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This question makes sense for any topological group $G$, but I'd particularly like to know the answer for $G$ a compact, connected Lie group.

$G$ acts on itself by conjugation. One has the equivariant singular cohomology $H^{\ast}_G(G) = H^{\ast}( (G\times EG)/G )$, with integer coefficients, say.

What is $H^{\ast}_G(G)$?

Concretely, $H^{\ast}_G(G)$ is the target of the Serre spectral sequence for the fibration $$G \hookrightarrow (G\times EG)/G \to BG.$$ When $G$ is path-connected, so that $BG$ is simply connected, this spectral sequence has $E_2^{p,q}=H^p(BG; H^q(G))$ (trivial local system). Does the spectral sequence always degenerate at $E_2$? It does when $G$ is abelian, because then the conjugation action is trivial.

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I don't know, but are you assuming field coefficients? Or is there another reason why $E_2^{p,q}$ is a tensor product? –  Paul Apr 8 '10 at 0:30
One other thought, your fibration has a section (e.g. the identity gauge transformation) and that might help collapse the spectral sequence. –  Paul Apr 8 '10 at 0:42
@Paul. No, that was carelessness; I will correct it. –  Tim Perutz Apr 8 '10 at 0:45
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5 Answers

up vote 22 down vote accepted

I asked Dan Freed, who gave a very clean general solution to this problem (as expected). Here it is (all mistakes in the transcription are mine of course).

The claim is that the equivariant cohomology of G acting on G is indeed the tensor product of cohomology of BG with cohomology of G - in other words the Leray spectral sequence for the fibration $G=\Omega BG \to G/G=LBG \to BG$ degenerates at E_2. To see this we will use the Leray-Hirsch theorem -- i.e. if we can show that every class on the fiber (G) extends to a class on the total space (LBG) then we will be done. Now the cohomology of G is generated (as an exterior algebra) by its primitive classes, and these all come from the generators of the cohomology of BG by transgression. So we just need to show that these transgressed classes actually lift to LBG.

But there is a nice direct construction of these classes on LBG. Namely we use the tautological correspondence $$LBG \leftarrow S^1\times LBG \rightarrow BG$$ where the right arrow is the evaluation map. Thus given a class on BG we can lift it to $S^1 \times LBG$ and then integrate along the circle to get a class on $LBG$. When we restrict these classes to a fiber, i.e. $G=\Omega BG$, we recover the usual transgression construction. (This can be seen very explicitly with differential forms.. the transgression involves the path fibration $$\Omega BG \to P(BG) \to BG$$ and is given by the same kind of tautological/evaluation construction for the map from the interval times the path space to BG, integrating over an interval.. when restricted to $\Omega BG$, ie closed paths, this integration becomes the integration over the circle we had above.)

So we've explicitly lifted all the generators of the cohomology of G to equivariant classes, ie to G/G, hence we're done. (Presumably this can also be seen concretely in the Cartan model, that the generators of H^*G lift to conjugation-equivariant classes..)

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Beautiful - thanks! –  Tim Perutz Apr 9 '10 at 17:13
Nice to see MO bringing UT math department together! –  HJRW Apr 9 '10 at 20:19
Does this decomposition into tensor product hold for arbitrary $G$ action on $G$, or only for the conjugation action? –  Gao 2Man Dec 28 '11 at 5:43
It doesn't hold for example for the left translation action - as for any free action, the equivariant cohomology in that case is the cohomology of the quotient, i.e., of a point, so very far from a tensor product of cohomology of $G$ and that of $BG$. –  David Ben-Zvi Dec 28 '11 at 16:51
Is the equivariant homology of $G$ acting on G by conjugation still the tensor product of homology of $BG$ with homology of $G$? Thanks. –  Colin Tan Dec 31 '11 at 3:33
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It is a well-known folk theorem that there is a homotopy equivalence between $(G\times EG)/G$, where $G$ acts by conjugation, and the free loop space of BG. My personal favorite proof is due to Kate Gruher, in the appendix to her thesis (Stanford, 2007). There's also a proof in the appendix to a recent paper of Klein, Schochet, and Smith (arXiv:0811.0771). The cohomology of the free loop space has been studied quite a bit in lots of cases. This KSS paper might have some useful information for you.

