Let $G$ be a (nice enough) topological group (actually a filtered colimit of compact Lie groups), and let $X$ be a manifold with an action (a proper one in fact) by a Lie group $H$. Let $X//H := (X\times EH)/H$ be the Borel construction.

Does the following claim appear in the literature:

Claim: The set of isomorphism classes of $H$-equivariant principal $G$-bundles is given by $[X//H,BG]$

For the case when $H$ is a compact Lie group and $G$ is an abelian Lie group, then this is proved in a 1983 paper by Lashof, May and Segal,

Equivariant bundles with Abelian structure group, Contemporary Math 19 (1983) 167--176. (pdf)

And some related cases are treated by May in

Some remarks on equivariant bundles and classifying spaces, Asterisque 191 (1990) 239--263 (pdf)

The claim seems intuitively true, but I can't seem to be able to find it, so my intuition may be off. If it is true, then surely May would have mentioned it (perhaps he does and for some reason I missed it).

EDIT: May also treats the topic in chapter VII of the book Equivariant homotopy and cohomology theory, but it seems a compilation of older work.

Another way to view the question is to ask if the categories of H-equivariant G-bundles on X and ordinary G-bundles of X//H are equivalent. There is some work in Lashof-May-Segal where they refrain from assuming G is abelian (there called A, however), I'm digesting it at the moment.

Also note I'm not asking for a classifying space for equivariant bundles - this is a much different question.

  • $\begingroup$ You at least need to take based maps. If you don't, then this doesn't even work for $X=*$ and finite groups if G is nonabelian. $\endgroup$ Mar 23, 2012 at 15:04
  • 1
    $\begingroup$ Probably no one cares anymore, but look at my answer to mathoverflow.net/q/156408/437 $\endgroup$ Feb 25, 2021 at 22:47
  • $\begingroup$ @CharlesRezk thanks for pointing it out. Indeed, this idea that I was pursuing was never going to work for what I was hoping it would, but it's good to collect related results that are interesting for others. $\endgroup$
    – David Roberts
    Feb 25, 2021 at 23:13

1 Answer 1


That guy that keeps getting mentioned here never claimed it in general because he does not believe it in general. Think about $G=U(n)$ and about equivariant $K$-theory. This is very close to the Atiyah-Segal completion theorem. The result for Abelian structure groups still seems somewhat surprising to him.

  • $\begingroup$ Hmm, ok. I was also thinking that there might be a conceptual explanation for why it might (or might not!) hold, coming from all this new-fangled higher category theory. Namely, can we compute hom-spaces between simplicial spaces (properly) by considering their geometric realisations. $\endgroup$
    – David Roberts
    Mar 23, 2012 at 3:14
  • $\begingroup$ Clearly here only taking simplicial spaces which are the nerves of groupoids. $\endgroup$
    – David Roberts
    Mar 23, 2012 at 3:15
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    $\begingroup$ Better you should get higher category theory off your brain: it only distracts here. Just look at representations versus bundles. $\endgroup$
    – Peter May
    Mar 23, 2012 at 3:18
  • $\begingroup$ It's not what I was going to use. incidentally, my group G is homotopic to an abelian group, does that help? $\endgroup$
    – David Roberts
    Mar 23, 2012 at 3:42
  • $\begingroup$ I don't think I'm going to get any other answers after this... :) $\endgroup$
    – David Roberts
    Mar 26, 2012 at 1:54

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