Latin America contains a wealth of UNESCO World Heritage Sites in countries from Mexico to Chile. Here are a few that we have managed to visit in our travels.
Cartagena Fortresses and Group of Monuments — The coastal city of Cartagena was threatened by pirates many times in its history. Spain built perhaps its most formidable fortifications to protect this beautiful city (Walking Tour available for $1.49 on website).
Pre-Hispanic City of Chichen-Itza — The Mayan complex at Chichen Itza is perhaps the best preserved and most impressive in Latin America. The crowning achievement of the ceremonial city is El Castillo, the huge pyramid which dominates the entire complex.
Historic Center of Mexico City — The world's largest urban area is loaded with American history — it was a significant Native American city prior to (and underneath) the modern city (Walking Tour available for $1.49 on website).
Guadalajara — Hospicio Cabanas — The Colonial city of Guadalajara has many fountains and a pedestrian-only central core that invites tourism (Walking Tour available for $1.49 on website).
Panama — Historic District of Panama City — Panama's Old City is a mixed bag. Some places are carefully and artfully restored and maintained; others are run down and seedy. But it is certainly a worthwhile place to visit (Walking Tour available for $1.49 on website).
This entry was posted in World Heritage Sites and tagged Latin American Travel, Walking Tours, World Heritage Sites, World Travel. Bookmark the permalink.


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text:
New clinic links pediatric sarcoma treatment services
Apr. 11, 2013, 8:23 AM
Scott Borinstein, M.D., left, and Jennifer Halpern, M.D., right, celebrate the debut of the Multidisciplinary Pediatric Sarcoma Clinic with patients, from left, Jennifer Simpson, J. Carlton Smith and Adrianna Crawford. (photo by John Russell)
J. Carlton Smith noticed severe pain in his knee five years ago during a lacrosse tournament. Smith, then 13 years old, never imagined hearing the diagnosis doctors would give him: osteosarcoma, a cancer of the bone.
He would get chemotherapy at the Monroe Carell Jr. Children's Hospital at Vanderbilt, and also see an orthopaedic oncology surgeon at the Orthopaedic Institute on another part of campus, plus blood work and X-rays in other locations. With a sports-career ending knee replacement, Smith and his parents were already overwhelmed.
To lessen some of the stress on patients and families, Children's Hospital launched the Multidisciplinary Pediatric Sarcoma Clinic April 4 as a one-stop location for patients with sarcoma to receive expedited therapy and more unified, convenient care. The clinic is located within the Hematology and Oncology Clinic on the sixth floor of the Doctors' Office Tower at Children's Hospital.
Jennifer Halpern, M.D., assistant professor of Orthopaedics and Rehabilitation, and Scott Borinstein, M.D., Ph.D., director of the Pediatric Sarcoma Program, joined forces to offer the joint clinic hours once a month — for now — from 1-4 p.m. As the program gets under way, they hope to expand days and hours available to patients being treated for sarcomas, including osteosarcoma, Ewing sarcoma, rhabdomyosarcoma and synovial sarcoma.
"The new clinic makes everything a lot easier. We're very fortunate to live in Nashville and have access to Vanderbilt. It definitely consolidates our time and makes it easier to get all our treatments and doctors' visits done in one location," said Smith, now an 18-year-old senior at Montgomery Bell Academy. He is in remission from his cancer.
His mother, Rebecca Smith, echoed his sentiments: "I couldn't imagine having a better experience or greater care, but the new clinic does make it easier having it all under one roof."
Until now, Halpern and Borinstein collaborated on and coordinated patient cases from their separate clinic locations – Halpern at the Orthopaedic Institute in Medical Center East and Borinstein at Children's Hospital. In addition to visiting both physicians, patients also might have scans and X-rays in different locations.
Borinstein treats the patients with sarcomas with chemotherapy, while Halpern's job is to perform surgery to resect and reconstruct the bone or soft tissues.
"Because we are a large referral center, we often have patients come from several hours away. Being able to create a clinic in which we work together is going to make it a whole lot easier for patients to get everything they need at one time," said Borinstein, assistant professor of Pediatrics in the Division of Hematology-Oncology.
"We can coordinate all their surveillance scans and their follow-up with Orthopaedic Surgery and Oncology. We can look at the X-rays in real time together and we can make a unified decision."
Fewer than 1,000 children and young adults (age 20 and younger) are diagnosed each year in the United States with sarcomas. Currently, Children's Hospital serves as a large referral center for the region, seeing about 40 active patients. Treatment lasts about six months to one year, with follow-up care for 10 years.
Jennifer Simpson, 24, an osteosarcoma survivor, was diagnosed at 19, and similar to Smith, went to multiple locations for her care. Now as a survivor for follow-up surveillance, she will be able to be seen at the new joint clinic.
"It's a lot easier, especially now that I have a job, I would have to schedule a whole day off to go to different places and have to wait hours in between. It's a lot easier, a lot faster," Simpson said.
Hyundai Hope on Wheels award boosts pediatric cancer research
Children's Hospital ED adds new rooms to speed treatment
Pediatric Pulmonary Hypertension program lauded
New clinic focuses on patients with rare ear malformations
Christina Echegaray · (615) 322-4747 ·
Reporter featured-Reporter jennifer halpern orthopaedics pediatric sarcoma Reporter April 12 2013 scott borinstein

redpajama_set_name: RedPajamaCommonCrawl


text:
NFL CBA Changes
Christopher Adams, Features Editor
The NFL Collective Bargaining Agreement (CBA) is planning on making some huge changes to future NFL seasons.
The new changes include expanding the regular season to 17 games, having 14 teams in the playoffs and changing in NFL roster sizes.
If the new rules are passed, the 17-game regular season won't go into effect this next season, but the 14 team playoff rule would start next season.
If the playoff rule was in effect last season of the NFL, the Pittsburgh Steelers and the Los Angeles Rams would have been in the playoffs as No. 7 seeds. The Super Bowl champions, the Kansas City Chiefs, would have had to play in the wild card round against the Steelers.
NFL players, including J.J. Watt, Russell Wilson, Aaron Rodgers and Richard Sherman, have expressed on social media that they don't agree with the new possible changes.
"Hard no on that proposed CBA," said Watt, a defensive end for the Houston Texans.
Not too many players have said anything positive about the new changes, but Detroit Lions linebacker Devon Kennard has.
"My vote was YES," Kennard said on Twitter.
I have a very mixed opinion on these new changes. Part of me likes the new changes but the other part of me doesn't like it.
As a huge NFL fan, it's hard to not like the new rule changes. We get to watch one extra regular season week of football and two extra wildcard games on wildcard weekend.
I can also see why some NFL players don't agree with the changes. It forces them to play an extra regular season game with no extra bye week and that is probably the reason why some players are voting no. I'm sure players are also concerned about getting injured in the extra game.
In my opinion, I hope the changes get passed and approved because it would be interesting to see an extra regular season game and an extra team in the playoffs. I won't be mad if the changes don't happen though. I'll still watch every game of the NFL that I can and still enjoy them.
Are you going out of town for spring break?
Yes (21%, 4 Votes)
Bonner Springs High School
Topeka West
Mar 3 / Varsity Boy's Basketball
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Jan 31 / Varsity Boy's Basketball
Keep up with BSHS
Tweets by bshspowwow
Athlete of the Week: Reyven McDonald
Braves swim team advances to State
Athlete of the Week: Rishaud Dockery
Athlete of the Week: Corrie Hassett
Athlete of the Week: Landen Hill
Gallery: Wrestling Senior Night
Keys to a Chiefs victory
Who will prevail? The Chiefs or Texans?
Small-town athletes make it big
Anthony Davis named new JV coach for boys basketball


text:
package org.aries.common.entity;

import java.util.ArrayList;
import java.util.Date;
import java.util.List;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import javax.persistence.Table;
import javax.persistence.Temporal;
import javax.persistence.TemporalType;
import org.aries.common.entity.EmailAccountEntity;
import org.aries.common.entity.EmailBoxEntity;
import org.aries.common.entity.EmailMessageEntity;
import org.hibernate.annotations.Cache;
import org.hibernate.annotations.CacheConcurrencyStrategy;

@Entity(name = "EmailBox")
@Table(name = "email_box")
@Cache(usage = CacheConcurrencyStrategy.READ_WRITE)
public class EmailBoxEntity {

	@Id
	@GeneratedValue
	private Long id;

	@Column(name = "type")
	private String type;

	@Column(name = "name")
	private String name;

	@Cache(usage = CacheConcurrencyStrategy.READ_WRITE)
	private EmailAccountEntity emailAccount;

	@Cache(usage = CacheConcurrencyStrategy.READ_WRITE)
	private EmailBoxEntity parentBox;

	@Cache(usage = CacheConcurrencyStrategy.READ_WRITE)
	private List<EmailMessageEntity> messageList;

	@Column(name = "creation_date")
	@Temporal(TemporalType.TIMESTAMP)
	private Date creationDate;

	@Column(name = "last_update")
	@Temporal(TemporalType.TIMESTAMP)
	private Date lastUpdate;

	public Long getId() {
		return id;
	}

	public void setId(Long id) {
		this.id = id;
	}

	public String getType() {
		return type;
	}

	public void setType(String type) {
		this.type = type;
	}

	public String getName() {
		return name;
	}

	public void setName(String name) {
		this.name = name;
	}

	public EmailAccountEntity getEmailAccount() {
		return emailAccount;
	}

	public void setEmailAccount(EmailAccountEntity emailAccount) {
		this.emailAccount = emailAccount;
	}

	public EmailBoxEntity getParentBox() {
		return parentBox;
	}

	public void setParentBox(EmailBoxEntity parentBox) {
		this.parentBox = parentBox;
	}

	public List<EmailMessageEntity> getMessageList() {
		if (messageList == null)
			messageList = new ArrayList<EmailMessageEntity>();
		return messageList;
	}

	public void setMessageList(List<EmailMessageEntity> messageList) {
		this.messageList = messageList;
	}

	public Date getCreationDate() {
		return creationDate;
	}

	public void setCreationDate(Date creationDate) {
		this.creationDate = creationDate;
	}

	public Date getLastUpdate() {
		return lastUpdate;
	}

	public void setLastUpdate(Date lastUpdate) {
		this.lastUpdate = lastUpdate;
	}

	@Override
	public boolean equals(Object object) {
		if (object == null)
			return false;
		if (!object.getClass().isAssignableFrom(this.getClass()))
			return false;
		EmailBoxEntity other = (EmailBoxEntity) object;
		if (this.getId() == null || other.getId() == null)
			return this == other;
		if (this.getId().equals(other.getId()))
			return true;
		return false;
	}

	@Override
	public int hashCode() {
		if (getId() != null)
			return getId().hashCode();
		return 0;
	}

}

redpajama_set_name: RedPajamaGithub


text:
\section{Introduction}

The \textit{Schottky problem} \cite{gru} concerns the characterization of Jacobians of genus $g$ curves
 among all abelian varieties of dimension $g$. The latter are parametrized by the
 \textit{Siegel upper-half space} $\mathfrak{H}_g$, i.e.~the set of complex symmetric $g\times g$
 matrices $\tau$ with positive definite imaginary part. The {\em Schottky locus}
 $ \mathfrak{J}_g$ is the subset of matrices $\tau$ in $\mathfrak{H}_g$ that represent Jacobians.
 Both sets are complex analytic spaces whose dimensions reveal that the inclusion is proper for $g \geq 4$:
\begin{equation}
\label{eq:dimensions}
  {\rm dim}(\mathfrak{J}_g) \,= \, 3g -3 \qquad \hbox{and} \qquad
 {\rm dim}(\mathfrak{H}_g) \,= \, \binom{g+1}{2}.
 \end{equation}
 For $g=4$, the dimensions in (\ref{eq:dimensions}) are $9$ and $10$,
 so $\mathfrak{J}_4$ is an analytic hypersurface in $\mathfrak{H}_4$.
 The equation defining this hypersurface is a polynomial of degree $16$ in
 the theta constants. First constructed by Schottky \cite{Schottky1888},
and further developed by Igusa \cite{igusa}, this modular form
embodies the theoretical solution (cf.~\cite[\S 3]{gru}) to the classical Schottky problem for $g=4$.

The Schottky problem also exists in tropical geometry \cite{mz}.
The {\em tropical Siegel space} $\mathfrak{H}_g^{\rm trop}$ is the
cone of positive definite $g \times g$-matrices, endowed with the fan structure given by the
second Voronoi decomposition. The {\em tropical Schottky locus}
$\mathfrak{J}_g^{\rm trop}$ is the subfan indexed by cographic matroids \cite[Theorem 5.2.4]{BMV}.
A detailed analysis for $g\leq 5$ is found in \cite[Theorem 6.4]{chan12}.
It is known, e.g.~by~\cite[\S 6.3]{BMV}, that the inclusion
$\mathfrak{J}_g^{\rm trop} \subset \mathfrak{H}_g^{\rm trop}$ correctly
tropicalizes the complex-analytic inclusion
$\mathfrak{J}_g \subset \mathfrak{H}_g$. However,
 it has been an open~problem (suggested in \cite[\S 9]{RSSS}) to find
a direct link between the equations that govern these two inclusions.

