text stringlengths 14 5.77M | meta dict | __index_level_0__ int64 0 9.97k ⌀ |
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Endermologie is the world's first patented technology to obtain market clearance in the USA for cellulite reduction.
The treatment uses therapeutic massagers with specific manoeuvres working on various levels of the skin. Over a period of time, the treatments will provide a significant reduction in cellulite and improve skin texture.
Recognised as the world's leading non-invasive treatment for cellulite and body shaping, Endermologie really does get you amazing results. Just book in for a free consultation and get all your questions answered by filling in the box on the right or by calling reception of 01179 500 500.
Here's to a cellulite-free, smooth skin future! | {
"redpajama_set_name": "RedPajamaC4"
} | 5,873 |
La Ville-sous-Orbais és un municipi francès, situat al departament del Marne i a la regió del Gran Est. L'any 2007 tenia 46 habitants.
Demografia
Població
El 2007 la població de fet de La Ville-sous-Orbais era de 46 persones. Hi havia 23 famílies, de les quals 11 eren unipersonals (7 homes vivint sols i 4 dones vivint soles), 4 parelles sense fills, 4 parelles amb fills i 4 famílies monoparentals amb fills.
La població ha evolucionat segons el següent gràfic:
Habitants censats
Habitatges
El 2007 hi havia 35 habitatges, 22 eren l'habitatge principal de la família, 10 eren segones residències i 3 estaven desocupats. Tots els 35 habitatges eren cases. Dels 22 habitatges principals, 14 estaven ocupats pels seus propietaris, 4 estaven llogats i ocupats pels llogaters i 5 estaven cedits a títol gratuït; 2 tenien una cambra, 6 en tenien tres, 5 en tenien quatre i 10 en tenien cinc o més. 12 habitatges disposaven pel capbaix d'una plaça de pàrquing. A 13 habitatges hi havia un automòbil i a 7 n'hi havia dos o més.
Piràmide de població
La piràmide de població per edats i sexe el 2009 era:
Economia
El 2007 la població en edat de treballar era de 30 persones, 20 eren actives i 10 eren inactives. De les 20 persones actives 19 estaven ocupades (12 homes i 7 dones) i 1 aturada (1 dona i 1 dona). De les 10 persones inactives 5 estaven jubilades, 4 estaven estudiant i 1 estava classificada com a «altres inactius».
Activitats econòmiques
Dels 2 establiments que hi havia el 2007, 1 era d'una empresa de construcció i 1 d'una empresa de serveis.
L'únic servei als particulars que hi havia el 2009 era un paleta.
L'any 2000 a La Ville-sous-Orbais hi havia 8 explotacions agrícoles que ocupaven un total de 696 hectàrees.
Poblacions més properes
El següent diagrama mostra les poblacions més properes.
Referències
Résumé statistique Fitxa resum de dades estadístiques de La Ville-sous-Orbais a l'INSEE.
Évolution et structure de la population Fitxa amb el detall de dades de La Ville-sous-Orbais a l'INSEE
France par commune Dades detallades de tots els municipis de França accessibles a partir del mapa.
Municipis del Marne | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 1,575 |
package me.itsrishi.exercisecounter.activities;
import android.content.DialogInterface;
import android.content.Intent;
import android.net.Uri;
import android.os.Build;
import android.os.Bundle;
import android.support.v7.app.AlertDialog;
import android.support.v7.app.AppCompatActivity;
import android.support.v7.widget.AppCompatButton;
import android.support.v7.widget.AppCompatEditText;
import android.support.v7.widget.SwitchCompat;
import android.support.v7.widget.Toolbar;
import android.view.Menu;
import android.view.MenuItem;
import android.view.View;
import android.widget.CompoundButton;
import android.widget.Toast;
import com.marcoscg.easylicensesdialog.EasyLicensesDialogCompat;
import java.util.ArrayList;
import java.util.Locale;
import butterknife.BindView;
import butterknife.ButterKnife;
import me.itsrishi.exercisecounter.R;
import me.itsrishi.exercisecounter.models.Exercise;
import me.itsrishi.exercisecounter.models.Session;
public class ExerciseCreateActivity extends AppCompatActivity implements CompoundButton.OnCheckedChangeListener {
@BindView(R.id.num_turns)
AppCompatEditText numTurns;
@BindView(R.id.time_per_turn)
AppCompatEditText timePerTurn;
@BindView(R.id.gap_between_turns)
AppCompatEditText gapBetweenTurns;
@BindView(R.id.exercise_submit)
AppCompatButton exerciseSubmit;
@BindView(R.id.exercise_name)
AppCompatEditText exerciseName;
@BindView(R.id.autoplay)
SwitchCompat autoplay;
@BindView(R.id.ex_create_activity_toolbar)
Toolbar toolBar;
private Exercise exercise;
private Session session;
private ArrayList<Session> sessions;
private int index;
private int position;
private boolean shouldAutoplay = true;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_exercise_create);
ButterKnife.bind(this);
setSupportActionBar(toolBar);
setTitle(R.string.title_activity_exercise_create);
if (savedInstanceState != null) {
sessions = savedInstanceState.getParcelableArrayList("sessions");
index = savedInstanceState.getInt("index");
position = savedInstanceState.getInt("position");
} else {
sessions = getIntent().getParcelableArrayListExtra("sessions");
index = getIntent().getIntExtra("index", 0);
position = getIntent().getIntExtra("position", 0);
}
session = sessions.get(index);
try {
exercise = session.getExercises().get(position);
} catch (IndexOutOfBoundsException ex) {
ex.printStackTrace();
} catch (NullPointerException ex) {
session.setExercises(new ArrayList<Exercise>());
}
if (exercise != null) {
setTitle(R.string.title_activity_session_edit);
exerciseName.setText(exercise.getName());
numTurns.setText(String.format(Locale.ENGLISH, "%d", exercise.getTurns()));
timePerTurn.setText(String.format(Locale.ENGLISH, "%f", exercise.getTimePerTurn()));
gapBetweenTurns.setText(String.format(Locale.ENGLISH, "%f", exercise.getGapBetweenTurns()));
if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.ICE_CREAM_SANDWICH) {
shouldAutoplay = exercise.getAutoplay();
autoplay.setChecked(exercise.getAutoplay());
}
}
autoplay.setOnCheckedChangeListener(this);
exerciseSubmit.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
if (updateSessions()) return;
Intent intent = new Intent(ExerciseCreateActivity.this, SessionCreateActivity.class);
intent.setFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP);
intent.putExtra("index", index);
intent.putExtra("sessions", sessions);
startActivity(intent);
}
});
ButterKnife.bind(this);
}
/**
* @return If an error occurs, this flag is raised
*/
private boolean updateSessions() {
String exName = exerciseName.getText().toString();
int turnCount;
float gapCount;
float timeCount;
try {
turnCount = Integer.valueOf(numTurns.getText().toString());
gapCount = Float.valueOf(gapBetweenTurns.getText().toString());
timeCount = Float.valueOf(timePerTurn.getText().toString());
} catch (NumberFormatException ex) {
Toast.makeText(ExerciseCreateActivity.this, "Enter valid values", Toast.LENGTH_SHORT).show();
return true;
}
if (exName.equals("")) {
Toast.makeText(ExerciseCreateActivity.this, "Enter valid name", Toast.LENGTH_SHORT).show();
return true;
}
exercise = new Exercise(exName, turnCount, timeCount, gapCount, -1, shouldAutoplay);
ArrayList<Exercise> exercises = session.getExercises();
if (position < exercises.size())
exercises.set(position, exercise);
else exercises.add(exercise);
session.setExercises(exercises);
sessions.set(index, session);
return false;
}
@Override
public void onBackPressed() {
AlertDialog.Builder alertBuilder = new AlertDialog.Builder(this);
alertBuilder.setMessage("Do you want to save the changes?");
alertBuilder.setTitle("Save");
alertBuilder.setNegativeButton(R.string.yes, new DialogInterface.OnClickListener() {
@Override
public void onClick(DialogInterface dialog, int which) {
exerciseSubmit.performClick();
}
});
alertBuilder.setPositiveButton(R.string.no, new DialogInterface.OnClickListener() {
@Override
public void onClick(DialogInterface dialog, int which) {
ExerciseCreateActivity.super.onBackPressed();
}
});
alertBuilder.show();
}
@Override
public void onCheckedChanged(CompoundButton buttonView, boolean isChecked) {
this.shouldAutoplay = isChecked;
}
@Override
public void onSaveInstanceState(Bundle outState) {
super.onSaveInstanceState(outState);
updateSessions();
outState.putParcelableArrayList("sessions", sessions);
outState.putInt("index", index);
outState.putInt("position", position);
}
@Override
public boolean onCreateOptionsMenu(Menu menu) {
getMenuInflater().inflate(R.menu.menu, menu);
return true;
}
@Override
public boolean onOptionsItemSelected(MenuItem item) {
Intent intent;
switch (item.getItemId()) {
case (R.id.action_favorite):
intent = new Intent(Intent.ACTION_VIEW,
Uri.parse("https://play.google.com/store/apps/details?id=" + "me.itsrishi.exercisecounter"));
startActivity(intent);
break;
case R.id.action_settings:
intent = new Intent(ExerciseCreateActivity.this, SettingsActivity.class);
intent.setFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP);
startActivity(intent);
break;
case R.id.action_stat:
intent = new Intent(ExerciseCreateActivity.this, StatsActivity.class);
intent.setFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP);
startActivity(intent);
break;
case R.id.action_about:
intent = new Intent(Intent.ACTION_VIEW, Uri.parse("https://rishiraj22.github.io"));
startActivity(intent);
break;
case R.id.action_license:
new EasyLicensesDialogCompat(this)
.setTitle("Open source licenses")
.setPositiveButton(android.R.string.ok, null)
.show();
break;
}
return true;
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 3,338 |
It's Friday again and Mumbai for Kids time…this week's Times of India article is about Nature Clubs in and around Mumbai.
It's about discovering heaven on earth for kids who venture out of the city and take the dirt track into the wild. And a growing number of NGOs and agencies are making this possible through camps, nature trails and other excursions into the great outdoors. They know an early appreciation of nature will not only sensitise children to vital issues like ecological balance and the value of biodiversity but will make a child's world view that much richer and more inclusive.
Another biggie in the field is the World Wildlife Fund (WWF), a pioneer in the country's Nature Clubs of India movement since 1976. With WWF, budding environmentalists can join trained guides for nature trails, habitat analysis, film shows, zoo visits and poster competitions. Children look forward to receiving issues of Panda Bulletin in English or Nisarg Vartta in Marathi, along with discounts on cool WWF products like sleeping bags, T-shirts and key chains. But for children who've glimpsed the wild, the forest experience itself is infinitely cooler. Education officer Goldin Quadros marvels at how instantly seasonal changes in nature get imprinted on young minds. "They are amazingly receptive to the slightest difference in sights and sounds on the forest floor,'' he says.
Sheathed in gloves and gumboots, families also love to get grubby at CampHarit, a 20-acre organic farm at Vikramgadh in Thane district, run by chartered accountant Vinod Harital. The programnme here includes getting down and dirty in the field, planting herbs (they have 150 varieties), sowing seeds, watering plants, de-weeding, gathering organic waste, plucking produce and drying and segregating harvests. The camp also has naturalist experts at hand to point out to children the range of flora and fauna around.
Another friend of nature is the aptly named SPROUTS (Society for Promotion of Research, Outdoors, Urbanity, Training and Social Welfare) which plans periodic trips on local soil, and slightly beyond, to places like Phansad Wildlife Sanctuary near Alibaug. Here kids absorb little known facts about beach and marine fauna, and spot rare giant squirrels and hornbills darting under the green cover of the northern Western Ghats. Outfits like Countryside Outdoors have come up with canny ways of keeping children hooked to the great outdoors for life.
Under the group's Free Spirit plan, historic fort ramparts, over gentle peaks or trekking to niches of the Sahyadri Hills—children are urged to sketch their outings. Winners of imaginative visuals will another round of adventurous excursions—free! And you thought only the city gave out good bargains.
This entry was tagged Education, Events, Mumbai for Kids Times of India. Bookmark the permalink.
The cost for GCE is low as compared to British Council. | {
"redpajama_set_name": "RedPajamaC4"
} | 6,418 |
\section*{Introduction}
The idea that the ground state of a frustrated spin 1/2 antiferromagnet
should be describable in terms of products of spin singlets was suggested
originally in the context of triangular spin $\frac{1}{2}$ antiferromagnets
\cite{PWA}. Products of singlet states are exact ground states of certain
frustrated spin 1/2 antiferromagnetic spin chains with special couplings
\cite{Majumbdar} and they have been used in variational approaches \cite
{variational} and in heuristic arguments.
In the present note we study the $O(N)$ Heisenberg antiferromagnet in the $%
N\rightarrow \infty $ limit by using a constraint free parametrisation of
the $O(N)$ Heisenberg antiferromagnet in terms of real fermions. We show
that the $N\rightarrow \infty $ limit is dominated by a mean field
configuration with small $\sim \frac{1}{\sqrt{N}}$ fluctuations. We are able
to locate the saddle point of lowest energy in the case of unfrustrated
antiferromagnets and for a subclass of frustrated ones. However, in many
cases we do not find a single saddle point, but instead an infinity of
saddle points that are all degenerate in energy. These saddle points are
essentially singlets on nearest neighbor dimers that cover the lattice and
correspond to the RVB singlets of Anderson.
Our results are similar to those obtained by D.S. Rokhsar on the $%
N\rightarrow \infty $ limit of the $SU(N)$ extension of the Heisenberg
antiferromagnet \cite{Rokhsar}.
\section*{Majorana representation of the O(N) Heisenberg model.}
It is known that operators that operate on spinors can be generalised from
the O(3) group to O(N) by using O(N) Dirac matrices. Somewhat less familiar
is the fact that these Dirac matrices can also be considered to be real
fermions \cite{Tsvelik}, thus providing us with the following representation
of $O(N)$ spin matrices:
\begin{eqnarray}
s_{k,l} &=&-\frac{i}{2}\left( \eta _{k}\eta _{l}-\eta _{l}\eta _{k}\right)
\text{, }k,l=1,2,..N \\
\text{ }\left[ \eta _{k,}\eta _{l}\right] _{+} &=&\delta _{kl}\text{, }\eta
_{l}^{+}=\eta _{l} \nonumber
\end{eqnarray}
The representations of $s_{k,l}$ can be characterised by the value of their
quadratic invariant
\begin{equation}
s^{2}\equiv \frac{1}{2}\sum_{i,j=1..N}s_{ij}s_{ij}=-\frac{1}{2}\sum_{i\neq
j=1..N}\eta _{i}\eta _{j}\eta _{i}\eta _{j}=\frac{1}{8}\left( N^{2}-N\right)
\end{equation}
For $N=3$ and $O(3)$ this representation reduces to
\begin{eqnarray}
s_{i} &=&-i\eta _{k}\eta _{l}\text{, }i,k,l=1,2,3\text{ cyclic} \\
s^{2} &=&\frac{1}{2}(\frac{1}{2}+1) \nonumber
\end{eqnarray}
a representation that has been discussed extensively in the first of ref\cite
{Tsvelik}. To define the $O(N)$ Heisenberg model, we associate a spin
represented by fermions with each point of a lattice:
\begin{eqnarray}
s_{x,kl} &=&-\frac{i}{2}\left( \eta _{xk}\eta _{yl}-\eta _{yl}\eta
_{xk}\right) \\
\left[ \eta _{xk}\text{,}\eta _{yl}\right] &=&\delta _{kl}\delta _{xy}
\nonumber
\end{eqnarray}
and use the invariant scalar product
\begin{equation}
(s_{x}s_{y})_{O(N)}=\frac{1}{2}\sum_{k,l=1}^{N}s_{x,kl}\cdot s_{y,kl})
\end{equation}
to define the interaction:
\begin{equation}
H_{Heisenberg}=\sum_{<x,y>}J_{xy}(s_{x}s_{y})_{O(N)}=\sum_{<x,y>}J_{xy}%
\left[ \frac{N}{8}+\frac{1}{2}\left( \sum_{k=1}^{N}\eta _{x,k}\eta
_{y,k}\right) ^{2}\right]
\end{equation}
The fermions $\eta _{x,k}$ are real as is appropriate in a situation without
electric charges, and there are no constraints on the physical Hilbert
space. To get familiar with the representation of spins in terms of real
fermions, we may calculate the energy of a pair of nearest neighbor points $%
x,y$ or a ''dimer '' that interact via their spins. We do this most easily
by introducing ordinary complex fermions:
\begin{eqnarray}
A_{\text{dimer},k} &=&\frac{1}{\sqrt{2}}\left( \eta _{x,k}+i\eta
_{y,k}\right) \text{, }A_{\text{dimer},k}^{+}=\frac{1}{\sqrt{2}}\left( \eta
_{x,k}-i\eta _{y,k}\right) \text{, }k=1,..N \\
\left( s_{x}s_{y}\right) _{O(N)} &=&\frac{N}{8}-\frac{1}{8}\left[
\sum_{k=1}^{N}(A_{\text{dimer,}k}^{+}A_{\text{dimer},k}-A_{\text{dimer,}k}A_{%
\text{dimer},k}^{+})\right] ^{2} \nonumber
\end{eqnarray}
and find
\begin{equation}
E_{\text{dimer}}\equiv <\left( s_{x}s_{y}\right) _{O(N)}>=-\frac{N^{2}}{8}+%
\frac{N}{8} \label{dimer}
\end{equation}
The ground state energy of a dimer is seen to vary smoothly as a function of
$N$.
\section*{A toy model}
To gain further insight into the nature of the $N\rightarrow \infty $
extrapolation, we calculate, for arbitrary $N$, the ground state energy of
four spins located at the corners of a square with interactions along the
edges and across the diagonals:
\begin{eqnarray}
H_{square} &=&s_{1}s_{2}+s_{2}s_{3}+s_{3}s_{4}+s_{4}s_{1}+\varepsilon \left(
s_{1}s_{3}+s_{2}s_{4}\right) \nonumber \\
&=&\frac{1}{2}\left[ \left( s_{1}+s_{3}\right) +\left( s_{2}+s_{4}\right)
\right] ^{2} \label{square} \\
&&+\frac{\varepsilon -1}{2}\left( \left( s_{1}+s_{3}\right) ^{2}+\left(
s_{2}+s_{4}\right) ^{2}\right) -2\varepsilon s^{2} \nonumber
\end{eqnarray}
Since one knows how to combine representations of O(3), one easily obtains
the spectrum of $H_{square}$ for $N=3$. One finds a singlet for the ground
state with an energy of
\begin{equation}
E_{square,3}=\left\{
\begin{array}{l}
\frac{\varepsilon }{2}-2\text{, }\varepsilon \leq 1 \\
-\frac{3\varepsilon }{2}\text{, }\varepsilon >1
\end{array}
\right\}
\end{equation}
We assume in our calculation for $O(N)$, with arbitrary $N\geq 3$, that the
ground state continues to be a singlet, so $\left[ \left( s_{1}+s_{3}\right)
+\left( s_{2}+s_{4}\right) \right] ^{2}=0$ in eq(\ref{square}). It remains
to find the spectra of $\left( s_{1}+s_{3}\right) ^{2}$ and $\left(
s_{2}+s_{4}\right) ^{2}$. We now use standard Dirac Gamma Matrices instead
of real fermions in our argument and reduce the tensor product of two $O(N)$
spinor representations, say $\xi ^{\alpha }\phi ^{\beta }$, of $2^{\left[
N/2\right] }$ dimensions each, to calculate $\left( s_{1}+s_{3}\right) ^{2}$
and $\left( s_{2}+s_{4}\right) ^{2}$. The reduction of $\xi ^{\alpha }\phi
^{\beta }$ is achieved via antisymmetrised products of Dirac $O(N)$ gamma
matrices \cite{Georgi}:
\begin{equation}
\xi ^{\alpha }\eta ^{\beta }\rightarrow \xi \eta \text{, }\xi \Gamma _{\mu
}\eta \text{, }\xi \Gamma _{\mu _{1}\mu _{2}}\eta \text{, ...,}\xi \Gamma
_{\mu _{1}...\mu _{N}}\eta \label{reduction}
\end{equation}
where $\Gamma _{\mu _{1}...\mu _{n}}$ denotes a product of $n$ gamma
matrices that are antisymmetrised in their indices. The $O(N)$ spin of an
antisymmetric tensor $\Gamma _{\mu _{1}..\mu _{n}}$ is most easily found by
treating it as Grassmann numbers acted upon by $O(N)$ generators $%
s_{ij}=\Gamma _{i}\frac{d}{d\Gamma _{j}}-\Gamma _{j}\frac{d}{d\Gamma _{i}}$.
In this way one can find without too much difficulty that
\begin{equation}
s^{2}\Gamma _{\mu _{1}..\mu _{n}}\equiv \frac{1}{2}%
\sum_{i,j=1..N}s_{ij}s_{ij}\Gamma _{\mu _{1}..\mu _{n}}=n(N-n)\Gamma _{\mu
_{1}..\mu _{n}} \label{eigen}
\end{equation}
with a minimum eigenvalue of $s^{2}$at $n=[\frac{N}{2}]$ $\equiv \frac{N}{2}-%
\frac{1}{2}\delta _{N,odd}$ and with maxima at $n=0,N$. This results in the
following ground state energy of $H_{square}$ for arbitrary $N$:
\begin{equation}
E_{square ,N}=\left\{
\begin{array}{l}
-\frac{1}{4}N^{2}+\frac{\varepsilon }{4}N-\frac{1}{4}\left( \varepsilon
-1\right) \delta _{N,odd}\text{, }\varepsilon \leq 1 \\
-\frac{\varepsilon }{4}N^{2}+\frac{\varepsilon }{4}N\text{,}\varepsilon >1
\end{array}
\right\} \label{groundstate}
\end{equation}
The ground state enery $E_{square,N}$ in eq(\ref{groundstate}) reduces
smoothly to its value at $N=3$, with a relative precision of order $\frac{1}{%
N^{2}}\sim 10\%$ for $N=3$, if one includes only the two leading terms, and
we conclude that the $O(N)$ extrapolation is satisfactory in this toy
example. By comparing with eq(\ref{dimer}) we notice that the leading $%
O(N^{2})$ term of $E_{square,N}$ can be interpreted in terms of the
formation of two dimers with coupling $\varepsilon $ or $1$, whichever is
larger. We shall see below that this is part of a more general pattern that
emerges at $N=\infty $.
\section*{Nature of the saddle point at N=$\infty $}
To use standard $N\rightarrow \infty $ techniques \cite{largeNpapers} we
rewrite the quartic interaction in terms of an auxiliary field \cite{Tsvelik}
\begin{eqnarray}
H &=&\sum_{<x,y>}J_{xy}(s_{x}s_{y}-\frac{N}{8})=\sum_{<x,y>}\frac{J_{xy}}{2}%
(\eta _{x}\eta _{y})^{2} \nonumber \\
Z &=&\int D\eta e^{-\int_{0}^{\beta }dt\left( \frac{1}{2}\eta _{\mu
}\partial _{t}\eta _{\mu }+H\right) }=const(\beta )\int DBD\eta e^{-S}
\nonumber \\
&=&const(\beta )\int DBe^{-\int_{0}^{\beta }dt(\frac{1}{4}\sum_{x,y}\frac{%
B_{xy}^{2}}{J_{xy}}-\frac{N}{2}\log (\det (\partial _{\tau }+iB))} \nonumber
\\
S &=&\int_{0}^{\beta }dt\left( \frac{1}{4}\sum_{x,y}\frac{B_{xy}^{2}}{J_{xy}}%
+\frac{1}{2}\sum_{x,\mu }\eta _{x\mu }\partial _{t}\eta _{x\mu }+\frac{i}{2}%
\sum_{x,y}B_{xy}\eta _{x\mu }\eta _{y\mu }\right) \label{aux}
\end{eqnarray}
With $J_{xy}$ scaling as $\frac{1}{N}$ the exponent is of order $N$ and the
integration over $B_{xy}$ is dominated by ''classical '' configurations of $%
B_{xy}$ plus fluctuations of $B_{xy}$ of order $O(1/\sqrt{N})$. To find the
optimal solution(s) of eq(\ref{aux}) at $T=0$ it is easiest to consider the
Hamiltonian that describes the saddlepoint of eq(\ref{aux}) and which is
given by
\begin{equation}
H_{\infty }=\frac{1}{4}\sum_{x,y}\frac{B_{xy}^{2}}{J_{xy}}+\frac{1}{2}%
\sum_{x,y,\mu }iB_{xy}\eta _{x\mu }\eta _{y\mu } \label{hinfinity}
\end{equation}
The optimal auxiliary fields $B_{xy}$ are those that give the lowest energy
and which minimise $<H_{\infty }>$. The matrix $iB_{x,y}$ is hermitean and
antisymmetric and its eigenvalues occur in complex conjugate pairs $(\xi
_{n},\lambda _{n}),(\xi _{n}^{*},-\lambda _{n})$, for $n=1,2..\frac{1}{2}%
\#(points)$. This property of $iB_{x,y}$ enables us to rewrite the kinetic
energy of the{\em \ real }fermions $\eta _{x\mu }$ in terms of {\em complex}
fermions with standard oscillator commutators:
\begin{eqnarray}
\frac{1}{2}\sum_{x,y,\mu }iB_{xy}\eta _{x\mu }\eta _{y\mu } &=&\frac{1}{2}%
\sum_{n=1}^{\frac{1}{2}\#(points)}\sum_{\mu =1}^{N}\lambda _{n}\left(
A_{n,\mu }^{+}A_{n,\mu }-A_{n,\mu }A_{n,\mu }^{+}\right) \label{oscillate}
\\
\text{with }A_{n,\mu } &=&\sum_{x=1}^{\#(points)}\xi _{nx}\eta _{x\mu }\text{%
, } \nonumber \\
\left[ A_{m,\mu }^{+},A_{n,\nu }\right] _{+} &=&\delta _{mn}\delta _{\mu \nu
}\text{; }\left[ A_{m,\mu },A_{n,\nu }\right] _{+}=0\text{; }m,n=1..\frac{1}{%
2}\#(points) \nonumber
\end{eqnarray}
We deduce from eq(\ref{oscillate}) that the zero point energy of the
fermions is given by
\begin{equation}
<\frac{1}{2}\sum_{x,y,\mu }iB_{xy}\eta _{x\mu }\eta _{y\mu }>=-\frac{N}{4}%
\sum_{\text{all }\lambda }|\lambda | \label{zeropoint}
\end{equation}
where one must sum over all $\#(points)$ eigenvalues. Combining eqs(\ref
{hinfinity},\ref{zeropoint}) we may rewrite the ground state energy at $%
N=\infty $ as
\begin{equation}
E=\frac{1}{4}\sum_{x,y}\frac{B_{xy}^{2}}{J_{xy}}-\frac{N}{4}\sum_{\text{all }%
\lambda }|\lambda | \label{start}
\end{equation}
In the case of frustrated magnetism we have, in general, distinct matrices $%
\frac{B}{\sqrt{J}}$ and $iB$ that do not commute. However, for a single
nonzero coupling $J_{xy}=J$ the energy associated with the auxiliary field
can be simplified and related to the spectrum of the matrix $iB$:
\begin{eqnarray}
\sum_{x,y}\frac{B_{xy}^{2}}{J_{xy}} &=&\frac{1}{J}\sum_{x,y}B_{xy}^{2}=-%
\frac{1}{J}\sum_{x,y}B_{xy}B_{yx}\text{, }B_{yx}=-B_{xy} \label{spectral} \\
&=&-\frac{1}{J}TrB^{2}=\frac{1}{J}Tr\left( iB\right) \left( iB\right) =\frac{%
1}{J}\sum_{\text{all }\lambda }\lambda ^{2} \nonumber
\end{eqnarray}
Combining eqs(\ref{hinfinity},\ref{zeropoint},\ref{spectral}) we obtain a
lower bound on the ground state energy:
\begin{equation}
E=\sum_{\text{all }\lambda }\left( \frac{\lambda ^{2}}{4J}-\frac{|\lambda |N%
}{4}\right) =\frac{1}{4J}\sum_{\text{all }\lambda }\left[ (|\lambda |-\frac{%
JN}{2})^{2}-\frac{(JN)^{2}}{4}\right] \geq -\frac{JN^{2}}{16}\text{\#(points)%
} \label{bound}
\end{equation}
We have used that the matrix $B_{x,y}$ has as many eigenvalues as there are
points on the lattice. We have thus obtained a lower bound for the energy of
an $O(\infty )$ antiferromagnet with a single coupling $J$.
For unfrustrated Heisenberg Hamiltonians on a lattice of points that can be
considered as a union of non overlapping pairs of points or ''dimers '' it
is easy to saturate this bound. For such Hamiltonians, we may choose a
configuration of $B_{xy}$ that is nonvanishing only on an arbitrary
collection of nonoverlapping dimers that cover the whole lattice. By
hypothesis, the couplings are the same on all these dimers. The matrix $%
B_{x,y}$ then decomposes into blocks, one for each dimer, and according to
eq(\ref{bound}) the energy can be rewritten as
\begin{eqnarray}
E &=&\frac{1}{4J}\sum_{\text{dimers}}\sum_{i=1,2}\left[ (|\lambda _{i}|-%
\frac{JN}{2})^{2}-\frac{(JN)^{2}}{4}\right] \stackrel{|\lambda |=\frac{JN}{2}%
}{\rightarrow }\frac{1}{4J}\sum_{\text{dimers}}\sum_{i=1,2}-\frac{(JN)^{2}}{4%
} \nonumber \\
&=&-\frac{JN^{2}}{16}\cdot 2\cdot \text{\#(dimers)}=-\frac{JN^{2}}{16}\cdot
\text{\#(points)} \label{saturated}
\end{eqnarray}
Here we have adjusted the block of $B$ that corresponds to each dimer in
such a way that its eigenvalues are $\lambda =\pm \frac{JN}{2}$. We may also
return to equation (\ref{dimer}) to see more directly that the minimal
energy of a collection of nonoverlapping dimers (in their singlet state)
saturates the inequality (\ref{bound}):
\begin{equation}
<\sum_{<x,y>}J(s_{x}s_{y})_{O(N)}>=-\frac{JN^{2}}{8}\cdot \text{\#(dimers)}%
+O(N) \label{simple}
\end{equation}
which visibly saturates eq(\ref{saturated}). Returning to the toy model of
eq(\ref{square}) we now understand why its ground state energy corresponds,
to leading order in $N$, to that of two dimers, but our general arguments
apply only to $\varepsilon =1$ and $\varepsilon =0$ where there is a single
coupling. The toy model suggests the stronger statement that the ground
state on lattices coverable by dimers is one of singlets on the ''strongest
'' dimers.
To follow the hint of the toy model, we reconsider eq(\ref{start}) and
derive a lower bound for the ground state energy of a frustrated
antiferromagnets at $N=\infty $:
\begin{equation}
E=\frac{1}{4}\sum_{x,y}\frac{B_{xy}^{2}}{J_{xy}}-\frac{N}{4}\sum_{\text{all }%
\lambda }|\lambda |\geq \frac{1}{4J_{\max }}\sum_{x,y}B_{xy}^{2}-\frac{N}{4}%
\sum_{\text{all }\lambda }|\lambda | \label{major}
\end{equation}
Here we have used the positivity of the couplings $J_{xy}$. We may now copy
word for word the arguments that lead to the lower bound of eq(\ref{bound})
in the unfrustated case, but with $J$ replaced by $J_{\max }$. We find:
\begin{equation}
E\geq -\frac{J_{\max }N^{2}}{16}\cdot \text{\#(points)} \label{frustlow}
\end{equation}
In some cases this bound can be saturated and lowest energy saddle points
can be found. Consider, for example, the square lattice spin $1/2$
antiferromagnet, with nearest neighbor couplings $J_{1}$ and next nearest
neighbor couplings (across the diagonals) $J_{2}$. In this case we chose the
stronger coupling $J_{\max }=\max (J_{1},J_{2})$, and any dimer covering of
the lattice by the ''stronger bonds '' that correspond to $J_{\max }$. As
before, we saturate the lower bound with these configurations. So the
infinite degeneracy persists in this particular frustrated antiferromagnet.
The $N=\infty $ saddle point of an antiferromagnet on a Kagome lattice\cite
{Kagome} is also infinitely degenerate, because a Kagome lattice can be
covered by dimers in an infinite number of ways.
However, there is only
finite degeneracy in the $N=\infty $ saddle point of
the frustrated spin chains of \cite{Majumbdar}.
Although we have found an infinite number of degenerate saddle points in
certain $O(\infty )$ antiferromagnets we cannot be sure to have found {\em %
all} the saddle points that saturate the lower bound. In particular, the
saddle point may exhibit continuous degeneracies. A one parameter continuous
degeneracy is indeed present in the $O(\infty )$ saddle points of the
frustrated square of eq(\ref{square}), and an analogous degeneracy was also
noted in \cite{Read} for the case of $SU(\infty )$.
\section*{Conclusions and open problems}
We have found that a large class of $O(N)$ Heisenberg models have $N=\infty $
saddle points consisting of products of singlets and which are infinitely
degenerate in many cases.
Because there is a gap at large $N$ but no gap in the unfrustrated model at $%
N=3$ and $N=4$ there should be a critical value of $N$ that separates the
two regimes in the case of unfrustrated antiferromagnets, while for
frustrated antiferromagnets there is no need for such a phase transition.
Our results closely parallel those obtained by D.S. Rokhsar\cite{Rokhsar} on
the $N\rightarrow \infty $ limit of an $SU(N)$ extension of the Heisenberg
antiferromagnet. In the latter case, the effects of $1/N$ corrections are
understood\cite{Read}, while in the $O(N)$ case, the effect of these
corrections is still unknown. Also, it is still an open problem whether the
correlations can be usefully organised in powers of $N$ at $N=3$.
\section*{Acknowledgements}
We acknowledge encouragement by the members of CPTMB and thank S.Meshkov and
Y.Meurdesoif for their help and useful comments and Y. Chen for a critical
reading of the manuscript. F.T. acknowledges support of ENS de Lyon. D.F.
acknowledges support by the ESPRIT / HCM / PECO - Workshop at the Institute
for Scientific Interchange Foundation, Torino, Italy, and discussions with
N.Schopohl.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 4,884 |
Scorched Earth (en anglais : « terre brûlée ») est un jeu vidéo d'artillerie. Développé pour DOS et édité en shareware, il s'agit d'un jeu où des tanks s'affrontent au tour par tour sur un terrain en deux dimensions, chaque joueur ajustant l'angle et la puissance de son tir.
Annexes
Liens internes
Jeu vidéo d'artillerie
Liens externes
Scorched Earth: The Mother of All Games
Scorched Earth (MobyGames)
An interview with the creator of Scorched Earth, Wendell T. Hicken (Ars Technica)
Jeu vidéo d'artillerie
Jeu vidéo sorti en 1991
Jeu DOS
Jeu vidéo développé aux États-Unis | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 5,379 |
ataract is the clouding of the natural crystalline lens and typically occurs because of ageing, although other reasons may be prolonged use of steroids, exposure to ultra violet and associated health problems like diabetes and eye injuries.
Cataract refers to the clouding of the natural lens of the eye. The lens of the eye is similar to the lens of a camera and is responsible for making a sharp image on the retina (light sensitive part at the back of the eye).
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The human eye has a natural lens, which is normally transparent and helps to focus the light on the retina and thus form a clear image. When this lens becomes cloudy, usually due to aging, it is called Cataract or Safed Motiya. This obstructs the light and causes vision to become hazy.
Apart from senile cataract, there are other types. Secondary cataract is due to an eye surgery, diabetes, use of steroids etc. Traumatic cataract follows an eye injury, sometimes many years later. Congenital cataract occurs often in both the eyes and may be due to metabolic diseases, consanguinity, viral infections etc.
A cataract is a progressive painless clouding of the lens inside the eye which leads to a decrease in vision, double vision, glare or halos around light.
Cataracts are the number one cause of blindness worldwide because many places do not have good access to modern care and technology.
Cataract is a clouding or opacity of the lens in the eyes that leads to a gradual loss of vision. It is one of the leading causes of blindness throughout the world especially in elderly people. In the asian sub-continent, millions of people are affected by cataract. More than 60% of those over 65 years of age and 90% of those over age of 85 have opacities of the lens.
Cataract is clouding of the normally transparent lens of the eye.With the increase in opacity of the lens, the light rays are prevented from focusing on retina, leading to blurring of vision. | {
"redpajama_set_name": "RedPajamaC4"
} | 4,326 |
\section{Introduction}
Within the last several years there has been a wealth of papers published in this journal regarding the motion of an orbiting body on a cylindrically-symmetric surface, such as a funnel or a circular spandex fabric, residing in a uniform gravitational field. Some of the interest in this topic can be attributed to fact that the orbital dynamics of an object on such a two-dimensional (2D) surface offers a non-trivial and interesting application of Lagrangian dynamics. With the help of a camera capable of generating short video clips and one of the aforementioned surfaces, one can easily extend this theoretical study to the experimental domain. This combined theoretical and experimental approach offers comprehensive undergraduate research possibilities as the relevant physics is accessible to any student who has taken an advanced dynamics course.\cite{weller}
Perhaps some of the interest in this topic arises from the fact that an orbiting body on a 2D surface is often used as a conceptual analogy when describing particle orbits about massive objects in general relativity. In Einstein's theory of general relativity (GR), gravity is described as the warping of space and time due to the presence of matter and energy. Although in Newtonian theory gravity is understood as the force of one massive object on another, in GR gravity emerges as a physical phenomenon due to the fact that a massive object warps the spacetime around it. According to GR, the planets move along geodesics, or free-particle orbits, in the non-Euclidean geometry about the sun. Hence, the planets move in their elliptical-like orbits aroung the sun not due to a `force' per se but rather due to the fact that the spacetime around the sun is curved by its presence.
In a typical introductory GR course, the spacetime external to a static, spherically-symmetric massive object is often extensively studied as it offers one of the simplest, exact solutions to the field equations of GR. When geodesics about this object are then encountered, the conceptual analogy of a marble rolling on a taut elastic fabric warped by a massive compact object, such as a bowling ball, is often embraced in the discussion. When the rolling marble is far from the massive object, the elastic fabric is flat and the marble moves in a straight line at a constant speed in agreement with Newton's first law. When the rolling marble approaches the massive object, as the analogy goes, it moves in a curved path not due to a force acting on the marble, but rather due to the fact that the space in which it resides (the 2D surface) is warped by the object.
This analogy offers the student of GR a way of visualizing a warped spacetime and should not be confused with embedding diagrams.\cite{embedding} However, with this being said, it has been shown that there exists no 2D cylindrically-symmetric surface residing in a uniform gravitational field that can generate the exact Newtonian orbits of planetary motion, except for the special case of circular orbits.\cite{English} This null result was shown to hold when considering the fully general relativistic treatment of planetary orbits around non-rotating, spherically-symmetric objects.\cite{me2}
White and Walker first explored marbles orbiting on a warped spandex surface by applying Newtonian mechanics to a marble residing on a cylindrically-symmetric spandex fabric subjected to a central mass. There they arrived at an expression describing the approximate shape of the spandex surface and found that a marble orbiting upon this surface obeys the relation $T^3\propto r^2$ for circular orbits in the small slope regime. This fascinating result is highly reminiscent of Kepler's third law for planetary motion, but with the powers transposed.\cite{GW} Lemons and Lipscombe then showed that the shape of the spandex surface can equivalently be found by minimizing the total potential energy of the spandex fabric through the use of the calculus of variations. The total potential energy considered includes the elastic potential energy of the spandex surface and the gravitational potential energy of the central mass. The calculus of variations approach proves advantageous over the Newtonian treatment as the results of White and Walker emerge as a special case of a more general treatment, where a pre-stretch of the fabric can be included in the analysis.\cite{DL} This work was later expanded upon by Middleton and Langston, where the authors considered circular orbits on the spandex fabric in \textit{both} the small and large slope regimes. There they generalized the aforementioned calculus of variations treatment by including the previously-neglected gravitational potential energy of the spandex surface to the total potential energy of the spandex/central mass system. They showed that the mass of the spandex fabric interior to the orbital path influences the motion of the marble and can even dominate the orbital characteristics. They also showed that the modulus of elasticity of the spandex fabric is not constant and is itself a function of the stretch.\cite{me2} Later, Nauenberg considered precessing elliptical-like orbits with small eccentricities on cylindrically-symmetric funnels with various shape profiles. By employing a perturbative method, an approximate solution to the orbital equation of motion was obtained where the precession parameter was found to be determined uniquely by the slope of the cylindrically-symmetric surface.\cite{Nauenberg} This perturbative method was then used to find the 2D surfaces that are capable of generating the stationary elliptical orbits of Newtonian gravitation and the precessing elliptical-like orbits of GR, for orbits with small eccentricities.\cite{me} Here, we're interested in understanding elliptical-like orbits of a marble on a warped spandex fabric.
Our paper is outlined as follows. In Sec. \ref{eqn}, we first consider the motion of a rolling marble on an arbitrary, cylindrically-symmetric surface and present the orbital equation of motion.
For elliptical-like orbits with small eccentricities, we solve this equation of motion perturbatively and find a valid solution when the precession parameter obeys an algebraic relation involving the slope of the surface. We then present the equation that describes the slope of an elastic surface, which arises from minimizing the total potential energy of the spandex/central mass system. In Sec. \ref{orbits} and Sec. \ref{orbits2}, we explore the elliptical-like orbits of the rolling marble in the small and large slope regimes, respectively. We arrive at an expression describing the angular separation between successive like-apsides in each regime and compare with our experimental results.\cite{apocenter} Lastly, in Appendix \ref{AdS}, we outline the general relativistic treatment of obtaining elliptical-like orbits with small eccentricities about a static, spherically-symmetric massive object in the presence of a vacuum energy. After arriving at an expression describing the angular separation between successive like-apsides, we compare it to its counterpart for the marble on the spandex surface in both the small and large slope regimes.
\section{Elliptical-like orbits of a marble rolling on a cylindrically-symmetric elastic surface} \label{eqn}
We begin this manuscript by presenting the equations of motion that describe a spherical marble of uniform mass density rolling on a cylindrically-symmetric surface in a uniform gravitational field. These equations can be obtained by first constructing the Lagrangian that describes a rolling marble, which is constrained to reside upon a cylindrically-symmetric surface, in cylindrical coordinates $(r,\phi,z)$. This Lagrangian includes both the translational and rotational kinetic energy of the rolling marble and the gravitational potential energy of the orbiting body. The resultant Lagrangian can then be subjected to the Euler-Lagrange equations, which yield the equations of motion of the form
\begin{eqnarray}
(1+z'^2)\ddot{r}+z'z''\dot{r}^2-r\dot{\phi}^2+\frac{5}{7}gz'&=&0,\label{roft}\\
\dot{\phi}&=&\frac{5\ell}{7r^2},\label{phi}
\end{eqnarray}
where a dot indicates a time-derivative, $z'\equiv dz/dr$, and $\ell$ is the conserved angular momentum per unit mass.\cite{Marion} It is noted that these equations of motion have been derived in full detail elsewhere.\cite{me2, English} In arriving at Eqs. (\ref{roft}) and (\ref{phi}), it is noted that the scalar approximation for an object that rolls without slipping was employed. A full vector treatment of the intrinsic angular momentum of a rolling object on a cylindrically-symmetric surface can be found elsewhere.\cite{white}
Notice that one can easily decouple the above differential equations by inserting Eq. (\ref{phi}) into Eq. (\ref{roft}). The resulting differential equation can then be solved for $r(t)$, at least in principle, once the equation describing the cylindrically-symmetric surface, $z(r)$, is specified. Here we are interested in arriving at an expression for the radial distance in terms of the azimuthal angle, $r(\phi)$. Using the chain rule and Eq. (\ref{phi}), we construct a differential operator of the form
\begin{equation}\label{oper}
\frac{d}{dt}=\frac{5\ell}{7r^2}\frac{d}{d\phi}.
\end{equation}
By employing Eqs. (\ref{phi}) and (\ref{oper}), Eq. (\ref{roft}) can be transformed into an orbital equation of motion of the form
\begin{equation}\label{rofphi}
(1+z'^2)\frac{d^2r}{d\phi^2}+(z'z''-\frac{2}{r}(1+z'^2))\left(\frac{dr}{d\phi}\right)^2-r+\frac{7g}{5\ell^2}\cdot z'r^4=0.
\end{equation}
This dynamical equation of motion describes the radial distance of the orbiting marble in terms of the azimuthal angle, for a given cylindrically-symmetric surface specified by $z(r)$, and equates to a non-linear differential equation that can be solved perturbatively. As here we are interested in studying elliptical-like orbits, we choose an approximate solution for the radial distance of the orbiting body to be of the form
\begin{equation}\label{sol}
r(\phi)=r_0(1-\varepsilon\cos(\nu\phi)),
\end{equation}
where $\varepsilon$ is the eccentricity of the orbit and $\nu$ is the precession parameter. For an orbiting body whose radial distance is described by Eq. (\ref{sol}), $r_0$ represents the average radial distance of the orbital motion. We note that in what follows, $\varepsilon$ will be treated as small and used as our expansion parameter, whereas the parameters $\nu$ and $r_0$ will be determined by Eq. (\ref{rofphi}).
The precession parameter, $\nu$, can be further understood through the relation
\begin{equation}\label{delphi}
\nu\equiv\frac{360^\circ}{\Delta\phi},
\end{equation}
where $\Delta\phi$ represents the angular separation between two like-apsides (the angle between two consecutive apocenters or two consecutive pericenters) of the orbital motion.\cite{apocenter} Notice that for $\nu=1$, Eq. (\ref{sol}) equates to the perturbative approximation of the \textit{exact} solution of Newtonian gravitation, whose orbits equate to the well-known conic sections, to first-order in the eccentricity.\cite{conic} When the azimuthal angle advances by $360^\circ$, the orbiting body will find itself at the same radial distance and will precisely retrace its preceding orbital path. Hence, for $\nu=1$, Eq. (\ref{sol}) approximately describes closed or stationary elliptical orbits with small eccentricities. When $\nu\neq 1$, $\Delta\phi\neq 360^\circ$ and Eq. (\ref{sol}) approximately describes precessing elliptical-like orbits with small eccentricities. Notice that for $\nu>1$, Eq. (\ref{delphi}) implies that the angular separation between two consecutive like-apsides will be \textit{less than} $360^\circ$ and the apsides will `march backwards' in the azimuthal direction over successive orbits. Contrarily, for $\nu<1$, the angular separation between two consecutive like apsides will be \textit{greater than} $360^\circ$ and the apsides will `march forward' in the azimuthal direction.
Inserting Eq. (\ref{sol}) into Eq. (\ref{rofphi}) and keeping terms up to first-order in $\varepsilon$, we find a valid approximate solution to the orbital equation of motion when the algebraic expressions
\begin{eqnarray}
\ell^2&=&\frac{7}{5}gr_0^3z_0'\label{ell}\\
\nu^2&=&\frac{3z_0'+r_0z_0''}{z_0'(1+z_0'^2)}\label{nu}
\end{eqnarray}
are satisfied, where $z_0'$ and $z_0''$ are radial derivatives of $z$ evaluated at the average radial distance, $r_0$. In arriving at the above expressions, it is noted that the slope and the radial derivative of the slope were expanded about the average radial distance of the respective orbit. The above expressions were presented in a previous work\cite{Nauenberg} and then later used to find the 2D surfaces that generate Newtonian and general relativistic orbits with small eccentricities.\cite{me} Equation (\ref{ell}) specifies the necessary angular momentum per unit mass needed, at a given average radius and for a given slope of the surface, for circular or elliptical-like orbits to occur. Equation (\ref{nu}) yields a prediction for the precession parameter in terms of the slope and the radial derivative of the slope of a given surface at a given average radial distance.
In the subsection that follows we present the equation that describes the slope of a cylindrically-symmetric spandex fabric, subjected to a central mass placed upon its surface, in a uniform gravitational field. This differential equation yields approximate expressions for the slope in the small and large slope regimes, which can be inserted into Eq. (\ref{nu}) to yield a theoretical prediction for the precession parameter in each regime. Using Eq. (\ref{delphi}), the approximate solution for the angular separation between two consecutive like-apsides for a marble orbiting on this spandex surface can be found in both the small and large slope regimes.
\subsection{The slope of an elastic fabric warped by a central mass}\label{Shape}
The equation that describes the slope of a cylindrically-symmetric spandex surface residing in a uniform gravitational field can be obtained through the use of the calculus of variations, as the surface will take the shape that \textit{minimizes} the total potential energy of the central mass/elastic fabric system. The total potential energy can be written as an integral functional, where we take into account the elastic potential energy of the spandex fabric and the gravitational potential energy of the surface of uniform areal mass density $\sigma_0$ and of the central object of mass $M$. This resultant functional can then be subjected to Euler's equation, which yields a differential equation that describes the slope of the spandex fabric.\cite{Marion}
This resultant differential equation can be integrated and yields the expression
\begin{equation}\label{shape}
rz'\left(1-\frac{1}{\sqrt{1+z'^2}}\right)=\alpha\left(M+\sigma_0\pi r^2\right),
\end{equation}
where we defined the parameter $\alpha\equiv g/2\pi E$, where $E$ is the modulus of elasticity of the elastic material. We note that this differential equation that describes the slope of the elastic surface was previously derived in an earlier work of one of the authors.\cite{me2}
Notice that Eq. (\ref{shape}) describes the slope of the spandex fabric at radial distances $r<R$, where $R$ is the radius of the spandex surface, and is dependent upon the modulus of elasticity, the mass of the central object, and the mass of the spandex fabric interior to this radial distance, which is explicitly seen through the $\sigma_0\pi r^2$ term. Equation (\ref{shape}) corresponds to a quartic equation, which can be solved exactly for the slope, $z'(r)$, but cannot be integrated analytically to yield the shape of the surface, $z(r)$. The exact expression for $z'$ is rather awkward to work with but can be conveniently approximated in the small and large slope regimes. These regimes correspond to where the surface is nearly horizontal and nearly vertical, respectively.
The theoretical work outlined in this section offers an interesting and non-trivial application of Lagrangian dynamics and of the calculus of variations for an advanced undergraduate student. For the student who wishes to forego these derivations and focus on the experimental aspects of this project, Eqs. \eqref{roft}, \eqref{phi}, and \eqref{shape} can easily serve as a starting point.
In the next section we explore the behavior of the slope of the spandex surface in the small slope regime and, using Eq. (\ref{nu}), arrive at an expression for the precession parameter. Interestingly, this theoretical prediction for the precession parameter can then easily be compared to experiment by directly measuring the angular separation between successive like-apsides and then comparing to the theoretical prediction of Eq. (\ref{delphi}).
\section{Elliptical-like orbits in the small slope regime}\label{orbits}
We first explore the behavior of the angular separation between consecutive like-apsides in the regime where the slope of the spandex surface is small and the surface nearly horizontal. This physically corresponds to the region near the perimeter of the spandex fabric when a small central mass (or no central mass) has been placed upon the surface. In this small slope regime where $z'\ll 1$, we can expand the square root in the denominator of Eq. (\ref{shape}) in powers of $z'^2$. Keeping only the lowest-order non-vanishing contribution and then solving Eq. (\ref{shape}) for $z'$, we obtain an approximate expression describing the slope of the elastic surface of the form
\begin{equation}\label{smallslope}
z'(r)\simeq\left(\frac{2\alpha}{r}\right)^{1/3}\left(M+\sigma_0\pi r^2\right)^{1/3}.
\end{equation}
Inserting Eq. (\ref{smallslope}) into Eq. (\ref{nu}) yields an approximate expression for the precession parameter when the slope of the surface is evaluated at the average radial distance, $r_0$, of the orbiting body. Plugging this expression for $\nu$ into Eq. (\ref{delphi}) and solving for $\Delta\phi$, we obtain an expression predicting the value of the angular separation between consecutive apocenters (or between consecutive pericenters), which takes the form
\begin{equation}\label{smallsol}
\Delta{\phi}=220^\circ\left[1+\left(2\alpha(M+\sigma_0\pi r_0^2)/r_0\right)^{2/3}\right]^{1/2}\left[1+\frac{1}{4}\frac{\sigma_0\pi r_0^2}{\left(M+\sigma_0\pi r_0^2\right)}\right]^{-1/2}.
\end{equation}
Upon inspection of Eq. (\ref{smallsol}), it is noted that the value of this angular separation between successive like-apsides is dependent upon the mass of the central object, $M$, and the mass of the spandex fabric interior to the average radial distance of the elliptical-like orbit, $\sigma_0\pi r_0^2$. These two competing terms are on equal footing when $M\simeq\sigma_0\pi r_0^2$. For our spandex fabric, we measured the uniform areal mass density to be $\sigma_0=0.172$ kg/m$^2$ and found that the average radial distance of the elliptical-like orbits in the small slope regime is $r_0\simeq 0.33$ m. Using these measured values, we find that these two terms are on equal footing for a central mass of about $M\simeq 0.059$ kg. Hence, when the mass of the central object is less than $0.059$ kg, the mass of the spandex fabric interior to the average radial distance of the elliptical-like orbit becomes the dominant factor in determining the angular separation between consecutive like-apsides, as described by Eq. (\ref{smallsol}). When the mass of the central object is much smaller than the mass term of the spandex fabric, Eq. (\ref{smallsol}) yields a limiting behavior for the angular separation of the form
\begin{equation}\label{smallM}
\lim_{M\ll\sigma_0\pi r_0^2}\Delta{\phi}\simeq 197^\circ\left[1+(2\alpha\sigma_0\pi r_0)^{2/3}\right]^{1/2}.
\end{equation}
Conversely, when the mass of the central object dominates over the mass term of the spandex fabric, Eq. (\ref{smallsol}) yields a limiting behavior for the angular separation of the form
\begin{equation}\label{smallsigma}
\lim_{M\gg\sigma_0\pi r_0^2}\Delta{\phi}\simeq 220^\circ\left[1+\left(\frac{2\alpha M}{r_0}\right)^{2/3}\right]^{1/2},
\end{equation}
where, physically, one must be careful to ensure that the elliptical-like orbits are still occurring in the small slope regime with this relatively large central mass.
Upon comparison of the above approximate expressions, notice that Eq. (\ref{smallM}) predicts that the angular separation increases with increasing average radial distance whereas Eq. (\ref{smallsigma}) predicts a decreasing angular separation with increasing average radial distance, for a given central mass. Further notice that Eq. (\ref{smallsol}) predicts a minimal angular separation of $\Delta\phi_{min}\simeq 197^\circ$, which occurs for orbits with vanishingly small average radial distances when the surface is void of a central mass.
\subsection{The experiment in the small slope regime}
To experimentally test our theoretical prediction for the angular separation between successive apocenters and pericenters in the small slope regime, we needed a cylindrically-symmetric spandex fabric, capable of supporting elliptical-like orbits of a marble upon its surface, and a means of precisely determining the polar coordinates of the orbiting marble. A 4-ft diameter mini-trampoline was stripped of its trampoline surface and springs. A styrofoam insert was fashioned to exactly fit the interior of the trampoline frame, which collectively formed a flat, table-like structure. With the styrofoam insert in place, a nylon-spandex fabric was draped over the trampoline frame, where we ensured that no pre-stretch or slack was introduced. The fabric was then securely fastened to the frame by means of a ratchet strap and the styrofoam insert was removed, allowing the fabric to hang freely.
\begin{figure}[!t]
\centering
\includegraphics[width=15cm]{Chad_MiddletonFig01.pdf}
\caption{Screenshot of elliptical-like orbits of a marble rolling on the spandex surface, here subjected to a central mass of $M=0.199$ kg, in the small slope regime. Each circle marks the position of the rolling marble, in intervals of 1/60 of a second, as it orbits the central mass. The three orbits imaged here were found to have angular separations between successive apocenters of $\Delta\phi=213.5^\circ, \;225.7^\circ,\; 223.9^\circ$ and eccentricities of $\varepsilon=0.31, \;0.29, \;0.31$, respectively.}
\label{fig:smallorbit}
\end{figure}
It is noted that the shape of the freely hanging spandex fabric can easily be obtained. When the spandex fabric is void of a central mass the slope of the surface is small, hence, the approximation for the slope given by Eq. (\ref{smallslope}) is valid everywhere on the surface. Setting $M=0$ in Eq. (\ref{smallslope}) and integrating we obtain the height of the surface, which is given by the expression
\begin{equation}\label{shape0}
z(r)_{M=0}=\frac{3}{4}(2\alpha\sigma_0\pi)^{1/3}(r^{4/3}-R^{4/3}),
\end{equation}
where $R=0.585$ m is the inner radius of the trampoline frame and we chose the zero point of our coordinate system to coincide with $z(R)=0$. By measuring the height of the spandex fabric at the center of the apparatus, which was found to be $z(0)=-0.098$ m, and by knowing the areal mass density of the spandex fabric, we experimentally obtain a value of $\alpha=0.018$ m/kg. Having an experimental value for $\alpha$ is necessary as we wish to compare our predicted values for the angular separation between any successive like-apsides to the experimentally measured values.
A camera capable of taking video clips at a rate of 60 frames per second was mounted above the spandex surface. A plumb line, strung from the camera's perched location, was used to align the trampoline frame below and to determine the precise center of the fabric where the central mass should be placed. A ramp was mounted near the perimeter of the trampoline and orientated in the tangential direction so elliptical-like orbits with varying eccentricities, and hence varying average radial distances, could be generated by simply varying the initial speed of the marble. Several short video clips of elliptical-like orbits were then generated for a range of central masses and eccentricities. The video clips were imported into Tracker,\cite{Tracker} a video-analysis and modeling software program, where the marble's radial and angular coordinates could readily be extracted one frame at a time (see Fig. \ref{fig:smallorbit}). The apocenters and pericenters of the orbiting marble were identified by scrolling through the instantaneous radii and determining the maximum and minimum values, along with the accompanying angular coordinates. To remain within the small slope regime, only the first five orbits of a given run were considered, which yielded a maximum of ten data points per run. These data were then imported into Excel for analysis.
\begin{figure}[!t]
\centering
\includegraphics[width=15cm]{Chad_MiddletonFig02.pdf}
\caption{Plot of the angular separation between two successive like-apsides, $\Delta\phi$, versus the average radial distance, $r_0$, of elliptical-like orbits in the small slope regime for central masses ranging from $M=0$ kg to $M=0.535$ kg. }
\label{fig:SmallSlope}
\end{figure}
The angular separation between successive like-apsides could easily be obtained from the measured angular coordinates. The average radial distance, $r_0$, was found by averaging over the measured instantaneous radii over one orbit; one data point was hence generated from these two quantities. It is noted that the radius of a successive apocenter or pericenter was found to decrease (dramatically in the small slope regime) due to frictional losses and drag. To account for this loss, the average radial distance of two successive apocenters or pericenters was calculated and used in the analysis in determining the eccentricity of an orbit.\cite{eccentricity} Elliptical-like orbits with eccentricities falling within the range of $0.15\leq\varepsilon<0.35$ were kept for analysis whereas orbits with eccentricities outside of this adopted range were systematically discarded.
Using our aforementioned measured value of $\alpha$, the theoretical curves of the angular separation between successive like-apsides, as described by Eq. (\ref{smallsol}), can be plotted versus the average radial distances, $r_0$, of an elliptical-like orbit for a given central mass. These theoretical curves can then be compared to the measured values of $\Delta\phi$, both are collectively displayed in Fig. \ref{fig:SmallSlope}.
Upon examination of Fig. \ref{fig:SmallSlope}, notice that the measured values of $\Delta\phi$ tend to reside below the theoretical curves, with larger discrepancies occurring for smaller central masses. The authors hypothesize that this is predominantly due to an increased frictional loss for an orbiting marble about smaller central masses. As the central mass decreases, the stretching of the spandex fabric decreases, and the warping of the spandex fabric due to the orbiting marble's weight increases. This generates a scenario were the marble is having to effectively climb out of its own gravitational well, more so then for a larger central mass, and encounters an increased frictional loss resulting in the marble reaching its apocenter earlier than it would without this loss. This can be compared to 2D projectile motion where the projectile is subjected to a retarding force. For a projectile subjected to a large retarding force, one finds that it reaches a smaller peak height in a shorter horizontal distance than it would for a smaller retarding force. Further note that the measured values for $\Delta\phi$ tend to increase for increasing average radial distance for zero central mass and decrease for increasing average radial distances for non-zero central mass, inline with the theoretical predictions.
In the next section, we explore the behavior of the angular separation between successive apocenters in the large slope regime. We find that within this regime, the angular separation takes on values \textit{greater than} 360$^\circ$ for very small radial distances and large central masses. This corresponds to the apsides marching forward in the azimuthal direction over successive orbits. Interestingly, elliptical-like orbits around non-rotating, spherically-symmetric massive objects in general relativity are also found to have angular separations between successive like-apsides taking on values greater than 360$^\circ$. A comparison between these GR orbits and the orbits on a spandex fabric will be discussed in the appendix.
\section{Elliptical-like orbits in the large slope regime}\label{orbits2}
We now wish to explore the behavior of the angular separation between successive apocenters in the large slope regime. This regime corresponds physically to the region deep inside the well, namely, on the spandex fabric where $r_0$ is small, when a large central mass has been placed upon the surface. In this large slope regime, where $z'\gg 1$, we can again expand the square root in the denominator of Eq. (\ref{shape}) by first pulling $z'$ out of the square root and then expanding in powers of $1/z'^2$. Keeping only the lowest-order contribution in this expression and then solving Eq. (\ref{shape}) for $z'$, the slope takes on the approximate form
\begin{equation}\label{largeslope}
z'(r)\simeq 1+\frac{\alpha M}{r},
\end{equation}
where we dropped the areal mass density term as it is insignificant in this regime. Now inserting Eq. (\ref{largeslope}) into Eq. (\ref{nu}) and then plugging this expression for $\nu$ into Eq. (\ref{delphi}), we arrive at an expression describing the angular separation between successive apocenters in the large slope regime, which takes the form
\begin{equation}\label{largesol}
\Delta\phi=360^\circ\sqrt{\frac{(1+\alpha M/r_0)}{(3+2\alpha M/r_0)}\left[1+\left(1+\frac{\alpha M}{r_0}\right)^2\right]},
\end{equation}
where we evaluated $r$ at the average radial distance, $r_0$, of the elliptical-like orbits.
Notice that this expression, like its counterpart in the small slope regime, also has two terms competing for dominance. However, as the above expression is only valid for $z'\gg 1$, with $z'$ approximately given by Eq. (\ref{largeslope}), Eq. (\ref{largesol}) isn't valid when $\alpha M/r_0\ll 1$.
For a large central mass and a \textit{very} small average radial distance, when the orbiting marble is deep within the well, Eq. (\ref{largesol}) yields a limiting behavior for the angular separation of the form
\begin{equation}\label{largesolapprox}
\lim_{\alpha M/r_0\gg 1}\Delta\phi\simeq\frac{360^\circ}{\sqrt{2}}\cdot\frac{\alpha M}{r_0}.
\end{equation}
Notice that when $\alpha M/r_0>\sqrt{2}$, Eq. (\ref{largesolapprox}) predicts that $\Delta\phi>360^\circ$. Hence, as $\alpha M/r_0$ approaches and then becomes larger than the $\sqrt{2}$, successive apocenters are predicted to transition from marching backwards to marching forwards in the azimuthal plane over successive orbits.
\subsection{The experiment in the large slope regime}
We now wish to compare our theoretical prediction for the angular separation in the large slope regime, given by Eq. (\ref{largesol}), to our experimentally measured values.\cite{apsides} If the modulus of elasticity was truly constant for the spandex fabric, then the experimentally obtained value of $\alpha=0.018$ m/kg in the small slope regime could be used here. This, however, is not the case and has been discussed extensively elsewhere.\cite{me2} Although the modulus of elasticity has been shown to be a function of the stretch, it is approximately constant in each regime. Therefore, the analysis presented in this manuscript is valid when applied to each regime separately.
\begin{figure}[!t]
\centering
\includegraphics[width=15cm]{Chad_MiddletonFig03.pdf}
\caption{Plots of the angular separation between two successive apocenters, $\Delta\phi$, versus the average radial distance, $r_0$, of elliptical-like orbits in the large slope regime for central masses that range from $M=5.274$ kg to $M=7.774$ kg. }
\label{fig:LargeSlope}
\end{figure}
To obtain $\alpha$ in this large slope regime, we note the slope of the spandex surface, given by Eq. (\ref{largeslope}), can be integrated, which yields the height of the spandex fabric as a function of the radius and the central mass. Performing this integration and then evaluating at a small fixed radius $R_B=0.0154$ m on the spandex fabric, we obtain the expression
\begin{equation}\label{alphaLS}
\frac{z(M)}{\ln(R_B)}=\alpha (M-M_0),
\end{equation}
where, here, we set $z(M_0)=0$ where $M_0$ corresponds to the smallest central mass probed. This expression gives a scaled height of the fabric as a function of the central mass, $M$, for a given radius of the fabric, $R_B$, in the large slope regime. As all of the quantities in Eq. (\ref{alphaLS}) save $\alpha$ can be measured directly, the slope of the best fit line of a plot $z(M)/\ln(R_B)$ vs $(M-M_0)$ yields the value of the parameter $\alpha$. By orientating our camera to obtain a side view of the fabric's surface, we subjected our spandex fabric to the same range of central masses used in generating particle orbits in this large slope regime. We then imported this video of the shape profile of the spandex fabric into Tracker and measured the heights of the fabric at the aforementioned fixed radius as a function of the central mass $M$.
The slope of this best fit line yields a value of $\alpha=0.0049$ m/kg.
The experimental setup and procedure for probing the large slope regime of the spandex fabric was quite similar to that of the small slope regime, with a few notable exceptions. As here we wish to probe small radial distances for large central masses, we positioned a small spherical object of radius $\sim1.5$ cm in the center of the fabric. A sling, capable of supporting several kilograms of mass, was fashioned and fastened to the spherical object from below the fabric with the help of an adjustable pipe clamp. This setup allowed us to probe a range of large central masses while maintaining a constant diameter circular well.
The average radii of successive orbits do not decrease as drastically when the orbiting body is deep within the well as compared to orbits in the small slope regime, hence many orbits can potentially be analyzed for any given run. The angular separations, $\Delta\phi$, were found and the corresponding average radial distances, $r_0$, were obtained. Again, ellliptical-like orbits with eccentricities found to lie within the range $0.15\leq\varepsilon<0.35$ were kept and further analyzed, those not in this defined range were systematically discarded.
Using our experimentally determined value of $\alpha=0.0049$ m/kg, the theoretical curves of the angular separation, $\Delta\phi$, described by Eq. (\ref{largesol}), can be plotted versus the average radial distance, $r_0$, of an elliptical-like orbit for a given central mass along with the measured values and are collectively displayed in Fig. \ref{fig:LargeSlope}. Notice that as the average radial distance decreases, the angular separation increases. The horizontal dashed line displayed in Fig. \ref{fig:LargeSlope} corresponds to $\Delta\phi=360^\circ$. Hence, for data points lying above this line, the apocenter marched forward in the azimuthal direction when compared to that of the previous orbits.
We note that when the average radial distances of the elliptical-like orbits were greater than $\sim10$ cm, the measured values of $\Delta\phi$ were found to become substantially smaller than those shown in Fig. \ref{fig:LargeSlope}. As our theoretical prediction for $\Delta\phi$ is only valid in the large slope regime, we neglected the data points beyond this radial distance. For radial distances smaller than $\sim3$ cm, the measured values of $\Delta\phi$ were found to be quite sporadic. For orbits with small average radial distances, the angular velocity becomes quite large and hence only a handful of particle locations are imaged ($\sim12$) for a given orbit; this yields an angular separation of $\sim30^\circ$ between imaged locations and gives rise to a fairly large uncertainty in the angular measurement. Here we display the data points with average radial distances $3.0\;\mbox{cm}\leq r_0\leq 10\;\mbox{cm}$.
\section{conclusion}
In this manuscript, we arrived at expressions describing the angular separation between successive like-apsides for a marble rolling on a cylindrically-symmetric spandex surface in a uniform gravitational field in both the small and large slope regimes. We found that a minimal angular separation of $\sim197^\circ$ was predicted for orbits with small radial distances when the surface is void of a central mass. We also showed that for a marble orbiting with a small radius about a large central mass, corresponding to when the marble is deep within the well, the angular separation between successive like-apsides takes on values greater than 360$^\circ$. For these orbits, successive apocenters `march forward' in the azimuthal direction in the orbital plane, reminiscent of particle orbits about spherically-symmetric massive objects as described by general relativity. These theoretical predictions were shown to be consistent with the experimental measurements.
Although the orbital characteristics of a marble on a warped spandex fabric fundamentally differ from those of particle orbits about a spherically-symmetric massive object as described by general relativity, the authors contend that embracing the conceptual analogy between the two is insightful for the student of GR. Hence, this proposed theoretical/experimental undergraduate research project can be extended into the realm of general relativity, where a study of the elliptical-like orbits about a spherically-symmetric massive object with small eccentricities can be compared and contrasted to the theoretical work contained within this manuscript.
In the appendix, we study the physics of the analogy and find that some striking similarities exist between these elliptical-like orbits. Although differing in functional form, we show that the angular separation between like-apsides for marbles orbiting in the large slope regime and for particles orbiting near the innermost stable circular orbit both diverge for vanishingly small radial distances in a somewhat similar manner. We also show that the areal mass density of the spandex fabric plays the role of the vacuum energy density in GR, in regards to their competing relationship with their respective central mass, in the expressions for the angular separation.
\begin{appendix}
\section{Elliptical-like orbits with small eccentricities in general relativity}\label{AdS}
In this section, we outline the general relativistic treatment of obtaining elliptical-like orbits with small eccentricities about a spherically-symmetric massive object in the presence of a constant vacuum energy density, or equivalently, a cosmological constant. We arrive at the necessary conditions for elliptical-like orbits to occur in this spacetime and compare and contrast the expression for the angular separation between consecutive like-apsides to that obtained for a rolling marble on the warped spandex fabric in \textit{both} the small and large slope regimes. Although these expressions differ in their functional forms, there are some striking similarities. Here we argue that a thorough study of the physics of the analogy, a marble orbiting on a spandex surface, can aid in the learning of free particle orbits, or geodesics, in general relativity.
The addition of the cosmological constant to the field equations of general relativity can be traced back to Einstein himself. One of his first applications of this new theory of gravitation, which describes gravity as the warping of space and time due to the presence of matter and energy, was to the field of cosmology. Einstein added the cosmological constant to the relativistic equations only after realizing that these equations fail to offer a static cosmological solution to describe the seemingly static universe of his time. When the universe was later discovered to be expanding, the cosmological constant became an unnecessary addition and Einstein missed the opportunity of predicting an expanding universe, as his cosmological constant-free equations seemed to be implying. With the much later discovery of a late-time \textit{accelerated} expansion of the universe, the importance of a cosmological constant has once again arisen, but with the cosmological constant now recast into the form of a vacuum energy density. Mathematically, a cosmological constant, $\Lambda$, which finds itself on the curvature side of the field equations of GR, can be equivalently reformulated into a constant vacuum energy density, $\rho_0$, which resides on the matter/energy side. These quantities are related through the algebraic relation
\begin{equation}\label{rholambda}
\rho_0=(c^2/8\pi G)\Lambda,
\end{equation}
where $G$ is Newton's universal constant and $c$ is the speed of light.\cite{hartle,carroll} The reason for the relabeling of the cosmological constant emerges in quantum field theory, where zero-point fluctuations, corresponding to particles and anti-particles coming into and out of existence, are predicted. These vacuum fluctuations give rise to a nonzero vacuum energy density. It is noted that the cosmological constant, and therefore the constant vacuum energy density, can be of either sign. Observational evidence, however, suggests that we live in a universe with a small but nonzero, \textit{positive} vacuum energy density
whose presence is needed to explain the observed accelerated expansion of the universe.
Using the Schwarzschild coordinates $(t,r,\theta,\phi)$, the equations of motion describing an object of mass $m$ orbiting about a non-rotating, spherically symmetric massive object of mass $M$, in the presence of a cosmological constant, in general relativity are of the form
\begin{eqnarray}
\ddot{r}+\frac{GM}{r^2}-\frac{\ell^2}{r^3}+\frac{3GM\ell^2}{c^2r^4}-\frac{1}{3}\Lambda c^2r&=&0,\label{releq}\\
\dot{\phi}&=&\frac{\ell}{r^2}\label{relphi},
\end{eqnarray}
where in this section a dot indicates a derivative with respect to proper time and $\ell$ is the conserved angular momentum per unit mass.\cite{Stuchlik,Hackmann,Cruz}
It is noted that the last two terms in Eq. (\ref{releq}) correspond to relativistic terms which are not present in Newtonian gravitation. By setting these two relativistic terms equal to zero, one obtains the equations of motion of Newtonian theory; these yield a solution for the radial distance that describes orbits that equate to the conic sections.\cite{conic} The $1/r^4$ term equates to a general relativistic correction term to Newtonian gravitation, present even for a vanishing cosmological constant, and offers a small-$r$ modification to Newtonian theory. This correction term demands that the closed elliptical orbits of Newtonian theory are replaced with precessing elliptical-like orbits, which will become apparent in the paragraphs that follow. For a nonzero cosmological constant, notice that the last term in Eq. (\ref{releq}) offers a large-$r$ modification to the orbits.
As in our treatment of elliptical-like orbits on a warped spandex surface, we are interested in obtaining the solution for the radial distance of the orbiting body in terms of the azimuthal angle. Using the chain rule and Eq. (\ref{relphi}), we construct a differential operator of the form
\begin{equation}\label{relop}
\frac{d}{d\tau}=\frac{\ell}{r^2}\frac{d}{d\phi},
\end{equation}
where $\tau$ is the proper time. By employing Eqs. (\ref{relphi}) and (\ref{relop}), Eq. (\ref{releq}) can be transformed into an orbital equation of motion of the form
\begin{equation}\label{releqphi}
\frac{d^2r}{d\phi^2}-\frac{2}{r}\left(\frac{dr}{d\phi}\right)^2+\frac{GM}{\ell^2}r^2-r+\frac{3GM}{c^2}-\frac{\Lambda c^2}{3\ell^2}r^5=0.
\end{equation}
The solution to this orbital equation of motion describes the possible particle orbits about a non-rotating, spherically-symmetric massive object in the presence of a constant vacuum energy density.
\subsection{$\Lambda=0$}
We wish to examine elliptical-like orbits with small eccentricities and first examine the case of a vanishing cosmological constant. Inserting the perturbative solution of the form
\begin{equation}\label{relsol}
r(\phi)=r_0(1-\varepsilon\cos(\nu\phi))
\end{equation}
into Eq. (\ref{releqphi}), setting $\Lambda=0$, and keeping terms up to first-order in $\varepsilon$, we find a valid solution when the constants and parameters obey the relations
\begin{eqnarray}
\ell^2&=&GMr_0\left(1-\frac{3GM}{c^2r_0}\right)^{-1},\label{relell2norho}\\
\nu^2&=&1-\frac{6GM}{c^2r_0}\label{relnu2norho}.
\end{eqnarray}
These algebraic expressions correspond to the conditions necessary for elliptical-like orbits with small eccentricities about a non-rotating, spherically-symmetric massive object, void of vacuum energy, and have been discussed elsewhere.\cite{me} It is noted that when the general relativistic term $GM/c^2r_0$ is set equal to zero, we recover the stationary elliptical orbits of Newtonian gravitation, characterized by $\nu=1$. Notice that $\nu$ becomes complex when $r_0<r_{ISCO}\equiv 6GM/c^2$; this implies that elliptical-like orbits are \textit{only} allowed for radii larger than this limiting value. This threshold radius, $r_{ISCO}$, is known as the innermost stable circular orbit. Further notice that for radii smaller than $r_0<3GM/c^2$, both the angular momentum per unit mass, $\ell$, and the parameter $\nu$ become complex. This implies that there are no circular orbits, stable or unstable, allowed for an object orbiting about a non-rotating, spherically-symmetric massive object for these small radii.
Using Eqs. (\ref{delphi}) and (\ref{relnu2norho}), the angular separation between successive apocenters takes the form
\begin{equation}\label{reldelphinorho}
\Delta\phi=360^\circ\left(1-\frac{ r_{ISCO}}{r_0}\right)^{-1/2},
\end{equation}
where we used our aforementioned definition of $r_{ISCO}$. Note that as the average radius, $r_0$, of an elliptical-like orbit with a small eccentricity takes on values that approach the innermost stable circular orbit, $r_{ISCO}$, the angular separation between successive apocenters increases without bound. To further explore this limiting behavior, we define
\begin{equation}\label{smallr}
r_0\equiv r_{ISCO}+r,
\end{equation}
where $r\ll r_{ISCO}$ is a vanishingly small radial distance. Inserting Eq. (\ref{smallr}) into Eq. (\ref{reldelphinorho}) and expanding for small $r$, we find the limiting behavior for the angular separation to be of the form
\begin{equation}\label{rll6GM}
\lim_{r\ll 6GM/c^2}\Delta\phi\simeq 360^\circ\sqrt{6}\cdot\sqrt{\frac{GM/c^2}{r}},
\end{equation}
near the innermost stable circular orbit. Recall that the functional form of the angular separation for a marble orbiting on the warped spandex fabric in the large slope regime took on a limiting behavior of the form
\begin{equation}\label{rllaM}
\lim_{r_0\ll \alpha M}\Delta\phi\simeq\frac{360^\circ}{\sqrt{2}}\cdot\frac{\alpha M}{r_0}.
\end{equation}
Upon comparison of Eqs. (\ref{rll6GM}) and (\ref{rllaM}), we note that both expressions diverge in the limit of vanishingly small distances $r_0$ and $r$, however, their functional forms differ. $\Delta\phi\sim 1/\sqrt{r}$ for the orbiting object about the non-rotating, spherically-symmetric massive object near the innermost stable circular orbit whereas $\Delta\phi\sim 1/r_0$ for the rolling marble on the spandex fabric in the large slope regime for small radial distances. In the vanishingly small $r_0$ regime the slope of the spandex surface diverges, which can be seen upon inspection of Eq. (\ref{largeslope}), when the central mass is assumed to be point-like. We further note that the parameter $\alpha\equiv g/2\pi E$ plays the role of $G/c^2$ in the particle orbits of general relativity. Both constants set the scale of their respective theories and determine the value of the angular separation for a given central mass and average radial distance.
\subsection{$\Lambda\neq 0$}
We wish to push the conceptual analogy further between particle orbits in general relativity and rolling marble orbits on a warped spandex fabric in the small slope regime by considering the functional form of the angular separation for a nonzero cosmological constant. Inserting Eq. (\ref{relsol}) into Eq. (\ref{releqphi}), with $\Lambda\neq 0$, and again keeping terms up to first-order in $\varepsilon$, we find that Eq. (\ref{relsol}) is a valid approximate solution when
\begin{eqnarray}
\ell^2&=&Gr_0\left(1-\frac{3GM}{c^2r_0}\right)^{-1}(M-2\cdot\frac{4}{3}\pi r_0^3\cdot\rho_0),\label{relell2}\\
\nu^2&=&1-\frac{6GM}{c^2r_0}-6\left(1-\frac{3GM}{c^2r_0}\right)\frac{ \frac{4}{3}\pi r_0^3\cdot\rho_0}{(M-2\cdot \frac{4}{3}\pi r_0^3\cdot\rho_0)}\label{relnu2},
\end{eqnarray}
where we eliminated $\Lambda$ in favor of $\rho_0$ by using Eq. (\ref{rholambda}).
Now inserting Eq. (\ref{relnu2}) into Eq. (\ref{delphi}) and solving for $\Delta\phi$, we obtain the relativistic expression for the angular separation between consecutive like-apsides, which takes the form
\begin{equation}\label{reldelphi}
\Delta\phi=360^\circ\left(1-\frac{6GM}{c^2r_0}\right)^{-1/2}\left[1-6\left(\frac{1-3GM/c^2r_0}{1-6GM/c^2r_0}\right)\frac{\frac{4}{3}\pi r_0^3\cdot\rho_0}{(M-2\cdot \frac{4}{3}\pi r_0^3\cdot \rho_0)}\right]^{-1/2}.
\end{equation}
Although it is tempting to identify $\frac{4}{3}\pi r_0^3\cdot\rho_0$ as the mass of the vacuum interior to the average radial distance of the elliptical-like orbit, as would be the case in classical physics, one must be careful when calculating volume in the non-Euclidean spacetime of general relativity.
Recall that the functional form of the angular separation between successive like-apsides for a marble orbiting on a warped spandex fabric in the small slope regime is of the form
\begin{equation}\label{smallsol2}
\Delta{\phi}=220^\circ\left[1+\left(2\alpha(M+\pi r_0^2\cdot\sigma_0)/r_0\right)^{2/3}\right]^{1/2}\left[1+\frac{1}{4}\frac{\pi r_0^2\cdot\sigma_0}{\left(M+\pi r_0^2\cdot\sigma_0\right)}\right]^{-1/2}.
\end{equation}
Comparing the denominators of the last terms in Eqs. (\ref{reldelphi}) and (\ref{smallsol2}), we find that the areal mass density, $\sigma_0$, of the spandex fabric plays the role of a \textit{negative} vacuum energy density, $-\rho_0$, of spacetime. In each respective theory, the mass of the central object competes for dominance with the mass term associated with the vacuum/spandex fabric.
As observational evidence implies that we live in a universe with a nonzero (albeit tiny), \textit{positive} vacuum energy density, one can envision a spandex fabric with a negative areal mass density when the conceptual analogy of particle orbits in general relativity and rolling marble orbits on a warped spandex fabric is employed. This fictional spandex fabric would feel a repulsive force from the uniform gravitational field of the earth and tend to hang \textit{up} towards the ceiling rather than down towards the floor.
This contribution to the total warping of the spandex surface gives rise to a repulsive force on the orbiting marble, effectively competing with the attractive force of the central mass contribution.
\vspace{-0.5cm}
\end{appendix}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 303 |
Is perianal Crohn's disease associated with intestinal fistulization?
Am J Gastroenterol. 2005 Jul;100(7):1547-9. doi: 10.1111/j.1572-0241.2005.40980.x.
David B Sachar 1 , Carol A Bodian, Eric S Goldstein, Daniel H Present, Theodore M Bayless, Michael Picco, Ruud A van Hogezand, Vito Annese, Judith Schneider, Burton I Korelitz, Jacques Cosnes; Task Force on Clinical Phenotyping of the IOIBD
1 Mount Sinai School of Medicine, New York 10029-6574, USA.
Background: When cases of Crohn's disease (CD) are described as "fistulizing," distinctions are often not drawn between perianal and intestinal fistulization. The question, therefore, remains open as to whether or not there is truly an association between perianal fistulization and intraabdominal intestinal fistulization in CD.
Aims: We have sought to determine the association between perianal and intestinal fistulization by analyzing the cases of CD recorded in databases from six international centers.
Patients: Six databases provided information on 5491 cases of CD in the United States, France, Italy, and The Netherlands. Of these cases, 1686 had isolated ileal disease and 1655 had Crohn's colitis.
Methods: An association between perianal disease and internal fistulae was sought by calculating relative risks for the chance of internal fistulae among patients with perianal fistulae relative to those without. Statistical significance was calculated by the Mantel-Haenszel procedure, stratifying on the separate centers. All statistical tests and estimates were implemented using SAS for the PC.
Results: Among the 1686 cases with isolated ileal disease, the evidence of an association between perianal disease and internal fistulization was not consistent across centers, with relative risks ranging from 0.8 to 2.2. For patients with Crohn's colitis (n = 1655), the association was much stronger and more consistent, with an estimated common relative risk of 3.4, 95% confidence interval (2.6-4.6, p < 0.0001).
Conclusions: We have found a statistically significant association between perianal CD and intestinal fistulization, much stronger and more consistent in cases of Crohn's colitis than in cases limited to the small bowel.
Anus Diseases / complications*
Colitis / complications
Colonic Diseases / complications
Crohn Disease / complications*
Ileal Diseases / complications
Intestinal Fistula / etiology* | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 9,152 |
Q: Splitting a string using a next line regex followed by a digit (JAVA) I have the following string which I need to split on space, commas, and next line character.
Sample string: "./homework2 3,phone3,desc3,brand4\n4,phone4,desc4,brand4\n5,phone5,desc5,brand5"
I am using split function as: .split("[,\\s+\\n+]"); but I am unable to get the required out put. The program keeps reading brand4\n4 as one substring.
Help would be appreciated.
A: Try this, it should work the way you want:
.split(" |,|\\n");
This splits the string at a space (" "), a comma, or a next line character. Use the or operator to add more entries if needed.
A: Try this code :
String str = "./homework2 3,phone3,desc3,brand4\\n4,phone4,desc4,brand4\\n5,phone5,desc5,brand5";
String[] arrOfStr = str.split("[\\s\\n,]+");
for (String a : arrOfStr)
System.out.println(a);
}
}
Out put :
./homework2
3
phone3
desc3
brand4\n4
phone4
desc4
brand4\n5
phone5
desc5
brand5
EDIT :
I found this function in the site : Link Here
for (int i = 0 ; i < arrOfStr.length ; i++) { //YOUR LOOP !!!
splitString(arrOfStr[i]);
}
}
static void splitString(String str)
{
StringBuffer alpha = new StringBuffer(),
num = new StringBuffer(), special = new StringBuffer();
for (int i=0; i<str.length(); i++)
{
if (Character.isDigit(str.charAt(i)))
num.append(str.charAt(i));
else if(Character.isAlphabetic(str.charAt(i)))
alpha.append(str.charAt(i));
else
special.append(str.charAt(i));
}
System.out.println(alpha);
System.out.println(num);
}
}
Combine the two functions and you get:
homework
2
3
phone
3
desc
3
brandn
44
phone
4
desc
4
brandn
45
phone
5
desc
5
brand
5
You can store the values in another array, or whatever you choose.
A: Check the below link to validate regex
https://regex101.com/r/2ZkRBp/1/
A: Instead of using \s+ and \n+ you could use a common quantifier. The below depicts the same
String s = "./homework2 3,phone3,desc3,brand4\n4,phone4,desc4,brand4\n5,phone5,desc5,brand5";
String[]s1 = s.split("[,\\s\\n]+");
for(String s2:s1) {
System.out.println(s2);
}
As it leads to
./homework2
3
phone3
desc3
brand4
4
phone4
desc4
brand4
5
phone5
desc5
brand5
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 76 |
Manas é um santuário de fauna aos pés dos Himalaias, no estado de Assam, Índia. O Santuário de Manas abriga grande variedade de fauna, incluindo muitas espécies em risco de extinção, como o tigre, o porco-pigmeu, o rinoceronte indiano e o elefante indiano.
A Unesco inclui o Santuário de Fauna de Manas na lista de Património Mundial da Humanidade em 1985 e, mais tarde, em 1992, na lista de Património Mundial em perigo. A decisão da Unesco de declarar o santuário Património Mundial em perigo foi tomada quando este foi invadido pelos militantes da tribo Bodo. Os danos no santuário foram estimados em dois milhões de dólares americanos. A infrastructura do sítio sofreu danos consideráveis entre 1992 e 1993. Uma missão de monitorizamento levada a cabo juntamente pela pelo Governo da Índia e pela Unesco em 1997 confirmou os extensos danos à infrastructura do parque e descida da população de algumas espécies, particularmente ao rinoceronte indiano de um só corno.
O Governo da Índia, Governo do estado de Assam e as autoridades do parque elaboraram um plano de reabilitação do parque com o custo de 2.35 milhões de dólares americanos que começou por ser implementado em 1997 e que progrede satisfatoriamente. Enquanto as condições de segurança melhoraram dentro e à volta de Manas, a ameaça de insurgência ainda se mantém no estado de Assam e os militantes atravessam ocasionalmente o santuário. No entanto, as condições do sítio - protecção e relações com os nativos da zona estão a melhorar.
Ligações Externas
Galeria de imagens da Unesco
Patrimônio Mundial da UNESCO na Índia | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 733 |
\section{Introduction}
\label{sec:intro}
There is ample theoretical evidence that it is useful to incorporate KL regularization into policy optimization in reinforcement learning. The most basic approach is to regularize towards a uniform policy, resulting in entropy regularization. More effective, however, is to regularize towards the previous policy.
By choosing KL regularization between consecutively updated policies, the optimal policy becomes a softmax over a uniform average of the full history of action value estimates \citep{vieillard2020leverage}.
This averaging smooths out noise, allowing for better theoretical results \citep{azar2012dynamic,kozunoCVI,vieillard2020leverage,Kozuno-2022-KLGenerativeMinMaxOptimal,abbasi-yadkori19b-politex}.
Despite these theoretical benefits, there are some issues with using KL regularization in practice.
It is well-known that the uniform average is susceptible to outliers; this issue is inherent to KL divergence \citep{Futami2018-robustDivergence}.
In practice, heuristics such as assigning vanishing regularization coefficients to some estimates have been implemented widely to increase robustness and accelerate learning \citep{grau-moya2018soft,haarnoja-SAC2018,Kitamura2021-GVI}.
However, theoretical guarantees no longer hold for those heuristics \citep{vieillard2020leverage,Kozuno-2022-KLGenerativeMinMaxOptimal}.
A natural question is what alternatives we can consider to this KL divergence regularization, that allows us to overcome some of these disadvantages while maintaining the benefits associate with restricting aggressive policy changes and smoothing errors.
In this work, we explore one possible direction by generalizing to Tsallis KL divergences. Tsallis KL divergences were introduced for physics \citep{TsallisEntropy,tsallis2009introduction}, using a simple idea: replacing the use of the logarithm with the deformed $q$-logarithm. The implications for policy optimization, however, are that we get quite a different form for the resulting policy. Tsallis \emph{entropy} with $q = 2$ has actually already been considered for policy optimization \citep{Nachum18a-tsallis,Lee2018-TsallisRAL}, by replacing Shannon entropy with Tsallis entropy to maintain stochasticity in the policy. The resulting policies are called \emph{sparsemax} policies, because they concentrate the probability on higher-valued actions and truncate the probability to zero for lower-valued actions. Intuitively, this should have the benefit of maintaining stochasticity, but only amongst the most promising actions, unlike the Boltzmann policy which maintains nonzero probability on all actions. Unfortunately, using only Tsallis entropy did not provide significant benefits, and in fact often performed worse than existing methods. We find, however, that using a Tsallis KL divergence to the previous policy does provide notable gains.
We first show how to incorporate Tsallis KL regularization into the standard value iteration updates, and prove that we maintain convergence under this generalization from KL regularization to Tsallis KL regularization. We then characterize the types of policies learned under Tsallis KL, highlighting that there is now a more complex relationship to past action-values than a simple uniform average. We then show how to extend Munchausen Value Iteration (MVI) \citep{vieillard2020munchausen}, to use Tsallis KL regularization, which we call MVI($q$). We use this naming convention to highlight that this is a strict generalization of MVI: by setting $q = 1$, we exactly recover MVI.
We first investigate the impact of choosing different $q$ in a small simulation environment. We then compare MVI($q = 2$) with MVI (namely the standard choice where $q = 1$), and find that we obtain significant performance improvements in Atari.
\section{Problem Setting}\label{sec:preliminary}
We focus on discrete-time discounted Markov Decision Processes (MDPs) expressed by the tuple $(\mathcal{S}, \mathcal{A}, d, P, r, \gamma)$, where $\mathcal{S}$ and $\mathcal{A}$ denote state space and finite action space, respectively.
$d(\cdot)$ denotes the initial state distribution.
$P(\cdot |s,a)$ denotes transition probability over the state space given state-action pair $(s,a)$,
and $r(s,a)$ defines the reward associated with that transition.
When time step $t$ is of concern, we write $r_t := r(s_t, a_t)$.
$\gamma \in (0,1)$ is the discount factor.
A policy $\pi(\cdot|s)$ is a mapping from the state space to distributions over actions.
We define the state value function following policy $\pi$ and starting from initial state $s_0 \sim d(\cdot)$ as $V_{\pi}(s) = \mathbb{E}\AdaRectBracket{\sum_{t=0}^{\infty} r_t | s_0=s }$.
Likewise, we define the state-action value function given initial action $a_0$ as $Q_{\pi}(s,a) = \mathbb{E}\AdaRectBracket{\sum_{t=0}^{\infty} r_t | s_0=s, a_0 = a }$,
where the expectation is with respect to the policy $\pi$ and transition probability $P$.
A standard approach to find the optimal value function $Q_*$ is value iteration. To define the formulas for value iteration, it will be convenient to write these functions as vectors, $V_{\pi} \in \mathbb{R}^{|\mathcal{S}|}$ and $Q_{\pi} \!\in\! \mathbb{R}^{|\mathcal{S}|\times|\mathcal{A}|}$.
For notational convenience, we define the inner product for any two functions $F_1,F_2\in\mathbb{R}^{|\mathcal{S}|\times|\mathcal{A}|}$ as $\AdaAngleProduct{F_1}{F_2} \in \mathbb{R}^{|\mathcal{S}|}$, where we only take the inner product over actions.
The Bellman operator acting upon any function $Q \!\in\! \mathbb{R}^{|\mathcal{S}|\times|\mathcal{A}|}$ can be defined as: $T_{\pi}Q := r + \gamma P_{\pi} Q$, where $P_{\pi}Q \!\in\! \mathbb{R}^{|\mathcal{S}|\times|\mathcal{A}|} \!:=\! \mathbb{E}_{s'\sim P(\cdot|s,a),a'\sim\pi(\cdot|s')}\!\AdaRectBracket{Q(s',a')} \!=\! P\!\AdaAngleProduct{\pi}{Q}$.
When $\pi$ is greedy with respect to $Q$, we have the Bellman optimality operator defined by $T_*Q := r + \gamma P_* Q$, $P_*Q \!=\! \mathbb{E}_{s'\sim P(\cdot|s,a)}\!\AdaRectBracket{\max_{a'}Q(s',a')}$.
The above definitions should be understood as component-wise.
Repeatedly applying the Bellman operator $T_{\pi}Q$ converges to the unique fixed point $Q_{\pi}$, and $T_*Q$ it converges to the optimal action value function $Q_*:=Q_{\pi^*}$:
\begin{align}
\begin{cases}
\pi_{k+1} = \greedy{Q_{k}}, & \\
Q_{k+1} = \AdaBracket{ T_* Q_k }^m , &
\end{cases}
\label{eq:PI}
\end{align}
where the integer $m \geq 1$ denotes repeated applications.
Choosing $m=1,\infty$ corresponds value iteration and policy iteration, respectively.
$1\!<\! m < \!\infty\!$ implies the use of approximate modified policy iteration \citep{scherrer15-AMPI}.
This basic recursion can be modified with the addition of a regularizer $\Omega(\pi)$:
\begin{align}
\begin{cases}
\pi_{k+1} = \greedyOmega{Q_{k}}, & \\
Q_{k+1} = \AdaBracket{T_{\pi_{k+1, \Omega}}Q_k}^m &
\end{cases}
\label{eq:regularizedPI}
\end{align}
where $T_{\pi_{k+1}, \Omega}Q_k \!=\! \BellmanOmega{k+1}{k}$ is the regularized Bellman operator \citep{geist19-regularized}. This modified recursion is guaranteed to converge if $\Omega$ is strongly convex in $\pi$. For example, the negative Shannon entropy $\Omega(\pi) = -\mathcal{H}\left( \pi \right) = \AdaAngleProduct{\pi}{\ln\pi}$ has the resulting optimal policy $\pi_{k+1} \propto \exp\AdaBracket{\tau^{-1}Q_k}$, where $\propto$ indicates \emph{proportional to} up to a constant not depending on actions.
Another popular choice is KL divergence $\Omega(\pi)=\KLany{\pi}{\mu} = \AdaAngleProduct{\pi}{\ln{\pi} \!-\!\ln{\mu}}$, which is more general since we recover Shannon entropy when we choose $\mu$ to be a uniform distribution, i.e. $\frac{1}{|\mathcal{A}|}$. In this work, when we say KL regularization, we mean the standard choice of setting $\mu = \pi_k$, the estimate from the previous update.
Therefore, $D_\text{KL}$ serves as a penalty to penalize aggressive policy changes.
The optimal policy in this case takes the form $\pi_{k+1}\propto\pi_k\exp\AdaBracket{\tau^{-1} Q_k}$.
By induction, we can show this KL-regularized optimal policy $\pi_{k+1}$ is a softmax over a uniform average over the history of action value estimates \cite{vieillard2020leverage}:
\begin{align}
\!\!\pi_{k+1} \!\propto \pi_k \exp\AdaBracket{\tau^{-1} \! Q_k} \!\propto\! \cdots \!\propto\! \exp\AdaBracket{\! \tau^{-1}\sumJK{1}Q_j}
.
\label{eq:average}
\end{align}
Using KL regularization has been shown to be theoretically superior to entropy regularization, in terms of error tolerance \cite{azar2012dynamic,vieillard2020leverage,Kozuno-2022-KLGenerativeMinMaxOptimal,chan2021-greedification}.
The definitions of $\mathcal{H}(\cdot)$ and $D_\text{KL}(\cdot||\cdot)$ rely on the standard logarithm and its inverse (the exponential) and both induce softmax policies as an exponential over action-values \citep{hiriart2004-convexanalysis,Nachum2020-RLFenchelRockafellar}.
Convergence properties of the resulting regularized algorithms have been well studied \citep{kozunoCVI,geist19-regularized,vieillard2020leverage}.
In this paper, we investigate Tsallis entropy and Tsallis KL divergence as the regularizer, which generalize Shannon entropy and KL divergence respectively.
\section{Generalizing to Tsallis Regularization}\label{sec:qlog}
We can easily incorporate other regularizers in to the value iteration recursion and maintain convergence, as long as those regularizers are strongly convex in $\pi$.
This property is satisfied by Tsallis entropy and the Tsallis KL divergence, for certain $q$. In this section, we define the Tsallis KL divergence and characterize the types of policies that arise from using this regularizer.
\subsection{Tsallis Regularization}
Tsallis regularization results from generalizing from the standard logarithm to the $q$-logarithm. The $q$-logarithm and its (unique) inverse function the $q$-exponential are defined as follows. For $q \in \mathbb{R}_{+} \! \setminus \! \{1\}$
\begin{align}
\logqstar\pi := \frac{\pi^{q-1} - 1}{q-1}, \quad \expqstar{\,Q} = \AdaRectBracket{1 + (q - 1)Q}^{\frac{1}{q-1}}_{+}
\label{eq:qlog_definition}
\end{align}
where $[\cdot]_{+} = \max\{\cdot, 0\}$.
When $q=1$, we define $\logqstar{ } \, , \expqstar$ be $\ln \, , \exp$ respectively,
because the above approaches these standard values as $q \rightarrow 1$.
Tsallis entropy is defined as $\tsallis{\pi}:= \AdaAngleProduct{-\pi}{\logqstar{\pi}}$.
The Tsallis KL divergence is defined as
$\qKLany{\pi}{\mu} := \AdaAngleProduct{\pi}{\ln_q\frac{\pi}{\mu}}$ \citep{TsallisRelativeEntropy}.
Tsallis KL is more \emph{mass-covering} than KL; i.e. its value is proportional to the $q$-th power of the ratio $\frac{\pi}{\mu}$, when $q>0$ is big, large values of $\frac{\pi}{\mu}$ are strongly penalized \citep{Wang2018-tailAdaptivefDiv}.
In the limit of $q\rightarrow 1$, Tsallis entropy recovers Shannon entropy and the Tsallis KL divergence recovers the KL divergence.
\footnote{The $q$-logarithm has been primarily used in physics.
It is worth noting that $q$ used in RL literature is different from the statistical physics (denote by $q^*$).
But the RL case can be recovered by leveraging the duality $q = 2-q^*$ \cite{Naudts2002DeformedLogarithm,Suyari2005-LawErrorTsallis,Lee2020-generalTsallisRSS}.}
We have a similar generalization in the form of the policy defined by the $q$-exponential.
When $\Omega(\pi) = -S_2(\pi)$, there is a simple
closed-form for the optimal policy called the sparsemax \citep{Martins16-sparsemax,Lee2020-generalTsallisRSS}:
\begin{align}
\begin{split}
&\pi_{k+1}(a | s) = \exp_2\AdaBracket{\frac{Q_k(s,a)}{2\tau} - \psi\AdaBracket{ \frac{Q_{k}(s,\cdot)}{2\tau}}}\\
&\psi\AdaBracket{\frac{Q_{k}(s,\cdot)}{2\tau}} = \frac{\sum_{a\in S(s)} \frac{Q_k(s,a)}{2\tau} - 1 }{|S{(s)}|} .
\end{split}
\label{eq:sparse_normalization}
\end{align}
$S(s)$ is the set of highest-valued actions, satisfying the relation $1 \!+\! i\frac{Q_k(s,a_{(i)})}{2\tau} \!>\! \sum_{j=1}^{i}\frac{Q_k(s,a_{(j)})}{2\tau}$, where $a_{(j)}$ indicates the action with $j$th largest action value.
This sparsemax sets the probability to zero on the lowest-valued actions,
$\pi_{k+1}(a_{(i)}|s)=0, i=z+1,\dots, |\mathcal{A}|$, where $Q_k(s,a_{(z)}) > 2\tau\AdaBracket{\psi\AdaBracket{\frac{Q_k(s,\cdot)}{2\tau}} - 1} > Q_k(s,a_{(z+1)})$.
This truncation is shown in the bottom of Figure \ref{fig:qkl_examples}.
It is worth noting that for general Tsallis entropy $\tsallis{\pi}, q\neq 1,2,\infty$, the normalization $\psi$ cannot be analytically solved \citep{Lee2020-generalTsallisRSS} and hence the policy does not have a closed-form expression.
We use first-order expansion to approximate the policy for those indices.
Note that $q = 1$ corresponds to Shannon entropy and $q = \infty$ to no regularization.
\begin{figure}
\centering
\includegraphics[width=.95\linewidth]{./new_figures/sns_GauBol_qkl.pdf}
\caption{(Top) Illustration of the components $\pi_1\ln\frac{\pi_1}{\pi_2}$ and $\pi_1\ln_q{\frac{\pi_1}{\pi_2}}$ between Gaussian policies and Boltzmann policies when $q=2$.
Tsallis KL divergence is more mass-covering than KL.
(Bottom) Illustration of sparsemax acting upon $\pi_1$ by truncating actions that have values lower than $\psi$ defined in \eq{\ref{eq:sparse_normalization}}.
}\label{fig:qkl_examples}
\end{figure}
\subsection{Convergence Results}
Now let us turn to formalizing when value iteration with Tsallis regularization converges.
Similar to logarithm, $q$-logarithm has the following properties\footnotemark\footnotetext{We prove these properties in Appendix \ref{apdx:basicfact_qlog}}.
\emph{Convexity: } $\logqstar{\pi}$ is convex for $q\leq 0$, concave for $q > 0$. When $q=2$, both $\logqstar\,\, ,\expqstar$ become linear.
\emph{Monotonicity: } $\ln_q\pi$ is monotonically increasing with respect to $\pi$.
The following similarity between Shannon entropy (reps. KL) and Tsallis entropy (resp. Tsallis KL) is highlighted:
\emph{Bounded entropy}: we have $ 0 \leq \mathcal{H}\left( \pi \right) \leq \ln |\mathcal{A}|$; and $\forall q, \, 0 \leq \tsallis{\pi} \leq -\ln_q \frac{1}{|\mathcal{A}|}$.
\emph{Generalized KL property}: $\forall q$, $\qKLany{\pi}{\mu} \geq 0$. $\qKLany{\pi}{\mu} = 0$ if and only if $\pi=\mu$ almost everywhere, and $\qKLany{\pi}{\mu} \rightarrow \infty$ whenever $\pi(a|s)>0$ and $\mu(a|s)=0$.
However, despite their similarity, a crucial difference is that $\ln_q$ is non-extensive, which means it is not additive \citep{TsallisEntropy}.
In fact, $\ln_q$ is only \emph{pseudo-additive}:
\begin{align}
\ln_q{\pi\mu} = \ln_q\pi + \ln_q \mu + (q-1)\ln_q\pi\ln_q\mu.
\label{eq:pseudo_add}
\end{align}
Pseudo-additivity complicates obtaining convergence results for \eq{\ref{eq:regularizedPI}} with $q$-logarithm regularizers, since the techniques used for Shannon entropy and KL divergence are generally not applicable to their $\ln_q$ counterparts.
Moreover, deriving the optimal policy may be nontrivial.
Convergence results have only been established for Tsallis entropy regularization with $q = 2$ \citep{Lee2018-TsallisRAL,Nachum18a-tsallis}.
Our first simple result is to show that \eq{\ref{eq:regularizedPI}} with $\Omega(\pi) = \qKLany{\pi}{\mu}$, for any $\mu$, converges at least for $q=2$. In fact, it converges for any $q$ where $\qKLany{\pi}{\mu}$ is strongly convex, but we are only able to show this for $q=2$.
\begin{theorem}\label{thm:converge}
The regularized recursion \eq{\ref{eq:regularizedPI}} with $\Omega(\pi) = \qKLany{\pi}{\cdot}$ when $q=2$ converges to the unique regularized optimal policy.
\end{theorem}
\vspace{-0.3cm}
\begin{proof}
See Appendix \ref{apdx:proof_qkl2} for the full proof. It simply involves proving that this regularizer is strongly convex.
\end{proof}
\subsection{TKL Regularized Policies Do More Than Averaging}
Our second result shows the optimal regularized policy under Tsallis KL regularization does more than uniform averaging: it can be seen as performing weighted average whose degree of weighting is controlled by $q$.
To be specific, we consider \eq{\ref{eq:regularizedPI}} with $m=1$ and $\Omega(\pi) = \qKLany{\pi}{\pi_k}$.
\begin{align}
\begin{cases}
\pi_{k+1} = \greedyTKL{\TsallisQ{k}}, & \\
\TsallisQ{k+1} = \BellmanTKL{k+1}{k}, &
\end{cases}
\label{eq:TsallisKLPI}
\end{align}
where we dropped the regularization coefficient $\tau$ for convenience.
\begin{proposition}\label{prop1}
The greedy policy $\pi_{k+1}$ in Equation \eqref{eq:TsallisKLPI} satisfies
\begin{align}
&\pi_{k+1}^{q-1} \propto \AdaBracket{ \expqstar{Q_1}\cdots\expqstar{Q_k} }^{q-1} \\
&=\expqstar\!{\AdaBracket{ \sum_{j=1}^{k}Q_j} }^{q-1} \!\!\!\!\!\!\!\!\!
+ \sum_{j=2}^{k}(q-1)^j \!\!\!\!\!\!\!\sum_{i_{1}=1 < \dots < i_j}^k \!\!\!\! Q_{i_1} \cdots Q_{i_j}.
\label{eq:pseudo_average}
\end{align}
When $q \!=\! 1$, \eq{\ref{eq:TsallisKLPI}} reduces to KL regularized recursion and hence \eq{\ref{eq:pseudo_average}} reduces to the KL-regularized policy \eq{\ref{eq:average}}.
When $q \!=\! 2$, \eq{\ref{eq:pseudo_average}} becomes:
\begin{align*}
\exp_{2}{Q_1}\cdots\exp_{2}{Q_k} \! =
\exp_{2}{\AdaBracket{ \sum_{j=1}^{k}Q_j \! } \!} + \!\!\!\!\!\!\! \sum_{\substack{j=2 \\ i_{1}=1 < \dots < i_j}}^{k} \!\!\!\!\!\!\! Q_{i_1} \cdots Q_{i_j}.
\end{align*}
i.e., Tsallis KL regularized policies average over the history of value estimates as well as computing the interaction between them $\sum_{j=2}^{k}\sum_{\substack{i_{1} < \dots < i_j}}^{k} Q_{i_1} \dots Q_{i_j}$.
\end{proposition}
\begin{proof}
See Appendix \ref{apdx:tsallis_kl_policy} for the full proof.
The proof comprises two parts: the first part shows $\pi_{k+1} \propto \expqstar{Q_1}\dots\expqstar{Q_k}$, and the second part establishes the \emph{more-than-averaging} property by using the generalized two-point property \cite{Yamano2004-properties-qlogexp} $\AdaBracket{\expqstar{x}\cdot \expqstar{y}}^{q-1} \!\!\! = \expqstar{\AdaBracket{x+y}}^{q - 1} \!\!+ (q -1)^2 xy $.
\end{proof}
\begin{figure}[t]
\includegraphics[width=0.925\linewidth]{new_figures/Acrobot-v1_barplot.png}
\caption{
End-learning scores of MVI($q$) for various entropic indices on \texttt{Acrobot-v1}, averaged over 50 seeds and $+100$ for visualization purpose.
Black lines indicate standard variations.
}
\label{fig:various_acrobot}
\end{figure}
The form of this policy is harder to intuit, than the Boltzmann policy for $q = 2$, but we can try to understand each component.
The first component actually corresponds to a weighted averaging by the property of the $\expqstar{\,}$:
\begin{align}
\expqstar{\AdaBracket{\sum_{i=1}^{k}Q_i}\!\!} \!=\! \expqstar{(Q_1)}\dots \expqstar{\AdaBracket{\!\frac{Q_k}{1 + (q-1)\sum_{i=1}^{k-1} Q_i}\! }}\!\!.
\label{eq:weighted_average}
\end{align}
\eq{\ref{eq:weighted_average}} states that any estimate $Q_j$ is weighted by the summation of estimates before $Q_j$ times $(q-1)$ and therefore the influence of outlier estimates can be largely reduced by larger $q$, and the weighting becomes more effective at the later stage of learning where degradation is likely to happen.
In fact, obtaining a weighted average is a typical goal in policy optimization. \citet{Futami2018-robustDivergence} proposed to leverage robust divergences to perform weighted averaging.
In RL, many heuristics coincide with weighted averaging \cite{grau-moya2018soft,haarnoja-SAC2018,Kitamura2021-GVI}.
Now let us consider the second term. Once again, we have a weighted average, with weighting $(q-1)^j$, but now on a term that multiplies rather than sums the previous action-values. It is not clear yet how this nonlinear relationship to past action-values benefits the policy. There is some intuition here that the above weighted average and this additional nonlinearity could improve the learned policy, but as yet there is no concrete theoretical justification. Instead, we focus primarily on motivating $q > 1$ empirically.
\subsection{Considering $q>1$ Is Beneficial}\label{sec:beneficial}
In this section, we demonstrate that selecting $q > 1$ with Tsallis KL regularization can be beneficial.
We test entropic indices from $[1.0, \infty) \cup \{\infty\}$ on \texttt{Acrobot-v1} from OpenAI gym \citep{brockman2016openai}.
Hyperparameters of each $q$ were independently fine-tuned.
The result is shown in Figure \ref{fig:various_acrobot}, with reward $+100$ for visualization.
For $q\neq 1,2,\infty$, we use the first-order Taylor expansion policy expressions detailed in Appendix \ref{apdx:approximate_policy}.
Recall that when $q=\infty$ corresponds to no regularization.
For each $q$ we run $5\times 10^5$ steps and average over 50 seeds.
It can be seen from Figure \ref{fig:various_acrobot} that the best performance was achieved by $q=2$.
The performance versus the choice of $q$ has some surprising outcomes.
The first is that performance drops drastically for $q=3, 4$.
Since the effect of Tsallis KL divergence is proportional to $q$, for large values such as $q\geq 10$ the regularization effect may well be approximating no regularization $q=\infty$.
In this regard, intermediate values between $2$ and $10$ may enforce too strong regularization or mass-covering property, hence constrain the policy from changing, leading to poor performance.
However, we are not yet able to fully explain this oscillatory behavior, but it is clear here that the choice of $q$ can have a significant impact on performance.
\section{A Practical Algorithm for Tsallis KL Regularization}\label{sec:Munchausen}
In this section we provide a practical algorithm for implementing Tsallis regularization.
We first explain why this is not straightforward to simply implement KL-regularized value iteration, even for $q = 2$, and how Munchausen Value Iteration ( MVI) overcomes this issue with a clever implicit regularization trick.
We then explain why we cannot directly use MVI, and then generalize MVI using a new interpretation of this trick as advantage learning.
\subsection{Implicit Regularization With MVI}
Even for the standard KL, it is difficult to implement KL-regularized value iteration with function approximation.
The difficulty arises from the fact that we cannot exactly obtain $\pi_{k+1} \propto \pi_k\exp\AdaBracket{Q_k}$. This policy might not be representable by our function approximator. For $q=1$, one could store all past $Q_k$ and use their average outputs; with neural networks, however, this is computationally infeasible.
An alternative direction has been to construct a different value function iteration scheme, which is equivalent to the original KL regularized value iteration \citep{azar2012dynamic,kozunoCVI}.
A recent method of this family is Munchausen VI ( MVI) \citep{vieillard2020munchausen}.
MVI implicitly enforces KL regularization using the recursion
\begin{align}
\begin{split}
\label{eq:mdqn_recursion}
\begin{cases}
\! \pi_{k+1} = \argmax_{\pi} \AdaAngleProduct{\pi}{Q_{k} - \tau\ln\pi} \\
\! Q_{k+1} \!=\! r \!+\! {\color{red}{\alpha\tau \ln\pi_{k+1} }} \!+\! \gamma P \AdaAngleProduct{\pi_{k+1}}{Q_k \!-\! {\color{black}{\color{blue}\tau\ln\pi_{k+1}}} }
\end{cases}
\end{split}
\end{align}
We see that \eq{\ref{eq:mdqn_recursion}} is \eq{\ref{eq:regularizedPI}} with $\Omega(\pi)=-\mathcal{H}\left( \pi \right)$ ({\color{blue}blue}) plus an additional {\color{red} red} \emph{Munchausen term}, with coefficient $\alpha$.
\citet{vieillard2020munchausen} showed that implicit KL regularization was performed under the hood, even though we still have tractable $\pi_{k+1}\propto\exp\AdaBracket{\tau^{-1}Q_k}$:
\begin{align}
&Q_{k+1} = r + {\color{red} \alpha\tau \ln{\pi_{k+1}} } + \gamma P \AdaAngleProduct{\pi_{k+1}}{Q_k - {\color{black} \tau\ln{\pi_{k+1}} } }\nonumber\\
&\Leftrightarrow Q_{k+1} - {\color{red}{\alpha\tau\ln{\pi_{k+1}}}} = r + \gamma P \big(\AdaAngleProduct{\pi_{k+1}}{Q_k - \alpha\tau\ln{\pi_k}} \nonumber\\
& \qquad -\AdaAngleProduct{\pi_{k+1}}{ \alpha\tau(\ln{\pi_{k+1}} - \ln{\pi_{k}}) - (1-\alpha)\tau\ln{\pi_{k+1}} } \big) \nonumber\\
& \Leftrightarrow Q'_{k+1} = r + \gamma P\big( \AdaAngleProduct{\pi_{k+1}}{Q'_k} - \alpha\tau\KLindex{k+1}{k} \nonumber \\
& \qquad \qquad \qquad \qquad + (1-\alpha)\tau \entropyIndex{k+1} \big)
\label{eq:mdqn}
\end{align}
where $Q'_{k+1}\!:=\! Q_{k+1} - \alpha\tau\ln{\pi_{k+1}}$ is the generalized action value function.
To extend this to $q > 1$, a natural choice seems to be to replace $\ln$ with $\ln_q$ in this recursion. Unfortunately, the $q$-logarithm is only pseudo-additive, rather than additive: $\ln (ab) = \ln a + \ln b$ but $\ln_q (ab) \neq \ln_q a + \ln_q b$. Consequently, we cannot use the same derivation above, because
$\qKLany{\pi}{\mu} \neq \AdaAngleProduct{\pi}{\logqstar{\pi} - \logqstar{\mu}}$. We show in Figure \ref{fig:MT_fail}
that this naive approach, that assumes additivity, indeed fails.
\subsection{MVI($q$) For General $q$}
We first discuss the alternative derivation for implicit KL regularization for $q > 1$, and then discuss
two practical approaches to approximation the recursion.
Instead of adding $\alpha\tau\logqstar{{\pi_{k+1}}}$, let us consider subtracting $\alpha\tau\logqstar{\frac{1}{\pi_{k+1}}}$. For $q = 1$, this is actually equivalent. For $q > 1$, we no longer have $\ln (a^{-1}) = -\ln a$, but we will see this gives us something much closer to implicit KL regularization. First, let us introduce $Q_{k+1}' = Q_{k+1} + \alpha\tau \logqstar{\frac{1}{\pi_{k+1}}}$ and write
\begin{align}
&Q_{k+1} + {\color{red} \alpha\tau\logqstar{\frac{1}{\pi_{k+1}}}} = \nonumber\\
& r + \gamma P \AdaAngleProduct{\pi_{k+1}}{Q_k + \alpha\tau\logqstar{\frac{1}{\pi_{k}}} - \alpha\tau\logqstar{\frac{1}{\pi_{k}}} - \tau\logqstar{\pi_{k+1}}} \nonumber\\
&{\Leftrightarrow} Q_{k+1}' = r + \gamma P\AdaAngleProduct{\pi_{k+1}}{Q_k'} \nonumber\\
&\quad\quad\quad - \gamma P\AdaAngleProduct{\pi_{k+1}}{ \alpha\tau\logqstar{\frac{1}{\pi_{k}}} + \tau\logqstar{\pi_{k+1}} }
\label{eq:tklvi_munchausen}
\end{align}
For $q = 1$, the term on the last line could be rearranged to produce the KL and entropy regularization.
But, for $q > 1$, because of pseudo-additivity (formula in Eq.\eqref{eq:pseudo_add}), we can only use the relationship
\begin{equation*}
\logqstar{\frac{1}{\pi_{k}}} = \logqstar{\frac{\pi_{k+1}}{\pi_{k}}} - \logqstar{\pi_{k+1}} -(q-1)\logqstar{\pi_{k+1}}\logqstar{\frac{1}{\pi_{k}}}
.
\end{equation*}
We can simplify the final term in Eq.\eqref{eq:tklvi_munchausen} as follows.
\begin{align*}
&\alpha\tau\AdaAngleProduct{\pi_{k+1}}{\logqstar{\frac{\pi_{k+1}}{\pi_{k}}} - \logqstar{\pi_{k+1}} -(q-1)\logqstar{\pi_{k+1}}\logqstar{\frac{1}{\pi_{k}}}}\\
& \quad\quad\quad\quad\quad\quad+\tau\AdaAngleProduct{\pi_{k+1}}{\logqstar{\pi_{k+1}}}\\
&= \alpha\tau\qKLindex{k+1}{k} - \tau\AdaBracket{1 - \alpha}\tsallis{\pi_{k+1}} \\
& \qquad \qquad -\alpha\tau(q-1)\AdaAngleProduct{\pi_{k+1}}{\logqstar{\frac{1}{\pi_k}}\logqstar{\pi_{k+1}} }
\end{align*}
Like with $q = 1$, we obtain an implicit regularization by adding $\alpha\tau\logqstar{\frac{1}{\pi_{k+1}}}$, though now with an additional term weighted by $q-1$.
\begin{figure}[t]
\centering
\includegraphics[width=0.925\linewidth]{new_figures/CartPole-MVI-TKLVI.png}
\caption{
MVI($q$) (proposed method) and MVI with $\pi$ being the sparsemax policy induced by $S_2(\pi)$.
}
\label{fig:MT_fail}
\end{figure}
\begin{algorithm}[tb]
\caption{MVI($q$) for $q=2$}
\label{alg:tklvi}
\begin{algorithmic}
\STATE {\bfseries Input:} number of iterations $K$, entropy coefficient $\tau$, Munchausen coefficient $\alpha$
\STATE Initialize $Q_0, \pi_0$ arbitrarily
\STATE Define $\texttt{Cond}(i, Q_k) = 1 \!+\! i\frac{Q_k(s,a_{(i)})}{2\tau} \!>\! \sum_{j=1}^{i}\frac{Q_k(s,a_{(j)})}{2\tau}$
\FOR{$k=1, 2 ,\dots, K$}
\STATE \# \texttt{Policy Improvement}
\FOR{$(s,a)\in \mathcal{(S, A)}$}
\STATE Sort $Q_{k}(s,a_{(1)}) > \dots > Q_{k}(s,a_{(|\mathcal{A}|)})$
\STATE $S(s) = \max\{i\in \{1, 2, \dots, |\mathcal{A}|\} \,\big| \texttt{Cond}(i, Q_k)\}$
\STATE Compute $\psi(\frac{Q_k(s,\cdot)}{2\tau}) = \frac{\sum_{a\in S(s)} \frac{Q_k(s,a)}{2\tau} - 1 }{|S{(s)}|}$
\STATE $\pi_{k+1}(a|s) = \exp_2\AdaBracket{\frac{Q_k(s,a)}{2\tau} - \psi\AdaBracket{ \frac{Q_k(s,\cdot)}{2\tau}} }$
\ENDFOR
\STATE \# \texttt{Policy Evaluation}
\FOR{$(s,a,s')\in \mathcal{(S, A)}$}
\STATE $Q_{k+1}(s, a) \!=\! r(s,a) \!+\! \alpha\tau \AdaBracket{Q_k(s,a) \!-\! \mathcal{M}_{2,\tau}{Q_k}(s)}$
\STATE \,\, $ + \gamma\sum_{b\in\mathcal{A}}\pi_{k+1}(b|s')\AdaBracket{Q_{k+1}(s',b) \!-\! \tau\ln_2{\pi_{k+1}(b|s')}\!}$
\ENDFOR
\ENDFOR
\end{algorithmic}
\end{algorithm}
This final term poses a problem. It involves both $k$ and $k+1$. If we try to construct a recursion that incorporates this term, the term that has to be added to $Q_{k+1}$ relies on the future policy, which is not yet available. One reasonable approximation, therefore, is to simply use $\alpha\tau\logqstar{\frac{1}{\pi_{k+1}}}$ and ignore this additional term.
There is another route to consider, to generalize MVI to $q > 1$. For the implementation of MVI, the red term is actually computed using $\alpha\tau\ln\pi_{k+1} = \alpha(Q_k - \mathcal{M}_{\tau}Q_k)$, where $\mathcal{M}_{\tau}Q_k:= \frac{1}{Z_k}\AdaAngleProduct{\exp\AdaBracket{\tau^{-1}Q_k}}{Q_k}, Z_k = \AdaAngleProduct{\boldsymbol{1}}{\exp\AdaBracket{\tau^{-1}Q_k}}$ is the Boltzmann softmax operator.\footnotemark\footnotetext{Using $\mathcal{M}_{\tau}Q$ is equivalent to the log-sum-exp operator up to a constant shift \citep{azar2012dynamic}.} They chose this route to improve numerical stability. We can generalize this $\mathcal{M}_{\tau}$ operator, to $q > 1$, and use $Q_k - \mathcal{M}_{q,\tau}{Q_k}$, where
\begin{equation}
\mathcal{M}_{q,\tau}{Q_k} = \AdaAngleProduct{\exp_q\AdaBracket{\frac{Q_k}{q\tau} - \psi\AdaBracket{ \frac{Q_k}{q\tau}}}}{ Q_k}
.
\end{equation}
Let us look more carefully at this operator.
Assume that $\tau$ absorbs $q$, and $Q(s,a_{(z)}) > \tau\AdaBracket{\psi\AdaBracket{\frac{Q(s,\cdot)}{\tau}} - \frac{1}{q-1}} > Q(s,a_{(z+1)})$, then $\exp_q\AdaBracket{\frac{Q_k(s,a_{(i)})}{\tau} - \psi\AdaBracket{ \frac{Q_k(s,\cdot)}{\tau}}}=0$ for $i=z+1,\dots,|\mathcal{A}|$.
Therefore, $\mathcal{M}_{q,\tau}{Q_k}$ is an expectation of $Q(s,a_{(1)}),\cdots,Q(s,a_{(z)})$ and $Q_k - \mathcal{M}_{q,\tau}{Q_k}$. The Tsallis policy has support only on these actions, and so this corresponds to the expected value under the Tsallis policy. This is the same property as $Q_k - \mathcal{M}_\tau Q_k$ for $q =1$, which also provides the advantage. An interesting thing to note is that this generalization also indicates an issue with using $\logqstar{\pi_{k+1}}$, which equals $\frac{Q_k}{\tau} - \psi\AdaBracket{\frac{Q_k}{\tau}}$. This term indicates the values compared to the normalization $\psi\AdaBracket{\frac{Q_k}{\tau}}$, which is not the same as the expected values, and so does not equal the advantage.
Unlike $q = 1$, however, adding the term $Q_k - \mathcal{M}_{q,\tau}{Q_k}$ for $ q > 1$ no longer has a clear connection to KL regularization. It is not the case that $-\alpha\tau\logqstar{\frac{1}{\pi_{k+1}}}$ equals $Q_k - \mathcal{M}_{q,\tau}{Q_k}$ (see Appendix \ref{apdx:adv_kl}).
Since $\alpha\tau\logqstar{\frac{1}{\pi_{k+1}}}$ is itself an approximation and one term is omitted, it is actually possible that $Q_k - \mathcal{M}_{q,\tau}{Q_k}$ more faithfully approximates KL regularization plus entropy regularization. However, this remains a big open question. We empirically tested both approximations---adding $\alpha\tau\logqstar{\frac{1}{\pi_{k+1}}}$ versus adding $Q_k - \mathcal{M}_{q,\tau}{Q_k}$---and found that adding the advantage term was generally more effective, and numerically stable. This result mirrors the original MVI, so the final algorithm we pursue for this extension uses $Q_k - \mathcal{M}_{q,\tau}{Q_k}$.
We summarize this MVI($q$) algorithm in Algorithm \ref{alg:tklvi} for $q = 2$, and provide the more general version in Algorithm \ref{alg:tklvi_general} in Appendix \ref{apdx:approximate_policy}.
When $q=1$, we get $\psi\AdaBracket{ \frac{Q_k}{\tau}} = \mathcal{M}_{\tau}{Q_k}$, recovering MVI.
For $q = \infty$, we get that $\mathcal{M}_{\infty,\tau}{Q_k}$ is $\max_{a}Q_k(s,a)$---no regularization---and we recover advantage learning \citep{Baird1999-gradient-actionGap}.
Similar to the original MVI algorithm, MVI($q$) enjoys tractable policy expression with $\pi_{k+1}\propto\exp_q\AdaBracket{\tau^{-1} Q_k}$.
\begin{figure*}[t]
\begin{minipage}[t]{0.245\textwidth}
\includegraphics[width=\textwidth]{new_figures/AmidarNoFrameskip.png}
\end{minipage}
\begin{minipage}[t]{0.245\textwidth}
\includegraphics[width=\textwidth]{new_figures/AsterixNoFrameskip.png}
\end{minipage}
\begin{minipage}[t]{0.245\textwidth}
\includegraphics[width=\textwidth]{new_figures/AssaultNoFrameskip.png}
\end{minipage}
\begin{minipage}[t]{0.245\textwidth}
\includegraphics[width=\textwidth]{new_figures/BeamRiderNoFrameskip.png}
\end{minipage}\\
\begin{minipage}[t]{0.245\textwidth}
\includegraphics[width=\textwidth]{new_figures/EnduroNoFrameskip.png}
\end{minipage}
\begin{minipage}[t]{0.245\textwidth}
\includegraphics[width=\textwidth]{new_figures/HeroNoFrameskip.png}
\end{minipage}
\begin{minipage}[t]{0.245\textwidth}
\includegraphics[width=\textwidth]{new_figures/MsPacmanNoFrameskip.png}
\end{minipage}
\begin{minipage}[t]{0.245\textwidth}
\includegraphics[width=\textwidth]{new_figures/SpaceInvadersNoFrameskip.png}
\end{minipage}
\caption{Learning curves of MVI($q$) and MVI on the selected Atari games.
On some environments MVI($q$) significantly improve upon MVI.
Quantitative improvements over MVI and Tsallis-VI are shown in Figures \ref{fig:evi_mdqn} and \ref{fig:evi_tsallis}.
}
\label{fig:learning_curves_atari}
\end{figure*}
\begin{figure}[t]
\vspace{-0.5cm}
\centering
\includegraphics[width=0.495\textwidth]{new_figures/MVIq-MVI-20atari.pdf}
\caption{ The percent improvement of MVI($q$) with $q = 2$ over standard MVI (where $q=1$) on the selected Atari games. The improvement is computed by subtracting the MVI scores from MVI($q$) and normalized by the MVI.}
\label{fig:evi_mdqn}
\end{figure}
\begin{figure}[t]
\vspace{-0.5cm}
\centering
\includegraphics[width=0.495\textwidth]{new_figures/MVIq-TVI-35atari.pdf}
\caption{
MVI($q$) improvement over Tsallis-VI on Atari environments, normalized with Tsallis-VI scores.
The maximum improvement for \texttt{RoadRunner} was capped at 1000\% to better visualize performance in the other games.
}
\label{fig:evi_tsallis}
\end{figure}
\section{Experiments}\label{sec:experiment}
In this section we investigate the utility of MVI($q$) in the Atari 2600 benchmark \citep{bellemare13-arcade-jair}.
We have seen in Figure \ref{fig:various_acrobot} that considering $q>1$ on simple environments could be helpful. We test whether this result holds in more challenging environments. Specifically, we compare to standard MVI ($q = 1$), which was already shown to have competitive performance on Atari \citep{vieillard2020munchausen}. We restrict our attention to $q=2$, which was generally effective in other settings and also allows us to contrast to previous work \citep{Lee2020-generalTsallisRSS} that only used entropy regularization with KL regularization. For MVI($q = 2$), we take the exact same learning setup---hyperparameters and architecture---as MVI($q=1$) and simply modify the term added to the VI update, as in Algorithm \ref{alg:tklvi}.
Hyperparameters and full results are provided in Appendix \ref{apdx:gym}.
We also compare MVI($q$) several advantage learning algorithms on another benchmark task, called MinAtar \citep{young19minatar}. Advantage learning has been under active investigation due to its appealing theory \citep{Farahmand2011-actionGap} and superior empirical performance \citep{Ferret2021-selfImitationAdvantage}. Performance and experimental settings are detailed in Appendix \ref{apdx:minatar}, due to space considerations. As a summary, we find that MVI($q$) with $q=1$ and $q=2$ are competitive with these methods, but that the difference between $q=1$ and $q=2$ is not nearly as stark in MinAtar as in Atari.
\subsection{Comparing MVI($q$) with $q=1$ to $q=2$}
We provide the overall performance of MVI versus MVI($q=2$) in Figure \ref{fig:evi_mdqn}. Using $q=2$ provides a large improvement in about 5 games, about double the performance in the next 5 games, comparable performance in the next 7 games and then slightly worse performance in 3 games (\texttt{PrivateEye}, \texttt{Chooper} and \texttt{Seaquest}). Both \texttt{PrivateEye} and \texttt{Seaquest} are considered harder exploration games, which might explain this discrepancy. The Tsallis policy with $q = 2$ reduces the support on actions, truncating some probabilities to zero. In general, with a higher $q$, the resulting policy is greedier, with $q = \infty$ corresponding to exactly the greedy policy. It is possible that for these harder exploration games, the higher stochasticity in the softmax policy from MVI whre $q = 1$ promoted more exploration. A natural next step is to consider incorporating more directed exploration approaches, into MVI($q=2$), to benefit from the fact that lower-value actions are removed (avoiding taking poor actions) while exploring in a more directed way when needed.
We examine the learning curves for the games where MVI($q$) had the most significant improvement, in Figure \ref{fig:learning_curves_atari}. Particularly notable is how much more quickly MVI($q$) learned with $q=2$, in addition to plateauing at a higher point. In \texttt{Hero}, MVI($q$) learned a stably across the runs, whereas standard MVI with $q=1$ clearly has some failures.
These results are quite surprising. The algorithms are otherwise very similar, with the seemingly small change of using Munchausen term $Q_k - \mathcal{M}_{q=2,\tau}{Q_k}$ instead of $Q_k - \mathcal{M}_{q=1,\tau}{Q_k}$ and using the $q$-logarithm and $q$-exponential for the entropy regularization and policy parameterization. Previous work using $q=2$ to get the sparsemax with entropy regularization generally harmed performance \citep{Lee2018-TsallisRAL,Lee2020-generalTsallisRSS}. It seems that to get the benefits of the generalization to $q > 1$, the addition of the KL regularization might be key. We validate this in the next section.
\subsection{The Importance of Including KL Regularization}
In the policy evaluation step of Algorithm \ref{alg:tklvi}, if we set $\alpha=0$ then we recover Tsallis-VI which uses regularization $\Omega(\pi)= - S_q(\pi)$ in \eq{\ref{eq:regularizedPI}}. In other words, we recover the algorithm that incorporates entropy regularization using the $q$-logarithm and the resulting sparsemax policy.
Unlike MVI, Tsallis-VI has not been comprehensively evaluated on Atari games, so
we include results for the larger benchmark set comprising 35 Atari games. We plot the percentage improvement of MVI($q$) over Tsallis-VI in Figure \ref{fig:evi_tsallis}.
The improvement from including the Munchausen term ($\alpha > 0$) is stark. For more than half of the games, MVI($q$) resulted in more than 100\% improvement. For the remaining games it was comparable. For 10 games, it provided more than 400\% improvement. Looking more specifically at which games there was notable improvement, it seems that exploration may again have played a role. MVI($q$) performs much better on \texttt{Seaquest} and \texttt{PrivateEye}. Both MVI($q$) and Tsallis-VI have policy parameterizations that truncate action support, setting probabilities to zero for some actions. The KL regularization term, however, likely slows this down. It is possible the Tsallis-VI is concentrating too quickly, resulting in insufficient exploration.
\section{Related Work}\label{sec:related_work}
There is a growing body of literature studying
generalizations of KL divergence in reinforcement learning \citep{Nachum2019-Algaedice,Zhang2020-GenDICE}.
\citet{Futami2018-robustDivergence} discussed the inherent drawback of KL divergence in generative modeling and proposed to use $\beta$- and $\gamma$-divergence to allow for weighted average of sample contribution.
A more commonly explored generalization is the $f$-divergence \citep{Sason2016-fDiverInequalities}, which is also being used in other machine learning areas including for generative modeling \citep{Nowozin2016-fGAN,Wan2020-fDivVI,Yu2020-EBMwithFDiv} and imitation learning \citep{Ghasemipour2019-divergenceMinPerspImitation,Ke2019-imitationLearningAsfDiv}.
In reinforcement learning, \citet{Wang2018-tailAdaptivefDiv} discussed using tail adaptive $f$-divergence to enforce the mass-covering property.
\citet{Belousov2019-AlphaDivergence} discussed a special family of $f$-divergence known as the $\alpha$-divergence.
Though the $f$-divergence offers great generality, it remains unknown how to find the environment-specific function $f$ since one cannot sweep functions naturally as we did for the scalar entropic indices in Figure \ref{fig:various_acrobot}.
The same holds true for another extension known as Bregman divergence \citep{Neu17-unified,Huang2022-bregmanPO}, which also requires specifying the generator function.
Exploring the generator-environment relationship is an interesting future direction.
\section{Conclusion and Discussion}\label{sec:conclusion}
We investigated the use of the more general $q$-logarithm for entropy regularization and KL regularization, instead of the standard logarithm ($q=1$), which gave rise to Tsallis entropy and Tsallis KL regularization. We extended several results previously shown for $q=1$, namely we proved (a) the form of the Tsallis policy in terms of the past action-values, (b) convergence of value iteration for $q=2$ and (c) a relationship between adding a $q$-logarithm to the action-value update, to provide implicit KL regularization and entropy regularization, mimicking Munchausen Value Iteration (MVI). We used these results to propose a generalization to MVI, which we call MVI($q$), because for $q=1$ we exactly recover MVI. We showed empirically that the generalization to $q >1$ can be beneficial, providing notable improvements in the Atari 2600 benchmark.
This is the first work to use Tsallis KL regularization for policy optimization, and naturally there are many open questions. One key question is to better understand the approximating in MVI($q$). When $q = 1$, the added Munchausen term is equivalent to adding explicit KL regularization. This perfect equivalence is lost for $q > 1$. It is important to better understand what the Munchausen term is really doing for $q > 1$, and when the approximation is very close to implicit KL regularization and when it is not. Another key question is to understand the role of the entropic index $q$ on performance. It was not that case that performance was very similar for all $q$, and in fact at times there seemed to be an oscillating pattern. For Acrobot, for example, $q=2$ performed very well, $q =3$ and $q=4$ performed very poorly, and then $q = 5$ had reasonable performance. Using $q = 2$ was generally effective, but in some cases we did find other $q$ were even better. To truly benefit from this generalization, it is important to better understand the role of $q$ on the policy and why certain $q$ are more effective in certain environments.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 5,418 |
namespace Envoy {
namespace {
class SocketInterfaceIntegrationTest : public BaseIntegrationTest,
public testing::TestWithParam<Network::Address::IpVersion> {
public:
SocketInterfaceIntegrationTest() : BaseIntegrationTest(GetParam(), config()) {
use_lds_ = false;
};
static std::string config() {
// At least one empty filter chain needs to be specified.
return absl::StrCat(echoConfig(), R"EOF(
bootstrap_extensions:
- name: envoy.extensions.network.socket_interface.default_socket_interface
typed_config:
"@type": type.googleapis.com/envoy.extensions.network.socket_interface.v3.DefaultSocketInterface
default_socket_interface: "envoy.extensions.network.socket_interface.default_socket_interface"
)EOF");
}
static std::string echoConfig() {
return absl::StrCat(ConfigHelper::baseConfig(), R"EOF(
filter_chains:
filters:
name: ratelimit
typed_config:
"@type": type.googleapis.com/envoy.config.filter.network.rate_limit.v2.RateLimit
domain: foo
stats_prefix: name
descriptors: [{"key": "foo", "value": "bar"}]
filters:
name: envoy.filters.network.echo
)EOF");
}
};
INSTANTIATE_TEST_SUITE_P(IpVersions, SocketInterfaceIntegrationTest,
testing::ValuesIn(TestEnvironment::getIpVersionsForTest()),
TestUtility::ipTestParamsToString);
TEST_P(SocketInterfaceIntegrationTest, Basic) {
BaseIntegrationTest::initialize();
const Network::SocketInterface* factory = Network::socketInterface(
"envoy.extensions.network.socket_interface.default_socket_interface");
ASSERT_TRUE(Network::SocketInterfaceSingleton::getExisting() == factory);
std::string response;
auto connection = createConnectionDriver(
lookupPort("listener_0"), "hello",
[&response](Network::ClientConnection& conn, const Buffer::Instance& data) -> void {
response.append(data.toString());
conn.close(Network::ConnectionCloseType::FlushWrite);
});
connection->run();
EXPECT_EQ("hello", response);
}
TEST_P(SocketInterfaceIntegrationTest, AddressWithSocketInterface) {
BaseIntegrationTest::initialize();
ConnectionStatusCallbacks connect_callbacks_;
Network::ClientConnectionPtr client_;
const Network::SocketInterface* sock_interface = Network::socketInterface(
"envoy.extensions.network.socket_interface.default_socket_interface");
Network::Address::InstanceConstSharedPtr address =
std::make_shared<Network::Address::Ipv4Instance>(
Network::Test::getLoopbackAddressUrlString(Network::Address::IpVersion::v4),
lookupPort("listener_0"), sock_interface);
client_ = dispatcher_->createClientConnection(address, Network::Address::InstanceConstSharedPtr(),
Network::Test::createRawBufferSocket(), nullptr);
client_->addConnectionCallbacks(connect_callbacks_);
client_->connect();
while (!connect_callbacks_.connected() && !connect_callbacks_.closed()) {
dispatcher_->run(Event::Dispatcher::RunType::NonBlock);
}
client_->close(Network::ConnectionCloseType::FlushWrite);
}
// Test that connecting to internal address will crash.
// TODO(lambdai): Add internal connection implementation to enable the connection creation.
TEST_P(SocketInterfaceIntegrationTest, InternalAddressWithSocketInterface) {
BaseIntegrationTest::initialize();
ConnectionStatusCallbacks connect_callbacks_;
Network::ClientConnectionPtr client_;
const Network::SocketInterface* sock_interface = Network::socketInterface(
"envoy.extensions.network.socket_interface.default_socket_interface");
Network::Address::InstanceConstSharedPtr address =
std::make_shared<Network::Address::EnvoyInternalInstance>("listener_0", sock_interface);
ASSERT_DEATH(client_ = dispatcher_->createClientConnection(
address, Network::Address::InstanceConstSharedPtr(),
Network::Test::createRawBufferSocket(), nullptr),
"panic: not implemented");
}
// Test that recv from internal address will crash.
// TODO(lambdai): Add internal socket implementation to enable the io path.
TEST_P(SocketInterfaceIntegrationTest, UdpRecvFromInternalAddressWithSocketInterface) {
BaseIntegrationTest::initialize();
const Network::SocketInterface* sock_interface = Network::socketInterface(
"envoy.extensions.network.socket_interface.default_socket_interface");
Network::Address::InstanceConstSharedPtr address =
std::make_shared<Network::Address::EnvoyInternalInstance>("listener_0", sock_interface);
ASSERT_DEATH(std::make_unique<Network::SocketImpl>(Network::Socket::Type::Datagram, address), "");
}
// Test that send to internal address will return io error.
TEST_P(SocketInterfaceIntegrationTest, UdpSendToInternalAddressWithSocketInterface) {
BaseIntegrationTest::initialize();
const Network::SocketInterface* sock_interface = Network::socketInterface(
"envoy.extensions.network.socket_interface.default_socket_interface");
Network::Address::InstanceConstSharedPtr peer_internal_address =
std::make_shared<Network::Address::EnvoyInternalInstance>("listener_0", sock_interface);
Network::Address::InstanceConstSharedPtr local_valid_address =
Network::Test::getCanonicalLoopbackAddress(version_);
auto socket =
std::make_unique<Network::SocketImpl>(Network::Socket::Type::Datagram, local_valid_address);
Buffer::OwnedImpl buffer;
Buffer::RawSlice iovec;
buffer.reserve(100, &iovec, 1);
auto result =
socket->ioHandle().sendmsg(&iovec, 1, 0, local_valid_address->ip(), *peer_internal_address);
ASSERT_FALSE(result.ok());
ASSERT_EQ(result.err_->getErrorCode(), Api::IoError::IoErrorCode::NoSupport);
}
} // namespace
} // namespace Envoy | {
"redpajama_set_name": "RedPajamaGithub"
} | 3,508 |
<constraint-mappings
xmlns="http://jboss.org/xml/ns/javax/validation/mapping"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://jboss.org/xml/ns/javax/validation/mapping validation-mapping-1.0.xsd">
<constraint-definition annotation="javax.validation.constraints.Future">
<validated-by include-existing-validators="true">
<value>org.opensmartgridplatform.domain.core.validation.joda.FutureValidator</value>
</validated-by>
</constraint-definition>
<constraint-definition annotation="javax.validation.constraints.Past">
<validated-by include-existing-validators="true">
<value>org.opensmartgridplatform.domain.core.validation.joda.PastValidator</value>
</validated-by>
</constraint-definition>
</constraint-mappings> | {
"redpajama_set_name": "RedPajamaGithub"
} | 1,261 |
Q: google maps api directions I am working with the google maps api. I have one issue that I was wondering if other people had and if it is fixable. When using the direction feature, sometimes the routes are unnecessarily long or complex. Like the picture in this link.
