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%% ================================================================
\title{Tensorama}
\author[J.\ Kerl]{John Kerl}
\institute[Arizona]{Department of Mathematics, University of Arizona\\
Graduate Student Colloquium}
\date{February 10, 2010}

\begin{document}

\begin{frame}
\titlepage
\end{frame}

%% ----------------------------------------------------------------
\begin{frame}
\frametitle{\emph{Polysemy:} the coexistence of many possible meanings for a
word or phrase.}

\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}

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'').  (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}

\end{frame}

%% ----------------------------------------------------------------
\begin{frame}
\frametitle{Freeness}

{\emcol Definition:}  An object (group, ring, module, etc.) $\mcO$ is
\emph{free} on a set $S \subset \mcO$ if for any other respective object $P$
and a morphism $f:S \to P$ there is a unique morphism $\tilde f$ from $\mcO$ to
$\mcP$ such that $\tilde f|_S = f$.

A commutative diagram is a bit more intuitive:

\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){$\mcO$}
\put(3.50, 0.05){$\mcP$}

\put(2.85, 0.95){inclusion}
\put(2.80, 0.40){$f$}
\put(3.65, 0.40){$\tilde f$}

\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}

\end{frame}

%% ----------------------------------------------------------------
\begin{frame}
\frametitle{Freeness, continued}

Still \ldots \emph{what}?  I claim that we all already know this well --- at
least, if $\mcO$ and $\mcP$ are vector spaces, $S$ is a basis for $\mcO$, and
$f$ is a linear transformation.  Specifying the images of the basis vectors
uniquely specifies the images of \emph{all} vectors, since any vector is a
linear combination of basis elements.

We have
\begin{align*}
	A(\vecv) &=  A\left(\sum_{j=1}^n v_j \vece_j\right)
	= \sum_{j=1}^n v_j A\left(\vece_j\right)
\end{align*}
where $A(\vece_j)$ is the $j$th column of $A$.

{\emcol Example:}
\begin{align*}
	\pcmatrix{ 1 & 2 \\ 3 & 4}
	\pcmatrix{
		\ucmatrix{1 \\ 0}
		\;\vline\;
		\ucmatrix{0 \\ 1}
	}
	&=
	\pcmatrix{
		\ucmatrix{1 \\ 2}
		\;\vline\;
		\ucmatrix{3 \\ 4}
	}
\end{align*}
i.e.  $\colvectwo{1}{0} \mapsto \colvectwo{1}{2}$
and   $\colvectwo{1}{0} \mapsto \colvectwo{3}{4}$.

\end{frame}

%% ----------------------------------------------------------------
\begin{frame}
\frametitle{Relations}

{\emcol Definition:}  a \emph{relation} is some non-trivial expression
(containing elements of an object) which is equal to the identity.

Group examples: $aba^{-1}b^{-1} = e$ (commutator relation) or $a^m=e$
(finite-order relation).

{\emcol Intuition:}  A set $S \subset \mcO$ is free on $S$ if the elements of
$S$ have only the minimal relations required for $\mcO$'s category.

Example:  If $G$ is a free group on $\{a,b,c\}$, then $ab$, $a^2$, $b^{-1}ac$,
etc. don't simplify.  But we must permit $aa^{-1}=e$ (which we wouldn't permit
for a free semigroup) since we want $G$ to \emph{be} a group.

Example:  $A$ is free abelian on $\{a,b,c\}$. Sample elements:
$3a+2b$, $-7a+4b-c$.

A free abelian group is a $\Z$-module.

\end{frame}

%% ----------------------------------------------------------------
\begin{frame}
\frametitle{Modules and vector spaces}

If we allow rational instead of integer coefficients, we have a free
$\Q$-module.  Sample element:  $\frac{2}{3}a + \frac{4}{7}b - 10c$.

Free means no non-trivial relations among basis elements: $K a + L b + M c = 0$
implies $K=L=M=0$.  For vector spaces $V$, a basis set $S$ is linearly
independent $\iff$ $V$ is free on $S$.

Vector spaces (free $\R$-modules) are typically obtained in one of two (very
different) ways:

(1) Start with a large $V$ and find a small $S$ inside.

(2) Start with (a smaller) $S$ and form a (larger) free $\R$-module $V$
generated by $S$.

