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Let $X$ be a reasonable topological space (let's say it has the homotopy type of a CW complex) and let $G$ be a topological group acting on that space. Let $E_G \rightarrow B_G$ be the universal bundle. Then the equivariant cohomology $H^*_G(X)$ is defined as $H^*(E_G \times_G X)$. We call the space $E_G \times_G X$ the homotopy quotient space. We always have a map $E_G \times_G X \rightarrow G \backslash X$, where $G \backslash X$ denotes the naive quotient space. This induces a map $H^* (G\backslash X) \rightarrow H^*_G(X)$.

I'm curious about the following proof. From now on, we will take cohomology with rational coefficients. Suppose the action of $G$ on $X$ has finite stabilizers. Then the map $H^* (G\backslash X) \rightarrow H^*_G(X)$ is in fact an isomorphism. This is proved in the following notes by Michel Brion (http://www-fourier.ujf-grenoble.fr/~mbrion/notesmontreal.pdf, see p. 4). The proof is as follows: the fibers of the map $E_G \times_G X \rightarrow G \backslash X$ are of the form $E_G/G_x$, where $G_x$ is the stabilizer of a point $x \in X$. Since $G_x$ is a discrete group, we have $\pi_1 (E_G/G_x) = G_x $ and vanishing of all higher homotopy groups. Since $G_x$ is finite, all the homotopy groups vanish when we tensor with $\mathbb{Q}$. I think this is what Brion means by "$\mathbb{Q}$-acyclic".

Then it is claimed that this is sufficient to prove that the map $H^* (G\backslash X) \rightarrow H^*_G(X)$ is an isomorphism. Why is this true? Could someone please explain this argument to me?

Even if the fibers were contractible, it seems to me that you would need this map to be a Serre fibration to prove a homotopy equivalence. Is it true that any surjective map with contractible fibers is a homotopy equivalence?

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3 Answers 3

up vote 8 down vote accepted

The argument in the quoted paper is a bit too sketchy. An actual proof will have two parts:

  1. Let $f:X \to Y$ be map. Assume that for each $y \in Y$, $\tilde{H}^{\ast} (f^{-1}(y);\mathbb{Q})=0$ (this is what the phrase ''$\mathbb{Q}$-acyclic'' means). Find conditions that guarantee that $f$ induces an isomorphism in cohomology. Essentially, one needs to provide assumptions that the Leray spectral sequence of the map is as nice as claimed in Marks answer.

  2. Provide conditions for group action of a topological group $G$ on a space $X$ such that $EG \times_G X \to X/G$ has the properties found in 1 (and the answer in 1 depends on the example you want to consider).

ad 1. Just having finite stabilizers is not enough. Consider the action of $G=\mathbb{Z}$ on $X=S^1$ by an irrational rotation. Then $EG \times_G S^1 \simeq S^1 \times S^1$. The quotient space is an uncountable set with the trivial topology. Because each map into this space is continuous, it is contractible and has trivial (Cech or singular) cohomology.

One needs criteria that show that a map $f:X \to Y$ with contractible/acyclic/$\mathbb{Q}$-acyclic fibres is itself a weak equivalence/acyclic/$\mathbb{Q}$-acyclic. Having contractible fibres does not suffice: look at the identity map $[0,1]^{\delta} \to [0,1]$, where $\delta$ means ''discrete topology''.

Such problems are solved by what I would call ''fibre theorems''. You wish to know how close the natural map $f^{-1}(y) \to hofib_f (y)$ from the point-preimages to the homotopy fibres is to be a weak homotopy equivalence. What one needs is a suitable local condition and then a local-to-global result. Many technical results are of this type: ''fibre bundles are Serre-fibrations'', ''local Hurewicz fibrations are Hurewicz fibrations'', Dold-Thom's result on quasifibrations, the Vietoris-begle mapping theorem, Quillens Theorems A and B and McDuff-Segals formulation of the ''group-completion'' theorem. A variation of the latter solves your problem:

Assume, as above, that for each $y \in Y$, $f^{-1}(y)$ is $\mathbb{Q}$-acyclic. Assume moreover that $Y$ is Hausdorff and for each $y \in U \subset Y$, there is $y \in V \subset U$ ($U$ and $V$ open) such that $V$ is contractible and $ f^{-1} (V)$ is $\mathbb{Q}$-acyclic (let us call these sets ''good''). Then, I claim, $f$ induces an isomorphism in rational cohomology.

Proof: Let $P:=\coprod_U U$, where $U$ runs through all good sets. This is a topological poset: $(U,x) \leq (V,y)$ iff $x=y$ and $U \subset V$. Take the nerve $|N_{\bullet} P|$ and consider the map $g:|N_{\bullet} P| \to Y$. It has contractible point-preimages: the preimage of $x$ is the poset of all good sets containing $x$, ordered by inclusion and the assumption says that this has contractible realization. In their recent paper arXiv:1201.3527, Galatius and Randal-Williams prove a lemma that under the present assumptions, $g$ is a Serre fibration with contractible fibres, in particular a weak homotopy equivalence.

