# Topological rigidity of tori and finite group actions on hypertoral
manifolds

Ignasi Mundet i Riera  Facultat de Matemàtiques i Informàtica
Universitat de Barcelona
Gran Via de les Corts Catalanes 585
08007 Barcelona
Spain<EMAIL_ADDRESS>
###### Abstract.

Let $X$ be a closed, connected and oriented topological $n$-dimensional
manifold. Suppose that:

  1. (1)

$X$ supports an effective action of $({\mathbb{Z}}/r)^{n}$ for arbitrarily
large values of $r$,

  2. (2)

the fundamental group $\pi_{1}(X)$ is virtually solvable, and

  3. (3)

there exists a map $X\to T^{n}$ of nonzero degree.

We then prove that $X$ is homeomorphic to $T^{n}$. Without using the
assumption on $\pi_{1}(X)$ we prove that the cohomology ring
$H^{*}(X;{\mathbb{Z}})$ is isomorphic to $H^{*}(T^{n};{\mathbb{Z}})$. We also
prove that any closed connected and oriented $n$-dimensional manifold $X$
admitting a map to $T^{n}$ of nonzero degree has Jordan homeomorphism group,
and we study for such manifolds the maximal number $k$ such that $X$ admits
effective actions of $({\mathbb{Z}}/r)^{k}$ for arbitrarily large $r$.
Finally, we prove that if $X$ is a compact and connected Kaehler manifold of
real dimension $n$ and $X$ supports effective holomorphic actions of
$({\mathbb{Z}}/r)^{m}$ for arbitrarily large $r$, then $m\leq n$, and if $m=n$
then $X$ is biholomorphic to a complex torus.

###### 2010 Mathematics Subject Classification:

57S17,54H15

This research was partially supported by the grant PID2019-104047GB-I00 from
the Spanish Ministeri de Ciència i Innovació.

##  1\. Introduction

###  1.1. Main theorems

Can one characterize a manifold by the collection of all finite groups that
act effectively on it? Of course not: there are plenty of examples of
asymmetric manifolds, i.e., manifolds which do not admit any effective action
of a finite group [17]. It is actually expected that in some sense most
manifolds are asymmetric [45]. But perhaps the question makes more sense if we
consider manifolds supporting actions of arbitrarily large finite groups. Here
”large” has to be understood in an appropriate sense and not merely as having
many elements, because assuming that a manifold supports an effective action
of the circle (which implies it supports actions of arbitrarily large finite
cyclic groups) is still very far from implying that the manifold can be
characterized by the finite groups which act effectively on it.

The $n$-dimensional torus $T^{n}$ supports a continuous effective action of
$({\mathbb{Z}}/r)^{n}$ for arbitrarily large integers $r$, given by rotations
of angle a multiple of $2\pi/r$ in each of the factors of
$T^{n}=S^{1}\times\dots\times S^{1}$. Our main result is a sort of converse to
this statement, that may be viewed as a partial positive answer to the
previous question. Suppose that a closed, connected and oriented manifold $X$
supports effective actions111Here and everywhere all actions of groups on a
topological manifold are implicitly assumed to be continuous. of
$({\mathbb{Z}}/r)^{n}$ for arbitrarily large integers $r$, meaning that there
exists a sequence of integers $r_{i}\to\infty$ such that $X$ admits an
effective action of $({\mathbb{Z}}/r_{i})^{n}$ for every $i$ (we emphasize
that we are not assuming any compatibility between the actions of
$({\mathbb{Z}}/r_{i})^{n}$ and $({\mathbb{Z}}/r_{j})^{n}$ for $i\neq j$).
Under this assumption, we prove that either $X$ is homeomorphic to $T^{n}$, or
$X$ is in some sense very different from $T^{n}$. We express this dichotomy in
terms of a strengthening of the topological rigidity of tori for maps from $X$
to $T^{n}$.

Topological rigidity of tori is the statement that if $X$ is a closed
connected manifold then any homotopy equivalence $X\to T^{n}$ is homotopic to
a homeomorphism. If $n\leq 2$ this is a consequence of the classification of
compact connected manifolds of dimensions at most $2$. It was proved for
$n\geq 5$ by Hsiang and Wall [29], for $n=4$ by Freedman [24, §11.5] (see also
[4]), and in dimension $n=3$ it is a consequence of Thurston’s geometrisation
conjecture proved by Perelman (see [6, 41] for proofs of the geometrisation
conjecture, and [33, §5] for the proof that geometrisation implies topological
rigidity of $T^{3}$). Topological rigidity of tori is a particular case of
Borel’s conjecture (see e.g. [37, §3] for a survey, [22, 23] for the case of
Riemannian manifolds with nonpositive curvature, and also the recent textbook
[15]).

In our main theorem, assuming a priori that $X$ supports effective actions of
$({\mathbb{Z}}/r)^{n}$ for arbitrarily large integers $r$, we replace homotopy
equivalence by the weaker requirement that the map has nonzero degree and that
$\pi_{1}(X)$ is virtually solvable. Without the assumption on $\pi_{1}(X)$ we
prove that $X$ is cohomologically indistinguishable from $T^{n}$.

######  Theorem 1.1.

Let $X$ be a closed, connected and oriented $n$-dimensional topological
manifold supporting an effective action of $({\mathbb{Z}}/r)^{n}$ for
arbitrarily large integers $r$. Suppose that there exists a map $\phi:X\to
T^{n}$ of nonzero degree. We have:

  1. (1)

There is an isomorphism of rings $H^{*}(X;{\mathbb{Z}})\simeq
H^{*}(T^{n};{\mathbb{Z}})$.

  2. (2)

If $\pi_{1}(X)$ is virtually solvable then $X$ is homeomorphic to $T^{n}$.

The following result complements the previous theorem in the smooth category.

######  Theorem 1.2.

Let $n\neq 4$ be a natural number. Let $X$ be a smooth manifold homeomorphic
to $T^{n}$. We have:

  1. (1)

$X$ supports effective smooth actions of $({\mathbb{Z}}/r)^{n}$ for
arbitrarily large values of $r$.

  2. (2)

There exists a number $\delta(n)$ (depending on $n$ but not on $X$) such that
if $X$ supports an effective smooth action of $({\mathbb{Z}}/m\delta(n))^{n}$
for some nonzero integer $m$ then $X$ is diffeomorphic to $T^{n}$.

Statement (2) of the previous theorem is related to many other existing
results in the literature showing that homeomorphic but not diffeomorphic
manifolds need not support smooth effective actions of the same finite or
compact groups (see [30] and the references therein for systematic results on
the case of the spheres; see also [13] for a recent paper studying analogous
questions).

Combining Theorems 1.1 and 1.2 we immediately obtain:

######  Corollary 1.3.

Let $n\neq 4$ be a natural number. Let $X$ be a closed, connected and oriented
$n$-dimensional smooth manifold supporting an effective action of
$({\mathbb{Z}}/r)^{n}$ for every natural number $r$. If there exists a
continuous map $\phi:X\to T^{n}$ of nonzero degree and $\pi_{1}(X)$ is
virtually solvable then $X$ is diffeomorphic to $T^{n}$.

A closed, connected and oriented $n$-dimensional manifold $X$ admitting a
continuous map $\phi:X\to T^{n}$ of degree $\pm 1$ is called hypertoral  in
[47, 48]. If the degree of $\phi$ is a nonzero integer, then we say that $X$
is rationally hypertoral. Equivalently, $X$ is rationally hypertoral if it
admits classes $\alpha_{1},\dots,\alpha_{n}\in H^{1}(X;{\mathbb{Z}})$ such
that $\alpha_{1}\cup\dots\cup\alpha_{n}\neq 0$, because $T^{n}=(S^{1})^{n}$
and $S^{1}$ is an Eilenberg-MacLane space $K({\mathbb{Z}},1)$. And similarly
$X$ is hypertoral if it admits classes $\alpha_{1},\dots,\alpha_{n}\in
H^{1}(X;{\mathbb{Z}})$ such that $\alpha_{1}\cup\dots\cup\alpha_{n}$ is a
generator of $H^{n}(X;{\mathbb{Z}})$. Using this characterisation it follows
from statement (1) in Theorem 1.1 that if a rationally hypertoral $n$-manifold
supports effective actions of $({\mathbb{Z}}/r)^{n}$ for arbitrarily large
values of $r$ then the manifold must be hypertoral.

There are plenty of examples of hypertoral manifolds. For example, if $Y$ is
any closed connected and oriented $n$-manifold, then the connected sum
$Y\sharp T^{n}$ is a hypertoral manifold, as the map $Y\sharp T^{n}\to T^{n}$
that identifies all points coming from $Y$ has degree $1$. There are also many
rationally hypertoral manifolds which are not hypertoral.

######  Example 1.4.

Let $\pi:Y\to T^{2}$ be a degree $2$ covering ramified at $2$ points. Then $Y$
is an orientable closed and connected surface of genus $2$. Let $\sigma:Y\to
Y$ be the nontrivial automorphism of $\pi$, and let $X$ be the mapping torus
of $\sigma$. In Section 14 we prove that $X$ is rationally hypertoral but not
hypertoral. Consequently, Theorem 1.1 implies the existence of some
$r_{0}\in{\mathbb{N}}$ such that if $X$ supports an effective action of
$({\mathbb{Z}}/r)^{3}$ for some integer $r$ then necessarily $|r|\leq r_{0}$.

Both Example 1.4 and the arguments in Section 14 can easily be generalized to
yield infinitely many examples in any dimension bigger than $2$ of rationally
hypertoral manifolds which are not hypertoral.

###  1.2. Jordan property, bounds on stabilizers, and almost asymmetric
manifolds

The tools used to prove Theorem 1.1 lead to other results of finite group
actions on rationally hypertoral manifolds. Recall that a group
${\mathcal{G}}$ is said to be Jordan if there exists a constant $C$ such that
every finite subgroup $G\leq{\mathcal{G}}$ has an abelian subgroup $A\leq G$
satisfying $[G:A]\leq C$. The following result extends to the topological
category the first part of [42, Theorem 1.4], and it also partially extends
[54, Corollary 1.7].

######  Theorem 1.5.

Let $X$ be a rationally hypertoral manifold. Then the homeomorphis group of
$X$ is Jordan.

Arguing as in [43, Corollary 4], the previous theorem has the following
implication, which was originally proved by Burghelea and Schultz in [12,
Theorem A].

######  Corollary 1.6.

Let $X$ be a rationally hypertoral manifold. If a compact connected Lie group
$G$ acts effectively and continuously on $X$ then $G$ is abelian.

If $X$ is a set supporting an action of a group $G$ we denote
$\operatorname{Stab}(X,G)=\\{G_{x}\mid x\in X\\}$ the set of stabilizers of
points in $X$. The following result gives a positive answer to [18, Question
1.8] for rationally hypertoral manifolds.

######  Theorem 1.7.

Let $X$ be a rationally hypertoral manifold. There exists a constant $C$ such
that every finite group $G$ acting on $X$ has a subgroup $G_{0}\leq G$
satisfying $[G:G_{0}]\leq C$ and $|\operatorname{Stab}(X,G)|\leq C$.

We say that a closed topological manifold $X$ is almost asymmetric if there
exist a number $C$ such that any finite group $G$ acting effectively on $X$
satisfies $|G|\leq C$. The following extends to the continuous category the
second part of [42, Theorem 1.4]:

######  Theorem 1.8.

If a rationally hypertoral manifold $X$ satisfies $\chi(X)\neq 0$ then $X$ is
almost asymmetric. More precisely, there exists a number $C$ depending only on
$\sum_{j}b_{j}(X)$, where $b_{j}(X)=\dim_{{\mathbb{Q}}}H_{j}(X;{\mathbb{Q}})$
is the $j$-th rational Betti number of $X$, such that any finite group acting
effectively on $X$ has at most $C$ elements.

Using the previous theorem one can give a simple construction showing that the
question at the beginning of the paper has a negative answer. Let
$X_{0}=T^{4}\sharp T^{4}\sharp T^{4}$. Then $X_{0}$ is hypertoral and
$\chi(X_{0})=-4$. Fix a nonzero class $c\in
H_{1}(X_{0};{\mathbb{Z}})\simeq{\mathbb{Z}}^{12}$ which is not a nontrivial
integral multiple. For any prime $p$ choose an embedded circle $\gamma\subset
X_{0}$ representing $p$ times the class $c$, and denote by $X_{p}$ the result
of performing surgery on $X_{0}$ along $\gamma$ (i.e., removing a tubular
neighborhood of $\gamma$, which will be diffeomorphic to $S^{1}\times D^{3}$,
and gluing back $D^{2}\times S^{2}$ using the identification of boundaries
$\partial(S^{1}\times D^{3})=S^{1}\times S^{2}=\partial(D^{2}\times S^{2})$).
We have $H_{1}(X_{p};{\mathbb{Z}})\simeq{\mathbb{Z}}/p\oplus{\mathbb{Z}}^{11}$
and $\chi(X_{p})=-2$. Furthermore, by Mayer–Vietoris the sum of the rational
Betti numbers of $X_{p}$ can be bounded by a constant independent of $p$. So
the previous theorem combined with the pigeonhole principle gives an infinite
sequence of pairwise non homeomorphic $4$-dimensional manifolds each of which
supports effective actions of the same collection of (isomorphism classes of)
finite groups.

###  1.3. Discrete degree of symmetry

For any closed topological manifold $X$ we denote by $\mu(X)$ the set of all
natural numbers $m$ such that $X$ supports effective actions of
$({\mathbb{Z}}/r)^{m}$ for arbitrarily large $r$. Note that $\mu(X)$ may be
empty. Define the discrete degree of symmetry of $X$ to be

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

This is always finite and hence well defined thanks to a theorem of Mann and
Su, see [38, Theorem 2.5]. This invariant is related to the previous notion in
the following terms:

######  Lemma 1.9.

For any closed manifold $X$ we have $\operatorname{disc-sym}(X)=0$ if and only
if $X$ is almost asymmetric.

###### Proof.

Clearly if $X$ is almost asymmetric then $\operatorname{disc-sym}(X)=0$.
Conversely, suppose that $\operatorname{disc-sym}(X)=0$. Then there is a
natural number $C$ such that any finite cyclic group acting effectively on $X$
has at most $C$ elements. Let $G$ be a finite group acting effectively on $X$.
For any prime divisor $p$ of $|G|$ there is a cyclic subgroup of $G$ with $p$
elements, so all prime divisors $p$ of $|G|$ satisfy $p\leq C$. By [38,
Theorem 2.5] there exists a natural number $d$ such that for any prime $p$ and
any natural number $m$ such that $({\mathbb{Z}}/p)^{m}$ acts effectively on
$X$ we have $m\leq d$. Fix some prime $p$ dividing $|G|$ and let $H$ be a
Sylow $p$-subgroup of $G$. Choose a maximal normal abelian subgroup $A\leq H$.
Then $A$ is isomorphic to the product of at most $d$ finite cyclic groups.
Each of these cyclic groups has at most $C$ elements, so $|A|\leq C^{d}$. By
[46, §5.2.3] the action of $H$ by conjugation on $A$ induces an injection
$H/A\hookrightarrow\operatorname{Aut}A$. Hence,
$|H|\leq|A|\cdot|\operatorname{Aut}A|\leq C^{d}(C^{d})!$. Repeating the
previous argument for all primes dividing $|G|$ we obtain the bound
$|G|\leq[C^{d}(C^{d})!]^{C}$. ∎

Recall that the rank of a finite group $G$ is the minimal size of a generating
subset of $G$; we denote it by $\operatorname{rk}G$. One can relate the
discrete degree of symmetry to actions of arbitrary finite abelian groups as
follows.

######  Lemma 1.10.

Let $X$ be a closed manifold and let $k$ be a natural number. The following
two conditions are equivalent.

  1. (1)

$\operatorname{disc-sym}(X)\leq k$;

  2. (2)

there exists some constant $C$ such that any finite abelian group acting
effectively on $X$ has a subgroup $A^{\prime}\leq A$ satisfying
$[A:A^{\prime}]\leq C$ and $\operatorname{rk}A^{\prime}\leq k$.

###### Proof.

If $\operatorname{disc-sym}(X)\leq k$ then there exists a natural number
$r_{0}$ such that if $({\mathbb{Z}}/r)^{k+1}$ acts effectively on $X$ then
$r\leq r_{0}$. Let $A$ be a finite abelian group acting effectively on $X$.
For any prime $p$ let $A_{p}\leq A$ denote the subgroup of elements whose
order is a power of $p$. Then $A\simeq\prod_{p}A_{p}$ and
$\operatorname{rk}A=\max_{p}\operatorname{rk}A_{p}$. For any prime $p$
satisfying $p>r_{0}$ we have $\operatorname{rk}A_{p}\leq k$, because if
$\operatorname{rk}A_{p}=s$ then the subgroup of $A_{p}$ consisting of elements
whose order divides $p$ is isomorphic to $({\mathbb{Z}}/p)^{s}$. Now suppose
that $p$ is a prime satisfying $p\leq r_{0}$ and let
$s=\operatorname{rk}A_{p}$. We are going to define a subgroup
$A_{p}^{\prime}\leq A_{p}$. If $s\leq k$ then we set $A_{p}^{\prime}:=A_{p}$.
Suppose now that $s>k$. By [38, Theorem 2.5] we have $s\leq C_{0}$ for some
number $C_{0}$ depending only on $X$. Choose an isomorphism
$\phi:A_{p}\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}{\mathbb{Z}}/p^{e_{1}}\times\dots\times{\mathbb{Z}}/p^{e_{s}}$,
where $e_{1},\dots,e_{s}$ are natural numbers written in nonincreasing order.
By the definition of $r_{0}$ we have $p^{e_{j}}\leq r_{0}$ for every $j>k$. So
if we set

$A_{p}^{\prime}=\phi^{-1}({\mathbb{Z}}/p^{e_{1}}\times\dots\times{\mathbb{Z}}/p^{e_{k}}\times\\{0\\}\times\dots\times\\{0\\})$

then $[A_{p}:A_{p}^{\prime}]\leq r_{0}^{C_{0}}$ and
$\operatorname{rk}A_{p}^{\prime}\leq k$. Let $A^{\prime}=\prod_{p\leq
r_{0}}A_{p}^{\prime}\times\prod_{p>r_{0}}A^{p}$. Then
$\operatorname{rk}A^{\prime}\leq k$ and $[A:A^{\prime}]\leq
C:=(r_{0}^{C_{0}})^{r_{0}}$. This proves (1) $\Rightarrow$ (2). The converse
implication follows from Lemma 6.1 below, the fact the rank does not increase
when passing from a finite abelian group to a subgroup, and the formula
$\operatorname{rk}({\mathbb{Z}}/a)^{b}=b$, which is valid for any natural
numbers $a\geq 2$ and $b$. ∎

###  1.4. Bounds on the discrete degree of symmetry

In the proof of statement (2) in Theorem 1.1 we will reduce the general case
to the case in which $\pi_{1}$ is solvable using the following result.

######  Theorem 1.11.

Let $X$ be a closed connected manifold and let $X^{\prime}\to X$ be a finite
covering. We have $\operatorname{disc-sym}(X^{\prime})\geq\operatorname{disc-
sym}(X)$.

The inequality in Theorem 1.11 can be strict in some cases, as the following
theorem proves. Here and in the rest of the paper we identify $T^{n}$ with
$({\mathbb{R}}/{\mathbb{Z}})^{n}$, so we use additive notation for the group
structure on $T^{n}$.

######  Theorem 1.12.

Fix natural numbers $k,n$ satisfying $1\leq k\leq n-1$. Consider the free
involution $\sigma:T^{n}\to T^{n}$ defined by
$\sigma(x_{1},\dots,x_{n})=(x_{1}+1/2,\dots,x_{k}+1/2,-x_{k+1},\dots,-x_{n})$.
Let $X^{\prime}=T^{n}$ and let $X=T^{n}/\sigma$. The natural projection
$\rho:X^{\prime}\to X$ is a covering map and we have $\operatorname{disc-
sym}X^{\prime}=n$ and $\operatorname{disc-sym}X=k$.

For example, setting $n=2$ and $k=1$ the manifold $X$ is the Klein bottle and
$X^{\prime}$, the $2$-torus, is the orientation $2$-cover of $X$.

The following theorem gives bounds on $\operatorname{disc-sym}(X)$ for
rationally hypertoral $X$.

######  Theorem 1.13.

Let $X$ be a rationally hypertoral manifold. We have:

  1. (1)

$\operatorname{disc-sym}(X)\leq\dim X$;

  2. (2)

suppose that $X$ is homeomorphic to $Y\times(Z\sharp Z^{\prime})$, where
$Y,Z,Z^{\prime}$ are closed connected topological manifolds satisfying $\dim
Z=\dim Z^{\prime}>1$; if neither $Z$ nor $Z^{\prime}$ are integral homology
spheres, then $\operatorname{disc-sym}(X)\leq\dim Y$.

