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How to prove that any vector bundle with a fiber metric g admits a metric connection? It seems I should use partition of unity, but I have no idea how to proceed.

Also it seems there are two definitions of connection on a vector bundle E, one is that $\nabla: \Gamma(TM) \times \Gamma(E) \to \Gamma$(E), and the other one is that $\nabla: \Gamma(E) \to \Gamma(E) \otimes\Gamma(T^\ast M)$, and they should be equivalent. But I don't see how to use this to show the following equations are equivalent:

If there is a fiber metric g on vector bundle E, then

$d(g(u,v))=g(\nabla u,v)+g(u,\nabla v)$ for all $u,v\in\Gamma(E)$

is equivalent to

$X(g(u,v))=g(\nabla_X u,v)+g(u,\nabla_X v)$ for all $X\in TM$,

is it because that $d(g(u,v))(X)=X(g(u,v))$? I am just not sure about how vector fields act on metrics.

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1 Answer

  1. Yes, if you put an arbitrary metric on each trivialized restriction and then glue them together using a partition of unity, you'll get a metric on the whole thing.

  2. To go from a map $\Gamma(E) \rightarrow \Gamma(E)\otimes \Gamma(T^*M)$ to a map $\Gamma(E)\otimes \Gamma(TM)\rightarrow \Gamma(E)$, if $s_0\mapsto s_1\otimes x$, then you can put $s \otimes v \mapsto s_1 \cdot x(v)$ -- basically, this all comes from the fact that $Hom_{\mathbb{R}}(V,W)\cong V^*\otimes W$.

  3. It depends which version of $\nabla$ you're using. From your notation it looks like either you're using the definition $\nabla:\Gamma(E)\rightarrow \Gamma(E)\otimes \Gamma(T^*M)$ or you're using the definition $\nabla:\Gamma(E)\otimes \Gamma(TM) \rightarrow \Gamma(E)$ and suppressing the vector field. My guess is that for $s_0\otimes x\in \Gamma(E)\otimes \Gamma(T^*M)$ you're extending the metric by $g(s_0\otimes x,s_1)=g(s_0,s_1)\cdot x \in \Gamma(T^*M)$; in this case, yes the fact is true by what you said, which can be taken as the definition of applying a 1-form to a vector.

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