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

Then $\ad x $ is nilpotent.

I know that if $\alpha, \beta\in H^*$ 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. enter image description here

Mark Murray
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3 Answers3

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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 $x\in L_\alpha$ and $y\in L_\beta$ then $(\text{ad}\, x)(y)=[x,y]\in L_{\alpha+\beta}$ and then $(\text{ad}\, x)^n(y)\in L_{n\alpha+\beta}$. If $\alpha\ne 0$ then there is $n$ large enough so that $L_{n\alpha+\beta}=0$ for all $\beta$ with $L_\beta\ne0$ since there are only finitely many root spaces. Then $(\text{ad}\, x)^n(y)=0$ for all $y\in L_\beta$ and so $(\text{ad}\, x)^n(y)=0$ for all $y\in L$.

Angina Seng
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  • I understand up until "for all $\beta$ with $L_\beta=0$" should this be $L_\beta\neq 0$? Also how do we go from "$(\text{ad}, x)^n(y)=0$ for all $y\in L_\beta$" to " $(\text{ad}, x)^n(y)=0$ for all $y\in L$". – Mark Murray Jul 24 '20 at 17:38
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    @MarkMurray $L$ is the sum of the $L_\beta$s (including $H=L_0$). – Angina Seng Jul 24 '20 at 18:12
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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 $N$.

Kenny Wong
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  • How do we know that we won't by looping through such $L_\gamma$'s – Mark Murray Jul 24 '20 at 17:47
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    @MarkMurray $\alpha$ and $\beta$ are vectors in a vector space!!! They are vectors in the dual of the Cartan subalgebra $H$. Since $\alpha \neq 0$, it must be the case that $\beta,\beta + \alpha,\beta + 2\alpha, \beta + 3\alpha, \dots $ are all distinct vectors in the vector space! – Kenny Wong Jul 24 '20 at 17:56
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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 $Y := \frac1r\mathop{ad}_Lx, H:=\mathop{ad}_L h \in End_k(L)$. 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).

Added: For more general statements, and exact reference to Jacobson, cf. If $A$ and $AB-BA$ commute, show that $AB-BA$ is nilpotent. User1551's answer to If $A=AB-BA$, is $A$ nilpotent? contains a minimal counterexample to the statement in positive characteristic.

Torsten Schoeneberg
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