For example, if $Z$ is a closed connected topological $l$-manifold and
$H_{*}(Z;{\mathbb{Z}})\not\simeq H^{*}(S^{l};{\mathbb{Z}})$ then
$\operatorname{disc-sym}(T^{k}\times(T^{l}\sharp Z))\leq k$. Since
$T^{k}\times(T^{l}\sharp Z))$ supports actions of $({\mathbb{Z}}/r)^{k}$ for
every $r$ (inherited from the actions on $T^{k}$) we conclude that
$\operatorname{disc-sym}(T^{k}\times(T^{l}\sharp Z))=k$. So among rationally
hypertoral $n$-manifolds we can find examples realizing any possible value of
$\operatorname{disc-sym}$ between $0$ and $n$.

We saw above how Theorem 1.1 implies that the manifold $X$ in Example 1.4
satisfies $\operatorname{disc-sym}(X)\leq 2$. The previous results lead to a
stronger inequality. The structure of mapping torus on $X$ defines a fibration
$\rho:X\to S^{1}$. Let $q:S^{1}\to S^{1}$ be a degree $2$ covering and let
$X^{\prime}\to S^{1}$ be the pullback of $\rho$ through $q$. More precisely,
$X^{\prime}=\\{(x,a)\in X\times S^{1}\mid\rho(x)=q(a)\\}$. The projetion
$X^{\prime}\to X$ sending $(x,a)$ to $x$ is an unramified degree $2$-covering,
so by Theorem 1.11 we have $\operatorname{disc-sym}(X)\leq\operatorname{disc-
sym}(X^{\prime})$. Now, $X^{\prime}\cong S^{1}\times Y$, where $Y$ is a closed
connected surface of genus $2$. So we may write $X^{\prime}\cong
T^{1}\times(T^{2}\sharp T^{2})$, and consequently Theorem 1.13 implies
$\operatorname{disc-sym}(X^{\prime})\leq 1$. This inequality is sharp. Indeed,
if $r$ is an odd integer and $q_{r}:S^{1}\to S^{1}$ is a degree $r$ covering
then the pullback $X_{r}$ of $X\to S^{1}$ via $q_{r}$ can be identified with
the mapping torus of $\sigma^{r}=\sigma$, and hence is homeomorphic to $X$.
But the projection $X_{r}\to X$ is a regular degree $r$ covering with group of
deck transformations ${\mathbb{Z}}/r$. Consequently, $X_{r}$ supports a free
action of ${\mathbb{Z}}/r$. Since $X_{r}\cong X$ we conclude that $X$ supports
a free action of ${\mathbb{Z}}/r$ for every odd integer $r$. It follows that
$\operatorname{disc-sym}(X)=1$.

Denote by $\nu(X)$ the set of natural numbers $m$ such that $X$ supports
effective actions of $T^{m}$, and define the abelian degree of symmetry of $X$
as

$\operatorname{ab-sym}(X)=\max(\\{0\\}\cup\nu(X)).$

It is a standard fact that if $T^{m}$ acts effectively on $X$ then $m\leq\dim
X$ (see e.g. the end of Subsection 13.2 for a proof), so $\operatorname{ab-
sym}(X)$ is well defined and satisfies $\operatorname{ab-sym}(X)\leq\dim X$.
In [48, Theorem 5.1] Schultz proves that if $Z$ is a closed connected
$l$-dimensional topological manifold then $\operatorname{ab-
sym}(T^{k}\times(T^{l}\sharp Z))\leq k$ unless $Z$ is a homotopy sphere.
Clearly, $\operatorname{disc-sym}(X)\geq\operatorname{ab-sym}(X)$ for every
$X$, so Theorem 1.13 is a strengthening of Schultz’s result, except that our
hypothesis that neither $Z$ nor $Z^{\prime}$ are integral homology spheres is
stronger than the assumption that neither of them is a homotopy sphere.

A construction due to Cappell, Weinberger and Yan provides exampes of
manifolds $X$ with $\operatorname{disc-sym}(X)>\operatorname{ab-sym}(X)$:

######  Theorem 1.14.

Let $X$ be any of the manifolds $T(h)\times H$ constructed in [14, §2]. We
have $\operatorname{disc-sym}(X)\geq 1$ and $\operatorname{ab-sym}(X)=0$.

This theorem is not new. The equality $\operatorname{ab-sym}(X)=0$ is the main
result in [14], while the inequality $\operatorname{disc-sym}(X)\geq 1$ stems
from the existence of regular self coverings $X\to X$ of degree $d$ for every
odd natural number $d$. The existence of such self coverings (for degree $3$,
and hence also every power of $3$) is stated without proof in [51, Remark
1.3], and we prove it (for any odd $d$) in Section 15. There are obvious
analogues of the invariants $\operatorname{disc-sym}$ and $\operatorname{ab-
sym}$ for locally linear actions and for smooth actions on smooth manifolds,
and in neither of these categories does one have the equality
$\operatorname{disc-sym}=\operatorname{ab-sym}$ in general. For locally linear
actions there are counterexamples in dimension $4$, by the work of Edmonds
[21] and Huck [31]. In the smooth category one may take $X=T^{n}\sharp\Sigma$,
where $\Sigma$ is an exotic $n$-sphere. Then $X$ is homeomorphic to $T^{n}$,
so $\operatorname{disc-sym}(X)=n$ by Theorem 1.2, but $\operatorname{ab-
sym}(X)=0$ by the main result in [1]. In contrast, for holomorphic actions on
compact Kaehler manifolds one does have $\operatorname{disc-
sym}=\operatorname{ab-sym}$ in general, as proved by Theorem 1.15 below.

Note that in contrast with the previous examples our main theorem proves that,
in some very particular situations, if $\operatorname{disc-sym}(X)$ is big
then $\operatorname{ab-sym}(X)$ is also big: namely if $X$ is an
$n$-dimensional rationally hypertoral manifolds with virtually solvable
$\pi_{1}$ then $\operatorname{disc-sym}(X)=n$ implies $\operatorname{ab-
sym}(X)=n$.

The proof we give of Theorem 1.1 uses crucially both hypothesis that $X$ is
rationally hypertoral and that $\pi_{1}(X)$ is virtually solvable (see Remark
10.5 for the latter). But it is tempting to believe that the result should be
true in greater generality. If the theorem were true for every closed
topological manifold $X$ (i.e., if one could prove that any closed topological
$n$-manifold $X$ satisfying $\operatorname{disc-sym}(X)\geq n$ is homeomorphic
to $T^{n}$) then a remarkable consequence would be that for any closed
topological $n$-manifold $X$ we have $\operatorname{disc-sym}(X)\leq n$, a
much stronger inequality than the elementary bound $\operatorname{ab-
sym}(X)\leq n$.

###  1.5. Holomorphic actions on Kaehler manifolds

We don’t know whether the previous statement is true, but we can prove the
following analogue for Kaehler manifolds:

######  Theorem 1.15.

Let $X$ be a compact connected Kaehler manifold of real dimension $n$. Suppose
that, for some natural number $m$, $X$ supports an effective holomorphic
action of $({\mathbb{Z}}/r)^{m}$ for arbitrarily large values of $r$. Then $X$
supports an effective holomorphic action of $T^{m}$. Furthermore, $m\leq n$,
and if $m=n$ then $X$ is biholomorphic to a complex torus.

We will prove Theorem 1.15 using a result of Fujiki [25] on automorphism
groups of compact Kaehler manifolds and the following result on Lie groups.

######  Theorem 1.16.

Let $G$ be a (finite dimensional) Lie group with finitely many connected
components. For every natural number $n$ the following properties are
equivalent:

  1. (1)

$G$ has a Lie subgroup isomorphic to $T^{n}$,

  2. (2)

for arbitrary large integers $r$ the group $G$ has a subgroup isomorphic to
$({\mathbb{Z}}/r)^{n}$.

The previous theorem does not extend to the infinite dimensional setting.
Indeed, if we take $G=\operatorname{Diff}(X)$ for some manifold $X$ then the
equivalence of statements (1) and (2) of the theorem applied to this $G$ is
the same thing as the equality of the smooth analogues of $\operatorname{disc-
sym}(X)$ and $\operatorname{ab-sym}(X)$, which as we explained previously is
false in general.

Theorem 1.15 seems to be new even for smooth projective varieties over the
complex numbers. One can also ask the analogous question for birational
transformation groups. Namely, if $X$ is an $n$-dimensional variety defined
over the complex numbers (or more generally a field of characteristic zero)
and the birational transformation group $\operatorname{Bir}(X)$ contains
subgroups isomorphic to $({\mathbb{Z}}/r)^{m}$ for arbitrarily large values of
$r$, does it follow that $m\leq 2n$? And if $m=2n$, does it follow that $X$ is
birational to an abelian variety?

A partial result on the first question, due to Prokhorov and Shramov, appears
in [44, Theorem 1.10]. An analogue of the second question for rationally
connected varieties has been recently proved by Xu [53, Theorem 1.3]: namely,
if $X$ is a rationally connected $n$-dimensional variety and
$\operatorname{Bir}(X)$ contains subgroups isomorphic to
$({\mathbb{Z}}/p)^{n}$ for sufficiently big primes $p$ then $X$ is rational.

Theorem 1.16 implies, when combined with the theorem of Myers–Steenrod, the
following Riemannian analogue of Theorem 1.15: if $(M,g)$ is a compact
$n$-dimensional Riemannian manifold whose isometry group contains subgroups
isomorphic to $({\mathbb{Z}}/r)^{m}$ for arbitrarily large values of $r$ then
$m\leq n$, and if $m=n$ then $(M,g)$ is isometric to a flat torus. One may
wonder whether an analogous result holds for Lorentz manifolds (note that the
isometry group of a Lorentz manifold, unlike the case of Riemannian manifolds,
may have infinitely many connected components — see e.g. [19, §1]).

###  1.6. Structure of the proofs and contents of the paper

To prove of Theorem 1.1 we use two main tools. The first one is geometric. In
the presence of an effective action of a finite group $G$ on a closed manifold
$X$, it allows to substitute any map of nonzero degree $\phi:X\to T^{n}$ by a
homotopic map that is $G$-equivariant with respect to an ”almost effective”
action of $G$ on $T^{n}$. By ”almost effective” we mean that the amount of
elements of $G$ acting trivially on $T^{n}$ is bounded above by a constant
depending only on $X$ and the degree of $\phi$. This fact is proved in Section
2. Immediate applications are Theorems 1.5 and 1.7 (proved in Section 3) and
Theorem 1.8 (proved in Section 4).

The second tool we use is algebraic. Let $M$ be a finitely generated
$A$-module, where $A={\mathbb{Z}}[t_{1}^{\pm 1},\dots,t_{n}^{\pm 1}]$. Suppose
that for every $1\leq i\leq n$ there exists a nonzero integer $d_{i}$, a
sequence of integers $(r_{i,j})_{j}$ satisfying $r_{i,j}\to\infty$ as
$j\to\infty$, and automorphisms $w_{i,j}\in\operatorname{Aut}_{A}M$ such that
$w_{i,j}^{r_{i,j}}$ coincides with multiplication by $t_{i}^{d_{i}}$. Then $M$
is finitely generated as a ${\mathbb{Z}}$-module. This is proved in Section 5
(see Corollary 5.3). Section 6 contains some elementary lemmas on finite
abelian groups, in Section 7 we prove Theorem 1.11, and in Section 8 we prove
Theorem 1.12.

Theorem 1.1 is proved in Section 10. The idea is the following: suppose that a
closed $n$-manifold $X$ supports effective actions of $({\mathbb{Z}}/r)^{n}$
for arbitrary large values of $r$ and that $\phi:X\to T^{n}$ is a map of
nonzero degree. Let $X_{\phi}\to X$ be the pullback of the principal
${\mathbb{Z}}^{n}$-bundle ${\mathbb{R}}^{n}\to{\mathbb{Z}}^{n}$ via $\phi$.
Then $X_{\phi}$ supports an action of ${\mathbb{Z}}^{n}$, so its homology
$H_{*}(X_{\phi};{\mathbb{Z}})$ is a module over
${\mathbb{Z}}[{\mathbb{Z}}^{n}]\simeq A$. We prove in Section 9 that
$H_{*}(X_{\phi};{\mathbb{Z}})$ is finitely generated as an $A$-module. Then we
use the geometric tool proved in Section 2 to deduce that
$H_{*}(X_{\phi};{\mathbb{Z}})$ satisfies the hypothesis allowing to apply the
algebraic tool, so that $H_{*}(X_{\phi};{\mathbb{Z}})$ is a finitely generated
${\mathbb{Z}}$-module. A simple argument using Serre’s spectral sequence
applied to the Borel construction of the ${\mathbb{Z}}^{n}$-space $X_{\phi}$
implies that $H_{*}(X_{\phi};{\mathbb{Z}})$ is zero in positive degrees. This
leads almost immediately to statement (1) in Theorem 1.1. As for statement
(2), if $\pi_{1}(X)$ is solvable then we conclude using Whitehead’s theorem
that the arcconnected components of $X_{\phi}$ are contractible, so $X$ is
homotopy equivalent to a torus. At this point we invoke the topological
rigidity of tori to deduce that $X\cong T^{n}$. Finally, the case of virtually
solvable $\pi_{1}(X)$ is reduced to the solvable case using Theorem 1.11.

In Section 11 we prove Theorem 1.2. Theorem 1.13 is proved in Section 12,
using the geometric tool introduced in Section 2, the finite generatedness
proved in Section 9, and statement (T1) above. In Section 13 we prove Theorems
1.16 and 1.15. Finally, in Section 14 we prove that the manifold in Example
1.4 is rationally hypertoral but not hypertoral, and in Section 15 we prove
Theorem 1.14.

###  1.7. Notation

For every finite set $S$ we denote by $|S|$ the cardinality of $S$.

###  1.8. Acknowledgements

I wish to thank Jordi Daura for pointing out to me reference [51], which led
to Theorem 1.14. Thanks also to Costya Shramov for comments on a preliminar
version of this paper and information on birational transformation groups.

##  2\. Equivariant maps to the torus: proof of Theorem 2.1

In all this section $X$ denotes a closed, connected and oriented
$n$-dimensional manifold. We identify $T^{n}$ with the quotient
${\mathbb{R}}^{n}/{\mathbb{Z}}^{n}$, and we use additive notation for the
group structure on $T^{n}$. Suppose that $G$ is a group, $\eta:G\to T^{n}$ is
a group homomorphism, and $G$ acts on a space $X$. A map $\phi:X\to T^{n}$
will be called $\eta$-equivariant if it satisfies $\phi(g\cdot
x)=\eta(g)+\phi(x)$ for every $x\in X$ and $g\in G$.

######  Theorem 2.1.

Let $X$ be a closed, connected and oriented $n$-dimensional topological
manifold. Let $\phi:X\to T^{n}$ be a continuous map of nonzero degree. Let $G$
be a finite group. Suppose that $X$ is endowed with an effective action of $G$
inducing the trivial action on $H^{1}(X;{\mathbb{Z}})$. Then there is a
morphism of groups $\eta:G\to T^{n}$ with these properties:

  1. (1)

the map $\phi$ is homotopic to an $\eta$-equivariant map $\psi:X\to T^{n}$,

  2. (2)

$|\operatorname{Ker}\eta|\leq|\deg\phi|$.

Before proving the theorem we prove three auxiliary lemmas. The first lemma is
a topological analogue of the construction at the beginning of [42, §2.1]. We
identify $S^{1}$ with ${\mathbb{R}}/{\mathbb{Z}}$ and accordingly we use
additive notation for the group structure on $S^{1}$.

######  Lemma 2.2.

Let $\alpha:X\to S^{1}$ be a continuous map. Let $\theta$ be a generator of
$H^{1}(S^{1};{\mathbb{Z}})$. Suppose that a finite group $G$ acts continuously
on $X$ preserving $\alpha^{*}\theta$. Let $r$ be the cardinal of $G$ and let
$\mu_{r}\subset S^{1}$ denote the group of $r$-th roots of unity. There exists
a morphism of groups $\xi:G\to\mu_{r}$ and a continuous map $\beta:X\to S^{1}$
homotopic to $\alpha$ such that $\beta(g\cdot x)=\xi(g)+\beta(x)$ for every
$x\in X$ and $g\in G$.

###### Proof.

Define $\zeta:X\to S^{1}$ by $\zeta(x)=\sum_{g\in G}\alpha(g\cdot x)$ for
every $x\in X$. Then $\zeta$ is continuous and constant on $G$-orbits. Let
$\rho_{g}:X\to X$ be the homeomorphism induced by the action of $g\in G$. By
assumption $\rho_{g}^{*}\alpha^{*}\theta=\alpha^{*}\theta$ for every $g\in G$.
We have $\zeta^{*}\theta=\sum_{g\in
G}\rho_{g}^{*}\alpha^{*}\theta=r\alpha^{*}\theta$.

Let $\Gamma=\\{(x,t)\in X\times S^{1}\mid\zeta(x)=rt\\}$. Let $\pi:\Gamma\to
X$ be the restriction of the projection map $X\times S^{1}\to X$. The action
of $\mu_{r}$ on $\Gamma$ given by $(x,t)\cdot\theta=(x,t\theta)$ endows
$\pi:\Gamma\to X$ with a structure of principal $\mu_{r}$-bundle. We claim
that it is a trivial principal bundle. This is equivalent to the triviality of
the monodromy of $\pi$, which we denote by $\nu:\pi_{1}(X,x_{0})\to\mu_{r}$,
where $x_{0}\in X$ is an arbitrary base point. If there existed some
$\lambda\in\pi_{1}(X,x_{0})$ such that $\nu(\lambda)\neq 0$ then the pairing
of $\zeta^{*}\theta$ with $[\lambda]\in H_{1}(X;{\mathbb{Z}})$ would not be
divisible by $r$, which contradicts the fact that
$\zeta^{*}\theta=r\alpha^{*}\theta$. Hence $\nu$ is trivial and consequently
the bundle $\pi:\Gamma\to X$ is trivial, so we may choose a section
$\sigma:X\to\Gamma$.

Define $\beta:X\to S^{1}$ by the condition that $\sigma(x)=(x,\beta(x))$. Then
$\beta$ is continuous and we have $r\beta(x)=\zeta(x)$ for every $x\in X$. For
any $g\in G$ define $\chi_{g}:X\to S^{1}$ by $\chi_{g}(x)=\beta(g\cdot
x)-\beta(x)$. We have $r\chi_{g}(x)=r\beta(g\cdot x)-r\beta(x)=\zeta(g\cdot
x)-\zeta(x)=0$ because $\zeta$ is $G$-invariant. Hence $\chi_{g}$ takes values
in $\mu_{r}$, and consequently, being continuous, it is a constant map. We may
thus define a map $\xi:G\to\mu_{r}$ by the condition that $\xi(g)=\chi_{g}(x)$
for every $x\in X$. Let $g,g^{\prime}\in G$ and let $x\in X$. We have

$\chi_{gg^{\prime}}(x)=\beta(gg^{\prime}\cdot
x)-\beta(x)=\beta(gg^{\prime}\cdot x)-\beta(g^{\prime}\cdot
x)+\beta(g^{\prime}\cdot x)-\beta(x)=\chi_{g}(g^{\prime}\cdot
x)+\chi_{g^{\prime}}(x),$

which proves that $\xi(gg^{\prime})=\xi(g)+\xi(g^{\prime})$, so $\xi$ is a
morphism of groups. Now the formula $\beta(g\cdot x)=\xi(g)+\beta(x)$ follows
immediately from the definition of $\xi$. To conclude the proof, note that
$r\beta^{*}\theta=\zeta^{*}\theta=r\alpha^{*}\theta$, so
$r(\beta^{*}\theta-\alpha^{*}\theta)=0$. Since $H^{1}(X;{\mathbb{Z}})$ has no
torsion we conclude that $\beta^{*}\theta=\alpha^{*}\theta$. Hence $\beta$ and
$\alpha$ are homotopic, because $S^{1}$ is a model for $K({\mathbb{Z}},1)$. ∎

######  Lemma 2.3.

Let $G$ be a finite group acting effectively, continuously and preserving the
orientation on $X$. Let $X^{*}\subseteq X$ be the set of points with trivial
stabilizer. Then $X^{*}$ is connected.

###### Proof.

For any $g\in G$ denote $X^{g}=\\{x\in X\mid g\cdot x=x\\}$. Let
$g_{1},\dots,g_{s}$ be the (nontrivial) elements of $G$ of prime order. Since
any nontrivial element of $G$ has some power belonging to the set
$\\{g_{1},\dots,g_{s}\\}$, we have $X^{*}=X\setminus\bigcup_{i}X^{g_{i}}$.
Define $X_{1}=X$ and $X_{i}=X\setminus(X^{g_{1}}\cup\dots\cup X^{g_{i-1}})$
for $2\leq i\leq s+1$. Then $X_{i}$ is an open subset of $X$ for each $i$. We
prove that $X_{i}$ is connected for every $1\leq i\leq s+1$, using ascending
induction on $i$. Since $X^{*}=X_{s+1}$, this will imply the lemma. Clearly
$X_{1}$ is connected. Now suppose that $1\leq i\leq s$ and that $X_{i}$ is
connected. Let $p_{i}$ be the order of $g_{i}$. Since $G$ acts on $X$
preserving the orientation, by [9, Chap V, Theorem 2.3] and [9, Chap V,
Theorem 2.5], $X^{g_{i}}$ is a ${\mathbb{Z}}/p_{i}$-cohomology manifold of
dimension $d_{i}$, where $n-d_{i}$ is even. Arguing as in the proof of [9,
Chap V, Theorem 2.6] we conclude that $d_{i}<n$, and consequently $d_{i}\leq
n-2$. Applying [9, Chap I, Corollary 4.7] we conclude that
$X_{i+1}=X_{i}\setminus(X_{i}\cap X^{g_{i}})$ is connected, so the lemma is
proved. ∎

The following result generalizes [20, Lemma 2.5].

