Cohomogeneity Two Nonsemisimple Isometric Actions

Abstract

We describe the orbits of a cohomogeneity two Riemannian G-manifold M from topological point of view, under the conditions that G is nonsemisimple and M decomposes as a product of negatively curved Riemannian manifolds.

1 Introduction

A manifold on which a group G acts is called a G-manifold. In this paper, we consider a complete Riemannian manifold M with the action of a closed and connected Lie subgroup G of isometries. Dimension of the orbit space is called the cohomogeneity of the action. Manifolds having actions of cohomogeneity zero are called homogeneous. A classic theorem about Riemannian manifolds of nonpositive curvature [22] states that a homogeneous Riemannian manifold M of nonpositive curvature is simply connected or it is diffeomorphic to a cylinder over a torus (i.e., it is diffeomorphic to $\mathbb {R}^{k} \times T^{s}$, $k+s=\text {dim} M$), and a theorem by S. Kobayashi [11] states that a homogeneous Riemannian manifold of negative curvature is simply connected. Therefore, it is diffeomorphic to $\mathbb {R}^{n}$, $n =$ dimM. There are many interesting theorems about topological properties of cohomogeneity one G-manifolds of nonpositive curvature [2, 17, 18, 21].

The authors of [21] studied cohomogeneity one G-manifolds of negative curvature. Among other results, they proved that if M is a nonsimply connected negatively curved cohomogeneity one Riemannian G-manifold and dim$(M) \ge 3$, then either M is diffeomorphic to $\mathbb {R}^{k} \times T^{s}$, $k+s=\text {dim} M$ or $\pi _{1}(M)=Z$ and the principal orbits are covered by $S^{n-2} \times \mathbb {R}$, $n=$dimM, and $\frac{M}{G}$ is homeomorphic to one of the spaces R and $[0, \infty )$. Also, topological properties of cohomogeneity one Riemannian manifolds of nonpositive curvature has been studied by many authors. But, classification of orbits and orbit spaces of cohomogeneity two Riemannian G-manifolds of nonpositive curvature is an open problem. This article follows previous papers [13,14,15,16], where we proved various results about topological properties of cohomogeneity two Riemannian G-manifolds. One of the main examples of Riemannian manifolds of nonpositive curvature is the product $M=M_{1} \times \cdots \times M_{k}$ of Riemannian manifolds such that $M_{i}$, $1\le i\le k$, has negative curvature. In the paper [15], we studied the orbits of cohomogeneity two G-manifolds of this kind, under the condition that $M^{G} \ne \emptyset$. In the present paper, we replace the condition $M^{G}\ne \emptyset$ by the condition that G is nonsemisimple and there is no non-principal orbit of positive dimension. Among other results, we show that if M is not simply connected then either it is homeomorphic to the product of a cohomogeneity one G-manifold with $\mathbb {R}$, or $\pi _{1}(M)=Z^{p}$, $p\ge 1$. Our main result is Theorem 3.2.

The paper is organized as follows: We recall some definitions and prove some statements about Riemannian manifolds of nonpositive curvature in Preliminaries. In the section of results, first we mention a remark about the relations of the orbits of a G-manifold and the orbits of its universal covering manifold, which is important in the proof of our theorem. Then, we give our main theorem and its proof.

2 Preliminaries

In what follows, M is a Riemannian manifold, G is a closed and connected subgroup of the isometries of M, and $M^{G}=\{ x \in M: G(x)=x \}$. All geodesics of M are considered to have unit speed.

A complete connected and simply connected Riemannian manifold of nonpositive curvature is called a Hadamard manifold. An isometry $\phi$ of a Hadamard manifold M is called elliptic if it has fixed point. $\phi$ is called hyperbolic (parabolic) if the function $d^{2}_{\phi }:M \rightarrow \mathbb {R}$ defined by $d^{2}_{\phi }(x)=d^{2}(x,\phi (x))$ has minimum point (has no minimum).

An isometry $\phi$ is called axial if there is a geodesic $\gamma$ translated by $\phi$ ( there exists a positive constant c such that $\phi (\gamma (t))=\gamma (t+c)$). $\gamma$ is called the axis of $\phi$.

