†† 2020 Mathematics Subject Classification. Primary 51F30; Secondary 53C23. ††
Key words and phrases. Non-positively curved spaces,

# A TOPOLOGICAL PRODUCT DECOMPOSITION OF BUSEMANN SPACE

Tomohiro Fukaya

###### Abstract.

We show that a Busemann space $X$ which is covered by parallel bi-infinite
geodesics is homeomorphic to a product of another Busemann space $Y$ and the
real line. We also show that a semi-simple isometry on $X$ preserving the
foliation by parallel geodesics canonically induces a semi-simple isometry on
$Y$.

T.Fukaya was supported by JSPS KAKENHI Grant Number JP19K03471

##  1\. Introduction

There are several generalization of the class of simply connected Riemannian
manifolds with nonpositive sectional curvature. One of the most studied class
is that of CAT(0) spaces. There are many results on the structure of CAT(0)
spaces. One of the most fundamental one is the following. We say that two
geodesics $\gamma,\eta\colon\mathbb{R}\to X$ on a metric space $X$ is parallel
if $d(\gamma(t),\eta(t))$ is bounded.

######  Theorem 1.1 (A Product Decomposition Theorem [2, II 2.14 ]).

Let $X$ be a CAT(0) space and let $c\colon\mathbb{R}\to X$ be a geodesic.

  1. $(1)$

Let $X_{c}$ be the union of the images of all geodesics
$\gamma\colon\mathbb{R}\to X$ parallel to $c$. Then $X_{c}$ is a convex
subspace of $X$.

  2. $(2)$

Let $p$ be the restriction to $X_{c}$ of the nearest point projection from $X$
to the image of the geodesic $c(\mathbb{R})$. Let $X_{c}^{0}:=p^{-1}(c(0))$.
Then, $X_{c}^{0}$ is convex, especially it is a CAT(0) space.

  3. $(3)$

$X_{c}$ is isometric to the product $X_{c}^{0}\times\mathbb{R}$.

The proof of the above decomposition theorem depends on the nice behaviour of
the nearest point projection onto a bi-infinite geodesic. It seems hard to
generalize Theorem 1.1 to more general class of metric spaces called Busemann
spaces, due to the luck of such nice property of the projections. So we
considered slightly weaker statement, and we obtained the following
topological decomposition theorem.

######  Theorem 1.2 (A Topological Product Decomposition Theorem).

Let $X$ be a Busemann space and let $\omega\colon\mathbb{R}\to X$ be a
geodesic.

  1. $(1)$

Let $X_{\omega}$ be the union of the images of all geodesics
$\gamma\colon\mathbb{R}\to X$ parallel to $\omega$. Then $X_{\omega}$ is a
convex subspace of $X$.

  2. $(2)$

Let $\mathbb{X}_{\omega}(\mathbb{R})$ be the set of the images of all
geodesics $\gamma\colon\mathbb{R}\to X$ parallel to $\omega$. We equip
$\mathbb{X}_{\omega}(\mathbb{R})$ with the Hausdorff metric. Then
$\mathbb{X}_{\omega}(\mathbb{R})$ is a Busemann space.

  3. $(3)$

$X_{\omega}$ is homeomorphic to the product
$\mathbb{X}_{\omega}(\mathbb{R})\times\mathbb{R}$.

In the above theorem, (1) is a direct consequence of the Strip Lemma (Lemma
3.5) by Bowditch [1]. Our essential contribution is to show (2) and (3) in
Theorem 1.2.

We also study the classification of isometries on
$\mathbb{X}_{\omega}(\mathbb{R})$ induced by those of on $X_{\omega}$. For a
map $\omega\colon\mathbb{R}\to X$, we denote by $\bar{\omega}$ the map defined
as $\bar{\omega}(t):=\omega(-t)$. We say that an isometry $g\colon
X_{\omega}\to X_{\omega}$ preserve the foliation by $\omega$ if $g\circ\omega$
or $g\circ\bar{\omega}$ is parallel to $\omega$.

######  Theorem 1.3.

Let $X$ be a Busemann space and let $\omega\colon\mathbb{R}\to X$ be a
geodesic. Let $g\colon X_{\omega}\to X_{\omega}$ be an isometry preserving the
foliation by $\omega$. Then $g$ induces an isometry
$g_{\omega}\colon\mathbb{X}_{\omega}(\mathbb{R})\to\mathbb{X}_{\omega}(\mathbb{R})$
by $g_{\omega}(L)=g(L)$ for $L\in\mathbb{X}_{\omega}(\mathbb{R})$. Moreover,
the following holds.

  1. $(1)$

If $g$ is elliptic, then so is $g_{\omega}$.

  2. $(2)$

If $g_{\omega}$ is elliptic, then $g$ is either elliptic or hyperbolic.

  3. $(3)$

Suppose $g$ is hyperbolic. Let $\xi\colon\mathbb{R}\to X$ be an axis of $g$.
If $\xi$ is parallel to either $\omega$ or $\bar{\omega}$, then $g_{\omega}$
is elliptic. Otherwise, $g_{\omega}$ is hyperbolic.

  4. $(4)$

If $g_{\omega}$ is hyperbolic, then so is $g$.

Especially, $g$ is semi-simple if and only if so is $g_{\omega}$.

The organization of the paper is as follows. In Section 2, we briefly review
the definition of Busemann spaces and some terminologies on geodesic
bicombings. We will remark a criterion that provide a sufficient condition for
a geodesic space to be a Busemann space. In Section 3, we construct a geodesic
bicombing on the space of parallel geodesics and we prove (2) of Theorem 1.2.
In Section 4, we introduce the Busemann functions and we prove (3) of Theorem
1.2. In Section 5, we review the standard notations on the classification of
isometries, and then we prove Theorem 1.3.