As Paul said, this fibration has a section, given by constant loops in the case of the free loop space model, or by sending $x\in BG$ to the pair $(1, \widetilde{x}) \in G\times EG$, where $\widetilde{x}$ is any point lying over $x$. This is well-defined since $1\in G$ is fixed under conjugation.

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I should say that some people attribute this result to a paper of Bokstedt-Hsiang-Madsen, but I've never really understood why. It's also sometimes said to be due to Waldhausen. And there's a proof for discrete groups in one of Dave Benson's books; Kate's argument is a clever extension of Benson's. –  Dan Ramras Apr 8 '10 at 0:48
Does this result deserve attribution? It seems to me equivalent to the fact that the inertia of the stack BG is the adjoint quotient stack G/G, which is almost definitional.. or more topologically, we know L(BG) is a Omega(BG)=G-bundle over BG, so comes from a G action on G, and we just need to check that it is the conjugation action.. –  David Ben-Zvi Apr 8 '10 at 3:27
More geometrically, free loops in BG are the same as G-local systems on the circle (here G considered as a homotopy type). Those are classified (via their monodromy) as elements in G, up to conjugation, i.e. the quotient stack G/G (aka the Borel construction (G x EG)/G). –  David Ben-Zvi Apr 8 '10 at 4:02
Sure, you can rephrase this in ways that make it sound like a tautology. But as far as I know, there is is no natural homotopy equivalence between these two objects, considered as topological spaces (not stacks). The proofs I've seen all make clever use of some third space, to which both of these spaces map (or vice versa). I don't think the abstract nonsense will help in writing down such a map, but I'd be quite interested to know if I'm wrong about that. I guess I'm saying that I think there's something more to the topological statement that just abstract nonsense. –  Dan Ramras Apr 8 '10 at 4:16
Probably I'm missing something basic, but I would have thought the statement on topological stacks is strictly stronger than one on spaces, which is obtained from it by a geometric realization. So we consider the top. stack of G-bundles on S^1, to which there's an obvious map from the stack G/G, and which has an obvious (classifying) map to the free loop space of BG, and that's our equivalence, no? –  David Ben-Zvi Apr 8 '10 at 13:45
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The cochains on G/G can be calculated as the Hochschild cochains of cochains on BG (this uses compactness of G - we'd get a kind of dual picture with Hochschild chains if we looked at free loops in a FINITE CW complex rather than BG). Now let me do something maybe evil and ignore gradings. Then we have a polynomial algebra, which is functions on the Chevalley vector space h/W. Its Hochschild cochains can be calculated as we would for any smooth affine variety as polyvector fields. Since the tangent bundle is trivial we get the tensor product of polynomials on h/W with the exterior algebra on the tangent space h/W. Now remembering the gradings we see precisely the cochains on BG tensor the cochains on G (this latter exterior algebra). (I'm assuming G is connected and simply connected just to be safe). I think this is all kosher algebraically - the crutch of using the HKR theorem for smooth affine varieties is just a way to avoid actually writing down the cyclic bar complex and calculating HH, but the result is the same..

EDIT: This discussion (in particular the appeal to HKR) is of rational cochains.. not sure what happens with torsion. Also I'm ignoring completion issues (related to the ignored grading), don't know how fatal they are. It's probably better to think of cochains on G/G as Hochschild cochains of the algebra of chains on G under convolution, but I think the answer comes out the same.

