0

Let $G$ be a complex semisimple connected Lie group with Lie algebra $\mathfrak{g}$. An element $x \in \mathfrak{g}$ is called ad-nilpotent if the operator $\text{ad} \ x : \mathfrak{g} \to \mathfrak{g}, \ y \mapsto [x,y]$ is nilpotent, i.e., $(\text{ad} \ x)^n = 0$ for some $n >0$. The group $G$ acts on $\mathfrak{g}$ by conjugation - this is also called the adjoint action and denoted by $\text{Ad}$.

My question is: why is ad-nilpotency preserved by the adjoint action of $G$? In other words, if $x \in \mathfrak{g}$ is ad-nilpotent, why is $gxg^{-1}$ also ad-nilpotent, for all $g \in G$?

Balerion_the_black
  • 2,060

1 Answers1

0

Let $Ad:G\rightarrow Aut(g)$ be the adjoint representation. For every $g\in G, x,y\in g$, $Ad_g([x,y])=[Ad_g(x),Ad_g(y)]$.

This implies that $(ad_{Ad_g(x)})^n=Ad_g((ad_x)^n(Ad^{-1}_g(y)))$.

To see this, we may use a recurrence: $ad_{Ad_g(x)}(y) =[Ad_g(x),y]=[Ad_g(x),Ad_g(Ad_g^{-1}(y)]=Ad_g([x,Ad_g^{-1}(y)])$.

Suppose $(ad_{Ad_g(x)})^n=Ad_g((ad_x)^n(Ad^{-1}_g(y)))$, then $ad_{Ad_g(x)}^{n+1}(y)=ad_{Ad_g(x)}(Ad_g((ad_x)^n(Ad^{-1}_g(y))))=[Ad_g(x),Ad_g((ad_x)^n(Ad_g^{-1}(y)))]=Ad_g[x,(ad_x)^n(Ad_g^{-1}(y))]=Ad_g((ad_x)^{n+1}(Ad_g^{-1}(y)))$

Since if $x$ is nilpotent, $(ad_x)^n=0$, this implies the result.

Tsemo Aristide
  • 87,475