If $\gamma$ is a geodesic in the Hadamard manifold M, we denote by $[\gamma ]$ the collection of all geodesics which are asymptotic to $\gamma$. The collection of all asymptotic classes of the geodesics of M is denoted by $M(\infty )$ and is called the ideal boundary of M (see [8] for details). In fact, we can imagine $M\bigcup M(\infty )$ as a manifold with boundary such that M is its interior and $M(\infty )$ is its boundary.

Remark 2.1

If $M_{1}$ and $M_{2}$ are Hadamard manifolds then their product $M=M_{1} \times M_{2}$ is also a Hadamard manifold. Consider unit speed geodesics $\gamma _{1}$ and $\gamma _{2}$ in $M_{1}$ and $M_{2}$ and put $\gamma (t)=(\gamma _{1}(\frac{\sqrt{2}}{2}t), \gamma _{2}(\frac{\sqrt{2}}{2}t))$. Then, $\gamma$ is a unit speed geodesic in M. Let $\eta _{1}$ and $\eta _{2}$ be other unit speed geodesics in $M_{1}$ and $M_{2}$ and $\eta$ be a geodesic in M, defined in the similar way as $\gamma$. If $[\gamma _{1}]=[\eta _{1}]$ and $[\gamma _{2}]=[\eta _{2}]$, then it is easy to show that $[\gamma ]=[\eta ]$. Consider the map $\rho : M_{1}(\infty ) \times M_{2}(\infty ) \rightarrow M(\infty )$, defined by $\rho ([\gamma _{1}],[\gamma _{2}])=[\gamma ]$. Clearly, $\rho$ is a one-to-one and well-defined map. Then, without loss of precision, we can denote $[\gamma ]$ by $([\gamma _{1}],[\gamma _{2}])$. In this way, $M_{1}(\infty ) \times M_{2}(\infty )$ can be considered as a subset of $M(\infty )$. We call a point $z \in M(\infty )$ a regular point at infinity, if z belongs to $M_{1}(\infty ) \times M_{2}(\infty )$ ( that is $z=(z_{1}, z_{2}), z_{1} \in M_{1}(\infty ), z_{2} \in M_{2}(\infty )$). A regular point in the infinity of the product of any finite number of Hadamard manifolds can be defined similarly.

Remark 2.2

A Hadamard manifold M satisfies Axiom 1 (see [8]) if for any distinct points $x, y \in M(\infty )$ there exists a geodesic $\gamma$ joining x to y.

Lemma 2.3

Let $\phi$ be a parabolic isometry on a Hadamard manifold M.

• (1) If z is a fixed point at infinity for $\phi$, then $\phi$ leaves each horosphere S centered at z invariant (see [8] for definition of the horosphere).

• (2) If M has strictly negative curvature, then there is a unique fixed point $z \in M(\infty )$ for $\phi$.

Proof

(1) is a direct consequence of [6, Lemma 3]. If M has strictly negative curvature then by [4, Lemma 9.10], M satisfies Axiom 1. Thus, by [8, Theorem 6.5], we get (2). $\square$

Lemma 2.4

If H is a closed and connected solvable subgroup of the isometries of a Riemannian manifold M of strictly negative curvature, then one of the following statements is true:

• (1) $M^{H} \ne \emptyset$.

• (2) H translates a unique geodesic.

• (3) There is a unique point in $M(\infty )$ fixed by H.

Proof

By [6, Theorem 5], one of the following statements is true:

• (1) $M^{H} \ne \emptyset$.

• (2) H translates a geodesic.

• (3) There is a point in $M(\infty )$ fixed by H.

Uniqueness of the geodesic in (2) comes from [4, Proposition 4.2]. If (1) and (2) are not true, then all elements of H are parabolic and uniqueness of the fixed point in (3) comes from Lemma 2.3(2). $\square$

Lemma 2.5

Let $M=M_{1} \times M_{2} \times \cdots \times M_{m}$ such that for each i, $M_{i}$ is a Hadamard manifold of negative curvature, and suppose that $\phi \in Iso(M)$ can be decomposed as the product $\phi =\phi _{1} \times \phi _{2} \times \cdots \times \phi _ {m}$ of non-elliptic isometries $\phi _{i} \in Iso(M_{i})$. If there is a geodesic $\beta =(\beta _{1},\ldots ,\beta _{m})$ such that $\phi (\beta )=\beta$, then $\beta$ is unique.

