MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given a general Lie algebra, is there a general procedure to find all its Casimir operator?

share|cite|improve this question
If the Lie algebra is semisimple, then the center of its universal enveloping algebra is polynomial, and lists of good generators (often "good" means "homogeneous with respect to some grading") are known. In general, I think there is the Duflo theorem that $Z(U(\mathfrak g)) = \operatorname{Sym}(\mathfrak g)^{\mathfrak g}$ as rings, but the map is somewhat nontrivial. And anyway, this just moves the problem: certainly I don't know how, for a general Lie algebra, to compute the ring structure on $\operatorname{Sym}(\mathfrak g)^{\mathfrak g}$ (GIT quotient of adjoint action). – Theo Johnson-Freyd Aug 27 '11 at 14:08

share|cite|improve this answer
The method in this paper is particularly suited for solvable Lie algebras, though, as it requires an explicit parametrisation of the adjoint group. Of course, the OP asks about "a general Lie algebra" so this goes some way towards answering the question. +1 – José Figueroa-O'Farrill Aug 27 '11 at 17:12
Due to "link rot", this answer is now useless :( Could anyone reproduce the author(s) & title of the linked article? – Danu Feb 10 at 11:14

I'm assuming you're thinking of some specific matrix representation $X_i \in \mathfrak{g}$ (let's assume it's the defining representation). Compute the Killing form, $\kappa_{ij} \doteq Tr (X_i\cdot X_j)$ (actually usually this is defined in the adjoint representation, but any faithful rep will do). The quadratic Casimir is then simply $ X_i \kappa_{ij} X_j$ (Einstein convention).

Other Casimirs can be obtained from the characteristic (secular) equation: define $X(\omega) = \omega^i X_i$. The characteristic equation is $\det\left( X(\omega) - \lambda I \right) = \sum\limits_{j} (-\lambda)^{N-j} \phi_j(\omega) \equiv 0 $ ($N$ is the matrix dimension, and/or the dimension of the Lie algebra if you're using the adjoint representation). If you now perform the substitution $\omega^i \to X_i$ in the coefficients $\phi_j (\omega)$, you get Casimir invariants $\phi_j (X)$!

It might seem like the higher the representation, the more invariants you get, but in fact all the invariants can be expressed in terms of the fundamental invariants of the defining representation. I can't think of many references right at the moment, but e.g. Gilmore: Lie groups, physics and geometry pp. 140 has a nice explanation. Also, google the boldface texts above.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.