##  2\. Combing and Busemann space

###  2.1. Busemann space

Let $(X,d)$ be a metric space. A map $\gamma\colon I\to X$ where
$I\subset\mathbb{R}$ is a closed interval is a geodesic if for all $t,s\in I$,
we have $d(\gamma(s),\gamma(t))=\left\lvert t-s\right\rvert$. We say that
$(X,d)$ is a geodesic space if for any pair $(x,y)\in X\times X$, there exists
a geodesic $\gamma\colon[0,d(x,y)]\to X$ with $\gamma(0)=x$ and
$\gamma(d(x,y))=y$. We also say that $(X,d)$ is a uniquely geodesic space if
it is a geodesic space and for any pair $(x,y)\in X\times X$, a geodesic
between $x$ and $y$ is unique. A linearly reparametrised geodesic is a map of
the form $[t\mapsto\gamma(\lambda t)]$ where $\gamma$ is a geodesic and
$\lambda$ is a positive real number.

######  Definition 2.1.

A geodesic space $(X,d)$ is a Busemann space if, for any two linearly
reparametrised geodesics $\gamma,\eta\colon[0,1]\to X$, the map $[t\mapsto
d(\gamma(t),\eta(t))]$ is convex in $t$.

It is easy to see that a Busemann space is a uniquely geodesic space.

###  2.2. Geodesic bicombing

A geodesic bicombing $\sigma$ on $X$ is a map

$\displaystyle\sigma\colon X\times X\times[0,1]\to X$

such that for every pair $(x,y)\in X\times X$, the map
$\sigma_{xy}:=\sigma(x,y,\cdot)\colon[0,1]\to X$ is a linearly reparametrised
geodesic with $\sigma_{xy}(0)=x$ and $\sigma_{xy}(1)=y$.

  1. (1)

We say that $\sigma$ is conical if for all $x,y,x^{\prime},y^{\prime}\in X$
and $t\in[0,1]$, we have

$\displaystyle
d(\sigma_{xy}(t),\sigma_{x^{\prime}y^{\prime}}(t))\leq(1-t)d(\sigma_{xy}(0),\sigma_{x^{\prime}y^{\prime}}(0))+td(\sigma_{xy}(1),\sigma_{x^{\prime}y^{\prime}}(1)).$

  2. (2)

We say that $\sigma$ is convex if the map $[t\mapsto
d(\sigma_{xy}(t),\sigma_{x^{\prime}y^{\prime}}(t))]$ is convex for all
$x,y,x^{\prime},y^{\prime}\in X$, that is, for any $a,b\in[0,1]$ and
$t\in[0,1]$, we have

$\displaystyle
d(\sigma_{xy}((1-t)a+tb),\sigma_{x^{\prime}y^{\prime}}((1-t)a+tb))$
$\displaystyle\leq(1-t)d(\sigma_{xy}(a),\sigma_{x^{\prime}y^{\prime}}(a))+td(\sigma_{xy}(b),\sigma_{x^{\prime}y^{\prime}}(b)).$

  3. (3)

We say that $\sigma$ is consistent if for all $x,y\in X$, $0\leq a\leq b\leq
1$, $p:=\sigma_{xy}(a)$, $q:=\sigma_{xy}(b)$, and $t\in[0,1]$, we have

$\displaystyle\sigma_{pq}(t)=\sigma_{xy}((1-t)a+tb).$

######  Example 2.2.

Let $(X,d)$ be a uniquely geodesic space. Then $X$ admits a geodesic bicombing
$\sigma$ such that $\sigma_{xy}$ is a unique linearly reparametrised geodesic
between $x$ and $y$. We call this the canonical bicombing. Obviously, the
canonical bicombing is consistent.

We remark that if a geodesic bicombing is conical and consistent, then it is
convex. Thus we have the following criterion.

######  Lemma 2.3.

Let $X$ be a uniquely geodesic space. If the canonical bicombing is conical,
then it is convex. Especially, $X$ is a Busemann space.

##  3\. Parallel geodesics in Busemann space

Let $(X,d)$ be a Busemann space.

######  Proposition 3.1.

Let $\gamma\colon\mathbb{R}\to X$ be a geodesic. For $x\in X$, there exists
unique $y\in\gamma(\mathbb{R})$ such that
$d(x,y)=d(x,\gamma(\mathbb{R}))=\inf\\{d(x,w):w\in\gamma(\mathbb{R})\\}$.

The following argument is due to Kato [4].

###### Proof..

Since $\mathbb{R}$ is proper, so is the image $\gamma(\mathbb{R})$. It follows
that there exists $y\in\gamma(\mathbb{R})$ such that
$d(x,y)=d(x,\gamma(\mathbb{R}))$.

Now we show that such $y$ is unique. Set $r:=d(x,\gamma(\mathbb{R}))$. Suppose
that there exists $y,z\in\gamma(\mathbb{R})$ with $y\neq z$ such that
$d(x,y)=d(x,z)=r$. Let $\sigma$ be the canonical bicombing of $X$. Set
$m=\sigma_{yz}(1/2)$. By the convexity, and the minimality, $d(x,m)=r$. By the
convexity of the map $[t\mapsto d(\sigma_{yx}(t),\sigma_{yz}(t))]$, we have
$d(\sigma_{yx}(1/2),m)\leq r/2$. By the triangle inequality,

$\displaystyle r=d(x,m)\leq d(x,\sigma_{yx}(1/2))+d(\sigma_{yx}(1/2),m)\leq
r.$

So $d(\sigma_{yx}(1/2),m)=r/2$. By the same argument, we have
$d(\sigma_{zx}(1/2),m)=r/2$. Then by the uniqueness of the geodesic between
$x$ and $m$, we have $\sigma_{yx}(1/2)=\sigma_{zx}(1/2)$. So set
$x(1):=\sigma_{yx}(1/2)=\sigma_{zx}(1/2)$. Replacing $x$ by $x(1)$, we apply
the same argument and we obtain $x(2)\in X$ such that
$d(x(2),y)=d(x(2),m)=d(x(2),z)=r/2^{2}$. Now we can apply inductively the same
arguments and obtain $x(n)\in X$ for $n\in\mathbb{N}$ such that
$d(x(n),y)=d(x(n),m)=d(x(n),z)=r/2^{n}$. It follows that $d(y,z)\leq
r/2^{n-1}\to 0$ as $n\to\infty$. This is a contradiction. ∎

######  Definition 3.2.

We say that two bi-infinite geodesics $\gamma,\eta\colon\mathbb{R}\to X$ are
parallel if $d(\gamma(t),\eta(t))$ is bounded. This define an equivalence
relation on the set of bi-infinite geodesics.