\end{frame}

%% ----------------------------------------------------------------
\begin{frame}
\frametitle{Equivalence classes and modding}

We {\emcol say} we work with equivalence classes.  But what we often {\emcol
mean} are coset representatives and transformation rules.

Example:  $\Z/5\Z$.  Do we really think
$$
	\{\ldots, -1, 4, 9, \ldots\}
	+
	\{\ldots, -2, 3, 8, \ldots\}
	=
	\{\ldots, -3, 2, 7, \ldots\}
$$
every time we think of addition mod 5?

No!  We think
$$
	4 + 3 = 7 \to 2.
$$
We choose canonical representatives $\{0,1,2,3,4\}$.  When we do modular
arithmetic, we lift to $\Z$ and add (e.g. 7).  Then we transform the answer,
taking it $\pm 5$ repeatedly, to get a canonical representative (e.g. 2).

\end{frame}

%% ----------------------------------------------------------------
\begin{frame}
\frametitle{Tensor products of vector spaces}

For vector spaces\footnote{In this talk, for simplicity, I take them to be
finite-dimensional.} $V$ and $W$, the tensor product $V \tensor W$ is the free
vector space\footnote{A free abelian group along with the scalar operations is
a vector space.}  on $V \oplus W$ --- where the $S$ is the (huge!) set of
\emph{all} $(\vecv,\vecw)$ pairs --- modulo the relations (for all scalars $r$
and vectors $\vecv$, $\vecw$, etc.):

\begin{itemize}

\item[(i)]   $r\vecv \tensor \vecw = \vecv \tensor r\vecw$,

\item[(ii)]  $(\vecv_1+\vecv_2) \tensor \vecw =
	\vecv_1\tensor \vecw + \vecv_2\tensor \vecw$,

\item[(iii)] $\vecv \tensor (\vecw_1+\vecw_2) =
	\vecv\tensor \vecw_1 + \vecv\tensor \vecw_2$.

\end{itemize}

So, elements $\vecv \tensor \vecw$ are \emph{equivalence classes} of
$(\vecv,\vecw)$ pairs with these transformation rules.  But here, unlike
modular arithmetic, there is no clear favorite representative for each
equivalence class.  And, the generating set $S$ is already large.

Example: in $\R^2 \tensor \R^2$,
\begin{align*}
	\colvectwo{2}{4} \tensor \colvectwo{3}{5}
	=
	\colvectwo{1}{2} \tensor \colvectwo{6}{10}.
\end{align*}

\end{frame}

%% ----------------------------------------------------------------
\begin{frame}
\frametitle{Tensor products of vector spaces in coordinates}

{\emcol Theorem:}
$\R \tensor V \cong V$ with $r \tensor \vecv \mapsto 1 \tensor r\vecv$.
\\
{\emcol Theorem:}
$(V_1 \oplus V_2) \tensor W \cong (V_1 \tensor W) \oplus (V_2 \tensor W)$
and likewise on the right.
\\
{\emcol Consequence:}
$\R^m \tensor \R^n \cong \R^{mn}$ with basis elements $\vece_i \tensor
\vecf_j$.  Specifically,
$$
	\vecv \tensor \vecw
	=
	\left(\sum_{i=1}^m v_i \vece_i \right) \tensor
	\left(\sum_{j=1}^n w_i \vecf_j \right)
	=
	\sum_{i=1}^m
	\sum_{j=1}^n
	v_i w_j (\vece_i \tensor \vecf_j).
$$
(Not all elements of $V \tensor W$ are of the pure form $\vecv \tensor \vecw$,
but all elements are linear combinations of such.)  Coefficients of $\vecv
\tensor \vecw$ are $v_i w_j$.  We can write these in a tableau:
$$
%	\vecv \tensor \vecw
%	=
	\colvectwo{v_1}{v_2} \tensor \colvectwo{w_1}{w_2}
	=
	\begin{tabular}{c|cc}
	             & $w_1$     & $w_2$     \\
	\hline $v_1$ & $v_1 w_1$ & $v_1 w_2$ \\
	       $v_2$ & $v_2 w_1$ & $v_2 w_2$ \\
	\end{tabular};
	\quad
%
%	\vece_1 \tensor \vecf_2
%	=
	\colvectwo{1}{0} \tensor \colvectwo{0}{1}
	=
	\begin{tabular}{c|cc}
	         & 0 & 1 \\
	\hline 1 & 0 & 1 \\
	       0 & 0 & 0 \\
	\end{tabular}.
$$
Then
$$
\colvectwo{2}{4} \tensor \colvectwo{3}{5}
=
\begin{tabular}{c|cc}
         &  3 &  5 \\
\hline 2 &  6 & 10 \\
       4 & 12 & 20 \\
\end{tabular}
%
\quad \textrm{and} \quad
%
\colvectwo{1}{2} \tensor \colvectwo{6}{10}
=
\begin{tabular}{c|cc}
         &  6 & 10 \\
\hline 1 &  6 & 10 \\
       2 & 12 & 20 \\
\end{tabular}.
$$
Much nicer!  Now we have 2D arrays in coordinates:  the link between guises (1)
and (3) from the beginning of the talk is now becoming clear.  Next, guise (2).