Proof

Let $\gamma =(\gamma _{1},\ldots , \gamma _{m})$ be another geodesic in M such that $\phi (\gamma )=\gamma$. Then, for each i, $\phi _{i}(\gamma _{i})=\gamma _{i}$ and $\phi _{i}(\beta _{i})=\beta _{i}$. We get from the uniqueness of the geodesic left invariant by non-elliptic isometry $\phi _{i}$ [4, Proposition 4.2] that $\gamma _{i}=\beta _{i}$. Thus, $\beta =\gamma$. $\square$

Remark 2.6

Let G be a connected solvable Lie subgroup of the isometries of $M=M_{1}\times M_{2}\times \cdots \times M_{m}$ such that for each i, $M_{i}$ is simply connected with strictly negative curvature. Since G is connected then each $g \in G$ can be decomposed as $g=g_{1}\times \cdots \times g_{m}$, $g_{i} \in Iso(M_{i})$ (see [10, Vol. 1, p. 240]). For each $i \in \{1,\ldots ,m\}$, let $P_{i}:G \rightarrow Iso(M_{i})$ be the map defined by $P_{i}(g_{1},\cdots g_{i},\ldots ,g_{n})=g_{i}$. Put $G_{i}=P_{i}(G)$. Clearly, $G_{i}$ is a closed and connected solvable subgroup of the isometries of $M_{i}$, and we have:

$$\begin{aligned} G\subset G_{1} \times \cdots \times G_{m} \ \ \ (*) \end{aligned}$$

Lemma 2.7

Under the assumptions of Remark 2.6, there is a unique subset $\Omega$ of M with the property that $G(\Omega )=\Omega$ and $\Omega$ is one of the following sets or a product of them.

• (1) $M^{G}$ or product of the fixed point set of some elements of $\{ G_{1},\ldots ,G_{m}\}$.

• (2) Image of a geodesic.

• (3) A regular point at infinity of the product of some elements of the set $\{ M_{1},\ldots ,M_{m}\}$.

Proof

Since $M_{i}$ has strictly negative curvature, then by Remark 2.4, one of the following statements is true:

• (I) For all i, $M_{i}^{G_{i}} \ne \emptyset$.

• (II) For each i, there is a unique geodesic $\gamma _{i}$ in $M_{i}$ such that $G_{i}(\gamma _{i})=\gamma _{i}$.

• (III) There is a unique point $\zeta _{i} \in M_{i}(\infty )$ for each i, such that $G_{i}(\zeta _{i})=\zeta _{i}$.

• (IV) $\{ G_{1}, \ldots , G_{m}\}$ is a collection of the groups with properties (I) or (II) or (III).

In the case (I), $(M_{1} \times \cdots \times M_{m})^{(G_{1} \times \cdots \times G_{m})} \ne \emptyset$, then $M^{G} \ne \emptyset$.

In the case (II), $\gamma =(\gamma _{1},\ldots ,\gamma _{m})$ is a geodesic such that $(G_{1} \times \cdots \times G_{m})(\gamma )=\gamma$.

Thus, $G(\gamma )=\gamma$ and by Lemma 2.5, $\gamma$ is unique. In a similar way if the case (III) is true, $\zeta =(\zeta _{1},\ldots ,\zeta _{m})$ will be the unique regular point in $M(\infty )$ fixed by G. Now, it is easy to show that if (IV) is true, then there is a unique set $\Omega$, which is a product of the sets similar to (I), (II), (III), such that $G(\Omega )=\Omega$. $\square$

3 Results

Remark 3.1

[5]. If M is a complete and connected Riemannian manifold and G is a connected subgroup of Iso(M), and if $\widetilde{M}$ is the universal Riemannian covering manifold of M with the covering map $\kappa : \widetilde{M} \rightarrow M$, then there is a connected covering $\widetilde{G}$ of G with the covering map $\pi : \widetilde{G} \rightarrow G$, such that $\widetilde{G}$ acts isometrically on $\widetilde{M}$ and

• (1) Each deck transformation $\delta$ of the covering $\kappa : \widetilde{M} \rightarrow M$ maps $\widetilde{G}$-orbits onto $\widetilde{G}$-orbits.

