Why the proof of existence of Riemannian metric does not apply to pseudo Riemannian metric?

Question

I just looked on the proof in Wikipedia Existence of Riemannian metric on smooth manifold, but I don’t see why this proof failed for pseudo-Riemannian metric, could anyone point out why this is not applicable for pseudo-Riemannian metric?

Answer 1

It fails even at the pointwise level: in the Riemannian case, $$\tag{1} \sum_\beta \tau_\beta g_\beta$$ is always positive definite since $g_\beta >0$, $\tau _\beta \ge 0$ and $\tau_{\beta_0} >0$ for some $\beta_0$. But if each $g_\beta$ is only non-degenerate, then (1) might give you a degenerate symmetric two tensor.

Related question about the existence of Puesdo Riemannian metric: here

Answer 2

Just an example: a sum of two metrics of signature $(1,1)$ can be the zero metric: $$\left(\begin{matrix}1 & 0 \\ 0 & -1 \end{matrix}\right)+ \left(\begin{matrix}-1 & 0 \\ 0 & 1 \end{matrix}\right)=\left(\begin{matrix}0 & 0 \\ 0 & 0 \end{matrix}\right)$$