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Let $G$ be a topological group and let $r \colon E \times G \to E$ be a continuous right-action on a topological space $X$.

If $p\colon E \to B$ is a continuous map into a topological space $B$ such that $(p, r)$ is a principal $G$-bundle, then it follows that $B \cong E/G$ where $E/G$ is the orbit space of the action, and that $p$ is essentially the projection map $\pi\colon E \to E/G$. It follows further that the action must be free.

I'm interested in the opposite direction, i.e. the question when the projection $E \to E/G$ determines a principal $G$-bundle.

In the beginning lines of this article I found the assertion that it is sufficient for the action of $G$ on $E$ to be free (this is obviously necessary) and proper. However, no proof is provided.

If $G$ is discrete, then one can show that the action is proper precisely if every $x \in E$ has a neighbourhood $U$ such that $Ua \cap U = \emptyset$ whenever $a \neq 1$, $a \in G$. Indeed, $V := \pi(U)$ is then the sought-after open neighbourhood of $xG \in E/G$ such that $\pi^{-1}(V) \cong V \times G$. It's not difficult to define a suitable $G$-homeomorphism.

If $G$ is locally compact, $X$ is Hausdorff, and the action is proper, then there is a similar result, for then every $x \in E$ has a neighbourhood $U$ such that $\{a \in G \mid Ua \cap U \neq \emptyset\}$ is contained in a compact set $K \subset G$.

I was wondering if this could be used for a proof. If anyone could help me with other ideas, I'd be very glad to hear them.

PS: I edited this post a bit as I didn't want to submit essentially the same question again.

Lisa
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1 Answers1

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The situation is, in fact, quite complicated.

  1. If $G$ is a Lie group, $X$ is a completely regular Hausdorff topological space (e.g. a metrizable space) and $G\times X\to X$ is a Palais-proper (see below) free continuous action, then the projection map $X\to X/G$ is indeed a principal $G$-bundle. This (and much more) is proven in (an unnumbered theorem on page 315, section 4.1)

R. Palais, On the existence of slices for actions of non-compact Lie groups. Ann. of Math. (2) 73 (1961) 295-323

although, according to Palais, the result was established earlier by Serre (notes from a Bourbaki seminar). (It takes some time to untangle the terminology used by Palais to understand what his theorem says.) The case when $G$ is a compact Lie group was proven earlier by Gleason.

Here an action is Palais-proper if for every pair of points $x, y\in X$ there exists a pair of their respective neighborhoods $U_x, U_y$ such that the subset $$ \{g\in G: gU_x \cap U_y \ne \emptyset\} $$ is relatively compact in $G$. If $X$ is locally compact, this definition is equivalent to the standard one.

  1. On the other hand, if $G$ is any locally compact metrizable group which is not a Lie group (e.g. the group of $p$-adic integers, which is compact and totally disconnected), then there exists a free Palais-proper continuous $G$-action $G\times X\to X$ on a metrizable space $X$, such that the projection $X\to X/G$ is not a principal fiber bundle. This (which I find quite astounding) is a corollary of Theorem 6 in

S. A. Antonian, Equivariant embeddings into G-AR's, Glasnik Matematički 22 (42) (1987), 503–533.

Namely, Antonian constructs a certain continuous linear actions $G\times B\to B$ on a Banach space and for some vector $v\in B$ with trivial $G$-stabilizer, observes nonexistence of an equivariant retraction of any neighborhood $X\subset B$ of $Gv$ to the orbit $Gv$.

Note that Antonian's examples are never locally compact.

Edit. In fact, there is an example of a compact 1-dimensional Hausdorff space $X$, a free action of a compact metrizable group $G\times X\to X$, such that $X/G$ has dimension 2. In particular, the projection $X\to X/G$ cannot be a locally trivial fiber bundle. The example is mostly due to Kolmogorov

A. Kolmogoroff, Über offene Abbildungen, Ann. of Math. (2) 38 (1937), 36-38

while the version for free action can be found, in English, here,

R. F. Williams, A useful functor and three famous examples in topology. Trans. Amer. Math. Soc. 106 (1963) 319–329.

Moishe Kohan
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  • Do you know if at least one can guarantee that $X/G$ has finite covering dimension when $G$ is a compact, finite dimensional group acting freely on a compact finite dimensional space $X$? – Shirly Geffen Mar 02 '18 at 12:18
  • @ShirlyGeffen: I do not have such examples but I expect negative answer. If you really need this, consider asking Alexandr Dranishnikov, he is at University of Florida. – Moishe Kohan Mar 02 '18 at 15:10
  • @MoisheCohen: Can you say a few more words about why $\dim X/G=2$ implies that the projection cannot be locally trivial? I don't know much about topological dimension, but my understanding is that the dimension of a product is not necessarily equal to the product of the dimensions. – Jack Lee Feb 05 '19 at 22:24
  • @JackLee: This is because you cannot embed a 2-dimensional space into a 1-dimensional space (I am not claiming additivity of dimensions here, of course): If $U_i\subset X/G$ were finitely many neighborhoods over which this were a trivial bundle, then one of these would have dimension $=2$ and then a local trivialization would embed it into $X$. – Moishe Kohan Feb 06 '19 at 00:14
  • Ah, I see. Thank you! – Jack Lee Feb 06 '19 at 00:22