I was interested in whether a manifold which admits a metric of constant sectional curvature can be homotopy equivalent to a product of non-contractible manifolds. Of course, there are three cases: positive, negative, and zero.


Positive
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If $M$ is a connected $n$-manifold which admits a metric with constant positive curvature, then it is finitely covered by $S^n$. Suppose $M$ were homotopy equivalent to $M\_1\times M\_2$ where $M\_1$ and $M\_2$ are manifolds. Then $S^n = \widetilde{M} = \widetilde{M\_1}\times\widetilde{M\_2}$ where the tilde is used to denote the universal cover. As $\pi\_k(S^n) = \pi\_k(\widetilde{M\_1})\oplus\pi\_k(\widetilde{M\_2})$, both $\widetilde{M\_1}$ and $\widetilde{M\_2}$ are $(n-1)$-connected. Moreover, as $\pi\_n(S^n) \cong \mathbb{Z}$, either $\pi\_n(\widetilde{M\_1}) \cong \mathbb{Z}$ and $\pi\_n(\widetilde{M\_2}) = 0$, or vice versa; without loss of generality, suppose $\pi\_n(\widetilde{M\_1}) \cong \mathbb{Z})$. It follows that $\widetilde{M\_1}$ is homotopy equivalent to $S^n$ and $\widetilde{M\_2}$ is contractible, so $M\_2$ is a $K(\pi\_1(M\_2), 1)$. As the cover $S^n \to M$ is finite, $\pi\_1(M)$, and hence $\pi\_1(M\_2)$, is finite. If $\pi\_1(M\_2) \neq 0$, then $M\_2$ has infinite cohomological dimension which contradicts the fact that it is a manifold; therefore $\pi\_1(M\_2) = 0$ and hence $M\_2$ is contractible. 


So a manifold which admits a metric with constant positive curvature cannot be homotopy equivalent to a product of non-contractible manifolds.


Negative
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If $M$ is a *compact* connected $n$-manifold which admits a metric with constant negative curvature (i.e. $M$ is hyperbolic), then by the Cartan-Hadamard Theorem, the universal cover of $M$ is isometric to $\mathbb{H}^n$; in particular, $M$ is aspherical. Suppose $M$ were homotopy equivalent to $M\_1\times M\_2$ where $M\_1$ and $M\_2$ are manifolds; as $M$ is aspherical, so are $M\_1$ and $M\_2$. Now by Priessman's theorem, every non-trivial abelian subgroup of $M$ is isomorphic to $\mathbb{Z}$. If both $M\_1$ and $M\_2$ are not contractible, then there are non-zero elements $\gamma\_i \in \pi\_1(M\_i)$ which individually generate subgroups isomorphic to $\mathbb{Z}$, and because $\gamma\_1$ and $\gamma\_2$ commute, $\langle\gamma\_1, \gamma\_2\rangle \cong \mathbb{Z}^2$. This contradicts Priessman's Theorem, so one of $M\_1$ and $M\_2$ must be contractible. 


So a compact manifold which admits a metric with constant negative curvature cannot be homotopy equivalent to a non-trivial product of manifolds.


What if we drop the compactness hypothesis?



> 
> **Question 1:** Let $M$ be a connected, non-compact, hyperbolic manifold. Can $M$ be homotopy equivalent to a product of non-contractible manifolds?
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> 
> 


If such a manifold exists, its fundamental group would be a direct sum of non-trivial groups. As such, $O^+(n, 1)$, the isometry group of $n$-dimensional hyperbolic space, contains such a subgroup. So we could also ask:



> 
> **Question 2:** Is there a subgroup of $O^+(n, 1)$ which is isomorphic to $G\_1\oplus G\_2$ where $G\_1$ and $G\_2$ are non-trivial groups?
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> 
> 


Note, question 2 is not equivalent to question 1 as the group may not act freely. However, a negative answer to question 2 provides a negative answer to the question 1.


Zero
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Flat manifolds can be products (e.g. tori). More generally, the product of flat manifolds is again flat.



Can a hyperbolic manifold be a product?
Question 1:


in $\mathrm{Isom}(\mathbf{H}^n)$, the centralizer of any loxodromic element preserves its axis, and hence is contained in a closed subgroup isomorphic to $\mathrm{O}(n-1)\times\mathrm{Isom}(\mathbf{R})$. In particular, discrete subgroups of the latter are virtually cyclic.


In addition, if a subgroup has no loxodromic element, either it has compact closure, or is horocyclic, in the sense that it fixes a unique point at infinity and is contained in the leafwise stabilizer of the corresponding horosphere foliation with respect to some point at infinity, which is isomorphic to $\mathrm{Isom}(\mathbf{R}^{n-1})$. The centralizer of a horocyclic subgroup fixes the same point at infinity, hence, in turn, is either relatively compact (pointwise fixing an axis), horocyclic, axial, or focal in the sense that it preserve (maybe not leafwise) the corresponding foliation. As regards discrete subgroups, they can't be focal.


Hence, if we have two discrete subgroups $G\_1,G\_2$ of $\mathrm{Isom}(\mathbf{H}^n)$ centralizing each other, we have one of the following, showing that there are fery few subgroups decomposing as direct products.


* both are virtually cyclic (each with a loxodromic element)
* one is virtually cyclic (with a loxodromic element) and the other one is virtually abelian (being horocyclic). This case is not possible if $G\_1\cap G\_2$ is finite and $G\_1G\_2$ is discrete (because the stabilizer of an axis does not contain a discrete $\mathbf{Z}^2$)
* both are virtually abelian (horocyclic)
* one is finite (and the other has to preserve its subspace of fixed points). In particular, if the other one does not preserve any proper geodesic subspace (e.g., is Zariski-dense), then it has a trivial centralizer.


As I said in a comment, Question 2 is too naive: just consider a cyclic subgroup of order 6. But also there are torsion-free subgroups decomposing as nontrivial direct products, for $n\ge 3$, such as a horocyclic copy of $\mathbf{Z}^2$. Geometrically this correspond to a 3-dimensional cusp, homeomorphic to $\mathbf{R}\times\mathbf{T}^2$, so well, this is indeed homeomorphic to a product of two non-contractible manifolds. By the previous description, this is essentially the only possibility, namely every real hyperbolic manifold that homotopically decomposes as a product has a finite covering homeomorphic to the product of a Euclidean space (of dimension $\ge 1$) and a torus (of dimension $\ge 2$), and moreover is "horocyclic" (cusp-like) in the sense that its $\pi\_1$ acts horocyclically on the universal cover.