Is there anything i could change in my directions code??
var directionsDisplay = new google.maps.DirectionsRenderer({
suppressMarkers: true
});
var directionsService = new google.maps.DirectionsService;
directionsService.route({
origin: marker.position,
destination: dest_end,
travelMode: google.maps.TravelMode.WALKING
}, function (response, status) {
if (status == google.maps.DirectionsStatus.OK) {
directionsDisplay.setDirections(response);
} else {
window.alert('Directions request failed due to ' + status);
}
});
A: I haven't enough credit to make comment so i am answering here according to google map they have three walking route to your destination and it shows best from that routs by default, in your case this rout has 0.4 mile walking distance and other two has 0.4 and 0.5 respectively so it take smallest distance with less time.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 8,113 |
MadMonday Hash
#### How to use?
```python
from MMHash import MMHash
hashsize = 256
print(MMHash('text', hashsize))
```
## MMCrypt
MadMonday Crypt
#### How to use?
```python
from MMCrypt import MMCrypt
crypted_text = MMCrypt().crypt('text')
encrypted_text = MMCrypt().encrypt(crypted_text)
```
| {
"redpajama_set_name": "RedPajamaGithub"
} | 764 |
{"url":"https:\/\/samacheer-kalvi.com\/samacheer-kalvi-12th-maths-guide-chapter-9-ex-9-9\/","text":"Tamilnadu State Board New Syllabus\u00a0Samacheer Kalvi 12th Maths Guide Pdf Chapter 9 Applications of Integration Ex 9.9 Textbook Questions and Answers, Notes.\n\nTamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.9\n\nQuestion 1.\nFind by integration, the volume of the solid generated by revolving about the x axis, the region enclosed by y = 2x\u00b2, y = 0 and x = 1\nSolution:\nThe region to be revolved is sketched\n\nSince revolution is made about the x axis, the volume of the solid generated is given by\n\nRequired volume = $$\\frac { 4\u03c0 }{ 5 }$$ cubic units\n\nQuestion 2.\nFind, by integration, the volume of the solid generated by revolving about the x axis, the region enclosed by y = e-2x, y = 0, x = 0 and x = 1.\nSolution:\nSince revolution is made about the x axis, the volume of the solid generated is given\n\nRequired volume = $$\\frac { \u03c0 }{ 4 }$$ [1 \u2013 e-4] cubic units\n\nQuestion 3.\nFind, by integration, the volume of the solid generated by revolving about the y axis, the region enclosed by x\u00b2 = 1 + y and y = 3.\nSolution:\nThe region to be revolved is sketched.\n\nSince revolution is made about the y axis, the volume of the solid generated is given by\n\nRequired volume = 8\u03c0 cubic units\n\nQuestion 4.\nThe region enclosed between the graphs of y = x and y = x\u00b2 is denoted by R. Find the volume generated when R is rotated through 360\u00b0 about x axis.\nSolution:\nThe region to be revolved is sketched.\n\nFind the intersecting point of y = x and y = x\u00b2\nx\u00b2 = x\nx\u00b2 \u2013 x = 0\nx (x \u2013 1) = 0 x = 0, x = 1\nIf x = 0, y = 0, x = 1, y = 1\n\u2234 Points of intersection are (0, 0), (1, 1)\n\nRequired volume = $$\\frac { 2\u03c0 }{ 15 }$$ cubic units\n\nQuestion 5.\nFind, by integration, the volume of the container which is in the shape of a right circular conical frustum as shown to figure.\n\nSolution:\nBy using integration we have to find the volume of the frustum. So first find the equation of the curve.\nLet A(0, 1) and B(2, 2) be two points. Line joining these two points form a straight line. That straight line revolves around x axis.\n\nVolume of the solid revolves around x axis\n\nVolume of the frustum = $$\\frac { 14 }{ 3 }$$ \u03c0 m\u00b3\n\nQuestion 6.\nA watermelon has an ellipsoid shape which can be obtained by revolving an ellipse with major axis 20 cm and minor axis 10 cm about its major axis. Find its volume using integration.\nSolution:\nA watermelon has an ellipsoid shape.\n2a = 20\na = 10\n2b = 10\nb = 5\n\n\u2234 Volume of the frustum = $$\\frac { 1000\u03c0 }{ 3 }$$ cubic units.","date":"2023-01-31 10:55:45","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.6273224949836731, \"perplexity\": 518.2568329637337}, \"config\": {\"markdown_headings\": false, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2023-06\/segments\/1674764499857.57\/warc\/CC-MAIN-20230131091122-20230131121122-00608.warc.gz\"}"} | null | null |
Mechery Louis Ouseppachan (born 13 September 1955), known mononymously as Ouseppachan, is an Indian film composer and singer who primarily works in Malayalam films. He is a recipient of National Film Award, Filmfare Award and Kerala State Film Awards for his numerous film soundtrack albums and background scores.
Early life
Ouseppachan was born to Mechery Louis and Mathiri Paliakkara on 13 September 1955 in Ollur, Thrissur, Kerala, India. He completed his formal education from St. Thomas College, Thrissur. He was interested in music and by the time he finished his formal education, he had become an expert violinist.
Career
He started his music career as violinist in some music troupes including Voice of Trichur. Later he got the chance to be the violinist in concerts of renowned playback singer Madhuri. He later became the violinist under music director Paravoor Devarajan master.
Then he moved to Madras, Tamil Nadu where most of the recording works of Malayalam films were done.
His first work in the film industry was the film Eenam, for which he set the background score. He also composed additional background music for the film Aaravam (1979) in which he also played the role of a fiddle player, however Kathodu Kathoram (1985), directed by Bharathan is considered as his debut. Ouseppachan's work was noted for the film and three songs from the film became super hits. These songs were noted also for the immense use of violin. This may be attributed to the facts that Ousephachan himself is a violinist and that the protagonist of the film played by Mammootty is also a violinist.
Ousephachan then went on to compose music for over 120 films. He also composed songs for the non-commercial Hindi film Freaky Chakra, directed by V.K. Prakash and background scores for various other Hindi films, notably directed by Priyadarshan. His non-film songs include popular albums like Onapoothalam, Vasanthageetangal and many Christian devotional songs. Aside from music direction, he had taken up a spot as a judge for Asianet's Idea Star Singer 2008 and also as a judge for Kairali TV's Gandharva Sangeetham 2012. He got his first state award in 1987 for the movie Unnikale Oru Kadha Parayam. His first national award was for ace director Shyamaprasad's Ore Kadal in 2007. All the songs in this film are composed in Shubhapantuvarali raga but all have their identity. Apparently, it is a challenge given to Ouseppachan by the director and he took up the challenge. His recent notable films are Ayalum Njanum Thammil, Nadan. For Nadan, he recreated the classical drama songs from the classical K.P.A.C era. He received a state award for this film for Best Music Director. Acclaimed music composers like A.R.Rahman, Vidyasagar, Harris Jayaraj, Gopi Sunder etc., has assisted him in various stages of his career and playback singer Franco Simon is his nephew.
Selected Discography
Awards
National Film Awards:
2007 – Best Music Direction: Ore Kadal
Filmfare Awards South:
2007 – Filmfare Award for Best Music Director – Malayalam: Ore Kadal
Kerala State Film Awards:
1987 – Best Music Director: Unnikale Oru Kadha Parayam
2007 – Best Background Music: Ore Kadal
2013 – Best Music Director: Nadan
References
External links
Ouseppachan at MSI
1954 births
Living people
Musicians from Thrissur
Kerala State Film Award winners
Malayalam film score composers
St. Thomas College, Thrissur alumni
Filmfare Awards South winners
Best Music Direction National Film Award winners
Film musicians from Kerala
20th-century Indian composers
21st-century Indian composers
People from Thrissur
Male actors from Thrissur
Male actors in Malayalam cinema
Indian male film actors | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 543 |
require 'spec_helper'
describe User do
it { should have_db_index(:email) }
it { should have_db_index(:remember_token) }
describe "When signing up" do
it { should validate_presence_of(:email) }
it { should validate_presence_of(:password) }
it { should allow_value("foo@example.co.uk").for(:email) }
it { should allow_value("foo@example.com").for(:email) }
it { should_not allow_value("foo@").for(:email) }
it { should_not allow_value("foo@example..com").for(:email) }
it { should_not allow_value("foo@.example.com").for(:email) }
it { should_not allow_value("foo").for(:email) }
it { should_not allow_value("example.com").for(:email) }
it "should store email in down case" do
user = Factory(:user, :email => "John.Doe@example.com")
user.email.should == "john.doe@example.com"
end
end
describe "When multiple users have signed up" do
before { Factory(:user) }
it { should validate_uniqueness_of(:email) }
end
describe "A user" do
before do
@user = Factory(:user)
@password = @user.password
end
it "is authenticated with correct email and password" do
(::User.authenticate(@user.email, @password)).should be
@user.should be_authenticated(@password)
end
it "is authenticated with correct uppercased email and correct password" do
(::User.authenticate(@user.email.upcase, @password)).should be
@user.should be_authenticated(@password)
end
it "is authenticated with incorrect credentials" do
(::User.authenticate(@user.email, 'bad_password')).should_not be
@user.should_not be_authenticated('bad password')
end
end
describe "When resetting authentication with reset_remember_token!" do
before do
@user = Factory(:user)
@user.remember_token = "old-token"
@user.reset_remember_token!
end
it "should change the remember token" do
@user.remember_token.should_not == "old-token"
end
end
describe "An email confirmed user" do
before do
@user = Factory(:user)
@old_encrypted_password = @user.encrypted_password
end
describe "who updates password" do
before do
@user.update_password("new_password")
end
it "should change encrypted password" do
@user.encrypted_password.should_not == @old_encrypted_password
end
end
end
it "should not generate the same remember token for users with the same password at the same time" do
Time.stubs(:now => Time.now)
password = 'secret'
first_user = Factory(:user, :password => password)
second_user = Factory(:user, :password => password)
second_user.remember_token.should_not == first_user.remember_token
end
describe "An user" do
before do
@user = Factory(:user)
@old_encrypted_password = @user.encrypted_password
end
describe "who requests password reminder" do
before do
@user.confirmation_token.should be_nil
@user.forgot_password!
end
it "should generate confirmation token" do
@user.confirmation_token.should_not be_nil
end
describe "and then updates password" do
describe 'with password' do
before do
@user.update_password("new_password")
end
it "should change encrypted password" do
@user.encrypted_password.should_not == @old_encrypted_password
end
it "should clear confirmation token" do
@user.confirmation_token.should be_nil
end
end
describe 'with blank password' do
before do
@user.update_password("")
end
it "does not change encrypted password" do
@user.encrypted_password.should == @old_encrypted_password
end
it "does not clear confirmation token" do
@user.confirmation_token.should_not be_nil
end
end
end
end
end
describe "a user with an optional email" do
before do
@user = User.new
class << @user
def email_optional?
true
end
end
end
subject { @user }
it { should allow_value(nil).for(:email) }
it { should allow_value("").for(:email) }
end
describe "a user with an optional password" do
before do
@user = User.new
class << @user
def password_optional?
true
end
end
end
subject { @user }
it { should allow_value(nil).for(:password) }
it { should allow_value("").for(:password) }
end
describe "user factory" do
it "should create a valid user with just an overridden password" do
Factory.build(:user, :password => "test").should be_valid
end
end
describe "when user exists before Clearance was installed" do
before do
@user = Factory(:user)
sql = "update users set salt = NULL, encrypted_password = NULL, remember_token = NULL where id = #{@user.id}"
ActiveRecord::Base.connection.update(sql)
@user.reload.salt.should be_nil
@user.reload.encrypted_password.should be_nil
@user.reload.remember_token.should be_nil
end
it "should initialize salt, generate remember token, and save encrypted password on update_password" do
@user.update_password('password')
@user.salt.should_not be_nil
@user.encrypted_password.should_not be_nil
@user.remember_token.should_not be_nil
end
end
describe "The password setter on a User" do
let(:password) { "a-password" }
before { subject.send(:password=, password) }
it "sets password to the plain-text password" do
subject.password.should == password
end
it "also sets encrypted_password" do
subject.encrypted_password.should_not be_nil
end
end
end
| {
"redpajama_set_name": "RedPajamaGithub"
} | 36 |
package io.quarkus.amazon.lambda.runtime;
import java.io.IOException;
import java.io.OutputStream;
import java.net.HttpURLConnection;
public interface LambdaOutputWriter {
void writeValue(OutputStream os, Object obj) throws IOException;
default void writeHeaders(HttpURLConnection conn) {
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 4,031 |
import * as assert from 'assert';
import 'vs/workbench/browser/parts/editor/editor.contribution'; // make sure to load all contributed editor things into tests
import { Promise, TPromise } from 'vs/base/common/winjs.base';
import { Event } from 'vs/base/common/event';
import { IQuickOpenService } from 'vs/platform/quickOpen/common/quickOpen';
import { Registry } from 'vs/platform/registry/common/platform';
import { QuickOpenHandlerDescriptor, IQuickOpenRegistry, Extensions as QuickOpenExtensions, QuickOpenAction, QuickOpenHandler } from 'vs/workbench/browser/quickopen';
export class TestQuickOpenService implements IQuickOpenService {
public _serviceBrand: any;
private callback: (prefix: string) => void;
constructor(callback?: (prefix: string) => void) {
this.callback = callback;
}
accept(): void {
}
focus(): void {
}
close(): void {
}
show(prefix?: string, options?: any): Promise {
if (this.callback) {
this.callback(prefix);
}
return TPromise.as(true);
}
get onShow(): Event<void> {
return null;
}
get onHide(): Event<void> {
return null;
}
public dispose() { }
public navigate(): void { }
}
suite('QuickOpen', () => {
class TestHandler extends QuickOpenHandler { }
test('QuickOpen Handler and Registry', () => {
let registry = (Registry.as<IQuickOpenRegistry>(QuickOpenExtensions.Quickopen));
let handler = new QuickOpenHandlerDescriptor(
TestHandler,
'testhandler',
',',
'Handler',
null
);
registry.registerQuickOpenHandler(handler);
assert(registry.getQuickOpenHandler(',') === handler);
let handlers = registry.getQuickOpenHandlers();
assert(handlers.some((handler: QuickOpenHandlerDescriptor) => handler.prefix === ','));
});
test('QuickOpen Action', () => {
let defaultAction = new QuickOpenAction('id', 'label', void 0, new TestQuickOpenService((prefix: string) => assert(!prefix)));
let prefixAction = new QuickOpenAction('id', 'label', ',', new TestQuickOpenService((prefix: string) => assert(!!prefix)));
defaultAction.run();
prefixAction.run();
});
}); | {
"redpajama_set_name": "RedPajamaGithub"
} | 742 |
{"url":"https:\/\/davidstutz.de\/multi-scale-context-aggregation-dilated-convolutions-yu-koltun\/","text":"# DAVIDSTUTZ\n\nCheck out our latest research on adversarial robustness and generalization of deep networks.\n18thMARCH2018\n\nFisher Yu, Vladlen Koltun. Multi-Scale Context Aggregation by Dilated Convolutions. CoRR, 2015.\n\nYu and Koltun use dilated convolutions, packaged in a so-called context module, to integrate multi-scale context into convolutional networks for semantic segmentation. Compared to regular, discrete convolution\n\n$(F\\ast k)(p) = \\sum_{s+t=p} F(s)k(t)$\n\nwhere $F$ and $k$ are discrete functions, $k$ defined on a grid of size $(2r + 1)^2$, dilated convolution introduces a dilation factor:\n\n$F \\ast_l k)(p) = \\sum_{s+lt=p} F(s)k(t)$\n\nIn the proposed context module, Yu and Koltun stack several layers of dilated convolution with exponentially increasing dilation factors:\n\n$F_{i+1} = F_i \\ast_{2^i}k_i$ for $i = 0,1,2,\\ldots,n-2$.$This allows to exponentially increase the receptive field while preserving the resolution (in contrast to pooling or strided convolution, which reduce resolution while increasing receptive field). The idea is illustrated in Figure 1. In practice, they also discuss initialization. Concretely, they were not able to improve semantic segmentation performance using standard initialization techniques. Instead they use an identity initialization:$k^b(t, a) = 1_{[t=0]} 1_{[a=b]}$where$a$is the index of the input feature map,$b$the index of the output feature map (assuming that the same number of output feature maps is computed);$t$indixes the kernel location (i.e. only the middle weight is set to 1. They also generalize this scheme to cases where input and output feature maps do not match. Let$c_i$and$c_{i+1}$be the number of feature maps in layer$i$and layer$i + 1$, then$k^b(t,a) = \\begin{cases}\\frac{C}{c_{i+1}} &\\quad t=0 \\text{ and } \\lfloor\\frac{aC}{c_i}\\rfloor = \\lfloor \\frac{bC}{c_{i+1}}\\rfloor\\\\\\epsilon &\\quad\\text{otherwise}\\end{cases}$with$\\epsilon$being a normal random variable with variance$\\sigma \\ll C\/c_{i + 1}\\$.\n\nWhat is your opinion on the summarized work? Or do you know related work that is of interest? Let me know your thoughts in the comments below or get in touch with me:","date":"2019-01-22 23:09:56","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.6464126706123352, \"perplexity\": 4769.724727865742}, \"config\": {\"markdown_headings\": true, \"markdown_code\": false, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-04\/segments\/1547583875448.71\/warc\/CC-MAIN-20190122223011-20190123005011-00065.warc.gz\"}"} | null | null |
Q: What to do if I don't have Context? I'm developing my first app in Android Studio. I have two classes. One which contains settings.
public class settings {
private int przed_termin;
private int przed_po_otwarciu;
private boolean powiadomienia;
private ustawienia_sql ustawienia_baza;
}
And a second one which is responsible for reading and saving them to SQLite database. And now I'm trying to create object settings_sql in settings, but to do that I need a Context. And is there possibility to get past that?
public class settings_sql extends SQLiteOpenHelper {
public settings_sql (Context context) {
super(context, "ustawienia.db", null, 1);
}
}
A: I'm going to give you a bit of understand then code. Context is a common frustration for new Android programmers.
There's actually 4 different types of Context objects. This important to know. Generally if you're doing user controls you'll want the Activity context the user control is hosted in. daylight's answer will work for that. Basically the context object is an Activity object. There's also Service instance objects that have a Service Context.
Another Context type is Application Context.
This is how you make sure any class in your app has access to the Application Context:
Create a class like this. That statically stores a reference to it's self with an accessor:
public class MyApp extends Application{
private static Context applicationContext;
public void onCreate() {
applicationContext = this;
}
public static Context getApplicationContext() {
return applicationContext;
}
}
Modify your ApplicationManifest.xml to include the name attribute in your application tag:
<application
android:name="MyApp"
android:allowBackup="true"
android:icon="@drawable/icon"
android:label="@string/app_name"
android:theme="@style/AppTheme">
<activity......
</application>
Then wherever you need a context object and don't have one convient use:
MyApp.getApplicationContext()
For example:
settings_sql(MyApp.getApplicationContext());
Because you create in the application class oncreate event you can be assured it'll never be null, while in your app.
A: I recommend passing the Context to classes that need it. You already have your settings_sql(Context context) constructor, which is a good start.
In your main application code, you can easily pass the Context from either your activity or application, depending on your needs and also on the actual lifetime of the settings_sql object. Does this object need to live beyond the duration of a single Activity?
If you then decide to write tests, you are free from depending on a static variable held in an arbitrary class that may not have much to do with your test. Why would a local unit test need to know about the Application class, for example?
You might not run into this problem early on, or at all, and a static variable may be a perfectly workable short-term solution, but just be aware that this can make things awkward and difficult to reason about when you're interacting with parts of your system from within limited scopes, particularly tests.
A: In your MainActivity do something like:
private static Context context;
public void onCreate() {
super.onCreate();
context = getApplicationContext();
}
Then when you make the call just pass the context in.
A: The simplest solution is to have an Application class in your APK and add a static method to it to retrieve the context, like this:
public class MyApplication extends Application {
private static MyApplication sInstance = null;
@Override
public void onCreate() {
super.onCreate();
sInstance = this;
}
public static Context getContext() {
return sInstance.getApplicationContext();
}
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 6,888 |
{"url":"http:\/\/www.dcode.fr\/primitive-integral","text":"Search for a tool\nPrimitives Functions\n\nTool to find primitives of a function. Integration of a function is the calculation of all its primitives, the inverse of the derivative.\n\nResults\n\nPrimitives Functions -\n\nTag(s) : Mathematics\n\ndCode and you\n\ndCode is free and its tools are a valuable help in games, puzzles and problems to solve every day!\nYou have a problem, an idea for a project, a specific need and dCode can not (yet) help you? You need custom development? Contact-me!\n\nTeam dCode read all messages and answer them if you leave an email (not published). It is thanks to you that dCode has the best Primitives Functions tool. Thank you.\n\n# Primitives Functions\n\nThis script has been updated, please report any problems.\n\n## Primitive Function Calculator\n\nAlso on dCode: Definite IntegralDerivative\n\nTool to find primitives of a function. Integration of a function is the calculation of all its primitives, the inverse of the derivative.\n\n### How to calculate a primitive\/integral?\n\nEnter the function and its variable to integrate and dCode do the computation of the primitive function.\n\nMathematicians talks about finding the function calculating the area under the curve.\n\ndCode knows all functions and their primitives.\n\n### What is the list of common primitives?\n\nFunctionPrimitive\n$$\\int \\,\\rm dx$$$$x + C$$\n$$\\int x^n\\,\\rm dx$$$$\\frac{x^{n+1}}{n+1} + C \\qquad n \\ne -1$$\n$$\\int \\frac{1}{x}\\,\\rm dx$$$$\\ln \\left| x \\right| + C \\qquad x \\ne 0$$\n$$\\int \\frac{1}{x-a} \\, \\rm dx$$$$\\ln | x-a | + C \\qquad x \\ne a$$\n$$\\int \\frac{1}{(x-a)^n} \\, \\rm dx$$$$-\\frac{1}{(n-1)(x-a)^{n-1}} + C \\qquad n \\ne 1 , x \\ne a$$\n$$\\int \\frac{1}{1+x^2} \\, \\rm dx$$$$\\operatorname{arctan}(x) + C$$\n$$\\int \\frac{1}{a^2+x^2} \\, \\rm dx$$$$\\frac{1}{a}\\operatorname{arctan}{ \\left( \\frac{x}{a} \\right) } + C \\qquad a \\ne 0$$\n$$\\int \\frac{1}{1-x^2} \\, \\rm dx$$$$\\frac{1}{2} \\ln { \\left| \\frac{x+1}{x-1} \\right| } + C$$\n$$\\int \\ln (x)\\,\\rm dx$$$$x \\ln (x) - x + C$$\n$$\\int \\log_b (x)\\,\\rm dx$$$$x \\log_b (x) - x \\log_b (e) + C$$\n$$\\int e^x\\,\\rm dx$$$$e^x + C$$\n$$\\int a^x\\,\\rm dx$$$$\\frac{a^x}{\\ln (a)} + C \\qquad a > 0 , a \\ne 1$$\n$$\\int {1 \\over \\sqrt{1-x^2}} \\, \\rm dx$$$$\\operatorname{arcsin} (x) + C$$\n$$\\int {-1 \\over \\sqrt{1-x^2}} \\, \\rm dx$$$$\\operatorname{arccos} (x) + C$$\n$$\\int {x \\over \\sqrt{x^2-1}} \\, \\rm dx$$$$\\sqrt{x^2-1} + C$$\n$$\\int \\sin(x)\\,\\rm dx$$$$-\\cos(x)+C$$\n$$\\int \\cos(x)\\,\\rm dx$$$$\\sin(x)+C$$\n$$\\int \\tan(x)\\,\\rm dx$$$$-\\ln|\\cos(x)|+C$$","date":"2016-12-10 12:39:42","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.4138864278793335, \"perplexity\": 2162.3046122945157}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2016-50\/segments\/1480698543170.25\/warc\/CC-MAIN-20161202170903-00304-ip-10-31-129-80.ec2.internal.warc.gz\"}"} | null | null |
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How to Live Stream on Kik [Full Guide]
Kik is a popular messaging app with over 300 million users. Kik offers a built-in live streaming feature that lets you share your screen with others in real time. You can use Kik to live stream anything from a game to a tutorial. Here's how to live stream on Kik.
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Kik Live Stream is a new feature on the popular messaging app Kik that allows users to stream live videos to their friends. It's a great way to stay connected with friends and family or to just show off your latest adventure. Here's everything you need to know about Kik Live Stream.
Read Also:How much do Starbucks holiday drinks cost this year?
To start using Kik Live Stream, simply open the app and tap on the "Live Stream" icon in the top right corner. Then, choose who you want to share your live stream with. You can either share it with all your contacts, or just specific friends. Once you've chosen who you want to share your stream with, start streaming! Your friends will be able to see your live video and chat with you in real time.
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Kik is a messaging app that has been gaining popularity in recent years. The app offers a unique feature called Kik Live Stream, which allows users to stream live videos to their contacts. Here's how to go live on Kik on Android and iOS:
To start, open the Kik app and tap on the "Live" icon at the bottom of the screen. On the next page, you'll need to provide a title for your live stream and select whether you want to allow comments from viewers. Once you've done that, tap on the "Start Live Stream" button.
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Download the Kik app and create an account.
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How To Cancel Live Stream on Kik?
If you're done with Kik and want to cancel your Live Stream, here's how.
First, open the app and go to your profile page. Tap the three dots in the top-right corner of the screen and select "Settings."
Under "Your Account," tap "Cancel Live Stream." You'll be prompted to confirm that you want to cancel. Once you do, your Live Stream will be canceled and you'll no longer be able to use the feature.
If you have any questions or need help, you can contact Kik's customer support team. They're available 24/7 and will be happy to assist you.
What are the Benefits Of Going Live On Kik?
Kik is a messaging app that allows users to connect with each other via text, video, and audio. The app also has a live-streaming feature that lets users share their lives with others in real time.
Going live on Kik can be a great way to connect with friends and family, as well as meet new people from all over the world. It's also a great way to share your life with others and let them into your world.
There are many benefits to going live on Kik. You can interact with your audience in real time, which can make for a more personal connection. You can also answer questions and give insights that you might not be able to do through text or video alone.
Going live on Kik can help you build an audience and following.
Kik is a messaging app that allows users to connect with friends and family. The app also has a live streaming feature that allows users to share their live stream with others. In order to start a live stream on Kik, follow the steps below.
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Takeaways from the 2022 midterm elections: The battle for control of... | {
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\section{Introduction and Summary}
In this work we study massive gravity in 2+1 spacetime dimensions.
A well-known theory of this is Topological Massive Gravity \cite {tmg}.
Here, we focus on the nonlinear dynamics of ``New Massive Gravity" (NMG)
proposed by Bergshoeff, Hohm, and Townsend, \cite{nmg}. Similar studies on the nonlinear
structure of massive gravity in {\it 3+1} dimensions have by now reached a certain level of maturity.
Since these developments are relevant for the motivation and
content of the present paper, we begin with a brief outline of the most important insights gained.
Fierz and Pauli (FP) \cite{fp} constructed a ghost-free and tachyon-less
linear theory of massive spin-2 in Minkowski space, describing 5 degrees of freedom in 4D
(2 degrees of freedom in 3D), that is consistent with the corresponding
representation of the Poincar\'e group. The FP theory has no continuous limit
to the massless theory \cite{vDVZ}, while the continuity can be restored in its
nonlinear extensions \cite{Vainshtein}. However, a generic nonlinear extension in 4D suffers
from the so-called Boulware - Deser (BD) ghost: the Hamiltonian constraint, that would
restrict the number of degrees of freedom to no more than 5 for any background, is
lost at the nonlinear level \cite{bd}. The consequences of this can be transparently
seen in the decoupling limit \cite{AGS}; here, the problem emerges
in the equation of motion for the helicity-0 mode that exhibits
ill-posed Cauchy problem due to nonlinear terms with
more than two time derivatives acting on a single field \cite{Creminelli,Deffayet}.
A toy example of this in 4D is the Lagrangian \cite {AGS}
\begin{eqnarray}
\mathcal{L}=\frac{1}{2}\pi\Box\pi+\frac{ (\Box\pi)^3 }{\Lambda^5_5} +\frac{1}{M_P}\pi T,
\label{pilagr}
\end{eqnarray}
where, $\Lambda_5$ is a certain scale (composed of the graviton mass and Planck
mass), and $T$ denotes the trace of the external energy-momentum tensor.
The equations of motion that follow from (\ref{pilagr}) involve nonlinear
terms with more than two time derivatives acting on $\pi$. This leads to an additional (sixth) mode
with negative kinetic term propagating on nontrivial backgrounds for the helicity-0 mode (e.g.,
that of a collapsing spherical source) \cite{GGruzinov,Creminelli,Deffayet};
this mode is identified with the BD ghost. This follows directly from the standard Ostrogradski
instability arguments and can be made manifest here by noting that the above theory can be equivalently
represented by the following lagrangian \cite{Deffayet},
\begin{eqnarray}
\mathcal{L'}=\frac{1}{2}\phi \Box\phi -\frac{1}{2}\psi\Box\psi- \epsilon \frac{2}{3\sqrt{3}}
\psi^{3/2}\Lambda^{5/2}_5+\frac{1}{M_P}\phi
T-\frac{1}{M_P}\psi T.
\label{psilagr}
\end{eqnarray}
The latter lagrangian is obtained from (\ref{pilagr}) in two steps.
First one introduces an auxiliary field $\lambda$ and a corresponding lagrangian
$\mathcal{L'}$, which involves no more than single derivatives per field.
The Lagrangian $\mathcal{L'}$ is defined so that integrating out $\lambda$, one recovers the original
$\pi$-lagrangian (\ref{pilagr}). On the other hand, making the further field
redefinitions $\lambda^2= \psi$ and $\pi=\phi-\psi$ brings $\mathcal{L'}$ to the form,
given in (\ref{psilagr}). Here $\epsilon=\pm 1$, depending on the sign of $\Box(\phi-\psi)$.
It is manifest from (\ref{psilagr}) that the additional
degree of freedom $\psi$ is inevitably a ghost; however, its mass is set by $\langle \psi \rangle ^{-1/2}$, and
although this mass is infinite when expanding around a Minkowski background, which explains why its presence
was not seen in the original FP analysis, it can drop down to physically accessible scales when expanding around a
non-trivial background, \cite{Creminelli}.
The simplest example of a nonlinear theory for a helicity-0 mode graviton that does not have
the above problem was found in the context of the DGP model
\cite{dgp} in Ref. \cite{lpr}. Its Lagrangian reads as follows:
\begin{eqnarray}
\mathcal{L}=\frac{1}{2} \pi\Box\pi+\frac{ (\partial\pi)^2\Box\pi }{\Lambda_3^3}+\frac{1}{M_P}\pi T.
\label{dgp}
\end{eqnarray}
The specific structure of the cubic term prevents the
appearance of more than two time derivatives in the equations of motion for $\pi$,
thereby rendering it free of the second, ghost-like, degree of freedom on any weak asymptotically flat
background, \cite{lpr}. The Lagrangian (\ref{dgp}) has been generalized to incorporate the quartic and quintic
self interactions with similar properties -- the so-called Galileon terms, \cite{nrt}.
These terms have been shown to naturally arise
in probe brane setups \cite{drt} and their extensions to multi-Galileon
theories have been studied in \cite{DeffayetGalileon,MultiGalileon1,MultiGalileon2}.
More importantly for the present work, Refs.~\cite{drg1,drg2}
showed that a certain class of theories of
4D massive gravity, contain all the Galileon terms
in the decoupling limit. In these theories, an infinite
number of nonlinear terms can be resummed and a Lagrangian containing just a few terms
can be obtained \cite{drgt}. Hence, the BD problem can be addressed in the full-fledged Hamiltonian
approach. So far the Hamiltonian has been calculated up to and including quartic order
in nonlinearities and beyond the decoupling limit,
showing that the BD problem does not arise at this level~\cite{drgt}\footnote{
A potentially dangerous term found in the quartic order of the Hamiltonian
construction in Ref. \cite {Creminelli} has a very special structure \cite {drgt},
and due to this, can be removed to higher orders by a consistent nonlinear field redefinition,
as shown explicitly in \cite {drgt}. The terms that could not have been
removed, do cancel out automatically in the quartic order
\cite {drgt}. We also note that the term of concern of Ref. \cite {Mukhanov}, found in that
work in the Lagrangian formalism, is nothing but the term already found in \cite {Creminelli} in the
Hamiltonian formalism, that was addressed above.}.
The purpose of the present work is to both study the nonlinear interactions
of the helicity-0 mode in the decoupling limit of NMG~\cite{nmg}, determining whether or
not it gives rise to the problematic nonlinear terms, as well as to address the
issue of the BD ghost in the full theory, away from the decoupling limit.
Since NMG is a 3D theory, one would expect to be able to do more than
in 4D, as well as gain some additional intuition about the 4D constructions.
The divergences arising in 3D being less severe, NMG can give some valuable insight on the renormalization of gravity, \cite{Oda:2009ys}. Furthermore, a DBI extension was proposed in Ref.~\cite{Gullu:2010pc}, and a remarkable connection with AdS/CFT was established, \cite{Skenderis:2009nt}. The existence of AdS$_3$ Black Hole solutions also makes this theory especially interesting, \cite{Clement:2009gq}.
In this work, we will focus instead on the stability of the theory and show that
the interaction Lagrangian for the helicity-0 in the decoupling limit of NMG
reduces to the three-dimensional version of Lagrangian ~(\ref {dgp}), supplemented by an
equation which determines the tensor modes. Hence, the theory
has a well posed Cauchy problem. The nonlinear term in (\ref {dgp}) seems to arise universally: It appears
in the context of the 5D DGP model, as found in \cite{lpr}, in 4D massive gravity \cite{drg1,drg2,drgt}, and now in 3D NMG.
In what follows we demonstrate that in NMG, the helicity-0 mode coincides with the conformal mode of the metric
in the decoupling limit. This motivates us to study the dynamics of the conformal mode
in the full theory. In ordinary GR, it is well known that the conformal mode has the wrong-sign kinetic term.
This is not a problem in GR since this mode is not dynamical (it can be removed via the gauge freedom and constraints).
In NMG the conformal mode is dynamical, but the
GR term itself has a wrong sign\footnote{This is acceptable as there are no propagating
degrees of freedom in 3D massless GR.}, hence, the conformal mode obtains a correct sign kinetic term.
Furthermore, the nonlinear terms of NMG give rise to interactions for the conformal mode that
are straightforward generalizations of the Galileon interactions, implying that they have a well-defined Cauchy problem.
Moreover, we give a non-perturbative counting of degrees of freedom that demonstrates that in NMG, even away from the
decoupling limit or from conformally
flat metric configurations, there are no extra modes beyond the two of a massive 3D graviton. This confirms that the
3D analogue of the BD ghost does not arise nonlinearly.
Last but not least, we present a class of 3D cubic order theories that generalize the NMG construction.
The paper is organized as follows. Section 2 serves as a review where we present some
well-known properties of massive gravity in 4D, adapted to three spacetime dimensions.
We illustrate that all known potential problems that arise when attempting to give a 4D graviton mass,
persist in generic three dimensional models as well. The reader who is well acquainted with the technology and
issues involved is encouraged to skip directly to section 3 where we investigate the nonlinear dynamics of NMG.
We start by studying the decoupling limit of the theory, and arrive at a well-defined,
ghost-free theory. We then move to the dynamics of the conformal mode, and find that it coincides
with the helicity-0 graviton in the decoupling limit. We show next that for conformally flat metrics the full
theory is free of any ghosts. Moreover, we give the non-perturbative arguments in favor of no-ghost propagation
beyond any limit or approximation. In section 4 we look at NMG from
the perspective of a generalization of the linear FP model, demonstrating how exactly the
cancellation of the BD ghost happens at the cubic level. Finally, we construct a class of generalizations
of the cubic theory, which exhibit similar properties.
\section{Ghosts and Strong Coupling in Massive Gravity}
In this section we review some of the known results from massive gravity in 4D and introduce the formalism used in this paper.
We begin by analyzing the Fierz-Pauli model at the linear level. We then extend the discussion to include nonlinear terms.
\subsection{Linear analysis}
The FP model is the unique theory at the linearized level, free of ghosts or tachyons \cite{Nieu} and propagating only the two degrees of freedom of a massive spin-2 particle in 3D. In terms of the metric perturbation $h_{\mu \nu}\equiv g_{\mu \nu}-\eta_{\mu \nu}$, the FP Lagrangian is given by,
\begin{eqnarray}
\mathcal{L}=-\frac{1}{2}M_P h^{\mu\nu}\mathcal{E}_{\mu\nu}^{\rho\sigma}h_{\rho\sigma}-\frac{1}{4}M_P m^2(h^{\mu\nu}h_{\mu\nu}-h^2)+\frac{1}{2}h^{\mu\nu}T_{\mu\nu},
\label{fp}
\end{eqnarray}
where $m$ sets the graviton mass, $T_{\mu \nu}$ denotes an external energy-momentum source, and $\mathcal{E}_{\mu\nu}^{\rho\sigma}$ is the linear Einstein operator
\begin{eqnarray}
\mathcal{E}_{\mu\nu}^{\rho\sigma}h_{\rho\sigma}=- {1\over 2} ( \square h_{\mu \nu} - \partial_\mu \partial^\alpha
h_{\alpha\nu} - \partial_\nu \partial^\alpha
h_{\alpha\mu} + \partial_\mu \partial_\nu h -
\eta_{\mu\nu} \square h + \eta_{\mu\nu}
\partial_\alpha \partial_\beta h^{\alpha\beta}),
\end{eqnarray}
with all indices contracted with the flat metric.
In order to single out the propagating helicity-1 and helicity-0 modes of the massive graviton, the metric perturbation can be conveniently decomposed at the linear order in the following way,
\begin{eqnarray}
h_{\mu\nu}=\frac{\bar h_{\mu\nu}}{\sqrt{M_P}}+\frac{\partial_\mu V_\nu}{\sqrt{M_P}m^2}+\frac{\partial_\nu V_\mu}{\sqrt{M_P}m^2}, \quad V_\mu= m A_\mu+\partial_\mu \pi,
\end{eqnarray}
where $\bar h_{\mu\nu}$, $A_\mu$ and $\pi$ encode the (canonically normalized) helicity-2, 1 and 0 components, respectively (note that in 3D the helicity-2 mode does not propagate). With this field content, the theory is invariant under the usual linearized diffeomorphisms
\begin{eqnarray} \bar h_{\mu\nu}\rightarrow \bar h_{\mu\nu}+\partial_\mu\xi_\nu+\partial_\nu\xi_\mu, \quad A_\mu\rightarrow A_\mu-m\,\xi_\mu~,
\end{eqnarray}
as well as an additional abelian U(1) symmetry, under which the vector and scalar modes transform as,
\begin{eqnarray}
A_\mu\rightarrow A_\mu+\partial_\mu \zeta, \quad \quad \pi\rightarrow \pi-m\,\zeta. ~~~~~~
\label{gt}
\end{eqnarray}
The latter invariance, which at this stage is introduced somewhat artificially, will turn out to provide a very convenient bookkeeping tool when studying different decoupling limits of nonlinear massive gravity.
As remarked in the introduction, the linear FP theory does not possess a continuous massless limit. To see this, we note that in terms of the different helicity components, the $m\rightarrow 0$ limit of the theory is given (up to a total derivative) by the following expression,
\begin{eqnarray}
\mathcal{L}_{m=0}=-\frac{1}{2}\bar h^{\mu\nu}\mathcal{E}_{\mu\nu}^{\rho\sigma}\bar h_{\rho\sigma}-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}-\bar h^{\mu\nu}(\partial_\mu\partial_\nu \pi-\eta_{\mu\nu}\Box\pi)+\frac{1}{2\sqrt{M_P}}\bar h^{\mu\nu}T_{\mu\nu},
\label{lim1}
\end{eqnarray}
where $F_{\mu\nu}$ denotes the usual abelian field strength for $A_\mu$. The mixing between the tensor and scalar modes in (\ref{lim1}) can be eliminated by a linear conformal redefinition
\begin{eqnarray}
\bar h_{\mu\nu}= \tilde h_{\mu\nu}+2\eta_{\mu\nu}\pi,
\label{cs}
\end{eqnarray}
which brings (\ref{lim1}) to the following form
\begin{eqnarray}
\mathcal{L}_{m=0}=-\frac{1}{2}\tilde h^{\mu\nu}\mathcal{E}_{\mu\nu}^{\rho\sigma}\tilde h_{\rho\sigma}-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+2 \pi\Box\pi+\frac{1}{2\sqrt{M_P}}\tilde h^{\mu\nu}T_{\mu\nu}+\frac{1}{\sqrt{M_P}}\pi T.
\label{lim2}
\end{eqnarray}
It is clear from the latter Lagrangian that the helicity-0 part of the massive graviton does not decouple from matter even in the $m\rightarrow 0$ limit, leading to the famous vDVZ discontinuity \cite{vDVZ}. This is an $\mathcal{O}(1)$ modification of the gravitational interactions as compared to GR, at least in the regime of validity of the linear approximation. However, as first pointed out in \cite{Vainshtein}, in nonlinear theories of massive gravity this approximation typically breaks down at a parametrically large distance $r_V$ from localized sources, allowing for their phenomenological viability. In fact, it is the nonlinear dynamics of the scalar mode itself that screens its contribution to the gravitational potential within the Vainshtein radius, $r_V$, restoring agreement with GR. In a generic nonlinear extension of the FP theory, however, the same nonlinear self-interactions of $\pi$ are responsible for a number of theoretical problems, such as the propagation of ghosts and the ill-posedness of the Cauchy problem, strong coupling at parametrically low scales, and etc. Before turning to the discussion of theories that avoid these problems, it is instructive to have a closer look at their origin by considering the minimal nonlinear extension of the FP model, to which we turn next.
\subsection{Fierz-Pauli model at the nonlinear level}
Let us ``naively" continue the theory of the previous subsection at the nonlinear level by keeping the graviton potential unchanged, while completing the derivative self-interactions to the full Einstein-Hilbert action,
\begin{eqnarray}
{\cal L} = M_{\rm P} \sqrt{-g} R - \frac{ M_{\rm P} m^2}{4}
\(h^2_{\mu\nu}-h^2\).
\label{PF}
\end{eqnarray}
Here index contractions on the metric perturbation are performed using the flat metric $\eta^{\mu\nu}$. In order to single out the high-energy degrees of freedom in (\ref{PF}), we employ a trick \textit{\`a la} St\"uckelberg \cite{Stueckelberg:1957zz}: For the purpose of dealing with the theory at cubic order, it is more convenient to restore general covariance by introducing the decomposition for $h_{\mu\nu}$, patterned after the nonlinear gauge transformation of the metric perturbation
\begin{eqnarray}
h_{\mu\nu}=\frac{\bar h_{\mu\nu}}{\sqrt{M_P}}+\frac{\partial_\mu V_\nu+\partial_\nu V_\mu}{\sqrt{M_P} m^2}+\frac{\partial_\mu V^\alpha \partial_\nu V_\alpha}{M_P m^4},
\label{nls}
\end{eqnarray}
where $\bar h_{\mu\nu}$ and $V_\mu$ are further decomposed as
\begin{eqnarray}
\bar h_{\mu\nu}=\hat h_{\mu\nu}+\frac{\partial_\mu V^\alpha \hat h_{\alpha\nu}+\partial_\nu V^\alpha \hat h_{\alpha\mu}+V^\alpha\partial_\alpha \hat h_{\mu\nu}}{\sqrt{M_P} m^2}, \quad V_\mu= m A_\mu+\partial_\mu \pi.
\label{dec'}
\end{eqnarray}
The fields $\tilde h_{\mu\nu}=\hat h_{\mu\nu}-2 \eta_{\mu\nu}\pi$, $A_\mu$ and $\pi$ will describe the canonically normalized tensor, vector, and scalar modes at high energies, respectively.
Plugging the decomposition (\ref{dec'}) into the Lagrangian (\ref{PF}), and considering the limit
\begin{eqnarray}
M_p\rightarrow \infty ,\quad m\rightarrow 0, \quad (M_P^{1/2} m^4)^{2/9}\equiv \Lambda_{9/2}=\text{finite},
\end{eqnarray}
one recovers a particular high-energy (``decoupling") limit of the theory. Performing the conformal shift (\ref{cs}) of the helicity-2 mode, (\ref{PF}) reduces up to a total derivative to the following expression,
\begin{eqnarray}
\mathcal{L}_{dec}=-\frac{1}{2}\tilde h^{\mu\nu}\mathcal{E}_{\mu\nu}^{\rho\sigma}\tilde h_{\rho\sigma}-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+2 \pi\Box \pi+\frac{1}{2 \Lambda ^{9/2}_{9/2}}\left((\Box\pi)^3-(\Box\pi)(\partial_\mu\partial_\nu\pi)^2\right)\nonumber \\
+\frac{1}{2\sqrt{M_P}}\tilde h^{\mu\nu} T_{\mu\nu}+\frac{1}{\sqrt{M_P}}\pi T.
\label{dll}
\end{eqnarray}
As noted in the introduction, the cubic decoupling limit of the FP theory is amended by the helicity-0 self-interactions of the form $(\partial^2\pi)^3$. They successfully screen the scalar contribution to the gravitational potential of a point source of mass $M$ inside the Vainshtein radius $r_V$ , given by
\begin{eqnarray}
r_V=\( \frac{M}{M_Pm^4} \)^{1/4},
\end{eqnarray}
which restores the agreement with General Relativity. This however happens at the expense of introducing a ghost in the theory: Although infinitely heavy on a flat background, the mass of the ghost becomes low enough to be disruptive around any reasonably localized source \cite{AGS, Creminelli, Deffayet}\footnote{Although the analysis in these references is performed in four spacetime dimensions, we confirmed that the conclusions remain intact in (2+1)D as well.}. Even if one does not consider an external source, the cubic self-interactions of the scalar field, containing more than two time derivatives, cause the Cauchy problem to be ill-posed for the theory at hand.
Recently, a class of theories of (four-dimensional) massive gravity that avoid these problems have been constructed \cite{drg1,drg2,drgt}. These theories modify the graviton potential order-by order, so as to cancel all potentially dangerous self-interactions of the scalar field. As a result, one obtains a sensible effective field theory for a massive graviton, with the cutoff given by $\Lambda_3=(M_P m^2)^{1/3}$ in four spacetime dimensions. Moreover, the decoupling limit of the theory, obtained by keeping $\Lambda_3$ finite while setting the graviton mass and Planck constant to zero and infinity respectively, is free of ghosts below the cutoff. Furthermore, the form of the decoupling limit is \textit{unique} at the quartic order; in other words, any nonlinearities in the potential of order higher than the quartic one have no effect on it\footnote{Interestingly enough, a recently proposed nonlinear completion of Fierz-Pauli massive gravity \cite{g, dr} automatically produces exactly the right structure of the potential at the cubic level, so as to fall into the category of such models, \cite{drg1}.}.
\section{New Massive Gravity}
In this section we analyze the nonlinear dynamics of NMG. As we show at the end of the section, viewed as a generalization of the FP model, NMG represents a curious example. In addition to introducing a potential for the graviton, it also modifies its derivative self-interactions beyond the ones originating from the Einstein-Hilbert action, leading to a well-defined decoupling limit similar to the four-dimensional case of \cite{drg1,drg2,drgt}. Most importantly, however, the full theory can be shown to be free of the BD ghost. We turn to the discussion of this remarkable property next.
One way to define NMG is through the following action
\begin{eqnarray}
S_{NMG}=M_P\int \,d^3x \ \sqrt{g} \left[-R-f^{\mu\nu}G_{\mu\nu}-\frac{1}{4}m^2(f^{\mu\nu}f_{\mu\nu}-f^2)\right].
\label{nmg}
\end{eqnarray}
Here $G_{\mu\nu}$ and $R$ are the usual Einstein tensor and Ricci scalar for the metric $g_{\mu\nu}$, respectively. All the indices are raised with the inverse metric $g^{\mu\nu}$ and $f_{\mu\nu}$ represents an auxiliary covariant tensor field ($f\equiv g^{\mu\nu}f_{\mu\nu}$) that can be integrated out to yield a particular higher-derivative extension of the `wrong sign' GR \cite{nmg}\footnote{Note the minus sign in front of the Einstein-Hilbert term in (\ref{nmg}), which is characteristic of Topologically Massive Gravity as well \cite{tmg}. }. The equations of motion, obtained by varying (\ref{nmg}) with respect to $g_{\mu\nu}$ and $f_{\mu\nu}$ (we keep the overall factors of $\sqrt{g}$ which will be essential for the perturbative analysis of the equations beyond the linear level) are given by the following expressions
\begin{align}
\sqrt{g}G^{\mu\nu}=\frac{1}{2}\sqrt{g}g^{\mu\nu}f^{\alpha\beta}G_{\alpha\beta}-\sqrt{g}f^\mu_\alpha G^{\alpha\nu}-\sqrt{g}f^\nu _\alpha G^{\alpha\mu}+ \frac{1}{2}\sqrt{g}(fR^{\mu\nu}-f^{\mu\nu}R)\nonumber \\+
\frac{1}{2}\sqrt{g}(\nabla^\alpha\nabla^\mu f^\nu_\alpha + \nabla^\alpha\nabla^\nu f^\mu_\alpha-\Box f^{\mu\nu}-g^{\mu\nu}\nabla^\alpha\nabla^\beta f_{\alpha\beta}-\nabla^\mu\nabla^\nu f+g^{\mu\nu}\Box f) \nonumber \\ +\frac{1}{8}m^2\sqrt{g}g^{\mu\nu}(f^{\alpha\beta}f_{\alpha\beta}-f^2)-\frac{1}{2}m^2\sqrt{g}(f^\mu_\alpha f^{\alpha\nu}-f^{\mu\nu}f)\, ,
\label{eq1}
\end{align}
\begin{align}
\sqrt{g}G^{\mu\nu}+\frac{1}{2}\sqrt{g}m^2(f^{\mu\nu}-g^{\mu\nu}f)=0,
\label{eq2}
\end{align}
and admit a Minkowski vacuum with $g_{\mu\nu}=\eta_{\mu\nu}$ and $f_{\mu\nu}=0$ (the Einstein and Ricci tensors are taken as functions of the metric $g_{\mu\nu}$ in the latter equations). From Eq.~(\ref{eq1}) we see that at the linear level the perturbations $h_{\mu\nu}$ and $f_{\mu\nu}$ of the metric and the auxiliary field satisfy
\begin{eqnarray}
G^{(1)}_{\mu\nu}(h-f)=0,
\nonumber
\end{eqnarray}
where $G^{(1)}_{\mu\nu}$ denotes the linearized Einstein tensor. Since it does not propagate any degrees of freedom in three dimensions, the only solution to the latter equation is (up to a gauge, which we choose to be zero) $h_{\mu\nu}=f_{\mu\nu}$. This, in combination with Eq.~(\ref{eq2}), yields the usual FP equations of motion for $f_{\mu\nu}$ in the linearized approximation.
\subsection{Exact decoupling limit of NMG}
Before turning to the full theory, we start by considering the decoupling limit of NMG, defined as,
$$M_P\to \infty, \quad m\to 0, \quad \Lambda_{5/2}\equiv (\sqrt{M_p}m^2)^{2/5}=\text{fixed}.$$
We consider the (\textit{\`a posteriori} justified) ansatz
\begin{eqnarray}
h_{\mu \nu}=\frac{\bar h_{\mu \nu}}{\sqrt{M_P}}, \quad f_{\mu \nu}=\frac{\bar f_{\mu \nu}}{\sqrt{M_P}}+\nabla_{\mu}V_{\nu}+\nabla_{\nu}V_{\mu},
\end{eqnarray}
where $V_\mu$ is a vector field which will encode the helicity-0 and -1 degrees of freedom of the massive three-dimensional graviton, and $\nabla$ denotes the covariant derivative associated with the metric $g_{\mu \nu}$. Plugging the latter decomposition into (\ref{nmg}), we obtain, up to a total derivative, the following expression for the Lagrangian of NMG,
\begin{eqnarray}
\mathcal{L}_{NMG} &=& M_P \sqrt{g} \left[-R-\frac{1}{\sqrt{M_P}}\bar{f}^{\mu\nu} G_{\mu \nu}-\frac{m^2}{4M_P} (\bar{f}^{\mu\nu}\bar{f}_{\mu \nu}-\bar{f}^2) \right. \nonumber \\&-& \left. \frac{m^2}{\sqrt{M_P}}\bar{f}^{\mu\nu}(\nabla_\mu V_\nu -g_{\mu \nu}\nabla^\alpha V_\alpha)
-\frac{m^2}{2}(\nabla^\mu V^\nu \nabla_\mu V_\nu-\nabla^\mu V_\mu \nabla^\nu V_\nu) \right. \nonumber \\ &+&\left. \frac{m^2}{2}R^{\mu\nu}V_\mu V_\nu \right].
\label{exactlagr}
\end{eqnarray}
In here, the last term comes from the commutator of covariant derivatives and the quantities $R$ and $G_{\mu \nu}$, as well as the covariant derivative itself, are associated with $g_{\mu \nu}$. Decomposing $V_\mu$ further into the canonically normalized helicity-1 and -0 modes $A_\mu$ and $\pi$,
\begin{eqnarray}
V_\mu=\frac{A_\mu}{\sqrt{M_P}m}+\frac{\nabla_\mu\pi}{\sqrt{M_P}m^2},
\end{eqnarray}
one arrives at the Lagrangian, which up to a total derivative can be written as,
\begin{eqnarray}
\mathcal{L}_{NMG}&=&\sqrt{g}\left[-M_P R-\sqrt{M_P} \bar f^{\mu\nu}G_{\mu \nu} -\frac{1}{4}m^2(\bar{f}^{\mu\nu}\bar{f}_{\mu \nu}-\bar{f}^2)-m \bar{f}^{\mu\nu}(\nabla_\mu A_\nu \right. \nonumber \\ &-& \left. g_{\mu \nu}\nabla^\alpha A_\alpha) -\bar{f}^{\mu\nu}(\nabla_\mu\nabla_\nu\pi-g_{\mu\nu}\Box\pi)-\frac{1}{2}(\nabla^\mu A^\nu\nabla_\mu A_\nu-\nabla^\mu A_\mu \nabla^\nu A_\nu) \right. \nonumber \\ &+& \left. \frac{1}{2}R^{\mu\nu}A_\mu A_\nu +\frac{2}{m}R^{\mu\nu}\nabla_\mu\pi A_\nu+\frac{1}{m^2}R^{\mu\nu}\nabla_\mu\pi \nabla_\nu\pi\right]. \nonumber
\end{eqnarray}
It is straightforward to check that at the linearized level the theory propagates the two degrees of freedom (encoded in $A_\mu$ and $\pi$) of the massive 3D graviton.
At this point we take the exact decoupling limit $M_{P} \rightarrow \infty$
defined by keeping the scale $\Lambda_{5/2}$ fixed, giving rise to the following Lagrangian,
\begin{eqnarray}
\mathcal{L}_{NMG}^{5/2}&=&\frac{1}{2}\bar h^{\mu\nu}(\mathcal{E}\bar h)_{\mu \nu}-\bar f^{\mu\nu}(\mathcal{E}\bar h)_{\mu \nu}-\bar f^{\mu\nu}(\partial_\mu\partial_\nu\pi-\eta_{\mu \nu}\Box\pi)-\frac{1}{4}F^2_{\mu \nu}\nonumber \\ &+&\frac{1}{\Lambda^{5/2}_{5/2}}\bar R^{(1)\mu\nu}\partial_\mu\pi\partial_\nu\pi,
\end{eqnarray}
where $\bar R^{(1)}_{\mu \nu}$ denotes the linearized Ricci tensor for $\bar h_{\mu \nu}$, $F_{\mu \nu}$ is the usual abelian field strength for $A_{\mu}$ and all indices are contracted with the flat metric. The auxiliary field $\bar f_{\mu \nu}$, being a Lagrange multiplier in the limit at hand, imposes the constraint
\begin{eqnarray}
(\mathcal{E}\bar h)_{\mu \nu}=-\partial_{\mu}\partial_{\nu}\pi+\eta_{\mu \nu}\Box\pi,
\end{eqnarray}
which is solved by,
\begin{eqnarray}
\bar h_{\mu \nu}=2\pi \, \eta_{\mu \nu} + \text{gauge transformation}.
\end{eqnarray}
Plugging the latter expression back into the Lagrangian and extracting a total derivative, one finally obtains the exact decoupling limit of the theory,
\begin{eqnarray}
\mathcal{L}_{NMG}^{5/2}=2 \pi\Box\pi-\frac{1}{2}(\partial\pi)^2\Box\pi-\frac{1}{4}F^2_{\mu \nu}.
\label{dlpi}
\end{eqnarray}
The decoupling limit of NMG therefore contains the helicity-0 mode with the cubic Galileon self-interaction, together with a free helicity-1 mode! The tensor modes $h_{\mu \nu}$ and $f_{\mu\nu}$ have both disappeared in this limit because they become massless, and by standard arguments massless spin 2 fields in 3D carry no propagating degrees of freedom (although in the presence of sources they will still contribute to the interaction energy).
Since $\bar h_{\mu \nu}=2\pi \, \eta_{\mu \nu} +$ gauge transformation, we automatically generate a coupling to matter that is of the form $\propto \pi \, T$ if external sources are considered. Remarkably, as we show below, this is an example of a theory of a massive 3D graviton with modified derivative self-interactions, free of ghosts to all orders in the decoupling limit. Even more importantly, we show next that the absence of the BD ghost persists even away from this limit.
\subsection{Dynamics of the conformal mode}
Before presenting arguments in support of this last statement, as a preliminary step it is very instructive to have a look at the dynamics of the conformal mode in NMG. Integrating out the auxiliary field $f_{\mu \nu}$ in (\ref{nmg}), one arrives at the original representation of the NMG action~\cite{nmg},
\begin{eqnarray}
\mathcal{L}_{NMG}=M_P \sqrt{g}\left[-R+\frac{1}{m^2}\left( R^{\mu\nu}R_{\mu \nu}-\frac{3}{8}R^2 \right) \right].
\label{nmg'}
\end{eqnarray}
Considering conformally flat configurations of the metric
\begin{eqnarray}
g_{\mu \nu}=\Omega^2 \eta_{\mu \nu}, \qquad \Omega=e^{\phi/\sqrt{M_P}},
\end{eqnarray}
the Lagrangian (\ref{nmg'}) can be rewritten in terms of the field $\phi$, as
\begin{eqnarray}
\mathcal{L}^{c}_{NMG}&=&e^{\phi/\sqrt{M_P}}\( 4\sqrt{M_P}~\Box\phi+2(\partial_\mu\phi)^2 \)+e^{-\phi/\sqrt{M_P}}~ [~\frac{1}{m^2} \( (\partial_\mu\partial_\nu\phi)^2-(\Box\phi)^2 \)\nonumber \\ &-&\frac{2}{\sqrt{M_P}m^2}\partial^\mu\partial^\nu\phi\partial_\mu\phi\partial_\nu\phi+\frac{1}{2M_P m^2}(\partial_\mu\phi)^2(\partial_\nu\phi)^2~],
\label{conformalnmg}
\end{eqnarray}
where all indices are contracted with the flat metric. Curiously enough, the $\Lambda_{5/2}$ decoupling limit of this theory coincides in form with the decoupling limit Lagrangian of the helicity-0 mode in Eq.~(\ref{dlpi})
\begin{eqnarray}
\mathcal{L}^{c}_{NMG} \supset 2 \phi\Box\phi-\frac{1}{2}(\partial\phi)^2\Box\phi,
\end{eqnarray}
identifying the helicity-0 part of the massive graviton with the conformal mode in this limit.
One can show that even away from the decoupling limit, the theory given by (\ref{conformalnmg}) does not propagate ghosts as one would naively assume from the presence of higher derivative interactions.
Indeed, up to a total derivative, Eq.~(\ref{conformalnmg}) can be rewritten as,
\begin{eqnarray}
\mathcal{L}^{c}_{NMG}&=&e^{\phi/\sqrt{M_P}}\( 4\sqrt{M_P}~\Box\phi+2(\partial_\mu\phi)^2 \)+\frac{1}{2m^2} e^{-\phi/\sqrt{M_P}}\( (\partial_\mu\partial_\nu\phi)^2-(\Box\phi)^2 \)\nonumber \\ &+& \frac{1}{2\sqrt{M_P}m^2}e^{-\phi/\sqrt{M_P}}\partial^\mu\phi\partial^\nu\phi\partial_\mu\partial_\nu\phi.
\label{conformalnmg'}
\end{eqnarray}
Naively, the second and third terms in the above Lagrangian, involving more than two time derivatives, might lead to the presence of ghosts (or equivalently, the ill-posedness of the Cauchy problem) in the theory. One can however show that the specific structure of these operators makes them harmless. The second term is conveniently expressed in terms of the Levi-Civita symbol as follows,
\begin{eqnarray}
\propto e^{-\phi/\sqrt{M_P}} \varepsilon^{\mu\alpha\rho\sigma}\varepsilon_{\nu\beta\rho\sigma}~\partial_\mu\partial^\nu\phi~\partial_\alpha\partial^\beta\phi.
\end{eqnarray}
The antisymmetry of the Levi-Civita symbol can then be used to show that no terms with more than two time derivatives are present in the equations of motion. Similarly, noticing that the third term in Eq.~(\ref{conformalnmg'}) includes a factor of a peculiar form $$\partial^\mu\phi\partial^\nu\phi\partial_\mu\partial_\nu\phi,$$ it becomes fairly straightforward to show that it does not cause the Cauchy problem to be ill-posed either. Indeed, this latter expression, being (up to a total derivative) equivalent to the DGP galileon $\Box\phi(\partial_\mu\phi)^2$, is well known to lead to no more than two time derivatives per field in the equation of motion. As can be straightforwardly checked, the factor of $e^{-\phi/\sqrt{M_P}}$ in front does not alter this property.
\subsection{No ghosts in New Massive Gravity}
Finally, we give an argument demonstrating the absence of the BD ghost in the full NMG. The argument we shall give is based on counting non-perturbatively the degrees of freedom utilizing the symmetries. The key observation is that there is a non-perturbative generalization of the $U(1)$ symmetry described in Eq.~(\ref{gt}) which combined with the existing 3D reparameterization invariance is sufficient to demonstrate that only the 2 propagating degrees of freedom remain non-perturbatively.
The Lagrangian (\ref{exactlagr}) is invariant under the usual reparametrizations of coordinates (3D diffeomorphisms), under which the metric $g_{\mu \nu}$ and the auxiliary field $\bar f_{\mu \nu}$ transform as tensors, while $V_\mu$ is a vector. One can use this symmetry to eliminate all potentially propagating degrees of freedom in the 3D metric $g_{\mu \nu}$.
By introducing the vector field $V_{\mu}$ we also introduce a new three-parameter local invariance (which we refer to as secondary linearized 3D diffeomorphisms), under which the metric $g_{\mu \nu}$ remains unchanged, while $\bar f_{\mu \nu}$ and $V_\mu$ transform as,
\begin{eqnarray}
\bar f_{\mu \nu}\to \bar f_{\mu \nu}+\sqrt{M_P}(\nabla_\mu \xi_\nu+\nabla_\nu \xi_\mu), \qquad V_\mu\to V_\mu-\xi_\mu.
\label{sym2}
\end{eqnarray}
It will prove helpful to slightly rewrite the Lagrangian (\ref{exactlagr}) in the following form,
\begin{eqnarray}
\mathcal{L}_{NMG}&=&\sqrt{g}~\left[-M_P R-\sqrt{M_P}\tilde f^{\mu\nu}G_{\mu \nu}-\frac{1}{4}m^2(\tilde{f}^{\mu\nu}\tilde{f}_{\mu \nu}-\tilde{f}^2)-\frac{1}{4}\bar F^{\mu\nu} \bar F_{\mu \nu} \right. \nonumber \\
&-&\left. \frac{1}{4}\frac{m^2}{\sqrt{M_P}}\tilde f^{\mu\nu}(\bar V_\mu \bar V_\nu+g_{\mu \nu}\bar V^\alpha \bar V_\alpha)-m\tilde f^{\mu\nu} (\nabla_\mu \bar V_\nu-g_{\mu \nu}\nabla^\alpha \bar V_{\alpha})\right. \nonumber \\ &-&\left. \frac{1}{2}\frac{m}{\sqrt{M_P}}\nabla^\mu \bar V^\nu (\bar V_\mu \bar V_\nu+g_{\mu \nu}\bar V^\alpha \bar V_\alpha)+\frac{1}{8}\frac{m^2}{M_P}(\bar V^\alpha \bar V_\alpha)^2~\right],
\label{lagr2}
\end{eqnarray}
where the following set of field redefinitions has been used,
\begin{eqnarray}
\bar V_\mu=\sqrt{M_P} m V_\mu, \qquad \tilde f_{\mu \nu}=\bar f_{\mu \nu}-\frac{1}{2\sqrt{M_P}}(V_\mu V_\nu-g_{\mu \nu} V^\alpha V_\alpha),
\end{eqnarray}
and $\bar F_{\mu \nu}$ denotes the abelian field strength for $\bar V_\mu$. The symmetry (\ref{sym2}) induces the corresponding transformation on $\tilde f_{\mu \nu}$, that leaves the action invariant. One can use this freedom to eliminate all potentially propagating d.o.f's in $\tilde f_{\mu \nu}$, leaving $V_\mu$ the only propagating field in the theory.
Generically, $\bar V_\mu$ would propagate three degrees of freedom in three dimensions; however, as we shall argue below, NMG is special in this sense, propagating only two degrees of freedom in the latter field.
The only place in the Lagrangian (\ref{lagr2}) where the time derivative of $\bar V_0$ appears, is the last term of the second line and the first term on the last line, which however include $\dot {\bar V}_0$ only \textit{linearly}. This means that the canonical momentum, conjugate to $\bar V_0$ is independent of $\dot{\bar V}_0$ itself, enforcing the primary constraint on the system. This reduces the number of propagating degrees of freedom in $\bar V_\mu$ to two, which, due to the arguments presented above, means that these are the only two dynamical degrees of freedom in the full theory of NMG. To make this argument clearer we can define
\begin{eqnarray}
\bar V_{\mu}= A_{\mu}+ \partial_{\mu} \pi ,
\end{eqnarray}
where $A_{\mu}$ is a vector under usual 3D diffeomorphisms, and transforms as $A_{\mu} \rightarrow A_{\mu} - \xi_{\mu}$ under the secondary linearized 3D diffeomorphisms, and $\pi$ is similarly a scalar under the former and invariant under the later.
Since the Lagrangian (\ref{lagr2}) depends only directly on $\bar V_{\mu}$, it is manifestly invariant under a new additional $U(1)$ symmetry
\begin{eqnarray}
A_\mu\rightarrow A_\mu+\partial_\mu \zeta, \quad \quad \pi\rightarrow \pi-\zeta,
\end{eqnarray}
generalizing the result in the linearized theory Eq.~(\ref{gt}). Since the above Lagrangian is only linear in $\dot {\bar V}_0$ it is in turn guaranteed to be only linear in $\ddot{\pi}$. Any such interaction will lead to second order equations of motion for $\pi$. However, one may worry about defining a conjugate momentum for $\pi$. That this can be done consistently,
can be seen by noting that it is always possible to integrate by parts to put the action in a form where there are no more than single time derivatives acting on the fields in the Lagrangian.
To see this explicitly we note that the only term in the above Lagrangian where the $\ddot{\pi}$ terms arise, is in `gauged' Galileon interaction on the last line
\begin{eqnarray}
\mathcal{L}_0=-\frac{1}{2}\sqrt{g}\, \nabla^\mu \bar V^\nu (\bar V_\mu \bar V_\nu+g_{\mu \nu}\bar V^\alpha \bar V_\alpha)\,,
\end{eqnarray}
as well as in
\begin{eqnarray}
\mathcal{L}_1=-\sqrt{g}\, \tilde f^{\mu\nu}\(\nabla_\mu \bar V_\nu-g_{\mu \nu} \nabla^\alpha\bar V_\alpha \)=-\sqrt{g}\(\nabla_\nu\bar V_\mu\)(\tilde f^{\mu\nu}-\tilde f g^{\mu\nu})\,.
\end{eqnarray}
On integration by parts we see that the first term is equivalent to
\begin{eqnarray}
\mathcal{L}_0=-\frac{1}{4}\sqrt{g} \, (\bar V^\alpha \bar V_\alpha) \nabla^\mu \bar V_\mu=-\frac 14 (\bar V^\alpha \bar V_\alpha) \partial_\mu(\sqrt{g}\, V^\mu)\,.
\end{eqnarray}
It is now easy to show that the would be problematic $\ddot{\pi}$ coming from the $\dot{\bar V}_0$ part is a total derivative and consequently can be removed by an integration by parts
\begin{eqnarray}
\mathcal{L}_0 &=& - \frac 14 \Big[-\frac{1}{N^2}( \bar V_0-N^i\bar V_i)^2+\gamma^{ij}\bar V_i \bar V_i\Big] \Big[ \partial_0 \(\frac{\sqrt{\gamma}}{N} (-\bar V_0+N^i \bar V_i)\)+\partial_i ( \dots)\Big] \nonumber \\
&=& - \frac{1}{12} \, \partial_0 \left( \sqrt{\gamma}\frac{(\bar V_0-N^i \bar V_i)^3}{N^3}\right) + \frac 14 \partial_0 \left(\frac{\sqrt{\gamma}}{N} ( \bar V_0-N^i \bar V_i)\gamma^{ij}\bar V_i \bar V_i \right) \nonumber \\
&+&\text{ terms with {\bf{no}} $\dot{\bar V}_0,\dot{N}$ or $\dot{N}^i$} ,
\end{eqnarray}
where $N$ is the ordinary lapse $g^{00}=-1/N^2$, $g_{0i}=N_i$ the associated shift and $g_{ij}=\gamma_{ij}$.
This is precisely for the same reason that the covariant DGP Galileon interaction $\Box \pi (\partial \pi)^2$ term gives well defined equations of motion. Indeed setting $A_{\mu}=0$ we see easily that this term is equivalent to the DGP interaction for $\pi$:
\begin{eqnarray}
(\bar V^\alpha \bar V_\alpha) \nabla^\mu \bar V_\mu|_{A_\mu=0}=\Box \pi (\partial \pi)^2\,.
\end{eqnarray}
Implicitly these integrations by parts determine the analogue of the Gibbons-Hawking boundary terms for NMG.
A similar argument can be employed for $\mathcal{L}_1$, where all potentially problematic terms can be seen to be a total derivative:
\begin{eqnarray}
\mathcal{L}_1&=&-\sqrt{\gamma}N_0 \(\frac{\dot N}{N}V_i N^i-V_0\dot N-V_i \dot N^i+\dot V_0+\cdots \)\frac{1}{N_0^2}\gamma^{ij}\tilde f_{ij}+\cdots\nonumber\\
&=&-\sqrt{\gamma}\gamma^{ij}\tilde f_{ij}\(\partial_0\(\frac{V_0}{N}\)-V_i\partial_0\(\frac{N^i}{N}\)\)+\cdots
\end{eqnarray}
so that this term is also free of any time derivative on the Lagrange multipliers,
\begin{eqnarray}
\mathcal{L}_1=-\partial_0\(\frac{\sqrt{\gamma}}{N}\gamma^{ij}\tilde f_{ij}\(V_0-V_iN^i\)\)+\text{ terms with {\bf{no}} $\dot{\bar V}_0,\dot{N}$, $\dot{N}^i$, $\dot{\tilde f}_{00}$ or $\dot{\tilde f}_{0i}$} .
\end{eqnarray}
Again the key feature to stress is that at the same time as removing the $\dot{V}_0$ terms we can also remove the $\dot{N}$, $\dot{N}^i$, $\dot{\tilde f}_{00}$ and $\dot{\tilde f}_{0i}$ terms ensuring that $N$, ${N}^i$, $\tilde f_{00}$ and $\tilde f_{0i}$ act as Lagrange multipliers for the normal 3D diffeomorphism Hamiltonian and momentum constraints. Similarly since there are by construction no $\dot{A}_0$ terms, then $A_0$ acts as the Lagrange multiplier for the $U(1)$ `Gauss law' constraint.
Putting this together, since all the double time derivative terms for $\pi$ can be removed by integration by parts, and by the same means the Lagrangian can be made independent of $\dot{N}, \dot{N^i},\dot{\tilde f}_{00}, \dot{\tilde f}_{0i},\dot{A}_0$, we are guaranteed that the equations of motion are well defined, the 7 constraints are preserved, and that there is a well-defined Hamiltonian formalism. The key point then is that following these steps the configuration space action can be written in a form in which it is a function of the 6+6+3+1=16 variables $g_{\mu\nu},f_{\mu\nu},A_{\mu},\pi$ each of which has well defined equations of motion. In addition there are 3+3+1=7 first class symmetries corresponding to 3D diffeomorphisms, secondary linearized 3D diffeomorphisms and the additional $U(1)$. Associated with each symmetry there is a gauge freedom and a constraint which allows us to remove 2 degrees of freedom per symmetry. Thus the total number of degrees of freedom non-perturbatively is 16 - 7 -7 =2. Since 2 is the correct number of physical polarizations of a massive spin 2 field in 3D this confirms the absence of the BD ghost to all orders, consistent with our previous arguments which were only valid in certain regimes.
\section{Explicit Computation of the Decoupling Limit at Nonlinear Order}
In this section we analyze the decoupling limit of NMG explicitly to cubic order. We show that one indeed recovers a ghost-free theory, governed by the scale $\Lambda_{5/2}$. We do so by explicitly demonstrating how the terms corresponding to the $\Lambda_{9/2}$ and $\Lambda_{7/2}$ all vanish. We then generalize the results and construct a new class of theories which are ghost-free to cubic order.
\subsection{NMG at Nonlinear Order}
To obtain the NMG Lagrangian to cubic order one must integrate out the metric perturbation $h_{\mu\nu}$ from the equations of motion of NMG (\ref{eq1})-(\ref{eq2}) beyond the quadratic order in the Lagrangian. This is a rather lengthy computation and it is outlined in the appendix. The resulting cubic effective Lagrangian for $f_{\mu\nu}$ can be written in the following form,
\begin{eqnarray}
\mathcal{L}^{(3)}_f &=& 2M_P(\sqrt{\gamma}R(\gamma))^{(3)}+\frac{1}{2}M_P f^{\mu\nu}\mathcal{E}_{\mu\nu}^{\rho\sigma} f_{\rho\sigma}
-\frac{M_P m^2}{4} ( f^2_{\mu\nu}-f^2 \nonumber \\ &+&\frac{5}{2}f^2_{\mu\nu}f- 2f^3_{\mu\nu}-\frac{1}{2}f^3).
\label{clag}
\end{eqnarray}
Here $\gamma_{\mu\nu}\equiv \eta_{\mu\nu}+f_{\mu\nu}$ is the analog of the full metric with $f_{\mu\nu}$ viewed as metric perturbation, and $M_P(\sqrt{\gamma}R(\gamma))^{(3)}$ denotes the cubic in $f_{\mu\nu}$ part of the corresponding Einstein-Hilbert action. At the quadratic level, we of course recover the FP action; a remarkable peculiarity of the cubic part, however, is that it modifies GR not just by the graviton's potential, but also by the deformation of the nonlinear Einstein-Hilbert part itself.
According to the discussion of section 2, in a generic nonlinear extension of the FP action we should expect $\Lambda_{9/2}$ to be the scale, governing the dynamics of the helicity-0 mode. On the other hand, we have shown in the previous section that $\Lambda_{5/2}$ represents the lowest scale, by which the Galileon self-interactions of the scalar graviton are suppressed in NMG. The theory must therefore possess a special structure, which eliminates the scale $\Lambda_{9/2}$, as well as the intermediate scale $\Lambda_{7/2}\equiv(\sqrt{M_P}m^3)^{2/7}$, from its dynamics. To show this, let us plug the decomposition (\ref{nls}) (with the obvious change $h\to f$, while keeping the same notation for the vector and scalar modes) into the cubic Lagrangian Eq.~(\ref{clag}). In the limit where all scales greater than $\Lambda_{7/2}$ are sent to infinity, one recovers the following expression for the action,
\begin{eqnarray}
\mathcal{L}^{dl}_{f} &=& -\frac{1}{2}\tilde{f}^{\mu\nu}\mathcal{E}^{\alpha\beta}_{\mu\nu} \tilde{f}_{\alpha\beta}-\frac{1}{4}F^{\mu\nu}F_{\mu \nu}+ 2\pi\Box\pi \nonumber \\ &+& \frac{1}{\Lambda_{9/2}^{9/2}}
\left(
2 \partial_\mu\partial_\nu\pi\partial^\mu\partial_\alpha\pi \partial^\nu\partial^\alpha\pi -
3 \Box\pi\partial_\mu\partial_\nu\pi\partial^\mu\partial^\nu\pi +
\left(\Box\pi\right)^3
\right)
\nonumber \\&+&
\frac{1}{\Lambda^{7/2}_{7/2}}(
7 \partial_\mu A_\nu \partial^\mu\partial^\alpha\pi \partial^\nu\partial_\alpha \pi -6 \partial_\mu A_\nu \partial^\mu\partial^\nu\pi \Box \pi
\nonumber \\ &-&4 \partial_\mu A^\mu \partial_\alpha\partial_\beta\pi \partial^\alpha\partial^\beta\pi +
3 \partial_\mu A^\mu \Box\pi \Box \pi
) \nonumber \\
&+&\(\frac{1}{\Lambda^{7/2}_{7/2}}\(\partial^\mu A^\rho \partial^\nu\partial_\rho \pi+\partial^\nu A^\rho \partial^\mu\partial_\rho \pi\)+\frac{1}{\Lambda^{9/2}_{9/2}}\partial^\mu\partial^\rho\pi\partial^\nu\partial_\rho\pi \)\mathcal{E}^{\alpha\beta}_{\mu\nu} \tilde{f}_{\alpha\beta}.
\label{lagr'''}
\end{eqnarray}
Here $\tilde f =\hat f-2\pi\eta$ represents the familiar conformally shifted tensor mode, so that in terms of the new fields the $U(1)$ gauge symmetry of (\ref{gt}) is given by,
\begin{eqnarray}
\tilde f_{\mu \nu}\to \tilde f_{\mu \nu}+2 m\xi \eta_{\mu \nu}, \quad A_{\mu}\to A_{\mu}+\partial_{\mu}\xi, \quad \pi\to\pi-m\xi.
\label{gi'}
\end{eqnarray}
Under these transformations, the Lagrangian (\ref{lagr'''}) is invariant up to terms that vanish in the decoupling limit,
\begin{eqnarray}
\delta_\xi \mathcal{L}^{dl}_{f} = \mathcal{O} \(\frac{\xi}{\Lambda^{5/2}_{5/2}}\).
\end{eqnarray}
The $\pi$ self-interactions in the second line of Eq.~(\ref{lagr'''}) combine to a total derivative and therefore drop out. Furthermore, extracting a total derivative\footnote{The corresponding total derivative is given by \\ $ 7 \( \partial^\mu A^\nu\partial_\mu\partial^\alpha\pi\partial_\nu\partial_\alpha\pi - \partial^\mu A^\nu\partial_\mu\partial_\nu\pi\Box\pi -\frac{1}{2} \partial^\mu A_\mu(\partial_\alpha\partial_\beta\pi)^2 +\frac{1}{2} \partial^\mu A_\mu(\Box\pi)^2 \)/\Lambda^{5/2}_{5/2}$.} from the $\partial A\partial\pi\partial\p\pi$-type operators and using the gauge invariance Eq.~(\ref{gi'}) to impose the condition $\partial^\mu A_\mu=0$, (\ref{lagr'''}) can be rewritten as,
\begin{eqnarray}
\mathcal{L}^{dl}_{f} &=& -\frac{1}{2}\tilde{f}^{\mu\nu}\mathcal{E}^{\alpha\beta}_{\mu\nu} \tilde{f}_{\alpha\beta}-\frac{1}{4}F^{\mu\nu}F_{\mu \nu}+ 2\pi\Box\pi +\frac{1}{\Lambda^{7/2}_{7/2}}\partial_\mu A_\nu \partial^\mu\partial^\nu\pi \Box \pi \nonumber \\ &+&\(\frac{1}{\Lambda^{7/2}_{7/2}}\(\partial^\mu A^\rho \partial^\nu\partial_\rho \pi+\partial^\nu A^\rho \partial^\mu\partial_\rho \pi\)+\frac{1}{\Lambda^{9/2}_{9/2}}\partial^\mu\partial^\rho\pi\partial^\nu\partial_\rho\pi \)\mathcal{E}^{\alpha\beta}_{\mu\nu} \tilde{f}_{\alpha\beta}.
\end{eqnarray}
The remaining interactions can be removed by a field redefinition of the helicity-0 and helicity-2 fields,
\begin{eqnarray}
\pi\to\pi-\frac{1}{4\Lambda^{7/2}_{7/2}}\partial_\mu A_\nu \partial^\mu\partial^\nu\pi; ~ \tilde{f}_{\mu\nu} \to \tilde{f}_{\mu\nu}&+&\frac{1}{\Lambda^{7/2}_{7/2}}\(\partial_\mu A^\rho \partial_\nu\partial_\rho \pi+\partial_\nu A^\rho \partial_\mu\partial_\rho \pi\)\nonumber \\ &+&\frac{1}{\Lambda^{9/2}_{9/2}}\partial_\mu\partial^\rho\pi\partial_\nu\partial_\rho\pi,
\end{eqnarray}
As promised, all the dangerous interactions associated with the $\Lambda_{9/2}$ and $\Lambda_{7/2}$ scales have dropped-out in this limit. This leaves $\Lambda_{5/2}$ to control the nonlinear dynamics of NMG, as shown in subsection 3.1.
It is important to note the significance of the $U(1)$ gauge invariance in the preceding discussion. In fact, precisely due to this freedom, any operator of the type $\partial A\partial^2\pi\partial^2\pi$ at the scale $\Lambda_{7/2}$ can be removed (up to a total derivative) by a redefinition of the helicity-0 graviton. This means that once the problematic $(\partial^2\pi)^3$-type operators at the scale $\Lambda_{9/2}$ are removed by a judicious choice of the coefficients of the different terms in the Lagrangian, the $\Lambda_{7/2}$ scale becomes automatically removable, leading to the decoupling limit, governed by $\Lambda_{5/2}$\footnote{Mathematically, this traces back to the fact that the $U(1)$ transformation of the scalar in (\ref{gt}) includes a factor of the mass $m$, connecting the scale $\Lambda_{9/2}$ to $\Lambda_{7/2}=m/\Lambda_{9/2}$. Because of this, tuning the theory so as to remove the former scale automatically results in the removal of the latter one, as was for instance encountered in \cite{drg2,drgt}.}.
\subsection{General Cubic-Order Ghost-Free Massive Gravity}
Finally, we extend NMG at cubic order considering a more general set of derivative self-interactions of the graviton. The generic type of cubic Lagrangians we wish to consider is written in terms of the Einstein-Hilbert term (with possibly a non-standard coefficient) together with reparametrisation invariance-violating contributions involving the metric perturbation $h_{\mu \nu}=g_{\mu \nu}-\eta_{\mu \nu}$,
\begin{eqnarray}
\label{eqn:generalLag}
\mathcal{L} &=& M_{P}~\left[\rho (\sqrt{g} R)^{(3)} - \frac{m^2}{4}\left(h_{\mu\nu}^2 - h^2 + c_1 h_{\mu\nu}^3 + c_2 h h_{\mu\nu}^2 + c_3 h^3 + \ldots \right) \right. \\\nonumber
&~~~& \left. + \alpha h^{\mu\nu}G_{\mu \nu}^{(1)} + \beta_1 h^{\mu\rho}h_\rho^\nu G_{\mu \nu}^{(1)} +\beta_2 h h^{\mu\nu}G^{(1)}_{\mu \nu}+ \beta_3 h_{\mu \nu} h^{\mu\nu}G + \beta_4 h^2G^{(1)} +\ldots \right]
\end{eqnarray}
where the ellipses denote possible higher order terms \footnote{All terms except the first one in the second line of (\ref{eqn:generalLag}) can in fact be removed at the cubic order by a \emph{nonlinear} redefinition of the metric perturbation $h_{_{\mu \nu}}$; in particular, setting $h_{_{\mu \nu}} \to h_{_{\mu \nu}}+ \beta_1 h_\mu^\rho h_{\rho\nu}+\beta_2 h h_{\mu\nu}+ \beta_3 h^{\alpha\beta} h_{\alpha\beta}\eta_{_{\mu \nu}}+ \beta_4 h^2\eta_{_{\mu \nu}}$, will eliminate the $hhG^{(1)}$-type terms, while generating additional $\beta$-dependent polynomial contributions. This leads to an alternative, equivalent formulation of the cubic theory; we will however choose to work with the original form of the action (\ref{eqn:generalLag}) below.}.
In order to investigate the relevant degrees of freedom of this theory, we employ again the St\"uckelberg decomposition, Eq.~(\ref{nls}).
We begin by investigating the $\Lambda_{9/2}$ - limit,
\begin{eqnarray}
\label{eqn:Lambda92Limit}
m \to 0, ~~~M_{\rm Pl} \to \infty, ~~~ {\rm keeping\ } \Lambda_{9/2} \equiv (m^4 \sqrt{M_{\rm Pl}} )^{2/9}
~~{\rm fixed}\,.
\label{declim3}
\end{eqnarray}
Let us for definiteness concentrate on the particular value $\rho=2$ for the coefficient in front of the Einstein-Hilbert term, since it corresponds to the case of NMG as shown above. Then, in order to recover the FP Lagrangian at the linearized level, $\alpha=1/2$ should hold. Expanding the Lagrangian of Eq.~(\ref{eqn:generalLag}) up to cubic order and performing the usual rescaling $\hat{h}_{\mu \nu}\rightarrow \tilde{h}_{\mu \nu} + 2\eta_{\mu \nu} \pi$, we obtain,
\begin{eqnarray}
\mathcal{L}_{9/2} &=& -\frac{1}{2}\tilde {h}^{\mu\nu}\mathcal{E}^{\alpha\beta}_{\mu\nu} \tilde{h}_{\alpha\beta} \nonumber \\&+& \frac{3}{2}\pi\Box\pi + \frac{1}{\Lambda_{9/2}^{9/2}}
\left(
\alpha_1 \partial_\mu\partial_\nu\pi\partial^\mu\partial_\alpha\pi \partial^\nu\partial^\alpha\pi +
\alpha_2 \Box\pi\partial_\mu\partial_\nu\pi\partial^\mu\partial^\nu\pi +
\alpha_3 \left(\Box\pi\right)^3
\right), \nonumber
\end{eqnarray}
where the corresponding coefficients are given as follows,
\begin{eqnarray}
\label{eqn:pipipiCoefficients}
\alpha_1 &=& -2-2c_1 -4\beta_1,\\\nonumber
\alpha_2 &=& 2 -2 c_2 + 4 \beta_1 - 4 \beta_2 + 8 \beta_3,\\\nonumber
\alpha_3 &=& -2 c_3 + 4 \beta_2 + 8 \beta_4.
\end{eqnarray}
In order to avoid the propagation of ghosts as discussed above, we choose these coefficients such that the three terms form a total derivative (or vanish) and therefore have no effect on the equations of motion. The relevant conditions therefore are,
\begin{eqnarray}
\alpha_1=-\frac{2}{3}\alpha_2=2\alpha_3.
\end{eqnarray}
NMG represents one example of a theory for which these conditions hold.
\vspace{10pt}
\noindent {\bf Acknowledgements:}
We would like to thank Gaston Giribet for useful correspondence. CdR is funded by the SNF, the work of GG was supported by NSF grant PHY-0758032, DP is supported by the Mark Leslie
Graduate Assistantship at NYU and IY is supported by the James Arthur fellowship. CdR and AJT would like to thank NYU for hospitality whilst part of this work was being completed.
\renewcommand{\theequation}{A-\Roman{equation}}
\setcounter{equation}{0}
\section*{Appendix A}
In this appendix we briefly outline the procedure of integrating out the metric perturbation $h_{\mu\nu}$ in (\ref{eq1})-(\ref{eq2}) in order to obtain the effective equations of motion, quadratic in $f_{\mu\nu}$ (or equivalently, the cubic effective Lagrangian for $f_{\mu \nu}$).
For this, we can use the expression for $\sqrt{g}G^{\mu\nu}$ found from (\ref{eq1}) in the equation for the auxiliary field (\ref{eq2}), and take into account the first-order relation $h_{\mu\nu}=f_{\mu\nu}$ to rewrite everything in terms of $f_{\mu\nu}$ only. Specifically, the latter relation is used for rewriting the (linearized) Einstein and Ricci tensors in terms of $f_{\mu\nu}$ in the first line of (\ref{eq1}), as well as the (linearized) quantities such as $\sqrt{g}$, $g^{\mu\nu}$ and the covariant derivatives in the second line of (\ref{eq1}) and in the second term of Eq. (\ref{eq2})\footnote{Note that we do not have to go beyond the first order in expressing $h_{\mu\nu}$ in terms of $f_{\mu\nu}$ for obtaining the second-order equations for the latter field.}. After a rather involved but straightforward computation, one arrives at the following effective equation for $f_{\mu\nu}$
\begin{eqnarray}
(\sqrt{\gamma}G^{\mu\nu}(\gamma))^{(1)}+2 (\sqrt{\gamma}G^{\mu\nu}(\gamma))^{(2)}+\frac{1}{2}m^2(f^{\mu\nu}-\eta^{\mu\nu}f)~~~~\nonumber \\ +m^2 \left( \frac{5}{8} \eta^{\mu\nu}f^{\alpha\beta}f_{\alpha\beta}-\frac{3}{8}\eta^{\mu\nu}f^2-\frac{3}{2}f^{\mu\alpha}f_\alpha^\nu+\frac{5}{4}f^{\mu\nu}f \right)=0,
\label{qe}
\end{eqnarray}
where $\gamma_{\mu\nu}\equiv\eta_{\mu\nu}+f_{\mu\nu}$ and $(\sqrt{\gamma}G^{\mu\nu}(\gamma))^{(n)}$ denotes the contribution to the corresponding quantity of order $n$ in $f_{\mu\nu}$ \footnote{The perturbative equation of motion for a massless GR graviton $h_{\mu\nu}$ is of the form
$(\sqrt{g}G^{\mu\nu}(g))^{(1)}+(\sqrt{g}G^{\mu\nu}(g))^{(2)}+ ... = 0$. It is crucial that the indices are raised in the latter equation; since $h_{\mu\nu}$ is defined as the perturbation of the metric $g_{\mu\nu}$ (and not its inverse), the variation of the Einstein-Hilbert action w.r.t $h_{\mu\nu}$ yields the Einstein's equations with \textit{upper} indices, expressed through the metric perturbation.}. More explicitly, the first two terms of (\ref{qe}) involve the following expressions,
\begin{eqnarray}
(\sqrt{\gamma}G^{\mu\nu}(\gamma))^{(1)}=\mathcal{E}^{\mu\nu}_{\rho\sigma} f^{\rho\sigma},\nonumber
\end{eqnarray}
\begin{eqnarray}
(\sqrt{\gamma}G^{\mu\nu}(\gamma))^{(2)}=\frac{1}{4}(\partial^\rho f \partial^\mu f^\nu_\rho+\partial^\rho f \partial^\nu f^\mu_\rho)-\frac{1}{4}\partial^\rho f \partial_\rho f^{\mu\nu}-\frac{1}{2}(\partial^\mu f^{\nu\rho} \partial^\sigma f_{\sigma\rho}+\partial^\nu f^{\mu\rho} \partial^\sigma f_{\sigma\rho})
\nonumber \\+ \frac{1}{2}\partial^\rho f _\rho^\sigma\partial_\sigma f^{\mu\nu}-\frac{1}{2}\partial^\rho f^{\mu\sigma} \partial_\sigma f^\nu_\rho+\frac{1}{2}\partial^\rho f^{\mu\sigma} \partial_\rho f^\nu_\sigma +\frac{1}{4}\partial^\mu f^{\rho\sigma} \partial^\nu f_{\rho\sigma}-\frac{1}{2}f^{\rho\sigma}(\partial_\rho\partial^\mu f^\nu_\sigma +\partial_\rho\partial^\nu f^\mu_\sigma) \nonumber \\
+\frac{1}{2}f^{\rho\sigma}\partial_\rho\partial_\sigma f^{\mu\nu}+\frac{1}{2}f^{\rho\sigma}\partial^\mu\partial^\nu f_{\rho\sigma}
-\frac{1}{2}\eta^{\mu\nu}[\partial^\rho f \partial^\sigma f_{\rho\sigma}-\frac{1}{4}\partial^\rho f\partial_\rho f -\partial^\rho f_\rho^\sigma\partial^\alpha f_{\alpha\sigma}-\frac{1}{2}\partial^\rho f^{\sigma\alpha}\partial_\sigma f_{\rho\alpha}\nonumber \\+\frac{3}{4}\partial^\alpha f^{\rho\sigma} \partial_\alpha f_{\rho\sigma}
-2 f^{\rho\sigma}\partial_\rho\partial^\alpha f_{\sigma\alpha}+f^{\rho\sigma}\partial_\rho\partial_\sigma f+f^{\rho\sigma}\Box f_{\rho\sigma}]+\frac{1}{2} f^{\mu\nu}\partial^\rho\partial^\sigma f_{\rho\sigma}-\frac{1}{2}f^{\mu\nu}\Box f+ \nonumber \\ \frac{1}{4} f (\partial^\mu\partial^\rho f_\rho^\nu +\partial^\nu\partial^\rho f_\rho^\mu -\partial^\mu\partial^\nu f -\Box f^{\mu\nu}+\eta^{\mu\nu}\Box f-\eta^{\mu\nu}\partial^\rho \partial^\sigma f_{\rho\sigma})-\frac{1}{2} f^{\mu\rho}(\partial_\rho\partial^\sigma f_\sigma^\nu +\partial^\nu\partial^\sigma f_{\sigma\rho} \nonumber \\ -\Box f^\nu_\rho-\partial^\nu\partial_\rho f)-\frac{1}{2} f^{\nu\rho}(\partial_\rho\partial^\sigma f_\sigma^\mu+\partial^\mu\partial^\sigma f_{\sigma\rho}-\Box f^\mu_\rho-\partial^\mu\partial_\rho f),\nonumber
\end{eqnarray}
where all indices are contracted with the flat metric.
It is now straightforward to obtain the cubic Lagrangian which yields (\ref{qe}) as the equation of motion for $f_{\mu \nu}$,
\begin{eqnarray}
\mathcal{L}^{(3)}_f&=& 2M_P(\sqrt{\gamma}R(\gamma))^{(3)}-M_P(\sqrt{\gamma}R(\gamma))^{(2)}
-\frac{M_Pm^2}{4} ( f^2_{\mu\nu}-f^2 \nonumber \\ &+& \frac{5}{2}f^2_{\mu\nu}f- 2 f^3_{\mu\nu}-\frac{1}{2}f^3 ).
\label{applagr}
\end{eqnarray}
Here, the notation $f^2_{\mu\nu}\equiv f^{\mu\nu}f_{\mu\nu}$ and $f^3_{\mu\nu}\equiv f^{\mu\alpha}f_{\alpha\beta}f^\beta_\mu$ has been used.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 8,552 |
\section{Introduction}\label{sec:introduction}
\begin{figure}[tb]
\centering
\subfigure[]{\label{fig:introa}\includegraphics[height=2.4cm]{introa}}\hspace{0.9cm}
\subfigure[]{\label{fig:introb}\includegraphics[height=2.4cm]{introb}}\hspace{0.9cm}
\subfigure[]{\label{fig:introc}\includegraphics[height=2.4cm]{introc}}\\
\subfigure[]{\centering\label{fig:mcubes}\includegraphics[width=0.8\columnwidth]{mcubes.jpg}}
\caption{The \subref{fig:introa} Ebbinghaus, \subref{fig:introb}~M\"{u}ller-Lyer, and \subref{fig:introc}~Ponzo illusions demonstrate that perception is not always veridical \cite{Goodale11}. For each, the orange elements are identical, although they appear unequal. \subref{fig:mcubes} In the \emph{size-weight illusion}, two objects that weigh the same feel as if the smaller one is heavier \cite{Charpentier91}.}
\label{fig:intro}
\end{figure}
\subsection{Motivation}
We like to believe that our sensory systems provide us with precise and accurate information about objects within the environment, but our perception is often subject to systematic errors, or {\em illusions} (Figure~\ref{fig:intro}). These can also occur between sensory modalities, often with visual information influencing haptic (touch) estimates of properties such as size or weight. For example, a curious experience occurs when we lift two objects of equal weight but different size; systematically and repeatably, the smaller object feels heavier than the larger. This \emph{size-weight illusion} (SWI) \cite{Charpentier91} cannot be explained by simple motor force error (i.e., it is not simply due to the production of more lifting or grip force for the larger object)~\cite{Flanagan00,Grandy06}, and so carries important implications for the dynamics of sensory integration between vision and haptics. Likewise, altered visual appearance of an object (e.g. through stereoscopic goggles~\cite{Ernst02} or optical distortion with prisms~\cite{Rock64}) can significantly impact haptically-judged estimates of its size. Simply put, when an object looks bigger than it really is, it feels bigger, too -- and any mismatch between vision and touch often goes completely unnoticed.
In order to establish a solid quantitative empirical assessment of these illusions, we have developed methodologies to examine the relationship between true size and perceived size. Previous investigations have uncovered evidence that the relationship between an object's true volume and its perceived volume often follows a power function with an average exponent of $0.704$ ($\sigma=0.08$)~\cite{Frayman81}. However, these prior investigations have predominantly used objects which are geometric, symmetrical, and convex -- properties which alone cannot adequately capture the range of objects regularly encountered in everyday environments. Thus, to systematically and comprehensively explore the relationship between true volume and perceived volume so as to better understand this percept's contribution to visual-haptic integration and, consequently, perception in general, we have developed dedicated methods to capture an ecologically valid set of stimuli.
\begin{figure}
\includegraphics[height=4cm]{cubes_all.jpg}\hspace{0.8cm}
\includegraphics[height=4cm]{cubes_rec}
\caption{We compare the performance of two scanning systems with the one proposed here at hand of a set of geometric primitives, whose volume can be easily measured by hand. Although shown as a collection, all items above have been reconstructed and texture-mapped individually.}\label{fig:cubes}
\end{figure}
Our goal is to build a data base of digital models of our specimens, which would allow us to infer volume and any other desired geometric properties. A method to create such models should meet the following list of criteria:
\begin{itemize}
\item It is mandatory that the sensing modality of choice be contactless: Cell phones and other hand-held consumer electronics, e.g., are among the classes of objects relevant to our psychological studies. They cannot simply be sunk in a fluid leveraging Archimedes' principle, neither can anything which is permeated by the fluid, as doing so would not provide the desired \emph{visible} volume.
\item The underlying sensor should be inexpensive and easy to use for non-experts; e.g., researchers in psychology and neuroscience. This rules out dedicated lab equipment, e.g., for white-light interferometry and the like.
\item The resulting models should exhibit some topological structure: For volume computations, it must at least be possible to distinguish interior and exterior. Therefore, point clouds alone are insufficient in this regard.
\item Given the complexity of everyday objects -- and that they frequently depart from the cubic, spherical, or cylindrical -- we target an improvement of accuracy over back-of-the-envelope estimates.
\item A rich spectrum of object classes is covered in terms of admissible geometry and reflectance properties. Specular surfaces, e.g., would need to be coated with powder to make them amenable to laser scanning. But this would contravene the first criterion and thus eliminates laser scanning from the list of candidates.
\end{itemize}
In light of these requirements, we opt for triangulation based on structured-light encoding as the primary sensing modality. The principle behind this method has been known for decades but has seen a renaissance in computer vision ever since Primesense introduced a fully functional color-range (``RGBD'') sensor unit in integrated-circuit design. This system-on-chip later became the core component of Microsoft's Kinect, which subsequently had an considerable impact in geometry reconstruction, tracking, occlusion detection, and action recognition, among other applications.
\subsection{Contribution and overview}
We develop a system for structured-light scanning of small- to medium-scale objects, which we dub \emph{Yet Another Scanner}, or YAS. Naturally, the question arises why we would need yet another scanner when several systems and commercial products are already available, e.g., Kinect Fusion~\cite{Newcombe2011}, ReconstructMe\footnote{\href{http://reconstructme.net/}{http://reconstructme.net/}}, Artec Studio\footnote{\href{http://www.artec3d.com/software/}{http://www.artec3d.com/software/}}, KScan3D\footnote{\href{http://www.kscan3d.com/}{http://www.kscan3d.com/}}, Scanect\footnote{\href{http://skanect.manctl.com/}{http://skanect.manctl.com/}}, Scenect\footnote{\href{http://www.faro.com/scenect/}{http://www.faro.com/scenect/}}, and Fablitec's 3d scanner\footnote{\href{http://www.fablitec.com/}{http://www.fablitec.com/}}; in particular, when most of these generate visually highly-pleasing results.
The main reason is that, albeit visually pleasing, the reconstructions provided by these methods are subject to biases that make them unsuitable for scientific investigation. The analysis, which is presented in Sect.~\ref{subsec:results}, compares the performance of YAS with that of two competing state-of-the-art implementations. While the reconstruction algorithm described in Sects.~\ref{subsec:alignment} and~\ref{subsec:poisson} itself is not novel, we carefully justify all choices to be made in its design w.r.t. above-listed requirements. Additionally, we address the issue that an aligned series of range images suffers from incompleteness precisely where the ground plane supports the object. Sect.~\ref{subsec:volumeestimation} proposes a strategy to circumvent this problem in volume estimation which avoids complicated hole-filling algorithms. We believe that a tool which outperforms commercial software but is accessible for further scientific development may be of interest to the computer vision community as well. Hence, as the final contribution, we will distribute the source through the repository at \href{https://bitbucket.org/jbalzer/yas}{https://bitbucket.org/jbalzer/yas}.
\section{System description}
\subsection{Data acquisition and calibration}
In all our experiments studies, we used Microsoft's Kinect and Primesense's Carmine 1.09. Both devices are shipped with a factory calibration of depth and RGB camera intrinsics as well as the coordinate transformation between their local reference frames. Initial visual assessment (by the naked human eye) approves of the default calibration simply because the point clouds computed from the range image seem to be accurately colored by the values of the RGB image. Extensive tests, however, have shown that -- in the spirit of our introductory remarks -- such an evaluation is misleading, and significant metric improvements through manual re-calibration are possible. For this purpose, we rely on the toolbox accompanying the paper~\cite{DanielHerreraC2012} to estimate all aforementioned parameters plus a depth uncertainty pattern, which varies both spatially and with depth itself.
\subsection{View alignment}\label{subsec:alignment}
A calibrated sensor immediately delivers physically plausible depth data. The integration of measurements from different vantage points into a common 3-d model can thus be seen as the core challenge here. A comprehensive overview of the state of the art in scan alignment is found in the recent survey~\cite{Tam2013}. Essentially, one can distinguish between two approaches: \emph{tracking} and \emph{wide-baseline matching}. The former is at the heart of Kinect Fusion~\cite{Newcombe2011} and the majority of commercially available software. Its main motivation stems from the fact that correspondence is easier to establish when two images haven been acquired closely in time -- supposing, of course, that the motion the camera has undergone between each image acquisition and the next meets certain continuity constraints. We believe, however, that for the purpose of small-scale object reconstruction, the disadvantages of tracking predominate. First and foremost, there is the question of redundancy: How do we deal
with the stream of depth data when operating an RGBD camera at frame rates up to $30~\mathrm{fps}$? On the one hand, redundancy is desirable because single depth images may not cover the entire surface of the unknown object, e.g., due to occlusions or radiometric disturbances of the projected infrared pattern. On the other hand, integration of overcomplete range data into a common 3-d model puts high demands on the quality of alignment. Most feature trackers operate on a reduced motion model\footnote{E.g., the Lucas-Kanade method assumes pure translational motion.} and are thus prone to drift. Such deviations in combination with the uncertainty in the raw depth data can lead to a \emph{stratification} of points in regions appearing in more than a single image. This effect is illustrated in Fig.~\ref{fig:misalign1}, which becomes more severe with higher numbers of processed images.
Kinect Fusion~\cite{Newcombe2011} deals with redundancy by instantaneously merging the depth stream into an implicit surface representation over a probabilistic voxel grid. The extra dimension, however, raises memory consumption -- even in implementations utilizing truncation or an efficient data structure such as an octree. Also, without a-priori knowledge of the specimen's size, it is difficult to gauge the interplay between the spatial resolutions of 3-d grid and raw depth data, the latter being left exploited only sub-optimally. Last but not least, a system based on tracking is little user-friendly: It requires the operator to move the sensor as steadily as possible. Otherwise, temporal under-sampling or motion blur can lead to a total breakdown of the alignment process.
\begin{figure}
\centering
\subfigure[Stratification]{\label{fig:misalign1}\includegraphics[height=4cm]{cubes_0_pcl_scenect_close}}\hspace{0.8cm}
\subfigure[Accumulation of tracking errors]{\label{fig:misalign2}\includegraphics[height=4cm]{free_lotion_pcl_scenect}}\\
\subfigure[Scenect]{\label{fig:misalign3}\includegraphics[height=4cm]{cubes_6_pcl_scenect}}\hspace{0.8cm}
\subfigure[YAS]{\label{fig:misalign4}\includegraphics[height=4cm]{cubes_6_pcl}}
\caption{We promote a sparse sampling of viewpoints for covering the object of interest with depth map measurements, for \subref{fig:misalign1} this reduces the impact of stratification and \subref{fig:misalign2} reduces the probability of drift or tracking failure. \subref{fig:misalign3}-\subref{fig:misalign4} The difference in alignment quality only becomes noticeable in cross-sections of the point clouds.}\label{fig:stratification}
\end{figure}
Here, we closely follow the wide-baseline matching procedure developed by the robotics community, notably in the work of Henry et al.~\cite{Henry2012}. It follows two quasi canonical steps: First, a set of local descriptors around interest points in each RGBD image is computed as well as a set of tentative matches between them. Such descriptors incorporate radiance and depth information either exclusively or in combination. Second, subsets of cardinality three are selected from all matches at random. Each subset admits a hypothesis about the rigid motion that transforms one of the point clouds into the other. The winning hypothesis is taken to be the one supporting the most matches, i.e., generating the highest number of \emph{inliers}~\cite{Fischler1981}. We review these initial two steps in the following sections.
\subsubsection{Sampling}\label{subsubsec:sampling}
Let us formally consider the case of two views $l=0,1$, i.e., we look for two rigid motions $g_l\in\SE(3)$, $g_l:\bm{x}\mapsto\mathbf{R}_l\bm{x}+\bm{t}_l$, with $\mathbf{R}_l\in\SO(3)$ and $\bm{t}_l\in\mathbb{R}^3$. Without loss of generality, one can assume that $g_0$ coincides with the world reference frame, i.e., $\mathbf{R}_0=\mathbf{I}$ and $\bm{t}_0=\bm{0}$, which leads to a simpler notation of the unknowns $\mathbf{R}_1=\mathbf{R}$ and $\bm{t}_1=\bm{t}$.
A number of interest points $\{\bm{p}^i_0\}, \{\bm{p}^j_1\}$ with high response is extracted from each of the two RGB images corresponding to $g_{0,1}$ by means of the SIFT detector. The points are combined into a set of putative correspondences $\mathcal{C}=\{(\bm{p}^{i_k}_0,\bm{p}^{j_k}_1)\;|\; k=1\ldots,n\in\mathbb{N}\}$ by thresholded forward-backward comparison of the distances between associated SIFT descriptors~\cite{Lowe1999}. The search for nearest neighbors can be sped up by a locality-sensitive hashing technique or similar. However, we found in all of our experiments that the time consumed by a brute-force search was within acceptable limits. Next, we repeatedly draw a sample of three matches from $\mathcal{C}$ and obtain a set of triples $\mathcal{H}=\{(k_1,k_2,k_3)\;|\; 1\leq k_1,k_2,k_3 \leq n,k_1 \neq k_2 \neq k_3 \}$.
\subsubsection{Consensus}\label{subsubec:consensus}
Implicitly, each of the elements of $\mathcal{H}$ determines a hypothesis about the transformation we are looking for: Suppose we already know $g_1\in\SE(3)$, then the geometric least-squares error for some $(k_1,k_2,k_3)\in\mathcal{H}$ is given by
\begin{equation}\label{eq:lse}
e(k_1,k_2,k_3)=\sum\limits_{k\in\{k_1,k_2,k_3\}}\frac{1}{2}\|\bm{x}^{i_{k}}_0-\mathbf{R}_k\bm{x}^{j_k}_1-\bm{t}_k\|^2.
\end{equation}
Here, the six points $\bm{x}_0^{i_{k}},\bm{x}_1^{j_{k}}\in\mathbb{R}^3$ equal the backprojections of the three matches $(\bm{p}^{i_k}_0,\bm{p}^{j_k}_1)$ forming the current hypothesis. Given the intrinsic camera parameters, they can be easily computed from the data delivered by the calibrated depth sensor. Conversely, given a triple $(k_1,k_2,k_3)\in\mathcal{H}$, we can find a global minimizer $g_1^*$ of the convex function~\eqref{eq:lse} in the following way: Denote by $\bar{\bm{x}}_0$ the mean of $\bm{x}_0^{i_{k}}$ over $k$ and define $\bar{\bm{x}}_0^{i_{k}}=\bm{x}_0^{i_{k}}-\bar{\bm{x}}_0$. The quantities $\bar{\bm{x}}_1$ and $\bar{\bm{x}}_1^{j_{k}}$ are defined analogously. A minimizer in the \emph{entire general linear group} of matrices $\GL(3)$ is
\[
\mathbf{H}=\sum\limits_{k\in\{k_1,k_2,k_3\}}\bar{\bm{x}}_0^{i_{k}}(\bar{\bm{x}}_1^{j_{k}})^{\top},
\]
cf.~\cite{Williams2001}. One needs to make sure that the optimal $g_1^*$ involves a genuine \emph{rotation matrix} by projecting $\mathbf{H}$ onto $\SO(3)$. This is commonly achieved by \emph{Procrustes analysis}, essentially consisting of a singular-value decomposition: Write $\mathbf{H}$ as the product $\mathbf{U}\mathbf{\Sigma}\mathbf{V}^{\top}$ with two orthogonal factors $\mathbf{U},\mathbf{V}\in\mathbb{R}^{3\times 3}$, then $\mathbf{R}_k^*=\mathbf{V}\mathbf{U}^{\top}$. Once $\mathbf{R}_k^*$ is known, the optimal translation vector can be computed as $\bm{t}_k^*=\bar{\bm{x}}_0-\mathbf{R}_k^*\bar{\bm{x}}_1$. The transformation for the element of $\mathcal{H}$ attaining the highest consensus among \emph{all} matches in $\mathcal{C}$ constitutes the solution to the global alignment problem~\cite{Fischler1981}. The result is refined based on the inlier correspondences with the iterative closest-point (ICP) method~\cite{Besl1992}. This also ensures a \emph{geometrically} continuous alignment, which is not guaranteed because $\mathcal{H}$ was generated
merely based on \emph{photometry}.
\subsection{Surface reconstruction}\label{subsec:poisson}
The common point cloud obtained after merging all aligned depth maps carries no information about the topological relationship between its elements, but as we will see shortly, such information plays a crucial part in volume estimation. There exists a wealth of algorithms for point-to-surface conversion, most of which depend on the signed or unsigned Euclidean distance field $\varphi:\mathbb{R}^3\to\mathbb{R}$ induced by the point cloud. Kinect Fusion, e.g., computes $\varphi$ directly. Alternatively, when the point cloud is oriented, i.e., each point $\bm{x}$ is endowed with an estimate of the normal vector $\bm{n}$ the surface should have at that location, one can search for the function $\varphi$ minimizing
\begin{equation}\label{eq:dirichlet}
\int\limits_D\frac{1}{2}\|\nabla\varphi-\bm{n}\|^2 \mathrm{d}\bm{x}\rightarrow\min.
\end{equation}
Here, with slight abuse of notation, $\bm{n}$ refers to an (arbitrary) continuation of the normal field from the point set to some sufficiently large rectangular domain $D\subset\mathbb{R}^3$. Since the gradient of any scalar function is orthogonal to its level sets, this gives a family of \emph{integral surfaces}
\begin{equation}\label{eq:isosurface}
\Gamma = \{\bm{x}\in\mathbb{R}^3\,|\,\varphi(\bm{x})=C\}
\end{equation}
of $\bm{n}$. A minimizer of~\eqref{eq:poisson} is found as the solution of the Euler-Lagrange equation
\begin{equation}\label{eq:poisson}
\Delta \varphi = \divv\bm{n}
\end{equation}
under natural boundary conditions (here of Neumann type). Eq.~\eqref{eq:poisson} is the well-known Poisson equation and eponymous for the \emph{Poisson reconstruction} algorithm proposed in~\cite{Kazhdan2006}.
The motivation for increasing the order of differentiation as compared to the direct approach (i.e., that followed in Kinect Fusion) is twofold: First, Eq.~\eqref{eq:dirichlet} is a variant of the Dirichlet energy, which implies that small holes in the point cloud, i.e., areas where $\bm{n}=\bm{0}$, will automatically be in-painted harmonically. Second, for a solution to exist in the strong sense $\nabla\varphi=\bm{n}$, the normal field must be \emph{integrable} or \emph{curl-free}. Noise, which is very common in RGBD images, is responsible for most of the non-integrability in a measured normal field $\bm{n}$. In the variational setting~\eqref{eq:dirichlet}, however, $\bm{n}$ is implicitly replaced by the next-best gradient (its so-called \emph{Hodge projection}, cf.~\cite{Cantarella2002}), which makes the approach very resilient to stochastic disturbances but at the same time destroys fine details. The smoothing effect can be mitigated by imposing Dirichlet conditions on~\eqref{eq:poisson} at a sparse set of salient points. This so-called \emph{screened Poisson reconstruction} has recently been introduced in~\cite{Kazhdan2013}.
The point cloud is easily oriented exploiting the known topological structure of the pixel lattice: Given a depth parametrization of the surface $z(x,y)$ over the two orthogonal camera coordinate directions $x$ and $y$, the normal can be written as $\bm{n}=(-\partial_x z, -\partial_y z,1)^{\top}$. The partial derivatives of $z$ w.r.t. camera and image coordinates $(x,y)^{\top}$ respectively $(u,v)^{\top}$ are related by the chain rule:
\begin{equation}\label{eq:normal}
\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial x}=\frac{f_u}{z}\frac{\partial z}{\partial u},\quad \frac{\partial z}{\partial y} = \frac{\partial z}{\partial v}\frac{\partial v}{\partial y}=\frac{f_v}{z}\frac{\partial z}{\partial v}.
\end{equation}
Here, $f_u, f_v$ are the focal lengths of the pinhole depth camera, and finite-differencing provides an approximation to the gradient of $z(u,v)$. For the details of numerically solving~\eqref{eq:poisson} and selecting the constant $C$ in~\eqref{eq:isosurface} appropriately, we refer the reader to the original paper~\cite{Kazhdan2006}.
\subsection{Volume estimation}\label{subsec:volumeestimation}
\subsubsection{Closed surfaces}\label{subsubsec:volcont}
Suppose for the moment that $\Gamma$ given by~\eqref{eq:isosurface} is compact and closed. The volume $V$ of the domain $\Omega\subset\mathbb{R}^3$ it encompasses is defined as the integral of the characteristic function $\chi_{\Omega}$ of $\Omega$:
\begin{equation}\label{eq:volume}
V=|\Omega|=\int\limits_{\mathbb{R}^3}\chi_{\Omega}\mathrm{d}\Omega=\int\limits_{\Omega}1 \mathrm{d}\Omega.
\end{equation}
Unfortunately, an evaluation of this integral is not very practical for two reasons. First, doing so would require a regular grid over $\Omega$, which introduces undesirable artefacts where it interacts with a discrete version of $\Gamma$. Second, an expensive nearest-neighbor problem would need to be solved\footnote{But could be remedied by re-visiting the signed distance function $\varphi$ of $\Gamma$ from Sect.~\ref{subsec:poisson}.} to determine whether a point is inside or outside of $\Omega$. The following trick is based on the classic Gauss divergence theorem, cf.~\cite{Mirtich1996}, which relates the flow of \emph{any} continuously differentiable vector field $\bm{v}:\mathbb{R}^3\to\mathbb{R}^3$ through the boundary $\Gamma=\partial\Omega$ of $\Omega$ with its source density or \emph{divergence} in the interior:
\begin{equation}\label{eq:gauss}
\int\limits_{\Omega}\divv\bm{v}\mathrm{d}\Omega=\int\limits_{\Gamma}\langle\bm{v},\bm{n}\rangle \mathrm{d}\Gamma.
\end{equation}
The left-hand side of this equation does not quite resemble the right-hand side of~\eqref{eq:volume}, yet. However, this can be achieved by a clever choice of $\bm{v}$, e.g., $\bm{v}:=(x,0,0)^{\top}$, but note that several variants will work equally well and that $\bm{v}$ is \emph{not} unitary. We will return to this point later in Sect.~\ref{subsubsec:holefilling}. We have $\divv \bm{v}=1$ so that combining~\eqref{eq:volume} and~\eqref{eq:gauss} yields
\begin{equation}\label{eq:volintegral}
V=\int\limits_{\Gamma}\langle\bm{v},\bm{n}\rangle \mathrm{d}\Gamma.
\end{equation}
\begin{figure}[tb]
\centering
\def0.8\columnwidth{0.8\columnwidth}
{\scriptsize\input{flow.pdf_tex}}
\caption{The volume of a compact object $\Omega$ equals the mean flow of the vector field $\bm{v}=(x,0,0)$ through its boundary surface $\Gamma$. If the support of the object is aligned with the $xy$- or $xz$-plane, it is not traversed by any stream line, and hence needs not be explicitly filled with mesh faces for a faithful volume estimate. The flow field is colored by increasing magnitude $\|\bm{v}\|$.}\label{fig:vol}
\end{figure}
Let us look at the discrete case: Here, a level set $\Gamma_h$ of~\eqref{eq:isosurface} is extracted by the marching cubes in the form of a triangular mesh~\cite{Lorensen1987}. Such a piece-wise linear representation of the geometry provides a likewise locally-linear approximation of any function $f$ whose values $f_i$ are known at the vertices $\bm{x}_i\in\Gamma_h$, $i\in\mathbb{N}$. A Gauss-Legendre quadrature rule for $f$ with linear precision defined over the triangle $T$ is
\[
\int\limits_{T}f\mathrm{d}\bm{x}\approx A(T)\sum_{i_k\in\mathcal{I}(T)}^3 \frac{1}{3}f_{i_k},
\]
where $A(T)$ equals the area of $T$, and $\mathcal{I}(T)$ enumerates its three corner vertices. Now substitute $f_i$ by the flow $\langle\bm{v}_i,\bm{n}_i\rangle$. The vertex normals $\bm{n}_i$ are usually taken to be the normalized mean of the face normals in a one-ring neighborhood of $\bm{x}_i$. Altogether, we finally obtain the following approximation of Eq.~\eqref{eq:volintegral}
\[
V\approx\sum_{T\in \Gamma_h}\frac{A(T)}{3}\sum_{i_k\in\mathcal{I}(T)}^3 \langle\bm{v}_{i_k},\bm{n}_{i_k}\rangle.
\]
\subsubsection{Surfaces with boundary}\label{subsubsec:holefilling}
As explained in Sect.~\ref{subsec:poisson}, Poisson reconstruction accounts for most smaller holes in the aligned point clouds. The support of the object, i.e., its ``bottom'' or the area where it is in contact with the ground plane, however, remains usually unfilled. At the beginning of Sect.~\ref{subsubsec:volcont}, we demanded that $\Gamma$ be compact and closed because only then the volume of $\Omega$ is well-defined. We can lift this assumption in parts simply by a coordinate transformation: Remember that we chose $\bm{v}$ to point in the direction of the $x$-axis of the world coordinate system. Consequently, the flow through any of the planes $y=\const$ or $z=\const$ vanishes. As shown in Fig.~\ref{fig:vol}, all we have to do is align the support of the model with one of these planes. Without loss of generality, we choose $\{(x,y,z)^{\top}\in\mathbb{R}^3\,|\, z=0 \}$. In our scanning scenario, it is reasonable to assume that the specimens to be measured are spatially isolated enough that the depth images capture a
significant portion of the ground plane surrounding the object, which can thus be detected fully automatically. To this end, we again invoke a RANSAC-type procedure, which samples triplets of points, calculates their common plane as a putative solution, and evaluates each such hypothetical plane by how many other points in the cloud it contains.
\section{Experiments}\label{sec:experiments}
\begin{figure}
\subfigure[YAS]{\label{fig:rec_vs_gt}\includegraphics[height=3cm]{cubes_0_rec_vs_gt}}\hfill
\subfigure[Scenect]{\label{fig:rec_vs_gt_scenect}\includegraphics[height=3cm]{cubes_0_rec_vs_gt_scenect}}\hfill
\subfigure[Kinect Fusion]{\label{fig:rec_vs_gt_kinfu}\includegraphics[height=3cm]{cubes_0_rec_vs_gt_kinfu}}
\caption{Reconstructions of item no. $1$ in the cube data set. The ground truth is shown in green. It becomes apparent that Kinect Fusion Explorer systematically overestimates the object volume.}\label{fig:rec_vs_gt}
\end{figure}
\subsection{Implementation}\label{subsec:implementation}
We created a C++ implementation of most of the reconstruction pipeline, including raw data acquisition, coarse and fine registration as well as detection of the ground plane. Ease of use is of premier priority in view of the interdisciplinary nature of this project. Therefore, the number of dependencies was kept as small as possible: The OpenCV library supplies us with all functionality for feature matching (Sect.~\ref{subsubsec:sampling}). Our implementation of the ICP method requires fast nearest-neighbor lookup which is based on the kd-tree data structure from the ANN library. We also created a graphical QT frontend which is showcased in the video included in the supplemental material. Poisson reconstruction (Sect.~\ref{subsec:poisson}) is currently done in Meshlab but will be integrated into our code in upcoming releases.
We compare our method to the Kinect Fusion algorithm and Scenect, which is one of the few commercial software packages without hindering functionality restraints in the trial version. To warrant a fair comparison, all participating systems should be operated with the same sensor and the same intrinsic calibration. This proved to be somewhat difficult: Both Scenect and YAS access devices through the OpenNI framework driver supporting all of Primesense's products and the Xtion by Asus among others. The Point Cloud Library provides an open-source version of the Kinect Fusion algorithm which could potentially function with the Carmine 1.09 as well, but our experiences with it were little encouraging. For this reason, we had to resort to Microsoft's own implementation Kinect Fusion Explorer, which works exclusively with proprietary hardware, the Kinect.
\subsection{Results}\label{subsec:results}
\begin{table}[tbp]
\centering
\setlength{\tabcolsep}{1.5pt}
\def 0.55cm{0.55cm}
\tiny
\begin{tabular}{lccccccc}\toprule
No. & $V_{\mathrm{manual}}\atop [10^{-3}m^3]$ & $V_{\mathrm{YAS}}\atop [10^{-3} m^3]$ & $E_{\mathrm{YAS}}\atop[\%]$ & $V_{\mathrm{Scenect}}\atop[10^{-3}m^3]$ & $E_{\mathrm{Scenect}}\atop[\%]$ & $V_{\mathrm{KinFu}}\atop[10^{-3}m^3]$ & $E_{\mathrm{KinFu}}\atop[\%]$ \\\midrule[\heavyrulewidth] %
1 & $1.26$ & $1.26$ & $0.11$ & $1.26$ & $0.6$ & $1.47$ & $16.6$ \\\midrule
2 & $2.82$ & $2.72$ & $-3.38$ & $2.68$ & $-5.1$ & $3.61$ & $28.4$\\\midrule
3 & $1.29$ & $1.21$ & $-6.07$ & $1.21$ & $-6.17$ & $1.63$ & $27.0$\\\midrule
4 & $3.07$ & $3.07$ & $-0.02$ & $2.83$ & $-7.57$ & $4.23$ & $38.0$\\\midrule
5 & $2.18$ & $2.06$ & $-5.44$ & $2.1$ & $-3.79$ & $2.85$ & $31.1$\\\midrule
6 & $6.26$ & $6.44$ & $2.86$ & $5.76$ & $-7.07$ & $7.69$ & $22.9$\\\midrule
7 & $1.66$ & $1.79$ & $8.22$ & $1.75$ & $5.65$ & $1.67$ & $0.55$\\\midrule
8 & $1.78$ & $1.89$ & $6.0$ & $1.67$ & $-6.28$ & $2.15$ & $20.3$\\\midrule
9 & $2.75$ & $2.75$ & $0.05$ & $2.68$ & $2.41$ & $3.84$ & $39.8$\\\midrule
10 & $24.6$ & $24.8$ & $1.05$ & $18.6$ & $-24.4$ & $28.8$ & $17.1$\\\bottomrule
$\mu$ & & & $-0.34$ & & $-5.74$ & & $24.2$\\\midrule
$\sigma$ & & & $4.59$ & & $7.76$ & & $11.5$ \\\bottomrule
\end{tabular}
\caption{Volume estimates and their relative error w.r.t. the value obtained by manual measurement.}
\label{tab:volumes}
\end{table}
\begin{figure}[tbp]
\subfigure[Volume]{\label{fig:volumes}\includegraphics[width=0.495\columnwidth]{volumes.pdf}}\hfill
\subfigure[Error]{\label{fig:errors}\includegraphics[width=0.495\columnwidth]{errors.pdf}}
\caption{Visualization of Tab.~\ref{tab:volumes}. The error bars in~\subref{fig:volumes} indicate the absolute deviation from the ground truth volume. From~\subref{fig:errors}, it seems like Kinect Fusion is biased towards excessive values.}\label{fig:plots}
\end{figure}
\begin{figure}[tbp]
\centering
\setlength{\tabcolsep}{0pt}
\def 2cm{2cm}
\subfigure[Object]{\label{fig:photos}
\begin{tabular}[b]{c}
\includegraphics[height=2cm]{statue}\\
\includegraphics[height=2cm]{tom}\\
\includegraphics[height=2cm]{robot}\\
\includegraphics[height=2cm]{statue_sa}\\
\includegraphics[height=2cm]{iron}\\
\includegraphics[height=2cm]{dog}\\
\includegraphics[height=2cm]{shoe}\\
\includegraphics[height=2cm]{lotion}\\
\includegraphics[height=2cm]{penguin}
\end{tabular}}
\subfigure[YAS]{\label{fig:ours}
\begin{tabular}[b]{c}
\includegraphics[height=2cm]{free_statue_rec_no_tex}\\
\includegraphics[height=2cm]{free_tom_rec_no_tex}\\
\includegraphics[height=2cm]{free_robot_rec_no_tex}\\
\includegraphics[height=2cm]{free_statue_sa_rec_no_tex}\\
\includegraphics[height=2cm]{free_iron_rec_no_tex}\\
\includegraphics[height=2cm]{free_dog_rec_no_tex}\\
\includegraphics[height=2cm]{free_shoe_rec_no_tex}\\
\includegraphics[height=2cm]{free_lotion_rec_no_tex}\\
\includegraphics[height=2cm]{free_penguin_rec_no_tex}
\end{tabular}}
\subfigure[Scenect]{\label{fig:scenect}
\begin{tabular}[b]{c}
\includegraphics[height=2cm]{free_statue_rec_scenect}\\
\includegraphics[height=2cm]{free_tom_rec_scenect}\\
\includegraphics[height=2cm]{free_robot_rec_scenect}\\
\includegraphics[height=2cm]{free_statue_sa_rec_scenect}\\
\includegraphics[height=2cm]{free_iron_rec_scenect}\\
\includegraphics[height=2cm]{free_dog_rec_scenect}\\
\includegraphics[height=2cm]{free_shoe_rec_scenect}\\
\includegraphics[height=2cm]{free_lotion_rec_scenect}\\
\includegraphics[height=2cm]{free_penguin_rec_scenect}
\end{tabular}}
\subfigure[Kinect Fusion]{\label{fig:kinfu}
\begin{tabular}[b]{c}
\includegraphics[height=2cm]{free_statue_rec_kinfu}\\
\includegraphics[height=2cm]{free_tom_rec_kinfu}\\
\includegraphics[height=2cm]{free_robot_rec_kinfu}\\
\includegraphics[height=2cm]{free_statue_sa_rec_kinfu}\\
\includegraphics[height=2cm]{free_iron_rec_kinfu}\\
\includegraphics[height=2cm]{free_dog_rec_kinfu}\\
\includegraphics[height=2cm]{free_shoe_rec_kinfu}\\
\includegraphics[height=2cm]{free_lotion_rec_kinfu}\\
\includegraphics[height=2cm]{free_penguin_rec_kinfu}
\end{tabular}}
\caption{Selection of reconstructed free-forms. Let us emphasize that -- to comply with the double-blind review policy -- the bust in the first row is \emph{not} taken from any one of the authors.}\label{fig:examplerec}
\end{figure}
The set of 20 specimens can be divided into two equally-sized groups: The first group contains objects of simple geometry like cubes and cylinders, whose basic dimensions can be measured manually with a tape measure or ruler, see Fig.~\ref{fig:cubes}. Free-forms of sizes ranging from just a few centimeters (fifth row of Fig.~\ref{fig:examplerec}) to the size of a human upper body (first row of Fig.~\ref{fig:examplerec}) make up the second group. The ``ground truth'' volumes for the first group are listed in the first column of Tab.~\ref{tab:volumes}. Needless to say these are afflicted by their own uncertainty, given that they were determined through measurements with a ruler. Our reconstructions of the cube data set are depicted in Fig.~\ref{fig:cubes}. We obtain the best average relative volume error $(V_{\mathrm{YAS}}-V_{\mathrm{Manual}})/V_{\mathrm{Manual}}$ of $-0.34$ $\%$. The performance of Scenect is comparable, which is somewhat surprising in view of Fig.~\ref{fig:scenect}. The meshes created by the Kinect Fusion explorer are of inferior topological quality: they contain a high number of non-manifold simplices. This, however, does not seem to affect the volume estimates negatively. Also it can be said that the sensitivity of volumes w.r.t. the ground plane parameters is relatively low.
As can be seen from the last two columns of Tab.~\ref{tab:volumes}, the Kinect Fusion explorer systematically overestimates the ground truth volume by a significant margin. We conjecture that the issue is rooted in calibration. In fact, an important lesson learned during our experimental studies was that a good calibration can make a difference in error of an order of magnitude. Indeed, Scenect provides a calibration program, but Kinect Fusion Explorer does not. A visualization of Tab.~\ref{tab:volumes} is plotted in Fig.~\ref{fig:plots}.
Results for the second group of objects are shown in Fig.~\ref{fig:examplerec}. Scenect performs worst among the three compared methods. It must be said, though, that Scenect does not offer mesh reconstruction feature, and the point clouds it exports are not oriented. Normals can be computed by singular value decomposition considering the nearest neighbors of a point, which is probably less accurate than the finite-differences approximations of~\eqref{eq:normal}. The lack of loop-closure whose effect can be better identified in the point cloud in Fig.~\ref{fig:misalign2} carries over to the triangular mesh in row of Fig.~\ref{fig:scenect}. Premature termination of the tracker is responsible for the poor reconstruction of the penguin in the last row of Fig.~\ref{fig:scenect}.
Although it behaved unreliably during volume estimation, Kinect Fusion produced high quality models in real-time, which justifies the tremendous success it had since its inception. Still, the scale bias shows, and a lot of details appear to be missing -- details which are present in our YAS reconstructions despite the fact that the Poisson algorithm is known to possess the characteristics of a low-pass filter (see the discussion in Sect.~\ref{subsec:poisson}). The precision setting in Kinect Fusion is quite rigid since the dimensions of the Cartesian grid have to be fixed \emph{before} the reconstruction process even starts. We found that the maximal resolution of $512\times 512\times 512$ voxels rendered the tracking unstable and/or led to incomplete meshes.
\subsection{Discussion}
We establish correspondence based on photometry, hence our approach fails in the absence of sufficiently exciting texture. This however does not necessarily need to be on the target object itself but can be found in the background, which may be ``enriched'' since after all, in our application, it is not of interest. The majority of competing approaches, including the two we compare against, can cope with the issue. The main reason is the trade-off between a sparse sampling of viewpoints and aforementioned need for sufficient texture: All previous method implicitly leverage on geometry in the motion estimation stage, which is only made possible by the small baseline between adjacent frames, i.e., through tracking, the disadvantages of which have been discussed in Sect.~\ref{subsec:alignment}. In fact, ICP measures the similarity between two points by their distance, hence endows each of them with a geometry descriptor, although a primitive one, too primitive to support wide-baseline matching. More suitable descriptors are available, but we are not yet considering them here, for already the process of salient point detection let alone the computation of informative geometry descriptors is extremely challenging on noisy, occlusion-ridden, and incomplete depth data such as from RGBD sensors.
The system has no real-time capabilities. We believe this is not necessary for our target application of small-scale object scanning (unlike e.g. navigation, map-building, or odometry). Considering the low resolution of RGB images, the matching process is a matter of seconds. This bottleneck could be removed by an approximate nearest-neighbors search. Our impression is that our system requires the same or even less overall acquisition time compared to e.g. Kinect Fusion, which requires slow and steady motion around the object (and sometimes even a complete reset).
\section{Conclusion}
We described a system for integrating a set of RGBD images of small-scale objects into a geometrically faithful 3-d reconstruction. The system is intended to support researchers in the field of cognitive neuroscience who will use it for acquiring ground truth data for their own experimental studies. To assess the suitability of our 3-d models and those obtained by comparable algorithms, we performed an in-depth analysis of theoretical and empirical kind. There are two main conclusions we would like to draw here: First, the quality of a set of calibration parameters or metric reconstruction can be deceptive. Only quantitative analysis enables veridical information. Second, while the uncertainties of hand-held structured-light scanners may be tremendous in the eyes of the optical metrologist, they help improving studies in the cognitive neurosciences, where manual measurement is still common practice.
\bibliographystyle{alpha}
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"redpajama_set_name": "RedPajamaArXiv"
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"redpajama_set_name": "RedPajamaC4"
} | 7,322 |
Mont Juic, suite of Catalan dances for orchestra és una obra escrita conjuntament per Lennox Berkeley i Benjamin Britten el 1937. Inspirada en la muntanya de Montjuïc, fou publicada com a Berkeley's op. 9 i Britten's op. 12.
Història
El 1936, Berkeley i Britten van assistir al XIV Festival de la Societat Internacional per la Música Contemporània a Barcelona. Berkeley havia estat vivint a l'estranger des de feia anys i mai abans havia conegut Britten. Aviat es van convertir en bons amics. Un altre amic de Berkeley, Peter Burra, també va assistir al festival, i també va esdevenir amic de Britten.
Al Festival, Britten va acompanyar el violinista Antoni Brosa en l'estrena de la seva Suite per a violí i piano, op. 6. El plat fort del Festival, al qual van assistir Britten, Berkeley i Burra, va ser l'estrena mundial pòstuma del Concert per a violí d'Alban Berg, "A la memòria d'un àngel", que es va realitzar el diumenge 19 d'abril, amb el solista Louis Krasner, sota la direcció de Hermann Scherchen.
L'endemà, el trio va visitar la muntanya de Montjuïc, el turó que domina el paisatge de Barcelona. Pocs dies després, el dimecres 22 d'abril van assistir a un Festival de Dansa Folklòrica a la zona de l'Exposició, on van escoltar diverses melodies populars catalanes.
Una mica més tard el mateix dia van fer alguns apunts sobre aquelles melodies en un cafè de Barcelona. L'any següent, de tornada a Anglaterra, van decidir escriure conjuntament una suite orquestral basat en algunes de les melodies del ball que havien sentit a Montjuïc. La van anomenar simplement Mont Juic, i la van dedicar "En memòria de Peter Burra", qui havia mort en un accident d'avió l'abril de 1937. L'obra va ser escrita entre el 6 d'abril i 12 de desembre de 1937.
Instrumentació
La instrumentació escollida van ser dues flautes (una que dobla el flautí), dos oboès, dos clarinets en si bemoll, un saxòfon alt (ad lib.), saxòfon tenor (ad lib.), dos fagots, quatre trompes, dues trompetes en si bemoll, tres trombons, tuba, timbals, glockenspiel, platerets, arpa i cordes.
Moviments
La suite consta de quatre moviments:
Andante maestoso
Allegro grazioso
Lament: Andante moderato ("Barcelona, july 1936")
Allegro molto
Els dos compositors van optar per no revelar qui havia escrit cada part de la música. El manuscrit que Benjamin Britten va presentar a l'editorial va ser escrit enterament a mà per ell mateix. El 1980, però, Lennox Berkeley va revelar a Peter Dickinson que ell havia escrit les dues primeres parts i Britten les dues últimes, tot i que conjuntament havien col·laborat en l'orquestració, la forma i altres detalls.
Homenatge a Catalunya
En el moment que l'obra va ser escrita, s'havia declarat la Guerra Civil espanyola, i el tercer Lament (en do menor), va ser escrit com un homenatge a Catalunya. S'inclou un saxòfon alt i solitari que es basa en el ritme de la sardana. Es va subtitular "Barcelona, juliol 1936", una clara referència a la guerra civil que havia esclatat el 18 de juliol.
Primera representació
L'estrena de l'obra va ser en un programa de BBC Radio 1 el 8 de gener de 1938, per l'Orquestra Simfònica de la BBC sota la direcció de Joseph Lewis. Després de l'actuació, Berkeley va escriure a Britten, dient-li: he de dir que vaig pensar que les teves dues peces són més efectives que les meves. Des de llavors Mont juïc ha estat representada en moltes actuacions i disposa de diversos enregistraments.
Referències
Suites per a orquestra
Composicions musicals del 1937
Obres de Benjamin Britten | {
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{"url":"https:\/\/thoughtstreams.io\/jtauber\/note-quantization\/4254\/","text":"Note Quantization\n\n48 thoughts\nlast posted April 25, 2014, 2:23 a.m.\n\n6 earlier thoughts\n\n0\n\nIn the case of pitch, we go from frequency to letter name + octave via a choice of tuning and temperament, then abstract away the octave and factor out the key to get an abstraction like \"the 3rd note of the scale\" or a \"IV64 chord\" or whatever.\n\n41 later thoughts","date":"2021-03-07 01:49:29","metadata":"{\"extraction_info\": {\"found_math\": false, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8308992385864258, \"perplexity\": 5096.61108112224}, \"config\": {\"markdown_headings\": false, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-10\/segments\/1614178376006.87\/warc\/CC-MAIN-20210307013626-20210307043626-00604.warc.gz\"}"} | null | null |
Der MHK Dubnica (Mestský hokejový klub Dubnica nad Váhom) ist ein slowakischer Eishockeyclub aus Dubnica nad Váhom, der 1942 gegründet wurde und momentan in der drittklassigen 2. Liga spielt, jedoch früher mehrere Spielzeiten in der Extraliga absolvierte. Seine Heimspiele trägt der Verein im Zimný štadión Dubnica nad Váhom aus, das 3.000 Zuschauer fasst.
Geschichte
Der MHK Dubnica wurde 1942 unter dem Namen ŠK Dubnica gegründet. Es folgten zahlreiche Namenswechsel, ehe der Verein 2007 den heutigen Namen wählte. Als Dubnica noch im tschechoslowakischen Ligensystem spielte, kam die Mannschaft nie über die zweite Spielklasse hinaus.
Nach Auflösung der Tschechoslowakei und der Teilung in die voneinander unabhängigen Staaten Tschechien und Slowakei wurde Dubnica in der Saison 1993/94 in die zweitklassige 1. Liga eingeteilt und erreichte auf Anhieb den Aufstieg in die Extraliga. Im dritten Jahr ihrer Zugehörigkeit zur Extraliga stieg die Mannschaft in der Saison 1996/97 jedoch wieder in die 1. Liga ab. Nach der Zweitligameisterschaft im Jahr 2004 gelang Dubnica noch einmal für eine Spielzeit die Rückkehr in die Extraliga. Mittlerweile tritt der Verein in der drittklassigen 2. Liga an.
Erfolge
Aufstieg in die 1. SNHL (zweitklassig): 1970
Meister der 1. SNHL 1979, 1980
Aufstieg in die Extraliga: 1995, 2004
Meister der 1. Liga 1994, 1999, 2004
Bekannte ehemalige Spieler
Pavol Demitra
Ivan Baranka
Ján Homér
Róbert Tomík
Tomáš Tatar
Rastislav Pavlikovský
Richard Pavlikovský
Weblinks
Offizielle Homepage (slowakisch)
Slowakischer Eishockeyclub | {
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{"url":"https:\/\/www.storyofmathematics.com\/fractions-to-decimals\/1-32-as-a-decimal\/","text":"# What Is 1\/32 as a Decimal + Solution With Free Steps\n\nThe fraction 1\/32 as a decimal is equal to 0.031.\n\nFractions are rational numbers expressed as the division of two numbers p and q as p\/q, with p being the numerator and q being the denominator. There are multiple types of fractions including proper (q > p), improper (q < p), mixed fractions, etc. The given fraction 1\/32 is a proper fraction because 32 > 1.\n\nHere, we are interested more in the types of division that results in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.\n\nNow, we introduce the method used to solve said fraction to decimal conversion, called Long Division which we will discuss in detail moving forward. So, let\u2019s go through the Solution of fraction 1\/32.\n\n## Solution\n\nFirst, we convert the fraction components i.e., the numerator and the denominator, and transform them into the division constituents i.e., the Dividend and the Divisor respectively.\n\nThis can be seen done as follows:\n\nDividend = 1\n\nDivisor = 32\n\nNow, we introduce the most important quantity in our process of division, this is the Quotient. The value represents the Solution to our division, and can be expressed as having the following relationship with the Division constituents:\n\nQuotient = Dividend $\\div$ Divisor = 1 $\\div$ 32\n\nThis is when we go through the Long Division solution to our problem.\n\nFigure 1\n\n## 1\/32 Long Division Method\n\nWe start solving a problem using the Long Division Method by first taking apart the division\u2019s components and comparing them. As we have 1, and 32 we can see how 1 is Smaller than 32, and to solve this division we require that 1 be Bigger than 32.\n\nThis is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. And if it is then we calculate the Multiple of the divisor which is closest to the dividend and subtract it from the Dividend. This produces the Remainder which we then use as the dividend later.\n\nIn this case, 1 multiplied by 10 gets us 10, which is still smaller than 32. Therefore, we again multiply by 10 to get 100, which is greater than 32. To indicate these two multiplications, we add a decimal \u201c.\u201d\u00a0and a\u00a00 to our quotient.\n\nNow, we begin solving for our dividend 1, which after getting multiplied by 100 becomes 100.\n\nWe take this 100 and divide it by 32, this can be seen done as follows:\n\n100 $\\div$ 32 $\\approx$ 3\n\nWhere:\n\n32 x 3 = 96\n\nWe add 3 to our quotient. This will lead to the generation of a Remainder equal to 100 \u2013 96 = 4, now this means we have to repeat the process by Converting the 4 into 40\u00a0and solving for that:\n\n40 $\\div$ 32 $\\approx$ 1\n\nWhere:\n\n32 x 1 = 32\n\nWe add 1 to our quotient. Finally, combining all the three pieces of the Quotient, we get\u00a00.031, with a Remainder equal to 8.\n\nImages\/mathematical drawings are created with GeoGebra.","date":"2022-09-27 21:21:35","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.7418448328971863, \"perplexity\": 395.6346289374324}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-40\/segments\/1664030335058.80\/warc\/CC-MAIN-20220927194248-20220927224248-00629.warc.gz\"}"} | null | null |
Q: Disable Wifi when battery voltage is 3.4 For my device I want disable wifi when battery voltage level comes down to 3.4.
