# Torsion groups of subexponential growth cannot act on finite-dimensional
CAT(0)-spaces without a fixed point

Hiroyasu Izeki111The first author was supported in part by JSPS Grants-in-Aid
for Scientific Research Grant Number JP20H01802. and Anders Karlsson222The
second author was supported in part by the Swiss NSF grants 200020-200400 and
200021-212864, and the Swedish Research Council grant 104651320.

(April 26, 2024)

###### Abstract

We show that finitely generated torsion groups of subexponential growth must
have a global fixed point whenever they act by isometry on a finite
dimensional complete CAT(0)-space. The same statement for simple groups of
subexponential growth is also shown. The method of proof uses ultralimits,
equivariant harmonic maps, subharmonic functions, horofunctions and random
walks.

##  1 Introduction

The existence of infinite, finitely generated, torsion groups has a long and
rich history which started with a problem posed by Burnside around the year
1900. The discovery that such groups actually could exist came in the 1960s by
work of Golod and Shafarevich. Since then other infinite torsion groups with
additional important features appeared, notably the papers of Novikov-Adian on
infinite Burnside groups and the Grigorchuk groups having intermediate growth,
see [CSD21, Os22] for recent discussions.

From the viewpoint of geometric group theory it is a natural question as to
whether such groups can act on CAT(0)-spaces, indeed this topic already
appeared in the foundational essay by Gromov [Gr87, sect. 4.5.C]. By a theorem
of Schur [S11] (or by Selberg’s lemma) it is known that any linear finitely
generated torsion group must be finite, thus one can say that every action by
a finitely generated torsion group on the classical symmetric spaces of
nonpositive curvature must fix a point by the Cartan fixed point theorem.
Another instance is the elegant argument in Serre’s book [Se80, sect. 6.5]
that finitely generated torsion groups acting on a tree must have a global
fixed-point. By the end of the 1990s it was a well-known open problem
formulated as follows (see [Sw99, Be00]):

_Can a finitely generated group that acts properly discontinuously (and
cocompactly) by isometry on a proper CAT(0)-space contain an infinite torsion
group?_

At least at that time the expected answer depended on whether the cocompact
assumption is included or not. It was again referred to as one of the key open
problems in [Ca14].

Grigorchuk constructed a continuum of finitely generated torsion groups of
growth strictly between polynomial and exponential (thus amenable) in the
early 1980s, see [G80, G84, dlH00, CSD21] and references therein. Amenable
groups admit proper actions on Hilbert spaces ([BCV95]). Both Grigorchuk
groups and Burnside groups, that is, infinite torsion groups of bounded
exponent, can act without a global fixed point on infinite-dimensional CAT(0)
cubical complexes (see [Os18] and [Sc22]). Also note that [Se80, Theorem 25]
shows that every infinitely generated group acts on a tree without a global
fixed-point, see also [BH99, II.7.11] for another infinitely generated torsion
group example. We refer to the recent paper [HO22] for a historical survey on
this problem, its variants and with references to the many previous partial
results. In [NOP22, HO22] the following conjecture is stated:

######  Conjecture 1.

([NOP22]) Every finitely generated group acting without a global fixed point
on a finite-dimensional CAT(0) complex contains an element of infinite order.

For CAT(0) cubical complexes this is known from a paper by Sageev [Sa95] (see
[LV20] for an explanation), and recently extended to CAT(0) cubical complexes
without infinite cubes in [GLU24]. Norin, Osajda, and Przytycki recently
proved the conjecture for two-dimensional CAT(0) complexes (with certain
additional minor assumptions) in their paper [NOP22].

We establish the following, that applies in particular to Grigorchuk’s torsion
groups of intermediate growth which seems previously not known:

######  Theorem 2.

Let $\Gamma$ be a finitely generated torsion group of subexponential growth.
Whenever $\Gamma$ acts by isometries on a complete CAT(0)-space $Y$ of finite
dimension, it must have a global fixed point in $Y$.

