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\begin{document}
\centerline{Math 372 lecture for Wednesday, Week 10}
\bigskip

\centerline{\bf The discrete Laplacian}
\bigskip

Let~$G=(V,E))$ be an undirected, connected, multigraph with vertex
set~$V=\left\{ v_1,\dots,v_n\right\}$.  For convenience, assume~$G$ has no
loops.  Last time, we defined the boundary mapping
\[
\partial\colon\Z E\to\Z V,
\]
determined by~$\partial(e)=e^{+}-e^{-}$ for each~$e\in E$.
%If~$f=\sum_{v\in V}a_vv\in\Z V$ for some~$a_v\in\Z$, define the {\em degree}
%of~$f$ to be
%\[
%  \deg(f):=\sum_{v\in V}a_v.
%\]
%We saw that there is an exact sequence
%\[
%  0\to\mathcal{C}\to\Z E\xrightarrow{\partial}\Z
%  V\xrightarrow{\mathrm{deg}}\Z\to0,
%\]
%where~$\mathcal{C}\to\Z E$ is just the inclusion of~$\mathcal{C}$ as a subset
%of~$\Z E$.
%By {\em exact}, we mean that the image of each mapping is the kernel of the
%following mapping.  In detail, the image of the mapping~$0\to\mathcal{C}$ is $\left\{ 0
%\right\}$, and that should be the kernel of~$\mathcal{C}\to\Z E$, which makes
%sense since the inclusion mapping is injective.  Next, the image of~$\mathcal{C}\to\Z
%E$ is just~$\mathcal{C}$ as a subset of~$\Z E$, and it is equal to the kernel
%of~$\partial$: 
%\[
%  \mathcal{C}=\ker\partial.
%\]
%Letting~$\Z V_0$ denote the elements of~$\Z V$ of degree~$0$, we have
%that~$\im\partial=\Z V_0$.  We thus get an induced isomorphism
%\[
%  \Z E/\mathcal{C}\xrightarrow[\partial]{\sim}\Z V_0.
%\]
Ordering the vertices and then using lexicographic ordering on the edges, as
usual, boundary mapping becomes a matrix, and we can take its transpose
\[
  \partial^t\colon\Z V\to \Z E.
\]
(More formally, define~$\Z E^*:=\Hom_{\Z}(\Z E^*,\Z)$ to be the set of
  all~$\Z$-linear mappings $\Z E\to\Z$, and identify~$\Z E$ with~$\Z E^*$
  by~$e\mapsto \chi_e$ for each~$e\in E$ where~$\chi_e$ is the characteristic function
  for~$e$.  Similarly, define~$\Z V^*$ and identify it with~$\Z V$.  Then we get
  a mapping
  \[
    \Z V^*\xrightarrow{\partial^*}\Z E^{*}
  \]
  defined by $\phi\mapsto \phi\circ\partial$ for each function~$\phi\colon\Z
  V\to\Z$. Fixing our usual bases for~$\Z V$ and~$\Z E$ turns~$\partial$ into a
matrix and~$\partial^*$ into the transpose~$\partial^t$.)

{\bf Definition.} The {\em Laplacian} of~$G$ is the mapping (or matrix)
\[
  L:=L(G):=\partial\circ\partial^t\colon\Z V\to\Z V.
\]
\medskip

{\bf Example.} Consider the {\em diamond graph}~$G$:
\begin{center}
\begin{tikzpicture}[scale=0.5]
  \node[ball,label={below:$v_1$}] (1) at (0,-1.732) {};
  \node[ball,label={left:$v_2$}] (2) at (-1,0) {};
  \node[ball,label={right:$v_3$}] (3) at (1,0) {};
  \node[ball,label={above:$v_4$}] (4) at (0,1.732) {};
  \draw (1)--(2)--(4)--(3)--(1);
  \draw (2)--(3);
\end{tikzpicture}
\end{center}
The Laplacian of~$G$ is
\[
L=\partial\partial^t
=
\left(\begin{array}{rrrrr}
    -1 & -1 & 0 & 0 & 0 \\
    1 & 0 & -1 & -1 & 0 \\
    0 & 1 & 1 & 0 & -1 \\
    0 & 0 & 0 & 1 & 1
\end{array}\right)
\left(\begin{array}{rrrr}
    -1 & 1 & 0 & 0 \\
    -1 & 0 & 1 & 0 \\
    0 & -1 & 1 & 0 \\
    0 & -1 & 0 & 1 \\
    0 & 0 & -1 & 1
\end{array}\right)
=
\left(\begin{array}{rrrr}
    2 & -1 & -1 & 0 \\
    -1 & 3 & -1 & -1 \\
    -1 & -1 & 3 & -1 \\
    0 & -1 & -1 & 2
\end{array}\right).
\]
\medskip

