# Actions of large finite groups on aspherical manifolds

Jordi Daura Serrano  Jordi Daura Serrano, Department de Màtematiques i
Informàtica, Universitat de Barcelona (UB), Gran Via de les Corts Catalanes
585, 08007 Barcelona (Spain<EMAIL_ADDRESS>
###### Abstract.

In this paper we study actions of finite groups on closed connected aspherical
manifolds. Under some assumptions on the outer automorphism group of the
fundamental group of a closed connected aspherical manifold $M$, we prove that
the homeomorphism group of $M$ is Jordan, we bound the discrete degree of
symmetry of $M$ and obtain a rigidity result, and we study the number of
stabilizers of finite group actions on $M$. Thereafter, we show that closed
connected aspherical locally homogeneous spaces $H\setminus G/\Gamma$ satisfy
the necessary hypothesis on the outer automorphism group of the fundamental
group.

2020 Mathematics Subject Classification: 57S17, 54H15, 22E40

##  1\. Introduction

Given a closed topological manifold $M$, can we determine which finite groups
act effectively on $M$? The answer of this question is completely out of reach
in the vast majority of cases with the current tools of finite transformation
groups theory. One way to modify this question to obtain a more tractable
problem is to consider actions of a finite group $F$ on $M$ and to study
properties not of the action of $F$ on $M$, but on the restriction to some
subgroup of $F$ of index bounded by a constant $C$ only depending on $M$. Let
us recall some problems that follow this philosophy from [MiR24a], a recent
survey on the topic.

We need the following definition to state the first problem:

######  Definition 1.1.

A group $\mathcal{G}$ is said to be Jordan if there exists a constant $C$ such
that every finite subgroup $F\leq\mathcal{G}$ has an abelian subgroup $A\leq
F$ such that $[F:A]\leq C$.

The name of this property is inspired by a classical theorem of Camille Jordan
which states that $\operatorname{GL}(n,\mathbb{R})$ is Jordan. Around 30 years
ago Étienne Ghys asked in a series of talks whether the diffeomorphism group
of a closed connected smooth manifold $M$ is Jordan. This question has been
answered affirmatively for a lot of different manifolds like closed flat
manifolds, integral homology spheres, closed connected manifolds with non-zero
Euler characteristics or closed connected manifolds up to dimension 3 (see
[MiR10, Zim14, Ye19, MiR19]), while it has been shown that there are closed
manifolds whose diffeomorphism group is not Jordan, like $S^{2}\times T^{2}$
(see [CPS14, MiR17]). More recently, the same question has been studied for
the homeomorphism group of closed topological manifolds, extending the results
obtained in the smooth case (see [MiR24a] and references therein). Note that
this property does not tell us anything for finite groups $F$ acting on $M$
with $|F|\leq C$, but it becomes relevant for large enough finite groups
acting effectively on $M$.

Another interesting problem (specially when $\operatorname{Homeo}(M)$ is
Jordan) is to study the rank of finite abelian groups (defined as the smallest
number of elements needed to generate the group) which can act effectively on
$M$. Here again, we follow the philosophy of looking for a statement that
eventually requires to replace the our group by a subgroup of bounded index.
Fix a natural number $k$. Does there exist a constant $C$ such that every
finite abelian group $A$ acting effectively on $M$ has an abelian subgroup $B$
such that $[A:B]\leq C$ and $\operatorname{rank}(B)\leq k$?. This question can
be reformulated in terms of the following invariant introduced in [MiR24b].

######  Definition 1.2.

Given a manifold $M$ let

$\mu(M)=\\{r\in\mathbb{N}:\text{ $M$ admits an effective action of
$(\mathbb{Z}/a)^{r}$ for arbitrarily large $a$}\\}.$

More explicitly, $r\in\mu(M)$ if there exists an increasing sequence of
natural number $\\{a_{i}\\}$ and effective group actions of
$(\mathbb{Z}/a_{i})^{r}$ on $M$ for each $i$.

The discrete degree of symmetry of a manifold $M$ is

$\operatorname{disc-sym}(M)=\max(\\{0\\}\cup\mu(M)).$

The above question is equivalent to asking whether $\operatorname{disc-
sym}(M)\leq k$. By a theorem of L.N.Mann and J.C.Su (see [MS63]) we know that
if $M$ is a closed connected manifold then $\operatorname{disc-sym}(M)$ is a
well-defined natural number, but finding the exact value of
$\operatorname{disc-sym}(M)$ is probably difficult in most cases. Note that a
manifold $M$ can admit group actions of abelian group of higher rank than
$\operatorname{disc-sym}(M)$. For example, for any natural numbers $a,b$ there
exists a closed surface $\Sigma_{g(a,b)}$ of genus $g(a,b)\geq 2$ such that
$(\mathbb{Z}/a)^{b}$ acts freely on $\Sigma_{g(a,b)}$. On the other hand
$\operatorname{disc-sym}(\Sigma_{g(a,b)})=0$ (see [FM11, Chapter 7]).

The definition of $\operatorname{disc-sym}(M)$ is analogous to the definition
of the toral degree of symmetry

$\operatorname{tor-sym}(M)=\max(\\{0\\}\cup\\{r\in\mathbb{N}:T^{r}\text{ acts
effectively on $M$}\\})$

studied in [Hsi12, Chapter VII. §2]. Recall that a classical result states
that if $M$ is a closed manifold of dimension $n$ then $\operatorname{tor-
sym}(M)\leq n$ and that $\operatorname{tor-sym}(M)=n$ if and only if $M\cong
T^{n}$. However, it is not known whether $\operatorname{disc-sym}(M)\leq n$,
and whether $\operatorname{disc-sym}(M)=n$ if and only if $M\cong T^{n}$.

The third problem is related to the number of stabilizers of a group action on
a manifold.

######  Definition 1.3.

Let $F$ be a finite group acting effectively on a manifold $M$. The set of
stabilizer subgroups of the action of $F$ on $M$ is denoted by

$\operatorname{Stab}(F,M)=\\{F_{x}:x\in M\\}.$

It is not possible to bound $|\operatorname{Stab}(F,M)|$ only depending on
$M$. For example, for all $n$ there is an effective action of the dihedral
group $D_{n}$ on $S^{1}$ such that $|\operatorname{Stab}(D_{n},S^{1})|\geq
n/2$. On the other hand, it is proven in [CPSMiR21, Theorem 1.3] that for any
closed connected manifold $M$ there exists a constant $C$ only depending on
$M$ such that any finite $p$-group $F$ acting effectively on $M$ has a
subgroup $H$ such that $[F:H]\leq C$ and $|\operatorname{Stab}(H,M)|\leq C$.
This result was crucial to prove a generalized version of the Jordan property
of homeomorphisms group of closed manifolds in [CPS22], which states that
there exists a constant $C$ such that every finite group $F$ acting on $M$ has
a nilpotent subgroup $N$ such that $[F:N]\leq C$. It is not known if we can
remove the hypothesis of $F$ being a $p$-group in [CPSMiR21, Theorem 1.3].

For the final question, we introduce the following definition:

######  Definition 1.4.

A group $\mathcal{G}$ is said to be Minkowski if there exists a constant $C$
such that every finite subgroup $F\leq\mathcal{G}$ fulfils that $|F|\leq C$.

######  Remark 1.5.

This name is motivated by a classical result of Hermann Minkowski which states
that $\operatorname{GL}(n,\mathbb{Z})$ is Minkowski. The Minkowski property
was studied in [Pop18, Gol24] under the name of bounded finite subgroups
property.

If $M$ is a closed manifold and $\operatorname{Homeo}(M)$ is Minkowski then
$M$ is said to be almost-asymmetric. In the particular case where $M$ does not
admit any effective finite group action we say that $M$ is asymmetric. This
case has been extensively studied (see [Pup07] and references therein).

In this paper we study these questions when $M$ is a closed connected
aspherical manifolds, generalizing some existing results in the literature.
For example, in [Ye19] it is proven that $\operatorname{Homeo}(M)$ is Jordan
when $M$ is closed flat manifold and in [JL10] a bound of the rank of
elementary abelian group acting on closed solvmanifolds is given. However, it
seems that these questions have not been studied in full generality on the
literature. They are particularly relevant for closed connected aspherical
manifolds, since tori are the only compact connected Lie groups that act
effectively on them [LR10, Thereom 3.1.16]. Therefore, $\operatorname{tor-
sym}(M)$ and $\operatorname{disc-sym}(M)$ become important invariants.

We recall some notation before stating the main results of this paper. Given a
group $\mathcal{G}$ we denote the automorphism group of $\mathcal{G}$ by
$\operatorname{Aut}(\mathcal{G})$. Given $g\in\mathcal{G}$, we denote by
$c_{g}:\mathcal{G}\longrightarrow\mathcal{G}$ the conjugation by $g$,
$c_{g}(h)=ghg^{-1}$. The normal subgroup $\\{c_{g}:g\in
G\\}\trianglelefteq\operatorname{Aut}(\mathcal{G})$ is denoted by
$\operatorname{Inn}(\mathcal{G})$. Recall that
$\operatorname{Inn}(\mathcal{G})\cong\mathcal{G}/\mathcal{Z}(\mathcal{G})$,
where $\mathcal{Z}(\mathcal{G})$ denotes the center of $\mathcal{G}$. The
outer automorphisms group of $\mathcal{G}$ is
$\operatorname{Out}(\mathcal{G})=\operatorname{Aut}(\mathcal{G})/\operatorname{Inn}(\mathcal{G})$.

Our first theorem is the following:

######  Theorem 1.6.

Let $M$ be a closed connected $n$-dimensional aspherical manifold such that
$\mathcal{Z}(\pi_{1}(M))$ is finitely generated and
$\operatorname{Out}(\pi_{1}(M))$ is Minkowski. Then:

  * 1.

$\operatorname{Homeo}(M)$ is Jordan.

  * 2.

$\operatorname{disc-sym}(M)\leq\operatorname{rank}\mathcal{Z}(\pi_{1}(M))\leq
n$, and $\operatorname{disc-sym}(M)=n$ if and only if $M$ is homeomorphic to
$T^{n}$.

  * 3.

If $\chi(M)\neq 0$ then $M$ is almost-asymmetric.

  * 4.

If $\operatorname{Aut}(\pi_{1}(M))$ is Minkowski, then there exists a constant
$C$ such that every finite group $F$ acting effectively on $M$ has a subgroup
$H$ such that $[F:H]\leq C$ and $|\operatorname{Stab}(H,M)|\leq C$.

This result is mainly a combination of several known results. Note that items
$2$ and $4$ partially answer affirmatively the questions [MiR24a, Question
3.4, Question 3.5] and [MiR24a, Question 12.2] respectively.

As an application of 1.6, we prove:

######  Proposition 1.7.

There exists a closed connected aspherical manifold $M$ such that
$\operatorname{Homeo}(M)$ is Jordan and $H^{*}(M)\cong H^{*}(T^{2}\times
S^{3})$.

Note that $\operatorname{Homeo}(T^{2}\times S^{3})$ is not Jordan by [MiR17].
1.7 provides the first example of two manifolds with the same cohomology with
integer coefficients such that one has Jordan homeomorphism group and the
other does not.

We also obtain a rigidity result for closed connected aspherical
$n$-dimensional manifolds when $\operatorname{disc-sym}(M)=n-1$. Let $K$
denote the Klein bottle and $SK$ denote the only non-trivial principal
$S^{1}$-bundle over $K$, then:

######  Proposition 1.8.

Let $M$ be a closed connected $n$-dimensional aspherical manifold such that
$\mathcal{Z}(\pi_{1}(M))$ is finitely generated and
$\operatorname{Out}(\pi_{1}(M))$ is Minkowski. Assume that
$\operatorname{Inn}\pi_{1}(M)$ has an element of infinite order. If
$\operatorname{disc-sym}(M)=n-1$ then $M\cong T^{n-2}\times K$ or $M\cong
T^{n-3}\times SK$

It is important to know when the hypothesis on the fundamental group of 1.6
are fulfilled. At the moment there are no known closed aspherical manifolds
where $\mathcal{Z}(\pi_{1}(M))$ is not finitely generated. Regarding the
second hypothesis, we prove:

######  Theorem 1.9.

Let $\Gamma$ be a lattice in a connected Lie group $G$. Then
$\operatorname{Out}(\Gamma)$ and $\operatorname{Aut}(\Gamma)$ are Minkowski.

The proof of this results uses several theorems of the theory of lattices of
Lie groups, like the Mostow-Prasad-Margulis rigidity theorem, the Borel
density theorem, Margulis superrigidity theorem and Margulis normal subgroup
theorem, as well as the results in [Mal02], which are used to compute the
outer automorphism group of a group extension.

If $G$ is a connected Lie group, $H$ is a maximal compact subgroup and
$\Gamma$ is a cocompact lattice of $G$ then the closed aspherical locally
homogeneous space $H\setminus G/\Gamma$ satisfies the hypothesis of 1.6. In
particular, flat manifolds, nilmanifolds, almost-flat manifolds,
solvmanifolds, infra-solvmanifolds and aspherical locally symmetric spaces
satisfy the hypothesis of 1.6. Note that if we remove the asphericity
hypothesis then closed locally homogeneous spaces do not necessarily have
Jordan homeomorphism group. Indeed, $T^{2}\times S^{2}$ is homogeneous (and
hence locally homogeneous) but $\operatorname{Homeo}(T^{2}\times S^{2})$ is
not Jordan.

1.9 generalizes [Gol24, Theorem 1.7], where it is proven that the outer
automorphism group of a cocompact lattices on connected complex Lie groups is
Minkowski. In the real case we need to be careful with the compact factors and
factors isomorphic to $\operatorname{PSL}(2,\mathbb{R})$ of the semisimple
part of $G$.

Note that 1.9 is also valid for non-cocompact lattices, although it cannot be
used to deduce properties of large finite groups actions on non-compact
aspherical locally homogeneous spaces, since the compactness hypothesis in 1.6
is essential.

To complement 1.9, we also prove that the bound on the discrete degree of
symmetry is reached for aspherical locally homogeneous space.

######  Theorem 1.10.

Let $H\setminus G/\Gamma$ be an aspherical locally homogeneous space. Then
$\operatorname{disc-sym}(H\setminus
G/\Gamma)=\operatorname{rank}\mathcal{Z}\Gamma$.

This result is a combination of the results in [LR10, Section 11.7] on the
toral degree of symmetry and the validity of the Borel conjecture for lattices
of connected Lie groups [KLR16].

Finally, using similar arguments used to prove 1.9 we can also prove:

######  Proposition 1.11.

Let $M=M_{1}\times\cdots\times M_{m}$, where $M_{i}$ are a closed aspherical
manifolds such that $\pi_{1}(M_{i})$ is hyperbolic and $\dim(M_{i})\geq 3$.
Then $\operatorname{Out}(\pi_{1}(M))$ is finite and
$\operatorname{Aut}(\pi_{1}(M))$ is Minkowski.

1.11 together with 1.6 and the fact that $\pi_{1}(M)$ is centreless implies
that $M$ is almost asymmetric.

The paper is divided as follows. In the second section we review the
construction in [Mal02] used to compute the outer automorphism group of a
group extension. We also make some observations on the Minkowski property that
will be used thereafter. In the third section we prove 1.6 and we make some
remarks on the relation between the discrete degree of symmetry and coverings
of closed connected aspherical manifolds. In section $4$ we deduce 1.9 for
solvable Lie groups and give some examples and in section 5 we prove 1.9 for
semisimple Lie groups and 1.11. Section 6 is devoted to combining the proofs
of section 4 and section 5 in order to complete the proof of 1.9 and we also
prove 1.10. Finally, in section 7 we prove 1.7 and in section 8 we prove 1.8.

Acknowledgements: First of all, I would like to thank Ignasi Mundet i Riera
for all the guidance given during this project and for the extensive revision
of the first draft of this paper. I would also like to thank László Pyber and
Endre Szabó for fruitful discussions, which helped to improve many results of
this paper, and for their hospitality during my stay at Álfred Rényi Institute
of Mathematics. This work was partially supported by the grant
PID2019-104047GB-I00 from the Spanish Ministry of Science and Innovation and
the Departament de Recerca i Universitats de la Generalitat de Catalunya (2021
SGR 00697).

##  2\. Preliminaries on outer automorphism group

The aim of this section is to briefly explain the constructions in [Mal02]
used to compute the outer automorphism group of a group extension.

Let

${1}$${K}$${G}$${Q}$${1}$$\scriptstyle{p}$

be a short exact sequence of groups. The extension is determined by the
morphism $\psi:Q\longrightarrow\operatorname{Out}(K)$ (called the abstract
kernel) such that $\psi(q)$ is the class of the conjugation
$c_{\sigma(q)|K}:K\longrightarrow K$, where $\sigma:Q\longrightarrow G$ is a
set-theoretic section, and a 2-cocycle $c\in H^{2}_{\psi}(Q,\mathcal{Z}K)$.
Let $\operatorname{Aut}(G,K)=\\{f\in\operatorname{Aut}(G):f(K)=K\\}$ and
$\operatorname{Out}(G,K)=\operatorname{Aut}(G,K)/\operatorname{Inn}G$.
Finally, recall that if $H$ is a subgroup of $G$, the centralizer $C_{G}(H)$
is $\\{g\in G:c_{h}(g)=g\text{ for all }h\in H\\}$.

######  Theorem 2.1.

[Mal02] There exist short exact sequences

${1}$${\mathcal{K}}$${\operatorname{Out}(G,K)}$${L_{1}}$${1,}$
${1}$${\overline{H}_{\psi}^{1}(Q,\mathcal{Z}K)}$${\mathcal{K}}$${L_{2}}$${1}$

and

${1}$${\overline{B}_{\psi}^{1}(Q,\mathcal{Z}K)}$${Z_{\psi}^{1}(Q,\mathcal{Z}K)}$${\overline{H}_{\psi}^{1}(Q,\mathcal{Z}K)}$${1}$

such that $L_{1}\leq\operatorname{Out}(Q)$, $L_{2}\leq
C_{\operatorname{Out}(K)}\psi(Q)/\psi(\mathcal{Z}Q)$ and
$B_{\psi}^{1}(Q,\mathcal{Z}K)\leq\overline{B}_{\psi}^{1}(Q,\mathcal{Z}K)$,
thus we have a surjective map
$H^{1}_{\psi}(Q,\mathcal{Z}K)\longrightarrow\overline{H}_{\psi}^{1}(Q,\mathcal{Z}K)$.

