Isometric group actions with vanishing rate of escape on
0 spaces
Hiroyasu Izeki
Department of Mathematics, Keio Univeristy,
Kohoku-ku, Yokohama 223-8522, Japan

[2010]53C23, 20F65
This work was supported by JSPS Grants-in-Aid for Scientific
Research Grant Number JP20H01802.

Let $Y=(Y,d)$ be a $\cat{0}$ space which is either proper or of finite
telescopic dimension, and $\Gamma$ a countable group equipped
with a symmetric and nondegenerate probability measure $\mu$.
Suppose that $\Gamma$ acts on $Y$ via a homomorphism
$\rho\colon \Gamma \rightarrow \isom{Y}$, where $\isom{Y}$ denotes the
isometry group of $Y$,
and that the action given by $\rho$ has finite second moment
with respect to $\mu$.
We show that if $\rho(\Gamma)$ does not fix a point in the boundary at
infinity $\partial Y$ of $Y$ and the rate of escape
$l_{\rho}(\Gamma)=l_{\rho}(\Gamma,\mu)$ associated to an action given
by $\rho$ vanishes, then there exists a flat subspace in $Y$ that is
left invariant under the action of $\rho(\Gamma)$.
Note that if the rate of escape does not vanish, then we know that
there exists an equivariant map from the Poisson boundary of
$(\Gamma,\mu)$ into the boundary at infinity of $Y$ by a result of
A. Karlsson and G. Margulis.
The key ingredient of the proof is $\mu$-harmonic functions on $\Gamma$
and $\mu$-harmonic maps from $\Gamma$ into $Y$.
We prove a result similar to the above for an isometric action of
$\Gamma$ on a locally finite-dimensional $\cat{0}$ space.

§ INTRODUCTION

Let $\Gamma$ be a countable group equipped with a probability
measure $\mu$.
We regard $\mu$ as the law of a random walk; for each $\gamma\in\Gamma$,
we regard $\mu(\gamma)$ as the transition probability of a move from the
identity element $e$ to $\gamma$.
Then the $n$-fold convolution $\mu^n$ of $\mu$ tells us the distribution
of the position of the random walk, generated by $\mu$ and starting from
$e$, at time $n$.

Suppose that $\Gamma$ acts on a metric space $Y=(Y,d_Y)$ via a
homomorphism $\rho\colon \Gamma \rightarrow \isom{Y}$, where $\isom{Y}$
denotes the isometry group of $Y$.
Take any point $p\in Y$ and consider an orbit map
$f_p\colon \gamma \mapsto \rho(\gamma)p$.
Then we can transplant the random walk generated by $\mu$ to
the orbit $f_p(\Gamma)=\rho(\Gamma)p\subset Y$.
The behavior of this transplanted random walk depends on how
$\Gamma$ acts on $Y$.  When $Y$ is a nonpositively curved
metric space, one can ask the following convergence question:
for which action do almost all sample paths of the transplanted random
walk converge to the “boundary at infinity” $\partial Y$ of $Y$?
If almost all sample paths converge to the boundary at infinity,
then this means that any orbit map induces a canonical equivariant map from
the Poisson boundary $\partial (\Gamma,\mu)$, a probability space that
describes the distribution of the position of a random walk on $\Gamma$
generated by $\mu$ at “time infinity”, into $\partial Y$.
Such a boundary map is known to be very useful tool in rigidity theory
as we can see, for example, in a pioneering work due to H. Furstenberg
discussed in [14], and a superrigidity result due to
G. Margulis [34], [35].

There have been many works related to the convergence question above.
Lie groups and their subgroups acting on associated symmetric spaces are
classical and well-understood ones (see [4]).
V. Kaimanovich and H. Masur [27] treated mapping class
groups acting on Teichmüller spaces and showed that, under certain
conditions, the Poisson boundary of a mapping class group can be
realized on the Thurston boundary of corresponding Teichmüller space.
Also a theorem due to Kaimanovich [26] says that it is the
case for hyperbolic-type groups acting on themselves, and that the
Poisson boundary of these groups
can be realized on the boundary at infinity of themselves.
Then, for example, F. Gautero and F. Mathéus [15]
treated certain groups acting on $\R$-trees, and J. Maher and G. Tiozzo
[33] treated groups acting on separable Gromov hyperbolic
spaces. Also groups acting on $\cat{0}$ cubical complexes were
considered by T. Fernós, J. Lécureux and F. Mathéus
Recently Chois [9] treated groups containing elements with
bounded geodesic image property acting on various nonpositively curved
spaces. The reader can find more works on the question in the references
of papers cited above.

On the other hand, there has been a known criterion for the
convergence to the boundary for groups acting on general nonpositively
curved metric spaces due to A. Karlsson and G. Margulis
Consider the rate of escape
$l_{\rho}(\Gamma)=l_{\rho}(\Gamma,\mu)$ of an action given by $\rho$
which turns out to be equal to

\begin{equation*}
l_{\rho}(\Gamma) = \lim_{n\to \infty}
\frac{\int_{\Gamma} d_Y(p,\rho(\gamma)p) d\mu^n(\gamma)}{n}
\end{equation*}

(see  <ref>).
This limit is known to exist, and since the numerator of the right-hand
side is the average of the distance between the starting point and the
position at time $n$ of the transplanted random walk, it measures the
speed of the transplanted random walk's escape to the boundary at
infinity $\partial Y$ of $Y$.
Indeed, for a nonpositively curved metric space $Y$,
A. Karlsson and G. Margulis proved in [28] that if
$l_{\rho}(\Gamma)>0$, then
almost all sample paths of the transplanted random walk
leave every bounded subset fast enough and go almost straight toward
$\partial Y$;
almost all sample paths converge to points in $\partial Y$.
However, it is usually difficult to check the positivity of
$l_{\rho}(\Gamma)$. For example, in
[12] and [33], the positivity
of $l_{\rho}(\Gamma)$ was obtained as a consequence of the convergence
to the boundary.

If $l_{\rho}(\Gamma)=0$, then the transplanted random
walk is rather wandering around in $Y$.
Thus one might think that information obtained from such a random
walk should contain lots of noise, and that it is hard to
Indeed, it seems that there has been no known efficient method to handle
this case in full generality.
However, under mild assumptions on $Y$ and $\rho$, our approach using
harmonic functions and harmonic maps associated to $\mu$ and $\rho$
enables us to specify such an action as stated in
Theorem <ref> below; an action given by $\rho$ with
$l_{\rho}(\Gamma)=0$ turns out to be the simplest one in the sense that
it is reduced to that on a Euclidean space.

Let $Y$ be a complete $\cat{0}$ space which is either proper or
of finite telescopic dimension,
and $\Gamma$ a countable group equipped with a symmetric and
nondegenerate probablity measure $\mu$.
Let $\Gamma$ act on $Y$ via a homomorphism
$\rho\colon \Gamma \rightarrow \isom{Y}$, and
suppose that $\rho(\Gamma)$ does not fix a point in
$\partial Y$ and that the action given by $\rho$ has finite
second moment with respect to $\mu$.
If  $l_{\rho}(\Gamma,\mu)=0$, then there exists a flat subspace $F$ in
$Y$, which is a convex subset of $Y$ isometric to $\R^n$ $(n\geq 0)$,
left invariant under the action of $\rho(\Gamma)$.

The case $n=0$ in the theorem above should be understood as that
$\rho(\Gamma)$ fixes a point in $Y$.
Let us briefly explain the assumptions in Theorem <ref>.
(See  <ref> for precise definitions and more
To obtain sufficient information of actions of $\Gamma$ from the
random walk generated by $\mu$, we assume that
$\mu$ is symmetric and nondegenerate.
Namely, $\mu$ satisfies $\mu(\gamma^{-1})=\mu(\gamma)$, and the support
$\supp{\mu}$ of $\mu$ generates $\Gamma$.
We also assume that the action given by $\rho$ has finite second
moment with respect to $\mu$:

\begin{equation*}
\int_{\Gamma} d_S(p,\rho(\gamma)p)^2 d\mu(\gamma)< \infty
\end{equation*}

holds for $p \in Y$;
it is easy to see that, by the triangle inequality
and the Schwarz inequality, this does not depend on the choice of
$p \in Y$.  If $\Gamma$ is finitely generated and $\mu$ has finite
second moment with respect to some (hence any) word metric on $\Gamma$,
then any $\rho \colon \Gamma \rightarrow \isom{Y}$ gives the action with
finite second moment (see  <ref>).
A metric space under consideration is a nonpositively curved metric
space called $\cat{0}$ space.
An important feature of $\cat{0}$ spaces on which our approach relies
is the convexity of distance functions.
We also assume that a $\cat{0}$ space $Y$ is either proper or of finite
telescopic dimension in order to provide $Y$ with a certain compactness
Here a $\cat{0}$ space $Y$ is called proper if every closed metric
ball $B(p,r)$, $p\in Y$, $r>0$, is compact, while a $\cat{0}$ space $Y$
is said to have telescopic dimension at most $n$
if every asymptotic cone of $Y$ has geometric dimension at most $n$;
the notion of geometric dimension and telescopic dimension was
introduced by B. Kleiner in [29] and by
P.-E. Caprace and A. Lytchak in [6], respectively.
(See  <ref> for more details.)
The assumption that $\rho(\Gamma)$ does not fix a point in $\partial Y$
guarantees the existance of a $\rho$-equiavariant $\mu$-harmonic map
$f \colon \Gamma \rightarrow Y$.
Here, a $\rho$-equivariant map $f\colon \Gamma \rightarrow Y$ is called
$\mu$-harmonic if $f$ minimizes the energy $\ene{\mu}$ defined by

\begin{equation*}
\ene{\mu}(f)=
\frac{1}{2} \int_{\Gamma} d_Y(f(e),f(\gamma))^2 d\mu(\gamma)
\end{equation*}

among all $\rho$-equivariant maps.
As equivariant maps from the Poisson boundary, equivariant harmonic
maps are also known to be useful tools in rigidity theory.
For example, K. Corlette [10],
N. Mok, Y.-T. Siu and S.-K. Yeung [36],
J. Jost and S.-T. Yau [24], [25], and
M. Gromov and R. Schoen [17] used the existence
of nonconstant equivariant harmonic maps from Riemannian symmetric
spaces in order to prove superrigidity results for lattices in Lie
groups, where the local geometric structure of the domains of harmonic
maps has key information to determine the images of these harmonic
maps and deduce superrigidity results.
Also discrete equivariant harmonic maps from the vertex set of a
simplicial complex or a finitely generated group itself were
treated by,
for example, M. Gromov [16], M.-T. Wang [39], and
H. Kondo, S. Nayatani and the author [22],
[20], [21],
[19], where various fixed-point theorems for isometric actions
of finitely generated groups on $\cat{0}$ spaces were obtained from the
existence of constant equivariant harmonic maps.
While in the present paper, we prove Theorem <ref> by using
the existence of a nonconstant equivariant harmonic map defined on a
countable group.
Typically, the image of our harmonic map is nowhere dense in the
target space or even in the convex hull of the image; the space we need
to specify its structure is far from being covered by the image.
In order to prove our theorem, however, we need to fill this huge
We achieve this by using the convexity of distance
functions of the target space effectively.

The outline of the proof of Theorem <ref> is as follows.
Since $Y$ is either proper or of finite telescopic dimension, under the
assumption that $\rho(\Gamma)$ does not fix a point in $\partial Y$,
one sees that there exists a $\rho$-equivariant $\mu$-harmonic map
$f\colon\Gamma \rightarrow Y$ (see  <ref> and
Proposition <ref>).
A $\mu$-harmonic map has a property that the pull-back of a convex
function $u$ on the target space $Y$ becomes a $\mu$-subharmomnic
function on $\Gamma$, that is, $-\Delta f^*u(\gamma)\geq 0$ holds for
any $\gamma \in \Gamma$.
Here, one may regard the value
$|-\Delta f^*u(\gamma)|$ as a quantity that measures the strength of
the convexity of a function $u$ around $f(\gamma)$.
We note here that a distance function $p\mapsto d(p,p_0)$ is a convex
function on $Y$, since $Y$ is a $\cat{0}$ space,
and that the strength of the convexity of a distance function should be
related to the strength of the nonpositivity (negativity) of the
“curvature” of $Y$.
Now, under the assumption that $l_{\rho}(\Gamma)=0$, we can
show that there exists a sequence $\{\gamma_j\}\subset \Gamma$ such that
the pull-back of $u_j\colon p \mapsto d(p, f(\gamma_j))$
approaches a $\mu$-harmonic function on $\Gamma$.
By our assumption that $Y$ is either proper or of finite telescopic
dimension, after a suitable normalization, we obtain a limit function
$u$ of $\{u_j\}$ which is either a distance function from a point in
$Y$ or a Busemann function associated to a point in $\partial Y$
(see  <ref> and
Then the pull-back $f^*u$ of the limit function $u$ is $\mu$-harmonic,
that is, $-\Delta f^*u(\gamma)=0$ holds for any $\gamma \in \Gamma$,
and hence $u$ has the weakest convexity on the convex hull
$\ch{f(\Gamma)}$ of $f(\Gamma)$.
This suggests that the nonpositivity of the curvature of
$\ch{f(\Gamma)}$ is the weakest one, namely, $\ch{f(\Gamma)}$ must be
a flat subspace in $Y$.

Combining Theorem <ref> with a theorem due to A. Karlsson
and G. Margulis mentioned above, we immediately obtain the following
theorem, which is a refinement of a part of a theorem due to U. Bader,
B. Duchesne, and J. Lécureux

Let $Y$ be a complete $\cat{0}$ space which is either proper or
of finite telescopic dimension, and $\Gamma$ a countable group
with a symmetric and nondegenerate probability measure $\mu$.
Let $\rho\colon \Gamma \rightarrow \isom{Y}$ be a homomorphism and
suppose that $\rho(\Gamma)$ does not fix a point in
$\partial Y$ and that the action given by $\rho$ has finite
second moment with respect to $\mu$.
Then either of the following is true.

$(1)$ An orbit map induces a canonical $\rho$-equivariant map from the
Poisson boundary $\partial (\Gamma,\mu)$ of $\Gamma$ into the boundary
$\partial Y$ of $Y$.

$(2)$ There exists a flat subsapce of $Y$ left invariant under the
action of $\rho(\Gamma)$.

For example, if $\Gamma$, equipped with the discrete topology,
has Kazhdan's property $(T)$, then
$\Gamma$ is finitely generated and any
isometric action of $\Gamma$ on a Hilbert space (a Euclidean space) has
a fixed point (see for example [3]).
Therefore we obtain the following.

Let $\Gamma$ be a discrete group with Kazhdan's
property $(T)$, and $\mu$ a symmetric and nondegenerate probability
measure $\mu$ on $\Gamma$ with finite second moment.
Let $Y$ be a complete $\cat{0}$ space which is either proper or
of finite telescopic dimension, and
$\rho\colon \Gamma \rightarrow \isom{Y}$ a homomorphism.
Suppose that $\rho(\Gamma)$ does not fix a point in $Y\cup \partial Y$.
Then any orbit map induces a canonical $\rho$-equivariant map from the
Poisson boundary $\partial (\Gamma,\mu)$ of $\Gamma$ into the boundary
$\partial Y$ of $Y$.

Since a group $\Gamma$ with Kazhdan's property $(T)$ is known to
have an “entropy gap”, the assertion of the corollary above says that
there is certain restrictions on isometric actions of $\Gamma$ on
$\cat{0}$ spaces as it has been suggested in many works from various

If $Y$ is locally finite-dimensional $\cat{0}$ space, we can show a
slightly weaker result.
We assume that $\rho(\Gamma)$ is
reductive in the sense of Jost
(see  <ref> for the definition), which is stronger
than the assumption that there exists a point in $\partial Y$ fixed by
$\rho(\Gamma)$; we need this assumption to deduce the existence of a
$\rho$-equivariant $\mu$-harmonic map from $\Gamma$ into $Y$.

Let $Y$ be a complete, locally finite-dimensional $\cat{0}$ space,
and $\Gamma$ a countable group equipped with a symmetric and
nondegenerate probablity measure $\mu$.
Let $\Gamma$ act on $Y$ via a homomorphism
$\rho\colon \Gamma \rightarrow \isom{Y}$, and
suppose that $\rho(\Gamma)$ is
reductive in the sense of Jost and that the action given by
$\rho$ has finite second moment with respect to $\mu$.
If  $l_{\rho}(\Gamma,\mu)=0$,  then there exists a flat subspace
of $Y$ left invariant under the action of $\rho(\Gamma)$.

It is clear that an analogue of Theorem <ref> holds for a
locally finite-dimensional $\cat{0}$ space $Y$.

Let $Y$ be a complete, locally finite-dimensional $\cat{0}$ space,
and $\Gamma$ a countable group
with a symmetric and nondegenerate probability measure $\mu$.
Let $\rho\colon \Gamma \rightarrow \isom{Y}$ be a homomorphism and
suppose that $\rho(\Gamma)$ is
reductive in the sense of Jost and that the action given by
$\rho$ has finite second moment with respect to $\mu$.
Then either of the following is true.

$(1)$ An orbit map induces a canonical $\rho$-equivariant map from the
Poisson boundary $\partial (\Gamma,\mu)$ of $\Gamma$ into the boundary
$\partial Y$ of $Y$.

$(2)$ There exists a flat subsapce of $Y$ left invariant under the
action of $\rho(\Gamma)$.

We also point out that, as a byproduct of our result, we obtain
a certain gap theorem on the energy growth of equivariant harmonic
For a $\rho$-equivariant map $f\colon \Gamma \rightarrow Y$,
$n$-step energy $\ene{\mu^n}(f)$ is defined by

\begin{equation*}
\ene{\mu^n}(f) =
\frac{1}{2} \int_{\Gamma} d_Y(f(e),f(\gamma))^2 d\mu^n(\gamma).
\end{equation*}

It is shown in [20] that,
for any $\rho$-equivariant map into a $\cat{0}$ space $Y$,
$\ene{\mu^n}(f)\leq Cn^2\ene{\mu}(f)$ holds for some $C>0$, and
that if $f$ is a $\rho$-equivariant $\mu$-harmonic map into a Hilbert
space (a Euclidean space), then

\begin{equation*}
\ene{\mu^n}(f) = n \ene{\mu}(f)
\end{equation*}

On the other, if $l_{\rho}(\Gamma)>0$, then the Schwarz inequality
tells us that

\begin{equation*}
\liminf_{n\to \infty} \frac{\ene{\mu^n}(f)}{n^2} > 0.
\end{equation*}

Therefore we obtain:

Let $Y$ be a $\cat{0}$ space which is either proper, of
finite telescopic dimension, or of locally finite dimension.
And let $\Gamma$ be a countable group with a symmetric and
nondegenerate probability measure $\mu$.
Suppose that the action given by a homomorphism
$\rho\colon \Gamma \rightarrow \isom{Y}$ has finite second moment with
respect to $\mu$ and that
there exists a nonconstant $\rho$-equivariant
$\mu$-harmonic map $f$.
Then either $\ene{\mu^n}(f)=n\ene{\mu}(f)$ or
$\ene{\mu^n}(f)\asymp n^2$ holds.

We note here that it was pointed out and proved in [16] and
[21] that if there exist $\varepsilon>0$ and a
natural number $n$ such that every $\rho$-equivariant map $f$ satisfies
$\ene{\mu^n}(f)\leq (n-\varepsilon)\ene{\mu}(f)$, then there exists a
constant $\rho$-equivariant harmonic map, and hence $\rho(\Gamma)$ fixes
a point in $Y$, which is used in [21] and
[19] to prove a fixed-point property for random groups.

This paper is organized as follows.
In  <ref>, we recall fundamental definitions and
facts on $\cat{0}$ spaces, and some facts on equivariant harmonic maps.
We first finish the proof of Theorem <ref> for proper
$\cat{0}$ spaces in  <ref>,
<ref>, and
In  <ref>, under the assumption
$l_{\rho}(\Gamma)=0$ together with the existence of an equivariant
$\mu$-harmonic map $f$, we show that the pull-back of a distance
function or a Busemannn function by $f$ becomes a (locally)
$\mu$-harmonic function; they are obtained as a limit of distance
functions from a sequence of points.
The properness of a $\cat{0}$ space guarantees that the limit
function can be regarded as either a distance function or a Busemann
Then, in  <ref>, we prove two
inequalities reflecting the convexity of metric balls and horoballs
respectively; the equality case plays a key role in the recognition of a
flat subspace in Theorem <ref>.
In  <ref>, using these results, we prove
Theorem <ref> for proper $\cat{0}$ spaces.

