Kostant's $G$-invariant part in the sym power ring of adjoint representation? Let $g$ be a Lie algebra, say $sl_n(\mathbb C)$. It is considered as the adjoint representation of $G=SL_n(\mathbb C)$. 
A famous theorem of Kostant from "Lie Group Representations on Polynomial Rings" states that $$S=I\otimes H,$$
where $S( g)=\oplus_{m=0}^\infty \vee^m {g}$ is a graded algebra of the symmetric powers of  the adjoint representation $g$, 
$H$ is $G$-harmonic part, and $I$ is $G$-invariant part.
I would like to see a formula for $I$, or the (degree of) generators of $I$. Is it true $I=\prod_{i=2}^n (1-q^{i})^{-1}$?
$q$ is the symbol for recording the grading.
 A: As Kostant himself says there, p. 330 (with $\mathfrak g$ complex reductive, $G$ the adjoint group, $J$ for your $I$):

Here the structure of $I$ is given by a theorem of Chevalley. This asserts that $I$ is a polynomial ring in $l$ (the rank of $\mathfrak g$) homogeneous generators $u_i$, $i = 1,2,\cdots, l$ with $\deg u_i= m_i+ 1$ where the $m_i$ are the exponents of $\mathfrak g$.

These exponents and their computation are discussed in his previous paper The principal three-dimensional subgroup..., e.g. for $sl_n$ they are $1,2,\cdots,n-1$. This is covered in Bourbaki, Lie Groups, Chap. VIII, §8, Thm 1 ((i): restriction $\mathrm S(\mathfrak g)\to\mathrm S(\mathfrak h)$ induces an isomorphism $I\to\mathrm S(\mathfrak h)^W$).
I am not sure how to interpret your proposed "formula for" $I$. It looks like a Poincaré series whereas Kostant gives the Poincaré polynomial of $G$ as $p_G(t)=\prod_{j=1}^l(1+t^{2m_j+1})$, after Hopf.
A: Unless I misunderstood something, in the case of $sl_n$ the Lie algebra is that of $n\times n$ matrices $M$ with zero trace. The invariants are given by traces of powers of $M$. Here, the algebra of invariants is generated by ${\rm tr}(M^k)$ for $k=2,\ldots,n$. These generators are algebraically independent, so your formula for the Cayley-Sylvester-Hilbert series of $I$, is I believe correct.
