Timeline for Annihilate a simple Lie algebra using two commutators
Current License: CC BY-SA 4.0
10 events
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| Feb 27, 2022 at 0:25 | comment | added | LSpice | @YCor, that is a good point. Interpreting 'simple' algebraically (i.e., as "Lie algebra of a simple group"), the worst that can happen is that one is dealing with a restriction of scalars, which doesn't change the argument in an essential way. If one takes a purely abstract definition of simplicity—which was surely intended—then you are right that things can be much worse in the positive-characteristic case (even in the algebraically closed case, I think). | |
| Feb 26, 2022 at 23:55 | comment | added | YCor | @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 $p$. | |
| Feb 26, 2022 at 23:29 | comment | added | LSpice | 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 $3$. | |
| Feb 26, 2022 at 23:26 | comment | added | LSpice | 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? | |
| Feb 26, 2022 at 23:25 | history | edited | LSpice | CC BY-SA 4.0 |
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| Feb 26, 2022 at 23:17 | history | edited | Peter McNamara | CC BY-SA 4.0 |
answered YCor's comment about closed orbits in the projectivisation of the Lie algebra
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| Feb 25, 2022 at 18:10 | comment | added | YCor | In any case this is a very nice and clever argument! | |
| Feb 25, 2022 at 16:30 | comment | added | YCor | 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 $\mathfrak{sl}_n$.) | |
| Feb 25, 2022 at 13:12 | history | edited | YCor | CC BY-SA 4.0 |
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| Feb 25, 2022 at 11:59 | history | answered | Peter McNamara | CC BY-SA 4.0 |