By Algebra, Chap. I, §8, no. 2, Prop. 2, any product of $m$ elements of $V^*$ is a $k$-linear combination of elements of the form $x^m$ where $x\in \Pi$. But $x^m=m$! pr$_m$(exp $x$). Q.E.D.

By transport of structure, every automorphism of V defines automorphisms of the algebras $\mathbf{S}(V)$ and $\mathbf{S}(V)^*$; this gives linear representations of $\mathbf{G}\mathbf{L}(V)$ on $\mathbf{S}(V)$ and $\mathbf{S}(V)^*$.

It is clear from the preceding that pr$_m(\psi (k[P]^W))\subset \mathbf{S}^m(\mathfrak{h}^*)^W$. Every element of $\mathbf{S}^m(\mathfrak{h}^*)$ is a $k$-linear combination of elements of the form

pr$_m$(exp $\lambda$ ) $=$ (pr$_m\circ \psi$ )$(e^{\lambda})$

where $\lambda \in P$ (Lemma 1). Hence every element of $\mathbf{S}^m(\mathfrak{h}^*)^W$ is a linear combination of elements of the form

$\sum_{w\in W}w$((pr$_m\circ \psi$ )$(e^{\lambda})$) $=$ (pr$_m\circ \psi$ )$(\sum_{w\in W}w(e^{\lambda}))$,

each of which belongs to pr$_m(\psi (k[P]^W))$.

Tr($u_E$) $=\langle \psi$(ch $E$)$, u\rangle$.

It suffices to treat the case in which $u=h_1. . . h_m$ with $h_1, . . . , h_m\in \mathfrak{h}$. For all $\lambda \in P$, let $d_{\lambda}=$ dim $E^{\lambda}$. Then ch $E =\sum_{\lambda}d_{\lambda}e^{\lambda}$, so $\psi$(ch $E$) $=\sum_{\lambda}d_{\lambda}$ exp($\lambda$ ) and hence

This is simply a reformulation of Prop. 2.

By Prop. 1, $\mathrm{S}^m(\mathfrak h^*)^W=(\mathrm{pr}_m\circ\psi)(k[P]^W)$. Now $\mathbf{Z}[P]^W=\mathrm{ch}\,\mathscr{R}(\mathfrak g)$ (§7, no. 7, Th. 2 (ii)). Thus, by Chap. VI, §3, no. 4, Lemma 3, $\psi(k[P]^W)$ is the $k$-vector subspace of $\mathrm{S}(\mathfrak h)^*$ generated by $\psi(\mathrm{ch}\,\mathscr{R}(\mathfrak g))=\zeta(\mathrm{Tr}\,\mathscr{R}(\mathfrak g))$. Consequently, $\mathrm{S}^m(\mathfrak h^*)^W$ is the vector subspace of $\mathrm{S}^m(\mathfrak h^*)$ generated by $(\mathrm{pr}_m\circ\zeta\circ\mathrm{Tr})(\mathscr{R}(\mathfrak g))$. But, if $\rho$ is a finite dimensional linear representation of $\mathfrak g$,

for all $x\in\mathfrak h$.

Put $g(x_1, . . . , x_m) =$ Tr($\rho (x_1). . . \rho (x_m)$) for $x_1, . . . , x_m\in \mathfrak{a}$. If $x\in \mathfrak{a}$, we have

$=\sum_{1\leq i\leq m}$ Tr($\rho (x_1). . . \rho (x_{i-1})[\rho (x), \rho (x_i)]\rho (x_{i+1}). . . \rho (x_m)$)

= Tr($\rho (x)\rho (x_1). . . \rho (x_m)$)$-$ Tr($\rho (x_1). . . \rho (x_m)\rho (x)$) $= 0$,

so $\theta^*(x)g= 0$. Let $h$ be the symmetric multilinear form defined by

$h(x_1, . . . , x_m) =m1$! $\sum g(x_{\sigma(1)}, . . . , x_{\sigma(m)})$.

For all $x\in \mathfrak{a}$, we have $\theta^*(x)h= 0$ and Tr($\rho (x)^m$) $=h(x, . . . , x)$, hence the lemma.

