% conjugation equals involution?

% action functional pf. -- type it up.

% Projected matrices with input & output indices marked.

% fdiff eqn's; z^k and z^k w.  Hmm ...

% S(.,.) = 0 is a SBF!  Very non-degenerate though.
% SBF (BFs in general really ...) as v^t Q v stuff?

% symm/non-degenerate/both BF -- which and where?



%% ----------------------------------------------------------------
% !!! wiki re dirac eq'n!!
% Levi-Civita connection ...
%   \nabla_{e_i} e_j = \sum_k \Gamma^k_{i,j} e_k.
% Christoffel symbols ...
% Green's fcn
% Dirac eqn is elliptic?

%% ================================================================
\section{Overview}

The Berezin integral is a concept arising in mathematical physics.  The term
``integral'' is a misnomer; it is a functional on functions of Grassmann
variables.  Such ``integrals'' are not measure-theoretic, and are not defined
for indefinite or infinite integrals (\cite{bib:rabin}).  Here we begin,
as guided by \cite{bib:palmer1}, to attempt to rigorize the Berezin integral
via the following scheme:
\begin{itemize}
\item Restrict the domain from $\R^3$ to $\R$.
\item Discretize the domain from $\R$ to $\Z$.
\item Replace continuous operators with range-one finite-difference operators.
\item Truncate the domain from $\Z$ to a finite, contiguous subset thereof.
\item Now working in the well-understood context of an exterior algebra over
	a finite-dimensional vector space, re-characterize the Berezin integral
	using Fock representations.
\item In future work, attempt to recover the continuum limit and multiple
	dimensions in a meaningful way.
\end{itemize}

This is a work in progress; not all proofs are complete as of this writing.

An outline of this paper is as follows:
\begin{itemize}
\item A brief, motivating discussion of physics terminology in section
	\ref{sec:physics}.
\item Notational conventions in section \ref{sec:notation}.
\item Algebraic preliminaries in section \ref{sec:la_mla}.
\item Motivation regarding linear operators in section
	\ref{sec:cts_disc_oprs}.
\item Discretization of the domain in section \ref{sec:discrete_vector_bundles}.
\item Discretized operators in section \ref{sec:discrete_difference_operators}.
\item The discrete Berezin integral in section \ref{sec:discrete_berezin}.
\item The Fock characterization of the discrete Berezin integral in section
	\ref{sec:fock_chzn}.
\item An explicit computation for a two-dimensional, constant-coefficient
	case is done in section \ref{sec:mma_rox}.
\item Directions for further research are sketched in section
	\ref{sec:further}.
\end{itemize}

%% ================================================================
\newpage
\section{Physical background}
\label{sec:physics}

The discussion in this section serves to motivate concepts, at the basic level
needed for this paper.  For detailed information, see \cite{bib:berezin},
\cite{bib:fs}, and \cite{bib:frankel}.

%% ----------------------------------------------------------------
\subsection{Probablity functions}

Quantum mechanics deals with elementary particles not as localized points, but
rather as \emphidx{probability distributions}.  Consider a function $\psi: \R^3
\to \C$ which is called the \bfidx{wave function} associated with a particle.
This is taken to mean that
$$
	\int_D \psi(x)\psibar(x) \,dx =
	\int_D |\psi(x)|^2 \,dx
$$
is the \emph{probability} of finding a particle in the region $D$ of
three-space.  (For example, if a particle is confined to a line, then $\int_a^b
|\psi(x)|^2\,dx$ is the probability of finding the particle between points $a$
and $b$.)  Only the \bfidx{probability function} $|\psi(x)|^2$ is observable;
the complex phase is not observable.  Why do we create a mathematical model
which uses unobservables?  The answer is that the underlying function $\psi(x)$
has linearity (superposition) properties which the probability function does
not.

Now, it is certain that the particle is \emph{somewhere} in space.  Thus, we
require
$$
	\int_{\R^3} |\psi(x)|^2\,dx = 1.
$$
That is, wave functions must be \emphidx{square integrable} and we have
$$
	\psi \in L^2(\R^3).
$$
In fact, historically such functions lead to the abstract notion of Hilbert
space.

Wave functions as presented thus far encode only the probability of a particle
being found in a given location:  the target space, $\C$, is one-dimensional.
Wave functions may also be constructed which encode not only position but also
\emphidx{momentum}, a quantity called \emph{spin}, etc.  Here we have functions
with values in $\C^N$ for some $N$.  In this case a wave function for a
particle is vector-valued, of the form
$$
	\psi(x) = (\psi_1(x), \ldots, \psi_N(x)).
$$

%% ----------------------------------------------------------------
\subsection{Fermions and bosons}

A system with two particles is described by a single function $\psi(x_1, x_2)$.
There exist in nature particles called \bfidx{bosons} which have
\bfidx{symmetric} wave functions.  That is, for all $x_1, x_2 \in \R^3$,
$$
	\psi(x_1,x_2) = \psi(x_2, x_1).
$$
Bosons include photons and gluons; these may be thought of as force carriers.
Likewise, there exist particles called \bfidx{fermions} which have
\bfidx{skew-symmetric} wave functions: for all $x_1, x_2 \in \R^3$,
$$
	\psi(x_1,x_2) = -\psi(x_2, x_1).
$$
Fermions include quarks and electrons; these may be thought of as the stuff of
matter.  (In fact, certain combinations of matter particles behave as bosons.
I am glossing this fact for simplicity of presentation.) Note that this
antisymmetry forces $\psi=0$ at $x_1=x_2$.  This is the \emphidx{Pauli
exclusion principle} which states that fermions do not occupy the same
location.  This repulsion may be thought of as comprising the ``stiffness'' of
matter.  When two automobiles approach each other head-on, their headlight
beams pass through one another but the vehicles themselves do not.

A system with $n$ particles is described by a function $\psi(x_1, \ldots,
x_n)$; skew-symmetry or symmetry of the particles is indicated by sign change,
or lack thereof, respectively, upon transposition of arguments.

%% ----------------------------------------------------------------
\subsection{Operators}

In physics, extensive use is made of \bfidx{operators} including the
differentiation operator.  In this paper, differentiation operators are briefly
treated in section \ref{sec:cts_disc_oprs}; the main treatment on Berezin
integrals deals with an abstract skew-symmetric operator.

See the above-cited references for more details.  Also see \cite{bib:lax}
for information on operators on function spaces.

%% ----------------------------------------------------------------
\subsection{Creation and annihilation operators}
\label{sec:phys_creat_annih}

An $n$-particle state is represented by a wave function
$$
	\psi(x_1, \ldots, x_n);
$$
an $n+1$-particle state is represented by a wave function
$$
	\phi(x_1, \ldots, x_{n+1}).
$$
Operators which map from the former to the latter, or vice versa, are called
\bfidxtwo{creation}{creation operator} and \bfidxtwo{annihilation
operators}{annihilation operator} respectively.  This topic is treated
beautifully in chapter 1 of \cite{bib:berezin}; discrete versions are presented
in section \ref{sec:creat_annih} of this paper.

%% ----------------------------------------------------------------
\subsection{Continuous Berezin integration}
\label{sec:cts_berezin_integral}

We attempt to define expressions of the form
$$
	\int F(\psi) e^{\mcL} \prod_{x,j} d\psi_j(x)
$$
where
$$
	\mcL = \int L \psi(x) dx
$$
is an \bfidx{action functional}.  The $\mcL$ integral is standard; it is the
outer integral which is the Berezin integral.  Here, $F(\psi)$ is a polynomial
in $\psi_1, \ldots, \psi_N$, and $L$ is an abstract skew-symmetric operator.
For the purposes of this paper, $F$ is ignored.  Thus we want to integrate
$$
	\int e^{\mcL} \prod_{x,j} d\psi_j(x).
$$
The rules for such integrals (\cite{bib:palmer1}) are
$$
	\int \psi_j \,d\psi_j = 1 \qqand
	\int 1 \,d\psi_j = 0
$$
with antisymmetric extension.  A more detailed description is provided in
\cite{bib:berezin}; see in particular the discussion of \emphidx{continual
integrals} on p. 38.

%% ================================================================
\newpage
\section{Notation and conventions}
\label{sec:notation}

$Y$ is used for a general finite-dimensional complex vector space; $V$ is used
for discrete bundles as discussed in section \ref{sec:discrete_vector_bundles}.

Given a set $S$, $2^S$ denotes the set of all subsets of $S$.  Recall that if
$S$ is a set containing $N$ elements, then $\#(2^S) = 2^N$ which justifies the
notation.

Given a complex vector space $Y$, write $Y^*$ for the \bfidx{dual space} of
$Y$, namely, the linear maps from $Y$ to $\C$.

The \bfidx{span} of a set of vectors $\vecv_1, \ldots, \vecv_k$ is written
$$
	\langle \left\{ \vecv_1, \ldots, \vecv_k \right\} \rangle.
$$

The \bfidx{tensor algebra} on $Y$ is written $\mcT(Y)$:
$$
	\mcT(Y) = \C \oplus Y \oplus (Y\tensor Y) \oplus (Y\tensor Y \tensor Y) \oplus \ldots .
$$

%% ================================================================
\newpage
\section{Linear and multilinear algebra}
\label{sec:la_mla}

%%% ----------------------------------------------------------------
%\subsection{Conjugate-pair bases}
%
%Let $Y$ be a complex vector space of dimension $2N$.
%
%$\{\phi_1, \ldots, \phi_{2N}\}$ vs.
%$\{\psi_1, \psibar_1, \ldots, \psi_N, \psibar_N\}$
%
%$\psi_1 = \phi_1 + i \phi_2$;
%$\psibar_1 = \overline{\phi_1 + i \phi_2}$.
%
%How does this affect skew-symmetry of linear operators on the space?
%
%Mention \bfidx{charge conjugation} and cite.

%%% ----------------------------------------------------------------
%\subsection{Orthonormal bases}
%
%\begin{lem}
%\label{lem:ortho_COB}
%Let $\Psi = \{\psi_1, \ldots, \psi_N\}$ and $\Phi= \{\phi_1, \ldots, \phi_N\}$
%be orthonormal bases for a complex vector space $Y$.  Then the change-of-basis
%matrix $Q$ from $\Phi$ to $\Psi$ is unitary.
%\begin{proof}
%TBD.
%\end{proof}
%\end{lem}

%% ----------------------------------------------------------------
\subsection{Bilinear forms, sesquilinear forms, and conjugations}

Throughout this section, let $Y$ be a complex vector space of dimension $N$.

\begin{defn}
A \bfidx{sesquilinear form} is a map
$$
	B: Y \times Y \to \C,
$$
such that for all $\vecu,\vecv,\vecu_1,\vecu_2,\vecv_1,\vecv_2 \in Y$ and all
$a,b \in \C$,
\begin{eqnarray*}
	B(a\vecu,\vecv) &=& \overline{a} B(\vecu,\vecv), \\
	B(\vecu,b\vecv) &=& bB(\vecu,\vecv), \\
	B(\vecu_1+\vecu_2,\vecv) &=& B(\vecu_1,\vecv) + B(\vecu_2,\vecv),
		\;\textrm{and} \\
	B(\vecu,\vecv_1+\vecv_2) &=& B(\vecu,\vecv_1) + B(\vecu,\vecv_2).
\end{eqnarray*}
Furthermore, a \bfidx{symmetric sesquilinear form} is a bilinear form such
that for all $\vecu,\vecv\in Y$,
$$
	B(\vecu,\vecv) = \overline{B(\vecv,\vecu)}.
$$
\end{defn}

\begin{defn}
A \bfidx{bilinear form} is a map
$$
	B: Y \times Y \to \C,
$$
such that for all $\vecu,\vecv,\vecu_1,\vecu_2,\vecv_1,\vecv_2 \in Y$ and all
$a,b \in \C$,
\begin{eqnarray*}
	B(a\vecu,\vecv) &=& aB(\vecu,\vecv), \\
	B(\vecu,b\vecv) &=& bB(\vecu,\vecv), \\
	B(\vecu_1+\vecu_2,\vecv) &=& B(\vecu_1,\vecv) + B(\vecu_2,\vecv),
		\;\textrm{and} \\
	B(\vecu,\vecv_1+\vecv_2) &=& B(\vecu,\vecv_1) + B(\vecu,\vecv_2).
\end{eqnarray*}
Furthermore, a \bfidx{symmetric bilinear form} is a bilinear form such
that for all $\vecu,\vecv\in Y$,
$$
	B(\vecu,\vecv) = B(\vecv,\vecu).
$$
A \bfidx{skew-symmetric bilinear form} is a bilinear form such that
that for all $\vecu,\vecv\in Y$,
$$
	B(\vecu,\vecv) = -B(\vecv,\vecu).
$$
\end{defn}

\begin{defn}
Let $B=\{\vecb_1, \ldots, \vecb_N\}$ a basis for $Y$.  Then $B$ is said to be
\bfidx{orthonormal} with respect to a symmetric sesquilinear or bilinear form
$F$ if $F(\vecb_i, \vecb_j) = \delta_{ij}$ for all $i,j = 1, \ldots, N$.
\end{defn}

\begin{defn}
A sesquilinear or bilinear form $F$ is said to be \bfidx{non-degenerate} if for
all $\vecv \in Y$, the kernel of $F(\cdot, \vecv)$ is the zero vector.
\end{defn}

\beginex
Given an orthonormal basis $\{\vece_1, \ldots, \vece_N\}$,
one may write a symmetric sesquilinear form $\ssfdot$
$$
	\ssf \vecu \vecv = \sum_{j=1}^N \overline{u}_i v_i
$$
or a symmetric bilinear form $\sbfdot$
$$
	\sbf \vecu \vecv = \sum_{j=1}^N u_i v_i.
$$
The following proposition shows that the converse is true.
\mendex

\begin{prop}
For any symmetric non-degenerate sesquilinear or bilinear form $F$ there is a
basis for $Y$ that is orthonormal with respect to $F$
\begin{proof}
Modify Gram-Scmidt.
\end{proof}
\end{prop}

%Conjugation of the first argument is the usual convention in
%mathematical physics; we follow that convention here.

There is another, equivalent notion of non-degeneracy: namely, a representation
theorem for the dual space.

\begin{prop}
A bilinear form $B$ is non-degenerate iff for all $\lambda \in
Y^*$, there exists a fixed $\vecu \in Y$ such that for all $\vecv \in Y$,
$\lambda(v) = \vecu^T \vecv$.
\begin{proof}
%First suppose that $B$ is non-degenerate.  Let $\lambda \in Y^*$.
TBD.
\end{proof}
\end{prop}

%xxx note that in the above example, $\ssf \vecu \vecv = \sbf{\overline\vecu} \vecv$.

\begin{defn}
A \bfidx{conjugation} on $Y$ is a linear transformation $K: Y \to Y$ such that
$K^2=1$, $K \ne 1$, and for all $\vecv \in Y$ and all $a \in \C$, $K(a\vecv) =
\overline{a} K(\vecv)$.
\end{defn}

\beginex
Let $Y$ be $N$-dimensional.  If $\vecv = (v_1, \ldots, v_N)$ in coordinates
with respect to some basis, then put $K(\vecv) = (\overline{v_1}, \ldots,
\overline{v_N})$.
\mendex

%\beginex
%Find another example.
%\mendex

%\beginex
%For $\vecu,\vecv$ in an $N$-dimensional complex vector space $Y$, define
%$$
%	B(\vecu,\vecv) = \sum_{k=1}^N u_i v_i.
%$$
%This is a bilinear form on $Y$.  Note that this bilinear form comes from
%the standard inner product $\langle \cdot, \cdot \rangle$
%and the standard complex conjugation $C$:
%$$
%	B(\vecu,\vecv) = \langle C\vecu,\vecv \rangle.
%$$
%\mendex

\begin{defn}
\label{defn:orthogonal_space}
An \bfidx{orthogonal space} is a complex vector space $Y$ with a distinguished
non-degenerate complex bilinear form.  We write that form simply as $(\cdot,
\cdot)$.
\end{defn}

%\begin{rem}
%xxx abstractly connect sesquilinear and bilinear forms via a conjugation.
%Uniqueness theorem (up to similiarity)?
%\end{rem}

%There are several bilinear forms used in this paper \ldots name and define all
%of them here, or xref forward.

%% ----------------------------------------------------------------
\subsection{Pairing and transpose}

\begin{defn}
Let $Y$ and $Z$ be vector spaces over the same base field.  A \bfidx{pairing}
is a bilinear map from $Y \times Z$ to the base field, written $\langle y, z
\rangle$, for $y\in Y$ and $z \in Z$.
\end{defn}
%(This is an unfortunate reuse of the angle brackets.)

\begin{defn}
A pairing $\langle \cdot, \cdot \rangle$ of $Y$ and $Z$ is said to be
\bfidxtwo{non-degenerate}{non-degenerate pairing} if for all $0\ne y \in Y$ and
all $0\ne z \in Z$, the linear maps $\langle y, \cdot\rangle:Z \to Z$ and
$\langle \cdot, z\rangle: Y \to Y$ have trivial kernels.
\end{defn}

Pairings generalize the notions of inner product, integral, and duality of
vector spaces.

\beginex
Any bilinear form on $Y$ is a pairing of $Y$ with itself.  In particular, given
a function space $Y$, we write
$$
	\langle f(x), g(x) \rangle = \int f(x) \,g(x) \,dx.
$$
\mendex

\beginex
One may pair a vector space $Y$ with its dual $Y^*$.  Then for
$\lambda \in Y^*$ and $\vecv \in Y$ we write
$$
	\langle \lambda, \vecv \rangle = \lambda(\vecv).
$$
\mendex

\begin{defn}
\label{defn:transpose}
Let $A:Y_1 \to Y_2$ be a linear transformation of vector spaces.  Further
suppose that we have non-degenerate pairings of $Y_1$ with $Z_1$, and of $Y_2$
with $Z_2$.  Then the \bfidx{transpose} of $A$ is the linear transformation
$A^\tau:Z_2 \to Z_1$ such that
$$
	\langle y_2, A z_1 \rangle = \langle A^\tau y_2, z_1 \rangle
$$
for all $y_2 \in Y_2$ and $z_1 \in Z_2$.  A diagram is as follows:
$$
\begin{array}{ccc}
	Y_1 & \stackrel{A}{\longrightarrow} & Y_2 \\
	\ssfdot &  & \ssfdot \\
	Z_1 & \stackrel{A^\tau}{\longleftarrow} & Z_2 \\
\end{array}
$$
\end{defn}

%xxx rmk:  non-degeneracy of the pairings implies uniqueness of the transpose.
%I should say at least something about existence and uniqueness of the
%transpose.

%\begin{defn}
%Let $Y$ be a vector space.  Let $T: Y \to Y$, $v \in Y$, and $\lambda \in Y^*$.
%We may form a \bfidx{pairing} of $Y^*$ and $Y$, i.e. a bilinear map from
%$Y\times Y^*$ to the base field, via
%$$
%	\langle \vecv, \lambda \rangle = \lambda(\vecv).
%$$
%Then we define the \bfidx{transpose} of $T$, a map
%$T^*: Y^* \to Y^*$, via
%$$
%	\langle T\vecv, \lambda \rangle = \langle \vecv, T^*\lambda \rangle.
%$$
%The use of angle brackets is standard in the literature, but is undesirable for
%this paper.  Thus, we instead write
%$$
%	\lambda(T\vecv) = (T^*\lambda)(\vecv).
%$$
%\end{defn}

%xxx how to compute $A^\tau$ (in coordinates!) with respect to a given bilinear
%form.  Do we need this?  We certainly do for the annihilation operator!  But do
%we need it for matrix operators?  It boils down to solving a linear system
%\ldots.  In the correct basis, it's just the usual transpose.  Prove this.

\begin{rem}
\label{rem:matrix_transpose_sbf}
Using an basis which is orthonormal with respect to the form, the transpose may
be computed easily by taking the usual matrix transpose.
\end{rem}

%% ----------------------------------------------------------------
\subsection{Quadratic forms}

Throughout this section, let $Y$ be a complex vector space.

\begin{defn}
A \bfidx{quadratic form} is a map
$$
	Q: Y \to \C
$$
such that for all $\vecu,\vecv \in Y$ and all $a \in \C$,
$$
	Q(a\vecu) = a^2 Q(\vecu),
$$
and such that the map
$$
	B: Y \times Y \to \C
$$
given by
$$
	B(\vecu,\vecv) = Q(\vecu+\vecv)-Q(\vecu)-Q(\vecv)
$$
is bilinear.
This map $B$ is called the \bfidx{associated bilinear form} $B$ for $Q$.  (Note
that $B$ is symmetric by construction.)
\end{defn}

\begin{rem}
The association may be done the other way as well:  given a symmetric bilinear
form $B$, we may create a quadratic form $Q$ via
$$
	Q(\vecu) = \frac{1}{2} B(\vecu,\vecu).
$$
\end{rem}

\begin{prop}
Given $B$, the associated $Q$ is in fact bilinear.
\begin{proof}
Let $a \in \C$ and $\vecu \in Y$.  Then
$$
	Q(a\vecu) = B(a\vecu,a\vecu) = a^2B(\vecu,\vecu) = a^2 Q(\vecu).
$$
For the second axiom, it suffices to show that, given $B$ and the associated
quadratic form $Q$, the bilinear form $B'$ associated to $Q$ is again $B$.
We have
\begin{eqnarray*}
	B'(\vecu,\vecv) &=& Q(\vecu+\vecv)-Q(\vecu)-Q(\vecv)\\
	&=& \frac{1}{2}\left[B(\vecu+\vecv,\vecu+\vecv)-B(\vecu,\vecu)-B(\vecv,\vecv)\right]\\
	&=& \frac{1}{2}\left[B(\vecu,\vecu)+B(\vecu,\vecv)+B(\vecv,\vecu)+B(\vecv,\vecv)-B(\vecu,\vecu)-B(\vecv,\vecv)\right]\\
	&=& \frac{1}{2}\left[B(\vecu,\vecv)+B(\vecv,\vecu)\right]\\
	&=& B(\vecu,\vecv)
\end{eqnarray*}
since $B$ is symmetric.
\end{proof}
\end{prop}

%xxx prove this?  I don't want to be leaving stuff ``as an exercise'' when it is
%I who am the student.

%% ----------------------------------------------------------------
\subsection{Exterior algebras}
\label{sec:ext_alg}

I take \index{exterior algebra}exterior algebras to be known to the reader.
One may consult any graduate algebra text, e.g. \cite{bib:df} or
\cite{bib:rotman}, or any graudate differential geometry text such as
\cite{bib:lee2}.  Here, I make note of my conventions and notation.  (Caution
is due:  in the literature, tensor algebras harbor an annoying diversity of
notation.)

\begin{defn}
\label{defn:ext_alg}
Recall $\vecu \wedge \vecv = -\vecv \wedge \vecu$, i.e.  $\vecu \wedge
\vecv + \vecv \wedge \vecu = 0$, and
$$
	\Lambda(Y) = \mcT(Y) \big/ \langle \vecu\vecv + \vecv\vecu \rangle.
$$
\end{defn}

\begin{defn}
\label{defn:wedge_scales}
Following \cite{bib:palmer2}, we define the \bfidx{scale factor} for the
wedge product of a $k$-vector and an $\ell$-vector as
$$
	\sqrt{\frac{(k+\ell)!}{k!|\ell!}}.
$$
This choice is perhaps non-standard, but is important as will be seen below.
\end{defn}

%xxx tabulate some scale factors.