We here solve this problem, and
develop computational tools for the Schottky problem,
both classically and tropically.
We distinguish between the {\em Schottky Decision Problem}
and the {\em Schottky Recovery Problem}. For the former, the input
is a matrix $\tau$ in $\mathfrak{H}_g$ resp.~$\mathfrak{H}_g^{\rm trop}$,
possibly depending on parameters, and we must decide whether
$\tau$ lies in $\mathfrak{J}_g$ resp.~$\mathfrak{J}_g^{\rm trop}$.
For the latter, $\tau$ already passed that test, and we
compute a curve whose Jacobian is given by~$\tau$.
The recovery problem also makes sense for $g=3$,
both classically \cite{bitangents} and tropically~\cite[\S 7]{BBC}.

This paper is organized as follows. In Section \ref{sec:classic} we tackle the
classical Schottky problem as a task in
numerical algebraic geometry \cite{DvH, HS, SD}. For $g=4$,
we utilize the software {\tt abelfunctions} \cite{abelfunctions} to
test whether the Schottky--Igusa modular form vanishes. In the affirmative case,
we use a numerical version of Kempf's method \cite{kempf} to compute a
canonical embedding into $\mathbb{P}^3$.
Our main results in Section \ref{sec:tropical}  are Algorithms \ref{alg:TSD} and \ref{alg:TSR}.
Based on the work in \cite{sikiric, sgsw, frank, voronoi},
these furnish a  computational solution to the tropical Schottky problem.
Key ingredients are  cographic matroids and the f-vectors of Voronoi polytopes.

Section \ref{sec:four} links the classical and tropical Schottky scenarios.
Theorem \ref{thm:recoverfromtheta} expresses the edge lengths of a
metric graph in terms of tropical theta constants, and
Theorem~\ref{thm:admissible} explains what happens to the
Schottky--Igusa modular form in the tropical limit.
We found it especially gratifying to discover how the cographic locus
is encoded in the classical theory.

\smallskip

The software we describe in this paper is made available at the supplementary website
\begin{equation}
\label{eq:url}
 \hbox{\url{http://eecs.berkeley.edu/~chualynn/schottky}}
 \end{equation}
This contains several pieces of code for the tropical Schottky problem,
as well as a more coherent \texttt{Sage} program
for the classical Schottky problem that makes calls to
{\tt abelfunctions}.

\section{The Classical Schottky Problem}\label{sec:classic}

We fix $g=4$, and review theta functions and
Igusa's construction \cite{igusa} of the equation that
cuts out~$\mathfrak{J}_4$.
For any vector $m\in\mathbb{Z}^{8}$ we write
 $m= (m',m'') $ for suitable $m',m''\in\mathbb{Z}^4$.
  The \textit{Riemann theta function}
with characteristic $m$~is the following function of  $\tau\in\mathfrak{H}_4$ and
$z\in\mathbb{C}^4$:
\begin{align}\label{eqn:thetafunc}
 \theta[m](\tau,z)\,\,=\,\,\,\sum_{n\in\mathbb{Z}^4}\exp\left[\pi \textnormal{i} (n+\frac{m'}{2})^t \tau (n+\frac{m'}{2})+2\pi\textnormal{i}(n+\frac{m'}{2})^t(z+\frac{m''}{2})\right].
\end{align}
For numerical computations of the theta function one has to make a good choice of lattice points to sum over in order for this series to converge rapidly  \cite{DHBvHS, SD}.
We use the software {\tt abelfunctions} \cite{abelfunctions}
to evaluate $\theta[m]$ for arguments $\tau$ and $z$
with floating point coordinates.

Up to a global multiplicative factor,
the definition (\ref{eqn:thetafunc}) depends only on the image of $m$ in
$ (\mathbb{Z}/2\mathbb{Z})^8$.
The sign of the characteristic $m$ is $\,e(m)=(-1)^{(m')^t m''}$.
Namely, $m$ is \textit{even} if $e(m)=1$ and \textit{odd} if $e(m) = -1$.
A triple $\{m_1,m_2,m_3\}\subset (\mathbb{Z}/2\mathbb{Z})^8$
 is called \textit{azygetic} if $e(m_1)e(m_2)e(m_3)e(m_1+m_2+m_3) = -1$.
Suppose that this holds. Then we choose a rank $3$ subgroup $N$ of $
(\mathbb{Z}/2\mathbb{Z})^8$ such that all elements of
 $(m_1{+}N) \cup (m_2{+}N) \cup (m_3{+}N)$ are even.

We  consider the following three products of eight {\em theta constants} each:
\begin{equation}
\label{eq:pidef}
  \pi_i \,\,\,\,=\prod_{m\in m_i+N} \! \theta[m](\tau,0) \qquad
  \hbox{for} \quad i=1,2,3.
\end{equation}

\begin{theorem}[Igusa \cite{igusa}]\label{thm:igusa}
 The function $\mathfrak{H}_4 \rightarrow \mathbb{C}$ that takes
 a symmetric $4 {\times} 4$-matrix $\tau$ to
\begin{equation}
\label{eq:schottkyigusa} \pi_1^2+\pi_2^2+\pi_3^2-2\pi_1\pi_2-2 \pi_1\pi_3-2 \pi_2\pi_3
\end{equation}
is independent of the choices above. It vanishes
if and only if $\tau $ lies in the closure of the Schottky locus $\mathfrak{J}_4$.
\end{theorem}

We refer to the expression (\ref{eq:schottkyigusa}) as the {\em Schottky--Igusa modular form}.
This is a polynomial of degree $16$ in the theta constants  $\theta[m](\tau,0)$. Of course, the
formula is unique only modulo the ideal that defines the embedding of the moduli space $\mathcal{A}_4$ in
the $\mathbb{P}^{15}$ of theta constants.

Our implementation uses the polynomial that is given by the following specific choices:
$$
 m_1=\begin{pmatrix}1 & 1 \\ 0 & 0 \\ 1 & 1 \\ 0 & 0     \end{pmatrix}\!,\,
m_2=\begin{pmatrix} 0 & 1 \\ 0 & 0 \\ 0 & 0 \\ 1 & 0    \end{pmatrix}\!,\,
m_3=\begin{pmatrix} 0 & 1 \\ 0 & 0 \\ 1 & 1 \\ 1 &  1  \end{pmatrix}\!,\,\,
  n_1=\begin{pmatrix} 0 & 1 \\ 0 & 1 \\ 0 &1 \\ 1 & 0  \end{pmatrix}\!,\,
  n_2=\begin{pmatrix} 0 & 0 \\ 0 & 0 \\ 1 & 0 \\ 1 & 1   \end{pmatrix}\!,\,
  n_3=\begin{pmatrix} 0 & 1 \\ 0 & 0  \\ 1 & 1\\ 0 & 1 \end{pmatrix}\!.
$$
The vectors $n_1,n_2,n_3$ generate the subgroup $N$ in $ (\mathbb{Z}/2\mathbb{Z})^8$.
One checks that the triple $\{m_1,m_2,m_3\}$ is azygetic
and that the three cosets $m_i + N$ consist of even elements only.
The computations to be described next were done with the {\tt Sage} library
{\tt abelfunctions} \cite{abelfunctions}.

The algorithm in \cite{DvH} finds the Riemann matrix $\tau \in \mathfrak{J}_g$ of a
plane curve in  $\mathbb{C}^2$. It is implemented in {\tt abelfunctions}.  We first check
that (\ref{eq:schottkyigusa})  does indeed vanish for such~$\tau$.

\begin{example}\label{exp:schottkycurve} \rm
The plane curve $\,y^5+x^3-1 = 0\,$ has genus four. Its
  Riemann matrix $\tau$~is
 \[
  \begin{pmatrix}   \,\,0.16913 + 1.41714\textnormal{i}  &  -0.81736 - 0.25138\textnormal{i} & \! -0.05626 - 0.44830\textnormal{i}  & \,\,0.24724 + 0.36327\textnormal{i}\\
\! -0.81736 - 0.25138\textnormal{i}  & -0.31319 + 0.67096\textnormal{i} &\! -0.02813 - 0.57155\textnormal{i}  & \,\,0.34132 + 0.40334\textnormal{i}\\
\!-0.05626 - 0.44830\textnormal{i} & -0.02813 - 0.57155\textnormal{i}  &  \,0.32393 + 1.44947\textnormal{i}  & -0.96494 - 0.63753\textnormal{i} \\
  \, \, 0.24724 + 0.36327\textnormal{i} &  \,\,\, 0.34132 + 0.40334\textnormal{i} & \! \! -0.96494 - 0.63753\textnormal{i}   & \,\,0.62362 + 0.73694\textnormal{i} \end{pmatrix} \! .
 \]
Evaluating the $16$ theta constants $\theta[m](\tau,0)$ numerically with {\tt abelfunctions}, we find that
\begin{eqnarray*}
 \pi_1^2+\pi_2^2+\pi_3^2&=&-5.13472888270289 + 6.13887870578982\textnormal{i},  \\
 2(\pi_1\pi_2+\pi_1\pi_3+\pi_2\pi_3)&=&-5.13472882638710 + 6.13887931435788\textnormal{i}.
\end{eqnarray*}
We trust that (\ref{eq:schottkyigusa}) is zero, and conclude that
$\tau$ lies in the Schottky locus $\mathfrak{J}_4$, as expected.
\end{example}

Suppose now that we are given a matrix  $\tau$ that depends on one or two parameters,
so it traces out a curve or surface in $\mathfrak{H}_4$. Then we can use our numerical
method to determine the Schottky locus  inside that curve or surface. Here is an illustration
for a surface in $\mathfrak{H}_4$.

 \begin{example}\label{exp:shimuracurve} \rm
 The following one-parameter family of genus $4$ curves is found in \cite[\S 2]{GM}:
 $$ y^6 \,\,=\,\, x(x+1)(x-t). $$
 This is both a Shimura curve and a Teichm\"uller curve.
Its Riemann matrix is  $\rho(t) = Z_2^{-1} Z_1$ where
$Z_1,Z_2$ are given in \cite[Prop.~6]{GM}.
Consider the following two-parameter family  in $\mathfrak{H}_4$:
\begin{equation}
\label{eq:taufamily}
 \tau(s,t)  \,= \, s \cdot {\rm diag}(2,3,5,7) \,\,+\,\, \rho(t).
\end{equation}
We are interested in the restriction of
the Schottky locus $\mathfrak{J}_4$ to the $(s,t)$-plane.
For our experiment, we assume that the two parameters satisfy
 $s \in [-0.5,0.5]$ and $ \lambda^{-1}(t) \in [\,\textnormal{i},\textnormal{i}+1]$,
 where $\lambda$ is the function  in \cite[Prop.~6]{GM}.
Using {\tt abelfunctions}, we computed the absolute value of
the modular form (\ref{eq:schottkyigusa})
at 6400 equally spaced rational points in the square
$[-0.5,0.5] \times [i,i+1]$. That graph
 is shown in Figure \ref{eq:taufamily}.
For $s$ different from zero, the smallest absolute value of
(\ref{eq:schottkyigusa}) is $4.3 \times 10^{-3}$.
For $s=0$, all absolute values are below $ 2.9 \times 10^{-8}$.
Based on this numerical evidence, we conclude that the
Schottky locus of our family is the line $s=0$.
\end{example}

\begin{figure}[h]
  \begin{center}
\includegraphics[height=8.4cm]{shimura.pdf}
\vspace{-0.2in}
\end{center}    \caption{\label{fig:eins}
Absolute value of the Schottky--Igusa modular form
on the 2-parameter family~(\ref{eq:taufamily}).
}
 \end{figure}

We now come to the Schottky Recovery Problem. Our input is a matrix
$\tau$ in $ \mathfrak{J}_4$. Our task is to compute a curve whose
Riemann matrix equals $\tau$. We use the following result
from Kempf's paper \cite{kempf}.
The {\em theta divisor} in the Jacobian
$\, \mathbb{C}^4/(\mathbb{Z}^4 + \mathbb{Z}^4 \tau )\,$
is the zero locus $\Theta^{-1}(0)$ of the Riemann theta function
$\,\Theta(z) := \theta[0](\tau,z)$. For generic $\tau$ this divisor is singular at
precisely two points. These represent $3$-to-$1$ maps from
the curve to $\mathbb{P}^1$. We compute a vector $z^* \in \mathbb{C}^4$ that is
a singular point of $\Theta^{-1}(0)$ by solving the system of five equations
\begin{equation}
\label{eq:findsingular}
\Theta(z) \, = \, \frac{\partial \Theta}{\partial z_1}(z) =
 \frac{\partial \Theta}{\partial z_2}(z) =
  \frac{\partial \Theta}{\partial z_3}(z) =
   \frac{\partial \Theta}{\partial z_4}(z) =  0.
\end{equation}
The Taylor series of the Riemann theta function
$\Theta$ at the singular point $z^*$ has the form
\begin{equation}
\label{eq:taylorseries}
\Theta(z^*+x) \,= \, f_2(x) \,+\, f_3(x) \, + \,f_4(x) \,+ \,\,\hbox{higher order terms},
\end{equation}
where $f_s$ is a homogeneous polynomial of degree $s$ in $ x = (x_1,x_2,x_3,x_4)$.

