Trivial line bundle

Question

Suppose that $L \to M$ is complex line bundle over a manifold $M$. One can therefore form the dual bundle $L^* \to M$. We can identify $L^* \otimes L$ with endomorphism bundle $End(L)$. Why it is true that $End(L)$ is trivial? Why the assumption of being line bundles is essential? Is it also true for other fields, for example $\mathbb{R}$?

Answer 1

If $L$ is a rank $1$ vector bundle, then $\operatorname{End}(L)$ is a rank $1$ vector bundle with a nowhere zero section: namely, the identity map. So it is trivial.

This doesn't work for higher-rank vector bundles; if $E$ has rank $k$, then $\operatorname{End}(E)$ has rank $k^2$, so we need $k^2$ linearly independent sections to trivialize it and they're nowhere to be found.

On the other hand, it doesn't in any way require that the ground field be $\Bbb{C}$.

Answer 2

The transition map for dual bundle $E^{*}$ is $(g_{\alpha \beta }^{t})^{-1}$, the inverse transpose of the original one $g_{\alpha \beta}$. In the line bundle case, this is ${g_{\alpha\beta}}^{-1}$. So the transition map for $End(E)\cong E \otimes E^{\ast}$ is nothing but $g_{\alpha \beta } {g_{\alpha \beta}}^{-1}=1$ which means triviality of $End(E)$.