The condition is clearly necessary. Assume now that it is satisfied. Let $a$ be a nilpotent element of $\mathfrak{g}$. There exists an integer $n$ such that $a^n_E= 0$. For all $t\in k$, we have

0 = exp($ta_E$)$.x-x=ta_Ex+$ 2!$1t^2a^2_Ex+\cdots +(n-1$ 1)! $t^{n-1}a^{n-1}_Ex$,

so $a_Ex= 0$. But the Lie algebra $\mathfrak{g}$ is generated by its nilpotent elements (§4, no. 1, Prop. 1). Hence $x$ is an invariant element of the $\mathfrak{g}$-module E. Q.E.D.

For any $\xi \in \mathbf{G}\mathbf{L}(\mathfrak{g})$, let $\mathbf{S}(\xi )$ be the automorphism of $\mathbf{S}(\mathfrak{g})$ that extends $\xi$, and $\mathbf{S}^*(\xi )$ the restriction to $\mathbf{S}(\mathfrak{g}^*)$ of the contragredient automorphism of $\mathbf{S}(\xi )$. Then $\mathbf{S}$ and $\mathbf{S}^*$ are representations of $\mathbf{G}\mathbf{L}(\mathfrak{g})$. If $a$ is a nilpotent element of $\mathfrak{g},\theta (a)$ is locally nilpotent on $\mathbf{S}(\mathfrak{g})$ and $\mathbf{S}$(exp ad $a$) $=$ exp$\theta (a)$, so

$\mathbf{S}^*$(exp ad $a$) $=$ exp$\theta^*(a)$. (2)

(i) $f\circ s=f$ for all $s\in$ Aut$_e(\mathfrak{g})$;

(ii) $f\circ s=f$ for all $s\in$ Aut$_0(\mathfrak{g})$;

(iii) $f$ is invariant.

The equivalence of (i) and (iii) follows from formula (2) and Lemma 3. It follows from this that (iii) implies (ii) by extension of the base field. The implication (ii) $=\Rightarrow$ (i) is clear.

Note carefully that, if $f$ satisfies the conditions of Prop. $3,f$ is not in general invariant under Aut($\mathfrak{g}$) (Exerc. 1 and 2).

(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.

This follows from Th. 2 (i) and Chap. V, §6, no. 2, Prop. 3.

(i) $\mathbf{S}(\mathfrak{g}) =\mathbf{S}(\mathfrak{h})\oplus J$.

(ii) Let $j$ be the homomorphism from the algebra $\mathbf{S}(\mathfrak{g})$ to the algebra $\mathbf{S}(\mathfrak{h})$ defined by the preceding decomposition of $\mathbf{S}(\mathfrak{g})$. Let $I(\mathfrak{g})$ be the set of invariant elements of $\mathbf{S}(\mathfrak{g})$. Let $\mathbf{S}(\mathfrak{h})^W$ be the set of elements of $\mathbf{S}(\mathfrak{h})$ invariant under the operation of W. Then $j|I(\mathfrak{g})$ is an isomorphism from $I(\mathfrak{g})$ to $\mathbf{S}(\mathfrak{h})^W$.

Assertion (i) is clear. The Killing form defines an isomorphism from the vector space $\mathfrak{g}^*$ to the vector space $\mathfrak{g}$, which extends to an isomorphism $\xi$ from the $\mathfrak{g}$-module $\mathbf{S}(\mathfrak{g}^*)$ to the $\mathfrak{g}$-module $\mathbf{S}(\mathfrak{g})$. We have $\xi (I(\mathfrak{g}^*)) = I(\mathfrak{g})$. The orthogonal complement of $\mathfrak{h}$ with respect to the Killing form is $\mathfrak{n}_++\mathfrak{n}_-($§2, no. 2, Prop. 1). If we identify $\mathfrak{h}^*$ with the orthogonal complement of $\mathfrak{n}_++\mathfrak{n}_-$ in $\mathfrak{g}^*$, then $\xi (\mathfrak{h}^*) =\mathfrak{h}$, so $\xi (\mathbf{S}(\mathfrak{h}^*)) =\mathbf{S}(\mathfrak{h})$ and $\xi (\mathbf{S}(\mathfrak{h}^*)^W) =\mathbf{S}(\mathfrak{h})^W$. Finally, $\xi^{-1}(J)$ is the set of polynomial functions on $\mathfrak{g}$ that vanish on $\mathfrak{h}$. This proves that $\xi$ transforms the homomorphism $i$ of Th. 1 into the homomorphism $j$ of Cor. 2. Thus assertion (ii) follows from Th. 1 (i).