We have an isomorphism
$\overline{B}_{\psi}^{1}(Q,\mathcal{Z}K)\cong(p^{-1}(\mathcal{Z}(Q))\cap
C_{G}K)/\mathcal{Z}G$.

There is also a version for the automorphism group.

######  Theorem 2.2.

[Mal02] There exist short exact sequences

${1}$${\mathcal{K}^{\prime}}$${\operatorname{Aut}(G,K)}$${L^{\prime}_{1}}$${1,}$

and

${1}$${Z_{\psi}^{1}(Q,\mathcal{Z}K)}$${\mathcal{K}^{\prime}}$${L^{\prime}_{2}}$${1}$

such that $L^{\prime}_{1}\leq\operatorname{Aut}(Q)$ and
$L^{\prime}_{2}\leq\operatorname{Aut}(K)$.

######  Remark 2.3.

If $K$ is a characteristic subgroup of $G$ then
$\operatorname{Aut}(G,K)=\operatorname{Aut}(G)$ and
$\operatorname{Out}(G,K)=\operatorname{Out}(G)$.

The relation between the Minkowski property and short exact sequences is
explained in the following lemma:

######  Lemma 2.4.

Let ${1}$${K}$${G}$${Q}$${1}$$\scriptstyle{p}$ be a short exact sequence of
groups. If $K$ and $Q$ are Minkowski, then $G$ is Minkowski. If $K$ is finite
and $G$ is Minkowski then $Q$ is Minkowski.

###### Proof.

Let us prove the first part of the statement. Let $F$ be a finite subgroup of
$G$. Then we have a short exact sequence $1\longrightarrow
F_{1}\longrightarrow F\longrightarrow F_{3}\longrightarrow 1$, where $F_{3}$
is the image of $F$ by the map $G\longrightarrow Q$ and $F_{1}=F\cap K$. If
$C_{1}$ and $C_{3}$ are the Minkowski constants of $K$ and $Q$ respectively
then $|F|\leq|F_{1}||F_{3}|\leq C_{1}C_{3}$. Therefore $G$ is Minkowski.

For the second part assume that $F$ is a finite subgroup of $Q$, then
$p^{-1}(F)$ is a finite subgroup of $G$ of order $|K||F|$. Since $G$ is
Minkowski we have that $|K||F|\leq C_{2}$ and therefore $|F|\leq C_{2}/|K|$.
Thus $Q$ is Minkowski. ∎

In particular, virtually torsion-free groups are Minkowski. A first
consequence of these results is the following corollary:

######  Corollary 2.5.

Let $\Gamma$ and $\Gamma^{\prime}$ be finitely generated groups with finitely
generated center such that $\Gamma^{\prime}\trianglelefteq\Gamma$ and
$\Gamma/\Gamma^{\prime}=F$ is a finite group. If
$\operatorname{Out}(\Gamma^{\prime})$ is a Minkowski, then
$\operatorname{Out}(\Gamma)$ is Minkowski.

###### Proof.

By [McC88, Lemma 1.(a)] we know that
$[\operatorname{Out}(\Gamma):\operatorname{Out}(\Gamma,\Gamma^{\prime})]<\infty$,
hence it is enough to prove that $\operatorname{Out}(\Gamma,\Gamma^{\prime})$
is Minkowski.

By 2.1 and 2.4, the group $\operatorname{Out}(\Gamma,\Gamma^{\prime})$ is
Minkowski if $\operatorname{Out}(F)$,
$C_{\operatorname{Out}(\Gamma^{\prime})}\psi(F)/\psi(\mathcal{Z}F)$ and
$H^{1}_{\psi}(F,\mathcal{Z}\Gamma^{\prime})$ are Minkowski. But
$\operatorname{Out}(F)$ and $H^{1}_{\psi}(F,\mathcal{Z}\Gamma^{\prime})$ are
Minkowski since $F$ is finite and $\mathcal{Z}\Gamma^{\prime}$ is finitely
generated. Lastly,
$C_{\operatorname{Out}(\Gamma^{\prime})}\psi(F)/\psi(\mathcal{Z}F)$ is
Minkowski since
$C_{\operatorname{Out}(\Gamma^{\prime})}\psi(F)\leq\operatorname{Out}(\Gamma^{\prime})$
is Minkowski by hypothesis and $\psi(\mathcal{Z}F)$ is a finite group, hence
we can use the second part of 2.4. ∎

##  3\. Finite group actions on aspherical manifolds: proof 1.6

Let $F$ be a finite group acting effectively on a connected manifold $M$.
Recall that if the action of $F$ has a fix point $x\in M$, then the group
action induces a group morphism
$F\longrightarrow\operatorname{Aut}(\pi_{1}(M,x))$. However, in the general
case were the action does not have a fix point this group morphism is only
well defined up to conjugation, so we have instead a group morphism
$\psi:F\longrightarrow\operatorname{Out}(\pi_{1}(M))$. This group morphism is
specially relevant when $M$ is a closed aspherical manifold, as the following
theorem of P.E. Conner and F. Raymond shows:

######  Theorem 3.1.

[LR10, Theorem 3.1.16] Let $M$ be a closed aspherical manifold such that
$\mathcal{Z}(\pi_{1}(M))$ is finitely generated. Assume that $F$ is a finite
group acting effectively on $M$. Then:

  * 1.

$\operatorname{Ker}\psi$ is isomorphic to a subgroup of the torus $T^{k}$,
where $k$ is the rank of $\mathcal{Z}(\pi_{1}(M))$.

  * 2.

If $F$ has a fixed point $x$ then the group morphism
$F\longrightarrow\operatorname{Aut}(\pi_{1}(M,x))$ is injective.

This theorem was used to find the first examples of asymmetric manifolds, by
constructing a closed aspherical manifold with a fundamental group with
trivial center and torsion-free outer automorphisms group. In our case, it is
the main tool to prove 1.6.

###### Proof of part 1. of 1.6.

Let $C$ denote the bound of the order of finite subgroups of
$\operatorname{Out}(\pi_{1}(M))$. If $F$ is a finite group acting effectively
on $M$, then $\operatorname{Ker}\psi$ is an abelian subgroup of $F$ and
$[F:\operatorname{Ker}\psi]=|F/\operatorname{Ker}\psi|\leq C$ since
$F/\operatorname{Ker}\psi\leq\operatorname{Out}(\pi_{1}(M))$. Thus
$\operatorname{Homeo}(M)$ is Jordan. ∎

To prove the second part we need the following group theoretic results (the
second one due to Schur).

######  Lemma 3.2.

[MiR24b, Lemma 2.1] Let $a,b,C$ be natural numbers and suppose that $F$ is a
subgroup of $(\mathbb{Z}/a)^{b}$ such that $[(\mathbb{Z}/a)^{b}:F]\leq C$.
Then there exists a subgroup $F^{\prime}\leq F$ isomorphic to
$(\mathbb{Z}/a^{\prime})^{b}$ such that $C!a^{\prime}\geq a$.

######  Lemma 3.3.

[Rob13, Theorem 4.12] Let $\Gamma$ be a finitely generated group such that
$[\Gamma:\mathcal{Z}\Gamma]<\infty$. Then the commutator subgroup
$[\Gamma,\Gamma]$ is finite.

###### Proof of part 2. of 1.6.

We start proving the inequality $\operatorname{disc-sym}(M)\leq k$. Let
$\\{a_{i}\\}_{i\in\mathbb{N}}$ be a sequence of natural numbers such that
$a_{i}\longrightarrow\infty$ and $(\mathbb{Z}/a_{i})^{b}$ acts effectively on
$M$ for some $b\in\mathbb{N}$. We have induced group morphisms
$\psi_{i}:(\mathbb{Z}/a_{i})^{b}\longrightarrow\operatorname{Out}(\pi_{1}(M))$
for each $i$ such that
$[(\mathbb{Z}/a_{i})^{b}:\operatorname{Ker}\psi_{i}]\leq C$. By 3.2, there
exists a sequence $\\{a^{\prime}_{i}\\}_{i\in\mathbb{N}}$ such that
$(\mathbb{Z}/a^{\prime}_{i})^{b}\leq\operatorname{Ker}\psi_{i}$. Since
$a_{i}^{\prime}C!\geq a_{i}$ we have that
$a^{\prime}_{i}\longrightarrow\infty$. Moreover,
$(\mathbb{Z}/a^{\prime}_{i})^{b}$ is a subgroup of $T^{k}$ for any $i$, so we
can conclude that $b\leq k$. In consequence $\operatorname{disc-sym}(M)\leq
k$. To prove the inequality $k\leq n$ we take the $n$-dimensional manifold
$\tilde{M}/\mathcal{Z}(\pi_{1}(M))$, where $\tilde{M}$ is the universal cover
of $M$. Since $\tilde{M}$ is contractible we have that
$H^{*}(\tilde{M}/\mathcal{Z}(\pi_{1}(M)))\cong H^{*}(T^{k},\mathbb{Z})$. The
fact that $\tilde{M}/\mathcal{Z}(\pi_{1}(M))$ has dimension $n$ implies that
$H^{i}(\tilde{M}/\mathcal{Z}(\pi_{1}(M)),\mathbb{Z})=0$ for $i>n$, hence
$k\leq n$.

Finally, we prove that $\operatorname{disc-sym}(M)=n$ if and only if $M\cong
T^{n}$. It is clear that $\operatorname{disc-sym}(T^{n})=n$ since
$\operatorname{Out}(\mathbb{Z}^{n})=\operatorname{GL}(n,\mathbb{Z})$ is
Minkowski. Conversely, assume that $\operatorname{disc-sym}(M)=n$. Since the
top cohomology $H^{n}(\tilde{M}/\mathcal{Z}(\pi_{1}(M)),\mathbb{Z})$ is non-
zero, we can conclude that $\tilde{M}/\mathcal{Z}(\pi_{1}(M))$ is a closed
connected manifold and that the map
$\tilde{M}/\mathcal{Z}(\pi_{1}(M))\longrightarrow M$ is a regular finite
cover. In consequence, $[\pi_{1}(M):\mathcal{Z}\pi_{1}(M)]<\infty$ and
$[\pi_{1}(M),\pi_{1}(M)]$ is trivial, by 3.3 and the fact that $\pi_{1}(M)$ is
torsion-free. Thus $\pi_{1}(M)\cong\mathbb{Z}^{n}$ and $M\cong T^{n}$, because
the Borel conjecture holds for $\mathbb{Z}^{n}$. ∎

The third part is a direct consequence of these two results:

######  Lemma 3.4.

[Got65, Theorem IV.1.] Let $X$ be a topological space with the homotopy type
of a compact connected aspherical CW-complex. If $\chi(X)\neq 0$, then
$\operatorname{rank}\mathcal{Z}\pi_{1}(X)=0$.

######  Lemma 3.5.

[MiR24a, Lemma 8.1] For any closed connected manifold $M$ we have
$\operatorname{disc-sym}(M)=0$ if and only if $M$ is almost asymmetric.

###### Proof of part 3. of 1.6.

If $M$ is a closed connected aspherical manifold, it has the homotopy type of
a compact connected aspherical CW-complex. Thus, if $\chi(M)\neq 0$ then
$\operatorname{disc-sym}(M)\leq\operatorname{rank}\mathcal{Z}(\pi_{1}(M))=0$.
In consequence $\operatorname{disc-sym}(M)=0$ and $M$ is almost asymmetric. ∎

Finally, to prove the last part we need the following general lemma:

######  Lemma 3.6.

Let $M$ be a closed manifold. Assume that $\operatorname{Homeo}(M)$ is Jordan
with constant $C$ and that there exists a constant $C^{\prime}$ such that
every finite group $F$ acting effectively on $M$ with a fixed point fulfils
that $|F|\leq C^{\prime}$. Then there exists a constant $D$ such that every
finite group $F$ acting on $M$ has a subgroup $H$ such that $[F:H]\leq D$ and
$|\operatorname{Stab}(H,M)|\leq D$.

###### Proof.

Let $F$ be a finite group acting effectively on $M$. Since
$\operatorname{Homeo}(M)$ is Jordan there exists an abelian subgroup $H$ such
that $[F:H]\leq C$. We will see that we can bound $\operatorname{Stab}(H,M)$
by a constant $C^{\prime\prime}$ for any finite abelian group acting
effectively on $M$ and therefore $D=\max\\{C,C^{\prime\prime}\\}$. Recall that
there exists a constant $r$ such that any finite abelian group $H$ which acts
effectively on $M$ fulfils that $\operatorname{rank}(H)\leq r$.

We have an inclusion $\operatorname{Stab}(H,M)\subseteq\\{L\leq H:|L|\leq
C^{\prime}\\}$, thus we will bound the number of subgroups of $H$ of order at
most $C^{\prime}$ instead of the number of stabilizers of the action of $H$.
Since $H$ is abelian, we have a decomposition $H=H_{1}\times\dots\times
H_{l}$, where each $H_{i}$ is a $p_{i}$-Sylow subgroup, thus is of the form
$H_{i}=\mathbb{Z}/p_{i}^{a_{i,1}}\times\dots\mathbb{Z}/p_{i}^{a_{i,r_{i}}}$,
where $a_{i,1}\leq\dots\leq a_{i,r_{i}}$ and $r_{i}\leq r$ for all $i$. Then
we have a bijective correspondence

$\\{L\leq H\\}\longleftrightarrow\prod_{i=1}^{l}\\{L_{i}\leq H_{i}\\}$

which induces an inclusion

$\\{L\leq H:|L|\leq C^{\prime}\\}\subseteq\prod_{i=1}^{l}\\{L_{i}\leq
H_{i}:|L|\leq C^{\prime}\\}.$

For a prime $p$ we define $e(p)=\max\\{e\in\mathbb{N}:p^{e}\leq C^{\prime}\\}$
and we denote the set of all primes such that $e(p)\neq 0$ by $\mathcal{P}$.
Note that $\mathcal{P}$ is precisely the set of primes which are equal or
smaller than $C^{\prime}$ and therefore $|\mathcal{P}|\leq C^{\prime}$.
Moreover, we have that

$\\{L_{i}\leq H_{i}:|L|\leq C^{\prime}\\}\subseteq S_{i}=\\{L_{i}\leq
H_{i}:\text{all elements of $L$ have order at most $e(p_{i})$}\\}$

for each $i$. Note that if $p_{i}\notin\mathcal{P}$ then $|S_{i}|=1$.
Therefore

$\\{L\leq H:|L|\leq
C^{\prime}\\}\subseteq\prod_{i=1,p_{i}\in\mathcal{P}}^{l}S_{i}.$

For each $H_{i}$ we choose an inclusion
$H_{i}\longrightarrow(\mathbb{Z}/p_{i}^{a_{i,r_{i}}})^{r}$, which induce an
inclusion

$S_{i}\subseteq\\{L\leq(\mathbb{Z}/p_{i}^{a_{i,r_{i}}})^{r}:\text{all elements
of $L$ have order at most $e(p_{i})$}\\}.$

Finally, we use that if $L$ is a subgroup of
$(\mathbb{Z}/p_{i}^{a_{i,r_{i}}})^{r}$ where all elements are of order at most
$e(p_{i})$ then
$L_{i}\leq(\mathbb{Z}/p_{i}^{e(p_{i})})^{r}\leq(\mathbb{Z}/p_{i}^{a_{i,r_{i}}})^{r}$.
In consequence, we have an inclusion

$\\{L_{i}\leq(\mathbb{Z}/p_{i}^{a_{i,r_{i}}})^{r}:\text{all elements of $L$
have order at most
$e(p_{i})$}\\}\subset\\{L_{i}\leq(\mathbb{Z}/p_{i}^{e(p_{i})})^{r}\\}.$

Since $p_{i}\leq C^{\prime}$ and $e(p_{i})\leq C^{\prime}$ we have that
$|\\{L_{i}\leq(\mathbb{Z}/p_{i}^{e(p_{i})})^{r}\\}|\leq
2^{|(\mathbb{Z}/p_{i}^{e(p_{i})})^{r}|}\leq 2^{(C^{\prime C^{\prime}})^{r}}$,
which does not depend on the prime $p_{i}$. Therefore, we have that
$|S_{i}|\leq 2^{(C^{\prime C^{\prime}})^{r}}$ if $p_{i}\in\mathcal{P}$ and
$|S_{i}|=1$ if $p_{i}\notin\mathcal{P}$. By using that $|\mathcal{P}|\leq
C^{\prime}$ we can conclude that $|\operatorname{Stab}(H,M)|\leq|\\{L\leq
H:|L|\leq
C^{\prime}\\}|\leq\prod_{i=1,p_{i}\in\mathcal{P}}^{l}|S_{i}|\leq(2^{(C^{\prime
C^{\prime}})^{r}})^{C^{\prime}}$, which completes the proof.

∎

###### Proof of part 4. of 1.6.

We have already seen that $\operatorname{Homeo}(M)$ is Jordan. Since a finite
group $F$ acting on $M$ with a fix point is a subgroup of
$\operatorname{Aut}(\pi_{1}(M))$ and $\operatorname{Aut}(\pi_{1}(M))$ is
Minkowski, the second condition of 3.6 is also fulfilled. Thus, we only need
to use 3.6 to finish the proof of the fourth part. ∎

This ends the proof of 1.6. There are some interesting and natural questions
that we summarise in the next remarks.

######  Remark 3.7.

(The hypothesis on the fundamental group) When are the two hypothesis on the
fundamental group fulfilled? No closed aspherical manifold with
$\mathcal{Z}(\pi_{1}(M))$ not finitely generated is known (see [LR10, Remark
3.1.19.]).

It is an interesting problem to find which closed connected aspherical
manifolds $M$ satisfy that $\operatorname{Out}(\pi_{1}(M))$ is Minkowski (see
[Gol24, Remark 7.1]). The next examples show some case where
$\operatorname{Out}(\pi_{1}(M))$ is Minkowski:

  * 1.

In this paper we prove that aspherical locally homogeneous spaces (or
classical aspherical manifolds following the terminology on [FJ90]) fulfil
that $\operatorname{Out}(\pi_{1}(M))$ is Minkowski.

  * 2.