\begin{proposition}[Kempf \cite{kempf}]
The canonical curve with Riemann matrix $\tau$ is the
degree $6$ curve in $\mathbb{P}^3$ that is defined by the quadratic equation $f_2 = 0$
and the cubic equation $ f_3 = 0$.
\end{proposition}

Thus our algorithm for the Schottky Recovery Problem consists of solving the
five equations (\ref{eq:findsingular}) for $z^* \in \mathbb{C}^4 $, followed by
extracting the polynomials $f_2$ and $f_3$ in the Taylor series (\ref{eq:taylorseries}). Both of
these steps can be done numerically using the software {\tt abelfunctions} \cite{abelfunctions}.

\begin{example}  \rm
Let $\tau \in \mathfrak{J}_4$ be the Riemann matrix of the genus $4$ curve $C = \{x^3 y^3 +x^3 + y^3 = 1\}$.
We obtain $\tau$ numerically using {\tt abelfunctions}. We want to recover $C$ from $\tau$. To be precise,
given only $\tau$,  we want to find defining equations
$f_2 = f_3 = 0$ in $\mathbb{P}^3$ of the canonical embedding of $C$. For that we use evaluations of $\Theta(z)$ and its derivatives in
{\tt abelfunctions}, combined with a numerical optimization routine in {\tt SciPy} \cite{spicy}.
We solve the equations (\ref{eq:findsingular}) starting from random
points $z = u + \tau v$ where $u,v \in \mathbb{R}^4$
with entries between $0$ and $1$. After several tries,
the local method in {\tt SciPy} converges to the following solution of our equations:
$$ z^* = \bigl(0.55517+0.69801\textnormal{i},  0.53678+0.26881\textnormal{i}, -0.50000-0.58958\textnormal{i},  0.55517+0.69801\textnormal{i}\bigr) . $$
Using (\ref{eq:taylorseries}), we computed the quadric $f_2$, which is nonsingular, as well as the cubic $f_3$:
$$
\begin{small}
 \begin{matrix}
  f_2(x) & =&
(-3.044822827 + 21.980542613 \textnormal{i})\cdot x_1^2   +( - 237.95207224  + 252.54744634 \textnormal{i})\cdot x_1 x_2 \\   & &
\,+(- 222.35552015  + 139.95612952 \textnormal{i})\cdot x_1 x_3   +(- 200.66932133  - 16.596272620 \textnormal{i})\cdot x_1 x_4 \\  & &
  + (- 191.16241727 - 85.22650070 \textnormal{i})\cdot x_2^2   + (- 429.11449060 + 167.32094535 \textnormal{i})\cdot x_2 x_3 \\  & &
+(  - 237.952072 + 252.54744632 \textnormal{i})\cdot x_2 x_4   +(- 206.75896934 + 27.364814282 \textnormal{i})\cdot x_3^2 \\ & &
+(  222.35552013 + 139.95612953 \textnormal{i})\cdot x_3 x_4   + (- 3.0448227745 + 21.980542601 \textnormal{i})\cdot x_4^2
 \end{matrix}
  \end{small} $$ $$ \begin{small}
\begin{matrix}
f_3(x) \quad =  \quad
(441.375966486 + 61.14097461986 \textnormal{i})\cdot x_1^3   +(  2785.727151434 + 2303.609067429 \textnormal{i})\cdot x_1^2 x_2 \\
+ \quad \cdots\,\,\cdots \quad   + \quad ( 441.3759668263 + 61.14097402189 \textnormal{i})\cdot x_4^3.
\end{matrix}
\end{small}
$$
As a proof of concept we also computed the $120$ tritangent planes numerically directly from $\tau$.
    These planes are indexed by the $120$ odd theta characteristics $m$.
    In analogy to the computation  in \cite[\S 5.2]{SD} of the $28$ bitangents for $g=3$,
    their defining equations are
$$
\frac{\partial \theta[m](\tau,z)}{\partial z_1}\biggl|_{z=0} \!\!\! \cdot \,x_1 \,\, + \,\,
\frac{\partial \theta[m](\tau,z)}{\partial z_2}\biggl|_{z=0} \!\!\! \cdot \,x_2 \,\, + \,\,
\frac{\partial \theta[m](\tau,z)}{\partial z_3}\biggl|_{z=0} \!\!\! \cdot \,x_3 \,\, + \,\,
\frac{\partial \theta[m](\tau,z)}{\partial z_4}\biggl|_{z=0} \!\!\! \cdot \,x_4 \,\,\, = \,\,\, 0.
$$
We verified numerically that each such plane
meets $\{f_2 = f_3 = 0\}$ in three double points.
\end{example}

\begin{remark} \rm
On our website (\ref{eq:url}), we offer a program in {\tt Sage} whose input is
a symmetric $4 \times 4$-matrix $\tau \in \mathfrak{H}_4$, given numerically.
The code decides whether $\tau $ lies in $\mathfrak{J}_4$ and, in the affirmative case, it computes
the canonical curve $\{f_2 = f_3 = 0\}$ and its 120 tritangent planes.
\end{remark}

\section{The Tropical Schottky Problem}\label{sec:tropical}

Curves, their Jacobians, and the Schottky locus have
natural counterparts in the combinatorial setting of tropical geometry.
We review the basics from \cite{BBC, BMV, chan12, mz}.
The role of a curve is played by a connected metric graph $\Gamma = (V,E,l,w)$.
This has vertex set $V$, edge set $E$, a length function $l: E \rightarrow \mathbb{R}_{> 0}$,
and a weight function $w : V \rightarrow \mathbb{Z}_{\geq 0}$.
The genus of $\Gamma$ is
\begin{equation}
\label{eq:graphgenus}
g \,\,= \,\, |E| -|V|  +1 \,+\, \sum_{v\in V} w(v).
\end{equation}
The moduli space $\mathcal{M}_g^{\rm trop}$ comprises all metric graphs of genus $g$.
This is a stacky fan of dimension $3g-3$.
See \cite[Figure 4]{chan12} for a colorful illustration.
The tropical Torelli map $\,\mathcal{M}_g^{\rm trop} \rightarrow \mathfrak{H}_g^{\rm trop}\,$
takes $\Gamma$ to its (symmetric and positive semidefinite)  Riemann matrix $Q_\Gamma$.

Fix a basis for the integral homology $H_1(\Gamma,\mathbb{Z}) \simeq \mathbb{Z}^g$.
Beside the usual cycles in $\Gamma$, this group has $w(v)$
generators for the virtual cycles at each vertex $v$.
Let $B$ denote the $g \times |E|$ matrix whose columns
record the coefficients of each edge in the basis vectors.
Let $D$ be the $|E| \times |E|$ diagonal matrix whose entries are the edge lengths.
The {\em Riemann matrix} of $\Gamma$~is
\begin{equation}
\label{eq:tropicalriemann} Q_\Gamma \,\, = \,\, B \cdot D \cdot B^t.
\end{equation}
One way to choose a basis is to fix an orientation
and a spanning tree of $\Gamma$.
Each edge not in that tree then determines a cycle
with $\pm 1$-coefficients.
See \cite[\S 4]{BBC} for details and an example.
Changing the basis of $H_1(\Gamma,\mathbb{Z})$
corresponds to the action of ${\rm GL}_g(\mathbb{Z})$ on
$Q_\Gamma$ by conjugation.

The matrix $Q_\Gamma$ has rank $g - \sum_{v\in V} w(v)$.
We defined $\mathfrak{H}_g^{\rm trop}$ with positive definite matrices.
Those have rank $g$.
For that reason, we now
restrict to graphs with
zero weights, i.e.~$w \equiv 0$.

The {\em tropical Schottky locus}  $\mathfrak{J}_g^{\rm trop}$
is the set of all matrices (\ref{eq:tropicalriemann}),
where $\Gamma = (V,E,l)$ runs over graphs
of genus $g$, and $B$ runs over their cycle bases.
This set is known as the {\em cographic locus}
in $\mathfrak{H}_g^{\rm trop}$, because
the $g \times |E|$ matrix $B$ is a representation
of the {\em cographic matroid} of~$\Gamma$.

The Schottky Decision Problem asks for a test of
membership in $\mathfrak{J}_g^{\rm trop}$. To be precise,
given a positive definite matrix $Q$, does there exist
a metric graph $\Gamma$ such that $Q = Q_\Gamma$?

To address this question, we need the polyhedral fan structures
on $\mathfrak{J}_g^{\rm trop}$ and  $\mathfrak{H}_g^{\rm trop}$.
Let $G = (V,E)$ be the graph underlying $\Gamma$,
with $E = \{e_1,e_2,\ldots,e_m\}$. Fix a cycle basis as above.
Let $b_1,b_2,\ldots,b_m$ be the column vectors of the $g \times m$-matrix $B$.
Formula (\ref{eq:tropicalriemann}) is equivalent~to
\begin{equation}
\label{eq:tropicalriemann2}
Q_\Gamma \,\,=\,\, l(e_1) b_1b_1^t + l(e_2) b_2 b_2^t +  \cdots +  l(e_m)b_m b_m^t.
 \end{equation}
The cone of all Riemann matrices for the graph $G$, allowing the edge lengths to vary, is
\begin{equation}  \label{eqn-cone}
\sigma_{G,B} \,\,=\,\, \mathbb{R}_{>0} \bigl\{ b_1b_1^t, \,b_2 b_2^t, \,\ldots\,,\, b_m b_m^t\bigr\}\,.
\end{equation}
This is a relatively open rational convex polyhedral cone,
 spanned by matrices of rank $1$.
The collection of all cones $\sigma_{G,B}$ is a polyhedral fan whose support
is the Schottky locus $\mathfrak{J}_g^{\rm trop}$.

This fan is a subfan of the
{\em second Voronoi decomposition} of the cone $\mathfrak{H}_g^{\rm trop}$ of positive
definite matrices.
 The latter fan is defined as follows. Fix
a Riemann matrix $Q \in \mathfrak{H}_g^{\rm trop}$ and
consider its quadratic form $\,\mathbb{Z}^g \rightarrow \mathbb{R},\,
x \mapsto x^t Q x$. The values of this quadratic form
define a regular polyhedral subdivision of $\mathbb{R}^g$
with vertices at $\mathbb{Z}^g$. This is denoted ${\rm Del}(Q)$ and
known as the {\em Delaunay subdivision} of $Q$.
Dual to ${\rm Del}(Q)$ is the {\em Voronoi decomposition} of
$\mathbb{R}^g$. The cells of the Voronoi decomposition of $Q$ are
the lattice translates of the \emph{Voronoi polytope}
\begin{equation}
\label{eq:voronoipolytope}
 \bigl\{\,  p\in\mathbb{R}^g \,:\, 2 p^t Q x \leq x^t Q x
 \,\,\,\hbox{for all} \,\,x\in\mathbb{Z}^g \,\bigr\}.
 \end{equation}
This is the set of points in $\mathbb{R}^g$ for which the origin is the closest lattice point,
in the norm given by $Q$.
If $Q$ is generic then
the Delaunay subdivision is a triangulation
and the Voronoi polytope (\ref{eq:voronoipolytope}) is simple.
It is dual to the link of the origin in the simplicial complex ${\rm Del}(Q)$.

The structures above represent principally polarized abelian varieties
in tropical geometry. A tropical abelian variety is the torus
$\mathbb{R}^g/\mathbb{Z}^g$ together with a quadratic form $Q \in \mathfrak{H}_g^{\rm trop}$.
The {\em tropical theta divisor} is given by the  codimension one cells in the
induced Voronoi decomposition of $\mathbb{R}^g/\mathbb{Z}^g$.
See \cite[\S 5]{BBC} for an introduction with many pictures and many references.