The {\em (weighted) adjacency matrix} for~$G$ is the~$n\times n$ matrix~$A$
where~$A_{ij}$ is the number of edges joining~$v_i$ to~$v_j$.  For each
vertex~$v\in V$, define~$\deg_G(v)$ to be the number of edges with an endpoint
equal to~$v$. Then define the diagonal
matrix~$D=\mathrm{diag}(\deg_G(v_1),\cdots,\deg_G(v_n))$.
\medskip

{\bf Proposition 1.} The Laplacian of~$G$ is $L=D-A$.
\smallskip

{\bf Proof.} Let~$m=|E|$.  For~$1\leq i<j\leq n$, we have
\begin{align*}
  (\partial\partial^t)_{ij}
  &= \sum_{k=0}^m\partial_{ik}\partial^t_{kj}\\
  &= \sum_{k=0}^m\partial_{ik}\partial_{jk}\\
  &= -A_{ij}=L_{ij}
\end{align*}
since
\[
  \partial_{ik}\partial_{jk}=
  \begin{cases}
    -1&\text{if~$\{v_i,v_j\}$ is the~$k$-th edge,}\\
    \hfill0\hfill&\text{otherwise.}
  \end{cases}
\]
Since ~$L$, ~$D$, and~$A$ are symmetric, we also
have~$L_{ij}=(\partial\partial^t)_{ij}$ for~$i>j$.  That leaves the case
where~$i=j$:
\begin{align*}
  (\partial\partial^t)_{ii}
  &= \sum_{k=0}^m\partial_{ik}\partial^t_{ki}\\
  &= \sum_{k=0}^m\partial_{ik}\partial_{ik}\\
  &=D_{ii}
\end{align*}
since
\[
\partial_{ik}\partial_{ik}=
  \begin{cases}
    1&\text{if~$v_i$ is on the~$k$-th edge,}\\
    \hfill0\hfill&\text{otherwise.}
  \end{cases}
\]
\hfill$\Box$
\medskip

As an immediate corollary, we get:

{\bf Corollary 2.} The sum of the rows and the sum of the columns of~$L$ are
both~$0\in\R^n$. 
\medskip

(Another way to see the corollary: $L=\partial\partial^t$ implies
  that~$\partial_0\circ L=0$ where~$\partial_0$ is the~$0$-th boundary map,
  represented by the matrix $[1\ 1\ \dots\ 1]$. (We will call~$\partial_0$ the
    {\em degree} mapping, below.)  This says the sum of the rows of
  $L$ is~$0$.  Since~$L$ is symmetric, the sum of the columns must then
be~$0$, also.)

{\bf Definition.} The {\em Picard group} of~$G$ is the abelian group
\[
  \Pic(G):=\Z V/\im(L).
\]
\medskip

Recall the exact sequence from last time:
\[
  0\to\mathcal{C}\to\Z E\xrightarrow{\partial}\Z
  V\xrightarrow{\mathrm{deg}}\Z\to0.
\]
Here, if~$f=\sum_{v\in V}a_v v$ for some~$a_v\in \Z$,
then~$\deg(f):=\sum_{v\in V}a_v$.  In particular,~$\deg(v)=1$, which would
usually not be equal to~$\deg_G(v)$, the number of edges incident on~$v$,
despite the similarity in notation.
Let~$\Z V_0$ denote the kernel of the degree mapping, i.e., the set of those
$f=\sum_{v\in V}a_v v$ such that~$\sum_{v\in V}a_v=0$.  The exact sequence says
that~$\im\partial\subseteq \Z V_0$.  Then, since~$L=\partial\circ\partial^t$, we
have
\[
  \im L\subseteq\im\partial\subseteq\Z V_0.
\]
\medskip

{\bf Definition.} The {\em Jacobian group} or {\em critical group} of~$G$ is
\[
  \Jac(G):=\Z V_0/\im L.
\]
\medskip

