%% ================================================================
% xxx row-vector vs. column-vector notation.

%% ================================================================
% Vector calc:

% * path integral = one form, one chain
% * surface integral = two form, two chain

% * potential function; poinc lemma.  ~1140 of Anton.  Proof??  Hildebrand ...
% * PDEs?  At least, solve dh/dx = f and dh/dy = g for h.
%   - Room for more PDE material in section 1?

% * d omega = 0 is a PDE?!?
% * ODEs <--> flows?

% cite Frankel ...

% Non-Eucl. vector calculus:
% * Domains w/ holes -- give real-world examples.
% * Configuration spaces of mech sys

%% ================================================================
% Analysis:

% mean-value thm
% Fubini ...

%% ================================================================
% AT!
% * A subspace of X
% * retraction: continuous projection (i.e. r(a) = a for all a in A).
% * retract: subspace with a retraction
% * deformation retract:  homotopy reaching the retraction, fixing the retract
% * def cov spc

% * de Rham thm
% * Poincar\'e duality
% * Poincar\'e-Hopf and index theory
% * HW 13 1ab:  scalar field & extrema -> vector field & zeroes, via gradient.

% * Brouwer

% ================================================================
% Gen topo:
% * def lebesgue #

% ================================================================
% 23.4.3
% 10.13.4

% * Put more specific mnfs in the mnfs section:  s2, t2, paraboloid,
%   quasi-paraboloid, hyperboloid of one sheet.

% spi2 p. 65 commutative diagram

% * def'n: or'ble mnf:  exists non-vanishing smooth m-form.
% * complexes
% * Is there a neat proof of the general tensor transformation formula?
%   Mine were all ad hoc for specific orders.  Frankel?
% * duoh pullback problem
% * Lie derivatives
% * See Cartan calculus entry in PM.  Cite it, use it, live and breathe it.
% * def vr fld
% * "it may be shown" in sec:loaf -> actually show it
% --
% * alt. notion of triv'n of vr bundle in terms of bundle map.
% * Tensors:
%   - Find AA ref for (lambda ox mu)(u,v) = lambda(u) mu(v).
%   - Explicit tensor examples
%   - For each structure type, note what the homs must preserve.
%   - Sym and Alt as functors.
% * Throughout:  I am conflating tangent *space* and tangent *bundle*.  Fix!
% * POUs ................
% * xxx note f dx means f ^ dx.
% * Note that uncited thms may be found in any reference.
% * for sym tensors, we don't always bother with the i<=j<=k business.
%   Cf. for alts, m=3, r=3, 27 vs. 1.  But for mtc tensor we often take sum ij.
% * forms: zero output at lin dep input, too.
% * row/col vectors througout.
% * triv'n section:  hair-combing ... moebius ...
% * def deriv'n
% * Lee DF vs. Pickrell final problem ... make sure I'm projecting down to \S^2
%   correctly ...  check Lee for examples.

% * fiber bundle in addition to vector bundle.
%   - Separate out the local-triviality condition.
%   - Example:  vector bundles.
%   - Example:  (geometric) tangent bundles.
%   - Example:  SO(n) over \S^{n-1}.  Here the fiber is a group.

% * \vecF throughout for maps of manifolds?  Vs. G for functions down to \R.

% Symmetric 2-tensor:  $g: \TqM \times \TqM \to \R$:
% $(\vecv_q, \vecw_q) \mapsto \vecv_q \cdot \vecw_q$.
% $$g = \sum_{i,j} g\left(\fracDx_i, \fracDx_j\right) dx_i \otimes dx_j.$$

% v ^ w = 1/2(v ox w - (v ox w)^t) = 1/2(v ox w - w ox v)
% Why:
%   (v ox w)ij = vi wj
%   (w ox v)ij = vj wi
%   ((v ox w)^t)ij = vj wi

% Q^t H Q: doubly covariant
% Q^-1 A Q: mixed
% similarity/congruence ... then, what about higher-order symmetries?

% * diffeom on mtc spaces in general?
% * what is the isom on metric spaces?

% What more DG HW problems to insert?  Be selective. :)

% Brauer and Nohel bib?  Esp. for flows.

% try to always have vectors as column vectors and functionals as row vectors
%   in this paper?  And/or carry around the $(\cdots)^t$?

% metric tensor vs metric space --- close, but no cigar
% alt: basis of b_i^* ^ ... w/ i<j
% sym: basis of b_i^* ^ ... w/ i<=j
% inf dim of Sigma^*
% fin dim of Lambda^*

% * Quack about coordinate invariance.  This is the main point, after all ... .

% * Why are only covariant fields pullbackable?  Note why it fails even for
%   vector fields.  If phi is a bijection (diffeom) then it's easy; relabelling.
%   Else problems w/ well-def'd or existing at non-1-1 or non-onto points.
%   But somehow even for non-diffeom phi, covts work OK.

% * Check Lang/Hung/DF re categorical freeness.
% * "Natural"
% * Pairing
%   - Examples:  inner product; (lambda, v) -> lambda(v)
%   - Adjoint vs. transpose?
%   - If V ox V is a v.s., what is its dual? (V ox V)^* = V^* ox V^*.
%   - V x V^* --> \R.  (v, \lambda) \mapsto <v, \lambda>
%   - Integ'n workshop notes re symmetric and symplectic pairings.
%   - Forms and chains as duals.  Cf. RA duals.
%   - Note integral is a linear functional.
%   - Stokes thm as simply saying that \D = adjoint of d:
%     <d omega, c> = <omega, \D c>
% * Adjoint
%   Given non-degen "form" (V --> V^*) <.,.>,
%   Ask about lin T: V -> V
%   Exists unique lin xfmn T^*: V -> V s.t. <Tv,w> = <v, T^*w> for all
%     v, w in W.  T^* is the "adjoint" of T.
%   - Given a pairing V x W \to \R, for all T:V \to V there exists T^*:W \to W
%     such that for all v, w: <Tv,w> = <v,T^*w>.
%     Then T^* is the ``adjoint'' of T.
%   - Do a simple 2x2 example to show that T^* = T^t --- in what cases?
%     Tu = sum_j a_{ij} u_j; T^* v = sum_j b_{ij} v_j
%     <Tu,v> = sum_{i,j} a_{ij} u_j v_i; <u,Tv> = sum_{i,j} b_{ij} u_i v_j.
%     So here T^* = T^t, at least as a matrix.
%   - This is the image of morphisms under the dualing functor.
% * Transpose:
%   Given T: V \to W, T^t: W^* \to V^* by T^t(lambda) = lambda T.
%   Then T^t is the ``transpose'' of T.

% Abelian categories
% multilin: V's and V*'s if necessary
% * Dual space as space of coordinate functions

% See Rudin ...
% * p. 260: d operator.

% Reproduce the *entire* magic sheet.  Make it obvious!  ``Cartan calculus''.
% * d(i_X dV) =? div(X) dV

% Misc:
% * Get eqn numbers to track thm, defn, etc.
% * the mnemonic ISRD
% * philosophical section about the recognition problem?
% * lots of hyperrefs for defns:  e.g. when "covering space" is used in the
%   stt of a thm, *href* to the defn of cov spc.
% * up top:  massey p. 149!?!  stokes etc.  *All* of this comes from vector
%   calculus.  *PLUS* modern development; rid of false starts; better notation
%   & use of algebra etc.  Focus on punctured potential?
% * have a section consisting solely of hrefs to the main thms?
% * section on "natural" isomorphisms.  examples too.
% * automorphisms of *anything* form a group.
% * sth big about commutativity of diagrams ............ along w/ categories,
%   sort of a unifying principle of modern math.
% * Disjoint union in Sets is categorically the same as direct sum in Modules.

% Pair alternating n-vector with alternating n-covector:
% * omega(v) = det(omega_i(v_j)).

% Tensors:
% * Multilinear fcns from V^k x V*^l to R.
% * *Linear* fcns from T^k_l(V) to R.

%% ================================================================
\section*{Preface}
\addcontentsline{toc}{section}{Preface}

\begin{center}
\emph{The art of doing mathematics consists in finding that special case which
contains all the germs of generality.}
--- David Hilbert\index{Hilbert, David} (1862-1943).
\end{center}

\begin{center}
\emph{
To Thales the primary question was not ``What do we know?'' but ``How do we
know it?''.
} \\
--- Aristotle\index{Aristotle}\index{Thales} (384-322 B.C.).
\end{center}

The following is a novice's guide (written by a novice, in mid-journey!) for
the often-twisting path from vector calculus to smooth manifolds.  The subject
matter is roughly that of the University of Arizona Mathematics Department's
first-year graduate course in geometry/topology, 534A-B.

This is a supplement, rather than a standalone reference.  One may consult
Lang's \emph{Algebra},
Spivak's \emph{Calculus on Manifolds},
Boothby's \emph{Introduction},
Frankel's \emph{The Geometry of Physics},
Lee's \emph{Introduction to Topological Manifolds} and
\emph{Introduction to Smooth Manifolds},
and other sources as cited throughout.

My intention is to gather together precise definitions and thorough examples in
a form that is concise and well-organized, or at least well-organized.  I
emphasize specific computations and examples that support and motivate theorems
which are proved elsewhere, spelling out important details which are all too
easily neglected in more formal treatments.  It is precisely these omitted,
taken-for-granted details which are the primary stumbling block for the
first-year graduate student.  I believe these stumbling blocks can be
identified and smoothed out, becoming paving stones on the student's road to
success.  (In part, this paper is an as-yet-incomplete attempt to introduce the
powerful techniques of reform calculus \cite{bib:hhgm}, including the Rule of
Four, into the graduate environment.)

In particular:
\begin{itemize}

\item This paper carefully revisits vector calculus, tying the known to the
unknown.

\item Substantial attention is given to tensors on finite-dimensional vector
spaces, before the development of bundles, sections and forms.

\item Topology is put in its proper place before geometry.

\item Geometry is emphasized; I have given geometrical motivation for symbolic
manipulations whenever I have found it possible to do so.

\item Exercises are worked through in full detail.  Rigor is the mouthwash of
mathematics, not the meat and bread:  when we are new to a subject, whether we
are six or sixty, we learn by imitation.

\end{itemize}

This document is a work perpetually in progress.

%% ----------------------------------------------------------------
\subsection*{Goals for the course}
\addcontentsline{toc}{subsection}{Goals for the course}

We have two main goals in Math 534A-B:

\begin{itemize}
\item[(1)] To put vector calculus on a more rigorous footing.  In particular,
	expressions of the form
	$$ df = \frac{\Df}{\Dx}\,dx + \frac{\Df}{\Dy}\,dy + \frac{\Df}{\Dz}\,dz$$
	were historically defined in terms of infinitesimals.  Our goal is to
	\emphidx{redefine} these familiar symbols in a rigorous way, in terms of
	linear approximations and vector spaces.  In particular, the grand goal is
	a generalized Stokes Theorem.
\item[(2)] To be able to do calculus on non-linear spaces, e.g. the surface
	of a sphere or torus.  In particular, the latter arise as configuration
	spaces of physical systems.
\end{itemize}

That is, we are re-thinking everything we thought we understood about vector
calculus, \emph{and} we doing it on a donut.  As a student, I can attest that
this dual goal gives students no small amount of difficulty.

Another irony is that in the undergraduate curriculum, calculus is one of the
first subjects taught, whereas our present re-working of calculus requires a
lot of math beyond calculus --- linear algebra, multilinear algebra, abstract
algebra, metric spaces, and quite a bit of new terminology.  The payoffs are
that (1) old concepts, which we once learned by rote, now become lucid and
elegant, and (2) we will be able to apply calculus far beyond the reach of the
freshman approach (see for example \cite{bib:penrose}).

% \begin{itemize}
% \item give examples of things we currently can compute
% \item give examples of things we will be able to compute
% \end{itemize}

%% ----------------------------------------------------------------
\subsection*{Goals for the paper}
\addcontentsline{toc}{subsection}{Goals for the paper}

The main, but not sole, purpose of this paper is to organize and annotate the
content of Math 534A-B.

If one wishes only to succeed on the geometry-topology qualifier, perhaps one
can simply learn examples by rote.  Since my goal is complete mastery of these
subjects, though, this paper takes a thorough approach toward the abstract
algebra which I claim is the majority of the content of the course.

Another goal of this paper is to illuminate and unify the various notions of
``tensor'' which appear in math and the sciences.  I am particularly interested
in understanding change of coordinates and ``index gymnastics''.  (That is, it
seems that mathematicians speak an entirely different language than scientists
do.  We can be bilingual; it will benefit all parties involved.) While this is
not qualifier material, it is a main goal of my graduate education.
Furthermore, this discussion is so intertwined with the main body of this paper
that it seemed to me to be foolish to separate it.  Non-qualifier sections of
this paper are clearly marked as such.

Do the contents of this paper form a redundant replication of pre-existing
textbooks?  Yes in part, and no:
\begin{itemize}
\item I want to explicitly state preliminaries (e.g. linear, abstract,
	tensor, and homological algebra, and category theory) which were going
	unstated, or were stated unclearly, in the course.
\item One learns by writing: the \emph{selection} of material winnows the
	important from the unimportant, and the \emph{organization} of material
	forces a clarity of thought which is otherwise unattainable.
\item I focus specifically on topics needed for UA's 534A-B course and
	geometry-topology qualifier.  This invites a best-of approach toward
	several different texts.
\item The contents of this paper are things I \emph{need} to know for the
	geometry-topology qual, and which are \emph{possible} to recall on qual
	day.  I am \emph{not} interesting in reciting technical details of proofs
	which I won't remember (details which authors of textbooks can and should
	include); I \emph{do}, however, want to have a single point of reference.
\end{itemize}

%% ================================================================
\subsection*{Structure of the paper}
\addcontentsline{toc}{subsection}{Structure of the paper}

\begin{itemize}

\item Necessary \bfidx{preliminary concepts} from calculus, algebra, analysis,
and topology are included in sections \ref{sec:prelim_geom} through
\ref{sec:prelim_alg}, rather than being interleaved throughout the
geometry-topology material per se.  The intents here are to summarize the key
points and to present unified notation.

\item \bfidx{Chains} (which generalize the paths, surfaces, and volumes that
appear in integrals in vector calculus) are a \emph{topological} concept.  Here
(section \ref{sec:topological_manifolds}) we need \bfidx{topological
manifolds}; we encounter homotopy, chains, and homology.

\item \bfidx{Forms} (which generalize the integrands and differentials that
appear in integrals in vector calculus) are a \emph{differential} concept.
Here (section \ref{sec:smooth_manifolds}) we \emph{add} a \bfidx{differentiable
structure}, or \bfidx{smooth structure}, to topological manifolds.  Here we
encounter tangent bundles, forms, and cohomology.

\item \bfidx{Integration} combines homology and cohomology.  Here (section
\ref{sec:duality_main_section}) we encounter \bfidx{duality} and
\bfidxtwo{pairings}{pairing}.

\item \bfidxtwo{Flows}{flows} (section \ref{sec:flows}) are the generalization
of ordinary differential equations to manifolds.

\item \bfidx{Lie derivatives} and the \bfidx{Cartan calculus} (section
\ref{sec:lie_cartan}) are, for this paper, a collection of notations and
manipulations which facilitate compuations.

\item Selected \bfidx{problem solutions} (section \ref{sec:problems}) are given
for homework and exam problems for Dr. Pickrell's 2005-2006 534A-B.

\item Solutions to some old qualifying exams are in a separate paper,
which you should find near this file as \texttt{prolrevqual.pdf}.

\end{itemize}

%% ================================================================
\subsection*{When I say \emph{map}, I mean \ldots}
\addcontentsline{toc}{subsection}{When I say \emph{map}, I mean \ldots}

In the context of group theory, one says merely \emph{homomorphism} in place of
\emph{group homomorphism}.  Likewise, in differential geometry, one simply says
\emph{map} in place of \emph{smooth map}.  With reference to section
\ref{sec:categories}, one says that one is \emph{working in the category of
smooth things}:  certain adjectives are omitted but understood.  Throughout
this paper, functions are assumed to be smooth (infinitely many times
differentiable) unless otherwise stated.

Moreover, in this paper I am \emph{working in the category of nice things}.  My
examples use Euclidean space or submanifolds thereof.  I will not discuss
infinite-dimensional vector spaces or non-Hausdorff topologies; I am not
interested in pathological examples.  Straightforward examples provide plenty
of work for a first-year course.

%% ================================================================
\subsection*{Specificity of dimension}
\addcontentsline{toc}{subsection}{Specificity of dimension}

I will happily prove a result for arbitrary $n$ by illustating only an $n=2$ or
$n=3$ case, as long as the $n=2$ or $n=3$ case contains all the elements of
generality.  Often, the general case adds no more detail, and in fact obscures
the simplicity of the presentation via its bric-a-brac of subscripts, \ldots's,
and summation signs.  We are low-dimensional creatures; we can see, touch, and
feel two and three dimensions.  We can use these abilities to help us rather
than hinder us.

%% ================================================================
\subsection*{The exemplar model}
\addcontentsline{toc}{subsection}{The exemplar model}

\begin{center}
\emph{I was born not knowing and have had only a little time to change that
here and there.} \\
--- Richard Feynman\index{Feynman, Richard} (1918-1988).
\end{center}

I am no math historian; in fact, as a second-year graduate student and a
teacher with only two years' classroom experience, I am only a budding
mathematician.  Yet this much seems to me to be a clear pattern in the
development of mathematics:
\begin{itemize}
\item Real-world problems are studied; mathematical models are developed.
\item After some time goes by, wherein mathematicians collaborate and
	communicate, similarities are discovered between problems which had seemed
	distinct.  (For example, permutations on symbols of letters and Galois'
	original ``substitutions'' \cite{bib:bma}; systems of linear equations and
	systems of differential equations.)
\item A system of axioms is developed, encapsulating and
	abstracting the common features of the once-disparate systems.  (In the
	above examples, we obtained abstract groups and vector spaces,
	respectively.)
\item We then study the abstract systems, proving general results
	and developing general methods which apply to all the original
	specific situations, \emph{and} any others which might apply
	in the future.  Herein lies much of the power of mathematics:  once we are
	familiar with, say, abstract groups, whenever we encounter a brand-new
	object, if we recognize it is a group then we instantly know much about
	it.
\item After enough time has gone by, we begin to teach the abstractions
	first.  In a pure-math environment, one might forget (or forget to
	mention) the origins entirely.
\end{itemize}

So much for the \emphidx{phylogeny} of mathematics --- its evolutionary
development across the lives of many thinkers.  What about its
\emphidx{ontogeny} --- the development of ideas within the mind of a single
thinker?  Like Feynman, none of us were born knowing calculus; every
mathematics student and future mathematician must painstakingly relearn these
ideas from scratch.

My claim is that \textbf{ontogeny recapitulates phylogeny}.  This is a
biological maxim\footnote{Attributed to the zoologist Ernst H\"ackel.  See also
the Wikipedia article on \emph{Recapitulation theory}.}, reflecting the fact
that the embryonic development of an individual retraces (albeit inexactly) the
evolutionary development of that individual's species.  When I was an embryo,
very early on I had gill slits --- and so did you.  One pair of slits became
our ears; the rest closed.  One might think that we, as modern humans, could
reproduce more humans directly, pole-vaulting over the intermediate stages ---
but we do not.

Likewise in the development of ideas.  When we are young, we experience many
specific situations.  We later learn to abstract and categorize: many
once-different objects begin to go by the common name of \emph{fork} or
\emph{dog}.  (See also the Boas quote in section \ref{sec:boas}.)  One might
think we could pole-vault over our early development, but we do not.

Few would argue that we could hand a two-year-old a dictionary and have him or
her skip over those early learning experiences.  But in the mathematics
classroom --- blinded by the power and beauty of modern methods --- we often do
just that.  In my eyes, the key trait distinguishing the mathematics of the
20th and 21st centuries from that of the centuries before is the
\emphidx{axiomatic approach}.  This is a rightful component of our discipline.
Yet we start the first day of an upper-division or graduate algebra course with
a mantra of the form ``A \emph{group} is a set endowed with a binary operation
\ldots''; we begin a geometry course with an abstract discussion of atlases and
manifolds.  This feels good and it feels right, especially to the teacher:  we
are going straight for the beauty and the elegance.  It is especially easy for
the teacher, who has already seen these concepts put to use in dozens of
concrete situtations:  the teacher is ready for abstraction.

%In our purity, we may never feel a desire to even mention why our subject
%exists in the first place.

But when we do this, we leave our students behind.  We attempt to leap over the
\emph{absolutely necessary} instantiation of ideas that we ourselves went
through (whether we did so consciously or not).  If we are to retain students
in mathematics, and if modern pure mathematics is to remain relevant to the
rest of the world, we must explicitly acknowledge that ontogeny recapitulates
phylogeny.  This does not mean that we need to retrace the full development of
every mathematical concept, with all its false starts and wrong turns.  It does
mean, however, that we \emph{first} need to tell our students \emph{why} a
discipline was invented in the first place.  We need to present the applied
problems which gave birth to the subject we are teaching.  We must give our
students the opportunity to generalize from concrete situations.

One raises the objection:  \emph{I don't have time for such trivialities.
Graduate students should be able to do this on their own.}  I reply, as a
graduate student and teacher:  It is hard enough for our students to learn what
we teach them; it is far harder for them to learn what we do not teach them.
The training of a new mind is something worth doing right:  it is better that
it happen \emph{because} of our teaching methods rather than in spite of them.

%A course in differential geometry ought to be more than a series of
%unmotivated definitions and theorems about differential symbology.

This is the one of the reasons I have found it not only entertaining but
necessary to write this paper.  As happened several times in my engineering
career (before graduate school), I could not obtain the kinds of explanations I
wanted (my questions were chiefly of the form ``\emph{But why?}'').  I found
myself lacking, and then writing, the references I wished I had had.  This in
turn led me to discover things I could not have found out otherwise.  I have,
as yet, found few historical references for the (phylogenetic) development of
differential forms.  Nonetheless, one of the primary themes of this paper is
the (ontogenetic) development of tensor algebra, tangent spaces, etc. from
fundamental concepts of area and measurement.  I have pieced together most of
these insights from my coursework; see also \cite{bib:bachman} who has a
similar approach.

%% ================================================================
\subsection*{Themes}
\addcontentsline{toc}{subsection}{Themes}

I encourage the reader to keep in mind the following central themes throughout
the paper:

\begin{itemize}

\item \bfidxtwo{Geometric}{geometric} vs. \bfidx{symbolic} methods\footnote{My
approach here is reminiscent of the Rule of Four as used in \cite{bib:hhgm},
although at this point it sounds like merely a Rule of Two.  Namely, the rule
is that mathematical ideas should be presented symbolically, graphically,
numerically, and verbally.  See section \ref{sec:matlab} for information about
numerics; however, I have much more work yet to do in this area.  (See also
\cite{bib:kerl}.)  Lastly, the very length of this paper constitutes a
non-empty verbal-description component.}.  We manipulate symbols because we can
do so efficiently, but the reason we invented them the way we did is often
geometric.  For example, the deteminant can be thought of symbolically in terms
of cofactor expansions of a square matrix, but it also has a purely geometric
existence in terms of the volume of a parallelepiped.  Likewise for the $d$ and
wedge operators of differential geometry: we can rattle off axioms for them,
but why were those axioms chosen?  What geometrical notions do these symbols
reflect?

\item \bfidxtwo{Coordinate-dependent}{coordinate-dependent} vs.
\bfidx{coordinate-free} definitions.  Early differential geometry [xxx cite
whom --- Spivak 2?] was done concretely, in terms of coordinate charts.  Any
coordinate-dependent definition needed to be accompanied by a proof that it was
invariant under change of coordinates.  The modern approach prefers
coordinate-independent definitions.  These are more elegant, but not as easy
for doing specific computations.  [xxx back that up or retract it.]

\item \bfidxtwo{Change of coordinates}{change of coordinates};
\bfidx{covariance} and \bfidx{contravariance}.  Different objects transform
differently on change of coordinates.  [xxx xref to some examples within this
paper.]  In the physics community [xxx see Frankel and Spivak 2?], the
transformation rules are taken to \emph{be} the defining characteristics.

\item The \bfidx{chain rule} and the \bfidx{Jacobian matrix}.  These are simply
all over the place in this subject.  Likewise, \bfidx{Riemann sums}.  (It is
ironic that in our first-year analysis class, we learn the beautiful subject of
measure theory; Riemann integration becomes a quaint relic.  Yet in the
geometry course, where everything is smooth or at worst piecewise smooth, the
latter works just fine.)

%\item \bfidxtwo{Intrinsic}{intrinsic} vs. \bfidx{extrinsic} definitions.  [xxx
%something about tangent spaces; incl. the usual pithy remark about living in
%curved 3-space and not being able to see out.]

\end{itemize}

%% ================================================================
\subsection*{Acknowledgements}
\addcontentsline{toc}{subsection}{Acknowledgements}

Doug Pickrell and Bill Faris provided numerous insights over the
course of several discussions.
Several problem solutions are due in full or in part to folks in the Monday
problem sessions --- Andy Lebovitz in particular --- as indicated in the
problems section of this paper.
Also, some ideas are due to the folks in the summer 2007 qualifier-preparation
session:  in particular, Victor Piercey and Dave Herzog.
Donna Harris proofread an early draft.
Dan Champion and Gabriela Dumitrascu provided useful feedback.
Thanks!

For constructive criticism and/or feedback please contact
\texttt{kerl at math dot arizona dot edu}.

%% ================================================================
%% ================================================================
%% ================================================================
\newpage
\section{Preliminaries from analytic geometry}
\label{sec:prelim_geom}

\begin{center}
\emph{To guess what to keep and what to throw away takes considerable
skill.  Actually it is probably merely a matter of luck, but it
looks as if it takes considerable skill.}
--- Richard Feynman\index{Feynman, Richard} (1918-1988).
\end{center}

%% ----------------------------------------------------------------
\subsection{Notation}
\label{sec:notn}

\begin{notn}
\label{notn:col_vec_row_vec}
Although I may occasionally slip, I endeavor to always write elements of $\R^n$
as \bfidx{column vectors}.  Also, I usually boldface their names.  For example,
$$
	\vecq = \colvecthree{x}{y}{z}
$$
in $\R^3$.  The \bfidx{transpose} of a column vector is a \bfidx{row vector},
written
$$
	\vecq^t = \rowvecthree{x,}{y,}{z},
$$
and vice versa.
\end{notn}

\begin{conv}
\label{conv:row_col_mx}
I will identify row vectors with $1 \times n$ matrices and column vectors with
$n \times 1$ matrices, respectively.
\end{conv}

\begin{notn}
When a row vector $\vecu^t$ is being thought of as a $1 \times n$ matrix, I
will write it as $\vecu^*$.  The reason for this will be seen in section
\ref{sec:VS_COB}.
\end{notn}

\begin{notn}
Throughout this paper, 
let
$$
	\xhat = \colvecthree{1}{0}{0}, \quad
	\yhat = \colvecthree{0}{1}{0}, \qand
	\zhat = \colvecthree{0}{0}{1}.
$$
Note that some writers call these $\ihat$, $\jhat$, and $\khat$.
When working in arbitrary dimensions (rather than 2 or 3), I will call them
$\ehat_1$ through $\ehat_n$, where $\ehat_j$ has a 1 in the $j$th slot and
zeroes elsewhere.
\end{notn}

%% ----------------------------------------------------------------
\subsection{Trigonometric functions}
\label{sec:trig}

xxx incl the wtrig diagram.  label and explain.

xxx the necessary trig and hyp-trig identities:  pyth, derivs, sum and difference.

\begin{align*}
	\sin^2(\alpha) + \cos^2(\alpha) &= 1 \\
	\sin(\alpha \pm \beta) &= \sin(\alpha)\cos\beta) \pm \cos(\alpha)\sin(\beta) &
	\cos(\alpha \pm \beta) &= \cos(\alpha)\cos\beta) \mp \sin(\alpha)\sin(\beta).
\end{align*}

Def'ns in terms of exponentials, which make anything possible (esp. choke
recovery).  [xxx perhaps a section/note on choke recovery \ldots it does very
much matter.] What else?  Check against the qual packet.

The key to the castle for spherical and hyperbolic trig functions is the following foursome:
\begin{align*}
	\cos(x)  &= \frac{e^{ix} + e^{-ix}}{2}  & \cosh(x) &= \frac{e^{x}  + e^{- x}}{2} \\
	\sin(x)  &= \frac{e^{ix} - e^{-ix}}{2i} & \sinh(x) &= \frac{e^{x}  - e^{- x}}{2}.
\end{align*}

%Using these identities we immediately obtain the following relationships
%between the spherical and hyperbolic trig functions:
%\begin{align*}
%	\cosh(ix) &=  \cos(x) &
%	\sinh(ix) &= i\sin(x).
%\end{align*}
%
%Using the sum and product identities we can find the complex sine and cosine
%in terms of the real functions:



xxx also inverse maps.

xxx mention graphs of cosh and sinh.

%% ----------------------------------------------------------------
\newpage
\subsection{The dot product and its applications}

Everything in this section is familiar from undergraduate math.  In this
section my goal is to connect these familiar terms with the new material to
come.

%% ----------------------------------------------------------------
\subsubsection{The dot product and projected length}
\label{sec:dot}

We are familar with the \bfidx{dot product} of two vectors $\vecu$ and
$\vecv$ in $\R^n$.  \emph{Algebraically}, we write
$$
	\vecu \cdot \vecv = \sum_{i=1}^n u_i v_i.
$$

What, though, is the \emph{geometric}
interpretation\footnote{See also the nice Java applet at
\texttt{www.falstad.com/dotproduct}.}?
It can be shown, using the law of cosines and the Pythagorean theorem, that
$$
	\vecu \cdot \vecv = \|\vecu\| \, \|\vecv\| \, \cos(\theta)
$$
where $\theta$ is the angle between the two vectors, with $\theta$ taken to be
from $0$ to $\pi$.  Now, the dot product is symmetric in $\vecu$ and $\vecv$.
But suppose that we distinguish, or fix, one vector --- say, $\vecu$ --- and
let the other one vary.  Then $\vecu \cdot \vecv$ is the length of
the \bfidx{projection} of $\vecv$ onto $\vecu$, times the length of $\vecu$
itself:

\begin{center}
\psfragscanon
\includegraphics{figures/dotuvcos.eps}
\end{center}

We can think of $\vecu$ as \emph{measuring length in a specified direction}.
[xxx xref fwd to fishbone plots in linear-algebra and form sections.]
Holding $\vecu$ fixed and varying $\vecv$, we will get different numbers.  We
can plot vectors $\vecv$ which all have the same projected length onto $\vecu$:

\begin{center}
\psfragscanon
\includegraphics{figures/dotuvs.eps}
\end{center}

[xxx xref fwd to contour plots]
Note in particular that if $\vecv$ is \bfidx{perpendicular} to $\vecu$, then
$\vecu \cdot \vecv$ is zero.  [xref to proj/perp; perp space; normal/tangent.]
In fact we can \emph{define} perpendicularity:  $\vecu$ is
\bfidx{perpendicular} to $\vecv$ if $\vecu \cdot \vecv$ is zero.
(This means that the zero vector is perpendicular to every vector.)

Observe that, using convention \ref{conv:row_col_mx}, the dot product
$\vecu \cdot \vecv$ is the same as the matrix product
$\vecu^t \vecv$.  For example, in 3 dimensions:
\begin{eqnarray*}
	\vecu \cdot \vecv &=&
		\colvecthree{u_1} {u_2} {u_3} \cdot
		\colvecthree{v_1} {v_2} {v_3}
		= u_1v_1 + u_2 v_2 + u_3 v_3 \\
	\vecu^t \cdot \vecv &=&
		\rowvecthree{u_1} {u_2} {u_3} 
		\colvecthree{v_1} {v_2} {v_3}
		= u_1v_1 + u_2 v_2 + u_3 v_3.
\end{eqnarray*}
This line of thought is continued in section \ref{sec:geom_dual}, where we
think of $\vecu^t$ as being a function on vectors $\vecv$.

Last, note the following:  the dot product is a function on two vectors $\vecu$
and $\vecv$ which is symmetric with respect to $\vecu$ and $\vecv$.  That is,
$\vecu\cdot\vecv = \vecv\cdot\vecu$.  This is the paradigm whose generalization
is the \bfidx{symmetric tensor}, discussed in more detail in section
\ref{sec:sym_alt}.

%% ----------------------------------------------------------------
\subsubsection{Projection and perpendicular operators}
\label{sec:proj_perp}

Let $\veca$ be a fixed vector in $\R^m$.  Often in what follows we will want to
\bfidx{decompose} a vector $\vecv$ into two \bfidx{components}, one parallel to
$\veca$ and one perpendicular to $\veca$:

%	\setlength{\unitlength}{1.0in}
%	\begin{picture}(6,1.0)
%	%\put(0,0){\framebox(6,1.0)}
%
%	\put(4.65, 0.75){$\vecv$}
%	\put(2.65, 0.75){$\vecv_\perp$}
%	\put(4.65, 0.10){$\vecv_\|$}
%	\put(3.55, 0.20){$\veca$}
%
%	%\put(2.55, 0.77){\vector(1,0){2.67}}
%	\put(2.55, 0.10){\vector(0,1){0.67}}
%	\put(2.55, 0.10){\vector(1,0){2.00}}
%	\put(2.55, 0.10){\vector(3,1){2.00}}
%	\thicklines
%	\put(2.55, 0.10){\vector(1,0){1.00}}
%	\thinlines
%	\end{picture}

\begin{center}
\psfragscanon
\includegraphics{figures/avparperp.eps}
\end{center}

That is, we want to write $$ \vecv = \vecv_{\|} + \vecv_\perp$$ where
$\vecv_{\|}$ is the component of $\vecv$ in the direction of $\veca$ and
$\vecv_\perp$ is the component of $\vecv$ perpendicular to $\veca$, but still
in the plane spanned by $\vecv$ and $\veca$.  (Note that if $\vecv$ is already
parallel to $\veca$, then $\vecv$ and $\veca$ don't span a plane.)  Since $$
\vecv_\perp = \vecv - \vecv_{\|},$$ it suffices to find the parallel component.

Given $\vecv$ and $\veca$, we need to construct a vector with magnitude equal
to $\|\vecv\| \cos \theta$ and direction $\veca$, where $\theta$ is the angle
between $\vecv$ and $\veca$.
Recall that the unit vector in the direction of
$\veca$ is $$\hat\veca = \frac{\veca}{\|\veca\|}.$$
So, we want
$$\vecv_{\|} = \|\vecv\| \cos \theta \, \frac{\veca}{\|\veca\|}.$$
Since
$$\vecv \cdot \veca = \|\vecv\| \, \|\veca\| \cos \theta,$$
and
$$\|\veca\| = \sqrt{\veca \cdot \veca},$$
we have
%$$
\begin{equation}
\label{eqn:par_perp}
\vecv_{\|} = \frac{\vecv \cdot \veca}{\veca \cdot \veca} \, \veca
\qqand
\vecv_{\perp} = \vecv - \frac{\vecv \cdot \veca}{\veca \cdot \veca} \, \veca.
\end{equation}
%$$
This looks familiar from the \plainidx{Gram-Schmidt} process.

This technique will be indispensable when we decompose vectors into components
perpendicular and parallel to the normal of a surface, namely, when working
with tangent spaces.  See for example section \ref{sec:fe_2}.

[xxx xref fwd to divergence/curl/etc.]

%% ----------------------------------------------------------------
\subsubsection{Point-normal form}
\label{sec:point_normal}

A (hyper)plane is defined by a point $\vecq_0$ on the plane, and a vector
$\vecn$ normal to the plane:

\begin{center}
\psfragscanon
\includegraphics{figures/ptnml.eps}
\end{center}

To get an equation for the plane, let $\vecq$ be an arbitrary point on the
plane.  Then $\vecq-\vecq_0$ is perpendicular to $\vecn$.  Algebraically,
\begin{equation}
\label{eqn:q_minus_q0_dot_n}
	(\vecq-\vecq_0) \cdot \vecn = 0.
\end{equation}
Suppose (in three dimensions) we have
$$
	\vecq_0 = \colvecthree{x_0}{y_0}{z_0},
	\vecq = \colvecthree{x}{y}{z}, \qand
	\vecn = \colvecthree{a}{b}{c}.
$$
Then equation \ref{eqn:q_minus_q0_dot_n} becomes
\begin{eqnarray*}
	\colvecthree{x-x_0}{y-y_0}{z-z_0} \cdot \colvecthree{a}{b}{c} &=& 0 \\
	a(x-x_0) + b(y-y_0) + c(z-z_0) &=& 0.
\end{eqnarray*}

xxx Include the many fwd xrefs.

%% ----------------------------------------------------------------
\newpage
\subsection{Coordinates}

\begin{center}
\emph{
I have an existential map. It has ``You are here'' written all over it.
} \\
--- Steven Wright\index{Wright, Steven} (1955 - ).
\end{center}

%% ----------------------------------------------------------------
\subsubsection{Coordinates, area, and the determinant}
\label{sec:coords_and_geom_det}

Here is a circle:

\begin{center}
\includegraphics{figures/circle1.eps}
\end{center}

How big is it?  What is its area?  What is its
curvature?  These things require \bfidx{rulers}\footnote{See \cite{bib:bma}
for the history of this key development, especially involving Descartes.}:

\begin{center}
\includegraphics{figures/circle_with_rulers.eps}
\end{center}

I have two points to make here:
(1) what is forced upon us when we consider reasonable notions of symmetry
and scaling for area, and
(2) what happens when we change coordinates.
I will address the former point in this section and the latter point in the
next section.
I'll switch from a circle to a square, or really any parallelogram, since
these are easier to measure:

\begin{center}
\psfragscanon
\includegraphics{figures/pargm1.eps}
\end{center}

(Parallelograms and parallelepipeds are particularly important since they are
regions swept out by systems of not-necessarily-orthonormal basis vectors.
[xxx xref fwd.]  Also, they are key for computing Riemann sums [xxx xref fwd].)

I'd like to think of the \bfidx{area} of the parallelogram as depending on the
lengths of two adjacent sides $\vecu$ and $\vecv$, and the angle $\theta$
between them.  In particular I'd want the area $A$ to be
$$
	A(\vecu,\vecv) = \|\vecu\|\,\|\vecv\|\,\sin(\theta).
$$  (In a moment I'll inquire about negative $\theta$.)

What properties do we want area to have?  Well, it must be \bfidx{additive}.
Two identical parallelograms have twice the area of one of them:

\begin{center}
\psfragscanon
\includegraphics{figures/pargm2.eps}
\end{center}

Algebraically,
$$A(2\vecu,\vecv) = 2 A(\vecu,\vecv).$$
More generally, we expect
$$A(c\vecu,d\vecv) = cd A(\vecu,\vecv)$$
for any zero or positive $c$ and $d$.

Next, the area of two unit squares should be one less than the area of three
unit squares, and so on.  From this we are forced to require $$A(c\vecu,d\vecv) =
cd A(\vecu,\vecv)$$ for any real $c$ and $d$, whether positive, zero, or
negative.  (This is the \bfidx{multilinearity} property of area, about which we
will have much more to say in section \ref{sec:tensors}.) That is, reasonable
arithmetic \emph{requires} that we allow the notion of \bfidx{signed area} ---
even though I have no idea how much paint it takes to cover a wall whose area
is $-100$ square feet.

Now what if I rotate the parallelogram?

\begin{center}
\psfragscanon
\includegraphics{figures/pargmrot.eps}
\end{center}

Its area is unchanged; area should be \bfidx{rotation invariant}.
Algebraically, this is nothing new:  $\theta$ is a \emph{relative} angle betwen
$\vecu$ and $\vecv$.

What if I pick up the parallelogram and flip it over?

\begin{center}
\psfragscanon
\includegraphics{figures/pargmflip.eps}
\end{center}

We declare that we want the area to change sign on flips:  it is
\bfidx{oriented}.  Algebraically,
$$
	A(\vecv,\vecu) = -A(\vecu,\vecv).
$$
This is the \bfidx{alternating} or \bfidx{skew symmetry} property of signed
area; we will have more to say about this in section \ref{sec:sym_alt}.  This
also makes sense of negative $\theta$.

These properties, namely multilinearity and skew symmetry, are the defining
characteristics of signed area.  They are what constitute the determinant.

\begin{defn}
The \index{determinant}\bfidx{geometric determinant} of two vectors $\vecu$ and
$\vecv$ (or three or more) is the oriented signed area of the parallelogram (or
parallepiped) spanned by them.  
\end{defn}

Now, there is a \bfidx{choice of orientation}:  \emph{after} I flip one edge,
then the sign of the area is reversed.  But what is the sign of the first area?
It's a choice and there is a conventional way to decide it.  [xxx Mention
right-handed and left-handed coordinate systems.]

Showing that this geometric definition is equivalent to the algebraic notion of
determinant [xxx below; perhaps type up handwritten notes for $2 \times 2$] is
a separate task; the geometric notion suffices for now.

%% ----------------------------------------------------------------
\subsubsection{Change of coordinates}

When we change coordinates, the object being measured is the same; only the
underlying measurement system changes.  But when we adopt a new measurement
system, it becomes the new reality we perceive:

\begin{center}
\psfragscanon
\includegraphics{figures/circle_shear.eps}
\end{center}

[xxx more here.]
This topic will be addressed further in section \ref{sec:vc_change_of_coords}.

%% ----------------------------------------------------------------
\subsubsection{Some coordinate systems}
\label{sec:coord_sys}

xxx to do: For $\R^2$ and $\R^3$; $\S^1$ and $\S^2$.  Rectangular, polar,
cylindrical, spherical with mnemonics.  Conventions.  Stereographic on both
poles with reciprocal transition function.

%% ----------------------------------------------------------------
\subsubsection{Stereographic projection on $\S^1$}
\label{sec:stereo_S1}

Questions involving \plainidx{stereographic coordinates} are popular on the
qualfiying exams.  The algebra involved is not too profound, but can throw one
for a loop if one has not worked through it at least once beforehand.

[xxx Draw the similar-triangles figures for $p=x/(1-y)$ and $q=x/(1+y)$.]

[xxx note the domains of definition for $p$ and $q$, and their intersection.]

\begin{prop}  One converts from $(x,y)$ coordinates on the circle to
$p$ or $q$ coordinates by
$$
	p = \frac{x}{1-y} \qand q = \frac{x}{1+y}.
$$
\begin{proof}
From the first figure, we can see that by similarity of triangles
$$
	\frac{p}{1} = \frac{x}{1-y}.
$$
Likewise, from the second figure,
$$
	\frac{q}{1} = \frac{x}{1+y}.
$$
\end{proof}
\end{prop}

\begin{prop}
The coordinates $p$ and $q$ are related by
$$ p = \frac{1}{q}.$$
\begin{proof}[Algebraic proof]
Recalling that $x^2+y^2=1$ on the circle, we have
\begin{eqnarray*}
	\frac{1}{q} &=& \frac{1+y}{x} \\
	&=& \left(\frac{1+y}{x}\right) \left(\frac{1-y}{1-y}\right) \\
	&=& \frac{1-y^2}{x(1-y)} \\
	&=& \frac{x^2}{x(1-y)} \\
	&=& \frac{x}{(1-y)} \\
	&=& p.
\end{eqnarray*}
\end{proof}
\begin{proof}[Geometric proof]
[xxx make figure; use similarity of triangles].
[xxx xref fwd to transition functions in the manifold section.]
\end{proof}
\end{prop}

\begin{prop}  One converts from $p$ or $q$ coordinates back to $(x,y)$
coordinates by
\begin{eqnarray*}
	x=\frac{2p}{p^2+1} &\qqand& y=\frac{p^2-1}{p^2+1} ; \\
	x=\frac{2q}{1+q^2} &\qqand& y=\frac{1-q^2}{1+q^2}.
\end{eqnarray*}
\begin{proof}
For the first, start with $p=x/(1-y)$.  We can solve for $y$ but we need to
eliminate $x$.  Our relation between $x$ and $y$ is $x^2+y^2=1$ so we will need
to square things.  We have
$$
	p^2 = \frac{x^2}{(1-y)^2}
	= \frac{1-y^2}{(1-y)^2}
	= \frac{(1-y)(1+y)}{(1-y)(1-y)}
	= \frac{1+y}{1-y}.
$$
This is an invertible rational function:  in general, if $ad-bc\ne 0$ in
$(ay+b)/(cy+d)$, then we can invert.  Doing as we instructed our freshman
students in college algebra, we get
\begin{eqnarray*}
	p^2 &=& \frac{1+y}{1-y} \quad \textrm{and likewise} \quad q^2   = \frac{1-y}{1+y} \\
	(1-y) p^2 &=& 1+y \\
	p^2 - p^2 y &=& 1+y \\
	(1+p^2) y &=& p^2-1 \\
	y &=& \frac{p^2-1}{p^2+1}.
\end{eqnarray*}
For $x$, use $p=x/(1-y)$ and solve for $p$:
\begin{eqnarray*}
	x &=& p(1 - y)
	= p\left(1 - \frac{p^2-1}{p^2+1}\right)
	= p\left(\frac{2}{p^2+1}\right)
	= \frac{2p}{p^2+1}.
\end{eqnarray*}

Similar algebra proves the formulas for $q$.
\end{proof}
\end{prop}

xxx move to the manifold section.  Establish and use the following to
illustrate covariance and contravariance:
\begin{eqnarray*}
	\frac{dp}{dq} &=& \frac{d}{dq} \left(\frac{1}{q}\right) = \frac{-1}{q^2} \\
	\frac{dq}{dp} &=& \frac{d}{dp} \left(\frac{1}{p}\right) = \frac{-1}{p^2}.
\end{eqnarray*}
Then (noting that we differentiate $x(p)$ and $y(p)$):
\begin{eqnarray*}
	\fracDp &=& \colvectwo{\slashDtwo{x}{p}}{\slashDtwo{y}{p}}
	= \colvectwo
		{\fracD{p}\left(\frac{2p}{p^2+1}\right)}
		{\fracD{p}\left(\frac{p^2-1}{p^2+1}\right)}
	= \colvectwo
		{\frac{2-2p^2}{(p^2+1)^2}}
		{\frac{4p}{(p^2+1)^2}} \\
	%
	\fracDq &=& \colvectwo{\slashDtwo{x}{q}}{\slashDtwo{y}{q}}
	= \colvectwo
		{\fracD{q}\left(\frac{2q}{1+q^2}\right)}
		{\fracD{q}\left(\frac{1-q^2}{1+q^2}\right)}
	= \colvectwo
		{\frac{2-2q^2}{(q^2+1)^2}}
		{\frac{-4q}{(q^2+1)^2}}.
\end{eqnarray*}
Then (noting that we differentiate the other way around, with $p(x)$):
\begin{eqnarray*}
	dp &=&
		\fracDx \left(\frac{x}{1-y}\right)\,dx +
		\fracDy \left(\frac{x}{1-y}\right)\,dy \\
	&=& \frac{dx }{1-y} + \frac{x\,dy}{(1-y)^2} \\
	dq &=&
		\fracDx \left(\frac{x}{1+y}\right)\,dx +
		\fracDy \left(\frac{x}{1+y}\right)\,dy \\
	&=& \frac{dx }{1+y} - \frac{x\,dy}{(1+y)^2}.
\end{eqnarray*}

[xxx describe what $dx$ and $dy$ mean here:  $(y^2, -xy)$ and $(-xy, x^2)$,
respectively.  They are scalar multiples of $d\theta$ which is $(-y,x)$.
And/or, use $x$ or $y$ as graph coordinates.]

[xxx important remark about how vector fields push forward but forms
do not.  xref to geometric-tangent-vector section.  More work for forms.
Edit and type up handwritten notes.]

Then [xxx do the algebra to show] (covariance and contravariance):
\begin{eqnarray*}
	dp &=& \frac{dp}{dq} \, dq. \\
	\fracDp &=& \frac{dq}{dp} \, \fracDq. \\
\end{eqnarray*}

%% ----------------------------------------------------------------
\subsubsection{Stereographic projection on $\S^n$}
\label{sec:stereo_Sn}

Here we generalize from section \ref{sec:stereo_S1}.  The pattern becomes
clear in $\S^3$.

For stereographic projection from the north pole which is
$(x,y,z,w)=(0,0,0,1)$, we want a coordinate chart and a parameterization, [xxx
xref fwd or reorg to charts/parameterizations] respectively:
$$
	\colvecfour{x}{y}{z}{w}
	\mapsto
	\colvecthree{r}{s}{t}
	\qqand
	\colvecthree{r}{s}{t}
	\mapsto
	\colvecfour{x}{y}{z}{w}.
$$

[xxx include figure and argument that this is done slotwise]

Then the desired coordinate chart is
$$
	r = \frac{x}{1-w}, \quad
	s = \frac{y}{1-w}, \quad
	t = \frac{z}{1-w}.
$$

For the parameterization, to use the relation $x^2+y^2+z^2+w^2=1$ we need to
square things; adding them up to collect the numerators we have
$$
	r^2+s^2+t^2 = \frac{x^2+y^2+z^2}{(1-w)^2}
	= \frac{1-w^2}{(1-w)^2} = \frac{1+w}{1-w}.
$$
Now in general, proceeding as in section \ref{sec:stereo_S1}, to invert such a
rational function we have
$$
	a = \frac{1+b}{1-b} \implies b = \frac{a-1}{a+1}.
$$
Here,
$$
	w = \frac{r^2+s^2+t^2-1}{r^2+s^2+t^2+1}.
$$
For $x$, use $x=r(1-w)$.  In general
$$
	1-\frac{c-1}{c+1} = \frac{2}{c+1}
$$
so here
$$
	x = \frac{2r}{r^2+s^2+t^2+1}, \quad
	y = \frac{2s}{r^2+s^2+t^2+1}, \quad
	z = \frac{2t}{r^2+s^2+t^2+1}.
$$

Generalizing to the $n$-sphere, one obtains the following coordinate chart and
paramerization for stereographic projection from the north pole:
\begin{eqnarray*}
	p_i &=& \frac{x_i}{1-x_{n+1}}, \; i=1, \ldots, n; \\
	x_i &=& \frac{2p_i}{\sum_{j=1}^{n}p_j^2 + 1}, \; i=1, \ldots, n, \qquad
	x_{n+1} = \frac{\sum_{j=1}^{n}p_j^2 - 1}{\sum_{j=1}^{n}p_j^2 + 1}.
\end{eqnarray*}
If one repeats the algebra for the south pole, one obtains
\begin{eqnarray*}
	q_i &=& \frac{x_i}{1+x_{n+1}}, \; i=1, \ldots, n; \\
	x_i &=& \frac{2q_i}{1+\sum_{j=1}^{n}q_j^2}, \; i=1, \ldots, n, \qquad
	x_{n+1} = \frac{1-\sum_{j=1}^{n}q_j^2}{1+\sum_{j=1}^{n}q_j^2}.
\end{eqnarray*}

It is evident from the figure [xxx] that the vectors $\vecp=(p_1, \ldots, p_n)$
and $\vecq=(q_1, \ldots, q_n)$ are collinear and on the same side of the
origin.  To see that $\|\vecp\|=1/\|\vecq\|$, it is sufficient to see that
$\|\vecp\|^2=1/\|\vecq\|^2$:
\begin{align*}
	\vecp &= \colvecthree{x_1}{\vdots}{x_n}\, \frac{1}{1-x_{n+1}} &
	\vecq &= \colvecthree{x_1}{\vdots}{x_n}\, \frac{1}{1+x_{n+1}} \\
	%
	\|\vecp\|^2 &= \frac{x_1^2+\ldots+x_n^2}{(1-x_{n+1})^2} &
	\|\vecq\|^2 &= \frac{x_1^2+\ldots+x_n^2}{(1+x_{n+1})^2} \\
	%
	&= \frac{1-x_{n+1}^2}{(1-x_{n+1})^2} &
	&= \frac{1-x_{n+1}^2}{(1+x_{n+1})^2} \\
	%
	&= \frac{1+x_{n+1}}{1-x_{n+1}} &
	&= \frac{1-x_{n+1}}{1+x_{n+1}}.
\end{align*}

[xxx xref fwd to transition functions in the manifold section.]

[xxx explicit transition functions.  They work out quickly and nicely.]

%% ----------------------------------------------------------------
\newpage
\subsection{Types of functions}
\label{sec:vector_calc_fcn_types}

We can have functions (linear or not) from $\R^m$ to $\R^n$.  However, special
cases and special terminology arise when $m=1$, $n=1$, or $m=n$.  In this
section I will discuss these types in the following order:
\begin{itemize}
\item paths from $\R \to \R^m$, then
\item maps from $\R^m \to \R^n$, then
\item scalar functions from $\R^n \to \R$.
\end{itemize}
This seemingly unimportant editorial decision reflects a natural order of
\plainidx{composition}:

\begin{itemize}
\item Given a path from $\R \to \R^m$, we can post-compose with
	a map from $\R^m \to \R^n$ to obtain another path.
\item Given a scalar function from $\R^n \to \R$, we can pre-compose with
	a map from $\R^m \to \R^n$ to obtain another scalar function.
\end{itemize}

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.0)
%\put(0,0){\framebox(6,1)}

\put(1.50, 0.10){$\R$}
\put(2.50, 0.70){$\R^m$}
\put(3.75, 0.70){$\R^n$}
\put(4.75, 0.10){$\R$}

\thicklines
\put(1.65, 0.22){\vector(3,2){.75}}
\put(2.75, 0.74){\vector(1,0){.90}}
\put(3.95, 0.70){\vector(3,-2){.75}}
\thinlines

\put(1.75, 0.15){\vector(4,1){1.80}}
\put(2.75, 0.63){\vector(4,-1){1.75}}

\put(1.50, 0.52){paths}
\put(3.00, 0.84){maps}
\put(4.50, 0.52){scalar functions}

\end{picture}

This points toward pushforwards and pullbacks as discussed in [xxx xref].  In
the differentiation section (section \ref{sec:vector_calc_diffn}), though, I
will discuss maps after functions.  This is simply because maps are made up of
functions.

xxx here or elsewhere: parameterized curves vs. graphs \ldots for VC we
prefer the latter and think of the former as a special case; here, the
other way around.  elaborate.

%% ----------------------------------------------------------------
\subsubsection{Paths: $\R$ to $\R^m$}
\label{sec:paths_R_to_Rm}

\begin{center}
\emph{
Reasoning draws a conclusion, but does not make the conclusion certain, unless the mind discovers it by the path of experience.}
--- Roger Bacon\index{Bacon, Roger} (c. 1214-1294).
\end{center}

\begin{defn}
A \bfidx{path}, or \bfidx{scalar-to-vector function}, is a function $\gamma:\R
\to \R^m$, which I will write for $m=3$ as
$$
\gamma(t) \quad\textrm{or}\quad
\begin{pmatrix} \kappa(t) \\ \lambda(t) \\ \mu(t) \end{pmatrix}.
$$
\end{defn}

A scalar-to-vector function has $m$ single-variable \bfidx{component
functions}.

Paths will lead us to \bfidx{tangent vectors} [xxx xref fwd] and [xxx xref to
topology, homotopy classes, first homology classes, etc.].

Visualizing paths is easy.  xxx include a figure with a starting point and an
arrow.

\beginex
\label{ex:gamma_circle}
Let
$$
	\gamma(t) = \colvectwo{\cos(t)}{\sin(t)}.
$$
[xxx matlab figure.]
\mendex

%% ----------------------------------------------------------------
\subsubsection{Maps: $\R^m$ to $\R^n$}
\label{sec:maps_Rm_to_Rn}

\begin{defn}
A \bfidx{map}, or \bfidx{vector-to-vector function}, is a function $\vecF:\R^m
\to \R^n$.  For $m=n=3$, I will often write such a function as
$$
\vecF(\vecq) \quad\textrm{or}\quad
\begin{pmatrix} f(x,y,z) \\ g(x,y,z) \\ h(x,y,z) \end{pmatrix}.
$$
\end{defn}

A vector-to-vector function has $n$ \bfidx{component functions} of
$m$ variables.  These component functions are vector-to-scalar functions
(section \ref{sec:scalar_fcns_Rn_to_R}).

Vector-to-vector functions will appear as
\bfidxtwo{diffeomorphisms}{diffeomorphism} [xxx xref fwd],
in [xxx xref fwd] \bfidx{change of variables},
and \bfidx{vector fields} and \bfidx{flows} [xxx xref fwd].

Maps are a bit harder to visualize than paths.  There are at least two
ways:

\begin{itemize}
\item \bfidxtwo{Grid images}{grid image}.  We think of each
	point $\vecq$ of $\R^m$ as being \bfidx{carried} by $\vecF$ to a point
	$\vecF(\vecq)$ of $\R^n$.
\item \bfidxtwo{Quiver plots}{quiver plot}.
	We think of $\vecF$ as \bfidx{attaching} to each point $\vecq$ of $\R^m$ a
	vector $\vecF(\vecq)$ which is \bfidxtwo{footed}{foot} not at the origin
	but at $\vecq$.
\end{itemize}

\beginex
\label{ex:map_bean}
Let $\vecF: \R^2 \to \R^2$ be given by
$$
	\colvectwo{x}{y} \mapsto \colvectwo{x-y^3}{x^3+2y}.
$$
A grid-image plot of $\vecF$ is as follows:
[xxx matlab figures].
The first figure is the original plane:  the positive $x$-axis is marked in
heavy red and the positive $y$-axis is marked in heavy blue.  The second figure
shows the image of the grid under $\vecF$. [xxx xref to matlab section:
section \ref{sec:matlab}.]

A quiver plot shows $\vecF$ as a vector field:
[xxx matlab figure]
\mendex

\begin{rem}
\label{rem:COV_func_nameless}
Generally we write vector-to-vector functions in the form
$$
	\vecx=\vecF(\vecs)
	\qqor
	\colvecthree{x}{y}{z} =
	\begin{pmatrix} f(s,t,u) \\ g(s,t,u) \\ h(s,t,u) \end{pmatrix}
$$
using separate letters for input, function name, and output.  A \bfidx{change
of coordinates} is nothing more than a vector-to-vector function (which we
generally want to be invertible), but we usually omit the function name.  (I
call these \bfidx{nameless functions}, although that terminology is
non-standard.) We have simply
$$
	\vecx=\vecx(\vecs)
	\qqor
	\colvecthree{x}{y}{z} =
	\begin{pmatrix} x(s,t,u) \\ y(s,t,u) \\ z(s,t,u) \end{pmatrix}.
$$
\end{rem}

%% ----------------------------------------------------------------
\subsubsection{Scalar functions: $\R^n$ to $\R$}
\label{sec:scalar_fcns_Rn_to_R}

\begin{defn}
A \bfidx{scalar function}, or \bfidx{vector-to-scalar function}, is a function
$G:\R^n \to \R$, written for $n=3$ as $$G(x,y,z) \quad\textrm{or}\quad
G(\vecq).$$
\end{defn}

Scalar functions will appear as [xxx write and xref fwd] [xxx
submanifolds] [xxx xref to regular-value theorem] [xxx coefficients in forms].

Scalar functions may be visualized in at least two ways:

\begin{itemize}
\item xxx \bfidx{surfaces} [xxx figure here]
\item xxx \bfidx{contours} [xxx figure here]
\end{itemize}

\beginex
The \bfidx{height function}
$$
	h(x,y,z) = z
$$
or
$$
	h(x,y) = y.
$$
[xxx elaborate.] [xxx figures.]
\mendex

\beginex
$G(x,y) = x^2+y^2$. [xxx elaborate]
\mendex

[xref fwd (not far!) to level sets]
   
%% ----------------------------------------------------------------
\subsubsection{Level sets}
\label{sec:level_sets}

\begin{defn}
Let $f: \R^3 \to \R$ be a \plainidx{vector-to-scalar function}.  The
\bfidx{level set} of $f$ and $c$ is the set of points
$$
	\{\vecq: f(\vecq) = c\}.
$$
\end{defn}

\beginex
Let $f(x,y) = x^2+y^2$.  It requires 3 dimensions to visualize this
(non-flat) 2D surface, with $z=f(x,y)$:

[xxx figure]

It requires 2 dimensions to visualize $f$ as a contour plot:

[xxx figure]

Now take the level set of $f$ and 1.  This result is a circle, which is
a (non-flat) 1D object in 2D space:

[xxx figure]

A single level set \emph{selects out} a one-dimensional space.  We will see
more of this in theorem \ref{thm:regular_value}.  In short, $f(x,y)=c$ is one
equation in two unknowns; there are two variables with one constraint and thus
one degree of freedom.
\mendex

xxx xref to tangent/normal dichotomy section.

%% ----------------------------------------------------------------
\subsubsection{Images of paths under maps: pushforwards}

Let $\gamma: \R \to \R^m$ and $\vecF: \R^m \to \R^n$.  Then $\vecF\circ\gamma$
is a path from $\R$ to $\R^n$.  The path $\gamma$ has been \bfidxtwo{pushed
forward}{pushforward} from $\R^m$ to $\R^n$.

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.0)
%\put(0,0){\framebox(6,1)}
\put(3.05, 0.80){$\R$}
\put(2.40, 0.20){$\R^m$}
\put(3.70, 0.20){$\R^n$}
\put(3.05, 0.05){$\vecF$}
\put(2.20, 0.60){$\gamma$}
\put(3.70, 0.60){$\vecF\circ\gamma$}
\put(3.00, 0.80){\vector(-1,-1){.45}}
\put(3.20, 0.80){\vector(1,-1){.45}}
\put(2.70, 0.25){\vector(1,0){.90}}
\end{picture}

\beginex
\label{ex:pf_bean}
Combining examples \ref{ex:map_bean} and
\ref{ex:gamma_circle}, let
$$
	\vecF(x,y) = \colvectwo{x-y^3}{x^3+2y}.
$$
and
$$
	\gamma(t) = \colvectwo{\cos(t)}{\sin(t)}.
$$
[xxx matlab plots].
\mendex

[xxx xref fwd to functors]

[xxx xref fwd to pushforward of tangent vectors]

[xxx xref fwd to $DF$ and $F_*$ section; 1st xref to v.c. pushf/deriv section.]

%% ----------------------------------------------------------------
\subsubsection{Images of maps under functions: pullbacks}
\label{sec:vc_pullback}

Let $G: \R^n \to \R$ and $\vecF: \R^m \to \R^n$.  Then $G\circ\vecF$ is a
function from $\R^m$ to $\R$.  The function $G$ has been \bfidxtwo{pulled
back}{pullback} from $\R^n$ to $\R^m$.

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.0)
%\put(0,0){\framebox(6,1)}
\put(2.40, 0.70){$\R^m$}
\put(3.70, 0.70){$\R^n$}
\put(3.05, 0.10){$\R$}
\put(3.10, 0.85){$\vecF$}
\put(2.20, 0.40){$G\circ \vecF$}
\put(3.60, 0.40){$G$}
\put(2.70, 0.75){\vector(1,0){.90}}
\put(2.60, 0.70){\vector(1,-1){.45}}
\put(3.60, 0.70){\vector(-1,-1){.45}}
\end{picture}

\beginex
TBD.  Make it punchy; make sure it connects to other example(s) later.
\mendex

[xxx xref fwd to functors]

[xxx xref fwd to pullback of forms]

%% ----------------------------------------------------------------
\subsection{A gallery of curves and surfaces, part 1}
\label{sec:gallery1}

xxx:
\begin{itemize}
\item $\S^1$ ($x^2+y^2=1$)
\item $\S^2$ ($x^2+y^2+z^2=1$)
\item $\T^2$ in coordinates
\item Paraboloids
\item Hyperboloid of one sheet ($x^2+y^2-z^2=1$)
\item Hyperboloid of two sheets ($z^2-x^2-y^2=1$)
\item Surface of revolution for $g(y)$
\item \ldots search 534 HW and quals for others.
\end{itemize}

%% ================================================================
%% ================================================================
%% ================================================================
\newpage
\section{Preliminaries from vector calculus}
\label{sec:prelim_vc}

\begin{center}
\emph{Hobbits delight in such things, if they are accurate: they like to have
books filled with things that they already know, set out fair and square with
no contradictions.}
--- J.R.R. Tolkien\index{Tolkien, J.R.R.}, \emph{The Fellowship of the Ring}.
\end{center}

Here we review notation which is old-fashioned in two senses: (1) it is
19th-century notation, and (2) it is what we were all taught as undergraduates.
Our aim in this course is to replace, generalize, and extend such things, while
still having the symbols \emph{look} largely the same.  Thus, it is worth
taking the time to review what it is that we want to replace.

Also, I have found that much of the supposedly new material in the
geometry-topology course is really vector calculus, being viewed in a new
light.  Viewing new things in new ways requires the learner to make a larger
mental leap, so to ease that leap I will discuss old things in new ways for a
while.

% xxx point out that this is all *Euclidean*.

%% ----------------------------------------------------------------
\subsection{Goals for vector calculus}
\label{sec:vector_calc_goals}

xxx list out classes of problems solved.  xxx note that many of these
arise in physics.

\begin{itemize}
\item Rate of change of functions.
\item Linear approximation of functions.
\item Optimization.
\item Volume, mass, center of mass.
\item Path and surface integrals:  used for computing \bfidx{work}, etc.
\item Find the evolution of a system given specified initial/boundary
conditions: ordinary differential equations, partial differential equations.
In particular, configuration space of mechanical systems.
(E.g. one-holed torus vs. double pendulum.)
\item Understand how results change when different coordinate systems
are introduced.
\end{itemize}

%% ----------------------------------------------------------------
\newpage
\subsection{Differentiation}
\label{sec:vector_calc_diffn}

\begin{center}
\emph{
Certainly he who can digest a second or third fluxion need not, methinks, be
squeamish about any point in divinity.
}
--- George Berkeley\index{Berkeley, George}, 1734.
\end{center}

We will see that almost everything in this differentiation section involves the
\bfidx{chain rule}, which in turn is encapsulated in the \bfidx{Jacobian
matrix}.  The Jacobian matrix will continue to reappear [xxx xref to COV for
integrals, and elsewhere].

%% ----------------------------------------------------------------
\subsubsection{Derivatives of functions from $\R$ to $\R$}
\label{sec:deriv_R_to_R}

Recall from single-variable calculus that we use the derivative function,
evaluated at a point $x_0$, to form a linear approximation to $f$ at $x_0$.
This is nothing more than \bfidx{point-slope} form:
\begin{eqnarray*}
	y &=& y_0 + m (x-x_0) \\
	f(x) &\approx& f(x_0) + f'(x_0)(x-x_0).
\end{eqnarray*}

Note that the derivative itself is not a linear transformation.  It is a
function; evaluated at a point, it gives a slope.  Linear transformations send
zero to zero and this approximation need not: consider $f(x)=\sin(x)$ at
$x_0=2\pi$.  Such approximations are \emph{affine}: a linear transformation
with translation.  But if we rearrange terms, then we do have a linear
transformation:
$$
	f(x) - f(x_0) \approx f'(x_0)(x-x_0)
$$
i.e.
$$
	\Delta f \approx f'(x_0) \Delta x.
$$
[xxx two figures here, the latter with translated origin.]

[xxx more about coord charts don't have the origin at each point but
the tangent spaces do.  xref to position/velocity.]

\begin{defn}
\label{defn:scalar_crit_pt}
Let $f:\R \to \R$.  We say that $x$ is a \bfidx{critical point} of $f$ if
$f'(x)$ is zero or undefined.  If $x$ is a critical point for $f$, we say that
$y=f(x)$ is a \bfidx{critical value} for $f$.
\end{defn}

\beginex
Let $f(x)=x^2+2x+3$.  We compute $f'(x)=2x+2$.  This is a polynomial and so is
not undefined anywhere; it is zero at $x=-1$.  Therefore $-1$ is the only
critical point for $f$.  The critical value is $f(-1)=2$.
\mendex

See definition \ref{defn:regular_point} in section \ref{sec:regular_value}
for the more general definition.

\begin{rem}
\label{rem:scalar_crit_pt_vs_extrema}
Remember that relative extrema happen at critical points, but not necessarily
vice versa.  The canonical examples are $f(x)=x^2$ and $g(x)=x^3$.  Both have
critical points at $x=0$; $f$ has a minimum there but $g$ does not.
\end{rem}

%% ----------------------------------------------------------------
\subsubsection{Derivatives of paths}
\label{sec:path_deriv}

Paths are simply differentiated componentwise.
\begin{defn}
Let
$$
	\gamma(t) = \colvecthree{\kappa(t)}{\lambda(t)}{\mu(t))}
$$
be a path.  Then
$$
	\gamma'(t) = \colvecthree{\kappa'(t)}{\lambda'(t)}{\mu'(t))}.
$$
\end{defn}

\begin{notn}
\label{notn:D_gamma}
Looking ahead to section \ref{sec:deriv_Rm_to_Rn}, I will write
$$\gamma' = D\gamma.$$
Note that $\gamma$ is from $\R^1 \to \R^m$ and $D\gamma$ is an $m \times 1$
matrix.
\end{notn}

[xxx figure]

We form tangent-line approximations for paths, much as in the single-variable
case as described in section \ref{sec:deriv_R_to_R}:
$$
	\gamma(t) \approx \gamma(t_0) + \gamma'(t_0)(t-t_0).
$$

\beginex
\label{ex:gamma_prime_circle}
Continuing example \ref{ex:gamma_circle}, let
$$
	\gamma(t) = \colvectwo{\cos(t)}{\sin(t)}.
$$
Then
$$
	\gamma'(t) = \colvectwo{-\sin(t)}{\cos(t)}.
$$
At $t_0=\pi/2$, we have an equation for the tangent line $\ell(t)$ given by
\begin{eqnarray*}
	\gamma(t) &\approx& \gamma(t_0) + \gamma'(t_0)(t-t_0) \\
	&=& \colvectwo{\cos(\pi/2)}{\sin(\pi/2)}
		+ \colvectwo{-\sin(\pi/2)}{\cos(\pi/2)}(t-\pi/2) \\
	&=& \colvectwo{0}{1} + \colvectwo{-1}{0}(t-\pi/2) \\
	&=& \colvectwo{\pi/2-t}{1}.
\end{eqnarray*}
[xxx matlab figure.]
\mendex

If a path $\gamma(t)$ represents \bfidx{position} of an object, then the
derivative $\gamma'(t)$ represents its \bfidx{velocity}.  When the velocity is
multiplied by the object's mass, we have \bfidx{momentum} (physicists write
$\vecp = m \vecv$).  [xxx elaborate:  a physicist might in fact think of a
tangent bundle [xxx xref fwd] as a space of positions and momenta.]

[xxx mnem: gamma as car driving on curve; gamma prime as the headlight beam.]

[xxx note: we lose the linear-transformation part of it when we
put it to work for the tangent-line approximation.  The match is at $t=t_0$
not $t=0$.]

See definition \ref{defn:regular_point} in section \ref{sec:regular_value} for
the definition of critical point.  For paths from $\R$ into higher-dimensional
spaces, critical points are not interesting.  (The derivative can never
be surjective.)

%% ----------------------------------------------------------------
\subsubsection{Derivatives of functions from $\R^n$ to $\R$}
\label{sec:deriv_Rn_to_R}

\begin{notn}
For single-variable functions we write the derivative as
$$
	\frac{dG}{dx} \qqor G'(x);
$$
for multi-variable functions we write
$$
	\frac{\DG}{\Dx} \qqor
	\frac{\DG}{\Dx}(x,y,z) \qqor
	G_x(x,y,z).
$$
When we evaluate at a point, we write
$$
	\frac{dG}{dx}\pickpipe{x_0} \qqor
	\frac{dG}{dx}(x_0) \qqor
	G'(x_0)
$$
in the single-variable case and
$$
	\frac{\DG}{\Dx}\pickpipe{\colvecthree{x_0}{y_0}{z_0}} \qqor
	\frac{\DG}{\Dx}(x_0,y_0,z_0) \qqor
	G_x(x_0,y_0,z_0)
$$
in the multivariable case.
\end{notn}

\begin{intuition}
We can use partial derivatives to form a \bfidx{linear approximation} to a
scalar function:
$$
	G(x,y) \approx G(x_0,y_0) + G_x(x_0,y_0)(x-x_0) + G_y(x_0,y_0)(y-y_0).
$$
The best way I know to motivate this is by using single-variable linear
approximations, one variable at at time.  Starting at $(x_0,y_0)$ we can make a
line to $(x,y_0)$ and from there to $(x,y)$:
\begin{eqnarray*}
	G(x,y_0) &\approx& G(x_0,y_0) + G_x(x_0,y_0)(x-x_0) \\
	G(x,y) &\approx& G(x,y_0) + G_y(x,y_0)(y-y_0) \\
	&\approx& G(x_0,y_0) + G_x(x_0,y_0)(x-x_0) + G_y(x,y_0)(y-y_0)
\end{eqnarray*}
Alternatively, starting at $(x_0,y_0)$ we can make a line to $(x_0,y)$ and from
there to $(x,y)$:
\begin{eqnarray*}
	G(x_0,y) &\approx& G(x_0,y_0) + G_y(x_0,y_0)(y-y_0) \\
	G(x,y) &\approx& G(x_0,y) + G_x(x_0,y)(x-x_0) \\
	&\approx& G(x_0,y_0) + G_y(x_0,y_0)(y-y_0) + G_x(x_0,y)(x-x_0) \\
	&=& G(x_0,y_0) + G_x(x_0,y)(x-x_0) + G_y(x_0,y_0)(y-y_0).
\end{eqnarray*}
Comparing these two expressions, we see that the difference is
$$
	G_x(x_0,y_0)(x-x_0) + G_y(x,y_0)(y-y_0)
$$
vs.
$$
	G_x(x_0,y)(x-x_0) + G_y(x_0,y_0)(y-y_0).
$$
If $G$ is linear, then
$$
	G_x(x_0,y_0) = G_x(x_0,y) \qqand G_y(x_0,y_0) = G_y(x,y_0);
$$
if $G$ is approximately linear then we can replace one with the other
to obtain the desired result
\begin{eqnarray*}
	G(x,y) &\approx& G(x_0,y_0) + G_x(x_0,y_0)(x-x_0) + G_y(x_0,y_0)(y-y_0) \\
	&=& G(x_0,y_0) + G_x(x_0,y_0) \Delta x + G_y(x_0,y_0) \Delta y.
\end{eqnarray*}
A careful use of the word ``approximately'' requires a limit proof; see
\cite{bib:rudin}, \cite{bib:spi1}, or the differentiation section of any
graduate analysis text.
\end{intuition}

xxx figure

xxx now note it looks like a plane form.

