Exercise 2.5 (Existence of a Riemannian metric)

Proof. Let ${\{(U_{\alpha },{\phi }_{\alpha })\}}_{\alpha \in A}$ be the smooth structure of $M$. Each smooth coordinate chart induces a Riemannian metric:

where $x_{\alpha }^1,\dots ,x_{\alpha }^n$ denote the local coordinates associated with the local coordinate map ${\phi }_{\alpha }$. Using a partition of unity ${\{{\psi }_{\alpha }\}}_{\alpha \in A}$ subordinate to ${\{U_{\alpha }\}}_{\alpha \in A}$, we define

Since ${\{{\psi }_{\alpha }\}}_{\alpha \in A}$ is a partition of unity, the family of supports ${\{\mathrm{supp}<!--\; FUNCTION\; APPLICATION\; -->{\psi }_{\alpha }\}}_{\alpha \in A}$ is locally finite, meaning that every point $p\in M$ has a neighborhood that intersects $\mathrm{supp}<!--\; FUNCTION\; APPLICATION\; -->{\psi }_{\alpha }$ only for finitely many values of $\alpha$. In other words, the sum in Eq. (2 ) has only finitely many non-zero elements at each point $p\in M$. Therefore, $g$ is

(i)

(a smooth tensor field)
Since a linear combination of smooth tensor fields is again a smooth tensor field, $g$ defines a smooth tensor field (LeeSM, p257).

(ii)

(symmetric)
Since a linear combination of symmetric tensors is again a symmetric tensor, $g$ is symmetric:

$$g{\text{|}}_p(v,w)=\underset{\alpha \in A}{\sum }{\psi }_{\alpha }{g}_{\alpha }{\text{|}}_p(v,w)=\underset{\alpha \in A}{\sum }{\psi }_{\alpha }{g}_{\alpha }{\text{|}}_p(w,v)=g{\text{|}}_p(w,v).$$

(iii)

(positive definite)
Let $v\in T_{p}M$ be any nonzero vector. We have

$$g{\text{|}}_p(v,v)=\underset{\alpha \in A}{\sum }{\psi }_{\alpha }{g}_{\alpha }{\text{|}}_p(v,v).$$

Each term in this sum is nonnegative since ${\psi }_{\alpha }\ge 0$ and $g_{\alpha }{\text{|}}_p(v,v)>0$; thus, $g(v,v)$ is nonnegative as well. But at least one of the functions ${\psi }_{\alpha }$ is strictly positive at $p$ (otherwise it would not sum up to 1). Thus, $g_p(v,v)$ must be positive as well. Whenever $v=0$, the assertion follows easily.