The proofs of Theorem <ref> for nonproper $\cat{0}$ spaces
and that of Theorem <ref> are given in
<ref> and  <ref>.
In order to apply our arguments to nonproper $\cat{0}$ spaces, we need a
sort of compactness theorem (Theorem <ref>) for
$\cat{0}$ spaces with certain dimension bounds.
In  <ref>, we present a quick review of the notion of
ultralimit and consider ultralimits of $\cat{0}$ spaces with certain
dimension bounds.
Assuming the compactness theorem mentioned above,
we finish the proof of Theorem <ref>.
The proof of Theorem <ref> is a straightforward variant of
that of Theorem <ref>
(see  <ref>).
Then  <ref> is devoted to the proof of
Theorem <ref>.
In the course of the proof of Theorem <ref>, we need to
replace a probability measure $\mu$ on $\Gamma$ with one whose support
covers all non-trivial elements of $\Gamma$.
It is achieved by taking a convex combination of the $n$-fold
convolution $\mu^n$'s of $\mu$.
In  <ref>, we show that this replacement does not affect
our assumption $l_{\rho}(\Gamma)=0$ in Theorem <ref>.

The author would like to thank Uri Bader, Atsushi Katsuda, Shin
Nayatani, Takefumi Kondo, Ryokichi Tanaka, Masato Mimura, and Tomohiro
Fukaya for their comments and interests in this work.

§ EQUIVARIANT HARMONIC MAPS INTO $\CAT{0}$ SPACES

In this section, we give some fundamental definitions and
facts on $\cat{0}$ spaces and equivariant harmonic maps from
countable groups into $\cat{0}$ spaces.

Let $Y$ be a metric space with metric $d_Y$, and $I\subset \R$ an
A map $c\colon I \rightarrow Y$ is called a geodesic if $c$ is
an isometric embedding, namely, $c$ satisfies $d_Y(c(t),c(t'))=|t-t'|$
for any $t,t' \in I$.
If $I$ is a bounded closed interval, we often call $c$ a
geodesic segment, while if $I=\ray{0,\infty}$, we call $c$ a
geodesic ray.
We call a geodesic defined on whole $\R$ a geodesic line.
If every pair of points in $Y$ is joined by a geodesic segment, then $Y$
is called a geodesic space.
In what follows, all metric spaces under consideration are assumed to be
complete as metric spaces.

A geodesic space $Y$ is called $\cat{0}$ space if any geodesic triangle
is thinner than that in the Euclidean plane in the following sense:
Let $p_1,p_2,p_3 \in Y$, and denote the image of geodesic joining $p_i$
and $p_j$ by $[p_i,p_j]$.
We call $[p_1,p_2]\cup [p_2,p_3]\cup [p_3,p_1]$ a geodesic
triangle, and denote by $\tri{p_1}{p_2}{p_3}$.
Note that we can take
a triangle $\tri{\overline{p_1}}{\overline{p_2}}{\overline{p_3}}$ in
$\R^2$ so that $d_Y(p_i, p_j)=d_{\R^2}(\overline{p_i},\overline{p_j})$,
since the side lengths of $\tri{p_1}{p_2}{p_3}$ satisfy the triangle
We call $\tri{\overline{p_1}}{\overline{p_2}}{\overline{p_3}}$ a
comparison triangle for $\tri{p_1}{p_2}{p_3}$.
And we call a point
$\overline{q}\in [\overline{p_i},\overline{p_j}]$
a comparison point for $q\in [p_i,p_j]$ if
$d_Y(p_i,q) = d_{\R^2}(\overline{p_i}, \overline{q})$.
Then we say that $\tri{p_1}{p_2}{p_3}$ satisfies the
$\cat{0}$ inequality
if $d_Y(q_1,q_2) \leq d_{\R^2}(\overline{q_1}, \overline{q_2})$ holds
for any pair of points $q_1, q_2$ on the sides of $\tri{p_1}{p_2}{p_3}$
and their comparison points $\overline{q_1}, \overline{q_2}$.
If every geodesic triangle in $Y$ satisfies the $\cat{0}$ inequality,
then $Y$ is called a $\cat{0}$ space.
A complete, simply
connected Riemannian manifold of nonpositive sectional curvature is a
typical example of $\cat{0}$ space.  Also trees and Hilbert spaces are
$\cat{0}$ spaces.
If one replaces the Euclidean plane, the present location of the
comparison triangle, with the $2$-dimensional space form of curvature
$\kappa$, we obtain the notion of $\cat{\kappa}$ space, which can be
regarded as a metric space whose curvature is bounded from above by
For a detailed exposition on $\cat{\kappa}$ spaces, we refer the reader
to [5].

If $Y$ is a $\cat{0}$ space, then it is clear that any pair of distinct
points is joined by a unique geodesic, and in particular, a $\cat{0}$
space turns out to be contractible.
Furthermore, it is not hard to see that the distance function on a
$\cat{0}$ space becomes a convex function, an important feature of a
$\cat{0}$ space:
For any geodesic $c_i\colon [0,1]\rightarrow Y$, $i=1,2$, parametrized
proportional to arc length (possibly a constant map), a function
$t \mapsto d(c_1(t),c_2(t))$ is a convex function, that is, for any
$t \in [0,1]$,

\begin{equation*}
d_Y(c_1(t), c_2((t)) \leq
(1-t)d_Y(c_1(0),c_2(0))  + td_Y(c_1(1),c_2(1))
\end{equation*}

holds (see <cit.>).
For a $\cat{0}$ space $Y$ the following is a part of the list of its
properties more or less directly derived from the convexity
above and the definition of $\cat{0}$ spaces.
We will frequently use them in what follows.

* Recall that $C \subset Y$ is called a convex subset if any
pair of points in $C$ can be joined by a geodesic lying in $C$.
For a closed convex subset $C$ of $Y$ and $p \in Y$, there exists
a unique point $q \in C$ such that $d_Y(p,C)=d_Y(p,q)$ holds.
Thus we can define the nearest point projection $\nrprj_C$
that sends $p \in Y$ to this $q \in C$; $\nrprj_C$ turns out to
be a distance nonincreasing map.
(See (<cit.>) for the

* Let $C$ be a closed convex subset of $Y$.
If $\nrprj_C$ restricted to $C'$ becomes an isometry onto its
image, then the convex hull of $\nrprj_C(C')\cup C'$ is isometric
to a product $C' \times [0,l]$.
(See <cit.>.)

* If two geodesic lines $c_i \colon \R \rightarrow Y$, $i=1,2$,
are parallel to each other, that is, $c_1(\R)$ stays in a bounded
distance from $c_2(\R)$ and vice versa, then the convex hull of
$c_1(\R)\cup c_2(\R)$ is isometric to $\R \times [0,l]$.
(See <cit.>.)

* Let $c\colon \R \rightarrow Y$ be a geodesic line.
The union of the images of geodesic lines parallel to $c$ forms a
closed convex subset $Y_c$ of $Y$, and $Y_c$ is isometric to a
product metric space $\R \times Y_0$ for some convex subset
$Y_0 \subset Y_c$.
(See <cit.>. )

From the viewpoint of the convexity of the distance function,
the Euclidean space $\R^n$, whose nonpositivity of the curvature is the
least among the $\cat{0}$ spaces, should be characterized as a $\cat{0}$
space with the “weakest” convexity.
In order to prove the existence of a flat subspace in
Theorem <ref>, we need to develope a quantitative
approach to the recognition of the subset with the weakest convexity.
It is plausible that, for example, the existence of solutions to a sort
of “equation” leads us to the goal;
indeed, it is achieved by considering harmonic maps and harmonic
functions as we will see.
In order to realize this, a tool like Riemannian metric provides us with
a great help; it enables us to produce various quantitative expressions
of the convexity (see, for example,
Propositions <ref> and
<ref>, and remarks following them).
An “inner product” defined below on the tangent cone
plays such a role in this paper.

Let $Y$ be a $\cat{0}$ space.

$(1)$  Let $c$ and $c'$ be two non-trivial geodesics in $Y$ starting
from $p \in Y$. The angle $\angle_p(c,c')$ between $c$ and $c'$
is defined by

\begin{equation*}
\angle_p(c,c')= \lim_{t,t' \rightarrow 0}
\angle_{\overline{p}}(\overline{c(t)},\overline{c'(t')}),
\end{equation*}

denotes the angle between the sides
$[\overline{p},\overline{c(t)}]$ and $[\overline{p},\overline{c'(t')}]$
of the comparison triangle
For $p,q \in Y$, denote by $\cc{p}{q}$ a geodesic starting from $p$ and
teminating at $q$.
We often use an expression $\angle_p(q,q')$ for the angle
$\angle_p(\cc{p}{q},\cc{p}{q'})$ between $\cc{p}{q}$ and $\cc{p}{q'}$.

$(2)$  Let $p \in Y$.
For a pair of non-trivial geodesics $c$, $c'$ starting from $p$, we
define a relation $\sim$ by $c\sim c'$ if and only if
Then $\sim$ becomes an equivalence relation on the set of
all non-trivial geodesics starting from $p$ denoted by
Then the angle $\angle_p$ induces a metric on the quotient
$(S_pY)^{\circ} = \widetilde{(S_pY)^{\circ}}/\sim$,
which we denote by the same symbol $\angle_p$.  The completion
$(S_pY,\angle_p)$ of the metric space $((S_pY)^{\circ}, \angle_p)$ is
called the space of directions  at $p$.

$(3)$ Let $TC_pY$ be the cone over $S_pY$, namely,

\begin{equation*}
TC_pY = (S_pY \times \R_{\geq 0}) / (S_pY \times \{0\}).
\end{equation*}

Let $W,W' \in TC_pY$. We may write $W=(V,t)$ and $W'=(V',t')$, where
$V,V' \in S_pY$ and $t,t' \in \R_{\geq 0}$.   Then

\begin{equation*}
d_{TC_pY}(W, W')= t^2 + {t'}^2 - 2tt'\cos \angle_p(V,V')
\end{equation*}

defines a metric on $TC_pY$. The metric space $(TC_pY, d_{TC_pY})$ is
known to be a $\cat{0}$ space and is called the tangent cone at
We define an “inner product”  on  $TC_pY$ by

\begin{equation*}
\inner{W}{W'} = tt'\cos \angle_p(V,V').
\end{equation*}

We often denote the length $t$ of $W$ by $|W|$; thus we have
$|W|=\sqrt{\langle W,W \rangle}=d_{TC_pY}(O_p,W)$, where $O_p$ denotes
the origin of $TC_pY$.
In what follows, we identify $S_pY$ with the set of elements
with the length equals one in $TC_pY$:
$S_pY = \{W \in TC_pY \mid |W|=1\}$.

$(4)$ Define a map $\tcprj_p \colon Y \longrightarrow TC_pY$ by
$\tcprj_p(q)=([c], d_Y(p,q))$, where $c$ is a geodesic starting from
$p$ and terminating at $q$, and $[c]\in S_pY$ is the equivalence class
of $c$. Then $\tcprj_p$ is distance nonincreasing.

For a $\cat{0}$ space $Y$, the space of direction $S_pY$ is known to be
a $\cat{1}$ space (<cit.>).
Note that, by definition, any isometry $\Phi\colon Y \rightarrow Y$
induces an isometry $\Phi_* \colon S_pY \rightarrow S_{\Phi(p)}Y$ and
$\Phi_*\colon TC_pY \rightarrow TC_{\Phi(p)}Y$.

While the tangent cone describes a local structure of a $\cat{0}$
space $Y$, the boundary at infinity defined below tells us an
asymptotic nature of $Y$; for example,
the boundary $\partial Y$ of $\cat{0}$ space $Y$
equipped with the angular metric (see below for the definition)
describes the distribution of the flat subspaces in $Y$.

Let $Y$ be a $\cat{0}$ space.

$(1)$ Two geodesic rays $c_i\colon \ray{0,\infty}\rightarrow Y$,
$i=1,2$, are asymptotic to each other if there exists
$M\in\ray{0,\infty}$ such that $d(c_1(t),c_2(t))\leq M$ for all
$t \in \ray{0,\infty}$.
Being asymptotic is easily seen to be an equivalence relation on the set
of geodesic rays. The boundary $\partial Y$ of $Y$ is defined to
be the set of equivalence classes (asymptotic classes) of geodesic
The topology on $Y \cup \partial Y$ naturally induced from
the uniform convergence of geodesics and geodesic rays on each bounded
interval is called the cone topology of $Y \cup \partial Y$.

$(2)$ Let $\xi_i \in \partial Y$.  Denote by $\cc{p}{\xi_i}$ a geodesic
ray starting from $p$ and representing $\xi_i$ (we will say
“terminating at $\xi_i$” and express as
Let us denote the angle between
$\cc{p}{\xi_1}$ and $\cc{p}{\xi_2}$ by $\angle_p(\xi_1,\xi_2)$.

\begin{equation*}
d_{\angle}(\xi_1,\xi_2)=\sup_{p \in Y} \angle_p(\xi_1,\xi_2)
\end{equation*}

gives a metric on $\partial Y$ called the angular metric.

$(3)$  Let $c\colon \ray{0,\infty}\rightarrow Y$ be a geodesic ray.
The Busemann function $b_c$ associated to $c$ is defined by

\begin{equation*}
b_c (x) =\lim_{t \to \infty} d(c(t),x)-t
\end{equation*}

The level set $S_c(r)=\{p \in Y \mid b_{c}(p)=r\}$ is called a
horosphere centered at $\xi=c(\infty)$ (see
Remark <ref> below).
The sublevel set $B_c(r)=\{p \in Y \mid b_{c}(p)\leq r\}$ is
called a horoball centered at $\xi=c(\infty)$.

Since any isometry preserves the relation asymptotic, the isometry
group $\isom{Y}$ of $Y$ acts on $\partial Y$.
The action is by homeomorphisms with respect to the cone topology.
On the other hand, it is clear, by definition, the action is by
isometry with respect to the angular metric on $\partial Y$.

A metric space $Y=(Y,d_Y)$ is called proper if every
closed metric ball is compact.
If a $\cat{0}$ space $Y$ is proper, then one sees that, by using
Arzelà-Ascoli theorem,
$Y\cup \partial Y$ equipped with the cone topology is compact.

For a $\cat{0}$ space $Y$, the boundary $(\partial Y, d_{\angle})$ with
the angular metric is a $\cat{1}$ spaces
(see <cit.>).
The topology on $\partial Y$ given by the angular metric is stronger
than the cone topology restricted to $\partial Y$ in general.
For example, for a hyperbolic $n$-space $\mathbb{H}^n$,
$\partial \mathbb{H}^n$ with the cone topology is homeomorphic to
$S^{n-1}$,  while $(\partial \mathbb{H}^n,d_{\angle})$ is easily seen
to be a discrete metric space, since every pair of points in
$\partial\mathbb{H}^n$ can be joined by a geodesic line.

Let $\xi, \xi' \in \partial Y$ with $d_{\angle}(\xi,\xi')<\pi$.
It is known that if there exists $p_0$ such that
$\angle_{p_0}(\xi,\xi')=d_{\angle}(\xi,\xi')$, then the convex hull of
$\cc{p_0}{\xi}(\ray{0,\infty})\cup \cc{p_0}{\xi'}(\ray{0,\infty})$ is
isometric to a flat sector, which is a sector in the Euclidean plane
bounded by a pair of half lines that meet at their endpoints with an
angle $\angle(\xi,\xi')$
On the other hand,
by <cit.>, one sees that
if the convex hull of $\ray{p,\xi}\cup \ray{p,\xi'}$ is isometric to a
flat sector, then $d_{\angle}(\xi,\xi')=\angle_p(\xi,\xi')$ holds.

Let $\xi \in \partial Y$ and $\{p_n\}\subset Y$ any sequence that
converges to $\xi$ in the cone topology. Fix $p_0 \in Y$.
Consider a function $p\mapsto b_{\xi}(p,p_0)$ defined by

\begin{equation}
\label{eq:def_of_busemann}
b_{\xi}(p,p_0)= \lim_n d(p_n,p)-d(p_n,p_0).
\end{equation}

The limit in the right-hand side exists and does not depend on the
choice of a sequence $\{p_n\}$ converging to $\xi$
(see <cit.>, see also
Lemma <ref>).
Therefore we see that $b_{\xi}(p)=b_c(p)$ for a geodesic ray $c$ with
$c(0)=p_0$ and $c(\infty)=\xi$;
we treat (<ref>) as the definition of the Busemann
function, and mainly use the symbol $b_{\xi}(\cdot,p_0)$ in what
Note that, by definition and the description above,
$\lim_{t\to \infty}b_{\xi}(c(t),p_0)\to -\infty$ for any geodesic ray
terminating at $\xi$.
Note also that the definition (<ref>) above shows that
the Busemann function has an additivity:

\begin{equation}
\label{eq:additivity}
b_{\xi}(p,p_0)=b_{\xi}(p,p_0')+ b_{\xi}(p_0',p_0)
\end{equation}

for any $p,p_0,p_0' \in Y$.
In particular, the change of the basepoint $p_0$ leaves the Busemann
function invariant up to a constant.
Thus we often denote the horosphere and the horoball centered at $\xi$
by $H_{\xi}(r)$ and $B_{\xi}(r)$ without specifying the basepoint.
Since a Busemann function $b_{\xi}(\cdot,p_0)$ is the limit of a
sequence of convex functions,  $b_{\xi}(\cdot,p_0)$ is also a convex
As a consequnece, a horoball $B_{\xi}(r)$, as well as a metric ball,
becomes a convex subset of $Y$.

Let $\Gamma$ be a countable group, and $\mu$ a probability
measure on $\Gamma$.
We assume that the support of $\mu$ generates $\Gamma$,
and that $\mu$ is symmetric, that is,
$\mu$ satisfies $\mu(\gamma)=\mu(\gamma^{-1})$ for any
$\gamma \in \Gamma$.
Suppose that $\Gamma$ acts on a metric space $Y$ isometrically via a
homomorphism $\rho\colon \Gamma \rightarrow \isom{Y}$, where $\isom{Y}$
is the isometry group of $Y$.
We assume that the action given by $\rho$ has
finite second moment with respect to $\mu$, namely, for $p\in Y$,

\begin{equation*}
\int_{\Gamma} d_Y(p, \rho(\gamma)p)^2 d\mu(\gamma) < \infty
\end{equation*}

Since the action given by $\rho$ is by isometry, using the triangle
inequality, we see that

\begin{equation*}
\begin{split}
d_Y(p', \rho(\gamma)p')^2
& \leq \left(d_Y(p',p) + d_Y(p,\rho(\gamma)p) + d_Y(\rho(\gamma)p,
\rho(\gamma)p')\right)^2 \\
& = \left(2d_Y(p',p) + d_Y(p,\rho(\gamma)p) \right)^2 \\
& = 4d_Y(p',p)^2 + 4d_Y(p',p)d_Y(p,\rho(\gamma)p) +
\end{split}
\end{equation*}

Applying the Schwarz inequality to the second term in the last
expression, we see that it is also integrable with respect to $\mu$; the
integrability does not depend on the choice of $p \in Y$.
A map $f \colon \Gamma \rightarrow Y$ is said to be
$\rho$-equivariant if $f(\gamma \gamma')=\rho(\gamma)f(\gamma')$
holds for any $\gamma, \gamma' \in \Gamma$.
The $\mu$-energy $\ene{\mu}(f)$ of a $\rho$-equivariant map $f$ defined

\begin{equation*}
\ene{\mu}(f) = \frac{1}{2} \int_{\Gamma} d_Y(f(e), f(\gamma))^2
\end{equation*}

is well-defined by the assumption that the action given by $\rho$ has
finite second moment with respect to $\mu$.

Now suppose that $\Gamma$ is finitely generated.
Then we have a finite generating set $S$ of $\Gamma$.
We assume that $S$ is symmetric; $s \in S$ implies that $s^{-1}\in S$.
The word length $\mathrm{length}_S(\gamma)$ of $\gamma \in \Gamma$
is defined as the length of the shortest expression of $\gamma$ by
generators belonging to $S$:

\begin{equation*}
\mathrm{length}_S(\gamma)= \min \{ k \in \N \mid \gamma=s_1 \dots s_k,\
s_i \in S\}.
\end{equation*}

Then $d_S(\gamma, \gamma')=\mathrm{length}_S(\gamma^{-1}\gamma')$
becomes a metric on $\Gamma$, and called the word metric on
$\Gamma$ with respect to a finite generating set $S$.
Note that, by definition, $d_S(\cdot,\cdot)$ is a left-invariant metric:
$d_S(\gamma \gamma', \gamma \gamma'')=d_S(\gamma',\gamma'')$.
For finite generating sets $S$ and $S'$, although $d_S$ and $d_{S'}$ may
be different, it is easy to see that they are Lipschitz equivalent.
We say that $\mu$ has finite second moment, if $\mu$ satisfies

\begin{equation*}
\int_{\Gamma} d_S(e,\gamma)^2 d\mu(\gamma) < \infty,
\end{equation*}

where $e \in \Gamma$ denotes the identity element in $\Gamma$, and
$d_S$ is a word metric on $\Gamma$ defined above.
Note that the property of having finite second monent does not depend on
the choice of a finite generating set $S$ ($S$ may not be contained in
Taking $u(\cdot)$ to be $d(f(e),\cdot)$ in the following lemma,
we see that if $\Gamma$ is finitely generated and
$\mu$ has finite second moment, then any isometric action of $\Gamma$ on
a metric space has finite second moment.