How to know the battery voltage is 3.4 programmatically?
A: You have to get the ACTION_BATTERY_CHANGED sticky intent and read the voltage from there:
IntentFilter ifilter = new IntentFilter(Intent.ACTION_BATTERY_CHANGED);
Intent batteryStatus = getContext().registerReceiver(null, ifilter);
int voltage = batteryStatus.getIntExtra(BatteryManager.EXTRA_VOLTAGE, -1);
if(voltage>0 && voltage<3400){
//disable wifi
}
| {
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\section{Introduction}
State-of-the-art crowd counting algorithms~\cite{Zhang15c,Zhang16s,Onoro16,Sam17,Xiong17,Sindagi17,Shen18,Liu18b,Li18f,Sam18,Shi18,Liu18c,Idrees18,Ranjan18,Cao18} rely on Deep Networks to regress a crowd density, which is then integrated to estimate the number of people in the image. Their application can have important societal consequences, for example when they are used to assess how many people attended a demonstration or a political event.
In the ``Fake News" era, it is therefore to be feared that hackers might launch adversarial attacks to bias the output of these models for political gain. While such attacks have been well studied for classification networks~\cite{Goodfellow2014a,Kurakin16,Madry18,Moosavi16,Moosavi17,Carlini17}, they remain largely unexplored territory for people counting and even for regression at large. The only related approach we know of~\cite{Ranjan19a} is very recent and specific to attacking optical flow networks, leaving the pixel-wise detection of attacks untouched.
In this paper, our goal is to blaze a trail in that direction. Our main insight is that if a two-stream network is trained to regress both the people density and the scene depth, it becomes very difficult to affect one without affecting the other. In other words, pixels that have been modified to alter the density estimate will also produce incorrect depths, which can be detected by estimating depth using unrelated means. As we will show, this can be done using a depth sensor, simple knowledge about the scene geometry, or even an unrelated deep model pre-trained to estimate depth from images. In other words, our approach works best for a static camera for which the depth of the scene it surveys can be accurately computed but remains applicable to mobile cameras that make such computation more difficult. This assumes that such reference depth maps can be kept safe from the attacker. When this cannot be done, we will show that using statistics from depth maps acquired earlier suffices to detect tampering at a later date.
\input{fig/deepsig.tex}
Fig.~\ref{fig:deepsig} depicts our approach, which we will refer to as the Density and Depth (DaD) model. We will show that it is robust to both unexposed attacks, in which the adversary does not know the existence of our adversarial detector, and exposed attacks, in which the adversary can access not only the density regression network but also our adversarial detector. In other words, even when our approach is exposed, a hacker cannot mount an effective attack while avoiding detection. This is largely because,
depth measurements in a crowded scene, are affected by the appearance and disappearance of people but the perturbations remain small whereas those of RGB pixels can be much larger due to illumination and appearance changes. Therefore, even if the attacker has access to the depth maps, it remains difficult to guarantee that they will be altered in a consistent and undetectable way.
Our contribution therefore is an effective approach to foiling adversarial attacks on people counting algorithms that rely on deep learning. Its principle is very generic and could be naturally extended to other image regression tasks. Our experiments on several benchmark datasets demonstrate that it outperforms heuristic-based and uncertainty-based attack detection strategies.
\section{Conclusion and Future Perspectives}
In this paper, we have shown that estimating density and depth jointly in a two-stream network could be leveraged to detect adversarial attacks against crowd-counting models at pixel level. Our experiments have demonstrated this to be the case even when the attacker knows our detection strategy, or has access to the reference depth map. In essence, our approach is an instance of a broader idea: One can leverage multi-task learning to detect adversarial attacks. In the future, we will therefore study the use of this approach for other tasks, such as depth estimation, optical flow estimation, and semantic segmentation.
\section{Related Work}
\label{sec:related}
\vspace{0.2cm}
\parag{Crowd Counting.}
Early crowd counting methods~\cite{Wu05,Wang11a,Lin10} tended to rely on {\it counting-by-detection}, that is, explicitly detecting individual heads or bodies and then counting them. Unfortunately, in very crowded scenes, occlusions make detection difficult and these approaches have been largely displaced by {\it counting-by-density-estimation} ones~\cite{Zhang15c,Xiong17,Liu18c,Sindagi17,Xu19a,Onoro16,Shen18,Zhang16s,Sam17,Cao18,Wang19a,Liu19c,Liu19e,Shi19a,Liu19a,Liu19b}. They rely on training a regressor to estimate people density in various parts of the image and then integrating. This trend began in~\cite{Chan09,Lempitsky10,Fiaschi12}, using either Gaussian Process or Random Forests regressors. Even though approaches relying on low-level features~\cite{Chen12f,Chan08,Brostow06,Rabaud06,Chan09,Idrees13} can yield good results, they have now mostly been superseded by CNN-based methods~\cite{Jiang19a,Zhao19a,Zhang19a,Wan19a,Lian19a,Liu19d,Yan19a,Ma19a,Liu19f,Xiong19a,Xu19a,Shi19b,Sindagi19a,Cheng19a,Wan19b,Zhang19b,Zhang19c}, a survey of which can be found in~\cite{Sindagi17}. In this paper, we therefore focus on attacks against these.
\parag{Defense against Adversarial Attacks.}
Deep networks trained to solve classification problems are vulnerable to adversarial attacks~\cite{Goodfellow2014a,Kurakin16,Madry18,Moosavi16,Moosavi17,Carlini17}. Existing attack strategies can be roughly categorized as optimization-based or gradient-based. The former~\cite{Moosavi16,Moosavi17,Carlini17} involve terms related to the class probabilities, which makes the latter~\cite{Goodfellow2014a,Kurakin16,Madry18} better candidates for attacks against deep regression models. The very recent work of~\cite{Ranjan19a} is the only one we know of that examines adversarial attacks against a regressor, specifically one that estimates optical flow. However, it does not propose defense mechanisms, which are the focus of this paper.
In the context of classification, one popular defense is adversarial training~\cite{Szegedy13}, which augments the training data with adversarial examples and has been shown to outperform many competing methods~\cite{Grosse17,Gong17a,Metzen17,Bhagoji17a,Li17b,Feinman17a}. However, it needs access to adversarial examples, which are often not available ahead of time and must be generated during training. As a consequence, several alternative approaches have been proposed. This includes training auxiliary classifiers, ranging from simple linear regressors to complex neural networks, to predict whether a sample is adversarial or not. However, as shown in~\cite{Carlini17b}, such detection mechanisms can easily be defeated by an adversary targeting them directly.
In any event, none of these methods are designed to detect attacks at the pixel-level. Even the few researchers who have studied adversarial attacks for semantic segmentation~\cite{Xiao18,Lis19}, which is a pixel-level prediction task, do not go beyond detection at the global image level.
A seemingly natural approach to performing pixel-level attack detection would be to rely on prediction uncertainty. In~\cite{Feinman17a}, the authors argue that Bayesian uncertainty~\cite{Gal16} is a reliable way to detect the adversarial examples because the perturbed pixels generally have much higher uncertainty values. Uncertainty can be computed using dropout, as in~\cite{Gal16}, learned from data~\cite{Kendall17}, or estimated using the negative log-likelihood of each prediction~\cite{Lakshminarayanan17}. In our experiments, we will extend this strategy to pixel-wise adversarial attack detection and show that our approach significantly outperforms it.
\section{Density and Depth Model}
\label{sec:model}
As discussed in Section~\ref{sec:related}, most state-of-the-art crowd counting algorithms rely on a deep network regressor $\mathcal{F}_{d}$, that takes an image $\mathbf{I}$ as input and returns $D_{\mathbf{I}}^{est} = \mathcal{F}_{d}(\mathbf{I},\Theta)$, an estimated density map, which should be as close as possible to a ground-truth one $D_{\mathbf{I}}^{gt}$ in $L^{2}$ norm terms. Here, $\Theta$ stands for the network's weights, that have been optimized for this purpose.
An adversarial attack then involves generating a perturbation $\mathbf{P} = \mathcal{F}_{p}(\mathbf{I},D_{\mathbf{I}}^{c})$, where $\mathcal{F}_{p}$ maps the input image $\mathbf{I}$ and the density $D_{\mathbf{I}}^{c}$ associated with the clean image to $\mathbf{P}$ in such a way that $\mathbf{A} = \mathbf{I} + \mathbf{P}$ is visually indistinguishable from $\mathbf{I}$ while yielding a crowd density estimate $\mathcal{F}_{d}(\mathbf{A},\Theta)$ that is as different as possible from the prediction obtained by the clean image $\mathbf{I}$.
We will review the best known ways to generate such attacks in Section~\ref{sec:attacks}. Here, our concern is to define $\mathcal{F}_{d}$ so as to defeat them by ensuring that they are easily detected. To this end, we leverage an auxiliary task, depth estimation, as discussed below.
\parag{Network Architecture.}
\input{fig/model.tex}
Instead of training a single regressor that predicts only people density, we train a two-stream network where one stream predicts people density and the other depth at each pixel. We write
\begin{align}
D_{\mathbf{I}}^{est} &= \mathcal{F}_{d}(\mathbf{I},\Theta) \; , \label{eq:regress} \\
Z_{\mathbf{I}}^{est} &= \mathcal{F}_{z}(\mathbf{I},\Theta) \;. \nonumber
\end{align}
where $D_{\mathbf{I}}^{est}$ and $Z_{\mathbf{I}}^{est}$ are the estimated densities and depths while $\mathcal{F}_{d}$ and $\mathcal{F}_{z}$ are two regressors parameterized by the weigths $\Theta$. The network that implements $\mathcal{F}_{d}$ and $\mathcal{F}_{z}$ comprises a single encoder and two decoders, one that produces the densities and the other the depths. It is depicted by Fig.~\ref{fig:model}, and we provide details of its architecture in the supplementary material. Note that some of the weights in $\Theta$ are specific to the first decoder and others to the second.
As the two decoders use the same set of features as input, it is difficult to tamper with the results of one without affecting that of the other, as we will see in Section~\ref{sec:compare}. More specifically, if a pixel is perturbed to change the local density estimate, the local depth estimate is likely to be affected as well. We therefore take the relative error in depth estimation with respect to a reference depth map $\mathbf{Z}^{ref}$
\begin{equation}
\frac{|\mathbf{Z}^{est}(j)-\mathbf{Z}^{ref}(j)|}{\mathbf{Z}^{ref}(j)}
\label{eq:indicator}
\end{equation}
to be an indicator of a potential disturbance. As will be shown in our experiments, the reference depth map can be either the ground-truth one, a rough estimate obtained using knowledge of the scene geometry, or the output of another monocular depth estimation network. In practice, we label a pixel as potentially tampered with if this difference is larger than 5\% of the largest difference in the training dataset and we will justify this choice in Section~\ref{sec:sensitivity}. In test sequences for which the reference depth map can also be tampered with by the attacker, we can use the statistics of the training depth maps to also detect such tampering, as will be shown in Section~\ref{sec:robust}.
\parag{Network Training.}
Given a set of $N$ training images $\{\mathbf{I}_{i}\}_{1 \leq i \leq N}$ with corresponding ground-truth density maps $\{\mathbf{D}_{i}^{gt}\}_{1 \leq i \leq N}$ and ground-truth depth maps $\{\mathbf{Z}_{i}^{gt}\}_{1 \leq i \leq N}$, we learn the weights $\Theta$ of the two regressors by minimizing the loss
\begin{small}
\begin{eqnarray}
L(\Theta) &=& L_d+\lambda \cdot L_z \;, \\ \nonumber
L_d &=& \frac{1}{2B}\sum_{i=1}^{B}\|\mathbf{D}^{gt}_{i}-\mathbf{D}^{est}_{\mathbf{I}_i}\|^{2}_{2} \;, \\ \nonumber
L_z &=& \frac{1}{2B}\sum_{i=1}^{B}\|\mathbf{Z}^{gt}_{i}-\mathbf{Z}^{est}_{\mathbf{I}_i}\|^{2}_{2}\;.
\label{eq:loss}
\end{eqnarray}
\end{small}
where $B$ is the batch size and $\lambda$ is a hyper-parameter that balances the contributions of the two losses. We found empirically that $\lambda=0.01$ yields the best overall performance, as will be shown in Section~\ref{sec:compare}.
To obtain the ground-truth density maps $D^{gt}_{i}$, we rely on the same strategy as previous work~\cite{Li18f,Sam17,Zhang16s,Sam18,Liu18a}. In each training image $\mathbf{I}_{i}$, we annotate a set of $c_{i}$ 2D points $O_{i}^{gt} = {\{O_{i}^j} \}_{1 \leq j \leq c_i}$ that denote the position of each human head in the scene. The corresponding ground-truth density map $\mathbf{D}_{i}^{gt}$ is obtained by convolving an image containing ones at these locations and zeroes elsewhere with a Gaussian kernel of mean $\mu$ and variance $\sigma$.
\section{Experiments}
\label{sec:results}
In this section, we first introduce the existing adversarial attack methods that can be used against a deep regressor and describe the evaluation metric and the benchmark datasets we used to assess their performance. We then use them against our approach to demonstrate its robustness, and conclude with an ablation study that demonstrates that our approach is robust to the hyper-parameter setting and works well when used in conjunction with several recent crowd density regressors~\cite{Zhang16s,Li18f,Liu19a} .
\subsection{Attacking a Deep Regressor}
\label{sec:attacks}
While there are many adversarial attackers~\cite{Goodfellow2014a,Kurakin16,Madry18,Moosavi16,Moosavi17,Carlini17}, their effectiveness has been proven mostly against classifiers but far more rarely against regressors~\cite{Ranjan19a}. As discussed in Section~\ref{sec:related}, the gradient-based methods~\cite{Goodfellow2014a,Kurakin16,Madry18} are the most suitable ones to attack regressors, and we focus on the so-called Fast Gradient Sign Methods ({\bf FGSM}s), which are the most successful and widely used ones. We will distinguish between unexposed attacks in which the attacker does not know that we use depth for verification purposes and exposed attacks in which they do.
\parag{Unexposed Attacks.} If the attacker is unaware that we use the depth map for verification purposes, they will only try to affect the density map. They might then use one of the following variants of {\bf FGSM}.
\vspace{0.1cm}
\sparag{\noindent (1) Untargeted FGSM (\FGSMU{n})~\cite{Goodfellow2014a,Kurakin16}.} It generates adversarial examples designed to increase the network loss as much as possible from that of the prediction obtained from the clean image, thereby preventing the network from predicting it. Given an input image $\mathbf{I}$, the density predicted from the clean image $\mathbf{D}$, and the regressor $\mathcal{F}_{d}$ of Eq.~\ref{eq:regress} parametrized by $\Theta$, the attack is performed by iterating
\begin{small}
\begin{align}
\mathbf{I}_{0}^{adv} & = \mathbf{I} \; , \label{eq:FGSMu} \\
\mathbf{I}_{i+1}^{adv} & = clip(\mathbf{I}_{i}^{adv} + \alpha \cdot sign(\nabla_{\mathbf{I}}L_d(\mathcal{F}_{d}(\mathbf{I}_{i}^{adv};\Theta),\mathbf{D})),\epsilon) \; . \nonumber
\end{align}
\end{small}
$n$ times. The adversarial example is then taken to be $\mathbf{I}_{n}^{adv}$, and we will refer to this as \FGSMU{n}. It is a single-step or multiple-step attack without target and $clip$ guarantees that the resulting perturbation is bounded by $\epsilon$.
For consistency with earlier work~\cite{Goodfellow2014a,Kurakin16}, when $n=1$, we reformulate this attack as
\begin{small}
\begin{align}
\mathbf{I}^{adv} & = \mathbf{I} + \epsilon \cdot sign(\nabla_{\mathbf{I}}L_d(\mathcal{F}_{d}(\mathbf{I};\Theta),\mathbf{D})). \label{eq:FGSMu1}
\end{align}
\end{small}
Unless otherwise specified we use $\epsilon=15$, $\alpha=1$, and $n=19$, as recommended in earlier work~\cite{Kurakin16}. These numbers are chosen to substantially increase the crowd counting error while keeping the perturbation almost imperceptible to the human eye. We will analyze the sensitivity of our approach to these values in Section~\ref{sec:sensitivity}. An example of this attack is shown in Fig.~\ref{fig:density}. By comparing Fig.~\ref{fig:density}(d) and (e), we can see that the attack made some people ``disappear''.
\input{fig/density.tex}
\vspace{0.1cm}
\sparag{\noindent (2) Targeted FGSM (\FGSMT{n})~\cite{Kurakin16}.} Instead of simply preventing the network from finding the right answer $\mathbf{D}$, we can target a specific wrong answer $\mathbf{D}_{t}$. This is achieved using the slightly modified iterative scheme
\begin{small}
\begin{align}
\mathbf{I}_{0}^{adv} & = \mathbf{I} \; , \label{eq:FGSMt}\\
\mathbf{I}_{i+1}^{adv} & = clip(\mathbf{I}_{i}^{adv} - \alpha \cdot sign(\nabla_{\mathbf{I}}L_d(\mathcal{F}_{d}(\mathbf{I}_{i}^{adv};\Theta),\mathbf{D}_{t})),\epsilon). \nonumber
\end{align}
\end{small}
Again, we take the adversarial example to be $\mathbf{I}_{n}^{adv}$ and use the same values as before for $\epsilon$ and $\alpha$. We will refer to this as \FGSMT{n}. In our experiments, we take the targets to be the original density values plus one, which creates an obvious error while yielding tampered images that are undistinguishable from the original ones.
\parag{Exposed Attacks.} If the attacker knows that we are using the depth maps and has access to both $\mathcal{F}_{d}$ and $\mathcal{F}_{z}$, the two regressors of Eq.~\ref{eq:regress}, their natural reaction will be to try to modify the density maps while leaving the depth maps as unchanged as possible. To this end, we propose the following {\it exposed} variations of the untargeted and targeted {\bf FGSM} attacks described above.
\sparag{\noindent (1) Untargeted Exposed FGSM (\FGSMUE{n}).} The iterative scheme becomes
\begin{small}
\begin{align}
\mathbf{I}_{0}^{adv} & = \mathbf{I} \; , \label{eq:FGSMue}\\
L_{i}^{all} & = L_d(\mathcal{F}_{d}(\mathbf{I}_{i}^{adv};\Theta),\mathbf{D})- \lambda \cdot L_z(\mathcal{F}_{z}(\mathbf{I}_{i}^{adv};\Theta),\mathbf{z}) \; , \nonumber \\
\mathbf{I}_{i+1}^{adv} & = clip(\mathbf{I}_{i}^{adv} + \alpha \cdot sign(\nabla_{\mathbf{I}}L_{i}^{all}),\epsilon) \; . \nonumber
\end{align}
\end{small}
where $\mathbf{z}$ is the depth map associated with clean image. When $n=1$, we reformulate the final line of Eq.~\ref{eq:FGSMue} as in Eq.~\ref{eq:FGSMu1} for consistency with earlier work. The additional term in the loss function aims to preserve the predicting power of $\mathcal{F}_{z}$ while compromising that of $\mathcal{F}_{d}$ as much as possible. We again use the same values as before for $\epsilon$ and $\alpha$, and $\lambda=0.01$ is the same balancing factor as in Eq.~\ref{eq:loss}.
\sparag{\noindent (2) Targeted Exposed FGSM (\FGSMTE{n}).} Similarly, the targeted attack iterative scheme becomes
\begin{small}
\begin{align}
\mathbf{I}_{0}^{adv} & = \mathbf{I} \; \label{eq:FGSMut}\\
L_{i}^{all} & = L_d(\mathcal{F}_{d}(\mathbf{I}_{i}^{adv};\Theta),\mathbf{D}_{t})+ \lambda \cdot L_z(\mathcal{F}_{z}(\mathbf{I}_{i}^{adv};\Theta),\mathbf{z}) \; , \nonumber \\
\mathbf{I}_{i+1}^{adv} & = clip(\mathbf{I}_{i}^{adv} - \alpha \cdot sign(\nabla_{\mathbf{I}}L_{i}^{all}),\epsilon) \; . \nonumber
\end{align}
\end{small}
When $n=1$, we again reformulate the final line of Eq.~\ref{eq:FGSMut} as in Eq.~\ref{eq:FGSMu1}.
\subsection{Evaluation Datasets}
We use three different datasets to evaluate our approach. The first two are RGB-D datasets with ground-truth depth values obtained from sensors. Since depth sensors may not always be available, we also evaluate our model on a third dataset that contains RGB images with accurate {\it perspective maps}, that is, maps inferred from the scene geometry without using depth sensors. Such maps only represent the scene to the exclusion of the people in it. This will let us show that
our approach not only works for RGB-D datasets but also achieves remarkable performance in RGB images if scene geometry is available. Furthermore, we will show that our approach also applies when the reference depth is obtained using a separate monocular depth estimation network.
\parag{ShanghaiTechRGBD~\cite{Lian19a}.} This is a large-scale RGB-D dataset with 2,193 images and 144,512 annotated heads. The valid depth ranges from 0 to 20 meters due to the limitation in depth sensors. The lighting condition ranges from very bright to very dark in different scenarios. We use the same setting as in~\cite{Lian19a}, with 1,193 images as training set and the remaining ones as test set, and normalize the depth values from [0,20] to [0,1] for both training and evaluation.
\parag{MICC~\cite{Bondi14a}.} This dataset was acquired by a fixed indoor surveillance camera. It is divided into three video sequences, named FLOW, QUEUE and GROUPS. The crowd motion varies from sequence to sequence. In the FLOW sequence, people walk from point to point. In the QUEUE sequence,
people walk in a line. In the GROUPS sequence, people move inside a controlled area. There are 1,260 frames in the FLOW sequence with 3,542 heads. The QUEUE sequence contains 5,031 heads in 918 frames, and the GROUPS sequence encompasses 1,180 frames with 9,057 heads.
We follow the same setting as in~\cite{Lian19a}, taking 20\% of the images of each scene as training set and using the remaining ones as test set.
\parag{Venice~\cite{Liu19a}.} The above two RGB-D datasets contain depth information acquired by sensors. Such information is hard to obtain in outdoor environments, particularly if the scene is far from the camera. Therefore, we also evaluate our approach on the \textit{Venice} dataset. This dataset contains RGB images and an accurate perspective map of each scene. It was inferred from the grid-like ground pattern that is provided in the supplementary material and without using a depth sensor. The dataset contains 4 different sequences for a total of 167 annotated frames with fixed 1,280 $\times$ 720 resolution. Our experimental setting follows that of~\cite{Liu19a,Liu19b}, with 80 images from a single long sequence as training data, and the images from the remaining 3 sequences for testing purposes.
\subsection{Metrics and Baselines}
In all our experiments, we partition the images into four parts and tamper with one while leaving the other three untouched. We then measure:
\begin{itemize}
\item How well we can detect the pixels that have been tampered with. We measure this in terms of the mean Intersection over Union
%
\begin{equation}
mIoU = \frac{1}{N} \sum_{i=1}^{N} \frac{ |v_{i} \cap \hat{v}_{i}|}{ | v_{i} \cup \hat{v}_{i} |}
\end{equation}
where $N$ is the number of images, $\hat{v}_{i}$ is 1 for the pixels in image $\mathbf{I}_i$ predicted to have been tampered with according to Eq.~\ref{eq:indicator} and $v_{i}$ is the ground-truth perturbation mask.
\item How well the modifications of the depth map correlate with those of the predicted density. As in many previous works~\cite{Zhang16s,Zhang15c,Onoro16,Sam17,Xiong17,Sindagi17,Liu19a,Liu19b}, we quantify these modifications in terms of the mean absolute error for densities and depths along with the root mean squared error for density. They are defined as
\begin{align}
DMAE & = \frac{1}{N}\sum_{i=1}^{N}|d_{i}-\hat{d_{i}}|\; , \nonumber \\
ZMAE & = \frac{1}{N}\sum_{i=1}^{N}\frac{ \sum_{j=1}^{M_{i}} |z_{i}^{j}-\hat{z_{i}^{j}}|}{M_{i}} \;, \\
RMSE & = \sqrt{\frac{1}{N}\sum_{i=1}^{N}(d_{i}-\hat{d_{i}})^{2}}\; . \nonumber
\end{align}
where $N$ is the number of test images, $d_{i}$ and $z_{i}^{j}$ denote the true number of people in the $i$th image and depth value at pixel $j$ of the $i$th image, and $\hat{d_{i}}$ and $\hat{z_{i}^{j}}$ are the estimated values. $M_{i}$ is the number of tampered pixels in the $i$th image. In practice $\hat{d_{i}}$ is obtained by integrating the predicted people densities.
\end{itemize}
In the absence of prior art on defenses against attacks of density estimation algorithms, we use the following baselines for comparison purposes.
\begin{itemize}
\item \rndh{} and \rndq{}. We randomly label either half or a quarter of the pixels as being under attack, given that we know {\it a priori} that exactly a quarter are. We introduced \rndh{} to show that using a random rate other than the true one does not help.
\item \uncertain{}. Since an adversarial attack is caused by modifying the input image, it can be seen as heteroscedastic aleatoric uncertainty~\cite{Kendall17}, which assumes that observation noise varies with the input, that is, uncertainty caused only by the input and invariant to the model. We threshold the uncertainty values to classify each pixel as perturbed or not and report the results obtained with the best threshold.
\item \simple{}. We use the approach of~\cite{Lakshminarayanan17} that relies on a scalable method for estimating predictive uncertainty from deep networks using a scoring rule as training criterion. The optional adversarial training process is not used as we do not know the potential attackers in advance. As before, we threshold the uncertainty values to obtain a pixel-wise classification map and report the best results.
\item \dropout{}. We further compare our model with Bayesian uncertainty~\cite{Gal16}, which uses dropout to
approximate model uncertainty. Again, we threshold the uncertainty value and report the results for the best threshold.
\end{itemize}
The baseline models are trained with the same backbone as our approach.
\subsection{Comparative Performance}
\label{sec:compare}
\paragraph{Using the CAN~\cite{Liu19a} architecture.}
CAN is an encoder-decoder crowd density estimation architecture that delivers excellent performance. We use it to implement $\mathcal{F}_{d}$ and duplicate its decoder to implement $\mathcal{F}_{z}$. Recall from Section~\ref{sec:model} that the hyper-parameter $\lambda$ of Eq.~\ref{eq:loss} balances the people density estimation loss and the depth estimation one while training $\mathcal{F}_{d}$ and $\mathcal{F}_{z}$. In Table~\ref{tab:lambda}, we report the errors of the two regressors as a function of $\lambda$. $\lambda=0.01$ yields the best performance overall, and we use regressors trained using this value in all our other experiments. Interestingly, training $\mathcal{F}_{d}$ and $\mathcal{F}_{z}$ jointly yields a better density regressor than training $\mathcal{F}_{d}$ alone, which is what we do when we set $\lambda$ to zero.
\input{table/lambda.tex}
\input{table/error.tex}
We report the counting and depth errors with/without attack in Table~\ref{tab:error} for the 3 datasets. All the attacks cause large increase in crowd counting errors, which always comes with a substantial increase in depth estimation error. The exposed methods reduce slightly this increase but at the cost of also making the attack less effective.
\input{table/miou_ShanghaiTechRGBD_new.tex}
\input{table/miou_micc_new.tex}
\input{table/miou_venice_new.tex}
In Tables~\ref{tab:miou_ShanghaiTechRGBD}, \ref{tab:miou_micc}, and \ref{tab:miou_venice}, we report the pixel-wise adversarial detection accuracy for the \textbf{ShanghaiTechRGBD}, \textbf{MICC} and \textbf{Venice} datasets. Our approach outperforms all the baseline models by a large margin for all the attacks. In Fig.~\ref{fig:visualization}, we show a qualitative result.
\input{fig/visualization.tex}
\input{fig/ablation_1.tex}
\input{table/ablation_combine.tex}
\input{fig/ablation_2.tex}
\parag{Using the CSRNet~\cite{Li18f} and MCNN~\cite{Zhang16s} architectures.}
To show that the above results are not tied to the CAN architecture, we re-ran our experiments using CSRNet~\cite{Li18f} and MCNN~\cite{Zhang16s}. As shown in Fig.~\ref{fig:backbones}, we obtain similar mIoU scores for all three architectures, with a slight advantage for the more recent CAN.
\subsection{Inference without Depth Map}
\label{sec:pretrain}
Using the scene geometry to infer a depth map, as we did in the \textit{Venice} dataset, is one way to avoid having to use a depth-sensor to create a reference depth map. Unfortunately, the strong geometric patterns we used to do this are not present in all images. In this section, we therefore explore the use of VNL~\cite{Yin19a}, a pre-trained deep model that can estimate depth from single images, to create the reference depth map. We report our results in
Table~\ref{tab:miou_pretrain}. As could be expected, there is a slight performance drop compared to the results we obtained using the ground-truth depth map, given in Tables~\ref{tab:miou_ShanghaiTechRGBD}, \ref{tab:miou_micc}, \ref{tab:miou_venice}. However, we still clearly outperform the baseline models. This shows that our approach does well even in the absence of a ground-truth depth map and is therefore applicable in a wide range of scenarios. In the supplementary material, we will show that our defense mechanism remains robust even when this pre-trained deep model is exposed to the attacker.
\subsection{Tampering with the Ground-Truth Depth Map}
\label{sec:robust}
The results of Sections~\ref{sec:compare} and~\ref{sec:pretrain} were obtained under the assumption that the reference depth map is safe from attack. In some scenarios, this might not be the case and the attacker might be able to tamper with the depth map. Fortunately, even if this were the case, the attack would still be detectable as follows. Given $N$ training depth maps $\{\mathbf{z}_{1},\mathbf{z}_{2},...,\mathbf{z}_{N}\}$ recorded by a fixed depth sensor, we can record the min and max depth values for each pixel $j$ as
\begin{eqnarray}
z_{min}^{j} &=& \min(z_{1}^{j},z_{2}^{j},...,z_{N}^{j})\;, \\ \nonumber
z_{max}^{j} &=& \max(z_{1}^{j},z_{2}^{j},...,z_{N}^{j})\;.
\end{eqnarray}
Given $M$ reference depth maps $\{\dot{\mathbf{z}}_{1},\dot{\mathbf{z}}_{2},...,\dot{\mathbf{z}}_{M}\}$ that are exposed and can be tampered with, the tampered depth value at pixel $j$ of the reference depth map $\dot{\mathbf{z}}_{k}$ can be written as
\begin{eqnarray}
t_{k}^{j} = \dot{z}_{k}^{j}+\beta \mu_z^{j} \; .
\end{eqnarray}
where $\mu_z^{j}$ is the mean depth value in the reference depth maps in the test dataset, and $\beta$ is a scalar that represents the perturbation strength of a potential attack. We take the perturbation $\beta \mu_z^{j}$ to be a function of $\beta$ because the reference depth map is not an input to our network. If the attacker were able to choose an appropriate $\beta$ for each pixel of the reference depth map, the tampering indicator of Eq.~\ref{eq:indicator} could be compromised. Fortunately, such an attack is very likely to be detected using the following simple but effective approach. If $t_{k}^{j} < z_{min}^{j}$ or $t_{k}^{j} > z_{max}^{j}$, we label pixel $j$ as potentially tampered with. As shown in Fig.~\ref{fig:depth_detection}, the pixel-wise detection accuracy is over 90\% even for extremely small perturbations with $\beta=0.01$ and quickly increases from there. This makes the attacker's task difficult indeed.
\subsection{Sensitivity Analysis}
\label{sec:sensitivity}
\input{table/indicator_all.tex}
\input{table/whitebox_error.tex}
We now quantify the influence of the three main hyper-parameters introduced in Section~\ref{sec:attacks} that control the intensity of the attacks.
\parag{Perturbation value.} We change the value of $\epsilon$ in Eq.~\ref{eq:FGSMu} and Eq.~\ref{eq:FGSMu1} from 1.0 to 35.0 for all attacks and plot the resulting mIoU in Fig.~\ref{fig:strength}. Our model can detect very weak attacks with $\epsilon$ down to 1.0 and its performance quickly increases for larger values. In the supplementary material, we will exhibit the monotonous relationship between $\epsilon$ and the people density estimation error. When $\epsilon=1.0$, there is already a small perturbation of the density estimates---around 6 in $DMAE$ for {\bf ShanghaiTechRGBD}---that then become much larger as $\epsilon$ increases. The number of iterations $n$ is set to $min(\epsilon+4,1.25\epsilon)$ as recommended in earlier work~\cite{Kurakin16}.
\parag{Threshold value.} In Table~\ref{tab:indicator_all}, we report mIoU values on each dataset as a function of the threshold we use to classify a pixel as tampered with or not, depending on the ratio of Eq.~\ref{eq:indicator}. 5\% gives the best answer across all attacks.
\parag{Strength of Exposed Attacks.}
To check the robustness of our model against exposed attacks, we evaluate different $\lambda$ values in the loss term of Eq.~\ref{eq:FGSMue}, whose role is to keep the depth estimate as steady as possible in {\bf ShanghaiTechRGBD}. We tested our approach for values of $\lambda$ ranging from 0.01 to 100.0 and report the detection accuracy results along with the crowd counting error and depth error in Table~\ref{tab:whitebox_error}. For larger values $\lambda$, both the crowd density error and the detection rate drop.
In other words, increasing $\lambda$ makes the attack harder to detect but also weaker. We show the same trend in the other datasets in the supplementary material.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 1,785 |
{"url":"https:\/\/www.experts-exchange.com\/questions\/26918511\/Get-last-record-for-each-student-in-database.html","text":"# Get last record for each student in database\n\nHi\n\nI have attached code below thats shows the first record for each student in my database which isn't correct\n\nThis code is outputing students who have missed 28 school days and I need it to show the last class that the student was absent not the last record in the database\n\nIf a register isnt marked the value is null\n\nHope this makes sense\n\nif object_id('tempdb..#absences','U') is not null drop table #absences\n\nselect dense_rank() over (partition by student_id order by week_no) as weekrn\n-- ,dense_rank() over (partition by student_id order by week_no,day_num) as dayrn -- not used\n,isnull(absence_code,'') as absence_ind\n,*\ninto #absences\nfrom nrc_eRegisters_allstudents_testing_tb\n\ncreate index idx_temp_absences_1 on #absences (student_id, weekrn, absence_ind) -- dayrn not used so removed from index\ncreate index idx_temp_absences_2 on #absences (student_id, date, absence_ind)\n\n;with absences_cte as\n( -- this will return the absence \"blocks\" of days - at the moment 4 school weeks using weekrn+4 indicates 4x7 = 28 \"school\" days not 28 calendar days\nselect a.student_id, a.date as start_date, b.date as end_date\nfrom #absences a\nouter apply (select top 1 date from #absences c where c.student_id = a.student_id and c.weekrn = a.weekrn+4 and c.day_num >= a.day_num ) b\nwhere a.absence_ind = 'O'\nand a.date < getdate()\nand b.date is not null\nand not exists (select NULL from #absences c where c.student_id = a.student_id and c.date between a.date and b.date and c.absence_ind <> 'O')\ngroup by a.student_id, a.date, b.date\n)\nSelect distinct ab.*\nfrom absences_cte c\nouter apply (select top 1 * from #absences na where na.student_id = c.student_id and na.date < c.start_date and na.absence_ind <> 'O' order by student_id, date) ab\n\nLVL 3\n###### Who is Participating?\n\nx\nI wear a lot of hats...\n\n\"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform.\" -Todd S.\n\nHi Lisa,\n\nhw are you??\n\nIt might helps.\n\n- Bhavesh\nselect c.*\nfrom absences_cte c, (select student_id, max(end_date)endDate\nfrom absences_cte\ngroup by student_id\n)d\nwhere c.student_id = D.Student_ID\nAnd c.end_date = d.end_date\n\n0\nAuthor Commented:\nhi brichsoft\n\nIm good how are you?\n\nI replaced line 25, 26, 27 with the code you supplied above and I have one error\n\nMsg 207, Level 16, State 1, Line 32\nInvalid column name 'end_date'.\n0\nhi,\n\ni'm fine....\n\npls check this.\n\nselect c.*\nfrom absences_cte c, (select student_id, max(end_date)endDate\nfrom absences_cte\ngroup by student_id\n)d\nwhere c.student_id = D.Student_ID\nAnd c.end_date = d.enddate\n0\nAuthor Commented:\nhi again\n\nTesting this for one student and I am getting\n\nstudent id \u00a0start_date end_date\n\nthis dates occur halfway through year and neither of them is the date of the last class that the student was absent\n0\nAuthor Commented:\noh sorry apologies\n\nthe end date is right\n0\nAuthor Commented:\ni just need the information for that student out as well\n\neg what module they missed, start_time, end time etc\n0\n*laughing* And I thought you wanted the last day present before all those absences :)\n\n(thats what the code does now)\n\nSo, to get the last day absent... that is something quite a bit different because there could be \"groups\" of absences, and assume you want the last one absent per group ?\n\ne.g.\n\nabsent weeks 4-8 = report last absent in week 8\nattended weeks 9-12\nabsent weeks 13-17 = report last absent in week 17\n\nis that correct ?\n\n0\nGreat Expert entered.. =)\n\nI'm out now....\n0\nHey Brichsoft - come back you chicken *laughing*\n0\nAuthor Commented:\nhi again mark\n\ndo you know something I am staring at this stuff for so long I have completly lost focus\n\nI do have to do the last class that they were present for but the code above only shows me the very fist class that that student had\n0\n\n0\nAuthor Commented:\neg the date that I am geting back is\n\n08\/09\/2010 which is the first class that student has\n0\n@lisa_mc\n\nYep the code at the top in your question is meant to be returning the last class ATTENDED prior to the absence\n\nas per \"would there be any way to show the last class each student attended \"\u00a0in : http:\/\/www.experts-exchange.com\/Microsoft\/Development\/MS-SQL-Server\/SQL-Server-2005\/Q_26884477.html#a35233763\n\nWill need to debug the code to make sure it is the last class present, rather than the first class. Reckon it will be the ORDER BY and should be date desc (not asc) but we probably have to add in start time and end time as well to that order by.\n\nGimme a moment...\n\n@brichsoft:\n*laughing* welcome back - where you been ?\n0\nAuthor Commented:\nok np thanks\n\n0\nOK, think it is the Order By...\n\nif object_id('tempdb..#absences','U') is not null drop table #absences\n\nselect dense_rank() over (partition by student_id order by week_no) as weekrn\n-- ,dense_rank() over (partition by student_id order by week_no,day_num) as dayrn -- not used\n,isnull(absence_code,'') as absence_ind\n,*\ninto #absences\nfrom nrc_eRegisters_allstudents_testing_tb\n\ncreate index idx_temp_absences_1 on #absences (student_id, weekrn, absence_ind) -- dayrn not used so removed from index\ncreate index idx_temp_absences_2 on #absences (student_id, date, absence_ind)\n\n;with absences_cte as\n( -- this will return the absence \"blocks\" of days - at the moment 4 school weeks using weekrn+4 indicates 4x7 = 28 \"school\" days not 28 calendar days\nselect a.student_id, a.date as start_date, b.date as end_date\nfrom #absences a\nouter apply (select top 1 date from #absences c where c.student_id = a.student_id and c.weekrn = a.weekrn+4 and c.day_num >= a.day_num ) b\nwhere a.absence_ind = 'O'\nand a.date < getdate()\nand b.date is not null\nand not exists (select NULL from #absences c where c.student_id = a.student_id and c.date between a.date and b.date and c.absence_ind <> 'O')\ngroup by a.student_id, a.date, b.date\n)\nSelect distinct ab.*\nfrom absences_cte c\nouter apply (select top 1 * from #absences na where na.student_id = c.student_id and na.date < c.start_date and na.absence_ind <> 'O' order by student_id, date desc, end_time desc, start_time desc) ab\n\n0\nAuthor Commented:\nim testing on e particular student who has been absent 2 periods of four weeks\n\nfirst absence is 15\/9\/2010 - 19\/11\/2010\nsecond absence is 07\/01\/2011 - 11\/03\/2011\n\nthe to records I am getting outputted for this student is 10\/09\/2010 and which seems to be the first date the student was absent\n\nand 17\/12\/2011 which is the date before the second group of four weeks absent\n\ndo you want me to send you this particular student would that make it easier\n0\n0\nAuthor Commented:\nsample student attached\ntest-student.xls\n0\nAuthor Commented:\nive been checking students who have only missed one block of 4 weeks and it seems to be working fine\n\nchecking a few more now\n0\n\nNot for Lisa_mc \u00a0 (secret expert only business)\n\n@brichsoft : If you wanna play with the spreadsheets, all I have been doing is importing the spreadsheet into a table and running the above code. e.g.\n\n-- definitely NOT for lisa\n\ndrop table nrc_eRegisters_allstudents_testing_tb\nSelect * into nrc_eRegisters_allstudents_testing_tb FROM OPENROWSET('Microsoft.ACE.OLEDB.12.0', 'Excel 12.0;Database=c:\\ee\\test-student.xls;HDR=Yes;imex=1', 'SELECT * FROM [sheet1\\$]') as a\n\n0\nOK, the challenge with the example student is the classes before the first absence have NULL as the absence_code\n\nAll I was checking was not \"O\", so we need to make sure that it is one of '\/', 'P','C' etc... Or, more simply put, checking the original absence_code for isnull(na.absence_code,'O') <>\u00a0'O'\n\nAnd (feeling slightly foolish) when I was importing the test data, the NULL value in the absence_code was importing as a four character string, not as NULL\n\nSo, with that in mind :) and a new check to eliminate NULL selection (which we need to do because of the outer apply anyway) then :\n\nif object_id('tempdb..#absences','U') is not null drop table #absences\n\nselect dense_rank() over (partition by student_id order by week_no) as weekrn\n-- ,dense_rank() over (partition by student_id order by week_no,day_num) as dayrn -- not used\n,isnull(absence_code,'') as absence_ind\n,*\ninto #absences\nfrom nrc_eRegisters_allstudents_testing_tb\n\ncreate index idx_temp_absences_1 on #absences (student_id, weekrn, absence_ind) -- dayrn not used so removed from index\ncreate index idx_temp_absences_2 on #absences (student_id, date, absence_ind)\n\n;with absences_cte as\n( -- this will return the absence \"blocks\" of days - at the moment 4 school weeks using weekrn+4 indicates 4x7 = 28 \"school\" days not 28 calendar days\nselect a.student_id, a.date as start_date, b.date as end_date\nfrom #absences a\nouter apply (select top 1 date from #absences c where c.student_id = a.student_id and c.weekrn = a.weekrn+4 and c.day_num >= a.day_num ) b\nwhere a.absence_ind = 'O'\nand a.date < getdate()\nand b.date is not null\nand not exists (select NULL from #absences c where c.student_id = a.student_id and c.date between a.date and b.date and c.absence_ind <> 'O')\ngroup by a.student_id, a.date, b.date\n)\nSelect distinct ab.*\nfrom absences_cte c\nouter apply (select top 1 * from #absences na where na.student_id = c.student_id and na.date < c.start_date and na.absence_code <> 'O' order by student_id, date desc, end_time desc, start_time desc) ab\nwhere ab.absence_code is not NULL\n\n0\n\nExperts Exchange Solution brought to you by\n\nFacing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.\n\nAuthor Commented:\nok thats great thats coming back with the date 17\/12\/2010 which is correct\n\ngoing to test a few more\n\none more thing I need to put a counter in there to say how many days they have been absent\n\neg for this particular student last class attended was 17\/12\/2010 but from their first absence on 07\/01\/2011 until the 11\/03\/2011 the period which cause them to breach 4 weeks absent they have missed 17 days\n\nmaybe I should post this as another question :-)\n\n0\nAuthor Commented:\nill post another question in a min\n\none more question mark\n\nsay I wanted to check students who have breached 3 weeks would I change the code\n\nweekrn + 4 to weekrn + 3\n0\nYep... So long as you are counting \"school weeks\" and not calendar periods then simply change the \"weekrn + 4\"\n\nOh, and you dont have to keep posting new questions.\n\nJust put all the details you need as output.\n0\nAuthor Commented:\ni have posted another question about the counter and would be very grateful for your help again :-)\n\n0\nAuthor Commented:\nsorry im posting another question as I think you deserve more than 500 points as you have been very good helping me\n\nif I wanted to say like above instead of 4 weeks off i now wanted >=3 weeks off but less than 4 weeks off is that possible\n0\nOK, I wont complain, but the points are only a small part of why I do this. Mainly interested in helping and getting the hard questions, so I dont pop up all the time, just for the good times :)\n\nNot so sure about a \"range\" because of the day element.\n\nDoing >= weekrn + 3 is easy enough, but doing >= 3 weeks and <\u00a04 weeks means we need to cut off the 4th week at the end of the (prior) day in that 4th week.\n\nSo if our first day absent was Week 3 Tue \u00a0then we want to extract everything up to Week 6 Tue or Week 7 Mon \u00a0 (being less than 4 weeks). Which we can do, but that \"where\" statement changes shape quite a bit when checking day_num. And then we will need a slightly different \"exists\" to accommodate an acceptable range, because we might have an attended day anywhere within the acceptable range.\n\nHmmmm.... would have to think about the \"range\" a fair bit me thinks...\n0\nAuthor Commented:\nwell I suppossed thats something to think about\n\nbelieve me you deserve the points as every part of that question was hard\n\nwould you be able to help me with my other question if poss\n\nID: 26918849\n\nthanks\n0\nAuthor Commented:\nsorry bit confused\n\n>=weekrn+3 - does this mean every student who has missed 3 or more weeks\nso if a student has missed 4 weeks they would show here\n\nthen what does weekrn+3 mean will this not also show people who have missed 3 or more weeks\n\n0\nOK, now posting in that other thread.\n\nHave we finished in this one ?\n0\nAuthor Commented:\nyeah just want to know what does weekrn+3\n\ndoess this mean 3 weeks and more or exactly 3 weeks\n\nthanks\n0\nOoops, didnt see the above question.\n\n>=weekrn+3 - does this mean every student who has missed 3 or more weeks\nso if a student has missed 4 weeks they would show here\n\nYes thats what I said, but thinking it is flawed. I wouldnt be doing this. More likely trying to show everything (which is wrong). That whole \"range\" thing requires a fair bit of thinking through.\n\nthen what does weekrn+3 mean will this not also show people who have missed 3 or more weeks\n\nYes that is correct. it will show at least 3 weeks (and it becomes a bit like a rolling period of 3 weeks)\n\n0\njust to clarify\n\ndoess this mean 3 weeks and more or exactly 3 weeks\n\nit means at least 3 weeks and could be more e.g.\n\nwk1 'O' \u00a0shows\nwk2 'O' \u00a0shows\nwk3 'O' \u00a0shows\nwk4 '\/'\nwk5 '\/'\nwk6 'O' \u00a0shows\nwk7 'O' \u00a0shows\nwk8 'O' \u00a0shows\nwk9 'O' \u00a0shows\nwk10 '\/'\nwk11 '\/'\n\n0\nAuthor Commented:\nthats just what i thought thanks again\n0","date":"2018-10-23 07:32:59","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.22793757915496826, \"perplexity\": 8733.831760175846}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2018-43\/segments\/1539583516117.80\/warc\/CC-MAIN-20181023065305-20181023090805-00062.warc.gz\"}"} | null | null |
Autism | Raising Children | Raising Special Needs Children | Raising Toddlers
Raising Autistic Twins: How I Became The Warrior Mommy
ByAmbreen Suhaib April 17, 2022 October 31, 2022
The first year of my life after becoming a mom to twin boys was beautiful, and to be honest very tiring as well. It was and is a lot of work raising autistic twins and managing them by myself since I was living in the UAE, in a home away from home and family.
The first symptoms
After their first birthday, I started noting that the twins were missing their milestones and that they seemed to be in their own world. I spoke about it with my husband and my mother. Both of them comforted me by telling me it's okay and that every child has his own pace. After six months, the symptoms became more clear. For instance, there was no eye contact, they were not responding to their name, and they weren't playing with their toys in a typical manner. Also, they had these really weird hand movements that were concerning me.
After their 2nd birthday, I was sure that my twins are different. They started walking at around 2 years and 3 months. Around this time, I had started reading about developmental delays and on top of my research, the word 'autism' was always there on Google.
Finally, at 2 years and 7 months, we went to a psychologist, and literally after 5-7 minutes of questions, she said they have very strong signs of ASD (Autism Spectrum Disorder). Since I was a first-time mother, I was in dismay and I cried rivers. My husband only said we will do whatever we can to make it better for them.
Then came the point of informing the families who at first completely denied it. It was after both families came and visited us in the UAE that they agreed with the diagnosis. But then, a list of different types of cures started coming from both sides of the family and that started annoying me at one point.
Schools and therapies
I sent them to Nursery in the beginning, but soon I realized that it was not the right place for them. I then started looking for therapy centers around me. After talking to other parents like us, I came to realize that therapy is really expensive here in the UAE, and mostly not even covered by insurance. In the beginning, you have absolutely no idea about this and you are just trying to do your best by doing so much as quickly as possible.
I went to one center and started ABA (Applied Behavior Therapy). It didn't work for my twins at that specific center. We changed the center because early intervention is key and I didn't want to waste time. This new center was amazing; they knew their work. They were newly opened so they offered us free therapy for one child. However, they were expensive, like really expensive. They worked amazingly with the twins and a lot of behaviors came under control. But slowly the improvement became stagnant.
Around this time I conceived our third child and was not feeling very well. It was after 1.5 years with them that they informed us that they can't offer us any kind of discount anymore and we that we now have to pay them double the amount that we were paying them already. We just couldn't afford that so we had to leave that center.
Finding the right center for us
Meanwhile, I had my third baby and it was a beautiful healthy baby boy. I started looking for a school readiness center around this time. I joined plenty of groups with the same journey as mine, which helped me a lot. Someone told me about this school readiness center that was providing therapies too. We met them and they were lovely. The issue with them was also a huge amount of fees. Somehow we managed to make it work after a lot of negotiations with them about sibling discount and all. Now they are going there and are very happy.
Twins on the spectrum
I sometimes wonder if it was one child with autism, would the financial strain have been less? But then as Muslims, I quickly remind myself that we believe in Allah's plans more than ours. Raising twins with autism is a tough journey, that too with no elder sibling to help you around. As time is passing by and they are growing older, it's getting more difficult. But I know that I have to make it work. I am their mother and I have no other choice.
Keeping house help or nannies doesn't help either because they don't understand. To be honest who wants to get bitten and punched. But as they say, life has to go on and I take it easy now. There are days when I just let them be and I take a break too because it's very important to remain sane.
Blogging about raising autistic twins as: The Warrior Mommy
I came up with the idea to make an Instagram account about my journey because I really want to help new parents who have just started out after receiving an autism diagnosis and want to read about real-life experiences. I want them to know that life doesn't end if you have a child with special needs. You need to be strong. You need to look at the positives and you need to focus on your strengths. No one is going to help you but you. This is not what you expected parenthood to be like, but then again, this can't ever be expected. But hey it's ok! You can't undo it. You can't undo their autism but you can help them live life better with autism.
A little advice
Don't look for bigger things to celebrate with them, enjoy their small advances. Love them for who they are. Accept them for who they are. Yes, it's not going to be easy, but you don't have any other choice. Also, it's ok to have emotional outbursts every other day. Yes, it's ok because we are humans. We are parents and we love our kids. Raising autistic twins hasn't been easy and watching them struggle is nothing less than painful. But as parents, we have to be there to help them get better and take care of ourselves at the same time. Don't be shy to look around for support. I'm telling you: you will find it and it will make you feel better.
Hang in there and don't let go of hope. You are truly warriors and for your autistic kids, you are the world.
Ambreen Suhaib
The writer is a housewife with three kids. She is working on the awareness and advocacy of autism and is known as Warrior Mommy on Instagram.
Latest posts by Ambreen Suhaib (see all)
Raising Autistic Twins: How I Became The Warrior Mommy - April 17, 2022
Post Tags: #autism#raising twins
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Raising readers: Your child's first step into reading
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Craniosynostosis | Raising Babies | Raising Children | Raising Special Needs Children | Raising Toddlers
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If you are a parent of a child with special needs, or if you just got to know that your child is differently-abled, and you are scared about how you're going to deal with it, then BINGO! This article is just for you. Be calm, have patience and try to make strategies on how to…
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You don't need to be strict to raise confident, loving and respectful kids. To be honest, there's no right or wrong way of parenting. It's what you feel will suit your household, you opt for that style. But, one thing I'm very clear about. Never ever EVER raise your hand on your child. I know…
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How are babies made? What is a belly button? How do we press it as a button? Is God a boy or a girl? Why does God give us pain? Why does poop have to be yucky? Are you even a parent if your child has never asked you awkward questions that stupified you for…
Ambreen says:
Jenna says:
Your kids are adorable and you sound like an amazing Mommy!
Thank you so much 💕😍
Loshane says:
Thank you for sharing this awareness to your readers. You are indeed a great warrior. I wish you the best of luck!
Thank you so much Loshane
You are a warrior! Thank you for sharing your journey! | {
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\section{Introduction and the Atiyah problem}\label{section-introduction.Atiyah}
In \cite{Ati} Atiyah (in collaboration with Singer) extended the Atiyah-Singer Index Theorem \cite{AS} to the non-compact setting; more concretely, to the arena of non-compact manifolds $M$ equipped with a free cocompact action of a discrete group $G \act M$ (i.e. with $M/G$ compact). In this situation, the space of solutions of the equation $Df = 0$, where $D$ is an elliptic differential operator, becomes an infinite-dimensional Hilbert space $\calH_D$, and the concept of analytic index is extended by using the theory of von Neumann algebras. This theory provides a well-defined notion of dimension for $\calH_D$, called the \textit{von Neumann dimension} of $\calH_D$. This new framework led Atiyah to define the so-called \textit{$\ell^2$-Betti numbers} as von Neumann dimensions of $\ell^2$-cohomology Hilbert modules \cite{Luck}.
In his paper \cite{Ati} Atiyah observed that, in case $G$ is a finite group, the $\ell^2$-Betti numbers coincide, \emph{modulo rescaling by $|G|$}, with the previously-known Betti numbers. Thus in this situation, they are in fact \textit{rational} numbers, and indeed Atiyah posed the following question, which is nowadays commonly known as the \textit{Atiyah Conjecture}.
\begin{question*}[Atiyah]\label{conjecture-atiyah}
Is it possible to obtain irrational values of $\ell^2$-Betti numbers?
\end{question*}
Although they were defined analytically, $\ell^2$-Betti numbers can be defined in an algebraic setting, purely in terms of the group $G$ and without explicit mention of the manifold $M$ (see Section \ref{subsection-Betti.numbers} for a detailed explanation). In this line, a real positive number $r$ is said to be an $\ell^2$-Betti number arising from $G$, with coefficients in a fixed subfield $K \subseteq \C$ closed under complex conjugation, whenever there exists a matrix operator $T \in M_n(KG)$ such that the von Neumann dimension of $\ker \, T$ is equal to $r$. Following this algebraic point of view, stronger questions --in relation with the original Atiyah question-- were formulated over the years. One of its strongest versions is the so-called \textit{Strong Atiyah Conjecture}, considered by Schick in \cite[Definition 1]{Schick}.
\begin{sac*}\label{conjecture-strong.atiyah}
The set of $\ell^2$-Betti numbers arising from $G$ with coefficients in $K$ is contained in the subgroup $\sum_H \frac{1}{|H|}\Z$ of $\mathbb Q$, where $H$ ranges over the finite subgroups of $G$.
\end{sac*}
The lamplighter is precisely the first counterexample to the SAC, as proved by R. I. Grigorchuk and A. \.{Z}uk \cite{GZ,GLSZ}, see also \cite{DiSc} for a more elementary proof. More recently, the original Atiyah's question has been solved in the negative, and some authors, including Austin \cite{Aus}, Grabowski \cite{Gra14,Gra16} and Pichot, Schick and Żuk \cite{PSZ} have found examples of groups having irrational values of $\ell^2$-Betti numbers. In particular, Grabowski shows in \cite{Gra16} that there are \textit{transcendental} numbers that appear as $\ell^2$-Betti numbers of the lamplighter group. Nevertheless, the following version of the SAC is still open.
\begin{bsac*}\label{conjecture-bounded.strong.atiyah}
Suppose that there exists an upper bound for the orders of the finite subgroups of $G$. Then the SAC holds for $G$.
\end{bsac*}
In particular it is open for torsion-free groups, which gives a generalization of the famous Kaplansky's zero-divisor Conjecture. For an extensive study of Atiyah's original question and strong versions of it, see \cite{DLMSY,Jaik-survey,Jaik,JL2,Lin91,Lin93,Lin08,LLS,LS07,LS12,Luck,LL18}.
During the last 40 years, the importance of both the SAC and BSAC has increased due to the wide variety of their consequences in several branches of mathematics. For instance, in differential geometry and topology the SAC has connections with the Hopf Conjecture on the possible sign of the Euler characteristic of a Riemannian manifold (\cite{Davis}, see also \cite[Chapters 10 and 11]{Luck}). In group theory, the SAC has implications relating group-theoretic properties of a group and its homological dimension. In particular, it is known that if a group has homological dimension one and satisfies the SAC, then it must be locally free \cite{KLL}.
One of the main problems (following these lines) is to actually compute the whole set $\calC(G,K)$ of $\ell^2$-Betti numbers arising from $G$ with coefficients in $K$. In this paper, we uncover a portion of this set for the lamplighter group $\Gamma$ with coefficients in $\Q$. The group $\Gamma$ is defined to be the semidirect product of $\Z$ copies of the finite group $\Z_2$ by $\Z$, i.e.
$$\Gamma = \Big( \bigoplus_{i \in \Z} \Z_2 \Big) \rtimes_{\rho} \Z,$$
whose automorphism $\rho$ implementing the semidirect product is the well-known Bernoulli shift. A precursor of our work here can be found in the paper \cite{AG} by Ara and Goodearl. In that article, the authors attack this problem algebraically, by trying to uncover the structure of what is called the \textit{$*$-regular closure} $\calR_{K \Gamma}$ of the lamplighter group algebra $K \Gamma$ inside $\calU(\Gamma)$, the algebra of unbounded operators affiliated to the group von Neumann algebra $\calN(\Gamma)$ or, more algebraically, the classical ring of quotients of $\calN(\Gamma)$. The precise connection between the $*$-regular closure $\calR_{K\Gamma}$ and the set $\calC(\Gamma,K)$ has been provided recently by a result of Jaikin-Zapirain \cite{Jaik}, which states that the rank function on $\calR_{K\Gamma}$, obtained by restricting the canonical rank function on $\calU(\Gamma)$, is completely determined by its values on matrices over $K\Gamma$. Since $\calR_{K\Gamma}$ is $*$-regular, this can be rephrased in the form
$$\phi(K_0(\calR_{K\Gamma})) = \calG(\Gamma,K),$$
where $\phi$ is the state on $K_0(\calR_{K\Gamma})$ induced by the restriction of the rank function on $K\Gamma$, and $\calG(\Gamma,K)$ is the subgroup of $\R$ generated by $\calC(\Gamma,K)$ \cite[Proposition 4.1]{AC2} (See e.g. \cite{Ros} for the definition of the $K_0$-group of a ring). The algebra $\calR_{K\Gamma}$, together with tight connections with $\calC(\Gamma,K)$, has been recently studied by the authors in \cite{AC2}.
A key observation in light of the development of the work presented in this paper is to realize the lamplighter group algebra as a $\Z$-crossed product $*$-algebra
$$K\Gamma \cong C_K(X) \rtimes_T \Z$$
through the Fourier transform. This method was first (somewhat implicitly) used in \cite{DiSc}, and very explicitly in \cite{Aus}. Here $X = \{0,1\}^{\Z}$ is the Pontryagin dual of the group $\bigoplus_{i \in \Z} \Z_2$, topologically identified with the Cantor set, $C_K(X)$ is the set of locally constant functions $f : X \ra K$, and $T : X \ra X$ is the homeomorphism of $X$ implemented by the Bernoulli shift. There is a natural measure $\mu$ on $X$, namely the usual product measure, having taken the $\big(\frac{1}{2},\frac{1}{2}\big)$-measure on each component $\{0,1\}$. This measure is ergodic, full and $T$-invariant, so we can apply the techniques developed in \cite{AC} to study the $\Z$-crossed product algebra $\calA := C_K(X) \rtimes_T \Z$ by giving `$\mu$-approximations' of the space $X$, which at the level of the algebra $\calA$ correspond to certain `approximating' $*$-subalgebras $\calA_n \subseteq \calA$ (see \cite[Section 4.1]{AC}, also \cite[Section 6]{AC2}). By using \cite[Theorem 4.7 and Proposition 4.8]{AC}, we obtain a canonical faithful Sylvester matrix rank function $\rk_{\calA}$ on $\calA$ which coincides, in case $K$ is a subfield of $\C$ closed under complex conjugation, with the rank function $\rk_{K\Gamma}$ on the group algebra $K\Gamma$ naturally inherited from the canonical rank function in the $*$-regular ring $\calU(\Gamma)$ \cite[Proposition 5.10]{AC2}. In light of this, one can define `generalized' $\ell^2$-Betti numbers in this more general setting, that is, arising from the $\Z$-crossed product algebra $\calA = C_K(X) \rtimes_T \Z$, for $K$ an \textit{arbitrary} field and $T$ an \textit{arbitrary} homeomorphism on a Cantor set $X$.
Turning back to the lamplighter group algebra $K\Gamma$, Graboswki has shown in a recent paper \cite{Gra16} the existence of irrational (in fact transcendental) $\ell^2$-Betti numbers arising from $\Gamma$, exhibiting a concrete example in \cite[Theorem 2]{Gra16}. With this result, the lamplighter group has become the simplest known example which gives rise to irrational $\ell^2$-Betti numbers. Very roughly, his idea is to compute $\ell^2$-Betti numbers by means of decomposing them as an infinite sum of (normalized) dimensions of kernels of finite-dimensional operators (i.e. matrices). He then realizes these matrices as adjacency-labeled matrices of certain graphs in order to determine the global behavior of the dimensions of their kernels. We use these ideas in this work, but applied to our construction. Our main result is the following:
\begin{theorem}[Theorem \ref{theorem-irrational.dimensions}]
Let $\Gamma$ be the lamplighter group.
Then for any family of natural numbers $\{d_i \mid 1 \leq i \leq n\}$, and any family of polynomials $\{p_i(x) \mid 0 \leq i \leq n\}$ of positive degrees and non-negative integer coefficients, such that the linear coefficient of $p_0(x)$ is non-zero, there exists a matrix operator $A$ over $\Q \Gamma$ such that
$$b^{(2)}(A) = q_0 + q_1 \sum_{k \geq 1} \frac{1}{2^{p_0(k) + p_1(k)d_1^k + \cdots + p_n(k)d_n^k}}$$
for some non-zero rational numbers $q_0,q_1$. If moreover the constant coefficient of $p_0(x)$ is $0$, then
$$\sum_{k \geq 1} \frac{1}{2^{p_0(k) + p_1(k)d_1^k + \cdots + p_n(k)d_n^k}} \in \calG(\Gamma,\Q).$$
\end{theorem}
Here $b^{(2)}(A)$ stands for the \textit{$\ell^2$-Betti number of $A$} (see Definition \ref{definition-l2bettinumber.grouprings}). Thus we obtain a whole family of irrational and even transcendental $\ell^2$-Betti numbers arising from $\Gamma$. Another source of irrational $\ell^2$-Betti numbers arising from $\Gamma$ comes from the fact that the $*$-regular closure $\calR_{K\Gamma}$ contains a copy of the algebra of non-commutative rational power series $K_{\text{rat}}\langle \pazX \rangle$ in infinitely many indeterminates, see \cite[Subsection 6.2]{AC2} and Subsection \ref{subsection-rationalseries}. Using this, we show that $\calG (\Gamma,\Q)$ contains for instance the irrational algebraic number $\frac{1}{4}\sqrt{\frac{2}{7}}$ (Theorem \ref{thm:irrat-algl2betti}).
We also apply our machinery to study a particular crossed product algebra known as the odometer algebra. It is defined as the $\Z$-crossed product algebra $\calO := C_K(X) \rtimes_T \Z$ where $X = \{0,1\}^{\N}$ is the one-sided shift space, and the automorphism $T : X \ra X$ implementing the crossed product is given by addition of the element $(1,0,0,...)$ with carry over. Although it is not possible to realize the odometer algebra as a group algebra, this example is interesting in its own right because we are able to fully determine the structure of its $*$-regular closure $\calR_{\calO}$ (see Theorem \ref{theorem-reg.closure.odometer}), thus giving a complete description of the set of $\calO$-Betti numbers (Theorem \ref{theorem-betti.numbers.odometer}). The algebra $\calO$ has also been studied by Elek in \cite{Elek16}, although the author does not exactly compute the $*$-regular closure $\calR_{\calO}$; instead, the author computes its rank-completion, showing that it must be isomorphic to the von Neumann continuous factor $\calM_K$, which is by definition the rank-completion of $\varinjlim_i M_{2^i}(K)$ with respect to its unique rank metric (cf. \cite[Proposition 4.2]{AC2}). In fact we study a general version of the classical odometer algebra, namely the dynamical system generated by ``addition of $1$'' for arbitrary profinite completions of $\Z$.
This work is structured as follows. In Section \ref{section-preliminaries} we provide essential background and preliminary concepts about Sylvester matrix rank functions and $\ell^2$-Betti numbers for general group algebras. We summarize, in Section \ref{section-approx.crossed.product}, a general approximation construction for $\Z$-crossed product algebras using measure-theoretic tools, which enables us to construct a canonical Sylvester matrix rank function over the crossed product algebra. Using this construction, we derive a formula for computing generalized $\ell^2$-Betti numbers (see Definition \ref{definition-vN.dim.cross.prod}) over this large class of algebras (Formula \eqref{equation-Betti.no.final}). In Section \ref{section-lamplighter.algebra} we focus on the particular case of the lamplighter group algebra. We explicitly realize it as a $\Z$-crossed product algebra, thus enabling us to apply the whole theory developed in \cite{AC} and \cite{AC2}. Our main result (Theorem \ref{theorem-irrational.dimensions}) provides a vast family of irrational, and even transcendental, $\ell^2$-Betti numbers arising from the lamplighter group.
We also explore another method to find $\ell^2$-Betti numbers arising from the lamplighter group, using the algebra of non-commutative rational series. In this direction, we obtain in Theorem \ref{thm:irrat-algl2betti} that the irrational algebraic number $\frac{1}{4}\sqrt{\frac{2}{7}}$ belongs to $\calG (\Gamma, \Q)$.
Finally we study, in Section \ref{section-odometer.alg}, the generalized odometer algebra $\calO(\ol{n})$ in great detail, and we are able to completely determine the algebraic structure of its $*$-regular closure (Theorem \ref{theorem-reg.closure.odometer}). We use this characterization in Theorem \ref{theorem-betti.numbers.odometer}, where we explicitly compute the whole set of $\calO(\ol{n})$-Betti numbers.
\section{Preliminaries}\label{section-preliminaries}
\subsection{\texorpdfstring{$*$}{}-regular rings and rank functions}\label{subsection-regular.rings.*.rank.functions}
A \textit{$*$-regular ring} is a regular ring $R$ endowed with a \textit{proper} involution $*$, that is, $x^*x = 0$ if and only if $x = 0$.
In a $*$-regular ring $R$, for every $x \in R$ there exist unique projections $e, f \in R$ such that $xR = eR$ and $Rx = Rf$. It is common to denote them by $e = \mathrm{LP}(x)$ and $f = \mathrm{RP}(x)$, and are termed the \textit{left} and \textit{right projections} of $x$, respectively.
We refer the reader to \cite{Ara87, Berb} for further information on $*$-regular rings.
For any subset $S \subseteq R$ of a unital $*$-regular ring, there exists a smallest unital $*$-regular subring of $R$ containing $S$ (\cite[Proposition 6.2]{AG}, see also \cite[Proposition 3.1]{LS12} and \cite[Proposition 3.4]{Jaik}). This $*$-regular ring is denoted by $\calR(S,R)$, and called the \textit{$*$-regular closure} of $S$ in $R$.
\hspace{0.1cm}
Let us denote by $M(R)$ the set of finite matrices over $R$ of arbitrary size, i.e. $\bigcup_{n \geq 1} M_n(R)$.
\begin{definition}\label{definition-sylvester.rank}
A \textit{Sylvester matrix rank function} on a unital ring $R$ is a map $\rk : M(R) \ra \R^+$ satisfying the following conditions:
\begin{enumerate}[a),leftmargin=1cm]
\item ${\rm rk} (0)= 0$ and ${\rm rk}(1)= 1$;
\item ${\rm rk} (M_1M_2) \leq \min\{{\rm rk}(M_1), {\rm rk}(M_2)\}$ for any matrices $M_1$ and $M_2$ of appropriate sizes;
\item ${\rm rk} \begin{pmatrix} M_1 & 0 \\ 0 & M_2 \end{pmatrix} = {\rm rk} (M_1) + {\rm rk}(M_2) $ for any matrices $M_1$ and $M_2$;
\item ${\rm rk} \begin{pmatrix} M_1 & M_3 \\ 0 & M_2 \end{pmatrix} \ge {\rm rk}(M_1) + {\rm rk}(M_2)$ for any matrices $M_1$, $M_2$ and $M_3$ of appropriate sizes.
\end{enumerate}
\end{definition}
The notion of Sylvester matrix rank function was first introduced by Malcolmson in \cite{Mal} on a question of characterizing homomorphisms from a fixed ring to division rings. Equivalent definitions exist, introduced by Malcolmson itself and Schofield \cite{Sch}. For more theory and properties about Sylvester matrix rank functions we refer the reader to \cite{Jaik} and \cite[Part I, Chapter 7]{Sch}.
In a regular ring $R$, any Sylvester matrix rank function $\rk$ is uniquely determined by its values on elements of $R$, see e.g. \cite[Corollary 16.10]{Goo91}. This is no longer true if $R$ is not regular.
Any Sylvester matrix rank function $\rk$ on a unital ring $R$ defines a pseudo-metric by the rule $d(x,y) = \rk(x-y)$. The rank function is called \textit{faithful} if the only element with zero rank is the zero element. In this case, $d$ becomes a metric on $R$.
The ring operations are continuous with respect to $d$, and $\rk$ extends uniquely to a Sylvester matrix rank function $\ol{\rk}$ on the completion $\ol{R}$ of $R$ with respect to $d$.
For a $*$-subring $S$ of a $*$-regular ring $R$, there are tight connections between the structure of projections of the $*$-regular closure $\calR(S,R)$ and possible values of a Sylvester matrix rank function defined on $R$, see for instance \cite{Jaik, AC2}.
\subsection{\texorpdfstring{$\ell^2$}{}-Betti numbers for group algebras}\label{subsection-Betti.numbers}
In this subsection we define $\ell^2$-Betti numbers arising from a group $G$ with coefficients in a subfield $K \subseteq \C$ closed under complex conjugation.
Let $G$ be a discrete, countable group. For any subring $R \subseteq \C$ closed under complex conjugation, let $RG$ denote the \textit{group $*$-algebra} of $G$ with coefficients in $R$, consisting of formal finite sums $\sum_{\gamma \in G} a_{\gamma} \gamma$ with $a_{\gamma} \in R$. The sum operation is defined pointwise, the product is induced by the group product and the $*$-operation is defined by linearity and according to the rule $(a_{\gamma}\gamma)^* = \ol{a_{\gamma}} \gamma^{-1}$, for $a_{\gamma} \in R$ and $\gamma \in G$. Let also $\ell^2(G)$ denote the Hilbert space of all square-summable functions $f : G \ra \C$ with obvious addition and scalar multiplication, and inner product defined by
$$\langle f,g \rangle_{\ell^2(G)} = \sum_{\gamma \in G} f(\gamma) \ol{g(\gamma)} \quad \text{ for } f,g \in \ell^2(G).$$
The space $\ell^2(G)$ has an orthonormal basis, naturally identified with $G$, consisting of indicator functions $\xi_{\gamma} \in \ell^2(G)$ for each $\gamma \in G$. Here $\xi_{\gamma}$ is defined to be $1$ over the element $\gamma$ and $0$ otherwise.
Observe that $G$ acts faithfully on $\ell^2(G)$ by left (resp. right) multiplication $\lambda : G \ra \calB(\ell^2(G))$ (resp. $\rho : G \ra \calB(\ell^2(G))$), defined by
$$\lambda_{\gamma}(f)(\delta) = f(\gamma^{-1}\delta) \quad \text{ and } \quad \rho_{\gamma}(f)(\delta) = f(\delta \gamma)$$
for $f \in \ell^2(G)$ and $\gamma,\delta \in G$. Either $\lambda$ or $\rho$ extend $R$-linearly to actions $RG \act \ell^2(G)$ by bounded operators, preserving the $*$-operation. We will identify $RG$ with the image of $\lambda$ inside $\calB(\ell^2(G))$.
We denote by $\calN(G)$ the weak-completion of $\C G \subseteq \calB(\ell^2(G))$, which is commonly known as the \textit{group von Neumann algebra of $G$}. An equivalent algebraic definition can be given: it consists exactly of those bounded operators $T : \ell^2(G) \ra \ell^2(G)$ that are $G$-equivariant, i.e. the relation $\rho_{\gamma} \circ T = T \circ \rho_{\gamma}$ is satisfied for every $\gamma \in G$. The algebra $\calN(G)$ is endowed with a normal, positive and faithful trace, defined as
$$\tr_{\calN(G)}(T) := \langle T(\xi_e),\xi_e \rangle_{\ell^2(G)} \quad \text{for } T \in \calN(G).$$
Note that for an element $T = \sum_{\gamma \in G} a_{\gamma} \gamma \in \C G$, its trace is simply the coefficient $a_e$.
All the above constructions can be extended to $k \times k$ matrices: the ring $M_k(RG)$ acts faithfully on $\ell^2(G)^k$ by left (resp. right) multiplication. We denote the extended actions by $\lambda_k$ and $\rho_k$, respectively. We identify $M_k(RG)$ with its image $M_k(RG) \subseteq \calB (\ell^2(G)^k)$ under $\lambda_k$. We denote by $\calN_k (G)$ the weak-completion of $M_k(\C G)$ inside $\calB (\ell^2(G)^k)$, which is easily seen to be equal to $M_k(\calN(G))$. The previous trace can be extended to an unnormalized trace over $M_k(\calN(G))$ by setting, for a matrix $T = (T_{ij}) \in M_k(\calN(G))$,
$$\Tr_{\calN_k (G)}(T) := \sum_{i=1}^k \tr_{\calN (G)}(T_{ii}).$$
A \textit{finitely generated Hilbert (right) $G$-module} is any closed subspace $V$ of $\ell^2(G)^k$, invariant with respect to the \textit{right} action $\rho^{\oplus k} := \rho \oplus \stackrel{k}{\cdots} \oplus \rho$. For $V \leq \ell^2(G)^k$ a finitely generated Hilbert $G$-module, the corresponding orthogonal projection operator $p_V : \ell^2(G)^k \ra \ell^2(G)^k$ onto $V$ belongs to $\calN_k(G)$. One then defines the \textit{von Neumann dimension of $V$} as the trace of $p_V$:
$$\dim_{\calN(G)}(V) := \Tr_{\calN_k (G)}(p_V).$$
In the particular case of matrix group rings $M_k(KG)$, being $K \subseteq \C$ a subfield closed under complex conjugation, every matrix operator $A \in M_k(KG)$ gives rise to an \textit{$\ell^2$-Betti number}, in the following way. Consider $A$ as an operator $A : \ell^2(G)^k \ra \ell^2(G)^k$ acting on the left, and take $p_A \in \calN_k (G)$ to be the projection onto $\ker \, A$, which is a finitely generated Hilbert $G$-module. One can then consider the von Neumann dimension of $\ker \, A$, which is simply the trace of the projection $p_A$.
\begin{definition}\label{definition-l2bettinumber.grouprings}
Let $A$ be a matrix operator in $M_k(KG)$ for some integer $k \geq 1$. We define the \textit{$\ell^2$-Betti number of $A$} by
$$b^{(2)}(A) := \dim_{\calN(G)}(\ker \, A) = \Tr_{\calN_k (G)}(p_A).$$
The set of all $\ell^2$-Betti numbers of operators $A \in M_k(KG)$ will be denoted by $\calC(G,K)$, and will be referred to as the set of all \textit{$\ell^2$-Betti numbers arising from $G$ with coefficients in $K$}. It should be noted that this set is always a subsemigroup of $(\R^+,+)$. We also write $\calG(G,K)$ for the subgroup of $(\R,+)$ generated by $\calC(G,K)$.
\end{definition}
It is also possible to define the von Neumann dimension by means of a \textit{Sylvester matrix rank function}, as follows. Let $\calU(G)$ be the algebra of unbounded operators affiliated to $\calN(G)$; equivalently, the classical ring of quotients of $\calN(G)$. It is a $*$-regular ring possessing a Sylvester matrix rank function $\rk_{\calU(G)}$ defined by
$$\rk_{\calU(G)}(U) := \Tr_{\calN_k(G)}(\text{LP}(U)) = \Tr_{\calN_k(G)}(\text{RP}(U))$$
for any matrix $U \in M_k(\calU(G))$, where $\text{LP}(U)$ and $\text{RP}(U)$ are the left and right projections of $U$ inside the $*$-regular algebra $M_k(\calU(G))$, respectively.
Notice that these projections actually belong to $\calN_k(G)$. In particular, we obtain by restriction a Sylvester matrix rank function $\rk_{KG}$ over $KG$. So for a matrix operator $A \in M_k(KG)$ we have $p_A= 1_k-\text{RP}(A)$ and we get the equality
\begin{equation}\label{equation-vN.dim.rank.UG}
b^{(2)}(A) = k - \rk_{KG}(A).
\end{equation}
Here $1_k$ stands for the identity matrix in $k$ dimensions.
\section{Approximating crossed product algebras through a dynamical perspective}\label{section-approx.crossed.product}
We recall the general construction used in \cite{AC} on approximating $\Z$-crossed product algebras.
Let $T : X \ra X$ be a homeomorphism of a totally disconnected, compact metrizable space $X$, which we also assume to be infinite (e.g. one can take $X$ to be the Cantor space). Let also $K$ be an arbitrary field endowed with a positive definite involution, that is, an involution $*$ such that for all $n\ge 1$ and $a_1,\dots, a_n\in K$, we have $\sum_{i=1}^n a_i^*a_i= 0 \implies a_i=0$ for each $i=1,\dots ,n$.
The algebras of interest are $\Z$-crossed product algebras of the form
$$\calA := C_K(X) \rtimes_T \Z,$$
where $C_K(X)$ denotes the algebra of locally constant functions $f : X \ra K$; equivalently, the algebra of continuous functions $f : X \ra K$ when $K$ is endowed with the discrete topology. For the approximation process, we choose a $T$-invariant, ergodic and full probability measure $\mu$ on $X$. We refer the reader to \cite[Section 3]{AC} for a detailed exposition of the construction.
For any clopen subset $\emptyset \neq E \subseteq X$ and any (finite) partition $\calP$ of the complement $X \backslash E$ into clopen subsets, let $\calB$ be the unital $*$-subalgebra of $\calA$ generated by the partial isometries $\{\chi_Z t \mid Z \in \calP\}$. Here $t$ denotes the generator of the copy of $\Z$ inside $\calA$, and $\chi_A$ denotes the characteristic function of the set $A$.
There exists a quasi-partition of $X$ (i.e. a countable family of non-empty, pairwise disjoint clopen subsets whose union has full measure) given by the $T$-translates of clopen subsets $W$ of the form
\begin{equation}\label{equation-Wexpression}
W = E \cap T^{-1}(Z_1) \cap \cdots \cap T^{-k+1}(Z_{k-1}) \cap T^{-k}(E)
\end{equation}
for $k \geq 1$ and $Z_i \in \calP$, whenever these are non-empty. In fact, if we write $|W| := k$ (the length of $W$) and $\V := \{W \neq \emptyset \text{ as above}\}$, then for a fixed $W \in \V$ and $0 \leq i < |W|$ the element $\chi_{T^i(W)}$ belongs to $\calB$, and moreover the set of elements
$$e_{ij}(W) = (\chi_{X \backslash E}t)^i \chi_W (t^{-1} \chi_{X \backslash E})^j, \quad 0 \leq i,j < |W|,$$
forms a set of $|W| \times |W|$ matrix units in $\calB$ (that is, they satisfy $e_{ik}(W) e_{lj}(W) = \delta_{k,l} e_{ij}(W)$ for all indices $0 \leq i,j,k,l < |W|$). In addition, by \cite[Proposition 3.11]{AC} the element $h_W := e_{00}(W) + \cdots + e_{|W|-1,|W|-1}(W)$ is central in $\calB$ and we have a $*$-isomorphism
$$h_W \calB \cong M_{|W|}(K).$$
In this way one obtains an injective $*$-representation $\pi : \calB \hookrightarrow \prod_{W \in \V} M_{|W|}(K) =: \gotR_{\calB}$ defined by $\pi(a) = (h_W \cdot a)_W$ \cite[Proposition 3.13]{AC}.
From now on the $*$-algebra $\calB$ corresponding to $(E,\calP)$ as above will be denoted by $\calA(E,\calP)$.
\hspace{0.06cm}
Take now $\{E_n\}_{n \geq 1}$ to be a decreasing sequence of clopen sets of $X$ together with a family $\{\calP_n\}_{n \geq 1}$ consisting of (finite) partitions into clopen sets of the corresponding complements $X \backslash E_n$, satisfying:
\begin{enumerate}[a),leftmargin=1cm]
\item the intersection of all the $E_n$ consists of a single point $y \in X$;
\item $\calP_{n+1} \cup \{E_{n+1}\}$ is a partition of $X$ finer than $\calP_n \cup \{E_n\}$;
\item $\bigcup_{n \geq 1} (\calP_n \cup \{E_n\})$ generates the topology of $X$.
\end{enumerate}
By writing $\V_n$ for the set of all the non-empty subsets $W$ of the form \eqref{equation-Wexpression} corresponding to the pair $(E_n,\calP_n)$, and setting $\calA_n := \calA(E_n,\calP_n)$ and $\gotR_n := \prod_{W \in \V_n} M_{|W|}(K)$, we get injective $*$-representations $\pi_n : \calA_n \hookrightarrow \gotR_n$, in such a way that the diagrams
\begin{equation*}
\vcenter{
\xymatrix{
\calA_n \ar@{^{(}->}[r]^{\iota_n} \ar@{^{(}->}[d]^{\pi_n} & \calA_{n+1} \ar@{^{(}->}[d]^{\pi_{n+1}} \ar@{^{(}->}[r]^{\iota_{n+1}} & \calA_{n+2} \ar@{^{(}->}[d]^{\pi_{n+2}} \ar@{^{(}->}[r] & \cdots \ar@{^{(}->}[r] & \calA_{\infty} \ar@{^{(}->}[d]^{\pi_{\infty}} \\
\gotR_n \ar@{^{(}->}[r]^{j_n} & \gotR_{n+1} \ar@{^{(}->}[r]^{j_{n+1}} & \gotR_{n+2} \ar@{^{(}->}[r] & \cdots \ar@{^{(}->}[r] & \gotR_{\infty}
}
}\label{diagram-An.Rn}
\end{equation*}
commute. Here $\iota_n$ is the natural embedding $\iota_n(\chi_Z t) = \sum_{Z'} \chi_{Z'}t$ where the sum is taken with respect to all the $Z' \in \calP_{n+1}$ satisfying $Z' \subseteq Z$, the maps $j_n : \gotR_n \hookrightarrow \gotR_{n+1}$ are the embeddings given in \cite[Proposition 4.2]{AC}, and $\calA_{\infty}, \gotR_{\infty}$ are the inductive limits of the direct systems $(\calA_n,\iota_n), (\gotR_n,j_n)$, respectively.
\begin{remark}\label{remark-algebra.Ainfty}
The algebra $\calA_{\infty}$ can be explicitly described in terms of the crossed product, as follows. For $U \subseteq X$ an open set, denote by $C_{c,K}(U)$ the ideal of $C_K(X)$ generated by the characteristic functions $\chi_V$, where $V$ ranges over the clopen subsets $V \subseteq X$ contained in $U$. By \cite[Lemma 4.3]{AC}, $\calA_{\infty}$ coincides with the $*$-subalgebra of $\calA = C_K(X) \rtimes_T \Z$ generated by $C_K(X)$ and $C_{c,K}(X \backslash \{y\}) t$. (Recall from a) above that $\{ y \}= \bigcap_{n\ge 1} E_n$.)
\end{remark}
One can define a Sylvester matrix rank function on each $\gotR_n$ by the rule
\begin{equation}\label{equation-rank.over.Rn}
\rk_n(M) = \sum_{W \in \V_n} \mu(W) \Rk(M_W) \quad \text{ for } M = (M_W)_W \in \gotR_n,
\end{equation}
being $\Rk$ the usual rank of matrices. These Sylvester matrix rank functions are compatible with respect to the embeddings $j_n$, so they give rise to a well-defined Sylvester matrix rank function $\rk_{\infty}$ on $\gotR_{\infty}$.
\begin{theorem}\cite[Theorem 4.7 and Proposition 4.8]{AC}\label{theorem-rank.function}
Following the above notation, if $\gotR_{\rk} := \ol{\gotR_{\infty}}$ denotes the rank-completion of $\gotR_{\infty}$ with respect to its Sylvester matrix rank function $\rk_{\infty}$ (see Subsection \ref{subsection-regular.rings.*.rank.functions}), then there exists an injective $*$-homomorphism $\pi_{\rk} : \calA \ra \gotR_{\rk}$ making the diagram
\begin{center}
\begin{tikzcd}[remember picture]
\calA \arrow[r,"\pi_{\rk}"] & \gotR_{\rk} \\
\calA_{\infty} \arrow[r,"\pi_{\infty}"] & \gotR_{\infty}\\
\end{tikzcd}
\begin{tikzpicture}[overlay,remember picture]
\path (\tikzcdmatrixname-1-1) to node[midway,sloped]{$\supseteq$}
(\tikzcdmatrixname-2-1);
\path (\tikzcdmatrixname-1-2) to node[midway,sloped]{$\supseteq$}
(\tikzcdmatrixname-2-2);
\end{tikzpicture}
\end{center}
\vspace{-0.6cm}
\noindent commutative, and sending the element $t$ to $\pi_{\rk}(t) = \lim_n \pi_n(\chi_{X \backslash E_n}t)$. Moreover, we obtain a Sylvester matrix rank function $\rk_{\calA}$ on $\calA$ by restriction of $\ol{\rk_{\infty}}$ (the extension of $\rk_{\infty}$ to $\gotR_{\rk}$) on $\calA$, which is extremal and unique with respect to the following property:
\begin{equation}\label{equation-unique.property}
\rk_{\calA}(\chi_U) = \mu(U) \quad \text{ for every clopen subset } U \subseteq X.
\end{equation}
Finally, the rank-completion of $\calA$ with respect to $\rk_{\calA}$ gives back $\gotR_{\rk}$, that is $\ol{\calA} = \gotR_{\rk}$.
\end{theorem}
In particular, since $\gotR_{\rk}$ has the structure of a $*$-regular algebra, we can consider the $*$-regular closure of $\calA$ inside $\gotR_{\rk}$, which we denote by $\calR_{\calA} := \calR(\calA,\gotR_{\rk})$, see \cite{AC2} for more details. The $*$-regular closure $\calR_{\calA}$ is extensively studied in \cite{AC2}. For our purposes, it will become important when computing $\calO(\ol{n})$-Betti numbers arising from the generalized odometer algebra $\calO(\ol{n})$ in Section \ref{section-odometer.alg}, and also in Subsection \ref{subsection-rationalseries}.
\subsection{A formula for computing \texorpdfstring{$\calA$}{}-Betti numbers inside \texorpdfstring{$\calB$}{}}\label{subsection-formula.compute.betti.no}
In this subsection we give a formula for computing $\calA$-Betti numbers of elements from matrix algebras over the approximation algebra $\calB = \calA(E,\calP)$. We will use ideas from \cite{Gra16}, although applied to our construction.
By Theorem \ref{theorem-rank.function}, the algebra $\calA = C_K(X) \rtimes_T \Z$ possesses a `canonical' Sylvester matrix rank function $\rk_{\calA}$, unique with respect to Property \eqref{equation-unique.property}. It is then natural to ask which is the set of positive real numbers reached by such a Sylvester matrix rank function.
\begin{definition}\label{definition-vN.dim.cross.prod}
Let $A \in M_k(\calA)$. We define the \textit{$\calA$-Betti number of $A$} by (cf. Definition \ref{definition-l2bettinumber.grouprings}, see also Equation \eqref{equation-vN.dim.rank.UG})
$$b^{\calA}(A) := k - \rk_{\calA}(A).$$
\end{definition}
The set consisting of all $\calA$-Betti numbers of elements $A \in M_k(\calA)$ will be denoted by $\calC(\calA)$. This set has the structure of a semigroup of $(\R^+,+)$ (cf. Definition \ref{definition-l2bettinumber.grouprings}). The subgroup of $(\R,+)$ generated by $\calC(\calA)$ will be denoted by $\calG(\calA)$. It is shown in Section \ref{section-lamplighter.algebra} that the definition of $\calA$-Betti number coincides with Definition \eqref{definition-l2bettinumber.grouprings} in case $\calA$ is the lamplighter group algebra $\Gamma$, and thus $\calC(\calA) = \calC(\Gamma,K)$.\\
We now focus on the approximation algebra $\calB = \calA(E,\calP)$, where recall that $E$ is any non-empty clopen subset of $X$, and $\calP$ a (finite) partition of the complement $X \backslash E$ into clopen subsets. Let $\pi : \calB \ra \gotR_{\calB}$ be the faithful $*$-representation on $\gotR_{\calB} = \prod_{W \in \V} M_{|W|}(K)$ given by $\pi(a) = (h_W \cdot a)_W$. We extend $\pi$ to a faithful $*$-representation, also denoted by $\pi$, over matrix algebras
$$\pi : M_k(\calB) \ra M_k(\gotR_{\calB}) \cong \prod_{W \in \V} M_k(K) \otimes M_{|W|}(K)$$
in a canonical way. Hence $\rk_{\calA}$ can be computed over elements $A \in M_k(\calB)$ by means of the formula (recall \eqref{equation-rank.over.Rn})
$$\rk_{\calA}(A) = \sum_{W \in \V} \mu(W) \Rk(\pi(A)_W),$$
where $\Rk$ is the usual rank of matrices. Note that $\pi(A)_W \in M_k(K) \otimes M_{|W|}(K) = M_{k|W|}(K)$. We thus obtain the following proposition.
\begin{proposition}\label{proposition-vNdim.computation}
With the above notation, for a given element $A \in M_k(\calB)$ we have the formula
$$b^{\calA}(A) = \sum_{W \in \V} \mu(W) \emph{dim}_K(\ker \, \pi(A)_W).$$
Here $\emph{dim}_K(\cdot)$ denotes the usual $K$-dimension of finite-dimensional $K$-vector spaces.
\end{proposition}
\begin{proof}
It is just a matter of computation, using that $\Rk(M) = m - \text{dim}_K(\ker \, M)$ for any matrix $M \in M_m(K)$ and that $\V$ forms a quasi-partition of $X$, so $\sum_{W \in \V} \mu(W) |W| = 1$
\end{proof}
Fix now $\{e_{ij} \mid 0 \leq i,j < k \}$ a complete system of matrix units for $M_k(K)$. For a fixed $W \in \V$, the family $\{e_{ij} \otimes e_{i'j'}(W) \mid 0 \leq i,j < k;\, 0 \leq i',j' < |W| \}$ is a complete system of matrix units for $M_k(K) \otimes M_{|W|}(K)$, where the $e_{i'j'}(W)$ are the matrix units for $h_W\calB = M_{|W|}(K)$ introduced in Section \ref{section-approx.crossed.product}. In particular $\pi(e_{ij} \otimes 1)_W = e_{ij} \otimes h_W$. Let now $M_k(K) \otimes M_{|W|}(K)$ act on the $K$-vector space $K^k \otimes K^{|W|}$, with $K$-basis $\pazB = \{e_i \otimes e_{i'}(W) \mid 0 \leq i < k;\, 0 \leq i' < |W| \}$, by left matrix multiplication:
$$(e_{ij} \otimes e_{i'j'}(W)) \cdot (e_a \otimes e_{a'}(W)) = \delta_{j,a} \delta_{j',a'} \text{ } e_i \otimes e_{i'}(W).$$
Let $\langle \cdot , \cdot \rangle$ be the scalar product on $K^k \otimes K^{|W|}$ rendering the basis $\pazB$ orthonormal.
Then for an element $A \in M_k(\calB)$, the entry of the matrix $\pi(A)_W$ corresponding to the $e_{ij} \otimes e_{i'j'}(W)$ component is given by
$$\langle \pi(A)_W \cdot (e_j \otimes e_{j'}(W)), e_i \otimes e_{i'}(W) \rangle.$$
One can think of the matrix $\pi(A)_W$ as the \textit{adjacency-labeled} matrix of an edge-labeled graph $E_A(W)$ defined as follows:
\begin{enumerate}[a),leftmargin=1cm]
\item The vertices are the pairs $(e_i, e_{i'}(W))$, for $0 \leq i < k$, $0 \leq i' < |W|$.
\item We have an arrow from $(e_j, e_{j'}(W))$ to $(e_i, e_{i'}(W))$ if and only if
$$\langle \pi(A)_W \cdot (e_j \otimes e_{j'}(W)), e_i \otimes e_{i'}(W) \rangle \neq 0.$$
In this case, we label the arrow as $d_{(j,j'),(i,i')} := \langle \pi(A)_W \cdot (e_j \otimes e_{j'}(W)), e_i \otimes e_{i'}(W) \rangle$.
\end{enumerate}
Here \textit{adjacency-labeled} matrix means that the coefficient of $\pi(A)_W$ corresponding to the $e_{ij} \otimes e_{i'j'}(W)$ component is exactly $d_{(j,j'),(i,i')}$. There is an example of such a graph in Figure \ref{figure-graph.EA.example}, corresponding to the matrix
$$\pi(A)_W = d_1 \cdot e_{11} \otimes e_{21}(W) + d_2 \cdot e_{k-2,k-2} \otimes e_{21}(W) + d_3 \cdot e_{k-1,k-2} \otimes e_{22}(W) + d_4 \cdot e_{11} \otimes e_{|W|-2,|W|-2}(W).$$
\begin{figure}[ht] \[
\xymatrix@=0.4cm{
& e_0(W) & e_1(W) & e_2(W) & \cdots & e_{|W|-2}(W) & e_{|W|-1}(W) \\
e_0 & \bullet & \bullet & \bullet & \cdots & \bullet & \bullet \\
e_1 & \bullet & \bullet \ar@[red][r]^{d_1} & \bullet & \cdots & \bullet \ar@[red]@(ul,ur)_{d_4} & \bullet \\
& \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
e_{k-2} & \bullet & \bullet \ar@[red][r]^{d_2} & \bullet \ar@[red][d]^{d_3} & \cdots & \bullet & \bullet \\
e_{k-1} & \bullet & \bullet & \bullet & \cdots & \bullet & \bullet \\
} \]
\caption{An example of a graph $E_A(W)$}\label{figure-graph.EA.example}
\end{figure}
If we denote by $\calG\text{r}_A(W)$ the set consisting of the connected components of the graph $E_A(W)$ and by $A_C$, for $C \in \calG\text{r}_A(W)$, the adjacency-labeled matrix of the corresponding connected component $C$, then the matrix $\pi(A)_W$ is similar to the block-diagonal matrix $\text{diag}(A_{C_1},A_{C_2},...,A_{C_r})$, being $\calG\text{r}_A(W) = \{C_1,...,C_r\}$.
As a consequence, and by using Proposition \ref{proposition-vNdim.computation}, we obtain the formula
\begin{equation}\label{equation-Betti.no.final}
\text{dim}_{\calA}(\ker \, A) = \sum_{W \in \V} \sum_{C \in \calG\text{r}_A(W)} \mu(W) \text{dim}_K(\ker \, A_C),
\end{equation}
which will be used in Section \ref{section-lamplighter.algebra} for explicit computations of $\ell^2$-Betti numbers.
The following lemma, whose proof is trivial, is an adaptation of \cite[Lemma 20]{Gra16} in our notation, and provides a graphical way for computing dimensions of kernels of finite matrices.
\begin{lemma}[Flow Lemma at $(e_i, e_{i'}(W))$]\label{lemma-flow.lemma}
An element $\alpha = \sum_{j,j'} \lambda_{(j,j')} e_j \otimes e_{j'}(W) \in K^k \otimes K^{|W|}$ belongs to the kernel of the matrix $\pi(A)_W$ if and only if, for every vertex $(e_i, e_{i'}(W))$,
$$\sum_{j,j'} \lambda_{(j,j')} d_{(j,j'),(i,i')} = 0.$$
\end{lemma}
One can think of this system of equations in a graphical way: if we think of the variable $\lambda_{(i,i')}$ as the label of the vertex $(e_i, e_{i'}(W))$, then the Flow Lemma at this vertex is telling us that
$$\sum_{\substack{\text{all arrows having} \\ \text{range } (i,i')}} (\text{label of the arrow}) \cdot (\text{label of the source of the arrow}) = 0.$$
To see how exactly the Flow Lemma works explicitly, we refer the reader to the appendix given in \cite{Gra16}, where some applications of it in concrete examples are described.
\section{The lamplighter group algebra}\label{section-lamplighter.algebra}
In this section we apply the construction given in Section \ref{section-approx.crossed.product} to the lamplighter group algebra. This algebra is of great relevance because, among other things, it gave the first counterexample to the Strong Atiyah Conjecture (SAC), see for example \cite{GZ}, \cite{DiSc}. We refer the reader to \cite[Section 5]{AC2}, where a further application of the construction from Section \ref{section-approx.crossed.product} to a wider class of group algebras is given.
\begin{definition}
The lamplighter group $\Gamma$ is defined to be the wreath product of the finite group of two elements, $\Z_2$, by $\Z$. In other words,
$$\Gamma = \Z_2 \wr \Z = \Big( \bigoplus_{i \in \Z} \Z_2 \Big) \rtimes_{\sigma} \Z,$$
where the action implementing the semidirect product is the well-known Bernoulli shift $\sigma$ defined by
$$\sigma_n(x)_i = x_{i+n} \quad \text{ for }x = (x_i) \in \bigoplus_{i \in \Z} \Z_2.$$
If we denote by $t$ the generator corresponding to $\Z$ and by $a_i$ the generator corresponding to the $i^{\text{th}}$ copy of $\Z_2$, we have the following presentation for $\Gamma$:
$$\Gamma = \langle t,\{a_i\}_{i \in \Z} \mid a_i^2 , a_ia_ja_ia_j, ta_it^{-1}a_{i-1} \text{ for } i,j \in \Z \rangle.$$
\end{definition}
Let now $K \subseteq \C$ be a subfield of $\C$ closed under complex conjugation, which will be the involution on $K$.
As in \cite[Section 6]{AC2}, the Fourier transform $\pazF$ gives a $*$-isomorphism $K\Gamma \cong C_K(X) \rtimes_T \Z =: \calA$ through the identifications
$$1 \mapsto \chi_X, \quad e_i := \frac{1 + a_i}{2} \mapsto \chi_{U_i}, \quad t \mapsto t,$$
where $X = \{0,1\}^{\Z}$ is the Cantor set, $T$ the shift map defined by $T(x)_i = x_{i+1}$ for $x = (x_i) \in X$, and $U_i$ is the clopen set consisting of all points $x \in X$ having a $0$ at the $i^{\text{th}}$ component.
Given $\epsilon_{-k},...,\epsilon_l \in \{0,1\}$, the cylinder set $\{ x=(x_i) \in X \mid x_{-k} = \epsilon_{-k},...,x_l = \epsilon_l \}$ will be throughout denoted by $[\epsilon_{-k} \cdots \underline{\epsilon_0} \cdots \epsilon_l]$. The collection of all the cylinder sets constitutes a basis for the topology of $X$.
We take $\mu$ to be the product measure on $X$, where the $\big( \frac{1}{2}, \frac{1}{2} \big)$-measure on each component $\{0,1\}$ is considered. It is well-known (see e.g. \cite[Example 3.1]{KM}) that $\mu$ is an ergodic, full and shift-invariant probability measure on $X$. Therefore by Theorem \ref{theorem-rank.function} we can construct a Sylvester matrix rank function $\rk_{\calA}$ on $\calA = C_K(X) \rtimes_T \Z$, as exposed in Section \ref{section-approx.crossed.product}.
\begin{proposition}\label{proposition-coincide.lamplighter.group.alg}
As in Subsection \ref{subsection-Betti.numbers}, let $\rk_{K\Gamma}$ be the Sylvester matrix rank function on $K\Gamma$ inherited from $\calU(\Gamma)$. Then $\rk_{K\Gamma}$ and $\rk_{\calA}$ coincide through the Fourier transform $\pazF : K\Gamma \cong \calA$. As a consequence, the $\ell^2$-Betti number of a matrix operator $A \in M_k(K\Gamma)$ equals its $\calA$-Betti number, that is,
\begin{equation}\label{equation-vN.dim.equality}
b^{(2)}(A) = b^{\calA}(\pazF(A)).
\end{equation}
Hence, $\calC(\calA) = \calC(\Gamma,K)$ and $\calG(\calA) = \calG(\Gamma,K)$.
\end{proposition}
\begin{proof}
Both $\rk_{K\Gamma}$ and $\rk_{\calA}$ coincide by Theorem \ref{theorem-rank.function}, as observed in \cite[Subsection 6.1]{AC2}. The result follows from Formula \eqref{equation-vN.dim.rank.UG}.
\end{proof}
\subsection{The approximating algebras \texorpdfstring{$\calA_n$}{} for the lamplighter group algebra}\label{subsection-approximating.algebras.lamplighter}
In this subsection we summarize the contents of \cite[Subsection 6.1]{AC2} in relation to the approximating algebras $\calA_n$ in the particular case of the lamplighter group algebra. We identify $K\Gamma = C_K(X) \rtimes_T \Z$ under the Fourier transform.
We take $E_n = [1 \cdots \underline{1} \cdots 1]$ (with $2n+1$ one's) for the sequence of clopen sets, whose intersection gives the point $y = (...,1,\underline{1},1,...) \in X$. We also take the partitions $\calP_n$ of the complements $X \backslash E_n$ to be
$$\calP_n = \{[0 0 \cdots \underline{0} \cdots 0 0], [0 0 \cdots \underline{0} \cdots 0 1], ..., [0 1 \cdots \underline{1} \cdots 1 1]\}.$$
Let $\calA_n := \calA(E_n, \calP_n)$ be the unital $*$-subalgebra of $\calA = C_K(X) \rtimes_T \Z$ generated by the partial isometries $\{\chi_Z t \mid Z \in \calP_n\}$.
The set $\V_n$ consists of the non-empty $W$ of the following types:
\begin{enumerate}[a),leftmargin=1cm]
\item $W_0 = [1 1 \cdots \underline{1} \cdots 1 1 1]$ of length $1$ (there are $2n+2$ ones);
\item $W_1 = [1 1 \cdots \underline{1} \cdots 1 1 0 1 1 \dots 1 \cdots 1 1]$ of length $2n+2$ (there are $4n+2$ ones, and a zero);
\item $W(\epsilon_1,...,\epsilon_l) = [1 1 \cdots \underline{1} \cdots 1 1 0 \epsilon_1 \cdots \epsilon_l 0 1 1 \cdots 1 \cdots 1 1]$ of length $(2n+3)+l$,
\end{enumerate}
where in the last type $l \geq 0$, and each $\epsilon_i \in \{0,1\}$, but with at most $2n$ consecutive ones in the sequence $(\epsilon_1,...,\epsilon_l)$. We understand $l = 0$ as having no $\epsilon_i$, so there are only $2$ zeros in the middle of the corresponding $W$.
We now recall the definition of the $m$-step Fibonacci sequence, as in \cite[Definition 6.2]{AC2}.
\begin{definition}\label{definition-fibonacci.numbers}
Fix an integer $m \geq 2$. For $k \in \Z^+$, we define the $k^{\text{th}}$ $m$-acci number, denoted by $\text{Fib}_m(k)$, recursively by setting
$$\text{Fib}_m(0) = 0, \quad \text{Fib}_m(1) = 1 \quad \text{ and } \quad \text{Fib}_m(l) = 2^{l-2} \text{ for } 2 \leq l \leq m-1,$$
and for $r \in \Z^+$,
$$\text{Fib}_m(r+m) = \text{Fib}_m(r+m-1) + \cdots + \text{Fib}_m(r).$$
\end{definition}
\begin{lemma}[Lemma 6.3 of \cite{AC2}]\label{lemma-fibonacci.numbers}
For integers $k,m \geq 2$, $\emph{Fib}_m(k)$ is exactly the number of possible sequences $(\epsilon_1,...,\epsilon_l)$ of length $l = k-2$ that one can construct with zeroes and ones, but having at most $m-1$ consecutive one's.
\end{lemma}
We thus have injective $*$-representations $\pi_n : \calA_n \hookrightarrow \gotR_n$ defined by the rule $\pi_n(x) = (h_W \cdot x)_W$, where we have the concrete realization $\gotR_n = K \times \prod_{k \geq 1}M_{2n+1+k}(K)^{\text{Fib}_{2n+1}(k)}$. We will use the sequence $\{\calA_n\}_{n \geq 0}$ in the next subsection for computing rational values of $\ell^2$-Betti numbers arising from $\Gamma$.
\subsection{Some computations of \texorpdfstring{$\ell^2$}{}-Betti numbers for the lamplighter group algebra}\label{subsection-computations.Betti.no.lamplighter}
Although it does not appear in our approximating sequence, we now introduce the algebra $\calA_{1/2}$, which gives the first non-trivial approximating algebra exhibiting irrational behavior of $\ell^2$-Betti numbers. Throughout this subsection $K$ will be any subfield of $\C$ closed under complex conjugation, which will be the involution on $K$.
We take as a non-empty clopen subset $E_{1/2} = [1 \underline{1}]$, and the partition $\calP_{1/2} = \{[0\underline{0}],[0\underline{1}],[1\underline{0}]\}$ of the complement $X \backslash E_{1/2}$. The algebra $\calA_{1/2} := \calA(E_{1/2},\calP_{1/2})$ is then the unital $*$-subalgebra of $\calA = C_K(X) \rtimes_T \Z$ generated by the partial isometries $\{\chi_{[0\underline{0}]}t, \chi_{[0\underline{1}]}t, \chi_{[1\underline{0}]}t\}$; equivalently, it is the unital $*$-subalgebra generated by $s_{-1} = e_{-1}t$ and $s_0 = e_0t$.
The set $\V_{1/2}$ consists of the non-empty $W$ of the form:
\begin{enumerate}[a),leftmargin=1cm]
\item either $W = [1 \underline{1}1]$ of length $1$, or
\item $W = [1\underline{1}0^{k_1}10^{k_2}1 \cdots 0^{k_r}11]$ of length $k = k_1 + \cdots + k_r + (r+1)$, with $r,k_i \geq 1$,
\end{enumerate}
where $0^{k_i}$ denotes $0 \stackrel{k_i}{\cdots} 0$. It is easily checked that
\begin{align*}
\sum_{W \in \V_{1/2}} |W| \mu(W) = \frac{1}{2^3} + \sum_{r \geq 1} \sum_{k_1,...,k_r \geq 1} \frac{k_1 + \cdots + k_r + (r+1)}{2^{k_1 + \cdots + k_r + (r+3)}} = \frac{1}{2^3} + \sum_{r \geq 1} \frac{3r+1}{2^{r+3}} = 1,
\end{align*}
so indeed the $T$-translates of the $W \in \V_{1/2}$ form a quasi-partition of $X$.
We are now ready to give examples of matrix operators with coefficients in $\Q$, i.e. $A \in M(\Q \Gamma)$, with irrational $\ell^2$-Betti number by using part $(i)$ of Proposition \ref{proposition-coincide.lamplighter.group.alg}, together with the Formula \eqref{equation-Betti.no.final}.
Our main result is the following.
\begin{theorem}\label{theorem-irrational.dimensions}
Fix $n$ a non-negative integer. For $0 \leq l \leq n$, take
$$p_l(x) = a_{0,l} + a_{1,l} x + \cdots + a_{m_l,l}x^{m_l}$$
polynomials of degrees $m_l \geq 1$ with non-negative integer coefficients, and we require that $a_{1,0} \geq 1$. Take also $d_1,...,d_n \geq 2$ natural numbers.
There exists then an element $A$ inside some matrix algebra over $\calA_{1/2}$ with rational coefficients such that
\begin{equation}\label{equation-irrational.vNdim}
b^{(2)}(A) = q_0 + q_1 \sum_{k \geq 1} \frac{1}{2^{p_0(k) + p_1(k)d_1^k + \cdots + p_n(k)d_n^k}} \in \calC(\Gamma,\Q),
\end{equation}
where $q_0, q_1$ are non-zero rational numbers which can be explicitly computed. Therefore we get a collection of irrational and even transcendental $\ell^2$-Betti numbers arising from the lamplighter group $\Gamma$. If moreover $a_{0,0} = 0$, then
$$\sum_{k \geq 1} \frac{1}{2^{p_0(k) + p_1(k)d_1^k + \cdots + p_n(k)d_n^k}} \in \calG(\Gamma,\Q).$$
\end{theorem}
\begin{proof}
Since the proof of the theorem is quite technical, we will start by giving some examples of elements whose $\ell^2$-Betti number behave as stated, first by just considering a single polynomial $p(k)$ and then by considering $p(k) d^k$. After that, we will present an explicit element having $\ell^2$-Betti number as in \eqref{equation-irrational.vNdim}.
We start by giving a concrete example, $p(x) = 2 + x + x^2$. Consider the element from $M_{11}(\calA_{1/2})$ given by
\begin{equation*}
\begin{split}
A = & - \chi_{[\underline{0}0]} t^{-1} \cdot e_{1,1} - \chi_{[\underline{0}]} \cdot e_{2,1} - \chi_{[\underline{0}0]} t^{-1} \cdot e_{2,2} \\
& - \chi_{[1\underline{0}]} \cdot e_{3,2} - \chi_{[0 \underline{0}]} t \cdot e_{3,3} - \chi_{[\underline{0}1]} \cdot e_{4,3} \\
& - \chi_{[\underline{0}0]} t^{-1} \cdot e_{4,4} - \chi_{[\underline{0}]} \cdot e_{5,4} - \chi_{[\underline{0}0]} t^{-1} \cdot e_{5,5} \\
& - \chi_{[1\underline{0}]} \cdot e_{6,5} - \chi_{[0\underline{0}]} t \cdot e_{6,6} - \chi_{[\underline{0}1]} \cdot e_{0,6} \\
& - \chi_{[01\underline{0}]} t^2 \cdot e_{8,1} - \chi_{[0\underline{0}]}t \cdot e_{8,8} + \chi_{[\underline{0}]} \cdot e_{9,8} \\
& - \chi_{[0\underline{0}]}t \cdot e_{9,9} - \chi_{[\underline{0}1]} \cdot e_{10,9} \\
& - \chi_{[01\underline{0}]}t \cdot e_{9,7} + \chi_{[\underline{0}10]}t^{-1} \cdot e_{0,7} \\
& + \chi_{[\underline{0}]} \cdot ( \text{Id}_{11} - e_{0,0} - e_{7,7} - e_{10,10} ) - \chi_{[\underline{0}1]} \cdot e_{1,1} \in M_{11}(\Q \Gamma),
\end{split}
\end{equation*}
\noindent being the family $\{e_{i,j} \mid 0 \leq i,j < 11 \}$ a complete system of matrix units for $M_{11}(\Q)$\footnote{It should be mentioned here that this example is similar to the one given in \cite[Section 4]{Gra16}, although some differences are present: apart from the fact that the element is not the same, our construction realizes the `levels' $e_i$ as a result of allowing matrix algebras over $\calA_{1/2}$.}. If we take $W = [1\underline{1}1]$ we observe that $\pi(A)_W$ equals the zero $11 \times 11$ matrix, so its kernel has dimension $11$. Figure \ref{figure-graph.EA.W} gives the prototypical graph $E_A(W)$ that appears in case one takes a $W$ of the second form $[1\underline{1}0^{k_1}10^{k_2}1 \cdots 0^{k_r}11]$ with length $k = k_1 + \cdots + k_r + (r+1)$. By having in mind the Formula \eqref{equation-Betti.no.final}, we study the different types of connected components of the graph $E_A(W)$. From now on, in the diagrams each straight line should be labeled with a $+1$, and each dotted line with a $-1$. Nevertheless, we will explicitly write down some of the labels when necessary in the diagrams, and we will use straight lines in such cases.
\begin{sidewaysfigure}
\centering
\myRule{0.67\textheight}{0.67\textheight}
\[
\xymatrix@R=0.5cm@C=0.3cm{
& \scriptstyle{e_0(W)} & \scriptstyle{e_1(W)} & \cdots & \scriptstyle{e_{k_1}(W)} & \scriptstyle{e_{k_1+1}(W)} & \scriptstyle{e_{k_1+2}(W)} & \cdots & \scriptstyle{e_{k_1+k_2+1}(W)} & \scriptstyle{e_{k_1 + k_2 + 2}(W)} & \cdots & \cdots & \scriptstyle{e_{k-k_r-1}(W)} & \cdots & \scriptstyle{e_{k-2}(W)} & \scriptstyle{e_{k-1}(W)} \\
e_0 & \bullet & \bullet & \cdots & \bullet & \bullet & \bullet & \cdots & \bullet & \bullet & \cdots & \cdots & \bullet & \cdots & \bullet & \bullet \\
e_1 & \bullet & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] \ar@/^1pc/@[red]@{.>}[dddddddrr] & \bullet & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] \ar@/^1pc/@[red]@{.>}[dddddddrr] & \bullet & \cdots & \cdots & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \bullet \\
e_2 & \bullet & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \bullet & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \bullet & \cdots & \cdots & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \bullet \\
e_3 & \bullet & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet & \cdots & \cdots & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \\
e_4 & \bullet & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \bullet & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \bullet & \cdots & \cdots & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \bullet \\
e_5 & \bullet & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \bullet & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \bullet & \cdots & \cdots \ar@/^0.5pc/@[red]@{.>}[dddr] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \bullet \\
e_6 & \bullet & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@/_2pc/@[red]@{.>}[uuuuuu] & \bullet & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@/_2pc/@[red]@{.>}[uuuuuu] & \bullet & \cdots & \cdots & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@/_2pc/@[red]@{.>}[uuuuuu] & \bullet \\
e_7 & \bullet & \bullet & \cdots & \bullet & \bullet \ar@[red]@{.>}[ddr] \ar@/_1pc/@[red][uuuuuuul] & \bullet & \cdots & \bullet & \bullet \ar@[red]@{.>}[ddr] \ar@/_1pc/@[red][uuuuuuul] & \cdots & \cdots \ar@[red]@{.>}[ddr] & \bullet & \cdots & \bullet & \bullet \\
e_8 & \bullet & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red][d] & \bullet & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red][d] & \bullet & \cdots & \cdots & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red][d] & \bullet \\
e_9 & \bullet & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet & \cdots & \cdots & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \\
e_{10} & \bullet & \bullet & \cdots & \bullet & \bullet & \bullet & \cdots & \bullet & \bullet & \cdots & \cdots & \bullet & \cdots & \bullet & \bullet
}
\]
\caption{The graph $E_A(W)$ for a $W = [1\underline{1}0^{k_1}10^{k_2}1 \cdots 0^{k_r}11]$ of length $k$. Each straight line should be labeled with a $+1$, and each dotted line with a $-1$.}
\label{figure-graph.EA.W}
\end{sidewaysfigure}
\noindent In this concrete example we have four different types of connected components $C$ for the graph, namely:
\begin{enumerate}[a),leftmargin=0.8cm]
\item $C_1$, given by the graphs with only one vertex $\bullet$. Note that in this case $\text{dim}_{\Q}(\ker \, A_{C_1}) = 1$, and we have $12 + 8r + 3(k_1+\cdots + k_r)$ connected components of this kind.
\item $C_2$, given by the graph
\vspace{0.2cm}
\begin{equation*}
\xymatrix @R=0.4cm @C=0.9cm {
\bullet \ar@[red]@(ul,ur) \ar@[red][d] \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \\
& & & & \bullet \\
}
\end{equation*}
Let us apply the Flow Lemma to compute $\text{dim}_{\Q}(\ker \, A_{C_2})$. We write $\lambda_{(i,j)}$ to denote the labels of the vertices of the $i^{\text{th}}$ row, for $1 \leq i \leq 3$ and $1 \leq j \leq k_1$. Here $k_1$ corresponds to the number of columns of the graph. We analyze the case where $k_1 \geq 2$, being the case $k_1=1$ completely analogous, and giving the same final result. Note that for $i = 3$ we only have the vertex corresponding to $j=k_1$. The Flow Lemma gives the following set of equations for the vertices:
\begin{equation*}
\begin{cases}
\lambda_{(1,1)} = 0, \\
\lambda_{(2,1)} + \lambda_{(1,1)} = 0, \\
\text{for } 1 \leq j \leq k_1-1, \\
\begin{cases}
\lambda_{(1,j+1)} - \lambda_{(1,j)} = 0, \\
\lambda_{(2,j+1)} + \lambda_{(1,j+1)} - \lambda_{(2,j)} = 0,
\end{cases} \\
- \lambda_{(2,k_1)} = 0.
\end{cases}
\end{equation*}
The solution of this system of equations is $\lambda_{(1,j)} = \lambda_{(2,j)} = 0$ for all $1 \leq j \leq k_1$, giving one degree of freedom, namely $\lambda_{(3,k_1)}$. Therefore $\text{dim}_{\Q}(\ker \, A_{C_2}) = 1$. We only have one connected component of this kind.
\item $C_3$, given by the graph
\vspace{-0.1cm}
\begin{equation*}
\xymatrix @R=0.4cm @C=0.9cm {
& & & & \bullet \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \bullet \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@/_2pc/@[red]@{.>}[uuuuuu] \\
}
\end{equation*}
Let us apply again the Flow Lemma to compute $\text{dim}_{\Q}(\ker \, A_{C_3})$. The labels of the vertices of the $i^{\text{th}}$ row will again be denoted by $\lambda_{(i,j)}$, for $1 \leq i \leq 7$ and $1 \leq j \leq k_r$. We restrict ourselves to the case where $k_r \geq 2$ (the case $k_r=1$ gives the same result for $\text{dim}_{\Q}(\ker \, A_{C_3})$). Note that for $i = 1$ we only have the vertex corresponding to $j=k_r$. The Flow Lemma gives the following set of equations for the vertices:
\begin{equation*}
\begin{cases}
\text{for } 1 \leq j \leq k_r-1, \\
\begin{cases}
\lambda_{(2,j)} - \lambda_{(2,j+1)} = 0, \\
\lambda_{(4,j+1)} - \lambda_{(4,j)} = 0, \\
\lambda_{(5,j)} - \lambda_{(5,j+1)} = 0, \\
\lambda_{(7,j+1)} - \lambda_{(7,j)} = 0,
\end{cases}
\end{cases}
\begin{cases}
\lambda_{(3,k_r)} - \lambda_{(2,k_r)} = 0, \\
\lambda_{(4,1)} - \lambda_{(3,1)} = 0, \\
\lambda_{(5,k_r)} - \lambda_{(4,k_r)} = 0, \\
\lambda_{(6,k_r)} - \lambda_{(5,k_r)} = 0, \\
\lambda_{(7,1)} - \lambda_{(6,1)} = 0, \\
- \lambda_{(7,k_r)} = 0,
\end{cases}
\begin{cases}
\text{for } 1 \leq j \leq k_r-1, \\
\begin{cases}
\lambda_{(3,j)} - \lambda_{(2,j)} - \lambda_{(3,j+1)} = 0, \\
\lambda_{(6,j)} - \lambda_{(5,j)} - \lambda_{(6,j+1)} = 0,
\end{cases}
\end{cases}
\end{equation*}
The first set of equations gives $\lambda_{(2,j)} = \lambda_{(2,1)}, \lambda_{(4,j)} = \lambda_{(4,1)}, \lambda_{(5,j)} = \lambda_{(5,1)}$, and $\lambda_{(7,j)} = \lambda_{(7,1)}$ for all $1 \leq j \leq k_r$. Analogously, the third set of equations gives $\lambda_{(3,j)} = \lambda_{(3,1)} - (j-1)\lambda_{(2,1)}$, and $\lambda_{(6,j)} = \lambda_{(6,1)} - (j-1)\lambda_{(5,1)}$ for all $2 \leq j \leq k_r$. For now, we are left with $\lambda_{(1,k_r)}$, and $\lambda_{(i,1)}$ for all $2 \leq i \leq 7$ as possible degrees of freedom. But the second set of equations then implies $\lambda_{(2,1)} = \lambda_{(3,1)} - (k_r-1)\lambda_{(2,1)}, \lambda_{(3,1)} = \lambda_{(4,1)} = \lambda_{(5,1)}, \lambda_{(5,1)} = -(k_r-1)\lambda_{(5,1)}$ and $\lambda_{(6,1)} = \lambda_{(7,1)} = 0$. Solving this gives $\lambda_{(i,1)} = 0$ for all $2 \leq i \leq 7$.
The conclusion is that we have only one degree of freedom, namely $\lambda_{(1,k_r)}$, and so $\text{dim}_{\Q}(\ker \, A_{C_3}) = 1$. We only have one connected component of this kind.
\item Finally $C_4$, given by the graphs
\vspace{-0.3cm}
\begin{equation*}
\xymatrix @R=0.6cm @C=0.9cm {
& & & & \bullet & & & & & & \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \bullet \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] \ar@/^2pc/@[red]@{.>}[dddddddrr] & & & & & & \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & & & & & & \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & & & & & & \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & & & & & & \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & & & & & & \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@/_2pc/@[red]@{.>}[uuuuuu] & & & & & & \\
& & & & & \bullet \ar@[red]@{.>}[ddr] \ar@/_1pc/@[red][uuuuuuul] & & & & & \\
& & & & & & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \\
& & & & & & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \\
& & & & & & & & & & \bullet \\
}
\end{equation*}
These connected components are the essential part of the whole graph since they give rise to the irrationality of $b^{(2)}(A)$. Note that we have $r-1$ connected components of this kind, and we fix one of them, corresponding to the lengths $k_l$ and $k_{l+1}$ respectively, for $1 \leq l \leq r-1$. Note that $k_l$ coincides with the number of columns of the top-left graph, and $k_{l+1}$ coincides with the number of columns of the bottom-right graph.
Once more, we apply the Flow Lemma to compute $\text{dim}_{\Q}(\ker \, A_{C_4})$. We write $\lambda_{(i,j)}$ to denote the labels of the vertices of the top-left graph, where $1 \leq i \leq 7$ are the rows, and $1 \leq j \leq k_l$ are the columns. Note that for $i=1$ we only have the vertex corresponding to $j=k_l$. Similarly, $\mu_{(i,j)}$ will denote the labels of the vertices of the bottom-right graph, where now $1 \leq i \leq 3$ and $1 \leq j \leq k_{l+1}$. As before, note that for $i=3$ we only have the vertex corresponding to $j=k_{l+1}$. We use $\rho$ to denote the label of the isolated vertex being in the middle of the graph. We have the following set of equations:
\begin{equation*}
\begin{cases}
\text{for } 1 \leq j \leq k_l-1, \\
\begin{cases}
\lambda_{(2,j)} - \lambda_{(2,j+1)} = 0, \\
\lambda_{(4,j+1)} - \lambda_{(4,j)} = 0, \\
\lambda_{(5,j)} - \lambda_{(5,j+1)} = 0, \\
\lambda_{(7,j+1)} - \lambda_{(7,j)} = 0,
\end{cases}
\end{cases}
\begin{cases}
\lambda_{(3,k_l)} - \lambda_{(2,k_l)} = 0, \\
\lambda_{(4,1)} - \lambda_{(3,1)} = 0, \\
\lambda_{(5,k_l)} - \lambda_{(4,k_l)} = 0, \\
\lambda_{(6,k_l)} - \lambda_{(5,k_l)} = 0, \\
\lambda_{(7,1)} - \lambda_{(6,1)} = 0, \\
\rho - \lambda_{(7,k_l)} = 0,
\end{cases}
\begin{cases}
\text{for } 1 \leq j \leq k_l-1, \\
\begin{cases}
\lambda_{(3,j)} - \lambda_{(2,j)} - \lambda_{(3,j+1)} = 0, \\
\lambda_{(6,j)} - \lambda_{(5,j)} - \lambda_{(6,j+1)} = 0,
\end{cases}
\end{cases}
\end{equation*}
\begin{equation*}
\begin{cases}
\mu_{(1,1)} - \lambda_{(2,k_l)} = 0, \\
\mu_{(2,1)} + \mu_{(1,1)} - \rho = 0, \\
\text{for } 1 \leq j \leq k_{l+1}-1, \\
\begin{cases}
\mu_{(1,j+1)} - \mu_{(1,j)} = 0, \\
\mu_{(2,j+1)} + \mu_{(1,j+1)} - \mu_{(2,j)} = 0,
\end{cases} \\
- \mu_{(2,k_{l+1})} = 0.
\end{cases}
\end{equation*}
The first set of equations gives $\lambda_{(2,j)} = \lambda_{(2,1)}, \lambda_{(4,j)} = \lambda_{(4,1)}, \lambda_{(5,j)} = \lambda_{(5,1)}$, and $\lambda_{(7,j)} = \lambda_{(7,1)}$ for all $1 \leq j \leq k_l$. Analogously, the third set of equations gives $\lambda_{(3,j)} = \lambda_{(3,1)} - (j-1)\lambda_{(2,1)}$, and $\lambda_{(6,j)} = \lambda_{(6,1)} - (j-1)\lambda_{(5,1)}$ for all $2 \leq j \leq k_l$. Now the fourth set of equations gives $\mu_{(1,j)} = \mu_{(1,1)}$, and $\mu_{(2,j)} = \mu_{(2,1)} - (j-1) \mu_{(1,1)}$ for all $2 \leq j \leq k_{l+1}$, together with $\mu_{(1,1)} = \lambda_{(2,k_l)} = \lambda_{(2,1)}$, $\mu_{(2,1)} + \mu_{(1,1)} = \rho$ and $\mu_{(2,k_{l+1})} = 0$.
For now, the possible degrees of freedom reduces to $\lambda_{(1,k_l)}$, $\lambda_{(i,1)}$ for all $2 \leq i \leq 7$, $\rho$, and $\mu_{(3,k_{l+1})}$. But imposing the restrictions coming from the second set of equations gives $\rho = \lambda_{(7,1)} = \lambda_{(6,1)} = k_l \lambda_{(5,1)} = k_l \lambda_{(4,1)} = k_l \lambda_{(3,1)} = k_l^2 \lambda_{(2,1)}$, with the condition $(k_l^2 - k_{l+1})\lambda_{(2,1)} = 0$. This implies that the possible degrees of freedom are now $\lambda_{(1,k_l)}$, $\lambda_{(2,1)}$, and $\mu_{(3,k_{l+1})}$, together with the restriction $(k_l^2-k_{l+1}) \lambda_{(2,1)} = 0$.
The conclusion is that, depending on whether $k_l^2 = k_{l+1}$ or not, we have either three degrees of freedom or two, respectively. Thus we get
\[ \text{dim}_{\Q}(\ker \, A_{C_4}) = \left\{ \begin{array}{lr}
3 & \text{ if $k_{l+1} = k_l^2$} \\
2 & \text{ otherwise}
\end{array}\right\} = 2 + \delta_{k_{l+1},k_l^2}. \]
As commented, we have $r-1$ connected components of this kind.
\end{enumerate}
We can now compute the $l^2$-Betti number of $A$ using Proposition \ref{proposition-vNdim.computation} and Formula \eqref{equation-Betti.no.final}:
\vspace{-0.4cm}
\begin{align*}
b^{(2)}(A) & = \sum_{W \in \V_{1/2}} \mu(W) \text{dim}_{\Q}(\ker \, \pi(A)_W) \\
& = \mu([1\ul{1}1]) \text{dim}_{\Q}(\ker \, \pi(A)_{[1\ul{1}1]}) + \sum_{r \geq 1} \sum_{k_1,...,k_r \geq 1} \mu([1\ul{1}0^{k_1}10^{k_2} \cdots 10^{k_r}11]) \Big( \sum_{i=1}^4 \dim_{\Q}(\ker \, A_{C_i}) \Big) \\
& = \frac{11}{2^3} + \sum_{r \geq 1} \sum_{k_1,...,k_r \geq 1} \frac{1}{2^{k_1+\cdots+k_r+(r+3)}} \Big[(12+8r+3(k_1+\cdots+k_r))+1+1+\sum_{l=1}^{r-1}(2 + \delta_{k_{l+1},k_l^2}) \Big] \\
& = \frac{11}{8} + \frac{1}{8} \sum_{r \geq 1} \frac{1}{2^r} \Big[ \sum_{k_1,...,k_r \geq 1} \frac{12+10r}{2^{k_1} \cdots 2^{k_r}} + 3 \sum_{k_1,...,k_r \geq 1} \frac{k_1 + \cdots + k_r}{2^{k_1} \cdots 2^{k_r}} + \sum_{k_1,...,k_r \geq 1} \sum_{l=1}^{r-1} \frac{\delta_{k_{l+1},k_l^2}}{2^{k_1} \cdots 2^{k_r}} \Big] \\
& = \frac{11}{8} + \frac{1}{8} \sum_{r \geq 1} \frac{1}{2^r} \Big[ (12+10r) \Big(\sum_{k \geq 1} \frac{1}{2^k} \Big)^r + 3r \Big( \sum_{k \geq 1} \frac{k}{2^k} \Big)\Big(\sum_{k \geq 1} \frac{1}{2^k}\Big)^{r-1} \\
& \hspace{6.52cm} + (r-1)\Big( \sum_{k \geq 1} \frac{1}{2^{k^2+k}} \Big) \Big( \sum_{k \geq 1} \frac{1}{2^k} \Big)^{r-2} \Big] \\
& = \frac{11}{8} + \frac{1}{8} \sum_{r \geq 1}\frac{12+16r}{2^r} + \frac{1}{8}\Big( \sum_{r \geq 1} \frac{r-1}{2^r} \Big) \Big( \sum_{k \geq 1}\frac{1}{2^{k^2+k}}\Big) \\
& = \frac{55}{8} + \frac{1}{2} \sum_{k \geq 1} \frac{1}{2^{k^2+k+2}} \approx 6.9082337619...
\vspace{-0.4cm}
\end{align*}
which is an irrational number (its binary expansion is clearly non-periodic).
With this example in mind, we now construct an element whose $\ell^2$-Betti number equals
$$\vspace{-0.1cm}\frac{12n+31}{8} + \frac{2^{a_0}}{8} \sum_{k \geq 1}\frac{1}{2^{p(k)}},\vspace{-0.1cm}$$
being $p(x) = a_0 + a_1x + \cdots + a_nx^n$ a polynomial of degree $n \geq 2$ with non-negative integer coefficients, whose linear coefficient satisfies $a_1 \geq 1$.
We consider the element from $M_{3n+5}(\calA_{1/2})$ given by
\begin{equation*}
\begin{split}
A = & - \chi_{[\underline{0}0]} t^{-1} \cdot e_{1,1} - \chi_{[\underline{0}]} \cdot e_{2,1} - \chi_{[\underline{0}0]} t^{-1} \cdot e_{2,2} \\
& - \chi_{[1\underline{0}]} \cdot e_{3,2} - \chi_{[0 \underline{0}]} t \cdot e_{3,3} - \chi_{[\underline{0}1]} \cdot e_{4,3} - (a_1 - 1) \chi_{[\underline{0}1]} \cdot e_{0,3} \\
& + \sum_{i=1}^{n-2} \Big( - \chi_{[\underline{0}0]} t^{-1} \cdot e_{3i+1,3i+1} - \chi_{[\underline{0}]} \cdot e_{3i+2,3i+1} - \chi_{[\underline{0} 0]} t^{-1} \cdot e_{3i+2,3i+2} \\
& \quad - \chi_{[1\underline{0}]} \cdot e_{3i+3,3i+2} - \chi_{[0 \underline{0}]} t \cdot e_{3i+3,3i+3} - \chi_{[\underline{0}1]} \cdot e_{3i+4,3i+3} - a_{i+1} \chi_{[\underline{0}1]} \cdot e_{0,3i+3} \Big) \\
& - \chi_{[\underline{0} 0]} t^{-1} \cdot e_{3n-2,3n-2} - \chi_{[\underline{0}]} \cdot e_{3n-1,3n-2} - \chi_{[\underline{0} 0]} t^{-1} \cdot e_{3n-1,3n-1} \\
& - \chi_{[1\underline{0}]} \cdot e_{3n,3n-1} - \chi_{[0 \underline{0}]} t \cdot e_{3n,3n} - a_n \chi_{[\underline{0}1]} \cdot e_{0,3n} \\
& - \chi_{[01\underline{0}]} t^2 \cdot e_{3n+2,1} - \chi_{[0 \underline{0}]}t \cdot e_{3n+2,3n+2} + \chi_{[\underline{0}]} \cdot e_{3n+3,3n+2} \\
& - \chi_{[0 \underline{0}]}t \cdot e_{3n+3,3n+3} - \chi_{[\underline{0}1]} \cdot e_{3n+4,3n+3} - \chi_{[01\underline{0}]}t \cdot e_{3n+3,3n+1} + \chi_{[\underline{0}10]}t^{-1} \cdot e_{0,3n+1} \\
& + \chi_{[\underline{0}]} \cdot \left( \text{Id}_{3n+5} - e_{0,0} - e_{3n+1,3n+1} - e_{3n+4,3n+4} \right) - \chi_{[\underline{0}1]} \cdot e_{1,1} \in M_{3n+5}(\Q \Gamma).
\end{split}
\end{equation*}
\vspace{0.2cm}
Again, $\pi(A)_W$ gives the zero $(3n+5) \times (3n+5)$ matrix for $W = [1 \underline{1}1]$, so its kernel has dimension $3n+5$. For a $W = [1\underline{1}0^{k_1}10^{k_2}1 \cdots 0^{k_r}11]$ of length $k = k_1 + \cdots + k_r + (r+1)$, its graph $E_A(W)$ has four different types of connected components $C$, namely:
\begin{enumerate}[a),leftmargin=0.8cm]
\item $C_1$, given by the graphs with only one vertex $\bullet$. Here $\text{dim}_{\Q}(\ker \, A_{C_1}) = 1$, and we have $(3n+6)+(3n+2)r + 3(k_1+\cdots + k_r)$ connected components of this kind.
\item $C_2$, given by the graph
\vspace{0.2cm}
\begin{equation*}
\xymatrix @R=0.4cm @C=0.9cm {
\bullet \ar@[red]@(ul,ur) \ar@[red][d] \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \\
& & & & \bullet \\
}
\end{equation*}
The Flow Lemma gives $\text{dim}_{\Q}(\ker \, A_{C_2}) = 1$. We only have one connected component of this kind.
\item $C_3$, given by the graph
\vspace{0.0cm}
\begin{equation*}
\xymatrix @R=0.45cm @C=0.9cm {
& & & & \bullet \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \bullet \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@/_4pc/@[red][uuu]|-{-(a_1-1)} \ar@[red]@{.>}[d] \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@/_3pc/@[red][uuuuuu]|-(0.45){-a_2} \ar@[red]@{.>}[d] \\
\vdots & \vdots & \ddots & \vdots & \vdots \ar@[red]@{.>}[d] \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@/_2pc/@[red][uuuuuuuuuu]|-(0.4){-a_n} \\
}
\end{equation*}
We get $\text{dim}_{\Q}(\ker \, A_{C_3}) = 1$. We only have one connected component of this type.
\item Finally $C_4$, given by the graphs
\vspace{-0.4cm}
\begin{equation*}
\xymatrix @R=0.4cm @C=0.9cm {
& & & & \bullet & & & & & & \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \bullet \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] \ar@/^1pc/@[red]@{.>}[dddddddddddrr] & & & & & & \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & & & & & & \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@/_4pc/@[red][uuu]|-{-(a_1-1)} \ar@[red]@{.>}[d] & & & & & & \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & & & & & & \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & & & & & & \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@/_3pc/@[red][uuuuuu]|-{-a_2} \ar@[red]@{.>}[d] & & & & & & \\
\vdots & \vdots & \ddots & \vdots & \vdots \ar@[red]@{.>}[d] & & & & & & \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & & & & & & \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & & & & & & \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@/_2pc/@[red][uuuuuuuuuu]|-(0.33){-a_n} & & & & & & \\
& & & & & \bullet \ar@[red]@{.>}[ddr] \ar@/_1pc/@[red][uuuuuuuuuuul] & & & & & \\
& & & & & & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \\
& & & & & & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \\
& & & & & & & & & & \bullet \\
}
\end{equation*}
Here
\[ \text{dim}_{\Q}(\ker \, A_{C_4}) = \left\{ \begin{array}{lr}
3 & \text{ if $k_{i+1} = p(k_i) - k_i - a_0$} \\
2 & \text{ otherwise}
\end{array}\right\} = 2 + \delta_{k_{i+1},p(k_i) - k_i - a_0}. \]
We have $r-1$ connected components of this kind.
\end{enumerate}
Putting everything together, we compute
\begin{align*}
b^{(2)}(A) & = \mu([1\ul{1}1]) \text{dim}_{\Q}(\ker \, \pi(A)_{[1\ul{1}1]}) + \sum_{r \geq 1} \sum_{k_1,...,k_r \geq 1} \mu([1\ul{1}0^{k_1}10^{k_2} \cdots 10^{k_r}11]) \Big( \sum_{i=1}^4 \dim_{\Q}(\ker \, A_{C_i}) \Big) \\
& = \frac{3n+5}{2^3} + \sum_{r \geq 1} \sum_{k_1,...,k_r \geq 1} \frac{1}{2^{k_1+ \cdots +k_r + (r+3)}} \Big[ (3n+6) + (3n+4)r + 3(k_1 + \cdots + k_r) \Big] \\
& \hspace{1.68cm} + \sum_{r \geq 1} \sum_{k_1,...,k_r \geq 1} \frac{1}{2^{k_1+ \cdots +k_r + (r+3)}} \Big[ \sum_{i=1}^{r-1} \delta_{k_{i+1},p(k_i) - k_i - a_0} \Big] \\
& = \frac{3n+5}{8} + \frac{1}{8} \sum_{r \geq 1} \frac{(3n+6)+(3n+10)r}{2^r} + \frac{2^{a_0}}{8} \Big(\sum_{r \geq 1} \frac{r-1}{2^r}\Big) \Big( \sum_{k \geq 1} \frac{1}{2^{p(k)}} \Big) \\
& = \frac{12n+31}{8} + \frac{2^{a_0}}{8} \sum_{k \geq 1} \frac{1}{2^{p(k)}},
\end{align*}
which is an irrational number since the degree of the polynomial is at least $2$. Note that, in the special case $a_0 = 0$, we obtain a number of the form $q_0 + \frac{1}{2^m} \alpha$, being $\alpha = \sum_{k \geq 1} \frac{1}{2^{p(k)}}$.
Let us do a step further: we now construct an element with $\ell^2$-Betti number exactly
$$\frac{12n+35}{8} + \frac{1}{8} \sum_{k \geq 1} \frac{1}{2^{k+p(k)d^k}},$$
with $p(x)$ a polynomial of degree at least $1$ with non-negative integer coefficients, and $d \geq 2$ a natural number. We consider the element from $M_{3n+6}(\calA_{1/2})$ given by
\begin{equation*}
\begin{split}
A = & - d \chi_{[1\underline{0}]} \cdot e_{1,2} - \chi_{[0 \underline{0}]} t \cdot e_{1,1} - a_0 \chi_{[\underline{0}1]} \cdot e_{0,1} \\
& - d \chi_{[\underline{0} 0]} t^{-1} \cdot e_{2,2} - \chi_{[\underline{0}]} \cdot e_{3,2} - d \chi_{[\underline{0} 0]} t^{-1} \cdot e_{3,3} \\
& - d \chi_{[1\underline{0}]} \cdot e_{4,3} - \chi_{[0 \underline{0}]} t \cdot e_{4,4} - \chi_{[\underline{0}1]} \cdot e_{5,4} - a_1 \chi_{[\underline{0}1]} \cdot e_{0,4} \\
& + \sum_{i=1}^{n-2} \Big( - \chi_{[\underline{0} 0]} t^{-1} \cdot e_{3i+2,3i+2} - \chi_{[\underline{0}]} \cdot e_{3i+3,3i+2} - \chi_{[\underline{0} 0]} t^{-1} \cdot e_{3i+3,3i+3} \\
& \quad - \chi_{[1\underline{0}]} \cdot e_{3i+4,3i+3} - \chi_{[0 \underline{0}]} t \cdot e_{3i+4,3i+4} - \chi_{[\underline{0}1]} \cdot e_{3i+5,3i+4} - a_{i+1} \chi_{[\underline{0}1]} \cdot e_{0,3i+4} \Big) \\
& - \chi_{[\underline{0} 0]} t^{-1} \cdot e_{3n-1,3n-1} - \chi_{[\underline{0}]} \cdot e_{3n,3n-1} - \chi_{[\underline{0} 0]} t^{-1} \cdot e_{3n,3n} \\
& - \chi_{[1\underline{0}]} \cdot e_{3n+1,3n} - \chi_{[0 \underline{0}]} t \cdot e_{3n+1,3n+1} - a_n \chi_{[\underline{0}1]} \cdot e_{0,3n+1} \\
& - \chi_{[01\underline{0}]} t^2 \cdot e_{3n+3,2} - \chi_{[0 \underline{0}]}t \cdot e_{3n+3,3n+3} + \chi_{[\underline{0}]} \cdot e_{3n+4,3n+3} \\
& - \chi_{[0 \underline{0}]} t \cdot e_{3n+4,3n+4} - \chi_{[\underline{0}1]} \cdot e_{3n+5,3n+4} - \chi_{[01\underline{0}]} t \cdot e_{3n+4,3n+2} + \chi_{[\underline{0}10]} t^{-1} \cdot e_{0,3n+2} \\
& + \chi_{[\underline{0}]} \cdot \left( \text{Id}_{3n+6} - e_{0,0} - e_{3n+2,3n+2} - e_{3n+5,3n+5} \right) - \chi_{[\underline{0}1]} \cdot e_{2,2} \in M_{3n+6}(\Q \Gamma).
\end{split}
\end{equation*}
\text{ }
\noindent For $W = [1 \underline{1}1]$, $\pi(A)_W$ gives the zero $(3n+6) \times (3n+6)$ matrix, so its kernel has dimension $3n+6$. For a $W = [1\underline{1}0^{k_1}10^{k_2}1 \cdots 0^{k_r}11]$ of length $k = k_1 + \cdots + k_r + (r+1)$, its graph $E_A(W)$ has four different types of connected components $C$, namely:
\begin{enumerate}[a),leftmargin=0.8cm]
\item $C_1$, given by the graphs with only one vertex $\bullet$. Here $\text{dim}_{\Q}(\ker \, A_{C_1}) = 1$, and we have $(3n+7) + (3n+3)r + 3(k_1+\cdots + k_r)$ connected components of this type.
