The problem:
Let $T^n=\mathbb{R}^n/\mathbb{Z}^n$ with arbitrary Riemannian metric $g$. Prove that there exists a point $p\in T^n$ and $v\in T_p T^n$ such that $\text{Ric}_p(v)\leq 0.$
My attempt:
By this corollary form do Carmos book Riemannian geometry
We get that for $n$ even, there is a point $p\in T^n$ where sectional curvature is not positive. And thus $$k(p)=\frac{1}{n}\sum \text{Ric}_p (X_i) <0,$$ for $\{X_i\}$ orthonormal basis in $T_p T^n.$
Is my answer in even dimension wrong? How can I solve the case of odd dimension?
Best Answer
Sectional curvature is not the right tool to use for this problem, because even if you have nonpositive sectional curvature on some $2$-planes at $p$ the averaging process in Ricci means it might end up being positive at $p$. (Also, you confused the sectional curvature with scalar curvature, which is the average of Ricci.)
Assume $\operatorname{Ric}>0$ at every point. Since $T^n$ is compact, we have $\operatorname{Ric}\geq\delta>0$ for some $\delta$. This is preserved when we lift to the universal cover $\mathbb{R}^n$, but that contradicts Bonnet-Myer's diameter bound.