We now fix an arbitrary Delaunay subdivision $D$ of $\mathbb{R}^g$.
Its {\em secondary cone} is defined as
\begin{align} \label{def-sec-cone}
\sigma_D \,\,=\,\, \bigl\{ \,Q\in \mathfrak{H}_g^{\rm trop}
\,\,|\,\, \Del(Q) = D \bigr\}\,.
\end{align}
This is a relatively open convex polyhedral cone. It consists of
positive definite matrices $Q$ whose Voronoi polytopes (\ref{eq:voronoipolytope})
have the same normal fan. The group ${\rm GL}_g(\mathbb{Z})$ acts on the
set of secondary cones. In his classical reduction theory for quadratic forms,
Voronoi \cite{voronoi} proved that the cones $\sigma_D$ form a polyhedral fan,
now known as the {\em second Voronoi decomposition} of $\mathfrak{H}_g^{\rm trop}$,
and that there are only finitely many secondary cones $\sigma_D$ up to the action
of ${\rm GL}_g(\mathbb{Z})$. The following summarizes  characteristic features for matrices
in the Schottky locus $\mathfrak{J}_g^{\rm trop}$.

\begin{proposition} \label{prop:features}
Fix a graph $G$ with metric $D$,
homology basis $B$, and Riemann matrix  $Q = BDB^t$.
The Voronoi polytope (\ref{eq:voronoipolytope}) is affinely isomorphic to the
zonotope $\sum_{i=1}^m [-b_i,b_i]$.
The secondary cone $\sigma_{{\rm Del}(Q)}$ is spanned
 by the rank one matrices $b_i b_i^t$: it equals $\sigma_{G,B}$ in
(\ref{eqn-cone}).
\end{proposition}

\begin{proof}
This can be extracted from Vallentin's thesis \cite{frank}.
The affine isomorphism is given by the invertible matrix
$Q$, as explained in item iii) of \cite[\S 3.3.1]{frank}.
The Voronoi polytope being the zonotope
 $\sum_{i=1}^m [-b_i,b_i]$ follows from
the discussion on cographic lattices in \cite[\S 3.5]{frank}.
The result for the secondary cone is derived from \cite[\S 2.6]{frank}.
See \cite[\S 4]{frank} for many examples.
\end{proof}

We now fix $g=4$. Vallentin \cite[\S 4.4.6]{frank} lists  all $52$ combinatorial types of Delaunay subdivisions
of $\mathbb{Z}^4$. His table contains the f-vectors of all $52$ Voronoi polytopes. Precisely
$16$ of these types are cographic, and these comprise the Schottky locus $\mathfrak{J}_4^{\rm trop}$.
These are described in rows 3 to 18 of the table in \cite[\S 4.4.6]{frank}. We reproduce
the relevant data in Table \ref{table-trop-schottky}. The following key lemma is found by
inspecting Vallentin's list of f-vectors.

\begin{lemma} The f-vectors of the $16$ Voronoi polytopes representing the Schottky locus
$\mathfrak{J}_4^{\rm trop}$ are distinct from the f-vectors of the other $36$ Voronoi polytopes,
corresponding to $\mathfrak{H}_4^{\rm trop} \backslash \mathfrak{J}_4^{\rm trop}$.
\end{lemma}

This lemma gives rise to the following method for the tropical Schottky decision  problem.

\begin{algorithm}[Tropical Schottky Decision]   \label{alg:TSD}
\underbar{Input}: $Q \in \mathfrak{H}_4^{\rm trop}\!$.
\underbar{Output}: Yes, if $\,Q \in \mathfrak{J}_4^{\rm trop}\!$. \\
1. Compute the Voronoi polytope in (\ref{eq:voronoipolytope}) for the quadratic form $Q$. \\
2. Determine the f-vector $(f_0,f_1,f_2,f_3)$ of this $4$-dimensional polytope. \\
3. Check whether this f-vector appears in our Table \ref{table-trop-schottky}.
Output ``Yes'' if this holds.
\end{algorithm}

\begin{longtable}{|c|c|c|c|c|c|c|}
  \hline
Graph $G$ & Riemann matrix $Q_\Gamma$ & $f_0$ & $f_1$ &
$f_2$ & $f_3$ & Dimension of $\sigma_D$ \\
\hline \rule{0pt}{1.4\normalbaselineskip}
\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}] \filldraw (0,0) circle (1pt) (0,1) circle (1pt)
  (0.5,0.5) circle (1pt) (1,0.5) circle (1pt) (1.5,0) circle (1pt) (1.5,1) circle (1pt); \draw (0,0)--(0,1)--(0.5,0.5)--(0,0) (0,1)--(1.5,1)--(1.5,0)--(0,0) (0.5,0.5)--(1,0.5) (1.5,0)--(1,0.5)--(1.5,1);
\end{tikzpicture}&
$\begin{psmallmatrix}3&1&-1&0\\1&4&1&1\\-1&1&4&-1\\0&1&-1&3\end{psmallmatrix}$
  & 96 & 198 & 130 & 28 & 9
\\[4mm] \begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
  \filldraw (0,0) circle (1pt) (0,1) circle (1pt) (1,0) circle (1pt) (2,0) circle (1pt)
  (1,0) circle (1pt) (1,1) circle (1pt) (2,1) circle (1pt);\draw
  (0,0)--(0,1) (0,0)--(1,1) (0,0)--(2,1) (1,0)--(0,1) (1,0)--(1,1) (1,0)--(2,1) (2,0)--(0,1) (2,0)--(1,1) (2,0)--(2,1);
\end{tikzpicture}& $\begin{psmallmatrix}4&2&-2&-1\\2&4&-1&-2\\-2&-1&4&2\\-1&-2&2&4\end{psmallmatrix}$
  & 102 & 216 & 144 & 30 & 9 \\[4mm] \hline \rule{0pt}{1.4\normalbaselineskip}
\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
\filldraw (0,0) circle (1pt) (0,1) circle (1pt) (1,0) circle (1pt)
(1,1) circle (1pt) (0.5,0.5) circle (1pt);
\draw (0,0)--(1,0) (1,1)--(0,1)--(0,0) (0,0)--(0.5,0.5)--(1,1)
(0.5,0.5)--(0,1) (1,0) to [bend left] (1,1) (1,0) to [bend right] (1,1);
\end{tikzpicture}
& $\begin{psmallmatrix}2&0&-1&0\\0&3&1&1\\-1&1&4&-1\\0&1&-1&3\end{psmallmatrix}$ & 72 & 150 & 102 & 24 & 8 \\[4mm]
\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
\filldraw (0,0) circle (1pt) (0,1) circle (1pt) (1,0) circle (1pt)
(1,1) circle (1pt) (0.5,0.5) circle (1pt);
\draw (0,0)--(1,0)--(1,1)--(0,1)--(0,0) (0,0)--(0.5,0.5)--(1,1) (1,0)--(0.5,0.5)--(0,1);
\end{tikzpicture}
& $\begin{psmallmatrix}3&2&1&-1\\2&4&2&-1\\1&2&4&1\\-1&-1&1&3\end{psmallmatrix}$ & 78 & 168 & 116 & 26 & 8 \\[4mm] \hline\rule{0pt}{1.4\normalbaselineskip}
\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
\filldraw (0,0) circle (1pt) (0,1) circle (1pt) (1,0) circle (1pt)
(1,1) circle (1pt);
\draw (0,0)--(1,0) (0,0)--(0,1) (0,0)--(1,1) (1,0) to [bend left]
(1,1) (1,0) to [bend right] (1,1) (0,1)--(1,1) (0,1)--(1,0);
\end{tikzpicture}
& $\begin{psmallmatrix}3&1&-1&-1\\1&3&1&1\\-1&1&3&2\\-1&1&2&3\end{psmallmatrix}$ & 60 & 134 & 98 & 24 & 7 \\[4mm]
 \begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
\filldraw (0,0) circle (1pt) (0,1) circle (1pt) (1,0) circle (1pt)
(1,1) circle (1pt);
\draw (0,0)--(1,0) (1,0) to [bend left] (1,1) (1,0) to [bend right]
(1,1) (0,1) to [bend left] (1,1) (0,1) to [bend right] (1,1)
(0,0) to [bend left] (0,1) (0,0) to [bend right] (0,1);
\end{tikzpicture}
& $\begin{psmallmatrix}2&0&-1&-1\\0&2&-1&-1\\-1&-1&4&3\\-1&-1&3&4\end{psmallmatrix}$ &
  54 & 116 & 84 & 22 & 7 \\[4mm]
\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
\filldraw (0,0) circle (1pt) (0,1) circle (1pt) (1,0) circle (1pt)
(1,1) circle (1pt);
\draw (0,0)--(1,0) (0,0)--(0,1) (0,0)--(1,1) (1,0) to [bend left]
(1,1) (1,0) to [bend right] (1,1) (0,1) to [bend left]
(1,1) (0,1) to [bend right] (1,1);
\end{tikzpicture}
& $\begin{psmallmatrix}2&0&-1&0\\0&2&0&-1\\-1&0&3&1\\0&-1&1&3\end{psmallmatrix}$
 & 54 & 114 & 80 & 20 & 7 \\[4mm]
\begin{tikzpicture}[baseline={([yshift=1ex]current bounding box.center)}]
\filldraw (0,0) circle (1pt) (0,0.8) circle (1pt) (0.8,0) circle (1pt)
(0.8,0.8) circle (1pt);
\draw (0,0)--(0.8,0) (0,0)--(0,0.8) (0,0)--(0.8,0.8) (0.8,0) (0.8,0)--(0.8,0.8)
(0,0.8)--(0.8,0.8) (0,0.8)--(0.8,0)
(0.8,0) to [in=30,out=-60,distance=8mm] (0.8,0);
\end{tikzpicture} & $\begin{psmallmatrix}3&1&-1&0\\1&3&1&0\\-1&1&3&0\\0&0&0&1\end{psmallmatrix}$ & 48 & 96 & 64 & 16 & 7 \\[-3mm] \hline \rule{0pt}{1.4\normalbaselineskip}
\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
\filldraw (0,1) circle (1pt) (1,1) circle (1pt) (0.5,0) circle (1pt);
\draw (0.5,0) to [bend left] (0,1) (0.5,0) to [bend right] (0,1)
(0.5,0) to [bend left] (1,1) (0.5,0) to [bend right] (1,1)
(0,1) to [bend left] (1,1) (0,1) to [bend right] (1,1);
\end{tikzpicture}
& $\begin{psmallmatrix}2&0&-1&-1\\0&2&1&1\\-1&1&3&2\\-1&1&2&3\end{psmallmatrix}$ & 46 & 108 & 84 & 22 & 6 \\[4mm]
\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
\filldraw (0,1) circle (1pt) (1,1) circle (1pt) (0.5,0) circle (1pt);
\draw (0.5,0) to [bend left] (0,1) (0.5,0) to [bend right] (0,1)
(0.5,0) to [bend left] (1,1) (0.5,0) to [bend right] (1,1)
(0,1)--(1,1) (0.5,0)--(1,1);
\end{tikzpicture}
& $\begin{psmallmatrix}2&-1&-1&-1\\-1&3&2&2\\-1&2&3&2\\-1&2&2&3\end{psmallmatrix}$ & 42 & 94 & 72 & 20 & 6 \\[4mm]
\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
\filldraw (0,0.6) circle (1pt) (0.6,0.6) circle (1pt) (0.3,0) circle (1pt);
\draw (0.3,0) to [bend left] (0,0.6) (0.3,0) to [bend right] (0,0.6)
(0.3,0) to [bend left] (0.6,0.6) (0.3,0) to [bend right] (0.6,0.6)
(0,0.6)--(0.6,0.6) (0.3,0) to [in=-30,out=-150, distance=8mm] (0.3,0);
\end{tikzpicture}
& $\begin{psmallmatrix}2&1&1&0\\1&3&2&0\\1&2&3&0\\0&0&0&1\end{psmallmatrix}$ & 36 & 74 & 52 & 14 & 6 \\[4mm]
\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
\filldraw (0,0) circle (1pt) (0.8,0) circle (1pt) (1.6,0) circle (1pt);
\draw (0,0) to [bend left=40] (0.8,0) (0,0) to [bend right=40] (0.8,0) (0,0)--(0.8,0);
\draw (0.8,0) to [bend left=40] (1.6,0) (0.8,0) to [bend right=40] (1.6,0) (0.8,0)--(1.6,0);
\end{tikzpicture}
& $\begin{psmallmatrix}2&1&0&0\\1&2&0&0\\0&0&2&1\\0&0&1&2\end{psmallmatrix}$ & 36 & 72 & 48 & 12 & 6 \\[4mm] \hline \rule{0pt}{1.4\normalbaselineskip}
\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
\filldraw (0,0) circle (1pt) (1,0) circle (1pt);
\draw (0,0) to [bend left=20] (1,0) (0,0) to [bend left=60] (1,0)
(0,0) to [bend right=20] (1,0) (0,0) to [bend right=60] (1,0) (0,0)--(1,0);
\end{tikzpicture}
& $\begin{psmallmatrix}2&1&1&1\\1&2&1&1\\1&1&2&1\\1&1&1&2\end{psmallmatrix}$ & 30 & 70 & 60 & 20 & 5 \\[3mm]
\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
\filldraw (0,0) circle (1pt) (1,0) circle (1pt);
\draw (0,0) to [bend left=20] (1,0) (0,0) to [bend left=60] (1,0)
(0,0) to [bend right=20] (1,0) (0,0) to [bend right=60] (1,0);
\draw (1,0) to[in=50,out=-50, distance=8mm,looseness=50] (1,0);
\end{tikzpicture}
& $\begin{psmallmatrix}2&1&1&0\\1&2&1&0\\1&1&2&0\\0&0&0&1\end{psmallmatrix}$ & 28 & 62 & 48 & 14 & 5 \\[-2mm]
\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
\filldraw (0.2,0) circle (1pt) (1,0) circle (1pt);
\draw (0.2,0) to [bend left=40] (1,0) (0.2,0) to [bend right=40] (1,0) (0.2,0)--(1,0);
\draw (1,0) to[in=90,out=0, distance=8mm] (1,0) (1,0) to[in=0,out=-90, distance=8mm] (1,0);
\end{tikzpicture}
 & $\begin{psmallmatrix}2&1&0&0\\1&2&0&0\\0&0&1&0\\0&0&0&1\end{psmallmatrix}$   & 24 & 48 & 34 & 10 & 5 \\[-2mm] \hline
\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
\filldraw (0,0) circle (1pt);
\draw (0,0) to[in=0, out=90,distance=8mm] (0,0) (0,0) to[in=90, out=180,distance=8mm](0,0) (0,0) to[in=180, out=270,distance=8mm] (0,0) (0,0) to[in=270, out=0,distance=8mm] (0,0);
\end{tikzpicture}
& $\begin{psmallmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{psmallmatrix}$ & 16 & 32 & 24 & 8 & 4 \\ \hline
\caption{The tropical Schottky locus for $g=4$}
\label{table-trop-schottky}
\end{longtable}

We implemented Algorithm \ref{alg:TSD} using existing software
for polyhedral geometry, namely the \texttt{GAP} package \texttt{polyhedral}
due to Dutour Sikiri\'c \cite{sikiric, sgsw}, as well as Joswig's
 {\tt polymake} \cite{polymake}.

