Proposition 8

Let $\mathfrak{n}$ be a Lie subalgebra of $\mathfrak{g}\mathfrak{l}(V)$ consisting of nilpotent elements, $\mathfrak{q}$ the normalizer of $\mathfrak{n}$ in $\mathfrak{g}\mathfrak{l}(V)$. There exists a finite dimensional vector space E, a representation $\rho$ of $\mathfrak{g}\mathfrak{l}(V)$ on E, and a vector subspace F of E, satisfying the following conditions:

(i) the image under $\rho$ of a homothety of V is diagonalizable;

(ii) F is stable under $\rho (\mathfrak{q})$;

(iii) $\mathfrak{n}$ is the set of $x\in \mathfrak{g}\mathfrak{l}(V)$ such that $\rho (x)(F) = 0$.

Let $n=$ dim V. By Engel’s theorem, V can be identified with $k^n$ in such a way that $\mathfrak{n}\subset \mathfrak{n}(n, k)$. Let P be the algebra of polynomial functions on $\mathfrak{g}\mathfrak{l}(n, k)$. For $i= 0,1, . .$., let $P_i$ be the set of elements of P homogeneous of degree $i$. Let N = exp $\mathfrak{n}$, which is a subgroup of the strictly lower triangular group T. Let J be the set of elements of P that are zero on N; this is an ideal in P. Let $N_J$ be the set of $x\in \mathfrak{g}\mathfrak{l}(n, k)$ such that $p(x) = 0$ for all $p\in J$. Then $N\subset N_J$. Conversely, let $x\in N_J$. Denote by $p_{ij}$ the polynomial functions giving the entries of an element of $\mathfrak{g}\mathfrak{l}(n, k)$. The ideal J contains the $p_{ij}$ (for $i < j$) and the $p_{ii}-1$; hence $x\in T$. On the other hand, if $u$ is a linear form on $\mathfrak{g}\mathfrak{l}(n, k)$ which is zero on $\mathfrak{n}$, there exists $p_u\in P$ such that $p_u(z) =u$(log $z$) for all $z\in T ($§3, no. 1, Lemma 1 (i)); we have $p_u\in J$, so $u$(log $x$) $= 0$. It follows that log $x$ belongs to $\mathfrak{n}$, so $x\in N$, proving that $N = N_J$.

For all $p\in P$ and $g\in \mathbf{G}\mathbf{L}_n(k)$, let $\lambda (g)p$ be the function $x \rightarrow p(g^{-1}x)$ on $\mathfrak{g}\mathfrak{l}(n, k)$; then $\lambda (g)p\in P,\lambda (g)$ is an automorphism of the algebra P, and $\lambda$ is a representation of $\mathbf{G}\mathbf{L}_n(k)$ on P which leaves each $P_i$ stable. We show that

If $x\in N, p\in J, y\in N$, then $(\lambda (x)p)(y) =p(x^{-1}y) = 0$ since $x^{-1}y\in N$; thus $\lambda (x)p\in J$, so $\lambda (x)J = J$. Let $x\in \mathbf{G}\mathbf{L}_n(k)$ be such that $\lambda (x)J = J$; let $p\in J$; then $p(x^{-1}) = (\lambda (x)p)(e) = 0$, so $x^{-1}\in N_J= N$ and $x\in N$. This proves (i).

The ideal J is of finite type (Commutative Algebra, Chap. III, §2, no. 10, Cor. 2 of Th. 2). Hence, there exists an integer $q$ such that, if $W = P_0+ P_1+\cdots + P_q$, then $J\cap W$ generates J as an ideal. Denote by $\lambda_j$ (resp. $\lambda '$) the subrepresentation of $\lambda$ defined by $P_J$ (resp. by W). By (1),

We show that, for all $j$, there exists a representation $\sigma_j$ of the Lie algebra $\mathfrak{g}\mathfrak{l}(n, k)$ on $P_j$ such that:

$\sigma_j|\mathfrak{n}(n, k)$ is compatible (§$3$, no$.1$) with $\lambda_j|T$. (3)

For all $x\in k.1_n, \sigma_j(x)$ is a homothety. (4)

Since $\lambda_j$ is the $j$th symmetric power of $\lambda_1$, it suffices to prove the existence of $\sigma_1$, cf. Lemma 1. Now $\lambda_1$ is the contragredient representation of the representation $\gamma$ of $\mathbf{G}\mathbf{L}_n(k)$ on $\mathfrak{g}\mathfrak{l}(n, k)$ given by

Let $c$ be the representation of the Lie algebra $\mathfrak{g}\mathfrak{l}(n, k)$ on $\mathfrak{g}\mathfrak{l}(n, k)$ given by

It is immediate that $c|\mathfrak{n}(n, k)$ and $\gamma |T$ are compatible, and that $c(x)$ is a homothety for all $x\in k.1_n$. Thus, it suffices to take for $\sigma_1$ the dual representation of $c$ (Chap. I, §3, no. 3).

Now let $\sigma '$ be the representation of $\mathfrak{g}\mathfrak{l}(n, k)$ on W given by the direct sum of the $\sigma_j, 0\leq j\leq q$. In view of (2) and the relations

$\lambda '$(exp($x$)) $=$ exp($\sigma '(x)$) and $\sigma '$(log($y$)) $=$ log($\lambda '(y)$)$, x\in \mathfrak{n}(n, k), y\in T$,

we have

Let $d=$ dim(J $\cap W$), and let $\tau =\bigwedge^d\sigma '$. Let $D =\bigwedge^d(J\cap W)$. By (5) and Lemma 2 (i),

But $\tau (\mathfrak{n}(n, k))$ consists of nilpotent endomorphisms, so (6) can also be written

Now let $E =\bigwedge^dW\oplus \bigwedge^1V\oplus \bigwedge^2V\oplus \cdots \oplus \bigwedge^nV$; let $\rho$ be the direct sum of $\tau$ and the canonical representations of $\mathfrak{g}\mathfrak{l}(n, k)$ on $\bigwedge^1V, . . . ,\bigwedge^nV$. Let $E_0\subset E$ be the sum of $D =\bigwedge^d(J\cap W)$ and the lines generated by $e_{n-j+1}\wedge \cdots \wedge e_n$ for $j= 1, . . . , n$. By (7) and Lemma 2 (ii),

It is immediate that, if $x\in k.1_n,\rho (x)$ is diagonalizable. Finally, if F is the set of elements of E annihilated by $\rho (\mathfrak{n}), F$ is stable under $\rho (\mathfrak{q})$ (Chap. I, §3, no. 5, Prop. 5), and by (8),