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Let $D=\mathop{Spec}(\mathbb C[[t]])$ be the algebraic small disk and let $\Delta= \{z\in \mathbb C: |z|<1\} $. Let $X_0$ be an algebraic surface over $\mathop{Spec}\mathbb C$

Suppose now that I have a deformation of $X_0\to\mathop{Spec}{\mathbb C}$ over $D$, that is a flat, proper (maybe projective) morphism $X\rightarrow D$ such that the special fibre is $X_0$.

Question: Under which conditions I can find a degeneration $Y\rightarrow \Delta$ with fibre at zero (central fibre) $X_0^{an}$ and with the property that we can recover $X\rightarrow D$ form $Y\rightarrow \Delta$ in the sense that if I start with the $$Y\rightarrow \Delta, \text{with central fibre }\ X_0^{an}$$ and we consider the composition $\Delta\rightarrow \mathbb C$, to get $Y\rightarrow \mathbb C$. If $Y$ is nice enough (projective, finite type etc) I can use GAGA to get an algebraic family $\mathcal X\rightarrow \mathop{Spec}\mathbb C[t]$ and if now I consider the inclusion $ \mathbb C[t]\rightarrow \mathbb C[[t]]$ and let $X$ be the base change I get a family $\mathcal X\otimes \mathop{Spec}(\mathbb C[[t]])\rightarrow D$ with special fibre $X_0$ and an isomorphism $\mathcal X\otimes \mathop{Spec}(\mathbb C[[t]])\simeq X$ over $D$?

If such a deformation exist, which properties are preserved? for example can I identify the Monodromy?

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    $\begingroup$ $D$ not being finite type is by no means the main problem, as there is no "global" GAGA for geometric objects over a non-proper base. Is your proper flat deformation over $D$ smooth over $D[1/t]$? It is also far too restrictive to expect an "algebraization" over $\mathbf{C}[t]_{(t)}$, since the completion cannot distinguish the genus of the curve arising from a given algebraic local ring. There is no "identification of $D$ and $\Delta$". You want results on openness of loci in the base for fibral properties in the scheme-theoretic and analytic space cases; see Exp. XII of SGA1. $\endgroup$
    – Marguax
    Oct 10, 2013 at 2:45
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    $\begingroup$ What is the meaning of $X(\mathbb{C}) = Y$? $Y$ is a complex analytic space, and $X(\mathbb{C})$ is a set. Are you looking for an analytification functor applied to $D$? $\endgroup$
    – S. Carnahan ♦
    Oct 10, 2013 at 9:45
  • $\begingroup$ By $X(\mathbb C)$ we understand the $\mathbb C$ points of $X$. We can give to it a structure of complex analytic space. So indeed $X_{an}=X(\mathbb C)$. $\endgroup$
    – Rogelio Yoyontzin
    Oct 12, 2013 at 0:44
  • $\begingroup$ Yes it is smooth outside the central fibre. I will take a look of SGA 1 to see if this is what I was looking for. Thanks. $\endgroup$
    – Rogelio Yoyontzin
    Oct 12, 2013 at 0:46
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    $\begingroup$ Yoyontzin: You are confused about something, since there is absolutely no way of turning a finite type $\mathbf{C}[[t]]$-scheme into an analytic space in general. It is meaningless to "specialize" a typical formal power series (which has radius of convergence 0) at $t \in \Delta - \{0\}$. The only meaningful definition of $X(\mathbf{C})$ as an analytic object is as $X_0^{\rm{an}}$, which is not what you want. Your statement "$X(\mathbf{C})=Y$" does not seem well-formed. Please think more carefully about the definitions involved. $\endgroup$
    – Marguax
    Oct 12, 2013 at 4:27

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