Let $\Gamma$ be a finitely generated group with a probability measure
$\mu$ with finite second moment. Suppose $\Gamma$ acts on a metric
space $Y$ via a homomorphism $\rho\colon \Gamma\rightarrow \isom{Y}$.
Suppose further that a function $u\colon Y \rightarrow \R$ satisfies
$|u(p)| \leq C_1 d(p_0,p)^2 + C_2$ for some $C_1,C_2>0$ and
$p_0 \in Y$. Then, for any $\rho$-equivariant map
$f\colon \Gamma \rightarrow Y$, the pull-back $f^*u$ of $u$ by $f$
belongs to $L^1(\Gamma,\mu)$.
In particular, $f^*u$ is a $\mu$-integrable function on $\Gamma$.

Fix a finite generating set $S$ of $\Gamma$.
Then, by setting $C=\max\{d_Y(f(e), f(s))\mid s \in S \}$,
for $\gamma = s_1 \dots s_k$ with $s_i \in S$
($i=1, \dots, k$), we have

\begin{equation*}
\begin{split}
& d_Y(f(e),f(s_1 \dots s_k)) \\
\leq & d_Y(f(e), f(s_1)) + d_Y(f(s_1), f(s_1 s_2))
+ \dots +
d_Y(f(s_1 \dots s_{k-1}), f(s_1 \dots s_k)) \\
= & d_Y(f(e), f(s_1)) + d_Y(f(e), f(s_2))
+ \dots + d_Y(f(e), f(s_k))
\leq  Ck,
\end{split}
\end{equation*}

because the action given by $\rho$ is isometric and $f$ is
This implies that $d_Y(f(e),f(\gamma)) \leq C d_S(e,\gamma)$.
Therefore $f^*u$ satisfies

\begin{equation*}
\begin{split}
& |f^*u(\gamma)| = |u(f(\gamma))| \\
\leq & C_1 \left(d(p_0, f(e)) + d(f(e),f(\gamma)) \right)^2 + C_2 \\
= & C_1 \left(d(f(e),f(\gamma))^2 +
2 d(p_0,f(e))d(f(e),f(\gamma)) + d(p_0,f(e))^2\right) + C_2 \\
\leq &  C_1'd_S(e,\gamma)^2 + C_1''d_S(e,\gamma) + C_2''
\end{split}
\end{equation*}

for some $C_1', C_1'', C_2''>0$.
Thus, by applying the Schwarz inequality to the second term in
the last expression and by our assumption of $\mu$ having
finite second moment, we see that

\begin{equation*}
\int_{\Gamma} |f^*u(\gamma)| d\mu(\gamma) < \infty.
\end{equation*}

This completes the proof.

A $\rho$-equivariant map $f$ is called a $\mu$-harmonic map if $f$
minimizes the $\mu$-energy $\ene{\mu}(f)$:
$\ene{\mu}(g)\geq \ene{\mu}(f)$ holds for any $\rho$-equivariant map
$g \colon \Gamma \rightarrow Y$.

Note that a $\rho$-equivariant map is completely determined by the image
of $e$; the set of $\rho$-equivariant map can be identified with $Y$.
We denote by $f_p$ a $\rho$-equivariant map defined by
Recall that a function $u \colon Y \rightarrow \R$ is said to be
convex if, for any geodesic $c\colon [0,l]\rightarrow Y$,
$t \mapsto u(c(t))$ is a convex function on $[0,l]$.
Take any geodesic $c \colon [0,1]\rightarrow Y$ parametrized
proportional to arc length.
Since the distance funtion of a $\cat{0}$ space $Y$ is nonnegative and
convex, its square $t \mapsto d_Y(c(t),\rho(\gamma)c(t))^2$ is also
convex, and hence

\begin{equation*}
\begin{split}
\ene{\mu}(f_{c(t)})
& =\frac{1}{2} \int_{\Gamma} d_Y(c(t),\rho(\gamma)c(t))^2 d\mu(\gamma) \\
& \leq \frac{1}{2}\int_{\Gamma} (1-t)d_Y(c(0),\rho(\gamma)c(0))^2
+ t d_Y(c(1),\rho(\gamma)c(1))^2 d\mu(\gamma) \\
& = (1-t)\ene{\mu}(f_{c(0)}) + t \ene{\mu}(f_{c(1)}),
\end{split}
\end{equation*}

which means that $p\mapsto \ene{\mu}(f_p)$ is a convex function on
$Y$.  Then we have the following.

Let $Y$ be a $\cat{0}$ space and
$\rho\colon \Gamma\rightarrow \isom{Y}$ a homomorphism.
Let $f\colon \Gamma \rightarrow Y$ be a $\rho$-equivariant map.
Then $f$ is $\mu$-harmonic if and only if $f$ satisfies the following
inequality$\,:$ For any $V \in S_{f(e)}Y$,

\begin{equation}
\label{eq:first_variation}
\int_{\Gamma} \inner{V}{\tcprj_{f(e)}(f(\gamma))} d\mu(\gamma)\leq 0.
\end{equation}

We use the following lemmas in the proof of
Proposition <ref>.

Let $Y$ be a $\cat{0}$ space and $p \in Y$. Take two non-trivial
geodesics $c$, $c'$ starting
from $p$, and fix a point $q$ on $c'$.  Then we have

\begin{equation*}
\cos \angle_p (c,c') = \lim_{t \rightarrow 0}
\frac{d_Y(p,q) - d_Y(c(t),q)}{t}.
\end{equation*}

Let $Y$ be a $\cat{0}$ space and $p,q \in Y$.  Let $c(t)$
$($resp. $c'(t)$ $)$ be a non-trivial geodesic starting from $p$
$($resp. $q$ $)$. Then we have

\begin{equation*}
\cos \angle_p(c,\cc{p}{q}) = \lim_{t\rightarrow 0}
\frac{ d_Y(p,c'(t))-d_Y(c(t),c'(t))}{t},
\end{equation*}

where $\cc{p}{q}$ denotes the geodesic starting from $p$ and
terminating at $q$.

Proof of Proposition <ref>.
Although the proof goes along the same line as that of
<cit.>, where the weights
corresoponding to our $\mu$ has finite support, we give a proof here
for the sake of completeness.

Suppose that $f$ is $\mu$-harmonic, that is, $f$ minimizes $E_{\mu}$.
Let $c\colon [0,\varepsilon]\rightarrow Y$ be a geodesic representing
$V \in S_{f(e)}Y$, and $f_t\colon \Gamma \rightarrow Y$ a
$\rho$-equivarinat map defined by
$f_t(\gamma)= \rho(\gamma) c(t)$.
By Lemmas <ref> and <ref> above, we see that

\begin{equation*}
\begin{split}
& \lim_{t\to +0} \frac{1}{t}
\left( d_Y(f_t(e),f_t(\gamma))- d_Y(f(e),f(\gamma)) \right) \\
= & \lim_{t\to +0} \frac{1}{t}\left[\big(d_Y(c(t),\rho(\gamma)c(t))
-d_Y(c(0),\rho(\gamma)c(t))\big) \right. \\
& \phantom{++}
+ \left. \big(d_Y(c(0),\rho(\gamma)c(t))
-d_Y(c(0),\rho(\gamma)c(0)) \big)\right] \\
=&  -\cos \angle_{f(e)}(V, \tcprj_{f(e)}(f(\gamma)))
-\cos \angle_{f(\gamma)}(\tcprj_{f(\gamma)}(f(e)),\rho(\gamma)V)
\end{split}
\end{equation*}

holds. Noting that
\angle_{f(e)}(\tcprj_{f(e)}(f(\gamma^{-1})),V)$,
we get

\begin{equation*}
\begin{split}
& \lim_{t\to +0} \frac{1}{t} \left(
d_Y(f_t(e),f_t(\gamma))^2-d_Y(f(e),f(\gamma))^2 \right) \\
= & -2 d_Y(f(e),f(\gamma)) \left(
\cos \angle_{f(e)}(V, \tcprj_{f(e)}(f(\gamma)))
+ \cos\angle_{f(e)}(\tcprj_{f(e)}(f(\gamma^{-1})),V)
\right) \\
= & -2\left(\inner{V}{\tcprj_{f(e)}(f(\gamma))}
+ \inner{V}{\tcprj_{f(e)}(f(\gamma^{-1})}\right).
\end{split}
\end{equation*}

\begin{equation*}
\begin{split}
|d_Y(f_t(e),f_t(\gamma)) - d_Y(f(e),f(\gamma))|
& \leq  d_Y(f(e),f_t(e)) + d_Y(f(\gamma),f_t(\gamma)) \\
& =  2d_Y(f(e),f_t(e))=2t,
\end{split}
\end{equation*}

and$\gamma\mapsto d_Y(f(e),f(\gamma))$ is $\mu$-integrable by the
Schwarz inequality, we see that

\begin{equation*}
\begin{split}
& \int_{\Gamma}\frac{|d_Y(f_t(e),f_t(\gamma))^2
-d_Y(f(e),f(\gamma))^2|}{t}   d\mu(\gamma) \\
\leq & \int_{\Gamma} \frac{1}{t }2d_Y(f(e),f_t(e))
\left(2d_Y(f(e),f(\gamma)) + d_Y(f(e),f_t(e))
+ d_Y(f(\gamma),f_t(\gamma))\right) d\mu(\gamma) \\
= & 4 \int_{\Gamma} \left(d_Y(f(e),f(\gamma)) + t\right)
d\mu(\gamma) < \infty.
\end{split}
\end{equation*}

Since $f$ is $\mu$-harmonic, by Lebesgue's dominated convergence
theorem and $\mu(\gamma)=\mu(\gamma^{-1})$, we obtain

\begin{equation}
\label{eq:differential}
\begin{split}
0 & \leq \lim_{t\to +0} \frac{\ene{\mu}(f_t)-\ene{\mu}(f)}{t} \\
&= \frac{1}{2}\int_{\Gamma} \lim_{t\to +0}
\frac{d_Y(f_t(e),f_t(\gamma))^2-d_Y(f(e),f(\gamma))^2}{t} d\mu(\gamma) \\
& = \int_{\Gamma} -2\inner{V}{\tcprj_{f(e)}(f(\gamma))} d\mu(\gamma).
\end{split}
\end{equation}

Now suppose $f$ satisfies (<ref>).
Suppose further that there exists a $\rho$-equivariant map $g$ with
Let $d=d_Y(f(e),g(e))$ and take a geodesic
$c\colon [0,d]\rightarrow Y$ joining $f(e)$ to $g(e)$.
Then, since $t \mapsto E_{\mu}(f_{c(t)})$ is a convex function,
together with the above computation, we see that

\begin{equation*}
\begin{split}
\int_{\Gamma}
-2\inner{V_c}{\tcprj_{f(e)}(f(\gamma))} d\mu(\gamma)
& = \lim_{t\to +0}\frac{\ene{\mu}(f_{c(dt)})-\ene{\mu}(f)}{dt} \\
& \leq \lim_{t\to +0} \frac{\left((1-t)\ene{\mu}(f)+t
\ene{\mu}(g)\right)-\ene{\mu}(f)}{dt}\\
& =\frac{\ene{\mu}(g)-\ene{\mu}(f)}{d} < 0,
\end{split}
\end{equation*}

where $V_c \in S_{f(e)}Y$ is the equivalence class of $c$.
A contradiction.
This completes the proof.

If $Y$ is isometric to a Riemannian manifold on a neighborhood of
$f(e)$, then the tangent cone at $f(e)$ is isometric to $\R^n$,
and $f_t$ can be extended uniquely and smoothly to $t<0$ with small
Then one sees that if $f$ is $\mu$-harmonic,

\begin{equation*}
\lim_{t\to 0} \frac{\ene{\mu}(f_t)-\ene{\mu}(f)}{t}=0
\end{equation*}

must holds, and we have an equality in (<ref>).
However we cannot replace the inequality with equality if there is a
singularity in $TC_{f(e)}Y$.

Proposition <ref> below asserts that a
$\mu$-harmonic map $f$ is characterized as a map that locates $f(e)$ at
the barycenter of $f_* \mu$ defined as follows, where $f_*\mu$ is
the push-forward of a measure $\mu$ by $f$.

Let $Y$ be a $\cat{0}$ space and $\nu$ a probability measure on $Y$
with finite second moment, that is,

\begin{equation*}
\int_Y d_Y(p_0, p)^2 d\nu(p) < \infty
\end{equation*}

for some (hence any) $p_0 \in Y$.  The barycenter $\bary{\mu}$ of
$\nu$ is defined to be a unique point that minimizes

\begin{equation*}
q \mapsto \int_Y d_Y(q, p)^2 d\nu(p).
\end{equation*}

The existence of the barycenter is a consequence of the convexity of a
$\cat{0}$ space
(see, for example,  <cit.> for the
Since $t \mapsto d_Y(c(t),p)$ is a convex function (regard $p$ as the
image of a geodesic with zero velocity),  we see that
$q \mapsto \int_Y d_Y(p,q)^2d\nu(p)$ is a convex function as in the
case of $p \mapsto \ene{\mu}(f_p)$.
A slight simplification of the proof of
Proposition <ref> leads us to the following
proposition;  since the support of $\nu$ is fixed, in contrast to
$f_t(\Gamma)$ in Proposition <ref>, we only need
Lemma <ref>.

Let $Y$ be a $\cat{0}$ space and $\nu$ a probability measrure with
finite second moment. Then $q$ is the barycenter of $\nu$
if and only if $q$ satisfies

\begin{equation*}
\int_{Y}\inner{V}{\tcprj_q(p)} d\nu(p) \leq 0
\end{equation*}

for any $V \in S_qY$.

Let $f$ be a $\rho$-equivariant $\mu$-harmonic map.
Then, since $f$ is $\rho$-equivariant, we see that $f(\gamma)$
is also the barycenter of $(f\circ \gamma)_* \mu$.

Let $\varphi \colon \Gamma \rightarrow \R$.
Suppose that
$\int_{\Gamma} \varphi(\gamma \gamma') d\mu(\gamma')<\infty$
holds for any $\gamma \in \Gamma$.
The $\mu$-Laplacian $\Delta=\Delta_{\mu}$ with respect to
$\mu$ is an operator defined by

\begin{equation*}
\Delta \varphi(\gamma)
= \varphi(\gamma) - \int_{\Gamma}
\varphi(\gamma \gamma') d\mu(\gamma').
\end{equation*}

A function $\varphi \colon \Gamma \rightarrow \R$ is called a
$\mu$-subharmonic function if $\Delta \varphi(\gamma)\leq 0$ holds
for any $\gamma \in \Gamma$.
If $\varphi$ satisfies $\Delta \varphi(\gamma)=0$ for any
$\gamma\in\Gamma$, then $\varphi$ is called a $\mu$-harmonic

The following fact plays an important role in the proof of
Theorem <ref>.

Let $Y$ be a $\cat{0}$ space and
$\rho\colon \Gamma\rightarrow \isom{Y}$ a homomorphism.
Let $f\colon \Gamma\rightarrow Y$ be a $\rho$-equivariant
$\mu$-harmonic map, and $u \colon Y \rightarrow \R$ a lower
semicontinuous convex function on $Y$ satisfying
$|u(p)| \leq C_1 d_Y(p_0,p)^2+C_2$ for some $p_0 \in Y$
and $C_1,C_2>0$.
Then the pull-back $f^*u$ of $u$ defined by $f^*u(\gamma)=u(f(\gamma))$
is a $\mu$-subharmonic function on $\Gamma$.

By Jensen's inequality for $\cat{0}$ spaces
(<cit.>), for a lower
semicontinuous convex function $u$
and a probability measure $\nu$ on $Y$, we have

\begin{equation*}
u(\bary{\nu}) \leq \int_Y u(p) d\nu(p).
\end{equation*}

Note that, since $\rho(\gamma)$ is an isometry,
$\bary{\rho(\gamma)_* \nu}=\rho(\gamma)\bary{\nu}$ holds by the
definition of the barycenter.
Therefore, by Propositions <ref> and
<ref>,  we see that a $\mu$-harmonic map $f$ satisfies
$f(\gamma)=\bary{(\rho(\gamma)\circ f)_*\mu}$ for any
$\gamma \in \Gamma$.
Thus, substituting $(\rho(\gamma)\circ f)_*\mu=f_*(\gamma_* \mu)$ to
$\nu$, we get

\begin{equation*}
u(f(\gamma)) \leq
\int_{\Gamma} u(p) d (f_*(\gamma_*\mu))(p)=
\int_{\Gamma} f^*u(\gamma') d\gamma_*\mu(\gamma')
= \int_{\Gamma} f^*u(\gamma\gamma'') d\mu(\gamma'')
\end{equation*}

for any $\gamma\in \Gamma$.
A computation similar to that in the proof of Lemma <ref>,
using the triangle inequality and the Schwarz inequality,
we see that $\gamma_*\mu$ also has finite second moment.
Thus, by our assumption on $u$ and Lemma <ref>,
integrals in the expression above make sense.
Thus we conclude that $f^*u$ is a $\mu$-subharmonic function.

According to Ishihara <cit.>, a smooth map
between Riemannian manifolds is harmonic if and only if the pull-back
of any (local) convex function on the target manifold is a (local)
subharmonic function.
A similar characterization is proved for a harmonic map from
a $1$-dimensional Riemannian polyhedron into a nonpositively curved
metric space <cit.>.
Although our situation is much simpler than that in
[11] as the proof above shows, we do not know if the
converse of Proposition <ref> is also true
or not.

It should be easy to imagine that, for a convex function $u$,
the quantity $|u(\bary{\nu})-\int_Y u(p) d\nu(p)|$ measures the
strength of the convexity of $u$.
Therefore, if we have a $\mu$-harmonic map
$f\colon \Gamma \rightarrow Y$ and a convex function
$u\colon Y\rightarrow \R$ that is $\mu$-harmonic, then
we may expect that $u$ has the “weakest possible convexity”
on the convex hull of $f(\Gamma)$.

In terms of probablity theory,
$\mu$-subharmonic functions correspond to submartingales associated to
a random walk generated by $\mu$.  See [38]
for more information on martingale theory and harmonic maps.

We list some known existence results of $\mu$-harmonic maps.
(See [16], [31],
[23], and [32] for related results.)
A homomorphism $\rho\colon \Gamma\rightarrow \isom{Y}$ is called
reductive in the sense of Jost [23] if there exists
a closed, convex $\rho(\Gamma)$-invariant subset $C$ of $Y$ satsifying
either of the following:

* $C$ is isometric to a finite-dimensional Euclidean space, or

* for any unbounded sequence $\{p_n\}\subset C$, there exists
$\gamma \in \Gamma$ such that $\{d(p_n, \rho(\gamma)p_n)\}$ is

It is well-known that if $\rho(\Gamma)$ is
reductive in the sense of Jost, then there exists a
$\rho$-equivariant $\mu$-harmonic map even when $Y$ is not proper (see
[23], [32], and Corollary <ref>).
Note that if $\rho(\Gamma)$ is not
reductive in the sense of Jost, then there exists an unbounded
sequence $\{p_n\} \subset Y$ such that, for each $\gamma \in \Gamma$,
$d(p_n,\rho(\gamma)p_n)$ is bounded.
If $Y$ is proper, by passing to a subsequence of $\{p_n\}$, we can find
a limit point $\xi \in \partial Y$, and see that $\xi$ is fixed by
$\rho(\Gamma)$; thus if $Y$ is proper and $\rho(\Gamma)$ does not fix a
point in $\partial Y$, then there exists a $\rho$-equivariant
$\mu$-harmonic map.
We will show that this is also true for nonproper $\cat{0}$ spaces with
certain dimension bound
(Proposition <ref>).