######  Lemma 2.4.

Suppose that a finite group $G$ acts effectively, continuously, and preserving
the orientation on $X$. Let $\pi:X\to X/G$ denote the quotient map. Let $r$
denote the cardinal of $G$. The image of the map
$\pi^{*}:H^{n}(X/G;{\mathbb{Z}})\to H^{n}(X;{\mathbb{Z}})$ is contained in
$rH^{n}(X;{\mathbb{Z}})$.

###### Proof.

Denote as in the previous lemma by $X^{*}$ the open subset of $X$ consisting
of points with trivial stabilizer. Since $X/G$ is endowed with the quotient
topology, $\pi(X^{*})=X^{*}/G$ is an open subset of $X/G$. Let $F=X\setminus
X^{*}$. Consider the following commutative diagram, where
$H_{c}^{*}(\cdot;{\mathbb{Z}})$ denotes cohomology with compact support, the
rows are portions of the long exact sequences for the inclusions
$X^{*}\hookrightarrow X\hookleftarrow F$ and $X^{*}/G\hookrightarrow
X/G\hookleftarrow F/G$ (see e.g. [9, Chap I, (2) in §1.1]), and the vertical
arrows are pullback morphisms induced by proper maps:

$\textstyle{H^{n}_{c}(X^{*};{\mathbb{Z}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{j}$$\textstyle{H^{n}_{c}(X;{\mathbb{Z}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{r}$$\textstyle{H^{n}_{c}(F;{\mathbb{Z}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{H^{n}_{c}(X^{*}/G;{\mathbb{Z}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi^{*}}$$\scriptstyle{j_{G}}$$\textstyle{H^{n}_{c}(X/G;{\mathbb{Z}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi^{*}}$$\scriptstyle{r_{G}}$$\textstyle{H^{n}_{c}(F/G;{\mathbb{Z}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0,}$

The morphism $j$ is an isomorphism because, by Lemma 2.3, $X^{*}$ is connected
(see [9, Chap I, Theorem 4.3]). By the exactness this implies that
$H^{n}_{c}(F;{\mathbb{Z}})=0$. The long exact sequence for
$X^{*}\hookrightarrow X\hookleftarrow F$ and the fact that $\dim X=n$ imply
that $H^{k}_{c}(F;{\mathbb{Z}})$ for every $k\geq n$ (here we are using [9,
Chap I, (3) in §1.2]). From [9, Chap III, Theorem 5.2] it follows that
$H^{k}_{c}(F/G;{\mathbb{Z}})$ for every $k\geq n$. Hence $j_{G}$ is
surjective. Consequently it suffices to prove that the image of the morphism
$\pi^{*}:H^{n}_{c}(X^{*}/G;{\mathbb{Z}})\to H^{n}_{c}(X^{*};{\mathbb{Z}})$ is
contained in $rH^{n}_{c}(X^{*};{\mathbb{Z}})$.

Denote $G^{*}=G\setminus\\{1\\}$. We are going to prove that there exists a
connected open subset $U\subset X^{*}$ such that $gU\cap U=\emptyset$ for
every $g\in G^{*}$. Fix some point $x\in X^{*}$. Since $X^{*}$ is Hausdorff,
for every $g\in G^{*}$ there exists disjoint open subsets $A_{g},B_{g}\subset
X^{*}$ such that $x\in A_{g}$ and $gx\in B_{g}$. Let $C=\bigcap_{g\in
G^{*}}A_{g}$. Then $gx\notin\overline{C}$ for every $g\in G^{*}$, because
$C\cap B_{g}=\emptyset$. This implies that $x$ belongs to the open set
$D:=C\setminus\bigcup_{g\in G^{*}}g\overline{C}$. Let $U\subset D$ be a
connected open subset containing $x$. Then for every $g\in G^{*}$ we have
$gU\subset gC$ because $U\subset C$, and consequently $gU\cap D=\emptyset$,
which implies that $gU\cap U=\emptyset$.

Let $V=\pi(U)$, so that $\pi^{-1}(V)=GU=\bigcup_{g\in G}gU$. Consider the
following commutative diagram:

$\textstyle{H_{c}^{n}(GU;{\mathbb{Z}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{j_{GU}}$$\textstyle{H_{c}^{n}(X^{*};{\mathbb{Z}})}$$\textstyle{H_{c}^{n}(V;{\mathbb{Z}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi_{V}^{*}}$$\scriptstyle{j_{V}}$$\textstyle{H_{c}^{n}(X^{*}/G;{\mathbb{Z}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces,}$$\scriptstyle{\pi^{*}}$

where $j_{GU}$ and $j_{V}$ are the covariant morphisms induced by open
embeddings and the vertical arrows are pullback morphisms induced by proper
morphisms. By [9, Chap I, Theorem 4.3] $j_{V}$ is an isomorphism, so it
suffices to prove that the image of $j_{GU}\circ\pi_{V}^{*}$ is contained in
$rH_{c}^{n}(X^{*};{\mathbb{Z}})$. Since $gU\cap U=\emptyset$ for every $g\in
G^{*}$, the open subset $GU$ contains $r=|G|$ connected components, which are
$\\{gU\mid g\in G\\}$. Denote by $i_{g}:gU\to GU$ the inclusion. The pullback
morphisms $i_{g}^{*}:H_{c}^{n}(GU;{\mathbb{Z}})\to H_{c}^{n}(gU;{\mathbb{Z}})$
combine to give an isomorphism
$H_{c}^{n}(GU;{\mathbb{Z}})\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}\bigoplus_{g\in
G}H_{c}^{n}(gU;{\mathbb{Z}})$. Let $j_{gU}:H_{c}^{n}(gU;{\mathbb{Z}})\to
H_{c}^{n}(X^{*};{\mathbb{Z}})$ be the morphism induced by the open embedding
$gU\hookrightarrow X^{*}$. We have $j_{GU}=\sum_{g\in G}j_{gU}\circ
i_{g}^{*}$, so if we prove that $j_{gU}\circ
i_{g}^{*}\circ\pi_{V}^{*}=j_{hU}\circ i_{h}^{*}\circ\pi_{V}^{*}$ for every
$g,h\in G$ then we will be done. Take two elements $g,h\in G$ and let
$\rho:X^{*}\to X^{*}$ be the map given by $\rho(x)=gh^{-1}\cdot x$. Then
$\rho$ is a homeomorphism and it restricts to a homeomorphism $\rho:hU\to gU$.
The induced morphism $\rho^{*}:H_{c}^{n}(X^{*};{\mathbb{Z}})\to
H_{c}^{n}(X^{*};{\mathbb{Z}})$ is the identity because $G$ acts on $X$ (and
hence on $X^{*}$) preserving the orientation. Now, the desired equality
follows from the commutativity of the following diagram:

|
---|---
$\textstyle{H_{c}^{n}(hU;{\mathbb{Z}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{j_{hV}}$$\textstyle{H_{c}^{n}(X^{*};{\mathbb{Z}})}$$\textstyle{H_{c}^{n}(V;{\mathbb{Z}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{i_{j}^{*}\circ\pi_{V}^{*}}$$\scriptstyle{i_{g}^{*}\circ\pi_{V}^{*}}$$\textstyle{H_{c}^{n}(gU;{\mathbb{Z}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho^{*}}$$\scriptstyle{j_{gU}}$$\textstyle{H_{c}^{n}(X^{*};{\mathbb{Z}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho^{*}=\operatorname{Id}}$

The triangle commutes because $\pi_{V}\circ i_{h}=\pi_{V}\circ i_{g}\circ\rho$
and the square commutes because $\rho$ is a homeomorphism and the inclusion
$hU\hookrightarrow X^{*}$ is equal to the composition of the inclusion
$gU\hookrightarrow X^{*}$ and $\rho$. ∎

We are now ready to prove Theorem 2.1. Let $\phi_{i}:X\to S^{1}$ be the
composition of $\phi$ with the projection to the $i$-th factor
$T^{n}=(S^{1})^{n}\to S^{1}$. Since $G$ acts trivially on
$H^{1}(X;{\mathbb{Z}})$, in particular it fixes $\phi_{i}^{*}\theta$, where
$\theta\in H^{1}(S^{1};{\mathbb{Z}})$ is any generator. Applying Lemma 2.2 to
$\phi_{i}$ we obtain the existence of a morphism of groups $\eta_{i}:G\to
S^{1}$ and a map $\psi_{i}:X\to S^{1}$ homotopic to $\phi_{i}$ such that
$\psi_{i}(g\cdot x)=\eta_{i}(g)+\psi_{i}(x)$ for every $x\in X$ and $g\in G$.
Define $\psi=(\psi_{1},\dots,\psi_{n}):X\to T^{n}$ and
$\eta=(\eta_{1},\dots,\eta_{n}):G\to T^{n}$. Then $\psi$ is homotopic to
$\phi$ and it is $\eta$-equivariant.

Let $G_{0}=\operatorname{Ker}\eta$. The map $\psi$ factors as a composition

$X\stackrel{{\scriptstyle\pi}}{{\longrightarrow}}X/G_{0}\stackrel{{\scriptstyle\psi_{0}}}{{\longrightarrow}}T^{n},$

where $\pi$ is the natural quotient map. Hence
$\psi^{*}=\pi^{*}\circ\psi_{0}$, so the image of
$\psi^{*}:H^{n}(T^{n};{\mathbb{Z}})\to H^{n}(X;{\mathbb{Z}})$ is contained in
the image of $\pi^{*}:H^{n}(X/G_{0};{\mathbb{Z}})\to H^{n}(X;{\mathbb{Z}})$.
By the definition of degree, the image of
$\psi^{*}:H^{n}(T^{n};{\mathbb{Z}})\to H^{n}(X;{\mathbb{Z}})$ is equal to
$(\deg\psi)H^{n}(X;{\mathbb{Z}})$, and by Lemma 2.4 the image of $\pi^{*}$ is
contained in $|G_{0}|\cdot H^{n}(X;{\mathbb{Z}})$. It then follows that
$|G_{0}|\leq|\deg\psi|$. But $|\deg\psi|=|\deg\phi|$ because $\psi$ and $\phi$
are homotopic. So the proof of the theorem is now complete.

######  Remark 2.5.

The results in this section have been independently proved by Csikós, Pyber
and Szabó, lifting $\phi$ to a map $\zeta$ from the universal cover of $X$ to
that of $T^{n}$, and defining $\psi$ as the average the translates of $\zeta$
by lifts of the action of elements of $G$ to the universal covers of $X$ and
$T^{n}$.

##  3\. Jordan property and bounds on stabilizers for rationally hypertoral
manifolds

The following is a classical result of Minkowski, originally published in
[40]. It follows from the elementary observation that the morphism
$\rho_{3}:\operatorname{GL}(k,{\mathbb{Z}})\to\operatorname{GL}(k,{\mathbb{Z}}/3)$
defined by taking residues mod $3$ componentwise is injective when restricted
to finite groups (equivalently: if $X\in\operatorname{GL}(k,{\mathbb{Z}})$ has
finite order and $\rho_{3}(X)=1$ then $X=1$). See [49] for a plethora of
generalisations of this fact.

######  Lemma 3.1.

For any $n$ there exists a constant $C_{n}$ with the property that any finite
subgroup of $\operatorname{GL}(n,{\mathbb{Z}})$ has at most $C_{n}$ elements.

If $X$ is a closed topological manifold then $H^{1}(X;{\mathbb{Z}})$ is a
finitely generated torsion free abelian group, hence isomorphic to
${\mathbb{Z}}^{b}$ for some $b\in{\mathbb{Z}}_{\geq 0}$. So the previous lemma
implies the following.

######  Lemma 3.2.

Let $X$ be a closed topological manifold. There exists a constant $C_{X}$ such
that for any finite group $G$ acting continuously on $X$ the kernel
$G^{\prime}$ of the induced morphism
$G\to\operatorname{Aut}H^{1}(X;{\mathbb{Z}})$ satisfies $[G:G^{\prime}]\leq
C_{X}$.

Let $X$ be a closed, connected and oriented $n$-dimensional manifold and let
$\phi:X\to T^{n}$ be a continuous map of nonzero degree satisfying

$|\deg\phi|=\min\\{|\deg\psi|\mid\psi:X\to T^{n},\,\deg\psi\neq 0\\}.$

Suppose that a finite group $G$ acts continuously on $X$. Let $G^{\prime}$ be
the kernel of the induced morphism
$G\to\operatorname{Aut}H^{1}(X;{\mathbb{Z}})$. By Lemma 3.2 we have
$[G:G^{\prime}]\leq C_{X}$ for some constant $C_{X}$ depending only on $X$. By
Theorem 2.1 there is a morphism of groups $\eta:G^{\prime}\to T^{n}$
satisfying $|\operatorname{Ker}\eta|\leq|\deg\phi|$ and an $\eta$-equivariant
map $\psi:X\to T^{n}$ homotopic to $\phi$. The existence of the
$\eta$-equivariant map $\psi$ implies that for every $x\in X$ we have
$G_{x}\leq\operatorname{Ker}\eta$ (if $g\in G_{x}$ then $\eta(g)=\psi(g\cdot
x)-\psi(x)=\psi(x)-\psi(x)=0$), so the previous bound implies that
$|\operatorname{Stab}(X,G)|\leq 2^{|\deg\phi|}$. This proves Theorem 1.7.

To prove Theorem 1.5 note that since $\eta(G^{\prime})$ is a subgroup of
$T^{n}$, it is abelian and can be generated by $n$ or fewer elements. If
$|\deg\phi|=1$ then $G^{\prime}\simeq\eta(G^{\prime})$, so $G^{\prime}$ is
abelian. This implies that $\operatorname{Homeo}(X)$ is Jordan in this case.
For other values of $|\deg\phi|$, we apply [42, Lemma 2.2] to the exact
sequence

$1\to\operatorname{Ker}\eta\to G^{\prime}\to\eta(G^{\prime})\to 1$

and since $|\operatorname{Ker}\eta|\leq|\deg\phi|$ we deduce the existence of
an abelian subgroup $G^{\prime\prime}\leq G^{\prime}$ such that
$[G^{\prime}:G^{\prime\prime}]$ is bounded above by a constant depending only
on $n$ and $\deg\phi$. Since $n$ and $\deg\phi$ only depend on $X$, it follows
that $[G:G^{\prime\prime}]$ is bounded above by a constant depending only on
$X$, so the proof that $\operatorname{Homeo}(X)$ is Jordan is now complete.

If $|\deg\phi|=1$ Theorem 1.5 can also be proved using [26, Theorem 2.5].

##  4\. Rationally hypertoral manifolds with $\chi\neq 0$ are almost
asymmetric

Here we prove Theorem 1.8. Let $X$ be a rationally hypertoral manifold
satisfying $\chi(X)\neq 0$. By hypothesis there exists a morphsm $\phi:X\to
T^{n}$ satisfying $d=|\deg\phi|\neq 0$. Let $b=\sum_{j}b_{j}(X)$ and let
$\operatorname{Tor}H^{*}(X;{\mathbb{Z}})$ denote the torsion of
$H^{*}(X;{\mathbb{Z}})$. Then
$H^{*}(X;{\mathbb{Z}})/\operatorname{Tor}H^{*}(X;{\mathbb{Z}})\simeq{\mathbb{Z}}^{b}$.
By Lemma 3.1 there exists a natural number $C$ such that no finite subgroup of
$\operatorname{GL}(b;{\mathbb{Z}})$ has more than $C$ elements. Let $G$ be a
finite group acting effectively on $X$. Let $G_{0}$ be the kernel of the
induced map

$G\to\operatorname{Aut}(H^{*}(X;{\mathbb{Z}})/\operatorname{Tor}H^{*}(X;{\mathbb{Z}}))\simeq\operatorname{GL}(b,{\mathbb{Z}}).$

Then $[G:G_{0}]\leq C$. Since $G_{0}$ acts trivially on
$H^{*}(X;{\mathbb{Z}})/\operatorname{Tor}H^{*}(X;{\mathbb{Z}})$, it also acts
trivially on $H^{*}(X;{\mathbb{Q}})$. Since $H^{1}(X;{\mathbb{Z}})$ is torsion
free, the action of $G_{0}$ on $H^{1}(X;{\mathbb{Z}})$ is also trivial. By
Theorem 2.1 there exists a morphism of groups $\eta:G_{0}\to T^{n}$ satisfying
$|\operatorname{Ker}\eta|\leq d$ and an $\eta$-equivariant map $\psi:X\to
T^{n}$. Let $g\in G_{0}$. Since the action of $g$ on $H^{*}(X;{\mathbb{Q}})$
is trivial and $\chi(X)\neq 0$, there exists some $x\in X$ such that $g\cdot
x=x$ (see e.g. [50, Chap III, Exercise (6.17) 3]; this applies to $X$ because
topological manifolds are Euclidean Neighborhood Retracts, see e.g. [27,
Corollary A.9]). It follows that every $g\in G_{0}$ belongs to
$\operatorname{Ker}\eta$, so $G_{0}=\operatorname{Ker}\eta$ and consequently
$|G_{0}|\leq d$, so $|G|\leq C\cdot d$.

##  5\. Finitely generated ${\mathbb{Z}}[t_{1}^{\pm 1},\dots,t_{1}^{\pm
1}]$-modules

######  Theorem 5.1.

Let $A$ be a Noetherian ring and let $M$ be a finitely generated
$A[z]$-module. Suppose that there exists a sequence of integers
$r_{j}\to\infty$ and $A[z]$-module morphisms $w_{j}:M\to M$ such that
$w_{j}^{r_{j}}$ coincides with multiplication by $z$. Then $M$ is finitely
generated as an $A$-module.

###### Proof.

Let $S\subset M$ be a finite $A[z]$-generating set. Let $M_{0}\subseteq M$ be
the $A$-submodule generated by $S$. Define an increasing sequence of
$A$-submodules of $M$

$M_{0}\subseteq M_{1}\subseteq\dots\subseteq M_{d}\subseteq\dots$

by the condition that $M_{d}=M_{d-1}+zM_{d-1}$ for every positive integer $d$.
Define also $M_{d}=0$ for negative integers $d$. For each $d$ the quotient
$M_{d}/M_{d-1}$ is a finitely generated $A$-module, because $M_{d}$ is a
finitely generated $A$-module.

For any $d\geq 1$, multiplication by $z$ gives a surjective morphism

$\mu_{d}:M_{d-1}/M_{d-2}\to M_{d}/M_{d-1}.$

Consider the composition

$\nu_{d}=\mu_{d}\circ\dots\circ\mu_{1}:M_{0}\to M_{d}/M_{d-1},$

and define $K_{d}=\operatorname{Ker}\nu_{d}$. Each $K_{d}$ is an $A$-submodule
of $M_{0}$, and there are inclusions

$K_{0}\subseteq K_{1}\subseteq K_{2}\subseteq\dots.$

Since $A$ is Noetherian and $M_{0}$ is a finitely generated $A$-module, there
exists some $d_{0}$ such that $K_{d}=K_{d-1}$ for $d\geq d_{0}$. If for some
$d$ the morphism $\mu_{d}$ fails to be an isomorphism, then
$\operatorname{Ker}\nu_{d}$ is strictly bigger than
$\operatorname{Ker}\nu_{d-1}$, because $\nu_{d}=\mu_{d}\circ\nu_{d-1}$ and
$\nu_{d-1}$ is surjective. It follows that $\mu_{d}$ is an isomorphism for any
$d\geq d_{0}$.

Let $N=M_{d_{0}}/M_{d_{0}-1}$. If $N=0$ then $M=M_{d_{0}-1}$, so $M$ is
finitely generated as an $A$-module and we are done.

