equivariant cohomology with respect to a loop group

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?

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It seems your question is answered here mathoverflow.net/questions/20671/… ? –  Daniel Pomerleano Apr 8 '11 at 8:04
Well a part of it anyways –  Daniel Pomerleano Apr 8 '11 at 8:31
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 '11 at 11:03
@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 '11 at 15:56

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.
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 '11 at 15:23