Given a complex vector space $Y$, write $\Lambda^0(Y) = \C$, $\Lambda^1(Y) =
Y$, $\Lambda^2(Y)$ for alternating 2-tensors on $Y$, etc.  Then write
$$
	\Lambda(Y) = \C \oplus Y \oplus \Lambda^2(Y) \oplus \ldots.
$$
for the graded alternating tensor algebra which is endowed with the wedge
product.
%[xxx still need to
%make it clear that these are covectors and \emph{not} linear functionals.]
%[xxx note zero is in all of them so we can't just say ``alternating
%$k$-tensors''.  Say it correctly as the space spanned by \ldots .]

\begin{rem}
\label{rem:biwedge_in_coords}
Recall that if $\vecu=(u_1, \ldots, u_N) \in Y$ and $\vecv=(v_1, \ldots, v_N)
\in Y$ then the \bfidx{tensor product} may be visualized in coordinates as an
\bfidx{outer product}
$$
	(\vecu \tensor \vecv)_{ij} = u_i v_j,
$$
i.e. the two-dimensional $N \times N$ array
$$
	\begin{pmatrix}
	u_1v_1 & \cdots & u_1 v_N \\
	\vdots &  & \vdots \\
	u_Nv_1 & \cdots & u_N v_N
	\end{pmatrix}.
$$
This may be computed by hand using the tableau
$$
	\begin{tabular}{r|rrrrr}
	{}       & $v_1$    & \ldots   & $v_N$ \\
	\hline
	$u_1$    & $u_1v_1$ & $\cdots$ & $u_1 v_N$ \\
	$\vdots$ & $\vdots$ &          &$\vdots$    \\
	$u_N$    & $u_Nv_1$ & $\cdots$ & $u_N v_N$. \\
	\end{tabular}
$$
Furthermore, since
$$
	\vecu \wedge \vecv = \frac{1}{\sqrt 2} \left(
		\vecu \tensor \vecv - \vecv \tensor \vecu \right),
$$
we have in coordinates
\begin{eqnarray*}
\vecu \wedge \vecv
&=&
		\frac{1}{\sqrt 2} \begin{pmatrix}
		u_1v_1 & \cdots & u_1 v_N \\
		\vdots &  & \vdots \\
		u_Nv_1 & \cdots & u_N v_N
		\end{pmatrix}
	-
		\frac{1}{\sqrt 2} \begin{pmatrix}
		u_1v_1 & \cdots & u_N v_1 \\
		\vdots &  & \vdots \\
		u_1v_N & \cdots & u_N v_N
		\end{pmatrix} \\
&=&
		\frac{1}{\sqrt 2} \begin{pmatrix}
		u_1v_1 - u_1 v_1 & \cdots & u_1 v_N - u_N v_1\\
		\vdots &  & \vdots \\
		u_Nv_1 - u_1v_N& \cdots & u_N v_N - u_N v_N
		\end{pmatrix} \\
&=&
		\frac{1}{\sqrt 2} \begin{pmatrix}
		0                     & u_1v_2-u_2v_1 & \cdots & u_1v_{N-1}-u_{N-1}v_1         & u_1v_N-u_Nv_1 \\
		u_2v_1-u_1v_2         & 0             &        &                       & \vdots \\
		\vdots                &               & \ddots &                       &     \\
		u_{N-1}v_1-u_1v_{N-1} &               &        & 0                     & u_{N-1}v_N-u_Nv_{N-1} \\
		u_N    v_1-u_1v_N     & \cdots        &        & u_Nv_{N-1}-u_{N-1}v_N & 0   \\
		\end{pmatrix}.
\end{eqnarray*}
\end{rem}

\begin{rem}
\label{rem:std_biwedge}
In particular, the wedge of two standard-basis vectors contains a single
1, a single -1, and the rest zeroes.  For example, with $N=4$,
$$
	\vece_3 \wedge \vece_4 =
	\frac{1}{\sqrt 2}
	\left[\begin{array}{rrrr}
		0 & 0 &  0 & 0 \\
		0 & 0 &  0 & 0 \\
		0 & 0 &  0 & 1 \\
		0 & 0 & -1 & 0 \\
	\end{array}\right].
$$
\end{rem}

\begin{rem}
\label{rem:multiwedge_in_coords}
A bi-vector as described in remark \ref{rem:biwedge_in_coords} has $N^2$
components when written in coordinates.  Similarly, a tri-vector, i.e. an
element of $\Lambda^3(Y)$, has $N^3$ components, and in general a $k$-vector
(an element of $\Lambda^k(Y)$) has $N^k$ components.  In particular, a
top-level form (an element of $\Lambda^N(Y)$) has $N^N$ components.  Recall,
however, that the dimension of $\Lambda^k(Y)$ is ${N \choose k}$ and in
particular $\Lambda^N(Y)$ is one-dimensional.  Thus, na\"ive calculations such
as those in remark \ref{rem:biwedge_in_coords} produce an overabundance of
coefficients; a more elegant computation extracts a single scalar from
a top-level form.
\end{rem}

%xxx that whole paragraph is awkward.  Re-phrase it somehow.  Top-level form is
%$$
%	\omega = \alpha \vece_1 \wedge \cdots \wedge \vece_N.
%$$
%The $\vece_1 \wedge \cdots \wedge \vece_N$ has $N^N$ components, but all of
%$\omega$ is specified solely by the single scalar $\alpha$.

%xxx include a $3\times 3$ example.

%xxx do the tensor pairing too.  xxx make sure the ext pairing follows from the
%tensor pairing along with the correct scale factors.

%xxx include an example computation of the wedge of a bivector with itself.  (Do
%$N=4$.)

\beginex
\label{ex:q_wedge_q}
Let $q$ be an arbitrary bivector on a 4-dimensional space:
$$
	q =
	\left[\begin{array}{rrrr}
	0  &  a &  b & c  \\
	-a &  0 &  d & e  \\
	-b & -d &  0 & f  \\
	-c & -e & -f & 0  \\
	\end{array}\right].
$$
To compute $q \wedge q$ from the definition, one could compute $q \tensor q$
and then compute the alternation of the result.  For the former we have
(disregarding the scale factor):
$$
	q \tensor q =
	\left[\begin{array}{rrrr|rrrr|rrrr|rrrr}
  % -    - -    - -    - ----    - -   - -    - -    - ----    - -    - -   - -    - ----    - -    - -    - -   -  -
  0 &    0 &    0 &    0    &    0 & a^2 &   ab &   ac    &    0 &   ab & b^2 &   bc    &    0 &   ac &   bc & c^2 \\
  0 &    0 &    0 &    0    & -a^2 &   0 &   ad &   ae    &  -ab &    0 &  bd &   be    &  -ac &    0 &   cd &  ce \\
  0 &    0 &    0 &    0    &  -ab & -ad &    0 &   af    & -b^2 &  -bd &   0 &   bf    &  -bc &  -cd &    0 &  cf \\
  0 &    0 &    0 &    0    &  -ac & -ae &  -af &    0    &  -bc &  -be & -bf &    0    & -c^2 &  -ce &  -cf &   0 \\
\hline
  % -    - -    - -    - ----    - -   - -    - -    - ----    - -    - -   - -    - ----    - -    - -    - -   -  -
  0 & -a^2 &  -ab &  -ac    &    0 &   0 &    0 &    0    &    0 &   ad &  bd &   cd    &    0 &   ae &   be &  ce \\
a^2 &    0 &  -ad &  -ae    &    0 &   0 &    0 &    0    &  -ad &    0 & d^2 &   de    &  -ae &    0 &   de & e^2 \\
 ab &   ad &    0 &  -af    &    0 &   0 &    0 &    0    &  -bd & -d^2 &   0 &   df    &  -be &  -de &    0 &  ef \\
 ac &   ae &   af &    0    &    0 &   0 &    0 &    0    &  -cd &  -de & -df &    0    &  -ce & -e^2 &  -ef &   0 \\
\hline
  % -    - -    - -    - ----    - -   - -    - -    - ----    - -    - -   - -    - ----    - -    - -    - -   -  -
  0 &  -ab & -b^2 &  -bc    &    0 & -ad &  -bd &  -cd    &    0 &    0 &   0 &    0    &    0 &   af &   bf &  cf \\
 ab &    0 &  -bd &  -be    &   ad &   0 & -d^2 &  -ed    &    0 &    0 &   0 &    0    &  -af &    0 &   df &  ef \\
b^2 &   bd &    0 &  -bf    &   bd & d^2 &    0 &  -fd    &    0 &    0 &   0 &    0    &  -bf &  -df &    0 & f^2 \\
 bc &   be &   bf &    0    &   cd &  de &   df &    0    &    0 &    0 &   0 &    0    &  -cf &  -ef & -f^2 &   0 \\
\hline
  % -    - -    - -    - ----    - -   - -    - -    - ----    - -    - -   - -    - ----    - -    - -    - -   -  -
  0 &  -ac &  -bc & -c^2    &    0 & -ae &  -be &  -ce    &    0 &  -af & -bf &  -cf    &    0 &    0 &    0 &   0 \\
 ac &    0 &  -dc &  -ce    &   ae &   0 &  -de & -e^2    &   af &    0 & -df &  -ef    &    0 &    0 &    0 &   0 \\
 bc &   cd &    0 &  -cf    &   be &  de &    0 &  -ef    &   bf &   df &   0 & -f^2    &    0 &    0 &    0 &   0 \\
c^2 &   ce &   cf &    0    &   ce & e^2 &   ef &    0    &   cf &   ef & f^2 &    0    &    0 &    0 &    0 &   0 \\
	\end{array}\right]
$$
However, this is beginning to get messy: there are 256 elements in this array,
and 24 elements in $\mcS_4$.  The point of this example is to convince the
reader of the benefits of alternative techniques: not all 256 elements are
independent, and a more efficient approach is possible.  Note from remark
\ref{rem:std_biwedge} that $q$ is simply
$$
\begin{array}{llllllll}
	q &=& a\,\vece_1\wedge\vece_2 &+& b\,\vece_1\wedge\vece_3 &+& c\,\vece_1\wedge\vece_4 \\
      & &                         &+& d\,\vece_2\wedge\vece_3 &+& e\,\vece_2\wedge\vece_4 \\
      & &                         & &                         &+& f\,\vece_3\wedge\vece_4.
\end{array}
$$
Then $q\wedge q$ is obtained by squaring this sum.  Note that any term in the product
which contains a repeated $\vece_i$ will be zero, so we have only
\begin{eqnarray*}
	q \wedge q
	&=& af\,\vece_1\wedge\vece_2 \wedge \vece_3\wedge\vece_4
	+	be\,\vece_1\wedge\vece_3 \wedge \vece_2\wedge\vece_4
	+	cd\,\vece_1\wedge\vece_4 \wedge \vece_2\wedge\vece_3 \\
	&+&	dc\,\vece_2\wedge\vece_3 \wedge \vece_1\wedge\vece_4
	+	eb\,\vece_2\wedge\vece_4 \wedge \vece_1\wedge\vece_3
	+	fa\,\vece_3\wedge\vece_4 \wedge \vece_1\wedge\vece_2 \\
%
	&=& 2af\,\vece_1\wedge\vece_2 \wedge \vece_3\wedge\vece_4
	+	2be\,\vece_1\wedge\vece_3 \wedge \vece_2\wedge\vece_4
	+	2cd\,\vece_1\wedge\vece_4 \wedge \vece_2\wedge\vece_3 \\
%
	&=& 2(af-be+cd)\,\vece_1\wedge\vece_2 \wedge \vece_3\wedge\vece_4.
\end{eqnarray*}
It may be shown that $\vece_1\wedge\vece_2 \wedge \vece_3\wedge\vece_4$
expands such that

$$
	q \wedge q =
	2(af-be+cd)
\left[\begin{array}{rrrr|rrrr|rrrr|rrrr}
	0 &  0 &  0 &  0 &  0 & 0 &  0 &  0 &  0 &  0 & 0 &  0 &  0 &  0 &  0 & 0 \\
	0 &  0 &  0 &  0 &  0 & 0 &  0 &  0 &  0 &  0 & 0 & -1 &  0 &  0 &  1 & 0 \\
	0 &  0 &  0 &  0 &  0 & 0 &  0 &  1 &  0 &  0 & 0 &  0 &  0 & -1 &  0 & 0 \\
	0 &  0 &  0 &  0 &  0 & 0 & -1 &  0 &  0 &  1 & 0 &  0 &  0 &  0 &  0 & 0 \\
	 \hline
	0 &  0 &  0 &  0 &  0 & 0 &  0 &  0 &  0 &  0 & 0 &  1 &  0 &  0 & -1 & 0 \\
	0 &  0 &  0 &  0 &  0 & 0 &  0 &  0 &  0 &  0 & 0 &  0 &  0 &  0 &  0 & 0 \\
	0 &  0 &  0 & -1 &  0 & 0 &  0 &  0 &  0 &  0 & 0 &  0 &  1 &  0 &  0 & 0 \\
	0 &  0 &  1 &  0 &  0 & 0 &  0 &  0 & -1 &  0 & 0 &  0 &  0 &  0 &  0 & 0 \\
	 \hline
	0 &  0 &  0 &  0 &  0 & 0 &  0 & -1 &  0 &  0 & 0 &  0 &  0 &  1 &  0 & 0 \\
	0 &  0 &  0 &  1 &  0 & 0 &  0 &  0 &  0 &  0 & 0 &  0 & -1 &  0 &  0 & 0 \\
	0 &  0 &  0 &  0 &  0 & 0 &  0 &  0 &  0 &  0 & 0 &  0 &  0 &  0 &  0 & 0 \\
	0 & -1 &  0 &  0 &  1 & 0 &  0 &  0 &  0 &  0 & 0 &  0 &  0 &  0 &  0 & 0 \\
	 \hline
	0 &  0 &  0 &  0 &  0 & 0 &  1 &  0 &  0 & -1 & 0 &  0 &  0 &  0 &  0 & 0 \\
	0 &  0 & -1 &  0 &  0 & 0 &  0 &  0 &  1 &  0 & 0 &  0 &  0 &  0 &  0 & 0 \\
	0 &  1 &  0 &  0 & -1 & 0 &  0 &  0 &  0 &  0 & 0 &  0 &  0 &  0 &  0 & 0 \\
	0 &  0 &  0 &  0 &  0 & 0 &  0 &  0 &  0 &  0 & 0 &  0 &  0 &  0 &  0 & 0 \\
\end{array}\right].
$$
Since 4-vectors are top-level for $N=4$, $\Lambda^4(Y)$ is 1-dimensional and
spanned by $\vece_1\wedge\vece_2 \wedge \vece_3\wedge\vece_4$.  Thus, all
information about the 256 elements of $q \wedge q$ is encapsulated in the
single scalar $2(af-be+cd)$.
\mendex

\begin{rem}
\label{rem:wedge_pairing}
For $\phi \in Y^*$ and $v \in Y$ we have the pairing
$$
	\langle \phi, v \rangle = \phi(v).
$$
Recall that $Y^*=\Lambda^1(Y^*)$ and $Y=\Lambda^1(Y)$.  Moreover,
for $\Phi=\phi_1\wedge\cdots\wedge\phi_k \in \Lambda^k(Y^*)$
and $\vecv=v_1\wedge\cdots\wedge v_k \in \Lambda^k(Y)$ we have the pairing
$$
	\langle \Phi, \vecv \rangle =
	\langle(\phi_1\wedge\cdots\wedge\phi_k),(v_1\wedge\cdots\wedge v_k)\rangle
	=
	\left|
		\begin{array}{ccc}
			\phi_1(v_1) & \cdots & \phi_1(v_k) \\
			\vdots      &        & \vdots      \\
			\phi_k(v_1) & \cdots & \phi_k(v_k) \\
		\end{array}
	\right|.
$$
The wedge scale factor in definition \ref{defn:wedge_scales} is chosen
precisely to make this statement work.  Details are probably in
\cite{bib:palmer2}; if not (or even if so), I should include them here.
\end{rem}

%% ----------------------------------------------------------------
\subsection{Clifford algebras}
\label{sec:cliff}

The geometric origin of Clifford and Grassmann algebras is presented
beautifully in sections 11.5 and 11.6 of \cite{bib:penrose}.  As well, the
skew-symmetry of Clifford algebras is used to mimic the skew-symmetry of
fermions; see \cite{bib:berezin}.
%xxx motivate by mentioning something about skew-symmetry of fermions and
%the CACR here \ldots .

\begin{defn}
\label{defn:cliff_alg}
Let $Y$ be a complex vector space, and let $S$ be a symmetric bilinear form on
$Y$.  (That is, $Y$ is an \plainidx{orthogonal space} in the sense of
definition \ref{defn:orthogonal_space}.) The \bfidx{Clifford algebra} of $Y$
and $S$, written $\Cliff(Y)$ without explicit mention of $S$, is the tensor
algebra on $Y$ modulo the \bfidxtwo{anticommutation relations}{anticommutation
relation} $\vecu\vecv+\vecv\vecu=S(\vecu,\vecv)$.  That is,
$$
	\Cliff(Y) = T(Y) \big/ \langle \vecu\vecv + \vecv\vecu - S(\vecu\vecv)\rangle
$$
\end{defn}

Note that the trivial (and highly degenerate) form $S(\vecu,\vecv)=0$ is
symmetric and bilinear.  The Clifford algebra with this form is precisely the
exterior algebra (definition \ref{defn:ext_alg}).  Thus, Clifford algebras
generalize the concept of exterior algebra.  However, we usually take the
symmetric bilinear forms for our Clifford algebras to be
\plainidx{non-degenerate}.

In the exterior algebra $\Lambda(Y)$, we compute with wedge products of vectors
and make use of the anticommutation relation $\vecv\wedge\vecu =
-\vecu\wedge\vecv$; we rarely need to mention a basis for $Y$ unless, say, we
are discussing the dimension of $\Lambda(Y)$ as a vector space.  Computation in
$\Cliff(Y)$ is a bit more subtle.  Definition \ref{defn:cliff_alg} tells us how
to formally juxtapose two vectors when we take their product.  Let $\{\vece_1,
\ldots, \vece_N\}$ be an \plainidx{orthonormal basis} for $Y$ with respect to
$S$, i.e.  $S(\vece_i, \vece_j) = \delta_{ij}$.  Then the anticommutation
relation specializes to
$$
	\vece_j \vece_i = -\vece_i \vece_j, \; i \ne j
$$
(i.e. for orthonormal vectors, we still have simple anticommutativity
as in the exterior algebra) and
$$
	\vece_i \vece_i = 1/2.
$$
%[xxx fix me \ldots $S$ as half of $Q$?  Need to get rid of this annoying factor
%of $1/2$.]
In fact, it is clear that with respect to $S$,
$$
	\vecu \perp \vecv \iff \vecu\vecv = -\vecv\vecu.
$$

It may be shown (but is not shown here) that the Clifford algebra has the same
dimension as the exterior algebra, namely, $2^N$ where $N$ is the dimension of
$Y$.

%xxx prove or cite:  basis and dimension \ldots note same as $\Lambda(Y)$.  V.S.
%dimension via commutators on orthonormal basis.

%Note non-homogeneous order of products.

%recall or at least mention the Q and B relationship;
%xxx need a quadratic form
%$Y$ and $B$ is enough to make a clifford algebra.

%xxx do some computational examples.

Clifford algebras play a key role in Fock representations\index{Fock
representation} (section \ref{sec:fock_rep}).

%% ----------------------------------------------------------------
\subsection{Creation and annihilation operators}
\label{sec:creat_annih}

In this section, let $Y$ be a finite-dimensional complex vector space.  We
present a discretized version of the creation and annihilation operators which
were briefly mentioned in section \ref{sec:phys_creat_annih}.

\begin{defn}
\label{defn:creation_operator}
For $u \in Y$ and $\vecv=v_1\wedge\cdots\wedge v_{k-1} \in \Lambda^{k-1}(Y)$,
define the \bfidx{creation operator}
$$
	c(u): \Lambda^{k-1}(Y) \to \Lambda^k(Y),
$$
via
$$
	c(u)\vecv = u \wedge \vecv.
$$
By varying $k$ we have, on the full alternating tensor algebra,
$c(u): \Lambda(Y) \to \Lambda(Y)$.  This means that also, for
$\lambda \in Y^*$ and $\Phi=\phi_1\wedge\cdots\wedge\phi_{k-1} \in
\Lambda^{k-1}(Y^*)$, we have a creation operator
$$
	c(\lambda): \Lambda^{k-1}(Y^*) \to \Lambda^k(Y^*)
$$
(and hence $c(\lambda): \Lambda(Y^*) \to \Lambda(Y^*)$) via
$$
	c(\lambda)\Phi = \lambda \wedge \Phi.
$$
\end{defn}

\beginex
We have
$$
	c(\vece_1) \vece_2 \wedge \vece_3 = \vece_1 \wedge\vece_2 \wedge \vece_3
$$
and
$$
	c(\vece_1) \vece_1 \wedge \vece_2 = \vece_1 \wedge\vece_1 \wedge \vece_2 = 0.
$$
Recall however that $0 \in \Lambda^3(Y)$, so the definition is intact.
\mendex

\begin{defn}
Formally, the \bfidx{annihilation operator} is the transpose of the creation
operator.  Concretely, let
$$
	\lambda \in Y^*,
$$
let
$$
	\Phi = \phi_1\wedge \cdots \wedge \phi_{k-1} \in \Lambda^{k-1}(Y^*),
$$
and let
$$
	\vecv = v_1 \wedge \cdots \wedge v_k \in \Lambda^k(Y).
$$
Recall from definition \ref{defn:creation_operator} that
$$
	c(\lambda): \Lambda^{k-1}(Y^*) \to \Lambda^k(Y^*)
$$
and recall from remark \ref{rem:wedge_pairing} that we have pairings
of $\Lambda^{k-1}(Y^*)$ with $\Lambda^{k-1}(Y)$
and
of $\Lambda^k(Y^*)$ with $\Lambda^k(Y)$.
From the definition of transpose (definition \ref{defn:transpose}), we have
$$
	a(\lambda) = c(\lambda)^\tau:
	\Lambda^k(Y) \to \Lambda^{k-1}(Y),
$$
specified by
$$
	\langle c(\lambda)\Phi, \vecv \rangle =
	\langle \Phi, a(\lambda)\vecv \rangle.
$$
Also note the following diagram:
$$
\begin{array}{ccc}
	\Lambda^{k-1}(Y^*) &
	\stackrel{c(\lambda)}{\longrightarrow} &
	\Lambda^k(Y^*) \\
	%
	\ssfdot &  & \ssfdot \\
	%
	\Lambda^{k-1}(Y) &
	\stackrel{c(\lambda)^\tau=a(\lambda)}{\longleftarrow} &
	\Lambda^k(Y) \\
\end{array}.
$$
\end{defn}

\begin{defn}
Let\index{hat} $\vecv = v_1 \wedge \cdots \wedge v_k$.  Write
$$
	\hat{\vecv}_j = v_1 \wedge \cdots \wedge v_{j-1} \wedge v_{j+1}
	\wedge \cdots \wedge v_k.
$$
That is, $\hat{\vecv}_j$ omits $v_j$ from $\vecv$.
Likewise,
$$
	\hat{\vecv}_{ij} = v_1 \wedge \cdots
	\wedge v_{i-1} \wedge v_{i+1} \wedge
	\cdots
	\wedge v_{j-1} \wedge v_{j+1} \wedge
	\cdots \wedge v_k,
$$
omitting $v_i$ and $v_j$ from $\vecv$.
\end{defn}

\begin{prop}
\label{prop:annih_formula}
The annihilation operator is given by the explicit formula:
\begin{eqnarray*}
	a(\lambda)\vecv &=& \sum_{j=1}^k (-1)^{j-1} \lambda(v_j) \hat{\vecv}_j.
\end{eqnarray*}
\begin{proof}
Consider the $k \times k$ determinant for wedge pairings, as discussed in
remark \ref{rem:wedge_pairing}:
\begin{eqnarray*}
	\langle c(\lambda)\Phi, \vecv \rangle
	&=& \langle \Phi, a(\lambda)\vecv \rangle \\
	\langle c(\lambda)\Phi, \vecv \rangle
	&=& \langle \lambda \wedge \phi_1 \wedge \cdots \wedge \phi_k,
		v_1 \wedge \cdots \wedge v_k \rangle \\
	&=& \left| \begin{matrix}
			\lambda(v_1) & \cdots & \lambda(v_k) \\
			\phi_1(v_1) & \cdots & \phi_1(v_k) \\
			\vdots      &        & \vdots \\
			\phi_k(v_1) & \cdots & \phi_k(v_k)
		\end{matrix} \right|.
\end{eqnarray*}
Now we simply expand along the top row, noting that the $j$th minor is the
$(k-1)\times(k-1)$ determinant
$$
\left| \begin{matrix}
\phi_1(v_1) &\cdots& \phi_1(v_{j-1}) &\phi_1(v_{j+1}) &\cdots& \phi_1(v_k) \\
\vdots      &      &  \vdots         &\vdots          &      & \vdots \\
\phi_k(v_k) &\cdots& \phi_k(v_{j-1}) &\phi_k(v_{j+1}) &\cdots& \phi_k(v_k) \\
\end{matrix} \right|.
$$
Writing the $j$th minor as $\langle \Phi, \hat{\vecv}_j \rangle$, we obtain
\begin{eqnarray*}
	\langle \Phi, a(\lambda)\vecv \rangle
	&=& \sum_{j=1}^k (-1)^{j-1} \lambda(v_j)
			\langle \Phi,\hat{\vecv}_j \rangle
	= \left\langle \Phi,
		\sum_{j=1}^k (-1)^{j-1} \lambda(v_j) \hat{\vecv}_j \right\rangle.
\end{eqnarray*}
Since this holds for all $\vecv$, we conclude (note: I need to elaborate here,
using non-degeneracy of the pairing) that
\begin{eqnarray*}
	a(\lambda)\vecv &=& \sum_{j=1}^k (-1)^{j-1} \lambda(v_j) \hat{\vecv}_j.
\end{eqnarray*}
\end{proof}
\end{prop}

\begin{prop}
\label{prop:creat_annih_linear}
The creation and annihilation operators are linear transformations on
$\Lambda(Y)$.
\begin{proof}
The linearity of the creation operator is obtained from arithmetic rules in
the exterior algebra:
$$
	c(au)\vecw = (au) \wedge \vecw = a(u \wedge \vecw)
$$
and
$$
	c(u+v)\vecw = (u+v)\wedge \vecw = u\wedge\vecw + v \wedge \vecw = c(u)\vecw + c(v)\vecw.
$$
The annihilation operator is linear since it is the transpose of a linear
operator (see definition \ref{defn:transpose}).
\end{proof}
\end{prop}

\beginex
Let $\lambda=\vece_2^*$ and $\vecv = \vece_1 \wedge \vece_2 \wedge \vece_3$.
Then
\begin{eqnarray*}
	a(\lambda)(\vecv) &=& a(\vece_2^*)(\vece_1 \wedge \vece_2 \wedge \vece_3) \\
	&=& \vece_2^*(\vece_1)(\vece_2 \wedge \vece_3)
	- \vece_2^*(\vece_2)(\vece_1 \wedge \vece_3)
	+ \vece_2^*(\vece_3)(\vece_1 \wedge \vece_2) \\
	&=& -\vece_1 \wedge \vece_3.
\end{eqnarray*}
Recall that when we compute determinants, we may omit the alternating signs
if we barrel-shift the minors.  E.g. we normally write
$$
	\left|\begin{matrix}
		a & b & c \\
		d & e & f \\
		g & h & i \\
	\end{matrix}\right|
	= a(ei-hf)-b(di-gf)+c(dh-ge)
$$
but instead we may write
$$
	\left|\begin{matrix}
		a & b & c \\
		d & e & f \\
		g & h & i \\
	\end{matrix}\right|
	= a(ei-hf)+b(fg-id)+c(dh-ge).
$$
Using this fact, we may write
\begin{eqnarray*}
	a(\lambda)(\vecv) &=& a(\vece_2^*)(\vece_1 \wedge \vece_2 \wedge \vece_3) \\
	&=& \vece_2^*(\vece_1)(\vece_2 \wedge \vece_3)
	+ \vece_2^*(\vece_2)(\vece_3 \wedge \vece_1)
	+ \vece_2^*(\vece_3)(\vece_1 \wedge \vece_2) \\
	&=& \vece_3 \wedge \vece_1
	= -\vece_1 \wedge \vece_3.
\end{eqnarray*}
\mendex

%xxx note why we WLOG assume the same set of $k$ indices.  The proof needs
%fixing.

