Corollary to Preissman's theorem

Question

Preismann's theorem states (ref. Petersen's "Riemannian Geometry", chapter 6):

On a compact manifold with negative sectional curvature, any abelian subgroup of the fundamental group is cyclic.

A corollary is stated:

No compact product manifold admits a metric with negative curvature.

I don't understand how the corollary follows from the thoerem. Of course it would if, on a product manifold, there are always non-cyclic abelian subgroups of the fundamental group. Is this true? (It seems like it shouldn't be, taking for example the product of simply connected manifolds.)

Answer 1

Recall Cartan-Hadamard: If $K\le 0$ and $M^n$ is complete, then the universal covering space of $M$ is $\Bbb R^n$.

Answer 2

Suppose that $M$ and $N$ are smooth compact manifolds. Suppose that $L=M\times N$ admits a metric of negative curvature. First, note that $\pi_1(L)$ is torsion-free: One can prove this by observing that, by Cartan's theorem, every finite group acting on a simply-connected complete manifold of nonpositive sectional curvature always has a fixed point (see Theorem 23, p. 164, in Peterson's book). Alternatively, notice that a finite nontrivial group $G$ cannot act freely on a contractible manifold as $G$ has infinite cohomological dimension, while a free action on a contractible manifold would yield a finite-dimensional classifying space $K(G,1)$. In either case, if $\pi_1(L)$ contains a nontrivial element of finite order, the fundamental group cannot act freely on the universal cover of $L$, which is a contradiction with the basic covering theory.

Thus, unless $\pi_1(L)$ contains ${\mathbb Z}^2$, either $M$ or $N$ is simply-connected. Assume that $\pi_1(M)$ is trivial; in particular, $M$ is orientable. Since $M$ is compact, there exists $k\in [2, dim(M)]$ such that $H_k(M)\ne 0$ (you can take $k=dim(M)$). Therefore, if $X=M\times \tilde{N}$, is the universal cover of $L$, then $H_k(X)\ne 0$ (use Kunneth formula). This contradicts Cartan-Hadamard theorem which says that the universal cover of a compact manifold of nonpositive curvature is diffeomorphic to a Euclidean space.

Answer 3

Consider a product $M\times N$ with negative sectional curvature.

If $M$ has nonnegative sectional curvature, it would be a contradiction.

So abelian subgroup of fundamental group of $M$ is $0$ or ${\bf Z}$, since it has no torsion.

Hence $\pi_1(M\times N)=\pi_1(M)\times \pi_1(N)$ has abelian subgroup in ${\bf Z}^2$.

From given theorem, so abelian subgroup of at least one factor of $M\times N$, say $M$, is $0$. Hence it is simply connected so that $M={\bf R}^n$. It contradicts to compactness.