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\usepackage{enumerate}

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% Miscellaneous
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% Permutation classes
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\newcommand{\gr}{\mathrm{gr}}
\newcommand{\Grid}{\mathsf{Grid}}
\newcommand{\Geom}{\mathsf{Geom}}
\newcommand{\Geomhash}{\Geom^\#}

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% Grid matrix
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% Grid class diagrams
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% 0/±1, M^{x2}, trace monoid of M
% ------------------------
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\title{\bf Growth rates of geometric grid\\ classes of permutations}

% input author, affiliation, address and support information as follows;
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% the street address

\author{David Bevan\\
\small Department of Mathematics and Statistics\\[-0.8ex]
\small The Open University\\[-0.8ex]
\small Milton Keynes, England\\
\small\tt David.Bevan@open.ac.uk
}

% \date{\dateline{submission date}{acceptance date}\\
% \small Mathematics Subject Classifications: comma separated list of
% MSC codes available from http://www.ams.org/mathscinet/freeTools.html}

\date{\dateline{Nov 17, 2014}{Nov 22, 2014}{Dec 4, 2014}\\
\small Mathematics Subject Classifications: 05A05, 05A16, 05C31}

\begin{document}

\maketitle

% E-JC papers must include an abstract. The abstract should consist of a
% succinct statement of background followed by a listing of the
% principal new results that are to be found in the paper. The abstract
% should be informative, clear, and as complete as possible. Phrases
% like "we investigate..." or "we study..." should be kept to a minimum
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% any other references to things in the paper that are not defined in
% the abstract. The abstract will be distributed without the rest of the
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\begin{abstract}
Geometric grid classes of permutations have proven to be
key in investigations of
classical permutation pattern classes.
By considering the representation of
gridded permutations as words in a trace monoid,
we prove that
every
geometric grid class
has a
growth rate
which is given by
the square of the largest root of the matching polynomial of
a related graph.
As a consequence, we characterise the set of growth rates of geometric grid classes in terms of the spectral radii of trees,
explore the influence of ``cycle parity'' on the growth rate, compare the growth rates of geometric grid classes against those of the corresponding
monotone grid classes, and present new results concerning the effect of edge subdivision on the largest root of the matching polynomial.
\end{abstract}

% ================================================================
\section{Introduction}

Following the proof by Marcus \& Tardos~\cite{MT2004} of the Stanley--Wilf conjecture, there has been particular interest in the growth rates of permutation classes.
Kaiser \& Klazar~\cite{KK2003} determined the possible growth rates less than $2$,
and then
Vatter~\cite{Vatter2011} characterised all the (countably many) permutation classes with growth rates below $\kappa\approx2.20557$ and
established that there are uncountably many permutation classes with growth rate $\kappa$.
Critical to these results has been the consideration of \emph{grid classes} of permutations, and particularly of \emph{geometric} grid classes.
Geometric grid classes have also been used to achieve the enumeration of some specific permutation classes~\cite{AAB2012,AAV2014,Pantone2013}.
Following initial work
on particular geometric grid classes
by Waton~\cite{WatonThesis},
Vatter \& Waton~\cite{VW2011b},
and Elizalde~\cite{Elizalde2011},
their general structural properties have been
investigated
in articles
by Albert, Atkinson, Bouvel, Ru\v{s}kuc \& Vatter~\cite{AABRV2011}
and Albert, Ru\v{s}kuc \& Vatter~\cite{ARV2012}.
We build on their work to establish the growth rate of any given geometric grid class.
Before we can state our result, we need a number of definitions.

\begin{figure}[t]
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\qquad\qquad\quad
 \raisebox{-.15in}{
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      \node[above]at(0,1.1){$r_2$};
      \node[above]at(1,1.1){$c_1$};
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$$
\caption[figGeomClass]{
At left: The standard figure for
$\begin{geommx}1&\pos0&-1\\ 1&-1&\pos1 \end{geommx}$,
showing two
plots
of the permutation $1527634$
with distinct griddings.
At right: Its row-column graph;
positive edges are shown as solid lines, negative edges are dashed.
}\label{figGeomClass}
\end{figure}
A geometric grid class is specified by a \opmo{} matrix which represents the shape of plots of permutations in the class.
To match the Cartesian coordinate system, we index these matrices from the lower left, by column and then by row.
If $M$ is such a matrix, then we say that the \emph{standard figure} of $M$, denoted  $\Lambda_M$, is the subset of $\mathbb{R}^2$ consisting of the
union of
oblique
open line segments $L_{i,j}$ with slope $M_{i,j}$ for each $i,j$ for which
$M_{i,j}$ is nonzero,
where $L_{i,j}$ extends
from $(i-1,j-1)$ to $(i,j)$
if $M_{i,j}=1$, and
from $(i-1,j)$ to $(i,j-1)$
if $M_{i,j}=-1$.
The \emph{geometric grid class} $\Geom(M)$ is
then defined to be
the set of
permutations $\sigma_1\sigma_2\ldots\sigma_n$ that can be plotted as a subset of the standard figure, i.e.~for which there exists a sequence of points $(x_1,y_1),\ldots,(x_n,y_n)\in\Lambda_M$
such that $x_1<x_2<\ldots<x_n$ and the sequence $y_1,\ldots,y_n$ is order-isomorphic to $\sigma_1,\ldots,\sigma_n$.
See Figure~\ref{figGeomClass} for an example.

If $g_n$ is the number of permutations of length $n$ in
$\Geom(M)$,
then the
\emph{growth rate}
of the class is given by
$\gr(\Geom(M))=\liminfty g_n^{\,1/n}$.
We will demonstrate that this limit exists for geometric grid classes\footnote{It is widely believed that all permutation classes have growth rates.
The proof of the Stanley--Wilf conjecture by
Marcus \& Tardos~\cite{MT2004} establishes only that each has an \emph{upper} growth rate ($\limsup g_n^{\,1/n}$).} and determine its value for any given \opmo{} matrix $M$.

Much of the structure of a geometric grid class is reflected in a
graph that
we associate with the underlying matrix.
If $M$ is a \opmo{} matrix of dimensions $t\ttimes u$, the \emph{row-column graph} $G(M)$ of $M$ is the bipartite graph
with vertices $r_1,\ldots,r_t,c_1,\ldots,c_u$ and an edge between $r_i$ and $c_j$ if and only if $M_{i,j}\neq0$.
We label each edge $r_ic_j$ with the value of $M_{i,j}$.
Edges labelled $+1$ are called \emph{positive}; edges labelled $-1$ are called \emph{negative}.
See Figure~\ref{figGeomClass} for an example.

\begin{figure}[ht]
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\caption[figDoubleRefinement]{At left: The standard figure of
$\begin{smallmx}1&\pos0&-1\\ 1&-1&\pos1 \end{smallmx}^{\!\times2}$,
with a consistent orientation marked. At right:
Its row-column graph.
}\label{figDoubleRefinement}
\end{figure}
We need one final definition related to geometric grid classes.
If $M$ is a \opmo{} matrix of dimensions $t\ttimes u$, we define the \emph{double refinement} $\drm$ of $M$ to be the \opmo{} matrix of dimensions $2\+t\ttimes 2\+u$ obtained from $M$ by replacing each $0$ with $\begin{smallmx}0&0\\0&0\end{smallmx}$, each $1$ with $\begin{smallmx}0&1\\1&0\end{smallmx}$, and each $-1$ with $\begin{smallmx}-1&\pos0\\ \pos0&-1\end{smallmx}$.
See Figure~\ref{figDoubleRefinement} for an example.
Note that the standard figure of $\drm$ is essentially a scaled copy of the standard figure of $M$, so we have:
\begin{observation}\label{obsGeomDouble}
$\Geom(\drm)=\Geom(M)$ for any \opmo{} matrix $M$.
\end{observation}
We will demonstrate a connection between the growth rate of
$\Geom(M)$ and the \emph{matching polynomial} of the graph $G(\drm)$, the row-column graph of the double refinement of~$M$.
A~$k$-\emph{matching} of a graph is a set of $k$ edges, no
pair of which have a vertex in common.
For example, the negative (dashed) edges in the graph in Figure~\ref{figDoubleRefinement} constitute a $4$-matching.
If, for each~$k$, $m_k(G)$ denotes the number of distinct $k$-matchings of a graph $G$
with $n$ vertices,
then the \emph{matching polynomial}
$\mu_G(z)$
of $G$ is defined to be
\begin{equation}\label{eqMatchingPolynomialDefn}
\mu_G(z)
\;=\;
\sum_{k\geqslant0}
(-1)^k\+ m_k(G)\+z^{n-2k}.
\end{equation}
Note that this \emph{is} a polynomial since $m_k(G)=0$ for $k>\floor{n/2}$.
Observe also that the exponents of the variable $z$ enumerate \emph{defects} in $k$-matchings: the number of vertices which are \emph{not} endvertices of an edge in such a matching.
If $n$ is even, $\mu_G(z)$ is an even function;
if $n$ is odd, $\mu_G(z)$ is an odd function.

With the relevant definitions complete, we can now state our theorem:

\begin{theorem}\label{thmGeomClassGrowthRate}
The growth rate of
geometric grid class
$\Geom(M)$
exists and
is equal to
the square
of the largest root of
the matching polynomial
$\mu_{G(\drm)}(z)$,
where
$G(\drm)$ is
the row-column graph of the double refinement of $M$.
\end{theorem}

In the next section, we prove this theorem by utilizing the link between geometric grid classes and trace monoids, and their connection to rook numbers and the matching polynomial.
Then, in Section~\ref{SectGeomImplications}
we investigate a number of implications of this result by utilizing properties of the matching polynomial, especially the fact that the moments of $\mu_G(z)$
enumerate certain closed walks on $G$.
Firstly, we characterise the growth rates of geometric grid classes in terms of the spectral radii of trees.
Then, we explore the influence of cycle parity on growth rates and relate
the growth rates of geometric grid classes to those of monotone grid classes.
Finally, we consider the effect
of subdividing edges in the row-column graph, proving some new results regarding how edge subdivision affects the largest root of the matching polynomial.

% ================================================================
\section{Proof of Theorem~\ref{thmGeomClassGrowthRate}}

In order to prove our result, we make use of the
connection between geometric grid classes and \emph{trace monoids}. This relationship was first used by Vatter \& Waton~\cite{VW2011} to establish certain structural properties of
grid classes, and was developed further in~\cite{AABRV2011} from where we use a number of results.
To begin with, we need to consider \emph{griddings} of permutations.

If $M$ has dimensions $t\ttimes u$, then an $M$-\emph{gridding} of a permutation $\sigma_1\ldots\sigma_n$ of length $n$ in $\Geom(M)$  consists of two sequences $c_1,\ldots,c_t$ and $r_1,\ldots,r_u$
such that
there is some plot $(x_1,y_1),\ldots,(x_n,y_n)$ of $\sigma$
for which
$c_i$ is the
number of points $(x_k,y_k)$ in column $i$
(with $i-1<x_k<i$),
and $r_j$ is the
number of points in row $j$
(with $j-1<y_k<j$).\footnote{This
definition
of an $M$-gridding
is equivalent to the traditional one
given
in terms
of the positions of the cell dividers
relative to the points
$(k,\sigma_k)$.}
Note that a permutation may have multiple distinct griddings in a given geometric grid class; see Figure~\ref{figGeomClass} for an example.
We call a permutation together with
one of its $M$-griddings
an
$M$-\emph{gridded permutation}.
We use $\Geomhash(M)$ to denote the set of all $M$-gridded permutations.

From an enumerative perspective,
it can be much easier working with $M$-gridded permutations than directly with the permutations themselves.
The following observation means that we can, in fact, restrict our considerations to $M$-gridded permutations:
\begin{lemma}[see Vatter~\cite{Vatter2011} Proposition~2.1]\label{lemmaGRGriddings}
If it exists, the growth rate of
$\Geom(M)$ is equal to the growth rate of the corresponding class of $M$-gridded permutations $\Geomhash(M)$.
\end{lemma}
\begin{proof}
Suppose that $M$ has dimensions $t\ttimes u$.
Each permutation in $\Geom(M)$ has at least one gridding in $\Geomhash(M)$, but no permutation of length $n$ in $\Geom(M)$ can have more than $\binom{n+t-1}{t-1}\binom{n+u-1}{u-1}$ griddings in $\Geomhash(M)$ because
that is the number of ways of choosing the number of points in each column and row.
Thus the number of $M$-gridded permutations of length $n$ is no more than a polynomial multiple of the number of $n$-permutations in $\Geom(M)$; the result follows immediately from the definition of the growth rate.
\end{proof}
To determine the growth rate of
$\Geomhash(M)$,
we relate $M$-gridded permutations to words in a trace monoid.
To achieve this, one additional concept is required, that of a \emph{consistent orientation} of a standard figure.
If $\Lambda_M=\bigcup\,\{L_{i,j}:M_{i,j}\neq0\}$ is the standard figure of a \opmo{} matrix $M$, then a \emph{consistent orientation} of $\Lambda_M$ consists of an orientation of each oblique line $L_{i,j}$ such that in each column either all the lines are oriented leftwards or all are oriented rightwards, and in each row either all the lines are oriented downwards or all are oriented upwards.\footnote{For
ease
of exposition, we
use
the concept of a consistent orientation
rather than the
approach used previously involving
partial multiplication matrices; results from \cite{AABRV2011} follow \emph{mutatis mutandis}.}
See Figures~\ref{figDoubleRefinement} and \ref{figPermWord} for examples.

It is not always possible to consistently orient a standard figure.
The ability to do so depends on the \emph{cycles} in the row-column graph.
We say that the \emph{parity} of a cycle in $G(M)$ is
the product of the labels of its edges,
a \emph{positive cycle} is one which has parity $+1$, and
a \emph{negative cycle} is one with parity $-1$.
The following result relates
cycle parity to
consistent orientations:
\begin{lemma}[see Vatter \& Waton~\cite{VW2011} Proposition~2.1]\label{lemmaNegativeCycles}
The standard figure
$\Lambda_M$
has a consistent orientation if and
only if its row-column graph $G(M)$ contains no negative cycles.
\end{lemma}
For example,
$\begin{rowcolmx}1&\pos0&-1\\ 1&-1&\pos1 \end{rowcolmx}$
contains a negative cycle so its standard figure
has no
consistent orientation (see Figure~\ref{figGeomClass}), whereas
$\begin{rowcolmx}-1&\pos0&-1\\ \pos1&-1&\pos1 \end{rowcolmx}$
has no negative cycles so its standard figure has a consistent orientation
(see Figure~\ref{figPermWord}).

On the other hand, we can always consistently orient the standard figure of the double refinement of a matrix
by orienting each oblique line towards the centre of its $2\times2$ block (as in Figure~\ref{figDoubleRefinement}).
So
we have the following:
\begin{lemma}[see \cite{AABRV2011} Proposition~4.1]\label{lemmaDoubleRefinementConsistentOrientation}
If $M$ is any \opmo{} matrix, then $\Lambda_{\drm}$ has a consistent orientation.
\end{lemma}
Thus, by Lemma~\ref{lemmaNegativeCycles}, the row-column graph of the double refinement of a matrix never contains a negative cycle.
Figure~\ref{figDoubleRefinement} shows a consistent orientation of the standard figure of the double refinement of a matrix whose standard figure (shown in Figure~\ref{figGeomClass}) doesn't itself have a consistent orientation.

\begin{figure}[ht]
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\node[right]at(0.5,1.47){${}^4$};
\node[right]at(1.375,0.595){${}^5$};
\node[right]at(2.75,0.72){${}_6$};
\node[above]at(2.925,1.125){${}_7$};
}
\gctwo[-1, -1,-1,3,12,-1,-1,-1, -1, -1,-1,5,-1,-1,-1,-1, -1, 15,14,-1,-1,-1,6,9]{3}{-2,0,-2}{2,-3,2}
\qquad\qquad
\setgcextra{
\node[right]at(0.125,0.095){${}_1$};
\node[right]at(2.25,1.72){${}^2$};
\node[right]at(1.625,0.345){${}^3$};
\node[right]at(2.5,1.47){${}^4$};
\node[right]at(2.625,0.595){${}_5$};
\node[right]at(0.75,1.22){${}^6$};
\node[above]at(2.925,1.125){${}_7$};
}
\gctwo[-1, 1,-1,-1,-1,-1,10,-1, -1, -1,-1,-1,-1,3,-1,-1, -1, -1,14,-1,12,5,-1,9]{3}{-2,0,-2}{2,-3,2}
}
$$
\caption[figPermWord]{The plots of permutation $1527634$
in $\begin{geommx}-1&\pos0&-1\\ \pos1&-1&\pos1 \end{geommx}$
associated with the
words $a_{32}a_{32}a_{11}a_{12}a_{21}a_{31}a_{32}$ and $a_{11}a_{32}a_{21}a_{32}a_{31}a_{12}a_{32}$.
Both plots correspond to the same gridding.
}\label{figPermWord}
\end{figure}
We are now in a position to describe the association between words and $M$-gridded permutations.
If $M$ is a \opmo{} matrix, then
we
let $\Sigma_M=\{a_{ij}:M_{i,j}\neq0\}$ be an alphabet of symbols, one for each nonzero cell in $M$.
If
we have a consistent orientation for $\Lambda_M$,
then
we can associate to each finite word $w_1\ldots w_n$ over $\Sigma_M$
a specific plot of
a permutation
in $\Geom(M)$
as follows: If $w_k=a_{ij}$, include the point at distance
$k\+\sqrt{2}/(n+1)$
along line segment
$L_{i,j}$
according to its orientation.
See Figure~\ref{figPermWord} for two examples.
Clearly, this induces a mapping
from
the set of all finite words over $\Sigma_M$ to $\Geomhash(M)$.
In fact, it can readily be shown that this map is surjective, every $M$-gridded permutation corresponding to some word over $\Sigma_M$ (\cite{AABRV2011} Proposition~5.3).

As can be seen in Figure~\ref{figPermWord}, distinct words may be mapped to the same gridded permutation.
This occurs because the order in which two consecutive points are included is immaterial if they occur in cells that
are neither in the same column nor in the same row.
From the perspective of the words, adjacent symbols corresponding to such cells may be interchanged without changing the
gridded permutation.
This corresponds to a structure known as a
{trace monoid}.

If we have a consistent orientation for standard figure $\Lambda_M$,
then
we define the \emph{trace monoid of~$M$},
which we denote by $\tmm$,
to be the set of equivalence classes of words over $\Sigma_M$ in which $a_{ij}$ and $a_{k\ell}$ commute (i.e.~$a_{ij}a_{k\ell}=a_{k\ell}a_{ij}$) whenever $i\neq k$ and $j\neq \ell$.
It is then relatively straightforward to show equivalence between gridded permutations and elements of the trace monoid:
\begin{lemma}[see \cite{AABRV2011} Proposition~7.1]\label{lemmaTraceMonoid}
If the standard figure $\Lambda_M$ has a consistent orientation, then gridded $n$-permutations in $\Geomhash(M)$ are in bijection with equivalence classes of words of length $n$ in
$\tmm$.
\end{lemma}
Hence, by combining Lemmas~\ref{lemmaGRGriddings}, \ref{lemmaDoubleRefinementConsistentOrientation} and \ref{lemmaTraceMonoid} with Observation~\ref{obsGeomDouble}, we know that the growth rate of $\Geom(M)$ is equal to the growth rate of $\mathbb{M}(\drm)$ if it exists.
All that remains
is to determine the growth rate of the trace monoid of a matrix.

Trace monoids
were first studied by Cartier \& Foata~\cite{CF1969}.
Using extended M\"obius inversion, they
determined the general form of the generating function, as follows:
\begin{lemma}[\cite{CF1969};
see also Flajolet \& Sedgewick~\cite{FS2009} Note~V.10]
The ordinary generating function for $\tmm$ is given by
$$
f_M(z) \;=\;
\frac{1}{
\sum_{k\geqslant0} (-1)^k \+ r_k(M) \+ z^k
}
$$
where $r_k(M)$ is the number of $k$-subsets of $\Sigma_M$ whose elements commute pairwise.
\end{lemma}
Since symbols in $\tmm$ commute if and only if they correspond to cells that are neither in the same column nor in the same row,
it is easy to see that
$r_k(M)$ is the number of distinct ways of placing $k$ chess rooks on the nonzero entries of $M$ in such a way that no two rooks attack each other by being in the same column or row.
The numbers $r_k(M)$ are known as the \emph{rook numbers} for $M$ (see Riordan~\cite{Riordan2002}).
Moreover, a matching
in the row-column graph $G(M)$ also corresponds to a set of cells
no pair of which share a column or row.
So
the rook numbers for $M$ are
the same as
the numbers of matchings in $G(M)$:
\begin{observation}\label{obsRookNumIsMatchingNum}
For all $k\geqslant0$, $r_k(M)=m_k(G(M))$.
\end{observation}
Now, by elementary analytic combinatorics, we know that the growth rate of $\tmm$ is given by the reciprocal of the root
of the denominator
of $f_M(z)$
that has least magnitude (see~\cite{FS2009} Theorem~IV.7).
The fact that this polynomial has a \emph{unique} root of smallest modulus
was proved by Goldwurm \& Santini in~\cite{GS2000}. It is real and positive by Pringsheim's Theorem.

But the \emph{reciprocal} of the \emph{smallest} root of a polynomial is the same as the \emph{largest} root of the \emph{reciprocal} polynomial (obtained by reversing the order of the coefficients). Hence,
if $M$ has dimensions $t\ttimes u$
and $n=t+u$,
then the growth rate of $\tmm$ is the largest (positive real) root of the polynomial
\begin{equation}\label{eqGMDefn}
g_M(z)
\;=\;
\frac{1}{
z^{\floor{n/2}} \+ f_M \!\left(\frac{1}{z}\right)\!
}
\;=\;
\sum_{k=0}^{\floor{n/2}} (-1)^k \+ r_k(M) \+ z^{\floor{n/2}-k}.
\end{equation}
Here,
$g_M(z)$ is the reciprocal polynomial of
$\left(f_M(z)\right)^{-1}$
multiplied by some nonnegative power of $z$,
since $r_k(M)=0$ for all $k>\floor{n/2}$.
Note also that $n$ is the number of vertices in $G(M)$.

If we now compare the definition of $g_M(z)$ in~\eqref{eqGMDefn}
with that of the matching polynomial $\mu_G(z)$ in~\eqref{eqMatchingPolynomialDefn}
and use Observation~\ref{obsRookNumIsMatchingNum}, then we see that:
$$
g_M(z^2) \;=\;
\begin{cases}
\phantom{z^{-1}\+}\mu_{G(M)}(z), & \text{if $n$ is even;} \\[3pt]
         z^{-1}\+ \mu_{G(M)}(z), & \text{if $n$ is odd.}
\end{cases}
$$
Hence,
the largest root of $g_M(z)$ is the square of the largest root of $\mu_{G(M)}(z)$.

We now have
all we need to prove Theorem~\ref{thmGeomClassGrowthRate}: The growth rate of $\Geom(M)$ is equal to the growth rate of $\mathbb{M}(\drm)$ which equals the square of the largest root of $\mu_{G(\drm)}(z)$.

In the above argument, we only employ the double
refinement $\drm$ to ensure that a consistent orientation is possible.
By
Lemma~\ref{lemmaNegativeCycles}, we know that if
$G(M)$
is free of negative cycles then $\Lambda_M$
can be consistently oriented.
Thus, we have the following special case of Theorem~\ref{thmGeomClassGrowthRate}:

\begin{corollary}\label{corNoNegCycles}
If
$G(M)$
contains no negative cycles,
then
the growth rate of
$\Geom(M)$ is
equal to
the square
of the largest root of
$\mu_{G(M)}(z)$.
\end{corollary}

% ================================================================
\section{Consequences}
\label{SectGeomImplications}

In this final section, we investigate some of the implications of Theorem~\ref{thmGeomClassGrowthRate}.
By considering properties of the matching polynomial, we characterise the growth rates of geometric grid classes in terms of the spectral radii of trees, prove a monotonicity result, and explore the influence of cycle parity on growth rates.
We then compare the growth rates of geometric grid classes with those of monotone grid classes.
Finally, we consider the effect
of subdividing edges in the row-column graph.

Let's begin by introducing some notation.
We use $G+H$ to denote the graph composed of two disjoint subgraphs $G$ and $H$.
The graph resulting from deleting
the vertex $v$
(and all edges incident to $v$)
from a graph $G$
is denoted $G-v$.
Generalising this,
if $H$ is a subgraph of~$G$, then $G-H$ is the graph obtained by deleting the vertices of $H$ from $G$.
In contrast, we use
$G\deledge e$ to denote
the graph resulting from deleting the edge $e$ from
$G$.
The number of connected components of $G$ is represented by $\mathrm{comp}(G)$.
The characteristic polynomial of a graph $G$ is denoted $\Phi_G(z)$. We use $\rho(G)$ to denote the spectral radius of $G$, the largest root of $\Phi_G(z)$.
Finally, we use $\lambda(G)$ for the largest root of the matching polynomial $\mu_G(z)$.

The matching polynomial was independently discovered a number of times, beginning with
Heilmann \& Lieb~\cite{HL1970} when investigating monomer-dimer systems in statistical physics.
It was first studied from a combinatorial perspective by Farrell~\cite{Farrell1979} and Gutman~\cite{Gutman1977}.
The theory was then further developed by Godsil \& Gutman~\cite{GG1981} and Godsil~\cite{Godsil1981}.
An introduction can be found in the books by Godsil~\cite{Godsil1993} and Lov\'asz \& Plummer~\cite{LP2009}.

The facts concerning the matching polynomial that we use are covered by three lemmas.
As a consequence of the first, we only need to consider connected graphs:
\begin{lemma}[Farrell~\cite{Farrell1979}, Gutman~\cite{Gutman1977}]
\label{lemmaMuProductComponents}
The matching polynomial of a graph is the product of the matching polynomials of its connected components.
\end{lemma}
Thus, in particular:
\begin{corollary}\label{corLambdaMaxComponents}
For any graphs $G$ and $H$, we have
  $\lambda(G+H)=\max(\lambda(G),\lambda(H))$.
\end{corollary}

The second lemma relates the matching polynomial to the characteristic polynomial.
\begin{lemma}[Godsil \& Gutman~\cite{GG1981}]
\label{lemmaMuCycleDef}
If $\CCC_G$ consists of all nontrivial subgraphs of $G$ which are unions of vertex-disjoint cycles (i.e., all subgraphs of $G$ which are regular of degree~2), then
  $$
  \mu_G(z) \;=\; \Phi_G(z) \:+\: \sum_{C\in\CCC_G} \!\!2^{\+\mathrm{comp}(C)} \+ \Phi_{G-C}(z),
  $$
where $\Phi_{G-C}(z)=1$ if $C=G$.
\end{lemma}
As an immediate consequence, we have the following:
\begin{corollary}[Sachs~\cite{Sachs1964}, Mowshowitz~\cite{Mowshowitz1972}, Lov\'asz \& Pelik\'an~\cite{LP1973}]
\label{corMuPhiTree}
The matching polynomial of a graph
is identical to its characteristic polynomial
if and only if the graph is acyclic.
\end{corollary}
In particular, their largest roots are identical:
\begin{corollary}\label{corLambdaRhoTree}
  If $G$ is a forest,
  then
$\lambda(G)=\rho(G)$.
\end{corollary}
Thus, using Corollaries~\ref{corNoNegCycles} and~\ref{corLambdaMaxComponents}, we have the following alternative characterisation for the growth rates of acyclic geometric grid classes:

\begin{corollary}\label{corForestRho}
  If $G(M)$ is a forest, then $\gr(\Geom(M))=\rho(G(M))^2$.
\end{corollary}

The last, and most important, of the three lemmas allows us to determine the largest root of the matching polynomial of a graph from
the spectral radius of a related tree.
It
is a consequence of the fact,
determined by Godsil in~\cite{Godsil1981},
that the moments (sums of the powers of the roots) of
$\mu_G(z)$ enumerate certain closed walks on~$G$, which he calls \emph{tree-like}.
This is analogous to the fact that the moments of
$\Phi_G(z)$
count \emph{all} closed walks on~$G$.
On a tree, all closed walks are tree-like.

\begin{lemma}[Godsil~\cite{Godsil1981}; see also~\cite{Godsil1993} and~\cite{LP2009}]
\label{lemmaExpandVertex}
  Let $G$ be a graph and let $u$ and $v$ be adjacent vertices in a cycle of $G$.
  Let $H$ be the component of $G-u$ that contains $v$.
  Now let $K$ be the graph constructed by taking a copy of $G\deledge uv$ and a copy of $H$ and joining the occurrence of $u$ in the copy of $G\deledge uv$ to the occurrence of $v$ in the copy of $H$
  (see Figure~\ref{figExpandingVertex}).
  Then $\lambda(G)=\lambda(K)$.
\end{lemma}
\begin{figure}[ht]
$$
\raisebox{12pt}{
\begin{tikzpicture}[scale=0.36]
    \draw [thick] (0,.3)--(1.7,1.5);
    \draw [thick] (0,.3)--(2.1,1.1);
    \draw [thick] (0,.3)--(3,0.75);
    \draw [fill=gray!30!white,thick] (3.4,1.4) circle [x radius=1.7, y radius=1];
    \draw [fill] (0,.3) circle [radius=0.15];
    \node[left]at(0,.3){$u$};
    \node[left]at(-1.5,.3){$G$:};
    \draw [fill] (1.7,1.5) circle [radius=0.15];
    \node[left]at(1.7,1.7){$v$};
    \node[]at(3.4,1.4){$H$};
    \draw [thick] (0,.3)--(3,-0.75);
    \draw [thick] (0,.3)--(2.33,-1);
    \draw [thick] (0,.3)--(2.17,-1.5);
    \draw [thick] (0,.3)--(2,-2);
    \draw [fill=gray!30!white,thick] (3.4,-1.35) circle [x radius=1.75, y radius=1.15];
    \node[]at(3.4,-1.35){$J$};
\end{tikzpicture}
}
\qquad\qquad\qquad
\begin{tikzpicture}[scale=0.36]
    \draw [fill=gray!30!white,thick] (3.4,3.8) circle [x radius=1.7, y radius=1];
    \draw [thick] (0,.3)--(1.7,3.9);
    \draw [thick] (0,.3)--(2,1);
    \draw [thick] (0,.3)--(3,0.75);
    \draw [fill=gray!30!white,thick] (3.4,1.4) circle [x radius=1.7, y radius=1];
    \draw [fill] (0,.3) circle [radius=0.15];
    \node[left]at(0,.3){$u$};
    \node[left]at(-1.5,1.4){$K$:};
    \draw [fill] (1.7,3.9) circle [radius=0.15];
    \node[left]at(1.7,4.1){$v$};
    \node[]at(3.4,1.4){$H$};
    \node[]at(3.4,3.8){$H$};
    \draw [thick] (0,.3)--(3,-0.75);
    \draw [thick] (0,.3)--(2.33,-1);
    \draw [thick] (0,.3)--(2.17,-1.5);
    \draw [thick] (0,.3)--(2,-2);
    \draw [fill=gray!30!white,thick] (3.4,-1.35) circle [x radius=1.75, y radius=1.15];
    \node[]at(3.4,-1.35){$J$};
\end{tikzpicture}
\qquad
$$
\caption{Expanding $G$ at $u$ along $uv$; $H$ is the component of $G-u$ that contains $v$}
\label{figExpandingVertex}
\end{figure}
The process that is described in Lemma~\ref{lemmaExpandVertex}
we will call
``\emph{expanding} $G$ \emph{at} $u$ \emph{along} $uv$''.
Each such expansion of a graph $G$ produces a graph with fewer cycles than $G$.
Repeated application of this process will thus eventually result in a forest $F$ such that $\lambda(F)=\lambda(G)$.
We shall say that $F$ results from \emph{fully expanding}~$G$.
Hence, by
Corollaries~\ref{corLambdaMaxComponents} and~\ref{corLambdaRhoTree},
the largest root of the matching polynomial of a graph equals the spectral radius of some tree:
for any graph~$G$, there is a tree $T$ such that $\lambda(G)=\rho(T)$.

It is readily observed that
every tree is the row-column graph of some geometric grid class. Thus
we have the following characterisation of geometric grid class growth rates.

\begin{corollary}
  The set of growth rates of geometric grid classes consists of the squares of the spectral radii of trees.
\end{corollary}

The spectral radii of connected graphs satisfy the following strict monotonicity condition:
\begin{lemma}[\cite{CRS2010} Proposition~1.3.10]\label{lemmaStrictMono}
  If $G$ is connected and $H$ is a proper subgraph of $G$, then we have $\rho(H)<\rho(G)$.
\end{lemma}
Lemma~\ref{lemmaExpandVertex} enables us to prove
the analogous fact for the largest roots of matching polynomials, from which we can deduce a monotonicity result for geometric grid classes:
\begin{corollary}\label{corStrictMono}
If $G$ is connected and $H$ is a proper subgraph of $G$, then $\lambda(H)<\lambda(G)$.
\end{corollary}
\begin{proof}
  Suppose we fully expand $H$
  (at vertices $u_1,\ldots,u_k$, say),
  then
  the result is a forest $F$ such that $\lambda(H)=\rho(F)$.
  Now suppose that we repeatedly expand $G$ analogously at $u_1,\ldots,u_k$, and then continue to fully expand the resulting graph. The outcome is a tree $T$ (since $G$ is connected)
  such that $F$ is a proper subgraph of $T$ and
  $\lambda(G)=\rho(T)$.
  The result follows from Lemma~\ref{lemmaStrictMono}.
\end{proof}
Adding a non-zero cell to a \opmo{} matrix $M$ adds an edge to $G(M)$.
Thus, geometric grid classes satisfy the following monotonicity condition:

\begin{corollary}\label{corAddCell}
If $G(M)$ is connected and $M'$ results from adding a non-zero cell to $M$
in such a way that $G(M')$ is also connected,
then $\gr(\Geom(M'))>\gr(\Geom(M))$.
\end{corollary}

% --------------------------------------------------------
\subsection{Cycle parity}

The growth rate of a geometric grid class
depends on the parity of its cycles.
Consider the case of $G(M)$ being a cycle graph $C_n$.
If $G(M)$ is a negative cycle, then $G(\drm)=C_{2n}$.
Now, by Lemma~\ref{lemmaExpandVertex}, we have $\lambda(C_n)=\rho(P_{2n-1})$, where $P_n$ is the path graph on $n$ vertices.
The spectral radius of a graph on $n$ vertices is $2\cos\frac{\pi}{n+1}$.
So,
\begin{equation}\label{eqGRGeomCycle}
\gr(\Geom(M)) \;=\;
\begin{cases}
4\cos^2\frac{\pi}{2n}, & \text{if $G(M)$ is a positive cycle;} \\[5pt]
4\cos^2\frac{\pi}{4n}, & \text{if $G(M)$ is a negative cycle.}
\end{cases}
\end{equation}
Thus the geometric grid class whose row-column graph is a
negative cycle has a greater growth rate than the class
whose row-column graph is a positive cycle.
As another example,
\begin{equation}\label{eqGRGeomExNegCyc}
  \gr\!\left(\begin{geommx}1&\pos0&-1\\ 1&-1&\pos1 \end{geommx}\right) \;=\; 3+\sqrt{2} \;\approx\; 4.41421,
\end{equation}
whereas
\begin{equation}\label{eqGRGeomExPosCyc}
  \gr\!\left(\begin{geommx}-1&\pos0&-1\\ \pos1&-1&\pos1 \end{geommx}\right) \;=\; 4.
\end{equation}
The former, containing a negative cycle,
has a greater growth rate than the latter, whose cycle is positive.
This is typical;
we will prove the following result:

\begin{corollary}\label{corNegateCell}
If $G(M)$ is connected and contains no negative cycles, and $\Msub$
results from
changing the sign of
a single entry of $M$ that is in a cycle (thus making one or more cycles in $G(\Msub)$ negative),
then
$\gr(\Geom(\Msub)) > \gr(\Geom(M))$.
\end{corollary}

In order to do this, we need to consider the structure of $G(\drm)$.
The graph $G(\drm)$ can be constructed from $G(M)$ as follows:
If $G(M)$ has vertex set $\{v_1,\ldots,v_n\}$, then we let $G(\drm)$ have vertices $v_1,\ldots,v_n$ and $v'_1,\ldots,v'_n$.
If $v_iv_j$ is a positive edge in $G(M)$, then in $G(\drm)$ we add an edge between $v_i$ and $v_j$ and also between $v'_i$ and $v'_j$.
On the other hand, if $v_iv_j$ is a negative edge in $G(M)$, then in $G(\drm)$ we join $v_i$ to $v'_j$ and $v'_i$ to $v_j$.
The correctness of this construction follows directly from the definitions of double refinement and of the row-column graph of a matrix.
For an illustration, compare the graph in Figure~\ref{figDoubleRefinement} against that in Figure~\ref{figGeomClass}.

Note that if $v_1,\ldots,v_k$ is a positive
$k$-cycle in $G(M)$, then $G(\drm)$ contains
two vertex-disjoint positive $k$-cycles, the union of whose vertices is $\{v_1,\ldots,v_k,v'_1,\ldots,v'_k\}$.
In contrast, if $v_1,\ldots,v_\ell$ is a negative
$\ell$-cycle in $G(M)$, then $G(\drm)$ contains a (positive) $2\+\ell$-cycle on $\{v_1,\ldots,v_\ell,v'_1,\ldots,v'_\ell\}$ in which $v_i$ is opposite $v'_i$ (i.e. $v'_i$ is at distance $\ell$ from $v_i$ around the cycle) for each $i$, $1\leqslant i\leqslant\ell$.
We make the following additional observations:
\begin{observation}\label{obsEvenx2}
  If $G(M)$ has no odd cycles, then $G(\drm)=G(M)+ G(M)$.
\end{observation}
\begin{observation}\label{obsOddx2}
  If $G(M)$ is connected and has an odd cycle, then $G(\drm)$ is connected.
\end{observation}
We now have all we require to prove our cycle parity result.

\begin{figure}[ht]
$$
\begin{tikzpicture}[scale=0.36]
    \draw [thick] (0,3)--(-2.5,0);
    \draw [thick] (0,3)--(-1.6,0);
    \draw [thick] (0,3)--(-0.7,0);
    \draw [thick] (0,3)--(-3,4);
    \draw [thick] (0,3)--(-3,3);
    \draw [thick] (0,3)--(-3,2);
    \draw [fill=gray!30!white,thick] (-1.7,0) circle [x radius=1.7, y radius=1];
    \node[]at(-1.7,0){$H$};
    \draw [fill=gray!30!white,thick] (-3,3) circle [x radius=1.7, y radius=1.15];
    \node[]at(-3,3){$J$};
    \draw [thick] (0,0)--(0,3);
    \draw [fill] (0,3) circle [radius=0.15];
    \draw [fill] (0,0) circle [radius=0.15];
    \node[above right]at(0,3){$\!\!u$};
    \node[below right]at(0,0){$\!v$};
    \draw [thick] (3,0)--(3.7,3);
    \draw [thick] (3,0)--(4.6,3);
    \draw [thick] (3,0)--(5.5,3);
    \draw [thick] (3,0)--(6,1);
    \draw [thick] (3,0)--(6,0);
    \draw [thick] (3,0)--(6,-1);
    \draw [fill=gray!30!white,thick] (4.7,3) circle [x radius=1.7, y radius=1];
    \node[]at(4.7,3){$H$};
    \draw [fill=gray!30!white,thick] (6,0) circle [x radius=1.7, y radius=1.15];
    \node[]at(6,0){$J$};
    \draw [thick] (3,0)--(3,3);
    \draw [fill] (3,3) circle [radius=0.15];
    \draw [fill] (3,0) circle [radius=0.15];
    \node[above left]at(3,3){$v'\!\!\!$};
    \node[below left]at(3,.27){$u'$};
    \node[]at(1.5,-3.1){$G(\drm)\+=\+G+ G$};
\end{tikzpicture}
\qquad\quad\;\:
\begin{tikzpicture}[scale=0.36]
    \draw [thick] (0,3)--(-2.5,0);
    \draw [thick] (0,3)--(-1.6,0);
    \draw [thick] (0,3)--(-0.7,0);
    \draw [thick] (0,3)--(-3,4);
    \draw [thick] (0,3)--(-3,3);
    \draw [thick] (0,3)--(-3,2);
    \draw [fill=gray!30!white,thick] (-1.7,0) circle [x radius=1.7, y radius=1];
    \node[]at(-1.7,0){$H$};
    \draw [fill=gray!30!white,thick] (3,0) circle [x radius=1.7, y radius=1];
    \node[]at(3,0){$H$};
    \draw [fill=gray!30!white,thick] (-3,3) circle [x radius=1.7, y radius=1.15];
    \node[]at(-3,3){$J$};
    \draw [thick] (0,3)--(1.3,0);
    \draw [fill] (0,3) circle [radius=0.15];
    \draw [fill] (1.3,0) circle [radius=0.15];
    \node[above right]at(0,3){$\!\!u$};
    \node[below left]at(1.3,0){$v\!$};
    \node[]at(0,-3.1){$\phantom{G_1}K\phantom{G_1}$};
\end{tikzpicture}
\qquad\quad\;\:
\begin{tikzpicture}[scale=0.36]
    \draw [fill=gray!10!white] (-5.2,4.65)--(6.9,4.65)--(6.9,1.5)--(1.25,1.5)--(1.25,-1.5)--(-5.2,-1.5)--(-5.2,4.65);
    \draw [thick] (0,3)--(-2.5,0);
    \draw [thick] (0,3)--(-1.6,0);
    \draw [thick] (0,3)--(-0.7,0);
    \draw [thick] (0,3)--(-3,4);
    \draw [thick] (0,3)--(-3,3);
    \draw [thick] (0,3)--(-3,2);
    \draw [fill=gray!30!white,thick] (-1.7,0) circle [x radius=1.7, y radius=1];
    \node[]at(-1.7,0){$H$};
    \draw [fill=gray!30!white,thick] (-3,3) circle [x radius=1.7, y radius=1.15];
    \node[]at(-3,3){$J$};
    \draw [thick] (0,0)--(3,0);
    \draw [fill] (0,3) circle [radius=0.15];
    \draw [fill] (0,0) circle [radius=0.15];
    \node[above right]at(0,3){$\!\!u$};
    \node[below right]at(0,-0.1){$\!v$};
    \draw [thick] (3,0)--(3.7,3);
    \draw [thick] (3,0)--(4.6,3);
    \draw [thick] (3,0)--(5.5,3);
    \draw [thick] (3,0)--(6,1);
    \draw [thick] (3,0)--(6,0);
    \draw [thick] (3,0)--(6,-1);
    \draw [fill=gray!30!white,thick] (4.7,3) circle [x radius=1.7, y radius=1];
    \node[]at(4.7,3){$H$};
    \draw [fill=gray!30!white,thick] (6,0) circle [x radius=1.7, y radius=1.15];
    \node[]at(6,0){$J$};
    \draw [thick] (0,3)--(3,3);
    \draw [fill] (3,3) circle [radius=0.15];
    \draw [fill] (3,0) circle [radius=0.15];
    \node[above left]at(3,3){$v'\!\!\!$};
    \node[below left]at(3,.17){$u'\!$};
    \node[]at(1.5,-3.1){$\Gsub\+=\+G({\Msub}^{\!\times2})$};
\end{tikzpicture}
\vspace{-10.5pt}
$$
\caption{Graphs used in the proof of Corollary~\ref{corNegateCell}}
\label{figNegCycle}
\end{figure}
\begin{proof}[Proof of Corollary~\ref{corNegateCell}]
Let $G=G(M)$ and $\Gsub=G({\Msub}^{\!\times2})$, and let
$uv$ be the edge in $G$ corresponding to the entry in $M$ that is negated to create $\Msub$.
Since $G$
contains no negative cycles, by Observation~\ref{obsEvenx2},
$G(\drm)=G+G$.
Thus,
since $G$ is connected, it has the form at the left of Figure~\ref{figNegCycle}, in which $H$ is the component of $G-u$ containing $v$.
Moreover,
we have
$\gr(\Geom(M))=\lambda(G)$.
(This also follows from Corollary~\ref{corNoNegCycles}.)
Now, if we expand $G$ at $u$ along $uv$, by Lemma~\ref{lemmaExpandVertex},
$\lambda(G)=\lambda(K)$, where $K$ is the graph in the centre of Figure~\ref{figNegCycle}.

On the other hand, $\Gsub$ is obtained from $G(\drm)$ by removing the edges $uv$ and $u'v'$, and adding $uv'$ and $u'v$, as shown at the
right of Figure~\ref{figNegCycle}.
It is readily observed that $K$ is a proper subgraph of $\Gsub$ (see the shaded box in Figure~\ref{figNegCycle}), and hence, by Corollary~\ref{corStrictMono}, $\lambda(K)<\lambda(\Gsub)$. Since $\gr(\Geom(\Msub))=\lambda(\Gsub)$, the result follows.
\end{proof}
Thus, making the first negative cycle increases the growth rate.
We suspect, in fact,
that the following stronger statement is also true:

\begin{conjecture}\label{conjNegateCell}
If $G(M)$ is connected and $\Msub$ results from negating a single entry of $M$
that is in one or more positive cycles but in no negative cycle,
then
we have
$\gr(\Geom(\Msub)) > \gr(\Geom(M))$.
\end{conjecture}

To prove this more general result seems to require some new ideas.
If $G(M)$ already contains a negative cycle, then $G(\drm)$ is connected, and, when this is the case,
there appears to be no obvious way to generate a subgraph of $G({\Msub}^{\!\times2})$
by expanding $G(\drm)$.

% --------------------------------------------------------
\subsection{Monotone grid classes}

In a recent paper~\cite{Bevan2013}, we established the growth rates of \emph{monotone} grid classes. If $M$ is a \opmo{} matrix, then the \emph{monotone grid class} $\Grid(M)$ consists of those permutations that can be plotted as a subset of some figure consisting of the union of \emph{any monotonic curves} $\Gamma_{i,j}$ with the same endpoints as the $L_{i,j}$ in $\Lambda_M$.
This permits greater flexibility in the positioning of points in the cells, so
$\Geom(M)$ is a subset of $\Grid(M)$
and we have $\gr(\Geom(M))\leqslant\gr(\Grid(M))$.
In fact, the geometric grid class
$\Geom(M)$ and the monotone grid class $\Grid(M)$
are identical
if
and only if
$G(M)$
is acyclic
(Theorem~3.2 in~\cite{AABRV2011}).
Hence, if $G(M)$ is a forest, %then
$\gr(\Geom(M))=\gr(\Grid(M))$.
We determined
in~\cite{Bevan2013}
that
the growth rate of monotone grid class $\Grid(M)$ is equal to the square of the spectral radius of $G(M)$.
For acyclic $G(M)$, this is consistent with
the growth rate of the geometric grid class as given by
Corollary~\ref{corForestRho}.

Typically,
the growth rate of a monotone grid class will be greater than
that
of the corresponding geometric grid class.
For example,
if $G(M)$ is a cycle then $\gr(\Grid(M))=4$, whereas from~\eqref{eqGRGeomCycle} we have $\gr(\Geom(M))<4$.
And we have
$$
\gr\!\left(\begin{gridmx}1&\pos0&-1\\ 1&-1&\pos1 \end{gridmx}\right)
\;=\;
\gr\!\left(\begin{gridmx}-1&\pos0&-1\\ \pos1&-1&\pos1 \end{gridmx}\right)
\;=\;
\thalf(5+\sqrt{17})
\;\approx\;
4.56155,
$$
which should be compared with~\eqref{eqGRGeomExNegCyc} and~\eqref{eqGRGeomExPosCyc}.

The fact that the growth rate of the monotone grid class is strictly greater
is a consequence of the fact that, if $G$ is connected and not acyclic, then
$\lambda(G)$ and $\rho(G)$ are distinct:
\begin{lemma}[Godsil \& Gutman~\cite{GG1981}]
\label{lemmaLambdaLessRhoCyclic}
If
$G$ is connected and contains a cycle, then we have
$\lambda(G)<\rho(G)$.
\end{lemma}
\begin{proof}
By Lemma~\ref{lemmaStrictMono},
if $C$ is a nonempty subgraph of $G$, then $\rho(G-C)<\rho(G)$.
So we have $\Phi_{G-C}(z)>0$ for all $z\geqslant\rho(G)$.
Moreover, $\Phi_G(z)\geqslant0$ for $z\geqslant\rho(G)$. So,
since $G$ contains a cycle, from Lemma~\ref{lemmaMuCycleDef} we can deduce that $\mu_G(z)>0$ if $z\geqslant\rho(G)$, and thus
$\lambda(G)<\rho(G)$.
\end{proof}
Note that,
analogously to Observation~\ref{obsGeomDouble},
$\Grid(\drm)=\Grid(M)$.
Hence it must be the case that $\rho(G(\drm))=\rho(G(M))$, the growth rate of a monotone grid class thus being independent of the parity of its cycles.
As a consequence, from Lemma~\ref{lemmaLambdaLessRhoCyclic} we can deduce that in the non-acyclic case
there is a strict inequality between the growth rate of a geometric grid class and the growth rate of the corresponding monotone grid class:

\begin{corollary}
  If $G(M)$ is connected,
  then $\gr(\Geom(M))<\gr(\Grid(M))$
  if and only if $G(M)$ contains a cycle.
\end{corollary}

% --------------------------------------------------------
\subsection{Subdivision of edges}

One particularly surprising result in~\cite{Bevan2013} concerning the growth rates of monotone grid classes is the fact that classes
whose row-column graphs
have longer internal paths or cycles exhibit \emph{lower} growth rates.
An edge $e$ of a graph $G$ is said to lie on an \emph{endpath} of $G$ if $G\deledge e$ is disconnected and one of its components is a (possibly trivial) path. An edge that does not lie on an endpath is said to be \emph{internal}.
The following result of Hoffman \& Smith states that the subdivision of an edge increases or decreases the spectral radius of the graph depending on whether the edge lies on an endpath or is internal:
\begin{lemma}[Hoffman \& Smith~\cite{HS1975}]\label{lemmaRhoSubdivision}
  Let $G$ be a connected graph and $G'$ be obtained from $G$ by subdividing an edge $e$. If $e$ lies on an endpath, then $\rho(G')>\rho(G)$. Otherwise (if $e$ is an internal edge), $\rho(G')\leqslant\rho(G)$, with equality if and only if $G$ is a cycle or has the following form (which we call an ``$H$~graph'':
$$
  \begin{tikzpicture}[scale=0.4]
    \draw [thick,dashed] (0,0)--(4,0);
    \draw [thick] (-.5,.866)--(0,0)--(-.5,-.866);
    \draw [thick] (4.5,.866)--(4,0)--(4.5,-.866);
    \draw [fill] (0,0) circle [radius=0.15];
    \draw [fill] (4,0) circle [radius=0.15];
    \draw [fill] (-.5,.866) circle [radius=0.15];
    \draw [fill] (-.5,-.866) circle [radius=0.15];
    \draw [fill] (4.5,.866) circle [radius=0.15];
    \draw [fill] (4.5,-.866) circle [radius=0.15];
  \end{tikzpicture}
$$
\end{lemma}
Thus for monotone grid classes, if $G(M)$ is connected,
and $G(M')$ is obtained from $G(M)$ by the subdivision of one or more internal edges, then $\gr(\Grid(M'))\leqslant\gr(\Grid(M))$.

As we will see, the situation is not as simple for geometric grid classes.
The effect of edge subdivision on the largest root of the matching polynomial does not seem to have been addressed previously.
In fact, the subdivision of an edge that is in a cycle may
cause $\lambda(G)$
to increase or decrease, or may leave it unchanged.
See Figures~\ref{figGeomSubdivisionIncr}--\ref{figGeomSubdivisionDecr} for illustrations of the three cases.
We investigate this further below.
However, if the edge being subdivided is not on a cycle in $G$, then the behaviour of $\lambda(G)$ mirrors that of $\rho(G)$, as we now demonstrate:
\begin{lemma}\label{lemmaSubdividing1}
  Let $G$ be a connected graph and $G'$ be obtained from $G$ by subdividing an edge $e$.
  If $e$ lies on an endpath, then $\lambda(G')>\lambda(G)$.
  However, if $e$ is an internal edge and not on a cycle, then $\lambda(G')\leqslant\lambda(G)$, with equality if and only if $G$ is an $H$ graph.
\end{lemma}

\begin{figure}[ht]
$$
\raisebox{12pt}{
\begin{tikzpicture}[scale=0.36]
    \draw [thick] (0,.3)--(1.7,1.5);
    \draw [thick] (0,.3)--(2.1,1.1);
    \draw [thick] (0,.3)--(3,0.75);
    \draw [thick,fill=gray!30!white] (3,1.4) circle [x radius=1.3, y radius=1];
    \draw [thick,fill=gray!30!white] (7.2,1.4) circle [x radius=1.2, y radius=.9];
    \draw [fill] (0,.3) circle [radius=0.15];
    \node[left]at(0,.3){$u$};
    \node[left]at(-1.5,.3){$G$:};
    \draw [fill] (1.7,1.5) circle [radius=0.15];
    \draw [fill] (4.3,1.4) circle [radius=0.15];
    \draw [fill] (6,1.4) circle [radius=0.15];
    \draw [ultra thick] (4.3,1.4)--(6,1.4);
    \node[above]at(5.15,1.4){$e$};
    \node[left]at(1.7,1.6){$v$};
    \node[]at(3,1.4){$\,H_1$};
    \node[]at(7.2,1.4){$H_2$};
    \draw [thick] (0,.3)--(3,-0.75);
    \draw [thick] (0,.3)--(2.33,-1);
    \draw [thick] (0,.3)--(2.17,-1.5);
    \draw [thick] (0,.3)--(2,-2);
    \draw [thick,fill=gray!30!white] (3.4,-1.35) circle [x radius=1.75, y radius=1.15];
    \node[]at(3.4,-1.35){$J$};
\end{tikzpicture}
}
\qquad\qquad\qquad
\begin{tikzpicture}[scale=0.36]
    \draw [thick,fill=gray!30!white] (3,3.8) circle [x radius=1.3, y radius=1];
    \draw [thick] (0,.3)--(1.7,3.9);
    \draw [thick] (0,.3)--(2.1,1.1);
    \draw [thick] (0,.3)--(3,0.75);
    \draw [thick,fill=gray!30!white] (3,1.4) circle [x radius=1.3, y radius=1];
    \draw [fill] (0,.3) circle [radius=0.15];
    \node[left]at(0,.3){$u$};
    \node[left]at(-1.5,1.4){$K$:};
    \draw [fill] (1.7,3.9) circle [radius=0.15];
    %
    \draw [thick,fill=gray!30!white] (7.2,3.8) circle [x radius=1.2, y radius=.9];
    \draw [fill] (4.3,3.8) circle [radius=0.15];
    \draw [fill] (6,3.8) circle [radius=0.15];
    \draw [ultra thick] (4.3,3.8)--(6,3.8);
    \node[above]at(5.15,3.8){$e_1$};
    %
    \draw [thick,fill=gray!30!white] (7.2,1.4) circle [x radius=1.2, y radius=.9];
    \draw [fill] (4.3,1.4) circle [radius=0.15];
    \draw [fill] (6,1.4) circle [radius=0.15];
    \draw [ultra thick] (4.3,1.4)--(6,1.4);
    \node[above]at(5.15,1.4){$e_2$};
    %
    \node[left]at(1.7,4){$v$};
    \node[]at(3,1.4){$\,H_1$};
    \node[]at(3,3.8){$\,H_1$};
    \node[]at(7.2,1.4){$H_2$};
    \node[]at(7.2,3.8){$H_2$};
    \draw [thick] (0,.3)--(3,-0.75);
    \draw [thick] (0,.3)--(2.33,-1);
    \draw [thick] (0,.3)--(2.17,-1.5);
    \draw [thick] (0,.3)--(2,-2);
    \draw [thick,fill=gray!30!white] (3.4,-1.35) circle [x radius=1.75, y radius=1.15];
    \node[]at(3.4,-1.35){$J$};
\end{tikzpicture}
\qquad
$$
\caption{Graphs used in the proof of Lemma~\ref{lemmaSubdividing1}}
\label{figExpandingVertexEInternal}
\end{figure}
\begin{proof}
If $e$ lies on an endpath, then $G$ is a proper subgraph of $G'$ and so
the result follows from Corollary~\ref{corStrictMono}.
On the other hand, if $e$ is internal and $G$ is acyclic, the conclusion is a consequence of Corollary~\ref{corLambdaRhoTree} and Lemma~\ref{lemmaRhoSubdivision}.
Thus, we need only consider the situation in which
$e$ is internal and $G$ contains a cycle. We proceed by induction on the number of cycles in $G$, acyclic graphs constituting the base case.
Let $uv$ be an edge in a cycle of $G$ such that $u$ is not an endvertex of~$e$.
Now, let $K$ be the result of expanding $G$ at $u$ along $uv$, and let $K'$, analogously, be the result of expanding $G'$ at $u$ along~$uv$.

We consider the effect of the expansion of $G$ upon $e$ and the effect of the expansion of $G'$ upon the two edges resulting from the subdivision of $e$.
If $e$ is in the component of $G-u$ containing~$v$, then $e$ is duplicated in $K$, both copies of $e$ remaining internal (see Figure~\ref{figExpandingVertexEInternal}).
Moreover, $K'$ results from subdividing both copies of $e$ in $K$.
Conversely, if $e$ is in a component of $G-u$ not containing~$v$, then $e$ is not duplicated in $K$ (and remains internal).
In this case, $K'$ results from subdividing $e$ in $K$.
In either case, $K'$ is the result of subdividing internal edges of $K$ (a graph with fewer cycles than $G$), and so
the result follows from the induction hypothesis.
\end{proof}
Now,
the subdivision of an edge
of a row-column graph
that is not on a cycle has no effect on the parity of the cycles.
Hence,
we
have the following conclusion
for the growth rates of geometric grid classes:

\begin{corollary}
If $G(M)$ is connected,
and $G(M')$ is obtained from $G(M)$ by the subdivision of one or more internal edges not on a cycle, then $\gr(\Geom(M'))\leqslant\gr(\Geom(M))$, with equality if and only if $G(M)$ is an $H$ graph.
\end{corollary}

\begin{figure}[ht]
$$
\raisebox{34.2pt}{
\begin{tikzpicture}[scale=0.36]
    \draw [thick] (0,3)--(-3,4);
    \draw [thick] (0,3)--(-3,3);
    \draw [thick] (0,3)--(-3,2);
    \draw [thick] (0,3)--(1.3,4.7);
    \draw [thick] (0,3)--(1.662,4.126);
    \draw [thick] (0,3)--(2.3,3.834);
    \draw [thick] (0,3)--(2.4,2.1);
    \draw [thick] (0,3)--(2.1,1.7);
    \draw [thick] (0,3)--(1.8,1.3);
    \draw [thick] (0,3)--(1.5,0.9);
    \draw [ultra thick] (5.3,1.3)--(5.3,4.7);
    \draw [fill=gray!30!white,thick] (3.3,4.7) circle [x radius=2, y radius=1];
    \node[]at(3.3,4.7){$\phantom{{}_.}H_1$};
    \draw [fill=gray!30!white,thick] (3.3,1.3) circle [x radius=2, y radius=1];
    \node[]at(3.3,1.3){$\phantom{{}_.}H_2$};
    \draw [fill=gray!30!white,thick] (-3,3) circle [x radius=1.7, y radius=1.1];
    \node[]at(-3,3){$J$};
    \draw [fill] (0,3) circle [radius=0.15];
    \draw [fill] (1.3,4.7) circle [radius=0.15];
    \draw [fill] (1.662,4.126) circle [radius=0.15];
    \draw [fill] (2.3,3.834) circle [radius=0.15];
    \draw [fill] (5.3,4.7) circle [radius=0.15];
    \draw [fill] (5.3,1.3) circle [radius=0.15];
    \node[above left]at(0,3.1){$u\!\!\!$};
    \node[right]at(5.3,3){$e$};
    \node[right]at(5.3,4.8){$x_1$};
    \node[right]at(5.3,1.2){$x_2$};
    \node[left]at(-5.4,3){$G$:};
    \node[below left]at(0,1.3){$\phantom{x_1}$};
\end{tikzpicture}
}
\qquad\qquad\qquad
\begin{tikzpicture}[scale=0.36]
    \draw [thick] (0,3)--(1.3,9.4);
    \draw [ultra thick] (5.3,9.4)--(7.7,9.4);
    \draw [fill=gray!30!white,thick] (3.3,9.4) circle [x radius=2, y radius=1];
    \node[]at(3.3,9.4){$\phantom{{}_.}H_1$};
    \draw [fill=gray!30!white,thick] (9.7,9.4) circle [x radius=2, y radius=1];
    \node[]at(9.7,9.4){$\phantom{{}_.}H_2$};
    \draw [fill] (1.3,9.4) circle [radius=0.15];
    \draw [fill] (5.3,9.4) circle [radius=0.15];
    \draw [fill] (7.7,9.4) circle [radius=0.15];
    \node[below left]at(8.2,9.4){$x_2$};
%
    \draw [thick] (0,3)--(1.662,6.126);
    \draw [ultra thick] (5.3,6.7)--(7.7,6.7);
    \draw [fill=gray!30!white,thick] (3.3,6.7) circle [x radius=2, y radius=1];
    \node[]at(3.3,6.7){$\phantom{{}_.}H_1$};
    \draw [fill=gray!30!white,thick] (9.7,6.7) circle [x radius=2, y radius=1];
    \node[]at(9.7,6.7){$\phantom{{}_.}H_2$};
    \draw [fill] (1.662,6.126) circle [radius=0.15];
    \draw [fill] (5.3,6.7) circle [radius=0.15];
    \draw [fill] (7.7,6.7) circle [radius=0.15];
    \node[below left]at(8.2,6.7){$x_2$};
%
    \draw [thick] (0,3)--(2.3,3.134);
    \draw [ultra thick] (5.3,4)--(7.7,4);
    \draw [fill=gray!30!white,thick] (3.3,4) circle [x radius=2, y radius=1];
    \node[]at(3.3,4){$\phantom{{}_.}H_1$};
    \draw [fill=gray!30!white,thick] (9.7,4) circle [x radius=2, y radius=1];
    \node[]at(9.7,4){$\phantom{{}_.}H_2$};
    \draw [fill] (2.3,3.134) circle [radius=0.15];
    \draw [fill] (5.3,4) circle [radius=0.15];
    \draw [fill] (7.7,4) circle [radius=0.15];
    \node[below left]at(8.2,4){$x_2$};
%
    \draw [thick] (0,3)--(-3,4);
    \draw [thick] (0,3)--(-3,3);
    \draw [thick] (0,3)--(-3,2);
    \draw [thick] (0,3)--(2.4,2.1);
    \draw [thick] (0,3)--(2.1,1.7);
    \draw [thick] (0,3)--(1.8,1.3);
    \draw [thick] (0,3)--(1.5,0.9);
    \draw [ultra thick] (5.3,1.3)--(7.7,1.3);
    \draw [fill=gray!30!white,thick] (3.3,1.3) circle [x radius=2, y radius=1];
    \node[]at(3.3,1.3){$\phantom{{}_.}H_2$};
    \draw [fill=gray!30!white,thick] (9.7,1.3) circle [x radius=2, y radius=1];
    \node[]at(9.7,1.3){$\phantom{{}_.}H_1$};
    \draw [fill=gray!30!white,thick] (-3,3) circle [x radius=1.7, y radius=1.1];
    \node[]at(-3,3){$J$};
    \draw [fill] (0,3) circle [radius=0.15];
    \draw [fill] (5.3,1.3) circle [radius=0.15];
    \draw [fill] (7.7,1.3) circle [radius=0.15];
    \node[above left]at(0,3.1){$u\!\:\!\!$};
    \node[below left]at(8.2,1.3){$x_1$};
    \node[left]at(-5.4,6.4){$K$:};
\end{tikzpicture}
$$
\caption{Graphs used in Lemma~\ref{lemmaSubdividing2}}
\label{figExpandingCycle}
\end{figure}
Let us now investigate the effect of subdividing an edge $e$ that lies on a cycle.
We restrict our attention to graphs
in which there is a vertex $u$ such that
the two endvertices of $e$ are in distinct
components of $(G\deledge e) - u$.
See the graph at the left of Figure~\ref{figExpandingCycle} for an illustration.
We leave the consideration of multiply-connected graphs that fail to satisfy this condition for future study.
\begin{lemma}\label{lemmaSubdividing2}
Let $G$ be a connected graph and $e=x_1x_2$ an edge on a cycle $C$ of $G$.
Let $u$ be a vertex on $C$, and let $H_1$ and $H_2$ be the distinct
components of $(G\deledge e) - u$ that contain $x_1$ and $x_2$ respectively.
Finally, let $G'$ be the graph obtained from $G$ by subdividing $e$.
  \begin{enumerate}[(a)]
    \item If, for $i\in\{1,2\}$, $H_i$ is a (possibly trivial) path of which $x_i$ is an endvertex, then $\lambda(G')>\lambda(G)$.
    \item If, for $i\in\{1,2\}$, $H_i$ is not a path or is a path of which $x_i$ is not an endvertex, then $\lambda(G')<\lambda(G)$.
  \end{enumerate}
\end{lemma}
\begin{proof}
Let $K$ be the result of repeatedly expanding $G$ at $u$ along every edge joining $u$ to $H_1$.
$K$~has the form shown at the right of Figure~\ref{figExpandingCycle}.
Also let $K'$ be the result of repeatedly expanding $G'$ ($G$~with edge $e$ subdivided) in an analogous way at $u$.
Clearly $K'$ is the same as the
graph that results from subdividing the
copies of $e$
in $K$.

Now, for part (a), since $H_1$ is a path with an end at $x_1$, and also $H_2$ is a path with an end at $x_2$, we see that $K'$ is the result of subdividing edges of $K$ that are on endpaths.
Hence, by the first part of Lemma~\ref{lemmaSubdividing1}, we have $\lambda(G')>\lambda(G)$ as required.

For part (b), since $H_1$ is not a path with an end at $x_1$, and nor is $H_2$ a path with an end at $x_2$,
we see that $K'$ is the result of subdividing internal edges of $K$.
Since $K$ is not an $H$ graph, by Lemma~\ref{lemmaSubdividing1}, we have $\lambda(G')<\lambda(G)$ as required.
\end{proof}
If the conditions for parts (a) and (b) of this lemma both fail to be satisfied (i.e. $H_1$ is a suitable path and $H_2$ isn't, or \emph{vice versa}), then
the proof fails.
This is due to the fact that
expansion leads to at least one copy of $e$ in $K$ being internal
and to another copy of $e$ in $K$ being on an endpath.
Subdivision of the former decreases $\lambda(G)$ whereas subdivision of the latter causes it to increase.
Sometimes, as in Figure~\ref{figGeomSubdivisionEq}, these effects balance exactly; on other occasions one or the other dominates.
We leave a detailed analysis of such cases for later study.