\beginex
\label{ex:annih_anticomm}
It will be shown below that the creation and annihilation operators
anticommute.  The proof of the latter wends through a dense thicket of
summation symbols, which will be illuminated by the following concrete example.
Let $\lambda,\mu\in Y^*$ and let $\vecw=w_1 \wedge w_2 \wedge w_3 \wedge w_4$.
Then, using the formula from proposition \ref{prop:annih_formula}, we have
\begin{eqnarray*}
	a(\lambda)a(\mu)\vecw &=& \begin{array}{rrrrrrl}
		 \lambda(w_1) &[&                      &  \mu(w_2)w_3\wedge w_4& -\mu(w_3)w_2\wedge w_4& +\mu(w_4)w_2\wedge w_3&] \\
		-\lambda(w_2) &[& \mu(w_1)w_3\wedge w_4&                       & -\mu(w_3)w_1\wedge w_4& +\mu(w_4)w_1\wedge w_3&] \\
		+\lambda(w_3) &[& \mu(w_1)w_2\wedge w_4& -\mu(w_2)w_1\wedge w_4&                       & +\mu(w_4)w_1\wedge w_2&] \\
		-\lambda(w_4) &[& \mu(w_1)w_2\wedge w_3& -\mu(w_2)w_1\wedge w_3& +\mu(w_3)w_1\wedge w_2&                       &] \\
	\end{array}
\end{eqnarray*}
and
\begin{eqnarray*}
	a(\mu)a(\lambda)\vecw &=& \begin{array}{rrrrrrl}
		 \mu(w_1) &[&                          &  \lambda(w_2)w_3\wedge w_4& -\lambda(w_3)w_2\wedge w_4& +\lambda(w_4)w_2\wedge w_3&] \\
		-\mu(w_2) &[& \lambda(w_1)w_3\wedge w_4&                           & -\lambda(w_3)w_1\wedge w_4& +\lambda(w_4)w_1\wedge w_3&] \\
		+\mu(w_3) &[& \lambda(w_1)w_2\wedge w_4& -\lambda(w_2)w_1\wedge w_4&                           & +\lambda(w_4)w_1\wedge w_2&] \\
		-\mu(w_4) &[& \lambda(w_1)w_2\wedge w_3& -\lambda(w_2)w_1\wedge w_3& +\lambda(w_3)w_1\wedge w_2&                           &]. \\
	\end{array}
\end{eqnarray*}
Here it is clear that $a(\mu)a(\lambda)\vecw = -a(\lambda)a(\mu)\vecw$, term by term.
\mendex

\begin{prop}
\label{prop:creat_annih_cacr}
The creation and annihilation operators satisfy the \bfidxtwo{anticommutation
relations}{anticommutation relation}
\begin{eqnarray*}
	c(u)\,c(v) + c(v)\,c(u) &=& 0, \\
	a(\lambda)\,a(\mu) + a(\mu)\,a(\lambda) &=& 0, \\
	a(\lambda)\,c(u) + c(u)\,a(\lambda) &=& \lambda(u).
\end{eqnarray*}
\begin{proof}
For the creation operator, let $\vecw=w_1 \wedge\cdots\wedge w_{k-2}
\in \Lambda^{k-2}(Y)$.  Then
\begin{eqnarray*}
	c(u)\,c(v)(\vecw)
	= u \wedge v \wedge \vecw = -v \wedge u \wedge \vecw = -c(v)\,c(u)(\vecw).
\end{eqnarray*}

For the annihilation operator, we simply generalize example
\ref{ex:annih_anticomm} using summation notation.  Let $\lambda,\mu \in Y^*$,
and let $\vecw=w_1 \wedge\cdots\wedge w_k \in \Lambda^k(Y)$.
Then
\begin{eqnarray*}
	a(\lambda)a(\mu)\vecw &=&
		\sum_{i=1}^k (-1)^{i+1}\lambda(w_i)
		\left[
		\sum_{j=1}^{i-1} (-1)^{j+1}\mu(w_j)\hat{\vecw}_{ij}
		+
		\sum_{j=i+1}^{k} (-1)^{j}\mu(w_j)\hat{\vecw}_{ij}
		\right] \\
	&=&
		\sum_{i=1}^k \sum_{j=1}^{i-1}
			(-1)^{i+j}
			\lambda(w_i) \mu(w_j)\hat{\vecw}_{ij}
		\quad+\quad
		\sum_{i=1}^k \sum_{j=i+1}^{k}
			(-1)^{i+j+1}
			\lambda(w_i) \mu(w_j)\hat{\vecw}_{ij} \\
	a(\mu)a(\lambda)\vecw &=&
		\sum_{j=1}^k (-1)^{j+1}\mu(w_j)
		\left[
		\sum_{i=1}^{j-1} (-1)^{i+1}\lambda(w_i)\hat{\vecw}_{ij}
		+
		\sum_{i=j+1}^{k} (-1)^{i}\lambda(w_i)\hat{\vecw}_{ij}
		\right] \\
	&=&
		\sum_{j=1}^k \sum_{i=1}^{j-1}
			(-1)^{i+j}
			\lambda(w_i) \mu(w_j)\hat{\vecw}_{ij}
		\quad+\quad
		\sum_{j=1}^k \sum_{i=j+1}^{k}
			(-1)^{i+j+1}
			\lambda(w_i) \mu(w_j)\hat{\vecw}_{ij}.
\end{eqnarray*}
Notice that in example \ref{ex:annih_anticomm} the diagonal entries are blank,
and we may split the double sum into the sum of above-diagonal entries and the
sum of below-diagonal entries.  Then
\begin{eqnarray*}
	a(\lambda)a(\mu)\vecw &=&
		\sum_{i<j} (-1)^{i+j+1} \lambda(w_i) \mu(w_j)\hat{\vecw}_{ij}
		\quad+\quad
		\sum_{i>j} (-1)^{i+j} \lambda(w_i) \mu(w_j)\hat{\vecw}_{ij} \\
	a(\mu)a(\lambda)\vecw &=&
		\sum_{j<i} (-1)^{i+j+1} \lambda(w_i) \mu(w_j)\hat{\vecw}_{ij}
		\quad+\quad
		\sum_{j>i} (-1)^{i+j} \lambda(w_i) \mu(w_j)\hat{\vecw}_{ij} \\
	&=&
		\sum_{i<j} (-1)^{i+j} \lambda(w_i) \mu(w_j)\hat{\vecw}_{ij}
		\quad+\quad
		\sum_{i>j} (-1)^{i+j+1} \lambda(w_i) \mu(w_j)\hat{\vecw}_{ij} \\
	&=&
		-\sum_{i<j} (-1)^{i+j+1} \lambda(w_i) \mu(w_j)\hat{\vecw}_{ij}
		\quad-\quad
		\sum_{i>j} (-1)^{i+j} \lambda(w_i) \mu(w_j)\hat{\vecw}_{ij} \\
	&=& -a(\lambda)a(\mu)\vecw
\end{eqnarray*}
which is what was to be shown.

For the anticommutation relation between the creation and annihilation
operators, we again use a test $k$-vector.  Let $\vecw=w_1 \wedge\cdots\wedge
w_k \in \Lambda^k(Y)$.  Then, using the formula from proposition
\ref{prop:annih_formula}, we have
\begin{eqnarray*}
a(\lambda)\,c(u)\,\vecw
	&=& a(\lambda)\,\left(u \wedge w_1 \wedge\cdots\wedge w_k \right) \\
	&=& \lambda(u) \, w_1 \wedge\cdots\wedge w_k
		+ u \wedge \left(\sum_{i=1}^k (-1)^i \lambda(w_i) \hat{\vecw}_i\right) \\
	&=& \lambda(u) \, w_1 \wedge\cdots\wedge w_k
		+ \sum_{i=1}^k (-1)^i \lambda(w_i) u \wedge \hat{\vecw}_i \\
\end{eqnarray*}
and
\begin{eqnarray*}
c(u)\,a(\lambda)\,\vecw
		&=& u \wedge \left(\sum_{i=1}^k (-1)^{i+1} \lambda(w_i) \hat{\vecw}_i \right) \\
	&=& \sum_{i=1}^k (-1)^{i+1} \lambda(w_i) u \wedge \hat{\vecw}_i \\
	&=& -\sum_{i=1}^k (-1)^i \lambda(w_i) u \wedge \hat{\vecw}_i \\
\end{eqnarray*}
so that
\begin{eqnarray*}
a(\lambda)\,c(u)\,\vecw + c(u)\,a(\lambda)\,\vecw
	&=& \lambda(u) \, w_1 \wedge\cdots\wedge w_k \\
	&=& \lambda(u) \vecw.
\end{eqnarray*}
Since $\vecw$ was arbitrary, we conclude that
$$
	a(\lambda)\,c(u) + c(u)\,a(\lambda) = \lambda(u)
$$
as functions.
\end{proof}
\end{prop}

%% ----------------------------------------------------------------
\subsection{Bra-ket notation}
\label{sec:bra_ket}

In this section we introduct a notation which is commonplace in the physics
literature.  (See any of the physics references in the appendix for more
information.)

\begin{defns}
\label{defns:bra_ket}
Let $Y$ be a complex vector space.  Then we write
\begin{eqnarray*}
	Y   \ni v       &=& \ket v \\
	Y^* \ni \lambda &=& \bra \lambda \\
	\langle \lambda, v \rangle &=& \braket{\lambda}{v}.
\end{eqnarray*}
One says that a functional (or covector) $\lambda$ is a \bfidx{bra} and a
vector $v$ is a \bfidx{ket}.  This notation is collectively referred to as
\bfidx{bra-ket} notation.  We also write
$$
	\Lambda(Y) \ni 1 \oplus 0 \oplus 0 \oplus \cdots \oplus 0 = \zvac.
$$
Note that this notation is a little misleading:  $\zvac$ is really 1.
\end{defns}

\begin{rem}
\label{rem:ca_vs_wedges}
Bra-ket notation, along with creation and annihilation operators (section
\ref{sec:creat_annih}) gives physicists an alternative way to write elements of
exterior algebras.  This is often used in place of the grading and wedges which
are preferred in the mathematical literature.  In particular, for $u,v \in Y$:
\begin{eqnarray*}
	c(u) \zvac &=& u \wedge 1 = u \\
	c(u) c(v) \zvac &=& c(u)(v) = u \wedge v \\
	a(u) \zvac &=& 0 \\
	A u \wedge v &=& c(Au) c(v) \zvac.
\end{eqnarray*}
\end{rem}

%% ----------------------------------------------------------------
\subsection{Pfaffians}
\label{sec:pfaffians}

The Pfaffian is a square root of the determinant which is defined on
skew-symmetric matrices.  It will be used for efficient computation of the
discrete Berezin integral in section \ref{sec:discrete_berezin}.

\begin{defn}
\label{defn:assocd_bivr}
Let $A$ be an $N \times N$ skew-symmetric matrix, with $N$ even.
Define the \bfidx{associated bivector} $b_A$ to be
$$
	b_A = \sum_{i<j} A_{ij} \vece_i \wedge \vece_j.
$$
\end{defn}

Recall that from remark \ref{rem:biwedge_in_coords} that a bivector, when
written in coordinates, is simply a two-dimensional array of numbers.  Also
recall that bivectors are doubly contravariant with respect to change of
coordinates, whereas matrices are of mixed variance.  Nonetheless, with respect
to a particular coordinate system, the skew-symmetric matrix $A$ and its
associated bivector $b_A$ contain precisely the same numbers in their
two-dimensional arrays.

\begin{prop}
With respect to a particular coordinate system, $A$ and $b_A$ as defined above
may be viewed as two-dimensional arrays containing the same elements.
\begin{proof}
Let
$$
	A = \begin{bmatrix}
		0          & A_{12} & \cdots &  A_{1,N-1} & A_{1,N}   \\
		-A_{12}    & 0      &        &  A_{2,N-1} & A_{1,N}   \\
		\vdots     &        & \ddots &            & \vdots    \\
		-A_{1,N-1} &        &        &  0         & A_{N-1,N} \\
		-A_{1,N}   &        &        & -A_{N-1,N} & 0         \\
	\end{bmatrix}
$$
be an arbitrary $N \times N$ matrix.  Recall from remark \ref{rem:std_biwedge}
that $\vece_i \wedge \vece_j$ may be viewed as an $N \times N$ array with
$+1$ at the $i,j$ element, $-1$ at the $j,i$ elements, and 0 elsewhere.
Summing as in definition \ref{defn:assocd_bivr} yields the array
$$
	b_A = \begin{bmatrix}
		0          & A_{12} & \cdots &  A_{1,N-1} & A_{1,N}   \\
		-A_{12}    & 0      &        &  A_{2,N-1} & A_{1,N}   \\
		\vdots     &        & \ddots &            & \vdots    \\
		-A_{1,N-1} &        &        &  0         & A_{N-1,N} \\
		-A_{1,N}   &        &        & -A_{N-1,N} & 0         \\
	\end{bmatrix}.
$$
\end{proof}
\end{prop}

\begin{ex}
\label{ex:4x4_assocd_bivr}
Consider a general $4 \times 4$ skew-symmetric matrix:
$$
	A = \left[\begin{array}{rrrrrr}
		0  &  a  & b  & c \\
		-a &  0  & d  & e \\
		-b & -d  & 0  & f \\
		-c & -e  & -f & 0 \\
	\end{array}\right].
$$
Recall from remark \ref{rem:std_biwedge} that $\vece_i \wedge \vece_j$ is a
two-dimensional array with (disregarding scale factors) $1$ at the $i,j$
element and $-1$ at the $j,i$ element, and zeroes elsewhere.  Thus, using
definition \ref{defn:assocd_bivr}, we have
\begin{eqnarray*}
	&&
	\begin{array}{llllllll}
	b_A &=& a \vece_1\wedge\vece_2 &+& b\,\vece_1\wedge\vece_3 &+& c\,\vece_1\wedge\vece_4  \\
    	& &                        &+& d\,\vece_2\wedge\vece_3 &+& e\,\vece_2\wedge\vece_4  \\
    	& &                         & &                        &+& f\,\vece_3\wedge\vece_4  \\
	\end{array}
\\
	&&
	b_A = \left[\begin{array}{rrrrrr}
		0  &  a  & b  & c \\
		-a &  0  & d  & e \\
		-b & -d  & 0  & f \\
		-c & -e  & -f & 0 \\
	\end{array}\right].
\end{eqnarray*}
\end{ex}

\begin{defn}
\label{defn:pfaffian}
Let $A$ be an $N \times N$ skew-symmetric matrix, with $N$ even.  Then the
\bfidx{Pfaffian} of $A$, written $\Pf(A)$, is specified by
$$
	\frac{1}{(N/2)!} \, b_A^{N/2} = \Pf(A) \, \vece_1 \wedge \cdots \wedge \vece_N.
$$
\end{defn}

\beginex
Use the $4 \times 4$ skew-symmetric matrix $A$ from example
\ref{ex:4x4_assocd_bivr}.  Working directly from definition
\ref{defn:pfaffian}, with $N=4$, we have
$$
	\frac{1}{2} \, b_A^2
	= \Pf(A) \, \vece_1 \wedge \vece_2 \wedge \vece_3 \wedge \vece_4.
$$
The product $b_A^2$ was computed in example \ref{ex:q_wedge_q}: it was found to
be $2(af-be+cd)\,\vece_1\wedge\vece_2 \wedge \vece_3\wedge\vece_4$.
Thus, we have
$$
	\Pf(A) = af-be+cd.
$$
Furthermore, a direct computation reveals that
$$
	\det(A) = (af-be+cd)^2,
$$
as expected.
\mendex

\begin{prop}
For any skew-symmetric matrix $A$,
$$
	\Pf^2(A) = \det(A)
$$
\begin{proof}
TBD.  Or cite \ldots.
\end{proof}
\end{prop}

\begin{cor}
$\Pf(\lambda A) = \lambda^{N/2}\Pf(A).$
\begin{proof}
Using the known $N$-linearity of the determinant, we have
$$
	\Pf^2 (\lambda A)
	= \det(\lambda A)
	= \lambda^N \det (A)
	= \lambda^N \Pf^2 (A).
$$
\end{proof}
\end{cor}

The computation of Pfaffians for matrices of larger dimensions is facilitated by
the following formulas.

\begin{prop}
\label{prop:pfaffian_of_odd_dim_mx}
The Pfaffian of an odd-dimensional skew-symmetric matrix is zero.
\begin{proof}
TBD.
\end{proof}
\end{prop}

\begin{defn}
Let $A$ be an $N \times N$ matrix.  Define $\hat{A}_{jk}$ to be the
$(N-2)\times(N-2)$ matrix with the $j$th and $k$th rows and the
$j$th and $k$th columns removed.
\end{defn}

\newcommand{\fooa}{\hspace{-0.10in}}
\newcommand{\foob}{\hspace{0.10in}}

\beginex
If
$$
	A = \left[\begin{array}{rrrr}
		0  &  a  & b  & c \\
		-a &  0  & d  & e \\
		-b & -d  & 0  & f \\
		-c & -e  & -f & 0 \\
	\end{array}\right]
$$
then
$$
	\hat{A}_{12} =
	\left[\begin{array}{r|r|rr}
			\vspace{-0.09in} \\
			\hline
			\vspace{-0.26in} \\
		0  \fooa &  \foob a  \fooa & \foob b  & c \\
			\vspace{-0.09in} \\
			\hline
			\vspace{-0.26in} \\
		-a \fooa & \foob  0  \fooa & \foob d  & e \\
		-b \fooa & \foob -d  \fooa & \foob 0  & f \\
		-c \fooa & \foob -e  \fooa & \foob -f & 0 \\
	\end{array}\right].
$$
\mendex

\begin{prop}[Reduction formula for Pfaffians]
\label{prop:pfaffian_reduction_formula}
Let\index{reduction formula for Pfaffians} $A$ be a skew-symmetric $N \times N$ matrix.
Then
$$
	\Pf(A) = \sum_{k=2}^n (-1)^{k} A_{1k} \Pf(\hat{A}_{1k}).
$$
\begin{proof}
See \cite{bib:palmer2}, appendix A.4.
\end{proof}
\end{prop}

\begin{rem}
This recursive formula requires a base case.
Directly from definition \ref{defn:pfaffian} we have
$$
	\Pf\left[\begin{array}{rr}
	0 & a \\ -a & 0
	\end{array}\right]
	= a.
$$
\end{rem}

\beginex
Again consider a general $4 \times 4$ skew-symmetric matrix
$$
	A = \left[\begin{array}{rrrr}
		0  &  a  & b  & c \\
		-a &  0  & d  & e \\
		-b & -d  & 0  & f \\
		-c & -e  & -f & 0 \\
	\end{array}\right].
$$
The reduction formula from proposition \ref{prop:pfaffian_reduction_formula}
gives
$$
	a \, \Pf\, \left[\begin{array}{rr}
		0  &  f  \\
		-f &  0  \\
	\end{array}\right]
	-
	b \, \Pf\, \left[\begin{array}{rr}
		0  &  e  \\
		-e &  0  \\
	\end{array}\right]
	+
	c \, \Pf\, \left[\begin{array}{rr}
		0  &  d  \\
		-d &  0  \\
	\end{array}\right]
=
	af-be+cd
$$
as expected.
\mendex

\begin{prop}[Block formula for Pfaffians]
\label{prop:pfaffian_block_formula}
Let\index{block formula for Pfaffians} $A$ be an $M \times M$ matrix; let $B$
be an $N \times N$ matrix which is also skew-symmetric and invertible.  Then
$$
	\Pf\begin{bmatrix}
		B & A \\ -A^\tau & 0 \\
	\end{bmatrix}
	= \Pf(-B) \Pf(-A^\tau B^{-1} A)
$$
\begin{proof}
TBD.  Or cite.
\end{proof}
\end{prop}

%% ----------------------------------------------------------------
\subsection{Exterior exponentials}
\label{sec:ext_exp}

\begin{defn}
Let $q \in \Lambda^2(Y)$ where $Y$ is an $N$-dimensional vector space.
Define\index{$\mathrm{exp}$}
$$
	\exp(q) = \sum_{k=0}^{N/2} \frac{q^k}{k!}
$$
where $q^k$ denotes $q$ wedged with itself $k$ times, and $q^0=1$.
Note that the sum terminates as the top level since the exterior algebra
is finite-dimensional.
\end{defn}

\begin{defns}
\label{defns:Y_proj_and_e_sigma}
Let $Y$ be a complex vector space with basis $\{\vece_1, \ldots, \vece_N\}$.
Let $\sigma \subseteq \{1, \ldots, N\}$, with $1 \le \sigma_1 < \ldots <
\sigma_k \le N$.
Define \index{$\mathbf{e}_\sigma$}
$$
	\vece_\sigma = \vece_{\sigma_1} \wedge \cdots \wedge \vece_{\sigma_k}.
$$
Also define \plainidx{$P_\sigma$} to be the projection from $Y$ onto the span
of $\{\vece_1, \ldots, \vece_N\}$.
\end{defns}

\beginex
For $N=4$ and $\sigma=\{1,2\}$, we have
$$
	\vece_\sigma = \vece_1 \wedge \vece_2 \qqand
	P_\sigma = \begin{bmatrix}
	1 & 0 & 0 & 0 \\
	0 & 1 & 0 & 0 \\
	\end{bmatrix}.
$$
\mendex

%xxx Write out the Pfaffian definition and point with glee.

%example:  $N=4$ or $6$ \ldots.

\begin{prop}
\label{prop:ext_exp_pfaffian}
Let $Y$ be an $N$-dimensional vector space with basis $\{\vece_1, \ldots,
\vece_N\}$.  Let $L$ be a skew-symmetric $N \times N$ matrix.  Then
$$
	\exp\frac{1}{2}\sum_{i,j} L_{ij} \vece_i \wedge \vece_j
	= \sum_{\sigma \subseteq 2^{\{1, \ldots, N\}}}
		\Pf(P_\sigma L P_\sigma) e_\sigma.
$$
\begin{proof}
TBD.
\end{proof}
\end{prop}

\beginex
\label{ex:ext_exp_pfaffian}
Let $N=4$ and consider the quadratic form
$$
	q = \vece_1\wedge\vece_2 + \vece_3\wedge\vece_4.
$$
One may verify by inspection that this is the bivector associated to the matrix
$$
	Q = \left[\begin{array}{rrrr}
	0  & 1 &  0 & 0 \\
	-1 & 0 &  0 & 0 \\
	0  & 0 &  0 & 1 \\
	0  & 0 & -1 & 0 \\
	\end{array}\right].
$$
Then
\begin{eqnarray*}
	\exp(q) &=& 1 + q + \frac{1}{2} \, q\wedge q \\
	&=& 1 + (\vece_1\wedge\vece_2 + \vece_3\wedge\vece_4) +
		\frac{1}{2}\, (\vece_1\wedge\vece_2 + \vece_3\wedge\vece_4) \wedge (\vece_1\wedge\vece_2 + \vece_3\wedge\vece_4) \\
	&=& 1 + (\vece_1\wedge\vece_2) + (\vece_3\wedge\vece_4) +
		(\vece_1\wedge\vece_2 \wedge \vece_3\wedge\vece_4).
\end{eqnarray*}
Recall from proposition \ref{prop:pfaffian_of_odd_dim_mx} that the Pfaffian of
an odd-dimensional matrix is zero, so we need to consider only $\sigma$'s
having 0, 2, or 4 elements.  These are
$$
	\{\}, \{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\}, \{1,2,3,4\}.
$$
From the matrix $Q$ we see that only the $\sigma$'s $\{\}$, $\{1,2\}$,
$\{3,4\}$, and $\{1,2,3,4\}$ contribute non-zero terms.
For $\sigma = \{1,2\}$ and $\{3,4\}$ we have
$$
	P_\sigma Q P_\sigma = \left[\begin{array}{rr}
	0 & 1 \\
	-1 & 0 \\
	\end{array}\right]
$$
which has a Pfaffian of 1.  The $4 \times 4$ matrix $Q$ has Pfaffian
$af-be+cd=1-0+0=1$.  Thus, we have
\begin{eqnarray*}
	\sum_{\sigma \subseteq 2^{\{1, \ldots, N\}}}
		\Pf(P_\sigma L P_\sigma) e_\sigma
	&=&
	1 +
	\vece_1 \wedge \vece_2 +
	\vece_3 \wedge \vece_4 +
	\vece_1 \wedge \vece_2 \wedge \vece_3 \wedge \vece_4
\end{eqnarray*}
as expected.
\mendex

This example shows that the Pfaffian serves here as a \bfidx{coefficient extractor}
as discussed in example \ref{ex:q_wedge_q}.

