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I am reading this paper of Teleman and Woodward.

On page 4, they say that $H^4(BG;\mathbb{R})$ can be identified with the space of invariant symmetric bilinear forms on $\mathfrak{g}_k$. Why is this true? Is there an easy way to see this?

$G$ is a complex reductive Lie group and $\mathfrak{g}_k$ is the Lie algebra of the compact form of $G$.

It seems like this must be a "standard fact", but I haven't seen it before.

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up vote 6 down vote accepted

You need to combine two classical theorems: 1. if $K$ denotes the compact form of $G$, then $K \to G$ is a homotopy equivalence; this is discussed in Bröcker-tom Diecks book "Representations of compact Lie groups", in the section with "complexification" in the title. So $H^{\ast}(BG)=H^{\ast}(BK)$. 2. The Chern-Weil homomorphism $CW : Sym^{\ast} (k^{\ast})^K \to H^{\ast} (BK;\mathbb{R})$ (which exists for an arbitrary Lie group $K$ with Lie algebra $k$) is an isomorphism for compact groups, this goes back to Borel or Cartan (?). This theorem is discussed and proven in Dupont's beautiful book "Curvature and characteristic classes" (there you find the construction of $CW$ as well, in case you do not now it already).

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Somewhere in this is an unbalanced TeX tag. – Ryan Reich Oct 19 '10 at 8:31
I have changed it, but MO does not save the changes (?) – Johannes Ebert Oct 19 '10 at 8:33
I fixed it. Using the asterisk sign (*) and the underscore (_) is always problematic because it gets parsed as an italics delimiter first and not as a LaTeX command. For the * sign, the solution is simply to replace it with \ast. For the underscore, putting the whole TeX formula between grave signs (...) does it. – darij grinberg Oct 19 '10 at 8:37
Oh, and my own comment is a perfect example of the problem... – darij grinberg Oct 19 '10 at 8:38
I see. Johannes – Johannes Ebert Oct 19 '10 at 8:39

One way to look at the invariant symmetric forms is by noting that they describe one dimensional central extensions of the loop algebra $L\mathfrak{g} = \operatorname{Maps}(S^1, \mathfrak{g})$. As a vector space, you have a direct sum $L\mathfrak{g} \oplus \mathbb{R}K$, and in order to get a Lie algebra structure that has a Lie algebra surjection to $L\mathfrak{g}$ with central kernel, it is necessary and sufficient that the bracket restricted to the $L\mathfrak{g}$ summand be given by: $$[f,g]_{\hat{\mathfrak{g}}} = [f, g]_{L\mathfrak{g}} - \phi(\operatorname{Res} fdg)K. $$ for some invariant form $\phi: \operatorname{Sym}^2 \mathfrak{g} \to \mathbb{R}$.

Given a pointed map $BG \to K(\mathbb{Z},4)$ (called the level), you can apply the loop space functor twice to get a two-fold loop map $\Omega G \to K(\mathbb{Z},2)$. Two-fold loop spaces (that are grouplike) are abelian groups up to homotopy, and two-fold loop maps are homotopy abelian homomorphisms. In this case, homotopy classes of two-fold loop maps classify the data of an $S^1$-bundle on the loop group, together with a multiplication that makes it a central extension, up to isomorphism. Delooping this twice gives you a correspondence with elements of $H^4(BG,\mathbb{Z})$.

Unfortunately, I am rather unfamiliar with the details of the remaining steps, namely switching to real coefficients, and passing from the loop group to its Lie algebra.

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Perhaps I can complement Johannes's answer. As he points out, one can work with $K$ ab initio, since $K$ and $G$ are homotopy equivalent.

Then $H(BK)$ is isomorphic to the $K$-equivariant cohomology $H_K(\mathrm{pt})$ of a point. Working over $\mathbb{R}$, you can use any of the infinitesimal models for equivariant cohomology. The isomorphism in the question is induced by the quasi-isomorphism of two $K$-DGAs: the ones in the Weil and Cartan models for equivariant cohomology.

Section 4 of Atiyah-Bott's The moment map and equivariant cohomology has a good treatment of the isomorphism.

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Sorry, but I am stupid: What are the Weil and Cartan models, exactly? – Kevin H. Lin Oct 19 '10 at 21:34
Or: Where can I look them up? – Kevin H. Lin Oct 19 '10 at 21:34
I learnt this from a paper of Atiyah and Bott: The moment map and equivariant cohomology. You can get the PDF for free here:… – José Figueroa-O'Farrill Oct 19 '10 at 21:44
Guillemin-Sternbergs "Equivariant de Rham theory" is another source. As far as I understand, the theory reduces to Chern-Weil theory if X is a point. – Johannes Ebert Oct 20 '10 at 18:31
I agree. I thought it might be helpful to phrase it as a quasi-iso of DGAs since Kevin has mentioned DGAs in previous MO questions/answers, if I recall correctly. – José Figueroa-O'Farrill Oct 20 '10 at 20:33

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