\item $C_2$, given by the graph
\vspace{-0.1cm}
\begin{equation*}
\xymatrix @R=0.4cm @C=0.9cm {
\bullet \ar@[red]@(ul,ur) \ar@[red][d] \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \\
& & & & \bullet \\
}
\end{equation*}
We compute $\text{dim}_{\Q}(\ker \, A_{C_2}) = 1$. We only have one connected component of this type.
\item $C_3$, given by the graph
\vspace{-0.3cm}
\begin{equation*}
\xymatrix @R=0.4cm @C=0.9cm {
& & & & \bullet \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red][u]|-{\scriptscriptstyle -a_0} \\
\bullet \ar@[red]@(ul,ur) \ar@[red][u]|-{\scriptscriptstyle -d} \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red][l]|-{\scriptscriptstyle (-d)} & \cdots \ar@[red][l]|-{\scriptscriptstyle (-d)} & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red][l]|-{\scriptscriptstyle (-d)} & \bullet \ar@[red]@{.>}[d] \ar@[red][l]|-{\scriptscriptstyle (-d)} \\
\bullet \ar@[red]@(ul,ur) \ar@[red][d]|-{\scriptscriptstyle -d} & \bullet \ar@[red]@(ul,ur) \ar@[red][l]|-{\scriptscriptstyle (-d)} & \cdots \ar@[red][l]|-{\scriptscriptstyle (-d)} & \bullet \ar@[red]@(ul,ur) \ar@[red][l]|-{\scriptscriptstyle (-d)} & \bullet \ar@[red]@(ul,ur) \ar@[red][l]|-{\scriptscriptstyle (-d)} \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@/_5pc/@[red][uuuu]|-{-a_1} \ar@[red]@{.>}[d] \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@/_4pc/@[red][uuuuuuu]|-(0.45){-a_2} \ar@[red]@{.>}[d] \\
\vdots & \vdots & \ddots & \vdots & \vdots \ar@[red]@{.>}[d] \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@/_3pc/@[red][uuuuuuuuuuu]|-(0.4){-a_n} \\
\end{equation*}
We get $\text{dim}_{\Q}(\ker \, A_{C_3}) = 1$. We only have one connected component of this type.
\item Finally $C_4$, given by the graphs
\begin{equation*}
\xymatrix @R=0.4cm @C=0.9cm {
& & & & \bullet & & & & & & \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red][u]|-{\scriptscriptstyle -a_0} & & & & & & \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red][u]|-{\scriptscriptstyle -d} & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red][l]|-{\scriptscriptstyle (-d)} & \cdots \ar@[red][l]|-{\scriptscriptstyle (-d)} & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red][l]|-{\scriptscriptstyle (-d)} & \bullet \ar@[red]@{.>}[d] \ar@[red][l]|-{\scriptscriptstyle (-d)} \ar@/^1pc/@[red]@{.>}[dddddddddddrr] & & & & & & \\
\bullet \ar@[red]@(ul,ur) \ar@[red][d]|-{\scriptscriptstyle -d} & \bullet \ar@[red]@(ul,ur) \ar@[red][l]|-{\scriptscriptstyle (-d)} & \cdots \ar@[red][l]|-{\scriptscriptstyle (-d)} & \bullet \ar@[red]@(ul,ur) \ar@[red][l]|-{\scriptscriptstyle (-d)} & \bullet \ar@[red]@(ul,ur) \ar@[red][l]|-{\scriptscriptstyle (-d)} & & & & & & \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@/_5pc/@[red][uuuu]|-{-a_1} \ar@[red]@{.>}[d] & & & & & & \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & & & & & & \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & & & & & & \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@/_4pc/@[red][uuuuuuu]|-{-a_2} \ar@[red]@{.>}[d] & & & & & & \\
\vdots & \vdots & \ddots & \vdots & \vdots \ar@[red]@{.>}[d] & & & & & & \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \ar@[red]@{.>}[l] & & & & & & \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \cdots \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[l] & & & & & & \\
\bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@/_3pc/@[red][uuuuuuuuuuu]|-(0.2){-a_n} & & & & & & \\
& & & & & \bullet \ar@[red]@{.>}[ddr] \ar@/_1pc/@[red][uuuuuuuuuuuul] & & & & & \\
& & & & & & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \\
& & & & & & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \\
& & & & & & & & & & \bullet \\
}
\end{equation*}
Here
\[ \text{dim}_{\Q}(\ker \, A_{C_4}) = \left\{ \begin{array}{lr}
3 & \text{ if $k_{i+1} = p(k_i)d^{k_i}$} \\
2 & \text{ otherwise}
\end{array}\right\} = 2 + \delta_{k_{i+1},p(k_i)d^{k_i}}. \]
We have $r-1$ connected components of this kind.
\end{enumerate}
In this case, we compute
\begin{align*}
b^{(2)}(A) & = \mu([1\ul{1}1]) \text{dim}_{\Q}(\ker \, \pi(A)_{[1\ul{1}1]}) + \sum_{r \geq 1} \sum_{k_1,...,k_r \geq 1} \mu([1\ul{1}0^{k_1}10^{k_2} \cdots 10^{k_r}11]) \Big( \sum_{i=1}^4 \dim_{\Q}(\ker \, A_{C_i}) \Big) \\
& = \frac{3n+6}{2^3} + \sum_{r \geq 1} \sum_{k_1,...,k_r \geq 1} \frac{1}{2^{k_1+ \cdots +k_r + (r+3)}} \Big[ (3n+7) + (3n+5)r + 3(k_1 + \cdots + k_r) \Big] \\
& \qquad \qquad \quad + \sum_{r \geq 1} \sum_{k_1,...,k_r \geq 1} \frac{1}{2^{k_1+ \cdots +k_r + (r+3)}} \Big[ \sum_{i=1}^{r-1} \delta_{k_{i+1},p(k_i)d^{k_i}} \Big] \\
& = \frac{3n+6}{8} + \frac{1}{8} \sum_{r \geq 1} \frac{(3n+7)+(3n+11)r}{2^r} + \frac{1}{8} \Big( \sum_{r \geq 1} \frac{r-1}{2^r} \Big) \Big( \sum_{k \geq 1} \frac{1}{2^{k + p(k)d^k}} \Big) \\
& = \frac{12n+35}{8} + \frac{1}{8} \sum_{k \geq 1} \frac{1}{2^{k+p(k)d^k}}
\end{align*}
which is again an irrational number, and even transcendental (see e.g. \cite{Tan}). Just as before, the rational number accompanying $\alpha = \sum_{k \geq 1} \frac{1}{2^{k+p(k)d^k}}$ is of the form $\frac{1}{2^m}$.
After these examples one can derive the pattern in order to obtain an exponent of the form $p_0(k) + p_1(k)d_1^k + \cdots + p_n(k) d_n^k$ by simply adding more levels, i.e. by considering matrices of higher dimension, and gluing the corresponding graphs in an appropriate way. We write down the corresponding element that gives rise to such a pattern. If we let $N = m_0 + \cdots + m_n$ to be the sum of the degrees of the polynomials $p_0,...,p_n$ respectively, with $p_i(x)=\sum_{j=0}^{m_i} a_{j,i}x^j$ for $0 \leq i \leq n$, then the element realizing the preceding pattern belongs to $M_{3N+n+5}(\calA_{1/2})$, and is given explicitly by
\hfill\tikzmark{right}
\begin{equation*}
\begin{split}
A = & \tikzmark{1} - \chi_{[\underline{0} 0]}t^{-1} \cdot e_{1,1} - \chi_{[\underline{0}]} \cdot e_{2,1} - \chi_{[\underline{0} 0]}t^{-1} \cdot e_{2,2} \\
& - \chi_{[1\underline{0}]} \cdot e_{3,2} - \chi_{[0 \underline{0}]} t \cdot e_{3,3} - \chi_{[\underline{0}1]} \cdot e_{4,3} - (a_{1,0} - 1) \chi_{[\underline{0}1]} \cdot e_{0,3} \\
& + \sum_{j=1}^{m_0-2} \Big( - \chi_{[\underline{0} 0]} t^{-1} \cdot e_{3j+1,3j+1} - \chi_{[\underline{0}]} \cdot e_{3j+2,3j+1} - \chi_{[\underline{0} 0]} t^{-1} \cdot e_{3j+2,3j+2} \\
& \qquad \qquad - \chi_{[1\underline{0}]} \cdot e_{3j+3,3j+2} - \chi_{[0 \underline{0}]} t \cdot e_{3j+3,3j+3} \\
& \qquad \qquad - \chi_{[\underline{0}1]} \cdot e_{3j+4,3j+3} - a_{j+1,0} \chi_{[\underline{0}1]} \cdot e_{0,3j+3} \Big) \\
& - \chi_{[\underline{0} 0]} t^{-1} \cdot e_{3m_0-2,3m_0-2} - \chi_{[\underline{0}]} \cdot e_{3m_0-1,3m_0-2} - \chi_{[\underline{0} 0]} t^{-1} \cdot e_{3m_0-1,3m_0-1} \\
& \tikzmark{2} - \chi_{[1\underline{0}]} \cdot e_{3m_0,3m_0-1} - \chi_{[0 \underline{0}]} t \cdot e_{3m_0,3m_0} - a_{m_0,0} \chi_{[\underline{0}1]} \cdot e_{0,3m_0} \\
& \quad \\
& \qquad \qquad \qquad \qquad \qquad - \chi_{[\underline{0}1]} \cdot e_{3m_0+2,1} \\
& \quad \\
& \tikzmark{3} - d_1 \chi_{[1\underline{0}]} \cdot e_{3m_0+1,3m_0+2} - \chi_{[0 \underline{0}]} t \cdot e_{3m_0+1,3m_0+1} - a_{0,1} \chi_{[\underline{0}1]} \cdot e_{0,3m_0+1} \\
& - d_1 \chi_{[\underline{0} 0]} t^{-1} \cdot e_{3m_0+2,3m_0+2} - \chi_{[\underline{0}]} \cdot e_{3m_0+3,3m_0+2} \\
& - d_1 \chi_{[\underline{0} 0]} t^{-1} \cdot e_{3m_0+3,3m_0+3} - d_1 \chi_{[1\underline{0}]} \cdot e_{3m_0+4,3m_0+3} \\
& - \chi_{[0 \underline{0}]} t \cdot e_{3m_0+4,3m_0+4} - \chi_{[\underline{0}1]} \cdot e_{3m_0+5,3m_0+4} - a_{1,1} \chi_{[\underline{0}1]} \cdot e_{0,3m_0+4} \\
& + \sum_{j=1}^{m_1-2} \Big( - \chi_{[\underline{0} 0]} t^{-1} \cdot e_{3m_0+3j+2,3m_0+3j+2} - \chi_{[\underline{0}]} \cdot e_{3m_0+3j+3,3m_0+3j+2} \\
& \qquad \qquad - \chi_{[\underline{0} 0]} t^{-1} \cdot e_{3m_0+3j+3,3m_0+3j+3} - \chi_{[1\underline{0}]} \cdot e_{3m_0+3j+4,3m_0+3j+3} \\
& \qquad \qquad - \chi_{[0 \underline{0}]} t \cdot e_{3m_0+3j+4,3m_0+3j+4} - \chi_{[\underline{0}1]} \cdot e_{3m_0+3j+5,3m_0+3j+4} \\
& \qquad \qquad - a_{j+1,1} \chi_{[\underline{0}1]} \cdot e_{0,3m_0+3j+4} \Big) \\
& - \chi_{[\underline{0} 0]} t^{-1} \cdot e_{3(m_0+m_1)-1,3(m_0+m_1)-1} - \chi_{[\underline{0}]} \cdot e_{3(m_0+m_1),3(m_0+m_1)-1} \\
& - \chi_{[\underline{0} 0]} t^{-1} \cdot e_{3(m_0+m_1),3(m_0+m_1)} - \chi_{[1\underline{0}]} \cdot e_{3(m_0+m_1)+1,3(m_0+m_1)} \\
& \tikzmark{4} - \chi_{[0 \underline{0}]} t \cdot e_{3(m_0+m_1)+1,3(m_0+m_1)+1} - a_{m_1,1} \chi_{[\underline{0}1]} \cdot e_{0,3(m_0+m_1)+1} \\
& \quad \\
& \qquad \qquad \qquad \qquad \qquad - \chi_{[\underline{0}1]} \cdot e_{3(m_0+m_1)+3,1} \\
& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \vdots \\
& \qquad \qquad \qquad \qquad \qquad - \chi_{[\underline{0}1]} \cdot e_{3(N-m_n)+n+1,1} \\
& \quad \\
& \tikzmark{5} - d_n \chi_{[1\underline{0}]} \cdot e_{3(N-m_n)+n,3(N-m_n)+n+1} - \chi_{[0 \underline{0}]} t \cdot e_{3(N-m_n)+n,3(N-m_n)+n} \\
& - a_{0,n} \chi_{[\underline{0}1]} \cdot e_{0,3(N-m_n)+n} - d_n \chi_{[\underline{0} 0]} t^{-1} \cdot e_{3(N-m_n)+n+1,3(N-m_n)+n+1} \\
& - \chi_{[\underline{0}]} \cdot e_{3(N-m_n)+n+2,3(N-m_n)+n+1} - d_n \chi_{[\underline{0} 0]} t^{-1} \cdot e_{3(N-m_n)+n+2,3(N-m_n)+n+2} \\
& \tikzmark{6} - d_n \chi_{[1\underline{0}]} \cdot e_{3(N-m_n)+n+3,3(N-m_n)+n+2} - \chi_{[0 \underline{0}]} t \cdot e_{3(N-m_n)+n+3,3(N-m_n)+n+3} \\
\end{split}
\end{equation*}
\begin{tikzpicture}[overlay, remember picture]
\node[anchor=base] (a) at (pic cs:1) {\vphantom{h}};
\node[anchor=base] (b) at (pic cs:2) {\vphantom{g}};
\draw [decoration={brace,amplitude=0.5em},decorate,ultra thick,black]
(a.north -| {pic cs:right}) -- (b.south -| {pic cs:right}) node[midway,below,rotate=90] {$1^{\text{st}}$ polynomial};
\end{tikzpicture}
\begin{tikzpicture}[overlay, remember picture]
\node[anchor=base] (a) at (pic cs:3) {\vphantom{h}};
\node[anchor=base] (b) at (pic cs:4) {\vphantom{g}};
\draw [decoration={brace,amplitude=0.5em},decorate,ultra thick,black]
(a.north -| {pic cs:right}) -- (b.south -| {pic cs:right}) node[midway,below,rotate=90] {$2^{\text{nd}}$ polynomial};
\end{tikzpicture}
\begin{tikzpicture}[overlay, remember picture]
\node[anchor=base] (a) at (pic cs:5) {\vphantom{h}};
\node[anchor=base] (b) at (pic cs:6) {\vphantom{g}};
\draw [decoration={brace,amplitude=0.5em},decorate,ultra thick,black]
(a.north -| {pic cs:right}) -- (b.south -| {pic cs:right}) node[midway,below,rotate=90] {$n^{\text{th}}$ polynomial};
\end{tikzpicture}
\begin{equation*}
\begin{split}
\text{ } & \tikzmark{7} - \chi_{[\underline{0}1]} \cdot e_{3(N-m_n)+n+4,3(N-m_n)+n+3} - a_{1,n} \chi_{[\underline{0}1]} \cdot e_{0,3(N-m_n)+n+3} \\
& + \sum_{j=1}^{m_n-2} \Big( - \chi_{[\underline{0} 0]} t^{-1} \cdot e_{3(N-m_n)+3j+n+1,3(N-m_n)+3j+n+1} \\
& \qquad \qquad - \chi_{[\underline{0}]} \cdot e_{3(N-m_n)+3j+n+2,3(N-m_n)+3j+n+1} \\
& \qquad \qquad - \chi_{[\underline{0} 0]} t^{-1} \cdot e_{3(N-m_n)+3j+n+2,3(N-m_n)+3j+n+2} \\
& \qquad \qquad - \chi_{[1\underline{0}]} \cdot e_{3(N-m_n)+3j+n+3,3(N-m_n)+3j+n+2} \\
& \qquad \qquad - \chi_{[0 \underline{0}]} t \cdot e_{3(N-m_n)+3j+n+3,3(N-m_n)+3j+n+3} \\
& \qquad \qquad - \chi_{[\underline{0}1]} \cdot e_{3(N-m_n)+3j+n+4,3(N-m_n)+3j+n+3} \\
& \qquad \qquad - a_{j+1,n} \chi_{[\underline{0}1]} \cdot e_{0,3(N-m_n)+3j+n+3} \Big) \\
& - \chi_{[\underline{0} 0]} t^{-1} \cdot e_{3N+n-2,3N+n-2} - \chi_{[\underline{0}]} \cdot e_{3N+n-1,3N+n-2} \\
& - \chi_{[\underline{0} 0]} t^{-1} \cdot e_{3N+n-1,3N+n-1} - \chi_{[1\underline{0}]} \cdot e_{3N+n,3N+n-1} \\
& \tikzmark{8} - \chi_{[0 \underline{0}]} t \cdot e_{3N+n,3N+n} - a_{m_n,n} \chi_{[\underline{0}1]} \cdot e_{0,3N+n} \\
& \qquad \qquad \quad - \chi_{[01\underline{0}]} t^2 \cdot e_{3N+n+2,1} \\
& \qquad \qquad \quad + \chi_{[\underline{0}10]} t^{-1} \cdot e_{0,3N+n+1} \\
& \qquad \qquad \quad - \chi_{[01\underline{0}]} t \cdot e_{3N+n+3,3N+n+1} \\
& \tikzmark{9} - \chi_{[0 \underline{0}]} t \cdot e_{3N+n+2,3N+n+2} + \chi_{[\underline{0}]} \cdot e_{3N+n+3,3N+n+2} \\
& \tikzmark{10} - \chi_{[0 \underline{0}]} t \cdot e_{3N+n+3,3N+n+3} - \chi_{[\underline{0}1]} \cdot e_{3N+n+4,3N+n+3} \\
& + \chi_{[\underline{0}]} \cdot \left( \text{Id}_{3N+n+5} - e_{0,0} - e_{3N+n+1,3N+n+1} - e_{3N+n+4,3N+n+4} \right) - \chi_{[\underline{0}1]} \cdot e_{1,1} \in M_{3N+n+5}(\Q \Gamma).
\end{split}
\end{equation*}
\begin{tikzpicture}[overlay, remember picture]
\node[anchor=base] (a) at (pic cs:7) {\vphantom{h}};
\node[anchor=base] (b) at (pic cs:8) {\vphantom{g}};
\draw [decoration={brace,amplitude=0.5em},decorate,ultra thick,black]
(a.north -| {pic cs:right}) -- (b.south -| {pic cs:right}) node[midway,below,rotate=90] {$n^{\text{th}}$ polynomial};
\end{tikzpicture}
\begin{tikzpicture}[overlay, remember picture]
\node[anchor=base] (a) at (pic cs:9) {\vphantom{h}};
\node[anchor=base] (b) at (pic cs:10) {\vphantom{g}};
\draw [decoration={brace,amplitude=0.5em},decorate,ultra thick,black]
(a.north -| {pic cs:right}) -- (b.south -| {pic cs:right}) node[midway,below,rotate=90] {last graph};
\end{tikzpicture}
\noindent The elements in between connect the different polynomials $p_i$, and the contributions (monomials) of the polynomials (that is, the sum $p_0(k) + p_1(k)d_1^k + \cdots + p_n(k)d_n^k$) are accumulated in the $\chi_{[\underline{0}1]} \cdot e_{0,0}$ component. A simplified schematic of a prototypical graph appearing here is as follows.
\begin{equation*}
\xymatrix@C=0.8cm@R=0.4cm{
& & & & \bullet & & & & & & \\
\bullet \ar@{-}@[red][rr] & & \cdots \ar@{-}@[red][rr] & & \bullet \ar@{-}@[red][d] \ar@/^3pc/@[red]@{.>}[ddddddddrr] & & & & & & \\
\vdots \ar@{-}@[red][u] \ar@{-}@[red][d] & & \text{$1^{\text{st}}$ polynomial} & & \vdots \ar@{-}@[red][d] \ar@/_1pc/@[red][uu] & & & & & & \\
\bullet \ar@{-}@[red][rr] & & \cdots \ar@{-}@[red][rr] & & \bullet \ar@/_2pc/@[red][uuu] & & & & & & \\
\vdots & \vdots & \ddots & \vdots & \vdots & & & & & & \\
\bullet \ar@{-}@[red][d] \ar@{-}@[red][rr] & & \cdots \ar@{-}@[red][rr] & & \bullet \ar@{-}@[red][d] & & & & & & \\
\vdots \ar@{-}@[red][d] & & \text{$(n+1)^{\text{th}}$ polynomial} & & \vdots \ar@{-}@[red][d] \ar@/_1pc/@[red][uuuuuu] & & & & & & \\
\bullet \ar@{-}@[red][rr] & & \cdots \ar@{-}@[red][rr] & & \bullet \ar@/_2pc/@[red][uuuuuuu] & & & & & & \\
& & & & & \bullet \ar@[red]@{.>}[ddr] \ar@/_2pc/@[red][uuuuuuuul] & & & & & \\
& & & & & & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red][d] \\
& & & & & & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \cdots \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[r] & \bullet \ar@[red]@(ul,ur) \ar@[red]@{.>}[d] \\
& & & & & & & & & & \bullet \\
}
\end{equation*}
Using the same procedure as before, a straightforward (but quite tedious) computation allows us to conclude the proof. We only write down the contribution corresponding to the components $C_4$ of the graph $E_A(W)$ where $W$ is of the second type, which are the ones that contribute to the irrationality of the $\ell^2$-Betti number. We get
\begin{align*}
\text{dim}_{\Q}(\ker \, A_{C_4})
= 2 + \delta_{k_{i+1},p_0(k_i) - k_i - a_{0,0} + p_1(k_i)d_1^{k_i} + \cdots + p_n(k_i)d_n^{k_i}}.
\end{align*}
We have $r-1$ connected components of this kind. Thus the contribution to $b^{(2)}(A)$ coming from these graphs is
\begin{align*}
& \sum_{r \geq 1} \sum_{k_1,...,k_r \geq 1} \frac{1}{2^{k_1 + \cdots + k_r + (r+3)}}\Big[ 2(r-1) + \sum_{i=1}^{r-1} \delta_{k_{i+1},p_0(k_i) - k_i - a_{0,0} + p_1(k_i)d_1^{k_i} + \cdots + p_n(k_i)d_n^{k_i}} \Big] \\
& \qquad \quad = \frac{2}{8} \sum_{r \geq 1}\frac{r-1}{2^r} + \frac{2^{a_{0,0}}}{8}\Big( \sum_{r \geq 1} \frac{r-1}{2^r} \Big) \Big( \sum_{k \geq 1} \frac{1}{2^{p_0(k) + p_1(k)d_1^{k} + \cdots + p_n(k)d_n^{k}}} \Big) \\
& \qquad \quad = \frac{1}{4} + \frac{2^{a_{0,0}}}{8} \sum_{k \geq 1} \frac{1}{2^{p_0(k) + p_1(k)d_1^{k} + \cdots + p_n(k)d_n^{k}}}.
\end{align*}
We leave the rest of the details to the reader. Note, again, that in the special case $a_{0,0} = 0$ the irrational number $\alpha = \sum_{k \geq 1} \frac{1}{2^{p_0(k) + p_1(k)d_1^{k} + \cdots + p_n(k)d_n^{k}}}$ is accompanied by a rational number of the form $\frac{1}{2^m}$.\\
For the second part of the theorem, note that under the assumption $a_{0,0} = 0$ we have always obtained irrational numbers of the form $q_0 + q_1 \alpha$, being $\alpha$ the irrational number
$$\alpha = \sum_{k \geq 1} \frac{1}{2^{p_0(k) + p_1(k)d_1^{k} + \cdots + p_n(k)d_n^{k}}},$$
and $q_0,q_1$ non-zero rational numbers with $q_1$ of the form $\frac{1}{2^m}$ for some $m \geq 1$. In particular, $q_0 + q_1 \alpha$ belongs to $\calG(\Gamma,\Q)$. By a result of Ara and Goodearl \cite[Corollary 6.14]{AG}, the group $\calG(\Gamma,\Q)$ contains the rational numbers, and hence the element $q_1 \alpha$ belongs to $\calG(\Gamma,\Q)$ too. Since $q_1 = \frac{1}{2^m}$, we get
$$\alpha = q_1 \alpha + \stackrel{2^m}{\cdots} + q_1 \alpha \in \calG(\Gamma,\Q).$$
This completes the proof of the theorem.
\end{proof}
\begin{remark}\label{remark-exotic.patterns}
By making use of the ideas and techniques developed in Theorem \ref{theorem-irrational.dimensions}, one can construct elements whose associated $\ell^2$-Betti number has a binary expansion following different kinds of exotic patterns. For example, by gluing in an appropriate way two graphs corresponding to different polynomials $p_1(x)$ and $p_2(x)$, that is, by constructing graphs of the form $C_3$ but substituting the bottom right graph (the one contributing to $k_{i+1}$) by another graph corresponding to the polynomial $p_2(x)$, we can obtain terms in the computation of the associated $\ell^2$-Betti number of the form
$$\sum_{k_1,k_2 \geq 1} \frac{\delta_{p_1(k_1),p_2(k_2)}}{2^{k_1+k_2}},$$
and more generally of the form
$$\sum_{k_1,k_2 \geq 1} \frac{\delta_{p_0(k_1) + p_1(k_1)d_1^{k_1} + \cdots + p_n(k_1) d_n^{k_1},q_0(k_2) + q_1(k_2)l_1^{k_2} + \cdots + q_m(k_2) l_m^{k_2}}}{2^{k_1+k_2}},$$
being $p_i(x),q_j(x)$ polynomials satisfying the hypotheses of the theorem, and $d_i,l_j \geq 2$ integers.
\end{remark}
To conclude this section, it may be instructive to compute some rational values of $\ell^2$-Betti numbers. In \cite{GZ} (cf. \cite{DiSc,AG,Gra14}) the authors compute the $\ell^2$-Betti number of the element $a_0 = e_0t + t^{-1} e_0 = \chi_{X \backslash E_0} t + t^{-1} \chi_{X \backslash E_0}$, which belongs to the first of our approximating algebras, $\calA_0$. We will compute, in general, the $\ell^2$-Betti number of the element $a_n = \chi_{X \backslash E_n} t + t^{-1} \chi_{X \backslash E_n} \in \Q \Gamma$ belonging to the $*$-subalgebra $\calA_n$. Under $\pi_n$ --and recalling that $\gotR_n = \Q \times \prod_{k \geq 1} M_{m+k}(\Q)^{\text{Fib}_m(k)}$ with $m = 2n+1$-- it gives
$$\pi_n(a_n) = ( 0, ( t_{m+k} + t_{m+k}^* , \stackrel{\text{Fib}_m(k)}{...}, t_{m+k} + t_{m+k}^*)_{k \geq 1}),$$
where $t_r$ is the $r \times r$ lower-triangular matrix given by
\[ t_r = \left( \begin{array}{@{}c@{}}
\begin{matrix}
0 & & & & \bigzero \\
1 & 0 & & & \\
& \ddots & \ddots & & \\
& & \ddots & 0 & \\
\bigzero & & & 1 & 0
\end{matrix}
\end{array} \right). \]
It is then straightforward to show that
$$\text{dim}_{\Q}(\ker \, (t_{m+k} + t_{m+k}^*)) = \begin{cases}
1 & \text{ if $k$ is even} \\
0 & \text{ otherwise}
\end{cases}$$
so
\begin{align*}
b^{(2)}(a_n) = \frac{1}{2^{m+1}} + \sum_{k \geq 1} \frac{\text{Fib}_m(2k)}{2^{2m+2k}}.
\end{align*}
This sum can be computed by using the summation rule
$$\sum_{k \geq 1} \frac{\text{Fib}_m(2k)}{2^{2k}} = 2^{m-1} \frac{2^{m+1}-1}{2^{m+2}+1},$
whose proof can be found in \cite[Lemma 3.2.11]{Claramunt}. We then have
$$b^{(2)}(a_n) = \frac{3}{1+2^{2n+3}}.$$
Note that $b^{(2)}(a_0) = \frac{1}{3}$, and we recover the result from \cite{GZ,DiSc}. Also, as $n \ra \infty$, this value tends to zero, as expected since $a_n \ra t + t^{-1}$ in rank, which is invertible inside $\gotR_{\rk}$.
\subsection{Rational series and \texorpdfstring{$\ell^2$}{}-Betti numbers}\label{subsection-rationalseries}
In this section we let again $K \subseteq \C$ be any field closed under complex conjugation. A more algebraic perspective to attack the problem of computing values from $\calG(\calA_{1/2}) \subseteq \calG(\Gamma,K)$ is through the determination of the $*$-regular closure $\calR_{1/2}$ of the algebra $\calA_{1/2}$ seen inside $\gotR_{1/2} = K \times \prod_{k \geq 1} M_{k+2}(K)^{\text{Fib}_2(k)}$ through the embedding $\pi_{1/2} : \calA_{1/2} \hookrightarrow \gotR_{1/2}$, and taking advantage of \cite[Proposition 4.1]{AC2} which states that the values of ranks of elements from $\calR_{1/2}$ are all included in $\calG(\Gamma,K)$.
In order to clarify the exact relationship of this approach with Atiyah's problem, it is convenient to introduce the following definitions for a general unital ring.
\begin{definition}\label{def:semigroupsCandCprime}
Let $R$ be a unital ring and let $\rk$ be a Sylvester matrix rank function on $R$. We denote by $\calC_{\rk} (R)$ the set of real numbers of the form $k- \rk(A)$ for $A\in M_k(R)$, $k\ge 1$. We denote by $\calC'_{\rk}(R)$ the set of real numbers of the form $\rk(A)$ for $A\in M_k(R)$, $k\ge 1$. Both $\calC_{\rk}(R)$ and $\calC_{\rk}'(R)$ are subsemigroups of $(\R^+,+)$. We denote by $\calG _{\rk}(R)$ the subgroup of $\R$ generated by $\calC_{\rk}(R)$. Clearly
$$\calG_{\rk}(R)= \calC _{\rk}(R) - \calC _{\rk}(R) = \calC _{\rk}'(R) - \calC_{\rk}'(R).$$
\end{definition}
\begin{remark}\label{rem:semigroupsCandCprime}
With this notation, \cite[Corollary 6.2]{Jaik} says that, if $\calU $ is a $*$-regular ring and $R$ is a $*$-subring of $\calU$ such that $\calU$ is the $*$-regular closure of $R$ in $\calU$, then we have
$$\calG_{\rk} (R) = \calG_{\rk}(\calU)$$
for every Sylvester matrix rank function $\rk$ on $\calU$. Thus the main object of study within this approach is the group $\calG_{\rk}(R)$. Of course $\calC (G,K) = \calC_{\rk}(KG)$ for any discrete group $G$, where $\rk$ is the canonical rank. It would be interesting to know conditions under which $\calC_{\rk}(R)= \calC_{\rk}'(R)$ and under which $\calC_{\rk}(R)= \calG_{\rk}(R)\cap \R^+$, or $\calC_{\rk}'(R)= \calG_{\rk}(R)\cap \R^+$. Note that
$$\forall g\in \calG_{\rk}(R)\, \exists t\in \Z^+ : g+t\in \calC_{\rk}(R) \iff
\forall g\in \calG_{\rk}(R)\, \exists t\in \Z^+ : g+t\in \calC_{\rk}'(R).$$
In case $R$ is von Neumann regular, we have that $\calC_{\rk}(R) = \calC_{\rk}'(R)$ and the above equivalent conditions hold automatically, for every Sylvester matrix rank function $\rk$.
\end{remark}
We briefly remind some concepts from \cite{AC2}. Write $\calA_{1/2} = \bigoplus_{i\in \Z} \calA_{1/2,i}t^i$, where $\calA_{1/2,0}= \calA_{1/2}\cap C_K(X)$ and
$$\calA_{1/2,i} = \chi_{X \setminus (E_{1/2} \cup T(E_{1/2}) \cup \cdots \cup T^{i-1}(E_{1/2}))} \calA_{1/2,0} \quad \text{and} \quad \calA_{1/2,-i} = \chi_{X \setminus (T^{-1}(E_{1/2}) \cup \cdots \cup T^{-i}(E_{1/2}))} \calA_{1/2,0}$$
for $i > 0$. Following \cite[Subsection 4.2]{AC2}, we consider a skew partial power series ring $\calA_{1/2,0}[[t;T]]$ by taking infinite formal sums
$$\sum_{i \geq 0} b_i (\chi_{X \backslash E_{1/2}} t)^i = \sum_{i \geq 0} b_i t^i , \text{ where } b_i \in \calA_{1/2,i} \text{ for all } i \geq 0.$$
Specializing \cite[Definition 4.17]{AC2} to our situation, we see that a {\it special term} of positive degree $i$ is a monomial $b_it^i$ in $\calA_{1/2,i}$, where $b_i = \chi_S$ with
$$S = [10^{i_1}10^{i_2}1\cdots 10^{i_r}\ul{1}],\quad \text{ where } r\ge 1, i_1,\dots ,i_r\ge 1 \text{ and } i= i_1+\cdots +i_r +r.$$
Moreover, since $E_{1/2}\cap T^{-1}(E_{1/2})\ne \emptyset$ and $S_0= T^{-1}(S_1)= [\ul{1}]$, we get from \cite[Definition 4.7(c)]{AC2} that the unique special term of degree $0$ is $\chi_{[\ul{1}]}$.
Next, consider the subset $\calS_{1/2}[[t;T]]$ of $\calA_{1/2,0}[[t;T]]$ consisting of those elements
$$\sum_{i \geq 0} b_i (\chi_{X \backslash E_{1/2}}t)^i = \sum_{i \geq 0} b_i t^i$$
such that each $b_it^i \in \calA_{1/2,i}t^i$ belongs to the $K$-linear span of the special terms of degree $i$. We denote by $\calS_{1/2} [t;T]$ the subspace of $\calS_{1/2} [[t;T]]$ consisting of elements with finite support.
Let $\pazX=\{x_1,x_2,\dots \}$ be an infinite countable set, and consider the algebras $K\langle \pazX \rangle $ and $K\langle \langle \pazX \rangle \rangle $ of non-commutative polynomials and non-commutative formal power series, respectively, with coefficients in $K$. We consider the degree in $K\langle \pazX \rangle $ given by $d(x_i) = i+1$ for all $i\ge 1$ and $d(1)=0$. With this degree function, $K\langle \pazX \rangle $ is a graded $K$-algebra, where $K$ has the trivial degree.
The following result is shown in \cite[Lemma 6.5 and Proposition 6.6]{AC2}. Note that \cite[Propositions 6.6 and 6.7]{AC2} are stated only for $n\ge 1$, but they hold, with the same proof, for $n=1/2$.
\begin{proposition}\label{prop:isospecial}
The linear subspaces $\calS_{1/2}[t;T]$ and $\calS_{1/2}[[t;T]]$ are indeed subalgebras of $\calA_{1/2}[[t;T]]$. There is an isomorphism of unital graded algebras $K\langle \pazX \rangle \cong \calS_{1/2} [t; T]$ sending the element $x_{i_r}\cdots x_{i_1}$ to $\chi_{[10^{i_1}10^{i_2}1\cdots 10^{i_r}\ul{1}]}t^{i}$, where $i=i_1+\cdots + i_r+r$, and $1$ to $\chi_{[\ul{1}]}$. This isomorphism extends naturally to an isomorphism between $K\langle \langle \pazX \rangle \rangle $ and $\calS_{1/2}[[t;T]]$.
\end{proposition}
The special terms of the form $\chi_{[10^i\ul{1}]}t^{i+1}$, for $i\ge 1$, which are the algebra generators of $\calS_{1/2}[t;T]$, are called {\it pure special terms}, see the proof of \cite[Proposition 6.6]{AC2}.
Let $\pazX^*$ denote the free monoid on $\pazX$. We now recall the concept of the {\it Hadamard product} $\odot$ on $K\langle \langle \pazX \rangle \rangle $, see \cite{BR} for more details.
If $a= \sum_{w\in \pazX^*}a_ww$ and $b= \sum_{w\in \pazX^*} b_ww$ are elements in $K\langle \langle \pazX \rangle \rangle $, set
$$a\odot b = \Big(\sum_{w\in \pazX^*}a_ww\Big)\odot \Big(\sum_{w\in \pazX^*} b_ww\Big) = \sum_{w\in \pazX^*} (a_wb_w)w \in K\langle \langle \pazX \rangle \rangle .$$
We will denote by $K\langle \langle \pazX \rangle \rangle ^{\circ}$ the algebra of non-commutative formal power series endowed with the Hadamard product.
Note that $K\langle \langle \pazX \rangle \rangle ^{\circ}$ is a $*$-regular ring with respect to the involution given by $\ol{\sum_{w\in \pazX^*}a_ww}= \sum _{w\in \pazX^*} \ol{a_w}w$.
The algebra of {\it non-commutative rational series}, denoted by $K_{\text{rat}}\langle \pazX \rangle $ is defined as the division closure of $K\langle \pazX \rangle $ in $K\langle \langle \pazX \rangle \rangle $, that is, $K_{\text{rat}}\langle \pazX \rangle $ is the smallest subalgebra of $K\langle \langle \pazX \rangle \rangle$ containing $K\langle \pazX \rangle $ and closed under inversion, see \cite{BR} for details. It is well-known that $K_{\text{rat}}\langle \pazX \rangle $ is closed under the Hadamard product \cite[Theorems 1.5.5 and 1.7.1]{BR}.
Let $p_{1/2} :=p_{E_{1/2}}= \pi_{1/2}(\chi_{E_{1/2}})$ be the projection in $\gotR_{1/2}= \prod_{W\in \mathbb V_{1/2}} M_{|W|}(K) = K\times \prod_{k \geq 1} M_{k+2}(K)^{\text{Fib}_2(k)}$ which has a $1$ in the left upper corner of each matrix factor $M_{|W|}(K)$, and $0's$ elsewhere, that is,
$(p_{1/2})_W= e_{00}(W)$ for each $W\in \mathbb V_{1/2}$. Recall from the beginning of Subsection \ref{subsection-computations.Betti.no.lamplighter} that the elements of $\mathbb V_{1/2}$ are all the clopen sets of the form $[1\ul{1}0^{i_1}10^{i_2}1\cdots 10^{i_r}11]$, where $r\ge 0$ and $i_1,\dots i_r\ge 1$ (we interpret this term to be $[1\ul{1}1]$ if $r=0$).
By \cite[Proposition 4.20]{AC2}, there is a $*$-isomorphism from $(\calS_{1/2}[[t;T]],\odot,-)$ onto $p_{1/2}\gotR_{1/2} p_{1/2}$. Composing the isomorphism $K\langle \langle \pazX \rangle \rangle \cong \calS_{1/2} [[t;T]]$ from Proposition \ref{prop:isospecial} (which respects the respective Hadamard products and involutions) with this isomorphism we obtain a $*$-isomorphism
$$\Delta \colon K\langle \langle \pazX \rangle \rangle^{\circ} \overset{\cong}{\longrightarrow} p_{1/2}\gotR_{1/2} p_{1/2}$$
such that
$$\Delta (f)_{[1\ul{1}0^{i_1}1\cdots 10^{i_r}11]} = (f, x_{i_r}\cdots x_{i_1})e_{00}([1\ul{1}0^{i_1}1\cdots 10^{i_r}11]) \qquad (f\in K\langle \langle \pazX \rangle \rangle )$$
for each $i_1,\dots ,i_r\ge 1$. Here, $(f, x_{i_r}\cdots x_{i_1})$ is the coefficient of $f$ corresponding to the monomial $x_{i_r}\cdots x_{i_1}$.
For each $n\ge 1$, set $\pazX_n=\{ x_1,\dots , x_n\}$. Let $L$ be a {\it language} over $\pazX_n$, i.e., a subset of the free monoid $\pazX_n^*$ on $\pazX_n$. We now give a formula for the rank of the element $\Delta(\ul{L})$, where
$$\ul{L} := \sum_{w\in L} w \in K\langle \langle \pazX_n \rangle \rangle $$
is the {\it characteristic series} of $L$ (\cite[p.8]{BR}). Note that, for $W=[1\ul{1}0^{i_1}10^{i_2}1\cdots 10^{i_r}11]\in \mathbb V_{1/2}$, we have
$$\Delta (\ul{L})_W= \begin{cases} e_{00}(W) & \text{ if } x_{i_r}\cdots x_{i_2}x_{i_1}\in L \\
0 & \text{ if } x_{i_r}\cdots x_{i_2}x_{i_1}\notin L\end{cases}.$$
Moreover, we have $\mu (W)= 2^{-(i_1+\cdots +i_r+r+3)}= 2^{-3}2^{-d(w)}$, where $w=x_{i_r}\cdots x_{i_2}x_{i_1}$. Let $s(L)= \sum_{j\ge 0} \alpha_j x^j$ be the series in $\Z^+[[x]]$ such that $\alpha_j$ is the number of words $w\in L$ such that $d(w)= j$. Then we have
\begin{equation}\label{equation-formula.for.rk.rat.series}
\rk(\Delta(\ul{L})) = \sum_{W \in \V_{1/2}} \mu(W) \Rk(\Delta(\ul{L})_W) = \frac{1}{2^3}\sum_{w\in L} \frac{1}{2^{d(w)}} = \frac{1}{8}s(L)\Big(\frac{1}{2}\Big).
\end{equation}
Here we make use of \eqref{equation-rank.over.Rn} for the first equality, of the above observations for the second, and of the definition of $s(L)$ for the third. We say that $s(L) \in \Z^+[[x]]$ is the {\it generating function} of $L$ with respect to $d$, and we denote by $\alpha_L$ the real number $\rk (\Delta (\ul{L}))$ described in \eqref{equation-formula.for.rk.rat.series}.
We now recall the notion of a rational language (see \cite[Chapter 3]{BR}).
\begin{definition}\label{definition-rational.language}
The set of \textit{rational languages} over a finite set $F$ is the smallest set of subsets of the free monoid $F^*$ containing all the finite subsets of $F^*$ and closed under the following operations:
\begin{enumerate}[a)]
\item union $L_1\cup L_2$;
\item product $L_1L_2 := \{w_1 w_2 \in F^* \mid w_1 \in L_1, w_2 \in L_2\}$;
\item $*$-product $L^* := \bigcup_{k \geq 0} L^k$.
\end{enumerate}
\end{definition}
A characterization of the rational languages is the following (\cite[Lemma 3.1.4]{BR}): a language over a finite alphabet $F$ is rational if and only if it is the support of some rational series $z \in \Z^+ \langle \langle F \rangle \rangle$. In fact, if $L$ is a rational language over $F$, its characteristic series $\ul{L} = \sum_{w \in L}w$ belongs to $R_{\mathrm{rat}}\langle F \rangle^{\circ}$ for any semiring $R$ (\cite[Proposition 3.2.1]{BR}).
It turns out that the set of rational languages over $F$, which we will denote by $\gotL(F)$, forms a Boolean subalgebra of $P (F^*)$, the power set of $F^*$ (\cite[Corollary 3.1.5]{BR}). However, the converse may not be true. Indeed, for a subfield $K$ of $\C$ closed under complex conjugation, given a rational series $z = \sum_{w \in F^*} \lambda_w w \in K_{\mathrm{rat}}\langle F \rangle$ it may be the case that its support $L = \text{supp}(z)$ is not a rational language. We put
$$\mathfrak K(F) := \{ \text{supp}(z) \mid z \in K_{\mathrm{rat}}\langle F \rangle^{\circ}\} \subseteq P(F^*),$$
and we let $\mathbf{B}_{\text{rat}}^K(F)$ be the Boolean algebra of subsets of $F^*$ generated by $\mathfrak K (F)$. Then by \cite[Proposition 6.10]{AC2} the $*$-regular closure of $K_{\text{rat}}\langle F\rangle ^{\circ}$ in $K\langle \langle F\rangle \rangle^{\circ}$ is contained in the set of formal power series whose support belongs to $\mathbf{B}_{\mathrm{rat}}^K(F)$, and for each set $L\in \mathbf{B}_{\mathrm{rat}}^K(F)$, the characteristic series $\ul{L}$ belongs to the $*$-regular closure.
We gather some of these facts in the following.
\begin{proposition}\label{proposition-numbers.from.languages}
Let $n\ge 1$ and let $\pazX_n=\{x_1,\dots , x_n\}$, endowed with the degree function $d(x_i)= i+1$. With the above notation, we have $\gotL(\pazX_n) \subseteq \mathfrak K (\pazX_n) \subseteq \textbf{\emph{B}}_{\mathrm{rat}}^K(\pazX_n)$. The sets $\gotL(\pazX_n)$ and $\textbf{\emph{B}}_{\mathrm{rat}}^K(\pazX_n)$ are Boolean subalgebras of $P(\pazX_n^*)$, but this is not the case in general for $\mathfrak K (\pazX_n)$. Moreover, if $L\in \mathbf{B}_{\mathrm{rat}}^K(\pazX_n)$, then $\alpha_L\in \calG (\Gamma, K)$.
\end{proposition}
\begin{proof}
See \cite[Chapter 3]{BR}. The fact that $\alpha_L\in \calG (\Gamma, K)$ for $L\in \mathbf{B}_{\mathrm{rat}}^K(\pazX_n)$ follows from \cite[Propositions 4.1, 4.26, 6.7 and 6.10]{AC2}.
\end{proof}
We now show that rational languages give rise to rational $\ell^2$-Betti numbers.
\begin{lemma}\label{lemma-generating.function.rational.language}
If a series $\sum_{j \geq 0} \alpha_j x^j \in \Z^+[[x]]$ is the generating function of a rational language $L \in \gotL(\pazX_n)$ with respect to $d$, then it is a rational series with constant term either $0$ or $1$. Consequently, $\alpha_L \in \Q$.
\end{lemma}
\begin{proof}
Suppose $\alpha_j = \#(L \cap d^{-1}(\{j\}))$, being $L \in \gotL(\pazX_n)$ a rational language. In particular the series $\ul{L}= \sum_{w \in L}w$ is rational. Define a map $\rho : \pazX_n \ra \Z^+[[x]]$ by setting $\rho(b) = x^{d(b)}$. By \cite[Proposition 1.4.2]{BR}, the map $\rho$ extends uniquely to a morphism $\rho : \Z^+\langle \langle \pazX_n \rangle \rangle \ra \Z^+[[x]]$ which induces the identity on $\Z^+$ and, moreover, preserves rationality. Then
$$\rho\Big( \sum_{w \in L}w \Big) = \sum_{w \in L}x^{d(w)} = \sum_{j \geq 0} \alpha_j x^j$$
is rational in $\Z^+[[x]]$, and clearly its constant term is either $0$ or $1$.
By \cite[Proposition 6.1.1]{BR}, there exist polynomials $p(x),q(x) \in \Z[x]$ such that the series $\sum_{j \geq 0}\alpha_j x^j$ is the power series expansion of the rational function $\frac{p(x)}{1-xq(x)}$, that is,
$$s(L) = \sum_{j \geq 0}\alpha_j x^j = \frac{p(x)}{1-xq(x)} = p(x) \Big[1 + \sum_{j \geq 0}(xq(x))^j\Big].$$
Consequently, if $L \in \gotL(\pazX_n)$, then $\alpha_L = \frac{1}{8}s(L)\big(\frac{1}{2}\big) \in \Q$.
\end{proof}
We obtain now a concrete irrational algebraic number of the form $\alpha_L$ for a suitable $L\in \mathbf{B}_{\text{rat}}^{\Q}(\pazX_2)$.
\begin{theorem}\label{thm:irrat-algl2betti}
There is $L\in \mathbf{B}_{{\rm rat}}^{\Q}(\pazX_2)$ such that $\alpha_L=\frac{1}{4}\sqrt{\frac{2}{7}}$. In particular, $\frac{1}{4}\sqrt{\frac{2}{7}}\in \calG(\Gamma, \mathbb Q)$.
\end{theorem}
\begin{proof}
Let $L$ be the language on $\pazX_2=\{x_1,x_2\}$ given by
$$L= \{ w\in \pazX_2^* \mid |w|_{x_1} =|w|_{x_2} \}.$$
Here $|w|_{x_i}$ is the number of appearances of $x_i$ in $w$, for $i=1,2$.
Then by \cite[Example 3.4.1]{BR}, $L\in \mathbf{B}_{\text{rat}}^{\Q}(\pazX_2)$ (although $L$ is {\it not} a rational language). If $w\in L$ then, since $d(x_1)= 2$ and $d(x_2)=3$, we get that $d(w)=5l$, where $l=|w|_{x_1}=|w|_{x_2}$. Therefore we get from Proposition \ref{proposition-numbers.from.languages} that
$$\alpha_L = \frac{1}{8} \sum_{j \geq 0} \frac{\alpha_j}{2^j} = \frac{1}{8} \sum_{l\ge 0} \binom{2l}{l} \frac{1}{2^{5l}} \in \calG (\Gamma, \Q).$$
Since
$$\sum_{l\ge 0} \binom{2l}{l} x^l = \frac{1}{\sqrt{1-4x}} \quad \text{ for } |x| < \frac{1}{4},$$
we obtain $\alpha _L = \frac{1}{4}\sqrt{\frac{2}{7}} \in \calG(\Gamma, \Q)$.
\end{proof}
Taking different choices of variables $x_r,x_s$, with $r\ne s$, we indeed obtain that $\calG (\Gamma, \Q)$ contains all the algebraic numbers
$$\frac{1}{8}\sqrt{\frac{2^t}{2^t-1}}, \quad \text{ for } t \geq 3.$$
Similar formulas can be obtained to compute the ranks of elements coming from the copy of the algebra $\Q_{\text{rat}}\langle \pazX\rangle$ in the $*$-regular closure $\calR_n$ of $\calA_n$ in $\mathfrak R_n$, where $\calA_n$ are the approximations of $\Q \Gamma$ described in Subsection \ref{subsection-approximating.algebras.lamplighter}. The only differences are that we have different degree functions on $\pazX$, depending on $n$, and that the factor $1/8$ must be substituted by the factor $2^{-2n-2}$.
\begin{remark}
By Theorem \ref{thm:irrat-algl2betti}, $\frac{1}{4}\sqrt{\frac{2}{7}}$ is a ``virtual'' $\ell^2$-Betti number arising from the lamplighter group $\Gamma $. This means that there are two $\ell^2$-Betti numbers arising from $\Gamma $, $\alpha_1$ and $\alpha_2$, such that $\frac{1}{4}\sqrt{\frac{2}{7}} = \alpha_1 - \alpha_2$. It would be interesting to know whether $\frac{1}{4}\sqrt{\frac{2}{7}}$ is itself an $\ell^2$-Betti number, and in that case, to determine a concrete matrix over $\mathbb Q \Gamma$ witnessing this fact. In particular, this would solve a question posed by Grabowski in \cite[p. 32]{Gra14}, see also \cite[Question 3]{Gra16}. We do not know any example of a group $G$ for which $\calC (G,\mathbb Q) \ne \calG (G,\mathbb Q)\cap \mathbb R^+$.
\end{remark}
We close this section with the following problem:
\begin{question}\label{quest:algebraics}
With the notation established before, is $\alpha_L$ always an algebraic number when $L \in \mathbf B_{{\rm rat}}^{\Q}(\pazX_n)$?
\end{question}
\section{The odometer algebra}\label{section-odometer.alg}
In this last section we focus on computing the whole set of values that the Sylvester matrix rank function constructed from Theorem \ref{theorem-rank.function} can achieve in the case of the generalized odometer algebra. We first recall its definition.
Fix a sequence of natural numbers $\ol{n} = (n_i)_{i \in \N}$ with $n_i \geq 2$ for all $i \in \N$, and consider $X_i$ to be the finite space $\{0,1,...,n_i-1\}$ endowed with the discrete topology. From these we form the topological space $X = \prod_{i \in \N}X_i$ endowed with the product topology, which is in fact a Cantor space. Let $T$ be the homeomorphism on $X$ given by the odometer, namely for $x = (x_i) \in X$, $T$ is given by
$$T(x) = \begin{cases}
(x_1+1,x_2,x_3,...) & \text{ if } x_1 \neq n_1-1, \\
(0,...,0,x_{m+1} + 1,x_{m+2},...) & \text{ if } x_j = n_j-1 \text{ for $1 \leq j \leq m$ and } x_{m+1} \neq n_m-1, \\
(0,0,...) & \text{ if } x_i = n_i-1 \text{ for all }i \in \N.
\end{cases}$$
Note that the odometer action is just addition of $(1,0,...)$ by carry-over.
Let $(K,-)$ be any field with a positive definite involution $-$. The \textit{generalized odometer algebra} is defined as the crossed product $*$-algebra $\calO(\ol{n}) := C_K(X) \rtimes_T \Z$. We obtain a measure $\mu$ on $X$ by taking the usual product measure, where we consider the measure on each component $X_i$ which assigns mass $\frac{1}{n_i}$ on each point in $X_i$. It is well-known (e.g. \cite[Section VIII.4]{Dav}) that $\mu$ is an ergodic, full and $T$-invariant probability measure on $X$, which in turn coincides with the Haar measure $\wh{\mu}$ on $X$ if one considers $X$ as an abelian group with addition by carry-over. We denote by $\rk_{\calO(\ol{n})}$ the Sylvester matrix rank function on $\calO(\ol{n})$ given by Theorem \ref{theorem-rank.function}.
\subsection{Characterizing the \texorpdfstring{$*$}{}-regular closure \texorpdfstring{$\calR_{\calO(\ol{n})}$}{}}\label{subsection-odometer.reg.closure}
We explicitly compute the $*$-regular closure $\calR_{\calO(\ol{n})}$ of the odometer algebra. Recall that it is defined as the $*$-regular closure of $\calO(\ol{n})$ inside $\gotR_{\rk}$ through the embedding $\calO(\ol{n}) \hookrightarrow \gotR_{\rk}$ given by Theorem \ref{theorem-rank.function}. We will use the notation $[a_1a_2 \dots a_r]$ for the cylinder sets $\{ (x_j)_{j\in \N}\mid x_i=a_i, 1 \leq i \leq r\}$, where $0\le a_i < n_i$, $1 \leq i \leq r$.
Define new integers $p_m := n_1 \cdots n_m$ for $m \in \N$. At each level $m \geq 1$, we take $E_m = [0 0 \cdots 0]$ (with $m$ zeroes) for the sequence of clopen sets, whose intersection gives the point $y = (0,0,...) \in X$. We take the partition $\calP_m$ of the complement $X \backslash E_m$ to be the obvious one, namely
$$\calP_m = \{[{1} 0 \cdots 0], ..., [{(n_1-1)} 0 \cdots 0],..., [{(n_1-1)} (n_2-1) \cdots (n_m-1)]\}.$$
We denote by $\calO(\ol{n})_m$ the unital $*$-subalgebra of $\calO(\ol{n})$ generated by the partial isometries $\{\chi_Z t \mid Z \in \calP_m\}$. The quasi-partition $\ol{\calP}_m$ is really simple in this case: write $Z_{m,l} = T^l(E_m)$ for $1 \le l < p_m$. Note that these clopen sets form exactly the partition $\calP_m$, and that $T(Z_{m,p_m-1}) = E_m$. Therefore there is only one possible $W \in \V_m$, which has length $p_m$ and is given by
$$W = E_m \cap T^{-1}(Z_{m,1}) \cap T^{-2}(Z_{m,2}) \cap \cdots \cap T^{-p_m+1}(Z_{m,p_m-1}) \cap T^{-p_m}(E_m) = E_m.$$
The representations $\pi_m : \calO(\ol{n})_m \to \gotR_m, x \mapsto (h_W \cdot x)_W$ become $*$-isomorphisms, where $\gotR_m = M_{p_m}(K)$.
Under this identification, the embeddings $\iota_m : \calO(\ol{n})_m \hookrightarrow \calO(\ol{n})_{m+1}$ become the block-diagonal embeddings
$$M_{p_m}(K) \hookrightarrow M_{p_{m+1}}(K), \quad x \mapsto \text{diag}(x,\stackrel{n_{m+1}}{\dots},x),$$
so that $\calO(\ol{n})_{\infty} := \varinjlim_m \calO(\ol{n})_m \cong \varinjlim_m M_{p_m}(K)$. Note that $\calO(\ol{n})_{\infty}$ is already $*$-regular, but it does not contain $\calO(\ol{n})$: in fact, it is contained in $\calO(\ol{n})$. Intuitively, in order to get containment of the whole algebra $\calO(\ol{n})$ we need to adjoin to $\calO(\ol{n})_{\infty}$ the element $t$.
\begin{definition}\label{definition-An.t}
For every $m \geq 1$, we denote by $\calO(\ol{n})^t_m$ the unital $*$-subalgebra of $\calO(\ol{n})$ generated by $\calO(\ol{n})_m$ and $t$.
\end{definition}
We can completely characterize these $*$-subalgebras.
\begin{lemma}\label{lemma-charac.An.t}
There exists a $*$-isomorphism $\calO(\ol{n})^t_m \cong M_{p_m}(K[t^{p_m}, t^{-p_m}])$.
\end{lemma}
\begin{proof}
For $0 \leq i,j < p_m$, the elements $e_{ij}^{(m)} := e_{ij}(E_m)$ form a complete system of matrix units inside $\calO(\ol{n})^t_m$, so there is an isomorphism $\calO(\ol{n})^t_m \cong M_{p_m}(T)$ with $T$ being the centralizer of the family $\{e_{ij}^{(m)} \mid 0 \leq i,j < p_m\}$ in $\calO(\ol{n})^t_m$. The isomorphism is given explicitly by
$$a \mapsto \sum_{i,j=0}^{p_m-1} a_{ij} e_{ij}^{(m)}, \quad \text{ with } a_{ij} = \sum_{k=0}^{p_m-1} e_{ki}^{(m)} \cdot a \cdot e_{jk}^{(m)} \in T,$$
which is also a $*$-isomorphism. Since $t^{p_m} e_{ij}^{(m)} t^{-p_m} = e_{ij}^{(m)}$ and
$$t = \sum_{i=0}^{p_m-1} t e_{ii}^{(m)} = \sum_{i=0}^{p_m-2} e_{i+1,i}^{(m)} + t^{p_m} e_{0,p_m-1}^{(m)} \in M_{p_m}(K[t^{p_m},t^{-p_m}]),$$
we deduce that $T = K[t^{p_m},t^{-p_m}]$, as desired.
\end{proof}
The inclusion $\calO(\ol{n})^t_m \subseteq \calO(\ol{n})^t_{m+1}$ translates to an embedding from $M_{p_m}(K[t^{p_m},t^{-p_m}])$ to \linebreak
$M_{p_{m+1}}(K[t^{p_{m+1}},t^{-p_{m+1}}])$ that extends the previous one $M_{p_m}(K) \hookrightarrow M_{p_{m+1}}(K)$, and sends the element $t^{p_m} \cdot \text{Id}_{p^m}$ to the element
$$\left( \begin{array}{@{}c@{}}
\begin{matrix}
\bigzero_{p_m} & & & \bigzero_{p_m} & t^{p_{m+1}} \cdot \text{Id}_{p_m} \\
\text{Id}_{p_m} & \bigzero_{p_m} & & & \bigzero_{p_m} \\
& \ddots & \ddots \stackinset{c}{-.1in}{c}{.15in}{\footnotesize$n_{m+1}$}{} & & \\
& & \ddots & \bigzero_{p_m} & \\
\hspace*{0.2cm} \bigzero_{p_m} & & & \text{Id}_{p_m} & \bigzero_{p_m}
\end{matrix}
\end{array}\right).$
\begin{lemma}\label{lemma-charac.An.t.2}
$\calO(\ol{n})$ is $*$-isomorphic to the direct limit $\varinjlim_m M_{p_m}(K[t^{p_m},t^{-p_m}])$ with respect to the previous embeddings.
\end{lemma}
\begin{proof}
By Remark \ref{remark-algebra.Ainfty} the algebra $\calO(\ol{n})_{\infty}$ contains $C_K(X)$, hence the $*$-subalgebra of $\calO(\ol{n})$ generated by $t$ and $\calO(\ol{n})_{\infty}$ is $\calO(\ol{n})$ itself. Now each $\calO(\ol{n})_m$ sits inside $\calO(\ol{n})^t_m$, hence $\calO(\ol{n})_{\infty} = \varinjlim_m \calO(\ol{n})_m \subseteq \varinjlim_m \calO(\ol{n})^t_m$. But $t \in \varinjlim_m \calO(\ol{n})^t_m$ already, so $\calO(\ol{n}) = \varinjlim_m \calO(\ol{n})^t_m \cong \varinjlim_m M_{p_m}(K[t^{p_m},t^{-p_m}])$ by Lemma \ref{lemma-charac.An.t}.
\end{proof}
We are now ready to compute $\calR_{\calO(\ol{n})}$.
\begin{theorem}\label{theorem-reg.closure.odometer}
There is a $*$-isomorphism $\calR_{\calO(\ol{n})} \cong \varinjlim_m M_{p_m}(K(t^{p_m}))$, where the maps $M_{p_m}(K(t^{p_m})) \hookrightarrow M_{p_{m+1}}(K(t^{p_{m+1}}))$ are induced by the inclusions of matrices over Laurent polynomial rings.
\end{theorem}
\begin{proof}
We have embeddings $\calO(\ol{n})^t_m \hookrightarrow \calO(\ol{n}) \hookrightarrow \calR_{\calO(\ol{n})} \hookrightarrow \gotR_{\rk}$. By \cite[Lemma 4.4]{AC2}, the rational function field $K(t^{p_m})$ sits inside $\calR_{\calO(\ol{n})}$. Hence there is, for each $m \geq 1$, a commutative diagram
\begin{equation*}
\xymatrix{
\calO(\ol{n})^t_m \ar@{^{(}->}[r] \ar@{^{(}->}[d] & M_{p_m}(K(t^{p_m})) \ar@{^{(}->}[r] & \calR_{\calO(\ol{n})} \ar@{}[d]|*=0[@]{=} \\
\calO(\ol{n})^t_{m+1} \ar@{^{(}->}[r] & M_{p_{m+1}}(K(t^{p_{m+1}})) \ar@{^{(}->}[r] & \calR_{\calO(\ol{n})}
}\label{diagram-comm.diag.5}
\end{equation*}
It is straightforward to prove that for any non-zero element $q(t) \in K[t^{p_m},t^{-p_m}]$, the corresponding matrix in $M_{p_{m+1}}(K[t^{p_{m+1}}, t^{-p_{m+1}}])$ becomes invertible in $M_{p_{m+1}}(K(t^{p_{m+1}}))$. Therefore the embedding $\calO(\ol{n})^t_m \hookrightarrow \calO(\ol{n})^t_{m+1}$ extends uniquely to an embedding $M_{p_m}(K(t^{p_m})) \hookrightarrow M_{p_{m+1}}(K(t^{p_{m+1}}))$, and
the previous commutative diagrams give embeddings
\begin{equation*}\label{equation-comm.diag.6}
\calO(\ol{n}) \cong \varinjlim_m \calO(\ol{n})^t_m \hookrightarrow \varinjlim_m M_{p_m}(K(t^{p_m})) \hookrightarrow \calR_{\calO(\ol{n})}.
\end{equation*}
But $\varinjlim_m M_{p_m}(K(t^{p_m}))$ is already $*$-regular and contains $\calO(\ol{n})$, so necessarily $\calR_{\calO(\ol{n})} \cong \varinjlim_m M_{p_m}(K(t^{p_m}))$, as required.
\end{proof}
In particular, since $\varinjlim_m M_{p_m}(K(t^{p_m}))$ has a unique Sylvester matrix rank function, it coincides with the one given by Theorem \ref{theorem-rank.function} under the previous $*$-isomorphism.
\subsection{Characterizing the set \texorpdfstring{$\calC'(\calO(\ol{n}))$}{}}\label{subsection-odometer.alg.betti.no}
We characterize the set $\calC'(\calO(\ol{n}))$ of all positive real values that the Sylvester matrix rank function $\rk_{\calO(\ol{n})}$ can achieve, and the subgroup $\calG(\calO(\ol{n}))$ it generates. First, let
$$\calC'(\calR_{\calO(\ol{n})}) := \rk_{\calR_{\calO(\ol{n})}}\Big( \bigcup_{k \geq 1} M_k(\calR_{\calO(\ol{n})}) \Big) \subseteq \R^+$$
and note that $\calC'(\calO(\ol{n})) \subseteq \calC'(\calR_{\calO(\ol{n})})= \calC(\calR_{\calO(\ol{n})}) $ (see Definition \ref{def:semigroupsCandCprime} and Remark \ref{rem:semigroupsCandCprime}). The following definition will be essential.
\begin{definition}\label{definition-supernatural.no}
For each sequence $\ol{n} = (n_1,n_2,...)$ of positive integers $n_i \geq 2$, one may associate to it the \textit{supernatural number}
$$n = \prod_{i \in \N} n_i = \prod_{q \in \P}q^{\varepsilon_q(n)},$$
where $\P$ is the set of prime numbers ordered with respect to the natural ordering, and each $\varepsilon_q(n) \in \{0\} \cup \N \cup \{\infty\}$. In more detail, if $n_i = \prod_{q \in \P}q^{\varepsilon_q(n_i)}$ is the prime decomposition of $n_i$, then, for each $q\in \P$, $\varepsilon_q (n)$ is defined by
$\varepsilon_q (n) = \sum_{i=1}^{\infty} \varepsilon _q (n_i) \in \{0\} \cup \N \cup \{\infty\}$, and $n$ is defined as the formal product $\prod_{q \in \P}q^{\varepsilon_q(n)}$.
\end{definition}
As in \cite[Definition 7.4.2]{LLR}, from any supernatural number $n$ one can construct an additive subgroup of $\Q$ containing $1$, denoted by $\Z(n)$, consisting of those fractions $\frac{a}{b}$ with $a \in \Z$, and $b \in \Z \backslash \{0\}$ being of the form
$$b = \prod_{q \in \P} q^{\varepsilon_q(b)},$$
where $\varepsilon_q(b) \leq \varepsilon_q(n)$ for all $q \in \P$, and $\varepsilon_q(b) = 0$ for all but finitely many $q$'s. If $n$ comes from a sequence $\ol{n} = (n_1,n_2,...)$ as above, $\Z(n)$ is exactly the additive subgroup of $\Q$ consisting of those fractions of the form
$$\frac{a}{n_1 \cdots n_r}, \quad \text{ with } a \in \Z \text{ and } r \geq 1.$$
Each group $\Z (n)$ gives rise to a subgroup $\mathbb G (n)=\Z(n)/\Z$ of the group $\Q/\Z$. Indeed the groups $\mathbb G (n)$ describe all the subgroups of $\Q/\Z$. These groups are called \textit{Pr\"ufer groups}.
\begin{theorem}\label{theorem-betti.numbers.odometer}
We have $\calC'(\calO(\ol{n})) = \calC(\calO(\ol{n})) = \calC'(\calR_{\calO(\ol{n})}) = \Z(n)^+$. In particular, $\calG(\calO(\ol{n})) = \Z(n)$.
\end{theorem}
\begin{proof}
The argument is similar to the one given in the proof of \cite[Proposition 4.1]{AC2}. Since $\calR_{\calO(\ol{n})}$ is a $*$-regular ring with positive definite involution, each matrix algebra $M_k(\calR_{\calO(\ol{n})})$ is also $*$-regular. Hence for each $A \in M_k(\calR_{\calO(\ol{n})})$ there exists a projection $P \in M_k(\calR_{\calO(\ol{n})})$ such that $\rk_{\calR_{\calO(\ol{n})}}(A) = \rk_{\calR_{\calO(\ol{n})}}(P)$. We conclude that $\calC'(\calR_{\calO(\ol{n})})$ equals the set of positive real numbers of the form $\rk_{\calR_{\calO(\ol{n})}}(P)$, where $P$ ranges over matrix projections with coefficients in $\calR_{\calO(\ol{n})}$. Now each such projection $P$ is equivalent to a diagonal projection \cite[Proposition 2.10]{Goo91}, that is of the form $\text{diag}(p_1,p_2,...,p_r)$ for some projections $p_1,...,p_r \in \calR_{\calO(\ol{n})}$,
so that $\rk_{\calR_{\calO(\ol{n})}}(P) = \rk_{\calR_{\calO(\ol{n})}}(p_1) + \cdots + \rk_{\calR_{\calO(\ol{n})}}(p_r)$. But since $\calR_{\calO(\ol{n})} \cong \varinjlim_m M_{p_m}(K(t^{p_m}))$ by Theorem \ref{theorem-reg.closure.odometer}, the set of ranks of elements in $\calR_{\calO(\ol{n})}$ is contained in $\Z(n)^+ \cap [0,1]$. Therefore $\rk_{\calR_{\calO(\ol{n})}}(P) \in \Z(n)^+$. This proves the inclusion $\calC'(\calR_{\calO(\ol{n})}) \subseteq \Z(n)^+$. The inclusion $\Z(n)^+ \subseteq \calC'(\calO(\ol{n}))$ is straightforward, since for $0 \leq i < p_m$ we have $\rk_{\calO(\ol{n})}(e_{ii}^{(m)}) = \frac{1}{p_m}$. The equality
$ \calC (\calO(\ol{n}))= \Z (n)^+$ also follows from the above.
The last part of the theorem is immediate.
\end{proof}
In a sense, Theorem \ref{theorem-betti.numbers.odometer} confirms the SAC for a class of crossed product algebras, as follows:
\begin{remark}\label{rem:fourierforX}
Let $n$ be a supernatural number and let $n= \prod_{i\in \N} n_i$ a decomposition of $n$, as above. Consider the presentation of $\mathbb G (n)$ given by generators $\{g_m \}_{m\in \N}$ and relations $g_1^{n_1}= 1$ and $g_{m+1}^{n_{m+1}} = g_m$ for all $m\ge 1$. The topological group $X(n)=\prod_{i\in \N} X_i$ is naturally isomorphic with the Pontryagin dual $\widehat{\mathbb G (n)}$ of the Pr\"ufer group $\mathbb G (n)$. Indeed, let $\xi_{p_m}$ denote the canonical primitive $p_m$-root of unity in $\C$, where $p_m= n_1\cdots n_m$, and observe that $\xi_{p_{m+1}}^{m_{n+1}}= \xi _{p_m}$ for each $m\in \N$. The map $\phi \colon X(n)\to \widehat{\mathbb G (n)}$ defined by
$$\phi_x (g_m)= \xi_{p_m}^{a_1+a_2n_1+\cdots + a_mn_{m-1}}, $$
where $x= (a_1,a_2,\dots )\in X(n)$, is a group isomorphism and a homeomorphism. Therefore if $K$ is a subfield of $\C$ closed under complex conjugation containing all $p_m^{\text{th}}$ roots of unity, Fourier transform defines an isomorphism
$$\mathcal F \colon K[\mathbb G(n)] \longrightarrow C_K(X(n)) ,$$
and we can pull back the automorphism induced by $T$ on $C_K(X(n))$ to an automorphism $\rho$ on $K[\mathbb G (n)]$ (which is {\it not} induced by an automorphism of $\mathbb G (n)$). Note that $\rho (g_m)= \xi_{p_m} g_m\in K\cdot \mathbb G (n)$ for all $m\ge 1$. Therefore we can interpret Theorem \ref{theorem-betti.numbers.odometer} as giving a positive answer to the SAC for the crossed product $\calA = K[\mathbb G (n)]\rtimes_{\rho}\Z\cong \calO (n)$: the set of $\calA$-Betti numbers is exactly the semigroup $\Z(n)^+$ generated by the inverses of the orders of the elements of $\mathbb G (n)$. Note finally that the groups $X(n)$, $n$ a supernatural, give all the profinite completions of $\Z$. With this view, the dynamical system considered on $X(n)$ is generated by addition by $1$.
\end{remark}
\section*{Acknowledgments}
The authors would like to thank the anonymous referees for their very careful reading of the manuscript and for their many suggestions, which have improved the exposition of the paper.
\section*{Data availability statement}
Data sharing not applicable to this article as no datasets were generated or analysed during the current study.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 4,011 |
\section{Introduction}
\label{sec:introduction}
Cryptocurrencies provide block rewards to incentivize miners in producing new blocks.
Transaction fees also contribute to the revenue of miners, motivating them to include transactions in blocks.
Miners maximize their profits by prioritizing the transactions with the highest fees.
In Bitcoin and its numerous clones~\cite{hum2020coinwatch}, the block reward is divided by two approx. every four years (i.e., after every 210k blocks), which will eventually result in a pure transaction-fee-based regime.
Before 2016, there was a belief that the dominant source of the miners' income does not impact the security of the blockchain.
However, Carlsten et al.~\cite{carlsten2016instability} pointed out the effects of the high variance of the miners' revenue per block caused by exponentially distributed block arrival time in transaction-fee-based protocols.
The authors showed that \emph{undercutting} (i.e., forking) a wealthy block is a profitable strategy for a malicious miner.
Nevertheless, Daian et al.~\cite{daian2020flash} showed that this attack is viable even in blockchains containing traditional block rewards due to front-running competition of arbitrage bots who are willing to extremely increase transaction fees to earn Maximum Extractable Value profits.
In this paper, we focus on mitigation of the undercutting attack in transaction-fee-based regime of PoW blockchains.
We also discuss related problems present (not only) in transaction-fee-based regime.
In particular, we focus on minimizing the mining gap~\cite{carlsten2016instability,tsabary2018gap}, (i.e., the situation, where the immediate reward from transaction fees does not cover miners' expenditures) as well as balancing significant fluctuations in miners' revenue.
To mitigate these issues, we propose a solution that splits transaction fees from a mined block into two parts -- (1) an instant reward for the miner and (2) a deposit sent into one or more fee-redistribution smart contracts ($\mathcal{FRSC}$s).
At the same time, these $\mathcal{FRSC}$s reward the miner of a block with a certain fraction of the accumulated funds over a fixed period of time (i.e., the fixed number of blocks).
This setting enables us to achieve several interesting properties that are beneficial for the stability and security of the protocol.
\subsubsection*{\textbf{Contributions}}
In detail, our contributions are as follows:
\begin{enumerate}
\item We propose an approach that normalizes the mining rewards coming from transaction fees by one or more $\mathcal{FRSC}$s that perform moving average on a certain portion of the transaction fees.
\item We evaluate our approach using various fractions of the transaction fees from a block distributed between a miner and $\mathcal{FRSC}$s.
We experiment with the various numbers and lengths of $\mathcal{FRSC}$s, and we demonstrate that usage of multiple $\mathcal{FRSC}$s of various lengths has the best advantages mitigating the problems we are addressing; however, even using a single $\mathcal{FRSC}$ is beneficial.
\item We demonstrated that with our approach, the mining gap can be minimized since the miners at the beginning of the mining round can get the reward from $\mathcal{FRSC}$s, which stabilizes their income.
\item We empirically demonstrate that using our approach the threshold of \textsc{Default-Compliant} miners who strictly do not execute undercutting attack is lowered from 66\% (as reported in the original work~\cite{carlsten2016instability}) to 30\%.
\end{enumerate}
\section{Problem Definition}
\label{sec:problemdefinition}
In transaction fee-based regime schemes, a few problems have emerged, which we can observe even nowadays in Bitcoin protocol~\cite{carlsten2016instability}.
We have selected three main problems and aim to lower their impact for protocols relying only on transaction fees.
In detail, we focus on the following problems:
\begin{enumerate}
\item \textbf{Undercutting attack.}
In this attack, a malicious miner attempts to obtain transaction fees by re-mining a top block of the longest chain, and thus motivates other miners to mine on top of her block~\cite{carlsten2016instability}.
In detail, consider a situation, where an honest miner mines a block containing transaction with extremely high transaction fees.
The malicious miner can fork this block while he leaves some portion of the ``generous'' transactions un-mined in the mempool.
These transactions motivate other miners to mine on top of the attacker's chain, and thus undercut the original block.
Such a malicious behavior might result in higher orphan rate, unreliability of the system, and double spending.
\item \textbf{The mining gap.}
As discussed in~\cite{carlsten2016instability}, the problem of mining gap arises once the mempool does not contain enough transaction fees to motivate miners in mining.
Suppose a miner succeeds at mining a new block shortly after the previous block was created, which can happen due to well known exponential distribution of block creation time in PoW blockchains.
Therefore, the miner might not receive enough rewards to cover his expenses because most of the transactions from the mempool were included in the previous block, while new transactions might not have yet arrived or have small fees.
Consequently, the miners are motivated to postpone mining until the mempool is reasonably filled with enough transactions (and their fees).
The mining gap was also analyzed by the simulation approach in the work of Tsabary and Eyal~\cite{tsabary2018gap}, who further demonstrated that this mining gap incentivizes larger mining coalitions, which impacts decentralization.
\item \textbf{Varying transaction fees over time.}
In the transaction-fee-based regime, any fluctuation in transaction fees directly affects the miners' revenue.
High fluctuation of transaction fees during certain time frames, e.g., in a span of a day or a week~\cite{b5}, can lead to an undesirable lack of predictability in rewards of miners and indirectly affect the security of the underlying protocol.
\end{enumerate}
\section{Proposed approach}
\label{sec:proposedapproach}
In this section, we describe our proposed solution in more detail.
\begin{figure}[t]
\includegraphics[width=0.45\textwidth]{images/explanation.png}
\vspace{-0.7cm}
\caption{Overview of our solution. }
\label{fig:overview}
\end{figure}
Our proposed solution collects a percentage of transaction fees in a native cryptocurrency from the mined blocks into one or multiple fee-redistribution smart contracts (i.e., $\mathcal{FRSC}$s).
Miners of the blocks who must contribute to these contracts are at the same time rewarded from them, while the received reward approximates a moving average of the incoming transaction fees across the fixed sliding window of the blocks.
The fraction of transaction fees (i.e., $\mathbb{C}$) from the mined block is sent to $\mathcal{FRSC}$ and the remaining fraction of transaction fees (i.e., $\mathbb{M}$) is directly assigned to the miner, such that $ \mathbb{C} + \mathbb{M} = 1.$
The role of $\mathbb{M}$ is to incentivize the miners in prioritization of the transactions with the higher fees while the role of $\mathbb{C}$ is to mitigate the problems of undercutting attacks and the mining gap.
\medskip\noindent
\subsubsection*{\textbf{Overview}}
We depict the overview of our approach in \autoref{fig:overview}, and it consists of the following steps:
\begin{enumerate}
\item Using $\mathcal{FRSC}$, the miner calculates the reward for the next block $B$ (i.e., $nextClaim(\mathcal{FRSC})$ -- see \autoref{eq:nextClaim}) that will be payed by $\mathcal{FRSC}$ to the miner of that block.
\item The miner mines the block $B$ using the selected set of the highest fee transactions from her mempool.
\item The mined block $B$ directly awards a certain fraction of the transaction fees (i.e., $B.fees ~*~ \mathbb{M}$) to the miner and the remaining part (i.e., $B.fees ~*~ \mathbb{C}$) to $\mathcal{FRSC}$.
\item The miner obtains $nextClaim$ from $\mathcal{FRSC}$.
\end{enumerate}
Our approach is embedded into the consensus protocol, and therefore consensus nodes are obliged to respect it in order to ensure that their blocks are valid.
It can be implemented with standard smart contracts of the blockchain platform or within the native code of the consensus protocol.
\subsection{Prioritization of High-Fee Transactions}
In the environment with constant transaction fees, a miner would receive the same amount with or without our solution.
However, in public blockchains (especially with transaction-fee based regime) there exist a mechanism to ensure prioritizaiton in processing of transactions with higher fees, which might result into fluctuations in rewards of the miners.
In our approach, we preserve the transaction prioritization since we directly attribute a part of the transaction fees to the miner (i.e., $\mathbb{M}$).
\subsection{Fee-Redistribution Smart Contracts}\label{sub:redistrib}
We define the fee-redistribution smart contract as
\begin{eqnarray}
\mathcal{FRSC} = (\nu, \lambda, \rho),
\end{eqnarray}
where
$\nu$ is the accumulated amount of tokens in the contract, $\lambda$ denotes the size of $\mathcal{FRSC}$' sliding window in terms of the number of preceding blocks that contributed to $\nu$,
and $\rho$ is the parameter defining the ratio for redistribution of incoming transaction fees among multiple contracts, while the sum of $\rho$ across all $\mathcal{FRSC}$s must be equal to 1:
\begin{eqnarray}\label{eq:frsc-redistrib-ratios}
\sum_{x ~\in~ \mathcal{FRSC}s} x.\rho &=& 1.
\end{eqnarray}
In contrast to a single $\mathcal{FRSC}$, multiple $\mathcal{FRSC}$s enable better adjustment of compensation to miners during periods of higher transaction fee fluctuations or in an unpredictable environment (we show this in \autoref{sec:exp3}).
\medskip
We denote the state of $\mathcal{FRSC}$s at the blockchain height $H$ as $\mathcal{FRSC}_{[H]}$.
Then, we determine the reward from $\mathcal{FRSC}_{[H]} \in \mathcal{FRSC}s_{[H]}$ for the miner of the next block with height $H+1$ as follows:
\begin{equation}
\partial Claim_{[H+1]}^{\mathcal{FRSC}_{[H]}} = \frac{\mathcal{FRSC}_{[H]}.\nu}{\mathcal{FRSC}_{[H]}.\lambda},
\end{equation}
while the reward obtained from all $\mathcal{FRSC}$s is
\begin{equation} \label{eq:nextClaim}
nextClaim_{[H+1]} = \sum_{\mathcal{X}_{[H]} ~\in~ \mathcal{FRSC}s_{[H]}^{}} \partial Claim_{[H+1]}^{\mathcal{X}_{[H]}}.
\end{equation}
\noindent
Then, the total reward of the miner who mined the block $B_{[H+1]}$ with all transaction fees $B_{[H+1]}.fees$ is
\begin{equation} \label{eq:rewardtotal}
rewardT_{[H+1]} = nextClaim_{[H+1]} + \mathbb{M} * B_{[H+1]}.fees.
\end{equation}
The new state of contracts at the height $H + 1$ is
\begin{eqnarray}
\mathcal{FRSC}s_{[H+1]} = \{\mathcal{X}_{[H+1]}(\nu, \lambda, \rho)\ ~|~
\end{eqnarray}
\begin{eqnarray}
\lambda &=& \mathcal{X}_{[H]}.\lambda,\\
\rho &=& \mathcal{X}_{[H]}.\rho,\\
\nu &=& \mathcal{X}_{[H]}.\nu - \partial Claim_{[H+1]} + deposit * \rho,\\
deposit &=& B_{[H+1]}.fees * \mathbb{C}\},
\end{eqnarray}
where $deposit$ represents the fraction $\mathbb{C}$ of all transaction fees from the block $B_{[H+1]}$ that are deposited across all $\mathcal{FRSC}$s in ratios respecting \autoref{eq:frsc-redistrib-ratios}.
\subsection{Example}
We present an example using Bitcoin~\cite{nakamoto2008bitcoin} to demonstrate our approach.
We assume that the current height of the blockchain is $H$, and we utilize only a single $\mathcal{FRSC}$ with the following parameters:
\[
\mathcal{FRSC}_{[H]} = (2016, 2016, 1).
\]
We set $\mathbb{M} = 0.4$ and $\mathbb{C} = 0.6$, which means a miner directly obtains 40\% of the $B_{[H+1]}.fees$ and $\mathcal{FRSC}$ obtains remaining 60\%.
\medskip
\noindent
Next, we compute the reward from $\mathcal{FRSC}$ obtained by the miner of the block with height $H+1$ as
\begin{equation*}
\partial Claim_{[H+1]} = \frac{\mathcal{FRSC}_{[H]}.\nu}{\mathcal{FRSC}_{[H]}.\lambda} = \frac{2016}{2016} = 1~\text{BTC},
\end{equation*}
resulting into
\begin{equation*}
nextClaim_{[H+1]} = \partial Claim_{[H+1]}~=~1~\text{BTC}.
\end{equation*}
\noindent
Further, we assume that the total reward collected from transactions in the block with height $H+1$ is $B_{[H+1]}.fees = 2$ BTC.
Hence, the total reward obtained by the miner of the block $B_{[H+1]}$ is
\begin{eqnarray*}
rewardT_{[H+1]} &=& nextClaim_{[H+1]} + \mathbb{M} * B_{[H+1]}.fees \\
&=& 1 + 0.4 * 2 \\
&=& 1.8 ~\text{BTC},
\end{eqnarray*}
and the contribution of transaction fees from $B_{[H+1]}$ to the $\mathcal{FRSC}$ is
\[
deposit = B_{[H+1]}.fees * \mathbb{C} = 1.2 ~\text{BTC}.
\]
Therefore, the value of $\nu$ in $\mathcal{FRSC}$ is updated at height H~+~1 as follows:
\begin{eqnarray*}
v_{[H+1]} &=& \mathcal{FRSC}_{[H]}.\nu - nextClaim_{[H+1]} + deposit \\
&=& 2016 - 1 + 1.2 ~\text{BTC} \\
&=& 2016.2 ~\text{BTC}.
\end{eqnarray*}
\noindent
\subsubsection*{\textbf{Traditional Way in Tx-Fee Regime}}
In traditional systems (running in transaction-fee regime) $rewardT_{[H+1]}$ would be equal to the sum of all transaction fees $B_{[H+1]}.fees$ (i.e., $2$ BTC); hence, using $\mathbb{M} = 1$.
In our approach, $rewardT_{[H+1]}$ can only be equal to the sum of all transaction fees in the block $B_{[H+1]}$, if:
\begin{eqnarray}
B_{[H+1]}.fees = \frac{nextClaim_{[H+1]}}{\mathbb{C}}.
\end{eqnarray}
In our example, a miner can mine the block $B_{[H+1]}$ while obtaining the same total reward as the sum of all transaction fees in the block if the transactions carry 1.66 BTC in fees:
\begin{equation*}
B_{[H+1]}.fees = \frac{1}{0.6} = 1.66 ~\text{BTC}.
\end{equation*}
\subsection{Initial Setup of $\mathcal{FRSC}$s Contracts}
To enable an even start, we propose to initiate $\mathcal{FRSC}$s of our approach by a genesis value.
The following formula calculates the genesis values per $\mathcal{FRSC}$ and initializes starting state of $\mathcal{FRSC}s_{[0]}$:
\begin{equation} \label{eq:setup}
\{\mathcal{FRSC}_{[0]}^{x}(\nu, \lambda, \rho)\ |\ \nu = \overline{fees} * \mathbb{C} * \rho * \lambda\},
\end{equation}
where $\overline{fees}$ is the expected average of incoming fees.
\section{Implementation}
\label{sec:implementation}
Our implementation is based on Bitcoin Mining Simulator~\cite{bitcoin_mining_simulator:kalodner}, introduced in~\cite{carlsten2016instability}, which was adjusted to our needs.
\subsubsection{Changes in Simulator}
We have created a configuration file to simulate custom scenarios of incoming transactions instead of the original design~\cite{carlsten2016instability} fees.
We added an option to switch simulation into a mode with a full mempool, and thus bound the total fees (and consequently the total number of transactions) that can be earned within a block --
this mostly relates to blocks whose mining takes longer time than the average time to mine a block.\footnote{Note that the original simulator~\cite{carlsten2016instability} assumes that the number of transactions (and thus the total fees) in the block is constrained only by the duration of a time required to mine the block, which was also criticized in~\cite{gong2022towards}.}
Next, we moved several parameters to arguments of the simulator with the intention to eliminate the need of recompilation the program frequently, and therefore simplify the process of running various experiments with the simulator.
Finally, we integrated our $\mathcal{FRCS}$-based solution into the simulator.
$\mathcal{FRSC}$s are initiated from a corresponding configuration file.
The source code of our modified simulator is available at anonymous link \url{https://www.dropbox.com/s/gwlbedq47csthqp/sources.zip?dl=0}.
\section{Evaluation}
\label{sec:evaluation}
We evaluated our proof-of-concept implementation of $\mathcal{FRCS}$s on a long term scenario that we designed to demonstrate significant changes in the total transaction fees in the mempool evolving across the time.
This scenario is depicted in the resulting graphs of most of our experiments, represented by the ``\textit{Fees in mempool}'' series -- see \autoref{sec:exp1} and \autoref{sec:exp2}.
We experimented with different parameters and investigated how they influenced the total rewards of miners coming from $\mathcal{FRSC}$s versus the baseline without our solution.
Mainly these included a setting of $\mathbb{C}$ as well as different lengths $\lambda$ of $\mathcal{FRSC}$s.
Note that we used the value of transaction fees per block equal to 50 BTC, alike in the original paper introducing undercutting attacks~\cite{carlsten2016instability}.
Also, in all our experiments but the last one (i.e., \autoref{sec:exp4}), we enabled the full mempool option to ensure more realistic conditions.
\begin{figure*}[t]
\vspace{-0.7cm}
\centering
\begin{subfigure}{0.40\textwidth}
\includegraphics[width=\textwidth]{images/fees-2016-1-50.png}
\vspace{-0.9cm}
\caption{$\mathcal{FRSC}^{1}$ and $\mathbb{C} = 0.5$.}
\end{subfigure}
\begin{subfigure}{0.40\textwidth}
\includegraphics[width=\textwidth]{images/fees-1-50.png}
\vspace{-0.9cm}
\caption{$\mathcal{FRSC}^{2}$ and $\mathbb{C} = 0.5$.}
\end{subfigure}
\begin{subfigure}{0.40\textwidth}
\includegraphics[width=\textwidth]{images/fees-2016-1-70.png}
\vspace{-0.9cm}
\caption{$\mathcal{FRSC}^{1}$ and $\mathbb{C} = 0.7$.}
\end{subfigure}
\begin{subfigure}{0.40\textwidth}
\includegraphics[width=\textwidth]{images/fees-1-70.png}
\vspace{-0.9cm}
\caption{$\mathcal{FRSC}^{2}$ and $\mathbb{C} = 0.7$.}
\end{subfigure}
\begin{subfigure}{0.40\textwidth}
\includegraphics[width=\textwidth]{images/fees-2016-1-90.png}
\vspace{-0.9cm}
\caption{$\mathcal{FRSC}^{1}$ and $\mathbb{C} = 0.9$.}
\end{subfigure}
\begin{subfigure}{0.40\textwidth}
\includegraphics[width=\textwidth]{images/fees-1-90.png}
\vspace{-0.9cm}
\caption{$\mathcal{FRSC}^{2}$ and $\mathbb{C} = 0.9$.}
\vspace{-0.4cm}
\end{subfigure}
\caption{Experiment I investigating various $\mathbb{C}s$ and $\lambda$s of a single $\mathcal{FRSC}$, where
$\mathcal{FRSC}^{1}.\lambda = 2016$ and $\mathcal{FRSC}^{2}.\lambda = 5600$.
\textit{Fees in mempool} show the total value of fees in the mined block (i.e., representing the baseline).
\textit{Block Value} is the reward a miner received in block $B$ as a sum of the fees he obtained directly (i.e. $\mathbb{M} * B.fees$) and the reward he got from $\mathcal{FRSC}$ (i.e., $nextClaim_{[H]}$).
\textit{Expected income from Contract} represents the reward of a miner obtained from $\mathcal{FRSC}$ (i.e., $nextClaim_{[H]}$).}\label{fig:50tocontract}
\end{figure*}
\subsection{Experiment I}
\label{sec:exp1}
\subsubsection{\textbf{Methodology}}
The purpose of this experiment was to investigate the amount of the reward a miner received with our approach versus the baseline (i.e., the full reward is based on all transaction fees).
In this experiment, we investigated how $\mathbb{C}$ influences the total reward of the miner and how $\lambda$ of the sliding window averaged the rewards.
In detail, we created two independent $\mathcal{FRSC}$s with different $\lambda$ -- one was set to 2016 (i.e., $\mathcal{FRCS}^1$), and the second one was se to 5600 (i.e., $\mathcal{FRCS}^2$).
We simulated these two $\mathcal{FRCS}$s with three different values of $\mathbb{C} \in \{0.5,~0.7,~0.9\}$.
\subsubsection{\textbf{Results}}
The results of this experiment are depicted in \autoref{fig:50tocontract}.
Across all runs of our experiment we can observe that $\mathcal{FRSC}^{2}$ adapts slower as compared to $\mathcal{FRSC}^{1}$, which leads to more significant averaging of the total reward paid to the miner.
Even though this is desired, we can see faster adaptation to changing environment by $\mathcal{FRSC}^{1}$, what a miner can expects from rising fees.
\begin{figure*}
\vspace{-0.5cm}
\centering
\begin{subfigure}{0.32\textwidth}
\includegraphics[width=\textwidth]{images/fees-4-50.png}
\caption{Scenario with 4 $\mathcal{FRSC}$s,\\ $\mathbb{C} = 0.5$.}
\end{subfigure}
\begin{subfigure}{0.32\textwidth}
\includegraphics[width=\textwidth]{images/contracts-4-50.png}
\caption{$\partial Claim$s and $nextClaim$, \\ $\mathbb{C} = 0.5$.}
\end{subfigure}
\begin{subfigure}{0.32\textwidth}
\includegraphics[width=\textwidth]{images/normalized-contracts-4-50.png}
\caption{$\partial Claim$s normalized by $\rho$, \\ $\mathbb{C} = 0.5$.}
\end{subfigure}
\begin{subfigure}{0.32\textwidth}
\includegraphics[width=\textwidth]{images/fees-4-70.png}
\caption{Scenario with 4 $\mathcal{FRSC}$s,\\ $\mathbb{C} = 0.7$.}
\end{subfigure}
\begin{subfigure}{0.32\textwidth}
\includegraphics[width=\textwidth]{images/contracts-4-70.png}
\caption{$\partial Claim$s and $nextClaim$, \\ $\mathbb{C} = 0.7$.}
\end{subfigure}
\begin{subfigure}{0.32\textwidth}
\includegraphics[width=\textwidth]{images/normalized-contracts-4-70.png}
\caption{$\partial Claim$s normalized by $\rho$, \\ $\mathbb{C} = 0.7$.}
\end{subfigure}
\begin{subfigure}{0.32\textwidth}
\includegraphics[width=\textwidth]{images/fees-4-90.png}
\caption{Scenario with 4 $\mathcal{FRSC}$s, \\ $\mathbb{C} = 0.9$.}
\end{subfigure}
\begin{subfigure}{0.32\textwidth}
\includegraphics[width=\textwidth]{images/contracts-4-90.png}
\caption{$\partial Claim$s and $nextClaim$, \\ $\mathbb{C} = 0.9$.}
\end{subfigure}
\begin{subfigure}{0.32\textwidth}
\includegraphics[width=\textwidth]{images/normalized-contracts-4-90.png}
\caption{$\partial Claim$s normalized by $\rho$, \\ $\mathbb{C} = 0.9$.}
\end{subfigure}
\caption{Experiment II investigating various $\mathbb{C}$s in the setting with multiple $\mathcal{FRSC}$s with their corresponding $\lambda$ = $\{1008, 2016, 4032, 8064\}$ and $\rho$ = $\{0.07, 0.14, 0.28, 0.51\}$. $\partial Claim$s represents contributions of individual $\mathcal{FRSC}$s to the total reward of the miner (i.e., its $nextClaim$ component).}\label{fig:exp-II}
\end{figure*}
\begin{figure*}
\centering
\begin{subfigure}{0.32\textwidth}
\includegraphics[width=\textwidth]{images/contracts-4-original.png}
\caption{$\rho$ correlating with $\lambda$.}
\end{subfigure}
\begin{subfigure}{0.32\textwidth}
\includegraphics[width=\textwidth]{images/contracts-4-equal.png}
\caption{$\rho$ equal for every $\mathcal{FRSC}$.}
\end{subfigure}
\begin{subfigure}{0.32\textwidth}
\includegraphics[width=\textwidth]{images/contracts-4-inverse.png}
\caption{$\rho$ negatively correlating with $\lambda$.}
\end{subfigure}
\caption{Experiment II -- multiple $\mathcal{FRSC}$s using various distributions of $\rho$ and their impact on $\partial Claim$, where $\mathbb{C}$ = 0.7.} \label{fig:percentages}
\end{figure*}
\begin{figure}
\centering
\begin{subfigure}{0.44\textwidth}
\includegraphics[width=\textwidth]{images/fees-1-percentage.png}
\caption{A custom fee scenario for Experiment III.}\label{fig:exp3-fees}
\end{subfigure}
\begin{subfigure}{0.44\textwidth}
\includegraphics[width=\textwidth]{images/percentage.png}
\caption{A relative difference in $nextClaim$ between 4 $\mathcal{FRSC}$s and a single $\mathcal{FRSC}$.}\label{fig:exp3-relative-diff}
\end{subfigure}
\caption{Experiment III comparing 4 $\mathcal{FRSC}$s and 1 $\mathcal{FRSC}$, both configurations having the same $\text{effective}\_\lambda$. }\label{fig:effective_length}
\end{figure}
\subsection{Experiment II}
\label{sec:exp2}
\subsubsection{\textbf{Methodology}}
In this experiment, we investigated how multiple $\mathcal{FRSC}$s dealt with the same scenario as before -- i.e., varying $\mathbb{C}$.
In detail, we investigated how individual $\mathcal{FRSC}$s contributed to the $nextClaim_{[H+1]}$ by their individual $\partial Claim^{\mathcal{FRSC}_{[H]}}_{[H+1]}$.
This time, we varied only the parameter $\mathbb{C} \in \{0.5, ~0.7, ~0.9\}$, and we considered four $\mathcal{FRSC}$s:
\begin{center}
$\mathcal{FRSC}s=\{$\\
$\mathcal{FRSC}^{1}(\_, 1008, 0.07), \mathcal{FRSC}^{2}(\_, 2016, 0.14),$\\
$\mathcal{FRSC}^{3}(\_, 4032, 0.28), \mathcal{FRSC}^{4}(\_, 8064, 0.51)\},$
\end{center}
where their lengths $\lambda$ were set to consecutive powers of two (to see differences in more intensive averaging spread across longer intervals), and their redistribution ratios $\rho$ were set to maximize the potential of averaging by longer $\mathcal{FRSC}$s (see details below in \autoref{sec:different-rhos}).
\subsubsection{\textbf{Results}}
The results of this experiment are depicted in \autoref{fig:exp-II}.
We can observe that the shorter $\mathcal{FRSC}$s quickly adapted to new changes and the longer $\mathcal{FRSC}$s kept more steady income for the miner.
In this sense, we can see that $\partial Claim^{4}$ held steadily over the scenario while for example $\partial Claim^{1}$ fluctuated more.
Since the scenarios of fees evolution in the mempool was the same across all our experiments (but \autoref{sec:exp3}), we can compare the $\mathcal{FRSC}$ with $\lambda$ = 5600 from \autoref{sec:exp1} and the current setup involving four $\mathcal{FRSC}$s -- both had some similarities.
This gave us intuition for replacing multiple $\mathcal{FRSC}$s with a single one, which we further investigated in \autoref{sec:exp3}.
\subsubsection{\textbf{Different $\rho$s}}\label{sec:different-rhos}
In \autoref{fig:percentages} we investigated different values of $\rho$ in the same set of four contracts and their impact on $\partial Claim$s.
The results show that the values of $\rho$ should correlate with $\lambda$ of multiple $\mathcal{FRSC}$s to maximize the potential of averaging by longer $\mathcal{FRSC}$s.
\subsection{Experiment III}\label{sec:exp3}
\subsubsection{\textbf{Methodology}}
In this experiment, we investigated whether it is possible to use a single $\mathcal{FRSC}$ setup to replace a multiple $\mathcal{FRSC}$s while preserving the same effect on the $nextClaim$.
To quantify a difference between such cases, we introduced a new metric of $\mathcal{FRSC}s$, called $\text{effective}\_\lambda$, which can be calculated as follows:
\begin{eqnarray} \label{eq:effective_length}
\text{effective}\_\lambda (\mathcal{FRSC}s) = \sum_{x ~\in~ \mathcal{FRSC}s} x.\rho* x.\lambda.
\end{eqnarray}
As the example, we were interested in comparing a single $\mathcal{FRSC}$ with 4 $\mathcal{FRSC}$s, both configurations having the equal $\text{effective}\_\lambda$.
The configurations of these two cases are as follow:
(1) $\mathcal{FRSC}(\_, 5292, 1)$ and (2)
\begin{center}
$\mathcal{FRSC}s=\{$\\
$\mathcal{FRSC}^{1}(\_, 1008, 0.07), \mathcal{FRSC}^{2}(\_, 2016, 0.19),$\\
$ \mathcal{FRSC}^{3}(\_, 4032, 0.28), \mathcal{FRSC}^{4}(\_, 8064, 0.46)\}$.
\end{center}
We can easily verify that the $\text{effective}\_\lambda$ of 4 $\mathcal{FRSC}$s is the same as in a single $\mathcal{FRSC}$ using \autoref{eq:effective_length}:
$0.07 * 1008 + 0.19 * 2016 + 0.28 * 4032 + 0.46 * 8064 = 5292$.
We conducted this experiment using a custom fee evolution scenario involving mainly linearly increasing/decreasing fees in the mempool (see \autoref{fig:exp3-fees}), and we set $\mathbb{C}$ to 0.7 for both configurations.
The new scenario of the fee evolution in mempool was chosen to contain extreme changes in fees, emphasizing possible differences.
\subsubsection{\textbf{Results}}
In \autoref{fig:exp3-relative-diff}, we show the relative difference in percentages of $nextClaim$ rewards between the settings of 4 $\mathcal{FRSC}$s versus 1 $\mathcal{FRSC}$.
It is clear that the setting of 4 $\mathcal{FRSC}$s in contrast to a single $\mathcal{FRSC}$ provided better reward compensation in times of very low fees value in the mempool, while it provided smaller reward in the times of higher values of fees in the mempool.
Therefore, we concluded that it is not possible to replace a setup of multiple $\mathcal{FRSC}$s with a single one.
\subsection{Experiment IV}\label{sec:exp4}
We focused on reproducing the experiment from Section 5.5 of~\cite{carlsten2016instability}.
We were searching for the minimal ratio of \textsc{Default-Compliant} miners, at which the undercutting attack is no longer profitable strategy.
\textsc{Default-Compliant} miners are honest miners in a way that they follow the rules of the consensus protocol such as building on top of the longest chain.
We executed several simulations, each consisting of multiple games (i.e., 300k as in~\cite{carlsten2016instability}) with various fractions of \textsc{Default-Compliant} miners.
From the remaining miners we evenly created \textit{learning miners}, who learn on the previous runs of games and switch with a certain probability the best mining strategy out of the following.
\begin{compactitem}
\item \textsc{PettyCompliant}: This miner behaves as \textsc{Default-Compliant} except one difference.
In the case of seeing two chains, he does not mine on the oldest block but rather the most profitable block.
Thus, this miner is not the (directly) attacking miner.
\item \textsc{LazyFork}:
This miner checks which out of two options is more profitable: (1) mining on the longest-chain block or (2) undercutting that block.
In either way, he leaves half of the mempool fees for the next miners, which prevents another \textsc{LazyFork} miner to undercut him.
\item \textsc{Function-Fork()}
The behavior of the miner can parametrized with a function f(.) expressing the level of his undercutting.
The higher the output number the less reward he receives and more he leaves to incentivize other miners to mine on top of his block.
This miner undercuts every time he forks the chain.
\end{compactitem}
\subsubsection{\textbf{Methodology}}
With the missing feature for difficulty re-adjustment (present in our work as well as in~\cite{carlsten2016instability}) the higher orphan rate occurs, which might directly impact our $\mathcal{FRSC}$-based approach.
If the orphan rate is around 40\%, roughly corresponding to~\cite{carlsten2016instability}, our blocks would take on average 40\% longer to be created, increasing the block creation time (i.e., time to mine a block).
This does not affect the original simulator, as there are no $\mathcal{FRSC}s$ that would change the total reward for the miner who found the block.
Nevertheless, this is not true for $\mathcal{FRSC}$-based simulations as the initial setup of $\mathcal{FRSC}s$ is calculated with $\overline{fees} = 50$ BTC (as per the original simulations).
However, with longer block creation time and transaction fees being calculated from it, the amount of $\overline{fees}$ also changes.
Without any adjustments this results in $\mathcal{FRSC}s$ initially paying smaller reward back to the miner before they are saturated.
To mitigate this problem, we increased the initial values of individual $\mathcal{FRSC}$s by the orphan rate from the previous game before each run.
This results in very similar conditions, which can be verified by comparing the final value in the longest chain of our simulation versus the original simulations.
We decided to use this approach to be as close as possible to the original experiment.
This is particularly important when the parameter for the full mempool is equal to $false$ (see \autoref{sec:implementation}), which means that the incoming transaction fees to mempool are calculated based on the block creation time.
In our simulations, we used the following parameters: 100 miners,
10 000 blocks per game,
300 000 games (in each simulation run),
exp3 learning model, and
$\mathbb{C} = 0.7$.
Modeling of fees utilized the same parameters as in the original paper~\cite{carlsten2016instability}: the full mempool parameter disabled, a constant inflow of 5 000 000 000 Satoshi (i.e., 50 BTC) every 600s.
For more details about the learning strategies and other parameters, we refer the reader to~\cite{carlsten2016instability}.
\subsubsection*{\textbf{Setup of $\mathcal{FRSC}s$}}
Since we have a steady inflow of fees to the mempool, we do not need to average the income for the miner.
Therefore, we used only a single $\mathcal{FRSC}$ defined as $\mathcal{FRSC}$(7 056 000 000 000, 2016, 1), where the initial value of $\mathcal{FRSC}.\nu$ was adjusted according to \autoref{eq:setup}, assuming $\overline{fees} = 50$ BTC.
In next runs of any game, $\mathcal{FRSC}.\nu$ was increased by the orphan rate from the previous runs, as mentioned above.
\begin{figure}
\centering
\begin{subfigure}{0.40\textwidth}
\includegraphics[width=\textwidth]{images/equilibrium.png}
\end{subfigure}
\caption{The number of \textsc{Default-Compliant} miners in our $\mathcal{FRSC}$ approach is $\sim$30\% (in contrast to $\sim66\%$ of~\cite{carlsten2016instability}).}\label{fig:improvement}
\end{figure}
\subsubsection{\textbf{Results}}
The results of this experiment are depicted in \autoref{fig:improvement}, and they demonstrate that with our approach using $\mathcal{FRSC}$s, we lowered down the number of \textsc{Default-Compliant} miners (i.e., purple meeting red) from the original $66\%$ to $30\%$.
This means that the profitability of undercutting miners is avoided with at least $30\%$ of \textsc{Default-Compliant} miners, indicating the more robust results.
\section{Security Analysis and Discussion}
\label{sec:discussion}
\subsection{Contract-Drying Attack}
This is a new attack that might potentially occur in our scheme; however, it is the abusive attack aimed at attacking the functionality of our scheme and not on maximizing the profits of the adversary.
In this attack, the adversary aims at getting his reward only from $\mathcal{FRSC}$s and does not include transactions in the block (or includes only a number of them).
This might result in slow drying of the funds from $\mathcal{FRSC}$s and would mean less reward for future honest miners.
Moreover, the attacker can mine in well times of higher saturation of $\mathcal{FRSC}$s and after some time decide to switch off the mining.
This might cause a deterioration in profitability for honest miners, consequently leading to deteriorated security w.r.t., undercutting attacks.
The attacker can keep repeating this behavior, which would likely lead to increased transaction fees in the mempool since the attacker keeps more un-mined transactions in the mempool than the honest miners.
Therefore, if an honest miner mines a block, he gets higher reward and at the same time deposits a higher amount from transaction fees to $\mathcal{FRSC}$s, which indicates a certain self-regulation of our approach, mitigating this kind of attack.
Additionally, we can think of lowering the impact of this attack by rewarding the miner with the full $nextClaim_{[H+1]}$ by $\mathcal{FRSC}$s only if the block contains enough transaction fees (e.g., specified by the network-regulated minimum threshold).
However, this assumes that there is always a reasonable amount of fees in the mempool, which might not be the case all the time and might result in a situation where the miners temporarily stop mining if there is not enough transactions to mine.
Nonetheless, it would require a more research to investigate this solution or a better solution mitigating this type of potential abusive attack, which we left for future work.
\subsection{Possible Improvements}
$\mathcal{FRSC}$ can contain the parameter enabling the interval of the possible change in the reward paid by $\mathcal{FRSC}^x$ (i.e., $nextClaim_{[H+1]}^{x}$) from the median of its value (computed over $\lambda$).
If $nextClaim_{[H+1]}^{x}$ would drastically increase from its median value as significantly more fees would come into the mempool, then $\mathcal{FRSC}_{[H]}^{x}$ would reward the miner with a certain value (specified by the parameter) from the interval $\langle average,nextClaim_{[H+1]}^{x} \rangle$ instead of the full $nextClaim_{[H+1]}^{x}$.
This would be particularly useful for $\mathcal{FRSC}$s with a small $\lambda$ parameter.
The parameter used for sampling might contain a stochastic function (e.g., exponential) attributing a higher likelihood of getting the values not far from the median.
However, we left the evaluation of this technique to future work.
\subsection{Adjustment of Mining Difficulty}
If the PoW blockchain with the longest chain fork-choice rule uses transaction-fee-based regime, the profitability of miners might be more volatile, which can further lead to decreased security w.r.t. undercutting attacks.
Although our solution with $\mathcal{FRSC}$s helps in mitigation of this problem, we propose another functionality that resides in adjusting the mining difficulty based on the total collected fees during the epoch.
In detail, the difficulty can be increased with higher fees collected from transactions during the epoch and vice versa.
Further research would be needed to evaluate this proposition.
\section{Related work}
\label{sec:relatedwork}
The work of Carlsten et al.~\cite{carlsten2016instability} is the inspiration for our paper, which for the first time describes undercutting attacks arising from the exponential distribution of block creation time and significant differences in transaction fees.
The authors simulated Bitcoin under transaction-fee-based regime and found that there exist the minimal threshold of \textsc{Default-Compliant} miners equal to $66\%$.
Gong et al.~\cite{gong2022towards} argue that using all accumulated fees in the mempool regardless of the block size limit is infeasible in practice and can inflate the profitability of undercutting that was originally described in~\cite{carlsten2016instability}.
Furthermore, Houy~\cite{houy2014economics} demonstrates that a constrain on the block size limit (thus the number of transactions) has economic importance and allows transaction fees not dropping to zero.
Therefore, Gong et al.~\cite{gong2022towards} model the profitability of undercutting with the block size limit presented, which bounds the claimable fees in a mining round.
The authors presented a countermeasure that selectively assembles transactions into the new block, while claiming fewer fees to avoid undercutting.
We argue that in contrast to our approach, this solution cannot be enforced by the consensus protocol, and thus might still enable undercutting to occur.
Zhou et al.~\cite{zhou2019robust} deal with the problem of a mining gap, which is more significant when the throughput of blockchain is high.
Therefore, the authors propose the self-adaptive algorithm to
adjust the block size every 1000 blocks and thus ensure that blocks have enough space to pack new transactions.
Even though Bitcoin-NG~\cite{eyal2016bitcoin} proposes a new consensus mechanism, it also contains an idea of splitting the transaction fees between two entities -- the current leader and the the miner of block -- which should incentivize the miner to include blocks created by the leader.
However, Bitcoin-NG is in some sense centralized and therefore, undercutting attacks are not its subject.
\section{Conclusion}
\label{sec:conclusion}
In this work, we focused on three problems related to transaction-fee-based regime of blockchains with the longest chain fork choice rule:
(1) the mining gap, (2) the possibility of undercutting attacks, and (3) the instability of mining rewards.
To mitigate these problems, we proposed the approach approximating a moving average based on the fee-redistributions smart contracts that accumulate a certain fraction of transaction fees and at the same time reward the miners from their reserves.
In this way, the miners are sufficiently rewarded even at the time of very low transaction fees, such as the beginning of the mining round, entering the mining protocol by new miners, market deviations, etc.
Besides, our approach brings a higher tolerance to undercutting attacks, and increases the minimal threshold of \textsc{Default-Compliant} miner that strictly do not perform undercutting attack from 66\% reported in state-of-the-art to 30\%.
\IEEEtriggeratref{6}
\subsection{Overview}
\subsubsection*{\textbf{Contributions}}
In detail, our contributions are as follows:
\begin{enumerate}
\item We propose an approach that normalizes the mining rewards coming from transaction fees by one or more $\mathcal{FRSC}$s that perform moving average on a certain portion of the transaction fees.
\item We evaluate our approach using various fractions of the transaction fees from a block distributed between a miner and $\mathcal{FRSC}$s.
We experiment with the various numbers and lengths of $\mathcal{FRSC}$s, and we demonstrate that usage of multiple $\mathcal{FRSC}$s of various lengths has the best advantages mitigating the problems we are addressing; however, even using a single $\mathcal{FRSC}$ is beneficial.
\item We demonstrated that with our approach, the mining gap can be minimized since the miners at the beginning of the mining round can get the reward from $\mathcal{FRSC}$s, which stabilizes their income.
\item We empirically demonstrate that using our approach the threshold of \textsc{Default-Compliant} miners who strictly do not execute undercutting attack is lowered from 66\% (as reported in the original work~\cite{carlsten2016instability}) to 30\%.
\end{enumerate}
\section{Acknowledgments}
\subsection{Overview}
Our proposed solution collects a percentage of transaction fees in a native cryptocurrency from the mined blocks into one or multiple fee-redistribution smart contracts (i.e., $\mathcal{FRSC}$s).
Miners of the blocks who must contribute to these contracts are at the same time rewarded from them, while the received reward approximates a moving average of the incoming transaction fees across the fixed sliding window of the blocks.
The fraction of transaction fees (i.e., $\mathbb{C}$) from the mined block is sent to $\mathcal{FRSC}$ and the remaining fraction of transaction fees (i.e., $\mathbb{M}$) is directly assigned to the miner, such that $ \mathbb{C} + \mathbb{M} = 1.$
The role of $\mathbb{M}$ is to incentivize the miners in prioritization of the transactions with the higher fees while the role of $\mathbb{C}$ is to mitigate the problems of undercutting attacks and the mining gap.
\medskip\noindent
\subsubsection*{\textbf{Overview}}
We depict the overview of our approach in \autoref{fig:overview}, and it consists of the following steps:
\begin{enumerate}
\item Using $\mathcal{FRSC}$, the miner calculates the reward for the next block $B$ (i.e., $nextClaim(\mathcal{FRSC})$ -- see \autoref{eq:nextClaim}) that will be payed by $\mathcal{FRSC}$ to the miner of that block.
\item The miner mines the block $B$ using the selected set of the highest fee transactions from her mempool.
\item The mined block $B$ directly awards a certain fraction of the transaction fees (i.e., $B.fees ~*~ \mathbb{M}$) to the miner and the remaining part (i.e., $B.fees ~*~ \mathbb{C}$) to $\mathcal{FRSC}$.
\item The miner obtains $nextClaim$ from $\mathcal{FRSC}$.
\end{enumerate}
Our approach is embedded into the consensus protocol, and therefore consensus nodes are obliged to respect it in order to ensure that their blocks are valid.
It can be implemented with standard smart contracts of the blockchain platform or within the native code of the consensus protocol.
\subsection{Prioritization of High-Fee Transactions}
In the environment with constant transaction fees, a miner would receive the same amount with or without our solution.
However, in public blockchains (especially with transaction-fee based regime) there exist a mechanism to ensure prioritizaiton in processing of transactions with higher fees, which might result into fluctuations in rewards of the miners.
In our approach, we preserve the transaction prioritization since we directly attribute a part of the transaction fees to the miner (i.e., $\mathbb{M}$).
\subsection{Fee-Redistribution Smart Contracts}\label{sub:redistrib}
We define the fee-redistribution smart contract as
\begin{eqnarray}
\mathcal{FRSC} = (\nu, \lambda, \rho),
\end{eqnarray}
where
$\nu$ is the accumulated amount of tokens in the contract, $\lambda$ denotes the size of $\mathcal{FRSC}$' sliding window in terms of the number of preceding blocks that contributed to $\nu$,
and $\rho$ is the parameter defining the ratio for redistribution of incoming transaction fees among multiple contracts, while the sum of $\rho$ across all $\mathcal{FRSC}$s must be equal to 1:
\begin{eqnarray}\label{eq:frsc-redistrib-ratios}
\sum_{x ~\in~ \mathcal{FRSC}s} x.\rho &=& 1.
\end{eqnarray}
In contrast to a single $\mathcal{FRSC}$, multiple $\mathcal{FRSC}$s enable better adjustment of compensation to miners during periods of higher transaction fee fluctuations or in an unpredictable environment (we show this in \autoref{sec:exp3}).
\medskip
We denote the state of $\mathcal{FRSC}$s at the blockchain height $H$ as $\mathcal{FRSC}_{[H]}$.
Then, we determine the reward from $\mathcal{FRSC}_{[H]} \in \mathcal{FRSC}s_{[H]}$ for the miner of the next block with height $H+1$ as follows:
\begin{equation}
\partial Claim_{[H+1]}^{\mathcal{FRSC}_{[H]}} = \frac{\mathcal{FRSC}_{[H]}.\nu}{\mathcal{FRSC}_{[H]}.\lambda},
\end{equation}
while the reward obtained from all $\mathcal{FRSC}$s is
\begin{equation} \label{eq:nextClaim}
nextClaim_{[H+1]} = \sum_{\mathcal{X}_{[H]} ~\in~ \mathcal{FRSC}s_{[H]}^{}} \partial Claim_{[H+1]}^{\mathcal{X}_{[H]}}.
\end{equation}
\noindent
Then, the total reward of the miner who mined the block $B_{[H+1]}$ with all transaction fees $B_{[H+1]}.fees$ is
\begin{equation} \label{eq:rewardtotal}
rewardT_{[H+1]} = nextClaim_{[H+1]} + \mathbb{M} * B_{[H+1]}.fees.
\end{equation}
The new state of contracts at the height $H + 1$ is
\begin{eqnarray}
\mathcal{FRSC}s_{[H+1]} = \{\mathcal{X}_{[H+1]}(\nu, \lambda, \rho)\ ~|~
\end{eqnarray}
\begin{eqnarray}
\lambda &=& \mathcal{X}_{[H]}.\lambda,\\
\rho &=& \mathcal{X}_{[H]}.\rho,\\
\nu &=& \mathcal{X}_{[H]}.\nu - \partial Claim_{[H+1]} + deposit * \rho,\\
deposit &=& B_{[H+1]}.fees * \mathbb{C}\},
\end{eqnarray}
where $deposit$ represents the fraction $\mathbb{C}$ of all transaction fees from the block $B_{[H+1]}$ that are deposited across all $\mathcal{FRSC}$s in ratios respecting \autoref{eq:frsc-redistrib-ratios}.
\subsection{Example}
We present an example using Bitcoin~\cite{nakamoto2008bitcoin} to demonstrate our approach.
We assume that the current height of the blockchain is $H$, and we utilize only a single $\mathcal{FRSC}$ with the following parameters:
\[
\mathcal{FRSC}_{[H]} = (2016, 2016, 1).
\]
We set $\mathbb{M} = 0.4$ and $\mathbb{C} = 0.6$, which means a miner directly obtains 40\% of the $B_{[H+1]}.fees$ and $\mathcal{FRSC}$ obtains remaining 60\%.
\medskip
\noindent
Next, we compute the reward from $\mathcal{FRSC}$ obtained by the miner of the block with height $H+1$ as
\begin{equation*}
\partial Claim_{[H+1]} = \frac{\mathcal{FRSC}_{[H]}.\nu}{\mathcal{FRSC}_{[H]}.\lambda} = \frac{2016}{2016} = 1~\text{BTC},
\end{equation*}
resulting into
\begin{equation*}
nextClaim_{[H+1]} = \partial Claim_{[H+1]}~=~1~\text{BTC}.
\end{equation*}
\noindent
Further, we assume that the total reward collected from transactions in the block with height $H+1$ is $B_{[H+1]}.fees = 2$ BTC.
Hence, the total reward obtained by the miner of the block $B_{[H+1]}$ is
\begin{eqnarray*}
rewardT_{[H+1]} &=& nextClaim_{[H+1]} + \mathbb{M} * B_{[H+1]}.fees \\
&=& 1 + 0.4 * 2 \\
&=& 1.8 ~\text{BTC},
\end{eqnarray*}
and the contribution of transaction fees from $B_{[H+1]}$ to the $\mathcal{FRSC}$ is
\[
deposit = B_{[H+1]}.fees * \mathbb{C} = 1.2 ~\text{BTC}.
\]
Therefore, the value of $\nu$ in $\mathcal{FRSC}$ is updated at height H~+~1 as follows:
\begin{eqnarray*}
v_{[H+1]} &=& \mathcal{FRSC}_{[H]}.\nu - nextClaim_{[H+1]} + deposit \\
&=& 2016 - 1 + 1.2 ~\text{BTC} \\
&=& 2016.2 ~\text{BTC}.
\end{eqnarray*}
\noindent
\subsubsection*{\textbf{Traditional Way in Tx-Fee Regime}}
In traditional systems (running in transaction-fee regime) $rewardT_{[H+1]}$ would be equal to the sum of all transaction fees $B_{[H+1]}.fees$ (i.e., $2$ BTC); hence, using $\mathbb{M} = 1$.
In our approach, $rewardT_{[H+1]}$ can only be equal to the sum of all transaction fees in the block $B_{[H+1]}$, if:
\begin{eqnarray}
B_{[H+1]}.fees = \frac{nextClaim_{[H+1]}}{\mathbb{C}}.
\end{eqnarray}
In our example, a miner can mine the block $B_{[H+1]}$ while obtaining the same total reward as the sum of all transaction fees in the block if the transactions carry 1.66 BTC in fees:
\begin{equation*}
B_{[H+1]}.fees = \frac{1}{0.6} = 1.66 ~\text{BTC}.
\end{equation*}
\subsection{Initial Setup of $\mathcal{FRSC}$s Contracts}
To enable an even start, we propose to initiate $\mathcal{FRSC}$s of our approach by a genesis value.
The following formula calculates the genesis values per $\mathcal{FRSC}$ and initializes starting state of $\mathcal{FRSC}s_{[0]}$:
\begin{equation} \label{eq:setup}
\{\mathcal{FRSC}_{[0]}^{x}(\nu, \lambda, \rho)\ |\ \nu = \overline{fees} * \mathbb{C} * \rho * \lambda\},
\end{equation}
where $\overline{fees}$ is the expected average of incoming fees.
\subsection{Contract-Drying Attack}
This is a new attack that might potentially occur in our scheme; however, it is the abusive attack aimed at attacking the functionality of our scheme and not on maximizing the profits of the adversary.
In this attack, the adversary aims at getting his reward only from $\mathcal{FRSC}$s and does not include transactions in the block (or includes only a number of them).
This might result in slow drying of the funds from $\mathcal{FRSC}$s and would mean less reward for future honest miners.
Moreover, the attacker can mine in well times of higher saturation of $\mathcal{FRSC}$s and after some time decide to switch off the mining.
This might cause a deterioration in profitability for honest miners, consequently leading to deteriorated security w.r.t., undercutting attacks.
The attacker can keep repeating this behavior, which would likely lead to increased transaction fees in the mempool since the attacker keeps more un-mined transactions in the mempool than the honest miners.
Therefore, if an honest miner mines a block, he gets higher reward and at the same time deposits a higher amount from transaction fees to $\mathcal{FRSC}$s, which indicates a certain self-regulation of our approach, mitigating this kind of attack.
Additionally, we can think of lowering the impact of this attack by rewarding the miner with the full $nextClaim_{[H+1]}$ by $\mathcal{FRSC}$s only if the block contains enough transaction fees (e.g., specified by the network-regulated minimum threshold).
However, this assumes that there is always a reasonable amount of fees in the mempool, which might not be the case all the time and might result in a situation where the miners temporarily stop mining if there is not enough transactions to mine.
Nonetheless, it would require a more research to investigate this solution or a better solution mitigating this type of potential abusive attack, which we left for future work.
\subsection{Possible Improvements}
$\mathcal{FRSC}$ can contain the parameter enabling the interval of the possible change in the reward paid by $\mathcal{FRSC}^x$ (i.e., $nextClaim_{[H+1]}^{x}$) from the median of its value (computed over $\lambda$).
If $nextClaim_{[H+1]}^{x}$ would drastically increase from its median value as significantly more fees would come into the mempool, then $\mathcal{FRSC}_{[H]}^{x}$ would reward the miner with a certain value (specified by the parameter) from the interval $\langle average,nextClaim_{[H+1]}^{x} \rangle$ instead of the full $nextClaim_{[H+1]}^{x}$.
This would be particularly useful for $\mathcal{FRSC}$s with a small $\lambda$ parameter.
The parameter used for sampling might contain a stochastic function (e.g., exponential) attributing a higher likelihood of getting the values not far from the median.
However, we left the evaluation of this technique to future work.
\subsection{Adjustment of Mining Difficulty}
If the PoW blockchain with the longest chain fork-choice rule uses transaction-fee-based regime, the profitability of miners might be more volatile, which can further lead to decreased security w.r.t. undercutting attacks.
Although our solution with $\mathcal{FRSC}$s helps in mitigation of this problem, we propose another functionality that resides in adjusting the mining difficulty based on the total collected fees during the epoch.
In detail, the difficulty can be increased with higher fees collected from transactions during the epoch and vice versa.
Further research would be needed to evaluate this proposition.
\subsubsection{Changes in Simulator}
We have created a configuration file to simulate custom scenarios of incoming transactions instead of the original design~\cite{carlsten2016instability} fees.
We added an option to switch simulation into a mode with a full mempool, and thus bound the total fees (and consequently the total number of transactions) that can be earned within a block --
this mostly relates to blocks whose mining takes longer time than the average time to mine a block.\footnote{Note that the original simulator~\cite{carlsten2016instability} assumes that the number of transactions (and thus the total fees) in the block is constrained only by the duration of a time required to mine the block, which was also criticized in~\cite{gong2022towards}.}
Next, we moved several parameters to arguments of the simulator with the intention to eliminate the need of recompilation the program frequently, and therefore simplify the process of running various experiments with the simulator.
Finally, we integrated our $\mathcal{FRCS}$-based solution into the simulator.
$\mathcal{FRSC}$s are initiated from a corresponding configuration file.
The source code of our modified simulator is available at anonymous link \url{https://www.dropbox.com/s/gwlbedq47csthqp/sources.zip?dl=0}.
\subsection{Experiment I}
\label{sec:exp1}
\subsubsection{\textbf{Methodology}}
The purpose of this experiment was to investigate the amount of the reward a miner received with our approach versus the baseline (i.e., the full reward is based on all transaction fees).
In this experiment, we investigated how $\mathbb{C}$ influences the total reward of the miner and how $\lambda$ of the sliding window averaged the rewards.
In detail, we created two independent $\mathcal{FRSC}$s with different $\lambda$ -- one was set to 2016 (i.e., $\mathcal{FRCS}^1$), and the second one was se to 5600 (i.e., $\mathcal{FRCS}^2$).
We simulated these two $\mathcal{FRCS}$s with three different values of $\mathbb{C} \in \{0.5,~0.7,~0.9\}$.
\subsubsection{\textbf{Results}}
The results of this experiment are depicted in \autoref{fig:50tocontract}.
Across all runs of our experiment we can observe that $\mathcal{FRSC}^{2}$ adapts slower as compared to $\mathcal{FRSC}^{1}$, which leads to more significant averaging of the total reward paid to the miner.
Even though this is desired, we can see faster adaptation to changing environment by $\mathcal{FRSC}^{1}$, what a miner can expects from rising fees.
\begin{figure*}
\vspace{-0.5cm}
\centering
\begin{subfigure}{0.32\textwidth}
\includegraphics[width=\textwidth]{images/fees-4-50.png}
\caption{Scenario with 4 $\mathcal{FRSC}$s,\\ $\mathbb{C} = 0.5$.}
\end{subfigure}
\begin{subfigure}{0.32\textwidth}
\includegraphics[width=\textwidth]{images/contracts-4-50.png}
\caption{$\partial Claim$s and $nextClaim$, \\ $\mathbb{C} = 0.5$.}
\end{subfigure}
\begin{subfigure}{0.32\textwidth}
\includegraphics[width=\textwidth]{images/normalized-contracts-4-50.png}
\caption{$\partial Claim$s normalized by $\rho$, \\ $\mathbb{C} = 0.5$.}
\end{subfigure}
\begin{subfigure}{0.32\textwidth}
\includegraphics[width=\textwidth]{images/fees-4-70.png}
\caption{Scenario with 4 $\mathcal{FRSC}$s,\\ $\mathbb{C} = 0.7$.}
\end{subfigure}
\begin{subfigure}{0.32\textwidth}
\includegraphics[width=\textwidth]{images/contracts-4-70.png}
\caption{$\partial Claim$s and $nextClaim$, \\ $\mathbb{C} = 0.7$.}
\end{subfigure}
\begin{subfigure}{0.32\textwidth}
\includegraphics[width=\textwidth]{images/normalized-contracts-4-70.png}
\caption{$\partial Claim$s normalized by $\rho$, \\ $\mathbb{C} = 0.7$.}
\end{subfigure}
\begin{subfigure}{0.32\textwidth}
\includegraphics[width=\textwidth]{images/fees-4-90.png}
\caption{Scenario with 4 $\mathcal{FRSC}$s, \\ $\mathbb{C} = 0.9$.}
\end{subfigure}
\begin{subfigure}{0.32\textwidth}
\includegraphics[width=\textwidth]{images/contracts-4-90.png}
\caption{$\partial Claim$s and $nextClaim$, \\ $\mathbb{C} = 0.9$.}
\end{subfigure}
\begin{subfigure}{0.32\textwidth}
\includegraphics[width=\textwidth]{images/normalized-contracts-4-90.png}
\caption{$\partial Claim$s normalized by $\rho$, \\ $\mathbb{C} = 0.9$.}
\end{subfigure}
\caption{Experiment II investigating various $\mathbb{C}$s in the setting with multiple $\mathcal{FRSC}$s with their corresponding $\lambda$ = $\{1008, 2016, 4032, 8064\}$ and $\rho$ = $\{0.07, 0.14, 0.28, 0.51\}$. $\partial Claim$s represents contributions of individual $\mathcal{FRSC}$s to the total reward of the miner (i.e., its $nextClaim$ component).}\label{fig:exp-II}
\end{figure*}
\begin{figure*}
\centering
\begin{subfigure}{0.32\textwidth}
\includegraphics[width=\textwidth]{images/contracts-4-original.png}
\caption{$\rho$ correlating with $\lambda$.}
\end{subfigure}
\begin{subfigure}{0.32\textwidth}
\includegraphics[width=\textwidth]{images/contracts-4-equal.png}
\caption{$\rho$ equal for every $\mathcal{FRSC}$.}
\end{subfigure}
\begin{subfigure}{0.32\textwidth}
\includegraphics[width=\textwidth]{images/contracts-4-inverse.png}
\caption{$\rho$ negatively correlating with $\lambda$.}
\end{subfigure}
\caption{Experiment II -- multiple $\mathcal{FRSC}$s using various distributions of $\rho$ and their impact on $\partial Claim$, where $\mathbb{C}$ = 0.7.} \label{fig:percentages}
\end{figure*}
\begin{figure}
\centering
\begin{subfigure}{0.44\textwidth}
\includegraphics[width=\textwidth]{images/fees-1-percentage.png}
\caption{A custom fee scenario for Experiment III.}\label{fig:exp3-fees}
\end{subfigure}
\begin{subfigure}{0.44\textwidth}
\includegraphics[width=\textwidth]{images/percentage.png}
\caption{A relative difference in $nextClaim$ between 4 $\mathcal{FRSC}$s and a single $\mathcal{FRSC}$.}\label{fig:exp3-relative-diff}
\end{subfigure}
\caption{Experiment III comparing 4 $\mathcal{FRSC}$s and 1 $\mathcal{FRSC}$, both configurations having the same $\text{effective}\_\lambda$. }\label{fig:effective_length}
\end{figure}
\subsection{Experiment II}
\label{sec:exp2}
\subsubsection{\textbf{Methodology}}
In this experiment, we investigated how multiple $\mathcal{FRSC}$s dealt with the same scenario as before -- i.e., varying $\mathbb{C}$.
In detail, we investigated how individual $\mathcal{FRSC}$s contributed to the $nextClaim_{[H+1]}$ by their individual $\partial Claim^{\mathcal{FRSC}_{[H]}}_{[H+1]}$.
This time, we varied only the parameter $\mathbb{C} \in \{0.5, ~0.7, ~0.9\}$, and we considered four $\mathcal{FRSC}$s:
\begin{center}
$\mathcal{FRSC}s=\{$\\
$\mathcal{FRSC}^{1}(\_, 1008, 0.07), \mathcal{FRSC}^{2}(\_, 2016, 0.14),$\\
$\mathcal{FRSC}^{3}(\_, 4032, 0.28), \mathcal{FRSC}^{4}(\_, 8064, 0.51)\},$
\end{center}
where their lengths $\lambda$ were set to consecutive powers of two (to see differences in more intensive averaging spread across longer intervals), and their redistribution ratios $\rho$ were set to maximize the potential of averaging by longer $\mathcal{FRSC}$s (see details below in \autoref{sec:different-rhos}).
\subsubsection{\textbf{Results}}
The results of this experiment are depicted in \autoref{fig:exp-II}.
We can observe that the shorter $\mathcal{FRSC}$s quickly adapted to new changes and the longer $\mathcal{FRSC}$s kept more steady income for the miner.
In this sense, we can see that $\partial Claim^{4}$ held steadily over the scenario while for example $\partial Claim^{1}$ fluctuated more.
Since the scenarios of fees evolution in the mempool was the same across all our experiments (but \autoref{sec:exp3}), we can compare the $\mathcal{FRSC}$ with $\lambda$ = 5600 from \autoref{sec:exp1} and the current setup involving four $\mathcal{FRSC}$s -- both had some similarities.
This gave us intuition for replacing multiple $\mathcal{FRSC}$s with a single one, which we further investigated in \autoref{sec:exp3}.
\subsubsection{\textbf{Different $\rho$s}}\label{sec:different-rhos}
In \autoref{fig:percentages} we investigated different values of $\rho$ in the same set of four contracts and their impact on $\partial Claim$s.
The results show that the values of $\rho$ should correlate with $\lambda$ of multiple $\mathcal{FRSC}$s to maximize the potential of averaging by longer $\mathcal{FRSC}$s.
\subsection{Experiment III}\label{sec:exp3}
\subsubsection{\textbf{Methodology}}
In this experiment, we investigated whether it is possible to use a single $\mathcal{FRSC}$ setup to replace a multiple $\mathcal{FRSC}$s while preserving the same effect on the $nextClaim$.
To quantify a difference between such cases, we introduced a new metric of $\mathcal{FRSC}s$, called $\text{effective}\_\lambda$, which can be calculated as follows:
\begin{eqnarray} \label{eq:effective_length}
\text{effective}\_\lambda (\mathcal{FRSC}s) = \sum_{x ~\in~ \mathcal{FRSC}s} x.\rho* x.\lambda.
\end{eqnarray}
As the example, we were interested in comparing a single $\mathcal{FRSC}$ with 4 $\mathcal{FRSC}$s, both configurations having the equal $\text{effective}\_\lambda$.
The configurations of these two cases are as follow:
(1) $\mathcal{FRSC}(\_, 5292, 1)$ and (2)
\begin{center}
$\mathcal{FRSC}s=\{$\\
$\mathcal{FRSC}^{1}(\_, 1008, 0.07), \mathcal{FRSC}^{2}(\_, 2016, 0.19),$\\
$ \mathcal{FRSC}^{3}(\_, 4032, 0.28), \mathcal{FRSC}^{4}(\_, 8064, 0.46)\}$.
\end{center}
We can easily verify that the $\text{effective}\_\lambda$ of 4 $\mathcal{FRSC}$s is the same as in a single $\mathcal{FRSC}$ using \autoref{eq:effective_length}:
$0.07 * 1008 + 0.19 * 2016 + 0.28 * 4032 + 0.46 * 8064 = 5292$.
We conducted this experiment using a custom fee evolution scenario involving mainly linearly increasing/decreasing fees in the mempool (see \autoref{fig:exp3-fees}), and we set $\mathbb{C}$ to 0.7 for both configurations.
The new scenario of the fee evolution in mempool was chosen to contain extreme changes in fees, emphasizing possible differences.
\subsubsection{\textbf{Results}}
In \autoref{fig:exp3-relative-diff}, we show the relative difference in percentages of $nextClaim$ rewards between the settings of 4 $\mathcal{FRSC}$s versus 1 $\mathcal{FRSC}$.
It is clear that the setting of 4 $\mathcal{FRSC}$s in contrast to a single $\mathcal{FRSC}$ provided better reward compensation in times of very low fees value in the mempool, while it provided smaller reward in the times of higher values of fees in the mempool.
Therefore, we concluded that it is not possible to replace a setup of multiple $\mathcal{FRSC}$s with a single one.
\subsection{Experiment IV}\label{sec:exp4}
We focused on reproducing the experiment from Section 5.5 of~\cite{carlsten2016instability}.
We were searching for the minimal ratio of \textsc{Default-Compliant} miners, at which the undercutting attack is no longer profitable strategy.
\textsc{Default-Compliant} miners are honest miners in a way that they follow the rules of the consensus protocol such as building on top of the longest chain.
We executed several simulations, each consisting of multiple games (i.e., 300k as in~\cite{carlsten2016instability}) with various fractions of \textsc{Default-Compliant} miners.
From the remaining miners we evenly created \textit{learning miners}, who learn on the previous runs of games and switch with a certain probability the best mining strategy out of the following.
\begin{compactitem}
\item \textsc{PettyCompliant}: This miner behaves as \textsc{Default-Compliant} except one difference.
In the case of seeing two chains, he does not mine on the oldest block but rather the most profitable block.
Thus, this miner is not the (directly) attacking miner.
\item \textsc{LazyFork}:
This miner checks which out of two options is more profitable: (1) mining on the longest-chain block or (2) undercutting that block.
In either way, he leaves half of the mempool fees for the next miners, which prevents another \textsc{LazyFork} miner to undercut him.
\item \textsc{Function-Fork()}
The behavior of the miner can parametrized with a function f(.) expressing the level of his undercutting.
The higher the output number the less reward he receives and more he leaves to incentivize other miners to mine on top of his block.
This miner undercuts every time he forks the chain.
\end{compactitem}
\subsubsection{\textbf{Methodology}}
With the missing feature for difficulty re-adjustment (present in our work as well as in~\cite{carlsten2016instability}) the higher orphan rate occurs, which might directly impact our $\mathcal{FRSC}$-based approach.
If the orphan rate is around 40\%, roughly corresponding to~\cite{carlsten2016instability}, our blocks would take on average 40\% longer to be created, increasing the block creation time (i.e., time to mine a block).
This does not affect the original simulator, as there are no $\mathcal{FRSC}s$ that would change the total reward for the miner who found the block.
Nevertheless, this is not true for $\mathcal{FRSC}$-based simulations as the initial setup of $\mathcal{FRSC}s$ is calculated with $\overline{fees} = 50$ BTC (as per the original simulations).
However, with longer block creation time and transaction fees being calculated from it, the amount of $\overline{fees}$ also changes.
Without any adjustments this results in $\mathcal{FRSC}s$ initially paying smaller reward back to the miner before they are saturated.
To mitigate this problem, we increased the initial values of individual $\mathcal{FRSC}$s by the orphan rate from the previous game before each run.
This results in very similar conditions, which can be verified by comparing the final value in the longest chain of our simulation versus the original simulations.
We decided to use this approach to be as close as possible to the original experiment.
This is particularly important when the parameter for the full mempool is equal to $false$ (see \autoref{sec:implementation}), which means that the incoming transaction fees to mempool are calculated based on the block creation time.
In our simulations, we used the following parameters: 100 miners,
10 000 blocks per game,
300 000 games (in each simulation run),
exp3 learning model, and
$\mathbb{C} = 0.7$.
Modeling of fees utilized the same parameters as in the original paper~\cite{carlsten2016instability}: the full mempool parameter disabled, a constant inflow of 5 000 000 000 Satoshi (i.e., 50 BTC) every 600s.
For more details about the learning strategies and other parameters, we refer the reader to~\cite{carlsten2016instability}.
\subsubsection*{\textbf{Setup of $\mathcal{FRSC}s$}}
Since we have a steady inflow of fees to the mempool, we do not need to average the income for the miner.
Therefore, we used only a single $\mathcal{FRSC}$ defined as $\mathcal{FRSC}$(7 056 000 000 000, 2016, 1), where the initial value of $\mathcal{FRSC}.\nu$ was adjusted according to \autoref{eq:setup}, assuming $\overline{fees} = 50$ BTC.
In next runs of any game, $\mathcal{FRSC}.\nu$ was increased by the orphan rate from the previous runs, as mentioned above.
\begin{figure}
\centering
\begin{subfigure}{0.40\textwidth}
\includegraphics[width=\textwidth]{images/equilibrium.png}
\end{subfigure}
\caption{The number of \textsc{Default-Compliant} miners in our $\mathcal{FRSC}$ approach is $\sim$30\% (in contrast to $\sim66\%$ of~\cite{carlsten2016instability}).}\label{fig:improvement}
\end{figure}
\subsubsection{\textbf{Results}}
The results of this experiment are depicted in \autoref{fig:improvement}, and they demonstrate that with our approach using $\mathcal{FRSC}$s, we lowered down the number of \textsc{Default-Compliant} miners (i.e., purple meeting red) from the original $66\%$ to $30\%$.
This means that the profitability of undercutting miners is avoided with at least $30\%$ of \textsc{Default-Compliant} miners, indicating the more robust results.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 8,035 |
{"url":"http:\/\/mathcentral.uregina.ca\/QQ\/database\/QQ.09.13\/h\/mary3.html","text":"SEARCH HOME\n Math Central Quandaries & Queries\n Question from Mary, a student: Solve the following trigonometric equation over 0 being less than or equal to A but less than 2pie 2cos(2A)=-2^1\/2 I have it partially worked out 2(2cos^2A) = -2^1\/2 4cos^2A = -2^1\/2 16cos^4A = 2 16cos^4A - 2 = 0 2(8cos^4A - 1) =0 I dont know where to go from here and I don't know if I am on the right track. If you could help me figure this out it would be much appreciated. thanks.\n\nHi Mary,\n\nYou state the equation as\n\n$2\\cos(2A)=-2^{1\/2}$\n\nand then you start to solve\n\n$2(2\\cos^2A) = -2^{1\/2}$\n\nWhich one is it?\n\nPenny\n\nMath Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.","date":"2017-10-17 20:28:03","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.4538786709308624, \"perplexity\": 537.668578320201}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2017-43\/segments\/1508187822488.34\/warc\/CC-MAIN-20171017200905-20171017220905-00511.warc.gz\"}"} | null | null |
// Code generated by protoc-gen-liverpc v0.1, DO NOT EDIT.
// source: v1/Room.proto
/*
Package v1 is a generated liverpc stub package.
This code was generated with go-common/app/tool/liverpc/protoc-gen-liverpc v0.1.