%% ----------------------------------------------------------------
\subsection{Isotropic subspaces and polarizations}
\label{sec:iso_sspcs}

This section provides machinery for Fock representations as presented in
section \ref{sec:fock_rep}.  Again suppose that $(\cdot,\cdot)$ is a
non-degenerate bilinear form on a complex vector space $Y$.

\begin{defn}
A subspace $W$ of $Y$ is \bfidx{isotropic} if $(\vecu,\vecv)=0$ for all
$\vecu,\vecv \in W$.
\end{defn}

\begin{rem}
In particular, this means $(\vecu,\vecu) = 0$ for all $\vecu \in W$.  Thus an
isotropic subspace can have dimension higher than 0 only when the bilinear form
$(\cdot,\cdot)$ is not positive-definite.
\end{rem}

\beginex
Let $Y=\C^2$, with bilinear form $(\vecu,\vecv) = u_1 v_1 + u_2 v_2$.  In an
attempt to find an isotropic subspace, let's first find a non-zero vector
$\vecw$ which is orthogonal to itself.  The condition $w_1^2+w_2^2 = 0$ forces
$\vecw=0$ over $\R$, but over $\C$ we have more freedom.  Take
$\vecw=\colvectwo{1}{i}$.  Then $(\vecw,\vecw)=1-1=0$.  Now take $W$ to be the
span of $\vecw$.  For any $\vecu=a\vecw$ and $\vecv=b\vecw$ in $W$ (with $a,b
\in \C$) we have
$$
	(\vecu,\vecv) =\left( \colvectwo{a}{ia}, \colvectwo{b}{ib} \right)
	= ab-ab = 0.
$$
Thus $W$ is an isotropic subspace for $Y$.  Note that the
non-positive-definiteness of the bilinear form was needed to construct a
non-trivial isotropic subspace.
\mendex

\begin{defn}
An \bfidx{isotropic splitting} or \bfidx{polarization} of $Y$ is a
decomposition of $Y$ into the direct sum of two subspaces $Y = A \oplus B$
where $A$ and $B$ are isotropic subspaces of $Y$.
\end{defn}

\begin{prop}
\label{prop:polar_dual_isom}
Suppose we have a polarization $Y=A \oplus B$ where the distinguished
bilinear form is non-degenerate.  Then $A \cong B^*$ (and vice versa)
with isomorphism $\veca \mapsto (\veca, \cdot)$.
\begin{proof}
The map $\veca \mapsto (\veca, \cdot)$ is linear due to the bilinearity of the
form.

For injectivity, let $\veca_1, \veca_2 \in A$ and suppose $(\veca_1, \cdot) =
(\veca_2,\cdot)$.  Then for all $\vecb\in B$, and recalling the bilinearity of
the form, we have
\begin{eqnarray*}
	(\veca_1,\vecb) &=& (\veca_2,\vecb) \\
	(\veca_1,\vecb) - (\veca_2,\vecb) &=& 0 \\
	(\veca_1-\veca_2,\vecb) &=& 0.
\end{eqnarray*}
Since the form is non-degenerate and $(\veca_1-\veca_2,\vecb) = 0$ for all
$\vecb$, $\veca_1-\veca_2=0$ i.e. $\veca_1=\veca_2$.

For surjectivity, recall that for finite-dimensional vector spaces
we have
$$
	\dim(A) = \dim(A^*) \qqand \dim(B) = \dim(B^*).
$$
The injectivity argument above, along with its obvious analogue for a map from
$B$ into $A^*$, shows that
$$
	\dim(A) \le \dim(B^*) \qqand \dim(B) \le \dim(A^*).
$$
From these facts we conclude that
$$
	\dim(A^*) \le \dim(B^*) \qqand \dim(B^*) \le \dim(A^*)
$$
which forces
$$
	\dim(A^*) = \dim(B^*) \qqand
	\dim(A) = \dim(B).
$$
\end{proof}
\end{prop}

\begin{cor}
\label{cor:polar_even_dim}
A polarized space is even-dimensional.
\end{cor}

\beginex
Continuing the previous example, we can seek a complementary isotropic subspace
$W'$ in addition to the isotropic subspace $W$ already found.  In an attempt to
satisfy the condition $Y=W\oplus W'$ --- which in particular requires that $W
\cap W' = \{0\}$ --- we might try looking for the space perpendicular to $W$
using the usual complex inner product.  We compute
\begin{eqnarray*}
	\colvectwo{1}{i} &\perp& \colvectwo a b \\
	\left\langle \colvectwo{1}{i} , \colvectwo{a}{b} \right\rangle &=& 0 \\
	a-ib&=& 0.
\end{eqnarray*}
Setting $b=-i$ as a parameter we have
$$
	\colvectwo a b = \colvectwo{1}{-i}.
$$
The subspace $W'$ is seen to be isotropic just as $W$ was.  Thus we have
the polarization $\C_2=Y=W\oplus W'$ where
$$
	W  = \left\langle \left\{ \colvectwo{1}{i}  \right\} \right\rangle
	\qqand
	W' = \left\langle \left\{ \colvectwo{1}{-i} \right\} \right\rangle.
$$
\mendex

Ad-hoc examples notwithstanding, one asks whether one may always obtain an
isotropic splitting for a given (even-dimensional by corollary
\ref{cor:polar_even_dim}) vector space $Y$.  We require a lemma before
answering this in the affirmative.

%[xxx perh. move this up]

\begin{lem}
\label{lem:gram_schmidt}
Given a non-degenerate bilinear form on a finite-dimensional complex vector
space $Y$, there exists a basis $\{\vecw_1, \ldots, \vecw_N\}$ for $Y$ which is
orthonormal with respect to the form.
\begin{proof}
The proof is not meaningfully different from that of the
\plainidx{Gram-Schmidt} process for non-degenerate sesquilinear forms.  See any
linear algebra text for details.
\end{proof}
\end{lem}

\begin{prop}
Let $\{\vecb_1, \ldots, \vecb_N\}$ be a basis for an $N$-dimensional complex
vector space $Y$, where $N$ is assumed to be even.  Further assume that the
basis is orthonormal (via lemma \ref{lem:gram_schmidt}) with respect to the
non-degenerate bilinear form.  Define
$$
	Y_+ = \langle \vecb_j + i \vecb_{j+N/2} \rangle \qand
	Y_- = \langle \vecb_j - i \vecb_{j+N/2} \rangle.
$$
Then $Y = Y_+ \oplus Y_-$ is a polarization of $Y$.
\begin{proof}
To see that $Y_+ \cap Y_- = \{0\}$, suppose
$\vecy \in Y_+ \cap Y_-$, i.e.
$$
	\vecy
	= \sum_{j=1}^{N/2} c_j(\vecb_j + i \vecb_{j+N/2})
	= \sum_{j=1}^{N/2} d_j(\vecb_j - i \vecb_{j+N/2}).
$$
Then
\begin{eqnarray*}
	\sum_{j=1}^{N/2} c_j(\vecb_j + i \vecb_{j+N/2}) &=& \sum_{j=1}^{N/2} d_j(\vecb_j - i \vecb_{j+N/2}) \\
	\sum_{j=1}^{N/2} (c_j - d_j)\vecb_j + i \sum_{j=1}^{N/2} (c_j + d_j)\vecb_{j+N/2} &=& 0.
\end{eqnarray*}
The linear independence of the $\vecb_j$'s forces $c_j=d_j$ and $c_j=-d_j$ for all
$j$, hence $c_j=d_j=0$ and thus $\vecy = 0$.

To see that $Y_+ + Y_- = Y$, \ldots .
let
$$
	\vecy = \sum_{j=1}^N c_j \vecb_j \in Y.
$$
Then
\begin{eqnarray*}
	\vecy = \sum_{j=1}^N c_j \vecb_j
	&=& \sum_{j=1}^{N/2} (c_j \vecb_j + c_{j+N/2} \vecb_{j+N/2}) \\
	&=& \sum_{j=1}^{N/2} \left(
		\frac{c_j - i c_{j+N/2}}{2}(\vecb_j + i \vecb_{j+N/2})
		+
		\frac{c_j + i c_{j+N/2}}{2}(\vecb_j - i \vecb_{j+N/2})
		\right).
\end{eqnarray*}

To see that $Y_+$ is an istropic subspace of $Y$, let
$$
	\vecy = \sum_{j=1}^{N/2} c_j (\vecb_j + i\vecb_{j+N/2}) \in Y_+.
$$
Then
\begin{eqnarray*}
	(\vecy,\vecy)
	&=& \left(
		\sum_{j=1}^{N/2} c_j (\vecb_j + i\vecb_{j+N/2}),
		\sum_{k=1}^{N/2} c_k (\vecb_k + i\vecb_{k+N/2}) \right) \\
	&=& \sum_{j=1}^{N/2} \sum_{k=1}^{N/2}
		c_j c_k
		\left(\vecb_j + i\vecb_{j+N/2}, \vecb_k + i\vecb_{k+N/2} \right) \\
	&=& \sum_{j=1}^{N/2} \sum_{k=1}^{N/2}
		c_j c_k \left[
			(\vecb_j, \vecb_k)
			+i(\vecb_j, \vecb_{k+N/2})
			+i(\vecb_{j+N/2}, \vecb_k)
			-(\vecb_{j+N/2}, \vecb_{k+N/2})
			\right] \\
	&=& \sum_{j=1}^{N/2} \sum_{k=1}^{N/2}
		c_j c_k \left[
			\delta_{jk} +0 +0 -\delta_{jk} \right] \\
	&=& 0.
\end{eqnarray*}
The proof that $Y_-$ is an istropic subspace of $Y$ is all but identical.
\end{proof}
\end{prop}

This orthonormalization process is one technique to construct a polarization of
a given vector space.  Another technique (using \emphidx{polarization
operators}) is shown in appendix A.2 of \cite{bib:palmer2}.

%Another technique is shown in section \ref{sec:polop}.

%%% ----------------------------------------------------------------
%\subsection{The polarization operator}
%\label{sec:polop}
%
%xxx emph:  Map from polarization to operator, OR from operator to polarization.
%
%xxx Given $Y=W_+ \oplus W_-$, define $Q$.  Show it has certain properties.
%
%xxx Given $Q$ with certain properties, construct a polarization.
%
%xxx cf. \cite{bib:palmer2} appendix A.2 (p. 292).
%
%xxx upshot:  $Q$ is skew-symm and $Q^2=I$.
%\begin{defn}
%An operator $Q$ from [xxx] to [xxx] is said to be a \bfidx{polarization
%operator} if $Q$ is skew-symmetric and $Q^2=I$.
%\end{defn}
%
%XXX do we need the polarization operator for this paper?!?

%% ----------------------------------------------------------------
\subsection{Fock representations}
\label{sec:fock_rep}

%Let $Y$ be an even-dimensional complex vector space.  Let $Q$ be a
%\plainidx{polarization operator} on $Y$.  Then, as in section
%\ref{sec:iso_sspcs}, we have a polarization

\begin{defn}
\label{defn:fock_Y_rep}
Let $Y$ be an even-dimensional complex vector space with a polarization
$$
	Y = Y_+ \oplus Y_-.
$$
In particular, for all $v \in Y$ there are $v_+ \in Y_+$ and $v_- \in Y_-$
such that
$$
	v = v_+ + v_-.
$$
Recall from proposition \ref{prop:polar_dual_isom} that we may identify $Y_-$
with $Y_+^*$ via
$$
	v_- \mapsto (\cdot, v_-).
$$
Define $F: \Cliff(Y) \to \Cliff(Y)$ by
$$
	F(v) = c(v_+) + a(v_-).
$$
\end{defn}

%\begin{rem}

Note that we have $v_+ \in Y_+$, $v_- \in Y_+^*$, and thus (recalling the
definitions of the creation and annihilation operators in section
\ref{sec:creat_annih}), we have
$$
	F(v): \Lambda(Y_+) \to \Lambda(Y_+).
$$
Since the creation and annihilation operators are linear transformations on
$\Lambda(Y_+)$ (see proposition \ref{prop:creat_annih_linear}), we have
$$
	F : Y \to \Hom(\Lambda(Y_+)).
$$
That is, $F$ is a \bfidx{representation} of $Y$ on the representation space
$\Lambda(Y_+)$.  We can extend $F$ to be a representation of $\Cliff(Y)$ on
$\Lambda(Y_+)$ if only the Clifford-algebra relations (see definition
\ref{defn:cliff_alg}) can be satisfied.  (We will re-use the letter $F$.)

%\end{rem}

\begin{prop}
The map $F:Y \to \Hom(\Lambda(Y_+))$ from definition \ref{defn:fock_Y_rep}
satisifies the Clifford-algebra relations from definition
\ref{defn:cliff_alg}), with composition as multiplication.  That is,
for all $u,v \in Y$,
$$
	F(u)F(v) + F(v)F(u) = (u,v).
$$
\begin{proof}
From the definition of $F$, we have
\begin{eqnarray*}
	F(u)F(v) + F(v)F(u)
	&=& (c(u_+)+a(u_-))\circ(c(v_+)+a(v_-))
	+   (c(v_+)+a(v_-))\circ(c(u_+)+a(u_-)) \\
	&=& c(u_+) \circ c(v_+)
	+   c(u_+) \circ a(v_-)
	+   a(u_-) \circ c(v_+)
	+   a(u_-) \circ a(v_-) \\
	&+& c(v_+) \circ c(u_+)
	+   c(v_+) \circ a(u_-)
	+   a(v_-) \circ c(u_+)
	+   a(v_-) \circ a(u_-).
\end{eqnarray*}
Now, using the first two anticommutation relations in proposition
\ref{prop:creat_annih_cacr}, we have $c(u_+)\circ c(v_+) +c(v_+)\circ c(u_+) =
0$ etc.  Thus
\begin{eqnarray*}
	F(u)F(v) + F(v)F(u)
	&=& c(u_+) \circ a(v_-)
	+   a(u_-) \circ c(v_+)
	+   c(v_+) \circ a(u_-)
	+   a(v_-) \circ c(u_+) \\
	&=& a(u_-) \circ c(v_+)
	+   c(v_+) \circ a(u_-)
	+   a(v_-) \circ c(u_+)
	+   c(u_+) \circ a(v_-).
\end{eqnarray*}
Using the third anticommutation relation in proposition
\ref{prop:creat_annih_cacr}, we have
\begin{eqnarray*}
	F(u)F(v) + F(v)F(u)
	&=& u_-(v_+) + v_-(u_+) \\
	&=& (u_-,v_+) + (v_-,u_+) \\
	&=& (u_-,v_+) + (u_+,v_-)
\end{eqnarray*}
by the symmetry of the bilinear form.
This is the same as $(u,v)$.
Using the fact that $u = u_+ + u_-$ and $v = v_+ + v_-$, along with the
bilinearity of the form, we have
$$
	(u,v)
	= (u_+ + u_-, v_+ + v_-)
	= (u_+, v_+)
	+ (u_+, v_-)
	+ (u_-, v_+)
	+ (u_-, v_-).
$$
Since $Y$ is polarized, $(u_+,v_+)$ and $(u_-,v_-)$ are both zero, and so
$$
	(u,v) = (u_+, v_-) + (u_-, v_+)
$$
which is the same as $F(u)F(v) + F(v)F(u)$, which is what was to be shown.
\end{proof}
\end{prop}

\begin{defn}
The map $F:\Cliff(Y)\to\Lambda(Y_+)$ as defined in this section is called the
\bfidx{Fock representation} of $\Cliff(Y)$ with respect to the polarization
$Y=Y_+ \oplus Y_-$ on the representation space $\Lambda(Y_+)$.
\end{defn}

%% ----------------------------------------------------------------
\subsection{Iterated commutators}

In this section, let $A$ and $B$ be linear transformations from a complex
vector space $Y$ to itself.

\begin{defn}
\label{defn:matrix_commutator}
Recall that we define the \bfidx{commutator} of $B$ with $A$ to be
$[A,B] = AB-BA$.
\end{defn}

\begin{lem}
\label{lem:commutator_bilinear}
The commutator is bilinear.  That is, for all $A,B,C,D$, and all scalars
$\lambda,\mu$, we have
$$
	[A+B,C+D] = [A,C] + [A,D] + [B,C] + [B, D].
$$
and
$$
	[\lambda A, \mu B] = \lambda \mu[A, B].
$$
\begin{proof}
For the first claim, we have
\begin{eqnarray*}
	[A+B,C+D] &=& (A+B)(C+D) - (C+D)(A+B) \\
	&=& AC + AD + BC + BD - CA - CB - DA - DB \\
	&=& AC - CA + AD - DA + BC - CB + BD - DB \\
	&=& [A,C] + [A,D] + [B,C] + [B, D].
\end{eqnarray*}
For the second claim, we have
\begin{eqnarray*}
	[\lambda A, \mu B] &=& \lambda A \mu B - \mu B \lambda A \\
	&=& \lambda \mu A B - \lambda \mu B A \\
	&=& \lambda \mu(AB - BA) \\
	&=& \lambda \mu[A, B].
\end{eqnarray*}
\end{proof}
\end{lem}


\begin{defn}
\label{defn:ad}
Define the $n$-fold \bfidx{iterated commutator} of $B$ with $A$, written
$\ad^n(A)B$, as
\begin{eqnarray*}
	\ad^0(A)B &=& B \\
	\ad^n(A)B &=& [A, \ad^{n-1}(A)B] = [A,[A, \ldots, [A, B]]].
\end{eqnarray*}
\end{defn}

\begin{rem}
\label{rem:ad0123}
In particular, we have the following:
\begin{eqnarray*}
	\ad^0(A)B &=& B \\
	\ad^1(A)B &=& AB-BA \\
	\ad^2(A)B &=& [A,[A,B]] = [A,AB-BA]
		= A(AB-BA) - (AB-BA)A \\
		&=& A^2B - 2ABA + BA^2 \\
	\ad^3(A)B &=& A(A^2B - 2ABA + BA^2) - (A^2B - 2ABA + BA^2)A \\
		&=& A^3B - 2A^2BA + ABA^2 - A^2BA + 2ABA^2 - BA^3 \\
		&=& A^3B - 3A^2BA + 3ABA^2 - BA^3.
\end{eqnarray*}
\end{rem}

These computations suggest the following proposition.
\begin{prop}
\label{prop:ad}
Using the notation from the above definition,
$$
	\ad^n(A)B = \sum_{k=0}^n (-1)^k {n \choose k} A^{n-k} B A^k.
$$
\begin{proof}
The proof is by induction on $n$.  The base case, $n=0$, is given by
$$
	\sum_{k=0}^0 (-1)^k {n \choose k} A^{n-k} B A^k = B = \ad^0(A)B.
$$
For the induction step, suppose the result holds for $n$.  Then we have
\begin{eqnarray*}
	\ad^{n+1}(A)B &=& [A, \ad^n(A)B] \\
	&=& \left[A, \sum_{k=0}^n (-1)^k {n \choose k} A^{n-k} B A^k \right]. \\
\end{eqnarray*}
Using the bilinearity of the commutator (lemma \ref{lem:commutator_bilinear}),
\begin{eqnarray*}
	\ad^{n+1}(A)B
	&=& \sum_{k=0}^n (-1)^k {n \choose k} \left[A, A^{n-k} B A^k \right] yy\\
	&=& \sum_{k=0}^n (-1)^k {n \choose k}
		\left( A(A^{n-k} B A^k) - (A^{n-k} B A^k) A \right) \\
	&=& \sum_{k=0}^n (-1)^k {n \choose k}
		\left( A^{n+1-k} B A^k - A^{n-k} B A^{k+1} \right).
\end{eqnarray*}
Expanding terms to find the pattern:
\begin{eqnarray*}
	&=& {n \choose 0} A^{n+1}B
		- {n \choose 1} A^n B A
		+ {n \choose 2} A^{n-1} B A^2
		- \ldots \\
	&&+   (-1)^{n-2} {n \choose {n-2}} A^3 B A^{n-2}
		+ (-1)^{n-1} {n \choose {n-1}} A^2 B A^{n-1}
		+ (-1)^{n}   {n \choose {n}}   A   B A^{n}
	\\
	&&- {n \choose 0} A^{n}B A
		+ {n \choose 1} A^{n-1} B A^2
		- {n \choose 2} A^{n-2} B A^3
		+ \ldots \\
	&&-   (-1)^{n-2} {n \choose {n-2}} A^2 B A^{n-1}
		- (-1)^{n-1} {n \choose {n-1}} A   B A^{n}
		- (-1)^{n}   {n \choose {n}}       B A^{n+1}.
\end{eqnarray*}
Collecting like terms:
\begin{eqnarray*}
	\ad^{n+1}(A)B
	&=&
		{n \choose 0} A^{n+1}B
		- \left({n \choose 0} + {n \choose 1}\right) A^{n}B A
		+ \left({n \choose 1} + {n \choose 2}\right) A^{n-1} B A^2 \\
	&&-   \left({n \choose 2} + {n \choose 3}\right) A^{n-2} B A^3
		+ \ldots \\
	&&+ (-1)^{n-1}\left({n \choose {n-2}} + {n \choose {n-1}}\right)
			A^2 B A^{n-1}
		+ (-1)^{n}\left({n \choose {n-1}} + {n \choose {n}}\right)
			A B A^{n} \\
	&&+ (-1)^n {n \choose n} B A^{n+1}.
\end{eqnarray*}

At this point we need a little combinatorial lemma.
\begin{lem}
Adjacent binomial coefficients obey
$$
	{n \choose k} + {n \choose {k+1}} = {{n+1} \choose {k+1}}.
$$
\begin{proof}
Using the factorial definition of binomial coefficients and acquiring a common
denominator, we obtain
\begin{eqnarray*}
	{n \choose k} + {n \choose {k+1}}
	&=& \frac{n!}{k!(n-k)!} + \frac{n!}{(k+1)!(n-k-1)!} \\
	&=& \frac{n!(k+1)}{(k+1)!(n-k)!} + \frac{n!(n-k)}{(k+1)!(n-k)!} \\
	&=& \frac{n!((k+1)-(n-k))}{(k+1)!(n-k)!}
	= \frac{(n+1)!}{(k+1)!((n+1)-(k+1))!} \\
	&=& {{n+1} \choose {k+1}}.
\end{eqnarray*}
\end{proof}
\end{lem}

Using the lemma, and recalling that
$$
	{n \choose 0} = {{n+1} \choose 0} =
	{n \choose n} = {{n+1} \choose {n+1}} = 1
$$
as well as $-(-1)^k = (-1)^{k+1}$, we have
\begin{eqnarray*}
	\ad^{n+1}(A)B
	&=&
		{{n+1} \choose 0} A^{n+1}B
		- {{n+1} \choose 1} A^{n}B A
		+ {{n+1} \choose 2} A^{n-1} B A^2
	 	- {{n+1} \choose 3} A^{n-2} B A^3
		+ \ldots \\
	&&+ (-1)^{n-1}{{n+1} \choose {n-1}} A^2 B A^{n-1}
		+ (-1)^{n} {{n+1} \choose n} A B A^{n}
	+ (-1)^{n+1} {{n+1} \choose {n+1}} B A^{n+1} \\
	&=& \sum_{k=0}^{n+1} (-1)^k {{n+1} \choose k}
		A^{n+1-k} B A^{k}
\end{eqnarray*}
as desired.
\end{proof}
\end{prop}

In the proof of theorem \ref{thm:vac_chzn} in section \ref{sec:vac_chzn} we
will use iterated commutators to evaluate an expression of the form $e^{\lambda
A} B e^{-\lambda A}$.  Writing out the matrix exponentials in Taylor series
about $\lambda=0$ we have:
\begin{eqnarray*}
	e^{ \lambda A}  &=& I + \lambda A  + \lambda^2 A^2/2!  + \lambda^3 A^3/3! + \ldots \\
	e^{\lambda A B} &=& B + \lambda AB + \lambda^2 A^2B/2! + \lambda^3 A^3B/3! + \ldots \\
	e^{-\lambda A}  &=& I - \lambda A  + \lambda^2 A^2/2!  - \lambda^3 A^3/3! + \ldots.
\end{eqnarray*}
FOILing out the product $e^{\lambda A} B e^{-\lambda A}$ in the following
tableau (freely re-arranging the terms of the sum) enables us to formulate a
conjecture:
$$
\begin{tabular}{l|llllll}
{} &
	$I$ &
	$-\lambda   A$ &
	$ \lambda^2 A^2/2!$ &
	$-\lambda^3 A^3/3!$ &
	$+\ldots$ \\
\hline
$B$ &
	$               B         $ &
	$-\lambda       B A       $ &
	$+\lambda^2     B A^2/  2!$ &
	$-\lambda^3     B A^3/  3!$ &
	$\ldots$ \\
$\lambda A B$ &
	$+\lambda   A   B         $ &
	$-\lambda^2 A   B A       $ &
	$+\lambda^3 A   B A^2/  2!$ &
	$-\lambda^4 A   B A^3/  3!$ &
	$\ldots$ \\
$\lambda^2 A^2 B/2!$ &
	$+\lambda^2 A^2 B    /2!  $ &
	$-\lambda^3 A^2 B A  /2!  $ &
	$+\lambda^4 A^2 B A^2/2!2!$ &
	$-\lambda^5 A^2 B A^3/2!3!$ &
	$\ldots$ \\
$\lambda^3 A^3 B/3!$ &
	$+\lambda^3 A^3 B    /3!  $ &
	$-\lambda^4 A^3 B A  /3!  $ &
	$+\lambda^5 A^3 B A^2/3!2!$ &
	$-\lambda^6 A^3 B A^3/3!3!$ &
	$\ldots$ \\
$+\ldots$ &. \\
\end{tabular}
$$
Collecting like powers of $\lambda$ on anti-diagonals suggests the following proposition.

