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Let $G$ be a compact connected simply connected Lie group. Let $LG$ be the corresponding loop group. What is the cohomology of its classifying space (i.e. what is the equivariant cohomology of a point with respect to $LG$?) I would like to express it in terms of the Lie algebra of $G$. Is the corresponding $dg$-algebra formal?

flag	asked Apr 8 2011 at 6:49 Alexander Braverman 3,333●6●25

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It seems your question is answered here mathoverflow.net/questions/20671/… ? – Daniel Pomerleano Apr 8 2011 at 8:04

Well a part of it anyways – Daniel Pomerleano Apr 8 2011 at 8:31

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Isn't $BLG$ just $LBG$? So you're really asking about the cohomology of $LBG$. Why is the question formulated in such a strange way? – John Klein Apr 8 2011 at 11:03

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@Daniel Pomerleano: Thank you. The question indeed is almost answered there -- the only thing which is not answered is the formality. @John Klein: You are right of course, $BLG=LBG$. The reason I am asking it this way is that I am thinking about equivariant cohomology of some other space with respect to $LG$ and I want to understand how $H^*_{LG}(pt)$ acts there. – Alexander Braverman Apr 8 2011 at 15:56

Craig Westerland · Answer 1 · 2011-04-08 23:40:19Z UTC

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You can compute $H^\ast(LBG)$ as the Hochschild cohomology

$$HH^\ast(C_\ast(G), C^\ast(G)),$$

where $C_\ast(G)$ is the singular chain complex of $G$, equipped with the Pontrjagin product, and $C^\ast(G)$ are the cochains, with the $C_\ast(G)$-module structure dual to the obvious one on $C_\ast(G)$. If you like, you can then replace $C^\ast(G)$ with the Chevalley-Eilenberg complex $K$ for the Lie algebra, and $C_\ast(G)$ with its dual $K^\ast$. At this point, though, it is perhaps not so obvious how to recover the ring structure on $K^*$ which lets you define the Hochschild cohomology.

answered Apr 8 2011 at 23:40

Craig Westerland
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	I have two questions, 1) you mean the answer to the last part of the question is "yes" since that ring isomorphism can be realized at the chain level? 2)Assuming you are taking rational or real coefficients, why don't we know that C_*(G) must be equivalent to an exterior algebra as a dg-Hopf-algebra? – Daniel Pomerleano Apr 10 2011 at 3:23
	I believe that the ring isomorphism can be realized at the chain level, since it comes from the fact that $LBG$ is the geometric realization of a simplicial space. I think that you're right that $C_(G)$ is generally going to be formal, if you pick the right base field to work over. And that will imply the spectral sequence converging to $HH^(C_(G), C^(G))$ with $E_2 = HH^(H_(G), H^(G))$ will collapse at $E_2$. But what about the Hochschild ($E_1$) differentials? If you use $Ext$ for $HH^$, you get a chain model which will have vanishing differential, but the multiplication is bad. – Craig Westerland Apr 10 2011 at 15:23

equivariant cohomology with respect to a loop group

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