Is it possible to put the a Riemannian metric on $n$-torus such that the sectional curvature is strictly lower than $0$ everywhere? If $n = 2$, I know that $T^2$ doesn't have a metric $g$ such that the sectional curvature $k < 0$ everywhere, otherwise,
$$0 = 2 \pi (2 - 2 * 1) = 2 \pi \chi(T^2) = \int_{T^2} k d\sigma < 0$$
from Gauss-Bonnet's theorem, which is a contradiction, but I don't know how to prove this for $n > 2$. I thought maybe use a corollary of Bonnet-Myers' theorem could be used to prove this once that it is used to prove that the $n$-torus doesn't have a metric $g$ such that the sectional curvature $k > 0$ everywhere. I would like to receive a hint about how to prove that the $n$-torus ($n > 2$) doesn't have a metric $g$ such that the sectional curvature $k < 0$ everywhere.
Thanks in advance!
 A: This is due to Preissmann that can be found in Lee's IRM (p. 357) or Petersen's RG (p. 247):

Theorem (Preissmann, 1943). If $(M, g)$ is a compact manifold of negative
curvature, then any nontrivial Abelian subgroup of the fundamental group is cyclic (i.e. is isomorphic to $\Bbb Z$.). In
particular, no compact product manifold $M\times N$ admits a metric with negative
curvature.

And this is from Jurgen Just book (Nonpositive Curvature: Geometric and Analytic Aspects):

Its fundamental group $\pi_1(\Bbb T^n)$ is $\Bbb Z^n$ . Actually, for any compact Riemannian
manifold $N$ of nonpositive sectional curvature, the rank of any Abelian subgroup
of $\pi_1(N)$ is at most $n$, and if the curvature is even negative, its rank is
precisely $1$ (if the Abelian subgroup is nontrivial) by Preissmann's theorem.
Since the fundamental group of a manifold of nonpositive curvature cannot
contain elements of finite order, in the negative curvature case therefore all
nontrivial Abelian subgroups of $\pi_1(N)$ are isomorphic to $\Bbb Z$.

I think one can use Bishop and O'Neill theorem to prove same statement but I don't know how!
A: You can prove this using Milnor's early paper A note on curvature and the fundamental group. The theorem there says that if $M$ is a compact Riemannian manifold whose sectional curvature is everywhere negative then the fundamental group $\pi_1 M$ has exponential growth.
But the fundamental group of the $n$-dimensional torus is isomorphic to $\mathbb Z^n$, which has polynomial growth of degree $n$.
