Lemma 3

Let E be a finite dimensional $\mathfrak{g}$-module, and $x\in E$. Then $x$ is an invariant element of the $\mathfrak{g}$-module E if and only if (exp $a_E$)$.x=x$ for every nilpotent element $a$ of $\mathfrak{g}$.

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)