# Representation of a Lie algebra from a representation of Lie group

Let $\pi$ be a unitary representation of a Lie group $G$ on a Hilbert space $H_{\pi}$. Thus $\pi:G \to B(H_{\pi})$. One can extend $\pi$ to a representation of $L^1(G)$ via the formula $\pi(f)=\int_{G}f(t)\pi(t)d\mu(t)$ where $\mu$ is the Haar measure. Thus $\pi$ may be viewed as a mapping $\pi: L^1(G) \to B(H_{\pi})$. I will describe two ways for obtaining a representation of the Lie algebra $\mathfrak{g}$ of $G$ from a representation $\pi$.

1. One way uses the description of $\mathfrak{g}$ as the Lie algebra of left invariant vector fields. For $X \in \mathfrak{g}$ (viewed as a left invariant vector fields) one defines an operator (in general unbounded) which we denote by $d\pi(X)$ as follows: it acts on vectors of the form $\pi(\varphi)\xi$ where $\varphi \in C^{\infty}_c(G)$ and $\xi \in H_{\pi}$ and the formula is $$d\pi(X)(\pi(\varphi)\xi)=\pi(X(\varphi))\xi$$ where $X(\varphi)$ is understood as the action of vector fields on smooth functions

2.The second way is as follows: for $X \in \mathfrak{g}$ viewed as an element in $T_eG$ we define $d\pi(X)$ as an operator which acts on smooth vectors for $\pi$ (i.e. those vectors $\xi \in H_{\pi}$ such that the map $G \ni t \mapsto \pi(t)\xi$ is smooth) by the formula $$d\pi(X)(\xi)=\frac{d}{dt}\Bigg|_{t=0}\pi(exp(tX))\xi$$

Are these two constructions equivalent?

• is $H_\pi$ a Hilbert space? are these unitary representations? – YCor May 6 '18 at 23:37
• Yes, I'm interested in the context of unitary representations on a Hilbert space. I also assume that all representations are strongly continuous. I've corrected my question – truebaran May 6 '18 at 23:48
• These definitions are equivalent where both are defined, that is on the space $\pi(C_c^\infty(G))H_\pi$. If $\xi$ lies in that space, it is a simple computation that the two definitions agree. The question remains whether $\pi(C_c^\infty(G))H_\pi=H_\pi^\infty$. This is true for irreducible reps of reductive Lie groups, but in general, I think the space of smooth vectors $H_\pi^\infty$ is bigger. – user1688 May 7 '18 at 9:51
• I don't see why these two definition agree on the common domain – truebaran May 7 '18 at 18:23

## 1 Answer

The equivalence of these two descriptions does depend on making both descriptions more precise. (Although, in some concrete cases, the potential ambiguity is pseudo-magically, but provably, dissipated by proving the "essential self-adjointness" of the relevant operator, that is, the fact that it has a unique self-adjoint extension, and that that unique extension is exactly its (graph-) closure.) Namely, an "unbounded operator" is not fully specified unless its domain is specified, and, quite disturbingly in contrast to the continuous/bounded operator case, being "symmetric" (in the sense that $\langle Tv,w\rangle=\langle v,Tw\rangle$ for $v,w$ in the domain of $T$) is not as strong as "self-adjoint" (meaning that the domain of the adjoint is exactly equal to the domain of $T$).

After such things are taken into account, then you would have a suitably qualified version of your equivalence.