Suppose from now on that $N\neq 0$. Since $A$ is Noetherian and $N$ is
finitely generated, there exists a filtration by $A$-submodules

(1) $0=N_{0}\subset N_{1}\subset\dots\subset N_{r}=N$

in such a way that $N_{j}/N_{j-1}\simeq A/{\mathfrak{p}}_{j}$ for primes
${\mathfrak{p}}_{1},\dots,{\mathfrak{p}}_{r}\in\operatorname{Spec}A$ (see [2,
Chap 7, Exercise 18] or [39, Theorem 6.4]). Let ${\mathfrak{p}}$ be a minimal
element of $\\{{\mathfrak{p}}_{1},\dots,{\mathfrak{p}}_{r}\\}$. Denote as
usual by $A_{{\mathfrak{p}}}$ the localisation of $A$ at ${\mathfrak{p}}$ and
by $k_{{\mathfrak{p}}}$ its residual field. For any $A$-module $R$ we denote
$R_{{\mathfrak{p}}}=R\otimes_{A}A_{{\mathfrak{p}}}$. Since
$A_{{\mathfrak{p}}}$ is a flat $A$-module, for any inclusion of $A$-modules
$R^{\prime}\subseteq R$ we have
$R_{{\mathfrak{p}}}/R_{{\mathfrak{p}}}^{\prime}\simeq(R/R^{\prime})_{{\mathfrak{p}}}$.
Since ${\mathfrak{p}}$ is a minimal element of
$\\{{\mathfrak{p}}_{1},\dots,{\mathfrak{p}}_{r}\\}$, for every $i$ we have

$(A/{\mathfrak{p}}_{i})_{{\mathfrak{p}}}\simeq\left\\{\begin{array}[]{cc}k_{{\mathfrak{p}}}&\qquad\text{if
${\mathfrak{p}}_{i}={\mathfrak{p}}$,}\\\ 0&\qquad\text{if
${\mathfrak{p}}_{i}\neq{\mathfrak{p}}$.}\end{array}\right.$

Hence, $(A/{\mathfrak{p}}_{i})_{{\mathfrak{p}}}$ is a simple
$A_{{\mathfrak{p}}}$-module for every $i$. So if we tensor by
$A_{{\mathfrak{p}}}$ the elements of the filtration (1) and we ignore the
resulting inclusions that are actually equalities, we get a composition series
for $N_{{\mathfrak{p}}}$ of length

$\lambda:=|\\{i\mid{\mathfrak{p}}_{i}={\mathfrak{p}}\\}|\geq 1$

(see e.g. the paragraph before Proposition 6.7 in [2]).

To conclude the proof of the theorem we are going to prove that there is no
$A[z]$-module morphism $w:M\to M$ satisfying $w^{r}=z$ for any $r>\lambda$.
Arguing by contradiction, let us assume that there exists an $A[z]$-module
morphism $w:M\to M$ satisfying $w^{r}=z$ for some $r>\lambda$.

Let $M_{0}^{\prime}=M_{0}$ and define recursively $M_{\delta}^{\prime}$ for
positive integers $\delta$ as
$M_{\delta}^{\prime}=M_{\delta-1}+wM_{\delta-1}$. Define also
$M_{\delta}^{\prime}=0$ for negative integers $\delta$. The action of $w$
defines a surjective $A$-module morphism
$\mu^{\prime}_{\delta}:M^{\prime}_{\delta-1}/M^{\prime}_{\delta-2}\to
M^{\prime}_{\delta}/M^{\prime}_{\delta-1}$. Arguing as we did for $M_{d}$, we
prove the existence of some $\delta_{0}$ such that $\mu^{\prime}_{\delta}$ is
an isomorphism for any $\delta\geq\delta_{0}$. Let
$N^{\prime}=M^{\prime}_{\delta_{0}}/M^{\prime}_{\delta_{0}-1}$.

Denote by $A[z]_{\leq d}$ (resp. $A[w]_{\leq\delta}$) the $A$-module of
polynomials in $z$ (resp. $w$) of degree at most $d$ (resp. $\delta$). We have

(2) $M_{d}=A[z]_{\leq d}M_{0},\qquad
M_{\delta}^{\prime}=A[w]_{\leq\delta}M_{0}.$

Since $w^{r}=z$, we have $A[z]_{\leq d}\subseteq A[w]_{\leq rd}$ for every
$d$. This implies that

$M_{d}\subseteq M^{\prime}_{rd}$

for every nonnegative $d$. Suppose that $S=\\{m_{1},\dots,m_{s}\\}$. Since
$m_{1},\dots,m_{s}$ generate $M$ as an $A[z]$-module, there exist polynomials
$P_{ijk}\in A[z]$ for $i=1,\dots,k-1$ and $j,k=1,\dots,s$ such that

$w^{i}m_{j}=P_{ij1}m_{1}+\dots+P_{ijr}m_{s}.$

Let $e=\max_{i,j,k}\deg P_{ijk}$. We have $w^{i}m_{j}\in M_{e}$ for every
$i=1,\dots,k-1$ and $j=1,\dots,s$, and this implies that for any $d$ we have
$A[w]_{\leq\delta}M_{0}\subseteq A[z]_{\leq[\delta/r]+e}M_{0}$, or
equivalently

$M_{\delta}^{\prime}\subseteq M_{[\delta/r]+e}.$

Following [2, Chap 6] (see the proof of [2, Proposition 6.7]) for any
$A_{{\mathfrak{p}}}$-module $R$ we denote by $l(R)$ the length of $R$. This is
defined to be $\infty$ if $R$ has no composition series of finite length and
it is equal to $n$ if $R$ has a composition series of lenght $n$. This is well
defined by [2, Proposition 6.7]. Furthermore, for any inclusion of
$A_{{\mathfrak{p}}}$-modules $R\subseteq R^{\prime}$ we have
$l(R^{\prime})=l(R)+l(R^{\prime}/R)$ (see [2, Proposition 6.9]). For example,
we have $l(N_{{\mathfrak{p}}})=\lambda$, and if $d\geq d_{0}$ then
$l((M_{d+k})_{{\mathfrak{p}}}/(M_{d})_{{\mathfrak{p}}})=k\lambda$ for every
$k$.

To simplify our notation we will denote
$M_{i,{\mathfrak{p}}}=(M_{i})_{{\mathfrak{p}}}$ and
$M^{\prime}_{i,{\mathfrak{p}}}=(M^{\prime}_{i})_{{\mathfrak{p}}}$ for every
$i$. Fix some value of $d$ satisfies both $d\geq d_{0}$ and
$[d/r]\geq\delta_{0}$. Let $k$ be a big number, to be specified later. The
inclusions $M_{d+k,{\mathfrak{p}}}\subseteq
M^{\prime}_{r(d+k),{\mathfrak{p}}}\subseteq M_{d+k+e,{\mathfrak{p}}}$ imply

$\displaystyle 0\leq
l(M_{d+k+e,{\mathfrak{p}}}/M^{\prime}_{r(d+k),{\mathfrak{p}}})$
$\displaystyle=l(M_{d+k+e,{\mathfrak{p}}}/M_{d+k,{\mathfrak{p}}})-l(M^{\prime}_{r(d+k),{\mathfrak{p}}}/M_{d+k,{\mathfrak{p}}})$
(3) $\displaystyle\leq
l(M_{d+k+e,{\mathfrak{p}}}/M_{d+k,{\mathfrak{p}}})=e\lambda.$

Using the additivity of $l$ and the filtration

$M_{d,{\mathfrak{p}}}=M^{\prime}_{rd,{\mathfrak{p}}}\subseteq
M^{\prime}_{rd+1,{\mathfrak{p}}}\subseteq\dots\subseteq
M^{\prime}_{r(d+k),{\mathfrak{p}}}\subseteq M_{d+k+e,{\mathfrak{p}}}$

we have

$\displaystyle(k+e)\lambda=l(M_{d+k+e,{\mathfrak{p}}}/M_{d,{\mathfrak{p}}})$
$\displaystyle=l(M^{\prime}_{r(d+k),{\mathfrak{p}}}/M^{\prime}_{rd,{\mathfrak{p}}})+l(M_{d+k+e,{\mathfrak{p}}}/M^{\prime}_{r(d+k),{\mathfrak{p}}})$
$\displaystyle=rk\,l(N^{\prime}_{{\mathfrak{p}}})+l(M_{d+k+e,{\mathfrak{p}}}/M^{\prime}_{r(d+k),{\mathfrak{p}}})$

and using (3) we have

$\frac{\lambda}{r}=\frac{(k+e)\lambda-e\lambda}{rk}\leq
l(N^{\prime}_{{\mathfrak{p}}})\leq\frac{(k+e)\lambda}{rk}.$

The lower bound for $l(N^{\prime}_{{\mathfrak{p}}})$ belongs to the interval
$(0,1)$, and if $k$ is big enough so that $(k+e)/k<r/\lambda$ then the upper
bound for $l(N^{\prime}_{{\mathfrak{p}}})$ also belongs to $(0,1)$. This
contradicts the fact that $l(N^{\prime}_{{\mathfrak{p}}})$ is an integer, so
the proof that $r$ cannot be bigger than $\lambda$ is now finished. ∎

######  Corollary 5.2.

Let $M$ be a finitely generated module over $A:={\mathbb{Z}}[t_{1}^{\pm
1},\dots,t_{n}^{\pm 1}]$. Suppose that for every $1\leq i\leq n$ there exists
a sequence of integers $(r_{i,j})_{j}$ satisfying $r_{i,j}\to\infty$ as
$j\to\infty$, and $A$-module automorphism $w_{i,j}:M\to M$ such that
$w_{i,j}^{r_{i,j}}$ coincides with multiplication by $t_{i}$. Then $M$ is
finitely generated as a ${\mathbb{Z}}$-module.

###### Proof.

Let $B={\mathbb{Z}}[z_{1},\dots,z_{2n}]$ and let $\phi:B\to A$ be the morphism
of rings defined by $\phi(z_{2i-1})=t_{i}$ and $\phi(z_{2i})=t_{i}^{-1}$. We
can look at $M$ as a finitely generated $B$-module via $\phi$, and the
automorphisms $w_{i,j}$ in the statement are $B$-module automorphisms
satisfying

$w_{i,j}^{r_{i,j}}=\text{multiplication by
$z_{2i-1}$},\qquad(w_{i,j}^{-1})^{r_{i,j}}=\text{multiplication by $z_{2i}$}.$

Let $B_{0}={\mathbb{Z}}$ and $B_{j}={\mathbb{Z}}[z_{1},\dots,z_{j}]$ for
$1\leq j\leq 2n$. Then $B_{2n}=B$, and we can prove that $M$ is finitely
generated as a $B_{j}$-module for any $0\leq j\leq 2n$ using descending
induction on $j$, applying Theorem 5.1 in the induction step. It follows that
$M$ is finitely generated as a $B_{0}$-module. ∎

######  Corollary 5.3.

Let $M$ be a finitely generated module over $A:={\mathbb{Z}}[t_{1}^{\pm
1},\dots,t_{n}^{\pm 1}]$. Suppose that for every $1\leq i\leq n$ there exists
a nonzero integer $d_{i}$, a sequence of integers $(r_{i,j})_{j}$ satisfying
$r_{i,j}\to\infty$ as $j\to\infty$, and $A$-module automorphisms $w_{i,j}:M\to
M$ such that $w_{i,j}^{r_{i,j}}$ coincides with multiplication by
$t_{i}^{d_{i}}$. Then $M$ is finitely generated as a ${\mathbb{Z}}$-module.

###### Proof.

Suppose that $M$ is generated as an $A$ module by $s_{1},\dots,s_{r}\in M$.
Let $A^{\prime}={\mathbb{Z}}[t_{1}^{\pm d_{1}},\dots,t_{1}^{\pm d_{n}}]$. Then
$M$ is generated as an $A^{\prime}$-module by the set

$\\{t_{1}^{a_{1}}t_{2}^{a_{2}}\dots t_{n}^{a_{n}}s_{i}\mid 1\leq i\leq
r,\,0\leq a_{i}\leq|d_{i}|-1\text{ for each $i$}\\},$

so $M$ is finitely generated as an $A^{\prime}$-module. Applying Corollary 5.2
to $M$ viewed as an $A^{\prime}$-module, we conclude that $M$ is a finitely
generated ${\mathbb{Z}}$-module. ∎

##  6\. Some lemmas on finite abelian groups

For any natural numbers $a,b$ we denote for convenience
$\Gamma_{a,b}=({\mathbb{Z}}/a)^{b}$. The first three lemmas below refer to
properties of the groups $\Gamma_{a,b}$.

######  Lemma 6.1.

Let $a,b,C$ be natural numbers and suppose that $\Gamma^{\prime}$ is a
subgroup of $\Gamma_{a,b}$ satisfying $[\Gamma_{a,b}:\Gamma^{\prime}]\leq C$.
There exists a subgroup $\Gamma^{\prime\prime}\leq\Gamma^{\prime}$ which is
isomorphic to $\Gamma_{a^{\prime},b}$ for some natural number $a^{\prime}$
satisfying $C!a^{\prime}\geq a$.

###### Proof.

Let $d$ be the greatest common divisor of $a$ and $C!$ and let
$a^{\prime}=a/d$. Note that $d\leq C!$, so $C!a^{\prime}\geq a$. For every
$1\leq j\leq a$ let
$F_{j}=\\{(\gamma_{1},\dots,\gamma_{b})\in\Gamma_{a,b}\mid\gamma_{i}=0\text{
for $i\neq j$}\\}$. Then $F_{j}\simeq{\mathbb{Z}}/a$. The set of subgroups of
$F_{j}$ can be put in bijection with the set of divisors of $a$ by assigning
to each subgroup its index in $F_{j}$, and given two subgroups
$F^{\prime},F^{\prime\prime}$ of $F_{j}$ we have $F^{\prime}\leq
F^{\prime\prime}$ if and only if $[F_{j}:F^{\prime\prime}]$ divides
$[F_{j}:F^{\prime}]$. Let $F_{j}^{\prime\prime}$ be subgroup of $F_{j}$
satisfying $[F_{j}:F_{j}^{\prime\prime}]=d$. Then $F_{j}^{\prime\prime}$ is a
cyclic group of $a^{\prime}$ elements. Let
$F_{j}^{\prime}=F_{j}\cap\Gamma^{\prime}$. We have
$[F_{j}:F_{j}^{\prime}]\leq[\Gamma_{a,b}:\Gamma^{\prime}]\leq C$, so
$[F_{j}:F_{j}^{\prime}]$ divides both $C!$ and $a$. Consequently
$[F_{j}:F_{j}^{\prime}]$ divides $d$ and hence $F_{j}^{\prime\prime}\leq
F_{j}^{\prime}$. Then
$\Gamma^{\prime\prime}:=F_{1}^{\prime\prime}\times\dots\times
F_{b}^{\prime\prime}$ is a subgroup of $\Gamma^{\prime}$ isomorphic to
$\Gamma_{a^{\prime},b}$. ∎

######  Lemma 6.2.

Let $\Gamma$ be a finite abelian group, and suppose that there exists a
surjection $f:\Gamma\to\Gamma_{a,b}$ for some natural numbers $a,b$. Then
there exists a subgroup of $\Gamma$ which is isomorphic to $\Gamma_{a,b}$.

###### Proof.

For any prime $p$ and any finite abelian group $A$ we denote by $A_{p}$ the
Sylow $p$-subgroup of $A$, which coincides with the set of elements in $A$
whose order is a power of $p$. Any morphism of finite abelian groups $g:A\to
B$ satisfies $g(A_{p})\leq B_{p}$, and $g$ is surjective if and only if
$g(A_{p})=B_{p}$ for every prime $p$. The group $A$ contains a subgroup
isomorphic to $B$ if and only there is a subgroup $A(p)\leq A$ isomorphic to
$B_{p}$ for every prime $p$, because in this case $\prod_{p}A(p)\simeq B$. So
it suffices to prove the lemma assuming that both $a$ and $|\Gamma|$ are
powers of the same prime $p$. Let $p$ be a prime, let $\Gamma$ be a finite
abelian $p$-group, and let $f:\Gamma\to\Gamma_{p^{e},b}$ be a surjection.
There is an isomorphism
$\phi:\Gamma\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}{\mathbb{Z}}/p^{e_{1}}\times\dots\times{\mathbb{Z}}/p^{e_{k}}$
for some natural numbers $e_{1}\geq e_{2}\geq\dots\geq e_{k}$. We have
$f(p^{e-1}A)=p^{e-1}f(A)\simeq({\mathbb{Z}}/p)^{b}$, so $e_{b}\geq e$. Then

$\phi^{-1}(p^{e_{1}-e}{\mathbb{Z}}/p^{e_{1}}\times\dots\times
p^{e_{b}-e}{\mathbb{Z}}/p^{e_{b}}\times\\{0\\}\dots\times\\{0\\})$

is a subgroup of $\Gamma$ which is isomorphic to $\Gamma_{p^{e},b}$. ∎

######  Lemma 6.3.

For any natural numbers $b$ and $C_{1}$ there exists a natural number $C_{2}$
with the following property. Suppose that $\Gamma$ is a finite group and that
$N$ is a normal subgroup of $\Gamma$ satisfying $|N|\leq C_{1}$. Suppose that
$\Gamma/N\simeq\Gamma_{a,b}$ for some natural number $a$. Then there is a
subgroup $\Gamma^{\prime}\leq\Gamma$ which is isomorphic to
$\Gamma_{a^{\prime},b}$ for some natural number $a^{\prime}$ satisfying
$C_{2}a^{\prime}\geq a$.

###### Proof.

By assumption there is a surjective morphism $\pi:\Gamma\to Q:=\Gamma_{a,b}$
whose kernel is $N$. Let $c:\Gamma\to\operatorname{Aut}N$ be the morphism
given by the conjugation action of $\Gamma$ on $N$. Let
$\Gamma_{0}=\operatorname{Ker}c$. Then
$[\Gamma_{a,b}:\pi(\Gamma_{0})]\leq[\Gamma:\Gamma_{0}]\leq|\operatorname{Aut}N|\leq
C_{1}!$. By Lemma 6.1 there is a subgroup $Q_{0}\leq\pi(\Gamma_{0})$ which is
isomorphic to $\Gamma_{a_{0},b}$ for some natural number $a_{0}$ satisfying
$(C_{1}!)!a_{0}\geq a$. The kernel of the restriction $\pi_{0}$ of $\pi$ to
$\Gamma_{0}$ coincides with the center $Z$ of $N$. Let
$\Gamma_{1}:=\pi_{0}^{-1}(Q_{0})$. We have an exact sequence

$0\to
Z\to\Gamma_{1}\stackrel{{\scriptstyle\pi_{0}}}{{\longrightarrow}}Q_{0}\simeq\Gamma_{a_{0},b}\to
0.$

Since this is a central extension, one can define a bilinear morphism
$\beta:Q_{0}\times Q_{0}\to Z$ by setting, for any two elements $u,v\in
Q_{0}$, $\beta(u,v)=[\widetilde{u},\widetilde{v}],$ where
$\widetilde{u},\widetilde{v}\in\Gamma_{1}$ are arbitrary lifts of $u,v$.
Define a morphism of groups

$\phi_{\beta}:Q_{0}\to\operatorname{Hom}(Q_{0},Z)$

setting $(\phi_{\beta}(u))(v)=\beta(u,v)$ for every $u,v\in Q_{0}$. Now,
$Q_{0}$ can be generated by $b$ elements because
$Q_{0}\simeq\Gamma_{a_{0},b}$, so we may bound

$|\operatorname{Hom}(Q_{0},Z)|\leq|Z|^{b}\leq|N|^{b}\leq C_{1}^{b}.$

Consequently, $Q_{1}:=\operatorname{Ker}\phi_{\beta}$ satisfies
$[Q_{0}:Q_{1}]\leq C_{1}^{b}$. By construction, $\pi_{0}^{-1}(Q_{1})$ is an
abelian subgroup of $\Gamma_{1}$. Using again Lemma 6.1 we deduce the
existence of a subgroup $Q_{2}\leq Q_{1}$ which is isomophic to
$\Gamma_{a^{\prime},b}$ for some natural number $a^{\prime}$ satisfying
$(C_{1}^{b})!a^{\prime}\geq a_{0}$. Then $\pi_{0}^{-1}(Q_{2})$ is a finite
abelian group surjecting onto $\Gamma_{a^{\prime},b}$, and hence by Lemma 6.2
there is a subgroup $\Gamma^{\prime}\leq\pi_{0}^{-1}(Q_{2})\leq\Gamma$ which
is isomorphic to $\Gamma_{a^{\prime},b}$. Setting
$C_{2}=(C_{1}^{b})!(C_{1}!)!$ we have

$C_{2}a^{\prime}=(C_{1}^{b})!(C_{1}!)!a^{\prime}\geq(C_{1}!)!a_{0}\geq a.$

This finishes the proof of the lemma. ∎

The next two lemmas refer to finite subgroups of tori. Recall that we use
additive notation for the group structure on tori.

######  Lemma 6.4.

Let $a,d$ be natural numbers and suppose that $\Gamma$ is a finite subgroup of
$T^{d}$ with the property that $a\gamma=0$ for every $\gamma\in\Gamma$. Then
$|\Gamma|\leq a^{d}$. In particular, if $T^{d}$ contains a subgroup isomorphic
to $\Gamma_{a,b}$ for some $a\geq 2$ then $b\leq d$.