\begin{prop}
\label{prop:exp_ad_prop}
Let $A$ and $B$ be linear transformations from a complex vector space $Y$ to itself.
Then
$$
	e^{\lambda A} B e^{-\lambda A} = \sum_{n=0}^\infty \frac{\lambda^n}{n!} \ad^n(A)B
$$
\begin{proof}
Writing out the Taylor expansions of $e^{\lambda A}$ and $e^{-\lambda A}$ about $\lambda=0$
and freely re-arranging terms gives
\begin{eqnarray*}
	(e^{\lambda A} B) e^{-\lambda A}
	&=&
		\left(\sum_{i=0}^\infty        \lambda^i A^i B/i!\right)
		\left(\sum_{j=0}^\infty (-1)^j \lambda^j A^j  /j!\right) \\
	&=& \sum_{i=0}^\infty \sum_{j=0}^\infty
		(-1)^j \lambda^{i+j} A^i B A^j / i!j! \\
	&=& \sum_{n=0}^\infty \sum_{i+j=n}
		(-1)^j \lambda^{n} A^i B A^j   / i!j!.
\end{eqnarray*}
Since the inner sum is over $i+j=n$, we may put $k=j$ and $i=n-k$ to obtain
\begin{eqnarray*}
	(e^{\lambda A} B) e^{-\lambda A}
	&=& \sum_{n=0}^\infty \sum_{k=0}^n
		(-1)^k \lambda^{n} \frac{A^{n-k} B A^k}{(n-k)!k!} \\
	&=& \sum_{n=0}^\infty \sum_{k=0}^n
		(-1)^k \lambda^{n} \frac{n! A^{n-k} B A^k}{n! (n-k)!k!} \\
	&=& \sum_{n=0}^\infty \sum_{k=0}^n
		(-1)^k \frac{\lambda^{n}}{n!} {n \choose k} A^{n-k} B A^k \\
	&=& \sum_{n=0}^\infty
			\frac{\lambda^{n}}{n!}
		\sum_{k=0}^n
			(-1)^k {n \choose k} A^{n-k} B A^k.
\end{eqnarray*}
By proposition \ref{prop:ad}, the inner sum is precisely $\ad^n(A)B$ and so
\begin{eqnarray*}
	(e^{\lambda A} B) e^{-\lambda A}
	&=& \sum_{n=0}^\infty \frac{\lambda^n}{n!} \ad^n(A)B
\end{eqnarray*}
as desired.
\end{proof}
\end{prop}

%% ================================================================
\newpage
\section{Continuous and discrete operators}
\label{sec:cts_disc_oprs}

%% ----------------------------------------------------------------
\subsection{Symmetric and skew-symmetric operators}

Let $X$ be a vector space, real or complex, and not necessarily
finite-dimensional.  In particular, we might consider $X$ to be $\R^n$, $\C^n$,
or a Hilbert function space such as (real- or complex-valued)
$L^2(-\infty,+\infty)$ or $L^2[0,1]$ or $L^2[a,b]$.

\begin{defn}
A \bfidx{linear operator} is simply a linear transformation from $X$ to $X$.
\end{defn}

\beginex
For finite-dimensional vector spaces, we usually think of operators as
matrices.
\mendex

\beginex
For function spaces, canonical examples include:
\begin{itemize}
\item $f(x) \mapsto df/dx$.
\item $f(x) \mapsto d^2f/dx^2$.
\item $f(x) \mapsto x\,f(x)$.
\item $f(x) \mapsto \int K(x,y)f(y)dy$.
\end{itemize}
\mendex

\begin{defn}
Now suppose $X$ has a distinguished bilinear form $\sbfdot$,
and let $L$ be a linear operator on $X$.  Then $L$ is said to be
\bfidx{symmetric} if
$$
	(Lf, g) = (f, Lg)
$$
for all $f,g \in X$.
Likewise, $L$ is said to be \bfidx{skew-symmetric} if
$$
	(Lf, g) = -(f, Lg)
$$
for all $f,g \in X$.
\end{defn}

\begin{rem}
On finite-dimensional vector spaces, there is no difference between symmetric
and self-adjoint operators.  See \cite{bib:lax} for more information.
\end{rem}

\beginex
Matrices are symmetric or skew-symmetric as operators if they are symmetric or
skew-symmetric, respectively, in coordinates: $A_{ij} = A_{ji}$ or $A_{ij} =
-A_{ji}$, respectively.
\mendex

\begin{defn}
For function spaces, define the symmetric bilinear form
$$
	(f,g) = \int f(x) g(x) dx.
$$
\end{defn}

\beginex
Let $X=L^2[a,b]$ and let
$$
	W = \{f \in X : f(a)=f(b)=0\}.
$$
The operator $L=d/dx$ is skew-symmetric on $W$, which is to say
$$
	(Lf,g)=-(f,Lg)
$$
i.e.
$$
	\int_a^b \frac{df}{dx} \,g(x) \,dx + \int_a^b f(x)\, \frac{dg}{dx} \,dx = 0.
$$
To see this, integrate by parts:
$$
	\int_a^b \frac{df}{dx} \,g(x) \,dx + \int_a^b f(x)\, \frac{dg}{dx} \,dx
	= f(b)g(b) - f(a)g(a) = 0
$$
due to the boundary conditions on $W$.
\mendex

%xxx temp:
%\begin{eqnarray*}
%	\int_a^b u v' dx + \int_a^b u'v dx &=& [uv]_a^b \\
%	\int_a^b u v' dx &=& [uv]_a^b - \int_a^b u'v dx \\
%	\int_a^b u'v dx &=& [uv]_a^b - \int_a^b u v' dx \\
%\end{eqnarray*}

%xxx temp:
%\begin{eqnarray*}
%	\int_a^b f''g dx &=& [f'g]_a^b - \int_a^b f' g' dx \\
%	\int_a^b fg'' dx &=& [fg']_a^b - \int_a^b f' g' dx \\
%\end{eqnarray*}

\beginex
Let $X$ and $W$ be as in the previous example.  Let $M=d^2/dx^2$.  Then
$M$ is symmetric on $W$, i.e.  $(Mf,g)-(f,Mg)=0$.  To see this, integrate by
parts on each term and apply the boundary conditions:
\begin{eqnarray*}
	(Mf,g) - (f,Mg)
	&=&\int_a^b f''(x)g(x) dx - \int_a^b f(x)g''(x) dx \\
	&=& \left([f'(x)g(x)]_a^b - \int_a^b f'(x) g'(x) dx\right) - \left([f(x)g'(x)]_a^b - \int_a^b f'(x) g'(x) dx\right) \\
	&=& [f'(x)g(x)]_a^b - [f(x)g'(x)]_a^b = 0.
\end{eqnarray*}
\mendex

\begin{rem}
\label{rem:d_dx_S_forms}
In the previous two examples, $L$ and $M$ are not skew-symmetric and symmetric,
respectively, on all of $X$.  One may think of the symmetric bilinear form
$$
	(f,g)_L = (Lf,g)+(f,Lg),
$$
which is identically zero on $W$, as measuring the lack of skew-symmetry of $L$
on $X$.  Likewise, the skew-symmetric bilinear form
$$
	(f,g)_M = (Mf,g)-(f,Mg),
$$
measures the lack of symmetry of $M$ on $X$.
\end{rem}

\begin{prop}
Let $X$, $L$, and $M$ be as above.  $L$ and $M$ are indeed bilinear forms.
\begin{proof}
Use the linearity of derivative and integral:
\begin{eqnarray*}
	(af_1+f_2, bg_1+g_2)_L
	&=& (af_1'+f_2', bg_1+g_2) + (af_1+f_2, bg_1'+g_2') \\
	&=& ab(f_1',g_1) + ab(f_1',g_2) + ab(f_2',g_1) + ab(f_2',g_2) \\
	&+& ab(f_1,g_1') + ab(f_1,g_2') + ab(f_2,g_1') + ab(f_2,g_2') \\
	&=& ab(f_1,g_1)_L + ab(f_1,g_2)_L + ab(f_2,g_1)_L + ab(f_2,g_2)_L
\end{eqnarray*}
as well as
\begin{eqnarray*}
	(af_1+f_2, bg_1+g_2)_M
	&=& (af_1''+f_2'', bg_1+g_2) - (af_1+f_2, bg_1''+g_2'') \\
	&=& ab(f_1'',g_1) + ab(f_1'',g_2) + ab(f_2'',g_1) + ab(f_2'',g_2) \\
	&-& ab(f_1,g_1'') - ab(f_1,g_2'') - ab(f_2,g_1'') - ab(f_2,g_2') \\
	&=& ab(f_1,g_1)_M + ab(f_1,g_2)_M + ab(f_2,g_1)_M + ab(f_2,g_2)_M.
\end{eqnarray*}
For symmetry and skew-symmetry:
\begin{eqnarray*}
	(f,g)_L
		&=& (f',g) + (f,g')
		= (g',f) + (g,f')
		= (g,f)_L,\\
	(f,g)_M
		&=& (f'',g) - (f,g'')
		= -(g'',f) + (g,f'')
		= -(g,f)_M.
\end{eqnarray*}
\end{proof}
\end{prop}

%%% ----------------------------------------------------------------
%\subsection{Multidimensional continuous operators}
%
%xxx quack briefly.  Maybe mention the Laplacian --- standard example.
%What's the canonical example of a skew-symmetric vector operator?

%% ----------------------------------------------------------------
\subsection{Discrete operators}

Recall that the derivatives of a real-valued function are often,
in practice, approximated numerically by finite differences:
\begin{eqnarray*}
	f'(x) &\approx& \frac{f(x+h)-f(x)}{h} \quad\textrm{(Right-difference rule)}\\
	f'(x) &\approx& \frac{f(x)-f(x-h)}{h} \quad\textrm{(Left-difference rule)}\\
	f'(x) &\approx& \frac{f(x+h)-f(x-h)}{2h} \quad\textrm{(Centered-difference rule)}.
\end{eqnarray*}
For this paper, the centered-difference rule will be used.  Likewise, for the
second derivative, we naturally use:
\begin{eqnarray*}
	f''(x)&\approx& \frac{f(x+h)-2f(x)+f(x-h)}{h^2}.
\end{eqnarray*}

The same approximation technique applies for directional derivatives, in the
direction of the vector $\vech$, of functions from $\R^n$ to $\R$:
\begin{eqnarray*}
	D_\vech f(\vecx) &\approx& \frac{f(\vecx+\vech)-f(\vecx-\vech)}{2\|\vech\|}\\
	D^2_\vech f(\vecx)&\approx& \frac{f(\vecx+\vech)-2f(\vecx)+f(\vecx-\vech)}{\|\vech\|^2}.
\end{eqnarray*}
%[xxx work through the second in terms of $\vech\cdot\nabla f$.]
In particular,
\begin{eqnarray*}
	\frac{df}{dx_i} &\approx& \frac{f(\vecx+h\vece_i)-f(\vecx-h\vece_i)}{2h}\\
	\frac{d^2f}{dx_i^2} &\approx& \frac{f(\vecx+h\vece_i)-2f(\vecx)+f(\vecx-h\vece_i)}{h^2}.
\end{eqnarray*}

Now consider $\vecf:\R \to \R^n$.  One approximates as before:
\begin{eqnarray*}
	\vecf'(x)  &\approx& \frac{\vecf(x+h)-\vecf(x-h)}{2h} \\
	\vecf''(x) &\approx& \frac{\vecf(x+h)-2\vecf(x)+\vecf(x-h)}{h^2}.
\end{eqnarray*}

For numerical work, one often replaces the continuous domain with a
\bfidx{mesh}, i.e. an evenly spaced sequence of $x_k$'s.  For the case
$\vecf:\R \to \R^n$, the estimated first and second derivatives become:
\begin{eqnarray*}
	\vecf'(x_k)  &\approx& \frac{1}{2h} \left(\vecf(x_{k+1})-\vecf(x_{k-1})\right) \\
	\vecf''(x_k) &\approx& \frac{1}{h^2} \left(\vecf(x_{k+1})-2\vecf(x_k)+\vecf(x_{k-1})\right).
\end{eqnarray*}
We may formalize these operations in terms of vector spaces and operators by
thinking of the input sequence $\{\vecf(x_k)\}$ as being in, say,
$X=\ell^\infty(\R^n)$.  Then we have output sequences $\{\vecy(x_k) \approx
\vecf'(x_k)\}$ and $\{\vecz(x_k) \approx \vecf''(x_k)\}$ also in $X$.  We obtain
linear operators $L$ and $M$ on $X$ via
\begin{eqnarray*}
	\vecy(x) &=& L \vecf(x) \\
	\left[\begin{array}{l}
		\vdots \\
		\vecy(x_{k-1}) \\
		\vecy(x_{k})   \\
		\vecy(x_{k+1}) \\
		\vdots
	\end{array}\right]
	&=&
	\frac{1}{2h}
	\left[\begin{array}{rrrrr}
		\ddots &      \ddots &      0      &      0      &      0      \\
		-1     &      0      &      1      &      0      &      0      \\
		0      &      -1     &      0      &      1      &      0      \\
		0      &      0      &      -1     &      0      &      1      \\
		0      &      0      &      0      &      \ddots &   \ddots    \\
	\end{array}\right]
	\left[\begin{array}{l}
		\vdots \\
		\vecf(x_{k-1}) \\
		\vecf(x_{k})   \\
		\vecf(x_{k+1}) \\
		\vdots
	\end{array}\right]
\end{eqnarray*}
and
\begin{eqnarray*}
	\vecz(x) &=& M \vecf(x) \\
	\left[\begin{array}{l}
		\vdots \\
		\vecz(x_{k-1}) \\
		\vecz(x_{k})   \\
		\vecz(x_{k+1}) \\
		\vdots
	\end{array}\right]
	&=&
	\frac{1}{h^2}
	\left[\begin{array}{rrrrr}
		\ddots &      \ddots &      0      &      0      &      0      \\
		1      &     -2      &      1      &      0      &      0      \\
		0      &      1      &     -2      &      1      &      0      \\
		0      &      0      &      1      &     -2      &      1      \\
		0      &      0      &      0      &      \ddots &   \ddots    \\
	\end{array}\right]
	\left[\begin{array}{l}
		\vdots \\
		\vecf(x_{k-1}) \\
		\vecf(x_{k})   \\
		\vecf(x_{k+1}) \\
		\vdots
	\end{array}\right]
\end{eqnarray*}
The central point of these examples is that $L=d/dx$ and $M=d^2/dx^2$ were
skew-symmetric and symmetric (respectively) as continuous operators, while
their discretized versions are skew-symmetric and symmetric (respectively) as
matrices.

%[xxx \emph{not} symm/skew-symm unless boundary conditions.  Work \emph{that}
%into the example.]

Discrete domains and discrete difference operators are discussed in more detail
in section \ref{sec:discrete_vector_bundles}, with this section having provided
a little motivation.

%xxx
%%% ----------------------------------------------------------------
%\subsection{Nullspaces and boundary values}
%
%Say something about $P_\Omega L$ and its nullspace; also $P_\Omega L P_\Omega$;
%discrete version; $h \to 0$; delta-function-ish things; note that
%for $d/dx$ we do not have $P_\Omega L P_\Omega$ invertible; etc.

%% ================================================================
\newpage
\section{Discrete vector bundles}
\label{sec:discrete_vector_bundles}

%% ----------------------------------------------------------------
\subsection{Infinite bundles}

\begin{defn}
Let $V$ be a finite-dimensional complex vector space.
Consider the set of functions
$$
	\tilde{V}(\Z) = \{f:\Z \to V\}.
$$
Define an \plainidx{inner product} $\ssfdot$ on $\tilde{V}(\Z)$ using the
standard inner product on $V$:
$$
	\langle f, g \rangle =
	\sum_{k\in\Z} \langle f(k), g(k) \rangle.
$$
This induces a norm on $\tilde{V}(\Z)$ via
$$
	\|f\|^2 = \langle f, f \rangle.
$$
Define $V(\Z)$ to be the \bfidx{square-summable functions}:
$$
	V(\Z) = \{ f: \Z \to V:
	\langle f, f \rangle < \infty.
	\}
$$
That is, $V(\Z) = \ell^2(V)$.
We call this a \bfidx{discrete vector bundle}\index{vector bundle};
we say that $V$ is the \bfidx{model} for the bundle.
\end{defn}

%verify \bfidx{vector bundle} axioms (with discrete topo on $\Z$).

%define \bfidx{fiber}

\begin{rem}
We assume we have an orthogonal space and with a basis of $\vece_i$'s which are
orthgonal with respect to the form.
\end{rem}

\begin{defn}
\label{defn:psijk}
We define $\psi_j(k)$ to be $\vece_j$ with support only on the $k$th fiber.
\end{defn}

\begin{defn}
\label{defn:VZ_sbf}
Define a \bfidx{symmetric bilinear form} on $V(\Z)$:
$$
	(f,g) = \sum_{k \in \Z} (f(k),g(k))
$$
where on the right-hand side we use the distinguished bilinear form on $V$.
\end{defn}

%vol form on single $V$ and bundle.

%bases \ldots .

%Discuss the wedge from different fibers.

%%% ----------------------------------------------------------------
%\subsection{Exterior algebra on bundles}

\begin{defn}
\label{defn:omega_Omega}
We define
$$
	\omega_k = \psi_1(k) \wedge \cdots \wedge \psi_N(k).
$$
and
$$
	\omega_\Omega = \prod_{k \in \Omega} \omega_k.
$$
\end{defn}

%xxx say something (proposition?) about $\omega_k$'s commuting with one another
%as long as the model fiber is even-dimensional.

%% ----------------------------------------------------------------
\subsection{Truncated bundles}
\label{sec:truncated_bundles}

\begin{defn}
For a finite subset $U$ of $\Z$, define \plainidx{$V(U)$} to be the subspace of
$V(\Z)$ consisting of functions with support in $U$.
\end{defn}

\begin{rem}
Due to the finiteness of $U$, \emph{all} functions with support on $U$ are
square-summable.
\end{rem}

A few particular finite subsets of $\Z$ are of interest for this paper.
Let $m<n \in \Z$.  We define
\index{$\Omega$}
\index{$\Omegabar$}
\index{$\Omegaubar$}
\index{$P$}
\index{$Q$}
%\index{$\dOmega$}
\begin{itemize}
\item $\Omega = \{m, m+1, \ldots, n-1, n\}$.
\item $\Omegabar = \{m-1, m, m+1, \ldots, n-1, n, n+1\}$.
\item $\Omegaubar = \{m+1, \ldots, n-1\}$.
\item $\Omegabar\setminus\Omega = \{m-1, n+1\}$.
\item $\Omega\setminus\Omegaubar = \{m+1, n-1\}$.
\item $\dOmega = \{m-1, m, n, n+1\}$.
\item $P = V(\Omega\setminus\Omegaubar)$.
\item $Q = V(\Omegabar\setminus\Omega)$.
\item $V(\dOmega) = P \oplus Q$.
\end{itemize}

These may be visualized as follows:

%%  - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
\setlength{\unitlength}{1.0in}
\begin{picture}(6.5,2.0)
%\put(0,0){\framebox(6.5,2.0)}

\put(0.80,1.80){$m-2$}
\put(1.30,1.80){$m-1$}
\put(1.95,1.80){$m$}
\put(2.30,1.80){$m+1$}
%
\put(3.35,1.80){$n-1$}
\put(3.95,1.80){$n$}
\put(4.35,1.80){$n+1$}
\put(4.85,1.80){$n+2$}

\put(5.90,1.60){$V(\Z)$}
\put(5.65,1.58){$\cdots$}
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\put(2.00,1.60){\circle*{.05}}
\put(2.50,1.60){\circle*{.05}}
	\put(2.90,1.60){\circle*{.05}}
	\put(3.00,1.60){\circle*{.05}}
	\put(3.10,1.60){\circle*{.05}}
\put(3.50,1.60){\circle*{.05}}
\put(4.00,1.60){\circle*{.05}}
\put(4.50,1.60){\circle*{.05}}
\put(5.00,1.60){\circle*{.05}}

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\put(5.65,1.38){$\cdots$}
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\put(4.50,1.40){\circle*{.05}}
\put(5.00,1.40){\circle{.05}}

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\put(5.65,1.18){$\cdots$}
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\put(2.50,1.20){\circle*{.05}}
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	\put(3.10,1.20){\circle*{.05}}
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\put(4.00,1.20){\circle*{.05}}
\put(4.50,1.20){\circle{.05}}
\put(5.00,1.20){\circle{.05}}

\put(5.90,1.00){$V(\Omegaubar)$}
\put(5.65,0.98){$\cdots$}
\put(0.15,0.98){$\cdots$}
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\put(4.50,1.00){\circle{.05}}
\put(5.00,1.00){\circle{.05}}

\put(5.90,0.80){$V(\Omegadiff)$}
\put(5.65,0.78){$\cdots$}
\put(0.15,0.78){$\cdots$}
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\put(3.25,0.80){\vector(1,0){2.25}}
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\put(5.00,0.80){\circle{.05}}

\put(5.90,0.60){$V(\dOmega)$}
\put(5.65,0.58){$\cdots$}
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\put(3.25,0.60){\vector(1,0){2.25}}
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	\put(3.10,0.60){\circle{.05}}
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\put(4.50,0.60){\circle*{.05}}
\put(5.00,0.60){\circle{.05}}

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\put(5.65,0.38){$\cdots$}
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\put(3.25,0.40){\vector(1,0){2.25}}
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\end{picture}
%%  - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

%% ================================================================
\newpage
\section{Discrete difference operators}
\label{sec:discrete_difference_operators}

%% ----------------------------------------------------------------
\subsection{Difference operators}

Accompany the discrete vector bundle by \bfidx{connection matrices}
$A(k+\frac{1}{2})$ and $B(k)$.  The former are indexed by half-integers, which
is intended to be reminiscent of connecting bonds between nodes.

\begin{defn}
Let $L:V(\Z) \to V(\Z)$ be given by
$$
	(Lf)(k) = A(k+\hlf)f(k+1) + B(k)f(k) - A^\tau(k-\hlf)f(k-1).
$$
We take the transpose $A^\tau$ with respect to the distinguished bilinear form.
(As discussed in remark \ref{rem:matrix_transpose_sbf}, this is trivial to
compute (using the standard matrix transpose) as long as we use a basis which
is orthonormal with respect to the form.)  Also, we assume $B(k) = -B^\tau(k)$.
As noted in \cite{bib:palmer1}, this $L$ is intended to be the most general
range-one finite-difference operator.
\end{defn}

Applying the definition to a few fibers yields:
$$
\begin{array}{lllllllllllllllllllll}
	&\vdots& \\
	(Lf)(k-1) &=& A_{k-\smfrac{1}{2}}&f(k)   &+& B(k-1)&f(k-1) &-& A^\tau(k-\smfrac{3}{2})&f(k-2)\\
	(Lf)(k)   &=& A_{k+\smfrac{1}{2}}&f(k+1) &+& B(k)  &f(k)   &-& A^\tau(k-\smfrac{1}{2})&f(k-1)\\
	(Lf)(k+1) &=& A_{k+\smfrac{3}{2}}&f(k+2) &+& B(k+1)&f(k+1) &-& A^\tau(k+\smfrac{1}{2})&f(k)\\
	&\vdots&.
\end{array}
$$
Writing an ordered basis for $V(\Z)$ as
$$
	\{\ldots,
	\psi_1(k-1), \ldots, \psi_N(k-1), \quad
	\psi_1(k),   \ldots, \psi_N(k), \quad
	\psi_1(k+1), \ldots, \psi_N(k+1),
	\ldots\},
$$
we may write $L$ as an infinite-dimensional tri-block-diagonal matrix:
$$
\left[ \begin{array}{l}
	\vdots    \\
	(Lf)(k-1) \\
	(Lf)(k)   \\
	(Lf)(k+1) \\
	\vdots \\
\end{array} \right]
=
%
\left[ \begin{array}{lllll}
\ddots & \ddots           & \ddots              &            &        \\
\ddots & B_{k-1}          & A_{k-\smfrac{1}{2}} & 0          &        \\
\ddots & \At_{k-\hlf} & B_{k}               & A_{k+\hlf} & \ddots \\
       & 0                & \At_{k+\hlf}    & B_{k+1}    & \ddots \\
       &                  & \ddots              & \ddots     & \ddots \\
\end{array} \right]
%
\left[ \begin{array}{l}
	\vdots \\
	f(k-1) \\
	f(k)   \\
	f(k+1) \\
	\vdots \\
\end{array} \right].
$$

\begin{prop}
$L$ is formally skew-symmetric with respect to the distinguished bilinear form.
\begin{proof}
This is visually obvious.
\end{proof}
\end{prop}