######  Lemma 3.3.

Let $\gamma,\eta$ be parallel geodesics. For any constant $C\in\mathbb{R}$,
the function

$\displaystyle\mathbb{R}\ni t\mapsto d(\gamma(t),\eta(t+C))\in\mathbb{R}$

is constant.

###### Proof..

Since the map $[t\mapsto d(\gamma(t),\eta(t+C))]$ is convex, if it is bounded,
it must be constant. ∎

For closed subsets $H,K\subset X$, we denote by $\mathcal{H}(H,K)$ the
Hausdorff metric between $H$ and $K$. For bi-infinite geodesics
$\gamma,\eta\colon\mathbb{R}\to X$, we abbreviate the Hausdorff metric of
their images $\mathcal{H}(\gamma(\mathbb{R}),\eta(\mathbb{R}))$ to
$\mathcal{H}(\gamma,\eta)$.

######  Corollary 3.4.

Let $\gamma,\eta$ be parallel geodesics. For any $t\in\mathbb{R}$, we have

$\displaystyle\mathcal{H}(\gamma,\eta)=d(\gamma(t),\eta(\mathbb{R}))=\inf_{s\in\mathbb{R}}d(\gamma(t),\eta(s)).$

Bowditch showed the following Strip Lemma.

######  Lemma 3.5 (Strip Lemma [1]).

Let $(X,d)$ be a Busemann space. Let $r>0$. We suppose that
$\gamma,\gamma^{\prime}\colon\mathbb{R}\to X$ are parallel geodesics such that
$d(\gamma(t),\gamma^{\prime}(t))=r$. For $t\in\mathbb{R}$, let
$\beta_{t}\colon[0,r]\to X$ be the geodesic from $\gamma(t)$ to
$\gamma^{\prime}(t)$. Define a map $\beta\colon\mathbb{R}\times[0,r]\to X$ by
$\beta(t,u):=\beta_{t}(u)$. Then:

  1. (1)

for each $u\in[0,r]$, the map $\gamma_{u}:=[t\mapsto\beta(t,u)]$ is a bi-
infinite geodesic parallel to $\gamma$ and $\gamma^{\prime}$;

  2. (2)

for each $t_{0}\in\mathbb{R}$ and $\lambda\in\mathbb{R}$, the maps
$[u\mapsto\beta(t_{0}+\lambda u,u)]$ is a linearly reparametrised geodesic.

######  Lemma 3.6.

Let $\gamma,\gamma^{\prime}\colon\mathbb{R}\to X$ be parallel geodesics such
that $d(\gamma(0),\gamma^{\prime}(0))=\mathcal{H}(\gamma,\gamma^{\prime})$.
Set $r=\mathcal{H}(\gamma,\gamma^{\prime})$. For $u\in[0,r]$, let $\gamma_{u}$
be the geodesic given by Lemma 3.5 which is parallel to $\gamma$ and
$\gamma^{\prime}$. Then for $u,v\in[0,r]$, we have

$\displaystyle\mathcal{H}(\gamma_{u},\gamma_{v})=\left\lvert u-v\right\rvert.$

###### Proof..

Let $\beta_{0}\colon[0,r]\to X$ be the geodesic from $\gamma(0)$ to
$\gamma^{\prime}(0)$. For $u\in[0,r]$, by the construction of $\gamma_{u}$, we
have $\gamma_{u}(0)=\beta_{0}(u)$, so $\beta_{0}(u)\in\gamma_{u}(\mathbb{R})$.
Then for $0\leq u\leq v\leq r$, we have

$\displaystyle\mathcal{H}(\gamma_{u},\gamma_{v})\leq
d(\beta_{0}(u),\beta_{0}(v))=\left\lvert u-v\right\rvert.$

Suppose $\mathcal{H}(\gamma_{u},\gamma_{v})<\left\lvert u-v\right\rvert$. Then

$\displaystyle r$
$\displaystyle=\mathcal{H}(\gamma,\gamma^{\prime})\leq\mathcal{H}(\gamma,\gamma_{u})+\mathcal{H}(\gamma_{u},\gamma_{v})+\mathcal{H}(\gamma_{v},\gamma^{\prime})$
$\displaystyle<d(\beta_{0}(0),\beta_{0}(u))+d(\beta_{0}(u),\beta_{0}(v))+d(\beta_{0}(v),\beta_{0}(r))=r.$

This is contradiction. Therefore we have
$\mathcal{H}(\gamma_{u},\gamma_{v})=\left\lvert u-v\right\rvert$. ∎

The following is the key to the uniqueness of geodesics in
$\mathbb{X}_{\omega}(\mathbb{R})$.

######  Lemma 3.7.

Let $\gamma_{1},\gamma_{2},\gamma_{3}$ be a pairwise parallel bi-infinite
geodesics in $X$ such that

$\displaystyle\mathcal{H}(\gamma_{1},\gamma_{3})=\mathcal{H}(\gamma_{1},\gamma_{2})+\mathcal{H}(\gamma_{2},\gamma_{3}).$

Let $x\in\gamma_{1}(\mathbb{R})$, $y\in\gamma_{2}(\mathbb{R})$ and
$z\in\gamma_{3}(\mathbb{R})$. If $d(x,y)=\mathcal{H}(\gamma_{1},\gamma_{2})$
and $d(x,z)=\mathcal{H}(\gamma_{1},\gamma_{3})$ holds, then we have
$d(y,z)=\mathcal{H}(\gamma_{2},\gamma_{3})$.

