Let $G$ be a connected reductive real Lie group with Lie algebra $\mathfrak{g}$. We denote by $\widehat{G}\_u$ the unitary dual, that is the set of isomorphism classes of unitary reprensentation of $G$. 


Let $C$ be the Casimir of $\mathfrak{g}$. It is in the center of the universal enveloping algebra $U(\mathfrak{g})$. Hence, it acts as a scalar $C\_\pi\in \mathbf{C}$ on $\pi\in \widehat{G}\_u$. 


Is it true that for a given $c\in \mathbf{C}$, there exits only finite $\pi\in \widehat{G}\_u$ such that $C\_\pi=c$?



Unitary representation with fixed Casimir
No, for higher-rank groups, such as $SL\_n(\mathbb R)$ with $n\ge 3$, it is straightforward to compute that the eigenvalue of Casimir on a unitary principal series is a quadratic polynomial in the parameters for the character, so there is a continuum of unitary principal series (generically irreducible) with the same eigenvalue of Casimir.


However Harish-Chandra proved that if the eigenvalues of *all* the center of the enveloping algebra are specified, there are only finitely-many irreducibles with those eigenvalues.


Edit: this is nearly a corollary of Harish-Chandra's sub-quotient theorem (and/or of Casselman's subrepresentation theorem), which asserts that irred admissibles are sub-quotients of principal series. The eigenvalues of the center of the enveloping algebra determine the principal series up to action of the Weyl group...


