Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose I have a surjective homomorphism of topological groups $f:E \to G$. Let K be the kernel of f. The topological group K acts on E in an obvious way. When is this a fiber bundle over G? (It will then be a K-principal bundle over G).

I'm writing a paper where I make a claim about when this holds. I thought I had a reference but when I went looking for it, my claim was not in the reference.

I don't want to consider examples that are too pathological so lets assume everything is Hausdorff and paracompact. (However if people are familiar with the more general setting, I'd be curious about that too!).

Clearly a necessary condition is that $G \cong E/K$, so let's assume this is the case. By homogeneity it is enough to show that f admits a local section in a neighborhood of the identity element of G. So my question is equivalent to asking if there are conditions I can impose on E and G which will ensure that f admit local sections near the identity of G.

I know by work of G. Segal ("Cohomology of Topological Groups" Symposia Math. Vol IV 1970 pg 377, in the appendix) that if G is abelian and locally contractible then the sequence $$G \to EG \to BG$$ is of this kind.

I want to know:

  1. Does this hold when K is locally contractible?
  2. What if K is globally contractible?
  3. Are there any simple (but not tautological) conditions I can put on K, E, or G to make this hold?

I'd also like to know some reasonable examples where this fails to be a principal bundle (if there are any).

share|improve this question
    
Is there perhaps a variant of Ehresmann's theorem (en.wikipedia.org/wiki/Ehresmann's_theorem) which contains the case you are interested in? –  José Figueroa-O'Farrill Apr 2 '10 at 19:54
1  
Does Ehresmann's theorem hold for infinite dimensional manifolds? The Wikipedia page is not specific. In any case, I would like a statement for locally contractible topological groups and would prefer to avoid assuming we are working with manifolds and smooth maps. –  Chris Schommer-Pries Apr 2 '10 at 20:12
    
Sorry -- I'm usually in the smooth (finite-dimensional) category and that's the version of Ehresmann I know, hence my question about the existence of suitable variants. Perhaps someone here knows. –  José Figueroa-O'Farrill Apr 2 '10 at 20:59
add comment

1 Answer 1

The following statement follows from results of Palais (see theorem 2.3.3 in "On the existence of slices for actions of non-compact Lie groups"). When $G$ is a Lie group, any principal $G$-bundle (in the sense of Husemoller's book, which does not assume local triviality) whose total space is completely regular is actually locally trivial.

It follows that when $H$ is a Lie group, and a closed subgroup of a Hausdorff topological group $G$, then $G\to G/H$ is a locally trivial principal $H$-bundle.

In a slightly different direction, it is a theorem of Skljarenko (theorem 15 of "On the topological structure of locally bicompact groups and their quotient spaces") that when $G$ is a locally compact Hausdorff group, and $H$ is a closed subgroup of $G$, then $G\to G/H$ is a Hurewicz fibration. In particular, if $G/H$ is (relatively) locally contractible then $G\to G/H$ is a locally trivial principal $H$-bundle. Again, by a result of Skljarenko, it is also enough to assume that $G/H$ is finite dimensional to conclude local triviality. See also Mostert's "Sections in principal fibre spaces".

share|improve this answer
    
Does the theorem of Palais also apply when G is an infinite dimensional Lie group? Unfortunately the cases of most interest for me are when G is something like a loop space or a classifying space or some hybrid. In particular the assumption of being locally compact is much too strong. –  Chris Schommer-Pries May 8 '12 at 14:16
    
The article of Palais unfortunately discusses only groups which are locally compact Hausdorff. Moreover, the results on local triviality seem to be only for Lie groups. –  Ricardo Andrade May 10 '12 at 22:41
add comment

Your Answer

 
discard

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.