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1. • CommentRowNumber1.
   • CommentAuthorDavidRoberts
   • CommentTimeMar 22nd 2012
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   Author: DavidRoberts
   Format: MarkdownItexThis is just to check if I have remembered something correctly. Let $G$ be a (nice enough) topological group, and let $X$ be a manifold with an action (proper, actually) by a Lie group $H$. Let $X//H := (X\times EH)/H$ be the Borel construction. Is it true that the set of isomorphism classes of $H$-equivariant principal $G$-bundles is given by $[X//H,BG]$? Where would I find a reference/proof of this?

   This is just to check if I have remembered something correctly.

   Let $G$ be a (nice enough) topological group, and let $X$ be a manifold with an action (proper, actually) by a Lie group $H$. Let $X//H := (X\times EH)/H$ be the Borel construction. Is it true that the set of isomorphism classes of $H$-equivariant principal $G$-bundles is given by $[X//H,BG]$? Where would I find a reference/proof of this?

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeMar 22nd 2012
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   Author: Urs
   Format: MarkdownItex* Peter May, _Some remarks on equivariant bundles and classifying spaces_ ([pdf](http://www.math.uchicago.edu/~may/PAPERS/65.pdf))

   • Peter May, Some remarks on equivariant bundles and classifying spaces (pdf)
3. • CommentRowNumber3.
   • CommentAuthorDavidRoberts
   • CommentTimeMar 23rd 2012
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   Author: DavidRoberts
   Format: MarkdownItexThanks, Urs. Ironically, I had found that just earlier, but didn't make the effort to read the poorly-scanned article past the first page. This is just a warmup. In fact, I need to replace the action groupoid $H \rtimes X$ with a more general Lie groupoid, but that shouldn't be too hard (I know what the Borel construction becomes in that case, it's the other part that needs doing). From an $(\infty,1)$-categorical point of view, we have two functors $LieGpd \to Top$, one being the Borel construction, the other being the nerve followed by geometric realisation. From the result I first mentioned, $[X//H,BG]$ gives the right connected components of the hom-space $Hom(H \rtimes X,\mathbf{B}G)$. But what about $[|H \rtimes X|,BG]$? It seems to me to be a little more difficult to show this is also the set of iso classes of equivariant bundles (well, certainly harder than the other one, which I think I could prove in a short paragraph).

   Thanks, Urs. Ironically, I had found that just earlier, but didn’t make the effort to read the poorly-scanned article past the first page.

   This is just a warmup. In fact, I need to replace the action groupoid $H \rtimes X$ with a more general Lie groupoid, but that shouldn’t be too hard (I know what the Borel construction becomes in that case, it’s the other part that needs doing).

   From an $(\infty,1)$-categorical point of view, we have two functors $LieGpd \to Top$, one being the Borel construction, the other being the nerve followed by geometric realisation. From the result I first mentioned, $[X//H,BG]$ gives the right connected components of the hom-space $Hom(H \rtimes X,\mathbf{B}G)$. But what about $[|H \rtimes X|,BG]$? It seems to me to be a little more difficult to show this is also the set of iso classes of equivariant bundles (well, certainly harder than the other one, which I think I could prove in a short paragraph).

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeMar 23rd 2012
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   Author: Urs
   Format: MarkdownItex> but didn't make the effort to read the poorly-scanned article Yeah, it takes a bit of effort. I also only now realized that the theorem 1 there assumes the structure group to be discrete. Which is probably not what you want.

   but didn’t make the effort to read the poorly-scanned article

   Yeah, it takes a bit of effort. I also only now realized that the theorem 1 there assumes the structure group to be discrete. Which is probably not what you want.

5. • CommentRowNumber5.
   • CommentAuthorUrs
   • CommentTimeMar 23rd 2012
   • (edited Mar 23rd 2012)
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   Author: Urs
   Format: MarkdownItex> But what about $[|H \rtimes X|,BG]$? I need to think about this. One special case which I can easily answer (but which will probably not be the case that you are interested in) is: $G$ discrete (as in the May article, in fact (where "$G$" is "$\Pi$")) and $H$ and $X$ both paracompact manifolds. In that case we may look at the situation inside [[ETop∞Grpd]] and then use the $(\Pi \dashv Disc)$-adjunction to compute $$ \begin{aligned} \pi_0 Hom(H \rtimes X, \mathbf{B}G) & \simeq \pi_0 Hom(H \rtimes X, Disc\mathbf{B}G) \\ & \simeq [\Pi(H \rtimes X), B G] \\ & \simeq [| H \rtimes X|, B G] \end{aligned} \,, $$ where in the last step I used the results of the discussion of geometric realization [here](http://ncatlab.org/nlab/show/Euclidean-topological+infinity-groupoid#GeometricHomotopy) at _[[ETop∞Grpd]]_ . I need to think about what one can say for non-discrete $G$. But probably not tonight...

   But what about $[|H \rtimes X|,BG]$?

   I need to think about this.

   One special case which I can easily answer (but which will probably not be the case that you are interested in) is: $G$ discrete (as in the May article, in fact (where “$G$” is “$\Pi$”)) and $H$ and $X$ both paracompact manifolds. In that case we may look at the situation inside ETop∞Grpd and then use the $(\Pi \dashv Disc)$-adjunction to compute

   $$\begin{aligned} \pi_0 Hom(H \rtimes X, \mathbf{B}G) & \simeq \pi_0 Hom(H \rtimes X, Disc\mathbf{B}G) \\ & \simeq [\Pi(H \rtimes X), B G] \\ & \simeq [| H \rtimes X|, B G] \end{aligned} \,,$$

   where in the last step I used the results of the discussion of geometric realization here at ETop∞Grpd .

