# Counting flat cycles in the homology of locally symmetric spaces

Daniel Studenmund and Bena Tshishiku

###### Abstract

For a prime $p$, let $\Gamma(p^{\ell})<\operatorname{SL}_{n+1}(\mathbb{Z})$ be
a principal congruence subgroup and let $Y(p^{\ell})$ denote its locally
symmetric space. We prove that the subspace of $H_{n}(Y(p^{\ell});\mathbb{Q})$
generated by flat cycles grows at least as fast as as a linear function of
$\operatorname{Vol}(Y(p^{\ell}))^{\frac{n+1}{n^{2}+2n}}$. We also prove
similar results for other locally symmetric spaces. This addresses a question
of Avramidi–Nguyen-Phan.

###### Contents

  1. 1 Introduction
  2. 2 Topological argument for homology growth in finite covers
  3. 3 Algebraic groups and geometric cycles
  4. 4 Explicit construction of periodic maximal flats
  5. 5 Periodic maximal flats in homology
  6. 6 Size of Lie groups over finite rings
  7. 7 Counting flat cycles

##  1 Introduction

Let $G<\operatorname{GL}_{d}$ be an algebraic group defined over $\mathbb{Q}$
whose real points $G(\mathbb{R})$ is a semisimple Lie group of noncompact
type. It is a basic problem to understand the (co)homology of the arithmetic
group $G(\mathbb{Z})$ and its principal congruence subgroups

$\Gamma(s):=\ker\big{[}G(\mathbb{Z})\rightarrow
G(\mathbb{Z}/s\mathbb{Z})\big{]}.$

######  Problem 1.1.

For a fixed prime $p$ and $k\geq 1$, determine the rate of growth of the
homology groups $H_{k}(\Gamma(p^{\ell});\mathbb{Q})$ as $\ell$ increases.

Let $Y(p^{\ell})=\Gamma(p^{\ell})\backslash G(\mathbb{R})/K$ denote the
locally symmetric space for $\Gamma(p^{\ell})$. Then

$H_{*}(\Gamma(p^{\ell});\mathbb{Q})\cong H_{*}(Y(p^{\ell});\mathbb{Q}),$

and by the theory of $L^{2}$-Betti numbers, if $k\neq\frac{1}{2}\dim
Y(p^{\ell})$, then $H_{k}(Y(p^{\ell});\mathbb{Q})$ grows sublinearly in terms
of the volume $\operatorname{Vol}(Y(p^{\ell}))$, i.e.

$\lim_{\ell\rightarrow\infty}\frac{\dim
H_{k}(Y(p^{\ell});\mathbb{Q})}{\operatorname{Vol}(Y(p^{\ell}))}=0.$ (1)

See e.g. [L0̈2, §0.6,§5.2] and [Gab02]. When $k=\frac{1}{2}\dim Y(p^{\ell})$
then $H_{k}(Y(p^{\ell});\mathbb{Q})$ grows linearly in the volume for certain
$G$ [L0̈2, op. cit.].

No other general results are known about the growth of
$H_{k}(Y(p^{\ell});\mathbb{Q})$ outside of Borel’s stable range. Some sporadic
results are known for $G(\mathbb{R})=\operatorname{SO}(n,1)$ [Xue92] and
$G(\mathbb{R})=U(n,m)$ [Mar14, MS19, GG20, Sto21].

Our main results provide lower bounds on the rate of growth of
$H_{n}(Y(p^{\ell});\mathbb{Q})$ for certain $G$, where $n$ is the real rank of
$G$. Precise statements follow.

######  Theorem A.

Fix $n\geq 2$ and let $G=\operatorname{SL}_{n+1}$. Denoting
$\Gamma(s)<\operatorname{SL}_{n+1}(\mathbb{Z})$ and
$Y(s)=\Gamma(s)\backslash\operatorname{SL}_{n+1}(\mathbb{R})/\operatorname{SO}(n)$
as above, there are infinitely many primes $p$ with the following property.
There exist constants $\mathcal{C},\mathcal{L}>0$ such that

$\dim
H_{n}(Y(p^{\ell});\mathbb{Q})\geq\mathcal{C}\cdot\operatorname{Vol}(Y(p^{\ell}))^{\frac{n+1}{n^{2}+2n}}$

for each $\ell\geq\mathcal{L}$.

######  Theorem B.

For $n=2m\geq 2$, let $Q_{n}$ be the unimodular, even integral symmetric
bilinear form with signature $(n,n)$. Let $G=\operatorname{SO}(Q_{n})$ be its
isometry group, and define $\Gamma(s)<\operatorname{SO}(Q_{n};\mathbb{Z})$ and
$Y(s)=\Gamma(s)\backslash G(\mathbb{R})/K$ as above. For any $\epsilon>0$,
there are infinitely many primes $p$ with the following property. There exist
constants $\mathcal{C},\mathcal{L}>0$ such that

$\dim
H_{n}(Y(p^{\ell});\mathbb{Q})\geq\mathcal{C}\cdot\operatorname{Vol}(Y(p^{\ell}))^{\frac{n-1-\epsilon}{n^{2}}}$

for each $\ell\geq\mathcal{L}$.

Theorems A and B are proved by giving a lower bound on the dimension of the
subspace of $H_{n}(Y(p^{\ell});\mathbb{Q})$ spanned by _flat cycles_ , i.e.
compact, totally geodesic, flat submanifolds of $Y(p^{\ell})$ of maximal
dimension $n$. Using the same technique we can prove a similar result for
Hilbert modular varieties.

######  Theorem C.

Fix $n\geq 2$ and fix a degree-$n$ totally real number field $F/\mathbb{Q}$
with ring of integers $\mathcal{O}_{F}$. For a nonzero ideal
$\mathfrak{s}\subset\mathcal{O}_{F}$, consider the congruence subgroup

$\Gamma(\mathfrak{s})=\ker\big{[}\operatorname{SL}_{2}(\mathcal{O}_{F})\rightarrow\operatorname{SL}_{2}(\mathcal{O}_{F}/\mathfrak{s})\big{]}$

and let
$Y(\mathfrak{s})=\Gamma(\mathfrak{s})\backslash(\mathbb{H}^{2})^{\times n}$.
There are infinitely many prime ideas $\mathfrak{p}\subset\mathcal{O}_{F}$
with the following property. There exist constants $\mathcal{C},\mathcal{L}>0$
such that if $\ell\geq\mathcal{L}$, then the subspace of
$H_{n}(Y(\mathfrak{p}^{\ell});\mathbb{Q})$ spanned by flat cycles has
dimension at least
$\mathcal{C}\cdot\operatorname{Vol}(Y(\mathfrak{p}^{\ell}))^{1/3}$.

In Theorem C, the real rank $n$ is equal to half the dimension of
$Y(\mathfrak{p}^{\ell})$, and it is known that
$H_{n}(Y(\mathfrak{p}^{\ell});\mathbb{Q})$ grows linearly in the volume, again
by results in $L^{2}$-cohomology [L0̈2, Thm. 5.12] and [Gab02]. In particular,
the homology we produce using flat cycles only accounts for a fraction of
$H_{n}(Y(\mathfrak{p}^{\ell});\mathbb{Q})$.

Avramidi–Nguyen-Phan [ANP15, §9] pose the problem of determining the growth
rate of the subspace of homology spanned by flat cycles in congruence covers.
Our results provide nontrivial lower bounds. We do not know of any upper bound
other than that given by (1). In particular, it would be interesting to
determine whether the exponent $\frac{1}{3}$ in Theorem C is sharp; if it is,
then the growth of $H_{n}(Y(\mathfrak{p}^{\ell});\mathbb{Q})$ is not fully
accounted for by flat cycles.

######  Remark 1.2.

Problem 1.1 is related to a conjecture of Sarnak–Xue. Recall that the
cohomology of an arithmetic group can be described representation-
theoretically by Matsushima’s formula, and computing
$H_{k}(Y(p^{\ell});\mathbb{C})$ reduces to computing the multiplicity
$m(\pi,p^{\ell})$ of certain representations $\pi$ occurring in
$L^{2}(\Gamma(p^{\ell})\backslash G(\mathbb{R}))$. Sarnak–Xue [SX91]
conjectured upper bounds on the growth of these multiplicities. Ultimately,
this has consequences for the growth of Betti numbers, but this conjecture has
been approached in only a small number of cases. See, for example, the work of
Marshall, Marshall–Shin, and Gerbelli-Gauthier [Mar14, MS19, GG20], where the
case $G=U(n,m)$ is studied using the theory of endoscopy.

#### About the proof techniques.

This paper combines several ingredients, as indicated in the table of
contents. The starting point is recent work [ANP15], [Tsh21], [Zsc21] that
proves that flat cycles give nontrivial homology classes in some congruence
cover $Y(p^{k})$ in the cases considered in Theorems A, B, and C,
respectively. We combine this with a simple topological argument of Xue
[Xue92] to estimate the number of linearly independent cycles in an orbit of
the deck group of a further cover $Y(p^{\ell})\rightarrow Y(p^{k})$. This
reduces the problem mostly to computing the size of the image of
$H(\mathbb{Z})\rightarrow H(\mathbb{Z}/m\mathbb{Z})$ for certain algebraic
groups $H$. A key difficulty is that we must consider the case when $H$ is an
algebraic torus, and these groups do not satisfy strong approximation. We
overcome this with ad hoc arguments that use classical results from number
theory about splitting primes in field extensions, including Chebotarev’s
density theorem and the Kronecker–Weber theorem; see e.g. Lemma 6.5 and
Proposition 4.15. The orthogonal case (Theorem B) is the most involved.

Theorems A–C remain true when their conclusions are strengthened from the
existence of “infinitely many primes” to “a set of primes of positive
density.” This is immediate from the proofs presented in §6 and §7, which
require the exclusion of finitely many primes and applications Chebotarev’s
density theorem and Dirichlet’s theorem on primes in arithmetic progressions.

######  Remark 1.3.

Examples of nonzero flat cycles in the homology of $Y=\Gamma\backslash
G(\mathbb{R})/K$ are constructed for $G=\operatorname{SL}_{n+1}$,
$G=\operatorname{SO}(Q_{n})$, and $G=R_{F/\mathbb{Q}}(\operatorname{SL}_{2})$
by [ANP15], [Tsh21], and [Zsc21], respectively. The same works [ANP15, Tsh21,
Zsc21] prove that flat cycles span subspaces in
$H_{n}(Y(p^{\ell});\mathbb{Q})$ of arbitrarily large dimension as $\ell$
increases. Our main theorems both quantify and give an alternate proof of this
latter fact, assuming that a single non-zero flat cycle exists (which is the
most involved part of [ANP15, Tsh21, Zsc21]).

######  Remark 1.4.

The cohomology of arithmetic groups has applications to the study of fiber
bundles. In particular, Theorem B implies a similar growth theorem for the
homology for certain connected covers of the classifying space
$B\operatorname{Diff}(W_{n})$ for bundles with fiber
$W_{n}=\\#_{n}(S^{2d}\times S^{2d})$ for $d\geq 2$. For more information, see
[Tsh21, Cor. 2] and Theorem 29 in the Appendix of [Tsh21] by M. Krannich.

#### Paper structure.

Sections 2 and 3 present the general approach used to prove each of Theorems
A–C. The details of the proofs appear in parallel in Sections 4–7.

#### Acknowledgements.

The second author is supported by NSF grant DMS-2104346. The authors thank
Jenny Wilson and Jeremy Miller for organizing of the Banff Workshop 21w5011,
where the idea for this project was formed. Thanks to Kevin Schreve for a
helpful email about L2 cohomology.

##  2 Topological argument for homology growth in finite covers

The main goal of this section is to explain a simple topological argument of
Xue [Xue92, §3] that we use to prove our main theorems. Xue explains the
argument in the context of arithmetic groups and congruence covers, but the
argument applies more generally. We explain this below. It could be of
independent interest.

Fix an oriented manifold $Y$ (possibly noncompact) with oriented, embedded
submanifolds $Y_{1},Y_{2}\subset Y$ of complementary dimension
$n_{1}+n_{2}=\dim Y$. Assume that $Y_{1}$ is compact and that $Y_{1},Y_{2}$
intersect transversely with nonzero algebraic intersection number $Y_{1}\cdot
Y_{2}\neq 0$.

The submanifold $Y_{1}\subset Y$ defines a homology class $[Y_{1}]\in
H_{n_{1}}(Y;\mathbb{Q})$. Since $Y_{2}$ is not assumed to be compact, it only
defines a homology class in Borel–Moore homology with closed supports
$H_{n_{2}}^{BM}(Y;\mathbb{Q})$. By Poincaré duality, the intersection pairing
between $H_{n_{1}}(Y;\mathbb{Q})$ and $H_{n_{2}}^{BM}(Y;\mathbb{Q})$ is a
perfect pairing. Note in particular, that $[Y_{1}]\neq 0$ in
$H_{n_{1}}(Y;\mathbb{Q})$ because $Y_{1}\cdot Y_{2}\neq 0$.

Fix a basepoint $y\in Y_{1}\cap Y_{2}$. Fix a finite quotient
$\rho:\pi_{1}(Y)\twoheadrightarrow W$, and let $\pi:\widehat{Y}\rightarrow Y$
be the corresponding covering space. Choose a component $\widehat{Y}_{i}$ of
$\pi^{-1}(Y_{i})$ for $i=1,2$, and set $W_{i}=\rho(\pi_{1}(Y_{i}))$. In
particular $\widehat{Y}_{i}\rightarrow Y_{i}$ is a regular cover with deck
group $W_{i}$.

The following result gives a lower bound on how many $W$-translates of
$\widehat{Y}_{1}$ are linearly independent in
$H_{n_{1}}(\widehat{Y};\mathbb{Q})$.

######  Proposition 2.1 (Xue).

With the preceding setup, the $W$-translates of $[\widehat{Y}_{1}]\in
H_{n_{1}}(\widehat{Y};\mathbb{Q})$ span a subspace of dimension at least
$\frac{1}{|Y_{1}\cap Y_{2}|}\cdot\frac{|W|}{|W_{1}|\cdot|W_{2}|}$.

The proof is an easy consequence of the following linear algebra fact, proved
in [Xue92, Lem. 6].

######  Lemma 2.2.

Let $V$ be a vector space with dual $V^{*}$, and let $\phi\cdot u$ denote the
pairing between $\phi\in V^{*}$ and $u\in V$. Let $S\subset V$ and
$S^{*}\subset V^{*}$ be finite sets and assume that $(1)$ for each $\phi\in
S^{*}$ there exists $u\in S$ such that $\phi\cdot u\neq 0$ and $(2)$ there
exists $t\geq 1$ such that for each $u\in S$, there are at most $t$ elements
in $S^{*}$ such that $\phi\cdot u\neq 0$. Then

$\dim_{\mathbb{Q}}\text{\emph{span}}(S)\geq\frac{|S^{*}|}{t}$

###### Proof of Proposition 2.1.

First we bound the number of translates of $\widehat{Y}_{2}$ that have
nontrivial algebraic intersection with $\widehat{Y}_{1}$.

$\begin{array}[]{rcl}\\#\\{w(\widehat{Y}_{2}):\widehat{Y}_{1}\cdot
w(\widehat{Y}_{2})\neq 0\\}&\leq&\\#\\{w(\widehat{Y}_{2}):\widehat{Y}_{1}\cap
w(\widehat{Y}_{2})\neq\operatorname{\varnothing}\\}\\\\[5.69054pt]
&\leq&|\widehat{Y}_{1}\cap\pi^{-1}(Y_{2})|\\\\[5.69054pt] &=&|Y_{1}\cap
Y_{2}|\cdot|W_{1}|.\end{array}$

The first inequality holds because disjoint submanifolds have zero algebraic
intersection number. The second inequality says that the number of components
of $\pi^{-1}(Y_{2})$ that intersect $\widehat{Y}_{1}$ is bounded by the total
number of intersections (so there is equality precisely when each component of
$\pi^{-1}(Y_{2})$ intersects $\widehat{Y}_{1}$ at most once). For the last
equality, observe that $\widehat{Y}_{1}\cap\pi^{-1}(Y_{2})$ is the preimage of
$Y_{1}\cap Y_{2}$ under the covering map $\widehat{Y}_{1}\rightarrow Y_{1}$.

