# Does an element in the center of universal enveloping algebra becomes a scalar in irreducible representations?

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.

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.
• @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. – S. Carnahan Aug 31 '15 at 9:03