%% ----------------------------------------------------------------
\subsection{Truncated difference operators}
\label{sec:P_L_orama}

\begin{defn}
For $U \subseteq \Z$, define \plainidx{$P_U$} to be the
\bfidxtwo{projection}{projection operator} of $V(\Z)$ onto $V(U)$.
(Note:  this extends definitions \ref{defns:Y_proj_and_e_sigma}.)
\end{defn}

We have the following matrices:
$$
P_\Omega =
\left[ \begin{array}{ll|llll|ll}
\cdots & 0      &1 &       &       &       &  \\
{} & {}     &  &1      &       &       &  \\
{} & {}     &  &       &\ddots &       &  \\
{} & {}     &  &       &       &       &  \\
{} & {}     &  &       &       &1      &0 &\cdots\\
\end{array} \right] \\
	\qand
P_\Omegadiff =
\left[ \begin{array}{l|llll|ll}
1      &0 &       &       &       &  &\textrm{(Output index $m-1$)}\\
\hline
{}     &  &       &       &0      &1 &\textrm{(Output index $n+1$)}\\
\end{array} \right] \\
$$

Note that $LP_U$ restricts input, making a vertical strip; $P_UL$ restricts
output, making a horizontal strip.  We have
$$
P_\Omega f =
\left[ \begin{array}{l}
	f(m)   \\
	f(m+1) \\
	\vdots \\
	f(n-1) \\
	f(n)   \\
\end{array} \right]
	\qqand
P_\Omegadiff f =
\left[ \begin{array}{l}
	f(m-1) \\
	f(n+1) \\
\end{array} \right].
$$
Likewise,
$$
P_\Omega Lf =
\left[ \begin{array}{l}
	(Lf)(m)   \\
	(Lf)(m+1) \\
	\vdots    \\
	(Lf)(n-1) \\
	(Lf)(n)   \\
\end{array} \right]
	\qqand
P_\Omegadiff Lf =
\left[ \begin{array}{l}
	(Lf)(m-1) \\
	(Lf)(n+1) \\
\end{array} \right].
$$
Thus we write $L$ as an infinite matrix\index{$L$}:
$$
L=
\left[ \begin{array}{ll|l|llll|l|ll}
{}              &               &(m-1)        &(m)          &\cdots       &\cdots       &(n)          &(n+1)      &\textrm{Input} &  \\
{}              &\ddots         &\ddots       &             &             &             &             &           &\textrm{index} &  \\
\hline
(m-1)           &\At_{m-\tlf}   &B_{m-1}      &A_{m-\hlf}   &             &             &             &           &               &  \\
\hline
(m)             &{}             &\At_{m-\hlf} &B_{m}        &A_{m+\hlf}   &             &             &           &               &  \\
(m+1)           &{}             &             &\At_{m+\hlf} &B_{m+1}      &A_{m+\tlf}   &             &           &               &  \\
\vdots          &{}             &             &             &\ddots       &\ddots       &\ddots       &           &               &  \\
(n-1)           &{}             &             &             &\At_{n-\tlf} &B_{n-1}      &A_{n-\hlf}   &           &               &  \\
(n)             &{}             &             &             &             &\At_{n-\hlf} &B_{n}        &A_{n+\hlf} &               &  \\
\hline
(n+1)           &{}             &             &             &             &             &\At_{n+\hlf} &B_{n+1}    &A_{n+\tlf}     &  \\
\hline
\textrm{Output} &\textrm{index} &             &             &             &             &             &\ddots     &\ddots         & \ddots \\
\end{array} \right] \\
$$
As well:
\index{$P_\Omega L$}
$$
P_\Omega L=
\left[ \begin{array}{l|llll|l}
0            &0            &             &             &           &  \\
\hline
\At_{m-\hlf} &B_{m}        &A_{m+\hlf}   &             &           &  \\
{}           &\At_{m+\hlf} &B_{m+1}      &A_{m+\tlf}   &           &  \\
{}           &             &\ddots       &\ddots       &\ddots     &  \\
{}           &             &\At_{n-\tlf} &B_{n-1}      &A_{n-\hlf} &  \\
{}           &             &             &\At_{n-\hlf} &B_{n}      & A_{n+\hlf} \\
\hline
{}           &             &             &             &0          &0 \\
\end{array} \right] \\
$$
and
\index{$L P_\Omega$}
$$
L P_\Omega =
\left[ \begin{array}{l|llll|l}
0            &A_{m-\hlf}   &             &             &           &  \\
\hline
0            &B_{m}        &A_{m+\hlf}   &             &           &  \\
{}           &\At_{m+\hlf} &B_{m+1}      &A_{m+\tlf}   &           &  \\
{}           &             &\ddots       &\ddots       &\ddots     &  \\
{}           &             &\At_{n-\tlf} &B_{n-1}      &A_{n-\hlf} &  \\
{}           &             &             &\At_{n-\hlf} &B_{n}      &0 \\
\hline
{}           &             &             &             &\At_{n+\hlf} &0 \\
\end{array} \right].
$$

Foreshadowing our attempt to do integrals which examine boundary values, we can
decompose the \bfidx{truncated operator} $L_\Omegabar$ into the sum of four
matrices:
\index{$P_\Omega LP_\Omega$}
$$
P_\Omega LP_\Omega=
\left[ \begin{array}{l|llll|l}
0      &0            &             &             &           &      \\
\hline
0      &B_{m}        &A_{m+\hlf}   &             &           &      \\
{}     &\At_{m+\hlf} &B_{m+1}      &A_{m+\tlf}   &           &      \\
{}     &             &\ddots       &\ddots       &\ddots     &      \\
{}     &             &\At_{n-\tlf} &B_{n-1}      &A_{n-\hlf} &      \\
{}     &             &             &\At_{n-\hlf} &B_{n}      &0     \\
\hline
{}     &             &             &             &0          &0     \\
\end{array} \right] \\
$$

\index{$P_\Omega LP_\Omegadiff$}
$$
P_\Omega LP_\Omegadiff =
\left[ \begin{array}{l|llll|l}
0             &0   &       &       &       &           \\
\hline
\At_{m-\hlf}  &0   &0      &       &       &           \\
{}            &0   &0      &0      &       &           \\
{}            &    &\ddots &\ddots &\ddots &           \\
{}            &    &0      &0      &0      &           \\
{}            &    &       &0      &0      &A_{n+\hlf} \\
\hline
{}            &    &       &       &0      &0          \\
\end{array} \right] \\
$$

\index{$P_\Omegadiff LP_\Omega$}
$$
P_\Omegadiff LP_\Omega=
\left[ \begin{array}{l|llll|l}
0        &A_{m-\hlf} &       &       &             &      \\
\hline
0       &0          &0      &       &             &      \\
{}      &0          &0      &0      &             &      \\
{}      &           &\ddots &\ddots &\ddots       &      \\
{}      &           &0      &0      &0            &      \\
{}      &           &       &0      &0            &0     \\
\hline
{}      &           &       &       &\At_{n+\hlf} &0     \\
\end{array} \right] \\
$$

\index{$P_\Omegadiff LP_\Omegadiff$}
$$
P_\Omegadiff LP_\Omegadiff=
\left[ \begin{array}{l|llll|l}
B_{m-1}      &0   &       &       &       &           \\
\hline
0             &0   &0      &       &       &           \\
{}            &0   &0      &0      &       &           \\
{}            &    &\ddots &\ddots &\ddots &           \\
{}            &    &0      &0      &0      &           \\
{}            &    &       &0      &0      &0          \\
\hline
{}            &    &       &       &0      &B_{n+1}    \\
\end{array} \right]
$$

\begin{defn}
\label{defn:L_Omegabar}
We can now define the matrix of the truncated skew-symmetric operator
$L_\Omegabar$ relative to the splitting $V(\Omegabar) = V(\Omega) \oplus
V(\Omegadiff)$:
\index{$L_\Omegabar$}
\begin{eqnarray*}
	L_\Omegabar &=& \begin{bmatrix}
		P_\Omega L P_\Omega     & P_\Omega L P_\Omegadiff \\
		P_\Omegadiff L P_\Omega & 0 \\
	\end{bmatrix} \\
	%
	&=&
\left[ \begin{array}{l|llll|l}
0      &A_{m-\hlf}   &       &       &             &      \\
\hline
\At_{m-\hlf} &B_{m}        &A_{m+\hlf}   &             &           &      \\
{}     &\At_{m+\hlf} &B_{m+1}      &A_{m+\tlf}   &           &      \\
{}     &             &\ddots       &\ddots       &\ddots     &      \\
{}     &             &\At_{n-\tlf} &B_{n-1}      &A_{n-\hlf} &      \\
{}     &             &             &\At_{n-\hlf} &B_{n}      &A_{n+\hlf} \\
\hline
{}      &           &       &       &\At_{n+\hlf} &0     \\
\end{array} \right]
\end{eqnarray*}
Note that we omit $P_\Omegadiff L P_\Omegadiff$.  This is intentional, and is
done solely to be able to make use, in section \ref{sec:vac_X}, of the Pfaffian
block formula which was presented in section \ref{prop:pfaffian_block_formula}.
\end{defn}

\begin{rem}
\label{rem:PWLPW_invertible}
It will be assumed below, but is not required by anything written so far, that
$P_\Omega L P_\Omega$ is invertible, but that $P_\Omega L$ has a nullspace.
\end{rem}

%xxx
%%% ----------------------------------------------------------------
%\subsection{$N(P_\Omega L)$}
%
%Mention the continuous problem --- 1st/2nd derivative operators and
%nullspace.  Then the discrete case.
%
%%% ----------------------------------------------------------------
%\subsection{$N(P_\Omega L)$ with constant matrices}

%Finite-difference problems \ldots.

%First consider the familiar 1D case, with $a,b,c$.  Then, multi-dimensional
%with $A,B,C$.  What about the \emph{dimension} of the solution space?!?
%
%Work out some examples e.g.
%$$
%	A=\left[\begin{array}{rr}a&0\\0&-a\end{array}\right] \qqand
%	B=\left[\begin{array}{rr}0&b\\-b&0\end{array}\right].
%$$
%
%Then quack a while about the inadequacy of this approach for higher dimensions;
%and then, of course, for non-constant matrices.

%% ----------------------------------------------------------------
\subsection{The action functional}
\label{sec:action_functional}

See \cite{bib:frankel} for more information about action functionals.

\begin{defn}
Given a skew-symmetric operator $L$ as above, we define
\index{$\mathcal{L}$}
$$
	\mcL = \frac{1}{2}
	\sum_{j=1}^N \sum_{k \in \Z} \left[L\psi_j(k)\right] \wedge \psi_j(k).
$$
\end{defn}

%\begin{rem}
%xxx mention the reason for $1/2$: the $j<k$ overcount.
%\end{rem}

\begin{rem}
This sum will not converge unless truncated.  This will be done below.
\end{rem}

\begin{prop}
$\mcL$ is formally independent of the choice of orthonormal basis for $V$.
\begin{proof}
TBD.
\end{proof}
\end{prop}

%Let $\Psi = \{\psi_1, \ldots, \psi_N\}$ and $\Phi= \{\phi_1, \ldots, \phi_N\}$
%be orthonormal bases.
%Let
%$$
%	\mcL = \sum_{j,k} \left[L\psi_j(k)\right] \wedge \psi_j(k)
%	\qqand
%	\mcM = \sum_{j,k} \left[L\phi_j(k)\right] \wedge \phi_j(k).
%$$
%It will be shown that $\mcL=\mcM$.  We may write
%$$
%	\psi_j(k) = Q \phi_j(k)
%$$
%where $Q$ is the change-of-basis matrix from $\Phi$ to $\Psi$.  Then
%$$
%	\mcL =
%		\sum_{j,k} \left[ A(k+\hlf)Q\phi_j(k+1) \right] \wedge Q\phi_j(k)
%		+
%		\sum_{j,k} \left[ B(k)Q\phi_j(k) \right] \wedge Q\phi_j(k)
%		-
%		\sum_{j,k} \left[ A^\tau(k-\hlf)Q\phi_j(k-1) \right] \wedge Q\phi_j(k)
%$$
%and
%$$
%	\mcM =
%		\sum_{j,k} \left[ A(k+\hlf)\phi_j(k+1) \right] \wedge \phi_j(k)
%		+
%		\sum_{j,k} \left[ B(k)\phi_j(k) \right] \wedge \phi_j(k)
%		-
%		\sum_{j,k} \left[ A^\tau(k-\hlf)\phi_j(k-1) \right] \wedge \phi_j(k).
%$$
%
%First I claim that for all vectors $\vecv(k)$,
%$$
%	\left[ B(k) Q\vecv(k) \right] \wedge Q\vecv(k) =
%	\left[ B(k) \vecv(k) \right] \wedge \vecv(k).
%$$
%By lemma \ref{lem:ortho_COB}, $Q$ is is unitary.  Since $B(k)$ is
%skew-symmetric, it commutes with $Q$ by proposition \ref{prop:BQ_QB} and so
%$$
%	\left[ B(k) Q\vecv(k) \right] \wedge Q\vecv(k) =
%	\left[ QB(k) \vecv(k) \right] \wedge Q\vecv(k);
%$$
%by proposition \ref{prop:Q_wedge},
%$$
%	\left[ QB(k) \vecv(k) \right] \wedge Q\vecv(k) =
%	\left[ B(k) \vecv(k) \right] \wedge \vecv(k).
%$$
%
%xxx now need to do the $A$ parts.

%%% ----------------------------------------------------------------
%\subsection{Truncated action}
%\label{sec:truncated_action}

\begin{defn}
Given $\mcL$ as above, we define the \bfidx{truncated action}
$$
	\mcL_\Omegabar = \frac{1}{2} \sum_{k \in \Omegabar} \sum_{j=1}^N
		[L_\Omegabar\psi_j(k)] \wedge \psi_j(k).
$$
\end{defn}

Note that $\mcL$ is the \bfidx{associated bivector} for the matrix $L$, in the
sense of definition \ref{defn:assocd_bivr}.

%$$
%\left[ \begin{array}{cc|cc|cc}
%\cdot&\cdot  &*&\cdot&  \cdot&\cdot \\
%\cdot&\cdot  &*&\cdot&  \cdot&\cdot \\
%\hline
%\cdot&\cdot  &*&\cdot&  \cdot&\cdot \\
%\cdot&\cdot  &*&\cdot&  \cdot&\cdot \\
%\hline
%\cdot&\cdot  &*&\cdot&  \cdot&\cdot \\
%\cdot&\cdot  &*&\cdot&  \cdot&\cdot \\
%\end{array} \right]
%\left[ \begin{array}{c}
%0 \\
%0 \\
%\hline
%1 \\
%0 \\
%\hline
%0 \\
%0 \\
%\end{array} \right]
%%
%=
%%
%\left[ \begin{array}{c}
%* \\
%* \\
%\hline
%* \\
%* \\
%\hline
%* \\
%* \\
%\end{array} \right]
%$$
%
%$$
%\left[ \begin{array}{cc|cc|cc}
%\ell_{11,11}&\ell_{11,12}  &\ell_{11,21}&\ell_{11,22}&  \ell_{11,31}&\ell_{11,32} \\
%\ell_{12,11}&\ell_{12,12}  &\ell_{12,21}&\ell_{12,22}&  \ell_{12,31}&\ell_{12,32} \\
%\hline
%\ell_{21,11}&\ell_{21,12}  &\ell_{21,21}&\ell_{21,22}&  \ell_{21,31}&\ell_{21,32} \\
%\ell_{22,11}&\ell_{22,12}  &\ell_{22,21}&\ell_{22,22}&  \ell_{22,31}&\ell_{22,32} \\
%\hline
%\ell_{31,11}&\ell_{31,12}  &\ell_{31,21}&\ell_{31,22}&  \ell_{31,31}&\ell_{31,32} \\
%\ell_{32,11}&\ell_{32,12}  &\ell_{32,21}&\ell_{32,22}&  \ell_{32,31}&\ell_{32,32} \\
%\end{array} \right]
%\left[ \begin{array}{c}
%0 \\
%0 \\
%\hline
%1 \\
%0 \\
%\hline
%0 \\
%0 \\
%\end{array} \right]
%%
%=
%%
%\left[ \begin{array}{c}
%\ell_{11,21} \\
%\ell_{12,21} \\
%\hline
%\ell_{21,21} \\
%\ell_{22,21} \\
%\hline
%\ell_{31,21} \\
%\ell_{32,21} \\
%\end{array} \right]
%$$

%So w/r/t this basis, $\mcL$ as a two-tensor \emph{IS} the same as $L_\Omega$!
%Same data in rows/columns.

%%% ----------------------------------------------------------------
%\subsection{Exponentiated action}
%\label{sec:exp_mcL}

\begin{defn}
Recalling section \ref{sec:ext_exp},

$$
	\exp(\mcL) = 1 + \mcL + \frac{\mcL\wedge\mcL}{2} + \ldots
		+ \frac{\mcL^{N/2}}{(N/2)!}
$$
%Recall from [xxx] that $\mcL$ is the \plainidx{associated bivector} for $L$ in
%the sense of definition \ref{defn:assocd_bivr}.
\end{defn}

As presented, computing this would be a mess of wedge products.
Pfaffians will be used below to faciliate the computation.

%xxx note this is a mess as-is and xref forward to pfaffian stuff.

%$$
%	\frac{1}{(N/2)!} \, b_A^{N/2}
%	= \Pf(A) \, \vece_1 \wedge \cdots \wedge \vece_N.
%$$

%% ================================================================
\newpage
\section{The discrete Berezin integral}
\label{sec:discrete_berezin}

%% ----------------------------------------------------------------
\subsection{The discrete Berezin integral and the vacuum vector}
\label{sec:dbivac}

The discrete Berezin integral on $\Lambda(V(\Omegabar))$ is defined by picking
out the coefficient of a top-level form in $\Lambda(V(\Omega))$.  This is not
an integral per se, but rather a functional.  Nonetheless, this boundary
behavior is vaguely reminiscent of the fundamental theorem of calculus.

\begin{defn}
Recall definition \ref{defn:omega_Omega} of $\omega_\Omega$.
Then we define
$$
	B_\Omega(\alpha \omega_\Omega + \textrm{lower-order terms}) = \alpha.
$$
and in particular for the action functional of section
\ref{sec:action_functional} we write
$$
	v_\Omega = \mcB_\Omega(\exp \mcL_\Omegabar).
$$
\end{defn}

\beginex
Let $N=2$ and
$$
	y =
	\psi_1(m-1)\wedge
	\Big[\psi_1(m)\wedge\psi_2(m)\Big]
	\wedge\cdots\wedge
	\Big[\psi_1(n)\wedge\psi_2(n)\Big]
	\wedge
	\psi_1(n+1)\wedge
	\psi_2(n+1).
$$
Then, since
$$
	\Big[\psi_1(m)\wedge\psi_2(m)\Big]
	\wedge\cdots\wedge
	\Big[\psi_1(n)\wedge\psi_2(n)\Big]
	=
	\omega_\Omega,
$$
we have
$$
	B_\Omega(y) =
	\psi_1(m-1)\wedge
	\psi_1(n+1)\wedge
	\psi_2(n+1).
$$
\mendex

It is clear that this discrete integral obeys the rules described in section
\ref{sec:cts_berezin_integral}.

%% ----------------------------------------------------------------
\subsection{Computation of the vacuum vector}
\label{sec:vac_X}

Pfaffians were developed in section \ref{sec:pfaffians} in order to facilitate
the following computation of the vacuum vector defined in section
\ref{sec:dbivac}.  First, some notation is presented.

In section \ref{sec:truncated_bundles} I presented a diagram of $V(\Omega)$
where each fiber was represented by a single bead.  Here I zoom in and think of
each fiber as an $N$-dimensional vector space.  I denote each one-dimensional subspace
of each fiber with a dot, gathering the one-dimensional subspaces of each fiber into
a box, and then gathering all of $V(\Omega)$ into a bigger box.  This leaves
$V(\Omegadiff)=Q$ on the left and right sides:

%%  - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
\setlength{\unitlength}{1.0in}
\begin{picture}(6.5,1.0)
%\put(0,0){\framebox(6.5,1.0)}

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\end{picture}
%%  - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Given this picture, the Berezin integrand is a form on $V(\Omegabar)$ which is
fully supported on all of $V(\Omega)$.  The result of the integral will be a
wedge product taken from among the $2N$ basis vectors for $Q$.  Since we are
integrating the exponential of a \emph{quadratic} form (the action functional
$\mcL$ as presented in section \ref{sec:action_functional}), the wedge products
in the result of the integral will consist only of even-degree terms.  Using
the work-avoidance principle which I exhorted in example \ref{ex:q_wedge_q}, I
would like to find the coefficients of each wedge term without needing to write
out high-dimensional arrays.

Here we can use proposition proposition \ref{prop:ext_exp_pfaffian}, as
illustrated by example \ref{ex:ext_exp_pfaffian}.  Since our integrand has full
support on $V(\Omega)$, the formula of proposition \ref{prop:ext_exp_pfaffian}
becomes
$$
	v_\Omega = \mcB_\Omega(\exp \mcL_\Omegabar) =
		\sum_\sigma \Pf
		\begin{bmatrix}
			P_\Omega L P_\Omega & P_\Omega L P_\sigma \\
			P_\sigma L P_\Omega & 0
		\end{bmatrix} e_\sigma
$$
where we let the $\sigma$'s range only over the $2N$ boundary values.

Now we apply the the Pfaffian block formula in proposition
\ref{prop:pfaffian_block_formula}: if $B$ is skew-symmetric and invertible,
then
$$
	\Pf\begin{bmatrix}
		B       & A \\
		-A^\tau & 0 \\
	\end{bmatrix}
	= \Pf(-B) \Pf(-A^\tau B^{-1} A).
$$

We defined $L_\Omegabar$, in definition \ref{defn:L_Omegabar}, to have zeroes
at the upper left and lower right corners.  We also assumed in remark
\ref{rem:PWLPW_invertible} that the matrix $P_\Omega L P_\Omega$ is invertible.
Both decisions were made precisely so that proposition
\ref{prop:pfaffian_block_formula} applies and we have, with $A$ being $P_\Omega
L P_\sigma$ and $B$ being $P_\Omega L P_\omega$:
\begin{eqnarray*}
	\Pf
	\begin{bmatrix}
		P_\Omega L P_\Omega & P_\Omega L P_\sigma \\
		P_\sigma L P_\Omega & 0
	\end{bmatrix}
	&=& \Pf(-P_\Omega L P_\Omega) \; \Pf(-(P_\Omega L P_\sigma)^\tau (P_\Omega L P_\Omega)^{-1} P_\Omega L P_\sigma) \\
	&=& \Pf(-P_\Omega L P_\Omega) \; \Pf(P_\sigma L P_\Omega (P_\Omega L P_\Omega)^{-1} P_\Omega L P_\sigma).
\end{eqnarray*}

Then, since clearly $P_\sigma P_\Omegadiff = P_\sigma$:
\begin{eqnarray*}
	v_\Omega &=& \mcB_\Omega(\exp \mcL_\Omegabar) =
		\sum_\sigma
		\Pf(P_\sigma L P_\Omega (P_\Omega L P_\Omega)^{-1} P_\Omega L P_\sigma)
		e_\sigma.
\end{eqnarray*}

\begin{defn}
\label{defn:X}
Define
$$
	X =
	P_\Omegadiff L P_\Omega (P_\Omega L P_\Omega)^{-1} P_\Omega L P_\Omegadiff.
$$
\end{defn}

\begin{prop}
$X$ is a skew-symmetric map on $Q$.
\begin{proof}
TBD.
\end{proof}
\end{prop}

Given this definition of $X$, we now have
\begin{eqnarray*}
	v_\Omega
	&=& \Pf(-P_\Omega L P_\Omega) \sum_\sigma \Pf(P_\sigma X P_\sigma) e_\sigma \\
	&=& \Pf(-P_\Omega L P_\Omega) \; \exp\left( \frac{1}{2} \sum [X \vece_j] \wedge \vece_j\right).
\end{eqnarray*}

%% ================================================================
\newpage
\section{The Fock characterization}
\label{sec:fock_chzn}

%xxx some pre-quack here.

%% ----------------------------------------------------------------
\subsection{The symmetric bilinear form}

We will construct a Clifford algebra (as presented in section \ref{sec:cliff})
using the symmetric bilinear form presented here.

