Timeline for When is a nilpotent Lie algebra isomorphic to the associated graded of its lower central series?
Current License: CC BY-SA 4.0
7 events
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| Mar 10, 2021 at 20:50 | comment | added | YCor | A sufficient condition, by the way, for $c$-step nilpotent $L$ to be Carnot, is that $L/L^c$ is that $L/L^c$ is free $(c-1)$-step nilpotent. This is, in particular, automatic if $c\le 2$. | |
| Mar 10, 2021 at 20:49 | comment | added | YCor | I largely survey/elaborate about the description of finite-dimensional Carnot Lie algebra (= those isomorphic to their associated graded) in this paper ("Gradings on Lie algebras, systolic growth, and cohopfian properties of nilpotent groups", Bull SMF 2016), see notably §3.2. | |
| Mar 10, 2021 at 20:28 | answer | added | Bugs Bunny | timeline score: 1 | |
| Mar 9, 2021 at 3:17 | comment | added | Irina | @LSpice: Thanks, I did not know that! I routinely use DeclareMathOperator when I'm writing papers, but didn't think of using it on MathOverflow. | |
| Mar 9, 2021 at 3:03 | comment | added | LSpice |
TeX note: for correct spacing, use \DeclareMathOperator, as in $\DeclareMathOperator\gr{gr}$$\gr L$ $\DeclareMathOperator\gr{gr}$$\gr L$ (or its one-shot version $\operatorname{gr} L$ \operatorname{gr} L) instead of $\text{gr} L$ \text{gr} L. I have edited accordingly.
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| Mar 9, 2021 at 3:02 | history | edited | LSpice | CC BY-SA 4.0 |
Inlined link to question; \DeclareMathOperator
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| Mar 9, 2021 at 2:56 | history | asked | Irina | CC BY-SA 4.0 |