Does every n-torus admit a semi-Riemannian metric?

Question

I know every smooth manifold admits a Riemannian metric, and I think I can state that a Riemannian metric is a semi-Riemannian metric,right? The former being just a special case of the later, so the answer should be yes. Is this reasoning correct? Otherwise how do I prove it?

Answer 1

Your reasoning is correct. Here is a proof that directly addresses the title of your question.

The $n$-dimensional torus $\Bbb T^n$ admits semi-Riemannian metrics of any signature $(k,n-k)$. Indeed, its cotangent bundle $T^*(\Bbb T^n)$ is trivial. Let $\{\alpha^1,\ldots,\alpha^n\}$ be a global trivialisation. Then $g_k = -\sum_{j=1}^k \alpha^j\otimes \alpha^j + \sum_{j=k+1}^n \alpha^j\otimes \alpha^j$ is a semi-Riemannian metric of signature $(k,n-k)$ for any $k\in \{0,\ldots,n\}$.