# When is a formal deformation convergent?

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?

• Is the $Spec$ in $Spec(\mathbb{C}\{t\})$ the usual spectrum of rings (or of $\mathbb{C}$-algebras) or some analytic variant? – Qfwfq Dec 5 '18 at 17:18

(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. $$xy(x+y)(x-\lambda y).$$

• 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. – Saal Hardali Dec 8 '18 at 9:25
• The answer still stands: take something like $E\colon y^2 z = x(x-z)(x-\lambda z)$ where $\lambda = \sum n! t^n$. – Piotr Achinger Dec 8 '18 at 10:55
• (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.) – Piotr Achinger Dec 8 '18 at 10:57
• 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 – Saal Hardali Dec 8 '18 at 11:18
• 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. $xy(x+y)(x-\lambda y)$. – Piotr Achinger Dec 8 '18 at 12:07

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).

• 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 $t$... – Piotr Achinger Dec 10 '18 at 11:42