Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm curious how one can classify the bundles over a given manifold. I recently read this paper on classifying $2$-sphere bundles over compact surfaces. A lot of the concepts went over my head since I'm still having trouble doing more "trivial" classifications.

To get used to harder classifications, I'm hoping to get used to doing easier ones first. Thus, I'm curious how one can classify the $2$-plane bundles over $S^2$. To be honest, I'm not sure how to even approach such a problem.

share|improve this question
add comment

1 Answer

up vote 10 down vote accepted

A vector bundle on $S^2$ can be constructed by gluing two trivial vector bundles over $S^2_+$ and $S^2_-$, the closed hemispheres. This is called the clutching construction; see, for example, Husemoller's book.

The «gluing instructions» are a map from the equator, a cicle $S^1$, to $\mathrm{GL}_n(\mathbb R)$, and the result depends only on the homotopy class of this map $S^1\to\mathrm{GL}_n(\mathbb R)$. Therefore, to do the classification you want, one needs to classify these up to homotopy.

Working this out, you end up with a bijection between isomorphism classes of $2$-dimensional vector bundles on $S^2$and elements of $\pi_1(\mathrm{GL}_2(\mathbb R))$. The latter group is isomorphic to $\mathbb Z$.

share|improve this answer
    
So given this clutching construction, we just consider all possible ways of gluing and then show which ones are the same up to homotopy? –  Landyn Oct 22 '11 at 21:55
    
Yes. This is explained in Husemoller's book and many other places —there is even a Wikipedia page on cluthing!— and they do a better jobs than what I could do here :) –  Mariano Suárez-Alvarez Oct 22 '11 at 21:56
    
This seems like a really useful construction! Does this construction work for classifying bundles on any (or most) manifolds? –  Landyn Oct 22 '11 at 21:57
    
I would imagine yes, provided you have a simple decomposition as a CW complex or a cover by charts with intersections you understand. The above argument works because the inthersection of the two hemispheres is a circle, but would fail in general with too complicated or unknown intersections. –  Olivier Bégassat Oct 22 '11 at 22:15
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.