###### Proof.

Let $\pi:{\mathbb{R}}^{d}\to T^{d}={\mathbb{R}}^{d}/{\mathbb{Z}}^{d}$ denote
the projection. Then $\pi^{-1}(\Gamma)$ is a discrete subgroup of
${\mathbb{R}}^{d}$, so it can be generated by $d$ or fewer elements, say
$g_{1},\dots,g_{d^{\prime}}$ with $d^{\prime}\leq d$, see e.g. [11, Chap I,
Lemma (3.8)]. Let $\gamma_{i}=\pi(g_{i})$. Then
$\gamma_{1},\dots,\gamma_{d^{\prime}}$ is a generating set of $\Gamma$, so the
morphism $\Gamma_{a,d^{\prime}}\to\Gamma$ sending
$(\overline{\alpha_{1}},\dots,\overline{\alpha_{d^{\prime}}})\in\Gamma_{a,d^{\prime}}$
to $\sum\alpha_{i}\gamma_{i}$ is surjective (here $\alpha_{i}\in{\mathbb{Z}}$
and $\overline{\alpha_{i}}$ is the class of $\alpha_{i}$ in ${\mathbb{Z}}/a$).
It follows that $|\Gamma|\leq|\Gamma_{a,d^{\prime}}|=a^{d^{\prime}}\leq
a^{d}$. ∎

######  Lemma 6.5.

Let $a,d$ be natural numbers and suppose that $\Gamma$ is a finite subgroup of
$T^{d}$ satisfying $a\gamma=0$ for every $\gamma\in\Gamma$. Let $1\leq j\leq
d$ be any integer and let

$S=\\{(\theta_{1},\dots,\theta_{d})\in T^{d}\mid\theta_{i}=0\text{ for $i\neq
j$}\\}.$

Then $|\Gamma\cap S|\geq|\Gamma|/a^{d-1}$.

###### Proof.

We can identify $\Gamma/(\Gamma\cap S)$ with a finite subgroup of
$T^{d}/S\simeq T^{d-1}$, all of whose elements have order dividing $a$. Hence,
by Lemma 6.4, we have $|\Gamma/(\Gamma\cap S)|\leq a^{d-1}$. The exact
sequence $0\to\Gamma\cap S\to\Gamma\to\Gamma/(\Gamma\cap S)\to 0$ then implies

$|\Gamma\cap S|=\frac{|\Gamma|}{|\Gamma/(\Gamma\cap S)|}\geq|\Gamma|/a^{d-1},$

as we wished to prove. ∎

##  7\. Proof of Theorem 1.11

Let $X$ be a closed connected manifold and let $n$ be a natural number. We say
that two degree $n$ coverings $\rho_{i}:X_{i}\to X$ ($i=1,2$) are isomorphic
if there is a homeomorphism $f:X_{1}\to X_{2}$ such that
$\rho_{1}=\rho_{2}\circ f$. Let $\operatorname{Cov}_{n}(X)$ denote the set of
isomorphism classes of degree $n$ coverings. We denote the isomorphism class
of a degree $n$ covering $\rho:X^{\prime}\to X$ as $[\rho:X^{\prime}\to X]$.
Fix a base point $x_{0}\in X$ and write $\pi_{1}=\pi_{1}(X,x_{0})$. Let
$S_{n}$ denote the symmetric group of permutations of $\\{1,\dots,n\\}$. The
group $\pi_{1}$ is finitely generated because $X$ is closed, so the set of
group homomorphisms $\operatorname{Hom}(\pi_{1},S_{n})$ is finite. The group
$S_{n}$ acts on $\operatorname{Hom}(\pi_{1},S_{n})$ by postcomposition, via
the conjugation action of $S_{n}$ on itself. Denote by
$\operatorname{Hom}(\pi_{1},S_{n})/S_{n}$ the orbit set of this action. Since
$\operatorname{Hom}(\pi_{1},S_{n})$ is finite, so is
$\operatorname{Hom}(\pi_{1},S_{n})/S_{n}$.

Let $\rho:X^{\prime}\to X$ be a degree $n$ covering. Taking a bijection
between $\rho^{-1}(x_{0})$ and $\\{1,\dots,n\\}$, the monodromy of $\rho$ can
be identified with an element in $\operatorname{Hom}(\pi_{1},S_{n})$, whose
class in $\operatorname{Hom}(\pi_{1},S_{n})/S_{n}$ is independent of the
chosen bijection. The resulting element in
$\operatorname{Hom}(\pi_{1},S_{n})/S_{n}$ only depends on the isomorphism
class of $\rho:X^{\prime}\to X$, so this construction gives a map
$\mu:\operatorname{Cov}_{n}(X)\to\operatorname{Hom}(\pi_{1},S_{n})/S_{n}.$ It
is an elementary fact that $\mu$ is a bijection. Hence, the set
$\operatorname{Cov}_{n}(X)$ is finite.

Let $G$ be a group acting continuously on $X$. For any $g\in G$ denote by
$h_{g}:X\to X$ the homeomorphism given by the action of $g$ on $X$. The group
$G$ acts on $\operatorname{Cov}_{n}(X)$ as follows: if $[\rho:X^{\prime}\to
X]\in\operatorname{Cov}_{n}(X)$, then $g\cdot[\rho:X^{\prime}\to
X]=[h_{g}^{*}\rho:h_{g}^{*}X^{\prime}\to X]$. Since
$\operatorname{Cov}_{n}(X)$ is finite, there exists a constant $C_{n}(X)$,
which only depends on $X$, and a subgroup $G_{0}\leq G$ whose action on
$\operatorname{Cov}_{n}(X)$ is trivial and which satisfies $[G:G_{0}]\leq
C_{n}(X)$. Let $\rho:X^{\prime}\to X$ be a finite covering of degree $n$.
Define

$G_{0}^{\prime}=\\{(g,f)\mid g\in G_{0},\,f:X^{\prime}\to X^{\prime}\text{
homeomorfism lifting $h_{g}$}\\}.$

Then $G_{0}^{\prime}$ is a group (with product defined componentwise) and it
acts on $X^{\prime}$ via the projection $(g,f)\mapsto f$. If the action of $G$
on $X$ is effective, then so is the action of $G_{0}$ on $X^{\prime}$. The
fact that the isomorphism class $[\rho:X^{\prime}\to X]$ is preserved by
$G_{0}$ implies that the projection $\pi:G_{0}^{\prime}\to G_{0}$ sending
$(g,f)$ to $g$ is surjective. Furthermore, the kernel of $\pi$ can be
identified with a subgroup of the permutation group of $\rho^{-1}(x_{0})$, and
hence has at most $n!$ elements.

Theorem 1.11 follows immediately from the next statement.

######  Lemma 7.1.

Given a finite covering $\rho:X^{\prime}\to X$ and a natural number $b$ there
is a constant $C$ such that, for every natural number $a$, if $\Gamma_{a,b}$
acts effectively on $X$ then there is a natural number $a^{\prime}$ satisfying
$Ca^{\prime}\geq a$ and an effective action of $\Gamma_{a^{\prime},b}$ on
$X^{\prime}$.

###### Proof.

Let $n$ be the degree of $\rho$, and assume that $\Gamma:=\Gamma_{a,b}$ acts
effectively on $X$. By the previous considerations, there is a subgroup
$\Gamma_{0}\leq\Gamma$ satisfying $[\Gamma:\Gamma_{0}]\leq C_{n}(X)$ and an
exact sequence

$1\to
F\to\Gamma_{0}^{\prime}\stackrel{{\scriptstyle\pi}}{{\longrightarrow}}\Gamma_{0}\to
1$

where $|F|\leq n!$ and $\Gamma_{0}^{\prime}$ acts effectively on $X^{\prime}$.
By Lemma 6.1 there is a subgoup $\Gamma_{1}\leq\Gamma_{0}$ isomorphic to
$\Gamma_{a_{1},b}$ for some integer $a_{1}$ satisfying $C_{n}(X)!a_{1}\geq a$.
Let $\Gamma_{1}^{\prime}=\pi^{-1}(\Gamma_{1})$. By Lemma 6.3 there is a
subgroup $\Gamma^{\prime\prime}\leq\Gamma_{1}^{\prime}$ isomorphic to
$\Gamma_{a^{\prime},b}$ for some natural number $a^{\prime}$ satisfying
$C^{\prime}a^{\prime}\geq a_{1}$, where $C^{\prime}$ only depends on $n$
(through the bound $|F|\leq n!$) and $b$. Setting $C=C_{n}(X)!C^{\prime}$ we
have $Ca^{\prime}\geq a$. Since $\Gamma_{0}^{\prime}$ acts effectively on
$X^{\prime}$, so does $\Gamma^{\prime\prime}$, so the proof is complete. ∎

##  8\. Proof of Theorem 1.12

First we introduce some notation. Let
$\operatorname{Aff}_{{\mathbb{Z}}^{n}}{\mathbb{R}}^{n}$ denote the group of
affine transformations of ${\mathbb{R}}^{n}$ that send the lattice
${\mathbb{Z}}^{n}$ to some translate of itself. There an exact sequence

$0\to{\mathbb{R}}^{n}\to\operatorname{Aff}_{{\mathbb{Z}}^{n}}{\mathbb{R}}^{n}\to\operatorname{GL}(n,{\mathbb{Z}})\to
1.$

The transformations in $\operatorname{Aff}_{{\mathbb{Z}}^{n}}{\mathbb{R}}^{n}$
descend to give affine diffeomorphisms of $T^{n}$. We denote the resulting
group of transformations of $T^{n}$ as $\operatorname{Aff}T^{n}$. This
coincides, as the notation suggests, with the group of affine transformations
of $T^{n}$. There is an exact sequence

$0\to
T^{n}\stackrel{{\scriptstyle\tau}}{{\longrightarrow}}\operatorname{Aff}T^{n}\stackrel{{\scriptstyle\mu}}{{\longrightarrow}}\operatorname{GL}(n,{\mathbb{Z}})\to
1,$

where $\tau$ sends $a\in T^{n}$ to the translation $b\mapsto a+b$. Using the
identification $T^{n}={\mathbb{R}}^{n}/{\mathbb{Z}}^{n}$ we may naturally
identify $H_{1}(T^{n};{\mathbb{Z}})\simeq{\mathbb{Z}}^{n}$. The morphism
$\sigma:\operatorname{GL}(n,{\mathbb{Z}})\to\operatorname{Aff}T^{n}$ induced
by the action of $\operatorname{GL}(n,{\mathbb{Z}})$ on ${\mathbb{R}}^{n}$
satisfies
$\mu\circ\sigma=\operatorname{Id}_{\operatorname{GL}(n,{\mathbb{Z}})}$. If
$A\in\operatorname{GL}(n,{\mathbb{Z}})$, the action of $\sigma(A)$ on
$H_{1}(T^{n};{\mathbb{Z}})$ coincides, via the isomorphism
$H_{1}(T^{n};{\mathbb{Z}})\simeq{\mathbb{Z}}^{n}$, with $A$. The following
lemma results from combining [36, Corollary to Lemma 1] and [36, Theorem 3].

######  Lemma 8.1.

Suppose that a finite group $\Gamma$ acts effectively on $T^{n}$. Using the
isomorphism $H_{1}(T^{n};{\mathbb{Z}})\simeq{\mathbb{Z}}^{n}$ the action of
$\Gamma$ on $H_{1}(T^{n};{\mathbb{Z}})$ gives a morphism
$\rho:\Gamma\to\operatorname{GL}(n,{\mathbb{Z}})$. Then there is an embedding
of groups $\eta:\Gamma\hookrightarrow\operatorname{Aff}T^{n}$ such that
$\mu\circ\eta=\rho$.

Fix natural numbers $k,n$ satisfying $1\leq k\leq n-1$. Recall that
$\sigma\in\operatorname{Aff}T^{n}$ is the involution defined by
$\sigma(x_{1},\dots,x_{n})=(x_{1}+1/2,\dots,x_{k}+1/2,-x_{k+1},\dots,-x_{n}),$
and that $X^{\prime}=T^{n}$, $X=T^{n}/\sigma$ and $\rho:X^{\prime}\to X$
denotes the projection.

Since $\Gamma_{r,n}$ acts effectively on $X^{\prime}$ for every $r$, we have
$\operatorname{disc-sym}X^{\prime}\geq n$. The action of $\Gamma_{r,k}$ on
$T^{n}$ given by

$(\overline{a_{1}},\dots,\overline{a_{k}})\cdot(x_{1},\dots,x_{n})=(x_{1}+a_{1}/r,\dots,x_{k}+a_{k}/r,a_{k+1},\dots,a_{n})$

commutes with $\sigma$, and hence defines an action of $\Gamma_{r,k}$ on $X$.
This action is effective if $r$ is odd, and hence we conclude that
$\operatorname{disc-sym}X\geq k$.

The inequality $\operatorname{disc-sym}X^{\prime}\leq n$ follows from (1) in
Theorem 1.13 (see Section 12). Let us prove that $\operatorname{disc-sym}X\leq
k$. We will need the following result.

######  Lemma 8.2.

Let $T_{\sigma}^{n}=\\{x\in T^{n}\mid\tau(x)\sigma=\sigma\tau(x)\\}$. There is
a short exact sequence of the form $0\to
T^{k}\stackrel{{\scriptstyle\iota}}{{\longrightarrow}}T_{\sigma}^{n}\to\Gamma_{2,n-k}\to
0.$

###### Proof.

Let $\iota:T^{k}\to T_{\sigma}^{n}$ be
$\iota((x_{1},\dots,x_{k}))=(x_{1},\dots,x_{k},0,\dots,0)$. Since
$(x_{1},\dots,x_{n})\in T^{n}$ belongs to $T_{\sigma}^{n}$ if and only if
$2x_{i}=0$ for every $i\geq k+1$, we have
$T_{\sigma}^{n}/\iota(T^{k})\simeq\Gamma_{2,n-k}$. ∎

Suppose that $\Gamma_{r,m}$ acts effectively on $X$. The arguments in the
proof of Theorem 1.11 imply the existence of a constant $C^{\prime}$ depending
only on $k$ and $n$, a subgroup $\Gamma_{0}\leq\Gamma_{r,m}$ satisfying
$[\Gamma_{r,m}:\Gamma_{0}]\leq C^{\prime}$, and a central extension of groups

$1\to Z\to\Gamma_{0}^{\prime}\to\Gamma_{0}\to 1$

such that $\Gamma_{0}^{\prime}$ acts effectively on $X$, and
$Z=\\{\operatorname{Id},\sigma\\}$. This implies that the order of every
element of $\Gamma_{0}^{\prime}$ is smaller than or equal to $2r$. By Lemma
8.1 there is an injective morphism
$\eta:\Gamma_{0}^{\prime}\to\operatorname{Aff}T^{n}$ satisfying
$\rho=\mu\circ\eta$, where
$\rho:\Gamma_{0}^{\prime}\to\operatorname{GL}(n,{\mathbb{Z}})$ is the morphism
induced by the action of $\Gamma_{0}^{\prime}$ on
$H_{1}(T^{n};{\mathbb{Z}})\simeq{\mathbb{Z}}^{n}$. The subgroup
$\Gamma_{0}^{\prime\prime}:=\operatorname{Ker}\rho$ satisfies
$[\Gamma_{0}^{\prime}:\Gamma_{0}^{\prime\prime}]\leq C$, where $C$ is given by
Lemma 3.1 applied to $\operatorname{GL}(n,{\mathbb{Z}})$ (so $C$ only depends
on $n$). Then $\Gamma_{0}^{\prime\prime}\leq\tau(T_{\sigma}^{n})$, so by Lemma
8.2 there is a subgroup
$\Gamma_{0}^{\prime\prime\prime}\leq\Gamma_{0}^{\prime\prime}$ satisfying
$[\Gamma_{0}^{\prime\prime}:\Gamma_{0}^{\prime\prime\prime}]\leq 2^{n-k}$ and
an embedding $\Gamma_{0}^{\prime\prime\prime}\hookrightarrow T^{k}$. By Lemma
6.4 we have $|\Gamma_{0}^{\prime\prime\prime}|\leq(2r)^{k}$ because
$\Gamma_{0}^{\prime\prime\prime}\leq\Gamma_{0}^{\prime}$, so

$r^{m}=|\Gamma_{r,m}|\leq
C^{\prime}|\Gamma_{0}|=\frac{C^{\prime}}{2}|\Gamma_{0}^{\prime}|\leq\frac{C^{\prime}}{2}C2^{n-k}(2r)^{k}=C^{\prime}C2^{n-1}r^{k}.$

Consequently, if $r>C^{\prime}C2^{n-1}$ then $m\leq k$.

##  9\. Finite generation of the cohomology of abelian covers

Denote by $\pi:{\mathbb{R}}^{k}\to T^{k}={\mathbb{R}}^{k}/{\mathbb{Z}}^{k}$
the quotient map. Given a topological space $X$ and a continuous map
$\phi:X\to T^{k}$ we denote by

$X_{\phi}=\\{(x,u)\in X\times{\mathbb{R}}^{k}\mid\phi(x)=\pi(u)\\}$

the pullback to $X$ of the covering ${\mathbb{R}}^{k}\to T^{k}$. The
projection

$\rho_{\phi}:X_{\phi}\to X,\qquad\rho_{\phi}(x,u)=x$

is an unramified covering map. We can also look at $\rho_{\phi}:X_{\phi}\to X$
as a principal ${\mathbb{Z}}^{k}$-bundle, where the action of
${\mathbb{Z}}^{k}$ on $X_{\phi}$ can be described as follows: if
$\nu\in{\mathbb{Z}}^{k}$ and $(x,u)\in X_{\phi}$ then
$\nu\cdot(x,s)=(x,u+\nu)$. Standard results on fiber bundles imply the
following.

######  Lemma 9.1.

Suppose that $X$ is paracompact. If two continuous maps $\phi,\psi:X\to T^{k}$
are homotopic then there is a ${\mathbb{Z}}^{k}$-equivariant homeomorphism
$\zeta:X_{\phi}\to X_{\psi}$ such that $\rho_{\phi}=\rho_{\psi}\circ\zeta$.

The action of ${\mathbb{Z}}^{k}$ on $X_{\phi}$ induces an action of
${\mathbb{Z}}^{k}$ on $H_{*}(X_{\phi};{\mathbb{Z}})$ by group automorphisms.
Let $e_{1},\dots,e_{k}$ denote the canonical basis of ${\mathbb{Z}}^{k}$, and
let $t_{i}:{\mathbb{Z}}^{k}\to{\mathbb{Z}}$ denote the characteristic function
of $\\{e_{i}\\}\subset{\mathbb{Z}}^{k}$. Then $t_{i}$ belongs to the group
ring ${\mathbb{Z}}[{\mathbb{Z}}^{k}]$, which we identify with the additive
group of finitely supported functions ${\mathbb{Z}}^{k}\to{\mathbb{Z}}$ with
ring structure given by convolution. Furthermore,
${\mathbb{Z}}[{\mathbb{Z}}^{k}]$ is isomorhic to ${\mathbb{Z}}[t_{1}^{\pm
1},\dots,t_{k}^{\pm 1}]$. The action of ${\mathbb{Z}}^{k}$ on
$H_{*}(X_{\phi};{\mathbb{Z}})$ defines a unique structure of module over the
group ring ${\mathbb{Z}}[{\mathbb{Z}}^{k}]\simeq{\mathbb{Z}}[t_{1}^{\pm
1},\dots,t_{k}^{\pm 1}]$ on $H_{*}(X_{\phi};{\mathbb{Z}})$.

######  Lemma 9.2.

Let $X$ be a closed topological manifold, and let $\phi:X\to T^{k}$ be a
continuous map. Then $H_{*}(X_{\phi};{\mathbb{Z}})$ is finite generated as a
${\mathbb{Z}}[t_{1}^{\pm 1},\dots,t_{k}^{\pm 1}]$-module.

###### Proof.

Since compact topological manifolds are Euclidean Neighborhood Retracts (see
e.g. [27, Corollary A.9]) we can identify homeomorphically $X$ with a closed
subset of some Euclidean space ${\mathbb{R}}^{N}$ in such a way that there
exists an open subset ${\mathcal{O}}\subset{\mathbb{R}}^{N}$ containing $X$
and a retraction $r:{\mathcal{O}}\to X$. For any $x\in X$ let
$B_{x}\subset{\mathcal{O}}$ be an open ball centered at $x$. By compactness we
may choose a finite set of points $x_{1},\dots,x_{s}\in X$ such that $X\subset
Y:=B_{x_{1}}\cup\dots\cup B_{x_{s}}$. Let $B_{i}=B_{x_{i}}$ and let
$\psi=\phi\circ r:Y\to T^{k}$.

Let $A:={\mathbb{Z}}[t_{1}^{\pm 1},\dots,t_{k}^{\pm 1}]$. For every $i$ the
space $B_{i}$ is contractible, so by Lemma 9.1 the principal
${\mathbb{Z}}^{k}$-bundle $(B_{i})_{\psi}\to B_{i}$ is trivial. Hence, for
every subspace $S\subseteq B_{i}$ we have an isomorphism of $A$-modules
$H_{*}(S_{\psi};{\mathbb{Z}})\simeq
H_{*}(S;{\mathbb{Z}})\otimes_{{\mathbb{Z}}}A$. So if $S\subseteq B_{i}$ has
the property that $H_{*}(S;{\mathbb{Z}})$ is a finitely generated abelian
group, then $H_{*}(S_{\psi};{\mathbb{Z}})$ is a finitely generated $A$-module.
It also follows that $H_{*}((B_{i})_{\psi};{\mathbb{Z}})$ is a free $A$-module
of rank $1$.

