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Let $\mathfrak{g}$ be a simple finite-dimensional Lie algebra over an arbitrary field $K$. For any nonzero $x\in\mathfrak{g}$ we must have $[\mathfrak{g},x]\neq\{0\}$, or else we violate simplicity.

Can it happen that there are two nonzero elements $x,y\in\mathfrak{g}$ such that $[[\mathfrak{g},x],y]=\{0\}$? If so, do $x,y$ have to satisfy some conditions? Are the previous questions easier to answer if we restrict to some $K$ (say algebraically closed, or of characteristic $0$) or to some $\mathfrak{g}$?

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  • $\begingroup$ (Assume char zero) Since passing from an element $z$ to its semisimple part (in the additive decomposition) reduces $[\mathfrak{g},z]$ and increases its centralizer, one can assume that each of $x$ and $y$ is either semisimple or nilpotent. And actually since for $x$ semisimple, $[\mathfrak{g},z]$ is a reductive subalgebra, one can assume that $x$ is nilpotent or both $x,y$ are semisimple. $\endgroup$
    – YCor
    Commented Feb 22, 2022 at 10:08
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    $\begingroup$ As a side note, it is very easy to see how to do it with 3 brackets. Applying $\mathrm{ad}_x$ 3 times for $x$ a root vector does the job. $\endgroup$
    – Callum
    Commented Feb 22, 2022 at 12:56
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    $\begingroup$ @LSpice Ah that's a very good point. I hadn't thought through my suggestion properly. We don't absolutely need the Lie algebra to be split but my idea certainly won't work for a compact Lie algebra. If we have a parabolic subalgebra (which is true for everything over $\mathbb{R}$, $\mathbb{C}$ except the compact real ones) we can find an element in the nilradical of the parabolic which has $\mathrm{ad}_x^3 = 0$. $\endgroup$
    – Callum
    Commented Feb 25, 2022 at 21:43
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    $\begingroup$ @Callum, a Lie algebra of an algebraic group $G$ will always have a (proper, as you of course meant) parabolic subalgebra unless it is 'compact', where the meaning of 'compact' over an arbitrary field is that $G$ contains no split torus (often called 'anisotropic'). That is, this isn't a special property of $\mathbb R$ or $\mathbb C$ (although the fact that we need not speak of algebraic groups is special to the charateristic-$0$ setting). Re, how do you find $x$ in the nilradical of an arbitrary proper parabolic subalgebra? $\endgroup$
    – LSpice
    Commented Feb 26, 2022 at 0:24
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    $\begingroup$ @LSpice I'm not sure how much of this carries over from the real/complex settings but there at least a parabolic subalgebra $\mathfrak{p}$ induces a filtration on the Lie algebra. Loosely we can see this by repeatedly applying $\operatorname{ad} \mathfrak{p}^\perp$ (i.e. the nilradical). This gives a filtration of the form $\mathfrak{p}^{n}\geq\mathfrak{p}^{n-1}\geq \cdots \leq \mathfrak{p}^{-n}$ with $[ \mathfrak{p}^{i}, \mathfrak{p}^{j}]\leq \mathfrak{p}^{i+j}$. Then any element $x\in\mathfrak{p}^{-n}$ will send $\mathfrak{g} \to \mathfrak{p}\to \mathfrak{p}^{-n} \to \{0\}$ $\endgroup$
    – Callum
    Commented Feb 26, 2022 at 8:12

2 Answers 2

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The special case $x=y$ is just the notion of a so-called sandwich element (that is, an element $x$ of $L$ such that $(\operatorname{ad}x)^2=0$). Such elements played an important role in the classification of finite-dimensional simple Lie algebras over fields of positive characteristic. For instance, over algebraically closed fields of characteristic $p>5$, in the paper

[A. A. Premet: Lie algebras without strong degeneration, Mat. Sb. (N.S.) 129(171(1))(1986), 140–153]

it is proved that the presence of sandwich elements characterizes finite-dimensional simple Lie algebras which are not classical.