\end{frame}

%% ----------------------------------------------------------------
\begin{frame}
\frametitle{Linear functionals}

{\emcol Definition:}  the dual space of $V$ is $V^* = \{\veclambda: V \to \R:
\veclambda \textrm{ linear}\}$.  It's the vector space of linear functionals on
$V$.

{\emcol Representation theorem} (trivial for finite-dimensional inner-product
spaces; Riesz-Fr\'echet for infinite-dimensional inner-product spaces):  Each
linear functional $\veclambda \in V^*$ has a unique $\vecu$ in $V$ so that
$\veclambda(\vecv) = \vecu^T \vecv = \vecu^* \vecv$ for all $\vecv$.
Specifically,
\begin{align*}
	\veclambda(\vecv) = \veclambda\left(\sum_{j=1}^n v_j \vece_j \right)
	= \sum_{j=1}^n \veclambda\left(\vece_j \right) v_j
\end{align*}
\emph{defines} the vector $\vecu$ with
$$
	u_j = \veclambda(\vece_j).
$$
Then
$$
	\veclambda(\vecv) = \vecu^* \vecv =
	\rowvecthree{u_1}{u_2}{u_3}
	\colvecthree{v_1}{v_2}{v_3}
	= u_1 v_1 + u_2 v_2 + u_3 v_3.
$$

A linear functional (in $V^*$) is a {\emcol row vector}.
A vector (in $V$) is a {\emcol column vector}.

\end{frame}

%% ----------------------------------------------------------------
\begin{frame}
\frametitle{Bases for linear functionals}

Given a basis $\{\vecb_1, \ldots, \vecb_n\}$ for $V$, the dual
basis
$\{\vecb^*_1, \ldots, \vecb^*_n\}$ for $V^*$ is such that
$$
	\vecb_i^*(\vecb_j) = \delta_{ij}.
$$
How to compute them?  Notation for matrices of row or column vectors
(e.g. $n=3$):
$$
\begin{pmatrix}
	\vecu \;\vline\; \vecv \;\vline\; \vecw
\end{pmatrix}
=
\begin{pmatrix}
	\ucmatrix{u_1 \\ u_2 \\ u_3} \;\vline\;
	\ucmatrix{v_1 \\ v_2 \\ v_3} \;\vline\;
	\ucmatrix{w_1 \\ w_2 \\ w_3}
\end{pmatrix}
\qqand
\begin{pmatrix}
	%\vecu^* \\ \hline \vecv^* \\ \hline \vecw^*
	\veclambda \\ \hline \vecmu \\ \hline \vecnu
\end{pmatrix}
=
\begin{pmatrix}
	%\ucmatrix{u^*_1 & u^*_2 & u^*_3} \\ \hline
	%\ucmatrix{v^*_1 & v^*_2 & v^*_3} \\ \hline
	%\ucmatrix{w^*_1 & w^*_2 & w^*_3}
	\ucmatrix{\lambda_1 & \lambda_2 & \lambda_3} \\ \hline
	\ucmatrix{\mu_1 & \mu_2 & \mu_3} \\ \hline
	\ucmatrix{\nu_1 & \nu_2 & \nu_3}
\end{pmatrix}.
$$
Then
$\vecb_i^*(\vecb_j) = \delta_{ij}$ means
$$
\begin{pmatrix}
	\vecb^*_1 \\ \hline \vdots \\ \hline \vecb^*_n
\end{pmatrix}
\begin{pmatrix}
	\vecb_1 \;\vline\; \cdots \;\vline\; \vecb_n
\end{pmatrix}
= I
\quad\textrm{i.e.}\quad
\begin{pmatrix}
	\vecb^*_1 \\ \hline \vdots \\ \hline \vecb^*_n
\end{pmatrix}
=
\begin{pmatrix}
	\vecb_1 \;\vline\; \cdots \;\vline\; \vecb_n
\end{pmatrix}^{-1}.
$$
Example:  if $\vecb_1=\colvectwo{2}{0}$ and $\vecb_2=\colvectwo{1}{1}$, then
$$
	\begin{pmatrix}
		\vecb^*_1 \\ \hline \vecb^*_2
	\end{pmatrix}
	=
	\begin{pmatrix}
		2 & 1 \\
		0 & 1 \\
	\end{pmatrix}^{-1}
	=
	\begin{pmatrix}
		1/2 & -1/2 \\
		0 & 1 \\
	\end{pmatrix}.
$$
So
$\vecb^*_1=\rowvectwo{1/2}{-1/2}$ and $\vecb^*_2=\rowvectwo{0}{1}$.