Apply the same argument to get a poset $Q$, formed not from good subsets of $Y$, but their preimages in $X$. Thus up to weak equivalence, we can replace $f:X \to Y$ by $|N_{\bullet} Q| \to |N_{\bullet} P|$. By assumption, on the $p$th simplicial stage, the map $N_p Q \to N_p P$ is a rational homology equivalence (because this is just a disjoint union of the maps $f^{-1}(U) \to U$). The spectral sequence of simplicial spaces gives then that $|N_{\bullet} Q| \to |N_{\bullet} P|$ is a rational homology equivalence.

ad 2. There are several instances where the above argument can be applied. Example 1: let $G$ be a discrete group that acts simplicially on a simplicial complex, with finite stabilizers. Example 2: let $G$ be a Lie group that acts properly on a smooth manifold with finite stabilizers. By some standard results, there is a $G$-invariant Riemannian metric and $G$-equivariant tubular neighborhoods of orbits $Gx\cong G/G_x$. The image of such a tubular neighborhood $U \subset M$ is a contractible subset of $M/G$. The preimage of $U/G$ under $EG \times_G M \to M/G$ is $EG \times_G U \simeq EG \times_G G/G_x \simeq BG_x$, which is $\mathbb{Q}$-acyclic.

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For ease of referencing I'll prove the stated isomorphism under the hypothesis that $G$ is a compact Lie group and $X$ is a finite dimensional compact topological $G$-manifold.

By the hypothesis on $G$ we can choose $EG$ to be a dimension-wise finite CW complex. The $(n+1)$-skeleton $E$ is compact, $n$-connected and for fixed $m$ and $n$ large enough, the inclusion $E \hookrightarrow EG$ induces an isomorphism $$H^m(EG\times_G X,\mathbb{Q}) \xrightarrow{\cong} H^m(E\times_G X,\mathbb{Q})$$ (that's 4) on p. 4 of the linked paper in the question). Hence it suffices to show that the map $E\times_G X \to X/G$ induces an isomorphism in rational cohomology in degree $m$.

But this follows from the Vietoris-Begle theorem [Q, Corollary A.7] that applies if we have shown that $f$ is closed and the fibres of $f$ are compact, n-acyclic and relative Hausdorff in $E\times_G X$:

Closedness of $f$ and compactness of the fibres $E/G_x$ are obvious, $n$-acyclity follows from $H^i(E/G_x,\mathbb{Q})=H^i(G_x,\mathbb{Q})=0$ for $0< i< n$. Relative Hausdorff means different points in the fibre have disjoint neighborhoods in $E \times_G X$. This holds since $E \times_G X$ is Hausorff. QED

[Q] Quillen: The Spectrum of an Equivariant Cohomology ring: I


The cited Vietoris-Begle theorem states:

If $f: X \to Y$ is closed and its fibres are compact, relatively Hausdorff in $X$ and $n$-acyclic, i.e. $H^i(f^{-1}(y),k)=H^i(\ast,k)$ for $i< n$ ($k$ any constant coefficients), then $f^\ast:H^i(Y,k) \xrightarrow{\cong} H^i(X,k)$ for $i \lt n$.

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1  
In the statement of the theorem you don't mean $H^i(X,k)$ the first time this appears, no? –  Mariano Suárez-Alvarez Jan 31 '13 at 4:45
    
Ooops, of course not. Thanks. –  Ralph Jan 31 '13 at 11:34

In fact "$\mathbb{Q}$-acyclic" means "having the same rational homology groups as a point". In particular, the classifying space of any finite group is $\mathbb{Q}$-acyclic, as can be seen by a simple argument using the transfer.

Once you know that the fibers of $f\colon\thinspace EG\times_G X\to G\backslash X$ are $\mathbb{Q}$-acyclic, you can conclude that its Leray spectral sequence collapses at the $E_2$-term, giving the desired isomorphism. (Any map $f\colon\thinspace X\to Y$ has a spectral sequence whose $E_2$-page is cohomology of $Y$ with coefficients in the sheaf given by $U\mapsto H^\ast(f^{-1}(U))$, of which the Leray--Serre spectral sequence of a fibration is a special case. See section 14 of Bott and Tu's "Differential forms in algebraic topology" for a nice introduction).

Edit: As others have noted, the spectral sequence does not collapse without additional assumptions. In my head when writing this answer I had $G$ a compact Lie group acting smoothly on a smooth manifold $X$. Then the spectral sequence does collapse (as noted by Johannes', since tubular $G$-neighbourhoods of orbits furnish $X/G$ with a good cover) but its probably easier to argue as in Ralph's answer, using a compact approximation to $E_G$ and the Vietoris-Begle theorem.

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2  
You cannot apply Vietoris-Begle here, properness of the map is a crucial assumption. In the present case, the fibres are spaces $BG_x$, and if $1 \neq G_x$, there is no compact model for $BG_x$. –  Johannes Ebert Jan 30 '13 at 21:02
    
@Johannes: You're right, thanks. I've removed the offending statement from my answer. –  Mark Grant Jan 30 '13 at 21:09
    
Why does the Leray spectral sequence collapse if the fibres are acyclic ? (I think the proof of the Vietoris-Begle theorem also uses the Leray spectral sequence, but requires in addition some compactness conditions). –  Demin Hu Jan 30 '13 at 21:32
    
@Demin: it does not collapse, unless some condition holds. –  Johannes Ebert Jan 30 '13 at 22:04

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