Non-invariant forms on loop Lie algebra of semisimple Lie group 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?
 A: 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.