\end{frame}

%% ----------------------------------------------------------------
\begin{frame}
\frametitle{Multilinear functionals}

{\emcol Definition:} $\phi:V \oplus W \to \R$ is \emph{bilinear} if it is
linear in each slot.  Multilinear functions $\phi: \bigoplus_{i=1}^k V_i \to
\R$ are defined similarly.

{\emcol Examples:}  linear functional ($k=1$), dot product ($k=2$), determinant
(as a function not on matrices but on $k$ vectors in a $k$-dimensional vector
space).

{\emcol Definition of tensor product of dual spaces:} 
for $\veclambda \in V^*$ and $\vecmu \in W^*$,
$$
	(\veclambda \tensor \vecmu): V \oplus W \to \R
	:= \veclambda(\vecv) \vecmu(\vecw).
$$
This is a bilinear map; similarly, tensor products of more than two dual spaces
consist of multilinear maps.

In coordinates:
$$
	(\veclambda \tensor \vecmu)(\vecv,\vecw)
	= \veclambda \tensor \vecmu \left(
		\sum_{i=1}^m v_i \vece_i,
		\sum_{j=1}^n w_j \vecf_j
	\right)
	= \sum_{i=1}^m \sum_{j=1}^n v_i w_j \veclambda(\vece_i) \vecmu(\vecf_j).
$$
Now, the $\veclambda(\vece_i)$'s and $\vecmu(\vecf_j)$'s uniquely specify row
vectors as before.  So now, for tensor products of row vectors, just as we had
with tensor products of column vectors, we have tensors as 2D arrays in
coordinates.  Guises (1), (2), and (3) have been unified!  Next:  guise (4), by
route of first going back and shedding more light on guises (2) and (3).

\end{frame}

%% ----------------------------------------------------------------
\begin{frame}
\frametitle{Change of basis}

We call elements of $V$, $V \tensor W$, etc. \emph{contravariant tensors}; we
call elements of $V^*$, $V^* \tensor W^*$, etc. \emph{covariant tensors}.  Why?

{\emcol Notation:}  $E$, $F$ are bases for $V$.  Write $[\vecv]_E$,
$[\veclambda]_E$ for column/row vectors in $E$ coordinates.  Similarly for
$[\vecv]_F$ and $[\veclambda]_F$.

