Timeline for ad-nilpotent degree of a nilpotent Lie Algebra
Current License: CC BY-SA 4.0
14 events
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| Jun 2, 2020 at 9:13 | comment | added | YCor | Actually the 4-layer of the free nilpotent Lie algebra on $k\ge 3$ generators splits into two irreducibles $V_1\oplus V_2$, with $V_1$ of dimension $(k-2)(k-1)k(k+1)/8$, generated by brackets $[[x,y],[z,t]]$ and $V_2$ of dimension $(k-1)k(k+1)(k+2)/8$, generated by brackets $[x,x,x,y]$ (in for 2 generators $V_1$ vanishes). So killing $V_2$ (and larger brackets) yields a Lie algebra of nilpotency class 4 and vanishing $[x,x,x,y]$. These facts about the free Lie algebras were initially observed by R.M. Thrall in 1942l. | |
| Jun 2, 2020 at 8:55 | comment | added | YCor | Actually this 17-dim Lie algebra is far from random (although the choice of basis is bit random!). It's Carnot and the first three layers are the same as those of the free 3-nilpotent Lie algebra on 3 generators (1st layer: 2,15,16; 2nd layer: 12,14,13, 3rd layer: 4,5,6,7,8,9,10,11). And 3 more dimensions from the 4th layer: 1,3,17. Actually, it's exactly the quotient of the 4-step free one by the relation $[x,x,x,y]$, and in particular has a canonical (grading-preserving) action of $\mathrm{GL}_3$. We can also kill two among $1,3,17$ to go down to dimension 15, but lose some symmetry. | |
| Jun 2, 2020 at 7:43 | history | edited | YCor | CC BY-SA 4.0 |
formatting to help readibility, added sentence from comment to introduce example
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| Jun 2, 2020 at 1:50 | history | edited | Ramiro Lafuente | CC BY-SA 4.0 |
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| Dec 19, 2019 at 18:10 | comment | added | LSpice | Would you post de Graaf's example that you mentioned? You can mark it community wiki so as not to get reputation from someone else's answer. | |
| May 25, 2011 at 9:19 | answer | added | Dietrich Burde | timeline score: 8 | |
| May 24, 2011 at 17:50 | comment | added | Ramiro Lafuente | Yesterday I was given an explicit counterexample by Willem de Graaf. It is a 17-dimensional Lie algebra over $\mathcal{Q}$ that is 3-Engel and of nilpotency class 4. I hope he posts it here. | |
| May 23, 2011 at 12:40 | comment | added | Jim Humphreys | @Torsten: Your approach looks attractively straightforward, but if it works it should be written down in more detail as an answer (preferably with a source in the literature). As it is, there seem to be two contradictory answers to the question asked. I've never worked directly with nilpotent Lie algebras and don't have any intuitive feeling about what is actually true here. But the usual polarisation technique seems to require commuting operators. | |
| May 20, 2011 at 13:36 | vote | accept | Ramiro Lafuente | ||
| May 23, 2011 at 18:15 | |||||
| May 20, 2011 at 12:51 | answer | added | user91132 | timeline score: 9 | |
| May 20, 2011 at 5:00 | comment | added | Torsten Ekedahl | As we are in characteristic 0 we may by polarisation express the product of linear operators as a linear combination of powers (of linear combinations of the operators). This implies that we have an equality. | |
| May 20, 2011 at 3:38 | history | edited | Ramiro Lafuente | CC BY-SA 3.0 |
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| May 19, 2011 at 21:18 | comment | added | Jim Humphreys |
Note that Engel's Theorem is not really involved here, but rather its more elementary converse: given a nilpotent Lie algebra, each operator $ad(x)$ is nilpotent. (On the other hand, combining Engel's Theorem with Ado's Theorem allows one to regard an abstract nilpotent Lie algebra as a subalgebra of the upper triangular matrices.) It might be helpful to recall more explicitly what is meant by $k$-step nilpotent.
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| May 19, 2011 at 12:15 | history | asked | Ramiro Lafuente | CC BY-SA 3.0 |