An identical bound holds for $\\#\\{w(\widehat{Y}_{2}):u(\widehat{Y}_{1})\cdot
w(\widehat{Y}_{2})\neq 0\\}$ for any $u\in W$.

The proposition now follows immediately by applying Lemma 2.2 to
$V=H_{n_{1}}(\widehat{Y};\mathbb{Q})$,
$V^{*}=H_{n_{2}}^{BM}(\widehat{Y};\mathbb{Q})$, and $S$ (resp. $S^{*}$) equal
to the $W$-translates of $[\widehat{Y}_{1}]$ (resp. $[\widehat{Y}_{2}]$). ∎

##  3 Algebraic groups and geometric cycles

Here we explain the general setup for studying geometric cycles in the
homology of arithmetic locally symmetric manifolds, following methods of
Millson [Mil76] and Millson–Raghunathan [MR80]. A good reference is [Sch10].

Let $G<\operatorname{GL}_{d}$ be an algebraic subgroup defined over
$\mathbb{Q}$ such that $G(\mathbb{R})$ is a semisimple Lie group. Fix
reductive $\mathbb{Q}$-subgroups $G_{1},G_{2}<G$.

For any subring $R\subset\mathbb{C}$, we define
$G(R):=G\cap\operatorname{GL}_{d}(R)$, and similarly for $G_{1},G_{2}$. If $G$
is actually defined over $\mathbb{Z}$ (as will hold for some of our examples),
then for any ring $R$ (e.g. $R=\mathbb{Z}/t\mathbb{Z}$), we can define
$G(R)=G\cap\operatorname{GL}_{d}(R)$. More generally, since $G$ is defined
over $\mathbb{Q}$, the group $G(\mathbb{Z}/t\mathbb{Z})$ is defined as long as
$t$ is relatively prime to the (finitely many) denominators appearing in the
defining equations of $G$.

For $r\geq 1$, let $\Gamma(r)<G(\mathbb{Z})$ denote the congruence subgroup

$\Gamma(r)=\ker\big{[}G(\mathbb{Z})\hookrightarrow\operatorname{GL}_{d}(\mathbb{Z})\rightarrow\operatorname{GL}_{d}(\mathbb{Z}/r\mathbb{Z})\big{]},$

and set $\Gamma_{i}(r)=G_{i}(\mathbb{Z})\cap\Gamma(r)$. When $G$ is defined
over $\mathbb{Z}$, we can equivalently define $\Gamma(r)$ as the kernel of the
reduction map $G(\mathbb{Z})\rightarrow G(\mathbb{Z}/r\mathbb{Z})$.

Suppose $K\subset G(\mathbb{R})$ is a maximal compact subgroup such that
$K_{i}:=G_{i}(\mathbb{R})\cap K$ is a maximal compact subgroup of
$G_{i}(\mathbb{R})$ for $i=1,2$. Then the orbit map
$G_{i}(\mathbb{R})\rightarrow G(\mathbb{R})/K$ defined by $g\mapsto gK$
descends to a totally geodesic embedding of $X_{i}:=G_{i}(\mathbb{R})/K_{i}$
into $X:=G(\mathbb{R})/K$. The embedding $X_{i}\hookrightarrow X$ descends to
an immersion of $Y_{i}(r):=\Gamma_{i}(r)\backslash G_{i}(\mathbb{R})/K_{i}$
into $Y(r):=\Gamma(r)\backslash G(\mathbb{R})/K$ for $i=1,2$.

######  Definition 3.1.

Let $G\leq\operatorname{GL}_{n}$ and $G_{1},G_{2}\leq G$ be as above. For a
maximal compact subgroup $K\subset G(\mathbb{R})$ and integer $r\geq 1$, say
that $(G,G_{1},G_{2},K,r)$ is a Millson–Raghunathan tuple if, using the above
notation, the following conditions hold:

  1. (A1)

$G_{1}(\mathbb{Z})$ is cocompact in $G_{1}(\mathbb{R})$.

  2. (A2)

$K_{i}\leq G_{i}(\mathbb{R})$ is a maximal compact subgroup for $i=1,2$.

  3. (A3)

$\dim X_{1}+\dim X_{2}=\dim X$ and $X_{1}\cap X_{2}=\\{eK\\}$.

  4. (A4)

$Y_{i}(r)$ is orientable, and $Y_{i}(r)\rightarrow Y(r)$ is an embedding.

  5. (A5)

$Y_{1}(r),Y_{2}(r)$ intersect transversely and each intersection has the same
sign.

Given a Millson–Raghunathan tuple $(G,G_{1},G_{2},K,r)$, for any natural
number $s$ that is a multiple of $r$, each component of the lift of $Y_{1}(r)$
over the covering map $Y(s)\to Y(r)$ represents a nonzero element of homology.
In particular, following the results of §2 we have the following bound on the
dimension of homology groups of covers of locally symmetric spaces arising
from Millson–Raghunathan tuples.

######  Lemma 3.2.

Suppose $(G,G_{1},G_{2},K,r)$ is a Millson–Raghunathan tuple and let
$d=\dim(X_{1})$. Given $s$ any multiple of $r$, let
$W(r,s):=\Gamma(r)/\Gamma(s)$ and $W_{i}(r,s)=\Gamma_{i}(r)/\Gamma_{i}(s)$ for
$i=1,2$. Then there is a constant $\mathcal{C}>0$ such that

$\dim(H_{d}(Y(s);\mathbb{Q}))\geq\mathcal{C}\cdot\frac{|W(r,s)|}{|W_{1}(r,s)|\cdot|W_{2}(r,s)|}$

###### Proof.

As discussed in §2, condition (A5) guarantees that the homology class
$[Y_{1}(r)]$ in $H_{d}(Y(r);\mathbb{Q})$ is nonzero, because it pairs
nontrivially with the class $[Y_{2}(r)]$ in the Borel–Moore homology of $Y(r)$
with closed supports.

The group $W(r,s)$ is the deck group of the cover $Y(s)\rightarrow Y(r)$. By
Proposition 2.1, the number of linearly independent translates of
$[Y_{1}(s)]\in H_{d}(Y(s);\mathbb{Q})$ under $W(r,s)$ is at least

$\frac{1}{|Y_{1}(r)\cap
Y_{2}(r)|}\cdot\frac{|W(r,s)|}{|W_{1}(r,s)|\cdot|W_{2}(r,s)|}$

∎

#### Construction of Millson–Raghunathan tuples.

We prove Theorems A–C by constructing Millson–Raghunathan tuples
$(G,G_{1},G_{2},K,r)$ with suitable properties. While constructing
Millson–Raghunathan tuples for general $G$ is challenging, many specific
examples are known. We begin in §4 by constructing $G_{1}<G$ as maximal
$\mathbb{R}$-split tori $G_{1}(\mathbb{R})\cong(\mathbb{R}^{\times})^{n}$ that
are anisotropic over $\mathbb{Q}$.

Given $G$ and $G_{1}$, it is not always possible to choose $G_{2}$ so that
condition (A3) holds. In §5 we build on the constructions of §4 to explicitly
construct $G_{2}<G$ and $K<G(\mathbb{R})$ satisfying (A2) and (A3).

For $G$, $G_{1}$, and $G_{2}$ satisfying (A1)–(A3), condition (A4) holds quite
generally. If $r$ is sufficiently large, then $Y_{i}(r)\rightarrow Y(r)$ is an
embedding by a result of Raghunathan; see [Sch10, Thm. E and its proof].
Similarly, one can find infinitely many primes $p$ so that if $r$ is a
multiple of $p^{k}$ for $k$ sufficiently large, then $Y_{i}(r)$ is oriented
for $i=1,2$; see [Sch10, Thm. F] and [RS93, Prop. 2.2 and its proof]. When
$G=\operatorname{SL}_{n+1}$, any prime $p$ admits such $k$, as argued in
[ANP15, §5].

In special cases, it is known that one can choose $r$ so that condition (A5)
holds. This can be shown using an argument that originated in work of Millson
[Mil76] and Millson–Raghunathan [MR80]. We apply these techniques in §5.

In light of Lemma 3.2, to prove Theorems A–C we are interested in bounding
$\frac{|W(r,s)|}{|W_{1}(r,s)|\cdot|W_{2}(r,s)|}$ below. This is done in §7. We
outline the approach here. By the definitions, there is a short exact sequence

$1\rightarrow W(r,s)\rightarrow G(\mathbb{Z})/\Gamma(s)\rightarrow
G(\mathbb{Z})/\Gamma(r)\rightarrow 1,$ (2)

and the group $G(\mathbb{Z})/\Gamma(t)$ is the image of
$G(\mathbb{Z})\hookrightarrow\operatorname{GL}_{d}(\mathbb{Z})\rightarrow\operatorname{GL}_{d}(\mathbb{Z}/t\mathbb{Z})$.
We will provide lower bounds on the size of the image of
$G(\mathbb{Z})\rightarrow\operatorname{GL}_{d}(\mathbb{Z}/s\mathbb{Z})$, and
upper bounds on the size of the images of
$G_{i}(\mathbb{Z})\rightarrow\operatorname{GL}_{d}(\mathbb{Z}/s\mathbb{Z})$
for $i=1,2$. For groups $G$ and $G_{2}$ we consider, these bounds will follow
from strong approximation. Our groups $G_{1}$ are algebraic tori, for which
strong approximation does not apply.

##  4 Explicit construction of periodic maximal flats

Let $X=G(\mathbb{R})/K$ be a symmetric space of noncompact type, and let
$\Gamma<G(\mathbb{R})$ be a lattice. A _flat_ is a geodesically embedded
Euclidean space $\mathbb{E}^{i}\hookrightarrow X$. The maximal dimension of a
flat is the real rank of $G(\mathbb{R})$, which we denote by $n$. A maximal
flat is preserved by a Cartan subgroup $H$ of $G(\mathbb{R})$, and we call the
flat _periodic_ if $\Gamma\cap H$ is a lattice in $H$. In this case,
$(\Gamma\cap H)\backslash H$ is a compact flat manifold and there is an
immersion $(\Gamma\cap H)\backslash H\looparrowright\Gamma\backslash X$.

There are general results of Prasad–Raghunathan [PR72] that guarantee the
existence of periodic maximal flats. These results, however, are not
constructive and are consequently difficult to apply in our setting. The
explicit construction we give below will facilitate the computation in §6 for
the proofs of Theorems A–C.

To construct periodic maximal flats in our cases, it suffices to find a
$\mathbb{Q}$-subgroup $G_{1}<G$ such that
$G_{1}(\mathbb{R})\cong(\mathbb{R}^{\times})^{n}$ and
$G_{1}(\mathbb{Z})<G_{1}(\mathbb{R})$ is cocompact (equivalently,
$G_{1}(\mathbb{Z})$ contains a free abelian group of rank $n$). For then
$G_{1}(\mathbb{R})$ has maximal compact subgroup $K_{1}\cong\\{\pm 1\\}^{n}$
and $G_{1}(\mathbb{R})/K_{1}\cong\mathbb{E}^{n}$ embeds in $G(\mathbb{R})/K$
as a maximal flat for any maximal compact subgroup $K\subset
G_{(}\mathbb{R})$, which is periodic because
$G_{1}(\mathbb{Z})<G_{1}(\mathbb{R})$ is cocompact.

######  Remark 4.1.

For $G=\operatorname{SL}_{n+1}$, it is easy to find $G_{1}$. Fix a totally
real number field $E/\mathbb{Q}$ of degree $n+1$, and let
$G_{1}=R_{E/\mathbb{Q}}^{1}(G_{m})$ be the norm torus in the restriction of
scalars group $R_{E/\mathbb{Q}}(G_{m})$. The group
$G_{1}(\mathbb{Z})\doteq\mathcal{O}_{E}$ is cocompact in
$G_{1}(\mathbb{R})\cong(\mathbb{R}^{\times})^{n}$ by Dirichlet’s unit theorem.
For the computation in §6, it is helpful to assume further that the ring of
integers $\mathcal{O}_{E}$ contains a unit
$\lambda\in\mathcal{O}_{E}^{\times}$ that is also a primitive element, i.e.
$E=\mathbb{Q}(\lambda)$. It may be well known that such fields exist for each
$n$, but we did not find a reference, so we give a construction below. We also
note that this simple construction for $G=\operatorname{SL}_{n+1}$ does not
seem to work as well for $G=\operatorname{SO}(Q_{n})$.

In each case, we define $G_{1}$ as the centralizer of a certain matrix $A$.
There are two basic constructions that will be useful for showing that
$G_{1}(\mathbb{Z})<G_{1}(\mathbb{R})$ is cocompact. Let $F/\mathbb{Q}$ be a
number field with ring of integers $\mathcal{O}_{F}$.

  * •

Fix a polynomial $\xi\in\mathcal{O}_{F}[x]$ that is irreducible in $F[x]$, and
let $\lambda$ be a root of $\xi$. View the number field $E=F[x]/(\xi)\cong
F(\lambda)$ as an $F$ vector space with basis
$1,\lambda,\ldots,\lambda^{d-1}$, where $d=\deg(\xi)$. Multiplication by $E$
on itself defines a ring homomorphism

$\rho:E\rightarrow M_{d}(F).$

The matrix $\rho(\lambda)$ belongs to $M_{d}(\mathcal{O}_{F})$ and has
characteristic polynomial $\xi$ (this is the so-called _companion matrix_ of
$\xi$). If $\xi(0)=1$, then
$\rho(\lambda)\in\operatorname{SL}_{d}(\mathcal{O}_{F})$. In this case
$\mathcal{O}_{F}[\lambda]^{\times}$ maps into the centralizer of
$\rho(\lambda)$ in $\operatorname{GL}_{d}(\mathcal{O}_{F})$.

  * •

Let $A\in\operatorname{SL}_{d}(\mathcal{O}_{F})$ be a matrix whose
characteristic polynomial $\chi$ is irreducible in $F[x]$, and let $\lambda$
be a root of $\chi$ (i.e. an eigenvalue of $A$). The field $F[x]/(\chi)$ can
be identified with

    1. (1)

the subalgebra of $M_{d}(F)$ generated by $A$ (the map $F[x]\rightarrow
M_{d}(F)$ defined by $x\mapsto A$ factors through $(\chi)$ by the
Cayley–Hamilton theorem, and this ideal is the kernel because $\chi$ is the
minimal polynomial of $A$), and

    2. (2)

the vector space $V=F^{d}$ viewed as a $F[x]$-module with $x$ acting by $A$
(for any nonzero $v\in V$, the vectors $v,Av,\ldots,A^{d-1}v$ are linearly
independent – $A$ acts irreducibly on $V$ because $\chi$ is irreducible).

###  4.1 Periodic maximal flats for $\operatorname{SL}_{n+1}$

Fix a polynomial

$\xi=x^{n+1}+a_{n}x^{n}+\cdots+a_{1}x+1$

with integer coefficients $a_{i}\in\mathbb{Z}$, and assume that $\xi$ is
irreducible in $\mathbb{Q}[x]$ and has only real roots. Consider the companion
matrix

$A=\left(\begin{array}[]{ccccc}0&0&\cdots&0&-1\\\ 1&0&\cdots&0&-a_{1}\\\
0&1&\cdots&0&-a_{2}\\\ \vdots&\vdots&\ddots&\vdots&\vdots\\\
0&0&\cdots&1&-a_{n}\end{array}\right)\in\operatorname{SL}_{n+1}(\mathbb{Z}).$

######  Lemma 4.2.

For $\xi$ and $A$ as above, define $G_{1}<\operatorname{SL}_{n+1}$ be the
centralizer of $A$. Then $G_{1}(\mathbb{R})\cong(\mathbb{R}^{\times})^{n}$,
and $G_{1}(\mathbb{Z})$ is cocompact in $G_{1}(\mathbb{R})$.

###### Proof.

Since $A$ has irreducible characteristic polynomial and real roots, $A$ is
diagonalizable over $\mathbb{R}$. This implies that
$G_{1}(\mathbb{R})\cong(\mathbb{R}^{\times})^{n}$.