Just as in section \ref{sec:deriv_R_to_R}, this approximation isn't a linear
transformation --- it only becomes one when we rearrange terms:
\begin{eqnarray*}
	f(x,y) - f(x_0,y_0)
		&\approx& f_x(x_0,y_0) \Delta x + f_y(x_0,y_0) \Delta y \\
	\Delta f &\approx&
		\rowvectwo{\fracDfx}{\fracDfy} \pickpipe{\colvectwo{x_0}{y_0}}
		\colvectwo{\Delta x}{\Delta y}.
\end{eqnarray*}
We can think of this product in two ways:
\begin{itemize}
\item As the dot product of two column vectors;
\item As the matrix product of a row vector by a column vector.
\end{itemize}
[xxx foreshadow ominously that this seemingly minor distinction leads into
covariance and contravariance.  numerous xrefs.]

\begin{defn}
Let $G: \R^n \to \R$.  Define the \bfidx{column vector}
$$
	\grad G = \begin{pmatrix}
		\frac{\DG}{\Dx_1}(x_1, \ldots, x_n) \\
		\vdots \\
		\frac{\DG}{\Dx_n}(x_1, \ldots, x_n) \\
	\end{pmatrix}.
$$
(This inverted triangle is pronounced ``grad'' or ``nabla''.
[xxx Anton ref w/ page number for the etymology.])
In particular, for $n=3$ we have
$$
	\grad G = \colvecthree{\DG/\Dx}{\DG/\Dy}{\DG/\Dz}.
$$
\end{defn}

Notice that we started with a vector-to-scalar function
$G: \R^n \to \R$.  Then we obtained a vector-to-vector function
$\grad G: \R^n \to \R^n$.  This suggests that the $\grad$ may be
thought of as a \bfidx{differential operator} $\grad$
which takes vector-to-scalar functions into vector-to-vector functions
[xxx xref to functors].

\begin{defn}
The \bfidx{gradient} is the following \bfidx{differential operator}, written as
a \bfidx{column vector}:
$$
	\grad = \colvecthree{\D/\Dx_1}{\vdots}{\D/\Dx_n}.
$$
For $n=3$ we have
$$
	\grad = \colvecthree{\D/\Dx}{\D/\Dy}{\D/\Dz}.
$$
\end{defn}

xxx figure: contour and quiver.

\begin{rem}
Since $\grad$ takes a scalar function $G$ into a vector field function
$\vecF=\grad G$, it is natural to ask if we can go the other way.  That is,
given $\vecF$, can we find $G$ such that $\grad G = \vecF$?  If so, we say that
$G$ is a \bfidx{potential function} for $\vecF$.  [xxx write and xref to
where we can answer that question.]
[xxx figures]
\end{rem}

\begin{notn}
\label{notn:grad_t_is_D}
Since the gradient of a scalar function is a column vector, its transpose is a
row vector.  Looking ahead to section \ref{sec:deriv_Rm_to_Rn}, and recalling
section \ref{sec:path_deriv}, I might write this as
$$
	\grad^t G = D G = \rowvecthree{\fracDGx}{\fracDGy}{\fracDGz}.
$$
Note that $G$ is from $\R^n \to \R^1$ and $DG$ is a $1 \times n$
matrix.  Then we have
$$
	\Delta f \approx DG \Delta \vecx.
$$
\end{notn}

xxx:
\begin{eqnarray*}
	f(x,y,z) - f(x_0,y_0,z_0) &\approx&
		\rowvecthree{\fracDfx}{\fracDfy}{\fracDfz}
			\pickpipe{\colvecthree{x_0}{y_0}{z_0}}
		\colvecthree{\Delta x}{\Delta y}{\Delta z}.
\end{eqnarray*}

[xxx dir'nal deriv, and grad as direction of greatest change?  figures.  this
leads to normals.]

\begin{defn}
\label{defn:fcn_crit_pt}
Let $G:\R^m \to \R$.  We say that $\vecq$ is a \bfidx{critical point} of $G$ if
$DG(\vecq)$ is zero or undefined.  If $\vecq$ is a critical point for $G$, we
say that $y=G(\vecq)$ is a \bfidx{critical value} for $G$.
\end{defn}

\begin{rem}
See definition \ref{defn:regular_point} in section \ref{sec:regular_value} for
the more general definition:  namely, for functions $\vecF:\R^m$ to $\R^n$, a
point $\vecq$ is critical if $D\vecF(\vecq)$ is not surjective.  Here, where
$n=1$, $DG$ is an $1 \times m$ matrix (which looks like a row vector).  For a
$1 \times m$ matrix not to be surjective means it must have rank less than 1.
But its rank can be at most 1.  For a $1 \times m$ matrix to have rank zero
means it must \emph{be} zero.  This is the same as the gradient (remember
$\grad^t G = DG$) being zero.
\end{rem}

\beginex
Let $G(x,y)=x^2-y^2+3$.  We compute $DG(x,y)=(2x,-2y)$.  This has polynomial
components and so is not undefined anywhere.  It is zero only when $2x=-2y=0$,
i.e. at the single point $x=0, y=0$.  The critical value is $G(0,0)=3$.
\mendex

\begin{rem}
Just as in the scalar case (see remark \ref{rem:scalar_crit_pt_vs_extrema}),
relative extrema occur at critical points but not necessarily vice versa.  For
the example just given, the point $x=0,y=0$ is a critical point, but it is a
saddle point rather than a relative minimum or relative maximum.  (If we had
defined $G(x,y)=x^2+y^2+3$ we would have had a minimum at the origin.)
\end{rem}

[xxx include figures here]

%% ----------------------------------------------------------------
\subsubsection{Derivatives of maps from $\R^m$ to $\R^n$}
\label{sec:deriv_Rm_to_Rn}

In section \ref{sec:maps_Rm_to_Rn} I wrote a vector-to-vector function as
$$
	\vecF(x,y,z) = \colvecthree{f(x,y,z)}{g(x,y,z)}{h(x,y,z)}.
$$
That is, a vector-to-vector function is simply a \emph{stack} of the component
vector-to-scalar functions.  To make a linear approximation for a
vector-to-vector function, then, we can simply apply the gradient as defined in
section \ref{sec:deriv_Rn_to_R} to each component function:
\begin{eqnarray*}
	f(x,y,z) - f(x_0,y_0,z_0) &\approx&
		\rowvecthree{\fracDfx}{\fracDfy}{\fracDfz}
			\pickpipe{\colvecthree{x_0}{y_0}{z_0}}
		\colvecthree{\Delta x}{\Delta y}{\Delta z} \\
	g(x,y,z) - g(x_0,y_0,z_0) &\approx&
		\rowvecthree{\fracDgx}{\fracDgy}{\fracDgz}
			\pickpipe{\colvecthree{x_0}{y_0}{z_0}}
		\colvecthree{\Delta x}{\Delta y}{\Delta z} \\
	h(x,y,z) - h(x_0,y_0,z_0) &\approx&
		\rowvecthree{\fracDhx}{\fracDhy}{\fracDhz}
			\pickpipe{\colvecthree{x_0}{y_0}{z_0}}
		\colvecthree{\Delta x}{\Delta y}{\Delta z}.
\end{eqnarray*}
But this is the same as saying
\begin{eqnarray*}
	\colvecthree
	{f(x,y,z) - f(x_0,y_0,z_0)}
	{g(x,y,z) - g(x_0,y_0,z_0)}
	{h(x,y,z) - h(x_0,y_0,z_0)}
		&\approx&
	\begin{pmatrix}
		\fracDfx & \fracDfy & \fracDfz \\
		\fracDgx & \fracDgy & \fracDgz \\
		\fracDhx & \fracDhy & \fracDhz \\
	\end{pmatrix}
			\pickpipe{\colvecthree{x_0}{y_0}{z_0}}
		\colvecthree{\Delta x}{\Delta y}{\Delta z}.
\end{eqnarray*}
Thus we have motivated the following definition.

\begin{defn}
\label{defn:jacobian}
Let $\vecF: \R^3 \to \R^3$ be
$$\vecF(x,y,z) =
\begin{pmatrix} f(x,y,z) \\ g(x,y,z) \\ h(x,y,z) \end{pmatrix}.$$
Then we define the \bfidx{Jacobian matrix} for $\vecF$ to be the matrix
with rows equal to the gradients of the component functions:
$$
D\vecF =
\begin{pmatrix}
	\fracDfx & \fracDfy & \fracDfz \\
	\fracDgx & \fracDgy & \fracDgz \\
	\fracDhx & \fracDhy & \fracDhz \\
\end{pmatrix}.
$$
In general, for $\vecF: \R^m  \to \R^m$ given by
$$\vecF(x_1, \ldots, x_m) =
\begin{pmatrix}
f_1(x_1, \ldots, x_m) \\ \vdots \\ f_m(x_1, \ldots, x_m)
\end{pmatrix}
$$
the Jacobian matrix has
entries $$\Df_i/\Dx_j.$$
\end{defn}

\begin{notn}
\label{notn:jac_multi_deriv}
In case we have a nameless function $\vecy=\vecy(\vecx)$, as in remark
\ref{rem:COV_func_nameless}, it is common to write the Jacobian as
either
$$
	\frac{\D(y_1,\ldots,y_n)}{\D(x_1,\ldots,x_m)}
$$
or
$$
	\frac{\D\vecy}{\D\vecx}.
$$
\end{notn}

\begin{rem}
Note that the Jacobian matrix, as presented, is a \bfidx{matrix-valued
function}.  Only when we evaluate it at a point, e.g.
$$
	D\vecF|_{\vecq},
$$
does it become a matrix.
\end{rem}

\beginex
\label{ex:jac_bean}
Let
$$
	\vecF(x,y) = \colvectwo{x-y^3}{x^3+2y}
$$
as in example \ref{ex:map_bean}.  Then
$$
	D\vecF(x,y) = \left(\begin{array}{cc}
		1 & -3y^2 \\
		3x^2 & 2 \\
	\end{array}\right).
$$
Evaluated at $x=0,y=1$, this is
$$
	D\vecF(0,1) = \left(\begin{array}{rr}
		1 & -3 \\
		0 &  2 \\
	\end{array}\right).
$$
We have
$$
	\vecF(0,1) = \colvectwo{-1}{2}
$$
so near $x_0=0$, $y_0=1$ we have
\begin{eqnarray*}
	\vecF(x,y) &\approx& \vecF(x_0,y_0) + D\vecF(x_0,y_0)
		\colvectwo{x-x_0}{y-y_0} \\
	%
	&\approx&
	\colvectwo{-1}{2}
	+
	\left(\begin{array}{rr}
		1 & -3 \\
		0 & 2 \\
	\end{array}\right)
	\colvectwo{x-0}{y-1}
	%
	= \colvectwo{x-3y+2}{2y}.
\end{eqnarray*}
[xxx figures: superimpose eucl grid on grid image]
\mendex

Note that $\gamma'=D\gamma$ and $\grad^tG=DG$ from notations \ref{notn:D_gamma}
and \ref{notn:grad_t_is_D} are both simply special cases of the Jacobian.
[xxx xref fwd to $DF$ section.]

%\begin{mnem}
%I know of two: the absurd \emph{Jac is stack}, and the
%more dignified \emph{the Jacobian
%is a matrix of partials}.
%\end{mnem}

See definition \ref{defn:regular_point} in section \ref{sec:regular_value} for
the definition of critical point.  The very definition requires linear algebra,
which is why we don't teach it at the vector-calculus level.

%% ----------------------------------------------------------------
\subsubsection{Derivatives for images of paths under maps}
\label{sec:deriv_image_path_map}

Let
$$
	\gamma(t) = \colvecthree{\kappa(t)}{\lambda(t)}{\mu(t)}
$$
be a path and let
$$
	\vecF(x,y,z) = \colvecthree{f(x,y,z)}{g(x,y,z)}{h(x,y,z)}.
$$
What is the derivative of the composition $\vecF\circ\gamma$?  Using the
\bfidx{chain rule}, we have
\begin{eqnarray*}
	\frac{d}{dt} \vecF(\gamma(t))
	&=& \begin{pmatrix}
		\frac{d}{dt} f(\kappa(t), \lambda(t), \mu(t)) \\
		\frac{d}{dt} g(\kappa(t), \lambda(t), \mu(t)) \\
		\frac{d}{dt} h(\kappa(t), \lambda(t), \mu(t)) \\
	\end{pmatrix} \\
	&&\\
	&=& \begin{pmatrix}
		\fracDfx(\kappa(t))\kappa'(t)+
		\fracDfy(\lambda(t))\lambda'(t)+
		\fracDfz(\mu(t))\mu'(t) \\
		%
		\fracDgx(\kappa(t))\kappa'(t)+
		\fracDgy(\lambda(t))\lambda'(t)+
		\fracDgz(\mu(t))\mu'(t) \\
		%
		\fracDhx(\kappa(t))\kappa'(t)+
		\fracDhy(\lambda(t))\lambda'(t)+
		\fracDhz(\mu(t))\mu'(t) \\
	\end{pmatrix} \\
	&&\\
	&=& \begin{pmatrix}
		\fracDfx & \fracDfy & \fracDfz \\
		\fracDgx & \fracDgy & \fracDgz \\
		\fracDhx & \fracDhy & \fracDhz \\
	\end{pmatrix}
	\pickpipe{\colvecthree{\kappa(t)}{\lambda(t)}{\mu(t)}}
	\begin{pmatrix}
		\kappa'(t) \\
		\lambda'(t) \\
		\mu'(t)
	\end{pmatrix}.
\end{eqnarray*}
That is to say,
$$
	\frac{d}{dt}(\vecF(\gamma(t)) = D\vecF(\gamma(t))\gamma'(t).
$$

We can use this rule to produce an equation for the tangent line to the image
of a path under a map.

\beginex
\label{ex:pf_deriv_bean}
Continuing examples
\ref{ex:gamma_circle},
\ref{ex:map_bean},
\ref{ex:pf_bean},
\ref{ex:gamma_prime_circle},
and
\ref{ex:jac_bean},
let
$$
	\vecF(x,y) = \colvectwo{x-y^3}{x^3+2y},
$$
$$
	\gamma(t) = \colvectwo{\cos(t)}{\sin(t)},
$$
and $t_0=\pi/2$.
Then from example \ref{ex:gamma_prime_circle} we have an equation for
the tangent line to the original path:
\begin{eqnarray*}
	\gamma(t) &\approx& \gamma(t_0) + \gamma'(t_0)(t-t_0) \\
	&\approx& \colvectwo{\pi/2-t}{1}.
\end{eqnarray*}
Then
\begin{eqnarray*}
	\vecF(\gamma(t))
	&\approx& \vecF(\gamma(t_0)) + D\vecF(\gamma(t_0))\gamma'(t_0)(t-t_0) \\
	&=& \vecF(0,1) + D\vecF(0,1)
		\colvectwo{-1}{0}
		(t-\pi/2) \\
	&=&
		\colvectwo{-1}{2}
		+
		\left(\begin{array}{rr}
			1 & -3 \\
			0 &  2 \\
		\end{array}\right)
		\colvectwo{-1}{0}
		(t-\pi/2) \\
	&=&
		\colvectwo{-1}{2}
		+
		\colvectwo{-1}{0}
		(t-\pi/2) \\
	&=&
		\colvectwo{\pi/2-1-t}{2}.
\end{eqnarray*}

[xxx matlab figures]
\mendex

Note:
\begin{itemize}
\item The not-necessarily-linear function $\vecF$ pushes forward the
not-necessarily-linear path $\gamma$.

\item The linear approximation $D\vecF$ pushes forward the tangent line defined
by $\gamma'$.
\end{itemize}

[xxx xref fwd to pushforward of tangent vectors]

%% ----------------------------------------------------------------
\subsubsection{Derivatives for images of maps under functions}

[xxx example]

[xxx xref fwd to pullback of forms]

%% ----------------------------------------------------------------
\subsubsection{The chain rule and the Jacobian}

xxx write out:

$D((B \circ A)(\vecq)) = DB(A(\vecq)) DA(\vecq)$.

xref back/fwd that prevs are special cases of this.

%% ----------------------------------------------------------------
\subsubsection{Divergence and curl}
\label{sec:div_and_curl}

xxx intuit first.  Follow \cite{bib:hhgm}

xxx the det formula seems quite arbitrary.  xref to $d$ section where it
falls out naturally.  also intuit \ldots .

\begin{defn}
Let $\vecF$ be a vector-to-vector function.  The \bfidx{curl} of $\vecF$,
written $\mathrm{curl}(\vecF)$ or $\nabla \times \vecF$, is the cross product
written as
$$ \nabla \times \vecF =
\left|
\begin{matrix}
\hat\vecx & \hat\vecy & \hat\vecz \\
\D/\Dx & \D/\Dy & \D/\Dz \\
f(x,y,z) & g(x,y,z) & h(x,y,z) \\
\end{matrix}
\right|
	=
\begin{pmatrix}
\Dh/\Dy - \Dg/\Dz \\
\Df/\Dz - \Dh/\Dx \\
\Dg/\Dx - \Df/\Dy
\end{pmatrix}
	=
\begin{pmatrix}
h_y - g_z \\
f_z - h_x \\
g_x - f_y
\end{pmatrix}
.$$
Note that the middle expression is just a silly mnemonic; it is not a
determinant in any formal sense.
\end{defn}

\begin{defn}
Let $\vecF$ be a vector-to-vector function.  The \bfidx{divergence} of $\vecF$,
sometimes written $\mathrm{div}(\vecF)$, is the dot product
$$ \nabla \cdot \vecF =
\begin{pmatrix} \D/\Dx \\ \D/\Dy \\ \D/\Dz \\ \end{pmatrix}
\cdot
\begin{pmatrix} f(x,y,z) \\ g(x,y,z) \\ h(x,y,z) \\ \end{pmatrix}
=
\Df/\Dx +
\Dg/\Dy +
\Dh/\Dz.
$$
\end{defn}

Note several things:
\begin{itemize}
\item The gradient takes vector-to-scalar functions to vector-to-vector
functions.
\item The curl takes vector-to-vector functions to vector-to-vector functions.
\item The divergence takes vector-to-vector functions to vector-to-scalar
functions.
\end{itemize}

Since the gradient of a vector-to-scalar function is a vector-to-vector
function, it makes sense to take the curl of it; also it makes sense to take
divergence of the curl of a vector-to-vector function.

\begin{prop}
As operators, $\grad \times \grad = 0$ and $\grad \cdot \grad \times = 0$.
\begin{proof}
For the first, let $G$ be a scalar function of three variables.
Then
\begin{eqnarray*}
	\grad \times \grad G
	&=& \grad \times \colvecthree{\DG/\Dx}{\DG/\Dy}{\DG/\Dz}
	= \left| \begin{matrix}
			\hat\vecx & \hat\vecy & \hat\vecz \\
			\D/\Dx & \D/\Dy & \D/\Dz \\
			\DG/\Dx & \DG/\Dy & \DG/\Dz \\
		\end{matrix}\right| \\
	&=& \begin{pmatrix}
			\D^2G/\Dy\Dz - \D^2G/\Dz\Dy \\
			\D^2G/\Dz\Dx - \D^2G/\Dx\Dz \\
			\D^2G/\Dx\Dy - \D^2G/\Dy\Dx
		\end{pmatrix}
	= 0.
\end{eqnarray*}
The last step is true because all our functions are assumed smooth: since the
first and second partials are continuous, the mixed partials are equal.

For the second, let $\vecF:\R^3 \to \R^3$.  Then
\begin{eqnarray*}
	\grad \cdot \grad \times \vecF
	&=& \grad \cdot \begin{pmatrix}
		\Dh/\Dy - \Dg/\Dz \\
		\Df/\Dz - \Dh/\Dx \\
		\Dg/\Dx - \Df/\Dy
	\end{pmatrix} \\
	%&=&
	%	   \D/\Dx(\Dh/\Dy - \Dg/\Dz) \\
	%	&+&\D/\Dy(\Df/\Dz - \Dh/\Dx) \\
	%	&+&\D/\Dz(\Dg/\Dx - \Df/\Dy) \\
	&=&
		   (\D^2h/\Dx\Dy - \D^2g/\Dx\Dz) \\
		&+&(\D^2f/\Dy\Dz - \D^2h/\Dy\Dx) \\
		&+&(\D^2g/\Dz\Dx - \D^2f/\Dz\Dy) \\
	&=& 0
\end{eqnarray*}
again due to equality of mixed partials.
\end{proof}
\end{prop}

Notes:
\begin{itemize}

\item Since $\nabla \times \vecF$ is vector-to-vector, it makes sense to
consider $\nabla \times \nabla \times \vecF$.  I am leaving this topic
untouched.  (See \cite{bib:hildebrand}, section 6.9.)

\item What do we do in higher dimensions (e.g.  phase spaces of complex
mechanical systems)?  The gradient and the divergence seem to make sense for
higher dimensions, but there is no clear notion here of how to extend the
concept of curl. [xxx xref to 2D and 4D $d^2=0$ complexes, when I write that.
Or to the $d$ section and let them do it themselves.]

\end{itemize}

%% ----------------------------------------------------------------
\newpage
\subsection{A first pass at tangent spaces}

[xxx s2tp]

circle.  level set and not.  two views of tangent line:
kernel of linear transformation.  show by example that we get the same
thing either way; question is how.

same with the two-sphere.  relate to plane forms \ldots.

main point: normal is perp to tangent.

%% ----------------------------------------------------------------
\newpage
\subsection{Integration}
\label{sec:vc_intgn}

\begin{center}
\emph{
But just as much as it is easy to find the differential of a given quantity, so
it is difficult to find the integral of a given differential. Moreover,
sometimes we cannot say with certainty whether the integral of a given quantity
can be found or not. 
}
--- Johann Bernoulli\index{Bernoulli, Johann} (1667-1748).
\end{center}

%% ----------------------------------------------------------------
\subsubsection{Integrals of paths}

If $\vecv(t)$ is an expression for the velocity of a particle, i.e.
$\vecv(t)=\gamma'(t)$ for some path $\gamma(t)$, then we can
recover $\gamma(t)$ via
$$
	\gamma(t) - \gamma(t_0) = \int_{t_0}^t \vecv(\tau)d\tau.
$$

\begin{defn}
The integral of a path is done componentwise:
if
$$
	\vecv(t) = \colvecthree{u(t)}{v(t)}{w(t)}
$$
then
$$
	\int_{t_0}^t \vecv(t) d\tau =
	\int_{t_0}^t\colvecthree{u(t)}{v(t)}{w(t)} d\tau
	= \begin{pmatrix}
		\int_{t_0}^t u(t) d\tau \\
		\int_{t_0}^t v(t) d\tau \\
		\int_{t_0}^t w(t) d\tau \\
	\end{pmatrix}
$$
\end{defn}

\beginex
Let
$$
	\vecv(t) = \colvectwo{-\sin(t)}{\cos(t)}
$$
and $t_0=0$.
Then
\begin{eqnarray*}
	\int_{t_0}^t \vecv(t) d\tau &=&
	\int_{0}^t\colvectwo{-\sin(t)}{\cos(t)} d\tau
	= \colvectwo
		{\int_{0}^t -\sin(t) d\tau}
		{\int_{0}^t \cos(t) d\tau} \\
	&=& \colvectwo{\cos(t)-1}{\sin(t)}
	= \colvectwo{\cos(t)}{\sin(t)} - \colvectwo{1}{0}
\end{eqnarray*}
which is $\gamma(t)-\gamma(0)$ for the path of example
\ref{ex:gamma_prime_circle}.
\mendex

%$\int_{a}^{b} \frac{df}{dx}dx = f(b)-f(a)$.

There is not much more to say here that we haven't already seen as
undergraduates; the graduate course is silent on this subject.

%% ----------------------------------------------------------------
\subsubsection{Integrals of scalar functions}

The integral of a scalar function $G(x)$, $G(x,y)$, or $G(x,y,z)$ is the
familiar area/volume under the curve:  if the circus grounds are on a region
$R$ of the $(x,y)$ plane and $G$ is an expression for the height of the circus
tent, then $\int_R G(x,y)\,dx\,dy$ is the volume enclosed by the tent.

[xxx figure]

Such integrals are easy to understand and (maybe) easy to compute.  Thus it is
perhaps no surprise that we have integration theorems which turn integrals of
vector-to-vector functions into integrals of scalar functions.  [xxx xref.]

%% ----------------------------------------------------------------
\subsubsection{Integrals of maps}

Line integrals: integral of a vector field over a path \ldots.

xxx

[xxx antideriv vs. solving differential equations?]

curl and div; xref fwd to the classical integration thms.  Point out the
interior/boundary pattern, including the FTC, and xref to DG stuff.

[xxx include $\vecF \cdot d\vecr$.]

\begin{defn}
Given a path
$$
	\gamma(t) = \colvectwo{x(t)}{y(t)}
$$
for $t\in [a,b]$ and a vector field
$$
	\vecF(x,y) = \colvectwo{f(x,y)}{g(x,y)}
$$
we write
\begin{eqnarray*}
	\int_\gamma f(x,y) dx + g(x,y) dy
		&=& \int_\gamma f(x,y) dx + \int_\gamma g(x,y) dy
\end{eqnarray*}
where
\begin{eqnarray*}
	\int_\gamma f(x,y)dx &=& \int_{t=a}^{t=b} f(x(t), y(t)) x'(t) dt
\end{eqnarray*}
and
\begin{eqnarray*}
	\int_\gamma g(x,y)dy &=& \int_{t=a}^{t=b} g(x(t), y(t)) y'(t) dt.
\end{eqnarray*}
\end{defn}

\begin{notn}
Writing
$$
	d\vecr = \colvectwo{dx}{dy}
$$
we have
$$
	\vecF \cdot d\vecr = f(x,y)dx + g(x,y)dy
$$
and so we write
$$
	\int_\gamma f(x,y) dx + g(x,y) dy = \int_\gamma \vecF\cdot d\vecr.
$$
\end{notn}

xxx rem about \bfidx{work} done by a \bfidx{force}, justifying the use of the
dot.

\begin{rem}
\label{rem:int_x_path_dy}
An expression of the form $$ \int_{x=0}^{x=1} f(x,y) dy $$ may appear shady.
However, in terms of the definition of the path integral, this is zero:  for
the parameterization path along the $x$ axis from 0 to 1, $y(t)$ is constant
and so $y'(t)$ is zero along that path.  Thus,
$$ \int_{x=0}^{x=1} f(x,y) dy = 0. $$
\end{rem}

%% ----------------------------------------------------------------
\newpage
\subsection{Change of coordinates}
\label{sec:vc_change_of_coords}

In this section we examine what happens to derivatives and integrals when we
change coordinates.  This foreshadows \bfidx{transition functions} on
manifolds, as discussed in section [xxx xref].

xxx xref to tensor stuff as well.  Several more xrefs \ldots.

%% ----------------------------------------------------------------
\subsubsection{Change of coordinates for derivatives}

Whether we change coordinates in the domain or the range of a function, we use
the chain rule as expected to compute the derivative in the new coordinates.

Let $\gamma(t):\R\to\R^m$ be a path.  Let $x=x(t)$ be a nameless
change-of-coordinate function, in the sense of remark
\ref{rem:COV_func_nameless}.  Then we get the derivative of the reparameterized
path using the \bfidx{chain rule}:
$$
	\frac{d}{dt}\,\gamma(x(t)) = \gamma'(x(t)) x'(t).
$$
Changing coordinates in the range is the same as section
\ref{sec:deriv_image_path_map}:
$$
	\frac{d}{dt}\, \vecF(\gamma(t)) =
		D\vecF(\gamma(t)) \gamma'(t).
$$

xxx note how paths change coordinates.  Use the jacobian.

xxx connect to pushforward.  The COC map is just an $\vecF$.

xxx note (cite frankel and who else?) that one can define a vector field as
\emph{anything} which changes coordinates in this way.

If $G(x,y)$ is a scalar function, with $x$ and $y$ reparameterized using $s$
and $t$, then the chain rule gives us
$$
	\frac{d}{ds}\,G(x(s,t),y(s,t))
	= \frac{\DG}{\Dx} \frac{\Dx}{\Ds}
	+ \frac{\DG}{\Dy} \frac{\Dy}{\Ds}
$$
etc.  Likewise for maps from $\R^m$ to $\R^n$.

%% ----------------------------------------------------------------
\subsubsection{Change of coordinates for single-variable integrals}

We all know how to do $u$-substitution, but let's take a fresh look at it.  We
can attach some new terminology which will help us in section [xxx xref].  I
will start with an absurdly simple example --- it contains (for me at least) a
few surprises.  Compute
$$
	\int_{x=0}^{x=1} x \,dx
	= \left[\frac{x^2}{2}\right]_{x=0}^{x=1} = \frac{1}{2}.
$$
A $u$-substitution is unnecessary but let's do one for the sake of discussion:
say, $u=2x$.  With reference to section [xxx xref], we are measuring the area
under the curve with a new ruler --- one which measures half-inches rather than
inches.  We expect the new ruler to overcount area, and so we expect to
compensate by dividing by four.  We have
$$
	u=2x,\quad du = 2\,dx
$$
so
$$
	\frac{du}{2} = dx
$$
and so
$$
	\int_{x=0}^{x=1} x \,dx =
	\int_{u=0}^{u=2} \frac{u}{2} \, \frac{du}{2} = 
	= \left[\frac{u^2}{8}\right]_{u=0}^{u=2} = \frac{1}{2}.
$$

I don't know about you, but when I do a $u$-substitution, I think of
$x$ as the ``old'' variable and $u$ as the ``new'' variable.  But let's
trace back through what we did in that example, using symbols this time.
We started with
$$
	\int_{x=a}^{x=b} f(x) \,dx.
$$
We chose a new variable as a function of the old,
$$
	u=u(x),
$$
then found
$$
	du = \frac{du}{dx}\,dx.
$$
But when we substituted in the integrand we wrote $x=x(u)$:
$$
	x=\frac{u}{2} \qand
	dx = \frac{dx}{du}\,du.
$$
That is, we mapped from $u$ to $x$ and obtained
$$
	\int_{x=a}^{x=b} f(x) \,dx =
	\int_{u=u(a)}^{u=u(b)} f(x(u)) \, \frac{dx}{du} \,du.
$$
Note that the integrand uses the function $x=x(u)$, while the limits of the
integral use the function $u=u(x)$.  In summary, when we transition from $u$
coordinates to $x$ coordinates, the behavior of the integrand
$f(x)\,dx$ looks reminiscent of the \bfidx{pullback} diagram in section
\ref{sec:vc_pullback}:

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.2)
%\put(0,0){\framebox(6,1.2)}
\put(1.70, 0.70){$f(x(u))\,\frac{dx}{du}\,du$}
\put(1.10, 0.40){$[x^{-1}(a),x^{-1}(b)]$}
\put(3.00, 0.90){$x(u)$}
\put(3.70, 0.70){$f(x)\,dx$}
\put(4.10, 0.40){$[a,b]$}
\put(3.05, 0.10){$\R$}
%\put(3.10, 0.85){$\vecF$}
%\put(2.20, 0.40){$G\circ \vecF$}
%\put(3.60, 0.40){$G$}
\put(2.70, 0.75){\vector(1,0){.90}}
\put(2.60, 0.70){\vector(1,-1){.45}}
\put(3.60, 0.70){\vector(-1,-1){.45}}
\end{picture}

We will revisit this picture in [xxx xref fwd. to transition functions,
$\omega=f(x)\,dx$ is a form, $[a,b]$ is a chain, integral is pairing,
$x^*(\omega) = \omega \circ x$ is a pullback of a form, etc. etc.]

%xxx notes to self on forms and chains::
%\begin{eqnarray*}
%	\omega &=& f(x)\,dx. \\
%	t &\mapsto& x(t). \\
%	x^*(\omega) &=& \omega \circ x. \\
%	&=& f(x(u))\,\frac{dx}{du}\,du. \\
%	c &\mapsto& x^{-1}(c).
%\end{eqnarray*}

\begin{center}* * *\end{center}

Here is another way to think of the change-of-coordinate formula for
single-variable integrals.  Again start with
$$
	\int_{x=a}^{x=b} f(x) \, dx.
$$

xxx picket figure (inkscape).

xxx form a Riemann sum and examine a single cell.

Old:  $x$ from $a$ to $b$.
Area:
$$
	(b-a)f(a).
$$

New:  $u$ from $c$ to $d$, where $a=x(c)$ and $b=x(d)$.
Area:
$$
	\left(\frac{b-a}{d-c}\right) \; (d-c) \; f(x(c))
$$
But
$$
	\frac{(b-a)}{(d-c)} = \frac{\Delta x}{\Delta u}
$$
so the area becomes
$$
	f(a) \;\Delta x
	= \frac{\Delta x}{\Delta u} \; (d-c) \; f(x(c))
	= \frac{\Delta x}{\Delta u} \; \Delta u \; f(x(c)).
$$
Returning to the limit we have
$$
	f(x) dx \approx f(x(u)) \;\frac{dx}{du} \; du.
$$

Result:
$$
	\int_{x=a}^{x=b} f(x) \, dx =
	\int_{u=x^{-1}(a)}^{u=x^{-1}(b)} f(x(u)) \,\frac{dx}{du}\, du.
$$

%% ----------------------------------------------------------------
\subsubsection{Change of coordinates for integrals of scalar functions}

For scalar functions of more than one variable, we encounter once again
the Jacobian matrix.

I want to show that
$$
	\int_R G(\vecx) \, d\vecx =
	\int_{\vecx^{-1}(R)} G(\vecx(\vecs))
		\left|\frac{\D\vecx}{\D\vecs}\right| d\vecs.
$$

Here is a picture of the function:

xxx tent figure here first.  Then:

We form a Riemann sum, as in [xxx xref], then reduce to consideration of a
single rectangle where $G$ is approximately linear.  We want to consider
what happens to the floor under this cell of $G$ in the new coordinates.
[xref back to section \ref{sec:coords_and_geom_det}.]  Then we have:

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.8)
%\put(0,0){\framebox(6,1.8)}

\put(0.50, 0.10){\vector(0,1){1.50}}
\put(0.30, 0.30){\vector(1,0){1.50}}

\put(1.60, 0.10){$s$}
\put(0.30, 1.40){$t$}
\put(1.25, 1.05){$\Delta s$}
\put(1.05, 1.35){$\Delta t$}
\put(0.70, 0.80){$(s_1,t_1)$}
\put(1.30, 0.80){$(s_2,t_1)$}
\put(0.70, 1.65){$(s_1,t_2)$}
\put(1.30, 1.65){$(s_2,t_2)$}
\put(1.00, 1.00){\line(1,0){0.50}}
\put(1.00, 1.50){\line(1,0){0.50}}
\put(1.00, 1.00){\line(0,1){0.50}}
\put(1.50, 1.00){\line(0,1){0.50}}
\thicklines
\put(1.00, 1.00){\vector(1,0){0.50}}
\put(1.00, 1.00){\vector(0,1){0.50}}
\thinlines


\put(2.50, 0.10){\vector(-1,3){0.50}}
\put(2.30, 0.30){\vector(4,1){1.50}}

\put(3.60, 0.40){$x$}
\put(1.85, 1.40){$y$}
\put(2.70, 0.80){$(x_{11},y_{11})$}
\put(3.30, 0.80){$(x_{21},y_{21})$}
\put(2.70, 1.65){$(x_{12},y_{12})$}
\put(3.30, 1.65){$(x_{22},y_{22})$}
\put(3.00, 1.00){\line(1,0){0.50}}
\put(3.00, 1.50){\line(1,0){0.50}}
\put(3.00, 1.00){\line(0,1){0.50}}
\put(3.50, 1.00){\line(0,1){0.50}}
\thicklines
\put(3.00, 1.00){\vector(1,0){0.50}}
\put(3.00, 1.00){\vector(0,1){0.50}}
\thinlines

\put(4.50, 0.10){\vector(0,1){1.50}}
\put(4.30, 0.30){\vector(1,0){1.50}}

\put(5.60, 0.10){$x$}
\put(4.30, 1.40){$y$}
\put(4.60, 0.80){$(x_{11},y_{11})$}
\put(5.20, 0.60){$(x_{21},y_{21})$}
\put(4.70, 1.65){$(x_{12},y_{12})$}
\put(5.30, 1.45){$(x_{22},y_{22})$}
\put(5.00, 1.00){\line(3,-1){0.50}}
\put(5.10, 1.40){\line(3,-1){0.50}}
\put(5.00, 1.00){\line(1,4){0.10}}
\put(5.50, 0.828){\line(1,4){0.10}}
\thicklines
\put(5.00, 1.00){\vector(3,-1){0.50}}
\put(5.00, 1.00){\vector(1,4){0.10}}
\thinlines

\end{picture}

As in section \ref{sec:coords_and_geom_det}, the area of the rectangle in the
original coordinates is the area of the parallelogram spanned by the
(perpendicular) vectors
$$
	\colvectwo{\Delta s}{0} \qand
	\colvectwo{0}{\Delta t}
$$
where
$$
	\Delta s = s_2 - s_1 \qqand \Delta t = t_2 - t_1.
$$
As in section \ref{sec:coords_and_geom_det}, this is the determinant
$$
	\det\begin{pmatrix}
		\Delta s & 0 \\
		0 & \Delta t \\
	\end{pmatrix}
$$
which simply the product
$$\Delta s \,\Delta t.$$
The question is, what is the area of the parallepiped
spanned by the images of $\Delta s$ and $\Delta t$, in $(x,y)$ coordinates?
For brevity, I will write
\begin{eqnarray*}
	x_{11} = x(s_1, t_1), && y_{11} = y(s_1, t_1), \\
	x_{12} = x(s_1, t_2), && y_{12} = y(s_1, t_2), \\
	x_{21} = x(s_2, t_1), && y_{21} = y(s_2, t_1), \qand \\
	x_{22} = x(s_2, t_2), && y_{22} = y(s_2, t_2).
\end{eqnarray*}
The images of the two edges in question are
$$
	\colvectwo{\Delta s}{0} \mapsto
	\colvectwo{x_{21}-x_{11}}{y_{21}-y_{11}}
	\qqand
	\colvectwo{0}{\Delta t} \mapsto
	\colvectwo{x_{12}-x_{11}}{y_{12}-y_{11}}.
$$
We have a nameless change-of-coordinate function $\vecx=\vecx(\vecs)$, with
Jacobian written as in notation \ref{notn:jac_multi_deriv} $\D\vecx/\D\vecs$.
We know from section \ref{sec:deriv_Rm_to_Rn} that this Jacobian
approximates differences:
$$
	\colvectwo{x_{21}-x_{11}}{y_{21}-y_{11}}
	\approx
	\begin{pmatrix}
		\Dx/\Ds & \Dx/\Dt \\
		\Dy/\Ds & \Dy/\Dt \\
	\end{pmatrix} \pickpipe{\colvectwo{s_0}{t_0}}
	\colvectwo{\Delta s}{0}
	= \colvectwo{\Dx/\Ds \Delta s}{\Dy/\Ds \Delta s}
$$
and
$$
	\colvectwo{x_{12}-x_{11}}{y_{12}-y_{11}}
	\approx
	\begin{pmatrix}
		\Dx/\Ds & \Dx/\Dt \\
		\Dy/\Ds & \Dy/\Dt \\
	\end{pmatrix} \pickpipe{\colvectwo{s_0}{t_0}}
	\colvectwo{0}{\Delta t}
	= \colvectwo{\Dx/\Dt \Delta t}{\Dy/\Dt \Delta t}.
$$
So, just as in section \ref{sec:coords_and_geom_det}, the area of the parallelogram
spanned by these two vectors is the determinant
$$
	\det\begin{pmatrix}
	\Dx/\Ds \Delta s & \Dy/\Ds \Delta s \\
	\Dx/\Dt \Delta t & \Dy/\Dt \Delta t \\
	\end{pmatrix}.
$$
But due to the multilinearity of the determinant, we can factor out $\Delta s$
from the first column and $\Delta t$ from the second to obtain
$$
	\Delta s \, \Delta t
	\det\begin{pmatrix}
	\Dx/\Ds & \Dy/\Ds \\
	\Dx/\Dt & \Dy/\Dt \\
	\end{pmatrix} \pickpipe{\colvectwo{s_0}{t_0}}
	%
	=
	%
	\Delta s \, \Delta t
	\left|\frac{\D\vecx}{\D\vecs}\right|.
$$
Then the ratio of the area of the image parallelogram to the area of
the original rectangle is
$$
	\frac{
		\Delta s \, \Delta t
		\left|\frac{\D\vecx}{\D\vecs}\right|
	}{
		\Delta s \, \Delta t
	}
	=
	\left|\frac{\D\vecx}{\D\vecs}\right|.
$$
Forming the Riemann sum of such cells and taking the limit as $\Delta s$ and
$\Delta t$ go to zero gives the result claimed at the top of this section.

Note that I have discussed this situation only for $G:\R^2\to\R$, for clarity
of presentation, but in fact nothing here is dependent on $n=2$.

%% ----------------------------------------------------------------
\newpage
\subsection{Integration theorems}
\label{sec:vc_intgn_thms}

Here we recall several theorems and constructions, stated here for ready
reference.  See \cite{bib:anton} or \cite{bib:hhgm} for more information.
These theorems will be proved here:  such questions occur from time to time on
the geometry qualifying exam.

We will see in section \ref{sec:stokes} that the generalized Stokes theorem,
which is the main point of the geometry course, subsumes all of the following
as special cases:

\begin{itemize}
\item The fundamental theorem of calculus (section \ref{sec:pf_ftc}).
\item Green's theorem (section \ref{sec:pf_green}).
\item Classical Stokes (section \ref{sec:pf_clstokes}).
\item Divergence theorem (section \ref{sec:pf_div}).
\item Cauchy's theorem (section \ref{sec:pf_cauchy_thm}).
\end{itemize}

There are two main points here:
\begin{itemize}
\item Most of the theorems in this section convert one kind of integral into
	another.
\item In particular, they follow the pattern of the fundamental theorem
	of calculus (section \ref{sec:pf_ftc}):  the integral of a derivative of a
	function over a region is equated to the integral of the original function
	over the boundary of the region.  We will discuss this situation in greater
	generality in section \ref{sec:duality_main_section}.
\end{itemize}

xref fwd: point out things we currently \emph{cannot} do.

%% ----------------------------------------------------------------
\subsubsection{Pieces for the integration theorems}

\begin{rem}
\label{rem:1D_deriv_approx}
Given a function $G:\R \to \R$, note that the left-hand approximation to
$G'(x)$ at $x_1$ is the same as the right-hand approximation to $G'(x)$ at
$x_2$, i.e.
$$
	G'(x_1) \approx \frac{G(x_2)-G(x_1)}{\Delta x} \approx G'(x_2)
$$
where $\Delta x = x_2 - x_1$:

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.3)
%\put(0,0){\framebox(6,1.3)}

\put(2.50, 0.10){\vector(0,1){1.00}}
\put(2.30, 0.20){\vector(1,0){1.25}}

\put(2.70, 0.20){\line(0,1){0.60}}
\put(3.40, 0.20){\line(0,1){0.78}}
\put(2.70, 0.80){\line(4,1){0.70}}
\put(2.70, 0.05){$x_1$}
\put(3.40, 0.05){$x_2$}
\put(2.60, 0.95){$G(x_1)$}
\put(3.30, 1.05){$G(x_2)$}

\end{picture}

This seems obvious, but it is one of the tricks to remember in the proof of
Green's theorem in section \ref{sec:pf_green}.
\end{rem}

\begin{rem}
\label{rem:trapezoid_1D_Riemann_cell}
I can re-use the same picture as in remark \ref{rem:1D_deriv_approx} for one
cell of a Riemman sum, using the \bfidx{trapezoid rule}:
$$
	\int_{x_1}^{x_2} f(x)\;dx \approx \frac{f(x_1)+f(x_2)}{2}\,\Delta x.
$$
\end{rem}

\begin{rem}
\label{rem:trapezoid_2D_Riemann_cell}
Likewise, we can use the trapezoid rule for a 2D cell:

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.3)
%\put(0,0){\framebox(6,1.3)}

\put(2.70, 0.20){\line(1,0){0.70}}
\put(2.70, 0.20){\line(0,1){0.60}}
\put(3.40, 0.20){\line(0,1){0.78}}
\put(2.70, 0.80){\line(4,1){0.70}}

\put(2.40, 0.30){\line(1,0){0.70}}
\put(2.40, 0.30){\line(0,1){0.60}}
\put(3.10, 0.30){\line(0,1){0.78}}
\put(2.40, 0.90){\line(4,1){0.70}}

\put(2.70, 0.20){\line(-3,1){0.30}}
\put(3.40, 0.20){\line(-3,1){0.30}}
\put(2.70, 0.80){\line(-3,1){0.30}}
\put(3.40, 0.98){\line(-3,1){0.30}}

\put(2.60, 0.05){$(x_1,y_1)$}
\put(3.70, 0.05){$(x_2,y_1)$}
\put(1.90, 0.25){$(x_1,y_2)$}
\put(3.50, 0.25){$(x_2,y_2)$}
\put(2.60, 1.10){$G$}
%\put(2.60, 0.95){$G(x_1)$}
%\put(3.30, 1.05){$G(x_2)$}

\end{picture}
\begin{eqnarray*}
	\int_{y_1}^{y_2} \int_{x_1}^{x_2} G(x,y)\;dx\;dy
	&=& \int_{y_1}^{y_2} \left[ \int_{x_1}^{x_2} G(x,y)\;dx \right] \;dy \\
	&\approx& \int_{y_1}^{y_2} \frac{G(x_1,y)+G(x_2,y)}{2} \;\Delta x \;dy \\
	&\approx& \frac{G(x_1,y_1)+G(x_2,y_1)+G(x_1,y_2)+G(x_2,y_2)}{4}
		\Delta x\; \Delta y.
\end{eqnarray*}
\end{rem}

\begin{notn}
On a cell with corners $(x_1,y_1)$, $(x_1,y_2)$, $(x_2,y_1)$, and $(x_2,y_2)$,
write $G_{11} = G(x_1, y_1)$ and so on.  Then we have the more streamlined
notation
$$
	\int_{y_1}^{y_2} \int_{x_1}^{x_2} G(x,y)\;dx\;dy
	\approx \frac{G_{11}+G_{12}+G_{21}+G_{22}}{4} \Delta x\; \Delta y.
$$
That is, we find the height of the cell by averaging the heights of the
four posts.
\end{notn}

\begin{rem}
\label{rem:trapezoid_3D_Riemann_cell}
Likewise, for a single cell of a 3D Riemann sum:
\begin{eqnarray*}
	\int_{z_1}^{z_2} \int_{y_1}^{y_2} \int_{x_1}^{x_2} G(x,y)\;dx\;dy\;dz
	\approx \frac{
	G_{111}+G_{112}+G_{121}+G_{122}+ G_{211}+G_{212}+G_{221}+G_{222}
	}{8} \;\Delta x\;\Delta y\;\Delta z.
\end{eqnarray*}
\end{rem}

%\begin{rem}
%xxx make this a definition.
%\cite{bib:hhgm} section 19.1:
%$\Delta\vec{A} = \vec n \Delta A$.
%\end{rem}

%\begin{rem}
%xxx elaborate:
%$\vecV = (P, Q, R)^t$.  Then $\vecV \cdot dx\;dy = R\;dx\;dy$.
%\end{rem}

%% ----------------------------------------------------------------
\subsubsection{The fundamental theorem of calculus}
\label{sec:pf_ftc}

\begin{thm}[Fundamental theorem of calculus]
\label{thm:first_fund}
If\index{fundamental theorem of calculus} $G$ is an antiderivative of $g$, then
$$ \int_a^b g(x) dx = G(b) - G(a).$$
\begin{proof}
Since $G$ is an antiderivative of $g$, $g$ is the derivative of $G$, i.e.
$g=G'$.  We form a Riemann sum for the left-hand side as
$$
	\int_a^b G'(x) dx \approx \sum_{k=0}^{n-1} G'(x_k) \Delta x_k
$$
where $a = x_0 < x_1 < \ldots < x_{n-1} < x_n = b$ and $\Delta x_k =
x_{k+1}-x_k$.  Then
$$
	G'(x_k) \approx \frac{G(x_{k+1})-G(x_k)}{\Delta x}
$$
so the sum becomes
$$
	\sum_{k=0}^{n-1} G'(x_k) \Delta x
	\approx
	\sum_{k=0}^{n-1} \frac{G(x_{k+1})-G(x_k)}{\Delta x} \Delta x
	=
	\sum_{k=0}^{n-1} G(x_{k+1})-G(x_k).
$$
But this is a \bfidx{telescoping sum}
$$
	G(x_1)-G(x_0)+
	G(x_2)-G(x_1)+
	G(x_3)-G(x_2)+
	\ldots+
	G(x_n)-G(x_{n-1})
	= G(x_n)-G(x_0)=G(b)-G(a).
$$

The lack of rigor here is in my use of the symbol $\approx$.
More precisely, one replaces $G'(x_k)$ with $G'(c_k)$ where $c_k$
is the point in the interval $(x_k,x_{k+1})$ whose existence is guaranteed
by the \bfidx{mean value theorem}, such that
$$
	G'(c_k) = \frac{G(x_{k+1})-G(x_k)}{\Delta x}.
$$
\end{proof}
\end{thm}

Regardless of the approach, the three key points are (1) \bfidx{linearization}
of $G$, (2) cancellation of the integral's numerator $\Delta x$'s by the
derivative's denominator $\Delta x$'s, and \bfidx{telescoping}.  These three
points all reappear in the proof of Green's theorem in section
\ref{sec:pf_green}.

[xxx note connected domain.  point out what happens if not.  xref/foreshadow.]

%% ----------------------------------------------------------------
\subsubsection{The second fundamental theorem of calculus}
\label{sec:pf_sftc}

\begin{thm}[Second fundamental theorem of calculus]
\label{thm:second_fund}
If\index{second fundamental theorem of calculus}
$$ F(x) = \int_a^x f(t) dt,$$
then
$$ F'(x) = f(x).$$
\begin{proof}[Proof heuristic]
The left-hand side is
\begin{eqnarray*}
	\frac{d}{dx}\, F(x) &=&
		\frac{d}{dx}\, \int_a^x f(t) \, dt \\
	&\approx& \frac{F(x+\Delta x)-F(x)}{\Delta x} \\
	&=& \frac{
		\int_a^{x+\Delta x} f(t)\, dt - \int_a^{x} f(t)\, dt
		}{\Delta x} \\
	&=& \frac{
		\int_x^{x+\Delta x} f(t)\, dt
		}{\Delta x} \\
	&\approx&
		\frac{f(x) \Delta x}{\Delta x} \\
	&=& f(x).
\end{eqnarray*}

[xxx make a picture w/ \bfidxtwo{linearized}{linearization}
extra strip at the right edge.]
\end{proof}
\end{thm}

%% ----------------------------------------------------------------
\subsubsection{Green's theorem}
\label{sec:pf_green}

Green's theorem:  Let $\sigma$ be a region of $\R^2$ with oriented
boundary $C$.  Let $f : \sigma \to \R$.
$$
\int_C f(x,y) dx + g(x,y) dy = \int\int_\sigma (\Dg /\Dx - \Df /\Dy) dA.
$$

\begin{thm}
Let $\vecF=(P,Q)$ [xxx stack vert] $: \R^2 \to \R^2$ with continuous partials.
Let $U$ be a simply connected [xxx define above and xref backward] region
of $\R^2$.  Let $C$ be a counterclockwise closed path in $U$, with
interior $D$.  Then
$$
	\oint_C P(x,y)\;dx + Q(x,y)\;dy =
	\int\int_D \left(\fracDtwo Q y - \fracDtwo P x \right)\;dx\;dy.
$$
\begin{proof}
xxx grids.  Mention \bfidx{linearization} and \bfidx{telescoping}.
xref back to proof of FTC.
Use $G_{12}$ notation.
\end{proof}
\end{thm}

%% ----------------------------------------------------------------
\subsubsection{Classical Stokes}
\label{sec:pf_clstokes}

\begin{thm}
Copy and paste into here.
\begin{proof}
{}[xxx goes here].
\end{proof}
\end{thm}

Classical Stokes:
Let $\sigma$ be a two-dimensional surface in $\R^3$ with oriented boundary $C$
which in turn has outward-pointing normal $\vecnhat$.
(To quote Ben Poletta, if $\sigma$ is a potato chip then $C$ is the peel on the
potato chip.)
Let $\vecF$ be a vector-to-vector function, i.e. $\vecF: \R^3 \to \R^3$,
defined on $\sigma$.
$$
	\int_C \vecF \cdot d\vecr
	= \int\int_\sigma (\nabla \times \vecF) \cdot \vecnhat \, dS.
$$
(Note that this reduces to Green's theorem when $\sigma$ is confined to the
plane,
since then
$$
	\vecF = \colvecthree{f}{g}{h},
	\qquad \vecnhat = \hat\vecz,
	\qqand d\vecr = dx + dy + dz.
$$
Alternatively, classical Stokes is the extension of Green's theorem when we
allow $R$ to levitate off the plane, and perhaps warp in the breeze.)

%% ----------------------------------------------------------------
\subsubsection{Divergence theorem}
\label{sec:pf_div}

\begin{thm}
Copy and paste into here.
\begin{proof}
Mimic proof of Green as much as possible.  Use $G_{122}$ notation.
Use $(\vecV \cdot \nhat) dA$ notation.
\end{proof}
\end{thm}

Divergence theorem:
Let $G$ be a solid region in $\R^3$ (e.g. a solid potato) with oriented
boundary $\sigma$ (the peel).  Let $\vecnhat$ be the outward-pointing normal.
Let $\vecF$ be a vector-to-vector function, i.e. $\vecF: \R^3 \to \R^3$, defined
on
$G$.
$$
	\int_\sigma (\vecF \cdot \vecnhat) \;dS =
	\int\int\int_G (\nabla \cdot \vecF) \;dV.
$$

%% ----------------------------------------------------------------
\subsubsection{Fundamental theorem of calculus for line integrals}
\label{sec:pf_ftc_line}

\begin{thm}
Goes here.
\begin{proof}
xxx Use Green's theorem / classical Stokes on closed loops to get path
independence.  Type up handwritten notes.
\end{proof}
\end{thm}

%% ----------------------------------------------------------------
\subsubsection{Cauchy's theorem}
\label{sec:pf_cauchy_thm}

\bfidx{Cauchy's theorem}

%% ----------------------------------------------------------------
\newpage
\subsection{Lagrange multipliers}
\label{sec:lagrange_multipliers}

\plainidx{Lagrange multipliers} are used to solve the following problem (stated
here for $\R^3$):  Maximimize the function $f(x,y,z)$, subject to the
constraint that $g(x,y,z) = 0$.

When I first took calculus, this seemed to be a quite arbitrary thing to want.
In differential geometry, though, it's the natural thing to use when
restricting a function from Euclidean space (e.g. $\R^3$) to a manifold (e.g.
$\S^2$), when that manifold is the level set of some function.  (E.g. $\S^2$ is
the subset of $\R^3$ subject to the constraint that $g(x,y,z) = x^2+y^2+z^2-1 =
0$.)

\begin{prop}[Anton's theorem 16.10.1]
Let $f, g:\R^3 \to \R$ with continuous first partials on some open set
containing the constraint surface $g(x,y,z) = 0$.  Further suppose that $g$ has
non-zero gradient on this surface.  If $f$ has a constrained relative extremum,
it occurs at a point $(x_0,y_0,z_0)$ where the gradients of $f$ and $g$ are
parallel, i.e.
$$
	\grad f(x_0, y_0, z_0) = \lambda \grad g(x_0, y_0, z_0)
$$
for some real number $\lambda$.
\end{prop}

\begin{rem}
Recall from remark \ref{rem:grad_F_is_normal} that $\grad g$ is a normal vector
to the surface defined by $g=0$.  Thus, the proposition is that if $f$ has an
extremum on the surface, it occurs when $f$'s gradient is normal to the
surface.
\end{rem}

\begin{rem}
Not all the points where $\grad f$ is parallel to $\grad g$ are necessarily
extrema for $f$ on $g$'s zero set.  We have the following:
\begin{itemize}
\item Solving $\grad f = \lambda \grad g$ will give you points at which
	the two gradients are parallel.
\item As explained in remark \ref{rem:lagr_mult_crit_pt} below, these will all
	be critical points.
\item Of the critical points, some may be extrema; others may be saddle points.
\item Remember that extrema of $f$ occur at critical points of $f$ or
	on the boundary (if any) of the intersection of the domains of $f$ and $g$.
\end{itemize}
\end{rem}

\beginex
Problem:  maximize $f(x,y,z) = y^2-z$ on $\S^2$, i.e. subject to the constraint
$g(x,y,z) = x^2+y^2+z^2-1 = 0$.

First compute the gradients:
$$
	\grad f = \colvecthree{0}{2y}{-1} \qqand
	\grad g = \colvecthree{2x}{2y}{2z}.
$$
Observe that $\grad g$ doesn't vanish on $\S^2$, since it only vanishes at
$(x,y,z) = (0,0,0)$ which is not a point on $\S^2$.  Thus the proposition
applies.  Next, find when $\grad f$ and $\grad g$ are parallel:
\begin{eqnarray*}
	&& \left\{\begin{matrix}
		0  &=& 2 \lambda x \\
		2y &=& 2 \lambda y \\
		-1 &=& 2 \lambda z \\
	\end{matrix}\right. \\
	&& \left\{\begin{matrix}
		\lambda x &=& 0\\
		(1-\lambda) y &=& 0\\
		\lambda z &=& -1/2\\
	\end{matrix}\right. \\
	&& \left\{\begin{matrix}
		\lambda = 0 &\qqor& x = 0 \\
		\lambda = 1 &\qqor& y = 0 \\
		\lambda z = -1/2\\
	\end{matrix}\right. \\
\end{eqnarray*}

\begin{itemize}
\item Case $\lambda=0$ and $\lambda=1$:  Absurd.

\item Case $\lambda=0$ and $y=0$:  The third equation cannot be satisfied.

\item Case $x=0$ and $\lambda=1$:  $x=0$ and $z=-1/2$, from which $y=\pm
\sqrt{1-x^2-z^2} = \pm\sqrt{3/4} = \pm \sqrt{3}/2$.

\item Case $x=0$ and $y=0$:  $z = \pm 1$, with $\lambda = \mp 1/2$.
\end{itemize}

This gives us four candidate extremum points for $f$ on $\S^2$:
$$
	\vvecthree{0}{\sqrt{3}/2}{-1/2},
	\vvecthree{0}{-\sqrt{3}/2}{-1/2},
	\vvecthree{0}{0}{1},
	\vvecthree{0}{0}{-1}.
$$
Evaluating $f$ at these four points yields, respectively,
$$
	\frac{5}{4}, \frac{5}{4}, -1, 1.
$$
Thus, the maxima of $f$ are at
$$
	\vvecthree{0}{\sqrt{3}/2}{-1/2} \qqand
	\vvecthree{0}{-\sqrt{3}/2}{-1/2},
$$
while the minimum of $f$ (which was not asked for) is at the north pole.
Note that $\S^2$ has no boundary so there are no extrema at boundaries.

[xxx incorporate Matlab figures from \texttt{y2z.m}.]
\mendex

\begin{rem}
\label{rem:lagr_mult_crit_pt}
It may appear at first blush that something is wrong.  Take the first point
(call it $\vecq$), for example:  If we evaluate $\grad f$ there, we don't get
zero:
$$
	\vecq          = \colvecthree{0}{\sqrt{3}/2}{-1/2}; \quad
	\grad f        = \colvecthree{0}{2y}{-1}; \quad
	\grad f(\vecq) = \colvecthree{0}{\sqrt{3}}{-1}.
$$
However, it is not the gradient of $f$ but rather the gradient of $f$
restricted to $\S^2$ which vanishes.  (We might write this $\grad^{\S^2} f$.)
How can we compute this restricted gradient?  We could do the following:
\begin{itemize}
\item Compute the tangent plane to the sphere at $\vecq$.
\item Use the projection operator (\ref{sec:proj_perp}) to project $\grad
	f(\vecq)$ down onto that plane.
\item Then see if that projected vector is zero.
\end{itemize}
Alternatively, we can simply recall [xxx xref bkwd] that the normal to the
sphere is given by $\grad g$.  The normal is perpendicular to the tangent
plane at each point, and we've found that $\grad f$ is parallel to $\grad g$,
i.e. normal to the tangent plane.  So we know that if we project $\grad
f(\vecq)$ down to the tangent plane at $\vecq$, we will get zero.  This sanity
check gives us reassurance that we've really found critical points using this
method.
\end{rem}

%cf. notes from discussion with Faris

%% ----------------------------------------------------------------
\newpage
\subsection{ODEs in one variable}
\label{sec:univt_ODEs}

Consult your favorite ODE text for full information, of course \ldots for
qualifier problems, the only types of single-variable ODEs you are likely to
need to solve are separable first-order ones, or linear homogeneous
second-order with constant coefficients.  That is, you don't need to re-learn a
thick volume worth of ODE techniques.

%Here is one example:
%\begin{eqnarray*}
%	\frac{dy}{dx} &=& x \\
%	dy &=& x\,dx \\
%	\int dy &=& \int x\,dx \\
%	y &=& x^2+C.
%\end{eqnarray*}
%Then use the initial conditions to determine $C$.

Here an example of a first-order equation:
\begin{eqnarray*}
	\frac{dy}{dx} &=& xy\\
	\frac{dy}{y} &=& dx \\
	\int \frac{dy}{y} &=& \int dx \\
	\ln(y) &=& x + C \\
	y &=& e^{x+C} = ke^x.
\end{eqnarray*}
Then use the initial conditions to determine $C$.

Here is the general technique for linear homogeneous second-order with constant
coefficients.  We start with
\begin{eqnarray*}
	a x'' + b x' + c &=& 0.
\end{eqnarray*}
Guess $x = e^{kt}$ for $k$ to be determined.  Then $x' = ke^{kt}$ and $x'' =
k^2 e^{kt}$.  Plug these back in and recall that $e^{kt}$ is non-vanishing:
\begin{eqnarray*}
	(a k^2 + bk + c) e^{kt} &=& 0 \\
	a k^2 + bk + c &=& 0 \\
	k &=& \frac{-b \pm \sqrt{b^2-4ac}}{2a}
\end{eqnarray*}
In the case there are two distinct $k$'s, say $k_1$ and $k_2$, write
$$
	x = A e^{k_1 t} + B e^{k_2 t}.
$$
Then apply the initial conditions and solve for $A$ and $B$.
If the quadratic formula gives one $k$, then use
$$
	x = A e^{k t} + B t e^{k t}.
$$

If $k$ is imaginary, then you can directly convert the complex exponentials to
sine and cosine using the identities in section \ref{sec:trig}.  Or, you can
directly take your solution to be in terms of sines and cosines.
This is shown by example:
\begin{eqnarray*}
	x'' + 4x &=& 0 \\
	k^2 + 4 &=& 0 \\
	k &=& \pm 2i \\
	x &=& A e^{2it} + B e^{-2it}.
\end{eqnarray*}
Now
$$
	\cos(2t) = \frac{e^{2it} + e^{-2it}}{2} \qqand
	\sin(2t) = \frac{e^{2it} - e^{-2it}}{2i}
$$
so
\begin{eqnarray*}
	x &=& A e^{2it} + B e^{-2it} \\
	&=& (A+iB) \frac{e^{2it} + B e^{-2it}}{2}
		\;+\; (A-iB) \frac{e^{2it} - B e^{-2it}}{2i} \\
	&=& C \cos(2t) \;+\; D \sin(2t).
\end{eqnarray*}
That is, since the constants are to be determined anyway, we can either write
down the solution as
$$
	x = A e^{2it} + B e^{-2it}
$$
or as
$$
	x = C \cos(2t) \;+\; D \sin(2t).
$$

If $k$ is real with distinct roots, you can again convert the exponentials to
sinh and cosh, or take the solution directly in terms of sinh and cosh as
follows:
\begin{eqnarray*}
	x'' - 4x &=& 0 \\
	k^2 - 4 &=& 0 \\
	k &=& \pm 2 \\
	x &=& A e^{2t} + B e^{-2t} \qquad \textrm{or} \\
	x &=& C \cosh(2t) + D \sinh(2t).
\end{eqnarray*}

If $k$ has distinct roots which aren't purely real or imaginary, proceed as follows:
\begin{eqnarray*}
	x'' - 4x' + 5x &=& 0 \\
	k^2 - 4k + 5 &=& 0 \\
	k &=& 2 \pm i \\
	x &=& A e^{2t}e^{it} + B e^{2t}e^{-it} \qquad \textrm{or} \\
	x &=& C e^{2t}\cos(t) + D e^{2t}\sin(t).
\end{eqnarray*}

%% ----------------------------------------------------------------
\subsection{ODEs in two variables}

\index{ordinary differential equations}
\index{ODEs}

As in previous sections, this material is review.  However, we want to be sure
that what we do later on, with new terminology, is consistent with what we
already know.

Systems of ODEs on the quals tend to be separable second-order.  One eliminates
variables, then uses techniques from section \ref{sec:univt_ODEs}.  This is best
illustrated by example.

In $\R^2$, consider the system of ODEs given by
$$
\begin{pmatrix} \dot x \\ \dot y \end{pmatrix}
=
\begin{pmatrix} -y \\ x \end{pmatrix}.
$$
Taking second derivatives, we get
$$
\begin{pmatrix} \ddot x \\ \ddot y \end{pmatrix}
=
\begin{pmatrix} -\dot y \\ \dot x \end{pmatrix}
=
\begin{pmatrix} -x \\ -y \end{pmatrix}.
$$
This gives two ODEs, each in one variable alone:
\begin{eqnarray*}
	\ddot x + x &=& 0 \\
	\ddot y + y &=& 0
\end{eqnarray*}
These have the solutions from section \ref{sec:univt_ODEs}, namely,
\begin{eqnarray*}
	x &=& a \cos t + b \sin t \\
	y &=& c \cos t + d \sin t.
\end{eqnarray*}
Putting initial conditions $(x(0), y(0)) = (x_0, y_0)$ gives
$$
	x(0) = a, \quad y(0) = c
$$
so we have
\begin{eqnarray*}
	x &=& x_0 \cos t + b \sin t \\
	y &=& y_0 \cos t + d \sin t.
\end{eqnarray*}
But these two equations are not independent; the original system was
\bfidx{coupled} in the sense that $\dot x$ depends on $y$ and vice versa:
$$
\begin{matrix}
	\dot x &=& -x_0 \sin t + b \cos t &=& -y &=& -y_0 \cos t + d \sin t\\
	\dot y &=& -y_0 \sin t + d \cos t &=&  x &=&  x_0 \cos t + b \sin t.
\end{matrix}
$$
from which
$$
\begin{matrix}
	\dot x &=& -x_0 \sin t + b \cos t &=& -y &=& -y_0 \cos t + d \sin t\\
	\dot y &=& -y_0 \sin t + d \cos t &=&  x &=&  x_0 \cos t + b \sin t.
\end{matrix}
$$
which gives
$$
	b = -y_0, \quad d = x_0
$$
and so the solution is
\begin{eqnarray*}
	x &=& x_0 \cos t - y_0 \sin t \\
	y &=& y_0 \cos t + x_0 \sin t
\end{eqnarray*}
or
$$
\begin{pmatrix} x \\ y \end{pmatrix}
=
\begin{pmatrix}
	\cos t & -\sin t \\
	\sin t &  \cos t \\
\end{pmatrix}
\begin{pmatrix} x_0 \\ y_0 \end{pmatrix}.
$$
Note that this matrix is the \bfidx{rotation matrix} for angle $t$.  That is,
any point $(x_0, y_0)$ will be revolved counterclockwise about the origin as
$t$ increases.  The origin itself stays fixed.

%xxx note $e^A$ for quick solutions.

%% ----------------------------------------------------------------
\newpage
\subsection{PDEs}

xxx TBD.  Lee makes a cryptic reference to $d\eta = \omega$ as a PDE \ldots.
Include some specific examples, with relevance.

\index{partial differential equations}
\index{PDEs}

%% ================================================================
\newpage
\subsection{Limitations of vector calculus}
\label{sec:vc_lim}

%% ----------------------------------------------------------------
\subsubsection{xxx 0}

FTC with disconnected domain.  Note this is no big deal:  it just
splits up.

%% ----------------------------------------------------------------
\subsubsection{xxx 1}

Cite and use Massey example:  non-existence of potential function,
given a 1-hole.

%% ----------------------------------------------------------------
\subsubsection{xxx 2}

Cite and use Massey example:  function with a 2-hole.

%% ================================================================
%% ================================================================
%% ================================================================
\newpage
\section{Preliminaries from topology and analysis}
\label{sec:prelim_ta}

\begin{center}
\emph{
Analysis does not set out to make pathological reactions impossible, but to
give the patient's ego freedom to decide one way or another.}
--- Sigmund Freud\index{Freud, Sigmund} (1856-1939).
\end{center}

xxx emph that much of this is \emph{used} in vector calculus, but not stated as
such.  Now, in graduate school, we use these things explicitly.

%% ----------------------------------------------------------------
\subsection{Connectedness?}
\label{sec:connectedness}

%% ----------------------------------------------------------------
\subsection{Compactness?}
\label{sec:compactness}

Warrants any separate discussion?

%% ----------------------------------------------------------------
\subsection{Homeomorphisms and diffeomorphisms}
\label{sec:homeom_diffeom}

Recall the following:

\begin{defn}
A function from one topological space to another is a \bfidx{homeomorphism} iff
it is a  bijection which is continuous with continuous inverse.
\end{defn}

This is the isomorphism in the category of topological spaces.  On Euclidean
spaces (more generally, on Banach spaces) we have the notion of
differentiability.

\begin{defn}
\label{defn:smooth}
A function from $\R^m$ to $\R^n$ is \plainidx{$C^k$} if it is $k$-times
differentiable.  A function which may be differentiated arbitrarily many times
is said to be \plainidx{$C^\infty$}, or \bfidx{smooth}.
\end{defn}

Most of the functions we deal with in this course are $C^\infty$.  This is a
convenience which frees us from having to track how many times a given function
is needs to be differentiable in each situation.

\begin{defn}
A function from $\R^m$ to $\R^n$ is a \bfidx{diffeomorphism} iff it is a
bijection which is smooth with smooth inverse.
\end{defn}

Note that this is distinct from a definition used in other contexts, namely,
that a diffeomorphism is a bijection from $\R^m$ to $\R^n$ which is
differentiable with differentiable inverse.  When need be, we may distinguish
these two concepts by saying that the latter is a $C^1$ diffeomorphism, whereas
the definition given above describes a $C^\infty$ diffeomorphism.

\beginex
The canonical example of a homeomorphism which is not a diffeomorphism is $f:\R
\to \R: x \mapsto x^3$, the inverse of which is not differentiable at 0.
\mendex

%% ----------------------------------------------------------------
\subsection{Implicit function theorem}

\begin{thm}[Implicit function theorem]
\label{thm:implicit_function}
Let\index{implicit function theorem} $E$ be an open subset of $\R^{m+n}$, and
let $\vecf:E \to \R^n$ be a vector-to-vector function which is continuously
differentiable, i.e. of type $C^2$.  Let
$$(\veca, \vecb) = (a_1, \ldots, a_m, b_1, \ldots, b_n) \in E$$
be such that
$$ \vecf(\veca, \vecb) = 0.$$
Let $D$ be the $n \times n$ submatrix of the Jacobian of $f$ given by
$\Df_i/\Dx_j$, for $i=1, \ldots, n$ and $j=m+1, \ldots, n$, evaluated at
$\vecb$.  If $\det(D) \ne 0$, then there exists a neighborhood $U$ of $\veca$
and a unique $C^2$ function $\vecg:U \to \R^n$ such that
$$
\vecg(\veca)=\vecb, \qie \vecf(\veca,\vecg(\veca))=0,
$$
and for all $\veca' \in U$,
$$
\vecf(\veca',\vecg(\veca')) = 0.
$$
That is, we can \emph{\bfidx{solve for}} the $\vecb$ variables at and near
$\veca$.
\begin{proof}
See any of the geometry texts in the bibliography.
% xxx include a sketch if it's informative.
\end{proof}
\end{thm}

\begin{rem}
The point is that, given a system of equations (for this course, usually a
single equation), we have an easy criterion for when we can solve for some
variables in terms of the others.  Note however that the implicit function
theorem ensures existence and uniqueness; actually \emph{finding} the function
$\vecg$ is another matter.
\end{rem}

\begin{rem}
The theorem, as stated, has the to-be-solved-for variable(s) written last.  In
practice, this may not be the case.  E.g. given a function $f(v,w,x,y,z):\R^5
\to \R^2$, we might want to solve for, say, $v$ and $x$.  In that case, we
would need to check the submatrix formed by the first and third columns of the
Jacobian of $f$.  Furthermore, we might not know ahead of time which variables
to solve for, until we apply the implicit function theorem to various
submatrices of the Jacobian.
\end{rem}

\begin{rem}
We make use of this theorem when we use \bfidx{graph coordinates} for a
manifold.  See section \ref{sec:coord_paramzn}.
\end{rem}

\beginex
\label{ex:impl_fcn_thm_S2}
Let
$$
	f:\R^3 \to \R: \colvecthree{x}{y}{z} \mapsto x^2+y^2+z^2-1.
$$
Here we have $m=2$ and $n=1$.
Then the kernel of $f$ is the sphere $\S^2$.
Consider the north pole, $(0,0,1)$, written as
$$
	(\veca,\vecb) = (a_1, a_2, b_1) = (x, y, z) = (0,0,1).
$$
The Jacobian of $f$ is
$$
	\begin{pmatrix}
		2x & 2y & 2z
	\end{pmatrix}
$$
which evaluated at the north pole is
$$
	\begin{pmatrix}
		0 & 0 & 1.
	\end{pmatrix}
$$
Now, there is only one $1 \times 1$ submatrix of this which is non-zero,
namely, the last.  So, there is a neighborhood $U$ of $(0,0)$ and a unique
function $g:U \to \R$ such that $z = g(x,y)$.  Here, it's clear what this is:
take
$$
	z = g(x,y) = \sqrt{1-x^2-y^2}.
$$
\mendex

Another example is in section \ref{sec:fe_1}.

%% ----------------------------------------------------------------
\subsection{Inverse function theorem}

\begin{thm}[Inverse function theorem]
\label{thm:inverse_function}
Let\index{inverse function theorem} $E$ be an open subset of $\R^{m}$, and
let $\vecf:E \to \R^m$ be a vector-to-vector function which is continuously
differentiable, i.e. of type $C^2$.  If the Jacobian of $\vecf$ is nonsingular
at a point $\vecq$ of $E$, then there exist open neighborhoods $U$ of
$\vecq$ and $V$ of $\vecf(\vecq)$ such that $\vecf$ is a diffeomorphsim
from $U$ into $V$.
\begin{proof}
See any of the geometry texts in the bibliography.
% xxx include a sketch if it's informative.
\end{proof}
\end{thm}

\begin{rem}
The point is that, even if such a function $\vecf$ is wildly non-linear, to
check for invertibility at a point it suffices to make the much simpler check
of the invertibility of the linearization of $\vecf$.
\end{rem}

% xxx example here.

%% ================================================================
%% ================================================================
%% ================================================================
\newpage
\section{Preliminaries from algebra}
\label{sec:prelim_alg}

\begin{center}
\emph{Structures are the weapons of the mathematician. }
--- Nicholas Bourbaki\index{Bourbaki, Nicholas}
\end{center}

%% ================================================================
\subsection{Algebraic axioms}
\label{sec:algebraic_axioms}

Here I gather together needed axioms.  This material is certainly review.  My
intention is not to insult the reader, but rather to ensure that terminology is
uniform throughout this paper.

%% ----------------------------------------------------------------
\subsubsection{Groups and semigroups}
\label{sec:groups}

Following \cite{bib:herstein}, I include closure as an axiom.  (This is worth
doing since it is often lack of closure which keeps a set from being a
subgroup.  In particular, the even integers are a subgroup of the integers with
respect to addition, whereas the odd integers fail to be closed under
addition.) Thus, I say that there are four group axioms, rather than the usual
three.  A \bfidx{group} $G$ is a set with a binary operation (written using
juxtaposition) satisfying the following four axioms:

\begin{itemize}
\item $G$ is \bfidx{closed} under the operation,
the operation is \bfidx{associative},
there is a (unique) \bfidx{identity},
and every element has a (unique) \bfidx{inverse}.
\item For an \bfidx{abelian group}, there is a fifth axiom: the operation is
\bfidx{commutative}.
\end{itemize}

\begin{defn}
A \bfidx{semigroup}, which will be referred to briefly in section
\ref{sec:free_groups}, lives up to the prefix \emph{semi} in that it is
\emph{halfway} to being a group:  it satisfies the first two of the four group
axioms.  Namely, it is closed under the semigroup operation, and that operation
is associative.
\end{defn}

\beginex
\label{ex:string_semigroup}
The canonical example of semigroup is strings of letters over some fixed
alphabet, with the operation being concatenation.  E.g. ``ab'' pasted together
with ``c'' is ``abc''.
\mendex

\begin{rem}
If we include the empty string as a letter, then we get an identity.  The
resulting structure is 3/4 of the way to being a group, and is called a
\bfidx{monoid}.  (Mnemonic: the Greek root \emph{monos} means single (or one),
and we often write an identity as 1 --- which is precisely what a monoid has
beyond a semigroup.)
\end{rem}

%% ----------------------------------------------------------------
\subsubsection{Normalizer and normal closure of a subgroup}

xxx re-do this with subsets, not subgroups.

\begin{defn}
Let $G$ be a group with subgroup $H$.  The \bfidx{normalizer} of $H$ in $G$,
written $N_G(H)$, or just $N(H)$, is the \emph{largest} subgroup of $G$ in
which $H$ is normal.
\end{defn}

\begin{defn}
Let $G$ be a group with subgroup $H$.  The \bfidx{normal closure} or
\bfidx{conjugate closure} of $H$ in $G$, written $\langle H \rangle_N$ or
$\overline H$, is the \emph{smallest} subgroup of $G$ in which $H$ is normal.
\end{defn}

\begin{rem}
In particular, if $H$ is already normal in $G$ (e.g. if $G$ is abelian, in
particular if $G$ is cyclic) then $N(H)$ is all of $G$.
\end{rem}

xxx examples

xxx How to actually \emph{compute} such things?

rmk: $H$ in center of $G$ implies $H$ normal in $G$.