As pointed out above, the finite generation and finite-dimensionality
assumptions are necessary, on the other hand, since we assume subexponential
growth, we do not answer the full general question nor settle all of
Conjecture 1. However, let us emphasize that our result is more general in the
sense that:

  * •

the space $Y$ is not necessarily proper or a CAT(0)-complex, and it is not
assumed to have a large isometry group.

  * •

no assumption on the action is made, such as being non-elementary (in the
sense of assuming no fixed-point in $\partial Y$), properly discontinuous, or
being part of a cocompact action.

It was previously shown by Caprace-Monod in [CM13, Corollary E] that finitely
generated groups of intermediate growth cannot be a _discrete_ subgroup of
$Isom(Y)$ under the assumption that the space $Y$ is a proper CAT(0)-space
having a _cocompact_ isometry group. On the other hand, Grigorchuk groups are
by construction subgroups of the automorphism groups of rooted binary trees,
thus they are subgroups of the isometry groups of certain proper CAT(0)-spaces
possessing a global fixed-point, the root. One should also keep in mind that
Caprace-Lytchak, [CL10, Theorem 1.6], extended theorems of Burger-Schroeder
and Adams-Ballmann to CAT(0) spaces of finite telescoping dimension showing
that amenable groups either fix a point at infinity or leave a flat invariant.
Our theorem by-passes the trouble with that there might be a fixed point on
the visual boundary at infinity. In this connection, we can mention a paper of
Papasoglu-Swenson [PS18, Theorem 3.17] which shows the following: Assume that
$G$ acts properly and cocompactly on a proper CAT(0)-space. If $\Gamma$ is an
infinite torsion subgroup of $G$ then it cannot fix a point in the limit set
of $\Gamma$. (Note that in our setting, that is assuming finite generation and
subexponential growth but no properness or cocompactness, the limit set is
empty since by Theorem 2 any orbit is bounded.) Groups which have a global
fixed point whenever they act on simplicial trees are said to have Serre’s
property FA. This is the 1-dimensional version of property $\mathrm{FA}_{n}$
in [F09], that is, having a global fixed point whenever acting by isometry on
$n$-dimensional CAT(0) cell complexes. The literature on these topics is vast,
and we cannot survey this here. For property FA and the groups of the type of
most immediate relevance for our paper, we can refer to [DG08, GS23] for
results and more information.

With a similar proof we also have:

######  Theorem 3.

Let $\Gamma$ be a finitely generated simple group of subexponential growth.
Whenever $\Gamma$ acts by isometries on a complete CAT(0)-space $Y$ of finite
dimension, it must have a global fixed point in $Y$.

Nekrashevych showed the existence of simple groups of intermediate growth in
[Ne18].

Our proof uses the technique of ultralimits of spaces and equivariant harmonic
maps, heavily inspired by Bourdon’s paper [Bo12], combined with substantial
arguments and results from the first author’s paper [Iz23] (much of what is
not repeated here, making our paper shorter than what it would be if all
needed details from [Iz23] were reproduced). The use of random walks might be
somewhat surprising. We think that the present method, in particular the
combination of ultralimits, equivariant harmonic maps, subharmonic functions
on the group, horofunctions and random walks, could have further applicability
in geometric group theory.

##  2 Ultralimits of metric spaces

For a detailed exposition on this topic we refer to [CSD21, Chapter 11] or
[Bo12, Iz23, Ly05], and for basics on CAT(0)-spaces we refer to [BH99]. Let
$\omega$ denote a non-principal ultrafilter on $\mathbb{N}$. Let $Y_{j}$ be
metric spaces and $p_{j}$ a point in each $Y_{j}.$ The associated ultralimit
metric space $(Y_{\infty},d)$ is defined as follows. First, let $Y^{\infty}$
be all (one-sided) sequences $y_{j}\in Y_{j}$ such that $d(y_{j},p_{j})$ is
bounded in $j$. Via the triangle inequality, one can therefore define the
following function on $Y^{\infty}$,

$d((x_{j}),(y_{j})):=\omega-\lim d(x_{j},y_{j}).$

This function is clearly symmetric and satisfies the triangle inequality. Now
we identify all sequences for which $d((x_{j}),(y_{j}))=0.$ The set of
equivalence classes, denoted by $Y_{\infty}$, is thus a metric space with the
same $d$ well-defined on the quotient. It is rather immediate that if each
$Y_{j}$ is a complete CAT(0)-space, then so is $(Y_{\infty},d)$.

