# Why is the trace of the Casimir on the irrep of a semisimple algebra nonzero?

A crucial step in the "purely algebraic" proof of Weyl's semisimplicity theorem is that the Casimir element $C\in U\mathfrak{g}$ acts by nonzero scalars on a nontrivial irrep $V$. However, at least two sources I have consulted assert that $\text{tr}_V(C)=\text{tr}(C)=\dim \mathfrak{g}$, i.e. using the fact that $C$ can be written as a sum of products dual basis elements and then asserting that the trace pairing on the irrep is the same as the Killing form. But this seems obviously false. Or at least, I don't see why it should be true.

What I can see is that if $\mathfrak{g}$ is simple, the trace pairing on the irrep is a scaling of the Killing form, so by picking a suitable basis, we get that the trace pairing on $V$ is a scaling of the sum of the Killing forms on the simple summands of semisimple $\mathfrak{g}$, so you get a linear combination of the dimensions of the simple summands. But this could very well be zero.

• I don't understand "But this seems obviously false. Or at least, I don't see why it should be true." The first sentence suggests that there is an obvious counterexample. The second sentence just says what it says, which is not the same.
– YCor
Nov 8, 2015 at 23:07
• That was essentially a way for me to say "I'm almost certain this is false but don't want to admit I was too lazy to actually craft a counterexample so am hedging my bets" without it sounding quite as self-damning. Nov 9, 2015 at 0:19
• @YCor: it is false because it is trivially false for $sl_2$ - and I really hope that anyone who asks such question on MO understands fully the $sl_2$ story - or else they should move to MSE... Dec 15, 2015 at 21:48

The Casimir operator lies in the center of the universal enveloping algebra of $\mathfrak{g}$. By Schur's lemma, it must act by a scalar multiple of identity on any irreducible representation. For a highest weight representation you can actually directly compute this scalar by considering action of the Casimir on the highest weight vector. The result is $\|\lambda + \rho\|^2 - \|\rho\|^2$, where $\lambda$ is the highest weight, $\rho$ is half the sum of positive roots and the norm is taken with respect to the Killing form. This is fairly standard material that can be found in many textbooks. See e.g. Peter Woit's notes on BRST.
I can't comment on your sources; if they claim, they get $\mathrm{dim}\,\mathfrak{g}$, then that is, indeed wildly false. For example, in $\mathfrak{sl}_2$, the representation with highest weight $n$ has Casimir eigenvalue proportional to $n(n-2)$ (let me not try to get the normalization right).
However, it's easy to see that the eigenvalue is non-zero: rather than consider the Casimir with respect to the the Killing form, consider the one constructed a dual basis according to the trace pairing of the representation $V$. This acts on a simple $V$ with eigenvalue $1$; if the algebra is simple, then we immediately see this has to be a non-zero multiple of the usual Casimir. The proof of semi-simplicity along these lines is written in Section 6.2-3 of "Introduction to Lie Algebras and Representation Theory" by @JimHumphreys.