[xxx xref fwd:  SvK uses normal closure; action of $\pi_1$ on fibers uses
normalizer.]

%% ----------------------------------------------------------------
\subsubsection{Group actions}

\begin{center}
\emph{
Vision without action is a daydream; action without vision is a nightmare.
} \\
--- Japanese proverb.
\end{center}

%\begin{defn}
%Let $G$ be a group and let $S$ be a set.  Let $e$ denote the identity
%element of $G$.  A map
%$
%	\mu: G \times S \to S,
%$
%written
%$
%	\mu(g, s) = g \cdot s
%$
%is said to be a \bfidx{left group action}\index{group action}\index{action,
%group} if the following two axioms (roughly speaking, identity and
%associativity) are satisfied:
%\begin{enumerate}
%\item For all $s \in S$, $e \cdot s = s,$ and
%\item For all $g_1, g_2 \in G$ and for all $s \in S$,
%	$ g_1 \cdot (g_2 \cdot s) = (g_1 g_2) \cdot s.$
%\end{enumerate}
%\end{defn}

xxx left actions most useful elsewhere, and probably are more familiar.  for
this course, though, we mostly need \emph{right} actions (xref fwd to $\pi_1$
acting on fibers).

xxx xref fwd to action of $\pi_1$ on fibers.

\begin{defn}
Let $G$ be a group and let $S$ be a set.  Let $e$ denote the identity
element of $G$.  A map
$
	\mu: S \times G \to S,
$
written
$
	\mu(s, g) = s \cdot g
$
or
$
	\mu(s, g) = s^g
$
is said to be a \bfidx{right group action}\index{group action}\index{action,
group} if the following two axioms (roughly speaking, identity and
associativity) are satisfied:
\begin{enumerate}
\item For all $s \in S$, $s \cdot e = s,$ and
\item For all $g_1, g_2 \in G$ and for all $s \in S$,
	$(s \cdot g_1) \cdot g_2 = s \cdot (g_1 g_2).$
\end{enumerate}
\end{defn}

standard examples.

xxx note the standard consequences and definitions (faithful and transitive).
Also the orbit-index formula.  xref to \cite{bib:grove}.

note that each $g \in G$ permutes $S$ (1-1), but that different group elements
need not give the \emph{same} permutation (not 1-1).

$G$ mod isotropy \emph{is} 1-1 and onto, if the action is transitive.

% isotropy sgr is same as stabilizer sgr?

note that the action axioms put some requirements on the structure (orbits,
total size) of what might otherwise seem an arbitrary set.

massey's ``$E$ is a homogenous $G$-space'' simply means that $G$ has
a transitive group action on $E$.

def auts:  homs are
set maps $\phi$ such that for all $g \in G$ and all $s \in S$,
$$
	\phi(s \cdot g) = \phi(s) \cdot g.
$$
write as commutative diagram w/ a nice picture.
then auts are bijective self-homs.

normalizer in group section?  own section?

%% ----------------------------------------------------------------
\subsubsection{Group action example}
\label{sec:boas}

\begin{center}
\emph{Suppose that you want to teach the ``cat'' concept to a very young child.
Do you explain that a cat is a relatively small, primarily carnivorous mammal
with retractible claws, a distinctive sonic output, etc.? I'll bet not. You
probably show the kid a lot of different cats, saying ``kitty'' each time,
until it gets the idea. To put it more generally, generalizations are best made
by abstraction from experience.} \\
--- R. P. Boas\index{Boas, R.P.}
\end{center}

make a big deal of this.  standard example.

$\Z$ acting on the vertices of an $n$-gon (take an octagon for specificity in
the picture).  (Kind of like $\mcD_n$ acting on the $n$-gon, but not a faithful
action, and no flips.)
\begin{itemize}
\item The group $G$ is $\Z \rho$ where $\rho$ is 1/$n$ of the way around.
\item The stability subgroup of a point is $H = n\Z$.
\item $G$ is cyclic hence abelian hence $H$ is normal.  So $N(H) = G$.
\item Auts of the action are $N(H)/H = \Z/n\Z$.
\end{itemize}

%% ----------------------------------------------------------------
\subsubsection{Rings}

A \bfidx{ring} is a set $R$ with addition and multiplication satisfying
the following 8 axioms:

\begin{itemize}
\item $R$ is an \bfidx{abelian group} with respect to addition.
\item $R$ is \bfidx{closed} with respect to multiplication.
\item Multiplication is \bfidx{associative}.
\item Addition and multiplication are related by left and right
\bfidx{distributivity}.
\end{itemize}

%% ----------------------------------------------------------------
\subsubsection{Fields}

A \bfidx{field} is a set $F$ with addition and multiplication satisfying
the following 11 axioms:

\begin{itemize}
\item $F$ is a \bfidx{ring}.
\item Multiplication is \bfidx{commutative}; there is a (unique)
\bfidx{multiplicative identity} for all non-zero elements; for each non-zero
element, there is a (unique) \bfidx{multiplicative inverse}.
\end{itemize}

Another way to clump the 11 field axioms is:

\begin{itemize}
\item $F$ is an \bfidx{abelian group} with respect to addition.
\item The non-zero elements of $F$ form an \bfidx{abelian group} with respect to
multiplication.
\item Addition and multiplication are related by left and right
\bfidx{distributivity}.
\end{itemize}

%% ----------------------------------------------------------------
\subsubsection{Modules and vector spaces}
\label{sec:modules}

There are 21 axioms for a vector space $V$ over a field $F$:

\begin{itemize}

\item $F$ is a \bfidx{field}.

\item $V$ is an \bfidx{abelian group} with respect to vector addition.

\item $V$ is \bfidx{closed} with respect to scalar multiplication (from either
side), scalar multiplication is \bfidx{associative} ($(ab)\vecv = a(b\vecv)$),
the field's \bfidx{multiplicative identity} $1_F$ satisfies $1_F\vecv = \vecv$
for all vectors, \bfidx{scalar distributivity} ($a(\vecu+\vecv) = a\vecu + a
\vecv$), \bfidx{vector distributivity} ($(a+b)\vecv = a\vecv + b\vecv$).

\end{itemize}

For a module $M$ over a ring $R$, here I mean a \bfidx{unital module}
\cite{bib:df}:  I require $R$ to have a multiplicative identity $1_R$ and I
require $1_R \vecv = \vecv$ for all $\vecv \in M$.  Otherwise, the axioms are
the same as those for a vector space over a field.  There are 19 axioms for
a unital module:

\begin{itemize}
\item $R$ is a \bfidx{ring} with multiplicative identity $1_R$.
\item $V$ is an \bfidx{abelian group} with respect to vector addition.
\item $V$ is \bfidx{closed} with respect to scalar multiplication (from either
side), scalar multiplication is \bfidx{associative} ($(ab)\vecv = a(b\vecv)$),
the field's \bfidx{multiplicative identity} $1_F$ satisfies $1_F\vecv = \vecv$
for all vectors, \bfidx{scalar distributivity} ($a(\vecu+\vecv) = a\vecu + a
\vecv$), \bfidx{vector distributivity} ($(a+b)\vecv = a\vecv + b\vecv$).
\end{itemize}

\begin{rem}
\label{rem:abn_gp_is_Z_module}
Any abelian group is already a $\Z$-module.  To see this, let $G$ be an abelian
group, and let $x,y \in G$.  When we write the group operation
multiplicatively, we write products as $xy$ and repeated multiplication as
$x^n$.  However, when we write the group operation additively, we write sums at
$x+y$ and repeated addition as $nx$.  Thus, scalar multiplication, by integer
scalar $n$, is nothing more than repeated addition $n$ times, using the
existing group operation.  You can check off the axioms for a $\Z$-module and
verify that they are in fact satisfied.
\end{rem}

%% ----------------------------------------------------------------
\subsubsection{Algebras}
\label{sec:algebras}

There are several ways to think of an algebra.  There are the same number of
axioms; the difference is only in the way in which those axioms are clumped
together.  One point of view is taken by \cite{bib:lang}.  Let $R$ be a
commutative ring with identity.  Then an $R$-algebra is a ring $A$ along with a
homomorphism $\phi:R \to A$ such that $\phi(R)$ is contained in the center of
$A$.  The other point of view, and the one I prefer here, is taken by
\cite{bib:hung}:

\begin{defn}
Let $R$ be a commutative ring with identity.  An unital $R$-module $A$ is an
\bfidx{algebra} if it is a ring and if $r(ab) = (ra)b = a(rb)$ for all $r \in
R$ and $a, b \in A$.
\end{defn}

When the base ring is a field, we can say the following (which is equivalent,
but phrased in more familiar terms of fields and vector spaces):

\begin{defn}
Let $F$ be a field.  An $F$-vector space $V$ is an \bfidx{algebra} if there is
a multiplication operation on $V$ making $V$ a \bfidx{ring}, and such that
$r(\vecu\vecv) = (r\vecu)\vecv = \vecu(r\vecv)$ for all $r \in F$ and $\vecu,
\vecv \in V$.
\end{defn}

There are three ideas here:

\begin{itemize}
\item[(i)] We already have a field.
\item[(ii)] We have a vector space and we want to be able to multiply vectors.
\item[(iii)] The old scalar multiplication and the new vector multiplication
should be compatible with each other.
\end{itemize}

There are 25 axioms for an algebra $A$ over a field $F$:

\begin{itemize}

\item $F$ is a field.

\item $A$ is a vector space.

\item $A$ is \bfidx{closed} under vector
multiplication, and the vector multiplication is \bfidx{associative}.

\item Vector multiplication \bfidxtwo{distributes}{distributivity} over vector
addition.

\item For all $r \in F$ and $\vecu, \vecv \in V$, $r(\vecu\vecv) =
(r\vecu)\vecv = \vecu(r\vecv)$.

\end{itemize}

Examples of algebras include:

\begin{itemize}

\item The ring of polynomials $\R[x]$ is an $\R$-algebra.  The vector-space
point of view is illustrated by the standard basis $\{1, x, x^2, x^3,
\ldots\}$.  When we add two polynomials, we add them coordinatewise, by
equating like terms.  Of course, we can multiply polynomials as well.

\item The ring $\R[x,y]$ of real bivariate polynomials
has basis $\{x^i y^j: i, j = 0, 1, 2, \ldots\}$.

\item The ring of polynomials $\R[x_1, \ldots, x_n]$ is likewise an
$\R$-algebra.

\item The ring of $m \times n$ matrices with entries in $\R$ is a vector space
over $\R$, but if $m=n$, then we can also multiply two such matrices.

\item Extension fields:  for example, take the base field $\Q$ and the
extension field $\Q(i)$, which is the rational complex numbers.  We can treat
$\Q(i)$ as a 2-dimensional vector space over $\Q$, with elements of the form
$(a, b)$ and $(c, d)$, but of course we can also multiply complex numbers using
$(a+bi)(c+di)$.

\item Likewise, the complex numbers $\C$ are just $\R^2$, along with a special
rule for multiplying vectors.  In a similar way, the real quaternions are an
$\R$-algebra, namely $\R^4$. Unlike for the complexes, though, the
vector-times-vector multiplication is not commutative.

\end{itemize}

Note that an algebra is a vector space with vector-vector multiplication.  So,
given an algebra we can obtain a vector space by forgetting about the
vector-vector multiplication.  In turn, a vector space is an abelian group with
multiplication by elements of a field.  So, given a vector space we can obtain
an abelian group by forgetting about the field.

%% ----------------------------------------------------------------
\subsubsection{Graded algebras}

\begin{defn*}
An abelian group $A$ is the \bfidx{direct sum} of abelian groups $B$ and $C$,
written $A = B \oplus C$, if:
\begin{enumerate}
\item $A = B + C$, i.e. for all $a \in A$, there are $b\in B$ and $c \in C$
	such that $a = b + c$.
\item $B \cap C = \{0\}$.
\end{enumerate}
\end{defn*}
Note in particular that this definition applies to rings and vector spaces,
which are simply abelian groups with some more structure added.

\begin{defn*}
A ring $S$ is \bfidx{graded} (\cite{bib:df}) if it is the direct sum of
additive subgroups $$S = S_0 \oplus S_1 \oplus S_2 \oplus \ldots$$ such that
for all $i, j \ge 0$, $S_i S_j \subseteq S_{i+j}$.  In particular, $s \in S_i$
and $t \in S_j$ implies $st \in S_{i+j}$.
\end{defn*}

\beginex
The ring of $\R[x]$ of real univariate polynomials is graded.  Each additive
subgroup $S_i$ consists of the zero polynomial along with all monomials with
degree $i$.
\mendex

The main examples for us (in fact the only reason for introducing graded rings
in this paper) are tensor algebras.  These are discussed in section
\ref{sec:tensor_algebras}.

%xxx def hom of graded algebras --- preserves grading.

%% ================================================================
\newpage
\subsection{Categories}
\label{sec:categories}

\begin{center}
\emph{Mathematics is the art of giving the same name to different things.}
--- Jules Henri Poincar\'e\index{Poincar\'e, Jules Henri} (1854-1912).
\end{center}

Like several sections of this paper, this section is certainly optional.  What
I want to do here is to give a thorough, Faris-esque grounding in fundamental
terminology.  The primary reason for presenting category theory in this paper
is as a mnemonic device:  for example, it helps us remember the difference
between $f_*$ and $f^*$ in differential geometry.  If you find that it only
adds confusion, you can certainly do differential geometry without it.

%% ----------------------------------------------------------------
\subsubsection{Definitions}

Category theory is sometimes (\cite{bib:lang}) referred to as \emphidx{abstract
nonsense}.  However, it is simply an abstraction of some very concrete things.
Namely, we are abstracting the concept of structures (groups, vector spaces,
topological spaces, etc.) and mappings (homomorphisms, linear transformations,
homeomorphisms, etc.).  One oddity is that the definitions below are
specifically designed to avoid mentioning the \emph{elements} of those
structures.

References are Appendix II of \cite{bib:df}, section I.11 of \cite{bib:lang},
and section I.7 and chapter X of \cite{bib:hung}.

\begin{defn}
A \bfidx{category} $\msC$ consists of the following (formally, it is an
ordered triple of the following three things):
\begin{itemize}
\item There is a class of \bfidx{objects}.

\item For each pair $(A, B)$ of objects, there is a set of
\bfidxtwo{morphisms}{morphism} or \bfidxtwo{arrows}{arrow} from $A$ to $B$.
This set is called a \bfidx{hom set} and is written $\Hom_\msC(A, B)$, or more
often simply $\Hom(A, B)$.  For an element $f$ of $\Hom(A, B)$, we often
write $f:A \to B$, or $A \stackrel{f}{\to} B$.

\item There is a \bfidxaux{law of composition}{composition}: if $f\in\Hom(A,
B)$ and $g\in\Hom(B, C)$ (which is to say $f:A\to B$ and $g:B \to C$), then
there is an element $g \circ f$ of $\Hom(A, C)$.
\end{itemize}

The objects, hom sets, and composition law must satisfy the following three
axioms:

\begin{itemize}

\item The hom sets are \bfidx{disjoint}:  If $A_1$ and $A_2$ are different, or
if $B_1$ and $B_2$ are different, then $\Hom(A_1, B_1)$ and $\Hom(A_2, B_2)$
have no morphisms in common.

\item Each object $A$ has an \bfidx{identity morphism}, written $1_A$.  Since
we are defining morphisms without reference to specific elements of an object,
we can't say something like ``\ldots such that $1_A(x)$ for all $x \in A$''.
Rather, an identity morphism on an object $A$ is characterized by the property
that for all other objects $B$, and for all $f:A \to B$ and all $g: B \to A$,
$$ f \circ 1_A = f \qand 1_A \circ g = g.$$

\item Composition of morphisms is \bfidx{associative}: if $f:A \to B$, $g:B \to
C$, and $h:C \to D$, then $h \circ(g \circ f) = (h\circ g) \circ f$.

\end{itemize}
\end{defn}

Category theory is carefully constructed to avoid references to elements of an
object.  This is a good and powerful thing.  However, for this paper, I use
categories mostly as a handy way to refer to concepts which apply equally well
to rings, vector spaces, modules, etc.  (This applies particularly to section
\ref{sec:sequences_and_complexes} on sequences of complexes.)  I always
\emph{do} have in mind categories which \emph{are} sets, and which \emph{do} at
least have a zero element.

\begin{defn}
For purposes of this paper, a \bfidx{concrete category} has objects which are
sets, each containing a zero element; the morphisms are homomorphisms.
\end{defn}

%xxx optional:  def monom, epim, endom, autom categorically

%xxx optional:  def monom, epim, endom, autom categorically

%% ----------------------------------------------------------------
\subsubsection{Examples of categories}

Some familiar examples of categories are:
\begin{itemize}
\item Objects are sets and morphisms are arbitrary functions.
\item Objects are groups and morphisms are group homomorphisms.
\item Objects are rings and morphisms are ring homomorphisms.
\item Objects are vector spaces and morphisms are linear transformations.
\end{itemize}

\beginex
A perhaps less familiar example is a partially ordered set.  Let $P$ be a
partially ordered set, with operation $\cle$.  Then the poset $P$ is itself the
entire category.  Objects of $P$ are the elements of $P$.  Morphisms of $P$ are
quite literally arrows:  there is an arrow from $a$ to $b$ if $a \cle b$.
Composition of arrows is nothing more than the transitivity property of posets.
The identity morphisms are provided by the reflexivity property of posets,
namely, $a \cle a$ for all $a \in P$.
\mendex

Here is a specific instance of that example:  the divisibility lattice on
factors of 12.  Here I am not showing all the arrows (morphisms):  (1) I am
omitting the arrows which may be obtained by composition, e.g. the arrow from 2
to 12 since $2 \mid 4$ and $4 \mid 12$.  (2) I am omitting arrows from objects
(integers) to themselves, e.g. $2\mid 2$.  I sometimes refer to these, the
identity morphisms which objects in any category must possess, as
\bfidx{self-loops}:  if we draw arrows as a graph, the identity morphisms
appear as loops from each object to itself.

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.4)
%\put(0,0){\framebox(6,1.4)}

\put(3.00, 0.05){1}
\put(2.50, 0.45){2}
\put(3.50, 0.45){3}
\put(2.50, 0.85){4}
\put(3.50, 0.85){6}
\put(2.95, 1.25){12}

\put(2.55, 0.60){\vector(0,1){.20}}
\put(3.55, 0.60){\vector(0,1){.20}}

\put(2.60, 0.60){\vector(4,1){.85}}

\put(2.90, 0.15){\vector(-1,1){.30}}
\put(3.15, 0.15){\vector( 1,1){.30}}

\put(2.60, 0.95){\vector( 1,1){.30}}
\put(3.45, 0.95){\vector(-1,1){.30}}
\end{picture}

\beginex
Another example is a single group $G$.  There is a single object which is all
of $G$; the hom set $\Hom_G(G, G)$ is all of $G$.  That is, the morphisms are
the elements of $G$.  The composition of two morphisms $x$ and $y$ is the usual
product $xy$.  The identity morphism is the identity element of $G$.
\mendex

%% ----------------------------------------------------------------
\subsubsection{Functors}
\label{sec:functors}

A \bfidx{functor} is a map between categories which takes objects to objects
and morphisms to morphisms.

\begin{defn}
Let $\msC$ and $\msD$ be categories.
A \bfidx{covariant functor} $T$ is a map from $\msC$ to $\msD$ such that:
\begin{itemize}

\item $T$ takes objects of $\msC$ to objects of $\msD$.

\item If $f: A \to B$ in $\msC$, then $T(f) : T(A) \to T(B)$ in $\msD$.
(This is the covariance property:  arrows point the same way.)

	\setlength{\unitlength}{1.0in}
	\begin{picture}(6,1.0)
	%\put(0,0){\framebox(6,1.0)}

	\put(2.00, 0.85){$A$}
	\put(2.00, 0.15){$B$}
	\put(2.20, 0.50){$f$}
	\put(2.05, 0.80){\vector(0,-1){.50}}

	\thicklines
	\put(2.50, 0.50){\vector(1,0){1.20}}
	\thinlines
	\put(3.00, 0.60){$T$}

	\put(4.00, 0.85){$T(A)$}
	\put(4.00, 0.15){$T(B)$}
	\put(4.30, 0.50){$T(f)$}
	\put(4.15, 0.80){\vector(0,-1){.50}}
	\end{picture}

\item Identity morphisms map to identity morphisms: for all objects $A$ of
$\msC$, $T(1_A) = 1_{T(A)}$.  A self-loop in $\msC$ must map to a self-loop in
$\msD$.

\item  Composition is respected:  $T(g \circ f) = T(g) \circ T(f)$.

\end{itemize}
\end{defn}

\begin{defn}
A \bfidx{contravariant functor} $T$ is almost the same, except for the direction
of arrows.  It is a map from $\msC$ to $\msD$ such that:
\begin{itemize}

\item $T$ takes objects of $\msC$ to objects of $\msD$.

\item If $f: A \to B$ in $\msC$, then $T(f) : T(B) \to T(A)$ in $\msD$.
(This is the contravariance property:  arrows point the other way.)

	\setlength{\unitlength}{1.0in}
	\begin{picture}(6,1.0)
	%\put(0,0){\framebox(6,1.0)}

	\put(2.00, 0.85){$A$}
	\put(2.00, 0.15){$B$}
	\put(2.20, 0.50){$f$}
	\put(2.05, 0.80){\vector(0,-1){.50}}

	\thicklines
	\put(2.50, 0.50){\vector(1,0){1.20}}
	\thinlines
	\put(3.00, 0.60){$T$}

	\put(4.00, 0.85){$T(A)$}
	\put(4.00, 0.15){$T(B)$}
	\put(4.30, 0.50){$T(f)$}
	\put(4.15, 0.30){\vector(0,1){.50}}
	\end{picture}

\item Identity morphisms map to identity morphisms: for all objects $A$ of
$\msC$, $T(1_A) = 1_{T(A)}$.  A self-loop in $\msC$ must map to a self-loop in
$\msD$.

\item Composition is respected (with arrows reversed):  $T(g \circ f) = T(f)
\circ T(g)$.

\end{itemize}
\end{defn}

\begin{notn}
\label{notn:functor_variance}
For a given functor $T$ and a morphism $f$, sometimes one writes \index{$f_*$}
\index{$f^*$}
\begin{eqnarray*}
	f_* &=& T(f), \quad \textrm{if $T$ is a covariant functor;}  \\
	f^* &=& T(f), \quad \textrm{if $T$ is a contravariant functor.}
\end{eqnarray*}
\end{notn}

Some simple examples are as follows.

\beginex
The \bfidx{identity functor} on any category, sending objects to
themselves and morphisms to themselves, is covariant.
\mendex

\beginex
The \bfidx{forgetful functor} is best illustrated by example.  Take $\msC$ to
be the category of groups, where morphisms are group homomorphisms, and take
$\msD$ to be the category of sets, where morphisms are arbitrary functions.
For any groups $G$ and $H$ and any homomorphism $\phi: G \to H$, send $G$ and
$H$ to the sets of elements of $G$ and $H$, respectively.  Have $\phi$ be the
same map sending elements of $G$ to elements of $H$, and simply forget the
homomorphism ploperty.
We also have a forgetful functor from vector spaces to abelian groups, which
forgets about scalar multiplication.
The forgetful functor is covariant.
\mendex

\beginex Let the category $A$ be the poset consisting of the elements 1, 2,
and 3 as shown.  There are arrows (morphisms) from 1 to 3 and 2 to 3, along
with the self-loops from 1 to 1, 2 to 2, and 3 to 3.  Next to that category in
the figure are two possible images of the category.

The first, say a functor named $f$, sends 3 to $c$ and collapses 1 and 2 into
$a$.  This functor is covariant since the arrows $1 \to 3$ and $2 \to 3$ are
mapped to the (identical) arrows $f(1) \to f(3)$ and $f(2) \to f(3)$, which are
both $a \to c$.

The second, say a functor named $g$, sends 1 to $a$ and collapses 2 and 3 into
$c$.  This functor is contravariant since the arrow $1 \to 3$ is mapped to
$g(3) \to g(1)$, which is $b \to a$, while the other non-identity arrow $2 \to
3$ is mapped to the self-loop $b \to b$.

\setlength{\unitlength}{1.0in}
\begin{picture}(6,0.8)
%\put(0,0){\framebox(6,0.8)}

\put(0.80, 0.35){$\mathbf{A}$}

\put(1.00, 0.15){$1$}
\put(1.90, 0.15){$2$}
\put(1.45, 0.55){$3$}
\put(1.10, 0.25){\vector( 1,1){.30}}
\put(1.85, 0.25){\vector(-1,1){.30}}

\put(2.30, 0.35){$\mathbf{B}$}

\put(3.00, 0.15){$a$}
\put(3.00, 0.55){$c$}
\put(3.02, 0.30){\vector( 0,1){.20}}

\put(3.80, 0.35){$\mathbf{C}$}

\put(4.00, 0.15){$a$}
\put(4.45, 0.55){$b$}
\put(4.40, 0.55){\vector(-1,-1){.30}}

\end{picture}
\mendex

%xxx More examples will be seen in subsequent sections, including the next
%section.

%% ----------------------------------------------------------------
\subsubsection{Hom functors}
\label{sec:hom_functors}

[xxx perhaps exclude this section]

\emph{Note: Perhaps this section introduces more abstraction than is
worthwhile.}

Here\index{hom functor} I want to formalize a familiar concept.  This will be
revisited in sections \ref{sec:vs_pullback} and \ref{sec:t_pullback}.  First, I
will consider vector spaces as a motivating example, followed by categories in
general.

Fix a vector space $V$.  For any other vector space $W$, the hom set
$\Hom(V,W)$ is what we call $\mcL(V,W)$, namely, the set of all linear
transformations from $V$ to $W$.  If we have a linear transformation $f$ from
vector spaces $W_1$ to $W_2$, and a linear transformation $g$ from $V$ to
$W_1$, then we can compose $f$ and $g$ to get a linear transformation $f \circ
g$ from $V$ to $W_2$:

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.0)
%\put(0,0){\framebox(6,1)}
\put(2.50, 0.70){$V$}
\put(3.70, 0.70){$W_1$}
\put(3.70, 0.10){$W_2$}
\put(3.10, 0.85){$g$}
\put(2.60, 0.30){$f \circ g$}
\put(3.85, 0.40){$f$}
\put(2.70, 0.75){\vector(1,0){.90}}
\put(2.70, 0.70){\vector(2,-1){.90}}
\put(3.75, 0.65){\vector(0,-1){.40}}
\end{picture}

We can use this to make a functor as follows:
\begin{itemize}

\item It will be a functor from the category of vector spaces to the category of
sets.  We have to say what it does to objects (vector spaces) and morphisms
(linear transformations).

\item A vector space $W$ is sent to $\mcL(V, W)$.  That is, a given $W$ is sent
to the collection of all linear transformations from the fixed vector space $V$
to that $W$.

\item A linear transformation $f:W_1 \to W_2$ is sent to a function
which converts a linear transformation $g:V \to W_1$ into
$f\circ g: V \to W_2$.  That is,
$$
(f: W_1 \to W_2) \mapsto
f_*: \mcL(V,W_1) \to \mcL(V,W_2)
\quad\textrm{where}\quad
f_*(g:V \to W_1) = (f\circ g: V \to W_2)
$$

\end{itemize}

This functor is written $$\Hom(V, \uscore).$$ It looks like this:

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.0)
%\put(0,0){\framebox(6,1)}
\put(1.70, 0.70){$W_1$}
\put(1.70, 0.10){$W_2$}
\put(1.85, 0.40){$f$}
\put(1.75, 0.65){\vector(0,-1){.40}}

\put(2.55, 0.55){$\Hom(V, \uscore)$}
\put(2.25, 0.45){\vector(1,0){1.10}}

\put(3.70, 0.70){$\mcL(V, W_1)$}
\put(3.70, 0.10){$\mcL(V, W_2)$}
\put(3.85, 0.40){$f_*$}
\put(3.75, 0.65){\vector(0,-1){.40}}
\end{picture}

Since it sends $W_1$ to $\mcL(V, W_1)$, $W_2$ to $\mcL(V, W_2)$, and $f:W_1 \to
W_2$ to $f_* : \mcL(V, W_1) \to \mcL(V, W_2)$, i.e. it preserves the direction
of the arrows, $\Hom(V, \uscore)$ is said to be a \bfidx{covariant hom
functor}.

\begin{center}
* * *
\end{center}

Since we have a covariant hom functor, it is natural to ask about a
contravariant hom functor.  Now fix a vector space $W$.  If we have a linear
transformation $f$ from vector spaces $V_1$ to $V_2$, and a linear
transformation $g$ from $V_2$ to $W$, then we can compose $g$ and $f$ to get a
linear transformation $g \circ f$ from $V_1$ to $W$:

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.0)
%\put(0,0){\framebox(6,1)}
\put(2.50, 0.70){$V_1$}
\put(2.50, 0.00){$V_2$}
\put(3.70, 0.00){$W$}
\put(3.10, 0.15){$g$}
\put(3.30, 0.50){$g \circ f$}
\put(2.35, 0.40){$f$}
\put(2.70, 0.05){\vector(1,0){.90}}
\put(2.55, 0.65){\vector(0,-1){.50}}
\put(2.70, 0.65){\vector(2,-1){.90}}
\end{picture}

This time, the functor associated to $W$ is written
$$ \Hom(\uscore, W).$$
It behaves as follows:

\begin{itemize}

\item A vector space $V$ is sent to $\mcL(V, W)$.  That is, a given $V$ is sent
to the collection of all linear transformations from that $V$ to the fixed
vector space $W$.

\item A linear transformation $f:V_1 \to V_2$ is sent to a function
which converts a linear transformation $g:V_2 \to W$ into
$g\circ f: V_1 \to W$.  That is,
$$
(f: V_1 \to V_2) \mapsto
f^*: \mcL(V_2,W) \to \mcL(V_1,W)
\quad\textrm{where}\quad
f^*(g:V_2 \to W) = (g\circ f: V_1 \to W)
$$

\end{itemize}

It looks like this:

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.0)
%\put(0,0){\framebox(6,1)}
\put(1.70, 0.70){$V_1$}
\put(1.70, 0.10){$V_2$}
\put(1.85, 0.40){$f$}
\put(1.75, 0.65){\vector(0,-1){.40}}

\put(2.55, 0.55){$\Hom(\uscore, W)$}
\put(2.25, 0.45){\vector(1,0){1.10}}

\put(3.70, 0.70){$\mcL(V_1, W)$}
\put(3.70, 0.10){$\mcL(V_2, W)$}
\put(3.85, 0.40){$f^*$}
\put(3.75, 0.25){\vector(0,1){.40}}
\end{picture}

Since $\Hom(\uscore, W)$ sends $V_1$ to $\mcL(V_1, W)$, $V_2$ to $\mcL(V_2,
W)$, and $f:V_1 \to V_2$ to $f^*: \mcL(V_2, W) \to \mcL(V_1, W)$, i.e. it
reverses the direction of the arrows, $\Hom(\uscore, W)$ is said to be a
\bfidx{contravariant hom functor}.

Now, here I used vector spaces as a concrete example, since they are familiar
to all of us.  However, the same goes for categories in general: given an
object $A$ of a category $\msC$, we have the covariant hom functor $\Hom(A,
\uscore)$; given an object $B$ of $\msC$, we have the contravariant hom functor
$\Hom(\uscore, B)$.

%% ================================================================
\newpage
\subsection{Freeness}

\begin{center}
\emph{
Everything that is really great and inspiring is created by the individual who
can labor in freedom.} \\
--- Albert Einstein\index{Einstein, Albert} (1879-1955).
\end{center}

The concept of freeness is central in algebra.  The formal definition --- which
is undeniably necessarily for mathematical rigor --- obscures the simple
intuition.  Thus, in this section I focus primarily on the intuition.

%% ----------------------------------------------------------------
\subsubsection{Definitions}

\begin{defn}
Let $\msC$ be a category.  An object $F$ in $\msC$ is \bfidx{free} with respect
to a set $S$ and with respect to a morphism $i$ from $S$ to $F$ if for any
other object $A$ of $\msC$ and a morphism $f: S \to A$ there is a unique
morphism $h$ from $F$ to $A$ such that $h \circ i = f$.
\end{defn}

This sounds too abstract to be of any use.  To rescue it, here is a
commutative diagram:

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.1)
%\put(0,0){\framebox(6,1.1)}

\put(2.50, 0.85){$S$}
\put(3.50, 0.85){$F$}
\put(3.50, 0.05){$A$}

\put(3.00, 0.95){$i$}
\put(2.80, 0.40){$f$}
\put(3.65, 0.40){$h$}

\put(2.65, 0.87){\vector(1,0){0.80}}
\put(2.65, 0.82){\vector(1,-1){0.70}}
\put(3.55, 0.77){\vector(0,-1){0.60}}
\end{picture}

This still looks overly abstract.  What the definition is saying, though, makes
sense when we consider the familiar example of vector spaces.

\beginex Let $\beta$ be a basis for a finite-dimensional vector space $V$.
Then $i$ is (as is always the case) the inclusion map.  Let $f$ be a
transformation sending the basis vectors to another vector space $W$.  The
existence and uniquess of $h$ simply means that when we specify the images of
the basis vectors under $f$, there is \emph{no more choice} in where to send
any other element of $V$.  This is precisely what we mean when we write a
linear transformation as a matrix: we are specifying the images of the
\emph{basis} vectors, and that gives us the image of \emph{any} vector in $V$
by linearity.  \mendex

This example motivates the following intuitive notion of freeness.

\begin{intuition} An object $F$ (group, abelian group, module, etc.) is
\bfidx{free} on a generating set $S$ (called the \bfidx{generating set} or
\bfidx{basis}) if, for a morphism (homomorphism, linear transformation, etc.),
specifying the image of the basis elements uniquely specifies the image of all
of $F$.
\end{intuition}

For the formal definition of generators and relations, see \cite{bib:grove}.
Here, the following suffices.

\begin{intuition} \label{defn:relation} In any algebraic structure with an
identity (group, abelian group, module, etc.) a \bfidx{relation} between two or
more elements is an expression which is equal to the identity element.  (When
there is more than one operation, e.g. in rings, we mean the additive
identity.) \end{intuition}

\beginex In an abelian group $G$, for all $a, b \in G$, we have
$aba^{-1}b^{-1}=1$.  The string $aba^{-1}b^{-1}$ is equal to the group
identity, so it is a relation.
\mendex

This gives us another way to think of freeness.

\begin{intuition} An algebraic structure $F$ (group, abelian group, module,
etc.) is \bfidx{free} on a generating set $S$ if the elements of $S$ have no
relations other than the minimal ones necessary for the operations appropriate
to the structure. \end{intuition}

\beginex
For vector spaces, this is the familiar notion of \bfidx{linear independence}:
a linear combination of basis vectors summing to zero (which is the additive
identity of the vector space) is simply a relation.  When the generating set is
linearly independent, any linear combination of basis vectors summing to zero
has all coefficients zero, which is the trivial relation $0=0$.
\mendex

More examples of free objects appear in the following sections.

%% ----------------------------------------------------------------
\subsubsection{Free groups}
\label{sec:free_groups}

We will need free groups in algebraic topology.  First, we have the definition
of free semigroup.  This is not needed for the rest of this paper (nor anywhere
in the geometry-topology course) but it helps motivate the concept of free
group.  Semigroups were defined in section \ref{sec:groups}.

\begin{intuition} A \bfidx{free semigroup} on a set $S$ is
the smallest semigroup containing $S$ in which there are no relations between
the elements of $S$.
\end{intuition}

The canonical example is the same as the semigroup in example
\ref{ex:string_semigroup}:

\beginex Let $S = \{a,b,c\}$.  Then sample elements of the free semigroup on
$S$ are the strings $a$, $abb$, $abbbbcccc$, $abab$, etc.  No string simplifies
into anything else.  \mendex

\begin{intuition} A \bfidx{free group} $F$ on a generating set $S$ is the
smallest group containing $S$ along with all inverses of elements of $S$.  As
well, elements of $F$ are completely distinct \emph{except} that inverses
cancel, and there must be an identity.  \end{intuition}

The formal construction of free groups, which makes this intuitive notion
precise, always take some fuss.  The shortest presentation I've seen is in
\cite{bib:grove}, and even there it takes several pages.  The simple concept
behind that fuss, though, is that a free group $F$ on a generating set $S$ is
as much like a free semigroup as possible:  elements of the free group are
strings composed of letters from $S$, with no simplification possible
\emph{except} for the cancelling of inverses.  We may think of the empty string
as the identity element.

\beginex
\label{ex:frg_abc}
Let $S=\{a,b,c\}$.  Then a sample element of the free group on $S$ is
$$abababccc.$$  Note however that $$ac = abb^{-1}c = ab^2b^{-2}c.$$
\mendex

%% ----------------------------------------------------------------
\subsubsection{Free abelian groups}

We need free abelian groups to construct tensor products, starting in section
\ref{sec:tensor_product_of_vs}, and we will use them to help define integrals
on manifolds in section \ref{sec:cubes_and_chains}.  Free abelian groups are
also used in free modules (coming next) which permeate this course.

\begin{intuition}
In the \bfidx{free abelian group} on a generating set $S$, all elements of $S$
are distinct as with a free group, but instead of forming strings of elements,
we keep track of how many elements we have.
\end{intuition}

\beginex
\label{ex:frab_abc}
Again take the generating set $S$ to be $\{a,b,c\}$.  When we form the free
abelian group $A$ on $S$, we have elements of the form $$3a - 2b + c.$$ Another
element might be $3a+0b+c$ (which we abbreviate as $3a+c$), and their sum would
be $6a-2b+2c$.  The sum $a+a$ simplifies to $2a$, but the sum $a+b$ stays
$a+b$: $a$ and $b$ are distinct; they are apples and oranges.  All we can do is
count how \emph{many} apples and how many oranges, etc.
\mendex

To make that last sentence a bit more precise, I note the following:

\begin{rem}
A free abelian group on a set $S$ is isomorphic to $\Z^r$ where $r$ is the
cardinality of $S$.  For each generator $g$ in $S$ there is one copy of $\Z$,
which counts the number of $g$'s.  For example, the free abelian group on
$\{a,b,c\}$ is isomorphic to $\Z^3$: $3a-2b+c$ maps to the triple $(3,-2,1)$.
\end{rem}

%% ----------------------------------------------------------------
\subsubsection{Free product of groups}
\label{sec:free_product_of_groups}

xxx stress intuition and examples, following the pattern above.
\index{free product of groups}

%% ----------------------------------------------------------------
\subsubsection{Free modules}
\label{sec:free_modules}

Next we have the notion of free module, which generalizes the previous concept
while also bringing us back to vector spaces.

\begin{intuition}
Let $R$ be a commutative ring with identity.  A \bfidxtwo{free $R$-module}{free
module} $M$ on a generating set $S$ is the smallest module containing all
$R$-linear combinations of elements of $S$.
\end{intuition}

\begin{rem}
A free $\Z$-module on a set $S$ is the precisely the same thing as a free
abelian group on $S$.  (Similarly, recall from remark
\ref{rem:abn_gp_is_Z_module} that any abelian group is a $\Z$-module.)
\end{rem}

We use this concept in two main ways for this course:

\begin{itemize}

\item We can start with a \emph{small} set of elements which, much as in free
groups and free abelian groups, are defined to be unmixable.  We then take all
$R$-linear combinations of elements of those basis elements, and we obtain a
\emph{big} module.

\item We can start with a \emph{big} $R$-module $M$, then look inside of it to
find a \emph{small} \bfidx{basis} $S$, or set of generators.  (We call the
cardinality of $S$ the dimension of $M$.)  Every element of $M$ is uniquely
expressible as an $R$-linear combination of basis elements.  This is precisely
what we do with vector spaces.  Every vector space is a free module, but we can
also form free modules when $R$ is not a field.  We can think of a free module
as being as close to a vector space as we can get when the base ring is not a
field.

\end{itemize}

\beginex
Here is an example of the first notion, i.e. starting with a generating set and
forming a free module on it.  As in example \ref{ex:frab_abc}, take $S$ to
consist of the letters $a,b,c$.  For the ring $R$, this time use the rational
numbers.  Then elements of the free $\Q$-module on $S$ are $\Q$-linear
combinations of $a,b,c$, e.g.  $$ \frac{3a}{4} - \frac{2c}{7}.$$
\mendex

Now for the second notion, i.e. finding a generating set inside a given module.
Just as a free abelian group is isomorphic to $\Z^r$, where $r$ is the
cardinality of the generating set, a similar statement holds for free modules.

\begin{rem} Let $R$ be a commutative ring with identity.  A free $R$-module $M$
is isomorphic with $R^m$ for some integer $m$.  The integer $m$ is called the
\bfidx{dimension} of $M$.  (For each generator $g$ of $M$ there is one copy of
$R$, which holds the coefficient on $g$.)  We are familiar with this for vector
spaces:  we know from linear algebra that every finite-dimensional real vector
space $V$ is isomorphic to $\R^m$, where $m$ is the size of any basis for $V$.
\end{rem}

%% ================================================================
\newpage
\subsection{Quotients, projections, and sections}
\label{sec:projections}

\begin{defn}
A function $f:X \to X$ from a set to itself is \bfidx{idempotent} if $f(f(x)) =
f(x)$ for all $x \in X$, i.e. if $f^2 = f$.
\end{defn}
This simply means that when we apply a function twice, we don't get anything
new, i.e. it acts like the identity on its image.

\begin{defn}
Let $E$ and $B$ be sets.  A \bfidx{projection} \index{$\pi$} $\pi: E \to B$ is
simply an \plainidx{epimorphism}.  If $E = B$, then we further require that
$\pi$ be idempotent.  Usually $E$ and $B$ are something more than sets; we
require that $\pi$ be a homomorphism for whatever category $E$ and $B$ belong
to (e.g. groups, rings, vector spaces).
\end{defn}

The set $E$ is called the \bfidx{total space} and $B$ is called the \bfidx{base
space}.

\begin{defn}
\label{defn:fiber}
Let $\pi:E \to B$ be a projection.  Let $b \in B$.  The \bfidx{fiber} of
$b$ is $\pi^{-1}(b)$.
\end{defn}
This is thought of as the \plainidx{preimage} of $b$, or everything
\bfidx{lying above} $b$.

\begin{defn}
\label{defn:section}
Let $\pi$ be a projection from a set $E$ to a set $B$.  A \bfidx{section} of
$\pi$ (or a \bfidx{cross-section}) is a function \index{$s$} $s: B \to E$ with
the property that $\pi \circ s$ is the identity on $B$.
\end{defn}

The $\pi \circ s = 1_B$ property simply means that a section maps elements of
$B$ somewhere into their \emph{own} fibers, and not into any other elements'
fibers.

Here is a picture of sets $E$ and $B$ with a projection $\pi$.  All the
elements of $E$ (black dots) are sent by $\pi$ down to some element of $B$
(also marked with black dots).  For each $b$ in $B$ (for each dot on the bottom
row) the elements of the fiber of $b$, i.e. $\pi^{-1}(b)$, are connected with a
line.
\input{fiberpic.tex}

Here is a picture of a section $s$ for the same $\pi: E \to B$.  Here, all the
elements $b$ of $B$ are darkened, but only one element from each fiber of $b$
is darkened.  The dark dot in each fiber indicates $s(b)$.  So, a section
begins to look a lot like a \bfidx{graph}, with $B$ as the abscissa and fibers
of $B$ as the ordinate.
\input{sectionpic.tex}

Here is another section of $\pi$:
\input{sectionpic2.tex}

For a given projection, there are in general many possible sections.  This
leads to the following definition:

\begin{defn}
\label{defn:gamma}
Let $\pi:E \to B$ be a projection.  Write $\Gamma(\pi,E,B)$ \index{$\Gamma$}
for the set of all sections, or \bfidx{space of sections}, of $\pi$.  (This is
what Pickrell calls $\Omega^0$; I choose to use a different symbol since
$\Omega$ is used for other things.  In fact, though, $\Gamma$ is rather common
in the literature.  We can think of $\Gamma$ for \emph{graph}.)
\end{defn}

\beginex
Let $E$ be the ring of integers $\Z$ and let $B$ be the integers mod $n$,
written $\Z/n/Z$.  Then $\pi$ is the map which reduces mod $n$.
\begin{itemize}
\item We write the elements of $\Z/n\Z$ as $\{ 0+\Z, 1+\Z, 2+\Z, \ldots,
	n-1+\Z\}$, or as $\{ \overline{0}, \overline{1}, \overline{2}, \ldots,
	\overline{n-1}\}$.
\item A section is a complete set of coset representatives.  Usually,
	we take $\{0,1,2,\ldots,n-1\}$ to be that set.  That is, $s(\overline{0}) =
	0$, $s(\overline{1}) = 1$, etc.  For example, with $n=5$, a section of
	$\pi$ is $\{0,1,2,3,4\}$.  But we could just as well take another section
	to be $\{4,8,12,16,20\}$.
\item The fiber above $\overline{k}$ is all numbers equivalent to $k$ mod $n$.
	For example, with $n=5$, the fiber above $\overline{2}$ is $\{\ldots, -8,
	-3, -2, 7, 12, 17, \ldots\}$.
\end{itemize}
\mendex

\beginex
Let $E$ be the symmetric group $\Sn$ and let $B =\{\pm 1\}$.  Let $\pi$ be the
sign function, which takes even permutations to $1$ and odd permutations to
$-1$.
\begin{itemize}
\item A section of $\pi$ is any set containing one even permutation and one odd
	permutation.
\item The fiber above $-1$ is the set of all odd permutations; the fiber above
	$1$ is the set of all even permutations, which is the alternating group
	$\An$.
\end{itemize}
\mendex

\beginex
Let $\pi: \R\times\R \to \R$ be given by $\pi(x,y) = x$, i.e. $\pi$ projects a
point onto its $x$ coordinate.
\begin{itemize}
\item A section is simply any real-valued function $s$: given $x$,
$s(x)$ gives a point $(x,y)$.
\item The set of all sections of $\pi$ is the set of all vector-to-scalar
functions.
\item If we further insist that $s(x)$ vary smoothly with $x$, then the set of
all such sections is the set of all smooth vector-to-scalar functions.
\item The fiber of $x_0 \in \R$ is the set of all $(x,y)$ pairs with $x=x_0$,
i.e. it is the vertical line through $x_0$.
\end{itemize}
\mendex

\beginex Let $\pi: \R^3\times\R \to \R$ be given by $\pi((x,y,z),u)) = u$.  A
section is simply any \bfidx{vector-to-scalar function}: $f(x,y,z)$ gives a
point $((x,y,z),u)$.  The set of all sections of $\pi$ is the set of all
vector-to-scalar functions.  Again, we may further insist that $u$ vary
smoothly with $(x,y,z)$.
\mendex

\beginex Let $\pi: \R^3\times\R^3 \to \R^3$ be given by
$\pi((x,y,z),(u,v,w))) = (u,v,w)$.  A section is any (smooth)
\bfidx{vector-to-vector function}: $\vecF(x,y,z)$ gives a point
$((x,y,z),(u,v,w))$.
\mendex

These examples recall the vector-calculus notions discussed in section
\ref{sec:vector_calc_diffn}.

xxx define quotient in two ways:

(1) If there is algebraic structure (e.g. groups, rings, vector spaces, etc.)
then in terms of cosets.  In particular, dimension of quotient is difference of
quotients of originals.

(2) Else, $A/B$ means contract $B$ to a point and leave all else the same.
Example: $\RP^n/\RP^m$.  Present in terms of cell complexes.  [Maybe this needs
to be moved forward and xref'ed to.  Or come up with a better example for
here.]  Here, the dimension-difference rule doesn't apply:  e.g.
$\RP^n/\RP^{n-1} = \S^n$ --- not dimension 1.

%% ================================================================
\newpage
\subsection{Sequences, complexes, and homological algebra}
\label{sec:sequences_and_complexes}

\begin{center}
\emph{In symbols one observes an advantage in discovery which is greatest when
they express the exact nature of a thing briefly and, as it were, picture it;
then indeed the labor of thought is wonderfully diminished.} \\
--- Gottfried Wilhelm Leibniz\index{Leibniz, Gottfried Wilhelm} (1646-1716).
\end{center}

Here we have some fancy terminology for what (keep in mind!) are abstractions
of down-to-earth concepts.  So, when confused by the terminology, invoke your
favorite example for concreteness.  Several examples are presented in the
following subsubsections.

%% ----------------------------------------------------------------
\subsubsection{Sequences}

Throughout math we have maps from one object to another, e.g. a group
homomorphism $G \to H$, a linear transformation $V \to W$, etc.  But we can as
well have a \bfidx{sequence} of maps, e.g.
$$ A \to B \to C $$
or
$$ A_1 \to A_2 \to A_3 \to A_4 $$
or perhaps an infinite sequence
$$ \ldots \to A_1 \to A_2 \to A_3 \to A_4 \to \ldots $$
etc.
We don't always name the maps, but we might, say, use the name $\phi_{i,i+1}$
for the map from the $i$th object to the $(i+1)$st object:
$$
\ldots \to A_1 \stackrel{\phi_{12}}{\to} A_2 \stackrel{\phi_{23}}{\to} A_3 \to
\ldots.
$$

There's not much more we can say about such sequences of maps, as stated here.
However, if we impose additional restrictions, we can obtain additional
structure and additional results.

For any group homomorphism $\phi: G \to H$, the kernel and image of $\phi$ are
always subgroups of $G$ and $H$, respectively.  Additionally, $\ker(\phi)$ is
normal in $G$, but $\im(\phi)$ is not necessarily normal in $H$.  Thus we can
always form the quotient group $G / \ker(\phi)$, but we can't always form the
quotient group $H / \im(\phi)$.  If $G$ and $H$ are abelian, then \emph{all}
subgroups are normal, and we can always form these quotients.  Note in
particular that rings and vector spaces have abelian additive groups.

Suppose we have a sequence
$$ A \stackrel{\phi}{\to} B \stackrel{\psi}{\to} C.$$
Then $\im(\phi)$ and $\ker(\psi)$ are both subobjects of $B$.  We can ask
about the relationship between the two, e.g. what additional results do
we obtain if
$\im(\phi) \subseteq\ker(\psi)$,
$\im(\phi) = \ker(\psi)$, or
$\im(\phi) \supseteq\ker(\psi)$?
To help address this question, I will first define some notation.

%% ----------------------------------------------------------------
\subsubsection{Trapezoid diagrams}

Let $A$ and $B$ \bfidx{objects} in a concrete category, with a homomorphism
$\phi_{AB}: A \to B$.  (For concreteness, perhaps think of them as being vector
spaces.) We will draw a \bfidx{trapezoid diagram} for this mapping.  Each
object has a zero element: $0_A$ and $0_B$, respectively.  Heavy vertical lines
represent all the elements of the sets $A$ and $B$, with the zero elements in
the middle:

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.5)
	%\put(0,0){\framebox(6,1.5)}

	\put(2.45, 1.25){$A$}
	\thicklines
	\put(2.50, 0.20){\line(0,1){1.00}}  % vert line at A
	\thinlines
	\put(2.50, 0.70){\circle*{0.05}}
	\put(2.25, 0.70){$0_A$}

	\put(3.45, 1.25){$B$}
	\thicklines
	\put(3.50, 0.20){\line(0,1){1.00}}  % vert line at B
	\thinlines
	\put(3.50, 0.70){\circle*{0.05}}
	\put(3.25, 0.70){$0_B$}

\end{picture}

Suppose we have a \bfidx{homomorphism} $\phi_{AB}: A \to B$.  Since
homomorphisms send 0 to 0, it makes sense to draw a line connecting the zero
elements.  Also, draw two nested trapezoids as follows:
\begin{itemize}
\item The outer trapezoid starts with all of $A$ and ends inside of $B$.  Its
right-hand edge represents the \bfidx{image} of $\phi_{AB}$, which is at most
all of $B$, but perhaps smaller.
\item The inner trapezoid ends with $0_B$.  Its left-hand edge represents
the \bfidx{kernel} of $\phi_{AB}$:  all of the elements of $A$ which map
to $0_B$.  This is at least $\{0_A\}$, but perhaps bigger.
\item By convention, I draw the outer-trapezoid lines parallel to the
	inner-trapezoid lines.
\end{itemize}

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.5)
	%\put(0,0){\framebox(6,1.5)}

	\put(2.45, 1.25){$A$}
	\thicklines
	\put(2.50, 0.20){\line(0,1){1.00}}  % vert line at A
	\thinlines
	\put(2.50, 1.20){\line(5,-1){1.00}}  % upper  horiz line A -> B
	\put(2.50, 0.90){\line(5,-1){1.00}}  % upper  horiz line A -> B
	\put(2.50, 0.70){\line(1,0){1.00}}  % middle horiz line A -> B
	\put(2.50, 0.50){\line(5,1){1.00}}  % upper  horiz line A -> B
	\put(2.50, 0.20){\line(5,1){1.00}}  % upper  horiz line A -> B
	\put(2.95, 1.25){$\phi_{AB}$}

	\put(3.45, 1.25){$B$}
	\thicklines
	\put(3.50, 0.20){\line(0,1){1.00}}  % vert line at B
	\thinlines

	\put(1.65, 0.70){$\ker(\phi_{AB}) \;\Big\{$}
	\put(3.85, 0.70){$\Bigg\}\;\im(\phi_{AB})$}

\end{picture}

Consequences:

\begin{itemize}
\item If the inner trapezoid disappears into the center line, that means the
	map is injective (i.e. has zero kernel).  Since I draw the outer-trapezoid
	lines parallel to the inner-trapezoid lines, one-to-one maps will have
	horizontal lines.
\item If the outer and inner trapezoid both disappear into the center line,
	that means $A = \{0\}$.
\item If outer trapezoid reaches all of $B$, then $\phi_{AB}$ is surjective.
\item If the inner trapezoid disappears into the center line, and if the
	outer trapezoid reaches all of $B$, then $\phi_{AB}$ is an isomorphism.
\end{itemize}

%% ----------------------------------------------------------------
\subsubsection{Non-complex, non-exact sequences}

Here are two sequences in which $\im(\phi_{AB}) \supseteq \ker(\phi_{BC})$.
(For a specific example of the latter, suppose $A=B=C$ and
$\phi_{AB}=\phi_{BC}$ are the identity map.)

[xxx include specific examples:  use $\Z/m\Z$, and $\R^n$.]

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.5)
	%\put(0,0){\framebox(6,1.5)}

	\put(0.45, 1.25){$A$}
	\thicklines
	\put(0.50, 0.20){\line(0,1){1.00}}  % vert line at A
	\thinlines
	\put(0.50, 1.20){\line(5,-1){1.00}}  % upper  horiz line A -> B
	\put(0.50, 0.90){\line(5,-1){1.00}}  % upper  horiz line A -> B
	\put(0.50, 0.70){\line(1,0){1.00}}  % middle horiz line A -> B
	\put(0.50, 0.50){\line(5,1){1.00}}  % upper  horiz line A -> B
	\put(0.50, 0.20){\line(5,1){1.00}}  % upper  horiz line A -> B
	\put(0.95, 1.25){$\phi_{AB}$}

	\put(1.45, 1.25){$B$}
	\thicklines
	\put(1.50, 0.20){\line(0,1){1.00}}  % vert line at B
	\thinlines
	\put(1.50, 1.20){\line(5,-1){1.00}}  % upper  horiz line B -> C
	\put(1.50, 0.90){\line(5,-1){1.00}}  % upper  horiz line B -> C
	\put(1.50, 0.70){\line(1,0){1.00}}  % middle horiz line B -> C
	\put(1.50, 0.50){\line(5,1){1.00}}  % upper  horiz line B -> C
	\put(1.50, 0.20){\line(5,1){1.00}}  % upper  horiz line B -> C
	\put(1.95, 1.25){$\phi_{BC}$}

	\put(2.45, 1.25){$C$}
	\thicklines
	\put(2.50, 0.20){\line(0,1){1.00}}  % vert line at C
	\thinlines

	%% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
	\put(3.45, 1.25){$A$}
	\thicklines
	\put(3.50, 0.20){\line(0,1){1.00}}  % vert line at A
	\thinlines
	\put(3.50, 1.20){\line(1,0){1.00}}  % upper  horiz line A -> B
	\put(3.50, 0.70){\line(1,0){1.00}}  % middle horiz line A -> B
	\put(3.50, 0.20){\line(1,0){1.00}}  % upper  horiz line A -> B
	\put(3.95, 1.25){$\phi_{AB}$}

	\put(4.45, 1.25){$B$}
	\thicklines
	\put(4.50, 0.20){\line(0,1){1.00}}  % vert line at B
	\thinlines
	\put(4.50, 1.20){\line(1,0){1.00}}  % upper  horiz line B -> C
	\put(4.50, 0.70){\line(1,0){1.00}}  % middle horiz line B -> C
	\put(4.50, 0.20){\line(1,0){1.00}}  % upper  horiz line B -> C
	\put(4.95, 1.25){$\phi_{BC}$}

	\put(5.45, 1.25){$C$}
	\thicklines
	\put(5.50, 0.20){\line(0,1){1.00}}  % vert line at C
	\thinlines

\end{picture}

For the right-hand sequence, any non-zero element of $A$ may be pushed through
the sequence, all the way to the right, with the result still being non-zero.
(This would still be true for a longer sequence of identity maps.) For the
left-hand sequence, there are also some elements of $A$ which map to non-zero
elements of $C$.

We don't have a term for these kinds of sequences.  However, do note that many
common maps (including isomorphisms!) are of this form.  Exact sequences and
complexes (to be discussed below) deal with the cases $\im(\phi_{AB}) =
\ker(\phi_{BC})$ and $\im(\phi_{AB}) \subseteq \ker(\phi_{BC})$, which are more
special.

%% ----------------------------------------------------------------
\subsubsection{Exact sequences}

Here\index{Rosetta stone} is a sequence in which $\im(\phi_{AB}) =
\ker(\phi_{BC})$:

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.5)
	%\put(0,0){\framebox(6,1.5)}

	\put(1.45, 1.25){$A$}
	\thicklines
	\put(1.50, 0.20){\line(0,1){1.00}}  % vert line at A
	\thinlines
	\put(1.50, 1.20){\line(4,-1){1.00}}  % upper  horiz line A -> B
	\put(1.50, 0.95){\line(4,-1){1.00}}  % upper  horiz line A -> B
	\put(1.50, 0.70){\line(1,0){1.00}}  % middle horiz line A -> B
	\put(1.50, 0.45){\line(4,1){1.00}}  % upper  horiz line A -> B
	\put(1.50, 0.20){\line(4,1){1.00}}  % upper  horiz line A -> B
	\put(1.95, 1.25){$\phi_{AB}$}

	\put(2.45, 1.25){$B$}
	\thicklines
	\put(2.50, 0.20){\line(0,1){1.00}}  % vert line at B
	\thinlines
	\put(2.50, 1.20){\line(4,-1){1.00}}  % upper  horiz line B -> C
	\put(2.50, 0.95){\line(4,-1){1.00}}  % upper  horiz line B -> C
	\put(2.50, 0.70){\line(1,0){1.00}}  % middle horiz line B -> C
	\put(2.50, 0.45){\line(4,1){1.00}}  % upper  horiz line B -> C
	\put(2.50, 0.20){\line(4,1){1.00}}  % upper  horiz line B -> C
	\put(2.95, 1.25){$\phi_{BC}$}

	\put(3.45, 1.25){$C$}
	\thicklines
	\put(3.50, 0.20){\line(0,1){1.00}}  % vert line at C
	\thinlines

\end{picture}

\begin{defn}
Let $\phi_{AB}: A \to B$
and $\phi_{BC}: B \to C$ be homomorphisms of concrete categories.
We say the sequence
$$
	A \stackrel{\phi_{AB}}{\longto}
	B \stackrel{\phi_{BC}}{\longto}
	C
$$
is \bfidx{exact at} $B$ if
$$
	\im(\phi_{AB}) = \ker(\phi_{BC}).
$$
\end{defn}

xxx specific examples

now define exact sequences.

%% ----------------------------------------------------------------
\subsubsection{Complexes}

Here is a sequence in which $\im(\phi_{AB}) \subseteq \ker(\phi_{BC})$:


%% ----------------------------------------------------------------
\subsubsection{[xxx merge w/ the above] Exact sequences}

\begin{defn}
A sequence
$$
\ldots \to A_1 \stackrel{\phi_{12}}{\to} A_2 \stackrel{\phi_{23}}{\to} A_3 \to
\ldots
$$
is said to be \bfidx{exact} if
$\im(\phi_{i-1,i}) = \ker(\phi_{i,i+1})$
for all $i$.
\end{defn}


Exact sequences have an important special case:

\begin{defn}
A sequence
$$
0
\to A
\stackrel{\phi}{\to} B
\stackrel{\psi}{\to} C
\to 0
$$
is said to be \bfidx{short exact}.
\end{defn}

Here is a graphical depiction of a short exact sequence:

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.5)
	%\put(0,0){\framebox(6,1.5)}

	\put(0.25, 0.70){$0$}
	\put(0.50, 0.70){\line(1,0){1.00}}  % horiz line 0 -> A
	\put(0.95, 0.80){$\phi_{0A}$}

	\put(1.45, 1.10){$A$}
	\thicklines
	\put(1.50, 0.40){\line(0,1){0.60}}  % vert line at A
	\thinlines
	\put(1.50, 1.00){\line(1,0){1.00}}  % upper  horiz line A -> B
	\put(1.50, 0.70){\line(1,0){1.00}}  % middle horiz line A -> B
	\put(1.50, 0.40){\line(1,0){1.00}}  % lower  horiz line A -> B
	\put(1.95, 1.20){$\phi_{AB}$}

	\put(2.45, 1.25){$B$}
	\thicklines
	\put(2.50, 0.20){\line(0,1){1.00}}  % vert line at B
	\thinlines
	\put(2.50, 1.20){\line(4,-1){1.00}} % upper  line B -> C
	\put(2.50, 0.70){\line(1,0){1.00}}  % middle horiz line B -> C
	\put(2.50, 0.20){\line(4,1){1.00}}  % lower  line B -> C
	\put(2.95, 1.20){$\phi_{BC}$}

	\put(3.45, 1.10){$C$}
	\thicklines
	\put(3.50, 0.45){\line(0,1){0.50}}  % vert line at C
	\thinlines
	\put(3.50, 0.95){\line(4,-1){1.00}} %
	\put(3.50, 0.70){\line(1,0){1.00}}  % middle horiz line C -> 0
	\put(3.50, 0.45){\line(4,1){1.00}}  %
	\put(3.95, 0.90){$\phi_{C0}$}

	\put(4.65, 0.70){$0$}

\end{picture}

In this diagram, the heavy vertical lines represent $A$, $B$, and $C$.  At the
middle of the vertical lines are the zero elements of $A$, $B$, and $C$,
respectively.  [xxx finish describing it.]

\begin{rem} Since the image of the (unnamed) map from $0$ to $A$ is 0 (it can't
be anything bigger!), and since that must be the same as $\ker(\phi)$, this
means that $\phi$ is necessarily \bfidx{injective}.  Likewise, since the kernel
of the (unnamed) map from $C$ to 0 is all of $C$ (it can't be anything
smaller!), and since this is the same as the image of $\psi$, $\psi$ is
necessarily \bfidx{surjective}.
Thus we can also think of short exact sequences as follows:
(1) $\phi$ is 1-1;
(2) $\psi$ is onto;
(3) $\im(\phi) = \ker(\psi)$.
\end{rem}

\beginex
Consider the sequence
$$
0
\to \Z
\stackrel{\phi}{\to} \Z
\stackrel{\psi}{\to} \Z/2\Z
\to 0
$$
where $\phi$ is the multiplication-by-$2$ map, $a \mapsto 2a$, and $\psi$ is
the reduction-mod-$2$ map.  The quotient $\Z/2\Z$ is just the two-element set
of evens and odds.  Multiplying by $2$ is 1-1, and reducing mod $2$ is
surjective.  Lastly, the image of $\phi$ is the even integers, which is
precisely the kernel of $\psi$.  (Also note that if we take all integers and
double them, then reduce them mod 2, we get only evens, which are the zero
element of $\Z/2\Z$.)
\mendex

\begin{rem}
Given objects $A$ and $B$ where the quotient $A/B$ is defined (for groups we
need $A$ to be a normal subgroup of $B$; for rings we need $A$ to be an ideal
of $B$; for modules we need $A$ to be a submodule of $B$; etc.), we can always
make an exact sequence
$$
	0 \to A \to B \to A/B \to 0
$$
where the left arrow is the inclusion of $A$ in $B$, and the right arrow is the
reduction of $B$ mod $A$.
\end{rem}

\begin{defn}
A \bfidx{long exact sequence} is one which is indexed by the natural numbers or
the integers, i.e. one which is countably infinite in one or both directions.
Note however that we can make any exact sequence countably infinite by putting
zeroes on either side.
\end{defn}

%% ----------------------------------------------------------------
\subsubsection{Techniques for exact sequences}

\begin{prop}
We have
$$
	0 \to A \to 0 \qquad \implies \qquad A =0.
$$
\begin{proof}
Let $a \in A$.  Then the right-hand map must send $a$ to 0, i.e. $a$ is in the
kernel of the right-hand map.  But the kernel of the right-hand map is equal to
the image of the left-hand map, which is 0, so $a=0$.
\end{proof}
\end{prop}

\begin{prop}
We have
$$
	0 \to A \stackrel{\phi}{\to} B \to 0 \qquad \implies \qquad A  \cong B.
$$
\begin{proof}
The 0 on the left means that $\phi$ is 1-1, and the 0 on the right means that
$\phi$ is onto.  Therefore $\phi$ is a bijection.  Since $\phi$ is also a
homomorphism, $\phi$ is an isomorphism.
\end{proof}
\end{prop}

Ways to \bfidx{disconnect a long exact sequence}:

(i) Show that an object is zero.  Then, its in-arrow must be surjective,
and its out-arrow must be injective.  For example, if we have
$$
	\ldots \to A \stackrel{\phi}{\to} B \stackrel{\psi}{\to} C \to \ldots
$$
and if $B$ is 0, then we obtain two long exact sequences:
\begin{eqnarray*}
	\ldots \to A \stackrel{\phi}{\to} &0& \\
	&0& \stackrel{\psi}{\to} C \to \ldots
\end{eqnarray*}

(ii) Show that an arrow is the zero map.  For example, if we have
$$
	\ldots \longto A
	\stackrel{\phi_{AB}}{\longto} B
	\stackrel{\phi_{BC}}{\longto} C
	\stackrel{\phi_{CD}}{\longto} D
	\stackrel{\phi_{DE}}{\longto} E
	\longto \ldots,
$$
and if $\phi_{BC}$ is the zero map, then its image is 0.  Then we
obtain two long exact sequences:
\begin{eqnarray*}
	\ldots \longto A
		\stackrel{\phi_{AB}}{\longto} B
		\stackrel{\phi_{BC}}{\longto} &0& \\
	&0&
		\stackrel{\phi_{CD}}{\longto} D
		\stackrel{\phi_{DE}}{\longto} E
		\longto \ldots \\
\end{eqnarray*}

(iii) Use the alternating-sum formula from p. 425 of \cite{bib:satan}.

%% ----------------------------------------------------------------
\subsubsection{[xxx merge w/ above] Complexes}

xxx discuss:  what if $\im \ne \ker$?

\begin{defn}
\bfidx{complex}:  $\phi^2=0$.  Show this pictorially via trapezoid diagrams.
\end{defn}

note exact implies complex.  prove with trapezoid diagrams.

note $\im \supseteq \ker$ implies non-complex.
prove with trapezoid diagrams.

example with properly complex.

examples: $\Z$, times 2, mod 2.
$\Z$, times 2, mod 10.
$\Z$, times 10, mod 2.
compute the homology when appropriate.

then, note we can take $\ker / \im$.

%% ----------------------------------------------------------------
\subsubsection{Zigzag lemma}

xxx better to prove this with or without chains?  If with, then move this
section.

%% ================================================================
\newpage
\subsection{Linear algebra}
\label{sec:linear_algebra}

\begin{center}
\emph{Halmos and I share a philosophy about linear algebra: we think
basis-free, we write basis-free, but when the chips are down we close the
office door and compute with matrices like fury.}
--- Irving Kaplansky\index{Kaplansky, Irving}.
\end{center}


Everything said in this section is applicable to finite-dimensional vector
spaces over an arbitrary field.  However, for this paper, the base field for
all vector spaces is $\R$.

%% ----------------------------------------------------------------
\subsubsection{Notation}
\label{sec:vs_notation}

% xxx row and column vector notation.

Given $m$ vectors $\vecv_1, \ldots, \vecv_m$ in a vector space $V$, written in
coordinates with respect to a basis, I will write
$$
\begin{pmatrix}
	\vecv_1 \\ \hline \vdots \\ \hline \vecv_m
\end{pmatrix}
\qqand
\begin{pmatrix}
	\vecv_1   \;\vline\; \cdots \;\vline\; \vecv_m
\end{pmatrix}
$$
for the matrix of row vectors and the matrix of column vectors, respectively.
In the case that $\dim V = m$, both of these matrices are square.

The standard basis for $\R^m$ is written
$$\{\vece_1, \ldots, \vece_m\}.$$
An arbitrary basis (not necessarily standard) will usually be written with
the letter \emph{b}, e.g.
$$\{\vecb_1, \ldots, \vecb_m\}.$$

%% ----------------------------------------------------------------
\subsubsection{To be filed}

\bfidx{Cramer's rule}: $$\mathrm{Adj}(A) A = \det(A) I.$$

\bfidxaux{Rank-nullity theorem}{rank-nullity theorem}:
Let $T: V \to W$ with finite dimensions.  Then
$$
	\dim(V) = \dim(\ker(T)) + \dim(\im(T)).
$$

%xxx Gram-Schmidt?

%% ----------------------------------------------------------------
\subsubsection{Basis and dual space}
\label{sec:basis_and_dual_space}

Let $V$ be a finite-dimensional vector space $V$ over $\R$, say of dimension
$m$.  Then $V$ is isomorphic with $\R^m$, which has many bases.  In particular,
it has a \bfidx{standard basis}
$$\{\vece_1, \ldots, \vece_m\}$$
with elements of the form
$$\vece_1 = (1, 0, 0, \ldots, 0),
\vece_2 = (0, 1, 0, \ldots, 0),
\ldots,
\vece_m = (0, 0, 0, \ldots, 1).$$

\begin{defn}
The \bfidx{dual space} of $V$ is the set $V^*$ of all linear transformations
from $V$ to $\R$.
\end{defn}

\begin{defn}
An element of a dual space is called a \bfidx{linear functional}.
\end{defn}

Recall that $V^*$ is a vector space over $\R$.  It has the same dimension,
namely $m$, as $V$.  To prove this, we can choose a basis for $V$, do the
following construction to obtain a basis for $V^*$, then use those bases to
construct an isomorphism between $V$ and $V^*$.

\begin{defn}
For any basis $\{\vecb_1, \ldots, \vecb_m\}$, there is a basis for the dual
space (i.e. a \bfidx{dual basis})
$$\{\vecb_1^*, \ldots, \vecb_m^*\}$$
where the $\vecb_i^*$ functions are defined to have value
$$ \vecb_i^*(\vecb_j) = \delta_{ij}$$
where the $\delta$ is \index{Kronecker delta} Kronecker's, i.e. $\delta_{ij} =
1$ when $i=j$, 0 otherwise.
\end{defn}

(See section \ref{sec:compute_dual_basis} for a technique to compute the
coefficients of the $\vecb_i^*$'s.)

In particular, the standard basis has a \bfidx{standard dual basis}
$$\{\vece_1^*, \ldots, \vece_m^*\} \;\textrm{with}\; \vece_i^*(\vece_j) =
\delta_{ij}.$$ To construct an explicit isomorphism between $V$ and $V^*$, we
can map each $\vece_i$ to the corresponding $\vece_i^*$.  The proof that this
map is in fact an isomorphism is a (brief) standard exercise in linear algebra.

\beginex
Let $\vecv = (v_1, \ldots, v_m) \in \R^m$.  Then $\vece_2^*(\vecv) = v_2$, i.e.
$\vece_2^*$ extracts the second coordinate of $\vecv$.  That is, the standard
dual basis may be thought of as consisting of \bfidxtwo{coordinate-selector
functions}{coordinate-selector function}.
\mendex

Let $\lambda: V \to \R$ and $\vecv \in V$.  Given the above basis,
we may write
$$ \lambda = \sum_{i=1}^m \ell_i \vece_i^* \qand
\vecv = \sum_{j=1}^m v_j \vece_j.
$$
Since the $e_i^*$'s are linear functions, and since $\vece_i^*(\vece_j) =
\delta_{ij}$, we have
$$
	\lambda(\vecv) =
		\sum_{i=1}^m \ell_i \vece_i^*\left( \sum_{j=1}^m v_j \vece_j\right)
	= \sum_{i=1}^m \sum_{j=1}^m \ell_i v_j \vece_i^*(\vece_j)
	= \sum_{i=1}^m \ell_i v_i.
$$

%% ----------------------------------------------------------------
\subsubsection{Geometric interpretation of dual}
\label{sec:geom_dual}

We can graph column vectors easily enough.  How, though, do we graph
functionals?  We will see that functionals are row vectors, then define the
notions of \emph{spine} and \emph{contour lines}.

First, a \plainidx{representation theorem}: Every linear functional on a
finite-dimensional vector space corresponds to a unique row vector.  More
precisely:

\begin{prop}
\label{thm:Rn_fcnal_rep}
Let $V$ be a finite-dimensional vector space.  For all $\lambda \in V^*$ there
is a unique $\vecu \in V$ such that for all $\vecv \in V$, $\lambda(\vecv) =
\vecu \cdot \vecv$.
\begin{proof}
The proof is by construction.  Since $V$ is finite-dimensional, it has a basis
$\{\vecb_1, \ldots, \vecb_m\}$.  Then $\vecv$ is written uniquely (which will
make $\vecu$ unique below) as a linear combination of the $\vecb_i$'s.  Since
$\lambda$ is a linear functional, we can move it through the sum, and since $V$
is finite-dimensional there is no question as to the convergence of the sum:
\begin{eqnarray*}
	\lambda(\vecv) &=& \lambda\left(\sum_{i=1}^m v_i \vecb_i \right)
	= \sum_{i=1}^m v_i \lambda(\vecb_i).
\end{eqnarray*}
Simply put
$$
	u_i = \lambda(\vecb_i)
$$
for each $i$ from 1 to $m$.  Then
\begin{eqnarray*}
	\vecu \cdot \vecv &=& \sum_{i=1}^m u_i v_i
	= \sum_{i=1}^m v_i \lambda(\vecb_i)
	= \lambda(\vecv).
\end{eqnarray*}
\end{proof}
\end{prop}

xxx xref to geom section; linear functionals.

xxx fishbone plots.

xxx define \bfidx{spine}

xxx xref back to inner-product stuff in the very first few sections.

%% ----------------------------------------------------------------
\subsubsection{Dual and double dual}
\label{sec:dual_and_double_dual}

For finite-dimensional vector spaces $V$, we saw in section
\ref{sec:basis_and_dual_space} that $V \cong V^*$.  But since $V^*$ is also a
finite-dimensional vector space, we also have $V^* \cong V^{**}$.  Since
isomorphisms are transitive, $V \cong V^*$ and $V^* \cong V^{**}$ imply $V
\cong V^{**}$.  The space $V^{**}$ is called the \bfidx{double dual} of $V$.

However, more can be said.  In section \ref{sec:basis_and_dual_space} we
obtained a basis for $V^*$ by first choosing a basis for $V$.  To construct an
isomorphism between $V$ and $V^{**}$, though, we don't need to choose a basis.
For each vector $\vecv \in V$ and each $\lambda \in V^*$, we can think of
$\vecv$ operating on $\lambda$ by $$ \vecv(\lambda) = \lambda(\vecv).$$ This
makes sense because $\lambda$ is in $V^*$, i.e. it is a linear function from
$V$ to $\R$.  Thus $\lambda(\vecv)$ is in $\R$, so we've taken $\lambda \in
V^*$ and used $\vecv$ to obtain a real number, which is precisely what it means
for something to be in $V^{**}$:  $v$ maps from $V^*$ to $\R$.  (The details of
the proof of the isomorphism are a quick linear-algebra exercise.  This leads
to a \bfidx{natural isomorphism} between $V$ and $V^{**}$, in the sense that it
works the same regardless of which basis we choose for $V$.)