It is generated from these files:
v1/Room.proto
*/
package v1
import context "context"
import proto "github.com/golang/protobuf/proto"
import "go-common/library/net/rpc/liverpc"
var _ proto.Message // generate to suppress unused imports
// Imports only used by utility functions:
// ==============
// Room Interface
// ==============
type RoomRPCClient interface {
// * 根据房间id获取房间信息
GetInfoById(ctx context.Context, req *RoomGetInfoByIdReq, opts ...liverpc.CallOption) (resp *RoomGetInfoByIdResp, err error)
// * 获取房间基本信息接口,前端/移动端房间页使用
GetInfo(ctx context.Context, req *RoomGetInfoReq, opts ...liverpc.CallOption) (resp *RoomGetInfoResp, err error)
}
// ====================
// Room Live Rpc Client
// ====================
type roomRPCClient struct {
client *liverpc.Client
}
// NewRoomRPCClient creates a client that implements the RoomRPCClient interface.
func NewRoomRPCClient(client *liverpc.Client) RoomRPCClient {
return &roomRPCClient{
client: client,
}
}
func (c *roomRPCClient) GetInfoById(ctx context.Context, in *RoomGetInfoByIdReq, opts ...liverpc.CallOption) (*RoomGetInfoByIdResp, error) {
out := new(RoomGetInfoByIdResp)
err := doRPCRequest(ctx, c.client, 1, "Room.get_info_by_id", in, out, opts)
if err != nil {
return nil, err
}
return out, nil
}
func (c *roomRPCClient) GetInfo(ctx context.Context, in *RoomGetInfoReq, opts ...liverpc.CallOption) (*RoomGetInfoResp, error) {
out := new(RoomGetInfoResp)
err := doRPCRequest(ctx, c.client, 1, "Room.get_info", in, out, opts)
if err != nil {
return nil, err
}
return out, nil
}
// =====
// Utils
// =====
func doRPCRequest(ctx context.Context, client *liverpc.Client, version int, method string, in, out proto.Message, opts []liverpc.CallOption) (err error) {
err = client.Call(ctx, version, method, in, out, opts...)
return
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 6,876 |
Many cosplayers would want wasp superhero costume cosplay and become the. first 1 is Richie (kinda and berserk Ryuko fight to. Animation Bump : Episode 3's x H): 70. Get the KItana cosplay costume 100 on materials per each most prepared for, especially.
This entry was posted in Final Fantasy Costumes on 12.11.2010 by Kigalkis . | {
"redpajama_set_name": "RedPajamaC4"
} | 5,026 |
Q: Recommended location to store intelliJ project files for project where the set of repos/modules it contains will change over time If I have an intelliJ project containing a number of git repos, the set of which will change over time, where/how should I store the project configuration files (/.idea folder) itself?
(For avoidance of doubt, what I mean above is that after some time, I will want to remove a repo/module from the project and add another of a different name.)
I have a project like this and have run into problems as I wanted to rename one of the repos, but the fact that the intelliJ project files (./idea folder) is contained within the repo/directory I want to rename, seems to be creating issues (specifically, when I change 'module name' in project settings, quit and reload, the change is not saved, and the old directory/module name reappears, which I don't want).
A: Create an empty project in any appropriate location:
It will create an .idea project configuration directory in the specified path. And then add other projects there as new modules from existing sources.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 5,418 |
{"url":"https:\/\/www.opuscula.agh.edu.pl\/om-vol38iss5art4","text":"Opuscula\u00a0Math.\u00a038,\u00a0no.\u00a05\u00a0(2018),\u00a0681-698\nhttps:\/\/doi.org\/10.7494\/OpMath.2018.38.5.681\n\nOpuscula Mathematica\n\n# Krein-von Neumann extension of an even order differential operator on a finite interval\n\nYaroslav I. Granovskyi\nLeonid L. Oridoroga\n\nAbstract. We describe the Krein-von Neumann extension of minimal operator associated with the expression $$\\mathcal{A}:=(-1)^n\\frac{d^{2n}}{dx^{2n}}$$ on a finite interval $$(a,b)$$ in terms of boundary conditions. All non-negative extensions of the operator $$A$$ as well as extensions with a finite number of negative squares are described.\n\nKeywords: non-negative extension, Friedrichs' extension, Krein-von Neumann extension, boundary triple, Weyl function.\n\nMathematics Subject Classification: 47A05.\n\nFull text (pdf)\n\n\u2022 Yaroslav I. Granovskyi\n\u2022 NAS of Ukraine, Institute of Applied Mathematics and Mechanics, Ukraine\n\u2022 Leonid L. Oridoroga\n\u2022 Donetsk National University, Donetsk, Ukraine\n\u2022 Communicated by A.A. Shkalikov.\n\u2022 Revised: 2017-12-15.\n\u2022 Accepted: 2017-12-20.\n\u2022 Published online: 2018-06-12.","date":"2018-09-23 17:49:47","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5517847537994385, \"perplexity\": 2806.3512484270245}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2018-39\/segments\/1537267159570.46\/warc\/CC-MAIN-20180923173457-20180923193857-00372.warc.gz\"}"} | null | null |
The Lost Interview Tapes Featuring Jim Morrison Volume Two sono delle interviste condotte da Salli Stevenson registrate il 13 ottobre 1970 nell'ufficio dei Doors situato nel West Hollywood per il Circus Magazine. Questo documento molto importante contiene le ultime dichiarazioni e discussioni di Jim Morrison del 1970, oltre a questo documento esiste l'ultima intervista catturata su nastro e pubblicata da Jim Morrison nel marzo 1971.
Tracce
Formazione
Jim Morrison – voce
Salli Stevenson - intervistatore
Note
Collegamenti esterni | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 4,906 |
Jan Ordoš (born September 18, 1996) is a Czech professional ice hockey player. He is currently playing for Bílí Tygři Liberec of the Czech Extraliga.
Ordoš made his Czech Extraliga debut playing with HC Bílí Tygři Liberec during the 2015-16 Czech Extraliga season.
References
External links
1996 births
Living people
HC Bílí Tygři Liberec players
Czech ice hockey forwards
Sportspeople from Ústí nad Labem
Czech expatriate ice hockey players in Sweden | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 6,620 |
\section{Introduction}
\label{sec:intro}
\IEEEPARstart{R}{ank} estimation is a important model order selection problem that arises in a variety of critical engineering applications, most notably those associated with statistical signal and array processing. Adaptive beamforming and subspace tracking provide two canonical examples in which one typically assumes $n$-dimensional observations that linearly decompose into a ``signal'' subspace of dimension $r \ll n$ and complementary ``noise'' subspace of dimension $n - r$. In many applications the goal is to enhance, null, or track certain elements of the signal subspace, based on observed array data.
In this context, we of course recover the classical statistical trade-offs between goodness of fit and model order; i.e., between system performance and complexity. In the rank selection case these trade-offs are particularly clear and compelling, as subspace rank estimation may well correspond to correctly identifying the number of interferers or signals of interest. We note that the goal of this article is a general understanding of the model order selection problem in this context, rather than an exhaustive application-specific solution.
To this end, we present here a collection of recent results from random matrix theory, and show how they enable a theoretical analysis of this instantiation of the model order selection problem. In Section~\ref{S:prob-state} we formulate the problem of model order selection in the context of the standard array observation model with additive white Gaussian noise, and in Section~\ref{S:null-alt-rmt} we present sample covariance asymptotics based on random matrix theory. In Section~\ref{S:rank-est} we bring these results to bear on the problem of optimal rank estimation by developing a decision-theoretic rank estimation framework, and associated algorithm whose asymptotic minimax optimality we prove. We then provide a brief simulation study in Section~\ref{S:emp-perf} to demonstrate the practical efficacy of our rank selection procedure, and conclude with a summary discussion in Section~\ref{s:summ}.
\section{Problem Formulation: Model Order Selection}
\label{S:prob-state}
Suppose at time $t$ we observe data vector $\vectorsymbol{x}(t) \in \mathbb{C}^n$
(or $\mathbb{R}^n$), a weighted combination of $r$ signal vectors corrupted
by additive noise. Using the notation $\mathcal{M}_r$ to denote an additive observation model of order $r$, we can express this as
\begin{equation}\label{E:Signal}
\mathcal{M}_r\!:\, \vectorsymbol{x}(t) = \sum_{i=1}^{r} s_{i}(t) \vectorsymbol{a}_{i} + \vectorsymbol{n}(t) = \matrixsymbol{A} \vectorsymbol{s}(t) + \vectorsymbol{n}(t),
\end{equation}
where
\(
\matrixsymbol{A}
=
\left(
\begin{matrix}
\vectorsymbol{a}_{1} & \vectorsymbol{a}_{2} & \cdots & \vectorsymbol{a}_{r}
\end{matrix}
\right)
\)
is an $n \times r$ mixing matrix and
\(
\vectorsymbol{s}(t)
=
\big(
s_1(t), s_2(t), \ldots, s_r(t)
\big)
\)
an $r \times 1$ signal vector. Here, we assume that $\matrixsymbol{A}$ is a deterministic matrix of weights and
$\vectorsymbol{s}(t)$ is a random source vector with nonsingular covariance matrix $\matrixsymbol{C}_\text{S}$.
If the noise $\vectorsymbol{n}(t)$ is independent of the source and white, with
all components having variance $\sigma^2$, then the covariance of $\vectorsymbol{x}(t)$ is
given by
\begin{equation}\label{E:C-decomp}
\matrixsymbol{C}
=
\mathbb{E}\left[ \vectorsymbol{x}(t) \vectorsymbol{x}(t)^* \right]
=
\matrixsymbol{A} \matrixsymbol{C}_\text{S} \matrixsymbol{A}^* + \sigma^2 \matrixsymbol{I}.
\end{equation}
In most array processing applications, it is generally desired to estimate or determine $\matrixsymbol{A}$ based on ``snapshot'' data vectors $\vectorsymbol{x}(t)$; however, note that the decomposition in~\eqref{E:C-decomp} is not statistically identifiable. We can render it identifiable up to a sign change by reparametrizing as
\begin{equation}\label{E:C-decomp-eig}
\matrixsymbol{C}
=
\matrixsymbol{W} \matrixsymbol{\Lambda} \matrixsymbol{W}^* + \sigma^2 \matrixsymbol{I},
\end{equation}
with
\(
\matrixsymbol{W}
=
\left(
\begin{matrix}
\vectorsymbol{w}_1 & \vectorsymbol{w}_2 & \cdots & \vectorsymbol{w}_r
\end{matrix}
\right)
\)
an $n \times r$ matrix having orthonormal columns and
\(
\matrixsymbol{\Lambda}
=
\diag(
\lambda_1, \lambda_2, \ldots, \lambda_r
)
\)
a diagonal matrix with
\(
\lambda_1 \geq \lambda_2 \geq \dots \geq \lambda_r > 0.
\)
Then, in many applications we can recover the parameters of interest from
$\matrixsymbol{W}$ using specialized knowledge of the structure of $\matrixsymbol{A}$ in conjunction with well-known algorithmic approaches
such as MUSIC~\cite{schmidt1986mel} or ESPRIT~\cite{roy1989ees}.
In general, then, the goal of subspace tracking is to estimate $\matrixsymbol{W}$ as well as
possible from observed array data $\vectorsymbol{x}$. Often, the associated data covariance $\matrixsymbol{C}$ will change in time; signals may change their physical characteristics or direction of arrival, with some ceasing while others appear. Such situations require \emph{adaptive} estimation of $\matrixsymbol{W}$. Over the past twenty years, a variety of algorithms
have been developed for recursively updating estimates of the dominant subspace
of a sample covariance matrix. Projection Approximation Subspace Tracking (PAST)~\cite{yang1995pas} is among the most popular, though many new algorithms continue to be proposed (see, e.g.,~\cite{badeau2008fas, bartelmaos2008fpc, doukopoulos2008fas}).
A deficiency of nearly all of these algorithms is that they require prior
knowledge of $r$, the rank of the desired signal subspace. In relative terms, little
attention has been devoted in the extant literature to estimating $r$ in an optimal manner. Kav\v{c}i\'c and Yang~\cite{kavcic1996are}, Rabideau~\cite{rabideau1996fra}, and Real et al.~\cite{real1999taf} separately address the problem, but ultimately all
rely on selection rules with problem-specific tuning parameters. More recently,
Shi et al.~\cite{shi2007aen} suggest a modification of~\cite{rabideau1996fra}; each of these approaches seeks to derive appropriate tuning parameters based on
user-specified false alarm rates.
In parallel to the above developments, the literature on random matrix theory
has progressed substantially over the past ten years. An important outgrowth of this theory is
a new set of rank-selection tools. To this end, Owen and Perry~\cite{owen2009bcv} suggest a
cross-validation-based approach that, for computational reasons, does not appear immediately applicable to real-time subspace tracking. Kritchman and
Nadler~\cite{kritchman2008dnc} provide a survey of other approaches, including those based on information-theoretic criteria~\cite{wax1985det, rao2008sam}, ultimately recommending a method based on eigenvalue thresholding.
Their threshold is determined from a specified false alarm rate using the
theory developed in~\cite{johansson2000sfa, johnstone2001dle, baik2005pto, baik2006els, paul2007ase, onatski2007adp}.
Even for the rank selection procedures with interpretable tuning parameters such as false alarm rate, it is not clear how these parameters should be chosen. In the sequel, we summarize recent sample covariance asymptotics and employ them to develop a decision-theoretic framework for estimation with specified costs for over- or under-estimating the rank. Within this framework we derive a selection rule based on eigenvalue thresholding, without the need for tuning parameters, that minimizes the maximum risk under a set of suitable alternate models.
\section{Sample Covariance Asymptotics
\label{S:null-alt-rmt}
Recall that the $r$th-order signal-plus-noise observation model $\mathcal{M}_r\!:\, \vectorsymbol{x}(t) = \matrixsymbol{A} \vectorsymbol{s}(t) + \vectorsymbol{n}(t)$ of~\eqref{E:Signal} gives rise to the covariance form $\matrixsymbol{C}$ defined in~\eqref{E:C-decomp} and~\eqref{E:C-decomp-eig}. The rank selection rule we derive is based on a sample covariance matrix $\widehat{\matrixsymbol{C}}$ comprised of $N$ array snapshots $\vectorsymbol{x}(t)$, indexed as a function of time $t$. For $N \leq t$, this empirical covariance estimator is defined as
\begin{equation}\label{E:C-empirical}
\widehat{\matrixsymbol{C}} = \frac{1}{N} \sum_{k=0}^{N-1} \vectorsymbol{x}(t-k) \vectorsymbol{x}(t-k)^*,
\end{equation}
and our appeal to random matrix theory will rely on properties of $\widehat{\matrixsymbol{C}}$ as the number of snapshots $N$ becomes large.
As is customary in the case of random matrix theory, we work in an asymptotic setting with $N$ and $n$ both tending to infinity, and their ratio $n/N$ tending to a constant $\gamma \in (0,\infty)$. Moreover, we suppose that the number of signals $r$ is fixed with respect to this asymptotic setting, although it is likely that this assumption can in fact be relaxed to $r = o(\sqrt{n})$. To simplify the presentation of the theory, we also suppose strict inequality in the ordering
$\lambda_1 > \lambda_2 > \cdots > \lambda_r > 0$ with respect to the decomposition of~\eqref{E:C-decomp-eig} of the true covariance $\matrixsymbol{C}$. Note that under the assumed model of~\eqref{E:Signal}, the actual eigenvalues of $\matrixsymbol{C}$ are given by $\{\lambda_i+\sigma^2\}_{i=1}^r \,\cup\, \{\sigma^2\}_{i=r+1}^n$.
To begin, we define the eigendecomposition of the empirical covariance $\widehat{\matrixsymbol{C}}$ of~\eqref{E:C-empirical} as
\begin{equation}\label{E:Cemp-decomp-eig}
\widehat{\matrixsymbol{C}} = \widehat{\matrixsymbol{W}} \matrixsymbol{L} \widehat{\matrixsymbol{W}}^* \text{,}
\end{equation}
where
\(
\widehat{\matrixsymbol{W}}
=
\left(
\begin{matrix}
\hat{\vectorsymbol{w}}_1 & \hat{\vectorsymbol{w}}_2 & \cdots & \hat{\vectorsymbol{w}}_n
\end{matrix}
\right)
\)
has orthonormal columns and
\(
\matrixsymbol{L}
=
\diag( \ell_1, \ell_2, \ldots, \ell_n )
\)
has $\ell_1 \geq \ell_2 \geq \cdots \geq \ell_{n} > 0$. Now consider the additive observation model $\mathcal{M}_0$ of~\eqref{E:Signal} corresponding to $r=0$; i.e., in the \emph{absence} of signal:
\begin{equation*}
\mathcal{M}_0\!:\, \vectorsymbol{x}(t) = \sum_{i=1}^{r} s_{i}(t) \vectorsymbol{a}_{i} + \vectorsymbol{n}(t), \quad r = 0 .
\end{equation*}
This $0$th-order case defines a natural null model for our rank estimation problem, in that $\vectorsymbol{x}(t) = \vectorsymbol{n}(t)$, and hence the observed snapshots $\vectorsymbol{x}(t)$ that comprise $\widehat{\matrixsymbol{C}}$ will consist entirely of noise. In the setting where $\vectorsymbol{x}(t)$ is white Gaussian noise, Johansson~\cite{johansson2000sfa} derives the distribution of $\ell_1$, the principal eigenvalue of $\widehat{\matrixsymbol{C}}$, for complex-valued data, and Johnstone~\cite{johnstone2001dle} gives the corresponding distribution in the real-valued case. These results are defined in terms of the density function of the celebrated Tracy-Widom law (illustrated in Fig.~\ref{F:tw-density}) and imply the following theorem.
\begin{figure}[t]
\centering
\includegraphics[width=\columnwidth]{plots/tw-density}
\caption{\label{F:tw-density}Density of the Tracy-Widom law $F_\beta(x)$ for real ($\beta=1$) and complex ($\beta=2$) cases, computed using the online software packages~\cite{dieng2006rml, perry2009rmtstat}.
}
\end{figure}
\begin{theorem}[Asymptotic Null Distribution]\label{T:null-value}
Let $\ell_1 > 0$ be the largest eigenvalue of an $n$-dimensional sample covariance matrix $\widehat{\matrixsymbol{C}}$ comprised of $N$ i.i.d.~observations $\vectorsymbol{x}$ according to~\eqref{E:C-empirical}, where each vector $\vectorsymbol{x}$ has i.i.d.~$\operatorname{Normal}(0,\sigma^2)$ entries. Defining the standardizing quantities
\begin{align*}
\mu_{N,n}
&=
\frac{\sigma^2}{N}
\left( \sqrt{n} + \sqrt{N} \right)^2 \\
\sigma_{N,n}
&=
\frac{\sigma^2}{N} \textstyle
\left( \sqrt{n} + \sqrt{N} \right)
\left( \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{N}} \right)^{1/3},
\end{align*}
we obtain as $n,N\to\infty$, with $n/N\to \gamma$, the result
\begin{equation*}
\mathbb{P} \left\{ \frac{\ell_1 - \mu_{N,n}}{\sigma_{N,n}} \leq x \right\}
\to
F_\beta ( x ),
\end{equation*}
where $F_\beta( x )$ is the distribution function for the Tracy-Widom law of
order $\beta$, with $\beta = 1$ if $\widehat{\matrixsymbol{C}} \in \mathbb{R}^n$ and $\beta = 2$ if $\widehat{\matrixsymbol{C}} \in \mathbb{C}^n$.
\end{theorem}
\begin{rem}
For complex $\widehat{\matrixsymbol{C}}$, El Karoui~\cite{elkaroui2006rcr} obtains a convergence rate of order $(n\wedge N)^{2/3}$ through small modifications of $\mu_{N,n}$ and $\sigma_{N,n}$; Ma~\cite{ma2008atw} treats the case of real $\widehat{\matrixsymbol{C}}$ similarly.
\end{rem}
Next, we consider properties of the empirical covariance estimator $\widehat{\matrixsymbol{C}}$ when at least one signal is present, by way of the following sequence of nested alternate models:
\begin{equation}\label{E:SignalAlt}
\left\{ \mathcal{M}_r\!:\, \vectorsymbol{x}(t) = \sum_{i=1}^{r} s_{i}(t) \vectorsymbol{a}_{i} + \vectorsymbol{n}(t) \right\}, \quad 0 < r < n.
\end{equation}
In this setting, Baik et al.~\cite{baik2005pto} discovered a phase-transition phenomenon in the size of the population eigenvalue. This work was further developed by Baik and Silversten~\cite{baik2006els}, Paul~\cite{paul2007ase}, and Onatski~\cite{onatski2007adp}. Paul derived the limiting distribution for the case $\widehat{\matrixsymbol{C}} \in \mathbb{R}^n$, with the ratio $n/N$ of sensors to snapshots tending to $\gamma < 1$. Onatski later generalized this to $\gamma \in (0,\infty)$. Finally, Bai and Yao~\cite{bai2008clt} derived these limiting distributions in the complex case. A simplified summary of the above work is given by the following theorem.
\begin{theorem}[Asymptotic Alternate Distribution]\label{T:spiked-value}
Consider a covariance $\matrixsymbol{C} = \mathbb{E}\left[ \vectorsymbol{x}(t) \vectorsymbol{x}(t)^* \right]$ under the model of~\eqref{E:SignalAlt}, with $r>0$ distinct principal eigenvalues $\{\lambda_i + \sigma^2\}$, and the corresponding
sample covariance matrix $\widehat{\matrixsymbol{C}}$, with $n$ ordered eigenvalues $\{\ell_i\}$.
Denoting by $\Phi( \cdot )$ the standard Normal distribution function, and defining the standardizing quantities
\begin{align*}
\mu_{N,n}(\lambda)
&= \left( \lambda + \sigma^2 \right)
\left(1 + \gamma \sigma^2/\lambda \right) \\
\sigma_{N,n}(\lambda)
&= \left( \lambda + \sigma^2 \right) \textstyle
\sqrt{
\frac{2}{\beta N}
\left( 1 - \gamma \sigma^4/\lambda^2 \right)
},
\end{align*}
we have that if $\lambda_{1}, \lambda_{2}, \ldots, \lambda_{q} > \sqrt{\gamma} \sigma^2$, then
\begin{equation*}
\mathbb{P} \left\{ \frac{\ell_i - \mu_{N,n}(\lambda_i)}
{\sigma_{N,n}(\lambda_i)} \leq x_i,
\quad
i = 1, \ldots, q \right\}
\to
\prod_{i=1}^q \Phi ( x_i ).
\end{equation*}
Otherwise, for any $\lambda_i\!:\! \lambda_i \leq \sqrt{\gamma} \sigma^2$, then $\ell_i \overset{a.s.}{\longrightarrow} \sigma^2 \left( 1 + \sqrt{\gamma} \right)^2$.
As in Theorem~\ref{T:null-value}, $\beta = 1$ for real data and
$2$ for complex data.
\end{theorem}
\begin{rem}
Theorem~\ref{T:spiked-value} yields a critical threshold $\sqrt{\gamma} \sigma^2$
below which any population eigenvalue is unrelated to its corresponding sample eigenvalue; sample eigenvalues corresponding to population eigenvalues above this threshold converge to a multivariate Normal with diagonal covariance.
\end{rem}
Paul and Onatski also give accompanying results linking the mutual information of population and corresponding sample eigen\emph{vectors} through the same critical threshold $\sqrt{\gamma} \sigma^2$, and moreover implying the general inconsistency of the latter as estimators of the former. For the case of real-valued data, they prove the following theorem.
\begin{theorem}[Sample Eigenvector Inconsistency]\label{T:spiked-vector}
Let $\vectorsymbol{w}_i$ denote the $i$th principal population eigenvector of $\mathbb{E}\left[ \vectorsymbol{x}(t) \vectorsymbol{x}(t)^T \right]$ under the model of~\eqref{E:SignalAlt}, and $\hat{\vectorsymbol{w}}_i$ its corresponding sample version via~\eqref{E:Cemp-decomp-eig}. Then we have that
\begin{equation*}
\langle \vectorsymbol{w}_i, \hat{\vectorsymbol{w}}_j \rangle \overset{a.s.}{\longrightarrow}
\begin{cases}
\sqrt{\frac{\lambda_i - \gamma \sigma^4/\lambda_i}
{\lambda_i + \gamma \sigma^2}} & \text{if $i = j$ and $\lambda_i > \sqrt{\gamma} \sigma^2$,} \\
0 & \text{otherwise.}
\end{cases}
\end{equation*}
\end{theorem}
\begin{rem}
Onatski gives a convergence rate of $\sqrt{N}$ for the quantities of Theorem~\ref{T:spiked-vector}.
\end{rem}
To conclude this section, we note that while the above results are asymptotic in nature, evidence suggests that they are achieved in practice for small sample sizes. In particular, Ma~\cite{ma2008atw} has catalogued empirical convergence rates for Theorem~\ref{T:null-value}, demonstrating that for $n$ ranging up to $500$, even with only $N=5$ samples the Tracy-Widom asymptotics remain a good approximation in the upper tail of the distribution---the setting of interest in the model selection problem posed here. In later simulations, we apply our results to the model selection regime of direction-of-arrival estimation~\cite{kavcic1996are}, and consider complex-valued data in $n=9$ dimensions using $N=45$ snapshots. Figure~\ref{F:null-alt-density} shows a comparison of empirical and asymptotic distributions in this scenario, with the generally good agreement providing further evidence for the practical utility of Theorems~\ref{T:null-value} and~\ref{T:spiked-value} above.
\begin{figure}[t]
\centering
\includegraphics[width=\columnwidth]{plots/rank-est-dists}
\caption{\label{F:null-alt-density}Example illustrating agreement of empirical and asymptotic distributions for the null and alternate settings of Theorems~\ref{T:null-value} and~\ref{T:spiked-value}, respectively. The latter case comprises a single signal eigenvalue $\lambda$ set at a signal-to-noise ratio of 3~dB, with all other parameter settings matched to the simulation study of Section~\ref{S:emp-perf} (complex-valued data with $n=9$ and $N=45$).
}
\end{figure}
\section{Minimax-Optimal Rank Estimation}
\label{S:rank-est}
The previous section has given us a relatively complete description of the behavior of $\widehat{\matrixsymbol{C}}$, our $n$-dimensional sample covariance comprised of $N$ array snapshots, with $n/N$ tending to $\gamma \in (0, \infty)$ as $n,N \to \infty$, and $\sigma^2$ the variance of additive white Gaussian noise in the observation model $\mathcal{M}_r$ of~\eqref{E:Signal}. With this information, we are ready to proceed to the task of estimating the model order $r$, corresponding to the number of signals present. In light of Theorems~\ref{T:spiked-value}~and~\ref{T:spiked-vector}, we need not consider the $r$th signal if its strength is below the critical threshold $\sqrt{\gamma} \sigma^2$, as the corresponding alternate model $\mathcal{M}_r$ will typically be indistinguishable from the null $\mathcal{M}_{r-1}$ in the asymptotic limit. In the sequel we thus restrict our attention to the case when $\lambda_i > \sqrt{\gamma} \sigma^2$ for $i = 1, 2, \ldots, r$.
\subsection{Derivation of Asymptotic Risk for Signal Absence/Presence}
To formulate our minimax-optimal rank estimation task, we adopt a classical decision-theoretic approach: we first define a loss function $L$ to measure the quality of a particular estimate of $r$, and then derive a decision rule $\delta$ that minimizes the risk $R = \mathbb{E} \, [ L(\delta) ]$; i.e., the expected loss under our assumed probability model.
Consider first the most basic problem to which Theorems~\ref{T:null-value} and~\ref{T:spiked-value} offer a solution: differentiating between observing no signal at all ($r=0$) and observing a single signal ($r=1$). When $r=0$, the snapshot $\vectorsymbol{x}(t)$ is assumed to be a zero-mean multivariate Normal with covariance $\sigma^2 \matrixsymbol{I}$. When the model order $r=1$, the population covariance matrix $\matrixsymbol{C}$ has one eigenvalue equal to $\lambda + \sigma^2$, and the rest equal to $\sigma^2$. We encode these models by noting that $\lambda = 0$ in the first and $\lambda > 0$ in the second, and next address the task of choosing between them according to the sample covariance $\widehat{\matrixsymbol{C}}$.
Using $\delta$ to denote a decision rule taking values in $\{ 0,1 \}$, we let $\delta = 0$ encode the decision $r = 0$. To this rule, we assign an ``inclusion'' penalty $c_\text{I} > 0$ for incorrectly overestimating the rank, and an ``exclusion'' penalty $c_\text{E} > 0$ for incorrectly underestimating it; when $\delta$ chooses the correct outcome we assign no penalty. We summarize this by introducing the loss function $L(\lambda, \delta)$, defined as
\begin{equation*}
L(\lambda, \delta)
=
\begin{cases}
c_\text{I} &\text{when $\lambda = 0$ and $\delta = 1$,} \\
c_\text{E} &\text{when $\lambda > 0$ and $\delta = 0$,} \\
0 &\text{otherwise.}
\end{cases}
\end{equation*}
Guided by the results of Section~\ref{S:null-alt-rmt}, we distinguish between the two cases above based on $\ell_1(\widehat{\matrixsymbol{C}})$, the principal eigenvalue of the observed sample covariance matrix $\widehat{\matrixsymbol{C}}$: If $\ell_1$ is larger than some fixed threshold, we estimate $r$ as $\hat r = 1$, and otherwise we choose $\hat r = 0$. For a threshold $T$, we thus define our decision rule $\delta$ as
\begin{equation}\label{E:delta}
\delta_T (\ell)
=
\begin{cases}
1 &\text{if $\ell > T$,} \\
0 &\text{otherwise.}
\end{cases}
\end{equation}
The risk associated with this rule is given by evaluating the expected loss $\mathbb{E} \, [ L( \lambda, \delta_T( \ell_1 ) )]$ associated with our chosen test statistic $\ell_1(\widehat{\matrixsymbol{C}})$, with respect to probabilities $\mathbb{P}_{\mathcal{M}_i}$ under the two competing models $\mathcal{M}_0$ and $\mathcal{M}_1$:
\begin{equation*}
R( \lambda, \delta_T )
= \begin{cases}
c_\text{I} \cdot \mathbb{P}_{\mathcal{M}_0} \! \left\{ \ell_1 > T \right\}
&\text{when $\lambda=0$,} \\
c_\text{E} \! \cdot \mathbb{P}_{\mathcal{M}_1} \! \left\{ \ell_1 \leq T \right\}
&\text{otherwise.}
\end{cases}
\end{equation*}
Theorem~\ref{T:null-value} in turn describes the asymptotic distribution of $\ell_1$ when
$r=0$, while Theorem~\ref{T:spiked-value} describes it when $r=1$. We thus obtain a precise asymptotic description of the risk $R$ as
\begin{equation}\label{E:asympt-risk}
R( \lambda, \delta_T )
\to \begin{cases}
c_\text{I} \cdot \left(1 - F_\beta \left( \frac{ T - \mu_{N,n} }
{ \sigma_{N,n}} \right) \right)
&\text{when $\lambda=0$,} \\
c_\text{E} \cdot \Phi \left( \frac{T - \mu_{N,n} (\lambda)}
{ \sigma_{N,n} (\lambda)} \right)
&\text{otherwise,}
\end{cases}
\end{equation}
where again $\Phi( \cdot )$ denotes the standard normal CDF.
Suppose we have knowledge that when a signal is present, its strength is at least equal to $\lambda_0$. We may then choose a threshold $T$ to minimize the maximum risk over all relevant scenarios. Specifically,
we seek to minimize
\begin{equation}\label{E:mm-risk}
\sup_{\lambda \in \{ 0 \} \cup [\lambda_0, \infty)} \!\!\!\! R( \lambda, \delta_T )
=
R(0, \delta_T) \vee R(\lambda_0, \delta_T)
.
\end{equation}
It is not hard to show that this occurs when
\(
R(0, \delta_T) = R(\lambda_0, \delta_T),
\)
and hence we conclude from~\eqref{E:asympt-risk} that $T$ must solve
\begin{equation}\label{E:T-eqn}
\textstyle
c_\text{I} \cdot \left(1 - F_\beta \left( \frac{ T - \mu_{N,n} }
{ \sigma_{N,n}} \right) \!\right)
\! =
c_\text{E} \cdot \Phi \left( \frac{T - \mu_{N,n} (\lambda_0)}
{ \sigma_{N,n} (\lambda_0)} \right) \!.
\end{equation}
\subsection{Asymptotic Analysis of Minimax Threshold Behavior}
While it is easy to compute the minimax-optimal $T$ in~\eqref{E:T-eqn} numerically using bisection, and therefore implement the decision rule $\delta_T( \ell_1 )$ of~\eqref{E:delta}, we know of no closed-form expression for $T$. Instead, we now present a brief asymptotic analysis of the minimax threshold behavior in order to gain insight as to how $T$ behaves as $\lambda_0$ varies. For clarity of presentation and without loss of generality, we assume $\sigma^2 = 1$ in the sequel.
Our first observation stems from a comparison of the mean standardization quantities in Theorems~\ref{T:null-value} and~\ref{T:spiked-value}: It is easily verified that $\mu_{N,n}(\lambda) \to \mu_{N,n}$ as
$\lambda \to \sqrt{\gamma}$, implying a need for analysis when
the minimal assumed signal strength $\lambda_0$ is close to $\sqrt{\gamma}$.
Indeed, for other values of $\lambda_0$ a threshold $T$ slightly above $\mu_{N,n}$ will yield minimax risk very close to $0$, since $\sigma_{N,n} \sim N^{-2/3}$ and $\sigma_{N,n}(\lambda_0) \sim N^{-1/2}$ in this case.
To study the variation of $T$ with $\lambda_0$ in this regime, we first parameterize $\lambda_0$ in $h$, with the restriction $h > 0$, as
\begin{equation}\label{E:Lh}
\lambda_0(h) = \sqrt{ \gamma } + h.
\end{equation}
The threshold behavior of interest occurs when $h$ is of size $N^{-1/3}$, and so we parametrize $T$ in $t$ for this case as
\begin{equation}\label{E:Tt}
T(t) = \mu_{N,n} + t \sigma_{N,n}.
\end{equation}
We summarize the behavior of $t$ for $h$ near $N^{-1/3}$ in the following two lemmas, whose proofs are given in Appendix~\ref{S:mm-asymp}.
\begin{lemma}\label{L:thresh-small-h}
Let $\lambda_0(h)$ and $T(t)$ be parameterized as in~\eqref{E:Lh} and~\eqref{E:Tt}, respectively, and fix $h = o\left(N^{-1/3}\right)$. The behavior of $t$ then depends on the ratio of costs $c_\text{E}$ and $c_\text{I}$ as follows:
\begin{enumerate}
\item If $c_\text{E} > (1 - F_\beta(0)) \cdot c_\text{I}$, then
\begin{multline*}
t
=
2
\beta^{-1/2}
\left(
\frac{h^3 N}{ \gamma^{1/4} + \gamma^{-1/4} }
\right)^{1/6}
\\
\cdot
\Phi^{-1} \! \left(
\frac{c_\text{I}}{c_\text{E}}
\left(
1 - F_\beta(0)
\right)
\right)
\left(
1 + o(1)
\right).
\end{multline*}
\item If $c_\text{E} < (1 - F_\beta(0)) \cdot c_\text{I}$, then for
\(
t_{\sqrt{\gamma}}
=
F_\beta^{-1} \!\left( 1 - c_\text{E}/c_\text{I} \right),
\)
we have that
\begin{multline*}
t
=
t_{\sqrt{\gamma}}
+
\left[
f_\beta \left( t_{\sqrt{\gamma}} \right)
\right]^{-1}
\cdot
\frac{c_\text{E}}{c_\text{I}}
\sqrt{\frac{2}{\beta \pi}}
\left(
\frac{ h^3 N }{\gamma^{1/4} + \gamma^{-1/4}}
\right)^{1/6}
\\
\cdot
t_{\sqrt{\gamma}}^{-1} \,
\exp\!\left(
-
\frac{\beta t_{\sqrt{\gamma}}^2}{8}
\left(
\frac{\gamma^{1/4} + \gamma^{-1/4}}{ h^3 N }
\right)^{1/3}
\right)
\left(
1 + o(1).
\right)
\end{multline*}
\item If $c_\text{E} = (1 - F_\beta(0)) \cdot c_\text{I}$, then $t$ solves
\begin{multline*}
t^2
=
\left[
f_\beta ( 0 )
\right]^{-1}
\cdot
\frac{c_\text{E}}{c_\text{I}}
\sqrt{\frac{2}{\beta \pi}}
\left(
\frac{ h^3 N }{\gamma^{1/4} + \gamma^{-1/4}}
\right)^{1/6}
\\
\cdot
\exp \!\left(
-
\frac{\beta t^2}{8}
\left(
\frac{\gamma^{1/4} + \gamma^{-1/4}}{ h^3 N }
\right)^{1/3}
\right)
\left(
1 + o\left( 1 \right)
\right).
\end{multline*}
\end{enumerate}
\end{lemma}
\begin{lemma}\label{L:thresh-med-h}
Suppose instead that $h = h_0 N^{-1/3}$ for some constant $h_0 > 0$. Then we have the result that
\begin{multline*}
c_\text{I}
\cdot
(1 - F_\beta ( t ) )
\sim
c_\text{E}
\cdot
\Phi \!\left(
\frac{\beta^{1/2} t}{ 2 \sqrt{ h_0 }}
\left( \gamma^{1/4} + \gamma^{-1/4} \right)^{1/6} \right.
\\
-
\left.
\frac{ \beta^{1/2} h_0^{3/2} }
{ \sqrt{ 2 \gamma }
\left( \gamma^{1/4} + \gamma^{-1/4} \right) }
\right).
\end{multline*}
Moreover, if it is also the case that $c_\text{E} = \omega( c_\text{I} )$, then
\[
t
\sim
\,-
\sqrt{
\frac{ 8 h_0 }{ \beta (\gamma^{1/4} + \gamma^{-1/4} )^{1/6} }
\log \frac{c_\text{E}}{c_\text{I}}
} \,;
\]
if instead we have that $c_\text{E} = o( c_\text{I} )$, then
$
t
\sim
\left(
\frac{3}{2 \beta}
\log \frac{c_\text{I}}{c_\text{E}}
\right)^{2/3}.
$
\end{lemma}
\subsection{Extension to the General Model Order Selection Problem}
In the above discussion we treated the basic model order selection problem of $\mathcal{M}_r: r = 0$ versus $\mathcal{M}_r: r = 1$, in order to develop our problem formulation and asymptotic results. In practice, of course, techniques are needed to address the general model order selection problem of estimating $r \geq 0$. The main result of this section is that an asymptotically minimax-optimal rank selection rule is in fact obtained through repeated application of the basic $\mathcal{M}_0$ vs.~$\mathcal{M}_1$ case.
To develop such a rule and verify its properties, we first extend the thresholding approach seen earlier to the case of arbitrary $r > 0$. Rather than specifying only a single cost $c_\text{E}$ for incorrectly excluding a term, we now require a sequence $\{c_\text{E}(i)\}_{i=1}^n$ of nonnegative costs, corresponding to exclusion of respective signal terms. To this end we define a cumulative exclusion cost $C_\text{E}(\cdot)$ as
\begin{equation}\label{E:Ce}
C_\text{E}(j) = \sum_{i=j}^n c_\text{E}(i), \quad 1 \leq j \leq n \text{.}
\end{equation}
One possible choice for the sequence $\{c_\text{E}(i)\}_{i=1}^n$ is simply to set all exclusion costs to be equal; alternatively, with prior knowledge that there are at most $r_\text{max}$ signals, one might well set $c_\text{E}(i) = 0$ for
$i > r_\text{max}$.
In a similar manner, we define a sequence of thresholds $\{T(i)\}_{i=1}^n$ associated to the sequence of ordered sample eigenvalues $\{\ell_i\}_{i=1}^n$, with each threshold $T(i)$ determined by an inclusion cost $c_\text{I}$ and the corresponding exclusion cost $C_\text{E}(i)$. An ordered rank selection procedure for $r$ is then given by Algorithm~\ref{A:rank-selection} below, and verified to be asymptotically minimax optimal by the theorem that follows.
\begin{algorithm}
\caption{\label{A:rank-selection}Minimax-optimal rank selection procedure}
\begin{enumerate}
\item Fix a threshold sequence $\{T(i)\}_{i=1}^n$ via~\eqref{E:T-eqn} and pre-assigned erroneous inclusion/exclusion costs $c_\text{I}, C_\text{E}(i)$;
\item Form the sample covariance $\widehat{\matrixsymbol{C}}$, set $i \leftarrow 1$, and test:
\begin{algorithmic}
\WHILE{ $\ell_{i}(\widehat{\matrixsymbol{C}}) > T( i )$ }
\STATE $i \leftarrow i + 1$
\ENDWHILE
\end{algorithmic}
\item Return $\hat r \leftarrow i - 1$ as the final estimate of rank $r$.
\end{enumerate}
\end{algorithm
\begin{theorem}[Minimax Optimality]\label{T:mm-opt} For fixed, nonnegative inclusion cost $c_\text{I}$ and exclusion
costs $\{C_\text{E}(i)\}$, the rank selection procedure of Algorithm~\ref{A:rank-selection} is asymptotically minimax.
\end{theorem}
\begin{IEEEproof}
Note that the $i$th iteration of Step~2 in Algorithm~\ref{A:rank-selection} tests $\mathcal{M}_r: r = i-1$ versus $\mathcal{M}_r: r \geq i$. By~\eqref{E:mm-risk}, the worst-case risk occurs when $r \to \infty$ and each signal has strength equal to the minimum assumed strength $\lambda_0$, in which case the cost for excluding them all is given by $C_\text{E}(i)$ in~\eqref{E:Ce}. Thus the optimal threshold at this stage is given by the corresponding $T(i)$ satisfying~\eqref{E:T-eqn}. Since Theorem~\ref{T:spiked-value} proves the asymptotic independence of the $r$ principal sample eigenvalues, we conclude in turn that considering only the $i$th eigenvalue $\ell_i$ at the $i$th iteration incurs no loss in power.
\end{IEEEproof}
According to~\eqref{E:T-eqn}, knowledge of the noise power $\sigma^2$ is required to determine optimal thresholds. In an adaptive setting, we note that $\sigma^2$ may be estimated at time $t$ by way of the residual variance from the $\hat r(t-1)$ signals at time index $t-1$~\cite{rabideau1996fra}. In a non-adaptive setting, Kritchman and Nadler~\cite{kritchman2008dnc} and Patterson et al.~\cite{patterson2006psa} suggest alternative approaches.
\section{Empirical Performance Examples}
\label{S:emp-perf}
Having derived a rank selection rule in the preceding section and investigated its theoretical properties, we now provide two brief simulation studies designed to demonstrate the empirical performance of this procedure. We report on an evaluation of Algorithm~\ref{A:rank-selection} by way of two simulations adopted from the direction-of-arrival estimation setting of~\cite{kavcic1996are}, shown in the top panels of Fig.~\ref{F:simulations}. The first simulation has signals of different strengths appearing and disappearing over time, leading to a varying rank, whereas the second simulation comprises a constant number of signals. In both settings, the snapshots are in $n = 9$ dimensions, and arrival directions vary over time.
\begin{figure*}
\includegraphics[width=2.1\columnwidth]{plots/doa-sim}
\caption{\label{F:simulations} Two simulations taken from a direction-of-arrival estimation setting, showing empirical performance of the minimax rank selection rule of Algorithm~\ref{A:rank-selection} and the hypothesis-testing-based rule of~\cite{kritchman2008dnc}. Signals of the form $C_0 ( 1, e^{j \omega}, \ldots, e^{j 8 \omega})$ appear and disappear over time $t \in [0,1000]$, with signal directions and strengths shown in the top row. One snapshot is sampled per unit time, and the subspace at time $t$ is estimated using the snapshots for times in the window $(t-45,t]$. The second row of the plot shows the true rank $r$ as a function of $t$, along with the estimated rank $\hat r$ according to the method of Algorithm~\ref{A:rank-selection}. The subspace approximation error in squared Frobenius norm, $\|\matrixsymbol{W} \matrixsymbol{W}^\ast - \widehat{\matrixsymbol{W}}_k \widehat{\matrixsymbol{W}}^\ast_k \|_F^2$, is plotted for $k = r, \hat r$ in the third row.}
\end{figure*}
Each simulation features a range of signal strengths, including some so low as to be indistinguishable from the background noise. In the first simulation, for example, there is one signal below the detection threshold $\sqrt{\gamma}$ between time indices $150$ and $400$; in the second simulation, the weakest signal is always below the detection threshold, implying that it will not in general be detected.
In these performance examples we employed a windowed covariance estimate with $N = 45$ observations, and set $c_\text{I} = c_\text{E}(1) = \cdots = c_\text{E}(n)$, with $\lambda_0 = \sqrt{\gamma} + N^{-1/3}$. We treat $\sigma^2$ as unknown, and estimate it for every time $t$ as the mean of the estimated noise eigenvalues at the previous time step $t-1$~\cite{rabideau1996fra}. The middle panels of Fig.~\ref{F:simulations} demonstrate the corresponding rank estimation results, from which we observe good empirical agreement with theoretical predictions; where we incur estimation errors, they tend to be as a result of signals whose strength falls below $\sqrt{\gamma}$, the detection limit. The bottom panels of Fig.~\ref{F:simulations} also show the error in squared Frobenius norm, $\| \matrixsymbol{W} \matrixsymbol{W}^\ast - \widehat{\matrixsymbol{W}}_k \widehat{\matrixsymbol{W}}^\ast_k \|_F^2$, of the corresponding subspace estimate
\(
\widehat{\matrixsymbol{W}}_k
=
\begin{pmatrix}
\hat{\vectorsymbol{w}}_1 & \hat{\vectorsymbol{w}}_2 & \cdots & \hat{\vectorsymbol{w}}_k
\end{pmatrix}
\)
for $k = r$ and $k = \hat r$, respectively the true and estimated ranks.
\begin{figure*}
\centering
\includegraphics[width=2.1\columnwidth]{plots/consistency-sim}
\caption{Simulations demonstrating that the performance of the rank estimators improves as the snapshot sampling frequency increases. With the same signal arrival patterns as in Fig.~\ref{F:simulations}, we vary the snapshot sampling rate and show how the rank estimation error $\frac{1}{1000} \int_0^{1000} |r(t) - \hat r(t)| \, dt$ behaves for the two rank estimators $\hat r$. The points show the mean behavior and the error bars show $1$ standard deviation, computed from $50$ replicates of the experiment.}\label{F:consistency-sim}
\end{figure*}
For purposes of comparison, we also show a hypothesis-testing-based approach adopted from Kritchman and Nadler~\cite{kritchman2008dnc}, based on Tracy-Widom quantiles and set at a fixed 0.5\% false alarm rate as suggested by those authors. We note however, that the variance estimation approach of~\cite{kritchman2008dnc} does not apply directly in the subspace tracking scenario, as it employs all eigenvalues of the sample covariance matrix; here we employed an estimate based on the residual from the previous time step. We see that in comparison to the proposed method of Algorithm~\ref{A:rank-selection}, this approach tends to provide a consistently more conservative estimate of rank, resulting in greater overall error in this simulation context.
In Fig.~\ref{F:consistency-sim}, we show how these rank estimators perform as the snapshot sampling frequency per unit time is increased. Since the number of sensors is held fixed, this corresponds to varying $\gamma$; the simulations of Fig.~\ref{F:consistency-sim} demonstrate that for higher sampling frequencies (lower values of $\gamma$), the rank estimation problem becomes easier and the error decreases. Note in the left-hand panel of Fig.~\ref{F:consistency-sim}, however, that because the rank is not constant within all time windows, the resultant error will not necessarily asymptote to zero. In contrast, that of the right-hand panel \emph{will} eventually reach zero when $\gamma$ becomes sufficiently small to render the weakest signal detectable (as per Theorem~\ref{T:spiked-value}).
\section{Discussion}
\label{s:summ}
In this article we have presented sample covariance asymptotics stemming from random matrix theory, and have in turn brought them to bear on the problem of optimal rank estimation. This task poses a classical model order selection problem that arises in a variety of important statistical signal and array processing systems, yet is addressed relatively infrequently in the extant literature. Key to our approach is the existence of a phase transition threshold in the context of the standard array observation model with additive white Gaussian noise, below which eigenvalues and associated eigenvectors of the sample covariance fail to provide any information on population eigenvalues. Using this and other results, we then developed a decision-theoretic rank estimation framework that led to a simple ordered selection rule based on thresholding; in contrast to competing approaches, this algorithm was shown to admit asymptotic minimax optimality and to be free of tuning parameters. We concluded with a brief simulation study to demonstrate the practical efficacy of our rank selection procedure, and plan to address a diverse set of rank estimation tasks as part of our future work.
\appendices
\section{Tracy-Widom Asymptotics}\label{S:tw-asymp}
A characterization of the tail behavior of $F_\beta(s)$, the distribution function of the Tracy-Widom law, is required to obtain the results of Lemmas~\ref{L:thresh-small-h} and~\ref{L:thresh-med-h}. In this appendix we derive the asymptotic properties of $F_\beta(s)$ for $\beta=1,2$ as
$|s| \to \infty$.
To begin, let $q(x)$ solve the Painlev\'e~II equation
\[
q''(x) = x q(x) + 2 q^3(x),
\]
with boundary condition $q(x) \sim \Ai(x)$ as $x \to \infty$ and $\Ai(x)$ the Airy function. Then it follows that
\begin{align*}
F_1(s)
&= \exp \left\{
-\frac{1}{2} \int_s^\infty q(x) + (x - s) q^2(x) dx
\right\}, \\
\intertext{and}
F_2(s)
&= \exp \left\{
- \int_s^\infty (x - s) q^2(x) dx
\right\}.
\end{align*}
As $x \to \infty$, the Airy function behaves as
\(
\Ai(x)
\sim \frac{1}{2\sqrt{\pi}} x^{-1/4}
\exp \left( -\frac{2}{3} x^{3/2} \right);
\)
asymptotic properties of $q$ as $x \to -\infty$ are studied by Hastings and McLeod \cite{hastings1980bvp}, who show that in this case,
\(
q(x) \sim \sqrt{ |x| / 2 }.
\)
Using these facts, we can compute for $s \to \infty$ the term
\begin{align*}
\int_s^\infty & q(x) dx \\
& \sim \frac{1}{2 \sqrt{\pi}}
\int_s^\infty
x^{-1/4} \exp\left( -\frac{2}{3} x^{3/2} \right) dx \\
&= \frac{1}{2 \sqrt{\pi}}
\int_0^\infty
\exp\left(
-\frac{2}{3} (x + s)^{3/2} - \frac{1}{4} \log (x + s)
\right) dx \\
&\sim \frac{1}{2 \sqrt{\pi}}
\int_0^\infty
\exp\left(
-\frac{2}{3} s^{3/2}
\left(
1 + \frac{3}{2} \frac{x}{s}
\right)
- \frac{1}{4} \log s
\right)
dx \\
&= \frac{1}{2 \sqrt{\pi}} s^{-1/4}
\exp\left( -\frac{2}{3} s^{3/2} \right)
\int_0^\infty \exp\left( -s^{1/2} x \right) dx \\
&= \frac{1}{2 \sqrt{\pi}} s^{-3/4}
\exp\left( -\frac{2}{3} s^{3/2} \right) ,
\end{align*}
and also
\begin{align*}
\int_s^\infty & (x - s) q^2(x) dx \\
& \sim \frac{1}{4 \pi}
\int_s^\infty
(x - s)
x^{-1/2}
\exp\left( -\frac{4}{3} x^{3/2} \right)
dx \\
&= \frac{1}{4 \pi}
\int_0^\infty
x
\exp\left(
-\frac{4}{3} (x + s)^{3/2}
-\frac{1}{2} \log( x + s )
\right)
dx \\
&\sim \frac{1}{4 \pi}
\int_0^\infty
x
\exp\left(
-\frac{4}{3}s^{3/2}
\left( 1 + \frac{3}{2} \frac{x}{s} \right)
- \frac{1}{2} \log s
\right)
dx \\
&= \frac{1}{4 \pi}
s^{-1/2}
\exp\left( -\frac{4}{3} s^{3/2} \right)
\int_0^\infty
x
\exp\left(
-2 s^{1/2} x
\right)
dx \\
&= \frac{1}{16 \pi}
s^{-3/2}
\exp\left( -\frac{4}{3} s^{3/2} \right).
\end{align*}
Likewise, for $s \to -\infty$ we have that
\begin{align*}
\int_s^\infty q(x) dx &\sim \frac{\sqrt{2}}{3} |s|^{3/2}, \\
\intertext{and}
\int_s^\infty (x - s) q^2(x) dx &\sim \frac{|s|^3}{12}.
\end{align*}
Now, we must have that as $s\to\infty$,
\begin{align*}
F_1(s)
&\sim \exp\left\{
-\frac{1}{2}
\left(
\frac{1}{2 \sqrt{\pi}}
s^{-3/4}
e^{-\frac{2}{3} s^{3/2}}
\right. \right.
\\ & \qquad \qquad \qquad \quad\,
\left. \left.
+
\frac{1}{16 \pi}
s^{-3/2}
e^{-\frac{4}{3} s^{3/2}}
\right)
\right\} \\
&\sim \exp\left\{
-
\frac{1}{4 \sqrt{\pi}}
s^{-3/4}
e^{-\frac{2}{3} s^{3/2}}
\right\} \\
&\sim 1
-
\frac{1}{4 \sqrt{\pi}}
s^{-3/4}
\exp\left( -\frac{2}{3} s^{3/2} \right), \\
\intertext{and similarly}
F_2(s) &\sim 1
-
\frac{1}{16 \pi}
s^{-3/2}
\exp\left( -\frac{4}{3} s^{3/2} \right).
\end{align*}
We also get that as $s\to -\infty$,
\begin{align*}
F_1(s) &\sim \exp\left( -\frac{|s|^3}{24} \right),
\intertext{and}
F_2(s) &\sim \exp\left( -\frac{|s|^3}{12} \right).
\end{align*}
In summary, then, for $\beta = 1,2$ we have that as $s \to -\infty$,
\begin{equation}\label{E:tw-s-neg}
F_\beta(s) \sim \exp\left( -\frac{\beta}{24} |s|^3 \right),
\end{equation}
while for $s \to \infty$ we have
\begin{equation}\label{E:tw-s-pos}
1 - F_\beta(s)
\sim
\left(
\frac{1}{16 \pi}
\right)^{\beta/2}
s^{-3 \beta/4 }
\exp\left(
-
\frac{2 \beta}{3}
s^{3/2}
\right).
\end{equation}
\section{Minimax Threshold Asymptotics}\label{S:mm-asymp}
In this appendix we prove Lemmas~\ref{L:thresh-small-h} and~\ref{L:thresh-med-h}. Recall that by~\eqref{E:Lh} we have the parameterization $\lambda_0(h) = \sqrt{\gamma} + h$ for some fixed $h > 0$, and we seek asymptotic properties of the associated minimax eigenvalue threshold $T$, parameterized according to~\eqref{E:Tt} as $T(t) = \mu_{N,n} + t \sigma_{N,n}$. To derive the asymptotic behavior of $T$ for small and large $h$, we first require the asymptotic
behaviors of $\mu_{N,n}( \sqrt{\gamma} + h )$ and $\sigma_{N,n}(\sqrt{\gamma} + h)$ for small $h$, along with the tail behaviors of $\Phi(x)$ and $F_\beta(x)$.
To this end, it is not hard to show that for small $h$,
\begin{align*}
\mu_{N,n}(\sqrt{\gamma} + h)
&= \mu_{N,n} + \frac{h^2}{\sqrt{\gamma}} + O(h^3), \\
\sigma_{N,n}(\sqrt{\gamma} + h)
&= 2 \beta^{-1/2}
(\gamma^{1/4} + \gamma^{-1/4}) \sqrt{\frac{h}{N}}
+ O \left( \frac{h}{\sqrt{N}} \right).
\end{align*}
Since
\(
\sigma_{N,n}
=
\left(
(\gamma^{1/4} + \gamma^{-1/4})/N
\right)^{2/3},
\)
for $h = O(N^{-1/2})$ we have that
\begin{multline*}
\frac{ T - \mu_{N,n}(\sqrt{\gamma} + h)}
{ \sigma_{N,n}(\sqrt{\gamma} + h) }
=
\beta^{1/2}
\frac{t}{2}
\left(
\frac{\gamma^{1/4} + \gamma^{-1/4}}{ h^3 N }
\right)^{1/6}
\\
-
\sqrt{\frac{\beta}{2 \gamma}}
\cdot
\frac{\sqrt{ h^3 N }}{\gamma^{1/4} + \gamma^{-1/4}}
+ O \left( h/\sqrt{N} \right) .
\end{multline*}
Using the notation $x \sim y$ to denote $x = y\left( 1 + o(1) \right)$, a
standard result \cite{abramowitz1970hmf} is that as $x \to \infty$, we have
\[
1 - \Phi(x)
\sim
\frac{1}{\sqrt{2 \pi}}
x^{-1}
\exp\left( -\frac{1}{2} x^2 \right).
\]
Therefore, as $\epsilon \to 0$,
\(
\Phi^{-1} (\epsilon)
\sim
-
\sqrt{ 2 \log \epsilon^{-1} }
\).
Tail properties of $F_\beta(x)$ are derived in Appendix~\ref{S:tw-asymp}, and given by~\eqref{E:tw-s-neg} and~\eqref{E:tw-s-pos}. From these, we have that as $\epsilon \to 0$,
\[
F_\beta^{-1} ( \epsilon )
\sim
-
\left( 24 \beta^{-1} \log \epsilon^{-1} \right)^{1/3}
\]
and
\[
F_\beta^{-1} ( 1 - \epsilon )
\sim
\left(
\frac{3}{2 \beta}
\log \epsilon^{-1}
\right)^{2/3}.
\]
Equipped with these results, we are now ready to give the proofs of Lemmas~\ref{L:thresh-small-h} and~\ref{L:thresh-med-h}.
\begin{IEEEproof}[Proof of Lemma~\ref{L:thresh-small-h}]
If $h = o\left(N^{-1/3}\right)$ then
\begin{multline*}
\Phi \!
\left(
\frac{ T - \mu_{N,n}(\sqrt{\gamma} + h)}
{ \sigma_{N,n}(\sqrt{\gamma} + h) }
\right)
=
\Phi \!
\left(
\frac{\beta^{1/2}t}{2}
\left(
\frac{\gamma^{1/4} + \gamma^{-1/4}}{ h^3 N }
\right)^{1/6}
\right)
\\
+
O\left(\sqrt{h^3 N} \right)
.
\end{multline*}
When $t = O\left((h^3 N)^{1/6}\right)$, the left-hand side of~\eqref{E:T-eqn} converges to $c_\text{I} ( 1 - F_\beta(0))$. Therefore,
\begin{multline*}
t
=
2
\beta^{-1/2}
\left(
\frac{ h^3 N }{\gamma^{1/4} + \gamma^{-1/4}}
\right)^{1/6}
\Phi^{-1}
\left(
\frac{c_\text{I}}{c_\text{E}}
\Big( 1 - F_\beta(0) \Big)
\right)
\\
+
O\left( (h^3 N)^{1/3} \right)
.
\end{multline*}
Of course, this only makes sense when $c_\text{E} > (1 - F_\beta(0)) \cdot c_\text{I}$.
Otherwise, we must have $t = \omega\left((h^3 N)^{1/6} \right)$. In this
case, using $\Phi(x) = 1 - (1/\sqrt{2 \pi}) x^{-1} \exp(-x^2/2)(1 + O(x^{-2}))$
as $x \to \infty$, we obtain for $t > 0$ the expression
\begin{multline*}
\Phi
\left(
\frac{ T - \mu_{N,n}(\sqrt{\gamma} + h)}
{ \sigma_{N,n}(\sqrt{\gamma} + h) }
\right)
=
1
-
\sqrt{\frac{2}{\beta \pi}}
\left(
\frac{ h^3 N }{\gamma^{1/4} + \gamma^{-1/4}}
\right)^{1/6}
\\
\cdot t^{-1}
\exp
\left(
-\frac{\beta t^2}{8}
\left(
\frac{\gamma^{1/4} + \gamma^{-1/4}}{ h^3 N }
\right)^{1/3}
\right)
\left( 1 + O \left(\sqrt{h^3 N}/t^3 \right) \right).
\end{multline*}
Consequently,
\begin{multline*}
t
= F_\beta^{-1}
\left\{
1 - \frac{c_\text{E}}{c_\text{I}}
+
\frac{c_\text{E}}{c_\text{I}}
\sqrt{\frac{2}{\beta \pi}}
\left(
\frac{ h^3 N }{\gamma^{1/4} + \gamma^{-1/4}}
\right)^{1/6}
\right.
\\
\cdot
\left . t^{-1}
\exp \!
\left(
-\frac{\beta t^2}{8}
\left(
\frac{\gamma^{1/4} + \gamma^{-1/4}}{ h^3 N }
\right)^{\!1/3}
\right) \!
\left(
1 + O\left(\sqrt{h^3 N}/t^3 \right)
\right) \!
\right\}
\\ \!\!\!
= t_{\sqrt{\gamma}}
+
\left[
f_\beta \left( t_{\sqrt{\gamma}} \right)
\right]^{-1}
\cdot
\frac{c_\text{E}}{c_\text{I}}
\sqrt{\frac{2}{\beta \pi}}
\left(
\frac{ h^3 N }{\gamma^{1/4} + \gamma^{-1/4}}
\right)^{1/6}
\\
\cdot t_{\sqrt{\gamma}}^{-1}
\exp
\left(
-\frac{\beta t_{\sqrt{\gamma}}^2}{8}
\left(
\frac{\gamma^{1/4} + \gamma^{-1/4}}{ h^3 N }
\right)^{1/3}
\right)
\left(
1 + O\left(\sqrt{h^3 N} \right)
\right) ,
\end{multline*}
where $t_{\sqrt{\gamma}} = F_\beta^{-1} \left( 1 - \frac{c_\text{E}}{c_\text{I}} \right)$.
Likewise, this expression only makes sense when
$c_\text{E} < (1 - F_\beta(0)) \cdot c_\text{I}$. The last case we need to consider is
when $c_\text{E} = (1 - F_\beta(0)) \cdot c_\text{I}$. In this case we have that $t$
solves
\begin{multline*}
t^2
=
\left[
f_\beta ( 0 )
\right]^{-1}
\cdot
\frac{c_\text{E}}{c_\text{I}}
\sqrt{\frac{2}{\beta \pi}}
\left(
\frac{ h^3 N }{\gamma^{1/4} + \gamma^{-1/4}}
\right)^{1/6}
\\
\cdot \exp
\left(
-\frac{\beta t^2}{8}
\left(
\frac{\gamma^{1/4} + \gamma^{-1/4}}{ h^3 N }
\right)^{1/3}
\right)
\left(
1 + O\left(\sqrt{h^3 N} \right)
\right).
\end{multline*}
\end{IEEEproof}
\begin{IEEEproof}[Proof of Lemma~\ref{L:thresh-med-h}]
We now suppose instead that $h = h_0 N^{-1/3}$, for some constant
$h_0 > 0$. In this case
\begin{multline*}
\frac{ T - \mu_{N,n}(\sqrt{\gamma} + h)}
{ \sigma_{N,n}(\sqrt{\gamma} + h) }
=
\frac{\beta^{1/2} t}{2 \sqrt{ h_0 } }
\left( \gamma^{1/4} + \gamma^{-1/4} \right)^{1/6}
\\
-
\frac{\beta^{1/2} h_0^{3/2}}
{ \sqrt{ 2 \gamma } \left( \gamma^{1/4} + \gamma^{-1/4} \right) }
+
O( h_0 N^{-5/6 }).
\end{multline*}
If $c_\text{E} = \omega( c_\text{I} )$, then we must have $t \to -\infty$ so that the
left-hand side of~\eqref{E:T-eqn} converges to $c_\text{I}$. Using the tail
behavior of $\Phi( \cdot )$, we have that
\[
\frac{\beta^{1/2} t}{2 \sqrt{h_0}}
\left(
\gamma^{1/4} + \gamma^{-1/4}
\right)^{1/6}
\sim
\Phi^{-1} \left( \frac{c_\text{I}}{c_\text{E}} \right)
\sim
-
\sqrt{
2 \log \frac{c_\text{E}}{c_\text{I}}
}
\]
so that
\[
t
\sim
-
\sqrt{
\frac{ 8 h_0 }{ \beta (\gamma^{1/4} + \gamma^{-1/4} )^{1/6} }
\log \frac{c_\text{E}}{c_\text{I}}
}.
\]
If, on the other hand, $c_\text{E} = o(c_\text{I})$, then the right-hand side of~\eqref{E:T-eqn}
must converge to $c_\text{E}$, and
\[
t
\sim
F_\beta^{-1} \left( 1 - \frac{c_\text{E}}{c_\text{I}} \right)
\sim
\left(
\frac{3}{2 \beta}
\log \frac{c_\text{I}}{c_\text{E}}
\right)^{2/3}.
\]
\end{IEEEproof}
\section*{Acknowledgment}
The authors wish to thank Art Owen for many helpful discussions.
\bibliographystyle{IEEEtran}%
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 1,732 |
3 people detained in Brussels in connection to Paris attacks
3 detained in Brussels related to Paris attack
Benjamin Hall reports from Belgium
Prosecutors say three people have been detained in Brussels for questioning in connection to the Paris attacks.
The Federal Prosecutor's Office said in a statement Tuesday that the three were detained following a morning search in Uccle, an upscale district of the Belgian capital.
It said a judge would decide Wednesday whether the three people should remain in custody. It said no further information would be made public about the search.
Brussels was home to many of the attackers who struck the French capital Nov. 13, killing 130 victims.
Brussels Terror Attacks: A Timeline of Events | Graphiq | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 2,794 |
Embarrassing mac and cheese incident spawns fundraisers
"A top-20 research university shouldn't have to be redeeming its name after one bad egg goes and ruins our reputation," said freshman Sadie Rumsey.
By PAT EATON-ROBB | Associated Press Monday, October 12, 2015
This undated photo provided by the University of Connecticut police department shows student Luke Gatti, 19, of Bayville, N.Y., who was arrested Sunday night, Oct. 4, 2015, following an altercation over purchasing macaroni and cheese at a market on the school's Storrs, Conn., campus. A 9-minute, obscenity-laced video clip went viral, showing Gatti arguing with and eventually shoving a manager at a food court inside the school's student union. University of Connecticut Police Department via AP
HARTFORD, Conn. (AP) — UConn students are trying to use an embarrassing video about macaroni and cheese to raise a little cheddar.
The video of a fellow student berating food service workers who refused to sell him jalapeno-bacon mac and cheese prompted a group of students to start an online fundraiser to give the beleaguered employees a well-deserved night out.
"A top-20 research university shouldn't have to be redeeming its name after one bad egg goes and ruins our reputation," said freshman Sadie Rumsey.
Rumsey and her friends set up a page on GoFundMe.com to show their support for the workers abused in the video captured inside the university's student union last week. As of Sunday, the page had collected more than $1,300.
The 9-minute, obscenity-laced video clip posted online shows freshman Luke Gatti arguing with and eventually shoving Dave Robinson, a food service supervisor. Police and the manager said Gatti was refused service on Oct. 4 for carrying an open alcohol container.
The video, which became fodder for late-night talk show hosts, shows the 19-year-old questioning why in America he can't have beer in the building. He uses a gay slur against Robinson and repeatedly demands, "Just give me some (expletive) bacon-jalapeno mac and cheese."
After shoving Robinson, Gatti is tackled by another employee, is arrested by a police officer and spits at the manager before being led out of the building.
Gatti, of Bayville, New York, has not returned phone calls or an email seeking comment. He is due in court Tuesday on charges of breach of peace and criminal trespass.
Kansas: No court order needed for same-sex birth certificate
Bus driver tells teenager his gay dads will go to hell | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 5,955 |
Stefan Parnicki-Pudełko (ur. 10 lipca 1914 r. w Majdanie Starym, zm. 16 kwietnia 1994 we Wrocławiu) – polski naukowiec, historyk i archeolog specjalizujący się w zakresie starożytności. Żołnierz AK i BCh. Honorowy obywatel Swisztowa.
Życiorys
Urodził się 10 lipca 1914 r. we wsi Majdan Stary na Lubelszczyźnie. W 1933 r. ukończył gimnazjum humanistyczne w Krasnymstawie. W okresie szkolnym był członkiem Związku Harcerstwa Polskiego.
W 1934 r. rozpoczął studia humanistyczne na Uniwersytecie Lwowskim (skupiając się głównie na archeologii śródziemnomorskiej i filologii klasycznej), które przerwał wybuch drugiej wojny światowej. W trakcie kampanii wrześniowej 1939 r. brał udział w obronie Lwowa. W okresie wojny pracując w gospodarstwie ojca, uczestniczył też w konspiracyjnym nauczaniu w Wojsławicach oraz działał w Armii Krajowej oraz Batalionach Chłopskich.
Po zakończeniu wojny kontynuował studia na Katolickim Uniwersytecie Lubelskim, uzyskując magisterium w 1946 r. na podstawie pracy Quo modo Homerus in poematibus suis colorem atque formam rerum exprimit. Następnie natychmiast podjął pracę w Katedrze Archeologii Klasycznej Uniwersytetu Wrocławskiego. W 1949 r. obronił tam pracę doktorską Udział czynników urbanistycznych w powstawaniu i rozwoju greckiej agory (promotor: Edmund Bulanda). W 1953 r. rozpoczął pracę w Zakładzie Archeologii Antycznej Instytutu Historii Kultury Materialnej Polskiej Akademii Nauk, gdzie w kwietniu 1956 r. awansowano go na stanowisko docenta.
W roku akademickim 1958/1959 przeniósł się na Uniwersytet im. Adama Mickiewicza w Poznaniu, zostając kierownikiem Katedry Archeologii Śródziemnomorskiej. W 1966 r. został profesorem nadzwyczajnym. W roku akademickim 1967-1968 gościnnie wykładał na Institute for Advanced Study w Princeton. Wobec likwidacji swojej katedry w 1969 r., został najpierw kierownikiem Katedry Historii Sztuki, a następnie w 1970 r. przeszedł do Zakładu Historii Starożytnej w Instytucie Historii UAM, którego objął kierownictwo.Tam pracował już do przejścia na emeryturę w 1984 r. W 1977 r. uzyskał nominację na profesora zwyczajnego.
W latach 1956-1957 brał udział w polsko-radzieckich (ukraińskich) badaniach archeologicznych w Olbii nad Morzem Czarnym. W 1960 r. odbył czteromiesięczne specjalistyczne studia w Grecji w zakresie architektury, urbanistyki i budownictwa starożytnej Hellady. W 1965 r. przebywał na trzymiesięcznym stypendium we Włoszech, badając budownictwo greckie w południowej części Włoch.
Od 1960 r. uczestnik wykopalisk w Novae, najpierw jako zastępca kierownika przy ekspedycji Uniwersytetu Warszawskiego, a od 1970 r. organizator samodzielnej Ekspedycji Archeologicznej w Novae Uniwersytetu im. Adama Mickiewicza. W wyniku jego prac odsłonięto m.in. fragmenty murów obronnych i pozostałości trzech bram wjazdowych do obozu.
Uczestniczył w organizowaniu i działał w Polskim Towarzystwie Archeologicznym. W latach 1982 i 1983 zorganizował międzynarodowe sympozja i konferencje naukowe na UAM-ie. Urządzał też wysoko oceniane wystawy archeologiczne zabytków z Novae w muzeach archeologicznych w Poznaniu, Inowrocławiu, Koszalinie i Gnieźnie. Współpracował też z poznańskim oddziałem Komisji Bałkanistycznej PAN.
Autor ok. 200 prac naukowych. W zakresie dydaktyki, prowadził rozmaite zajęcia dla studentów filologii klasycznej, historii, historii sztuki i archeologii.
Odznaczenia
Naukowe
Nagroda Ministra Szkolnictwa Wyższego i Techniki II stopnia – dwukrotnie (1976, 1981)
Nagroda Ministra Szkolnictwa Wyższego i Techniki I stopnia (1984)
Nagroda Sekretarza Naukowego PAN (1977)
Nagroda Rektora UAM (?)
Medal 100-lecia Bułgarskiej Akademii Nauk (1979)
Medal XX-lecia IHKM PAN (1978)
Państwowe
Krzyż Kawalerski Orderu Odrodzenia Polski (1974)
Medal Komisji Edukacji Narodowej (1979)
Medal Prezydium Miejskiej Rady Narodowej za Zasługi dla Miasta Swisztowa (1982)
Honorowe obywatelstwo Swisztowa (1982)
Publikacje
Plac w budownictwie greckim a współczesne planowanie, "Politechnika", 1947, nr 9-10, s. 247-254.
Z problemów planowania miast w starożytnej Grecji, "Archeologia", IV, 1950-1951 (1953), s. 27-38
Próba rekonstrukcji agory w Kynaitha, "Archeologia", V, 1952-1953, s. 76-82, 429-431
Gospodarcza rola agory greckiej, "Archeologia", VI, 1954 (1956), s. 90-115, 300-302.
Agora. Geneza i rozwój rynku greckiego, Wrocław 1957
Uwagi do genezy kolumny doryckiej [w:] Księga pamiątkowa ku czci Władysława Podlachy, Wrocław 1957, s. 74-83, 201-205.
Dom mieszkalny a świątynia w Grecji archaicznej, "Archeologia", VIII, 1, 1956 (1958), s. 49-64
Budownictwo w starożytnej Grecji, I, "Prace Zakładu Archeologii Antycznej ZHKM PAN", z. 9, Warszawa 1958
Budownictwo i topografia Olbii [w:] Olbia. Teksty źródłowe i badania archeologiczne. Prace Zakładu Archeologii Antycznej IHKM PAN, z. 6, Warszawa 1957, s. 121-194.
Budownictwo starożytnej Grecji od okresu archaicznego do rzymskiego, Wrocław-Warszawa 1962
Olimpia i olimpiady, Poznań 1964
The Western Gate of Novae, "Archeologia Polona", XIV, 1974, s. 297-314
Architektura starożytnej Grecji, ? 1975.
Befestigungslagen von Novae [w:] Ars historica. Prace z dziejów powszechnych i Polski, Poznań 1976, s. 179-192
Budowle po zachodniej stronie forum w Novae. Building at Western Side of Forum in Novae, "Sprawozdanie Komisji Archeologicznej nr 94 za 1976 r. Poznańskiego Towarzystwa Przyjaciół Nauk. Wydział Historii i Nauk Społecznych"
Budownictwo greckie [w:] Kultura materialna starożytnej Grecji. Zarys, t. II, Wrocław 1977, s. 345-475.
Castra Novae [w:] Novae – Sektor zachodni 1974, I, Poznań 1978, s. 113-117
Krepostnite porti na Novae, "Arheologija", Sofija, XXIII, 4, 1981 (1982), s. 9-21
Wczesnochrześcijańska bazylika katedralna w Novae, Bułgaria, "Poznańskie Towarzystwo Nauk. Wydział Nauk o Sztuce, Sprawozdanie nr 99 za 1981 r.", Poznań 1983, s. 68-73
The Earl-christian Episcopal Basilica in Novae, "Archeologia Polona", s. XXI-XXII, 1983, s. 241-261
Das Heizungssystem der römischen Festung von Novae, Nord-Bulgarien, "Jahreschefte aus August und Kaiseraugust", 3, Liestal 1983, s. 147-155 (wspólnie z L. Press)
Wczesnochrześcijańska bazylika episkopalna w Polsce, "Balcanica Posnaniensia", I, Poznań 1984, s. 147-153
Ambona wczesnochrześcijańskiej bazyliki biskupiej w Novae, "Balcanica Posnaniensia", V, Poznań 1989 [1990], s. 319-341.
The Fortifications in the Western Sector of Novae, Poznań 1990
Bibliografia
Małgorzata Biernacka-Lubańska, Stefan Parnicki-Pudełko [w:] Novensia, t. 9, Warszawa 1997, s. 97-101.
Absolwenci Katolickiego Uniwersytetu Lubelskiego Jana Pawła II
Odznaczeni Medalem Komisji Edukacji Narodowej
Odznaczeni Krzyżem Kawalerskim Orderu Odrodzenia Polski (Polska Ludowa)
Pochowani na Cmentarzu Grabiszyńskim we Wrocławiu
Polscy archeolodzy
Polscy historycy starożytności
Urodzeni w 1914
Wykładowcy Wydziału Historycznego Uniwersytetu im. Adama Mickiewicza w Poznaniu
Zmarli w 1994
Pracownicy Institute for Advanced Study w Princeton | {
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The Konebada Petroleum Park Authority – A 'golden bowl' for who?
by Dr Kristian Lasslett*
01 Jun 201710 min read
Part 1 - KPPA 2006-08
Part II - KPPA 2008-17
Part III - One Final Twist
Data maps
PNG-NRI researchers recently raised concerns over the Lands Minister's decision to place 23,000+ hectares of land – which includes substantial customary holdings – under the stewardship of the Konebada Petroleum Park Authority (KPPA).
They were right to.
Evidence extracted from internal company records, the Investment Promotion Authority corporate registry, and a Public Accounts Committee inquiry, point to a range of irregular transactions and governance weaknesses within the KPPA, that require immediate investigation.
This trail of evidence prompting concern begins in the 2006-2008 period, when KPPA business was conducted through a private, limited liability company.
It continues today.
Despite now being on a statutory footing, the KPPA continues to conduct business through private, limited liability companies. These companies are connected to foreign businessmen implicated in a number of notable state-corporate scandals, including the Paga Hill Estate development and the UBS loan affair.
We have also just discovered that KPPA's Managing Director has abruptly 'left employment without notice'.
* Dr Kristian Lasslett is Director of the Institute for Research in Social Sciences, Ulster University. He specialises in anti-corruption research and forensic methodologies.
Part I – The Konebada Petroleum Park Authority Limited 2006-2008
Initially KPPA business was conducted through a private, limited liability vehicle, KPPA Limited.
Investment Promotion Authority (IPA) records reveal that KPPA Limited was incorporated on 7 May 2004, as Magico Limited. Magico was owned and managed by Debra Varpiam.
Then in 2005, the company was renamed KPPA Limited through a resolution signed by Debra Varpiam, and witnessed by John Beattie, a partner at the Pacific Legal Group:
Debra Varpiam was replaced as company Director by Kila Ai and Joseph Gabut.
Her single share passed to Kila Ai, who held it on trust for the Independent State of Papua New Guinea.
A Company Secretary was appointed – Filipino national, Ofelia Raymundo Carlos.
The company documents ratifying these changes were all submitted by the Pacific Legal Group.
It appears KPPA Limited's primarily role was to act as a government business vehicle, with a mandate to maximize downstream enterprise opportunities emerging from PNG's gas and oil extraction industries.
A Public Accounts Committee report published in 2010 alleges that KPPA Limited's core activities were managed by two consultancy companies.
Nikita Consulting Limited
IPA records show that FIRMS is a registered business name, under which two individuals operate:
Filipino national, Ofelia Raymundo Carlos.
Australian national, Peter Nicholls OBE.
Nikita Consulting Limited is jointly owned by:
Kila Ai (50%).
Inez Ai (50%).
The Public Accounts Committee claims that KPPA Limited's operations were in part funded 'illegally' out of public moneys held in the Konebada Petroleum Park Authority Working Group Trust Account.
K8,529,634.24 was spent on consultants, the Public Accounts Committee maintains, which represented 83% of the total Trust Account expenditure (K10,252,962.83) during 2006-2008.
The Public Accounts Committee alleges that Kila Ai and Peter Nicholls were two core beneficiaries of these payments, presumably via FIRMS and Nikita Consulting Limited.
In total, payments to these individuals evidently constituted '28.01% of all expenditure undertaken in the Konebada Petroleum Park Project' for the years 2006-2008.
Figure 1.1: KPPA Limited (Source: Investment Promotion Authority and the Public Accounts Committee)
The Public Accounts Committee also points to a series of other payments, it deems irregular.
In particular, it alleges, public money from the Trust Account was spent on:
Dining expenses at the Golden Bowl restaurant, Daikoku restaurant, Roundhouse restaurant, Asia Aromas restaurant and others.
Bilums, gold jeweller (K4,500), and considerable overseas travel made without records or reports, as required by the Financial Instructions.
Renovations of an apartment owned by Avonmore Limited. IPA records suggest Avonmore Limited is jointly owned by Peter Nicholls, and Filipino national, Arlene Nicholls.
Lunches for cleaners and security (K20,850)
In particular, K164,320 was expended on 'TA for travel to Australia for meeting'. The Public Accounts Committee notes 'of that amount … K100,562.21 was apparently used for "USD notes" and the balance was sent by TT to Australia for unknown purposes'.
Cash payments. In total, K545,124.03 in cash payments were made during 2006-07, 88.13% of which were for casual workers
The committee concludes:
'The Project has received and expended a very considerable amount of public money for very little result'. It further found that the KPPA 'Trust Account was intentionally and deliberately hidden from the Auditor General and from Government scrutiny'.
In addition to logging the alleged abuse of public moneys perpetrated by KPPA Limited and linked personnel, the Public Accounts Committee also examines the diligence with which the company's senior management team was appointed.
'No tendering procedures were conducted. "Consultants" seem to have been chosen by the individuals concerned with the management of the Project with no lawful competitive selective procurement conducted'.
The Public Accounts Committee adds:
'The Project Manager, Mr. Kila Ai, was appointed directly by the NEC without any competitive applications and solely upon the recommendation of Sir Moi Avei the then Minister for Petroleum and Energy'.
Sir Moi Avei reportedly told the NEC, 'to publicly advertise the position will cause an undue delay and may diminish professional services offered by the successful applicant'.
Examination of IPA records reveal that KPPA Limited's Company Secretary, Ofelia Carlos, and Sir Moi Avei, share a registered residential address: Unit 306, Pacific View Apartments.
Figure 1.2: Sir Moi Avei and Ofelia Carlos' shared residential address (Source: Investment Promotion Authority)
A Leadership Tribunal later recommended Sir Moi Avei's dismissal, for 'serious misconduct in public office', after it was found he had misapplied District Support Grant money.
Sir Moi Avei now holds senior positions in state owned enterprises, including:
Chairman of Ok Tedi Limited
Chairman and Director of Kumul Petroleum Holdings (he recently replaced outgoing Chairman Frank Kramer, who was notable for his opposition to the acquisition of a 10% stake in Oil Search Limited, through the controversial UBS loan).
Director of Bougainville Copper Limited.
Part II – Konebada Petroleum Park Authority 2008-2017
KPPA Limited it appears was set up through a NEC directive. The company's operations seem to have been largely shut down in late 2008.
In its place a statutory authority has been established through the Konebada Petroleum Park Authority Act 2008. Its mission echoes that of KPPA Limited, except it does so now with a statutory footing.
According to KPPA's website**, the authority was established as a 'vehicle for developing land and related assets of the PNG government in key petroleum areas'.
The core team standing behind this new statutory body includes:
Board: Raho Kevau (Chairman), Ralph Saulep (Deputy Chairman), Sisia Morea (Director)
Senior Management Team: Donald Valu (CEO), Joel Oli (Managing Director – currently MIA)
Figure 1.3: Konebada Petroleum Park Authority (Source: Investment Promotion Authority)
However, company records suggest KPPA business is still being conducted, in part, through private, limited liability corporate vehicles, a practice that was previously criticised by the Public Accounts Committee.
In particular, KPPA's senior management team own Petroleum Parks Holding Limited on trust for KPPA.
A recent Board meeting for Petroleum Parks Holdings, was held at KPPA's office. KPPA's Chairman, and its senior managers, were all in attendance. This indicates how closely entwined, in practice, both organisations are.
Figure 1.4: Company Resolution, Petroleum Parks Holding Limited, 3 February 2017 (Source: Investment Promotion Authority)
The minutes also reveal that KPPA's Managing Director, Joel Oli, has evidently left employment without notice. This could be interpreted as a warning sign things are not well within KPPA.
Oli has been replaced by Samuel Pepena as Shareholder/Director at Petroleum Park Holdings Limited.
If we continue to follow the corporate chain, it is apparent that Petroleum Park Holdings Limited is in business with foreign commercial figures, who have been implicated in a series of state-corporate scandals.
The circuit board for this relationship takes place through the company Pacific (PNG) Oil & Gas Limited.
Incorporated on 4 June 2014, the company initially was solely owned by Able Wain (sometimes spelt Abel Wain). Its Directors were Able Wain and Gudmundur Fridriksson.
Gudmundur Fridriksson is currently CEO of Paga Hill Development Company (PNG) Limited. To date, Fridriksson has run businesses at the centre of transactions censured in 1 x Commission of Inquiry, 4 x Public Accounts Committee inquiries, and 2 x Auditor General reports.
Table 1.1 Fridriksson associated entities featured in public inquiries. (Source: Auditor General's Office, Investment Promotion Authority, Public Accounts Committee, Commission of Inquiry into the Department of Finance).
KPPA's formal connection with Pacific (PNG) Oil & Gas Limited was established at a Board meeting on 21 November 2016, whose minutes were signed and witnessed by Gudmundur Fridriksson.
At the meeting:
200 shares were issued to Petroleum Park Holdings Limited.
100 shares were issued to Pertusio Capital Partners Limited.
As a result, the company was then owned by Able Wain (25%), Petroleum Park Holdings Limited (50%) and Pertusio Capital Partners Limited (25%).
Figure 1.5 Company Resolution, Pacific (PNG) Oil & Gas Limited, 21 November 2016 (Source: Investment Promotion Authority)
It appears that Fridriksson left the company on 10 April this year. However, it is notable that Able Wain, shares both a residential and postal address with Gudmundur Fridriksson.
Figure 1.6: Able Wain and Gudmundur Fridriksson's shared residential and postal address (Source: Investment Promotion Authority)
This residential and postal address has also been employed by:
Stanley Liria, Paga Hill Development Company (PNG) Limited's lawyer and sole shareholder.
Andayap No. 1, a subsidiary of Paga Hill Development Company (PNG) Limited.
Neyapu Investments Limited. Its shareholder and director is Zhongqin Shi, who is closely linked to the Paga Hill Estate project.
A large number of Chinese nationals, and Chinese owned companies.
South Pacific (PNG) Oil & Gas Limited, a company owned by Able Wain. Gudmundur Fridriksson was Director from 2014-2017.
The other major partner involved in Pacific (PNG) Oil & Gas Limited is Pertusio Capital Partners. Pertusio Capital Partners is jointly owned by Australian national, Nathan Chang, and Norwegian national, Lars Mortensen.
Fairfax journalist, John Garnaut, claims that both men advised the Government of Papua New Guinea on its controversial Oil Search Limited acquisition, secured through a UBS loan – a transaction that is currently under investigation by the Ombudsman Commission.
Figure 1.7: The KPPA corporate network (Source: Investment Promotion Authority)
** According to domain registry information, the registrar for this website is Australian national, Joshua Coughran. Coughran claims to be a Director at Kindi PNG Limited.
There is one final twist in this story. On 5 May 2017 – after this report had been drafted – a share transfer document was lodged with the IPA for Pacific (PNG) Oil & Gas Limited by Samuel Pepena, Petroleum Park Holdings' Managing Director. Attached to the share transfer form are meeting minutes also dated 5 May 2017, signed by Pepena, and Chairman of Petroleum Park Holdings, Donald Valu.
Through Petroleum Park Holdings Limited, both men hold 200 shares in Pacific (PNG) Oil & Gas' on trust for KPPA. At this fifteen minute meeting, Pepena and Valu elected to sell 160 of Petroleum Park Holding's 200 shares in Pacific (PNG) Oil & Gas Limited. In effect they relinquished 80% of the company's stake in the latter venture.
The meeting minutes wrongly indicate they are only divesting a 40% stake.
The minutes also state that the shares have been purchased by Pertusio Advisors Limited for the amount of K5 million (Petusio Advisors is a subsidiary of Pertusio Capital Partners Limited).
The meeting record goes on to declare that the shareholders have 'resolved to accept part-payment of K500,000 on transfer of the above shares, and the balance of K1.5 million within 30 days hence, and the remaining balance in equal instalments over 10 years with inbuilt interest component'.
Figure 1.8: Petroleum Park Holdings Limited, Meeting Minutes, 5 May 2017 (Source: Investment Promotion Authority)
It is not clear from the meeting minutes, or share transfer form, whether this sale of shares by two trustees, on behalf of KPPA, was preceded by a professional valuation of the asset price, ministerial approval, KPPA board approval, or independent legal review of the proposed 10 year agreement. These are questions perhaps best answered by Samuel Pepena.
To summarise the key facts:
The KPPA initially conducted its business through a private limited company, which suffered serious and sustained criticism from the Public Accounts Committee for its alleged, opaque spending, irregular transactions, illegal use of trust account moneys, and notable lack of impact.
A new statutory body was set up in 2008, through the Konebada Petroleum Park Authority Act.
The statutory authority still conducts its business through private limited companies, including Petroleum Park Holdings Limited and Pacific (PNG) Oil & Gas Limited.
Through this corporate network, KPPA has developed close links with businessmen implicated in a range of state-corporate scandals.
KPPA, through its senior management team, have disposed of share assets held in Pacific (PNG) Oil & Gas Limited. It is not clear from the documents submitted to the IPA, whether Ministerial approval was acquired for this sale or what sort of professional vetting was employed.
Given the historical track record of KPPA Limited, and those individuals connected to the new statutory authority – through Pacific (PNG) Oil & Gas Limited – PNGNRI researchers were right to raise the alarm.
Their diligence and bravery demonstrates the critical role researchers and public intellectuals can play in keeping government and business to account in PNG.
Evidence to date suggests the initial findings of the Public Accounts Committee have gone largely unheeded. This again is a timely reminder that the Independent Commission Against Corruption (ICAC) remains unactioned.
PNG has no substantive, independent body to thoroughly investigate irregular conduct, and known governance weak-points – even though the former Supreme Court Judge appointed to head the interim ICAC has declared his willingness to begin work at the earliest available opportunity. As a result of this inaction, the public is largely reliant on the executive to police itself.
Given the hostile response of the Minister for Lands & Physical Planning to the issues raised by PNGNRI researchers, there is little evidence to suggest that the executive branch is motivated to respond judiciously when such governance concerns are raised by reputable bodies.
PPHL Meeting Minutes, 05.05.2017
POG Company Resolution, 21.11.2016
PPHL Company Resolution, 3.02.2017 | {
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{"url":"http:\/\/blogs.msdn.com\/b\/murrays\/archive\/2007\/07\/14\/automatic-arguments.aspx","text":"The arguments of math display objects are either optional or essential. Examples of essential arguments are the numerator and denominator of a fraction. For a legitimate fraction, both are nonempty. Hence if either or both are empty, they should be represented by a dotted square box, which indicates that an essential argument is missing.\n\nOptional arguments are like limits of a summation or integral, the degree of a square root, or some elements of a matrix. It\u2019s entirely legitimate to omit these arguments, but later on you may want to enter text into them. Word 2007 treats optional arguments by explicit user choice: if the user doesn\u2019t want the argument to appear (and take up space), it isn\u2019t displayed at all and the left\/right arrow keys cannot navigate into the argument. The user can use a right-mouse context menu to unhide a hidden optional argument or hide an unhidden one. When empty, an unhidden optional argument displays a dotted box.\n\nRichEdit has the simpler and more powerful approach of \u201cautomatic arguments.\u201d We figured this style out too late in the Office 2007 cycle to get it into Word 2007. When an optional argument is automatic, it doesn\u2019t display the dotted box, unless the user moves the insertion point into the argument by using left or right arrow keys. With the insertion point inside the empty argument, the dotted box appears and the user can enter text. Similarly the user can delete the contents of an automatic argument and no dotted box appears unless the insertion point is inside the argument.\n\nFor example, the linear-format entry \\sqrt(a+b) creates a square root of the quantity a+b with no degree displayed. This implies a degree of 2, i.e., a square root. With automatic arguments if the insertion point immediately precedes the variable a, typing the left arrow key moves the insertion point into the radical\u2019s empty degree argument, which then displays the dotted box. The user can then type something, for example, n for an nth root.\n\nThis also works with n-ary operator limits, empty matrix elements, etc. It\u2019s a simple, globalized method for handling optional arguments in a general way without menus.","date":"2014-07-30 13:08:41","metadata":"{\"extraction_info\": {\"found_math\": false, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8010814785957336, \"perplexity\": 1141.0260320895065}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2014-23\/segments\/1406510270528.34\/warc\/CC-MAIN-20140728011750-00105-ip-10-146-231-18.ec2.internal.warc.gz\"}"} | null | null |
{"url":"http:\/\/www.bobbydurrettdba.com\/2015\/04\/29\/ddl_lock_timeout-to-sneak-in-change-on-active-system\/","text":"# DDL_LOCK_TIMEOUT to sneak in change on active system\n\nI need to change a view and an index on an active production system. \u00a0I\u2019m concerned that the change will fail with a \u201cORA-00054: resource busy\u201d error because I\u2019m changing things that are in use. \u00a0I engaged in a twitter conversation with\u00a0@FranckPachot and\u00a0@DBoriented and they gave me the idea of using\u00a0DDL_LOCK_TIMEOUT with a short timeout to sneak in my changes on our production system. \u00a0Really, I\u2019m more worried about backing out the changes since I plan to make the change at night when things are quiet. \u00a0If the changes\u00a0cause a problem it will be during the middle of the next day. \u00a0Then I\u2019ll need to sneak in and make the index invisible or drop it and put the original view text back.\n\nI tested setting\u00a0DDL_LOCK_TIMEOUT to one second at the session level. \u00a0This is the most conservative setting:\n\nalter session set DDL_LOCK_TIMEOUT=1;\n\nI created a test table with a bunch of rows in it and ran a long updating transaction against it like this:\n\nupdate \/*+ index(test testi) *\/ test set blocks=blocks+1;\n\nThen I tried to alter the index invisible with the lock timeout:\n\nalter index testi invisible\n*\nERROR at line 1:\nORA-00054: resource busy and acquire with NOWAIT specified or timeout expired\n\nSame error as before. \u00a0The update of the entire table took a lot longer than 1 second.\n\nNext I tried the same thing with a shorter running update:\n\nupdate \/*+ index(test testi) *\/ test set blocks=blocks+1 where owner='SYS' and table_name='DUAL';\ncommit;\nupdate \/*+ index(test testi) *\/ test set blocks=blocks+1 where owner='SYS' and table_name='DUAL';\ncommit;\n... lots more of these so script will run for a while...\n\nWith the default setting of\u00a0DDL_LOCK_TIMEOUT=0 my alter index invisible statement usually exited with an\u00a0ORA-00054 error. \u00a0But, eventually, I could get it to work. \u00a0But, with\u00a0DDL_LOCK_TIMEOUT=1 in my testing my alter almost always worked. \u00a0I guess in some cases my transaction exceeded the 1 second but usually it did not.\n\nHere is the alter with the timeout:\n\nalter session set DDL_LOCK_TIMEOUT=1;\n\nalter index testi invisible;\n\nOnce I made the index invisible the update started taking 4 seconds to run. \u00a0So, to make the index visible again I had to bump the timeout up to 5 seconds:\n\nalter session set DDL_LOCK_TIMEOUT=5;\n\nalter index testi visible;\n\nSo, if I have to back out these changes at a peak time setting\u00a0DDL_LOCK_TIMEOUT to a small value should enable me to make the needed changes.\n\nHere is a zip of my scripts if you want to recreate these tests: zip\n\nYou\u00a0need Oracle 11g\u00a0or later to use DDL_LOCK_TIMEOUT.\n\nThese tests were all run on Oracle 11.2.0.3.\n\nAlso, I verified that I studied DDL_LOCK_TIMEOUT for my 11g OCP test. \u00a0I knew it sounded familiar but I have not been using this feature. \u00a0Either I just forgot or I did not realize how helpful it could be for production changes.\n\n\u2013 Bobby","date":"2017-03-28 19:43:01","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.4382893443107605, \"perplexity\": 4188.374717048494}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2017-13\/segments\/1490218189884.21\/warc\/CC-MAIN-20170322212949-00008-ip-10-233-31-227.ec2.internal.warc.gz\"}"} | null | null |
Los miserables or Los Miserables may refer to:
Los miserables (1973 TV series)
Los miserables (2014 TV series)
Los Miserables (band), a Chilean punk rock band | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 4,794 |
This is a work of fiction. Names, characters, places, and incidents either are the product of the author's imagination or are used fictitiously. Any resemblance to actual persons, living or dead, events, or locales is entirely coincidental.
Text copyright © 2016 by Ginger Garrett
Cover art and interior illustrations copyright © 2016 by Dinara Mirtalipova
All rights reserved. Published in the United States by Delacorte Press, an imprint of Random House Children's Books, a division of Penguin Random House LLC, New York.
Delacorte Press is a registered trademark and the colophon is a trademark of Penguin Random House LLC.
Visit us on the Web
Educators and librarians, for a variety of teaching tools, visit us at RHTeachersLibrarians.com
_Library of Congress Cataloging-in-Publication Data_
Garrett, Ginger.
The last monster / Ginger Garrett. — First edition.
pages cm
Summary: "Thirteen-year-old cancer survivor Sofia has been chosen as the next Guardian of a book called The Bestiary, an ancient text. Drawn into violent and unpredictable mysteries, Sofia learns that these misunderstood monsters from the book are in danger and she is the only one who can save them" —Provided by publisher.
ISBN 978-0-553-53524-2 (hc) — ISBN 978-0-553-53525-9 (glb) — ISBN 978-0-553-53526-6 (ebook)
[1. Monsters—Fiction. 2. Cancer—Fiction. 3. Amputees—Fiction. 4. People with disabilities—Fiction.] I. Title.
PZ7.1.G375Las 2016
[Fic]—dc23
2015003381
eBook ISBN 9780553535266
Cover design by Sarah Hokanson
Random House Children's Books supports the First Amendment and celebrates the right to read.
v4.1
ep
# Contents
Cover
Title Page
Copyright
Dedication
Chapter One
Chapter Two
Chapter Three
Chapter Four
Chapter Five
Chapter Six
Chapter Seven
Chapter Eight
Chapter Nine
Chapter Ten
Chapter Eleven
Chapter Twelve
Chapter Thirteen
Chapter Fourteen
Chapter Fifteen
Chapter Sixteen
Chapter Seventeen
Chapter Eighteen
Chapter Nineteen
Chapter Twenty
Chapter Twenty-one
Chapter Twenty-two
Chapter Twenty-three
Acknowledgments
About the Author
To every reader who
still checks under the bed at night
•••
Friday, February 21
I had to pick up a new leg after school.
Dark storm clouds hung low over the Children's Cancer Center of Atlanta. Mom took the last turn into the parking lot, and all the old familiar fears coiled around me. I groaned under their weight.
Mom glanced at me, worried.
"It won't stop people from staring," I mumbled.
"Is that really what's bothering you?" she asked softly.
I glanced at her and realized her knuckles were white. She was gripping the steering wheel as if it were a life preserver. Coming here was just as hard for her as it was for me.
"I'm fine," I lied. "The worst is over, isn't it?"
She slowed the car as we approached the entrance. "I'll let you off at the front door, in case it starts raining," she said. "But wait for me before you go back."
I nodded as I glanced out the car window. A hawk circled overhead, hunting some small, frightened thing that hid in the shrubs by the entrance. The bird screamed when it saw me, a shrill welcome.
Mom stopped in front of the double doors before leaning over to give me a quick peck on the cheek. Her lips were cold. This was our last big appointment for four weeks. We'd basically lived here for two months last year, from the end of October until just after New Year's; then we had switched to outpatient visits. After a month of rest to build my strength, I had started physical therapy. So this place had been my second home for almost four months. Now the idea of freedom felt surreal. "Free" was a strong word, though. It was more like I'd gotten extra length on my leash. I still needed to come back. Cancer was a part of my life that I would never escape.
The wind moaned as I opened the car door. I swallowed hard, my stomach in knots, as if this place were the nightmare I would never wake up from.
Inside, the TV in the waiting room was tuned to a twenty-four-hour health channel. I settled into a vinyl-cushioned chair that wheezed when I sat. The volume was so loud no one noticed. A special feature called _Meet Your Meat_ was playing, a behind-the-scenes look at how meat products are manufactured. In the corner, a coffee machine sat undisturbed, its electric red eye watching us all.
_"Sausage casings are made from the intestines of pigs, horses, or sheep. Machines remove the fat and mucus before extruding the final casing, which will give some lucky sausage that satisfying snap."_
My eyebrows shot up in horror. Sausages were like babies; I had a general idea of how they were made, but I didn't want the clinical details. I glanced behind me, praying Mom would find a parking spot soon. We didn't do the "free" valet parking anymore. Tipping required cash, and we rarely had any now.
I noticed a boy about seven or eight years old, hunched over, holding a beach sand pail in front of his mouth in case he vomited. A little bit of drool clung to his chin. He still had his hair, though, so he was probably a new patient, trapped between terror and denial. He cringed as images of greasy pink sausages flashed above his head.
The TV was mounted on the wall. If there was a remote control, I had never found it. The only way to change the channel would be to reach way up and press the buttons, and somebody needed to do that fast before the poor little guy hurled again. He glanced at me and nodded weakly in the direction of his mother. She was standing by the windows, chatting furiously on a cell phone, her back to us. Fat drops of sleet began pelting the tinted lobby windows. A bright whip of lightning ripped across the sky and split in five directions, like a giant claw. Strange, even for a city known for its weird weather.
_"And now we'll take a closer look at ground beef. The name itself is misleading, since it includes skin, connective tissue, fat, and even bone...."_
The kid put his head down and retched loudly into the bucket. I stood and walked to the TV, praying the other kids didn't notice how one of my legs moved differently from the other one. The prosthesis I wore was too big, and it made my gait clumpy. The new prosthesis would make my gait very natural, but of course everyone at school would stare anyway, trying to figure out which leg was real. They got them mixed up.
I pressed the channel button repeatedly, but every other station was running a special bulletin on the weather. A bad storm was coming. Lightning in the northern suburbs of Atlanta was especially dangerous; we had huge blocks of granite under the ground that attracted lightning. When a storm came, it didn't just give us a pretty show; someone usually got hurt.
Finally, I found the cartoon channel and released the button. I turned around to check on the boy and saw that his mom was still on the phone, unaware that he would never eat sausage again. The thunder boomed and a fragile-looking baby cradled in its mother's arms began to wail.
"Thanks," the boy said to me. "You're my hero."
I nodded and quickly looked away, my cheeks getting warm and probably red. My face had always been like a giant mood ring. "Hang in there," I said. "It's not always this bad, I promise."
If he could tell I was lying, he didn't show it. The boy rested his face on the side of the pail and sighed, eyes closed. I felt a little bit better too, which surprised me. All I had done was change a TV channel.
The baby's crying suddenly grew louder, as if it knew something we didn't.
Lightning flashed again, illuminating the room like a sinister X-ray. Thunder chased it immediately, and the walls shook. It sounded like the hospital had been hit.
The TV blanked out with a hiss of static as all the lights shut off and the hum of the heater died away. There was nothing but darkness and cold silence. The baby whimpered; I could hear its breath coming in little stutters.
The lightning must have knocked out the lobby's power. The hospital's power grid was designed to protect patient rooms and equipment first, so it didn't surprise me that the power down here had gone out. Still, if lightning struck again, maybe the lights wouldn't come back at all. I got nervous about it. I knew a girl whose mom had parked their car under a bank's drive-through awning and lightning had struck their car, melting all their tires. Lightning ruled the skies here.
Everyone began talking loudly or crying, but all I could hear was the drumbeat of my own heart. What was taking Mom so long? The cancer center shared a parking lot with the children's emergency room and hospital, so maybe the lots were all full. Fridays weren't normally this busy, though.
Through the chaos, I noticed a little girl illuminated in the red glow from an overhead Exit sign on the other side of the waiting room. She had one hand wrapped around her IV pole; the other was dragging a pillowcase with something heavy inside.
Her face was gaunt and her long hair hung lifeless. Instead of a blue hospital gown, she wore a white nightgown that skimmed the floor. Bloodstains dotted the hem and there were rips in the fabric, as if she had been attacked. The _Walking Dead_ television series filmed here in Atlanta; maybe she had wandered off the set and was lost.
A shadow rose from the floor beside the girl, like smoke from a fire. Goose bumps spread across my arms, as if my body recognized a change in the atmosphere, the warning of a lightning strike gathering energy before choosing its victim. In the blood-red glow of the exit light, the dark smoke blossomed, unfurling into the form of a woman dressed in flowing robes, her braided blond hair pinned under a crown of round gray stones, each with what looked like a blinking eye at its center. Every stone swiveled like an eyeball, each moving in a different direction, until they all came to rest on me.
She glared at me like I was her worst enemy, but I had never seen her before.
Dread surged through my body, an icy river of fear that burned down into my stomach, making me clutch it in pain. The woman sneered as she rested a paper-white hand with thin, spidery fingers on the girl's shoulder, making her flinch. The woman's fingers stretched down and around, covering the girl like the roots of a growing tree.
The girl shuddered as if the touch was cold, but her eyes never left mine. She needed to tell me something; I could feel her sense of urgency. She nodded once, slowly, willing me to come and help her.
I froze, panic overtaking my body. I didn't know her. I didn't _want_ to know either of them. I pushed myself backward in my seat, sinking down. My heart pounded against my ribs, like it was trying to escape. Was this a nightmare? Or some sort of dream? I wasn't even asleep.
An emergency alarm blasted through the lobby. "Code Amber. Code Amber. All personnel report."
The power snapped back on, light exploding inside the room.
I turned to check on the vomiting boy to make sure he was safe. He was. When I looked back at the hallway, the girl and the woman were gone.
I exhaled in relief, then started breathing too fast, pushing the air out and sucking it back in until I felt light-headed. Oh man, if I tried to tell someone about what I had just seen, they'd keep me overnight for tests, certain I was suffering a late side effect from chemo. It was possible. Chemo was just carefully measured poison, a Pandora's box of dreadful surprises.
A hand grabbed my shoulder and I lurched around, ready to scream, my hands balled into fists. It was Barnes, my prosthetic technician. He curled his hands into fists too. "You want to hit old Barnes? You want much of this?"
He broke into a huge grin and I could see all the metal bits that attached his dentures. I grinned back, more from relief than anything else. Barnes was probably about seventy or eighty years old, but he said he needed to keep up with his slang for all the young patients. I think he did it to make us laugh, and it worked on me every time. I felt the dread and fear of a moment ago melt away as I giggled. My nerves had made me slightly nuts. Everything was back to normal.
Mom walked in, waving to me before stopping at the reception desk. I stood and took Barnes by the hand. He didn't notice mine was shaking. "You're supposed to say 'You want _some_ of this?' "
"What- _ever,_ " he said, hitting the inflection just right. I had taught him well. His knuckles were red and swollen today, so I held his hand extra carefully, willing my own to soften into a steady grip.
"Code Amber is a missing kid, right?" I asked.
He didn't answer me, just took a few steps in the direction of his lab and turned back to wave to Mom. She nodded back, meaning she would follow in just a second. I knew she looked forward to talking to Barnes during our visits. He never used slang with her. They had had several disagreements and he was still endlessly sweet to her. A lot of men treated my mom that way.
"Maybe they should look down that hallway," I added, pointing to the right. I didn't want to sound too certain, but if the girl was real, that was where they'd find her.
An impatient wave was his way of saying my words were falling on deaf ears. And maybe they were; he had to tilt his head in my direction if I talked in a quiet voice. Plus, his hair was thinning at an alarming rate. He'd be as bald as me before long. I wondered if it bothered him to look in the mirror the way it bothered me.
"Let security do their jobs. You stay out of that."
"But, Barnes," I started to protest. "I just—"
He cut me off. "No arguing." Barnes had been in the navy a long time ago and still liked to bark a command now and then. He started down the hallway on the left and I followed, soft instrumental music greeting us through the speakers.
My hands started to shake again, but I couldn't shove them in my pockets; I needed loose arms for balance. I hadn't been here as an inpatient since January, but the silent terror I felt when I walked the halls of this hospital was hard to explain to anyone except another survivor.
When you leave the hospital for the last time, they call it being discharged. That's the same term the military uses when they release a soldier. Veterans say that once you've been in combat, you never completely forget it, or get over it. The pain stays embedded, like invisible shrapnel.
I think veterans of any war probably understand each other.
They would understand how all the scary, sad moments in here flooded my mind: the burnt smell of antiseptic, the quiet moans that echoed through the heating ducts at night, the dark stains in the bathrooms' white grout, the endless predawn hours I spent watching Animal Planet, witnessing predators take down the weak and frightened. The doctors had fought the cancer while I had clung to life. I didn't feel brave or strong, not here, not in this place where the enemy won so many battles and did not spare the weak.
My head throbbed. My awkward gait made one side of my neck muscles tight, which caused the headaches. It felt like my head was being squeezed in the grip of some angry god. I was grateful that my mom had insisted on the new prosthesis. We had tried to go the conventional route, using one that was too big so I could grow into it, but I hated the way I walked with it, and she didn't like it either. We shouldn't have been getting a new prosthesis just after picking the other one out, but Barnes had allies all over the hospital system. He was a veteran too, after all.
Pain, by the way, is the reason that hospitals don't play music that has voices and words. Words fail everyone in here. Every day when I was a patient, I had to point to a chart that showed happy and frowny faces. I had to point to the expression that represented how much pain I felt. With all the modern technology and medicine, we have to communicate through symbols about what we feel inside, like we're still drawing on cave walls.
Lightning flashed in the window behind us.
I realized Barnes had slowed down to wait for my mom. He was still at the end of the hall, near the lobby. No one else was in this corridor, so I raised my voice. "I thought maybe I saw a girl in the other hallway by the waiting room."
"Security is working on it," he called back. "Don't let it ruin your big day. You are one lucky kid. Wait till you see this leg."
I was thirteen, bald, and had one leg, and yet everyone told me how lucky I was. Life is a precious gift, people said. No one ever mentioned that it was a gift you didn't get to pick out. Apparently, God and I had very different tastes.
Mom finally caught up to Barnes and they walked together toward me. His lab was at the end of a very long passageway. He said he liked to work in solitude, but I wondered if his lab was so remote because other people freaked out when they saw what he had inside. He held the door for me and Mom.
"It looks so real," Mom gasped. "It's...beautiful."
My new leg rested on the table in front of us. The paperwork under it had a bar code and I could read my name printed in block letters. On the far end of the workbench was a bag with my name on it too, like a carrying case. Next to the case was a pair of crutches for when I didn't have the leg on.
Mom ran her fingers lightly over the leg.
"Mom," I groaned. "Don't pet it." She looked offended, so I shrugged. "It's weird."
"Oh, look, Sofia!" Mom giggled as she pointed. "He even got your cute little freckle just right." My real leg, the one they amputated, had had a heart-shaped freckle on the calf. Mom must have told Barnes about it, but I was hurt that she hadn't consulted me on something as personal as a freckle placement.
Barnes sat at his workbench, which was lit with two bright lights on either end, just like the adjustable kind we used in art class. He put on his jeweler's headlamp, which had a magnifying glass over one eye.
"Try it on, kid," he said. "Barnes has worked a minor miracle for his favorite patient." He looked like a grinning cyclops. "Feel how soft it is too. New material."
"Oh!" Mom said. "I left my phone in the car. I wanted to get a picture of you two. Do you mind?"
Barnes shook his head.
"I'll be back in a flash," Mom said as she bolted out of the room. I was jealous, and not just because she could move so fast.
It's hard to describe the sensation of staring at your leg on a table while you're standing at the other end of the room. This was probably why most patients weren't allowed in Barnes's lab at all but were ushered into a separate room with comfortable chairs. Maybe Barnes had decided he could trust me not to freak out in here because he knew I never tried to cheer people up. In a hospital setting, cheerful people are usually the most unstable.
I grabbed the carrying case and pretended to inspect it.
"What are you working on today?" I asked.
He didn't look up but motioned for me to hand him the tiny screwdriver that was just out of his reach. I did, glad to do anything but put that leg on. This one would be mine for a while. In January, Barnes had given me a temporary prosthesis that I had to practice walking on while he and Mom decided on a final model. It was like learning to ride a bike using training wheels. They wanted me to get the general idea before I graduated to the final leg.
After four weeks with that first clunker, I had another couple of weeks working with a newer design while Mom and Barnes argued about the final fit. This was the leg I'd have for at least a year, until I started growing again.
No one was sure when that would happen, because chemo can delay puberty. Cancer sucked on so many levels, and not just because of what it did to my hair. Since money was tight, we had agreed to do what everyone else did: get a bigger leg, put up with the gait issues for a year. Then we'd gone in to discuss my post-treatment plan of regular bloodwork and CT scans, and the doctor kept talking about the miracle of my cancer being caught early and entirely removed, because survival rates for bone cancer, especially once it spreads, haven't improved much in years. Mom and I realized then that maybe waiting a year to get a better leg wasn't a good idea.
Maybe I wouldn't have another year. No one was guaranteeing anything.
Anyway, most of the time, Barnes built the legs too big so the patients could grow into them over time, kind of like how a mom buys her kids shoes that are just a little too big so they don't outgrow them right away. Prostheses were incredibly expensive, and making them big helped them last longer, but Barnes knew I hated how the big one fit. He hadn't been all that surprised when I told him we wanted a new one.
What Barnes really wanted to build for me was a running blade, the kind of prosthesis that has no fake flesh and looks like a weapon instead of a leg. I hated them, but Barnes thought I would want to run again soon. I used to consider myself a major track star; as in, I'd been famous for majorly vomiting after every race, and sometimes beforehand too. I had run harder than anyone but never won a race. I got a lot of awards for participating, though. At least my best friend, Alexis, had thought I had potential.
"Barnes, you know what's sad? No one ever sees your best work," I said. "Your real talent, all that complicated stuff inside? The fake skin hides your genius."
He looked at me with his huge magnified eye. It blinked, the fringed eyelid lowering and lifting like a garage door. "Your mother should be back any minute. She'll want to see how this looks."
I refused to take the hint. I knew he wasn't ashamed of his work, so which side was he on? Did he think prostheses should help patients blend in or stand out? Which one did he think was more important?
"People don't see how gifted you are," I continued. "Doesn't that bother you?"
He stood without turning back to me and opened the door. "I'm going to get you a soda pop," he called. "And then we're going to talk about the next genius prosthesis I'd like to create for you. No getting out of it today."
I couldn't risk following Barnes out, because he would talk about the blade. I didn't want a freaky prosthesis that everyone would stare at. Besides, right now it felt good to be alone in a quiet room at the end of a hallway where no one needed to be rescued from sausage documentaries. I looked at the leg again. When I put it on, it would look as if nothing had ever happened, as if the past had just performed a vanishing act.
"I want her to look completely natural, just like any other girl her age, and I will hold a thousand bake sales to pay for it, if that's what it takes," Mom had insisted during one of my last rehab sessions with the old leg. "My daughter deserves to look in the mirror and like what she sees."
It wasn't a good time to mention it, but I had _never_ liked what I saw in the mirror, and I had certainly never felt just like any other girl my age. At school, it seemed like every other girl was pretty or popular. I was neither. I was timid and mousy, and had fallen through the rungs of the social ladder in elementary school. When I looked in the mirror, it wasn't to see what I looked like. It was to see what I could still fix. Truth was, there wasn't a lot left. On my best days, I had a vague resemblance to a snapping turtle: sharp pointed nose, squinty eyes, thin lips. That was what I saw, anyway.
Sighing, I slipped off my pants and then the stocking that went over my current prosthesis. I had zero energy after my first full day at school today, and it took me a couple of minutes to get the old prosthesis off.
A fast little squeaking sound caught my attention, but I ignored it. It was hard work wiggling around. At least I had remembered to wear boy shorts instead of underwear.
The noise grew louder, closer. I finally looked over. The girl from before stood framed in the doorway, breathing hard, one hand on her IV pole, one hand dragging the sack. Why hadn't security found her already? She was attached to a pole; how hard could it be?
I spotted a tear in the fabric of the pillowcase; inside was a big book, like the kind they made ages ago, with an engraved leather cover and tons of thick ivory-colored pages. She had been dragging around an old book? Most kids carried a teddy bear. What was wrong with this girl? I craned my neck to try to get a view of the hall, but I couldn't see anyone—or anything—else out there.
I pushed against the floor with my one foot, scooting myself back in the chair. Barnes or my mother would be here any minute, but for now I was stuck in my boy shorts and only one leg.
"Everyone is looking for you," I said.
She stared at me, her face impassive, not a muscle moving. Only her eyes danced with happiness and relief, as if she had just finished a long race.
"Now I know why you were chosen," she whispered. "I saw you take care of that boy. You are a kind person and you'll be brave for them."
I had no idea what she meant, but she was starting to scare me. "We need to call security," I said. "We have to let someone know you're safe."
Tears formed in her eyes. They rose, spilling down her cheeks as her chin quivered. "I will be. He promised."
The girl glanced down the hall fearfully and licked her upper lip, catching a tear. I immediately grabbed the pair of crutches and forced myself up. "Please don't cry," I said softly. "Let's just call someone, okay?"
One more big, fat tear fell from her cheek onto her nightgown. She stared at the spreading wet spot, as if mystified. Her gaze moved to the bottom of her hem and then to the bloodstains.
"I'm not like you," she sighed. "I tried to be brave and kind, but they scared me." She paused, and looked at the ground as if ashamed. "Plus it's hard to keep secrets. It's lonely."
A gray mist snaked around her feet. The girl's eyes widened in terror, and she slung the pillowcase forward, hitting me in the shin.
Thrown off balance, I stumbled back, landing on my butt, hard.
She reached for the tape that held the IV line inside her other arm. "She doesn't want me to give it to you. She's going to be very angry."
I scrambled to grab the crutches again and stand back up, but the floor was too slippery. The mist was growing thicker and rising. I vowed to write a strongly worded letter to the hospital about the decision to wax the floor of a prosthetics lab.
"Don't!" I yelled, shoving the pillowcase behind me to get it out of the way before I tried to stand again. "You have to leave your IV in until a nurse takes it out."
"Tell them all I'm sorry, especially the Golem." With that, she ripped the IV line out as the mist rose over her head and the outline of the same woman began to form. Blood bubbled up from the place where the IV had punctured her skin.
"Mom! Barnes!" I yelled. "Somebody!"
She took off running, her IV pole crashing to the floor. Her bare feet padded hard against the blue floors as blood dripped from the IV line and trailed behind her.
"Help!" I screamed. "Stop her!" I crawled into the hallway.
Barnes was rounding the corner carrying two soda cans when she crashed into him and fell. The cans flew out of his hands and burst against the walls. A shower of dark cola erupted, and Barnes shielded himself. The girl scrambled back to her feet and ran into the waiting room, then directly out into the storm.
I caught one last glimpse of her through the window at the end of the hall. Her white nightgown blew in the wind as she ran. Above her, lightning split, and in that brief flash, the terrifying shadow woman was illuminated across the sky. She pointed one long finger at me, her eyes blazing until her face dissolved into the night rain.
I took a deep breath, forcing myself to slow my racing heart. The drops of blood and the girl's tears and the evil shadow flashing through a dark sky shredded by lightning...Black dots gathered in the edges of my vision. My thoughts were a blur and so was the hallway, the floor tiles swimming in shifting shapes.
My mom's high heels clacked rapidly toward me. I couldn't breathe or speak as she began calling my name more loudly. Security guards in blue uniforms swarmed the entrance; flashing blue and red police lights danced outside.
The black dots in my vision got thicker, my own dark storm.
Suddenly, I knew why they called it blacking out.
Monday, February 24
"Your test scores on evolution were disappointing," Ms. Kerry said with a clap of her hands, waking up a kid in the front row. "And contrary to what several of you implied, our principal, Mr. Reeves, is _not_ proof of the missing link."
Two boys from class who wore eyeliner and dyed their hair black laughed quietly.
"However, one student in particular had a fascinating answer to the extra-credit question on the next stage of human evolution. Sofia, would you care to read your answer?" Ms. Kerry spoke slowly whenever she addressed me, as if losing a leg had affected my hearing. People were nice to me in the most extraordinarily stupid ways.
I shook my head. "No." Then, to be polite, I added, "I would not care to do that." I said it fast, hoping she'd get the point. I had come back to school last week for half days from Monday to Thursday so I could build stamina and rest in the afternoons if I needed to. Friday had been my first full day of seventh-grade reality. When you have a prosthetic leg, you burn more calories than other people, even just doing ordinary stuff. The body looks whole, but invisibly, it's always working to make up for the loss. If my doctors thought it was my body that needed to adjust, they were wrong.
I was suddenly receiving a ton of attention. People who didn't know my name last year had signed a Welcome Back, Sofia banner, as if my absence had left a big hole at the school. We all knew it hadn't, but everyone pretended it was a big deal that I was back. I knew people were reacting to the cancer, not me. They were terrified of it, so they celebrated that it hadn't killed me, because everyone's worst fear is that it could kill them.
The banner wasn't really for me, even if it had my name all over it. No one would understand that, and I didn't try to explain. But it felt wrong, even dishonest, to be called brave or have anyone want to be friends with me now.
I felt more invisible than ever, yet overwhelmed with attention. How weird was that?
Right now everyone was staring anyway, including my former best friend, Alexis, who was sitting two rows up, wearing her track team jersey. Before class started, she had motioned for me to take the seat next to her, but I had pretended not to see her. We had said hi to each other a couple of times, but there hadn't been a chance to say anything else, and I was grateful for that. I had no idea what to say yet. We hadn't really talked since my life fell apart, just a few quick, awkward phone calls, but with the winter break and all the excitement of January's huge snowstorms, maybe she didn't miss me that much anymore. I hoped she didn't hurt inside like I did.
Alexis's long, curly brown hair was pulled up in a high pony. She must have run with the team that morning. I envied her for that, and for all that hair. I wondered if you could envy someone so much you actually hated them. I wished we could be friends again, but that wasn't fair to her. We had spent most of our time together at track or cross-country practice. Now I couldn't run. Was she supposed to rearrange her whole life just to spend time with me? A real friend would let her go. She needed to run because it made her happy, and she needed some happy in her life. There was too much drama at home for Alexis; running was her only refuge.
"Oh, I don't mind, then." Ms. Kerry pulled my paper from the stack on her desk. "I'll read it for you."
I lowered my face into my hands. This was going to be painful, and I knew pain.
Ms. Kerry began. "The extra-credit question was, 'Pretend that you have discovered the next step in human evolution. How will humans change, and why?' Here's Sofia's answer:
_"Humans will lose their eyesight, because most of what everyone sees now is fake. Like the people on TV. They look real, like they're really there, but they're not. When people want to know the weather? We look at a screen instead of the sky. People don't even see what's real about the ones they love. Instead, they see what they wish we were, or what they wanted to be themselves. And it hurts if you know that the image will never be real. So humans were made to see real things, but no one does. So we'll all go blind soon. Or maybe we already are."_
Everyone snuck glances at me while she read my answer. I was the weird, quiet girl who used to hide behind stringy brown hair that fell into my eyes and covered my face as I worked, the girl who always had a book in her hand instead of a phone. They didn't know much about me except that I stayed off the social radar, but they forgot that I had been pushed off a long time ago. Besides, teachers never embarrassed kids in class for reading, not like they did for passing notes and texting. No teacher had ever grabbed my book and read a passage to the class just to teach me a lesson. So my personal thoughts and interests were kind of a curiosity to everyone, especially since my name was now on a banner near the front office. I was the bald mystery girl with a fake leg ("Is it the left?" "No, it's the right."), and now I had just accused all of them of being blind. (It actually was my left.)
Alexis turned and stared at me, probably wondering if my answer was a cry for help, a sign that I was finally reaching out. My face grew hot from frustration and embarrassment. Honestly, I was just trying to score some extra points, because Mom would do her happy dance when my next report card had all As.
At least I hadn't written about my latest theory on global warming: that global warming was really caused by the Earth getting bigger every year, and therefore closer to the sun. Millions of people die annually and get buried in the ground. Millions and millions, year after year, decade after decade...shouldn't the Earth be getting bigger all the time?
Ms. Kerry was passing out the tests, instructing us to correct any wrong answers before the end of the period. Above me, a thin black spider slipped down from its web, floating on an unseen current. No one else noticed it. My pencil kept slipping from my hands as I worked on my test. I hunched over my paper so no one would see. I had finished my chemo three months ago, but it had a lot of side effects that took months longer to go away, if they ever did, including how my muscles felt like rubber bands. Even holding a pencil was challenging at times. I dropped it repeatedly while I corrected my answer on natural selection. "The weak get weeded out" took me two whole minutes to write. Finished with that sentence, I shook out my hand and let it rest.
I squinted up at the spider, trying to decide what species it was. I had seen a documentary on them on Animal Planet. Spiders looked horrifying, but they were delicate and even beautiful in a strange way. Plus they had eight legs. If they lost one, it probably wasn't a big deal.
The spider slipped farther down and I leaned back in my chair and cupped my hands, hoping it would land in them so I could get a closer look.
I accidentally locked eyes with the new guy, who was staring at the spider too. Technically, he wasn't new anymore, since he had transferred here over the winter break, but I'd overheard girls gossiping about him nonstop since I'd been back. Apparently he was wild and unpredictable, but the stories they told sounded far-fetched. I doubted he had really meant to superglue a sixth grader to the bench in the cafeteria. Plus, no one had seen him put the plastic roaches in the teachers' lunch bags, so it wasn't fair to assume he had done that either.
He grinned at me and nodded. This was my second full day back, so to him, _I_ was the new kid. I looked away immediately, but my cheeks fired up again. My face felt like a kiln. He was incredibly cute.
"Don't forget to turn in your permission slips for the field trip," Ms. Kerry said.
Natalie giggled from the back row. Ms. Kerry walked over and lifted Natalie's cell phone from underneath her paper. The screen was lit up with an incoming text. Candy and Natalie both glanced over at the new guy and blushed.
"Billy, the girls in class are wondering if you have a girlfriend yet," Ms. Kerry said.
The class froze, waiting for his response. I tried not to stare. Hot guys were an alien species to me.
"I'm currently accepting applications," he said, and the rest of the class started laughing. He looked right at me when he said it. I buried myself back in my test while Ms. Kerry got everyone under control.
A few minutes later, I stole a quick peek to check on the spider, but it was gone.
Suddenly, Candy squealed. Jumping out of her seat, she pointed at the floor. Everyone saw the spider. Matt, a football player who was shaped like a giant meat loaf with ankles, stood up and used his sneaker to grind it into a smudge.
I made a sad noise, a little choking sound, but no one noticed.
"I hate spiders," he said, and sat back down. "And I'm not afraid to smash things." He glared at Billy. Billy probably didn't know yet that Matt had had a crush on Candy since the fifth grade. Candy enjoyed the attention but not him, if that makes sense. Attention was like money to her. She wanted all that she could get.
The bell rang, and everyone shuffled their papers and books into their bags for the next class.
"Oh, Sofia, I am so sorry," Ms. Kerry called over the commotion. The halls were now swarming with kids. Last week she had insisted on giving me a five-minute head start, but I didn't want or need the special treatment. I had tried to tell her I didn't need it, but she'd just given me a big smile and promised me that it wasn't a problem at all, really. She was glad to do it for me. Ugh. "This is not about you" was what I wanted to say, but I couldn't, because she was only trying to help. The harder everyone tried to make my life easier, the worse I felt.
Everything was supposed to get back to normal, although I did have a couple of new teachers this term. I'd always been in advanced classes, but when I got my diagnosis in October, I had dropped back to on-level classes, which made keeping up with my classwork easier. _School_ should have been easier too, if school was only about learning facts and figures. But it's not. Middle school is about learning your label, locating your pack, and avoiding humiliation. Why didn't teachers acknowledge that? School was like the Eagle Cam that had run a live camera feed of an eagle's nest last summer. One of the little babies had been in danger of falling out of the nest, and everyone saw it but none of the scientists helped it. The baby bird fell and died and people watching over the Internet were furious. It's not okay to just let it fall, they said, and the scientists argued that it was wrong to interfere. Nature had to run her course, like Mother Nature was a sweet old lady who could be trusted.
For the record, Mother Nature is a total thug. Teachers should at least admit that school is a dangerous social experiment, not a serene learning environment.
I braced myself to face the hormonal hive out there, wishing I had time to write a goodbye note to Mom. Even if I made it through the crush of humanity, the toxic cloud of body spray might disorient me, and I'd wander the school for days. I'd be like the class hamster they found two years after it escaped from the science lab, living off a forgotten bag of Skittles in the school library with a crazed look in its eyes.
Alexis was waiting for me in the hall. I tried to move past and pretend I didn't see her.
"Hey," she said. "Want me to walk with you?"
To avoid her, I walked right into the path of a sixth grader zigzagging through the crowd. The sixth graders had a separate hall, but there was a shortcut through the seventh-grade hall that led straight to the gym. He was probably racing to dress for gym before any bullies walked into the locker room.
The collision knocked my book bag off my shoulder, sending my papers flying out all over the hall.
People trampled over them, dragging papers underfoot in every direction. "Oh, come on!" I snapped.
Alexis froze, as if she knew it was her fault. And it was. She assumed we could be best friends again just because my mom finally let me come back. She wasn't seeing the situation clearly.
Then Billy appeared, rescuing my papers one by one. I was so surprised that I didn't speak or move. My Language Arts essay fluttered through the air and landed in front of Matt's locker. Matt sauntered over to it, unaware that he was now standing on my essay. Billy grabbed him by the shoulder and nudged him off the paper. "Watch where you're standing," Billy said.
Matt lowered his head like he was going to headbutt Billy. "Touch me again and you'll be dead meat!" Matt replied.
Billy grabbed the paper, then stood up and placed his hand back on Matt's shoulder with a sad look on his face. "All meat is dead. I thought you knew."
Matt shoved Billy's hand off, then stomped away.
"Come back!" Billy called after him. "You shouldn't be alone right now!"
He walked over to me and held out my paper. I hesitated before reaching to take it. Then Billy changed his mind and held the paper up to look for something.
"Sofia," he said. "With an 'f,' not a 'ph.' " He lowered the paper and grinned at me. "I like that name. And I liked your answer on that science test. I wish I had written it."
I was rendered mute by the sheer force of his hotness plus this bizarre circumstance. I mean, he was talking to _me._ Willingly. It wasn't like a teacher had assigned us to be lab partners.
My hand fluttered up to touch the bandana on my head. He took a breath like he was about to ask me something.
"Sofia, hi." The voice calling out behind us was like a needle snapping off in my vein. I turned to see Candy walk up to us, her eyes never leaving Billy's face. She didn't really want to talk to me, of course. I was a human placeholder, taking up space until others had something interesting to say.
"Hello, Candy," I sighed.
Candy was an unfortunate name for such a sour girl. Maybe her parents had been confused when they'd first seen her as a baby, pink and soft. But all babies look like that, including rats and vultures.
She gave Billy her brightest smile, the one reserved for when she was running for class office. "You know each other?" she gushed. "That's great. I've been so busy organizing the school dance I haven't had a chance to catch up with Sofia." She didn't even look at me when she said my name.
Candy's friends circled around us. Predators hunted in packs.
"I'm the student council president," she said to Billy, "but I'm sure you knew that already."
Billy extended his hand and shook hers. "It's an honor to know an elected official."
The Winter Gala was the Saturday after next, but how much work could it be to hang crepe paper in the gym and ask the PTA to donate snacks? Stale cookies do not a gala make.
Besides, the dance was already doomed. It had originally been scheduled for January 29, the day the infamous Snowpocalypse hit the city, trapping kids taking school buses on the road and stranding people overnight at stores and offices. The teachers weren't happy at all about giving up a Saturday evening to chaperone us, even if the dance would end at eight p.m., but I guess Mr. Reeves wasn't willing to schedule it any sooner and this was the only good date. Anyway, lots of people were complaining about the change. School functions were usually held during the week, but this had been an unusual year. For all of us.
Change makes people hostile. I don't know why.
The bell was going to ring any minute. I mentally mapped out the quickest route to my next class. I hated being late and walking in alone. Moving with the herd offered protection from people who stared.
Candy's manicured talons grasped her perfect midnight-black hair and swept it into place. Billy's eyes darted side to side. He was looking for an exit route too. He caught my eye and I nodded slightly to the left, indicating where I was going. My stomach rumbled. I had just accomplished something extraordinary. I had made contact with the hot guy, tuned in to his wavelength. We were _communicating._ It made me a little dizzy.
Candy's smile stayed plastered on. I could tell she was trying to decide if he was mocking her. But that wouldn't make sense, because Candy was, to be fair, completely gorgeous. No one humiliated the pretty people.
"So, Sofia," Candy said, turning to me with teeth bared in a full-wattage smile, "I was planning my wardrobe for the week and a crazy idea popped into my head. I couldn't believe I hadn't thought of it before."
I raised my eyebrows. Candy had never said hello to me in the halls before today, and suddenly she was acting like we were friends.
"We need to talk this week, Sofia. I think you're going to love this."
Her best friend, Natalie, interrupted. "A group of us are going to get pedicures after school. Want to come?"
Candy glared at Natalie. "Sorry about that," she said.
My prosthesis had toenails that could be painted, but there was no way I was going to say that in front of Billy. He could see that something was wrong with me, but I didn't want to give him the whole picture yet. We had only just met. It would be like trying to hang a tire swing from a sapling. Too much too soon. Besides, I'm sure he had seen the Welcome Back, Sofia banner and someone would connect the dots for him.
Natalie huffed in her own defense. "Why is that a bad thing to say? Plus, Sofia, yours would be half off!" She grinned at me, thrilled to have just thought of this.
Mother Nature should let us submit names for natural selection.
Billy frowned, obviously confused, and stared at me.
It made sense. No one had talked to him about me while I was gone, because no one talked to me when I was here. Alexis was probably the only person who missed me, and she wasn't the type to gossip.
"Wait. Why would yours be half off?" Billy asked. No one moved. Candy had picked the worst possible moment to start being nice to me. "Nice" was going to ruin my life.
Mr. Reeves walked around the corner. He frowned when he saw our group standing there. "Save it for after school."
Candy and her friends scattered, but Billy stayed where he was. Did he think I would stick around and explain? Not a chance. I walked away without a word.
"Wait!" he called, but that only made me walk faster. "You never answered the question! I'm so confused. Why would yours be half off?"
All around me, people were rushing to class, and the sounds of the lockers slamming made my head throb. The doors swung and snapped shut like rows of hungry metal jaws. The bell rang, a shrill and piercing alarm, but there was no waking up from this nightmare. When would I learn that?
On Tuesday, October 8, of last year, a month into seventh grade, I fell during a cross-country training run. Which wasn't unusual, but the pain afterward was. My left leg had been bothering me for weeks, but I assumed it was from training so much. I loved the long training runs because there was no pressure to win. Alexis and I could run side by side, our feet finding the same steady rhythm, and then our breathing too, until I was certain that even our hearts were beating in time. We plodded through the wooded trails behind the school, the world becoming ours in a way no one but a runner could understand. School drifted from our thoughts as we sailed through waving branches and down the winding path.
Autumn in the South is a long, gorgeous goodbye, so I didn't want to miss any of it. Instead, I popped ibuprofen every night, willing the pain to go away. I loved how the birds and butterflies swooped carelessly close as we ran, distracted by their preparations. Most of them migrated in the fall as the flowers wilted and curled brown and then the leaves above burst into orange and red flames, like Mother Nature was cremating what remained of the year. The colors, the noise, the rush of wings overhead...all leading to the long gray silence of winter.
I wanted to see every moment, and a stupid bum leg was not going to stop me.
So I popped some painkillers and ignored it...until that run.
After I fell, I couldn't stand back up. The pain was unbearable, "disproportionate," the paramedic said, as if my problem were mathematical. In a way it was, because tumors are caused by cells that have a fatal error and spend all their energy multiplying the mistake. Their miscalculation wove its way through my bone with ferocious enthusiasm. A simple X-ray at the ER clinic showed the presence of a shadowy mass. Nearly two weeks later, we had the results of the biopsy: primary bone cancer. The good news was that the doctors didn't think it had spread. The bad news was that the tumor had my nerves and blood vessels in a bear hug. Removing it would be tricky, even if everything went well.
It didn't.
I was supposed to have three rounds of chemo before surgery to shrink the tumor, but I kept getting all kinds of infections. The biopsy might have caused one of them, but no one really knew. Every round of chemo took nearly three weeks. Finally, the doctors canceled the last round because they worried I wouldn't survive it. Everything that happened after that came so quickly it was like hitting the fast-forward button.
On December 7, I turned thirteen.
I celebrated my thirteenth birthday by having my leg amputated.
There was a certain irony to that.
Germs were a constant threat. It was cold and flu season, and the news was full of stories about strange and lethal viruses floating around. Alexis tried to visit me in the hospital at first, but she had a cough and the nurses sent her home. Mom was freaked out and didn't want me to have any more visitors until I was absolutely safe. She assumed that after my surgery, I'd be healthy enough to see Alexis.
But I didn't want Alexis to see me like that, ghostly pale with tubes sticking out of me. I was always dozing off in the middle of a sentence. I was also incredibly skinny, and I wanted to put on weight before Alexis saw me. People think recovery is a simple process that moves in a straight line, but it's more like trying to build a house of cards with a deck that is bent and torn in weird places. It's frustrating and complicated, and I didn't want Alexis to see how weak I was. Just staying awake to watch a whole episode of something on TV took effort.
I had always wanted to be just like Alexis, strong and brave. I was in awe of her. I didn't know why she had chosen me as her best friend last year at the beginning of sixth grade. She just ran with me one day at practice, before she even knew my name. We were getting to know each other. And then, before our first official race, I heard her crying in a bathroom stall in the locker room, so I crawled under the door because she refused to open it. "My sister is sick again" was all she would say. We were only track buddies, not best friends yet, so I wasn't sure how hard to push her to talk.
She shook her head, like her sister disgusted her, and then she stared at me for a long time. "You have toilet paper on your forehead," she finally said.
I pulled it off and felt my cheeks get hot from embarrassment, but then Alexis doubled over in laughter. I laughed too, and she stood up and hugged me. From that point on, I knew we were best friends and not just running buddies, because if you crawl under a bathroom stall to talk to someone and they hug you for it, that's pretty special.
After that we ran together every chance we got, because we didn't have any classes together that year and I could never come over after school to hang out because our moms worked and we rode different buses. Alexis liked that I wasn't on social media very much, because she said that meant I wasn't a drama queen, constantly comparing myself to everyone. It actually meant we couldn't afford a computer or smartphone, but that was just one way that Alexis could take something bad about me and decide it was one of my best features.
Then summer came.
We both signed up for summer cross-country practice with the high school girls. The program was free, and the older girls got community service credit. We ran twice a week. After practice, usually around lunchtime, Alexis and I would get a ride back to one or the other's house and hang out until our moms got home from work. But those hours together were sometimes harder than running.
It was just us, and we weren't running side by side. We had to face each other when we talked, and that was suddenly weird. We started to talk about the serious things, like why I didn't have a dad and how much I wished I could cry at night about it but I couldn't find the tears and so what was wrong with me? Then one day we talked about Alexis's frustration at being so thin. Formless, she called it, as if she were more raw material than finished work.
"I wish I was as skinny as you." That's what I actually said. We were sitting on the couch eating popcorn. Movie credits were rolling on the TV screen. Alexis's face turned dark with anger.
"Don't you ever say that again!"
I swallowed, too shocked to say anything else. Alexis's chin trembled and she cleared her throat a couple of times.
"Remember when I told you my sister was sick?" She wasn't looking at me, but I nodded anyway.
"That's why she's not here this summer. She's spending it at a treatment facility for anorexics. All she ever wanted to be was skinny. It's like she wants to evaporate. She thinks that if her body doesn't take up any space, people will like her more." Alexis kicked the carpet with her foot in disgust. "I hate being skinny, and you know what? I think she hates me for it too. Maybe my whole family does."
We talked about it a few more times over the summer. Alexis was terrified she would lose her sister, and her whole family, to an enemy she couldn't see, one that had its own peculiar laws and punishments. Alexis didn't want to be thin. She was terrified that somehow she would fall prey to the same disease. But she didn't want to give up running, because it made her feel so good, and she rarely ate junk food, except ice cream. Even then, she just liked it...she didn't love it. She couldn't help being skinny. It wasn't her choice. She was just made that way. She had the body her sister wanted, and she was worried that her sister hated her for it.
Alexis wasn't motivated by food or looks or even popularity. She lived more in her head, which I thought was cool, even if she said she wanted to live more from her heart. She sighed that afternoon when we first talked about her sister, and then put her head on my shoulder. "I just want to be able to run and eat and live, without feeling like everything I do depresses someone else. I'm tired of feeling like I only remind my sister of everything she'll never be."
Our conversations scared me at those moments, like we were testing a patch of ice, unsure how strong it was. And by the start of seventh grade, when we had said pretty much everything we had ever wanted to say to a best friend, there were little invisible cracks all around us. There was nowhere to hide. I knew she saw them, because she hesitated before we talked about the deep stuff. I hesitated too, not wanting to go forward until I knew for sure what those cracks meant, because cracks can mean different things. Like, before a baby chick is born, a crack appears in the egg, and that's a good thing. But before glass shatters, a crack runs across it. So which was it? Had we destroyed our friendship by sharing too much, or was something beautiful about to be born?
When school started, her sister came home and I started to shy away a little. I wanted to give Alexis space to spend time with her family, and we were both so busy with homework and running that having some distance between us felt inevitable. I began to move a little too fast between classes, not sure of the right things to say. I was trying to make it easy on Alexis, to give her plenty of room to heal and be certain, so she would know that being best friends had been her choice. That I was here for her when she was ready to catch up, but I wasn't a burden.
Turns out the tumor caught up to me before she did.
So how could we be friends now? I couldn't run with her, and I didn't want to hold her back from the only thing that made her happy. What if she stopped running to hang out with me and she started to hate me for forcing her to make that choice? Maybe she didn't want to be around another person who was sick. What if I was that last straw people talk about, and she stopped fighting this plague of perfection that had already claimed her sister? If she had to choose between running and me, I wanted her to choose what made her happy.
The tumor had been in _my_ leg, but it affected everyone. I saw what it was doing to my mom. Cancer didn't make victims; it took hostages. So I decided. It couldn't have Alexis.
She had to be free, because I wasn't.
Tuesday, February 25
I hit the snooze button and snuggled deeper into my warm bed. Downstairs, the dryer beeped softly three times as the furnace hummed its one long, low note. I loved waking up alone, listening to the sounds of our home, without strangers with coffee breath standing over me reading a chart or adjusting a tube.
I heard Mom in the kitchen. She fixed breakfast only on school days, like a consolation prize. The microwave door slammed closed, and she turned up the volume on the television's traffic report.
I swung my leg off the side of the bed and leaned forward to grab the nightstand. Grunting, I hoisted myself up and hopped over to part the curtains, inspecting the neighborhood. Our house was just two doors down from the elementary school bus stop, and the kids got picked up super early. Caitlyn, a kindergartner with long blond pigtails, waved furiously up at me and lost her balance, tipping over backward. Her backpack was larger than her body, making her look like a turtle in its shell. I waved back as she stood and then ran to the bus stop. We'd been friends ever since I had first seen her trying to teach her cat, Newman, to walk on a leash. Caitlyn was an optimist.
It was time to get my prosthesis on, and that took forever. I groaned just thinking about it.
The light from the hallway made the cracks around my bedroom door glow and cast an outline of my head on the window. I had a pencil-thin neck and a big round head with short, stiff bristles of new hair. My shadow looked like the outline of a toilet scrubber. My eyes followed the shadow down to the pajama pant leg hanging limp below my left hip, like a balloon with no air, the kind you find behind your dresser six months after the party. Shadows could lie and distort what was real, but this one was all true. My body was strange and frightening, even to me. Especially to me.
I grabbed a bandana for my head, watching my outline move with me.
Mom knocked before she swung my door open. "You're still not dressed?" she gasped.
