Theorem 1

Let $I(\mathfrak{g}^*)$ be the algebra of invariant polynomial functions on $\mathfrak{g}$. Let $i:\mathbf{S}(\mathfrak{g}^*)\rightarrow \mathbf{S}(\mathfrak{h}^*)$ be the restriction homomorphism.

(i) The map $i|I(\mathfrak{g}^*)$ is an isomorphism from the algebra $I(\mathfrak{g}^*)$ to the algebra $\mathbf{S}(\mathfrak{h}^*)^W$.

(ii) For any integer $n\geq 0$, let $I^n(\mathfrak{g}^*)$ be the set of homogeneous elements of $I(\mathfrak{g}^*)$ of degree $n$. Then $I^n(\mathfrak{g}^*)$ is the set of linear combinations of functions on $\mathfrak{g}$ of the form $x \rightarrow$ Tr($\rho (x)^n$), where $\rho$ is a finite dimensional linear representation of $\mathfrak{g}$.

(iii) Let $l=$ rk($\mathfrak{g}$). There exist $l$ algebraically independent homogeneous elements of $I(\mathfrak{g}^*)$ that generate the algebra $I(\mathfrak{g}^*)$.

a) Let $f\in I(\mathfrak{g}^*)$ and $w\in W$. There exists $s\in$ Aut$_e(\mathfrak{g},\mathfrak{h})$ such that $s|\mathfrak{h}=w($§2, no. 2, Cor. of Th. 2). Since $f$ is invariant under $s$ (Prop. $3$)$,i(f)$ is invariant under $w$. Hence $i(I(\mathfrak{g}^*))\subset \mathbf{S}(\mathfrak{h}^*)^W$.

b) We prove that, if $f\in I(\mathfrak{g}^*)$ is such that $i(f) = 0$, then $f= 0$. Extending the base field if necessary, we can assume that $k$ is algebraically closed. By Prop. $3,f$ vanishes on $s(\mathfrak{h})$ for all $s\in$ Aut$_e(\mathfrak{g})$. Hence $f$ vanishes on every Cartan subalgebra of $\mathfrak{g}$ (Chap. VII, §3, no. 2, Th. 1), and in particular on the set of regular elements of $\mathfrak{g}$. But this set is dense in $\mathfrak{g}$ for the Zariski topology (Chap. VII, §2, no. 2).

c) Let $n$ be an integer $\geq 0$. Let $L^n$ be the set of linear combinations of functions of the form $x \rightarrow$ Tr($\rho (x)^n$) on $\mathfrak{g}$, where $\rho$ is a finite dimensional linear representation of $\mathfrak{g}$. By Lemma $2, L^n\subset I^n(\mathfrak{g}^*)$. Thus

By Cor. 2 of Prop. $2,\mathbf{S}^n(\mathfrak{h}^*)^W\subset i(L^n)$. Hence $i(I^n(\mathfrak{g}^*)) =\mathbf{S}^n(\mathfrak{h}^*)^W$, which proves (i), and $i(L^n) =i(I^n(\mathfrak{g}^*))$ so $L^n= I^n(\mathfrak{g}^*)$ by b). Thus (ii) is proved.

d) Assertion (iii) follows from (i) and Chap. V, §5, no. 3, Th. 3.