The key point is that while we think of vectors $V$ as \bfidx{data}, they also
can be thought of as \bfidx{functions} as well, operating on the elements of
the dual space.

%% ----------------------------------------------------------------
\subsubsection{Explicit computation of dual basis}
\label{sec:compute_dual_basis}

A linear functional $\lambda \in V^*$ is a linear transformation from $V$ to
$\R$.  With respect to a basis, say the standard basis, we can think of a
general linear transformation from an $m$-dimensional space to an
$n$-dimensional space as an $n \times m$ matrix.  So, this means $\lambda$ can
be represented as a $1 \times m$ matrix.  For example:
$$
\lambda(\vecv) = \begin{pmatrix}a_1 a_2 \cdots a_m\end{pmatrix}
\begin{pmatrix}v_1 \\ v_2 \\ \vdots \\ v_m\end{pmatrix}.
$$
Thus, a linear functional may be explicitly represented by a row vector.  In
section \ref{sec:basis_and_dual_space} we saw that each $\lambda \in V^*$ is a
linear combination of the $\vecb_i^*$'s, the basis functionals corresponding to
the basis vectors $\vecb_j$'s, where $\vecb_i^*(\vecb_j) = \delta_{ij}$.  So,
we immediately ask, given a basis of $\vecb_j$'s, how do we compute the
coefficients of the $\vecb_i^*$'s?

Recall the notation for row and column vectors shown in section
\ref{sec:vs_notation}.  Use the fact that $\vecb_i^*(\vecb_j) = \delta_{ij}$.
Now, $\vecb_i^*(\vecb_j)$ is the product of the $1 \times m$ array $\vecb_i^*$
and the $m \times 1$ array $\vecb_j$.  Matrix multiplication is nothing more
than a collection of such products:  i.e. given any matrices $A$ and $B$, the
$ij$th cell of $C=AB$ is the dot product of the $i$th row of $A$ with the $j$
column of $B$.  For example, in the product of two $4 \times 4$ matrices, the
$2,3$ cell of the product is the dot of the 2nd row of the first matrix with
the 3rd column of the second matrix:
$$
\begin{pmatrix}
. & . & . & .  \\
. & . & \circ & .  \\
. & . & . & .  \\
. & . & . & .  \\
\end{pmatrix}
=
\begin{pmatrix}
. & . & . & .  \\
\circ & \circ & \circ & \circ  \\
. & . & . & .  \\
. & . & . & .  \\
\end{pmatrix}
\begin{pmatrix}
. & . & \circ & .  \\
. & . & \circ & .  \\
. & . & \circ & .  \\
. & . & \circ & .  \\
\end{pmatrix}
$$
Since the identity matrix $I$ has $\delta_{ij}$ in its $i,j$th cell, this means
that we can collect all the $\vecb_i^*(\vecb_j)$'s into a matrix product:
$$
	I =
	\begin{pmatrix}
		\vecb_1^* \\ \hline \vdots \\ \hline \vecb_m^*
	\end{pmatrix}
	\begin{pmatrix}
		\vecb_1   \;\vline\; \cdots \;\vline\; \vecb_m
	\end{pmatrix}
	\quad\implies\quad
	\begin{pmatrix}
		\vecb_1^* \\ \hline \vdots \\ \hline \vecb_m^*
	\end{pmatrix} =
	\begin{pmatrix}
		\vecb_1   \;\vline\; \cdots \;\vline\; \vecb_m
	\end{pmatrix}^{-1}.
$$
See section \ref{sec:VS_COB} for an example.  Also see section
\ref{sec:TS_COB}, for change of coordinates for tensors.

[xxx perhaps note:  If the basis is orthonormal then the primaries and their
respective dual spines coincide.  If it is merely orthogonal then they are
collinear with reciprocal lengths.  Else they are skewy.]

%% ----------------------------------------------------------------
\subsubsection{Change-of-basis matrices; covariance and contravariance}
\label{sec:VS_COB}

Here, by means of a carefully worked example, I relate the new (and important)
concepts of \bfidx{covariance} and \bfidx{contravariance} to the familiar
language of \bfidx{change-of-basis matrices}.  I also show the geometric
significance of change of basis for dual vectors.

Let
$$
	\vece_1 = \colvectwo{1}{0} \qand \vece_2 = \colvectwo{0}{1}
$$
as usual, and let
$$
	\vecb_1 = \colvectwo{2}{1} \qand \vecb_2 = \colvectwo{1}{3}.
$$
I want to see what happens to $\vecb_1^*$ and $\vecb_2^*$ when we change from
$\{\vece_1, \vece_2\}$ coordinates to $\{\vecb_1, \vecb_2\}$ coordinates.
Using the reasoning in section \ref{sec:compute_dual_basis}, I have
$$
	\begin{pmatrix}
		\vecb_1^* \\ \hline \vecb_2^*
	\end{pmatrix}
	= \begin{pmatrix}
		\vecb_1 \; \vline \; \vecb_2
	\end{pmatrix}^{-1}
	= \begin{pmatrix}
		2 & 1 \\
		1 & 3 \\
	\end{pmatrix}^{-1}
	= \begin{pmatrix}
		3/5 & -1/5 \\
		-1/5 & 2/5 \\
	\end{pmatrix}
$$
so
$$
	\vecb_1^* = \rowvectwo{3/5}{-1/5} \qand
	\vecb_2^* = \rowvectwo{-1/5}{2/5}.
$$
(Note that these are not simply the \plainidx{transpose} of $\vecb_1$ and
$\vecb_2$.  Hence my use of $\vecb_i^*$ rather than $\vecb_i^t$, as mentioned
in section \ref{sec:notn}.)

Now I want to look at the change-of-basis matrix.
Since $\vece_1$ maps to $\vecb_1$ and $\vece_2$ maps to
$\vecb_2$, we write down the matrix
$$
	Q = \begin{pmatrix}
		\vecb_1 \; \vline \; \vecb_2
	\end{pmatrix}
	= \begin{pmatrix}
		2 & 1 \\
		1 & 3 \\
	\end{pmatrix}.
$$
This sends the original basis vectors to the new basis vectors:
$$
	Q
	\begin{pmatrix}
		\vece_1 \; \vline \; \vece_2
	\end{pmatrix}
	= QI
	= 
	\begin{pmatrix}
		\vecb_1 \; \vline \; \vecb_2
	\end{pmatrix}.
$$
For the dual-basis vectors, we have
$$
	\begin{pmatrix}
		\vecb_1^* \\ \hline \vecb_2^*
	\end{pmatrix}
	=
	\begin{pmatrix}
		\vecb_1 \; \vline \; \vecb_2
	\end{pmatrix}^{-1}
	I
	= Q^{-1}
	\begin{pmatrix}
		\vece_1^* \\ \hline \vece_2^*
	\end{pmatrix}
$$
or
$$
	\begin{pmatrix}
		\vece_1^* \\ \hline \vece_2^*
	\end{pmatrix}
	=
	Q \begin{pmatrix}
		\vecb_1^* \\ \hline \vecb_2^*
	\end{pmatrix}.
$$
In summary, we used the change-of-basis matrix $Q$ as follows:
\begin{eqnarray*}
	\begin{pmatrix}
		\vece_1 \; \vline \; \vece_2
	\end{pmatrix}
	&\stackrel{Q}{\longrightarrow}&
	\begin{pmatrix}
		\vecb_1 \; \vline \; \vecb_2
	\end{pmatrix} \\
%
	\begin{pmatrix}
		\vece_1^* \\ \hline \vece_2^*
	\end{pmatrix}
	&\stackrel{Q}{\longleftarrow}&
	\begin{pmatrix}
		\vecb_1^* \\ \hline \vecb_2^*
	\end{pmatrix}.
\end{eqnarray*}
That is:
\begin{itemize}
\item The coefficients of a linear functional vary together with $Q$,
	i.e. \bfidxtwo{covariantly}{covariant} with $Q$.
	(Now is as good a time as any to note that linear functionals are sometimes
	called \bfidxtwo{covectors}{covector}.)
\item The coefficients of a vector vary opposite from $Q$,
	i.e. \bfidxtwo{contravariantly}{contravariant} with $Q$.
\end{itemize}
We will see more about this subject in sections \ref{sec:TS_COB} and [xxx write
and xref --- incl. $dx\ne dx$ section(s)].

\begin{center}* * *\end{center}

So much for the symbolic manipulation.  From a geometric point of view, why is
it that the dual basis must transform in this way?  Here is a plot of
the basis vectors and the dual-basis vectors:

[xxx figure here]

I am using fishbone plots for the dual-basis vectors, as defined in section
\ref{sec:geom_dual}.  Recall also the discussion in section \ref{sec:dot}.
From the figure it's now clear that $\vecb_1^*$ has its spine pointing
perpendicular to $\vecb_2$, and vice versa.  Also, the magnitude of $\vecb_1^*$
is such that $\vecb_1^*(\vecb_1)$ is still 1.

%%% ----------------------------------------------------------------
%\subsubsection{Overwrought section:  revise me}
%
%Here, by means of a carefully worked example, I relate the new (and important)
%concepts of \bfidx{covariance} and \bfidx{contravariance} to the familiar
%language of \bfidx{change-of-basis matrices}.
%
%Let
%$$
%A = \{\veca_1, \veca_2\} = \left\{\colvectwo{1}{0}, \colvectwo{0}{1} \right\}
%\qand
%B = \{\vecb_1, \vecb_2\} = \left\{\colvectwo{2}{0}, \colvectwo{1}{1}\right\}
%$$
%be two ordered bases for $\R^2$.  (I am intentionally making the second basis
%non-orthonormal so that we can see the full range of behavior.)
%Also consider a vector and a linear functional, namely,
%$$
%	\vecv = \colvectwo{3}{4}, \qquad
%	\lambda = (5\;6).
%$$
%I want to see how the coefficients of $\vecv$ and $\lambda$ change when we switch from
%the $A$ basis to the $B$ basis.  First, some notation.
%
%\begin{defn}
%Following \cite{bib:fis}, I write
%$$
%	{}[\vecv]_A,\quad {}[\lambda]_A \qqand
%	{}[\vecv]_B,\quad {}[\lambda]_B
%$$
%for the arrays of coefficients of $\vecv$ and $\lambda$ with respect to the $A$
%and $B$ bases.  (Of course, when we write $\vecv$ and $\lambda$ without
%brackets, we think of them as being with respect to the standard basis.)
%Likewise, I write
%$$
%	{}[v_j]_A,\quad {}[\lambda_i]_A \qqand
%	{}[v_j]_B,\quad {}[\lambda_i]_B
%$$
%for individual coordinates of $\vecv$ and $\lambda$ with respect to the $A$
%and $B$ bases.
%\end{defn}
%
%I will look at the basis and dual basis elements, $\lambda$, and $\vecv$ in
%both coordinate systems.  Let's first write down what we already know.
%
%\begin{itemize}
%
%\item With respect to the $A$ basis, the basis elements $\veca_i$ have a 1 in
%the $i$th row, and the dual basis elements $\veca_i^*$ have a 1 in the $i$th
%column.
%
%\item With respect to the $B$ basis, the basis elements $\vecb_i$ have a 1 in
%the $i$th row, and the dual basis elements $\vecb_i^*$ have a 1 in the $i$th
%column.
%
%\item We can use the formula from section \ref{sec:compute_dual_basis} to
%compute $\vecb_i^*$'s in $A$ coordinates.  That is, in $A$ coordinates,
%	$$
%		\begin{pmatrix}
%		\vecb_1^* \\ \hline \vecb_2^*
%		\end{pmatrix}
%		=
%		\begin{pmatrix}
%		\vecb_1   \;\vline\; \vecb_2
%		\end{pmatrix}^{-1}
%		=
%		\begin{pmatrix}
%			2 & 1 \\
%			0 & 1 \\
%		\end{pmatrix}^{-1}
%		=
%		\begin{pmatrix}
%			1/2 & -1/2 \\
%			0 & 1 \\
%		\end{pmatrix}
%	$$
%	so
%	$$
%		[\vecb_1^*]_A = \left(\frac{1}{2} \;-\frac{1}{2}\right), \qquad [\vecb_2^*]_A = (0 \; 1).
%	$$
%
%\item We expect $\lambda(\vecv) \cdot 5 + 4 \cdot 6 = 39$ to have the same
%value when we write $\lambda$ and $\vecv$ in $B$ coordinates.
%
%\item You are probably aware of change-of-basis matrices, but my intention here
%is to work through the derivation.  So, let's say that we don't know the
%coordinates $\alpha, \beta$ of $\vecv$ and $\gamma, \delta$ of $\lambda$ with
%respect to the $B$ basis.
%
%\end{itemize}
%
%Organizing these facts, we have:
%
%\begin{tabular}{|lll||lll|}
%\hline
%\multicolumn{3}{|c||}{$A$ coordinates} & \multicolumn{3}{|c|}{$B$ coordinates} \\
%\hline
%
%%%%
%	$[\veca_1]_A   = \colvectwo{1}{0}$ &
%	$[\veca_2]_A   = \colvectwo{0}{1}$ &
%	$[\vecv]_A     = \colvectwo{3}{4}$ &
%
%	$[\veca_1]_B   = \colvectwo{?}{?}$ &
%	$[\veca_2]_B   = \colvectwo{?}{?}$ &
%	$[\vecv]_B     = \colvectwo{\alpha}{\beta}$ \\
%&&&&&\\
%
%	$[\veca_1^*]_A = (1\;0)$&
%	$[\veca_2^*]_A = (0\;1)$&
%	$[\lambda]_A   = (5 \; 6)$ &
%
%	$[\veca_1^*]_B = (?\;?)$&
%	$[\veca_2^*]_B = (?\;?)$&
%	$[\lambda]_B   = (\gamma \; \delta)$ \\
%&&&&&\\
%
%	$[\vecb_1]_A   = \colvectwo{2}{0}$ &
%	$[\vecb_2]_A   = \colvectwo{1}{1}$ &
%	$[\lambda]_A([\vecb_1]_A) = 10$ &
%
%	$[\vecb_1]_B   = \colvectwo{1}{0}$ &
%	$[\vecb_2]_B   = \colvectwo{0}{1}$ &
%	$[\lambda]_B([\vecb_1]_B) = 10$ \\
%&&&&&\\
%
%	$[\vecb_1^*]_A = \left(\frac{1}{2}\;-\frac{1}{2}\right)$&
%	$[\vecb_2^*]_A = (0\;1)$&
%	$[\lambda]_A([\vecb_2]_A) = 11$ &
%
%	$[\vecb_1^*]_B = (1\;0)$&
%	$[\vecb_2^*]_B = (0\;1)$&
%	$[\lambda]_B([\vecb_2]_B) = 11$ \\
%&&&&&\\
%
%	&
%	&
%	$[\lambda]_A([\vecv]_A) = 39$ &
%
%	&
%	&
%	$[\lambda]_B([\vecv]_B) = 39$ \\
%
%\hline
%\end{tabular}
%
%To find $[\vecv]_B$, note that $\vecv$ must be a linear combination of the
%$B$ basis vectors:
%\begin{eqnarray*}
%	\vecv &=& \alpha \vecb_1 + \beta \vecb_2 \\
%	\colvectwo{3}{4}
%		&=& \alpha \colvectwo{2}{0} + \beta \colvectwo{1}{1}
%		= \begin{pmatrix}
%			2\cdot\alpha & 1\cdot\beta \\
%			0\cdot\alpha & 1\cdot\beta
%			\end{pmatrix}
%		= \begin{pmatrix} 2 & 1 \\ 0 & 1 \\ \end{pmatrix}
%			\colvectwo{\alpha}{\beta} \\
%	\colvectwo{\alpha}{\beta}
%		&=& \begin{pmatrix} 2 & 1 \\ 0 & 1 \\ \end{pmatrix}^{-1}
%			\colvectwo{3}{4}
%		= \begin{pmatrix} 1/2 & -1/2 \\ 0 & 1 \\ \end{pmatrix}
%			\colvectwo{3}{4}
%		= \colvectwo{-1/2}{4}.
%\end{eqnarray*}
%So, writing $Q$ for the matrix of column vectors of the $B$ basis, we have
%$$
%	[\vecv]_A = Q [\vecv]_B, \qqor
%	[\vecv]_B = Q^{-1} [\vecv]_A.
%$$
%Now we can use the same technique to find the coordinates of the standard
%basis vectors in $B$ coordinates:
%$$
%	[\veca_1]_B = \colvectwo{1/2}{0}, \quad
%	[\veca_2]_B = \colvectwo{-1/2}{1}.
%$$
%
%To find the coefficients of $\lambda$ in $B$ coordinates, I can use the fact
%that the values $\lambda(\vecb_1)$ and $\lambda(\vecb_2)$ should be independent
%of choice of coordinates.  Since in $A$ coordinates we have
%\begin{eqnarray*}
%	&& \left\{ \begin{matrix}
%		[\lambda]_A([\vecb_1]_A) &=& 10 \\
%		[\lambda]_A([\vecb_2]_A) &=& 11 \\
%	\end{matrix} \right.
%\end{eqnarray*}
%in $B$ coordinates we must have
%\begin{eqnarray*}
%	&& \left\{ \begin{matrix}
%		[\lambda]_B([\vecb_1]_B) &=& 10 \\
%		[\lambda]_B([\vecb_2]_B) &=& 11 \\
%	\end{matrix} \right. \\
%	&& \left\{ \begin{matrix}
%		(\gamma \; \delta) \colvectwo{1}{0} &=& 10 \\
%		(\gamma \; \delta) \colvectwo{0}{1} &=& 11 \\
%	\end{matrix} \right.
%\end{eqnarray*}
%which is to say we can just read off $\gamma=10$, $\delta=11$.
%We can use the same technique to obtain
%$$
%	[\veca_1^*]_B = \rowvectwo{2}{1}, \quad
%	[\veca_2^*]_B = \rowvectwo{0}{1}.
%$$
%
%Now the blanks have been filled in:
%
%\begin{tabular}{|lll||lll|}
%\hline
%\multicolumn{3}{|c||}{$A$ coordinates} & \multicolumn{3}{|c|}{$B$ coordinates} \\
%\hline
%
%%%%
%	$[\veca_1]_A   = \colvectwo{1}{0}$ &
%	$[\veca_2]_A   = \colvectwo{0}{1}$ &
%	$[\vecv]_A     = \colvectwo{3}{4}$ &
%
%	$[\veca_1]_B   = \colvectwo{1/2}{0}$ &
%	$[\veca_2]_B   = \colvectwo{-1/2}{1}$ &
%	$[\vecv]_B     = \colvectwo{-1/2}{4}$ \\
%&&&&&\\
%
%	$[\veca_1^*]_A = (1\;0)$&
%	$[\veca_2^*]_A = (0\;1)$&
%	$[\lambda]_A   = (5 \; 6)$ &
%
%	$[\veca_1^*]_B = (2\;1)$&
%	$[\veca_2^*]_B = (0\;1)$&
%	$[\lambda]_B   = (10 \; 11)$ \\
%&&&&&\\
%
%	$[\vecb_1]_A   = \colvectwo{2}{0}$ &
%	$[\vecb_2]_A   = \colvectwo{1}{1}$ &
%	$[\lambda]_A([\vecb_1]_A) = 10$ &
%
%	$[\vecb_1]_B   = \colvectwo{1}{0}$ &
%	$[\vecb_2]_B   = \colvectwo{0}{1}$ &
%	$[\lambda]_B([\vecb_1]_B) = 10$ \\
%&&&&&\\
%
%	$[\vecb_1^*]_A = \left(\frac{1}{2}\;-\frac{1}{2}\right)$&
%	$[\vecb_2^*]_A = (0\;1)$&
%	$[\lambda]_A([\vecb_2]_A) = 11$ &
%
%	$[\vecb_1^*]_B = (1\;0)$&
%	$[\vecb_2^*]_B = (0\;1)$&
%	$[\lambda]_B([\vecb_2]_B) = 11$ \\
%&&&&&\\
%
%	&
%	&
%	$[\lambda]_A([\vecv]_A) = 39$ &
%
%	&
%	&
%	$[\lambda]_B([\vecv]_B) = 39$ \\
%
%\hline
%\end{tabular}
%
%My key point here is how we used the $Q$ matrix to find the coordinates of
%$\vecv$ and $\lambda$ with respect to the new $B$ basis.  We had
%$$
%	[\lambda]_B = [\lambda]_A Q \qqand
%	[\vecv]_A = Q [\vecv]_B
%$$
%so we used the change-of-basis matrix $Q$ as follows:
%\begin{eqnarray*}
%	{}[\lambda]_A &\stackrel{Q}{\longrightarrow}& [\lambda]_B \\
%	{}[\vecv]_A   &\stackrel{Q}{\longleftarrow}&  [\vecv]_B.
%\end{eqnarray*}
%That is:
%\begin{itemize}
%\item The coefficients of a linear functional vary together with $Q$,
%	i.e. \bfidxtwo{covariantly}{covariant} with $Q$.
%	(Now is as good a time as any to note that linear functionals are sometimes
%	called \bfidxtwo{covectors}{covector}.)
%\item The coefficients of a vector vary opposite from $Q$,
%	i.e. \bfidxtwo{contravariantly}{contravariant} with $Q$.
%\end{itemize}
%We will see more about this subject in sections \ref{sec:TS_COB} and [xxx write
%and xref].

%% ----------------------------------------------------------------
\subsubsection{Pullbacks}
\label{sec:vs_pullback}

Now let $f:V \to W$ be a linear transformation, and let $\lambda: W \to \R$ be
a linear functional, i.e. $\lambda \in W^*$.  We can map from $V$ to $\R$ by
going through $W$ first.  That is, for $\vecv \in V$, $f(\vecv)$ is in $W$, and
so we can apply $\lambda$ to it:

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.0)
%\put(0,0){\framebox(6,1)}
\put(2.50, 0.70){$V$}
\put(3.70, 0.70){$W$}
\put(3.05, 0.10){$\R$}
\put(3.10, 0.85){$f$}
\put(2.40, 0.40){$\lambda\circ f$}
\put(3.60, 0.40){$\lambda$}
\put(2.70, 0.75){\vector(1,0){.90}}
\put(2.60, 0.70){\vector(1,-1){.45}}
\put(3.60, 0.70){\vector(-1,-1){.45}}
\end{picture}

Formally, $\lambda \circ f: V \to \R$.  Given $\lambda \in W^*$, we have
obtained $\lambda \circ f \in V^*$.  Post-composing $\lambda$ by $f$ is said to
be a \bfidx{pullback} of $\lambda$ from $W$ to $V$.  Another way to look at
this is that, given $f:V \to W$, we have\index{Rosetta stone} a map from $W^*$
to $V^*$:
\begin{eqnarray*}
	f:&&V \to W \\
	f^*:&& W^* \to V^* \\
	f^*(\lambda) &=& \lambda\circ f \\
	(f^*\lambda)(\vecv) &=& \lambda(f\vecv).
\end{eqnarray*}

In this way, $f^*$ takes functionals to functionals:

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.0)
%\put(0,0){\framebox(6,1)}
\put(2.50, 0.70){$V$}
\put(3.70, 0.70){$W$}
\put(3.10, 0.85){$f$}
\put(2.70, 0.75){\vector(1,0){.90}}
\put(2.50, 0.20){$V^*$}
\put(3.70, 0.20){$W^*$}
\put(3.10, 0.35){$f^*$}
\put(3.60, 0.25){\vector(-1,0){.90}}
\end{picture}

There are two key points here:
\begin{itemize}

\item[(1)] We have a \bfidx{contravariant} functor from vector spaces to vector
spaces.  It is contravariant in the categorical sense, as in section
\ref{sec:functors}.  This is because the arrows are reversed.  (See
\ref{sec:hom_functors} or any of the cited algebra texts for more information;
$f^*$ is called a contravariant \bfidx{hom functor}.)

\item[(2)] To compute the pullback $f^*\lambda$ of $\lambda$ by $f$, just write
$\lambda\circ f$.

\end{itemize}

%Is it true that $f^*(\phi) = \phi\circ f$ and $f_*(\phi) = f \circ \phi$?
%Check Lang.  or at least, warn.

% ----------------------------------------------------------------
\subsubsection{Pairings}
\label{sec:pairings}

\begin{defn}
Let $V$ and $W$ be real vector spaces.  A \bfidx{pairing} is a bilinear
function from $V \times W$ to $\R$.  Given $\vecv \in V$ and $\vecw \in W$, we
write
$$
	\pig{\vecv, \vecw}.
$$
\end{defn}

\beginex
The usual dot product is a pairing of $V$ with itself:
$$\langle\vecv_1,\vecv_2\rangle = \vecv_1 \cdot \vecv_2.$$
\mendex

\beginex
We can pair $V$ and $V^*$ via
$$\langle\lambda,\vecv\rangle = \lambda(\vecv).$$
\mendex

%\beginex
%Let the first vector space (not finite-dimensional) be the set $C$ of smooth
%vector-to-scalar functions $f$ from the unit interval; let the second vector
%space be $C^*$, the set of integrals $\mu$ from $C$ to $\R$.  (Here, the
%integrals play the role of linear functionals, while the functions are
%vectors.) Just as with the previous example, we pair functions with integrals
%by integrating the functions.
%\mendex

See section \ref{sec:gen_stokes} for another example of a pairing.

\begin{defn}
Let $V$ and $W$ be paired vector spaces.  For a linear transformation
$T:V \to V$, a linear transformation $T^*: W \to W$ is the \bfidx{adjoint}
of $T$ if
$$
	\pig{T\vecv, \vecw} = \pig{v, T^* \vecw}
$$
for all $\vecv \in V$ and all $\vecw \in W$.
\end{defn}

The adjoint exists and is unique \cite{bib:fis}.

\beginex
In particular, for the usual pairing of $V$ and $V^*$, we have
$$ \pig{T\vecv, \lambda} = \pig{v, T^* \lambda} = \pig{\vecv, \lambda T^t}.$$
[xxx elaborate.]
\mendex

%% ----------------------------------------------------------------
\subsubsection{Symmetric and skew-symmetric matrices}
\label{sec:mx_symms}

%[xxx perhaps motivate a bit via the adjoint kind of transpose.]

\begin{defn}
Let $A$ be an $m \times m$ matrix.
Then $A$ is said to be \bfidx{symmetric}
if $a_{ij} = a_{ji}$ for all $i,j$ from 1 to $m$.
Likewise, $A$ is said to be \bfidx{skew-symmetric}
if $a_{ij} = -a_{ji}$ for all $i,j$ from 1 to $m$.
\end{defn}

\begin{rem}
It is an easy exercise to show that an arbitrary $m \times m$ matrix $A$ may be
written as the sum of symmetric and skew-symmetric matrices.  This is done
explicitly by forming the \bfidx{symmetrization} and
\bfidx{skew-symmetrization} of $A$, namely,
$$
	B = \frac{1}{2}\left(A + A^t\right),\quad
	C = \frac{1}{2}\left(A - A^t\right),\quad
	A = B+C.
$$
These are appropriately named:  we take the entries of $A$ and make matrices
which are, \emph{by construction}, symmetric and skew-symmetric, respectively.
\end{rem}

\beginex
Let $$A=\begin{pmatrix}1 & 2 \\ 3 & 4 \end{pmatrix}.$$
Then the symmetrization of $A$ is
$$
\frac{1}{2}
\left[
\begin{pmatrix}1 & 2 \\ 3 & 4 \end{pmatrix} +
\begin{pmatrix}1 & 3 \\ 2 & 4 \end{pmatrix}
\right]
=
\begin{pmatrix}1 & 5/2 \\ 5/2 & 4 \end{pmatrix}
$$
while the skew-symmetrization of $A$ is
$$
\frac{1}{2}
\left[
\begin{pmatrix}1 & 2 \\ 3 & 4 \end{pmatrix} -
\begin{pmatrix}1 & 3 \\ 2 & 4 \end{pmatrix}
\right]
=
\begin{pmatrix}0 & -1/2 \\ 1/2 & 0 \end{pmatrix}.
$$
\mendex

It is easy to show that the set of symmetric $m \times m$ matrices is a real
vector space of dimension $m(m+1)/2$; the set of skew symmetric matrices is a
real vector space of dimension $m(m-1)/2$; the set of all $m \times m$ matrices
is a real vector space of dimension $m^2$, and is the direct sum of the former
two spaces.

Let $A$ be an $m \times m$ symmetric matrix with entries $a_{ij}$.  Each
$a_{ij}$ must be the same as $a_{ji}$.  Thus, to specify a symmetric matrix it
suffices to specify the entries $a_{ij}$ for $i \le j$.  There are $m(m+1)/2$
choices for values $a_{ij}$ with $i\le j$.  Thus, for example, a basis for the
vector space of symmetric $3 \times 3$ matrices is
\begin{eqnarray*}
	&&
%
	\begin{pmatrix}
	1 & 0 & 0 \\
	0 & 0 & 0 \\
	0 & 0 & 0 \\
	\end{pmatrix}, \quad
%
	\begin{pmatrix}
	0 & 0 & 0 \\
	0 & 1 & 0 \\
	0 & 0 & 0 \\
	\end{pmatrix}, \quad
%
	\begin{pmatrix}
	0 & 0 & 0 \\
	0 & 0 & 0 \\
	0 & 0 & 1 \\
	\end{pmatrix}, \\
%
	&&
	\begin{pmatrix}
	0 & 1 & 0 \\
	1 & 0 & 0 \\
	0 & 0 & 0 \\
	\end{pmatrix}, \quad
%
	\begin{pmatrix}
	0 & 0 & 1 \\
	0 & 0 & 0 \\
	1 & 0 & 0 \\
	\end{pmatrix}, \quad
%
	\begin{pmatrix}
	0 & 0 & 0 \\
	0 & 0 & 1 \\
	0 & 1 & 0 \\
	\end{pmatrix}.
\end{eqnarray*}
Likewise, for the skew-symmetric matrices, all the $a_{ii}$'s must be 0; there
are $m(m-1)/2$ choices for the remaining coefficients.  We can specify the
values of a skew-symmetric matrix by specifying the coefficients $a_{ij}$ for
$i<j$.  A basis is
\begin{eqnarray*}
	&&
	\begin{pmatrix}
	0 & 1 & 0 \\
	-1 & 0 & 0 \\
	0 & 0 & 0 \\
	\end{pmatrix}, \quad
%
	\begin{pmatrix}
	0 & 0 & 1 \\
	0 & 0 & 0 \\
	-1 & 0 & 0 \\
	\end{pmatrix}, \quad
%
	\begin{pmatrix}
	0 & 0 & 0 \\
	0 & 0 & 1 \\
	0 &-1 & 0 \\
	\end{pmatrix}.
\end{eqnarray*}

%% ================================================================
\newpage
\subsection{Tensors}
\label{sec:tensors}

\begin{center}
\emph{
Let us be patient! These severe afflictions \\
Not from the ground arise; \\
But oftentimes celestial benedictions \\
Assume this dark disguise.
} \\
--- Henry Wadsworth Longfellow\index{Longfellow, Henry Wadsworth} (1807-1882).
\end{center}

%\begin{center}
%\emph{
%The bottom line for mathematicians is that the architecture has to be right.
%%In all the mathematics that I did, the essential point was to find the right
%%architecture.
%\ldots
%It's like building a bridge. Once the main lines of the structure
%are right, then the details miraculously fit. The problem is the overall
%design.} --- Freeman Dyson (1923-).
%\end{center}

Tensors are encountered throughout math and physics.  They are presented in at
least four seemingly distinct guises:

\begin{itemize}

\item[(1)] Tensor products from abstract algebra.

\item[(2)] Tensors as $k$-linear functions $T:V^k \to \R$.

\item[(3)] Tensors as $k$-dimensional arrays: a 0-tensor is related to a
scalar, a 1-tensor is related to an array, a 2-tensor is related to a matrix, a
3-tensor is related to a 3-dimensional array, etc.

\item[(4)] Old-fashioned tensors from physics (``transform according to
\ldots'', e.g. \cite{bib:pdm}).  (In this guise, tensors appear particularly
foreign.  You will see lots of superscripts, subscripts, Einstein summation,
etc.  Why bother?  Well, this is the way tensors are usually viewed in
applications, so this is the language that your scientific collaborators will
be speaking.)

\end{itemize}

[xxx introduce a new section on the zeroth guise:  geometry.  Functional is
$\vecu^*$.  Bivector $\vecu\wedge\vecv$ is the equivalence class of all vector
pairs coplanar with $\vecu$ and $\vecv$, with the same signed area.  For the
symmetric product, all pairs coplanar to $\vecu$ and $\vecv$ with the same
inner product.  Bifunctional $\vecu^*\wedge\vecv^*$ measures area spanned
by two other vectors, projected onto the plane spanned by $\vecu$ and $\vecv$.
Also equivalence classes using
$$
	(\lambda\wedge\mu)(\vecu,\vecv) =
	(\lambda\wedge\mu)(\vecu\wedge\vecv) =
	\det(\lambda|\mu)\det(\vecu|\vecv).
$$
Generalizations to 3 and more dimensions.  xref back frequently
to the geometry section; xref forward to the axiomatic tensor definitions.
Suggest that, although I don't know the history well enough, I \emph{suspect}
that it was precisely these geometric notions which led to the axioms we have
today.]

These four guises are all equivalent, as we will see.  Here, the first guise is
taken to be the base definition.  Note that \cite{bib:satan} takes the second
guise to be the base definition.  Following \cite{bib:conlon} and
\cite{bib:abr}, I take the first guise to be the base definition; the second
guise follows as a consequence as shown in section
\ref{sec:tensor_product_of_duals}.  There are two reasons for this: (1) we can
draw on the abstract algebra that we possess as first-year graduate students in
math; (2) we will be able to view vector fields as a special case of tensor
fields.  This will lead to a more elegant view of Lie derivatives later on.

%% ----------------------------------------------------------------
\subsubsection{Tensor products of vector spaces}
\label{sec:tensor_product_of_vs}

The tensor product of modules (here, it suffices to consider tensor products of
vector spaces) is discussed in any graduate algebra text, e.g.  \cite{bib:df},
\cite{bib:grove}, \cite{bib:hung}, \cite{bib:lang}.  There is some amount of
fuss about the existence and uniqueness of the tensor product; one loses the
forest for the trees.  The essence of the tensor product is very simple, as
follows.

\begin{itemize}

\item[(1)] Let $V$ and $W$ be vector spaces.  We can form the \bfidx{Cartesian
product} $V \oplus W$.  Elements of this product are of the form $(\vecv,
\vecw)$ for $\vecv \in V$ and $\vecw \in W$.

\item[(2)] Given any set we can form a \bfidx{free abelian group} on that set,
as discussed in section \ref{sec:free_modules}.  That is, for the free abelian
group on $V \oplus W$, all of the $(\vecv, \vecw)$ pairs are distinct.

\item[(3)] The \bfidx{tensor product} of $V$ and $W$, written $$ V \tensor W,$$
with elements written $$ \vecv \tensor \vecw $$ for $\vecv \in V$ and $\vecw
\in W$, is the free abelian group \bfidx{modded out} by the following
\bfidx{relations} (see definition \ref {defn:relation}):

\begin{itemize}
\item[(a)] For any $r\in \R$,
$$r\vecv \tensor \vecw = \vecv \tensor r\vecw.$$
\item[(b)] We can distribute:
\begin{eqnarray*}
	(\vecv_1 + \vecv_2) \tensor \vecw &=&
		\vecv_1 \tensor \vecw + \vecv_2 \tensor \vecw\\
	\vecv \tensor (\vecw_1 + \vecw_2) &=&
		\vecv \tensor \vecw_1 + \vecv \tensor \vecw_2.
\end{eqnarray*}
\end{itemize}

That is, all $(\vecv,\vecw)$ pairs are distinct, \emph{except} when we can do
some of the above three operations to make them equal.  (This is to say that
all elements of a tensor product are equivalence classes, with equivalence
given by the above relations.) In property (a), we permit ourselves to move
\emph{real} numbers back and forth.  For this reason, in an algebra text, $V
\tensor W$ would be written $V \tensor_\R W$.
\end{itemize}

\beginex
\label{tensor_r2_example_1}
Let $V=W=\R^2$.  Then $$ (2,4)\tensor(3,5) = (1,2)\tensor(6,10) $$ since we can
move the 2 from the left-hand component to the right-hand component.
\mendex

%% ----------------------------------------------------------------
\subsubsection{Explicit representation of the tensor product; tensor product
as vector space}
\label{sec:explicit_tensor_v}

It is proved in the above-cited algebra texts (see \cite{bib:df} section 10.4
for a particularly lucid discussion) that for vector spaces, $$\R^m \tensor
\R^n \cong \R^{mn}$$ as an isomorphism of vector spaces.  What this means for
us is that, given
\begin{eqnarray*}
	\dim(V) = m, && \dim(W)=n\\
	\vecv=(v_1, \ldots, v_m),&& \vecw=(w_1, \ldots, w_n)
\end{eqnarray*}
where coefficients are taken with respect to the standard basis, we can think
of $\vecv \tensor \vecw$ as the \bfidx{outer product}
$$
\begin{pmatrix}
u_1v_1 & \cdots & u_1 v_n \\
\vdots &  & \vdots \\
u_mv_1 & \cdots & u_m v_n
\end{pmatrix}
$$
which is to say that $\vecv \tensor \vecw$ is an $m \times n$ \bfidx{array}
with $i,j$th entry $$ u_i v_j.$$

\beginex
\label{ex:outer_product}
Outer products may be done with pencil and paper as follows.  Write $\vecu$
down a column and $\vecv$ across a row, then fill in products of $u_i v_j$.
For example, let $\vecu = (1, 2)$ and $\vecv = (3, 4)$.  Then we have
$$
\begin{tabular}{c|cc}
         & 3 & 4 \\
\hline 1 & 3 & 4 \\
       2 & 6 & 8 \\
\end{tabular}
$$
\mendex

These are not matrices in the regular sense: such things are added
componentwise just as for matrices, but we'll see in section
\ref{sec:tensor_algebras} that they are multiplied using a different rule.

Of course, we can take the tensor product of several vector spaces.  Sample
elements of $U\tensor V \tensor W$ would be of the form $\vecu \tensor \vecv
\tensor \vecw$, with $i,j$th entry $u_i v_j w_k$.  For this course we are
interested in $V$ tensored with itself an arbitrary number of times.  In fact,
we have the following definitions:

\begin{defn}
Take $V$ tensored with itself $r$ times.  This is called the \bfidx{$r$-fold
tensor product} of $V$.
\end{defn}

\begin{rem}
Let $m=\dim(V)$.
By the above isomorphism, $\dim(V \tensor V) = m^2$, and in general the
$r$-fold tensor product of $V$ with itself has dimension $m^r$ over $\R$.
\end{rem}

Notation will be introduced in section \ref{sec:mixed_tensors}.  We will write
the $r$-fold tensor product of $V$ with itself as \index{$\mathcal{T}^r_0(V)$}
$\mathcal{T}^r_0(V)$, where the subscript 0 will be explained in section
\ref{sec:tensor_product_of_duals}.

%\begin{defn} An element of $\mcT^r_0(V)$, of the form $\vecv_1 \tensor \cdots
%\tensor \vecv_r$, is called an \bfidx{$r$-tensor}.  The number $r$ is called
%the \bfidx{degree} of the tensor $\vecv_1 \tensor \cdots \tensor \vecv_r$.
%\end{defn}

Given the above isomorphism, we can think of $V \tensor W$ as a \bfidx{vector
space} over $\R$.  Let $m=\dim V$, $n=\dim W$, and
let
$$
\{\vece_1, \ldots, \vece_m\},
\{\vecf_1, \ldots, \vecf_n\}
$$
be the standard bases for $V$ and $W$, respectively.  Basis elements for $V
\tensor W$ are of the form $$ \vece_i \tensor \vecf_j$$ where $1 \le i \le m$
and $1 \le j \le n$.

\beginex
\label{ex:r2_ox_r3_basis}
Let $V = \R^2$ and $W = \R^3$.  Then we have bases
$$
\{ \vece_1=(1,0), \vece_2=(0,1) \}, \qquad
\{ \vecf_1=(1,0,0), \vecf_2=(0,1,0), \vecf_3=(0,0,1) \}
$$
for $\R^2$ and $\R^3$.
Computing as in example \ref{ex:outer_product},
a basis for $\R^2 \tensor \R^3$ is
\begin{eqnarray*}
&&
\vece_1 \tensor \vecf_1 = \begin{pmatrix}
	1 & 0 & 0 \\
	0 & 0 & 0 \\
\end{pmatrix},
\vece_1 \tensor \vecf_2 = \begin{pmatrix}
	0 & 1 & 0 \\
	0 & 0 & 0 \\
\end{pmatrix},
\vece_1 \tensor \vecf_3 = \begin{pmatrix}
	0 & 0 & 1 \\
	0 & 0 & 0 \\
\end{pmatrix}, \\
&&
\vece_2 \tensor \vecf_1 = \begin{pmatrix}
	0 & 0 & 0 \\
	1 & 0 & 0 \\
\end{pmatrix},
\vece_2 \tensor \vecf_2 = \begin{pmatrix}
	0 & 0 & 0 \\
	0 & 1 & 0 \\
\end{pmatrix},
\vece_2 \tensor \vecf_3 = \begin{pmatrix}
	0 & 0 & 0 \\
	0 & 0 & 1 \\
\end{pmatrix}.
\end{eqnarray*}
\mendex

\beginex
\label{ex:u_outer_v}
Let $V=W=\R^2$.  In example \ref{tensor_r2_example_1} we saw that from
the definition of the tensor product we could set
$$(2,4)\tensor(3,5) = (1,2)\tensor(6,10)$$
by moving the scalar 2 from one side to the other.  What happens when
we think of these as explicit outer products?  The left-hand side is
$$
(2,4) \tensor(3,5)
= \begin{pmatrix} 2\cdot 3 & 2 \cdot 5 \\ 4 \cdot 3 & 4 \cdot 5 \end{pmatrix}
= \begin{pmatrix} 6 & 10 \\ 12 & 20 \\\end{pmatrix}
$$
whereas the right-hand side is
$$
(1,2) \tensor(6,10)
= \begin{pmatrix} 1\cdot 6 & 1 \cdot 10 \\ 2 \cdot 6 & 2 \cdot 10 \end{pmatrix}
= \begin{pmatrix} 6 & 10 \\ 12 & 20 \\\end{pmatrix}.
$$
This example drives home the fact that while the tensor product is associative,
it is certainly not commutative.  Also note that, in contrast to the
representation used in example \ref{tensor_r2_example_1}, arrays give us
\emph{unique} representations for tensors.
\mendex

%% ----------------------------------------------------------------
\subsubsection{Decomposability of tensors}
\label{sec:decompose}

In example \ref{ex:u_outer_v} we wrote an element of $V \tensor V$ as the outer
product of two vectors $\vecu$ and $\vecv$ in $V$.  Given the vector-space
representation, it is clear that we can add a pair of two-tensors
componentwise, e.g. if $\vecu_1, \vecu_2, \vecv_1, \vecv_2 \in V$, then $$
\vecu_1 \tensor \vecv_1 + \vecu_2 \tensor \vecv_2 \in V \tensor V.$$ When we do
that, will the result be representable as the outer product of two vectors?
Probably not.  To see this, note the following: consider $\vecu \tensor \vecv$.
With respect to the standard basis, we wrote this as an $m \times m$ array
(where $m = \dim V$), with coefficients $u_i v_j$.  Now, the $i$th row of
$\vecu \tensor \vecv$ is just the vector $u_i \vecv$.  So, $\vecu \tensor
\vecv$ has row rank 1 (or 0, if $\vecu$ or $\vecv$ is zero) when viewed as a
2-dimensional array:  every row is a multiple of the same vector $\vecv$.
(Likewise, the $j$th column of $\vecu \tensor \vecv$ is $\vecu v_j$.) So, if an
element $\psi$ of $V \tensor V$ is of the form $\vecu \tensor \vecv$, then it
will have row rank at most 1, with respect to any basis we choose.  Likewise,
if an element of $V \tensor V$ has row rank greater than 1, it could not be of
the form $\vecu \tensor \vecv$.  Now, we can certainly write down lots of $m
\times m$ arrays which have row rank greater than 1.  This means most elements
of $V \tensor V$ aren't expressible in the form $\vecu \tensor \vecv$.  This
motivates the following definition.

\begin{defn}
\label{defn:decompose}
A tensor $\psi \in \mcT^r(V)$ is said to be \bfidx{simple} or
\bfidx{decomposable} if it can be written as a tensor product of $r$ vectors in
$V$.
\end{defn}

\beginex
Let $V=\R^2$.  With respect to the standard basis,
$$
\alpha = \begin{pmatrix} 1 & 2 \\ 3 & 6 \\ \end{pmatrix}
$$
is an element of $V \tensor V$ which is decomposable into
$$ \alpha = (1,3) \tensor (1,2).$$
On the other hand,
$$
\beta = \begin{pmatrix} 1 & 2 \\ 3 & 4 \\ \end{pmatrix}
$$
is indecomposable.
\mendex

Note that using the vector-space representation as shown in section
\ref{sec:explicit_tensor_v}, while we cannot decompose an arbitrary tensor as a
tensor product of vectors, we can write an arbitrary tensor as the \emph{sum}
of simple tensors.  (We can do this because of the isomorphism in section
\ref{sec:explicit_tensor_v}.)

\beginex
Take $\beta$ from the previous example.  Proceeding as in example
\ref{ex:r2_ox_r3_basis}, write $\beta$ with respect to the basis $\{\vece_i
\tensor \vecf_j\}$:
\begin{eqnarray*}
	\beta
	&=& \begin{pmatrix} 1 & 2 \\ 3 & 4 \\ \end{pmatrix}
	= 1 \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix}
	+ 2 \begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix}
	+ 3 \begin{pmatrix} 0 & 0 \\ 1 & 0 \\ \end{pmatrix}
	+ 4 \begin{pmatrix} 0 & 0 \\ 0 & 1 \\ \end{pmatrix} \\
	&=& \vece_1 \tensor \vecf_1
	+ 2 \vece_1 \tensor \vecf_2
	+ 3 \vece_2 \tensor \vecf_1
	+ 4 \vece_2 \tensor \vecf_2.
\end{eqnarray*}
\mendex

%% ----------------------------------------------------------------
\subsubsection{Tensor products of dual spaces; tensors as $k$-linear functions}
\label{sec:tensor_product_of_duals}

In section \ref{sec:tensor_product_of_vs} we saw how to form the tensor product
$V \tensor W$ of vector spaces $V$ and $W$, or the $r$-fold tensor product
$\mcT^r_0(V)$.  However, given a vector space $V$, the dual $V^*$ is also a
vector space so it makes sense to tensor $V^*$ with itself.  Here, however, we
get an additional feature:  we can think of such tensors as \emph{functions}.
We could write $V^* \tensor V^*$ as $\mcT^2_0(V^*)$, but we will instead write
$V^* \tensor V^*$ as $\mcT^0_2(V)$.  (This will be explained in section
\ref{sec:mixed_tensors}.)  In general, the $r$-fold tensor product of $V^*$
with itself will be written $\mcT^0_r(V)$.

Given linear functionals $\lambda, \mu : V \to \R$, we can certainly write down
the tensor product $\lambda \tensor \mu.$ But what can this mean as a function?
Given $\vecv, \vecw \in V$, we think of this as
$$
(\lambda \tensor \mu)(\vecv \tensor \vecw) = \lambda(\vecv) \mu(\vecw)
$$
where we simply multiply the outputs $\lambda(\vecv)$ and $\mu(\vecw)$ in $\R$.
This means that $V^* \tensor V^*$ consists of functions from $V \oplus V$ to
$\R$.

However, more can be said about such functions:  they are \bfidx{bilinear},
i.e. they are linear in each slot if we hold the other slot fixed.

\beginex
For example, let $\vecu_1, \vecu_2, \vecv_1, \vecv_2 \in V$ and let $a, b \in
\R$.  Given $\lambda \tensor \mu$ as above,
\begin{eqnarray*}
(\lambda \tensor \mu)(\vecu_1+\vecu_2, \vecv)
	&=& \lambda(\vecu_1 + \vecu_2) \mu(\vecv) \\
	&=& [\lambda(\vecu_1) + \lambda(\vecu_2)] \mu(\vecv) \\
	&=& \lambda(\vecu_1) \mu(\vecv) + \lambda(\vecu_2) \mu(\vecv) \\
	&=& (\lambda \tensor \mu)(\vecu_1, \vecv)
	+ (\lambda \tensor \mu)(\vecu_2, \vecv).
\end{eqnarray*}
Likewise, we have
\begin{eqnarray*}
	(\lambda \tensor \mu)(\vecu, \vecv_1+\vecv_2)
		&=& (\lambda \tensor \mu)(\vecu, \vecv_1) + \lambda(\vecu, \vecv_2) \\
%
	(\lambda \tensor \mu)(a \vecu, \vecv)
		&=& a (\lambda\tensor\mu)(\vecu, \vecv)
		= (\lambda \tensor \mu)(\vecu, a \vecv)
\end{eqnarray*}
These things work precisely because $\lambda$ and $\mu$ are linear.
As a consequence:
\begin{eqnarray*}
\lambda(a \vecu, b \vecv) &=& ab \lambda(\vecu, \vecv) \\
\lambda(\vecu_1+\vecu_2, \vecv_1+\vecv_2) &=&
\lambda(\vecu_1, \vecv_1) +
\lambda(\vecu_1, \vecv_2) +
\lambda(\vecu_2, \vecv_1) +
\lambda(\vecu_2, \vecv_2)
\end{eqnarray*}
and so on.
\mendex

Likewise, given $r$ linear functionals $\lambda_1, \ldots, \lambda_r \in V^*$,
and $r$ vectors $\vecv_1, \ldots, \vecv_r \in V$, we can write
$$
(\lambda_1 \tensor \cdots \tensor \lambda_r)
(\vecv_1 \tensor \cdots \tensor \vecv_r)
=
\lambda_1(\vecv_1) \cdots \lambda_r(\vecv_r).
$$
The following notation is handy:

\begin{defn}
Write $V^r$ for the $r$-fold \plainidx{Cartesian product} of $V$ with itself.
\end{defn}

Then we can say that $\mcT^0_r(V)$ consists of \bfidx{$r$-linear} functions
from $V^r$ to $\R$.  Note that elements of $V^r$ are just $r$-tuples of
vectors.  For example, if $V$ has dimension $m$, we might represent an element
of $V^r$ using $mr$ scalars, with respect to some basis for $V$:
$$
	(\vecv_1, \ldots, \vecv_r) =
	((v_{11}, \ldots, v_{1m}),
	(v_{21}, \ldots, v_{2m}),
	\ldots,
	(v_{r1}, \ldots, v_{rm})).
$$

Note that, just as in section \ref{sec:decompose}, an element of $\mcT^0_r(V)$
may or may not be decomposable into a tensor product of linear functionals.
However, as discussed in section \ref{sec:decompose}, we can decompose an
element of $\mcT^0_r(V)$ into a sum of simple tensors, using a basis of $V^*$.
An example appears in the next section.

%\begin{defn}
%Write $V^k$ for the $k$-fold cartesian product $V \oplus V \oplus \ldots \oplus
%V \oplus V$.
%\end{defn}
%
%For example, if $V$ has dimension $m$, we might represent an element
%of $V^k$ using $mk$ scalars, with respect to some basis for $V$:
%$$
%	(\vecv_1, \ldots, \vecv_k) =
%	((v_{11}, \ldots, v_{1m}),
%	(v_{21}, \ldots, v_{2m}),
%	\ldots,
%	(v_{k1}, \ldots, v_{km}))
%$$

%% ----------------------------------------------------------------
\subsubsection{Explicit representation of multilinear functions}
\label{sec:explicit_rep_multilin_fcns}

In section \ref{sec:explicit_tensor_v} we saw that, given a basis, we can view
elements of $$\mcT^s_0(V) = \overbrace{V\tensor \cdots \tensor V}^{s \;
\textrm{times}}$$ as multidimensional arrays.  The same is true for elements of
$$\mcT^0_r(V) = \underbrace{V^* \tensor \cdots \tensor V^*}_{r
\;\textrm{times}}.$$

\beginex
\label{ex:decompose_three_tensor}
Let $\psi \in V^* \tensor V^* \tensor V^*$ where $m=\dim V$. (I do not need to
assume that $\psi$ is decomposable into the form $\lambda \tensor \mu \tensor
\nu$ for linear functionals $\lambda$, $\mu$, and $\nu$.)  Then for $u,v,w\in
V$, using multilinearity (here, trilinearity) we have:
\begin{eqnarray*}
	\psi(\vecu, \vecv, \vecw)
	&=& \psi(
		u_1 \vece_1 + u_2 \vece_2 + u_3 \vece_3,
		v_1 \vece_1 + v_2 \vece_2 + v_3 \vece_3,
		w_1 \vece_1 + w_2 \vece_2 + w_3 \vece_3) \\
	&=& \sum_{i=1}^m \sum_{j=1}^m \sum_{k=1}^m u_i v_j w_k
		\psi(\vece_i, \vece_j, \vece_k)
	= \sum_{i=1}^m \sum_{j=1}^m \sum_{k=1}^m u_i v_j w_k a_{ijk}
\end{eqnarray*}
where the $a_{ijk}$'s are real numbers which depend on $\psi$.  All the
information about what $\psi$ does is contained in the 3-dimensional $m\times m
\times m$ array with entries $a_{ijk}$.  In particular, all the $a_{ijk}$'s are
\emph{constants}.
\mendex

Here is the key \bfidx{application formula} for tensors.  That is, we can think
of an $r$-tensor as an $r$-dimensional array, while we can also think of it as
an $r$-linear function operating on $r$ vectors.  How do we apply the former to
the latter?  This is illustrated by example.

\beginex Let $T$ be a 3-tensor on $V$; for this example, use the standard
basis for $V$.  As discussed in section \ref{sec:explicit_tensor_v}, $T$ is a
linear combination of the $m^3$ basis tensors $\vece_i^* \tensor \vece_j^*
\tensor \vece_k^*$.  Likewise, vectors $\vecu$, $\vecv$, and $\vecw$ are also
linear combinations of the basis vectors $\vece_i$:
\begin{eqnarray*}
	T &=& \sum_{ijk} T(\vece_i, \vece_j, \vece_k) \vece_i^* \otimes \vece_j^* \otimes \vece_k^*
	= \sum_{ijk} t_{ijk} \vece_i^* \otimes \vece_j^* \otimes \vece_k^* \\
	\vecu &=& \sum_{i'} u_{i'} \vece_{i'}, \qquad
	\vecv = \sum_{j'} v_{j'} \vece_{j'}, \qquad
	\vecw = \sum_{k'} w_{k'} \vece_{k'} \\
	T\left(\vecu,\vecv,\vecw\right) &=& \sum_{ijk} t_{ijk} \vece_i^* \otimes \vece_j^* \otimes \vece_k^*\left(
		\sum_{i'} u_{i'} e_{i'},
		\sum_{j'} v_{j'} e_{j'},
		\sum_{k'} w_{k'} e_{k'}\right)
	= \sum_{ijk} t_{ijk} u_i v_j w_k.
\end{eqnarray*}
\mendex

\begin{rem}
When doing computations by hand, another expression is useful.  This is
illustrated by a 2-tensor example, with $\dim(V)=2$.  Suppose that, with
respect to the standard basis, we have the following 2-tensor $\psi$ and vectors $\vecu,
\vecv$.  We might write
$$
	\psi(\vecu,\vecv) =
	\begin{pmatrix}
		1 & 2 \\
		3 & 4 \\
	\end{pmatrix}
	\begin{pmatrix}
		5 \\ 6 \\
	\end{pmatrix}
	\begin{pmatrix}
		7 \\ 8 \\
	\end{pmatrix}
	=
	1\cdot 5 \cdot 7 +
	2\cdot 5 \cdot 8 +
	3\cdot 6 \cdot 7 +
	4\cdot 6 \cdot 8
	= 433.
$$
However, it is easy to mistake whether the 2 goes with the 5 and 8, or with the
6 and 7.  It is less error-prone to observe that in the expression
$$
	\psi(\vecu, \vecv) = \sum_{ij} t_{ij} u_i v_j
$$
we have $u_i v_j$ which are precisely the components of $\vecu \tensor \vecv$.
Define $\odot$ on multi-dimensional arrays to be \emph{componentwise}
multiplication, e.g.  if $A$ and $B$ have entries $a_{ij}$ and $b_{ij}$, then
$A \odot B$ has entries $a_{ij}b_{ij}$.  Then
$$
	\psi(\vecu, \vecv) = \psi \odot (\vecu \tensor \vecv).
$$
In this example, multiplying out $\vecu \tensor \vecv$ as in example
\ref{ex:outer_product}, we have
$$
	\psi(\vecu,\vecv)
%
	= \begin{pmatrix}
		1 & 2 \\
		3 & 4 \\
	\end{pmatrix}
	\odot \left[
	\begin{pmatrix}
		5 \\ 6 \\
	\end{pmatrix}
	\tensor
	\begin{pmatrix}
		7 \\ 8 \\
	\end{pmatrix}
	\right]
%
	= \begin{pmatrix}
		1 & 2 \\
		3 & 4 \\
	\end{pmatrix}
	\odot
	\begin{pmatrix}
		35 & 40 \\
		42 & 48 \\
	\end{pmatrix}
%
	= 35 + 80 + 126 + 192 = 433.
$$
In general, for an $r$-tensor, write out the tensor product of the argument
vectors, then do a componentwise product with the tensor coefficients, then sum
up.
\end{rem}

\begin{rem}
In this paper, I do many tensor computations with respect to coordinates ---
usually the standard basis.  However, change of coordinates is a key part of
tensor algebra, as well as differential geometry.  In fact, coordinates are
just an expression of an abstract object with respect to specific coordinates.
$$
\framebox[6.0in][c]{
	\textbf{A multidimensional array is to a tensor
	as a matrix is to a linear transformation.}
}
$$
Change of coordinates will be discussed further in sections [xxx write and
xref].
\end{rem}
%whereas
%\begin{eqnarray*}
%	\vecu \otimes \vecv \otimes \vecw &=& (u_i v_j w_k)_{ijk}.
%\end{eqnarray*}
%So, applying $T$ is the first formula, or equivalently we form
%$\vecu \otimes \vecv \otimes \vecw$, then sum up the elements
%of that array using $t_{ijk}$'s as weighting coefficents.

%\beginex
%xxx numerical example goes here
%\mendex

%% ----------------------------------------------------------------
\subsubsection{Mixed tensors}
\label{sec:mixed_tensors}

\begin{defn}
We write
$$
	\mcT^s_r(V) =
	\overbrace{V\otimes \cdots \otimes V}^{s \;\textrm{times}}
	\otimes
	\underbrace{V^* \otimes \cdots \otimes V^*}_{r \;\textrm{times}}.
$$
The integer $s$ is called the \bfidx{contravariant degree}; $r$ is called the
\bfidx{covariant degree}.  There are $s$ \bfidx{contravariant
indices}\index{indices} and $r$ \bfidx{covariant indices}.  An element of
$\mcT^s_r(V)$ is said to be of \bfidx{type} $s, r$.  If $s>0$ and $r>0$,
elements of $\mcT^s_r(V)$ are said to be of \bfidx{mixed type}.  When we refer
to an $r$-tensor, we mean a tensor of type $0,r$, i.e.  an $r$-linear function
on $V$.  An $r$-tensor is said to have \bfidx{degree} $r$, or sometimes
\bfidx{order} $r$.  We write
$$
	\ord(\omega)
$$
for the order\index{$\mathrm{ord}$} of an $r$-tensor $\omega$.
\end{defn}

Note the following in particular:

\begin{itemize}
\item $\mcT^0_0(V) = \R$.
\item $\mcT^1_0(V) = V$; $\mcT^0_1(V) = V^*$.
\item $\mcT^2_0(V) = V \tensor V$; $\mcT^1_1(V) = V \tensor V^*$;
	$\mcT^0_2(V) = V^* \tensor V^*$.
\item A sample decomposable element of $\mcT^2_0(V)$ is $\vecu \otimes \vecv$ for
$\vecu, \vecv \in V$.
\item A sample decomposable element of $\mcT^0_2(V)$ is $\lambda \otimes \mu$ for
$\lambda, \mu: V \to \R$ i.e. $\lambda, \mu \in V^*$.
\end{itemize}

Note that $\R$ is a subspace of $V$ and $V^*$, and so on.  So we have the
following\index{Rosetta stone}:

\setlength{\unitlength}{1.0in}
\begin{picture}(6,2.0)
%\put(0,0){\framebox(6,2.0)}

\put(2.90, 0.05){scalars}
\put(2.70, 0.25){$\mcT^0_0(V) = \R$}
	\put(2.70, 0.45){\vector(-1,1){.20}}
	\put(3.35, 0.45){\vector( 1,1){.20}}

\put(2.30, 0.75){vectors}
\put(3.50, 0.75){covectors}
\put(2.10, 0.95){$\mcT^1_0(V) = V$}
\put(3.30, 0.95){$\mcT^0_1(V) = V^*$}
	\put(2.25, 1.15){\vector(-1,1){.20}}
	\put(2.50, 1.15){\vector( 1,1){.20}}
	\put(3.55, 1.15){\vector(-1,1){.20}}
	\put(3.80, 1.15){\vector( 1,1){.20}}

\put(1.50, 1.45){$\mcT^2_0(V) = V\tensor V$}
\put(2.70, 1.45){$\mcT^1_1(V) = V \tensor V^*$}
\put(3.90, 1.45){$\mcT^0_2(V) = V^*\tensor V^*$}
	\put(1.65, 1.65){\vector(-1,1){.20}}
	\put(1.90, 1.65){\vector( 1,1){.20}}
	\put(2.95, 1.65){\vector(-1,1){.20}}
	\put(3.20, 1.65){\vector( 1,1){.20}}
	\put(4.15, 1.65){\vector(-1,1){.20}}
	\put(4.40, 1.65){\vector( 1,1){.20}}

\put(1.30, 1.95){\ldots}
\put(2.10, 1.95){\ldots}
\put(2.60, 1.95){\ldots}
\put(3.40, 1.95){\ldots}
\put(3.80, 1.95){\ldots}
\put(4.60, 1.95){\ldots}

\end{picture}

%xxx Show by example why this is backward from the functorial idea of covariance
%and contravariance.  Sadly bemoan the sullying influence of historical physics.
%Point out that this difference cannot be glossed over:  one sense of
%co/contravariant is frequently encountered; the other sense is also frequently
%encountered.

\begin{rem}
\label{rem:mixed_tensors_also_multilinear}
As noted in section \ref{sec:dual_and_double_dual}, we can identity $V$ with
$V^{**}$.  So, we can view an arbitrary mixed tensor as a multilinear function
from $(V^*)^s \oplus V^r$ to $\R$:
\begin{eqnarray*}
	\psi \in \mcT^s_r(V) &=&
	\overbrace{V^{**}\otimes \cdots \otimes V^{**}}^{s \;\textrm{times}}
	\otimes
	\underbrace{V^* \otimes \cdots \otimes V^*}_{r \;\textrm{times}} \\
	\psi: (V^*)^s \oplus V^r &\to& \R.
\end{eqnarray*}
\end{rem}

\begin{rem}
\label{rem:100r}
For this course we focus on tensors of type $0,r$, which we call
\bfidx{$r$-tensors}, and tensors of type $1,0$, which we call \bfidx{vectors}.
Mixed tensors perhaps seem like too much generality.  However, they have the
following benefits:
\begin{itemize}
\item In sections \ref{sec:metric_tensor} and
	\ref{sec:forms} we will extend this notion, and obtain \bfidx{forms} and
	\bfidx{vector fields}, respectively, over \bfidx{manifolds}.  Mixed tensors
	unify these concepts.
\item Fundamental concepts such as linear transformations (section
	\ref{sec:TS_COB}) and curvature (\cite{bib:gug}, \cite{bib:lee3}) turn out
	to be mixed tensors.
\end{itemize}
\end{rem}

% xxx write down examples.  xxx copy the nice tree diagram from section xxx to
% here.

%% ----------------------------------------------------------------
\subsubsection{Examples of tensors}
\label{sec:examples_of_tensors}

\beginex
Let $\lambda \in V^*$.  As discussed in section \ref{sec:compute_dual_basis},
with reference to a given basis, $\lambda$ is represented by a $1 \times m$
matrix, or \bfidx{row vector}.  The action $\lambda(\vecv)$ is the product of
$\lambda$'s row vector with $\vecv$'s \bfidx{column vector}, which is the dot
product of $\lambda$ and $\vecv$.  For example:
$$
\lambda(v) = \begin{pmatrix}a_1 a_2 \cdots a_m\end{pmatrix}
\begin{pmatrix}v_1 \\ v_2 \\ \vdots \\ v_m\end{pmatrix}
= a_1 v_1 + \ldots + a_m v_m.
$$
Another way to see this is that linearity of $\lambda$ gives
\begin{eqnarray*}
	\lambda(v) &=& \lambda\left(\sum_{i=1}^m v_i \vece_j \right)
	= \sum_{i=1}^m v_i \lambda(\vece_j)
	= \sum_{i=1}^m v_i a_i
\end{eqnarray*}
The 1-dimensional array $\veca$, of length $m$, gives the information needed to
compute $\lambda$ of any $\vecv \in V$.
\mendex

\beginex
An example of a 2-tensor is the standard dot product.  Let $\vecu, \vecv \in V$
with $m=\dim(V)$.  Holding either $\vecu$ or $\vecv$ fixed and varying the
other gives a linear transformation, so this makes sense.  With respect
to the standard basis, we can think of $\vecu \cdot \vecv$ as
\begin{eqnarray*}
	\vecu \cdot \vecv &=& \left(\sum_{i=1}^m u_i \vece_i\right) \cdot \vecv \\
	&=& \sum_{i=1}^m u_i \left(\vece_i \cdot \vecv\right) \\
	&=& \sum_{i=1}^m u_i \vece_i \cdot \left(\sum_{j=1}^m v_j \vece_j \right) \\
	&=& \sum_{i=1}^m \sum_{j=1}^m u_i v_j (\vece_i \cdot \vece_j) \\
	&=& \sum_{i=1}^m \sum_{j=1}^m u_i v_j \delta_{ij}.
\end{eqnarray*}
Here, the $m \times m$ array is the \bfidx{identity matrix}, with entries
$\delta_{ij}$.  These $m^2$ numbers indicate how to form the inner product with
respect to the standard basis in Euclidean space.  (Since the $m \times m$
identity matrix has rank $m$, if $m>1$ then the dot-product tensor is
\bfidx{indecomposable} by the reasoning in section \ref{sec:decompose}.)
\mendex

\begin{rem}
Using matrices which are not necessarily the identity leads to \bfidx{bilinear
forms}; see \cite{bib:fis} for more information.  These two-tensors generalize
the inner product:  in fact, an inner product is nothing more than a symmetric
positive-definite bilinear form.
\end{rem}

\beginex
Recall that the determinant on $m \times m$ matrices is an $m$-linear function,
either on rows or on columns.  Here, think of it as an $m$-linear function on
rows.  (More can be said about the determinant:  see section
\ref{sec:sym_alt_algebras}.) Let $\vecu=(u_1, u_2), \vecv = (v_1, v_2) \in
\R^2$.  Because of the bilinearity of the determinant (xref backward after
marking the equation), we have
\begin{eqnarray*}
	\det(\vecu,\vecv) &=& \det\left(
		u_1 \vece_1 + u_2 \vece_2,
		v_1 \vece_1 + v_2 \vece_2
		\right) \\
	&=& \det\left(u_1 \vece_1, v_1 \vece_1\right)
		+ \det\left(u_1 \vece_1, v_2 \vece_2\right)
		+ \det\left(u_2 \vece_2, v_1 \vece_1\right)
		+ \det\left(u_2 \vece_2, v_2 \vece_2\right) \\
	&=& u_1 v_1 \det\left(\vece_1, \vece_1\right)
		+ u_1 v_2 \det\left(\vece_1, \vece_2\right)
		+ u_2 v_1 \det\left(\vece_2, \vece_1\right)
		+ u_2 v_2 \det\left(\vece_2, \vece_2\right).
\end{eqnarray*}
It remains to find the values of $\det(\vece_i, \vece_j)$.  But since we
are thinking of matrices as lists of row vectors, these are
\begin{eqnarray*}
\det(\vece_1, \vece_1)=\det\begin{pmatrix} 1 & 0 \\ 1 & 0 \\ \end{pmatrix}=0&&
\det(\vece_1, \vece_2)=\det\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}=1\\
\det(\vece_2, \vece_1)=\det\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}=-1&&
\det(\vece_2, \vece_2)=\det\begin{pmatrix} 0 & 1 \\ 0 & 1 \\ \end{pmatrix}=0
\end{eqnarray*}
So, the determinant tensor on rows of $2 \times 2$ matrices may be thought
of as
$$ \det(\vecu, \vecv) = \sum_{i=1}^2 \sum_{j=1}^2 u_i v_j d_{ij}$$
where $d_{ij}$ are the entries of the $2 \times 2$ array
$$ \begin{pmatrix}0 & 1 \\ -1 & 0 \\ \end{pmatrix}.$$
\mendex

\beginex
Similarly, for $3 \times 3$ matrices:
\begin{eqnarray*}
	\det(\vecu,\vecv,\vecw) &=& \det\left(
		\sum_{i=1}^3 u_i \vece_i,
		\sum_{j=1}^3 v_j \vece_j,
		\sum_{k=1}^3 w_k \vece_k
		\right). \\
	&=& \sum_{i=1}^3 \sum_{j=1}^3 \sum_{k=1}^3
		\det\left(u_i \vece_i, v_j \vece_j, w_k \vece_k \right) \\
	&=& \sum_{i=1}^3 \sum_{j=1}^3 \sum_{k=1}^3
		u_i v_j w_k \det\left(\vece_i, \vece_j, \vece_k \right).
\end{eqnarray*}
It remains to find the values of $\det(\vece_i, \vece_j, \vece_k)$.
There are 27 of them, but only 6 with non-zero values.  They are
\begin{eqnarray*}
\det(\vece_1, \vece_2, \vece_3)=\det\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}=1,&&
\det(\vece_1, \vece_3, \vece_2)=\det\begin{pmatrix}1&0&0\\ 0&0&1\\ 0&1&0 \end{pmatrix}=-1,\\
\det(\vece_2, \vece_3, \vece_1)=\det\begin{pmatrix}0&1&0\\ 0&0&1\\ 1&0&0 \end{pmatrix}=1,&&
\det(\vece_3, \vece_2, \vece_1)=\det\begin{pmatrix}0&0&1\\ 0&1&0\\ 1&0&0 \end{pmatrix}=-1,\\
\det(\vece_3, \vece_1, \vece_2)=\det\begin{pmatrix}0&0&1\\ 1&0&0\\ 0&1&0 \end{pmatrix}=1,&&
\det(\vece_2, \vece_1, \vece_3)=\det\begin{pmatrix}0&1&0\\ 1&0&0\\ 0&0&1 \end{pmatrix}=-1.\\
\end{eqnarray*}
The determinant tensor on rows of $3 \times 3$ matrices may be thought
of as
$$ \det(\vecu, \vecv, \vecw) = \sum_{i=1}^3 \sum_{j=1}^3 \sum_{k=1}^3 u_i v_j w_k d_{ijk}$$
where $d_{ijk}$ are the entries of the following $3 \times 3 \times 3$ array
where $i$ indexes the face (front, middle, back), $j$ indexes rows, and $k$
indexes columns:
$$
\begin{matrix}
\textrm{front} &
\textrm{middle} &
\textrm{back} \\
\left(
\left(\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 &-1 & 0 \\
\end{matrix}\right)
\right.
&
\left.
\left(\begin{matrix}
0 & 0 &-1 \\
0 & 0 & 0 \\
1 & 0 & 0 \\
\end{matrix}\right)
\right.
&
\left.
\left(\begin{matrix}
0 & 1 & 0 \\
-1 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix}\right)
\right)
\end{matrix}
.$$
\mendex

% ----------------------------------------------------------------
\subsubsection{Pullbacks}
\label{sec:t_pullback}

%Let $\lambda$ be a $k$-tensor, i.e. $\lambda: V^k \to \R$, and let $\vecv \in
%V$.  Given the above basis, we may write
%xxx fix me for tensors
%$$ \lambda = \sum_{i=1}^m a_i \vece_i^* \qand
%\vecv = \sum_{j=1}^m b_j \vece_j.
%$$
%Since the $e_i^*$'s are linear functions, and since $\vece_i^*(\vece_j) =
%\delta_{ij}$, we have
%$$
%	\lambda(\vecv) =
%		\sum_{i=1}^m a_i \vece_i^*\left( \sum_{j=1}^m b_j \vece_j\right)
%	= \sum_{i=1}^m \sum_{j=1}^m a_i b_j \vece_i^*(\vece_j)
%	= \sum_{i=1}^m a_i b_i.
%$$

This example is from \cite{bib:spi1}, p. 77.

Let $f:V \to W$ be a linear transformation, and let $\lambda: W^k \to \R$ be a
$k$-tensor, i.e. $\lambda \in \mcT^k(W)$.  We can map from $V^k$ to $\R$ by
going through $W^k$ first.  That is, for $(\vecv_1, \ldots, \vecv_k) \in V^k$,
$(f(\vecv_1), \ldots, f(\vecv_k))$ is in $W^k$, and so we can apply $\lambda$
to it:

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.0)
%\put(0,0){\framebox(6,1)}
\put(0.20, 0.80){$V^k$}
\put(0.25, 0.75){\vector(0,-1){.50}}
\put(0.10, 0.50){\small{$f$}}
\put(0.20, 0.10){$W^k$}
\put(0.35, 0.75){\vector(2,-1){.40}}
\put(0.35, 0.25){\vector(2,1){.40}}
\put(0.55, 0.70){\small{$\lambda \circ f$}}
\put(0.65, 0.20){\small{$\lambda$}}
\put(0.80, 0.50){$\R$}
\end{picture}

Formally, $\lambda \circ f: V^k \to \R$.  Given $\lambda \in \mcT^k(W)$, we
have obtained $\lambda \circ f \in \mcT^k(V)$.  Post-composing $\lambda$ by
$f$ is said to be a \bfidx{pullback} of $\lambda$ from $W^k$ to $V^k$.  Another
way to look at this is that, given $f:V \to W$, we have a map from $\mcT^k(W)$
to $\mcT^k(V)$:
\begin{eqnarray*}
	f:&&V \to W \\
	f^*:&& \mcT^k(W) \to \mcT^k(V) \\
	f^*(\lambda) &=& \lambda\circ f \\
	(f^*\lambda)(\vecv_1, \ldots, \vecv_k) &=& \lambda(f(\vecv_1), \ldots,
		f(\vecv_k)).
\end{eqnarray*}

In this way, $f^*$ takes $k$-tensors to $k$-tensors:

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.0)

%\put(0,0){\framebox(6,1)}
\put(0.20, 0.80){$V^k$}
\put(0.25, 0.75){\vector(0,-1){.50}}
\put(0.10, 0.50){\small{$f$}}
\put(0.20, 0.10){$W^k$}
\put(0.35, 0.75){\vector(2,-1){.40}}
\put(0.35, 0.25){\vector(2,1){.40}}
\put(0.55, 0.70){\small{$\lambda \circ f$}}
\put(0.65, 0.20){\small{$\lambda$}}
\put(0.80, 0.50){$\R$}

\put(2.20, 0.80){$\mcT^k(V)$}
\put(2.25, 0.25){\vector(0,1){.50}}
\put(2.10, 0.50){\small{$f^*$}}
\put(2.20, 0.10){$\mcT^k(W)$}

\end{picture}

%% ----------------------------------------------------------------
\subsubsection{Tensor algebras}
\label{sec:tensor_algebras}

In section \ref{sec:explicit_tensor_v} we saw that $\mcT^r_s(V)$ is a vector
space over $\R$, with dimension $\dim(V)^{r+s}$.  This permits addition of
tensors.  We can think of this in two ways, both of which are illustrated by
example.  First, since tensors are multilinear functions, we can pick two
tensors and $r$ vectors, then see what happens to the sum of tensors.  Doing
this uses the vectors as something external, though, so this is less than
satisfying.  Second, we can see what happens to the coefficients in the tensors
themselves.  The former turns into the latter as follows, for a pair of
2-tensors $\phi$ and $\psi$:
\begin{eqnarray*}
	\phi &=& \sum_{ij} \phi(\vece_i, \vece_j) \vece_i^* \tensor \vece_j^*, \quad
		\phi(\vece_i, \vece_j) = a_{ij} \\
	\psi &=& \sum_{ij} \psi(\vece_i, \vece_j) \vece_i^* \tensor \vece_j^*, \quad
		\psi(\vece_i, \vece_j) = b_{ij} \\
	(\phi + \psi)(\vecu, \vecv)&=& \phi(\vecu, \vecv) + \psi(\vecu, \vecv) \\
	&=& \sum_{ij} a_{ij} u_i v_j + \sum_{ij} b_{ij} u_i v_j \\
	&=& \sum_{ij} \left[a_{ij} + b_{ij}\right] u_i v_j.
\end{eqnarray*}
So, by applying our pair of tensors to an arbitrary pair of vectors, we see
that we didn't need them after all:  namely, the coefficients of $\phi+\psi$
are the sums of coefficients of $\phi$ and $\psi$.  This is just vector-space
addition, as we would expect.

