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Edit : There are several satisfying proofs in the case each $L^i$ is finite-dimensional. It is proven (for example, Hinich DG coalgebras as formal stacks) that for $A$ : local Artin ring then $\operatorname{Def}_L(A)=\operatorname{Hom}(A^*, H^0(\mathscr{C}(L)))$. So when everything is finite-dimensional, just use basic linear algebra and duality to get $\operatorname{Hom}(H^0(\mathscr{C}(L))^*, A)$. How could one reduce to that case ? If one can not, what happens ? How to obtain pro-representability ?


I work with deformation theory and DG Lie algebras, motivation coming from Goldman-Millson theory. Here DG Lie algebras are always assumed to have $L^i=0$ for $i<0$ and $H^i(L)$ finite-dimensional.

Context : To such a DG Lie algebra $L$ over a field $\mathbf{k}$ of characteristic zero is associated a functor $\operatorname{Def}_L$ on local Artin rings, sending $(A,\mathfrak{m}_A)$ so the set of solutions of the Maurer-Cartan equation $dx+\frac{1}{2}[x,x]=0$ in $L^1\otimes\mathfrak{m}_A$ modulo the exponential action of $L^0\otimes\mathfrak{m}_A$ by gauge transformations. It is known by classical methods in deformation theory that $\operatorname{Def}_L$ is pro-representable if $H^0(L)=0$ (Manetti, Deformation theory via differential graded Lie algebras, section 4 on the Kuranishi map).

On the other hand I'm reading about more modern deformation theories (Kontsevich, Hinich, Manetti...) They describe a functor which has several different names: Quillen $\mathscr{C}$ functor, bar construction, Chevalley-Eilenberg complex... obtained as follows : take the symmetric algebra on $L$ shifted by $1$, $\operatorname{Sym}(L[1])$, and turn it into a DG coalgebra $\mathscr{C}(L)$ (the codifferential has one part coming from the differential in $L$ and one part coming from the Lie bracket). Its dual $\mathscr{C}(L)^*$ is a DG complete local algebra.

Question : Now I would like the following theorem to be true but I never find it stated so explicitly and each time I try to extract it from the existing litterature (Kontsevich, Hinich, Manetti) there seems to be some subtle points (about duality algebras-coalgebras, finite-dimensionality, or algebras up to quasi-isomorphism) that I may not understand.

Let $L$ be a DG Lie algebra ($L^i=0$ for $i<0$ and $H^i(L)$ finite-dimensional). If $H^0(L)=0$ then $\operatorname{Def}_L$ is pro-represented by $H^0(\mathscr{C}(L))^*$.

Is this theorem true ? I would like some help to understand it (I work at the most "concrete" possible level of deformation theory, with the least possible amount of $\infty$ / simplicial / derived techniques). I'm suprised I never see this so stated nor used. It means, that $\operatorname{Def}_L$ is pro-represented by some very explicit and functorial object, which is much better than choosing a non-canonical splitting of $L^1$ as is done in Manetti's lecture notes.

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  • $\begingroup$ Do you work with dg Artin rings or just Artin rings concentrated in degree $0$? I'm think all the results you stated fail in the dg case, but there is a good chance that everything works out more or less as you want in the non-dg case. $\endgroup$ Commented Mar 7, 2017 at 13:23
  • $\begingroup$ Exactly I want to work in the non-DG case ("classical deformation theory" à la Schlessinger) and it seems that to get this theorem, one has to go through higher deformation theories (DG Artin rings, coalgebras, simplicial sets...) That's why I'm having such a hard time coming back to the classical case ! $\endgroup$ Commented Mar 7, 2017 at 13:34
  • $\begingroup$ I think I have something close to what you would like, but not exactly that. I can write it as an answer in the next couple of days if you want. I just need to know: how familiar are you with the language of operads? $\endgroup$ Commented Mar 7, 2017 at 15:18
  • $\begingroup$ I'm sorry I'm not familiar with operads. But anyway I need to understand this theorems and what could go wrong. What result do you obtain ? $\endgroup$ Commented Mar 14, 2017 at 16:12
  • $\begingroup$ Never mind, I would basically have recovered Hinich's result you put in your edit (using more general results). I really think you need some kind of finite-dimensionality assumption if you want to dualize. I would advise working with coalgebras. It can be a bit hard to get used to it, but once you're familiar with the objects then everything becomes very natural. $\endgroup$ Commented Mar 16, 2017 at 1:18

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Edit: The answer below only works for the case where all the $L^i$'s are finite dimensional.

The statement is true. In fact, under the assumptions of the question, it is also true that ${\scr C}(L)^*$ pro-represents the associated derived moduli problem for commutative artinian dg-algebras. Recall that the derived deformation problem determined by $L$ associates to a commutative artinian dg-algebra $A$ the derived mapping space ${\rm Map}_{{\rm dg-Lie}}({\scr D}A,L)$, where ${\scr D}A$ is the dg-lie algebra Koszul dual to $A$. On the other hand, the commutative dg-algebra ${\scr C}(L)^*$ is the Koszul dual of $L$. Furthermore, under the given assumptions $L$ is also the Koszul dual of ${\scr C}^*(L)$ (see Proposition 13.3.1.1 of http://www.math.harvard.edu/~lurie/papers/SAG-rootfile.pdf). One then obtains a natural equivalence ${\rm Map}_{{\rm dg-Lie}}({\scr D}A,L) \simeq {\rm Map}_{{\rm dg-Lie}}({\scr D}A,{\scr D}{\scr C}(L)^*) \simeq {\rm Map}_{{\rm dg-com}}({\scr C}(L)^*,A)$. To get the pro-representability in the classical setting use the fact if $A$ is concentrated in degree 0 then ${\rm Map}_{{\rm dg-com}}({\scr C}(L)^*,A) \simeq {\rm Map}_{{\rm com}}(H_0({\scr C}(L)^*),A)$. Note that if the cohomology groups $H^i(L)$ are not finite dimensional but $H^0(L)$ is still 0 then then the moduli problem associated to $L$ is still pro-representable (see Proposition 13.3.3.1 of http://www.math.harvard.edu/~lurie/papers/SAG-rootfile.pdf), but just not usually by ${\scr C}(L)^*$, since in this case $L$ is generally not the Koszul dual of ${\scr C}(L)^*$.

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    $\begingroup$ First, thank you for answering and confirming this is true. However, Lurie's theory is not an easy thing ! I would like to understand it with more down-to-earth arguments. In particular in section 13.3 you quote, Koszul duality seems very abstract but it is written that it can be be more explicitly obtained via $L^\infty$ algebras.. ? By the way the theorem you quote is for $L^i$ finite-dimensional. In that case I already have a satisfying proof (use Hinich + duality algebras-coalgebras). But I have $H^i(L)$ finite-dimensional, how to reduce to that ? $\endgroup$ Commented Mar 10, 2017 at 14:23
  • $\begingroup$ You're right, there is a gap here. I'm not sure how to reduce from $H^i(L)$ finite to $L^i$ finite. Maybe one can try to argue by comparing the lie-commutative Koszul duality with the associative-associative Koszul duality (and then use Corollary 14.1.3.3.of SAG). $\endgroup$ Commented Mar 11, 2017 at 20:49

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