Given compact Lie groups $G$ and $\Pi$, there is a notion of "$G$-equivariant principal $\Pi$-bundle", and a corresponding notion of classifying space, often denoted $B\_G\Pi$, so that $G$-equivariant maps $X\to B\_G\Pi$ (up to equivariant homotopy) correspond to $G$-equivariant $\Pi$-bundles on $X$ (up to equivalence), for nice $X$.


There is a natural map of $G$-spaces
$$ f\colon B\_G\Pi \to \mathrm{Map}(EG, B\Pi), $$
which in some sense encodes the fact that a $G$-equivariant $\Pi$-bundle $P\to X$ gives rise to a principal $\Pi$-bundle of the form $P\times\_G EG\to X\times\_G EG$.


It's natural to ask whether $f$ is a $G$-equivariant weak homotopy equivalence. This amounts to showing that for any closed subgroup $H\leq G$, the induced map on fixed points
$$ f^H\colon (B\_G\Pi)^H \to \mathrm{Map}(EG,B\Pi)^H\approx\mathrm{Map}(BH, B\Pi)$$
is a weak equivalence. In general, this is not the case, but it is known to be true for arbitrary compact $G$ when either


1. $\Pi$ is finite, or
2. $\Pi$ is compact abelian.


The reference for both of these is: [JP May, Some remarks on equivariant bundles and classifying spaces.
Astérisque No. 191 (1990)](http://www.math.uchicago.edu/~may/PAPERS/65.pdf).
Case 1 follows from covering space theory, while case 2 needs a bit of work and is proved in [Lashof, May, Segal,
Equivariant bundles with abelian structural group.
Contemp. Math. 19, (1983)](http://www.math.uchicago.edu/~may/PAPERS/45.pdf).


The obvious guess is that there is a common generalization, where 1. or 2. is replaced by 


* $\Pi$ is an extension of a finite group by a compact torus (or more cleanly, $\Pi$ is a compact Lie group with contractible simply connected cover).


Is this true? Has someone proved it? 



Equivariant classifying spaces from classifying spaces
**Added Aug 2016:** I've written this up, available at <https://arxiv.org/abs/1608.02999>


$\def\Hom{\mathrm{Hom}} \def\Map{\mathrm{Map}} \def\ad{\mathrm{ad}}$


I think this is true. I'll sketch a possible proof here. I haven't
carefully checked everything, and there are things that need checking. Feel free to do that.


First, we can assume $H=G$: we want to show that $(B\_G\Pi)^G\to
\mathrm{Map}(BG,B\Pi)$ is an equivalence if $G$ is
compact Lie and $\Pi$ is compact Lie and a 1-type. 


We might as well
consider the induced map on homotopy fibers over the maps to $B\Pi$ (induced by
evaluating at the basepoint of $BG$.) That is, we want to show
$$
\Hom(G,\Pi) \to \Map\_\*(BG,B\Pi)
$$
is a weak equivalence. Here $\Hom(G,\Pi)$ is topologized as a
subspace of $\Map(G,\Pi)$. We know that this is homeomorphic to
$\coprod\_{[\phi]} \Pi/C\_\Pi(\phi(G))$, a coproduct over conjugacy
classes of homomorphisms $G\to \Pi$ (see
[Nearby homomorphisms from compact Lie groups are conjugate](https://mathoverflow.net/questions/123624/nearby-homomorphisms-from-compact-lie-groups-are-conjugate)). 


Given this, it is already clear we get an equivalence when $G$ is
compact and *connected* (reduce to the case where $\Pi$
is a torus). It's not so easy to see why this is so for general $G$:
although we can "compute" both sides, the accounting is different and
hard to match up. 


Here's an attempt at a general proof, based on the ideas which work
when $\Pi$ is a torus (which involve the idea of continuous cochains
as in Graeme Segal, "Cohomology of topological groups"). It should fit into some already-known technology (cohomology of topological groups with coefficients in a topological 2-group?), but I don't want to bother to figure out what or how.


Consider data consisting of 


* a group $\Pi$ (a compact Lie 1-type as above), with connected
component $\Pi\_0$ (which is abelian),
* a vector space $V$,
* a group homomorphism $\exp\colon V\to \Pi$,
* an action $\ad\colon \Pi/\Pi\_0\to \mathrm{Aut} V$,
* such that $\exp(\ad(\pi)v)=\pi \exp[v] \pi^{-1}$.


(It's a kind of crossed module.) Given this, define $E(G, (\Pi,V))$
to be the space of pairs $(f,v)$ where $f\colon G\to \Pi$ and $v\colon
G\times G\to V$ are continuous maps, satisfying


* $f(g\_1)f(g\_2)=\exp[ v(g\_1,g\_2)] f(g\_1g\_2)$,
* $v(g\_1,g\_2)+v(g\_1g\_2,g\_3)= \ad(f(g\_1))v(g\_2,g\_3) + v(g\_1,g\_2g\_3)$.