###### Proof..

We choose $z^{\prime}\in\gamma_{3}(\mathbb{R})$ such that
$d(y,z^{\prime})=\mathcal{H}(\gamma_{2},\gamma_{3})$. Then we have

$\displaystyle\mathcal{H}(\gamma_{1},\gamma_{3})$ $\displaystyle\leq
d(x,z^{\prime})\leq d(x,y)+d(y,z^{\prime})$
$\displaystyle=\mathcal{H}(\gamma_{1},\gamma_{2})+\mathcal{H}(\gamma_{2},\gamma_{3})=\mathcal{H}(\gamma_{1},\gamma_{3}).$

It follows that $d(x,z^{\prime})=\mathcal{H}(\gamma_{1},\gamma_{3})$. Since
such $z^{\prime}\in\gamma_{3}(\mathbb{R})$ is unique, we have $z=z^{\prime}$,
and so $d(y,z)=\mathcal{H}(\gamma_{2},\gamma_{3})$. ∎

######  Definition 3.8.

Let $\gamma\colon\mathbb{R}\to X$ be a geodesic. We denote by
$\mathbb{X}_{\omega}$ the set of all geodesic $\gamma\colon\mathbb{R}\to X$
which is parallel to $\omega$. We also denote by
$\mathbb{X}_{\omega}(\mathbb{R})$ the set of images of geodesic in
$\mathbb{X}_{\omega}$, and denote by $X_{\omega}$ the union of the images of
geodesic in $\mathbb{X}_{\omega}$. Thus we have

$\displaystyle\mathbb{X}_{\omega}(\mathbb{R})=\\{\gamma(\mathbb{R}):\gamma\in\mathbb{X}_{\omega}\\},\quad
X_{\omega}=\bigcup_{\gamma\in\mathbb{X}_{\omega}}\gamma(\mathbb{R})=\bigcup_{L\in\mathbb{X}_{\omega}(\mathbb{R})}L.$

It follows from Lemma 3.5 that $X_{\omega}$ is a convex subspace of $X$. We
equip $\mathbb{X}_{\omega}(\mathbb{R})$ with a Hausdorff metric $\mathcal{H}$
so that $(\mathbb{X}_{\omega}(\mathbb{R}),\mathcal{H})$ is a metric space. Now
we construct a geodesic bicombing on $\mathbb{X}_{\omega}(\mathbb{R})$.

######  Definition 3.9.

Let $\gamma,\eta\in\mathbb{X}_{\omega}$. We suppose that
$r:=d(\gamma(0),\eta(0))=\mathcal{H}(\gamma,\eta)$. Let
$\beta\colon\mathbb{R}\times[0,r]\to X$ be a map given in Lemma 3.5. For
$u\in[0,r]$, set $\gamma_{u}=[t\mapsto\beta(t,u)]$ as in Lemma 3.5. Then a map
$[u\mapsto\gamma_{u}(\mathbb{R})]$ is a geodesic from $\gamma(\mathbb{R})$ to
$\eta(\mathbb{R})$ by Lemma 3.6. Therefore we define
$\Sigma(\gamma,\eta,u)=\Sigma_{\gamma\eta}(u)=[u\mapsto\gamma_{ru}(\mathbb{R})]$
for $u\in[0,1]$.

######  Remark 3.10.

Let $\gamma,\eta\in\mathbb{X}_{\omega}$. We suppose that
$r:=d(\gamma(0),\eta(0))=\mathcal{H}(\gamma,\eta)$. Let $a\in\mathbb{R}$. We
define $\gamma^{a},\eta^{a}\colon\mathbb{R}\to X$ by
$\gamma^{a}(t):=\gamma(t+a)$ and $\eta^{a}(t):=\eta(t+a)$ for
$t\in\mathbb{R}$. Then we have
$\Sigma_{\gamma\eta}=\Sigma_{\gamma^{a}\eta^{a}}$.

It follows that $\mathbb{X}_{\omega}(\mathbb{R})$ is a geodesic space. In
fact, we have the following.

######  Proposition 3.11.

$\mathbb{X}_{\omega}(\mathbb{R})$ is a uniquely geodesic space. Therefore
$\Sigma$ is the canonical bicombing.

###### Proof..

Let $\gamma,\eta\in\mathbb{X}_{\omega}$. We suppose that
$r:=d(\gamma(0),\eta(0))=\mathcal{H}(\gamma,\eta)$. Let
$\Gamma\colon[0,r]\to\mathbb{X}_{\omega}(\mathbb{R})$ be a geodesic such that
$\Gamma(0)=\gamma(\mathbb{R})$ and $\Gamma(r)=\eta(\mathbb{R})$. Now we define
a map $\xi\colon[0,r]\to X$ as follows. Set $\xi(0):=\gamma(0)\in\Gamma(0)$.
For $u\in[0,r]$, we choose unique $\xi(u)\in\Gamma(u)$ such that
$d(\xi(0),\xi(u))=\mathcal{H}(\Gamma(0),\Gamma(u))$. Then for $s,t\in[0,r]$,
by Lemma 3.7, we have

$\displaystyle d(\xi(s),\xi(t))=\mathcal{H}(\Gamma(s),\Gamma(t))=\left\lvert
s-t\right\rvert.$

Especially, $d(\gamma(0),\xi(r))=r=d(\gamma(0),\eta(0))$ and
$\xi(r)\in\Gamma(r)=\eta(\mathbb{R})$. So $\xi(r)=\eta(0)$. Therefore $\xi$ is
a geodesic between $\gamma(0)$ and $\eta(0)$. Since $X$ is a uniquely geodesic
space, $\xi$ is equal to the geodesic $[u\mapsto\beta(0,u)]$ where $\beta$ is
a map given by Lemma 3.5 for $\gamma,\eta$. It follows that $\Gamma$ is equal
to the geodesic $\Sigma_{\gamma\eta}$. ∎

######  Lemma 3.12.

Let $\sigma$ be the canonical bicombing of $X$, and let $\Sigma$ be that of
$\mathbb{X}_{\omega}(\mathbb{R})$. For
$\gamma_{1},\gamma_{2}\in\mathbb{X}_{\omega}$, and
$x_{1}\in\gamma_{1}(\mathbb{R}),\,x_{2}\in\gamma_{2}(\mathbb{R})$, we have

$\displaystyle\sigma_{x_{1}x_{2}}(t)\in\Sigma_{\gamma_{1}\gamma_{2}}(t)\quad(t\in[0,1]).$

###### Proof..

We can assume that $x_{1}=\gamma_{1}(0)$ and
$r:=d(\gamma_{1}(0),\gamma_{2}(0))=\mathcal{H}(\gamma_{1},\gamma_{2})$. Then
$x_{2}=\gamma_{2}(\lambda)$ for some $\lambda\in\mathbb{R}$. Let
$\beta\colon\mathbb{R}\times[0,r]\to X$ be a map given by Lemma 3.5 for
parallel geodesics $\gamma_{1},\gamma_{2}$. Then the map
$[u\mapsto\beta(\lambda u,ru);u\in[0,1]]$ is a unique linearly reparametrised
geodesics between $x_{1}$ and $x_{2}$, so it coincides with
$\sigma_{x_{1}x_{2}}$. It follows that $\sigma_{x_{1}x_{2}}(t)=\beta(\lambda
t,rt)\in\Sigma_{\gamma_{1}\gamma_{2}}(t)$. ∎

######  Theorem 3.13.

$\mathbb{X}_{\omega}(\mathbb{R})$ is a Busemann space.

