# Unitary representation with fixed Casimir

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 $C_\pi=c$?

• Isn't this equivalent to saying that the Laplacian on $G$ doesn't have infinite dimensional eigenspaces? – Joonas Ilmavirta Feb 28 '15 at 19:54
• Dear Joonas Ilmavirta, when $G$ is non compact, the Casimir is not elliptic. Only if restict to $C^\infty(G/K)$, Casimir is the Laplacian. – shu Feb 28 '15 at 20:00
• @shu: Treatments of infinitesimal characters related to unitary representations of real Lie groups usually involve the complexified Lie algebra and its universal enveloping algebra. For a Casimir-type operator to act by a scalar may require an algebraically closed field. – Jim Humphreys Feb 28 '15 at 20:44
• @JimHumphreys, Thank you for the comment. I modified the field. – shu Feb 28 '15 at 20:58

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.