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    $\begingroup$ Just for completeness: in char. zero the result of non-existence of sandwich element follows from the Jacobson-Morozov theorem (which ensures that in a semisimple Lie algebra for every $x$ such that $\mathrm{ad}(x)$ is nilpotent and nonzero, there exists $y$ such that $[x,[x,y]]=x$, so that $\mathrm{ad}(x)^2$ is nonzero as well). $\endgroup$
    – YCor
    Commented Feb 22, 2022 at 23:21
  • $\begingroup$ Huh, so it is indeed possible, thanks! I'll go through the related literature, whose existence I ignored. About my second question, it is natural for me to ask then: is this the only case in which it happens, or are there examples when $x\neq y$ (or rather $x\notin Ky$)? $\endgroup$
    – D. Dona
    Commented Feb 23, 2022 at 8:59
  • $\begingroup$ @YCor: Sure, I agree: thanks for the remark. $\endgroup$
    – Salvatore Siciliano
    Commented Feb 23, 2022 at 11:41
  • $\begingroup$ @D. Dona: I would expect that there exist also examples with $x$ and $y$ non-proportional. $\endgroup$
    – Salvatore Siciliano
    Commented Feb 23, 2022 at 11:46
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    $\begingroup$ OK LSpice, thank you. $\endgroup$
    – Salvatore Siciliano
    Commented Feb 25, 2022 at 18:14
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I will sketch the proof that over the complex numbers, the answer is no.

The set $$\{x\in \mathbb{P}(\mathfrak{g}) \mid \exists y\neq 0, [[\mathfrak{g},x],y]=0\}$$ is closed and $G$-invariant. Therefore it suffices to assume that $x$ lies in a closed $G$-orbit in $\mathbb{P}(\mathfrak{g})$.

So we can assume that $x\in \mathfrak{g}_\alpha$ for some root $\alpha$, where we have also fixed a Cartan subalgebra $\mathfrak{h}$ to talk about root spaces. [I'll give a proof of this fact at the end.]

Now write $y=h+\sum_\beta c_\beta X_\beta$, where $h\in \mathfrak{h}$ and $X_\beta\in \mathfrak{g}_\beta$. Then $[y,X_\gamma]=0$ for all $\gamma$ with $\gamma-\alpha$ a root or zero.

The set of possible $\gamma$ does not lie in a hyperplane, which forces $h=0$. For every root $\beta$ there exists such a $\gamma$ with $\gamma+\beta$ a root or zero, which forces $c_\beta=0$, QED.

Proof of the classification of closed $G$-orbits on $\mathbb{P}(\mathfrak{g})$: If $x$ is in a closed $G$-orbit, then its stabiliser contains a Borel $B$. To be stable under the torus $T$ implies that $x$ lies in a single weight-space. To be stable under the unipotent radical implies that that weight space must be the highest weight. (An alternative approach to this result is the argument in the proof of Theorem 4.3.3 in Collingwood and McGovern's "Nilpotent Orbits in Semisimple Lie Algebras".)

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    $\begingroup$ You seem to use the fact that a nonzero element of $\mathfrak{g}$ defines a closed $G$-orbit in $\mathbb{P}(\mathfrak{g})$ if and only if in the $G$-orbit of some nonzero element of $\mathfrak{g}_\alpha$ for some root $\alpha$. Do you have a reference for this fact? (I could check the fact inside $\mathfrak{sl}_n$.) $\endgroup$
    – YCor
    Commented Feb 25, 2022 at 16:30
  • $\begingroup$ In any case this is a very nice and clever argument! $\endgroup$
    – YCor
    Commented Feb 25, 2022 at 18:10
  • $\begingroup$ Sorry to ask a foolish follow-up, but why must a line (that's what $\mathbb P(\mathfrak g)$ means, right?) in a closed $G$-orbit be stabilised at least by a Borel? Is it because the closed orbit is a projective variety, and the Borels are the minimal subgroups giving projective quotients? $\endgroup$
    – LSpice
    Commented Feb 26, 2022 at 23:26
  • $\begingroup$ Also, your argument doesn't really use that we're working over $\mathbb C$, does it? Since you show non-existence, we may harmlessly pass to an algebraic closure. It seems to me that the only place you even use characteristic $0$ is to say that stability under the unipotent radical of a Borel implies belonging to the highest-root space, and even that surely is a problem at worst in characteristics $2$ and $3$. $\endgroup$
    – LSpice
    Commented Feb 26, 2022 at 23:29
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    $\begingroup$ @LSpice beware that "simple" is not preserved by passing to an algebraic closure (and the result is false in non-simple semisimple Lie algebra, taking $x,y$ in distinct factors). So in char 0 one needs a little additional argument to treat simple Lie algebras that are not absolutely simple. I'm even less sure what simple yields when passing to al algebraic closure, in char $p$. $\endgroup$
    – YCor
    Commented Feb 26, 2022 at 23:55

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