vector bundles and their cross-sections

408 Views Asked by At

Let $B$ be a compact Hausdorff space and $C^0(B)$ be the ring of continuous real valued functions on $B$. For any vector bundle $\xi$ over $B$, let $\Gamma(\xi)$ denote the $C^0(B)$-module consisting of all cross-sections of $\xi$.

(1). Prove that $\xi\cong\eta$ if and only if $\Gamma(\xi)\cong\Gamma(\eta)$ as $C^0(B)$-modules.

(2). Prove that $\xi$ is trivial if and only if $\Gamma(\xi)$ is a free $C^0(B)$-module.

How to do this? I have no way to solve it...

Moreover, if $B=M$ is a smooth manifold and $\xi$ is a vector bundle over $M$, if we let $\Gamma^\infty(\xi)$ be the module of smooth cross-sections over $C^\infty(M)$, is it true that:

(1). $\xi\cong\eta$ if and only if $\Gamma^\infty(\xi)\cong\Gamma^\infty(\eta)$ as $C^\infty(M)$-modules?

(2). $\xi$ is trivial if and only if $\Gamma^\infty(\xi)$ is a free $C^\infty(M)$-module?

Thank you a lot.

1

There are 1 best solutions below

1
On BEST ANSWER

This is not a complete answer, since this is worth working through on your own.

Instead, here is the main idea for all of these questions: if you have a $U \subset B$ with a trivialization of $\xi|_U$, you now have a way of constructing $C^0$ sections ($C^\infty$ sections if $B$ is smooth) by taking "constant" sections times a cut-off function. This is why knowledge of the module $\Gamma(\xi)$ is sufficient to recover the bundle.