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Show that if a manifold $M$ is contractible, then every vector bundle over $M$ is equivalent to the trivial bundle.

Got this as homework but I'm kind of lost in the hole vector bundle subject, so if anyone could give some hints to this one as well as to understanding the subject, it would be highly appreciated.

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3 Answers 3

up vote 7 down vote accepted

Here are a couple of things that I found helpful when thinking about vector bundles:

  1. Over nice spaces they they can be classified up to isomorphism. So by nice here we mean compact manifold, or CW complex. The idea is that every finite dimensional vector bundle $\xi: E \to X$ is isomorphic to the pull back of the universal bundle $EG \to BG$ where $G$ is some particular structure group. (for $\xi$ a real $n$ dimensional bundle $G=GL_n(\mathbb{R})$, note though that $O(n)$ is homotopy equivalent to $GL_n(\mathbb{R})$). You can learn a lot about a vector bundle by examining the map that classifies it. For example, $\xi: E \to X$ is orientable if and only if the map that classifies it can be lifted over the fibration $BSL_n(\mathbb{R}) \to BGL_n(\mathbb{R})$.

  2. Vector bundles are really nicely behaved families of vector spaces over $X$. We usually use them to keep track of some important data about $X$. In fact, most operations on vector spaces extend to operations on vector bundles over $X$. Things such as tensor products, direct sums, exterior and symmetric powers pop up all the time.

  3. Surprisingly enough every vector bundle $\xi: E \to X$ is a homotopy equivalence since the fibers are contractible. It seems like this makes vector bundles sort of silly, but it illustrates that they are inherently geometric objects. While we use them for certain constructions in homotopy theory, they are really about the geometry of the base space. That being said, homotopy theory can say a lot about vector bundles, consider Chern-Weil Theory which relates cohomology classes (a gadget that can only see homotopy theory) to things like curvature.

A big help for learning about bundles for me was thinking about them like objects, like when you think about a manifold you want to think about the manifold not its parametrizations.

Mariano's answer is dead on, I am just adding some helpful facts. Let me know if you want me to clarify something.

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Show that if $f,g:X\to Y$ are two maps between manifolds which are homotopic, and if $E$ is a vector bundle on $Y$, then $f^{-1}(E)$ and $g^{-1}(E)$ are isomorphic vector bundles on $X$. (You don't really need that both $X$ and $Y$ be manifolds: you can get by if they are just spaces and $X$ is paracompact...)

(You should really provide information about your background, otherwise it is more or less impossible to come up with a useful answer, you see!)

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Considering the OP is new to vector bundles, perhaps you should define the pullback bundle? Or at least give this link: en.wikipedia.org/wiki/Pullback_bundle –  Dylan Wilson Jan 30 '11 at 22:40
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@Dylan, that is why I suggested that the OP provide information about his background. Rather than Wikipedia, in the case he is so new, I'd recommend Husemoller's book! –  Mariano Suárez-Alvarez Jan 30 '11 at 22:50

This may be a repetition of what Mariano and Sean were suggesting: you will most likely ultimately need to either: produce a bundle morphism between your bundle and the trivial bundle, or produce a global trivialization of your top space.

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