# The Casimir invariant of an irreducible representation of a compact Lie group

Let $G$ be a compact Lie group (not necessarily connected) and $\rho:G\to \mathrm{End}(V)$ an irreducible (hence finite-dimensional) unitary representation of $G$. Let $\mathfrak{g}$ be the Lie algebra of $G$, $\mathfrak{U}(\mathfrak{g})$ the universal enveloping algebra, and write also $\rho$ for the representation $\mathfrak{U}(\mathfrak{g})\to \mathrm{End}(V)$ induced by $\rho$. Let $B$ be some non-degenerate $\mathrm{Ad}(G)$-invariant Bilinear form on the Lie algebra $\mathfrak{g}$ of $G$ and let $\Omega_B\in \mathfrak{U}(\mathfrak{g})$ be the Casimir element associated to $B$, i.e. one has $$\Omega_B=\sum_{i=1}^{\mathrm{dim}\,\mathfrak{g}}X_iX^i,$$ where $\{X_i\}$ is any basis of $\mathfrak{g}$ and $\{X^i\}$ the dual basis of $\mathfrak{g}$ w.r.t. $B$. Then, by Schur's Lemma, the endomorphism $$\rho(\Omega_B):V\to V$$ is a scalar multiple of the identity: $$\rho(\Omega_B)=\lambda(\rho,B)\,\mathrm{Id}_V,\qquad \lambda(\rho,B)\in \mathbb{C}.$$ Let us call $\lambda(\rho,B)$ the Casimir invariant of the representation $\rho$ with respect to the bilinear form $B$.

In this setting, I have the following questions:

• How is the Casimir invariant related to the character $\chi(\rho):G\to \mathbb{C}$ given by $g\mapsto \mathrm{tr}\,\rho(g)$ ?
• Is there a simple direct formula to compute the Casimir invariant without using the character?

Of course, my hope is that to an expert these questions seem very easy and there is a standard reference giving the answers that I have been unable to find.

I have found a ton of literature about Casimir operators of semisimple Lie groups (where $B$ is the Killing form), expressing their eigenvalues via highest weights, and computations of Casimir invariants in special cases, that is, for certain concretely given representations. But in the compact group case I could not find any good resources and in particular none on the relation to the character.

• Bourbaki, Lie Groups and Lie Algebras, Chapter 9, §7, no. 6: "Casimir elements". – abx May 15 '17 at 9:50

[I started to type this as a comment but it took like ten comments to fit, so I paste it into an answer. It is an answer as it explains how you can use the references you already have to answer your question, but it does not answer your question directly.]

[EDIT: as pointed out in the comments, the answer below is only true when the center of $G$ is discrete, hence finite (by compactness). In the general case you can compute the Casimir action for $G$ from the character of the center $Z$ of $G$ together with the Casimir of $G/Z$ which is what my answer below deals with, but right now I'm too tired to see whether this pasting together the two elements into one big Casimir is really straightforward or requires some deeper insight into the group structure. Hopefully I or someone else can say something about that later.]

The semi-simple case covers the compact case, so the reference you already have is - in principle - all you need.

In more detail: let $\mathfrak{g}_0$ be the Lie algebra of $G$ (the tangent space to $G$ at the identity so in particular a real vector space) and let $\mathfrak{g}$ be the complexification of $\mathfrak{g}_0$ (so the thing you get when you pick any basis of $\mathfrak{g}_0$ and consider the Lie algebra of all complex linear combination of these basis elements, with the Lie bracket defined by making it complex linear.)

Now since $G$ acts on $V$, so does $\mathfrak{g}_0$ and, assuming $V$ is a vector space over the complex numbers, we get for free that this action extends to an action of all of $\mathfrak{g}$ on $V$. This has numerous advantages. For instance the complex Lie algebra has much nicer bases, giving rise to the whole business of highest weight vectors and stuff which enable you to write down a Casimir for which computations are relatively easy. In general, most of these computationally nicer basis elements do not lie in the (real) subalgebra $\mathfrak{g}_0$ of $\mathfrak{g}$, but we need not worry about that since $V$ is a complex space anyway and if we were to use a basis of $\mathfrak{g}_0$ instead we would (eventually, after more annoying computations) find the same action of the Casimir.

Now the reason I know that we can find a really nice basis of the complex Lie algebra $\mathfrak{g}$ giving rise to highest weight theory is that the compactness of the real group $G$ implies the semi-simpleness of the complex algebra $\mathfrak{g}$. (Even though the associated complex algebraic group is (as a manifold) far from compact and there are many non-compact real Lie groups having lie algebras $\mathfrak{g}_1$ non-isomorphic to $\mathfrak{g}_0$ whose complexification also equals this same $\mathfrak{g}$.)

This result: compact group implies semi-simple complexified Lie-algebra really is an example of the magic of Lie theory where topological properties imply algebraic properties and vice versa.

Finally, to see why compact groups imply semi-simple complexified Lie algebra, take the definition of semi-simple that reads 'every finite dimensional representation is the direct sum of irreducible representations'. (The equivalence of various definitions of semi-simple takes place entirely inside the world of complex Lie algebras without any role for the group so I won't discuss that here.) This same statement (any finite dimensional rep is a direct sum of irreducibles) is easy to prove for compact groups, as below.

Take any inner product $( , )$ on $V$ and turn in into a $G$-equivariant inner product $\langle , \rangle$ by letting $\langle v, w \rangle = \int_G (gv, gw) d\mu$ where $\mu$ is the Haar measure, renormalized to have total measure 1. (This is where we use compactness.)

Now that we have an inner product satisfying $\langle hv, hw \rangle = \langle v, w \rangle$ for all $v, w \in V$ we see that whenever $W$ is a $G$-invariant subspace of $V$, so is the orthocomplement to $W$ under $\langle , \rangle$.

In other words: whenever $V$ is not irreducible it decomposes as a direct sum of two subrepresentations, and continuing in this fashion we see that it decomposes as a sum of irreducibles.

• O where I say 'covers' I mean it in the every day world sense of the world, not in the topological sense. – Vincent May 15 '17 at 21:34
• There's something I don't understand here. Take a torus. This is a compact group. Being commutative, its (complexified) Lie algebra is abelian and hence not semisimple. – José Figueroa-O'Farrill May 15 '17 at 22:26
• O that's right. I totally forgot about the existence of Abelian groups! – Vincent May 15 '17 at 22:29
• Ok, so the answer needs some serious rewriting, dealing with the center of the group (and Lie algebra). It is a bit late here, hopefully I get to it tomorrow (or you can do it if you want). Let me just grab the opportunity to quickly say that I really enjoy reading the appendix to your paper about triple systems! Even if I am progressing slowly. (See also my MSE question about it: math.stackexchange.com/q/2279785/101420. I believe the given answer is correct but I want to so some computations before accepting.) – Vincent May 15 '17 at 22:42
• Sorry, I didn't see the question on MSE. The answer is correct. I'll make a comment there. – José Figueroa-O'Farrill May 15 '17 at 22:46