Timeline for Annihilate a simple Lie algebra using two commutators
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Feb 26, 2022 at 8:12 | comment | added | Callum | @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 $\mathfrak{g} \to \mathfrak{p}\to \mathfrak{p}^{-n} \to \{0\}$ | |
| Feb 26, 2022 at 0:24 | comment | added | LSpice | @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? | |
| Feb 25, 2022 at 21:43 | comment | added | Callum | @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 $\mathrm{ad}_x^3 = 0$. | |
| Feb 25, 2022 at 14:30 | comment | added | LSpice | @Callum, that only works for a split Lie algebra (e.g., for $K$ algebraically closed), I think. | |
| Feb 25, 2022 at 11:59 | answer | added | Peter McNamara | timeline score: 4 | |
| Feb 22, 2022 at 17:14 | history | became hot network question | |||
| Feb 22, 2022 at 12:56 | comment | added | Callum | 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. | |
| Feb 22, 2022 at 12:54 | answer | added | Salvatore Siciliano | timeline score: 16 | |
| Feb 22, 2022 at 10:08 | comment | added | YCor | (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. | |
| Feb 22, 2022 at 9:11 | history | asked | D. Dona | CC BY-SA 4.0 |