   I need to think about what one can say for non-discrete $G$. But probably not tonight…

6. • CommentRowNumber6.
   • CommentAuthorDavidRoberts
   • CommentTimeMar 23rd 2012
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   Author: DavidRoberts
   Format: MarkdownItexThat's cool, no need to jump on it. I'm quite happy to work completely with the Boral construction version, provided it will work for non-discrete $G$

   That’s cool, no need to jump on it. I’m quite happy to work completely with the Boral construction version, provided it will work for non-discrete $G$

7. • CommentRowNumber7.
   • CommentAuthorUrs
   • CommentTimeMar 23rd 2012
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   Author: Urs
   Format: MarkdownItexWe need an $n$Lab entry on the Borel construction and how it essentially is just the geometric realization of the action groupoid.

   We need an $n$Lab entry on the Borel construction and how it essentially is just the geometric realization of the action groupoid.

8. • CommentRowNumber8.
   • CommentAuthorDavidRoberts
   • CommentTimeMar 23rd 2012
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   Author: DavidRoberts
   Format: MarkdownItexAh, silly me, of course they're the same. And May's paper is actually not very helpful, it assumes that the structure group of the bundle and the group acting on the space are related. I'm not assuming that. Hmm, MO time?

   Ah, silly me, of course they’re the same. And May’s paper is actually not very helpful, it assumes that the structure group of the bundle and the group acting on the space are related. I’m not assuming that. Hmm, MO time?

9. • CommentRowNumber9.
   • CommentAuthorUrs
   • CommentTimeMar 23rd 2012
   • (edited Mar 23rd 2012)
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   Author: Urs
   Format: MarkdownItexI started an entry _[[Borel construction]]_, indicating how to show that it is $\vert X // G\vert$. > it assumes that the structure group of the bundle and the group acting on the space are related. Not necessarily. If you take $\Gamma = \Pi \times G$ then the theory just gives plain $G$-equivariant $\Pi$-principal bundles. But it can also handle more general cases.

   I started an entry Borel construction, indicating how to show that it is $\vert X // G\vert$.

   it assumes that the structure group of the bundle and the group acting on the space are related.

   Not necessarily. If you take $\Gamma = \Pi \times G$ then the theory just gives plain $G$-equivariant $\Pi$-principal bundles. But it can also handle more general cases.

10. • CommentRowNumber10.
    • CommentAuthorDavidRoberts
    • CommentTimeMar 23rd 2012
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    Author: DavidRoberts
    Format: MarkdownItexAh, that's true. Duh...

    Ah, that’s true. Duh…

11. • CommentRowNumber11.
    • CommentAuthorDavidRoberts
    • CommentTimeMar 23rd 2012
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    Author: DavidRoberts
    Format: MarkdownItexI have asked this on MO: <http://mathoverflow.net/questions/91973/>

    I have asked this on MO: http://mathoverflow.net/questions/91973/

12. • CommentRowNumber12.
    • CommentAuthorDavidRoberts
    • CommentTimeMar 23rd 2012
    • (edited Mar 23rd 2012)
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    Author: DavidRoberts
    Format: MarkdownItexHmm, Peter May doesn't think it is true. The problem is, I know that my group G is homotopically very simple (actually homotopic to an abelian topological group, but certainly not as a topological group). But at this stage my proof (of a bigger theorem) won't allow me to use that fact.

    Hmm, Peter May doesn’t think it is true.

    The problem is, I know that my group G is homotopically very simple (actually homotopic to an abelian topological group, but certainly not as a topological group). But at this stage my proof (of a bigger theorem) won’t allow me to use that fact.

13. • CommentRowNumber13.
    • CommentAuthorjim_stasheff
    • CommentTimeMar 23rd 2012
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    Author: jim_stasheff
    Format: Text@12 my group G is homotopically very simple (actually homotopic to an abelian topological group, but certainly not as a topological group) assume you mean homotopy equivalent with neither map of the equivalence being multiplicative? not even up to homotopy?

    @12 my group G is homotopically very simple (actually homotopic to an abelian topological group, but certainly not as a topological group)
    
    assume you mean homotopy equivalent with neither map of the equivalence being multiplicative? not even up to homotopy?
14. • CommentRowNumber14.
    • CommentAuthorDavidRoberts
    • CommentTimeMar 23rd 2012
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    Author: DavidRoberts
    Format: MarkdownItexThat is true. it is a product, up to homotopy, of Eilenberg-MacLane spaces. Unless I can reconstruct the Postnikov tower in the category of topological groups, or at least H-spaces, I can't assume multiplicativity.

    That is true. it is a product, up to homotopy, of Eilenberg-MacLane spaces. Unless I can reconstruct the Postnikov tower in the category of topological groups, or at least H-spaces, I can’t assume multiplicativity.

15. • CommentRowNumber15.
    • CommentAuthorDavidRoberts
    • CommentTimeMar 23rd 2012
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    Author: DavidRoberts
    Format: MarkdownItexTyler Lawson pointed out, no pun intended, that one needs pointed maps, or it is clearly false.

    Tyler Lawson pointed out, no pun intended, that one needs pointed maps, or it is clearly false.