In deformation theory and the nearby areas of maths, the central object of study is almost always a generic differential graded Lie (dgLie) algebra $\mathfrak{g}$ respecting some finiteness assumption (in order to make some formulae converge, mainly). Now, I've principally seen two approaches to these finiteness assumptions:
- The first, conceptually easier possibility is to assume that $\mathfrak{g}$ is nilpotent, i.e. that there exists some integer $n$ such that any $n+1$ iterated bracket gives zero. This is done e.g. by Getzler in Lie theory for nilpotent $L_\infty$ algebras (well, in the context of $L_\infty$ algebras, which generalizes dgLie).
- Take $\mathfrak{g}$ an arbitrary dgLie algebra. Let $(R,\mathfrak{m})$ be an (artinian) parameter algebra, i.e. a complete noetherian local commutative ring with maximal ideal $\mathfrak{m}$ such that $R/\mathfrak{m}$ gives the base field $k$. The mosti important example of this i deformation theory is the parameter algebra of formal power series in one variable $k[[\hbar]]$, coding formal deformations. Now, instead of using $\mathfrak{g}$, we consider instead the $\mathfrak{m}$-adic completion of $\mathfrak{m}\otimes\mathfrak{g}$, that is the dgLie algebra $$\mathfrak{m}\widehat{\otimes}\mathfrak{g} := \lim_{\leftarrow j}\left(R/m^{j+1}\otimes_R(\mathfrak{m}\otimes\mathfrak{g})\right)$$ with the $\mathfrak{m}$-adic topology (the necessary formulae will then be convergent with respect to that topology). This will be pronilpotent, and nilpotent if $R$ is artinian. This approach is used, among others, by Yekutieli in MC elements in pronilpotent dg Lie algebras.
Now, my question is:
Is it possible to go in a natural way from one approach to the other?
For example, given a nilpotent dgLie algebra $\mathfrak{g}$, is it always possible to find another dgLie algebra $\mathfrak{g}'$ and an artinian parameter algebra $(R,\mathfrak{m})$ such that $\mathfrak{g}\cong\mathfrak{m}\widehat{\otimes}\mathfrak{g}'$?
There is the following easy case: If $\mathfrak{g}$ is abelian (i.e. the Lie bracket is zero), then take $R = k[t]/(t^2)$ the algebra of dual numbers, with $\mathfrak{m} = (t)$ the ideal generated by $t$. Then $t$ is $1$-dimensional (as a $k$-vector space), and we have $$\mathfrak{g}\cong\mathfrak{m}\widehat{\otimes}\mathfrak{g}.$$