3
$\begingroup$

Fix a nilpotent Lie algebra $L$ over some char 0 field $k$ which is naturally graded, i. e. isomorphic to graded algebra $\bar L$ associated to lower central filtration.

I'm interested in some reasonable description of $M \subset Hom(L \otimes L, L)$ consisting of algebras $M$ with an algebra isomorphism $\phi: \bar M \to L$. Reasonable description includes action of $GL(L)$ and some invariant compactification.


Maybe someone knows a lot more and have already found a way to describe sheaves of nilpotent algebras for which PBW morphism is a deformation of coalgebra map in some sense — like in Lefevre-Hasegawa thesis; I guess this remark needs some elaboration, which is best suited as separate question. So, references to any papers about this kind of "Lie algebra sheaves with connection" are welcome.

$\endgroup$
  • $\begingroup$ It would be natural to fix the given Carnot grading on $L$, consider the set $B_L$ of bilinear maps $b:L\otimes L\to L$ such that $b(L^i\otimes L^j)\subset L^{i+j+1}$ for all $i,j$ such that $[\cdot,\cdot]+b$ is a Lie algebra law on $L$. And then mod out by the action of the graded automorphism group of $(L,[\cdot,\cdot])$. $\endgroup$ – YCor Feb 5 at 1:39
  • $\begingroup$ But if we consider only +1 degree part of bilinear maps, this will give only a germ of $M$ at $L$, right? And considering all positive degrees breaks this approach, because quotening by homogenous automorphisms does not give anything meaningful in that case. $\endgroup$ – Denis T. Feb 5 at 1:46
  • $\begingroup$ Sorry, I meant this where $L^i$ is the lower central series. In terms of grading, it means $b(L_i\otimes L_j)\subset \bigoplus_{k>i+j}L_k$ for all $i,j$. $\endgroup$ – YCor Feb 5 at 2:30
  • $\begingroup$ I do not understand why it should give the right answer anyways, sorry. $\endgroup$ – Denis T. Feb 5 at 8:12
  • $\begingroup$ Classification of nilpotent Lie algebras (even under some restrictions) is probably hopeless, with arbitrary families occuring as soon as the dimension gets large. $\endgroup$ – F. C. Feb 8 at 19:05

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.