For any $1\leq j\leq s$ we denote $B_{\leq j}=B_{1}\cup\dots\cup B_{j}$. We
claim that $H_{*}((B_{\leq j})_{\psi};{\mathbb{Z}})$ is a finitely generated
$A$-module for every $1\leq j\leq s$. The case $j=1$ has been already been
proved. For general $j$ we use ascending induction on $j$. Suppose that $j>1$
and that the claim is true for $j-1$. The Mayer–Vietoris (MV) exact sequence:

$\dots\to H_{k}((B_{\leq j-1})_{\psi}\cap(B_{j})_{\psi};{\mathbb{Z}})\to
H_{k}((B_{\leq j-1})_{\psi};{\mathbb{Z}})\oplus
H_{k}((B_{j})_{\psi};{\mathbb{Z}})\to\\\ \to H_{k}((B_{\leq
j})_{\psi};{\mathbb{Z}})\to H_{k-1}((B_{\leq
j-1})_{\psi}\cap(B_{j})_{\psi};{\mathbb{Z}})\to\dots{}$

is actually an exact sequence of $A$-modules. This follows from the naturality
of the MV exact sequence and the fact that the $A$-module structure on each
term comes from an action of ${\mathbb{Z}}^{k}$ on each of the spaces
commuting with the inclusions

$(B_{\leq j-1})_{\psi}\hookleftarrow(B_{\leq
j-1})_{\psi}\cap(B_{j})_{\psi}\hookrightarrow(B_{j})_{\psi}$

and

$(B_{\leq j-1})_{\psi}\hookrightarrow(B_{\leq
j})_{\psi}\hookleftarrow(B_{j})_{\psi}.$

By the induction hypothesis $H_{k}((B_{\leq j-1})_{\psi};{\mathbb{Z}})\oplus
H_{k}(B_{j})_{\psi};{\mathbb{Z}})$ is a finitely generated $A$-module. We have
$(B_{\leq j-1})_{\psi}\cap(B_{j})_{\psi}=(B_{\leq j-1}\cap B_{j})_{\psi}$,
$B_{\leq j-1}\cap B_{j}$ is obviously a subset of $B_{j}$, and $H_{*}(B_{\leq
j-1}\cap B_{j};{\mathbb{Z}})$ is finitely generated (because $B_{\leq j-1}\cap
B_{j}$ is the union of finitely many convex subsets of ${\mathbb{R}}^{N}$).
Hence, $H_{k}((B_{\leq j-1})_{\psi}\cap(B_{j})_{\psi};{\mathbb{Z}})$ is a
finitely generated $A$-module.

Since $A$ is Noetherian, the previous considerations and the exactness of the
sequence imply that $H_{k}((B_{\leq j})_{\psi};{\mathbb{Z}})$ is a finitely
generated $A$-module for every $k$. Finally, $(B_{\leq j})_{\psi}$ is an
$N$-dimensional topological manifold, so its homology vanishes in dimensions
bigger than $N$. It follows that the entire homology $H_{*}((B_{\leq
j})_{\psi};{\mathbb{Z}})$ is a finitely generated $A$-module, so the proof of
the claim is complete.

To conclude the proof note that the inclusion $\iota:X\hookrightarrow Y$ and
the retraction $r:Y\to X$ induce ${\mathbb{Z}}^{k}$-equivariant maps
$\iota^{\prime}:X_{\phi}\hookrightarrow Y_{\psi}$ and $r^{\prime}:Y_{\psi}\to
X_{\phi}$ satisfying
$r^{\prime}\circ\iota^{\prime}=\operatorname{Id}_{X_{\phi}}$. It follows that
$H_{*}(X_{\phi};{\mathbb{Z}})$ is an $A$-submodule of
$H_{*}(Y_{\psi};{\mathbb{Z}})$. Since $A$ is Noetherian and
$H_{*}(Y_{\psi};{\mathbb{Z}})$ is finitely generated, it follows that
$H_{*}(X_{\phi};{\mathbb{Z}})$ is also finitely generated. ∎

##  10\. Proof of Theorem 1.1

Let $X$ be a rationally hypertoral $n$-dimensional manifold satisfying
$\operatorname{disc-sym}(X)\geq n$, so that $X$ supports effective actions of
$\Gamma_{r,n}=({\mathbb{Z}}/r)^{n}$ for arbitrarily large integers $r$. Fix a
continuous map $\phi:X\to T^{n}$ of nonzero degree. Let $d$ denote the
absolute value of the degree of $\phi$. Define the principal
${\mathbb{Z}}^{n}$-bundle $X_{\phi}\to X$ as in the previous section. The
action of ${\mathbb{Z}}^{n}$ on $X_{\phi}$ induces an action of
${\mathbb{Z}}^{n}$ on $H_{*}(X_{\phi};{\mathbb{Z}})$ by isomorphisms of
${\mathbb{Z}}$-modules. Hence $H_{*}(X_{\phi};{\mathbb{Z}})$ is a module over
the ring ${\mathbb{Z}}[{\mathbb{Z}}^{n}]={\mathbb{Z}}[t_{1}^{\pm
1},\dots,t_{n}^{\pm 1}]$.

The following lemma describes how certain homeomorphisms of $X$ lift to
homeomorphisms of $X_{\psi}$ for suitable maps $\psi:X\to T^{n}$. Recall that
for any $a\in T^{n}$ we denote by $\tau(a):T^{n}\to T^{n}$ the translation
$\tau(a)(t)=t+a$.

######  Lemma 10.1.

Let $f:X\to X$ be a homeomorphism of order $r$, and suppose that
$\tau(a)\circ\psi=\psi\circ f$ for some $a\in T^{n}$, which necessarily
satisfies $ra=0$. Let $v\in{\mathbb{R}}^{n}$ satisfy $\pi(v)=a$. Then there
exist a lift $g:X_{\psi}\to X_{\psi}$ satisfying $g^{r}(x,u)=(x,u+rv)$ for
every $(x,u)\in X_{\psi}$.

###### Proof.

Recall that $X_{\psi}=\\{(x,u)\in
X\times{\mathbb{R}}^{k}\mid\psi(x)=\pi(u)\\}$. Define $g:X_{\psi}\to X_{\psi}$
by $g(x,u)=(f(x),u+v)$. The equality $\tau(a)\circ\psi=\psi\circ f$ guarantees
that this is indeed a well defined homeomorphism of $X_{\psi}$. The definition
immediately implies that $g^{r}(x,u)=(x,u+rv)$ for every $(x,u)\in X_{\psi}$.
∎

######  Lemma 10.2.

For every $1\leq j\leq n$ there exists a nonzero integer $d_{j}$ and a
sequence of natural numbers $o_{i,j}$ satisfying $o_{i,j}\to\infty$ as
$i\to\infty$, together with isomorphisms of groups
$w_{i,j}:H_{*}(X_{\phi};{\mathbb{Z}})\to H_{*}(X_{\phi};{\mathbb{Z}})$, such
that $w_{i,j}^{o_{i,j}}$ coincides with multiplication by $t_{j}^{d_{j}}$.

###### Proof.

By Lemma 3.2 there exists a number $C$ such that for every action of a finite
group $G$ on $X$ the kernel $G^{\prime}$ of the natural map
$G\to\operatorname{Aut}H^{1}(X;{\mathbb{Z}})$ satisfies $[G:G^{\prime}]\leq
C$. Let $0<r_{1}<r_{2}<\dots$ be the infinite sequence of integers such that
$X$ supports an effective action of $G_{i}:=\Gamma_{r_{i},n}$ for every $i$.
For every $i$ let
$G_{i}^{\prime}=\operatorname{Ker}(G_{i}\to\operatorname{Aut}H^{1}(X;{\mathbb{Z}}))$.
Then $[G_{i}:G_{i}^{\prime}]\leq C$. By Lemma 6.1 there is a subgroup
$G_{i}^{\prime\prime}\leq G_{i}$ such that
$G_{i}^{\prime\prime}\simeq\Gamma_{s_{i},n}$ for a natural number $s_{i}$
satisfying $C!s_{i}\geq r_{i}$. In particular, $s_{i}\to\infty$ as
$i\to\infty$.

Applying Theorem 2.1 to the action of $G_{i}^{\prime\prime}$ on $X$ we obtain
a morphism of groups

$\eta_{i}:G_{i}^{\prime\prime}\to T^{n}$

and an $\eta_{i}$-equivariant map $\psi_{i}:X\to T^{n}$ which is homotopic to
$\phi$. Also, $|\operatorname{Ker}\eta_{i}|\leq d$, so
$|\eta(G_{i}^{\prime\prime})|\geq s_{i}^{n}/d$. For every $1\leq j\leq n$ let
$S_{j}=\\{(\theta_{1},\dots,\theta_{n})\in T^{n}\mid\theta_{k}=0\text{ for
$k\neq j$}\\}$.

By Lemma 6.5 we have

$\overline{o}_{i,j}:=|\eta(G_{i}^{\prime\prime})\cap
S_{j}|\geq\frac{s_{i}^{n}}{d\cdot s_{i}^{n-1}}=\frac{s_{i}}{d}.$

The element $e_{i,j}=\pi(e_{j}/\overline{o}_{i,j})\in T^{n}$ is a generator of
$\eta(G_{i}^{\prime\prime})\cap S_{j}$. Let $f_{i,j}:X\to X$ be the
homeomorphism given by the action of an element of $G_{i}^{\prime\prime}$
which is sent to $e_{i,j}$ by the morphism $\eta$. Then we have $\psi_{i}\circ
f_{i,j}=\tau(e_{j}/\overline{o}_{i,j})\circ\psi_{i}$. Furthermore, the order
of $f_{i,j}$ is a number of the form

$o_{i,j}=d_{i,j}\overline{o}_{i,j},$

where $d_{i,j}$ is a natural number satisfying $1\leq d_{i,j}\leq d$, because
$|\operatorname{Ker}\eta_{i}|\leq d$. Passing to a subsequence and relabelling
accordingly we may assume that all natural numbers $d_{1,j},d_{2,j},\dots$ are
equal to the same number $d_{j}$. By Lemma 10.1 there is a homeomorphism
$g_{i,j}:X_{\psi_{i}}\to X_{\psi_{i}}$ such that
$g_{i,j}^{o_{i,j}}:X_{\psi}\to X_{\psi}$ coincides with the action of
$d_{j}e_{j}$ on $X_{\psi}$ given by the structure of principal
${\mathbb{Z}}^{n}$-bundle on $X_{\psi}$.

Since $\psi_{i}$ is homotopic to $\phi$, by Lemma 9.1 there is a
${\mathbb{Z}}^{n}$-equivariant homeomorphism $\zeta_{i}:X_{\phi}\to
X_{\psi_{i}}$. Let $w_{i,j}:H_{*}(X_{\phi};{\mathbb{Z}})\to
H_{*}(X_{\phi};{\mathbb{Z}})$ be the isomorphism induced by the homeomorphism

$\zeta_{i}^{-1}\circ g_{i,j}\circ\zeta_{i}:X_{\phi}\to X_{\phi}.$

Then $w_{i,j}^{o_{i,j}}$ coincides with multiplication by $t_{j}^{d_{j}}$.
Since ${\overline{o}}_{i,j}\geq s_{i}/d$ and $s_{i}\to\infty$ as $i\to\infty$,
we conclude that $o_{i,j}\to\infty$ as $i\to\infty$. ∎

Combining the previous lemma with Corollary 5.3 it follows that
$H_{*}(X_{\phi};{\mathbb{Z}})$ is a finitely generated ${\mathbb{Z}}$-module.
(Note that we may apply Corollary 5.3 to the ${\mathbb{Z}}[t_{1}^{\pm
1},\dots,t_{n}^{\pm 1}]$-module $H_{*}(X_{\phi};{\mathbb{Z}})$ thanks to Lemma
9.2.)

######  Lemma 10.3.

We have $H_{k}(X_{\phi};{\mathbb{Z}})=0$ for every $k>0$.

###### Proof.

Suppose that $H_{k}(X_{\phi};{\mathbb{Z}})\neq 0$ for some $k>0$. By the
universal coefficient theorem, there exists some prime $p$ such that
$H_{k}(X_{\phi};{\mathbb{Z}}/p)\neq 0$ for some $k>0$. The action of
${\mathbb{Z}}^{n}$ on $X_{\phi}$ induces a morphism
$\alpha_{p}:{\mathbb{Z}}^{n}\to\operatorname{Aut}(H_{*}(X_{\phi};{\mathbb{Z}}/p))$.
Since $H_{*}(X_{\phi};{\mathbb{Z}})$ is a finitely generated
${\mathbb{Z}}$-module, $H_{*}(X_{\phi};{\mathbb{Z}}/p)$ is a finite group
(again by the universal coefficient theorem). Consequently, the group
$\Lambda=\operatorname{Ker}\alpha_{p}$ has finite index in ${\mathbb{Z}}^{n}$.

Consider the action of $\Lambda$ on $X_{\phi}\times{\mathbb{R}}^{n}$ given by
$\lambda\cdot((x,u),v)=(\lambda\cdot(x,u),v-\lambda)=((x,u+\lambda),v-\lambda)$,
and let $X_{\phi}\times_{\Lambda}{\mathbb{R}}^{n}$ denote the quotient of
$X_{\phi}\times{\mathbb{R}}^{n}$ under this action. There are natural maps

$\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
20.3976pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{X_{\phi}\times_{\Lambda}{\mathbb{R}}^{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx<EMAIL_ADDRESS>0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-2.39166pt\hbox{$\scriptstyle{\Pi}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 0.0pt\raise-29.76665pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
26.67537pt\raise 5.39166pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-2.39166pt\hbox{$\scriptstyle{\Theta}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 44.3976pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
44.3976pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{X_{\phi}/\Lambda}$}}}}}}}{\hbox{\kern-15.44452pt\raise-40.80551pt\hbox{\hbox{\kern
0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{{\mathbb{R}}^{n}/\Lambda}$}}}}}}}\ignorespaces}}}}\ignorespaces,$

where $\Pi$ is induced by the projection
$X_{\phi}\times{\mathbb{R}}^{n}\to{\mathbb{R}}^{n}$ and $\Theta$ is induced by
the projection $X_{\phi}\times{\mathbb{R}}^{n}\to X_{\phi}$. The map $\Pi$ is
a fibration with fiber $X_{\phi}$, and $\Theta$ is a fibration with fiber
${\mathbb{R}}^{n}$ and hence is a homotopy equivalence. (This is of course a
general phenomenon: $X_{\phi}\times_{\Lambda}{\mathbb{R}}^{n}$ is the Borel
construction for the action of $\Lambda$ on $X_{\phi}$, and the fact that
$X_{\phi}\times_{\Lambda}{\mathbb{R}}^{n}$ is homotopy equivalent to the
quotient $X_{\phi}/\Lambda$ is a consequence of the fact that $\Lambda$ acts
freely on $X_{\phi}$.) Since $X_{\phi}/\Lambda$ is an $n$-dimensional
manifold, we have

(4)
$H_{k}(X_{\phi}\times_{\Lambda}{\mathbb{R}}^{n};{\mathbb{Z}}/p)=H_{k}(X_{\phi}/\Lambda;{\mathbb{Z}}/p)=0\qquad\text{for
every $k>n$}.$

The monodromy action of $\pi_{1}({\mathbb{R}}^{n}/\Lambda)\simeq\Lambda$ on
the homology over ${\mathbb{Z}}/p$ of any given fiber of the fibration $\Pi$
coincides with the action of $\Lambda$ on $H_{*}(X_{\phi};{\mathbb{Z}}/p)$,
and hence is trivial. Consequently, the homology Serre spectral sequence for
the fibration $\Pi$ takes the form

$H_{a}({\mathbb{R}}^{n}/\Lambda;H_{b}(X_{\phi};{\mathbb{Z}}/p))\simeq
H_{a}({\mathbb{R}}^{n}/\Lambda;{\mathbb{Z}}/p)\otimes_{{\mathbb{Z}}/p}H_{b}(X_{\phi};{\mathbb{Z}}/p)\Longrightarrow
H_{a+b}(X_{\phi}\times_{\Lambda}{\mathbb{R}}^{n};{\mathbb{Z}}/p).$

Let $l=\max\\{k\mid H_{k}(X_{\phi};{\mathbb{Z}}/p)\neq 0\\}$. By our choice of
$p$, we have $l>0$. Then
$H_{n}({\mathbb{R}}^{n}/\Lambda;{\mathbb{Z}}/p)\otimes_{{\mathbb{Z}}/p}H_{l}(X_{\phi};{\mathbb{Z}}/p)$
is a nonzero entry in the second page of the spectral sequence, and for
dimension reasons it is contained in the kernel of every differential and none
of its elements is killed by any differential; consequently,
$H_{n}({\mathbb{R}}^{n}/\Lambda;{\mathbb{Z}}/p)\otimes_{{\mathbb{Z}}/p}H_{l}(X_{\phi};{\mathbb{Z}}/p)$
can be identified with a subquotient of
$H_{n+l}(X_{\phi}\times_{\Lambda}{\mathbb{R}}^{n};{\mathbb{Z}}/p)$. Hence (4)
implies that $l=0$, which is a contradiction. ∎

######  Lemma 10.4.

$\pi_{0}(X_{\phi})$ is finite, and the action of ${\mathbb{Z}}^{n}$ on
$X_{\phi}$ induces a transitive action on $\pi_{0}(X_{\phi})$.

###### Proof.

Recall that $\rho_{\phi}:X_{\phi}\to X$ is the projection
$\rho_{\phi}(x,p)=x$, and that $X_{\phi}$ has a structure of principal
${\mathbb{Z}}^{n}$-bundle over $X$ with projection $\rho_{\phi}$. Fix some
base point $x_{0}\in X$. Since $X$ is arcconnected, each arconnected component
of $X_{\phi}$ intersects $\rho^{-1}(x_{0})$ nontrivially. This implies that
the action of ${\mathbb{Z}}^{n}$ on $\pi_{0}(X_{\phi})$ is transitive.

Again because $X$ is arcconnected, $\pi_{0}(X_{\phi})$ can be identified with
the set ${\mathcal{O}}$ of orbits of the monodromy action of
$\pi_{1}(X,x_{0})$ on $\rho_{\phi}^{-1}(x_{0})$. Since the bundle
$\rho_{\phi}:X_{\phi}\to X$ is the pullback of $\pi:{\mathbb{R}}^{n}\to T^{n}$
by $\phi$, the set ${\mathcal{O}}$ can be identified with the set
${\mathcal{O}}^{\prime}$ of orbits of the action of
$\phi_{*}\pi_{1}(X,x_{0})\leq\pi_{1}(T^{n},\phi(x_{0}))$ on
$\pi^{-1}(\phi(x_{0}))$. Since $\pi^{-1}(\phi(x_{0}))$ is a torsor over
$\pi_{1}(T^{n},\phi(x_{0}))$ (via the monodromy action), the choice of any
element $y\in\pi^{-1}(\phi(x_{0}))$ gives a bijection between
${\mathcal{O}}^{\prime}$ and
$\pi_{1}(T^{n},\phi(x_{0}))/\phi_{*}\pi_{1}(X,x_{0})$. Summarizing, we have
proved that $\pi_{0}(X_{\phi})$ can be put in bijection with
$\pi_{1}(T^{n},\phi(x_{0}))/\phi_{*}\pi_{1}(X,x_{0})$.

We thus need to prove that
$\pi_{1}(T^{n},\phi(x_{0}))/\phi_{*}\pi_{1}(X,x_{0})$ is finite. Since
$\pi_{1}(T^{n},\phi(x_{0}))$ is abelian, the previous quotient is isomorphic
to $H_{1}(T^{n};{\mathbb{Z}})/\phi_{*}H_{1}(X;{\mathbb{Z}})$. Now suppose that
the index of $\phi_{*}H_{1}(X;{\mathbb{Z}})$ in $H_{1}(T^{n};{\mathbb{Z}})$ is
infinite. Then
$\phi_{*}H_{1}(X;{\mathbb{Z}})\otimes_{{\mathbb{Z}}}{\mathbb{Q}}=\phi_{*}H_{1}(X;{\mathbb{Q}})$
is a proper subspace of
$H_{1}(T^{n};{\mathbb{Z}})\otimes_{{\mathbb{Z}}}{\mathbb{Q}}=H_{1}(T^{n};{\mathbb{Q}})$.
Hence there exists some nonzero $\beta\in H^{1}(T^{n};{\mathbb{Q}})$ inducing
a morphism $H_{1}(T^{n};{\mathbb{Q}})\to{\mathbb{Q}}$ whose kernel contains
$\phi_{*}H_{1}(X;{\mathbb{Q}})$. This implies that $\phi^{*}\beta=0.$

Since $\beta\neq 0$ there exist classe
$\beta_{1}=\beta,\beta_{2},\dots,\beta_{n}\in H^{1}(T^{n};{\mathbb{Q}})$ such
that $\Omega=\beta_{1}\cup\dots\cup\beta_{n}$ is a generator of
$H^{n}(T^{n};{\mathbb{Q}})$. Then $\phi^{*}\beta=0$ implies
$\phi^{*}\Omega=0$, so $\phi^{*}:H^{n}(T^{n};{\mathbb{Q}})\to
H^{n}(X;{\mathbb{Q}})$ is the zero map. This contradicts the assumption that
$\deg\phi\neq 0$.