Recall from section \ref{sec:algebras} that we can turn this vector space into
an algebra if only we can multiply tensors.  As with addition, we can do this
\bfidx{argumentwise} (with reference to vectors) or \bfidx{coefficientwise}
(with reference only to the tensors themselves).  That is, knowing that tensors
are $r$-linear functions, we can think of what tensor multiplication would
\emph{have} to be.  Again, this is illustrated by example.  Here, let $\phi$ be
a 2-tensor, and let $\psi$ be a 1-tensor.
\begin{eqnarray*}
	\phi &=& \sum_{ij} \phi(\vece_i, \vece_j) \vece_i^* \tensor \vece_j^*, \quad
		\phi(\vece_i, \vece_j) = a_{ij} \\
	\psi &=& \sum_{k} \psi(\vece_k) \vece_k^*, \quad
		\psi(\vece_k) = b_{k} \\
	(\phi \tensor \psi)(\vecu, \vecv, \vecw)&=& \phi(\vecu, \vecv) \psi(\vecw) \\
	&=& \left[\sum_{ij} a_{ij} u_i v_j\right] \left[\sum_{k} b_{k} w_k \right] \\
	&=& \sum_{ijk} a_{ij} b_k u_i v_j w_k.
\end{eqnarray*}
As above, we get the coefficients of the product tensor $\phi\tensor\psi$,
namely $a_{ij}b_k$, as an object of its own.

I chose to do a specific example to keep the notation from becoming
frightening.  But we can extend this example to mixed tensors in general.

\begin{defn}
Let $\phi$ be a tensor of type $s_1, r_1$;
let $\psi$ be a tensor of type $s_2, r_2$.
Then the \bfidx{tensor product} $\phi \tensor \psi$ is of type
$s_1+s_2, r_1+r_2$.  It is defined by the property
\begin{eqnarray}
	&&(\phi\tensor\psi)(\lambda_1, \ldots, \lambda_{s_1}, \mu_1, \ldots, \mu_{s_2},
		\vecu_1, \ldots, \vecu_{r_1}, \vecv_1, \ldots, \vecv_{r_2}) \\
	&&=
	\phi(\lambda_1, \ldots, \lambda_{s_1}, \vecu_1, \ldots, \vecu_{r_1})
	\psi(\mu_1, \ldots, \mu_{s_2}, \vecv_1, \ldots, \vecv_{r_2}).
\end{eqnarray}
The coefficients of $\phi\tensor\psi$ are obtained by taking the outer product
of the coefficients of $\phi$ and $\psi$, as in section
\ref{sec:explicit_tensor_v}.
\end{defn}

Note however that unless $\phi$ or $\psi$ is of type 0,0 --- that is, a scalar
--- then the vector space $\mcT^s_r(V)$ is not closed under multiplication:
orders increase when tensors are multiplied.

\begin{defn}
Let $V$ be a vector space.  The \bfidx{full tensor algebra}, written $\mcT(V)$,
is
$$
	\mcT(V) = \bigoplus_{s,r=0}^\infty \mcT^s_r(V).
$$
\end{defn}

\begin{rem}
Using categorical terminology, $\mcT^r_s$ is a functor from vector spaces to
vector spaces; $\mcT$ is a functor from vector spaces to graded rings.
\end{rem}

% xxx if I'm going to say that, I need to say where objects and morphisms go.

%xxx Note that $(V \otimes V)^* = V^* \otimes V^*$, i.e. $\mcT^r_s(V)^* = \mcT^s_r(V)$.
%So, perhaps think of $(\cdot)^*$ as the ``dualing functor''?  Maybe overkill.
% Requires adjoint for image of arrows.

%xxx Include several specific examples.
%
%Include as much computation as possible.
%
%$f:V \to W$, $\mcT^k(f) = (f, \ldots, f)$
%
%categorical associativity implies there is a bilinear map:
%$\mcT^r(V) \oplus \mcT^s(V) \to \mcT^{r+s}(V)$.

%$f: V \to W$:  $\mcT^k(f) : \mcT^k(V) \to \mcT^k(W)$.

%xxx Pickrell notes:
%
%$V^* \otimes V^* \cong (V \otimes V)^* \cong \mcT^2(V^*)$.  Since a lin fcn on $V
%\oplus V$ is a bilin fcn on $V$?
%
%Categories: VS $\to$ assoc. algebras.
%All these are functors:
%$V \mapsto T(V)$.
%$V \mapsto S(V)$.
%$V \mapsto \Lambda(V)$.
%
%$\Lambda(V^*) = T(V^*)$ mod $\phi\otimes\psi + \psi\otimes\phi$
%
%$S(V^*) = T(V^*)$ mod $\phi\otimes\psi - \psi\otimes\phi$
%
%VS is an abelian category?
%$V \oplus W \mapsto T(V \oplus W) = T(V) \oplus T(W)$.
%
%Looks exponential, and in fact it's often important to consider $\vecv
%\mapsto e^\vecv = 1 + \vecv + \vecv^2/2 + \ldots$.

%% ----------------------------------------------------------------
\subsubsection{Permutations acting on multilinear functions}
\label{sec:pmtns_on_tensors}

It is natural to ask about the symmetries of a tensor.
For example, if $\psi$ is a 2-tensor, then we might ask if
$$
	\psi(\vecu, \vecv) =  \psi(\vecv, \vecu) \qquad\textrm{(symmetry)}
$$
or
$$
	\psi(\vecu, \vecv) = -\psi(\vecv, \vecu) \qquad\textrm{(skew symmetry)}
$$
for all vectors $\vecu$ and $\vecv$.  For example, in section
\ref{sec:examples_of_tensors} we saw that the inner product is symmetric, while
the determinant is skew symmetric.  In order to generalize this concept, and in
order to define forms in section \ref{sec:sym_alt}, we will need a notion of
how permutations act on tensors.  Proceeding as in section
\ref{sec:tensor_algebras}, we can first do this argumentwise, using test
vectors, then find out what the effect is on the tensor itself.

This works for all tensors of non-mixed type, i.e. contravariant or covariant
tensors.  (One could define a notion of symmetry for mixed tensors, but this
appears \cite{bib:lee2} not to be useful in practice.)  Here, I will use
covariant tensors for my examples.  Let $\psi$ be an $r$-tensor; let $\vecu_1,
\ldots, \vecu_r \in V$.  Let $\sigma \in \mcS_r$.  Then we want
$$
	\psi\left((\vecu_1, \ldots, \vecu_3)^\sigma \right)
	=
	\psi\left(\vecu_{\sigma^{-1}(1)}, \ldots, \vecu_{\sigma^{-1}(r)}
	\right).
$$
This is completely reasonable.  The only question is why we use the inverse of
the permutation on the right-hand side.  [xxx discuss]

As we did for tensor addition and tensor multiplication, let's see if we can
get rid of the need for test vectors.  Taking the specific permutation
$\sigma=(1\;2\;3)$, we have
\begin{eqnarray*}
	\psi &=& \sum_{ijk} \psi(\vece_i, \vece_j, \vece_k)
		\vece_i^* \tensor \vece_j^* \tensor \vece_k^*; \quad
		a_{ijk} = \psi(\vece_i, \vece_j, \vece_k) \\
	%\psi &=& \sum_{ijk} a_{ijk} \vece_i^* \tensor \vece_j^* \tensor \vece_k^* \\
	\psi(\vecu, \vecv, \vecw)
		&=& \sum_{ijk} a_{ijk} u_i v_j w_k \\
	\psi(\vecw, \vecu, \vecv)
		&=& \sum_{ijk} a_{ijk} w_i u_j v_k
		= \sum_{ijk} a_{ijk} u_j v_k w_i
		= \sum_{ijk} a_{kij} u_i v_j w_k.
\end{eqnarray*}
Generalizing from this example, we can define
$$
	\psi^\sigma(\vecu_1, \ldots, \vecu_r) =
	\psi\left((\vecu_1, \ldots, \vecu_r)^\sigma\right)
$$
where on the right-hand side, we permute the \emph{arguments} to $\psi$, and
on the left-hand side we permute the \emph{coefficients}.

This generalizes the concept of \plainidx{transpose} discussed in section
\ref{sec:mx_symms}.

%% ----------------------------------------------------------------
\subsubsection{Symmetric and alternating tensor spaces}
\label{sec:sym_alt}

Above we said that a 2-tensor $\psi$ is symmetric if $\psi(\vecv,\vecu) =
\psi(\vecu, \vecv)$ for all $\vecu, \vecv$, and likewise $\psi$ is
skew-symmetric or alternating if $\psi(\vecv,\vecu) = -\psi(\vecu, \vecv)$ for
all $\vecu, \vecv$.  We can now generalize this as follows.

\begin{defn}
A non-mixed tensor $\psi$ of order $r$ is \bfidx{symmetric} if
$$
	\psi\left((\vecv_1, \ldots, \vecv_r)^\tau\right) =
	\psi\left(\vecv_1, \ldots, \vecv_r\right)
$$
for all transpositions $\tau \in \mcS_r$.
Likewise, $\psi$ of order $r$ is \bfidx{alternating} if
$$
	\psi\left((\vecv_1, \ldots, \vecv_r)^\tau\right) =
	-\psi\left(\vecv_1, \ldots, \vecv_r\right)
$$
for all transpositions $\tau$.  An alternating tensor is also called a
\bfidx{form}.
\end{defn}

\begin{rem}
The definition was done argumentwise, but following the discussion in section
\ref{sec:pmtns_on_tensors}, we can also look at it coefficientwise.
That is, for symmetric $r$-tensors $\psi$ and transpositions $\tau \in \mcS_r$,
$$
	\psi^\tau = \psi.
$$
Likewise, for alternating $\psi$,
$$
	\psi^\tau = -\psi.
$$
Note in particular that the \bfidx{symmetric} $r$-tensors are held fixed by the
\bfidx{symmetric group} $\mcS_r$, while the \bfidx{alternating} $r$-tensors are
held fixed only by the \bfidx{alternating group} $\mcA_r$.
\end{rem}

\begin{rem}
All tensors of order 0 or 1 are (trivially) symmetric and alternating.
\end{rem}

\begin{rem}
\label{rem:alt_coeffs_diag_zero}
Let $\psi$ be an alternating $3$-tensor.  The alternation property forces
$\psi(\vecu, \vecu, \vecv) = 0$.  In general, for an arbitrary alternating
tensor $\psi$, $\psi$ has value zero when evaluated on $r$ vectors any time two
or more of those vectors are equal.  As discussed in section
\ref{sec:explicit_rep_multilin_fcns}, the coefficients $a_{ijk}$ of a 3-tensor
$\psi$ with respect to a basis
$\{\vecb_1, \ldots, \vecb_m\}$ are given by
$
	a_{ijk} = \psi(\vecb_i, \vecb_j, \vecb_k)
$
and so the $a_{ijk}$ coefficients must be zero whenever $i=j$, $i=k$, or $j=k$.
In general, the coefficients of an arbitrary alternating $r$-tensor are zero
when any index is repeated.  (This is reminiscent of the fact that the diagonal
entries of a skew-symmetric matrix must be zero.  The diagonal entries $a_{ii}$
have equal first and second indices.)
\end{rem}

\begin{rem}
\label{rem:dup_wedge_zero_output}
Just as 1-forms have zero output when vector arguments are repeated, they have
zero output when wedge components are repeated.  Let $\lambda$ be a linear
functional, i.e. a 1-tensor, on $V$.  By the definition of wedge product, for
any $\vecu, \vecv \in V$, we have
\begin{eqnarray*}
	\lambda\wedge\lambda(\vecu,\vecv)
		&=& \lambda\tensor\lambda(\vecu,\vecv)
		- \lambda\tensor\lambda(\vecv,\vecu) \\
	&=& \lambda(\vecu)\lambda(\vecv) - \lambda(\vecv)\lambda(\vecu)
	= \lambda(\vecu)\lambda(\vecv) - \lambda(\vecu)\lambda(\vecv)
	= 0.
\end{eqnarray*}
Note that this remark applies to one-forms.  See remark
\ref{prop:tensor_semicomm} for full details.
\end{rem}

It is easy to check that the sets of symmetric and alternating $r$-tensors are
closed under scalar multiplication and vector addition, and thus that they are
vector spaces.

\begin{defn}
More or less following \cite{bib:lee2} and Pickrell, I write
$$
\Sigma^r(V^*), \qquad \Lambda^r(V^*)
$$
for the spaces of symmetric and alternating $r$-tensors, respectively.
\end{defn}

One naturally asks about the dimension of these vector spaces.  Just as for
matrices in section \ref{sec:mx_symms}, we note that to specify the
coefficients of a symmetric tensor, it suffices to consider those with
non-decreasing indices.  For example, for a 3-tensor $\psi$, if we know what
the coefficient $a_{112}$ is, then we know that $a_{121}$ and $a_{211}$ must be
the same; for the any of the coefficients of a symmetric 3-tensor, it suffices
to specify the coefficients $a_{ijk}$ for $i \le j \le k$.  For a symmetric
tensor of arbitrary order $r$, it suffices to specify the coefficients $a_{i_1,
\ldots, i_r}$ for $i_1 \le \ldots \le i_r$.

For an alternating 3-tensor $\psi$, we know by remark
\ref{rem:alt_coeffs_diag_zero} that the coefficients $a_{ijk}$ are zero for any
repeated indices.  For this reason, along with the reasons from the previous
paragraph, it suffices to specify the values of $a_{ijk}$ for $i<j<k$.  For an
alternating tensor of arbitrary order $r$, it suffices to specify the
coefficients $a_{i_1, \ldots, i_r}$ for $i_1 < \ldots < i_r$.

To find the dimensions of the spaces of symmetric and alternating $r$-tensors
on a vector space of dimension $m$, we only need to count how many ways we can
choose coefficients as just described --- that is, non-decreasing and
increasing indices.  A counting argument shows that
$$
	\dim(\Sigma^r(V^*)) = {{m+r-1}\choose{r}} \qqand
	\dim(\Lambda^r(V^*)) = {{m}\choose{r}}.
$$

\begin{rem}
Note that the above formulas may be written as
$$
	{{m+r-1}\choose r} = \frac{m(m+1)\cdots(m+r-2)(m+r-1)}{r(r-1)\cdots2\cdot 1}
	, \qquad
	{{m}\choose r} = \frac{m(m-1)\cdots(m-r+2)(m-r+1)}{r(r-1)\cdots2\cdot 1}.
$$
For $m=2$, we have
$$
	\frac{m(m+1)}{2} + \frac{m(m-1)}{2} = m^2
$$
and so an arbitrary 2-tensor may be written as the sum of symmetric and
alternating components.  However, for $r\ge 3$, ${(m+r-1)\choose r} + {m
\choose r} < m^r$ and so we cannot write an arbitrary $r$-tensor in terms of
symmetric and alternating components.
\end{rem}

%[xxx examples for $r=2$ and $r=3$, leading into exam questions.]

%% ----------------------------------------------------------------
\subsubsection{Symmetric and alternating tensor algebras}
\label{sec:sym_alt_algebras}

The usual tensor product of two symmetric or alternating tensors need not be
symmetric or alternating, respectively.  (Since all 1-tensors are both
alternating and symmetric, we can use example \ref{ex:outer_product} to see
this.)  So, in order to turn the vector spaces of symmetric and alternating
tensors into algebras, we need to define multiplications on them which preserve
the symmetries.

First, much as in section \ref{sec:mx_symms}, we define techniques for
computing the symmetrization and the alternation of a given tensor.

\begin{defn}
Let \plainidx{$\mathrm{Sym}$} map from $r$-tensors to $r$-tensors by
$$
	\Sym(\psi) = \frac{1}{r!}\sum_{\sigma \in \mcS_r} \psi^\sigma.
$$
Likewise, let \plainidx{$\mathrm{Alt}$} map from $r$-tensors to $r$-tensors by
$$
	\Alt(\psi) = \frac{1}{r!}\sum_{\sigma \in \mcS_r} \sgn(\sigma) \psi^\sigma.
$$
where $\sgn(\sigma)$ is $+1$ for even permutations and $-1$ for odd
permutations.
\end{defn}

See any of the geometry texts in the bibliography for proofs that the
symmetrization and alternation do in fact produce symmetric and alternating
tensors, respectively.  Moreover, these are both \bfidx{projections} from
$\mcT^0_r(V)$ onto $\Sigma^r(V^*)$ and $\Lambda^r(V^*)$, respectively, in the
sense of section \ref{sec:projections}: they are surjective as well as
idempotent.

Now we can define symmetry-preserving multiplications.  Just as with the tensor
product as described in section \ref{sec:tensor_algebras}, though, the
product of tensors won't have the same order as the multiplicands.

\begin{defn}
We write
$$
	\Sigma(V^*) = \bigoplus_{r=0}^\infty \Sigma^r(V^*) \qqand
	\Lambda(V^*) = \bigoplus_{r=0}^\infty \Lambda^r(V^*).
$$
\end{defn}

\begin{defn}
Let $\phi \in \Sigma^k(V^*)$ and $\psi \in \Sigma^\ell(V^*)$.  That is, suppose
$\phi$ has order $k$ and $\psi$ has order $\ell$.  The \bfidx{symmetric
product} of $\phi$ and $\psi$, written using juxtaposition, is
$$
	\phi\psi = \Sym(\phi\tensor\psi).
$$
Likewise, if $\phi \in \Lambda^k(V^*)$ and $\psi \in \Lambda^\ell(V^*)$, then
the \bfidx{alternating product}, or \bfidx{wedge product}, of $\phi$ and $\psi$
is
\begin{eqnarray*}
	\phi \wedge \psi &=& \frac{(k+\ell)!}{k!\ell!} \Alt(\phi \tensor \psi) \\
	&=& \frac{(k+\ell)!}{k!\ell!} \frac{1}{(k+\ell)!}
		\sum_{\sigma\in\mcS_{k+\ell}} \sgn(\sigma)(\phi \tensor \psi)^\sigma \\
	&=& \frac{1}{k!\ell!} \sum_{\sigma\in\mcS_{k+\ell}}
		\sgn(\sigma)(\phi \tensor \psi)^\sigma.
\end{eqnarray*}
Note that in both cases the product has order $k+\ell$.  The factorial
multiplier is included so that the wedge of $m$ basis functionals, where
$m=\dim(V)$, will be exactly the determinant.  This is because the $1/k!\ell!$
in the rightmost expression above is 1 whenever we are wedging 1-tensors, which
have orders $k=\ell=1$.  We will see more about this in example
\ref{ex:det_uvw}.
\end{defn}

With these multiplications, $\Sigma(V^*)$ and $\Lambda(V^*)$ are both
\bfidx{algebras}.  That is, we can add, subtract, and multiply tensors.  In
particular, all the usual arithmetic rules for (non-commutative) rings apply,
such as distributivity.  But what about commutativity?

\begin{prop}
\label{prop:tensor_semicomm}
Let $\phi$ and $\psi$ be covariant symmetric tensors of orders $k$
and $\ell$, respectively; let $\omega$ and $\eta$ be covariant alternating
tensors of orders $k$ and $\ell$.  Then it can be shown that
$$
	\phi\psi = \psi\phi
$$
and
$$
	\omega \wedge \eta = (-1)^{k\ell} \eta \wedge \omega.
$$
\end{prop}

\begin{mnem}
For the symmetric case, take
$$\phi=\lambda_1 \lambda_2$$
and
$$\psi=\lambda_3 \lambda_4 \lambda_5$$
where each $\lambda_i$ has degree 1.
Then
$$\phi\psi =
(\lambda_1 \lambda_2) (\lambda_3 \lambda_4 \lambda_5) =
\lambda_1 (\lambda_3 \lambda_4 \lambda_5) \lambda_2 =
(\lambda_3 \lambda_4 \lambda_5) \lambda_1 \lambda_2 =
\psi\phi
$$
since we can transpose degree-1 symmetric terms.
Likewise, for the alternating case, take
$$\omega=\theta_1 \wedge\theta_2$$
and
$$\eta=\theta_3 \wedge\theta_4 \wedge\theta_5$$
where each $\theta_i$ has degree 1.
Then
$$\omega\wedge\eta =
(\theta_1 \wedge\theta_2) \wedge(\theta_3 \wedge\theta_4 \wedge\theta_5) =
-\theta_1 \wedge(\theta_3 \wedge\theta_4 \wedge\theta_5) \wedge\theta_2 =
(\theta_3 \wedge\theta_4 \wedge\theta_5) \wedge\theta_1 \wedge\theta_2 =
\eta\wedge\omega.
$$
This is because we get three minus signs when we move
$\theta_2$ past $\theta_3,$ $\theta_4$, and $\theta_5$, then
three more minus signs when we move
$\theta_1$ past $\theta_3,$ $\theta_4$, and $\theta_5$.
If either $k$ or $\ell$ is even, there is an even number of transpositions.

The proof simply generalizes this specific example from $2$ to $k$ and $3$ to
$\ell$, along with $\phi,\psi,\omega,\eta$ being linear combinations of
elementary terms of the form shown here:  use the distributive property of
symmetric and alternating algebras to complete the proof.
\end{mnem}

\begin{cor}
\label{cor:omega_odd_wedge_self_is_zero}
Let $\omega$ be a $k$-form.  If $k$ is odd, then $\omega \wedge \omega = 0$.
\end{cor}

\beginex
Let $\omega$ and $\eta$ both be 1-forms.  Applying the definition of
wedge product to test vectors $\vecu$ and $\vecv$ gives
\begin{eqnarray*}
	\omega \wedge \eta(\vecu, \vecv) &=&
		\omega(\vecu)\eta(\vecv) - \omega(\vecv)\eta(\vecu) \\
	\eta \wedge \omega(\vecu, \vecv) &=&
		\eta(\vecu)\omega(\vecv) - \eta(\vecv)\omega(\vecu) \\
	&=& \omega(\vecv) \eta(\vecu) - \omega(\vecu) \eta(\vecv) \\
	&=& -\omega\wedge\eta(\vecu, \vecv).
\end{eqnarray*}
Since the test vectors are arbitrary,
$$
	\omega\wedge\eta = -\eta\wedge\omega.
$$
Now suppose $\omega$ is a 1-form and $\eta$ is a 2-form.  Again applying the
definition of wedge product, recalling that $\eta$ is alternating, and pulling
the factorial coefficient over to the left, we have
\begin{eqnarray*}
	2 \omega\wedge\eta(\vecu,\vecv,\vecw)
	&=& \begin{matrix}
			& \omega\tensor\eta(\vecu,\vecv,\vecw) &-& \omega\tensor\eta(\vecv,\vecu,\vecw) \\
			+&\omega\tensor\eta(\vecv,\vecw,\vecu) &-& \omega\tensor\eta(\vecw,\vecv,\vecu) \\
			+&\omega\tensor\eta(\vecw,\vecu,\vecv) &-& \omega\tensor\eta(\vecu,\vecw,\vecv) \\
		\end{matrix} \\
	&=& \begin{matrix}
			& \omega(\vecu)\eta(\vecv,\vecw) &-& \omega(\vecv)\eta(\vecu,\vecw) \\
			+&\omega(\vecv)\eta(\vecw,\vecu) &-& \omega(\vecw)\eta(\vecv,\vecu) \\
			+&\omega(\vecw)\eta(\vecu,\vecv) &-& \omega(\vecu)\eta(\vecw,\vecv) \\
		\end{matrix} \\
	&=& \begin{matrix}
			& 2\omega(\vecu)\eta(\vecv,\vecw) \\
			+&2\omega(\vecv)\eta(\vecw,\vecu) \\
			+&2\omega(\vecw)\eta(\vecu,\vecv) \\
		\end{matrix} \\
	2 \eta\wedge\omega(\vecu,\vecv,\vecw)
	&=& \begin{matrix}
			& \eta\tensor\omega(\vecu,\vecv,\vecw) &-& \eta\tensor\omega(\vecv,\vecu,\vecw) \\
			+&\eta\tensor\omega(\vecv,\vecw,\vecu) &-& \eta\tensor\omega(\vecw,\vecv,\vecu) \\
			+&\eta\tensor\omega(\vecw,\vecu,\vecv) &-& \eta\tensor\omega(\vecu,\vecw,\vecv) \\
		\end{matrix} \\
	&=& \begin{matrix}
			& \eta(\vecu,\vecv)\omega(\vecw) &-& \eta(\vecv,\vecu)\omega(\vecw) \\
			+&\eta(\vecv,\vecw)\omega(\vecu) &-& \eta(\vecw,\vecv)\omega(\vecu) \\
			+&\eta(\vecw,\vecu)\omega(\vecv) &-& \eta(\vecu,\vecw)\omega(\vecv) \\
		\end{matrix} \\
	&=& \begin{matrix}
			& 2\eta(\vecu,\vecv)\omega(\vecw) \\
			+&2\eta(\vecv,\vecw)\omega(\vecu) \\
			+&2\eta(\vecw,\vecu)\omega(\vecv) \\
		\end{matrix} \\
	&=& 2 \omega\wedge\eta(\vecu,\vecv,\vecw).
\end{eqnarray*}
Since the test vectors are arbitrary,
$$
	\omega\wedge\eta = \eta\wedge\omega.
$$
\mendex

\begin{defn}
The alternating tensor algebra $\Lambda(V^*)$ is also called the
\bfidx{exterior algebra} or \bfidx{Grassmann algebra} on $V$.
\end{defn}

\begin{rem} One may write down explicit bases for $\Sigma^r(V^*)$ and
$\Lambda^r(V^*)$, just as for matrices in section \ref{sec:mx_symms}.  In fact,
letting $\{\vecb_1^*, \ldots, \vecb_r^*\}$ be a basis for $V^*$, one obtains
$$
	\left\{\vecb^*_{i_1}\cdots \vecb^*_{i_r} :
		1 \le i_1 \le \ldots \le i_r \le m
	\right\}
$$
as a basis for $\Sigma^r(V^*)$ and
$$
	\left\{\vecb^*_{i_1}\wedge \ldots \wedge \vecb^*_{i_r} :
		1 \le i_1 < \ldots < i_r \le m \right\}
$$
as a basis for $\Lambda^r(V^*)$.
\end{rem}

\beginex
Let $m=3$.  Then a basis for $\Sigma^2(V^*)$
is
$$
	\left(
	\vecb^*_1\vecb^*_1, \quad
	\vecb^*_1\vecb^*_2, \quad
	\vecb^*_1\vecb^*_3, \quad
	\vecb^*_2\vecb^*_2, \quad
	\vecb^*_2\vecb^*_3, \quad
	\vecb^*_3\vecb^*_3.
	\right\}
$$
\mendex

\beginex
Let $m=3$.
Bases for $\Lambda^r(V^*)$, for $r=0,1,2,3$, respectively, are
\begin{eqnarray*}
	&&\left\{
	1
	\right\}, \\
%
	&&\left\{
	\vecb^*_1, \quad
	\vecb^*_2, \quad
	\vecb^*_3
	\right\}, \\
%
	&&\left\{
	\vecb^*_1\wedge\vecb^*_2, \quad
	\vecb^*_1\wedge\vecb^*_3, \quad
	\vecb^*_2\wedge\vecb^*_3
	\right\}, \\
%
	&&\left\{
	\vecb^*_1\wedge\vecb^*_2\wedge\vecb^*_3
	\right\}.
\end{eqnarray*}
\mendex

\begin{rem}
We saw that $\Lambda^r(V^*)$ has dimension ${m \choose r}$.  This means that
$\Lambda^m(V^*)$ has dimension ${m \choose m} = 1$.  Now, the determinant is an
alternating $m$-linear function on $m$ vectors of length $m$, so it is
necessarily a scalar multiple of the single basis vector for $\Lambda^m(V^*)$.
In fact, that scalar multiple is 1, and the determinant may be
characterized\index{determinant, characterization of} as the unique alternating
$m$-linear function whose value on the standard basis is 1.
\end{rem}

\beginex
\label{ex:det_uvw}
Here we see what $\vece_1^* \wedge \vece_2^* \wedge \vece_3^*$ does to three
vectors on a 3-dimensional space.
\begin{eqnarray*}
	\vece_1^* \wedge \vece_2^* \wedge \vece_3^*(\vecu, \vecv, \vecw)
	&=& \sum_{\sigma\in \mcS_3}
		\vece_1^* \tensor \vece_2^* \tensor \vece_3^*(\vecu, \vecv, \vecw)^\sigma \\
	&=&
		\vece_1^* \tensor \vece_2^* \tensor \vece_3^*(\vecu, \vecv, \vecw)
		+ \vece_1^* \tensor \vece_2^* \tensor \vece_3^*(\vecv, \vecw, \vecu)
		+ \vece_1^* \tensor \vece_2^* \tensor \vece_3^*(\vecw, \vecu, \vecv) \\
	&-&	\vece_1^* \tensor \vece_2^* \tensor \vece_3^*(\vecu, \vecw, \vecv)
		- \vece_1^* \tensor \vece_2^* \tensor \vece_3^*(\vecw, \vecv, \vecu)
		- \vece_1^* \tensor \vece_2^* \tensor \vece_3^*(\vecv, \vecu, \vecw) \\
	&=& u_1 v_2 w_3
	+	v_1 w_2 u_3
	+	w_1 u_2 v_3
	-	u_1 w_2 v_3
	-	w_1 v_2 u_3
	-	v_1 u_2 w_3 \\
	&=& u_1 v_2 w_3
	-	u_1 v_3 w_2
	+	u_2 v_3 w_1
	-	u_2 v_1 w_3
	+	u_3 v_1 w_2
	-	u_3 v_2 w_1 \\
	&=& \left| \begin{matrix}
		u_1 & u_2 & u_3 \\
		v_1 & v_2 & v_3 \\
		w_1 & w_2 & w_3 \\
		\end{matrix}\right| \\
	&=& \det(\vecu, \vecv, \vecw).
\end{eqnarray*}
\mendex

\begin{rem} \label{rem:foreshadow_metric_tensor_and_forms} While our most
important example of an alternating tensor is the determinant, the key
symmetric example is the dot product.  Starting in section \ref{sec:loaf}, we
will use varying coefficients rather than constant coefficients.  Then, in
sections \ref{sec:forms} and \ref{sec:metric_tensor}, the determinant and the
dot product will become the \bfidx{volume form} and the \bfidx{metric tensor},
respectively. \end{rem}

%\beginex
%\label{ex:metric_tensor}
%The \bfidx{metric tensor} or \bfidx{Euclidean metric tensor} is a symmetric
%two-tensor
%$$ g = dx\tensor dx + dy\tensor dy + dz\tensor dz $$
%which is sometimes written
%$$ g = dx^2 + dy^2 + dz^2.$$
%[xxx need to move this example, or re-word it so that it can belong here.]
%[xxx express this in terms of the dot product w/ COB.  Then xref forward.]
%\mendex

%% ----------------------------------------------------------------
\subsubsection{Some explicit computations with tensors}
\label{sec:decompose_tensors}

In example \ref{ex:decompose_three_tensor} we decomposed a three-tensor into a
linear combination of its value on basis functions.  Here we will do something
similar for a one-tensor.  This kind of computation comes up very often in
exams and qualifiers, where the tensors in question are forms on a manifold.
[xxx xref forward to section xxx, once I write it.]  See for example the
problems in section \ref{sec:ex_1_6} and \ref{sec:ex_2_2}.

\beginex
\label{ex:decompose_alt_one_tensor}
Let $\eta$ be a one-tensor on $V$ and let $m=\dim(V)$.  Let $V$ have basis
$\{\vecb_1,\ldots, \vecb_m\}$ with corresponding dual basis
$\{\vecb_1^*,\ldots, \vecb_m^*\}$.  Then we know that $\eta$ is a linear
combination of the one-tensors, namely,
$$ \eta = \sum_{i=1}^m n_i \vecb_i^*.$$
How do we find out what the coefficients $n_i$ are?  We can apply $\eta$
to each of the basis elements $\vecb_j$.  We get
$$
	\eta(\vecb_j) = \sum_{i=1}^m n_i \vecb_i^*(\vecb_j) = n_j
$$
since $\vecb_i^*(\vecb_j) = \delta_{ij}$.  So, we have the useful result
\begin{equation*}
%\label{eqn:decompose_one_tensor}
\framebox[4.0in][c]{
	$\displaystyle \eta = \sum_{i=1}^m \eta(\vecb_i) \vecb_i^*.$
}
\end{equation*}
That is, to write our one-tensor with respect to a certain coordinate system,
we just need to find its value on the basis vectors for that coordinate system.
Note that a one-tensor is trivially alternating, so this formula also applies
to one-forms as well.
\mendex

\beginex
\label{ex:decompose_alt_two_tensor}
A similar computation shows that for an alternating two-tensor $\eta$, we
have
\begin{equation*}
%\label{eqn:decompose_alt_two_tensor}
\framebox[4.0in][c]{
	$\displaystyle
	\eta = \sum_{i<j} \eta(\vecb_i, \vecb_j) \vecb_i^* \wedge \vecb_j^*.$
}
\end{equation*}
\mendex

\beginex
\label{ex:decompose_sym_two_tensor}
For a symmetric two-tensor $\phi$, we
have
\begin{equation*}
%\label{eqn:decompose_sym_two_tensor}
\framebox[4.0in][c]{
	$\displaystyle
	\phi = \sum_{i\le j} \phi(\vecb_i, \vecb_j) \vecb_i^* \vecb_j^*.$
}
\end{equation*}
\mendex

\beginex
\label{ex:decompose_two_tensor}
Likewise, for a general two-tensor $\psi$, we
have
\begin{equation*}
%\label{eqn:decompose_two_tensor}
\framebox[4.0in][c]{
	$\displaystyle
	\psi = \sum_{i,j} \psi(\vecb_i, \vecb_j) \vecb_i^* \tensor \vecb_j^*.$
}
\end{equation*}
\mendex

%\beginex
%\label{ex:decompose_alt_two_tensor}
%Let $\omega \in \Lambda^2(V)$ and suppose $\dim(V)=2$.  Then for $u,v\in V$,
%using multilinearity (here, bilinearity) we have:
%\begin{eqnarray*}
%	\vecu &=& \sum_{i=1}^m u_i \vece_i, \quad
%		\vecv = \sum_{j=1}^m v_j \vece_j \\
%	\omega(\vecu, \vecv) &=& \omega\left(
%		\sum_{i=1}^m u_i \vece_i,
%		\sum_{j=1}^m v_j \vece_j \right) \\
%	&=& \sum_{i=1}^m \sum_{j=1}^m u_i v_j
%		\omega(\vece_i, \vece_j) \\
%	&=&
%		u_1 v_1 \omega(\vece_1, \vece_1) +
%		u_1 v_2 \omega(\vece_1, \vece_2) +
%		u_2 v_1 \omega(\vece_2, \vece_1) +
%		u_2 v_2 \omega(\vece_2, \vece_2).
%\end{eqnarray*}
%But $\omega$ is alternating, so we have
%$$
%	\omega(\vecu,\vecv) = (u_1 v_2 - u_2 v_1) \omega(\vece_1, \vece_2).
%$$
%\mendex

%% ----------------------------------------------------------------
\subsubsection{Change of basis; more about covariance and contravariance}
\label{sec:TS_COB}

\emph{Note:  This section is not qualifier material.}

In section \ref{sec:VS_COB} we saw how the coefficients of linear functionals
and vectors transform when we change bases.  We now know these to be tensors of
type 0,1 and 1,0, respectively, so one may ask how the coefficients of general
tensors transform with respect to change of basis.

First, some notation.  Let
$$
	A = \{\veca_1, \ldots, \veca_m\} \qand
	B = \{\vecb_1, \ldots, \vecb_m\}
$$
be ordered bases for an $m$-dimensional real vector space $V$, and for
concreteness assume here that $A$ is the standard basis.  (As usual, my
examples use $V=\R^m$.) As in section \ref{sec:VS_COB} as well as
\cite{bib:fis}, let $Q$ be the change-of-basis matrix which converts vectors
from $B$ coordinates to $A$ coordinates.  Recall that $Q$ simply has $j$th
column equal to $\vecb_j$ in $A$ coordinates.  Write $q_{ij}$ for the elements
of $Q$, and write $p_{ij}$ for the elements of $Q^{-1}$.

We saw in section \ref{sec:VS_COB} that the coefficients of a covariant tensor
$\lambda$ and a contravariant tensor $\vecv$ transform according to
$$
	{}[\lambda]_B =        [\lambda]_A Q \qqand
	{}[\vecv]_B   = Q^{-1} [\vecv]_A.
$$
Written in terms of coefficients, these are
\begin{eqnarray*}
	\begin{pmatrix}
		[\lambda_1]_B & \ldots & [\lambda_m]_B
	\end{pmatrix}
	&=&
	\begin{pmatrix}
		[\lambda_1]_A & \ldots & [\lambda_m]_A
	\end{pmatrix}
	\begin{pmatrix}
		q_{11} & \ldots & q_{1m} \\
		\vdots &        & \vdots \\
		q_{m1} & \ldots & q_{mm} \\
	\end{pmatrix}, \\
%
	\begin{pmatrix}
		[v_1]_B \\ \vdots \\ [v_m]_B
	\end{pmatrix}
	&=&
	\begin{pmatrix}
		p_{11} & \ldots & p_{1m} \\
		\vdots &        & \vdots \\
		p_{m1} & \ldots & p_{mm} \\
	\end{pmatrix}
	\begin{pmatrix}
		[v_1]_A \\ \vdots \\ [v_m]_A
	\end{pmatrix}
\end{eqnarray*}
which is to say
$$
	{}[\lambda_j]_B = \sum_{j=1}^m q_{ij} [\lambda_i]_A \qqand
	{}[v_i]_B       = \sum_{j=1}^m p_{ij} [v_j]_A.
$$
Note in particular that the covariant 1-tensor uses $q_{ij}$, and the
entries of $[\lambda]_A$ sum along the first (row) index of $Q$, namely, $i$.
Likewise,
the contravariant 1-tensor uses $p_{ij}$, and the
entries of $[\lambda]_A$ sum along the $j$, or column, index of $Q^{-1}$.

I will discuss two more examples, as motivation for the general principle.

The \bfidx{dot product} is a bilinear function on $V$, i.e. a covariant
2-tensor.  It will be convenient to name it, so write
$$
	\mu(\vecu, \vecv) = \vecu \cdot \vecv.
$$
Given $\vecu, \vecv$ written in coordinates with respect to the standard basis,
we are accustomed to writing
$$
	\vecu \cdot \vecv = \sum_{i=1}^m u_i v_i.
$$
In the spirit of section \ref{sec:decompose_tensors}, though,
we can also write
$$
	\mu = \sum_{ij} \mu(\vece_i, \vece_j) \vece_i^* \tensor \vece_j^*.
$$
Recalling that $\vece_i^*$ is nothing more than the $i$th
\plainidx{coordinate-selector function}, we have
\begin{eqnarray*}
	\mu(\vecu,\vecv) &=&
		\sum_{ij} \mu(\vece_i, \vece_j)
		\vece_i^* \tensor \vece_j^*(\vecu,\vecv)
	= \sum_{ij} (\vece_i \cdot \vece_j) u_i v_j
	= \sum_{ij} \delta_{ij} u_i v_j
	= \sum_{i} u_i v_i
\end{eqnarray*}
which is what we expect.  So, with respect to the standard basis, the
coefficients of $\mu$ are $\delta_{ij}$, which is the same as the identity
matrix.  These coefficients $\vece_i \cdot \vece_j$ of $\mu$ are equal to
$\delta_{ij}$ precisely for \plainidx{orthonormal coordinates}.

With respect to another basis $\{\vecb_1, \ldots, \vecb_m\}$, we have
\begin{equation}
	\mu = \sum_{ij} \mu(\vecb_i, \vecb_j) \vecb_i^* \tensor \vecb_j^*.
\end{equation}
Given the discussion in section \ref{sec:VS_COB}, I want to state this in terms
of the change-of-basis matrix $Q$.  Since the covariant 1-tensor above had
$$
	{}[\lambda_j]_B = \sum_{j=1}^m q_{ij} [\lambda_i]_A,
$$
and guided by my readings elsewhere, I want the coefficients of the covariant
2-tensor $\mu$ to transform by using $Q$ twice, summing along the first index
of $Q$ in each case.  That is, for the $m \times m$ array of coefficients
$[\mu]_B$, I want
\begin{equation}
	{}[\mu_{ij}]_B = \sum_{k,\ell=1}^m q_{ki} q_{\ell j} [\mu_{k\ell}]_A.
\end{equation}
It remains to show that
$$
	\mu(\vecb_i, \vecb_j) = \sum_{k,\ell=1}^m q_{ki} q_{\ell j} [\mu_{k\ell}]_A.
$$
But this is true since the $i$th and $j$th columns of $Q$ are precisely
$\vecb_i$ and $\vecb_j$ in $A$ coordinates.  That is, in $A$ coordinates
we would write
$$
	\mu(\vecb_i, \vecb_j) = \sum_{k,\ell=1}^m
	\mu(\veca_k, \veca_\ell) (\veca_k^* \tensor \veca_\ell^*)(\vecb_i, \vecb_j)
	= \sum_{k,\ell=1}^m [\mu_{k\ell}]_A q_{ki} q_{\ell j}.
$$

\begin{center}
* * *
\end{center}

The next example after the dot product is a tensor of mixed variance.  It might
seem that we don't know of any such thing.  However, we do: in fact, a linear
transformation is a mixed tensor.  (This becomes guessable when one uses the
\plainidx{Einstein summation convention}, as in \cite{bib:lee2} and
\cite{bib:gug}, where we write $y^i = A^i_j x^j$ with superscripts being
contravariant and subscripts being covariant.)

Let $T:V \to V$.  Then for $\vecy = T \vecx$, taking coordinates with respect
to a basis $A$, for the $i$th component of $\vecy$ we write
$$
	{}[y_i]_A = \sum_{j=1}^m [T_{ij}]_A [x_j]_A.
$$
Writing all the components of $\vecy$ at once, we can say
$$
	{}[\vecy]_A = \sum_{ij} [T_{ij}]_A \veca_i \tensor \veca_j^*
		\left( \sum_k [x_k]_A \veca_k\right)
$$
Cavalierly manipulating symbols (I should justify this using tensor
contraction, which I've not defined here), and recalling that
$\veca_j^*(\veca_k) = \delta_{jk}$, this is
$$
	{}[\vecy]_A = \sum_{ijk} [T_{ij}]_A [x_k]_A
		\veca_i \tensor \veca_j^* (\veca_k)
	= \sum_{ijk} [T_{ij}]_A [x_j]_A \veca_i
$$
which is what we would expect.

The question at hand is how the coefficients of $T$ transform when we change
bases.  Let $Q$ be the change-of-basis matrix as above.  It is shown in
\cite{bib:fis} that
$$
	{}[T]_B = Q^{-1} [T]_A Q.
$$
This is in fact easy to see.  Above we saw that for vectors $\vecv$, $Q$
converts from $B$ coordinates to $A$ coordinates.  So, when we compute
$(Q^{-1}[T]_A Q)([\vecv]_B)$, we are doing the following:
\begin{itemize}
\item Convert $\vecv$ from $B$ coordinates to $A$ coordinates.
\item Apply the linear transformation in $A$ coordinates.
\item Convert the result back to $B$ coordinates.
\item So, the net effect is to apply the linear transformation in $B$
coordinates.
\end{itemize}
As with the covariant 2-tensor example, trying to generalize the pattern I've
been seeing so far, I would want to write
$$
	{}[T_{ij}]_B = \sum_{k\ell} p_{ik} q_{\ell j} [T_{k\ell}]_A
$$
which looks a bit more familiar as
$$
	{}[T_{ij}]_B = \sum_{k\ell} p_{ik} [T_{k\ell}]_A q_{\ell j}.
$$
This is precisely what the triple matrix product $Q^{-1}[T]_A Q$ is:

$$
	\begin{matrix}
		  &   & k \\
		  &   & \longrightarrow \\
		i & \Bigg\downarrow &
		\begin{pmatrix}
			p_{11} & \ldots & p_{1m} \\
			\vdots &        & \vdots \\
			p_{m1} & \ldots & p_{mm} \\
		\end{pmatrix}
	\end{matrix}
	\quad
	\begin{matrix}
		  &   & \ell \\
		  &   & \longrightarrow \\
		k & \Bigg\downarrow &
		\begin{pmatrix}
			{}[T_{11}]_A & \ldots & [T_{1m}]_A \\
			\vdots       &        & \vdots \\
			{}[T_{m1}]_A & \ldots & [T_{mm}]_A \\
		\end{pmatrix}
	\end{matrix}
	\quad
	\begin{matrix}
		  &   & j \\
		  &   & \longrightarrow \\
		\ell & \Bigg\downarrow &
		\begin{pmatrix}
			q_{11} & \ldots & q_{1m} \\
			\vdots &        & \vdots \\
			q_{m1} & \ldots & q_{mm} \\
		\end{pmatrix}
	\end{matrix}.
$$

\begin{center}
* * *
\end{center}

Following this pattern, we have a general rule for change of coordinates.  To
recap, let $T$ be a tensor of type $s,r$ on an $m$-dimensional real vector
space $V$.  Let $A$ and $B$ be ordered bases for $V$.  Let $Q$ be the
change-of-basis matrix with entries $q_{ij}$, and write $p_{ij}$ for the
entries of $Q^{-1}$.  Then
$$
	{}[T^{i_1, \ldots, i_s}_{j_1, \ldots, j_r}]_B
	= \sum_{k_1, \ldots, k_s, \ell_1, \ldots, \ell_r}
	p_{i_1,k_1} \cdots p_{i_s,k_s}
	q_{\ell_1,j_1} \cdots q_{\ell_r,j_r}
	{}[T^{k_1, \ldots, k_s}_{\ell_1, \ldots, \ell_r}]_A.
$$
In section \ref{sec:VS_COB} and this section I've proved this for tensors of
type 1,0, type 0,1, type 1,1, and type 0,2.  I will not prove this general
formula, but the examples here should make it plausible and rememberable.

[xxx xref forward to varying-coefficients section.]

%% ----------------------------------------------------------------
\subsubsection{Contractions}
\label{sec:vs_i_X}

\begin{defn}
\label{defn:contraction_vs}
Let $V$ be an $m$-dimensional real vector space.  Let $\omega$ be a $k$-form,
i.e. $\omega \in \mcT^k(V^*)$.  Let $\vecu \in V$.
Define the \index{$X \ctt \vecu$} \index{$i_X$} \bfidx{contraction} $\vecu \ctt
\omega$ (also written $i_\vecu(\omega)$) to be
$$
	\vecu\ctt\omega(\vecv_1, \ldots, \vecv_{k-1}) =
	\omega(\vecu, \vecv_1, \ldots, \vecv_{k-1}).
$$
\end{defn}
Thus, if $\omega$ is a $k$-form, then $\vecu\ctt\omega$ is a $k-1$ form.

\begin{mnem}
We have an $i$ in $i_\vecu$, and it inserts $\vecu$ into a form's argument
list:  \emph{i} for \emph{insert}.
\end{mnem}

%\beginex(\cite{bib:spi1}, pp. 83-84.)
%Let $\vecu \in \R^3$ and consider the 3-form which is the determinant on $\R^3
%\oplus \R^3 \oplus \R^3$.  Then
%$$ i_\vecu\det(\vecv, \vecw) = \det(\vecu, \vecv, \vecw).$$
%
%If we fix $\vecv$ and $\vecw$, then this is a 1-form with argument $\vecu$.
%
%If we fix $\vecu$, then this is a 2-form with arguments $\vecv$ and $\vecw$.
%
%xxx discuss
%\mendex

\begin{defn}
\label{defn:derivn_and_antiderivn}
Let $\phi$ be a function from the alternating tensor algebra $\mcT^k(V^*)$ to
itself.  Suppose that $\phi$ is linear, i.e. for all forms $\omega$ and $\eta$
and for all $s \in \R$,
$$
	\phi(\omega+\eta) = \phi(\omega)+\phi(\eta)
	\qqand
	\phi(s\omega) = s\phi(\omega).
$$
Let $k$ and $\ell$ be the degrees of $\omega$ and $\eta$.
We say that $\phi$ is a \bfidx{derivation} if in addition to linearity,
$$
	\phi(\omega \wedge \eta) = \phi(\omega) \wedge \eta +
	\omega \wedge \phi(\eta).
$$
Likewise, we say that $\phi$ is an \bfidx{antiderivation} if in addition to
linearity,
$$
	\phi(\omega \wedge \eta) = \phi(\omega) \wedge \eta
		+ (-1)^k \omega \wedge \phi(\eta).
$$
\end{defn}

\begin{rem}
These two properties are also called the\index{Leibniz rule} \bfidx{signed
Leibniz rule} and the \bfidx{unsigned Leibniz rule}, respectively, due to the
analogy with the product rule in calculus.  Also note that the \emph{anti} in
\emph{antiderivation} has only to do with the alternating sign --- it has
nothing to do with antidifferentiation.  (We will see in section [xxx xref]
that the differentiation and antidifferentiation operators $d$
and $\vecu\ctt$ are both antiderivations on tangent bundles.)
\end{rem}

\begin{prop}
The contraction $\vecu\ctt$ is an antiderivation.
\begin{proof}
\ldots goes here.
\end{proof}
\end{prop}

xxx examples.

%% ----------------------------------------------------------------
\subsubsection{Tensor products over free modules}

Everything said about linear algebra and tensor algebra in sections
\ref{sec:linear_algebra} and \ref{sec:tensors} was stated in terms of
finite-dimensional vector spaces over $\R$.  I did this for the sake of
familiarity: we've all seen vector spaces before.  Since this is our first year
of graduate school and since we'll be getting modules in the core algebra
course only in the second semester, I delayed generalizing the presentation.
But everything said above for vector spaces is in fact true for \bfidxtwo{free
modules}{free module} (see section \ref{sec:modules}) over commutative rings
with identity as well.  This will be discussed in section \ref{sec:loaf}, after
some more terminology is defined.

%% ================================================================
%% ================================================================
%% ================================================================
\newpage
\section{Topological manifolds}
\label{sec:topological_manifolds}

\begin{center}
\emph{
All art is at once surface and symbol. Those who go beneath the surface do so
at their peril.} \\
--- Oscar Wilde\index{Wilde, Oscar} (1854-1900).
\end{center}

Algebraic topology is an extensive subject.  See \cite{bib:hatcher},
\cite{bib:massey}, and/or \cite{bib:lee1} for more information.  The minimum
needed for our purposes is enough for the student to understand and use the
following:
\begin{itemize}
\item classification of \bfidx{surfaces},
\item the \bfidx{lifting lemmas},
\item the \bfidx{Seifert-van Kampen} theorem,
\item the \bfidx{Mayer-Vietoris} theorem, and
\item cubes and chains which are needed for \bfidx{integration}
	on manifolds (section \ref{sec:duality_main_section}).
\end{itemize}
The essential idea is that section \ref{sec:vc_lim} showed us that traditional
vector calculus fails when domains have \bfidx{holes} in them.  Here, we simply
want to understand what holes are.

%xxx include as many problem solutions as possible.

%% ================================================================
\subsection{CW complexes}

CW complexes early on, following Hatcher?  Is there any reason \emph{not} to?
It would make it possible to do classification of surfaces all at once, without
having to come back.

%% ================================================================
\subsection{Classification of surfaces}

$$
\begin{matrix}
	\S^2 &\qquad& \T &\qquad& \T^2=\T\#\T &\qquad& \T^3 &\qquad& \ldots \\
	P=\P^2 &\qquad& \mathrm{KB}=\P^2 \# \P^2 = P^2 &\qquad&
		P^3 &\qquad& P^4 &\qquad&
	\ldots
\end{matrix}
$$

[xxx alg section:  quotients / relations:  we do NOT use equiv classes in most
actual work.  rather, pick reps and use xfmn rules.]

give edge presentations (multiple ones!) and stress they have different uses.
S2 with/without the edge.  In each case, stress careful counting of ID'd
vertices and edges.  Mention free groups on strings, mod canonicalizing
relations.  Pound operator as concat op on edge strings.

[xxx have a revisit section at the end of topo --- incl various info including
$\chi$'s, univ covs ($\S^2$ over $\S^2$ and $\P^2$, $\R^2$ over everything else);
fundamental groups; what else \ldots.]

to be filed: genus has been soft-pedaled in this course \ldots
$\chi = 2-2g$ \ldots.

%% ================================================================
\subsection{Homotopy and the fundamental group}

%% ----------------------------------------------------------------
\subsection{Homologies}

Cellular, simplicial, singular.  Note I prefer singular since it gels nicely
with the elementary proof of Stokes using the FTC, as in little Spivak.

%% ----------------------------------------------------------------
\subsection{Singular homology}

Crucial\index{singular homology}\index{homology} example: latitudinal paths
along $\S^2$: $c_1 - c_2 = \partial B$ where [insert figure with cancelling
arrows]  $B$ is the image of the unit square as shown.

%% ----------------------------------------------------------------
\subsection{Cellular homology}

Work\index{cellular homology}\index{homology} some examples.  This makes quick
work out of $\R\P^n$, $\C\P^n$, and $\T^n$.

xxx note it's a homology theory for CW complexes so I need to have first said
something about CW complexes.

%% ----------------------------------------------------------------
\subsection{Homotopy equivalence and homotopy invariance}

what, and why (xxx xref fwd).  Makes it easier to do SvK and MV.

\begin{defn}
[xxx cite wiki]

Topological spaces $X$ and $Y$ are \bfidx{homotopy equivalent} (or of the same
\bfidx{homotopy type}) if there exist continuous $f:X\to Y$ and $g:Y\to X$ such
that $g \circ f$ is homotopic to $\mathrm{id}_X$ and $f \circ g$ is homotopic
to $\mathrm{id}_Y$.  The functions $f$ and $g$ are called
\bfidx{homotopy equivalences}.
\end{defn}

\begin{rem}
A homeom is a h.e. but not v.v.  find a ctrex.
\end{rem}

\bfidx{homotopy invariance}:
If $X$ and $Y$ are homotopy equivalent, then:
\begin{itemize}
\item $X$ path conn iff $Y$ is
\item $X$ simply conn iff $Y$ is
\item sing homol and coho groups are isom
\item $\pi_k$ (incl fund grps) are isom as long as $X$, $Y$ path-conn
\end{itemize}

%% ----------------------------------------------------------------
\subsection{Deformation retracts}

what, and why (xxx xref fwd).  Makes it easier to do SvK and MV.

intuition: ctsly shrink space into subspace.

%% ----------------------------------------------------------------
\subsection{Seifert-van Kampen}

stt of thm (lee not'n)

torus example

claim:  for surfaces, just read off the edge presentation.

%% ----------------------------------------------------------------
\subsection{Mayer-Vietoris for homology}

xxx include diagrams from Notation TBD section.  Connections with SvK.
Emphasize importance of choice of $U$, $V$, and $X$ --- be sure their
homologies are known.  Xref bkwd to def retracts and homotopy invariance.
Tabulate known homologies.  Also note that we needn't always solve for $X$:
sometimes it works out more nicely to solve for $U$.  Include example from
August 2006 qual.

%% ================================================================
\subsection{Covering spaces}

\begin{defn}
Let $X$ and $\tilde X$ be topological spaces, with $p$ a continuous surjection
of $\tilde X$ onto $X$.  Then the map $p$ is said to be a \bfidx{cover}, and
the space $\tilde X$ is said to be a \bfidx{covering space} of $X$, if each $x
\in X$ has an open neighborhood $U$ such that $p^{-1}(U)$ is the union is
disjoint open sets in $\tilde X$, each of which is homemorphic to $U$.
Furthermore, Pickrell requires that $\tilde X$ and $X$ be connected.
\end{defn}

\begin{mnem}
You can recover all of the items of the definition by the following example:
send the real line $\tilde X = \R$ to $X = \S^1$ (viewed as the unit circle in
$\C$) via $p(t) = e^{2\pi it}$.  The point 1 on the unit circle has a
neighborhood with preimages $(n-\veps, n+\veps)$ for each integer $n$.  For
other points of $\S^1$, just shift around the circle.
\end{mnem}

\bfidxaux{Lift criterion}{lift criterion}:  nice diagram here.  Massey V.~5.
Note the intuitive obviousness of it (at least if \emph{both} uppers are
covering spaces), given the monomorphism property of covering paths.

monomorphism property:  make a nice picture of a spoked wheel over a
single-dotted wheel.

Def \bfidx{automorphism} in this context.  Make intuitive sense of Massey's
commutative diagram:  homomorphisms must preserve something; what is it that a
homomorphism of cov spcs needs to preserve.

no-fixed-points --- state thm and find examples.  Or, just as a consequence
of group actions.

def univ cov spc\index{universal cover}.

\bfidx{action}\index{group action} of $\pi_1(X, x)$ on $p^{-1}(x)$ --- Massey
is most thorough here.

Let $\phi \in \Aut(\tilde X, X)$ and $\alpha \in \pi_1(X, x_0)$.
Let $\tilde x \in p^{-1}(x_0)$.  Then
$$
	\phi(\tilde x \cdot \alpha) = (\phi \tilde x) \cdot \alpha
$$

[xxx note this is a compatibility between left and right group actions, namely,
the aut group and $\pi_1$ of the base both acting on fibers.  xref to the
beautiful Wikipedia article.]

Massey's cor. 7.3 --- central.

$$
	\Aut(\tilde X, X) \cong
	\frac{N[p_*(\pi_1(\tilde X, \tilde x))]}{p_*(\pi_1(\tilde X, \tilde x))}.
$$

\emph{natural} isom?  Also belabor the group-action lemma in Massey's
appendix B.  Phrase $p_*(\pi_1)$ as an \emph{isotropy subgroup}.

\bfidx{normalizer} notation.  How to actually \emph{compute} such things?  Note
that the normalizer of $H$ in $G$ is the \emph{largest} normal subgroup of $G$
in which $H$ is normal.  In particular, if $H$ is already normal in $G$ (e.g.
abelian, in particular if cyclic) then $N(H)$ is all of $G$.

Find a good example space and covering for this normalizer business: draw up a
nice picture (or set of pictures) for 8-to-1 covering of $\S^1$.  ``Stop
sign''.  See handwritten AT final-review notes from 5-10-06.

\begin{defn}
A covering space $\tilde X$ of $X$ is said to be a \bfidx{regular covering} if
$$
	p_*(\pi_1(\tilde X, \tilde x)) \lhd \pi_1(X, x).
$$
\end{defn}

(Massey 7.4)
$\tilde X/X$ a regular covering implies that for all $x \in X$ and for all
$\tilde x \in p^{-1}(x)$,
$$
	\Aut(\tilde X, X) \cong \frac{\pi_1(X,x)}{p_*(\pi_1(\tilde X, \tilde x))}.
$$
since the normalizer is taken inside the parent group which is $\pi_1(X,x)$.

(Massey 7.5)
Let $U$ be a universal covering space of $X$.  Then
$$
	\Aut(U, X) \cong \pi_1(X,x)
$$
and
$$
	\# \pi_1(X,x) = \# p^{-1}(x)
$$

(Massey 8.1) Aut group is \bfidx{transitive on fibers} iff the covering is
regular.

$\tilde X/X$ a regular covering implies $X$ is homeom to $\tilde X /
\Aut(\tilde X, X)$.

\begin{defn}
The action of a group $G$ on a topological space $X$ is \bfidx{free} (we say
$G$ acts \bfidx{freely} on $X$) if the stabilizer subgroup of each $x$ in $X$
is trivial, i.e. if all non-identity elements of $G$ move all points of $X$.
\end{defn}

\begin{defn}
The action of a group $G$ on a topological space $X$ is \bfidx{properly
discontinuous} if for all $x \in X$, there is a neighborhood $U$ of $x$ such
that for all $g \in G$ other than the identity of $G$, $gU \cap U = \emptyset$.
\end{defn}

%% ================================================================
\subsection{Topo TBD}

xxx in particular, \emph{lift recognition}:  many problems turn out to be an
application of the lifting theorem; the task is to \emph{see} them as such.

\begin{itemize}
\item Retract and deformation retract:  emphasize recognition and computation.
\item Seifert-van Kampen
\item Universal covering
\item $\pi_1(X)$ acting on the covering space $\tilde X$.
	(This is an application of the lifting theorem.)
\item $\Aut(\tilde X/X)$ acting on the covering space $\tilde X$.
\item Mayer-Vietoris
\end{itemize}

Also look at the same topics in this order:

\begin{itemize}
\item What options do we have for computing a fundamental group?
\item What options do we have for computing the automorphism group
	of a covering space?
\end{itemize}

For MV:  disconnect via (a) showing a \emph{group} is 0, or (b) showing a
\emph{map} has zero \emph{image}, or (c) showing a \emph{map} has zero
\emph{kernel}.

%% ----------------------------------------------------------------
\subsection{Cubes and chains}
\label{sec:cubes_and_chains}

\begin{defn}
$k$-cube \ldots .
\end{defn}

\begin{defn}
$k$-chain \ldots .
\end{defn}

%% ----------------------------------------------------------------
\subsection{The boundary operator $\D$}

Make sure to emphasize the sign which alternates with variable index.

%% ----------------------------------------------------------------
\subsection{Homology}

%% ================================================================
%% ================================================================
%% ================================================================
\newpage
\section{Smooth manifolds}
\label{sec:smooth_manifolds}

\begin{center}
\emph{
Life is a short affair; we should try to make it smooth, and free from strife.
} \\
--- Euripides\index{Euripides} (c. 480-406 B.C.).
\end{center}

%% ================================================================
\subsection{Manifolds}

%% ----------------------------------------------------------------
\subsubsection{Manifolds}

\begin{defn} A \bfidx{manifold} is a \plainidx{metric space} with the property
that for all $q \in M$, there is some neighborhood $U$ of $q$ and some
non-negative integer $m(q)$ such that $U$ is homeomorphic to $\R^m$.  If $m(q)$
is the same for all $q \in M$, then we say that $M$ has \bfidx{dimension} $m$.
\end{defn}

For this course we consider only connected manifolds.  In particular, all our
manifolds have a dimension.  (For a counterexample, consider the plane $z=1$ in
$\R^3$ along with the line $y=z=0$.  This has two disconnected components: the
plane with dimension 2 and the line with dimension 1.)

%% ----------------------------------------------------------------
\subsubsection{Coordinates and parameterizations}
\label{sec:coord_paramzn}

\begin{defn}
Let $M$ be an $m$-dimensional manifold.  A \bfidx{coordinate chart} is a
homeomorphism $\phi$ from an open subset $U$ of $M$ to $\R^m$.  (Formally, the
chart is the ordered pair $U, \phi$.)
\end{defn}

A manifold by definition has such homeomorphisms for every point in the
manifold, so we are just naming them.  Also, homeomorphisms are invertible by
definition, so it also makes sense to name their inverses.

\begin{defn}
\label{defn:paramzn} Let $M$ be an $m$-dimensional manifold.  A
\bfidx{parameterization} is the inverse of a coordinate chart.
\end{defn}

\begin{defn}
A \bfidx{transition function} on a manifold $M$ is of the form $y\circ x^{-1}$
where $x$ and $y$ are coordinate charts on a manifold $M$.
\end{defn}

Note that a transition function is a homeomorphism from $\R^m$ to $\R^m$.

\beginex
There are several different coordinate charts we can put on the unit sphere
$\S^2$.
\begin{itemize}

\item We use \bfidx{graph coordinates} when we solve for one rectangular
coordinate in terms of the others.  E.g. off the equator we can solve for $z$
in terms of $x$ and $y$.  (Here we are implicitly using the \plainidx{implicit
function theorem}, theorem \ref{thm:implicit_function}.)
An example coordinate chart around the north pole is
$$
	\colvecthree{x}{y}{z} \mapsto \colvectwo{x}{y}
$$
with parameterization
$$
	\colvectwo{x}{y} \mapsto \colvecthree{x}{y}{\sqrt{1-x^2-y^2}}.
$$

\item We use \bfidx{spherical coordinates} which work everywhere except the
north and south poles.  (Why?  Well, a homeomorphism must be one-to-one.
Points with $\phi=0$ and \emph{any} value of $\theta$ all correspond to the
north pole.  Likewise for the south pole and $\phi=\pi$.) Spherical coordinates
for $\R^3$ are
$$
	\colvecthree{x}{y}{z} =
	\colvecthree{\rho\cos\theta\sin\phi}{\rho\sin\theta\sin\phi}{\rho\cos\phi}.
$$
So, when we restrict to $\S^2$ (where $\rho=1$), the coordinate chart is
$$
	\colvecthree{x}{y}{z} \mapsto \colvectwo{\theta}{\phi}
$$
with parameterization
$$
	\colvectwo{\theta}{\phi} \mapsto
	\colvecthree{\cos\theta\sin\phi}{\sin\theta\sin\phi}{\cos\phi}.
$$
The domain of this coordinate chart is any open set as big as the open set
consisting of all of $\S^2$ except the poles and some meridian.  If we were to
include a meridian, then the resulting open set would not be homeomorphic to a
subset of $\R^m$.

\item \bfidxtwo{Stereographic coordinates}{stereographic coordinates} are
discussed in section [xxx homework problem early on].  (That section also gives
an example of transition functions.) Here, we have two coordinate charts, each
of which cover all but \emph{one} point of $\S^2$.

\end{itemize}
\mendex

One would want the transition maps $y\circ x^{-1}$ to be smooth.  The notions
of \bfidx{atlas}, \bfidx{maximal atlas}, and \bfidx{differentiable structure}
are important foundationally, but do not arise in practice in problems given in
this course.  See any of the geometry texts in the bibliography for
definitions.

%% ----------------------------------------------------------------
\subsubsection{Maps of manifolds}

In section \ref{defn:smooth}, we defined smoothness\index{smooth} for functions
from $\R^m$ to $\R^n$.  To apply this notion to manifolds, we can use
coordinate charts and parameterizations.  Let $M$ and $N$ be manifolds of
dimension $m$ and $n$, respectively.  Let $x$ and $y$ be coordinate charts on
$M$ and $N$, respectively.  If $f$ is a function from $M$ to $N$, then we want
to say that $f$ is smooth in the manifold sense if $y\circ f \circ x^{-1}:\R^m
\to \R^n$ is smooth in the Euclidean-space sense.  Of course, we have to
consider what happens if the range of $f\circ x^{-1}$ doesn't intersect the
domain of $y$, and whether this manifold smoothness is independent of the
choice of $x$ and $y$.  This can in fact be done.

\begin{defn} Let $f: M \to N$ be a map of manifolds.  A \bfidx{coordinate
expression} for $f$ is $y \circ f \circ x^{-1}$, where $x$ and $y$ are
coordinate functions on $M$ and $N$, respectively.  \end{defn}

\begin{defn} A map of manifolds $f:M \to N$ is \bfidx{differentiable} if for
all coordinates $(x, U)$ of $M$ and $(y, V)$ of $N$, $y \circ f \circ x^{-1}$
is a differentiable function from $\R^m$ to $\R^n$.  \end{defn}

\begin{defn} A map of manifolds $f:M \to N$ is \bfidx{smooth} if for all
coordinates $(x, U)$ of $M$ and $(y, V)$ of $N$, $y \circ f \circ x^{-1}$ is a
smooth function from $\R^m$ to $\R^n$.  \end{defn}

\begin{defn} A map of manifolds $f:M \to N$ is a \bfidx{diffeomorphism} if it
is a smooth map of manifolds which is also bijective.  \end{defn}

Of course we could say that diffeomorphisms need only be bijective
differentiable (i.e. $C^1$) maps, but for this course we use the same
convention as in section \ref{sec:homeom_diffeom}.

%% ----------------------------------------------------------------
\subsubsection{Immersion and embedding}

\begin{defn}
Let $f:M \to N$ be a map of manifolds.  We say $f$ is an \bfidx{immersion} at
$\vecq\in M$ if $Df_\vecq$ is 1-1 (i.e.  has zero kernel).  If $f$ is an
immersion at all points $\vecq \in M$, then we simply say that $f$ is an
immersion.
\end{defn}

\begin{defn}
Let $f:M \to N$ be a map of manifolds.
We say that $f$ is an \bfidx{embedding} if:
\begin{itemize}
\item It is an immersion, and
\item it is a homemorphism onto its image.
\end{itemize}
\end{defn}

\begin{rem}
An immersion is locally 1-1.  An embedding is globally 1-1.
\end{rem}

\beginex
\ldots
\mendex

\beginex
\ldots
\mendex

\begin{defn} Let $M \subseteq N$ be manifolds.  Then the inclusion map $i$ is a
map of manifolds.  We say that $M$ is an \bfidx{immersed submanifold} of $N$ if
the inclusion map $i$ is an immersion.  Likewise, we say that $M$ is an
\bfidx{embedded submanifold} of $N$ if the inclusion map is an embedding.
\end{defn}

%% ----------------------------------------------------------------
\subsubsection{Regular and critical values}
\label{sec:regular_value}

\begin{defn}
\label{defn:regular_point}
Let $f:M \to N$ be a map of manifolds.  Then $\veca \in M$ is said to be a
\bfidx{regular point} of $f$ if the local derivative of $f$ at $\veca$ is
surjective.
[xxx draw a picture:  $x$ is a coordinate function on $M$ and $y$ is
a coordinate function on $N$.]
That is, $f$ is regular at $\veca$ if
$D(yfx^{-1})|_{x(\veca)}$ has rank $n$ where $n = \dim(N)$.
[xxx note that we need to prove (or cite) that this is independent of the
choice of coordinates.  here or elsewhere:  soap-box about the fact that
much of the fuss in differential geometry is showing that this or that is
independent of choice of coordinates.  For this paper, such fuss will at most
be deferred to an appendix or a citation.]

We then define the following related terms:
\begin{itemize}
\item A point in $M$ is a \bfidx{critical point} of $f$ if it is not
regular.
\item A point $\vecc \in N$ is a \bfidx{critical value} of $f$ if any of the
points in $f^{-1}(\vecc)$ are critical points of $f$, i.e. if $\vecc$
is the image of a critical point.
\item A point $\vecc \in N$ is a \bfidx{regular value} of $f$ if all the points
in $f^{-1}(\vecc)$ are regular points of $f$.
\end{itemize}
\end{defn}

Note that the \emph{points} are in the source manifold $M$; the \emph{values}
are in the destination manifold $N$.

[xxx draw a nice picture here.]

\begin{thm}[regular value theorem]
\label{thm:regular_value}
\index{regular value theorem}
Let $M$ and $N$ be manifolds of dimension $m$ and $n$, respectively, and let
$f:M \to N$ be a map of manifolds.  If $\vecc \in N$ is a regular value of $f$,
then either $f^{-1}(\vecc) = \emptyset$ or $f^{-1}(\vecc)$ has the structure of
an embedded submanifold of $M$.  In the latter case $f^{-1}(\vecc)$ has
dimension $m-n$.
\end{thm}

\begin{rem}
For the last sentence we can intuitively think of $m$ as the number of
variables and $n$ as the number of constraints, leaving $m-n$ degrees of
freedom.
\end{rem}

\begin{rem}
Note that \cite{bib:lee2} refers to this as the \emphidx{regular level set
theorem}.
\end{rem}

\begin{rem}
See section \ref{sec:fr_3} for an example which shows that the converse
of the regular value theorem does not hold.
\end{rem}

\begin{rem}
Manifolds are often presented as embedded submanifolds of Euclidean space (e.g.
$\S^2$ inside $\R^3$).  The $D$ in definition \ref{defn:regular_point} must be
restricted to the manifold.  Either (1) compute $D$ in the parent space and
restrict as in section \ref{sec:proj_perp}, or (2) use \plainidx{Lagrange
multipliers} as described in section \ref{sec:lagrange_multipliers}.  For an
example using both techniques, see section \ref{sec:fe_2}.
\end{rem}

zzzz

[xxx needs an example.]

xxx define codimension.

xxx move this to another section further on:

\begin{rem}
\label{rem:rvt_short_exact}
At a regular point $\vecq\in M$ of $f$, $f_*$ has full rank $n$.  The
linearization $Df|_\vecq$ (which is $f_*|_\vecq$) is a map from the
$m$-dimensional space $TM|_\vecq$ to the $n$-dimensional space $TN|_\vecc$.
Therefore, by the rank-nullity theorem, $Df|_\vecq$ has kernel dimension $m-n$.
[xxx need to fix this b0rk3n statement.]
Let $f:M \to N$ be a map of manifolds.
Let $\vecq \in M$ and $\vecc \in N$ be such that $\vecc = f(\vecq)$.
Assume that $\vecc$ is a regular value of $f$ but do not assume that $f$
is 1-1.
$$
	0 \quad\longto\quad \ker(f_*) \cong Tf^{-1}(\vecc)
	\quad\longto\quad TM|_\vecq
	\stackrel{f_*}{\quad\longto\quad} TN_\vecc
	\quad\longto\quad 0.
$$
\end{rem}

xxx move this to the linear-algebra section:

\begin{prop}[adjugate criterion]
\label{prop:adjugate_criterion}
An\index{adjugate criterion} $n \times m$ matrix (with $m > n$) has full rank
$n$ iff not all $n \times n$ determinants\index{adjugate
criterion}\index{determinant} are zero.
\begin{proof}
This is an exercise.  See section \ref{sec:hw_1_2d}.
[This can't be quite right --- the exercise is for square matrices; this is for
non-square.  How do we generalize it?]
\end{proof}
\end{prop}

xxx note how cramer's rule is related to exterior algebras.  The proof of
Cramer's rule is elementary and easy \emph{if} you already have the formula ---
but how would a person have thought of it?  Appeal to exterior algebra?

\beginex
Insert an example here, or xref.
\mendex

%% ================================================================
\newpage
\subsection{Vector bundles}

%% ----------------------------------------------------------------
\subsubsection{Varying coefficients}
\label{sec:loaf}

All the vectors, linear functionals, tensors, etc. discussed up until now have
had \bfidx{constant coefficients}, i.e. coefficients in $\R$.  Now picture the
Euclidean space $V=\R^m$ evolving with time $t$.  (Or, for $V = \R^2$, one may
think of a loaf of bread, where each slice of bread is a vector space and the
parameter $t$ indexes the slices of bread.  Of course, the loaf of bread is
infinitely long, with slices that are infinitely thin and infinitely wide.)

\begin{itemize}

\item In this collection of vector spaces, you could imagine a vector evolving
in time.  Using the other metaphor, this would be a collection of vectors, one
vector in each slice, with \bfidx{varying coefficients}, depending smoothly on
$t$.  We could label the vector in the $t$th slice $\vecv(t)$, or, looking
ahead to the notation used in differential geometry, $\vecv_t$.  For example,
in $\R^2$ we might have $\vecv_t = (\cos t, \sin t)$.  We could label the
collection $\vecv$.

\item If, in the sense of section \ref{sec:projections}, we form a
\bfidx{projection} $\pi$ from $V \times \R$ to $\R$ by $\pi(\vecv, t) = t$,
then we can think of $\vecv$ as a \bfidx{section} of $\pi$.  (We will be
calling it a \bfidx{vector field}).

\item We can write this as $V \times \R$, but it is not the same as $\R^{m+1}$.
It doesn't make sense to add one time's $\vecu_{t_1}$ to another time's
$\vecv_{t_2}$, but it does make sense to add vectors in the same time slice.

\item Here the coefficients are not just in $\R$; they're smooth functions,
i.e.  elements of $C^\infty(\R)$.  We might think of this $V \times \R$ as a
vector space over $\R$, but it sure feels like it ought to be some sort of
space over $C^\infty(\R)$ as well.  (We will be calling it a \bfidx{vector
bundle} over $\R$.)

