# The analytic and the algebraic “small disc”

I would like to understand the relation between an analytic object (the so called "small disc") and an algebraic one (the spectrum of a DVR). The framework is that of one-parameter families of complex curves $X\to S$.

Analysis: $S_0=\{z\in \mathbb C:|z|<\epsilon\}$.

Algebra: $S_1=\textrm{Spec }\mathbb C[[t]]$.

I understand (better: I accept!) that, under some GAGA equivalence of categories, the arrow $X\to S$ can be thought of as either a holomorphic map in the category of complex manifolds, or as a morphism of complex algebraic varieties. But why is $S_1$ the "translation" of $S_0$?

I figure $S_1$ as a two point space where the unique closed point is $(t)$ and $\eta=(0)$ is the generic point. When I look at $S_0$, I can guess that the origin $0\in S_0$ plays the role of $(t)\in S_1$. Can anyone tell me how I should interpret $S_0$ and $S_1$ in terms of each other? (I am sorry because I know this is a vague question.)

The main difficulty is to deal with the fact that there are only two points in $S_1$ (so a family $X\to S_1$ consists of two curves), while there are infinitely many points in $S_0$. But I understand that $\eta$, being an open point, plays the role of a neighborhood of $(t)$ exactly as every open neighborhood of the origin in $S_0$.

Thanks in advance.

• My knowledge of the matter (especially on the analytic side) is rather skeletal, but in general the process of a completion (in the simplest case: of a local ring at a point) is the algebraic analogue to taking "infinitesimal neighbourhood". For projective varieties, the stalk of the structure sheaf at a given point remembers the whole variety up to birational equivalence, so contains a lot of information that we would not consider "local". On the other hand, the completion of a local ring at a non-singular point is always $\mathbb{C}[x_{1}, \ldots, x_{n}]$, where $n$ is the dimension. – Piotr Pstrągowski Apr 23 '13 at 20:08
• I'm not sure what happens on the analytic side, but on the algebraic you can think of this map as encoding the behaviour of $X \rightarrow S$ in some infinitesimal neighbourhood of your point. I hope someone with actual expertise will add something. – Piotr Pstrągowski Apr 23 '13 at 20:10
• Dear @PiotrPstragowski, I think you forgot some double square brackets near the end of your first comment. – Andrew Apr 23 '13 at 20:13
• @Andrew, you're right, of course! Although I'm not sure if I can edit it now. – Piotr Pstrągowski Apr 23 '13 at 20:15
• Dear @PiotrPstragowski, thanks for your comments; understanding the correspondence between completion and "taking infinitesimal neighborhood" was exactly the aim of my question :) – Brenin Apr 26 '13 at 9:01

## 3 Answers

There are a few things going on here.

First, strictly speaking, I don't think GAGA applies here, since $S_1$ is not finite type (e.g. here).

But your interpretation of the roles of $(t)$ and $(0)$ are correct in the case we are studying a degeneration of a family of curves over a DVR. In this case, the family $X\to S_1$ is determined by a homomorphism $\mathbb C[[t]]\to\Gamma(X,\mathcal O_X)$, and there are two fibres to consider: the fibre over the closed point $(t),$ and the fibre over the generic point $(0).$ Over the closed point we will get an honest curve over $\mathbb C,$ while over the generic point, we get a "family" which is just a curve over $\mathbb C((t)).$

What Qiaochu mentions in his answer is actually an arc of the curve $V(f)\in\mathbb C^2.$ You should actually think of this as an infinitesimal analytic approximation of $V(f)$ at $(x(0),y(0)),$ rather than as a family of curves degenerating to $V(f).$

Edit

Let us consider a particular example, just for the sake of having something concrete to look at. We can see what happens if $X$ is something relatively easy, like $\operatorname{Spec}(\mathbb C[[t]][x])=\mathbb A^1_{\mathbb C[[t]]}.$ The morphism $X\to S_1$ is determined by its dual $\mathbb C[[t]]\hookrightarrow\mathbb C[[t]][x].$ Let's check the fibres.