The dimension of a CAT(0)-space $Y$ is defined as follows (see [Kl99] for
various equivalent definitions of dimension for CAT(0)-spaces). Given a set of
points $x_{0},x_{1},...,x_{n}$ in $Y$, one has a barycentric simplex
$\sigma:\Delta_{n}\rightarrow Y$ defined by mapping the point
$(\alpha_{0},...,\alpha_{n})$ in the standard simplex to the unique minimizer
(the _barycenter_), which exists by standard CAT(0)-space theory, see [BH99,
Ch. II.2], of the uniformly convex function

$f(y)=\sum\alpha_{i}d(x_{i},y)^{2}.$

The dimension of $Y$ is the greatest number $D$ such that for some points
$x_{0},x_{1},...,x_{D}$ the corresponding barycentric simplex is non-
degenerate. A simplex $\sigma$ is degenerate if
$\sigma(\Delta)=\sigma(\partial\Delta)$, see [Kl99, Definition 4.7]. We repeat
the argument by Lytchak in [Ly05, Lemma 11.1] (alternatively see [Iz23,
Proposition 6.10]) to show:

######  Proposition 4.

If the dimensions of $Y_{j}$ is at most $D<\infty,$ then the dimension of
$Y_{\infty}$ is at most $D$.

###### Proof.

Let $x_{0},x_{1},...,x_{n}$ be points in $Y_{\infty}$ such that the associated
barycentric simplex $\sigma$ is not degenerate. Represent these classes by
actual sequences of points in $Y$, denoted $z_{i,j}$, here
$j\rightarrow\infty$. By the continuity of barycenters in CAT(0)-spaces and
the definition of barycentric simplices implies that $\sigma$ is the
ultralimit of the simplices defined by $z_{i,j}$ in $Y_{j}$. If all of these
simplices would be degenerate then so would $\sigma$. ∎

##  3 Existence of equivariant harmonic maps with positive energy when passing
to certain ultralimits

Let $\Gamma$ be a group generated by a finite set $S$. Assume that $\Gamma$
acts by isometry on a complete metric space $Y$, that is, there is a
homomorphism $\rho:\Gamma\rightarrow Isom(Y)$. A _( $\rho$-)equivariant map
_is a map $f:\Gamma\rightarrow Y$ such that

$f(g)=\rho(g)f(e)$

for all $x\in Y$ and $g\in\Gamma.$ These maps are determined by $f(e)$ and
hence are just the orbit maps, $g\mapsto\rho(g)f(e)$.

Consider the following _( $S$-)energy _of maps $f:\Gamma\rightarrow Y$,

$E(f)=\sum_{S}d(f(e),f(s))^{2},$

which clearly is finite. We are looking to find a map with minimal energy
among the maps which are equivariant. Such maps are called _harmonic._
Intuitively we want to place our orbit in a minimal position, which is a
generalization of the consideration of the infimal displacement for single
isometries.

When $Y$ is a complete CAT(0)-spaces it is well-known, and already used in the
previous section about barycenters, that a convex function such as

$x\mapsto\sum_{S}d(x,\rho(s)x)^{2},$

which is the energy of the map defined by $f(e)=x$, either has a minimum in
$Y$ or a minimizing sequence $p_{j}$ that escapes to infinity, that is
$d(x,p_{j})\rightarrow\infty$ for any $x\in Y$.

In order to deal with the second scenario, we assume that $\Gamma$ acts on a
finite-dimensional CAT(0)-space $Y$ without a global fixed point. We now
follow Bourdon’s techniques for $L^{p}$-spaces in [Bo12]. Let

$\delta(x)=\max_{s\in S}d(x,s(x)),$

where we from now on suppress $\rho$ in the notation. Later on, we want
$\inf_{x\in Y}\delta(x)>0$, which clearly would imply that

$\inf E(f)>0$

over all equivariant maps $f$, see below. Since $S$ is a generating set, the
action has a global fixed point in $Y$ if and only if there exists an
equivariant map $f$ such that $E(f)=0$.