\item Non-zero real numbers are invertible, so $\R$ is a field.  Not all
non-zero elements of $C^\infty(\R)$ can be reciprocated:  for a function $f \in
C^\infty(\R)$ to be non-zero, it must have a non-zero value for \emph{some}
$t$; to be reciprocable, it would have to be non-zero for \emph{all} $t$.
However, $C^\infty(\R)$ is a commutative ring with identity.  Furthermore, it
may be shown (see section \ref{sec:def_vb}) that this $V \times \R$ is a
\bfidx{free module} over $C^\infty(\R)$.  Remember from section
\ref{sec:modules} that every element of a free module $M$ over a ring $R$ is
uniquely expressible as an $R$-linear combination of basis elements.  So, we
get everything we would want from a vector space, except division by scalars.

\item We can also form mixed tensors here:  the tensor product, which we
examined for vector spaces, can also be formed on modules.  (This kind of
construction will be the subject of most of the rest of this paper.) Here, the
coefficients of the tensors won't be real numbers; instead, they will be smooth
functions of the parameter $t$.

\item If we select a particular value of $t$, then we do have a plain old
vector space over $\R$, where all the vectors, tensors, etc. have constant
coefficients which are simply the coefficient functions evaluated at $t$.

\end{itemize}

In this picture, constructions on $V$ (i.e.  the space, its dual, higher-order
tensors, etc.) have coefficients in $\R$, while constructions in $V \times \R$
have coefficients in $C^\infty(\R)$:

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.0)
%\put(0,0){\framebox(6,1)}
\put(2.20, 0.80){$V$}
\put(2.25, 0.75){\line(0,-1){.50}}
\put(2.20, 0.10){$\R$}
\put(3.20, 0.80){$V \times \R$}
\put(3.30, 0.25){\line(0,1){.50}}
\put(3.20, 0.10){$C^\infty(\R)$}
\end{picture}

Note also instead of parameterizing by a single variable $t$, we could attach a
copy of $V$ to every point $(x,y)$ of $\R^2$.  In that case, scalars
(coefficients) would be of the form $f(x,y)$ for $f \in C^\infty(\R^2, \R)$.

%% ----------------------------------------------------------------
\subsubsection{Definition of vector bundle}
\label{sec:def_vb}

Motivated by the preceding $V \times \R$ example, what properties would we want
a vector bundle to have?

\begin{itemize}
\item Every time $t$ should have a copy of $V$ attached to it.
\item Even though we took $V$ to be $\R^m$, our $V \times \R$ should not be the
same as $\R^{m+1}$.  It should only make sense to add vectors from the same
time slice $t$.
\item Coefficients on vector fields $\vecv$ should vary smoothly with time.
\item The vector spaces $V_t$ themselves should be connected to one another
smoothly with time.
\end{itemize}

Here is the formal definition (following \cite{bib:lee2}).

\begin{defn} Let $E$ and $M$ be manifolds, and let $\pi:E \to M$ be a
surjective smooth map of manifolds.  (Note that $\pi$ is a \bfidx{projection}
in the sense of section \ref{sec:projections}.)
Then $E$, $M$, and $\pi$ form a \bfidx{vector bundle} of \bfidx{rank}
$k$ if the following conditions are satisfied:

\begin{itemize}

\item For each $\vecq \in M$, the \bfidx{fiber} (see definition
\ref{defn:fiber}) $\pi^{-1}(\vecq)$ is a $k$-dimensional real vector space.
Write $E_\vecq$ for the fiber $\pi^{-1}(\vecq)$.  Note that since $\pi$ is
surjective, every point $\vecq$ of $M$ has a fiber over it.  We call $M$ the
\bfidx{base space}.

\item For each $\vecq \in M$, there exist an open subset $U$ of $M$ and a
diffeomorphism $\Phi: \pi^{-1}(U) \to U \times \R^k$.  This means that the
fibers vary smoothly as we move around open sets in the base space.

\item For the $\vecq$, $U$, and $\Phi$ in the previous item, the following
diagram commutes:

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.1)
%\put(0,0){\framebox(6,1.1)}
\put(2.75, 0.90){$\Phi$}
\put(2.10, 0.80){$\pi^{-1}(U)$}
\put(2.25, 0.75){\vector(1,-1){.50}}
\put(2.20, 0.40){$\pi$}
\put(3.20, 0.80){$U\times \R^k$}
\put(3.35, 0.75){\vector(-1,-1){.50}}
\put(3.30, 0.40){$\pi_1$}
\put(2.60, 0.85){\vector(1,0){.55}}
\put(2.75, 0.10){$U$}
\end{picture}

where $\pi_1$ is the projection onto the first component.  If $\vece$ is a
point of $E$ lying above $\vecq$, then $\pi(\vece) = \vecq$, whereas if we take
$\Phi(\vece)$ and then project down to $M$, we have to get the same base point
$\vecq$.  So, all this means is that $\Phi$ respects which fiber lies over
which base point.

\item Furthermore, for each $\vecq' \in U$ as in the previous item, $\Phi$
restricted to the fiber $E_{\vecq'}$ is a vector-space isomorphism between
$E_{\vecq'}$ and $\{\vecq'\} \times \R^k$, which in turn is isomorphic to
$\R^k$.  Sometimes we state this condition by saying that $\Phi$ is
\bfidx{linear on fibers}.

\end{itemize}
\end{defn}

Some more terminology:

Recall from definition \ref{defn:section} that a \bfidx{section} of $\pi$ is a
function \index{$s$} $s: M \to E$ such that $\pi \circ s$ is the identity on
$M$, which simply means that a section maps base points into their own fibers.
As in section \ref{defn:section}, whenever we have projections and sections, we
want those maps to have whatever properties are appropriate for the category in
question.  Here, this means that projections and sections are taken to be
\bfidx{smooth}.

\begin{defn}
A \bfidx{vector field} is a section of vector bundle.
\end{defn}

That is, a section is something of the form described in section
\ref{sec:loaf}, i.e. a vector whose coefficients depend smoothly on the base
point $\vecq$.

\begin{defn}
Given a vector field $\vecv$ on a manifold $M$, for each point $\vecq$ of $M$
there is a vector $\vecv_\vecq$.  We say that $\vecv_\vecq$ has its
\bfidx{foot} at $\vecq$.
\end{defn}

\begin{defn} Sometimes we write $\vecv|_\vecq$ instead of $\vecv_\vecq$.  We do
this particularly when the thing being subscripted (or the subscript itself, or
both) is a complicated expression, has other subscripts, etc.  See also
notation \ref{notn:v_comma_q}. \end{defn}

Recall from definition \ref{defn:gamma} that we write $\Gamma(E)$ for all the
sections of $E$, or the \bfidx{space of sections} of $E$.  This is actually
makes $\Gamma(E)$ a \bfidx{module} over $C^\infty(M,\R)$, as was alluded to in
section \ref{sec:loaf}:

\begin{itemize}

\item Given two sections, or vector fields, $\vecv_1, \vecv_2:M \to E$, we have
vector-plus-vector addition of the form $(\vecv_1 + \vecv_2)|_\vecq =
\vecv_1|_\vecq + \vecv_2|_\vecq$.

\item Given a ``scalar'' $f \in C^\infty(M, \R)$ and $\vecv \in \Gamma(E)$, we
have multiplication by the scalar of the form $(f\vecv)|_\vecq = f|_\vecq
\vecv|_\vecq$.

\end{itemize}

Thus we have\index{Rosetta stone} the following analogy between vector spaces
and vector bundles:

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.0)
%\put(0,0){\framebox(6,1)}
\put(2.20, 0.80){$V$}
\put(2.25, 0.75){\line(0,-1){.50}}
\put(2.20, 0.10){$\R$}
\put(3.15, 0.80){$\Gamma(E)$}
\put(3.30, 0.25){\line(0,1){.50}}
\put(3.00, 0.10){$C^\infty(M,\R)$}
\end{picture}

%% ----------------------------------------------------------------
\subsubsection{Examples of vector bundles}

Let $M$ be a manifold.

\beginex
\label{ex:product_bundle}
For any $k$, we can attach a copy of $\R^k$ at each point $\vecq$ of $M$.  This
is called a \bfidx{product bundle}.
\mendex

\beginex For a manifold which is, say, the level set of some function from
$\R^m$ to $\R$, we can define a \bfidx{normal bundle}, which at each point
$\vecq$ of $M$ consists of all the vectors in the \plainidx{ambient space}
$\R^m$ which are normal to $M$ at $\vecq$.  For example, for the sphere $\S^2$,
the normal bundle has rank 1, where the fiber above a point $\vecq$ of $\S^2$ is
the line normal to the sphere at $\vecq$.  \mendex

\beginex Similarly, in the same context, we can also define a \bfidx{tangent
bundle}, where the vector space attached to each point $\vecq$ of $M$ consists
of the space perpendicular to the normal space.  For example, on $\S^2$, this is
of rank 2 where fibers are tangent planes.  \mendex

Tangent bundles are central to the geometry course.  In section \ref{sec:tb},
they will be defined in a way that is independent of an ambient space.

%% ----------------------------------------------------------------
\subsubsection{Trivializations of vector bundles}
\label{sec:triv_vb}

\begin{defn} A diffeomorphism $\Phi$, defined on an open subset $U$ of $M$ as
in section \ref{sec:def_vb}, is called a \bfidx{local trivialization} of the
vector bundle.  If there is such a diffeomorphism $\Phi$ defined on all of $M$,
then it is said to be a \bfidx{global trivialization} of the vector bundle.  If
a vector bundle has a global trivialization, the bundle is said to be
\bfidx{trivial}.  \end{defn}

\begin{notn}
\label{notn:v_comma_q}
Pickrell adopts the convention that $\vecv_\vecq$ may be written as the ordered
pair $(\vecv, \vecq)$ only for trivial bundles.
\end{notn}

There is a criterion to determine whether a vector bundle is trivial.  It
requires a few definitions.

\begin{defn} \label{defn:frame} Let $(E,M,\pi)$ be a rank-$k$ vector bundle.
Given an open subset $U$ of $M$, a \bfidx{local frame}\index{frame} is an
ordered set $\{\vecb_1, \ldots, \vecb_k\}$ of $k$ (smooth) sections of the
vector bundle such that for all $\vecq \in U$, the set $\{\vecb_1|_\vecq,
\ldots, \vecb_k|_\vecq\}$ forms a basis for the fiber $E_\vecq$.  If there is
such an ordered set defined on all of $M$, then it is called a \bfidx{global
frame}.  \end{defn}

Lee proves the following in chapter 5 of \cite{bib:lee2}.  Pickrell uses this
criterion implicitly:  when he asks you to compute an explicit trivialization,
what he wants is a global frame.

\begin{prop}[Global frame criterion]
A vector bundle is trivial iff it admits a global frame.
\end{prop}

For an example, see section \ref{sec:hw_6_1}.  [xxx that problem uses
terminology from sections downstream from here.]

Note that the \plainidx{product bundle} (example \ref{ex:product_bundle}) is
certainly trivial: take the standard basis for each $\R^k$ to be the global
frame.

%\begin{defn}
%A vector bundle is \bfidx{trivial} if it is isomorphic to the direct product of
%the base space and a fiber.  A \bfidx{trivialization} of a vector bundle is
%such an isomorphism.
%\end{defn}

%% ================================================================
\newpage
\subsection{The tensor bundle}
\label{sec:tb}

In this section we define tangent vectors, then build a tensor algebra on top
of that.  This will encapsulate many concepts, including derivatives,
functions, functionals, area forms, and dot products.

%% ----------------------------------------------------------------
\subsubsection{Notions of tangent vectors}

There are many different definitions of tangent space given in
\cite{bib:satan}, \cite{bib:conlon}, and \cite{bib:abr}.  If I count correctly,
\cite{bib:satan} alone presents no less than 5 definitions, and proves them to
be equivalent.  This overabundance of definitions is a great source of
confusion for the learner.

For Pickrell's course, two notions are actually used reguarly in computations,
with a third notion appearing briefly early on.  (Namely, this third notion is
tangent vectors as equivalence classes of curves.  This is discussed in section
\ref{sec:tang}).  As well, these are precisely the two notions that
\cite{bib:lee2} emphasizes.  I will follow that approach.  Namely:

\begin{itemize}

\item If we have a surface explicitly embedded in $\R^{k}$ for some $k$, then
we can think of the tangent space in the usual way from vector calculus.
Namely, we compute the normal vector to the surface at each point, then find
the space of vectors perpendicular to the normal.  These are called
\bfidx{geometric tangent vectors}.  Such tangent vectors are \bfidx{extrinsic}
in that they stick off the manifold: for example, consider tangent planes on a
sphere.  This works fine when our manifold is already embedded into a
higher-dimensional Euclidean space in some obvious way.

\item The standard example of why geometric tangent vectors are insufficient is
the spacetime of the universe itself:  regardless of whether it \emph{could} be
visualized as embedded in some higher-dimensional space, we might not
\emph{want} to do so, particularly when our manifold is already
three-dimensional.  (Trying to draw three-dimensional pictures on a chalkboard
is hard enough!) The more general way to think of the tangent space is as the
space of \bfidx{directional derivatives}.  These are \bfidx{intrinsic} in that
we don't need to visualize anything sticking off the manifold.

\end{itemize}

%% ----------------------------------------------------------------
\subsubsection{Geometric tangent vectors}
\label{sec:geom_tang}

Geometric tangent vectors are just what you would expect from vector calculus
(\cite{bib:anton}, \cite{bib:hhgm}).  In the case that our manifold $M$ is the
level set of a single equation, we can get the tangent space as a two-step
process:

\begin{itemize}
\item[(1)] Compute the normal space to the surface.  If the surface is the
level set of a function $F$, then the normal vector to a point $\vecq \in M$ is
$\nabla(F)|_\vecq$.
\item[(2)] Do linear algebra to compute the perpendicular space, which is
the tangent space.
\end{itemize}

This technique is time-consuming; a more efficient technique is presented in
section \ref{sec:compute_tang}.  An example comparing both methods appears in
section \ref{sec:fr_5}.

%% ----------------------------------------------------------------
\subsubsection{Tangent vectors}
\label{sec:tang}

Here I want to accomplish two things:  make some sense of Pickrell's choice of
assigned problems, and intuitively connect geometric tangent vectors with
tangent vectors as directional derivatives.  The latter is made quite clear in
chapter 3 of \cite{bib:lee2}, so I will confine myself to presenting motivation
and examples.

\textbf{Preliminary \#1.} Suppose $M$ is the Euclidean space $\R^m$ and let $f:\R^m \to
\R$.  Then, given a point $\vecq$ of $M$, it is reasonable to equate a vector
$\vecv_\vecq$ footed at $\vecq$ with the \bfidx{directional derivative}, taken
on functions from $\R^m$ to $\R$, in the direction of $\vecv$.  This directional
derivative is written
$$
	Df|_\vecq(\vecv) = \frac{d}{dt}\pickpipe{t=0}f(\vecq+t\vecv)
$$
The left-hand side is the derivative of $f$ at $\vecq$ in the direction of
$\vecv$.  One can show that there is in fact an isomorphism identifying each
$\vecv_\vecq$ with the directional derivative at $\vecq$ in the direction of
$\vecv$.

\textbf{Preliminary \#2.} Let $\vecx = \vecq + t\vecv$, i.e. for $i=1, \ldots, m$, $x_i
= q_i + t v_i$, and let $f \in C^\infty(\R^m, \R)$.  Then $f(\vecx)$ is a
function of $t$.  By the chain rule, we have
$$
	\frac{d}{dt}\pickpipe{t=0}f(\vecq+t\vecv)
	= \sum_{i=1}^m \fracDtwo{f}{x_i} \frac{dx_i}{dt}
	= \sum_{i=1}^m \fracDtwo{f}{x_i} v_i.
$$
In particular, take $\vecv$ to be a standard basis vector $\vece_j$.  Then
the above gives
$$
	\frac{d}{dt}\pickpipe{t=0}f(\vecq+t\vece_j)
	= \sum_{i=1}^m \fracDtwo{f}{x_i} \frac{dx_i}{dt}
	= \fracDtwo{f}{x_j}.
$$

\textbf{Preliminary \#3.}  Now let $M$ be a manifold not necessarily $\R^m$.
Note that in the previous step, any other curve through $\vecq$ in the
direction of $\vecv$ which is equal to $\vecq+t\vecv$ up to first order would
have produced the same result.  The intuitive example is $M=\S^2$.  Then tangent
lines of the form $\vecq+t\vecv$ stick off $\S^2$, but we can have a curve
through $\vecq$, heading off in the direction of $\vecv$, but staying on the
sphere.  (This notion makes an appearance in some problems early on in the
course.  See for example section \ref{sec:hw_1_2d}.)  We could use this
intuition to motivate a definition of tangent vectors as equivalence classes of
curves which agree up to first order.  However, we do not in fact define them
that way.

\begin{defn}
We now define \bfidxtwo{tangent vectors}{tangent vector} as follows.
\begin{itemize}
\item Since we have an $m$-dimensional manifold $M$, at each point $\vecq$ of
$M$ there is an open set $U$ with a \plainidx{coordinate chart} $\phi$ which is
a homeomorphism from $M$ to $\R^m$.

\item From step 1 we know that in $\R^m$ there is an isomorphism between
$\vecv_\vecq$'s and directional derivatives at $\vecq$ in the direction of
$\vecv$.

\item In particular, using that isomorphism we can identify each $\vece_j$ in
$\R^m$ with the directional derivative $\slashD{x_j}$.

\item We use that isomorphism followed by the parameterization $\phi^{-1}$ (the
inverse of the coordinate chart $\phi$) to map the standard basis for $\R^m$
back to an open subset $U$ of $M$.  (See Lee's proposition 3.6 for a technical
detail.)

\end{itemize}
\end{defn}

In summary, the tangent space at a point $\vecq$ of $M$ has a basis, in
coordinates, which is the set
$$
	\left\{ \fracD{x_1}\pickpipe\vecq, \ldots, \fracD{x_m}\pickpipe{\vecq} \right\}
$$

\begin{defn}
Let $M$ be a manifold, and let $\vecq \in M$.  The \bfidx{tangent space} of $M$
at $\vecq$, written $T_\vecq M$ or $TM|_\vecq$, consists of all
\bfidx{directional derivatives}, applied to $C^\infty(M) = \mcT^0_0(M)$.
\end{defn}

\begin{defn}
The \bfidx{tangent bundle} of $M$ is the disjoint union
$$
	TM  = \coprod_{\vecq \in M} T_\vecq M.
$$
\end{defn}

\begin{intuition}
Disjointification is a way to make otherwise equal things unequal.  For
example, take two copies of the integers, one colored red and one colored blue.
Then red 3 is unequal to blue 3.  Two copies of $\Z$ have been disjointified by
the color label.  Here, we have many tangent spaces, which are disjointified by
the foot label.
\end{intuition}

%% ----------------------------------------------------------------
\subsubsection{Explicit computations of tangent vectors}
\label{sec:compute_tang}

For problems appearing in Pickrell's course, we start with the
directional-derivative notion of tangent vector and end up with a geometric
tangent vector.  An example will help illustrate.

\beginex
\label{ex:s2_tang}
Let $M=\S^2$.  A point $\vecq$ of $\S^2$ is (in rectangular and spherical
coordinates, respectively)
$$
	\vecq = \colvecthree{x}{y}{z} =
	\colvecthree{\cos\theta\sin\phi}{\sin\theta\sin\phi}{\cos\phi}.
$$
The tangent space is spanned by the two (orthogonal) basis vectors
$\slashD\theta$ and $\slashD\phi$.  Now, we make the key observation that
$\slashD\theta$ and $\slashD\phi$ are directional derivative operators,
acting on $C^\infty(\S^2)$, while the $x$, $y$, and $z$ coordinates are in
fact smooth functions of the point $\vecq$.  So, we \emph{apply}
$\slashD\theta$ and $\slashD\phi$ to the coordinate functions $x$, $y$, and
$z$:
\begin{eqnarray*}
\fracD\theta\colvecthree x y z &=&
	\fracD\theta \colvecthree{\cos\theta\sin\phi}{\sin\theta\sin\phi}{\cos\phi}
	= \colvecthree{-\sin\theta\sin\phi}{\cos\theta\sin\phi}{0}; \\
\fracD\phi\colvecthree x y z &=&
	\fracD\phi \colvecthree{\cos\theta\sin\phi}{\sin\theta\sin\phi}{\cos\phi}
	= \colvecthree{\cos\theta\cos\phi}{\sin\theta\cos\phi}{-\sin\phi}.
\end{eqnarray*}
Observe that when we do this, the results no longer look like a
directional-derivative operators:  they are just geometric tangent vectors.
The isomorphism of section \ref{sec:tang} goes unspoken; we write
$$
\fracD\theta =
	\colvecthree{-\sin\theta\sin\phi}{\cos\theta\sin\phi}{0} \qqand
\fracD\phi =
	\colvecthree{\cos\theta\cos\phi}{\sin\theta\cos\phi}{-\sin\phi}.
$$
\mendex

Here is another example.

\beginex[Surface of revolution.]
\label{ex:gy_tang}
Let $g(y) > 0$ be a smooth function.  Let $M$ be the surface of revolution
obtained by revolving $g(y)$ about the $y$ axis.  Then a point $\vecq$ of
$M$ has rectangular coordinates
$$
	\vecq = \colvecthree{g(y) \cos \theta}{y}{g(y) \sin \theta}.
$$
Now, we can parameterize this surface using $y$ and $\theta$, where $\theta$ is
the angle from the $x$ axis to $\vecq$.  Thus, the tangent space will be
spanned by $\slashDy$ and $\slashD\theta$.  Proceeding as in example
\ref{ex:s2_tang}, we apply $\fracDy$ and $\fracD\theta$ to the $x$, $y$,
and $z$ coordinates of $\vecq$ to obtain
$$
	\fracDy = \colvecthree{g'(y) \cos \theta}{1}{g'(y) \sin \theta}, \quad
	\fracD \theta = \colvecthree{-g(y) \sin \theta}{0}{g(y) \cos \theta}.
$$
\mendex

%% ----------------------------------------------------------------
\subsubsection{$F_*$ and $DF$}

Let\index{$F_*$}\index{$DF$} $F:M \to N$ be a map of manifolds of dimensions
$m$ and $n$, respectively.  Then we would like to define a
\bfidx{linearization}\index{$F_*$} of $F$ at $\vecq\in M$ as a function on the
respective tangent spaces, i.e. a linear approximation to $F$ at $\vecq$.  How
can we do this?  Well, we saw in section \ref{sec:tang} that the tangent space
$T_\vecq M$ is spanned by the $\D/\Dx_i$'s.  An arbitrary element $X$ of the
tangent space is just a linear combination of those.  A map on the tangent
spaces is defined by what it does to the $X$'s.  And what do the $X$'s do?
They operate on functions from a manifold to $\R$.  So (\cite{bib:lee2},
chapter 3) we make the following definition.

\begin{defn}
Let $F:M \to N$ be a map of manifolds of dimensions $m$ and $n$, respectively.
Let $\vecq \in M$.  Let $X \in T_\vecq M$ and $f:N \to \R$.  Recall that the
pullback $F^*f$ is $f\circ F$.  Then
$$F_*:T_\vecq M \to T_{F(\vecq)}N$$ is defined by
$$
\framebox[2.0in][c]{
	$(F_* X)(f) =
	%X(f \circ F) =
	X(F^*f).$
}
$$
\end{defn}

Note the following diagram\index{Rosetta stone} of the situation:

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.6)
%\put(0,0){\framebox(6,1.6)}

\put(2.20, 1.30){$T_\vecq M$}
\put(4.20, 1.30){$T_{F(\vecq)}N$}
\put(2.20, 0.80){$M$}
\put(4.20, 0.80){$N$}
\put(3.20, 0.10){$\R$}

\put(1.20, 1.30){$X$}
\put(5.20, 1.30){$F_*X$}
\put(1.20, 0.80){$\vecq$}
\put(5.20, 0.80){$F(\vecq)$}

\put(3.10, 1.50){$F_*$}
\put(2.55, 1.35){\vector(1,0){1.50}}
\put(3.10, 1.00){$F$}
\put(2.55, 0.85){\vector(1,0){1.50}}
\put(4.10, 0.40){$f$}
\put(4.00, 0.75){\vector(-1,-1){0.55}}
\put(2.10, 0.40){$F^*f$}
\put(2.50, 0.75){\vector(1,-1){0.55}}
\end{picture}

There are several points:

\begin{itemize}
\item A map $f:N \to \R$ has been \bfidxtwo{pulled back}{pullback}, via $F$,
from $N$ to $M$.

\item A tangent vector $X$ has been \bfidxtwo{pushed forward}{pushforward}, via
$F_*$, from $T_\vecq M$ to $T_{F(\vecq)}N$.

\item This is a \bfidx{coordinate-free} definition of $F_*$.

\item As shown in \cite{bib:lee2}, whenever we do use coordinates, this map
$F_*$ is represented by the Jacobian matrix
$$
	DF = \left(\fracDtwo{F_i}{x_j}\right)_{i,j=1, \ldots, m}.
$$
%Hence the following statement in \cite{bib:satan}:
%$$
%	F_*: T\R^m \to T\R^n:  \vecv_q \mapsto [DF|_\vecq(\vecv)]_{F(\vecq)}.
%$$
%That is, we take a tangent vector $\vecv$ footed at $\vecq$.  We compute the
%Jacobian and evaluate its coefficients at $\vecq$.  We apply that to $\vecv$,
%and foot the result at $F(\vecq)$.

\end{itemize}

%% ----------------------------------------------------------------
\subsubsection{The cotangent bundle}

\begin{defn}
Let $M$ be a manifold, and let $\vecq \in M$.  The \bfidx{cotangent space} of
$M$ at $\vecq$, written $T_\vecq^*M$ or $T^*M|_\vecq$, is the dual space of
$T_\vecq M$.  It consists of all \plainidx{linear functionals}, or
\bfidxtwo{covectors}{covector}, on $T_\vecq M$.  The \bfidx{cotangent bundle}
of $M$ is the disjoint union
$$
	T^*M  = \coprod_{\vecq \in M} T^*_\vecq M.
$$
\end{defn}

Just as in section \ref{sec:basis_and_dual_space}, given a basis we can form a
\bfidx{dual basis}.  We saw in section \ref{sec:tang} that a basis for $T_\vecq
M$, in coordinates, is given by
$$
	\left\{
		\fracD{x_1}\pickpipe{\vecq}, \ldots, \fracD{x_m}\pickpipe{\vecq}
	\right\}.
$$
The dual basis elements are
$$
	\left\{ d\,x_1\pickpipe\vecq, \ldots, d\,x_m\pickpipe{\vecq} \right\}
$$
and they are defined, just as in section \ref{sec:basis_and_dual_space}, by
$$dx_i\left(\fracD{x_j}\right) = \delta_{ij}.$$  That is, these are nothing
more than coordinate-selector functions, in the sense of section
\ref{sec:basis_and_dual_space}, on the tangent space.

Just as for frames (see definition \ref{defn:frame}), we can have
\bfidx{coframes} of the form
$$
	\left\{ d\,x_1, \ldots, d\,x_m \right\}.
$$

%% ----------------------------------------------------------------
\subsubsection{The differential $df$}
\label{sec:df}

\begin{defn}
\label{defn:differential}
Let $f:M \to \R$.  Let $\vecq \in M$ and $X_\vecq \in T_\vecq M$.  Then
\plainidx{$df$}, the \bfidx{differential} of $f$, is defined (in a
\bfidx{coordinate-free} way) as
$$
	df|_\vecq(X_\vecq) = X_\vecq f
$$
\end{defn}

Note that $df$ is a linear functional on each tangent space:  it takes a
tangent vector and produces a scalar, and it is linear.

\begin{rem}
In coordinates (\cite{bib:lee2}), $df$ is written
$$
	df = \sum_{i=1}^m \fracDtwo{f}{x_i} \,dx_i.
$$
This looks a lot like the gradient, but written as a covector rather than a
vector.  Why?  Well, the Jacobian matrix for a vector-to-scalar function is a
$1 \times m$ matrix.  This looks like a \bfidx{row vector}, and we've been
equating row vectors with linear functionals (see section
\ref{sec:compute_dual_basis}).  Recall from the discussion in section
\ref{sec:VS_COB} that if we were to treat it as a \bfidx{column vector} (plain
old vector) rather than as a row vector (linear functional or covector) then it
wouldn't transform correctly on change of basis.
\end{rem}

\begin{rem}
In addition to asking what kind of map $df$ is (we saw that it's a 1-form), we
can also ask what kind of map $d$ is.  Here, we started with $f$, which is a
scalar, or 0-form.  Then $df$ is a 1-form.  One might guess that $d$ will in
general send a $k$-form to a $(k+1)$-form.  That is indeed the case, as is
discussed in section \ref{sec:d_operator}.
\end{rem}

% xxx include Lee's intuition here.

%% ----------------------------------------------------------------
\subsubsection{The tensor bundle}

Given a manifold $M$, we now have a tangent bundle $TM$ and a cotangent bundle
$T^*M$.  We also have the respective spaces of smooth sections\index{section},
$\Gamma(TM)$ and $\Gamma(T^*M)$.

\begin{defn} Elements of $\Gamma(TM)$ and $\Gamma(T^*M)$ are called
\bfidx{vector fields} and \bfidx{covector fields}, respectively.  \end{defn}

As discussed in section \ref{sec:def_vb}, $\Gamma(TM)$ and $\Gamma(T^*M)$ are
free modules over $C^\infty(M, \R)$.  We can appropriate the discussion of
tensors in section \ref{sec:tensors} wholesale, replacing constant coefficients
with coefficients that vary smoothly over the manifold.  This includes forming
the full tensor algebra as well as defining symmetric and alternating tensors.

\begin{notn}
We write $\mcT^s_r(M)$ in place of $\mcT^s_r(TM)$.
\end{notn}

Just as in section \ref{sec:mixed_tensors}, $C^\infty(M,\R)$ is a subspace
of $\Gamma(TM)$ and so on.  So\index{Rosetta stone} we have:

\setlength{\unitlength}{1.0in}
\begin{picture}(6,2.0)
%\put(0,0){\framebox(6,2.0)}

\put(2.90, 0.05){scalars}
\put(2.50, 0.25){$\mcT^0_0(M) = C^\infty(M,\R)$}
	\put(2.70, 0.45){\vector(-1,1){.20}}
	\put(3.35, 0.45){\vector( 1,1){.20}}

\put(2.10, 0.75){tangent space}
\put(3.30, 0.75){cotangent space}
\put(2.05, 0.95){$\mcT^1_0(M) = TM$}
\put(3.25, 0.95){$\mcT^0_1(M) = T^*M$}
	\put(2.25, 1.15){\vector(-1,1){.20}}
	\put(2.50, 1.15){\vector( 1,1){.20}}
	\put(3.55, 1.15){\vector(-1,1){.20}}
	\put(3.80, 1.15){\vector( 1,1){.20}}

\put(1.70, 1.45){$\mcT^2_0(M)$}
\put(2.90, 1.45){$\mcT^1_1(M)$}
\put(4.10, 1.45){$\mcT^0_2(M)$}
	\put(1.65, 1.65){\vector(-1,1){.20}}
	\put(1.90, 1.65){\vector( 1,1){.20}}
	\put(2.95, 1.65){\vector(-1,1){.20}}
	\put(3.20, 1.65){\vector( 1,1){.20}}
	\put(4.15, 1.65){\vector(-1,1){.20}}
	\put(4.40, 1.65){\vector( 1,1){.20}}

\put(1.30, 1.95){\ldots}
\put(2.10, 1.95){\ldots}
\put(2.60, 1.95){\ldots}
\put(3.40, 1.95){\ldots}
\put(3.80, 1.95){\ldots}
\put(4.60, 1.95){\ldots}

\end{picture}

as well as:

\setlength{\unitlength}{1.0in}
\begin{picture}(6,2.65)
%\put(0,0){\framebox(6,2.65)}

\put(2.70, 0.05){\textbf{scalar fields}}
\put(2.30, 0.25){$\Gamma(\mcT^0_0(M)) = \Gamma(C^\infty(M,\R))$}
	\put(2.70, 0.45){\vector(-1,1){.20}}
	\put(3.35, 0.45){\vector( 1,1){.20}}

\put(2.05, 0.75){\textbf{vector fields}}
\put(3.35, 0.75){\textbf{covector fields}}
\put(1.80, 0.95){$\Gamma(\mcT^1_0(M)) = \Gamma(TM)$}
\put(3.30, 0.95){$\Gamma(\mcT^0_1(M)) = \Gamma(T^*M)$}
	\put(2.25, 1.15){\vector(-1,1){.20}}
	\put(2.50, 1.15){\vector( 1,1){.20}}
	\put(3.55, 1.15){\vector(-1,1){.20}}
	\put(3.80, 1.15){\vector( 1,1){.20}}

\put(1.20, 1.65){contravariant 2-tensor}
\put(2.70, 1.65){mixed 2-tensor}
\put(3.80, 1.65){\textbf{covariant 2-tensor}}
\put(1.80, 1.45){fields}
\put(2.90, 1.45){fields}
\put(4.10, 1.45){\textbf{fields}}
\put(1.60, 1.85){$\Gamma(\mcT^2_0(M))$}
\put(2.80, 1.85){$\Gamma(\mcT^1_1(M))$}
\put(4.00, 1.85){$\Gamma(\mcT^0_2(M))$}
	\put(1.65, 2.05){\vector(-1,1){.20}}
	\put(1.90, 2.05){\vector( 1,1){.20}}
	\put(2.95, 2.05){\vector(-1,1){.20}}
	\put(3.20, 2.05){\vector( 1,1){.20}}
	\put(4.15, 2.05){\vector(-1,1){.20}}
	\put(4.40, 2.05){\vector( 1,1){.20}}

\put(1.30, 2.35){\ldots}
\put(2.10, 2.35){\ldots}
\put(2.60, 2.35){\ldots}
\put(3.40, 2.35){\ldots}
\put(3.80, 2.35){\ldots}
\put(4.17, 2.55){\textbf{higher-order covariant}}
\put(4.50, 2.35){\textbf{tensor fields}}

\end{picture}

In the second diagram, I've boldfaced the parts of the full tensor bundle which
are of interest for the geometry course:  namely, covariant tensor fields (the
right-hand linear branch of the tree), as well as contravariant 1-tensor
fields, i.e. vector fields (the left-hand spot).

%% ----------------------------------------------------------------
\subsubsection{The metric tensor}
\label{sec:metric_tensor}

\begin{defn}
As foreshadowed in remark \ref{rem:foreshadow_metric_tensor_and_forms}, the
\bfidx{metric tensor} is nothing more than the dot product in tangent spaces.
That is, the metric tensor $g \in \Gamma(\mcT^0_2(M))$ is defined by
$$
	g(\vecv_q, \vecw_q) = \vecv \cdot \vecw.
$$
\end{defn}

Explicit computations are much as in section \ref{sec:decompose_tensors},
except that the basis elements $\vecb_1, \ldots, \vecb_m$ are replaced by
$\slashDx_1, \ldots, \slashDx_m$, and likewise the dual basis elements
$\vecb_1^*, \ldots, \vecb_m^*$ are replaced by $dx_1, \ldots, dx_m$.

\begin{rem}
\label{rem:g_dx2_dy2_dz2}
In Euclidean coordinates,
$$
	\vecv \cdot \vecw = \sum_{i=1}^m v_i w_i.
$$
Recall that $dx_i$ is the $i$th coordinate-selector function, so we can
write this in tensor lingo as
$$
	g(\vecv_q, \vecw_q) = \sum_{i=1}^m (dx_i \tensor dx_i)(\vecv, \vecw).
$$
Treating $g$ as a 2-tensor, we can write
$$
	g = \sum_{i=1}^m dx_i \tensor dx_i.
$$
In particular, for Euclidean 3-space, we have
$$
	g = dx \tensor dx + dy \tensor dy + dz \tensor dz
$$
which we sometimes abbreviate as
$$
	g = dx^2 + dy^2 + dz^2.
$$

\end{rem}

See section \ref{sec:ex_1_6} for an example computation.

%% ----------------------------------------------------------------
\subsubsection{Forms}
\label{sec:forms}

\begin{center}
\emph{The human mind has first to construct forms, independently, before we can
find them in things.} \\
--- Albert Einstein\index{Einstein, Albert} (1879-1955).
\end{center}

xxx to do.

%\ref{sec:ex_2_2}
%\ref{sec:ex_2_3}

% forms
% note df is a one-form
% note that d takes this 0-form to 1-form
% volume form

somewhere (perhaps not here):  \bfidx{pullback of forms}.
xxx incl. a diagram too?
\begin{eqnarray*}
	f: z &\mapsto& z^n \\
	\omega &=& dz \\
	f^*(\omega) &=& \omega \circ f \\
	f^*(dz) &=& d(z^n) = n z^{n-1}dz.
\end{eqnarray*}

incl/xref to $i^*$ example too.

xxx diagram with coords/parameterizations and $dp$ vs. $\slashDp$ --- which
goes which way.

%% ================================================================
\newpage
\subsection{The exterior derivative}
\label{sec:d_operator}

\begin{center}
\emph{
Mathematicians are like Frenchmen: whatever you say to them they translate into
their own language and forthwith it is something entirely different.
}
--- Johann Wolfgang von Goethe\index{Goethe, Johann Wolfgang von} (1749-1832).
\end{center}

%% ----------------------------------------------------------------
\subsubsection{Definition of $d$}

\begin{center}
\emph{
And what are these fluxions? The velocities of evanescent increments. And what
are these same evanescent increments? They are neither finite quantities, nor
quantities infinitely small, nor yet nothing. May we not call them the ghosts
of departed quantities...?
} \\
--- Bishop Berkeley\index{Berkeley, George}, in \emph{The Analyst: A Discourse
Addressed to an Infidel Mathematician} (1734).
\end{center}

Both \cite{bib:lee2} and \cite{bib:satan} define the exterior derivative
in coordinates.  Namely, if we have an $r$-form
$$
	\omega = f(x_1, \ldots, x_m) dx_{i_1} \wedge \ldots \wedge dx_{i_r}
$$
then we write
$$
	d\omega = df(x_1, \ldots, x_m) dx_{i_1} \wedge \ldots \wedge dx_{i_r}.
$$
where the $df$ is taken on coefficient functions in the sense of section
\ref{sec:df}.  The $d$ operator takes $r$-forms to $(r+1)$-forms.  It can be
shown that the following properties hold:

\begin{itemize}
\item The $d$ operator is unique, and is independent of change of coordinates.

\item Given forms $\omega$ and $\eta$, $d(\omega + \eta) = d\omega + d\eta$.
\item $d(s\omega) = s d\omega$ for $s \in \R$.
\item $d(\omega \wedge \eta) = d\omega \wedge \eta + (-1)^{\ord(\omega)}\omega
\wedge d\eta$.  (This is the \bfidx{signed Leibniz rule}\index{Leibniz rule}.)
\item (Note that the previous three items satisfy the definition of an
	\bfidx{antiderivation} on the alternating tensor algebra.)
\item $d^2=0$.  [xxx xref to complex section.]
\end{itemize}

The following computational rules apply:

\begin{itemize}

\item We simply write out the differentials, then simplify.

\item From corollary \ref{cor:omega_odd_wedge_self_is_zero}, if $\omega$ is an
odd-degree form then $\omega \wedge \omega = 0$.  This applies in particular to
the 1-forms $dx$ et al.: $dx \wedge dx = 0$.

\item We can replace $dy \wedge dx$ with $-dx\wedge dy$.

\item Note that $f\;dx$ is shorthand for $f \wedge dx$.

\item We are working in the algebra of alternating tensors, with the wedge as
multiplication.  So, the usual rules of arithmetic apply, e.g. the distributive
property.  However, see immediately below for the commutativity rule.

\end{itemize}

\beginex
\begin{eqnarray*}
	\omega &=&
		f(x,y,z) dy \wedge dz +
		g(x,y,z) dz \wedge dx +
		h(x,y,z) dx \wedge dy \\
	d\omega &=&
		df \wedge dy \wedge dz +
		dg \wedge dz \wedge dx +
		dh \wedge dx \wedge dy \\
	&=& (f_x dx + f_y dy + f_z dz)  \wedge dy \wedge dz \\
		&+& (g_x dx + g_y dy + g_z dz)  \wedge dz \wedge dx \\
		&+& (h_x dx + h_y dy + h_z dz)  \wedge dx \wedge dy \\
	&=& f_x dx \wedge dy \wedge dz
		+ g_y dy \wedge dz \wedge dx
		+ h_z dz  \wedge dx \wedge dy \\
	&=& (f_x + g_y + h_z) dx \wedge dy \wedge dz.
	%&=& \nabla \vecF \cdot dV.
\end{eqnarray*}
Furthermore, if
$$
	\vecF = (f, g, h)
$$
then we can write
$$
	(f_x + g_y + h_z) dx \wedge dy \wedge dz = (\nabla \cdot \vecF) dV.
$$
\mendex

%\begin{defn}
%A \bfidx{derivation} on an $\R$-algebra $A$ is function $f:A \to A$
%such that:
%\begin{itemize}
%\item $f$ is $\R$-linear, i.e. for all $\vecu, \vecv \in A$ and all
%	$r \in \R$,
%	$$f(\vecu + \vecv) = f(\vecu) + f(\vecv) \qand f(r\vecu) = r f(\vecu).
%	$$
%\item $f$ satisfies an \bfidx{unsigned Leibniz rule} \index{Leibniz rule}:
%	for all $\vecu, \vecv \in A$ and all
%	$$ f(\vecu \vecv) = f(\vecu) \vecv + \vecu f(\vecv).$$
%\item (xxx where to say) $f(c) = 0$ for ``constant'' functions $c$.
%	Does this make sense in the context of general algebras, or do we need this
%	defined only for tensor algebras?  Or only for forms?  Or only for bundles?
%\end{itemize}
%\end{defn}

% xxx xref back to contractions section.  $i_X$ for vector fields $X$.
% Later!

[xxx:  major lack --- geometric view of the $d$ operator.  Without this it's
hard to come up with a geometric notion of when two forms are cohomologous.
Perhaps it suffices to work locally.  Can I claim that then all $k$ forms are
linear combinations of $dx_1 \wedge \cdots \wedge dx_k$?  Then does it reduce
to obtaining a geometric view of $df$?  If so:
$$
	df = \frac{\Df}{\Dx}\,dx + \frac{\Df}{\Dy}\,dy.
$$
Use $dx$, $dy$ as spines.  This is just the gradient (covariantly of course) .
How to extend to higher-level forms?
\begin{eqnarray*}
	\omega &=& f(x,y)\,dx + g(x,y)\,dy \\
	d\omega &=& \left(\frac{\Df}{\Dx}\,dx+\frac{\Df}{\Dy}\,dy\right)\wedge dx
		\;+\; \left(\frac{\Dg}{\Dx}\,dx+\frac{\Dg}{\Dy}\,dy\right)\wedge dy \\
	&=& \left(\frac{\Dg}{\Dy} - \frac{\Df}{\Dx}\right) \, dx \wedge dy.
\end{eqnarray*}
First, plot a contour of $f$.  Then, superimpose gradient arrows.  Second, draw
a quiver of arrows, with $f$ and $g$ components in a pair of adjacent dotted
triangles.  Then, what does $\Dg/\Dy-\Df/\Dx$ look like?
\ldots
]

%% ----------------------------------------------------------------
\subsubsection{Forms in 3-space TBD}
\index{Rosetta stone}

\setlength{\unitlength}{1.0in}
\begin{picture}(6,2.5)
\put(0,0){\framebox(6,2.5)}
\put(0.05,0.05){\framebox(5.90,2.40)}

\put(0.20, 1.30){$\Omega^0(\R^3)$}
\put(0.25, 0.45){\vector(0,1){.75}}
\put(0.20, 0.30){$C^\infty(\R^3,\R)$}

	\put(0.65, 1.33){\vector(1,0){0.90}}
	\put(0.95, 0.33){\vector(1,0){0.60}}
	\put(1.05, 1.43){$d$}
	\put(1.00, 0.40){grad=$\nabla$}

\put(1.70, 1.30){$\Omega^1(\R^3)$}
\put(1.80, 0.45){\vector(0,1){.75}}
\put(1.70, 0.30){$\mathrm{vect}(\R^3)$}

	\put(2.15, 1.33){\vector(1,0){0.90}}
	\put(2.25, 0.33){\vector(1,0){0.80}}
	\put(2.55, 1.43){$d$}
	\put(2.35, 0.40){$\mathrm{curl}=\nabla\times$}

\put(3.20, 1.30){$\Omega^2(\R^3)$}
\put(3.30, 0.45){\vector(0,1){.75}}
\put(3.20, 0.30){$\mathrm{vect}(\R^3)$}

	\put(3.65, 1.33){\vector(1,0){0.90}}
	\put(3.75, 0.33){\vector(1,0){0.80}}
	\put(4.05, 1.43){$d$}
	\put(3.85, 0.40){$\mathrm{div}=\nabla\cdot$}

\put(4.70, 1.30){$\Omega^3(\R^3)$}
\put(4.80, 0.45){\vector(0,1){.75}}
\put(4.70, 0.30){$C^\infty(\R^3,\R)$}

	\put(0.20, 0.10){$f$}
	\put(0.20, 1.70){$f$}

	\put(1.60, 0.10){$\vecF=(f,g,h)$}
	\put(1.60, 1.90){$\vecF\cdot d\vecx =$}
	\put(1.30, 1.70){$fdx + gdy + hdz$}

	\put(3.00, 0.10){$\vecF=(f,g,h)$}
	\put(3.00, 2.30){$\vecF\ctt dV = $}
	\put(3.00, 2.10){$f\,dy\wedge dz+$}
	\put(3.00, 1.90){$g\,dz\wedge dx+$}
	\put(3.00, 1.70){$h\,dx\wedge dy$}

	\put(4.75, 0.10){$f$}
	\put(4.70, 1.70){$f dV$}

	\put(0.40, 0.80){$\mathrm{Id}$}
	\put(1.90, 0.80){$\vecF\cdot d\vecx$}
	\put(3.40, 0.80){$\vecF\ctt dV$}
	\put(4.90, 0.80){$f\,dV$}

\end{picture}

Include the cor. 12 stuff here.  The property $d^2=0$ means:

\begin{itemize}
\item curl of grad is 0
\item div of curl is 0
\end{itemize}

incl. $X\ctt$ in the diagram.

$\vecnhat\ctt dV = dA$.

need to show $\vecF\ctt dV = (\vecF \cdot \vecnhat) dS$.
Lee lemma 13.25.

xxx here or somewhere else:  closed, exact.  $dx$ is exact;
$d \theta$ is not; $dA$ is exact iff $M$ is non-compact.
Explain more ............

%% ----------------------------------------------------------------
\subsection{Computations with tangent vectors and forms}

%% ----------------------------------------------------------------
\subsubsection{When $dx \ne dx$:  charts, embeddings, projections, and variance}

also graph coordinates on, say, upper sheet of hyperboloid of two sheets.
different meanings for $\slashDx$ etc.

xxx crucial distiction between \emph{enough coordinates} (i.e. a chart) and
\emph{too many coordinates} (i.e. an embedding).  Neither is ``wrong''; both
have their uses.  We simply need to be careful about what the symbols mean.

xxx peril:  illustrate by example.  On $\S^2$, the function $f=x^2+y^2+z^2-1$
is 0 --- obviously.  Less obviously, $y \,dy \wedge dz - x \,dz \wedge dx$ is
also zero.  Show this using spherical coordinates.

\begin{center}* * *\end{center}

In the plane, we convert \emph{from} polar coordinates \emph{to} rectangular
coordinates as follows:
%$$
%	\colvectwo{r}{\theta} \mapsto \colvectwo{x}{y} 
%$$
%by
$$
	\colvectwo{x}{y} = \colvectwo{r \cos\theta}{r \sin\theta}.
$$
Then [xxx point out directions] the following two pop out easily:
\begin{eqnarray}
	\label{eqn:polar_DDr}
	\fracDr &= \colvectwo{\cos\theta}{\sin\theta}
		= \colvectwo{x/r}{y/r}
		= \colvectwo{\frac{x}{\sqrt{x^2+y^2}}}{\frac{y}{\sqrt{x^2+y^2}}} \\
	\label{eqn:polar_DDtheta}
	\fracDtheta &= \colvectwo{-r \sin\theta}{r \cos\theta}
		= \colvectwo{-y}{x}
\end{eqnarray}
[xxx footing caveats] and
\begin{eqnarray}
	\label{eqn:polar_dx}
	dx	&=& \frac{dx}{dr} \,dr + \frac{dx}{d\theta}\,d\theta
		= \cos\theta\, dr - r \sin\theta\, d\theta \\
	\label{eqn:polar_dy}
	dy	&=& \frac{dy}{dr} \,dr + \frac{dy}{d\theta}\,d\theta
		= \sin\theta\, dr + r \cos\theta\, d\theta.
\end{eqnarray}

For the other four, it takes a bit more manipulation.  [xxx disclaim about the
arctan nonsense.]  First, $dr$ and
$d\theta$.
\begin{eqnarray*}
	\colvectwo{dx}{dy}
	&=& \mxrsqtwo
			{\cos\theta}{-r\sin\theta}
			{\sin\theta}{ r\cos\theta}
		\colvectwo{dr}{d\theta}
	= \mxrsqtwo
			{\frac{x}{\sqrt{x^2+y^2}}}{-y}
			{\frac{y}{\sqrt{x^2+y^2}}}{ x}
		\colvectwo{dr}{d\theta} \\
	\colvectwo{dr}{d\theta}
	&=& {\mxrsqtwo
			{\frac{x}{\sqrt{x^2+y^2}}}{-y}
			{\frac{y}{\sqrt{x^2+y^2}}}{ x}}^{-1}
		\colvectwo{dx}{dy} \\
	&=& \mxcsqtwo
			{x} {y}
			{-\frac{y}{\sqrt{x^2+y^2}}} {\frac{x}{\sqrt{x^2+y^2}}}
			\;\left(\frac{1}{\sqrt{x^2+y^2}}\right)\;
		\colvectwo{dx}{dy} \\
	&=& \mxcsqtwo
			{\frac{x}{\sqrt{x^2+y^2}}} {\frac{y}{\sqrt{x^2+y^2}}}
			{-\frac{y}{x^2+y^2}} {\frac{x}{x^2+y^2}}
		\colvectwo{dx}{dy}.
\end{eqnarray*}

Second, $\slashDx$ and $\slashDy$:
\begin{eqnarray*}
	\colvectwo{\slashDr}{\slashDtheta}
		&=& \mxcsqtwo{\frac{x}{\sqrt{x^2+y^2}}}{\frac{y}{\sqrt{x^2+y^2}}}{-y}{x}
		\colvectwo{\slashDx}{\slashDy} \\
	\colvectwo{\slashDx}{\slashDy}
		&=& \mxcsqtwo
			{x}{\frac{-y}{\sqrt{x^2+y^2}}}
			{y}{\frac{ x}{\sqrt{x^2+y^2}}}
		\;\left(\frac{1}{\sqrt{x^2+y^2}}\right)\;
		\colvectwo{\slashDr}{\slashDtheta} \\
	&=& \mxcsqtwo
		{\frac{x}{\sqrt{x^2+y^2}}}{\frac{-y}{x^2+y^2}}
		{\frac{y}{\sqrt{x^2+y^2}}}{\frac{ x}{x^2+y^2}}
	\colvectwo{\slashDr}{\slashDtheta}
	= \mxcsqtwo
		{\frac{r\cos\theta}{r}}{\frac{-r\sin\theta}{r^2}}
		{\frac{r\sin\theta}{r}}{\frac{ r\cos\theta}{r^2}}
	\colvectwo{\slashDr}{\slashDtheta} \\
	&=& \mxcsqtwo
		{\cos\theta}{\frac{-\sin\theta}{r}}
		{\sin\theta}{\frac{ \cos\theta}{r}}
	\colvectwo{\slashDr}{\slashDtheta}.
\end{eqnarray*}

So, as a final result,
\begin{align*}
%
	\fracDx &= \cos\theta\, \fracDr
		\;-\; \frac{\sin\theta}{r}\, \fracDtheta &
	dx	&= \cos\theta\, dr - r \sin\theta\, d\theta  \\
%
	\fracDy &= \sin\theta\, \fracDr
		\;+\; \frac{\cos\theta}{r}\, \fracDtheta &
	dy	&= \sin\theta\, dr + r \cos\theta\, d\theta
\end{align*}
and
\begin{align*}
%
	dr      &= \frac{x\, dx \;+\; y\,dy}{\sqrt{x^2+y^2}} &
	\fracDr &=
		\frac{x}{\sqrt{x^2+y^2}}\,\fracDx \;+\;
		\frac{y}{\sqrt{x^2+y^2}}\,\fracDy \\
%
	d\theta     &= \frac{-y\, dx \;+\; x\,dy}{x^2+y^2} &
	\fracDtheta &= -y\,\fracDx    \;+\; x\,\fracDy.
\end{align*}
[xxx compute norms and show, and plot:
	length of $\slashDr$ is 1,
	length of $\slashDtheta$ is $r$;
	spine length of $dr$ is 1,
	spine length of $d\theta$ is $1/r$.
Thus it's no surprise that $d\theta$ is undefined at zero.
]

\begin{center} * * * \end{center}

When we restrict to $\S^1$, i.e. set $x^2+y^2=1$ and project onto the
tangents, then we have
\begin{align*}
%
	\fracDx &= -\sin\theta\, \fracDtheta &
	dx	&= -\sin\theta\, d\theta  \\
%
	\fracDy &= \cos\theta\, \fracDtheta &
	dy	&= \cos\theta\, d\theta
\end{align*}
and
\begin{align*}
%
	d\theta     &= -y\, dx \;+\; x\,dy &
	\fracDtheta &= -y\,\fracDx    \;+\; x\,\fracDy
\end{align*}

[xxx then, use this to drive home the point about when $dx \ne dx$.
Spine plots will make this quite clear.]

\begin{center} * * * \end{center}

Note that [xxx $\vece_1$ with $\slashDx$ and $\vece_2$ with $\slashDy$]
equations \ref{eqn:polar_DDr} and \ref{eqn:polar_DDtheta}
are the same as
\begin{align*}
	\fracDr     &=\fracDtwo{x}{r}\,\fracDx + \fracDtwo{y}{r}\,\fracDy \\
	\fracDtheta &=\fracDtwo{x}{\theta}\,\fracDx + \fracDtwo{y}{\theta}\,\fracDy.
\end{align*}
Compare to equations \ref{eqn:polar_dx} and \ref{eqn:polar_dy}:
\begin{eqnarray}
	dx	&=& \frac{dx}{dr} \,dr + \frac{dx}{d\theta}\,d\theta \\
	dy	&=& \frac{dy}{dr} \,dr + \frac{dy}{d\theta}\,d\theta.
\end{eqnarray}

Graphically, we see
$$
\xymatrix{
	{\colvectwo{r}{\theta}}\ar[rr]^{F} & {} & {\colvectwo{x}{y}}
		= F(r,\theta)
		= \colvectwo{r\cos\theta}{r\sin\theta} \\
%
	F^*(dx)=d(r\cos\theta) = \cos\theta \,dr - r\sin\theta \,d\theta
		\ar@{<-|}[rr]^{F^*}
		& {} & dx\\
%
	F^*(dy)=d(r\sin\theta) = \sin\theta\,dr + r\cos\theta \,d\theta
		\ar@{<-|}[rr]^{F^*}
		& {} & dy\\
%
	\fracDr\ar@{|->}[rr]^{F_*} & {} &
		F_*\left(\fracDr\right)
		= \frac{\Dx}{\Dr}\fracDx + \frac{\Dy}{\Dr}\fracDy
		%= {\colvectwo{\Dx/\Dr}{\Dy/\Dr}}
		= {\colvectwo{\cos\theta}{\sin\theta}} \\
%
	\fracDtheta\ar@{|->}[rr]^{F_*} & {} &
		F_*\left(\fracDtheta\right)
		= \frac{\Dx}{\D\theta}\fracDx + \frac{\Dy}{\D\theta}\fracDy
		%= {\colvectwo{\Dx/\D\theta}{\Dy/\D\theta}}
		= {\colvectwo{-r\sin\theta}{r\cos\theta}}.
}
$$

Two points:
\begin{itemize}
\item This echoes the discussion in section \ref{sec:VS_COB}, wherein
	I noted that column vectors (e.g. $\slashDx$), which are contravariant, and
	row vectors (e.g. $dx$), which are covariant, change coordinates
	differently.
\item This justifies our use of $\slashDx$ for vectors and $dx$ for
	functionals:  namely, the $\partial$'s and $d$'s cancel correctly
	in the numerators and denominators in the above formulas, in accordance
	with their respective contravariance and covariance.
\end{itemize}

%% ----------------------------------------------------------------
\subsubsection{Stereographic coordinates on $\S^1$}

xxx $dp$, $dq$, etc.  In terms of (known) $d\theta$.

%% ----------------------------------------------------------------
\subsubsection{A gallery of curves and surfaces, part 2}
\label{sec:gallery2}

xref back to gallery1, and have it xref here.  what coordinates to use.
computing partials.  Make sure to do upper sheet of two-sheeted hyperboloid;
compare and contrast $\slashDx$ with $\slashDx$.

%% ----------------------------------------------------------------
\subsubsection{asdfasdfasdf}

Formalize the following mnemonic: with $i : M \to \R^m$:  $M = \ker F$, $TM =
\ker DF$.

xxx xref to the SL problem from the final review.

xxx Include here some references to vector calculus.  Common surfaces such as
the sphere are a level set of some $F: \R^3 \to \R$.  The gradient points in
the direction of greatest change; it is the field of vectors normal to the
surface.  Perpendicular to that, $F$ is (to linear approximation) constant.
Now, $\vecv \in \ker DF$ means $DF \cdot \vecv = 0$ which simply means that the
dot of $\vecv$ with $DF$ is zero, i.e. $\vecv$ is perpendicular to the normal.
This is exactly what we would expect.

%% ================================================================
\newpage
\subsection{Cohomologies}

xxx cellular, simplicial, singular, de Rham w/ xref back to homologies sectoin.
xxx note prefer singular for the same reason as there.

xxx put the de Rham thm right up front.  This is a representation theorem,
equating singular cohomology with de Rham cohomology.  So, it suffices to work
with the latter.  Note that qual questions only ever ask about de Rham
cohomology.

xxx def or xref to closed, exact.

xxx in VS section, have already done sequences of mappings including exact and
short exact ones.  Then, note $d$ as a sequence of mappings.  Then, closed mod
exact as ker mod im.

%% ----------------------------------------------------------------
\subsubsection{de Rham cohomology}
\label{sec:de_rham_cohomology}

\begin{defn}
\bfidx{de Rham cohomology}\index{cohomology} \ldots
\end{defn}

\begin{thm}[Poincar\'e lemma]
A contractible (technically, \emphidx{star-shaped}) open subset $U$ of $\R^n$
has trivial cohomology $H^k_{dR}(U)$ for $k>0$.
\index{Poincar\'e lemma}
\end{thm}

\begin{rem}
This means that \emph{if} a form is closed on a contractible open set, then it
is exact.  If a form is not closed, then the Poincar\'e lemma does not apply.
\end{rem}

long exact seqs, complexes, homology and cohomology.  where to put this in the
right order vs. $d$ and $\D$.

%% ================================================================
\newpage
\subsection{Section title goes here}

\emph{This section is not qualifier material.}

% ----------------------------------------------------------------
\subsubsection{Old-fashioned tensors}

From \cite{bib:pdm}:

\begin{itemize}

\item Let $\myvecx$ and $\vecy$ be coordinate functions (xxx for what).  Thus,
a point $q$ (of what) has coordinates $(x^1(q), \ldots, x^m(q))$ in the first
coordinate system, and coordinates $(y^1(q), \ldots, y^m(q))$ in the second
coordinate system.

\item A set of $m$ components, written $A^i$, that are functions of the first
set of coordinates will become a set of $m$ components, denoted $B^i$, after
transformation to the second set of coordinates.

\item A set of $m^2$ components, written $A^{ij}$, that are functions of the
first set of coordinates will become a set of $m^2$ components, denoted
$B^{ij}$, after transformation to the second set of coordinates.

\item A set of $m^3$ components, written $A^{ijk}$, that are functions of the
first set of coordinates will become a set of $m^3$ components, denoted
$B^{ijk}$, after transformation to the second set of coordinates.

\item A tensor is a set of components that obeys some transformation law.  The
number of suffixes indicates the \bfidx{rank} of the tensor; their position
indicates the type (\plainidx{covariant}, \plainidx{contravariant}, or
\plainidx{mixed}) of the tensor.

\item A covariant 1-tensor is a set (?) $A_i$ satisfying, for each $i$,
$$ B_i = \sum_{r=1}^n \frac{\Dx^r}{\Dy^i} A_r.$$

\item A contravariant 1-tensor is a set (?) $A^i$ satisfying, for each $i$,
$$ B^i = \sum_{r=1}^n \frac{\Dy^i}{\Dx^r} A^r.$$

\item A covariant 2-tensor is a set (?) $A_{ij}$ satisfying, for each $i$
and $j$, $$ B_{ij} = \sum_{r=1}^n \sum_{s=1}^n
\frac{\Dx^r}{\Dy^i}
\frac{\Dx^s}{\Dy^j} A_{rs}.$$

\item A contravariant 2-tensor is a set (?) $A^{ij}$ satisfying, for each $i$
and $j$,
$$ B^{ij} = \sum_{r=1}^n \sum_{s=1}^n \frac{\Dy^i}{\Dx^r}
\frac{\Dy^j}{\Dx^s} A^{rs}.$$

\item A mixed 2-tensor is a set (?) $A^i_j$ satisfying, for each $i$
and $j$,
$$ B^i_j = \sum_{r=1}^n \sum_{s=1}^n \frac{\Dy^i}{\Dx^r}
\frac{\Dx^s}{\Dy^j} A^r_s.$$

\end{itemize}

xxx several examples here.  First, a simple rotation in $\R^2$.  Show how
linear functional, dot, and determinant transform.

Try to express this in terms of concepts used in the course.

Is this just using the COB matrix??  Try a couple specific examples.

\begin{eqnarray*}
	\lambda(\vecu, \vecv, \vecw)
	&=& \lambda(
		u_1 \vece_1 + u_2 \vece_2 + u_3 \vece_3,
		v_1 \vece_1 + v_2 \vece_2 + v_3 \vece_3,
		w_1 \vece_1 + w_2 \vece_2 + w_3 \vece_3) \\
	&=& \sum_i \sum_j \sum_k u_i v_j w_k \lambda(\vece_i, \vece_j, \vece_k) \\
	&=& \sum_i \sum_j \sum_k u_i v_j w_k L_{ijk}
\end{eqnarray*}

xxx xref back to section \ref{sec:VS_COB}:  here we are generalizing that
concept, which was done there for vectors and functionals (i.e. tensors of type
1,0 and type 0,1).

%% ================================================================
%% ================================================================
%% ================================================================
\newpage
\section{Duality}
\label{sec:duality_main_section}

\begin{center}
\emph{The union of the mathematician with the poet, fervor with measure,
passion with correctness, this surely is the ideal.}
--- William James\index{James, William} (1842-1910), \emph{Collected Essays}.
\end{center}

xxx quack about chains and forms; homology and cohomology.

%% ================================================================
\subsection{Integration}

xxx xref to chapter 10 of \cite{bib:rudin}.

\begin{defn}
Integral over a $k$-chain \ldots .
\end{defn}

\begin{prop}
Well-definedness with respect to choice of pullback \ldots.
\end{prop}

%% ================================================================
\newpage
\subsection{Stokes' theorem}
\label{sec:stokes}

%First of all I am blamed for crossing the boundary of experimental evidence.
%This, I reply, is the habitual action of the scientific mind.at least of that
%portion of it which applies itself to physical investigation. Our theories of
%light, heat, magnetism, and electricity, all imply the crossing of this
%boundary.
%--- Lord Kelvin

Here we discuss a \index{Generalized Stokes}\index{Stokes'
theorem}\plainidx{generalized Stokes' theorem}.  This modern formulation
includes several classical integral theorems as special cases.

%% ----------------------------------------------------------------
\subsubsection{Stokes theorem for chains}

steps (xxx make a picture):
\begin{itemize}
\item $c$ is a single $k$-cube and $w$ is a simple $(k-1)$-form.  Moreover
	$c$ is the inclusion of $[0,1]^k$ into $\R^k$.  Write out using
	definitions of $\D c$, $dw$, Fubini, and the FTC on one variable.
\item Show independence of parameterization on $[0,1]^k$ (source).
\item Let $c$ be an arbitrary $k$-cube (destination).
\item Sums of $c_i$'s and $\omega_j$'s.
\end{itemize}

%% ----------------------------------------------------------------
\subsubsection{Stokes theorem for manifolds}
\label{sec:gen_stokes}

xxx need to define:

\begin{defn}
\bfidx{manifold with boundary}
\end{defn}

\begin{defn}
\bfidx{boundary} of a manifold.
\end{defn}

This is written $\DM$.  If $M$ has dimension $m$,then $\DM$ has dimension
$m-1$.

\begin{defn}
\bfidx{integral}.
\end{defn}

\begin{defn}
\bfidx{induced orientation}.
\end{defn}

Let $M$ be a compact, oriented manifold.  Suppose $\omega \in \Omega^{m-1}(M)$.
Then $$ \int_{M} d \omega = \int_{\partial M} \omega.$$

xxx express in terms of the pairing of forms and chains:  $\D$ is the adjoint
of $d$:

$$ \langle d\omega, c\rangle = \langle \omega, \D c \rangle.$$

%% ----------------------------------------------------------------
\subsubsection{Proof of Stokes}

[xxx first use POUs to say why it suffices to work within a single chart.]

[xxx next use chart invariance (cancelling dets) to say why it now remains to
work in $\R^n$.]

Sketch (needs a picture w/ transition function):
$$
	\int_U (y^{-1})^* \omega =
	\int_U (yx^{-1})^* (y^{-1})^* \omega =
	\int_U (x^{-1})^* y^* (y^{-1})^* \omega =
	\int_U (x^{-1})^* \omega.
$$

%For the proof in $\R^n$, first we do 1-chains.

%\includegraphics*{two_holed_torus.eps}
%\input{two_holed_torus.pstex_t}
%\includegraphics[bb=0 0 6 2]{two_holed_torus.pdf}
%\includegraphics{two_holed_torus.pdf}

%\includegraphics{foo.jpg}
%\includegraphics[length=1.0in,width=1.0in]{foo.jpg}
%\includegraphics*[length=1.0in,width=1.0in]{foo.jpg}
%\includegraphics*[bb=0 0 1 1,scale=1.0]{foo.jpg}
%\scalebox{1.0}{ \includegraphics*{foo.jpg} }

Let\index{pullback}\index{fundamental theorem of calculus}\index{Stokes'
theorem} $f$ be a 0-form; let $c$ be a 1-chain.  That is, $c$ is a smooth map
from $[0,1]$ to $X$.
\begin{eqnarray*}
	\int_c df &=& \int_0^1 c^*(df) \quad\textrm{(pull $df$ back to $[0,1]$)}\\
	&=& \int_0^1 dc^*(f) \quad\textrm{($d$ commutes with pullback)}\\
	&=& (c^*f)(1) - (c^*f)(0)
		\quad\textrm{(fundamental theorem of calculus)}\\
	&=& f(c(1)) - f(c(0)) \quad\textrm{(definition of pullback)}
\end{eqnarray*}
Meanwhile,
\begin{eqnarray*}
	\D c &=& 1\cdot c(1) - 1 \cdot c(0) \\
	\int_{\D c} f &=& f(c(1)) - f(c(0)) \\
	&=& \int_c df.
\end{eqnarray*}

See section \ref{sec:fr_13} for the two-dimensional case.
[xxx merge that in here, or eliminate.]

Next, we will look at how the classical integration theorems of section
\ref{sec:vc_intgn_thms} are special cases of Stokes' theorem.  We will pay
particular attention to how the classical notation translates into the
generalized notation.

%% ----------------------------------------------------------------
\subsubsection{The fundamental theorem of calculus}

Let $M$ be the closed interval $[a,b]$ for some $a, b \in \R$.  Then $\partial
M = \{a, b\}$.  Let $f(x)$ be a a differentiable function on $[a,b]$, i.e.  $f
\in \Omega^0(M)$.  For an antiderivative $F(x)$ of $f$, i.e.  $dF/dx = f$, we
are accustomed to thinking of the \plainidx{fundamental theorem of calculus} as
$$\int_a^b f(x) dx = F(b) - F(a).$$  Using our current terminology, we can
observe that the integrand $f dx$ is a 1-form.  This is $d$ of something,
namely, $f dx = dF$.  Then we can write the left-hand side of the Stokes
equation as $$ \int_a^b f dx = \int_a^b \fracdtwo{F}{x} dx = \int_a^b dF, $$
and we can write the right-hand side as $$ \int_{\D[a,b]} F = F(b) - F(a).$$

%% ----------------------------------------------------------------
\subsubsection{Line integrals independent of path}

\index{line integral}
\index{path}

xxx xref to the figure in section \ref{sec:d_operator}.

Let $M$ be a curve (i.e. a manifold of dimension 1) of $\R^2$ between points
$\vecp$ and $\vecq$.  Then $\DM = \{\vecp, \vecq\}$.  Given $f, g: \R^2 \to
\R$, we write the line integral $$ \int_M f(x,y) dx + g(x,y)dy.$$ The integrand
is a 1-form, i.e.  $$ f(x,y) dx + g(x,y)dy\in \Omega^1(M).$$ To use generalized
Stokes, we need to write the integrand as $d$ of something.  If $f$ and $g$ are
smooth (which we assume in this course) then in fact (\cite{bib:anton}, section
xxx) there is a \plainidx{vector-to-scalar function} $H : \R^2 \to \R$ such
that $$ \fracDtwo{H}{x} = f, \fracDtwo{H}{y} = g.$$ Then $$ fdx + gdy =
\fracDtwo{H}{x} dx + \fracDtwo{H}{y} dy = \grad H \cdot d\vecx$$ where $d\vecx$
is the vector $(dx, dy)$.  Note that we can also write $$ dH = \fracDtwo{H}{x}
dx + \fracDtwo{H}{y} dy$$ so $$ \int_M f dx + g dy = \int_M dH.$$ Now we can
use Stokes' theorem to say $$ \int_M dH = \int_{\DM} H = H(\vecq) - H(\vecp).$$
Hence the classical result that $$ \int_M f dx + g dy = H(\vecq) - H(\vecp).$$

Note that the right-hand side is independent of the curve $M$, as we would
expect from xxx cite.

%% ----------------------------------------------------------------
\subsubsection{Green's theorem}

\index{Green's theorem} Let $M$ be a connected blob, i.e. a bounded manifold of
dimension 2, in $\R^2$.  (One might think of $M$ as a pancake, with $\R^2$ as
the griddle.)  Let $C$ be a closed curve around the perimeter of $M$.  Use the
positive induced orientation (xxx put in a picture here) with tangent vector
$\vecnhat$.  Let $$ \omega = f(x,y) dx + g(x,y) dy \in \Omega^1(M).$$ Then we
want to compute $$ \oint_C f(x,y) dx + g(x,y) dy.$$ Note that we have $C =
\DM$.  In the previous examples, we had to obtain the integrand as $d$ of
something.  Here, we are given the other side of a Stokes equation: we are
given $\omega$ and all we need to do is to compute $d\omega$.  We have (xxx
xref to above section)
\begin{eqnarray*}
	d\omega &=& d(f dx + g dy)
	= d(f dx) + d(g dy) \\
	&=& \left(\fracDtwo{f}{x} dx + \fracDtwo{f}{y} dy \right) \wedge dx
	+ \left(\fracDtwo{g}{x} dx + \fracDtwo{g}{y} dy \right) \wedge dy \\
	&=&
	\fracDtwo{f}{x} dx \wedge dx +
	\fracDtwo{f}{y} dy \wedge dx +
	\fracDtwo{g}{x} dx \wedge dy +
	\fracDtwo{g}{y} dy \wedge dy \\
	&=&
	\fracDtwo{f}{y} dy \wedge dx +
	\fracDtwo{g}{x} dx \wedge dy \\
	&=&
	\left(\fracDtwo{g}{x} - \fracDtwo{f}{y}\right) dx \wedge dy.
\end{eqnarray*}
Thus
$$
\int_M d\omega
= \int_M \left(\fracDtwo{g}{x} - \fracDtwo{f}{y}\right) dx \wedge dy
= \int_M \left(\fracDtwo{g}{x} - \fracDtwo{f}{y}\right) dA
$$
and we obtain the classical result
$$
\int_M \left(\fracDtwo{g}{x} - \fracDtwo{f}{y}\right) dA
= \oint_C f dx + g dy.
$$

xxx need to write this in terms of the normal vector as well.

%% ----------------------------------------------------------------
\subsubsection{Classical Stokes}

\index{Classical Stokes}
\index{Stokes' theorem}

Let $M$ be a 2-dimensional submanifold of $\R^3$.  That is, $M$ is a surface,
where the canonical example is a potato chip.  Let $\vecnhat$ be an
outward-pointing normal vector.  Start with a 1-form $$\omega = \vecB \cdot
d\vecx = B_1(x,y,z) dx + B_2(x,y,z) dy + B_3(x,y,z) dz.$$

xxx incomplete.

xxx think of: the work done by a magnetic field around a wire equals
flux of field through interpolating surface?

%% ----------------------------------------------------------------
\subsubsection{Divergence theorem}

\index{divergence theorem}

[xxx elaborate on this sketch]

$$
	\int_M (\grad \cdot \vecF) dV = \int_M d(\vecF\ctt dV)
	= \int_{\D M} \vecF\ctt dV = \int_{\D M} \vecF \cdot \nhat \; dA.
$$

xxx more:

\begin{eqnarray*}
	L_\vecF(dV) &=& d(\vecF\ctt dV) + \vecF\ctt (d(dV)) \\
	L_\vecF(dx \wedge dy \wedge dz) &=&
		 (L_\vecF(dx)) \wedge dy \wedge dz
		+dx \wedge (L_\vecF(dx)) \wedge dz
		+dx \wedge dy \wedge (L_\vecF(dx)) \\
	&=& d(L_\vecF(x)) \wedge dy \wedge dz
		+dx \wedge d(L_\vecF(x)) \wedge dz
		+dx \wedge dy \wedge d(L_\vecF(x)) \\
	L_\vecF(x) &=& (F_1 \D/\Dx + F_2 \D/\Dy + F_3 \D/\Dz) x = F_1 \\
	&& \textrm{etc.} \\
	L_\vecF(dx \wedge dy \wedge dz) &=&
		 dF_1 \wedge dy   \wedge dz
		+dx   \wedge dF_2 \wedge dz
		+dx   \wedge dy   \wedge dF_3 \\
	&=& (\grad \cdot \vecF) dV.
\end{eqnarray*}

%% ----------------------------------------------------------------
\subsubsection{Cauchy's theorem}

\index{Cauchy's theorem}

%% ================================================================
\newpage
\subsection{Duality theorems}

\bfidxtwo{Pairing}{pairing} (see section \ref{sec:pairings}) of
homology and cohomology (cycles and forms).
This:
$$
	\Omega^k(M) \times C_k(M) \to R
$$
via
$$
	(\omega, c) \mapsto \int_c \omega
$$
Induces a pairing of homology and cohomology:
$$
	H^k_{dR}(M) \times H_k(M) \to R
$$
via
$$
	([\omega], [c]) \mapsto \int_c \omega
$$
which is well-defined.
Type up 4-13:  pull back to $I^m$ and Riemann integrals.

\begin{prop}[Poincar\'e duality]
If\index{Poincar\'e duality} $M$ is an orientable smooth manifold, then
$$
	H^{m-k}_{dR}(M) \cong H^k_{dR}(M)^*
$$
via
$$
	[\omega^{(m-k)}] \quad \longmapsto \quad
		\left( [\eta^{(k)}] \mapsto \int_M \omega \wedge \eta \right).
$$
\end{prop}

\begin{rem}
We can remember that this theorem has orientability as a hypothesis --- else
the integral of a top form over all of $M$ would be zero.
\end{rem}

\begin{rem}
Also (via what map? ``cap product'' (?!?) with ``fundamental class'' $[M]$):
$$
	H^{m-k}(M) \cong H_k(M).
$$
\end{rem}

\begin{prop}[de Rham Theorem]
\label{thm:de_Rham}
Let\index{de Rham Theorem} $M$ be a smooth manifold.  Then
$$
	H^k_{dR}(M) \cong H_k(M)^*
$$
via
$$
	[\omega^{(k)}] \quad \longmapsto \quad
		\left( [c^{(k)}] \mapsto \int_c \omega \right).
$$
\begin{rem}
Note that orientability is not needed here --- the integrals are over chains,
which are oriented.
\end{rem}
\begin{proof}
See \cite{bib:lee2} for the full proof, which is non-trivial.  Here, though, I
will note why it is that the above map is well-defined on cohomology (the
$\omega$'s) and homology (the $c$'s).