To show $G_{1}(\mathbb{Z})$ is cocompact in $G_{1}(\mathbb{R})$, we show that
$G_{1}(\mathbb{Z})$ contains an abelian subgroup of rank $n$. Consider the
subalgebra $\mathbb{Q}[A]\subset M_{n+1}(\mathbb{Q})$ generated by $A$. As
mentioned above, $\mathbb{Q}[A]$ is isomorphic to the field
$E=\mathbb{Q}[x]/(\xi)$. The group $\mathbb{Z}[A]^{\times}$ is contained in
$\operatorname{GL}_{n+1}(\mathbb{Z})$ and centralizes $A$. The group
$\mathbb{Z}[A]^{\times}\cap\operatorname{SL}_{n+1}(\mathbb{Z})$ has finite
index in $\mathbb{Z}[A]^{\times}$ and is contained in $G_{1}(\mathbb{Z})$, so
it suffices to show $\mathbb{Z}[A]^{\times}$ has rank $n$. For this, note that
$\mathbb{Z}[A]^{\times}\cong\mathbb{Z}[x]/(\xi)$ has the same rank as
$\mathcal{O}_{E}^{\times}$, which is $n$ by Dirichlet’s unit theorem; cf.
[Neu99, Ch. I, Thms. 7.4 and 12.12]. ∎

######  Remark 4.3 (Irreducible is necessary).

If $\xi$ is reducible and square-free with real roots, then again
$G_{1}(\mathbb{R})=(\mathbb{R}^{\times})^{n}$, but now $G_{1}(\mathbb{Z})$ has
rank $<n$ and hence is not cocompact. For example, suppose $A$ is a block
matrix $B\oplus C$, where $B,C$ have irreducible characteristic polynomials
which are relatively prime. Then $A$ can be diagonalized over $\mathbb{R}$
using a block diagonal matrix, and this implies that its centralizer is also
block diagonal. Now one computes $\operatorname{rank}G_{1}(\mathbb{Z})<n$
using Dirichlet’s unit theorem similar to the proof of Lemma 4.2.

Next we show that for each $n\geq 0$, there is a degree-$(n+1)$ monic
polynomial $\xi\in\mathbb{Z}[x]$ that is irreducible in $\mathbb{Q}[x]$, has
only real roots, and satisfies $\xi(0)=1$. There are likely several arguments
for this, but we were not able to find a reference. Below we make a few
remarks, and then give explicit examples in Lemma 4.7.

######  Remark 4.4.

The characteristic polynomial of a “random” element in
$\operatorname{SL}_{n+1}(\mathbb{Z})$ is irreducible by Rivin [Riv08], but
typically one expects the characteristic polynomial will have non-real roots.
It would be nice to show that a random symmetric matrix in
$\operatorname{SL}_{n+1}(\mathbb{Z})$ has irreducible characteristic
polynomial (since then the roots are necessarily real by the spectral
theorem). There seem to be difficulties to adapting Rivin’s argument [Riv08,
Thm. 5.2] to show this.

######  Remark 4.5.

If $E/\mathbb{Q}$ is a number field and $\alpha\in\mathcal{O}_{E}^{\times}$ is
a unit and also $\alpha$ is a primitive element ($E=\mathbb{Q}(\alpha)$), then
the minimal polynomial of $\alpha$ has the desired properties. However, not
every number field has a primitive element that’s also a unit [ABZ16].

######  Remark 4.6.

Kedlaya [Ked12] constructs for each $n=r+2s$, monic, irreducible polynomials
$\xi\in\mathbb{Z}[x]$ so that the number field $F=\mathbb{Q}[x]/(\xi)$ has $r$
real embeddings and $s$ complex embeddings, and such that
$\mathcal{O}_{F}=\mathbb{Z}[x]/(\xi)$. The polynomials $\xi$ are somewhat
complicated, and we did not check if Kedlaya’s construction includes examples
with $\xi(0)=1$. In any case, the examples we give below are quite simple.

######  Lemma 4.7.

Fix $n\geq 6$, and let $a_{0},a_{1},\ldots,a_{n}$ be integers such that

$0=a_{0}<a_{1}<a_{2}<\cdots<a_{n}\>\>\>\text{ and }\>\>\>a_{i+1}-a_{i}>2.$

Then the degree-$(n+1)$ polynomial

$\xi=1+x(x-a_{1})\cdots(x-a_{n})\in\mathbb{Z}[x]$ (3)

has only real roots and is irreducible in $\mathbb{Q}[x]$.

Lemma 4.7 only gives examples for $n\geq 6$, but it is easy to construct ad
hoc examples for $n\leq 5$. The proof of Lemma 4.7 uses the following
irreducibility criterion, which can be found in [Pra04, Thm. 2.2.8].

######  Theorem 4.8 (Polya irreducibility criterion).

Let $\xi$ be a degree-$d$ polynomial and let $k=\lfloor\frac{d+1}{2}\rfloor$.
If there are distinct integers $b_{0},\ldots,b_{d-1}$ such that

$0<|\xi(b_{i})|<\frac{k!}{2^{k}}\>\>\>\text{ for }\>\>\>0\leq i\leq d-1,$

then $\xi$ is irreducible in $\mathbb{Q}[x]$.

###### Proof of Lemma 4.7.

We sketch the argument. To see that $\xi$ has $n+1$ real roots, let
$m_{0}=-2$, $m_{i}=\frac{a_{i-1}+a_{i}}{2}$ for $i=1,\dotsc,n$, and
$m_{n+1}=a_{n}+2$. Then the assumption that $a_{i+1}-a_{i}>2$ implies that the
sign of $\xi(m_{i})$ is $(-1)^{n-i}$, and so $\xi$ has $n+1$ real roots by the
intermediate value theorem.

Next we explain why $\xi$ is irreducible. Observe that
$\xi(0)=\xi(a_{1})=\cdots=\xi(a_{n})=1$. Since $1<\frac{k!}{2^{k}}$ for $k\geq
4$, we can apply Polya’s Theorem 4.8 as long as
$\lfloor\frac{n+2}{2}\rfloor\geq 4$, i.e. $n\geq 6$, to conclude that $\xi$ is
irreducible. ∎

###  4.2 Periodic maximal flats for $R_{F/\mathbb{Q}}(\operatorname{SL}_{2})$

This case can be treated similar to the $\operatorname{SL}_{n+1}$ case. Let
$F/\mathbb{Q}$ be a totally real number field of degree $n\geq 2$.

Let $\xi=x^{2}+ax+1$ be a polynomial in $\mathcal{O}_{F}[x]$ that is
irreducible in $F[x]$ and has real roots (equivalently, $a^{2}-4$ is positive
and is not a square in $F$). Define

$A=\left(\begin{array}[]{cc}0&-1\\\
1&-a\end{array}\right)\in\operatorname{SL}_{2}(\mathcal{O}_{F}),$

which has characteristic polynomial $\xi$.

Consider $\mathcal{H}=\operatorname{SL}_{2}$ viewed as an algebraic group over
$F$, and define $\mathcal{H}_{1}<\mathcal{H}$ to be the centralizer of $A$.
Set $G=R_{F/\mathbb{Q}}(\mathcal{H})$ and
$G_{1}=R_{F/\mathbb{Q}}(\mathcal{H}_{1})$.

######  Lemma 4.9.

The group $G_{1}(\mathbb{R})$ is isomorphic to $(\mathbb{R}^{\times})^{n}$ and
is embedded diagonally in
$G(\mathbb{R})=\big{(}\operatorname{SL}_{2}(\mathbb{R})\big{)}^{n}$. Also,
$G_{1}(\mathbb{Z})<G_{1}(\mathbb{R})$ is cocompact.

###### Proof.

The proof is similar to the proof of Lemma 4.2. Since $A$ is diagonalizable in
$\mathcal{H}_{1}(\mathbb{R})$, its centralizer is isomorphic to
$\mathbb{R}^{\times}$. Applying restriction of scalars gives the first
statement.

For the second statement, we prove that $G_{1}(\mathbb{Z})$ contains a
rank-$n$ abelian group. The groups $G_{1}(\mathbb{Z})$ and
$\mathcal{H}_{1}(\mathcal{O}_{F})$ are commensurable, so it suffices to work
with $\mathcal{H}_{1}(\mathcal{O}_{F})$.

Similar to the proof of Lemma 4.2, consider the subalgebra $F[A]\subset
M_{2}(F)$ generated by $A$, which is isomorphic to the field $E=F[x]/(\xi)$.
The group $(\mathcal{O}_{F}[A])^{\times}$ is contained in
$\operatorname{GL}_{2}(\mathcal{O}_{F})$ and centralizes $A$. Consequently,
$\mathcal{H}_{1}(\mathcal{O}_{F})$ contains the kernel of the composition

$\big{(}\mathcal{O}_{F}[A])^{\times}\rightarrow\operatorname{GL}_{2}(\mathcal{O}_{F})\xrightarrow{\det}\mathcal{O}_{F}^{\times}.$

This kernel has the same rank as the kernel of the norm map

$\mathcal{O}_{E}^{\times}\rightarrow\operatorname{GL}_{2}(\mathcal{O}_{F})\xrightarrow{\det}\mathcal{O}_{F}^{\times},$

which is $n=(2n-1)-(n-1)$ by Dirichlet’s unit theorem. (Note that the norm map
restricted to $\mathcal{O}_{F}^{\times}$ is the map
$\lambda\mapsto\lambda^{2}$, so the map
$\mathcal{O}_{E}^{\times}\rightarrow\mathcal{O}_{F}^{\times}$ is virtually
surjective.) ∎

######  Remark 4.10 (Example).

Consider $F=\mathbb{Q}(\sqrt{2})$ and $\xi=x^{2}-4x+1$. Then

$A=\left(\begin{array}[]{cc}0&-1\\\
1&4\end{array}\right)\in\operatorname{SL}_{2}(\mathcal{O}_{F}),$

which has $\lambda=2+\sqrt{3}$ as an eigenvalue. Here
$\mathcal{O}_{F}=\mathbb{Z}\\{1,\sqrt{2}\\}$ and
$E\cong\mathbb{Q}(\sqrt{2},\sqrt{3})$. The ring $\mathcal{O}_{F}[A]$ is
isomorphic to

$\mathcal{O}_{F}[\lambda]=\mathbb{Z}\\{1,\sqrt{2},\sqrt{3},\sqrt{6}\\},$

which has finite index in
$\mathcal{O}_{E}=\mathbb{Z}\\{1,\sqrt{2},\sqrt{3},\frac{\sqrt{2}+\sqrt{6}}{2}\\}$.
We can see in this case that the centralizer
$\mathcal{H}_{1}(\mathcal{O}_{F})$ has rank 2 since $\lambda$ and
$(5-4\sqrt{2})+(2\sqrt{2})\lambda$ belong to
$\mathcal{O}_{F}[\lambda]^{\times}$ and are linearly independent. The
corresponding commuting matrices $A$ and $(5-4\sqrt{2})I+(2\sqrt{2})A$ in
$\operatorname{SL}_{2}(\mathcal{O}_{F})$ are

$\left(\begin{array}[]{cc}0&-1\\\ 1&4\end{array}\right)\>\>\>\text{ and
}\>\>\>\left(\begin{array}[]{cc}5-4\sqrt{2}&-2\sqrt{2}\\\
2\sqrt{2}&5+4\sqrt{2}\end{array}\right).$

###  4.3 Periodic maximal flats in $\operatorname{SO}(Q_{n})$

Define

$Q_{n}=\left(\begin{array}[]{cc}0&I_{n}\\\ I_{n}&0\end{array}\right)$ (4)

where $I_{n}$ is the $n\times n$ identity matrix. The bilinear form defined by
$Q_{n}$ is the unique unimodular, even integral symmetric bilinear form with
signature $(n,n)$. For a ring $R$, we define

$\operatorname{SO}(Q_{n};R)=\\{g\in\operatorname{SL}_{2n}(R):g^{t}Q_{n}g=Q_{n}\\}.$

Explicitly constructing periodic maximal flats in this case is more subtle
than the preceding cases because there is no known simple characterization of
the characteristic polynomials of elements in $\operatorname{SO}(Q_{n})$. See
e.g. [GM02] and [BF15]. The main result we use follows.

######  Lemma 4.11.

Fix $n\geq 2$, and fix a rational symmetric bilinear form $Q$ of signature
$(n,n)$. Assume that $A\in\operatorname{SO}(Q;\mathbb{Z})$ has characteristic
polynomial $\chi$ that’s irreducible in $\mathbb{Q}[x]$ and has only real
roots. Let $G_{1}<\operatorname{SO}(Q)$ be the centralizer of $A$. Then
$G_{1}(\mathbb{R})\cong(\mathbb{R}^{\times})^{n}$ and $G_{1}(\mathbb{Z})$ is
cocompact in $G_{1}(\mathbb{R})$.

######  Remark 4.12.

We do not actually know if there always exists
$A\in\operatorname{SO}(Q;\mathbb{Z})$ with irreducible characteristic
polynomial and only real roots. Generically one expects that the
characteristic polynomial is irreducible by an argument similar to [Riv08,
Thm. 5.2]. (For $Q=Q_{n}$, one can use [GM02] to prove that the map from
(orthogonal) matrices to characteristic polynomials is dominant as a map of
varieties.) But similar to Remark 4.4, we are unaware of an argument that
ensures that there are examples with only real roots.

######  Remark 4.13.

From the proof we will see that we do not need to assume that $\chi$ is
irreducible. For example, we have the same conclusion if
$\chi=\chi_{1}\cdots\chi_{j}$ is a product of irreducible polynomials with
real roots. We will use this observation to construct explicit examples in
each $\operatorname{SO}(Q_{n})$ after the proof of Lemma 4.11.

######  Remark 4.14.

The statement of Lemma 4.11 can be generalized to $Q$ of signature $(n,m)$
with minor changes. For example, if $Q^{\prime}=Q\oplus P$ where $Q$ has
signature $(n,n)$ and $P$ is positive (or negative) definite, then a
conclusion similar to that of Lemma 4.11 holds for
$A\in\operatorname{SO}(Q^{\prime})$ of the form
$A=A^{\prime}\oplus\operatorname{Id}$, where
$A^{\prime}\in\operatorname{SO}(Q)$ has irreducible characteristic polynomial
with real roots. In this case the characteristic polynomial is not irreducible
and the centralizer of $A$ is not a Cartan subgroup, rather
$G_{1}(\mathbb{R})$ is the product of $(R^{\times})^{n}$ with the compact
group $\operatorname{SO}(P;\mathbb{R})$.

###### Proof of Lemma 4.11.

By assumption, the action of $A$ on $\mathbb{R}^{2n}$ has $2n$ distinct real
eigenvalues, which occur in $(\lambda,1/\lambda)$ pairs with $\lambda\neq\pm
1$. From this one deduces that there is an eigenbasis
$u_{1},\ldots,u_{n},v_{1},\ldots,v_{n}\in\mathbb{R}^{2n}$ consisting of
isotropic vectors whose intersection matrix is $Q_{n}$. Then it is not hard to
see that $G_{1}(\mathbb{R})$ is isomorphic to
$\operatorname{SO}(1,1)^{n}\cong(\mathbb{R}^{\times})^{n}$.

Now we show that $G_{1}(\mathbb{Z})$ is cocompact in $G_{1}(\mathbb{R})$ by
showing that $G_{1}(\mathbb{Z})$ contains a rank-$n$ free abelian group. For
this, we use some structural results of Milnor [Mil69]. Let
$V\cong\mathbb{Q}^{2n}$ denote the standard representation of
$\operatorname{SO}(Q;\mathbb{Q})$. For any nonzero vector $v\in V$, the
vectors $v,Av,\ldots,A^{2n-1}v$ are linearly independent (otherwise the action
of $A$ on $V$ would be reducible, contradicting the fact that $\chi$ is
irreducible). This allows us to identify $E:=\mathbb{Q}[x]/(\chi)$ with $V$ as
$\mathbb{Q}[x]$-modules, with multiplication by $x$ on $\mathbb{Q}[x]/(\chi)$
corresponding to the action of $A$ on $V$. According to [Mil69, Lem. 1.1],
there exists a unique bilinear form $h:E\times E\rightarrow E$ that is
Hermitian with respect to the involution $e\mapsto\bar{e}$ on $E$ defined by
$\bar{x}=x^{-1}$ and such that

$(u,v)\mapsto\operatorname{Tr}_{E/\mathbb{Q}}(h(u,v))$

is equal to the given form $Q$ on $V$ (with respect to the identification of
$E$ and $V$). From this we conclude that the action of $e\in E$ on $V$
preserves the form $Q$ if and only if $e\bar{e}=1$.