Suppose instead that $\inf_{x\in Y}\delta(x)=0$. We will now do the initial
scaling of the action using ultralimits.

First, we rewrite a lemma that Bourdon attributes to Shalom [Sh00, Lemma 6.3].

######  Lemma 5.

Assume that $\inf_{x\in Y}\delta(x)=0$ but there is no global fixed point.
Then for any integer $n>0$, there exists $r_{n}>0$ and $v_{n}\in Y$ such that
$\delta(v_{n})\leq r_{n}/n$ and for all $v$ with $d(v_{n},v)<r_{n},$ it holds
that

$\delta(v)\geq r_{n}/2n.$

###### Proof.

Let $n$ be a positive integer. Take $w_{1}\in Y$ such that $\delta(w_{1})\leq
1/2n.$ If $\delta(v)\geq 1/4n$ for every $v$ with $d(v,w_{1})<1/2$, we let
$v_{n}=w_{1}$ and $r_{n}=1/2$. Otherwise, there is $w_{2}$ with
$d(w_{1},w_{2})<1/2$ and $\delta(w_{2})\leq 1/4n$. If $\delta(v)\geq 1/8n$ for
every $v$ with $d(v,w_{2})<1/4$, then we let $v_{n}=w_{2}$ and $r_{n}=1/4$.
This procedure terminates, because otherwise we would get a sequence $w_{k}$
such that $d(w_{k},w_{k+1})<2^{-k}$. By completeness of $Y$ this Cauchy
sequence converges to a point $w\in Y$. But note that

$\left|d(w,s(w))-d(w_{k},s(w_{k}))\right|\leq 2d(w,w_{k}),$

which implies that $\delta(w)=0$ since $\delta(w_{k})\leq 1/(2^{k}n)$. This
contradicts the assumption that the orbit did not have a global fixed point. ∎

Using this lemma we can now show, what essentially is part of [Bo12, Prop.
3.1]:

######  Proposition 6.

Let $\Gamma$ be a group generated by a finite set $S$. Assume that it acts on
a finite-dimensional complete CAT(0)-space $Y$, with

$\inf_{x\in Y}\max_{s\in S}d(x,s(x))=0$

but there is no global fixed point. Then by passing to a certain rescaled
ultralimit action, there is a finite-dimensional complete CAT(0)-space
$Y_{\infty}$ on which $\Gamma$ acts without a global fixed point and

$\inf_{x\in Y_{\infty}}\max_{s\in S}d(x,s(x))>0.$

###### Proof.

Let $\lambda_{n}=n/r_{n}$ with $r_{n}$ and $v_{n}$ from Lemma 5 with
$\delta(x)=\max_{s\in S}d(x,s(x))$ whose infimum is zero. Consider the metric
spaces $\lambda_{n}Y=(Y,\lambda_{n}d)$, which of course again is a complete
CAT(0) on which $\Gamma$ acts by isometry via the same action. Take the
ultralimit space $Y_{\infty}$ of the sequence of pointed spaces
$(\lambda_{n}Y,v_{n})$ as in the previous section. Thanks to Proposition 4 it
is still finite-dimensional. Let

$\delta_{n}(x)=\max_{s\in S}\lambda_{n}d(x,s(x)),$

so $\delta_{n}(v_{n})\leq 1$ and $\delta_{n}(v)\geq 1/2$ for all $v$ in the
$\lambda_{n}d$-ball of radius $n$ around $v_{n}$. Thanks to the condition
$\delta_{n}(v_{n})\leq 1$ we have that $\Gamma$ still acts (by isometry) on
the ultralimit and thanks to the inequality $\delta_{n}(v)\geq 1/2$ on larger
and larger balls around $v_{n}$ we have that this new action of $\Gamma$ on
$Y_{\infty}$ satisfy $\inf_{x\in Y_{\infty}}\delta(x)>0.$ ∎

In summary, possibly by replacing $Y$ with an ultralimit we have that $\Gamma$
acts on a finite dimensional CAT(0)-space $Y$ such that