(I might want to additionally require a normalization: $f(e)=e$. Or not.)


The examples I have in mind are $E:= E(G, (\Pi, T\_e\Pi))$ and $E^0:=
E(G, (\Pi,0))$. The claims are as follows.


**$E$ is weakly equivalent to $\Map\_\*(BG,B\Pi)$.** To compute the
mapping space, you need to climb the cosimplicial space $[k] \mapsto
\Map\_\*(G^k, B\Pi)$. Because $B\Pi$ is a 2-type, you don't need to
climb very far. The idea is that if you do this, and you keep in mind
facts such as:


* $\Pi$ is equivalent to $\Omega B\Pi$, and
* the fibration $(v,\pi)\mapsto (\pi, \exp[v]\pi) \colon V\times\Pi\to \Pi\times \Pi$ is equivalent to the free path fibration $\Map([0,1],\Pi)\to \Pi\times \Pi$,


you see that you get an equivalence. (I came up with the definition
of $E$ exactly by doing this.) 


That's kind of sketchy. 


*More concretely:* $\Map\_\*(BG,B\Pi)$ can be identified with the space of maps between pointed simplicial spaces, from $G^\bullet$ to $S\_\bullet:=\bigl([n]\mapsto \Map\_\*(\Delta^n/\mathrm{Sk}\_0\Delta^n, B\Pi)\bigr)$. The space $E$ is also a space of maps between such, from $G^\bullet$ to $N\_\bullet$, where $N\_\bullet$ is a simplicial space built from the crossed module $(\Pi,V)$ (the nerve of the crossed module, as in <https://mathoverflow.net/q/86486> ) with $N\_n \approx \Pi^n\times V^{\binom{n}{2}}$. It's not to hard to show that $N\_\bullet$ and $S\_\bullet$ are weakly equivalent Reedy fibrant simplicial spaces; they both receive a map from $\Pi^\bullet$, which exhibits the equivalence. (But note: showing that $N\_\bullet$ is Reedy fibrant relies crucially on the fact that $\exp$ is a *covering map*.)


**$E^0$ is homeomorphic to $\Hom(G,\Pi)$.** Yup.


**The inclusion $E^0\subseteq E$ is a weak equivalence.**


To see this, let $C^1:=\Map(G,V)$, as a topological group under
pointwise addition. There is an action $C^1\curvearrowright E$, by
$u\cdot (f,v)=(f',v')$ where


* $f'(g) := \exp[u(g)] f(g)$,
* $v'(g\_1,g\_2) := u(g\_1)-u(g\_1g\_2) + \ad(f(g\_1))u(g\_2) + v(g\_1,g\_2)$.


It's useful to note that for any $(f,v)\in E$, the resulting map
$G\xrightarrow{f} \Pi\to \Pi/\Pi\_0$ is a homomorphism. Thus we write
$E=\coprod E\_\gamma$ for $\gamma\in \Hom(G,\Pi/\Pi\_0)$, and $C^1$ acts
on each $E\_\gamma$. 


Consider $(f,0)\in E\_\gamma^0= E\_\gamma\cap E^0$. Note that $u\cdot (f,0)$ has the form $(f',0)$ for some $f'$ if and
only if $u\in Z^1\_\gamma$, where this is the set of $u\colon G\to V$
such that


* $u(g\_1)-u(g\_1g\_2) + \ad\gamma(g\_1) u(g\_2)=0$.


So the action passes to an injective map $C^1\times\_{Z^1\_\gamma} E^0\_\gamma\to
E\_\gamma$. In fact, it should be a homeomorphism. To see that it's
surjective, fix 
$(f,v)\in E\_\gamma$; we need to solve for $u\in C^1$ such that 


* $u(g\_1)-u(g\_1g\_2)+\ad\gamma(g\_1)u(g\_2)=v(g\_1,g\_2)$.


This amounts to the vanishing of $H^2$ in the complex $C^\bullet\_\gamma$ of
continuous 
cochains: $C^t\_\gamma:=\Map(G^{t}, V\_\gamma)$ (where the differential uses
the action $\ad\gamma\colon G\to\mathrm{Aut}(V)$). The
vanishing is because $G$ is compact, so we can "average" over Haar
measure to turn a non-equivariant contracting homotopy on
$D^\bullet\_\gamma=\Map(G^{\bullet+1}, V\_\gamma)$ into a contracting homotopy
on $C^\bullet\_\gamma = (D^\bullet\_\gamma)^G$. 


Given this, since both $C^1$ and $Z^1\_\gamma$ are
contractible groups, (in fact, $Z^1\_\gamma=V/V^{\gamma(G)}$ by $H^1=0$), we
should have that $C^1\times\_{Z^1\_\gamma} E^0\_\gamma$ is weakly
equivalent to $E^0\_\gamma$. 


Note: in the case that $\Pi$ is abelian, we simply get a homeomorphism
$C^1\times E^0\approx E$. 