Let $\lambda$ be a root of $\chi$. Under the map
$E\cong\mathbb{Q}(\lambda)\rightarrow M_{2n}(\mathbb{Q})$ sending $\lambda$ to
$A$, the subgroup of $\mathbb{Z}[\lambda]^{\times}$ of elements satisfying
$e\bar{e}=1$ maps into $G_{1}(\mathbb{Z})$. We will show this subgroup has
rank $n$. Since $\mathbb{Z}[\lambda]^{\times}$ has finite index in
$\mathcal{O}_{E}^{\times}$, it suffices to prove the same statement for
$\mathcal{O}_{E}^{\times}$.

Define $\Omega^{\pm}=\\{e\in\mathcal{O}_{E}^{\times}:\bar{e}=e^{\pm 1}\\}$,
each of which is a subgroup of $\mathcal{O}_{E}^{\times}$. We want to compute
the rank of $\Omega^{-}$. The product $\Omega^{+}\times\Omega^{-}$ is a
finite-index subgroup of $\mathcal{O}_{E}^{\times}$ (this is a general fact
about integral representations of $\mathbb{Z}/2\mathbb{Z}$). The group
$\Omega^{+}$ coincides with $\mathcal{O}_{F}$, where $F=\\{e\in
E:\bar{e}=e\\}$ is the fixed field of the involution. Now $F/\mathbb{Q}$ has
degree $n$ because $E/F$ has degree 2. Then
$\operatorname{rank}\Omega^{+}=\operatorname{rank}\mathcal{O}_{F}^{\times}=n-1$.
Since $\operatorname{rank}\mathcal{O}_{E}^{\times}=2n-1$, this implies that
$\operatorname{rank}\Omega^{-}=n$. This completes the proof. ∎

Next we construct explicit examples $A\in\operatorname{SO}(Q_{n})$ for every
even $n\geq 2$. Start with the case $n=2$ and consider the matrix

$B=\left(\begin{array}[]{rrrr}1&1&-2&2\\\ 1&2&-4&2\\\ -2&-4&10&-5\\\
2&2&-5&5\end{array}\right).$ (5)

This matrix preserves the bilinear form $Q_{2}$ and has characteristic
polynomial

$\chi=1-18x+43x^{2}-18x^{3}+x^{4}.$ (6)

This polynomial has real roots (note $B$ is symmetric) and is irreducible in
$\mathbb{Q}[x]$ (according to Mathematica).111The matrix $B$ was basically
constructed as $U^{t}U$ with $U$ a random upper-triangular unipotent matrix in
$\operatorname{SO}(Q_{2})$.

Applying Lemma 4.11 to the matrix $B$ above, we obtain periodic maximal flats
for $G(\mathbb{Z})\backslash G(\mathbb{R})/K$ when
$G=\operatorname{SO}(Q_{2})$ and $K\subset G(\mathbb{R})$ is any maximal
compact subgroup. (Note that $\operatorname{SO}(2,2)$ is isogenous to
$\operatorname{SL}_{2}(\mathbb{R})\times\operatorname{SL}_{2}(\mathbb{R})$, so
this case is related to §4.2.)

We can obtain periodic maximal flats for any even $n\geq 2$ as follows. Fixing
$n=2m$, consider the block diagonal matrix

$A=B\oplus B^{2}\oplus\ldots\oplus B^{m},$ (7)

and let $G_{1}<\operatorname{SO}(Q_{n})$ be the centralizer of $A$. Then by
the proof of Lemma 4.11, we conclude that
$G_{1}(\mathbb{R})\cong(\mathbb{R}^{\times})^{n}$ and $G_{1}(\mathbb{Z})$ is
cocompact in $G_{1}(\mathbb{R})$. A key point here is that $A$ has no repeated
eigenvalues; otherwise its centralizer is larger than a Cartan subgroup and
the symmetric space for $G_{1}(\mathbb{R})$ is not flat.

The following proposition will be used in the proof of Theorem B in §7.3.

######  Proposition 4.15.

For $\chi=1-18x+43x^{2}-18x^{3}+x^{4}$ and $E=\mathbb{Q}[x]/(\chi)$, there are
infinitely many primes $p\equiv 3\pmod{4}$ that split completely in $E$.

###### Proof.

First we note that $E=\mathbb{Q}[x]/(\chi)$ is a Galois extension of
$\mathbb{Q}$ and has Galois group
$\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$. We checked this using
Sage.

By the Kronecker–Weber theorem [Neu99, Ch. V, Thm. 1.10], every abelian
extension of $\mathbb{Q}$ is contained in a cyclotomic field
$\mathbb{Q}(\zeta_{m})$, where $\zeta_{m}$ a primitive $m$-th root of unity.
In fact, $E\subset\mathbb{Q}(\zeta_{40})$. To see this, first compute that the
roots of $\chi$ are

$\begin{array}[]{c}\frac{1}{2}\big{(}9-6\sqrt{2}-3\sqrt{5}+2\sqrt{10}\big{)}\\\\[5.69054pt]
\frac{1}{2}\big{(}9-6\sqrt{2}+3\sqrt{5}-2\sqrt{10}\big{)}\\\\[5.69054pt]
\frac{1}{2}\big{(}9+6\sqrt{2}-3\sqrt{5}-2\sqrt{10}\big{)}\\\\[5.69054pt]
\frac{1}{2}\big{(}9+6\sqrt{2}+3\sqrt{5}+2\sqrt{10}\big{)}\end{array}$

This implies that $E\cong\mathbb{Q}(\sqrt{2},\sqrt{5})$, and since

$\sqrt{2}=\zeta_{8}+\zeta_{8}^{-1}\>\>\>\text{ and
}\>\>\>\sqrt{5}=(\zeta_{5}+\zeta_{5}^{-1})-(\zeta_{5}^{2}+\zeta_{5}^{-2})$

we conclude that
$E\subset\mathbb{Q}(\zeta_{5},\zeta_{8})\subset\mathbb{Q}(\zeta_{40})$.

The primes $p\nmid 40$ that split completely in $E$ are characterized as
follows. The Galois group of $\mathbb{Q}(\zeta_{40})/\mathbb{Q}$ is
$(\mathbb{Z}/40\mathbb{Z})^{\times}$ and contains
$\operatorname{Gal}(\mathbb{Q}(\zeta_{40})/E)$ as a subgroup. If $p\nmid 40$
then $p$ splits completely if and only if the image of $p$ in
$(\mathbb{Z}/40\mathbb{Z})^{\times}\cong\operatorname{Gal}(\mathbb{Q}(\zeta_{40})/\mathbb{Q})$
lies in $\operatorname{Gal}(\mathbb{Q}(\zeta_{40})/E)$; [Neu99, Ch. I, §10 and
Ch. VII, Lem. 13.5]. One computes

$\operatorname{Gal}(\mathbb{Q}(\zeta_{40})/E)=\\{1,9,31,39\\}\subset(\mathbb{Z}/40\mathbb{Z})^{\times}.$

Therefore, $\chi$ splits completely in $\mathbb{Z}/p\mathbb{Z}$ whenever $p$
is a prime in one of the following arithmetic progressions.

$40x+1,\>\>\>40x+9,\>\>\>40x+31,\>\>\>40x+39$

By Dirichlet’s theorem, each of these arithmetic progressions contain
infinitely many primes. In particular, since $40x+31\equiv 3\pmod{4}$, there
are infinitely many primes congruent to $3\pmod{4}$ so that $\chi$ splits
completely in $\mathbb{Z}/p\mathbb{Z}$. ∎

##  5 Periodic maximal flats in homology

In this section we complete our construction of Millson–Raghunathan tuples
that will be used in the proofs of Theorems A–C. For the algebraic tori
$G_{1}$ constructed in §4, we construct $G_{2}$ and $K$ and identify primes
$p$ with the property that $(G,G_{1},G_{2},K,p^{k})$ is a Millson–Raghunathan
tuple for sufficiently large $k$. To do this, we apply the main results of
[ANP15], [Zsc21], and [Tsh21] about flat cycles in homology.

###  5.1 Flats and the homology of $\operatorname{SL}_{n+1}(\mathbb{Z})$

Let $G=\operatorname{SL}_{n+1}$. As in §4.1, we define $G_{1}$ as the
centralizer of $A_{1}\in\operatorname{SL}_{n+1}(\mathbb{Z})$, where $A_{1}$
has real eigenvalues and whose characteristic polynomial is irreducible in
$\mathbb{Q}[x]$. By Lemma 4.2, the group $G_{1}$ is an $\mathbb{R}$-split
torus $G_{1}(\mathbb{R})\cong(\mathbb{R}^{\times})^{n}$, and
$G_{1}(\mathbb{Z})<G_{1}(\mathbb{R})$ is cocompact.

There is a unique maximal compact $K<G(\mathbb{R})$ such that $K_{1}:=K\cap
G_{1}(\mathbb{R})$ is a maximal compact subgroup of $G_{1}(\mathbb{R})$.
Specifically, let $v_{0},\ldots,v_{n}\in\mathbb{R}^{n+1}$ be an eigenbasis for
$A_{1}$, fix the inner product for $\mathbb{R}^{n+1}$ where
$v_{0},\ldots,v_{n}$ is an orthonormal basis, and let
$K\cong\operatorname{SO}(n+1)$ be the group of isometries of
$\mathbb{R}^{n+1}$ with respect to this inner product. Then $K$ is a maximal
compact subgroup of $G(\mathbb{R})$ and it contains the maximal compact
subgroup $K_{1}\cong(\mathbb{Z}/2\mathbb{Z})^{n}$ of $G_{1}(\mathbb{R})$.

We will choose $G_{2}$ as follows. Fix a vector $v\in\mathbb{Q}^{n+1}$, set
$L=\mathbb{R}\\{v\\}$, and let $P$ be the orthogonal complement of $L$ in
$\mathbb{R}^{n+1}$ (with respect to our fixed inner product). Let
$A_{2}\in\operatorname{GL}_{n+1}(\mathbb{Q})$ be the matrix that preserves the
decomposition $\mathbb{R}^{n+1}=P\oplus L$, and acts on $P$ and $L$ by $-1$
and $+1$, respectively. Define $G_{2}$ to be the centralizer in
$\operatorname{SL}_{n+1}$ of $A_{2}$. By construction, $K_{2}:=K\cap
G_{2}(\mathbb{R})$ is the maximal compact subgroup of $G_{2}(\mathbb{R})$, and
$K_{2}$ is isomorphic to $\operatorname{Isom}(P)\times\operatorname{Isom}(L)$.

The group $G_{2}(\mathbb{Q})$ is conjugate in
$\operatorname{SL}_{n+1}(\mathbb{Q})$ to

$\left\\{\left(\begin{array}[]{cc}A&0\\\
0&\det(A)^{-1}\end{array}\right):A\in\operatorname{GL}_{n}(\mathbb{Q})\right\\}\subset\operatorname{SL}_{n+1}(\mathbb{Q}).$

We will impose further condition on the matrix $A_{2}$ below to ensure that
$X_{1}$ and $X_{2}$ meet transversely in a single point.

#### Intersection of $X_{1}$ and $X_{2}$.

Recall that $X_{i}:=G_{i}(\mathbb{R})/K_{i}$, where $K_{i}=K\cap
G_{i}(\mathbb{R})$.

We explain how to choose $A_{2}$ so that the $X_{1}$ and $X_{2}$ meet
transversely in a single point. By construction, $X_{1}$ and $X_{2}$ intersect
in the basepoint $eK\in G/K=X$, so the point is to show that the intersection
is no larger. Avramidi–Nguyen-Phan [ANP15] give a projective geometry
criterion for $X_{1}$ and $X_{2}$ to intersect transversely in a single point.
Instead we use Lemma 5.1 below, which is simpler for our purposes.

Fix $v\in\mathbb{Q}^{n+1}$ such that if we write
$v=a_{0}v_{0}+\cdots+a_{n}v_{n}$, then $a_{i}\neq 0$ for each $i$. Set
$L=\mathbb{R}\\{v\\}$ and $P$ be the orthogonal complement of $L$ in
$\mathbb{R}^{n+1}$ (with respect to our chosen inner product). Let
$A_{2}\in\operatorname{GL}_{n+1}(\mathbb{Q})$ be the matrix that acts
trivially on $L$ and by $-1$ on $P$. Observe that $K\cap
G_{2}(\mathbb{R})=\operatorname{Isom}(P)\times\operatorname{Isom}(L)$ is a
maximal compact subgroup of $G_{2}(\mathbb{R})$.

######  Lemma 5.1.

For $G_{1}$ and $G_{2}$ as above, $G_{1}(\mathbb{R})\cap
G_{2}(\mathbb{R})=\\{\pm I\\}$. Consequently, $X_{1}\cap X_{2}=\\{eK\\}$.

###### Proof.

Fix $g\in G_{1}(\mathbb{R})\cap G_{2}(\mathbb{R})$. Since $g$ commutes with
$A_{1}$ and $A_{2}$, there exist nonzero constants $c_{0},\ldots,c_{n},c$ so
that $g(v_{i})=c_{i}v_{i}$ and $g(v)=cv$. On the one hand,

$g(v)=cv=c(a_{0}v_{0}+\cdots+a_{n}v_{n})$

and on the other hand,

$g(v)=g(a_{1}v_{1}+\cdots+a_{n}v_{n})=a_{0}c_{0}v_{0}+\cdots+a_{n}c_{n}v_{n}.$

Combining these we conclude that $c_{i}=c$ for each $i$. This shows that
$g=cI$ is a scalar matrix, and $c\in\\{\pm 1\\}$ because
$g\in\operatorname{SL}_{n+1}(\mathbb{R})$.

To prove the last statement of the lemma, first observe that $X_{1}\cap X_{2}$
is connected (because $X_{1}$ and $X_{2}$ are totally geodesic in $X$), so if
$X_{1}\cap X_{2}$ has more than one point, then $\dim\big{(}X_{1}\cap
X_{2}\big{)}>0$. This implies that $\dim\big{(}G_{1}(\mathbb{R})\cap
G_{2}(\mathbb{R})\big{)}>0$. ∎

#### Intersection of $Y_{1}(r)$ and $Y_{2}(r)$.

Recall that $Y_{i}(r)=\Gamma_{i}(r)\backslash G_{i}(\mathbb{R})/K_{i}$. Fix a
prime $p$. The main theorem of [ANP15] proves the following.

######  Theorem 5.2 (Avramidi–Nguyen-Phan).

Let $G,G_{1},G_{2},K$ be as above and fix a prime $p$. There exists
$\mathcal{K}$ such that if $k\geq\mathcal{K}$, then
$Y_{1}(p^{k}),Y_{2}(p^{k})$ are orientable embedded submanifolds of $Y(p^{k})$
that intersect transversely and such that every intersection has the same
sign.

Combined with the general observations in §3, we conclude the following.

######  Corollary 5.3.

Let $G,G_{1},G_{2},K$ be as above. For any prime $p$ there exists
$\mathcal{K}$ such that if $k\geq\mathcal{K}$, then $(G,G_{1},G_{2},K,p^{k})$
is a Millson–Raghunathan tuple.