First suppose $[\omega] = [\eta]$.  Recall that cohomology classes are closed
forms mod exact forms, and homology classes are cycles mod boundaries.  This
means $\omega-\eta$ is exact:  say $\omega-\eta = d\alpha$.  Then
$$
	\int_c(\omega-\eta) = \int_c d\alpha = \int_{\D c} \alpha = 0
$$
by Stokes theorem, and because $c$ is a cycle so $\D c = 0$.
Thus
$$
	\int_c \omega = \int_c \eta.
$$

Second suppose $[c]=[b]$.  Then $c-b$ is the boundary of a chain:  say
$c-b=\D a$.  Then
$$
	\int_{c-b} \omega = \int_{\D a} \omega = \int_a d\omega = 0
$$
by Stokes theorem, and because $\omega$ is closed so $d\omega = 0$.
Thus
$$
	\int_c \omega = \int_b \omega.
$$
\end{proof}
\end{prop}

\begin{rem}
One direction is obvious:  given any $k$ form $\omega$ we can certainly turn it
into a functional as shown above.  The payload is the other direction.  This is
really a \bfidx{representation theorem}, much like proposition
\ref{thm:Rn_fcnal_rep}.  Namely, \emph{all} functionals on (real) homology come
from forms in this way.
\end{rem}

xxx \emph{nature} of the explicit maps for M-V on cohomology: $j_U^* \oplus
j_V^*$ and $i_U^* - i_V^*$, with $j_U^*(\omega)$ being just the
\emphidx{restriction} of $\omega)$ to $U$, etc.  Make a nice picture.  What
about the connecting map $\delta$?

%% ================================================================
\newpage
\subsection{Notation TBD}
\index{Rosetta stone}

\begin{center}
\emph{By relieving the brain of all unnecessary work, a good notation sets it
free to concentrate on more advanced problems, and, in effect, increases the
mental power of the race.}
--- Alfred North Whitehead\index{Whitehead, Alfred North} (1861-1947).
\end{center}

[xxx needs elaboration]

\begin{itemize}
\item $C_k(M)$:  $k$-chains.
\item $Z_k(M)$:  $k$-cycles ($\partial c = 0$).
\item $B_k(M)$:  $k$-boundaries ($c = \partial b$).
\item $H_k(M)$:  $Z_k(M) / B_k(M)$ ($k$th homology is $k$-cycles mod
$k$-boundaries).
\end{itemize}

\begin{itemize}
\item $\Omega^k(M)$:  $k$-forms.
\item $Z^k(M)$:  closed $k$-forms ($d\omega = 0$).
\item $B^k(M)$:  exact $k$-forms ($\omega = d\eta$).
\item $H^k(M)$:  $Z^k(M) / B^k(M)$.  ($k$th cohomology is closed $k$ forms
	mod exact $k$-forms).
\end{itemize}

and quotients (since $\partial^2=0$ and $d^2=0$):
\begin{eqnarray*}
	&&C_k(M) \supseteq Z_k(M) \supseteq B_k(M); \qquad H_k(M) = Z_k(M) / B_k(M) \\
	&&\Omega^k(M) \supseteq Z^k(M) \supseteq B^k(M); \qquad H^k(M) = Z^k(M) / B^k(M)
\end{eqnarray*}

%% ----------------------------------------------------------------
Here are the inclusions and their induced maps on chains and homology (used,
with varying connectivity hypotheses, in Seifert-van Kampen and
Mayer-Vietoris).  Note that the subscript $\#$ and $*$ are covariant functors
from manifolds to chain complexes and homology groups:  they preserve the
directions of the arrows (see notation \ref{notn:functor_variance}).

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.0)
%\put(0,0){\framebox(6,1)}

\put(0.70, 0.90){$M$}
\put(0.20, 0.50){$U$}
\put(1.20, 0.50){$V$}
\put(0.60, 0.10){$U\cap V$}
\put(0.35, 0.60){\vector( 1, 1){.30}}
\put(1.15, 0.60){\vector(-1, 1){.30}}
\put(0.65, 0.22){\vector(-1, 1){.30}}
\put(0.85, 0.22){\vector( 1, 1){.30}}
\put(0.25, 0.75){$j_U$}
\put(1.15, 0.75){$j_V$}
\put(0.25, 0.25){$i_U$}
\put(1.15, 0.25){$i_V$}

\put(2.60, 0.90){$C_k(M)$}
\put(1.90, 0.50){$C_k(U)$}
\put(3.20, 0.50){$C_k(V)$}
\put(2.40, 0.10){$C_k(U\cap V)$}
\put(2.35, 0.60){\vector( 1, 1){.25}}
\put(3.15, 0.60){\vector(-1, 1){.25}}
\put(2.65, 0.22){\vector(-1, 1){.30}}
\put(2.85, 0.22){\vector( 1, 1){.30}}
\put(2.15, 0.75){$j_{U\#}$}
\put(3.15, 0.75){$j_{V\#}$}
\put(2.15, 0.25){$i_{U\#}$}
\put(3.15, 0.25){$i_{V\#}$}

\put(4.60, 0.90){$H_k(M)$}
\put(3.90, 0.50){$H_k(U)$}
\put(5.20, 0.50){$H_k(V)$}
\put(4.40, 0.10){$H_k(U\cap V)$}
\put(4.35, 0.60){\vector( 1, 1){.25}}
\put(5.15, 0.60){\vector(-1, 1){.25}}
\put(4.65, 0.22){\vector(-1, 1){.30}}
\put(4.85, 0.22){\vector( 1, 1){.30}}
\put(4.15, 0.75){$j_{U*}$}
\put(5.15, 0.75){$j_{V*}$}
\put(4.15, 0.25){$i_{U*}$}
\put(5.15, 0.25){$i_{V*}$}

\end{picture}

%% ----------------------------------------------------------------
Here are the inclusions and their induced maps on forms and cohomology.  Note
that the superscript $\#$ and $*$ are contravariant functors from manifolds to
cochain complexes and cohomology groups: they reverse the directions of the
arrows (see notation \ref{notn:functor_variance}).

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.0)
%\put(0,0){\framebox(6,1)}

\put(0.70, 0.90){$M$}
\put(0.20, 0.50){$U$}
\put(1.20, 0.50){$V$}
\put(0.60, 0.10){$U\cap V$}
\put(0.35, 0.60){\vector( 1, 1){.30}}
\put(1.15, 0.60){\vector(-1, 1){.30}}
\put(0.65, 0.22){\vector(-1, 1){.30}}
\put(0.85, 0.22){\vector( 1, 1){.30}}
\put(0.25, 0.75){$j_U$}
\put(1.15, 0.75){$j_V$}
\put(0.25, 0.25){$i_U$}
\put(1.15, 0.25){$i_V$}

\put(2.60, 0.90){$\Omega^k(M)$}
\put(1.90, 0.50){$\Omega^k(U)$}
\put(3.20, 0.50){$\Omega^k(V)$}
\put(2.40, 0.10){$\Omega^k(U\cap V)$}
\put(2.65, 0.85){\vector(-1,-1){.30}}
\put(2.85, 0.85){\vector( 1,-1){.30}}
\put(2.35, 0.52){\vector( 1,-1){.30}}
\put(3.15, 0.52){\vector(-1,-1){.30}}
\put(2.15, 0.75){$j_{U}^{\#}$}
\put(3.15, 0.75){$j_{V}^{\#}$}
\put(2.15, 0.25){$i_{U}^{\#}$}
\put(3.15, 0.25){$i_{V}^{\#}$}

\put(4.60, 0.90){$H^k(M)$}
\put(3.90, 0.50){$H^k(U)$}
\put(5.20, 0.50){$H^k(V)$}
\put(4.40, 0.10){$H^k(U\cap V)$}
\put(4.65, 0.85){\vector(-1,-1){.30}}
\put(4.85, 0.85){\vector( 1,-1){.30}}
\put(4.35, 0.52){\vector( 1,-1){.30}}
\put(5.15, 0.52){\vector(-1,-1){.30}}
\put(4.15, 0.75){$j_{U}^{*}$}
\put(5.15, 0.75){$j_{V}^{*}$}
\put(4.15, 0.25){$i_{U}^{*}$}
\put(5.15, 0.25){$i_{V}^{*}$}
\end{picture}

Behavior of induced maps:
\begin{itemize}
\item Induced maps on chains are \emph{inclusions}.  E.g. a chain $c$
	in $U \cap V$ is a chain in $U$ by inclusion: $i_{U\#}(c) = c$.
\item Induced maps on forms are \emph{restrictions}.  E.g. a form $\omega$
	on $M$ is a form on $U$ by restriction: $j_{U}^{\#}(\omega) = \omega|_U$.
\end{itemize}

%% ----------------------------------------------------------------
Short exact sequences of chain maps:
$$
\begin{matrix}
	&
	&\vdots&
	&\vdots&
	&\vdots&
		\\
	0 &\longto &
	C_k(U\cap V)
		&\stackrel{i_{U\#} \oplus i_{V\#}}{\longto}&
	C_k(U) \oplus C_k(V)
		&\stackrel{j_{U\#} - j_{V\#}}{\longto}&
	C_k(M) &\longto & 0
		\\
	&
	&\downarrow \partial&
	&\downarrow \partial&
	&\downarrow \partial&
		\\
	0 &\longto &
	C_{k-1}(U\cap V)
		&\stackrel{i_{U\#} \oplus i_{V\#}}{\longto}&
	C_{k-1}(U) \oplus C_{k-1}(V)
		&\stackrel{j_{U\#} - j_{V\#}}{\longto}&
	C_{k-1}(M) &\longto& 0
		\\
	&
	&\vdots&
	&\vdots&
	&\vdots&
		\\
\end{matrix}
$$

Long exact sequence of homology:

\xymatrix{
	&\vdots
		& &
	\vdots
		& &
	\vdots
		& \\
%
	&H_k(U\cap V)
		&\stackrel{i_{U*} \oplus i_{V*}}{\longto}&
	H_k(U) \oplus H_k(V)
		&\stackrel{j_{U*} - j_{V*}}{\longto}&
	H_k(M)
	% Long arrow starts in this cell
	\ar
		`r[d]
		`_l
		^{\partial_*}
		`^d[lllll]
		`^r[dlllll]
		[rdlllll]
%
		&\\
%
	&H_{k-1}(U\cap V)
		&\stackrel{i_{U*} \oplus i_{V*}}{\longto}&
	H_{k-1}(U) \oplus H_{k-1}(V)
		&\stackrel{j_{U*} - j_{V*}}{\longto}&
	H_{k-1}(M)
		&\\
%
	&\vdots
		& &
	\vdots
		& &
	\vdots
		&\\
}

%% ----------------------------------------------------------------
Short exact sequences of form maps:
$$
\begin{matrix}
	&
	&\vdots&
	&\vdots&
	&\vdots&
		\\
	0 &\longfrom &
	\Omega^k(U\cap V)
		&\stackrel{i_{U}^{\#} - i_{V}^{\#}}{\longfrom}&
	\Omega^k(U) \oplus \Omega^k(V)
		&\stackrel{j_{U}^{\#} \oplus j_{V}^{\#}}{\longfrom}&
	\Omega^k(M) &\longfrom & 0
		\\
	&
	&\uparrow d&
	&\uparrow d&
	&\uparrow d&
		\\
	0 &\longfrom &
	\Omega^{k-1}(U\cap V)
		&\stackrel{i_{U}^{\#} - i_{V}^{\#}}{\longfrom}&
	\Omega^{k-1}(U) \oplus \Omega^{k-1}(V)
		&\stackrel{j_{U}^{\#} \oplus j_{V}^{\#}}{\longfrom}&
	\Omega^{k-1}(M) &\longfrom& 0
		\\
	&
	&\vdots&
	&\vdots&
	&\vdots&
		\\
\end{matrix}
$$

Long exact sequence of cohomology:

\xymatrix{
	&\vdots
		& &
	\vdots
		& &
	\vdots
		& \\
%
	&H^k(U\cap V)
		&\stackrel{i_{U}^{*} - i_{V}^{*}}{\longfrom}&
	H^k(U) \oplus H^k(V)
		&\stackrel{j_{U}^{*} \oplus j_{V}^{*}}{\longfrom}&
	H^k(M)
		&\\
%
	% Long arrow starts in this cell
	\ar
		`l[u]
		`_r
		^\delta
		`^u[rrrrrr]
		`^l[urrrrr]
		[lurrrrrr]
%
	&H^{k-1}(U\cap V)
		&\stackrel{i_{U}^{*} - i_{V}^{*}}{\longfrom}&
	H^{k-1}(U) \oplus H^{k-1}(V)
		&\stackrel{j_{U}^{*} \oplus j_{V}^{*}}{\longfrom}&
	H^{k-1}(M)
		&\\
%
	&\vdots
		& &
	\vdots
		& &
	\vdots
		&\\
}

%% ================================================================
\newpage
\section{Flows}
\label{sec:flows}

\begin{center}
\emph{
Flow with whatever may happen and let your mind be free. Stay centered by
accepting whatever you are doing. This is the ultimate.
}
--- Chuang Tzu\index{Chuang Tzu} (c. 370-301 B.C.).
\end{center}


xxx define in terms of diffeomorphisms

xxx define \bfidx{flow}, \bfidx{complete flow}, \bfidx{one-parameter family}.

xxx make sure to xref back to the ODE section.

1-parameter group of diffeomorphisms:
$$\phi_s \circ \phi_t = \phi_{s+t}.$$

To convert a diffeomorphism into a vector field, we have the following
formula:
$$
	\vecv\pickpipe{\vvecthree{x}{y}{z}}
	= \frac{\D \phi_t}{\D t}\pickpipe{t=0}\left(\vvecthree{x}{y}{z}\right).
$$
[xxx How to convert a vector field into a diffeomorphism?  Solve ODE?
Include an example or two.  xref to Lee2 thms about existence.]

If
$$
	X = f(x,y,z)\,\fracDx
	+ g(x,y,z)\,\fracDy
	+ h(x,y,z)\,\fracDz
$$
then solve
\begin{eqnarray*}
	\dot x &=& f(x,y,z) \\
	\dot y &=& g(x,y,z) \\
	\dot z &=& h(x,y,z).
\end{eqnarray*}

\beginex (Fall 2001 qualifying exam, \#4.)
For each $t \in \R$, let $\phi_t$ denote the map of $\S^2$ into itself which is
defined by
$$
	\phi_t: \S^2 \to \S^2: \colvecthree{x}{y}{z} \mapsto
	\colvecthree{\cos(t)x - \sin(t) y}{\sin(t)x + \cos(t) y}{z}.
$$
Show that $\phi_t$ is a one-parameter group of diffeomorphisms, and compute and
graph the vector field $\vecv$ on $\S^2$ for which $\phi_t$ is the corresponding
flow.

Remark:  In defining the vector field $\vecv$, please specify how you are
viewing the tangent bundle of $\S^2$.

\emph{Solution.}  To show that $\phi_t$ is a one-parameter group of
diffeomorphisms, we need to show:
\begin{itemize}
\item Each $\phi_t$ is bijective on $\S^2$.
\item Each $\phi_t$ is smooth with smooth inverse.
\item For each $s$ and $t$ we have $\phi_s \circ \phi_t = \phi_{s+t}$.
\end{itemize}
First note that the map $\phi_t$ may be written in matrix form as
$$
	\colvecthree{x}{y}{z} \stackrel{\phi_t}{\to}
	\mxrsqthree
		{\cos(t)}{-\sin(t)}{0}
		{\sin(t)}{ \cos(t)}{0}
		{0}{0}{1}
	\colvecthree{x}{y}{z}.
$$
The matrix has determinant $\cos^2(t) + \sin^2(t)=1 \ne 0$ so it is invertible,
hence bijective.  In particular its inverse is
$$
	\colvecthree{x}{y}{z} \stackrel{\phi_t^{-1}}{\to}
	\mxrsqthree
		{\cos(t)}{\sin(t)}{0}
		{-\sin(t)}{\cos(t)}{0}
		{0}{0}{1}
	\colvecthree{x}{y}{z}.
$$

This much shows that $\phi_t$ is bijective on $\R^3$.  To show bijectivity on
$\S^2$, we need to show that $\phi_t(\vecq) \in \S^2$ for all $q \in \S^2$, and
likewise for $\phi_t^{-1}$.  Now, we've already shown that $\phi_t^{-1}$ is
rotation by $-t$ (since $\cos(-t)=\cos(t)$ and $\sin(-t)=\sin(t)$) so it
suffices to show this for $\phi_t$ only.  Namely, the coordinates of $(u, v, w)
= \phi_t(x,y,z)$ should satisfy $u^2+v^2+w^2=1$.  Check:
\begin{eqnarray*}
	u^2 + v^2 + w^2 &=& 
		(\cos(t)x - \sin(t)y)^2 + (\sin(t)x + \cos(t)y)^2 + z^2 \\
&&\\
	&=& \cos^2(t) x - \cos(t)\sin(t)xy + \sin^2(t)y^2 \\
	&+& \sin^2(t)x^2 + \cos(t)\sin(t)xy + \cos^2(t)y^2 \\
	&+& z^2 \\
&&\\
	&=& x^2 + y^2 + z^2 = 1.
\end{eqnarray*}

The transformation is linear, hence it is its own derivative and thus eminently
differentiable; likewise for the inverse.  (Remember we're checking smoothness
for fixed $t$.  Smoothness with respect to $t$ is a different question.) For
the composition property, we compute (using the sum formulas for sine and
cosine)
\begin{eqnarray*}
	\phi_{s+t}
	\colvecthree{x}{y}{z}
	%
	&=&
	\mxrsqthree
		{\cos(s+t)}{-\sin(s+t)}{0}
		{\sin(s+t)}{ \cos(s+t)}{0}
		{0}{0}{1}
	\colvecthree{x}{y}{z} \\
	%
	&=&
	\mxrsqthree
		{\cos(s)\cos(t)-\sin(s)\sin(t),}{-\sin(s)\cos(t)-\cos(s)\sin(t),}{0}
		{\sin(s)\cos(t)+\cos(s)\sin(t),}{ \cos(s)\cos(t)-\sin(s)\sin(t),}{0}
		{0,}{0,}{1}
	\colvecthree{x}{y}{z}.
\end{eqnarray*}
On the other hand,
\begin{eqnarray*}
	\phi_s\circ\phi_t
	\colvecthree{x}{y}{z}
	&=&
	\mxrsqthree
		{\cos(s)}{-\sin(s)}{0}
		{\sin(s)}{ \cos(s)}{0}
		{0}{0}{1}
	\mxrsqthree
		{\cos(t)}{-\sin(t)}{0}
		{\sin(t)}{ \cos(t)}{0}
		{0}{0}{1}
	\colvecthree{x}{y}{z} \\
	&=&
	\mxrsqthree
		{\cos(s) \cos(t)-\sin(s)\sin(t),}{-\cos(s)\sin(t)-\sin(s)\cos(t),}{0}
		{\sin(s) \cos(t)+\cos(s)\sin(t),}{-\sin(s)\sin(t)+\cos(s)\cos(t),}{0}
		{0}{0}{1}
	\colvecthree{x}{y}{z}
\end{eqnarray*}
which is the same as desired.  This was the last item needed to prove
that the $\phi_t$'s form a one-parameter group of diffeomorphisms.

Recalling the necessary formula for the corresponding vector field, we have
\begin{eqnarray*}
	\vecv\pickpipe{\colvecthree{x}{y}{z}}
	&=& \frac{\D \phi_t}{\D t}\pickpipe{t=0}\colvecthree{x}{y}{z} \\
	&=& \frac{\D }{\D t}\pickpipe{t=0}
		\colvecthree{\cos(t)x - \sin(t) y}{\sin(t)x + \cos(t) y}{z} \\
	&=& \colvecthree{-\sin(t)x - \cos(t) y}{\cos(t)x - \sin(t) y}{0}
		\pickpipe{t=0} \\
	&=& \colvecthree{-y}{x}{0}\pickpipe{\colvecthree{x}{y}{z}} \\
	&=& -y\,\slashDx + x\,\slashDy.
\end{eqnarray*}
This is of course rotation about the $z$-axis.  [xxx insert figure here.] Here,
we are viewing the tangent bundle as directional derivatives.

This is a vector field on $\R^3$.  [xxx to do:  convert to graph coordinates, or spherical.]
\mendex


\beginex
\label{ex:pb_rot_S2_vol}
Here is a variation of the previous example.  Let $\omega = x\,dy\wedge dz +
y\,dz\wedge dx + z\, dx\wedge dy$.  Compute $\phi_t^*(\omega)$.  (In example
\ref{ex:Lie_rot_S2_vol} we will revisit this problem using the Lie derivative.)

\emph{Solution.}  Let $c=\cos(t)$ and $s=\sin(t)$.  Then, replacing $x$ with
$cx-sy$ and $y$ with $sx+cy$, we have
\begin{eqnarray*}
	\phi_t^*(\omega)
		&=& (cx-sy)\, d(sx+cy)\wedge dz \\
		&+& (sx+cy)\, dz\wedge d(cx-sy) \\
		&+& z\, d(cx-sy)\wedge d(sx+cy) \\
	&&\\
		&=& (cx-sy)\, (sdx\wedge dz +c dy\wedge dz) \\
		&+& (sx+cy)\, (c dz\wedge dx - s dz \wedge dy) \\
		&+& z\, (cdx-sdy)\wedge (sdx+cdy) \\
	&&\\
		&=& (cx-sy)\, (-sdz\wedge dx +c dy\wedge dz) \\
		&+& (sx+cy)\, (c dz\wedge dx + s dy \wedge dz) \\
		&+& z\,(c^2+s^2) dx \wedge dy \\
	&&\\
		&=& -cs\,x\, dz\wedge dx
		+c^2\,x\, dy \wedge dz
		+s^2\,y\, dz \wedge dx
		-cs\,y\, dy\wedge dz \\
	&+& cs\,x\, dz\wedge dx
		+s^2\,x\, dy \wedge dz
		+c^2\,y\, dz \wedge dx
		+cs\,y\, dy\wedge dz \\
		&+& z\, dx \wedge dy \\
	&&\\
		&=& x\, dy \wedge dz +y\, dz \wedge dx + z\, dx \wedge dy \\
	&=& \omega.
\end{eqnarray*}
Note that this is a bit messy, only because I chose a form with three terms in
it.
\mendex

% xxx GT.F01.4 example w/ attrib to Dan.

%% ================================================================
\newpage
\section{Lie derivatives and the Cartan calculus}
\label{sec:lie_cartan}

\begin{center}
\emph{
If you do not change the direction in which you are going, you will end up
where you are headed.
} \\
--- Confucius\index{Confucius} (551-479 B.C.).
\end{center}

%% ================================================================
\subsection{The $X\ctt$ operator}
\label{sec:i_X}

xxx note we are revisiting this for tensor fields; xref back to section
\ref{sec:vs_i_X}.

xxx xref back to $d$ section as well.

xxx collect rules here.
\begin{eqnarray*}
	X \ctt f &=& 0 \\
	X \ctt \omega^{(1)} &=& \omega(X) \\
	X \ctt (\omega \wedge \eta) &=& (X \ctt \omega) \wedge \eta + (-1)^{\ord(\omega)}\, \omega \wedge (X \ctt \eta).
\end{eqnarray*}

Rules for $X \ctt \omega$:
\begin{eqnarray*}
	(f \slashDx)\ctt(g\, dx\wedge dy \wedge dz) &=& fg\, dy \wedge dz \\
	(f \slashDy)\ctt(g\, dx\wedge dy \wedge dz) &=& -fg\, dx \wedge dz \\
	&=& g\, dz \wedge dx \\
	(f \slashDz)\ctt(g\, dx\wedge dy \wedge dz) &=& fg\, dx \wedge dy.
\end{eqnarray*}
In general,
$$
	f\slashDx_k\ctt(g\, dx_1 \wedge \cdots \wedge dx_k \wedge \cdots \wedge dx_m)
	= (-1)^{k-1} fg\, dx_1 \wedge \cdots \wedge \hat{dx}_k \wedge \cdots \wedge dx_m)
$$
where the overhat indicates that that term is omitted.

\begin{mnem}
Put your left forefinger over the $\slashDx_k$ and your right forefinger over
the $dx_k$.  Write down what remains, toggling the minus sign by the number of
$d$'s between your fingers.
\end{mnem}

%% ================================================================
\newpage
\subsection{Lie derivatives}

Let $X$ be a vector field on a manifold $M$, and $T$ be a tensor field on $M$.
The \bfidx{Lie derivative}\index{$L_X$} $L_X(T)$ is, intuitively, the
\bfidx{rate of change} of $T$ in the direction of $X$.  We will see that $L_X$
takes tensors to tensors of the same rank.  This is in contrast to $d$ and
$X\ctt$ which are rank-raising and rank-lowering operators.

%% ================================================================
\subsubsection{Lie derivatives on general tensor fields}

\begin{defn}
\label{defn:lie_deriv_flow_pullback}
Let $X$ be a vector field on a manifold $M$, with flow $\phi^X_t$.  Let $T$ be
a tensor on $M$.  Then
$$L_X(T) = \frac{d}{dt}\bigg|_{t=0}(\phi^X_t)^* T.$$
\end{defn}

Meaning:  How $T$ changes under the flow in the direction of $X$: (1) Pull
back (w/r/t flow --- need to have defined pullback in terms of the
diffeomorphism map), then (2) differentiate with respect to time.

xxx make a better diagram (and xref to pullback section(s)):

$$
\begin{matrix}
M & \stackrel{\phi_t}{\to}& M \\
T \circ \phi_t \downarrow & & T \downarrow \\
\end{matrix}
$$

%\beginex
%If $T$ is a 0-form, then
%$$(\phi_t^X)^* \omega\bigg|_\vecq = \omega \circ \phi^X_t\bigg|_\vecq
%= \omega(\phi^X_t(\vecq)).$$
%\mendex
%xxx explain that!!

%\beginex
%If $\omega$ is a 1-form, then
%$$
%(L_X \omega)(\vecv_\vecq) = \frac{d}{dt}\bigg|_{t=0}[\omega((\phi^X_t)^*(\vecv_\vecq)])
%$$
%\mendex
%xxx also needs some explaining!

%% ================================================================
\subsubsection{Lie derivatives on 0-forms}

\begin{prop}
\label{prop:lie_deriv_on_0_forms}
Let $f: M \to \R$, and let $X$ be a vector field on $M$.
Then
$$
	L_X(f) = Xf
$$
where the action of $X$ on $f$ is the directional derivative as usual.
\begin{proof}
\cite{bib:lee2}, proposition 18.9.
\end{proof}
\end{prop}

Note that $f$ and $Xf$ are both functions from $M$ to $\R$, i.e. 0-forms on
$M$.

%% ================================================================
\subsubsection{Lie derivatives on vector fields}

\begin{defn}
\label{prop:lie_deriv_on_vr_flds}
For smooth vector fields $X$ and $Y$, the \bfidx{Lie bracket} of $X$ and $Y$
is
$$
	[X,Y]=XY-YX
$$
via
$$
	[X,Y]f=XYf-YXf
$$
for a smooth function $f$.
\end{defn}

Since $Xf$ and $Yf$ are 0-forms, just like $f$, it makes sense to compute $YXf$
and $XYf$.

Vector fields on $M$ become a \bfidx{Lie algebra}: Addition is as usual and
multiplication is $[X, Y]$.

\begin{prop}
$L_X(Y) = [X,Y]$.
\begin{proof}
See \cite{bib:lee2}, proposition 18.9.
\end{proof}
\end{prop}

%% ================================================================
\subsubsection{Properties of $L_X$ and relationships between $X\ctt$, $L_X$,
and $d$}

Definition \ref{defn:lie_deriv_flow_pullback} is unwieldy.  Propositions
\ref{prop:lie_deriv_on_0_forms} and \ref{prop:lie_deriv_on_vr_flds} (the proofs
of which also use definition \ref{defn:lie_deriv_flow_pullback}) give us better
ways to compute Lie derivatives of 0-forms and vector fields.  For forms in
general, using a couple more propositions along with the properties we've seen,
we can derive some more handy properties that allow us to manipulate $L_X$'s,
$X\ctt$'s, and $d$'s directly.

[xxx xref:]
\begin{prop}
\label{prop:i_and_d_are_antiderivns}{$\,$}
\begin{eqnarray*}
	d(\omega \wedge \eta) &=& d(\omega)\wedge \eta + (-1)^{\ord(\omega)} \; \omega \wedge d\eta \\
	X\ctt(\omega \wedge \eta)
		&=& X\ctt(\omega)\wedge \eta + (-1)^{\ord(\omega)} \; \omega \wedge X\ctt(\eta)
\end{eqnarray*}
\end{prop}

\begin{prop}
\label{prop:lie_deriv_is_derivn}
$L_X(\omega \wedge \eta) = L_X(\omega) \wedge \eta + \omega \wedge L_X(\eta)$.
\begin{proof}
This is \cite{bib:lee2}, proposition 18.9.
\end{proof}
\end{prop}

\begin{rem}
That is, while $X\ctt$ and $d$ are \bfidxaux{antiderivations}{antiderivation},
the Lie derivative is a \bfidx{derivation}.  (See definition
\ref{defn:derivn_and_antiderivn}.)
\end{rem}

\begin{prop}
$L_X(Y\ctt(\omega)) = {L_X Y}\ctt(\omega) + Y\ctt(L_X \omega)$.
\end{prop}

%\begin{rem}
%xxx needs more info; intuition at least.  Does it need to be here at all?
%See \cite{bib:lee2} prop 18.9.
%\end{rem}

\begin{prop}
\label{prop:lie_deriv_commutes_with_d}
For any smooth vector field $X$ and any smooth differential form $\omega$,
$$
	L_X(d\omega) = d(L_X\omega).
$$
\begin{proof}
This is \cite{bib:lee2}, corollaries 18.11 (for 1-forms) and 18.14 (for
$k$-forms).
\end{proof}
\end{prop}

What does it mean to contract a one-form?

\begin{prop}
\label{prop:iXdf_is_Xf}
For a smooth function $f$ and a vector field $X$ on a manifold $M$,
$$
	X\ctt(df) = df(X) = Xf.
$$
\begin{proof}
The first equality is by definition of the $X\ctt$ operator (definition
\ref{defn:contraction_vs}):  we insert $X$ into the argument list of the
form $df$.  Since $df$ is a 1-form, we need no more arguments:
$$
	X\ctt(df) = df(X).
$$

For the second equality, the only way I know how to do this is using
coordinates.  The vector field $X$ is a linear combination of the basis vectors
$\slashDx_i$, with coefficients being smooth functions, say, $v_i$.  Then,
since the $dx_i$'s are the dual basis to the $\slashDx_j$'s,
\begin{eqnarray*}
	df(X) &=& \sum_{i=1}^m \fracDfx_i dx_i \left(
		\sum_{j=1}^m v_j \fracDx_j
		\right)
	= \sum_{i=1}^m \fracDfx_i v_i \\
	&=& \left(\sum_{i=1}^m  v_i\fracDx_i\right)(f)
	= Xf.
\end{eqnarray*}
\end{proof}
\end{prop}

Given these facts, we can now justify a formula which connects $L_X$, $X\ctt$,
and $d$ --- all using simple algebraic manipulations.

\begin{prop}[Cartan's magic formula]
\label{prop:cartan_magic_formula}
For\index{magic formula}\index{Cartan's magic formula} any smooth vector field
$X$ and any smooth differential form $\omega$,
$$
	L_X \omega = X\ctt(d\omega) + d(X\ctt(\omega)).
$$
That is,
$$
	L_X = X\ctt \circ d + d \circ X\ctt.
$$
\begin{mnem}
The $d$ operator raises the degree of forms, the $X\ctt$ operator lowers degree,
and the $L_X$ operator preserves degree.  So, with regard to degrees, we have
$0 = (1-1) + (-1+1)$.
Or,
$$
\begin{matrix}
	        &   & \nearrow & \searrow &   &          &          \\
	\longto & = &          &          & + &          &          \\
	        &   &          &          &   & \searrow & \nearrow \\
\end{matrix}.
$$
\end{mnem}
\begin{proof}
The proof is in three steps:  (1) for 0-forms; (2) for 1-forms; (3) for
$k$-forms, using induction.

First suppose $\omega$ is a 0-form.  Rename $\omega$ to $f$ for clarity.
Then
$$
	L_X(f) = Xf
$$
whereas
$$
	X\ctt(df) + d(X\ctt(f)) = Xf + 0 = Xf.
$$
The first term is $Xf$ because of proposition \ref{prop:iXdf_is_Xf}.  The
second term is 0 because $X\ctt$ is degree-lowering and $f$ is already of degree
0; our convention [xxx write and xref backward] is that $X\ctt$ of 0-forms is 0.

Now suppose $\omega$ is a 1-form.  Then [xxx write and xref backward] $\omega$
is a linear combination of 1-forms of the form $u\;dv$, where $u$ and $v$ are
0-forms.  Recall that by convention, we omit the wedge in wedge products when
one factor is a scalar --- e.g. $u\;dv$ means $u \wedge dv$.  Using the above
propositions, the left-hand side of the magic formula is
\begin{eqnarray*}
	L_X(u\;dv) &=& L_X(u)\wedge dv + u \wedge L_X(dv) \\
	&=& Xu\;dv + u\;L_X(dv) \\
	&=& Xu\;dv + u\;dL_X(v) \\
	&=& Xu\;dv + u\;d(Xv).
\end{eqnarray*}
Recall that while the Lie derivative is a derivation, the $X\ctt$ and $d$
operators are antiderivations.  For the right-hand side, we have
\begin{eqnarray*}
	d(u\;dv) &=& du \wedge dv \\
	X\ctt(d(u\;dv)) &=& X\ctt(du \wedge dv) \\
		&=& X\ctt(du) \wedge dv - du \wedge X\ctt(dv) \\
		&=& Xu\;dv - du\;Xv \\
	X\ctt(u\;dv) &=& X\ctt(u) \wedge dv + u\;X\ctt(dv) \\
		&=& u\;Xv \\
	d(X\ctt(u\;dv)) &=& d(u\;Xv) \\
		&=& du\;Xv + u\;d(Xv) \\
	X\ctt(d(u\;dv)) + d(X\ctt(u\;dv))
		&=& Xu\;dv - du\;Xv + du\;Xv + u\;d(Xv) \\
		&=& Xu\;dv + u\;d(Xv) \\
		&=& L_X(u\;dv).
\end{eqnarray*}

[xxx finish up the induction part.]
\end{proof}
\end{prop}

%% ================================================================
\subsection{Lie derivatives and flows}

xxx attribute:

\begin{prop}
Let $V$ and $W$ be vector fields with associated flows $\phi_s$ and
$\psi_t$.  The following are equivalent:
\begin{itemize}
\item $L_V(W) = 0$.
\item $L_W(V) = 0$.
\item $[V,W]=0$
\item $W$ is \plainidx{invariant under the flow} of $V$.
\item $V$ is \plainidx{invariant under the flow} of $W$.
\item $\psi_t \circ \phi_s = \phi_s \circ \psi_t$.
\end{itemize}
\end{prop}

xxx examples here, or xref to prolrevqual.

%% ================================================================
\newpage
\subsection{Computations using Lie derivatives}
\label{sec:lie_deriv_computations}

\begin{center}
\emph{
An idea which can be used once is a trick. If it can be used more than once it
becomes a method.} \\
--- George Polya\index{Polya, George} and Gabor Szego\index{Szego, Gabor}
\end{center}

Let's summarize the results of the previous sections.  If an exam question asks
you if a form $\omega$ is ``\bfidx{invariant} under the flow'' of a vector
field $X$, or asks you whether the ``flow of $X$ \bfidx{preserves}'' $\omega$,
what can you do?  Compute the Lie derivative of $\omega$ or $Y$ with respect to
$X$: the Lie derivative $L_X(\omega)$ is zero if and only if $\omega$ is
invariant under the flow of $X$ (proposition 18.16 of \cite{bib:lee2}).  Here
are our options for carrying that out that computation:

\begin{itemize}
\item[(i)] For a vector field $Y$, $L_X(Y) = [X,Y]$.  [xxx type up qual problem
	and xref to it.]
\item[(ii)] For a 0-form $f$, $L_X(f) = X(f)$.
\item[(iii)] For an exact form $d\omega$, $L_X(d\omega) = dL_X(\omega)$.
\item[(iv)] For a wedge of forms $\omega\wedge\eta$,
	$L_X(\omega\wedge\eta) = L_X(\omega)\wedge\eta + \omega \wedge L_X(\eta)$.
\end{itemize}

This allows us to split up a lot of things.  Perhaps not everything imaginable,
but hopefully anything we will see on an exam in our first year.

Common qualifier questions are:  is a given form $\omega$ \bfidx{invariant}
under the flow corresponding to a given vector field $X$, or does a given flow
\bfidx{preserve} a given form.  We can answer these questions in the
affirmative by showing that the Lie derivative $L_X(\omega)$ is zero.

\beginex
On $\S^2$, does the flow of $\D/\D\theta$ preserve the area form $dA$?

Using spherical coordinates, the area form is $dA=\sin\phi \,d\theta\wedge
d\phi$.  Then we need to compute $$L_{\D/\D\theta}(\sin\phi\,d\theta\wedge
d\phi).$$ Remember from [xxx write and xref] that there is an implicit wedge
between 0-forms and higher-level forms, so this is really
$$L_{\D/\D\theta}(\sin\phi\wedge d\theta\wedge d\phi).$$  Using
rule (iv), this splits up as
$$
	L_{\D/\D\theta}(\sin\phi)\wedge d\theta\wedge d\phi \quad+\quad
	\sin\phi\wedge L_{\D/\D\theta}(d\theta)\wedge d\phi \quad+\quad
	\sin\phi\wedge d\theta \wedge L_{\D/\D\theta}(d\phi).
$$
Using rule (ii) we can set the first term to zero.  Can we use rule (iii) for
the remaining two terms?  Just because a form traditionally is written with a
$d$ in front doesn't mean it is exact.  Locally, at least, $\theta$ and $\phi$
are 0-forms.  Then we have
$$
	0 \quad+\quad
	\sin\phi\wedge d L_{\D/\D\theta}(\theta)\wedge d\phi \quad+\quad
	\sin\phi\wedge d\theta \wedge d L_{\D/\D\theta}(\phi) = 0.
$$
Since this is identically zero, $\D/\D\theta$ preserves the area form $dA$.

Suppose the question asked about the flow of $\D/\D\phi$.  The second and third
terms would still go away, but the first would not.
\mendex

\beginex
\label{ex:Lie_rot_S2_vol}
(This example revisits example \ref{ex:pb_rot_S2_vol}, using the Lie derivative.)
Show that the two-form $\omega = x\,dy\wedge dz + y\,dz\wedge dx + z\, dx\wedge
dy$ is invariant under the flow of $\phi_t$.

\emph{Solution.}  We will compute the Lie derivative [xxx for prolrev, xref
forward or reorg] of $\omega$ in the direction of $\vecv$ and [xxx].  Since
$$
	\vecv = -y\, \slashDx + x\, \slashDy,
$$
we have
$$
	L_\vecv(x) = -y, \quad
	L_\vecv(y) =  x, \quad
	L_\vecv(z) =  0.
$$
Recall that $L_\vecv$ is a derivation, i.e. it follows the product rule through
wedges.  Also, it commutes with $d$ for exact forms.  For the first term of
$\omega$, namely $x\, dy\wedge dz$, we then have
\begin{eqnarray*}
	L_\vecv(x\, dy \wedge dz)
		&=& L_\vecv(x) dy \wedge dz
		\;+\; x \, d(L_\vecv(y)) \wedge dz
		\;+\; x \, dy \wedge d(L_\vecv(z)) \\
	%
		&=& -y\, dy \wedge dz
		\;+\; x\, dx \wedge dz
		\;+\; x \, dy \wedge 0 \\
	%
		&=& -y\, dy \wedge dz \;-\; x\, dz \wedge dx.
\end{eqnarray*}
For the second term:
\begin{eqnarray*}
	L_\vecv(y\, dz \wedge dx)
		&=& L_\vecv(y) dz \wedge dx
		\;+\; y \, d(L_\vecv(z)) \wedge dx
		\;+\; y \, dz \wedge d(L_\vecv(x)) \\
	%
		&=& x\, dz \wedge dx
		\;+\; y \, 0 \wedge dx
		\;-\; y \, dz \wedge dy \\
	%
		&=& x\, dz \wedge dx \;+\; y \, dy \wedge dz.
\end{eqnarray*}
For the third term:
\begin{eqnarray*}
	L_\vecv(z\, dx \wedge dy)
		&=& L_\vecv(z) dx \wedge dy
		\;+\; z \, d(L_\vecv(x)) \wedge dy
		\;+\; z \, dx \wedge d(L_\vecv(y)) \\
	%
		&=& 0\, dx \wedge dy
		\;-\; z \, dy \wedge dy
		\;+\; z \, dx \wedge dx \\
	%
		&=& 0.
\end{eqnarray*}
Combining these, we have
\begin{eqnarray*}
	L_\vecv(\omega)
	&=& -y\, dy \wedge dz \;-\; x\, dz \wedge dx \\
	&+& x\, dz \wedge dx \;+\; y \, dy \wedge dz \\
	&=& 0.
\end{eqnarray*}
which shows that $\omega$ is invariant under the flow of $\phi_t$.
\mendex

[xxx do it again using the magic formula.  Much nicer.]

[xxx type up the August 06 qual problem and xref to it.]

%% ================================================================
\newpage
\section{Problems}
\label{sec:problems}

\begin{center}
\emph{Each problem that I solved became a rule which served afterwards to solve
other problems.}

--- Ren\'e Descartes\index{Descartes, Ren\'e} (1596-1650).
\end{center}

Here are various problems, along with solutions.  The problems are taken from
homework assignments, exam reviews, and exams.

%I am not including solutions to old qualifiers, since each of us should do
%that on our own.

%% ================================================================
\subsection{Geometry homework 1}

%%% ----------------------------------------------------------------
%\subsubsection{Geometry homework 1 problem 2a}
%
%Find $Df$ where $f: \mcL(\R^n) \to \mcL(\R^n)$ is given by $A \mapsto A^m$,
%for positive integer $m$.
%
%\begin{answer}
%\end{answer}

%%% ----------------------------------------------------------------
%\subsubsection{Geometry homework 1 problem 2b}
%
%Find $Df$ where $f: \mcL(\R^n) \to \mcL(\R^n)$ is given by $A \mapsto A^{-1}$.
%
%\begin{answer}
%\end{answer}

%%% ----------------------------------------------------------------
%\subsubsection{Geometry homework 1 problem 2c}
%
%Find $Df$ where $f: \mcL(\R^n) \to \GL(\R^n)$ is given by $A \mapsto e^A$.
%
%\begin{answer}
%\end{answer}

%% ----------------------------------------------------------------
\subsubsection{Geometry homework 1 problem 2d}
\label{sec:hw_1_2d}

Find $Df$ where $f: \mcL(\R^n) \to \R$ is given by $A \mapsto \det(A)$.

Hint: First consider the case when $A$ is invertible, i.e.  $\det(A) \ne 0$.
Then by the multiplicativity of the determinant,
$$
\frac{d}{dt}\pickpipe{t=0} \det(A+tB) =
\frac{d}{dt}\pickpipe{t=0} \det(I+tA^{-1}B)\det(A).
$$
Next, recall that the exponential map from $\mcL(\R^n)$ to $\GL(\R^n)$
has the form
$$ e^A = I + A + A^2/2! + \ldots $$
and so
$$ e^{tA^{-1}B} = I + tA^{-1}B + \textrm{(higher-order terms)}.
$$
Thus
$$
	\frac{d}{dt}\pickpipe{t=0} \det(I+tA^{-1}B)\det(A) =
	\frac{d}{dt}\pickpipe{t=0} \det(e^{tA^{-1}B})\det(A).
$$

\begin{answer}[Answer (due largely to Andy Lebovitz).]
Recall that $\det(e^X) = e^{\tr(X)}$.  So,
$$
	D\det|_A(B) = \frac{d}{dt}\pickpipe{t=0} e^{\tr(tA^{-1}B} \det(A)
	= \det(A) \frac{d}{dt} \pickpipe{t=0} e^{\tr(tA^{-1}B)}.
$$
Since the trace is linear, $\tr(tA^{-1}B = t \cdot \tr(A^{-1}B)$.  Also note
that $A^{-1}B$ is constant with respect to $t$.  Then
\begin{eqnarray*}
	D\det|_A(B)
	&=& \det(A)\frac{d}{dt} \pickpipe{t=0} e^{t\cdot \tr(A^{-1}B)} \\
	&=& \det(A) \tr(A^{-1}B) \frac{d}{dt}\pickpipe{t=0} e^t \\
	&=& \det(A) \tr(A^{-1}B).
\end{eqnarray*}

Continuing to follow the hint, consider non-invertible $A$.  Recall that
when $A$ is invertible [xxx have defined adjugate somewhere above],
$$ A^{-1} = \frac{\adju(A)}{\det(A)}.$$
From above,
$$
	D \det|_A(B) = \tr\left(\frac{\adju(A) B}{\det(A)}\right) \det(A).
$$
Since the trace is linear,
$$
	D \det|_A(B) = \tr(\adju(A) B).
$$
This formula applies for non-singular $A$.  Note that $\tr$ is linear, the
adjugate is polynomial in the coefficients of $A$, and $\GL(\R^n)$ is dense in
$\mcL(\R^n)$.  Thus by continuity, this formula holds for all $A$.
\end{answer}

%% ================================================================
\newpage
\subsection{Geometry homework 6}

%% ----------------------------------------------------------------
\subsubsection{Geometry homework 6 problem 1}
\label{sec:hw_6_1}

Suppose $z=g(y)$ is a positive function, and let $M$ denote the surface of
revolution obtained by revolving the graph of $g$ around the $y$ axis in
$\R^3$.  Show that the tangent bundle of $M$ is trivial by finding an
explicit trivialization.

\begin{answer}
It suffices to apply the \bfidx{global frame criterion} as defined in section
\ref{sec:triv_vb}.  Let $\vecq \in M$.  Then in rectangular coordinates for
$\R^3$ we can write
$$
	\vecq = \colvecthree{g(y) \cos\theta}{y}{g(y)\sin\theta}.
$$
We can parameterize this surface using $y$ and $\theta$ coordinates, where
$\theta$ is the angle from the $x$ axis to $\vecq$.  As in section
\ref{sec:compute_tang}, we compute
$$
	\fracDy      = \colvecthree{g'(y) \cos\theta}{1}{g'(y)\sin\theta}, \quad
	\fracD\theta = \colvecthree{-g(y) \sin\theta}{0}{g(y)\cos\theta}.
$$
It suffices to show that the sections $\slashDy$ and $\slashD\theta$ form a
global frame, i.e. that they are linearly independent and globally defined on
$M$.  Clearly, they are globally defined on $M$.  (Unlike, say, [xxx xref]
$\S^2$, where our various coordinates leave some part of the sphere uncovered.)
It is also clear that $\slashDy$ and $\slashD\theta$ are smooth on $M$, as long
as $g(y)$ is smooth on $\R$.

For linear independence, we can show that (1) both sections are non-zero, and
(2) their dot product is identically zero.  For the first claim, $\fracDy$ is
nowhere zero since $g(y)$ is assumed non-zero, and since sine and cosine have
no common zeroes.  Likewise, $\slashD\theta$ is nowhere zero due to the 1 in
the $y$ component.  For the second claim, we compute the dot product
$$
	\fracDy \cdot \fracD\theta =
	\colvecthree{g'(y) \cos\theta}{1}{g'(y)\sin\theta} \cdot
	\colvecthree{-g(y) \sin\theta}{0}{g(y)\cos\theta}
	= -g(y) g'(y) \sin\theta \cos\theta + g(y) g'(y) \sin\theta \cos\theta
	= 0.
$$
\end{answer}

%% ================================================================
\newpage
\subsection{Geometry exam 1}

%% ----------------------------------------------------------------
\subsubsection{Geometry exam 1 problem 6}
\label{sec:ex_1_6}

Fix $0<r<R$.  Consider the manifold $M$ obtained by revolving the circle
$(y-R)^2+z^2=r^2$ around the $z$-axis in $\R^3$.  As we discussed in class,
there is an explicit diffeomorphism
$$
	\S^1 \times \S^1 \to M:
	(e^{i\theta}, e^{i\phi}) \longmapsto
	(R+r\cos\phi)) \colvecthree{\cos\theta}{\sin\theta}{0}
	+ r\sin\phi\colvecthree{0}{0}{1}.
$$
Express the Euclidean inner product (or metric), viewed as a 2-tensor field on
$M$, in terms of the coordinates $\theta, \phi$.

\begin{answer}
See example \ref{ex:decompose_two_tensor}.  Here we are doing the same thing,
but with varying coefficients in the sense of sections \ref{sec:loaf} and
\ref{rem:g_dx2_dy2_dz2}.  A point $\vecq$ of $M$ is
$$
	\vecq = \colvecthree{x}{y}{z} =
	(R+r\cos\phi) \colvecthree{\cos\theta}{\sin\theta}{0}
	+ r\sin\phi\colvecthree{0}{0}{1}.
$$
so we can write
$$
	\fracD\theta = (R + r\cos\phi)\colvecthree{-\sin\theta}{\cos\theta}{0}
	\qand
	\fracD\phi = (-r\sin\phi) \colvecthree{\sin\theta}{\cos\theta}{0}
		+ r\cos\phi \colvecthree{0}{0}{1}.
$$
As in example \ref{ex:decompose_two_tensor}, we want to write the metric
tensor $g$ as
$$
	g = g\left(\fracD\theta, \fracD\theta\right) d\theta\tensor d\theta
	+ g\left(\fracD\phi, \fracD\phi\right) d\phi\tensor d\phi
$$
where we can omit the diagonal terms $g(\slashD\theta,\slashD\phi)$ etc. since
$g$ is diagonal and $\{\slashD\theta, \slashD\phi\}$ is an orthogonal basis.
First we find $g(\slashD\theta,\slashD\theta)$.  Recalling that $dx$ extracts
the $x$ coordinate, etc., this is
\begin{eqnarray*}
	g\left(\fracD\theta,\fracD\theta\right) &=&
	(dx\tensor dx + dy\tensor dy + dz\tensor dz)
		\left(\fracD\theta,\fracD\theta\right) \\
	&=&
		dx\tensor dx\left(\fracD\theta,\fracD\theta\right) +
		dy\tensor dy\left(\fracD\theta,\fracD\theta\right) +
		dz\tensor dz\left(\fracD\theta,\fracD\theta\right) \\
	&=&
		(R + r\cos\phi)^2\sin^2\theta +
		(R + r\cos\phi)^2\cos^2\theta +
		\left(0\right)^2 \\
	&=& (R + r\cos\phi)^2.
\end{eqnarray*}
Next we find $g(\slashD\phi,\slashD\phi)$:
\begin{eqnarray*}
	g\left(\fracD\phi,\fracD\phi\right)
	&=&
		dx\tensor dx\left(\fracD\phi,\fracD\phi\right) +
		dy\tensor dy\left(\fracD\phi,\fracD\phi\right) +
		dz\tensor dz\left(\fracD\phi,\fracD\phi\right) \\
	&=& r^2 \sin^2\theta \sin^2\phi
		+ r^2 \cos^2\theta \sin^2\phi
		+ r^2 \cos^2\phi \\
	&=& r^2 (\sin^2\theta + \cos^2\theta \sin^2\phi) \sin^2\phi
		+ r^2 \cos^2\phi \\
	&=& r^2 \sin^2\phi + r^2 \cos^2\phi \\
	&=& r^2.
\end{eqnarray*}
So, we have
\begin{eqnarray*}
	g &=& g\left(\fracD\theta, \fracD\theta\right) d\theta\tensor d\theta
		+ g\left(\fracD\phi, \fracD\phi\right) d\phi\tensor d\phi \\
	&=& (R + r\cos\phi)^2 d\theta\tensor d\theta + r^2 d\phi\tensor d\phi.
\end{eqnarray*}
\end{answer}

%% ================================================================
\newpage
\subsection{Geometry homework 11}

%% ----------------------------------------------------------------
\subsubsection{Geometry homework 11 problem 2}
\label{sec:hw_11_2}

Let $x,y,z$ denote linear coordinates for $\R^3$.  Let
$$
	\nabla = \colvecthree{\slashDx}{\slashDy}{\slashDz}, \quad
	\vecx = \colvecthree{x}{y}{z}, \quad
	\vecA = \colvecthree{A}{B}{C}, \qand
	dV = dx \wedge dy \wedge dz.
$$
Show that for the exterior derivative
\begin{eqnarray*}
	(a) \qquad d(\vecA \cdot d\vecx) &=& (\nabla\times \vecA)\ctt dV \\
	(b) \qquad d(\vecA \ctt dV) &=& \div(\vecA) dV.
\end{eqnarray*}

%\begin{answer}[Inefficient answer.]
%This solution is brute-force and obvious, and sufficient for a homework
%assignment, but time-consuming on an exam.  I present this solution because it
%is nonetheless illustrative.
%
%For part (a), we can simply write out the left-hand and right-hand sides.
%Using the abbreviated notation $ A_x = \fracD{A}{x},$ etc., we have on the
%left-hand side
%\begin{eqnarray*}
%	d(\vecA \cdot d\vecx) &=& dA \wedge dx + dB \wedge dy + dC \wedge dz \\
%	\\
%	&=& [A_x dx + A_y dy + A_z dz] \wedge dx \\
%	&+& [B_x dx + B_y dy + B_z dz] \wedge dy \\
%	&+& [C_x dx + C_y dy + C_z dz] \wedge dz \\
%	\\
%	&=& A_y dy \wedge dx + A_z dz \wedge dx \\
%	&+& B_x dx \wedge dy + B_z dz \wedge dy \\
%	&+& C_x dx \wedge dz + C_y dy \wedge dz \\
%	\\
%	&=&-A_y dx \wedge dy + A_z dz \wedge dx \\
%	&+& B_x dx \wedge dy - B_z dy \wedge dz \\
%	&-& C_x dz \wedge dx + C_y dy \wedge dz \\
%	\\
%	&=& (B_x - A_y) dx \wedge dy
%	+   (C_y - B_z) dy \wedge dz
%	+   (A_z - C_x) dz \wedge dx.
%\end{eqnarray*}
%
%On the right-hand side, recall that we defined the contraction operator in
%sections \ref{sec:vs_i_X} and \ref{sec:i_X}.  Let
%$$
%	\vecv = \colvecthree{v_1}{v_2}{v_3} \qand
%	\vecw = \colvecthree{w_1}{w_2}{w_3}.
%$$
%Then by definition of the contraction operator we have
%$$
%	i_{\nabla \times \vecA} dV(\vecv, \vecw) =
%	dx \wedge dy \wedge dz(\nabla \times \vecA, \vecv, \vecw).
%$$
%Recall that $dx(\vecv) = v_1$, etc. since $dx$, $dy$, and $dz$ are dual to the
%standard basis in $\R^3$; they are coordinate-selector functions.  as mentioned
%in section \ref{sec:basis_and_dual_space} and [xxx].  Abbreviate
%$\vecu = \nabla \times \vecA$.
%Then
%\begin{eqnarray*}
%	dx \wedge dy \wedge dz(\vecu, \vecv, \vecw)
%	&=& dx(\vecu) dy(\vecv) dz(\vecw) - dx(\vecu) dy(\vecw) dz(\vecv) \\
%	&+& dx(\vecw) dy(\vecu) dz(\vecv) - dx(\vecw) dy(\vecv) dz(\vecu) \\
%	&+& dx(\vecv) dy(\vecw) dz(\vecu) - dx(\vecv) dy(\vecu) dz(\vecw) \\
%	\\
%	&=& u_1 v_2 w_3 - u_1 w_2 v_3
%	+ u_1 v_2 w_3 - u_1 w_2 v_3
%	+ u_1 v_2 w_3 - u_1 w_2 v_3.
%\end{eqnarray*}
%But
%$$
%	u_1 = dx(\nabla \times \vecA)
%	= dx\colvecthree{C_y - B_z}{A_z-C_x}{B_x - A_y} = C_y - B_z,
%$$
%etc.  So
%\begin{eqnarray*}
%	dx \wedge dy \wedge dz(\nabla \times \vecA, \vecv, \vecw)
%	&=& (C_y - B_z) v_2 w_3 - (C_y - B_z) v_3 w_2 \\
%	&+& (A_z - C_x) v_3 w_1 - (B_x - A_y) v_2 w_1 \\
%	&+& (B_x - A_y) v_1 w_2 - (A_z - C_x) v_1 w_3 \\
%	\\
%	&=& (B_x - A_y) (v_1 w_2 - v_2 w_1) \\
%	&+& (C_y - B_z) (v_2 w_3 - v_3 w_2) \\
%	&+& (A_z - C_x) (v_3 w_1 - v_1 w_3) \\
%	\\
%	&=& (B_x - A_y) dx \wedge dy (\vecv, \vecw) \\
%	&+& (C_y - B_z) dy \wedge dz (\vecv, \vecw) \\
%	&+& (A_z - C_x) dz \wedge dx (\vecv, \vecw).
%\end{eqnarray*}
%Now, $\vecv$ and $\vecw$ were arbitrary, so as 3-linear functions,
%$$
%	i_{\nabla \times \vecA} dV = (B_x - A_y) dx \wedge dy
%		+ (C_y - B_z) dy \wedge dz
%		+ (A_z - C_x) dz \wedge dx
%	= d(\vecA \cdot d\vecx).
%$$
%
%For part (b), [need to type this in.]
%\end{answer}

\begin{answer}
The key is that $dV$ is the \bfidx{volume operator}, and as such is represented
by a determinant.  (Without this observation this problem can be become a
time-consuming mess.)

%time-consuming mess, as the previous solution demonstrates.)

For part (a), first we compute $\vecA \cdot d\vecx$.  Using the
shorthand notation $A_x = \DA / \Dx$, etc., we have
$$
	\vecA \cdot d\vecx = A dx + B dy + C dz.
$$
Next we find $d$ of this.  Recalling that [xxx cite]
$$
	dA = A_x dx + A_y dy + A_z dz,
$$
etc., we have
\begin{eqnarray*}
	d(\vecA \cdot d\vecx)
	&=& (A_x dx + A_y dy + A_z dz) \wedge dx \\
	&+& (B_x dx + B_y dy + B_z dz) \wedge dy \\
	&+& (C_x dx + C_y dy + C_z dz) \wedge dz \\
	\\
	&=& (A_y dy + A_z dz) \wedge dx \\
	&+& (B_x dx + B_z dz) \wedge dy \\
	&+& (C_x dx + C_y dy) \wedge dz \\
	\\
	&=& A_y dy \wedge dx + A_z dz \wedge dx \\
	&+& B_x dx \wedge dy + B_z dz \wedge dy \\
	&+& C_x dx \wedge dz + C_y dy \wedge dz \\
	%\\
	%&=&-A_y dx \wedge dy + A_z dz \wedge dx \\
	%&+& B_x dx \wedge dy - B_z dy \wedge dz \\
	%&-& C_x dz \wedge dx + C_y dy \wedge dz \\
	\\
	&=& (B_x - A_y) dx \wedge dy
	+ (C_y - B_z) dy \wedge dz
	+ (A_z - C_x) dz \wedge dx.
\end{eqnarray*}

We also need to write out $\nabla \times \vecA$.  This is
$$
	\nabla \times \vecA
	= \left|\begin{matrix}
		\hat\vecx & \hat\vecy & \hat\vecz \\
		\slashDx  & \slashDy  & \slashDz \\
		A         & B         & C
		\end{matrix}\right|
	= \colvecthree{C_y - B_z}{A_z - C_x}{B_x - A_y}
$$
Let $\vecv, \vecw \in \R^3$.  Then, since $dV$ of $m$ vectors in
$\R^m$ is the volume of the parallepiped spanned by them, and using the
definition of the contraction operator from [xxx cite], we have
\begin{eqnarray*}
	(\nabla\times\vecA)\ctt dV(\vecv, \vecw) = \det\begin{pmatrix}
		\nabla\times\vecA \\ \hline \vecv \\ \hline \vecw
		\end{pmatrix}
	&=& \left|\begin{matrix}
		C_y - B_z & A_z - C_x & B_x - A_y \\
		v_1 & v_2 & v_3 \\
		w_1 & w_2 & w_3 \\
	\end{matrix}\right| \\
	&=& (C_y - B_z)(v_2 w_3 - v_3 w_2) \\
	&+& (A_z - C_x)(v_3 w_1 - v_1 w_3) \\
	&+& (B_x - A_y)(v_1 w_2 - v_2 w_1).
\end{eqnarray*}
At this point we should simply \emph{see} that
$$
	v_2 w_3 - v_3 w_2 = dy \wedge dz (\vecv, \vecw),
$$
etc. so
\begin{eqnarray*}
	(\nabla\times\vecA)\ctt dV
	&+& (B_x - A_y)dx \wedge dy \\
	&+& (C_y - B_z)dy \wedge dz \\
	&=& (A_z - C_x)dz \wedge dx.
\end{eqnarray*}
as desired.

\begin{center}* * *\end{center}

For part (b), first we will find out what $\vecA \ctt dV$ is, then compute $d$ of
it, then compare that to $\div(A) dV$.

As in part (a), let $\vecv, \vecw \in \R^3$.  Then
\begin{eqnarray*}
	\vecA \ctt dV(\vecv, \vecw) = \det\begin{pmatrix}
		\vecA \\ \hline \vecv \\ \hline \vecw
		\end{pmatrix}
	&=& \left|\begin{matrix}
		A & B & C \\
		v_1 & v_2 & v_3 \\
		w_1 & w_2 & w_3 \\
	\end{matrix}\right| \\
	&=& A(v_2 w_3 - v_3 w_2)
	+ B(v_3 w_1 - v_1 w_3)
	+ C(v_1 w_2 - v_2 w_1).
\end{eqnarray*}
Again observe that
$$
	v_2 w_3 - v_3 w_2 = dy \wedge dz (\vecv, \vecw),
$$
etc., so
$$
	\vecA \ctt dV(\vecv, \vecw)
	= A\, dy \wedge dz + B\, dz \wedge dx + C\, dx \wedge dy.
$$
Then, it is easy to compute $d$ of this [xxx cite justifcations for steps]:
\begin{eqnarray*}
	d(\vecA \ctt dV(\vecv, \vecw))
	&=& d(A\, dy \wedge dz + B\, dz \wedge dx + C\, dx \wedge dy) \\
	&=& d(A\, dy \wedge dz) + d(B\, dz \wedge dx) + d(C\, dx \wedge dy)
\end{eqnarray*}
Now [xxx cite], and using the shorthand $A_x = \DA/\Dx$, etc.,
$$
	dA = A_x dx + A_y dy + A_z dz,
$$
etc.  When we FOIL this out, we'll get 9 terms, but 6 of them are zero since
$dx \wedge dx = 0$ etc.  So,
\begin{eqnarray*}
	d(\vecA \ctt dV(\vecv, \vecw))
	&=& A_x dx \wedge dy \wedge dz
	+ B_y dx \wedge dy \wedge dz
	+ C_z dx \wedge dy \wedge dz \\
	&=& (A_x + B_y + C_z) dx \wedge dy \wedge dz \\
	&=& \div(\vecA) dV.
\end{eqnarray*}
\end{answer}

%% ================================================================
\newpage
\subsection{Geometry exam 2}

%% ----------------------------------------------------------------
\subsubsection{Geometry exam 2 problem 1}

Let $\vecv$ denote the vector field on $\R^2$ given by
$$
\vecv\pickpipet{\vvectwo{x}{y}} =
\vvectwo{x^2 y}{-y}\pickpipet{\vvectwo{x}{y}}.
$$
Compute the flow of this vector field.
\begin{answer}
We have
$$
\vvectwo{\dot x}{\dot y} =
\vvectwo{x^2 y}{-y}
$$
which immediately gives $\dot y = y$ with solution $y = y_0 e^t$.
Plugging this into the first equation gives
\begin{eqnarray*}
	 \fracdtwo{x}{t} &=& x^2 y_0 e^t \\
	 x^{-2}dx &=& y_0 e^t dt \\
	 \int x^{-2}dx &=& y_0 e^t + C \\
	 -1/x &=& y_0 e^t + C \\
\end{eqnarray*}
\ldots finish typing up later.
\end{answer}

%% ----------------------------------------------------------------
\subsubsection{Geometry exam 2 problem 2}
\label{sec:ex_2_2}

Recall that we can identify the tangent space of $\S^2$ at a point $\vecq$ with
$\{(\vecv,\vecq): \vecv \in \R^3, \vecq \perp \vecv\}$.  Let $\eta$ denote the
1-form (or 1-tensor) which is given by
$$
\eta((\vecq, \vecv)) = v_1
$$
where $v_1$ is the first component of $\vecv$.

(a) Express $\eta$ in terms of spherical coordinates $\phi, \theta$ for $\S^2$,
where $0 < \phi < \pi$ and $0 < \theta < 2\pi$.

(b) Find two reasons that $d\eta = 0$.

\begin{answer}
See example \ref{ex:decompose_alt_one_tensor}.  Here we are doing the same
thing, but with varying coefficients in the sense of section \ref{sec:loaf}.
Using that approach, we have
$$
\eta = \eta\left(\fracD \theta\right) d\theta
+ \eta\left(\fracD \phi\right) d\phi.
$$
The point $\vecq$ is
$$ \vecq=\begin{pmatrix}
	\cos\theta \sin\phi \\
	\sin\theta \sin\phi \\
	\cos\phi
\end{pmatrix}.$$
Thus
$$
\D/\D\theta =
\vecq=\begin{pmatrix}
	-\sin\theta \sin\phi \\
	\cos\theta \sin\phi \\
	0
\end{pmatrix},
\qquad
\D/\D\phi =
\vecq=\begin{pmatrix}
	\cos\theta \cos\phi \\
	\sin\theta \cos\phi \\
	-\sin\phi
\end{pmatrix}.
$$
Now, $\eta$ extracts the first coordinate, so
$$\eta = -\sin\theta \sin\phi d\theta + \cos\theta \cos\phi d\phi.$$

For part (b), we might like to say that $d^2=0$.  However, it is an unfortunate
fact [xxx xref to the appropriate section once I write it] that not everything
that begins with $d$ is exact.  [in the xref, say whether $d\theta$ and $d\phi$
are in fact exact.  I guess that's the same as asking whether $\theta$ and
$\phi$ are 0-forms.  Also in the xref, talk about $dA$, etc.] Let's compute
$d\eta$ using the above expression.  We use the fact that $d$ is an
antiderivation, so it is linear and has a product rule.  We get
\begin{eqnarray*}
d\eta &=& -d(\sin\theta \sin\phi d\theta) + d(\cos\theta \cos\phi d\phi)\\
&=& -d(\sin\theta) \sin\phi d\theta
	-\sin\theta d(\sin\phi) d\theta
	-\sin\theta \sin\phi dd\theta \\
&&+ d(\cos\theta) \cos\phi d\phi
	+ \cos\theta d(\cos\phi) d\phi
	+ \cos\theta \cos\phi dd\phi\\
\end{eqnarray*}
Now, $dd$ of anything is zero, and $d\theta \wedge d\theta$ = 0 and likewise for
$d\phi \wedge d\phi$, so we have
\begin{eqnarray*}
d\eta
&=& -\sin\theta \cos\phi d\phi \wedge d\theta
	- \sin\theta d\theta \cos\phi \wedge d\phi
\end{eqnarray*}
But $d\phi \wedge d\theta = -d\theta \wedge d\phi$ so this is zero.
\end{answer}

%% ----------------------------------------------------------------
\subsubsection{Geometry exam 2 problem 3}
\label{sec:ex_2_3}

Consider the manifold $M = \{(x,y,z) \in \R^3: z = x^2 + y^2\}$ with the
upward-pointing orientation.

(a) Compute the area form of $M$ in $x, y$ coordinates.

(b) Let
$$
	h: M \to \R^2 : \colvecthree x y z \longmapsto \colvectwo u v
	= \colvectwo{xy}{z}.
$$
If $du \wedge dv$ denotes the standard area form on $\R^2$, then
$h^*(du \wedge dv)$ is a two-form on $M$.  Express this two-form in $x,y$
coordinates.

\begin{answer}[Answer to part (a).]
Proceeding as in [xxx xref], we note that $M$ is a warp of the plane, so the
graph coordinates $x,y$ are appropriate.  Then a point $\vecq \in M$ is, in the
$x,y$ parameterization,
$$
	\vecq = \colvecthree{x}{y}{x^2+y^2}.
$$
Since we are using $x,y$ coordinates, a basis for the tangent space of $M$ is
$\{\slashDx, \slashDy\}$, and a basis for the cotangent space of $M$ is $\{dx,
dy\}$.  As in [xxx], we compute
$$
	\fracDx = \colvecthree{1}{0}{2x} \qand
	\fracDy = \colvecthree{0}{1}{2y}.
$$
As in example \ref{ex:decompose_alt_two_tensor}, for the area form $dA$ we have
$$
	dA = dA\left(\fracDx, \fracDy\right) dx ^ \wedge dy.
$$
so it remains to find out what $dA(\slashDx, \slashDy)$ is.

Unlike in example [xxx], here we need a little trick: from vector calculus we
know that the area of the parallelogram spanned by two vectors is the signed
magnitude of their cross product.  The cross product, in turn, is just the
normal vector to the surface:
$$
	\vecn = \fracDx \times \fracDy =
		\left| \begin{matrix}
			\hat\vecx & \hat\vecy & \hat\vecz \\
			1 & 0 & 2x \\
			0 & 1 & 2y \\
		\end{matrix}\right|
	= \colvecthree{-2x}{-2y}{1}
$$
The magnitude of this is
$$
	\|\vecn\| = \sqrt{4x^2 + 4y^2 + 1}.
$$
To get the sign correct, use the right-hand rule:  moving from $\slashDx$ to
$\slashDy$ we would expect $\vecn$ to point upward, so we want the positive
sign, so we can leave the sign as is.  In conclusion, we have
$$
	dA = (4x^2+4y^2+1)^{1/2} dx ^ \wedge dy.
$$
\end{answer}

\begin{answer}[Answer to part (b).]
We have
$$
	h: M \to \R^2 : \colvecthree x y z \longmapsto \colvectwo u v
	= \colvectwo{xy}{z}.
$$
Then using [xxx write and xref; also include a pullback diagram], we have
\begin{eqnarray*}
	h^*(du\wedge dv) &=& h^*(du) \wedge h^*(dv)
		= d(u\circ h) \wedge d(v\circ h) \\
	&=& d(xy) \wedge dz = (y\,dx + x\,dy) \wedge dz
		= (y\,dx + x\,dy) \wedge d(x^2+y^2) \\
	&=& (y\,dx + x\,dy) \wedge(2x\,dx + 2y\,dy)
		= 2y^2 dx\wedge dy - 2x^2 dx\wedge dy \\
	&=& (2y^2-2x^2)dx\wedge dy.
\end{eqnarray*}

Note that the key to this problem is pulling back correctly; the rest is
straightforward computation.
\end{answer}

%% ================================================================
\newpage
\subsection{Geometry final review}

%% ----------------------------------------------------------------
\subsubsection{Geometry final review problem 1}

(a) Calculate the derivative of $\det: \mcL(\R^n) \to \R$.

(b) Determine the set of critical points for $\det$.

\begin{answer}[Answer to part (a).] This was done in homework 1, section
\ref{sec:hw_1_2d}.  I remark that the derivation of the formula required
several non-obvious tricks, so for an exam one would perhaps be advised to
simply memorize the formulas
$$ D\det|_A(B) = \det(A) \tr(A^{-1}B) \qquad (\det(A) \ne 0)$$
and
$$ D\det|_A(B) = \tr(\adju(A)B) \qquad \textrm{(general $A$)}.$$
\end{answer}

\begin{answer}[Answer to part (b).] Critical points $A$ of $\det$ occur when
$D\det|_A$ is not surjective.  Since the target space $\R$ has dimension 1 over
$\R$, the rank of $D\det|_A$ is either 0 or 1.  We need to find points $A$ such
that $D\det|_A$ is zero.  This is precisely when $\adju(A)$ is 0.  That is, if
$A$ is $n\times n$, then $\adju(A)$ is zero when all $(n-1) \times (n-1)$
submatrices of $A$ are singular.

This result is perhaps surprising:  we might have guessed that the critical
points of $\det$ would be all the singular matrices, i.e. those of rank less
than $n$.  However, the above tells us that the matrices of rank $n-1$ are also
regular points for the $\det$ function.  It is the matrices of rank $n-2$ or
less which are critical.  \end{answer}

%% ----------------------------------------------------------------
\subsubsection{Geometry final review problem 2}

(a) State the regular value theorem.

(b) Prove that $\rSL(n,\R) = \{A \in \mcL(\R^n) : \det(A) = 1\}$ is an embedded
submanifold of $\mcL(\R^n)$.  What is its dimension?

(c) Calculate the tangent space to $\rSL(\R^n)$ at the identity matrix.

\begin{answer}[Answer for part (a)]
See theorem\index{regular value theorem} \ref{thm:regular_value} in section
\ref{sec:regular_value}: if $f:M \to N$ is a map of manifolds, and if $\vecc
\in N$ is a regular value of $f$, then $f^{-1}(\vecc)$ is either the empty set,
or an embedded submanifold of $M$.  In the latter case, $f^{-1}(\vecc)$ has
dimension $m-n$.
\end{answer}

\begin{answer}[Answer for part (b)] Apply the regular value theorem, with
$M=\mcL(\R^n)$, $f=\det$, $N=\R$, and $\vecc=1$.  It remains only to show that
$1$ is a regular value of $\det$.  In turn, $1$ is a regular value of $\det$ if
all $A$ in $\det^{-1}(A)$ are regular points.  This would mean $D\det|_A$ is
surjective when $\det(A) = 1$.  Since $\R$ is a one-dimensional vector space
over $\R$, it suffices to show that $D\det|_A$ is not identically zero when
$\det(A) = 1$.  From the previous problem, we can use
$$
	D\det|_A(B) = \det(A) \tr(A^{-1}B).
$$
Since $\det(A) = 1$, $D\det|_A(B)$ won't vanish for that reason.  The only
remaining task is to argue that for fixed $A$, the $B$-linear function
$\tr{A^{-1}B}$ is not identically zero.  If it were identically zero, it would
be true for $B=A$.  But $\tr{A^{-1}A} = \tr{I} = n \ne 0$ where $n$ is the
dimension of $A$.  This is a contradiction, so $D\det|_A$ is not identically
zero.

