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I just want to how much people know about this at the moment? I thought this is elementary and may be of execrise level, but a quick google search showed serious papers written on this subject (like this). Maybe I asked this prematurely and more research on my side is needed.

If some expert at here knows the answer already, please provide a hint instead of a straightforward answer. This way I can probably learn something myself.

Bombyx mori
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    What other dimension can a surface be? – Qiaochu Yuan Apr 12 '13 at 06:35
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    To be specific so not to be confused with an algebraic surface. – Bombyx mori Apr 12 '13 at 06:36
  • I think what is much better understood is the case of a surface bundle over some other space, and this is related to representations into the mapping class group. –  Apr 12 '13 at 07:37
  • It't not exactly what you looking for, but a text I recently found very useful that seems related, is http://www.math.columbia.edu/~scautis/math428/notes-bundles.pdf. Considering the amount of work needed there, I don't expect your question to be elementary, nor of exercise level. – HSN Apr 12 '13 at 11:37
  • Is this what you're looking for? http://en.wikipedia.org/wiki/Principal_bundle#Classification_of_principal_bundles – user17786 Apr 12 '13 at 15:23
  • @SteveD: Thank you! I thought this primarily in terms of how $G$ may define a covariant derivative on $M$, and if the connection is flat whether I can conclude certain things. But I never thought this way about it. – Bombyx mori Apr 12 '13 at 19:19
  • @user17786: The classification of principal bundles over the sphere is done via the classifying space, see for example Streerod. I just do not know how it works in the case of the surface. – Bombyx mori Apr 12 '13 at 19:20
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    At the level of generality at which you posed your question there is little more to be said apart from recalling the fact that $G$-principal bundles over a (paracompact!) space $X$ are classified by homotopy classes of maps $X\to BG$. If you have a specific group in mind, one might be able to do better. – Mariano Suárez-Álvarez Apr 12 '13 at 19:30
  • @MarianoSuárez-Alvarez: So if I limit my groups to be classicial Lie groups. Then I might hope for a deeper classification theorem? – Bombyx mori Apr 12 '13 at 19:33
  • Let me think about this for a while. – Bombyx mori Apr 12 '13 at 19:35
  • No, he answer would be pretty much the same. If you want something explicit, pick a group :-) – Mariano Suárez-Álvarez Apr 12 '13 at 19:59
  • @MarianoSuárez-Alvarez: I see. Let me pick $SU(3)$ (my favoriate Lie group) and see how it goes... – Bombyx mori Apr 12 '13 at 20:27
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    Well, now find a description of $BSU(3)$ good enough to compute the set of homotopy classes $[X,BSU(3)]$. – Mariano Suárez-Álvarez Apr 12 '13 at 21:02
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    @user32240: $\mathrm{SU}(3)$ is simply connected, and hence $B\mathrm{SU}(3)$ is $2$-connected. Since a surface is $2$-dimensional, it follows that $[\Sigma, B\mathrm{SU}(3)] = 0$ for any surface $\Sigma$, i.e. up to isomorphism the only $\mathrm{SU}(3)$ bundle over a surface $\Sigma$ is the trivial bundle. – Henry T. Horton Apr 13 '13 at 02:37
  • @HenryT.Horton - If $M$ is 2-dimensional and $N$ is 2-connected, why is $[M,N]=0$? (I'm new to homotopy theory.) – Daan Michiels Apr 13 '13 at 20:39
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    @DaanMichiels: The surface can be think as a 2-dimensional CW-complex, you extend the map over each skeleton and since $BSU(3)$ is 2-connected any such extension would be homotopically equivalent. This means ther eis only one bundle (the trivial one) up to homotopy. – Bombyx mori Apr 26 '13 at 05:56

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