3
$\begingroup$

Some context for my question. Given a vector space $V$, $\mathbb{C}[V]$ is defined to be the ring of polynomials on the vector space $V$. In other words, we can make the identification $\mathbb{C}[V] = Sym V^*$, where $V^*$ is the dual vector space.

I wish to know whether there is a nice description for $\mathbb{C}[\mathfrak{g}]^G$.

An explicit description of the invariants in the case of $G = A_2$ would be enlightening and helpful too.

$\endgroup$
2
  • 3
    $\begingroup$ It is a polynomial ring. The degrees of the generators are given here: en.wikipedia.org/wiki/Harish-Chandra_isomorphism $\endgroup$ Dec 9, 2014 at 6:59
  • 1
    $\begingroup$ There are quite a few questions on MO close to this one, such as mathoverflow.net/questions/132455 (try searching for Chevalley restriction theorem, Weyl group invariants, etc.) Basic invariants in the Weyl group framework are often discussed in these questions (and supporting literature), but lifting these explicitly to the Lie algebra framework isn't trivial. $\endgroup$ Dec 9, 2014 at 18:22

1 Answer 1

7
$\begingroup$

Spivak, Comprehensive Introduction to Differential Geometry, volume 5, chapter 13, explains how to calculate these using classical invariant theory for all of the classical groups and gives the computations explicitly for $SO(n)$: the Pfaffian (if $n$ is even) and the even degree elementary symmetric functions of the eigenvalues. For $A_2$, the invariant polynomials are the determinant and the trace of the square of each matrix, as the trace is zero. More generally for $A_n$ you get the elementary symmetric functions of the eigenvalues, except for the trace (which is zero), as in the theory of Chern classes.

I think you find all of the invariants written out pretty explicitly in M. L. Mehta, Basic sets of invariant polynomials for finite reflection groups, Comm. in Algebra, 16 (5), 1988, p. 1083-1098. The idea is that any invariant polynomial must restrict to any Cartan subalgebra to become invariant under the Weyl group, and conversely any Weyl group invariant polynomial extends to an invariant polynomial on the Lie algebra.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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