9
$\begingroup$

Two questions (more details below):

  1. Let $G$ be a compact Lie group and $X$ a $G$-space such that all stabilizer subgroups are conjugate to a fixed $H \leq G$. Denote by $\pi: X \to X/G$ the quotient map. Under which conditions on $X$ is $\pi$ a Serre fibration?

  2. Let $G$ be as above and $F$ a free $G$-space. Under which conditions is the canonical map $F_{hG} \to F/G$ from the homotopy quotient to the quotient a weak equivalence?

All spaces are assumed to be CGWH spaces.

Relation between 1. and 2.:

I am most interested in 2.. However, if $F$ is a free $G$-space, $EG$ denotes Bar construction, such that $(F \times EG)/G$ models $F_{hG}$, and both $F$ and $F \times EG$ are as in 1., i.e., the respective quotient maps are Serre fibrations, then it is not hard to see that $F_{hG} \to F/G$ is a weak equivalence, using the long exact sequence of homotopy groups.

So, any condition for 1. that holds for $EG$ and is stable under products provides a condition for 2.

Results known to me:

  • A sufficient condition for 1. is being a completely regular Hausdorff space (also known as a Tychonoff space) by a result in Bredon's "Introduction to Compact Transformation groups." He shows that the quotient map is a fiber bundle in this case. However, the Tychonoff property does not seem to be preserved by CGWH products, so it will not hold for many constructions which one would like to have results as in 1. or 2. for.

  • A sufficient condition for 2. should be that $F$ is a (retract of a) free $G$-CW-complex by using model category arguments. However, this also a rather severe condition that can be hard or impossible to check in practice.

Thus, I would like to know if there any other known results, preferably with a reference, that improve the sufficient conditions outlined above. Any results for more arbitrary topological groups will also be appreciated.

Improve this question
$\endgroup$
3
  • $\begingroup$ Are you familiar with the condition (for (2)) that the group action "has slices" (any point $x$ has a set $U$ containing $x$ such that the map $G \times U \to X$ is a homeomorphism onto a neighborhood)? E.g. this is always true for a smooth, free action of a Lie group on a smooth manifold (the slice theorem in differential geometry). This condition is strong enough to make $X \to X/G$ into a fiber bundle with fiber $G$. $\endgroup$
    – Tyler Lawson
    Dec 4, 2016 at 4:35
  • $\begingroup$ Thank you for your comment. As far as I know, the concept of slices only appears in the context of Tychonoff spaces throughout the literature (Bredon, Palais, tom Dieck …). One of the key steps in Bredon's result is the existence of slices for actions on a Tychonoff space. Manifolds are Tychonoff (no smoothness required), so, unfortunately, your condition does not add anything to my list of known results above. $\endgroup$
    – Alexander Körschgen
    Dec 8, 2016 at 2:36
  • $\begingroup$ On a side note: while the Tychonoff property answers 1. (as described in the OP), it is also sufficient for 2. because it is stable under products and EG is Tychonoff (it is normal and Hausdorff). The distinction between the Kelley product and the classical product does not make a difference for the weak homotopy type of $(F \times EG)/G$. $\endgroup$
    – Alexander Körschgen
    Dec 8, 2016 at 2:41

1 Answer 1

Reset to default
3
$\begingroup$

Assertions 1. and 2. hold for spaces that are (compactly generated and) Hausdorff (we still have to require $G$ to be compact Lie). This is Theorems A.10 and A.8 of my recent preprint https://arxiv.org/abs/1612.04267v3. An application, where the "results known to me" from above are not sufficient, can be found there as well.

(It's hard to find out what happens for spaces that are CGWH but not Hausdorff. This is related to the following question: Compact Lie group action on non-Hausdorff (but CGWH) space with Hausdorff quotient)

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.