• (2) If $x \in M$ and $\widetilde{x}\in \widetilde{M}$ such that $\kappa (\widetilde{x})=x$ then $\kappa (\widetilde{G}(\widetilde{x}))=G(x)$.

• (3) $\widetilde{M}^{\widetilde{G}}= \kappa ^{-1}( M^{G}).$

• (4) If G is non-semisimple then $\widetilde{G}$ is non-semisimple.

• (5) Deck transformation group, which we denote it by $\Delta$, centralizes $\widetilde{G}$ (i.e., for each $\delta \in \Delta$ and $\widetilde{g}\in \widetilde{G} , \delta \widetilde{g} = \widetilde{g}\delta$).

Remark 3.2

Let $\widetilde{M}$ be a Hadamard manifold and S be a horosphere in $\widetilde{M}$ related to the asymptotic class of geodesics $[\gamma ]$ (i.e., all elements of $[\gamma ]$ intersect S perpendicularly). The function $f: \widetilde{M} \rightarrow \mathbb {R}$, $f(p)=lim_{t \rightarrow \infty } d(p, \gamma (t))-t$, is called a Bussmann function. For each point $p \in \widetilde{M}$, there is a point $\eta _{_{S}}(p)$ in S which is the unique point of S nearest p, and the following map is a homeomorphism ([8, pp. 47, 58]):

$$\begin{aligned} \phi : \widetilde{M} \rightarrow S \times \mathbb {R}, \ \ \phi (p)=( \eta _{_{S}}(p), f(p)). \end{aligned}$$

Theorem 3.3

(see [15, Theorem 3.5]). Let $M^{n+2}$ be a nonsimply connected Riemannian manifold which is of cohomogeneity two under the action of a closed and connected subgroup G of isometries, and suppose that M can be decomposed as a product of Riemannian manifolds of negative curvature and $M^{G}\ne \emptyset$. Then,

• (a) M is diffeomorphic to $S^{1} \times \mathbb {R}^{n+1}$ or $B \times \mathbb {R}^{n}$, where B is the Moebius band.

• (b) Each principal orbit is diffeomorphic to $S^{n}.$

As we mentioned in Introduction, classification of orbits of cohomogeneity two Riemannian manifolds of nonpositive curvature is an open problem and seems to be a difficult problem in general case. In the following theorem, we consider an important category of Riemannian manifolds of nonpositive curvature, containing products of negatively curved Riemannian manifolds. In direction of [15], we give a description of the manifold and its orbits under the condition that the acting group is non-semisimple.

Theorem 3.4

Let $M^{n+2}$, be a nonsimply connected Riemannian manifold such that it can be decomposed as a product of Riemannian manifolds of strictly negative curvature of dimension bigger than two, and let G be a non-semisimple closed and connected subgroup of the isometries of M. If M is of cohomogeneity two under the action of G, without non-principal orbits of positive dimension, then one of the following statements is true:

• (a) M is a parabolic manifold homeomorphic to $\frac{S}{\pi _{1}(M)}\times \mathbb {R}$. Where, S is a horosphere in the universal Riemannian covering of M, and $\frac{S}{\pi _{1}(M)}$ is a cohomogeneity one G-manifold.

• (b) $\pi _{1}(M)=Z^{p}$ for some positive integer p, and all orbits are diffeomorphic to $\mathbb {R}^{n-p}\times T^{p}.$

• (c) M is diffeomorphic to $S^{1} \times \mathbb {R}^{n+1}$ or $B \times \mathbb {R}^{n}$, where B is the Moebius band, and each principal orbit is diffeomorphic to $S^{n}$.

• (d) dimM=3 and M has negative curvature. $\pi _{1}(M)=Z$ and each orbit is diffeomorphic to $S^{1}$.