Example:
\begin{align*}
	E &= \left\{\vece_1, \vece_2\right\}
	= \left\{\colvectwo{1}{0}, \colvectwo{0}{1}\right\}, &
	%
	F &= \left\{\vecf_1, \vecf_2\right\}
	= \left\{\colvectwo{2}{0}, \colvectwo{1}{1}\right\}; \\
	[\vecv]_E &= \colvectwo{3}{4}, &
	[\veclambda]_E &= \rowvectwo{3}{4}.
\end{align*}
What are $[\vecv]_F$ and $[\veclambda]_F$?
\begin{align*}
	\colvectwo{v_{1,E}}{v_{2,E}} &=
	\colvectwo{3}{4}
		= v_{1,F} \colvectwo{2}{0}
		+  v_{2,F}  \colvectwo{1}{1}
	= \pcmatrix{2 & 1 \\ 0 & 1}
		\colvectwo{v_{1,F}}{v_{2,F}}; \\
	%
	\colvectwo{v_{1,F}}{v_{2,F}}
	&= \pcmatrix{2 & 1 \\ 0 & 1}^{-1}\colvectwo{3}{4}
	= \pcmatrix{1/2 & -1/2 \\ 0 & 1}^{-1}\colvectwo{3}{4}
	= \colcvectwo{-1/2}{4}.
\end{align*}

\end{frame}

%% ----------------------------------------------------------------
\begin{frame}
\frametitle{Explicit change of basis for column and row vectors, continued}

{\emcol Definition:}  The change-of-basis matrix $Q$ from basis $E$ to basis
$F$, so that $[\vecv]_E = Q [\vecv]_F$, is
$$
	Q =
	\pcmatrix{
		\vecf_1
		\;\vline\;
		\cdots
		\;\vline\;
		\vecf_n
	}_E.
$$
What about $[\veclambda]_F$?  Regardless of basis used, $\veclambda(\vecv)$
should be the same number:
\begin{align*}
	[\veclambda]_E [\vecf_1]_E &= \rowvectwo{3}{4} \colvectwo{2}{0} = 6; &
	[\veclambda]_E [\vecf_2]_E &= \rowvectwo{3}{4} \colvectwo{1}{1} = 7.
\end{align*}
So,
\begin{align*}
	[\veclambda]_F [\vecf_1]_F &= \rowvectwo{\lambda_1}{\lambda_2} \colvectwo{1}{0} = 6; &
	[\veclambda]_F [\vecf_2]_F &= \rowvectwo{\lambda_1}{\lambda_2} \colvectwo{0}{1} = 7.
\end{align*}
So $[\veclambda]_F = \rowvectwo{\lambda_1}{\lambda_2} = \rowvectwo{6}{7}$.
Stacking these equations together and generalizing for arbitrary $n$, we have
\begin{align*}
	[\veclambda]_E
	\pcmatrix{
		\vecf_1 \;\vline\; \cdots \;\vline\; \vecf_n
	}_E
	&=
	[\veclambda]_F
	\pcmatrix{
		\vecf_1 \;\vline\; \cdots \;\vline\; \vecf_n
	}_F
	= [\veclambda]_F I,
	= [\veclambda]_F,
\end{align*}
i.e.
\begin{align*}
	[\veclambda]_E \; Q &= [\veclambda]_F.
\end{align*}
In summary, {\emcol column vectors and row vectors transform differently on
change of coordinates}:
\begin{align*}
	[\vecv]_E   \stackrel{Q}{\longleftarrow} [\vecv]_F; & \qquad \qquad
	[\veclambda]_E \stackrel{Q}{\longrightarrow} [\veclambda]_F.
\end{align*}

\end{frame}

%%% ----------------------------------------------------------------
%\begin{frame}
%\frametitle{Explicit change of basis for column and row vectors, continued}
%
%What about
%$[\vece^*_1]_F = \rowvectwo{2}{1}$ and $[\vece^*_2]_F = \rowvectwo{0}{1}$?
%Again, $[\vece_i^*]_F [\vece_j]_F = \delta_{ij}$:
%\begin{align*}
% 	\pcmatrix{
%		\vece_{11}^* & \vece_{12}^* \\ \hline
%		\vece_{21}^* & \vece_{22}^*
%	}_F
% 	\pcmatrix{
%		\ucmatrix{\vece_{11} \\ \vece_{12}} \;\vline\;
%		\ucmatrix{\vece_{21} \\ \vece_{22}}
%	}_F
%	=
%	\pcmatrix{
%		1 & 0 \\
%		0 & 1 \\
%	}.
%\end{align*}
%But the second matrix is just the inverse change-of-basis matrix, $Q^{-1}$.  So
%(for general $n$) we have
%\begin{align*}
% 	\pcmatrix{ \vece_1^* \\ \hline \vdots \\ \hline \vece_n^*}_F
%	= QI = Q.
%\end{align*}

%\end{frame}

%% ----------------------------------------------------------------
\begin{frame}
\frametitle{Tensors in physics}

Instead of vector spaces $V$ and $V^*$ with real coefficients, use tangent
bundles/cobundles $TM$ and $T^*M$ over a manifold $M$.  The coefficients are
$C^\infty(M)$ functions.  (There are vector spaces attached to each point on the
manifold, and the coefficients vary smoothly as you move from one point to
another.)