###### Proof..

We denote by $\Sigma$ the canonical bicombing of
$\mathbb{X}_{\omega}(\mathbb{R})$. By Proposition 3.11,
$\mathbb{X}_{\omega}(\mathbb{R})$ is a uniquely geodesic space. So by Lemma
2.3, it is enough to show that the canonical bicombing $\Sigma$ is conical.

Let $\gamma_{1},\gamma_{2},\eta_{1},\eta_{2}\in\mathbb{X}_{\omega}$. We choose
$x_{i}\in\gamma_{i}(\mathbb{R}),\,y_{i}\in\eta_{i}(\mathbb{R})$ for $i=1,2$
such that

$\displaystyle d(x_{1},y_{1})=\mathcal{H}(\gamma_{1},\eta_{1}),\quad
d(x_{2},y_{2})=\mathcal{H}(\gamma_{2},\eta_{2}).$

Let $\sigma$ be the canonical bicombing of $X$. For $t\in[0,1]$, we have

$\displaystyle\sigma_{x_{1}x_{2}}(t)\in\Sigma_{\gamma_{1}\gamma_{2}}(t),\quad\sigma_{y_{1}y_{2}}(t)\in\Sigma_{\eta_{1}\eta_{2}}(t).$

Thus we have

$\displaystyle\mathcal{H}(\Sigma_{\gamma_{1}\gamma_{2}}(t),\Sigma_{\eta_{1}\eta_{2}}(t))$
$\displaystyle\leq d(\sigma_{x_{1}x_{2}}(t),\sigma_{y_{1}y_{2}}(t))$
$\displaystyle\leq(1-t)d(x_{1},y_{1})+td(x_{2},y_{2})$
$\displaystyle=(1-t)\mathcal{H}(\gamma_{1},\eta_{1})+t\mathcal{H}(\gamma_{2},\eta_{2}).$

Therefore $\Sigma$ is conical and this completes the proof. ∎

##  4\. Busemann function and topological splitting

First we introduce the Busemann function.

######  Definition 4.1.

Let $X$ be a metric space and $\eta\colon\mathbb{R}\to X$ be an geodesic. We
define a function $b_{\eta}\colon X\to\mathbb{R}$ by

$\displaystyle b_{\eta}(x)=\lim_{t\to\infty}(d(x,\eta(t))-t),$

and call it the Busemann function associated to $\eta$.

######  Remark 4.2.

Since $d(x,\eta(t))-t$ is non-increasing on $t$ and bounded below, the limit
exists, and $b_{\eta}$ is a 1-Lipschitz function.

######  Lemma 4.3.

Let $X$ be a Busemann space and $\eta\colon\mathbb{R}\to X$ be a geodesic. For
a geodesic $\xi\colon\mathbb{R}\to X$, the composite
$b_{\eta}\circ\xi\colon\mathbb{R}\to\mathbb{R}$ is convex. Moreover, if $\xi$
and $\eta$ are parallel, it is strictly decreasing.

###### Proof..

Since the canonical bicombing is convex, the function $s\mapsto d(x,\xi(t))$
is convex, and so are its translates. Since the Busemann function $b_{\eta}$
is the pointwise limit of the function $[s\mapsto d(\xi(s),\eta(t))-t]$, it is
convex.

Suppose $\sup_{t\in\mathbb{R}}d(\eta(t),\xi(t))<\infty$. Then

$\displaystyle\lim_{s\to\infty}b_{\eta}(\xi(s))=-\infty,\quad\lim_{s\to-\infty}b_{\eta}(\xi(s))=\infty,$

it follows that $b_{\eta}\circ\xi$ is strictly decreasing. ∎

Let $X$ be a Busemann space and $\omega\colon\mathbb{R}\to X$ be a geodesic.
We define a subset $\mathbb{X}_{\omega}^{b}$ of $\mathbb{X}_{\omega}$ and a
subset $\mathbb{X}_{\omega}^{b}(t)$ of $X$ where $t\in\mathbb{R}$ as follows:

$\displaystyle\mathbb{X}_{\omega}^{b}:=$
$\displaystyle\left\\{\gamma\in\mathbb{X}_{\omega}:b_{\omega}(\gamma(0))=0\right\\},$
$\displaystyle\mathbb{X}_{\omega}^{b}(t):=$
$\displaystyle\\{\gamma(t):\gamma\in\mathbb{X}_{\omega}^{b}\\}.$

There exists a bijection between $\mathbb{X}_{\omega}^{b}$ and
$\mathbb{X}_{\omega}(\mathbb{R})$. Therefore we identify
$\mathbb{X}_{\omega}^{b}$ with $\mathbb{X}_{\omega}(\mathbb{R})$, and we
regard $(\mathbb{X}_{\omega}^{b},\mathcal{H})$ as a metric space.

######  Lemma 4.4.

For $t\in\mathbb{R}$, define a map
$\mathrm{Ev}_{t}\colon\mathbb{X}_{\omega}^{b}\to\mathbb{X}_{\omega}^{b}(t)$ by
$\mathrm{Ev}_{t}(\gamma):=\gamma(t)$. Then $\mathrm{Ev}_{t}$ is continuous.

###### Proof..

It is enough to show that $\mathrm{Ev}_{0}$ is continuous. Indeed, for
$\gamma,\gamma^{\prime}\in\mathbb{X}_{\omega}^{b}$, by Lemma 3.3, we have
$d(\gamma(t),\gamma^{\prime}(t))=d(\gamma(0),\gamma^{\prime}(0))$. Therefore,
if $\mathrm{Ev}_{0}$ is continuous, so is $\mathrm{Ev}_{t}$ for all
$t\in\mathbb{R}$.