The first column of Table \ref{table-trop-schottky} shows all relevant graphs $G$ of genus $4$.
The second column gives a representative Riemann matrix.
Here all edges have length $1$ and a cycle basis $B$ was chosen.
Using (\ref{eqn-cone}), we also  precomputed
the secondary cones  $\sigma_{G,B}$ for the $16$
 representatives.

\begin{example}\label{ex-decision} \rm
 Using the \texttt{GAP} package \texttt{polyhedral} \cite{sikiric} we compute the Voronoi polytope of
$$ \begin{small}
 Q\,\,=\,\,\begin{pmatrix}14&-9&11&0\\-9&11&-2&1\\11&-2&21&11\\0&1&11&14\end{pmatrix}.
 \end{small}
 $$
Its $f$-vector is $(62,142,104,24)$. This does not appear in
Table \ref{table-trop-schottky}.
Hence $Q$ is not in  $\mathfrak{J}_4^{\rm trop}$.
\end{example}

We now address the Schottky Recovery Problem.
The input is a matrix $Q \in \mathfrak{J}_4^{\rm trop}$.
From Algorithm \ref{alg:TSD} we know the f-vector
of the Voronoi polytope. Using Table \ref{table-trop-schottky},
this uniquely identifies the graph $G$.
Note that our graphs $G$ are dual to those in \cite[\S 4.4.4]{frank}.
From our precomputed list, we also know the secondary cone $\sigma_{G,B}$
for some choice of basis~$B$.

\begin{algorithm}[Tropical Schottky Recovery]   \label{alg:TSR}
\underbar{Input}: $Q \in \mathfrak{J}_4^{\rm trop}$. \\
\underbar{Output}: A metric graph $\Gamma $ whose Riemann matrix $Q_\Gamma$ equals $Q$. \\
1. Identify the underlying graph $G$  from Table \ref{table-trop-schottky}.
Retrieve the basis $B$ and the cone $\sigma_{G,B}$. \\
2. Let $D = {\rm Del}(Q)$ and compute the secondary cone $\sigma_D$
as in (\ref{def-sec-cone}). \\
3. The cones $\sigma_D$ and $\sigma_{G,B}$ are related by a linear transformation
$X \in {\rm GL}_4(\mathbb{Z})$.~Compute $X$. \\
4. The matrix $ X^t Q X$ lies in $\sigma_{G,B}$. Compute
$\ell_1,\ldots,\ell_m$ such that
$X^t Q X = \sum_{i=1}^m \ell_i b_i b_i^t$. \\
5. Output the graph $G$ with length $\ell_i$ for its $i$-th edge,
corresponding to the column $b_i$ of $B$.
\end{algorithm}

We implemented this algorithm as follows.
Step 2  can be done using \texttt{polyhedral} \cite{sikiric}.
This code computes the secondary cone  $\sigma_D$ containing
a given positive definite matrix $Q$.
The matrix $X \in {\rm GL}_4(\mathbb{Z}) $ in Step 3 is also
found by \texttt{polyhedral}, but with external calls to the package \texttt{isom}
due to Plesken and Souvignier \cite{ps95}.
We refer to \cite[\S 4]{sgsw} for details.
For Step 4 we note that the rank $1$ matrices $b_1 b_1^t, \ldots, b_m b_m^t$ are linearly
independent \cite[\S 4.4.4]{frank}. Indeed, the two $9$-dimensional secondary cones $\sigma_{G,B}$ at the top of
Table \ref{table-trop-schottky} are simplicial, and so are their faces.
Hence the multipliers $\ell_1,\ldots,\ell_m$ found in Step 4 are unique and positive.
These $\ell_i$ must agree with the desired edge lengths $l(e_i)$, by the formula
for $Q= Q_\Gamma$ in (\ref{eq:tropicalriemann2}).

\begin{example}\label{ex-recov} \rm
Consider the Schottky Recovery Problem for the matrix
\begin{equation}
\label{eq:ex36matrix}
\begin{small}
Q \,\,=\,\,\begin{pmatrix}
17 & 5 & 3 & 5\\
5 & 19 & 7 & 11\\
3 & 7 & 23 & 16 \\
5 & 11 & 16 & 29
\end{pmatrix}.
 \end{small}
\end{equation}
Using \texttt{polyhedral}, we find that the f-vector of its
Voronoi polytope is $(96,198,130,28)$. This matches the first row in
Table \ref{table-trop-schottky}. Hence $Q \in \mathfrak{J}_4^{\rm trop}$,
and $G$ is the triangular prism.
Using \texttt{polyhedral} and \texttt{isom}, we find a matrix that
 maps $Q$ into our preprocessed secondary cone:
$$
\begin{small}
X = \begin{pmatrix}
  0 & 0 & 0 & 1 \\
  1 & 0 & 0 & 0 \\
  0 & 1 & 1 & 0 \\
  -1 & -1 & 0 & 0
\end{pmatrix}\,\in\, {\rm GL}_4(\mathbb{Z}) \qquad \hbox{gives} \qquad
Q' = X^tQX = \begin{pmatrix}
  26 & 9 & -9 & 0 \\
  9 & 20 & 7 & -2 \\
  -9 & 7 & 23 & 3 \\
  0 & -2 & 3 & 17
\end{pmatrix}\, \in\, \sigma_{G,B}.
\end{small}
$$
This $Q'$ is the Riemann matrix of the metric graph in Figure
\ref{fig-recover-trop-schottky}, with basis cycles
 $e_2+e_6-e_3$, $-e_1-e_4+e_7+e_2$, $-e_1-e_5+e_8+e_3$, and
$e_4+e_9-e_5$. These are the rows of the $4 \times 9$-matrix $B$.
In Step 4 of Algorithm \ref{alg:TSR} we compute
$D = {\rm diag}(\ell_1,\ldots,\ell_9) = {\rm diag}(
7,9,9,2,3,8,2,4,12)$.
 In Step 5 we output the metric graph in
Figure \ref{fig-recover-trop-schottky}.
Its Riemann matrix equals $Q = BDB^t$.

\begin{figure}[h]\label{fig-graph1}
\begin{center}
\begin{tikzpicture}[scale=1.5]
\usetikzlibrary{decorations.markings}
\begin{scope}[thick, decoration={markings, mark=at position 0.5 with {\arrow{latex}}}]
\filldraw (0,0) circle (1pt) (0,2) circle (1pt)
  (1,1) circle (1pt) (2,1) circle (1pt) (3,0) circle (1pt)
(3,2) circle (1pt);
\draw[postaction={decorate}] (1,1)--node[above,red]{7}node[below]{$e_1$}(2,1);
\draw[postaction={decorate}] (1,1)--node[above,red]{9}node[below]{$e_2$}(0,2);
\draw[postaction={decorate}] (1,1)--node[above,red]{9}node[below]{$e_3$}(0,0);
\draw[postaction={decorate}] (2,1)--node[above,red]{2}node[below]{ $e_4$}(3,2);
\draw[postaction={decorate}] (2,1)--node[above,red]{3}node[below]{$e_5$}(3,0);
\draw[postaction={decorate}] (0,2)--node[right,red]{8}node[left]{$e_6$}(0,0);
\draw[postaction={decorate}] (0,2)--node[above,red]{2}node[below]{$e_7$}(3,2);
\draw[postaction={decorate}] (0,0)--node[above,red]{4}node[below]{$e_8$}(3,0);
\draw[postaction={decorate}] (3,2)--node[right,red]{12}node[left]{$e_9$}(3,0);
\end{scope}
\end{tikzpicture}
\caption{Metric graph with edge lengths in red. Its
Riemann matrix matches (\ref{eq:ex36matrix}).
}
\label{fig-recover-trop-schottky}
\end{center}
\end{figure}
\end{example}

It is instructive to compare Algorithms \ref{alg:TSD} and
\ref{alg:TSR} with Section~\ref {sec:classic}.
Our classical solution is not just the abstract
Riemann surface but it consists of a canonical embedding into $\mathbb{P}^3$.

Canonical embeddings also exist for metric graphs $\Gamma$, as
explained in \cite[\S 7]{HMY}. However, even computing the
ambient space $|K|$, that plays the role of $\mathbb{P}^3$, is non-trivial
in that setting. For $g=4$ this is solved in \cite{BoLin}.
An alternative approach is to construct a
classical curve over a non-archimedean field that tropicalizes to $\Gamma$.
See \cite[\S 7.3]{BBC} for first steps in that direction.

Example \ref{exp:shimuracurve} explored the Schottky locus in a
two-parameter family of Riemann matrices.
In the tropical setting, it is natural to intersect
$ \mathfrak{H}_g^{\rm trop}$ with an
affine-linear space $L$ of symmetric matrices.
The intersection  $ \mathfrak{H}_g^{\rm trop} \cap L$
is a {\em spectrahedron}.
By the {\em Schottky locus of a spectrahedron}
we mean $ \mathfrak{J}_g^{\rm trop} \cap L$.
This is an infinite periodic polyhedral complex
inside the spectrahedron. For
quartic spectrahedra \cite{ORSV}, when $g=4$,
this locus has codimension one.

\begin{example}[The Schottky locus of a quartic spectrahedron] \rm
We consider the matrix
$$
\begin{small}
\! Q =
 \begin{bmatrix}1589 - 2922 s + 960 t &   789 - 1322 s & -820 + 660 s - 1350 t & \!\! -820 + 3260 s +    2550 t \\ 789 - 1322 s &   1589 - 2922 s - 960 t \! &\!\!\! -820 + 3260 s - 2550 t \!& -820 + 660 s +  1350 t \\ -820 + 660 s - 1350 t & \!-820 + 3260 s - 2550 t \! &   1665 + 450 s + 3120 t & -25 - 2930 s \\ \!-820 + 3260 s +    2550 t & -820 + 660 s + 1350 t & -25 - 2930 s &   1665 + 450 s - 3120 t\end{bmatrix} \!.
 \end{small}
$$
Here $s$ and $t$ are parameters. This defines a plane $L$ in the
space of symmetric $4 \times 4$-matrices. The left diagram in  Figure \ref{fig:spec}
shows the hyperbolic curve $\{{\rm det}(Q)=0\}$.
The spectrahedron $ \mathfrak{H}_g^{\rm trop} \cap L$ is bounded by its inner oval.
The right diagram shows the second Voronoi decomposition.
The Schottky locus  $ \mathfrak{J}_g^{\rm trop} \cap L$
is a proper subgraph
of its edge graph.
It is shown in red. Note that the graph has infinitely many edges and regions.
\end{example}

\begin{remark} \rm
We described some computations in {\tt GAP} and in {\tt polymake}
that realize Algorithms~\ref{alg:TSD} and \ref{alg:TSR}.
The code for these implementations  is made available on our website~(\ref{eq:url}).
\end{remark}
\begin{figure}[h]
  \begin{center}
  \vspace{-0.1in}
\includegraphics[width=7.5cm]{spectrahedron.pdf}   \quad
 \includegraphics[width=7.8cm]{spectrahedron_yr.pdf}
\vspace{-0.3in}
\end{center}    \caption{\label{fig:spec}
A quartic spectrahedron (left) and its second Voronoi decomposition (right).
The Schottky locus of that spectrahedron consists of those edges that are highlighted in red.
}
 \end{figure}

\section{Tropical Meets Classical}
\label{sec:four}

In this section we present a second solution to the tropical Schottky problem.
It is new and different from the one in Section \ref{sec:tropical}, and it
links directly to  the classical solution in Section~\ref{sec:classic}.

