Timeline for Nilpotency of Lie Algebra from Structure Constants
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Mar 1, 2016 at 9:15 | comment | added | Bulois Michael | An idea in this direction: for a Lie algebra L to be nilpotent, it is sufficient to show that, for a given basis B of L, all the iterated brackets of elements of B have to be nilpotent, see: arxiv.org/abs/1010.0821 but this is far from being effective due to the number of nilpotent matrices to consider | |
| Feb 29, 2016 at 11:37 | comment | added | YCor | About the edited answer, while at first sight it does not seem to be an algorithm, it works at it just mean that $(t_1A_1+\dots+t_nA_n)^n=0$, where $t_i$ are formal variables. (Of course, it is only an algorithm relying on the assumption that we can compute in the field.) | |
| Feb 28, 2016 at 18:55 | history | edited | Ben McKay | CC BY-SA 3.0 |
fixed error
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| Feb 28, 2016 at 13:31 | comment | added | Robert Bryant | @YCor: It used to be true (I don't know whether it is still true) that you couldn't delete an answer that had been accepted as correct by the OP. The best you could (still can?) do is edit the answer, pointing out the error and, if you can't correct it, then putting up a warning that the answer has been found to be in error. | |
| Feb 27, 2016 at 15:20 | comment | added | Ben McKay | @YCor: Thanks. I should have required that every linear combination of the matrices $A_k$ is nilpotent. | |
| Feb 27, 2016 at 11:28 | comment | added | YCor | Seems like saying that when there's a basis $(e_i)$ such that each $ad(e_i)$ is nilpotent, then the Lie algebra is nilpotent. This is false in $sl_2$, e.g. with a basis $(s,x,y)$ with law $[x,y]=s-x+y$, $[s,x]=s+x+y$, $[s,y]=s-x-y$. | |
| Feb 27, 2016 at 9:11 | history | edited | Ben McKay | CC BY-SA 3.0 |
grammar
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| Feb 27, 2016 at 9:08 | vote | accept | Jjm | ||
| Feb 27, 2016 at 9:04 | history | answered | Ben McKay | CC BY-SA 3.0 |