# Kapranov's Theorem for tropical hypersurfaces: Understanding closure


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!

• This seems to be equivalent to showing $\{\w \in \mathbb R^n : \text{in}_\w(f) \text{ is not a monomial } \}$ is closed. It surely is necessary; and if it's true then closure of intersection with a dense set is the set itself. – Max Apr 5 '17 at 21:02
• On the other hand $\{\w \in \mathbb R^n : \text{in}_\w(f) \text{ is not a monomial } \}$ is closed as finite union (over all pairs of monomial exponents $u$ with non zero $c_u$) of closed sets $K_{ij}$ of $w$s for which $f(u)=val(c_u)+w\cdot u$ has $f(u_i)=f(u_j)$ (linear equation) and this is $\leq f(u_k)$ for all other $u_k$ (finite set of non-strict linear inequalities). – Max Apr 5 '17 at 21:36


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 $\implies B_\delta(w) \subset C \implies C \text{ open } \implies B \text{ closed}$.