Let $Q \in \mathfrak{H}_g^{\rm trop}$ be a positive definite matrix for arbitrary $g$.
Mikhalkin and Zharkov \cite[\S 5.2]{mz} define the following analogue
to the Riemann theta function in the max-plus algebra:
\begin{equation}
\label{eq:troptheta}
\Theta(Q,x)\,\,:=\,\,\max_{\lambda\in\mathbb{Z}^g}\{\lambda^t Q x-\frac{1}{2} \lambda^t Q \lambda\}.
\end{equation}
This {\em tropical theta function} describes the asymptotic behavior of the
classical Riemann theta function with Riemann matrix $t\cdot\tau$ when $t$ goes to infinity, as long as there are no cancellations.   This is made precise in Proposition \ref{lem:tropicallimit}. Here,
the real matrix $Q$ is the imaginary part of $\tau$.

  Analogously, for $u\in\mathbb{Z}^g$, we define the \textit{tropical theta constant with characteristic} $u$ to be
  \begin{equation}
  \label{eq:tropthetaconst}
 \Theta_{u}(Q)\,\,:=\,\,2\cdot\Theta(Q,\frac{u}{2})-\frac{1}{4} u^t Q u.
\end{equation}
  In the classical case, characteristics are vectors $m = (m',m'')$ in $\mathbb{Z}^{2g}$. But, only
  $u=m'$  contributes to the aforementioned asymptotics.
  Note that $\Theta_u(Q)$ depends only on $u$ modulo~$2$.

\begin{definition} \rm
  For any $v\in\mathbb{Z}^g$ consider the following signed sum of tropical theta constants:
\begin{equation}
\label{eq:vartheta}
 \vartheta_v(Q) \,\,\,:=\sum_{u\in(\mathbb{Z}/2\mathbb{Z})^g} \!\! (-1)^{u^t v}\cdot \Theta_u(Q) .
 \end{equation}
 The \textit{theta matroid} $M(Q)$ is the binary matroid represented by the collection of vectors
 \begin{equation}
 \label{eq:magicmatroid}
\bigl\{v \in(\mathbb{Z}/2\mathbb{Z})^g:\, \vartheta_v(Q)\neq 0 \bigr\}.
 \end{equation}

\end{definition}

The tropical theta constants and the theta matroid are invariant under basis changes
$S\in\textnormal{GL}_g(\mathbb{Z})$. We have
$\, \vartheta_u(Q)=\vartheta_{S^{-1} u}(S^tQ S)\,$ for all
$u\in\mathbb{Z}^g$, and therefore
$M(Q) = M(S^tQS)$.

Here is the promised new approach to the  Schottky problem.
If $Q$ lies in the tropical Schottky locus then
 $M(Q)$ is the desired cographic matroid
 and (\ref{eq:vartheta}) furnishes edge lengths.

\begin{theorem}\label{thm:recoverfromtheta}
 If $\,Q \in \mathfrak{J}_g^{\rm trop}$ then the matroid $M(Q)$ is cographic.
 In that graph, we assign the length
$\,2^{3-g} \cdot \vartheta_v(Q)\,$ to the edge labeled $v$.
The resulting metric graph has Riemann matrix~$Q$.
\end{theorem}

This says, in particular, that $\vartheta_v(Q)$ is non-negative when $Q $ comes from a metric graph.

\begin{proof}
Since $Q \in \mathfrak{J}_g^{\rm trop}$, there exists a unimodular matrix
$B = (b_1,\ldots,b_m) \in \{-1,0,+1\}^{g \times m}$
and a diagonal matrix $D = {\rm diag}(\ell_1,\ldots,\ell_m)$ such that $Q = B D B^t
= \sum_{i=1}^m \ell_i b_i b_i^t$. We claim
\begin{equation}
\label{eq:weclaim}
   \Theta_u (Q)\,\,=\,\,-\,\frac{1}{4}\,\cdot \! \sum_{b_i^t u \textnormal{ is odd}} \!\!\ \ell_i \qquad \quad
    \hbox{for all $\,u\in \mathbb{Z}^g$}.
    \end{equation}
 Here the $\ell_i$ are positive real numbers. First, we note that
 $$
  \Theta_u(Q)\,\,=\,\,\max_{\lambda\in\mathbb{Z}^g} \,\bigl\{ -(\lambda+\frac{u}{2})^t Q(\lambda+\frac{u}{2})\,\bigr\}
\,    \,\,\leq \,
  \,\,\sum_{i=1}^m -\ell_i\cdot \min_{\lambda\in\mathbb{Z}^g}\biggl\{ \bigl(\,b_i^t\cdot (\lambda+\frac{u}{2})\,\bigr)^2 \biggr\}.
  $$
 If $\,b_i^t u\,$ is even, then $b_i^t\cdot (\lambda+\frac{u}{2})=0$ for some  $\lambda\in\mathbb{Z}^g$.
     Otherwise, the absolute value of $b_i^t\cdot (\lambda+\frac{u}{2})$ is at least $1/2$.
     This shows that $\Theta_u(Q)\leq-\frac{1}{4}\cdot\sum_{b_i^t u\textnormal{ is odd}} \ell_i$.
To derive the reverse inequality, let $\,I = \bigl\{ i \,:\,
u_i \,\hbox{is odd} \bigr\} \subset \{1,\ldots,g\}$.
 By a result of Ghouila-Houri \cite{houri} on unimodular matrices,
 we can find $w\in\mathbb{Z}^g$ with $w_i=\pm1$ if $i\in I$ and $w_i=0$ otherwise,
 such that $b_i^t\cdot w\in\{0,\pm1\}$ for all $1\leq i\leq m$.
 The vector $\lambda_0 = \frac{1}{2}(w-u)$ lies in $\mathbb{Z}^g$.
 One checks that
  \[-(\lambda_0+\frac{u}{2})^t Q (\lambda_0+\frac{u}{2})
  \,=\,
  \sum_{i=1}^m -\ell_i\cdot (b_i^t\cdot (\lambda_0+\frac{u}{2}))^2
  \,=\,-\frac{1}{4}\sum_{i=1}^m \ell_i\cdot (b_i^t\cdot w)^2
  \,=\,-\frac{1}{4}\cdot\sum_{b_i^t u\textnormal{ is odd}} \!\!\ell_i.\]
  Therefore, we also have $\,\Theta_u(Q)\geq-\frac{1}{4}\cdot\sum_{b_i^t u\textnormal{ is odd}} \ell_i$.
  This establishes the assertion in (\ref{eq:weclaim}).

We next claim that, under the same hypotheses as above,
the function in (\ref{eq:vartheta})  satisfies
\begin{equation}
\label{eq:nextclaimthat}
  \vartheta_v(Q)\,\,=\,\,\,2^{g-3}\!\sum_{b_i\equiv v \,{\rm mod} \,2} \!\! \ell_i
 \qquad \hbox{for all}\,\, v \in \mathbb{Z}^g .
\end{equation}
Indeed, substituting the right hand side of (\ref{eq:weclaim}) for $\Theta_u(Q)$ into (\ref{eq:vartheta}), we find that
  $$
 \vartheta_v(Q)\,\,=\,\,-\frac{1}{4}\cdot\sum_{u\in(\mathbb{Z}/2\mathbb{Z})^g}\sum_{b_i^t u \textnormal{ is odd}}
 (-1)^{u^t v}\cdot  \ell_i \,\,=\,\,-\frac{1}{4}\cdot\sum_{i=1}^m \ell_i \cdot (|E_i|-|O_i|),
  $$
  where $E_i=\{u\in(\mathbb{Z}/2\mathbb{Z})^g: b_i^t u \,\,{\rm odd},\, u^t v \, \,{\rm even}\}$ and
  $O_i=\{u \in(\mathbb{Z}/2\mathbb{Z})^g: b_i^t u \,\,{\rm odd}, \, u^t v \,\,{\rm odd}\}$.
    If $b_i\equiv v \,{\rm mod}\,2$ then $E_i=\emptyset$ and $|O_i|=2^{g-1}$. Otherwise, $|E_i|=|O_i|=2^{g-2}$.
    This proves~(\ref{eq:nextclaimthat}).

Since $Q \in \mathfrak{J}_g^{\rm trop}$, this matrix comes from a graph $G$.
We may assume that $G$ has no $2$-valent vertices. This ensures
that any pair is independent in the cographic matroid of $G$.

The column $b_i$ of the matrix $B$ records the coefficients of the $i$-th edge
in a cycle basis of the graph $G$. The residue class of $b_i$ modulo $2$ is unique.
For $v \in \mathbb{Z}^g$ with $b_i \equiv v \,{\rm mod} \, 2 $, the sum in
(\ref{eq:nextclaimthat}) has only term $\ell_i$, and we have
$\ell_i = 2^{3-g}\vartheta_v(Q)$. If $v \in \mathbb{Z}^g$ is not congruent to $b_i$
for any $i$ then $\vartheta_v(Q) = 0$. This proves that the theta matroid
$M(Q)$ equals the cographic matroid of $G$, and the edge lengths $\ell_i$
are recovered from $Q$ by the rule in Theorem \ref{thm:recoverfromtheta}.
\end{proof}

By Theorem \ref{thm:recoverfromtheta}, the
non-negativity of $\vartheta_v(Q)$ is a necessary condition for $Q$ to be in
 $\mathfrak{J}_g^{\rm trop}$.

\begin{example} \rm
For the matrix $Q$ in Example \ref{ex-decision}, we find $\vartheta_{0001}(Q)=-\frac{1}{2}$.
Hence $\,Q \not\in \mathfrak{J}_4^{\rm trop}$.
\end{example}

This necessary (but not sufficient) condition translates into the following algorithm:

\begin{algorithm}[Tropical Schottky Recovery]   \label{alg:TSR2}
\underbar{Input}: $Q \in \mathfrak{J}_g^{\rm trop}$. \\
\underbar{Output}: A metric graph $\Gamma $ whose Riemann matrix $Q_\Gamma$ equals $Q$. \\
1. Compute the theta matroid $M(Q)$. It is cographic and determines a unique graph $G$. \\
2. Compute all edge lengths using the formula $\ell_i = 2^{3-g}\vartheta_v(Q)$. Set
 $\,D = {\rm diag}(\ell_1,\ldots,\ell_m)$. \\
 3. Output the metric graph $(G,D)$.  \\
4. (Optional) As in Algorithm \ref{alg:TSR}, find a basis $B$ such that $BDB^t = Q$.
\end{algorithm}

\begin{example} \rm
Let $Q$ be the matrix  in Example \ref{ex-recov}.
For each $u \in (\mathbb{Z}/2\mathbb{Z})^4$, we list
the theta constant $\Theta_u(Q)$, the weight $2^{-1}\vartheta_u(Q)$
and the label of the corresponding edge in Figure \ref{fig-graph1}:
\setcounter{MaxMatrixCols}{20}
$$
\begin{small}
\begin{matrix}
  u &  \! 0001 \! & \! 0010 \! & \! 0011 \!& \! 0100 \! & \! 0101 \! & \! 0110 \! & \! 0111 \! &
  \! 1000 \! & \! 1001 \! &\! 1010 \! &\! 1011 \! &\! 1100\! &\! 1101 \! &\! 1110 \! &\!1111 \smallskip \\
-\Theta_u &\frac{29}{4} & \frac{23}{4} &5& \frac{19}{4} & \frac{13}{2} & 7 & \frac{31}{4} & \frac{17}{4} & 9 & \frac{17}{2} & \frac{33}{4} & \frac{13}{2} & \frac{43}{4} & \frac{41}{4}& \frac{21}{2} \smallskip \\
   2^{-1}\vartheta_u  &9&7&9&8&2&0&4&12&0&0&0&0&2&0&3 \\
   {\rm Edge} & e_2 & e_1 & e_3 & e_6 & e_7 &-& e_8 & e_9 &-&-&-&- & e_4 &-& e_5    \\
\end{matrix}
\end{small}
$$
\end{example}

We now explain the connection between the classical
and tropical theta functions. In particular, we will show how
the process of tropicalization relates
Theorems \ref{thm:igusa} and \ref{thm:recoverfromtheta}.