{\bf Proposition 3.} Fix any~$q\in V$.  Then there is an isomorphism
\begin{align*}
  \phi\colon\Pic(G)&\to\Z\oplus\Jac(G)\\
  f&\mapsto (\deg(f),f-\deg(f)q).
\end{align*}
\smallskip

{\bf Proof.}  First notice that~$\phi$ is well-defined: if~$f,f'\in \Z V$
and~$f=f'\bmod \im L$, write~$f'=f+h$ where~$h\in\im L$. We saw above
that everything in the image of~$L$ has degree~$0$.  Therefore, 
\[
  \deg(f')=\deg(f+h)=\deg(f)+\deg(h)=\deg(f).  
\]
The inverse to~$\phi$ is the mapping defined by $(d,g)\mapsto g+dq$.
\hfill$\Box$
\medskip

{\bf Definition.} Fix~$q\in V$.  The {\em reduced Laplacian} of~$G$ (with respect
to~$q$) is the matrix~$\tilde{L}$ obtained by removing the row and column
indexed by~$q$ from the Laplacian~$L$.
\medskip

{\bf Proposition 4.} Fix~$q=v_j\in V=\left\{ v_1,\dots,v_n \right\}$.  Then there is a~$\Z$-linear isomophism
\begin{align*}
 \psi\colon\Jac(G)&\to \Z^{n-1}/\im(\tilde{L})\\
  \textstyle
  \sum_{i=1}^na_iv_i\bmod\im(L)&\mapsto\textstyle(a_1,\dots,\hat{a_j},\dots,a_n)\bmod\im(\tilde{L}).
  \end{align*}
\smallskip

{\bf Proof.} Without loss of generality, take~$j=1$. We will first verify
that~$\psi$ is well-defined. Think of~$L$ as an~$n\times n$ matrix. For each~$i$
let the~$i$-th column of~$L$ be denoted by~$\ell_i$ and let~$\tilde{\ell}_i$ be
the vector~$\ell_i$ with its first entry removed.  Thus, the columns of the
reduced Laplacian are~$\tilde{\ell}_2,\dots,\tilde{\ell}_n$.  We need to show
that~$\psi(\ell_i)\in\im(\tilde{L})$ for each~$i$.  That follows immediately
for~$i>1$, since~$\psi(\ell_i)=\tilde{\ell}_i$.  For~$i=1$, use Corollary~2,
which says that~$\sum_{i=1}^n\ell_i=0$.  It follows
that~$\ell_1=-\sum_{i=2}^n\ell_i$ and thus
\[
  \psi(\ell_i)=\tilde{\ell}_1=-\sum_{i=2}^n\tilde{\ell_i}\in\im(\tilde{L}).
\]

maps to~$0$ under~$\psi$ if~$i=j$ and maps to a column of~$\tilde{L}$,
otherwise.  Therefore,~$\psi$ is well-defined.  

To show~$\psi$ is an isomorphism, we exhibit its inverse:
\begin{align*}
 \rho\colon\Z^{n-1}/\im(\tilde{L})\colon&\to\Jac(G)\\
  \textstyle
  (a_2,\dots,a_n)\bmod\im(\tilde{L})
  &\mapsto(-\textstyle\sum_{i=2}^na_i,a_2,\dots,a_n)\bmod\im(L).
\end{align*}
Reasoning as above, we see~$\rho(\tilde{\ell}_i)=\ell_i$ for~$i=2,\dots,n$.
Thus,~$\rho$ is well-defined.  Recalling that the sum of the coefficients of any
element of~$\Jac(G)$ is~$0$, it is clear that~$\rho$ is inverse to~$\psi$.
\hfill$\Box$

\medskip

{\bf Example.} Let~$G$ be the diamond graph from the first example.  Then
\[
  \Pic(G)\simeq\Z^4/\Span_{\Z}\left\{
    (2,-1,-1,0),(-1,3,-1,-1),(-1,-1,3,-1),(0,-1,-1,2) \right\}.
\]
and
\[
  \Jac(G)\simeq\Z^3/\Span_{\Z}\left\{ (3,-1,-1),(-1,3,-1),(-1,-1,2)
    \right\},
\]
choosing~$q=v_1$ in Proposition~3.

\medskip
\end{document}

