1-dim Vector Bundle sufficient condition to be trivial

Question

I'm a physics student studying differential geometry. I'm trying to understand how vector bundles work, I have the following exercise.

Let be $ L $ a $1$-dim vector bundle on $M$. Prove that if $\exists\sigma: M \to L$, a section of $L$, so that $\sigma(P) \neq 0 \; \forall P \in M$ then $L$ is trivial.

This is how I tried to proceed. $L$ is a vector bundle so $ \forall P \in M, \; \exists U_P (\ni P) \subset M $ open

$\exists\chi$ a diffeomorphism so that $\chi: L_U = \pi^{-1}(U) \mapsto U \times R$ where $\pi: L \to M$ is the projection ($\pi^{-1}(P) = L_P$ is a 1-dim vector space).

I have to prove that $L = M \times \mathbb{R}$ so $U = M$.

Let be $P \in M$, then $\chi \circ \sigma (U_P) = U_P \times \mathbb{R}$. But I can't figure out how the value $0$ is important.

Are there any propositions or theorems to prove that a vector bundle is trivial? And is it useful to know that is trivial?

Answer 1

Here's a theorem that might help you.

Let $\pi:E\to M$ be a rank $k$ vector bundle. Then $E$ is trivial if and only if it admits a global frame.

Admitting a global frame means that there exists $k$ linearly independent sections. That is, there exists sections $s_1,\dots,s_k$ such that $\{s_1(p),\dots,s_k(p)\}$ form a basis for $E_p$.

Now look at your question where you are considering a line bundle. What does it mean to have a global frame in this case?

Answer 2

It has been awhile since this question was asked, but I thought I give a physicsy answer which may help others.

To prove that the vector bundle is trivial you need to show that it is a product of the fiber (i.e. a 1d vector space) and the base space $M$. Now since there exists a global nonzero section (in physics terms you can think of it as a field configuration which does not vanish anywhere) $\sigma(p)$, it gives you a basis for the fiber at every point $p$ of the base-space. I.e. any vector in that 1d vector space can be represented as $t\sigma(p)$ if $\sigma(p)\ne0$, where $t$ takes values in the field $F$ over which the vector space is defined (say a real number for real vector spaces). That means that a vector bundle which allows a global nonzero section can be completely described by a pair $(t,p)$, which means that the bundle is a product space.

To see how this fails for a nontrivial bundle, consider a Möbius strip. There any section would necessarily go through zero somewhere. To see, think of the section as a line on the Möbius strip. You want to draw the line so that it doesn't cross the mid-point of the strip. You cannot. I put a link of the drawing of this scenario below.

Fig: The Möbius strip does not allow a global nonzero section