Over $(t)\subseteq\mathbb C[[t]]$ we get $\kappa((t))=\mathbb C$ for the residue field, so the fibre is the spectrum of $\mathbb C[[t]][x]\otimes_{\mathbb C[[t]]}\mathbb C\cong\mathbb C[x].$ Thus, the fibre over $(t)$ is $\mathbb A^1_{\mathbb C},$ which is an honest curve over $\mathbb C.$ Analytically, we think of this as exactly the fibre over $0,$ since $t$ vanishes at $0\in\mathbb C.$

Over $(0)\subseteq\mathbb C[[t]]$ we get $\kappa((0))=\mathbb C((t)),$ and we can compute easily that the fibre is $\mathbb A^1_{\mathbb C((t))},$ which is now a curve over a transcendental field extension of $\mathbb C.$ The fact that this is the fibre over the generic point tells us to consider this curve as being a general member of the "family of curves" $X$, in exactly the same way as the generic point of $\mathbb C[[t]]$ is considered to be a general point in the "infinitesimal neighbourhood" of the origin represented by $S_1$, while the closed point $(t)\subseteq\mathbb C[[t]]$ refers to exactly the origin.

In order to really make a tight connection with the complex analysis picture, I think one has to consider the family $X\to S_1$ as a deformation of the closed fibre over the ring $\mathbb C[[t]].$ In general, once we have a small deformation of an object, the question of deformation theory is whether we can extend the deformation to larger (than $\mathbb C[[t]]$ in this case) base rings, eventually finding the "algebraic" deformations (i.e., over non-local rings) of the given object. As long as certain obstructions vanish, we will be able to compute extensions, though there is a huge theory behind this, and even showing that obstructions vanish can be tough. But, if we have found an algebraic deformation, over something like $\mathbb C[t],$ then we can really use GAGA, or simply the fact that $\mathbb A^1_{\mathbb C}$ is naturally a complex manifold, to find the right correspondence.

• :thank you for your answer. In what sense can I view the generic fiber (which is one fiber) as a "family", as you wrote? – Brenin Apr 26 '13 at 8:52
• Dear @atricolf, I have edited my answer to add some comments which (I hope!) will answer your question. – Andrew Apr 26 '13 at 12:54
• Dear @Andrew, your answer is great. Thank you very much! – Brenin Apr 26 '13 at 21:24

Don't think in terms of points. It will be easier for me to talk about morphisms $S_1 \to X$ first. If $X = \text{Spec } R$ then this is a homomorphism $R \to \mathbb{C}[[t]]$. For example, if $R = \mathbb{C}[x, y]/f(x, y)$, then this is a pair of power series $x(t), y(t)$ such that $f(x(t), y(t)) = 0$. If these power series have some finite radius of convergence, then we really do get an honest (analytic) curve parameterizing some small neighborhood of $(x(0), y(0))$ by setting $t$ to be sufficiently small, but in the algebraic picture we don't care about radii of convergence; we get a "curve" parameterizing some "infinitesimal neighborhood" of $(x(0), y(0))$.

Perhaps it is helpful to first think about homomorphisms $R \to \mathbb{C}$, which are points, and then homomorphisms $R \to \mathbb{C}[t]/t^2$, which are points together with tangent vectors (exercise), and then homomorphisms $R \to \mathbb{C}[t]/t^3$, which are...

• ......... two-jets. – Andrew Apr 23 '13 at 20:17

This is one idea.

If you consider your family $X\rightarrow S_0$ with central fibre $X_0$ (the fibre at the origin) and the inclusion $S_0\to \mathbb C$ then the composition gives you $f:X\rightarrow \mathbb C$. If $X$ is nice, then you can use GAGA to get an algebraic family $\mathcal X\to \mathbb A^1_{\mathbb C}$ such that $f_{an}:\mathcal{X_{an}} \rightarrow \mathbb A^1_{\mathbb{C},an}$ is a family with coincide with your family after restriction to the open disc $S_0$. In particular the central fibre is the same.

Now considering the morphism $\mathbb C[x]\rightarrow \mathbb C[[x]]$ and base change we get a family over $S$: $$\mathfrak f: \mathfrak X\rightarrow S$$ we get an algebraic familly with special fibre $\mathfrak X_0\rightarrow \mathop{Spec}\mathbb C$ such that $$\mathfrak X_{0,an} = X_0 \rightarrow \mathbb C.$$

That is to an analytic family over the open disc you can associate an algebraic familly over the algebraic open disc with the "same" central fibre. Moreover we must have also that coherent sheaves, Hodge structures etc.. translate… I think.