###  5.2 Flats and the homology of $\operatorname{SL}_{2}(\mathcal{O}_{F})$

Fix a totally real number field $F/\mathbb{Q}$ of degree $n\geq 2$. As in
§4.2, consider $\mathcal{H}=\operatorname{SL}_{2}(\mathcal{O}_{F})$ and
$G=R_{F/\mathbb{Q}}(\mathcal{H})$, and let
$A_{1}\in\operatorname{SL}_{2}(\mathcal{O}_{F})$ be the companion matrix of a
polynomial $\xi=x^{2}+ax+1$ in $\mathcal{O}_{F}[x]$ that is irreducible in
$F[x]$ and has real roots. Let $\mathcal{H}_{1}<\mathcal{H}$ be its
centralizer and $G_{1}=R_{F/\mathbb{Q}}(\mathcal{H}_{1})$. The group
$G_{1}(\mathbb{R})$ is isomorphic to $(\mathbb{R}^{\times})^{n}$ and is
embedded in $G(\mathbb{R})\cong(\operatorname{SL}_{2}(\mathbb{R}))^{n}$
diagonally by Lemma 4.9. The (unique) maximal compact subgroup
$K_{1}<G_{1}(\mathbb{R})$ maps to the subgroup $\\{(\pm I,\ldots,\pm I)\\}$ in
$(\operatorname{SL}_{2}(\mathbb{R}))^{n}$. Since $\\{\pm I\\}$ is central in
$\operatorname{SL}_{2}(\mathbb{R})$, every maximal compact subgroup of
$G(\mathbb{R})$ contains $K_{1}$. As such, we fix any maximal compact subgroup
$K<G(\mathbb{R})$, and consider the corresponding maximal flat
$X_{1}:=G_{1}(\mathbb{R})/K_{1}\hookrightarrow G(\mathbb{R})/K=:X$.

######  Lemma 5.4.

Let $A_{1}\in\operatorname{SL}_{2}(\mathcal{O}_{F})$ as above. There exists
$g\in\operatorname{SL}_{2}(F)$ such that the flats $X_{1}$ and $g(X_{1})$
intersect transversely in a single point.

###### Proof.

First we recall a few facts which make the lemma easy.

Parameterizing the space of flats. Any flat
$\mathcal{F}\subset\operatorname{\mathbb{H}}^{2}\times\cdots\times\operatorname{\mathbb{H}}^{2}$
is a product of geodesics
$\mathcal{F}\cong\gamma_{1}\times\cdots\times\gamma_{n}$. Since a geodesic in
$\operatorname{\mathbb{H}}$ is determined uniquely by its endpoints in
$\partial\operatorname{\mathbb{H}}^{2}\cong S^{1}$, a flat is uniquely
determined by a tuple $(\theta_{1},\ldots,\theta_{2n})\in(S^{1})^{\times 2n}$.
Since $\operatorname{SL}_{2}(\mathbb{R})$ acts transitively on geodesics in
$\operatorname{\mathbb{H}}^{2}$, it is obvious that $G(\mathbb{R})$ acts
transitively on flats. Furthermore, since
$G(\mathbb{Q})\cong\operatorname{SL}_{2}(F)$ is dense in $G(\mathbb{R})$, the
$G(\mathbb{Q})$-orbit of a flat is dense in the space of flats (where the
space of flats is topologized as a subset of $(S^{1})^{\times 2n}$).

Intersection of two flats. Two flats $\mathcal{F}$ and $\mathcal{F}^{\prime}$
with corresponding tuples $(\theta_{1},\ldots,\theta_{2n})$ and
$(\theta_{1}^{\prime},\ldots,\theta_{2n}^{\prime})$ intersect transversely in
a single point if and only if the pairs $(\theta_{2i-1},\theta_{2i})$ and
$(\theta_{2i-1}^{\prime},\theta_{2i}^{\prime})$ are linked in
$\partial\operatorname{\mathbb{H}}^{2}$ for each $i=1,\ldots,n$.

Now we prove the lemma. Choose a flat $\mathcal{F}$ that intersects $X_{1}$
transversely in a point. By the preceding paragraphs, we can choose
$g\in\operatorname{SL}_{2}(F)$ so that $g(X_{1})$ is close enough to
$\mathcal{F}$ in the space of flats so that $g(X_{1})$ and $X_{1}$ also
intersect transversely in a point. ∎

Using Lemma 5.4, choose $g\in\operatorname{SL}_{2}(F)$ so that $X_{1}$ and
$X_{2}:=g(X_{1})$ intersect transversely in a single point. Set
$A_{2}=gA_{1}g^{-1}$, define $\mathcal{H}_{2}\cong g\mathcal{H}_{1}g^{-1}$ to
be the centralizer of $A_{2}$, and define
$G_{2}=R_{F/\mathbb{Q}}(\mathcal{H}_{2})$.

For an ideal $\mathfrak{r}\subset\mathcal{O}_{F}$, we consider the principal
congruence subgroup

$\Gamma(\mathfrak{r}):=\ker\big{[}\operatorname{SL}_{2}(\mathcal{O}_{F})\rightarrow\operatorname{SL}_{2}(\mathcal{O}_{F}/\mathfrak{r})\big{]}.$

Accordingly, we write $Y(\mathfrak{r})=\Gamma(\mathfrak{r})\backslash
G(\mathbb{R})/K$ and similarly define $\Gamma_{i}(\mathfrak{r})$ and
$Y_{i}(\mathfrak{r})$ for $i=1,2$.

######  Theorem 5.5 (Zschumme).

Let $G,G_{1},G_{2},K$ be as above. For any prime ideal
$\mathfrak{p}\subset\mathcal{O}_{F}$ there exists $\mathcal{K}$ such that if
$k\geq\mathcal{K}$, then $Y_{1}(\mathfrak{p}^{k}),Y_{2}(\mathfrak{p}^{k})$ are
oriented submanifolds of $Y(\mathfrak{p}^{k})$ that intersect transversely,
and each intersection of $Y_{1}(\mathfrak{p}^{k})$ and
$Y_{2}(\mathfrak{p}^{k})$ has the same sign.

Theorem 5.5 has stricter hypotheses and a slightly stronger conclusion than
what is proved by Zschumme [Zsc21]. Zschumme considers more general lattices
$R_{F/\mathbb{Q}}(\operatorname{SL}_{D})$, where $D$ is a quaternion algebra,
and only concludes that there is _some_ ideal
$\mathfrak{r}\subset\mathcal{O}_{F}$ for which the conclusion above holds.

To reach the stronger conclusion in our case, note that a special feature of
the $G=R_{F/\mathbb{Q}}(\operatorname{SL}_{2})$ case is that
$G_{1}(\mathbb{R})$ automatically preserves the orientations on $X$ and
$X_{1}$, and similarly for $G_{2}(\mathbb{R})$. Therefore, once we choose $k$
large enough so that $Y_{1}(\mathfrak{p}^{k})$ and $Y_{2}(\mathfrak{p}^{k})$
are embedded, the conclusions about transversality and signs of the
intersections are satisfied by general arguments. See [Tsh21, §2.2 and Prop.
6].

Combined with the general observations in §3, we conclude the following.

######  Corollary 5.6.

Let $G,G_{1},G_{2},K$ be as above. For any prime $p$ there exists
$\mathcal{K}$ such that if $k\geq\mathcal{K}$, then $(G,G_{1},G_{2},K,p^{k})$
is a Millson–Raghunathan tuple.

###  5.3 Flats and the homology of $\operatorname{SO}(Q_{n};\mathbb{Z})$

Continue with the notation of §4.3, so that $G=\operatorname{SO}(Q_{n})$, and
$G_{1}$ is the centralizer of the matrix
$A\in\operatorname{SO}(Q_{n};\mathbb{Z})$ that was constructed in §4.3
equation 7. Then $G_{1}(\mathbb{R})\cong(\mathbb{R}^{\times})^{n}$ and
$G_{1}(\mathbb{Z})<G_{1}(\mathbb{R})$ is cocompact by Lemma 4.11.

To define $G_{2}$, decompose $\mathbb{R}^{2n}=U_{1}\oplus\cdots\oplus U_{n}$
into 2-dimensional summands, where each $U_{j}$ is spanned by two by a
$(\lambda,\frac{1}{\lambda})$-eigenspace pair for $A$. Fix a rational vector
$v\in\mathbb{R}^{2n}$ such that $Q_{n}(v,v)<0$. Let $L$ be the span of $v$,
and set $P=L^{\perp}$. Assume that $P\cap U_{i}$ is a positive line for each
$i$. Define $G_{2}$ as the centralizer of
$A_{2}\in\operatorname{GL}_{n+m}(\mathbb{Q})$, where $A_{2}$ acts by $-1$ on
$P$ and acts trivially on $L$. With this choice of $G_{1},G_{2}$, there is a
unique maximal compact subgroup $K\subset G(\mathbb{R})$ such that
$K_{i}=G_{i}(\mathbb{R})\cap K$ is a maximal compact subgroup of
$G_{i}(\mathbb{R})$ for $i=1,2$ (maximal compact subgroups of $G(\mathbb{R})$
correspond to maximal positive-definite subspaces of $\mathbb{R}^{2n}$; here
the unique choice for $K$ is the subspace spanned by the lines $P\cap U_{i}$).

With these choices, $G_{1}\cap G_{2}=\\{\operatorname{Id}\\}$ and
$X_{1},X_{2}\subset X$ meet transversely in a single point, as explained in
[Tsh21, §3]. Furthermore, from [Tsh21] we have the following analogue of
Theorem 5.2 of Avramidi–Nguyen-Phan [ANP15].

######  Theorem 5.7.

Let $G,G_{1},G_{2},K$ be as above. Let $p$ be prime and assume $p\equiv
3\pmod{4}$. There exists $\mathcal{K}$ so that if $k\geq\mathcal{K}$, then
$Y_{1}(p^{k}),Y_{2}(p^{k})$ are orientable embedded submanifolds of $Y(p^{k})$
that intersect transversely and so that every intersection has the same sign.

Theorem 5.7 is a special case of what is proved in [Tsh21, Thm. 1] and has a
stronger conclusion. We explain this below.

###### Proof sketch of Theorem 5.7.

Using general theory, we can find $\mathcal{K}_{0}$ so that if
$k\geq\mathcal{K}_{0}$, then $Y_{1}(p^{k})$ and $Y_{2}(p^{k})$ are oriented,
embedded in $Y(p^{k})$, and intersect transversely. Fixing
$k\geq\mathcal{K}_{0}$, there is an embedding $\iota:Y_{1}(p^{k})\cap
Y_{2}(p^{k})\hookrightarrow\Gamma_{2}(p^{k})\backslash\Gamma(p^{k})/\Gamma_{1}(p^{k})$.
Fix $z\in Y_{1}(p^{k})\cap Y_{2}(p^{k})$ and write $[\gamma]=\iota(z)$. To
show that $Y_{1}(p^{k})$ and $Y_{2}(p^{k})$ intersect positively222Here we
have fixed orientations on $X_{1},X_{2}$ so that their intersection is
positive. at $z$ it suffices to show that we can write $\gamma=g_{2}g_{1}$
where $g_{i}\in G_{i}(\mathbb{R})$ preserves the orientation on $X_{i}$ and on
$X$. As explained in [Tsh21, Prop. 6], it is possible to write
$\gamma=g_{2}g_{1}$, where $g_{i}\in G_{i}(\mathbb{Q})$. To show that the
orientations are preserved, we will show that the spinor norm
$\theta(g_{i})\in\mathbb{R}^{\times}/(\mathbb{R}^{\times})^{2}\cong\\{\pm
1\\}$ is trivial; cf. [Tsh21, §3.2]. (For more on the spinor norm, see e.g.
[HO89, §7.2].) On $\operatorname{SO}(Q_{n};\mathbb{Z})$ the spinor norm
factors through a surjection
$\theta:\operatorname{SO}(Q_{n};\mathbb{Z})\rightarrow\mathbb{Z}^{\times}/(\mathbb{Z}^{\times})^{2}\cong\\{\pm
1\\}$; therefore, $\eta\in\operatorname{SO}(Q_{n};\mathbb{Z})$ has trivial
spinor norm if and only if it is in the kernel of the map

$\operatorname{SO}(Q_{n};\mathbb{Z})\rightarrow\operatorname{SO}(Q_{n};\mathbb{F}_{p})\rightarrow\mathbb{F}_{p}^{\times}/(\mathbb{F}_{p}^{\times})^{2},$
(8)

and in particular the spinor norm is trivial on $\Gamma(p)$. The preceding
statement uses the assumption $p\equiv 3\pmod{4}$, which implies that the map
in (8) is surjective, since then $-1$ is not a square in
$\mathbb{F}_{p}^{\times}$.

Since $\gamma\in\Gamma(p)$, we know $\theta(\gamma)=1$, and so
$\theta(g_{1})=\theta(g_{2})$. To finish the proof, we will show that the
spinor norm is trivial on $G_{1}(\mathbb{Q})$. Over $\mathbb{R}$, a matrix
$g\in G_{1}(\mathbb{Q})$ can be diagonalized into the form
$\left(\begin{array}[]{cc}D_{g}&0\\\ 0&D_{g}^{-1}\end{array}\right)$, and the
spinor norm is given by $\theta(g)=\det(D_{g})\mod(\mathbb{R}^{\times})^{2}$
(as can be easily checked). It is easy to see that $\det(D_{g})>0$ for each
$g\in G_{1}(\mathbb{Q})$ (the main observation here is that for $g\in
G_{1}(\mathbb{Q})$ the number of eigen-lines on which $g$ acts by $-1$ must be
a multiple of 4; this is because the matrix $B$ used to define $A_{1}$ is a
$4\times 4$ matrix with irreducible characteristic polynomial, so the only
diagonal matrix in $G_{1}(\mathbb{Q})$ that commutes with $B$ is $-I$).
Therefore $\theta$ vanishes on $G_{1}(\mathbb{Q})$. ∎

Combined with the general observations in §3, we conclude the following.

######  Corollary 5.8.

Let $G,G_{1},G_{2},K$ be as above. For any prime $p\equiv 3\pmod{4}$, there
exists $\mathcal{K}$ such that if $k\geq\mathcal{K}$, then
$(G,G_{1},G_{2},K,p^{k})$ is a Millson–Raghunathan tuple.

##  6 Size of Lie groups over finite rings

The Lie groups of finite type that are of interest in this paper fall into two
groups: (1) $\operatorname{SL}_{d}(\mathcal{O}_{F}/\mathfrak{p}^{j})$, where
$F$ is a number field and $\mathfrak{p}\subset\mathcal{O}_{F}$ is a prime
ideal, and (2) the orthogonal groups $O(Q;\mathbb{Z}/p^{j}\mathbb{Z})$ for a
prime number $p\neq 2$. In this section we will provide estimates of the size
of these groups as functions of $j$, considering all other variables fixed.

Given two functions $f,g$ of $j$, we write $f\sim g$ if there is a constant
$C$ (independent of $j$) such that $\frac{1}{C}\cdot g(j)\leq f(j)\leq C\cdot
g(j)$ for all $j\geq 1$.

###  6.1 Size of $\operatorname{SL}_{d}(\mathcal{O}_{F}/\mathfrak{p}^{j})$

Fix a number field $F/\mathbb{Q}$. Throughout this section we consider an
ideal $\mathfrak{t}\subset\mathcal{O}_{F}$ of the form
$\mathfrak{t}=\mathfrak{p}^{j}$ for some prime ideal $\mathfrak{p}$. Set
$\mathbb{F}=\mathcal{O}_{F}/\mathfrak{p}$.

######  Proposition 6.1.

Fix $j\geq 1$, and let $\mathfrak{t}=\mathfrak{p}^{j}$, where
$\mathfrak{p}\subset\mathcal{O}_{F}$ is a prime ideal. For $d\geq 2$,

$|\operatorname{SL}_{d}(\mathcal{O}_{F}/\mathfrak{t})|\sim
N(\mathfrak{t})^{d^{2}-1},$

where $N(\mathfrak{t})=|\mathcal{O}_{F}/\mathfrak{t}|$ is the ideal norm.

Proposition 6.1 is surely well known. We include a proof for completeness and
also because it is a good warm-up for the orthogonal case in the next section.

###### Proof of Proposition 6.1.

Set $R=\mathcal{O}_{F}/\mathfrak{t}$. Given the short exact sequence

$1\rightarrow\operatorname{SL}_{d}(R)\rightarrow\operatorname{GL}_{d}(R)\rightarrow
R^{\times}\rightarrow 1,$

we may break up our computation into computing $|\operatorname{GL}_{d}(R)|$
and $|R^{\times}|$.

