$M$ simply connected $\implies$ every line bundle $E\to M$ is trivial

Question

Let $M$ be simply connected smooth manifold. Prove that every real line bundle $\pi:E\to M$ is trivial.

Here is my attempt: let $\pi:E\to M$ be a line bundle. Using the fact that there is a metric $<\cdot,\cdot>$ on $E$, we can take orthonormal local sections, so that the transition functions $g_{\alpha\beta}:U_\alpha\cap U_\beta\to GL(1)$ take values on the orthogonal group $O(1)=\{1,-1\}$.

In order to prove $E$ is trivial, we need to prove that for every transition function we have $g_{\alpha\beta}\equiv 1$. I guess here is where the simply connected hypothesis comes in.

When I try to draw concrete examples, I can see that this is very intuitive, but I still can't find a way to formalize the idea that every closed path being homotopic to a point implies all transitions are the identity.

I feel like this has something to do with orientation, but I also don't know how to make this precise.

Answer 1

You mean real line bundle surely? Assuming, as you do, there is a reduction to structure group $O(1)$, then taking the two unit vectors in each fibre gives a two-sheeted covering map $E'\to M$. But if $M$ is simply connected then this covering is trivial: the direct product of $M$ with $\{1,-1\}$. Picking one of the sheets gives a trivialisation of $E$.

Answer 2

Hint

A real line bundle $L$ on a topological space $X$ is trivial if and only if its first Stiefel-Whitney class $w_1(L)\in H^1(X, \mathbb Z/2\mathbb Z)$ is zero.

(there is also an argument that shows that if $X$ is a closed smooth manifold whose Stiefel-Whitney and Pontryagin classes vanish, then $X$ is stably parallelizable)

Answer 3

I know this is a late reply, but I guess I can add a different approach here. Let's consider a simply connected space $X$. Then, we can consider the universal line bundle $G_1(\mathbb{R}^{\infty})$ - the $1$-dimensional Grassmannian(which I'll simply denote as $G_1$). The universal cover of $G_1$ is $S^{\infty}$ - the direct limit of spheres. Any map $f:X \rightarrow G_1$ can be lifted to a map $\tilde{f}:X \rightarrow S^{\infty}$ (as $X$ is simply connected). We denote the set of homotopy classes of maps from $X$ to $Y$ as $[X, Y]$. Since $S^{\infty}$ is contractible, $[X, S^{\infty}]$ is singleton. This implies $[X, G_1]$ is singleton as a homotopy in a cover gives a homotopy in the base space. Now, by the definition of the universal bundle, there's a bijection between $[X, G_1]$ and $Vect_1(X)$ (this is the set of all real line bundles of $X$ identified up-to isomorphism). Hence, all real line bundles of $X$ are trivial.