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 $\mathbb{R}$?
[Math] Trivial line bundle
vector-bundles
Related Solutions
Apparently, Mario Serna has produced pictures of $U(1)$-bundles on his webpage and in his paper "Riemannian Gauge Theory and Charge Quantization". Here an example
The image represents a trivial $\mathbb R^3$ over some rectangular base manifold. The $U(1)$ bundle which we want to visualize is shown as an $\mathbb R^2$-sub-bundle: the disks indicate the 2-dimensional fibers at each point, to be understood as subspaces of small 3-dimensional boxes at each point (not shown).
It seems that the disks are also meant to give an impression of the connection, but I don't fully understand how parallel transport is supposed to work here.
He cites a result by Narasimhan and Ramanan which says that every $U(1)$ bundle can be embedded into a trivial $(2d+1)$-dimensional complex vector bundle where $d = \text{dim} M$ is the dimension of the base manifold. Fortunately, the dimension is lower in the cases drawn.
The difference between a rank $2k$ real bundle and a rank $k$ complex bundle is that in a complex bundle one is using $\mathbb C$ as the scalar field, whereas in a real bundle one is only using $\mathbb R$ as the scalar field.
Now let's suppose we are given a complex $n$-dimensonal bundle $E \mapsto X$.
It is true that if this bundle is trivial over $\mathbb C$, i.e. if there is a $\mathbb C$-linear isomorphism of bundles $E \mapsto X \times \mathbb C^n$ over $X$, then one immediately obtains an $\mathbb R$ linear isomorphism of bundles $E \mapsto X \times \mathbb R^{2n}$ over $X$: just restrict the scalar field from $\mathbb C$ to $\mathbb R$.
But the converse does not hold: if there is an $\mathbb R$ linear isomorphism of bundles $E \mapsto X \times \mathbb R^{2n}$ over $X$, you don't know how to extend this to a $\mathbb C$-linear isomorphism; roughly speaking, you don't know how to define multiplication by $i$ in $X \times \mathbb R^{2n}$ in such a way that makes the given map of bundles into a $\mathbb C$-linear isomorphism.
Best Answer
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 $\Bbb{C}$.