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
13 events
| when toggle format | what | by | license | comment | |
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| Mar 10, 2021 at 22:59 | comment | added | LSpice | (Excuse me, I forgot what we were calling what. My comment about conditions on $L$ was meant to refer to conditions on $\exp(L)$, which seems to be the goal of the question. But of course it is up to @Irina to say what counts as an answer. My point was sort of that the generality of this answer means that it's not really saying anything specifically about nilpotent algebras.) | |
| Mar 10, 2021 at 21:25 | comment | added | Bugs Bunny | Yes, I meant with coefficients in the adjoint representation. | |
| Mar 10, 2021 at 21:24 | comment | added | YCor | Oh, by the way it seems this 2-cohomology class rather lies in $H^2(L,L)$ than $H^2(L)$. However, it's probably true that $H^2(L,L)$ is nonzero for every nilpotent Lie algebra of dimension $>1$ (in dimension $2,3,4,5,6,7$, it has dimension $=2$, $\ge 8$, $\ge 15$, $\ge 24$, $\ge 34$, $\ge 48$ by classification). | |
| Mar 10, 2021 at 21:00 | comment | added | Irina | @YCor: I wasn't expressing an opinion as to the answer since I hadn't worked out any examples. It was mostly an attempt to probe what Bugs Bunny was saying in his answer. I have never thought about deformation theory, so I don't have strong intuitions for how this stuff should work. | |
| Mar 10, 2021 at 20:56 | comment | added | YCor | @Irina for your last question, this sounds like a quite random expectation; the 2-dimensional abelian Lie algebra is already a counterexample. | |
| Mar 10, 2021 at 20:54 | comment | added | YCor | It's quite easy: assume that the Lie algebra $L$ is $c$-step nilpotent and has abelianization of rank $d\ge 2$. Write it as quotient $L=G/H$ of $G$ free $(c+1)$-step nilpotent of rank $d$. Let $J$ be an ideal in $G$ that has codimension 1 in $H$, $G'=G/J$, $H'=H/J$. Then $L=G'/H'$ and $H'$ has dimension 1; in addition by construction $H'\subset [G',G']$. So the central extension $0\to H'\to G'\to L\to 0$ defines a nonzero 2-cohomology class. | |
| Mar 10, 2021 at 20:50 | comment | added | Bugs Bunny | @YCor I did not know. That is depressing. | |
| Mar 10, 2021 at 20:46 | comment | added | YCor | @BugsBunny $H^2(L)=0$ for a finite-dimensional nilpotent Lie algebra over a field implies $\dim(L)\le 1$. | |
| Mar 10, 2021 at 20:39 | comment | added | Irina | Is it possibly the case that given a finite-dimensional positively graded Lie algebra $G$ over $\mathbb{Q}$ (the finite-dimensional and positively graded assumptions force $G$ to be nilpotent), the set of isomorphism classes of nilpotent Lie algebras with associated graded $G$ are in bijection with $H^2(G)$? Maybe you have to talk about filtered Lie algebras with filtrations that aren't necessarily the LCS. That would be a cool result. | |
| Mar 10, 2021 at 20:32 | comment | added | Bugs Bunny | $H^2=0$ is sufficient. Vanishing of a natural cohomology class is necessary and sufficient. Sorry, I am not good with spirits. | |
| Mar 10, 2021 at 20:31 | history | edited | LSpice | CC BY-SA 4.0 |
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| Mar 10, 2021 at 20:30 | comment | added | LSpice | This doesn't seem in the spirit of the question, which asks for conditions on $L$ (not general cohomological obstructions). | |
| Mar 10, 2021 at 20:28 | history | answered | Bugs Bunny | CC BY-SA 4.0 |