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?

  • 1
    $\begingroup$ If the connection is smooth, then so will be the function $(x,v) \mapsto \tau_x(v)$. This follows from standard facts about ODEs, for example, although that might be overkill. $\endgroup$ Nov 14, 2013 at 17:36
  • $\begingroup$ But it has to depend on the paths $p_x$, otherwise a vector bundle over a path-connected manifold would be always trivial. $\endgroup$
    – Seirios
    Nov 15, 2013 at 16:32
  • $\begingroup$ Hi Seirios: I'm not sure if your comment is in response to mine. My comment was just about your final question "how to show that $(x,v) \mapsto \tau_x(v)$ is smooth" — certainly it needs to be shown, but my preference is to prove the general fact that a smooth ODE defines a smooth "flow" function: the output depends smoothly both in the time variable and the initial-condition variable. Anyway, on a higher dimensional manifold, the first problem to repeating your proof is that the connection might not be flat. (Flatness is trivial in one dimension.) ... $\endgroup$ Nov 16, 2013 at 3:27
  • $\begingroup$ ... If the connection is flat, then homotopic paths give the same isomorphism of fibers. But of course non-homotopic paths might give different isomorphisms of fibers. Although not obvious, it is nevertheless true that every vector bundle admits a flat connection. It follows that vector bundles are trivial on manifolds with vanishing $\pi_1$, but not in general. $\endgroup$ Nov 16, 2013 at 3:33
  • 2
    $\begingroup$ Possible duplicate of A fiber bundle over Euclidean space is trivial. $\endgroup$
    – Alex M.
    Oct 5, 2018 at 6:10

2 Answers 2


Theo says in the comments:

Although not obvious, it is nevertheless true that every vector bundle admits a flat connection. It follows that vector bundles are trivial on manifolds with vanishing $\pi_1$, but not in general.

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.


Following Theo Johnson-Freyd's comments, I found:

Let $E \to B$ be a vector bundle and $\nabla$ be a connection. Let $x_0 \in B$ and, for every $x \in B$, let $\gamma_x : [0,1] \to B$ be a path from $x$ to $x_0$. A candidate for a global trivialization is $$\Phi : \left\{ \begin{array}{ccc} E & \to & B \times E_{x_0} \\ (x,v) & \mapsto & (x,\tau_x(v)) \end{array} \right.$$ where $\tau_x$ is the parallel transport along $\gamma_x$. The only obstruction so that $\Phi$ be a global trivialization is that $(x,v) \mapsto \tau_x(v)$ can be non-smooth.

In fact, $\tau_x(v)=\tilde{\gamma}_x(1)$ where $\tilde{\gamma}_x : [0,1] \to E$ is a parallel path (along $\gamma_x$) satisfying $\tilde{\gamma}_x(0)=(x,v)$, that is $\tilde{\gamma}_x$ is solution to $$\left\{ \begin{array}{l} \nabla_{\dot{\gamma}_x} \dot{\tilde{\gamma}}_x=0 \\ \tilde{\gamma}_x(0)=(x,v) \end{array} \right..$$

If locally $(X_1, \dots, X_r)$ is a basis of $E$, $\dot{\gamma}_x=a^i\partial_i$ and $\dot{\tilde{\gamma}}_x=b^iX_i$, the equation above can be written as $$(b^i)'X_i+b^ia^j \nabla_{\partial_j}X_i=0.$$

It is a linear ODE in $b^i$, and we now that the solution is smooth if the coefficients so are.

Therefore, if the path $\gamma_x$ depends smoothly on $x$ (in the sense that the $a^i$ are smooth), then $\Phi$ is a global trivialization. For the closed ball $B(0,1) \subset \mathbb{R}^n$, we may choose $x_0=0$ and $\gamma_x(t)=(1-t)x$ so that $\tilde{\gamma}_x=-x$; thus, any vector bundle over $[0,1]^n$ is trivial.


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