Let $M$ be the smooth manifold,prove there exist some Riemann metric on it.
This is the standard application of partition of unity thoerem:First find the local $g_\alpha$ on each chart by pullback the standard metric on Euclidean space.
since the chart of the manifold covers $M$,appling partition of unity to glue those $g_\alpha$ togother as :
$$g = \sum\rho_\alpha g_\alpha$$
To prove it's a Riemann metric,we need to check two things :
first it's smooth,second at each point it satisfy the inner product axioms.
Here is a bit detail that confuses me is how to define $g_\alpha$ outside original domain $U_\alpha$?It may depend on the existence of smooth extension of $g_\alpha$ outside $U_\alpha$.Does this result holds?To make it more clear I make the proposition below:
Let $M$ be a smooth manifold,prove given closed subset $A \subset M$,and a smooth tensor field on $A$ ,there exist a smooth extension of this tensor field to $M$.
Best Answer
It doesn't matter because $\rho_\alpha$ is supported on $U_\alpha$, which means that $\rho_\alpha g_\alpha$ is zero outside $U_\alpha$.