Let $\gamma\in\mathbb{X}_{\omega}^{b}$. We fix $\epsilon>0$. Set
$\delta^{\prime}:=\min\\{\left\lvert
b_{\omega}(\gamma(\epsilon/2))\right\rvert,\,b_{\omega}(\gamma(-\epsilon/2))\\}$.
By Lemma 4.3, we have $\delta^{\prime}>0$. Set
$\delta:=\min\\{\delta^{\prime}/2,\epsilon/2\\}$. Here we use Lemma 4.3 again,
we have

(1) $\displaystyle\left\lvert
b_{\omega}(\gamma(t))\right\rvert>\delta^{\prime}\quad\text{for all
}t\in\mathbb{R}\text{ with }\left\lvert t\right\rvert>\epsilon/2.$

Now let $\gamma^{\prime}\in\mathbb{X}_{\omega}^{b}$ such that
$\mathcal{H}(\gamma,\gamma^{\prime})<\delta$. There exists $t\in\mathbb{R}$
such that $d(\gamma(t),\gamma^{\prime}(0))<\delta$. We suppose that
$d(\gamma(0),\gamma^{\prime}(0))>\epsilon$. Then we have

$\displaystyle\left\lvert t\right\rvert=d(\gamma(0),\gamma(t))\geq
d(\gamma(0),\gamma^{\prime}(0))-d(\gamma^{\prime}(0),\gamma(t))>\epsilon-\delta\geq\epsilon/2.$

Since $b_{\omega}$ is 1-Lipschitz, using (1), we have

$\displaystyle\left\lvert
b_{\omega}(\gamma^{\prime}(0))\right\rvert>\left\lvert
b_{\omega}(\gamma(t))\right\rvert-\delta\geq\delta^{\prime}/2>0.$

This contradicts that $\gamma^{\prime}\in\mathbb{X}_{\omega}^{b}$. Thus we
have $d(\gamma(0),\gamma^{\prime}(0))\leq\epsilon$. Therefore
$\mathrm{Ev}_{0}$ is continuous. ∎

######  Theorem 4.5.

The map

$\displaystyle\mathrm{Ev}\colon\mathbb{X}_{\omega}^{b}\times\mathbb{R}\to
X_{\omega},\quad\mathrm{Ev}(\gamma,t)=\gamma(t)$

is homeomorphism.

###### Proof..

First we show that $\mathrm{Ev}$ is continuous. By Lemma 4.4, for
$(\gamma,t)\in\mathbb{X}_{\omega}^{b}\times\mathbb{R}$ and $\epsilon>0$, there
exists $\delta>0$ such that for all
$\gamma^{\prime}\in\mathbb{X}_{\omega}^{b}$ with
$\mathcal{H}(\gamma,\gamma^{\prime})<\delta$, we have
$d(\mathrm{Ev}_{t}(\gamma),\mathrm{Ev}_{t}(\gamma^{\prime}))=d(\gamma(t),\gamma^{\prime}(t))<\epsilon/2$.
Then for
$(\gamma^{\prime},t^{\prime})\in\mathbb{X}_{\omega}^{b}\times\mathbb{R}$ with
$\mathcal{H}(\gamma,\gamma^{\prime})<\delta$ and $\left\lvert
t-t^{\prime}\right\rvert<\epsilon/2$, we have

$\displaystyle
d(\mathrm{Ev}(\gamma,t),\mathrm{Ev}(\gamma^{\prime},t^{\prime}))=d(\gamma(t),\gamma^{\prime}(t^{\prime}))\leq
d(\gamma(t),\gamma^{\prime}(t))+d(\gamma^{\prime}(t),\gamma^{\prime}(t^{\prime}))<\epsilon.$

Therefore $\mathrm{Ev}$ is continuous.

Now we will see that the inverse is also continuous. We define a map $q\colon
X_{\omega}\to\mathbb{X}_{\omega}^{b}$ such that for $x\in X_{\omega}$, we have
$x\in q(x)(\mathbb{R})$. Since for $x\in X_{\omega}$ there exists unique
$\gamma\in\mathbb{X}_{\omega}^{b}$ with $x\in\gamma(\mathbb{R})$, this map is
well-defined. For $x,x^{\prime}\in\mathbb{X}_{\omega}(\mathbb{R})$, by
Corollary 3.4, we have $\mathcal{H}(q(x),q(x^{\prime}))\leq d(x,x^{\prime})$.
So the map $q$ is continuous. We also define a map $\theta\colon
X_{\omega}\to\\{1,-1\\}$ by

$\displaystyle\theta(x)=\begin{cases}1&\text{if }x\in q(x)([0,\infty))\\\
-1&\text{else}\end{cases}$

Then the map $x\mapsto\theta(x)d(x,q(x)(0))$ is continuous. Since the inverse
$\mathrm{Ev}^{-1}$ is given by
$\mathrm{Ev}^{-1}(x)=(q(x),\theta(x)d(x,q(x)(0)))$, it is continuous. ∎

######  Corollary 4.6.

The map $\mathrm{Ev}_{t}$ defined in Lemma 4.4 is a homeomorphism.

Using the fact from the dimension theory [3, Theorem 5.11.], we have the
following.

######  Corollary 4.7.

If the topological dimension of $X$ is finite, we have

$\displaystyle\mathop{\mathrm{dim}}\nolimits\mathbb{X}_{\omega}(\mathbb{R})\leq\mathop{\mathrm{dim}}\nolimits
X-1.$

######  Lemma 4.8.

Let $\gamma$ be a geodesic which is parallel to $\omega$. For any
$u,s\in\mathbb{R}$,

$\displaystyle b_{\omega}(\gamma(u+s))=b_{\omega}(\gamma(u))-s.$

###### Proof..

By Lemma 3.3,

$\displaystyle b_{\omega}(\gamma(u+s))$
$\displaystyle=\lim_{t\to\infty}\left\\{d(\gamma(u+s),\omega(t))-t\right\\}$
$\displaystyle=\lim_{t\to\infty}\left\\{d(\gamma(u),\omega(t-s))-(t-s)-s\right\\}$
$\displaystyle=b_{\omega}(\gamma(u))-s.$

∎

######  Corollary 4.9.

For $\gamma\in\mathbb{X}_{\omega}^{b}$ and $t\in\mathbb{R}$, we have
$\mathrm{Ev}(\gamma,t)\in b_{\omega}^{-1}(-t)$.

