Edit: I realize the mathematics below is lacking a precise phrasing. I hope that the intuitiion behind the question is clear enough that a reader will understand the question and provide guidance. The question is essentially to what extent a degeneration of tropical curves reflects an actual degeneration of complex tropical curves in $(\mathbb{C}^*)^2$ More precisely, On page 10 of his paper,


Mikhalkin discusses the degeneration of a smooth tropical curve to a nodal tropical curve. Given such a local degeneration, which we assume occurs in a one dimensional real family paramaterized by $t$, via some reconstruction process, we can associate for each $t$ an actual hypersurface in $(\mathbb{C}^*)^2$ (either a curve in some degenerated complex structure or a complex tropical curve), whose tropicalization is the tropical curve $\Pi_t$. Is it true that the limiting curve over the nodal tropical curve is nodal?

To be more demanding, can one associate to this degeneration in a canonical way a fibration of curves in $(\mathbb{C}^*)^2$,

$H_t \to Spec(\mathbb{C}[\tau,\tau^{-1}])$

such that one fiber has a nodal curve and the rest of the fibers are smooth? The following example makes me believe this may be possible: In the local model corresponding to Mikhalkin's example, we may consider the family of hypersurfaces:

$$\tau+x+y+xy $$

when $\tau$ is not $0,1$, the smooth tropical curve is a smooth deformation retract of the amoebaas of curves in this family. The singular tropical is the (tropical) amoeba of:

$$ 1+x+y+xy=(1+x)(1+y) $$


If you ask is it true that a tropical nodal curve can be presented such a way as a limit of nodal curves then the answer is no.

What is true:

Curves in a family which tropicalises to tropical curve of genus g has genus at least g.

So, tropical nonsingular curve may be presented as a limit of nonsingular curves and a tropical limit of singular curves is singular too.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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