Let us consider a Lie group $G$ with Lie algebra $\mathfrak{g}$ and let $L\mathfrak{g} = C^\infty(S^1, \mathfrak{g})$ the Lie algebra of the loop group $LG$.
My question is about continuous Lie algebra 2-cocycles on $L\mathfrak{g}$.


It is well-known (see, e.g., Prop. 4.2.4 in Pressley-Segal "Loop groups") that if $G$ is semisimple and compact, the only continuous $G$-invariant 2-cocycles are of the form
$$\omega(X, Y) = \int\_{S^1} b(X(t), Y^\prime(t))dt, $$
where $b$ is some (necessarily symmetric) $G$-invariant bilinear map on $\mathfrak{g}$.


If I understand correctly, the proof in Pressley-Segal does not use the compactness assumption (which is a general assumption throughout the entire chapter there), but the compactness is used in the proof that all continuous 2-cocycles on $L\mathfrak{g}$ are cohomologous to a $G$-invariant one (this follows from averaging over $G$, which only works in the compact case).


**Question: What is known if $G$ is a non-compact semisimple Lie group?**


Still, all continuous $G$-invariant cocycles are of the form given above, but now there may be non-trivial cohomology classes that are not represented by a $G$-invariant one.


**More precisely: What are examples of non-trivial classes in $H^2\_c(L\mathfrak{g}, \mathbb{R})$ that are not represented by a $G$-invariant one, where $G$ is some semisimple Lie group?**



Non-invariant forms on loop Lie algebra of semisimple Lie group
Thanks to Yves Cornulier, for suggesting to look at the paper of Neeb and Wagemann. After reading Example 6.2 of that paper (arxiv version), I think the answer to my question is that in fact all 2-cycles have a representative of the form in my original post. In other words, the assumption of $G$-invariance in the proposition from the book of Pressley-Segal can be dropped (at least up to replacing the cocycle with a cohomologous one).


Explicitly, set $A = C^\infty(S^1)$.
By the results of Neeb and Wagemann, any cocycle $\omega$ on $L \mathfrak{g} = A \otimes \mathfrak{g}$ is defined by two continuous linear maps, $$f\_1 :\Lambda^2 (A) \otimes \mathrm{Sym}^2(\mathfrak{g}) \to \mathbb{R}, \qquad f\_2 : A \otimes Z\_2(\mathfrak{g}) \to \mathbb{R}.$$
(In general, there is also $f\_3$, but that does not occur in this special case.)
Moreover, we can also ignore $f\_2$, because $H^2(\mathfrak{g}) = 0$ in the semisimple case, hence the $f\_2$ part corresponds to a coboundary.
The map $f\_1$ corresponds to a map
$$ \tilde{f}\_1: A \times A \to \mathrm{Sym}(\mathfrak{g})^{\mathfrak{g}}$$
and the corresponding cocycle $\omega$ is then given by
$$ \omega(X, Y) = \sum\_{ij=1}^n \tilde{f}\_1(X^i, Y^j)(b\_i, b\_j),$$
where $b\_1, \dots, b\_n$ is a basis for $\mathfrak{g}$, and we expanded $X = \sum\_i X^i b\_i$, $Y = \sum\_j Y^j b\_j$.
Now, the condition on $f\_1$ is that there exist a continuous linear map $f\_1^\flat : \Omega^1(S^1) \to \mathrm{Sym}^2(\mathfrak{g})^{\mathfrak{g}}$ with
$$ \tilde{f}\_1(a, b) = f\_1^\flat(a \,db - b \,da) $$
and
$$ \Gamma(f\_1^\flat(da)) = 0$$
for all $a \in A$ (in general there is a term depending on $f\_2$ on the right hand side and such cocycles are called "coupled", but this term vanishes in our case).
Here $\Gamma: \mathrm{Sym}^2(\mathfrak{g})^{\mathfrak{g}} \to Z\_3(\mathfrak{g})$ is the Koszul map, which is injective for semisimple Lie algebras, hence $f\_1^\flat(da) = 0$.
Hence we are looking for 1-currents $T$ on $S^1$ that are co-closed, $\partial T = 0$, and it is well-known that the space of such is one-dimensional; up to a scalar, they are of the form
$$ T(\alpha) = \int\_{S^1} \alpha. $$
Hence $f\_1^\flat = T \otimes b$ for some $b \in \mathrm{Sym}^2(\mathfrak{g})^{\mathfrak{g}}$ and assembling $\omega$ for this $f\_1^\flat$ yields the cocycle from the original post.