$\delta(x)=\max_{s\in S}d(x,s(x))$

is bounded away from $0$, which of course in particular excludes the existence
of a global fixed-point. This implies that the energy of any equivariant map
$f$ is also bounded away from $0$, since

$E(f)=\sum_{S}d(f(e),f(s))^{2}=\sum_{S}d(f(e),sf(e))^{2}\geq\delta(f(e))^{2}.$

Now we pass to the other part of [Bo12, Prop. 3.1]:

######  Proposition 7.

Let $\Gamma$ be a group generated by a finite set $S$. Assume that it acts on
a finite-dimensional complete CAT(0)-space $Y$ without a global fixed point.
Then there is a finite-dimensional complete CAT(0)-space $Z$ on which $\Gamma$
acts by isometry, such that there is a harmonic equivariant map
$h:\Gamma\rightarrow Z$ with strictly positive energy.

###### Proof.

In view of Proposition 6, by passing to a certain ultralimit if necessary, we
may assume that $\delta(x)=\max d(x,s(x))$ bounded away from $0$ and hence the
energy $E(f)$ of equivariant maps $f$ is also bounded away from $0$ in view of
the above displayed inequality.

Let $p_{j}$ be a sequence such that the equivariant maps with
$f_{j}(e):=p_{j}$ have energy approaching the minimum. As remarked above in
case the the $p_{j}$ stay bounded there is an equivariant harmonic map because
of the uniform convexity property of complete CAT(0)-spaces. In any case, we
can argue as follows, consider the $\omega$-ultralimit of $(Y,p_{j})$. By the
above displayed inequality we have $\delta(p_{j})$ is bounded from above, say
by twice the square root of the minimal energy. This guarantees that the
$\Gamma$ action extends to this limit space by isometry, and as before this
limit is a complete finite dimensional CAT(0)-space (in particular in view of
Proposition 4). Likewise we may also define the map $h:\Gamma\rightarrow
Y_{\infty}$ by $h(g)=\omega-\lim f_{j}(g)$. Since for each $j$ this is
equivariant so is $g$. We now claim it is moreover harmonic. Clearly,
$E(h)=\omega-\lim E(f_{j})=\inf_{f}E(f).$ To show that this value of the
energy is also minimal in the limit space, take $h_{1}$ an equivariant map
$\Gamma\rightarrow Y_{\infty}$ and consider its energy. Take a sequence
$w_{j}$ representing $h_{1}(e)$. Define the corresponding equivariant maps
$f_{1,j}(e)=w_{j}.$ For a given $g\in\Gamma$ let $z_{j}$ be a sequence
representing $g(w_{j})$, by equivariance we now have

$0=\omega-\lim d(gw_{j},z_{j})=\omega-\lim d(f_{1,j}(g),z_{j})$

which shows that we can represent $h_{1}$ by the maps $f_{1,j}$, but then it
follows that

$E(h_{1})=\omega-\lim E(f_{1,j})\geq\inf_{f}E(f)=E(h)$

. ∎

##  4 Proofs of the main theorems

Let $\Gamma$ be a group generated by a finite set $S$, which we take to be
symmetric, $S^{-1}=S$. Assume that the balls in the word metric defined by $S$
grow subexponential in the radius (we refer to [dlH00] for more information
about growth in groups). This is independent of the finite generating set. We
denote by $\mu$ the uniform probability measure on $S$. In view of the
subexponential growth, a simple count implies that the asymptotic entropy of
the associated random walk must be $0$. It then follows from the standard
theory that the linear drift in the word length of this symmetric random walk
is zero. All this is very well-known, we refer to the book [La23] for details,
proofs and historical references. For any isometric action
$\rho:\Gamma\rightarrow Isom(X,d)$, the orbit map $g\mapsto\rho(g)x$ from the
group with its word metric to the metric space $(X,d)$ is Lipschitz. Taken
together this implies that

$\ell=\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{g\in\Gamma}d(x,\rho(g)x)\mu^{*n}(g)=0,$

where $\mu^{*n}$ are the convolution powers giving the distribution of the
random walk at time $n$. We emphasize that this holds for any isometric action
of $\Gamma$ on a metric space.