Proof

Following Remark 3.1, let $\widetilde{M}$ be the universal Riemannian covering manifold of M with the deck transformation group $\Delta$ and let $\widetilde{G}$ be the corresponding connected covering of G which acts isometrically and by cohomogeneity two on $\widetilde{M}$. Since by assumptions of the theorem, M can be decomposed as the product of Riemannian manifolds of strictly negative curvature, then $\widetilde{M}$ can be decomposed as $\widetilde{M}=M_{1} \times M_{2} \times \cdots \times M_{m}$ such that for all i, $M_{i}$ has strictly negative curvature, and each $\delta \in \Delta$ decomposes as $\delta =\delta _{1} \times \delta _{2} \times \cdots \times \delta _{m}$, $\delta _{i} \in Iso(M_{i})$. If $M^{G}\ne \emptyset$, then we get part (c) of the theorem from Theorem 3.3. Thus, in the rest of the proof we suppose that $M^{G}= \emptyset$ which implies $\widetilde{M}^{\widetilde{G}} = \emptyset$. Then, according to the assumptions of the theorem, all orbits must be regular and of dimension n. Since G is non-semisimple, $\widetilde{G}$ is non-semisimple. Let H be a connected solvable normal subgroup of $\widetilde{G}$. By Lemma 2.7, there exists a unique subset $\Omega$ of $\widetilde{M}$ such that $H(\Omega )=\Omega$ and $\Omega$ is one of the following sets or product of them.

A: a totally geodesic submanifold of $\widetilde{M}$ or a totally geodesic submanifold of the product of some elements of the set $\{M_{1}, \ldots , M_{m}\}$ (because, fixed point sets of connected subgroups of isometries are totally geodesic submanifolds).

$\gamma$: image of a geodesic of $\widetilde{M}$ or a geodesic in the product of some elements of the set $\{M_{1}, \ldots , M_{m}\}$ .

$\zeta$: a regular point at infinity of $\widetilde{M}$ or a regular point at infinity for product of some elements of the set $\{M_{1}, \ldots , M_{m}\}$.

That is

$$\begin{aligned} \Omega \in \{ \gamma , \zeta , A, A \times \gamma , A \times \zeta , \zeta \times \gamma , A \times \zeta \times \gamma \} \ \ \ \ \ (*) \end{aligned}$$

Since H is normal in $\widetilde{G}$, for each $g \in \widetilde{G}$ we have:

$$\begin{aligned} g^{-1}Hg=H \Rightarrow g^{-1}Hg(\Omega )=H(\Omega ) \Rightarrow g^{-1}Hg(\Omega )=\Omega \Rightarrow H(g\Omega )=g\Omega . \end{aligned}$$

Now, from the uniqueness of $\Omega$ with the property $H(\Omega )=\Omega$, we get that $g\Omega =\Omega$, then

$$\begin{aligned} \widetilde{G}(\Omega )=\Omega \ \ \ \ \ (**) \end{aligned}$$

Since the elements of $\Delta$ commute with the elements of $\widetilde{G}$, for each $\delta \in \Delta$ we have $\widetilde{G}( \delta \Omega )= \delta \widetilde{G}(\Omega )= \delta \Omega$. Uniqueness of $\Omega$ implies that $\delta (\Omega )=\Omega$. Thus,

$$\begin{aligned} \Delta (\Omega )=\Omega \ \ \ \ (***) \end{aligned}$$

We consider now the seven cases of $\Omega$ in $(*)$.

Case 1. $\Omega = \gamma$.

$\widetilde{G}(\gamma )=\gamma$ and By $(**)$, $\Delta (\gamma )= \gamma$. Since $\widetilde{M}^{\widetilde{G}} = \emptyset$, then $\gamma$ is a $\widetilde{G}$-orbit. Since the cohomogeneity is two and all orbits are of dimension n, then $n=1$. Since by assumptions of the theorem, for all i, dim$M_{i}\ge 3$, then $\widetilde{M}=M_{1}$. Thus, M has strictly negative curvature. All $\widetilde{G}$-orbits in $\widetilde{M}$ are diffeomorphic to $\gamma$, then they are diffeomorphic to R. By Lemma 3.5 in chapter 12 of [7], $\Delta = Z$. Thus, each G-orbit in M will be diffeomorphic to $\frac{R}{\Delta }= \frac{R}{Z}=S^{1}$. This is part (d) of the theorem.

