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I am looking for a geometric and topological way to make a visualization of higher dimensional berkovich spaces, starting with the berkovich plane. Of course, this is just a collection of bounded semi-norms, but the question remains:

Is there a visualization possible for $\mathbb A^2_{\text{Berk}}$ like the infinite branched tree for $\mathbb A^1_{\text{Berk}}$?

(For $\mathbb A^1_{\text{Berk}}$ see for example Baker and Rumely's Potential Theory and Dynamics on the Berkovich projective line, Chapters 1 - 2.)

I think you get a simplicial complex, but I don't know exactly how. On the one hand (reading Favre and Johnsson's The valuative tree), you have this list of valuations (thus, also of seminorms, allthough this book discusses seminorms on $\mathbb{C}^2$). On the other hand we have Berkovich's theory of Type I - Type IV points. I guess there just more Type I - Type IV points in a plane (i.e. more seminorms that occur as it were type I points), and some can be only represented by faces (two dimensional simplices), only I don't know how.

Are there any references on the visualization part?

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    $\begingroup$ I don't think you get a simplicial complex, but you may get a metric $\mathbb{R}^2$-building. This is similar to the situation with the projective line. You get some Type I points from maximal ideals, and Type II and III points from neighborhoods, but you also get extra points from analytic arcs. The points of the plane may be rather messy to describe in full, since I think they include all of the points in all possible blow-ups. $\endgroup$
    – S. Carnahan ♦
    Commented Jul 22, 2010 at 12:14
  • $\begingroup$ @Scott: Do you know a reference (unpublished notes or whatnot are ok) for a classification of points of the Berkovich plane? $\endgroup$
    – David Zureick-Brown
    Commented Aug 3, 2010 at 16:42
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    $\begingroup$ 'm not sure if it is possible to really vizualize the Berkovich plane. Of course, you can always project it onto the line and fibers will then be lines over the residue fields. But I'm not sure how much this helps. Actually, I already lack a satisfactory answer for schemes. Can you vizualize the scheme $\mathbb{A}^2_{\mathbb{Q}_p}$? And the Berkovich plane surjects to this scheme (through the map that sends a seminorm on $\mathbb{Q}_p[X,Y]$ to the prime ideal that consists of elements of seminorm 0)... $\endgroup$
    – Jérôme Poineau
    Commented Sep 2, 2010 at 12:56
  • $\begingroup$ @David: No. What I said above was bold ansatz. $\endgroup$
    – S. Carnahan ♦
    Commented Sep 3, 2010 at 9:39
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    $\begingroup$ Related to S. Carnahan's comment: You might be interested in some of the papers of Bertrand Rémy (math.univ-lyon1.fr/~remy/maths.html) related to compactifications of Bruhat-Tits buildings in Berkovich spaces. It doens't describe the full Berkovich space, but I guess that is about as good as you can get. $\endgroup$
    – Tom De Medts
    Commented Apr 29, 2011 at 6:25

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Since I don't seem to be able to edit my own question. I found s link which might be useful here:

  • http://users.math.yale.edu/~sp547/pdf/Anayltification-tropicalizations.pdf, in which a homeomorphism is constructed between $X^{an}$ (or $X_{Berk}$, if that is you cup of coffee) and an inverse limit of tropicalizations of embeddings of (toric) subvarieties of $X(K)$, where $K$ is the basefiel. Payne first takes the field $\mathbb{C}((t^{\mathbb{R}}))$ (in which, for example in $\mathbb{P}^1_{Berk,K}$ there are only Type 1 and Type 2 points), then switches to fields with trivial norms.
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In fact you get more type of points, I believe. Points of type I are just k-point, i.e. points such that H(x)=k, you have a general fact that if l is algebraic over the completion l' of trascendence degree n over k then the sum of the trancendence degree of the reduction of l over the reduction of k and the rank of |l*| over |k*| is less or equal than n.

In the projective line, as in all the 1-dimentionals analytics spaces, this implies that you have only four types of points (H(x)=k, OR |H(x)*| increase of one degree of trascendence, OR the reduction of H(x) increase of "one rank", OR H(x) it is a non trivial immediate extention of k). When n>1 you may have then a lot more types of points, and the number of types increase when n grows.

If you want to find those generals informations there's a paper of Temkin wich is very intresting.

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    $\begingroup$ What you say about type I points is only true when k is algebraically closed. If k is, say, $\mathbb{Q}_p$, all points in $\bar{\mathbb{Q}}_p$ are of type I. Actually, all points in $\mathbb{C}_p$ are of type I too, which can be rather misleading (and probably indicates that this classification only really suits the algebraically closed case). $\endgroup$
    – Jérôme Poineau
    Commented May 19, 2011 at 15:52
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@David. The best reference I was able to find (of some relevance) is "The Valuative Tree" by Matthias Johnsson, which discusses norms on $\mathbb{C}[x,y]$. The only thing which isn't clear from this is, which points are Type I, ..., Typer IV.

You don't get more types (so no Type V or something like that) [this is proven somewhere in Berkovich's Analysis on Non-archimedean spaces], but I think you get more Type I, more Type II, etc points.

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    $\begingroup$ I'm not sure I understand your last sentence. Berkovich's classification only holds for points of the line. What do you mean by a type I point on a surface? By the way, Favre and Jonsson work over $\mathbf{C}$ with trivial valuation. Things would get more complicated for nontrivially valued fields. $\endgroup$
    – Jérôme Poineau
    Commented Sep 2, 2010 at 13:08

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