In order to tropicalize the Schottky--Igusa modular form, we must study the order of growth of the theta constants when the entries of the Riemann matrix grow. This information is captured by the tropical theta constants.
 The following proposition makes that precise.

\begin{proposition} \label{lem:tropicallimit}
Fix $Q \in \mathfrak{H}_g^{\rm trop}$, and let $P(t)$ be any real symmetric
$g \times g$-matrix that depends on a parameter $t\in\mathbb{R}$.
  For every $m\in(\mathbb{Z}/2\mathbb{Z})^{2g}$ there is a constant $C \in\mathbb{R}$ such that
\begin{equation}
\label{eq:thetaratio}
 0 \,\,\leq \,\,\frac{| \,\theta[m](P(t)+t\cdot \textnormal{i} Q, 0)\, |}{|\, \exp (t \cdot\pi\cdot  \Theta_{m'}(Q)) \,|}\leq
 \,\, C\,\, \quad  \hbox{for all $\,t\geq0$}.
\end{equation}
Moreover, we can choose $P(t)$ such  that the ratio above does not approach zero for $t\to\infty$.
\end{proposition}

  Here $ \theta[m](\tau,0)$ is the classical theta constant from
   (\ref{eqn:thetafunc}), and    $\Theta_{m'}(Q)$ is the tropical theta constant
   defined  in (\ref{eq:tropthetaconst}).
We use the notation $m = (m',m'')$ for vectors in $\mathbb{Z}^{2g}$ as in Section~\ref{sec:classic}.

\begin{proof}
Consider the lattice points $\lambda$ where the maximum in
(\ref{eq:troptheta}) for $x = m'/2$ is attained.
The corresponding summands in (\ref{eqn:thetafunc}) with $\lambda = n$
have the same asymptotic behavior as $\exp (t \cdot\pi  \Theta_{m'}(Q))$ for $t\to\infty$.
 The sum over the remaining exponentials tends to zero since it can be bounded by a sum of finitely many Gaussian integrals with variance going to zero for $t\to\infty$. We can choose the
 real symmetric matrix $P(t)$
 in such a way that no cancellation of highest order terms happens. Then the expression
 in (\ref{eq:thetaratio}) is bounded away from zero.
 \end{proof}

\begin{remark}
 On the Siegel upper-half space $\mathfrak{H}_g$ we have an action by the symplectic group $\textnormal{Sp}_{2g}(\mathbb{Z})$. Two matrices from the same orbit under this action correspond to the same abelian variety. However their tropicalizations may vary drastically. Consider for example the case $g=1$: $\begin{pmatrix}
               0&-1\\
               1&k                                                                                \end{pmatrix}\in\textnormal{Sp}_{2}(\mathbb{Z})$
 sends $\tau=\textnormal{i}$ to a complex number with imaginary part $\frac{1}{1+k^2}$.
\end{remark}

We now assume that $g=4$.
For any subset $M\subset (\mathbb{Z}/2\mathbb{Z})^{8}$ we write $M'=\{m':\, m\in M\}$ and similarly for $M''$.
The following lemma concerns the possible choices
 for Theorem \ref{thm:igusa}.

\begin{lemma}\label{cor:conj} For any azygetic triple $\{m_1,m_2,m_3\}$ and any matching
subgroup $N \subset (\mathbb{Z}/2\mathbb{Z})^8$,
\begin{itemize}
\item[(1)] there exist indices $1 \leq i < j \leq 3$ such that $(m_i+N)'=(m_j+N)'$, and \vspace{-0.1in}
\item[(2)] if $\dim N' =3$ and $(m_1+N)'=(m_2+N)'\neq (m_3+N)'$, then $m_1',m_2'\in N'$.
\end{itemize}
\end{lemma}

\begin{proof}
This purely combinatorial statement can be proved by exhaustive computation.
\end{proof}

For instance, consider the specific choice of $m_1,m_2,m_3,N$ made prior to Example \ref{exp:schottkycurve}.
This has $\dim N'=2$, and Lemma \ref{cor:conj} (1) holds with $i= 2$, $j= 3$.
If we exchange the first four coordinates with the last four coordinates, then
$\dim N'=3$, $m_1',m_3'\in N'$ and $m_2'\not\in N'$.

Recall from Theorem \ref{thm:igusa} that a
 matrix $\tau\in\mathfrak{H}_4$ is in the Schottky locus if and only if $\pi_1^2+\pi_2^2+\pi_3^2-2(\pi_1\pi_2+\pi_1\pi_3+\pi_2\pi_3)$ vanishes. The tropicalization of this expression equals
 \begin{equation}
 \label{eq:tropicalschottkyigusa}
  \max_{i,j=1,2,3}( \pi^{\textnormal{trop}}_i+\pi^{\textnormal{trop}}_j),
  \end{equation}
 where $\pi^{\textnormal{trop}}_i=\sum_{m\in m_i+N} \Theta_{m'}(Q)$
 is the tropicalization of the product (\ref{eq:pidef}), with $Q=
 {\rm im}(\tau)$.

 The {\em tropical Schottky--Igusa modular form} (\ref{eq:tropicalschottkyigusa})
defines a piecewise-linear convex function $\mathfrak{H}_4^{\rm trop} \rightarrow \mathbb{R}$.
Its breakpoint locus  is the set of  Riemann matrices $Q$ for which the maximum in (\ref{eq:tropicalschottkyigusa})
  is attained twice. That set depends on our choice of $m_1,m_2,m_3,N$.
That choice is called {\em admissible} if $N\subset (\mathbb{Z}/2\mathbb{Z})^8$ has rank three, the triple
$\{m_1,m_2,m_3\} \subset (\mathbb{Z}/2\mathbb{Z})^8$ is azygetic, all elements of $m_i+N$ are even,
\underbar{and} the group $N' \subset (\mathbb{Z}/2\mathbb{Z})^4$ also has rank three.
We define the {\em tropical Igusa locus}
in $\mathfrak{H}_g^{\rm trop}$ to be the intersection,
over all admissible choices $m_1,m_2,m_3,N$, of
the breakpoint loci of the tropical modular forms $(\ref{eq:tropicalschottkyigusa})$.

\begin{theorem}
\label{thm:admissible}
A matrix $Q \in \mathfrak{H}_4^{\rm trop}$ lies in the tropical Igusa locus
if and only if   $\vartheta_v(Q)\geq0$ for all $v\in\mathbb{Z}^4$.
That locus contains
the tropical Schottky locus $\mathfrak{J}_4^{\rm trop}\!$,
but they are not equal.
\end{theorem}

\begin{proof}
We are interested in how the maximum in (\ref{eq:tropicalschottkyigusa}) is attained.
By Lemma \ref{cor:conj} (1), after relabeling, $\,\pi^{\textnormal{trop}}_1=\pi^{\textnormal{trop}}_2$.
   The maximum is attained twice  if and only if $\pi^{\textnormal{trop}}_1\geq \pi^{\textnormal{trop}}_3$.
   By Lemma \ref{cor:conj} (2),    this is equivalent to
\begin{equation}
\label{eq:inout}
\sum_{u\in N'} \Theta_{u}(Q)\,\,\geq \,\,\sum_{u\not\in N'} \Theta_{u}(Q).
\end{equation}
Let $v $ be the non-zero vector in $(\mathbb{Z}/2\mathbb{Z})^{4}$ that is orthogonal to $N'$.
Then (\ref{eq:inout}) is equivalent to
$$ \vartheta_v(Q) \,\,\, =
\sum_{u\in(\mathbb{Z}/2\mathbb{Z})^{4}}\!\!(-1)^{u^t v} \Theta_u(Q)
\,\,\, \geq \,\,\, 0 . $$
This proves the first assertion, if we knew that
 every $v$ arises from some admissible choice.

We saw in Theorem \ref{thm:recoverfromtheta}
that $\vartheta_v(Q) \geq 0 $  for all $v$
whenever $Q \in \mathfrak{J}_4^{\rm trop}$.
Hence the tropical Schottky locus $\mathfrak{J}_4^{\rm trop}$ is contained in the tropical Igusa locus.
The two loci are not equal because the latter contains the
{\em zonotopal locus} of $\mathfrak{H}_4^{\rm trop}$. This consists of matrices
$Q = BDB^t$ where $B$ represents any unimodular matroid, not necessarily
cographic. By  \cite[\S 4.4.4]{frank},  the second Voronoi decomposition
of $\mathfrak{H}_4^{\rm trop}$ has a non-cographic $9$-dimensional cone in its zonotopal locus.
It is unique modulo ${\rm GL}_4(\mathbb{Z})$. We
verified that all $16$ tropical modular forms $\vartheta_v$ are non-negative on that cone.
This establishes the last assertion in Theorem \ref{thm:admissible}.

To finish the proof, we still need that every
$v \in (\mathbb{Z}/2\mathbb{Z})^{4} \backslash \{0\}$ is orthogonal to $N'$
for some admissible choice $m_1,m_2,m_3,N$.
By permuting coordinates,
  it suffices to show this~for
  $$v \,\in \, \{(1,0,0,0)^t, (1,1,0,0)^t, (1,1,1,0)^t, (1,1,1,1)^t\}.$$
  For $v=(1,0,0,0)^t$ we take
$$
\begin{small}
 m_1=\begin{pmatrix}0 & 1 \\ 0 & 0 \\ 0 & 0 \\ 0 & 1     \end{pmatrix}\!,\,
m_2=\begin{pmatrix} 1 & 1 \\ 1 & 0 \\ 1 & 1 \\ 0 & 1    \end{pmatrix}\!,\,
m_3=\begin{pmatrix} 0 & 1 \\ 0 & 0 \\ 1 & 0 \\ 1 &  0  \end{pmatrix}\!,\,\,
  n_1=\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ 0 &1 \\ 0 & 1  \end{pmatrix}\!,\,
  n_2=\begin{pmatrix} 0 & 1 \\ 0 & 0 \\ 1 & 0 \\ 1 & 1   \end{pmatrix}\!,\,
  n_3=\begin{pmatrix} 0 & 0 \\ 0 & 1  \\ 1 & 0\\ 0 & 0 \end{pmatrix}\!.
  \end{small}
$$
For $v=(1,1,0,0)^t$ we take
$$
\begin{small}
 m_1=\begin{pmatrix}0 & 1 \\ 0 & 0 \\ 0 & 0 \\ 0 & 1     \end{pmatrix}\!,\,
m_2=\begin{pmatrix} 1 & 1 \\ 1 & 0 \\ 1 & 1 \\ 0 & 0    \end{pmatrix}\!,\,
m_3=\begin{pmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 0 \\ 0 &  1  \end{pmatrix}\!,\,\,
  n_1=\begin{pmatrix} 1 & 1 \\ 1 & 1 \\ 0 &1 \\ 1 & 0  \end{pmatrix}\!,\,
  n_2=\begin{pmatrix} 0 & 0 \\ 0 & 1 \\ 1 & 0 \\ 0 & 0   \end{pmatrix}\!,\,
  n_3=\begin{pmatrix} 0 & 0 \\ 0 & 1  \\ 0 & 0\\ 1 & 1 \end{pmatrix}\!.
\end{small}
$$
For $v=(1,1,1,0)^t$ we take
$$
\begin{small}
 m_1=\begin{pmatrix}1 & 0 \\ 0 & 0 \\ 0 & 0 \\ 1 & 0     \end{pmatrix}\!,\,
m_2=\begin{pmatrix} 0 & 1 \\ 0 & 1 \\ 0 & 1 \\ 1 & 0    \end{pmatrix}\!,\,
m_3=\begin{pmatrix} 0 & 0 \\ 0 & 0 \\ 0 & 0 \\ 1 &  0  \end{pmatrix}\!,\,\,
  n_1=\begin{pmatrix} 0 & 0 \\ 0 & 0 \\ 0 &0 \\ 1 & 1  \end{pmatrix}\!,\,
  n_2=\begin{pmatrix} 1 & 0 \\ 0 & 0 \\ 1 & 0 \\ 1 & 1   \end{pmatrix}\!,\,
  n_3=\begin{pmatrix} 1 & 0 \\ 1 & 0  \\ 0 & 0\\ 0 & 0 \end{pmatrix}\!.
  \end{small}
$$
For $v=(1,1,1,1)^t$ we take
$$
\begin{small}
 m_1=\begin{pmatrix}1 & 0 \\ 0 & 1 \\ 0 & 1 \\ 0 & 0     \end{pmatrix}\!,\,
m_2=\begin{pmatrix} 0 & 1 \\ 0 & 1 \\ 1 & 0 \\ 1 & 0    \end{pmatrix}\!,\,
m_3=\begin{pmatrix} 0 & 1 \\ 0 & 0 \\ 0 & 1 \\ 0 &  1  \end{pmatrix}\!,\,\,
  n_1=\begin{pmatrix} 1 & 0 \\ 1 & 1 \\ 0 &0 \\ 0 & 1  \end{pmatrix}\!,\,
  n_2=\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \\ 1 & 0   \end{pmatrix}\!,\,
  n_3=\begin{pmatrix} 1 & 1 \\ 0 & 1  \\ 1 & 1\\ 0 & 0 \end{pmatrix}\!.
  \end{small}
$$
This completes the proof of Theorem \ref{thm:admissible}.
\end{proof}

We have shown that the tropicalization of the classical Schottky locus
satisfies the constraints coming from the tropical Schottky--Igusa modular forms
in (\ref{eq:tropicalschottkyigusa}). However, these constraints are not yet tight.
The tropical Igusa locus, as we have defined it, is strictly larger than the tropical
Schottky locus. It would be desirable to close this gap, at least for $g=4$.
One approach might be a more inclusive definition of which choices are ``admissible''.

\begin{question}
Can the tropical Schottky locus $\,\mathfrak{J}_4^{\rm trop}$ be cut out by
additional tropical modular forms, notably those obtained in  (\ref{eq:tropicalschottkyigusa})
by allowing choices $\,m_1,m_2,m_3,N\,$ with $\dim N' \leq 2 $?
\end{question}

The next question concerns arbitrary genus $g$.
We ask whether just computing the theta matroid $M(Q)$
solves the Tropical Schottky Decision problem. Note that we
did not address this subtle issue in Algorithm \ref{alg:TSR2} because
we had assumed that the input $Q $ lies in $ \mathfrak{J}_g^{\rm trop}$.

\begin{question}\label{qu:thetamatroid}
 Let $Q$ be a positive definite $g\times g$ matrix such that the
 matroid $M(Q)$ is cographic with positive weights. Does this imply that $Q$ is in the tropical Schottky locus?
\end{question}

If the answer is affirmative then we can use Tutte's classical algorithm \cite{tutte} as
a subroutine for Schottky Decision. That algorithm can decide whether the
matroid $M(Q)$ is cographic.
We close with a question that pertains to classical Schottky Reconstruction  as in Section~\ref{sec:classic}.

\begin{question}\label{qu:riccardo}
How to generalize the results in {\rm \cite{bitangents}} from $g=3$ to $g=4$?
Is there a nice {\em tritangent matrix}, written explicitly in theta constants, for canonical curves of genus four?
\end{question}

\bigskip
\medskip

\noindent
{\bf Acknowledgments.}
We thank Riccardo Salvati Manni for telling us about Kempf's article \cite{kempf}.
We also had helpful conversations with Christian Klein and Emre Sert\"oz.
Lynn Chua was supported by a UC Berkeley University Fellowship and the Max Planck Institute for Mathematics in the Sciences, Leipzig.
Bernd Sturmfels received funding from the US National Science
Foundation (DMS-1419018) and the Einstein Foundation Berlin.

