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

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

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?

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?

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

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