Another source of closed connected aspherical manifolds is the strict
hyperbolization processes (see [CD95] and references therein). Given a closed
oriented manifold $M^{\prime}$ of dimension $n\geq 3$ we can construct a
closed oriented aspherical manifold $M$ and a non-zero degree map
$f:M\longrightarrow M^{\prime}$ such that $\pi_{1}(M)$ is a hyperbolic group.
These groups fulfils that $\operatorname{Out}(\pi_{1}(M))$ is finite (see
[Gro87, 5.4.A] or [Pau91]). Moreover $\mathcal{Z}(\pi_{1}(M))=0$ and therefore
$M^{\prime}$ is almost-asymmetric.

  * 3.

If $M$ is a closed connected aspherical 3-dimensional manifold, then
$\operatorname{Out}(\pi_{1}(M))$ is Minkowski (see [Koj84]). This fact was
used in [Zim14] to prove that $\operatorname{Diff}(M)$ is Jordan when $M$ is a
closed smooth 3-manifold.

  * 4.

If $\operatorname{Out}(\pi_{1}(M))$ is a finitely generated virtually abelian
group then $\operatorname{Out}(\pi_{1}(M))$ is Minkowski. This is the case for
piecewise linear locally symmetric spaces, a new type of closed aspherical
manifold defined in [TNP11] by compactifying non-compact symmetric spaces and
using the reflection trick on its corners. If $M$ is a piecewise linear
locally symmetric space then $\operatorname{Out}(\pi_{1}(M))$ is finitely
generated and virtually abelian by [TNP11, Theorem 4]. Moreover,
$\mathcal{Z}\pi_{1}(M)=\\{e\\}$ by [TNP11, Lemma 7], hence piecewise linear
locally symmetric spaces are almost asymmetric.

######  Remark 3.8.

(Groups whose (outer) automorphism group is not Minkowski) There exists groups
$\Gamma$ such that $B\Gamma$ is a finite CW-complex and
$\operatorname{Out}(\Gamma)$ is not Minkowski. For example, let $\Gamma$ be
the Baumslag-Solitar group $B(m,ml)=\langle a,b|ba^{m}b^{-1}=a^{ml}\rangle$
with $m,l\geq 2$. The space $BB(m,ml)$ is a finite aspherical $2$-dimensional
CW-complex since it is a torsion-free one relator group (see [LSLS77]).
Moreover $\operatorname{Out}(B(m,ml))$ and $\operatorname{Aut}(B(m,ml))$ have
elements of order $l^{t}(l-1)$ for arbitrarily large $t$ (see [CL83, Lemma
3.8] or [Lev07]) and thus they are not Minkowski.

######  Remark 3.9.

(The discrete degree of symmetry versus the toral degree of symmetry) The
question of when $\operatorname{tor-sym}(M)$ is equal to
$\operatorname{rank}\mathcal{Z}(\pi_{1}(M))$ has been extensively studied for
closed connected aspherical manifolds $M$. We also note that any effective
torus action on a closed aspherical manifold $M$ is almost-free (see [LR10,
Corollary 3.1.12]), therefore $\operatorname{tor-sym}(M)$ is equal to the
toral rank of $M$ (see [FOT08, §7.3]).

If $\operatorname{Out}(\pi_{1}(M))$ is Minkowski and $\operatorname{tor-
sym}(M)=\operatorname{rank}\mathcal{Z}\pi_{1}(M)$ then $\operatorname{disc-
sym}(M)=\operatorname{rank}\mathcal{Z}\pi_{1}(M)$, since $\operatorname{tor-
sym}(M)\leq\operatorname{disc-
sym}(M)\leq\operatorname{rank}\mathcal{Z}(\pi_{1}(M))$. The equality holds for
infra-solvmanifolds or some aspherical locally homogeneous spaces (see [LR10,
Section 11.7]). In 1.10 we prove that the equality is valid for closed
aspherical locally homogeneous spaces. On the other hand, there exist closed
aspherical manifolds such that $\operatorname{tor-sym}(M)=0$ and
$\operatorname{disc-sym}(M)\geq 1$ (see [CWY13, MiR24b]). It is an interesting
question whether all closed connected aspherical manifolds satisfy
$\operatorname{disc-sym}(M)=\operatorname{rank}\mathcal{Z}\pi_{1}(M)$.

######  Remark 3.10.

(Euler characteristic and asymmetry) The converse of part 3 of 1.6 is not
true. There exist asymmetric flat manifolds (see [Szc12]). The fundamental
group of a closed flat manifold fulfils the hypothesis of 1.6 and the manifold
Euler characteristic is $0$ since they are finitely covered by a torus.

######  Remark 3.11.

(Generalizations of aspherical manifolds) 3.1 has been generalized to closed
connected manifolds $M$ satisfying that the only periodic self-homeomorphisms
of the universal cover $\tilde{M}$ commuting with the deck of transformation
groups $\pi_{1}(M)$ are elements of $\mathcal{Z}\pi_{1}(M)$ (following the
terminology of [LR10, Theorem 3.2.2] we will say that $M$ is admissible). This
class of manifolds includes, for example, manifolds which admits a non-zero
degree map to a closed aspherical manifold.

If $M$ is a closed connected admissible manifold such that
$\operatorname{Out}(\pi_{1}(M))$ is Minkowski and $\mathcal{Z}\pi_{1}(M)$ is
finitely generated then $\operatorname{Homeo}(M)$ is Jordan and
$\operatorname{disc-
sym}(M)\leq\operatorname{rank}(\mathcal{Z}\pi_{1}(M)/\operatorname{Torsion}(\mathcal{Z}\pi_{1}(M)))$.
For example, if $M$ is a closed aspherical manifold satisfying the hypothesis
of 1.6 and $N$ is a closed simply-connected manifold of the same dimension as
$M$, then $M\\#N$ is a closed admissible manifold such that
$\operatorname{Homeo}(M\\#N)$ is Jordan and $\operatorname{disc-
sym}(M\\#N)\leq\operatorname{rank}\mathcal{Z}\pi_{1}(M)$. This last inequality
is usually strict. For example, assume that $N$ is an odd dimensional manifold
such that $\chi(N)\neq 0$. Then $\chi(T^{n}\\#N)\neq 0$ and hence
$\operatorname{disc-
sym}(T^{n}\\#N)=0<n=\operatorname{rank}\mathcal{Z}(\pi_{1}(T^{n}\\#N))$ (see
[MiR24b]).

However, there exists a closed admissible manifold $M$ such that
$\operatorname{Homeo}(M)$ is Jordan but $\operatorname{Out}(\pi_{1}(M))$ is
not Minkowski. In [Blo75, Theorem 1.1.1] it is proved that if $\Gamma$ and
$\Lambda$ are not free groups then
$\operatorname{Out}(\Gamma*\Lambda)\cong\operatorname{Aut}(\Gamma)\times\operatorname{Aut}(\Lambda)$
if $\Gamma\ncong\Lambda$ and
$\operatorname{Out}(\Gamma*\Lambda)\cong(\operatorname{Aut}(\Gamma)\times\operatorname{Aut}(\Lambda))\times\mathbb{Z}/2$
if $\Gamma\cong\Lambda$. In particular, if $M$ is a closed 4-manifold with
$\pi_{1}(M)\cong B(m,ml)$ with $m,l\geq 2$ then
$\operatorname{Out}(\pi_{1}(M\\#T^{4}))\cong\operatorname{Aut}(B(m,ml))\times\operatorname{GL}(4,Z)$
is not Minkowski (see 3.7), but $M\\#T^{4}$ is hypertoral and hence
$\operatorname{Homeo}(M\\#T^{4})$ is Jordan by [MiR24b, Theorem 1.15].

######  Remark 3.12.

(Non-compact aspherical manifolds) If $M$ is a non-compact connected
aspherical manifold then $\operatorname{Homeo}(M)$ is not necessarily Jordan,
even when $\operatorname{Out}(\pi_{1}(M))$ is Minkowski. For example, since
$\mathbb{R}^{3}$ admits effective actions by $\operatorname{SO}(3)$ then
$\operatorname{Homeo}(T^{2}\times\mathbb{R}^{3})$ is not Jordan by [MiR17,
Theorem 1].

We end this section by exposing some facts on the relation between large
finite group actions on manifolds and covering maps.

######  Lemma 3.13.

Let $M$ and $M^{\prime}$ be closed connected manifolds and
$p:M^{\prime}\longrightarrow M$ be a finite covering. Then:

  * 1.

If $\operatorname{Homeo}(M^{\prime})$ is Jordan, then
$\operatorname{Homeo}(M)$ is Jordan.

  * 2.

$\operatorname{disc-sym}(M^{\prime})\geq\operatorname{disc-sym}(M)$.

  * 3.

Assume that there exists a constant $D^{\prime}$ such that any finite group
$F^{\prime}$ acting effectively on $M^{\prime}$ with a fix point satisfies
$|F^{\prime}|\leq D^{\prime}$. Then there exists a constant $D$ such that any
finite group $F$ acting effectively on $M$ with a fix point satisfies $|F|\leq
D$

###### Proof.

The first two parts are proven in [MiR10, §2.3] and [MiR24b, Theorem 1.12].
The proof of the third part follows the same arguments as the proofs of the
first two parts.

Assume that $p:M^{\prime}\longrightarrow M$ is a $n$-sheeted covering and $F$
is a finite group acting effectively on $M$. Then $F$ also acts on
$\operatorname{Cov}_{n}(M)$, the set of $n$-sheeted coverings of $M$, by pull-
backs. On the other hand
$\operatorname{Cov}_{n}(M)\cong\operatorname{Hom}(\pi_{1}(M),S_{n})/\sim$
where $S_{n}$ is the $n$-th symmetric group and the equivalence relation is
given by conjugation of elements of $S_{n}$. Therefore
$\operatorname{Cov}_{n}(M)$ is finite, which implies that there exists a
constant $C$ only depending on $M$ and $n$ such that any finite group $F$
acting effectively on $M$ has a subgroup $F_{0}$ which acts trivially on
$\operatorname{Cov}_{n}(M)$ and $[F:F_{0}]\leq C$. Then there exists a finite
group $F^{\prime}_{0}$ acting effectively on $M^{\prime}$ and a surjective
group morphism $\pi:F^{\prime}_{0}\longrightarrow F_{0}$ which makes the
covering map $p:M^{\prime}\longrightarrow M$ $\pi$-equivariant and
$|\operatorname{Ker}\pi|\leq n!$.

Let $F$ be a finite group acting effectively on $M$ with a fix point $x\in M$.
Then $x$ is also fixed by the action of $F_{0}$ and $F^{\prime}_{0}$ acts on
$p^{-1}(x)$. Given $x^{\prime}\in p^{-1}(x)$, we have that the orbit of
$x^{\prime}$ by $F^{\prime}_{0}$ is
$F_{0}^{\prime}/{F^{\prime}_{0}}_{x^{\prime}}\subseteq p^{-1}(x)$, which
implies that $|F_{0}^{\prime}/{F^{\prime}_{0}}_{x^{\prime}}|\leq n$. Finally,
${F^{\prime}_{0}}_{x^{\prime}}$ acts effectively on $M^{\prime}$ with a fixed
point, therefore $|{F^{\prime}_{0}}_{x^{\prime}}|\leq D^{\prime}$. Since $\pi$
is surjective, we can take $D=C\cdot D^{\prime}\cdot n$. ∎

In consequence, if $p:M^{\prime}\longrightarrow M$ is a covering of closed
connected aspherical manifolds and $M^{\prime}$ fulfils the hypothesis of 1.6
then all the conclusions of 1.6 also hold for $M$. This fact can also be
deduced for regular coverings using 2.5.

Given a regular covering $p:M^{\prime}\longrightarrow M$, an interesting
problem is to determine when $\operatorname{disc-sym}(M)=\operatorname{disc-
sym}(M^{\prime})$. The next proposition will be used to give a partial answer
to this question for the case of closed connected aspherical manifolds, which
will be given in 3.15.

######  Proposition 3.14.

Let $M$ be a closed connected aspherical manifold such that
$\mathcal{Z}(\pi_{1}(M))$ is finitely generated. Assume that $F$ is a finite
group acting freely on $M$. Then
$k=\operatorname{rank}(\mathcal{Z}(\pi_{1}(M)))=\operatorname{rank}(\mathcal{Z}(\pi_{1}(M/F)))$
if and only if the map
$\psi^{\prime}:F\longrightarrow\operatorname{Out}(\mathcal{Z}(\pi_{1}(M)))$ is
trivial.

###### Proof.

Recall that the free action of $F$ on $M$ induces a commutative diagram

${1}$${1}$${1}$${1}$${\mathcal{Z}\pi_{1}(M)}$${C_{\tilde{F}}(\pi_{1}(M))}$${\operatorname{Ker}\psi}$${1}$${1}$${\pi_{1}(M)}$${\tilde{F}}$${F}$${1}$${1}$${\operatorname{Inn}\pi_{1}(M)}$${\tilde{\psi}(\tilde{F})}$${\psi(F)}$${1}$${1}$${1}$${1}$$\scriptstyle{\tilde{\psi}}$$\scriptstyle{p}$$\scriptstyle{\psi}$

where $\tilde{\psi}(\tilde{F})\leq\operatorname{Aut}(\pi_{1}(M))$ and
$\psi(F)\leq\operatorname{Out}(\pi_{1}(M))$. We also note that
$\mathcal{Z}\tilde{F}\trianglelefteq C_{\tilde{F}}(\pi_{1}(M))$ and
$\tilde{F}=\pi_{1}(M/F)$ is torsion-free. Recall also that
$\tilde{\psi}(\tilde{f})=c_{\tilde{f}|\pi_{1}(M)}$ and that
${\psi}(f)=[c_{\sigma{f}|\pi_{1}(M)}]$ where
$\sigma:F\longrightarrow\tilde{F}$ is a set-theoretic section of $p$.

Since $\mathcal{Z}\pi_{1}(M)$ is a characteristic subgroup of $\pi_{1}(M)$, we
have maps
$\tilde{\psi}^{\prime}:\tilde{F}\longrightarrow\operatorname{GL}(k,\mathbb{Z})$
and $\psi^{\prime}:F\longrightarrow\operatorname{GL}(k,\mathbb{Z})$ by
restricting (outer) automorphisms to $\mathcal{Z}\pi_{1}(M)$. Moreover,
$\tilde{\psi}^{\prime}=\psi^{\prime}\circ p$ and since $p$ is surjective we
can conclude that $\tilde{\psi}^{\prime}$ is trivial if and only if
$\psi^{\prime}$ is trivial.

Assume now that $\psi^{\prime}$ is trivial and hence $\tilde{\psi}^{\prime}$
is trivial too. This implies that $[\tilde{f},z]=e$ for any
$\tilde{f}\in\tilde{F}$ and any $z\in\mathcal{Z}\pi_{1}(M)$. Thus
$\mathcal{Z}\pi_{1}(M)\trianglelefteq\mathcal{Z}\tilde{F}\trianglelefteq
C_{\tilde{F}}(\pi_{1}(M))$. But
$\operatorname{rank}(\mathcal{Z}\pi_{1}(M))=\operatorname{rank}(C_{\tilde{F}}(\pi_{1}(M)))$
since the first row of the commutative diagram is a central exact sequence and
$C_{\tilde{F}}(\pi_{1}(M))$ is torsion-free. In consequence,
$\operatorname{rank}(\mathcal{Z}\pi_{1}(M))=\operatorname{rank}(\mathcal{Z}{\tilde{F}})$.

If
$\operatorname{rank}(\mathcal{Z}\pi_{1}(M))=\operatorname{rank}(\mathcal{Z}{\tilde{F}})$
then $[C_{\tilde{F}}(\pi_{1}(M)):\mathcal{Z}\tilde{F}]<\infty$. This implies
that $\tilde{\psi}^{\prime}(\tilde{F})\leq\operatorname{GL}(r,\mathbb{Z})$
fixes a sublattice of $\mathcal{Z}\pi_{1}(M)$ and therefore
$\tilde{\psi}^{\prime}$ is trivial. Thus $\psi^{\prime}$ is also trivial, as
desired. ∎

######  Corollary 3.15.

Let $M$ be a closed connected aspherical manifold such that
$\mathcal{Z}\pi_{1}(M)$ is finitely generated,
$\operatorname{Out}(\pi_{1}(M))$ is Minkowski and $\operatorname{disc-
sym}(M)=\operatorname{rank}\mathcal{Z}(\pi_{1}(M))$. Assume that $F$ is a
finite group acting freely on $M$ such that $\psi^{\prime}$ is not trivial.
Then $\operatorname{disc-sym}(M/F)<\operatorname{disc-sym}(M)$.

See 3.9 for examples where the hypothesis of the corollary holds true.

##  4\. Solvmanifolds

Recall that a manifold $M$ is a solvmanifold if $M$ admits transitive group
action of a connected solvable Lie group $R$. If $R$ is nilpotent then we will
say that $M$ is a nilmanifold. The structure of nilmanifolds and solvmanifolds
and their fundamental groups have been vastly studied (see [Mal51, Gor09,
Aus73] and [Rag12, Chapter II,III,IV]). The fundamental group of any
solvmanifold is polycyclic. Recall that $\Gamma$ is a polycyclic group if
there exists a sequence of
$\Gamma\trianglerighteq\Gamma_{1}\trianglerighteq\dots\trianglerighteq\Gamma_{r}=\\{e\\}$
such that each quotient $\Gamma_{i-1}/\Gamma_{i}$ is cyclic. Equivalently, a
group is polycyclic if it is solvable group and all its subgroups are finitely
generated. In particular, the center of a polycyclic group is finitely
generated. Recall that $\Gamma$ is a virtually polycyclic group if it contains
a polycyclic subgroup of finite index.

######  Theorem 4.1.

[Weh94] Let $\Gamma$ be a virtually polycyclic group, then
$\operatorname{Out}(\Gamma)$ is isomorphic to a subgroup of
$\operatorname{GL}(n,\mathbb{Z})$ for some $n$.

The proof of this theorem uses the analogue statement for the automorphism
group.

######  Theorem 4.2.

[Meb70] Let $\Gamma$ be a virtually polycyclic group, then
$\operatorname{Aut}(\Gamma)$ is isomorphic to a subgroup of
$\operatorname{GL}(n,\mathbb{Z})$ for some $n$.

