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Let $X$ be a finite type scheme over $\mathbb{C}$ and let $ \mathcal{X} \to Spf(\mathbb{C}[[x]])$ be a formal deformation of $X$. Which of the following assumptions (or combinations thereof) are sufficient to imply that this deformation is convergent? (i.e. that it comes from some flat analytic family $\tilde{\mathcal{X}} \to \mathcal{D}$ - where $\mathcal{D}$ means a closed analytic disk of some non-zero radius).

  1. $X$ quasi-projective.

  2. $X$ proper.

  3. $X$ projective (equivalently, both (1) and (2)).

  4. $X$ affine.

  5. $X$ smooth.

Are there simple (preferably low dimensional) counter examples to convergence of formal defomrations?

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  • $\begingroup$ Is the $Spec$ in $Spec(\mathbb{C}\{t\})$ the usual spectrum of rings (or of $\mathbb{C}$-algebras) or some analytic variant? $\endgroup$ – Qfwfq Dec 5 '18 at 17:18
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EDIT. New version, addressing questions in the comments.

(1) Affine and smooth implies what you want.

Indeed, suppose $\mathcal{X}$ is smooth and that $H^1(X_0, T_{X_0/\mathbb{C}}) = 0$ where $X_0/\mathbb{C}$ is the special fiber and where $T_{X_0/\mathbb{C}}$ is the tangent bundle. This is of course satisfied if $X_0$ is affine.

I claim that in this case $\mathcal{X}$ is actually constant, i.e. $\mathcal{X}$ is the $t$-adic completion of $X_0 \times {\rm Spec}\, \mathbb{C}[[t]]$; if this is true then the constant family $X_0 \times (\text{unit disc})$ is the desired extension.

The claim follows by usual deformation theory: one shows by induction on $n$ that $X_n = \mathcal{X}\otimes_{\mathbb{C}[[t]]} \mathbb{C}[t]/(t^{n+1})$ is isomorphic over $\mathbb{C}[[t]]$ to $X_0 \otimes_\mathbb{C} \mathbb{C}[t]/(t^{n+1})$. For the induction step, basic deformation theory (e.g. Fantechi's and Illusie's articles in "FGA explained") tells you that liftings of $X_n \cong X_0 \otimes_\mathbb{C} \mathbb{C}[t]/(t^{n+1})$ to $\mathbb{C}[t]/(t^{n+2})$ are a torsor under $H^1(X_0, T_{X_0/\mathbb{C}})$. Since this group vanishes, there is only one such lifting, up to a non-canonical isomorphism. In the limit, we obtain the desired isomorphism.

See also the results of Renee Elkik Solutions d’équations à coefficients dans un anneau hensélien Annales scientifiques de l’É.N.S. 4e série, tome 6, no 4 (1973), p. 553-603. She shows (see Theorem 6, p. 580) that a smooth algebra over a ring $R_0= R/I$ can always be lifted (not only formally) over $R$ if $R$ is noetherian and henselian along $I$.

(2) Smooth affines really are the only case when the answer is positive.

Indeed, take an elliptic curve over $\mathbb{C}[[t]]$ with divergent $j$-invariant, for example $$ E\colon y^2 z = x(x-z)(x-\lambda z), \quad \lambda = \sum_{n\geq 0} n! t^n. $$ Then the $j$-invariant $$ j(\lambda) = 258 \frac{(1-\lambda(1-\lambda))^3}{(\lambda(1-\lambda))^2} \in \mathbb{C}[[t]] $$ is likely not convergent. If $E$ was the completion of an analytic family (I am identifying $E$ with the corresponding formal scheme which does not make much difference since $E$ is proper), the $j$-invariant would have to have positive radius of convergence.

The same example works in the affine but singular case by taking the affine cone, i.e. $$ \mathcal{X} = {\rm Spf}\, \mathbb{C}[[t]]\{x,y,z\} / (y^2 z - x(x-z)(x-\lambda z), $$ or the same with $[[x,y,z]]$.

Perhaps the simplest example is the plane curve singularity with four lines meeting at a point whose cross-ratio $\lambda$ is divergent, i.e. $$ xy(x+y)(x-\lambda y).$$

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  • $\begingroup$ I changed the question in hope that it would be more precise so that I could understand the answer. I'm not sure how an elliptic curve can have a divergent $j$-invariant if it comes from a formal deformation. In particular I think both singular affine curves and complete non-singular curves can't be counterexamples. Could you be more precise please? Sorry about the abrupt edit of the question. $\endgroup$ – Saal Hardali Dec 8 '18 at 9:25
  • $\begingroup$ The answer still stands: take something like $E\colon y^2 z = x(x-z)(x-\lambda z)$ where $\lambda = \sum n! t^n$. $\endgroup$ – Piotr Achinger Dec 8 '18 at 10:55
  • $\begingroup$ (I didn't check that in this case $j(\lambda) = 256 (1-\lambda(1-\lambda)^3) \lambda^{-2} (1-\lambda)^{-2}$ is divergent, but most likely it is.) $\endgroup$ – Piotr Achinger Dec 8 '18 at 10:57
  • $\begingroup$ Yes, I understand now, of course you are correct, sorry. This covers the proper-smooth case. Do you know a simple example of a surface singularity with a divergent formal deformation? Just to make this answer complete $\endgroup$ – Saal Hardali Dec 8 '18 at 11:18
  • $\begingroup$ Of course! Take $y^2z =x(x-z)(x-\lambda z)$ but in $\mathbb{C}[[x,y,z]]$. There are also plane curve singularity examples, e.g. $xy(x+y)(x-\lambda y)$. $\endgroup$ – Piotr Achinger Dec 8 '18 at 12:07
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A sufficient condition comes from well known results: Grothendieck's effectiveness theorem and Artin's approximation theorem. Grothendieck's result shows that if $X$ is projective and there is a closed embedding of formal schemes of $\mathcal X$ into formal projective space over $Spf(\mathbb C[t])$ then the formal deformation is effective (meaning the formal deformation comes from a deformation over the $Spec$ of a local noetherian $\mathbb C$--algebra). If $X$ is projective and $h^2(X, \mathcal O_X ) = 0$, then every formal deformation of $X$ is effective. Artin's theorems assures that any effective formal VERSAL deformation of $X$ is algebraizable (meaning that it comes from a deformation over the $Spec$ of a $\mathbb C$--algebra of finite type).

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    $\begingroup$ Yes, this is also very useful, but does not give the desired convergence. For example, the elliptic curve in my answer is of course defined over the finitely generated subalgebra $\mathbb{C}[\lambda] \subseteq \mathbb{C}[[t]]$, but $\lambda$ is a divergent series in $t$... $\endgroup$ – Piotr Achinger Dec 10 '18 at 11:42

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