**Background:**

Let $G$ be a cscsc¹ Lie group, and let $\widetilde{LG}$ be the universal central extension (center = $S^1$) of $LG:=Map_{C^\infty}(S^1,G)$, with the topology inherited from the $C^\infty$ topology on $LG$.

Let $\mathfrak g$ be the complexified Lie algebra of $G$, and let $\widetilde{L\mathfrak g}$ be the universal central extension of $L\mathfrak g:=\mathfrak g[t,t^{-1}]$ (center = $\mathbb R$).

There is a (non-faithful) action of $\mathbb R$ on $\widetilde{LG}$ given by reparametrizing the loop, and there is an induced action of $\mathbb R$ by derivations of $\widetilde{L\mathfrak g}$. We denote the respective semidirect products $\widetilde{LG}^e$ and $\widetilde{L\mathfrak g}^e$. We also write $\widetilde{L\mathfrak g}^e=\widetilde{L\mathfrak g}\oplus \mathbb C L_0$.

Definition:Apositive energy representationof $\widetilde{LG}$ on a Hilbert space $H$ is a continuous homomorphism $\widetilde{LG}\to U(H)$ for which there exists an extension $\widetilde{LG}^e\to U(H)$ such that the spectrum of $L_0$ is bounded from below. Here, $U(H)$ is equipped with the strong operator topology.

If the action of $L_0$ on $H$ has only point spectrum, then the algebraic direct sum of its eigenspaces is called the space of finite energy vectors, and denoted $H_0$.

**The following is something that I know how to prove.²
I would however prefer to find a reference in the literature. Is this written anywhere?
Alternatively, does someone have an easy proof of this fact?**

Let $H$ be a positive energy representation of $\widetilde{LG}$, and let us assume that $L_0$ can be chosen so that it only has point spectrum ($\Rightarrow H_0$ is dense in $H$), then there is an induced Lie algebra action of $\widetilde{L\mathfrak g}$ on $H_0$.

The difficulty lies in the fact that the action of $\widetilde{LG}$ is only assumed to be continuous, and the it is therefore difficult to know that differentiation makes sense.

¹: "cscsc" = compact, simple, connected, simply connected

²: to be precise, I only know how to prove it if $G$ ≠ $SU(2)$.

Since I said that I know how to prove the above claim, let me include a proof. This will also make it clear why the case $G=SU(2)$ is special.

**Proof:** Let $\tilde T$ be the maximal torus of $\widetilde{LG}$, let $T^e:=\mathbb R\times \tilde T\subset \widetilde{LG}^e$, and let $\Lambda=Hom(T^e,S^1)$ be its character group. Every positive energy representation $H$ of $\widetilde{LG}$ decomposes as a direct sum of weight spaces
$$
H=\bigoplus_{\lambda\in\Lambda} H_\lambda.
$$
Let $P\subset \Lambda$ be the set of $\lambda$ for which $H_\lambda\not =0$.

Let $\widetilde W= NT^e/T^e$ be the affine Weyl group. This group acts on $\Lambda$, and preserves $P$. By the positive energy condition, we know that $P$ is contained in a "half-space" of $\Lambda$, and combining this with $\widetilde W$-invariance, we learn that $P$ is contained inside a paraboloid.

Let $D$ be the affine Dynkin diagram associated to $G$. Every node $a\in D$ corresponds to an embedding $SU(2)\hookrightarrow \widetilde{LG}^e$ whose image we shall denote $SU(2)_a$. Let $T_a\subset T^e$ be the centralizer of $SU(2)_a$, and let $\Lambda_a$ be the corresponding quotient of $\Lambda$, with projection map $p_a:\Lambda\to \Lambda_a$.

Since $P$ is contained in a paraboloid, we know that for each $\sigma\in \Lambda_a$, the set $\Lambda(\sigma):=\{\lambda\in P\,|\,p_a(\lambda)=\sigma\}$ is finite. It follows that for every $\sigma$, the representation of $SU(2)_a$ on $\bigoplus_{\lambda\in\Lambda(\sigma)} H_\lambda$ contains only finitely many isomorphism classes of $SU(2)_a$-irreps. In particular, the action of $\mathfrak{su}(2)_a$ on $\bigoplus_{\lambda\in\Lambda(\sigma)} H_\lambda$ is by bounded operators (i.e., everywhere defined). This way, one obtains Lie algebra actions of $\mathfrak{su}(2)_a$ on the space of finite energy vectors of $H$.

Those Lie algebras contain all the generators $\{E_a, F_a, H_a\}$ of the Serre presentation of $\widetilde{L\mathfrak g}^e$. We still need to check the Serre relations.

We now consider rank two subgroups of $\widetilde{LG}^e$ corresponding to pairs of vertices of $D$. To every such pair of vertices corresponds a **compact** subgroup $G_{ab}$ of $\widetilde{LG}^e$ (this is where we use that $G$ is not $SU(2)$). Applying the same sort of argument as above, we see that there is an action of its Lie algebra $\mathfrak g_{ab}$ on the space of finite energy vectors.

Every Serre relation is detected in one of the Lie algebras $\mathfrak g_{ab}$, so the generators $\{E_a, F_a, H_a\}$ satisfy the Serre relations. $\square$

neversmooth (even if the group is compact). – André Henriques Jun 23 '13 at 8:32