Hence, if $\Gamma$ is virtually polycyclic, then $\operatorname{Out}(\Gamma)$
and $\operatorname{Aut}(\Gamma)$ are Minkowski. In consequence, by using 1.6
we can conclude that a closed solvmanifold $M$ has Jordan homeomorphism group,
$\operatorname{disc-sym}(M)\leq\operatorname{rank}(\mathcal{Z}(\pi_{1}(M)))$
(and indeed $\operatorname{disc-
sym}(M)=\operatorname{rank}(\mathcal{Z}(\pi_{1}(M)))$ by the results in [LR10,
Section 11.7]) and there exists a constant $C$ such that any finite group $F$
acting effectively on $M$ has a subgroup $H$ such that
$|\operatorname{Stab}(H,M)|\leq C$ and $[F:H]\leq C$. These properties are
also satisfied by any manifold finitely covered by a solvmanifold. This
includes flat manifolds, almost-flat manifolds and infra-solvmanifolds.

######  Example 4.3.

Let us show a low dimensional example in order to illustrate some consequences
of 1.6 on nilmanifolds. The 3-dimensional Heisenberg group

$H=\\{(x,y,z)=\left(\begin{array}[]{ccc}1&x&z\\\ 0&1&y\\\
0&0&1\end{array}\right):x,y,z\in\mathbb{R}\\}$

is a simply connected nilpotent Lie group. Any lattice of $H$ is isomorphic to
a lattice of the form

$\Gamma_{k}=\\{(x,y,z)=\left(\begin{array}[]{ccc}1&x&\tfrac{1}{k}z\\\ 0&1&y\\\
0&0&1\end{array}\right):x,y,z\in\mathbb{Z}\\}$

where $k$ is a positive integer. Note that
$\mathcal{Z}\Gamma_{k}=\langle(0,0,1)\rangle$ and two lattices
$\Gamma_{k}\cong\Gamma_{l}$ are isomorphic if and only if $k=l$. A possible
presentation of these lattices is $\Gamma_{k}=\langle
a,b,c|[c,a]=[c,b]=1,[a,b]=c^{k}\rangle$ where $a=(1,0,0)$, $b=(0,1,0)$ and
$c=(0,0,1)$. Thus $\mathcal{Z}\Gamma_{k}=\langle c\rangle\cong\mathbb{Z}$ for
all $k$. In [CR06, §8] it is shown that
$\operatorname{Aut}(\Gamma_{k})\cong\mathbb{Z}^{2}\rtimes\operatorname{GL}(2,\mathbb{Z})$
and
$\operatorname{Out}(\Gamma_{k})\cong(\mathbb{Z}/k)^{2}\rtimes\operatorname{GL}(2,\mathbb{Z})$.
Note that as an abstract group, $\operatorname{Aut}(\Gamma_{k})$ does not
depend on $k$, but $\operatorname{Out}(\Gamma_{k})$ does. Both of them are
Minkowski and therefore the conclusions of 1.6 are valid for $H/\Gamma_{k}$.
In particular, $\operatorname{Homeo}(H/\Gamma_{k})$ is Jordan and
$\operatorname{disc-sym}(H/\Gamma_{k})=1$. However, not all finite subgroups
of $\operatorname{Out}(\Gamma_{k})$ can be realized by a group action on
$H/\Gamma_{k}$ (see [RS]).

If a $n$-dimensional flat manifold such that $\operatorname{disc-
sym}(M)=n=\operatorname{disc-sym}(T^{n})$ then $M$ is homeomorphic to $T^{n}$.
On the other hand, an almost-flat manifold $M$ finitely covered by a
nilmanifold $N/\Gamma$ satisfying that $\operatorname{disc-
sym}(M)=\operatorname{disc-sym}(N/\Gamma)$ is not necessarily isomorphic to
$N/\Gamma$. The 3-dimensional Heisenberg manifold can be used to construct an
almost-flat manifold $M$ which is finitely covered by $H/\Gamma_{2}$ and
$\operatorname{disc-sym}(M)=\operatorname{disc-sym}(H/\Gamma_{2})=1$ but $M$
is not homeomorphic to $H/\Gamma_{2}$ ($M$ is not even a nilmanifold). There
is a free action of $\mathbb{Z}/2$ on $H/\Gamma_{2}$ such that its orbit space
$M$ is an almost-flat manifold with fundamental group $\pi_{1}(M)=\langle
a,b,c,\alpha|[c,\alpha]=[c,a]=[c,b]=1,[a,b]=c^{2},\alpha a=a^{-1}\alpha,\alpha
b=b^{-1}\alpha,\alpha^{2}=c\rangle$ (see [Dek06, pg. 160]). It is clear that
$\langle c\rangle\leq\mathcal{Z}\pi_{1}(M)$ and therefore
$1\leq\operatorname{disc-sym}(M)\leq\operatorname{disc-sym}(H/\Gamma_{2})=1$.
In consequence, we have an equality $\operatorname{disc-
sym}(M)=\operatorname{disc-sym}(H/\Gamma_{2})=1$. Note that $\alpha
c\alpha^{-1}=c$, hence the morphism
$\psi^{\prime}:\mathbb{Z}/2\longrightarrow\operatorname{Out}(\mathcal{Z}\Gamma_{2})$
is trivial, as we expected from 3.15.

Let $M$ be a closed aspherical manifold satisfying the hypothesis of 1.6. The
fact that $\operatorname{disc-
sym}(M)=\operatorname{rank}(\mathcal{Z}(\pi_{1}(M)))$ does not imply that
there are no effective actions of $(\mathbb{Z}/k)^{r}$ with
$r>\operatorname{disc-sym}(M)$ for some $k$. Indeed, if $M$ is a compact
solvmanifold $R/\Gamma$ then [JL10, Corollary 3.3] asserts that if
$(\mathbb{Z}/p)^{r}$ acts freely on $R/\Gamma$ then $r\leq\dim R/\Gamma$. This
bound is sharp even when $R/\Gamma$ is not a torus. For an integer $k\leq 2$
we consider the subgroup

$\Gamma^{\prime}_{k}=\\{(x,y,z)=\left(\begin{array}[]{ccc}1&\tfrac{1}{k}x&\tfrac{1}{k^{3}}z\\\
0&1&\tfrac{1}{k}y\\\ 0&0&1\end{array}\right):x,y,z\in\mathbb{Z}\\}$

which is a lattice of $H$ isomorphic to $\Gamma_{k}$. An straightforward
computation shows that $\Gamma_{k^{2}}$ is a normal subgroup of
$\Gamma^{\prime}_{k}$ and
$\Gamma^{\prime}_{k}/\Gamma_{k^{2}}\cong(\mathbb{Z}/k)^{3}$. Therefore
$(\mathbb{Z}/k)^{3}$ acts freely on $H/\Gamma_{k^{2}}$ even though
$\operatorname{disc-sym}(H/\Gamma_{k^{2}})=1$ (see [CS05] for a classification
of all finite abelian group actions on Heisenberg manifolds). Therefore the
bound of [JL10, Corollary 3.3] is sharp.

##  5\. Locally symmetric spaces

A locally symmetric space is a double coset space $H\setminus G/\Gamma$ where
$G$ is a connected semisimple Lie group, $H$ is a maximal compact subgroup of
$G$ and $\Gamma$ is a torsion-free lattice of $G$. Since $H\setminus
G\cong\mathbb{R}^{n}$ for some $n$ (see [Hel01, Chapter VI, Theorem 1]), then
$H\setminus G/\Gamma$ is a connected aspherical manifold with fundamental
group $\Gamma$. If we assume that $H\setminus G/\Gamma$ is compact then
$G/\Gamma$ is also compact and therefore $\Gamma$ is a cocompact lattice.

The aim of this section is to prove the following proposition.

######  Proposition 5.1.

Let $\Gamma$ be a lattice of a connected semisimple Lie group $G$ without
compact factors. Then $\operatorname{Out}(\Gamma)$ Minkowski.

The objective of the first part of the section is to recall the results of the
theory of lattices of semisimple Lie groups and the theory of relatively
hyperbolic groups needed to prove 5.1. We start stating some well-known facts
and theorems on lattices of semisimple Lie groups and hyperbolic groups. We
refer to [Mor01] for an introduction to this topic.

######  Theorem 5.2.

Let $G$ be a connected semisimple linear group without compact factors and
$\Gamma$ a lattice of $G$. Then:

  * 1.

(Selberg’s lemma, [Mor01, (4.2.8) Theorem]) Any lattice $\Gamma\leq G$
contains a normal torsion-free lattice $\Gamma^{\prime}$ such that
$[\Gamma:\Gamma^{\prime}]<\infty$.

  * 2.

(Reducible lattices, [Mor01, (4.3.3) Proposition]) Assume that $G$ is
centreless. Then there exist simple subgroups $G_{1},...,G_{r}$ of $G$ and
lattices $\Gamma_{i}\leq G_{i}$ for all $1\leq i\leq r$ such that
$G=G_{1}\times\cdots G_{r}$ and $\Gamma_{1}\times\cdots\times\Gamma_{r}$ is a
normal finite index subgroup of $\Gamma$.

  * 3.

(Consequences of Borel density theorem I, [Mor01, (4.5.3) Corollary]) The
group $N_{G}(\Gamma)/\Gamma$ is finite and $C_{G}(\Gamma)=\mathcal{Z}G$
(equivalently, $\mathcal{Z}\Gamma=\Gamma\cap\mathcal{Z}G$).

  * 4.

(Consequences of Borel density theorem II, [Mor01, (4.5.2) Corollary]) Let $H$
be a connected closed subgroup of $G$. Assume that $\Gamma$ normalizes $H$,
then $H$ is normal in $G$.

  * 5.

(Mostow-Prasad-Margulis rigidity theorem, [Mor01, (15.1.2) Thereom]) Let
$G^{\prime}$ be another connected semisimple linear Lie group without compact
factors, and let $\Gamma^{\prime}$ be a lattice of $G^{\prime}$. Assume that
$G$ and $G^{\prime}$ have trivial center. Finally, assume that there does not
exists any simple factor $H$ of $G$ such that
$H\cong\operatorname{PSL}(2,\mathbb{R})$ and $H\cap\Gamma_{1}$ is a lattice in
$H$. Then any isomorphism from $\Gamma$ to $\Gamma^{\prime}$ extends to a
unique continuous isomorphism from $G$ to $G^{\prime}$.

  * 6.

(Margulis superrigidity, [Mor01, §16.1]) Assume that
$\operatorname{rank}_{\mathbb{R}}G\geq 2$ and that $\Gamma$ is an irreducible
lattice. Given a representation
$\rho:\Gamma\longrightarrow\operatorname{GL}(n,\mathbb{R})$, let
$\overline{\rho(\Gamma)}^{o}$ be the identity component of the Zariski closure
of $\rho(\Gamma)$ in $\operatorname{GL}(n,\mathbb{R})$ and set
$\Gamma_{0}=\rho^{-1}(\overline{\rho(\Gamma)}^{o})$. Then there exists a
representation $\tilde{\rho}:G\longrightarrow\overline{\rho(\Gamma)}^{o}$ such
that $\tilde{\rho}_{|\Gamma_{0}}=\rho_{|\Gamma_{0}}$.

  * 7.

(Margulis normal theorem, [Mor01, (17.1.1) Theorem]) Assume that $G$ has
$\operatorname{rank}_{\mathbb{R}}G\geq 2$ and finite center, and that $\Gamma$
is an irreducible lattice. If $N$ is a normal subgroup of $\Gamma$, then
either $N\leq\mathcal{Z}(G)$ or $\Gamma/N$ is finite.

From 5.2(3) we can deduce:

######  Corollary 5.3.

Let $G$ be a connected semisimple linear group without compact factors and
$\Gamma$ a torsion-free lattice of $G$. Then $\Gamma$ does not contain normal
solvable subgroups.

###### Proof.

Assume on the contrary, that $\Gamma$ contains a solvable normal subgroup
$\Lambda$. Then, the connected component of the identity of the Zariski
closure $\overline{\Lambda}^{o}$ is a connected solvable Lie subgroup of $G$.
Since $\Lambda$ is normal in $\Gamma$ then $\overline{\Lambda}^{o}$ is
normalized by $\Gamma$. By 5.2(3), $\overline{\Lambda}^{o}$ is a connected
solvable normal subgroup of $G$, which contradicts the semisimplicity of $G$.
∎

We now introduce the results we need from the theory of relatively hyperbolic
groups. For an introduction to relatively hyperbolic groups we refer to
[Bow97, Osi06].

Let us recall the definition of relative hyperbolicity from [Osi06, MO10].
Given a group $H$ and a collection of proper subgroups $\\{H_{i}\\}_{i\in I}$,
a subset $X$ of $G$ is a relative generating set of $H$ with respect to
$\\{H_{i}\\}_{i\in I}$ if $X$ together with the union of all $H_{i}$ generates
$H$. In this situation $H$ can be written as a quotient group of a free group
$\mathcal{F}=(*_{i\in I}H_{i})*F(X)$, where $F(X)$ denotes the free group with
basis $X$. If the kernel $\mathcal{F}\longrightarrow H$ is the normal closure
of a subset $R$ of $\mathcal{F}$ then we say that $H$ has a relative
presentation $\langle X,H_{i},i\in I|R\rangle$. If $X$ and $R$ are finite sets
then we say that $H$ is finitely presented relative to $\\{H_{i}\\}_{i\in I}$.

Set $\mathcal{H}=\bigsqcup_{i\in I}(H_{i}\setminus\\{e\\})$. Given a word $W$
in the alphabet $X^{\pm}\cup\mathcal{H}$ representing the trivial element $e$
in $H$ there exists an element of the form $\prod_{j=1}^{k}f_{j}R_{j}^{\pm
1}f_{j}^{-1}$ which is equal to $W$ in the group $\mathcal{F}$, where
$R_{j}\in R$ and $f_{j}\in\mathcal{F}$. The smallest possible $k$ for which
$W$ is equal to an element of the form $\prod_{j=1}^{k}f_{j}R_{j}^{\pm
1}f_{j}^{-1}$ is called the relative area of $W$ and denoted by
$Area^{rel}(W)$. If $||W||$ denotes the length of $W$ in the alphabet
$X^{\pm}\cup\mathcal{H}$ then:

######  Definition 5.4.

[MO10, Definition 2.1] A group $H$ is hyperbolic relative to a collection of
proper subgroup $\\{H_{i}\\}_{i\in I}$ if $H$ is finitely presented relative
to $\\{H_{i}\\}_{i\in I}$ and there is a constant $C>0$ such that any word $W$
in $X^{\pm}\cup\mathcal{H}$ representing the identity in $H$ satisfies that

$Area^{rel}(W)\leq C||W||.$

The groups $H_{i}$ are called peripheral subgroups. We say that a group $H$ is
relatively hyperbolic if there exists a collection of subgroups
$\\{H_{i}\\}_{i\in I}$ such that $H$ is hyperbolic relative to
$\\{H_{i}\\}_{i\in I}$.

######  Remark 5.5.

The definition is independent of the choice of $X$ and $R$ (see [Osi06]).

######  Remark 5.6.

There are other definitions of relative hyperbolicity of a group $H$ with
respect a collection of subgroups $\\{H_{i}\\}_{i\in I}$ in the literature. If
$H$ is torsion-free and finitely presented and $H_{i}$ is finitely presented
for all $i$, then all the definitions are equivalent (see [BSB08, Definition
1.1] and references therein). This will happen in our setting.

######  Remark 5.7.

A group $H$ is hyperbolic if it is hyperbolic relative to the collection of
subgroups which only contains the trivial subgroup.

We need to introduce relative hyperbolicity because of the next theorem:

######  Theorem 5.8.

([Far98] and [Gro87, §0.2(F)]) A lattice $\Gamma$ in a connected semisimple
Lie group without compact factors $G$ is relatively hyperbolic if and only if
$\operatorname{rank}_{\mathbb{R}}G=1$. More precisely, $\Gamma$ is hyperbolic
relative to the collection of all its cusp subgroups associated to the cusps
of the symmetric space $H\setminus G/\Gamma$. The cusp subgroups are virtually
nilpotent.

Let $H$ be a relatively hyperbolic group to $\\{H_{i}\\}_{i\in I}$. Recall
that $h\in H$ is said to be parabolic if its is conjugate to an element of
$H_{i}$ for some $i$. An element $h\in H$ which is not parabolic is said to be
hyperbolic.

######  Lemma 5.9.

Let $H$ be torsion-free a relative hyperbolic group to $\\{H_{i}\\}_{i\in I}$.
Then:

  * 1.

[MO10, Lemma 2.4] If $h\in H$ is hyperbolic then $C_{H}(h)$ is cyclic.

  * 2.

[MO10, Proposition 3.3] The set of hyperbolic elements generates $H$.

  * 3.

[MO10, Lemma 2.2] Given $i\in I$ and $h\in H\setminus H_{i}$, we have that
$H_{i}\cap hH_{i}h^{-1}=\\{e\\}$.

We are interested in the following corollary of item 3.

######  Corollary 5.10.

Let $h$ be a non-trivial element of $H_{i}$, then $C_{H}(h)\leq H_{i}$.

###### Proof.

Suppose that $h^{\prime}\in C_{H}(h)$. Then $h^{\prime}hh^{\prime-1}=h$,
therefore $H_{i}\cap h^{\prime}H_{i}h^{\prime-1}\neq\\{e\\}$. By 5.9,
$h^{\prime}\in H_{i}$. Thus, $C_{H}(h)\leq H_{i}$. ∎

We are ready to prove 5.1. Firstly, let us state the following well-known
result and give a proof of it for the sake of completeness.

######  Lemma 5.11.

Let $G$ be connected linear semisimple Lie groups with trivial center and no
compact factors, let $\Gamma$ be a lattice of $G$ and assume that there does
not exist any simple factor $H$ of $G$ such that
$H\cong\operatorname{PSL}(2,\mathbb{R})$ and $H\cap\Gamma_{1}$ is a lattice in
$H$. Then the group $\operatorname{Out}(\Gamma)$ is finite.