#### Computing $|R^{\times}|$.

Let $M\subset R$ be the unique maximal ideal, which is the image of
$\mathfrak{p}\subset\mathcal{O}_{F}$ under
$\mathcal{O}_{F}\rightarrow\mathcal{O}_{F}/\mathfrak{p}^{j}\equiv R$. Note
that $R/M\cong\mathbb{F}$.

Claim $1$. The natural map
$R^{\times}\rightarrow(R/M)^{\times}=\mathbb{F}^{\times}$ gives a short exact
sequence

$1\rightarrow(1+M)\rightarrow
R^{\times}\rightarrow\mathbb{F}^{\times}\rightarrow 1.$ (9)

Given the claim, we conclude that

$|R^{\times}|=|M|\cdot|\mathbb{F}^{\times}|=N(\mathfrak{t})\>\frac{|\mathbb{F}^{\times}|}{|\mathbb{F}|}.$
(10)

Here we use that $|M|=\frac{|R|}{|R/M|}$ and $R/M=\mathbb{F}$.

Proof of Claim $1$. We need to show that $R^{\times}\rightarrow(R/M)^{\times}$
is surjective and has kernel $1+M$.

The kernel of $R^{\times}\rightarrow(R/M)^{\times}$ is equal to $(1+M)\cap
R^{\times}$. In fact, $1+M$ is contained in $R^{\times}$: for each $m\in M$,
the element $1+m$ is invertible in $R$ because the ideal $M\subset R$ is
nilpotent ($M^{j}=0$). Therefore, $1+M$ is the kernel of
$R^{\times}\rightarrow(R/M)^{\times}$.

To show that $R^{\times}\rightarrow(R/M)^{\times}$ is surjective, fix $x\in R$
whose image in $R/M$ is invertible. To prove surjectivity, it suffices to show
that $x$ is invertible. By assumption there exists $y\in R$ such that $xy=1+m$
for some $m\in M$. Since $1+m$ is invertible, this implies $xy$ is invertible;
hence $x$ is also invertible, as desired.

#### Computing $|\operatorname{GL}_{d}(R)|$.

Similar to the preceding computation, we reduce to the following claim.

Claim $2$. The natural map
$\operatorname{GL}_{d}(R)\rightarrow\operatorname{GL}_{d}(R/M)=\operatorname{GL}_{d}(\mathbb{F})$
gives a short exact sequence

$1\rightarrow\big{[}I+M_{d}(M)\big{]}\rightarrow\operatorname{GL}_{d}(R)\rightarrow\operatorname{GL}_{d}(\mathbb{F})\rightarrow
1,$

where $I$ is the $2\times 2$ identity matrix and $M_{d}(M)$ is the group of
$d\times d$ matrices with entries in $M$.

The claim implies that

$|\operatorname{GL}_{d}(R)|=|M_{d}(M)|\cdot|\operatorname{GL}_{d}(\mathbb{F})|=N(\mathfrak{t})^{d^{2}}\>\frac{|\operatorname{GL}_{d}(\mathbb{F})|}{|\mathbb{F}|^{d^{2}}}.$
(11)

Combining Equations (10) and (11), we conclude that

$|\operatorname{SL}_{d}(R)|=\frac{|\operatorname{GL}_{d}(R)|}{|R^{\times}|}=N(\mathfrak{t})^{d^{2}-1}\cdot\frac{|\operatorname{GL}_{d}(\mathbb{F})|}{|\mathbb{F}^{\times}|\cdot|\mathbb{F}|^{d^{2}-1}},$

which proves the proposition. Therefore, it remains only to prove Claim 2.

Proof of Claim $2$. The kernel of
$\operatorname{GL}_{d}(R)\rightarrow\operatorname{GL}_{d}(R/M)$ is the
intersection of $I+M_{d}(M)$ with $\operatorname{GL}_{d}(R)$. But in fact
$I+M_{d}(M)$ is contained in $\operatorname{GL}_{d}(R)$ because each $\mu\in
M_{d}(M)$ is nilpotent ($\mu^{j}=0$). Furthermore,
$\operatorname{GL}_{d}(R)\rightarrow\operatorname{GL}_{d}(R/M)$ is surjective:
given $A\in M_{d}(R)$ whose image in $M_{d}(R/M)$ is invertible, there exists
$B$ so that $AB=I+\mu$ for some $\mu\in M_{d}(M)$. Since $1+\mu$ is
invertible, this implies $AB$ and hence $A$ is invertible.

This finishes the proof of Claim 2 and the proof of the proposition. ∎

###  6.2 Size of finite orthogonal groups

Fix a prime $p\neq 2$ and $d\geq 2$. Let $Q\in M_{d}(\mathbb{Z})$ be the
matrix of a nondegenerate symmetric bilinear form
$\langle\cdot,\cdot\rangle:\mathbb{Z}^{d}\times\mathbb{Z}^{d}\rightarrow\mathbb{Z}$,
and assume that $Q$ remains nondegenerate after reducing mod $p$, i.e.
$\det(Q)\not\equiv 0\pmod{p}$. Define the orthogonal group

$O(Q;\mathbb{Z}/p^{j}\mathbb{Z})=\\{g\in\operatorname{GL}_{d}(\mathbb{Z}/p^{j}\mathbb{Z}):g^{t}Qg=Q\\}.$

The reduction map

$\rho_{j}:\mathbb{Z}/p^{j}\mathbb{Z}\rightarrow\mathbb{Z}/p^{j-1}\mathbb{Z}$

induces a homomorphism

$\pi_{j}:O(Q;\mathbb{Z}/p^{j}\mathbb{Z})\rightarrow
O(Q;\mathbb{Z}/p^{j-1}\mathbb{Z}),$

which we will use to inductively compute $|O(Q;\mathbb{Z}/p^{j}\mathbb{Z})|$.
For each $i\geq 2$, set $M_{i}:=\ker(\rho_{i})$, and define

$S_{i}(Q)=\\{\mu\in M_{d}(M_{i}):\mu^{t}Q+Q\mu=0\\}.$

######  Proposition 6.2.

With the preceding setup,

$|O(Q;\mathbb{Z}/p^{j}\mathbb{Z})|=|O(Q;\mathbb{Z}/p\mathbb{Z})|\cdot|S_{2}(Q)|\cdots|S_{j}(Q)|.$

###### Proof.

First we show that $\pi_{j}$ is surjective for each $j\geq 2$. For this, we
use that $O(Q;\mathbb{Z}/p^{j}\mathbb{Z})$ is generated by reflections

$r_{x}:v\mapsto v-2\frac{\langle v,x\rangle}{\langle x,x\rangle}\>x,$

where $x\in(\mathbb{Z}/p^{j}\mathbb{Z})^{d}$ is such that $\langle x,x\rangle$
is a unit in $\mathbb{Z}/p^{j}\mathbb{Z}$; see [MH73, Ch. I, Cor. 4.3]. Given
this, it is straightforward to check that if
$x\in(\mathbb{Z}/p^{j-1}\mathbb{Z})^{d}$ is such that $\langle x,x\rangle$ is
a unit and $\hat{x}\in(\mathbb{Z}/p^{j}\mathbb{Z})^{d}$ is a lift of $x$, then
$\langle\hat{x},\hat{x}\rangle$ is a unit in $\mathbb{Z}/p^{j}\mathbb{Z}$, and
$\pi_{j}(r_{\hat{x}})=r_{x}$. Thus $\pi_{j}$ is surjective.

It remains to understand $\ker(\pi_{j})$. Similar to the proof of Proposition
6.1, there is a short exact sequence

$1\rightarrow\big{[}I+M_{d}(M_{j})\big{]}\rightarrow\operatorname{GL}_{d}(\mathbb{Z}/p^{j}\mathbb{Z})\rightarrow\operatorname{GL}_{d}(\mathbb{Z}/p^{j-1}\mathbb{Z})\rightarrow
1,$

so $\ker(\pi_{j})$ is the intersection of $I+M_{d}(M_{j})$ with
$O(Q;\mathbb{Z}/p^{j}\mathbb{Z})$. If $I+\mu$ belongs to $I+M_{d}(M_{j})$ and
$O(Q;\mathbb{Z}/p^{j}\mathbb{Z})$, then $(1+\mu)^{t}Q(1+\mu)=Q$, which implies
that $\mu^{t}Q+Q\mu=0$ (note that $\mu^{t}Q\mu$ is 0 because $M_{j}^{2}=0$ in
$\mathbb{Z}/p^{j}\mathbb{Z}$). Therefore $|\ker(\pi_{j})|=|S_{j}(Q)|$, and the
Proposition follows by induction on $j$. ∎

#### Example: $Q=Q_{n}$.

When $Q=Q_{n}$ as in §4.3, one computes

$|S_{i}(Q_{n})|=|M_{i}|^{n^{2}}+2|M_{i}|^{n(n-1)/2}=p^{n^{2}}+2p^{n(n-1)/2}.$

for each $i\geq 2$. Then Proposition 6.2 implies that if $p\neq 2$ and we set
$t=p^{j}$, then
$|O(Q_{n};\mathbb{Z}/t\mathbb{Z})|\sim(p^{n^{2}}+2p^{n(n-1)/2})^{j}$.

For a ring $R$, define $\Omega(Q_{n};R)$ as the kernel of the spinor
homomorphism

$\Omega(Q_{n};R):=\ker\big{[}\operatorname{SO}(Q_{n};R)\xrightarrow{\theta}R^{\times}/(R^{\times})^{2}\big{]}.$
(12)

When $R=\mathbb{Z}/p^{j}\mathbb{Z}$, the subgroups
$\Omega(Q_{n};R)\subset\operatorname{SO}(Q_{n};R)$ and
$\operatorname{SO}(Q_{n};R)\subset O(q_{n};R)$ each have index at most 2, so
we conclude the following corollary.

######  Corollary 6.3.

Fix a prime $p\neq 2$, and set $t=p^{j}$ for $j\geq 1$. Then

$|\Omega(Q_{n};\mathbb{Z}/t\mathbb{Z})|\sim\left(p^{n^{2}}+2p^{n(n-1)/2}\right)^{j}.$

#### Example: $Q=\widetilde{Q}_{n}$.

Define

$\widetilde{Q}_{n}=\left(\begin{array}[]{cc}Q_{n}&0\\\
0&1\end{array}\right)\in M_{2n+1}(\mathbb{Z}).$ (13)

Here one computes that
$|S_{i}(\widetilde{Q}_{n})|=|S_{i}(Q_{n})|+|M_{i}|^{n}=p^{n^{2}}+2p^{n(n-1)/2}+p^{n}$
for each $i\geq 2$.

######  Corollary 6.4.

Fix a prime $p\neq 2$, and set $t=p^{j}$ for $j\geq 1$. Then

$|\Omega(\widetilde{Q}_{n};\mathbb{Z}/t\mathbb{Z})|\sim\left(p^{n^{2}}+2p^{n(n-1)/2}+p^{n}\right)^{j}.$

###  6.3 Finite tori

For the proof of the main theorems, we will need to know the size of the
centralizer of certain elements in certain finite groups of Lie type.

######  Lemma 6.5.

Let $F/\mathbb{Q}$ be a number field. Fix
$A\in\operatorname{SL}_{d}(\mathcal{O}_{F})$ and assume that its
characteristic polynomial $\chi$ is irreducible in $F[x]$. There are
infinitely many prime ideals $\mathfrak{p}\subset\mathcal{O}_{F}$ such that
for each $j\geq 1$,

  1. (i)

$\chi$ is square-free and splits completely in
$(\mathcal{O}_{F}/\mathfrak{p}^{j})[x]$, and

  2. (ii)

$A$ is diagonalizable in
$\operatorname{GL}_{d}(\mathcal{O}_{F}/\mathfrak{p}^{j})$.

Consequently, for these primes, the centralizer of $A$ in
$\operatorname{GL}_{d}(\mathcal{O}_{F}/\mathfrak{p}^{j})$ has size
$\mathcal{C}^{d}$, where
$\mathcal{C}=N(\mathfrak{p}^{j})\>\frac{N(\mathfrak{p})-1}{N(\mathfrak{p})}$
and $N(\mathfrak{t})=|\mathcal{O}_{F}/\mathfrak{t}|$ is the ideal norm.

For the last statement, compare with (10). As we will explain, (i) is well-
known, and (i) implies (ii).

######  Remark 6.6.

There is reason to be cautious when trying to diagonalize a matrix
$A\in\operatorname{GL}_{d}(R)$ when $R$ is a commutative ring that is not a
field because in this generality, (1) a given eigenspace $\\{v\in
R^{d}:Av=\lambda v\\}$ need not have a basis, (2) the sum of eigenspaces need
not be a direct sum, and (3) the sum of eigenspaces need not span $R^{d}$.
These phenomena are illustrated in [RW90]. These issues can be avoided if the
characteristic polynomial of $A$ is square-free and splits completely in
$\big{(}\mathcal{O}_{F}/\mathfrak{p}\big{)}[x]$, as will be explained in the
proof of Lemma 6.5.

###### Proof of Lemma 6.5.

We begin with (i). Fix a root $\lambda$ of $\chi$, and let $E=F(\lambda)\cong
F[x]/(\chi)$.

First we recall that there are infinitely many primes
$\mathfrak{p}\subset\mathcal{O}_{F}$ so that $\chi$ is square-free and splits
completely in $\big{(}\mathcal{O}_{F}/\mathfrak{p}\big{)}[x]$. By Chebotarev’s
density theorem [Neu99, Ch. VII, Cor. 13.6 and Ch. I, §8, Exer. 4], there are
infinitely many primes $\mathfrak{p}$ that split completely
$\mathfrak{p}\mathcal{O}_{E}=\mathfrak{q}_{1}\cdots\mathfrak{q}_{d}$. If
$\mathfrak{p}$ is such a prime, and if we assume further that $\mathfrak{p}$
is not one of the (finitely many) divisors of the conductor ideal
$\mathfrak{c}\subset\mathcal{O}_{F}$ of the ring $\mathcal{O}_{F}[\lambda]$,
then $\chi$ is square-free and splits completely in
$\big{(}\mathcal{O}_{F}/\mathfrak{p}\big{)}[x]$; see [Neu99, Ch. I, Prop.
8.3].

If $\chi$ is square-free and splits completely in
$\big{(}\mathcal{O}_{F}/\mathfrak{p}\big{)}[x]$, then the same also holds in
$\big{(}\mathcal{O}_{F}/\mathfrak{p}^{j}\big{)}[x]$ for each $j\geq 1$ by
Hensel’s lemma [Neu99, Ch. II, Lem. 4.6]. Furthermore, the uniqueness part of
Hensel’s lemma implies that $\chi$ has exactly $n$ roots in
$\mathcal{O}_{F}/\mathfrak{p}^{j}$, one lying over each root in
$\mathcal{O}_{F}/\mathfrak{p}$. This proves (i).

Next we prove (ii). Fixing $j\geq 1$, we argue that $A$ is diagonalizable in
$\operatorname{GL}_{d}(\mathcal{O}_{F}/\mathfrak{p}^{j})$ by showing $A$ has a
basis of eigenvectors. Let
$\lambda_{1},\ldots,\lambda_{d}\in\mathcal{O}_{F}/\mathfrak{p}^{j}$ be the
roots of $\chi$, and let
$E_{1},\ldots,E_{d}\subset(\mathcal{O}_{F}/\mathfrak{p}^{j})^{n}$ be the
corresponding eigenspaces. Each $E_{i}$ contains a linearly independent vector
$v_{i}$ [RW90, Lem. 2] (here it is important that $\chi(\lambda_{i})$ is zero
and not merely a zero-divisor). Furthermore, the standard proof from linear
algebra shows that $v_{1},\ldots,v_{n}$ are linearly independent; cf. [RW90,
Lem. 3] (here it is important that for $i\neq j$, $\lambda_{i}-\lambda_{j}$ is
not a zero-divisor in $\mathcal{O}_{F}/\mathfrak{p}^{j}$, which holds because
$\lambda_{i}$ and $\lambda_{j}$ map to distinct elements in
$\mathcal{O}_{F}/\mathfrak{p}$). Finally, $v_{1},\ldots,v_{n}$ span
$(\mathcal{O}_{F}/\mathfrak{p}^{j})^{n}$, as can be seen by counting since the
span of $v_{1},\ldots,v_{n}$ has the same cardinality as
$(\mathcal{O}_{F}/\mathfrak{p}^{j})^{n}$. This shows that $v_{1},\ldots,v_{n}$
are an eigenbasis for $A$, so $A$ is diagonalizable in
$\operatorname{GL}_{d}(\mathcal{O}_{F}/\mathfrak{p}^{j})$. ∎

We prove a similar theorem for the orthogonal case.

