We are working over $$ \mathbb{K} = \bigcup_m \mathbb{C}\lbrace\lbrace t^{1/m} \rbrace\rbrace. $$ Roughly speaking, tropicalization is taking $$ f(x,y) = \sum_{i,j} a_{ij} x^i y^i \to \sum_{i,j} v(a_{ij}) x^i y^i = \max_{i,j} (xi + yj + v(a_{ij})). $$ But the tropical limit will have more structure than this. Given a polynomial $f$ defining an algebraic curve, the tropical limit of $f$ keeps track of
- the image under the valuation map,
- some collection of algebraic curves “as $t \to 0$.”
The setup was that $\Delta$ is a rational polytope, and this gives rise to a toric variety $\operatorname{Tor}(\Delta)$. We are then looking at a polynomial of the form $$ f(z,w) = \sum_{(i,j) \in \Delta} a_{i,j}(t) z^i w^j \in \mathbb{K}[z,w]. $$ Then $f$ defines a curve $C \subset \operatorname{Tor}_ \mathbb{K}(\Delta)$. Without loss of generality let us assume that all the coefficients lie in $\mathbb{C}\lbrace\lbrace t \rbrace\rbrace$ by rescaling $f$ and reparametrizing $t$. Then there exists a $\delta \gt 0$ such that if $0 \lt \lvert t \rvert \le \delta$ then we obtain a family of curves $C^{(t)} \subset \operatorname{Tor}_ \mathbb{C}(\Delta) = \operatorname{Tor}(\Delta)$.
Lemma 1. The family $C^{(t)}$ is equi-singular, and moreover topologically the singularities of $C^{(t)}$ correspond to singularities of $C \subset \operatorname{Tor}_ \mathbb{K}(\Delta)$.
Consider the tropical polynomial $N_f(x, y) = \sum_{(i,j) \in \Delta} v(a_{ij}) x^i y^j$, which gives rise to a tropical curve on the one hand and also a “dual function” $\nu_f(i, j)$ by taking the convex hull of $(i, j, -v(a_{ij}))$. Then $\nu = \nu_f \colon \Delta \to \mathbb{R}$ gives rise to a subdivision $\Delta = \Delta_1 \cup \dotsb \cup \Delta_N$. Let us also write $$ f(z,w) = \sum_{(i,j) \in \Delta} (a_{i,j}^{(0)} + O(t)) t^{\nu(i,j)} z^i w^j. $$
Definition 2. The tropicalization of $f$ consists of
- the tropical curve $T_f$ defined by $N_f(x, y)$,
- the curves $C_m \subset \operatorname{Tor}(\Delta_m)$ defined by $f_m(z, w) = \sum_{(i,j) \in \Delta_m} a_{i,j}^{(0)} z^i w^j$.
For $\nu \colon \Delta \to \mathbb{R}$, let us define the overgraph of $\nu$ as $$ \lbrace (x, y) \colon x \in \Delta, \nu(x) \le y \le M \rbrace $$ for $M$ large enough. If $\tilde{\Delta}$ is the overgraph, we can consider the toric variety for $\tilde{\Delta}$.
Fact 3. There is a map $\operatorname{Tor}(\tilde{\Delta}) \to \mathbb{A}^1$ where the generic fiber is $\operatorname{Tor}(\Delta)$ and there is some degeneration at $0$.
Tropicalization of nodal curves
Recall that $\operatorname{Sev}_ n(\Delta)$ is the variety parametrizing nodal curves passing through $n$ nodes. We want to study the degree of $\operatorname{Sev}_ n(\Delta)$ or $\operatorname{Sev}_ n^\mathrm{irr}(\Delta)$. This amounts to asking how many such curves pass through a generic set of points.
Theorem 4. Pick points $p_1, \dotsc, p_\zeta$ such that $x_i = \operatorname{val}(p_i)$ are $\Delta$-generic. Consider a curve $C \in \operatorname{Sev}_ n(\Delta)$ passing through $p_i$. Then the tropicalization of $C$ consists of
- a simple (in particular, nodal) tropical curve of rank $\zeta$,
- a curve $C_m \in \operatorname{Tor}(\Delta_m)$ such that
- if $\Delta_m$ is a triangle then $C_m$ is an irreducible rational nodal curve, meeting $\operatorname{Tor}(\partial \Delta_m)$ in precisely three points, all of them unibranch,
- if $\Delta_m$ is a parallelogram then $C_m$ arises from a product of monomials and binomials.
Recall that given a tropical curve $T$ with Newton polygon $\Delta$, we obtained subdivision $S$ of $\Delta$. Then
- components of $\mathbb{R}^2 - T$ correspond to vertices of $S$,
- edges correspond to edges,
- vertices of $T$ correspond to polygons of $S$,
- the valence of a vertex in $T$ corresponds to the number of sides of a polygon in $S$.
Then we were saying that a tropical curve is simple when the subdivision consists of triangles and parallelograms, and moreover all the edge vertices are used.
Recall that a genus $g(T)$ of a tropical curve $T$ is the minimum of $b_1(\Gamma)$ for all $\Gamma$ parametrizing $T$. We first show that the genus $g(C)$ agrees with $g(T)$ in a generic situation.
Lemma 5. If $T$ is a nodal tropical curve, then $$ \operatorname{rank}(T) = \lvert \operatorname{Ends}(T) \rvert + g(T) - 1. $$
Recall that for $C \in \operatorname{Sev}_ n^\mathrm{irr}(\Delta)$ we have $g(C) = \zeta + 1 - \lvert \partial\Delta \cap \mathbb{Z}^2 \rvert$. Then $$ \operatorname{rank}(T) - \zeta \ge 0 = \lvert \operatorname{Ends}(T) \rvert - \lvert \partial\Delta \cap \mathbb{Z}^2 \rvert $$ shows that $g(T) \ge g(C)$. In the other direction, we can roughly imagine that a cycle in $T$ gives rise to a cycle in $C$, and this shows that $g(T) \le g(C)$. This can be made precise.
Finally, if $g(T) = g(C) = g$, why does $T$ have to be nodal? We have $$ \operatorname{rank}_ \mathrm{expected}(T) = \lvert \operatorname{Vert}(S) \rvert - 1 - \sum_{k=1}^N (\lvert \operatorname{Vert}(\Delta_k) \rvert - 3) \le \operatorname{rank}(T). $$ But this has be an equality, so the curve has to be nodal.