The nicest definition of $L_\infty$-algebras ---which I will call a "Chevalley-Eilenberg" style definition after the obvious analogy with the Chevalley-Eilenberg differential of Lie algebras--- is the following:

Let $z^a$ be formal variables of integer degrees $\deg z^a\in\mathbb Z$. They are a basis of some graded vector space $V$ over the reals or complexes. Form the free graded-commutative algebra $S(V)$ as the quotient $$ S(V):=T[z^a]\mod [z^a \otimes z^b - (-1)^{(\deg z^a)(\deg z^b)} z^b\otimes z^a] $$ of the tensor algebra in the variables $z^a$ by the ideal in brackets; in other words odd $z^a$s anticommute with each other. Then the following differential (where $\frac{\partial}{\partial z^b}$ is a left derivative) $$ Q=\sum_{n=1}^{\infty}\frac{1}{n!} C^b_{a_1a_2\dots a_n} z^{a_1}z^{a_2}\cdots z^{a_n} \frac{\partial}{\partial z^b}\,,\qquad C^b_{a_1a_2\dots a_n}\in\mathbb R\, \forall n $$ defines an $L_\infty$-algebra structure on the degree-shifted dual $W=(V[-1])^\star$ iff $Q^2=0$.

Clearly if $\deg z^a=1$ we obtain the definition of a Lie algebra in terms of some set of structure constants $C^b_{a_1a_2}$, and $Q^2=0$ implies the Jacobi identities. In fact $Q$ is the Chevalley-Eilenberg differential in that case. There are other definitions of $L_\infty$-algebras, involving either coalgebras or an infinite number of higher products, neither of which are particularly nice.

In this language, a minimal $L_\infty$-algebra is one with $C^b_{a}=0$ i.e. no linear term in $Q$ (this term defines the dual to the $L_\infty$-algebra unary bracket). The minimal model theorem claims that every $L_\infty$-algebra is quasiisomorphic to a minimal $L_\infty$-algebra (its minimal model); for a construction see e.g. the review in the recent paper 1809.09899.

What I would like to know is if the minimal model has been explicitly constructed in terms of a "Chevalley-Eilenberg" style definition anywhere.

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    $\begingroup$ I don't understand how your definition is "nicer" than the usual definition as a square-zero coderivation on the cofree cocommutative coalgebra, considering it's completely identical, except that 1. you have dualized it (meaning trouble in characteristic $\neq 0$ or for infinite-dimensional stuff), 2. you have expressed the (co)differential explicitly in terms of coefficients in a basis, the choice of which is unnatural... Anyway, take any reference which "constructs" minimal models in terms of cocommutative coalgebras (for the notion of "construct" that you have in mind), dualize, and voilà. $\endgroup$ Nov 19, 2018 at 15:56
  • $\begingroup$ In fact, if you don't dualize, then you get Chevalley--Eilenberg chains (a quasi-cofree cocommutative coalgebra) instead of cochains (quasi-free commutative coalgebra). I guess it's a matter of taste. $\endgroup$ Nov 19, 2018 at 16:02
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    $\begingroup$ @NajibIdrissi Thanks for looking at this. Well, niceness is subjective; you can substitute "familiar to Alex specifically" instead if you'd like. Point 1. is irrelevant/can be dealt with in the context I need this for; point 2. you can interpret the above expressions in abstract index notation if you want, and be rid of arbitrary choices. $\endgroup$ Nov 19, 2018 at 16:05
  • $\begingroup$ Let me restate my point 1 more clearly then. If you have found any reference which deals with the usual definition of $L_\infty$-algebras, then you have found one which deals with your definition, just by putting duals everywhere (and exchanging subscripts/superscripts, I guess). // Moreover, I'd like to know what you means by "construct", exactly... Writing it down is nontrivial in general. For a Koszul dual example, try to write down the (Sullivan) minimal model of a wedge of two spheres. In Lie algebras, I guess this would correspond to the product $S^n \times S^m$. It's not easy. $\endgroup$ Nov 19, 2018 at 16:11
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    $\begingroup$ @NajibIdrissi: There is a more or less explicit construction in the paper I cite, page 24. However that construction is in terms of the definition involving the various products, and is moreover recursive, so writing down the differential (or indeed the codifferential) is hard. $\endgroup$ Nov 19, 2018 at 16:30


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