§ THE RATE OF ESCAPE AND EQUIVARIANT HARMONIC MAPS

Let $\Gamma$ be a countable group with a symmetric probability
measure $\mu$ whose support generates $\Gamma$.
This measure $\mu$ can be regarded as a transition probability of a move
starting from the identity element $e$ in $\Gamma$, and hence generates
a random walk on $\Gamma$.
We define $(\Omega, \mathbb{P})$ to be
the set of increments of this random walk:

\begin{equation*}
(\Omega, \mathbb{P})=(\Gamma,\mu)^{\N}=
(\Gamma, \mu) \times (\Gamma, \mu) \times \dots.
\end{equation*}

Then, for each $n \in \N$ and a walk $\omega \in \Omega$,
the position of $\omega$, starting from $e$, at time $n$ is given by a

\begin{equation*}
\gamma_n \colon \Omega \rightarrow \Gamma; \
\omega \mapsto \omega_1 \dots \omega_n,
\end{equation*}

where $\omega=(\omega_1, \omega_2, \dots, \omega_n, \dots) \in \Omega$.
Let $\mu^n$ be a probability measure on $\Gamma$ obtained as the
push-forward of $\mathbb{P}$ by a map $\gamma_n$, that is,
$\mu^n ={\gamma_n}_* \mathbb{P}$.
Then we see that $\mu^n$ describes the distribution of the position of a
random walk generated by $\mu$ at time $n$,
and, for $\mu^n$-integrable function
$\varphi\colon \Gamma \rightarrow \R$, we have

\begin{equation*}
\begin{split}
\int_{\Gamma} \varphi(\gamma) d\mu^n(\gamma)
& = \int_{\Omega} \varphi(\gamma_n(\omega)) d\mathbb{P}(\omega) \\
& = \int_{\Gamma}\dots \left(\int_{\Gamma}\left(
\int_{\Gamma} \varphi (\omega_1 \dots \omega_{n-1}\omega_n)
d\mu(\omega_{n-1})\right) \dots d\mu(\omega_1);
\end{split}
\end{equation*}

$\mu^n$ is the $n$-fold convolution of $\mu$.
Suppose that $\Gamma$ acts on a metric space $Y=(Y,d)$ via a
homomorphism $\rho\colon \Gamma \rightarrow \isom{Y}$ and that the
action given by $\rho$ has finite second moment with respect to $\mu$.
Then the action also has finite second moment with respect to
$\mu^n$, according to the following computation and induction:
for $p \in Y$,

\begin{equation*}
\begin{split}
&\int_{\Gamma} d(p,\rho(\gamma)p)^2 d\mu^n(\gamma)  \\
= & \int_{\Gamma}\left(
\int_{\Gamma} d(p,\rho(\gamma'\omega_n)p)^2 d\mu(\omega_n)\right)
d\mu^{n-1}(\gamma') \\
\leq & \int_{\Gamma}\left( \int_{\Gamma}
\left(d(p,\rho(\gamma')p)+
d\mu^{n-1}(\gamma') \\
= & \int_{\Gamma}\left( \int_{\Gamma}
+ d(p,\rho(\omega_n)p)^2
d\mu^{n-1}(\gamma') \\
\leq &  \left(
1+ 2 \left(\int_{\Gamma} 1+d(p,\rho(\omega_n)p)^2\
\int_{\Gamma}1+ d(p,\rho(\gamma')p)^2\ d\mu^{n-1}(\gamma') \\
& \phantom{aaa} + \int_{\Gamma} d(p,\rho(\omega_n)p)^2 d\mu(\omega_n),
\end{split}
\end{equation*}

where we use the invariance of $d$ under $\rho(\gamma')$ to deduce the
second equality, and
$d(p,q)\leq 1+d(p,q)^2$ to deduce the last inequality.

Now we introduce the quantity called the rate of escape and see
what happens when it vanishes.
It is well-known that, for almost all $\omega \in \Omega$,

\begin{equation*}
\lim_n \frac{d(p, \rho(\gamma_n(\omega))p)}{n}
\end{equation*}

exists and the function

\begin{equation*}
\omega \mapsto \lim_n \frac{d(p, \rho(\gamma_n(\omega))p)}{n}
\end{equation*}

converges to a constant function in $L^1(\Omega, \mathbb{P})$
(see, for example, <cit.>).
The value of this constant function is called the rate of escape
(or the drift) of $\rho$ and denoted by
It is easy to check, by using the triangle inequality,
$l_{\rho}(\Gamma)$ does not depend on the choice of $p \in Y$.

Let $Y=(Y,d)$ be a proper $\cat{0}$ space, and $\Gamma$ a
countable group equipped with a symmetric probability measure $\mu$
whose support generates $\Gamma$.
Suppose that $\Gamma$ acts on $Y$ via a homomorphism
$\rho\colon \Gamma \rightarrow \isom{Y}$ and that
there exists a $\rho$-equivariant $\mu$-harmonic map
$f\colon \Gamma \rightarrow Y$.
$($Note that the existence of a $\rho$-equivariant $\mu$-harmonic
map means that the action given by $\rho$ has finite second moment.$)$
If $l_{\rho}(\Gamma)=0$, then, for almost all $\omega \in \Omega$,
there exists
$\xi(\omega) \in Y\cup \partial Y$ such that a function
$\varphi_{\omega} \colon \Gamma \rightarrow \R$ defined by

\begin{equation*}
\varphi_{\omega}(\gamma) =
\begin{cases}
d(\xi(\omega), f(\gamma))-d(\xi(\omega), f(e))
& \text{if } \xi(\omega) \in    Y \\
b_{\xi(\omega)}(f(\gamma), f(e))
& \text{if } \xi(\omega) \in \partial Y
\end{cases}
\end{equation*}

satisfies $\Delta \varphi_{\omega}(e)=0$.

If one assumes that $\mu(\gamma)\not= 0$ for any
$\gamma \in \Gamma\setminus\{e\}$, then one can show that there exists
$\omega \in \Omega$ such that $\varphi_{\omega}$ in the proposition
is $\mu$-harmonic, that is, $\Delta \varphi_{\omega}(\gamma)=0$ for any
$\gamma \in \Gamma$.
This is true even if $Y$ is not proper, although $\xi(\omega)$ may not
be a point in $Y\cup \partial Y$
(Proposition <ref>).
We will prove this in  <ref>.

In order to prove Proposition <ref>,
let us consider a function $L^n(f)$ of a $\rho$-equivariant map
$f \colon \Gamma \rightarrow Y$ defined as

\begin{equation*}
L^n(f) = \int_{\Gamma} d(f(e), f(\gamma)) d\mu^n(\gamma)
= \int_{\Gamma} d(f(e),\rho(\gamma_n(\omega))f(e))
\end{equation*}

Take $\varphi$ to be a function
$\gamma \mapsto d(f(e), f(\gamma))$ and set

\begin{equation*}
\alpha_n = \int_{\Gamma} -\Delta \varphi(\gamma) d\mu^n(\gamma)
= \int_{\Omega} -\Delta \varphi(\gamma_n(\omega)) d\mathbb{P}(\omega).
\end{equation*}

Then there is the following relation between this $\alpha_n$ and
the rate of escape $l_{\rho}(\Gamma)$:

If $\liminf_n \alpha_n >0$, then $l_{\rho}(\Gamma)>0$.

Notice that

\begin{equation*}
\begin{split}
& L^n(f)+ \alpha_n =
L^n(f) - \int_{\Gamma} \Delta \varphi(\gamma)
d\mu^n(\gamma) \\
= & \int_{\Gamma} d(f(e), f(\gamma)) d\mu^n(\gamma)
- \int_{\Gamma} \left(d(f(e),f(\gamma)) -
\int_{\Gamma} d(f(e), f(\gamma \gamma')) d\mu(\gamma') \right)
= & \int_{\Gamma} \left(\int_{\Gamma} d(f(e), f(\gamma \gamma'))
d\mu(\gamma')\right) d\mu^n(\gamma) \\
= & L^{n+1}(f).
\end{split}
\end{equation*}

Now suppose that $\liminf_n \alpha_n = c>0$.
Then there exists $n_0 \in \N$ such that

\begin{equation*}
\inf_{n \geq n_0} \alpha_n \geq c/2.
\end{equation*}

Therefore, by taking $c'=\min\{ L^{n_0}(f)/n_0, c/2 \}$, we see that

\begin{equation*}
L^n (f) \geq L^{n_0}(f) + \frac{c(n-n_0)}{2} \geq c'n
\end{equation*}

for any $n\geq n_0$.

\begin{equation*}
\omega \mapsto \frac{d(f(e), f(\gamma_n(\omega)))}{n}
\end{equation*}

converges to $l_{\rho}(\Gamma)$ in $L^1(\Omega, \mathbb{P})$,
we have

\begin{equation*}
= \lim_{n\to \infty} \int_{\Omega}\frac{d(f(e),f(\gamma_n(\omega)))}{n}
\ d\mathbb{P}(\omega)
= \lim_{n\to \infty} \frac{L^n(f)}{n} \geq c'>0.
\end{equation*}

This completes the proof.

Proof of Proposition <ref>.
Suppose that there exists a
$\rho$-equivariant $\mu$-harmonic map $f \colon \Gamma \rightarrow Y$.
Since $f$ is harmonic and $y \mapsto d(f(e),y)$ is a continuous convex
function, by Proposition <ref>, we see that
$\varphi(\gamma)=d(f(e),f(\gamma))$ becomes a
$\mu$-subharmonic function, that is,

\begin{equation*}
\Delta \varphi(\gamma)
= \varphi(\gamma) - \int_{\Gamma}
\varphi(\gamma \gamma') d\mu(\gamma') \leq 0
\end{equation*}

holds for any $\gamma \in \Gamma$.

Since we are assuming $l_{\rho}(\Gamma)=0$, we see that
$\liminf_n \alpha_n\leq 0$ by Lemma <ref>.
On the other hand, since $\varphi$ is subharmonic, we have
$-\Delta \varphi(\gamma)\geq 0$ for any $\gamma \in \Gamma$,
and hence we conclude that

\begin{equation*}
\liminf_{n} \int_{\Omega} -\Delta \varphi(\gamma_n(\omega))
d\mathbb{P}(\omega) = 0.
\end{equation*}

Therefore, by taking a subsequence $\{n_j\}\subset \N$, we have

\begin{equation*}
\lim_{j} \int_{\Omega} -\Delta \varphi(\gamma_{n_j}(\omega))
d\mathbb{P}(\omega) = 0.
\end{equation*}

Since $-\Delta \varphi(\gamma) \geq 0$ for any $\gamma \in \Gamma$, this means
$\{-\Delta \varphi(\gamma_{n_j}(\cdot))\colon \Omega \rightarrow \R\}_{j\in \N}$
converges to $0$ in $L^1(\Omega,\mathbb{P})$.
Thus, by taking a subsequence if necessary, we may assume that
$-\Delta \varphi(\gamma_{n_j}(\cdot)) \to 0$ almost everywhere on
Take $\omega \in \Omega$ so that
$-\Delta\varphi(\gamma_{n_j}(\omega))\to 0$.

\begin{equation*}
\varphi_{\omega, n_j}(\gamma)
= d(f(\gamma_{n_j}(\omega)^{-1}), f(\gamma)) -
d(f(\gamma_{n_j}(\omega)^{-1}), f(e)),
\end{equation*}

we have

\begin{equation*}
\Delta \varphi_{\omega, n_j} (\gamma)
= d(f(\gamma_{n_j}(\omega)^{-1}), f(\gamma))
- \int_{\Gamma}
d(f(\gamma_{n_j}(\omega)^{-1}), f(\gamma \gamma')) d\mu(\gamma').
\end{equation*}

In particular, we see that

\begin{equation*}
\begin{split}
\Delta \varphi_{\omega, n_j} (e)
& = d(f(\gamma_{n_j}(\omega)^{{-1}}), f(e))
- \int_{\Gamma}
d(f(\gamma_{n_j}(\omega)^{-1}), f(\gamma')) d\mu(\gamma') \\
& = d(f(e), f(\gamma_{n_j}(\omega))) -
\int_{\Gamma}d(f(e), f(\gamma_{n_j}(\omega))\gamma')d \mu(\gamma') \\
& = \Delta \varphi(\gamma_{n_j}(\omega)).
\end{split}
\end{equation*}

Passing to a subsequence if necessary, since $Y$ is proper, we may
assume that the sequence $\{\gamma_{n_j}(\omega)^{-1}\}$ converges to a
point $\xi(\omega) \in Y\cup \partial Y$.
If  $\xi(\omega) \in Y$, we have

\begin{equation*}
\varphi_{\omega}(\gamma) := \lim_j \varphi_{\omega, n_j}(\gamma)
= d(\xi(\omega), f(\gamma)) - d(\xi(\omega), f(e)),
\end{equation*}

while in the case $\xi(\omega) \in \partial Y$ we get

\begin{equation*}
\varphi_{\omega}(\gamma):= \lim_j \varphi_{\omega, n_j}(\gamma)
= b_{\xi(\omega)}(f(\gamma), f(e)),
\end{equation*}

where $b_{\xi(\omega)}(\cdot, \cdot)$ denotes the
Busemann function associated to $\xi(\omega)$
(see Remark <ref>).
Note that

\begin{equation*}
|\varphi_{\omega,n_j}(\gamma)| =
|d(f(\gamma_{n_j}(\omega)^{-1}),f(\gamma)) -
d(f(\gamma_{n_j}(\omega)^{-1}),f(e))| \leq d(f(e),f(\gamma))
\end{equation*}

and that
the action given by $\rho$ has finite second moment with respect
to $\mu$.
Then we see that, by Lebesgue's dominated convergence theorem,

\begin{equation*}
\lim_j \int_{\Gamma} \varphi_{\omega,n_j}(\gamma) d\mu(\gamma)
= \int_{\Gamma} \varphi_{\omega}(\gamma) d\mu(\gamma),
\end{equation*}

and hence we get $\Delta \varphi_{\omega} (e)=0$.
This completes the proof of Proposition <ref>.

§ SOME CONSEQUENCES OF THE CONVEXITY OF DISTANCE FUNCTIONS

In this section, we show two propositions which reflect the
convexity of distance functions on $\cat{0}$ spaces.
These propositions, together with Proposition <ref>
above,  play a key role in the proof of Theorem <ref>.

Let $Y$ be a $\cat{0}$ space,  and $p, q \in Y$.
Let $c\colon [0,\infty)\longrightarrow Y$ be a geodesic ray starting
from $p$.
Take $r_0>0$ and let $p_0=c(r_0)$.
Denote by $V_c$ the element in $S_pY$ corresponding to the geodesic
segment $t \mapsto c(t)$.
Then we have

\begin{equation*}
\inner{V_c}{\tcprj_p(q)}  \geq d(p_0,p)-d(p_0,q).
\end{equation*}

Suppose that $\angle (\tcprj_p(p_0),\tcprj_p(q))=0$.
Then $\tcprj_p(q)$ lies in a geodesic ray $\{tV_c\mid t\geq 0\}$ in
$TC_pY$ and

\begin{equation*}
\inner{V_c}{\tcprj_p(q)} = d_{T_p}(O_p,\tcprj_p(q))
= d(p,q) \geq d(p_0,p)-d(p_0,q)
\end{equation*}

by the triangle inequality.

Suppose $\angle (\tcprj_p(p_0),\tcprj_p(q))\not=0$ and
$\angle (\tcprj_p(p_0),\tcprj_p(q))< \pi$.
Set  $W_q = \tcprj_p(q)/\Vert \tcprj_p(q) \Vert \in S_pY$.
Since $0< \angle (V_c, W_q) =\angle (\tcprj_p(p_0),\tcprj_p(q)) < \pi$,
there is a non-trivial geodesic segment $\bar c$ in $S_pY$ starting
from $V_c$ and terminating at $W_q$.
By attaching a segment of length $\pi-\angle (V_c, W_q)$ to the
endpoint $W_q$, we obtain a $\cat{1}$ space $S_p$, containing
$S_pY$ as a convex subset in which $\bar c$ can be extended up to
length $\pi$
Denote by $-V_c \in S_p$ the endpoint of this extended
geodesic $\bar c$.  Then we have

\begin{equation*}
\angle(V_c, W_q) + \angle(W_q,-V_c) = \pi,
\end{equation*}

in particular,

\begin{equation}
\label{eq:minus_vector}
-\inner{-V_c}{\tcprj_p(q)} %= -\cos \left(\pi-\angle(W_q,V_c)\right)
% = \cos \left(\angle(V_c, W_q)\right)
= \inner{V_c}{\tcprj_p(q)}.
\end{equation}

Let $T_p$ be the Euclidean cone over $S_p$.
Then $T_p$ contains $TC_pY$ as a convex subset.
Note that $\{tV_c\mid t\geq 0\} \cup \{t(-V_c) \mid t\geq 0\}$
forms a geodesic line in $T_p$ since $\angle(V_c,-V_c)=\pi$.
Let $\tilde c$ be a geodesic defined by

\begin{equation*}
\tilde c (t)=
\begin{cases}
tV_c & t\geq 0, \\
\vert t \vert (-V_c) & t \leq 0.
\end{cases}
\end{equation*}

Then $\tilde c$ satisfies
$\tilde c(\R) = \{tV_c\mid t\geq 0\} \cup \{t(-V_c) \mid t\geq 0\}$,
$\tilde c(0)=O_p$ and $\inner{V_c}{\tilde c(t)} = t$ for
$t \in \R$.
Set $d_o=d_{T_p}(\tcprj_p(p_0),O_p)$ and
$d_q=d_{T_p}(\tcprj_p(p_0),\tcprj_p(q))$, and let
$q'=\tilde c(d_o-d_q)$.
Then $q'$ is a point on $\tilde c(\R)$ satisfying

\begin{equation*}
\inner{V_c}{q'} = d_o-d_q \geq d(p_0,p)-d(p_0,q),
\end{equation*}

where the last inequality holds because
$d_q \leq d(p_0,q)$ while $d_o=d(p_0,p)$.
Let $\bar q$ be the image of $\tcprj_p(q)$ by the nearest point
projection onto the image $\tilde c(\R)$ of $\tilde c$  in
$T_p$. Since the nearest point projection does not increase the
$d_{T_p}(\tcprj_p(p_0),\bar q) \leq d_q$,
and hence

\begin{equation*}
\bar q \in \tilde
c\left(\left[d_o-d_q,\  d_o+d_q\right]\right).
\end{equation*}

\begin{equation*}
\inner{V_c}{\bar q} \geq d_o-d_q
= \inner{V_c}{q'}\geq d(p_0,p)-d(p_0,q).
\end{equation*}

By the definition of the distance of a Euclidean cone, we have

\begin{equation}
\label{eq:distance_from_tilde_c}
\begin{split}
& d_{T_p}(\tilde c(t), \tcprj_p(q))^2
\\ & =
\begin{cases}
t^2 + |\tcprj_p(q)|^2 -2t |\tcprj_p(q)| \cos \angle(V_c,\tcprj_p(q))
& t \geq 0, \\
t^2 + |\tcprj_p(q)|^2 -2|t| |\tcprj_p(q)| \cos \angle(-V_c,\tcprj_p(q))
& t \leq 0.
\end{cases}
\end{split}
\end{equation}

Take $t_0$ to be $t_0V_c=\tilde c(t_0)=\bar q$, and note that the
left-hand side of (<ref>) is minimized at
If $t_0\not= 0$, then the differentiation of
$d_{T_p}(\tilde c(t), \tcprj_p(q))^2$ at $t=t_0$ exists and it
must be equal to $0$, that is,

\begin{equation*}
\begin{cases}
t_0 - |\tcprj_p(q)| \cos \angle(V_c,\tcprj_p(q)) = 0 & t_0 > 0, \\
t_0 + |\tcprj_p(q)| \cos \angle(-V_c, \tcprj_p(q)) = 0 & t_0 < 0. \\
\end{cases}\end{equation*}

\begin{equation*}
\begin{cases}
\inner{V_c}{\tcprj_p(q)} =
|\tcprj_p(q)|\cos \angle (V_c,\tcprj_p(q)) =t_0 = \inner{V_c}{\bar q}.
& t_0 >0, \\
\inner{-V_c}{\tcprj_p(q)} =
|\tcprj_p(q)|\cos \angle (-V_c,\tcprj_p(q)) =-t_0 =
-\inner{V_c}{\bar q}. & t_0 <0. \\
\end{cases}
\end{equation*}

Since $q'$ lies in $\tilde c(\R)$, we have
$\inner{-V_c}{q'} = -\inner{V_c}{q'}$.
Together with (<ref>), we get

\begin{equation*}
\begin{cases}
\inner{V_c}{\tcprj_p(q)} = \inner{V_c}{\bar q}
\geq d(p_0,p)-d(p_0,q) & t_0 >0 \\
\inner{V_c}{\tcprj_p(q)} = -\inner{-V_c}{\tcprj_p(q)}
= \inner{V_c}{\bar q}
\geq d(p_0,p)-d(p_0,q)
& t_0 <0.
\end{cases}
\end{equation*}

Thus we obtain the desired inequality in the case $t_0 \not= 0$.

If  $t_0=0$, then, from (<ref>),
the right (resp. the left) differentiation of
$d_{T_p}(\tilde c(t), \tcprj_p(q))^2$ at $t=0$ exists and it is
nonnegative (resp. nonpositive).  Noting (<ref>) again,
we have

\begin{equation*}
\begin{cases}
- |\tcprj_p(q)| \cos \angle(V_c,\tcprj_p(q))
= -\inner{V_c}{\tcprj_p(q)} \geq 0  \\
- |\tcprj_p(q)| \cos \angle(-V_c, \tcprj_p(q))
= -\inner{-V_c}{\tcprj_p(q)}
= \inner{V_c}{\tcprj_p(q)} \geq 0,  \\
\end{cases}\end{equation*}

which implies $\inner{V_c}{\tcprj_p(q)} = 0$.
On the other hand, since the nearest point projection does not increase
the distance, $t_0=0$ means that $d_q \geq d_o$;

\begin{equation*}
\inner{V_c}{\tcprj_p(q)}
\geq d_o - d_q
\geq d(p_0,p)-d(p_0,q)
\end{equation*}

Now suppose $\angle(V_c, \tcprj_p(q)) = \pi$.  In this case,
$V_c$ and $\tcprj_p(q)$ can be joined by a geodesic segment in $T_p$
passing through $O_p$, hence

\begin{equation*}
\begin{split}
\inner{V_c}{\tcprj_p(q)}
& = d_{T_p}(O_p, \tcprj_p(q)) \cos \angle (V_c, \tcprj_p(q)) \\
& = -d_{T_p}(O_p, \tcprj_p(q))\\
& = d_o - d_q.
\end{split}
\end{equation*}

Since $d_o=d(p_0,p)$ and $d_q \leq d(p_0,q)$, we get

\begin{equation*}
\inner{V_c}{\tcprj_p(q)} \geq d(p_0,p) -d(p_0,q).
\end{equation*}

This completes the proof.

