Kapranov's Theorem for tropical hypersurfaces: Understanding closure

Question

I am working with the book Introduction to Tropical Geometry by Sturmfels and Maclagan, and am trying to understand Kapranov's Theorem (Theorem 3.1.3 in the book), which states (I only mention the relevant parts): $\newcommand{\u}{\textbf{u}} \newcommand{\w}{\textbf{w}}$

Let $K$ be algebraically closed field with a non-trivial valuation $\text{val}: K \to \mathbb{R} \cup \infty$.

Fix a Laurent polynomial $f = \sum_{\u \in \mathbb{Z}^n}{c_{\u}x^\u}$ in $K[x_1^{\pm1},\dots, x_n^{\pm1}]$. The following sets coincide:

(1) the tropical hypersurface $\text{trop}(V(f))$

(2) the closure in $\mathbb{R}^n$ of the set $\{\w \in \Gamma^n_{\text{val}} : \text{in}_\w(f) \text{ is not a monomial } \}$

(3) ...

Here $\Gamma_\text{val}$ denotes the value group which is dense in $\mathbb{R}$, and $\text{in}_\w(f)$ is the initial form with respect to vector $\w$. Topology on $\mathbb R^n$ is the Euclidean topology.

Then the book says without explanation:

Note that the closure in Set 2 equals $\{\w \in \mathbb R^n : \text{in}_\w(f) \text{ is not a monomial } \}$

This is something I don't understand (although I know that the value group is dense in $\mathbb{R}$, so the closure of $\Gamma_\text{val}^n$ is $\mathbb R ^n $).

Could someone explain to me why this set is exactly the closure?

Thanks a lot!

Answer 1

$\def\valgroup{\Gamma_{\text{val}}} \def\trop{\text{trop}} \def\Z{{\mathbb Z}} \def\Q{{\mathbb Q}} \def\R{{\mathbb R}} \newcommand{\val}[1]{\text{val}(#1)} \newcommand{\vect}[1]{\text{\bf #1}}$

I will answer this myself. Maybe it will help someone sometime. Here is a detailed proof:

We want to show that the closure $\bar A$ with regard to the Euclidean topology (let $\langle \cdot, \cdot \rangle$ denote inner product) of the set $A := \{ w \in \valgroup^n : \text{in}_{w}(f) \text{ is not a monomial} \}$ is $$\bar A = B := \{ w \in \R^n : \text{in}_{w}(f) \text{ is not a monomial} \}$$

As $K$ is algebraically closed and its valuation is non-trivial, we know that $\valgroup$ is dense in $\R$. Thus every open subset in $B$ contains a point in $A$, and we have $B \subset \bar A$. So it remains to show that $B$ is closed, i.e. show that $\R^n \setminus B = \{ w \in \R^n : \text{in}_{w}(f) \text{ is a monomial} \} =: C$ is open.

Let $w \in C$. Then the minimum in $$\trop(f)(w) = \underset{u \in \Z^n : c_{u} \neq 0 }{\min}\{ \val{c_u} + \langle u, w \rangle \}$$ is achieved only once, i.e. $\exists v \in \Z^n$ such that $c_v \neq 0$ and $\forall u \neq v$ we have $\val{c_v} + \langle v, w \rangle < \val{c_u} + \langle u, w \rangle$.

So, given any $u \in \Z^n$ with $c_u \neq 0$, define the oviously continuous function $$g_u : \R^n \to \R, x \mapsto \val{c_u} + \langle u, x \rangle$$

Let $U := \{u \in \Z^n : c_{u} \neq 0 \}$ which is finite. Have $g_{v}(w) < g_{u}(w) \ \forall u \in U \setminus \{v\}$. Set $\epsilon := \underset{u \in U \setminus \{ v \} }{\min}\{g_u(w) - g_v(w)\} > 0$. Because of continuity $\forall u \in U$ get $\delta_u > 0$ s.t. $\lvert g_u(w) - g_u(\tilde w) \rvert < \frac{\epsilon}{2} \ \forall \tilde w$ in the open ball $B_{\delta_u}(w)$ around $w$. Set $\delta := \underset{u \in U}{\min}\{\delta_u\}$ and let $x \in B_\delta(w)$.

Then $\forall u \in U \setminus \{v\}$ we have

\begin{equation*} \begin{split} \underbrace{g_u(x)}_{> g_u(w) - \frac{\epsilon}{2}} - \underbrace{g_v(x)}_{< g_v(w) + \frac{\epsilon}{2}} > g_u(w) - \frac{\epsilon}{2} - g_v(w) - \frac{\epsilon}{2} \underset{g_u(w) - g_v(w) \geq \epsilon}{\geq} \epsilon - \epsilon = 0 \end{split} \end{equation*}

Thus the minimum in $\trop(f)(x)$ is also achieved only once and $\text{in}_x(f)$ is monomial $\implies B_\delta(w) \subset C \implies C \text{ open } \implies B \text{ closed}$.