Consequently, the index of $\phi_{*}H_{1}(X;{\mathbb{Z}})$ in
$H_{1}(T^{n};{\mathbb{Z}})$ is finite, so $\pi_{0}(X_{\phi})$ is finite. ∎

Fix an arcconnected component $X_{\phi}^{0}\subseteq X_{\phi}$. Since
$H_{k}(X_{\phi};{\mathbb{Z}})=0$ for every $>0$, $X_{\phi}^{0}$ has the same
integral homology as the point:

(5) $H_{0}(X_{\phi}^{0};{\mathbb{Z}})\simeq{\mathbb{Z}}\qquad\text{and}\qquad
H_{k}(X_{\phi}^{0};{\mathbb{Z}})=0\text{ for $k>0$}.$

Let $V\leq{\mathbb{Z}}^{n}$ be the subgroup consisting of those elements of
${\mathbb{Z}}^{n}$ whose action of $X_{\phi}$ maps $X_{\phi}^{0}$ to itself.
By the previous lemma, $V$ has finite index in ${\mathbb{Z}}^{n}$ and we have
$X=X_{\phi}/{\mathbb{Z}}^{n}=X_{\phi}^{0}/V$. Since $V$ has finite index in
${\mathbb{Z}}^{n}$, the quotient ${\mathbb{R}}^{n}/V$ is homeomorphic to
$T^{n}$. Arguing as in the definition of
$X_{\phi}\times_{\Lambda}{\mathbb{R}}^{n}$ we prove that $X=X_{\phi}^{0}/V$ is
homotopy equivalent to $X_{\phi}^{0}\times_{V}{\mathbb{R}}^{n}$, where the
latter is defined exactly as $X_{\phi}\times_{\Lambda}{\mathbb{R}}^{n}$ but
replacing $X_{\phi}$ resp. $\Lambda$ with $X_{\phi}^{0}$ resp. $V$. The fibers
of the projection
$Q:X_{\phi}^{0}\times_{V}{\mathbb{R}}^{n}\to{\mathbb{R}}^{n}/V$ can be
identified with $X_{\phi}^{0}$, and hence are acyclic by (5); it follows that
$Q$ induces an isomorphism in integral homology. It follows that
$H_{*}(X;{\mathbb{Z}})=H_{*}(X_{\phi}^{0}/V;{\mathbb{Z}})$ is isomorphic to
$H_{*}({\mathbb{R}}^{n}/V;{\mathbb{Z}})\simeq H_{*}(T^{n};{\mathbb{Z}})$,
which proves statement (1) of Theorem 1.1.

To prove statement (2) of Theorem 1.1 we momentarily assume that $\pi_{1}(X)$
is solvable. Since the projection $X_{\phi}^{0}\to X$ is a covering space, it
identifies $\pi_{1}(X_{\phi}^{0})$ with a subgroup of $\pi_{1}(X)$, so
$\pi_{1}(X_{\phi}^{0})$ is solvable. The abelianization of
$\pi_{1}(X_{\phi}^{0})$ can be identified with $H_{1}(X_{\phi}^{0})$, which we
know is $0$. A solvable group with trivial abelianization is the trivial
group, so $X_{\phi}^{0}$ is simply connected. By the Hurewicz theorem,
$X_{\phi}^{0}$ is contractible (see e.g. [27, Corollary 4.33]).

######  Remark 10.5.

This is the only point where we use that $\pi_{1}(X)$ is (virtually) solvable.
Note that an acyclic manifold can perfectly be noncontractible. Indeed, there
are plenty of examples of finitely generated acyclic groups (see e.g. [5]) and
by a result of Kervaire [32, Theorem 1] any such group is the fundamental
group of an integral smooth homology sphere of dimension $n>4$. Removing a
point from such manifold we obtain an acyclic manifold with the same
fundamental group. It is not clear, however, whether a non simply connected
acyclic $n$-dimensional manifold can support a free action of
${\mathbb{Z}}^{n}$ with compact quotient (let alone that its quotient by
${\mathbb{Z}}^{n}$ supports free actions of $({\mathbb{Z}}/r)^{n}$ for
arbitrarily large $r$). So it could be the case that our assumption that
$\pi_{1}(X)$ is virtually solvable is unnecessary.

It follows that
$Q:X_{\phi}^{0}\times_{V}{\mathbb{R}}^{n}\to{\mathbb{R}}^{n}/V$ is a homotopy
equivalence. Precomposing it with a homotopy inverse of the projection
$X_{\phi}^{0}\times_{V}{\mathbb{R}}^{n}\to X_{\phi}^{0}/V=X$ we obtain a
homotopy equivalence $X\to{\mathbb{R}}^{n}/V$. It then follows from the
topological rigidity of tori (see the Introduction for references) that $X$ is
homeomorphic to $T^{n}$.

To conclude, let us prove statement (2) of Theorem 1.1 in the general case.
Suppose that $X$ is a rationally hypertoral manifold satisfying
$\operatorname{disc-sym}(X)\geq n$ and that $\pi_{1}(X)$ is virtually
solvable. Then there is a finite covering $r:X^{\prime}\to X$ such that
$\pi_{1}(X^{\prime})$ is solvable. Let $\phi:X\to T^{n}$ be a map of nonzero
degree and let $\phi^{\prime}=\phi\circ r$. Then
$\deg\phi^{\prime}=\deg\phi\cdot\deg r\neq 0$ and we have a Cartesian diagram

$\textstyle{X^{\prime}_{\phi^{\prime}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{r_{\phi}}$$\scriptstyle{\rho_{\phi^{\prime}}}$$\textstyle{X_{\phi}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho_{\phi}}$$\textstyle{X^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{r}$$\textstyle{X.}$

In particular, $r_{\phi}$ is a finite (unramified) covering space. By Theorem
1.11 we have $\operatorname{disc-sym}(X^{\prime})\geq n$. Applying the
previous discussion to $X^{\prime}$ we conclude that the connected components
of $X^{\prime}_{\phi^{\prime}}$ are contractible. Let $X_{\phi}^{0}$ be any
connected component of $X_{\phi}$ and denote by
$(X^{\prime}_{\phi^{\prime}})^{0}$ its preimage under $r_{\phi}$. Let
$\pi=\pi_{1}(X_{\phi}^{0})$, so that
$X_{\phi}^{0}=(X^{\prime}_{\phi^{\prime}})^{0}/\pi$, where $\pi$ acts freely
on $(X^{\prime}_{\phi^{\prime}})^{0}$. Note that $\pi$ is finite, because
$r_{\phi}$ is a finite covering. The freeness of the action implies that $\pi$
acts freely on the set of connected components of
$(X^{\prime}_{\phi^{\prime}})^{0}$ because by Smith’s theory a homeomorphism
of primer order of a contractible manifold has necessarily some fixed point
(see e.g. [9, Chap III, Corollary 4.6]). This implies that $X_{\phi}^{0}$ is
also contractible, so the same argument as in the case of solvable fundamental
group allows to finish the proof of statement (2) of Theorem 1.1 in the
general case.

##  11\. Proof of Theorem 1.2

In dimensions up to three any topological manifold has a unique smooth
structure, so we will assume from now on in this section that $n\geq 5$.
According to [52, §15A], for any smooth $n$-manifold $X$ and any simple
homotopy equivalence $h:X\to T^{n}$ one can define a ”characteristic class”

$c(h:X\to T^{n})\in A_{n}:=H^{3}(T^{n};{\mathbb{Z}}/2)\oplus\bigoplus_{i\leq
n}H^{i}(T^{n};\pi_{i}(PL/O))$

with the property that if $h^{\prime}:X^{\prime}\to T^{n}$ is another simple
homotopy equivalence and $c(h:X\to T^{n})=c(h^{\prime}:X^{\prime}\to T^{n})$
then $X$ and $X^{\prime}$ are diffeomorphic. The piece of the characteristic
class in $H^{3}(T^{n};{\mathbb{Z}}/2)$ accounts for the PL structure of $X$,
whereas that in $H^{i}(T^{n};\pi_{i}(PL/O))$ accounts for the different
choices of smooth structure compatible with the given PL structure. The
identity map $\operatorname{Id}:T^{n}\to T^{n}$ has trivial characteristic
class, so if $c(h:X\to T^{n})=0$ then $X$ is diffeomorphic to the standard
torus. In addition, if $\pi:T^{n}\to T^{n}$ is a covering and
$\pi^{*}h:\pi^{*}X\to T^{n}$ is the pullback of $h$, then

$c(\pi^{*}h:\pi^{*}X\to T^{n})=\pi^{*}c(h:X\to T^{n})$

(note that $\pi^{*}h:\pi^{*}X\to T^{n}$ is also a simple homotopy
equivalence). The homotopy groups $\pi_{i}(PL/O)$ are finite and hence so is
the group $A_{n}$.

Let $X$ be a smooth $n$-manifold and suppose that $h:X\to T^{n}$ is a
homeomorphism. Then $h$ is a simple homotopy equivalence by Chapman’s theorem
(see e.g. the Appendix in [16]), so we have a well defined characteristic
class $c(h:X\to T^{n})\in A_{n}$. Let $k$ be any natural number and let
$r=k|A_{n}|+1$. Multiplication by $r$ is the identity map on $A_{n}$, so if
$\pi_{r}:T^{n}\to T^{n}$ is the covering space defined by
$\pi_{r}(x_{1},\dots,x_{n})=(rx_{1},\dots,rx_{n})$ (where
$x_{i}\in{\mathbb{R}}/{\mathbb{Z}}$) then $\pi_{r}^{*}c(h:X\to
T^{n})=c(\pi_{r}^{*}h:\pi_{r}^{*}X\to T^{n})=c(h:X\to T^{n})$. Hence there
exists a diffeomorphism $\phi_{r}:X\to\pi_{r}^{*}X$. The manifold
$\pi_{r}^{*}X$ has a free and smooth action of $({\mathbb{Z}}/r)^{n}$ given by
deck transformations of the covering $\pi_{r}^{*}X\to X$. This action can be
transported via $\phi_{r}$ to a free action of $({\mathbb{Z}}/r)^{n}$ on $X$.
This proves statement (1) of Theorem 1.2.

Let us now prove (2). Let $X$ be a smooth manifold homeomorphic to $T^{n}$,
and fix a homotopy equivalence $h:X\to T^{n}$. By Lemma 3.2 and Theorem 2.1
there exists a constant $C$ such that for any finite group $\Gamma$ acting
continuously on $T^{n}$ there is a subgroup $\Gamma_{0}\leq\Gamma$ satisfying
$[\Gamma:\Gamma_{0}]\leq C$, a map $\psi:X\to T^{n}$ homotopic to $h$, and a
monomorphism $\eta:\Gamma_{0}\to T^{n}$ such that $\psi$ is
$\eta$-equivariant. In particular, the action of $\Gamma_{0}$ on $T^{n}$ is
free.

We next define $\delta(n)$. By Lemma 6.1 there exists a constant $C^{\prime}$
such that, if $\Gamma^{\prime}$ is any subgroup of $\Gamma_{r,n}$ satisfying
$[\Gamma_{r,n}:\Gamma^{\prime}]\leq C$, then $\Gamma^{\prime}$ contains a
subgroup isomorphic to $\Gamma_{s,n}$, where $C^{\prime}s\geq r$. Suppose that
$|A_{n}|=p_{1}^{e_{1}}\dots p_{k}^{e_{k}}$, where $p_{1},\dots,p_{k}$ are
pairwise distinct prime numbers and the exponents $e_{i}$ are natural numbers.
For each $i$ let $f_{i}$ be the smallest natural number such that
$p_{i}^{f_{i}}\geq C^{\prime}$. Define $\delta(n)=p_{1}^{e_{1}+f_{1}}\dots
p_{k}^{e_{k}+f_{k}}.$

If $r$ is divisible by $\delta(n)$ and $\Gamma^{\prime}$ is a subgroup of
$\Gamma_{r,n}$ satisfying $[\Gamma_{r,n}:\Gamma^{\prime}]\leq C$ then there
exists a subgroup $\Gamma^{\prime\prime}\leq\Gamma^{\prime}$ isomorphic to
$\Gamma_{s,n}$ for some $s$ satisfying $C^{\prime}s\geq r$. Since
$\Gamma_{r,n}$ has a subgroup isomorphic to $\Gamma_{s,n}$,
$s^{n}=|\Gamma_{s,n}|$ divides $r^{n}=|\Gamma_{r,n}|$, which implies that $s$
divides $r$. We next prove that $s$ is divisible by $p_{i}^{e_{i}}$ for every
$i$. Let $p_{i}^{g_{i}}$ (resp. $p_{i}^{h_{i}}$) be the biggest power of
$p_{i}$ dividing $s$ (resp. $r$). We have $h_{i}\geq e_{i}+f_{i}$ because $r$
is divisible by $\delta(n)$, and $h_{i}\geq g_{i}$ because $s$ divides $r$.
The quotient $r/s$ is divisible by $p_{i}^{h_{i}-g_{i}}$. Since $r/s\leq
C^{\prime}$, we have $h_{i}-g_{i}\leq f_{i}$, so $g_{i}\geq h_{i}-f_{i}\geq
e_{i}$. Hence $s$ is divisible by $|A_{n}|$.

If the group $\Gamma_{r,n}$ acts smoothly and effectively on $X$ then there is
a monomorphism $\eta:\Gamma_{s,n}\to T^{n}$ and an $\eta$-equivariant map
$\psi:X\to T^{n}$ homotopic to $h$. The quotient $T^{n}\to
T^{n}/\eta(\Gamma_{s,n})$ is a covering map and $T^{n}/\eta(\Gamma_{s,n})$ is
homeomorphic to $T^{n}$. So the map $\psi$ descends to a continuous map
$\zeta:X/\Gamma_{s,n}\to T^{n}/\eta(\Gamma_{s,n})\cong T^{n}$. The projection
map identifies $\pi_{1}(X)$ (resp. $\pi_{1}(T^{n})$) with a subgroup of
$\pi_{1}(X/\Gamma_{s,n})$ (resp. $\pi_{1}(T^{n}/\eta(\Gamma_{s,n})$), and via
these identifications
$\zeta_{*}:\pi_{1}(X/\Gamma_{s,n})\to\pi_{1}(T^{n}/\eta(\Gamma_{s,n}))$ can be
seen as an extension of $\psi_{*}:\pi_{1}(X)\to\pi_{1}(T^{n})$. Since
$\psi_{*}$ is an isomorphism,
$[\pi_{1}(X/\Gamma_{s,n}):\pi_{1}(X)]=[\pi_{1}(T^{n}/\eta(\Gamma_{s,n})):\pi_{1}(T^{n})]$
(both are equal to $|\Gamma_{s,n}|$), and
$\pi_{1}(X/\Gamma_{s,n})\simeq\pi_{1}(T^{n}/\eta(\Gamma_{s,n}))\simeq{\mathbb{Z}}^{n}$,
it follows that $\zeta_{*}$ is an isomorphism. Both $X/\Gamma_{s,n}$ and
$T^{n}/\eta(\Gamma_{s,n})$ are aspherical spaces, so $\zeta$ is a homotopy
equivalence. By the topological rigidity of tori, $\zeta$ is homotopic to a
homeomorphism $\xi:X/\Gamma_{s,n}\to T^{n}$. Since $\xi$ is homotopic to
$\zeta$, it can be lifted to a homeomorphism $\theta:X\to T^{n}$ that makes
the following diagram commutative:

$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\theta}$$\textstyle{T^{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{q}$$\textstyle{X/\Gamma_{s,n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\xi}$$\textstyle{T^{n}.}$

Since $s$ is divisible by $|A_{n}|$ and $q$ can be written as
$q(x_{1},\dots,x_{n})=(sx_{1},\dots,sx_{n})$, it follows that $c(\theta:X\to
T^{n})=0$, so $X$ is diffeomorphic to $T^{n}$.

##  12\. Proof of Theorem 1.13

Statement (1) of Theorem 1.13 follows from combining Lemma 3.2, Theorem 2.1
and Lemma 6.4. For the proof of statement (2) we will need the following two
lemmas.

######  Lemma 12.1.

Let $X,Y$ be two closed connected topological manifolds. Then $X\times Y$ is
rationally hypertoral if and only if both $X$ and $Y$ are rationally
hypertoral.

###### Proof.

Denote by $\pi_{X},\pi_{Y}$ the projections from $X\times Y$ to $X,Y$
respectively. Let $m=\dim X$ and $n=\dim Y$. If $X$ and $Y$ are rationally
hypertoral then there exist classes $\alpha_{1},\dots,\alpha_{m}\in
H^{1}(X;{\mathbb{Z}})$ and $\beta_{1},\dots,\beta_{n}\in
H^{1}(Y;{\mathbb{Z}})$ such that $\alpha_{1}\cup\dots\cup\alpha_{m}\neq 0$ and
$\beta_{1}\cup\dots\cup\beta_{n}\neq 0$. Then
$\pi_{X}^{*}\alpha_{1}\cup\dots\cup\pi_{X}^{*}\alpha_{m}\cup\pi_{Y}^{*}\beta_{1}\cup\dots\cup\pi_{Y}^{*}\beta_{n}\neq
0$, because the map $H^{m}(X;{\mathbb{Z}})\otimes H^{n}(Y;{\mathbb{Z}})\to
H^{m+n}(X\times Y;{\mathbb{Z}})$ given by pullback and cup-product is an
isomorphism, so $X\times Y$ is rationally hypertoral. Conversely, suppose that
there exist classes $\gamma_{1},\dots,\gamma_{m+n}\in H^{1}(X\times
Y;{\mathbb{Z}})$ such that $\gamma_{1}\cup\dots\cup\gamma_{m+n}\neq 0$. By
Künneth we may write $\gamma_{i}=\delta_{i}+\epsilon_{i}$ where
$\delta_{i}\in\pi_{X}^{*}H^{1}(X;{\mathbb{Z}})$ and
$\epsilon_{i}\in\pi_{Y}^{*}H^{1}(Y;{\mathbb{Z}})$. Expanding the product
$\gamma_{1}\cup\dots\cup\gamma_{m+n}=(\delta_{1}+\epsilon_{1})\cup\dots\cup(\delta_{m+n}+\epsilon_{m+n})$
the only terms that are not automatically zero are equal (after reordering
terms) to terms of the form
$\delta_{i_{1}}\cup\dots\cup\delta_{i_{m}}\cup\epsilon_{j_{1}}\cup\dots\cup\epsilon_{j_{n}}$
where $\\{i_{1},\dots,i_{m}\\}\cup\\{j_{1},\dots,j_{n}\\}=\\{1,\dots,m+n\\}$.
Some of these monomials is different from zero, and hence both $X$ and $Y$ are
rationally hypertoral. ∎

######  Lemma 12.2.

Let $X,Y$ be closed connected topological manifolds of the same dimension.
Suppose that $X\sharp Y$ is rationally hypertoral. Then at least one of the
manifolds $X,Y$ is also rationally hypertoral.

###### Proof.

If $n=1$ the statement is obvious, and if $n=2$ it follows from the
classification of closed connected surfaces. Let us suppose from now on that
$n\geq 3$ and that $Z:=X\sharp Y$ is rationally hypertoral. Let $X_{0},Y_{0}$
be the complementaries of open balls in $X,Y$ respectively, so that $\partial
X_{0}\cong S^{n-1}\cong\partial Y_{0}$. We identify $Z$ with
$X_{0}\cup_{\partial X_{0}\cong\partial Y_{0}}Y_{0}$. Denote by
$i_{X}:X_{0}\hookrightarrow Z$ and $i_{Y}:Y_{0}\hookrightarrow Z$ the natural
inclusions.

