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

I often hear people speaking of the many connections between algebraic varieties and tropical geometry and how geometric information about a variety can be read off from the associated tropical variety. Although I have seen some concrete examples of this, I am curious about how much we can get out of this correspondence in general. More precisely, my question is the following:

Which information of $X=V(I)$ can be read off its tropicalization $\mbox{Trop(X)}=\bigcap_{f\in I}\mbox{trop}(f)$?

As a very basic example, it is known that $\dim(X)=\dim_{\mathbb{R}}\mbox{Trop}(X)$.

share|improve this question
add comment

4 Answers 4

By work of Matt Baker, the dimension of a linear system on a curve is bounded above by the dimension of the corresponding tropical linear system on the corresponding tropical curve -- see Lemma 2.8 of his paper, "Specialization of linear systems from curves to graphs."

share|improve this answer
    
I have no idea whether anything like this is known in higher dimension, by the way, or even whether it's completely clear how to articulate the correct question! –  JSE Jan 26 '11 at 15:26
    
Actually, I find Dustin Cartwright's work on the higher-dimensional case very convincing. –  JSE Feb 24 at 14:24
add comment

I hope others will have lots more to say, but one nice property is $$g(X) \geq b_1(\mathrm{Trop}(X))$$ when $X$ is a (plane) curve, where $g$ is the genus and $b_1$ is the betti number (=number of cycles) of the graph. (Thanks to quim for the correcting the inequality.)

share|improve this answer
2  
Actually, $g(x)\ge b_1(Trop(X))$. For elliptic curves, the Betti number of the tropicalization will be zero or one depending on the sign of the valuation of the j-invariant (see Speyer arXiv:0711.2677 and Katz-Markwig-Markwig DOI:10.1112/S1461157000001522) –  quim Jan 26 '11 at 10:19
    
Oops! Fixed it, but please leave your comment up with the references. –  Dave Anderson Jan 26 '11 at 17:07
add comment

For general subvarieties of an algebraic torus, the tropicalization knows about the class of the subvariety in a suitable toric compactification of the algebraic torus. So you can compute intersection products of subvarieties of a torus tropically. See the recent preprint of Osserman-Payne for the state of the art.

For subvarieties X that are schon (which is a natural smoothness condition), you can say a lot more. There is a natural dualizing complex $\Gamma_X$ which maps to Trop(X) whose homology reflects the lowest bit of the weight filtration on X. From this fact, you can get the natural generalization of $g(X)\geq b_1(\Gamma)$ (it is not in general true that $g(X)\geq b_1(\operatorname{Trop}(X))$ because the tropicalization map may have disconnected fibers (as was pointed out by Speyer)).

There are two special cases where you can say a lot more: when $X$ is a schon hypersurface and when Trop(X) is smooth (smoothness here means Trop(X) is locally modeled on matroid fans). In this case, you can say things about the Hodge numbers of X. See my paper with Stapledon for details. Warning: the results of that paper require compactifying X by completing the algebraic torus to a toric variety. We have a sequel in the works that will use more sophisticated Hodge theory to get around that problem.

share|improve this answer
    
To JSE: higher dimensional analogs of Matt Baker's specialization lemma should hold. I've worked out some special cases for surfaces using tropical intersection theory. I think the right approach to the higher-dimensional analog should use rigid analytic geometry. –  Eric Katz Feb 6 '11 at 4:06
add comment

In arxiv.org/0805.1916 Sam Payne shows that one can reconstruct the analytification of a quasiprojective variety over a nonarchimedean field as the inverse limit of its tropicalisations.

share|improve this answer
add comment

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.