######  Lemma 6.7.

Fix $A\in\operatorname{SO}(Q_{n};\mathbb{Z})$ and assume that its
characteristic polynomial factors $\chi=\chi_{1}\cdots\chi_{m}$, where the
$\chi_{i}$ are distinct and irreducible in $\mathbb{Q}[x]$. There exist
infinitely many primes $p$ such that for each $j\geq 1$,

  1. (i)

$\chi$ is square-free and splits completely in
$(\mathbb{Z}/p^{j}\mathbb{Z})[x]$, and

  2. (ii)

$A$ is diagonalizable in
$\operatorname{SO}(Q_{n};\mathbb{Z}/p^{j}\mathbb{Z})$.

Consequently, for these primes, the centralizer of $A$ in
$\operatorname{SO}(Q_{n};\mathbb{Z}/p^{j}\mathbb{Z})$ has size
$\mathcal{C}^{n}$, where $\mathcal{C}=p^{j}\left(\frac{p-1}{p}\right)$.

The proof is similar to the proof of Lemma 6.5, but with a few key
differences. Specifically we need to deal with the fact that $\chi$ is not
assumed to be irreducible and we need to show that $A$ is diagonalizable in
$\operatorname{SO}(Q_{n};\mathbb{Z}/p^{j}\mathbb{Z})$ rather than just
$\operatorname{GL}_{2n}(\mathbb{Z}/p^{j}\mathbb{Z})$.

###### Proof of Lemma 6.7.

$None$ Let $\lambda_{i}$ be a root of $\chi_{i}$, and set
$E_{i}=F(\lambda_{i})\cong F[x]/(\chi_{i})$ for $1\leq i\leq m$. Set
$\widetilde{E}=F(\lambda_{1},\ldots,\lambda_{m})$. By Chebotarev’s theorem,
there are infinitely many primes that split completely in $\widetilde{E}$, and
hence also in each $E_{i}$ (see [Neu99, §13]). As in the proof of Lemma 6.5,
for such a prime $p$, if $p$ does not divide the conductor
$c_{i}\in\mathbb{Z}$ of $\mathbb{Z}[\lambda_{i}]$, then $\chi_{i}$ splits
completely in $(\mathbb{Z}/p^{j}\mathbb{Z})[x]$ for each $j\geq 1$. To finish
the proof of (i), we argue that after imposing finitely many more constraints
on $p$, we may assume that $\chi$ is squarefree in
$(\mathbb{Z}/p^{j}\mathbb{Z})[x]$. By assumption, for each $i<i^{\prime}$,
$\chi_{i}$ and $\chi_{i^{\prime}}$ are relatively prime in $\mathbb{Q}[x]$, so

$a_{i,i^{\prime}}\chi_{i}+b_{i,i^{\prime}}\chi_{i^{\prime}}=1$ (14)

for some $a_{i,i^{\prime}},b_{i,i^{\prime}}\in\mathbb{Q}[x]$. If $p$ does not
divide the numerator or denominator of any coefficient of $a_{i,i^{\prime}}$
or $b_{i,i^{\prime}}$ (for $1\leq i<i^{\prime}\leq m$), then the relation (14)
holds in $(\mathbb{Z}/p^{j}\mathbb{Z})[x]$ (with both $a_{i,i^{\prime}}$ and
$b_{i,i^{\prime}}$ nonzero), so $\chi_{i},\chi_{i^{\prime}}$ are relatively
prime in $(\mathbb{Z}/p^{j}\mathbb{Z})[x]$ (for each $i<i^{\prime}$), which
implies that they have no common roots. Hence $\chi$ is square-free.

$None$ Fix a prime $p$ as in the preceding paragraph and fix $j\geq 1$. By
Lemma 6.5, $A$ is diagonalizable in
$\operatorname{GL}_{2n}(\mathbb{Z}/p^{j}\mathbb{Z})$. Let
$v_{1},\ldots,v_{2n}$ be an eigenbasis for $A$.

We choose $p$ so that each eigenvector is isotropic. If $v$ is not isotropic,
then its eigenvalue is $\pm 1\pmod{p}$ (since $\langle v,v\rangle=\langle
Av,Av\rangle=\lambda^{2}\langle v,v\rangle$). Therefore, if we assume $p$ does
not divide $\chi(1)$ or $\chi(-1)$, then $\pm 1$ are not roots of $\chi$ in
$\mathbb{Z}/p^{j}\mathbb{Z}$, which implies $v_{1},\ldots,v_{2n}$ are
isotropic in $(\mathbb{Z}/p^{j}\mathbb{Z})^{2n}$.

Vectors in our eigenbasis come in pairs whose eigenvalues are inverses, so we
write instead $v_{1},\ldots,v_{n},u_{1},\ldots,u_{n}$, where $v_{i},u_{i}$
have eigenvalues $\lambda_{i},\frac{1}{\lambda_{i}}$. For each $i$, the
vectors $v_{i},u_{i}$ span a hyperbolic summand; furthermore, up to scaling
$v_{i}$, we can assume that the intersection matrix of $v_{i},u_{i}$ is

$\left(\begin{array}[]{cc}0&1\\\ 1&0\end{array}\right).$

Therefore there is an isometry of $(\mathbb{Z}/p^{j}\mathbb{Z})^{2n}$ taking
the standard basis $\\{e_{i},f_{i}\\}$ to $\\{v_{i},u_{i}\\}$. This shows that
$A$ is diagonalizable in
$\operatorname{SO}(Q_{n};\mathbb{Z}/p^{j}\mathbb{Z})$. ∎

##  7 Counting flat cycles

In this section we complete the proofs of Theorem A–C.

###  7.1 $\operatorname{SL}_{n+1}$ case (Theorem A)

Fix $G,G_{1},G_{2}$ as in §5.1. For any prime $p$, let $r=p^{k}$ for some
fixed $k\gg 0$ large enough that Corollary 5.3 applies. For any $\ell>k$, let
$s=p^{\ell}$. The main remaining task is to estimate the sizes of the groups
$W(r,s)=\Gamma(r)/\Gamma(s)$ and $W_{i}(r,s)=\Gamma_{i}(r)/\Gamma_{i}(s)$ for
$i=1,2$ as functions of $\ell$ for infinitely many primes $p$.

#### Size of $W(r,s)$.

We show that $|W(r,s)|\sim(p^{n^{2}+2n})^{\ell}$. Recall the following short
exact sequence (2) from §3.

$1\rightarrow W(r,s)\rightarrow G(\mathbb{Z})/\Gamma(s)\rightarrow
G(\mathbb{Z})/\Gamma(r)\rightarrow 1,$

where $G(\mathbb{Z})/\Gamma(t)$ is isomorphic to the image of

$\operatorname{SL}_{n+1}(\mathbb{Z})\rightarrow\operatorname{SL}_{n+1}(\mathbb{Z}/t\mathbb{Z}).$

Because $r$ is fixed, it follows that
$|W(r,s)|\sim|\operatorname{SL}_{n+1}(\mathbb{Z}/s\mathbb{Z})|$.

It is well-known that
$\operatorname{SL}_{n+1}(\mathbb{Z})\rightarrow\operatorname{SL}_{n+1}(\mathbb{Z}/t\mathbb{Z})$
is surjective for each $t$, e.g. because these groups are generated by
elementary matrices [HO89, Thm. 4.3.9]. Since
$|\operatorname{SL}_{n+1}(\mathbb{Z}/t\mathbb{Z})|\sim
t^{(n+1)^{2}-1}=t^{n^{2}+2n}$, we conclude that
$|W(r,s)|\sim\left(p^{n^{2}+2n}\right)^{\ell}$.

#### Size of $W_{2}(r,s)$.

We show that $|W_{2}(r,s)|\sim\left(p^{n^{2}-1}\right)^{\ell}$ for all but
finitely many $p$.

Recall that $G_{2}(\mathbb{Q})$ is conjugate to the block diagonal subgroup
$\operatorname{GL}_{n}(\mathbb{Q})\subset\operatorname{SL}_{n+1}(\mathbb{Q})$.
Fix $\epsilon\in\operatorname{SL}_{n+1}(\mathbb{Q})$ so that

$\epsilon\>G_{2}(\mathbb{Q})\>\epsilon^{-1}=\operatorname{GL}_{n}(\mathbb{Q}).$

Observe that there is a subgroup $\Lambda<G_{2}(\mathbb{Z})$ whose index in
$G_{2}(\mathbb{Z})$ is at most 2 and such that the Zariski closure of
$\epsilon\Lambda\epsilon^{-1}$ is

$\operatorname{SL}_{n}=\left\\{\left(\begin{array}[]{cc}A&0\\\
0&1\end{array}\right)\right\\}\subset\operatorname{SL}_{n+1}(\mathbb{Q}).$

Specifically, if $\mathbb{Q}^{n}=P\oplus L$ is the decomposition preserved by
$G_{2}(\mathbb{Q})$, then $G_{2}(\mathbb{Z})$ preserves
$L\cap\mathbb{Z}^{n}\cong\mathbb{Z}$, and we can take $\Lambda$ to be the
kernel of the homomorphism
$G_{2}(\mathbb{Z})\rightarrow\operatorname{Aut}(L\cap\mathbb{Z}^{n})$. The
Zariski closure of $\epsilon\Lambda\epsilon^{-1}$ is certainly contained in
$\operatorname{SL}_{n}$, and it is equal because
$\epsilon\Lambda\epsilon^{-1}$ contains a finite-index subgroup of
$\operatorname{SL}_{n}(\mathbb{Z})$.

Let $G_{2}^{\prime}$ be the Zariski closure of $\Lambda$. To estimate the size
of $W_{2}(r,s)$, it suffices to compute the image of
$G_{2}^{\prime}(\mathbb{Z})\rightarrow G_{2}^{\prime}(\mathbb{Z}/t\mathbb{Z})$
for $t=p^{j}$. In fact, we will show the following.

#### Claim.

For all but finitely many $p$, the map $G_{2}^{\prime}(\mathbb{Z})\rightarrow
G_{2}^{\prime}(\mathbb{Z}/t\mathbb{Z})$ is surjective and the group
$G_{2}^{\prime}(\mathbb{Z}/t\mathbb{Z})$ is isomorphic to
$\operatorname{SL}_{n}(\mathbb{Z}/t\mathbb{Z})$.

From the claim we deduce (similar to the computation for $W(r,s)$) that

$|W_{2}(r,s)|\sim\left(p^{n^{2}-1}\right)^{\ell}.$

###### Proof of Claim.

First assume that $p$ does not divide the denominator of any entry of
$\epsilon$ (this excludes finitely many $p$). Then $\epsilon$ descends to an
invertible matrix in $\operatorname{SL}_{n+1}(\mathbb{Z}/t\mathbb{Z})$.
Observe that there is a finite index subgroup
$\Lambda^{\prime}<G_{2}^{\prime}(\mathbb{Z})$ so that
$\epsilon\Lambda^{\prime}\epsilon^{-1}$ is contained in
$\operatorname{SL}_{n}(\mathbb{Z})$. Then the images of the maps

$\Lambda^{\prime}\hookrightarrow G_{2}^{\prime}(\mathbb{Z})\rightarrow
G_{2}^{\prime}(\mathbb{Z}/t\mathbb{Z})\>\>\>\text{ and
}\>\>\>\epsilon\Lambda^{\prime}\epsilon^{-1}\hookrightarrow\operatorname{SL}_{n}(\mathbb{Z})\rightarrow\operatorname{SL}_{n}(\mathbb{Z}/t\mathbb{Z})$

are conjugate in $\operatorname{SL}_{n+1}(\mathbb{Z}/t\mathbb{Z})$. This
proves the claim, since the map on the right is surjective for almost every
$p$ by strong approximation [LS03, Window 9, Cor. 3].333More precisely, strong
approximation implies that the inclusion
$\epsilon\Lambda\epsilon^{-1}\hookrightarrow\operatorname{SL}_{n}(\mathbb{Z})\hookrightarrow\operatorname{SL}_{n}(\mathbb{Z}_{p})$
has dense image. Therefore $\epsilon\Lambda\epsilon^{-1}$ surjects onto the
image of
$\operatorname{SL}_{n}(\mathbb{Z}_{p})\rightarrow\operatorname{SL}_{n}(\mathbb{Z}/t\mathbb{Z})$,
and this latter map is surjective because
$\operatorname{SL}_{n}(\mathbb{Z})\rightarrow\operatorname{SL}_{n}(\mathbb{Z}/t\mathbb{Z})$
is surjective, as observed above. ∎

#### Size of $W_{1}(r,s)$.

We show that $|W_{1}(r,s)|\sim\left(p^{n}\right)^{\ell}$.

By the definition of $G_{1}$, the $R$-points $G_{1}(R)$ is the centralizer of
$A_{1}$ in $\operatorname{SL}_{n+1}(R)$, for any ring $R$. Let
$\chi\in\mathbb{Z}[x]$ be the characteristic polynomial of $A_{1}$.

By Lemma 6.5, there are infinitely many primes $p$ so that $\chi$ splits
completely in $(\mathbb{Z}/p^{j}\mathbb{Z})[x]$ and $A_{1}$ is diagonalizable
in $\operatorname{GL}_{n+1}(\mathbb{Z}/p^{j}\mathbb{Z})$ for each $j\geq 1$.
Fix such a prime $p$, and set $R=\mathbb{Z}/p^{j}\mathbb{Z}$. Since $A_{1}$ is
diagonalizable with distinct eigenvalues, the centralizer of $A_{1}$ in
$\operatorname{GL}_{n+1}(R)$ is conjugate to the group of diagonal matrices;
hence this centralizer has size

$|(\mathbb{Z}/p^{j}\mathbb{Z})^{\times}|^{n+1}\sim(p^{j})^{n+1}.$

Next we show that $G_{1}(\mathbb{Z})$ surjects to the centralizer of $A_{1}$
in $\operatorname{SL}_{n+1}(\mathbb{Z}/p^{j}\mathbb{Z})$. The centralizer of
$A_{1}$ in $M_{n+1}(\mathbb{Z})$ contains a subring isomorphic to
$\mathbb{Z}[t]/(\chi)$ (see §4). The same statement holds with $\mathbb{Z}$
replaced by $R$. Since $\chi$ splits completely in $R$ by our choice of $p$,
there is a ring isomorphism $R[x]/(\chi)\cong R^{n+1}$. Up to conjugation the
map $R[x]/(\chi)\rightarrow M_{n+1}(R)$ has image equal to the diagonal
matrices. From this we deduce that the invertible elements of $R[x]/(\chi)$
surject onto the diagonal subgroup of $\operatorname{GL}_{n+1}(R)$ and that
$G_{1}(\mathbb{Z})$ surjects onto the centralizer of $A_{1}$ in
$\operatorname{SL}_{n+1}(\mathbb{Z}/p^{j}\mathbb{Z})$. From this we conclude
that $|W_{1}(r,s)|\sim\left(p^{n}\right)^{\ell}$.

#### An expression in terms of volume.

So far, we have shown that

$\frac{|W(r,s)|}{|W_{1}(r,s)|\cdot|W_{2}(r,s)|}\sim\left(p^{n^{2}+2n-(n^{2}-1)-n}\right)^{\ell}=\left(p^{n+1}\right)^{\ell}$

for infinitely many primes $p$.