Note that, according to the proof above, we have

\begin{equation*}
\inner{V_c}{\tcprj_p(q)} = \inner{V_c}{\bar q}
\geq \inner{V_c}{q'}
\end{equation*}

in any case.
This can be regarded as an expression of
the convexity of a metric ball centered at $\tcprj_p(p_0)$ (or $p_0$).
And if the equality in the inequality above holds,
then it should be understood that the points in the statement are
arranged in a convex set with the “weakest” convexity, which is
an interval.
See the next remark.

In the proof above, when $\angle (V_c, \tcprj_p(q))$ is equal to either
$0$ or $\pi$, the geodesic
segment joining $V_c$ and $\tcprj_p(q)$ extends or passes through the
origin $O_p$, and hence $\tcprj_p(q)=\bar q$.
If $\angle (V_c, \tcprj_p(q))\not=0, \pi$, then the equality in the
proposition holds if and only if $\bar q=q'$.
In this case, since
$d(\tcprj_p(p_0), \tcprj_p(q))=d(\tcprj_p(p_0),q')=d(\tcprj_p(p_0),\bar q)$,
the nearest point projection from $[\tcprj_p(p_0),\tcprj_p(q)]$ to
$\tilde c(\R)$ turns out to be an isometry.
By <cit.>,
the distance between $\tilde c(\R)$ and each point of
$[\tcprj_p(p_0),\tcprj_p(q)]$ must be constant.
Both sets share the same point $\tcprj_p(p_0)$, the distance between these
set must be $0$.  Therefore $\tcprj_p(q)=\bar q$.

Letting $p_0\to \xi=c(\infty)$ in
Proposition <ref>, we obtain the
following proposition, which can be viewed as a sort of
quantitative expression of the convexity of the horoball.

Let $Y$ be a $\cat{0}$ space, and $p, q \in Y$.
Let $c\colon [0,\infty)\longrightarrow Y$ be a geodesic ray starting
from $p$ and terminating at $\xi \in \partial Y$.
Denoting the element corresponding to $t \mapsto c(t)$ in $S_pY$ by
$V_c$, we have

\begin{equation*}
\inner{V_c}{\tcprj_p(q)}
\geq b_{\xi}(p,q).
\end{equation*}

If the equality holds in the inequality above, then the convex hull of
$[p, \xi)\cup [q, \xi)\cup [p, q]$ is
isometric to a half-infinite parallelogram in the Euclidean plane.

Let $p_0 = c(t_0)$ and recall
Proposition <ref>. Then we have

\begin{equation*}
\inner{V_c}{\tcprj_p(q)} \geq d(c(t_0),p)-d(c(t_0),q)
\end{equation*}

for any $t_0 \in \R$. Letting $t_0 \to \infty$ yields the desired

Take $S_p$ and $T_p$ as in the proof of
Proposition <ref>.
Note that
$O_p$ and $q'$ in $\tilde c(\R)$ are located on the same side of
$\tilde c(\R)$ with respect to $\tcprj_p(p_0) = \tilde c(t_0)$ for
sufficiently large $t_0$.  Recalling the definition of $q'$ and
Remark <ref>, although $q'$ may depend on $t_0$, we have

\begin{equation*}
\begin{split}
b_{\tilde c(\infty)}(O_p,\tcprj_p(q))
& = \lim_{t_0\to \infty}d_{T_p}(\tilde c(t_0),O_p)
- d_{T_p}(\tilde c(t_0), \tcprj_p(q)) \\
& = \lim_{t_0\to \infty}d_{T_p}(\tilde c(t_0),O_p)
- d_{T_p}(\tilde c(t_0), q') \\
& = \lim_{t_0\to \infty}\inner{V_c}{q'} \\
& \leq \inner{V_c}{\tcprj_p(q)}.
\end{split}
\end{equation*}

On the other hand, since $\tcprj_p$ does not increase the distance while
$d(c(t_0),p)=d_{TC_pY}(\tilde c(t_0),O_p)$ and
$\tcprj_p(c(t_0))=\tilde c(t_0)$ for $t_0 \geq 0$, we have

\begin{equation*}
\begin{split}
b_{c(\infty)}(p,q) & = \lim_{t_0 \to \infty}
d(c(t_0),p) -d(c(t_0),q) \\
& \leq \lim_{t_0 \to \infty}
d_{TC_pY}(\tilde c(t_0), O_p) -
d_{TC_pY}(\tilde c(t_0), \tcprj_p(q)) \\
& = b_{\tilde c(\infty)}(O_p, \tcprj_p(q)).
\end{split}
\end{equation*}

Therefore the equality
$\inner{V_c}{\tcprj_p(q)} =b_{c(\infty)}(p,q)$ means that

\begin{equation*}
b_{\tilde c(\infty)}(O_p,\tcprj_p(q))=b_{c(\infty)}(p,q)
\end{equation*}

which implies that

\begin{equation*}
\lim_{r_0 \to \infty}
\left(d(c(t_0),q) -d_{TC_pY}(\tilde c(t_0), \tcprj_p(q)) \right)
= b_{\tilde c(\infty)}(O_p,\tcprj_p(q))-b_{c(\infty)}(p,q)
= 0.
\end{equation*}

Then, by Lemma <ref> below, we see that
$d(c(t_0),q)=d_{TC_pY}(\tilde c(t_0), \tcprj_p(q))$ for any
$t_0 \in [0,\infty)$.  Note that, by the definition of the distance on
$TC_pY$,  any triangle in $TC_pY$ having a vertex $O_p$ is isometric to
a triangle in $\R^2$. Thus
$\tri{\tilde c(t_0)}{O_p}{\tcprj_p(q)}$ can be regarded as the
comparison triangle for $\tri{c(t_0)}{p}{q}$ $t_0 \in [0,\infty)$.
Since $\angle c(t_0)pq = \angle c(t_0) O_p \tcprj_p(q)$ by definition,
<cit.> tells us that
$\tri{c(t_0)}{p}{q}$ is a flat triangle for any $t_0 \in [0,\infty)$.
Therefore the convex hull of $[p, \xi)\cup [q, \xi) \cup [p, q]$ is
isometric to a region obtained as the limit of a increasing sequence of
flat triangles isometric to $\tri{c(t_0)}{p}{q}$, $t_0 \in [0,\infty)$,
in $\R^2$, which gives us a half-infinite parallelogram.
This completes the proof.

Let $c\colon [0,\infty) \rightarrow Y$ be a geodesic ray and $q \in Y$.
Set $\tilde c (t):=\tcprj_{c(0)}(c(t))$. If

\begin{equation*}
\lim_{t\to \infty}
d(c(t),q)-d_{TC_{c(0)}Y}(\tilde c(t),\tcprj_{c(0)}(q))=0
\end{equation*}

holds, then, for any $t \in [0,\infty)$,

\begin{equation*}
d(c(t),q)=d_{TC_{c(0)}Y}(\tilde c(t),\tcprj_{c(0)}(q))
\end{equation*}

Let $\tri{\overline{c(0)}}{\overline{c(t)}}{\overline{q}}$ be a
comparison triangle of $\tri{c(0)}{c(t)}{q}$, and denote the
angle $\angle_{\overline{c(0)}} (\overline{c(t)},\overline{q})$
by $\theta(t)$.  Let
$\tilde \theta = \angle_{O_{c(0)}}(\tilde c(t),\tcprj_{c(0)}(q))$.
Note that $\tilde \theta$ is a constant independent of $t$, and
$\theta(t)\to \tilde \theta$ as $t \to 0$ by the definition of the
angle, and that
$\theta(t)$ is nondecreasing as $t$ getting larger
by the definition of $\cat{0}$ space
(or see, for example, <cit.>).

\begin{equation*}
d(t) = d(c(t),q), \quad
\tilde d(t) =d_{TC_{c(0)}Y}(\tilde c(t),\tcprj_{c(0)}(q)), \quad
P= d(c(0),q),
\end{equation*}

then by the cosine rule,
we see that

\begin{equation*}
\cos \theta(t) = \frac{t^2 + P^2 - d(t)^2}{2tP}, \quad
\cos \tilde \theta = \frac{t^2 + P^2 - \tilde d(t)^2}{2tP}.
\end{equation*}

Since  $\theta(t)$ is nondecreasing, if $\theta(t_0) > \tilde \theta$
for some $t_0$, then by putting $C=\theta(t_0)-\tilde \theta$ we have
$\theta(t)-\tilde \theta\geq C$ for any $t\geq t_0$.
Therefore, there exists $C'>0$ such that

\begin{equation*}
\cos \tilde \theta - \cos \theta(t)
= \frac{d(t)^2-\tilde d(t)^2}{2tP} \geq C'
\end{equation*}

for any $t \geq t_0$.

\begin{equation*}
d(t)=\sqrt{t^2 + P^2 -2tP\cos \theta(t)}, \quad
\tilde d(t)= \sqrt{t^2+P^2 - 2tP\cos \tilde \theta},
\end{equation*}

we get

\begin{equation*}
\frac{d(t)^2-\tilde d(t)^2}{2tP}
= \frac{d(t)-\tilde d(t)}{2P}
\left(\sqrt{1 + \frac{P^2}{t^2} -2\frac{P\cos \theta(t)}{t}}
+ \sqrt{1 + \frac{P^2}{t^2} -2\frac{P\cos \tilde \theta}{t}}
\right).
\end{equation*}

On the other hand, for $t\geq t_0$, we have

\begin{equation*}
\sqrt{1 + \frac{P^2}{t^2} -2\frac{P\cos \theta(t)}{t}},
\sqrt{1 + \frac{P^2}{t^2} -2\frac{P\cos \tilde \theta}{t}}
\leq \sqrt{\left(1+\frac{P}{t} \right)^2}
= 1 + \frac{P}{t} \leq 1 + \frac{P}{t_0},
\end{equation*}

\begin{equation*}
C' \leq \frac{d(t)^2-\tilde d(t)^2}{2tP}
\leq
2\left(1 + \frac{P}{t_0}\right)
\frac{d(t)-\tilde d(t)}{2P}.
\end{equation*}

Thus, for $t\geq t_0$,

\begin{equation*}
\frac{t_0C'P}{(t_0 + P)} \leq
d(t)-\tilde d(t) .
\end{equation*}

Thus $\lim_{t\to \infty} d(t)-\tilde d(t)=0$ means that $C'=0$, that
is, $\theta(t)=\tilde \theta$ for any $t \geq 0$.
Therefore we get $d(t)=\tilde d(t)$ for any $t \geq 0$.

§ PROOF OF THEOREM <REF>: WHEN $Y$ IS PROPER

In this section, we give the proof of Theorem <ref> in the
case that a $\cat{0}$ space $Y$ is proper.
Suppose that a countable group $\Gamma$ with a random walk
$\mu$ acts on $Y$ via $\rho \colon \Gamma \rightarrow \isom{Y}$
with finite second moment,
and we are assuming that $\rho(\Gamma)$ does not fix a point in
$\partial Y$.
Then as we mentioned in  <ref>, there exists
$\rho$-equivariant $\mu$-harmonic map $f$.
If $l_{\rho}(\Gamma)=0$, then by Proposition <ref>,
for almost all $\omega \in \Omega$, there exists
$\xi(\omega) \in Y \cup \partial Y$ such that

\begin{equation*}
\varphi_{\omega}(\gamma) =
\begin{cases}
d(\xi(\omega), f(\gamma))-d(\xi(\omega), f(e)) & \xi(\omega) \in
Y \\
b_{\xi(\omega)}(f(\gamma), f(e)) & \xi(\omega) \in \partial Y
\end{cases}
\end{equation*}

satisfies $\Delta \varphi_{\omega}(e)=0$.

First we note that we may assume, by replacing $\mu$ with a convex
combination of $\mu^n$, that the support $\supp\mu$ of
a probability measure $\mu$ contains $\Gamma \setminus \{e\}$.
Indeed, since we are assuming that the support of our original $\mu$
generates $\Gamma$, we see that, for any $\gamma$, there exists $n$ such
that $\mu^n(\gamma)\not=0$.
Thus we may assume that $\sum_{n=1}^{\infty }a_n \mu^n(\gamma)\not=0$
for any $\gamma \in \Gamma \setminus \{e\}$, where $a_n\geq 0$ and
It is easy to see that, by taking $a_n$ suitably,
the action given by $\rho$ has finite second moment with respect to
$\sum_{n=1}^{\infty}a_n \mu^n$, since it is true for each
$\mu^n$ as we have seen in  <ref>.
Furthermore, according to <cit.>,
\left(\sum_{n=1}^{\infty}n a_n \right) l_{\rho}(\Gamma,\mu)$
holds, and hence we may assume
$l_{\rho}(\Gamma,\sum_{i=n}^{\infty}a_n\mu^n)=0$, which is our
assumption on $\mu$ in Theorems <ref> and <ref>.
(Although the outline of the proof of <cit.> is
suggested there, for the sake of completeness, we will give a proof of a
weaker result in  <ref>, which is sufficient for our
Note also that, since we are assuming that $\rho(\Gamma)$
fixes no point in $\partial Y$ and $Y$ is proper,
the existence of $\mu$-harmonic map is not
affected by this change of a probability measure $\mu$.

The following lemma summarizes the core argument used in the
proof of Theorem <ref> and Theorem <ref>.

Let $Y$ be a $\cat{0}$ space $($we do not assume $Y$ to be any one of  proper,
of finite telescopic dimension, or of locally finite dimension here$)$,
and $\Gamma$ a countable group with
symmetric and nondegenerate probability measure $\mu$.
Further assume that $\supp \mu \supset \Gamma \setminus \{e\}$.
Suppose that $\Gamma$ acts on $Y$ via a homomorphism
$\rho\colon \Gamma \rightarrow \isom{Y}$ and that there exists a
$\rho$-equivariant $\mu$-harmonic map $f\colon \Gamma \rightarrow Y$.

$(1)$ Let $\xi \in Y$ and
$\varphi\colon \gamma \mapsto d(\xi,f(\gamma))-d(\xi,f(e))$.
If $\Delta \varphi(e)=0$ holds, then there exists a point in $Y$ fixed
by $\rho(\Gamma)$.

$(2)$ Let $\xi \in \partial Y$ and
$\varphi\colon \gamma \mapsto b_{\xi}(f(\gamma),f(e))$.
If $\Delta \varphi(e)=0$ holds, then, for any
$\gamma \in \Gamma\setminus \{e\}$, the convex hull of
$\ray{f(e),\xi}\cup [f(e),f(\gamma)]\cup \ray{f(\gamma),\xi}$
is isometric to a half-infinite parallelogram in $\R^2$.

$(3)$ Suppose that the assumption of $(2)$ is satisfied and that there
exists a sequence $\{\gamma_j\}\subset \Gamma$ such that
$\{f(\gamma_j)\}$ converges to $\xi$ in $(2)$ in the cone topology.
If further $\{\rho(\gamma_j^{-1})\xi\}$ and $\{f(\gamma_j^{-1})\}$
converge to $\xi^+, \xi^{-}\in \partial Y$ in the cone topology
then, for each $\gamma \in \Gamma$, there exists a geodesic
$c_{\gamma}\colon \R \rightarrow Y$ such that $c_{\gamma}(0)=f(\gamma)$
and $c_{\gamma}(\pm \infty)=\xi^{\pm}$.

$(4)$ Suppose that
Under the assumptions of $(2)$ and $(3)$,
the convex hull $\ch{f(\Gamma)}$ splits as $H\times Z$, where
$H$ is a complete Hilbert space and $Z$ does not admit any splitting
$\R \times Z'$.
Furthermore, $\rho$ splits along this splitting
as $\rho=\rho_1 \times \rho_2$, where
$\rho_1 \colon \Gamma \rightarrow \isom{H}$ and
$\rho_2 \colon \Gamma \rightarrow \isom{Z}$.
There also exists $\rho_2$-equivariant $\mu$-harmonic map
$f_2\colon \Gamma \rightarrow Z$, and $l_{\rho_2}(\Gamma)=0$

$(1)$ We show that $f$ is a constant map under the assumption.
Then we see that $f(\Gamma)=\rho(\Gamma)f(e)$ is a point, and this point
is fixed by $\rho(\Gamma)$.

If $\\xi=f(e)$, then
and $\Delta \varphi(e)=0$ means that

\begin{equation*}
\Delta \varphi(e)=
\varphi(e)- \int_{\Gamma}\varphi(\gamma)d\mu(\gamma)
= - \int d(f(\gamma), f(e)) d\mu(\gamma)= 0,
\end{equation*}

namely $f(\Gamma)=f(e)$; $f$ is a constant map.

Now assume $\xi\not= f(e)$.
Take any $\gamma \in \Gamma\setminus \{e\}$, and note that
$\mu(\gamma)\not=0$ by our assumption.
Let $p=f(e)$, $q=f(\gamma)$ and $p_0=\xi$ in
Proposition <ref>,
and denote by $V_c \in S_{f(e)}Y$ the element corresponding to the
geodesic $\cc{f(e)}{\xi}$ joining $f(e)$ and $\xi$.
Then we get

\begin{equation*}
\inner{V_c}{\tcprj_{f(e)}(f(\gamma))}
\geq d(\xi, f(e)) - d(\xi, f(\gamma))= -\varphi(\gamma).
\end{equation*}

By Proposition <ref> and
$\varphi(e)=0$, we see that

\begin{equation*}
\begin{split}
0 & \geq \int_{\Gamma} \inner{V_c}{\tcprj_{f(e)}(f(\gamma))}
d\mu (\gamma)
\geq \int_{\Gamma} -\varphi(\gamma) d\mu (\gamma) \\
& = \left(
\varphi(e) - \int_{\Gamma} \varphi(\gamma) d\mu(\gamma)
\right) \\
& = \Delta (\varphi(e))=0.
\end{split}
\end{equation*}

Therefore we get

\begin{equation*}
\int_{\Gamma} \inner{V_c}{\tcprj_{f(e)}(f(\gamma))}
\end{equation*}

Moreover the equality in
Propostion <ref> must hold
for each $\gamma \in \Gamma\setminus \{e\}$:

\begin{equation}
\label{eq:on_the_same_side}
\inner{V_c}{\tcprj_{f(e)}(f(\gamma))} = d(\xi,f(e))-d(\xi,f(\gamma)).
\end{equation}

By Remark <ref> this implies that
$\tcprj_{f(e)}(f(\gamma)) \in \tilde c(\R)$ for every
$\gamma \in \Gamma\setminus \{e\}$, where $\tilde c$ is a geodesic in
$T_{f(e)}$ passing through $O_{f(e)}$ and $\tcprj_{f(e)}(\xi)$.
Hence we have

\begin{equation*}
\begin{split}
d(\xi,f(\gamma)) &=
d(\xi,f(e))-\inner{V_c}{\tcprj_{f(e)}(f(\gamma))} \\
& = d_{T_{f(e)}}(\tcprj_{f(e)}(\xi), \tcprj_{f(e)}(f(\gamma))).
\end{split}
\end{equation*}

which becomes a comparison triangle of $\tri{f(e)}{f(\gamma)}{\xi}$,
degenerates to a segment and contained in $\tilde c(\R)$.
This means that $f(\gamma)$ is contained in a geodesic segment passing through
$f(e)$ and $\xi$ for every $\gamma \in \Gamma\setminus \{e\}$.
Note that, although such a geodesic on $Y$ may not be unique, all such
geodesics share a geodesic segment $[\xi, f(e)]$ in common.

\begin{equation*}
a(\gamma):=  \inner{V_c}{\tcprj_{f(e)}(f(\gamma))}
= d(\xi,f(e))-d(\xi,f(\gamma)),
\end{equation*}

and $b:=d(\xi,f(e))$.
Then we have

\begin{equation*}
\begin{cases}
d(f(e),f(\gamma))  &
\inner{V_c}{\tcprj_{f(e)}(f(\gamma))} \geq 0 \\
-d(f(e),f(\gamma))  &
\inner{V_c}{\tcprj_{f(e)}(f(\gamma))} \leq 0,
\end{cases}
\end{equation*}

and, by definition, we get $b \geq a(\gamma)$
for all $\gamma \in \Gamma \setminus \{e\}$.
Suppose that, for $\gamma \in \Gamma$, $f(\gamma)$ lies on a
geodesic segment starting from $f(e)$ and passing through $\xi$.
Then we have $\inner{V_c}{\tcprj_{f(e)}(f(\gamma))}\geq 0$,
and $b\geq a(\gamma)$ means that
$d(\xi,f(e))\geq d(f(\gamma),f(e))$, in other words,
$f(\gamma) \in [\xi,f(e)]$.
Therefore, for any $\gamma \in \Gamma$,
$f(\gamma)$ must lie on a geodesic segment starting from $\xi$
and passing through $f(e)$; it cannot lie somewhere beyond $\xi$.