Case 2. $\Omega =\zeta$.

Let $[\gamma ]$ be the asymptotic class of the geodesics related to $\zeta$. We have $\widetilde{G}(\zeta )=\zeta$ and by $(**)$, $\Delta (\zeta )=\zeta$(i.e, $\widetilde{G}([\gamma ])=[\gamma ]$ and $\Delta ([\gamma ])=[\gamma ]$). First, suppose that there is an axial element $\delta \in \Delta$ and let $\lambda$ be the unique geodesic such that $\delta \lambda =\lambda$ ( uniqueness of $\lambda$ comes from Lemma 2.5). If $g\in \widetilde{G}$ then $\delta (g\lambda )=g\delta \lambda =g\lambda$. Thus, we get from the uniqueness of $\lambda$ that $g\lambda =\lambda$. Then, $\lambda$ is a $\widetilde{G}$-orbit, and we get part (d) of the theorem as like as Case 1.

Now, suppose that all elements of $\Delta$ are non-axial. Non-identity elements of $\Delta$ are without fixed points, then they must be parabolic and M will be a parabolic manifold. By Lemma 2.3, for each $\delta \in \Delta$ and each horosphere S related to the asymptotic class $[\gamma ]$, $\delta S = S$. Fix a horosphere S related to $[\gamma ]$. Put $M_{1}=\frac{S}{\Delta }$ and let $\eta _{s}$ and f be the maps defined in Remark 3.2. The homeomorphism $\phi :\widetilde{M}\rightarrow S\times \mathbb {R}$ mentioned in Remark 3.2, induces a homeomorphism $\phi _{1}:\frac{\widetilde{M}}{\Delta }=M \rightarrow \frac{S}{\Delta }\times \mathbb {R}$, such that $\phi _{1}(x)=(\kappa \eta _{s}(\widetilde{x}),f(\widetilde{x})), \widetilde{x}\in \kappa ^{-1}(x)$.

If there is a $g \in \widetilde{G}$ which is axial and $\lambda$ is its unique axis, then we get from the fact that the elements of $\Delta$ and g commute that for all $\delta \in \Delta$, $\delta (\lambda )$ is also an axis for g. Since the axis is unique then $\delta (\lambda )=\lambda$. Now, if $g' \in \widetilde{G}$, then again we get from the uniqueness of the axis $\lambda$ for $\delta$ that $g' \lambda =\lambda$, thus $\widetilde{G}(\lambda )=\lambda$, and we get part (d) of the theorem as like as case 1.

Now, Suppose that all elements of $\widetilde{G}$ are non-axial. If for some $g \in \widetilde{G}$ and $x \in \widetilde{M}$, $gx=x$, then for the geodesic $\lambda$ in $[\gamma ]$ which passes from x, we have $g\lambda =\lambda$, and g must be axial which is contradiction. Thus, we can assume that the elements of $\widetilde{G}$ are parabolic and by Lemma 2.3, $\widetilde{G}(S)=S$. Thus, S is a cohomogeneity one $\widetilde{G}$-manifold and $\frac{S}{\Delta }$ is a cohomogeneity one G-manifold. This is part (a) of the theorem.

Case 3. $\Omega =A$.

Similar to the previous cases, we have $\widetilde{G}(A)=A$ and $\Delta (A)=A$. Put $A_{1}=\kappa (A)$. A is a nontrivial totally geodesic submanifold of $\widetilde{M}$, thus $A_{1}$ is a totally geodesic submanifold of M. Since $\Delta (A)=A$, then $\Delta$ is equal to deck transformation group of $A_{1}$, and $\pi _{1}(A_{1})=\Delta =\pi _{1}(M)$. Since all orbits are of dimension n, we have for all $x \in A$:

$$\begin{aligned} n = \text {dim} \widetilde{G}(x) \le \text {dim} A < \text {dim} \widetilde{M} = n + 2 \Rightarrow \text {dim} A = n orn + 1 \end{aligned}$$

Now, we consider dim $A = n$, dim $A = n + 1$, separately.

I) dim$A = n$.

