Timeline for Characterisation of parabolic subalgebras: reference sought
Current License: CC BY-SA 2.5
16 events
| when toggle format | what | by | license | comment | |
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| Aug 5, 2019 at 5:34 | answer | added | Alexander Premet | timeline score: 3 | |
| Mar 21, 2012 at 13:54 | answer | added | ulnor | timeline score: 4 | |
| Jul 2, 2010 at 21:09 | comment | added | Fran Burstall | Jim: many thanks for yr kind attention! | |
| Jun 26, 2010 at 14:58 | comment | added | Jim Humphreys | Fran: I've just looked at your proof (but not gone through the inductions carefully); it looks OK, but reinforces my impression that the result itself is rather technical and a bit negative by itself (no interesting new subalgebras appear). Presumably it's needed in a further argument of yours. The Bourbaki result cited by Jose covers your Lemma B attributed to Grothendieck, but for them it's just part of their broader discussion of maximal subalgebras. Your converse might have come up in papers by Dixmier or others on polarizations, etc. Including your proof in a paper is safe, I guess. | |
| Jun 25, 2010 at 13:14 | history | edited | Fran Burstall | CC BY-SA 2.5 |
Added link to pdf of proof
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| Jun 25, 2010 at 13:11 | comment | added | Fran Burstall | Jim: Here is a sketch of the proof. Start with the case where $\mathfrak{p}^\perp$ lies inside $\mathfrak{p}$. Here the central descending series of $\mathfrak{p}^\perp$ together with its orthocomplements makes $\mathfrak{g}$ into a filtered Lie algebra with $\mathfrak{p}$ in weight $0$ and $\mathfrak{p}^\perp$ in weight $-1$. In particular, $\mathfrak{p}^\perp$ consists of nilpotent elements and the Bourbaki/Grothendieck result applies. For the general case, apply the preceding to $\mathfrak{q}=\mathfrak{p}+\mathfrak{p}^\perp$. I will add a link to details in the question. | |
| Jun 24, 2010 at 13:53 | comment | added | Jim Humphreys | Fran: This back-and-forth has made me more curious about what your own undisclosed proof involves. What you've stated in terms of "nilpotent subalgebra" rather than "subalgebra consisting of nilpotent elements" isn't really the converse of the familiar statement you start with about the orthocomplement of a parabolic subalgebra (?) | |
| Jun 23, 2010 at 18:29 | comment | added | Fran Burstall | Victor: thank you for drawing the Ozeki-Wakimoto paper to my attention. It is indeed interesting but does not, as far as I can see, prove my statement. In fact, their result seems much deeper and uses 'non-algebraic' considerations: they look at the analytic subgroup corresponding to a w-polarisable subalgebra and see that the resulting coset space is compact whence the subgroup and so, eventually, the subalgebra is parabolic. | |
| Jun 23, 2010 at 0:09 | comment | added | Victor Protsak | A more "forward" way is to say that if $\mathfrak{n}$ is a nilpotent Lie subalgebra of $\mathfrak{g}$ such that its orthocomplement $\mathfrak{p}:=\mathfrak{n}^\perp$ is a Lie subalgebra then $\mathfrak{p}$ is a parabolic subalgebra (and hence $\mathfrak{n}$ is its nilradical). I can't help thinking that Ozeki and Wakimoto paper, which proves that any polarizable subalgebra is parabolic, is somehow relevant; at least, it gives the right conclusion. | |
| Jun 22, 2010 at 22:16 | comment | added | Jim Humphreys | Maybe I understand better: the essential statement not already implied by the Bourbaki theorem is that the orthocomplement of a nilpotent subalgebra containing nonzero semisimple elements is never a Lie subalgebra (since if it were, it would have to be parabolic and thus its orthocomplement in turn would be a nil algebra)? This is somewhat roundabout to state though probably true. I haven't seen it in print, however. | |
| Jun 22, 2010 at 18:01 | history | edited | Fran Burstall | CC BY-SA 2.5 |
Clarified what I mean by nilpotent subalgebra
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| Jun 22, 2010 at 17:58 | comment | added | Fran Burstall | I think I am being careful, Jim: I only require that $\mathfrak{p}^\perp$ is nilpotent in the usual sense that the central descending series terminates. However, I am definitely requiring that the orthocomplement be a subalgebra, else, as you say, a CSA would provide a counterexample. I will edit my question to make my meaning clearer. | |
| Jun 22, 2010 at 12:35 | comment | added | Jim Humphreys |
I think you need to be more careful about what you mean by nilpotent subalgebra here, since you are implicitly requiring that it consist of "nilpotent" elements in the sense of the abstract Jordan decomposition in $\mathfrak{g}$. A Cartan subalgebra is also nilpotent, for example, but consists of "semisimple" elements. An arbitrary nilpotent subalgebra could involve both types. Unless you assume $\mathfrak{n}$` consists of nilpotent elements, the discussion gets more subtle (and the orthocomplement need not even be a subalgebra of $\mathfrak{g}$)
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| Jun 22, 2010 at 1:06 | history | edited | Q.Q.J. |
edited tags
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| Jun 21, 2010 at 23:45 | answer | added | José Figueroa-O'Farrill | timeline score: 4 | |
| Jun 21, 2010 at 22:46 | history | asked | Fran Burstall | CC BY-SA 2.5 |