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Compact connected simply-connected Lie groups have so much structure that you can calculate their cohomology from their Lie algebras using Lie algebra cohomology (certain Ext-groups) and similarly their homology from their Lie algebras using Lie algebra homology (certain Tor-groups).

Is there similar theorem that gives the (co)homology of the loop space of a Lie group in terms of its Lie algebra?

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  • $\begingroup$ Are you looking for the cohomology of the pointed loop space ΩG or the free loop space LG? $\endgroup$
    – Reid Barton
    Commented Oct 22, 2009 at 19:32
  • $\begingroup$ Preferably both, because I want to do some calculations of the loop algebra structure of Chas-Sullivan and both appear in a spectral sequence converging to the homology with the loop algebra structure. But one would be fine. $\endgroup$
    – skupers
    Commented Oct 22, 2009 at 20:08

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Yes.

At least, rationally.

The result that you want starts on p68 of "Loop Groups" by Pressley and Segal. There, they prove surjectivity of H*(L𝔤;ℝ) → H*(LG;ℝ). The basic idea of the argument is as follows: for reasonably simple reasons, the cohomology of LG is easily obtainable from that of G. This yields specific formulae for generators of the de Rham cohomology of LG. Although these forms are not themselves left invariant, they are cohomologous to rational multiples of left invariant forms, and thus come from the cohomology of the Lie algebra, L𝔤. The proof of surjectivity is thus not hard.

The proof that it is an isomorphism is a little trickier and they defer that to section 14.6 (p299). That this is quite some way through the book is a good indication of how much you need to absorb to understand it.

Amazingly, parts of Loop Groups (including pages 68, 69, and 299, but not 300) are available on Google books. It says that it is a "Limited preview" but whether or not that is just limited in pages or limited in time, I do not know. However, Loop Groups is a great book for anyone interested in Lie Groups and infinite dimensional "stuff".

(Incidentally, this is all for G simply connected.)

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    $\begingroup$ What is 𝔤? It doesn't display correctly for me. $\endgroup$
    – Anton Geraschenko
    Commented Oct 21, 2009 at 22:13
  • $\begingroup$ I would guess it's fraktur g. $\endgroup$
    – Kevin H. Lin
    Commented Oct 22, 2009 at 0:11
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    $\begingroup$ Yes, that's right. Even without using MathML, the STIX fonts (or equivalent) do add a certain amount of functionality. $\endgroup$
    – Andrew Stacey
    Commented Oct 22, 2009 at 7:02
  • $\begingroup$ Thanks for the answer. Is there anything known about what happens over $\mathbb{Z}/p\mathbb{Z}$ or the integers? $\endgroup$
    – skupers
    Commented Sep 6, 2010 at 19:00
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Well, couldn't you use the fibration \Omega G\to PG\to G, note that PG is always contractible, and use the spectral sequence of a fibration? You know the cohomology of PG and G from contractibility and from the lie algebra, and then you just need to do the spectral sequence yoga to compute it for \Omega G.

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  • $\begingroup$ Yes, that would work, but I was looking for a method that doesn't refer to the Lie group. Something along the lines of defining a 'loop Lie algebra' for the loop space of the Lie algebra and relating that to the original Lie algebra. $\endgroup$
    – skupers
    Commented Oct 21, 2009 at 14:56
  • $\begingroup$ Ahh. I don't know anything like that. I figured that this spectral sequence, which only requires knowledge of the cohomology of the Lie Group, and which implies that the cohomology of all the loop spaces is determined by the lie algebra, was the sort of thing you were looking for. $\endgroup$
    – Charles Siegel
    Commented Oct 21, 2009 at 15:14

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