###### Proof.

Let $F:\operatorname{Aut}(\Gamma)\longrightarrow\operatorname{Aut}(G)$ be the
morphism sending an automorphism of $\Gamma$ to its unique extension on $G$ by
5.2(5). Clearly, $F(\operatorname{Inn}(\Gamma))\leq\operatorname{Inn}(G)$, so
$F$ descends to a group morphism
$f:\operatorname{Out}(\Gamma)\longrightarrow\operatorname{Out}(G)$. Then, the
claim follows from the fact that $\operatorname{Out}(G)$ is finite and that
$\operatorname{Ker}f=N_{G}(\Gamma)/\Gamma$ is also finite by 5.2(4). ∎

######  Lemma 5.12.

Let $G$ be a connected semisimple Lie groups without compact factors and let
$\Gamma$ be a lattice of $G$. If
$\operatorname{Out}(\Gamma/\mathcal{Z}\Gamma)$ is Minkowski then
$\operatorname{Out}(\Gamma)$ is Minkowski.

###### Proof.

We consider the central short exact sequence
$1\longrightarrow\mathcal{Z}\Gamma\longrightarrow\Gamma\longrightarrow\Gamma/\mathcal{Z}\Gamma\longrightarrow
1$. The center is a characteristic subgroup, hence
$\operatorname{Out}(\Gamma)=\operatorname{Out}(\Gamma,\mathcal{Z}\Gamma)$.

On the other hand, $\operatorname{Out}(\Gamma/\mathcal{Z}\Gamma)$ is Minkowski
by hypothesis, $\operatorname{Out}(\mathcal{Z}\Gamma)$ is Minkowski since
$\mathcal{Z}\Gamma$ is a finitely generated abelian group and
$\overline{H}^{1}(\Gamma/\mathcal{Z}\Gamma,\mathcal{Z}\Gamma)$ is Minkowski
since it is finitely generated and abelian. Therefore by 2.1 we obtain that
$\operatorname{Out}(\Gamma)$ is Minkowski. ∎

In view of the preceding lemma, to prove 5.1 it remains to show that
$\operatorname{Out}(\Gamma/\mathcal{Z}\Gamma)$ is Minkowski. Note that
$\Gamma/\mathcal{Z}\Gamma$ is a lattice in $G/\mathcal{Z}G$ which is a
centreless connected semisimple linear Lie group without compact factors. By
5.2 (1) and (2), $G/\mathcal{Z}G=G_{1}\times\cdots\times G_{n}$ and
$\Gamma/\mathcal{Z}\Gamma$ has a normal torsion-free finite index subgroup
$\Lambda$ of the form $\Lambda=\Lambda_{1}\times\cdots\times\Lambda_{n}$ where
each $\Lambda_{i}$ is an irreducible lattice of $G_{i}$. Note that each of
this groups is centreless. We will use the decomposition
$\Lambda=\Lambda^{(1)}\times\Lambda^{(\geq 2)}$ where $\Lambda^{(1)}$ is the
product of all lattices of real rank $1$ and $\Lambda^{(\geq 2)}$ is the
product of all lattices of real rank greater or equal than $2$. After
reordering we can assume that
$\Lambda^{(1)}=\Lambda_{1}\times\cdots\Lambda_{m}$ and $\Lambda^{(\geq
2)}=\Lambda_{m+1}\times\cdots\Lambda_{n}$.

######  Lemma 5.13.

The group $\Lambda^{(\geq 2)}$ is characteristic in $\Lambda$.

###### Proof.

We will prove that given any $f\in\operatorname{Aut}(\Lambda)$ and any
$m+1\leq j\leq n$ we have $f(\Lambda_{j})\leq\Lambda^{(\geq 2)}$.

Let $\pi_{i}:\Lambda\longrightarrow\Lambda_{i}$ and
$\iota_{i}:\Lambda_{i}\longrightarrow\Lambda$ denote the natural projection
and inclusion morphisms. We take the group morphism $\pi_{i}\circ
f\circ\iota_{j}:\Lambda_{j}\longrightarrow\Lambda_{i}$ with $1\leq i\leq m$
and $m+1\leq j\leq n$. By Margulis normal subgroup theorem (see 5.2(7)) we
have that $N=\operatorname{Ker}(\pi_{i}\circ
f\circ\iota_{j})\trianglelefteq\Lambda_{j}$ is either finite or has finite
index in $\Lambda_{j}$. If $N$ is finite then it is trivial since
$\Lambda_{j}$ is torsion-free and $\pi_{i}\circ f\circ\iota_{j}$ is injective.
If it has finite index then $\Lambda_{j}/N$ is a finite subgroup of
$\Lambda_{i}$ and therefore it is trivial since $\Lambda_{i}$ is torsion-free.
In this case, $\pi_{i}\circ f\circ\iota_{j}$ is trivial.

The morphisms $\pi_{i}\circ
f\circ\iota_{j}:\Lambda_{j}\longrightarrow\Lambda_{i}$ cannot be injective for
any $i$ and $j$. Indeed, since $G_{i}$ is centreless (hence
$G_{i}\leq\operatorname{GL}(n_{i},\mathbb{R})$ for some $i$), we can construct
a representation
$\rho:\Lambda_{j}\longrightarrow\operatorname{GL}(n_{i},\mathbb{R})$ given by
the composition $\pi_{i}\circ f\circ\iota_{j}$ with the inclusions
$\Lambda_{i}\leq G_{i}\leq\operatorname{GL}(n_{i},\mathbb{R})$. By Margulis
superrigidity (see 5.2(6)) there exists a finite index subgroup
$\Lambda_{j0}\leq\Lambda_{j}$ and a representation
$\tilde{\rho}:G_{j}\longrightarrow\overline{\rho(\Lambda_{j})}^{o}$ such that
$\tilde{\rho}_{|\Lambda_{j0}}={\rho}_{|\Lambda_{j0}}$. Since
$\overline{\rho(\Lambda_{j})}^{o}\leq G_{i}$, $\tilde{\rho}$ induces an group
morphism $G_{j}\longrightarrow G_{i}$, which is trivial since $G_{i}$ and
$G_{j}$ are simple, $\operatorname{rank}_{R}G_{j}\geq 2$ and
$\operatorname{rank}_{R}G_{i}=1$. Therefore
$\Lambda_{j0}\leq\operatorname{Ker}\pi_{i}\circ f\circ\iota_{j}$. Since
$\pi_{i}\circ f\circ\iota_{j}$ is not injective then $\pi_{i}\circ
f\circ\iota_{j}$ is trivial, as desired. ∎

Note that $\operatorname{Out}(\Lambda^{(\geq 2)})$ is finite by 5.11.

######  Lemma 5.14.

The group $\operatorname{Out}(\Lambda^{(1)})$ is Minkowski.

###### Proof.

Let
$H=\\{f\in\operatorname{Aut}(\Lambda^{(1)}):f(\Lambda_{i})=\Lambda_{i}\text{
for all $i$}\\}$. Firstly, we will show that $H$ has finite index in
$\operatorname{Aut}(\Lambda^{(1)})$. Since
$\operatorname{Inn}(\Lambda^{(1)})\trianglelefteq H$, this will imply that
$[\operatorname{Out}(\Lambda^{(1)}):H/\operatorname{Inn}(\Lambda^{(1)})]<\infty$.

We denote by $r(\lambda)$ the number of non-trivial entries of an element
$\lambda\in\Lambda^{(1)}$. Let $S$ be the set of elements of $\Lambda^{(1)}$
whose centralizer $C_{\Lambda^{(1)}}(\lambda)$ is isomorphic as an abstract
group to $\mathbb{Z}\times\Lambda_{\lambda}$, where $\Lambda_{\lambda}$ is a
product of lattices in $\mathbb{R}$-rank one semisimple Lie groups (which
depend on the element $\lambda\in\Lambda$).

Firstly, we note that if $\lambda\in S$ then $r(\lambda)=1$. Given
$\lambda=(\lambda_{1},\dots,\lambda_{m})\in\Lambda^{(1)}$, its centralizer is
$C_{\Lambda^{(1)}}(\lambda)=\prod_{i=1}^{m}C_{\Lambda_{i}}(\lambda_{i})$. In
addition, $C_{\Lambda_{i}}(\lambda_{i})=\Lambda_{i}$ if and only if
$\lambda_{i}=e_{i}$. If $\lambda_{i}$ is not trivial then
$C_{\Lambda_{i}}(\lambda_{i})$ is virtually nilpotent. Indeed, if
$\lambda_{i}$ is hyperbolic then $C_{\Lambda_{i}}(\lambda_{i})\cong\mathbb{Z}$
by 5.9(1) and if $\lambda_{i}$ is parabolic then
$C_{\Lambda_{i}}(\lambda_{i})$ is virtually nilpotent by 5.10 and 5.8. If
$r(\lambda)\neq 1$ and $\lambda\in S$ then $C_{\Lambda^{(1)}}(\lambda)$ would
contain at least two virtually nilpotent factors and one of this factors would
be a normal subgroup of a lattice in a centreless semisimple Lie group. This
contradicts 5.3. Thus, we can conclude that $\lambda$ is not in $S$. However,
note that it is possible for an element $\lambda$ with $r(\lambda)=1$ to not
be in $S$. If $\lambda=(e_{1},\dots,\lambda_{i},\dots,e_{m})$, then
$C_{\Lambda^{(1)}}(\lambda)\cong
C_{\Lambda_{i}}(\lambda_{i})\times\Lambda_{\lambda}$ and
$C_{\Lambda_{i}}(\lambda_{i})$ is not necessarily isomorphic to $\mathbb{Z}$
if $\lambda_{i}$ is not hyperbolic.

Clearly $S$ is preserved by automorphisms of $\Lambda^{(1)}$. Finally, $S$
generates $\Lambda^{(1)}$ by 5.9(2), since $S$ contains all elements whose
only non-trivial entry is hyperbolic. Consequently, any
$f\in\operatorname{Aut}(\Lambda^{(1)})$ permutes the factors of
$\Lambda^{(1)}$ and we can construct a group morphism
$\phi:\operatorname{Aut}(\Lambda^{(1)})\longrightarrow S_{m}$ such that
$H=\operatorname{Ker}\phi$. Consequently,
$[\operatorname{Aut}(\Lambda^{(1)}):H]<m!$.

We are ready to prove that $\operatorname{Out}(\Lambda^{(1)})$ is Minkowski.
We proceed by induction on the number of factors $m$. If $m=1$ then
$\Lambda^{(1)}$ is an irreducible lattice in a centreless semisimple Lie group
$G_{1}$. If $G_{1}\ncong\operatorname{PSL}(2,\mathbb{R})$ then we can use 5.11
to conclude that $\operatorname{Out}(\Lambda^{(1)})$ is finite. If
$G_{1}\cong\operatorname{PSL}(2,\mathbb{R})$ then $\Lambda^{(1)}$ is a
Fuchsian group and therefore $\operatorname{Out}(\Lambda^{(1)})$ is virtually
torsion-free [MS06, Corollary 2.6] and hence Minkowski.

Assume now that
$\operatorname{Out}(\Lambda_{1}\times\cdots\times\Lambda_{m-1})$ is Minkowski.
By 2.1, $\operatorname{Out}(\Lambda^{(1)},\Lambda_{m})$ is Minkowski if
$\operatorname{Out}(\Lambda_{m})$,
$\operatorname{Out}(\Lambda_{1}\times\cdots\times\Lambda_{m-1})$ and
$H^{1}(\Lambda_{1}\times\cdots\times\Lambda_{m-1},\mathcal{Z}\Lambda_{m})$ are
Minkowski. But they are Minkowski by induction hypothesis and the fact that
$\mathcal{Z}\Lambda_{m}$ is trivial. Since
$H\leq\operatorname{Aut}(\Lambda^{(1)},\Lambda_{m})\leq\operatorname{Aut}(\Lambda^{(1)})$
and $[\operatorname{Aut}(\Lambda^{(1)}):H]<\infty$ we have that
$[\operatorname{Out}(\Lambda^{(1)}):\operatorname{Out}(\Lambda^{(1)},\Lambda_{m})]<\infty$,
which implies that $\operatorname{Out}(\Lambda^{(1)})$ is Minkowski. ∎

We are ready to prove that $\operatorname{Out}(\Gamma/\mathcal{Z}\Gamma)$ is
Minkowski and finish the proof of 5.1.

######  Lemma 5.15.

The group $\operatorname{Out}(\Gamma/\mathcal{Z}\Gamma)$ is Minkowski.

###### Proof.

Since $[\Gamma/\mathcal{Z}\Gamma:\Lambda]<\infty$, by 2.5 it is enough to
prove that $\operatorname{Out}(\Lambda)$ is Minkowski. We have seen that
$\operatorname{Out}(\Lambda^{(\geq 2)})$, $\operatorname{Out}(\Lambda^{(1)})$
and $H^{1}(\Lambda^{(1)},\mathcal{Z}\Lambda^{(\geq 2)})$ are Minkowski. Thus,
by 2.1 $\operatorname{Out}(\Lambda)$ is Minkowski, as we wanted to see. ∎

The rest of this section is devoted to prove 1.11. Recall that if $H$ is a
torsion-free hyperbolic group, then $C_{H}(h)$ is cyclic for all non-trivial
$h\in H$. In addition, if $H$ contains a normal abelian subgroup then $H$ is
cyclic (see [BH13, Part III $\Gamma$.3]). Finally, if $M$ is a closed
connected aspherical manifold of dimension $n\geq 3$ and $\pi_{1}(M)$ is
hyperbolic then $\operatorname{Out}(\pi_{1}(M))$ is finite (see 3.7(2)).

The arguments used in 5.14 can be used to prove the next statement.

######  Proposition 5.16.

Let $M=M_{1}\times\cdots\times M_{m}$, where $M_{i}$ are closed aspherical
manifolds such that $\pi_{1}(M_{i})$ is hyperbolic and $\dim(M_{i})\geq 3$.
Then $\operatorname{Out}(\pi_{1}(M))$ is finite.

###### Proof.

Firstly, note that
$\pi_{1}(M)=\pi_{1}(M_{1})\times\cdots\times\pi_{1}(M_{m})$. Let
$H=\\{f\in\operatorname{Aut}(\pi_{1}(M)):f(\pi_{1}(M_{i}))=\pi_{1}(M_{i})\text{
for all $i$}\\}$. As in 5.14, we will show that $H$ has finite index in
$\operatorname{Aut}(\pi_{1}(M))$. Since
$\operatorname{Inn}(\pi_{1}(M))\trianglelefteq H$, this will imply that
$[\operatorname{Out}(\pi_{1}(M)):H/\operatorname{Inn}(\pi_{1}(M))]<\infty$.

Let $e_{i}$ denote the trivial element of $\pi_{1}(M_{i})$. We know that for
every $\lambda=(\lambda_{1},\dots,\lambda_{m})\in\pi_{1}(M)$ we have
$C_{\pi_{1}(M)}(\lambda)=\prod_{i=1}^{m}C_{\pi_{1}(M_{i})}(\lambda_{i})$. In
addition, $C_{\pi_{1}(M_{i})}(\lambda_{i})=\pi_{1}(M_{i})$ if
$\lambda_{i}=e_{i}$ and $C_{\pi_{1}(M_{i})}(\lambda_{i})=\mathbb{Z}$
otherwise. As before, let $r(\lambda)$ denote the number of non-trivial
entries of $\lambda$. We claim that that if $r(\lambda)=1$ and
$f\in\operatorname{Aut}(\pi_{1}(M))$ then $r(f(\lambda))=1$.

Assume on the contrary, that $\lambda=(e_{1},\dots,\lambda_{i},\dots,e_{m})$
and that $r({f(\lambda)})>1$. Since
$f(C_{\pi_{1}(M)}(\lambda))=C_{\pi_{1}(M)}(f(\lambda))$ we can take the
inverse morphism $f^{-1}:C_{\pi_{1}(M)}(f(\lambda))\longrightarrow
C_{\pi_{1}(M)}(\lambda)$ and restrict it to $\mathbb{Z}^{r(f(\lambda))}\leq
C_{\pi_{1}(M)}(f(\lambda))$. The morphism $\pi_{i}\circ
f^{-1}_{|\mathbb{Z}^{r(f(\lambda))}}:\mathbb{Z}^{r(f(\lambda))}\longrightarrow\mathbb{Z}$
cannot be injective. If $a$ is a non trivial element of
$\operatorname{Ker}\pi_{i}\circ f^{-1}_{|\mathbb{Z}^{r(f(\lambda))}}$, then
there exists a $j\neq i$ such that $\pi_{j}\circ f^{-1}_{|\langle
a\rangle}:\langle a\rangle\longrightarrow\Lambda_{j}$ is injective. Since
$\langle a\rangle\trianglelefteq C_{\pi_{1}(M)}(f(\lambda))$ and
$\pi_{j}:C_{\pi_{1}(M)}(\lambda)\longrightarrow\pi_{1}(M_{j})$ is surjective
we can conclude that $\mathbb{Z}\cong\pi_{j}\circ f^{-1}_{|\langle
a\rangle}(\langle a\rangle)\trianglelefteq\pi_{1}(M_{j})$. But
$\pi_{1}(M_{j})$ is hyperbolic, so from the fact that it contains an abelian
normal subgroup we can conclude $\pi_{1}(M_{j})\cong\mathbb{Z}$. This is a
contradiction with the fact that $\pi_{1}(M_{j})$ is the fundamental group of
a closed aspherical manifold of dimension $\dim(M_{j})\geq 3$.

In conclusion, any $f\in\operatorname{Aut}(\pi_{1}(M))$ permutes the factors
of $\pi_{1}(M)$ and thus we can construct a group morphism to the permutation
group of $m$ letters, $\phi:\operatorname{Aut}(\pi_{1}(M))\longrightarrow
S_{m}$ such that $H=\operatorname{Ker}\phi$. Consequently,
$[\operatorname{Aut}(\pi_{1}(M)):H]<m!$.