Let $\xi'$ be the point in the closure of $f(\Gamma)$ closest to $\xi$.
Thus, for any $\varepsilon >0$, there exists a $\gamma_j \in \Gamma$
with $d(f(\gamma_j), \xi')< \varepsilon$.
Suppose that there is $\gamma_0 \in \Gamma$ such
that  $f(\gamma_0)=\xi'$.
If $\gamma_0=e$, then we have $d(f(\gamma),\xi)\geq d(f(e),\xi)$ for
any $\gamma \in \Gamma$.
Hence $\Delta \varphi(e)=0$ implies that
$f(\Gamma)$ must become the point $f(e)$, that is, $f$ is a constant
Suppose that $\gamma_0\not= e$.
For every $\gamma \in \Gamma$,
$f(\gamma)$ lies in a geodesic segment starting from
$\xi'=f(\gamma_0)$ and passing through $f(e)\not= \xi'$.
Then there is $V_0 \in S_{f(\gamma_0)}Y$ representing these geodesics,
and we get

\begin{equation*}
\begin{split}
\int_{\Gamma} \inner{V_0}{\tcprj_{f(\gamma_0)} (f(\gamma))}
\ d{\gamma_0}_*\mu(\gamma)
= & \int_{\Gamma} d(f(\gamma_0), f(\gamma_0(\gamma_0^{-1}\gamma))) \
d\mu(\gamma_0^{-1}\gamma) \\
\geq & d(f(\gamma_0),f(e)) \mu(\gamma_0^{-1}) >0.
\end{split}
\end{equation*}

Recalling Remark <ref>, we see that this
contradicts the harmonicity of $f$.
Therefore $f(\gamma)\not=\xi'$ for any $\gamma \in \Gamma$.
Note that, for sufficiently small $\varepsilon$, $f(\gamma_j)$ must lie
in the interior of a geodesic segment $[\xi', f(e)]$.
where $V_{c,f(\gamma_j)}$ denotes the element in $S_{f(\gamma_j)}Y$
corresponding to the geodesic $\cc{f(\gamma_j)}{\xi'}$.

Note that, since $f(\Gamma)$ lies in a union of geodesic
having $[\xi',f(e)]$ in common, we may regard that $f(\Gamma)$
lies in a $\cat{0}$ space locally isometric to $\R$ around
$f(\gamma_j) \in (\xi',f(e))$, and hence
$\tcprj_{f(\gamma_j)}(f(\Gamma))$ lies in a geodesic line in
Then, by Remark <ref> and Remark <ref>,
we see that $f$ satisfies

\begin{equation}
\label{eq:mu-harmonic}
\int_{\Gamma}  a_j(\gamma) d{\gamma_j}_*\mu(\gamma) = 0.
\end{equation}

Since each $f(\gamma)$ lies in a geodesic segment starting from
$\xi'$ and passing through $f(\gamma_j)$, we have

\begin{equation*}
a_j(\gamma) =
\begin{cases}
d(f(\gamma_j), f(\gamma)) & a_j(\gamma)\geq 0 \\
-d(f(\gamma_j), f(\gamma)) & a_j(\gamma) < 0,
\end{cases}
\end{equation*}

and $a_j(\gamma) \leq d(f(\gamma_j),\xi') < \varepsilon$.
Thus we have

\begin{equation*}
\int_{\Gamma} \max\{a_j(\gamma), 0 \}
d{\gamma_j}_*\mu (\gamma) <  \varepsilon.
\end{equation*}

Therefore, by (<ref>),

\begin{equation*}
\begin{split}
\int_{\Gamma} d(f(e),f(\gamma')) d\mu(\gamma')
& = \int_{\Gamma} d(f(\gamma_j), f(\gamma_j \gamma'))
d{\gamma_j}_* \mu(\gamma_j \gamma') \\
& = \int_{\Gamma} |a_j(\gamma)| d{\gamma_j}_* \mu(\gamma)
< 2\varepsilon.
\end{split}
\end{equation*}

Since $\varepsilon$ is arbitrary, this means that $f$ is a constant

Take any $\gamma \in \Gamma \setminus \{e\}$.
Let $p=f(e)$, $q=f(\gamma)$ in Proposition <ref>.
Then we get

\begin{equation*}
\inner{V_c}{\tcprj_{f(e)}(f(\gamma))}
\geq b_{\xi}(f(e),f(\gamma))
= -\varphi(\gamma).
\end{equation*}

As in the proof of $(1)$ above, by
Proposition <ref> and $\varphi(e)=0$, we see that

\begin{equation*}
\begin{split}
0 & \geq \int_{\Gamma} \inner{V_c}{\tcprj_{f(e)}(f(\gamma))}
d\mu (\gamma)
\geq \int_{\Gamma} -\varphi(\gamma) d\mu (\gamma) \\
& = \left(
\varphi(e) -
\int_{\Gamma} \varphi(\gamma) d\mu(\gamma)
\right) \\
& = \Delta (\varphi(e))=0.
\end{split}
\end{equation*}

Therefore we get

\begin{equation}
\label{eq:harmonic_at_e2}
\int_{\Gamma} \inner{V_c}{\tcprj_{f(e)}(f(\gamma))}
\end{equation}

Moreover the equality in
Propostion <ref> must hold
for all $\gamma \in \Gamma \setminus \{e\}$:

\begin{equation*}
\inner{V_c}{\tcprj_{f(e)}(f(\gamma))}
= b_{\xi}(f(e),f(\gamma)).
\end{equation*}

By Proposition <ref>, for all
$\gamma \in \Gamma$, the convex hull of
$[f(e),\xi(\omega)) \cup [f(e),f(\gamma)] \cup [f(\gamma),\xi(\omega))$
is isometric to a half-infinite parallelogram in $\R^2$.

Denote by $F_{\gamma}$ a half-infinite parallelogram bounded by
$[f(e),\xi) \cup [f(e),f(\gamma)] \cup [f(\gamma),\xi)$
obtained in $(2)$.
Take any $\gamma \in \Gamma$ and fix it for a while.  Note that
$\{f(\gamma_j\gamma)\}=\{\rho(\gamma_j)f(\gamma)\}$ also converges to
$\xi$ by our assumption.
Since $F_{\gamma_j\gamma}$'s are half-infinite parallelograms,
$[f(e),\xi)$ and $[f(\gamma_j\gamma),\xi)$ are
parallel to each other.
Thus by taking $q_j \in [f(e), \xi)$ so that
$d([f(e), \xi), f(\gamma_j\gamma))=d(q_j, f(\gamma_j\gamma))$, for
sufficiently large $j$, we may assume

\begin{equation*}
d(q_j, f(\gamma_j\gamma))
=d([f(e), \xi), [f(\gamma_j\gamma), \xi)),
\end{equation*}

\begin{equation}
\label{eq:flat_angle_sum}
\begin{split}
& \angle_{f(\gamma_j\gamma)}(\xi, f(e))
= \angle_{f(\gamma_j\gamma)}(\xi, q_j) +
\angle_{f(\gamma_j\gamma)}(q_j, f(e)), \\
& \angle_{f(\gamma_j\gamma)}(\xi, q_j) =
\angle_{q_j} (\xi, f(\gamma_j\gamma))=\frac{\pi}{2} .
\end{split}
\end{equation}

$\theta_{\gamma_j\gamma}= \angle_{f(\gamma_j\gamma)}(q_j, f(e))$.
Again noting that $F_{\gamma_j\gamma}$ is flat, we get

\begin{equation*}
\begin{split}
\cos \theta_{\gamma_j\gamma}& =
\frac{d(q_j, f(\gamma_j\gamma))}{d(f(e),f(\gamma_j\gamma))}=
\frac{d([f(e), \xi),
f(\gamma_j\gamma))}{d(f(e),f(\gamma_j\gamma))} \\
& = \sin \angle_{f(e)}(\xi, f(\gamma_j\gamma))\to 0,
\end{split}
\end{equation*}

since $f(\gamma_j\gamma) \to \xi$.
Therefore $\theta_{\gamma_j\gamma}\to \pi/2$.
By (<ref>), we see that the angle
$\angle_{f(\gamma_j\gamma)}(\xi(\omega), f(e))$ tends to $\pi$.

Consider the union of a geodesic segment and a geodesic ray

\begin{equation*}
\rho(\gamma_j^{-1})
\left( [f(e), f(\gamma_j\gamma)] \cup [f(\gamma_j\gamma), \xi)
\right)
= [f(\gamma_j^{-1}), f(\gamma)]\cup [f(\gamma),
\rho(\gamma_j^{-1})\xi),
\end{equation*}

for each $j$.
Recall that
$\{\rho(\gamma_j^{-1})\xi\}$ and $\{f(\gamma_j^{-1})\}$ converge
to points $\xi^+, \xi^-\in \partial Y$ respectively by our
Therefore we may assume that a sequence of geodesic rays
$\{\cc{f(\gamma)}{\rho(\gamma_j^{-1})\xi}\}$ converges to
$\cc{f(\gamma)}{\xi^+}$ unifomly on each bounded interval,
where $\cc{f(\gamma)}{\rho(\gamma_j^{-1})\xi}$ denotes the
geodesic starting from $f(\gamma)$ and terminating at
$\rho(\gamma_j^{-1})\xi$ as before.
On the other hand, setting
$\tilde c_j \colon \ray{0,\infty}\rightarrow Y$ to be

\begin{equation*}
\tilde c_j(t)=
\begin{cases}
\cc{f(\gamma)}{f(\gamma_j^{-1})} (t) & t \leq
d(f(\gamma),f(\gamma_j^{-1})) \\
f(\gamma_j^{-1}) & t \geq d(f(\gamma),f(\gamma_j^{-1})),
\end{cases}
\end{equation*}

we see that $\{\tilde c_j\}$ converges to $\cc{f(\gamma)}{\xi^-}$
uniformly on each bounded interval.
Recalling that letting $j\to \infty$ gives

\begin{equation*}
\angle_{f(\gamma)}(\rho(\gamma_j^{-1})\xi,f(\gamma_j^{-1}))
= \angle_{f(\gamma_j\gamma)} (\xi, f(e)) \to \pi,
\end{equation*}

we see that the images of $\cc{f(\gamma)}{\xi^+}$ and
$\cc{f(\gamma)}{\xi^-}$ form a straight geodesic line passing through
$f(\gamma)$ whose endpoints in $\partial Y$ are $\xi^+$ and $\xi^-$.

$(4)$ Now let $\tilde S$ be the union of the images of all geodesics
parallel to $c(\R)$, where $c$ is a geodesic satisfying $c(0)=f(e)$,
$c(-\infty)=\xi^-$, $c(\infty)=\xi^+$.
Then, by the discussion above, there exists a geodesic
joining $\xi^+$ and $\xi^-$ passing through $f(\gamma)$ for each
$\gamma \in \Gamma$, and thus
$f(\Gamma)\subset \tilde S$.
By <cit.>, we see that
$\tilde S$ is convex, and isometric to a product metric
space $\R \times S$.
We may take $S$ to be a
convex subset of the form $\nrprj_{c}^{-1}(c(0))$,
where $\nrprj_{c} \colon Y \rightarrow c(\R)$ is the nearest point
projection onto $c(\R)$ and $c$ is a geodesic satisfying $c(0)=f(e)$,
$c(-\infty)=\xi^-$, and $c(\infty)=\xi^+$ as above.
Let $\{q_j\} \subset S$ and $q_j \to q$ in $Y$. Then
there is a geodesic $c_j$ satisfying $c_j(0)=q_j$, $c_j(-\infty)=\xi^-$,
and $c_j(\infty)=\xi^+$ for each $j$.  Since $c_j$'s are parallel to each
other, $d(c_j(t), c_{j'}(t))=d(c_j(0),c_{j'}(0))$ for any $t \in \R$ and
$j, j' \in\N$, and hence $\{c_j(t)\}$ is a Cauchy sequence for each $t$.
Since $Y$ is complete, $\{c_j(t)\}$ converges and forms a geodesic
joining $\xi^-$ and $\xi^+$ and passing through $q$.  Therefore $S$ is
closed, and $\tilde S=\R \times S$ is also closed.

Now we show that the convex hull $\ch{f(\Gamma)}$ of $f(\Gamma)$
splits as $\R \times S_0$.
By definition, $\ch{f(\Gamma)}$ is the closure of the minimal convex
subset of $Y$ containing $f(\Gamma)$.
Since $\tilde S$ is a closed convex subset and
$f(\Gamma) \subset \tilde S$, $\ch{f(\Gamma)}$ is contained in
$\tilde S$.
Note that $f(\Gamma)$ is invariant under the action of $\rho(\Gamma)$,
hence so is $\ch{f(\Gamma)}$.
Recall that $\xi(\omega)=\lim f(\gamma_j)$.
Then for any $q \in \ch{f(\Gamma)}$, we have
$[q,f(\gamma_j)]\subset \ch{f(\Gamma)}$ and, as a set
$[q,f(\gamma_j)]$ converges to $[q,\xi(\omega))$.  Since
$\ch{f(\Gamma)}$ is closed, we see that
$[q,\xi(\omega)) \subset \ch{f(\Gamma)}$.  The same is true for
$\xi^- =\lim f(\gamma_j^{-1})$.  Since $\ch{f(\Gamma)}$ is
$[q, \rho(\gamma_j^{-1})\xi(\omega)) \subset \ch{f(\Gamma)}$ for
any $q \in \ch{f(\Gamma)}$ and $j \in \N$. Since
$[q, \rho(\gamma_j^{-1})\xi(\omega))$ converges to $[q, \xi^+)$ by the
definition of $\xi^+$, we see that $[q, \xi^+)$ is also contained in

Let $S_0 = \ch{f(\Gamma)} \cap S$.  Then $S_0$ is a closed convex
subset of $Y$.
As we have seen above, for each $\gamma \in \Gamma$,
$[f(\gamma),\xi^+)$ (resp. $[f(\gamma),\xi^-)$) is contained in
Since there is a geodesic $c_{\gamma}$ joining $\xi^+$ and $\xi^-$ passing
through $f(\gamma)$, $[f(\gamma),\xi^+)\cup [f(\gamma),\xi^-)$ must
coincide with the image $c_{\gamma}(\R)$ of $c_{\gamma}$.
Therefore $c_{\gamma}(\R) \subset \ch{f(\Gamma)}$ and,
by reparametrizing $c_{\gamma}$ so that $c_{\gamma}(0) \in S$, we see
that $c_{\gamma}(0) \in S \cap \ch{f(\Gamma)}=S_0$.
Hence $f(\gamma) \in \R \times S_0 \subset \tilde S$ for every
$\gamma \in \Gamma$.
Since $S_0$ is closed and convex, so is $\R \times S_0$, and we see that
$\ch{f(\Gamma)} \subset \R \times S_0$.
On the other hand, for any $s \in S_0$, since
$s \in \ch{f(\Gamma)}$,
$[s,\xi^+)$ and $[s,\xi^-)$ are contained in $\ch{f(\Gamma)}$ as we
have seen above.
Since there is a geodesic line joining $\xi^+$ and $\xi^-$ passing through
$s$, $[s,\xi^+) \cup [s,\xi^-)$ must coincide with the image
$\R \times \{s\}$ of this geodesic line.
Therefore $\R \times S_0 \subset \ch{f(\Gamma)}$.
We conclude that $\ch{f(\Gamma)}=\R \times S_0$.

Note that there is an element $g_a \in \isom{\ch{f(\Gamma)}}$ given by

\begin{equation*}
g_a \colon (r,s)\mapsto (r+a,s),
\ (r,s) \in \ch{f(\Gamma)}=\R \times S_0, \ a \in \R,
\end{equation*}

and recall that an isometry satisfying

\begin{equation*}
d(g p, p)=d(g q,q)
\end{equation*}

for any $p,q \in Y$ is called a Clifford tranlation; $g_a$ is a
Clifford translation.
Let $H \subset \isom{\ch{f(\Gamma)}}$ be the set of all Clifford
translation on $\ch{f(\Gamma)}$.
Then, by
$H$ is an abelian group and it can be identified with a Hilbert space,
and $\ch{f(\Gamma)}$ splits as $\ch{f(\Gamma)}=H\times Z$;
each fiber isometric to $H$ is given as the $H$-orbit $Hx$ of
$x \in \ch{f(\Gamma)}$.
Since we have $g_a$, $H$ is a non-trivial Hilbert space.
Note that there is no Clifford translation in $\isom{Z}$, and hence $Z$
does not admit any splitting $Z=\R \times Z'$.

According to the proof of
for any $x \in \ch{(f(\Gamma))}$ and $g \in \isom{\ch{f(\Gamma)}}$,
$g$ maps $Hx$ to $Hgx$.
Setting $x=(h,y) \in H \times Z$ and $y'=\nrprj_Z(gx)$,  we see that
$g (H\times \{y\}) = H\times \{y'\}$.
Thus, by <cit.>,
we see that $g$ splits as $g=g_1\times g_2$, where
$g_1 \in \isom{H}$ and $g_2 \in \isom{Z}$.
Then it is easy to see that, along this splitting,
$\rho \colon \Gamma \rightarrow \isom{f(\Gamma)}$ splits
as $\rho=\rho_1 \times \rho_2$, where
$\rho_1 \colon \Gamma \rightarrow \isom{H}$ and
$\rho_2 \colon \Gamma \rightarrow \isom{Z}$.

Note that, along the splitting $\rho=\rho_1\times \rho_2$, a
$\rho$-equivariant $\mu$-harmonic map $f$ also splits as
$f=f_1\times f_2$; setting $f(e)=(f_1(e), f_2(e))$ and
$f_i(\gamma)=\rho_i(\gamma)f_i(e)$, $i=1,2$, we see that
In particular, each $f_i$
is a $\rho_i$-equivariant map for $i=1,2$.
If there is a $\rho_i$-equivariant map $f'_i$ with strictly smaller
$\mu$-energy than $f_i$, then replacing $f_i$ by $f_i'$ and taking
$f'=(f_1', f_2')$ decreases the energy of $f$ strictly.
This contradicts the fact that $f$ is $\mu$-harmonic.
Hence $f_i$ is a $\rho_i$-equivariant $\mu$-harmonic map.
Since the projection onto $Z$ does not increase the distance,
$l_{\rho_2}(\Gamma)=0$ follows from the assumption

Now we finish the proof of Theorem <ref> for proper $Y$.

Proof of Theorem <ref>.
Under our assumption, by Proposition <ref>,
there exists $\omega \in \Omega$ such that
a subsequence $\{f(\gamma_j)\}$ of $\{f(\gamma_n(\omega)^{-1})\}$
converges to $\xi(\omega) \in \partial Y$, and

\begin{equation*}
\varphi_{\omega}(\gamma)=
\begin{cases}
d(\xi(\omega),f(\gamma)) - d(\xi(\omega),f(e)) & \text{if }\xi(\omega)
\in Y \\
b_{\xi(\omega)}(f(\gamma), f(e))  & \text{if }\xi(\omega) \in \partial Y
\end{cases}
\end{equation*}

satisfies $\Delta \varphi_{\omega}(e)=0$.
If $\xi(\omega) \in Y$, then Lemma <ref> $(1)$ tells us that
$\rho(\Gamma)$ fixes a point in $Y$.

Suppose $\xi(\omega) \in \partial Y$.
Since $Y$ is proper, taking further subsequence if necessary, we may
assume that $\rho(\gamma_j^{-1})\xi(\omega)$ and $\{f(\gamma_j^{-1})\}$
converge to points $\xi^+, \xi^- \in Y\cup \partial Y$
Since $d(f(e),f(\gamma^{-1}))=d(f(\gamma),f(e))\to \infty$,
$\xi^- \in \partial Y$ as well as $\xi^+$.
Then by Lemma <ref> $(4)$, we see that we have
a splitting $\ch{f(\Gamma)}=H \times Z$,
$\rho_2$-equivariant $\mu$-harmonic map
$f_2 \colon \Gamma \rightarrow Z$ and $l_{\rho_2}(\Gamma)=0$.
Note that $H=\R^k$ for some $k \in \N$ and that $Z$ is proper, since
$Y$, and hence $\ch{f(\Gamma)}$, is proper.
Therefore, by Proposition <ref>, we have
$\xi \in Z \cup \partial Z$ such that

\begin{equation*}
\varphi(\gamma) =
\begin{cases}
d(\xi,f(\gamma))-d(\xi,f(e)) & \text{if } \xi \in Z \\
b_{\xi}(f(\gamma),f(e)) & \text{if } \xi \in \partial Z
\end{cases}
\end{equation*}

satisfies $\Delta \varphi(e)=0$ again.
If $\xi \in \partial Z$, then Lemma <ref> $(4)$ tells us that
$Z$ must admit further splitting $Z = \R^{k'}\times Z'$.
However, by our choice of $Z$ in Lemma <ref> $(4)$, this is
Thus $\xi$ must lie in $Z$, and Lemma <ref> $(1)$ implies that
there exists a point $p \in Z$ fixed by $\rho_2(\Gamma)$.
Then we see that $\R^k \times \{p\}$ is left invariant under the action
of $\rho(\Gamma)=\rho_1(\Gamma)\times \rho_2(\Gamma)$, which means the
existence of $\rho(\Gamma)$-invariant flat subspace.
This completes the proof of Theorem <ref> for proper $Y$.

