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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?

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    $\begingroup$ 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. $\endgroup$
    – LSpice
    Commented Mar 9, 2021 at 3:03
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    $\begingroup$ @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. $\endgroup$
    – Irina
    Commented Mar 9, 2021 at 3:17
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    $\begingroup$ 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. $\endgroup$
    – YCor
    Commented Mar 10, 2021 at 20:49
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    $\begingroup$ 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 $c\le 2$. $\endgroup$
    – YCor
    Commented Mar 10, 2021 at 20:50

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$\DeclareMathOperator\gr{gr}$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 $\mu=d\theta$.

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

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