By the regular value theorem, since the source space $\mcL(\R^n)$ has dimension
$n^2$ and the destination space $\R$ has dimension 1, $\rSL(\R^n)$ has
dimension $n^2-1$. \end{answer}

\begin{answer}[Answer for part (c) (Pickrell)]
As discussed in remark \ref{rem:rvt_short_exact}, the tangent space to
$\rSL(\R^n)$ at $I$ is $\ker(D\det|_I)$.  Now, $$D\det|_I(B) = tr(B)$$ so the
tangent space at $I$ consists of the matrices with zero trace.
\end{answer}

%% ----------------------------------------------------------------
\subsubsection{Geometry final review problem 3}
\label{sec:fr_3}

(a) Determine the critical points and critical values for the function
$$ F:\R^3 \to \R^2: \colvecthree x y z \mapsto \colvectwo{z^2-y^2-x^2}{z-y}.$$

(b) Show that \emph{all} of the level sets of $F$ are embedded submanifolds
(of various dimensions).

\begin{answer}[Answer for part (a)]
From section \ref{sec:regular_value} we know that $F$ is critical at all points
$\vecq \in \R^3$ such that $DF$ has rank less than 2.
We can use the adjugate criterion (proposition \ref{prop:adjugate_criterion})
to find such points.  The Jacobian of $F$ is
$$
	DF = \begin{pmatrix}
		-2x & -2y & 2z \\
		0 & -1 & 1 \\
	\end{pmatrix}.
$$
The determinants of $2 \times 2$ submatrices are
$$ 2x, -2x, \qand 2z - 2y.$$
The critical points of $F$ are where these are simultaneously zero, i.e. when
$$ x=0 \qand z=y.$$
Inserting these into the expression for $F$ gives
$$
F\colvecthree{0}{y}{y} = \colvectwo{0}{0}.
$$
\end{answer}

\begin{answer}[Answer for part (b)] By the regular value theorem, the level
sets for regular values are embedded submanifolds.  It remains to show that the
level set of the critical value $(0,0)$ is also an embedded submanifold.  But
that level set is just the line $z=y$.  This is certainly an embedded
submanifold of $\R^3$, since the inclusion map is globally 1-1.  \end{answer}

\begin{rem}[Pickrell] This does not contradict the regular value theorem.  The
regular value theorem says that if $\vecc$ is a regular value of the map $f:M
\to N$, then $f^{-1}(\vecc)$ is an embedded submanifold of $M$.  Here we have
an example of a critical value, the inverse image of which also happens to be
an embedded submanifold.  This shows that the converse to the regular value
theorem does not hold.
\end{rem}

%% ----------------------------------------------------------------
\subsubsection{Geometry final review problem 4}

Determine whether the function $$ \chi: \S^2 \to \R^2 : \colvecthree x y z
\mapsto \colvectwo{x^2-y^2}{2xy+z}$$ is a coordinate for $\S^2$ in the
neighborhood of the north pole.

\begin{answer} It would be tempting to compute the rank of $\chi$ by taking the
$2 \times 3$ Jacobian, but note that we have $\S^2 \into \R^3 \to \R^2$, and
$\chi$ as shown above is a map from $\R^3$ to $\R^2$.  We need to somehow
restrict $\chi$ to $\S^2$.

My approach is that a coordinate chart on $\S^2$ does give a two-dimensional
surface.  Consider a known-good coordinate chart, namely the graph coordinates
$x,y$.  If $\chi$ were also a coordinate chart, then the transition function
from graph coordinates to $\chi$ would be a 1-1 function.  [xxx include a
picture here, with two coordinate charts mapping out of the upper hemisphere.]
Let $\psi$ be the transition function:
\begin{eqnarray*}
	\psi\colvectwo{x}{y} &=& \colvectwo{x^2 - y^2}{2xy + (1-x^2-y^2)^{1/2}}
\end{eqnarray*}
Now, $\psi$ is a non-linear function, but we can use the inverse function
theorem.  That is, by linearizing, we can use the Jacobian test:
\begin{eqnarray*}
	D\psi &=&
		\begin{pmatrix}
			2x & -2y \\
			2y - \frac{x}{(1-x^2-y^2)^{1/2}} &
			2x - \frac{y}{(1-x^2-y^2)^{1/2}}
		\end{pmatrix}
\end{eqnarray*}
At the point $x=y=0$, this is the zero matrix.  So, $\psi$ is not 1-1
at the north pole, which is all that remained to be shown.
\end{answer}

%% ----------------------------------------------------------------
\subsubsection{Geometry final review problem 5}
\label{sec:fr_5}

(a) Compute the tangent bundle of the manifold
$$ N  = \{(x,y,z) \in \R^3 : z^2 - x^2 - y^2 = 1\}.$$

(b) What is the dimension of $TN$, as a manifold?

\begin{rem*}{} [xxx insert picture here.] To get an idea for what this looks
like,  fix $y=0$ and graph that cross-section on the plane.  Then by rotational
symmetry you'll see that near the origin $x=y=0$, it looks like a paraboloid,
but away from the origin $x=y=0$ it straightens out.  So, it's like a cone with
a rounded end.  \end{rem*}

\begin{answer}[First answer to part a] This solution uses the technique shown
in section \ref{sec:geom_tang}.  This uses vector calculus and linear algebra,
and is a bit tedious.  The next, more efficient, solution uses the technique of
section \ref{sec:compute_tang}.

The tangent plane is perpendicular to the normal vector.  To get the normal
vector, in turn, we can observe that $N$ is the level set of 1 and $F(x,y,z) =
z^2-x^2-y^2$.  As in section \ref{sec:level_sets}, the equation for the normal
is given by $\grad F$.  This is $$ \grad F = \begin{pmatrix}-2x \\ -2y \\
2z\end{pmatrix}.$$

Let $\vecq=(x,y,z)$ be an arbitrary point on $N$.  Then the normal vector
footed at $\vecq$ is $\vecnq$, as given above.  Any element $\vecvq$ of the
tangent space $TM_\vecq$ is perpendicular to $\vecnq$, i.e.  $$ \vecnq \cdot
\vecvq = 0.$$ This means
$$
\begin{pmatrix} n_1 &  n_2 &  n_3 \end{pmatrix}
\begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}
=
\begin{pmatrix} -2x &  -2y &  2z \end{pmatrix}
\begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}
= (-2x)v_1 + (-2y)v_2 + (2z)v_3 = 0.
$$
This is a linear system, where $\vecvq$ is the variable we want to solve for.
The coefficients are $(x,y,z)$.  These look like variables, but they're
constant for each point $\vecq = (x,y,z)$.  Since the matrix has rank 1, and we
are computing its kernel, from the rank-nullity theorem we expect kernel
dimension 2.  So, since there are two degrees of freedom, we first choose
$v_1=1$ and $v_2=0$, then $v_1=0$ and $v_2=1$.  It remains to compute what the
$v_3$ coordinate must be in each case.  The factor of $-2$ is irrelevant, so
the equation to be solved is
$$ x v_1 + y v_2 - z v_3 = 0.$$
For the first and second choices, respectively, we have
$$ x - z v_3 = 0 \qand y - z v_3 = 0.$$  These give $v_3 = x/z$ and
$y/z$, respectively, and so a basis for the tangent space at $\vecq$
is
$$
\left\{ (1, 0, x/z), (0, 1, y/z) \right\}
\qqor
\left\{ (z, 0, x), (0, z, y) \right\}.
$$

Let's do a quick sanity check.  Pick $x=0$ and $y=1$ on $N$.  Then
$z^2=x^2+y^2+1$, i.e. $z^2 = 2$.  Take the upper point $\vecq=(x,y,z) =
0,1,\sqrt 2$.  Then the tangent space at $(0,1,\sqrt 2)$ should be spanned by
$(\sqrt 2, 0, 0)$ and $(\sqrt 0, 2, 1)$.  This makes sense: [xxx need a picture
here.]
As a second check, take $x=y=0$ and $z=1$.  Then the basis vectors for the
tangent space are just $(1,0,0)$ and $(0,1,0)$, which is what we would
intuitively expect.
\end{answer}

\begin{answer}[Second answer to part a] This solution uses the technique shown
in section \ref{sec:compute_tang}.  From the graph, we know that $M$ looks like
a cone with a rounded end.  In particular, it looks like a deformation
of the plane.  So, the key observation is to parameterize $M$ using the
graph coordinates $x$ and $y$.  (How do we know to use these coordinates?
We have to intuitively know, although we haven't had algebraic topology
until the next semester, that this manifold looks like a plane.)
Since we are parameterizing using $x$ and $y$, the tangent space will be
spanned by $\slashDx$ and $\slashDy$.  A point $\vecq \in M$ has coordinates
$$
	\vecq = \colvecthree{x}{y}{\sqrt{x^2+y^2+1}}.
$$
As in section \ref{sec:compute_tang}, we apply $\slashDx$ and $\slashDy$
to the rectangular coordinates of $\vecq$ to obtain
$$
	\fracDx = \colvecthree{1}{0}{\frac{x}{\sqrt{x^2+y^2+1}}}
	= \colvecthree{1}{0}{x/z}, \quad
	\fracDy = \colvecthree{0}{1}{\frac{y}{\sqrt{x^2+y^2+1}}}
	= \colvecthree{1}{0}{x/z}.
$$
We can (and should) do the sanity check as in the first solution.  Note that
the $z$ in the denominator is non-singular, since all points on this surface
have non-zero $z$ (in fact, $z \ge 1$).
\end{answer}

\begin{center} * * *\end{center}

\begin{answer}[Answer to part b]
We have found that, at all points $\vecq$ of $N$, the tangent space $TN_\vecq$
has dimension 2, i.e. it is a tangent plane.  Thus, $TN$ has dimension 2.
Furthermore, the dimension of the tangent space is the same as the dimension of
the manifold [xxx write and xref].
\end{answer}

%% ----------------------------------------------------------------
\subsubsection{Geometry final review problem 6}

(a) What is a trivial vector bundle?  Give an example of a vector bundle which
is not trivial.

(b) Suppose that $M$ is an embedded submanifold of a manifold $X$.  How is the
normal bundle of $M$ inside of $X$ defined?  What is its dimension as a
manifold (in terms of the dimensions of $X$ and $M$)?

(c) Consider $\S^n \subseteq \R^{n+1}$.  Show that the normal bundle is a
trivial bundle.

\begin{answer}[Answer to part (a).] xxx move up to main body of paper: I
believe the following to be correct: Let $E$ be a $k$-dimensional bundle over a
manifold $M$.  A (global) trivialization is a diffeomorphism $E \cong M \times
\R^k$ which is linear on fibers.  I believe that, in practice, it suffices to
construct smoothly varying, linearly independent sections which are defined
globally on $M$.

xxx include the additional commuting-diagram criterion.

xxx move up to main body: surface of revolution and $\S^1$ are examples of
manifolds with trivial tangent bundles.  $\S^2$, on the other hand, has a
non-trivial tangent bundle.  Proof of this requires appealing to the
sphere-combing theorem, but we can provide some motivation by noting that
spherical coordinates are not defined at the poles, graph coordinates are only
defined on hemispheres, stereographic projections are defined everywhere except
\emph{one} pole, etc.  \end{answer}

\begin{answer}[Answer to part (b).]
\end{answer}

\begin{answer}[Answer to part (c).]
\end{answer}

%% ----------------------------------------------------------------
\subsubsection{Geometry final review problem 7}

Compute the flow for the vector field on $\S^2$ given by
$$ \vecv \pickpipet{\colvecthree{x}{y}{z}} = \colvecthree{-z}{0}{x}.$$

\begin{answer}
We are given
$$
	\colvecthree{\dot x}{\dot y}{\dot z}
	= \colvecthree{-z}{0}{x}.
$$
Ideally we would want univariate differential equations.  Clearly we already
have $\dot y = 0$ from which $y = y_0$ for all $t$.  Differentiating twice
gives
$$
	\colvecthree{\ddot x}{\ddot y}{\ddot z}
	= \colvecthree{-\dot z}{0}{\dot x}
	= \colvecthree{-x}{0}{-z}.
$$
The univariate ODEs $\ddot x + x = 0$ and $\ddot z + z = 0$ have solutions
\begin{eqnarray*}
	x &=& a \cos t + b \sin t \\
	z &=& c \cos t + d \sin t.
\end{eqnarray*}
Putting in initial conditions gives $x(0) = a$ and $z(0) = c$,
so
\begin{eqnarray*}
	x &=& x_0 \cos t + b \sin t \\
	z &=& z_0 \cos t + d \sin t.
\end{eqnarray*}
Doing this required only looking at $x$ and $z$ independently.  Now, to find
$b$ and $d$, we can use the coupling of $x$ and $z$ from the original equation
to obtain
$$
\begin{matrix}
	\dot x &=& -x_0 \sin t + b \cos t &=& -z &=& -z_0 \cos t - d \sin t \\
	\dot z &=& -z_0 \sin t + d \cos t &=&  x &=&  x_0 \cos t + b \sin t
\end{matrix}
$$
Equating like terms (namely, $\sin$ and $\cos$) gives $d = x_0$
and $b = -z_0$, so
\begin{eqnarray*}
	x &=& x_0 \cos t - z_0 \sin t \\
	y &=& y_0 \\
	z &=& z_0 \cos t + x_0 \sin t.
\end{eqnarray*}
or
$$
	\begin{pmatrix} x \\ y \\ z \end{pmatrix}
	=
	\begin{pmatrix}
	\cos t & 0 & -\sin t \\
	0 & 1 & 0 \\
	\sin t & 0 &  \cos t \\
	\end{pmatrix}
	\begin{pmatrix} x_0 \\ y_0 \\ z_0 \end{pmatrix}.
$$
Note that this motion is rotating about the $y$ axis.
\end{answer}

%%% ----------------------------------------------------------------
%\subsubsection{Geometry final review problem 8}

%Draw various tori embedded in $\R^3$.  Draw the trajectories for the negative
%gradient flow for the function $f=z$ on your tori, paying special attention to
%the critical points.

%\begin{answer}
%[xxx insert drawings here]
%\end{answer}

%% ----------------------------------------------------------------
\subsubsection{Geometry final review problem 9}

(a) Suppose that $M=\R^3$.  Explain the sense in which $d:\Omega^k(M) \to
\Omega^{k+1}(M)$ (the exterior derivative) is equivalent to the gradient, curl,
and divergence.

(b) Suppose that $\vecA$ is a vector field on $\R^m$.  Show that
$$
d(\vecA \ctt dx_1 \wedge \ldots \wedge dx_m) = \div(\vecA) dx_1 \wedge \ldots
\wedge dx_m.$$
(Hence the exterior derivative on $m-1$ forms in $\R^m$ is equivalent to the
divergence.)

\begin{answer}[Answer to part (a).]
\end{answer}

\begin{answer}[Answer to part (b).]
This is the same as problems \ref{sec:hw_11_2} and \ref{sec:hw_11_2} part (b),
except generalized from $3$ to $m$ dimensions.  Proceeding as in the $m=3$
case, we let
$$
	\vecA = \colvecthree{A_1}{\vdots}{A_m}.
$$
Then, again observing that $dV$ of $m$ vectors $\vecv_1, \ldots, \vecv_m$
is equal to the area of the parallelepiped spanned by them, we have
$$
	\vecA ctt dx_1 \wedge \ldots \wedge dx_m =
	\left|\begin{matrix}
		A_1    & \cdots & A_m    \\
		v_{21} & \cdots & v_{2m} \\
		\vdots &        & \vdots \\
		v_{m1} & \cdots & v_{mm} \\
	\end{matrix}\right|
	= \sum_{i=1}^m A_i
		dx_1 \wedge\ldots\wedge \widehat{dx_i}\wedge\ldots\wedge \ldots dx_m
$$
where the\index{hat symbol $\widehat\cdot$} $\widehat\cdot$ notation
indicates omission.  Recall [xxx cite] that
$$
	dA_i = \sum_{j=1}^m \frac{\DA_i}{\Dx_j} dx_j.
$$
Then using the linearity of the derivative,
\begin{eqnarray*}
	d(\vecA \ctt dx_1 \wedge \ldots \wedge dx_m)
	&=& \sum_{i=1}^m d\left(A_i
		dx_1 \wedge\ldots\wedge \widehat{dx_i}\wedge\ldots\wedge \ldots dx_m
		\right) \\
	&=& \sum_{i=1}^m \left[
			\sum_{j=1}^m \frac{\DA_i}{\Dx_j} dx_j
		\right]
		dx_1 \wedge\ldots\wedge \widehat{dx_i}\wedge\ldots\wedge \ldots dx_m.
\end{eqnarray*}
Now, $dx_j \wedge dx_j = 0$ so, as in the $m=3$ case above, these terms are
zero except when $i=j$, where the $dx_i$ from the inner sum fills the empty
spot in the outer sum, making up the full $dV$ form each time.  We now have
\begin{eqnarray*}
	d(\vecA \ctt dx_1 \wedge \ldots \wedge dx_m)
	&=& \left(\sum_{i=1}^m \frac{\DA_i}{\Dx_i}\right)
		dx_1 \wedge \ldots \wedge dx_m.
\end{eqnarray*}
which is precisely $\div(\vecA) dV$.
\end{answer}

%% ----------------------------------------------------------------
\subsubsection{Geometry final review problem 10}

Let $\omega$ denote the two-form defined on $\R^3$ by $$\omega = y^2 dx \wedge
dz.$$  Let $i:\S^2 \to \R^3$ denote the inclusion.

(a) Calculate $i^*\omega$ in spherical coordinates.

(b) Orient $\S^2$ using the outward-pointing normal vector.  Set up an explicit
integral expression for $$ \int_{(\S^2)^+} i^*\omega$$ in spherical coordinates.

%\begin{rem}
%\label{rem:non_closed_non_integrable}
%xxx $d^2=0$ means exact implies closed.  Thus, a non-closed form is non-exact.
%Note (here or later) that we can only integrate over exact forms.  Thus, a
%non-closed form (i.e. $\omega$ such that $d\omega \ne 0$) is non-integrable.
%\end{rem}

%\begin{rem*}[Pickrell]
%%Can we integrate this form over $\R^3$?
%Can we apply Stokes?
%As discussed in remark \ref{rem:non_closed_non_integrable}, to test for
%non-integrability we can compute $d$ of this form (in rectangular coordinates):
%$$
%	d \omega = 2y \, dy \wedge dx \wedge dz = 2y \, dV \ne 0.
%$$
%\end{rem*}

\begin{answer}[Answer to part (a).]
Convert directly to spherical coordinates for $\S^2$ (i.e. use spherical
coordinates for $\R^3$ with $\rho=1$):
\begin{eqnarray*}
	i^*(\omega) &=& (i^*y)^2 d(i^*x) \wedge d(i^*z) \\
	&=& (i^*y)^2 d(i^*x) \wedge d(i^*z) \\
	&=& (\sin\theta \sin\phi)^2 d(\cos \theta \sin \phi) \wedge d(\cos \phi) \\
	&=& \sin^2\theta \sin^2\phi
		\left(-\sin\theta \sin\phi d\theta + \cos\theta\cos\phi d\phi\right)
		\wedge \left(-\sin\phi d\phi\right) \\
	&=& \sin^2\theta \sin^2\phi
		\left(\sin\theta \sin\phi d\theta - \cos\theta\cos\phi d\phi\right)
		\wedge \left(\sin\phi d\phi\right) \\
	&=& \sin^3\theta \sin^4\phi d\theta \wedge d\phi.
\end{eqnarray*}
\end{answer}

\begin{answer}[Alternate answer to part (a).]
We want to write $\omega=y^2 \, dx \wedge \dz$ in spherical coordinates on
$\S^2$.  As in problem \ref{sec:ex_2_2}, we should be able to write
$$ \omega = \omega\left(\fracD\theta, \fracD\phi\right) d\theta \wedge d\phi.$$
We know that a point $\vecq$ of $\S^2$ has coordinates
$$
	\vecq = \colvecthree{\cos\theta\sin\phi}{\sin\theta\sin\phi}{\cos\phi}
$$
so
$$
	\fracD\theta =
		\colvecthree{-\sin\theta\sin\phi}{\cos\theta\sin\phi}{0}, \quad
	\fracD\phi =
		\colvecthree{\cos\theta\cos\phi}{\sin\theta\cos\phi}{-\sin\phi}.
$$
Then, recalling that $dx$ and $dz$ are the coordinate-extractor functions
for $x$ and $z$, respectively, we have
\begin{eqnarray*}
	\omega\left(\fracD\theta, \fracD\phi\right)
		&=& y^2 dx \wedge dz\left(\fracD\theta, \fracD\phi\right) \\
	&=& \left(\sin\theta\sin\phi\right)^2 \left[
		dx\left(\fracD\theta\right) \, dz\left(\fracD\phi\right) -
		dz\left(\fracD\theta\right) \, dx\left(\fracD\phi\right)
		\right] \\
	&=& \sin^2\theta\sin^2\phi \left[\sin\theta\sin^2\phi - 0 \right] \\
	&=& \sin^3\theta\sin^4\phi
\end{eqnarray*}
which is in agreement with the previous result.
\end{answer}

\begin{answer}[Answer to part (b).]
%By the above remark we can't integrate globally (xxx discuss the restriction
%from $\R^3$ to $\S^2$ \ldots) (the remark was that we can't reduce this to an
%integral on the boundary \ldots) (something about how $\S^2$ is incontractible
%but the octant is contractible), but we can integrate over a portion of the
%sphere.
It turns out that what Pickrell means by ``explicit integral
expression'' is a Riemann integral of the form
$$
	\int_{\phi=0}^{\pi/2} \int_{\theta=0}^{2\pi}
		\sin^3\theta \sin^4\phi
		d\theta \,d\phi.
$$
It remains to get the sign right.  We have $\D/\D\theta$ going latitudinally
and counterclockwise (i.e. west to east) and $\D/\D\phi$ going longitudinally
toward the north pole.    Thus to get an outward-pointing normal, using the
right-hand rule, we would need to order our basis for $T\S^2$ as
$\{\D/\D\theta, \D/\D\phi\}$.  So, no sign change is necessary.
\end{answer}
%[xxx insert a nice picture here.]

%% ----------------------------------------------------------------
\subsubsection{Geometry final review problem 11}

Let $\D/\D\rho$ denote the outward-pointing vector field in
$\R^n\setminus\{0\}$.

(a) Express this vector field in Euclidean coordinates.

(b) Compute $w = (\D/\D\rho)\ctt dx_1\wedge\ldots\wedge dx_n$ in Euclidean
coordinates.

(c) Explain why the $(n-1)$-form $\omega$ restricts to the standard volume
form on $\S^{n-1}$.

\begin{answer}[Answer to part (a).]
The vector pointing outward from the origin at a point $\vecq$ of $\R^n$
is simply $\vecq$ itself.  Thus we can get an outward-pointing normal
vector by normalizing $\vecq$:
$$
	\D/\D\rho = \hat\vecq = \frac{\vecq}{\|\vecq\|} =
		\frac{
			(x_1, \ldots, x_n)
		}{
			\sqrt{x_1^2 + \ldots + x_n^2}
		}.
$$
Here we think of $x_i$ as the coordinates of $\D/\D\rho$ with respect to the
basis $\{\D/\Dx_1, \ldots, \D/\Dx_n\}$ for $T\R^n$, which are the same $n$ real
numbers as the coordinates of $\vecq$ with respect to the basis $\{\vece_1,
\ldots, \vece_n\}$ for $\R^n$.
\end{answer}

\begin{answer}[Answer to part (b).]
If I am understanding this correctly, I can use problem 9b.
Recall that the divergence is defined in $\R^n$, not just $\R^3$:
just dot the gradient with the vector in question.
The $i$th component of $\D/\D\rho$ is
$$\frac{x_i}{(x_1^2 + \ldots + x_n^2)^{1/2}}.$$
Then $\D/\Dx_i$ of that is
\begin{eqnarray*}
	\frac{\D}{\Dx_i} \left( \frac{x_i}{(x_1^2 + \ldots + x_n^2)^{1/2}} \right)
&=&
	\frac{1}{(x_1^2 + \ldots + x_n^2)^{1/2}} +
	\frac{x_i \cdot 2 x_i \cdot (\frac{-1}{2})}{(x_1^2 + \ldots + x_n^2)^{3/2}}
	\\
&=&
	\frac{1}{(x_1^2 + \ldots + x_n^2)^{1/2}} -
	\frac{x_i^2}{(x_1^2 + \ldots + x_n^2)^{3/2}} \\
&=&
	\frac{x_1^2 + \ldots + x_{i-1}^2 + x_{i+1}^2 + \ldots + x_n^2}
	{(x_1^2 + \ldots + x_n^2)^{3/2}} \\
\end{eqnarray*}
\end{answer}

\begin{answer}[Answer to part (c).]
This is clear geometrically:  think of the area form on $\S^2$, which along with
$d\rho$ gives the volume form on $\R^3$.  To be more formal, though, \ldots .
[xxx type up Pickrell's remarks from the review session.]
\end{answer}

%% ----------------------------------------------------------------
\subsubsection{Geometry final review problem 12}

Let $M$ be a manifold.

(a) Define $H^*_{dR}(M)$.

(b) Attempt to define a product (the \emph{cup product}) on $H^*_{dR}(M)$ by $$
[\omega] \cupprod [\eta] = [\omega \wedge \eta].$$ Show that this is
well-defined, and hence cohomology is an associative algebra.  (An example:
Next semester we will see that in the case of the torus $M=\S^1 \times \S^1$,
$H^*(M)$ is the exterior algebra generated by $[d\theta_1]$ and $[d\theta_2]$.

\begin{answer}[Answer for part (a)] For any given $r$, the $rt$th de Rham
cohomology of $M$, written $H^r_{dR}(M)$, is the quotient space of closed
$r$-forms mod exact $r$-forms.  (How can one remember this?  There are two
ways:  (1) \emph{c} comes alphabetically before \emph{e}; (2) Since $d^2=0$, an
exact form is closed.  That is, if $\omega = d\eta$, then $d\omega = d^2\eta =
0$.  So, taking exact forms mod closed forms wouldn't make sense.)

Now, the $r$ forms, for each $r$, form a real vector space.  The $r$th de Rham
cohomology is a quotient of vector spaces, but we sometimes forget (in the
sense of section \ref{sec:algebras}) about scalar multiplication and think of
this as a quotient of abelian groups.  The set of all $r$ forms, for varying
$r$, form a tensor algebra.  When we form the quotient, we lose the
multiplication, but this can be restored by considering the cup product (see
below).

Thus $H^*_{dR}(M)$ is the union of $H^r_{dR}(M)$ for all $r$.
\end{answer}

\begin{answer}[Answer for part (b)] To show that the cup product is
well-defined, we need to show that it is independent of choice of coset
representative.  By the symmetry of the problem it suffices to show that the
cup product is independent of the choice of $\omega$; showing that it is
independent of the choice of $\eta$ is essentially the same problem.

Suppose
$$
	\omega_1 \sim \omega_2, \qqie [\omega_1] = [\omega_2]
$$
and fix $\eta$.  We need to show that
$$
	[\omega_1 \wedge \eta] = [\omega_2 \wedge \eta].
$$
Since $H^*$ is the quotient of closed forms by exact forms, this means (see
also section \ref{sec:de_rham_cohomology}) that $\omega_1 - \omega_2$ is exact.
That is, there is a form $\alpha$, of order one less than the orders of the
$\omega_i$'s, such that
$$
	\omega_1 - \omega_2 = d\alpha.
$$
We need to show that
$$
	[\omega_1 \wedge \eta] = [\omega_2 \wedge \eta].
$$
For this to be true, we would need
$$
	\omega_1 \wedge \eta - \omega_2 \wedge \eta = d\beta
$$
for some $\beta$.  But recall that the set of forms is an algebra, in
particular a ring, with the wedge operator as its multiplication.  So, we can
undistribute:
$$
	(\omega_1 - \omega_2) \wedge \eta = d\alpha \wedge \eta = d\beta.
$$
So our remaining task is to find such a form $\beta$.  One might suggest $\beta
= \alpha \wedge \eta$.  Then $$ d\beta = d\alpha \wedge \eta + (-1)^k \alpha
\wedge d\eta$$ where $k = \ord(\alpha)$.  All we need is $\alpha \wedge d\eta$
is 0.  But $d\eta=0$ since $\eta$ is closed: recall that homology is
\emph{closed} forms mod exact ones.
\end{answer}

%% ----------------------------------------------------------------
\subsubsection{Geometry final review problem 13}
\label{sec:fr_13}

(a) State and prove Stokes' theorem for the standard cube in $\R^n$.  (On the
exam I might ask you to do this for $n=2$, which amounts to Green's theorem.)

(b) Explain why the classical divergence and Stokes' theorems are special cases
of our Stokes' theorem.

\begin{answer}[Answer to part (a)]
First, I will do $n=2$.  (The solution for the $n=2$ case is due largely to Dr.
Pickrell and Yuliya Gorlina.)

We need to prove
$$ \int_c d\omega = \int_{\Dc} \omega$$
where $c$ is the unit square $I^2$.
Since $n=2$, $n-1=1$, $\omega$ is a 1-form and hence [xxx write and xref to
basis statement] is of the form
$$
	\omega = f(x,y) dx + g(x,y) dy.
$$
To find the left-hand side of the equation to be proved, we need $d\omega$.
This is
\begin{eqnarray*}
	d(\omega) &=& d(f\,dx + g\,dy)
	= d(f\,dx) + d(g\,dy) \\
	&=& \left(\fracDfx dx + \fracDfy dy\right) \wedge dx
	+ \left(\fracDgx dx + \fracDgy dy\right) \wedge dy \\
	&=& \fracDfy dy \wedge dx + \fracDgx dx \wedge dy \\
	&=& \left( \fracDgx -\fracDfy\right) dx \wedge dy.
\end{eqnarray*}
Now we can integrate this.  At this point we switch [xxx write and xref] from
the chains-and-forms notation to the Riemann integral.  Once the double
integral is in that form, we can switch the order of integration, which is
justifiable by \bfidx{Fubini's theorem}.  We have
\begin{eqnarray*}
	\int_c d\omega
	&=& \int_{y=0}^{y=1} \int_{x=0}^{x=1}
		\left(\fracDgx - \fracDfy\right) dx \, dy \\
	&=&
	\int_{y=0}^{y=1} \left[ \int_{x=0}^{x=1} \fracDgx dx \right] dy -
	\int_{x=0}^{x=1} \left[ \int_{y=0}^{y=1} \fracDfy dy \right] dx.
\end{eqnarray*}
By the fundamental theorem of calculus (using it in reverse), this is
\begin{eqnarray*}
	\int_c d\omega
	&=&
	\int_{y=0}^{y=1} \left[ g(1,y) - g(0, y) \right] dy -
	\int_{x=0}^{x=1} \left[ f(x,1) - f(x, 0) \right] dx.
\end{eqnarray*}
Since I'm not sure how to continue with this, it seems like a good time to
leave it as is, and work on the right-hand side for a while.

For the right-hand side we have $\int_{\Dc} \omega$ where $\Dc$ is the
counterclockwise path around the unit square.  I have labeled the four pieces
of the path path $c_1$, $c_2$, $c_3$, and $c_4$:

\setlength{\unitlength}{1.0in}
\begin{picture}(6,1.5)
%\put(0,0){\framebox(6,1.5)}

\thicklines
\put(3.00, 0.10){\vector(1,0){0.50}}
\put(3.50, 0.10){\line(1,0){0.50}}

\put(4.00, 1.10){\vector(-1,0){0.50}}
\put(3.50, 1.10){\line(-1,0){0.50}}

\put(4.00, 0.10){\vector(0,1){0.50}}
\put(4.00, 0.60){\line(0,1){0.50}}

\put(3.00, 1.10){\vector(0,-1){0.50}}
\put(3.00, 0.60){\line(0,-1){0.50}}
\thinlines

\put(2.90, 0.10){\vector(1,0){1.50}}
\put(3.00, 0.00){\vector(0,1){1.50}}

\put(3.45, 0.15){$c_1$}
\put(3.85, 0.55){$c_2$}
\put(3.45, 0.95){$c_3$}
\put(3.05, 0.55){$c_4$}

\put(4.50, 0.05){$x$}
\put(3.10, 1.40){$y$}

\put(3.05, 0.02){\tiny{0}}
\put(4.00, 0.02){\tiny{1}}
\put(2.90, 0.15){\tiny{0}}
\put(2.90, 1.05){\tiny{1}}
\end{picture}

Again switching to the Riemann integral, we have
\begin{eqnarray*}
	\int_{\Dc} \omega
	&=& \int_{c_1} (f(x,y) dx + g(x,y) dy)
	+   \int_{c_2} (f(x,y) dx + g(x,y) dy) \\
	&+& \int_{c_3} (f(x,y) dx + g(x,y) dy)
	+   \int_{c_4} (f(x,y) dx + g(x,y) dy) \\
%
	&=& \int_{x=0}^{x=1} (f(x,0) dx + g(x,0) dy)
	+   \int_{y=0}^{y=1} (f(1,y) dx + g(1,y) dy) \\
	&+& \int_{x=1}^{x=0} (f(x,1) dx + g(x,1) dy)
	+   \int_{y=1}^{y=0} (f(0,y) dx + g(0,y) dy)
\end{eqnarray*}
Half of these vanish as discussed in remark \ref{rem:int_x_path_dy}.  Then
\begin{eqnarray*}
	\int_{\Dc} \omega
	&=& \int_{x=0}^{x=1} f(x,0) dx
	+   \int_{y=0}^{y=1} g(1,y) dy
	+   \int_{x=1}^{x=0} f(x,1) dx
	+   \int_{y=1}^{y=0} g(0,y) dy \\
	&=& \int_{x=0}^{x=1} f(x,0) dx
	+   \int_{y=0}^{y=1} g(1,y) dy
	-   \int_{x=0}^{x=1} f(x,1) dx
	-   \int_{y=0}^{y=1} g(0,y) dy) \\
	&=& \int_{x=0}^{x=1} (f(x,0) - f(x,1)) dx
	+   \int_{y=0}^{y=1} (g(1,y) - g(0,y)) dy \\
	&=&
	\int_{y=0}^{y=1} \left[ g(1,y) - g(0, y) \right] dy -
	\int_{x=0}^{x=1} \left[ f(x,1) - f(x, 0) \right] dx
\end{eqnarray*}
which is what we wanted to show.
\end{answer}

%% ----------------------------------------------------------------
\subsubsection{Geometry final review problem 14}

(a) Let $M$ be an oriented manifold.  Prove that there exists a volume form on
$M$ which is positive at all points of $M$.

(b) Suppose that $M$ is an oriented compact $n$-dimensional manifold without
boundary.  Explain why the map
$$ H^n_{dR}(M, \R) \to \R: [\omega] \mapsto \int_M \omega$$
is well-defined.  Use (a) to explain why, in this case,
$H^n_{dR}(M,\R)$ is not zero.

(c) It turns out that the map in part (b) is an isomorphism.  This has the
following consequence:  if $\omega$ is an $n$-form on $M$, then there is an
$n-1$ form $\eta$ on $M$ such that $d\eta = \omega$ iff $\int_M \omega = 0$.
Explain why this is true for $M=\S^1$.

%% ----------------------------------------------------------------
\subsubsection{Geometry final review problem 15}

Calcluate the area of $\S^2$ (the 2-sphere with radius $r$), for the Euclidean
area form.  Calculate the volume of the ball of radius $r$ in $\R^3$, for the
standard volume form.  How are these two quantities abstractly related?  (Hint:
consider a derivative with respect to $r$ of something, and use units.  This is
a first-semester calculus question related to the second fundamental theorem of
calculus.)

\begin{answer}
Let $B^2$ be the filled unit ball in $\R^3$ (i.e. $\S^2 = \D B^2$).
Recall that
$$A = \int_{r\S^2} dA = 4 \pi r^2$$
and
$$V = \int_{rB^2} dV = \frac{4}{3} \pi r^3.$$
This gives us $V/A = r/3$, from which $V = Ar/3$.  I will use this fact to
check my work.

To use the second fundamental theorem of calculus (theorem
\ref{thm:second_fund}), though, we need to write something of the form
$$ F(r) = \int_0^r f(t) dt$$
which by the SFTC will tell us
$$ F'(r) = f(r).$$
So, think of the ball as the union of nested shells, and take the limit as the
number of shells goes to infinity.  The volume of a thin shell from radius $a$
to radius $b$ may be approximated by
$$
	4\pi a^2 (b-a) = 4\pi a^2 h
$$
where $h = b - a$.  Summing up $n$ such shells, from radius $r_i$ to $r_{i+1}$,
with $\Delta r = r_{i+1} - r_i$, gives
$$ V_n = \sum_{i=0}^{n-1} 4\pi r_i^2 \Delta r.$$
Taking the limit of this finite sum gives
$$ V = \int_0^r 4\pi t^2 dt = \frac{4\pi r^3}{3}.$$
Since $A(t) = 4\pi t^2$, we can also think of this integral as
$$ V = \int_0^r A(t) dt.$$
The SFTC says
$$ V'(r) = A(r),$$
which we can verify by writing
$$
\frac{dV}{dr} = \frac{d}{dr}\left(\frac{4}{3}\pi r^3 \right) = 4 \pi r^2 = A(r).
$$
\end{answer}

\begin{answer}[Alternative answer (Pickrell).]
Start with
$$
	V(r) = \frac{4\pi r^3}{3}.
$$
Use the definition of derivative to obtain
\begin{eqnarray*}
	V'(r) &=& \lim_{h\to 0} \frac{V(r+h) - V(r)}{h}
	= \lim_{h\to 0} \frac{4\pi}{3} \frac{(r+h)^3 - r^3}{h} \\
	&=& \left(\frac{4\pi}{3}\right) \lim_{h\to 0}
		\left(\frac{r^3+3r^2h+3rh^2 + h^3 - r^3}{h}\right)
	= \frac{4\pi}{3} \, 3r^2
	= 4 \pi r^2.
\end{eqnarray*}
\end{answer}

\begin{rem*}
I'm a bit puzzled by both of these solutions.  Pickrell asked for how the
volume and area are ``abstractly related'', but both solutions are quite
concrete.
\end{rem*}

%% ----------------------------------------------------------------
\subsubsection{Geometry final review problem 16}

(a) State a change-of-variables theorem for integration.

(b) Explain why it is true for a constant integrand, the standard cube as
domain, and a linear change of variables.

\begin{answer} For part (a), see \cite{bib:lee2}, p. 593.  [xxx move to main
body of paper and xref.]  Let $A, B$ be compact domains of integration in
$\R^m$.  Let $f:B \to \R$ be continuous, and let $G:A \to B$ be a smooth
bijection with smooth inverse.  Then
$$
	\int_{G(A)} f dV = \int_A (f \circ G) |\det(DG)|dV.
$$

xxx include a nice pullback diagram here:
$G:A \to B$,
$f:B \to \R$,
$f \circ G: A \to \R$.

For part (b), I will first do a particular example.  The emphasis here is on
having a quick computation we can do in our heads, or on scratch paper, to make
sure we have the $\det$ on the correct side of the equation.

Let $x,y$ be coordinates for $\R^2$.  Let $I$ be the unit square in $\R^2$ and
let $u,v$ be given by
$$
	\colvectwo{u}{v} = \colvectwo{2x}{3y}
	= \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} \colvectwo{x}{y}
	= G \colvectwo{x}{y}.
$$
Note that $G$ is already linear, so $DG = G$, and $\det(G) = 6$.  Integrate the
constant function 1.  From freshman calculus we have
$$
	\int_I dA = \int_{y=0}^{y=1} \int_{x=0}^{x=1} dx\,dy = 1.
$$
Using the change of variables $u=2x$ and $v=3y$, we have
$$
	\int_{v=0}^{v=3} \int_{u=0}^{u=2} du \,dv = 6.
$$
So, we may safely remember that the $\det(G)$ goes on the left-hand side, and
we have
$$
	\int_I \det(G) dA = \int_{G(I)} dA.
$$

[xxx replace $v=3x$ with $v=-3x$ and check that the signs are correct.]

Now for part (b) per se.  We are integrating a constant function $c(\vecx)$
over the standard cube $I$ of $\R^m$, with linear change of variables $G: \R^m
\to \R^m$.  Again, $G$ is already linear so $DG = G$.  The function $c$ is from
$G(I)$ to $\R$, so to get a map from $I$ to $\R$ we need to pull $c$ back as $c
\circ G$.  This helps us get the $c$ and the $c\circ G$ in the right places.
Then the left-hand side is
$$
	\int_I (c \circ G) \det(G) dV
$$
while the right-hand side is
$$
	\int_{G(I)} c dV.
$$
The composition $c \circ G$ is again a constant function.  (I.e.  $c$ is not
the identity function.)  Thus the constants pull out of both sides, and
the constant $\det(G)$ pulls out of the left-hand side.  We have
$$
	\det(G) \int_I dV, \qquad \int_{G(I)} dV.
$$
All we need to do to prove these two integrals are equal is recognize that
$$
	\int_{G(I)} dV = \det(G).
$$
\end{answer}

%% ================================================================
\newpage
\subsection{Geometry final}

%% ----------------------------------------------------------------
\subsubsection{Geometry final problem 1}
\label{sec:fe_1}

(a) Show that
$$
	M = \left\{\colvecthree{x}{y}{z} \in \R^3 :
		x^2 + xy + y^2 + xz + z^2 = 1
	\right\}
$$
is an embedded submanifold of $\R^3$.

(b)  The point $\vecq = (1,-1,-1)$ is in $M$.  Find a parameterization for $M$
in a neighborhood of $\vecq$.

\begin{answer}[Answer to part (a).] When we hear \emph{embedded submanifold},
we should think \emph{regular value theorem}.  In particular, let $f:\R^3 \to
\R$ be given by $f(x,y,z) = x^2 + xy + y^2 + xz + z^2$.  Then $M$ is the level
set of $f$ and 1, i.e.  $M = f^{-1}(1)$.  By the \plainidx{regular value
theorem} (theorem \ref{thm:regular_value}), if 1 is a regular value of $f$,
then $f^{-1}(1)$ is either empty, or is an embedded submanifold of $\R^3$.
Since part (b) gives us a point on $M$, $f^{-1}(1)$ is not empty.  So, it
remains to show that 1 is a regular value of $f$.  This is the case if all
preimages of 1, i.e. all points $\vecp$ of $M$, are regular points of $f$.
This in turn is the case if $Df|_\vecp$ is surjective for each $p$.  Since
$f:\R^3 \to \R$, $Df|_\vecp$ has full rank when it is non-zero.  We compute
$$
	Df = \begin{pmatrix} 2x + y + z, & x + 2y, & x + 2z \end{pmatrix}.
$$
For all of these to simultaneously vanish at a point $\vecp=(x,y,z)$, we would
have
$$
	x=-2y, x=-2z, 2x+y+z=0.
$$
The first two force $y=z$, so the last becomes $2x+2y=0$.  This combined with
$x=-2y$ gives $x=0$, and so we have $\vecp=(0,0,0)$.  But this point is not on
$M$.  Since any critical point lies off $M$, all points $p$ on $M$ are regular.
\end{answer}

\begin{answer}[Answer to part (b).]
When we hear \emph{parameterization in a neighborhood}, we should think
\emphidx{implicit function theorem}.  We know by the regular value theorem that
$M$ is two-dimensional, since it is defined by one equation in three unknowns.
As in example \ref{ex:impl_fcn_thm_S2}, we can use the theorem to find which
variable to solve for in terms of the other two.  We compute $Df$ as above, and
substitute $\vecq=(1,-1,-1)$:
$$
	Df|_\vecq = \begin{pmatrix} 0, & -1, & -1\end{pmatrix}.
$$
So, we can solve for $y$ or $z$.  Let's use $z$.  We have a quadratic and a
linear term in $z$ in the equation
$$
		x^2 + xy + y^2 + xz + z^2 = 1,
$$
suggesting that we will need to complete the square.  Solving for the $z$
terms gives
\begin{eqnarray*}
	z^2 + xz &=& 1 - x^2 - xy - y^2 \\
	z^2 + zx + \frac{x^2}{4}&=& 1 - x^2 - xy - y^2 +\frac{x^2}{4}\\
	\left(z + \frac{x}{2}\right)^2 &=& 1 - xy - y^2 -\frac{3x^2}{4}\\
	z &=& - \frac{x}{2} - \sqrt{1 - xy - y^2 -\frac{3x^2}{4}}\\
\end{eqnarray*}
where we select the negative square root so that $z$ will be $-1$
at $x=1$, $y=-1$.

The problem asked for a parameterization (see definition \ref{defn:paramzn}).
So, to finish up, we write
$$
	\colvectwo{x}{y} \longmapsto
	\colvecthree{x}{y}{
		- \frac{x}{2} - \sqrt{1 - xy - y^2 -\frac{3x^2}{4}}
	}.
$$
\end{answer}

%% ----------------------------------------------------------------
\subsubsection{Geometry final problem 2}
\label{sec:fe_2}

Consider the function
$$
	f: \S^2 \to \R: \colvecthree{x}{y}{z} \mapsto xy.
$$

(a) Find the critical points and critical values for this function.

(b) Consider the point $\vecq=(1,0,0)$.  Find the direction at this point in
which $f$ increases most rapidly.

Here I give two solutions, one using projections as discussed in section
\ref{sec:proj_perp}, and the other using \plainidx{Lagrange multipliers} as
discussed in section \ref{sec:lagrange_multipliers}.

\begin{answer}[Answer to part (a) using projections.]
For part (a), let $\vecq=(x,y,z)$ vary over $\S^2$.  On $\R^3$, we have
$$
	Df_\vecq = (y, x, 0)_\vecq.
$$
We need to restrict this to $\S^2$.  Proceeding as in section
\ref{sec:proj_perp}, we recognize that $\vecnhat = \vecq$ is normal to $\S^2$.
So, we can \bfidx{decompose} $Df_\vecq$ into components parallel and
perpendicular to $\vecnhat$.  Writing $\vecu$ for $Df_\vecq$, and
recognizing that $\vecnhat$ already has unit length to we don't need to
scale by $\vecnhat \cdot \vecnhat$,
the parallel part (which we need first) is
$$
	\vecu_\|
	= (\vecu \cdot \vecnhat) \vecnhat
	=
	\left[\colvecthree{y}{x}{0} \cdot
	\colvecthree{x}{y}{z}\right]
	\colvecthree{x}{y}{z}
	=
	\colvecthree{2x^2 y}{2xy^2}{2xyz}.
$$
Then
$$
	Df|_{T\S^2} = \vecu_\perp = \vecu - \vecu_\|
	= \colvecthree{y-2x^2 y}{x - 2xy^2}{-2xyz}.
$$

The critical points of $f$ are those for which $Df|_{T\S^2}$ has rank less than
1, i.e. zero.  For this to happen, all three entries must be zero.  So, the
following must be simultaneously true:
$$
	\left\{
	\begin{matrix}
	y(1-2x^2) &=& 0, \\
	x(1-2y^2) &=& 0, \\
	-2xyz &=& 0.
	\end{matrix}
	\right.
$$
That is,
$$
	\begin{matrix}
\left[y = 0 \right.&\textrm{or}& \left.x = \pm 1/\sqrt 2\right],&\textrm{and}\\
\left[x = 0 \right.&\textrm{or}& \left.y = \pm 1/\sqrt 2\right],&\textrm{and}\\
\left[x = 0 \right.&\textrm{or}& y = 0 &\textrm{or}& \left.z = 0\right].&
	\end{matrix}
$$
Enumerating the possibilities, we obtain the 6 critical points
$$
	\colvecthree{0}{0}{\pm 1},\quad
	\colvecthree{\pm 1/\sqrt 2}{\pm 1/\sqrt 2}{0}.
$$
The critical values of $f$ are the images of these 6 points, which are
$$
	0, \quad \pm 1/2.
$$
\end{answer}

\begin{answer}[Answer to part (a) using Lagrange multipliers.]
The manifold $\S^2$ is the zero set of $g(x,y,z) = x^2+y^2+z^2-1$; the function
to be maximized is $f(x,y,z) = xy$.  We put
\begin{eqnarray*}
	\grad f &=& \lambda \grad g \\
	\vvecthree{y}{x}{0} &=& \vvecthree{2 \lambda x}{2 \lambda y}{2 \lambda z}.
\end{eqnarray*}
If $\lambda = 0$, then $x=y=0$ and $z=\pm 1$.  If $\lambda\ne 0$, then $z=0$
and we can solve for $\lambda$ in the first two equations to find $x$ and $y$.
Note that if $x=0$ then $y=0$ and we have $x=y=z=0$ which is not a point on
$\S^2$.  So, we may divide by $x$ and $y$.  We obtain
\begin{eqnarray*}
	\lambda &=& y/2x\\
	\lambda &=& x/2y\\
	2x^2 &=& 2y^2.
\end{eqnarray*}
Since $z=0$ and $x^2+y^2=1$, we obtain for $\lambda \ne 0$ the same
four points as in the previous solution.  These four, along with the
north and south poles for $\lambda=0$, give us the critical points
$$
	\colvecthree{0}{0}{\pm 1},\quad
	\colvecthree{\pm 1/\sqrt 2}{\pm 1/\sqrt 2}{0}.
$$
as before.  Of course, the critical values are the same as well.
\end{answer}

\begin{answer}[Answer to part (b).]
The direction of greatest change of $f$ on $\S^2$ is simply the gradient of $f$
restricted to $\S^2$, which is what we just computed.  So all we need to do
is evaluate $Df_{T\S^2}$ at $\vecq=(1,0,0)$.  We obtain
$$
	Df|_{T\S^2} = \colvecthree{y-2x^2 y}{x - 2xy^2}{-2xyz}
	= \colvecthree{0}{1}{0}.
$$
\end{answer}

%%% ----------------------------------------------------------------
%\subsubsection{Geometry final problem 3}
%
%Consider the vector field on $\R^2$ given by
%$$
%	\vecv\pickpipe{\colvectwo{x}{y}} = \colvectwo{y}{x}.
%$$
%
%(a) Compute the flow exactly.
%
%(b) Graph the trajectories of the flow.
%
%\begin{answer}[Answer to part (a).]
%\end{answer}
%
%\begin{answer}[Answer to part (b).]
%\end{answer}

%%% ----------------------------------------------------------------
%\subsubsection{Geometry final problem 4}
%
%\emph{Last-minute cancellation.}

%% ----------------------------------------------------------------
\subsubsection{Geometry final problem 5}

Let
$$
	\eta = \frac{-y\,dx + x\,dy}{x^2+y^2}
$$
on $\R^2 \setminus \{0\}$.

(a) Show that $\eta$ is closed.

(b) Calculate $\int_{\S^1}\eta$, where $\S^1$ has the counterclockwise
orientation.

(c) Why does your answer to (b) imply that there does not exist a function $f$
on $\R^2 \setminus \{0\}$ such that $df = \eta$?

\begin{answer}[Answer to part (a).]
\end{answer}

\begin{answer}[Answer to part (b).]
\end{answer}

\begin{answer}[Answer to part (c).]
\end{answer}

%% ----------------------------------------------------------------
\subsubsection{Geometry final problem 6}

Let $X$ denote the upper hemisphere of $\S^2$ with the orientation induced
by the upward-pointing normal vector.  Compute the integral
$$
	\int_X z \, dx \wedge dy.
$$

\begin{answer}[Answer, due in part to Tommy Occhipinti.]
My (the author's) natural inclination is to treat $X$ as the surface, with
perimeter $\DX$ being the unit circle in the $x,y$ plane.  Then Stokes' theorem
would give us
$$
	\int_X d\omega = \int_{\DX} \omega.
$$
In this scenario, the given form $z \, dx \wedge dy$ is $d\omega$, and we need
to find out what $\omega$ is.  But will this search be futile?  That is, is $z
\, dz \wedge dy$ exact?  Recall that exact forms are closed, since $d^2=0$: if
$d\omega = z\,dx\wedge dy$ then $d^2\omega = 0$.  But this is
$$
	d(z\,dx \wedge dy) = dz \wedge dz \wedge dy = dx \wedge dy \wedge dz = dV,
$$
which is not identically zero.  So, $z\,dx \wedge dy$ is not closed, and
therefore is not exact.

Since I want to invoke Stokes' theorem, I can't have $X$ being the surface and
the perimeter being the boundary.  Instead, I must think of $X$ as being the
boundary.  This requires a little leap of insight:  we have to also include the
closed disk on the $x,y$ plane, the floor of the dome if you will, in the
boundary.  (One hopes this will not affect the value of the integral, or at
least that the contribution will easily be taken back out.  However, the
integral over the floor of the dome of $z\,dx\wedge dy$ is zero, since
$z=0$ on the floor.)  Here, the volume to be integrated is the upper half of
the filled sphere:  call it $H$.  If $$\omega = z \, dx \wedge dy,$$ then (as
above) $$ d\omega = dV.$$ Then this problem is easy, since $$ \int_X \omega =
\int_H d\omega = \int_H dV.$$ Now, the integral of the volume form over the
whole filled unit sphere is $4\pi/3$, so half that is $2\pi/3$.
\end{answer}

%%% ----------------------------------------------------------------
%\subsubsection{Geometry final problem 7}
%
%\emph{Last-minute cancellation.}

%%% ----------------------------------------------------------------
%\subsubsection{Geometry final problem 8}
%
%State the classical Stokes' theorem.  Explain why this theorem is a
%consequence of our abstract Stokes' theorem.

%% ----------------------------------------------------------------
\subsubsection{Geometry final problem 9}

Suppose that $z=g(y)$ is a smooth positive function, defined for $y \in [a,b]$.
Consider the surface $M$ of revolution with boundary, obtained by revolving the
graph of $g$ around the $y$ axis in $\R^3$.  Orient $M$ using the
outward-pointing normal vector $\vecnhat$, and consider the coordinates
$\theta,y$ for $M$.  [xxx orientation of $\theta$.  or xref to prev sec.]

(a) Briefly explain why the oriented area form of $M$ is given by the
expression
$$
	dA = g(y) \sqrt{1+g'(y)^2} \, d\theta \wedge dy.
$$

(b) Suppose that $G(y)$ is an antiderivative of
$g(y) \sqrt{1+g'(y)^2}$ for $y\in[a,b]$.  Show that
$$
	\eta =  G(y)\,d\theta
$$
is a well-defined one-form on $M$, and that $d\eta = dA$.

[xxx rmk and xref that not everything starting with a $d$ is exact.]

(c) Use part (b) and Stokes' theorem to compute the area of $M$ in terms of
$G$.  (Remark:  this just recovers a standard formula for the area from
first-year calculus.)

%% ----------------------------------------------------------------
\subsubsection{Geometry final problem 10}

Determine whether the following statements are true or false.
Briefly explain your answers.

(a) The form $\omega_1 = y^2\,dx \wedge dz$ is exact in $\R^3$, by the
Poincar\'e Lemma, because $\R^3$ is contractible.

(b) If $\omega$ is exact in $\R^2 \setminus\{0\}$, then for any closed oriented
curve $c$ in $\R^2\setminus\{0\}$, $\int_c\omega = 0$.

(c) On $\S^2$, $L_{\D/\D\theta}dA = 0$.

(d) If $X$ is a vector field on $M$ and $f:M \to N$ is a map, then we can use
$f$ to push $X$ forward to a vector field on $N$.


%% ----------------------------------------------------------------
\newpage
\subsection{Topology homework 1}

%%% ----------------------------------------------------------------
%\subsubsection{Topology homework 1 problem 2}
%
%For what $n$ is $\R\P^n$ orientable?
%
%\begin{answer}
%\ldots
%\end{answer}

%% ----------------------------------------------------------------
\subsection{Topology exam 1 review}

%% ----------------------------------------------------------------
\subsubsection{Topology exam 1 review problem 5}

For each real number $\alpha$, define a group action
$$
	\Z \times \S^1 \longto \S^1:
		(n, e^{2\pi i \theta}) \mapsto e^{2\pi i(\theta + n \alpha )}.
$$

\begin{itemize}
\item[(a)] Determine the values of $\alpha$ for which this is a free
	action, i.e. the stabilizers are all trivial.
\item[(b)] Fix a value of $\alpha$ as in (a) such that $\Z$ is acting
	freely.  Show that $\Z$ is not acting properly discontinuously.
\end{itemize}

\begin{answer}[Answer to part (a).]
Either $\alpha$ is rational, or it is not.  First consider the former
case.  Put $\alpha=a/b$ in lowest terms.  Then for $n=b$, we have
$$
	(b, e^{2\pi i \theta}) \mapsto e^{2\pi i(\theta + b(a/b))}
		= e^{2\pi i(\theta + a)}
		= e^{2\pi i\theta}
$$
since $a$ is an integer.  Thus the stabilizer of $\alpha$ is non-trivial.
(Note in particular that for $\alpha = 0$, the stabilizer is all of $\Z$.)

Now consider the case when $\alpha$ is irrational.  For
$$
	e^{2 \pi i \theta} = e^{2 \pi i (\theta + n \alpha)}
$$
requires
$$
	e^{2 \pi i (\theta + n \alpha)} e^{-2 \pi i \theta} = 0
	= e^{2 \pi i 0}
	= e^{2 \pi i n \alpha}
$$
which puts $n \alpha \equiv 0 \pmod 1$, i.e. $n \alpha$ is an integer.  This of
course happens when $n=0$.  Now suppose (seeking a contradiction) that there is
some integer $m$ such that $n \alpha = m$ and $n\ne 0$.  Then we may divide to
obtain $\alpha = n/m$.  This shows that an irrational number is equal to a
rational number, which is the desired contradiction.

Thus we have shown that the action has trivial stabilizer if and only if
$\alpha$ is irrational.
\end{answer}

\begin{answer}[Answer to part (b).]
\end{answer}

% Tommy's sol'n for homot of topo group stuff:
% (a.b)(t) ab(st) [(a.b)(st)]^-1?

%% ----------------------------------------------------------------
\subsection{Topology final}

%\setlength{\unitlength}{1.0in}
%\begin{picture}(6,1.6)
%\put(0,0){\framebox(6,1.6)}
%\put(2.70, 0.05){$\mcT^0_0(V) = C^\infty(M)$}
%	\put(2.90, 0.25){\vector(-1,1){.20}}
%	\put(3.15, 0.25){\vector( 1,1){.20}}
%\put(2.10, 0.55){$\mcT^1_0(V) = TM$}
%\put(3.30, 0.55){$\mcT^0_1(V) = T^*M$}
%	\put(2.55, 0.70){\vector(-1,1){.20}}
%	\put(2.80, 0.70){\vector( 1,1){.20}}
%	\put(3.35, 0.70){\vector(-1,1){.20}}
%	\put(3.60, 0.70){\vector( 1,1){.20}}
%\put(2.10, 1.05){$\mcT^2_0(M)$}
%\put(2.90, 1.05){$\mcT^1_1(M)$}
%\put(3.70, 1.05){$\mcT^0_2(M)$}
%	\put(2.15, 1.25){\vector(-1,1){.20}}
%	\put(2.40, 1.25){\vector( 1,1){.20}}
%	\put(2.95, 1.25){\vector(-1,1){.20}}
%	\put(3.10, 1.25){\vector( 1,1){.20}}
%	\put(3.75, 1.25){\vector(-1,1){.20}}
%	\put(3.90, 1.25){\vector( 1,1){.20}}
%\put(1.70, 1.55){$\ldots$}
%\put(2.50, 1.55){$\ldots$}
%\put(3.30, 1.55){$\ldots$}
%\put(4.10, 1.55){$\ldots$}
%\end{picture}
%

%\item For this course, we consider vector-to-scalar and vector-to-vector
%functions which are \bfidx{smooth} or \bfidx{infinitely differentiable}.
%Given a metric space $X$, we write $$C^\infty(X, \R)$$ for the
%set\index{$C^\infty$} of such functions.  Since our functions are smooth,
%mixed partials are equal and so we do have $\nabla \times \nabla = 0$ and
%$\nabla \cdot \nabla \times = 0$ as operators, i.e. curl of grad and div of
%curl are both zero.

%% ================================================================
\newpage
\section{About the references}

As mentioned in the preface, this is not a standalone work.  My focus is on
providing intuition and examples that have been short-changed elsewhere; I
offer little rigor in this paper since there is no lack of that in the
literature --- as described below.  The following annotations are based on my
experiences and the experiences of my fellow graduate students.  This is an
admittedly small sample.

\begin{itemize}

\item Calculus:
	\begin{itemize}
	\item \cite{bib:anton} is a standard approach to calculus.
	\item \cite{bib:hhgm} is a reform approach to calculus; the present
		work is in part inspired by this text.
	\item \cite{bib:hildebrand} is advanced calculus.
	\end{itemize}

\item Linear algebra:
	\begin{itemize}
	\item \cite{bib:fis} A senior undergraduate text.  The determinant
		is done entirely without exterior algebra.
	\end{itemize}

\item Abstract algebra:
	\begin{itemize}
	\item \cite{bib:herstein}:  A senior undergraduate text.
	\item \cite{bib:df},
	\cite{bib:grove},
	\cite{bib:hung},
	\cite{bib:lang}:
	Graduate texts.  In particular, you can find information here about
	sequences, homology, and cohomology.
	\end{itemize}

\item Analysis:
	\begin{itemize}
	\item \cite{bib:rudin}.  Contains some information about differential
		forms.
	\end{itemize}

\item Physics approach to geometry/topology:
	\begin{itemize}
	\item \cite{bib:abr}, \cite{bib:frankel}:
		I find the latter more readable; also, the former is more rigorous.
	\item \cite{bib:penrose}:  Really a pop-science book, but with
		plenty of mathematical content.
	\end{itemize}

\item Differential geometry:
	\begin{itemize}
	\item \cite{bib:spi1} is a slim, senior undergraduate text which manages
		to include much of the content of the geometry course.
		It includes chains, forms, and duality, ending in a proof of
		generalized Stokes.
	\item \cite{bib:boothby} and \cite{bib:lee2} are textbooks for
		a geometry/topology course.  I have received positive reviews
		of the former and mixed reviews of the latter.  The latter
		is fussy with proofs, which some readers find obscures the big
		picture.  \cite{bib:lee2} is intended to be preceded by \cite{bib:lee1}
		(below).
	\item I have received very few if any positive student reviews
		of \cite{bib:conlon} and \cite{bib:satan}.  Fans of the latter
		tend to be faculty members.
	\item \cite{bib:lee3} is a follow-on to a geometry/topology course.
	\end{itemize}

\item Algebraic topology:
	\begin{itemize}
	\item \cite{bib:hatcher} and \cite{bib:massey}:  standalone sources
		for algebraic topology.
	\item \cite{bib:lee1} is intended to precede \cite{bib:lee2}.
	\item All three are textbooks for a graduate geometry/topology course.
	\end{itemize}

\item Classical differential geometry:
	\begin{itemize}
	\item \cite{bib:gug}:  Focuses on concepts such as curvature and
		torsion which, for reasons I don't understand, we no longer
		consider in a first-year geometry course.
	\end{itemize}

\item Math history:
	\begin{itemize}
	\item \cite{bib:bma}:  history of math from ancient times.
		For this reason, information on the development of calculus
		is limited.
	\item \cite{bib:dunham}:  initial development and rigorization of
		single-variable calculus.
	\item \cite{bib:crowe}:  the advent of the vectorial system.
	\item I am still looking for a good reference on the historical development
		of differential forms.
	\end{itemize}

\item Works similar to this one:
	\begin{itemize}
	\item \cite{bib:lamb}: succinct notes for qualifier preparation.
	\item \cite{bib:bachman}: a work very similar to the present one,
		although less ambitious.  Intended for an undergraduate audience
		(immediately post-calculus) although there is plenty of information
		for graduate students.  I have discovered this work only recently;
		its geometrical approach to forms is very similiar to mine.
	\end{itemize}

\item Encyclopediae:
	\begin{itemize}
	\item \cite{bib:crc}, \cite{bib:pdm}, \cite{bib:pm}, and
		of course Wikipedia.
	\end{itemize}

\end{itemize}

%% ================================================================
\newpage
\section{Typesetting and computation}

\begin{center}
\emph{
The soul cannot think without a picture.
}
--- Aristotle\index{Aristotle} (384-322 B.C.).
\end{center}

This section presents software-related issues related to learning differential
geometry and the preparation of this paper.  Such topics may seem ephemeral:
software changes quickly.  However, central concepts and algorithms, happily, do
not.

%% ----------------------------------------------------------------
\subsection{Files and tools}

The source for this paper is (as of this writing) located at

\texttt{http://math.arizona.edu/{\tat}kerl/doc/prolrev/}.

The top-level file is \texttt{prolrev.tex}; most of the contents are in
\texttt{body.tex}.  Figures and artwork are in the \texttt{figures/}
subdirectory.  The index was prepared using my Perl script \texttt{kmkidx}: see 

\texttt{http://math.arizona.edu/{\tat}kerl/index.php?v=software}

for more information.  PostScript was created using \texttt{dvips}; The PDF was
created using \texttt{ps2pdf}.  The \texttt{psfrag} package was used to include
{\LaTeX} math symbols in the figures.  (Thanks to David Bachman for the tip!)

%% ----------------------------------------------------------------
\subsection{Figures and artwork}

The figures in this document were created using Inkscape and Matlab.  The
latter is discussed in more detail in section \ref{sec:matlab}.

The cover art suggests that we can approach geometry/topology using the
familiar circle as a starting point, as discussed in section [write and xref].

The back art (page \pageref{fig:black_flag}) depicts the following:

\begin{itemize}

\item The large and imposing edifice of geometry-topology is constructed of
simple building blocks.

\item The black flag is the flag of anarchy:  this paper was written by a
graduate student, not a faculty member.  This paper emphasizes issues of
interest and/or difficulty \emph{for the learner}.

\item The flag is flapping and warped by the breeze --- the same breeze which
turns Green's theorem into classical Stokes.

\item The flag is a 2-dimensional manifold with boundary, with rank-one
fundamental group.

\item Although the edifice may appear well-armed and impenetrable, the central
feature is an entryway, through which we are invited.

\item The stones of the archway are reminiscent of a chain of singular 3-cubes.

\item Inscribed on the keystone is the generalized Stokes theorem, in pairing
notation.

\end{itemize}

%% ----------------------------------------------------------------
\subsection{Matlab}
\label{sec:matlab}

xxx paste in some examples here.  Include gradient, surface, contour,
and quiver.

How-to's.

\texttt{http://math.arizona.edu/{\tat}kerl/doc/prolrev/figures}

list.

See \cite{bib:kerl}.

%%% ----------------------------------------------------------------
%\newpage
%\subsection{2D, 3D, and 4D complexes TBD}
%\label{sec:complexes_2D_3D_4D}
%
%In 2D: we have \bfidx{scalar curl}.  What do we have in 4D?
%
%First 2D.
%
%\begin{itemize}
%\item A general 0-form is
%$$
%	G(x,y).
%$$
%Then $d$ of that is a 1-form:
%$$
%	dG(x,y) = \fracDGx\,dx + \fracDGy\,dy.
%$$
%This is the \bfidx{differential} form $dG$.
%
%\item A general 1-form is
%$$
%	f(x,y)\,dx + g(x,y)\,dy.
%$$
%Then $d$ of that is a 2-form:
%\begin{eqnarray*}
%	d[f(x,y)\,dx + g(x,y)\,dy]
%	&=& \left(\fracDfx\,dx + \fracDfy\,dy\right) \wedge dx +
%		\left(\fracDgx\,dx + \fracDgy\,dy\right) \wedge dy \\
%	&=& \left(\fracDgx-\fracDfy\right)\,dx\wedge dy.
%\end{eqnarray*}
%This is the \bfidx{scalar curl} form.
%\end{itemize}
%
%Now 3D.
%
%\begin{itemize}
%\item A general 0-form is
%$$
%	G(x,y,z).
%$$
%Then $d$ of that is a 1-form:
%$$
%	dG(x,y,z) = \fracDGx\,dx + \fracDGy\,dy + \fracDGz\,dz.
%$$
%This is the \bfidx{differential} form $dG$.
%
%\item A general 1-form is
%$$
%	f(x,y,z)\,dx + g(x,y,z)\,dy + h(x,y,z)\,dz.
%$$
%Then $d$ of that is a 2-form:
%\begin{eqnarray*}
%	d[f(x,y,z)\,dx + g(x,y,z)\,dy + h(x,y,z)\,dz]
%	&=& \left(\fracDfx\,dx + \fracDfy\,dy + \fracDfz\,dz\right) \wedge dx \\
%	&+& \left(\fracDgx\,dx + \fracDgy\,dy + \fracDgz\,dz\right) \wedge dy \\
%	&+& \left(\fracDhx\,dx + \fracDhy\,dy + \fracDhz\,dz\right) \wedge dz \\
%	&&  \\
%	&=& \left(\fracDhy - \fracDgz\right)\,dy\wedge dz \\
%	&+& \left(\fracDfz - \fracDhx\right)\,dz\wedge dx \\
%	&+& \left(\fracDgx - \fracDfy\right)\,dx\wedge dy.
%\end{eqnarray*}
%This is the \bfidx{curl} form $(\grad \times \vecF) \cdot \vecnhat \,dA$.
%See also section \ref{sec:div_and_curl}.
%[xxx to do:  explain more about the $\vecnhat$ and the $dA$.]
%
%\item A general 2-form is
%$$
%	f(x,y,z)\,dy\wedge dz + g(x,y,z)\,dz\wedge dx + h(x,y,z)\,dx\wedge dy.
%$$
%Then $d$ of that is a 3-form:
%\begin{eqnarray*}
%	&&d[
%	f(x,y,z)\,dy\wedge dz + g(x,y,z)\,dz\wedge dx + h(x,y,z)\,dx\wedge dy]\\
%	&=& \left(\fracDfx\,dx + \fracDfy\,dy + \fracDfz\,dz\right) \wedge dy\wedge dz \\
%	&+& \left(\fracDgx\,dx + \fracDgy\,dy + \fracDgz\,dz\right) \wedge dz\wedge dx \\
%	&+& \left(\fracDhx\,dx + \fracDhy\,dy + \fracDhz\,dz\right) \wedge dx\wedge dy \\
%	&&  \\
%	&=& \left(\fracDfx + \fracDgy + \fracDhz \right)\,dx\wedge dy \wedge dz.
%\end{eqnarray*}
%This is the \bfidx{divergence} form $\grad\cdot \vecF dV$.
%See also section \ref{sec:div_and_curl}.
%\end{itemize}
%
%Last, 4D.
%
%\begin{itemize}
%\item A general 0-form is
%$$
%	G(w,x,y,z).
%$$
%Then $d$ of that is a 1-form:
%$$
%	dG(w,x,y,z) = \fracDGw\,dw + \fracDGx\,dx + \fracDGy\,dy + \fracDGz\,dz.
%$$
%This is the \bfidx{differential} form $dG$.
%
%\item A general 1-form is
%$$
%	e(w,x,y,z)\,dw + f(w,x,y,z)\,dx + g(w,x,y,z)\,dy + h(w,x,y,z)\,dz.
%$$
%Then $d$ of that is a 2-form:
%\begin{eqnarray*}
%	&&d[e(w,x,y,z) + f(w,x,y,z)\,dx + g(w,x,y,z)\,dy + h(w,x,y,z)\,dz] \\
%	&=& \left(e_w\,dw + e_x\,dx + e_y\,dy + e_z\,dz \right) \wedge dw \\
%	&+& \left(f_w\,dw + f_x\,dx + f_y\,dy + f_z\,dz \right) \wedge dx \\
%	&+& \left(g_w\,dw + g_x\,dx + g_y\,dy + g_z\,dz \right) \wedge dy \\
%	&+& \left(h_w\,dw + h_x\,dx + h_y\,dy + h_z\,dz \right) \wedge dz \\
%	&&  \\
%	&+& (f_w - e_x)\,dw\wedge dx \\
%	&+& (g_x - f_y)\,dx\wedge dy \\
%	&+& (h_y - g_z)\,dz\wedge dy \\
%	&+& (e_z - h_w)\,dz\wedge dw \\
%	&+& (g_w - e_y)\,dw\wedge dy \\
%	&+& (h_x - f_z)\,dx\wedge dz.
%\end{eqnarray*}
%What do we call this?
%
%\item A general 2-form is
%$$
%	f(x,y,z)\,dy\wedge dz + g(x,y,z)\,dz\wedge dx + h(x,y,z)\,dx\wedge dy.
%$$
%Then $d$ of that is a 3-form:
%\begin{eqnarray*}
%	&&d[
%	f(x,y,z)\,dy\wedge dz + g(x,y,z)\,dz\wedge dx + h(x,y,z)\,dx\wedge dy]\\
%	&=& \left(\fracDfx\,dx + \fracDfy\,dy + \fracDfz\,dz\right) \wedge dy\wedge dz \\
%	&+& \left(\fracDgx\,dx + \fracDgy\,dy + \fracDgz\,dz\right) \wedge dz\wedge dx \\
%	&+& \left(\fracDhx\,dx + \fracDhy\,dy + \fracDhz\,dz\right) \wedge dx\wedge dy \\
%	&&  \\
%	&=& \left(\fracDfx + \fracDgy + \fracDhz \right)\,dx\wedge dy \wedge dz.
%\end{eqnarray*}
%This is the \bfidx{divergence} form $\grad\cdot \vecF dV$.
%See also section \ref{sec:div_and_curl}.
%\end{itemize}
%
%Last, 4D.


%% ================================================================
%% ================================================================
%% ================================================================
\newpage
\section{Under construction and to be merged}

%% ----------------------------------------------------------------
\subsection{temp}

xxx rm2r:

\begin{defn}
Let $f$ be a vector-to-scalar function and let $\vecu=(u, v, w) \in \R^3$.
The \bfidx{directional derivative} of $f$ in the direction of $\vecu$
is the dot product
$$
D_\vecu(f) = \vecu \cdot \nabla f =
u \frac{\Df}{\Dx} + v \frac{\Df}{\Dy} + w \frac{\Df}{\Dz}.
$$
\end{defn}
The directional derivative $\vecu \cdot \grad(f)$ is the \bfidx{rate of change}
of $f$ in the arbitrary direction $\vecu$.  Note that the partial derivative
with respect to $x$ is the directional derivative of $f$ in the direction of
$\hat \vecx$ and likewise for $y$ and $z$.  That is, the partial derivative is
the fundamental notion; the directional derivative is the more general notion
which has the partial derivative as a special case.

Now let $\vecu$ vary over the unit sphere, i.e. consider all $\vecu$ such that
$\|\vecu\| = 1$.  Then $\vecu \cdot \nabla f$ is biggest when $\vecu$ points in
the same direction as $\nabla f$.  This is true because
$$
\vecu \cdot \nabla f = \|\vecu\| \, \|\nabla f\| \cos \theta
= \|\nabla f\| \cos \theta
$$
where $\theta$ is the angle between $\vecu$ and $\nabla f$.  This is biggest
when $\theta = 0$, i.e. when $\vecu$ and $\nabla f$ point in the same
direction.  This means that the gradient of $f$ points in the \bfidx{direction
of greatest change} \index{greatest change} of $f$.

%% ----------------------------------------------------------------
\subsection{Jacobian matrix TBD}

xxx ways to motivate:
\begin{itemize}

\item Normal to surface (gradient as column vector)

\item Tangent hyperplane to embedded submanifold: kernel of gradient.

\item Regular-value thm (express in vr. calc. terms here)

\item Impl. fcn. thm. (cf. Frankel)

\item Inv. fcn. thm. (cf. Frankel p. 29)

\end{itemize}

%% ----------------------------------------------------------------
\newpage
\subsection{Level sets TBD}

\begin{defn}
Let $f: \R^3 \to \R$ be a \plainidx{vector-to-scalar function}.  The
\bfidx{level set} of $f$ and $c$ is the set of points $$\{\vecq: f(\vecq) =
c\}.$$
\end{defn}

\beginex
If $f(x,y,z) = x^2+y^2+z^2$, then the level set of $f$ and 1 is the unit sphere
\bfidx{$\S^2$}.
\mendex

\begin{rem}
\label{rem:grad_F_is_normal}
Recall that $\nabla f$ points in the direction of greatest change of $f$.  When
evaluated at $\vecq$, it is also the normal vector to the level sets at
$\vecq$, written $\vecnhat_\vecq$.  Perpendicular to the normal is along the
direction of the surface.
\end{rem}

\begin{rem}
\label{rem:point_normal_form_for_tangent_plane}
The \bfidx{point-normal form} gives an equation for the \bfidx{tangent plane}
to the level-set surface.  That is, if
$$
	\vecq=(x_0, y_0, z_0)
$$
then the point-normal form of the tangent plane at $\vecq$ is
$$
	\fracDtwo{f}{x}(x-x_0) + \fracDtwo{f}{y}(y-y_0) + \fracDtwo{f}{z}(z-z_0) = 0.
$$
Write this [xxx to do] in terms of (1) dot product, when $\nabla f$ is thought
of as a vector, and (2) kernel of linear transformation $Df = \grad f$, when
$\nabla f$ is thought of as a $1 \times 3$ matrix.  Conclude that the tangent
space at $q$ is the kernel of $Df|q$.
\end{rem}

xxx to do:  header style!!

xxx rcol/rowvectwo/three macros.

xxx really need to understand $(\grad \times \vecF) \cdot \vecnhat dA$.

If $\vecnhat = (a, b, c)^t$ then what is $dA$?  Try to work in point-normal form
somehow \ldots.

xxx notation: col/row stack.

%xxx foo fig start:
%\begin{center}
%\psfragscanon
%\includegraphics{figures/test_psfrag1.eps}
%\end{center}
%xxx foo fig end:
