# Almost sure GOE fluctuations of energy levels for hyperbolic surfaces of
high genus

Zeév Rudnick and Igor Wigman  School of Mathematical Sciences, Tel Aviv
University, Tel Aviv 69978, Israel<EMAIL_ADDRESS>Department of
Mathematics, King’s College London, UK<EMAIL_ADDRESS>
###### Abstract.

We study the variance of a linear statistic of the Laplace eigenvalues on a
hyperbolic surface, when the surface varies over the moduli space of all
surfaces of fixed genus, sampled at random according to the Weil-Petersson
measure. The ensemble variance of the linear statistic was recently shown to
coincide with that of the corresponding statistic in the Gaussian Orthogonal
Ensemble (GOE) of random matrix theory, in the double limit of first taking
large genus and then shrinking size of the energy window. In this note we show
that in this same limit, the energy variance for a typical surface is close to
the GOE result, a feature called “ergodicity” in the random matrix theory
literature.

This research was supported by the European Research Council (ERC) under the
European Union’s Horizon 2020 research and innovation programme (grant
agreement No. 786758) and by the Israel Science Foundation (grant No.
1881/20).

##  1\. Introduction

###  1.1. Statement of results

Let $X$ be a compact hyperbolic surface of genus $g\geq 2$,
$\lambda_{j}=1/4+r_{j}^{2}$ the Laplace eigenvalues on $X$. Let $f$ be a test
function so that its Fourier transform $\widehat{f}$ is even, smooth and
compactly supported, $L\geq 1$, $\tau>0$, and define the smooth linear
statistic

$N_{L,\tau}(X):=\sum\limits_{j\geq 0}f(L(r_{j}-\tau))+f(L(r_{j}+\tau)),$

effectively counting eigenvalues in a window of size $\approx\tau/L$ around
$\tau^{2}$, as in [10, 11, 7, 12]. We write $N_{f,L,\tau}(X)$ as a sum of a
smooth and fluctuating terms

$N_{L,\tau}(X)=\overline{N}+\widetilde{N}_{L,\tau}(X),$

where

$\overline{N}:=2(g-1)\int\limits_{-\infty}^{\infty}f(L(r-\tau))r\tanh(\pi
r)dr\sim\frac{2\tau(g-1)}{L}\int\limits_{-\infty}^{\infty}f(x)dx.$

In [11] it was shown that when averaged over the moduli space
$\mathcal{M}_{g}$ of surfaces of genus $g$ with respect to the Weil-Petersson
measure, the variance of $\widetilde{N}_{L,\tau}(X)$ matches that of the
corresponding statistic in the Gaussian Orthogonal Ensemble (GOE) of random
matrix theory, in the double limit, taking $g\to\infty$, and then
$L\to\infty$:

(1.1) $\lim_{L\to\infty}\lim_{g\to\infty}\mathbb{E}^{\rm
WP}_{g}\left(\left|\widetilde{N}_{L,\tau}-\mathbb{E}^{\rm
WP}_{g}(\widetilde{N}_{L,\tau})\right|^{2}\right)=\Sigma^{2}_{\rm GOE}(f)$

where $\Sigma^{2}_{\rm
GOE}(f)=2\int_{-\infty}^{\infty}|x|\widehat{f}(x)^{2}dx$.

We now consider the energy average. Let $\omega$ be a non-negative, even
weight function, normalized by $\int_{-\infty}^{\infty}\omega(x)dx=1$, with
Fourier transform $\widehat{\omega}$ smooth and supported in $[-1,1]$. For
$T>0$ define an averaging operator

(1.2)
$\operatorname{\mathbb{E}}_{T}[F]:=\frac{1}{T}\int_{-\infty}^{\infty}F(\tau)\omega\left(\frac{\tau}{T}\right)dt$

and a corresponding variance

(1.3)
$\operatorname{Var}_{T}(F)=\operatorname{\mathbb{E}}_{T}\left[\left|F-\operatorname{\mathbb{E}}_{T}(F)\right|^{2}\right].$

Denote by $\mathcal{V}_{T,L}(X)$ the energy variance of
$\widetilde{N}_{L,\tau}(X)$, thought of as a random variable on
$\mathcal{M}_{g}$:

$\mathcal{V}_{T,L}(X):=\operatorname{Var}_{T}(\widetilde{N}_{L,\tau}(X)).$

Following Berry [1], it is believed that for generic surfaces111Arithmetic
surfaces are exceptional, see [2, 6]. $X\in\mathcal{M}_{g}$, for fixed
genus222Genus $g=2$ requires a modification due to the presence of the
hyperelliptic involution. $g>2$, the energy variance $\mathcal{V}_{T,L}(X)$
will converge to the GOE variance $\Sigma^{2}_{\rm GOE}(f)$ when $T\to\infty$,
and $L\to\infty$ but $L=o(T)$. Our principal result supports this if we first
take the large genus limit $g\to\infty$, and then the high energy limit
$T\to\infty$, while restricting to energy windows $L\to\infty$ with $L=o(\log
T)$, more precisely that in this limit the random variable
$\mathcal{V}_{T,L}(X)$ converges in distribution to the constant
$\Sigma^{2}_{\rm GOE}(f)$:

######  Theorem 1.1.

For every $\epsilon>0$,

(1.4) $\lim\limits_{\begin{subarray}{c}L,T\rightarrow\infty\\\
L=o(\log{T})\end{subarray}}\limsup\limits_{g\rightarrow\infty}\operatorname{Prob}^{WP}_{g}\left(\left|\mathcal{V}_{L,T}-\Sigma_{GOE}^{2}(f)\right|>\epsilon\right)=0.$

This may be viewed as an analogue of a feature called “ergodicity” in random
matrix theory: In the limit of large matrix size, the energy variance equals
the ensemble variance for almost all matrices, see [9].

###  1.2. Method of proof

Using the Selberg trace formula, we have [11]

$N_{L,\tau}(X)=\overline{N}+\widetilde{N}_{L,\tau}(X)$

with

$\overline{N}=2(g-1)\int\limits_{-\infty}^{\infty}f(L(r-\tau))r\tanh(\pi
r)dr\sim\frac{2\tau(g-1)}{L}\int\limits_{-\infty}^{\infty}f(x)dx,$

and

(1.5)
$\widetilde{N}_{L,\tau}(X)=\frac{2}{L}\sum\limits_{\ell_{\gamma}\in\mathcal{L}_{g}(X)}\sum\limits_{k\geq
1}\frac{\ell_{\gamma}}{\sinh(k\ell_{\gamma}/2)}\widehat{f}\left(\frac{k\ell_{\gamma}}{L}\right)\cos(\tau
k\ell_{\gamma})$

where the summation is over the primitive length spectrum

$\mathcal{L}_{g}(X)\subset(0,\infty),$

the set of lengths of primitive non-oriented closed geodesics of $X$.

We view the primitive length spectrum as a random point set $\mathcal{L}_{g}$,
parameterized by the random variable $X\in\mathcal{M}_{g}$, that is a random
point process, and the variance $\mathcal{V}_{T,L}(X)$ is an application
$\varphi(\mathcal{L}_{g})$ of a functional $\varphi$ on $\mathcal{L}_{g}$,
which is continuous in a suitable topology. A fundamental result of Mirzakhani
and Petri [8] shows that the point processes $\mathcal{L}_{g}$ converge, in
distribution as $g\to\infty$, to a Poisson point process, denoted
$\mathcal{L}_{\infty}$ on the positive reals, with intensity measure

$d\nu_{MP}(x)=\frac{2\sinh^{2}x}{x}dx.$

From the theory of point processes, it follows that
$\mathcal{V}_{T,L}(X)=\varphi(\mathcal{L}_{g})$ converges, in distribution as
$g\rightarrow\infty$ to the random variable
$\mathcal{V}_{T,L}^{\infty}=\varphi(\mathcal{L}_{\infty})$, see Proposition
2.2. To show Theorem 1.1, we argue that it suffices to prove that
$\mathcal{V}_{T,L}^{\infty}$, converges in distribution to the constant
$\Sigma^{2}_{\rm GOE}(f)$ as $L,T\to\infty$ with $L=o(\log T)$ (see Theorem
2.3). The proof of the latter result takes up the bulk of this paper, and is
done in § 4.

##  2\. Outline of the proof of the main result

We rewrite (1.5) as

(2.1)
$\widetilde{N}_{L,\tau}(X)=\sum_{\ell\in\mathcal{L}_{g}}H_{L,\tau}(\ell),$

where

(2.2) $H_{L,\tau}(x)=2\frac{x}{L}\sum_{k\geq
1}\frac{\widehat{f}(\frac{kx}{L})\cos(kx\tau)}{\sinh(kx/2)}$

that we think of as a functional of the length spectrum, that is considered as
a random point process, as in the following notation.

######  Notation 2.1.

  1. i.)

For $g\geq 1$ let $\mathcal{L}_{g}=\mathcal{L}_{g}(X)=\\{\ell_{j}\\}_{j\geq
1}$ be the (random) primitive unoriented length spectrum of
$X\in\mathcal{M}_{g}$, considered as a random point process on
${\mathbb{R}}_{>0}$.

  2. ii.)

Let $\mathcal{L}_{\infty}=\\{\ell_{j}\\}_{j\geq 1}$ be the Poisson point
process of intensity

(2.3) $\nu_{MP}(dt):=\frac{2\sinh(t/2)^{2}}{t}dt.$

It was proved by Mirzakhani-Petri [8], that $\mathcal{L}_{g}$ converges, as a
sequence of point processes, to $\mathcal{L}_{\infty}$, see § 3 below. It is
then natural to consider the analogue of $\widetilde{N}_{L,\tau}(X)$, with
$\mathcal{L}_{g}$ replaced by $\mathcal{L}_{\infty}$. That is, define the
random variable

(2.4) $S_{L,\tau}=\sum_{\ell\in\mathcal{L}_{\infty}}H_{L,\tau}(\ell),$

where $H_{L,\tau}(x)$ is given by (2.2). We recall the averaging operator
$\operatorname{\mathbb{E}}_{T}[\cdot]$, as in (1.2), and apply the
corresponding variance operator $\operatorname{Var}_{T}(\cdot)$ as in (1.3),
on $S_{L,\tau}$:

$\mathcal{V}_{T,L}^{\infty}:=\operatorname{Var}_{T}(S_{L,\tau})=\operatorname{\mathbb{E}}_{T}\left[\left|N_{L,\tau}(X)-\operatorname{\mathbb{E}}_{T}\left(N_{L,\tau}\left(X\right)\right|\right)^{2}\right].$

We aim to express $\mathcal{V}_{T,L}^{\infty}$ directly in terms of
$\mathcal{L}_{\infty}$.

To this end, squaring (2.4), after some simple manipulations with the
resulting expression, gives

(2.5)
$\mathcal{V}_{T,L}^{\infty}=\left(\frac{2}{L}\right)^{2}\sum_{\ell_{1},\ell_{2}\in\mathcal{L}_{\infty}}\sum_{k_{1},k_{2}\geq
1}\frac{H_{L}(k_{1}\ell_{1})H_{L}(k_{2}\ell_{2})}{k_{1}k_{2}}U_{T}(k_{1}\ell_{1},k_{2}\ell_{2})$

with

(2.6) $H_{L}(x)=\frac{x\widehat{f}\left(\frac{x}{L}\right)}{\sinh(x/2)}$

and

(2.7)
$\begin{split}U_{T}(x,y)&=\operatorname{\mathbb{E}}_{T}\left[\cos\left(\tau
x\right)\cos\left(\tau
y\right)\right]-\operatorname{\mathbb{E}}_{T}\left[\cos\left(\tau
x\right)\right]\cdot\operatorname{\mathbb{E}}_{T}\left[\cos\left(\tau
y\right)\right]\\\
&=\frac{1}{2}\widehat{w}\left(T\left(x-y\right)\right)+\frac{1}{2}\widehat{w}\left(T\left(x+y\right)\right)-\widehat{w}\left(Tx\right)\cdot\widehat{w}\left(Ty\right).\end{split}$

Note that $U_{T}(x,y)$ is real valued since $w(x)$ is real valued and even.
Starting from (2.1) in place of (2.4), and likewise yields

(2.8)
$\mathcal{V}_{T,L}(X)=\left(\frac{2}{L}\right)^{2}\sum_{\ell_{1},\ell_{2}\in\mathcal{L}_{g}}\sum_{k_{1},k_{2}\geq
1}\frac{H_{L}(k_{1}\ell_{1})H_{L}(k_{2}\ell_{2})}{k_{1}k_{2}}U_{T}(k_{1}\ell_{1},k_{2}\ell_{2}).$

Below we will be able to deduce Theorem 1.1 from the following two results:
Proposition 2.2, proved at the end of § 3, asserting that the random variables
$\mathcal{V}_{T,L}(X)$ converge in distribution to
$\mathcal{V}_{T,L}^{\infty}$, and Theorem 2.3, proved along § 4, asserting
that, in the regime of Theorem 1.1, $V_{T,L}^{\infty}$ converge, in
distribution, to the constant $\Sigma^{2}_{\rm GOE}(f)$.

######  Proposition 2.2.

For every $T,L>0$, the random variables
$\mathcal{V}_{T,L}=\mathcal{V}_{T,L}(X)$, with $X\in\mathcal{M}_{g}$ random
uniform w.r.t. the WP measure, converge, in distribution as
$g\rightarrow\infty$, to the random variable $\mathcal{V}_{T,L}^{\infty}$.

######  Theorem 2.3.

One has

$\lim_{\begin{subarray}{c}L,T\to\infty\\\ L=o(\log
T)\end{subarray}}\operatorname{\mathbb{E}}_{\operatorname{Pois}}\left[\left|\mathcal{V}_{T,L}^{\infty}-\Sigma^{2}_{\rm
GOE}(f)\right|\right]=0.$

Theorem 2.3 with Markov’s inequality implies that the probability that
$\operatorname{Var}_{T}(S_{L,\tau}^{2})$ is far from $\Sigma^{2}_{\rm GOE}(f)$
vanishes: for every $\epsilon>0$,

(2.9) $\lim_{\begin{subarray}{c}L,T\to\infty\\\ L=o(\log
T)\end{subarray}}\operatorname{Prob}_{\operatorname{Pois}}\left(\left|\mathcal{V}_{T,L}^{\infty}-\Sigma^{2}_{\rm
GOE}(f)\right|>\epsilon\right)=0.$

##  3\. Background on point processes

Let $\mathcal{N}$ be the space of measures $\mu$ on ${\mathbb{R}}_{\geq 0}$
s.t. $\mu(0)=0$, for every $t\geq 0$, $\mu(t)\in{\mathbb{Z}}_{\geq 0}$, where
$\mu(t):=\mu((0,t])$, which is assumed to be right-continuous at $0$. A
measure $\mu\in\mathcal{N}$ has a representation

(3.1) $\mu=\sum\limits_{j=1}^{k}\delta_{\xi_{j}},$

where $0\leq k\leq+\infty$, and $0<\xi_{1}\leq\xi_{2}\leq\ldots$, and
sometimes it is convenient to regard $\mu$ as a discrete multi-set

(3.2) $\mu=\\{\xi_{j}\\}_{j\geq 1}\subseteq{\mathbb{R}}$

Given a sequence $\\{\mu_{n}\\}\subseteq\mathcal{N}$ and $\mu\in\mathcal{N}$,
we say that $\mu_{n}$ vaguely converges to $\mu$, denoted
$\mu_{n}\rightarrow\mu$, if for all $t\in{\mathbb{R}}_{\geq 0}$ so that
$\mu(\cdot)$ is continuous at $t$,

(3.3) $\mu^{n}(t)\rightarrow\mu(t).$

Equivalently, for every non-negative continuous test function of compact
support $f:{\mathbb{R}}_{\geq 0}\rightarrow{\mathbb{R}}$, one has

$\int\limits_{0}^{\infty}f(x)\mu^{n}(dx)\rightarrow\int\limits_{0}^{\infty}f(x)\mu(dx).$

A point process (in a somewhat restrictive sense, sufficient for our purposes)
is a random element $\mu\in\mathcal{N}$ (w.r.t. the vague topology on
$\mathcal{N}$, generating the Borel $\sigma$-algebra on $\mathcal{N}$).
(Equivalently, it is a probability measure on $\mathcal{N}$.) We will regard a
point process as a random element of $\mathcal{N}$.

We say that a point process $\mathbf{N}$ is Poisson with intensity measure
$\nu$ on ${\mathbb{R}}_{>0}$, if for every Borel set $B\subseteq{\mathbb{R}}$,
the distribution of $\mathbf{N}(B)$ is Poisson with parameter $\nu(B)$, and
the random variables $\mathbf{N}(B_{1}),\ldots,\mathbf{N}(B_{k})$ are
independent for every $k\geq 2$, and any choice of disjoint Borel sets
$B_{1},\ldots,B_{k}$. A sequence of point processes $\mathbf{N}_{n}$ on
${\mathbb{R}}_{>0}$ is said to converge (in distribution) to a point process
$\mathbf{N}$ (written $\mathbf{N}_{n}\overset{d}{\rightarrow}\mathbf{N}$) if
the sequence of random vectors
$(\mathbf{N}_{n}(B_{1}),\dots,\mathbf{N}_{n}(B_{k}))$ converges in
distribution to the random vector
$(\mathbf{N}(B_{1}),\dots,\mathbf{N}(B_{k}))$ for all $k\geq 1$ and Borel sets
$B_{i}$ with boundaries satisfying $\mathbf{N}(\partial B_{i})=0$ almost
surely for all $i$. The convergence of $\mathbf{N}_{n}$ to $\mathbf{N}$
implies [3, Theorem 23 on p. 521] that the random element $\mathbf{N}_{n}$
converges, in distribution, to $\mathbf{N}$.

For a measure $\mu\in\mathcal{N}$ as in (3.1) with $k\geq 1$ finite, and
$1\leq m\leq k$, one defines [5, p. 28] the $m$’th factorial measure
$\mu^{(m)}$ of $\mu$ on ${\mathbb{R}}_{\geq 0}^{(m)}$ as

$\mu^{(m)}=\sum\limits_{i_{1},\ldots i_{m}\text{
distinct}}\delta_{(x_{i_{1}},\ldots x_{i_{m}})}.$

Next, for a point process $\eta$ we define [5, Definition 4.9] the $m$’th
factorial moment measure as

$\alpha_{m}(C)=\alpha_{\eta;m}(C):=\operatorname{\mathbb{E}}\left[\eta^{(m)}(C)\right]$

for $C\subseteq{\mathbb{R}}_{\geq 0}^{m}$ measurable (Borel). Then the $m$’th
factorial measure of a Poisson point process $\eta$ with intensity $\lambda$
is [5, Corollary 4.10] a Poisson point process with the intensity
$\lambda^{m}$, the $m$’th product measure of $\lambda$ on ${\mathbb{R}}_{\geq
0}^{m}$, that is also the $m$’th factorial moment measure of $\eta$.

In what follows use the identification (3.2) of a measure $\mu\in\mathcal{N}$.
Thus the factorial moment measure of a proper point process $\eta$ (that
includes any Poisson point process) computes the correlations of
$\\{\xi_{j}\\}$ in the sense that for a function $f:{\mathbb{R}}_{\geq
0}^{m}\rightarrow{\mathbb{R}}$,

(3.4)
$\operatorname{\mathbb{E}}\Big{[}\sum_{\begin{subarray}{c}(\xi_{1},\dots,\,\xi_{m})\in\eta^{m}\\\
\text{
distinct}\end{subarray}}f(\xi_{1},\dots,\xi_{m})\Big{]}=\int\limits_{{\mathbb{R}}_{\geq
0}^{m}}f(\xi_{1},\ldots,\xi_{m})d\alpha_{m}(\xi_{1},\dots,\xi_{m}),$

see [5, Definition 4.9 and immediately after]. In particular, for $m=1$, this
is Campbell’s formula [5, Proposition 2.7].

We conclude this section with the following observation. By comparing (2.5) to
(2.8), we may express both as

$\mathcal{V}_{T,L}(X)=\varphi_{T,L}(\mathcal{L}_{g});\;\;\mathcal{V}_{T,L}^{\infty}=\varphi_{T,L}(\mathcal{L}_{\infty}),$

with the functional $\varphi_{T,L}:\mathcal{N}\rightarrow{\mathbb{R}}$,
defined as

(3.5)
$\varphi_{T,L}(\mu):=\left(\frac{2}{L}\right)^{2}\sum_{\ell_{1},\ell_{2}\in\mu}\sum_{k_{1},k_{2}\geq
1}\frac{H_{L}(k_{1}\ell_{1})H_{L}(k_{2}\ell_{2})}{k_{1}k_{2}}U_{T}(k_{1}\ell_{1},k_{2}\ell_{2}).$

This is easily seen to be continuous, see [12, Lemma 2.1].

###### Proof of Proposition 2.2.

We interpret the result of Mirzakhani-Petri [8, Theorem 4.1] as the
convergence, as $g\rightarrow\infty$, of the point processes $\mathcal{L}_{g}$
to the Poisson point process $\mathcal{L}_{\infty}$ (Notation 2.1), and recall
that it implies that the random element $\mathcal{L}_{g}\in\mathcal{N}$
converges, in distribution, to $\mathcal{L}_{\infty}\in\mathcal{N}$. Hence, an
application of the Continuous Mapping theorem, with the continuous functional
$\varphi:\mathcal{N}\rightarrow{\mathbb{R}}$ as in (3.5), yields that, as
$g\rightarrow\infty$, the random variables

$\mathcal{V}_{T,L}(X)=\varphi(\mathcal{L}_{g})$

converge, in distribution, to

$\mathcal{V}_{T,L}^{\infty}=\varphi(\mathcal{L}_{\infty}),$

that is, the statement of Proposition 2.2.

∎

##  4\. GOE fluctuations for the Poisson model: Proof of Theorem 2.3

We need to compute the expected value of
$\left|\mathcal{V}_{T,L}^{\infty}-\Sigma^{2}_{\rm GOE}(f)\right|$ with
$\mathcal{V}_{T,L}^{\infty}$ given by (2.5) with $\ell_{1},\ell_{2}$ selected
according to the Poisson point process. We write (2.5) according to whether
$\ell_{1}=\ell_{2}$ (the diagonal) and $\ell_{1}\neq\ell_{2}$:

$\mathcal{V}_{T,L}^{\infty}=\sum_{\ell_{1}=\ell_{2}}+\sum_{\ell_{1}\neq\ell_{2}}=\operatorname{Diag}+\operatorname{Off}$

and then we have

$\operatorname{\mathbb{E}}_{\operatorname{Pois}}\left[\left|\mathcal{V}_{T,L}^{\infty}-\Sigma^{2}_{\rm
GOE}(f)\right|\right]\leq\operatorname{\mathbb{E}}_{\operatorname{Pois}}\left[\left|\operatorname{Diag}-\Sigma^{2}_{\rm
GOE}(f)\right|\right]+\operatorname{\mathbb{E}}_{\operatorname{Pois}}\left[\left|\operatorname{Off}\right|\right].$

We will show that both terms tend to zero as $L,T\to\infty$, $L=o(\log T)$.

###  4.1. Bounding
$\operatorname{\mathbb{E}}_{\operatorname{Pois}}\left[\left|\operatorname{Off}\right|\right]$

######  Lemma 4.1.

There is some $C_{0}=C_{0}(f)>0$ so that

(4.1)
$\operatorname{\mathbb{E}}_{\operatorname{Pois}}\left[\left|\operatorname{Off}\right|\right]\ll\frac{e^{2C_{0}L}}{TL},$

the constant involved in the ‘$\ll$’-notation absolute.

###### Proof.

We apply the correlation formula (3.4) with $m=2$ to obtain

$f(\xi_{1},\xi_{2})=\frac{4}{L^{2}}\sum\limits_{k_{1},k_{2}\geq
1}\frac{|H_{L}(k_{1}\xi_{1})||H_{L}(k_{1}\xi_{1}))|}{k_{1}k_{2}}|U_{T}(k_{1}\xi_{1},k_{1}\xi_{1})|$

and intensity $d\nu_{MP}(x)=2\sinh^{2}(x/2)/x\;dx$, to yield that

(4.2)
$\operatorname{\mathbb{E}}_{\operatorname{Pois}}\left[\left|\operatorname{Off}\right|\right]=\int\limits_{0}^{\infty}\int\limits_{0}^{\infty}f(x,y)d\nu_{MP}(x)d\nu_{MP}(y)\\\
=\frac{4}{L^{2}}\sum\limits_{k_{1},k_{2}\geq
1}\int\limits_{0}^{\infty}\int\limits_{0}^{\infty}\frac{|H_{L}(k_{1}x)H_{L}(k_{2}y)|}{k_{1}k_{2}}|U_{T}(k_{1}x,k_{2}y)|d\nu_{MP}(x)d\nu_{MP}(y)\\\
=\frac{16}{L^{2}}\sum\limits_{k_{1},k_{2}\geq
1}\int\limits_{0}^{\infty}\int\limits_{0}^{\infty}\widehat{f}\left(\frac{k_{1}x}{L}\right)\widehat{f}\left(\frac{k_{2}y}{L}\right)\cdot\frac{\sinh(x/2)^{2}\cdot\sinh(y/2)^{2}}{\sinh\left(\frac{k_{1}x}{2}\right)\cdot\sinh\left(\frac{k_{2}y}{2}\right)}\times\\\
\times|U_{T}(k_{1}x,k_{2}y)|dxdy$

by (2.6). In what follows we show that when $L=o(\log{T})$, this expression
vanishes.

We note that $\widehat{f}$ being compactly supported forces that in the range
of the integral one has $0<x\ll\frac{L}{k_{1}}$, so that $\sinh(x/2)\leq
e^{C_{0}\cdot L}$ for some absolute $C_{0}>0$, and likewise $\sinh(y/2)\leq
e^{C_{0}\cdot L}$. Since for $k\geq 1$ and $t\in{\mathbb{R}}$

$\frac{\sinh(t)}{\sinh(kt)}\leq\frac{1}{k},$

we have

$\frac{\sinh(x/2)^{2}\cdot\sinh(y/2)^{2}}{\sinh\left(\frac{k_{1}x}{2}\right)\cdot\sinh\left(\frac{k_{2}y}{2}\right)}\leq
e^{2C_{0}\cdot L}\cdot\frac{1}{k_{1}k_{2}}.$

Given $x$ in the domain of the integration (as above, $x\ll\frac{L}{k_{1}}$),
the compact support of $\widehat{\omega}$ forces that $y$ is contained in an
interval of length $\frac{1}{k_{2}T}$ (from (2.7)). Therefore, taking into
account the boundedness of both $\widehat{f}$ and $\widehat{w}$ (hence of
$U_{T}$), we have

(4.3)
$\begin{split}\frac{1}{L^{2}}\int\limits_{0}^{\infty}\int\limits_{0}^{\infty}&\widehat{f}\left(\frac{k_{1}x}{L}\right)\widehat{f}\left(\frac{k_{2}y}{L}\right)\frac{\sinh(x/2)^{2}\sinh(y/2)^{2}}{\sinh\left(\frac{k_{1}x}{2}\right)\sinh\left(\frac{k_{2}y}{2}\right)}|U_{T}(k_{1}x,k_{2}y)|dxdy\\\
&\ll\frac{1}{L^{2}}\cdot\frac{e^{2C_{0}L}}{k_{1}k_{2}}\cdot\frac{1}{Tk_{2}}\cdot\frac{L}{k_{1}}=\frac{e^{2C_{0}L}}{TL}\cdot\frac{1}{k_{1}^{2}k_{2}^{2}}.\end{split}$

The bound (4.1) of Lemma 4.1 finally follows upon summing up (4.3) for
$k_{2},k_{1}\geq 1$ and substituting it into (4.2). ∎

###  4.2. Bounding the diagonal term

We now want to bound the term
$\operatorname{\mathbb{E}}_{\operatorname{Pois}}\left[\left|\operatorname{Diag}-\Sigma^{2}_{\rm
GOE}(f)\right|\right]$. Recall

$\operatorname{Diag}=\frac{4}{L^{2}}\sum_{x}\sum_{k_{1},k_{2}\geq
1}\frac{H_{L}(k_{1}x)H_{L}(k_{2}x)}{k_{1}k_{2}}U_{T}(k_{1}x,k_{2}x)$

We separate out the contribution of the pair $(k_{1},k_{2})=(1,1)$ and the
rest and use

(4.4)
$\operatorname{\mathbb{E}}_{\operatorname{Pois}}\left[\left|\operatorname{Diag}-\Sigma^{2}_{\rm
GOE}(f)\right|\right]\leq\operatorname{\mathbb{E}}_{\operatorname{Pois}}\left[\left|\frac{4}{L^{2}}\sum_{x}H_{L}(x)^{2}U_{T}(x,x)-\Sigma^{2}_{\rm
GOE}(f)\right|\right]\\\
+\frac{4}{L^{2}}\operatorname{\mathbb{E}}_{\operatorname{Pois}}\left[\sum_{x}\sum_{k_{1}+k_{2}\geq
3}\frac{|H_{L}(k_{1}x)H_{L}(k_{2}x)|}{k_{1}k_{2}}|U_{T}\left(k_{1}x,k_{2}x\right)|\right].$

####  4.2.1. The sum $k_{1}+k_{2}\geq 3$

We set

$D(k_{1},k_{2}):=\frac{4}{L^{2}}\operatorname{\mathbb{E}}_{\operatorname{Pois}}\left[\sum_{x}\frac{|H_{L}(k_{1}x)H_{L}(k_{2}x)|}{k_{1}k_{2}}|U_{T}\left(k_{1}x,k_{2}x\right)|\right]$

and want to show that

$\sum_{k_{1}+k_{2}\geq 3}D(k_{1},k_{2})\ll\frac{\log L}{L^{2}}.$

We apply Campbell’s formula ((3.4) with $m=1$)

(4.5)
$\operatorname{\mathbb{E}}_{\operatorname{Pois}}\left[\sum_{x}h(x)\right]=\int_{0}^{\infty}h(x)d\nu_{MP}(x)$

with
$h(x)=\frac{|H_{L}(k_{1}x)H_{L}(k_{2}x)|}{k_{1}k_{2}}|U_{T}\left(k_{1}x,k_{2}x\right)|$
to find

$\begin{split}D(k_{1},k_{2})&=\frac{4}{L^{2}}\int_{0}^{\infty}\frac{|H_{L}(k_{1}x)H_{L}(k_{2}x)|}{k_{1}k_{2}}|U_{T}\left(k_{1}x,k_{2}x\right)|d\nu_{MP}(x)\\\
&\ll\int_{0}^{C_{0}L}\frac{\sinh^{2}(Lx/2)}{\sinh(Lk_{1}x/2)\sinh(Lk_{2}x/2)}xdx\end{split}$

on using boundedness of $U_{T}$ and $\widehat{f}$ being supported in
$[-C_{0},C_{0}]$. It is easy to show (see the proof of [11, Lemma 5.2]) that
this last integral is bounded by

$\ll\min\left(\frac{1}{L^{2}(k_{1}+k_{2}-2)^{2}},\frac{1}{k_{1}k_{2}^{3}}\right)$

(assuming $k_{2}\geq k_{1}$) and from this to show that

$\sum_{k_{1}+k_{2}\geq 3}D(k_{1},k_{2})\ll\frac{\log L}{L^{2}}$

which is our claim.

####  4.2.2. The term $k_{1}=k_{2}=1$

We are left with showing that the first term on the RHS of (4.4) tends to
zero, and using Cauchy-Schwartz it suffices to bound the second moment

$\lim_{L\to\infty}\operatorname{\mathbb{E}}_{\operatorname{Pois}}\left[\left|\frac{4}{L^{2}}\sum_{x}H_{L}(x)^{2}U_{T}(x,x)-\Sigma^{2}_{\rm
GOE}(f)\right|^{2}\right]=0$

for $T\geq 1$. Squaring out, we want to show

$\operatorname{\mathbb{E}}_{\operatorname{Pois}}\left[\left(\frac{4}{L^{2}}\sum_{x}H_{L}(x)^{2}U_{T}(x,x)\right)^{2}\right]\\\
-2\Sigma^{2}_{\rm
GOE}(f)\operatorname{\mathbb{E}}_{\operatorname{Pois}}\left[\frac{4}{L^{2}}\sum_{x}H_{L}(x)^{2}U_{T}(x,x)\right]\to(\Sigma^{2}_{\rm
GOE}(f))^{2}.$

We have

(4.6)
$\operatorname{\mathbb{E}}_{\operatorname{Pois}}\left[\frac{4}{L^{2}}\sum_{x}H_{L}(x)^{2}U_{T}(x,x)\right]=\Sigma^{2}_{\rm
GOE}(f)+O\left(\frac{1}{TL}\right).$

Indeed, by Campbell’s formula (4.5),

$\operatorname{\mathbb{E}}_{\operatorname{Pois}}\left[\frac{4}{L^{2}}\sum_{x}H_{L}(x)^{2}U_{T}(x,x)\right]=\frac{4}{L^{2}}\int_{0}^{\infty}H_{L}(x)^{2}U_{T}(x,x)d\nu_{MP}(x)\\\
=\frac{4}{L^{2}}\int_{0}^{\infty}\frac{x^{2}\widehat{f}\left(\frac{x}{L}\right)}{\sinh^{2}(x/2)}\frac{1}{2}\left(1+\widehat{w}(2Tx)-2\widehat{w}(Tx)^{2}\right)2\frac{\sinh^{2}(x/2)}{x}dx\\\
=4\int_{0}^{\infty}\widehat{f}(y)^{2}ydy+4\int_{0}^{\infty}\widehat{f}(y)^{2}y\left(\widehat{w}(2TLy)-2\widehat{w}(TLy)^{2}\right)dy,$

using

$U_{T}(x,x)=\frac{1}{2}+\frac{1}{2}\widehat{w}(2Tx)-\widehat{w}(Tx)^{2}.$

The first term is
$2\int_{-\infty}^{\infty}\widehat{f}(y)^{2}|y|dy=\Sigma^{2}_{\rm GOE}(f)$. The
second term is treated by observing that since
$\operatorname{Supp}\widehat{w}\subseteq[-1,1]$, the integral is bounded by

$\int_{0}^{\infty}\widehat{f}(y)^{2}y\left(|\widehat{w}(2TLy)|+2|\widehat{w}(TLy)^{2}|\right)dy\ll\int_{0}^{1/2TL}\widehat{f}(y)^{2}ydy\ll\frac{1}{TL},$

hence we obtain (4.6)

Therefore it suffices to show

$\operatorname{\mathbb{E}}_{\operatorname{Pois}}\left[\left(\frac{4}{L^{2}}\sum_{x}H_{L}(x)^{2}U_{T}(x,x)\right)^{2}\right]\to(\Sigma^{2}_{\rm
GOE}(f))^{2}.$

Again square out, obtain a diagonal sum and an off-diagonal sum.

We bound the diagonal sum using Campbell’s formula (4.5)

$\operatorname{\mathbb{E}}_{\operatorname{Pois}}\left[\frac{16}{L^{4}}\sum_{x}H_{L}(x)^{4}U_{T}(x,x)^{2}\right]=\frac{16}{L^{4}}\int_{0}^{\infty}H_{L}(x)^{4}U_{T}(x,x)^{2}d\nu_{MP}(x)\\\
=\frac{16}{L^{4}}\int_{0}^{\infty}\frac{x^{4}\widehat{f}\left(\frac{x}{L}\right)^{4}}{\sinh^{4}(x/2)}\frac{2\sinh^{2}(x/2)}{x}U_{T}(x,x)^{2}dx\\\
\ll\frac{1}{L^{4}}\int_{0}^{cL}\frac{x^{3}}{\sinh^{2}(x/2)}dx\ll\frac{1}{L^{4}}.$

We evaluate the off-diagonal term using the correlation formula (3.4) with
$m=2$ for the Poisson process

$\operatorname{\mathbb{E}}_{\operatorname{Pois}}\left[\frac{16}{L^{4}}\sum_{x\neq
y}H_{L}(x)^{2}U_{T}(x,x)H_{L}(y)^{2}U_{T}(y,y)2\right]\\\
=\frac{16}{L^{4}}\iint
H_{L}(x)^{2}U_{T}(x,x)H_{L}(y)^{2}U_{T}(y,y)d\nu_{MP}(x)d\nu_{MP}(y)\\\
=\left(\frac{4}{L^{2}}\int_{0}^{\infty}H_{L}(x)^{2}U_{T}(x,x)d\nu_{MP}(x)\right)^{2}=\left(\Sigma^{2}_{\rm
GOE}(f)\right)^{2}+O\left(\frac{1}{TL}\right)$

by (4.6), so we are done.

##  5\. Almost sure GOE fluctuations: Concluding the proof of Theorem 1.1

###### Proof.

Let $\epsilon>0$ be given, and let us consider for a moment the random
variables $\mathcal{V}_{T,L}=\mathcal{V}_{T,L}(X)$ and
$\mathcal{V}_{T,L}^{\infty}$ for sufficiently large, but fixed $T,L$ within
the allowed range $L=o(\log{T})$. It would follow that, as
$g\rightarrow\infty$,

(5.1)
$\operatorname{Prob}\left(\left|\mathcal{V}_{T,L}-\Sigma_{GOE}^{2}(f)\right|>\epsilon\right)\rightarrow\operatorname{Prob}\left(\left|\mathcal{V}_{T,L}^{\infty}-\Sigma_{GOE}^{2}(f)\right|>\epsilon\right),$

provided that $\Sigma_{GOE}^{2}(f)\pm\epsilon$ are not atoms of the
distribution of $\mathcal{V}_{T,L}^{\infty}$. Otherwise, apply (5.1) with

$\epsilon/2<\epsilon^{\prime}=\epsilon^{\prime}(T,L)<\epsilon$

so that, for the given $T,L$, the distribution of $\mathcal{V}_{T,L}^{\infty}$
does not have an atom at $\Sigma_{GOE}^{2}(f)\pm\epsilon^{\prime}$, and use
that, for sufficiently large admissible $T,L$, the probability of

$\left\\{\left|\mathcal{V}_{T,L}^{\infty}-\Sigma_{GOE}^{2}(f)\right|>\epsilon^{\prime}\right\\}\subseteq\left\\{\left|\mathcal{V}_{T,L}^{\infty}-\Sigma_{GOE}^{2}(f)\right|>\frac{\epsilon}{2}\right\\}$

is arbitrarily small, by (2.9), a corollary from Theorem 2.3. Finally, the
double limit statement (1.4) follows, as

$\begin{split}&\operatorname{Prob}\left(\left|\mathcal{V}_{T,L}-\Sigma_{GOE}^{2}(f)\right|>\epsilon\right)\leq\operatorname{Prob}\left(\left|\mathcal{V}_{T,L}-\Sigma_{GOE}^{2}(f)\right|>\epsilon^{\prime}\right)\\\
&=\operatorname{Prob}\left(\left|\mathcal{V}_{T,L}^{\infty}-\Sigma_{GOE}^{2}(f)\right|>\epsilon^{\prime}\right)\\\
&+\left(\operatorname{Prob}\left(\left|\mathcal{V}_{T,L}-\Sigma_{GOE}^{2}(f)\right|>\epsilon^{\prime}\right)-\operatorname{Prob}\left(\left|\mathcal{V}_{T,L}^{\infty}-\Sigma_{GOE}^{2}(f)\right|>\epsilon^{\prime}\right)\right)\\\
&\leq\operatorname{Prob}\left(\left|\mathcal{V}_{T,L}^{\infty}-\Sigma_{GOE}^{2}(f)\right|>\frac{\epsilon}{2}\right)\\\
&+\left(\operatorname{Prob}\left(\left|\mathcal{V}_{T,L}-\Sigma_{GOE}^{2}(f)\right|>\epsilon^{\prime}\right)-\operatorname{Prob}\left(\left|\mathcal{V}_{T,L}^{\infty}-\Sigma_{GOE}^{2}(f)\right|>\epsilon^{\prime}\right)\right),\end{split}$

with the first term small by (2.9), and the difference tends to zero as
$g\to\infty$ since $\Sigma_{GOE}^{2}(f)\pm\epsilon^{\prime}$ is not an atom of
$\mathcal{V}_{T,L}^{\infty}$. ∎

## References

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# Symmetric bilinear forms, superalgebras and integer matrix factorization

Dan Fretwell, Jenny Roberts

###### Abstract.

We construct and investigate certain (unbalanced) superalgebra structures on
$\text{End}_{K}(V)$, with $K$ a field of characteristic $0$ and $V$ a finite
dimensional $K$-vector space (of dimension $n\geq 2$). These structures are
induced by a choice of non-degenerate symmetric bilinear form $B$ on $V$ and a
choice of non-zero base vector $w\in V$. After exploring the construction
further, we apply our results to certain questions concerning integer matrix
factorization and isometry of integral lattices.

##  1\. Introduction

Let $K$ be a field and $A$ be a $K$-algebra. We say that $A$ is a superalgebra
over $K$ if it is equipped with a $\mathbb{Z}/2\mathbb{Z}$-grading, i.e. a
vector space decomposition $A=A^{(0)}\oplus A^{(1)}$ satisfying
$A^{(i)}A^{(j)}\subseteq A^{(i+j\bmod 2)}$ for all $i,j\in\\{0,1\\}$. Every
$K$-algebra gives a trivial superalgebra structure by setting $A^{(0)}=A$ and
$A^{(1)}=\\{0\\}$, but non-trivial examples of superalgebras exist, e.g.
exterior algebras, Clifford algebras and polynomial rings (decomposition given
by symmetric and antisymmetric polynomials). Superalgebras appear throughout
Mathematics but are also of direct importance to Physics, since they provide
robust mathematical models for supersymmetry (e.g. see [2] and [7]).

Let $K$ be a field and $V$ be a $K$-vector space of dimension $n\geq 2$. The
endomorphism algebra $\text{End}_{K}(V)$ is then a $K$-algebra, and one asks
whether there are any interesting superalgebra structures on this algebra. One
particular well known example comes to mind. If a basis for $V$ is fixed then
$\text{End}_{K}(V)\cong M_{n}(K)$, and a superalgebra structure is found by
letting $A^{(0)}$ and $A^{(1)}$ be the subspaces of symmetric and
antisymmetric matrices respectively. It is natural to wonder, as we have,
whether there are other interesting superalgebra structures to be found beyond
this example.

In this paper we provide a new class of superalgebra structures on
$\text{End}_{K}(V)$ for any field $K$ of characteristic zero and any finite
dimensional $K$-vector space $V$ (of dimension $n\geq 2$). These are induced
from a choice of non-degenerate symmetric bilinear form $B$ on $V$ and a fixed
choice of non-zero vector $w\in V$. More precisely, we show in Theorem 2.8
that there is a superalgebra decomposition:

$\text{End}_{K}(V)=E^{(0)}(B,w)\oplus E^{(1)}(B,w),$

where

$\displaystyle E^{(0)}(B,w)$
$\displaystyle=\\{\phi\in\text{End}_{K}(V)\,|\,B(u,\phi(w))=B(w,\phi(u))=0,\forall\
u\in\\{w\\}_{B}^{\perp}\\},$ $\displaystyle E^{(1)}(B,w)$
$\displaystyle=\\{\phi\in\text{End}_{K}(V)\,|\,B(u,\phi(v))=0,\forall\
u,v\in\\{w\\}_{B}^{\perp},B(w,\phi(w))=0\\}.$

Our recipe is inspired by results coming from the papers [3], [4] and [5],
which correspond to the special case with $B$ the standard symmetric bilinear
form and $w=(1,1,...,1)$ (after choosing a basis for $V$). In this case we
have that $E^{(0)}(B,w)$ is the subspace of semi-magic squares and
$E^{(1)}(B,w)$ is the subspace of matrices satisfying a vertex cross-sum
property. We give a similar interpretation of the subspaces $E^{(i)}(B,w)$ in
more generality, although we have tried to present our results in a basis
independent way where possible.

Unlike the superalgebra structures given by Clifford algebras, the
superalgebra structure given above is in general unbalanced. If
$\text{dim}(V)=n$ then we have that $\text{dim}(E^{(0)}(B,w))=n^{2}-2n+2$ and
$\text{dim}(E^{(1)}(B,w))=2n-2$ (which are not equal if $n\geq 3$). Also, in
general the above superalgebra structure is not equivalent to a symmetric and
antisymmetric matrix decomposition (e.g. if $n\neq 4$ then we cannot have
$(n^{2}-2n+2,2n-2)=\left(\frac{n(n+1)}{2},\frac{n(n-1)}{2}\right)$).

After establishing the above result, we study potential implications towards
the isometry problem for quadratic spaces. If $B^{\prime}$ is another non-
degenerate symmetric bilinear form on $V$ then $(V,B)$ and $(V,B^{\prime})$
are isometric if $B^{\prime}=B_{\phi}$ for some $\phi\in\text{GL}(V)$ (here
$B_{\phi}(x,y)=B(\phi(x),\phi(y))$). Expanding $\phi$ into its “odd” and
“even” parts according to the superalgebra structure induces a decomposition
of $B_{\phi}$ into a sum of symmetric bilinear forms (Proposition 4.1).
Careful substitution then provides us with a set of necessary equations that
must be satisfied by $\phi$ in order for the equality $B^{\prime}=B_{\phi}$ to
hold (Theorem 3.2). In practice, these equations can either be used to prove
the non-existence of $\phi$ or to provide information about it.

Certain long standing problems in number theory concern the topic of integer
matrix factorization. One particular family of such problems considers
factorizations of the form $\mathcal{B}^{\prime}=M^{T}M$ with
$\mathcal{B}^{\prime},M\in\text{GL}_{n}(\mathbb{Z})$ and
$\mathcal{B}^{\prime}$ symmetric of determinant $1$. Indeed, classical results
(e.g. discussed in the paper of Mordell [6]) show that if $n\leq 7$ and
$\mathcal{B}^{\prime}$ is positive definite then such a factorization is
always guaranteed to exist (for $n=8$ a counterexample is given by any Gram
matrix of the $E_{8}$ lattice under the standard Euclidean inner product).

The above problem can be instead viewed as an isometry problem for integral
lattices. Interpreting $\mathcal{B}^{\prime}$ as the Gram matrix of a non-
degenerate symmetric bilinear form $B^{\prime}$ on $V=\mathbb{Q}^{n}$, we see
that the lattice $\Lambda=\mathbb{Z}^{n}$ is integral with respect to
$B^{\prime}$. The existence of $M$ in the matrix factorization is then
equivalent to the equality $B^{\prime}=B_{\phi}$ for some
$\phi\in\text{GL}(\Lambda)\subset\text{GL}(V)$, where $B$ is the standard
symmetric bilinear form (corresponding to Gram matrix $\mathcal{B}=I_{n}$ with
respect to the standard basis). It is then clear that a special case of the
above decomposition results (or at least their integral counterparts) can then
be of use in first determining whether such an isometry $\phi$ exists and then
determining precise information about it. This strategy was adopted by the
authors in the papers [3] and [5] to produce new insights on this problem
(although a matrix superalgebra decomposition was used, as opposed to the
analogous isometry decomposition that we consider). In particular, the authors
use a superalgebra decomposition of $M_{4}(\mathbb{Q})$ to classify integer
(and rational) matrix factorizations of the Wilson matrix:

$W=\begin{pmatrix}5&7&6&5\\\ 7&10&8&7\\\ 6&8&10&9\\\ 5&7&9&10\end{pmatrix},$

a symmetric integral positive definite matrix of determinant $1$ (that is
mildly ill-conditioned).

More generally, one could ask for integer matrix factorizations of the form
$\mathcal{B}^{\prime}=M^{T}\mathcal{B}M$ for an arbitrary
$\mathcal{B},\mathcal{B}^{\prime},M\in\text{GL}_{n}(\mathbb{Z})$ with
$\mathcal{B},\mathcal{B}^{\prime}$ symmetric and of equal determinant. As
above, our more general results let us consider this question from the point
of view of decompositions of lattice isometries (again, we require
$B^{\prime}=B_{\phi}$ for some $\phi\in\text{GL}(\Lambda)$, but now $B$ is not
necessarily the standard symmetric bilinear form). We give a precise set of
integral equations that must be satisfied for such an isometry $\phi$ to
exist, and demonstrate via examples how these equations can be used to prove
that certain integer matrix factorizations are impossible (alternatively
proving that certain families of integral symmetric bilinear forms are non-
isometric).

##  2\. Symmetric bilinear forms and superalgebras

Let $K$ be a field with $\text{char}(K)=0$ and let $V$ be a finite dimensional
$K$-vector space equipped with a non-degenerate symmetric bilinear form
$B:V\times V\rightarrow K$. We assume that $\text{dim}_{K}(V)=n\geq 2$ and we
fix a choice of vector $w\in V\backslash\\{0\\}$. Denote by
$\\{w\\}_{B}^{\perp}$ the orthogonal complement of $w$ with respect to $B$.

The aim of this section is to use the above data to obtain a superalgebra
structure on $\text{End}_{K}(V)$, this being a decomposition of $K$-vector
spaces

$\text{End}_{K}(V)=E^{(0)}(B,w)\oplus E^{(1)}(B,w)$

satisfying $E^{(i)}(B,w)E^{(j)}(B,w)\subseteq E^{(i+j\bmod 2)}(B,w)$ for all
$i,j\in\\{0,1\\}$ (i.e. a $\mathbb{Z}/2\mathbb{Z}$-graded algebra structure).

###  2.1. The subspace $E^{(0)}(V)$

We define the following subspace of $\text{End}_{K}(V)$:

$E^{(0)}(B,w)=\\{\phi\in\text{End}_{K}(V)\,|\,B(u,\phi(w))=B(w,\phi(u))=0,\forall\
u\in\\{w\\}_{B}^{\perp}\\}.$

######  Lemma 2.1.

$\phi\in E^{(0)}(B,w)$ if and only if there exists $\lambda\in K$ such that

$\displaystyle B(w,\phi(\cdot))$ $\displaystyle=\lambda B(w,\cdot),$
$\displaystyle B(\cdot,\phi(w))$ $\displaystyle=\lambda B(\cdot,w).$

###### Proof.

The reverse implication is clear. To prove the forward implication we first
note that since $B(u,\phi(w))=0$ for all $u\in\\{w\\}_{B}^{\perp}$, we have
that $\phi(w)\in\\{w\\}_{B}^{\perp\perp}=Kw$, so that

$\phi(w)=\lambda w$

for some $\lambda\in K$. From here, we see that

$B(\cdot,\phi(w))=\lambda B(\cdot,w).$

For the second equation, let $\phi^{\dagger}$ be the adjoint of $\phi$ with
respect to $B$. Then we have that $\phi^{\dagger}\in E^{(0)}(B,w)$, since:

$\displaystyle B(u,\phi^{\dagger}(w))$
$\displaystyle=B(\phi^{\dagger}(w),u)=B(w,\phi(u))=0$ $\displaystyle
B(w,\phi^{\dagger}(u))$ $\displaystyle=B(\phi^{\dagger}(u),w)=B(u,\phi(w))=0$

for all $u\in\\{w\\}_{B}^{\perp}$. It follows that:

$B(\cdot,\phi(w))=B(\phi(w),\cdot)=B(w,\phi^{\dagger}(\cdot))=\lambda^{\prime}B(w,\cdot)=\lambda^{\prime}B(\cdot,w),$

for some $\lambda^{\prime}\in K$.

Finally, the fact that $\lambda=\lambda^{\prime}$ follows by evaluating both
maps at $w$:

$\lambda
B(w,w)=B(w,\phi(w))=B(\phi(w),w)=B(w,\phi^{\dagger}(w))=\lambda^{\prime}B(w,w),$

and noting that $B(w,w)\neq 0$ by non-degeneracy. ∎

Choosing a basis $\\{e_{1},e_{2},...,e_{n}\\}$ for $V$, we get a matrix $M\in
M_{n}(K)$ associated to $\phi$ and a Gram matrix
$\mathcal{B}=(B(e_{i},e_{j}))$ associated to $B$. The conditions above then
mean that all row/column sums of the matrix $\mathcal{B}M$ (weighted with
respect to the coordinates of $w$) are a fixed scaling of those of the matrix
$\mathcal{B}$. In the particular case of $w=(1,1,...,1)$ and
$\mathcal{B}=I_{n}$, we recover the conditions for $M$ to be a semi-magic
square (as observed in [3] and [5]).

######  Definition 2.2.

If $\phi\in E^{(0)}(B,w)$ then we define the weight of $\phi$ as follows:

$\text{wt}(\phi)_{B,w}=\frac{B(w,\phi(w))}{B(w,w)}$

(i.e. the constant $\lambda\in K$ corresponding to $\phi\in E^{(0)}(B,w)$).

In [3] and [5], a similar notion of weight was given to matrices $M\in
M_{n}(K)$:

$\text{wt}(M)=\frac{1}{n^{2}}\sum_{i}\sum_{j}m_{ij}=\frac{1}{n^{2}}1_{n}^{T}M1_{n}.$

where $1_{n}=(1,1,...,1)\in K^{n}$. By comparison, when $B$ is the standard
symmetric bilinear form and $w=(1,1,...,1)$ (after choosing a basis for $V$)
then our definition gives $\text{wt}(M)_{B,w}=n\,\text{wt}(M)$, a simple
scaling of the above definition.

We can use the weight map to decompose the subspace $E^{(0)}(B,w)$ further.
For $u,v\in V$ let $\phi_{B,u,v}\in\text{End}_{K}(V)$ be given by

$\phi_{B,u,v}(x)=B(v,x)u,$

the endomorphism corresponding to the outer product of $u$ and $v$ with
respect to $B$ (i.e. in coordinates this is given by the matrix
$uv^{T}\mathcal{B}$).

The following Lemma will be useful later.

######  Lemma 2.3.

There exists a decomposition of $K$-vector spaces:

$E^{(0)}(B,w)=E^{(0)}_{0}(B,w)\oplus K\phi_{B,w,w},$

where $E^{(0)}_{0}(B,w)$ is the subspace consisting of weight $0$
endomorphisms.

###### Proof.

Note first that the weight is a linear map $\text{End}_{K}(V)\rightarrow K$.
The decomposition will then follow from the fact that $\phi_{B,w,w}\in
E^{(0)}(B,w)$ has weight $B(w,w)\neq 0$. This is a simple consequence of Lemma
2.1, since:

$\displaystyle B(w,\phi_{B,w,w}(\cdot))$
$\displaystyle=B(w,B(w,\cdot)w)=B(w,w)B(w,\cdot),$ $\displaystyle
B(\cdot,\phi_{B,w,w}(w))$ $\displaystyle=B(\cdot,B(w,w)w)=B(w,w)B(\cdot,w).$

∎

The above implies that every $\phi\in E^{(0)}(B,w)$ can be uniquely written in
the form

$\phi=\phi_{0}+\frac{\text{wt}(\phi)_{B,w}}{B(w,w)}\phi_{B,w,w}$

for some $\phi_{0}$ of weight $0$.

###  2.2. The subspace $E^{(1)}(B,w)$

We define another subspace of $\text{End}_{K}(V)$:

$E^{(1)}(B,w)=\\{\phi\in\text{End}_{K}(V)\,|\,B(u,\phi(v))=0,\forall\
u,v\in\\{w\\}_{B}^{\perp},B(w,\phi(w))=0\\}.$

######  Lemma 2.4.

$\phi\in E^{(1)}(B,w)$ if and only if $\phi=\phi_{B,w,a}+\phi_{B,b,w}$ for
some $a,b\in\\{w\\}_{B}^{\perp}$.

###### Proof.

First note that if $\phi\in\text{End}_{K}(V)$ is of the above form then
$\phi\in E^{(1)}(B,w)$, since for all $u,v\in\\{w\\}_{B}^{\perp}$ we have

$\displaystyle B(u,\phi(v))$
$\displaystyle=B(u,\phi_{B,w,a}(v))+B(u,\phi_{B,b,w}(v))=0,$ $\displaystyle
B(w,\phi(w))$ $\displaystyle=B(w,\phi_{B,w,a}(w))+B(w,\phi_{B,b,w}(w))=0.$

For the other direction, let $\phi\in E^{(1)}(B,w)$. Then for
$v\in\\{w\\}_{B}^{\perp}$, we have that
$\phi(v)\in\\{w\\}_{B}^{\perp\perp}=Kw$. So, there exists a linear map
$f:\\{w\\}_{B}^{\perp}\rightarrow K$ such that $\phi(v)=f(v)w$ for all
$v\in\\{w\\}_{B}^{\perp}$. Now, by the Riesz representation theorem there
exists a vector $a\in\\{w\\}_{B}^{\perp}$ satisfying $f(v)=B(a,v)$ for all
$v\in\\{w\\}_{B}^{\perp}$. Thus $\phi(v)=\phi_{B,w,a}(v)$ for all
$v\in\\{w\\}_{B}^{\perp}$.

To provide the necessary extension to $V$, let $b=\frac{1}{B(w,w)}\phi(w)\in
V$ (well defined by non-degeneracy of $B$). Then $b\in\\{w\\}_{B}^{\perp}$,
since $\phi\in E^{(1)}(B,w)$ and so

$B(w,b)=\frac{1}{B(w,w)}B(w,\phi(w))=0.$

It follows that

$\phi(w)=B(w,w)b=\phi_{B,b,w}(w),$

so that $\phi=\phi_{B,w,a}+\phi_{B,b,w}$.

∎

###  2.3. The endomorphism superalgebra induced by $B$.

We are now ready to show that there is an orthogonal decomposition
$\text{End}_{K}(V)=E^{(0)}(B,w)\oplus E^{(1)}(B,w)$, and that this defines a
superalgebra structure on $\text{End}_{K}(V)$.

######  Definition 2.5.

For $\phi_{1},\phi_{2}\in\text{End}_{K}(V)$ we define:

$B(\phi_{1},\phi_{2})=\text{Tr}(\phi_{1}^{\dagger}\phi_{2}),$

where $\phi_{1}^{\dagger}$ is the adjoint of $\phi_{1}$ with respect to $B$.

We check that this is a symmetric bilinear form. Bilinearity follows
immediately since the adjoint and trace maps are $K$-linear. Symmetry follows
since:

$B(\phi_{1},\phi_{2})=\text{Tr}(\phi_{1}^{\dagger}\phi_{2})=\text{Tr}(\phi_{2}\phi_{1}^{\dagger})=\text{Tr}((\phi_{2}^{\dagger}\phi_{1})^{\dagger})=\text{Tr}(\phi_{2}^{\dagger}\phi_{1})=B(\phi_{2},\phi_{1}).$

The above is an analogue of the Frobenius inner product, but induced by $B$
instead of the standard bilinear form. Indeed, in matrix form we have
$B(M_{1},M_{2})=\text{Tr}(\mathcal{B}^{-1}M_{1}^{T}\mathcal{B}M_{2})$, and if
$K\subseteq\mathbb{R}$ and the quadratic form corresponding to $B$ is positive
definite then this bilinear form can be shown to be positive definite.

######  Lemma 2.6.

Let $i\in\\{0,1\\}$. If $\phi\in E^{(i)}(B,w)$ then $\phi^{\dagger}\in
E^{(i)}(B,w)$.

###### Proof.

We already saw the case $i=0$ in the proof of Lemma 2.1. For the case $i=1$,
write $\phi=\phi_{B,w,a}+\phi_{B,b,w}$ for some $a,b\in\\{w\\}_{B}^{\perp}$.

Note that $\phi_{B,w,a}^{\dagger}=\phi_{B,a,w}$ since:

$B(\phi_{B,w,a}(x),y)=B(B(a,x)w,y)=B(a,x)B(w,y)=B(B(w,y)a,x)=B(\phi_{B,a,w}(y),x)$

for all $x,y\in V$. Similarly $\phi_{B,b,w}^{\dagger}=\phi_{B,w,b}$ and so it
follows that:

$\phi^{\dagger}=\phi_{B,w,a}^{\dagger}+\phi_{B,b,w}^{\dagger}=\phi_{B,a,w}+\phi_{B,w,b},$

and we notice that $a,b$ have simply switched roles. Thus $\phi^{\dagger}\in
E^{(1)}(B,w)$. ∎

######  Lemma 2.7.

There exists an orthogonal decomposition

$\text{End}_{K}(V)=E^{(0)}(B,w)\oplus E^{(1)}(B,w)$

with respect to the symmetric bilinear form in Definition 2.5.

###### Proof.

Let $\phi_{1}\in E^{(0)}(B,w)$ and $\phi_{2}\in E^{(1)}(B,w)$. Choose
$a,b\in\\{w\\}_{B}^{\perp}$ such that $\phi_{2}=\phi_{B,w,a}+\phi_{B,b,w}$.
Then:

$\displaystyle B(\phi_{1},\phi_{2})=\text{Tr}(\phi_{1}^{\dagger}\phi_{2})$
$\displaystyle=\text{Tr}(\phi_{1}^{\dagger}\phi_{B,w,a})+\text{Tr}(\phi_{1}^{\dagger}\phi_{B,b,w})$
$\displaystyle=B(w,\phi_{1}^{\dagger}(a))+B(b,\phi_{1}^{\dagger}(w))$
$\displaystyle=0$

since $\phi_{1}^{\dagger}\in E^{(0)}(B,w)$ (by Lemma 2.6). It follows that
$E^{(0)}(B,w)$ and $E^{(1)}(B,w)$ are orthogonal.

Since $B$ is non-degenerate, $\text{dim}(E^{(0)}(B,w))=n^{2}-2n+2$ and
$\text{dim}(E^{(1)}(B,w))=2n-2$, we see that
$E^{(0)}(B,w)^{\perp}=E^{(1)}(B,w)$ and so the decomposition follows. ∎

We are now in a position to prove the main result. We will do this in a way
that also highlights certain extra features of the decomposition (e.g.
multiplicativity of the weight).

######  Theorem 2.8.

The decomposition in Lemma 2.7 defines a superalgebra structure on
$\text{End}_{K}(V)$, i.e. $E^{(i)}(B,w)E^{(j)}(B,w)\subseteq E^{(i+j\bmod
2)}(B,w)$ for all $i,j\in\\{0,1\\}$.

###### Proof.

First let $\phi_{1},\phi_{2}\in E^{(0)}(B,w)$ have weights
$\lambda_{1},\lambda_{2}\in K$ respectively. Then $\phi_{1}\phi_{2}\in
E^{(0)}(B,w)$ has weight $\lambda_{1}\lambda_{2}$ since:

$\displaystyle
B(\cdot.(\phi_{1}\phi_{2})(w))=B(\phi_{1}^{\dagger}(\cdot),\phi_{2}(w))=\lambda_{2}B(\phi_{1}^{\dagger}(\cdot),w)=\lambda_{1}\lambda_{2}B(\cdot,w),$
$\displaystyle
B(w,(\phi_{1}\phi_{2})(\cdot))=B(\phi_{1}^{\dagger}(w),\phi_{2}(\cdot))=\lambda_{1}B(w,\phi_{2}(\cdot))=\lambda_{1}\lambda_{2}B(w,\cdot).$

Now let $\phi_{3},\phi_{4}\in E^{(1)}(B,w)$ and suppose that
$a_{1},b_{1},a_{2},b_{2}\in\\{w\\}_{B}^{\perp}$ are such that

$\displaystyle\phi_{3}$ $\displaystyle=\phi_{B,w,a_{1}}+\phi_{B,b_{1},w},$
$\displaystyle\phi_{4}$ $\displaystyle=\phi_{B,w,a_{2}}+\phi_{B,b_{2},w}.$

From this one checks that
$(\phi_{3}\phi_{4})(w)=(\phi_{4}^{\dagger}\phi_{3}^{\dagger})(w)=B(a_{1},b_{2})B(w,w)w$,
and so:

$\displaystyle B(\cdot,(\phi_{3}\phi_{4})(w))$
$\displaystyle=B(a_{1},b_{2})B(w,w)B(\cdot,w),$ $\displaystyle
B(w,(\phi_{3}\phi_{4})(\cdot))$
$\displaystyle=B((\phi_{4}^{\dagger}\phi_{3}^{\dagger})(w),\cdot)=B(a_{1},b_{2})B(w,w)B(w,\cdot)$

Since $B(a_{1},b_{2})B(w,w)\neq 0$ we see that $\phi_{3}\phi_{4}\in
E^{(0)}(B,w)$ has weight $B(a_{1},b_{2})B(w,w)$.

We now show that $\phi_{1}\phi_{3}$ and $\phi_{3}\phi_{1}$ lie in
$E^{(1)}(B,w)$. First note that:

$\displaystyle B(w,(\phi_{1}\phi_{3})(w))$
$\displaystyle=\lambda_{1}B(w,\phi_{3}(w))=0,$ $\displaystyle
B(w,(\phi_{3}\phi_{1})(w))$
$\displaystyle=B(\phi_{3}^{\dagger}(w),\phi_{1}(w))=\lambda_{1}B(\phi_{3}^{\dagger}(w),w)=0.$

Finally, if $u,v\in\\{w\\}_{B}^{\perp}$, we have:

$\displaystyle B(u,(\phi_{1}\phi_{3})(v))$
$\displaystyle=B(a_{1},v)B(u,\phi_{1}(w))=\lambda_{1}B(a_{1},v)B(u,w)=0,$
$\displaystyle B(u,(\phi_{3}\phi_{1})(v))$
$\displaystyle=B((\phi_{1}^{\dagger}\phi_{3}^{\dagger})(u),v)=B(b_{1},u)B(\phi_{1}^{\dagger}(w),v)=\lambda_{1}B(b_{1},u)B(w,v)=0.$

∎

Note that the even and odd pieces of this superalgebra decomposition have
different dimensions if $n\geq 3$. This is in contrast with the Clifford
algebra coming from the corresponding quadratic form on $V$. This
decomposition is also not equivalent to ones given by symmetric and
antisymmetric matrices in general (e.g. if $n\neq 4$ then we cannot have
$(n^{2}-2n+2,2n-2)=\left(\frac{n(n+1)}{2},\frac{n(n-1)}{2}\right)$).

We pause to give a simple example of the above superalgebra decomposition.

######  Example 2.9.

Let $n=3$ and $K=\mathbb{Q}$. Choose the standard basis for
$V=\mathbb{Q}^{3}$. The following symmetric bilinear form on $V$ is non-
degenerate:

$B(u,v)=u_{1}v_{1}+2u_{2}v_{2}+3u_{3}v_{3}.$

For the choice $w=(1,0,0)$ we get the superalgebra decomposition
$\text{End}_{K}(V)\cong M_{3}(\mathbb{Q})=E^{(0)}(B,w)\oplus E^{(1)}(B,w)$
with:

$\displaystyle E^{(0)}(B,w)$ $\displaystyle=\left\\{\begin{pmatrix}a&0&0\\\
0&e&f\\\ 0&h&i\end{pmatrix}\,\Bigg{|}\,a,e,f,h,i\in\mathbb{Q}\right\\},$
$\displaystyle E^{(1)}(B,w)$ $\displaystyle=\left\\{\begin{pmatrix}0&b&c\\\
d&0&0\\\ g&0&0\end{pmatrix}\,\Bigg{|}\,b,c,d,g\in\mathbb{Q}\right\\}.$

However, for the choice $w=(1,1,1)$ we get the very different looking
superalgebra decomposition with components:

$\displaystyle E^{(0)}(B,w)$ $\displaystyle=\left\\{\begin{pmatrix}a&b&c\\\
d&e&a+b+c-d-e\\\
\frac{b+c-2d}{3}&\frac{2a+b+2c-2e}{3}&\frac{a+b+2d+2e}{3}\end{pmatrix}\,\Bigg{|}\,a,b,c,d,e\in\mathbb{Q}\right\\},$
$\displaystyle E^{(1)}(B,w)$ $\displaystyle=\left\\{\begin{pmatrix}f&g&h\\\
i&-2f+g+2i&-3f+h+3i\\\
\frac{4f-g-h-2i}{3}&\frac{2f+g-2h-4i}{3}&f-g-2i\end{pmatrix}\,\Bigg{|}\,f,g,h,i\in\mathbb{Q}\right\\}.$

A natural question to ask in general is whether the above recipe produces
isomorphic superalgebras as $(B,w)$ vary. The theorem below addresses this
question. First recall that if $\phi_{1},\phi_{2}\in\text{End}_{K}(V)$ then we
can construct a new symmetric bilinear form on $V$ via

$B_{\phi_{1},\phi_{2}}(x,y)=\frac{1}{2}(B(\phi_{1}(x),\phi_{2}(y))+B(\phi_{1}(y),\phi_{2}(x))).$

When $\phi_{1}=\phi_{2}=\phi$ we will simply write $B_{\phi}$.

######  Theorem 2.10.

Suppose that $\phi\in\text{GL}(V)$. Then conjugation by $\phi$ gives a
superalgebra isomorphism:

$E^{(0)}(B,w)\oplus E^{(1)}(B,w)\cong E^{(0)}(B_{\phi},\phi^{-1}(w))\oplus
E^{(1)}(B_{\phi},\phi^{-1}(w)).$

###### Proof.

We first note that the map $\psi\mapsto\phi^{-1}\psi\phi$ clearly defines an
algebra isomorphism $\text{End}_{K}(V)\rightarrow\text{End}_{K}(V)$. It then
suffices to prove that this map preserves the grading.

Suppose that $\psi\in E^{(0)}(B,w)$. Then $\phi^{-1}\psi\phi\in
E^{(0)}(B_{\phi},\phi^{-1}(w))$ since for any
$u\in\\{\phi^{-1}(w)\\}^{\perp}_{B_{\phi}}$ we have (using Lemma 2.1):

$\displaystyle B_{\phi}(u,(\phi^{-1}\psi\phi)(\phi^{-1}(w)))$
$\displaystyle=B(\phi(u),\psi(w))=\lambda B(\phi(u),w)=\lambda
B_{\phi}(u,\phi^{-1}(w))=0,$ $\displaystyle
B_{\phi}((\phi^{-1}\psi\phi)(u),\phi^{-1}(w))$
$\displaystyle=B(\psi(\phi(u)),w)=\lambda B(\phi(u),w)=\lambda
B_{\phi}(u,\phi^{-1}(w))=0.$

Similarly, if $\psi\in E^{(1)}(B,w)$ then $\phi^{-1}\psi\phi\in
E^{(1)}(B_{\phi},\phi^{-1}(w))$ since for any
$u,v\in\\{\phi^{-1}(w)\\}^{\perp}_{B_{\phi}}$ we have:

$\displaystyle B_{\phi}(u,(\phi^{-1}\psi\phi)(v))$
$\displaystyle=B(\phi(u),\psi(\phi(v)))=0,$ $\displaystyle
B_{\phi}(\phi^{-1}(w),(\phi^{-1}\psi\phi)(\phi^{-1}(w))$
$\displaystyle=B(w,\psi(w))=0$

(noting that $u,v\in\\{\phi^{-1}(w)\\}^{\perp}_{B_{\phi}}$ implies that
$\phi(u),\phi(v)\in\\{w\\}^{\perp}_{B}$).

∎

##  3\. Further results

In this section we study some simple consequences of the superalgebra
decomposition. In particular, we will see how it can be used to deduce
information about the corresponding isometry problem for the non-degenerate
symmetric bilinear space $(V,B)$.

To recall, the isometry problem for $(V,B)$ is the problem of detecting
whether a given non-degenerate symmetric bilinear form $B^{\prime}$ on $V$ can
be written as $B^{\prime}=B_{\phi}$ for some $\phi\in\text{GL}(V)$ (and
further to construct such a map $\phi$ if it exists).

First, we consider the decomposition of $B_{\phi}$ for an arbitrary
$\phi\in\text{End}_{K}(V)$, induced by the decomposition of $\phi$ given by
Lemma 2.7.

######  Proposition 3.1.

Let $\phi\in\text{End}_{K}(V)$ have decomposition

$\phi=\phi_{0}+\frac{\text{wt}(\phi)_{B,w}}{B(w,w)}\phi_{B,w,w}+\phi_{B,w,a}+\phi_{B,b,w},$

for some $a,b\in\\{w\\}_{B}^{\perp}$ and $\phi_{0}\in E^{(0)}(B,w)$ of weight
$0$. Then

$B_{\phi}=B_{\phi_{0}}+\frac{\text{wt}(\phi)_{B,w}^{2}}{B(w,w)^{2}}B_{\phi_{B,w,w}}+B_{\phi_{B,w,a}}+B_{\phi_{B,b,w}}+2\left(B_{\phi_{0},\phi_{B,b,w}}+\frac{\text{wt}(\phi)_{B,w}}{B(w,w)}B_{\phi_{B,w,w},\phi_{B,w,a}}\right).$

###### Proof.

Writing $\phi$ as above and expanding gives

$\displaystyle B_{\phi}$
$\displaystyle=B_{\phi_{0}}+\frac{\text{wt}(\phi)_{B,w}^{2}}{B(w,w)^{2}}B_{\phi_{B,w,w}}+B_{\phi_{B,w,a}}+B_{\phi_{B,b,w}}$
$\displaystyle+2\frac{\text{wt}(\phi)_{B,w}}{B(w,w)}(B_{\phi_{0},\phi_{B,w,w}}+B_{\phi_{B,w,w},\phi_{B,w,a}}+B_{\phi_{B,w,w},\phi_{B,b,w}})$
$\displaystyle+2(B_{\phi_{0},\phi_{B,w,a}}+B_{\phi_{0},\phi_{B,b,w}}+B_{\phi_{B,w,a},\phi_{B,b,w}}).$

The seventh and tenth terms vanish since $a,b\in\\{w\\}_{B}^{\perp}$, whereas
the fifth and eighth terms vanish since $\phi_{0}\in E^{(0)}(B,w)$ has weight
$0$. ∎

The above Lemma allows us to either extract information about potential
isometries $\phi$ satisfying the equality $B^{\prime}=B_{\phi}$, or to give
potential obstructions to their existence.

######  Theorem 3.2.

Let $B^{\prime}$ be an arbitrary non-degenerate symmetric bilinear form on
$V$. If $B^{\prime}=B_{\phi}$ for some $\phi\in\text{End}_{K}(V)$ then the
following holds for any $w\in V\backslash\\{0\\}$ and
$z_{1},z_{2}\in\\{w\\}_{B}^{\perp}$:

$\displaystyle B^{\prime}(w,w)$
$\displaystyle=B(w,w)(\text{wt}(\phi)_{B,w}^{2}+B(w,w)B(b,b)),$ $\displaystyle
B^{\prime}(w,z_{1})$
$\displaystyle=B(w,w)(\text{wt}(\phi)_{B,w}B(a,z_{1})+B(b,\phi_{0}(z_{1}))),$
$\displaystyle B^{\prime}(z_{1},z_{2})$
$\displaystyle=B(\phi_{0}(z_{1}),\phi_{0}(z_{2}))+B(a,z_{1})B(a,z_{2})B(w,w).$

Here $a,b\in\\{w\\}_{B}^{\perp}$ are as in Proposition 3.1.

###### Proof.

If $B^{\prime}=B_{\phi}$ then Proposition 3.1 tells us that:

$B^{\prime}=B_{\phi_{0}}+\frac{\text{wt}(\phi)_{B,w}^{2}}{B(w,w)^{2}}B_{\phi_{B,w,w}}+B_{\phi_{B,w,a}}+B_{\phi_{B,b,w}}+2\left(B_{\phi_{0},\phi_{B,b,w}}+\frac{\text{wt}(\phi)_{B,w}}{B(w,w)}B_{\phi_{B,w,w},\phi_{B,w,a}}\right).$

The claim follows by evaluating both sides at each pair of vectors, recalling
that $\phi_{0}$ has weight $0$ (so that $B(\phi_{0}(w),x)=B(w,\phi_{0}(x))=0$
for any $x\in V$) and that $a,b,z_{1},z_{2}$ are orthogonal to $w$. ∎

In the special case of $B$ being the standard symmetric bilinear form and
$w=(1,1,...,1)$, we find that equation one of Theorem 3.2 is equivalent to
equation $(18)$ of [5] and equation $(3.1)$ of [3] (both of which proved
useful in determining integer matrix factorizations). Theorem 3.2 allows us to
consider the case of arbitrary non-degenerate symmetric bilinear form, and
fills in the missing equations that must be solved to determine an isometry
$\phi$ (if it exists).

######  Remark 3.3.

When $K\subseteq\mathbb{R}$ and $B^{\prime}$ is an inner product then we can
apply the Cauchy-Schwarz inequality to give following constraint for any
$z\in\\{w\\}_{B}^{\perp}$:

$|B^{\prime}(w,z)|^{2}\leq B^{\prime}(w,w)B^{\prime}(z,z).$

This implies a corresponding inequality between the quantities on the RHS in
Theorem 3.2.

##  4\. Application to integer matrix factorization

In this section we let $K=\mathbb{Q}$ and fix a lattice $\Lambda\subset V$
(i.e. a $\mathbb{Z}$-submodule satisfying $\mathbb{Q}\Lambda=V$). Further, we
assume that $B(v,v^{\prime})\in\mathbb{Z}$ for all $v,v^{\prime}\in\Lambda$,
so that $\Lambda$ is an integral lattice with respect to $B$. In particular,
this implies that $\Lambda\subseteq\Lambda^{*}=\\{v\in
V\,|\,B(u,v)\subseteq\mathbb{Z},\forall u\in\Lambda\\}$.

We are interested in studying the isometry problem for $(\Lambda,B)$. This is
the problem of detecting whether a given non-degenerate symmetric bilinear
form $B^{\prime}$ on $V$ can be written as $B^{\prime}=B_{\phi}$ for some
$\phi\in\text{GL}(\Lambda)=\\{\phi\in\text{GL}(V)\,|\,\phi(\Lambda)=\Lambda\\}$
(and further to construct such a map if it exists). Fixing a basis for
$\Lambda$, this is equivalent to giving an integer matrix factorization of the
form $\mathcal{B}^{\prime}=M^{T}\mathcal{B}M$ for some
$M\in\text{GL}_{n}(\mathbb{Z})$ (where $\mathcal{B},\mathcal{B}^{\prime}$ are
Gram matrices of $B,B^{\prime}$ respectively, with respect to this basis for
$\Lambda$).

Such problems have been studied in number theory for a long time. For example,
in the case $K=\mathbb{Q}$ and $\mathcal{B}=I_{n}$ (i.e. the standard
symmetric bilinear form on $\Lambda=\mathbb{Z}^{n}\subset V=\mathbb{Q}^{n}$),
it is a well known theorem (e.g. see the paper of Mordell [6]) that such a
factorization is always possible if $n\leq 7$,
$\text{det}(\mathcal{B}^{\prime})=1$ and $\mathcal{B}^{\prime}$ is positive
definite. In other words, every symmetric positive definite matrix
$\mathcal{B}^{\prime}\in\text{SL}_{n}(\mathbb{Z})$ has a factorization
$\mathcal{B}^{\prime}=M^{T}M$ for some $M\in\text{GL}_{n}(\mathbb{Z})$
whenever $n\leq 7$. In dimension $n=8$ there may not be such a factorization.
Indeed, any Gram matrix coming from a basis of the $E_{8}$ lattice (under the
usual Euclidean inner product) provides a counter-example. As the rank $n$
grows, the number of counter-examples grows rapidly.

As mentioned previously, in the papers [3] and [5], the authors utilised a
superalgebra decomposition of $M_{n}(\mathbb{Q})$ in order to provide further
understanding of this special case of the isometry problem. In doing so, they
were able to provide potential obstructions to the existence of such
isometries by relating the problem with integer points on certain quadratic
forms. When such an isometry exists, information about the isometry is learned
from the corresponding integer points. The results of the previous sections
allow us to extend the results in the above papers to study more general
isometry problems through this lens.

######  Proposition 4.1.

If $w\in\Lambda$ and $\phi\in\mathrm{GL}(\Lambda)$ decomposes as

$\phi=\phi^{(0)}+\phi^{(1)}=\phi_{0}+\frac{\text{wt}(\phi)_{B,w}}{B(w,w)}\phi_{B,w,w}+\phi_{B,w,a}+\phi_{B,b,w}$

with $a,b\in\\{w\\}_{B}^{\perp}$, then $a\in\frac{1}{B(w,w)^{2}}\Lambda^{*}$
and $b\in\frac{1}{B(w,w)^{2}}\Lambda$.

###### Proof.

We first claim that $\phi^{(0)}(w)\in\frac{1}{B(w,w)}\Lambda$ and
$\phi^{(1)}(w)\in\frac{1}{B(w,w)}\Lambda$. The first inclusion follows from
the proof of Lemma 2.1 since:

$\displaystyle\phi^{(0)}(w)=\frac{B(w,\phi^{(0)}(w))}{B(w,w)}w=\frac{B(w,\phi^{(0)}(w)+\phi^{(1)}(w))}{B(w,w)}w=\frac{B(w,\phi(w))}{B(w,w)}w\in\frac{1}{B(w,w)}\Lambda,$

and the second follows from the fact that
$\phi^{(1)}(w)=\phi(w)-\phi^{(0)}(w)\in\frac{1}{B(w,w)}\Lambda$. Note that the
proof of Lemma 2.4 gives $b=\frac{1}{B(w,w)}\phi^{(1)}(w)$, and so
$b\in\frac{1}{B(w,w)^{2}}\Lambda$. For the claim about $a$ we recall that by
Lemma 2.6 we have that

$\displaystyle\phi^{\dagger}=(\phi^{\dagger})^{(0)}+(\phi^{\dagger})^{(1)}=(\phi^{(0)})^{\dagger}+(\phi^{(1)})^{\dagger},$

since $(\phi^{(0)})^{\dagger}\in E^{(0)}(B,w)$ and $(\phi^{(1)})^{\dagger}\in
E^{(1)}(B,w)$. Note that $\phi(\Lambda)=\Lambda$ implies that
$\phi^{\dagger}(\Lambda^{*})=\Lambda^{*}$. We also have
$w\in\Lambda\subseteq\Lambda^{*}$, and so we may apply the same argument as
above to conclude that
$(\phi^{(0)})^{\dagger}(w)\in\frac{1}{B(w,w)}\Lambda\subseteq\frac{1}{B(w,w)}\Lambda^{*}$,
hence that
$(\phi^{(1)})^{\dagger}=\phi^{\dagger}(w)-(\phi^{(0)})^{\dagger}(w)\in\frac{1}{B(w,w)}\Lambda^{*}$.
Finally, since the adjoint swaps the roles of $a$ and $b$ in the
$E^{(1)}(B,w)$ component we learn that
$a=\frac{1}{B(w,w)}(\phi^{(1)})^{\dagger}(w)\in\frac{1}{B(w,w)^{2}}\Lambda^{*}$.
∎

######  Corollary 4.2.

Suppose that $B^{\prime}$ is a non-degenerate symmetric bilinear form on $V$.
If $B^{\prime}=B_{\phi}$ for some $\phi\in\text{GL}(\Lambda)$ then the vectors
$\tilde{a}=B(w,w)^{2}a\in\Lambda^{*}$, $\tilde{b}=B(w,w)^{2}b\in\Lambda$ from
Proposition 4.1 satisfy:

$\displaystyle B(w,w)^{2}B^{\prime}(w,w)$
$\displaystyle=B(w,w)B(w,\phi(w))^{2}+B(\tilde{b},\tilde{b}),$ $\displaystyle
B(w,w)^{3}B^{\prime}(w,z)$
$\displaystyle=B(w,w)B(w,\phi(w))B(\tilde{a},z_{0})+B(\tilde{b},\tilde{\phi}_{0}(z_{0})),$
$\displaystyle B(w,w)^{4}B^{\prime}(z,z)$
$\displaystyle=B(\tilde{\phi}_{0}(z_{0}),\tilde{\phi}_{0}(z_{0}))+B(w,w)B(\tilde{a},z_{0})^{2},$

for any $z_{0}\in\Lambda$ (with
$z=z_{0}-\frac{B(z_{0},w)}{B(w,w)}w\in\\{w\\}^{\perp}$ and
$\tilde{\phi}_{0}(z_{0})=B(w,w)^{2}\phi_{0}(z_{0})\in\Lambda$).

###### Proof.

By Theorem 3.2 we have that

$B^{\prime}(w,w)=B(w,w)(\text{wt}(\phi)_{B,w}^{2}+B(w,w)B(b,b)).$

Clearing denominators gives the first equation. Also note that

$\displaystyle\phi(z)$ $\displaystyle=\phi_{0}(z)+B(a,z)w$
$\displaystyle=\phi_{0}(z_{0})+\frac{B(z_{0},w)}{B(w,w)}\phi_{0}(w)+\left(B(a,z_{0})+\frac{B(z_{0},w)}{B(w,w)}B(a,w)\right)w$
$\displaystyle=\phi_{0}(z_{0})+B(a,z_{0})w.$

Since $\phi(z)\in\frac{1}{B(w,w)}\Lambda$ and
$B(a,z_{0})w=\frac{1}{B(w,w)^{2}}B(\tilde{a},z_{0})w\in\frac{1}{B(w,w)^{2}}\Lambda$,
we find that $\phi_{0}(z_{0})\in\frac{1}{B(w,w)^{2}}\Lambda$. Again, by
Theorem 3.2 we have:

$\displaystyle B^{\prime}(w,z_{1})$
$\displaystyle=B(w,w)(\text{wt}(\phi)_{B,w}B(a,z_{1})+B(b,\phi_{0}(z_{1}))),$
$\displaystyle B^{\prime}(z_{1},z_{2})$
$\displaystyle=B(\phi_{0}(z_{1}),\phi_{0}(z_{2}))+B(a,z_{1})B(a,z_{2})B(w,w).$

Inserting $z_{1}=z_{2}=z$ and clearing denominators gives the second and third
equations. ∎

Corollary 4.2 tells us that information about potential lattice isometries can
be determined by solving certain Diophantine equations. For example, the first
and third equations define integral quadratic forms, and if the underlying
quadratic space is positive definite then these quadratic forms are also
positive definite (allowing for a finite search). If there are no solutions to
any of these equations then we immediately deduce that no such isometry
exists.

As mentioned above, after choosing a basis for $\Lambda$ we see that the above
equations give insight into the solution of integer matrix factorization
problems of the form $\mathcal{B}^{\prime}=M^{T}\mathcal{B}M$ with
$M,\mathcal{B},\mathcal{B}^{\prime}\in\text{GL}_{n}(\mathbb{Z})$ and
$\mathcal{B},\mathcal{B}^{\prime}$ symmetric (Gram matrices).

In the examples below we consider the space $V=\mathbb{Q}^{n}$ and lattice
$\Lambda=\mathbb{Z}^{n}$ (i.e. $\mathbb{Z}$-span of the standard basis). In
all cases we find that the first equation of Corollary 4.2 suffices to prove
that no isometry exists.

######  Example 4.3.

If $n=2$ then

$\displaystyle B(u,v)$ $\displaystyle=u_{1}v_{1}+5u_{2}v_{2},$ $\displaystyle
B^{\prime}(u,v)$
$\displaystyle=2u_{1}v_{1}+u_{1}v_{2}+u_{2}v_{1}+3u_{2}v_{2},$

are non-degenerate symmetric bilinear forms on $V$, and we see that $\Lambda$
is integral with respect to both forms.

Suppose that $B^{\prime}=B_{\phi}$ for some
$\phi\in\text{GL}(\Lambda)\cong\text{GL}_{2}(\mathbb{Z})$. Choosing $w=(1,0)$
gives $\tilde{b}=b=(0,t)\in\mathbb{Z}^{2}$, $B(w,w)=1$ and
$B^{\prime}(w,w)=2$. The first equation of Corollary 4.2 then implies that
there are integer solutions to the equation:

$2=B(w,\phi(w))^{2}+5t^{2}.$

This is a contradiction, hence $(\Lambda,B^{\prime})$ and $(\Lambda,B)$ are
not isometric (despite these forms having Gram matrices of the same
determinant, a necessary condition for isometry).

We can also use Corollary 4.2 to prove that certain infinite families of pairs
of forms in rank 2 and rank 3 lattices cannot be isometric, despite their
corresponding Gram matrices having the same determinant.

######  Example 4.4.

If $n=2$ then

$B=(u,v)=m^{2}u_{1}v_{1}+n^{2}u_{2}v_{2}$

is a non-degenerate symmetric bilinear form on $V$ whenever
$m,n\in\mathbb{Q}\backslash\\{0\\}$. Suppose that
$m,n\in\mathbb{Z}\backslash\\{0\\}$. Then $\Lambda$ is an integral lattice
with respect to $B$.

For any choice of $\alpha,\beta,\gamma\in\mathbb{Z}$ such that
$\alpha\gamma-\beta^{2}=(mn)^{2}$, we have that

$B^{\prime}(u,v)=\alpha u_{1}v_{1}+\beta u_{1}v_{2}+\beta u_{2}v_{1}+\gamma
u_{2}v_{2}$

is another non-degenerate symmetric bilinear form on $V$ with the same
determinant as $B$. Also, $\Lambda$ is integral with respect to this form.

Suppose that there exists a prime $p\equiv 3\bmod 4$ dividing $\alpha$ to an
odd power. Then $B^{\prime}=B_{\phi}$ for some $\phi\in\text{GL}(\Lambda)$ is
impossible, since choosing $w=(1,0)$ in the first equation of Corollary 4.2
gives:

$\alpha\cdot
m^{4}=m^{2}B(w,\phi(w))^{2}+n^{2}t^{2}=(mB(w,\phi(w))^{2}+(nt)^{2}.$

This has no integer solutions by Fermat’s theorem on sums of two squares (e.g.
Theorem $13.3$ of [1] and the Corollary that follows). It follows that
$(\Lambda,B)$ and $(\Lambda,B^{\prime})$ are not isometric.

######  Example 4.5.

If $n=3$ then

$\displaystyle B(u,v)$
$\displaystyle=(2m^{2}+1)u_{1}v_{1}-u_{1}v_{2}-u_{2}v_{1}+u_{2}v_{2}+2m^{2}u_{3}v_{3},$
$\displaystyle B^{\prime}(u,v)$
$\displaystyle=4m^{3}u_{1}v_{1}+mu_{2}v_{2}+u_{3}v_{3},$

are non-degenerate symmetric bilinear forms on $V$ for any
$m\in\mathbb{Q}\backslash\\{0\\}$. Suppose that
$m\in\mathbb{Z}\backslash\\{0\\}$. Then $\Lambda$ is an integral lattice with
respect to $B$ and also with respect to $B^{\prime}$. As in previous examples,
these forms have the same determinant and so there is a chance that
$(\Lambda,B)$ and $(\Lambda,B^{\prime})$ are isometric.

Suppose that $B^{\prime}=B_{\phi}$ for some $\phi\in\text{GL}(\Lambda)$.
Choosing $w=(1,1,1)$ gives $\tilde{b}=(b_{1},b_{2},-b_{1})$ for some
$b_{1},b_{2}\in\mathbb{Z}$, since $B(w,\tilde{b})=0$ if and only if
$2m^{2}(b_{1}+b_{3})=0$. It is then readily calculated that

$\displaystyle B(w,w)$ $\displaystyle=4m^{2},$ $\displaystyle B^{\prime}(w,w)$
$\displaystyle=(1+m+4m^{3}),$ $\displaystyle B(\tilde{b},\tilde{b})$
$\displaystyle=(4m^{2}+1)b_{1}^{2}-2b_{1}b_{2}+b_{2}^{2}=(2mb_{1})^{2}+(b_{1}-b_{2})^{2},$

so that equation one of Corollary 4.2 gives:

$\displaystyle 16m^{4}(4m^{3}+m+1)$
$\displaystyle=4m^{2}B(w,\phi(w))^{2}+(2mb_{1})^{2}+(b_{1}-b_{2})^{2},$
$\displaystyle=(2mB(w,\phi(w))^{2}+(2mb_{1})^{2}+(b_{1}-b_{2})^{2}.$

By Legendre’s theorem on sums of three squares (e.g. Theorem $13.5$ of [1]),
there are no integer solutions to this equation whenever $16m^{4}(4m^{3}+m+1)$
is of the form $4^{t}(8k+7)$ for some $t,k\geq 0$. It follows that in this
case there is no such isometry between $(\Lambda,B)$ and
$(\Lambda,B^{\prime})$.

Since any odd integer satisfies $m^{4}\equiv 1\bmod 8$, the above condition is
equivalent to $4m^{3}+m+1=4^{t}(8k+7)$ for some $t,k\geq 0$. By a tedious case
by case check, one finds that asymptotically this happens for $\frac{1}{6}$ of
the possible $m$ values. Thus our method has proved that at least
$\frac{1}{6}$ of the pairs of forms in this family cannot be isometric over
$\mathbb{Z}$. In fact, there are no values of $m$ for which these pairs of
forms are isometric, and this can be seen by similar calculations for other
choices of vector $w$.

Similarly, for any choice of
$\alpha,\beta,\gamma\in\mathbb{Z}\backslash\\{0\\}$ satisfying
$\alpha\gamma-\beta^{2}=4m^{4}$ the form:

$B^{\prime\prime}(u,v)=\alpha u_{1}v_{1}+\beta u_{1}v_{2}+\beta
u_{2}v_{1}+\gamma u_{2}v_{2}+u_{3}v_{3}$

is a non-degenerate symmetric bilinear form on $V$ such that $\Lambda$ is
integral with respect to $B^{\prime\prime}$ (and also has the same determinant
as $B$).

Suppose again that $w=(1,1,1)$ and that
$B^{\prime\prime}(w,w)=\alpha+2\beta+\gamma+1=4^{t}(8k+7)$ for some $t,k\geq
0$. Then the same argument as above shows that $(\Lambda,B)$ and
$(\Lambda,B^{\prime\prime})$ are not isometric either, by Legendre’s theorem
on sums of three squares.

Numerical examples of the above are given by $m=3$ and
$(\alpha,\beta,\gamma)=(60,6,6)$. Since we have that

$\displaystyle 4m^{3}+m+1$ $\displaystyle=28=4\cdot 7,$
$\displaystyle\alpha\gamma-\beta^{2}$ $\displaystyle=4m^{4}=324,$
$\displaystyle\alpha+2\beta+\gamma+1$ $\displaystyle=79\equiv 7\bmod 8,$

we conclude that the following forms are such that $(\Lambda,B)$ and
$(\Lambda,B^{\prime})$ are not isometric and $(\Lambda,B)$ and
$(\Lambda,B^{\prime\prime})$ are not isometric:

$\displaystyle B(u,v)$
$\displaystyle=19u_{1}v_{1}-u_{1}v_{2}-u_{2}v_{1}+u_{1}v_{2}+18u_{3}v_{3},$
$\displaystyle B^{\prime}(u,v)$
$\displaystyle=108u_{1}v_{1}+3u_{2}v_{2}+u_{3}v_{3},$ $\displaystyle
B^{\prime\prime}(u,v)$
$\displaystyle=60u_{1}v_{1}+6u_{1}v_{2}+6u_{2}v_{1}+6u_{2}v_{2}+u_{3}v_{3}.$

## References

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