2
$\begingroup$

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?

$\endgroup$
4
  • 2
    $\begingroup$ is $H_\pi$ a Hilbert space? are these unitary representations? $\endgroup$
    – YCor
    May 6, 2018 at 23:37
  • $\begingroup$ 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 $\endgroup$
    – truebaran
    May 6, 2018 at 23:48
  • $\begingroup$ 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. $\endgroup$
    – user1688
    May 7, 2018 at 9:51
  • $\begingroup$ I don't see why these two definition agree on the common domain $\endgroup$
    – truebaran
    May 7, 2018 at 18:23

1 Answer 1

3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.