# When is a nilpotent Lie algebra isomorphic to the associated graded of its lower central series?

All Lie algebras in this question are finite-dimensional and defined over a field $$k$$ of characteristic $$0$$ which I'm happy to take to be $$\mathbb{R}$$ or $$\mathbb{C}$$.

$$\DeclareMathOperator\gr{gr}$$Let $$L$$ be a nilpotent Lie algebra. It is then filtered by its lower central series, and we have an associated graded nilpotent Lie algebra $$\gr L$$. It is definitely not the case that $$L$$ and $$\gr L$$ have to be isomorphic; see Malcev Lie algebra and associated graded Lie algebra for some examples.

Question: what kinds of conditions can I put on $$L$$ that ensure that it is isomorphic to $$\gr L$$? E.g. if the field is $$\mathbb{R}$$ are the there geometric/topological/algebraic conditions on the associated simply-connected nilpotent Lie group that ensure this?

• 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. Mar 9, 2021 at 3:03
• @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 3:17
• 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.
– YCor
Mar 10, 2021 at 20:49
• 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 $c\le 2$.
– YCor
Mar 10, 2021 at 20:50

$$\DeclareMathOperator\gr{gr}$$Too tired to think clearly, but it looks like a standard Deformation Theory thingy.

We have natural linear maps $$\gamma_n (L)\rightarrow \gr_n L$$. Split them as linear maps. We get a bijective linear map $$L\rightarrow \gr L$$.

Use it to equip $$\gr L$$ with a second Lie algebra structure $$[,]^{\prime}$$, coming from $$L$$. Now consider the difference $$\mu:L\otimes L \rightarrow L, \ \ \mu (x\otimes y) = [x,y]-[x,y]^{\prime}.$$ I claim that $$\mu$$ is a 2-cocycle on $$\gr L$$ and that finding an isomorphism $$\gamma_n (L)\cong \gr_n L$$ is equivalent to finding a 1-cochain $$\theta$$ such that $$\mu=d\theta$$.

• This doesn't seem in the spirit of the question, which asks for conditions on $L$ (not general cohomological obstructions). Mar 10, 2021 at 20:30
• 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 20:39
• @BugsBunny $H^2(L)=0$ for a finite-dimensional nilpotent Lie algebra over a field implies $\dim(L)\le 1$.
– YCor
Mar 10, 2021 at 20:46
• @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 21:00
• 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).
– YCor
Mar 10, 2021 at 21:24