Linear Algebra – Eigenvalue of a Casimir Element in Representation Theory

Question

For a simple Lie algebra $\mathfrak{g}$ which contains a subalgebra isomorphic to $\mathfrak{sl}(2,\mathbb{R})$, let $\pi:\mathfrak{g}\to\mathrm{End}(V)$ be a non-trivial irreducible representation. Define the trace bilinear form $B_V$ on $\mathfrak{g}$ by $B(X,Y)=\mathrm{tr}(\pi(X)\pi(Y))$. Then there exists a Casimir element $\Delta$ in the center of $U(\mathfrak{g})$ (by universal property). Now I'm trying to show that the eigenvalue of $\Delta$ on $V$ is $\frac{1}{\dim(V)}$.

I've proved that $\pi$ is faithful and $B_V$ is nondegenerate. Taking the trace, I concluded that
$$\mathrm{tr}(\Delta)=\sum\limits_{i}\mathrm{tr}(\pi(x_i)\pi(y_i))=\sum\limits_iB(x_i,y_i)=\dim\mathfrak{g}$$
after fixing a basis $(x_1,\cdots,x_n)$ of $\mathfrak{g}$ and its determined dual basis $(y_1,\cdots,y_n)$. What to do next to compute and verify the eigenvalue of $\Delta$ on $V$?

Answer 1

There might be something incorrect in the original question. Review a proposition in Bump's Lie Groups.

Proposition 10.2. Let $\pi:\mathfrak{g}\to\mathrm{End}(V)$ be an irreducible representation of the Lie algebra $\mathfrak{g}$. If $c$ is in the center of $U(\mathfrak{g})$, then there exists a scalar $\lambda$ such that $\pi(c)=\lambda I_V$.

Back to this question, it immediately follows from the above proposition that $\pi(\Delta)=\lambda I_V$ for a certain scalar $\lambda$, which can be regarded as the eigenvalue of $\Delta$. Taking traces on both sides, we obtain $$\mathrm{tr}(\Delta)=\dim\mathfrak{g}=\lambda\dim V,$$ i.e., the eigenvalue $\lambda=\frac{\dim\mathfrak{g}}{\dim V}$.