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Thanks! - Let me see if I followed: viewing $G/G$ as $L(BG)$. But by Jones, cochains on a loop-space are homotopy equivalent to dg Hochschild chains of cochains on the base. Now $H^\ast(BG)$ is (say in the simply connected case?) just $S^\ast(\mathfrak{g})^G$. By Chevalley's theorem, $G$-invariant polynomials on $\mathfrak{g}$ are the same as Weyl-invariant polys on the Cartan $\mathfrak{h}$, and now one can calculate Hochschild homology of $H^\ast(BG)$ via HKR. –  Tim Perutz Apr 8 '10 at 13:25
We actually wanted dg Hochschild homology of cochains, which in principle could be smaller than ordinary Hochschild hom of cohomology. But maybe you used a cochain-level model for $BG$, or a formality statement for it? All this seems quite close to the Cartan model for equivariant diff forms; I wonder whether one can say something directly in that model. –  Tim Perutz Apr 8 '10 at 13:27
I was thinking via Jones as well but you have to be careful - the theorem you cite (using Hochschild chains) is about loops of finite CW complex (cf his Theorem A -- well that's the cyclic version, I think the Hochschild is much older). But what we need here is for loops in a classifying space, to which his Theorem B applies -- we really want Hochschild COchains there.. –  David Ben-Zvi Apr 8 '10 at 13:47
Tim - yes I was using formality to identify cochains on BG with the symmetric algebra --- it's better as you say to work with cochains but in this case they're the same.. –  David Ben-Zvi Apr 8 '10 at 15:24
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With regards to some of the comments towards the bottom from David Ben-Zvi and Tim Perutz: you can get around some finite dimensional restrictions. Specifically, there is an interpretation of the Hochschild cohomology of $C^*(BG)$ in terms of string topology. Namely, it's an inverse limit of the homology of a pro-object that approximates the free loop space of $BG$ by finite dimensional manifolds. As hinted at by Dan Ramras' comments, a lot of this comes from Kate Gruher's work. This comment is explained in detail in a paper of hers and mine: arXiv:0710.1445.

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For any interested latecomers who somehow discover this question in the future, I've found a lower-tech answer, bootstrapping from the low-tech answer to Is a Lie group equivariantly formal under conjugation by a maximal torus?. I post this because I still don't understand the current answers.

Anyway, once you believe that the conjugation action upon a compact Lie group $G$ of its maximal torus $T$ is equivariantly formal, it follows that the action of $G$ by conjugation is as well. This argument will actually work for any reasonably good space $M$ on which $G$ acts and the restricted $T$-action is equivariantly formal, because one has $$H_G(M) = H_T(M)^W = (H(M) \otimes H(BT))^W = H(M) \otimes H(BT)^W = H(M) \otimes H(BG)$$ as $H(BT)$-modules, where $W = N_G(T)/T$ is the Weyl group of $G$.

If, say, you don't buy that the action of $W$ on $H(M)$ is trivial, a longer proof goes like this.


The homotopy quotient $M_G$ is a further quotient of $M_T$, and the projection $EG \times M \to EG$ then induces a commutative diagram

\begin{CD} M @= M\\ @VVV @VVV\\ M_T @>>> M_G\\ @VVV @VVV\\ BT @>>> BG \end{CD}

where the upper vertical maps are fiber inclusions.

The projection $BT = EG/T \to EG/G = BG$ induces an inclusion $H(BG) \cong H(BT)^W \hookrightarrow H(BT)$ in cohomology, and there are induced maps both in cohomology and on the Serre spectral sequences for the equivariant cohomologies, starting with this $E_2$ page:

\begin{CD} H(M) @= H(M)\\ @AAA @AAA\\ H(M) \otimes H(BT) @<<< H(M) \otimes H(BG)\\ @AAA @AAA\\ H(BT) @<<< H(BG) \end{CD}

Because the top and bottom horizontal maps are injective, so is the middle one, so the differentials for the spectral sequence converging to $H_G(M)$ are restrictions of those for $H_T(M)$. But the differentials for $H_T(M)$ are all zero, by equivariant formality, so the spectral sequence for $H_G(M)$ collapses as well.

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I don't know how to fix the diagrams, unfortunately. I looked up how to put diagrams in, and apparently that doesn't work here. –  jdc Apr 9 at 21:48
You just need to put your "\require{AMScd}" between dollar signs. –  Francois Ziegler Apr 9 at 21:57
Ah, it doesn't apply globally. That makes sense, actually. Thanks! –  jdc Apr 10 at 18:55
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