I adjusted the knot at the back of the bandana without turning around. She always wanted to check the progress of my hair growth, but I wouldn't let her. If you've ever heard of shocking a pool with chlorine, that's sort of like what the drugs did to my body. My hair had thinned and fallen out in wispy handfuls, until a nurse helped me shave what was left one night when Mom went downstairs to the hospital cafeteria for a soda. I thought it would be easier for us both that way.
Mom got really upset when she came back. I didn't know shaving my head was something we were supposed to do together, like shopping for a dress or getting a manicure. Even now, it's hard to explain why I did it that way, except that I wanted to face my new reality, and it wasn't entirely about my hair.
When a brief surge of courage hit—which may have actually been a fresh dose of morphine, I can't be sure—I decided to accept the inevitable and go bald. Whether that was an example of optimism or fatalism would be a great subject for a Language Arts essay. Either way, I knew my hair wasn't my best feature.
So why not get it over with?
Now my once-brown hair was growing back as ghost-white bristles. If you think something can't get any worse, you don't have all the facts.
"I don't feel so good," I said, not turning around.
Mom ignored me and reached for the new leg. She carefully set it next to me on the bed, handling it like fine china. I shuddered when I thought about how much it cost.
The price wasn't just money; it was Mom's dream of going to night school to finish her degree. We had serious debt now, and she couldn't afford the classes. My dad had left before I was born and didn't pay child support, so we didn't have any help with the medical bills that insurance wouldn't cover. Drive-throughs, department stores, and cancer centers: they all take Visa and MasterCard. But we weren't entirely on our own; her coworkers loved her. They had taken up a collection and bought me some gift cards when I was in the hospital. I got to pick out some new clothes and a soft blue chenille bedspread that I adored.
Mom lightly ran her fingers over different outfits in my closet. "Well, this one will draw the eye up," she said, holding a tunic out to me. I held my nose, pretending I was trying not to retch. Mom bit her lip and turned back around.
Mom spied another outfit to show me, and slid the clothes down the bar to get it. In the far corner, on the floor, sat the pillowcase from the girl at the hospital. It was tucked inside a bigger plastic bag with the hospital's logo on the side. Mom didn't even notice it.
I blinked in confusion. That pillowcase shouldn't be here; it wasn't mine. I didn't want it.
We hadn't talked much about the incident at the hospital. It made Mom uneasy. The news made a brief mention of a girl brought into the emergency room with wounds from a suspected animal attack. Police were investigating, but the girl had used a fake name and they didn't have any leads. They didn't even know what had attacked her. The nurse had begun to clean the wounds and hook up an IV, but when she stepped out of the room for a moment, the girl slipped off the gurney and ran. It was all very peculiar.
When I had revived from passing out, my prosthesis, the carrying case, and the paperwork had already been loaded into the car by hospital staff. The pillowcase had been with me in the room, so someone must have assumed it was mine and packed it. Mom never realized it wasn't part of my personal stuff.
I sucked the inside of my cheek, thinking. I doubted the news story. If a dangerous animal really was on the loose, someone would know. A homeless man had reported seeing a freakishly large dog that night, but no one else had seen it. The Atlanta area had quite a few Bigfoot sightings every year, plus a persistent rumor of a pterodactyl-like bird hiding in the woodlands, so police didn't usually investigate reports of bizarre animals.
Mom was holding up a long sweater and cargo pants, wiggling her eyebrows at me.
"Yeah, whatever," I said, my chest tight. I cleared my throat.
"You loved it when you saw it online," Mom said.
"I was on drugs," I reminded her. I forced myself to breathe and look right at her, not at the pillowcase. Why had that girl shoved it at me like she wanted to get away from it? I tried to remember what she had said, something about me being brave and kind and... _chosen._
The girl had been completely crazy. Crazy enough to—
"Put it on," Mom said, laying the outfit on the bed beside me. "And hurry up. I've got your favorite breakfast ready."
"I didn't know McDonald's delivered," I murmured.
Mom paused at the door and gave me a bright smile. "See? We _are_ back to normal." Her voice cracked a tiny bit on the word "normal."
I stared at the pillowcase and then, without moving, examined the entire floor of my closet for any sign that the girl was there too. I braced myself to see tiny blood-splattered feet. A pair of pants had fallen off the hanger when Mom had shuffled my closet around, but she hadn't noticed. I stared at the lump on the floor, expecting it to move.
It didn't.
Mom must have been waiting for me to reply, because I heard her finally shut the door softly. I hadn't meant to be rude. I'd make up for it later.
Besides, before last year, she never shut my door. I had complained about that all the time. But now she shut it, and I knew why. It was hard to see that some things would never be like they were. We had each lost our hopes for the future, and neither of us knew where to find new ones.
I didn't have my prosthesis on yet, so no matter what happened next, I wouldn't be able to run away. I slid my body to the floor, sitting on my butt, preparing myself to scoot closer to the closet to inspect the pile of clothes. I'd be able to nudge the pile with my leg, and if it moved, I'd scream.
Pushing both palms flat against the carpet, I scooted a few inches, eyes focused on the lumpy pile in the right corner.
In the left corner, the pillowcase shifted.
I stopped, my hands frozen in midair.
It sat in the center of the closet floor now, not the corner. It had to be a trick of the morning light moving across my carpet. Or maybe Mom had moved it and I hadn't seen her. The book inside was clearly visible.
Slowly, I reached for my prosthesis, grateful to feel how heavy it was. I could swing it like a club if I had to.
I pretended to be wiping imaginary lint from the hollowed cup area of the leg, but from the corner of my eye, I continued to spy on the book. I had the strangest feeling it was spying on me too.
"Get moving up there!" Mom yelled.
Adrenaline shocked me at the sound of her voice, and I jabbed my prosthesis into the pile of clothes.
Nothing was there.
The book didn't move either.
I was an idiot.
Pulling the leg back, I used it to whack the closet door. The door slammed shut, sending a burst of air back at me. It had a strange but familiar smell, like the air after a storm. But this storm had brought something with it. It smelled like an animal. And leather. And old churches where candles burned all night.
The latch on the closet door popped and the door swung open.
Slowly.
The book was out of the pillowcase, against the door. As the door opened farther, the book fell forward, its cover facing me.
I climbed onto the bed, my heart hammering in my ears. This book was moving itself, which was impossible.
Books were lifeless.
The thick, cracked cover bore embossed swirls and something written in a strange language—the title, probably. The book was about a foot wide, and thick like two dictionaries stacked together. Deep cracks ran up and down the spine. There were two badly scratched brass buckles across the sides, and leather belt straps with black grimy edges ran through them.
A shiver shot down my back and into my belly.
Mom pounded on my bedroom door and I flung my hand over my mouth to keep from screaming. "Let's go!" she called. I heard her footsteps as she kept moving down the hall.
"Coming!" I yelled. I didn't stir.
Outside my window, tiny snowflakes began to drift down. I sensed that the earth had suddenly tipped, and its axis, so deep and hidden from all our eyes, was shifting. It rarely snowed in Atlanta, and we had already survived two bad winter storms. Snow in late February, even a light dusting like this, was rare. It made me feel like anything was possible, and I wasn't sure that was good.
I moved to one side of the bed and quickly put on my prosthesis and then my clothes, my eyes never leaving the book. It didn't move again.
My bedroom felt warmer, like I wasn't alone. Standing up, I took a deep breath, kicked the book back into the closet, and shut the door, fast.
Something nagged at me, like a name you can't remember. Maybe it wasn't a storm that had crept up on us on the way to the hospital.
Maybe it was something much, much bigger.
My stomach still churned like a rock tumbler every time I opened the front doors to the school. I was glad I hadn't eaten much this morning. Forcing my shoulders back, I made myself take a deep breath. If I took small steps, my gait was perfect. The new prosthesis felt smooth. Barnes was brilliant.
I maneuvered close to the lockers, my left shoulder grazing the metal. This way it was less likely that anyone would run into me and knock me down. It was too awkward trying to get back up, and then everyone went nuts with the apologies. But being careful meant being slow, and I hated being slow.
In the wild, a predator would already have me by the throat.
I didn't want to think about that right now. I didn't want to think at all, especially about what I was going to do with a possessed book in my room. I would have thrown it in the trash, except I was afraid to even touch it.
Mr. Reeves stood in the doorway to his office. Everyone on staff knew about my treatment schedule, including the new leg. He caught my eye and grinned, giving me a thumbs-up. I smiled politely, resisting the urge to roll my eyes.
The point of a prosthesis is to lead a normal life. Normal people don't get a thumbs-up for having all four limbs, do they? And having a prosthesis didn't automatically make me brave or a hero either. Heroes choose their paths. I hadn't chosen anything, not even where to put that stupid freckle.
I passed the counselor's office and noticed that the lights were on. Mr. Reeves had informed my mom and me that our counselor had retired this past Christmas because his wife was ill, so I was surprised that anyone was in there. Maybe they had hired someone over the winter break. A woman with long blond hair swept into a messy chignon sat at the desk. I loved that hairstyle. I had tried a chignon once, back when I had hair, but the only thing I got right was the messy part.
"Sofia," she called. Her voice was low and raspy, like a movie star's.
I stopped and ducked my head in. "Uh, yes?"
She smiled and stood, extending her hand to me. I entered, adjusted my backpack, and shook her hand. It was ice cold.
"Do you have a minute?"
"Uh..." I glanced down the hall.
"I'm sure your teachers won't mind if you're a little late." She didn't ask if I minded.
I shrugged and sat in the empty chair she gestured to. She hadn't put anything in her office yet: no posters or framed photos, not even a pencil cup on her desk. The office was lifeless. I shivered; it was freezing cold in here.
"Your backpack looks heavy," she said, a smile spreading again across her face.
"It is," I replied. "Lost of books."
She leaned forward. "And what are you reading, Sofia?" She placed her elbows on the desk and wove her long fingers together, making a steeple. Her nails were clipped short and painted black. She rested her chin on her hands as she stared at me. I shifted in the chair, suddenly uncomfortable. Her eyes were light gray, almost blue, like a thin frost covered the irises. I swallowed nervously.
"Are you reading anything _interesting_?"
The bell rang and I jumped without meaning to. She stared but didn't say anything else. Something about this conversation didn't feel right.
"I really should get to class," I said, and stood.
"They say only two things can change the course of your life," she said. "The books you read and the people you meet. Don't throw your life away, Sofia. Not for a book. And not for the fool who wrote it."
I stepped back into the hallway and noise flooded my ears. Kids brushed past me from all directions and I was swept back into the morning hallway rush, revived by the heat of all the bodies in motion.
When I glanced back, the office was dark and completely empty.
I got to homeroom with Ms. Forester. Her son was a marine serving overseas. She had raised a soldier, so she frowned on us acting like weenies. Sitting in my chair, I stuck my hands under my thighs until they stopped shaking. A few minutes later, I walked up to her desk while everyone was doing their work.
"Do we have a new counselor?" I asked.
Ms. Forester glanced up from grading papers. "Not yet." She looked back down at her papers, then paused, like she realized who was asking. "Do you need to talk to someone? I can ask Mr. Reeves about finding someone for you."
"No!" I smiled like that was a silly suggestion. "I was just curious."
She stared at me with a confused smile.
"Because I noticed the counselor's office is empty....It is empty, right? No one is in there?"
She frowned, her head cocked to one side.
I couldn't tell anyone about the book and I couldn't tell anyone about the disappearing woman, so I decided to keep my head down and my mouth shut for the rest of the day. I managed to avoid Alexis and Billy in science, although Billy made that easy by getting called to the principal's office at the beginning of class. Apparently he'd had detention yesterday and something bad had happened.
After class, I summoned the old power of blending in, hiding in plain sight, fading away, which is also known as sitting in a bathroom stall for as long as possible between classes.
Mom's car inched through traffic all the way home while she stole glances at me every few minutes. I counted six churches as she drove. One had a cross that stood taller than all the other roofs, like a stern reminder to be on our best behavior.
"This city is God-struck," a neighbor once said when I had asked why we needed so many churches. I wasn't sure what that meant, but it sounded right. They all seemed to have the same basic beliefs, about being kind and honest, but some also thought you should get baptized by being dunked underwater, while others thought the water ought to be sprinkled, while some thought the real secret was to go door to door and interrupt people's dinners.
"Did your leg hurt today?" Mom asked, veering around a slower-moving car. She honked the horn, frowning. Speed limits in Atlanta were more like baseline readings. They gave you a general idea where you were starting from. I counted one more church.
"Not much." Sometimes I had phantom pains, sharp flashes that came and went for no obvious reason. Part of my brain didn't believe my leg was gone and misinterpreted nerve impulses. In the hospital, when chatty doctors lowered themselves to sit on the side of the bed during my daily checks, I'd instinctively try to make room for them and move my left leg out of the way.
Then I'd remember it already was.
So phantom sensations sounded crazy but felt real, just like everything else right now: a strange girl who told me I'd been chosen, a book that moved by itself, and disappearing school counselors. I had an impulse to stop at one of those churches and ask someone inside how they knew for sure what to believe. How did they know what was real?
I was making this too complicated. Mom would believe me without any proof. She believed in crazy-sounding things like phantom sensations. She might also believe in books that moved or counselors that disappeared. It was okay to doubt what had happened, but I shouldn't doubt her. Even if this was all a strange side effect from treatment, I could tell her.
I drummed my fingers on my knee, trying to think of the best way to start the conversation.
Her hand rested on my arm. "I'm so proud of you," she said softly. "You've done an incredible job holding it all together."
I then realized how exhausted she looked. Her eyeliner was smudged, leaving dark circles under her eyes, and her face was pale and thin.
"Thanks, Mom." I tried to smile. "My leg didn't hurt that much. I promise. It was a pretty good day, overall. I forgot to tell you: I even got an A on my science test. I rocked the extra credit."
"That's my girl," she said.
Once we got home, Mom lugged her black leather briefcase to the kitchen table. The other customer service reps just used totes or grocery bags, but Mom said it was important to look the part of a successful businesswoman. Mom was stuck in a job that didn't pay much, and she was hoping to break out of it.
"Dinner in twenty," she said. She went to a peg by the back door and grabbed her apron. I heard her sigh with exhaustion.
Upstairs, I opened the door to my quiet bedroom. The closet door was still shut. Tomorrow I would definitely tell Mom about the strange things I had seen, and she would tell me it was a side effect. We might even laugh. I would show her the book and she'd take it to the trash, shaking her head and laughing.
Sitting on the edge of the bed, I began the long process of removing my leg. Even considering my hair, I really had been lucky. After the amputation, my oncologist told us that the margins around the tumor were clear and all tests were negative for cancer, which meant I didn't have to do more chemo. The only time Mom and I had let go and cried together was when the doctor told us the news. Most of the time, Mom tried to cry in the bathroom so she wouldn't upset me, as if I didn't already know that trick. Everyone knows that trick. It made me feel worse, actually, when she did that. I cried under my blanket at night. I hoped she didn't know that one.
I wiggled the leg off and then bent down to peel away the stocking that protected my stump. "Stump" was an ugly word but better than "residual limb." That made it sound like a hygiene problem. Anyway, I was done with all of that...at least until my next appointment with Barnes. Before that, I had another round of blood work, but the lab techs never talked about the leg. I'm not even sure if they knew I had a prosthesis.
Outside, the sun was setting. Shadows shifted on my walls, illuminating dozens of little jagged holes where I had ripped off my track and cross-country ribbons. Only my _Doctor Who_ poster remained. I had hidden the ribbons and team jersey in a box under the bed. Someday, I'd be strong enough to throw them out. For now, it was like that wistful feeling you get when you hang up your Christmas stocking. Some part of you still wants to believe in Santa, and some part of me still wanted to believe that none of this was real.
The crack under my closet door glowed. The light was green, like someone had just turned on a string of Christmas lights.
My closet didn't have a light inside.
I stared at the light, my heart beating rabbit-fast in my chest.
Something thunked against my bedroom door and I gasped.
"You left your crutches in the bathroom," Mom called. "Dinner in ten!"
Resting my head on my hands, I tried to tell myself to get a grip. It wasn't fair to my mom to freak out over an unexpected side effect like this hallucination. She'd have to put me back in the hospital, and she had been through so much already.
I didn't need her help to ignore this. I gritted my teeth, ready to get this over with. Once I knew it was all in my mind, I'd feel so much better.
The closet doorknob twisted, first to the left, just slightly. I exhaled once, then sucked in a breath and held it, counting to ten. The knob returned to center.
I let my breath out slowly. My shoulders contracted like someone had punched me, slow motion, in the chest.
It twisted again, right.
The door swung open, sending goose bumps across my skin.
The book stood upright, like it was staring at me.
Green light rolled from the book like fog, making a faint hissing sound as it crossed the carpet. The fog felt warm as it touched my ankle, probing it. Then it wrapped itself around me, as if it was making sure I was real. I couldn't breathe; my eyes watered from holding my breath, this time by accident. Hallucinations weren't supposed to touch you, were they?
As the book moved across the floor toward me, the cracked leather corners caught on the carpet. I tried to raise a hand to warn it to stop, but my arm was deadweight. I managed to exhale, a little shrill wheeze of panic.
The book's cover opened slightly, revealing a strange gold and green and blue etching inside. It might have been of an animal; I wouldn't get closer to check.
It seemed to be waiting for me. I was so polite even my hallucinations had manners.
There was one way to know for sure if I was imagining this. If I couldn't grab it, it wasn't real.
With a fast shallow breath, I flung my hands down and grabbed the book's spine. It wriggled against my grip, like a big fat animal that had to be dragged from its den.
The book was real.
My physical therapist said I had weak core muscles. She was right. I had to adjust my grip twice before I could lift the book up next to me on the bed, and I was panting from the effort. Or maybe from anxiety.
I edged one finger under the cover and paused.
Nothing happened.
Biting my lip, I flipped the cover open. Green light flooded my room, but it wasn't a scary, venomous green light. No, this was the green of spring, of grass and leaves and iridescent hummingbirds, a green I had almost forgotten, one I desperately needed to see again. When they released me from the hospital, it had been a shock to find that it was winter and the whole world was dead and gray. Now I longed to see life, and spring, return to the cold world.
The next page was made of leather. It was thick, with ragged edges and stitching on the sides.
In the center were words in fancy black script.
Beneath this was a beautiful image of a great winged bull. I had no idea what a bestiary or Xeno was, but I had heard of Aristotle. I had once thought they named a clothing line after him, but that, Ms. Kerry had told me gruffly, was a company called Aeropostale. Aristotle, she said, was a famous ancient philosopher who loved to document every animal he could find. He wrote endless descriptions of all the lives he found in nature, and many scientists consider him the first biologist. He founded his own school and loved teaching.
I turned the page again. The next one felt thinner. It wasn't made of leather, but it wasn't exactly paper either. It was like a thin fabric. Every page seemed to be different, as if added at different times.
I didn't recognize the language, but thankfully someone had scribbled a translation beneath the original wording.
> Year 436, Ab urbe condita
>
> "To Aristotle's disciples, to the healers, and to all who love Truth,
>
> Though our Master, Aristotle, is now dead, murdered at the hand of our enemy, I carry on.
>
> Herein are the secrets about the oddities of this world, the forgotten, the misused, and the feared. You and I are called as witnesses to their Truth, which is our Truth, and the Truth of the human condition. For we know ourselves best only when we see through the eyes of the strangers among us.
>
> Do not misunderstand. These are not playthings, pets, or wild animals that may be tamed.
>
> These are monsters.
>
> Xeno
Outside my window, a mournful howl rose from the belly of some awful beast. I shivered and snapped the book shut.
The room went dark.
Wednesday, February 26
I huddled on the school steps in the morning cold, my teeth clacking. Mom had dropped me off early at my request. She was shocked that I'd asked, of course. And since I'd never gone at that time before, neither of us realized the school wouldn't actually be open.
Last night she'd popped into my room unexpectedly and practically dragged me down the stairs for dinner. She thought I was sitting on my bed with the lights out because I was depressed. She hadn't noticed the book next to me, maybe because my room was dark. The book had stopped glowing when I shut it.
When I finally climbed into bed at nine-thirty, the book was nowhere to be seen. Mom tucked me in and I lay there, eyes wide open. Nothing moved in the darkness, and I fell asleep working through the details of a plan.
Thankfully, my school had a good library. If I was hallucinating, it meant my subconscious wanted to tell me something. Or I was crazy. Either way, I needed to know if there was any truth in the book. I didn't believe in monsters, but this was Atlanta, and it was a city with very old secrets.
People forget that Georgia was one of the original thirteen colonies, and when the English arrived, they discovered that the Native Americans believed this area to be a hive of supernatural activity. To the northeast of Atlanta, there is a canyon with a name that translates as "terrible," because tribes believed evil spirits lived there. It isn't far from Blood Mountain, which the tribes fought a horrible war for control of. It was believed to be the home of powerful and good spirits called the Immortals.
Plus, compared with other American cities, especially the ones we studied in school, like Boston and Philadelphia, all the buildings around Atlanta appear new. You have to look hard for a glimpse of the past. That's because General Sherman and his men burned the city during the Civil War. The city's history is in the ashes buried under our feet. When you're in Atlanta, you're always walking on the dead.
So Atlanta is a mysterious city with weird weather.
This winter especially had been unexpectedly brutal. The morning cold snuck in between the tiny stitches of my sweater. My skin stung all over, even on my head. If I had worn a wig instead of my favorite bandana, my head wouldn't feel like an ice cube. Mom had bought me a wig, but apparently she thought hairstyles were like wingspans, and that a larger size offered an evolutionary advantage. I refused to wear it. I didn't like hats either, because when you pulled a hat off, the bandana came off with it.
Now, with nothing to do but wait, I remembered the last time I had been this cold and alone. Instinctively I reached down and touched my left hip. The only place that was colder and lonelier than an operating room was the hallway just outside it. I had been left there while they prepped the room, drowsy from the pre-op drugs, when I heard a child wailing in pain somewhere in the hospital. My blanket had slipped from the bed and I had shivered in fear, drifting in and out of consciousness.
Suddenly, the track team burst around the corner of the school, moving in a pack. They always left the track and ran one last lap around the school before hitting the showers. Alexis was easy to spot, with her ponytail swinging side to side as she ran. I ducked my head and pretended I didn't see her.
The metal bolt slid back behind me and I turned to see Mr. Reeves opening the door. "Sofia! Come in! It's freezing out there!"
"Thank you," I mumbled, wondering if my lips were blue.
Mr. Reeves wanted to say something more. I saw his mouth open and heard him take a long breath. People did that around me, like they were about to dive into another world where they couldn't breathe the air.
"I have to get to the media center," I said before he could find his voice.
I hustled through the foyer and turned down the hall to my left. A basketball whizzed past my head. I ducked to avoid it and lost my balance, stumbling right into the path of two girls from the basketball team who were chasing the ball. The girls fell and I got slammed back into the lockers.
The team was using the long, empty hallway for passing and dribbling practice. The coach blew on her whistle, hard, screeching at the girls to get back up and continue the drill. The girls went right back to practicing, leaving me standing there, stunned. They were so focused on the ball that they hadn't even noticed me. That was exactly what I wanted, so even if my back was bruised and stinging, I should have been happy. The tear rolling down my cheek was just misinformed.
Thankfully, the media center doors were unlocked. Ms. Hochness, our librarian, was already at work sorting returned books. She was a really curvy woman the boys had secretly nicknamed Ms. Hotness. Her wispy blond hair swirled around her head like a personal cloud.
"Good to see you, Sofia. What do you need today?" She didn't hyperventilate when she saw me, which I appreciated. She glanced up to make sure I had heard her, smiled, and went back to work sorting a stack of books about sweet girls who loved shy ponies.
"I need to know everything about Aristotle," I said. He was my starting point, since I had already heard of him. The rest I would Google.
"He's dead," she said flatly. "I hope you can account for your whereabouts."
I blinked, not really getting the joke.
"Just a little librarian humor," she sighed. Then she nodded and crooked her finger at me, motioning for me to follow, and walked toward the nonfiction bookshelves.
"How's your mother?" she asked, not looking back, just walking and expecting me to keep up.
I took my first deep breath of the morning.
"She's good."
"Still nervous about you being back at school?" Ms. Hochness asked. "With all these germy kids?"
"Completely," I laughed. Ms. Hochness made it okay. I didn't know how. She just had that odd magic some adults have. "Thanks again for those books you sent me at the hospital. They were great."
I wondered if she knew I was lying. I had stopped reading right after my diagnosis. No one here knew.
I used to love books, and carried one everywhere. I enjoyed anything except the kind written especially for girls. That was weird, I know, but those books seemed to be softer and sweeter, and I couldn't be soft, not when I was in middle school, where bullies might try to slam your fingers in your locker or snap pictures with their cell phones when you changed for gym class. That hadn't happened to me yet, but I had to be ready since I wasn't even remotely popular.
After the diagnosis, when Mom and I spent every day together, she finally noticed that I had stopped reading but blamed the meds and the nausea and exhaustion. It wasn't cancer's fault, though. I stopped reading because I had never felt so alone, and books weren't helping. Books are like messages in a bottle washing up on a deserted island. It horrified me to know how many people were struggling to be heard, like they were stranded and calling out for help.
The realization had started back in September, in science class. Nothing ruins your life like the facts.
Let's start with color. What I call the color blue—that may not be what other people see. If you're color-blind, you don't see the same colors. You literally don't see the same world. And some people have extra color receptors in their retinas; they see about ninety million additional colors. The colors really do exist, but _no one else can see them._
I remember sitting in class and wondering, do I see what no one else sees or do they see what I can't? We'll never know.
And then there's taste. When Alexis and I split a bag of chips, were we tasting the same thing? No one knows, because no one has the same number of taste buds. Some people have a ton, so they can taste flavors other people can't. Some people don't have many, so they can't taste much; oatmeal tastes about the same as a chocolate chip cookie, which I find deeply disturbing. Alexis likes broccoli because it tastes good to her, but to me it tastes like a dirty diaper left in a hot car. So when Alexis and I split a bag of chips, we thought we were sharing the experience but we weren't.
I will never know what anyone else tastes, or sees or hears or feels. No one will. Everyone is alone in the way they experience the world. My reality isn't your reality, because we're different in a basic, cellular way. That's frightening. I didn't want to be alone. And books, I realized, were just one way people tried to communicate their reality. But it felt hopeless.
The social workers warned Mom that depression was common after cancer, especially once the intense treatments were over. They warned her to look for the telltale signs. But the depression doesn't come from having cancer; it comes from knowing how alone you really are. There are no words, and even hospitals know that. All my life I've loved words and reading, and then I realized there was no book written in my native language. No book could heal the sadness of being so alone, because _everyone was alone._ That's why I had stopped reading. Books depressed me. What good did it do me to know that everyone else was alone too? No book could save me from that truth.
Ms. Hochness gestured to a long metal shelf. "Histories are in the middle, essays about him to the left, and his personal notes are to the right."
"Can I take them all?" I asked. "And I need to look something up online." The media center had the best computers in the building.
She nodded and started pulling books off the shelf. "Help yourself to the computers. I'll grab these for you and put them on a table."
Sitting in a cubicle, I pulled up an Internet search engine and typed in "bestiary." I learned that it meant a book of mythical animals. I typed in the other name in the book: "Xeno." Google told me it was an ancient Greek male name meaning "strange voice." Nothing else was recorded about him and monsters, of course.
Finally I typed in "the year 436 Ab urbe condita." I learned that it meant Xeno wrote his letter 436 years from the founding of the city of Rome, also known as 318 BCE.
I chewed my lower lip, staring at the screen. There was no way that book was so old. But if it wasn't a hallucination, what was it? A joke, or a prop that got lost from one of those dragon and wizard games or some TV series that filmed in Atlanta?
Maybe the books on Aristotle would have some answers.
I made my way back to the table and could barely see Ms. Hochness over the growing pile of books she had pulled. "Aristotle wrote dozens of books. Maybe hundreds," she said.
I was going to have to read hundreds of books? My eyes widened.
Ms. Hochness raised one finger as if to warn me. "But they were destroyed after his death." She lowered her voice like she was telling me a juicy secret. "Consider this: He was the greatest natural scientist the world has ever known. Yet no one really knows how he died; some blame a mysterious stomach ailment, but others say Aristotle had powerful enemies. I've always wondered if there's more to his story than we know."
"Yeah, apparently there's a lot more," I said. "I mean..."
Ms. Hochness cocked her head to one side.
I shrugged, the universal body language of thirteen-year-old girls.
She led me to a table closest to the one good heating vent in the room, past a shapeless form slumped over a desk.
"A new kid. Transferred in from a fancy private school in the city," she whispered. "He was already asleep there when I arrived this morning. I don't even know how he got in."
"Serious student," I whispered back.
Something about the look she gave me—one eyebrow arched up and a wry smile—told me I was wrong. His records must have told a different story.
A glittering pool was spreading out from his mouth across the table. A badge that read "I Love Hot Chicks," with a chicken nugget roasting over a flame, was displayed on his jean jacket. And he smelled bad, like the-last-day-of-the-state-fair bad, when they're loading the animals from the petting zoo and trashing the old corn dogs that didn't sell.
An open book rested under his arms, and Ms. Hochness gently pulled it from under the drool-hole.
The guy woke up with a start, then looked around, dazed. He and I both did a double take.
It was Billy.
She turned the book around. "What have you been drawing, Billy?"
He wiped his eyes with the back of his hands and winced at the harsh overhead lights.
"I was making the cats anatomically correct," he said. "I'd think that would be important to a school deeply committed to academic excellence."
"You want the picture to be more realistic?" she asked.
He nodded.
"Then this poor cat needs a support garment," Ms. Hochness said. "Because she's going to have back trouble later."
Billy cringed.
"Here's another dose of reality." Ms. Hochness smiled sweetly. "Defacing school property carries an automatic penalty of detention."
"I'd love detention in the library," Billy countered. "Detention in the computer lab is so boring. Books are where the real action is."
The corners of Ms. Hochness's mouth twitched. She was trying not to smile.
"Anyway," Billy said, "it's my dad's book. He's a vet. I draw the pictures to let him know I have a keen interest in the family business."
I had never heard anyone my age use the word "keen" in a sentence before. I liked it. It was undeniably polite, with a sneaky touch of snark.
He glanced up at the clock before stuffing everything into his backpack and walking toward me. Something fluttered out from all his stuff. I buried myself in my work and prayed he would keep walking.
Instead, he sat down at my table. "Tornado season is coming."
I cleared my throat and kept reading.
"You watch TV during storms?" he asked.
I didn't answer. I wished I had put on lip gloss this morning, even though I never wore it. I was so pale. Plus this baggy sweater made me look like a cat trapped inside a parachute. Maybe I should have worn that wig instead of the bandana. He wouldn't have recognized me.
He sat back and sighed. "Why won't you talk to me?"
Ironic, because the chatter in my head was deafening.
He sighed and continued. "So, when you watch TV during a storm, the station sometimes stops the show and you hear this random male voice say, 'We interrupt this program to bring you an emergency weather report....' And you're mad at first, really mad, and you have to change channels to find something else to watch while a reporter blathers on and on about some stupid storm, right?"
I nodded even though I never did that. Secretly, I always worried it would hurt the reporter's feelings somehow.
"But then you really like the new show. You're glad you got interrupted. 'Cause you would have missed out. And maybe now it will become your favorite show. You can't wait to see what happens next."
Was he still talking about TV? I had a feeling the subject had changed.
He picked up one of my books. "Aristotle!" he yelled.
Ms. Hochness gave us an arched-eyebrow stare.
Billy scooted lower in his seat and looked at me like I had passed another secret test. "You are the most interesting girl I've ever met. You're not afraid of spiders, you read Aristotle, and you don't want to hang out with the popular girls, even though they want to hang out with you. Plus your extra-credit answer put into words everything I've been thinking for the last year."
I wished he had said I was pretty, not interesting. Scabs were interesting...and so were turtles and mildew on bread. Ugh.
He kept talking. "And I'm sorry about Natalie embarrassing you. Somebody filled me in during detention. I want you to know, I don't care. I mean, I care. But not enough to talk about it. Unless you need to. And I hope you don't." He turned red and cleared his throat. "What I'm trying to say is, don't be embarrassed around me."
But it clearly embarrassed him to say that. He was strange...and maybe wonderful. But he was definitely too much for me. I had gone from lowly outcast to class pet on a pedestal so fast that it was a wonder I didn't have a nosebleed. Hot guys would only add to the social chaos.
"I really need to read now," I said. His sea-green eyes made it hard to concentrate.
He ignored my hint, pushing the book out of my reach.
"At one of my other schools, we saw a documentary on the ancient philosophers." Billy scooted closer to the table. "Back then, parents worried that the truth would corrupt young minds. Some people were even put to death for seeking the truth." He lowered his voice and leaned even farther toward me. I pulled back, worried about my breath. "Ask too many hard questions and..."
He made a slicing motion across his neck with one finger.
I glanced at Ms. Hochness, who was refilling the candy jar on her desk. "I think we're safe in here," I said.
He sat back suddenly and folded his arms behind his head. Little white hairs clung all over his shirt. "I better stick around while you read," he said. "For your own good."
A shrill, rattling noise, like one made by an animal, came from the hallway, and we both jumped in our seats.
Then Billy stood. "Almost forgot," he said, and it sounded like an apology. "Look, if I don't see you again, Sofia, know this: you've restored my faith in humanity."
He left so fast I didn't even have a chance to point out the paper he had dropped. He was definitely a rare combination of strange and sweet, like a popcorn-flavored jelly bean.
When I was sure he was gone, I walked over and picked up the paper. It was a picture of his family, his mom and dad and him as a little kid. The photo had been torn and then taped back together. It seemed private, like it was more than just a picture.
"Something wrong?" Ms. Hochness asked. She stood in her office doorway.
I didn't want to show her the picture for some reason. "Nope," I replied. I tucked the picture in my bag. Later I would find his locker and slip it in through the crack.
"Need more time?" Ms. Hochness asked.
"Yeah, thanks," I replied.
Ms. Hochness ducked back into her office to call Ms. Forester and let her know that I was staying a few more minutes after the bell. They'd probably agree it was a good idea to let the hallways clear out. So occasionally, I admit, having a prosthesis worked in my favor.
I went back to the books, eager to find answers. There wouldn't be any about a student called Xeno, because he didn't exist, at least according to the Internet.
But plenty had been written about his "master," Aristotle. "Master" was an ancient title for teachers, which was a disturbing discovery I vowed to keep to myself.
Aristotle, I learned, began his career as a medical doctor but wanted to document everything in nature, so he looked for truth under every leaf and in every eye.
He believed we were two things at once: what the world saw on the outside and the thing hidden inside that we would become. He tutored Alexander the Great, the warlord with fabulous hair who conquered most of the ancient world, including the Persian empire. The Persians were the first people to decide that men should wear pants instead of flowing skirts.
I groaned and looked up from the pile of books. This was not the information I needed, but the last piece of my plan was the most embarrassing.
"I need books on monsters," I told Ms. Hochness.
"Novels?" she asked.
"Nonfiction." My cheeks got really hot. I might as well have asked about the Tooth Fairy and Santa.
She brought me a few books on folklore and urban legends. Each of them featured a ghoul on the cover. One creature was dripping with blood; the other had a severed arm in its mouth. I nodded and she scanned them; then I threw them in my book bag, not wanting to see the covers. Was that what monsters really looked like? I was hoping more for the Muppets.
Then I heard the first scream.
Outside the library, teachers were yelling for help. I hustled to the door, right behind Ms. Hochness, to the sounds of pounding footsteps and shouting. People were slipping on the floors because they were running too fast, or else they were pressed flat against the lockers in horror. A terrified goat raced past and nailed Ms. Hochness on the foot. She cried out, bending down to grab her toes, which gave me a good view of the hall.
The hit-and-run goat had a huge number three painted on its side. His beady eyes were wide with fear. Mr. Bronson, our beefy gym teacher, sprinted after it, his sneakers giving him traction. The goat bleated in panic and went down in a whirlwind of hooves and biceps. Mr. Bronson grabbed it in a wrestling hold, and its thick pink tongue stuck out and waved like a flag of surrender as dark pebbles shot from its rear end in every direction. This made some of the sixth graders scream even louder before the goat broke free and ran off.
Another goat had cornered Mr. Reeves by the boys' restroom. This one had a big number one painted on its side. Number One lowered its head and rammed Mr. Reeves right in the no-no zone. The principal fell to his knees, his mouth open in a silent scream. Number One charged down the hall toward the administration offices just as a math teacher pulled the fire alarm.
Once the fire department arrived, Number Three was caught within thirty minutes. We got updates from firefighters whenever they walked back to the truck. The older guys had to clear their throats a lot to keep from laughing.
Number One was found standing on a table in the room for in-school suspensions. He had braced his front hooves against the wall and was eating the posters.
We stood in the cold, waiting an hour more for them to catch Number Two, but Number Two was never found, despite a thorough search. The decision was made to let us back inside, but we were instructed to travel in pairs for the remainder of the day. The teachers rushed us from class to class and made us eat lunch at our desks so we could all get back on schedule.
When the final bell rang, I knew Mom would be waiting in the carpool line. I hoped she hadn't heard about the goats. She didn't need anything else to worry about.
Billy brushed past me, heading toward a silver pickup. Mr. Reeves was watching him from a distance, a notepad open and pen in hand.
"Hey!" I called.
Billy stopped and turned back. When he saw me, he grinned like he had waited all day to talk to me.
Mr. Reeves made a note. I had never been in his notes before.
Billy sauntered over. "I guess I'm still here."
"Where else would you be?" I asked.
He cast a glance at Mr. Reeves, but didn't answer me.
"You dropped something in the library," I said, and passed the picture to Billy facedown, like it was a secret.
Billy looked at the picture, then back at me.
My stomach flipped. I liked his eyes. I liked his _whole face._
He ran a hand through his thick brown hair and nodded thoughtfully. I had passed yet another test I didn't even know about. He leaned in but I pulled back, terrified that he might try to kiss me. I didn't know how those things worked. He stepped closer and tried again.
"Can you keep a secret?" he whispered, his breath tickling my ear.
"Yes," I said. I sounded like a dork, and not just because of my one-syllable answer.
He took a breath, then whispered again. "There was no Number Two." He pulled back to watch my reaction as I figured out what that meant.
My mouth fell open.
"Mr. Reeves made me clean the bathrooms yesterday during detention, even though I told him I wasn't the one who sent the email, but he said—"
I was frowning, so Billy backtracked. "Apparently, after my detention in the computer lab, the sixth graders all received emails from the school that said they had to dissect a human head in science class. If they refused, they'd be held back a year. Parents were furious. Mr. Reeves said cleaning someone else's mess would teach me not to make my own. But I'm glad I didn't get caught today."
"Because you would have been expelled," I said, trying not to laugh with Mr. Reeves watching.
"Because I wouldn't get to see what happens next."
Mom tore the grocery list in half. I took the bottom portion and she took the top before we both checked our watches. It was our usual grocery store challenge: We each had to get an equal number of items, at the best price, and then meet back at the registers. Whoever got there first and scored the best deal controlled the TV remote after dinner.
Tonight's grocery challenge was a spaghetti dinner. I had to find pasta, sauce, and bread. Mom was going to hit the produce department and get everything to create an epic salad. Frankly, I think she decided to lose the contest just to cheer me up. Pasta was shelved next to the sauce, so I had a clear advantage. Plus, I was kind of squeamish about the produce department. You had to touch everything and sometimes even smell it. It always felt inappropriate to do that in public.
Six minutes later, I was waiting at the registers. The produce department was to the left, around the corner from the pharmacy. Mom didn't emerge. I shifted my weight and worked out my plan. Since I was going to control the remote, we would watch a short show, and then I'd make an excuse to head to bed early and read.
I had a very particular book in mind. I was pretty sure it was safe to explore it, and I wanted to know what else was in there.
When Mom still didn't come around the corner, I groaned and went to find her. She was spoiling my victory, but at least we wouldn't watch any boring British dramas tonight. Turning the corner, I spied her talking to a woman wearing yoga pants and a pullover, as if she had just left the gym. She had long blond hair swept into a perfect ponytail, like a model on a magazine cover.
"Mom!" I called, and took a step toward them.
The woman standing next to my mom turned around, and a dark smile slowly spread across her face. It was the woman from the counselor's office. My breath caught in my throat.
"Sofia, be there in a sec," Mom called. "I'm just talking to someone."
I moved closer. The woman stood between us.
"Your mother and I were just talking about all the difficult struggles that girls your age must face. It's so important for us moms to know how to support our girls."
She placed an odd emphasis on the word "our." I nodded but said nothing. Mom put a red onion in her basket and stuck out her hand to the woman.
"Again, it was lovely to meet you. Thank you for your advice."
"Well, I hope I didn't overstep my bounds," the woman replied in her rich, soothing voice. "I guess I just know from experience that once the wrong door gets opened, it can take forever to shut."
At the registers, when I was sure the woman wasn't following us, I grabbed Mom's arm.
"What advice did that woman give you?"
Mom shrugged slightly. "Just how important it is that girls your age feel like they fit in. Being the odd girl out is a sign of trouble, she said. She's right about one thing: this is a hard age."
I focused on emptying my basket for the cashier. The advice didn't sound sinister. But then, I didn't get the sense that woman was a fairy godmother.
"Yeah, but, Mom," I said, carefully placing my last item, the jar of spaghetti sauce, on the belt. "I just don't fit in. I've tried. Does that mean I'm headed for trouble?"
"Not unless you count my cooking as trouble." Mom shook her head and set her bag of lettuce on the belt. "And you fit in just fine."
I shrugged like I was debating that, and she smiled at me.
"You know what's weird?" Mom asked a minute later. "She said she just got out of a spin class, but she wasn't sweaty at all."
I glanced behind me with a frown.
"Some women are just perfect," Mom sighed. The cashier handed her the receipt and Mom tucked it neatly into her wallet. She had more receipts than money.
The television switched on downstairs. Mom was watching the news while she did the dishes. The loud clanks and clinks were accented by a man's deep, authoritative voice. The only man's voice we ever heard inside these walls came from the TV. I wished Mom would get married. But she had survived two broken hearts, so even though I wanted a dad, I never pressured her to risk a third heartbreak. Sometimes hiding what I felt seemed like the only way to show that I loved her.
Mom didn't like to talk about my dad. He left when he found out she was pregnant with me. Mom said it was a complicated situation and he wasn't ready to be a dad, so she offered him a deal: he never had to pay child support if he relinquished all parental rights in the divorce. He took it.
Then, when I was in third grade, my mom was engaged to the man of her dreams. He was going to be my first real father. But four weeks before the wedding, he was killed in a motorcycle accident. Mom's wedding dress hung in her closet for a year before she finally gave it away. I wish there were a ribbon for people who survived broken hearts. I'd wear one for my mom. I was thirteen now, and maybe she thought it was too late for us both to have a happily-ever-after as a family, but I still wanted her to have one.
I opened my closet door and didn't see the book. Panicking, I looked in each corner, then around the room. I knew Mom hadn't moved it when she hung my laundry up. She would have asked me about it. Finally, I bent down and peeked under my bed. The book sat in the shadows. It looked like it had hidden itself.
I pulled back, surprised. But if that was what it had done, that was a good thing. Keeping the book hidden was one thing I didn't have to worry about. If it was real, which it wasn't, exactly.
Definitely.
Maybe.
I sat on the edge of the bed, reached for the book, and tried to pull it out, but my body had the muscle tone of a wet noodle. I grunted and tried two more times before I was able to get the thing into my lap.
I opened the book and flipped past the dedication. On the left side of the next page was a picture of a young girl with long white hair wearing a golden yellow dress. The girl stood in a room with a wooden floor and a window with bars on it, like a prison. In her hands she clutched a bottle. I couldn't tell if it was supposed to be poison or medicine. At her feet crouched a strange creature with long white teeth, red eyes, and floppy dog ears. It had lizard feet and a dragon's sharp, pointed tail. I didn't understand what the picture meant. The lines were delicate and exact. Someone must have spent hours making it.
Around the edges of the picture were gold and blue swirls, with tiny scrolls of leaves and ocean waves. I ran a finger over the picture. It had a slight texture—not bumpy, like an oil painting, but wavy and just a little warped, like paper that's been wet, then dried. And it was definitely paper of some kind, not leather.
On the opposite page was a similar frame of blue and gold, but it had a giant eye staring right at me, front and center. I didn't like that.
I turned the page.
The next one was blank except for two words in an old, fancy script, each letter fat and thick, with blurs of ink around the edges.
Leaning down to look more closely, I read:
> Welcome, Sofia.
I slammed the book closed. The freaky girl from the hospital must have written that before she gave it to me.
No, that was impossible. She didn't know my name.
Then it had to be the creepy counselor who had warned me not to throw my life away for a book, because she _did_ know my name. But how? She wasn't a real counselor. And how would she have gotten access to the book or know I would open it?
I started to feel that irresistible temptation to do what I knew I shouldn't, like digging a fingernail under the edge of a scab.
I opened the book and flipped right back to the same page.
It was blank.
As I continued to stare in confusion, tiny scratches appeared, like someone was trapped inside an Etch A Sketch. I covered my mouth with one hand and watched.
Whoever or whatever it was scratched out one letter at a time. It was torture, guessing what the letter would be, waiting for each letter to be complete, guessing at the word it would make. A question formed.
> Shall we begin?
I jerked back and blinked to clear my vision. When I opened my eyes again, the four words were still there. This was really happening.
The newscaster's voice floated up the stairs. _"Tonight, we look into the threat of meningitis in schools: what you can do to keep your child safe."_
Mom would be glued to the television.
I stared at the book. Who was talking to me? And how were they doing it? Was I supposed to write my answer? And what were we going to begin?
"I don't exactly know...," I said. "I mean, I've never talked through a book. I'm not exactly sure how this works."
Immediately the words were rubbed away, one by one, fast. Someone was magically erasing them. Tiny scratches appeared in their place, faster now, like someone on the other side was excited. I watched, half frozen with disbelief.
> Neither am I.
>
> I am trapped, alone, between the realms of life and afterlife. This book is my window between our worlds.
>
> I can only speak through these pages.
>
> I am human, even now, and am bound by human limitations.
>
> I can hear what you say and watch what you do, wherever you are. But I cannot read your mind, and I do not know the future.
>
> However, I do—
"Seriously? You can hear me?" I gasped, leaning forward. "I'm talking to a _person_?"
> You interrupted me.
"Sorry," I murmured. It was probably an adult.
> There are many believers but only one Guardian. Though the truth speaks to everyone who listens, I myself can only speak to the one who holds this book. The Guardian has almost always been a child, because children are not often suspected of wielding great power. But know this: The power is not in the book. The power is in you, and in your choices.
>
> Please understand: this is life and death. Do everything I say, exactly as I say. No Guardian who obeyed me has died of her injuries. No matter what happens, trust me. I have walked countless roads and seen the rise and fall of many empires.
When no more words appeared, I cleared my throat. "May I ask a question now?" I said. The page remained blank. I assumed the answer was yes.
"The girl who gave me the book—is she okay? What happened to her?"
> Her name is Claire, and she is well. Claire did not want this life. She is not strong like we are.
"Who are you?" I whispered.
> The last student of the Master of Them That Know, Aristotle. My name is Xeno.
"Xeno, if that were really true, you'd be dead by now," I said.
> NOT DEAD
"Okay, trapped," I answered. "Between two worlds. And you're talking to me through a book, a book about monsters and someone called a Guardian. Xeno, if that's even your real name, I don't believe you."
> I did not ask you to believe. I asked you to trust. Will you do that and help me protect this last, great secret?
Xeno thought I could guard a last, great secret? My mom couldn't even leave me alone with an open bag of potato chips. Stalling for time before I answered, I glanced around my room, at the pockmarked walls, at the closet full of clothes to hide what made me different, and I felt that aching hole in my heart from missing my best friend even though I had to act like I didn't. I was pretty good at keeping secrets, but mine all sucked.
Maybe I needed a new secret. I was so tired of being me. Reading and answering this book was like playing a game of dress-up. Xeno and I were pretending. He was offering me the chance to be someone else. None of this was real.
And I hated real.
"I'd love to," I said.
> Excellent.
>
> The book was lost for generations and has only recently returned to the light. I am anxious to begin. Many friends have wandered and been lost.
>
> If you are ready, press your hand to the page.
So far, this was interesting. A little scary, maybe, but a definite improvement over staring at my blank walls.
I pressed my palm flat against the page. It was soft and warm, like an animal's belly. The black letters beneath my palm pressed back against my hand. I could feel their ridges and sharp lines tickling me. Then the letters turned a warm and deep gold color, seeping together in one puddle that stretched itself into a long thin line and surrounded my hand with the outline of a much bigger hand. It was a man's hand holding mine, and it radiated heat.
In the distance, a shrill scream split through the night air. Animals all over the neighborhood reacted, dogs barking madly and cats screeching as if they'd seen an enemy. But softer, farther away still, a different creature sang. It was not a song with notes of growls or cries, but a song in a language of long ago, like something brought up from the ocean floor.
My heart seemed to expand and open. Something in my bones remembered this song, as if this were music I had made when I took my first breath as a newborn. For the very first time, I heard the world as I was meant to.
The television went quiet downstairs. Mom would be coming up for bed. I shut the book with a sharp snap. Musty air swirled up at my face from between the pages. Something familiar was in that smell again. Something warm and salty and...It reminded me of the way Alexis's dad smelled when he hugged me after a track meet. Comforting and loving. What was happening to me?
I heard Mom's footsteps on the stairs. I only had a split second left before I needed to hide the book. I flipped it back open.
"But wait...who's the Guardian?" I whispered.
> YOU ARE.
Thursday, February 27
I ate breakfast while Mom pouted, loading our dishes into the dishwasher with a lot of unnecessary clanging and banging.
"Mom—" I tried again.
"You didn't want to talk last night. You just rushed upstairs to your room, and every time I checked on you, you were just sitting there in the dark and you wouldn't let me come in. Did I do something? Is there a reason you suddenly need to be alone?"
I opened my mouth to answer but nothing came out. _Yes,_ I wanted to say, but that would hurt her, and if I tried to explain, she'd think I was having emotional problems like the doctors warned.
They couldn't have prepared either of us for this. I had looked up the word "paranormal" at school, and it meant life outside the normal senses, so it was a perfect description. I was dealing with something I couldn't explain. I had tried to stay awake last night so I could open the book once she fell asleep, because there was so much to ask and learn and school sure wouldn't cover any of it.
The last thing I remembered was hiding the book under my pillow when I heard Mom coming down the hall to check on me again. I had nodded off, still fully clothed.
A glass shattered when she slammed it into the rack. She didn't stop but just reached for the next one.
"I still get tired," I said. "And I need time to think."
"If something is going on, you can talk to me about it. You know you can tell me anything."
I could, but I didn't. I always wanted to tell her everything, but there were so many little things, overheard comments and embarrassing moments and daily hassles that I kept to myself. Changing for gym class had embarrassed me in sixth grade so badly, but Mom thought it was no big deal. I thought it sucked. Stripping to your underwear in front of someone like Candy, who was perfect, was humiliating. One time, Candy had brought the whole changing room to a stop when she pointed out my underwear.
"Sofia, are you seriously still wearing underwear with pictures on it?"
I had on my Polly Pocket panties I bought in fifth grade. They weren't cool even when I bought them, but they were on clearance and Mom loved a bargain. They had stretched out a lot, but they still fit, and I hadn't found anything else clean in the laundry that morning.
In sixth grade, girls are not supposed to wear underwear with cartoon characters on it. No one had told me. Candy made sure I knew. She was special that way.
I started changing in the bathroom stalls to avoid being seen.
So if Mom didn't understand some issues about ordinary life, even the stuff that happened last year, how could I explain this?
Fresh tears made my throat tighten, and I sniffed to chase them away.
Mom heard me and lowered her head, still holding an empty coffee mug. She didn't set it down when I walked over to her. I could tell she was still upset that I hadn't wanted to talk to her last night. I took the mug and set it in the sink, then picked up her arms, wrapping them around my waist. I rested my head on her chest.
"Are you angry with me?" she whispered. "I worry that we're not communicating."
"What? No."
She pushed me back so she could see my face.
I raised my head and kissed her on the cheek. "You wanted to get back to normal, right? Well, I'm a moody, difficult teenager. That's normal."
She rolled her eyes in mock dismay. My attempt at humor had worked.
"Let's get moving," she said. "Don't want to be late for school."
Grateful for the out, I went upstairs, but before I reached the top I glanced back to be sure she wasn't watching. It was hard to move my prosthetic leg like a real one when going up the stairs. I had to angle my torso sharply to one side and swing the prosthesis up.
When I got to my room, I dumped my book bag onto the bedspread before I packed it for the day. It was too heavy to drag around with all the books from the media center. I chose three to keep in my backpack for now and put the rest in a pile on the floor next to my bed.
A business card fell out as I picked up the books.
> Hamby Animal Clinic
>
> Andy Hamby, DVM
>
> After-Hours Emergencies Accepted
>
> Call Our Home Number at...
Billy must have slipped that into my bag while we were in the library.
I remembered seeing a new vet clinic open really close to our house. Billy had told Ms. Hochness the truth; his dad really was a vet. I suspected that was where Billy had gotten the goats.
Something scratched at my window.
The media center books about monsters were spread out on my bed. I hadn't had a chance to read them yet. I froze, glancing between them and the window. Could a monster really be out there?
Could they be real?
Not a chance. People imagine monsters, that's all.
My pillow began to glow. I guessed Xeno wanted to tell me something.
Wind rattled my window. I glanced at it, hoping I was wrong about the noise. A patch of wet fog grew on the other side of the glass. Something big had just exhaled.
I yanked Xeno's book out and turned my back to the window, feeling dizzy, like I was standing on a very high ledge looking down.
I flipped the book to the blank page where words had appeared last night. How did this work, exactly? "What's outside my window?" I asked.
> Throw away those worthless books! Will I not reveal all?
I glanced at the books on my bed, then back at the window. "I just needed more information to—"
> Once, I did not believe in monsters either. But the Master sent me into new lands to document every creature I found. And I found creatures beyond explanation. If I had not seen them myself, I would not have believed.
>
> You must see them for yourself too.
A shock like the touch of cold metal shot up my spine. I wasn't ready. I thought I was, but I wasn't. It was just like the moment when they took the bandages off my stump for the first time.
Outside my window, something huffed loudly. It sounded impatient.
"Wait!" I said.
I heard a clacking like rapid-fire gunshots or teeth being gnashed in frustration.
"You haven't even told me what a Guardian is," I pointed out to Xeno. "Or does."
> I can only show you what is out there, beyond your window. How you respond determines who you are. When you know that, you will also know what you must do.
The thing outside scratched at my window again. It sounded like it was testing the glass, looking for a weakness.
I heard Mom coming up the stairs.
I closed the book and shoved it under my pillow. "Oh my gosh!" I whispered. "School! I have to go to school. I can't do this right now."
Mom paused outside my room and knocked on the frame of my door. "Ten minutes, kiddo." Then she hustled down the hall toward her bedroom. Right before we left the house, she always ironed her clothes, then brushed her teeth with an electric toothbrush that was so loud and sucked up so much energy that I think it had been banned in thirteen countries.
My window slid open, just a notch.
I leaped toward it to clamp it back down, but my arm bumped the lamp shade on my nightstand, knocking it over. It tipped and fell straight to the floor, the lightbulb shattering into tiny milk-colored shards across my carpet.
I listened for my mom, but she didn't call my name. She hadn't heard anything from her bedroom.
My hands rested on the windowsill as my breath thundered in my ears.
Two red eyes glowed in the morning's darkness, framed by the rising sun. My window continued to slide open in tiny, jerky movements.
Veins instantly pinched tight all over my body, shocked by a burst of adrenaline. Huge, bumpy gray fingers slid through the opening. This was real. Xeno was real.
_Monsters_ were real.
A hand followed, a massive square hand with fingers thick as corn dogs. The hand smeared a gray paste against the white wood trim. The hand turned, palm up, and raised the window the rest of the way. The sunrise on the horizon made a soft pink glow.
I didn't want to look. The nurses had said it wouldn't be scary the first time I saw my mutilated leg, but it was.
It was.
I looked up and saw two red eyes burning brightly, watching me frozen and alone.
The window was completely open.
I would never be able to undo this moment.
The monster growled, a gritty sound, like it had gargled rocks. It came through the window all at once, an enormous gray creature, so tall it had to stoop to avoid hitting its head on my ceiling. Its skin glistened like wet clay, and its body was lumpy with mounds of strange flesh bulging out in odd places. It looked like a rough, unfinished sculpture, as if the beautiful part was yet to be revealed, still hidden beneath all the clay.
Goose bumps pricked my skin. The awe of seeing a creature so bizarre, so _monstrous_ and strange, and the shock of standing in the same room, breathing the same air, made my knees go weak. We were _both_ real, living in the same world, mangled girl and monster. How was that possible?
The creature had an enormous square head with strange symbols written across the forehead and hollow places gouged out for its eyes. It didn't have eyeballs, just a dark hole in the clay, with a red fire burning inside. Where lips should have been, a gray slash created two lips stretched so thin they were almost white. The lips revealed two rows of giant teeth, each tooth the size of a Post-it note. The creature watched me, cocking its head to one side, as if I were the curiosity.
Tears formed in my eyes. I carried too many secrets in my ordinary, everyday life. I couldn't handle one more. Why had I agreed to this? I would never be able to tell anyone what I had seen. This secret would make me lonelier than ever.
It growled again and I grimaced. The monster reached its hand out to me as if to reassure me, but I shrank back. Its skin looked wet and dirty and I didn't want to touch it.
Huge tears like marbles welled in the monster's dark eyes and fell. They rolled across my carpet and burst like water balloons. Its chin trembled, and with a pitiful cry, it jumped out the window.
From my mother's bathroom I heard the whir of her electric toothbrush at full speed.
I swallowed again and breathed through my mouth, slow and steady. Every nerve in my body was on fire.
I had to shut my window and lock it. That thing was still out there. It might come back and I wasn't ready for this. I didn't want this secret.
I leaned forward to grab my nightstand for support, and a slurping noise caught my attention. It was a sound like a toddler sucking its thumb. I looked down and saw the monster, its fingers in its mouth, curled up in a ball, rocking itself right there in our shrubbery. It reached up to wipe the tears streaming down its face, then popped its fingers back in its mouth.
I started to say something, on instinct. I had hurt its feelings and I was supposed to be sorry, but I wanted to slam the window shut. It was a freaking _monster._ How did it expect me to react? Ten minutes ago I hadn't even believed they were real, and now I wished they weren't.
It looked up at me just then. Anger, pain, and sorrow emanated from its fiery red eyes. I had the unmistakable feeling I was seeing myself through them. _I_ was the monster. I had hurt something because it was strange and different and ugly.
_"How you respond determines who you are."_ Xeno's words came back to me. I felt sick.
"I'm sorry," I whispered. I really was. "I'm sorry I hurt your feelings."
The monster scowled at me and ran away through the bushes. I lost sight of it as Mom called my name. It was time to get in the car.
The first bell had just rung when Billy approached me in the hall.
"Whoa, didn't you sleep last night?" he asked, falling into step beside me. "You look tired."
I kept moving so he wouldn't know how much he was embarrassing me. I wanted to be ignored, but he stayed as close as an IV pole.
"What's wrong?" he asked. I kept walking. There was no way to answer that question.
"Are you sick?"
My heart raced as sweat broke out on my upper lip. In the hospital, a nurse would have rushed in to check my vitals at this point. Diagnosis? I think I had a crush. Plus I was still reeling from an encounter with something I shouldn't have seen and would never forget. My body felt all wrong, like my brain had to rewire itself after this morning to make room for the new information and it didn't have the energy to remind me of how to move properly.
I didn't know how to act around Billy. He was so smart and quick that anything I said sounded stupid. In the movies, the pretty girls said cute stuff and flung their hair around. I had bristles. If I swung my head, he might lose an eye.
"Look, I love that you're as fierce as I am," Billy added, "and I thought that was why we were going to be friends. But friends _talk._ " He got in front of me and stopped, forcing me to as well.
He thought I was fierce? Because of an answer I wrote on a test? I wrote that because I felt sorry for myself. It was not exactly a battle cry.
I had to get to class. "I have a lot on my mind." I chose my first words carefully; then my mouth picked up speed. "And I didn't sleep well last night. I feel half dead, like a zombie."
"Don't be a pessimist," he said. "Zombies aren't half dead. They're half alive."
I grinned, and he moved back to my side. We resumed the walk to class, and a couple of girls said hi to me, loudly, while they smiled at Billy.
We passed Alexis, who stood at her locker getting a book out for homeroom. She watched us, her forehead wrinkling like when she didn't understand a question. How could I explain my friendship with Billy? Every girl flirted with him, but I didn't. Alexis knew that wasn't my style. When it came to guys, I had no style. So what was I doing walking down the hall with the best-looking guy in our grade? She might even be wondering if he was the reason I wasn't talking to her.
"So the dance Candy talked about?" Billy asked. "Will it be any good?"
I had to change the subject or be forced to admit I hadn't been to one yet. Our whole school district had a winter dance every year, even for the elementary kids. No one had ever asked me, and I didn't have the courage to go alone. Alexis hated wearing dresses—she said they made it more obvious that she had the curves of a toothpick—so it wasn't her idea of a fun night either.
"Well...," I murmured.
What if Billy asked me to go? My stomach bounced down to my knees and back up. How would I know if he meant as friends or as a date? He paused, looking at me, obviously waiting for a complete sentence.
I ducked into the girls' restroom.
Once inside, I pushed the door closed and turned. Candy was there, looking at herself in the mirror. And beyond her stood the strange woman with light blue-gray eyes and blond hair. She was wearing a long black gown of a style I'd only seen in history books. The sleeves of the gown moved as if an imaginary breeze were ruffling them.
A grim smile slowly curled her lips as she glided forward, glancing from Candy to me. Clearly Candy had no idea she was there. My body jerked back, reacting before my brain could process what I was seeing, or imagining. It was the strangest feeling, like being pulled apart, my inside and outside working at different speeds. I reached out to warn Candy, my face still turned up toward the specter hovering between us.
Then her image dissolved into smoke, disappearing through the cracks of the door and out into the hallway.
"That's weird. I was just thinking about you," Candy said. She glanced at me, then went back to brushing her long straight hair in sweeping, elegant strokes, as if she hadn't noticed anything at all. Of course she hadn't. I was the only one seeing things these days. Just when I thought being the odd girl out couldn't get any lonelier. I realized my hand was still extended toward her and I dropped it to my side.
Even under the fluorescent lights of the bathroom, her hair was gorgeous, flowing under the brush like liquid.
I wanted my old hair back so badly. When Candy looked in the mirror, did she see how beautiful she was? Did perfect-looking people _feel_ perfect? I almost started to ask. It would be weird if being perfect and feeling perfect weren't the same thing at all. Still, I kept my distance to avoid seeing our reflections side by side in the mirror.
Every stall door was closed, but there was no sound in the restroom except the soft whish of the brush through her hair. I refused to sit in a stall with her in here, so I pretended I needed to wash my hands.
Candy caught me stealing another glance. She shrugged as she sighed. "You're giving me the silent treatment, aren't you? Is it because of what Natalie said?"
The paper towels were on her side of the sink, so I either needed to reach over for them or walk around her. Out of the corner of my eye, I glimpsed something else in the mirror. Two brown scaly legs with claws on each toe began to slowly descend to the floor inside a stall, as if a monster had been crouched on a toilet the whole time I had been in here.
I swallowed and forced myself to breathe. Blood pounded through my veins like a freight train. I had never seen a monster at school before. Did that mean they were following me, or had they been here all along and I was only now able to see them?
Candy set her brush down with a loud clank and faced me. I glanced back at the stall as she cleared her voice, expecting me to pay attention. "Natalie isn't the brightest girl, but she didn't mean to hurt your feelings." She reached for a paper towel and handed it to me. The monster legs slid out from underneath the stall door, slithering soundlessly across the tile toward Candy. I noticed they were covered in bumps, not scales, with a layer of fine white feathers.
I wiped my hands and crumpled the paper towel. Chances were, the monster was here because I was the Guardian, even though I still wasn't sure what that meant. I only knew that it probably wouldn't eat me. Not that Xeno had promised they wouldn't, but it was a hopeful theory. That didn't exempt Candy. I needed to get her out of here. I started to tell her we both had to go, when she interrupted me.
"Just listen." Candy rested her hand on my arm. Surprisingly, her skin was warm. I had assumed she was cold-blooded, like all reptiles.
"I know we've never been friends. But don't judge me."
I picked her hand off of me and she blinked. "We should get to class," I said. One claw scraped lightly against the tile floor right next to Candy's foot. She didn't notice. I froze, not knowing what it was going to do next.
She crossed her arms. "I know you don't like me because you think I'm superficial."
One leg extended even farther, like the monster had elastic limbs. It began slipping around the back of Candy's legs.
"Maybe I just know something you don't." She leaned forward as if she was about to spill a major secret. "Girls are judged for our looks in a way that boys aren't." She waited for me to agree, one eyebrow raised.
I nodded.
"Girls who make an effort get treated better. Life is easier for them. So being pretty is really more like being smart."
"Candy, I—"
"No! Don't argue." She met my eyes with an intense gaze. "You may not like me, but at least I won't lie to you. This is an ugly thing to say, but the truth is, life hurts less when you look good and you fit in with everyone. And, Sofia, I can't help but think that you've had enough pain already."
She moved her hand to rest on my shoulder. "The point is, you have such potential if you would just try. Let me help you."
The toilet flushed and we both jumped. The monster's two legs were sucked backward toward the bowl and disappeared.
Candy moved to the stall and used her fist to bang the door open. It was empty. The red light on the automatic toilet flashed off and on. "I hate those things," she muttered. "They always flush without any warning."
I exhaled in relief, although I felt a pang of guilt for whatever monster had just gotten unexpectedly sucked away. Monsters were old and automatic flushing toilets were new.
"Candy, we better get to class. The bell is going to ring any minute," I said.
She looked annoyed that I hadn't fallen at her feet in gratitude. "You don't even want to talk about this?"
I shook my head. I didn't know if that monster could swim. He might be back. "You might get in trouble if you're late," I reminded her. "You're on the student council. The teachers don't mind if I'm late."
"Exactly!" she said. "Because everyone feels sorry for you. But that won't last forever. You have to decide who you want to be before it's too late. You'll either go back to being the sad girl without a lot of friends, or you can use this opportunity to become everything you ever wanted to be. The whole school would get behind you! You have to decide what you want, and if you do want to change, you'll need a plan. You'll need me."
I grabbed for the door, biting my lip so I wouldn't cry. The door swung open, hard. Someone on the other side pushed as I pulled. I fell backward and landed on my rear end. Alexis stood over me, looking horrified upon seeing what she had done. Billy peeked over her shoulder. He must have been waiting for me.
Candy moved first. She grabbed me from behind, lifting me from under my shoulders, and helped me to my feet.
"I thought you two were supposed to be friends," she snapped at Alexis. Alexis inhaled like she was about to explode.
"No one here is my friend, okay?" I said. It came out all wrong. Alexis winced as if I had slapped her. Billy frowned, as if I had hurt him too.
"Think about my offer," Candy said. Then she brushed past the three of us on her way to class. "I've already started a list of things you'll need to do."
I didn't know what else to say, so I just followed her. It was the right direction to walk but felt like the wrong way to go.
I dropped my book bag on the floor when I got to the kitchen. Mom was right behind me, and picked up one of the books that fell out.
"Ewww," she said, staring at the book cover. Mom wrinkled her nose like she smelled something rotten. On the cover was a snarling monster with a body hanging limp from its jaws. Before I could stop her, she dug out my other assorted titles. I had planned on returning them all, since Xeno thought they were useless.
"I didn't even read them," I said, trying to grab the book from her hands. She lifted her hand higher, staring at me with a frown.
"Well, good, because this is not appropriate reading material," she said. "Besides, I don't want you worried about things that aren't even real."
"Why is everyone on my case today?" I groaned. _And who knows what's real, anyway?_ I wanted to shout.
Mom stared at me in silence for a long moment before she replied. "I am not 'everyone.' I am your mother." She suddenly sounded tired. "And if there's something going on, we need to talk about it this weekend, okay?"
I had forgotten it was a long weekend, since tomorrow was a teacher in-service day. Normally I'd have been excited about relaxing and watching TV for three whole days, but I was heartsore from a bad day at school; plus I was nervous about what the weekend would bring. All I seemed to do was accidentally hurt people's feelings. Maybe monsters were unsafe, but apparently, so was I.
Forty minutes later, Mom moved to answer the doorbell when the pizza arrived. "Let me get it," I offered. "You relax."
I wanted her to think I was trying to make up for being moody, but in reality I had to make sure it wasn't a monster ringing the doorbell. I still had no idea how this Guardian thing worked. I needed to get upstairs as fast as I could without hurting Mom's feelings and figure all this out, because maybe next time there wouldn't be an automatic toilet flush to save me.
I paid the guy and carried the pizza into the kitchen while Mom stood in front of the TV. She had been searching to find a movie for us to watch but was transfixed by a commercial. As she watched, text rolled across the screen like credits after a film, listing every dangerous germ and virus that lives in the average home. The only defense was a new disinfectant conveniently priced at $29.95.
"Mom, seriously," I said. She worried all the time about me getting sick. More than once I'd caught her in the store spraying Lysol in the air and sniffing it. I worried she would make me use that instead of perfume.
She snapped off the TV and came into the kitchen to set out paper plates and napkins.
"How's Alexis?" she asked, then cleared her throat. I could tell the commercial had upset her.
I rummaged in the spice cabinet, glad to have my back to her. "Fine." When I found the red pepper flakes, I turned around and tried to think of a way to change the subject.
"I told you she was welcome to come over," Mom said, her voice sounding a little too bright. She was scared about my health, but I knew she really wanted my life to go back to the way it was, even if that meant catching colds and stomach bugs from other kids.
"She's busy with track," I replied, hoping Mom would make the connection. Alexis still had her normal life, and I no longer fit into it.
"You could call her anyway," Mom said.
She was not going to let this go. She was going to keep asking questions about Alexis, because she thought I needed a gentle push to talk about it. I didn't need a gentle push. That was like giving a nervous woman on a building ledge a gentle push to get the conversation going.
Some things are hard to explain, especially to your mom. And with my mom and me being so close, I shouldn't have had to explain anything at all. But it was just the opposite.
Maybe I just sucked at relationships.
"Can I eat later?" I asked. "I'm not that hungry. I just want to go upstairs and rest."
I tried not to look her in the eye as I said it.
"Should I wait for you?" Mom asked.
"No," I said softly.
But as hard as I resisted, I finally looked up and saw the hurt. There was a wall between us. I walked upstairs slowly, half hoping she would insist I stay downstairs and talk, but she didn't. I had put the wall up, so she was waiting for me to pull it down. I needed to find a way to do that, but it's hard to pull down a wall when you're not sure what needs to remain protected. Xeno and his monsters were strangely like chemo. I had agreed to be the Guardian, but I hadn't expected all these other problems.
In cancer centers, there are two versions of the word "unexpected." You can have an unexpected benefit, and that's wonderful, but most people get unexpected outcomes, and those are all different and usually bad. I still didn't know which version of "unexpected" Xeno's book and his monsters were. But he had asked my permission before we began. Since last October, my world had been defined by one word, and it wasn't cancer. It was "no."
Alone in the dark, I didn't really want to open his book. Monsters were the strangest, scariest, ugliest things I had ever seen, and I couldn't trust myself not to hurt them. I didn't think I was going to be a good Guardian, especially if I screamed every time I saw a monster.
But being the Guardian was the first chance I'd had to say yes to anything. Since my diagnosis, the whole world had become one big no. I couldn't walk unassisted, I had no appetite, no energy, and no guarantees about the future. Whatever I wanted, whatever I would normally do, the answer was no.
I had said yes to Xeno for selfish reasons, for a chance to escape, but the kick in the pants was that being the Guardian just brought me right back to the basic problem.
Me.
Downstairs, the heater turned on. Warm air blew from the vent over my bed, but I was still cold when it tickled the hair on my arms. I pulled the bedspread up around me and scooted down under the covers.
Dogs all over the neighborhood began to bark with a sudden urgency. Just like the other night. Had I never noticed them before, or did I now hear animals in a powerful new way?
I reached out for my lamp but then remembered the bulb had broken. Two red eyes were staring at me through the window.
It was back. My eyes adjusted to the darkness around him. In the soft moonlight, I could see the huge square outline of his head, with a face like cement that had hardened in a thick puddle, rough and uneven. The eyes were like two huge dark caves.
We just stared at each other.
He had tears streaming down his face. Again. One huge tear after the other as his breath fogged up the window.
Was he still upset about my reaction when I first saw him? There wasn't anything I could do except try to apologize again. Under my bed, the book began to glow.
If he had the courage to come back, I should find the courage to try one more time, I thought. I owed him that. I opened the window.
Like a frightened, badly scolded child, the monster crawled into my room and stood, his head hanging low.
The book's light grew brighter.
A stray snowflake swirled in through the open window. The monster and I watched it flutter in the space between us. It landed softly on the carpet, a hint of sparkle in the moonlight. The monster reached out with one foot, each of his toes as big as my fist. He nudged the snowflake. His eyes were wide with wonder, like the snowflake was a small miracle. In Atlanta, snow was always a miracle, so I grinned, but then pressed my lips together so he wouldn't think I was teasing him.
I pointed to the night sky.
It whirled with snowflakes, each one falling and landing in a dark place, like a forgotten star.
The monster stared, then looked back at the snowflake between us. I wondered why he would think that just one was special. The whole neighborhood was full of them. Why care about a single snowflake when there were thousands, maybe millions more?
The snowflake melted into the carpet and disappeared. The monster inhaled sharply, witnessing this magic. Its chin began to tremble, like a child who had lost his balloon to the wind.
I pointed to the sky again.
He stepped toward me and one huge arm wrapped itself around my waist. His flesh was warm and hard, like a mug just out of the dishwasher. Before I could scream, he was climbing out the window. I clung to his massive body, my arms barely able to span his chest. He climbed down the side of our house like it was as easy as walking. With two great steps that shook my body, we stood under the streetlight by my house. Snow fell all around us, twinkling like fairy lights. He set me down and nodded like a child waiting for me to open a present.
I didn't move. It was so quiet. When the neighborhood kids woke up to snow, they'd go nuts. It never snowed this much here. The kids would be dizzy with the realization that impossible things still happened in the world. They had not yet learned to hate the unexpected.
The monster extended his enormous hand, palm up, catching a snowflake to give me. I opened mine and cupped them, trying to catch more. He sucked in his breath, delighted. He was so goofy I couldn't help but laugh. Maybe I could learn to love the world again too.
"Watch this," I whispered. Tilting my head back, I opened my mouth wide and stuck out my tongue. Snowflakes swirled around my face, landing on my lashes and falling into my mouth.
He tried it, but his mouth was just a black hole. Still, after swallowing at least a dozen, he stamped his feet in pleasure. I wrapped my arms around my body; the cold had started to hurt me. Without a sound, he wrapped an arm around me and we climbed back through my window.
I didn't have any magic, but he did. And he didn't just have magic; he saw it too, in the simplest, most ordinary things. I had never believed in monsters, but now I realized it wasn't just monsters I needed to believe in. I needed to believe in the extraordinary. This monster believed in a world that still had a little bibbidi-bobbidi-boo in its back pocket.
Inside my room, I shook off the snow, still smiling, then reached toward him to brush him off.
He shook his head.
"Why not?" I asked. The monster glanced at me, embarrassed, before hanging his head again. Then he turned, slowly. At first I was afraid he was going to leave, but then I understood. He wanted to show me his back. His raw red back was littered with dozens of little black-edged holes scattered all over his flesh. Each hole was no bigger than a piece of confetti.
Biting my lip for courage, I touched one of the holes. The monster whimpered. He had been so kind to me, and I hadn't even known he was in pain. He wasn't just strong in his body. He had a strong heart.
"I know what these are," I whispered.
I turned the monster around to face me. "You were shot, weren't you? Was it—a shotgun? A pellet gun?"
The monster opened its mouth and the saddest sound I have ever heard, the sound of every pain, of all pain, the pain of being alive and the pain of wishing you were dead, poured out.
"I know," I whispered.
I did.
I held my breath. Mom did not call my name. The television volume was too high for her to hear anything.
The book glowed so brightly it almost pulsed, so I dug it out from under the bed. I was moving faster than I ever had in physical therapy, but I noticed this as if from a distance. Right then, it didn't feel like I was even in my body. I heaved the book out and up in one try. It fell open to the page I needed. Elaborate edging highlighted bold black letters painted long ago. A smudged map was drawn alongside the monster's image. The names and boundaries on the map had been changed many times.
> # THE GOLEM
>
> Origins: Created by a Jewish rabbi in the 1500s for defense against enemies. First documented in Prague, the largest city in the Czech Republic. Sightings now range all over the world. His forehead is marked with an ancient language and the Guardian must never touch the symbols, for they contain the mystery of creation and death.
>
> Nature: Fiercely protective of those he loves, the Golem may be injured if humans see him and lash out in fear. If he is wounded by humans, you may use human first aid to care for his wounds. A good and kind Guardian will also do what she can for his heart too, which is the most vulnerable part of the Golem.
>
> Of all creatures, in fact.
The page fluttered and turned over to the next. Here were more detailed notes: what he ate, where he roamed, what kinds of injuries he was prone to and how to treat them. Running my finger down the page, I found what Xeno wanted to show me:
> For minor flesh wounds, do not use water. Use alcohol and be most gentle.
"Stay right here, okay?" I whispered to the Golem. "I need to get something."
He nodded and sat on my bed. It sank deep in the middle, and I hoped the frame wouldn't break. I had to move fast.
I made my way downstairs. Mom was on the couch, her face red and swollen. She saw me standing there and muted the television.
"Sad movie," she said, but I knew she was lying. Maybe she didn't tell me everything either.
"I just need something from the kitchen," I said. That's where we kept the bandages and ointments for cuts and burns. With my mom's cooking skills, the kitchen was the most dangerous room in the house.
"You hungry?" she asked. "Want some pizza?" She started to stand up and I overreacted.
"No! Just watch your movie, okay?" If she saw me getting into the first-aid kit, she would want answers. She'd demand to see the wound.
Her eyes widened and she sat back down. "Okay, then."
I promised myself I would make up for this in the morning. First, though, I had to take care of the Golem.
Once the movie was back on, I made my way to the cabinet above our mixer, where Mom kept our first-aid supplies. I grabbed a handful of bandages, Neosporin, and a pair of tweezers, stuffing them in my pocket. There was no rubbing alcohol, though, and Xeno said I needed it, but then I spied a few bottles of alcohol. The adult-beverage variety, not the kind used for first aid.
Mom rarely drank, so the bottles were full. I wasn't sure if I could use them as a substitute, and even if I could, which one? I read the labels. One hundred proof? Eighty proof? Proof of what? The labels didn't say.
"Whatcha in the mood for?" Mom chirped. I whirled around, a guilty look plastered across my face. She was standing at the kitchen counter. Her eyes narrowed when she saw which cabinet I had open.
I faked a smile. "You know. Just anything."
"You're looking at the liquor bottles?" she said, her voice deceptively flat, like a sinkhole.
"Well, obviously I'm not looking for alcohol. Why would I be looking for alcohol?"
Mom didn't reply, and I realized she was waiting for me to answer my own question. "Where do you hide the chocolate from yourself these days?"
She stared at me, her face set and blank, before she walked over and reached for the cabinet door on the opposite side of the alcohol. Inside was a new bag of chocolates.
"Busted," she said, and then laughed. The smile didn't make it to her eyes, though. They were still suspicious.
I grabbed a chocolate and pecked her on the cheek. "We'll hang out tomorrow. Promise."
I made my way back to my bedroom with a burning lump in my throat. Mom couldn't talk to me and I wouldn't talk to her, and in the middle of all this not-talking, I had to take care of an injured monster.
I sat behind Golem on the bed, my back against the wall, so I would be closer to eye level with the wounds. I had to check each little hole and remove the pellet. The Golem was quiet as I worked. I didn't want to stick the tweezers in and poke around; that would hurt. I gently pressed with my fingers on either side of the hole, hoping to ease the pellet up to the surface, where I could easily grab it with the tweezers. I did that, one pellet hole after another, until I had removed forty-three pellets and dabbed ointment on each wound.
He whimpered but I kept working, the television blaring the whole time. When I thought the pain was too much for him, I rested my hand on his shoulder and let him gather his courage. I needed some too. What made people so mean? Why would they hurt someone just for being different?
After the pellets were removed, I eased myself to sit beside him and took his hand in mine. It was like holding a big canned ham. The Golem bowed his head, unwilling to look at me. I rested my head against his shoulder and sighed, and even though I couldn't see it and he technically didn't have lips, I think he kissed the top of my head.
Then he jumped up and ran to the window, leaping into the darkness.
I hopped up and followed, sticking my head out to look for him, but he was gone. Glancing up, I saw the bright full moon above me, illuminating the street below. A woman stood under the streetlight, face turned in my direction, her long blond hair blowing in the night wind.
I ducked my head back in and shut the window, fast. The night noises grew louder—the barks and growls and howls and slithering things. I looked back out the window, but the woman was gone.
The book was open and glowing, so I turned to the blank page Xeno used for communicating. "Is that what a Guardian does? Takes care of monsters? The ones who get hurt?" I asked. The page remained blank. "Monsters need people sometimes, right?"
> In truth, it is people who need monsters. How do you feel?
I chewed my lip for a second before answering. "Better, I think. It felt good to help him. And it felt good to play in the snow with him." I laughed softly, thinking about his delight in one little snowflake. "But there's a strange lady following me and there are so many questions I have for you. I have to know how everything works."
> I've been alive for over two thousand years and I still don't know that.
>
> Ask me your first question.
A hundred filled my head at once. It was hard to know which to ask. How did I know which one was the most important? Mom might be coming to bed at any moment. I would run out of time before I asked everything.
"Monsters aren't like animals—your book says that. People invented them, but they aren't just stories. They're real. How is that possible?"
> What is underneath you right now?
I was sitting on my bed, but didn't he know that? "Uh, my bedspread."
> Your bedspread began as a thought. Someone imagined it before it came to be. The home you live in? Someone first imagined that too. Imagination made the world. Why, then, would you think that imagination couldn't make a monster too? Did you think the boundaries of imagination stop at bedspreads and buildings? Imagination is the most practical and powerful force on earth.
"But how did a monster go from being in someone's imagination to walking around on the earth?"
> The entire world pulses with wonder and creation and magic. How did your bedspread go from the mind to the mill? How does the painting go from the soul to the canvas? Every one of us does some bit of magic like that every day. Do not be surprised, then, that some people can make their imagination walk out into the sun or howl under the moon. Some people perfect their magic.
"How?" I asked.
> I never learned that secret. I studied what was already here, not what could be. That was what the Master asked of me.
Something still didn't make sense. "But we're frightened of monsters," I said. "Why would we make something we're scared of?"
> By creating monsters, we chose what we would be most afraid of. We set limits to it, told it where it could live and what it would look like. We gave fear a place of its own and we were free to enjoy our lives once more.
Mom turned off the television. The sudden silence made me jump. Time was running out for the night, and I had only asked four questions. There was still so much I needed to know!
"How do they find me?" I whispered urgently.
> You made a decision to love what is unlovable. They will always be able to find you. You are a light in a dark world, Sofia. But do not trust the darkness. Monsters are not the only things that dwell in the shadows.
I heard my mom on the stairs, so I closed the book, wincing. I needed more time, but it wouldn't be tonight. Mom opened my door to say good night, and I stood up to hug her. If she felt my heart pounding through my thin shirt, she didn't say anything, and so neither did I.
Friday, February 28
When I peeked out the window early the next morning, there was no blanket of snow on the ground, but only a few glistening flakes clinging to the dead grass. Atlanta weather was a real heartbreaker. I got my leg on and went downstairs in my pajamas and fluffy robe for a glass of orange juice. Mom was working in the backyard, so I lugged myself outside to sit in a patio chair and join her. There was always so much to be done in the yard every fall, but last year Mom had been too overwhelmed to care about pulling dead flowers, freshening the mulch, and pruning bushes. Now that the worst of winter was hopefully over, she had a brief second chance to get it done. Everything dead had to be cut off and cleared away. If not, when spring came we wouldn't have any blossoms.
Spring was the most beautiful season in Atlanta, but preparing for it was hard work. I was glad Mom felt strong enough to do it.
Besides, I wasn't much of a morning person. I felt like my head had been swaddled in a thick cotton blanket, making it hard to think, maybe because there was too much to think about. I was glad Mom had gotten the day off to stay home with me. Her office friends had pooled their sick and vacation days to give to her, which not every company allows but hers did. They valued her.
I understood why.
I liked watching Mom putter around in the backyard, nothing burdening her except boring, everyday decisions. Boring, everyday decisions were a luxury we missed.
Spring wouldn't be here for at least another month, so we still had mostly cold days, but the sun shone bright and warm. Mom attacked the dead flowers from last year with a vengeance. I saw a row of loose dirt. She must have planted some bulbs.
"Hey, Mom," I called.
She straightened up, replacing her frown with a halfhearted smile, then grabbed the garden hose and soaked the bulbs she had just planted.
"Isn't it too late to plant those?" I called. Bulbs had to be planted in fall, not a few weeks before spring.
She didn't respond even though I knew she'd heard me. She concentrated on watering the dirt. I watched, and waited for her to say something.
In the cascade of water leaping from the hose, a rainbow shimmered. Above her, in our giant oak tree, still bare from winter, I saw a perfect, round bird's nest from last spring. A family of baby birds had lived right in my backyard. I hadn't even known they were there.
I closed my eyes for a moment and lifted my face to the morning sun. A bird sang in the distance, and dead leaves rustled in the breeze. I could hear a squirrel scamper across the dry grass by the fence.
I loved our backyard. I loved our little world.
Then Mom screamed, the noise followed by a hard thud. Something else was here with us.
I jumped up and lost my balance. Falling face forward onto the concrete patio, I scraped the palms of my hands. Hot tears gouged my eyes as I scrambled to get up. Then Mom screamed again. Another thud, then another. I grabbed the patio table, pulling myself up.
Moving fast, I found her around the corner by the shed. She raised the shovel over her head, then slammed it to the ground. I didn't see any monsters. I didn't see anything at all. Just Mom, a shovel, and the empty shed.
"Mom!" I yelled. "Stop! What's wrong?"
She looked at me like her mind was somewhere far away. "It was a fuzzy caterpillar. They can sting."
We stared at each other for a long moment, and my mind flooded with questions. Caterpillars didn't appear here till summer, so something must have brought one from a warmer climate like a dog carries fleas. Which meant something had been hanging around in my backyard.
Mom wasn't focusing on me; she seemed to be remembering something terrible.
I didn't feel too good about any of this.
"But, Mom?" I said. "If you kill all the caterpillars, there won't be any butterflies in the spring."
She shook her head as she stared at the pulverized blob on the ground. She set the shovel in the shed and walked toward the house.
"That's the price we have to pay," she muttered.
A dull-colored cloud rolled over the sun as she walked back inside. I hated winter and wanted it to be over.
I stayed outside because I didn't know what to do. My life was changing but hers wasn't, and I knew it was my fault. Maybe if I could explain what I was seeing and feeling and learning, she wouldn't hate all the pain we had been through anymore. Because I could see now that pain isn't a death, but a door. Everything beautiful has a violent birth: cocoons tear, seeds crack, babies come into the world with their mothers screaming and panting and sweating and people call it a miracle. All my pain, the grief of feeling so alone, the torture of feeling like a freak...maybe it had been not a death but a door. And who knew what was going to come through it? Doors work both ways, you know. I might find a whole new life, or a new life might come find me. And pain had been the only price. Was that really so bad? It was time to stop hating the past for tearing a giant hole in our lives. It was time to take a step into the dark passage and see where it led.
I had to try and tell her one more time about Xeno's book and the monsters.
I heard the clangs and dings of Mom in the kitchen, setting out bowls and spoons for our breakfast. As I walked back into the house, something rustled in the bushes behind me and hissed for my attention. I had no idea what it was, and I didn't want to know. Not right now. I kept moving. My mom was hurting and she needed me.
Inside the kitchen, I sat at the table, watching her. I wasn't sure how to begin the conversation. As she set out the cereal and milk, I took a breath for courage and tried to decide what to say first. She began to pour the cereal for me but then stopped, holding the box in midair.
"The night you were born, I stayed awake all night long, just looking at you," she said. I would find a way to tell her when she finished. "I was afraid that if I closed my eyes, if I even blinked, God would change his mind and ask for you back." She set the box on the table and sat down, smiling at the memory. "You were just that perfect. I was sure God had made a mistake by giving me the most perfect baby ever."
I grinned without even meaning to. I loved this story. She told it to me a lot.
A shadow passed over the sun and the kitchen lost all of its morning light. Her face darkened. "And then, when you got sick, we were back in the hospital, with you asleep and me afraid to close my eyes. One night, I heard a nurse tell another mother that it was time to let go."
She droned on, like she wasn't even aware I was in the room. This was a story she had never told. "The mom let out this wail, this howl like she was the one dying. I climbed into your bed while you were sleeping, and held on to you until my arms shook. I've never been so afraid in all my life."
Her eyes filled with tears, but she blinked fast. She pushed the cereal box in front of me and wiped her eyes. I wanted to tell her it was all going to be okay, that maybe everything had happened so that I could become the Guardian and protect the abandoned creatures of the night, but all I could think of was that caterpillar she had just smashed.
"I don't know how to protect you," she whispered, her face pale. "All I want is for you to be happy, but this world is not a safe place."
I reached my hand across the table and held hers. "It was never safe. We just didn't know that before." Taking a breath, I paused to tell her that I had found something better than being safe. "I can never be safe again. Because—"
She squeezed my hand and gave me a sharp look. "I know you've been depressed. But don't talk like that. We'll find a way. You will have a long, normal life, I promise."
I didn't want that. I didn't want to settle for any kind of normal, because that meant I might never have anything better. I needed something more.
Mom stood and grabbed her empty cereal bowl. Putting it in the sink, she ran water over it even though she hadn't used it. I think she was wrestling with big questions too, the kind of questions that have lots of easy answers but no great ones. I wanted to share my secret with her, but the answer, of course, was no. Or _not now,_ which was almost the same thing.
We spent a quiet day together. She gave me some uninterrupted alone time in my room to work on homework, but I used most of the time to study the Bestiary. It wasn't just monsters I had to understand; it was geography and history and science and medicine. I swear my skull hurt from the rapid expansion of brain cells. At least I was studying important subjects, and that could help my grades. I had always liked being an A student, but recently I had begun to understand that I _needed_ to be an A student. Without a scholarship, college wasn't going to happen for me.
Finally, we said an early goodnight at the top of the stairs. Mom went to run the water for a bath, a magazine tucked under her arm. She shut her bedroom door behind her; then I heard the bathroom door shut too. I had a two-door warning system.
Alone in my room, I grabbed the book for more studying. Mom could never, ever know about its existence. She had enough to handle. Normal girls didn't talk to dead men or give first aid to nature's nightmares. At least Golem was sweet. Maybe I would ask Xeno if I could just be the Guardian for Golem. I flipped through the book. Some of these monsters were terrifying; plus, they ate people. I also wanted to ask Xeno why there wasn't an entry for Bigfoot or the chupacabra. Did that mean they weren't real monsters or that no Guardian had ever treated one? It would be incredible, I thought, if I got to add a page of my own to this book.
As I turned another page, a horrid smell hooked me by the nose. On instinct, I dropped the book and threw one hand over my face to shield it from the stench.
It smelled like fish. Dead fish. Dead, rotting, stinking fish.
My room suddenly grew cold, and tiny goose bumps popped up along my arms. I exhaled slowly, trying to control my breathing and stop myself from gulping in the putrid air. White mist floated out from between my fingers, like I was breathing outside in winter.
Something wet began to slither up my calves toward my stomach. A slimy, barrel-shaped tentacle, heavy and cold, smashed against my chest, pinning me to the bed. In the moonlight, I watched another tentacle wave. It crept across my chest like a snake and struck my forehead. I couldn't move.
Little bursts of light danced in my vision. The tentacle on my chest had cut off my air. The book lay on the floor, glowing green and urgent, but I couldn't reach it.
A wet scraping sound, like a sack of wet clothes being dragged across my floor, broke the silence. A face emerged from the darkness and loomed over mine. It hissed softly, watching me. The monster had almond-shaped eyes with pupils that opened wide and black against its stark white eyeballs. It didn't have control of them, though. Each eye roved independently around the room and spun whenever the monster hissed. I didn't get the feeling this monster was anything like Golem.
I tried to turn my head. The monster smelled like it had digested something awful and dead, something that might come back up right in my face.
Thousands of scales overlapped along its skin. When it turned its head to look out my window, I saw gills along its neck, and a mouth like a fish's, with two thick pieces of rubbery cartilage for lips, which opened and closed silently. Dead leaves clung to its body. This must have been what was in the garden this morning, and I had ignored it. Now it was mad.
The monster turned its head back to me, lips opening again, revealing row after row of tiny white triangles, each with a jagged edge. The teeth looked like they were made for tearing meat away from bone. A low giggle escaped from its open mouth as its eyes spun like loose marbles.
It was going to eat me.
Xeno would know what to do. I had to reach the book. I tried to move my right arm, but the monster pinned me tight. It looked me up and down, as if trying to decide where to start eating first. All I could do was wiggle the fingers on my left hand. I struggled to move my thighs, to dislodge the monster, but they were trapped. Had I survived cancer just to be eaten by a filthy monster with wonky eyes?
Its mouth opened, teeth emerged, and, whipping around, the monster sank them into my prosthesis. The teeth ripped through the fake skin and then scraped against the metal rod inside.
The monster snapped its head back up in a snarl of protest. I took my chance and slid to the floor. Reaching fast, I grabbed the book. It had flipped open to the blank page, glowing bright white. The words burned black.
> TELL IT YOU HAVE A CUCUMBER.
"That makes no sense!" I said.
The monster tensed its muscles, preparing to pounce, its teeth like tombstones in the moonlight.
"I have a cucumber," I whispered.
It stopped.
Settling back on its haunches, if you could call them that, the creature cocked its head like a puppy, a wide grin splitting its face. Clear juice dripped from the monster's teeth. I got a good look at the whole of it. The creature had the body of a hairy man, or gorilla maybe, and a bunch of tentacles that braided together to function as arms and legs, or released and went off in different directions, like a remarkable combination of fish and man and loose marbles.
"Rrreowr?" it asked, the sound lilting up at the end like a question.
The book glowed so bright I could read the words out of the corner of my eye.
> GET THE CUCUMBER.
"My mother is up here," I protested. "I'm not leaving her alone."
> GET THE CUCUMBER.
Did we even have one? I tried to remember what I had seen in the fridge.
> NOW.
I struggled to stand. My real knee felt like jelly.
"Please, please let us have a cucumber," I whispered. I made it to the bedroom door, expecting to be pounced on and eaten at any moment. When I turned back, the monster was crouched on my bed, waiting. It seemed very happy.
The house was dark and no light was visible under my mom's door, but I heard water running in her bathroom. I made my way down the stairs in the darkness, cautiously testing each step. I could not trip or stumble. Mom might hear and come out of her room to check on me, and the monster would devour her. Whatever this thing was, I didn't like it.
I thought back to the girl in the hospital, her nightgown stained in blood. What had happened to her? Did she not have a cucumber? Why hadn't Xeno mentioned that produce would be involved? And why did I always have more questions than answers?
I made it to the fridge, breathing heavily. I grasped the cold metal handle and pulled. White light flooded over me like I was at the gates of heaven. I reached for the vegetable drawer with trembling hands, holding my breath.
A cucumber sat on a bed of wilted lettuce.
I was never so happy to see a vegetable in all my life. I took it out and held it above my head in triumph. "Yes!"
A growl floated down from my room. I shut the fridge quietly and quickly started my way back up. The fridge light had blinded me. I worked my way slowly up the stairs. I heard the radio playing from behind Mom's door and the water still running.
I opened my bedroom door to see the monster bouncing up and down on my bed. It seemed to like the spring action of the mattress.
It snatched the cucumber from my hands and cradled it like a baby. Then the monster lifted it and bit one end off. It stopped every few bites and held the cucumber up to the light, clicking its teeth in delight before resuming. While it ate, I noticed a fishhook embedded in its lower lip. The wound was swollen and oozy. It was infected. The poor thing had been in pain when I refused to help this morning.
I remembered when I was about five or six and went on a field trip to the lake with my summer day care. A little boy had gotten a fishhook caught in his finger when he was playing near the shore. Our teacher called the park rangers for help. I could still recall exactly what the guy had to do to remove it. When you're a little kid, scary medical procedures sear themselves into your brain.
I moved slowly, trying not to disturb the monster. It didn't have a lot of the cucumber left. In my dresser I kept my jewelry-making tools, a leftover from craft days at the hospital. I pulled out a pair of wire clippers and promised myself I wouldn't hurl if this got nasty.
As I approached the monster, its eyes swirled faster, watching me. It hugged the cucumber tightly and growled.
"No," I whispered, shaking my head. "I don't want your cucumber." I pointed to my lip and then to its lip.
The monster uncurled one tentacle, reaching up to touch the hook. It whimpered.
"I can take that out," I said. I pantomimed how I would pull it free, leaving out the part with the contorted face of anguish.
The monster looked at the cucumber sadly, then set it on the bed and nodded.
It had already tried to eat me, and it was not going to like what I had to do, but the wound looked bad. I knew how dangerous infected wounds were, and how they made doctors really nervous. If I didn't help, the monster might get sick or die.
I just wished I understood why that was a bad thing. This monster didn't seem to have any redeeming qualities.
Lifting the clippers, I reached for its lip. The monster's eyes stopped swirling and came to rest on mine. Beneath the thin green flesh, I could now see a blue heart beating inside its chest, shooting blood through the abdomen in messy squirts.
The creature's lip felt like cold rubber. Rust had begun to eat through the hook, so I hoped the next part would work. I snipped off the eye end of the hook—the end where the fishing line had been tied to it. The barbed end was still inside the lip.
I pushed with one hand and held the lip taut with the other. The monster cried as it squirmed, but I forced the barbed end of the hook all the way through and out the other side. Because I'd made the top a clean edge, the rest of the metal slipped through the hole smoothly.
I held the hook out to show the monster. It opened its mouth, flashing serrated teeth. The eyes were spinning so fast they looked like they were in a blender. It was furious.
> OPEN THE WINDOW.
I obeyed Xeno immediately. The monster's tentacles reached for me.
"You're forgetting your cucumber," I whispered.
It stopped and glanced back to the bed. Grabbing and cradling it out of my view, the monster flew out the window. I slid the window shut and locked it too. I did not want any more of those things in my room. Ever. Its smell still lingered.
The letters erased on the page and new words formed as I watched.
> That was a Kappa. It eats children.
Cold sweat broke out across my forehead. "What?" I said. "It was going to eat me? And you let it come to my bedroom?"
> If you follow my directions, you are in little danger.
"Can you define 'little'?" I snapped. "That thing bit me."
> Kappas are from Japan and have only one love greater than the flesh of children.
"Cucumbers?" I guessed. "But how did you know we had a cucumber? And what if we didn't?"
> Sofia?
"What?" I said.
> Close your eyes and listen.
Outside, the world was alive. I heard insects singing and cats meowing to each other through the bushes across the street. Dogs barked over fences, passing messages down the block, and a wood frog kept time with deep, happy bellows. I had spent thousands of nights in this same room, with these same noises. But only just now, tonight, my heart beat loudly, wanting to join in their song.
It was a strange feeling, one that I recognized but couldn't quite remember, sort of like when you see a friend from summer camp a year or two later. You remember them, but it takes a minute to remember from where exactly.
And then it hit me.
I was part of a bigger, mysterious, and beautiful world. It wasn't a cold and gray place unless you chose to live in the shadows. Had I only forgotten that, or were these nightmarish creatures teaching me something new?
I inhaled and exhaled in slow, deep breaths. A smile seemed to bubble all the way up from my toes and stretch across my cheeks. I opened my eyes and the blank page swirled with new letters.
> You are changing, Sofia. And you are becoming a very good Guardian.
"I learned how to do that a long time ago," I said, thinking he was talking about the fishhook.
> You were kind to the Golem because you realized he was sweet. But you were kind to the Kappa simply because I asked you to be. You have great potential, Sofia. A Guardian must watch over all, the terrible and the good.
"But why?" I asked. "Why protect all of them? Some of them eat people. Why not just protect the ones who are nice? It's not like we need all of them."
> Throughout history, every culture created its own monsters. Why?
Social studies wasn't my best subject, but I tried to remember what Xeno had taught me. "Because we wanted to choose what we were afraid of."
> Yes. A common enemy, a common dread, is a strong bond between people. It forces the most unlikely individuals to work together. It gives fear an honorable focus. Monsters made us better people.
"But no one believes in them anymore," I pointed out.
> We have chosen, instead, to fear each other. Perhaps one day, people will remember the wisdom of governing through imagination instead of fear. You must protect the monsters, Sofia, because one day the world will need them. When we grow tired of war and suspicion and hate, we will need them.
Would the world really be a better place if everyone was terrified of monsters? I thought about school. If we had monsters to fight and fear, I might be thankful for the kids who were mean and strong. Bullies would actually be useful.
Maybe monsters had to be bad, I thought, so we could learn to be good to each other.
"Xeno, there's something else I need to tell you," I said. "A woman has been following me. I think she knows about the book. I didn't tell her, I promise."
> I know.
"Who is she?" I asked. "What does she want?"
> It is late, and that is not a story to be told at bedtime.
Saturday, March 1
As soon as I heard Mom stirring the next morning, I got up. All night I had worried about the Kappa's bite to my prosthetic. How would I explain the damage? Mom was going to freak. I got my leg on and pulled on some sweats before making my way to the bathroom across the hall.
Wincing from the glare of the bathroom lights, I moved to sit on the covered toilet seat and bent over to inspect the prosthesis.
It looked pretty good, actually. The imitation-flesh material had sealed back over the teeth marks, and if you didn't look too closely, the wound wasn't obvious. Out of curiosity, I pried the bite marks apart to look inside. A glint of surgical steel winked at me. I let the wound snap closed, then ran my finger along it. I'd almost forgotten how strangely beautiful the inner workings of the prosthesis were, like a watch.
All that incredible work inside the leg, I had told Barnes, and no one would ever see it. Why had he worked so hard to give me this leg, I wondered, if no one would ever really know what he had created? His best efforts meant that when people looked at me, nothing caught their eye. How could he be happy with pouring his genius into his work for people who only wanted to hide it? Was that what I really wanted, and what Mom had paid so dearly for? The privilege of being ignored? The prosthesis fit perfectly, but somehow...it was beginning to feel just a tiny bit off.
I wished I could talk to him about it, but I wasn't due to see him again for weeks, and I couldn't even think about getting a new prosthesis until I outgrew this one.
I sat back on the toilet seat and sighed. We didn't study Aristotle in school, and I'd found nothing online about Xeno or his monsters. Last year in social studies, we'd studied Alexander the Great, but only to debate whether it was right to invade and conquer other countries. Alexander had wanted to rule the world and be worshipped as a god. We debated whether he was right to force his beliefs onto everyone else. One thing we all agreed on: when countries didn't have the same beliefs and values, they usually ended up fighting each other.
Adults went to war for reasons I didn't understand, but it seemed far-fetched that monsters had anything to do with world peace. Xeno thought they held an important key and were worth saving, but I didn't know yet what I believed.
After I'd showered and dressed for the day, I went to check our vegetable drawer.
"Whatcha looking for?"
I jumped, not knowing Mom had come into the kitchen behind me. "We're out of cucumbers."
"You hate cucumbers," she said, with the tiniest trace of a frown.
"I know, right?" I tried to sound cheerful. "But I want to eat healthy. Can you buy some?"
"I guess." She didn't sound convinced.
"Like, today?" I didn't want to pressure her, but neither of us deserved to die for lack of produce.
Mom sighed. "Add it to the list. We'll try to swing by the store later."
I clenched my jaw to keep from groaning. We really needed cucumbers. Now. But if I acted too desperate, she'd wonder why. I pasted on a smile and grabbed a Toaster Strudel out of the freezer.
Mom snatched it from my hand. She threw it in the trash and handed me a mushy brown banana.
"Eating healthy," she said.
"Right. Thank you." I wanted to stab myself in the eye with the banana.
"I need to get caught up on bill paying and laundry," Mom said. "Do you mind if I take over the kitchen table?"
"No problem," I replied, sounding like the perfect daughter. In reality, this was a lucky break. "I have a bunch of homework left to do upstairs." Which was true, even though it wasn't schoolwork. I needed to study the Bestiary.
"So today will be a workday for both of us. But tomorrow the weather is supposed to be gorgeous. If you really want to be healthy," Mom said, "we should take a walk around the park."
"Sounds great." With the price of a peck on the cheek, I had just bought myself hours of uninterrupted study time.
Sunday, March 2
On our walk the next day, I wore a scarf to make Mom happy, but the sun had come out and both of us took our winter coats off after a while. We walked slowly, enjoying the tiny signs of life creeping back. The tulips had sprouted thick green stalks, but because of the crazy winter we'd had, they wouldn't flower for several more weeks. The pear trees were heavy with tight green buds. Everyone in Atlanta was worried about them because one more freeze would destroy the blossoms. The tail end of winter could kill whatever was tender and new.
A woman passed us wearing dark wraparound glasses, her hair tucked under a baseball cap. She had in earbuds and was pushing a baby carriage at a fast pace. Mom let her pass and the woman waved one hand to thank us.
A half hour later, we sat on a bench, watching squirrels dart back and forth across the path. One squirrel even put on a show for us. He chattered angrily at something in the trees before dashing through the bushes. Every few minutes, he would return, his tail full and swishing, barking in protest at whatever was preventing him from climbing that tree. I didn't worry about monsters. I doubted they were out in broad daylight with so many people around. They seemed to prefer the shadows as much as they needed them. The woman with the stroller passed us again. She must have made three laps around the park by now, compared to our one.
Mom stood and pointed to the restrooms behind us. "I need to hit the ladies' room. You?"
"Nah," I said, and tipped my head back to let the sun wash over me. Squinting one eye, I smiled at her. "This was a good idea, coming here today."
Mom grinned and turned for the restrooms. Above me, cranes circled, honking loudly at each other. Mom stopped and pointed up at them, and I nodded. Cranes flew in circles, blaring like party horns, waiting for other cranes to find the right air current and join the loop before forming a V shape and flying north. Cranes migrated up from the southern beaches through Atlanta at this time of year, heading back to Canada for the spring. They were so loud that people always stopped whatever they were doing and watched them, because it was usually too hard to hear anything else.
As the V took shape and flew, the woman pushing the baby stroller walked toward me. I whipped my head to the right and looked down the path. Hadn't she just passed us? The birds must have disoriented me.
The woman slowed as she got closer to the bench.
Mom had just gone inside the ladies' room when the woman pushed her baby stroller to the side and sat next to me. I looked inside the stroller.
It was empty. She took off her glasses, but I already knew who she was. The temperature felt ten degrees colder.
My chest tightened with anxiety. "Why are you doing this?" I asked. My imagination leaped to the worst possible answer. "You're going to kill me, aren't you?"
The woman tilted her head back as she laughed. She was perfect, even up close. "What use would I have for a body? I don't want your body, Sofia." She leaned in to me and pointed to my chest with one long finger. "I want your heart."
I recoiled, and she laughed again.
"Not literally." She reached for my arm next, her fingers extending like a rake, as if to scratch me, but her hand disappeared through my arm, like a ghost. Her mouth pursed in sadness, then she looked at me and her eyes narrowed. "I'm not here to hurt you. I've come to help you. Xeno is letting you make a terrible mistake."
"I don't believe that."
We sat in silence for a moment. Then she pointed to the scarf I was wearing. "Have you ever seen anyone knit?"
"Yes."
"Our worlds, the natural and the supernatural, are like two knitting needles," she said. "We collide constantly. All that is left behind is the pattern, what we wove together when we met. The supernatural is woven into the pattern of this world. People either refuse to see that reality or they become hopelessly distracted by it." She pointed one long finger back at my face. "You do not have the luxury of denial or distraction. I need you to focus."
The invisible wind rustled like music through the trees as birds soared above. There were things in this world that I had to accept on faith, like wind. I would never see it, but I knew it was real and it affected everything. I blinked once and focused on the woman's eyes, ignoring her finger still pointed at my face. Her irises never moved as she watched me. She was alive in a vague sense of the word but with something essential missing.
"Xeno is filling your head with ideas that will only lead to heartbreak." Her voice held an icy edge. "Just as his master, Aristotle, did to him. You must understand the difference between reality and delusion before you destroy any hope you have of happiness. You are making a pattern that remains. It cannot be undone."
I looked at the restroom. "My mom will be back any minute," I said. "You should go."
"I am Olympias, the mother of Alexander the Great. I do not take orders from children."
My eyebrows shot up. I knew that name. She smiled at my reaction, her teeth bright and glistening.
"Thousands of years later and the whole world still knows who he is," she said. "Unlike Xeno, I might add. My son Alexander showed us what a truly great man can do. He would have ended all war, forever, and given us one government, with one ruler, a true and good god. There was no need for tribes and monsters and separate histories, all those identities that clashed and provoked. With my son as god and king, everyone could have lived in peace." She shook her head. "But Xeno promoted the lie that what made us different could make us strong. He even wanted monsters kept alive and recruited a child to guard them."
I glanced at the ladies' room again. Mom stepped out, cell phone pressed to her ear, and held one finger up to signal she needed a minute or two. She rarely chatted on the phone anymore. Mom needed to reconnect with her friends, but this was spectacularly bad timing.
"The monsters are in hiding," I said. "Why can't you just leave them alone?"
"But I do," she said, sounding offended. "Xeno has told you so little. I rather like monsters, my dear. If I can find the raw materials, I'll create one, to serve my cause. I made a rather special one just recently."
She sighed deeply, as if remembering a delicious memory. "Ah, yes, the old magic. So few practice it anymore. Everyone's minds are so... _sterile._ Except yours."
I didn't know what she meant by that, but I was more confused about something else. "So you _don't_ want to kill them?"
"Monsters? No. There was an age, long ago, when that should have been done, of course, but time has passed and done the work for me. They are as good as dead now." She turned and grinned at me. A shiver crept up my spine.
Her eyes were lifeless. They were like glass marbles, hard and shiny.
"Monsters do not trouble me, Sofia. You do."
"Me?" I asked.
"Sofia, people can have their identity and their beliefs, or they can have peace. No one can have both. Not in this world." Olympias neatly folded her sunglasses. "Watch the news. The bombings, the wars, the beheadings...all because we believe different things."
"We call it diversity," I said.
"I call it madness," she snapped.
Her tone was shrill, and I winced at the sound of her voice. But what really bothered me was that she was right. No matter what a person believed, it seemed like someone else wanted to kill them for it.
She continued, softening her tone. "And so what if a Guardian, more powerful than all the others, was born in this strange, soulless age? What if her courage woke a sleeping generation? What if she convinced people that their real enemy was fear? That an individual should be celebrated instead of silenced? All my work, all Alexander's dreams, would be undone. And there would be much suffering."
Mom laughed at something her friend must have said, and her nose crinkled just like it used to, when we didn't have so much trouble.
Olympias tapped my prosthetic leg with her glasses. "But what if this girl used her strength to save the world? What if she used her gifts to convince others that only in unity can we survive? Conformity is not defeat. It is wisdom. Sofia, it is salvation."
Mom had her back turned to me now, obviously relieved that I was chatting with someone as safe as a mom and her new baby.
"Take the first step," Olympias said. "Let the monsters go. I'm not asking you to kill them or do anything cruel. I'm only asking you to stop caring for them. They are individual, personal creations, and individuality is what we're fighting. We cannot have separate identities and beliefs, because that always leads to conflict. We must become one so that wars will end. This is a dangerous time to be distracted by our imaginations. But if we let go of our beliefs, all the nations could join together, under one government, with one ruler."
She looked up at the sun for a moment, and I noticed she did not blink, not once, even though the sun was bright. Then she turned back to me and I watched as a tear slipped down one of her cheeks. It was like seeing a doll cry.
"It has been more than two thousand years since I have beheld my son's face. I long to return to him, Sofia, but I must finish what he began. Help me. Don't save the monsters. Save the world."
Then she leaned in and whispered, "Save yourself."
I pulled away, desperate for a deep breath of air, as if Olympias carried an infection, a bacteria waiting to thaw and spoil all the oxygen. Closing my eyes, I inhaled, then let my breath out slowly.
Mom must have finished her call, because she was walking toward us now, a big smile on her face. Today had been a really good day for her.
Olympias stood and loomed over me, blocking out my mother and the sun. In her shadow, the cold crept back over my body.
"Aristotle—and Xeno and the rest of his students—believed that the search for truth was a good thing. But you know, Sofia, what happens when people see the truth. How kind are people to those who are weak or different? Knowing the truth about the world, about ourselves, will never lead to peace. Xeno wants you to seek the truth, but he never explained what truth costs."
Frowning, I tried to catch a glimpse of my mom. Olympias smiled at my nervousness.
"I'm not evil, Sofia, and I am not your enemy. But I have made one last monster who will take everything you hold dear and smash it to bits, piece by piece, until you realize that what I offer the world is far better than truth."
When I got home, I locked my window and didn't take the book out from under the bed. It glowed once, briefly, but I looked away, pretending not to see it. Two weeks ago, I hadn't even believed in monsters. Now I did, but Olympias wanted to challenge everything else I believed.
Mom had dumped some laundry on my bed, so I got to work separating the pile.
Talking to Olympias had made me sick to my stomach, like that feeling you get when you realize too late that an assignment is due. When I opened the Bestiary, I hadn't known that anyone from the ancient past was going to come looking for me, especially someone like her.
I found a pair of socks and rolled them together. I hated wearing one sock; it made my phantom foot cold.
Grabbing a shirt, I replayed everything Olympias had said, and one thing stuck with me, a thought that lodged itself in my brain as unwelcome as a seed caught in my teeth: _What does truth cost?_
I knew part of the answer: the price of truth was pain. Billy had said that too. Even I avoided it sometimes. That didn't make me a bad person. It just meant that I really loved my mom and I guess I loved Alexis too. I loved her enough to refuse to tell her the truth about why we couldn't be friends yet. I did want to be friends once I was stronger and she didn't have to worry about me. When I could be happy to see her running with someone new, I'd know it was time. Right now, it would be a lie.
And Alexis knew when I was lying. Sometimes she laughed at me for it and sometimes it made her mad. Alexis was like salt. She usually made everything better, but if her words touched on some wound I was trying to ignore, they stung like crazy. She was never mean; she was just always right.
I opened my dresser drawer and tucked my underwear inside. The socks went in next; then I'd hang the shirts and be done.
Why were relationships so complicated?
I grabbed a hanger and started the last little bit of my chore.
Was I happier now without a best friend? Could I be happy without any friends at all? I wouldn't have to worry about making mistakes and hurting people. I used to be completely alone at school. Maybe I had been happier then, even if I didn't think so at the time. Grabbing a shirt to hang up, I flinched at the memory of the day it all started.
In first grade, my mom bought a shirt for me at a garage sale for a quarter. I loved it. It had a lion made out of gold glitter, but everywhere I went, I left specks of glitter behind. When I wore it to school, Candy pretended that the glitter was germs, and all the girls took big steps to avoid touching me and the glitter germs. I kept my head down all day so no one would see the tears brimming in my eyes, and tried to avoid calling attention to myself.
Mom washed my shirt that night and hung it back in my closet, but I had vowed to never wear it again. Being the odd girl out, even for just a day, had been the most painful thing I'd ever experienced.
In the wash, the glitter got all over my other clothes. For months after that, Candy would point to a speck of glitter on me, and then everyone would talk to each other behind hands cupped at their mouths. Of course, glitter was used everywhere in elementary school, so I never escaped the teasing.
I eventually just learned to keep to myself. But if it sucked not having friends, at least I didn't have to worry all the time about hurting someone's feelings, or getting hurt either.
The last shirt was hung and I could finally sit on my bed.
So maybe truth hurt, but what hurt worse: being hurt by the truth or by someone you loved? It was the kind of question a philosopher would ask. I grinned a little and made a note to tell Xeno.
Which was more painful, truth or love?
I inhaled sharply because I had this weird thought.
What if they were the same thing?
Monday, March 3
Mom's alarm clock buzzed without end. She was already downstairs and refused to come up to turn it off.
She was pretty mad when I finally made it into the kitchen. I was moving slowly since I was tired; plus I couldn't find a head scarf or bandana to match my outfit and it had taken me forever to get dressed. I wasn't sure what the rule was for matching. I only knew nail polish should match on both fingertips and toes, and not to wear white bras after Labor Day or something. I made a mental note to read a fashion magazine and figure this stuff out once and for all.
Mom was stuffing papers from the table into her briefcase and tried to grab one that fluttered away. She quickly read it before she stuffed it in her bag. "Sofia, when were you going to mention this?"
I could tell it was from school by the logo at the top. My brain raced to think of what else I had recently neglected to tell her.
"The school dance?" she said. "In sixth grade, you told me that more than anything, you wanted to go and dance with a cute boy."
"No one asked me in sixth grade," I reminded her.
"So?"
"No one asked me this year either." I faked a big bright smile. "Maybe life really is getting back to normal."
At lunch, I loaded up my tray with double the amount of cookies and a little bitty salad. Mom was making me eat plenty of vegetables at home. A lunch lady grabbed one of the cookies and put an apple in its place, giving me a wink. "We all want to keep you healthy, dear."
I walked outside to our butterfly garden, a big area in the middle of the school with a bunch of picnic tables. All the plants had been chosen by the Science Club to attract butterflies, and when it was butterfly season and the place was filled with dozens of them, it was amazing. Right now everything was dead.
I spied Alexis sitting alone at a table. I used to always sit next to Alexis on her right, because she's a leftie. My body moved toward her by instinct, seeing her with no one on her right. That was my place, ever since the beginning of sixth grade last year.
I pulled back, reminding myself that our friendship was a tangled mess and I couldn't unravel it in one day. Even if I wanted to, I would need time and space to concentrate and get my words right. I had to make sure she didn't feel guilty, and I needed to be strong enough to be happy for her that she still had her leg, her hair, and her whole life.
_Happy, happy, happy,_ I said under my breath. So much happiness. Like stabbing a fork in your eye.
Maybe someday she would understand that I had to do it this way because I loved her. Tears blurred my vision and I blinked hard to clear them. I just wasn't ready yet.
The track team flooded outside as I stood there, and a loud-mouthed brunette grabbed the seat to Alexis's right. My spot was gone. Alexis sat with her back to me, so she hadn't seen me hesitating. I hung my head and moved toward an empty table.
I lowered myself with a soft groan, my butt sore from the hard plastic chairs in class. Why did they make desks and chairs so cold and hard? Were they afraid we might start to actually enjoy learning?
Above me, the sky was ice blue with thick white clouds. A sliver of the moon was visible, a pale shimmer beyond the clouds.
"It's a waxing crescent," Billy said, setting his tray across from mine. " 'Waxing' means 'growing.' " I noticed he took a package of carrots from his tray and stuffed them in his pocket. "I looked it up. But no more snow in the forecast. I think winter is almost over."
People were starting to claim the tables all around us. There was no place I could sit without pretending I wanted to join someone's clique.
I opened my milk and took a sip, not looking at Billy. His knee grazed mine under the table. The butterflies I was longing for suddenly appeared in my stomach. I shifted to the right to avoid touching him in case he didn't want to touch me again.
"I looked for you all morning," he said, unfolding his napkin and setting it on his lap. "We never had a chance to catch up last week. You should give me your number."
I had never given my number to a guy. My mind went blank, like the blue screen of death on a computer. What was my number? I had no idea.
He placed his plastic fork on the left and spoon on the right, then began scooting the cheese off his mystery slop with his knife. "So anyway, first, don't let Candy get to you again. And second, I wanted to apologize."
"For what?"
"Well, being the new guy sucks, because the girls always chase me and the other guys don't like it. So I end up alone and just try to keep things interesting until I get kicked out again. But you didn't chase me, and I thought it was because you were cool. But then I realized, maybe you just don't like me. Not that I really care, but I'm sorry if you actually hate me and I got it all wrong."
"No," I said. "I like you."
He scowled and narrowed his eyes at me.
"A lot!" I added.
His face brightened. "That's what I thought."
I felt like an idiot and changed the subject. "Why do you do it?"
He raised one eyebrow, so I clarified the question. "Why do you want to be the bad kid and get in trouble all the time?"
He bit his bottom lip, thinking. "I think...I'm trying to make a point."
"To who?" I asked.
He frowned, anger flashing in his eyes. "If you have to ask, then it's obviously not you." Shaking his head to dismiss the conversation, he picked up his fork and pointed it at me. "Anyway, we should go to the dance."
"What about the goats?" I said, pretending I hadn't heard him.
"They already turned me down," Billy said.
I struggled not to laugh. "No, I mean, what did you do with them?"
"The fire department returned them to the owner, and I gave my dad's intern a twenty for the help. Everyone went home happy."
His fork hovered over the salad that he hadn't even looked at. He was too busy staring at me.
"It drives me crazy when people get food they know they aren't going to eat," I blurted. "Why not just say 'No, thank you' to the lunch lady? Why pretend you're going to eat the salad if you know you're really going to throw it away?" My voice rose up a little at the end, and I wanted to run for the nearest exit in embarrassment.
He wiped his mouth with the napkin before answering. "You're kinda weird." He took another bite of his meat slop and jabbed his fork back in my direction. "But I admire your sense of justice. I mean, who else would stick up for a defenseless side salad? You're a vegetable vigilante."
I nodded, unable to speak. It's too bad they don't make EpiPens for episodes of life-threatening embarrassment.
"I'd like to take you to the dance," he said, loading another spoonful of cheesy glob. "If you're not too busy saving helpless produce."
"For one thing, it's called a _dance,_ " I said. "And I don't dance." Why did I sound angry when I was scared? I didn't mean to. I just needed to cut him off before this got out of control. I had enough problems, like Xeno, Alexis, monsters, and my mom. I didn't want a new one called Billy and another one called dancing.
Billy opened his eyes wide, like I was missing something painfully obvious. "No, actually, it's called a date, and you should try it. You've never been very nice to me, and I'm your only boy friend." He took another bite, then paused. "I am your only boy friend, right?"
"Yes!" I said. "No!"
He grinned.
"I meant no. You're not my boyfriend. I don't have a boyfriend. I've never had a boyfriend." Why did I have to say that? I clenched my fists and considered leaping up from the table, then remembered it wouldn't be much of a leap. I'd only embarrass myself more if I tried to run.
"I am a boy and I am your friend, right? Don't make it complicated. Look, if you're scared about the dancing, I can teach you a move, if you want," he said. "I learned it from the Internet. All you have to do is bend over a little at the waist and pretend to write your name in the air with your butt. Boom! You're dancing."
"Stop," I said. "I'm not who you think I am. I'm not fierce! I'm..." I looked down at my lap. "Do you know what an insurance company once called me?" I asked softly. "A liability."
Billy stared at me for a long time. I could feel it by the growing burn in my cheeks.
"Maybe I understand that better than you know," he said; then he slapped his palm on the table.
"You don't need to learn to dance," he said, and with the other hand grabbed the edge of my lunch tray. "You need to learn to fight back."
"I feel like all I do is fight."
"Well, not for the right things, then. Sofia, you may need me more than you realize."
I looked up and glared at him. Why did everyone want to make me their pet project?
"I'll teach you. Let's start with your lunch." He pushed my tray a couple of inches to the left. I didn't know if he meant to do it or not, but that was still my weak side. It just took time, the doctors said, to relearn balance and counterweight, and all those daily exercises I was supposed to do at home were going to help.
I didn't move. I hadn't been doing the daily exercises.
"Come on, stop me," he said, pushing my tray a couple of inches more. "Make me sorry I messed with you."
The tray was getting dangerously close to the edge of the table. His eyes locked with mine and chills went through my whole body, even my prosthesis, I swear. I wasn't sure if I wanted to kiss him really badly or just slap him repeatedly.
Bit by bit, he pushed my tray farther off the edge. I kept my arms folded, trying not to watch. It made me crazy. When it tilted at last, I refused to look, but then, at the last possible second, I lunged.
My lunch splattered all over the ground. I slowly sat back in defeat.
"Too late," he said, and shook his head. He sounded disappointed. "I'll get you a new one. One cookie, an apple, and some wilted lettuce that you're going to eat just on principle, right?"
Everyone stared at me as he left and I got busy picking up the tray.
Sneaking a glance in his direction, I saw Billy swing the door open, but before he went into the cafeteria, he reached into his pocket and bent down.
Looking back down, I reached for a clump of lettuce. A new Sperry loafer stepped on it and I looked up.
"What is his deal?" Candy stood over me, glaring at Billy's back as he disappeared.
I motioned for her to move as I finished scooping up the remainders of my lunch and she watched.
"He's cute but psycho," Candy sighed. "I called him and he didn't even call me back, can you believe that? He told Natalie he already had a girlfriend, even though the only girl he ever talks to is you, so I know that's a lie."
I stood, keeping the tray between us for extra space. She must have applied her perfume like she was crop dusting.
"Anyway," she said, stroking her hair absentmindedly with one manicured hand, "if we're going to work on our plan, we have to start right away. I'm assuming you're going to the dance, so I'll go dress shopping with you. My aunt owns a great boutique at the mall."
More help I didn't want, though I realized that the only way to stop her was by confessing that we were broke. I took a deep breath.
Candy grabbed my arm. "I know what you're thinking, but here comes the best part. She wanted me to tell you that if you let a photographer take your picture for the newspaper, you'll get the dress for free."
"Ice cream!" Alexis stood up at her table, her voice easily carrying across the small courtyard. "Who's in?" None of her tablemates raised their hands. Only Alexis would want ice cream on a chilly day.
"Hello?" Candy said, sounding exasperated with me.
I turned my attention back to her. How could I possibly tell her everything wrong with that idea? When Candy saw a camera, her whole face lit up. When I saw a camera, I tried to shrink my body inward and turn my face away.
Candy and I lived in parallel universes and it was just too hard to explain mine.
"We're going Thursday," she said. "My mom's signing us out early, but your mom has to send in a permission slip."
Alexis stood and turned around, catching me talking with Candy. Her face darkened.
"You're so lucky," Candy said, resting her hand on my forearm. "My aunt doesn't give just anybody a free dress."
Alexis rolled her eyes and stomped off toward the cafeteria.
I jerked my arm away. "I never agreed to this!"
Candy gasped, like I had just spit on a kitten. "I'm offering you my help, and a free dress from the best store in town. How does that make me the bad guy?" Then she took a deep breath and closed her eyes, like this was hard on _her._ I wanted to scream.
"Do you know the definition of insanity, Sofia?" She gestured to my outfit. "It's doing the same thing over and over and expecting different results. We have to upgrade you. The first step is getting you a fabulous dress for the dance. People will remember it for weeks. That'll buy us some time to work on your other...choices." She made a face like she was trying to avoid looking directly at my clothes.
Natalie bounced up, pulling on Candy's arm. "Isabel just walked in!" she hissed in Candy's ear. "Let's go ignore her."
"Please do," I sighed under my breath.
Candy gave me a confused, searching look. I was clearly a mystery to her, but she walked off with Natalie to go ruin someone else's day.
Grabbing my tray to return it, I turned to leave through the media center doors to avoid seeing anyone else, especially Billy if he really was going to come back with another lunch. As I swung the door open, the bushes rustled beside me. I stood still, listening, but the movement was too soft and gentle to be a monster. Bending down, I peered into the gap between the shrubs. A skinny brown rabbit was chewing on a pile of carrots, its nose twitching back and forth. Poor guy looked starved from the long winter months. On impulse, I reached for the spot where my left leg had once been.
Mother Nature was nobody's friend.
After dinner that night, I tried to make up for my moodiness by tidying up the kitchen while Mom prepped for an online software class. Everyone in her office had to take it, but Mom was probably the only one excited about it. She grabbed her pajamas from the dryer while I took out the trash. Outside, dusk was settling, the night's darkness creeping in as birds with long black wings glided overhead. Several trees in the neighborhood were still bare, their spidery branches frozen in their final pose from last fall. The tight green buds on the branches were still only promises.
I looked up and down the neighborhood, at the line of lifeless trees, the brown grass, the flower beds with collapsed, rotting leaves. The world still looked dead. Who knew what would come back and what was lost forever?
I dumped the trash in the can and dropped the lid. The can shook in response, a huge dent appearing in the right side. Something was in there. Something big.
Dumb, fat possums got trapped in garbage bins all the time. Raccoons got trapped too, but they always figured out how to escape. Possums were hopeless. Even though my hands shook at just the thought of a possible bite from one of those nasty oversized rats, I popped off the lid and jumped back.
Nothing happened. Nothing moved. Maybe the possum was playing dead.
I turned to walk back into the house, when I saw two huge yellow eyes glaring at me, like twin moons side by side. With my peripheral vision, I saw tufts of brown fur, tinged at the end with white. Beneath the eyes was a snout that ended with thick black nostrils, which flared as the beast sniffed the air.
I carefully took one step back.
An enormous wolf stood before me on four huge paws with knuckles the size of golf balls and long black claws as long as my pinkie.
I knew I was seeing an extraordinary creature, one that had gone undetected for hundreds of years, maybe more. My heart beat faster.
I was alone. If I died, there would be no witnesses, except perhaps the birds that stirred in the dark trees.
Pins and needles shot up and down my arms. Maybe this was what I would feel after my first kiss, overwhelmed and unsteady. When you're terrified, odd thoughts float up from your subconscious. Why did I associate terror with my first kiss? I made a mental note to consider that at a more convenient time.
I calmed my nerves by taking a few deep breaths, but the deeper I breathed, the more I started to hyperventilate. I had to focus on the wolf, not myself. I extended one hand to reach out and touch it to make certain it was real. My hand began to shake and the wolf's gold eyes targeted it as if I were shaking a toy at a dog. Dropping my hand back to my side, I tried to take inventory, the way Xeno might have when he recorded his own sightings.
The fur on its body was a dirty brown with light tan streaks like swirls of dirt, and it smelled like old socks and blood. The skin of its thick, wide black nose was pebbly. My gaze dipped down to its mouth. Flecks of a dirty paper towel and old candy bar wrapper clung to its lips. Saliva mixed with what looked like blood, hanging in juicy thick ropes. Bits of meat, perhaps a fast-food hamburger, stuck out between its teeth. Its breath was hot in my face and stank.
Without warning, the beast lurched to the right and snapped furiously at his side. I jumped too, even though I didn't know what had spooked him. His eyes went wide and he began panting. Then he whipped toward the left, snapping at his other side, and stumbled when he landed. Something inside his stomach was punching outward, making his side expand. He growled in frustration and snapped again, but the thing inside him moved to the other side and punched. The wolf continued to whip his body back and forth. Something was at war inside his stomach.
"Wait here!" I whispered. "I'll be right back."
I got upstairs just as Mom was walking to her bedroom with her company laptop. She saw me walking as fast as I could to my room and frowned. "Where's the fire?" she asked.
"Uh..." My brain searched for a quick excuse. "I thought you would be on the computer already."
"Do I need to cancel my class?" she said, an ominous tone to her voice. "Are you up to something?"
"No!" I forced a smile. "Do your class. I'm not going to get into any trouble. Promise."
Mom stared at me as if deciding whether to trust me or her gut. I guess I won, because she disappeared into her room and I was only able to see the top of her feet as she stretched out on her bed to watch the video.
Once in my room, I grabbed Xeno's book and flipped until I found the page with the monster's picture on it. It was at the end, and the paper and handwriting were different from the rest of the book, as if someone had made this entry later. How many Guardians had there been? Did they quit, like Claire, or did they get eaten?
Bending over closely to read, I kept glancing at my window, straining to listen for new sounds, any howl or growl that would alert the neighbors.
> # THE BEAST OF GEVAUDAN
>
> ## Origins: Gevaudan territory, France
>
> The Beast terrorized France in the 1700s, eating over one hundred villagers. He first appeared on a spring morning in 1734, attacking a young girl tending to her sheep. He ate the sheep too. The King of France sent the army after it, but the Beast made his way across Europe to England, killing as it went, evading the troops. Finally, it boarded a boat in the fog, hoping to sleep and digest its last meal, which was a rather fat baker. Aboard this ship, the Beast ate two passengers and several crew members. To avoid panic, their deaths were attributed to sickness.
>
> The Colonists had brought the Beast of Gevaudan to the New World.
I studied the picture carefully. This was definitely the monster in our driveway. If any of our neighbors saw him, Mom was going to get a strongly worded letter from our Homeowners Association. Unless he ate them all first.
> The Beast has poor digestion because he does not chew thoroughly.
I was beginning to get a bad feeling. I read the rest of the entry, which contained a map of the Beast's natural habitat and a few notes about his behavior.
A low growl erupted outside. I opened my window and peeked down at our driveway. There was no sign of him, but I knew he was there. The dogs in our neighborhood were barking furiously.
"Shhh!" I whispered. "You're going to get us both caught!"
A moth flew toward my bedroom light. The Beast lunged out from the bushes, his white fangs glistening in the moonlight as he snapped up the moth before falling back to the pavement with a crash. Mrs. Cranston across the street turned on her porch light and I panicked.
Sliding my window open, I motioned to the Beast below to climb inside, fast. He made the jump to the second story easily, like a dog leaping onto a couch.
Mrs. Cranston opened her front door and squinted into the dusky evening. She had rollers in her hair and was wearing a white robe that made her look like a giant marshmallow.
The Beast whined and lifted his paw, knocking my bed against the wall with it as he clawed at his mouth frantically.
Why had I opened my window? Now I had Mrs. Cranston outside and a Beast inside. I didn't know which one was scarier.
The Beast couldn't fully close his mouth and he panted heavily, his saliva dripping onto my carpet.
"What do I do?" I asked, and flipped to the blank page Xeno used. The page remained empty. I tried shaking the book, like it was an Etch A Sketch, but still nothing happened.
The Beast tried to circle me like I was prey, but he barely fit in my room. He scooted his butt across my wall as he turned and knocked the _Doctor Who_ poster off center.
The book began to glow at last. I exhaled, my knee going weak with relief.
> The Beast will bite you if you are not careful. You might not survive.
"Isn't it your job to keep me safe?" I snapped.
> I do. I give you advice.
I scowled. "I don't want advice! I thought being the Guardian would give me superpowers or something."
> Following good advice is a superpower.
I rolled my eyes. "So how do I help him?" Maybe I shouldn't help at all, I thought. The world needed monsters, but this Beast had eaten lots of people.
> He must be restrained. It will require a special rope.
"A special rope? What kind?"
> A rope made from the beard of a woman.
"You're not helping," I groaned. _And you're not funny,_ I wanted to add.
> Are you sure you trust me? Remember, it is the only thing I asked of you.
"I didn't realize what that meant," I said, my voice getting high-pitched. "I'm trying."
> I don't want you to try. I want you to trust. This is not about monsters. This is about you. Answer one question and I will help.
The Beast whined, a pitiful sound like someone who's had the stomach flu for two days and just wants it to be over.
"Okay," I said, "I trust you. I have a million questions of my own and you never give me all the answers, but I trust you."
The Beast belched, causing my hair to blow back. I clasped one hand over my mouth as a shield against the smell; it was worse than fish-stick farts.
> To survive, monsters must hide. You do not have to hide, yet you do. Why?
The Beast retched, and a squirrel flew out of his mouth still wearing a wide-eyed look of shock. The squirrel ran toward my bed, then back toward my closet, then back to my bed. The Beast and I watched as it made one more zigzag, before the Beast ate it again. It made a little barking noise as it went down the Beast's neck and into his stomach.
The Beast swallowed animals whole! There were live animals inside his stomach, fighting to get out! I had to get the Beast out of my room. Whatever came up next might be harder to handle than a squirrel.
I glanced back at the book. "I can't answer that right now! I have to help the Beast."
> And I have to help you. You are not meant to live in the shadows. And because you are a rather difficult student, I must try a new method of teaching. I will refuse to say anything else tonight. Your help will not come from me.
I gasped. I wasn't stubborn!
The Beast's stomach changed shape again when the squirrel hit the bottom. It looked like animals fighting inside a pillowcase. He slammed his side against my wall and the fighting stopped for a moment.
"Sofia! Keep it down!" Mom hollered from her bedroom. "I'm trying to concentrate!"
"You've got to get out of here," I whispered to the Beast, panicked.
The Beast slowly turned away from me. He knew I had failed him. I saw it in the way he moved his legs. Sadness made them heavy. I couldn't help him.
"What do I do?" I asked Xeno.
The page erased itself and remained blank. I waited for as long as I could and realized Xeno meant what he had said: he was not going to help me. I had the world's most dangerous wolf in my bedroom, the one who had started the legend about the existence of werewolves.
I couldn't handle this alone.
Billy's dad was a vet. A veterinarian would know what to do. But what if I called and Billy answered? I had left the butterfly garden before he could replace my lunch and ask me about the dance again.
We only had one phone in the house, not counting Mom's cell phone. It was the landline downstairs.
I reached for my book bag and pulled out the business card tucked in the front pocket. I had fished it out of the trash earlier, feeling guilty about tossing it. Billy deserved a good friend even if I thought he had chosen the wrong person.
The Beast moved to the window, a groan escaping his closed lips as his stomach rumbled again.
"Wait!" I said to him. "I can help you. Just follow me downstairs and do not make a sound. Not even one. Do you understand? Not a peep!"
He lifted his lips to reveal huge canine teeth. They were as long as steak knives. I don't think he liked my tone. I stepped back carefully and crooked my finger, indicating that he should follow me.
Adrenaline was making me light-headed, and I took each stair as carefully and as quickly as I could without risking a fall, his snout edging along me next to the wall.
We made it into the kitchen, where the landline was. I left the lights off and grabbed the phone, sitting on the couch before I dialed.
The Beast stayed in the kitchen, his snout wrinkling as he sniffed loudly. I whirled around, one finger to my mouth in warning. He snapped his teeth at me. I gulped once and dialed fast.
In the dim light, I could barely read the business card. It listed an office phone and a number for after-hour emergencies. This was an emergency.
Billy picked up right away. I closed my eyes in agony and every other painful, delicious feeling that came from hearing his voice. A cereal box flew out from the pantry, ripped to shreds and empty, as it skidded across the floor. Then I heard another box being ripped open.
"Billy, it's Sofia," I whispered.
Nothing but silence.
"Billy?"
"Why are we whispering?" he whispered.
An empty box of crackers hit me in the head. I swirled around and glared at the Beast. He had one paw poised over a can of tomatoes, preparing to use his claw to pierce it open. I shook my head.
"Billy, what should I do if my...dog...has swallowed something he can't digest?"
"That's why we're whispering?" Billy said loudly, sounding disappointed.
The Beast punctured the can of tomatoes. He jerked his head back, then leaned in and sniffed the can. Knocking it to the side, he grabbed another.
"I swear, Billy, this is important," I pleaded. "I need your help right now."
"You should talk to my dad, not me," he said. "He's the vet."
"So get him," I said.
"Say you'll go to the dance with me."
It was my turn to be silent. The Beast was now on his third can. He was opening each one just to smell what was inside. I held one hand over the phone and hissed at him, making that little hissy-snappy sound that the Dog Whisperer uses on bad dogs. It had no effect.
Billy repeated the demand, slowly. "I'll get my dad if you'll be my date to the dance."
The Beast turned from the pantry and began using his giant teeth to pull open a cabinet. He grabbed a coffee mug and dropped it to the floor, then sniffed the shards before moving to open the next cabinet.
"Okay! I'll be your date. Now get your dad."
I heard him drop the phone and yell for his dad.
"This is Dr. Hamby." A deep, fatherly voice came on the line.
"This is Sofia, a...friend of Billy's from school. Sort of. I mean, definitely from school." I giggled nervously. "Anyway, my...dog...has really bad indigestion."
"Did he get into anything he shouldn't have?" Dr. Hamby asked.
"No, just the usual, I think. But too much of it."
"Have your mother bring him over. Can I speak to her? She'll need our address."
"No! I can't. I mean, she can't. Talk to you. She's too upset. And the dog can't come over. Never."
A bright light temporarily blinded me. The Beast had opened the refrigerator. I heard him make a happy noise in the back of his throat, in between a bark and a whine.
Oh, no, no, no.
"Because?" Dr. Hamby asked, sounding patient. He was probably a really good dad. Not that I would know how to judge one.
"Because...he...hates men. He, like..." I tried to think of the absolute worst thing a dog would do. "He bites crotches. He's a really horrible crotch biter. So...maybe you could tell me what to do? I can do it here, by myself."
"Well..." He sounded unsure.
"I'm sure I'll be fine. He never bites my crotch or anything." I rolled my eyes. Had I really just said that?
The Beast poked his snout around in our fridge, using his teeth to open our cheese drawer.
Dr. Hamby cleared his throat, like maybe he was trying not to laugh at me, then spoke slowly. I could tell he was thinking this through in his head. "Is he acting normal right now?"
I watched as the Beast swallowed a block of cheddar whole, the wrapper still intact.
"It's hard to define normal for him," I replied. "But he seems okay."
"Is he drinking normally?"
The Beast now had his head tilted back, drinking straight from the milk carton. Mom didn't even let me do that. I hoped he didn't leave any hairs.
"Yep."
"No vomiting or diarrhea?" Dr. Hamby asked.
I hadn't realized that was a possibility. "Not yet," I said, my voice getting weaker.
"Well," Dr. Hamby said, "I'd watch him for a little while. You might try withholding any more food until tomorrow to give his system a chance to settle down. If he's not better by then, bring him by. I'm not afraid of a cranky patient."
"You should be," I whispered under my breath. Louder, I said, "Okay. Thanks for your help. Bye." What I wanted to say was, _Thanks for nothing,_ because I had no useful information and I had just agreed to be his son's date.
Could this night get any worse?
The Beast lifted his tail and expelled a big cloud of gas that shimmered like a rainbow in the refrigerator light. Lifting my shirt over my mouth and nose, I marched into the kitchen and grabbed the Beast by the scruff of the neck, just like I had seen mother wolves do on Animal Planet. I had to stand on my toes to do it, and stretch my arms until they burned, but it worked.
The Beast whined and cowered, his big yellow eyes blinking at me innocently.
"To the backyard," I said through gritted teeth. "Right now."
I opened the back door and he followed me obediently. In the moonlight, protected from the neighbors' view by our fence, I collapsed onto a patio chair to think, my muscles burning. The Beast sat on his haunches beside me. He was so big he blocked out the light of the moon, but I could still hear strange and awful noises coming from inside him. I stroked the Beast's fur, trying to think of a plan. A Heimlich maneuver wouldn't work, since I couldn't wrap my arms around him. It wasn't safe to get him human medicine either.
The Beast's breathing slowed. Under my palms, his muscles released and elongated. A little burp escaped his mouth.
"You shouldn't swallow things whole," I scolded. He pressed his body against my hand, and it gave me an idea. Didn't mothers burp their babies by patting them on the back? I pushed myself up and stood next to him, running my palms down his side, trying to position them near his stomach. When I felt the trapped animal push back, I began gently slapping his body in that spot. The Beast turned and lifted an eyebrow, curious but unaffected and slightly condescending. He was French, after all.
I took a deep breath, made two fists, then punched him twice in the side. His eyebrows shot straight up in surprise. I did it again. This time I remembered to suck in my abs, because the physical therapists had told me that whenever I needed physical strength I should move from my core. I landed another one-two punch, and his mouth popped open, so I let more punches fly, until I was working up a sweat.
