Chevalley-Eilenberg complexes of curved dg Lie algebras Let $k$ be a field of characterisitic $0$.
A curved dg Lie algebra (curved dgla) is a triple $(\mathfrak{g},d,w)$ where $\mathfrak{g}$ is a graded Lie algebra, $d$ is a derivation with degree $1$ and $w \in \mathfrak{g} _2$ such that $d(w) = 0$ and $d^2(x) = [w,x]$. The element $w$ s called the curvature.
For any curved dgla $(\mathfrak{g}, d, w)$, we can construct a cochain complex CE$(\mathfrak{g},d,w)$ called Chevalley-Eilenberg complex.
CE$(\mathfrak{g},d,w)$ is isomorphic to Sym$(\mathfrak{g}[1])^*$ as a graded algebra.
The differential $d'$ of CE$(\mathfrak{g},d,w)$ is a sum $d' = d'_0 + d'_1 +d'_2$ where the restriction is given by

*

*$d'_0$: $(\mathfrak{g}[1])^* \rightarrow k$ is given by the evaluation on $w$ the curvature of $\mathfrak{g}$.


*$d'_1$: $(\mathfrak{g}[1])^* \rightarrow (\mathfrak{g}[1])^*$ is given by the differential on $g$.


*$d'_2$ : $(\mathfrak{g}[1])^* \rightarrow S^2(\mathfrak{g}[1])^*$ is given by the bracket on $\mathfrak{g}$.
It seems that the evaluation of $w$ can be defined only on an element of $(\mathfrak{g}[1]_1)^*$ for me.
How is define the evaluation of $w$ on an element of $(\mathfrak{g}[1]_i)^*$ $(i \neq 1)$ ?
And, is $d_0$ of degree $-1$ ?
 A: $\newcommand{\g}{\mathfrak g}$
Summary: the construction takes $S(s\g)$ with the usual CE differential $\partial$ and the right multiplication $R_w$ by $s\omega$. Then $(\partial+R_w)^2 = 0$ because $R_w^2=0$ and $\partial R_w + R_w\partial + \partial^2=0$.

I think $\g$ needs to have degree $-1$ differential and $w$ of degree $-2$ for things to work as you want, and you also want $dw=0$, i.e. the curvature is always closed.
As you write, let $V = s\mathfrak g$ and form $S^c(V)$. This is the free conilpotent unital coalgebra generated by $V$. Instead of defining the algebra $S(V)^* \simeq S(V^*)$ I will define this coalgebra, which is easier. Namey, a coderivation on $S^c(s\g)$ is completely determined by the resulting map
$$S^c(s\g) \longrightarrow s\g$$
obtained by projecting onto the cogenerators. We will define it to vanish everywhere except in three sumands: the unit, $s\g$ and $S^2(s\g)$. As in the previous version of this post:

*

*The bracket $s\g\otimes s\g\longrightarrow s\g$ is now of degree $-1$, and defines the map on the summand $S^2(s\g) \to s\g$.


*The map $\mathbb k\to s\g$ sends $1$ to $sw$. This is also of degree $-1$. You must extend this to $d_0(\omega) = (-1)^\omega\omega\wedge sw$, i.e. multiplication by $sw$.


*The map $s\g\longrightarrow  s\g$ sends $sg\to -sd(g)$, so it is still of degree $-1$.
Thus the unique coderivation extending this map acts like the usual CE-differential in positive degree, and satisfies $d(1) = sw$. As you mentioned, this differential has three components $d_0+d_1+d_2$ which I am indexing by the weight, and then $d^2=0$ follows from:

*

*$d_1d_0+d_0d_1=0$ since $dw=0$

*$d_0^2$ since $sw$ is odd.

*$d_2^2=0$ by the Jacobi identity, as usual.

*$d_1d_2+d_2d_1=0$ by the Leibniz rule, again as usual.

*$d_1^2+d_2d_0+d_0d_2=0$ by definition of the curvature.

Proof. We compute:

*

*For any form $\omega$, you get $d_1d_0(\omega) = d_1(\omega\wedge sw) = (-1)^{|\omega|} d\omega \wedge sw$ while $d_0d_1(\omega) =  (-1)^{|\omega|-1} d\omega \wedge sw$ is its opposite.


*As mentioned before, $sw\wedge sw=0$.


*This is the same computation as usual, and the same with 4.


*Take $\omega = v_1\cdots  v_n$. Then $d_1^2(\omega)$ is equal to a signed sum of $v_1\cdots d^2v_i \cdots v_n$. On the other hand, $d_0d_2(\omega)$ is equal to a sum of the form $v_1\cdots [v_i,v_j]\cdots v_n w$ while in $d_2d_0(\omega)$ you have the same terms (with opposite sign) and also terms where you take the bracket with $w$, i.e. $v_1\cdots [v_i,w]\cdots v_n$, which cancel precisely with $v_1\cdots d^2v_i \cdots v_n$.
When you dualize this construction, as you want to, the argument is the same. For more details see this preprint, around page 34.