(i) I and $I'$ are graded polynomial algebras (Chap. V, §5, no. 1) of transcendance degree $l$.

(ii) Z is isomorphic to the algebra of polynomials in $l$ indeterminates over $k$.

The canonical filtration of the enveloping algebra of $\mathfrak{a}$ induces a filtration of Z. By Chap. I, §2, no. 7, Th. 1 and p. 25, gr Z is isomorphic to $I'$. In view of Commutative Algebra, Chap. III, §2, no. 9, Prop. 10, it follows that (i) $=\Rightarrow$ (ii).

On the other hand, Th. 1 and its Cor. 2 show that (i) is true whenever $\mathfrak{a}$ is split. The general case reduces to that case in view of the following lemma:

We have ${A'}^0=k'$, so $A^0=k$. Put $A_+=\bigoplus_{n\geq 1}A^n$ and $P = A_+/A^2_+$. Then P is a graded vector space, and there is a graded linear map $f: P\rightarrow A_+$ of

$^2$ In Lemmas 4, 5 and $6,k$ can be any (commutative) field. degree zero such that the composite with the canonical projection $A_+\rightarrow P$ is the identity on P. Give $\mathbf{S}(P)$ the graded structure induced by that of P (Algebra, Chap. III, p. 506). The homomorphism of $k$-algebras $g:\mathbf{S}(P)\rightarrow A$ that extends $f$ (Algebra, Chap. III, p. 497) is a graded homomorphism of degree 0; an immediate induction on the degree shows that $g$ is surjective.

If P is finite dimensional, $\mathbf{S}(P)$ is clearly a graded polynomial algebra, and so is A if $g$ is bijective. Conversely, assume that A is generated by algebraically independent homogeneous elements $x_1, . . . , x_m$ of degrees $d_1, . . . , d_m$. Let $\overline{x}_i$ be the image of $x_i$ in P. It is immediate that the $\overline{x}_i$ form a basis of P; since $\overline{x}_i$ is of degree $d_i$, it follows that $\mathbf{S}(P)$ and A are isomorphic; in particular, dim $\mathbf{S}(P)^n=$ dim $A^n$ for all $n$. Since $g$ is surjective, it is necessarily bijective.

Lemma 4 is now immediate. Indeed, Lemma 5, applied to the $k'$-algebra $A'$, shows that $g\otimes 1 :\mathbf{S}(P)\otimes k'\rightarrow A\otimes k'$ is bijective, and hence so is $g$.

Extending the base field if necessary, we can assume that $\mathfrak{a}=\mathfrak{g}$ is splittable. Assume that $x$ is nilpotent. For any finite dimensional linear representation $\rho$ of $\mathfrak{g}$, and any integer $n\geq 1$, we have Tr($\rho (x)^n$) $= 0$, so $f(x) = 0$ for all homogeneous $f\in I(\mathfrak{g}^*)$ of degree $\geq 1$ (Th. 1 (ii)), and hence $f(x) = 0$ for all $f\in \mathfrak{p}$. Conversely, if $f(x) = 0$ for all $f\in \mathfrak{p}$, then Tr((ad $x$)$^n$) $= 0$ for all $n\geq 1$ (Th. 1 (ii)), so $x$ is nilpotent.

Indeed, for each $s\in G$ there exists a polynomial function $g_s$ on V equal to 1 at $v$ and to 0 at $sv'$. Then the function $g= 1-\prod_{s\in G}g_s$ is equal to 0 at

$v$ and to 1 on $Gv'$. The polynomial function $f=\prod_{t\in g}t.g$ is G-invariant, equal to 0 at $v$ and to 1 at $v'$.