Let $W:=i_{X}(X_{0})\cap i_{Y}(Y_{0})$, choose a homeomorphism $\xi:W\to
S^{n-1}$, and let $E$ be the result of attaching to $Z$ an $n$-disk $D^{n}$
along its boundary $\partial D^{n}=S^{n-1}$, using $\xi$. Let $\phi:Z\to
T^{n}$ be a map of nonzero degree. Since $\pi_{n-1}(T^{n})=0$, the restriction
of $\phi$ to $W$ is homotopically trivial, so $\phi$ extends to a continuous
map $\psi:E\to T^{n}$. By contracting $D^{n}\subset E$ we obtain a map $c:E\to
X\vee Y$ that is a homotopy equivalence. Composing $\psi$ with a homotopy
inverse of $c$ we obtain a map $\zeta:X\vee Y\to T^{n}$. The map $\phi$
coincides up to homotopy with the composition

$Z\stackrel{{\scriptstyle c|_{Z}}}{{\longrightarrow}}X\vee
Y\stackrel{{\scriptstyle\zeta}}{{\longrightarrow}}T^{n}.$

Since the degree of $\phi$ is nonzero, it follows that the restriction of
$\zeta$ to one of the two summands $X\subset X\vee Y$ or $Y\subset X\vee Y$
has to be nonzero, so the corresponding manifold is rationally hypertoral. ∎

We now prove statement (2) in Theorem 1.13. Let $X=Y\times(Z\sharp
Z^{\prime})$ be a rationally hypertoral manifold, where $\dim Y=k$ and $\dim
Z=\dim Z^{\prime}=n-k$ for some integer $0\leq k<n$. Since $X$ is orientable,
the three manifolds $Y,Z,Z^{\prime}$ are orientable. By Lemma 12.1 both $Y$
and $Z\sharp Z^{\prime}$ are hypertoral, and by Lemma 12.2 at least one of the
manifolds $Z,Z^{\prime}$ (say, $Z$) is rationally hypertoral. Assume that
$H^{*}(Z^{\prime};{\mathbb{Z}})$ is not isomorphic to
$H^{*}(S^{n-k};{\mathbb{Z}})$. Choose maps of nonzero degree $\phi_{Y}:Y\to
T^{k}$, $\phi_{Z}:Z\to T^{n-k}$ and let
$\phi=(\phi_{Y},\phi_{Z}^{\prime}):X=Y\times(Z\sharp Z^{\prime})\to
T^{k}\times T^{n-k}=T^{n}$, where $\phi_{Z}^{\prime}=\phi_{Z}\circ
c_{Z^{\prime}}:Z\sharp Z^{\prime}\to T^{n-k}$ and $c_{Z^{\prime}}:Z\sharp
Z^{\prime}\to Z$ is the map collapsing $Z^{\prime}$. Then $d:=\deg\phi$ is
nonzero.

We prove that $\operatorname{disc-sym}(X)\leq k$ by contradiction. Assume that
there exists a sequence of natural numbers $r_{i}\to\infty$ such that
$\Gamma_{i}:=\Gamma_{r_{i},k+1}$ acts effectively on $X$. By Lemma 3.2,
Theorem 2.1 and Lemma 6.1, there exists a constant $C$ and, for every $i$, a
subgroup $\Gamma_{i}^{\prime}\leq\Gamma_{i}$ isomorphic to
$\Gamma_{s_{i},k+1}$, with $s_{i}$ satisfying $Cs_{i}\geq r_{i}$, a morphism
$\eta_{i}:\Gamma_{i}^{\prime}\to T^{n}$ satisfying
$|\operatorname{Ker}\eta_{i}|\leq d$, and an $\eta_{i}$-equivariant map

$\phi_{i}:X\to T^{n}$

homotopic to $\phi$.

Let $\lambda:T^{n}\to T^{k}$ be the projection that forgets the last $n-k$
coordinates. By Lemma 6.4 we have
$|\lambda(\eta_{i}(\Gamma_{i}^{\prime}))|\leq s_{i}^{k}.$ Consequently

$\Gamma_{i,0}^{\prime}:=\eta_{i}(\Gamma_{i}^{\prime})\cap\operatorname{Ker}\lambda$

satisfies

(6)
$|\Gamma_{i,0}^{\prime}|=\frac{|\eta_{i}(\Gamma_{i}^{\prime})|}{|\lambda(\eta_{i}(\Gamma_{i}^{\prime}))|}\geq\frac{s_{i}^{k+1}/d}{|\lambda(\eta(\Gamma_{i}^{\prime}))|}\geq
s_{i}/d.$

Let $\mu:T^{n}\to T^{n-k}$ be the projection that forgets the first $k$
coordinates, and consider the sequence of projections

$T^{n}\stackrel{{\scriptstyle\mu}}{{\longrightarrow}}T^{n-k}\stackrel{{\scriptstyle\mu_{1}}}{{\longrightarrow}}T^{n-k-1}\to\dots\stackrel{{\scriptstyle\mu_{n-k-1}}}{{\longrightarrow}}T^{1}\stackrel{{\scriptstyle\mu_{n-k}}}{{\longrightarrow}}\\{1\\},$

where each $\mu_{j}$ forgets the first coordinate. Denote the compositions
$\nu_{j}=\mu_{j}\circ\dots\circ\mu_{1}$. Define
$G_{i,0}:=\mu(\Gamma_{i,0}^{\prime})$ and, for every $1\leq j\leq n-k$,
$G_{i,j}=\nu_{j}(G_{i,0})$. Consider the set

$B=\\{j\in\\{0,1,\dots,n-k-1\\}\mid|G_{i,j}\cap\operatorname{Ker}\mu_{j+1}|\text{
is not bounded as $i\to\infty$}\\}.$

For every $j$ there is a short exact sequence

$0\to G_{i,j}\cap\operatorname{Ker}\mu_{j+1}\to G_{i,j}\to G_{i,j+1}\to 0,$

and since $|G_{i,n-k}|=1$, we have

$|G_{i,0}|=\prod_{j=0}^{n-k-1}|G_{i,j}\cap\operatorname{Ker}\mu_{j+1}|.$

Since $|G_{i,0}|\to\infty$ as $i\to\infty$, it follows that the set $B$ is
nonempty. Let $j_{0}:=\min B$. Replacing the sequence $(\Gamma_{i})$ by a
subsequence, we may assume that
$|G_{i,j_{0}}\cap\operatorname{Ker}\mu_{j_{0}+1}|\to\infty$ as $i\to\infty$.
By the choice of $j_{0}$, there is a constant $C$ such that
$|G_{i,0}\cap\operatorname{Ker}\nu_{j_{0}}|\leq C$ for every $i$. Define

$\Gamma_{i}^{\prime\prime}=\eta_{i}^{-1}(\Gamma_{i,0}^{\prime})$

and let

$\widetilde{\eta}_{i}:=\nu_{j_{0}}\circ\eta_{i}|_{\Gamma_{i}^{\prime\prime}}:\Gamma_{i}^{\prime\prime}\to
T^{n-k-j_{0}}.$

Then the fibers of $\widetilde{\eta}_{i}|_{\Gamma_{i}^{\prime\prime}}$ have at
most $\delta C$ elements.

Let $S=\operatorname{Ker}\nu_{j_{0}+1}\leq T^{n-k-j_{0}}$. This can be
identified with the elements of $T^{n-k-j_{0}}$ whose coordinates are all zero
except possibly the first one. Let
$\Gamma_{i}^{\prime\prime\prime}=\widetilde{\eta}_{i}^{-1}(S)$. By the
definition of $j_{0}$, we have
$|\widetilde{\eta}_{i}(\Gamma_{i}^{\prime\prime\prime})|\to\infty$ as
$i\to\infty$. Let $\xi:T^{n-k-j_{0}}\to S$ be the projection that forgets all
coordinates except the first one, and let

$\psi_{i}=\xi\circ\nu_{j_{0}}\circ\mu\circ\phi_{i}:X\to S.$

Then $X_{\psi_{i}}\to X$ is a ${\mathbb{Z}}$-principal bundle and
$H_{*}(X_{\psi_{i}};{\mathbb{Z}})$ is a finitely generated
${\mathbb{Z}}[{\mathbb{Z}}]={\mathbb{Z}}[t^{\pm 1}]$ module by Theorem 5.1.
The map $\psi_{i}$ is homotopic to the map

$\zeta=\xi\circ\nu_{j_{0}}\circ\mu\circ\phi:X\to S,$

so there exists a ${\mathbb{Z}}$-equivariant homeomorphism $X_{\zeta}\cong
X_{\psi_{i}}$. We are next going to prove that $H_{*}(X_{\zeta};{\mathbb{Z}})$
is finitely generated as a ${\mathbb{Z}}$-module, using an argument similar to
the proof of Lemma 10.2.

Then for any $\gamma\in\Gamma_{i}^{\prime\prime\prime}$ and $x\in X$ we have

$\psi_{i}(\gamma\cdot x)=\widetilde{\eta}_{i}(\gamma)+\psi_{i}(x).$

Let $o_{i}$ be the order of
$\widetilde{\eta}_{i}(\Gamma_{i}^{\prime\prime\prime})$. We can identify $S$
with ${\mathbb{R}}/{\mathbb{Z}}$, and we denote by $\pi:{\mathbb{R}}\to S$ the
projection. There exists an element
$\gamma_{i}\in\Gamma_{i}^{\prime\prime\prime}$ satisfying
$\widetilde{\eta}_{i}(\gamma_{i})=\pi(1/o_{i})$. Then the order of
$\gamma_{i}$ is equal to $C_{i}$, where $C_{i}$ is a natural number satisfying
$1\leq C_{i}\leq\delta C$. Passing to a subsequence of $(\Gamma_{i})$ we may
assume that all numbers $C_{i}$ are equal to a natural number $C^{\prime}$.
Then the homeomorphism $f_{i}:X\to X$ given by the action of $\gamma_{i}$ has
order $C^{\prime}o_{i}$. By Lemma 10.1 the map $f_{i}$ admits a lift
$g_{i}:X_{\psi_{i}}\to X_{\psi_{i}}$ such that the action of
$g_{i}^{C^{\prime}o_{i}}$ on $H_{*}(X_{\psi_{i}};{\mathbb{Z}})$ coincides with
multiplication by $t^{C^{\prime}}$. Since $o_{i}\to\infty$, Corollary 5.3
implies that $H_{*}(X_{\zeta};{\mathbb{Z}})$ is a finitely generated
${\mathbb{Z}}$-module, as we wanted to see.

The map $\zeta:X\to S$ factors through the composition

$X=Y\times(Z\sharp Z^{\prime})\to Z\sharp Z^{\prime}\stackrel{{\scriptstyle
c_{Z^{\prime}}}}{{\longrightarrow}}Z,$

where the first map is the projection to the second factor. Let $\kappa:X\to
Z$ be this composition. Define $\psi_{Z}=\xi\circ\nu_{j_{0}}\circ\phi_{Z}$.
Then $\zeta=\psi_{Z}\circ\kappa$, so $X_{\zeta}$ can be identified with
$Y\times(Z_{\psi_{2}}\sharp(Z^{\prime}\times{\mathbb{Z}}))$, the product of
$Y$ with the connected sum of $Z_{\psi_{Z}}$ with countably many copies of
$Z^{\prime}$. Since $H_{*}(Z^{\prime};{\mathbb{Z}})\not\simeq
H^{*}(S^{n};{\mathbb{Z}})$, making connected sum with $Z^{\prime}$ increases
the size of the homology, so $H_{*}(X_{\zeta};{\mathbb{Z}})$ is not finitely
generated over ${\mathbb{Z}}$, contradicting our previous conclusion. This
finishes the proof that $\operatorname{disc-sym}(X)\leq k$.

##  13\. Holomorphic finite group actions on Kaehler manifolds

###  13.1. Proof of Theorem 1.16

The implication (1)$\Rightarrow$(2) is immediate, as the $r$-torsion of
$T^{n}$ is isomorphic to $({\mathbb{Z}}/r)^{n}$. To prove the converse
implication (2)$\Rightarrow$(1) it is enough to consider the case in which $G$
is compact. Indeed, if $G$ is an arbitrary Lie group with finitely many
connected components then the existence and uniqueness up to conjugation of
maximal compact subgroups (see e.g. [28, Theorem 14.13]) implies the existence
of a compact subgroup $K\leq G$ with the property that any compact (in
particular, any finite) subgroup of $G$ is conjugate to a subgroup of $K$.
Hence, replacing $G$ by $K$ we assume from now on that $G$ is a compact Lie
group.

Let ${\mathfrak{g}}$ denote the Lie algebra of $G$. For every
$\alpha\in{\mathfrak{g}}$ we denote by $e^{\alpha}$ the image of $\alpha$
under the exponential map ${\mathfrak{g}}\to G$. We claim that there exists a
neighborhood of zero $U\subset{\mathfrak{g}}$ with the property that for any
$\alpha,\beta\in U$ such that the elements $e^{\alpha},e^{\beta}\in G$ commute
we have $[\alpha,\beta]=0$. To prove the claim, pick an
$\operatorname{Ad}$-invariant norm on ${\mathfrak{g}}$ and choose $\epsilon>0$
small enough so that the exponential map sends the ball
$U:=B_{\epsilon}(0)\subset{\mathfrak{g}}$ diffeomorphically to a neighborhood
of the identity in $G$. Choose elements $\alpha,\beta\in U$ such that
$e^{\alpha}$ and $e^{\beta}$ commute. Then

$e^{\beta}=e^{\alpha}e^{\beta}(e^{\alpha})^{-1}=e^{\operatorname{Ad}(e^{\alpha})(\beta)},$

so by the injectivity of $\exp|_{U}$ we have
$\beta=\operatorname{Ad}(e^{\alpha})(\beta)$. Since
$\operatorname{Ad}(e^{\alpha})$ is a linear automorphism of ${\mathfrak{g}}$,
it follows that $t\beta=\operatorname{Ad}(e^{\alpha})(t\beta)$ for every
$t\in{\mathbb{R}}$ and consequently $e^{\alpha}$ commutes with $e^{t\beta}$
for every $t\in[-1,1]$. Repeating the same argument with $t\beta$ resp.
$\alpha$ instead of $\alpha$ resp. $\beta$ we conclude that for every
$s,t\in[-1,1]$ we have $e^{s\alpha}e^{t\beta}=e^{t\beta}e^{s\alpha}$. This
implies $[\alpha,\beta]=0$.

Choose a neighborhood of the identity $V\subset G$ such that $V^{-1}V\subseteq
e^{U}$. Since $G$ is compact, we can cover $G$ with finitely many translates
of $V$, say $G=g_{1}V\cup\dots\cup g_{m}V$ for some elements
$g_{1},\dots,g_{m}\in G$. Suppose that $\Gamma$ is a finite group of $G$.
Since $\Gamma=(g_{1}V\cap\Gamma)\cup\dots\cup(g_{m}V\cap\Gamma)$, there exists
some $1\leq i\leq m$ such that $|g_{i}V\cap\Gamma|\geq|\Gamma|/m$. Take any
element $\gamma_{0}\in g_{i}V\cap\Gamma$ and let
$S_{U}=\\{\gamma_{0}^{-1}\gamma\mid\gamma\in g_{i}V\cap\Gamma\\}$. Then
$S_{U}\subseteq V^{-1}V\cap\Gamma\subseteq e^{U}\cap\Gamma$ and
$|S_{U}|\geq|\Gamma|/m$, so the subgroup $\Gamma_{U}\leq\Gamma$ generated by
$S_{U}$ satisfies $[\Gamma:\Gamma_{U}]\leq m$. Let
${\mathfrak{s}}_{U}=\\{\alpha\in U\mid e^{\alpha}\in S_{U}\\}$. By the
construction of $U$, ${\mathfrak{s}}_{U}$ spans an abelian subalgebra
${\mathfrak{t}}\subseteq{\mathfrak{g}}$. Let $T$ be the closure of
$e^{{\mathfrak{t}}}$. Then $T$ is a torus and $S_{U}\subset T$, so
$\Gamma_{U}\subset T$. We have thus proved that for any finite subgroup
$\Gamma$ of $G$ there is a torus $T\subseteq G$ such that
$[\Gamma:T\cap\Gamma]\leq m$, where $m$ is independent of $\Gamma$.

Suppose that $n$ is a natural number such that $G$ contains subgroups
isomorphic to $\Gamma_{r,n}=({\mathbb{Z}}/r)^{n}$ for arbitrary large numbers
$r$. By Lemma 6.1 there is a constant $C$ such that any subgroup
$\Gamma^{\prime}\leq\Gamma_{r,n}$ satisfying
$[\Gamma_{r,n}:\Gamma^{\prime}]\leq m$ contains a subgroup
$\Gamma^{\prime\prime}\leq\Gamma^{\prime}$ such that
$\Gamma^{\prime\prime}\simeq\Gamma_{s,n}$ with $Cs\geq r$. By assumption,
there exists some $r\geq 2C$ and some subgroup $\Gamma$ of $G$ isomorphic to
$\Gamma_{r,n}$. By our previous result, there exists a torus $T\leq G$ such
that $[\Gamma:T\cap\Gamma]\leq m$. Then there is a subgroup
$\Gamma^{\prime\prime}\leq\Gamma^{\prime}:=T\cap\Gamma$ such that
$\Gamma^{\prime\prime}\leq\Gamma_{s,n}$ for some $s$ satisfying $Cs\geq r\geq
2C$, so $s\geq 2$. Then Lemma 6.4 implies that $\dim T\geq n$, so the proof of
Theorem 1.16 is complete.

###  13.2. Proof of Theorem 1.15

Let $X$ be a compact connected Kaehler manifold of real dimension $n$. Let
$\operatorname{Aut}X$ denote the group of biholomorphisms of $X$. A theorem of
Bochner and Montgomery [8, §9] states that $\operatorname{Aut}X$ has a natural
structure of Lie group. Let $\omega$ denote the Kaehler form of $X$ and let
$\overline{\omega}$ denote the cohomology class in $H^{2}(X;{\mathbb{R}})$
represented by $\omega$ through the de Rham isomorphism. Let
$\operatorname{Aut}_{\overline{\omega}}X$ denote the subgroup of
$\operatorname{Aut}X$ consisting of those biholomorphisms fixing the class
$\overline{\omega}$. According to a theorem of Fujiki [25, Theorem 4.8],
$\operatorname{Aut}_{\overline{\omega}}X$ has finitely many connected
components.

By Lemma 3.1 there exists a constant $C$ such that any finite subgroup acting
effectively and by group automorphisms on
$H^{2}(X;{\mathbb{Z}})/\operatorname{Tor}H^{2}(X;{\mathbb{Z}})\simeq{\mathbb{Z}}^{b_{2}(X)}$
has at most $C$ elements. This implies that if $\Gamma$ is a finite group
acting on $X$ then there is a subgroup $\Gamma^{\prime}\leq\Gamma$ satisfying
$[\Gamma:\Gamma^{\prime}]\leq C$ and whose action on
$H^{2}(X;{\mathbb{Z}})/\operatorname{Tor}H^{2}(X;{\mathbb{Z}})$ is trivial.
Equivalently, the action of $\Gamma^{\prime}$ on $H^{2}(X;{\mathbb{R}})$ is
trivial. In particular, if $\Gamma$ acts effectively on $X$ by biholomorhic
transformations, so that we can identify $\Gamma$ with a subgroup of
$\operatorname{Aut}X$, then
$[\Gamma:\operatorname{Aut}_{\overline{\omega}}X\cap\Gamma]\leq C$.

Now assume that for some natural number $m$ the group $\operatorname{Aut}X$
contains subgroups isomorphic to $\Gamma_{r,m}$ for arbitrary large values of
$r$. Arguing as in the preceding subsection (applying the existence of the
constant $C$ and using Lemma 6.1) we may conclude that
$\operatorname{Aut}_{\overline{\omega}}X$ contains subgroups isomorphic to
$\Gamma_{s,m}$ for arbitrary large values of $s$. This implies, by Theorem
1.16, that $\operatorname{Aut}_{\overline{\omega}}X$ contains a torus $T$ of
satisfying $\dim T\geq m$.

Now, if an $m$-dimensional torus $T$ acts effectively on an $n$ dimensional
connected topological manifold $X$ then $m\leq n$, and if $m=n$ then $X$ is
homeomorphic to a torus. This is probably well known, but since we did not
find a reference with a proof, we sketch it here. For any nontrivial closed
subgroup $K\leq T$ the fixed point set $X^{K}$ is closed and has empty
interior (see [10, Chap III, Theorem 9.5]). The set of closed subgroups of $T$
is countable and topological manifolds are Baire spaces, so
$\bigcap_{\\{e\\}\neq K\leq T}X\setminus X^{K}\neq\emptyset$. Hence, there
exists a point $x\in X$ whose stabilizer is trivial, so the map $f:T\ni
t\mapsto t\cdot x\to X$ is continuous and injective. By the theorem of
invariance of domain it follows that $\dim T\leq\dim X$ and that, if $\dim
T=\dim X$, then $f$ is an open map. In that case, $f(T)$ is closed (because
$T$ is compact) and open in $X$ and since $X$ is connected it follows that
$f(T)=X$. Since $f$ is open and injective, it gives a homeomorphism $T\cong
f(T)$, so we conclude that $T\cong X$.

Finally, a Kaehler manifold homeomorphic to a torus is biholomorphic to a
complex torus (see e.g. [3, Theorem B] for a very nice exposition of a more
general result), so the proof of Theorem 1.15 is now finished.

##  14\. The manifold in Example 1.4 is rationally hypertoral but not
hypertoral

Recall that $X$ is the mapping torus of $\sigma:Y\to Y$, the involution of a
degree $2$ covering $\pi:Y\to T^{2}$ ramified at two points. So
$X=([0,1]\times Y)/\sim$, where $\sim$ identifies $(0,y)$ with $(1,\sigma(y))$
for every $y\in Y$. The projection $[0,1]\times Y\to[0,1]$ induces a fibration
$\rho:X\to S^{1}$. The map $\operatorname{Id}\times\pi:[0,1]\times
Y\to[0,1]\times T^{2}$ descends to a map $X\to T^{3}$, where we view $T^{3}$
as the mapping torus of the identity on $T^{2}$. So $X$ is rationally
hypertoral.

We prove that $X$ is not hypertoral using the following fact, whose proof is