MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there a way to classify (and build) the principal SU(2) bundles over a given topological surface up to homeomorphism? In the end, I would like to examine the associated bundle whose fiber is a given irreducible representation of SU(2). Perhaps this construction appears in the physics literature.

share|cite|improve this question
Principal $G$-bundles are classified up to isomorphisms by homotopy classes of mappings to $BG$ or by elements of a non-abelian cohomology set. I've tried to summarize this here:… – algori Feb 22 '10 at 1:47

By "surface" you mean a 2-real-dimensional manifold $F$? If so, there is only one principal $SU(2)$ bundle up to bundle isomorphism, the trivial bundle: $F\times SU(2)\to F$. This is because the 3 skeleton of $BSU(2)$ is a point. So the associated bundles to any rep $SU(2)\to GL(V)$ are also trivial: $F\times V$.

If by "surface" you mean smooth complex projective variety of complex dimension 2, then there are an integer's worth of bundles (topologically), determined entirely by the second Chern class.

share|cite|improve this answer
Although there is only one bundle up to isomorphism, it's wrong to say that it "is" the trivial bundle: there's no canonical isomorphism between a given SU(2) bundle and the trivial bundle, but rather the space of such isomorphisms is a torsor over the group of sections of the trivial bundle. – Theo Johnson-Freyd Feb 22 '10 at 4:55
I don't see what is wrong with the statement that the trivial bundle is the only bundle up to bundle isomorphism. But (as everywhere in mathematics) the existence of non-trivial automorphisms implies that there is ambiguity in choosing an identification of your object with a fixed object. Actually, in this case (SU(2) bundle over a 2-manifold), the group of bundle automorphisms is path connected, and so any two identifications of your bundle with the trivial bundle are homotopic, so you get a little bit more "canonicalness" than in general. – Paul Feb 22 '10 at 15:10

Your Answer


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.