##  5\. isometries

In this section, we consider a classification of isometries induced on the
first factor of the decomposition in Theorem 1.2.

First we review standard terminologies. Let $X$ be a metric space and $f\colon
X\to X$ be an isometry. The displacement function $d_{f}\colon X\to\mathbb{R}$
of $f$ is defined as $d_{f}(x)=d(x,f(x))$ for $x\in X$. The minimal
displacement of $f$, denoted by $\left\lvert f\right\rvert$, is the infimum of
$d_{f}$, that is,

$\displaystyle\left\lvert f\right\rvert=\inf_{x\in X}d_{f}(x).$

The minimal set of $f$, denoted by $\mathrm{Min}(f)$, is the subset defined as

$\displaystyle\mathrm{Min}(f)=\\{x\in X:d_{f}(x)=\left\lvert
f\right\rvert\\}.$

We say that an isometry $f\colon X\to X$ is

  1. (i)

parabolic if $\mathrm{Min}(f)=\emptyset$,

  2. (ii)

elliptic if $\mathrm{Min}(f)\neq\emptyset$ and $\left\lvert f\right\rvert=0$,

  3. (iii)

hyperbolic if $\mathrm{Min}(f)\neq\emptyset$ and $\left\lvert
f\right\rvert>0$,

  4. (iv)

semi-simple if $f$ is elliptic or hyperbolic.

It is easy to see that $f$ is elliptic if and only if $f$ has a fixed point.

Let $f\colon X\to X$ be an isometry. We say that $f$ is axial if $f$ has no
fixed point and there exists a geodesic $\gamma\colon\mathbb{R}\to X$ whose
image is setwise invariant by $f$. We call such $\gamma$ an axis of $f$.

######  Proposition 5.1 ([5, Proposition 11.2.10]).

Let $f\colon X\to X$ be an axial isometry and $\gamma\colon\mathbb{R}\to X$ be
an axis of $f$. Set $b_{\gamma}(f)=-b_{\gamma}(f(\gamma(0)))$. Then we have
$\left\lvert b_{\gamma}(f)\right\rvert=\left\lvert f\right\rvert$ and for all
$t\in\mathbb{R}$, we have

$\displaystyle f\circ\gamma(t)=\gamma(t+b_{\gamma}(f)).$

Axial isometries are hyperbolic ([5, Corollary 11.2.9]). The converse holds if
$X$ is a Busemann space.

######  Proposition 5.2 ([5, Corollary 11.4.4]).

Let $X$ be a Busemann space and $f\colon X\to X$ be an isometry. Then $f$ is
hyperbolic if and only if $f$ has an axis.

Let $X$ be a Busemann space and let $\omega\colon\mathbb{R}\to X$ be a
geodesic. We define a geodesic $\bar{\omega}\colon\mathbb{R}\to X$ by
$\bar{\omega}(t)=\omega(-t)$.

######  Definition 5.3.

We say that an isometry $g\colon X_{\omega}\to X_{\omega}$ preserve the
foliation by $\omega$ if $g\circ\omega$ or $g\circ\bar{\omega}$ is parallel to
$\omega$.

Let $g\colon X_{\omega}\to X_{\omega}$ be an isometry preserving the foliation
by $\omega$. For $\gamma,\eta\in\mathbb{X}_{\omega}$, we have
$g\circ\gamma(\mathbb{R}),g\circ\eta(\mathbb{R})\in\mathbb{X}_{\omega}(\mathbb{R})$,
and
$\mathcal{H}(\gamma(\mathbb{R}),\eta(\mathbb{R}))=\mathcal{H}(g\circ\gamma(\mathbb{R}),g\circ\eta(\mathbb{R}))$.
Thus we define an isometry
$g_{\omega}\colon\mathbb{X}_{\omega}(\mathbb{R})\to\mathbb{X}_{\omega}(\mathbb{R})$
by $g_{\omega}(\gamma(\mathbb{R})):=g\circ\gamma(\mathbb{R})$.

######  Proposition 5.4.

Let $g\colon X_{\omega}\to X_{\omega}$ be an isometry preserving the foliation
by $\omega$. If $g$ is elliptic, then so is $g_{\omega}$.

###### Proof..

If $g$ is elliptic, then there exists $x\in X_{\omega}$ such that $g(x)=x$.
Then for $L\in\mathbb{X}_{\omega}(\mathbb{R})$ with $x\in L$, we have
$g_{\omega}(L)=L$. It follows that $g_{\omega}$ is elliptic. ∎

######  Proposition 5.5.

Let $g\colon X_{\omega}\to X_{\omega}$ be an isometry preserving the foliation
by $\omega$. If $g_{\omega}$ is elliptic, then $g$ is either elliptic or
hyperbolic.

###### Proof..

If $g_{\omega}$ is elliptic, then $g_{\omega}$ has a fixed point, that is,
there exists $\gamma\in\mathbb{X}_{\omega}$ such that
$g\circ\gamma(\mathbb{R})=\gamma(\mathbb{R})$. So $\gamma$ is an invariant
geodesic of $g$. Therefore $g$ is elliptic or hyperbolic. ∎

######  Proposition 5.6.

Let $g\colon X_{\omega}\to X_{\omega}$ be a hyperbolic isometry preserving the
foliation by $\omega$. Let $\xi\colon\mathbb{R}\to X$ be an axis of $g$. If
$\xi$ is parallel to either $\omega$ or $\bar{\omega}$, then $g_{\omega}$ is
elliptic. Otherwise, $g_{\omega}$ is hyperbolic.

###### Proof..

Suppose that $\xi$ is parallel to either $\omega$ or $\bar{\omega}$, Then
$g\circ\omega(\mathbb{R})=\omega(\mathbb{R})$. So $g_{\omega}$ is elliptic.