%xxx recall (xref backward) motivation of the construction of $S$ via $d/dx$
%example \ldots remark
%\ref{rem:d_dx_S_forms}.

\begin{defn}
\label{defn:S}
Recall the definitions of $P_\Omega L$ and $L P_\Omega$ from section \ref{sec:P_L_orama}.  Then\index{$S$}
$$
S = P_\Omega L - L P_\Omega =
\left[ \begin{array}{l|llll|l}
0            &-A_{m-\hlf}  &             &             &           &  \\
\hline
\At_{m-\hlf}  &0   &0      &       &       &           \\
{}            &0   &0      &0      &       &           \\
{}            &    &\ddots &\ddots &\ddots &           \\
{}            &    &0      &0      &0      &           \\
{}            &    &       &0      &0      &A_{n+\hlf} \\
\hline
{}           &             &             &             &A^\tau_{n+\hlf} &0 \\
\end{array} \right].
$$
This form is designed to measure the lack of skew-symmetry of $P_\Omega L$, as
motivated in remark \ref{rem:d_dx_S_forms}.
\end{defn}

On $V(\dOmega)$, $S$ is:
$$
\left[ \begin{array}{llll}
	(Sf)(m-1) &\in& Q \\
	(Sf)(m)   &\in& P \\
	(Sf)(n)   &\in& P \\
	(Sf)(n+1) &\in& Q \\
\end{array} \right]
=
\left[ \begin{array}{llll}

0            &-A_{m-\hlf} &0               &0          \\
\At_{m-\hlf} &0           &0               &0          \\
0            &0           &0               &A_{n+\hlf} \\
0            &0           &A^\tau_{n+\hlf} &0          \\

\end{array} \right]
\left[ \begin{array}{llll}
	f(m-1) &\in& Q \\
	f(m)   &\in& P \\
	f(n)   &\in& P \\
	f(n+1) &\in& Q \\
\end{array} \right]
$$

\begin{prop}
$S$ as given in definition \ref{defn:S} is a symmetric linear transformation on
$V(\dOmega)$.
\begin{proof}
This is obvious from the picture.
\end{proof}
\end{prop}

\begin{prop}
\label{prop:S_PQ_QP}
$S: P \to Q$ and $S: Q \to P$.
\begin{proof}
This is obvious from the picture.
\end{proof}
\end{prop}

\begin{defn}
\label{defn:sbf_S}
Define a form on $P \oplus Q$ via
$$
	(f,g)_S = (Sf, g)
$$
using the bilinear form $\sbfdot$ on $V$ defined in definition
\ref{defn:VZ_sbf}.
\end{defn}

\begin{prop}
\label{prop:sbf_S}
The form in definition \ref{defn:sbf_S} is symmetric and bilinear.
\begin{proof}
Let $f,g \in P \oplus Q$.
For symmetry, observe that
\begin{eqnarray*}
	(f,g)_S &=& (Sf, g) \qquad\textrm{(definition of $S$)}\\
	&=& (g,Sf)  \qquad\textrm{(symmetry of $\sbfdot$)}\\
	&=& (Sg,f)  \qquad\textrm{(symmetry of $S$)}\\
	&=& (g,f)_S \qquad\textrm{(definition of $S$)}.
\end{eqnarray*}
For bilinearity, we have
\begin{eqnarray*}
	(f_1+f_2, g)_S &=& (S(f_1+f_2),g) \qquad\textrm{(definition of $S$)} \\
	&=& (S(f_1)+S(f_2),g) \qquad\textrm{(linearity of $S$)} \\
	&=& S(f_1),g + S(f_2),g \qquad\textrm{(bilinearity of $\sbfdot$)} \\
	&=& (f_1,g)_S + (f_2,g)_S \qquad\textrm{(definition of $S$)}.
\end{eqnarray*}
Right-linearity and left/right scalar multiplicativity are all but identical
and are omitted.
\end{proof}
\end{prop}

%xxx written out:
%$$
%\begin{array}{lllll}
%	(Sf,g)
%	&=& (-A_{m-\hlf}      &f(m),   &g(m-1)) \\
%	&+& (-A^\tau_{m-\hlf} &f(m-1), &g(m)) \\
%	&+& (A_{m-\hlf}       &f(n+1), &g(n)) \\
%	&+& (A^\tau_{m-\hlf}  &f(n),   &g(n+1)).
%\end{array}
%$$

\begin{rem}
\label{rem:S_tau}
Since $S$ is symmetric, i.e. $S = S^\tau$, we have
$$
	(Sf,g) = (f,S^\tau g) = (f,Sg).
$$
for all $f,g \in P \oplus Q$.
\end{rem}

%xxx written out:
%\begin{eqnarray*}
%&&
%\begin{array}{lllll}
%	(Sf,g)
%	&=& (-A_{m-\hlf}      &f(m),   &g(m-1)) \\
%	&+& (-A^\tau_{m-\hlf} &f(m-1), &g(m)) \\
%	&+& (A_{m-\hlf}       &f(n+1), &g(n)) \\
%	&+& (A^\tau_{m-\hlf}  &f(n),   &g(n+1)).
%\end{array} \\
%%%
%&&
%\begin{array}{lllll}
%	(f,Sg)
%	&=& (-A_{m-\hlf}      &f(m),   &g(m-1)) \\
%	&+& (-A^\tau_{m-\hlf} &f(m-1), &g(m)) \\
%	&+& (A_{m-\hlf}       &f(n+1), &g(n)) \\
%	&+& (A^\tau_{m-\hlf}  &f(n),   &g(n+1)).
%\end{array}
%\end{eqnarray*}

\begin{prop}
$S$ invertible implies $\sbfdot_S$ is non-degenerate.
\begin{proof}
TBD.
\end{proof}
\end{prop}

\begin{rem}
\label{rem:S_invble}
We assume (see \cite{bib:palmer1}) that $S$ is invertible, so that
$(\cdot,\cdot)_S$ is non-degenerate.
\end{rem}

%% ----------------------------------------------------------------
\subsection{The Fock representation}
\label{sec:bundle_fock_rep}

Recall from section \ref{sec:fock_rep} that we may form a Fock representation
of $V(\dOmega)=P\oplus Q$ if the vector space is even-dimensional with a
non-degenerate symmetric bilinear form and a polarization.  Since $P$ and $Q$
are of the same dimension, the first criterion is satisfied; the second
criterion is satisfied by $\sbfdot_S$ (proposition \ref{prop:sbf_S}).

\begin{prop}
The decomposition $V(\dOmega)=P\oplus Q$ is a polarization with respect
to $\sbfdot_S$.
\begin{proof}
$V(\dOmega)=P \oplus Q$ is already a direct-sum decomposition.  It remains to
show that $\sbfdot_S$ is zero on $P$ and $Q$.  Let $f_P \in P \oplus Q$ be
supported only in $P$.  Then
$$
	(f_P,f_P)_S = (Sf_P, f_P).
$$
Recall from proposition \ref{prop:S_PQ_QP} that $Sf_P \in Q$.  From definition
\ref{defn:VZ_sbf}, the form evaluated on an element supported solely in $Q$ and
an element supported solely in $P$ yields zero.  The proof for the case when
$f_Q$ is supported only in $Q$ is analogous.
\end{proof}
\end{prop}

\begin{defn}
\label{defn:Psi_PQ}
Let $f \in V(\dOmega) = P \oplus Q$.  Then there exist $f_P \in P$ and $f_Q \in
Q$ such that $f=f_P+f_Q$.
View $f_Q$ as an element of $Q^*$ via $f_Q \mapsto (\cdot,f_Q)_S$.
Define $\Psi: P \oplus Q \to \Hom(\Lambda(Q))$ by
$$
	\Psi(f) = \Psi(f_P + f_Q) = c(Sf_P) + a(f_Q).
$$
\end{defn}

\begin{rem}
Since $S:P \to Q$ and $S:Q \to P$ (proposition \ref{prop:S_PQ_QP}), we have
$$
	c(Sf_P) \in \Hom(\Lambda(Q))
$$
and
$$
	a(f_Q) \in \Hom(\Lambda(Q)).
$$
In section \ref{sec:fock_rep} we identified $v_- \in Y_-$ with an element of
$Y_+^*$, where here $P$ and $Q$ play the roles of $Y_-$ and $Y_+$,
respectively.  Here, though, we identify $f_Q$ with an element of $Q^*$ rather
than $P^*$.  This is because $(f_P,g)_S = (Sf_P,g)$ sends $f_P$ from $P$ to $Q$
before the bilinear form is applied.
\end{rem}

\begin{prop}
\label{prop:Psi_Cliff_PQ}
The map $\Psi$ of definition \ref{defn:Psi_PQ} satisfies the Clifford-algebra
relations of definition \ref{defn:cliff_alg}), i.e. for $f,g \in P\oplus Q$,
$$
	\Psi(f)\Psi(g) + \Psi(g)\Psi(f) = (f,g)_S.
$$
\begin{proof}
Let $f=f_P+f_Q$ and $g=g_P+g_Q$.  Then
\begin{eqnarray*}
	\Psi(f)\Psi(g) + \Psi(g)\Psi(f)
	&=& (c(Sf_P) + a(f_Q)) \circ (c(Sg_P) + a(g_Q)) \\
	&+& (c(Sg_P) + a(g_Q)) \circ (c(Sf_P) + a(f_Q)) \\
	&=& c(Sf_P) \circ c(Sg_P)
	+   c(Sf_P) \circ a( g_Q)
	+   a( f_Q) \circ c(Sg_P)
	+   a( f_Q) \circ a( g_Q) \\
	&+& c(Sg_P) \circ c(Sf_P)
	+   c(Sg_P) \circ a( f_Q)
	+   a( g_Q) \circ c(Sf_P)
	+   a( g_Q) \circ a( f_Q).
\end{eqnarray*}
Using the first two anticommutation relations from proposition
\ref{prop:creat_annih_cacr}, we have
\begin{eqnarray*}
	\Psi(f)\Psi(g) + \Psi(g)\Psi(f)
	&=& c(Sf_P) \circ a( g_Q)
	+   a( f_Q) \circ c(Sg_P) \\
	&+& c(Sg_P) \circ a( f_Q)
	+   a( g_Q) \circ c(Sf_P) \\
	&=& a( f_Q) \circ c(Sg_P)
	+   c(Sg_P) \circ a( f_Q) \\
	&+& a( g_Q) \circ c(Sf_P)
	+   c(Sf_P) \circ a( g_Q).
\end{eqnarray*}
Using the third anticommutation relation from proposition
\ref{prop:creat_annih_cacr}, we have
\begin{eqnarray*}
	\Psi(f)\Psi(g) + \Psi(g)\Psi(f) &=& (f_Q, Sg_P) + (g_Q, Sf_P).
\end{eqnarray*}

On the other hand,
\begin{eqnarray*}
	(f,g)_S &=& (f_P+f_Q, g_P+g_Q)_S
		\qquad\textrm{(given)} \\
	&=& (f_P, g_P)_S + (f_P, g_Q)_S + (f_Q, g_P)_S + (f_Q, g_Q)_S
		\qquad\textrm{(bilinearity of $S$)} \\
	&=& (Sf_P, g_P) + (Sf_P, g_Q) + (Sf_Q, g_P) + (Sf_Q, g_Q)
		\qquad\textrm{(definition of $S$)} \\
	&=& (Sf_P, g_Q) + (Sf_Q, g_P)
		\qquad\textrm{(polarization of $P \oplus Q$)} \\
	&=& (f_P, Sg_Q) + (f_Q, Sg_P)
		\qquad\textrm{(symmetry of $S$)} \\
	&=& \Psi(f)\Psi(g) + \Psi(g)\Psi(f)
\end{eqnarray*}
which is what was to be shown.
\end{proof}
\end{prop}

Thus, as in section \ref{sec:fock_rep},
$$
	\Psi:\Cliff(P \oplus Q) \to \Lambda(Q)
$$
is a \bfidx{Fock representation} associated with the symmetric form
$(\cdot,\cdot)_S$.

%% ----------------------------------------------------------------
\subsection{Vacuum characterization}
\label{sec:vac_chzn}

\begin{thm}
\label{thm:vac_chzn}
The vacuum vector
$$
	v_\Omega = \mcB_\Omega(e^{\mcL_\Omegabar})
$$
is projectively characterized (i.e. up to a constant multiplier)
as the unique vector in $\Lambda(Q)$ that is annihilated by all
elements $\Psi(f)$ where $f$ is the boundary value in $P\oplus Q$
of a function $F \in V(\Omegabar)$ in the nullspace of $P_\Omega L$.
That is,
$$
	\Psi(f) v_\Omega = 0
$$
whenever $f = F|_{\dOmega}$ and
$$
	P_\Omega L F = 0.
$$
\begin{proof}
(Note: The presentation here will complete the sketch given in \cite{bib:palmer1}.)

\textbf{Step 1.}
Recall from section \ref{sec:vac_X} that the vacuum vector we are trying to
characterize is
$$
	v_\Omega =
	\exp\left( \frac{1}{2} \sum_{j=1}^{2N} X\vece_j \wedge \vece_j \right).
$$
Using the bra-ket notation from section \ref{sec:bra_ket}, and in particular
using remark \ref{rem:ca_vs_wedges}, this may be written as
$$
	v_\Omega
	= \exp\left(\frac{1}{2}\sum_{j=1}^{2N} c(X\vece_j)c(\vece_j) \zvac \right)
	= \exp\left(\frac{1}{2}\sum_{j=1}^{2N} c(X\vece_j)c(\vece_j) \right) \zvac
$$
%[xxx justify moving the zvac]
where the exterior arithmetic is done in $\Lambda(Q)$.

\textbf{Step 2.}
Now suppose $f \in Q$.  Use [do that!] proposition
\ref{prop:exp_ad_prop} to show that
$$
	\exp\left( \frac{1}{2} \sum_{j=1}^{2N} c(X\vece_j)c(\vece_j) \right)
	a(f)
	\exp\left(-\frac{1}{2} \sum_{j=1}^{2N} c(X\vece_j)c(\vece_j) \right)
	= c(Xf) + a(f).
$$
\textbf{Step 3.}
From this it follows [show that!] that
\begin{equation}
\label{eqn:cXf_afv}
	(c(Xf)+a(f))v_\Omega =
	\exp\left(\frac{1}{2} \sum_{j=1}^{2N} c(X\vece_j)c(\vece_j)\right)
	a(f) \zvac = 0
\end{equation}
for $f \in Q$, since $a(f)\zvac = 0$.

\textbf{Step 4.}
Now recall the representation of the Clifford algebra for $(V_\dOmega,S)$ is
given by
$$
	\Psi(f) = c(S f_P) + a(f_Q).
$$
Suppose that $f \in V_\dOmega$ is the boundary value of a function $F \in
V(\Omegabar)$ (i.e. $f=F|_{\dOmega}$) with
\begin{equation}
\label{eqn:PLf=0}
	P_\Omega L f = 0.
\end{equation}
Write $N_\dOmega$ for the subspace of $V_\dOmega$ which consists of boundary
values of functions $F \in V(\Omegabar)$ which satisfy equation
\ref{eqn:PLf=0}.  Now suppose that $F$ satisfies \ref{eqn:PLf=0} and write
$$
	f = F|_\dOmega = f_P + f_Q
$$
and
$$
	F = F_\Omega + F_Q.
$$
The equation which $F$ must satisfy becomes
$$
	P_\Omega L P_\Omega F_\Omega + P_\Omega L f_Q = 0,
$$
from which it follows that
\begin{equation}
\label{eqn:FW_fQ}
	F_\Omega = -(P_\Omega L P_\Omega)^{-1} P_\Omega L f_Q.
\end{equation}

\textbf{Step 5.}
Recall from remark \ref{rem:PWLPW_invertible} that we suppose $P_\Omega L
P_\Omega$ is invertible.  Now apply $P_\Omegadiff S$ to both sides of this last
equality and recall the definitions of $S$ and $X$ to find
$$
	S f_P = X f_Q.
$$
Thus we see that for $f \in N_\dOmega$ we have
$$
	\Psi(f) = c(Sf_P) + a(f_Q) = c(X f_Q) + a(f_Q).
$$
From equation \ref{eqn:cXf_afv} we see that
\begin{equation}
\label{eqn:Psi_f_vW}
	\Psi(f) v_\Omega = 0
\end{equation}
for $f \in N_\dOmega$.

\textbf{Step 6.}
It remains to check that $v_\dOmega$ is projectively characterized by this
condition.  This turns out to be equivalent to showing that $N_\dOmega$ is a
maximal isotropic subspace of $(V_\dOmega, S)$.  To see this, suppose that $f,g
\in N_\dOmega$ are boundary values of $F,G$ satisfying equation
\ref{eqn:PLf=0}.  Then
$$
	S(f,g) = (P_\Omega L F, G) + (F, P_\Omega L G) = 0.
$$
This shows that $N_\dOmega$ is isotropic.  Because $S$ interchanges $Q$ and
$P$, and because the subspaces are orthogonal to one another with respect to
the symmetric form $(\cdot,\cdot)$, it follows that $Q$ and $P$ are
complementary istropic subspaces for the symmetric form $(\cdot,\cdot)_S$.
Since $S$ is supposed non-degenerate, this means that both $Q$ and $P$ are
maximal isotropic subspaces.  However, equation \ref{eqn:FW_fQ} shows that
$N_\dOmega$ is a graph over $Q$.  Thus it has the same dimension as $Q$ and so
must be a maximal isotropic subspace.

\textbf{Step 7.}
To finish the proof that equation \ref{eqn:Psi_f_vW} projectively characterizes
$v_\Omega$, first show that the vector
$$
	\zvac = 1 \oplus 0 \oplus \cdot \oplus 0 \in \Lambda(V_+)
$$
is projectively characterized by the condition
$$
	\Psi(f)\zvac = 0
$$
for $f \in V_-$, in the Fock representation of the Clifford relations $\Psi$
associated with the polarization $V_+ \oplus V_-$.  Then use the equivalence of
representations.
\end{proof}
\end{thm}

%% ================================================================
\newpage
\section{Example of vacuum vector}
\label{sec:mma_rox}

To help fix ideas, we may explicitly compute some of the quantities described
in this paper.
Take a simple case of $N=2$ and constant coefficients:
$$
	A = \left[\begin{array}{rr} a & 0 \\ 0 & -a \end{array}\right], \quad
	B = \left[\begin{array}{rr} 0 & b \\ -b & 0 \end{array}\right], \quad
	m=1, n=4.
$$
Then recall from section \ref{sec:P_L_orama} that we have
$$
L_\Omegabar =
\left[\begin{array}{rr|rr|rr|rr|rr|rr}
	 0 & 0 &  a &  0 &  0 &  0 &  0 &  0 &  0 &  0 & 0 &  0 \\
	 0 & 0 &  0 & -a &  0 &  0 &  0 &  0 &  0 &  0 & 0 &  0 \\
	 \hline
	-a & 0 &  0 &  b &  a &  0 &  0 &  0 &  0 &  0 & 0 &  0 \\
	 0 & a & -b &  0 &  0 & -a &  0 &  0 &  0 &  0 & 0 &  0 \\
	 \hline
	 0 & 0 & -a &  0 &  0 &  b &  a &  0 &  0 &  0 & 0 &  0 \\
	 0 & 0 &  0 &  a & -b &  0 &  0 & -a &  0 &  0 & 0 &  0 \\
	 \hline
	 0 & 0 &  0 &  0 & -a &  0 &  0 &  b &  a &  0 & 0 &  0 \\
	 0 & 0 &  0 &  0 &  0 &  a & -b &  0 &  0 & -a & 0 &  0 \\
	 \hline
	 0 & 0 &  0 &  0 &  0 &  0 & -a &  0 &  0 &  b & a &  0 \\
	 0 & 0 &  0 &  0 &  0 &  0 &  0 &  a & -b &  0 & 0 & -a \\
	 \hline
	 0 & 0 &  0 &  0 &  0 &  0 &  0 &  0 & -a &  0 & 0 &  0 \\
	 0 & 0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  a & 0 &  0 \\
\end{array}\right]
$$
and
%By [xxx xref here], $\mcL_\Omega$ is a 2D-array with the very same components.
$$
P_\Omega L P_\Omega =
\left[\begin{array}{rr|rr|rr|rr}
	 0 &  b &  a &  0 &  0 &  0 &  0 &  0 \\
	-b &  0 &  0 & -a &  0 &  0 &  0 &  0 \\
	 \hline
	-a &  0 &  0 &  b &  a &  0 &  0 &  0 \\
	 0 &  a & -b &  0 &  0 & -a &  0 &  0 \\
	 \hline
	 0 &  0 & -a &  0 &  0 &  b &  a &  0 \\
	 0 &  0 &  0 &  a & -b &  0 &  0 & -a \\
	 \hline
	 0 &  0 &  0 &  0 & -a &  0 &  0 &  b \\
	 0 &  0 &  0 &  0 &  0 &  a & -b &  0 \\
\end{array}\right].
$$
One computes
$$
\Pf(-P_\Omega L P_\Omega) = a^4 + 3 a^2b^2 + b^4.
$$

One may compute $(P_\Omega L P_\Omega)^{-1}$ using a tool such as Mathematica,
but it is too messy to fit on this page.  This illustrates the need for further
cleverness --- we don't need all of this matrix, but rather only its boundary
components in $X$.

%$$
%(P_\Omega L P_\Omega)^{-1} =
%\left(a^4+3 b^2 a^2+b^4\right)^2
%\left[\begin{array}{rr|rr|rr|rr}
%    0 & -b^7-5 a^2 b^5-7 a^4 b^3-2 a^6 b & -a^7-4 b^2 a^5-4 b^4 a^3-b^6 a & 0
%      & 0 & -b a^6-3 b^3 a^4-b^5 a^2 & -a^7-3 b^2 a^5-b^4 a^3 & 0 \\
%    b^7+5 a^2 b^5+7 a^4 b^3+2 a^6 b & 0 & 0 & a^7+4 b^2 a^5+4 b^4 a^3+b^6 a &
%      b a^6+3 b^3 a^4+b^5 a^2 & 0 & 0 & a^7+3 b^2 a^5+b^4 a^3 \\
%	\hline
%    a^7+4 b^2 a^5+4 b^4 a^3+b^6 a & 0 & 0 & -b^7-4 a^2 b^5-4 a^4 b^3-a^6 b &
%      -a b^6-3 a^3 b^4-a^5 b^2 & 0 & 0 & -b a^6-3 b^3 a^4-b^5 a^2 \\
%    0 & -a^7-4 b^2 a^5-4 b^4 a^3-b^6 a & b^7+4 a^2 b^5+4 a^4 b^3+a^6 b & 0 & 0
%      & a b^6+3 a^3 b^4+a^5 b^2 & b a^6+3 b^3 a^4+b^5 a^2 & 0 \\
%	\hline
%    0 & -b a^6-3 b^3 a^4-b^5 a^2 & a b^6+3 a^3 b^4+a^5 b^2 & 0 & 0 & -b^7-4
%      a^2 b^5-4 a^4 b^3-a^6 b & -a^7-4 b^2 a^5-4 b^4 a^3-b^6 a & 0 \\
%    b a^6+3 b^3 a^4+b^5 a^2 & 0 & 0 & -a b^6-3 a^3 b^4-a^5 b^2 & b^7+4 a^2
%      b^5+4 a^4 b^3+a^6 b & 0 & 0 & a^7+4 b^2 a^5+4 b^4 a^3+b^6 a \\
%	\hline
%    a^7+3 b^2 a^5+b^4 a^3 & 0 & 0 & -b a^6-3 b^3 a^4-b^5 a^2 & a^7+4 b^2 a^5+4
%      b^4 a^3+b^6 a & 0 & 0 & -b^7-5 a^2 b^5-7 a^4 b^3-2 a^6 b \\
%    0 & -a^7-3 b^2 a^5-b^4 a^3 & b a^6+3 b^3 a^4+b^5 a^2 & 0 & 0 & -a^7-4 b^2
%      a^5-4 b^4 a^3-b^6 a & b^7+5 a^2 b^5+7 a^4 b^3+2 a^6 b & 0
%\end{array}\right]
%$$

%$P_\sigma X P_\sigma$ is just a row/column selector; xref.

