Timeline for Replacement for Lie-algebra complements
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Feb 26, 2016 at 0:53 | comment | added | LSpice | Oh, got it. (Of course, a direct sum of $\mathfrak{sl}_2$'s won't be adjoint, but I assume that's just the first example that came to mind.) Thanks! I'll have a look at Strade's book. | |
| Feb 25, 2016 at 23:51 | comment | added | David Stewart | The issue would be if you took a direct sum of 5 copies of sl_2 in characteristic 5, then the extra derivations may magically appear on top in the normaliser. | |
| Feb 24, 2016 at 20:54 | comment | added | LSpice | I put that book on hold at the library, but haven't got it yet. These weird semisimple algebras won't be algebraic (i.e., Lie algebras of algebraic groups), right? I'm only concerned with the algebraic case. | |
| Feb 21, 2016 at 20:26 | comment | added | David Stewart | It also occurs to me that there might be more examples coming from weird semisimple algebras which turn up in characteristic $p$. You can read about these in Strade's book but the basic point is that you can tensor any simple $S$ with a truncated polynomial ring $O_1=k[X]/X^p$ to get $S\otimes O_1$ and act by derivations $1\otimes W_1$ where $W_1=Der O_1$. Then the semidirect product of these two things is semisimple but not the direct sum of simples. I don't know if these would also give counterexamples. | |
| Feb 21, 2016 at 20:22 | comment | added | David Stewart | Yes. $\mathfrak{psl}_{rp}$ is $\mathfrak{sl}_{rp}/z$ for $z=k.I_{pr}$ the centre generated by the identity matrix and this is simple if $pr>2$. It is not the Lie algebra of anything, but clearly very close. | |
| Feb 21, 2016 at 19:02 | vote | accept | LSpice | ||
| Feb 21, 2016 at 18:58 | comment | added | LSpice | (For 'diagonal' in the last sentence of my last comment please read 'scalar'.) | |
| Feb 21, 2016 at 18:49 | comment | added | LSpice | Your rephrasing of my question is just right, and probably more enlightening; I want specifically only to include those outer derivations that are realised in a super-Lie algebra. (I mean super- as opposed to sub-, not in the $\mathbb Z_2$-graded sense.) It will take some time for this non-GAP user to unpack the rest of your answer, but it looks like it answers my question. What is $\mathfrak{psl}_n$? Is it $\mathfrak{sl}_n$ modulo the subtorus of diagonal matrices (of which there are none unless $p \mid n$)? | |
| Feb 21, 2016 at 15:52 | history | edited | David Stewart | CC BY-SA 3.0 |
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| Feb 21, 2016 at 15:29 | history | edited | David Stewart | CC BY-SA 3.0 |
Sorry for all these edits.
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| Feb 21, 2016 at 14:53 | history | edited | David Stewart | CC BY-SA 3.0 |
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| Feb 21, 2016 at 14:38 | history | edited | David Stewart | CC BY-SA 3.0 |
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| Feb 21, 2016 at 14:16 | history | edited | David Stewart | CC BY-SA 3.0 |
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| Feb 21, 2016 at 14:03 | history | edited | David Stewart | CC BY-SA 3.0 |
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| Feb 21, 2016 at 13:08 | history | answered | David Stewart | CC BY-SA 3.0 |