§ PROOFS OF
THEOREM <REF> FOR NONPROPER $\CAT{0}$
SPACES AND THOREM <REF>

In what follows, we consider the case that $Y$ is not proper.
Before proceed, recall that we have used the properness of $Y$ to

* $\mu$-harmonic map exists when $\rho(\Gamma)$ does not fixes a
point in $\partial Y$,

* $\{f(\gamma_j(\omega)^{-1})\}$ has a convergent subsequence

* $\{f(\gamma_j^{-1})\}$ has a convergent subsequence

* $\{\rho(\gamma_j')\xi(\omega)\}$ has a convergent subsequence.

We have to make a detour or find a substitute for these arguments.
It is certainly possible when $Y$ is of finite telescopic dimension, and
almost possible when $Y$ is locally finite-dimensional as we will see in
<ref> and <ref>.

§.§ A quick review of ultralimits

In what follows, we will frequently use the ultralimit of a sequence of
metric spaces. We start with a quick review of ultralimits.

Let $\lambda$ be a nonprincipal ultrafilter on $\N$, that is, a finitely
additive probablity measure on $\N$ satisfying

* every $S \subset \N$ is $\lambda$-measurable,

* for any $S \subset \N$ we have $\lambda (S)\in \{0,1\}$,

* if $\# S <\infty$, then $\lambda (S)=0$.

See, for example, <cit.> for the
existence of a nonpricipal ultrafilter.
For any bounded sequence $\{a_j\}$ of real numbers, there exists
$l \in \R$ such that, for any $\varepsilon>0$,
$\lambda (\{j \in \N \mid |a_j-l|<\varepsilon\})=1$, which we express
as $l=\ulim_j a_j$.

Let $Y_j$ be a metric space, and $p_j \in  Y_j$, $j\in \N$.
Let $Y^{\infty}$ be a set of sequences $\{x_j\}$, $j \in \N$, such that
$d_{Y_j}(p_j,x_j)$ is bounded independent of $j \in \N$.
We set $\{x_j\}\sim \{y_j\}$ if $\{x_j\}$ and $\{y_j\}$ satisfy
$\ulim_j d(x_j,y_j)=0$.  Then $\sim$ becomes an equivalence relation on
$Y^{\infty}$, and we denote the set of equivalence classes by
$Y_{\infty}=\ulim_j (Y_j,p_j)$.
We also denote $x=\{x_j\}\in Y_{\infty}$ by $x=\ulim_j x_j$.
It is obvious that $d_{\lambda}(\{x_j\},\{y_j\})=\ulim_j d(x_j,y_j)$
becomes a metric on $Y_{\infty}$.
Let $q_j$ be a point in $Y_j$ whose distance from $p_j$ is bounded
independently of $j \in \N$, and $c_j\colon [0,d_j]\rightarrow Y_j$ a
geodesic with $c_j(0)=q_j$.
Then, for each $t \in [0,\ulim_j d_j]$, $d(c_j(t),p_j)$ is
bounded independently of $j\in \N$, and hence we get
$c_{\infty}(t)=\ulim_j c_j(t)$.
Since $d(c_j(t),c_j(t'))=|t-t'|$ for any $j \in N$, we see that
$c_{\infty} \colon [0,\ulim_j d_j]\rightarrow Y_{\infty}$ turns out to
be a geodesic.
Then it is easy to see that if each $Y_j$ is a complete $\cat{0}$
space, then $Y_{\infty}=\ulim_j (Y_j,p_j)$ becomes a complete $\cat{0}$
Note that, even when $d_j\nearrow\infty$,  we can define a geodesic ray
$c_{\infty}\colon \ray{0,\infty}\rightarrow Y_{\infty}$ as an
ultralimit of a sequence of  geodesics
$\{c_j\colon [0,d_j]\rightarrow Y_j\}$.
In this case, we have a point
$\xi=c_{\infty}(\infty) \in \partial Y_{\infty}$.
We often express this $\xi$ as $\xi=\ulim_j c_j(d_j)$.

In what follows, we mainly consider an ultralimit of the form
$Y_{\infty}=\ulim (Y,o)$ for fixed $Y=(Y,d)$ and $o \in Y$, and
we use the same symbol $d$ for the metric on $Y_{\infty}$ instead of
$d_{\lambda}$ for the sake of simplicity.
It can be shown that if $Y=(Y,d)$ is proper, then
$Y_{\infty}=(Y_{\infty},d)$ is isometric to $Y=(Y,d)$.
However, this is no longer true for nonproper metric space $Y$.
In that case, $Y$ becomes a convex closed proper subset of

Let $\lambda$ be a nonprincipal ultrafilter.
Consider a metric space $Y$ obtained by attaching a half line $L_n$ to
$1/n \in \ray{0,\infty}$ for $n=1,2,\dots$.
Then $Y$ becomes a $\cat{0}$ space.
Let $t>0$ and take a point $p_n^t \in L_n$ so that
Choose a basepoint
$o$ to be the endpoint $0$ of $\ray{0,\infty}$ to which we
attach $L_n$'s, and consider an ultralimit
$Y_{\infty}=\ulim_n (Y,o)$.
Then there must be a point $p^t=\ulim_n p_n^t$ in $Y_{\infty}$.
However, since, for any $q \in Y$,  $d(p_n^t,q)\geq t$ holds unless
$q \in L_n$, we see that
$\lambda( \{n \in \N \mid d(p_n^t,q)\geq t\})=1$, and hence $p^t$
cannot be a point in $Y$.
It is clear that, for each $t>0$, $d(p^t,o)$ must be equal to
$t$,  and, for $t\not=t'$, $d(p^t,p^{t'})=|t-t'|$ must hold.
Therefore $\{p^t\mid t \in \ray{0,\infty}\}$ is a half line $L$
attached to $o$, the endpoint of $\ray{0,\infty}$, and $L$
does not exist in $Y$.
Thus we see $Y \not=Y_{\infty}$.
Take an ultralimit one more time and consider
$(Y_{\infty})_{\infty}=\ulim_n (Y_{\infty},o)$.
Looking at $\{p_n^t\}$ again, we see that, since
$d(p_n^t,p^t)\geq 2t$ for any $n$, an ultralimit of this sequence
cannot be $p^t$.
Indeed, we see that $\ulim_n p_n^t$ cannot be any point in $Y_{\infty}$
as before, and there must be another half line $L'$ containing
$\ulim_n p_n^t$ and attached to $o$, which does not
exist in $Y_{\infty}$.
Therefore we get $(Y_{\infty})_{\infty}\not= Y_{\infty}$ again.
Repeating this, we get larger and larger metric spaces.
This observation leads us to an argument often used in the proof of
our theorems in what follows.

Although $Y_{\infty}$ can be larger than $Y$ as the example above
shows,  $Y_{\infty}$ inherits a certain dimension bound as we will see
in Proposition <ref>,
which plays an important role in the proof of Theorem

Let $\rho\colon \Gamma \rightarrow \isom{Y}$ be a homomorphism.
For $p=\ulim_j p_j \in Y_{\infty}=\ulim (Y,o)$, we see that

\begin{equation*}
d(\rho(\gamma)p_j,o) \leq d(\rho(\gamma)p_j, \rho(\gamma)o) +
d(\rho(\gamma)o,o) = d(p_j,o)+d(\rho(\gamma)o,o)
\end{equation*}

is bounded independent of $j\in \N$,  and hence we have
$\ulim_j \rho(\gamma)p_j \in Y_{\infty}$.
Therefore,  the action of $\Gamma$ on $Y$ given by
$\rho$ can be extended to that on $Y_{\infty}$ by setting
$\rho(\gamma)(\ulim_j p_j)=\ulim_j \rho(\gamma)p_j$.
It is clear that this gives a homomorphism from $\Gamma$ into
$\isom{Y_{\infty}}$, which we denote by the same symbol $\rho$.

Also, as one easily imagine,  $\mu$-energy $\ene{\mu}$ behaves well under
taking ultralimit.
Although the proof is standard for experts, we give a proof of the
following proposition for the sake of completeness.

Let $Y$ be a $\cat{0}$ space, $o\in Y$, and
$\lambda$ a nonprincipal ultrafilter.
Suppose that a countable group $\Gamma$ acts on $Y$ via a
homomorphism $\rho\colon \Gamma \rightarrow \isom{Y}$, and that the
action given by $\rho$ has finite second moment with respect to $\mu$.
Let $f_q \colon \Gamma \rightarrow Y$ be a $\rho$-equivariant map
defined by $f_q(\gamma)=\rho(\gamma)q$, and
$Y_{\infty}=\ulim (Y,o)$.
Suppose that $p= \ulim_j p_j$ for a bounded sequence
$\{p_j\}\subset Y$.  Then we have

\begin{equation*}
\ene{\mu} (f_p) = \ulim_j \ene{\mu}(f_{p_j}).
\end{equation*}

\begin{equation*}
\begin{split}
d(p_j, \rho(\gamma)p_j) & \leq d(o, p_j) + d(o, \rho(\gamma)o)
+ d(\rho(\gamma)o, \rho(\gamma)p_j) \\
& = 2d(o, p_j) + d(o, \rho(\gamma)o),
\end{split}
\end{equation*}

and $\{p_j\}$ is bounded, we have

\begin{equation*}
d(p_j, \rho(\gamma)p_j)^2 \leq C_1 d(o, \rho(\gamma)o)^2 + C_2
\end{equation*}

for some positive constant $C_1$ and $C_2$.
Since $\Gamma$ is countable, we can take an increasing sequence
$D_1 \subset D_2 \subset \dots \subset D_n \subset \dots$ of finite
subsets of $\Gamma$ so that $\bigcup_{n=1}^{\infty}D_n=\supp \mu$.
Then, by the inequality above, we have

\begin{equation*}
\int_{\Gamma \setminus D_n} d(p_j, \rho(\gamma)p_j)^2 d\mu(\gamma)
\leq \int_{\Gamma \setminus D_n} C_1 d(o, \rho(\gamma)o)^2 + C_2
\ d\mu(\gamma)
\end{equation*}

for any $j \in \N$, and also

\begin{equation*}
\int_{\Gamma \setminus D_n} d(p, \rho(\gamma)p)^2 d\mu(\gamma)
\leq \int_{\Gamma \setminus D_n} C_1 d(o, \rho(\gamma)o)^2+ C_2
\ d\mu(\gamma).
\end{equation*}

Note that the right-hand side tends to $0$ as
$n\to \infty$ since the action given by $\rho$ has finite second
moment with respect to $\mu$.
Therefore, for any $\varepsilon>0$,
there exists $n$ such that

\begin{equation*}
\frac{1}{2} \int_{\Gamma \setminus D_n} d(p_j, \rho(\gamma)p_j)^2
d\mu(\gamma) < \frac{\varepsilon}{4}
\end{equation*}

for any $j \in \N$, and

\begin{equation*}
\frac{1}{2} \int_{\Gamma \setminus D_n} d(p, \rho(\gamma)p)^2
d\mu(\gamma) < \frac{\varepsilon}{4}
\end{equation*}

On the other hand, since
$\ulim_j d(p_j, \rho(\gamma)p_j)^2=d(p,\rho(\gamma)p)^2$ and $\lambda$ is
a nonprincipal ultrafilter, letting

\begin{equation*}
A_{\gamma}=\left\{k \in \N \mid
\right\},
\end{equation*}

we have $\lambda(A_{\gamma})=1$ for each $\gamma \in \Gamma$.
Since $D_n$ is a finite set, we obtain
$\lambda (\bigcap_{\gamma \in D_n} A_{\gamma}) =1$.
Note that, for any $j \in \bigcap_{\gamma \in D_n} A_{\gamma}$,

\begin{equation*}
\left|\frac{1}{2}\int_{D_n} d(p_j,\rho(\gamma)p_j)^2 d\mu(\gamma)
- \frac{1}{2}\int_{D_n} d(p,\rho(\gamma)p)^2 d\mu(\gamma) \right|
< \frac{\varepsilon}{2}
\end{equation*}

holds.  Therefore, for any $\varepsilon>0$, by taking $n \in \N$ as
above, we see that

\begin{equation*}
\begin{split}
\left|\ene{\mu}(f_{p_j}) - \ene{\mu}(f_p)\right|
& = \frac{1}{2} \left|
\int_{\Gamma \setminus D_n} d(p_j, \rho(\gamma)p_j)^2 d\mu(\gamma)
+ \int_{D_n} d(p_j, \rho(\gamma)p_j)^2 d\mu(\gamma) \right. \\
& \phantom{==}   \left.
- \int_{\Gamma \setminus D_n} d(p, \rho(\gamma)p)^2 d\mu(\gamma)
- \int_{D_n}d(p, \rho(\gamma)p)^2 d\mu(\gamma)
\right| \\
& \leq  \frac{1}{2} \int_{\Gamma \setminus D_n}
d(p_j, \rho(\gamma)p_j)^2 d\mu(\gamma)
+ \frac{1}{2}\int_{\Gamma \setminus D_n}
d(p, \rho(\gamma)p)^2 d\mu(\gamma)  \\
& \phantom{==}
+ \frac{1}{2}
\left|\int_{D_n} d(p_j, \rho(\gamma)p_j)^2 d\mu(\gamma)
- \int_{D_n}d(p, \rho(\gamma)p)^2 d\mu(\gamma)
\right|\\
& < \frac{\varepsilon}{4} + \frac{\varepsilon}{4} +
\frac{\varepsilon}{2}= \varepsilon
\end{split}
\end{equation*}

holds for any $j \in \bigcap_{\gamma \in D_n}A_{\gamma}$.
This completes the proof.

Recall that $\rho(\Gamma) \subset \isom{Y}$ is said to be
reductive in the sense of Jost
if there exists a closed, convex $\rho(\Gamma)$-invariant subset $C$ of
$Y$ satisfying either

* $C$ is isometric to a finite-dimensional Euclidean space, or

* for any unbounded sequence $\{p_n\} \subset C$ , there exists
$\gamma \in \Gamma$ such that $\{d(p_n, \rho(\gamma)p_n)\}$ is

If $C$ is isometric to a finite-dimensional Euclidean space, then we can
regard $C$ as $Y$.
The existence of $\rho$-equivariant harmonic map in this case follows
from, for example, an argument similar to that in [32].
Suppose that we have the second case in the definition of reductiveness
in the sense of Jost.
Take any sequence $\{f_{p_j}\}$ that minimizes $\mu$-energy.
Suppose that $\{p_j\}$ is unbounded.
Then there exists $\gamma \in \Gamma$ such that
$\{d(p_j,\rho(\gamma)p_j)\}$ is unbounded.
Since $\supp \mu$ generates $\Gamma$, there exists
$s_1, \dots, s_k \in \supp \mu$ such that $\gamma=s_1\dots s_k$.

\begin{equation*}
d(p_j, \rho(\gamma)p_j) \leq
d(p_j, \rho(s_1)p_j) + \dots + d(p_j,\rho(s_k)p_j)
\end{equation*}

as we have seen in the proof of Lemma <ref>,
$\{d(p,\rho(s_j)p)\}$ is unbounded for some $j$, and we conclude that
$\ene{\mu}(f_{p_j})$ is unbounded.
A contradiction.
Now since $\{p_j\}$ is bounded, by the proposition above,  we see that
$f_p$ for $p=\ulim p_j$ minimizes $\mu$-energy, and hence it is
a $\mu$-harmonic map into $Y_{\infty}$.
Note that $Y$ is a convex closed subset in $Y_{\infty}$.
Then the nearest point projection $\nrprj_Y$ onto $Y$ is
$\rho$-equivariant, and does not increase
the distance, and hence the energy.
Therefore $\nrprj\circ f_{p}$ is also
$\mu$-harmonic; we may assume $f_{p}$ is a map into $Y$.
Summarizing this gives the following.

If $\rho(\Gamma)$ is
reductive in the sense of Jost, then there is a $\rho$-equivariant
harmonic map $f \colon \Gamma \rightarrow Y$.

The corollary above also follows from
<cit.> without using ultralimits.

The following lemma tells us that Busemann functions also behave
well under taking ultralimits.

Take $o \in Y$.
Suppose that a sequence $\{x_j\} \subset Y$ satisfies
Then we have $\ulim_j x_j = \xi \in \partial Y_{\infty}$, where
$Y_{\infty}=\ulim (Y,o)$.
Furthermore,  we have, for any $z \in Y$,

\begin{equation*}
b_{\xi}(z,o)=\ulim \left(d(x_j,z)-d(x_j,o)\right).
\end{equation*}

Take $z \in Y$.
Consider a geodesic ray $c$ with $c(0)=o$.
Applying <cit.> by setting
we see that, for any $\varepsilon/3>0$, there exists $r>0$ such that

\begin{equation*}
d(c(r),z) -d(c(t),z) + (t-r) < \varepsilon/3
\end{equation*}

holds for any geodesic $c$ starting from $o$ and $t\geq r$.
This inequality can be rewritten as

\begin{equation*}
\left(d(c(r),z)-r\right) -
\left(d(c(t),z)-t\right) < \varepsilon/3.
\end{equation*}

Since $t\geq r$ is arbitrary and $d(c(t),z)-t$ is easily seen to be
nonincreasing by the triangle inequality
we see that

\begin{equation*}
|\left(d(c(r),z)-r\right) - b_{c(\infty)}(z, o)|
\leq \varepsilon/3
\end{equation*}

for any geodesic ray $c$ starting from $o$.
Let $c_j$ be a geodesic joining $o$ and $x_j$ with $c_j(d_j)=x_j$, and

\begin{equation*}
D(r)= \{j \in \N \mid d_j \geq r\}.
\end{equation*}

By our assumption, $d_j \to \infty$, and hence $\lambda(D(r))=1$.
Note that, as we have seen before, we have an ultralimit
$c_{\infty}\colon \ray{0,\infty}\rightarrow Y_{\infty}$ of $\{c_j\}$,
and $\xi=c_{\infty}(\infty)=\ulim_j x_j \in \partial Y_{\infty}$.

\begin{equation*}
C(r,\varepsilon)  = \{j \in \N \mid
< \varepsilon/3\}.
\end{equation*}

Then $\lambda(C(r,\varepsilon))=1$ for any $r$ and $\varepsilon$.
Thus $\lambda(D(r) \cap C(r,\varepsilon))=1$,
and for any $j \in D(r) \cap C(r,\varepsilon)$,

\begin{equation*}
\begin{split}
& |d(c_j(d_j), z)-d_j -
b_{\xi}(z,o)| \\
\leq &
|\left(d(c_j(d_j), z)-d_j\right)
- \left(d(c_j(r), z)-r\right)| \\
& +|\left(d(c_j(r),z)-r\right)-
\left(d(c_{\infty}(r), z)-r\right)| \\
& + |\left(d(c_{\infty}(r), z)-r\right) -
b_{\xi}(z,o)| \\
< &  \varepsilon/3  +
|d(c_j(r), z)-d(c_{\infty}(r), z)| +
\varepsilon/3 < \varepsilon
\end{split}
\end{equation*}

holds.  Since $\varepsilon>0$ is arbitrary, we get

\begin{equation*}
\ulim  \left(d(c_j(d_j), z) - d_j\right)
= b_{\xi}(z,o),
\end{equation*}

and this completes the proof.

§.§ Existence of a $\mu$-harmonic function

We have seen that there is a function $\varphi_{\omega}$ which
satisfies $\Delta \varphi_{\omega}(e)=0$ in
<ref> when $Y$ is proper.
In this section, we show that, even when $Y$ is not proper,
there does exist a function which is $\mu$-harmonic on whole $\Gamma$
under the assumption that $\mu(\gamma)\not= 0$ for any
$\gamma \in \Gamma \setminus \{e\}$.
We will use this fact in  <ref>.