%PW_L_PWdiff:
% -a   0      0   0
%  0   a      0   0
%
%  0   0      0   0
%  0   0      0   0
%
%  0   0      0   0
%  0   0      0   0
%
%  0   0      a   0
%  0   0      0  -a

%PWdiff_L_PW:
%  a   0      0   0      0   0      0   0
%  0  -a      0   0      0   0      0   0
%
%  0   0      0   0      0   0     -a   0
%  0   0      0   0      0   0      0   a

%PW_L_PW_inv:
% -0.000   0.455      1.091  -0.818     -0.364   0.455      1.091  -1.000
% -0.455   0.000     -0.545   0.273     -0.000  -0.091     -0.364   0.273
%
% -1.091   0.545      0.000   0.273      0.364  -0.273     -0.364   0.455
%  0.818  -0.273     -0.273   0.000     -0.182   0.091     -0.000  -0.091
%
%  0.364  -0.000     -0.364   0.182      0.000   0.273      1.091  -0.818
% -0.455   0.091      0.273  -0.091     -0.273   0.000     -0.545   0.273
%
% -1.091   0.364      0.364  -0.000     -1.091   0.545      0.000   0.455
%  1.000  -0.273     -0.455   0.091      0.818  -0.273     -0.455   0.000

%X = PWdiff_L_PW * PW_L_PW_inv * PW_L_PWdiff:
%  0.000   0.909     -1.000  -1.091
% -0.909   0.000     -3.909  -4.000
%
%  1.000   3.909     -0.000   0.909
%  1.091   4.000     -0.909  -0.000

%det(A)       = -2.000000e+00
%det(B)       = 1.000000e+00
%det(S)       = 1.600000e+01
%det(PW_L_PW) = 1.210000e+02
%det(X)       = 1.190083e+00
%
%A            is neither symmetric nor skew-symmetric
%B            is skew-symmetric
%S            is symmetric
%X            is skew-symmetric
%PW_L_PW      is skew-symmetric
%PW_L_PW_inv  is skew-symmetric

Recall that
$$
P_\Omegadiff L P_\Omega =
\left[
\begin{array}{rr|rr|rr|rr}
 a  &  0  &  0  &  0  &  0  &  0  &  0  &  0 \\
 0  & -a  &  0  &  0  &  0  &  0  &  0  &  0 \\
 \hline
 0  &  0  &  0  &  0  &  0  &  0  & -a  &  0 \\
 0  &  0  &  0  &  0  &  0  &  0  &  0  &  a
\end{array}
\right]
$$
and
%$$
%PWLPWinv =
%   \left[
%   \begin{array}{llllllll}
%    0 & \frac{-b^7-5 a^2 b^5-7 a^4 b^3-2 a^6 b}{a^8+6 b^2 a^6+11 b^4 a^4+6 b^6
%      a^2+b^8} & \frac{-a^7-4 b^2 a^5-4 b^4 a^3-b^6 a}{a^8+6 b^2 a^6+11 b^4
%      a^4+6 b^6 a^2+b^8} & 0 & 0 & \frac{-b a^6-3 b^3 a^4-b^5 a^2}{a^8+6 b^2
%      a^6+11 b^4 a^4+6 b^6 a^2+b^8} & \frac{-a^7-3 b^2 a^5-b^4 a^3}{a^8+6 b^2
%      a^6+11 b^4 a^4+6 b^6 a^2+b^8} & 0 \\
%    \frac{b^7+5 a^2 b^5+7 a^4 b^3+2 a^6 b}{a^8+6 b^2 a^6+11 b^4 a^4+6 b^6
%      a^2+b^8} & 0 & 0 & \frac{a^7+4 b^2 a^5+4 b^4 a^3+b^6 a}{a^8+6 b^2 a^6+11
%      b^4 a^4+6 b^6 a^2+b^8} & \frac{b a^6+3 b^3 a^4+b^5 a^2}{a^8+6 b^2 a^6+11
%      b^4 a^4+6 b^6 a^2+b^8} & 0 & 0 & \frac{a^7+3 b^2 a^5+b^4 a^3}{a^8+6 b^2
%      a^6+11 b^4 a^4+6 b^6 a^2+b^8} \\
%    \frac{a^7+4 b^2 a^5+4 b^4 a^3+b^6 a}{a^8+6 b^2 a^6+11 b^4 a^4+6 b^6
%      a^2+b^8} & 0 & 0 & \frac{-b^7-4 a^2 b^5-4 a^4 b^3-a^6 b}{a^8+6 b^2
%      a^6+11 b^4 a^4+6 b^6 a^2+b^8} & \frac{-a b^6-3 a^3 b^4-a^5 b^2}{a^8+6
%      b^2 a^6+11 b^4 a^4+6 b^6 a^2+b^8} & 0 & 0 & \frac{-b a^6-3 b^3 a^4-b^5
%      a^2}{a^8+6 b^2 a^6+11 b^4 a^4+6 b^6 a^2+b^8} \\
%    0 & \frac{-a^7-4 b^2 a^5-4 b^4 a^3-b^6 a}{a^8+6 b^2 a^6+11 b^4 a^4+6 b^6
%      a^2+b^8} & \frac{b^7+4 a^2 b^5+4 a^4 b^3+a^6 b}{a^8+6 b^2 a^6+11 b^4
%      a^4+6 b^6 a^2+b^8} & 0 & 0 & \frac{a b^6+3 a^3 b^4+a^5 b^2}{a^8+6 b^2
%      a^6+11 b^4 a^4+6 b^6 a^2+b^8} & \frac{b a^6+3 b^3 a^4+b^5 a^2}{a^8+6 b^2
%      a^6+11 b^4 a^4+6 b^6 a^2+b^8} & 0 \\
%    0 & \frac{-b a^6-3 b^3 a^4-b^5 a^2}{a^8+6 b^2 a^6+11 b^4 a^4+6 b^6
%      a^2+b^8} & \frac{a b^6+3 a^3 b^4+a^5 b^2}{a^8+6 b^2 a^6+11 b^4 a^4+6 b^6
%      a^2+b^8} & 0 & 0 & \frac{-b^7-4 a^2 b^5-4 a^4 b^3-a^6 b}{a^8+6 b^2
%      a^6+11 b^4 a^4+6 b^6 a^2+b^8} & \frac{-a^7-4 b^2 a^5-4 b^4 a^3-b^6
%      a}{a^8+6 b^2 a^6+11 b^4 a^4+6 b^6 a^2+b^8} & 0 \\
%    \frac{b a^6+3 b^3 a^4+b^5 a^2}{a^8+6 b^2 a^6+11 b^4 a^4+6 b^6 a^2+b^8} & 0
%      & 0 & \frac{-a b^6-3 a^3 b^4-a^5 b^2}{a^8+6 b^2 a^6+11 b^4 a^4+6 b^6
%      a^2+b^8} & \frac{b^7+4 a^2 b^5+4 a^4 b^3+a^6 b}{a^8+6 b^2 a^6+11 b^4
%      a^4+6 b^6 a^2+b^8} & 0 & 0 & \frac{a^7+4 b^2 a^5+4 b^4 a^3+b^6 a}{a^8+6
%      b^2 a^6+11 b^4 a^4+6 b^6 a^2+b^8} \\
%    \frac{a^7+3 b^2 a^5+b^4 a^3}{a^8+6 b^2 a^6+11 b^4 a^4+6 b^6 a^2+b^8} & 0 &
%      0 & \frac{-b a^6-3 b^3 a^4-b^5 a^2}{a^8+6 b^2 a^6+11 b^4 a^4+6 b^6
%      a^2+b^8} & \frac{a^7+4 b^2 a^5+4 b^4 a^3+b^6 a}{a^8+6 b^2 a^6+11 b^4
%      a^4+6 b^6 a^2+b^8} & 0 & 0 & \frac{-b^7-5 a^2 b^5-7 a^4 b^3-2 a^6
%      b}{a^8+6 b^2 a^6+11 b^4 a^4+6 b^6 a^2+b^8} \\
%    0 & \frac{-a^7-3 b^2 a^5-b^4 a^3}{a^8+6 b^2 a^6+11 b^4 a^4+6 b^6 a^2+b^8}
%      & \frac{b a^6+3 b^3 a^4+b^5 a^2}{a^8+6 b^2 a^6+11 b^4 a^4+6 b^6 a^2+b^8}
%      & 0 & 0 & \frac{-a^7-4 b^2 a^5-4 b^4 a^3-b^6 a}{a^8+6 b^2 a^6+11 b^4
%      a^4+6 b^6 a^2+b^8} & \frac{b^7+5 a^2 b^5+7 a^4 b^3+2 a^6 b}{a^8+6 b^2
%      a^6+11 b^4 a^4+6 b^6 a^2+b^8} & 0
%   \end{array}
%   \right]
%$$
$$
P_\Omega L P_\Omegadiff =
\left[
\begin{array}{rr|rr}
-a  &  0  &  0  &  0 \\
 0  &  a  &  0  &  0 \\
 \hline
 0  &  0  &  0  &  0 \\
 0  &  0  &  0  &  0 \\
 \hline
 0  &  0  &  0  &  0 \\
 0  &  0  &  0  &  0 \\
 \hline
 0  &  0  &  a  &  0 \\
 0  &  0  &  0  & -a
\end{array}
\right].
$$
Then, using definition \ref{defn:X} of $X$ and the omitted
messy $(P_\Omega L P_\Omega)^{-1}$, we obtain
$$
X =
   \left[
   \begin{array}{ll|ll}
    0 & -\frac{a^2 \left(b^3+2 a^2 b\right)}{a^4+3 b^2 a^2+b^4} &
      -\frac{a^5}{a^4+3 b^2 a^2+b^4} & 0 \\
    \frac{a^2 \left(b^3+2 a^2 b\right)}{a^4+3 b^2 a^2+b^4} & 0 & 0 &
      \frac{a^5}{a^4+3 b^2 a^2+b^4} \\
	&&&\\
\hline
	&&&\\
    \frac{a^5}{a^4+3 b^2 a^2+b^4} & 0 & 0 & -\frac{a^2 \left(b^3+2 a^2
      b\right)}{a^4+3 b^2 a^2+b^4} \\
    0 & -\frac{a^5}{a^4+3 b^2 a^2+b^4} & \frac{a^2 \left(b^3+2 a^2
      b\right)}{a^4+3 b^2 a^2+b^4} & 0
   \end{array}
   \right].
$$

Since $N=2$, $\sigma$'s include the $1,2$ slots of node $m=1$ and the $1,2$
slots of node $n=4$.  Here are all 16 $\sigma$'s, represented schematically
with indicator functions:
$$
\begin{tabular}{|l|l|l|l|}
\hline $\vece_{00,00}$ & $\vece_{00,01}$ & $\vece_{00,10}$ & $\vece_{00,11}$ \\
\hline $\vece_{01,00}$ & $\vece_{01,01}$ & $\vece_{01,10}$ & $\vece_{01,11}$ \\
\hline $\vece_{10,00}$ & $\vece_{10,01}$ & $\vece_{10,10}$ & $\vece_{10,11}$ \\
\hline $\vece_{11,00}$ & $\vece_{11,01}$ & $\vece_{11,10}$ & $\vece_{11,11}$ \\
\hline
\end{tabular}
$$

Only even $\sigma$'s contribute to the Pfaffian.  Here are all the even
$\sigma$'s and their corresponding wedges:
$$
\begin{tabular}{|l|l|l|l|}
\hline
	$\vece_{00,00} = 1$ &
	$\vece_{00,11} = \psi_1(5) \wedge \psi_2(5)$ \\
\hline
	$\vece_{01,01} = \psi_2(0) \wedge \psi_2(5)$ &
	$\vece_{01,10} = \psi_2(0) \wedge \psi_1(5)$ \\
\hline
	$\vece_{10,01} = \psi_1(0) \wedge \psi_2(5)$ &
	$\vece_{10,10} = \psi_1(0) \wedge \psi_1(5)$ \\
\hline
	$\vece_{11,00} = \psi_1(0) \wedge \psi_2(0)$ &
	$\vece_{11,11} =
		\psi_1(0) \wedge \psi_2(0) \wedge \psi_1(5) \wedge \psi_2(5)$ \\
\hline
\end{tabular}
$$

%$$
%(P_\Omega L P_\Omega)^{-1} =
%   \left[
%   \begin{array}{llllllll}
%    0 & -\frac{b^3+2 a^2 b}{a^4+3 b^2 a^2+b^4} & -\frac{a
%      \left(a^2+b^2\right)}{a^4+3 b^2 a^2+b^4} & 0 & 0 & -\frac{a^2 b}{a^4+3
%      b^2 a^2+b^4} & -\frac{a^3}{a^4+3 b^2 a^2+b^4} & 0 \\
%    \frac{b^3+2 a^2 b}{a^4+3 b^2 a^2+b^4} & 0 & 0 & \frac{a
%      \left(a^2+b^2\right)}{a^4+3 b^2 a^2+b^4} & \frac{a^2 b}{a^4+3 b^2
%      a^2+b^4} & 0 & 0 & \frac{a^3}{a^4+3 b^2 a^2+b^4} \\
%    \frac{a \left(a^2+b^2\right)}{a^4+3 b^2 a^2+b^4} & 0 & 0 & -\frac{b
%      \left(a^2+b^2\right)}{a^4+3 b^2 a^2+b^4} & -\frac{a b^2}{a^4+3 b^2
%      a^2+b^4} & 0 & 0 & -\frac{a^2 b}{a^4+3 b^2 a^2+b^4} \\
%    0 & -\frac{a \left(a^2+b^2\right)}{a^4+3 b^2 a^2+b^4} & \frac{b
%      \left(a^2+b^2\right)}{a^4+3 b^2 a^2+b^4} & 0 & 0 & \frac{a b^2}{a^4+3
%      b^2 a^2+b^4} & \frac{a^2 b}{a^4+3 b^2 a^2+b^4} & 0 \\
%    0 & -\frac{a^2 b}{a^4+3 b^2 a^2+b^4} & \frac{a b^2}{a^4+3 b^2 a^2+b^4} & 0
%      & 0 & -\frac{b \left(a^2+b^2\right)}{a^4+3 b^2 a^2+b^4} & -\frac{a
%      \left(a^2+b^2\right)}{a^4+3 b^2 a^2+b^4} & 0 \\
%    \frac{a^2 b}{a^4+3 b^2 a^2+b^4} & 0 & 0 & -\frac{a b^2}{a^4+3 b^2 a^2+b^4}
%      & \frac{b \left(a^2+b^2\right)}{a^4+3 b^2 a^2+b^4} & 0 & 0 & \frac{a
%      \left(a^2+b^2\right)}{a^4+3 b^2 a^2+b^4} \\
%    \frac{a^3}{a^4+3 b^2 a^2+b^4} & 0 & 0 & -\frac{a^2 b}{a^4+3 b^2 a^2+b^4} &
%      \frac{a \left(a^2+b^2\right)}{a^4+3 b^2 a^2+b^4} & 0 & 0 & -\frac{b^3+2
%      a^2 b}{a^4+3 b^2 a^2+b^4} \\
%    0 & -\frac{a^3}{a^4+3 b^2 a^2+b^4} & \frac{a^2 b}{a^4+3 b^2 a^2+b^4} & 0 &
%      0 & -\frac{a \left(a^2+b^2\right)}{a^4+3 b^2 a^2+b^4} & \frac{b^3+2 a^2
%      b}{a^4+3 b^2 a^2+b^4} & 0
%   \end{array}
%   \right]
%$$

One computes
\begin{eqnarray*}
P_{11,11} X P_{11,11} &=& X \\
%
\Pf(-P_\Omega LP_\Omega) &=& a^4+3 b^2 a^2+b^4 \\
%
\Pf(-P_\Omega L P_\Omega) \Pf(P_{11,00} X P_{11,00}) &=&
-a^2 \left(b^3+2 a^2 b\right) \\
%
\Pf(-P_\Omega L P_\Omega) \Pf(P_{10,10} X P_{10,10}) &=&
-a^5 \\
%
\Pf(-P_\Omega L P_\Omega) \Pf(P_{10,01} X P_{10,01}) &=& 0 \\
%
\Pf(-P_\Omega L P_\Omega) \Pf(P_{01,10} X P_{01,10}) &=& 0 \\
%
\Pf(-P_\Omega L P_\Omega) \Pf(P_{01,01} X P_{01,01}) &=& a^5 \\
%
\Pf(-P_\Omega L P_\Omega) \Pf(P_{00,11} X P_{00,11}) &=&
-a^2 \left(b^3+2 a^2 b\right) \\
%
\Pf(-P_\Omega L P_\Omega) \Pf(X) &=& a^4 \left(a^2+b^2\right).
\end{eqnarray*}

Using the formula in section \ref{sec:vac_X}, we have
\begin{eqnarray*}
	v_\Omega &=&
	\Pf(-P_\Omega L P_\Omega) \sum_\sigma \Pf(P_\sigma X P_\sigma) e_\sigma \\
	&&\\
	&=& \Pf(-P_\Omega LP_\Omega) \vece_{00,00}  \\
	&+& \Pf(-P_\Omega L P_\Omega) \Pf(P_{11,00} X P_{11,00}) \vece_{11,00}  \\
	&+& \Pf(-P_\Omega L P_\Omega) \Pf(P_{10,10} X P_{10,10}) \vece_{10,10}  \\
	&+& \Pf(-P_\Omega L P_\Omega) \Pf(P_{10,01} X P_{10,01}) \vece_{10,01}  \\
	&+& \Pf(-P_\Omega L P_\Omega) \Pf(P_{01,10} X P_{01,10}) \vece_{01,10}  \\
	&+& \Pf(-P_\Omega L P_\Omega) \Pf(P_{01,01} X P_{01,01}) \vece_{01,01}  \\
	&+& \Pf(-P_\Omega L P_\Omega) \Pf(P_{00,11} X P_{00,11}) \vece_{00,11}  \\
	&+& \Pf(-P_\Omega L P_\Omega) \Pf(X)  \\
	&&\\
	&=& (a^4+3 b^2 a^2+b^4) \\
	&+& a^2 \left(b^3+2 a^2 b\right) \psi_1(0) \wedge \psi_2(0)  \\
	&-& a^5 \psi_1(0) \wedge \psi_1(5) \\
	&+& (a^5) \psi_2(0) \wedge \psi_2(5)  \\
	&-& a^2 \left(b^3+2 a^2 b\right) \psi_1(5) \wedge \psi_2(5) \\
	&+& (a^4 \left(a^2+b^2\right)  \psi_1(0) \wedge \psi_2(0) \wedge \psi_1(5) \wedge \psi_2(5).
\end{eqnarray*}

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

This should be equal, up to a constant multiplier, to the vector described in
theorem \ref{thm:vac_chzn}.  To begin to compute this, we first need to find
the nullspace of $P_\Omega L$.  Recall from the theory of finite difference
equations with constant coefficients (such as the Fibonacci sequence) that we
guess a solution of the form
$$
	f(k) = c^k u.
$$
We write down
%A={{a,0},{0,-a}};
%B={{0,b},{-b,0}};
%TeXForm[Expand[Det[-Transpose[A] + c*B + c^2*A]]]
%b^2 c^2-a^2 \left(c^2-1\right)^2
\begin{eqnarray*}
	-A^\tau f(k-1) + Bf(k) + Af(k+1) &=& 0 \\
	-A^\tau c^{k-1}u + Bc^k u + Ac^{k+1}u &=& 0 \\
	\left(-A^\tau + Bc + Ac^2\right) c^{k-1}u &=& 0.
\end{eqnarray*}

Ignore trivial solutions:
\begin{eqnarray*}
	\left(-A^\tau + Bc + Ac^2\right) u &=& 0 \\
	\det\left(-A^\tau + B c + A c^2\right) &=& 0 \\
%
-A^\tau + Bc + Ac^2 &=&
\left[
\begin{array}{ll}
a c^2-a & b c \\
-b c & a-a c^2
\end{array}
\right] \\
%
	\det\left(-A^\tau + B c + A c^2\right) &=& -a^2 c^4+2 a^2 c^2+b^2 c^2-a^2 = 0 \\
	a^2 c^4-2 a^2 c^2-b^2 c^2+a^2 &=& 0.
\end{eqnarray*}
Solve for $c$ in terms of $a$ and $b$, using the quadratic formula:
\begin{eqnarray*}
	a^2 c^4 - (2 a^2 + b^2) c^2 + a^2 &=& 0 \\
	c^2 &=& \frac{1}{2} a^2 \left(2 a^2+b^2\pm\sqrt{b^4+4 a^2 b^2}\right) \\
\end{eqnarray*}
We have four possibilities for $c$:
\begin{eqnarray*}
	c_1 &=& \frac{\sqrt{\frac{2 a^2+b^2-\sqrt{b^4+4 a^2 b^2}}{a^2}}}{\sqrt{2}} \\
	c_2 &=& \frac{\sqrt{\frac{2 a^2+b^2+\sqrt{b^4+4 a^2 b^2}}{a^2}}}{\sqrt{2}} \\
	c_3 &=& -\frac{\sqrt{\frac{2 a^2+b^2-\sqrt{b^4+4 a^2 b^2}}{a^2}}}{\sqrt{2}} \\
 	c_4 &=& -\frac{\sqrt{\frac{2 a^2+b^2+\sqrt{b^4+4 a^2 b^2}}{a^2}}}{\sqrt{2}}.
\end{eqnarray*}

To compute the eigenvector $u$ in each case, we examine the matrices $-A^\tau +
B c + A c^2$:

$$
M_1 = 
   \left[
   \begin{array}{ll}
    \frac{b^2-\sqrt{b^4+4 a^2 b^2}}{2 a} & \frac{b \sqrt{\frac{2
      a^2+b^2-\sqrt{b^4+4 a^2 b^2}}{a^2}}}{\sqrt{2}} \\
    -\frac{b \sqrt{\frac{2 a^2+b^2-\sqrt{b^4+4 a^2 b^2}}{a^2}}}{\sqrt{2}} &
      \frac{\sqrt{b^4+4 a^2 b^2}-b^2}{2 a}
   \end{array}
   \right]
$$
$$
M_2 = 
   \left[
   \begin{array}{ll}
    \frac{b^2+\sqrt{b^4+4 a^2 b^2}}{2 a} & \frac{b \sqrt{\frac{2
      a^2+b^2+\sqrt{b^4+4 a^2 b^2}}{a^2}}}{\sqrt{2}} \\
    -\frac{b \sqrt{\frac{2 a^2+b^2+\sqrt{b^4+4 a^2 b^2}}{a^2}}}{\sqrt{2}} &
      -\frac{b^2+\sqrt{b^4+4 a^2 b^2}}{2 a}
   \end{array}
   \right]
$$
$$
M_3 = 
   \left[
   \begin{array}{ll}
    \frac{b^2-\sqrt{b^4+4 a^2 b^2}}{2 a} & -\frac{b \sqrt{\frac{2
      a^2+b^2-\sqrt{b^4+4 a^2 b^2}}{a^2}}}{\sqrt{2}} \\
    \frac{b \sqrt{\frac{2 a^2+b^2-\sqrt{b^4+4 a^2 b^2}}{a^2}}}{\sqrt{2}} &
      \frac{\sqrt{b^4+4 a^2 b^2}-b^2}{2 a}
   \end{array}
   \right]
$$
$$
M_4 = 
   \left[
   \begin{array}{ll}
    \frac{b^2+\sqrt{b^4+4 a^2 b^2}}{2 a} & -\frac{b \sqrt{\frac{2
      a^2+b^2+\sqrt{b^4+4 a^2 b^2}}{a^2}}}{\sqrt{2}} \\
    \frac{b \sqrt{\frac{2 a^2+b^2+\sqrt{b^4+4 a^2 b^2}}{a^2}}}{\sqrt{2}} &
      -\frac{b^2+\sqrt{b^4+4 a^2 b^2}}{2 a}
   \end{array}
   \right]
$$

We may then use row reduction to find the eigenvectors.

%$$
%\det M_1 = 0
%$$
%
%$$
%\det M_2 = 0
%$$
%
%$$
%\det M_3 = 0
%$$
%
%$$
%\det M_4 = 0
%$$

%Row reds:
%
%$$
%   \left[
%   \begin{array}{ll}
%    1 & \frac{\sqrt{2} a b \sqrt{\frac{2 a^2+b^2-\sqrt{b^4+4 a^2
%      b^2}}{a^2}}}{b^2-\sqrt{b^4+4 a^2 b^2}} \\
%    0 & 0
%   \end{array}
%   \right]
%$$
%
%$$
%   \left[
%   \begin{array}{ll}
%    1 & \frac{\sqrt{2} a b \sqrt{\frac{2 a^2+b^2+\sqrt{b^4+4 a^2
%      b^2}}{a^2}}}{b^2+\sqrt{b^4+4 a^2 b^2}} \\
%    0 & 0
%   \end{array}
%   \right]
%$$
%
%$$
%   \left[
%   \begin{array}{ll}
%    1 & \frac{\sqrt{2} a b \sqrt{\frac{2 a^2+b^2-\sqrt{b^4+4 a^2
%      b^2}}{a^2}}}{\sqrt{b^4+4 a^2 b^2}-b^2} \\
%    0 & 0
%   \end{array}
%   \right]
%$$
%
%$$
%   \left[
%   \begin{array}{ll}
%    1 & -\frac{\sqrt{2} a b \sqrt{\frac{2 a^2+b^2+\sqrt{b^4+4 a^2
%      b^2}}{a^2}}}{b^2+\sqrt{b^4+4 a^2 b^2}} \\
%    0 & 0
%   \end{array}
%   \right]
%$$

%[1 x] [u1] = [0]
%[0 0] [u2]   [0]
%
%[1 x] [u1] = [0]
%[0 0] [u2]   [0]
%
%u1 + xu2 = 0
%u1 = -xu2
%
%u2 = 1
%u1 = -x
%
%[1 x] [-x] = [0]
%[0 0] [ 1]   [0]

%xxx do $c^{-k}$'s too \ldots.

%% ================================================================
\newpage
\section{Further directions}
\label{sec:further}

The work presented here is a starting point for much further investigation.
Possibilities include:

\begin{itemize}
\item Compare and contrast symmetric (bosonic) integrals with the
	skew-symmetric (fermionic) integrals presented here.

\item Consider the join of multiple intervals $\Omega_1$ and $\Omega_2$.
	This may entail examination of the space of all skew-symmetric boundary
	conditions for the intersection node, using line bundles.

\item Restore the continuum limit for the truncated, discretized integral,
	obtaining an integral on an interval; extend that integral to the real line.
	This may involve zeta-function regularization for determinants.

\item Extend the one-dimensional discussion to two or three
	dimensions.
\end{itemize}

%%% ================================================================
%%% ================================================================
%%% ================================================================
%\appendix
%\newpage
%
%%% ================================================================
%\section{History}
%
%To be filed:  somewhere, a section with some history (Dirac).