In this case, A is a $\widetilde{G}$-orbit and $A_{1}$ must be a G-orbit of nonpositive curvature. Thus, we get part (b) of the theorem ( Because, a homogeneous Riemannian manifold of nonpositive curvature is diffeomorphic to product of a torus and a euclidean space [22].

II) dim $A = n + 1$.

$A_{1}$ is a cohomogeneity one G-manifold of non-positive curvature, without singular orbits. Consider the following two cases separately:

(II-1) For all $\delta \in \Delta$, $d^{2}_{\delta }$ has no minimum point.

(II-2) There is a $\delta \in \Delta$ such that $d^{2}_{\delta }$ has minimum point.

(II-1): In this case by [17, Lemma 3.2 ], $\Delta$ maps each orbit of A onto itself. Thus by a similar way in the proof of [17, Lemma 3.6 ], all orbits of $A_{1}$ are diffeomorphic to $T^{p} \times \mathbb {R}^{n-p}$ for some nonnegative integer p, thus we get part (b) of the theorem.

(II-2): By [4, Proposition 4.2], the minimum point set of $d^{2}_{\delta }$ is equal to the image of all geodesics translated by $\delta$. But by Lemma 2.5, there is at most one geodesic translated by $\delta$. Let $\gamma$ be the unique geodesic such that $\delta (\gamma )=\gamma$. Since the elements of $\widetilde{G}$ and $\Delta$ commute, then we get from the uniqueness of $\gamma$ that $\widetilde{G}(\gamma )=\gamma$, and we get part (d) of the theorem in the similar way as Case 1.

Case 4. $\Omega \in \{ A \times \gamma , A \times \zeta , \zeta \times \gamma , A \times \zeta \times \gamma \}$.

By dimensional reasons, this cases can not occur. We give the proof for $\Omega = A \times \gamma$, other cases are similar. We can assume (after a possible rearrangement of the indices) that

$$\begin{aligned} \widetilde{M}= & {} M_{1} \times M_{2} \times \ldots M_{k} \times M_{k+1} \times M_{k+2} \times \cdots \times M_{m} \\&A \subset M_{1} \times M_{2} \times \ldots M_{k}, \ \ \ \ \gamma \subset M_{k+1} \times M_{k+2} \times \cdots \times M_{m} \end{aligned}$$

By $(**)$, $\widetilde{G}(A \times \gamma )=A \times \gamma$. Thus, $A \times \gamma$ is a union of $\widetilde{G}$-orbits. Since A and $\gamma$ are nontrivial then the codimension of A in $M_{1} \times M_{2} \times \ldots M_{k}$ is at least 1, and because for all i, dim$M_{i}\ge 3$, then the codimension of $\gamma$ in $M_{k+1} \times M_{k+2} \times \cdots \times M_{m}$ is at least 2. Thus, the codimension of $A \times \gamma$ in $\widetilde{M}$ will be at least 3. This is contradiction (because, $A \times \gamma$ is union of orbits which have codimension two in $\widetilde{M}$). $\square$

Remark 3.5

In Theorem 3.4, decomposability of M to the product of negatively curved manifolds can be replaced by the weaker condition of decomposability of the universal covering manifold $\widetilde{M}$ to negatively curved manifolds and decomposability of $\Delta$ ( see [17], definition of UND-manifolds and examples).

Remark 3.6

In case (a) of Theorem 3.4, M is homeomorphic to the product of a cohomogeneity one manifold $\frac{S}{\Delta }$ with $\mathbb {R}$. By proof of the theorem, in this case, there is no singular orbit. Since the orbit space of cohomogeneity one manifolds with no singular orbit are homeomorphic to $S^{1}$ or R, then the orbit space of M under the action of G will be homeomorphic to $S^{1} \times R$ or $R^{2}$. Study of the orbits in this case reduces to the study of the orbits of cohomogeneity one actions on horospheres. In the special case when M has constant negative curvature (decomposition of M has one factor of constant negative curvature), the horospheres of $\widetilde{M}$ are isometric to $R^{n+1}$, and from the known results about cohomogeneity one actions on flat Riemannian manifolds, orbits of M are diffeomorphic to $R^{k} \times T^{n-k}$, for some positive integer k.