I would like to show the followoing result:
Every vector bundle over $[0,1]^n$ is trivial
First, I consider the case $n=1$, so let $E$ be a vector bundle over $[0,1]$. If $\nabla$ is a connexion for $E$, let $\tau_x : E_x \to E_0$ be the parallel transport along the path $p_x : t \mapsto (1-t)x$. Now, I want to show that the map
$$ \left\{ \begin{array}{ccc} E & \to & [0,1] \times E_0 \\ (x,v) & \mapsto & (x, \tau_x(v)) \end{array} \right.$$
is an isomorphism of vector bundles. The only non-trivial point seems to be to show that the previous map is smooth, so my question is: how to show that $(x,v) \mapsto \tau_x(v)$ is smooth?
Best Answer
Theo says in the comments:
This is false. I wouldn't post this as an answer except that it's a bit long for a comment.
Let $G$ be, say, a compact Lie group. Recall that Chern-Weil theory allows us to compute the real characteristic classes of a principal $G$-bundle in terms of the curvature of a connection on the bundle. In particular, if a principal $G$-bundle admits a flat connection, then all of its real characteristic classes vanish. E.g. if a real vector bundle ($G = O(n)$) admits a flat connection then its real Pontryagin classes vanish (equivalently its Pontryagin classes are torsion), and if a complex vector bundle ($G = U(n)$) admits a flat connection then its real Chern classes vanish (equivalently its Chern classes are torsion). Since most $G$-bundles don't have this property, most $G$-bundles don't admit a flat connection, and most $G$-bundles aren't trivial even on spaces with vanishing $\pi_1$.
The simplest examples are complex line bundles over $S^2$, which are classified by their first Chern class in $H^2(S^2, \mathbb{Z}) \cong \mathbb{Z}$; in particular there are nontrivial complex line bundles over $S^2$. Moreover the cohomology group above injects into $H^2(S^2, \mathbb{R})$, so a nontrivial complex line bundle over $S^2$ necessarily does not admit a flat connection. In particular, despite the fact that $S^2$ is simply connected, it is not possible to write down a smooth choice of paths from a basepoint of $S^2$ to any other point to carry out the argument in Seirios' answer. (There is an obvious choice of such path for most points, which is to take a geodesic, but that prescription fails badly for the point opposite the basepoint.)
More abstractly, while principal $G$-bundles are classified by $BG$, flat principal $G$-bundles are classified by $BG_{\delta}$ where $G_{\delta}$ is $G$ equipped with the discrete topology. The problem of equipping a principal $G$-bundle with a flat connection then becomes the problem of lifting a classifying map $X \to BG$ to a classifying map $X \to BG_{\delta}$, or equivalently of reducing the structure group from $G$ to $G_{\delta}$, and there are general machines you can run on this problem to find out what the obstructions to doing this are.