Let $\Gamma$ act by isometry on a finite-dimensional complete CAT(0)-space
$Y$. Assume that $\Gamma$ has no global fixed point in $Y$. Then the
ultralimits, in two steps as explained in the previous section, Propositions 6
and 7, give another finite-dimensional complete CAT(0) space $Z$ on which
$\Gamma$ acts by isometry and with the existence of an equivariant harmonic
map with positive energy. Since the energy is strictly positive there is no
fixed point.

As is explained in [Iz23, sect. 4] we may modify the measure $\mu$ so that we
can assume that its support contains all non-identity elements of $\Gamma$
keeping the drift $0$. The notion of horofunctions $b_{\xi}$ with respect to a
base point $o$ also extends to ultralimits in natural way ([Iz23, Lemma 6.6]).
We recall the important Proposition 6.9 in [Iz23] slightly tailored to fit the
present notations:

######  Proposition 8.

([Iz23, Prop. 6.9]) Let $Z$ be a complete CAT(0)-space. Take $o\in Z$ and set
$Z_{\infty}=\omega-\lim(Z,o).$ Let $\Gamma$ be a countable group equipped with
a symmetric probability measure $\mu$ whose support contains
$\Gamma\setminus\\{e\\}.$ Suppose that $\Gamma$ acts by isometry on $Z$ and
that there exists an equivariant harmonic map $f:\Gamma\rightarrow Z$. If the
drift $\ell=0$, then for almost all random walk trajectories $\sigma$ , there
exists $\xi(\sigma)\in Z_{\infty}\cup\partial Z_{\infty}$such that a function
$\varphi_{\sigma}:\Gamma\rightarrow\mathbb{R}$ defined by

$\varphi_{\sigma}(\xi)=\begin{cases}d(\xi(\sigma),f(g))-d(\xi(\sigma),f(e)),&\text{if
$\xi(\sigma)\in Z_{\infty}$}\\\ b_{\xi(\sigma)}(f(g),f(e)),&\text{if
$\xi(\sigma)\in\partial Z_{\infty}$}\end{cases}$ (1)

is $\mu$-harmonic.

The last notion of $\mu$-harmonic means that the Laplacian defined by $\mu$
annihilates $\varphi_{\sigma}.$ The proof requires a few pages and since it
can be used in a form directly applicable in the present paper, we refer to
[Iz23] for the proof.

Using this proposition one can show the following:

######  Proposition 9.

Let $Z$ be a complete CAT(0)-space of finite dimension and $\Gamma$ be a
countable group equipped with a symmetric probability measure $\mu$ whose
support contains $\Gamma\setminus\\{e\\}.$ Suppose that $\Gamma$ acts on $Z$
by isometry and that there exists an equivariant harmonic map
$f:\Gamma\rightarrow Z$. If the drift $\ell=0$, then there is a
$\Gamma$-invariant convex subset $F$ isometric to a finite dimensional
(possibly zero-dimensional) Euclidean space.

###### Proof.

Although this assertion is not explicitly stated in [Iz23], it can be proved
by following the arguments in [Iz23, §6.4] appropriately in our setting. We
sketch the outline here.

Suppose $\xi(\sigma)\in Z_{\infty}$ in Proposition 8. Then, as in the first
part of [Iz23, §6.4], there is a point in $Z_{\infty}$ fixed by $\Gamma$, and
hence every orbit of $\Gamma$ in $Z_{\infty}$ is bounded. In particular, any
orbit of $\Gamma$ in $Z$ is bounded, and this implies the existence of a
global fixed point, an invariant zero-dimensional Euclidean space, in $Z$ via
the Bruhat-Tits fixed point theorem, [BH99, II.2.8].