Suppose that $\Gamma$ acts on $\cat{0}$ space $Y$ via a homomorphism
$\rho\colon \Gamma \rightarrow \isom{Y}$ and that the action given by
$\rho$ has finite second moment.
Fix $o \in Y$.
Let $\varphi\colon \Gamma \rightarrow \R$ be a function satisfying
$\varphi(\gamma) \leq C_1 d(o,\rho(\gamma)o)^2+C_2$ for some
Since the action give by $\rho$ has finite second moment with respect
to $\mu$, it is also true for $\mu^2$ as we have seen
in  <ref>.
Noting that $d(p,q) \leq 1+d(p,q)^2$ holds for any $p,q \in Y$,  we see

\begin{equation*}
\begin{split}
|\varphi(\gamma \gamma')| &\leq C_1 d(o,\rho(\gamma \gamma')o)^2 + C_2 \\
& \leq C_1
\left(d(o,\rho(\gamma)o)+d(\rho(\gamma)o,\rho(\gamma\gamma')o)\right)^2
+ C_2 \\
& \leq C_1'd(o,\rho(\gamma')o)^2 + C_2'
\end{split}
\end{equation*}

for some positive constant $C_1', C_2'$ which depend on $\gamma$.
Thus, fixing $\gamma \in \Gamma$,
we see that $\gamma' \mapsto \varphi(\gamma \gamma')$ is
$\mu^2$-integrable by our assumption.
In the rest of this section, a function $\varphi$ under consideration
is always assumed to satisfy
$|\varphi(\gamma)| \leq C_1 d(o,\rho(\gamma)o)^2 + C_2$ for some

Let us denote by $\Delta_2$ the $\mu^2$-Laplacian, which is defined by

\begin{equation*}
\Delta_2 \varphi (\gamma) =
\varphi(\gamma)- \int_{\Gamma} \varphi(\gamma\gamma') d\mu^2(\gamma').
\end{equation*}

Then we have

\begin{equation*}
\begin{split}
-\Delta_2 \varphi(\gamma) & =
\int_{\Gamma} \varphi(\gamma \gamma') - \varphi(\gamma) d\mu^2(\gamma') \\
& = \int_{\Gamma} \left(\int_{\Gamma}
\varphi(\gamma \gamma_1 \gamma_2) - \varphi(\gamma) d\mu(\gamma_2)\right)
d\mu(\gamma_1) \\
& = \int_{\Gamma} \left(\int_{\Gamma}
\left(\varphi(\gamma \gamma_1 \gamma_2) - \varphi(\gamma \gamma_1)\right)
+ \varphi(\gamma \gamma_1) - \varphi(\gamma)
d\mu(\gamma_2)\right) d\mu(\gamma_1) \\
& = \int_{\Gamma}
\left(
-\Delta \varphi(\gamma \gamma_1) + \varphi(\gamma \gamma_1) -
\varphi(\gamma)\right)
d\mu(\gamma_1) \\
& =\int_{\Gamma}-\Delta \varphi(\gamma\gamma_1) d\mu(\gamma_1)
-\Delta \varphi(\gamma),
\end{split}
\end{equation*}

and hence we get the following:

Let  $\varphi\colon \Gamma \rightarrow \R$ be a $\mu$-subharmonic

$(1)$ $\varphi$ is $\mu^2$-subharmonic$;$ for any $\gamma \in \Gamma$,
$-\Delta_2 \varphi(\gamma)\geq 0$ holds.

$-\Delta_2 \varphi(\gamma) \geq -\Delta \varphi(\gamma)\geq 0$

$(3)$ If $-\Delta_2 \varphi(\gamma)=0$ holds for $\gamma \in \Gamma$,
then $-\Delta \varphi(\gamma)=0$ holds.

Suppose that $\mu(\gamma)\not= 0$ holds for any
$\gamma \in \Gamma \setminus \{e\}$.
Let $\varphi \colon \Gamma \longrightarrow \R$ be a $\mu$-subharmonic
If  $-\Delta_2 \varphi(e)=0$ holds, then $\varphi$ is a $\mu$-harmonic
function, that is, $-\Delta \varphi(\gamma)=0$ holds for any
$\gamma \in \Gamma$.

By the computation above and $-\Delta_2\varphi(e)=0$, we get

\begin{equation*}
0= -\Delta_2 \varphi(e)
= -\Delta \varphi(e) +
\int_{\Gamma}-\Delta \varphi (\gamma)  d\mu(\gamma).
\end{equation*}

On the other hand, since $\varphi$ is $\mu$-subharmonic, we have
$-\Delta\varphi(\gamma)\geq 0$.
Therefore, by our assumption that $\mu(\gamma)\not= 0$ for any
$\gamma \in \Gamma$, we see that $-\Delta \varphi(\gamma)=0$ for any
$\gamma \in \Gamma$.
This completes the proof.

Now take a $\mu$-harmonic map $f$ and let
$\varphi(\gamma) = d(f(e), f(\gamma))$.
Then $\varphi$ is a $\mu$-subharmonic function by
Proposition <ref>.

\begin{equation*}
\alpha_{2,n}:= \int_{\Omega} -\Delta_2 \varphi(\gamma_{2n}(\omega))
\ d\mathbb{P}(\omega)
= \int_{\Gamma} -\Delta_2 \varphi(\gamma) d\mu^{2n}(\gamma).
\end{equation*}

Then, by the same computation as in
<ref>,  we see that

\begin{equation*}
L^{2(n+1)}(f)= L^{2n}(f) + \alpha_{2,n}
\end{equation*}

Thus if $l_{\rho}(\Gamma)=0$, then we have
$\liminf_n\alpha_{2,n}=0$, and by Lebesgue's convergence theorem and
$-\Delta_2 \varphi(\gamma)\geq 0$ which follows from
Lemma <ref> $(1)$, we get

\begin{equation}
\label{eq:0_for_each_k}
\liminf_n -\Delta_2 \varphi(\gamma_{2n}(\omega)) = 0
\quad \text{a.e.}
\end{equation}

\begin{equation*}
A =
\{ \omega \in \Omega \mid \
\liminf_n -\Delta_2 \varphi(\gamma_{2n}(\omega))=0 \},
\end{equation*}

then we see that $\mathbb{P}(A)=1$.
Take $\omega \in A$.
Then, for any $j$, we can find $n_j \in \N$ such that

\begin{equation*}
-\Delta_2 \varphi(\gamma_{n_j}(\omega)) \leq 1/j.
\end{equation*}

By Lemma <ref> $(2)$, we also have

\begin{equation*}
-\Delta \varphi(\gamma_{n_j}(\omega)) \leq 1/j.
\end{equation*}

Letting $j\to \infty$, we see that

\begin{equation}
\label{eq:converges_to_harmonic}
\lim_{j\to \infty} -\Delta_k \varphi(\gamma_{n_j}(\omega))=0
\end{equation}

for $k=1,2$.  Let us define

\begin{equation*}
\varphi_{\omega,n_j} (\gamma)
= d(f(\gamma_{n_j}(\omega)^{-1}), f(\gamma))
- d(f(\gamma_{n_j}(\omega)^{-1}), f(e)).
\end{equation*}

Note that, by the triangle inequality, $\varphi_{\omega,n_j}$ satisfies
$|\varphi_{\omega,n_j}(\gamma)|\leq d(f(e),f(\gamma))$ for each
$j \in \N$ and $\gamma \in \Gamma$.
Thus $\{\varphi_{\omega,n_j}\}$ can be viewed as a sequence in the
product space
$\prod_{\gamma \in \Gamma}[-d(f(e),f(\gamma)),d(f(e),f(\gamma))]$,
which is sequentially compact.
Since the topology of this product space agrees with that given
by the pointwise convergence of real valued functions on $\Gamma$,
by taking a subsequence if necessary, we may
assume that $\{\varphi_{\omega,n_j}\}$ converges pointwise to
Since, by fixing $j$, the second term in the right-hand side of the
definition of $\varphi_{\omega,n_j}$ is independent of
$\gamma\in \Gamma$, we have

\begin{equation*}
\begin{split}
\Delta_2 \varphi_{\omega,n_j} (\gamma)
& = d(f(\gamma_{n_j}(\omega)^{-1}), f(\gamma))
- \int_{\Gamma}
d(f(\gamma_{n_j}(\omega)^{-1}), f(\gamma \gamma')) d\mu^2(\gamma') \\
& = d(f(e), f(\gamma_{n_j}(\omega)\gamma)) -
\int_{\Gamma}d(f(e), f(\gamma_{n_j}(\omega)\gamma\gamma'))d \mu^2(\gamma') \\
& = \Delta_2 \varphi(\gamma_{n_j}(\omega)\gamma),
\end{split}
\end{equation*}

in particular, we get

\begin{equation*}
\Delta_2 \varphi_{\omega, n_j} (e)
= \Delta_2 \varphi(\gamma_{n_j}(\omega)).
\end{equation*}

\begin{equation*}
=|d(f(\gamma_j), f(\gamma)) - d(f(\gamma_j), f(e))|
\leq d(f(e), f(\gamma)),
\end{equation*}

where $\gamma_j=\gamma_{n_j}(\omega)^{-1}$,
and $\int_{\Gamma} d(f(e),f(\gamma)) d\mu^2(\gamma) < \infty$,
we see that

\begin{equation*}
\lim_{j\to \infty} \int_{\Gamma} \varphi_{\omega,n_j}(\gamma)
= \int_{\Gamma} \varphi_{\omega}(\gamma) d\mu^2(\gamma)
\end{equation*}

by Lebesgue's dominated convergence theorem.
Combining with (<ref>), we get

\begin{equation*}
\Delta_2 \varphi_{\omega}(e)
=\lim_j \Delta_2 \varphi_{\omega,n_j}(e) = 0.
\end{equation*}

Then, by Proposition <ref>, we see
that $\Delta \varphi_{\omega}(\gamma)=0$ for any $\gamma \in \Gamma$.
Now let us denote $\xi(\omega)=\ulim_j f(\gamma_j)$.
If $\xi(\omega) \in Y_{\infty}$, then, since $\varphi_{\omega,n_j}$
converges pointwise to $\varphi_{\omega}$ while
we see that
On the other hand, if $\xi(\omega)\in \partial Y_{\infty}$, then
Lemma <ref> tells us that
Thus we get a $\mu$-harmonic function of the form

\begin{equation*}
\varphi_{\omega}(\gamma)=
\begin{cases}
d(f(\gamma),\xi(\omega)) - d(f(e),\xi(\omega)) & \text{if }
\xi(\omega) \in Y_{\infty}, \\
b_{\xi(\omega)}(f(\gamma),f(e)) & \text{if }
\xi(\omega) \in \partial Y_{\infty}.
\end{cases}
\end{equation*}

What we have shown is the following generalization of
Proposition <ref>.

Let $Y=(Y,d)$ be a $\cat{0}$ space.
Take $o \in Y$ and set $Y_{\infty}=\ulim (Y,o)$.
Let $\Gamma$ be a countable group equipped with a symmetric
probability measure $\mu$ whose support generates $\Gamma$.
Suppose that $\Gamma$ acts on $Y$ via a homomorphism
$\rho\colon \Gamma \rightarrow \isom{Y}$ and that
there exists a $\rho$-equivariant $\mu$-harmonic map
$f\colon \Gamma \rightarrow Y$.
If $l_{\rho}(\Gamma)=0$, then, for almost all $\omega \in \Omega$,
there exists
$\xi(\omega) \in Y_{\infty}\cup \partial Y_{\infty}$ such that a
$\varphi_{\omega} \colon \Gamma \rightarrow \R$ defined by

\begin{equation*}
\varphi_{\omega}(\gamma) =
\begin{cases}
d(\xi(\omega), f(\gamma))-d(\xi(\omega), f(e))
& \text{if } \xi(\omega) \in    Y_{\infty} \\
b_{\xi(\omega)}(f(\gamma), f(e))
& \text{if } \xi(\omega) \in \partial Y_{\infty}
\end{cases}
\end{equation*}

is $\mu$-harmonic$;$ the function $\varphi_{\omega}$ satisfies
$\Delta \varphi_{\omega}(\gamma)=0$ for any $\gamma\in \Gamma$.

§.§ $\cat{0}$ spaces with certain dimension bounds

Let us denote the metric ball of radius $r$ centered at $p$ by
The radius $\rad{Y}{B}$ of $B \subset Y$ is defined to be the
infimum of $r$ such that $B\subset B(p,r)$ for some $p\in Y$.
It is known that, for any bounded subset $B$ of a $\cat{0}$ space $Y$,
there exists a unique point $p_0$ such that
$B\subset B(p_0, \rad{Y}{B})$
(see <cit.>).
Such a point $p_0$ is called the circumcenter of $B$.

The geometric dimension of a metric space with curvature bounded
from above is defined in an inductive manner in [29] and
studied there in detail.
For a $\cat{0}$ space $Y$,
according to <cit.>, $Y$
has geometric dimension at most $n$ if and only if any subset $B$ of
finite diameter satisfies

\begin{equation}
\label{eq:geom_dim}
\rad{Y}{B} \leq \sqrt{\frac{n}{2(n+1)}} \diam{B},
\end{equation}

where $\diam{B}$ denotes the diameter of $B$.
Note that the equality is achieved by a regular Euclidean $n$-simplex.
We say $Y$ is locally finite-dimensional if, for each $R>0$ and
$p \in Y$, there exists $n \in \N$ such that $B(p,R) \subset Y$ has
geometric dimension at most $n$.  (Note that a convex subset of a
$\cat{0}$ space is again a $\cat{0}$ space, and that it inherits the
upper bound on the geometric dimension from the ambient
$\cat{0}$space.  Thus the choice of a point $p$ is irrelevant
A $\cat{0}$ space $Y$ is said to have telescopic dimension at most
$n$ if every asymptotic cone of $Y$ has geometric dimension at most
For example, a $\cat{0}$ space $Y$ has telescopic
dimension at most $1$
if and only if $Y$ is hyperbolic in the sense of Gromov
Again <cit.> tells us that $Y$ has telescopic
dimension at most $n$ if and only if, for any $\delta>0$, there exists
$D>0$ such that every bounded subset $B$ with $\diam{B} > D$

\begin{equation}
\label{eq:tele_dim}
\rad{Y}{B} \leq
\left(\delta +\sqrt{\frac{n}{2(n+1)}} \right) \diam{B}.
\end{equation}

The first part of the following proposition has already been pointed out
in <cit.>; P.-E. Caprace and A. Lytchack used the
term ultracompletion for the ultralmit of a constant sequence of
pointed metric space $(Y,p_0)$ there.
We include, however,  the proof of the result here for the sake of

Let $Y$ be a $\cat{0}$ space and $o \in Y$.  Denote by
$Y_{\infty}$ an ultralimit $\ulim (Y,o)$ with respect to a
nonprincipal ultrafilter $\lambda$.

$(1)$ If $Y$ has telescopic dimension at most $n$, so does

$(2)$ Suppose that, for any $R>0$, there exsits $n \in \N$ such that
any $B \subset B(o, R)$ has geometric dimensin at most $n$.
Then $Y_{\infty}$ has the same property.

$(1)$ Assume that $Y$ has telescopic dimension at most $n$.
We first show that, for any $\delta>0$, there exists $D>0$ such that any
finite subset $B \subset Y_{\infty}$ with $\diam{B}>D$ satisfies

Take any finite subset $B =\{p^{(1)}, \dots, p^{(k)}\}$ in
$Y_{\infty}$ with $\diam{B}>D$,
and fix an expression $p^{(i)}=\ulim_j p^{(i)}_j$ with $p^{(i)}_j\in Y$
for each $p^{(i)} \in B$.
Let $B_j=\{p^{(1)}_j, \dots, p^{(k)}_j\}$ be the set of $j$th elements
in the expression of each $p^{(i)} \in B$.
Since $B_j$ is a finite set, there is $1\leq j(0)< j(1) \leq k$ such
that $d(p^{(j(0))}_{j}, p^{(j(1))}_{j})=\diam{B_j}$.
For $m=0,1$ there exists unique $l_m \in \{1, \dots, k\}$ such that
$\lambda(\{i \in \N \mid  j(m)=l_m\}) = 1$,
since $\lambda$ is an ultrafilter.
Then we see that  $\ulim_i p^{(j(m))}_{i}=p^{(l_m)}$ and

\begin{equation*}
\ulim \diam{B_j}=\ulim
\leq \diam{B}.
\end{equation*}

On the other hand, since, for any
$p=\ulim_j p_j,\ p'= \ulim_j p_j' \in B$,
we have

\begin{equation*}
d(p,p') = \ulim_j d(p_j, p_j') \leq \ulim_j \diam{B_j},
\end{equation*}

which implies $\diam{B} \leq \ulim_j \diam{B_j}$.
Hence we get $\diam{B}= \ulim_j \diam{B_j}$.
In particular, we may assume that $\diam{B_j}> D$ for $\lambda$-almost
all $j \in \N$. Let $p^{(0)}_j$ be a circumcenter of $B_j$.
Take $R>0$ so that $B(o, R)$ contains $B$.  Then we may assume that
$B_j \subset B(o, R+1)$, and that $p^{(0)}_j$'s lie in $B(o,R+1)$.
Thus we obtain a point
$p^{(0)}= \ulim_j p^{(0)}_j \in Y_{\infty}$.
Since, for any $p_j \in B_j$, we have
$d(p^{(0)}_j, p_j) \leq \rad{Y}{B_j}$, we see that
$d(p^{(0)},p) \leq \ulim_j \rad{Y}{B_j}$ for any
$p \in B$, which implies that
$\rad{Y_{\infty}}{B} \leq  \ulim_j \rad{Y}{B_j}$.
Since each $B_j$ is a subset of $Y$ with $\diam{B_j}>D$, we

\begin{equation*}
\ulim \rad{Y}{B_j}
\leq \left( \delta + \sqrt{\frac{n}{2(n+1)}}\right)
\ulim \diam{B_j}
\end{equation*}

by (<ref>), hence we get

\begin{equation}
\label{eq:finite_set}
\rad{Y_{\infty}}{B}
\leq \left( \delta + \sqrt{\frac{n}{2(n+1)}}\right) \diam{B}.
\end{equation}

Now suppose that $Y_{\infty}$ has telescopic
dimension greater than $n$.  Then, by <cit.>,
there exsits a sequence of pointed subsets $\{(Y_j, p_j)\}$ of
$Y_{\infty}$ such that $\ulim_j (Y_j,p_j)$ contains $\R^{n+1}$.
Therefore we can find
$S=\{q^{(0)}, \dots, q^{(n+1)}\}\subset \ulim_j (Y_j,p_j)$ that forms a
regular Euclidean $(n+1)$-simplex with arbitrarily large diameter.
Take $D>0$ for $\delta$  satisfying

\begin{equation}
\label{eq:delta}
0< \delta < \sqrt{\frac{n+1}{2(n+2)}} - \sqrt{\frac{n}{2(n+1)}},
\end{equation}

and choose $S$ so that $\diam{S}>D$.
Expressing $q^{(i)}=\ulim_j q^{(i)}_j$
with $q^{(i)}_j \in Y_j \subset Y_{\infty}$, and setting
$S_j=\{q^{(0)}_j, \dots, q^{(n+1)}_j\}$, we see that

\begin{equation*}
\rad{\ulim_j(Y_j,p_j)}{S}\leq \ulim \rad{Y_j}{S_j},
\quad
\diam{S} = \ulim_j \diam{S_j},
\end{equation*}

and we may assume $\diam{S_j}>D$ as before.
Together with (<ref>) and (<ref>), we get

\begin{equation*}
\rad{\ulim_j (Y_j,p_j)}{S}
\leq \left( \delta + \sqrt{\frac{n}{2(n+1)}}\right) \diam{S}
< \sqrt{\frac{n+1}{2(n+2)}} \diam{S}.
\end{equation*}

This contradicts our assumption that $S$ is a regular Euclidean
Therefore we conclude that $Y_{\infty}$ has telescopic
dimension at most $n$.

$(2)$ Now assume that $Y$ is locally finite-dimensional.
Fix $R>0$ and take any finite subset
$B=\{p^{(1)}, \dots, p^{(k)}\} \subset Y_{\infty}$ so
that $B \subset B(o,R)$.
Fix an expression $p^{(i)}=\ulim_j p^{(i)}_j$ with $p^{(i)}_j \in Y$,
and define $B_j$ as before.
We may assume $p^{(i)}_j \in B(o, R+1) \subset Y$, and hence
there exists $n \in \N$ such that $B_j$ satisfies (<ref>)
with $n$. Thus $B$ also satisfies (<ref>) with the same
$n$ as we have seen above.

Suppose that $B(o,R)\subset Y_{\infty}$ has geometric dimension greater
than $n$.
Then, by <cit.>, there are sequences
$r_j \to \infty$, $S_j \subset B(o,R)$, and $p \in B(o,R)$ such that
$d(S_j,p)\to 0$ and $r_j S_j$ converges to the unit ball in $\R^{n+1}$
in the Gromov-Hausdorff topology.
Then we can find a sequence of $n+2$ points in $B(o,R)$ which converges
to the vertices of a regular Euclidean $(n+1)$-simplex by scaling them
with $r_j$'s.
Then this contradicts the fact that any finite set of
$B(o,R)$ satisfies (<ref>). Thus $B(o,R)$ has
geometric dimension at most $n$.

If $Y$ is proper and $\rho(\Gamma)$ does not fix a point in
$\partial Y$, then
$\rho(\Gamma)$ is reductive in the sense of Jost as we have seen
in the end of  <ref>,
and hence $\{f_j(e)\}$ is bounded for any minimizing sequence $\{f_j\}$.
In particular, we can find a harmonic map, by passing to a subsequence,
as a limit map of $\{f_j\}$ (see also Corollary <ref>).
This is not true in general for nonproper $\cat{0}$ space; we cannot
conclude that $\rho(\Gamma)$ is
reductive in the sense of Jost,
assuming that $\rho(\Gamma)$ does not fix a point in $\partial Y$.
However we can show the following, which is related to

Let $\Gamma$ be a countable group equipped with a symmetric
and nondegenerate probability measure $\mu$.
Let $Y$ be a $\cat{0}$ space with finite telescopic dimension
and $\rho \colon \Gamma \rightarrow \isom{Y}$ a homomorphism.
Assume that the action given by $\rho$ has finite second
moment with respect to $\mu$.