Preissmann's theorem

Remarks

The idea of the proof: by the Cartan–Hadamard theorem, the universal cover $\tilde M$ is diffeomorphic to $\R^n$ and $\pi_1(M)$ acts on it by deck isometries. Each nontrivial element has infinite order and, by compactness, moves points along a unique invariant geodesic, its axis. Negative curvature forces two commuting elements to have the same axis. So an abelian subgroup acts by translations on one geodesic line, and a discrete group of translations of a line is infinite cyclic.

Example: let $M_1$ and $M_2$ be compact manifolds whose fundamental groups contain elements of infinite order, such as $M_1 = M_2 = S^1$. Then $\pi_1(M_1 \times M_2)$ contains $\Z^2$, so $M_1 \times M_2$ carries no metric of negative sectional curvature.

Negative curvature is needed: the flat torus $\R^2 / \Z^2$ has curvature $0$ and fundamental group $\Z^2$.

Going deeper

For a student

Preissmann's theorem is a group-theoretic obstruction to negative curvature. It rules out, for example, every product of two compact manifolds whose fundamental groups have elements of infinite order. It is an early instance of the link between negative curvature and the algebra of the fundamental group that later led to Gromov's hyperbolic groups.