Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I wonder if someone could explain briefly what is the relation between these 3 formal models, of a Berkovich space, a rigid analytic space and a formal scheme?

I have been working with formal schemes in the last few years, but I know virtually nothing about these other models.

What are the relations between these categories? is one contained in the other?

Are there any recommended references about this comparison ?

Many thanks.

share|improve this question
4  
You might also add Huber's theory of adic spaces. In any case, all of these models have their respective merits. A large part of Berkovich's ``red book'' is devoted to comparison theorems between his theory and those of Tate (rigid spaces) and Raynaud (formal schemes modulo admissible blow ups). –  ACL Jul 31 '13 at 18:29
2  
Just in case someone actually wants to check the references, I am pretty sure that there are no formal schemes in the red book. –  Jérôme Poineau Aug 5 '13 at 11:10

2 Answers 2

There are different answers to your questions depending on what you have in mind. If you want to compare the spaces themselves, then Berkovich spaces and rigid analytic spaces are very close. I would say that it is similar to the relation between schemes of finite type over $\mathbb{C}$ and complex varieties in a naive sense, i.e. with only $\mathbb{C}$-points.

You can have a look at Berkovich's book "Spectral theory etc.", section 3.3, for precise statements. Fix a non-archimedean complete valued field $k$ with non-trivial valuation. Then, there is a functor from the category of strictly $k$-analytic spaces (Berkovich spaces) to rigid spaces with good properties: it preserves cohomology, for instance. At the level of points, the functor is easy to describe: you keep the points whose residue field is a finite extension of $k$.

As for the essential image of the functor, I am not sure what to say, but you can at least build a Berkovich space from a quasi-compact and quasi-separated rigid space in a satisfactory way (see Conrad's notes mentioned in Colin McLarty's answer).

(There are technical points that show that Berkovich's theory is actually more general. First, trivial valuation on $k$ is allowed. Second, the basic algebras of rigid geometry are the Tate algebras $k\{T_1,\dots,T_n\}$ that contain power series with radius of convergence at least 1, whereas Berkovich allows any radii.)

So the spaces are more or less the same, but I think the way you think about them is different. For example, Berkovich spaces are true topological spaces (as opposed to the Grothendieck topology of rigid spaces), which makes it easier to adapt the local methods from complex analytic geometry. In my mind, rigid analytic spaces look more algebraic.

As regards formal schemes, they are quite different objects. The relation is that the rigid spaces and Berkovich spaces are their generic fibers. So a formal scheme has more information. Raynaud's theory (see his paper in Mémoires de la S.M.F 39-40 (1974), p.319-327) actually tells you how to pass from a model to another (by blowing-ups ideals supported on the special fiber), which makes it possible to recover the category of rigid spaces purely in terms of formal schemes (by a suitable localization). This makes the analytic theory very much algebraic! A big gain is that you may now use everything that has been developed in the algebraic setting and push it on the analytic side. For example, you can show the coherence of proper direct images of coherent sheaves this way (see Lütkebohmert, Math. Ann. 286 (1990), p.341-171).

I hope this helps. If you have more specific questions, please ask!

share|improve this answer

A discussion of Berkovich spaces and rigid analytic spaces, with references is at http://ncatlab.org/nlab/show/Berkovich+space. For all three, also with references, see http://en.wikipedia.org/wiki/Rigid_analytic_space. The references include notably a long chapter, "Several approaches to non-archimedean geometry," by Brian Conrad, http://math.stanford.edu/~conrad/papers/aws.pdf

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.