# Reconstructing a nilpotent Lie algebra from its cohomology with $A_{\infty}$-structure

Let $L$ be a nilpotent Lie algebra (over a field of char 0) and $CE^{\bullet}(L)$ be its Chevalley-Eilenberg dg-algebra. By homotopy transfer, there exists a structure of an $A_{\infty}$-algebra on the cohomology $H^{\bullet}(L)$ of $CE^{\bullet}(L)$. I'm asking for the more or less direct way of reconstructing $L$ from this $A_{\infty}$-structure.

One way of do this is to take the bar-cobar construction of $H^{\bullet}(L)$, obtaining the dg-algebra which is quasi-isomorphic to the $H^{\bullet}(L)$, and then take its Sullivan minimal model. By uniqueness of the minimal model, we'll get precisely $CE^{\bullet}(L)$, and the reconstruction of a Lie algebra from its Chevalley-Eilenberg complex is a triviality.

However, bar-cobar construction is huge, so this is more of a proof that the reconstruction is possible, than a reconstruction itself. I'm looking for probably not so canonical, but simpler ways of describing the Lie algebra with given cohomology.

Here is the simplest case of the problem: suppose we are given the graded algebra with the basis $\{1,p,q,ph,qh,pqh\}$, the obvious multiplication and the $m_3$ operation given by $m_3(p,p,q)=ph$ and so on. It is very easy to guess that the Lie algebra with this cohomology is the 3-dimensional Heisenberg, but I do not know how to obtain this answer algorithmically.

UPD: As Pavel said in comments, since $CE^{\bullet}(L)$ is commutative, one could consider the Harrison construction (the free Lie algebra on $H^{\bullet}$ with the differential depending on $m_i$'s) instead of the full bar-construction. Since we have the a posteriori knowledge of what the cohomology of $Har^{\bullet}(H^{\bullet})$ is --- it should be $L$ in degree zero and 0 in higher degrees --- and since $Har^{<0}(H^{>0})$ is zero, we have a very simple description of lower central factors. Concretely, the factor $L^{k-1}/L^k \otimes \mathbb{Q}$ is isomorphic (or dual) to the kernel of the operation $m_k: Lie^k(H^1) \rightarrow H^2$. Maybe I'm making a mistake with how the grading in the Harrison complex is defined, but at least for the Heisenberg and some other simple nilpotent algebras this gives the right answer.
1. This is only true when $k \le n+1$, where $n$ is the nilpotency depth of $L$. It seems that from carefully inspecting homotopy transfer formulas one can conclude that the highest arity of a non-zero operation is equal to $n+1$ if and only if the nilpotency depth is equal to $n$. I haven't checked this enough times to be sure, though.
2. Since the natural map from the group $G$ to its Malcev completion $G \otimes \mathbb{Q}$ induces the isomorphism on the first rational cohomology $H^1$ and the surjection on the $H^2$, it seems that the same procedure gives the right answer for lower central factors of all groups, not only nilpotent ones.
• It seems you actually have a $C_\infty$ structure on $H^\bullet(CE^\bullet(L))$. Can't you take the Harrison complex of that to recover the Lie algebra? Commented Mar 24, 2016 at 19:53
• Actually, in the case when $L$ is concentrated in the degree $0$, the Harrison complex becomes much more tractable. It seems that in order to get the $k$th lower central factor, I have to take just the kernel of $m_k: Lie^k(H^1) \mapsto H^2$. Commented Jul 2, 2016 at 12:35