The Beast looked up at the moon and his body trembled. I realized, almost too late, what he was about to do, and I scrambled to move away. His jaws opened wide, his giant pink tongue flopping to the side as he dry heaved once; then the squirrel popped back out, followed by Newman the cat, and then a coyote. The Beast opened his mouth wider, his head jerking back and forth, until at last a huge hawk, wings spread out in either direction, emerged and flew into the sky.
The terrified animals looked around, then fled into the night.
The Beast sighed and lay down, obviously exhausted. He coughed once more, and a can of Cheese Whiz rolled across our patio.
The light came on above our stairs.
I froze. The Beast leaped into the darkness and disappeared.
I stood up just as Mom turned on the kitchen lights. Our eyes met through the glass of the patio door; then she turned her head and surveyed the damage.
I walked into the kitchen without saying anything. Opened cans, shredded boxes, and torn wrappers were everywhere. Cabinet doors stood ajar. The sweat on my face was cold as it ran down in little drops from my forehead.
"Well" was all Mom said. I started to clean but she waved me off, the heat of her anger and frustration rolling off her in waves. I went upstairs and got into bed, not even bothering to open the book. Xeno had said he was done for the night. My mom clearly was too. I wasn't worried about that.
I was worried about tomorrow.
Tuesday, March 4
"I'm in," I said to Candy. She was sitting with her little coven of friends at the center lunch table. The source of so much frustration was now the answer to my problems. It's too bad that colleges don't give out scholarships for irony. I was in the accelerated program.
Candy nodded in smug satisfaction, then took a long sip from her water bottle, her eyes never leaving mine. Xeno probably wanted me in this situation. He was using monsters to push me out into the open and ask others for help. Everyone around Candy smiled blankly, waiting to resume their private conversations.
Walking away, I could hear them gossiping about Billy asking me to the dance. News spread fast in this school. I tossed my cookie wrappers into the trash on the way out the door.
The intercom blasted my name. "Sofia Calloway, come to the office for early checkout."
My stomach lurched, and not just because I had eaten three oatmeal cream pies. There was only one reason I had ever gotten checked out early: the doctor.
Mom was waiting in the office but refused to give me any details. She was amazed that I had been hungry for breakfast that morning, and even a little suspicious that I was feeling so good. Plus, I wasn't due for blood work for another month and I'd already had my teeth cleaned a few weeks ago. I started to go a little nuts in the car, but Mom's only response was reaching over to pat my leg, except she patted my prosthesis, not me.
When we arrived, she parked under a big magnolia tree that cast a dark shadow across the lot. Gray clouds hovered above. I had another odd feeling, as if something were not right. I glanced around, expecting to catch a glimpse of Olympias. Xeno had been so wrapped up with me asking questions that he hadn't told me what to do about her.
We walked toward a plain one-story building. The lobby was brightly lit by a harsh blue light that buzzed overhead. In front of the elevators, Mom pulled out a piece of paper from her purse, checking it, then walked down the hall to the left. She kept looking from her paper to the signs on the doors. Finally she stuffed the paper back in her purse, straightened her posture, and took a deep breath.
> MARIE INEZ CAPISTRANO, MD, PhD
The sign on the door was thick gray plastic with black letters.
Mom was already turning the knob before I could stop her.
"Who is that?" I asked, pointing to the sign.
Mom ignored me and walked to a receptionist's desk with a frosted glass window made to slide open. Nobody was there.
"All my labs came back normal," I protested again. "I'm not due for any more tests until the end of the month."
An old woman, with white hair swept all around her head into a big fluffy bun, opened the door into the waiting room from the inner office. Her hairstyle looked like it had been shot out of a can of whipping cream.
"Sofia?" she asked. She had on an embroidered shawl with fringe that swayed when she moved, and big doughy cheeks that swelled up around her eyes. "Come with me. Your mom can stay out here." Her body moved from side to side as she turned to walk down the hall. I followed, only stopping once to turn back and glare at Mom.
The woman led me into an office. A movie poster from the old black-and-white _Frankenstein_ film was on the wall behind the desk. I counted four diplomas hanging on either side of it. A plastic dragon sat next to a box of tissues. I leaned forward to examine it closely.
"Like it?" she asked, waddling around to her chair.
"It's kind of dumb," I replied, without thinking. What I wanted to say was that I was wondering how accurate it was. I hadn't seen one yet.
"Good," she laughed. "You're honest."
I felt my cheeks get hot. I hadn't meant to insult her statue. The rest of the office was decorated in normal adult fashion: practical, uncomfortable, and beige.
She settled down into her chair with a wiggle. A pile of mail sat unopened on her desk, including a newspaper. The front page had a photo of the girl from the hospital next to an illustration of odd-looking bite marks. I hadn't had time to ask Xeno what had happened to her or which beast had attacked, but the marks didn't look like they'd come from anything normal. Beneath the photo I read the words "Animal Control Warns the Public: Unknown Animal Still on the Loose."
She caught me staring at the article and I glanced away. We took turns stealing glances at each other, sizing one another up. This definitely did not feel like a normal doctor's appointment. Looking anywhere but at that paper, I noticed she had a big cooler next to her on the floor. This lady packed some extremely large lunches, but there was no medical equipment in sight. She had several black-and-white movie posters: _Dracula, The Wolf Man, The Mummy._ I looked back and caught her staring at me.
"What are you?" I asked. "I mean, obviously you're a doctor. A doctor of what, though?"
She glanced out the window to my left, craning her neck and frowning, as if checking something. Turning her attention back to me, she smiled politely. "I'm a psychiatrist." She tapped her forehead as if I didn't know what a psychiatrist was. "I'm on staff with the hospital. I teach at the university. I'm a specialist in childhood fears and post-traumatic stress disorder. Your mother thought you might benefit from my services. She found a paper online that I wrote on the role of monsters in the history of mental health."
It was hard to pay attention. She had huge white eyebrows that jumped and batted each other like kung fu caterpillars when she spoke.
"I don't need a psychiatrist."
Dr. Capistrano shrugged. "Your mother thinks you do. And the only other psychiatrist here who treats children works with functional constipation."
It was my turn to shrug.
"Children who refuse to poop." She glanced out the window, then scowled slightly. She turned back, smiling at me next as she folded her hands, resting them on her big pudding-bowl belly. "Don't be embarrassed by your interest in monsters, my dear. It's all very normal."
It was anything but normal, I wanted to yell.
"You see, by fearing monsters, we stay in control of our emotions. We use our fear of monsters to distract us from what we cannot manage ourselves. Monsters are a beautiful illusion, a dream that protects us from feeling pain."
The hair along my arms stood up, the same uneasy feeling creeping back: something definitely wasn't right.
"But what if you're wrong?" I asked.
The air around me turned cold.
"What if they're not an illusion?"
The doctor was rummaging through the cooler, grumbling beneath her breath, not listening to me. Did no one ever listen to what I was trying to say?
A monster hovered outside the window, watching us both. It had huge red feathered wings, each feather three or four feet long. Its body looked like a man's, but with bulging muscles. Its skin shimmered like gold. The skin on its face was black and lumpy and hung in folds. Instead of a mouth, the creature had a long beak, so white that it made its black eyes sparkle like a scalpel. I searched my mind for images from the book. What was it? What did it eat? I hope it didn't eat people, but if it did, my odds of survival were pretty good with Dr. Capistrano sitting there.
The monster floated outside the window as it watched us.
Nodding as if she was satisfied with something, Dr. Capistrano sat back and picked up a pen. Didn't she see the monster? She scribbled thoughtfully in my patient file, her head down. And how could she diagnose me so fast? She had no idea what the real problem was.
I looked back at the window. She might be about to find out.
"I have a patient waiting," the doctor said, smiling to herself as she recapped her pen. I knew she was lying; the waiting room had been empty. "This was a brief consultation, but I think I have all I need to make a recommendation for therapy." She snapped the folder closed and pointed to the door. "Let's go find your mom."
This monster must have followed me here. It needed me. I couldn't leave it floating outside, but I couldn't go out into the parking lot with my mom either.
"Go on." Dr. Capistrano motioned with her head for me to leave. "I'll be right behind you."
I stood still, goose bumps popping on my arms, my stomach flipping in every direction. I couldn't let it get inside. What if it tried to eat her? If a monster attacked a human, whose side was I on?
"I can't leave," I said weakly. "You might need me."
She paused and her eyes grew wide. "You see him, don't you?"
"Of course," I said, then looked back at her in surprise when it hit me. " _You_ see him too?"
"Cover your eyes, quickly," she demanded. I did, just as the creature crashed through the plate-glass window. I tilted my head down as glass flew everywhere, followed by a sharp cold wind. I opened my eyes to see its red wings spread and open, creating a dark screen that blocked out the sun. Fear rooted me in place.
"That's the third time in a month," she muttered. "By the saints, those windows are expensive."
The monster chirped and then clicked his razor-sharp beak at her.
"Excuse us a moment," she said, then pulled a human foot from the cooler and tossed it to the monster. He caught it in midair and took a step toward her, chirping again.
"Who's a good boy?" she cooed.
She pulled out another foot and tossed it. The creature craned its head, catching it easily and swallowing it whole.
Then it took a step toward me, talons on each foot scraping against the carpet.
"Don't be afraid, dear," she said. "He's not bad. He's just hungry."
The monster cocked its head and bobbed side to side, making clicking sounds in its throat.
I tucked my good foot under the chair to keep it safe. I was beginning to get a little protective of it after the Kappa. I used my prosthetic foot to scoot away from the creature. Without the Bestiary, I couldn't be sure, but it looked like the Native American monster the Thunderbird. A Jesuit priest had written of its existence in America back in the 1600s.
The monster danced from foot to foot, coming closer. I held my breath, my spine pressed against the back of the chair. It cawed loudly right into my face, the fetid burst causing my bandana to slip backward on my head. The smell of nasty, dead feet made my eyes water. I reached up to adjust the bandana, blinking back tears. The monster nudged my hand with its beak, like it wanted to be petted. I shrank back.
Turning, it looked out the window before flying out, eclipsing the gray sky with its red feathers.
"So you're a believer too." Dr. Capistrano clapped her hands together and studied me with delight. She closed the curtains to disguise the shattered window and sat back down. "Well, this changes everything."
"You're treating it like a pet," I said quietly. "You shouldn't do that."
She paused, studying me. "But I'm not afraid of it. _We're_ not afraid, right?"
I wasn't sure how to answer that. I helped monsters, but I didn't trust them. They were shadowy dreams and nightmares, but it was my job to protect them, not make friends.
"You keep body parts in there?" I pointed to the cooler.
She pulled the cooler closer to her knees and opened it. I stood and peeked inside.
"Most monsters are carrion eaters, and they like roadkill. A few prefer human flesh, so I buy cadaver parts from the medical college, or I just steal medical waste. You wouldn't believe what an arm and a leg cost."
I didn't laugh.
She frowned, obviously disappointed that I shared her secret but not her sense of humor. With a sigh, she set the arm back in the cooler. I sat back in my chair.
"We used to live in perfect harmony, you know," she said. "All over the world. Every monster had a tribe, every tribe lived in peace. Monsters were created to be scapegoats for the human race, and they accepted their job with grace."
She sighed as if the story hurt her to tell.
"Then medicine killed the monsters, in a manner of speaking. People don't have to feel fear anymore; we could take a pill. But science can never change the human heart. Without monsters, we're cruel to each other. We still need these creatures, Sofia. And they need us, because what could be worse than knowing you are invisible? That no one sees you?"
She leaned back in her chair and opened the desk drawer. Pulling out a bag of M&M's, she tore them open and tilted her head back, pouring a few in. I watched her chew, and then she spoke. "We'll save the manatees and protect forest snails, but no one gives a rip about monsters."
"I give a rip," I said, leaning toward her desk.
"When did you first see one?" she asked, her eyes narrowing. She was testing me, I think. But how much was I allowed to reveal?
"Recently." Vague answers might be best until I knew if I could trust her.
She jerked her head back. "Recently? That's odd. Little children sometimes see a monster, or the telltale signs that one is near, perhaps even under the bed, but no one takes them seriously. That's why adults rarely witness these miraculous creatures. Adults are not willing to believe, so they are not able to see." She clasped her hands together as if she were praying. "As a doctor, I was trained to set aside my personal prejudices, to listen and observe. Medical school made me curious about mysteries, not afraid. I had been a doctor for eight years when I saw my first monster." She lowered her voice. "So what happened recently? Something must have changed you."
I looked down at my knees. Xeno said not to share his secret, but Dr. Capistrano already knew it, didn't she? And what else did she know? I needed a lot of answers. Relief washed over me. I hadn't realized how hard it had been to do this alone.
"I was given a book," I said. "It belonged to Aristotle's last student, Xeno, and he uses it to communicate with me. He understands the monsters and helps me take care of them. But I'm not the only one who knows about the book, or the monsters. There's a ghost of a woman following me, or the book, but I don't know why." My words spilled out faster. "I don't think she wants the book or she would have taken it by now. I don't know what she wants, and I never have enough time..."
Dr. Capistrano's eyes were as wide as saucers. "It exists?" she whispered breathlessly. "The Bestiary? And you have it?" She paused. "You're the Guardian, aren't you? Their very survival depends on you." She began furiously scribbling notes in my file.
"I don't do much for them," I said, embarrassed.
"It's not what you do, it's who you are. This is extraordinary."
In the distance, a scream split the sky. I turned toward the window, once again frightened. The doctor didn't flinch; she must not have heard it.
"I'll help you in any way I can," Dr. Capistrano said, suddenly smiling. "You don't have to do this alone."
But I was alone. For reasons I couldn't quite explain yet, I knew that.
The door swung open. "I am so sorry," my mom said. "I was reading a magazine and looked at my watch and was shocked by how the time had flown. I know you said twenty minutes, tops, since you had to work us in between your other appointments. You should really hire a receptionist."
Dr. Capistrano stood and maneuvered herself to block my mom's view of the broken glass on the carpet, then turned to smile at Mom. "It's hard to keep good help." She took my mom by the arm and steered her back toward the door. "You have a wonderful daughter. Her mental health is excellent, but I do believe she would benefit from a few sessions."
"To talk about her recovery from cancer?" Mom asked.
Dr. Capistrano cocked her head to one side and looked back at me. We hadn't talked about my other problems at all. We'd both forgotten about them entirely once the monster had shown up.
"Yes, of course," the doctor said, then focused on me. "Sofia, I want you to consider me a friend. Call me anytime, day or night."
Mom treated me to dinner at the cheap Italian place we both loved. It had an aquarium in the middle of the room. There were huge fish in brilliant hues of green and blue, but my favorite had always been a yellow fish with bulging eyes and big white lips, because he looked startled every time he finished another lap around the tank.
On the way to our table, I checked all the windows and exits. I glanced toward the front door before sliding into the booth. I picked the side with a great view of the aquarium and studied the fish while Mom went to the buffet. She returned with two fully loaded plates.
"Look, Sofia, about the doctor, let me explain," she began.
A monster floated in the aquarium. He looked like a swollen, colorless frog, with huge gills that opened and closed while he swam, all six eyes watching me. He must have recognized me as the Guardian, because he merrily waved with his fin.
I frowned, trying to remember his classification. An Afanc, maybe—a British water monster. Before I could decide, he opened his mouth and ate my yellow fish. My jaw fell open in outrage. The Afanc continued to swim in lazy circles around the tank, selecting his next fish to eat.
Mom shook her head and talked faster. "No, no, let me explain. I'm worried about the changes in your behavior, that's all. I thought a neutral third party could help you open up about what's been bothering you."
"I have a date," I blurted. Ugh. What made my mouth disconnect from my brain like that? But it worked.
She set down her fork and changed the subject. "What? Where? Who?"
The Afanc hovered next to the sucker fish on the wall and opened its gaping maw. The sucker fish clung to the glass and didn't budge.
"The school dance on Saturday. A new guy named Billy asked me." The Afanc's lips made a sucking motion and the sucker fish flew in, along with most of the gravel along the bottom of the tank and a few more fish. The hostess stood in front of the aquarium checking her phone, oblivious.
I buried my face in the plate of spaghetti.
Mom sat back, a big smile fighting to bust out across her face. Her mouth twitched and she tried to stop it. "Is that what you didn't want to tell me? You have your first boyfriend?" She grinned. "That's wonderful."
This was so awkward it hurt. "No, not my first boyfriend."
She grinned even wider. "Do tell."
I hung my head. "The new guy asked and I said yes and I guess it's kind of a big deal. Everyone's going."
"Alexis?" Mom asked. I reached for a bread roll and tore it into four pieces, ignoring the question.
Mom sighed. "Well, we can talk about her later. I guess we need to go shopping for something to wear. I get paid on Friday. Maybe we can find something before then and ask the store to put it on hold."
"You won't have to pay for a dress," I said. "Candy's aunt owns a boutique at the mall. She's going to give me a dress in exchange for letting the paper take a few pictures. Candy really wants to help me with my new look. It's all she can talk about."
Mom frowned. "Candy?" She chewed her lip. "You've been hanging around Candy, instead of Alexis?"
"No, Mom, this has nothing to do with Alexis."
Mom looked confused. She had heard my stories about Candy for years. I could tell she wanted to ask more, so I took another big bite of spaghetti, and reached for another roll too. I hadn't eaten like this since before the diagnosis. She always worried that I didn't eat much. She looked happy and confused at the same time.
"I don't know how I feel about the paper taking pictures of you," she said finally.
"No one's ever asked to take a picture of me before."
"That woman I talked to in the grocery store," Mom began, her voice drifting as she thought about it, "she said something that stuck with me. She said girls who make an effort to look pretty and fit in with their peers have an easier time in life. I don't think that's always good, but I think it's true. If you're going to the dance, you need a nice dress."
"So you think I should take Candy up on her offer?" I asked.
Mom groaned. It was a tough decision. "So much has changed. Maybe Candy has too. All I know for certain is that I want you to be happy. You may have to try new things to make that happen."
Neither of us had anything to say to that. To avoid talking any more, I polished off my spaghetti, plus a salad and two breadsticks.
Mom paid the check and tucked the receipt into her wallet with all the others. When we stood to leave, I saw the hostess throw one hand over her mouth, having finally noticed the fish tank. It was completely empty: no fish, no plants, no gravel. The Afanc was gone. The hostess glanced around the restaurant wildly, like one of us had done it. I kept my head down and made for the front door. Mom was busy checking her cell phone and didn't look up.
On the way home, I had to move the seat belt away from my abdomen. I felt like I'd swallowed a bowling ball. This made Mom laugh. I kept my eyes closed, pretending to concentrate on my digestion.
I had never been a girl anyone wanted to take pictures of. I didn't think Candy was going to embarrass me or do anything cruel; she genuinely seemed like she wanted to help me. Her logic was probably twisted but right. People were nicer to me now because they felt sorry for me, but that wouldn't last forever. The medical term for it was "compassion fatigue." Soon everyone would be ready to forget my trauma and focus on their own problems, and they'd need a place to put me, a label for me to wear just like everyone else. I would go back to being the outsider...unless I changed.
I knew we couldn't afford a nice dress, but Mom wouldn't have to worry about the money. She'd be excited to see me dressed up to join my classmates for a fun night out. Mom would be relieved and happy and hopeful.
But what would _I_ be?
We stopped for gas on the way home. I tilted my head back and closed my eyes, glad for one quiet moment alone. The acrid smell of gasoline stung my nose as the clicking noise of the pump kept ticking on. I hated the sound of money flying out of Mom's wallet. I wished now that she hadn't bought me dinner, but I was glad I had made the right decision about the dress.
I certainly wasn't going to tell her that a monster from an ancient bestiary was largely responsible for my first date. There's only so much a mom can take.
Mom was having trouble with the pump. I heard her arguing with it and opened my eyes to watch her pressing the button for a receipt over and over. No receipt came out. She marched off toward the cashier inside the store. I closed my eyes again.
Her door opened. I opened my eyes to tell her how unbelievably fast she was when a smell hit me, worse than the gas smell before, if that was even possible. It was the Kappa. It slid into the driver's seat, eyes swirling fast in every direction like two pinwheels. Tentacles slithered out, then braided together as it reached for me.
I pushed back in my seat as far as I could go. "I don't have any cucumbers," I whispered. "Go away!"
Its tentacles were cold and wet as it grabbed my arms, pulling me toward its face.
"Please, no!" I glanced around and saw my mom waiting in line to talk to the cashier about the stupid receipt. "I'll get you a cucumber, okay? Just let me go."
The Kappa's jaws opened wide, saliva stretching like gummy rivers between its lips. Its mouth opened wider and wider, revealing a darkness with the stench of dead fish. Fog burst from its nostrils, and little tendrils of slime ran down its face. Its eyes slowly came to rest on me and I recognized something in them, my face only inches away from that awful mouth.
I tried to see why it had come back. Another fishhook? An accident? A craving for human flesh?
The Kappa whimpered, but I was the one helplessly trapped. Then I realized what it was. It was fear.
The Kappa was afraid.
"What is it?" I asked. I needed Xeno here. These beasts couldn't talk, at least not to me.
The Kappa whimpered again; then it released me and used its tentacles to cover its eyes. It wasn't here to eat me. It was here to be saved, but from what? I took a deep breath for courage and gagged. Kappas really stank. Steadying my nerves, I took another look at it. The body looked normal, at least for a Kappa. Nothing was wrong there. The face was still horrendous, which seemed like a good sign. So what was wrong?
It opened its mouth wider.
I peered inside. The smell was worse than spoiled shrimp left in the car trunk. My stomach closed like a hard fist, threatening to send my dinner spewing out.
Inside its mouth were hundreds of tiny Kappa babies. They squealed in terror when they saw my gigantic face looming in front of them, sounding like a hundred tiny balloons deflating at once. The Kappa snapped its jaws shut, nearly giving me that perky nose job I had always wanted.
This thing was a mother? Suddenly mine didn't seem so bad.
A low growl tore across the sky. People at other pumps looked up, as if they had just heard thunder. I knew it was something much worse.
I heard the car door close and turned back around. The Kappa was gone.
My heart was thumping twice as fast when Mom got back in the car. She wrinkled her nose and looked at me. "Whew! Garlic bread does not agree with you," she muttered, putting her finger on the button to roll down her window.
"No!" I said. "Don't." I didn't know what was happening or what was outside the car. "It's cold outside."
"Sorry," Mom said. "Turn on the heater if you want. You won't freeze to death."
"It isn't the weather that I'm afraid of," I muttered under my breath, but the night air was already whooshing into the car as we left the gas station. The stench of the Kappa faded away, and something new was being carried on the wind, vaguely familiar, a stale musky scent, one I remembered from visits to the pet store.
It smelled like a snake.
"Wow, you're getting good on the stairs," Mom said, trailing behind me. I hadn't realized I was moving so fast or that she would notice the difference. Being the Guardian was paying off in unexpected ways: as I pushed my body to work and move faster, it responded with grace.
"I work my core every day," I said, omitting the part about monsters being responsible for that too. Pausing at my bedroom, one hand on my doorknob, I smiled awkwardly. "Um, well, good night, then."
"You understand why I did it, don't you?" Mom asked, taking a step toward me. I didn't want her to come in my room for a late-night heart-to-heart. Something was wrong, and I had no idea what it was. Something awful was out there, something that scared even the Kappa.
Mom tried again. "You understand why I wanted you to talk to Dr. Capistrano?"
"I'm not mad at you," I said. "But I am really tired."
She opened her mouth, then shut it and smiled. "Good night, then. Sleep well."
The book was glowing dark green when I got inside the room and shut the door.
"I get to ask the questions tonight," I said. "And I need answers. About the woman who's been following me, Olympias, and a doctor who sees monsters too. And the Kappa, because it has babies in its mouth and it's scared and I think there's something awful lurking out there. Something that's not in your book. Something you didn't tell me about."
Flipping to the blank page, I saw that it was empty. He was there but he wasn't speaking. "I've done everything you asked, haven't I?"
> The Kappa is a mouth brooder. Mothers keep their young inside it when a bigger predator is near.
"A bigger predator? A monster that other monsters are afraid of?"
Xeno was silent. Did he not know the answer or was he afraid to tell me? Letters etched slowly onto the page as Xeno responded with small, frightened marks.
> Yes.
"What kind of monster? How can there be a new monster? I thought your book listed all of them. How could a brand-new monster suddenly appear?"
> Your birthday wish—do you remember it?
I did. On my thirteenth birthday, the nurses had put a candle in a hospital cafeteria cupcake and brought it to me. Mom was asleep in the chair by my bed. I leaned forward, and just before I pursed my lips to blow, I made a wish. I didn't want to seem ungrateful, because being alive was, after all, the best possible way to start another year, but birthday candles are made for big, selfish wishes.
"I don't know if I'll ever have another birthday, or another chance to make a wish," I whispered. "There are a lot of things I want to experience, and if I get to grow up, I will. But there's one thing that more time isn't going to give me. I've never been pretty. I've never been the girl who could star in the story. I've always just faded into the background of a crowd. So if I could have one wish, it's to look in the mirror and smile. To see myself as strong and powerful and... _exquisite._ " That word surprised me as it tumbled out, but it was perfect. It made my imaginary self seem beautiful and rare. That's what I wanted to be, really, not just some model in a magazine.
Opening my eyes, I blew out the candle and made the wish.
The flame flickered as if deciding, and just before it went out, the lights dimmed in my room, like my wish had interrupted some unseen electrical current.
The next morning I looked in the mirror, sighed in defeat, and decided to shave the rest of my hair off.
> That moment shook the supernatural realm. In the age of science and doubt, how could any child wish with such power? Every cell in your body vibrated and pulsed with the desire for what existed only in your imagination. You are so different from every Guardian of the past. You are a Guardian that neither Olympias nor I knew could be born in an age of unbelief.
I picked up the book and shook it. I had to grab his attention. "I don't understand what you're saying! Who created the new monster?"
> You did.
My body revolted; my mouth dried up and my stomach tensed as his words sank deeper into my brain like some poisonous arrow. It couldn't be true; this couldn't be my fault.
> His dark shadow has haunted many girls. They feel him near when they look in the mirror, when they walk alone down the school hallways. He is the voice which whispers that they will never belong, that they are unloved. His presence is often mistaken for the human emotion of shame, but it is something much worse: it is a monster begging to be born into the world, waiting for the one person who still remembers how to call fear into existence.
>
> If you have read about Olympias, you know she practiced a terrible and cruel magic. Like me, Olympias felt the power of your imagination when you made your wish, and knew you would make a strong ally. You would convince people to see the world through her eyes. Your fears, which were many and dark, gave the monster its form, and when you chose to become the Guardian, Olympias granted him life. He was created not to kill you, but to convince you.
"Convince me of what?" I asked.
> That you cannot be at peace with yourself, that you will never see beauty when you look in the mirror, that you will see only what you lack. You will see a void. He whispers that you must hide the wound because the world is not a safe place. Those who do not belong suffer.
My tongue felt like sandpaper in my mouth, and it was hard to speak. "Olympias used my fears against me."
The words erased slowly and a new message appeared, as if Xeno was carefully thinking as he wrote.
> She calls him Entropion.
"En- _tro-_ pee-on?" I repeated. "What kind of name is that?"
Xeno didn't reply. I guessed he wanted me to figure that part out on my own.
I thought about what Olympias had said in the park about differences leading to war. If everyone was alike, there would be no more suffering. Nobody likes different. The only way to stay safe in this world is to look like everyone else, to act like them and want the same things.
It reminded me of how weird it was to go with my mom to vote. She had to stand inside this box with privacy screens all around so no one knew who she voted for. And we were a country founded on the idea of freedom. So how come adults had to hide who they voted for?
But what Olympias didn't understand was that no one wanted to be like me, so I had no power or influence. It wasn't like I could convince anyone that being more like me was a good thing.
What good would I be as an ally?
"She said she was going to take things away from me, though." I paused. "Xeno? I'm scared."
> I do not know what waits for you out there. But, Sofia? I know what lives within you. I will put my hope in that. Now, you have work to do.
"What work?" I asked.
> First, you must create one last entry in the Bestiary. Entropion must be documented for future generations. And then you must find a way to overcome him and the power of his lies.
I suddenly realized I was now part of history. The thing about history, though?
Everyone's dead.
Wednesday, March 5
Ms. Hochness was sipping a cup of coffee when I hustled into the library the next morning. I had fifteen minutes until the first bell. Nothing else had come to my window last night, not even a moth, and Xeno hadn't said any more either. I had begged Mom to do me a favor later today, but I didn't know yet if she would come through.
I threw my book bag on a table and went to the row of books Ms. Hochness had shown me earlier.
"Whoa, where's the fire?" she asked.
"I think I missed something," I called to her. "I need to know more about Alexander the Great and his mother."
I grabbed the dictionary that sat next to the computers and looked up the _E_ s while she gathered a few books and carried them to a table. "Entropion: from the Greek word _'entropia,'_ meaning 'a turning inward against oneself.' " "Entropion" also described a medical condition affecting the eyes. So Entropion was created by the part of me that had turned against myself, the way I saw myself.
"You're working on a special project?" she called as she pulled books for me.
I looked up. Our eyes met and I nodded. "Very special."
She got my meaning. This was important.
She motioned for me to stand, and when I did, she took my place in front of the computer.
"I'll rummage around in the university's database. Sometimes they upload research articles. It's good stuff."
"Thanks," I said. I immediately sat at the table, opened one of the books, and dug in.
Fifteen minutes later, I had discovered nothing new. I dropped my head onto the book, groaning. There had to be something on how to stop a monster like Entropion, and Olympias probably held the only clue. Sucking in a deep breath, I sat up and went at it again. If I knew what sort of rituals she used, or magic she practiced, maybe I could find her weakness too. But it seemed like people had hated and feared her so much they'd been afraid to write about her.
By the time the bell rang, nothing new had turned up, and my shoulders hurt from bending over the books.
"Cool beans!" Ms. Hochness yelled.
I jumped up and made my way over to her. "No one says that anymore, Ms. Hochness."
"I just did." She smirked, hitting a button on her keyboard with a flourish. I arched one eyebrow.
She grinned. "An academic piece that someone at the teaching hospital was working on," she said. Behind her, the printer whirred to life. "It's not even finished," she continued. "I'm shocked that a professional would upload anything before it's finished."
I didn't care about any of that. I just wanted it.
The tardy bell rang.
I grabbed my book bag, then snatched the two pages from the printer. The second hand on the wall clock swept around again.
"Last one," Ms. Hochness said.
"Thank you!" I yelled, grabbing it from her as I hustled, making it to homeroom just in front of Ms. Forester, who was carrying a steaming cup of coffee. She tsked at me for brushing past her so fast. I apologized, but secretly I was pleased. It meant I was getting faster.
After I settled into my seat, I pulled out my textbook and laid the papers from the library over it. The little writing at the top of the page didn't catch my eye at first.
I had to reread the front page twice before it really sank in.
> The Mysterious Death of Aristotle and His Last Student
>
> By Dr. Inez Capistrano
I stared at the name in disbelief, vaguely aware that Ms. Forester was passing out a pop quiz on bathroom etiquette. Apparently, the janitors were refusing to clean the boys' bathrooms unless the boys learned some basic skills, like how to put paper towels in a waste basket. And aiming, I suppose.
The tiny time stamp at the top of Dr. Capistrano's paper showed when it was uploaded into the system. It read 4:36 a.m.
And today's date. So at 4:36 a.m. this morning, just a few hours ago, in the middle of the night, Dr. Capistrano had uploaded it, before it was even finished. But why?
The paper began,
> In investigating the history of Aristotle, in hopes of verifying the legend of his last student, Xeno, and reported sightings of a book popularly referred to as the Bestiary, a scholar must be able to separate fact from fiction. Fiction, however, would surely offer scholars a better story, because fiction holds the promise of a happy ending.
>
> As I have tonight learned, there can be no happy ending for the student who possesses the Bestiary. The book is cursed and should never have been opened again.
>
> I am sorry I could be of no further help, but I have chosen to move on to safer subjects.
>
> May the one who reads this take every precaution.
Ms. Forester grabbed my book and shut it. She then tapped her finger on the pop quiz now in front of me on my desk.
I was going to pass this quiz, but the bigger question was, what else was I going to fail, and who would pay the price?
After school, Mom turned out of the parking lot and headed in the direction of Dr. Capistrano's office.
I cleared my throat so I wouldn't sound too anxious. "Thanks again for getting an appointment."
"I didn't. No one answered the phone," she replied. "We'll just stop by and see if she can fit you in."
It was hard to tell if she was still frustrated with me or just really tired. Either way, this was not the moment to confess that I had helped create a super-monster that just might try to eat us both. _Sometimes,_ I told myself, _good relationships are all about timing._
Mom turned up the volume on the radio, listening to a report on a new variety of drug-resistant, flesh-eating bacteria. _Besides,_ I thought, _she has enough of her own monsters to deal with._
Mom glanced into the rearview mirror and pulled to the right. A police car zoomed past, lights flashing and sirens screaming.
We watched it take the same turn we were about to make toward Dr. Capistrano's office. A bad feeling deep in my bones grew stronger. The doctor had uploaded that paper at 4:36 a.m., but it hadn't been finished. Why couldn't she wait? What had she discovered? Had Entropion interrupted her?
I reviewed the paper in my mind one more time. The facts were simple. But the meaning escaped me. I knew enough from history to remember that Alexander the Great was still considered a military hero. But not many people knew about his mother. Bile flooded my mouth as I read on; Olympias was believed to practice a dark, evil magic, and snakes were said to be her closest companions. In fact, she told Alexander that he had been fathered by Zeus, who came to her one night in the form of a snake. That's why Alexander was worshipped as a god. And during his travels, he sent letters to Aristotle, detailing all he had discovered, so he had contact with his former teacher for years. Dr. Capistrano added one last note:
> Here the historical record ends, and we must imagine what other letters Aristotle received. If indeed a student had discovered fierce and terrible creatures in the new world, Alexander would not be able to force his troops to march upon them. The men would have been too terrified. And if Alexander as a god was not able to defeat a monster, he would be revealed as a fraud and failure. Alexander would have faced a difficult decision, to either send his men to their death or he himself die in dishonor.
>
> And then suddenly, Alexander the Great, the god, was dead. Some believe he was poisoned. Some believe that he had epilepsy and tried unsuccessfully to hide it, lest his men see him as weak.
>
> Olympias went mad with grief and surely must have craved revenge. Some wonder if Olympias blamed Aristotle for her son's death. We will never know, but within a year, Aristotle was found dead, then all his books mysteriously disappeared. Forever.
>
> Legend holds that Aristotle's last student, Xeno, saved a book documenting the existence of monstrous creatures, but no trace of him or the book has ever been seen. The last student who so dearly loved the truth has become a legend.
>
> And legends do not die. They are extinguished, often violently, in the name of progress.
Dr. Capistrano's paper ended abruptly with that detail. The sirens continued to race past us as an ambulance rushed toward the highway exit ramp.
Crime scene tape stretched across the entrance to Dr. Capistrano's building. A police van blocked my view of the doors, and four police cars were parked in front, their lights silently swirling. Dozens of people in uniform walked around with walkie-talkies.
Mom slowed the car and reached for my hand, finally stopping at the end of the street. Neither of us spoke.
A news van sped around the corner and raced right to the edge of the tape. A woman with a perfectly styled helmet of hair jumped out, a cameraman following. She tried to get a police officer to talk to her, but he gestured for her to back up and get out.
The police didn't want anyone knowing what had happened inside.
"Wait here," Mom said, getting out and walking away before I could argue.
Pressing a hand to my mouth, I rocked back and forth in the seat.
When Mom got back to the car, she opened the door and slid in, locking the doors and checking the mirrors. She cleared her throat and put the car in reverse.
"What happened?" I asked.
"Dr. Capistrano is dead," Mom said. She glanced in the rearview mirror, then sideways at me.
It felt like a block of ice had crashed into my chest.
"What happened?" I asked, my voice hoarse.
"Let's go home," she said.
"What happened?" I asked again.
She stopped at a red light, knuckles white, wrapped around the steering wheel.
"Mom. Please."
"It's better not to know."
"You can't protect me from this," I said.
She turned to look at me, and frowned. "Yes, I can," she said.
The light turned green but she didn't press the gas pedal. Her eyes were glazed over.
The car behind us honked twice, startling us both. Mom floored it and I jerked back in my seat. The city became a smudge, a blur racing across my window, trees whipping past, houses looming and then shrinking, people weaving in and out of focus.
Mom shook her head, as if giving up, and spoke quickly. "Dr. Capistrano's first patient this morning found the office door torn off its hinges. The police found a sweater that had been slashed to bits," she said. "They think, when the doctor came in to work, her murderer was waiting."
I swallowed down the sick feeling in the pit of my stomach. Entropion had been there.
Mom shook her head. "Plus...the police found a cooler with a leg in it." She glanced at me again and reached a hand out to rest it on my thigh. "Not a body. Just a leg. I am so sorry to have to tell you this."
I sighed in relief. That leg didn't mean Dr. Capistrano was dead. She'd uploaded that paper to warn me that she was getting out of this before Entropion got to her.
I settled down into the seat to think. Mom looked angry. "Sofia, you've watched too much TV. This is reality, not make-believe. A woman has lost her life."
"I'm just trying to understand it," I said, and it wasn't a lie.
Dr. Capistrano was the only person who knew my secret, and now she had fled, probably forever, because of the dangers. I should have obeyed Xeno and kept my mouth shut. Thankfully, I hadn't said anything to my mom. Now I was certain I had to carry this burden alone.
All I could do was wait for Olympias's next move. If I asked anyone for help, they would get hurt. Olympias wanted me to give up being the Guardian and let the monsters die. Xeno wanted me to keep the monsters alive until the world needed them. But neither of them had really answered the most basic question: why me? Did they think I was powerful because I survived cancer? That had been a battle, true, but I hadn't really won. Cancer had taken my leg and could still return for the rest of me. Every day, it took other kids, ones who really were brave and strong. Heroes are supposed to defeat the enemy, not just stay alive. If Xeno and Olympias expected the most powerful Guardian in history, they had picked the wrong kid. I knew how to fight, but I didn't know how to win.
Only Entropion knew I wasn't a hero. He knew the shame and fear I carried, because he was born from those foul thoughts, a secret truth that he and I shared. But Xeno was teaching me about the other truths that shape the world. One thing was clear: the truth could be a tool to build or a weapon to destroy.
Everything depended on the one who wielded it.
Thursday, March 6
Facing my own possible death, I knew the only logical next step was to go to the mall.
Mom had insisted I keep my shopping date with Candy the next day. Because that was "normal" and would be good for me, she promised, after the shock of hearing about Dr. Capistrano. I wasn't too worried about the doctor. I hoped she had made her escape safely; I just didn't know if I would.
I left fifth period early as planned and headed down the empty hallway to the front office to meet Candy. In the music room, the chamber choir was practicing a classical song with lots of long, high notes that gave me the chills. As I glanced down the opposite hall, a door at the end of it slowly closed, hinges squeaking like a haunted-house sound track. The choir continued chanting.
Something small and white floated past my head, and I brushed at it instinctively. A little clump of Styrofoam clung to my bandana, so I picked it off for inspection. I gazed at it, frowning, until I realized it was a piece of the ceiling tile. I snapped my head up to see the tiles shake and warp as something crawled above them.
Panicking, I walked as fast as I could to the office. My stump burned as it chafed against the prosthesis, but I didn't care. Moments like this were the reason I didn't feel guilty for skipping my rehab exercises. I got a good workout just trying to stay alive each day.
Candy smiled when she saw me swing the door open.
"Ready?" she asked. Before I could reply, she handed me a huge stack of books. "Can you put these in your bag? I only brought a purse today."
"I didn't know you studied so much," I said, struggling to adjust my stance under the load, the top book threatening to spill over.
"I don't," she said. "I just have to look like I do, to keep my dad off my back."
She turned to finish signing her check-out card. Her mom waved to us through the office windows as she stood beside her SUV.
The top book slid to the right, and I thrust my hips to the side to catch it. I couldn't carry all these without dropping one. I decided to adjust the stack, and set everything on the counter.
I opened my book bag to see if there was any room.
"Hey, Sofia. Hey, Candy." Alexis stood there watching me, a bottle of Pamprin in her hand. Girls could keep it in the nurse's office for their period if their parents approved.
Alexis stuffed the Pamprin in her pocket. "What are you doing?" she asked.
I just stood there, mute.
"We're going shopping," Candy said, stepping closer to me. "I'm giving Sofia a makeover."
Alexis looked at me like I had just kicked her grandmother in the crotch. "Seriously?" she said.
"Sofia, we better go," Candy said. "My aunt is waiting. I can't wait to try on the dresses she picked out." She acted like Alexis wasn't even in the room.
Alexis glanced again between me and Candy, then frowned. She was judging me.
Alexis always used to worry that she might not ever get her period, especially because of all the running she did. What if she never got curves, she said, and was always as skinny as a stick? Getting her period was hugely important to her. She wanted her body to change so maybe her sister wouldn't resent her anymore.
Turns out Alexis kept secrets from me too. She had gotten her period and never told me, and who knew when it had happened? Maybe when we were still talking, even. Everyone kept something hidden, even Alexis, but it felt like I was the only one being pushed out of the shadows. It wasn't fair.
"Sofia," Candy snapped, "we're wasting valuable shopping time."
Alexis's face clouded over again, darker this time.
"I can't wait," I replied, and followed Candy out.
As for Alexis? Our friendship was officially dead. I knew I should be sad, maybe even devastated, but mainly I was relieved. It's weird how someone can be so important to you that sometimes it's easier to let go than to hold on. If everything had to change, I guessed my friends would too.
Everywhere I looked, headless women were frozen in unnatural poses. Some had heads, but then their faces were all white, no eyes or mouth, nothing to tell any of them apart. I wanted to enjoy the mall, especially since I didn't have to pay for the dress I was going to get. Everyone loves free, right? But I had never noticed how creepy mannequins were, maybe because I had always been double-checking how much cash I had. The worst feeling in the world was not having enough money when you were at the register and adults were in line behind you, tapping their feet impatiently.
Candy's aunt, Mrs. Baker, had a whole rack of dresses already selected for us to try on. They were gorgeous. I lifted the tag on one that caught my eye. Black, mid-length, form-fitting to create some serious curves. It was $355. I dropped the tag.
"Don't be silly, Sofia, that one is for me," Candy said. She pushed it farther down on the rack, separating the dresses into two sections. "It wouldn't work for you anyway. Not with...just trust me."
I looked down at myself, then shrugged. It was hard to ignore the stinging in my throat, the warning that I was going to cry. But Candy was probably right. What did I know about fashion?
The dresses for me were all long and straight, like Popsicle wrappers. I had never worn a long dress, except maybe a nightgown when I was little.
Mrs. Baker stood in front of the rack and studied my body. Then my face. When she pointed to my bandana, I shook my head.
"That's all right," she said. "You can remove it when you're ready."
"I won't be ready today," I countered, frowning to make sure she took me seriously.
"Well," she said, turning back to the dresses, "let's take this one step at a time."
I didn't hate her, which was a surprise since she was related to Candy. She was sweet and soft-spoken, and I liked how she wore loose black pants with a fuzzy sweater. She also wore little pearl earrings and a plain silver necklace. Mrs. Baker was like an elegant librarian. It seemed to be an achievable fashion goal, even for someone like me.
Candy grabbed her selections and disappeared into a dressing room.
Mrs. Baker chose the first dress for me to try. It was shapeless. Just a straight gray tube of shimmering fabric. But gray could be a good color, I tried to tell myself. Gray was the color of useful things, like knives and scissors and surgical instruments.
No, I corrected myself. Instruments like the flute. Shrill and annoying when badly played. Good grief, why was I even here? I was a thrift-store shopper. This was not my world.
Mrs. Baker followed me to the dressing room, making polite chitchat about how my day was and whether I wanted a soda or bottle of water. If she could tell how nervous I was, she didn't show it.
Outside the dressing room was a table with several trays on top. Inside two of them were different pieces of jewelry. The other three held cosmetics and hair products and accessories like ribbons and barrettes.
My stomach bounced down to my knees. I didn't have enough hair to use ribbons or barrettes. I would look like an idiot if she decorated my head.
Mrs. Baker waited outside the door while I stepped into the dressing room.
The inside was pretty. It had striped wallpaper, a chair, but no mirror. Was that good or bad? I wondered. Did she give me this room on purpose?
"The photographer called," she said.
I had almost forgotten about that horror. In real life I could get away and escape from people staring at me, but in a photograph I'd be frozen. I wouldn't be able to hide.
I struggled to get into the dress. The zipper was in back. Sweat broke out on my upper lip as I writhed and wriggled and yanked at it. If I braced against the wall with one hand to keep my balance, I couldn't quite get the zipper to hold still.
"He's stuck at the airport in Boston," Mrs. Baker continued. "Would you mind if we took the pictures at the dance instead? The lighting won't be as good, but I'll make him promise to take an extra shot of you and Candy together."
Wonderful. Something for the fridge.
I didn't answer, because I knew it was a rhetorical question. After a few more embarrassing grunts, I defeated the zipper. Before I could celebrate, though, I realized I had to use the big three-way mirror at the end of the dressing-room hallway next.
Maybe in these fancy stores, it wasn't your own opinion that counted. Other people had to approve too. I ran my hands down the front of the dress, feeling the weak spot where the Kappa had bit me.
Opening the door slightly, I motioned for her to come in. Instead, she reached for my hand and pulled. I felt like a turtle getting yanked from its shell. Mrs. Baker was surprisingly strong.
Standing there in a six-hundred-dollar dress that covered me from neck to ankle, I had never felt more exposed.
Mrs. Baker slapped her hands to her cheeks.
"I'll change," I said quickly. "It probably looks bad."
"No," she gasped. "It's perfect! Just wait." She disappeared for a minute, then reappeared with two boxes of shoes. When she opened both boxes, I saw that the two pairs were the exact same style. She grinned and pulled one shoe out of each box. "You're not my first customer with a prosthesis," she whispered.
Squatting at my feet, she helped me into the shoes. "Since you probably haven't worn heels yet, let me tell you a little secret. Sometimes you'll need a different size for each foot when you buy high heels. A prosthesis has a different fit in heels, even if it's the same size as your natural foot."
If I wanted to wear anything but tennis shoes, that meant I'd have to buy two pairs. We could never afford two pairs of new shoes at one time. I'd have to cry later; Mrs. Baker had already started on my face. Soft brushes tickled my chin and nose. Then a cold metal necklace circled my neck. Next, she reached for my bandana.
My hand caught hers in midair.
"Time to take a leap of faith," she whispered.
I didn't know if that was a joke but I let go of her hand and closed my eyes.
The bandana slid off. A light comb pulled through my short stubble up top. A cold spray followed, then more combing. The combing felt good, like my mom's fingernails on my back.
"Ready?" she asked.
I took a deep breath and opened my eyes.
"I want to see!" Candy squealed from her dressing room. I heard her door handle turning.
Mrs. Baker stepped to my side. Candy stood in between me and the mirror.
Candy stared at me. She didn't move. I had never seen that expression on anyone's face when they looked at me. I didn't know what it meant.
I walked around her to the mirror, holding my breath. There in front of me stood someone I had never met.
She was beautiful.
_I_ was beautiful.
The long dress hid everything I hated about my body. My toilet-scrubber hair was slicked back and stylish. Mrs. Baker had draped me with blue and silver jewelry, and my eyes reflected the glittery shine.
"I look like a model in a magazine," I said softly. "Exquisite, even."
"I know," Candy said. "You look amazing. You don't even look real."
I turned side to side, admiring myself. This was what I had wished for, wasn't it? I wouldn't blend in and be forgotten, ever again.
I used to hate that no one noticed me. Then I came back after I got sick, and it was like I was doubly invisible, because people still didn't see me. They just saw a new label: cancer. That was the loneliest feeling in the world, to realize that despite everything I was going through, and all the ways I was changing, people just saw a different label. But now I realized that on my birthday, I had wished for the wrong thing. I had wanted to look in the mirror and like what I saw. I should have wished to look in the mirror and see someone else.
That was what Candy was trying to help me understand. I didn't have to show anyone the real me, ever.
I could show them something _better._
I thought of Billy suddenly and blushed. What would he see? Would he think I was beautiful too?
"All you had to do was trust me," Candy said. She moved to stand side by side in the mirror. The air around us turned cold. "I knew you'd like what you saw."
I suddenly thought of Entropion. If I wasn't ashamed of myself, his lies held no power. Maybe this was the way to defeat him, by beating him at his own game. I didn't have to like the real me. I only had to be someone different, and I had just learned that it wasn't really that hard.
Mrs. Baker walked up behind me on the other side and put her arm around my shoulders. She shivered and glanced at the thermostat on the wall.
"Well, thank you, Mrs. Baker," I finally said. "And...thanks, Candy."
"This is going to change everything," Candy replied.
A rumbling noise echoed in the air-conditioning vent above the dressing rooms.
"We should get going," I said, glancing up. "I have a ton of homework." I didn't know what that noise was, but I didn't want to find out. Had something followed me from school, maybe the thing that had been hiding in the ceiling?
After I changed, Mrs. Baker took the dress and put it in a bag with the hanger at the top, and she put the shoes in a velvet bag too. I had never owned anything this expensive. At the store where I shopped, all the clothes got thrown together into one plastic bag. Walking out, I passed by the faceless mannequins again. One head swiveled slowly as I passed, and out of the corner of my eye, I saw it watching me with light gray eyes, just like Olympias.
I walked so fast Candy was breathing hard when we got to the car. I took a moment to tie a bandana back on my head, even though Candy rolled her eyes when I did. I wasn't ready for anyone else to see my hair.
Candy talked the whole way home about everything she still had to do before the dance. I tried to listen, since she had just scored me a dress and shoes, and her aunt had thrown in a bunch of makeup samples. Thankfully, we had a short drive to my mom's office.
I opened the car window to allow cold air to blow onto my face. A bunch of plain brown sparrows sat on a telephone wire. We had studied the life cycle of sparrows in science last year. The birds looked at us when we were stopped at a light. Not one of them was beautiful. No one went to the zoo to see sparrows; they weren't special. My science teacher said most sparrows don't even survive after they hatch. If they do, they only live a few years.
"So if you're one of them, what's the point of living?" the teacher had asked, obviously hoping that a student would have some profound insight about the circle of life. No one did. And the teacher didn't even answer her own question, so it hung in the air all period like a dark cloud. We didn't want to try and explain the meaning of death. Most of us were still trying to figure out if we needed to start wearing deodorant.
I was still staring out the window at the sparrows when a shiny black SUV drove past. A woman with long blond hair and sunglasses was driving. She didn't look at me or turn her head, but somehow I knew she was smiling. And I knew exactly who she was.
Friday, March 7
I slept fitfully and awakened early to the sound of weeping. I pushed myself to sit up, groaning with the effort. I hopped to the window and looked down to where the noise was coming from. Caitlyn, my next-door kindergartner neighbor, stood at the edge of the street, in tears. Her father said something to comfort her that I couldn't hear.
"But I told you!" she wailed. "I told you that we shouldn't let Newman go outside last night. There was a monster in the bushes!" When he shook his head, she stomped her foot. "I saw it!"
He wrapped his arm around her shoulders, trying to nudge her home. Behind his back, he held a collar with a dangling tag. It looked like it had been ripped from the cat's neck.
Caitlyn refused to move, and stood with her head bent down to her chest. Her father finally had to pick her up and carry her home. That's when I saw it.
A huge red bloodstain was smeared across the road. It must have been Newman...or what was left of him.
I glanced up and down the street but saw nothing. If a car had hit Newman, there would have been a body. But Caitlyn wasn't crying over a body. Newman was dead, and it wasn't a car that got him.
The little hairs on my neck rose. If Entropion was out there, why had he attacked Newman? Cats weren't monsters. I mean, most cats weren't.
"Sofia! Breakfast!" Mom called from downstairs.
I grabbed an outfit and got dressed. "I need to see if I left something in the car," I called when I got downstairs. The street was quiet and empty. I walked to the stain in the road and bent down to inspect it. It didn't look like blood—at least, not animal blood, not if you really stared at it. It had a greenish-blue tint. It looked like...
Newman meowed at me from the bushes.
Turning, I made my way to him, relief flooding through me. He pranced out, looking offended that he had been forced to hide. Poor Newman; he had been eaten and vomited up by the Beast of Gevaudan, and now this. I reached to scoop him up and carry him home to Caitlyn. I couldn't wait to ring the doorbell and see her face.
As I reached for him, I heard a little squish. I lifted my right foot and looked down. Part of a tentacle clung to the sole of my shoe.
Entropion hadn't killed a cat. Entropion had killed a Kappa.
He wasn't going to let any of the monsters get to me. Even if they needed help, he'd stop them. I would be useless as the Guardian.
Newman hissed when he saw the bloodstain, and leaped from my arms. There weren't going to be any happy endings today.
> Hijacking Darwin: The Science of Evil
The banner in front of the Natural History Museum hung lifeless in the morning light. I'd forgotten all about the science field trip, just like I'd forgotten my paper on evolution was due in less than a week. I needed this. My assigned topic was on how evolution impacted the balance of power in any ecosystem. I wasn't even sure what that meant. The bus ride here had been fairly miserable, with Mr. Reeves having us sing the school song twice to fire up the track team for their meet on Saturday.
I avoided looking anyone in the eyes.
When the bus doors opened, Billy called out my name and told me to wait. Mr. Reeves had assigned Billy a seat with a chaperone in the back. I pretended I didn't hear him and hustled down the steps, scanning the area for Entropion. I didn't know what he looked like yet, but there was nothing else I could think of to do.
Inside the museum, the exhibit was cold and dark. There were black-and-white pictures on the walls, and war relics from all over the world in square glass boxes. A lot of the exhibit had to do with Nazi Germany. I saw pictures of children with Down syndrome sitting on an examination bench without any clothes on, so skinny their ribs were visible under the skin. They were smiling, as if the bright camera flash were a sparkling promise that these men in uniform were humans too, and weren't all people basically good?
The Golem came to mind, his back riddled with buckshot. I dropped my head, my heart suddenly hurting. Had he been shot out of fear or for fun? Would either answer make it right? Not all humans were good. And even good people could do bad things. The monsters were gone from our lives, but we had never felt less safe.
An old guy in a crisp museum-guide uniform clapped his hands to signal the beginning of our tour. He had white hair trimmed into sharp clean lines at the temples, and a deep sadness that haunted his eyes. When he turned his head to one side, I saw a red scar, old and flat, that ran from ear to mouth.
Our first stop was an exhibit called "Survival of the Fittest?" As we filed in, Candy caught my eye and smiled. "What did your mom think of the dress?" she asked. "I told everyone you're going to look amazing in it."
The museum guide was talking, so I turned to listen.
"As you see, the natural world is extraordinary in its diversity. Whether we are looking at a lowly insect or an apex predator, every creature is wonderful in its own way. Some would have you believe that these creatures"—here he waved to all the pictures—"must destroy each other to survive. Survival of the fittest, they call it. I call it utter nonsense."
We all glanced at each other, confused. Wasn't that what evolution was all about? The strong gobbling up the weak?
The museum guide continued. "Competition does _not_ determine survival, not in the way you may think. Species survive by being _different,_ not stronger. The dinosaurs? So strong the earth trembled beneath them." He lowered his voice as he spoke and made fists, imitating a strong beast. Then he shrugged. "And today, of course, quite dead."
A gloomy feeling settled among us. He pointed a finger, sweeping it past all our faces, then jabbing it at the pictures of the strange creatures from nature.
"To survive, one must adapt, change, do the unexpected."
Candy nudged Natalie and whispered something to her. Natalie just shrugged.
He cleared his throat. "The mistake that Hitler's men made was in not understanding the true power of evolution. Survival does not go to the strong or the beautiful, not to the fast or smart. Only the _unique_ live on, those with the rarest of strengths: the ability to stand alone for the sake of the greater good." He pulled out a linen handkerchief and dabbed at his mouth. The corner of his mouth drooped when he talked.
"He's saying it's better to be a freak," Candy whispered, but loud enough for us all to hear. A bunch of people giggled. "Good luck with that."
The guide stared at Candy. Something in his eyes told me that he was too tired to argue with her, like maybe he had already fought that fight a thousand times and knew how it ended.
The guide waved his hands, motioning for us to line up to walk to another exhibit. He made a wide path around Candy.
Without warning, Billy was right next to me, his shoulder grazing mine. I pretended not to notice. I worried he might sense the way I had been thinking about him when I looked at myself in the dress.
"Good grief, I can't leave you alone for a minute," he said. "I just heard you went shopping with Candy for a dress."
"I had to," I said. "Long story."
"We have the whole bus ride back."
Thankfully, the guide continued the lecture. "The Nazis tried to create a superior human race. Those inferior were identified by what made them different: religion, disabilities, artistic gifts, sexual orientation, being left-handed or just noticeably strange."
Several people snuck glances at me. Instinctively, I reached up and touched my bandana to make sure it hadn't slipped.
"Hitler wanted to control evolution to gain power," the guide said. He pointed to a picture of a soldier shooting a mother cradling an infant in her arms. Some of the girls in our class started crying. Hitler was a real monster, I thought, the kind even Xeno would be afraid of.
I looked down to where my leg should have been.
How could anyone believe they could control nature?
The exhibit had more photographs, but my brain shut down. It was an eternity before we got the signal to head back to the tour buses. I was the first one out the door, and that was when I saw it.
The first flower of spring.
The first real, growing flower I had seen up close since last fall, when they admitted me to the hospital. The world had been feeling gray for so long. But here was a perfect tulip shooting straight up from a bulb buried in the cold dirt, its thick red petals a signal to the others to do the same. None of the flowers were blooming yet, but there's always one that can't wait to feel the sun.
Mr. Reeves made me keep moving, but I turned just before getting on the bus for one last look at it.
Candy hovered, casting a shadow over the tulip. Then she reached down and, using her fingernails, severed the blossom from the stalk. She held it in her hand, showing it off as her friends gathered. A teacher yelled at them to get moving. Candy tossed it aside as she walked onto her bus, and the first bloom of spring was trampled by the rest of the seventh grade.
I swallowed hard, trying to keep my tears from spilling.
Billy climbed onto the bus and was just about to sit down next to me, but with a shake of his head, Mr. Reeves shot his long arm out and pointed to the back of the bus. Billy huffed in protest but the kids in line pushed him down the aisle. He took his seat about six or seven rows behind me.
Mr. Reeves sat in the row across from mine and turned toward me, giving me a kind smile. He almost looked like he felt guilty for making me see that exhibit. I turned to look out the window. Everyone knew I would have been chosen for extinction. Why did evil people pick on people who were different? Why were they so afraid of us? I was different because I was less than perfect. How could that be a threat to anyone?
I was directly behind the driver, with his seat blocking most of my view. Mr. Reeves had plenty of extra space, with no rows in front of him. He was busy making notes in his notebook.
Mr. Reeves got up to borrow the driver's intercom mike. "Traffic may be a little heavy as we head back to the school. Please talk quietly so we don't disturb our driver. If you brought a snack, you may eat it, but remember to pick up your trash."
I looked back at the museum as the windows darkened in the afternoon sun. No one was safe. Not if we were different, out of order...or just the first to sense a change coming. My eyes began to fill up again.
Billy cleared his throat loudly. I refused to look back.
"Sofia!"
I jerked my head around to face him.
"What's wrong?" he asked.
"I don't want to talk about it." The last thing I wanted was for everyone on the bus to know why I was crying.
The bus rumbled forward and I settled farther down in my seat, resting my head against the window. The sun warmed the glass and my eyelids grew heavy.
"Sofia!"
I sat up and turned around again. "What!" I snapped at Billy.
"Is it what I said in the museum?" he called back.
In the rearview mirror, I saw the bus driver raise his eyebrows, irritated with us. Mr. Reeves turned around and warned Billy not to get out of his seat.
Facing forward, I refused to look back again. Then I realized I could spy on Billy by looking in the driver's mirror, and snuck a glance.
Billy caught me. Before I could look away, he held up an orange. He had used a Sharpie to draw a smiley face in what I think was supposed to represent me.
I turned around. "Am I supposed to be flattered?"
"You're supposed to laugh. I'm trying to cheer you up."
We were pulling into the school lot when Billy sent the orange rolling, fast, straight up the center of the aisle toward me.
A dad on his cell phone drove his SUV too close to the bus, and our driver had to swerve to avoid him. Instead of landing in my hand, the orange shot forward and lodged itself under the brake pedal just as the bus driver slammed his foot on the brakes.
The orange exploded.
The bus skidded to a stop, fleshy bits of orange pulp dripping down the driver's face. He pulled the lever to open the doors and stumbled down the steps, collapsing like someone had kicked him in the back of the knees. When he caught his breath, the driver shouted an impressive list of four-letter words at us.
I heard Billy gulp.
No one else made a sound.
No one even dared to breathe.
Carefully, quietly, Mr. Reeves bent down and retrieved the burst orange, holding it up for us all to see. If I had ever wondered what my face would look like if it was orange and completely smashed in, now I knew.
I was standing by myself as the rest of our class gossiped, their backs turned, eyes darting back and forth between me and Billy. Alexis was listening to her friends from the track team. A few of them had been on my bus, and they were filling her in on the incident.
Parents started to arrive. I could tell the story was traveling fast by the adults' expression when they looked back at me. As if this were my fault.
Alexis turned away from her friends and stormed toward Billy.
"What is wrong with you?" Alexis snapped at him. Mr. Reeves raised his eyebrows, but he made no move to stop Alexis. I think he was hoping she would punch him.
"What, because I'm her friend?" Billy asked, like Alexis had asked the most obvious, ridiculous question ever. "Something's wrong with me because I care how she feels?"
"Something's wrong with you because you won't leave her alone! She doesn't want you bothering her! Anyone can see that, you moron! Look at her!" Alexis shouted as she pointed at me.
I looked away.
Billy took a step toward her. He got right in her face.
"Who cares what you think?" Billy shot back.
Alexis shoved him, both hands on his chest. "I'm her _best friend_!" she shouted.
My heart felt like someone was squeezing it. Billy looked down at the ground. I could see his jaw muscles flexing as he cleared his throat.
"I didn't know you already had a best friend," Billy said to me when he finally looked up. His eyes were hard and cold.
Why did it matter if she was a best friend instead of just a friend? Was it against the rules to have more than one?
"You should have said something," Billy said. "I don't want to get caught in the middle. I don't want to be a _liability._ "
My toes curled in my shoe. If I had a zipper on my back, I would have unpeeled myself from my body and run.
Alexis held her ground, arms crossed. They both looked at me like I owed them an explanation.
"She _was_ my best friend," I finally said. I wanted to look at Alexis, but I couldn't. "It's complicated."
"Why are you doing this to me?" Alexis uncrossed her arms and moved closer. She moved like an athlete, strong and confident, even in short strides. "Just give me one good reason."
"To save you the trouble," I said. "You don't want to be friends with me anymore."
"Yes, I do," she said, and stomped one foot. "Why don't you believe that?"
"When we were friends," I said, "and we would run together..." I had to stop. My throat burned from the sobs I kept swallowing back. Saying each word was like chewing razors. I took a breath and tried again. "We'd be racing, and your shoes would be rubbing a blister, and then I'd have to stop and vomit because I always ate the wrong thing for breakfast, but you never let us stop or give up. If there's one thing you really hate in life, it's pain. You said there was too much of it already. You always said you refused to let pain win."
"Because I'm not a quitter," she said. "I don't quit on races...or friends."
A tear burned my cheek and I wiped it away, fast. Billy frowned in concentration as he listened and watched.
At our very first cross-country practice, Alexis and I hadn't even known each other's names. The coach had divided the girls into two groups to encourage teamwork, and Alexis and I weren't in the same group. The average pace for a middle school runner is twelve minutes and fourteen seconds per mile, so our goal was to run a 5K in under thirty-eight minutes. That would qualify us as average, our coach said, but he wanted better-than-average runners, so he would be standing on the track, clipboard in hand. We had twelve laps to impress him.
I knew I was going to do that.
I hadn't trained much; I wasn't a great runner, so I didn't see the point of training really hard. I sprinted from the start line and ran until my sides burned in agony. Before the first lap was even done, Alexis blew past me in the lead for her group. On the third lap she saw me: my hands on my knees, breathing hard and crying. I knew I wasn't a great runner, but I hadn't known that I would humiliate myself in front of everyone. I couldn't even finish all twelve laps unless I walked.
Other girls passed by but Alexis circled back for me. "We'll run together," she had said. "If you run with a friend, it doesn't hurt as bad."
We introduced ourselves when I stopped to stretch out a cramp in my calf.
Alexis never left my side after that. She was strong enough for both of us. Even when it meant she had to run extra portions of a race just to keep me company. I didn't know why she wanted to be friends with me. She kind of dazzled me, but after I found her crying in the bathroom, I knew she had issues at home so she needed a good friend. I decided to keep my mouth shut and hope that a little of her strength would rub off on me. She really only had one rule: it didn't matter if we lost, as long as the pain didn't win.
Remembering it all, I took a deep breath that made me shudder. "You tried to visit me in the hospital, right?"
Alexis glared at me, but I saw tears in her eyes. "You know I did."
"But you didn't come back after that. Maybe you knew I wasn't strong enough. The pain won, Alexis. Look at me!" I gestured with one hand to my body. "How could I be your friend like this? You'd feel guilty that you can run and I can't, and then you might even stop. I would ruin the one thing you love most."
"Are you crazy?" she yelled. She looked like she was trying not to explode. Billy stepped toward her, raising his arms in case he needed to grab her. I shook my head at him angrily. This was my fight.
Alexis chewed her lip, rolling the bottom one under her front teeth, mentally working something through before she said it. She started three separate times before she finally spoke.
"When we raced, there were rules," she said, holding her hands out, palms up, for emphasis. "If we followed the rules, we might win. But if we broke the rules, we got disqualified. _No matter how hard we ran._ Your cancer? That was a race that you had to run without me." Tears slipped down her cheeks. "Yeah, I love running. But I love you too."
My own tears were building. Big swallows of air kept me from completely disintegrating when she continued.
"And I didn't want to hurt your chances," she went on. "I left you alone in the hospital, because that's what your mom wanted me to do! She said I might give you another infection. And then suddenly you didn't want to be my friend anymore. I didn't know why and you wouldn't even talk to me to explain it, but I would never, ever quit on you. So I waited for you to circle back for me this time. And you never did!"
"Because I can't!" I screamed. "You have to go on without me. You _get_ to go on! And I can't be happy for you, not yet. I'm angry! It's not fair, not to either one of us. Every time I look at you, I feel sad and angry, and every time you look at me, you feel guilty."
She briefly looked away, then shook her head, but we both knew I was right. I knew how she felt when her sister looked at her and how much Alexis hated having the one thing her sister wanted more than anything else. I knew the truth, and it cost us our friendship.
So what hurts worse, truth or love?
Both, because they end up being the same thing.
I buried my face in my hands and bawled.
Alexis held out her arms to wrap me up in a hug, but instinctively I took a step back. I knew that the cracks around me weren't the good kind. If she touched me, I would break into a thousand sharp pieces.
Billy stepped between us, holding his hands out to keep Alexis back, like he knew what it meant to be shattered.
"Leave us alone," Alexis said to Billy. "Go find someone else to stalk."
Billy's face turned red and I watched Mom's car turn in to the lot.
"You abandoned her," Billy said. "I would never do that, no matter who told me to. A _real_ friend stays."
Alexis pulled one arm back to shove him, just as I tried to twist around and push Billy out of her path.
Mom got out of the car and waved just as my hand hit Billy's face. He fell, hands covering his face, blood seeping out between his fingers. I had accidentally smashed his nose with the flat of my palm.
"Wow," Alexis whispered. "Just...wow." She slowly walked off and glanced over her shoulder at me, once.
I stood there, my head swimming. Everything about the moment had gone wrong. I had finally spoken my truth, and it wasn't a beautiful thing, not at all. My truth was not something to write poetry about or frame on a poster. It was ugly and small and screaming in pain, but it was mine.
So what was I going to do about it?
I watched from the passenger seat of our car as Mom and Mr. Reeves talked. Billy was holding an ice pack to his face, and a girl was laughing at something he had just said. She flung her hair and must have said something witty, because he laughed next. It hurt to watch them. Why had he completely shut down when he learned that Alexis and I were best friends? And were we best friends, even now that she knew how I felt? She'd have to be either strong or crazy to be my best friend after today.
My attention snapped back to the adults deciding my fate. Their discussion involved a lot of finger-pointing. Mom straightened her back and went toe-to-toe with Mr. Reeves to stare him down.
But then she said something and his whole face changed. He softened and reached out to touch her arm. He pulled out his wallet and showed her a picture. She smiled and nodded, and touched his arm in the same spot. I had never seen my mom flirt. I rubbed my eyes, trying to stop the damage, but it was no use. The image was seared on my retinas forever.
She walked slowly back to the car. The expression etched on her face was a work of art, a delicately balanced composition of shock and fury.
My life was a constant loop of misunderstanding.
"Well, you had a memorable field trip," Mom chirped after a few minutes of silence. She merged the car onto the highway. Apparently there had also been some confusion as to whether I had tried to stage a mock explosion of the driver's head, and whether I had meant to punch Billy. The car was cold and she reached for the heater, but accidentally turned it to AC.
Her knuckles were white but she didn't notice the drop to subzero temperature and I didn't dare move. Or speak.
"I should probably blame myself," she said. "I pushed you to be a normal kid again. Perhaps I should have been more specific. A normal _law-abiding_ kid with good grades and nice friends!"
She hit the wheel with the palm of her fist. "Mr. Reeves felt sorry for me," she sputtered, like the words left a bad taste in her mouth. "Like I need pity, since I'm a single mom."
Pity rubbed us both the wrong way. We didn't speak again until she pulled into the driveway and turned off the engine.
"What is going on with you?" she demanded.
I shook my head and looked at my lap. If I said what I was thinking—that she was only getting a little taste of my everyday existence—we'd end up in a huge fight. Should I tell her that the boy I had just injured might be my first boyfriend? That I had recently been a beacon of hope for terrifying monsters from all over the world? That what I wanted more than anything right now was to be able to talk to a dead man? Or that I had accidentally helped create a new predator?
"Where is the Sofia I know?" Her voice got quiet. "I want that girl back. I miss her."
"If I told you the truth, you wouldn't believe me."
"Try me," she said.
I folded my arms. "I didn't like the old me. No one liked her, Mom."
"That's not true."
"I'm not saying I was bullied, not exactly. But no one paid attention to me, except to make sure I wasn't in their clique," I said. "No one wanted me around, and just because people feel sorry for me now doesn't mean they suddenly like me. I still don't belong anywhere."
"But you can't live like this. I can't live like this! Not one more day," she said. "Something has got to change. I just want you to be happy."
I turned my face to the window. "I'm trying," I whispered.
My mind went back to everything in the museum exhibit. If all the animals in a species were exactly like each other, eventually they all died. But if one was born different? It died first.
And that wasn't the worst part. I had learned the worst part in the last exhibit.
Change was useless unless it was the right adaption for what was about to happen—and no one ever knew what that would be.
Evolution was so unfair.
We pulled into the driveway a few minutes later, and I glanced up at my window. It was open, the curtains fluttering in the early evening breeze. I was sure I hadn't left my window open. Mom got out, lugging her briefcase from the car, and I followed.
She turned the doorknob and nudged the door open with her shoulder. A small brown package sat on our doorstep with a return label that read "Germs-B-Gone." Mom bent down and tried to hide it behind her briefcase, knowing I would probably say something snarky.
That was why I saw the inside of our house first.
The couch was standing on its end against the living room wall, where huge chunks of drywall had been ripped out. Our pictures were facedown on the floor. Papers and books had been thrown everywhere and torn to shreds.
Mom saw my face and turned slowly. She dropped her package and briefcase as she slumped against the door. I stepped over her, picking my way through the debris until I was in the kitchen. The phone base was empty but still plugged in, which seemed like a miracle. Instinctively, I hit the page button and listened for the beeping. I found the phone inside a soup pan resting by the TV. The TV was lying on its side by the back door.
I held up the phone. _We still have this,_ I was trying to say, _if that helps._
Mom's face went blank, the way people look in the movies just before they pass out. She clutched her cheek with one hand.
I dialed 911.
The emergency operator stayed on the line with me and told me what to do, which was not much. I propped Mom's head up and fanned her. If she had been a monster, I might actually have had a clue as to how to treat her.
Police and paramedics arrived seven minutes and eighteen seconds later. One of the paramedics checked Mom's vitals, then, after a few tests, smiled brightly at us both. I rested my head in my hands, relieved. Mom was okay. They use a very different smile when it's serious.
A police officer tried to get me to answer questions while the paramedic wrote on a clipboard. It was hard for me to form words. My brain was like pudding.
"Do you have any valuables that a thief would want to steal?" he asked again.
I shook my head. "We don't have anything valuable."
I suddenly remembered: we did have a valuable book, one of the greatest treasures in history. Dr. Capistrano was the only one who had known about it, but she had wanted to help, not stop me. Was she so worried about the curse on the Bestiary that she'd try to steal it?
I looked around the place again, seeing it for the first time: ragged edges, scrapes and tears covering everything.
This wasn't the work of a thief.
And it wasn't a monster who had come here looking for help either.
This was Entropion.
Olympias had told me she would destroy everything I held dear. She was angry, maybe because I had weakened Entropion's power when I looked in the mirror and liked what I saw. She was punishing me for fighting back.
A shudder rolled through my body like a wave. Right behind it something else rose too: anger. Entropion had no right to come into my house.
I made my way to the stairs as fast as I could, stumbling twice over the debris. A policeman caught me gently by the arm when I tripped for the third time.
His name badge said "Officer Lopez."
I jerked my arm free. "I need to check my room!"
"Let my partner go first," he said, his eyes searching mine. He had soft brown eyes, with a few deep wrinkles around the edges, the kind from laughing a lot. His warm, steady hand was still on my arm.
I nodded.
After his partner had gone up and declared it safe, Officer Lopez helped me up the rest of the stairs. At the top, he placed one hand on my shoulder, forcing me to pause for a moment. I needed it. Each breath hurt, like my lungs were filled with glass shards.
"I'm sorry this happened to you," Officer Lopez said. "People do ugly things sometimes."
All my clothes had been ripped from the hangers, torn to shreds. Even my track trophies, which had been hidden in their box, were in bits and pieces. The bedspread was in ribbons, but hanging neatly in the closet was the gray dress.
Why hadn't Entropion destroyed that?
Looking around, I saw a scrap of shredded purple-and-gold fabric on the floor in front of my nightstand. Disbelief drowned out all my other thoughts.
These were the remains of my track jersey. It looked staged, like I was meant to find them only after I had seen everything else around me destroyed. The edges glistened with saliva.
I sat on my bed, focused on taking one breath at a time.
That was the jersey I had worn on my very last run, the one when I fell, literally, into another world called Cancer. Alexis had injured her knee that week, so she was using our run-a-minute/walk-a-minute technique. She had urged me to run ahead. "Not without you!" I argued. "You have to!" she yelled back. So I ran, _fast._ Running with Alexis had made me strong. I had been right: a little bit of her really had rubbed off. My legs were like springs attached at my hips. I bounded, I leaped, I _flew._
For the last time.
Entropion had shredded the memory, torn apart any last hope of wearing that jersey again someday, and almost killed my mother from shock.
Olympias had won.
I never wanted to see my mother collapse from shock and pain again. I never knew I could hurt like this again either. The track jersey was a scar that Entropion had split wide open. Unless you've spent weeks watching your body slowly knit itself back together in ragged seams and thick red scars, you take healing for granted. You get a paper cut and don't worry. A few days later, without any effort or thought, you notice that your body has quietly healed itself, one more little magic trick for an audience that forgets to applaud.
But sometimes healing is like being taken hostage in a violent siege. You have to survive each hour passing on the clock. You pray to make it until the next one, or the next pill, without losing your mind, telling yourself that relief can't be far.
I couldn't survive another open wound. I didn't have the strength to wait for it to close.
Officer Lopez stood in the doorway while I began searching through the rubble. He was short but really muscular and had thick, dark black hair. "Can I help you find something?" he asked. I shook my head. "Are you sure? You look pretty upset."
What could I say? I was looking for a two- or three-thousand-year-old book? That it was the connection to a dead guy and all the monsters in the world? Careful to keep my back turned to him, I bent down to check under the bed. My breath caught in my chest.
"How about you take a break and I'll get you a glass of water?" he asked. "Or a Coke if your mom says it's okay."
The book was gone. It really was over. Entropion had everything he wanted. He'd granted me a tiny bit of mercy, though, because he'd left me that gray dress.
I sat on the bed and cried, opening my mouth to take in big gulps of air. I wanted to be angry, but strangely, I felt as if a big weight had been lifted off my shoulders. I hadn't been a great Guardian, not like Xeno wanted, but at least I hadn't quit. Someone stronger had beaten me, and that was different. I wished, for one second, to be the kind of girl who would get angry and vow revenge, but that type of girl would never really be me. At least I didn't have to try anymore.
Officer Lopez offered me a fresh tissue. When Mom came up to see the damage, he rested a hand on her arm and was really nice to her too. She looked like she wanted to collapse against him. He seemed like he could handle it. She held the phone in her hands, as if she needed to call someone but couldn't remember who it was.
I wiped the tears from my cheeks and took the phone, setting it on my bed. I wanted her to scoop me up in her arms like she used to when I was little. But she didn't look at me, not really. She picked her way through the stuff on the floor, looking around in dismay.
"At least they didn't get this," she said, touching the gray dress and stroking the soft fabric with her fingertips. Empty hangers swung on either side of it, but she seemed mesmerized as it sparkled in the ruins of my closet. She didn't act like she knew I was even in the room, so I didn't reply, neither of us understanding why the dress hadn't been destroyed. It didn't make sense. Olympias had promised to destroy everything I held dear, and I loved that dress. It made me look flawless.
"You are going to be so beautiful. I can't tell you how happy I am that you still have this." She turned to me and smiled, tears shimmering in her eyes. "We still have that to look forward to."
"I'm so sorry," Officer Lopez said again. "Strange things have been happening ever since this little girl showed up at the hospital with bites from an animal she couldn't—or wouldn't—describe. Then an animal attacked an office near the hospital. Weird, even by Atlanta standards."
Mom dabbed at her eyes. She wasn't listening, but I was. They thought an animal had broken into Dr. Capistrano's office? No animal could tear a door off its hinges.
"Do you think we should get a dog?" Mom asked. She wrung her hands, looking around like she was lost, even though she was standing in my room. Officer Lopez stepped closer and gently nudged her toward the stairs before he gave me one last reassuring smile.
"Please let me get you a glass of water or something," he said to her. "You should sit down and take all this real slow." They walked out of the room, speaking in hushed tones.
I kicked one of my broken trophies with my prosthetic leg. A piece of it cut through the material and got stuck right in my big toe.
The fake stuff is never as strong as it looks.
Neither of us wanted dinner. Mom poured herself a glass of wine and wrapped herself up in a blanket to sit outside on the patio. I don't think she wanted to look at the house anymore.
It was strangely quiet. Without monsters, the world felt dead.
I grabbed another blanket off the tattered couch and followed Mom outside. I scanned the night sky.
"What a day," she muttered. She was standing in front of our patio swing, staring straight ahead.
I nodded.
"The locksmiths will be here in a few hours. It's nice that they agreed to come out so late." She didn't sound like she was really talking to me.
"Yeah," I said. "Officer Lopez was a big help." He had called in a favor to get them to agree to do the work after hours—it was already past dinnertime.
Neither of us said anything for a while. Things had been so awkward between us lately. Mom didn't understand why going right back to our "normal life" had felt like a death sentence to me. And all I knew was that the old me was gone, and I had been trying really hard to find the new me.
Mom broke the silence. "Girls your age push their moms away. It's normal," she said. "Healthy, the experts say."
"I'm not pushing you away."
She laughed under her breath and sat down on our patio bench.
I groaned. "I'm really not."
"I can't understand everything you've been through," she said. "I know that."
I touched my prosthesis, without even meaning to. She and I locked eyes.
"I just want to try," she said softly.
"I'm not pushing you away, I promise." I sighed. "I just can't explain everything to you right now. Some things I can't even explain to myself."
She gently waved a hand to dismiss the conversation and took a sip of her wine.
"It's okay," she said. "I'll wait till you're ready. That's what moms do. We give you everything we ever wanted for ourselves." She took another sip. "We give you the life we once wished for, and then we discover it's not what you needed after all. My worst fear is that you're nearly grown and I've wasted all my chances to be a good mom."
I sat beside her as she opened her arms to wrap the blanket around me too. She never really talked about her mother or childhood. I wondered if she had her own secrets.
"You've been through so much," she whispered. "You deserve a Cinderella ending to your story."
I giggled softly. I had been Cinderella for a while, hadn't I? Except that songbirds flocked to her window, but man-eating monsters slithered through mine.
An owl flew above us, hunting something we couldn't see. The bats were still hibernating, so the owls currently ruled the skies at night.
The phone rang, but Mom didn't move. She just stared at the sky, so I went inside to answer it.
"You broke my nose."
I gasped. "Billy?"
"Who else did you punch today?"
"Did I really break it?" I cringed. "I didn't punch you. I smushed you. Accidentally."
"That's why I'm doing this over the phone. In case of another accident."
"Doing what?" I asked.
"Officially breaking up."
"You can't break up with me. We aren't dating."
"Not anymore we aren't. My dad won't let me take you to the dance tomorrow night," Billy said. "He thinks you're a bad influence. Seeing as how you nearly killed the entire student body on the bus today and then broke my nose. You were supposed to catch the orange, you know."
"But I got a dress!" I blurted. "I went shopping with _Candy_ for it!" He couldn't back out now. "We made a deal," I added.
"The deal was you would agree to go with me. And you did."
"Is this because of the fight with Alexis?" I said. My body tensed. I wanted to punch him for real now. "You _told_ me to fight for myself, and now you don't want to go to the dance with me?"
"Try to pay attention. I _do_ want to go with you," he said. "My dad won't let me. I have to stay out of trouble at this school. My mom is threatening to send me to military school if I get expelled again. I think she's only mad that I've ruined the illusion of having a perfect son."
"I didn't know you had a mom," I muttered. That came out wrong, like everything else I said. I had seen his family picture, but he never talked about her.
"Did you think I was hatched? Of course I have a mom."
"I just didn't know if she was still around." I moved to sit on the couch, searching for a cushion that was still intact.
"She's not. But at least now she might feel guilty about it."
Neither of us said anything for a long time. Silence made talking again even harder. A thousand thoughts went through my mind all at once. Nothing I wanted to say sounded right. The longer we were silent, the more important it seemed to say something perfect.
"On Christmas morning two years ago," Billy said, "I opened a present that didn't have a label. I just assumed it was mine. My mom got this funny expression on her face. She hadn't meant to put it under the tree. It wasn't for me. It wasn't for my dad either. When New Year's rolled around a week later, she was gone. She had a whole new family."
"I'm sorry," I said weakly.
"That's not the worst part," he said. "She always told me I was the ideal son, and we had a picture-perfect Christmas that year; she even said so. Everything in our house looked like it came out of a magazine. We had all this amazing food, we went to a candlelight church service, we sang Christmas carols at the neighborhood party...and it was all a lie. Maybe she thought she could have both, but in the end she had to choose." He sighed. "The funny thing is, it wasn't her lies that nearly killed me. It was finding out the truth."
"So you wanted a friend who saw what was real." I thought about my test answer and why he loved it so much. "So you wouldn't get hurt again."
"You were supposed to be my early warning system," Billy replied. He sounded exhausted. "As long as I could see the truth about people, I wouldn't get hurt."
"You know what's funny?" I asked. "The only part of me that doesn't hurt is the fake part. It doesn't feel anything at all."
I could hear him exhale.
"Billy, I think that, sometimes, getting hurt is the only way we know what's real."
"Tell that to my nose."
I laughed so hard I snorted.
"My dad says I have to go," he said, "but remember, I said you had to learn to fight for the _right_ things." He paused before going on, like what he was going to say next would cost him something. "Alexis is one of the right things. You need to fight to hold on to your friendship. Don't worry about me right now, okay?"
Then he said good night.
We had just broken up, and now we were closer than ever. How was that even possible? Boys were the strangest discovery of all.
The locksmiths left a little before ten. Every hour or so, a police cruiser silently rolled past our house, and the walls lit up in red and blue. When Mom and I hugged good night, she held on for a long time. Or maybe I held on to her. She wanted me to sleep in her bed, but I was worried that I might start crying if I thought about the house, and I didn't want to upset her. She found an old quilt for my bed and then a couch pillow downstairs that was still in one piece.
An hour later, I lay there, unable to sleep, waiting for the next wave of lights, staring at the silver-gray dress hanging in my closet. The price tag twirled whenever the heat vent turned on.
A noise startled me. I turned my face to the window.
Thick fingers gripped the windowsill, followed by a hand the size of a dinner plate, the skin rough and gray.
"Golem!" I whispered, sitting up and reaching to open the window the rest of the way. My heart bounced in my chest. I scanned the street below. No police car was in sight, no stirring in the trees or bushes. Entropion wasn't there, but he had already made his point. Maybe he didn't think I would be strong enough to care about anything but myself now.
One massive foot, which looked like a big gray brick, came through the window first. Then the leg followed. He repeated the process slowly with the other leg, and I tried to see what was making him move so carefully.
When the Golem was completely in the room, he wasn't able to stand up to his full height, but instead hung his head. He was holding his left hand against his chest. I looked him over quickly but saw no wounds or marks.
"If you're hurt, I won't be able to help you, Golem," I said. "I'm not the Guardian anymore. The book is gone."
I braced myself to stand up by grabbing his arm. With one hand, I lifted his chin up so I could look at his face. A tear from one of his dark eyes splashed near my lip. On impulse, I licked away the tear with my tongue. It tasted salty, just like every other tear.
"Oh, Golem, what is it? What's happened?" I asked.
He lowered his left hand, opening it to let me see.
A baby sparrow lay in his palm. It was bald and raw, and then I saw what made Golem weep.
A sharp white bone protruded through its leg. The leg was swollen twice the size of the other one and bright red, filled with blood. The baby had its eyes closed, taking shallow rapid breaths.
I could feel its tiny heart beating under my fingers. It was strong. The bird refused to give up, even if the battle had already been decided.
An ant crawled through its feathers and I picked it off, crushing it between my fingers, angry. Outside, a mother bird called, a sharp _chirrup_ sound. The bird was so small, so fragile. It had never even flown and it wanted so badly to live.
Golem and I sat together on my bed. Golem held the baby bird in his palm as I rested my hand in his, cupping mine around the bird to keep it warm. We kept vigil over it as the heartbeats slowed, then became irregular and soft...and then stopped.
Forever.
A star shot past my window, the darkest hours of the night still unfolding. I rested my head against Golem's massive arm and went right to sleep.
When I woke, I was lying in bed with the covers tucked neatly around me. Golem and his sparrow were gone.
The sun rose in hot pink and orange colors above the dead landscape. Winter was lifting; out of the cold, dark earth, millions of forgotten seeds were breaking open. New life would finally rise through the darkness.
The dawn illuminated the frost on my window. In it, Golem had etched a heart. I stood and used my finger to trace its outline. I remembered how when I had first seen him, I was repulsed by his appearance. Then I began to see his heart, and I realized that the heart was all that really mattered.
I placed my hand over my chest, feeling my own heartbeat, hoping it would tell me whether I had changed in the right ways, whether I had finally made the right choices. It just kept its rhythm. Nothing had slowed it: not the tumor, not Entropion, not my own stupid fears and flaws and mistakes. Nothing had convinced it to stop beating.
It was strong, and, maybe, so was I.
Saturday, March 8
Mom had gotten up to work on the house while I slept in. It looked a lot better. Big chunks of drywall were still missing, but at least all the broken things were picked up. We went out to eat fast food for lunch, which was unusual for us, but I think she wanted to get out of the house for a little while. We'd need to do a lot of work to get it looking good, and more work wasn't what we needed right now.
Alexis was at her track meet, so I couldn't talk to her yet, but I was feeling a little better. Maybe telling the truth, as ugly and selfish as it sounded, had opened a wound that desperately needed air and light before it could heal. In the hospital, the nurses had always said sunlight was good medicine as they opened the curtains in my room every day. So maybe there was hope.
I still wasn't sure how I was going to coordinate hanging out with her and Candy, though. I owed Candy for the dress, even if it was free. Billy was right about one thing: sometimes sharing didn't work.
Late in the afternoon, after she had taken a long, hot shower, Mom pulled a box out of the freezer and rubbed the frost off with her palm to read the instructions. It was what she called "cooking from scratch." She scratched off the frost and read the directions.
I excused myself to get ready. I couldn't eat anyway. My stomach was a mess from nerves. I didn't tell Mom that Billy had canceled. I just said I wanted her to drive. She thought that was a good idea for a school dance with a date she hadn't even met yet. I think she wanted to add "and one that you assaulted." She had already set out all my stuff for tonight, plus some of her own cosmetics and a fragrance called Eternity. I wondered if anything would really change when I slipped into that dress and did my hair and makeup. All I knew was that once people saw me like that, there was no going back. I had to keep up the look. No one would want the other me. Was that why Entropion had spared my dress? Olympias had promised to destroy everything I held dear....Did that include my real self, the part of me that shone through my heart instead of the mirror?
And was evolution always this terrifying?
"Set out the plates anyway," Mom said, waving toward the dining table. She had chewed her fingernails around the edges. I wished we had time to talk. She handed me two glasses. "You're going to be beautiful. I can't wait to see you come down those stairs."
She turned on the radio and moved her head to the rhythm of an old song. I watched her, a soft smile spreading on my face. We needed music in this house again.
I knew what I had to do.
I shut my bedroom door softly, then rested my forehead against the cold wood. Finally I made myself let go of the doorknob. I turned and faced the dress.
It was still perfect. The silver-gray threads winked at me in the late afternoon light.
Was I supposed to get dressed and then do my makeup? Or the other way around? I had no clue. Part of me didn't want to do anything at all. I looked down at my feet, willing one of them to take a small step forward. I wasn't sure which would move first, the fake part of me or the real.
A bright flash blinded me. I winced and started blinking to clear my vision. From under the bed a fierce white light blazed.
Goose bumps spread over my skin, each of them a sharp protest. A thousand thoughts exploded inside my head.
The book scooted out from under the bed, approaching in slow stutters across the carpet, just like on our first morning.
I looked at the dress, and a wave of panic hit me, like I was being yanked out of a really good dream.
"Entropion stole you!" I said. "And you can't come back now. I already made up my mind."
There was a burst of light and a ripping noise, like fabric being torn. I was again momentarily blinded, as if someone had flashed a camera in my face.
"I have made up my mind too," said an unmistakably male voice.
I rubbed my eyes fast. Who was talking? None of the monsters had talked. "The book was stolen. I checked under the bed."
"As it happens," the voice said, "being stuck between worlds provides for an excellent hiding spot."
My vision cleared. The book was on the carpet, but it wasn't glowing. Instead, the letters flowed into the air. I watched, transfixed, as black letters hovered around me, some of them grazing my hair, flattening little strands as they flew past. Then streams of gold and red and blue, every paint color I had seen in the book, poured out in shimmering rivers, circling the letters in curls.
I reached out a finger to touch a letter hovering near my nose. It stung me.
"Ow!" I said, and stuck my finger in my mouth. The letters softened their shapes as if they were melting, until I was looking at an inky black cloud surrounded by the rivers of color. It was like watching a painter assemble their paints before starting on a canvas. Then, with a noise like someone snapping their fingers, the ink and colors flew into an image that took shape and began to move.
An old man made of moving pixels sat on my bed. I could see light between each of the dots as they moved. His hair was pure white, and long, almost as long as his beard, and he had a mustache so big and bushy it reached across each cheek to his ears.
"Xeno?" I whispered. Only my gaze had moved; my body was still pinned against the door.
He smiled. "You were expecting someone else?" His clear blue eyes sparkled. The pixels around the edge of his body were softer than the rest, more like smoke or mist, and a few drifted in my direction. I inhaled and caught the scent of pine and old books and the sea. "The world will go on without its monsters," Xeno said, crossing his legs and smoothing his robe. "But, oh, my dear, how will it go on without the real you?"
It was, I realized, the way I thought a father should smell. I smiled softly, staring at him in wonder.
Then I suddenly remembered what night it was.
"Do you not see that dress?" I blurted. "I'm starting over with a _new_ me. You're interrupting the most important night of my life!"
He held up a hand to stop me from talking. "There is no time to thank me."
My mouth dropped open. I wanted to throw something at him, but he wasn't really there, was he? I could still see the outline of my nightstand through the dots that made up his body.
"I thought being the Guardian would help you to discover who you really are," he said. "You would find all those wonderful seeds of the bizarre and beautiful, and grow into all you were meant to be. But you are afraid, Sofia, much more than anyone realized."
His words burned and I knew they were true. I also knew they didn't matter anymore. "I'm not the Guardian now," I said.
Xeno's face softened into a smile. "You will always be the Guardian. I'm not here to convince you of that."
"Then why are you here?" I asked. "I told you I've already made my decision."
"Because you do not understand the choice you are making," he said. "Aristotle gave me more than one chance to choose the correct answer, and so I will do that for you."
"I'm just putting on a dress and going to a dance," I said.
"You do not understand what you are throwing away." He looked out my window at the darkening sky. "And you do not understand what you are inviting in."
A cloud rolled over the moon.
"When I first discovered monsters, I thought everyone would be as thrilled as I. I had much to learn about the human heart. Olympias used fear to destroy courage, and suspicion replaced hope. People forgot what they were capable of."
He shook his head, as if lost in memories. "And then you became the Guardian. You were so very different from the others. You are one of the extraordinary ones, and yet you hate yourself for that. You want to hide what makes you different. You are ashamed. Olympias worried that you would become her greatest enemy, and then she realized the truth: you could be her most powerful ally. Everyone's eyes are upon you right now. If you conform, the others will follow." I got the feeling he was talking about more than just the kids at school.
A bitter taste rose in the back of my throat. His words burned like matches held too long. Maybe that was what real truth felt like. I swallowed hard, trying to push the pain away.
"Olympias wants me to fit in," I said, my voice hoarse. "It's better than killing me."
Then it hit me.
Candy's sudden interest in me; the disappearance of Dr. Capistrano, the only adult who knew my secrets and could help me; the loss of Alexis, the best friend who had loved my real self; the shredding of my old track jersey. Nothing was left but a gray dress that hid all my imperfections, every feature that made me different. Olympias didn't care what I saw in the mirror after tonight, as long as I didn't see the real me. She was afraid of who I really was. So was I.
Xeno glanced at the dress on the bed between us.
My cheeks grew hot, as if I was guilty of something.
He pointed to the dress. "If this is who you want to be, put it on. Just understand the choice you are making." His chin quivered. "You are the only one who can decide who you will be."
I had to know the truth about one thing, but I needed time to think too. "The other Guardians, the ones in the past?" I asked. "I bet they were afraid sometimes. They didn't always like who they were or what they looked like. No one is perfect."
"Of course."
"And you let Claire out of being the Guardian."
Xeno sighed, and a little puff of pixels floated from his mouth. "Of all my daughters, you were the only one who had already faced death, and it held no fear for you. I thought that meant you would be safe from Olympias. I thought you might become the most powerful Guardian of all. But I was wrong! You, Sofia, do not fear death. You are terrified of life."
I sank down to the floor. His words had finally burned through me and I was as light as ash.
Everything Xeno said was awful and true. I had always been the odd girl out, and hated myself for that. I thought I was the sparrow that no one cared about.
Xeno crossed the room and held the dress out to me. I couldn't lift my arms. I didn't want the dress anymore. I shook my head and a teardrop landed on my thigh.
"Then become who you were born to be!" he yelled. I gasped and looked up. The edges of Xeno's body burst into a white flame. A mist rose from his feet, and he dissolved into a swirling cloud; the book fluttered open to draw the funnel into its pages.
The book closed and all went dark.
The dress rested on the floor at my feet.
I thought of Golem and his little brown bird. He needed me. The weak and the forgotten and the ugly, they needed me, the real me. Whatever it cost. One person had to be alive to see and love the forgotten things in this world, the sparrows and the monsters and the freaks. They mattered, even if I couldn't always explain why.
My fingers rested against my prosthesis. They ran back and forth over the uneven bite mark as I sat still, staring at my legs, thinking. I knew how much tonight meant to my mom. It was a new beginning for us both.
My index finger dug into the wound. Without thinking, I ripped off a piece of the fake flesh. It was soft and rubbery. The metal frame underneath it was hard and shiny and beautiful in a strange way. I stared at it, mesmerized; then my fingers began to frantically pick off the fake stuff.
I shuddered, trying to catch a deep breath, and realized that pieces of fake flesh were all over my carpet.
> You should hurry. You have a dance to go to, Cinderella.
I laughed out loud, but I didn't know if that was my conscience or Xeno speaking to me inside my head.
"But..." Words failed me. I looked at the metal of my leg. Why had I done that? It was like my body knew something my brain didn't.
I got this weird feeling that Xeno was smiling, watching me. Instinct made me reach for the phone Mom had left in my room, without even knowing what I was going to say. I punched in the old familiar numbers.
"Hello?" Alexis answered, breathless.
"It's me," I said. She was silent, and I didn't know how to continue. I guessed she hadn't checked caller ID.
"Sorry, I was busy changing," she said. "I thought you were someone else."
"That was the problem. I wanted to be _anyone_ else," I said, then paused. "I'm so sorry."
"Um...," she said, "okay."
Closing my eyes, I prayed I was doing the right thing. "Listen, I know we need to talk more," I said, the idea forming in my mind even while I spoke. I needed something, but if I went looking, Mom would get suspicious. "But first...can you come over right now? And...bring scissors?"
Alexis and her mom arrived fifteen minutes later. When Mom opened the front door, Alexis hesitated, looking at the ground and chewing her lip. My mom immediately pulled her into the house and into a big bear hug.
They hugged for a while. I had to clear my throat a few times at the top of the stairs for Mom to let Alexis go. Alexis looked up at me and frowned.
"You're not even dressed yet!"
"You look amazing!" I said at the same time.
We both broke into big grins.
She was wearing a fitted pink dress that made her muscles pop and her boobs look huge. I tried not to stare, but I hadn't known she had boobs. Since illness had interrupted puberty, Mom said I had to wait for the Boob Fairy to visit. With my luck, the fairy would be eaten by a monster.
"Come up and help me, okay?" I asked.
Mom and Alexis's mom wandered over to the couch and sat down, talking like they were afraid of running out of oxygen before they said everything.
"Don't take too long," Mom called up to us. "I am so excited to see you in that dress!"
Alexis carried a cute beaded purse up the stairs with her. Once we were in my room, she opened it, pulled out the scissors, and held them out to me. I reached for them, but she didn't let go.
"Not yet," she said. "First you have to tell me what you're going to do."
I smiled, my hand on top of hers. The metal of the scissors winked at us.
"I'm _not_ going to hurt myself, but you just need to trust me right now, okay?"
She considered that for a minute and then released her grip. I took the scissors and held them.
There would be no turning back.
Ever.
I took the blade's edge and lowered it to the prosthesis, then looked up at Alexis. Our eyes met and she nodded. Somehow, she knew. Best friends are like that. A little lump formed in my throat; I was so grateful to have her with me for this.
Working fast, we cut and pulled and picked all the fake flesh away from the leg, revealing the metal skeleton underneath. Alexis ran and got a washcloth from the bathroom and rubbed the metal down, removing the last bits of flesh material.
I stared at my leg, transfixed. "Now the dress," Alexis said. She grabbed it while I shimmied out of my shirt and skirt.
"Oh my gosh!" Alexis yelled, seeing the price tag.
"Candy," I muttered.
"Say no more," she replied. She snipped it off and the tag fluttered to the floor. "Good riddance," she said. Next she handed me the dress. I slipped it on and smoothed it down.
"Wow" was all she could say. I think she meant it in a good way this time. She set the scissors down on the nightstand.
"Not yet," I said.
I grabbed the hem of the dress, so long and lovely, the perfect length to disguise what made me feel different from everyone else. I looked at Alexis. Her eyes went wide.
People were going to see. They were going to stare at me, just like she was doing now. I narrowed my eyes at her to tell her I was ready.
Alexis gave me a serious nod.
"I'll do it," she said. Taking the scissors back, she approached me and knelt down. Grabbing the hem, she stopped and looked back up.
"You want to know a secret?" she asked.
I nodded, too nervous to speak.
"When we ran together, I was always jealous of you," she said.
I frowned. "Why? You're a better runner."
"Yeah, I know," she said. "But I never had to fight for it. I knew I would finish every race. You had to fight for each one." She plunged the scissors into the fabric and cut.
I inhaled, holding my breath as she worked and talked. "I always watched your face when you crossed the finish line. I was jealous of how happy you were." She cut and motioned for me to turn. I did. "It's why I pushed you so hard, because I knew I would never get to feel anything like that. I loved running because it made me feel free and I never wanted the runs to end. But you? You weren't like me. You weren't running away from anything. You were running _toward_ something. I guess that secretly, I hoped I would get to do that someday too."
After two more cuts, she stood up. We grabbed hands and each took a few deep breaths. There was still so much to get caught up on, but I knew this: when we faced life together, it hurt less.
I turned toward the door.
She grabbed my arm. "Wait," she said, then swiped off my bandana and gasped.
Alexis used her fingers to tousle the fringe on my head. "Wow. Just...wow."
I giggled.
"I—I'm sorry," she stammered, waving her hands to make me stop. "I know I need a new word. But...wow!"
"Come on!" Mom yelled from downstairs. "You girls don't want to miss the dance!"
Alexis pointed a finger at me to stay right where I was. She went to the top of the stairs and hollered sweetly down at my mom, "Down in a sec!" When she returned, she was all business. I closed my eyes as she fluffed and brushed and sprayed.
When we walked out of the bedroom, Alexis nudged me toward my mother's room. Mom had the only full-length mirror in the house, fastened to the back of her bedroom door. Alexis leaned against the doorjamb and smiled. I took a deep breath and walked.
Everything led to this private moment between me and the mirror. The cold, familiar twist of fear snaked through my body. I had always hated the mirror because it reminded me that I was the odd girl out, and didn't nature prove that the weak were only safe in a pack? Those who got separated were eaten alive. Olympias had made sure I understood that my only chance at a happy life was to hide my true self and go with the herd.
But she was wrong. The difference between humans and animals is that any human can become an alpha. Being pushed out of the pack isn't a death sentence. It's an invitation.
I opened the bedroom door and stepped inside, ready to look in the mirror with fresh eyes.
I had finally figured out what was worth standing alone for.
I can't lie, though; at first glance I felt a tiny, familiar lurch of disappointment. I had wanted to see perfection. It was an old habit, I guess. But really, what _is_ perfection?
Then that feeling was replaced slowly, sweetly, by a strange awe for this new creation. I reached one hand out and gently grazed my reflection with my fingertips. The cocoon had torn and here I was. There had never been anyone like me in a magazine or on a billboard. I had learned that the world is full of people who are both unusual and stunning, but they have to find the courage to see themselves with new eyes. Because when they do, when they see who they truly are, they catch a glimpse of what they might yet become. And the world needs a new generation that is unafraid to change.
Breathing quietly, I studied my reflection and listened as the din of old lies faded and the howl of the alpha began to rise. This was not who I had once so desperately wanted to be.
This was _far_ better.
And then, yeah, I bent over at the waist, just a little, and wrote my name in the air with my butt. It totally worked.
Minutes later, at the top of the stairs, Alexis whispered for me to go first. I descended and waited for Mom to glance up. When she did, her mouth fell open, her eyes wide and unblinking.
Alexis's mom turned to look up at us too, and froze.
I wanted to go back upstairs, but Alexis kept her hand on the small of my back. The tiniest of nudges told me I had to keep going. This was me, the real me, on display.
"Your dress," Mom whispered.
I nodded. Sliced off at an angle, it deliberately exposed every inch of my prosthesis, and that was all that caught your eye...at first.
"Your hair," she murmured.
I reached up and tried to smooth it a little. Alexis hissed at me. "Don't you touch it!"
She had moussed it into white, spiky perfection. It resembled a medieval weapon.
Then Mom stood and screamed, "Your _leg_!"
I guess her brain had random disconnects too. She'd just realized that I had destroyed all the fake flesh.
I hurried down the last three steps before she could keel over or yell at me again.
"Mom, calm down! It's only a fake leg anyway. You let them cut off my real one, remember?"
I was the only one who found that even a little funny.
Mom was opening and closing her mouth like a fish that's been yanked out of the river and thrown into a chest of ice.
"Anyway, it still works," I said. I did a little jig to prove my point. She sank back onto the couch, shaking her head.
I took a deep breath, knowing that what I had to say next was the most important thing I might ever say to her. "I just can't hide anymore. I can't hide who I am. I don't want to be afraid when I look in the mirror or when people look at me."
"I don't want you to be afraid," she whispered, sitting forward on the couch. "All I ever wanted was for you to be happy."
Neither of us moved.
"So..." Alexis came down the rest of the stairs. "Maybe we should go."
Her mom stood and rested a hand on my mom's shoulder.
"I've never seen you look more beautiful," Mom said finally. Every word seemed to cost her something, as if a little piece of her ached as she said it, but the bigger part, the deeper part, kept pushing those words to the surface.
"I forgot one thing upstairs," I said. "Be right back."
I made it up there in record time. I was free, and free felt fast and light.
"Xeno?" The room was dark, the sun outside gone. The moon had claimed her place in the sky, a delicate silver crescent.
"Thank you," I said. "For helping me choose something better. For helping me choose myself."
Silence.
The book was open, but no words appeared. Maybe even philosophers struggled for words sometimes.
"If Entropion shows up tonight, what do I do?" I asked.
"Do I have to come up there?" Alexis yelled. "Let's go!"
I stood closer to the book, speaking quickly but keeping my voice down. "Can the Guardian kill a monster if she has to?"
"Sofia, hurry up," Mom called.
Xeno's words came fast.
> Entropion was made from fear, and fear does not die. But nature teaches us that even things that cannot die can still be done away with. For example, an apple cannot be killed. It can only be transformed, forced into another state of being.
"How do you transform an apple?" I asked.
> It must become part of something or someone else.
"Like possession?" I said. "When a ghost takes over your body?"
> Like digestion.
The problem with getting advice from a philosopher is that most of it doesn't make sense. Plus I had a vague suspicion he was mocking me. But we'd deal with that later.
If I survived.
Alexis's mom offered to drive. I said goodbye to Mom with one last hug. Alexis opened the front door. The night air took our breath away.
It was sharp and brittle, a mean-spirited cold, like winter was hanging on, forcing spring back into hiding. The full moon cast a ghostly pallor on our faces, making us look like corpses.
"Is it too cold?" Mom called from the doorway. She sounded hopeful, like maybe I wouldn't go through with this.
I shook my head, not wanting to look her in the eye.
I sat in the backseat next to Alexis. My foot kept tapping the floor. What had Xeno meant by "digestion"? My heart beat too fast, and I willed it to keep time with my foot.
Alexis smiled at me.
If anything went wrong tonight...I groaned a little, and Alexis's mom glanced at me in her rearview mirror.
"Just nervous," I said to everyone.
Alexis laughed. "Don't be nervous, Sofia. Enjoy it. Think of this as a finish line. I'll be right beside you to cheer you on."
She reached for my hand, and I let her take it. We clutched hands and nodded to one another; then she realized how sweaty and cold my palm was. She giggled and let go, wiping her hand on the seat between us.
We were still giggling when her mom made the final turn. We were almost there. White clouds huddled together in the sky, covering the moon. Floodlights lit the school entrance, and we saw police barricades set up around the perimeter. Uniformed police officers stood in front of the school, holding leashes.
Two German shepherds, one of them smaller than the other, paced in front of the main entrance. Everyone had to walk past these two animals to get inside.
"Whoa," Alexis said. I looked at her, one eyebrow cocked. She shook her head. "It's a different word entirely," she insisted.
Her mom slowed the car and rolled down her window. "Since there have been reports of strange animal attacks recently," she said, "the PTA voted to pay for extra security."
She pulled closer to the barricade, but a police officer held up his hand.
The dogs paced and panted, watching us.
The smaller shepherd was probably a female. She tilted her head back and sniffed the air to read my scent. She did not blink. Or bark. She just held my gaze, and something in my soul understood.
It was a warning.
The police officer was explaining to Alexis's mom that only students and faculty were allowed past this checkpoint.
Alexis's mom hesitated, but the officer nodded at her warmly. "Ma'am, I have a daughter in there too. Nothing is going to hurt those kids on my watch."
With that, Alexis and I got out of the car and walked in. In the bustle and confusion of the unexpected drop-off, I forgot about my appearance.
The disco ball and dimmed lights disoriented me for a moment, and Alexis grabbed my arm to steady me. People were staring. Slowly, one by one, everyone on the dance floor stopped moving. Friend nudged friend, and girl whispered to boy, until every set of eyes was glued to me. Only the boy taking our tickets didn't seem to notice me; he was too busy staring at Alexis's boobs.
I took my first step forward.
Everyone looked different tonight, not just me, and they all looked nervous. Change wasn't easy for any of us. As people caught sight of me, they stared, eyes wide, a few of them actually letting their jaws drop open.
And then I understood.
Xeno hadn't chosen me. My peers had. They sensed I was different and pushed me outside the protective circle of our little herd, but my mistake—and maybe theirs—had been assuming that my difference was a weakness. I was suddenly thankful for the heartbreak. I might never have become this strong. For a split second, I pitied everyone who had stayed in the circle, protected and popular. Maybe one day they would discover this same strength, if they could endure the loneliness that comes before the understanding.
Music boomed all around us, sending vibrations through my whole body. I made my way over to the punch bowl, with Alexis behind me. Walking on heels took a ton of concentration because my center of gravity shifted forward slightly. I wouldn't feel so triumphant if I face-planted on the floor. A couple of guys moved to get out of our way, one of them looking at my leg, the other looking at Alexis's chest.
A cluster of girls surrounded the refreshment table, their backs to us.
"So I was at my grandmother's and I wanted some milk, but she was out, right? So I went home and opened my fridge and there was no milk! I was like, noooo! And my mom just said, 'Sorry, we're out.' I was so sad, I was about to cry. Really. That's how bad I wanted a drink of milk. Because on this diet, liquids are all I can have." Candy gestured to the cup of punch she was holding and sighed.
"Uh, Candy...," the other girls murmured. They must not have seen me enter, but they saw me now. Alexis grabbed my hand protectively. Everyone stared, mouths open. Candy put her hands on her hips, ready to scold them for not taking her problem more seriously, but first she glanced back to see what had distracted them.
She looked at my leg. The disco ball spun over us, reflecting little circles of light that washed over their heads. Candy's face morphed from confusion to disgust to anger.
I dropped Alexis's hand. I wasn't going to hide anymore, not from myself and not from Candy.
"Do you know how expensive that dress was?" she hissed. "Is this how you show your gratitude? By turning this into some kind of joke? Who's going to want to see a picture of that?" She gestured at my dress, or my leg, or maybe both.
"Candy, let me explain," I said.
"What happened?" Natalie asked breathlessly. "I mean, to your leg?"
"Everyone knew it was fake," I said with a shrug, keeping my eyes on Candy. Her face was a swirling vortex of anger.
Several football players, including Matt, swaggered up to us. Candy handed Matt her punch and he drank it in one gulp. I guessed he was her date.
"Her leg looks like a vegetable peeler," Matt whispered to his friends, but we all heard it.
Natalie's face brightened. "Ohmygosh, my mom just got one of those automatic ones. It can shred a carrot, like, really fast."
Billy saw me from across the gym. He was alone, wearing a dark shirt and nice khakis with a belt, his hair combed back neatly. He looked scrubbed and fresh and nervous.
His eyes met mine and he didn't look away. He didn't just see the leg, or the dress, or the hair...he saw _me._ A warm, bright joy surged from my heart into a smile.
He pushed his way through the crowd. Candy noticed him and smiled too, maybe because he looked so good. Matt watched her, then glared at Billy. When Billy lightly touched Candy's arm to nudge her out of the way, Matt shoved him on the shoulder.
"Not tonight, okay?" Billy said, raising his hands like a shield.
Matt pushed him again, but Billy stepped back just in time, causing Matt to trip and ram an eighth grader in the back. The eighth grader turned, punch running down his chin onto his shirt. He shoved Matt into Billy. In a flash, it looked like a swirling mass of fists and hair and legs.
Everyone was yelling, but a shrill scream pierced my ears. I whipped my head around, searching for the sound. No one else heard it. Sweat beaded my upper lip.
Candy was hyperventilating, glancing wide-eyed between the pile of boys and me. "I can't believe you ruined that dress," she said. "After everything I did for you."
"Look, I'm sorry," I said, backing away. I needed to find out where that sound had come from. What if Entropion was here and he attacked the wrong person? "I appreciate your help, but that wasn't the real me."
"That was the whole point," Candy said.
Mr. Reeves ran up and called for more teachers. They separated the boys, and I tried to make my exit. Strange new vibrations moved up both my legs. Something heavy was moving through the hallway outside, coming closer. I cut across the dance floor through the couples. Candy and Alexis shouted at me, but I didn't turn around.
I burst through the gym doors and looked in either direction, listening for the sound again. Two teachers and one security guard stood in the hallway. I turned to the left.
"That hallway is closed," one of them called to me.
"Just need a bathroom!" I hollered back. "Gotta adjust my leg."
Neither knew how to respond to that in a politically correct way. They didn't try to stop me.
Around the corner, I paused to listen, but all I could hear was the pounding of my heart.
A low moan, like the sound of an injured animal, rose from somewhere deep within the school.
I closed my eyes and strained to hear with everything inside me, everything that had gone into creating this last monster: my heart, my mind, my fears. I wanted to face Entropion at last.
He was in the library.
An officer was down.
I opened the main door to the library, but Entropion was already dragging his victim out the door on the opposite side of the room. The body disappeared through the door, two limp hands dragging across the floor. I heard the wet thud of the body being dropped. Carefully, I made my way to the door and glanced out of the windows built into the top.
In the hallway, only the emergency lights were on. The little bulbs gave a flickering, unreliable light, casting strange moving shadows. I stepped out of the library and looked in either direction.
The dog was on her side, panting and whimpering. Blood seeped out from underneath her body.
A thick, musky odor hung in the air. I took a deep breath anyway. I could see the dark shadow of Entropion looming.
A low, soft laugh echoed in the darkness as the air around me turned bone-numbingly cold, like a door to a giant freezer had been opened. Olympias stood before me while her body separated in two, dividing down the middle like curtains being pulled apart.
And there was Entropion. Blood froze in my veins. He looked like a giant serpent with a misshapen human head. He had no lips, only thick pink gums mottled with blood, like the mouth of a great white shark. His arms and legs were short and scaled, like a crocodile's, with clawed hands and feet. It was as if someone had tried to create a monster, taking the worst parts of every creature imaginable and setting them at odd angles. Then I remembered that that someone was me. He was all that I had feared, the misshapen, the broken, and the scarred.
Vomit crept up the back of my throat. My vision dissolved at the edges; the sight of him disoriented me, like I was looking at a face inside a kaleidoscope. His eyes were the eyes of a wolf, bright yellow with a hollow black center that glowed with rage when he looked at my dress. In the flickering shadows I saw rows of shiny white teeth as he growled. I fled back inside the library and shoved the door closed.
Two cold hands curled around my shoulders.
I spun around, arms flailing.
Candy grabbed my shoulders and pushed me against the metal door handle. It crunched against my spine.
"Let go," I begged. "You've got to get out of here!"
"I don't have to do anything," she replied. She looked around the library in disgust, as if I had only retreated here to hide from the dance. "I didn't have to help you, but I made a plan and you agreed to it. This is how you repay me?"
Natalie was behind her, shaking her head.
"This isn't about you," I replied. I couldn't hear anything moving in the darkness around us. "Let me go."
"A newspaper reporter is waiting in the gym to take your picture. You will smile and pose, just like we agreed. Natalie and I will stand just a little in front and on either side so no one can see what you did to the dress."
"You go," I said. "I _promise,_ I'll be there as soon as I can."
Candy tried to pull me, but I planted my good foot and braced against her.
"Like I would ever listen to you again," she said.
"You never listened to me at all!" I yelled, slapping her hands away. "You have no idea who I am!" A terrible realization hit me. "You have no idea who _you_ are either."
She sneered, but a flicker of sadness in her eyes told me I was right.
My voice hardened; she _had_ to get this. "We're not packages, or illusions, or magic tricks. We're not just girls...we're _humans,_ and we're supposed to want things that can't be found in a mirror."
She shook her head, refusing to listen.
"Candy, some people can't change how they look, or disguise who they are. They're just born different, and this world is not a safe place for people who don't fit in. And I'm one of them. You think that makes me a loser, but I think that might make me a hero. I'm not afraid anymore that people won't love me; I'm afraid of hurting the ones who really do. So you can keep the limelight, because I've learned that not everyone is like me. _Not everyone is strong enough to face the shadows._ "
"You don't know anything about life in the shadows!" Natalie snapped, pushing both of us to the side. She swung the door open, expecting Candy to force me out into the hall. Candy and I stood face to face.
A flickering shadow moved across Candy's pupils.
Natalie's body flew backward, crashing into a bookshelf. Books scattered around her, pages flapping. Candy and I screamed at the same time.
Entropion stood in the doorway. His lips lifted to show rows of white teeth as he reached his arms out wide to either side, claws scraping down the metal doorframe. I winced from the sound. Olympias stood behind him, delight etched across her face as she watched her creation at work.
Entropion lowered his head and hissed.
My good foot was wet and warm. Glancing over, I saw a dark stain spread across Candy's dress. She had peed all over us both.
Entropion shifted his head back and screamed, the sound piercing my ears. I shook my head, trying to stop the pain.
Above us, the ceiling tiles shuddered, flakes of insulation fluttering down. Something else was up there. I had seen this a thousand times on Animal Planet. It was a hunting strategy. One animal bared its teeth and scared the prey while the other circled behind and attacked.
Xeno was wrong. I hadn't made just one monster.
I'd made two.
I grabbed Candy and shoved her to the ground with all my strength. I was the one they wanted, not her.
But it was Golem who crashed through the ceiling. He opened that beautiful, thick dark mouth and roared. Windows shattered and books flew through the air. I hadn't created Golem, had I? No, but I had loved him, and that was maybe close enough. He was mine, and I was his.
Entropion lowered himself on his haunches and lunged.
Golem caught him in midair before he reached us. One of Entropion's claws shot out for me from under Golem's arm, slicing my skin. Blood bubbled up beneath the long mark.
Golem had Entropion pinned down, his mouth open, thick teeth snapping in fury. Every second, Golem seemed to grow bigger and bigger. Entropion thrashed, lifting his knees to knock Golem off. Golem flew up and back, crashing into the lockers outside the doors. They burst open, papers flying like a snowstorm.
Free now, Entropion turned toward me again and jumped, throwing me on my back, one clawed hand around Candy next to me, the other around my prosthesis. I bent my good leg and kicked him. He clung to my prosthesis, trying to regain control. I tried another tactic and spun onto my stomach. The prosthesis came off in his hand.
Confusion flashed in Entropion's eyes.
Behind him, Golem rose. He grabbed Entropion by the waist and lifted him into the air, and then Golem's eyes met mine. A softness shone within them, something like love.
He slammed Entropion to the floor, and Olympias released a bloodcurdling scream.
Entropion didn't move again.
My breath came out in spurts like short laughs. Golem had done it. He had killed Entropion. Golem looked at me and pointed to his heart.
With a trembling hand, I pointed to my own.
Olympias raised her arms like a puppeteer commanding her creation.
Entropion rose and shifted, sinking his claws into Golem's chest. The tender monster slowly fell backward with a mighty sigh.
"Golem!" I screamed. "No!"
Entropion was on top of him now, claws twisting inside Golem's chest, sunk to the knuckle. Golem lifted one shaking arm and rested a finger against his own forehead. Then he opened his mouth wide, inhaling. Locker doors swung in his direction, books and papers careening toward him. He inhaled harder, chest expanding, mouth gaping wider and wider.
Entropion trembled and yanked at his arm to free it, but Golem had caught it with his other arm, pinning Entropion to himself. Golem inhaled so powerfully that Candy and I began to move across the floor toward him.
Olympias collapsed, her hands covering her face as she shrieked.
With Golem's final mighty inhale, Entropion exploded into a cloud of ash that was sucked into Golem's mouth. Golem snapped it shut and looked toward me.
With one raised finger, he smudged the last symbol on his forehead.
Golem's head rolled to one side, limp, his eyes blank holes.
He began to fade, and the air between us shimmered and danced like snowflakes falling softly at midnight. Then he was gone. I pressed my hand to my heart, expecting it to break, but instead, it was stronger. Golem was still with me somehow. He was not lost, not as long as he lived inside me. I knew I would see him again.
My focus shifted to Olympias. Underneath her hands, her face was shifting, the skin rippling between her fingers. She turned her back to me and stood.
Then she looked over her shoulder at me, and I could see her profile. I didn't recognize her anymore; if I passed her on the street now, I wouldn't know who she was.
"The first chapter of our story is finished," she whispered. "But oh, what surprises await!" Like smoke floating up from a candle, she vanished into the night.
"What happened?" Natalie sat up, rubbing her head.
At some point, Candy had fainted, and she was still flat on the ground. Natalie stumbled over to her and shook her awake. Candy pushed herself up, her eyes glazed over. I grabbed my prosthesis and wrestled it on, feeling the suction that secured it into place, then wiggled the stretchy support bands over the top to secure it.
Mr. Reeves burst in from the opposite set of doors, and his mouth fell open when he saw the mess of papers and books lying everywhere.
"Don't be angry, Mr. Reeves," I said. "They saved my life."
"We did?" Natalie asked.
Candy nudged her, hard. "We did," she said.
"I ran away from the dance," I said. "Candy and Natalie followed me. They were worried because I was so upset. We found the library like this, and a big animal—a dog or wolf, I think—jumped out and Candy shoved me to the floor to protect me."
Candy looked at me, suspicion lingering in her eyes. I didn't know whether she doubted my story or her sanity.
Mr. Reeves tapped his notebook. "Is what you're telling me true?" he asked. I nodded.
Billy and Matt ran into the room and froze. Billy looked at me first, checking me up and down, then exhaled so hard his upper body caved forward. Matt ran over to Candy and she let him put his arm around her.
Resting his hands on his thighs to catch his breath, Billy looked up at Mr. Reeves.
"Somebody called for an ambulance and it just pulled up," Billy said. "Who's it for?"
Outside the school doors, the familiar blue and red lights were flashing in all directions. Every police car and ambulance in the county must have been there. Parents were calling their kids' names and jumping out of cars with the motors still running. Cell phone rings came from all directions. Teachers were shouting about the strange animal that had been sighted in the school.
A beehive of EMTs and police officers swirled around the fallen officer. He was alive—Entropion had only knocked him out—but he had a nasty gash on his hand. One of the paramedics said it looked like a defensive wound; another one said it was a bite mark. Either way, something was wrong. A hand injury couldn't cause the amount of blood I had spied on the floor.
I pushed past the kids peering into the library and found a new trail of blood drops leading toward another hallway.
Billy saw it too. He grabbed my arm and propelled me forward, his strength carrying me along. The trail led to the drinking fountain. The smaller, female German shepherd had collapsed in the recessed area under it. Red slashes crisscrossed her fur. She must have tried to protect the officer when Entropion attacked.
"Is she dead?" I asked, hand over my mouth. "Please don't let her be dead."
Billy gently rested two fingers on the inside of her thigh. He was still, waiting for a pulse. And I swear, in the silence, my own heart stopped and listened for hers.
"She's alive."
"Call your dad," I said.
"There's no time." He scooped his hands under the dog's body, and she whimpered when he lifted her.
I was already moving toward the school's front doors, where all the adults were.
"We can't," Billy said. "It's crazy out there with all the cars. They won't move fast enough."
We had to get help. We couldn't handle this alone.
Billy turned and staggered toward an emergency exit. "We'll take my golf cart."
He glanced back at me. "It was the only way I could get here. I snuck out."
I kept moving, following behind Billy and the dog faster than I thought possible. My brain was screaming, but all my thoughts were jumbled. Billy pressed his back against the exit doors and burst through them to the outside, carefully cradling the dog.
"The key's in my pocket," he said, and turned to present his rear end to me. With no time to argue, I grimaced and stuck my hand in his pocket, feeling for the key. Billy had a gift for putting me in awkward situations.
"You'll have to drive," he said, while carefully lowering himself into the passenger seat. A golf cart only went about ten miles per hour, I knew, but we were desperate. I put the key in the ignition and the engine roared to life. It sounded like a jet engine had been dropped under the hood.
"Take Main Street down to Milton," Billy yelled above the noise.
I hit the gas and the cart burst out of the lot, gravel flying behind us. There was no speedometer, so I had no idea how fast we were going, except that it was somewhere between Crazy and Dangerous. I glanced at him in amazement.
"My dad let me tinker with it. I upgraded the motor. But be careful!"
I narrowly dodged an elderly woman setting out her trash can. She yelled at me, shaking a fist in the air.
I made the turn onto Milton Street, smoking past a middle-aged guy in a Prius. The cart straightened onto the road.
"Clinic is one mile ahead on your right!" Billy yelled. "Almost there!"
Flashing blue and red appeared behind us. We didn't have time to stop for a cop. I jammed the gas pedal to the floor. The engine burst into flames and the cart rolled to a stop. The officer jumped out of his car and caught up to us on foot, his flashlight blazing into our faces.
"Sofia?"
I looked up and squinted. "Officer Lopez?"
"Sofia, there's a bottle of Betadine next to the table," Dr. Hamby said. "Grab it and pour it wherever I tell you, okay? It's the dark orange stuff."
"I know what it looks like," I said.
Dr. Hamby had already grilled us about the dog's injuries. We were in the operating room of his clinic now. Officer Lopez stood to the side, busy on his radio. All the other officers were at the school, but everything was calming down there. My story about the animal had everyone panicked, but no one had seen anything. Officer Grant, the one injured at the school, had regained consciousness on the way to the hospital. He didn't remember anything except his dog growling, and a flash of pain when something hit him on the head.
Dr. Hamby poured Betadine over the dog's leg before he started an IV. I watched the needle go in, and she lifted her head, licking Dr. Hamby on his hand. Tears sprang up in my eyes, hot and stinging. I hated needles.
I tried to hide my face from Billy, who was standing on the opposite side of the table, covered in blood and dog hair. He walked over and put his arm around me. He didn't have to say anything.
Dr. Hamby inspected the dog's scalp and head and seemed to find nothing serious. Then he peeled back the skin on either side of an obvious chest wound. I could see the white of the bone. Billy left my side and wheeled a big metal box over and positioned it above her.
"Wait!" Dr. Hamby told Billy. Then he looked at me. "No chance you're pregnant, right?"
"No." It was the most awkwardly spoken single word ever.
Dr. Hamby nodded and Billy flipped the switch. A low buzz told us the X-ray machine had snapped the picture. They turned to a computer screen on a table behind them. Blurry black-and-white images came to life. Billy exhaled hard.
"What?" I asked, my voice rising.
"No broken ribs. Lungs look good too. It's a bad cut. That's all," his dad said.
Mom burst through the clinic's front doors, yelling my name. Alexis was with her.
Dr. Hamby didn't wait for introductions. He plunged a syringe into the IV and depressed it, sending a light pink medicine into the line. The dog's body relaxed immediately, her breathing becoming more even.
"Your mom was freaking out when she got to the school," Alexis whispered to me. "She called Officer Lopez. He told her where you were."
"Billy, suture kit," Dr. Hamby said.
"Your daughter did a brave thing," Officer Lopez said. "She found an injured police dog and got help."
Dr. Hamby agreed. "And thankfully, Officer Lopez got the dog here in time."
I exchanged glances with Billy.
I reached out toward the sleeping dog and stroked her fur. The monitor registered a change immediately as the dog's heart rate finally pushed back into the normal range. Everyone watched, fascinated.
"She just needs to know that she's not alone," I said. "I think they feel more than we know."
Billy excused himself and walked out.
I found him sitting outside on the curb.
A yellow streetlight beamed down like a low moon hanging in the dark sky.
I managed to sit beside him rather gracefully. Alexis started to come out the door, but I held a finger up to stop her. She stepped back inside, nodding like she understood.
"What'll you tell your dad about the golf cart?" I asked.
"As little as possible," he replied. "I wasn't supposed to be at the dance."
"Or with me," I said.
"You're the only reason I showed up," he said. "I had to see what kind of dress was worth the pain of being friends with Candy."
I looked down at the dress. The fabric hadn't held up very well in all the excitement; the side seams were splitting and tiny threads poked out of them. It was cheaper than it looked.
"Well, it did look a little better on the hanger," I admitted.
Then Mom came outside and cleared her throat. I nodded in her direction. Alexis stood in the doorway behind her.
Billy grabbed my hand before I could stand up. "My dad won't let us date. So we can't be boyfriend and girlfriend. And we can't be best friends, because you already have one."
"Is everything always so black-and-white with you?" I asked.
"I'm willing to create a new category. I still want to hang out. I'm just not sure what we would call that."
"It's called being friends," I said.
"No, it's something more," he said.
"Okay," I said. "You're my Something More."
Mom cleared her throat, this time really loudly.
"Why did you really show up tonight?" I asked.
"I decided that there's only one thing worse than feeling pain, and that's feeling nothing at all."
He leaned over and kissed me on the cheek. I glanced at my mom to make sure she wasn't watching.
This time, she blushed.
"I better go," I said.
The soft smile on his lips made my stomach flutter.
Mom and Alexis were waiting. I glanced down at my shadow and saw the strange outline of spiky hair and mismatched legs, then forced myself to look back up at the dark blue horizon. The full moon was bright silver, and the stars around it burned gold. I never realized how much color hid in the darkness. Wind rustled the leaves of a nearby tree, and something flew from its branches into the night.
Now I knew that a book really could save you, if it was the right book. And if I had to carry this secret alone, it didn't mean I had to be lonely. It meant I had to burn brightly, like the stars. I had to reclaim what was lost in the darkness.
I am an outpost, a lighthouse, a planet silently gliding past yours. Gravity has set us all in motion and chosen our path. Maybe we don't feel or taste or even believe the same things, and maybe that's not the answer. We have to stop fighting what we see, especially what we see in the mirror, because it is the unseen that truly matters, and I've made my decision.
I will hold my course.
The world around us is full of monsters and freaks and mysteries.
And me?
I am the Guardian who will watch over them all.
# ACKNOWLEDGMENTS
I'm grateful to so many people for getting this book into your hands. First, my beloved husband, Mitch, urged me to take the time I needed to write the book I needed to write. He even took over the laundry. My kids, James, Elise, and Lauren, tried very hard to let me write in peace, even when a tree crashed in our yard, blocking the road and unleashing a nest of angry hornets. ("Don't worry, Mom, we have knives!" will always be a memorable line from you guys.)
My own mom and dad contributed a lot to this book, starting a few decades ago, when they brought me to my first horror movie. And while I still hate seeing people's heads explode, I did develop a passion for monsters. Thank you for letting me host my tea parties late at night in case Dracula and the Wolf Man could come. And thanks to my brother, Steve, for taking me on my first Sasquatch hunt. We'll find him next time.
My dear friend Dr. Ana Adams, DVM, gave me advice about the Beast's medical predicament. Dr. Adams is a well-respected vet, and I am grateful my monster had such good care. If you need to treat a giant wolf of your own, please don't rely on this book but consult a qualified cryptozoologist.
My literary agent, Melissa Jeglinski, is a gem. Melissa, you thought this book had heart, and you certainly have mine for all you've done. (Thank you, Cecil Murphy, for introducing us, and thanks to the wonderful Julie Garmon for cheering us all on while we ate gluten-free French food together.) My film agent, Andy Cohen, asked to represent the book and gave me a vote of confidence that meant the world to me. I do love writing, but the people I meet are the real gifts.
And my critique group, the Y-Nots: Johnna Stein, Sharon Pegram, and Sandra Havriluk. I am so indebted to each of you for your encouragement and feedback, but most of all for your friendship, which means so much to me. Please say something extra nice about me in your acknowledgments. I also owe a debt to my favorite nemesis, Jim Reinoehl, and his wonderful wife, Louise. I love you both!
And finally, the team at Delacorte Press, especially Krista Vitola. Thank you for believing in the book and then showing me how we could make it even stronger. You truly have an alchemist's gifts, and it was my honor to work with you. I hope our work impacts a lot of readers.
Designer Sarah Hokanson and illustrator Dinara Mirtalipova, I'm thrilled you were chosen to create the "welcome mat" for the readers. You captured the heart of the book with your pencil, which is rare magic indeed.
And, finally, my copy editor, Deb Dwyer. Your attention to detail made this book shine. Although I expressed my gratitude a bazillion times, let me say it once more: thank you!
# ABOUT THE AUTHOR
Ginger Garrett graduated from Southern Methodist University in Dallas with a major in theatre arts and a focus on playwriting. Although she applied to the CIA to become an international master of espionage, she had to settle for selling pharmaceuticals for a global corporation. She eventually traveled the world on her own dime and without a fake mustache.
Ginger now lives in Atlanta with her husband, three children, and two rescue dogs. She spends her time baking gluten-free goodness for her friends and family and mentoring middle school students who want to become working writers. Passionate about science, history, and women's studies, Ginger loves exploring new ideas and old secrets. She especially loves good books read late at night.
Ginger is a popular speaker and a frequent guest on radio and television shows. She has been featured in and on media outlets across the country, including Fox News, _USA Today,_ _Library Journal,_ and 104.7 The Fish Atlanta. You can learn more at gingergarrett.com.
1. Cover
2. Title Page
3. Copyright
4. Contents
5. Dedication
6. Chapter One
7. Chapter Two
8. Chapter Three
9. Chapter Four
10. Chapter Five
11. Chapter Six
12. Chapter Seven
13. Chapter Eight
14. Chapter Nine
15. Chapter Ten
16. Chapter Eleven
17. Chapter Twelve
18. Chapter Thirteen
19. Chapter Fourteen
20. Chapter Fifteen
21. Chapter Sixteen
22. Chapter Seventeen
23. Chapter Eighteen
24. Chapter Nineteen
25. Chapter Twenty
26. Chapter Twenty-one
27. Chapter Twenty-two
28. Chapter Twenty-three
29. Acknowledgments
30. About the Author
1. Cover
2. Cover
3. Title Page
4. Contents
5. Start
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| {
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\section{Introduction and the statement of the main result}
In this paper we consider parabolic equations of the type
\begin{align}\label{maineq}
F(D^2u, Du, x, t)-u_t=0,
\end{align}
where $F$ is assumed to be uniformly elliptic in the Hessian and has a certain nonlinear growth in the gradient variable. More precisely, we assume that there exist constants $0 < \lambda \leq \Lambda$ such that
\begin{align*}
\lambda Tr(N) \leq F(M+N,p,x,t)-F(M,p,x,t) \leq \Lambda Tr(N)
\end{align*}
$\forall\ N \geq 0$ and for every $(p,x,t) \in \mathbb R^n \times Q_2$. We furthermore assume that $F$ has the following growth dependence in the gradient variable,
\begin{align}\label{nonlin}
|F(0,p,x,t)| \leq \phi(|p|)
\end{align}
for every $(p,x,t) \in \mathbb R^n \times Q_2$ and where $\phi : [0,\infty) \rightarrow [0,\infty)$ is of the form $\phi(t)=\eta(t)t$ and satisfies the following conditions analogous to that in \cite{j}:
\begin{enumerate}
\item[(P1)] $\phi : [0,\infty) \rightarrow [0,\infty)$ is increasing, locally Lipschitz continuous in $(0,\infty)$ and $\phi(t)\geq t$ for every $t\geq0$. Moreover, $\eta : [0,\infty) \rightarrow [1,\infty)$ is nonincreasing on $(0,1)$ and nondecreasing on $[1,\infty]$;
\item[(P2)] $\eta$ satisfies
\begin{equation*}
\underset{t\rightarrow\infty}{\text{lim}}\frac{t\eta^{'}(t)}{\eta(t)}\text{log}(\eta(t))=0;
\end{equation*}
\item[(P3)] There is a constant $\Lambda_0$ such that
\begin{align*}
\eta(st)\leq \Lambda_0\eta(s)\eta(t);
\end{align*}
for every $s$, $t\in (0,\infty)$.
\end{enumerate}
Following the ideas of Caffarelli \cite{Ca}, we will replace the equation (\ref{maineq}) by two extremal inequalities which takes into account the ellipticity assumption and the growth condition on the drift term. In other words, we assume that $u \in C(Q_2)$ is a viscosity supersolution of
\begin{equation}\label{eq1}
P^-(D^2u)-u_t\leq\phi(|Du|)
\end{equation}
and a viscosity subsolution of
\begin{equation}\label{eq2}
P^+(D^2u)-u_t\geq-\phi(|Du|),
\end{equation}
where $P^{\pm}$ correspond to the extremal Pucci operators as defined in \eqref{pucci}. We refer to Section \ref{s:n} for the precise notion of viscosity sub/supersolutions.
Before proceeding further, we make the following discursive remark.
\begin{rmrk}
Throughout this paper, by a universal constant, we refer to a constant $C$ which depends only on the ellipticity constants, the nonlinearity $\phi$ and the dimension $n$.\end{rmrk}
\subsection*{Statement of the main result}
We now state the main result of the paper which is regarding the validity of a parabolic Harnack type inequality for solutions to \eqref{maineq} in a suitable intrinsic geometry corresponding to the nonlinearity.
\begin{thrm}\label{mthm}
Let $u \in C(Q_2)$ be a positive viscosity supersolution of (\ref{eq1}) and viscosity subsolution of (\ref{eq2}). There is a universal constant $C>0$ such that
\begin{align}\label{Harnack}
\underset{A}{\text{sup}}\hspace{0.8mm} u(a_0 x, a_0^2 t) \leq C u(0,0) \hspace{2mm}\text{for} \hspace{2mm} a_0=\frac{u(0,0)}{C(\phi(u(0,0))+u(0,0))},
\end{align}
where $A=\Big\{(x,t):|x|_{\infty} \leq \frac{c_n}{2}, -1+ \frac{c_n^2}{4} \leq t \leq -1+\frac{c_n^2}{2}\Big\}$. Here $c_n$ depends only on $n$.
\end{thrm}
\begin{rmrk}
Theorem \ref{mthm} roughly ensures that in a cube of size $a_0 \sim \frac{u(0,0)}{\phi(u(0,0))+u(0,0)}$, the parabolic Harnack inequality holds. It is to be noted that when $\phi(t) \equiv t$, we have that $a_0 \approx 1$ and consequently \eqref{Harnack} reduces to the scale invariant Krylov-Safonov type parabolic Harnack inequality as in \cite{KS}. We would also like to mention that in view of a counterexample in \cite{j}, the standard Harnack inequality doesn't hold for solutions to \eqref{maineq}.
\end{rmrk}
\begin{rmrk}
In section \ref{s:n}, we provide an explicit example of a function satisfying the extremal inequalities in \eqref{eq1} -\eqref{eq2} above which vanishes in finite time. Such an example would thus demonstrate that the time lag in the Harnack inequality for positive solutions to \eqref{eq1}-\eqref{eq2} has to depend on the solution in general. \end{rmrk}
Now in order to provide a proper perspective to our work, we mention that in the time independent case, Julin in \cite{j} showed that nonnegative viscosity solutions to
\begin{equation}
\begin{cases}
P^-(D^2u)\leq\phi(|Du|)
\\
P^+(D^2u)\geq-\phi(|Du|),
\end{cases}
\end{equation}
in $B_2$ satisfy the following generalized Harnack inequality,
\begin{equation}\label{eHarnack}
\int_{\inf_{B_1} u}^{\sup_{B_1} u} \frac{dt}{ \phi(t) + t} \leq C.
\end{equation}
Such a Harnack inequality as in \eqref{eHarnack} above quantifies the strong maximum principle. Moreover as mentioned above, a counterexample in \cite{j} shows that the standard Harnack inequality cannot hold in this setting. The proof of the Harnack inequality in \cite{j} involves a fairly delicate adaptation of the Vitali type covering argument which utilizes the slow growth assumption on $\eta$ in a very crucial way. It is to be noted that the classical proof of the Krylov-Safonov type weak Harnack estimate as in \cite{Ca} for elliptic equations and \cite{W1} for the parabolic equations is based on a basic measure estimate which uses the Alexandrov-Bakeman-Pucci (ABP) type maximum principle following which an iterative argument is set up on the super level sets of supersolutions using the Calderon-Zygmund decomposition. In the parabolic case, an additional ``stack of cubes" covering argument is required which takes into account the time lag between the measure and the pointwise information in the basic measure estimate. The switch from the Calderon-Zygmund type decomposition to the more elementary Vitali argument is relatively well known in the elliptic setting ( see for instance \cite{IS1}, \cite{Sa}). However in the parabolic case, because of such a time lag in the measure estimate, adapting the Vitali type covering which takes into account the parabolic geometry appears like a serious obstruction. Thus it is not clear to us as to whether the methods in \cite{j} can be generalized to the parabolic case in order to obtain an analogous parabolic Harnack inequality as \eqref{eHarnack} above. Therefore in this present work, we obtain a new intrinsic framework where a parabolic Harnack inequality can be established for \eqref{maineq}. Moreover in the course of the proof, we also show that in such a framework, the weak Harnack estimate for positive supersolutions also remains valid. This makes our work somewhat different from \cite{j} where instead a conditional weak Harnack estimate is proven ( see Lemma 4.8 in \cite{j}) which however is optimal in the extrinsic Euclidean geometry and suffices for the Harnack estimate \eqref{eHarnack}.
\medskip
We proceed as in the classical case as in \cite{W1}, but similar to that in \cite{j}, the difficulty arises because of the fact that if $u$ solves (\ref{eq1}) and $M>0$, then
\begin{align*}
v(x,t)=\frac{u(x,t)}{M}
\end{align*}
is a solution of $$P^-(D^2u)-u_t\leq \Lambda_0\eta(M){\phi}(|Du|),$$ and thus the growth of the nonlinearity gets altered. However after further rescaling,
\begin{align*}
v(x,t)=u(rx,r^2t)
\end{align*}
we observe that $v$ solves $$P^-(D^2v)-v_t\leq \Lambda_0 r \eta(1/r){\phi}(|Dv|).$$ At this point, by noting that $r \eta(1/r) \to 0$ as $r \to 0$, we obtain that in a small enough region dictated by the nonlinearity, the growth conditions on the nonlinear term remains comparable to the initial assumptions and thus one can normalize and use the equation in such a region.
Subsequently, the basic measure estimate is obtained via the ABP type comparison principle which relies on a somewhat subtle computation of an appropriate barrier function ( see the proof of Lemma \ref{barrier}). The $L^{\epsilon}$-estimate is then obtained by applying the measure estimate repeatedly on appropriate normalized solutions using a delicate stack of cubes covering argument. Over here, we would like to emphasize that the stack of cubes defined in \cite{W1} ( see also \cite{is}) doesn't work in our inhomogeneous situation. Therefore, we define a different stack of cubes which is tailor-made for our nonlinearity. This constitutes the key novelty of this work. Moreover, in our entire analysis, the precise growing nature of $\eta$ crucially comes into play at various steps. Following the $L^{\epsilon}$-estimate, the subsolution estimate is then obtained by adapting the ideas in \cite{j} to our parabolic situation. We refer to the subsquent work \cite{AJ} where a related boundary Harnack inequality in $C^{1,1}$ domains has been established for such inhomogeneous elliptic equations.
