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Let $G$ be a simple algebraic group over the complex numbers. Is it true that the Lie algebra cohomology $H^*(L\mathfrak{g}, \mathbb{C})$ of the loop Lie algebra $L\mathfrak{g}=\mathfrak{g} \otimes \mathbb{C}((z))$ is isomorphic to the singular cohomology $H^*(LG, \mathbb{C})$ of the loop group of $G$?

I know that the corresponding statement is true for the real form of $G$. I.e. if we want to compute the Lie algebra cohomology for $\mathfrak{g}$ we can just compute the de Rham cohomology for $G$. However, the argument I know of this fact uses the averaging process on $G$ to get a left-invariant differential form from an ordinary differential form. Is there some other way of getting the isomorphism?

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    $\begingroup$ In defining $H^*(L\mathfrak{g},\mathbf{C})$, could you be more specific? Do you really mean Lie algebra cohomology, or do you have in mind some topological restriction using the topology of $\mathbf{C}((z))$? Indeed $H_2(L\mathfrak{g},\mathbf{C})$ is known (due to Bloch for $\mathfrak{sl}_n$) to be isomorphic to $\mathrm{HC}_1(A)=\Omega^1_{A/\mathbf{C}}/dA$, where $A=\mathbf{C}((z))$. Abstractly, this is a space of dimension continuum, so the dual $H^2$ has dimension $2^c$. But restricting to the continuous part of $\mathrm{HC}_1(A)$, i.e. continuous cocycles, makes it, I think, 1-dimensional. $\endgroup$
    – YCor
    Commented Sep 16, 2019 at 6:48
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    $\begingroup$ So he's using a continuity assumption, or his statement is just wrong. Write $K=\mathbf{C}((z))$, let $W_K$ be the space of $\mathbf{C}$-linear maps $f:K\otimes K\to K$ such that $f(ab\otimes c+bc\otimes a+ca\otimes b)=f(a\otimes bc+b\otimes ca+c\otimes ab)$ for all $a,b,c$. If $u,v$ are $\mathbf{C}$-linear derivations of $K$, then $(a,b)\mapsto u(a)v(b)$ belongs to $W_K$, and it's easy to deduce that $W_K$ has at least continuum dimension (if no continuity assumption is made). More precisely is $B$ is a transcendence basis of $K$ over $\mathbf{C}$ (...) $\endgroup$
    – YCor
    Commented Sep 16, 2019 at 7:48
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    $\begingroup$ (...) and $a\neq b\in B$ then there exists $f=f_{a,b}\in W_K$ such that $f(a\otimes b)=1$ and $f(c\otimes d)=0$ for all $(c,d)\in B^2-\{(a,b)\}$. Now map $\Lambda^2(\mathfrak{g}\otimes K)$ into $K$ by $ux\wedge vy\mapsto f(u\otimes v-v\otimes u)\langle x,y\rangle$, where $\langle .,.\rangle$ is the Killing form. This vanishes on 2-boundaries. Hence it defines a 2-cohomology class $q(f)$. Then, evaluating on the 2-cycles $ax\wedge by-bx\wedge ay$, one checks that the family $(q(f_{a,b}))$, $(a,b)\in B^2$, generates a subspace of continuum dimension in $H^2(\mathfrak{g}\otimes K)$. $\endgroup$
    – YCor
    Commented Sep 16, 2019 at 7:48
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    $\begingroup$ PS: in this context $\mathfrak{g}\otimes\mathbf{C}[t,1/t]$ is sometimes also called the loop Lie algebra. In this case the statement about cohomology is correct: $H^2$ is 1-dimensional. PPS: in my previous two comments, $u,v$ mean two derivations in the first one, and two scalars in the second one, sorry. $\endgroup$
    – YCor
    Commented Sep 16, 2019 at 8:20
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    $\begingroup$ I haven't checked that the continuous $H^2$ is 1-dimensional. Still, I guess it should follow with no major difficulty from the fact that $\mathbf{C}[z,1/z]$ is dense in $\mathbf{C}((z)$. $\endgroup$
    – YCor
    Commented Sep 16, 2019 at 13:50

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