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

If $H$ is a closed subgroup of a topological group $G$, then the orbit map $G\to G/H$ is a principal bundle, yet somewhat surprisingly, it need not be locally trivial. In the wikipedia article on fiber bundles it is claimed that if $H$ is a Lie group, then $G\to G/H$ is locally trivial. Is the claim true, and if so, what is the reference?

Remarks:

  1. That $G\to G/H$ is a principal bundle is explained e.g. in Husemoller's "Fiber bundles", example 2.4 in the 3rd edition. In the same section one can also find a definition of a principal bundle (which does not require local triviality).

  2. A simple example when $G\to G/H$ is not locally trivial can be found in the paper of Karube [On the local cross-sections in locally compact groups, J. Math. Soc. Japan 10 1958 343–347]. In the example $G$ is the product of infinitly many circles, and $H$ is the product of their order $2$ subgroups; there can be no cross-section because $G$ is locally-connected and $H$ is not, so $G$ is not even locally homeomorphic to $H\times G/H$.

  3. In the same paper Karube proves that $G\to G/H$ is locally trivial in a number of cases, including when $G$ is locally compact, and $H$ is a Lie group.

UPDATE: If $H$ is a Lie group, Palais's paper mentioned in his answer actually characterises the principal $H$-bundles that are locally trivial; details are below.

For a topological group $H$ acting freely and by homeomorphisms on a space $X$, we let $X^\ast$ be the subsets of $X\times X$ consisting of pairs $(x,hx)$ where $x\in X$ and $h\in H$. Since $H$ acts freely, there is a map $t: X^\ast\to H$ given by $t(x,hx)=h$.

Theorem 4.1 of Palais's paper says that if the space $X$ is completely regular, and if $H$ is a Lie group, then the free $H$-space $X$ is locally trivial if and only if the map $t$ is continuous.

Note that in the terminology of Husemoller's "Fiber bundles" book continuity of $t$ is assumed in the definition of a $H$-principal bundle, thus Husemoller's $H$-principal bundles are all locally trivial (provided $H$ is a Lie group and $X$ is completely regular).

If $X$ is a topological group and $H$ is a subgroup, then continuity of $t$ follows from continuity of multiplication and inverse in $X$. It is fun to see why Palais's result doesn't show that the $\mathbb Z$-action on $S^1$ by irrational rotation is a principal bundle: here $X=S^1$, and $H$ is the subgroup $\{e^{in}: n\in \mathbb Z\}$ with the subspace topology. The map $t$ is continuous, but $H$ is not a Lie group.

share|improve this question
    
(just edited the title to show it is about G/H) –  Qfwfq Mar 1 '11 at 18:36
    
I asked this question on mathstackexchange and got a nice answer and comment: math.stackexchange.com/questions/23739/… –  Eric O. Korman Mar 1 '11 at 19:11
    
@Eric, the answer/comment you got at stackexchange did not really address your question there, I think. –  Igor Belegradek Mar 1 '11 at 19:20
add comment

1 Answer 1

up vote 19 down vote accepted

...if $H$ is a Lie group, then $G \to G/H$ is locally trivial. Is the claim true, and if so, what is the reference?

Yes, it is true. See the Corollary in section 4.1 of: "On the Existence of Slices for Actions of Non-compact Lie Groups", which you can download here: http://vmm.math.uci.edu/ExistenceOfSlices.pdf

This is a paper originally published in the March 1961 Annals of Math.

The Corollary says that "If X is a topological group and G is a closed Lie subgroup of X then the fibering of X by left G cosets is locally trivial."

share|improve this answer
1  
wonderful! Thank you. –  Igor Belegradek Mar 1 '11 at 19:04
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.