(i) $s\in$ Aut$_0(\mathfrak{a})$;

(ii) for any invariant polynomial function $f$ on $\mathfrak{a}$, we have $f\circ s=f$.

By extending scalars if necessary, we can assume that $k$ is algebraically closed. The implication (i) $=\Rightarrow$ (ii) follows from Prop. 3. We assume that condition (ii) is satisfied and prove (i). In view of Prop. 3, and §5, no. 3, Cor. 1 of Prop. 5, we can assume that $s\in$ Aut($\mathfrak{g},\mathfrak{h}$) and that $s$ leaves stable a Weyl chamber C. Let $x\in C\cap \mathfrak{h}_{\mathbf{Q}}$. We have $sx\in C$. If $g$ is a W-invariant polynomial function on $\mathfrak{h}$, we have $g(x) =g(sx)$ (Th. 1 (i)). By Lemma 6, it follows that $sx\in Wx$. Since $sx\in C$, we have $x=sx$ (Chap. V, §3, no. 3, Th. 2). Then $s|\mathfrak{h}=$ Id$_{\mathfrak{h}}$, and $s\in$ Aut$_0(\mathfrak{g},\mathfrak{h}) ($§5, no. 2, Prop. 4).

Prop. 6 shows that Aut$_0(\mathfrak{a})$ is closed. Let $\overline{k}$ be an algebraic closure of $k$. The group Aut($\mathfrak{a}\otimes \overline{k}$)$/$Aut$_0(\mathfrak{a}\otimes \overline{k})$ is finite (§5, no. 3, Cor. 1 of Prop. 5); a fortiori, the group Aut($\mathfrak{a}$)$/$Aut$_0(\mathfrak{a})$ is finite. Since the cosets of Aut($\mathfrak{a}$) in Aut($\mathfrak{a}$) are closed, it follows that Aut$_0(\mathfrak{a})$ is open in Aut($\mathfrak{a}$).

Let $S(\mathfrak h)^W$ be the set of elements of $S(\mathfrak h)$ invariant under the action of $W$. Then $(\delta\circ\varphi)|_Z$ is an isomorphism from $Z$ to $S(\mathfrak h)^W$, independent of the choice of $B$.

a) Let $P_{++}$ be the set of dominant weights of $R$, $w\in W$, $\lambda\in P_{++}$, $\mu=w\lambda$. Then $Z(\mu-\rho)$ is isomorphic to a submodule of $Z(\lambda-\rho)$ (§6, no. 3, Cor. 2 of Prop. 6), and $\varphi(u)(\lambda-\rho)=\varphi(u)(\mu-\rho)$ for all $u\in Z$ (§6, no. 4, Prop. 7). Thus, the polynomial functions $(\delta\circ\varphi)(u)$ and $(\delta\circ\varphi)(u)\circ w$ on $\mathfrak h^*$ coincide on $P_{++}$. But $P_{++}$ is dense in $\mathfrak h^*$ in the Zariski topology: this can be seen by identifying $\mathfrak h^*$ with $k^B$ by means of the basis consisting of the fundamental weights $\varpi_\alpha$, and by applying Prop. 9 of Algebra, Chap. IV, §2, no. 3. Hence

which proves that $(\delta\circ\varphi)(Z)\subset S(\mathfrak h)^W$.

b) Let $\eta$ be the isomorphism from $I(\mathfrak g)$ to $S(\mathfrak h)^W$ defined in no. 3, Cor. 2 of Th. 1. Consider the canonical isomorphism from the $\mathfrak g$-module $U$ to the $\mathfrak g$-module $S(\mathfrak g)$ (Chap. I, §2, no. 8), and let $\theta$ be its restriction to $Z$. Then $\theta(Z)=I(\mathfrak g)$. Let $z$ be an element of $Z$ with filtration $\leqslant f$ in $U$.

Introduce the notations of §6, no. 4, and put