9

I am trying to understand the notion of a principal $G$-bundle versus a vector bundle. Here $G$ is a Lie group.

Supposedly, principal $G$-bundles are a generalization of vector bundles. My problem here is that most sources, for example the wikipedia page, talks about bundles over $GL_n(\mathbb{R})$ or some other such matrix group. But the fibers of vector bundles are of the form $\mathbb R^n$ and not $GL_n(\mathbb{R})$. So, how are principal bundles the generalization of vector bundles?

On the other hand, is there some kind of correspondence between vector bundles and $GL_n(\mathbb{R})$-bundles, so that principal bundles are in some indirect way a generalization?

Thomas
  • 211

1 Answers1

5

On the other hand, is there some kind of correspondence between vector bundles and $\text{GL}_n(\mathbb{R})$-bundles, so that principal bundles are in some indirect way a generalization?

Yes. Given a principal $G$-bundle and a linear representation $\rho : G \to \text{Aut}(V)$, you get an associated vector bundle whose fibers look like $V$ instead of $G$. This gives you a functor from principal $\text{GL}_n(\mathbb{R})$-bundles to $n$-dimensional vector bundles (taking the standard $n$-dimensional representation) which is an equivalence of categories (the inverse functor is given by taking the frame bundle).

Qiaochu Yuan
  • 419,620
  • 1
    For the benefit of the OP: Given a principal $G$-bundle and any (say, smooth) $G$ action on a manifold $M$, you get an associated bundle with fiber $M$. – Jason DeVito - on hiatus Feb 27 '13 at 19:54
  • For a general G is (not necessarily $GL_n(\mathbb R)$) is it a equivalence of categories? What if we take G is a subgroup of $GL_n(\mathbb R)$? Can we give a reference of the proof of the equivalence in case of G=$GL_n(\mathbb R)$? – Babai Jul 19 '14 at 20:05
  • @Susobhan: is what an equivalence of categories? – Qiaochu Yuan Jul 19 '14 at 20:06
  • The functor which takes an Principal G-bundle and an action of G into V to a Vector bundle with fibre V and Structure group G. – Babai Jul 19 '14 at 20:07
  • @Susobhan: yes (it's not true if you don't add "structure group $G$," which is why I asked), provided that the representation of $G$ on $V$ is faithful. This follows directly from e.g. the Cech cocycle characterization of either bundles with a structure group or principal bundles respectively. – Qiaochu Yuan Jul 19 '14 at 20:10
  • I think the frame bundle is associated only to a vector bundle. So we need the structure group to be $GL_n(\mathbb R)$. Otherwise, the inverse functor is not defined. – Babai Jul 19 '14 at 20:15
  • @Susobhan: off the top of my head, I don't know a clean description of the inverse functor in general, but it's more complicated than just taking the frame bundle in general. For example, if $n = 2k$ and $G = \text{GL}_k(\mathbb{C})$ then we are taking about complex vector bundles and instead of the frame bundle we can take the complex frame bundle. – Qiaochu Yuan Jul 19 '14 at 20:18
  • Okay. I have another doubt. We can associate to Fiber Bundle with Fiber F and structure group G a Principal Bundle G. Also to a principal G-bundle and a action of G into F we can associate a Fiber bundle with F and structure group G. Will this functor will define an equivalence of categories between Principal G bundles and Fiber bundles with Fiber F and structure group G? – Babai Jul 19 '14 at 20:27
  • @Susobhan: yes, again assuming that the action of $G$ on $F$ is faithful, and again it follows directly from the Cech cocycle description of either. – Qiaochu Yuan Jul 19 '14 at 20:33
  • To be an equivalence of categories the functor has to be faithful. But the discussion here " http://math.stackexchange.com/questions/73256/is-the-functor-associating-a-bundle-with-a-structure-group-to-a-principal-bundle " says the functor is faithful only when G is identity. – Babai Jul 19 '14 at 20:37
  • @Susobhan: ah, I left out an important detail, my apologies; on the fiber bundle side one must take morphisms to be isomorphisms. Otherwise "frame bundle" isn't even a functor. – Qiaochu Yuan Jul 19 '14 at 20:44
  • You mean if we take morphisms as isomorphisms of fiber bundles then it will be equivalence of categories even though G is arbitrary group acting faithfully on F? – Babai Jul 19 '14 at 20:47
  • @Susobhan: yes, that's what I'm claiming. – Qiaochu Yuan Jul 19 '14 at 20:50
  • Okay. Thank you for the clarification. Can you give some reference for the proof? – Babai Jul 19 '14 at 20:55
  • 1
    @Susobhan: nope. I would look e.g. in Steenrod's Topology of Fibre Bundles. – Qiaochu Yuan Jul 19 '14 at 21:05