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I am reading through Humphrey's Introduction to Lie Algebras and Representation Theory on my own and I am currently stumped by one of the exercises, namely Exercise 2 from Section 15.

Let $L$ be a semisimple Lie algebra over an algebraically closed field of characteristic zero. We want to prove that the only solvable Engel subalgebras of $L$ are necessarily Cartan subalgebras. The hint is to use a previous exercise, in which we proved that given a Cartan subalgebra $H\subseteq L$ and $h\in H$, the centralizer $C_L(h)$ of $h$ is reductive.

My idea was to show that our Engel subalgebra $K$ contains a Cartan subalgebra $H$. This is clear. I then tried to show that $K=C_L(h)$ for some $h\in H$, which I can't quite prove. I am only able to prove this if $K=L_0(\text{ad }x)$ for $x\in K$ semisimple. If I can show this in general however then I know where to go from here.

My question can be distilled to the following: is every Engel subalgebra of the form $C_L(h)$? I suspect not but it would be nice if true. If it's not true, I don't know how I should tackle this exercise.

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I am not completely sure about this, but I give you my thoughts.

Let $H\subseteq K\subseteq L$ be as in the question. Let $B$ be a Borel subalgebra containing $K$. We can choose such a $B$ since $K$ is sovable. Let $R^+$ be the positive roots w.r.t. $H$ and $B$. Then we can find $\Phi\subseteq R^+$ such that $K=H\oplus\bigoplus_{\alpha\in\Phi}\mathfrak{g}_\alpha$.

Let $x\in K$ be such that $K$ is the kernel of $(\mathrm{ad}x)^m$ for some $m\geq 1$. Write $x=h+\sum_{\alpha\in\Phi}x_\alpha$ where $h\in H$ and $x_\alpha\in\mathfrak{g}_\alpha$. Suppose for a contradiction that $\Phi\neq\emptyset$. Let $\gamma\in\Phi$ and $y\in\mathfrak{g}_\gamma\setminus\{0\}$. Since $K$ is Engel, we must have $(\mathrm{ad}x)^m y=0$. This is only possible if $h=0$. But then it follows that $x$ is nilpotent. And if $x$ is nilpotent, we have $L=K$ by definition of $K$. Since $L$ is semisimple and $K$ is solvable, this implies that $L=0$ - a contradiction. (By convention, semisimple Lie algebras are $\neq 0$.)

NB. I don't know how to use the previous exercise from the book as suggested by the hint. Maybe, I am using it implicitly.

NB. Once you have established $K=H$, it is clear that $K=C_L(h)$ for some $h\in H$. At least in this setting: solvable Engel subalgebra of semisimple Lie algebra.

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  • $\begingroup$ Looks good! One quick question though. How do we know that $K$ decomposes as expressed in the last line of your first paragraph? $\endgroup$ Jun 27, 2017 at 2:24
  • $\begingroup$ I think the same arguments which give the root space decomposition of semisimple Lie algebras $L$ (cf. page 35 of Humphreys), give you also the root space decomposition of $K$. Indeed, $H$ acts on $K$ and $\mathrm{ad}_K H$ is simultaneously diagonalizable. You also have $C_L(H)=C_K(H)=H=\mathfrak{g}_0$ because $L$ is semisimple. $\endgroup$ Jun 27, 2017 at 11:05
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Since answer to exercise is given above, this answer proceeds the hint Humphreys stated. I tried to prove it in the following; the proof is not so elegant as above since it is very much set theoretic. But, it is by the way Humphreys suggested. May be some dropping/jumping possible.

(0) Let $H$ be Engel and solvable. So $H=L_0(ad_x)$ for some $x\in L$.

(1) Write $x=x_s+x_n$ (Jordan decomposition); then $L_0(ad_{x_s})\subseteq L_0(ad_x)$ [Cor. of 15.3].

(2) Since $x_s$ is semisimple so $L_0(ad_{x_s})=C_L(x_s)$ [see Cor. of 15.3].

(3) Since $x_s$ is semisimple part of $x$, so $C_L(x)\subseteq C_L(x_s)$. We have got $$C_L(x)\subseteq C_L(x_s)\subseteq H.$$

(4) $x_s$ is semisimple, let $K$ a maximal toral containing $x_s$. Then $K$ is abelian, so $K\subseteq C_L(x_s)$.

(5) $L$ is semisimple so $C_L(x_s)$ is reductive; so $C_L(x_s)=Z(C_L(x_s))+ \underbrace{[C_L(x_s), C_L(x_s)]}_{semisimple}$

(6) Since $C_L(x_s)$ is reductive and contained in solvable $H$, its semisimple part in (5) is zero. So $$C_L(x_s)=Z(C_L(x_s)).$$ (7) Since Center of Centralizer=Centralizer of Centralizer, so $$Z(C_L(x_s))=C_L(C_L(x_s))\subseteq C_L(K)=K.$$ where inclusion is by inclusion in (4) and last equality is property of maximal toral in semisimple $L$.

(8) By (3), (6) and (7) we get $C_L(x)\subseteq C_L(x_s)\subseteq K $; so $x\in K$ i.e. $x$ is semisimple. Thus $$H=L_0(ad_x)=L_0(ad_{x_s})=C_L(x_s)\subset K \,\,\,\,\Rightarrow\,\,\,\, H=K.$$

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    $\begingroup$ I quite like this proof! However, I am wondering why it is the case that $C_L(x)\subseteq C_L(x_s)$? If you could shed some light on this, it would be very helpful! Thanks. $\endgroup$ Oct 22, 2020 at 23:55
  • $\begingroup$ @ET-phone-homology I also don't see why $C_L(x) \subseteq C_L(x_s)$ should follow. But since $[x_s,x_n] =0$, then $x = x_s+x_n \in C_L(x_s) \subseteq K$ shows that $x$ is semisimple. The argument can proceed now. $\endgroup$ Dec 15, 2020 at 20:30
  • $\begingroup$ This is because $ad\,x_s$ is a zero constant polynomial function of $ad \,x$. $\endgroup$ May 24, 2021 at 2:54

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