26
$\begingroup$

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). 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).

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?

$\endgroup$
  • 2
    $\begingroup$ I'm traveling and only have my phone, so I can't check this, but the answer might be in a recent paper of Luck and Uribe which I remember seeing on the arxiv within the past six months. The references point to a very complete survey by Luck which might serve as a modern update to the work of May which you cite. Hope that helps. $\endgroup$ – David White Feb 1 '14 at 21:56
  • $\begingroup$ @DavidWhite You mean arxiv.org/abs/1304.4862 ? The closest I see is section 14.2, which proves the case of $\Pi$ compact abelian (in a much more general context). $\endgroup$ – Charles Rezk Feb 1 '14 at 23:02
  • $\begingroup$ Yes, that's the paper I meant, plus Luck's 2004 survey which I found from the references. I'm sorry to hear the answer wasn't in there. I haven't seen much about G equivariant pi bundles elsewhere so that was my best guess $\endgroup$ – David White Feb 2 '14 at 0:53
10
$\begingroup$

Charles, thanks for asking. This is not an answer, but it is too long for a comment. Like you, I encourage others to pursue the question and closely related ones I'll raise here. The paper of mine you cite is miserably written and hard to parse for what is proven in what generality. It is written more generally in terms of extensions $\Gamma$ of $G$ by $\Pi$ rather than just products $G\times \Pi$, and much of it does not restrict to compact Lie groups. Your (2) is its Theorem 10 (which has a typo $\Gamma$ for $G$), and that is just a reinterpretation of the result in the Lashof-May-Segal paper you cite. Its Theorem 5 specializes to say that your map $f$ is an equivalence for any topological group $G$ and discrete group $\Pi$. As you say, that is just an exercise in covering space theory.

More substantially, its Theorem 9 uses consequences of the Sullivan conjecture due to Dwyer and Zabrodsky and Notbohm to partially answer a kind of opposite question to the one you ask. When $G$ (not $\Pi$) is an extension of a finite $p$-group by a torus and $\Pi$ is a compact Lie group it specializes to show that your map $f$ induces an isomorphism on mod $p$ homology, with a stronger statement when $G$ is finite.

The related question goes as follows. One can ask for categorical models of equivariant classifying spaces. Let $G$ be a topological group and let $\tilde G$ denote the ``chaotic'' topological category with object space $G$ and morphism space $G\times G$, so that there is a unique morphism $g\to h$ for each pair of elements. Think of the topological group $\Pi$ as a topological category with a single object. (I'm actually more interested in the more general situation where $G$ acts on $\Pi$.) Then the classifying space of the topological category $Cat(\tilde G,\Pi)$ is another candidate for $B_G\Pi$, and there is a natural map $BCat(\tilde G,\Pi) \to Map(EG,B\Pi)$. At least if $G$ is discrete and $\Pi$ is either discrete or a compact Lie group, $B_G\Pi$ is equivalent to $BCat(\tilde G,\Pi)$ by a result of Guillou, Merling, and myself http://arxiv.org/pdf/1201.5178.pdf. The related questions then are

(1) How generally is $BCat(\tilde G,\Pi)$ equivalent to $B_G\Pi$?

(2) How generally is $BCat(\tilde G,\Pi)\longrightarrow Map(EG,B\Pi)$ a $G$-equivalence?

A key point in Charles' question is that $Map(EG,E\Pi)$ is a universal principal $(G,\Pi)$-bundle in complete generality (by Theorem 5 in my early paper he cites), so that $B_G\Pi$ is $Map(EG,E\Pi)/\Pi$. There is an evident natural map $$ Map(EG,E\Pi)/\Pi \longrightarrow Map(EG,B\Pi).$$ It is a $G$-equivalence if $\Pi$ is discrete, and a rephrasing of Charles' question is to ask ask how generally that map is a $G$-equivalence. An obvious diagram (Section 5 of the GMM paper) relates this question to my questions.

Edit: Maybe worth further clarifying the context of Charles' original question and his own answer below. In the early papers that Charles cites, I study general extensions of $G$ by $\Pi$, at first with no restrictions on the topological groups $G$ and $\Pi$. Charles is taking $G$ to act trivially on $\Pi$ and in that case he is proving that the natural map $ B_G \Pi = Map(EG,\Pi)/\Pi \to Map(EG,B\Pi)$ is a weak $G$-equivalence when $G$ and $\Pi$ are compact Lie groups with $\Pi$ a $1$-type, so a split extension of a torus by a finite group. The early work had only shown that when $\Pi$ is abelian or when $\Pi$ is discrete (in which case the map is a homeomorphism). The related questions (1) and (2) above are about categorical models for equivariant classifying spaces in the intermediate generality of split extensions of $G$ by $\Pi$ (so with $G$ acting non-trivially on $\Pi$), as studied with Guillou and Merling in http://arxiv.org/pdf/1201.5178.pdf.

$\endgroup$
10
$\begingroup$

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).

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 $C^1\times E^0\approx E$.

$\endgroup$
  • 2
    $\begingroup$ Charles, I see that your paper about this has appeared in print: msp.org/agt/2018/18-1/p15.xhtml Congratulations! I'll accept your answer. $\endgroup$ – Charles Rezk Mar 3 '19 at 5:05
  • $\begingroup$ Thanks! :) :) :) $\endgroup$ – Charles Rezk Jun 1 '19 at 17:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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