We are now ready to prove that $\operatorname{Out}(\pi_{1}(M))$ is finite. We
proceed by induction on the number of factors $m$. If $m=1$ then
$\operatorname{Out}(\pi_{1}(M))$ is finite (3.7(2)). Assume now that
$\operatorname{Out}(\pi_{1}(M_{1})\times\cdots\times\pi_{1}(M_{m-1}))$ is
finite. By 2.1, $\operatorname{Out}(\pi_{1}(M),\pi_{1}(M_{m}))$ is finite if
$\operatorname{Out}(\pi_{1}(M_{m}))$,
$\operatorname{Out}(\pi_{1}(M_{1})\times\cdots\times\pi_{1}(M_{m-1}))$ and
$H^{1}(\pi_{1}(M_{1})\times\cdots\times\pi_{1}(M_{m-1}),\mathcal{Z}\pi_{1}(M_{m}))$
are finite. But they are finite by induction hypothesis and the fact that
$\mathcal{Z}\pi_{1}(M_{m})$ is trivial. Since
$H\leq\operatorname{Aut}(\pi_{1}(M),\pi_{1}(M_{m}))\leq\operatorname{Aut}(\pi_{1}(M))$
and $[\operatorname{Aut}(\pi_{1}(M)):H]<\infty$ we have that
$[\operatorname{Out}(\pi_{1}(M)):\operatorname{Out}(\pi_{1}(M),\pi_{1}(M_{m}))]<\infty$,
which implies that $\operatorname{Out}(\pi_{1}(M))$ is finite. ∎

Note that the hyperbolic groups are centreless. Consequently,
$\mathcal{Z}\pi_{1}(M)$ is trivial, which implies that $M$ is almost
asymmetric by 1.6 and 5.16. In addition:

######  Corollary 5.17.

The group $\operatorname{Aut}(\pi_{1}(M))$ is Minkowski.

###### Proof.

Note that $\mathcal{Z}\pi_{1}(M)$ is trivial, therefore there is a short exact
sequence

$1\longrightarrow\pi_{1}(M)\longrightarrow\operatorname{Aut}(\pi_{1}(M))\longrightarrow\operatorname{Out}(\pi_{1}(M))\longrightarrow
1.$

Since $\operatorname{Out}(\pi_{1}(M))$ is finite and $\pi_{1}(M)$ is torsion-
free, we conclude that $\operatorname{Aut}(\pi_{1}(M))$ is virtually torsion-
free and hence Minkowski. ∎

##  6\. Combining the two cases: Aspherical locally homogeneous spaces

The aim of this section is to finish the proof of 1.9 and to prove 1.10. For
the first task, the strategy is to combine the results on solvable Lie groups
and semisimple Lie groups obtained in the previous sections in a similar way
we proved 5.12.

Given a connected Lie group $G$, we would like to use the Levi decomposition
$G=R\rtimes S$, where $S$ is the semisimple part and $R$ is the solvable
radical, the maximal normal connected solvable Lie subgroup of $G$. However,
$\Gamma\cap R$ is not a lattice in $R$ in general (see [Gen15]).

Instead of using the solvable radical we will use the amenable radical, the
maximal normal connected amenable Lie subgroup of $G$. Let us recall some
basic properties of amenable groups.

######  Lemma 6.1.

[Zim13, Proposition 4.1.6] Let $1\longrightarrow G_{1}\longrightarrow
G_{2}\longrightarrow G_{3}\longrightarrow 1$ be a short exact sequence of
groups. Then $G_{2}$ is amenable if and only if $G_{1}$ and $G_{3}$ are
amenable.

Solvable Lie groups and compact Lie groups are amenable. In particular, finite
groups and $\mathbb{Z}$ are amenable.

The lemma implies that virtually polycyclic groups are amenable.

We can decompose $G=A\rtimes S_{nc}$ where $S_{nc}$ is semisimple with no
compact factors and $A$ is the amenable radical. Moreover, $A=R\rtimes S_{c}$
where $S_{c}$ is compact and semisimple. If $\Gamma$ is a lattice in $G$ then
$A\cap\Gamma$ is a lattice in $A$ (see [GS20, Gen15]). In addition, a lattice
$\Gamma$ of $G$ is amenable if and only if $G$ is amenable (see [Zim13,
Proposition 4.1.11]). We have the short exact sequence

$1\longrightarrow\Gamma\cap
A\longrightarrow\Gamma\longrightarrow\Gamma/\Gamma\cap A\longrightarrow 1,$

where the group $\Gamma\cap A$ is a lattice in the amenable radical $A$ and
$\Gamma/\Gamma\cap A$ is a lattice in the semisimple Lie group without compact
factors $S_{nc}$. We write $\Gamma\cap A=\Gamma_{A}$ and $\Gamma/\Gamma\cap
A=\Gamma_{nc}$.

Our first goal is to see that $\Gamma_{A}$ is virtually polycyclic. 6.2 is
probably well-known to experts, but it is difficult to find a proof in the
literature. Hence, we provide a proof for the sake of completeness.

######  Lemma 6.2.

Let $\Gamma$ be a lattice in an amenable group $A$. Then $\Gamma$ is virtually
polycyclic.

###### Proof.

Denote by $\pi_{A/R}:A\longrightarrow A/R$ the quotient map and define
$L=\overline{\pi_{A/R}(\Gamma)}^{o}$. It is connected solvable Lie group by
[Rag12, 8.24]. Moreover, $L$ is abelian since it is a connected solvable Lie
subgroup of the compact Lie group $A/R$. Then, $\tilde{R}=\pi_{A/R}^{-1}(L)$
is a connected solvable group since $\tilde{R}/R$ is abelian and $R$ is
connected.

$\tilde{R}\cap\Gamma$ is a lattice in $\tilde{R}$ (see [GS20, Claim 2.2]). We
claim that $\tilde{R}\cap\Gamma$ is polycyclic. Indeed, we have that
$\tilde{R}/\tilde{R}\cap\Gamma$ is a compact solvmanifold. Then
$\tilde{R}/\tilde{R}\cap\Gamma\cong R^{\prime}/\Gamma^{\prime}$, where
$R^{\prime}$ is the universal cover of $\tilde{R}$ and $\Gamma^{\prime}$ is a
lattice in a simply connected solvable Lie group thus it is polycyclic.
Moreover, using the long exact sequence of homotopy groups for a fibration we
obtain the short exact sequence

$1\longrightarrow\pi_{1}(\tilde{R})\cong\mathbb{Z}^{n}\longrightarrow\Gamma^{\prime}\longrightarrow\tilde{R}\cap\Gamma\longrightarrow
1.$

Since $\tilde{R}\cap\Gamma\cong\Gamma^{\prime}/\mathbb{Z}^{n}$ we can conclude
that $\tilde{R}\cap\Gamma$ is polycyclic.

We want to show that $[\Gamma:\tilde{R}\cap\Gamma]<\infty$. Let
$H=N_{A}(\tilde{R})$ and let $H^{o}$ be its connected component (then
$R\leq\tilde{R}\leq H^{o}\leq H\leq A$). Thus $\Gamma\leq H$ and $|H/H^{o}|$
is bounded by [GS20, Corollary $2.6$]. In addition, $H^{o}/\tilde{R}$ is a
compact Lie group.

In consequence, $[\Gamma:\Gamma\cap\tilde{R}]=[\Gamma:\Gamma\cap
H^{o}][\Gamma\cap H^{o}:\Gamma\cap\tilde{R}]$. The first term is finite
because $\Gamma\leq H$ and $|H/H^{o}|$ is bounded. The second term is finite
because
$\Gamma/\Gamma\cap\tilde{R}\cong\Gamma\tilde{R}/\tilde{R}=\pi_{H^{o}/\tilde{R}}(\Gamma)$
is a discrete subgroup in a compact Lie group and thus it is finite. ∎

Now that we know the structure of $\Gamma_{A}$ we are interested in the
relation between $\Gamma_{A}$ and $\Gamma$. Since we want to study the
automorphisms and outer automorphisms of $\Gamma$, it is natural to study
whether $\Gamma_{A}$ is a characteristic subgroup of $\Gamma$.

If $\mathcal{Z}(S_{nc})\neq\\{e\\}$ then we consider the subgroup
$\pi^{-1}(\mathcal{Z}(\Gamma_{nc}))=\Gamma_{A}^{\prime}$, which is a virtually
polycyclic group since it fits in the short exact sequence
$1\longrightarrow\Gamma_{A}\longrightarrow\Gamma_{A}^{\prime}\longrightarrow\mathcal{Z}(\Gamma_{nc})\longrightarrow
1$. In consequence, there is a short exact sequence
$1\longrightarrow\Gamma_{A}^{\prime}\longrightarrow\Gamma\longrightarrow\overline{\Gamma}_{nc}\longrightarrow
1$, where $\overline{\Gamma}_{nc}$ is a lattice of the centreless connected
semisimple Lie group $S_{nc}/\mathcal{Z}(S_{nc})$. By 5.2(1), there exists a
normal finite index torsion-free lattice
$\Gamma_{nc}^{\prime}\leq\overline{\Gamma}_{nc}$. Then we can consider the
short exact sequence
$1\longrightarrow\Gamma_{A}^{\prime}\longrightarrow\Gamma^{\prime}\longrightarrow\Gamma_{nc}^{\prime}\longrightarrow
1$. We will later prove that $\operatorname{Out}(\Gamma^{\prime})$ is
Minkowski. This fact, together with the previous short exact sequence, will
imply that $\operatorname{Out}(\Gamma)$ is Minkowski by 2.5.

######  Lemma 6.3.

$\Gamma_{A}^{\prime}$ is a characteristic subgroup of $\Gamma^{\prime}$.

###### Proof.

Assume that $\Gamma_{A}^{\prime}$ is not characteristic. Thus, there exists
$f\in\operatorname{Aut}(\Gamma^{\prime})$ such that $f(\Gamma_{A}^{\prime})$
has a non trivial projection $\Lambda$ in $\Gamma_{nc}^{\prime}$. The group
$\Lambda$ is a torsion-free virtually polycyclic subgroup and hence it has a
characteristic subgroup $\Lambda^{\prime}$ which is polycyclic (see [LR10,
§9.5]). In particular, $\Lambda^{\prime}$ is solvable and normal in the
semisimple centreless lattice $\Gamma^{\prime}_{nc}$, contradicting 5.3.
Consequently, $\Gamma_{A}^{\prime}$ is characteristic. ∎

In consequence,
$\operatorname{Out}(\Gamma^{\prime})=\operatorname{Out}(\Gamma^{\prime},\Gamma_{A}^{\prime})$.
2.1 shows that to prove that $\operatorname{Aut}(\Gamma^{\prime})$ and
$\operatorname{Out}(\Gamma^{\prime})$ are Minkowski it is enough to prove that
$\operatorname{Aut}(\Gamma_{A}^{\prime})$,
$\operatorname{Aut}(\Gamma_{nc}^{\prime})$,
$Z^{1}_{\psi}(\Gamma_{nc}^{\prime},\mathcal{Z}(\Gamma_{A}^{\prime}))$,
$\operatorname{Out}(\Gamma_{A}^{\prime})$,
$\operatorname{Out}(\Gamma_{nc}^{\prime})$ and
$\overline{H}^{1}_{\psi}(\Gamma_{nc}^{\prime},\mathcal{Z}(\Gamma_{A}^{\prime}))$
are Minkowski. We already know that $\operatorname{Aut}(\Gamma_{A}^{\prime})$
and $\operatorname{Out}(\Gamma_{A}^{\prime})$ are Minkowski because
$\Gamma_{A}^{\prime}$ is virtually polycyclic (4.1), and
$\operatorname{Out}(\Gamma_{nc}^{\prime})$ is Minkowski by 5.12. Thus we only
need to check that $\operatorname{Aut}(\Gamma_{nc}^{\prime})$,
$Z^{1}_{\psi}(\Gamma_{nc}^{\prime},\mathcal{Z}(\Gamma_{A}^{\prime}))$,
$\operatorname{Out}(\Gamma^{\prime}_{A})$ and
$\overline{H}^{1}_{\psi}(\Gamma_{nc}^{\prime},\mathcal{Z}(\Gamma_{A}^{\prime}))$
are Minkowski.

######  Lemma 6.4.

The group $\operatorname{Aut}(\Gamma_{nc}^{\prime})$ is Minkowski.

###### Proof.

Since $\Gamma_{nc}^{\prime}$ is centreless, we have a short exact sequence

$1\longrightarrow\Gamma_{nc}^{\prime}\longrightarrow\operatorname{Aut}(\Gamma_{nc}^{\prime})\longrightarrow\operatorname{Out}(\Gamma_{nc}^{\prime})\longrightarrow
1.$

Since $\Gamma_{nc}^{\prime}$ is virtually torsion-free and
$\operatorname{Out}(\Gamma_{nc}^{\prime})$ is Minkowski, then
$\operatorname{Aut}(\Gamma_{nc}^{\prime})$ is Minkowski. ∎

######  Lemma 6.5.

The group
$\overline{H}^{1}_{\psi}(\Gamma_{nc}^{\prime},\mathcal{Z}\Gamma_{A}^{\prime})$
is Minkowski.

###### Proof.

Since $H^{1}_{\psi}(\Gamma_{nc}^{\prime},\mathcal{Z}\Gamma_{A}^{\prime})$ is a
finitely generated abelian group and there is a surjective morphism
$H^{1}_{\psi}(\Gamma_{nc}^{\prime},\mathcal{Z}\Gamma_{A}^{\prime})\longrightarrow\overline{H}^{1}_{\psi}(\Gamma_{nc}^{\prime},\mathcal{Z}\Gamma_{A}^{\prime})$,
the group
$\overline{H}^{1}_{\psi}(\Gamma_{nc}^{\prime},\mathcal{Z}\Gamma_{A}^{\prime})$
is finitely generated and abelian, hence it is Minkowski. ∎

######  Lemma 6.6.

The group
$Z^{1}_{\psi}(\Gamma^{\prime}_{nc},\mathcal{Z}(\Gamma^{\prime}_{A}))$ is
Minkowski.

###### Proof.

There is a short exact sequence

$1\longrightarrow\overline{B}^{1}_{\psi}(\Gamma^{\prime}_{nc},\mathcal{Z}(\Gamma^{\prime}_{A}))\longrightarrow
Z^{1}_{\psi}(\Gamma^{\prime}_{nc},\mathcal{Z}(\Gamma^{\prime}_{A}))\longrightarrow\overline{H}^{1}_{\psi}(\Gamma_{nc}^{\prime},\mathcal{Z}\Gamma_{A}^{\prime})\longrightarrow
1.$

We have that
$\overline{B}^{1}_{\psi}(\Gamma^{\prime}_{nc},\mathcal{Z}(\Gamma^{\prime}_{A}))\cong\mathcal{Z}(\Gamma^{\prime}_{A})/\mathcal{Z}(\Gamma^{\prime})$
(see 2.1), therefore
$\overline{B}^{1}_{\psi}(\Gamma^{\prime}_{nc},\mathcal{Z}(\Gamma^{\prime}_{A}))$
is a finitely generated abelian group and hence it is Minkowski. Since
$\overline{H}^{1}_{\psi}(\Gamma_{nc}^{\prime},\mathcal{Z}\Gamma_{A}^{\prime})$
is also a finitely generated abelian group, we use 2.4 to conclude that
$Z^{1}_{\psi}(\Gamma^{\prime}_{nc},\mathcal{Z}(\Gamma^{\prime}_{A}))$ is
Minkowski. ∎

These three lemmas complete the proof of 1.9.

######  Remark 6.7.

Note that we can use the proof of 1.9 to conclude that if $\Gamma$ is a
torsion-free group which fits in a short exact sequence
$1\longrightarrow\Gamma_{A}\longrightarrow\Gamma\longrightarrow\Gamma_{nc}\longrightarrow
1$ where $\Gamma_{A}$ is virtually polycyclic and $\Gamma_{nc}$ is a lattice
in a centreless semisimple Lie group then $\operatorname{Out}(\Gamma)$ is
Minkowski. Not all groups of this form are lattices in connected Lie groups,
see [BK18, Theorem 7.5]. Baues and Kamishima introduced in [BK18] the notion
of the closed aspherical manifolds with large symmetry. From the definition of
these manifolds [BK18, Definition 1.1] one can deduce that their fundamental
group fulfils the hypothesis of this remark and hence the outer automorphism
group of their fundamental group is Minkowski.

######  Remark 6.8.

The homeomorphism group of a coset space $G/\Gamma$ where $G$ is a connected
Lie group and $\Gamma$ is a cocompact lattice is not necessarily Jordan.
Indeed, we can take a non-compact semisimple Lie group $G^{\prime}$ and a
cocompact lattice $\Gamma^{\prime}$ such that a connected compact maximal
subgroup $K^{\prime}$ containing $\operatorname{SU}(2)$ acts effectively on
$G^{\prime}/\Gamma^{\prime}$. By [MiR17] we know that
$\operatorname{Homeo}(G^{\prime}/\Gamma^{\prime}\times T^{2})$ is not Jordan.
Hence, taking $G=G^{\prime}\times\mathbb{R}^{2}$ and
$\Gamma=\Gamma^{\prime}\times\mathbb{Z}^{2}$ we obtain a coset space
$G/\Gamma$ whose homeomorphism group is not Jordan. On the other hand, A.
Golota proved in [Gol24] that if $G$ is a complex Lie group and $\Gamma$ is a
cocompact lattice, then the group of automorphisms
$\operatorname{Aut}(G/\Gamma)$ is Jordan.

The end of this section is devoted to prove 1.10. We will need the following
two results from [LR10].

######  Theorem 6.9.

[LR10, Theorem 11.7.29] Let $M$ be a closed connected aspherical manifold
whose fundamental group fits into a short exact sequence
$1\longrightarrow\Gamma_{A}\longrightarrow\pi_{1}(M)\longrightarrow\Gamma_{nc}\longrightarrow
1$, where $\Gamma_{A}$ is virtually polycyclic and $\Gamma_{nc}$ is a
cocomapct lattice in a semisimple connected Lie group without compact factors.
Then there exists a closed connected aspherical manifold $M^{\prime}$ which is
homotopically equivalent to $M$ and satisfies $\operatorname{tor-
sym}(M^{\prime})=\operatorname{rank}\mathcal{Z}\pi_{1}(M)$.

######  Theorem 6.10.

[LR10, Theorem 11.7.28] Let $H\setminus G/\Gamma$ be a closed aspherical
locally homogeneous space with $G$ simply connected and let $R$ be the
solvable radical of $G$. Assume that $R\cap\Gamma$ is a lattice in $R$ such
that any automorphism of $R\cap\Gamma$ can be extended to $R$, and that
$\exp:\mathcal{L}(R)\longrightarrow R$ is surjective. Then $\operatorname{tor-
sym}(H\setminus G/\Gamma)=\operatorname{rank}\mathcal{Z}\Gamma$.