Now we suppose that $\xi$ is parallel to neither $\omega$ nor $\bar{\omega}$.
We define a map $\Gamma\colon\mathbb{R}\to\mathbb{X}_{\omega}(\mathbb{R})$
such that $\xi(t)\in\Gamma(t)$. Since $\xi$ is parallel to neither $\omega$ or
$\bar{\omega}$, such $\Gamma(t)$ in $\mathbb{X}_{\omega}(\mathbb{R})$ uniquely
exists, so $\Gamma$ is well-defined. Then $\Gamma$ is a geodesic. Indeed, let
$\sigma$ be the canonical bicombing of $X$, and let $\Sigma$ be that of
$\mathbb{X}_{\omega}(\mathbb{R})$. For $u,v\in\mathbb{R}$ with $u\leq v$ and
for $t\in[u,v]$, by the construction of $\Sigma$, we have

(2)
$\displaystyle\xi(t)=\sigma_{\xi(u)\xi(v)}((t-u)/(v-u))\in\Sigma_{\Gamma(u)\Gamma(v)}((t-u)/(v-u)).$

It follows that $\Gamma(t)=\Sigma_{\Gamma(u)\Gamma(v)}((t-u)/(v-u))$.
Therefore $\Gamma$ is a geodesic.

Since $g(\xi(\mathbb{R}))=\xi(\mathbb{R})$, we have
$g_{\omega}(\Gamma(\mathbb{R}))=\Gamma(\mathbb{R})$. Thus $\Gamma$ is an
invariant bi-infinite geodesic on which $g_{\omega}$ acts as a translation.
Therefore $g_{\omega}$ is hyperbolic. ∎

The following Lemma is a direct consequence of Lemma 3.5 and the uniqueness of
geodesic in $\mathbb{X}_{\omega}(\mathbb{R})$ (Proposition 3.11).

######  Lemma 5.7 (Flat plane).

Let $\Gamma\colon\mathbb{R}\to\mathbb{X}_{\omega}(\mathbb{R})$ be a bi-
infinite geodesic in $\mathbb{X}_{\omega}(\mathbb{R})$. We choose
$x\in\Gamma(0)$ as a base point. For $u\in\mathbb{R}$, we choose
$\gamma_{u}\in\mathbb{X}_{\omega}$ such that, $\gamma_{0}(0)=x$, and for all
$u\in\mathbb{R}$, $\gamma_{u}(\mathbb{R})=\Gamma(u)$ and

$\displaystyle
d(x,\gamma_{u}(0))=d(\gamma_{0}(0),\gamma_{u}(0))=\mathcal{H}(\gamma_{0},\gamma_{u})=u.$

We define a map $\beta\colon\mathbb{R}^{2}\to X_{\omega}$ by
$\beta(t,u):=\gamma_{u}(t)$. Then a map $[u\mapsto\beta(t_{0}+\lambda u,u)]$
is a linearly reparametrised geodesic where $t_{0},\lambda\in\mathbb{R}$ are
constants.

######  Proposition 5.8.

Let $g\colon X_{\omega}\to X_{\omega}$ be an isometry preserving the foliation
by $\omega$. If $g_{\omega}$ is hyperbolic, so is $g$.

###### Proof..

Suppose $g_{\omega}$ is hyperbolic. For simplicity, we assume that
$g\circ\omega$ is parallel to $\omega$.

Let $\Gamma\colon\mathbb{R}\to\mathbb{X}_{\omega}(\mathbb{R})$ be an axis of
$g_{\omega}$. We choose a base point $x\in\Gamma(0)$. For $u\in\mathbb{R}$, we
choose $\gamma_{u}\in\mathbb{X}_{\omega}$ as in Lemma 5.7 and define the map
$\beta$ by $\beta(t,u):=\gamma_{u}(t)$. Let $r>0$ be the translation length of
$g_{\omega}$. Then we have $g(\gamma_{0}(\mathbb{R}))=\gamma_{r}(\mathbb{R})$.
Since $g(x)=g(\gamma_{0}(0))\in\gamma_{r}(\mathbb{R})$ there uniquely exists
$s_{0}\in\mathbb{R}$ such that $g(x)=\gamma_{r}(s_{0})$. By Lemma 3.3,

$\displaystyle
d(g(\gamma_{0}(0)),\gamma_{0}(s_{0}))=d(\gamma_{r}(s_{0}),\gamma_{0}(s_{0}))=d(\gamma_{r}(0),\gamma_{0}(0))=\mathcal{H}(\gamma_{r},\gamma_{0})=r.$

We show inductively that for $n\in\mathbb{Z}$, we have

$\displaystyle g^{n}(x)=g^{n}\circ\gamma_{0}(0)=\gamma_{rn}(s_{0}n).$

We suppose that $g^{n-1}\circ\gamma_{0}(0)=\gamma_{r(n-1)}(s_{0}(n-1))$ for
positive $n$. Then $g^{n-1}\circ\gamma_{0}(s_{0})=\gamma_{r(n-1)}(s_{0}n)$. So
we have

$\displaystyle d(g^{n}\circ\gamma_{0}(0),\gamma_{r(n-1)}(s_{0}n))$
$\displaystyle=d(g^{n}\circ\gamma_{0}(0),g^{n-1}\circ\gamma_{0}(s_{0}))$
$\displaystyle=d(g\circ\gamma_{0}(0),\gamma_{0}(s_{0}))$ $\displaystyle=r$
$\displaystyle=\mathcal{H}(\gamma_{nr},\gamma_{(n-1)r}).$

Then we have $g^{n}\circ\gamma_{0}(0)=\gamma_{rn}(s_{0}n)$. For negative $n$,
we can prove in the same way.

Now by Lemma 5.7, a map $[u\mapsto\beta(s_{0}u,ru)=\gamma_{ru}(s_{0}u)]$ is a
linearly reparametrised geodesic on which the orbit of $x$ by $g$ lies. It
follows that this is an axis of $g$, and so $g$ is hyperbolic. ∎

Propositions 5.4, 5.5, 5.6 and 5.8 implies Theorem 1.3.

## References

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Tomohiro Fukaya

Department of Mathematical Sciences, Tokyo Metropolitan University, Minami-
osawa Hachioji, Tokyo, 192-0397, Japan

E-mail address:

E-mail address<EMAIL_ADDRESS>