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Let $k$ be an algebraically closed field, e.g. $k=\mathbb C$. Let $Art_k$ be the category of local Artin $k$-algebras with residue field $k$. A deformation functor is a functor $D:Art_k\to Sets$ such that $D(k)$ has one element.

Let $D:Art_k\to Sets$ be a prorepresentable deformation functor, i.e. $D\cong h_R=\hom_k(R,-)$ for some local $k$-algebra $(R,m)$ with residue field $k$ and finite-dimensional tangent space $T=(m/m^2)^\vee$. This is equivalent to the existence of an obstruction theory $(T_1,T_2)$ on $D$ such that for every small extension $\xi:0\to I\to B\to A\to 0$ there is an exact sequence of sets $$0\to T_1\otimes I\to DB\to DA\overset{ob_\xi}{\to}T_2\otimes I.$$ Now, $T_1$ is naturally isomorphic to $T$, which is finite-dimensional, say $\dim T=d$. If $\dim T_2=e$, then it is known that $$d\geq \dim R\geq d-e.$$

The vanishing $T_2=0$ would assure an isomorphism $R\cong k[[t_1,\dots, t_d]]$.

Question. In what situations, different from $T_2=0$, do we get $R\cong k[[t_1,\dots, t_d]]$? For example, what about if $ob_\xi=0$ for all small extensions $\xi$?

Basically I am interested in understanding when a point $p$ on a moduli space $X$ is smooth. So for instance if $X$ is a fine moduli space then the deformation functor associated to $p$, i.e. the functor $$D_{X,p}:A\mapsto \{g:\textrm{Spec }A\to X\textrm{ such that }g|_{\textrm{Spec k}}=p\},$$ is isomorphic to $h_R$ where $R=\hat{\mathscr O}_{p}$. So we are in the above situation: we find some obstruction theory $(T,T_2)$ and if $T_2=0$ then $p$ is smooth. Actually if $R=k[[t_1,\dots, t_d]]/J$ then we can take $T_2=(J/\mathfrak nJ)^\vee$, where $\mathfrak n$ is the maximal ideal of $k[[t_1,\dots, t_d]]$. What else than $J=\mathfrak nJ$ can make $p$ a smooth point of $X$?

Thank you for any help!

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I'm not completely sure I understand your question. If $p$ is a smooth point on a scheme of finite type over $k$ (for instance your moduli space), then by Cohen's structure theorem you get that the completion of the local ring at $p$ is isomorphic to $P:=k[[t_{1},...,t_{d}]]$. So $J$ is trivial in this case.

In fact you can use formal deformation theory to prove this part of the structure theorem. If $R$ is smooth over $k$, by the infinitesimal lifting property of smooth morphisms you get an obstruction theory taking values in the trivial vector space.

Also, to answer your question, if the obstruction vanishes for any small extension, it will vanish for any thickening because any surjective morphism of Artin-k-algebras can be factored into successive small extensions. So, again, you can take $T_{2}=0$.

Note also that, in general you have a canonical obstruction theory for $hom(R,-)$, with $R=k[[t_{1},...,t_{d}]]/I$ taking values in $V=Hom(I/I^{2},k)/Hom(\hat{\Omega},k)$. Here $\hat{\Omega}$ is the free $k[[t_{1},...,t_{d}]]$-module generated by $dt_{i}$. $V$ has dimension $s$, where $s$ is the minimal number of generators of $I/I^{2}$. This obstruction theory induces any other obstruction theory for $hom(R,-)$.

Ultimately you see that in your case you have: The moduli space $X$ is smooth at $p$$\iff$ The completion of the local ring at $p$ is $k[[t_{1},...,t_{d}]]$$\iff$ The canonical obstruction theory for the prorepresentable local deformation functor at $p$ takes values in the trivial vector space.

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  • $\begingroup$ I see your point. So is your last paragraph saying that smoothness is equivalent to formal smoothness? So you say that smoothness is equivalent to the fact that the canonical obstruction theory on $D_{X,p}$ takes values in $T_2=0$. That confuses me, because there are thousands of smooth moduli spaces which are obstructed. But that probably refers to another obstruction theory, one which is induced by the specific moduli problem. Does it make sense? $\endgroup$
    – Brenin
    May 19, 2015 at 17:30
  • $\begingroup$ Not really. As I said, when you have a smooth point, the local deformation functor at that point will be unobstructed, in the very sense that the obstruction for lifting deformations along ANY surjection of Artin-k-algebras will vanish. So I don't really see which examples you could mean. Of course, it's not impossible that I'm the one missing out on something here. $\endgroup$
    – Rieux
    May 28, 2015 at 16:43

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