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 | |
|---|---|---|---|---|---|
| Feb 28, 2016 at 12:38 | answer | added | Dietrich Burde | timeline score: 3 | |
| Feb 27, 2016 at 17:01 | comment | added | YCor | ... and the most obvious algorithm is to compute all the $n^n$ $n$-fold brackets ($n$ is the dimension) $[e_{i_1},\dots,e_{i_n}]$ and check whether they are all zero. It's not the most efficient: computing inductively the lower central series requires a polynomial number of operations in terms of $n$ (the coefficients can grow in the process, but this depends on how we compute in the ground field anyway). | |
| Feb 27, 2016 at 11:30 | comment | added | YCor | The lower central series can be computed "computationally". | |
| Feb 27, 2016 at 9:08 | vote | accept | Jjm | ||
| Feb 27, 2016 at 9:04 | answer | added | Ben McKay | timeline score: 2 | |
| Feb 27, 2016 at 9:02 | history | edited | Jjm | CC BY-SA 3.0 |
Focusing on Structure Constants
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| Feb 27, 2016 at 8:54 | comment | added | Jjm | @BenMcKay Yes, of course I may use the standard definition, but I wonder whether there exists a specially refined (preferrably algebraic) condition for those coefficients. For instance, checking that a matrix is nilpotent is equivalent to the characteristic polynomial being $x^n$, which is an easy-to-use algebraic condition on the coefficients. | |
| Feb 27, 2016 at 8:48 | comment | added | Ben McKay | The structure constants are right there; take a dual basis and see that the structure constants are these $a_{ijk}$. Then just check as usual. | |
| Feb 27, 2016 at 8:42 | history | asked | Jjm | CC BY-SA 3.0 |