Since the volume satisfies
$\operatorname{Vol}(Y(s))=|W(r,s)|\cdot\operatorname{Vol}(Y(r))$, we can write
$\operatorname{Vol}(Y(s))\sim\left(p^{n^{2}+2n}\right)^{\ell}$. Therefore,

$\frac{|W(r,s)|}{|W_{1}(r,s)|\cdot|W_{2}(r,s)|}\sim\left(p^{n+1}\right)^{\ell}\sim\operatorname{Vol}(Y(s))^{\frac{n+1}{n^{2}+2n}}.$

Combined with Lemma 3.2 and Corollary 5.3, this proves Theorem A. ∎

###  7.2 $R_{F/\mathbb{Q}}(\operatorname{SL}_{2})$ case (Theorem C)

In this section we prove Theorem C following the same outline as the preceding
section.

Fix $F\subset\mathbb{R}$ a totally real number field. Write
$\mathcal{H}=\operatorname{SL}_{2}$, viewed as an algebraic group over $F$,
and consider the restriction of scalars $G=R_{F/\mathbb{Q}}(\mathcal{H})$.

Given a prime ideal $\mathfrak{p}\subset\mathcal{O}_{F}$, let
$\mathfrak{r}=\mathfrak{p}^{k}$ for some fixed $k\gg 0$ large enough that
Corollary 5.6 applies. For any $\ell>k$, let
$\mathfrak{s}=\mathfrak{p}^{\ell}$. Next we estimate the size of the groups
$W(\mathfrak{r},\mathfrak{s})=\Gamma(\mathfrak{r})/\Gamma(\mathfrak{s})$ and
$W_{i}(\mathfrak{r},\mathfrak{s})=\Gamma_{i}(\mathfrak{r})/\Gamma_{i}(\mathfrak{s})$
for $i=1,2$ as functions of $\ell$ for infinitely many primes $\mathfrak{p}$.

#### Size of $W(\mathfrak{r},\mathfrak{s})$.

First we show that $|W(\mathfrak{r},\mathfrak{s})|\sim N(\mathfrak{s})^{3}$,
where $N(\mathfrak{t})=|\mathcal{O}_{F}/\mathfrak{t}|$ is the ideal norm.
Similar to (2), we have a short exact sequence

$1\rightarrow
W(\mathfrak{r},\mathfrak{s})\rightarrow\operatorname{SL}_{2}(\mathcal{O}_{F})/\Gamma(\mathfrak{s})\rightarrow\operatorname{SL}_{2}(\mathcal{O}_{F})/\Gamma(\mathfrak{r})\rightarrow
1.$

By [HO89, Thm. 4.3.9], the reduction map
$\operatorname{SL}_{2}(\mathcal{O}_{F})\rightarrow\operatorname{SL}_{2}(\mathcal{O}_{F}/\mathfrak{t})$
is surjective, so there is a short exact sequence

$1\rightarrow\Gamma(\mathfrak{t})\rightarrow\operatorname{SL}_{2}(\mathcal{O}_{F})\rightarrow\operatorname{SL}_{2}(\mathcal{O}_{F}/\mathfrak{t})\rightarrow
1.$ (15)

Using (15) and Proposition 6.1, we conclude that

$W(\mathfrak{r},\mathfrak{s})=\frac{|\operatorname{SL}_{2}(\mathcal{O}_{F}/\mathfrak{s})|}{|\operatorname{SL}_{2}(\mathcal{O}_{F}/\mathfrak{r})|}\sim
N(\mathfrak{s})^{3}.$

#### Size of $W_{i}(\mathfrak{r},\mathfrak{s})$.

Recall that $\mathcal{H}_{i}$ is the centralizer of
$A_{i}\in\operatorname{SL}_{2}(F)$, and we have assumed that $A_{1},A_{2}$ are
conjugate in $\operatorname{SL}_{2}(F)$. As such, the computation of
$|W_{i}(\mathfrak{r},\mathfrak{s})|$ is essentially the same for $i=1,2$. To
simplify exposition, we present the proof for $i=1$ and show that
$|W_{1}(\mathfrak{r},\mathfrak{s})|\sim N(\mathfrak{s})$.

We start by assuming $\mathfrak{p}$ is not one of the finitely many factors of
denominators of entries of $A_{1}$. This allows us to consider $A_{1}$ as an
element of $\operatorname{SL}_{2}(\mathcal{O}_{F}/\mathfrak{p}^{k})$ for each
$k\geq 1$.

By Lemma 6.5, there are infinitely many prime ideals
$\mathfrak{p}\subset\mathcal{O}_{F}$ such that the characteristic polynomial
of $A_{1}$ splits completely in $(\mathcal{O}_{F}/\mathfrak{p}^{j})[x]$ and
$A_{1}$ is diagonalizable in
$\operatorname{GL}_{2}(\mathcal{O}_{F}/\mathfrak{p}^{j})$ for each $j\geq 1$.
Fixing such a prime, the centralizer of $A_{1}$ in
$\operatorname{GL}_{2}(\mathcal{O}_{F}/\mathfrak{p}^{j})$ is conjugate to the
group of diagonal matrices, so the centralizer of $A_{1}$ in
$\operatorname{SL}_{2}(\mathcal{O}_{F}/\mathfrak{p}^{j})$ is isomorphic to
$(\mathcal{O}_{F}/\mathfrak{p}^{j})^{\times}$. Denoting
$\mathfrak{t}=\mathfrak{p}^{j}$, this group has order
$N(\mathfrak{t})\cdot\frac{N(\mathfrak{p})-1}{N(\mathfrak{p})}$, cf. Equation
(10).

Next we show that $\mathcal{H}_{1}(\mathcal{O}_{F})$ surjects to the
centralizer of $A_{1}$ in
$\operatorname{SL}_{2}(\mathcal{O}_{F}/\mathfrak{p}^{j})$. The centralizer of
$A_{1}$ in $M_{2}(\mathcal{O}_{F})$ contains a subring isomorphic to
$\mathcal{O}_{F}[x]/(\chi)$, where $\chi$ is the characteristic polynomial.
The same statement holds with $\mathcal{O}_{F}$ replaced by
$R:=\mathcal{O}_{F}/\mathfrak{p}^{j}$. Since $\chi$ splits completely in $R$,
there is a ring isomorphism $R[x]/(\chi)\cong R^{2}$. Up to conjugation the
map $R[t]/(\chi)\rightarrow M_{2}(R)$ has image equal to the diagonal
matrices. From this we deduce that the invertible elements of $R[t]/(\chi)$
surject onto the diagonal subgroup of $\operatorname{GL}_{2}(R)$ and that
$\mathcal{H}_{1}(\mathcal{O}_{F})$ surjects onto the centralizer of $A_{1}$ in
$\operatorname{SL}_{2}(R)$.

From this we conclude that $|W_{1}(\mathfrak{r},\mathfrak{s})|\sim
N(\mathfrak{s})$. By the same argument, we have
$|W_{2}(\mathfrak{r},\mathfrak{s})|\sim N(\mathfrak{s})$.

#### An expression in terms of volume.

So far, we have shown that

$\frac{|W(\mathfrak{r},\mathfrak{s})|}{|W_{1}(\mathfrak{r},\mathfrak{s})|\cdot|W_{2}(\mathfrak{r},\mathfrak{s})|}\sim\frac{N(\mathfrak{s})^{3}}{N(\mathfrak{s})\cdot
N(\mathfrak{s})}=N(\mathfrak{s})$

for infinitely many primes $p$.

Since the volume satisfies
$\operatorname{Vol}(Y(\mathfrak{s}))=|W(\mathfrak{r},\mathfrak{s})|\cdot\operatorname{Vol}(Y(\mathfrak{r}))$,
we can write $\operatorname{Vol}(Y(s))\sim N(\mathfrak{s})^{3}$. Therefore,
$\frac{|W(\mathfrak{r},\mathfrak{s})|}{|W_{1}(\mathfrak{r},\mathfrak{s})|\cdot|W_{2}(\mathfrak{r},\mathfrak{s})|}\sim\operatorname{Vol}(Y(\mathfrak{s}))^{1/3}$.
Combined with Lemma 3.2 and Corollary 5.6, this proves Theorem C. ∎

###  7.3 Orthogonal case (Theorem B)

In this section we prove Theorem B. Fix $n\geq 2$ and let $G$ denote the
algebraic group

$\operatorname{SO}(Q_{n})=\\{g\in\operatorname{SL}_{2n}:g^{t}Q_{n}g=Q_{n}\\},$

where $Q_{n}$ is the matrix defined in (4).

Take $G_{1},G_{2}<G$ as in §5.3. For any prime $p\equiv 3\pmod{4}$, let
$r=p^{k}$ for some fixed $k\gg 0$ large enough that Corollary 5.8 applies. For
any $\ell>k$, let $s=p^{\ell}$. Below we estimate the size of the groups
$W(r,s)=\Gamma(r)/\Gamma(s)$ and $W_{i}(r,s)=\Gamma_{i}(r)/\Gamma_{i}(s)$ for
$i=1,2$ as functions of $\ell$ for infinitely many primes $p$.

#### Size of $W(r,s)$.

First show that $|W(r,s)|\sim\left(p^{n^{2}}\right)^{\ell}$. Recall the
following short exact sequence (2) from §3:

$1\rightarrow W(r,s)\rightarrow G(\mathbb{Z})/\Gamma(s)\rightarrow
G(\mathbb{Z})/\Gamma(r)\rightarrow 1.$

For $t=p^{j}$, the group $G(\mathbb{Z})/\Gamma(t)$ is isomorphic to the image
of

$\operatorname{SO}(Q_{n};\mathbb{Z})\rightarrow\operatorname{SO}(Q_{n};\mathbb{Z}/t\mathbb{Z}).$

By strong approximation and Hensel’s lemma, the image of this map contains the
group $\Omega(Q_{n};\mathbb{Z}/t\mathbb{Z})$ defined in (12) for almost every
prime $p$. This implies that
$|W(r,s)|\sim\left(p^{n^{2}}+2p^{n(n-1)/2}\right)^{\ell}$ by Corollary 6.3.

Our application of these results is routine, so we only give a brief
explanation. Similar to §7.1, strong approximation is used to conclude that
the image of the inclusion $G(\mathbb{Z})\hookrightarrow G(\mathbb{Z}_{p})$ is
dense. One difference is that the algebraic group $\operatorname{SO}(Q_{n})$
is not simply connected, so in order to apply strong approximation, one needs
to first pass to the 2-fold spin cover; see [LS03, Window 9, §1 and §4] and
[Pin00]. Hensel’s lemma in algebraic geometry implies that the reduction map
$G(\mathbb{Z}_{p})\rightarrow G(\mathbb{Z}/p^{j}\mathbb{Z})$ is surjective for
each $j\geq 1$; see [Gro67, §18.5] and [Stu20, Thm. 4.17]. In order to apply
Hensel, one needs $G$ to be a smooth $\mathbb{Z}_{p}$-scheme. The smoothness
of $G=\operatorname{SO}(Q_{n})$ as a group scheme is discussed in [Con14,
C.1.5].

#### Size of $W_{2}(r,s)$.

For all but finitely many primes $p$ (depending on the choice of the negative
vector), the group $G_{2}(\mathbb{Z}/t\mathbb{Z})$ is isomorphic to
$\operatorname{SO}(\widetilde{Q}_{n-1},\mathbb{Z}/t\mathbb{Z})$, where
$\widetilde{Q}_{n}$ is defined in (13). Again by strong approximation, the
image of $G_{2}(\mathbb{Z})\rightarrow G_{2}(\mathbb{Z}/t\mathbb{Z})$ is
isomorphic to $\Omega(\widetilde{Q}_{n-1};\mathbb{Z}/t\mathbb{Z})$ for almost
every prime $p$. This implies that
$|W_{2}(r,s)|\sim\left(p^{(n-1)^{2}}+2p^{(n-1)(n-2)/2+p^{n-1}}\right)^{\ell}$
by Corollary 6.4.

#### Size of $W_{1}(r,s)$.

We show that $|W_{1}(r,s)|\sim\left(p^{n}\right)^{\ell}$.

By the definition of $G_{1}$, the group $G_{1}(R)$ is the centralizer of
$A_{1}$ in $\operatorname{SO}(Q_{n};R)$, for any ring $R$. Let
$\chi\in\mathbb{Z}[x]$ be the characteristic polynomial of $A_{1}$.

By Proposition 4.15 and Lemma 6.7, there are infinitely many primes $p\equiv
3\pmod{4}$ so that $\chi$ splits completely in
$(\mathbb{Z}/p^{j}\mathbb{Z})[x]$ and $A_{1}$ is diagonalizable in
$\operatorname{SO}(Q_{n};\mathbb{Z}/p^{j}\mathbb{Z})$ for each $j\geq 1$. Fix
such a prime $p$, and set $R=\mathbb{Z}/p^{j}\mathbb{Z}$. Since $A_{1}$ is
diagonalizable with distinct eigenvalues, the centralizer of $A_{1}$ in
$\operatorname{SO}(Q_{n};R)$ is conjugate to the group of diagonal matrices in
$\operatorname{SO}(Q_{n};R)$; hence this centralizer has size

$|(\mathbb{Z}/p^{j}\mathbb{Z})^{\times}|^{n}\sim(p^{j})^{n}.$

It remains to show that $G_{1}(\mathbb{Z})$ surjects to the centralizer of
$A_{1}$ in $\operatorname{SO}(Q_{n};\mathbb{Z}/p^{j}\mathbb{Z})$. The
centralizer of $A_{1}$ in $M_{2n}(\mathbb{Z})$ contains a subring isomorphic
to $\mathbb{Z}[t]/(\chi)$ (see §4). The same statement holds with $\mathbb{Z}$
replaced by $R$. Since $\chi$ splits completely in $R$ by our choice of $p$,
there is a ring isomorphism $R[x]/(\chi)\cong R^{n+1}$. Up to conjugation the
map $R[x]/(\chi)\rightarrow M_{2n}(R)$ has image equal to the diagonal
matrices. From this we deduce that the invertible elements of $R[x]/(\chi)$
surject onto the diagonal subgroup of $\operatorname{GL}_{2n}(R)$ and that
$G_{1}(\mathbb{Z})$ surjects onto the centralizer of $A_{1}$ in
$\operatorname{SO}(Q_{n};\mathbb{Z}/p^{j}\mathbb{Z})$. From this we conclude
that $|W_{1}(r,s)|\sim\left(p^{n}\right)^{\ell}$ (for an isometry, the
diagonal entries occur in $(\lambda,\frac{1}{\lambda})$ pairs).

#### An expression in terms of volume.

Combining the above estimates, we have

$\frac{|W(r,s)|}{|W_{1}(r,s)|\cdot|W_{2}(r,s)|}\sim\left(\frac{p^{n^{2}}+p^{n(n-1)/2}}{p^{n^{2}-n+1}+2p^{(n^{2}-n+2)/2}+p^{2n-1}}\right)^{\ell}$

for infinitely many primes $p\equiv 3\pmod{4}$.

Since the volume satisfies
$\operatorname{Vol}(Y(s))=|W(r,s)|\cdot\operatorname{Vol}(Y(r))$, we can write
$\operatorname{Vol}(Y(s))\sim\left(p^{n^{2}}+p^{n(n-1)/2}\right)^{\ell}$
Therefore,

$\frac{|W(r,s)|}{|W_{1}(r,s)|\cdot|W_{2}(r,s)|}\sim\operatorname{Vol}(Y(s))^{\kappa},$

where

$\kappa=1-\frac{\log(p^{n^{2}-n+1}+2p^{(n^{2}-n+2)/2}+p^{2n-1})}{\log(p^{n^{2}}+p^{n(n-1)/2})}.$

For given $p$ and $n$, the numerator of the above fraction is bounded above by
$\log(3p^{n^{2}-n+1})$ while the denominator is bounded below by
$\log(p^{n^{2}})$, so there is an inequality

$\kappa\geq\frac{n-1-\log_{p}(3)}{n^{2}}.$

Combined with Lemma 3.2 and Corollary 5.8, this proves Theorem B. ∎

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Daniel Studenmund
Department of Mathematical Sciences, Binghamton University
<EMAIL_ADDRESS>
Bena Tshishiku
Department of Mathematics, Brown University
<EMAIL_ADDRESS>