###### Proof of 1.10.

Note that $\operatorname{Out}(\Gamma)$ is Minkowski by 1.9, hence
$\operatorname{disc-sym}(H\setminus
G/\Gamma)\leq\operatorname{rank}\mathcal{Z}\Gamma$ by 1.6. On the other hand,
it is clear that $\operatorname{tor-sym}(H\setminus
G/\Gamma)\leq\operatorname{disc-sym}(H\setminus G/\Gamma)$. Therefore, it
suffices to show that $\operatorname{tor-sym}(H\setminus
G/\Gamma)=\operatorname{rank}\mathcal{Z}\Gamma$ to conclude that
$\operatorname{disc-sym}(H\setminus
G/\Gamma)=\operatorname{rank}\mathcal{Z}\Gamma$.

By 6.2 $\Gamma$ satisfies the conditions of 6.9, hence there exists a closed
connected aspherical manifold $M^{\prime}$ homotopically equivalent to
$H\setminus G/\Gamma$ such that $\operatorname{tor-
sym}(M^{\prime})=\operatorname{rank}\mathcal{Z}\Gamma$. On the other hand, the
Farrel-Jones conjecture is true for cocompact lattices in connected Lie groups
(see [BL12, KLR16]), which implies the Borel conjecture for closed aspherical
locally homogeneous space of dimension equal or greater than 5 (see [BL12,
Proposition 0.3 (ii)]). In consequence, if $\dim(H\setminus G/\Gamma)\geq 5$
then $H\setminus G/\Gamma\cong M^{\prime}$ and hence $\operatorname{tor-
sym}(H\setminus G/\Gamma)=\operatorname{rank}\mathcal{Z}\Gamma$. Thus, it
remains to study the cases where $\dim(H\setminus G/\Gamma)\leq 4$.

If $\dim(H\setminus G/\Gamma)=1$ or $2$ then $\operatorname{tor-
sym}(H\setminus G/\Gamma)=\operatorname{rank}\mathcal{Z}\Gamma$ by classical
results and if $\dim(H\setminus G/\Gamma)=3$ then $\operatorname{tor-
sym}(H\setminus G/\Gamma)=\operatorname{rank}\mathcal{Z}\Gamma$ by [Gab92,
Corollary 8.3] and [CJ94, Theorem 1.1]. Thus, it only remains to study the
case where $\dim(H\setminus G/\Gamma)=4$.

First, assume that the noncompact semisimple part $S_{nc}$ of $G$ is trivial.
Then $\Gamma$ is virtually polycyclic. Virtually polycyclic groups are
Freedman good (this property was called "good" and it was introduced in
[Fre83]). If the fundamental group is Freedman good, the Farrel-Jones
conjecture implies the Borel conjecture in dimension $4$ (see [BL12,
Proposition 0.3 (ii)]) and hence $\operatorname{tor-sym}(H\setminus
G/\Gamma)=\operatorname{rank}\mathcal{Z}\Gamma$.

In consequence, it only remains to study the case where $S_{nc}\neq\\{e\\}$.
We can assume two extra hypothesis without losing generality. Firstly, let
$q:\tilde{G}\longrightarrow G$ be the universal cover of $G$. Let $\tilde{H}$
be a maximal compact connected subgroup of $\tilde{G}$ inside $q^{-1}(H)$.
Then $q^{-1}(\Gamma)=\tilde{\Gamma}$ is a cocompact lattice of $\tilde{G}$ and
$H\setminus G/\Gamma\cong\tilde{H}\setminus\tilde{G}/\tilde{\Gamma}$. Thus, we
can assume that $G$ is simply connected and hence its solvable radical $R$ and
its semisimple part $S$ are also simply connected. Secondly, note that if $K$
is a compact connected normal subgroup of $G$ and $p:G\longrightarrow G/K$ is
the quotient map then $\Gamma\cap K=\\{e\\}$ since $\Gamma$ is torsion-free.
This implies that $H\setminus G/\Gamma\cong p(H)\setminus p(G)/\Gamma$.
Therefore, we can assume that $G$ has no compact factors.

Note that $\dim(R)+\dim(S)-\dim(H)=4$ and since $S_{nc}\neq\\{e\\}$ then
$\dim(S)-\dim(H)\geq 2$. Hence, $\dim(R)\leq 2$. There are four simply
connected solvable groups of dimension less or equal than $2$: $\dim(R)=0$ and
$R=\\{e\\}$, $\dim(R)=1$ and $R\cong\mathbb{R}$, and $\dim(R)=2$ and
$R\cong\mathbb{R}^{2}$ or
$R\cong\operatorname{Aff}(\mathbb{R})^{0}\cong\mathbb{R}\rtimes\mathbb{R}$. To
construct the desired torus action on $H\setminus G/\Gamma$ with $R$ belonging
to one of this four cases we will use 6.10. We need to check that the
hypothesis of 6.10 are satisfied in these four cases. Note that in all the
cases the exponential map is surjective and any lattice automorphism extends
uniquely to an automorphism of the group. This is because $R$ is either
abelian or solvable of type (R) (see [LR10, §6.3]). It only remains to see
that $\Gamma\cap R$ is a lattice in $R$.

Assume that $R$ is abelian, then if $C$ is a compact factor of $S$ acting
trivially on $R$ then $C$ is normal on $G$ and therefore $C=\\{e\\}$.
Therefore no compact factor of $S$ acts trivially on $R$. This implies by
[Gen15, Theorem 1.3 (i)] that $R\cap\Gamma$ is a lattice in $R$. Thus,
$\operatorname{tor-sym}(H\setminus
G/\Gamma)=\operatorname{rank}\mathcal{Z}\Gamma$ by 6.10.

If $R\cong\operatorname{Aff}(\mathbb{R})^{0}$ then $\operatorname{Aut}(R)\cong
R$ and hence any compact factor $C$ of $S$ needs to act trivially on $R$.
Consequently, $C\trianglelefteq G$ and therefore $C=\\{e\\}$. By [Gen15,
Theorem 1.3 (i)] if $\Gamma$ were a lattice of $G$ then $R\cap\Gamma$ would be
a lattice of $R$. But $R$ does not admit any lattice (see [Boc16, pg. 82]).
Thus, a closed aspherical locally homogeneous 4-manifold with solvable radical
$R$ does not exists. ∎

##  7\. Jordan property and cohomology

This section is devoted to proving 1.7. We start with the following general
lemma.

######  Lemma 7.1.

Let $\Gamma$ be a finitely generated group and let $\mathcal{Z}_{i}\Gamma$ be
the $i$-th term of the upper central series. Assume that
$\Gamma/\mathcal{Z}_{i}\Gamma$ is finitely generated, centreless and that
$\operatorname{Out}(\Gamma/\mathcal{Z}_{i}\Gamma)$ is Minkowski for some $i$.
Then $\operatorname{Out}(\Gamma)$ is Minkowski.

###### Proof.

Recall that the upper central series
$\\{e\\}=\mathcal{Z}_{0}\Gamma\trianglelefteq\mathcal{Z}_{1}\Gamma\trianglelefteq\mathcal{Z}_{2}\Gamma\trianglelefteq\cdots$
of a group $\Gamma$ is a subnormal series where
$\mathcal{Z}_{i}\Gamma/\mathcal{Z}_{i-1}\Gamma\cong\mathcal{Z}(\Gamma/\mathcal{Z}_{i-1})$
for all $i\geq 0$. The group $\Gamma$ is nilpotent if and only if
$\Gamma\cong\mathcal{Z}_{i}\Gamma$ for some $i$. Moreover, the groups
$\mathcal{Z}_{i}\Gamma$ are characteristic and each group morphism
$\psi_{i}:\Gamma/\mathcal{Z}_{i}\Gamma\longrightarrow\operatorname{Out}(\mathcal{Z}_{i}\Gamma)$
is trivial.

Consequently, the group $\operatorname{Out}(\Gamma)$ is Minkowski if
$\operatorname{Out}(\mathcal{Z}_{i}\Gamma)$,
$\operatorname{Out}(\Gamma/\mathcal{Z}_{i}\Gamma)$ and
$H^{1}(\Gamma/\mathcal{Z}_{i}\Gamma,\mathcal{Z}\Gamma_{i}/\mathcal{Z}\Gamma_{i-1})$
are Minkowski. But they are Minkowski by hypothesis and 4.1, obtaining the
desired conclusion. ∎

If we assume that $\Gamma$ is torsion-free and $\Gamma/\mathcal{Z}_{i}\Gamma$
acts properly on a contractible manifold $\tilde{X}$ then we can use the
Seifert fiber construction in [LR10, Chapter 7] to construct closed connected
aspherical manifolds $M$ such that $\pi_{1}(M)\cong\Gamma$. In addition, if
$\tilde{X}/(\Gamma/\mathcal{Z}_{i}\Gamma)$ is a closed aspherical manifold,
then $M$ can be seen as an iterated principal torus bundle over
$\tilde{X}/(\Gamma/\mathcal{Z}_{i}\Gamma)$. A special case of 7.1 and 1.6 is
the next corollary.

######  Corollary 7.2.

If $M$ is a closed connected aspherical manifold such that
$\mathcal{Z}\pi_{1}(M)$ is trivial and $\operatorname{Out}(\pi_{1}(M))$ is
Minkowski, then $\operatorname{Homeo}(M\times T^{n})$ is Jordan.

Finally, there exists closed hyperbolic $3$-manifolds $N$ which are integral
homology spheres (see [BP92, Thu22]). We have that $\mathcal{Z}(\pi_{1}(N))$
is trivial and $\operatorname{Out}(\pi_{1}(N))$ is finite by [Gro87, 5.4 A].
Thus, $H^{*}(N\times T^{2})\cong H^{*}(S^{3}\times T^{2})$,
$\operatorname{Homeo}(N\times T^{2})$ is Jordan by 7.2 and
$\operatorname{Homeo}(S^{3}\times T^{2})$ is not Jordan by [MiR17]. This
completes the proof of 1.7.

Another example can be produced using Brieskorn manifolds $\Sigma(p,q,r)$.
Recall that
$\Sigma(p,q,r)=\\{(z_{1},z_{2},z_{3})\in\mathbb{C}^{3}:z_{1}^{p}+z_{2}^{q}+z_{3}^{r}=0,\sum_{i=1}^{3}|z_{i}|^{2}=1\\}$.
If $p$, $q$ and $r$ are relatively prime and
$\tfrac{1}{p}+\tfrac{1}{q}+\tfrac{1}{r}<1$ then $\Sigma(p,q,r)$ is a closed
connected aspherical integral homology $3$-sphere (see [Mil75]). Moreover,
$\Sigma(p,q,r)$ are Seifert manifolds with
$\mathcal{Z}\pi_{1}(\Sigma(p,q,r))\cong\mathbb{Z}$ (see [LR10, §14.11]), hence
there exists a $S^{1}$-action on $\Sigma(p,q,r)$ inducing the short exact
sequence

$1\longrightarrow\mathbb{Z}\longrightarrow\pi_{1}(\Sigma(p,q,r))\longrightarrow\operatorname{Inn}\pi_{1}(\Sigma(p,q,r))\longrightarrow
1$

where $\operatorname{Inn}\pi_{1}(\Sigma(p,q,r))$ is centreless. In addition,
$\operatorname{Inn}\pi_{1}(\Sigma(p,q,r))$ is a subgroup of isometries of the
hyperbolic plane. Hence, it contains a centreless torsion-free Fuchsian
subgroup $Q$ of finite index. Since $\operatorname{Out}(Q)$ is virtually
torsion-free then
$\operatorname{Out}(\operatorname{Inn}\pi_{1}(\Sigma(p,q,r)))$ is also
virtually torsion-free (see [MS06, Lemma 2.4, Corollary 2.6]) and therefore
$\operatorname{Out}(\operatorname{Inn}\pi_{1}(\Sigma(p,q,r)))$ is Minkowski.

The manifold $\Sigma(p,q,r)\times T^{2}$ is a closed connected aspherical
manifold such that $H^{*}(\Sigma(p,q,r)\times T^{2})\cong H^{*}(S^{3}\times
T^{2})$. Moreover, since $\mathcal{Z}(\pi_{1}(\Sigma(p,q,r)\times
T^{2}))\cong\mathbb{Z}^{3}$ we have a central extension

$1\longrightarrow\mathbb{Z}^{3}\longrightarrow\pi_{1}(T^{2}\times\Sigma(p,q,r))\longrightarrow\operatorname{Inn}\pi_{1}(\Sigma(p,q,r))\longrightarrow
1.$

By 7.1, the group $\operatorname{Out}(\pi_{1}(T^{2}\times\Sigma(p,q,r)))$ is
Minkowski. Therefore, 1.6 implies that
$\operatorname{Homeo}(T^{2}\times\Sigma(p,q,r)))$ is Jordan.

######  Remark 7.3.

A lot of the results about the Jordan property on $\operatorname{Homeo}(M)$ of
a closed connected manifold $M$ rely on the cohomology of $M$, for example in
[MiR19] its is proven that $\operatorname{Diff}(M)$ is Jordan if $M$ is an
integral homology sphere or $\chi(M)\neq 0$. However, 1.7 shows that the
Jordan property on $\operatorname{Homeo}(M)$ and $\operatorname{Diff}(M)$ does
not only depend on the cohomology of $M$ in general.

######  Remark 7.4.

We also note that $\operatorname{disc-sym}(N\times T^{2})<\operatorname{disc-
sym}(\Sigma(p,q,r)\times T^{2})<\operatorname{disc-sym}(S^{3}\times T^{2})$,
thus all three manifolds have different discrete degree of symmetry.

We have $\operatorname{disc-sym}(N\times T^{2})=2$, since $\operatorname{disc-
sym}(N\times T^{2})\leq 2$ by 1.6 and $N\times T^{2}$ admits a $T^{2}$ free
action. Similarly, $\operatorname{disc-sym}(\Sigma(p,q,r)\times T^{2})=3$,
since $\operatorname{disc-sym}(\Sigma(p,q,r)\times T^{2})\leq 3$ by 1.6 and
$\Sigma(p,q,r)\times T^{2}$ admits an action of $T^{3}$. Finally, $T^{4}$ acts
effectively on $S^{3}\times T^{2}$, hence $\operatorname{disc-sym}(S^{3}\times
T^{2})\geq 4$.

##  8\. Remarks when the discrete degree of symmetry is close to the dimension
of the aspherical manifold: Proof of 1.8

We have seen that if $M$ is a closed $n$-dimensional aspherical manifold with
$\mathcal{Z}\pi_{1}(M)$ finitely generated and
$\operatorname{Out}(\pi_{1}(M))$ Minkowski then $\operatorname{disc-sym}(M)=n$
if and only if $M\cong T^{n}$. An interesting question is whether there exist
similar rigidity results when $\operatorname{disc-sym}(M)$ is close to $n$,
for example $\operatorname{disc-sym}(M)=n-1$. The aim of this section is to
answer this question provided we have an additional hypothesis on
$\pi_{1}(M)$.

We start by recalling what happens in low dimensions. Let $M$ be a
$2$-dimensional closed connected aspherical manifold. If $M$ is orientable
then $M$ is either a torus $T^{2}$ and $\operatorname{tor-sym}(M)=2$ or $M$ is
a surface $\Sigma_{g}$ of genus $g\geq 2$ and $\operatorname{tor-sym}(M)=0$.
If $M$ is not orientable, then $M$ is either the Klein bottle $K$ and
$\operatorname{tor-sym}(M)=1$ or $M$ has a surface $\Sigma_{g}$ of genus
$g\geq 2$ as an orientable 2-cover, hence $\operatorname{tor-sym}(M)=0$.

If $M$ is a closed connected $3$-dimensional aspherical manifold with an
effective $S^{1}$ action, then $M$ falls in one of the following $4$ cases
(see [LR10, §14.4]):

  * 1.

$M\cong T^{3}$.

  * 2.

$M$ is homeomorphic to $K\times S^{1}$ or $SK$, the non-trivial principal
$S^{1}$-bundle over $K$.

  * 3.

$M\cong H/\Gamma$, where $H$ is the $3$-dimensional Heisenberg group and
$\Gamma$ is a lattice of $H$.

  * 4.

$\mathcal{Z}\pi_{1}(M)\cong\mathbb{Z}$ and
$\operatorname{Inn}\pi_{1}(M)\cong\pi_{1}(M)/\mathcal{Z}\pi_{1}(M)$ is
centreless.

Note that in all cases, we have a central extension

$1\longrightarrow\mathbb{Z}\longrightarrow\pi_{1}(M)\longrightarrow
Q\longrightarrow 1$

where $Q$ acts effectively, properly and cocompactly on $\mathbb{R}^{2}$.

In the first case $\operatorname{tor-sym}(M)=3$ and in the third and fourth
cases $\operatorname{tor-sym}(M)=1$. In the second case, it is clear that
$\operatorname{tor-sym}(K\times S^{1})=2$, so we focus on $SK$. One can see
that $\pi_{1}(SK)\cong\mathbb{Z}^{2}\rtimes_{\phi}\mathbb{Z}$, where
$\phi:\mathbb{Z}\longrightarrow\operatorname{GL}(2,\mathbb{Z})$ satisfies that

$\phi(1)=\left(\begin{matrix}0&1\\\ 1&0\end{matrix}\right).$

Since $\phi(1)^{2}=Id$, we have that $SK$ is a flat solvmanifold and therefore
$\operatorname{tor-sym}(SK)=\operatorname{rank}\mathcal{Z}\pi_{1}(SK)$.

######  Lemma 8.1.

Let $\phi:\mathbb{Z}\longrightarrow\operatorname{GL}(n,\mathbb{Z})$ be group
morphism such that $\phi(1)$ has finite order $a$. Then
$\mathcal{Z}(\mathbb{Z}^{n}\rtimes_{\phi}\mathbb{Z})=\operatorname{Fix}(\phi)\times
a\mathbb{Z}$, where
$\operatorname{Fix}(\phi)=\\{v\in\mathbb{Z}^{n}:\phi(1)v=v\\}$.