The change-of-basis matrix in coordinates is $Q = \partial x_i / \partial y_j$,
whose inverse is $Q^{-1} = \partial y_i / \partial x_j$.  Then
\begin{align*}
	[\lambda_i]_F
		&= \sum_{j=1}^n \frac{\partial x^j}{\partial y^i} [\lambda_j]_E
		& \quad \textrm{covariant}; \\
	[v^i]_F
		&= \sum_{j=1}^n \frac{\partial y^i}{\partial x^j} [v^j]_E
		& \quad \textrm{contravariant}.
\end{align*}

Conventions:  subscripts outside derivative denominators, and superscripts
inside derivative denominators, are called {\emcol covariant indices}.
Superscripts outside derivative denominators, and subscripts inside derivative
denominators, are called {\emcol contravariant indices}.

Physicists say that a covariant/contravariant tensor, respectively, is
\emph{anything} which transforms this way on change of coordinates.

\end{frame}

%% ----------------------------------------------------------------
\begin{frame}
\frametitle{Aside:  Tensor algebras and mixed tensors}

The tensor product of vector spaces is another vector space.  You can {\emcol
add} two tensors together, or {\emcol multiply} one tensor by a scalar.  But if
you can moreover define some way $\vecu * \vecv$ to {\emcol multiply} one
tensor $\vecu$ by another tensor $\vecv$, you have an {\emcol algebra}.  The
product is typically of higher dimension than either of the operands:  if
$\vecu \in V \tensor V$ and $\vecv \in V$, then $\vecu * \vecv \in V \tensor V
\tensor V$.  This leads to a {\emcol graded algebra} which I would discuss more
today, if there were time.

If $\vecu * \vecv = \pm \vecv * \vecu$ then we get {\emcol symmetric} and
{\emcol alternating} tensor algebras, respectively:
$\vecu \vecv$ and $\vecu \wedge \vecv$.

One can tensor together $V \tensor V^*$, etc.  Such tensors are of {\emcol
mixed valence}, not purely covariant or contravariant.  In fact, plain old
linear transformations may be viewed as being of this form.


\end{frame}

%% ----------------------------------------------------------------
\begin{frame}
\frametitle{There's more \ldots}

Tensors in all four of their guises have been unified --- they really are all
the same thing.  There is a {\emcol fifth guise}, which I (unfortunately)
stopped thinking about shortly after (on the second attempt) I (high-)passed my
geometry-topology qualifying exam:  the {\emcol geometrical guise}.

Namely:
\begin{itemize}

\item The bivector $\vecu\wedge\vecv$ is the equivalence class of all vector
pairs coplanar with $\vecu$ and $\vecv$, with the same signed area.

\item The symmetric product $\vecu\vecv$ is all pairs coplanar to $\vecu$ and
$\vecv$ with the same inner product.

\item The bifunctional $\vecu^*\wedge\vecv^*$ measures the area spanned by two
other vectors, projected onto the plane spanned by $\vecu$ and $\vecv$.

\item We can also form equivalence classes using things like
$$
	(\veclambda\wedge\vecmu)(\vecu,\vecv) =
	(\veclambda\wedge\vecmu)(\vecu\wedge\vecv) =
	\det(\veclambda|\vecmu)\det(\vecu|\vecv).
$$
\end{itemize}

These generalize to three and more dimensions.  Although I don't know the
history well enough, I suspect that it was precisely these geometric notions
which led to the axioms we have today.

\end{frame}

%% ----------------------------------------------------------------
\begin{frame}

\begin{center}
{\emcol Thanks for attending!}
\end{center}

\end{frame}

%%% ================================================================

\end{document}
