# Rootspaces are $mathop{ad}$ nilpotent

Mathematics Asked on January 1, 2022

$$DeclareMathOperator{ad}{ad}$$
Let $$L$$ be a semisimple Lie algebra with root space $$L=H oplus bigoplus_{alpha in Phi}L_alpha$$. Let $$xin L_alpha$$ with $$alphaneq 0$$. I want to show

Then $$ad x$$ is nilpotent.

I know that if $$alpha, betain H^*$$ then $$[L_alpha,L_beta]subset L_{alpha+beta}$$. I think I should make if we could show that there are only finitely many non-zero $$L_alpha$$, we could use this fact to push $$x$$ into a trivial root space. Then it should follow that $$(ad x)$$ is nilpotent? I am slightly confused as to what it means for $$(ad x)$$ to be nilpotent, is it that $$(ad x)^n$$ is zero, or is it that $$ad^n x$$ is zero?

Note this is a Proposition in Humphreys book but I do not see how it follows directly.

The other two answers give a quick argument for the case at hand. I'd like to point out the following far more general statement and proof (due to N. Jacobson, as far as I know):

Let $$L$$ be a finite-dimensional Lie algebra over a field $$k$$ with $$mathop{char}(k)=0$$, and let $$x in L$$ such that there exists $$h in L$$ with $$[h,x] = rx$$, $$r in k^ast$$. Then $$mathop{ad}_Lx$$ is nilpotent.

Proof: Abbreviate $$Y := frac1rmathop{ad}_Lx, H:=mathop{ad}_L h in End_k(L)$$. Then $$mathop{ad}_{L}x =HY-YH$$, and iteratively (using that $$mathop{ad}_Lx$$ commutes with $$Y$$),

$$(ad_{L} x)^n=(ad_Lx)^{n-1}(HY-YH) = ((ad_Lx)^{n-1}H)Y- Y((ad_Lx)^{n-1}H)$$

for all $$n ge 1$$. So we've written all $$(ad_{L} x)^n in End_k(L)$$ as commutators, which have trace $$0$$; but this famously implies that $$ad_{L} x$$ is nilpotent (for this step we need the restriction on the characteristic though).

Answered by Torsten Schoeneberg on January 1, 2022

I presume we are in the finite-dimensional case, when there are only finitely many non-zero $$L_alpha$$.

$$text{ad}, x$$ is a map from $$L$$ to $$L$$ and then its $$n$$-th power $$(text{ad}, x)^n$$ is also a map from $$L$$ to $$L$$. If $$xin L_alpha$$ and $$yin L_beta$$ then $$(text{ad}, x)(y)=[x,y]in L_{alpha+beta}$$ and then $$(text{ad}, x)^n(y)in L_{nalpha+beta}$$. If $$alphane 0$$ then there is $$n$$ large enough so that $$L_{nalpha+beta}=0$$ for all $$beta$$ with $$L_betane0$$ since there are only finitely many root spaces. Then $$(text{ad}, x)^n(y)=0$$ for all $$yin L_beta$$ and so $$(text{ad}, x)^n(y)=0$$ for all $$yin L$$.

Answered by Angina Seng on January 1, 2022

I would interpret the statement that $${rm ad }x$$ is nilpotent as saying that there exists an $$N$$ such that the linear map $$({rm ad }x)^N : L to L$$ is the zero map.

As you say, that fact that $${rm ad}(x)$$ maps from each $$L_beta$$ to $$L_{alpha + beta}$$, and the fact that there finitely many non-zero $$L_{gamma}$$'s, implies that $$({rm ad }x)^N$$ is zero for sufficiently large $$N$$.

Answered by Kenny Wong on January 1, 2022

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