8
$\begingroup$

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.

1 Answers 1

13

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 $\mathbb Z$.

  • 0
    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.2011-10-22