Homotopy equivalence vs gauge equivalence

Let $(\mathfrak{g},[-,-])$ be a pronilpotent Lie algebra (considered of degree zero). We can consider $(\mathfrak{g},[-,-])$ as a differential graded Lie algebra endowed with the $0$-differential. Let $A$ be a commutative non-negatively graded differential algebra. Then $A\widehat{\otimes }\mathfrak{g}$ is a differential graded Lie algebra whit bracket

$[a\otimes V,b\otimes W]:=ab\otimes[V,W]$

and where the differential is the obvious differential in the tensor product. We denote such a differential with $d$. Let $MC(A\widehat{\otimes}\mathfrak{g})$ be the set of maurer-Cartan elements, i.e solutions of the equations

$d\alpha +\frac{1}{2}[\alpha,\alpha]=0$.

There are two type of equivalences on this differential graded algebra.

1) gauge equivalences

The set $(A\widehat{\otimes }\mathfrak{g})^{0}$ acts on the set of maurer-Cartan elements via the action

$d+\alpha\mapsto e^{u}(d+\alpha)e^{-u}=d+e^{ad_{u}}\alpha+\frac{1-e^{ad_{u}}}{ad_{u}}du$.

Two maurer-Cartan elements $\alpha_{0},\alpha_{0}$ are gauge equivalent if for there exists an $u$ as above connecting them.

2)homotopy equivalence Let $\Omega(1)$ be the differential graded algebra of polynomial forms on $[0,1]$. Then $(\Omega(1)\otimes A)\widehat{\otimes }\mathfrak{g}$ is again a differential graded algebra. Two maurer-Cartan elements $\alpha_{0},\alpha_{0}$ are homotopy equivalent if there exists a Maurer-Cartan element $\alpha$ in $(\Omega(1)\otimes A)\widehat{\otimes }\mathfrak{g}$ such that $\alpha(0)=\alpha_{0}$ and $\alpha(1)=\alpha_{1}$.

Here my question: Under which condition of $A$ is 1 equivalent to 2? Do I need some special condition on $A$?

Edit: here some observations: Notice that if $A$ is finite dimensional then $A\widehat{\otimes }\mathfrak{g}$ is again pronilpotent and the two definitions are equivalent. On the other hand if $A$ is not finite dimensional, then $A\widehat{\otimes }\mathfrak{g}$ can be written as the projective limit of non necessarlily finite dimensional Lie algebras. So my question can be writtens as: what happen if $A$ is not finite-dimensional?

• I guess you mean $A$ is a commutative algebra rather than a Lie algebra. What do you mean by $\hat{\otimes}$? Is it just $A\hat{\otimes} \lim \mathfrak{g}_n = \lim (A\otimes \mathfrak{g}_n)$? Given any cdga $A$ and a nilpotent Lie algebra $\mathfrak{g}_n$, $A\otimes \mathfrak{g}_n$ is nilpotent. – Pavel Safronov Jun 5 '17 at 0:12
• Ups is a typo. it is exaktaly what I mean. I agree but $A\widehat{\otimes}\mathfrak{g}$ may be not pronilpotent since the Lie algebras $A\widehat{\otimes}\mathfrak{g}_{n}$ are not finite dimensional. – Cepu Jun 5 '17 at 7:32

1 Answer

The two conditions are always equivalent, since $A\widehat{\otimes}\mathfrak{g}$ is complete. This is present in the literature, I think a possible reference is A tale of three homotopies by Dotsenko and Poncin, as well as my own article Representing the Deligne-Hinich-Getzler ∞-groupoid, where this fact is a direct consequence of Corollary 5.3.