\begin{figure}[t]
    $$
    \gctwo{4}{1,1,1}{0,1,1,1}
    \quad
    \raisebox{-0.09in}{\begin{tikzpicture}[scale=0.4]
    \draw [thick] (0,0)--(1,0)--(1.707,0.707)--(2.414,0)--(1.707,-0.707)--(1,0);
    \draw [thick] (2.414,0)--(3.414,0);
    \draw [fill] (0,0) circle [radius=0.15];
    \draw [fill] (1,0) circle [radius=0.15];
    \draw [fill] (2.414,0) circle [radius=0.15];
    \draw [fill] (3.414,0) circle [radius=0.15];
    \draw [fill] (1.707,0.707) circle [radius=0.15];
    \draw [fill] (1.707,-0.707) circle [radius=0.15];
    \end{tikzpicture}}
    \quad\quad\quad\quad
    \gcthree{5}{1,1,0,1}{0,1,1}{0,0,1,1,1}
    \quad
    \raisebox{-0.13in}{\begin{tikzpicture}[scale=0.4,rotate=30]
    \draw [thick] (0,0)--(1,0)--(1.5,0.866)--(2.5,0.866)--(3,0)--(2.5,-0.866)--(1.5,-0.866)--(1,0);
    \draw [thick] (2.5,-0.866)--(3,-1.732);
    \draw [fill] (0,0) circle [radius=0.15];
    \draw [fill] (1,0) circle [radius=0.15];
    \draw [fill] (3,0) circle [radius=0.15];
    \draw [fill] (3,-1.732) circle [radius=0.15];
    \draw [fill] (1.5,0.866) circle [radius=0.15];
    \draw [fill] (2.5,0.866) circle [radius=0.15];
    \draw [fill] (1.5,-0.866) circle [radius=0.15];
    \draw [fill] (2.5,-0.866) circle [radius=0.15];
    \end{tikzpicture}}
    \quad\quad\quad\quad
    \gcfour{6}{1,1,0,0,1}{0,1,1}{0,0,1,1}{0,0,0,1,1,1}
    \quad
    \raisebox{-0.22in}{\begin{tikzpicture}[scale=0.4,rotate=45]
    \draw [thick] (0,0)--(1,0)--(1.383,0.924)--(2.307,1.307)--(3.23,0.924)--(3.613,0)--(3.23,-0.924)--(2.307,-1.307)--(1.383,-0.924)--(1,0);
    \draw [thick] (2.307,-1.307)--(2.307,-2.307);
    \draw [fill] (0,0) circle [radius=0.15];
    \draw [fill] (1,0) circle [radius=0.15];
    \draw [fill] (3.613,0) circle [radius=0.15];
    \draw [fill] (1.383,0.924) circle [radius=0.15];
    \draw [fill] (1.383,-0.924) circle [radius=0.15];
    \draw [fill] (3.23,0.924) circle [radius=0.15];
    \draw [fill] (3.23,-0.924) circle [radius=0.15];
    \draw [fill] (2.307,1.307) circle [radius=0.15];
    \draw [fill] (2.307,-1.307) circle [radius=0.15];
    \draw [fill] (2.307,-2.307) circle [radius=0.15];
    \end{tikzpicture}}
    \vspace{-9pt}
    $$
\caption{Standard figures and row-column graphs of geometric grid classes whose growth rates increase
from left to right}
\label{figGeomSubdivisionIncr}
    $$
    \gctwo{5}{1,1,1,1}{0,0,1,1,1}
    \quad
    \raisebox{-0.09in}{\begin{tikzpicture}[scale=0.4]
    \draw [thick] (0.293,0.707)--(1,0)--(1.707,0.707)--(2.414,0)--(1.707,-0.707)--(1,0);
    \draw [thick] (0.293,-0.707)--(1,0);
    \draw [thick] (2.414,0)--(3.414,0);
    \draw [fill] (0.293,0.707) circle [radius=0.15];
    \draw [fill] (0.293,-0.707) circle [radius=0.15];
    \draw [fill] (1,0) circle [radius=0.15];
    \draw [fill] (2.414,0) circle [radius=0.15];
    \draw [fill] (3.414,0) circle [radius=0.15];
    \draw [fill] (1.707,0.707) circle [radius=0.15];
    \draw [fill] (1.707,-0.707) circle [radius=0.15];
    \end{tikzpicture}}
    \quad\quad\quad\:\;
    \gcthree{6}{1,1,1,0,1}{0,0,1,1}{0,0,0,1,1,1}
    \quad
    \raisebox{-0.16in}{\begin{tikzpicture}[scale=0.4,rotate=30]
    \draw [thick] (0.234,0.643)--(1,0)--(1.5,0.866)--(2.5,0.866)--(3,0)--(2.5,-0.866)--(1.5,-0.866)--(1,0);
    \draw [thick] (2.5,-0.866)--(3,-1.732);
    \draw [thick] (0.234,-0.643)--(1,0);
    \draw [fill] (0.234,0.643) circle [radius=0.15];
    \draw [fill] (0.234,-0.643) circle [radius=0.15];
    \draw [fill] (1,0) circle [radius=0.15];
    \draw [fill] (3,0) circle [radius=0.15];
    \draw [fill] (1.5,0.866) circle [radius=0.15];
    \draw [fill] (2.5,0.866) circle [radius=0.15];
    \draw [fill] (3,-1.732) circle [radius=0.15];
    \draw [fill] (1.5,-0.866) circle [radius=0.15];
    \draw [fill] (2.5,-0.866) circle [radius=0.15];
    \end{tikzpicture}}
    \quad\quad\quad\:\;
    \gcfour{7}{1,1,1,0,0,1}{0,0,1,1}{0,0,0,1,1}{0,0,0,0,1,1,1}
    \quad
    \raisebox{-0.23in}{\begin{tikzpicture}[scale=0.4,rotate=45]
    \draw [thick] (0.207,0.609)--(1,0)--(1.383,0.924)--(2.307,1.307)--(3.23,0.924)--(3.613,0) --(3.23,-0.924)--(2.307,-1.307)--(1.383,-0.924)--(1,0);
    \draw [thick] (2.307,-1.307)--(2.307,-2.307);
    \draw [thick] (0.207,-0.609)--(1,0);
    \draw [fill] (0.207,0.609) circle [radius=0.15];
    \draw [fill] (0.207,-0.609) circle [radius=0.15];
    \draw [fill] (1,0) circle [radius=0.15];
    \draw [fill] (3.613,0) circle [radius=0.15];
    \draw [fill] (1.383,0.924) circle [radius=0.15];
    \draw [fill] (1.383,-0.924) circle [radius=0.15];
    \draw [fill] (3.23,0.924) circle [radius=0.15];
    \draw [fill] (3.23,-0.924) circle [radius=0.15];
    \draw [fill] (2.307,1.307) circle [radius=0.15];
    \draw [fill] (2.307,-1.307) circle [radius=0.15];
    \draw [fill] (2.307,-2.307) circle [radius=0.15];
    \end{tikzpicture}}
    \vspace{-9pt}
    $$
\caption{Standard figures and row-column graphs of geometric grid classes whose growth rates are all the same
(equal to 5)
}
\label{figGeomSubdivisionEq}
    $$
    \gctwo{6}{1,1,1,1}{0,0,1,1,1,1}
    \quad
    \raisebox{-0.09in}{\begin{tikzpicture}[scale=0.4]
    \draw [thick] (0.293,0.707)--(1,0)--(1.707,0.707)--(2.414,0)--(1.707,-0.707)--(1,0);
    \draw [thick] (0.293,-0.707)--(1,0);
    \draw [thick] (3.121,0.707)--(2.414,0)--(3.121,-0.707);
    \draw [fill] (0.293,0.707) circle [radius=0.15];
    \draw [fill] (0.293,-0.707) circle [radius=0.15];
    \draw [fill] (1,0) circle [radius=0.15];
    \draw [fill] (2.414,0) circle [radius=0.15];
    \draw [fill] (3.121,0.707) circle [radius=0.15];
    \draw [fill] (3.121,-0.707) circle [radius=0.15];
    \draw [fill] (1.707,0.707) circle [radius=0.15];
    \draw [fill] (1.707,-0.707) circle [radius=0.15];
    \end{tikzpicture}}
    \quad\quad\quad
    \gcthree{7}{1,1,1,0,1}{0,0,1,1}{0,0,0,1,1,1,1}
    \quad
    \raisebox{-0.16in}{\begin{tikzpicture}[scale=0.4,rotate=30]
    \draw [thick] (0.234,0.643)--(1,0)--(1.5,0.866)--(2.5,0.866)--(3,0)--(2.5,-0.866)--(1.5,-0.866)--(1,0);
    \draw [thick] (2.326,-1.851)--(2.5,-0.866)--(3.44,-1.208);
    \draw [thick] (0.234,-0.643)--(1,0);
    \draw [fill] (0.234,0.643) circle [radius=0.15];
    \draw [fill] (0.234,-0.643) circle [radius=0.15];
    \draw [fill] (1,0) circle [radius=0.15];
    \draw [fill] (3,0) circle [radius=0.15];
    \draw [fill] (1.5,0.866) circle [radius=0.15];
    \draw [fill] (2.5,0.866) circle [radius=0.15];
    \draw [fill] (2.326,-1.851) circle [radius=0.15];
    \draw [fill] (3.44,-1.208) circle [radius=0.15];
    \draw [fill] (1.5,-0.866) circle [radius=0.15];
    \draw [fill] (2.5,-0.866) circle [radius=0.15];
    \end{tikzpicture}}
    \quad\quad\quad
    \gcfour{8}{1,1,1,0,0,1}{0,0,1,1}{0,0,0,1,1}{0,0,0,0,1,1,1,1}
    \quad
    \raisebox{-0.23in}{\begin{tikzpicture}[scale=0.4,rotate=45]
    \draw [thick] (0.207,0.609)--(1,0)--(1.383,0.924)--(2.307,1.307)--(3.23,0.924)--(3.613,0) --(3.23,-0.924)--(2.307,-1.307)--(1.383,-0.924)--(1,0);
    \draw [thick] (1.698,-2.1)--(2.307,-1.307)--(2.916,-2.1);
    \draw [thick] (0.207,-0.609)--(1,0);
    \draw [fill] (0.207,0.609) circle [radius=0.15];
    \draw [fill] (0.207,-0.609) circle [radius=0.15];
    \draw [fill] (1,0) circle [radius=0.15];
    \draw [fill] (3.613,0) circle [radius=0.15];
    \draw [fill] (1.383,0.924) circle [radius=0.15];
    \draw [fill] (1.383,-0.924) circle [radius=0.15];
    \draw [fill] (3.23,0.924) circle [radius=0.15];
    \draw [fill] (3.23,-0.924) circle [radius=0.15];
    \draw [fill] (2.307,1.307) circle [radius=0.15];
    \draw [fill] (2.307,-1.307) circle [radius=0.15];
    \draw [fill] (1.698,-2.1) circle [radius=0.15];
    \draw [fill] (2.916,-2.1) circle [radius=0.15];
    \end{tikzpicture}}
    \vspace{-9pt}
    $$
\caption{Standard figures and row-column graphs of geometric grid classes whose growth rates decrease
from left to right}
\label{figGeomSubdivisionDecr}
\end{figure}
To conclude, we state the consequent result for the growth rates of geometric grid classes.
To simplify its statement and avoid having to concern ourselves directly with cycle parities, we
define $\grm(M)$ to be $G(M)$ when $G(M)$ has no odd cycles and $\grm(M)$ to be $G(\drm)$ otherwise.

\begin{corollary}
  Suppose $\grm(M)$ is connected.
  \begin{enumerate}[(a)]
    \item If $\grm(M')$ is obtained from $\grm(M)$ by subdividing one or more edges that satisfy the conditions of part (a) of Lemma~\ref{lemmaSubdividing2}, then $\gr(\Geom(M'))>\gr(\Geom(M))$.
    \item If $\grm(M')$ is obtained from $\grm(M)$ by subdividing one or more edges that satisfy the conditions of part (b) of Lemma~\ref{lemmaSubdividing2}, then $\gr(\Geom(M'))<\gr(\Geom(M))$.
  \end{enumerate}
\end{corollary}

Figure~\ref{figGeomSubdivisionIncr} provides an illustration of part (a) and Figure~\ref{figGeomSubdivisionDecr} an illustration of part (b).

% ================================================================
\subsection*{Acknowledgements}
The author would like to thank Penny Bevan
and Robert Brignall
for reading earlier drafts of this paper.
Their
feedback led to significant improvements in its presentation.
Thanks are also due to an anonymous referee whose comments resulted in a much expanded final section,
and to Brendan McKay, whose helpful response to a question on \emph{MathOverflow}~\cite{McKay2014} alerted the author to the significance of Godsil's Lemma~\ref{lemmaExpandVertex}.
\emph{S.D.G.}

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\end{document}

