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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.

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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: mathoverflow.net/questions/15224/triviality-of-fibre-bundles/… –  algori Feb 22 '10 at 1:47

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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.

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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

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