If $\xi(\sigma)\in\partial Z_{\infty}$ in Proposition 8, then the horofunction
$\varphi_{\sigma}:g\mapsto b_{\xi(\sigma)}(f(g),f(e))$ is $\mu$-harmonic.
Since horofunctions on CAT(0) spaces are convex, the pull-back of them by a
harmonic map are $\mu$-subharmonic functions on $\Gamma$ in general (see for
example [Iz23, Prop. 2.17]). Here a function $u$ on $\Gamma$ is
$\mu$-subharmonic if the Laplacian of $u$ defined by $\mu$ is nonpositive. If
$u$ is the pull-back of a convex function on a CAT(0) space by a harmonic map,
then the absolute value of its Laplacian measures the strength of the
convexity of the function around the image of the harmonic map. Thus the fact
$\varphi_{\sigma}:g\mapsto b_{\xi(\sigma)}(f(g),f(e))$ is $\mu$-harmonic means
that the horofunction has the weakest convexity around the image $f(\Gamma)$
of $f$. This suggests that the convex hull of $f(\Gamma)$ is flat, and it can
be proved as follows.

Since we assume that $Z$ is finite-dimensional, a compactness principle [Iz23,
Theorem 6.12] tells us that $\xi(\sigma)$ actually lies in $\partial Z$. The
compactness principle and [Iz23, Lemma 5.1] imply that there exist
$\xi\in\partial Z$ and a bi-infinite geodesic $c_{g}$ passing through $f(g)$
and terminating at $\xi$ for each $g\in\Gamma$, and all the $c_{g}$ geodesics
are parallel to each other (in order to prove this, we use the weakest
convexity of the horofunction). Then an argument involving the splitting
theorem, see [BH99, Ch. II.2], leads us to see that there is a splitting of
the convex hull of $f(\Gamma)$ one of whose factor is a Hilbert space, and the
action of $\Gamma$ also splits along this splitting. Taking a maximal flat
factor in this splitting, and applying the argument above to the other factor,
we see that the action of $\Gamma$ on the other factor must fix a point; the
action leaves the product $F$ of the maximal flat factor and the fixed point
in the other factor invariant. Since $Z$ is finite-dimensional, so is $F$, and
it becomes the desired flat subset in $Z$. ∎

Recall now that we have excluded the existence of a global fixed point in $Z$
in the preceding section. Therefore, by Proposition 9, we obtain a
representation of $\Gamma$ into the group of affine isometries of a Euclidean
space of finite and strictly positive dimension. Thus we obtain a linear
representation of $\Gamma$ over the real numbers.

In the case of a finitely generated torsion group, by Schur’s theorem this
representation must have a finite image. Hence by the Cartan-Bruhat-Tits fixed
point theorem it must fix a point in this flat. This leads to a contradiction
to the construction of the space $Z$ from Proposition 7. The conclusion is
then that there must be a global fixed point for the original action,
$\rho:\Gamma\rightarrow Isom(Y)$ which concludes the proof of Theorem 2.

In the case $\Gamma$ is a simple group of subexponential growth, then by the
Tits alternative it must be virtually nilpotent. Virtually nilpotent groups
have a finite index normal subgroup, which must be trivial since the group is
simple. Hence it follows that the image of any linear representation is
trivial or a cyclic group of prime order (which must be $\Gamma$ itself) and
again we have a fixed point in $Z$, leading to the desired contradiction, that
is, there must be a global fixed point for the original action,
$\rho:\Gamma\rightarrow Isom(Y)$.

######  Remark 10.

Suppose $\Gamma=\mathbb{Z}$ then the drift is surely $0$ for the simple
symmetric random walk. This group may act without a fixed point, precisely
when being a parabolic or hyperbolic isometry of $Y$. Note that the proof here
applies all the way until the point that it preserves a flat in $Z$ so it has
an axis (and the hypothesis of only possessing finite linear representations
is obviously not satisfied). This clearly does not contradict the no fixed-
point assumption, and is also consistent with the case when $\Gamma$ acts
parabolically (i.e. without an axis) on the original space $Y$.

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Department of Mathematics, Keio University, Kohoku-ku, Yokohama 223-8522,
Japan.

Section de mathématiques, Université de Genève, Case postale 64, 1211 Genève,
Switzerland; Mathematics department, Uppsala University, Box 256, 751 05
Uppsala, Sweden.