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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.
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3 Answers

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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