6
$\begingroup$

I'm asking a question about Lie group representation. Let $G$ be a Lie group, not necessarily connected. Let $\Omega$ be an element in the center of the universal enveloping algebra $U(\mathfrak{g})$ of the Lie algebra $\mathfrak{g}$ of $G$, so $\Omega$ commutes with every vector $X\in\mathfrak{g}$. Let $\rho$ be a unitary representation of $G$ on a complex Hilbert space $H$. Then $\rho$ induces an algebra homomorphism (which will be denoted as just $\rho$) from $U(\mathfrak{g})$ into the linear endomorphism algebra of some dense subspace of $H$, the space of smooth vectors. If $\rho$ is irreducible, meaning that there is no proper nontrivial closed subspace of $H$ that is invariant under all of $\rho(g)$'s, then can we say that $\rho(\Omega)$ is a scalar multiple of the identity map on $H$?

I've been convinced that the Schur's lemma is still true for unbounded operators; more precisely, if $T$ is a densely-defined closable unbounded operator on $H$ commuting with every $\rho(g)$ (so in particular the domain $D(T)$ of $T$ is invariant under $\rho(g)$'s) then $T$ should be a scalar. But it is not clear to me that whether or not $\rho(\Omega)$ commutes with $\rho(g)$'s. Perhaps the answer to my question is true when $\Omega$ is invariant under the adjoint action of $G$, but this is also not clear.

So my question is:

  1. Is it true that $\rho(\Omega)$ should be a scalar?
  2. Should $\Omega$ be invariant under $\mathrm{Ad}_{g}$?
  3. If not, when those are true?

    (1) What if $H$ is finite-dimensional?

    (2) What if $G$ is connected, or simply-connected?

Thank you.

$\endgroup$
7
$\begingroup$

No. Let $G$ be $O_2(\mathbb{R})$, so the Lie algebra is one dimensional, and the center of the universal enveloping algebra is the symmetric algebra of the Lie algebra. Then the usual 2-dimensional representation (which is irreducible) does not have elements of the Lie algebra acting by scalars.

You need $\Omega$ to be $Ad$ invariant.

$\endgroup$
  • 1
    $\begingroup$ Wow, nice and simple counterexample! Thank you. So, what if G is connected? Is ad-invariance implies Ad-invariance for that case? $\endgroup$ – Junekey Jeon Aug 30 '15 at 2:26
  • 1
    $\begingroup$ @JunekeyJeon Yes, if $G$ is connected, then $G$ is generated by exponentials of elements in the Lie algebra. This implies the two versions of invariance are the same. $\endgroup$ – S. Carnahan Aug 31 '15 at 9:03

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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