\bigskip

\begin{small}

redpajama_set_name: RedPajamaArXiv


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Digital health records soon available in Principality
Patients in Monaco will soon benefit from having their medical history stored in e-health records. EHRs, as they are otherwise known, are real-time, patient-centered records that make information available instantly and securely to authorised users.
The Government of Monaco and the French Ministry of Solidarity and Health recently met in Paris at the Delegation for European and International Affairs to discuss the future of e-health in the Principality.
The meeting, led by Deputy Delegate for European and International Affairs of the French Ministry of Health Elvire Aronica and Minister of Social Affairs and Health Didier Gamerdinger, was dedicated to discussions on shared medical files and the positive impact on caring for patients in future.
E-health records will soon be available in Monaco and will contain pertinent information on any health-related act carried out in Monaco, except in the case of explicit refusal by the patient, allowing both authorised medical professionals and patients to have records available at their fingertips, saving time and possibly lives.
The ideal set-up is to have an open exchange system with health care providers outside of the Principality, particularly in France, to ensure the best care is possible by allowing medical teams access to instant patient data, and opening the doors to the newest telemedical practices.
"The shared medical file that will be set up in the Principality of Monaco will have to be interoperable with foreign healthcare systems and more particularly with the French system, to be truly comprehensive and at the service of the patient," said Didier Gamerdinger.
Monaco is dedicated to the implementation of digital health records within its borders, but is also taking measures to educate health professionals abroad on the benefits of using technology in patient assessment.
In October, Monaco hosted the e-HealthWorld Conference at the Monte-Carlo Bay Hotel, bringing together health care workers to learn more about connected health care options, notably the burgeoning telemedicine field.
Telemedicine, sometimes called telehealth, is the distribution of health related services and information via email, phone or video conference. It allows long-distance clinician and patient interactions with regard to care, advice, diagnosis, reminders, monitoring, education and even more advanced situations such as remote robotic surgeries, physical therapy at a distance or test interpretation.
The advent of e-health can allow patients, particularly the elderly or those with limited mobility, almost instant access to health care as well as creating a system whereby patients are not required to be physically present in a hospital or doctor's office to receive information.
Top photo: ©DR


Rise Above Performing Arts is thrilled to announce the cast of The Addams Family!
The Addams Family will be performed May 2-5, 2019 at the Glenridge Performing Arts Center 7333 Scotland Way Sarasota, FL 34238. Get your tickets today at http://riseabovearts.com/upcoming-events/.


text:
Quiz: Place These One-Hit Wonders in Heaven or Hell and We'll Guess How Old You Are: HowStuffWorks
Place These One-Hit Wonders in Heaven or Hell and We'll Guess How Old You Are
By: Kennita Leon
Image: SBK Records
There's something about music that can soothe even the grumpiest of folk. What many believe to be magic, has the power to heal others, make some of us move and give bravery to those who don't think they have the strength. Of course, there are particular songs that we listen to when we're in a certain mood and oftentimes, these songs are the really popular ones to which everyone knows the words.
But even popular songs can sometimes have flaws. Sometimes the artists singing those songs were blessed with just that one song- which is why we call them one-hit wonders. Can you imagine having an entire career spanning years in the music industry, to only have one really great song? But we're not here to discuss that. Instead, we want you to rate some pretty unforgettable one-hit wonders for us.
Tell us whether these songs belong in heaven or hell and then we'll tell you how old we think you really are. If you agree to our terms, let's stop wasting time and get started right away!
Hopefully, our answer isn't wrong because today is your birthday!
Where does "I'm Too Sexy" belong?
Which option works for Willa Ford's "I Wanna Be Bad"?
What about Ylvis' "The Fox (What Does the Fox Say?)"?
Where would you put "Cotton Eye Joe"?
Where do you think "The Ketchup Song" should go?
Can you place "Barbie Girl" by Aqua for us?
Tell us where you'd put Hinder's "Lips of an Angel."
Where should "Ice Ice Baby" go?
What about "Don't Drop That Thun Thun" by The Finatticz?
Where would you put Billy Ray Cyrus' "Achy Breaky Heart"?
Which place does "This Is Why I'm Hot" belong?
Remember "Harlem Shake" by Baauer? Where would you put it?
What's your opinion on where "One of Us" by Joan Osborne should go?
Where are you going with for "Blue (Da Ba Dee)"?
And what about "Who Let the Dogs Out?"
Where does "Kung Fu Fighting" belong?
Which option best suits D4L"s "Laffy Taffy"?
What about Magic!'s "Rude"?
Where would you put "Gangnam Style"?
Where do you really think "Funkytown" should go?
Can you place Tag Team's "Whoomp! (There It Is)" for us?
Tell us where you'd put "Mickey" by Toni Basil?
Where should "Crank That" by Soulja Boy go?
And "Juju On That Beat"?
Where would you put "Play that Funky Music"?
Where does "Nothing Compares 2 U" belong?
Remember Sir Mix-a-Lot's "Baby Got Back"? Where should it go?
What's your opinion on Debby Boone's "You Light Up My Life"?
Where are you going with for "Macarena"?
And what about "Take On Me"?
Can We Guess How Old You Are Based on What You Find Attractive?
How Old Is Your Spirit?
Can We Guess How Old You Are and How Old You Act Based on These Strange Questions?
Can We Guess How Old You Are and How Old You Act Based on These Random Questions?
Order a Bunch of Food From Olive Garden and We'll Guess How Old You Are
Walk Us Through Your Day and We'll Guess How Old You Are
Build a Taco Bell Order and We'll Guess How Old You Are
🎸 Pick 30 of Your Favorite Classic Rock Songs and We'll Guess Your Age and Gender
Place These Cars in Heaven or Hell and We'll Guess What Truck You Own
What's Your Dark Power?


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Bev Benda: My Coach Bev — When Money Doesn't Grow on Trees, Must We Cut Them Down?
Six years ago, my childhood home was bulldozed after the new owners had a fire and the city condemned it. When I learned that it took less than a half-hour to crush our huge old house into a pile of rubble and lay pavement in its place, I felt like my parents and seven sisters were also crushed into that pile.
Our home was one of the original homesteads of Wahpeton, N.D. My mother painted it Bubblegum Pink for as many years as I lived in it. In her words, "Who couldn't smile looking at this lovely pink?" My brother-in-law, Jim, nicknamed it The Wad. However, to sell it after my mother died, the Realtor suggested we paint it a drab gray.
It was hard to look at it after that. I avoided it even more once I heard it was torn down. Yet when I finally mustered up the courage to see the property, I discovered grace had been granted.
Driving up to my old address on a sunny July day, I discovered thick green grass and the majestic trees of my youth still standing. Yes, there was the big old tree that served as first base in hundreds of backyard softball games (until the night my twin sister knocked a homer through the glass porch window.) The white picket fence separating our property from the neighbors was shedding paint, but it was still ready to serve as "steady catcher." The rose of China and flowering crab trees were in full bloom, fully pinkifying my world again. The banana tree out back was still shedding "baby bananas," sustenance for our dolls and stuffed animals.
In the years since, I was willing to go back and visit my childhood home of grass and trees, plop down under the big old tree and bask in a whirlwind of memories.
Oh, if those trees could talk, they would tell of Kick the Can, Ante—I-Over and Starlite Moonlight games, snowball fights, crab apple battles, productive gardens, family reunions, animal funerals and burials and even a Wimbledon grass court in celebration of our favorite tennis event. They would tell of how my mother fed hobos picnic meals as they jumped off the trains near our backyard. The trees heard her explain to us why she did it: "God says to be kind to strangers, for you never know when you are entertaining angels." From their mighty stature, they saw so much, including her last breath. Life, death, it all took place here.
The city owned the land after the fire and eventually put it up for sale. Because it was next to the railroad tracks, it wasn't very marketable, so I had this feeling I could visit it as my own for many years. Maybe it will become a little city park, just green grass and trees, where people (like me) can enjoy the greenery and the fragrant trees in the summertime?
To sell it, the city rezoned my old address from residential to commercial and sold it recently to a computer business. Without any input from the neighbors, the computer gurus promptly cut down all the trees. Our beloved trees!
I don't begrudge any business for buying property that is inexpensive. However, maintaining the beautiful trees of one of the original homesteads in Wahpeton, rather than impulsively cutting them down, is a decision that could have been made with more consideration.
You may say it's not my home anymore, so it's none of my business. If this computer business wants trees, it can easily create trees on computer with a few clicks of a mouse. Never mind that computer-generated trees don't provide shade or permeate an incredible fragrance blocks away — with eyes fixed to computer screens, these business owners are oblivious to the loss to the community. They aren't concerned either about longtime neighbors who are mourning the loss of the trees. Money was the issue. And money did not grow on our trees.
Fortunately, no tree-chomping truck will ever take those away from us.
I enjoyed reading about your love of trees and your memories. We bought our house in EGF in an old neighborhood because of the big, old trees and the view of more in the distance from our windows. We both grew up with trees and needed them in our yard. We are moving soon and know it will be a selling feature for the new owners. I am sure the whole neighborhood is missing those trees. So glad you had a chance to visit them and remember your childhood before they were cut. I write poems for fun and once wrote a poem for a friend's 40th birthday party. She was complaining about getting old so the first line was: I thought that I would never see, a friend as old as a tree. I do not recall the rest but we had a good laugh and nobody laughed more than she did. I think I will go hug a tree.
Thank you, Helen, for your kind and comforting words. Bless you, and keep hugging those trees!


text:
Intervals of clouds and sunshine. High 102F. Winds NW at 10 to 15 mph..
Partly cloudy. A stray shower or thunderstorm is possible. Low near 80F. WSW winds shifting to SSE at 10 to 15 mph.
editor's pick spotlight top story
Logan Burtch-Buus, Tucson Local Media
As one would expect after the narrative of the 2018 election in Oro Valley, there are those within the community who believe the future operations of the town-owned 45 golf holes and the community center are of the utmost importance to the town's financial future. At least, that's what it seems like in our Voices section.
And while I am by no means discounting the importance of that fateful 2014 acquisition of the facilities, the ongoing operation or the importance of the community center and its many amenities to the town—there are near-countless decisions to be made from the dais if a town is to flourish. While each choice in their own right may not alter a significant portion of a municipality's future (though many do), it is the collective sum of those decisions that creates a community.
While each local legislative body reigns over a different portion of the region, the community center and its amenities are in Oro Valley—and within the control of its seven-member council.
Mayor Joe Winfield, Vice Mayor Melanie Barrett and Councilmembers Joyce Jones-Ivey and Josh Nicolson each put the community center and golf courses front and center in their campaigns last year, though it's been a relatively quiet stretch on the subject from the entire council since the four new members took office.
There are plenty of reasons the golf courses and community center hasn't received the lion's share of the limelight in recent months, as anything involving government takes time. And though it's completely reasonable that a decision hasn't yet been made, it's clear some people in the community are interested in accelerating the process and introducing more public involvement.
That's at the heart of this week's update from Kathleen Kunz, who recently delved into the efforts of the Men's Golf Association at the El Conquistador and the Cañada Hills Homeowners Association. Members of the two community groups have met with the mayor and town staff this year to discuss their plan to introduce new revenue and cost saving measures.
As is the way with bureaucracy, that proposal hasn't gained too much traction with the council, and it hasn't seen much in the way of public attention, either. With the newly-formed Budget and Finance Commission and the town council both meeting on the community center this month, there's sure to be plenty of time for residents to voice their opinion before any decisions are made.
To read Kathleen's update, click here.
Police Beat: Taser Chase


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