###### Proof.

A straightforward computation shows that given
$(v,t),(w,s)\in\mathbb{Z}^{n}\rtimes_{\phi}\mathbb{Z}$ we have
$(v,t)(w,s)=(w,s)(v,t)$ if and only if $(Id-\phi(s))v=(Id-\phi(t))w$. If
$(v,t)\in\mathcal{Z}(\mathbb{Z}^{n}\rtimes_{\phi}\mathbb{Z})$ then we have
$(Id-\phi(t))w=0$ for all $w\in\mathbb{Z}^{n}$, by taking $s=0$. Thus,
$\phi(t)=Id$ and hence $t\in a\mathbb{Z}$.

If we assume that $w=0$, then $(Id-\phi(s))v=0$ for all $s\in\mathbb{Z}$.
Thus, $v\in\operatorname{Fix}(\phi)$. Consequently,
$\mathcal{Z}(\mathbb{Z}^{n}\rtimes_{\phi}\mathbb{Z})\subseteq\operatorname{Fix}(\phi)\times
a\mathbb{Z}$. The other inclusion follows from the fact that if
$(v,t)\in\operatorname{Fix}(\phi)\times a\mathbb{Z}$ then
$(Id-\phi(s))v=0=(Id-\phi(t))w$ for any $s\in\mathbb{Z}$ and
$w\in\mathbb{Z}^{n}$. ∎

Applying the previous lemma to $\pi_{1}(SK)$, we obtain
$\operatorname{Fix}(\phi)=\langle(1,1)\rangle$ and therefore
$\mathcal{Z}\pi_{1}(SK)\cong\mathbb{Z}^{2}$ and $\operatorname{tor-
sym}(SK)=2$. Thus, $K$ and $SK$ are the two only $3$-dimensional aspherical
manifolds such that where $\operatorname{tor-sym}(M)=\dim(M)-1$. The next
proposition generalizes the previous fact to arbitrary dimension. It is
probably well-known to experts but we could not find a proof in the
literature, so we provide it for the sake of completeness.

######  Proposition 8.2.

Let $M$ be a closed aspherical manifold of dimension $n$. If
$\operatorname{tor-sym}(M)=n-1$ then $M\cong K\times T^{n-2}$ or $M\cong
T^{n-3}\times SK$. In particular, $M$ is always a non-orientable flat
solvmanifold.

###### Proof.

Let $H\leq T^{n-1}$ be the isotropy subgroup of the principal orbit of the
action. Since $T^{n-1}$ is abelian and its action on $M$ is effective we can
conclude that $H$ is trivial. Therefore, the principal orbits of the action
have dimension $n-1$. In this case we say that we have a cohomogenity one
action. The next theorem describes the cohomogenity one actions.

######  Theorem 8.3.

[GGZ18, Theorem A] Let $M$ be a $n$-dimensional closed connected manifold with
a cohomogenity one action of a compact Lie group $G$. Let $H$ be the isotropy
subgroup of a principal orbit. Then we have one of these two options:

  * 1.

The quotient $M/G\cong S^{1}$. Then $M$ is equivariantly homeomorphic to the
total space of a fiber bundle $\pi:M\longrightarrow S^{1}$ is a fiber bundle
with fiber $G/H$. The action does not have exceptional orbits.

  * 2.

The quotient $M/G\cong[-1,1]$. Then $M$ is the union of two fiber bundles over
the two singular orbits with isotropy subgroups $K_{+}$ and $K_{-}$ whose
fibers are cones over spheres or the Poincaré homology sphere. More
explicitly,

$M=G\times_{K_{-}}C(K_{-}/H)\cup_{G/H}G\times_{K_{+}}C(K_{+}/H)$

where $C(K_{\pm}/H)$ denotes the cone over $K_{\pm}/H$, which are spheres or
Poincaré homology spheres. The exceptional orbits $G/K_{\pm}$ correspond to
the preimages of $\pm 1\in[-1,1]$.

If the cohomogenity one action of $T^{n-1}$ on $M$ does not have exceptional
orbits, then the action is free and the orbit map $\pi:M\longrightarrow S^{1}$
is a principal $T^{n-1}$-bundle. Since the base is $S^{1}$ the principal
bundle is trivial and therefore $M\cong T^{n}$ and $\operatorname{tor-
sym}(M)=n$, which is not possible. Therefore the action has exceptional
orbits.

Since $M$ is aspherical, the evaluation map $ev_{x}:T^{n-1}\longrightarrow M$
such that $ev_{x}(g)=gx$ for all $g\in T^{n-1}$ induces an injective group
morphism ${ev_{x}}_{*}:\pi_{1}(T^{n-1})\longrightarrow\pi_{1}(M)$ for any
$x\in M$ ([LR10, Lemma 3.1.11]). Consequently, all the isotropy subgroups of
the action are discrete. The quotients $K_{\pm}/H\cong K_{\pm}$ are
homeomorphic to a sphere or a Poincaré homology sphere. Since they are also
discrete we obtain that $K_{\pm}\cong S^{0}\cong\mathbb{Z}/2$. Then

$M=T^{n-1}\times_{\mathbb{Z}/2}I_{+}\cup_{T^{n-1}}T^{n-1}\times_{\mathbb{Z}/2}I_{-},$

where $I_{+}=[0,1]$ and $I_{-}=[-1,0]$. Let
$Z_{+}=\pi_{1}(T^{n-1}\times_{\mathbb{Z}/2}I_{+})$ and
$Z_{-}=\pi_{1}(T^{n-1}\times_{\mathbb{Z}/2}I_{-})$. The principal
$\mathbb{Z}/2$-bundles $T^{n-1}\times I_{\pm}\longrightarrow
T^{n-1}\times_{\mathbb{Z}/2}I_{\pm}$ induce two short exact sequences of
fundamental groups

$1\xrightarrow{}\mathbb{Z}^{n-1}\xrightarrow{i_{\pm}}Z_{\pm}\xrightarrow{}\mathbb{Z}/2\xrightarrow{}1.$

Note that $Z_{+}$ and $Z_{-}$ are isomorphic to $\mathbb{Z}^{n-1}$. By the
Seifert-van Kampen theorem, $\pi_{1}(M)=Z_{+}*_{\mathbb{Z}^{n-1}}Z_{-}$ where
the amalgamated product is induced by the inclusions
$i_{\pm}:\mathbb{Z}^{n-1}\longrightarrow Z_{\pm}$, which we are going to
describe explicitly.

Firstly, there exist primitive elements $\alpha_{\pm}\in Z_{\pm}$ such that
$\alpha_{\pm}\notin i_{\pm}(\mathbb{Z}^{n-1})$. Furthermore, there exist two
$x_{\pm}\in\mathbb{Z}^{n-1}$ such that $i_{\pm}(x_{\pm})=2\alpha_{\pm}$. We
have two possibilities, that $x_{+}=x_{-}$ or that $x_{+}\neq x_{-}$.

Firstly, assume that $x_{+}=x_{-}$. If we denote $x_{+}=x_{-}=x$ then we can
choose generators $\\{x,y_{1},...,y_{n-2}\\}$ of $\mathbb{Z}^{n-1}$ such that
$\\{\alpha_{+},i_{+}(y_{1}),...,i_{+}(y_{n-2})\\}$ generates $Z_{+}$ and
$\\{\alpha_{-},i_{-}(y_{1}),...,i_{-}(y_{n-2})\\}$ generates $\mathbb{Z}_{-}$.
Thus,

$\pi_{1}(M)\cong\langle\alpha_{+},\alpha_{-}|\alpha_{+}^{2}=\alpha_{-}^{2}\rangle\times\mathbb{Z}^{n-2}\cong\pi_{1}(K)\times\mathbb{Z}^{n-2}=\pi_{1}(K\times
T^{n-2}).$

Since $\pi_{1}(M)$ is the fundamental group of a flat manifold and the Borel
conjecture holds for these groups (see [BL12]) we conclude that $M\cong
K\times T^{n-2}$.

We now assume that $x_{+}\neq x_{-}$. As before, we can choose a generator set
$\\{x_{+},x_{-},y_{1},...,y_{n-3}\\}$ of $\mathbb{Z}^{n-1}$ such that
$\\{\alpha_{+},i_{+}(x_{-}),i_{+}(y_{1}),...,i_{+}(y_{n-3})\\}$ generates
$Z_{+}$ and $\\{i_{-}(x_{+}),\alpha_{-},i_{-}(y_{1}),...,i_{-}(y_{n-3})\\}$
generates $Z_{-}$. Thus,

$\pi_{1}(M)\cong\langle\alpha_{+},i_{+}(x_{-}),i_{-}(x_{+}),\alpha_{-}|\alpha_{+}^{2}=i_{-}(x_{+}),\alpha_{-}^{2}=i_{+}(x_{-}),[\alpha_{+},i_{+}(x_{-})]=1,[\alpha_{-},i_{-}(x_{+})]=1\rangle\times\mathbb{Z}^{n-3}.$

The presentation of the first factor $\Lambda$ can be rearranged to obtain a
new presentation

$\displaystyle\Lambda\cong$ $\displaystyle\langle
a,b|[a,b^{2}]=1,[a^{2},b]=1\rangle$ $\displaystyle\cong$ $\displaystyle\langle
a,b,c,d|c=ab,d=ba,ac=bd,cb=da\rangle$ $\displaystyle\cong$
$\displaystyle\langle a,c,d|[c,d]=1,aca^{-1}=d,ada^{-1}=c\rangle.$

With the last presentation, we can define an isomorphism
$f:\Lambda\longrightarrow\mathbb{Z}^{2}\rtimes_{\phi}\mathbb{Z}$ such that
$f(a)=(0,0,1)$, $f(b)=(1,0,0)$ and $f(c)=(0,1,0)$. Consequently,
$\pi_{1}(M)\cong\pi_{1}(SK\times T^{n-3})$. Since $\pi_{1}(M)$ is the
fundamental group of a flat manifold and the Borel conjecture holds for these
groups we conclude that $M\cong SK\times T^{n-3}$.

∎

######  Corollary 8.4.

Let $M$ be a closed connected aspherical manifold of dimension $n$. If
$M\ncong T^{n}$ and $M$ is orientable then $\operatorname{tor-sym}(M)\leq
n-2$.

######  Remark 8.5.

The hypothesis of $M$ being aspherical is essential. For example
$\operatorname{tor-sym}(S^{2})=1$ and $S^{2}\ncong K\times
T^{n-2},T^{n-3}\times SK$.

######  Theorem 8.6.

Let $M$ be a closed connected aspherical manifold such that
$\mathcal{Z}(\pi_{1}(M))$ is finitely generated and
$\operatorname{Out}(\pi_{1}(M))$ is Minkowski. Assume that
$\operatorname{Inn}(\pi_{1}(M))$ has an element of infinite order. Then
$\operatorname{disc-sym}(M)=n-1$ if and only if $M\cong K\times T^{n-2}$ or
$M\cong T^{n-3}\times SK$.

###### Proof.

Because of the hypothesis on $\pi_{1}(M)$ we know that
$\mathcal{Z}(\pi_{1}(M))\cong\mathbb{Z}^{n-1}$. Let
$x\in\operatorname{Inn}\pi_{1}(M)$ be an element of infinite order and
consider the commutative diagram

${1}$${\mathcal{Z}\pi_{1}(M)}$${p^{-1}(\langle x\rangle)}$${\langle
x\rangle}$${1}$${1}$${\mathcal{Z}\pi_{1}(M)}$${\pi_{1}(M)}$${\operatorname{Inn}\pi_{1}(M)}$${1}$$\scriptstyle{Id}$$\scriptstyle{p}$

where the vertical arrows are inclusion maps. The upper short exact sequence
is also central and therefore it is classified by
$H^{2}(\mathbb{Z},\mathbb{Z}^{n-1})=0$. In consequence, $p^{-1}(\langle
x\rangle)\cong\mathbb{Z}^{n}$. We consider now the covering
$q:\tilde{M}/p^{-1}(\langle x\rangle)\longrightarrow M$. Note that
$H^{*}(\tilde{M}/p^{-1}(\langle x\rangle),\mathbb{Z})\cong
H^{*}(\mathbb{Z}^{n},\mathbb{Z})$ and therefore
$H^{n}(\tilde{M}/p^{-1}(\langle x\rangle),\mathbb{Z})$ is not trivial.
Consequently, $\tilde{M}/p^{-1}(\langle x\rangle)$ is a closed connected
aspherical manifold and $\tilde{M}/p^{-1}(\langle x\rangle)\cong T^{n}$. Since
$M$ and $\tilde{M}/p^{-1}(\langle x\rangle)$ are compact, we obtain that $q$
is a finite covering map and therefore $p^{-1}(\langle
x\rangle)\leq\pi_{1}(M)$ has finite index and $\pi_{1}(M)$ is the fundamental
group of a flat manifold. Thus, we have that $\operatorname{disc-
sym}(M)=\operatorname{tor-sym}(M)$ by 1.10 and the result follows from 8.2. ∎

It is an interesting question to know whether $\operatorname{Inn}\pi_{1}(M)$
always has an element of infinite order when $M$ is closed aspherical manifold
not homeomorphic to a torus and such that $\mathcal{Z}\pi_{1}(M)$ is finitely
generated. Note that $\operatorname{Inn}\pi_{1}(M)$ cannot be finite by 3.3
unless $M$ is homeomorphic to a torus. Thus, if $\operatorname{Inn}\pi_{1}(M)$
does not contain elements of infinite order then $\pi_{1}(M)$ is abelian or
$\operatorname{Inn}\pi_{1}(M)$ is infinite periodic.

We present some evidences that support that the answer to the question is
affirmative. Firstly, let $\Gamma$ and $\Lambda$ be two groups. Assume that we
have a surjective group morphism $p:\Gamma\longrightarrow\Lambda$. Then there
is an induced surjective group morphism
$p^{\prime}:\operatorname{Inn}\Gamma\longrightarrow\operatorname{Inn}\Lambda$
which sends a conjugation $c_{\gamma}$ to $c_{p(\gamma)}$. Thus, if
$\operatorname{Inn}\Lambda$ has an element of infinite order, then
$\operatorname{Inn}\Gamma$ also has an element of infinite order.

Assume now that we have an inclusion $i:\Gamma\longrightarrow\Lambda$ instead.
Then there is an induced injective group morphism
$i^{\prime}:\operatorname{Inn}\Gamma\longrightarrow\operatorname{Inn}\Lambda$
which sends a conjugation $c_{\gamma}$ to $c_{i(\gamma)}$. In particular, if
$\operatorname{Inn}\Gamma$ has an element of infinite order, then
$\operatorname{Inn}\Gamma$ also has an element of infinite order. We can
deduce the following corollary from these observations:

######  Corollary 8.7.

Let $M$ be a closed aspherical manifold, then:

  * (1)

If $M^{\prime}\longrightarrow M$ is a covering and
$\operatorname{Inn}\pi_{1}(M^{\prime})$ has an element of infinite order then
$\operatorname{Inn}\pi_{1}(M)$ has an element of infinite order.

  * (2)

Suppose that we have a fibration of closed connected aspherical manifolds
$M^{\prime}\longrightarrow M\longrightarrow M^{\prime\prime}$. If
$\operatorname{Inn}\pi_{1}(M^{\prime})$ or
$\operatorname{Inn}\pi_{1}(M^{\prime\prime})$ have an element of infinite
order, then $\operatorname{Inn}\pi_{1}(M)$ has an element of infinite order.

###### Proof.

In the first case, we use that $\pi_{1}(M^{\prime})\leq\pi_{1}(M)$ and in the
second case the short exact sequence
$1\longrightarrow\pi_{1}(M^{\prime})\longrightarrow\pi_{1}(M)\longrightarrow\pi_{1}(M^{\prime\prime})\longrightarrow
1$ together with the observations above. ∎

Note that if $\Gamma$ is a non-abelian polycyclic group or $\Gamma$ is
torsion-free centreless then $\operatorname{Inn}\Gamma$ has elements of
infinite order. Thus, closed aspherical locally homogeneous spaces and closed
aspherical manifolds whose fundamental group is hyperbolic have an infinite
order element in the inner automorphism group of their fundamental group.
Moreover, any closed aspherical manifold $M$ not homeomorphic to a torus
constructed using fibrations of these two classes of aspherical manifolds will
have an element of infinite order in $\operatorname{Inn}\pi_{1}(M)$.

There are cohomological restrictions to $\operatorname{Inn}\pi_{1}(M)$ being
infinite periodic. Assume that $M$ is a closed connected aspherical manifold
such that $\operatorname{Inn}\pi_{1}(M)$ is infinite periodic. Then
$H^{1}(\operatorname{Inn}\pi_{1}(M),\mathbb{Z})=\operatorname{Hom}(\operatorname{Inn}\pi_{1}(M),\mathbb{Z})$
is trivial and the inflation-restriction exact sequence becomes

$1\longrightarrow H^{1}(\pi_{1}(M),\mathbb{Z})\longrightarrow
H^{1}(\mathcal{Z}\pi_{1}(M),\mathbb{Z})\longrightarrow
H^{2}(\operatorname{Inn}\pi_{1}(M),\mathbb{Z})\longrightarrow
H^{2}(\pi_{1}(M),\mathbb{Z})$

where we see $\mathbb{Z}$ as a trivial $\pi_{1}(M)$-module. In particular, if
$\operatorname{rank}H^{1}(M,\mathbb{Z})>\operatorname{rank}\mathcal{Z}\pi_{1}(M)$
then $\operatorname{Inn}\pi_{1}(M)$ has elements of infinite order.

Moreover, since $H^{2}(\pi_{1}(M),\mathbb{Z})\cong H^{2}(M,\mathbb{Z})$ and
$H^{1}(\mathcal{Z}\pi_{1}(M),\mathbb{Z})\cong
H^{1}(T^{\operatorname{rank}\mathcal{Z}\pi_{1}(M)},\mathbb{Z})$, the group
$H^{2}(\operatorname{Inn}\pi_{1}(M),\mathbb{Z})$ needs to be finitely
generated. Using the results in [AA18], we can conclude that
$\operatorname{Inn}\pi_{1}(M)$ cannot be isomorphic to a free Burnside group
$B(a,b)$ with $b\geq 665$ odd. The results in [Che21] imply that
$\operatorname{Inn}\pi_{1}(M)$ cannot be an infinite periodic 2-group of
bounded exponent. Finally, we note that $\operatorname{Inn}\pi_{1}(M)$ is a
finitely presented group and it is an open question if there exists finitely
presented infinite periodic groups (this question is the Burnside problem for
finitely presented groups, see [Sap07, pg. 3])

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