Lemma 2

$^2$ (i) Let W be a vector subspace of V of dimension $d$, D the line $\bigwedge^dW\subset \bigwedge^dV,\theta$ the canonical representation of $\mathfrak{g}\mathfrak{l}(V)$ on $\bigwedge V$ (Chap. III, App.). Let $x\in \mathfrak{g}\mathfrak{l}(V)$. Then $x(W)\subset W$ if and only if $\theta (x)(D)\subset D$.

(ii) Let $(e_1, . . . , e_n)$ be the canonical basis of $k^n,\theta$ the canonical representation of $\mathfrak{g}\mathfrak{l}(n, k)$ on $\bigwedge(k^n)$, and $x\in \mathfrak{g}\mathfrak{l}(n, k)$. Then $x\in \mathfrak{n}(n, k)$ if and only if

for $1\leq d\leq n$.

(i) If $x(W)\subset W$, it is clear that $\theta (x)D\subset D$. Conversely, assume that $\theta (x)D\subset D$. Let $u$ be a non-zero element of D and let $y\in W$. Then $y\wedge u= 0$. Since $\theta (x)$ is a derivation of $\bigwedge V$, this implies

Now $\theta (x)u\in ku$, so $y\wedge \theta (x)u= 0$ and consequently $\theta (x)y\wedge u= 0$. By Algebra, Chap. III, §7, no. 9, Prop. 13, this implies that $\theta (x)y\in W$, i.e. $x(y)\in W$, which proves that $x(W)\subset W$.

(ii) The condition stated in (ii) is clearly necessary for $x\in \mathfrak{n}(n, k)$. Assume that it is satisfied. By (i), $x$ leaves

$^2$ In this lemma, $k$ can be an arbitrary (commutative) field.

stable, and since this holds for $d= 1, . . . , n,x$ is lower triangular. Put

We have $0 =x(e_n) =x_{nn}e_n$, so $x_{nn}= 0$. Let $i < n$, and assume that we have proved that $x_{jj}= 0$ for $j > i$. Then

so $x_{ii}= 0$. Thus, $x\in \mathfrak{n}(n, k)$.