\medskip
In closing, we would like to mention that various types of intrinsic Harnack inequalities for divergence form parabolic equations modelled on
\[
\operatorname{div}(|Du|^{p-2} Du) = u_t,
\]
have been obtained in a series of fundamental works by DiBenedetto, Gianazza and Vespri (see \cite{DGV1, DGV2, DGV3}) which in part has also inspired our present work. See also \cite{Ku} where an intrinsic weak Harnack inequality has been obtained for such equations. It remains to be seen whether our techniques can be extended to such structures. We also refer to a recent interesting work \cite{PV} where intrinsic Harnack inequality for
\[
|Du|^{\gamma} \Delta_p u= u_t,\ \gamma> 1-p,
\]
has been established by comparison with explicit Barenblatt solutions which in turn is inspired by an approach of DiBenedetto for the parabolic p-Laplacian. ( see \cite{Di}).
We would also like to mention some other works \cite{IS1, Sa, Wa} which introduces new techniques
for proving regularity estimates for nondivergence form elliptic and parabolic equations. These works study
equations which are homogeneous but which fail to be uniformly elliptic. The key idea is to
bypass the classical ABP estimate by directly touching the solution from below by paraboloids (or
by other suitable functions).\\
The paper is organized as follows. In Section \ref{s:n}, we introduce some basic notations and notions and gather some preliminary results. In Section \ref{s:m}, we prove our main result. Finally in the appendix, we provide proofs of Lemma \ref{barrier} and Lemma \ref{measureuniform} which involves a somewhat delicate and long computation.\\
\\
\textbf{Acknowledgments}
We would like to thank Agnid Banerjee for various helpful discussions and suggestions. We would also like to thank the editor for the kind handling of the paper and also the reviewer for various valuable comments and suggestions which has substantially improved the presentation of this article. Especially, we are grateful to the reviewer for suggesting the possibility of finding an example of a solution which vanishes in finite time which we have now put in section \ref{s:n}. Such an example demonstrates that the intrinsic nature of our Harnack inequality in Theorem \ref{mthm} is unavoidable.
\section{Notations and Preliminaries}\label{s:n}
A point in space time will be denoted by $(x,t)$ where $(x,t) \in \mathbb R^n \times \mathbb R.$
We will denote a point in space by $x$, $y$ etc.. We will use $0$ for both origin of $\mathbb R^n$ and real number. We will denote Euclidean norm of $x$ in $\mathbb R^n$ by $|x|$.
$Df$ and $D_{x,t}f$ will denote the gradient of $f$ in $x$ variable and gradient of $f$ in $x$ and $t$ both variables respectively. $f_t$ will denote partial derivative of $f$ with respect to $t$. $D^2f$ will represent the Hessian matrix of $f$ with respect to $x$.
For a given set $A$, $|A|$ will denote the Lebesgue measure of $A$.
A cube of radius $\rho$ and center $(x_0,t_0)$ is defined as following
\begin{align*}
Q_\rho(x_0,t_0) = \{x \in \mathbb R^n:|x-x_0|_{\infty} < \rho \}\times (t_0-\rho^2,t_0),
\end{align*}
where $|x|_{\infty} = \text{max}\{ |x_1|, |x_2|,..., |x_n|\}$.
We define
\begin{align}\label{tildq}
\tilde {Q}_{\rho}(x_0,t_0) := \{(x,t) \hspace{1mm}: \hspace{1mm}|x-x_0|_{\infty} < 3{\rho},\hspace{1mm} t_0 <t< t_0+9{\rho}^2\}.
\end{align}
We denote $Q_{\rho}(0,0)$ by $Q_{\rho}$ and $\tilde {Q}_{\rho}(0,0)$ by $\tilde {Q}_{\rho}$.
For $(\Omega\times (t_1,t_2)) \subset \mathbb R^n \times \mathbb R$,
$\partial_p (\Omega\times (t_1,t_2))$ will denote the parabolic boundary of $\Omega \times (t_1,t_2)$ and defined as:
\begin{align*}
\partial_p(\Omega\times (t_1,t_2))&=\Omega\times \{t_1\}\cup\partial\Omega\times[t_1,t_2].
\end{align*}
Let $S$ be the space of real $n\times n$ symmetric matrices. We recall the definition of Pucci's extremal operators (for more detail see \cite{cc}). For $M\in S$, Pucci's extremal operators with ellipticity constant $0<\lambda\leq\Lambda$ are defined as
\begin{align}\label{pucci}
P^-(M,\lambda,\Lambda)=P^-(M)=\lambda\sum_{e_i>0}e_i+\Lambda\sum_{e_i<0}e_i,
\\
P^+(M,\lambda,\Lambda)=P^+(M)=\Lambda\sum_{e_i>0}e_i+\lambda\sum_{e_i<0}e_i\notag,
\end{align}
where $e_i$'s are the eigenvalues of $M$.
We recall the definition of a viscosity supersolution of (\ref{eq1}) a viscosity subsolution of (\ref{eq2}).
\begin{dfn}
A function $u :Q_{\rho} \rightarrow \mathbb R$ is a viscosity supersolution of (\ref{eq1}) in $Q_\rho$ if it is continuous and the following holds: if $(x_0,t_0)\in Q_\rho$ and $\varphi \in C^2(Q_\rho)$ is such that $\varphi\leq u$ and $\varphi(x_0,t_0)=u(x_0,t_0)$ then
\begin{align*}
P^-(D^2\varphi(x_0,t_0))-\varphi_t(x_0,t_0)\leq\phi(|D\varphi(x_0,t_0)|).
\end{align*}
\end{dfn}
\begin{dfn}
A function $u :Q_{\rho} \rightarrow \mathbb R$ is a viscosity subsolution of (\ref{eq2}) in $Q_\rho$ if it is continuous and the following holds: if $(x_0,t_0)\in Q_\rho$ and $\varphi \in C^2(Q_\rho)$ is such that $\varphi\geq u$ and $\varphi(x_0,t_0)=u(x_0,t_0)$ then
\begin{align*}
P^+(D^2\varphi(x_0,t_0))-\varphi_t(x_0,t_0)\geq-\phi(|D\varphi(x_0,t_0)|).
\end{align*}
\end{dfn}
We will state some more properties of $\eta$. See Proposition 2.3 in \cite{j}.
\begin{lemma}\label{etaprop}
Let $\eta$ satisfy the conditions $(P1)-(P3)$. Then the following hold:
\begin{itemize}
\item[(i)] For every $c>0$, we have
\begin{equation*}
\underset{t\rightarrow\infty}{\text{lim}}\frac{\eta(ct)}{\eta(t)}=1.
\end{equation*}
\item[(ii)] For every $\gamma>0$, we have
\begin{equation*}
\underset{t\rightarrow\infty}{\text{lim}}\frac{\eta(t)}{t^\gamma}=0.
\end{equation*}
\item[(iii)] There is a constant $\Lambda_1$ such that for every $t>0$ it holds that
\begin{align*}
\eta(\eta(t)t)\leq \Lambda_1 \eta(t).
\end{align*}
\item[(iv)] There is a constant $\Lambda_2$ such that for every $t>0$ and $0<r<s$ it holds that
\begin{align*}
r\eta(t/r)\leq \Lambda_2s\eta(t/s).
\end{align*}
\end{itemize}
\end{lemma}
Since the equation (\ref{eq1}) is not scaling invariant, we need the following lemma. Proof is same as in the elliptic case, see Lemma $4.4$ in \cite{j}.
\begin{lemma}\label{scaling}
Let $u\in C(Q_2)$ be a viscosity supersolution of (\ref{eq1}) in $Q_2$. There exists a universal constant $L_2$ such that if $A\in (0, \infty)$ then for every $r\leq r_A$, where
\begin{align*}
r_A=\frac{A}{L_2(\phi(A)+A)}=\frac{1}{L_2(\eta(A)+1)}
\end{align*}
the rescaled function
\begin{align*}
\tilde{u}(x,t):=\frac{u(rx,r^2t)}{A},
\end{align*}
is a supersolution of (\ref{eq1}) in its domain.
\end{lemma}
\begin{dfn}\label{medef}
\textbf{(Monotone envelope of a function)}. The monotone envelope of a lower semi-continuous function $u : Q_\rho(x,t) \rightarrow \mathbb R$ is the largest function $v : Q_\rho(x,t) \rightarrow \mathbb R$ lying below $u$ which is convex with respect to $x$ and non-increasing with respect to $t$. It is denoted by $\Gamma_u.$
\end{dfn}
We need the following lemma to prove Lemma \ref{measureuniform}. See Lemma $4.4$ in \cite{is}.
\begin{lemma}\label{glem}
If $u$ is $C^{1,1}$ with respect to $x$ and Lipschitz continuous with respect to $t$,
then the function $G : \Omega \times (a,b) \rightarrow \mathbb R^{n+1}$ defined as follows:
\begin{align*}
Gu(x,t)=G(u)(x,t)=(Du(x,t),u(x,t)-x\cdot Du(x,t)),
\end{align*}
is Lipschitz continuous in $(x,t)$ and for a.e $(x,t) \in \Omega \times (a,b) $,
\begin{align*}
\text{det}\hspace{1mm} D_{x,t}G(u) = u_t\hspace{1mm} \text{det}\hspace{1mm}D^2u.
\end{align*}
\end{lemma}
We need the following lemma to prove Lemma \ref{measureuniform}. See Lemma $4.13$ in \cite{is}.
\begin{lemma}\label{setmeas}
If $u \in C(Q_1)$ with $u \geq 0$ on $\partial_p(Q_1)$ and $\text{sup}_{Q_1} u^- =1,$ then
\begin{align*}
\Big\{(\xi,h) \in \mathbb R^n \times \mathbb R:|\xi| \leq \frac{1}{4}, \frac{5}{8} \leq -h \leq \frac{6}{8}\Big\} \subset {G\Gamma_u(Q_1 \cap C_u)}.
\end{align*}
where $C_u=\{u=\Gamma_u\}$ and $\Gamma_u$ represents the monotone envelope of min$\{u,0\}$ extended by $0$ to $Q_2$.
\end{lemma}
We will now introduce parabolic setting of Calderon-Zygmund decomposition. (See \cite{is})
Consider a cube $Q$ of the form $(x_0,t_0)+(-s,s)^n \times (0,s^2)$. A \textbf{dyadic cube} $K$ of $Q$ is obtained by repeating a finite number of times the following iterative process: $Q$ is divided into $2^{n+2}$ by considering all translations of the form $(0,s)^n \times (0,s^2/4)$ by vectors of the form $(ks,l(s^2/4))$ with $k\in \mathbb{Z}^n$ and $l \in \mathbb{Z}$ included in $Q$. $\bar{K}$ is called the \textbf{predecessor} of $K$ if $K$ is one of the $2^{n+2}$ cubes obtained from dividing $\bar{K}$.\\
In figure below cube $ABCD$ is the predecessor of $K$.
\begin{center}
\begin{tikzpicture}
\fill[color=gray!20] (0,1) rectangle (2,2);
\draw (0,0) rectangle (4,4);
\draw[-] (0,1)--(4,1);
\draw[-] (0,2)--(4,2);
\draw[-] (0,3)--(4,3);
\draw[-] (2,0)--(2,4);
\draw (1,1.5) node{$K$};
\draw (0,-0.2) node{$A$};
\draw (4,-0.2) node{$D$};
\draw (0,4.2) node{$B$};
\draw (4,4.2) node{$C$};
\end{tikzpicture}
\end{center}
Let $m$ be a natural number. For a dyadic cube $K$ of $Q$, the set $\bar{K}^m$ is obtained by ``stacking" $m$ copies of its predecessor $\bar{K}$. More precisely, if $\bar{K}$ is of the form $\Omega \times (a,b)$ then $\bar{K}^m$ is $\Omega \times (b,b+m(b-a)).$
\begin{center}
\begin{tikzpicture}[scale=0.35]
\fill[color=gray!10] (0,8) rectangle (4,12);
\fill[color=gray!10] (0,4) rectangle (4,8);
\fill[color=gray!20] (0,1) rectangle (2,2);
\draw (0,0) rectangle (4,4);
\draw (0,4) rectangle (4,8);
\draw (0,8) rectangle (4,12);
\draw (2,8) node{$\bar{K}^2$};
\draw[-] (0,1)--(4,1);
\draw[-] (0,2)--(4,2);
\draw[-] (0,3)--(4,3);
\draw[-] (2,0)--(2,4);
\draw (1,1.5) node{$K$};
\draw (0,-0.4) node{$A$};
\draw (4,-0.4) node{$D$};
\draw (-0.5,4.2) node{$B$};
\draw (4.4,4.2) node{$C$};
\end{tikzpicture}
\end{center}
We need the following covering lemma. See Lemma $4.27$ in \cite{is}.
\begin{lemma}\label{cald}
Let $m$ be a natural number. Consider two subsets $A$ and $B$ of a cube $Q$. Assume that $|A|\leq \delta |Q|$ for some $\delta \in (0,1)$. Assume also the following: for any dyadic cube $K \subset Q$,
\begin{align*}
|K \cap A|>\delta |K| \implies \Bar{K}^m \subset B.
\end{align*}
Then $|A| \leq \delta\frac{m+1}{m}|B|.$
\end{lemma}
As mentioned in the introduction, we now provide an example of a function satisfying \eqref{eq1}-\eqref{eq2} which vanishes in finite time.
\subsection*{Example.}
Consider for $(x,t) \in Q_2(0,1),$
\begin{align*}
u(x,t)=
\begin{cases}
e^{1/t}(x+3) \hspace{2mm}&t<0\\
0 & t\geq 0.
\end{cases}
\end{align*}
Also let $\phi(s)=5s(|\operatorname{ln}(s)|+4)^2.$
Then for $t<0$,
\begin{align}\label{x1}
u_{xx} -u_t=\frac{e^{1/t}(x+3)}{t^2} \geq 0 \geq -\phi(|u_x|)
\end{align}
and moreover,
\begin{align}\label{x2}
u_{xx} -u_t&=\frac{e^{1/t}(x+3)}{t^2}\\
&=|u_x|(x+3)\left(|\operatorname{ln}(e^{1/t})|\right)^2\notag\\
&\leq |u_x|(x+3)\left(|\operatorname{ln}(|u_x|)|+4\right)^2\notag\\
&\leq 5|u_x|\left(|\operatorname{ln}(|u_x|)|+4\right)^2\notag\\
&=\phi(|u_x|)\notag.
\end{align}
Moreover since all the derivatives of $u$ decay as $t \to 0^{-}$, therefore \eqref{x1} and \eqref{x2} continues to remain valid for $t>0$. Thus $u(x,t)$ solves (\ref{eq1}) and (\ref{eq2}) corresponding to $\phi(s)=5s(|\operatorname{ln}(s)|+4)^2.$
Now we show that such a $\phi$ satisfies (P1), (P2) and (P3). Note that for $s<1,$
\begin{align*}
\phi '(s)&=5(4-\operatorname{ln}s)^2 +10 s (4-\operatorname{ln}(s))(-1/s)=5(4-\operatorname{ln}s)(2-\operatorname{ln}s)>0,\\
\eta '(s)&=\frac{-10}{s}(4-\operatorname{ln}s)<0.
\end{align*}
For $s>1,$ we instead have,
\begin{align*}
\phi '(s)&=5(4+\operatorname{ln}s)(6+\operatorname{ln}s))>0,\\
\eta '(s)&=10(4+\operatorname{ln}s)(1/s)>0.
\end{align*}
Also $\phi(s) \geq s.$ Thus $\phi$ satisfies (P1). Now (P2) is satisfied as
\begin{align*}
\underset{s \rightarrow \infty}{\text{lim}}\frac{s \eta'(s)}{\eta(s)}\operatorname{ln}(\eta(s))=\underset{s \rightarrow \infty}{\text{lim}}\frac{10s (4+\operatorname{ln}s)(1/s)}{25(4+\operatorname{ln}s)^2}\operatorname{ln}((4+\operatorname{ln}s))= \underset{s \rightarrow \infty}{\text{lim}} \frac{10\operatorname{ln}(4+\operatorname{ln}s)}{25(4+\operatorname{ln}s)}=0.
\end{align*}
Observe that (P3) is also satisfied as
\begin{align*}
\eta(s_1)\eta(s_2)&=25(4+|\operatorname{ln}s_1|)^2(4+|\operatorname{ln}s_2|)^2\\
&=25(16+4|\operatorname{ln}s_1|+4|\operatorname{ln}s_2|+|\operatorname{ln}s_1||\operatorname{ln}s_2|)^2\\
& \geq 25(16+4|\operatorname{ln}s_1|+4|\operatorname{ln}s_2|)^2\\
& \geq 25(16+4|\operatorname{ln}s_1s_2|)^2= 80 \eta(s_1 s_2).
\end{align*}
\section{Proof of Main Theorem}\label{s:m}
We will follow the same approach as in \cite{W1} but in the intrinsic setting. \\
First, we will construct a barrier function as Lemma $4.16$ in \cite{is}. Proof is in Appendix \ref{a}.
We define the following subsets of $Q_1$.
\begin{align*}
{K_1}&:=(-c_n,c_n)^n \times (-1,-1+c_n^2),\\
{K_2}&:=(-3c_n,3c_n)^n \times (c_n^2-1,10c_n^2-1),\\
{K_3}&:=(-3c_n,3c_n)^n \times (-1+c_n^2,0),
\end{align*}
where $c_n=(10n)^{-1}.$
\begin{center}
\begin{tikzpicture}{Scale=1.3}
\draw (2,0) node{\tiny{$\bullet$}};
\draw (2,-0.25) node{$(0,-1)$};
\draw (0,0) rectangle (4,4);
\draw (1.6,0) rectangle (2.4,0.8);
\draw (0.8,0.8) rectangle (3.2,4);
\draw (2,0.4) node{$K_1$};
\draw (2,2) node{$K_3$};
\draw (2,4) node{\tiny{$\bullet$}};
\draw (2,4.25) node{$(0,0)$};
\draw[<-] (2,3.8)--(2.4,3.8);
\draw (2.7,3.8) node{$3c_n$};
\draw[->] (3,3.8)--(3.2,3.8);
\draw[<->] (2.6,0)--(2.6,0.8);
\draw (2.8,0.4) node{$c_n^2$};
\draw[<->] (2,0.9)--(2.4,0.9);
\draw (2.2,1.1) node{$c_n$};
\end{tikzpicture}
\end{center}
\begin{lemma}\label{barrier}
There exist a universal constant $r_0>0$ and a nonpositive Lipschitz function $h:Q_{1}$ $\rightarrow$ $\mathbb R$, which is $C^2$ with respect to $x$ on the set where $h$ is negative and solves (in viscosity sense)
\begin{align*}
P^+(D^2h)-h_t+\tilde{\phi}(|Dh|)\leq g
\end{align*}
for some continuous, bounded function $g:$ $Q_{1}$ $\rightarrow$ $\mathbb R$ and supp $g$ $\subset$ ${K_1}$,\\
where $\tilde{\phi} = {\Lambda_{0}}{r} \eta \big(\frac{1}{r}\big) \phi$ for $r \leq r_0$.
Also, $h\leq-2$ in ${K_3}$ and $h=0$ on $\partial_pQ_1$.
\end{lemma}
We will use previous barrier function to obtain the following basic measure estimate.
\begin{lemma}\label{measureuniform}
There exist universal constants $L>1$, $\mu \in (0,1)$ and $r_1 \in (0,1)$ such that for any nonnegative supersolution of
\begin{equation}\label{211}
P^-(D^2u) - u_t \leq \tilde{\phi}(|Du|) \hspace{5mm} \text{in} \hspace{2mm}Q_1(0,0),
\end{equation}
where $\tilde{\phi} = {\Lambda_{0}}{r_1} \eta \big(\frac{1}{r_1}\big) \phi$,
the followings holds:\\
If ${\inf}_{K_3}u \leq 1$ then,
\begin{align*}
|\{(x,t) \in K_1:u(x,t) \leq L\}| \geq \mu |K_1|.
\end{align*}
\end{lemma}
The proofs of both Lemma \ref{barrier} and Lemma \ref{measureuniform} are given in the Appendix \ref{a} because they involve long computations.
\begin{rmrk}\label{rmu}
The above lemma is true for the case when $u$ is viscosity supersolution of (\ref{211}) in $ D=\{(x,t):|x|_{\infty}<1,-1<t\leq c_n^2+{\delta}^2\}$ for some $\delta >0$.\\
More precisely, if ${\inf}_{K_3 \cap D}u \leq 1$ then,
\begin{align*}
|\{(x,t) \in K_1:u(x,t) \leq L\}| \geq \mu |K_1|.
\end{align*}
\end{rmrk}
\textbf{Stack of cubes.} As we mentioned in introduction stack of cubes defined in \cite{W1} ( also \cite{is}) does not work in our inhomogeneous situation. So, we will define stack of cubes which will work in our situation. First, we will explain it in general setting, later we will specify required quantities.\\
Let's say we are given a cube $Q_\rho(x_0,t_0)$, for some $\rho \in (0,1)$ and $z_1$, $z_2$, . . ., $z_m$ such that $z_i(\leq 3)$. We will define stack of cubes, denoted by $Q^i(Q_\rho(x_0,t_0))$ for $i=0,1,2,...,m$, corresponding to $Q_\rho(x_0,t_0)$ and $z_1$, $z_2$, . . ., $z_m$. We will define $Q^i(Q_\rho(x_0,t_0))$ inductively as follows:\\
For $i=0$, define $Q^0(Q_\rho(x_0,t_0)):=Q_\rho(x_0,t_0)$.\\
Now,assume $Q^i(Q_\rho(x_0,t_0))$ is defined for $i=k$. We will define a cube for $i=k+1$, i.e., $Q^{k+1}(Q_\rho(x_0,t_0))$.
Let $Q^k(Q_\rho(x_0,t_0))$ be a cube of radius $r_k$ and center $(x_k,t_k)$, i.e., $Q^k(Q_\rho(x_0,t_0))=Q_{r_k}(x_k,t_k)$.\\
Take a cube with the following properties:
\begin{itemize}
\item Radius of cube is $z_{k+1}{r_k}$.
\item Center of cube is $(x_{k+1},t_k+(z_{k+1}{r_k})^2)$, where $x_{k+1}$ is such that $|x_{k+1}-x_k|_{\infty} \leq 3r_k-z_{k+1}{r_k} $.
\item Cube is closest to the line $\{(0,t):t \in \mathbb R \}$.
\end{itemize}
This will be our $Q^{k+1}(Q_\rho(x_0,t_0))$. Thus, we have defined our stack of cubes.
Let $Q^k(Q_\rho(x_0,t_0))$ be a cube of radius $r_k$ and center $(x_k,t_k)$, i.e., $Q^k(Q_\rho(x_0,t_0))=Q_{r_k}(x_k,t_k)$. Then
\begin{align}\label{tildqk}
\tilde{Q}^k(Q_\rho(x_0,t_0)):=\tilde{Q}_{r_k}(x_k,t_k),
\end{align}
where $\tilde{Q}_{r_k}(x_k,t_k)$ is defined in (\ref{tildq}).
In the figure below, $Q^k$ denotes $Q^{k}(Q_\rho(x_0,t_0))$. The bigger cube represents $\tilde{Q}^k(Q_\rho(x_0,t_0))$. For instance cube filled with dots is $\tilde{Q}^3$.
\noindent
\begin{center}
\begin{tikzpicture}[scale=1.36]
\fill[color=gray!20] (8,1) rectangle (8.75,1.5);
\fill[color=gray!20] (7.3,1.5) rectangle (8.2,2.1);
\fill[color=gray!20] (6.2,2.1) rectangle (7.4,2.75);
\fill[color=gray!20] (5.1,2.75) rectangle (6.8,3.53);
\draw (01,0) rectangle (11,9);
\draw (8,1) rectangle (8.75,1.5);
\draw (8.5,1.2) node{$Q_{\rho}$};
\draw[<-] (6,1) -- (6.8,1);
\draw[->] (7.1,1)--(8,1);
\draw (6.95,0.95) node{$d$};
\draw (7.3,1.5) rectangle (9.45,2.65);
\draw (7.3,1.5) rectangle (8.2,2.1);
\draw[<->] (7.75,2.2)--(8.2,2.2);
\draw (7.75,2.2) node{\tiny{$\bullet$}};
\draw (8,2.35) node{$z_1\rho$};
\draw[<-] (7.3,1.35) -- (7.7,1.35);
\draw[->] (8.1,1.35) -- (8.37,1.35);
\draw (7.9,1.3) node{$3\rho$};
\draw (7.8,1.75) node{$Q^1$};
\draw (8.37,1.5) node{\tiny{$\bullet$}};
\draw[<->] (6,1.6) -- (7.3,1.6);
\draw (6.6,1.75) node{$d-2\rho$};
\draw (6.2,2.1) rectangle (9.2,3.6);
\draw (6.2,2.1) rectangle (7.4,2.75);
\draw[<->] (6,2.4)--(6.2,2.4);
\draw[->] (5.5,1.5)--(6.1,2.35);
\draw (5,1.3) node{$d-2\rho-2z_1\rho$};
\draw (6.7,2.4) node{$Q^2$};
\draw (5,2.75) rectangle (8.7,4.75);
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\end{center}
\begin{rmrk}\label{stackrmrk}
We will mention some properties of stack of cubes.
\begin{itemize}
\item $Q^{k}(Q_\rho(x_0,t_0))$ may not be unique.
\item Radius of $Q^{k}(Q_\rho(x_0,t_0))$ is $z_k z_{k-1}...z_{1}\rho$.
\item $Q^{k+1}(Q_\rho(x_0,t_0)) \subset \tilde{Q}^{k}(Q_\rho(x_0,t_0))$. We would like to remind the reader that $\tilde{Q}^{k}(Q_\rho(x_0,t_0))$ corresponding to $Q^{k}(Q_\rho(x_0,t_0))$ as defined in (\ref{tildqk}) is the "shifted in time predecessor" and $Q^{k+1}(Q_\rho(x_0,t_0))$ is the next member in the family of cubes.
\item If $d$ is the distance between $\{(0,t):t \in \mathbb R \}$ and $Q^{k}(Q_\rho(x_0,t_0))$, then the distance between $\{(0,t):t \in \mathbb R \}$ and $\tilde{Q}^{k}(Q_\rho(x_0,t_0))$ is at most max$\{0,d-2r_k\}$. Hence, the distance between $\{(0,t):t \in \mathbb R \}$ and $Q^{k+1}(Q_\rho(x_0,t_0))$ is at most max$\{0,d-2r_k\}$.
\end{itemize}
\end{rmrk}
Now, we will specify $z_1$, $z_2$, . . ., $z_m$ for defining our stack of cubes corresponding to the given nonlinearity. To do this, we will first
define, for $k \geq 1$,
\begin{align*}
a_k=\frac{1}{kL_2(\eta(L^k)+1)},
\end{align*}
where $L$ is from Lemma \ref{measureuniform}.
Then, we have
\begin{align*}
\frac{a_k}{a_{k+1}}=\frac{(k+1)(\eta(L^{k+1})+1)}{k(\eta(L^k)+1)}\leq\frac{(k+1)\eta(LL^{k})}{k\eta(L^k)}+\frac{k+1}{k(\eta(L^k)+1)}\leq\frac{(k+1)\eta(LL^{k})}{k\eta(L^k)}+\frac{k+1}{2k},
\end{align*}
where last inequality is a consequence of $\eta(t)\geq1$ for all $t\geq0$.\\
Using the fact that $(k+1)/k$ goes to $1$ as $k\rightarrow\infty$ and (i) of Lemma \ref{etaprop} with $c=L$, we can choose $k_1 \geq 4$ such that
\begin{align}\label{k1}
\underset{k\geq k_1}{\text{sup}}\frac{a_k}{a_{k+1}}\leq\underset{k\geq k_1}{\text{sup}}\frac{(k+1)\eta(LL^{k})}{k\eta(L^k)}+\frac{k+1}{2k}\leq2.
\end{align}
We also define
\begin{align*}
m_{k_1}:=3,\hspace{2mm}m_{k}:=\frac{a_{k-1}}{a_k} \hspace{2mm} \text{for} \hspace{2mm} k>k_1.
\end{align*}
Now, we are ready to define $z_i$'s.\\ If $Q_\rho(x_0,t_0)$ is such that $|Q_\rho(x_0,t_0) \cap \{{u} > L^{l}\}| > (1-\mu)|Q_\rho(x_0,t_0)|,$ then
\begin{align*}
z_i=m_{l-i} \hspace{2mm} \text{for} \hspace{2mm} 1 \leq i \leq l-{k_1}.
\end{align*}
Note that $z_i$'s are depending on super level set at which measure density of cube has crossed. To emphasize this fact,
we will denote $Q^{k}(Q_\rho(x_0,t_0))$ by $Q^{k}(Q_\rho(x_0,t_0),l)$.
To prove Theorem \ref{mthm} we require a couple of lemmas. In the lemmas, we will assume $u$ is a solution of following equations:
\begin{equation}\label{meq1}
P^-(D^2u)-u_t\leq \tilde{\phi}(|Du|)
\end{equation}
and
\begin{equation}\label{meq2}
P^+(D^2u)-u_t\geq-\tilde{\phi}(|Du|)
\end{equation}
where $\tilde{\phi} = {\Lambda_{0}}{r_1} \eta \big(\frac{1}{r_1}\big) \phi$ and $r_1$ is from Lemma \ref{measureuniform}.
Now, we will prove the lemma which states how measure information will propagate.
\begin{lemma}\label{prop}
Let $u\in C(Q_2)$ be a viscosity supersolution of (\ref{meq1}) in $Q_2$.
Let $ Q_\rho(x_0,t_0)$ satisfy the following conditions:
\begin{itemize}
\item$|Q_\rho(x_0,t_0) \cap \{\tilde{u} > L^{k+1}\}| > \big( 1-\mu \big) |Q_\rho(x_0,t_0)|$,
\item $ \rho \leq \frac{c_n{a_k}}{r}$,
\end{itemize}
where $\tilde{u}(x,t)=u(rx,r^2t).$
Then, the following hold:
\begin{itemize}
\item We have
\begin{align*}
\tilde{u}(x,t)> L^{k} \hspace{2mm}
\text{in} \hspace{2mm} \{(x,t): |x-x_0|_{\infty}<3\rho,t_0<t<(c_n^{-1}\rho)^2+t_0-\rho^2\}\cap\{-1 < t < 0\},
\end{align*}
which in particular implies,
\begin{align*}
\tilde{u}(x,t)> L^{k} \hspace{2mm}
\text{in} \hspace{2mm} \widetilde{Q}^0(Q_{\rho}(x_0,t_0),k+1)\cap\{-1 < t < 0\}.
\end{align*}
\item For $k > k_1$, we have $\tilde{u}> L^{k-i}$ in $\widetilde{Q}^i(Q_{\rho}(x_0,t_0),k+1)\cap\{-1 < t < 0\}$ for all $i$ such that $0 \leq i < k-k_1$.
\end{itemize}
where ${Q^i}(Q_{\rho}(x_0,t_0),k+1)$ denotes the $i^{th}$ member of stacks of cube corresponding to $Q_{\rho}(x_0,t_0)$.
\end{lemma}
\begin{rmrk}
We can rephrase the first assertion of above lemma as follows: Take $(\hat{x},\hat{t} )$ such that $Q_{\rho}(x_0,t_0) =(\hat{x},\hat{t})+(c_n^{-1}\rho)K_1$, then
\begin{align*}
\tilde{u} > L^k \hspace{2mm} \text{in} \hspace{2mm} ((\hat{x},\hat{t})+(c_n^{-1}\rho)K_3) \cap \{-1 <t <0\}.
\end{align*}
\end{rmrk}
\begin{proof}
We will prove the result by induction on $i$.
For the case $i=0$, if
\begin{align*}
\{(x,t):|x-x_0|_{\infty}<3\rho, t_0 <t<(c_n^{-1}\rho)^2+t_0-\rho^2\}\cap\{-1 < t < 0\}=\emptyset,
\end{align*}
then there is nothing to prove. So, we will assume
\begin{align}\label{200}
\{(x,t):|x-x_0|_{\infty}<3\rho, t_0<t<(c_n^{-1}\rho)^2+t_0-\rho^2\}\cap\{-1< t < 0\} \not= \emptyset.
\end{align}
Define
\begin{align*}
v(\,x, t )\, &=\frac{\tilde{u}(c_n^{-1}\rho x + x_0, {(\,c_n^{-1} \rho)\,}^2 t +t_0-\rho^2+{(\,c_n^{-1} \rho)\,}^2)\,}{L^k} ;\\
&=\frac{u(c_n^{-1}\rho r x + x_0, {(\,c_n^{-1} \rho)\,}^2 r^2t +t_0-\rho^2+{(\,c_n^{-1} \rho)\,}^2)\,}{L^k} ;
\end{align*}
where $(x,t) \in D:=\{(x,t):|x|_{\infty}<1,-1<t\leq -1+c_n^2-t_0{(\,c_n^{-1} \rho)\,}^2\}$.\\
From (\ref{200}) we have $t_0 <0$, which implies $D$ contains $K_1$.
It is given that $\rho \leq \frac{c_na_k}{r}$, i.e., $c_n^{-1}\rho r \leq a_k$. So, by Lemma \ref{scaling}, $v$ is a supersolution in $D$.\\ Clearly, $(x,t) \in K_1$ iff $(c_n^{-1}\rho x + x_0, {(\,c_n^{-1}\rho)\,}^2t +t_0-\rho^2+{(\,c_n^{-1} \rho)\,}^2) \in Q_\rho(x_0,t_0)$, so
\begin{align*}
\frac{|K_1 \cap \{v > L\}|}{|K_1|}=\frac{|Q_\rho(x_0,t_0) \cap \{\tilde{u} > L^{k+1}\}|}{|Q_\rho(x_0,t_0)|}.
\end{align*}
Now use first hypothesis of lemma to get $|K_1 \cap \{v > L\}| > \big( 1-\mu \big) |K_1|$, which using Remark \ref{rmu} gives ${\inf }_{K_3 \cap D} v > 1$.
By rescaling back, we get the first part of the lemma. In particular, it gives the conclusion for $i=0.$
Assume result holds for $i\leq l$. We will prove it for $i=l+1 < k-k_1$.\\
If $\widetilde{Q}^{l+1}(Q_{\rho}(x_0,t_0),k+1)\cap\{-1 < t < 0\}= \emptyset$ then there is nothing to prove and we are done.\\
So, we will proceed for the case where $\widetilde{Q}^{l+1}(Q_{\rho}(x_0,t_0),k+1)\cap\{-1< t < 0\}\not= \emptyset$. In this case,
${Q}^{l+1}(Q_{\rho}(x_0,t_0),k+1)\subset Q_{2}$.
By induction hypothesis, we have
\begin{equation}\label{1}
\tilde{u} > L^{k-l} \hspace{2mm}\text{in} \hspace{2mm} {Q}^{l+1}(Q_{\rho}(x_0,t_0),k+1).
\end{equation}
By definition, ${Q}^{l+1}(Q_{\rho}(x_0,t_0),k+1)$ is a cube of radius $z_{l+1}z_{l}...z_1\rho$, which is the same as $m_{k-l}m_{k+1-l}...m_{k}\rho$ (we will call it $r_{l+1}$) and we call its center $(\,x_1,t_1)\,$ i.e., ${Q}^{l+1}(Q_{\rho}(x_0,t_0),k+1)=Q_{r_{l+1}}(\,x_1,t_1)\,$.\\
Define
\begin{align*}
v(\,x, t )\,&=\frac{\tilde{u}(c_n^{-1}r_{l+1} x + x_1, {(\,c_n^{-1}r_{l+1})\,}^2t +t_1-{r_{l+1}}^2+{(\,c_n^{-1}r_{l+1})\,}^2)\,}{L^{k-l-1}}\\
&= \frac{u(c_n^{-1}r_{l+1}r x + x_1, {(\,c_n^{-1}r_{l+1}r)\,}^2t +t_1-{r_{l+1}}^2+{(\,c_n^{-1}r_{l+1})\,}^2)\,}{L^{k-l-1}},
\end{align*}
where $(x,t)\in D:=\{(x,t):|x|_{\infty}<1,-1<t\leq -1+c_n^2-t_1{(\,c_n^{-1}r_{l+1})\,}^2\}$.\\
Note that $\widetilde{Q}^{l+1}(Q_{\rho}(x_0,t_0),k+1)\cap\{-1 < t < 0\}\not= \emptyset$ implies $t_1 < 0$. Hence, $D$ contains $K_1$.
In view of Lemma \ref{scaling}, we know $v$ will be a supersolution in its domain if
\begin{align*}
c_n^{-1}r_{l+1}r &\leq a_{k-l-1},\\
\text{i.e.,} \hspace{2mm} c_n^{-1} z_{l+1}z_{l}...z_1\rho r &\leq a_{k-l-1},\\
\text{i.e.,} \hspace{2mm} c_n^{-1}m_{k-l}m_{k-l+1}...m_{k}\rho r &\leq a_{k-l-1},\\
\text{i.e.,} \hspace{2mm} c_n^{-1}\frac{a_{k-l-1}}{a_{k-l}}\frac{a_{k-l}}{a_{k-l+1}}...\frac{a_{k-1}}{a_{k}}\rho r &\leq a_{k-l-1},\\
\text{i.e.,} \hspace{2mm} c_n^{-1} \rho r &\leq a_{k}.
\end{align*}
This is true by the second hypothesis of lemma. Hence, $v$ is a supersolution in its domain. Also, we have $(x,t) \in K_1$ iff $(c_n^{-1}r_{l+1} r x + x_1, {(\,c_n^{-1}r_{l+1} r)\,}^2t +t_1-{r_{l+1}}^2+{(\,c_n^{-1}r_{l+1})\,}^2))\, \in Q_{r_{l+1}}(\,x_1,t_1)\,$ which gives
\begin{align*}
\frac{|K_1 \cap \{v > L\}|}{|K_1|}=\frac{|Q_{r_{l+1}}(\,x_1,t_1)\, \cap \{\tilde{u} > L^{k-l}\}|}{|Q_{r_{l+1}}(\,x_1,t_1)\,|}.
\end{align*}
The above equation using \eqref{1} gives $|K_1 \cap \{v > L\}| = |K_1| > \big( 1-\mu \big) |K_1|$. Now, use Remark \ref{rmu} to get $\underset{K_3 \cap D}{\inf v} > 1$, which in particular implies $\tilde{u}>L^{k-l-1}$ in $\widetilde{Q}^{l+1}(Q_{\rho}(x_0,t_0),k+1)\cap\{-1 < t < 0\}$. This proves the induction step and hence the lemma.
\end{proof}
In the above lemma, measure information was propagating provided initial cube's size is small. We will now prove the lemma which will ensure smallness of cube.
\begin{lemma}\label{decay}
Let $u\in C(Q_2)$ be a viscosity supersolution of (\ref{meq1}) in $Q_2$ such that
$u(\,0,0)\, = 1$.
Let there exist $Q_\rho(x_1,t_1) \subset K_1$ such that
\begin{align}\label{aa}
| Q_\rho(x_1,t_1) \cap \{\tilde{u} > L^k\}| > \big( 1-\mu \big) |Q_\rho(x_1,t_1)|,
\end{align}
where $\tilde{u}=u(a_{k_1}x,a_{k_1}^2t)$.
Then,
$\hspace{2mm} \rho < \frac{ c_n a_{k-1}}{3 a_{k_1}}$ for all $k \geq k_1 +1$.
\end{lemma}
\begin{proof}
We will prove it by induction on $k$.\\
Step 1: For $k=k_1+1$.
Since $Q_\rho(x_1,t_1) \subset K_1$ so $\rho \leq c_n$, which is the same as $\rho \leq \frac{c_n a_{k_1}}{a_{k_1}}$.
Then by Lemma \ref{prop}, we have
\begin{align}\label{ak2}
\tilde{u}(x,t) &>L^{k_1} \hspace{2mm} \text{in} \hspace{2mm}\tilde{Q}_{\rho}(x_1,t_1).
\end{align}
We claim that $3\rho < c_n$.\\
Assume this is not true, i.e., $3\rho \geq c_n$.
Recall radius of $\tilde{Q}_{\rho}(x_1,t_1)$ is $3\rho$, which is greater than or equal to $c_n$, so it contains some translated version of $K_1$. So, we can
take $(\,x_0,t_0)\,$ such that the following hold:
\begin{itemize}
\item[(i)] $(\,x_0,t_0)\, +K_1 \subset \tilde{Q}_{\rho}(x_1,t_1)$,
\item[(ii)]$\{(0,s):s<0\} \cap (\,x_0,t_0)\, +K_1 \not= \emptyset.$
\end{itemize}
(we can assure the condition (ii) by a simple application of triangle inequality to the fact that $Q_\rho(x_1,t_1) \subset K_1
$) \\
Also, note that condition (ii) will imply $(0,0) \in (\,x_0,t_0)\, + K_3$. Note that (\ref{ak2}) gives
\begin{align*}
\frac{|(x_0,t_0)+K_1 \cap \{\tilde{u} > L^{k_1}\}|}{|(x_0,t_0)+K_1|}=1>(1-\mu).
\end{align*}
Since $(a_k)$ is a decreasing sequence, $c_n < \frac{c_n a_{k_1-1}}{a_{k_1}}$. Hence, by Lemma \ref{prop}, we get $\tilde{u} > L^{k_1-1}$ in $((x_0,t_0)+K_3) \cap Q_1$. In particular, we get $u(0,0) > 1$, which is a contradiction to $u(0,0)=1$.
Hence we get our claim, i.e., $3\rho < c_n$.
This proves the lemma for $k=k_1+1.$\\
Step 2: Assume result holds for $k=l \geq k_1+1$, i.e., if there exists $Q_\rho \subset K_1$ such that
\begin{align}\label{aa1}
| Q_\rho(x_1,t_1) \cap \{\tilde{u} > L^l\}| > \big( 1-\mu \big) |Q_\rho(x_1,t_1)|
\end{align}
then $\hspace{2mm} \rho < \frac{ c_n a_{l-1}}{3 a_{k_1}}$.\\
Take $Q_\rho(x_1,t_1) \subset K_1$ such that
\begin{align}\label{aa2}
| Q_\rho(x_1,t_1) \cap \{\tilde{u} > L^{l+1}\}| > \big( 1-\mu \big) |Q_\rho(x_1,t_1)|.
\end{align}
Clearly, (\ref{aa2}) implies (\ref{aa1}). So, by induction hypothesis, we have
\begin{align*}
\rho < \frac{ c_n a_{l-1}}{3 a_{k_1}}.
\end{align*}
Using $\underset{k\geq k_1}{\text{sup}}\frac{a_k}{a_{k+1}} \leq 2$ and $l-1 \geq k_1$, We get
\begin{align}\label{aa3}
\rho < \frac{ c_n a_{l}}{ a_{k_1}}.
\end{align}
Now (\ref{aa2}) and (\ref{aa3}) make us eligible to apply Lemma \ref{prop}, and we get
\begin{align}\label{aa500}
\tilde{u}> L^{l-i} \hspace{2mm}\text{in}\hspace{2mm} \widetilde{Q}^i(Q_{\rho}(x_1,t_1),l+1)\cap\{-1 < t < 0\} \hspace{2mm}\text{for all}\hspace{2mm} i\hspace{2mm} \text{such that}\hspace{2mm} 0 \leq i < l-k_1,
\end{align}
which implies,
\begin{align}\label{aa501}
\tilde{u}> L^{l+1-i} \hspace{2mm}\text{in}\hspace{2mm} {Q}^i(Q_{\rho}(x_1,t_1),l+1)\cap\{-1 < t < 0\} \hspace{2mm}\text{for all}\hspace{2mm} i\hspace{2mm} \text{such that}\hspace{2mm} 0 \leq i \le l-k_1.
\end{align}
If $\widetilde{Q}^{l-k_1}(Q_{\rho}(x_1,t_1),l+1)\cap\{-1 < t < 0\} \not= \emptyset$, then ${Q}^{l-k_1}(Q_{\rho}(x_1,t_1),l+1) \subset Q_2.$\\
Also, note that the radius of ${Q}^{l-k_1}(Q_{\rho}(x_1,t_1),l+1)$ is $z_{l-k_1}z_{l-k_1-1}...z_1\rho$, i.e., $m_{k_1+1}m_{k_1+2}...m_l\rho$. Using (\ref{aa3}) we get
\begin{align*}
\text{radius of ${Q}^{l-k_1}(Q_{\rho}(x_1,t_1),l+1)$}=\frac{a_{k_1}}{a_l}\rho<c_n
\leq \frac{c_n a_{k_1}}{a_{k_1}}.
\end{align*}
Now, apply Lemma \ref{prop} ( for $k=k_1$ and cube ${Q}^{l-k_1}(Q_{\rho}(x_1,t_1),l+1)$ ) to get
\begin{align}\label{aa502}
\tilde{u}>L^{k_1} \hspace{2mm} \text{in} \hspace{2mm}\widetilde{Q}^{l-k_1}(Q_{\rho}(x_1,t_1),l+1)\cap\{-1 < t < 0\}.
\end{align}
Hence, from (\ref{aa500}), (\ref{aa501}) and (\ref{aa502}), we have
\begin{align}\label{aa4}
\tilde{u}> L^{l-i} \hspace{2mm}\text{in}\hspace{2mm} \widetilde{Q}^i(Q_{\rho}(x_1,t_1),l+1)\cap\{-1 < t < 0\} \hspace{2mm}\text{for all}\hspace{2mm} i\hspace{2mm} \text{such that}\hspace{2mm} 0 \leq i \leq l-k_1,
\end{align}
which implies,
\begin{align}\label{aa5}
\tilde{u}> L^{l+1-i} \hspace{2mm}\text{in}\hspace{2mm} {Q}^i(Q_{\rho}(x_1,t_1),l+1)\cap\{-1 < t < 0\} \hspace{2mm}\text{for all}\hspace{2mm} i\hspace{2mm} \text{such that}\hspace{2mm} 0 \leq i \le l-k_1+1.
\end{align}
\textbf{Claim(1)}: $\rho < \frac{ c_n a_{l}}{3 a_{k_1}}$.\\
\textbf{Proof of Claim(1)}: We will prove it by contradiction. Assume Claim(1) is not true, i.e., $\rho \geq \frac{ c_n a_{l}}{3 a_{k_1}}$, which is the same as saying that, $\frac{3 a_{k_1}}{ a_{l}}\rho \geq c_n$, which we can write as $3\frac{a_{k_1}}{a_{k_1+1}}\frac{a_{k_1+1}}{a_{k_1+2}}...\frac{a_{l-1}}{a_{l}}\rho \geq c_n$, which is $m_{k_1}m_{k_1+1}...m_{l}\rho \geq c_n$, i.e., radius of ${Q}^{l+1-k_1}(Q_{\rho}(x_1,t_1),l+1) \geq c_n.$ Choose first $i_0 $ such that radius of ${Q}^{i_0}(Q_{\rho}(x_1,t_1),l+1) \geq c_n.$
Clearly, $i_0 \leq l+1-k_1$. \\
There are three possible situations:
\begin{itemize}
\item[1.] ${Q}^{i_0}(Q_{\rho}(x_1,t_1),l+1) \subset \mathbb R^n\times \{t<0\}$.
\item[2.] ${Q}^{i_0}(Q_{\rho}(x_1,t_1),l+1) \cap \mathbb R^n\times \{t=0\} \not= \emptyset$.
\item[3.] ${Q}^{i_0}(Q_{\rho}(x_1,t_1),l+1) \subset \mathbb R^n\times \{t>0\}$.
\end{itemize}
Now, we will rule out all three situations to get a contradiction.
\textbf{Case 1}: ${Q}^{i_0}(Q_{\rho}(x_1,t_1),l+1) \subset \mathbb R^n\times \{t<0\}$.
Note that radius of ${Q}^{i_0}(Q_{\rho}(x_1,t_1),l+1) \geq c_n$, so we can take $(x_0,t_0)$ such that
\begin{itemize}
\item[(A)] $(x_0,t_0)+K_1 \subset {Q}^{i_0}(Q_{\rho}(x_1,t_1),l+1)$ and
\item[(B)] $(x_0,t_0)+K_1$ intersects $\{0\} \times \mathbb R$.
\end{itemize}
Note that radius of ${Q}^{i_0}(Q_{\rho}(x_1,t_1),l+1) \geq c_n$ and $Q_{\rho}(x_1,t_1) \subset K_1$. With an easy application of triangle inequality to these facts, we can ensure condition (B). Condition (B) will further imply $(0,0) \in (x_0,t_0)+K_3$.
Note that, from (\ref{aa5}) and $i_0 \leq l+1-k_1$, we have $\tilde{u}>L^{k_1}$ in $(x_0,t_0)+K_1$. Also, radius of $(x_0,t_0)+K_1=c_n \leq \frac{c_n a_{k_1-1}}{a_{k_1}}.$ Therefore, from Lemma \ref{prop}, we get $\tilde{u}>L^{k_1-1}$ in $(x_0,t_0)+K_3 \cap Q_1$, which in particular implies $u(0,0)=\tilde{u}(0,0)>L$, but this is not true as $u(0,0)=1$. Hence, this situation is ruled out.
\textbf{Case 2}: ${Q}^{i_0}(Q_{\rho}(x_1,t_1),l+1) \cap \mathbb R^n\times \{t=0\} \not= \emptyset$.
By definition of $i_0$, we have ${Q}^{i_0}(Q_{\rho}(x_1,t_1),l+1) \cap \{0\}\times \mathbb R \not= \emptyset$. Hence, we get $(0,0) \in {Q}^{i_0}(Q_{\rho}(x_1,t_1),l+1).$ Use (\ref{aa5}) and $i_0 \leq l+1-k_1$ to get $u(0,0)=\tilde{u}(0,0)>L^{k_1}$, which is false as $u(0,0)=1$. Hence, this situation is ruled out.\\
\textbf{Case 3}: ${Q}^{i_0}(Q_{\rho}(x_1,t_1),l+1) \subset \mathbb R^n\times \{t>0\}$.
\textbf{Claim(2)}: There exists $i<i_0$ such that $(0,0) \in {Q}^{i}(Q_{\rho}(x_1,t_1),l+1)$.
Assume Claim(2) has been established, then by (\ref{aa5}) we have $u(0,0) > L^{l+1-i}$. Now, $i \leq l-k_1$ implies $u(0,0) > L^{k_1+1}$, which contradicts $u(0,0)=1$. Hence, we will rule this situation out provided we could establish the Claim(2).
Now we will establish Claim(2).\\
Choose first $j_0$ such that ${Q}^{j_0}(Q_{\rho}(x_1,t_1),l+1)\cap \{0\}\times \mathbb R \not= \emptyset$. Clearly, $j_0 \leq i_0$.
Note the following observations:
\begin{itemize}
\item
For $j_0 \geq 2$, we have
\begin{align*}
r_0+r_1+...+r_{j_0-2}<\frac{c_n\sqrt{n}}{2}.
\end{align*}
The reason is as follows. If $d$ is the distance between $Q_{\rho}(x_1,t_1)$ and $\{0\}\times \mathbb R$, then by Remark \ref{stackrmrk}, distance between ${Q}^{j_0-1}(Q_{\rho}(x_1,t_1),l+1)$ and $\{0\}\times \mathbb R$ is at most max$\{d-2r_0-2r_1-...-2r_{j_0-2},0\}$. Now,
\begin{align*}
{Q}^{j_0-1}(Q_{\rho}(x_1,t_1),l+1)\cap \{0\}\times \mathbb R = \emptyset\\
\implies d-2r_0-2r_1-...-2r_{j_0-2} > 0\\
\implies 2r_0+2r_1+...+2r_{j_0-2}<d.
\end{align*}
Also, $Q_{\rho}(x_1,t_1) \subset K_1$ implies $d<c_n\sqrt{n}$. Hence, we get
\begin{align*}
r_0+r_1+...+r_{j_0-2}<\frac{c_n\sqrt{n}}{2}.
\end{align*}
\item For $j_0 \geq 1$, $r_{j_0-1} < c_n$ and $r_{j_0}< 3c_n$.\\
($r_{j_0-1} < c_n$ follows from definition of $j_0$. And $r_{j_0}< 3c_n$ follows from $r_{j_0} \leq 3r_{j_0-1}$ )
\end{itemize}
Now, we will show ${Q}^{j_0}(Q_{\rho}(x_1,t_1),l+1) \subset \mathbb R^n \times \{t<0\}$. For $j_0=0$, we are done because $Q_{\rho}(x_1,t_1) \subset K_1$. Also, for $j_0 \geq 1$, we have $(x,t) \in {Q}^{j_0}(Q_{\rho}(x_1,t_1),l+1)$ implies
\begin{align*}
t
\leq &\hspace{1mm}t_1+r_1^2+...+r_{j_0-2}^2+r_{j_0-1}^2+r_{j_0}^2\\
\leq&-1+ c_n^2+r_1^2+...+r_{j_0-2}^2+r_{j_0-1}^2+r_{j_0}^2\\
<&-1+ c_n^2+r_0^2+r_1^2+...+r_{j_0-2}^2+r_{j_0-1}^2+r_{j_0}^2\\
\leq&-1+ c_n^2+(r_0+r_1+...+r_{j_0-2})^2+r_{j_0-1}^2+r_{j_0}^2\\
\leq&-1+ c_n^2+\frac{nc_n^2}{4}+c_n^2+9c_n^2\\
\leq&-1+12nc_n^2\\
\leq&-1+\frac{12}{100n}\\
<&0.
\end{align*}
Therefore we are done. Thus, we get $j_0<i_0$,
\begin{align*}
{Q}^{j_0}(Q_{\rho}(x_1,t_1),l+1) \subset \mathbb R^n\times \{t<0\},\\
{Q}^{j_0}(Q_{\rho}(x_1,t_1),l+1)\cap \{0\}\times \mathbb R \not= \emptyset.
\end{align*}
Also, we have ${Q}^{i_0}(Q_{\rho}(x_1,t_1),l+1) \subset \mathbb R^n\times \{t>0\}$. Therefore, the way we have defined the stack of cubes will give $i$ such that $j_0<i<i_0$ and $(0,0) \in {Q}^{i}(Q_{\rho}(x_1,t_1),l+1)$. This establishes the Claim(2).
\end{proof}
Now, we are ready to prove the $L^{\epsilon}$-estimate.
\begin{lemma}\label{wh}
Let $u\in C(Q_2)$ be a viscosity supersolution of (\ref{meq1}) in $Q_2$ such that
$u(\,0,0)\, = 1$. Then, there exist universal constants $C$ and $\epsilon >0$ such that the following hold:
\begin{align}
|\{\tilde{u} >\tau\} \cap \hat{K}| \leq C\tau^{-{\epsilon}}
\end{align}
and
\begin{align}\label{lepsilon}
\bigg(\int_{\hat{K}}\tilde{u}^{\epsilon}\bigg)^{\frac{1}{\epsilon}} \leq C,
\end{align}
where $\tilde{u}(x,t)=u(a_{k_1}x,a_{k_1}^2t),$ and
$\hat{K} = \{(x,t):|x|_{\infty} < c_n, -1 < t < -1+c_n^2/2\}$.
\end{lemma}
\begin{proof}
Choose $m$ large enough so that $\big(1-\frac{\mu}{2}\big) \leq \frac{m+1}{m}(1-\mu)$, where $\mu$ is from Lemma \ref{measureuniform}. Now, choose $k_{2} >k_1$ large enough such that
\begin{equation}
\sum_{k\geq k_2} \frac{(m+1) a_k^2}{a_{k_1}^2}= \frac{m+1}{a_{k_1}^2} \sum_{k\geq k_2}a_k^2 \leq \frac{m+1}{a_{k_1}^2} \sum_{k\geq k_2} \frac{1}{k^2} \leq \frac{c_n^2}{2}.
\end{equation}
Define, for $k$ $\geq$ $k_2$,
\begin{equation}
A_k = \bigg\{(x,t):\hspace{1mm} \tilde{u}(x,t) > L^{km},\hspace{1mm} |x|_{\infty} < c_n \hspace{2mm} \& \hspace{2mm} -1 < t < -1+c_n^2- \frac{m+1}{a_{k_1}^2}\sum_{j=k_2}^{k} \frac{1}{j^2} \bigg\}.
\end{equation}
We claim that, for $k \geq k_2$, the following holds:
\begin{align*}
| A_{k+1} | \leq \Big( 1- \frac{\mu}{2} \Big) | A_{k} |.
\end{align*}
To prove this, we want to apply the Lemma \ref{cald} with
\begin{align*}
A&=A_{k+1},\\
B&=A_{k},\\
Q&=K_1=(-c_n,c_n)^n \times (-1,-1+c_n^2),\\
\text{and} \hspace{2mm} \delta&=1-\mu.
\end{align*}
We have $\tilde{u}(0,0)=u(0,0)=1$, so $\text{inf}_{K_3}\tilde{u}\leq 1$. Hence, by Lemma \ref{measureuniform}, we have
\begin{align*}
|A| \leq |\{\tilde{u} > L\} \cap K_1| < (1-\mu)|K_1|=(1-\mu)|Q|.
\end{align*}
Take any dyadic cube $Q_\rho(x_1,t_1) \subset Q$ such that
\begin{equation*}
| Q_\rho(x_1,t_1) \cap A | > ( 1-\mu ) |Q_\rho(x_1,t_1)|,
\end{equation*}
i.e.,
\begin{equation}\label{11}
| Q_\rho(x_1,t_1) \cap A_{k+1} | > ( 1-\mu ) |Q_\rho(x_1,t_1)|.
\end{equation}
From now onwards in this proof we will call $Q_\rho(x_1,t_1)$ by $Q_\rho$.
In view of Lemma \ref{cald}, we have to show $ \Bar{Q_{\rho}}^m \subset B$,
i.e., we have to show $\Bar{Q_{\rho}}^m \subset A_{k}$.\\
First we will show
\begin{align}\label{145}
\Bar{Q_{\rho}}^m \subset \{|x|_{\infty} < c_n\} \times \bigg\{-1 < t < -1+c_n^2- \frac{m+1}{a_{k_1}^2}\sum_{j=k_2}^{k} \frac{1}{j^2} \bigg\}.
\end{align}
Because of (\ref{11}), we have
\begin{align*}
Q_{\rho} \cap \{|x|_{\infty} < c_n\} \times \bigg\{-1 < t < -1+c_n^2- \frac{m+1}{a_{k_1}^2}\sum_{j=k_2}^{k+1} \frac{1}{j^2} \bigg\} \not= \emptyset.
\end{align*}
Hence
\begin{align*}
\Bar{Q_{\rho}}^m \subset \{|x|_{\infty} < c_n\} \times \bigg\{-1 < t < -1+c_n^2- \frac{m+1}{a_{k_1}^2}\sum_{j=k_2}^{k+1} \frac{1}{j^2} + \text{height}(\Bar{Q_{\rho}})+\text{height}(\Bar{Q_{\rho}}^m) \bigg\},
\end{align*}
where height($L$) = sup$\{t:\exists\hspace{1mm} x,\hspace{1mm}(x,t)\in L\}$ - inf$\{t:\exists\hspace{1mm} x,\hspace{1mm}(x,t)\in L\}$.
Moreover,
\begin{align*}
\text{height}({Q_{\rho}})&=\rho^2,\\
\text{height}(\Bar{Q_{\rho}})&=4\hspace{1mm} \text{height}({Q_{\rho}})=4\rho^2,\\
\text{height}(\Bar{Q_{\rho}}^m)&=m\hspace{1mm}\text{height}(\Bar{Q_{\rho}})=4m\rho^2.
\end{align*}
To show (\ref{145}), it is enough to show
\begin{align*}
-1+c_n^2- \frac{m+1}{a_{k_1}^2}\sum_{j=k_2}^{k+1} \frac{1}{j^2} + \text{height}(\Bar{Q_{\rho}})+\text{height}(\Bar{Q_{\rho}}^m) \leq -1+c_n^2- \frac{m+1}{a_{k_1}^2}\sum_{j=k_2}^{k} \frac{1}{j^2},
\end{align*}
which is the same as showing
\begin{align*}
\text{height}(\Bar{Q_{\rho}})+\text{height}(\Bar{Q_{\rho}}^m) \leq \frac{m+1}{a_{k_1}^2} \frac{1}{(k+1)^2},
\end{align*}
which is the same as showing
\begin{align}\label{1452}
4(m+1)\rho^2 \leq \frac{m+1}{a_{k_1}^2} \frac{1}{(k+1)^2}.
\end{align}
By Lemma \ref{decay}, (\ref{11}) implies
\begin{align*}
\rho \leq \frac{c_n}{3a_{k_1}} a_{(k+1)m-1}.
\end{align*}
$\text{Using}\hspace{2mm} \underset{k\geq k_1}{\text{sup}}\frac{a_k}{a_{k+1}}\leq 2$, we get
\begin{align*}
\rho \leq \frac{c_n}{a_{k_1}} a_{(k+1)m}.
\end{align*}
Since $a_k \leq \frac{1}{k}$, we have
\begin{align*}
\rho \leq \frac{c_n}{a_{k_1}} \frac{1}{(k+1)m}.
\end{align*}
Using $4 c_n^2 =4 (10n)^{-2}\leq 1$, we get
\begin{align*}
4(m+1)\rho^2 \leq \frac{m+1}{a_{k_1}^2} \frac{1}{((k+1)m)^2}.
\end{align*}
Since $m \geq 1$, we get (\ref{1452}). Hence we get (\ref{145}).
Now we will show
\begin{align*}
\tilde{u}>L^{km}\hspace{2mm}\text{in} \hspace{2mm} \Bar{Q_{\rho}}^m.
\end{align*}
Since $Q_\rho(x_1,t_1) \subset K_1$ satisfies (\ref{11}), by Lemma \ref{decay}, we get $\rho < \frac{c_n}{3a_{k_1}} a_{(k+1)m-1}$.
Hence, we can apply Lemma \ref{prop} to get
\begin{align*}
\tilde{u}>L^{(k+1)m-1} \hspace{2mm}\text{in} \hspace{2mm} B_{3\rho}(x_1) \times (t_1,t_1+(c_n^{-1}\rho)^2-\rho^2) \cap \{-1<t<0\},
\end{align*}
where $B_{3\rho}(x_1):=\{x \in \mathbb R^n:|x-x_1|_{\infty}<3\rho\}$.
If \[\Bar{Q_{\rho}}^m \subset B_{3\rho}(x_1) \times (t_1,t_1+(c_n^{-1}\rho)^2-\rho^2) \cap \{-1<t<0\},\] then we are done.
If not, then note that (\ref{145}) gives
\[D:=\{x:|x-x_1|_{\infty}<\rho\} \times (t_1+(c_n^{-1}\rho)^2-2\rho^2,t_1+(c_n^{-1}\rho)^2-\rho^2) \subset Q_1(0,0),\]
Also, we have
\begin{align*}
\rho &< \frac{c_n}{3a_{k_1}} a_{(k+1)m-1}\\
& < \frac{c_n}{3a_{k_1}} a_{(k+1)m-2}\hspace{5mm} \text{(since $(a_k)$ is a decreasing sequence).}
\end{align*}
Now, apply Lemma \ref{prop} ( with $k=(k+1)m-2$ and cube $D$) to get
\begin{align*}
\tilde{u}>L^{(k+1)m-2} \hspace{2mm}\text{in} \hspace{2mm} B_{3\rho}(x_1) \times (t_1+(c_n^{-1}\rho)^2-\rho^2,t_1+2[(c_n^{-1}\rho)^2-\rho^2]) \cap \{-1<t<0\}.
\end{align*}
Then, we have
\begin{align*}
\tilde{u}>L^{(k+1)m-2} \hspace{2mm}\text{in} \hspace{2mm} B_{3\rho}(x_1) \times (t_1,t_1+2[(c_n^{-1}\rho)^2-\rho^2]).
\end{align*}
If $\Bar{Q_{\rho}}^m \subset B_{3\rho}(x_1) \times (t_1,t_1+2[(c_n^{-1}\rho)^2-\rho^2]) \cap \{-1<t<0\}$ then we are done.\\
If not, then continue like before and repeat the process.
Also, note that
\begin{align*}
m(c_n^{-2}-1)\rho^2 =m(100n^2-1) \geq 99m\rho^2 > 4(m+1)\rho^2.
\end{align*}
Therefore we get
\begin{align*}
\Bar{Q_{\rho}}^m \subset B_{3\rho}(x_1) \times (t_1,t_1+j[(c_n^{-1}\rho)^2-\rho^2]) \cap \{-1<t<0\}
\end{align*}
for some $j \in \{1,2,3,...,m\}$ and
\begin{align*}
\tilde{u}>L^{(k+1)m-j} \hspace{2mm}\text{in} \hspace{2mm} B_{3\rho}(x_1) \times (t_1,t_1+j[(c_n^{-1}\rho)^2-\rho^2]),
\end{align*}
which implies,
\begin{align*}
\tilde{u}>L^{km} \hspace{2mm}\text{in} \hspace{2mm} B_{3\rho}(x_1)\times (t_1,t_1+j[(c_n^{-1}\rho)^2-\rho^2]).
\end{align*}
In particular, we have
\begin{align}\label{1453}
\tilde{u}>L^{km}\hspace{2mm}\text{in} \hspace{2mm} \Bar{Q_{\rho}}^m.
\end{align}
Thus, from (\ref{145}) and (\ref{1453}) we get $\Bar{Q_{\rho}}^m \subset A_k$.
Hence, by Lemma \ref{cald}, we get
\begin{align*}
|A_{k+1}| &\leq (1-\mu)\frac{m+1}{m}|A_{k}|\\
& \leq \big(1-\frac{\mu}{2}\big)|A_{k}|.
\end{align*}
Thus, for $k\geq k_2$, we have
\begin{align*}
|A_k| \leq \bigg(1-\frac{\mu}{2}\bigg)^{k-k_2}|A_{k_2}|.
\end{align*}
In particular, for any natural number $k$, we have
\begin{align*}
\bigg|\bigg\{(x,t): \tilde{u}(x,t) > L^{km}, |x|_{\infty} < c_n \hspace{2mm} \& \hspace{2mm} -1 < t < -1+\frac{c_n^2}{2} \bigg\}\bigg| &\leq \bigg(1-\frac{\mu}{2}\bigg)^{k-k_2}|A_{k_2}|\\
&\leq C_1\bigg(1-\frac{\mu}{2}\bigg)^{k},
\end{align*}
where $C_1$ is a universal constant.\\
This implies
\begin{align*}
|\{\tilde{u} >\tau\} \cap \hat{K}| \leq C_1\tau^{-{\epsilon}_1},
\end{align*}
where ${\epsilon}_1=-\frac{\text{ln}(1-\mu/2)}{m (\text{ln} L)}$.\\
Now, for ${\epsilon}=-\frac{\text{ln}(1-\mu/2)}{2m (\text{ln} L)}$ we have
\begin{align*}
\int_{\hat{K}}\tilde{u}^{\epsilon}dxdt&=\epsilon\int_0^{\infty}\tau^{\epsilon-1}|\{\tilde{u} >\tau\} \cap \hat{K}|d\tau\\
&\leq \epsilon|\hat{K}|\int_0^{1}\tau^{\epsilon-1}d\tau+\epsilon\int_1^{\infty}\tau^{\epsilon-1}|\{\tilde{u} >\tau\} \cap \hat{K}|d\tau\\
&\leq \epsilon|\hat{K}|\int_0^{1}\tau^{\epsilon-1}d\tau+\epsilon C_1\int_1^{\infty}\tau^{\epsilon-1}\tau^{-{\epsilon}_1}d\tau\\
&\leq \epsilon|\hat{K}|\int_0^{1}\tau^{\epsilon-1}d\tau+\epsilon C_1\int_1^{\infty}\tau^{\frac{-{\epsilon}_1}{2}-1}d\tau\\
&\leq C_2,
\end{align*}
where $C_2$ is a universal constant.\\
Thus, there exist universal constants $C$ (=max$\{C_1,C_2\}$) and $\epsilon >0$ such that
\begin{align*}
|\{\tilde{u} >\tau\} \cap \hat{K}| \leq C\tau^{-{\epsilon}}
\end{align*}
and
\begin{align*}
\bigg(\int_{\hat{K}}\tilde{u}^{\epsilon}\bigg)^{\frac{1}{\epsilon}} \leq C.
\end{align*}
\end{proof}
\begin{lemma}\label{super}
Let $u \in C(Q_2)$ be a solution of (\ref{meq1}) and (\ref{meq2}) with $u(0,0)=1$. Then, there exist universal constants $L_0$ and $\sigma$ such that for $\nu=\frac{L_0}{L_0-1/2}$, the following holds:\\
If \[(x_0,t_0) \in \{(x,t): |x|_{\infty}<c_n/2, -1+c_n^2/4< t < -1+c_n^2/2\}\] and $l$ is a natural number such that
\begin{align*}
\tilde{u}(x_0,t_0)\geq \nu^lL_0,
\end{align*}
where $\tilde{u}(x,t)=u(a_{k_1}x,a_{k_1}^2t)$.
Then,
\begin{align*}
\text{sup}_{Q_{r_l}(x_0,t_0)} \tilde{u}(x,t)>\nu^{l+1}L_0,
\end{align*}
where $r_l=\sigma \nu^{-(l+1)\frac{\epsilon}{n+2}}\big(\frac{L_0}{2}\big)^{-\frac{\epsilon}{n+2}} < 1$. Here, $\epsilon >0$ is a universal constant in Lemma \ref{wh}.
\end{lemma}
\begin{proof}
Choose $\sigma$ such that $\sigma^{n+2} > \frac{C}{\mu c_n^{n+2}}$, where $C$ is a constant as in Lemma \ref{wh}.
Assume, on the contrary, $\text{sup}_{Q_{r_l}}\tilde{u}(x,t)\leq \nu^{l+1}L_0$. We will choose the universal constant $L_0$ later. We define
\begin{align*}
G=\{(x,t):|x-x_0|_{\infty}<r_l c_n,t_0-r_l^2<t<t_0+r_l^2(c_n^2-1))\}.
\end{align*}
\begin{center}
\begin{tikzpicture}[scale=0.8];
\draw (2,-0.25) node{$(x_0,t_0-r_l^2)$};
\draw (-0.5,0) rectangle (4.5,4);
\draw (2,3) node{$Q_{r_l}$};
\draw (1,0) rectangle (3,2);
\fill[color=gray!20] (1,0) rectangle (3,2);
\draw (2,1) node{$G$};
\draw (2,4) node{\tiny{$\bullet$}};
\draw (2,4.25) node{$(x_0,t_0)$};
\draw (3.5,1.0) node{$(r_l c_n)^2$};
\draw[->] (3.2,1.2)--(3.2,2);
\draw[<-] (3.2,0)--(3.2,.8);
\draw (2.6,2.3) node{$r_l c_n$};
\draw[<->] (2,2.1)--(3,2.1);
\draw[<-] (2,3.8)--(3,3.8);
\draw[->] (3.5,3.8)--(4.5,3.8);
\draw (3.3,3.8) node{$r_l$};
\draw [<-] (4.7,0)--(4.7,1.8);
\draw [->] (4.7,2.2)--(4.7,4);
\draw (4.75,2) node{$r_l^2$};
\draw (2,0) node{\tiny{$\bullet$}};
\end{tikzpicture}
\end{center}
Now, we will estimate measure of $G$ using $L^{\epsilon}$-estimate and basic measure estimate.\\
First, we will estimate
\begin{align*}
\bigg|\bigg\{(x,t)\in G:\tilde{u}(x,t)\leq\nu^{l+1}\frac{L_0}{2}\bigg\}\bigg|.
\end{align*}
Define $v:Q_1\rightarrow\mathbb R$ as
\begin{align*}
v(x,t)&=\frac{\nu}{\nu-1}-\frac{\tilde{u}(r_lx+x_0,r_l^2t+t_0)}{(\nu-1)\nu^lL_0}\\
&=\frac{\nu}{\nu-1}-\frac{u(a_{k_1}r_lx+a_{k_1}x_0,a_{k_1}^2r_l^2t+a_{k_1}^2t_0)}{(\nu-1)\nu^lL_0}
\end{align*}
Now we have to show that $v$ is a supersolution. By Lemma \ref{scaling}, we will be done if
\begin{align*}
a_{k_1}r_l\leq \frac{1}{L_2\eta((\nu-1)\nu^lL_0)+L_2}.
\end{align*}
Since $a_{k_1}\leq1$, we will be done if
\begin{align*}
r_l\leq \frac{1}{L_2\eta((\nu-1)\nu^lL_0)+L_2}.
\end{align*}
Since $\eta(t)\geq 1$ for all $t$, we will be done if
\begin{align*}
2L_2\eta((\nu-1)\nu^lL_0)&\leq r_l^{-1},\\
\text{i.e.,} \hspace{2mm} 2L_2\eta((\nu-1)\nu^lL_0)&\leq {\sigma}^{-1} \nu^{(l+1)\frac{\epsilon}{n+2}}\big(\frac{L_0}{2}\big)^{\frac{\epsilon}{n+2}}
\end{align*}
Since $\eta(st)\leq {\Lambda}_0\eta(s)\eta(t)$ so we will be done if
\begin{align*}
2L_2{\Lambda}_0\eta({\nu}^l)\eta((\nu-1)L_0)&\leq {\sigma}^{-1} \nu^{(l+1)\frac{\epsilon}{n+2}}\big(\frac{L_0}{2}\big)^{\frac{\epsilon}{n+2}}\\
\text{i.e.,}\frac{2L_2{\Lambda}_0\eta((\nu-1)L_0)}{{\nu}^\frac{\epsilon}{n+2}\big(\frac{L_0}{2}\big)^{\frac{\epsilon}{n+2}}}\frac{\eta({\nu}^l)}{\nu^{l\frac{\epsilon}{n+2}}}&\leq {\sigma}^{-1}.
\end{align*}
Now from part (ii) of Lemma \ref{etaprop}, we have $\underset{t\rightarrow\infty}{\text{lim}}\frac{\eta(t)}{t^{\frac{\epsilon}{n+2}}}=0$, which implies that for all $t\geq 1$, we have $\frac{\eta(t)}{t^{\frac{\epsilon}{n+2}}} \leq C_1$ for some universal constant $C_1$.\\
Also, for $L_0 \geq 2$, we have $1<\nu<\frac{3}{2}$ and $\frac{1}{2}<(\nu-1)L_0<\frac{3}{4}$.\\
We will be done if
\begin{align*}
\frac{C_2}{\big(\frac{L_0}{2}\big)^{\frac{\epsilon}{n+2}}} \leq {\sigma}^{-1},
\end{align*}
where $C_2$ is universal constant.\\
We will be done if $L_0 \geq \text{max}\{L,2\}$ is large enough to satisfy above inequality; but $L_0$ is in our hand, so we are done. Also, We will choose $L_0$ large enough to make sure that $r_l < 1$.
Thus $v$ is a supersolution in $Q_1$.
Note that $\tilde{u} \leq \nu^{l+1}L_0$ in $Q_{r_l}$, which implies $v \geq 0.$ Also, $\tilde{u}(x_0,t_0) \geq \nu^l L_0$ implies $v(0,0) \leq 1$.\\
So, by Lemma \ref{measureuniform},
we have
\begin{align*}
|\{v>L\}\cap K_1| &\leq (1-\mu)|K_1|.
\end{align*}
Since $L_0>L$, above inequality in particular implies,
\begin{align*}
|\{v\geq L_0\}\cap K_1| &\leq (1-\mu)|K_1|,
\end{align*}
which by rescaling and the fact that, $\tilde{u} \leq \nu^{l+1}\frac{L_0}{2}$ iff $v\geq L_0$, gives
\begin{align*}
|\{\tilde{u}\leq \nu^{l+1}\frac{L_0}{2}\} \cap G| \leq (1-\mu)|G|.
\end{align*}
Now, by Lemma \ref{wh}
we have
\begin{align*}
\bigg|\bigg\{(x,t)\in G:\tilde{u}(x,t) > \nu^{l+1}\frac{L_0}{2}\bigg\}\bigg| \leq C\bigg(\nu^{l+1}\frac{L_0}{2}\bigg)^{-\epsilon}.
\end{align*}
Hence,
\begin{align*}
|G| \leq C\bigg(\nu^{l+1}\frac{L_0}{2}\bigg)^{-\epsilon} + (1-\mu)|G|,
\end{align*}
which implies
\begin{align*}
\mu |G| &\leq C\bigg(\nu^{l+1}\frac{L_0}{2}\bigg)^{-\epsilon},\\
\text{i.e.,}\hspace{2mm}\mu\sigma^{n+2}\nu^{-(l+1)\epsilon}\bigg(\frac{L_0}{2}\bigg)^{-\epsilon} c_n^{n+2} &\leq C\bigg(\nu^{l+1}\frac{L_0}{2}\bigg)^{-\epsilon}.
\end{align*}
Then, we have
\begin{align*}
\sigma^{n+2} \leq \frac{C}{\mu c_n^{n+2}},
\end{align*}
which is a contradiction to $\sigma^{n+2} > \frac{C}{\mu c_n^{n+2}}$.
This proves the lemma.
\end{proof}
\begin{lemma}\label{final}
Let $u \in C(Q_2)$ be a solution of (\ref{meq1}) and (\ref{meq2}) with $u(0,0)=1$. Then, there exists a universal constant $l_0 $ such that
\begin{align}\label{aaa}
\text{sup}_A \tilde{u}(x,t) \leq \nu^{l_0}L_0,
\end{align}
where $L_0$ is universal constant from Lemma \ref{super}, $\tilde{u}(x,t)=u(a_{k_1}x,a_{k_1}^2t)$, and\\ $A=\Big\{(x,t):|x|_{\infty} \leq \frac{c_n}{2}, -1+ \frac{c_n^2}{4} \leq t \leq -1+\frac{c_n^2}{2}\Big\}$.
\end{lemma}
\begin{center}
\begin{tikzpicture}[scale=1.35]
\draw (0,0) rectangle (4,4);
\draw (1,0) rectangle (3,2);
\draw (1.5,1) rectangle (2.5,2);
\fill[color=gray!10] (1.5,1) rectangle (2.5,2);
\draw (2,1.5) node{$A$};
\draw (2,4) node{\tiny{$\bullet$}};
\draw (2,4.25) node{$(0,0)$};
\draw (2,-0.25) node{$(0,-1)$};
\draw (2,0) node{\tiny{$\bullet$}};
\draw[<-] (1.5,0.8)--(1.85,0.8);
\draw[->] (2.1,0.8)--(2.5,0.8);
\draw (2,0.75) node{$c_n$};
\draw[<->] (2.7,1)--(2.7,2);
\draw (2.85,1.6) node{$\frac{c_n^2}{4}$};
\draw[<-] (1,0.2)--(1.75,0.2);
\draw[->] (2.2,0.2)--(3,0.2);
\draw (2,0.2) node{$2c_n$};
\draw[<-] (3.2,0)--(3.2,0.6);
\draw[->] (3.2,1.4)--(3.2,2);
\draw (3.2,1) node{$\frac{c_n^2}{2}$};
\end{tikzpicture}
\end{center}
\begin{proof}
Choose $l_0$ such that
\begin{align*}
\sum_{j=l_0}^{\infty}r_j \leq \frac{c_n}{2} \hspace{3mm}\text{and} \hspace{3mm} \sum_{j=l_0}^{\infty}r_j^2 \leq \frac{c_n^2}{8},
\end{align*}
where $r_j$ is as in Lemma \ref{super}. We can choose such an $l_0$ because $\nu >1$.\\
Assume (\ref{aaa}) is not true. Then, there exists $(x_{l_0},t_{l_0}) \in A$ such that
\begin{align*}
\tilde{u}(x_{l_0},t_{l_0}) > \nu^{l_0}L_0.
\end{align*}
By Lemma \ref{super}, there exists $(x_{l_0+1},t_{l_0+1}) \in Q_{r_{l_0}}(x_{l_0},t_{l_0})$ such that
\begin{align*}
\tilde{u}(x_{l_0+1},t_{l_0+1}) > \nu^{l_0+1}L_0,
\end{align*}
$|x_{l_0+1}-x_{l_0}|_{\infty} <r_{l_0}$ and $t_{l_0}\geq t_{l_0+1} > t_{l_0}-r_{l_0}^2$.\\
By repeating this process we get a sequence $(x_{l},t_{l})$ for $l >l_0$ such that
\begin{align}\label{12}
\tilde{u}(x_{l},t_{l}) > \nu^{l}L_0,
\end{align}
$|x_{l+1}-x_{l}|_{\infty} <r_{l}$ and $t_{l} \geq t_{l+1} > t_{l}-r_{l}^2$.\\
Note that, for $l>l_0$,
\begin{align*}
|x_{l}|_{\infty} &\leq |x_{l_0}|_{\infty} +\sum_{j=l_0}^{l-1}|x_{j+1}-x_{j}|_{\infty}\\
&\leq \frac{c_n}{2} +\sum_{j=l_0}^{\infty}r_j\\ &\leq \frac{c_n}{2} + \frac{c_n}{2}\\
&=c_n.
\end{align*}
Also,
\begin{align*}
t_{l_0} \geq t_l &\geq t_{l_0} -\sum_{j=l_0}^{l-1}r_j^2\\ &\geq t_{l_0} -\sum_{j=l_0}^{l-1}r_j^2\\ &\geq -1 + \frac{c_n^2}{4} -\frac{c_n^2}{8}\\ &\geq -1+\frac{c_n^2}{8}.
\end{align*}
Thus, $(x_{l},t_{l})$ is a bounded sequence in $K_1$. Now, by continuity of $u$ in $Q_2$, $\tilde{u}(x_{l},t_{l})$ is a bounded sequence, which is a contradiction to (\ref{12}). This proves the lemma.
\end{proof}
Now, we will prove our main theorem Theorem \ref{mthm}, which is a direct consequence of the previous lemma.
\begin{proof}[Proof of Theorem \ref{mthm}]
Let $u \in C(Q_2)$ be a solution of (\ref{eq1}) and (\ref{eq2}).
Define
\begin{align*}
v(x,t):=\frac{u(a_0x,a_0^2t)}{u(0,0)}.
\end{align*}
Then, by Lemma \ref{scaling}, $v$ is a solution of (\ref{eq1}) and (\ref{eq2}).\\
Now, define
\begin{align*}
v_1(x,t):=v(r_1x,r_1^2t).
\end{align*}
Then, $v_1$ is a solution of (\ref{meq1}) and (\ref{meq2}) with $v_1(0,0)=1$.\\
Then, by Lemma \ref{final}, we get
\begin{align*}
\text{sup}_A \tilde{v_1}(x,t) \leq C,
\end{align*}
where $C=\nu^{l_0}L_0$ is a universal constant, $\tilde{v_1}(x,t)=v_1(a_{k_1}x,a_{k_1}^2t)$, and\\ $A=\Big\{(x,t):|x|_{\infty} \leq \frac{c_n}{2}, -1+ \frac{c_n^2}{4} \leq t \leq -1+\frac{c_n^2}{2}\Big\}$.\\
Hence, we have
\begin{align*}
\text{sup}_A u(a_0 a_{k_1} r_1 x,(a_0 a_{k_1} r_1)^2t) \leq C u(0,0),
\end{align*}
i.e.,
\begin{align*}
\text{sup}_A u(a_0 \gamma x,(a_0 \gamma)^2 t) \leq C u(0,0),
\end{align*}
where $\gamma=a_{k_1} r_1$.
This proves the result.
\end{proof}
\section{Appendix}\label{a}
\begin{proof}[Proof of Lemma \ref{barrier}]
We will work with the same barrier function which is constructed in \cite{is}. In our case, we have to take care of nonlinearity as well. So, here we will do calculations more carefully. Our idea of scheme is to make $P^+(D^2h)-h_t$ strictly negative instead of nonpositive as in \cite{is}. With the help of this negative quantity and favourable scaling we will take care of nonlinearity.\\ First we will define $h$ in the region
\begin{align*}
D= \bigg\{(x,t):|x|<1 \hspace{2mm}\&\hspace{2mm} \frac{c_n^2}{36n}\leq t \leq \frac{1}{36n} \bigg\}
\end{align*}
as
\begin{align*}
h(x,t)=t^{-p}H\Big(\frac{x}{\sqrt{t}}\Big),
\end{align*}
where $H:\mathbb R^n\rightarrow\mathbb R$ is defined as
\[H(y)=\begin{cases}
\text{some smooth and bounded function between $-((6\sqrt{n})^q(2^q-1)^{-1})$ and $-1$} &\quad \text{if}\hspace{1.5mm} |y|\leq3\sqrt{n},\\
((6\sqrt{n})^{-q}-|y|^{-q}) &\quad \text{if}\hspace{1.5mm}3\sqrt{n}\leq|y|\leq6\sqrt{n},\\
0 &\quad \text{if}\hspace{1.5mm}|y|\geq6\sqrt{n}.
\end{cases}
\]
To calculate $P^+(D^2h)-h_t$, we will do some calculations:
\begin{align*}
h_i&=t^{-p}H_i\bigg(\frac{x}{\sqrt{t}}\bigg);
\hspace{5mm} h_{ij}=t^{-p}H_{ij}\bigg(\frac{x}{\sqrt{t}}\bigg)\frac{1}{t};\\
h_t&=-pt^{-p-1}H\bigg(\frac{x}{\sqrt{t}}\bigg)+t^{-p}\nabla H\bigg(\frac{x}{\sqrt{t}}\bigg)\cdot\frac{x}{t^{-3/2}}\bigg(-\frac{1}{2}\bigg)\\
&=-p t^{-p-1}H\bigg(\frac{x}{\sqrt{t}}\bigg)-\frac{1}{2}t^{-p-1}\nabla H\bigg(\frac{x}{\sqrt{t}}\bigg)\cdot\frac{x}{\sqrt{t}}.
\end{align*}
Here, $h_i$ represents the partial derivative of $h$ with respect to $x_i$.\\
Thus, we have
\begin{align}\label{H}
P^+(D^2h)-h_t=t^{-p-1}P^+\bigg(D^2H\bigg(\frac{x}{\sqrt{t}}\bigg)\bigg)+p t^{-p-1}H\bigg(\frac{x}{\sqrt{t}}\bigg)+\frac{1}{2}t^{-p-1}\nabla H\bigg(\frac{x}{\sqrt{t}}\bigg)\cdot\frac{x}{\sqrt{t}}.
\end{align}
We will calculate the derivative of $H$ in the region $ \{y\in \mathbb R^n:3\sqrt{n} \leq |y| \leq 6\sqrt{n}\}$ as
\begin{align*}
H_{i}(y)&=q|y|^{-q-1}\frac{y_i}{|y|}=q|y|^{-q-2}{y_i},\\
H_{ij}(y)&=-q(q+2)|y|^{-q-4}{y_i}{y_j}+q|y|^{-q-2} \delta_{ij}.
\end{align*}
Hence, in the region $ \{y \in \mathbb R^n:3\sqrt{n} \leq |y| \leq 6\sqrt{n}\}$, eigenvalues of $D^2H(y)$ are $q|y|^{-q-2}$ with multiplicity $n-1$ and $-q(q+1)|y|^{-q-2}$ with multiplicity 1 so,
\begin{align*}
P^+(D^2H(y))&=q(\Lambda(n-1)-\lambda(q+1))|y|^{-q-2}.
\end{align*}
Then, we have
\begin{align}\label{star}
P^+(D^2h)=q(\Lambda(n-1)-\lambda(q+1))t^{-p-1}\bigg(\frac{|x|}{\sqrt{t}}\bigg)^{-q-2}.
\end{align}
Note that
\begin{align*}
\nabla H\bigg(\frac{x}{\sqrt{t}}\bigg)\cdot\frac{x}{\sqrt{t}}&=q \bigg(\frac{|x|}{\sqrt{t}}\bigg)^{-q}.
\end{align*}
Thus, in the region $\Big\{(x,t)\in D:3\sqrt{n}\leq\frac{|x|}{\sqrt{t}}\leq 6\sqrt{n}\Big\}$, we have
\begin{align*}
P^+(D^2h)-h_t\leq q(\Lambda(n-1)-\lambda(q+1))t^{-p-1}\bigg(\frac{|x|}{\sqrt{t}}\bigg)^{-q-2}+p t^{-p-1}H\bigg(\frac{|x|}{\sqrt{t}}\bigg)+\frac{q}{2}t^{-p-1} \bigg(\frac{|x|}{\sqrt{t}}\bigg)^{-q}.
\end{align*}
By definition $H(y)\leq 0$ for $y \in \mathbb R^n$. So, we have
\begin{align*}
P^+(D^2h)-h_t\leq q(\Lambda(n-1)-\lambda(q+1))t^{-p-1}\bigg(\frac{|x|}{\sqrt{t}}\bigg)^{-q-2}+\frac{q}{2}t^{-p-1} \bigg(\frac{|x|}{\sqrt{t}}\bigg)^{-q}.
\end{align*}
Using $\frac{|x|}{\sqrt{t}} \le 6\sqrt{n}$, we get
\begin{align*}
P^+(D^2h)-h_t\leq q(\Lambda(n-1)-\lambda(q+1) +18n)t^{-p-1}\bigg(\frac{|x|}{\sqrt{t}}\bigg)^{-q-2}.
\end{align*}
Choose $q$ large enough such that $\Lambda(n-1)-\lambda(q+1) +18n \leq -1$. Then, we have
\begin{align}\label{112}
P^+(D^2h)-h_t\leq -qt^{-p-1}\bigg(\frac{|x|}{\sqrt{t}}\bigg)^{-q-2}
\end{align}
in $\Big\{(x,t)\in D:3\sqrt{n}\leq\frac{|x|}{\sqrt{t}}\leq 6\sqrt{n}\Big\}$.\\
\\
Now we will estimate $ P^+(D^2h)-h_t$ in the region $\Big\{(x,t)\in D:\frac{|x|}{\sqrt{t}}\leq 3\sqrt{n}\Big\}$.\\
Note that $H(y)$ is smooth in $|y|\leq3\sqrt{n}$, so there exists $M$ such that
\begin{align*}
P^+\Big(D^2H\Big(\frac{x}{\sqrt{t}}\Big)\Big)+\frac{1}{2}\nabla H\Big(\frac{x}{\sqrt{t}}\Big)\cdot\frac{x}{\sqrt{t}} \leq M.
\end{align*}
Choose large $p$ such that $-p+M\leq-1$. Then, from (\ref{H}) and $H(y) \le -1$ in $|y| \le 3 \sqrt{n}$, we have
\begin{align*}
P^+(D^2h)-h_t\leq -t^{-p-1} \hspace{2mm} \text{in} \hspace{2mm} \Big\{(x,t)\in D:\frac{|x|}{\sqrt{t}}\leq 3\sqrt{n}\Big\}.
\end{align*}
Thus, in $D \cap \{(x,t):h(x,t)\not=0\}$, we get
\begin{align}\label{113}
P^+(D^2h)-h_t\leq -t^{-p-1}.
\end{align}
Now we will extend the definition of $h$ in $\{(x,t):|x|\leq 1,1/(36n) \leq t \leq 1\} $ as
\begin{align*}
h(x,t)=e^{(\kappa (t-{1/{36n}}))}h(x,1/36n),
\end{align*}
where $\kappa$ is chosen such that
$\underset{\{|x|<1\}}{\text{inf}}\frac{P^+(D^2h(x,1/36n))}{h(x,1/36n)}\geq \kappa+1$.
(We can choose $\kappa <0$, i.e., inf exists because
\begin{itemize}
\item For $|x| \leq \frac{1}{2}$, $h(x,1/36n) \leq -(36n)^p$ and $h$ is smooth so $P^+(D^2h(x,1/36n))$ is bounded. Hence, $\frac{P^+(D^2h(x,1/36n))}{h(x,1/36n)}$ is bounded below.
\item For $1>|x| \geq \frac{1}{2}$, from (\ref{star}), we have $P^+(D^2h(x,1/36n)) < 0$ and by definition $h < 0$ so $\frac{P^+(D^2h(x,1/36n))}{h(x,1/36n)}$ is bounded below by $0$.)
\end{itemize}
Now, in $\{(x,t):|x|\leq 1,1/(36n) \leq t \leq 1\} $,
\begin{align*}
P^+(D^2h)-h_t=e^{(\kappa (t-{1/{36n}}))}(P^+(D^2h)-\kappa h(x,1/36n)).
\end{align*}
Take $m >0$ such that $-m=\underset{\{{1/2}<|x|<1\}}{\text{min}}P^+(D^2h(x,1/36n))$. Note that $h(x,1/36n)=0$ for $|x|=1$. Hence, by continuity of $h$, we can choose $\delta >0$ such that \[-m-\kappa h(x,1/36n) \leq \frac{-m}{2}\] in $\{1-\delta <|x| \le 1\}$,
which gives us
\begin{align}
P^+(D^2h)-h_t\leq \frac{-m}{2}e^{(\kappa (t-{1/{36n}}))} \hspace{2mm} \text{in} \hspace{2mm} \{(x,t):1-\delta < |x| \le 1,1/(36n) \leq t \leq 1\}.
\end{align}
Now, using the definition of $\kappa$, we have
\begin{align*}
P^+(D^2h)-h_t&=e^{(\kappa (t-{1/{36n}}))}(P^+(D^2h)-\kappa h(x,1/36n))\\
&\leq e^{\kappa (t-{1/{36n}})} h(x,1/36n).
\end{align*}
Thus, in $\{(x,t):1-\delta < |x| \le 1,1/(36n) \leq t \leq 1\}$, we have
\begin{align*}
P^+(D^2h)-h_t \leq e^{\kappa (t-{1/{36n}})} \underset{\{|x| \leq 1-\delta\}}{\text{max}}h(x,1/36n).
\end{align*}
Thus, in $\{(x,t):|x|\leq 1,1/(36n) \leq t \leq 1\} $, we have
\begin{align}\label{114}
P^+(D^2h)-h_t &\leq \text{max}\{-m/2,\underset{\{|x| \leq 1-\delta\}}{\text{max}}h(x,1/36n)\}e^{\kappa (t-{1/{36n}})}\\
&\leq \text{max}\{-m/2,\underset{\{|x| \leq 1-\delta\}}{\text{max}}h(x,1/36n)\}\label{114}.
\end{align}
Hence, from (\ref{112}),(\ref{113}) and (\ref{114}) we get
\begin{align*}
P^+(D^2h)-h_t&\leq \text{max}\{-m/2,\underset{\{|x| \leq 1-\delta\}}{\text{max}}h(x,1/36n),-(36n)^{p+1},-q(6\sqrt{n})^{q+2}c_n^{-2p-2}(36\sqrt{n})^{p+1}\}\\
&=:-c.
\end{align*}
Note that $h(x,1/36n) <0$ in $|x| \leq 1-\delta$, implies $\underset{\{|x| \leq 1-\delta\}}{\text{max}}h(x,1/36n) <0$. Hence, $c>0$ depends only on dimension and ellipticity constants.
We will define the following subsets of $Q_1(0,1).$
\begin{align*}
\hat{K_1}&:=(-c_n,c_n)^n \times (0,c_n^2),\\
\hat{K_3}&:=(-3c_n,3c_n)^n \times (c_n^2,1),
\end{align*}
where $c_n=(10n)^{-1}.$
Note that the way we constructed $h$, it vanishes only in $\{(x,t):|x| \geq 6\sqrt{nt},t<(36n)^{-1}\}$, and $\hat{K}_3$ does not intersects with it, so
\begin{align}\label{al}
\alpha:=\text{sup}_{\hat{K}_3}h(x,t)<0.
\end{align}
Now we will define function $\tilde{h}:Q_1(0,1) \rightarrow \mathbb R$ as
\[\tilde{h}(x,t)=\begin{cases}
\frac{-2h(x,t)}{\text{sup}_{\hat{K}_3}h} &\quad \text{whenever $h$ is defined,}\\
0 &\quad \text{otherwise}.
\end{cases}
\]
Take a smooth function $\bar{h}$ defined in $\hat{K}_1$ such that $\bar{h} = \tilde{h}$ on boundary of $\hat{K}_1$ (set theoretic boundary).
Define $h:Q_1 \rightarrow \mathbb R$ (this $h$ is not same as the function we were defining earlier) as
\[h(x,t)=\begin{cases}
\tilde{h}(x,t+1) &\quad \text{if} (x,t)\in Q_1\setminus {K}_1\\
\bar{h}(x,t+1) &\quad \text{if} (x,t)\in {K}_1.
\end{cases}
\]
Note that $h$ is a Lipschitz function so $\phi(|D h|)$ is bounded, say by $M_1$.\\
Now, choose $r_0>0$ such that
\begin{align}
\frac{2c}{\alpha}+\Lambda_0 r_0 \eta \bigg(\frac{1}{r_0}\bigg) M_1 \leq 0,
\end{align}
where $\alpha$ is define in (\ref{al}). This will imply for $r \leq r_0$ that,
\begin{align}\label{121}
\frac{2c}{\alpha}+\Lambda_0 r \eta \bigg(\frac{1}{r}\bigg) M_1 \leq 0.
\end{align}
Then, we have (in viscosity sense)
\begin{align*}
P^+(D^2h)-h_t+\tilde{\phi}(|D h|)\leq g,
\end{align*}
and support of $g$ lies in ${K}_1$.
Hence, $h$ is required barrier function.
\end{proof}
\noindent
\begin{proof}
[Proof of Lemma \ref{measureuniform}]
By approximating with inf-convolution
\begin{align*}
u_{\epsilon}(x,t)=\underset{(y,s) \in Q_2}{\text{inf}}\bigg\{u(y,s)+\frac{1}{\epsilon}(|x-y|^2+|t-s|^2)\bigg\}
\end{align*}
We may assume $u$ is semiconcave in $x$ variable and Lipschitz in $x$ and $t$ variable.\\
Define $v(x,t)=u(x,t)+h(x,t)$ in $Q_1$, where $h$ is the barrier function constructed in Lemma \ref{barrier}.
We will choose $r_1$ later such that $r_1 \leq r_0$.\\
Now, using properties of Pucci operators, increasing nature of $\phi$, $r_1 \le r_0$, and properties of barrier function, we obtain $v$ is a viscosity supersolution of
\begin{align}\label{100}
P^-(D^2v)-v_t \leq \tilde{\phi}(|Dv|+|Dh|)-\tilde{\phi}(|Dh|)+g.
\end{align}
Now, extend $v$ by zero in $Q_2$, and denote the convex envelope of $\text{min}\{v,0\}$ by $\Gamma_v$.\\
Using (\ref{100}), we get (in viscosity sense)
\begin{align*}
P^-(D^2v)-v_t \leq M
\end{align*}
for some $M$.
Then, $\Gamma_v$ is $C^{1,1}$ with respect to $x$ and Lipschitz with respect to $t$ in $\{v=\Gamma_v\}$ ( see Corollary $3.17$ in \cite{W1}).
Since $v$ is a viscosity supersolution of equation (\ref{100}) and $\Gamma_v \leq v$ we have
\begin{align}\label{eqgammav}
P^-(D^2(\Gamma_v))-(\Gamma_v)_t \leq \tilde{\phi}(|D\Gamma_v|+|Dh|)- \tilde{\phi}(|Dh|) +g \hspace{2mm} \text{a.e} \hspace{2mm} \text{in} \hspace{2mm} \{v=\Gamma_v\}.
\end{align}
Define $E:=\{v=\Gamma_v\} \cap \{(x,t) \in Q_1: |D\Gamma_v(x,t)| \leq 1\}$.
\textbf{Claim:} There exists a constant $b,$ depending on $\phi$ and $n$ such that for a.e. in $E$ the following holds:
\begin{align}\label{eqfinal}
P^-(D^2(\Gamma_v))-(\Gamma_v)_t \leq b |D\Gamma_v|+ \tilde{\phi}(2) +g.
\end{align}
Proof of claim: We will prove this in two cases depending on the value of $|Dh|$ on that point.\\
Case 1: We assume $(x,t) \in E$ such that $|Dh|(x,t)\leq 1$. In this case, we can bound $\tilde{\phi}(|D\Gamma_v|+|Dh|)- \tilde{\phi}(|Dh|)$ by $\tilde{\phi}(2)$ because $\phi\geq0$ and we have $|D\Gamma_v|\leq 1$ in $E$.\\
Case 2: We assume $(x,t) \in E$ such that $|Dh(x,t)|\geq 1$.
Since $h$ is Lipschitz, $\text{sup}_{Q_1}$$|Dh|$ exists and it depends only on $n$ owing to the way we constructed $h$. Since $\phi$ is locally Lipschitz, we have
\begin{align}
\phi(|D\Gamma_v|+|Dh|)- \phi(|Dh|) \leq b |D\Gamma_v|,
\end{align}
Where
\begin{align}\label{defb}
b= \text{max}\{|D\phi(y)|:1 \leq y \leq \text{sup}_{Q_1}|Dh|+1\}.
\end{align}
Then, we have
\begin{align}
\tilde{\phi}(|D\Gamma_v|+|Dh|)- \tilde{\phi}(|Dh|) \leq b |D\Gamma_v|.
\end{align}
From both the cases and (\ref{eqgammav}) we get our claim.\\
Recall for given $f$, $Gf(x,t)=(Df(x,t), f(x,t)-x\cdot Df(x,t))$.
Since $\Gamma_v$ is $C^{1,1}$ with respect to $x$ and Lipschitz with respect to $t$ in $\{v=\Gamma_v\}$, by Lemma \ref{glem}, $G{\Gamma}_v$ is Lipschitz continuous in $E$. Now,
Coarea formula and Lemma \ref{glem} gives
\begin{align*}
\int_{G\Gamma_v(E)}\frac{d\xi dh}{|\xi|^{n+1} + \delta}
&\leq \int_{E}\frac{|\text{det}(G\Gamma_v)|}{|D\Gamma_v|^{n+1} + \delta}dxdt\\
&=\int_{E}\frac{|\partial_t\Gamma_v|\hspace{1mm} |\text{det}D^2(\Gamma_v)|}{|D\Gamma_v|^{n+1} + \delta}dxdt,
\end{align*}
where $\delta >0$ is a small universal constant which will be chosen later.
Now, by definition of Monotone envelope (\ref{medef}), we have $\Gamma_v$ is convex with respect to $x$ variable so $\text{det}D^2(\Gamma_v)\geq0$. Also, $\Gamma_v$ is non-increasing with respect to $t$ so $\partial_t\Gamma_v\leq0$. Hence, we get
\begin{align*}
\int_{G\Gamma_v(E)}\frac{d\xi dh}{|\xi|^{n+1} + \delta}
&\leq \int_{E}\frac{-\partial_t\Gamma_v\hspace{1mm} \text{det}D^2(\Gamma_v)}{|D\Gamma_v|^{n+1} + \delta}dxdt\\
&=\frac{\lambda^n}{\lambda^n} \int_{E}\frac{-\partial_t\Gamma_v\hspace{1mm} \text{det}D^2(\Gamma_v)}{|D\Gamma_v|^{n+1} + \delta}dxdt\\
&\leq\frac{\lambda^k\Lambda^{n-k}}{\lambda^n} \int_{E}\frac{-\partial_t\Gamma_v\hspace{1mm} \text{det}D^2(\Gamma_v)}{|D\Gamma_v|^{n+1} + \delta}dxdt,
\end{align*}
where $k$ is number of positive eigenvalues of $D^2(\Gamma_v)$.\\
Now, write determinant as the product of eigenvalues, club positive eigenvalues with $\lambda$ and negative eigenvalues with $\Lambda$, and use AM-GM inequality to get
\begin{align*}
\int_{G\Gamma_v(E)}\frac{d\xi dh}{|\xi|^{n+1} + \delta}
&\leq\frac{n+1}{\lambda^n} \int_{E}\frac{(-\partial_t\Gamma_v\hspace{1mm} + P^-(D^2\Gamma_v))^{n+1}}{|D\Gamma_v|^{n+1} + \delta}dxdt.
\end{align*}
Now, using (\ref{eqfinal}), we get
\begin{align*}
\int_{G\Gamma_v(E)}\frac{d\xi dh}{|\xi|^{n+1} + \delta}
&\leq\frac{n+1}{\lambda^n} \int_{E}\frac{( b |D\Gamma_v|+ \tilde{\phi}(2) +g)^{n+1}}{|D\Gamma_v|^{n+1} + \delta}dxdt\\
&\leq C b^{n+1} + C \frac{(\tilde{\phi}(2))^{n+1}}{\delta} + C\frac{|K_1\cap E|}{\delta},
\end{align*}
where $C$ depends on $n$ and ellipticity constant.\\
Now, $\text{inf}_{K_3}u\leq 1$ and $h \le -2$ in $K_3$ imply
\begin{align*}
\text{inf}_{K_3} v \leq -1,
\end{align*}
which is the same as saying
\begin{align*}
\text{sup}_{K_3} (\text{min}\{v,0\})^-\geq 1\\
\implies \text{sup}_{Q_1} (\text{min}\{v,0\})^-\geq 1.
\end{align*}
Also, $u \geq 0$ and $h=0$ on $\partial_p Q_1$ imply $v \geq 0$ on $\partial_p Q_1$.
Hence, by Lemma \ref{setmeas}, we get
\begin{align*}
\Big\{(\xi,h) \in \mathbb R^n \times \mathbb R:|\xi| \leq \frac{1}{4}, \frac{5}{8} \leq -h \leq \frac{6}{8}\Big\} \subset {G\Gamma_v(E)}.
\end{align*}
Thus, we have
\begin{align}\label{eqcon}
\frac{1}{8} \int_{|\xi|<1/4}\frac{d\xi dh}{|\xi|^{n+1} + \delta}
\leq C b^{n+1} + C \frac{(\tilde{\phi}(2))^{n+1}}{\delta} + C\frac{|K_1\cap E|}{\delta}.
\end{align}
Since $b$ and $C$ are universal constants, and
\begin{equation*}
\int_{|\xi|<1/4}\frac{d\xi}{|\xi|^{n+1} + \delta} \rightarrow\infty \hspace{2mm}\text{as} \hspace{2mm} \delta \rightarrow 0,
\end{equation*}
we can choose a universal constant $\delta >0$ such that
\begin{align}
\frac{1}{8} \int_{|\xi|<1/4}\frac{d\xi dh}{|\xi|^{n+1} + \delta}
\geq C b^{n+1} +2.
\end{align}
Chosen $\delta>0$, choose $r_1 \leq r_0$ universal such that
\begin{align*}
C(r_1\eta(1/r_1)\phi(2))^{n+1}\leq \delta, \hspace{2mm}\text{which is the same as}\hspace{2mm} C\frac{(\tilde{\phi}(2))^{n+1}}{\delta}\leq1.
\end{align*}
Hence (\ref{eqcon}) becomes
\begin{align*}
1\leq C\frac{|K_1\cap E|}{\delta}.
\end{align*}
Take $\mu_1=\frac{\delta}{C}$, which is a universal constant. By definition of $E$, we have
\begin{align*}
\mu_1 &\leq |K_1\cap E|
\leq |K_1 \cap \{v=\Gamma_v\}|.
\end{align*}
Using $\Gamma_v$ is monotone envelope of $\text{min}\{v,0\}$ we get,
\begin{align*}
\mu_1
&\leq |K_1 \cap \{v\leq 0\}|.
\end{align*}
Substitute $v=u+h$ to get
\begin{align*}
|K_1 \cap \{u\leq -h\}| \geq \mu_1.
\end{align*}
Take $L=\text{max}_{Q_1}(-h)$. Then, we have
\begin{align*}
|K_1 \cap \{u\leq L\}| \geq \mu_1,
\end{align*}
which implies,
\begin{align*}
|\{(x,t) \in K_1 : u(x,t) \leq L\}| \geq |K_1|-\mu_1.
\end{align*}
Take $\mu=\frac{|K_1|-\mu_1}{|K_1|}$ and get conclusion.
\end{proof}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,844 |
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\begin{document}
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\title[Optimal Novikov-type criteria]{Optimal Novikov-type criteria
for local martingales with jumps}
\author{Alexander Sokol}
\address{Alexander Sokol: Institute of Mathematics, University of
Copenhagen, 2100 Copenhagen, Denmark}
\email{alexander@math.ku.dk}
\urladdr{http://www.math.ku.dk/$\sim$alexander}
\subjclass[2000] {Primary 60G44; Secondary 60G40}
\keywords{Martingale, Exponential martingale, Uniform integrability,
Novikov, Optimal, Poisson process}
\begin{abstract}
We consider local martingales $M$ with jumps larger than $a$ for some
$a$ larger than or equal to $-1$, and prove Novikov-type criteria for
the corresponding exponential local martingale
to be a uniformly integrable martingale. We obtain criteria using both
the quadratic variation and the predictable quadratic variation. We
prove optimality of the coefficients in the criteria. As a corollary,
we obtain a verbatim extension of the classical Novikov criterion for
continuous local martingales to the case of local martingales with nonnegative jumps.
\end{abstract}
\maketitle
\noindent
\section{Introduction}
\label{sec:intro}
The motivation of this paper is the question of when an
exponential local martingale is a uniformly integrable
martingale. Before introducing this problem, we fix our notation and
recall some results from stochastic analysis.
Assume given a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},P)$
satisfying the usual conditions, see \cite{PP} for the definition
of this and other probabilistic concepts such as being a local martingale, locally
integrable, locally square-integrable, and for the quadratic variation
and quadratic covariation et cetera. For any local martingale
$M$, we say that $M$ has initial value zero if $M_0=0$. For any local
martingale $M$ with initial value zero, we denote by $[M]$ the
quadratic variation of $M$, that is, the unique increasing adapted
process with initial value zero such that $M^2-[M]$ is a local martingale. If $M$ furthermore is locally square
integrable, we denote by $\langle M\rangle$ the predictable quadratic
variation of $M$, which is the unique increasing predictable process
with initial value zero such that $[M]-\langle M\rangle$ is a local
martingale.
For any local martingale with initial value zero,
there exists by Theorem 7.25 of \cite{HWY} a unique decomposition
$M=M^c+M^d$, where $M^c$ is a continuous local martingale and $M^d$ is
a purely discontinuous local martingale, both with initial value
zero. Here, we say that a local martingale with initial value zero is
purely discontinuous if it has zero quadratic covariation with
any continuous local martingale with initial value zero. We refer to
$M^c$ as the continuous martingale part of $M$, and refer to $M^d$ as
the purely discontinuous martingale part of $M$.
Let $M$ be a local martingale with
initial value zero and $\Delta M\ge-1$. The exponential martingale of $M$, also known as the
Dol{\'e}ans-Dade exponential of $M$, is the unique c\`{a}dl\`{a}g
solution in $Z$ to the stochastic differential equation $Z_t = 1 +
\int_0^t Z_{s-}\df{M}_s$, given explicitly as
\begin{align}
\mathcal{E}(M)_t &= \exp\left(M_t-\frac{1}{2}[M^c]_t\right)\prod_{0<s\le t}(1+\Delta M_s)\exp(-\Delta M_s),
\end{align}
see Theorem II.37 of \cite{PP}. Applying Theorem 9.2 of
\cite{HWY}, we find that $Z$ always is a local martingale with initial
value one. Also, $\mathcal{E}(M)$ is always nonnegative. We wish to
understand when $\mathcal{E}(M)$ is a uniformly integrable martingale.
The question of when $\mathcal{E}(M)$ is a uniformly integrable
martingale has been considered many times in the litterature, and is
not only of theoretical interest, but has several applications in connection
with other topics. In particular, exponential martingales are of use
in mathematical finance, where checking uniform integrability of
a particular exponential martingale can be used to prove absence of
arbitrage and obtain equivalent martingale measures for option
pricing. For more on this, see \cite{PS} or chapters 10
and 11 of \cite{TB}. Also, exponential martingales arise naturally in
connection with maximum likelihood estimation for stochastic
processes, where the likelihood viewed as a stochastic process often
is an exponential martingale which is a true martingale, see for example the likelihood for
parameter estimation for Poisson processes given in (3.43)
of \cite{KA} or the likelihood for parameter estimation for diffusion
processes given in Theorem 1.12 of \cite{YK}. Finally, exponential martingales which are true martingales can be used in the
explicit construction of various probabilistic objects, for example
solutions to stochastic differential equations, as in Section 5.3.B of
\cite{KS2}.
Several sufficient criteria for $\mathcal{E}(M)$ to be a uniformly integrable
martingale are known. First results in this regard were obtained by \cite{AAN}
for the case of continuous local martingales. Here, we are interested in
the case where the local martingale $M$ is not necessarily
continuous. Sufficient criteria for $\mathcal{E}(M)$ to be a uniformly
integrable martingale in this case have been obtained by
\cite{LM}, \cite{ISS}, \cite{TO}, \cite{JAY} and \cite{KS}.
We now explain the particular result to be obtained in this paper. In
\cite{AAN}, the following result was obtained: If $M$ is a continuous
local martingale with initial value zero and $\exp(\frac{1}{2}[M]_\infty)$ is integrable, then
$\mathcal{E}(M)$ is a uniformly integrable martingale. This criterion is
known as Novikov's criterion. We wish to understand whether
this result can be extended to local martingales which are not
continuous.
In the case with jumps, another process in addition to the
quadratic variation process is relevant: the predictable quadratic
variation. As noted earlier, the predictable quadratic variation is defined for any
locally square-integrable local martingale $M$ with initial value zero, is denoted $\langle
M\rangle$, and is the unique predictable, increasing and locally
integrable process with initial value zero such that $[M]-\langle M\rangle$ is a local
martingale, see p. 124 of \cite{PP}. For a continuous local
martingale $M$ with initial value zero, we have that $M$ always is locally square integrable
and $\langle M\rangle =[M]$.
Using the predictable quadratic variation, the following result is
demonstrated in Theorem 9 of \cite{PS}. Let
$M$ be a locally square integrable local martingale with initial value
zero and $\Delta M\ge-1$. It then holds that if
$\exp(\frac{1}{2}\langle M^c\rangle_\infty+\langle M^d\rangle_\infty)$ is
integrable, then $\mathcal{E}(M)$ is a uniformly integrable martingale. This is an extension of the classical Novikov criterion
of \cite{AAN} to the case with jumps. \cite{PS} also argue in Example
10 that the constants in front of $\langle M^c\rangle$ and
$\langle M^d\rangle$ are optimal, although their argument contains a
flaw, namely that the formula (28) in that paper does not hold.
In this paper, we specialize our efforts to the case where
$M$ has jumps larger than or equal to $a$ for some $a\ge-1$ and prove
results of the same type, requiring either that $M$ is a locally square
integrable local martingale and that $\exp(\frac{1}{2}\langle
M^c\rangle+\alpha(a)\langle M^d\rangle)$ is integrable for some
$\alpha(a)$, or that $\exp(\frac{1}{2}[M^c]+\beta(a)[M^d])$ is
integrable for some $\beta(a)$. For all $a\ge-1$, we identify the optimal
value of $\alpha(a)$ and $\beta(a)$, in particular giving an argument circumventing
the problems of Example 10 in \cite{PS}. Our results are stated as
Theorem \ref{theorem:OptimalNovikov} and Theorem
\ref{theorem:OptimalNovikov2}. In particular, we obtain that for local
martingales $M$ with initial value zero and $\Delta M\ge0$, $\mathcal{E}(M)$ is a uniformly
integrable martingale if $\exp(\frac{1}{2}[M]_\infty)$ is integrable
or if $M$ is locally square integrable and $\exp(\frac{1}{2}\langle
M\rangle_\infty)$ is integrable, and we obtain that both the constants
in the exponents and the requirement on the jumps of $M$ are
optimal. This result is stated as Corollary
\ref{coro:NovikovExtension} and yields a verbatim extension of the
Novikov criterion to local martingales $M$ with initial value zero and
$\Delta M\ge0$.
\section{Main results and proofs}
\label{sec:main}
In this section, we apply the results of \cite{LM} to obtain optimal
constants in Novikov-type criteria for local martingales with jumps. For $a>-1$ with $a\neq0$, we define
\begin{align}
\alpha(a)&=\frac{(1+a)\log(1+a)-a}{a^2}\quad\textrm{ and }\\
\beta(a)&=\frac{(1+a)\log (1+a)-a}{a^2(1+a)},
\end{align}
and put $\alpha(0)=\beta(0)=\frac{1}{2}$ and $\alpha(-1)=1$. The
functions $\alpha$ and $\beta$ will yield the optimal constants in the
criteria we will be demonstrating. Before proving our main results, Theorem
\ref{theorem:OptimalNovikov} and Theorem
\ref{theorem:OptimalNovikov2}, we state three lemmas.
\begin{lemma}
\label{lemma:alphaFun}
The functions $\alpha$ and $\beta$ are continuous, positive and
strictly decreasing. Furthermore, $\beta(a)$ tends to infinity as $a$ tends to
minus one.
\end{lemma}
\begin{proof}
We first prove the result on $\alpha$.
Define $h(a)=(1+a)\log(1+a)-a$ for $a>-1$ and $h(-1)=1$. Note that $h$ is differentiable with
$h'(a)=\log(1+a)$. By the l'H{\^o}pital
rule, we have
\begin{align}
\lim_{a\to -1}h(a)
&=1+\lim_{a\to-1}\frac{\log(1+a)}{(1+a)^{-1}}
=1-\lim_{a\to-1}\frac{(1+a)^{-1}}{(1+a)^{-2}}=1,
\end{align}
which yields that $h$ and $\alpha$ are continuous at $-1$. Similarly,
\begin{align}
\lim_{a\to0}\alpha(a)
&= \lim_{a\to0}\frac{\log(1+a)}{2a}
= \lim_{a\to0}\frac{1}{2(1+a)}=\frac{1}{2},
\end{align}
so $\alpha$ is continuous at $0$. As $h$ is zero at zero, $h(a)$ is positive for
$a\neq0$, from which is follows that $\alpha$ is positive. It remains
to show that $\alpha$ is strictly decreasing. For $a\ge-1$ with
$a\notin\{-1,0\}$, we have that $\alpha$ is differentiable with
\begin{align}
\alpha'(a)
&=\frac{a^2\log(1+a)-2((1+a)\log(1+a)-a)a}{a^4}\notag\\
&=\frac{2a^2-a(2+a)\log(1+a)}{a^4}.
\end{align}
By the l'H{\^o}pital rule, we obtain
\begin{align}
\lim_{a\to0}\alpha'(a)
&=\lim_{a\to0}\frac{4a-2(1+a)\log(1+a)-a(2+a)(1+a)^{-1}}{4a^3}\notag\\
&=\lim_{a\to0}\frac{a(2+a)(1+a)^{-2}-2\log(1+a)}{12a^2}\notag\\
&=-\lim_{a\to0}\frac{2a(2+a)(1+a)^{-3}}{24a}
=-\frac{1}{12}\lim_{a\to0}\frac{2+a}{(1+a)^3}
=-\frac{1}{6},
\end{align}
so defining $\alpha'(0)=-\frac{1}{6}$, we obtain that $\alpha'$ is
a continuous mapping on $(-1,\infty)$, and as $\alpha'$ is the
derivative of $\alpha$ for $a\ge-1$ with $a\notin\{-1,0\}$, $\alpha'$
is also the derivative of $\alpha$ for $(1,\infty)$. In order to show
that $\alpha$ is strictly decreasing, it then suffices to show that
that $2a^2-a(2+a)\log(1+a)$ is negative for $a>-1$ with $a\neq0$. Now,
for $a\neq0$, note that
\begin{align}
\frac{\df{}}{\df{a}}(2a-(2+a)\log(1+a))
&=2-\log(1+a)-\frac{2+a}{1+a}\quad\textrm{ and }\\
\frac{\df{}^2}{\df{a}^2}(2a-(2+a)\log(1+a)) &=\frac{1}{(1+a)^2}-\frac{1}{1+a}
=-\frac{a}{(1+a)^2}.
\end{align}
From this, we conclude that $a\mapsto 2a-(2+a)\log(1+a)$
is positive for $-1<a<0$ and negative for $a>0$. Therefore, $a\mapsto 2a^2-a(2+a)\log(1+a)$ is negative for $a>-1$ with
$a\neq0$. As a consequence, $\alpha$ is strictly decreasing. As
$\beta(a)=\alpha(a)/(1+a)$, the results on $\beta$ follow from those
on $\alpha$.
\end{proof}
\begin{lemma}
\label{lemma:PoissonCompMart}
Let $N$ be a standard Poisson process, let $b$ and $\lambda$ be in $\mathbb{R}$, and define $f_b(\lambda)=\exp(-\lambda)+\lambda(1+b)-1$. With
$L^b_t=\exp(-\lambda(N_t-(1+b)t)-tf_b(\lambda))$, $L^b$ is a nonnegative
martingale with respect to the filtration induced by $N$.
\end{lemma}
\begin{proof}
Let $\mathcal{G}_t=\sigma(N_s)_{s\le t}$. Fix $0\le s\le t$. As $N_t-N_s$ is
independent of $\mathcal{G}_s$ and follows a Poisson distribution with
parameter $t-s$, we obtain
\begin{align}
E(\exp(-\lambda(N_t-N_s))|\mathcal{G}_s)
&=\exp((t-s)(\exp(-\lambda)-1)),
\end{align}
which implies
\begin{align}
E(L^b_t|\mathcal{G}_s)
&=E(\exp(-\lambda(N_t-N_s))|\mathcal{G}_s)
\exp(-\lambda N_s)\exp(\lambda(1+b)t-tf_b(\lambda))\notag\\
&=\exp((t-s)(\exp(-\lambda)-1))\exp(-\lambda N_s)\exp(\lambda(1+b)t-tf_b(\lambda))\notag\\
&=\exp(-\lambda(N_s-(1+b)s)-sf_b(\lambda))=L^b_s,
\end{align}
proving the lemma.
\end{proof}
\begin{lemma}
\label{lemma:EMUI}
Let $M$ be a local martingale with initial value zero and $\Delta
M\ge-1$. Then $E\mathcal{E}(M)_\infty\le 1$, and $\mathcal{E}(M)$ is a uniformly
integrable martingale if and only if $E\mathcal{E}(M)_\infty=1$.
\end{lemma}
\begin{proof}
This follows from the the optional sampling theorem for nonnegative
supermartingales.
\end{proof}
In the proof of Theorem \ref{theorem:OptimalNovikov}, note that for a
standard Poisson process $N$, it holds that with $M_t = N_t-t$,
$\langle M\rangle_t = t$, since $[M]_t = N_t$ by Definition VI.37.6 of
\cite{RW2} and since $\langle M\rangle$ is the unique predictable and
locally integrable increasing process making $[M]-\langle M\rangle$ a
local martingale.
\begin{theorem}
\label{theorem:OptimalNovikov}
Fix $a\ge -1$. Let $M$ be a locally square integrable local martingale
with $\Delta
M1_{(\Delta M\neq0)}\ge a$. If $\exp(\frac{1}{2}\langle
M^c\rangle_\infty+\alpha(a)\langle M^d\rangle_\infty)$ is integrable,
then $\mathcal{E}(M)$ is a uniformly integrable martingale. Furthermore, for all
$a\ge-1$, the coefficients $\frac{1}{2}$ and $\alpha(a)$ in front of
$\langle M^c\rangle$ and $\langle M^d\rangle$ are optimal in the sense
that the criterion is false if any of the coefficients are reduced.
\end{theorem}
\begin{proof}
\textit{Sufficiency.} With $h(x)=(1+x)\log(1+x)-x$, we find by Lemma \ref{lemma:alphaFun} that
for $-1\le a\le x$, $\alpha(a)\ge \alpha(x)$, which implies $h(x)\le
\alpha(a)x^2$. Letting $a\ge-1$ and letting $M$ be a locally square
integrable local martingale with initial value zero, $\Delta M1_{(\Delta M\neq0)}\ge a$ and $\exp(\frac{1}{2}\langle
M^c\rangle_\infty+\alpha(a)\langle M^d\rangle_\infty)$ integrable, we
obtain for all $t\ge0$ the inequality $(1+\Delta M_t)\log(1+\Delta
M_t)+\Delta M_t\le \alpha(a) (\Delta M_t)^2$, and so Theorem III.1 of
\cite{LM} shows that $\mathcal{E}(M)$ is a
uniformly integrable martingale. Thus, the condition is sufficient.
As regards optimality of the coefficients, optimality of the
coefficient $\frac{1}{2}$ in front of $\langle M^c\rangle$ is
well-known, see \cite{AAN}. It therefore suffices to to prove optimality of the
coefficient $\alpha(a)$ in front of $\langle M^d\rangle$. To do so, we
need to show the following: That
for each $\eps>0$, there exists a locally square
integrable local martingale with initial value zero and $\Delta M\ge a$ such that
$\exp(\frac{1}{2}\langle M^c\rangle_\infty+(1-\eps)\alpha(a)\langle M^d\rangle_\infty)$ is
integrable, while $\mathcal{E}(M)$ is not a uniformly integrable martingale.
\textit{The case $a>0$.} Let $\eps,b>0$, put $T_b = \inf\{t\ge0\mid N_t -
(1+b)t=-1\}$ and define $M_t = a(N^{T_b}_t-t\land T_b)$. We claim that
we may choose $b>0$ such that $M$ satisfies the requirements stated
above. It holds that $M$ is a locally square
integrable local martingale with initial value zero and $\Delta M1_{(\Delta M\neq0)}\ge a$, and $M$ is purely
discontinuous by Definition 7.21 of \cite{HWY} since it is of locally
integrable variation. In particular, $M^c=0$, so it suffices to show that $\exp((1-\eps)\alpha(a)\langle M\rangle_\infty)$ is
integrable while $\mathcal{E}(M)$ is not a uniformly
integrable martingale. To show this, we first argue that $T_b$ is almost surely finite. To this end, note that since
$t\mapsto N_t-(1+b)t$ only has nonnegative jumps, has initial value
zero and decreases between jumps, the process hits $-1$ if and only if it is less than or equal to
$-1$ immediately before one of its jumps. Therefore, with $U_n$ denoting the $n$'th jump time of
$N$, we have
\begin{align}
P(T_b=\infty)
&=P(\cap_{n=1}^\infty (N_{U_n-}-(1+b)U_n>-1))\notag\\
&=P(\cap_{n=1}^\infty (n>U_n(1+b)))\notag\\
&\le P(\limsup_{n\to\infty}U_n/n\le (1+b)^{-1}),
\end{align}
which is zero, as $\lim_{n\to\infty}U_n/n=1$ almost surely by the law
of large numbers, and $(1+b)^{-1}<1$ as $b>0$. Thus, $T_b$ is almost surely
finite, and by the path properties of $N$, $N_{T_b} = (1+b)T_b-1$
almost surely. We then obtain
\begin{align}
\mathcal{E}(M)_\infty
&=\exp(a(N_{T_b}-T_b)+N_{T_b}(\log(1+a)-a))\notag\\
&=\exp(N_{T_b}\log (1+a)-aT_b)\notag\\
&=\exp(((1+b)T_b-1)\log (1+a)-aT_b)\notag\\
&=(1+a)^{-1}\exp(T_b((1+b)\log(1+a)-a)).
\end{align}
Recalling Lemma \ref{lemma:EMUI}, we wish to choose $b>0$ such that $E\mathcal{E}(M)_\infty<1$ and $E\exp((1-\eps)\alpha\langle
M\rangle_\infty)<\infty$ holds simultaneously. Note that $\langle
M\rangle_\infty = a^2T_b$. Therefore, we need to select a positive $b$
with the properties that
\begin{align}
E\exp(T_b((1+b)\log(1+a)-a))&< 1+a\textrm{ and }\label{eqn:PrimaryFirstReq}\\
E\exp(T_ba^2(1-\eps)\alpha(a))&<\infty.\label{eqn:PrimarySecondReq}
\end{align}
Consider some $b>0$ and let $f_b$ be as in Lemma \ref{lemma:PoissonCompMart}. By that same lemma, the process $L^b$ defined by
putting $L^b_t=\exp(-\lambda(N_t-(1+b)t)-tf_b(\lambda))$ is a
martingale. In particular, it is a nonnegative supermartingale with
initial value one, so Theorem II.77.5 of \cite{RW1} yields $1\ge EL^b_{T_b}
=E\exp(\lambda-T_bf_b(\lambda))$,
and so $E\exp(-T_bf_b(\lambda))\le \exp(-\lambda)$. Note that $f_b'(\lambda)
=-\exp(-\lambda)+1+b$, such that $f_b$ takes its minimum at $-\log(1+b)$. Therefore, $-f_b$ takes
its maximum at $-\log(1+b)$, and we find that the maximum is
$h(b)$. In particular, $E\exp(T_bh(b))$ is
finite. Next, define a function $\lambda$ by putting $\lambda(b) = -\log((1+a)\frac{b}{a})$, we then have $E\exp(-T_bf_b(\lambda(b)))\le
(1+a)\frac{b}{a}$, which is strictly less than $1+a$ whenever $b<a$. Thus, if we can choose $b\in(0,a)$ such that
\begin{align}
(1+b)\log(1+a)-a&\le -f_b(\lambda(b))\textrm{ and }\label{eqn:FirstReq}\\
a^2(1-\eps)\alpha(a)&\le h(b),\label{eqn:SecondReq}
\end{align}
we will have achieved our end, since (\ref{eqn:FirstReq}) implies
(\ref{eqn:PrimaryFirstReq}) and (\ref{eqn:SecondReq}) implies
(\ref{eqn:PrimarySecondReq}). To this end, note that
\begin{align}
-f_b(\lambda(b))
&=-\exp(\log((1+a)\tfrac{b}{a}))+\log((1+a)\tfrac{b}{a})(1+b)+1\notag\\
&=1-(1+a)\tfrac{b}{a}+(1+b)\log((1+a)\tfrac{b}{a})\notag\\
&=1-(1+a)\tfrac{b}{a}+(1+b)\log(1+a)+(1+b)\log \tfrac{b}{a},
\end{align}
such that, by rearrangement, (\ref{eqn:FirstReq}) is equivalent to
\begin{align}
0\le 1+a-(1+a)\tfrac{b}{a}+(1+b)\log \tfrac{b}{a},
\end{align}
and therefore, as $1-\frac{b}{a}>0$ for $0<b<a$, equivalent to
\begin{align}
\label{eqn:FirstReqReduction}
(1+b)\frac{\log \tfrac{b}{a}}{\tfrac{b}{a}-1} &\le 1+a,
\end{align}
which, as $\log x\le x-1$ for $x>0$, is satisfied for all
$0<b<a$. Thus, it suffices to choose $b\in(0,a)$ such that
(\ref{eqn:SecondReq}) is satisfied, corresponding to choosing $b\in(0,a)$
such that $(1-\eps)h(a)\le h(b)$. As $h$ is positive and continuous on $(0,\infty)$, this is
possible by choosing $b$ close enough to $a$. With this choice of $b$, we now obtain $M$ yielding an
example proving that the coefficient $\alpha(a)$ is optimal. This
concludes the proof of optimality in the case $a>0$.
\textit{The case $a=0$.} Let $\eps>0$. To prove optimality, we wish to
identify a locally square integrable local martingale
$M$ with initial value zero and $\Delta M1_{(\Delta M\neq0)}\ge0$ such that $\exp((1-\eps)\alpha(0)\langle M\rangle_\infty)$ is integrable while
$\mathcal{E}(M)$ is not a uniformly integrable martingale. Recalling that
$\alpha$ is positive and continuous, pick $a>0$ so
close to zero that $(1-\eps)\alpha(0)\le
(1-\frac{1}{2}\eps)\alpha(a)$. By what was already shown, there
exists a locally square integrable local
martingale $M$ with initial value zero and $\Delta
M1_{(\Delta M\neq0)}\ge a$ such that $\exp((1-\frac{1}{2}\eps)\alpha(a)\langle
M\rangle_\infty)$ is integrable while $\mathcal{E}(M)$ is not a uniformly
integrable martingale. As $\exp((1-\eps)\alpha(0)\langle
M\rangle_\infty)$ is integrable in this case, this shows that $\alpha(0)$ is optimal.
\textit{The case $-1<a<0$.} Let $\eps>0$, let $-1<b<0$, let $c>0$ and
define a stopping time $T_{bc}$ by putting $T_{bc} = \inf\{t\ge0\mid N_t -
(1+b)t\ge c\}$. Also define $M$ by $M_t = a(N^{T_{bc}}_t-t\land T_{bc})$. We claim that
we can choose $b\in(-1,0)$ and $c>0$ such that $M$ satisfies the
requirements to show optimality. Similarly to the case $a>0$, $M$ is a
purely discontinuous locally square
integrable local martingale with initial value zero and $\Delta M1_{(\Delta M\neq0)}\ge a$, so it suffices to
show that $\exp((1-\eps)\alpha(a)\langle M\rangle_\infty)$ is
integrable while $\mathcal{E}(M)$ is not a uniformly integrable
martingale. We first investigate some properties
of $T_{bc}$. As $t\mapsto N_t-(1+b)t$ only has nonnegative jumps, has initial value
zero and decreases between jumps, the process advances beyond $c$ at
some point if and only it advances beyond $c$ at one of its jump
times. Therefore, with $U_n$ denoting the $n$'th jump time of $N$,
\begin{align}
P(T_{bc}=\infty)
&=P(\cap_{n=1}^\infty (N_{U_n}-(1+b)U_n<c))\notag\\
&=P(\cap_{n=1}^\infty (n-c<U_n(1+b)))\notag\\
&\le P(\liminf_{n\to\infty}U_n/n\ge (1+b)^{-1}),
\end{align}
which is zero, as $U_n/n$ tends to one almost surely and $(1+b)^{-1}>1$. Thus, $T_{bc}$ is almost surely
finite. Furthermore, by the path properties of $N$, $N_{T_{bc}}
\ge (1+b)T_{bc}+ c$ and $N_{T_{bc}}\le (1+b)T_{bc}+c+1$ almost surely. Since $\log(1+a)\le
0$, we in particular obtain $N_{T_{bc}}\log(1+a)\le ((1+b)T_{bc}+c)\log(1+a)$ almost
surely. From this, we conclude that
\begin{align}
\mathcal{E}(M)_\infty
&=\exp(a(N_{T_{bc}}-T_{bc})+N_{T_{bc}}(\log(1+a)-a))\notag\\
&=\exp(N_{T_{bc}}\log (1+a)-aT_{bc})\notag\\
&\le\exp(((1+b)T_{bc}+c)\log (1+a)-aT_{bc})\notag\\
&=(1+a)^c\exp(T_{bc}((1+b)\log(1+a)-a)).
\end{align}
We wish to choose $-1<b<0$ and $c>0$ such that $E\exp((1-\eps)\alpha(a)\langle
M\rangle_\infty)<\infty$ and $E\mathcal{E}(M)_\infty<1$ holds
simultaneously. As $\langle M\rangle_\infty = a^2T_{bc}$, this is
equivalent to choosing $-1<b<0$ and $c>0$ such that
\begin{align}
E\exp(T_{bc}((1+b)\log(1+a)-a))&< (1+a)^{-c}\textrm{ and }\label{eqn:PrimaryFirstReqMinus}\\
E\exp(T_{bc}a^2(1-\eps)\alpha(a))&<\infty.\label{eqn:PrimarySecondReqMinus}
\end{align}
Let $f_b$ and $L^b$ be as in Lemma \ref{lemma:PoissonCompMart}. The
process $L^b$ is then a
nonnegative supermartingale. As $N_{T_{bc}}\le (1+b)T_{bc}+c+1$, the
optional stopping theorem allows us to conclude that for $\lambda\ge0$,
\begin{align}
1&\ge EL^b_{T_{bc}}
=E\exp(-\lambda(N_{T_{bc}}-(1+b)T_{bc})-T_{bc}f_b(\lambda))\notag\\
&\ge E\exp(-(c+1)\lambda-T_{bc}f_b(\lambda)),
\end{align}
so that $E\exp(-T_{bc}f_b(\lambda))\le \exp((c+1)\lambda)$. As in the case $a>0$, $-f_b$ takes
its maximum at $-\log(1+b)$, and the maximum is $h(b)$, leading
us to conclude that $E\exp(T_{bc}h(b))$ is finite. Put
$\lambda(b,c)=(c+1)^{-1}\log((1+a)^{-c}\frac{b}{a})$. For all $b\in(a,0)$,
$\frac{b}{a}<1$, leading to $E\exp(-T_{bc}f_b(\lambda(b,c)))\le
(1+a)^{-c}\frac{b}{a}<(1+a)^{-c}$. Therefore, if we can choose
$b\in(a,0)$ and $c>0$ such that
\begin{align}
(1+b)\log(1+a)-a&\le -f_b(\lambda(b,c)) \textrm{ and }\label{eqn:FirstReqMinus}\\
a^2(1-\eps)\alpha(a)&\le h(b),\label{eqn:SecondReqMinus}
\end{align}
we will have obtained existence of a local maringale yielding the
desired optimality of $\alpha(a)$. We first note that $a^2(1-\eps)\alpha(a)\le h(b)$
is equivalent to $(1-\eps)h(a)\le h(b)$. As $h$ is continuous and
positive on $(-1,0)$, we find that (\ref{eqn:SecondReqMinus}) is satisfied
for $a<b<0$ with $b$ close enough to $a$. Next, we turn our
attention to (\ref{eqn:FirstReqMinus}). We have
\begin{align}
-f_b(\lambda(b,c))
&=-\exp\left(-\frac{1}{c+1}\log\left((1+a)^{-c}\frac{b}{a}\right)\right)-\frac{1+b}{c+1}\log\left((1+a)^{-c}\frac{b}{a}\right)+1\notag\\
&=1-(1+a)^{\frac{c}{c+1}}\left(\frac{b}{a}\right)^{-\frac{1}{(c+1)}}-\frac{1+b}{c+1}\log\left((1+a)^{-c}\frac{b}{a}\right)\notag\\
&=1-(1+a)^{\frac{c}{c+1}}\left(\frac{a}{b}\right)^{\frac{1}{c+1}}+\frac{c(1+b)}{c+1}\log(1+a)+\frac{1+b}{c+1}\log\frac{a}{b},
\end{align}
such that (\ref{eqn:FirstReqMinus}) is equivalent to
\begin{align}
\label{eqn:FirstReqReductionMinus}
0&\le
1+a-(1+a)^{\frac{c}{c+1}}\left(\frac{a}{b}\right)^{\frac{1}{c+1}}+\frac{1+b}{c+1}\left(\log\frac{a}{b}-\log(1+a)\right).
\end{align}
Fixing $a<b<0$, we wish to argue that for $b$ close enough to $a$,
(\ref{eqn:FirstReqReductionMinus}) holds for $c$ large enough. To this
end, let $\rho_b(c)$ denote the right-hand side of
(\ref{eqn:FirstReqReductionMinus}). Then
$\lim_{c\to\infty}\rho_b(c)=0$. We also note that
$\frac{\df{}}{\df{c}}\frac{1}{c+1}=-\frac{1}{(c+1)^2}$ and
$\frac{\df{}}{\df{c}}\frac{c}{c+1}=\frac{1}{(c+1)^2}$, yielding
\begin{align}
&\frac{\df{}}{\df{c}}(1+a)^{\frac{c}{c+1}}\left(\frac{a}{b}\right)^{\frac{1}{c+1}}\notag\\
&=\frac{\df{}}{\df{c}}\exp\left(\frac{c}{c+1}\log(1+a)+\frac{1}{c+1}\log\frac{a}{b}\right)\notag\\
&=\left(\frac{\log(1+a)}{(c+1)^2}-\frac{\log\frac{a}{b}}{(c+1)^2}\right)\exp\left(\frac{c}{c+1}\log(1+a)+\frac{1}{c+1}\log\frac{a}{b}\right)\notag\\
&=\frac{\log(1+a)-\log\frac{a}{b}}{(c+1)^2}\exp\left(\frac{c}{c+1}\log(1+a)+\frac{1}{c+1}\log\frac{a}{b}\right)
\end{align}
and
\begin{align}
\frac{\df{}}{\df{c}}\frac{1+b}{c+1}\left(\log\frac{a}{b}-\log(1+a)\right)
&=-\frac{1+b}{(c+1)^2}\left(\log\frac{a}{b}-\log(1+a)\right)\notag\\
&=(1+b)\frac{\log(1+a)-\log\frac{a}{b}}{(c+1)^2},
\end{align}
which leads to
\begin{align}
\rho_b'(c)
&=\frac{\log(1+a)-\log\frac{a}{b}}{(c+1)^2}\left(1+b-\exp\left(\frac{c}{c+1}\log(1+a)+\frac{1}{c+1}\log\frac{a}{b}\right)\right).
\end{align}
Now note that for $a<b$, we obtain
\begin{align}
\lim_{c\to\infty}1+b-\exp\left(\frac{c}{c+1}\log(1+a)+\frac{1}{c+1}\log\frac{a}{b}\right)
&=1+b-(1+a)>0,
\end{align}
and for $b$ close enough to $a$, $\log(1+a)-\log\frac{a}{b}<0$,
since $a<0$. Therefore, for all $c$ large enough, $\rho_b'(c)<0$. Consider such
a $c$, we then obtain
\begin{align}
\rho_b(c)&=\lim_{y\to\infty}\rho_b(c)-\rho_b(y)
=-\lim_{y\to\infty}\int_c^y \rho_b'(z)\df{z}>0.
\end{align}
Thus, we conclude that for $b$ close enough to $a$, it holds that $\rho_b(c)>0$ for $c$ large enough.
We now collect our conclusions in order to obtain $b\in(a,0)$ and $c>0$
satisfying (\ref{eqn:FirstReqMinus}) and
(\ref{eqn:SecondReqMinus}). First choose $b\in(a,0)$ so close to
$a$ that $(1-\eps)h(a)\le h(b)$ and
$\log(1+a)-\log\frac{a}{b}<0$. Pick $c$ so large that $\rho_b(c)>0$. By our deliberations, (\ref{eqn:FirstReqMinus}) and
(\ref{eqn:SecondReqMinus}) then both hold, demonstrating the existence of
a locally square integrable local martingale $M$ with initial value
zero and $\Delta M1_{(\Delta M\neq0)}\ge a$ such that $\exp((1-\eps)\alpha(a)\langle M\rangle_\infty)$ is
integrable while $\mathcal{E}(M)$ is not a uniformly integrable
martingale.
\textit{The case $a=-1$.} Let $\eps>0$. We wish to identify a purely
discontinous locally square integrable local martingale $M$ with
$\Delta M1_{(\Delta M\neq0)}\ge-1$ such that integrability of
$\exp((1-\eps)\alpha(-1)\langle M\rangle_\infty)$ holds while
$\mathcal{E}(M)$ is not a uniformly integrable martingale. We proceed as in
the case $a=0$. By positivity and continuity of $\alpha$, take $a>0$ so
close to $-1$ that $(1-\eps)\alpha(-1)\le
(1-\frac{1}{2}\eps)\alpha(a)$. By what was shown in the previous case, there
exists a purely discontinuous locally square integrable local
martingale $M$ with initial value zero and $\Delta
M1_{(\Delta M\neq0)}\ge a$ such that $\exp((1-\frac{1}{2}\eps)\alpha(a)\langle
M\rangle_\infty)$ is integrable while $\mathcal{E}(M)$ is not a uniformly
integrable martingale. As $\exp((1-\eps)\alpha(-1)\langle
M\rangle_\infty)$ then also is integrable, this shows that $\alpha(-1)$ is optimal.
\end{proof}
\begin{theorem}
\label{theorem:OptimalNovikov2}
Fix $a>-1$. Let $M$ be a local martingale with $\Delta M1_{(\Delta M\neq0)}\ge a$. If $\exp(\frac{1}{2}[M^c]_\infty+\beta(a)[M^d]_\infty)$ is integrable,
then $\mathcal{E}(M)$ is a uniformly integrable martingale. Furthermore, for all
$a>-1$, the coefficients $\frac{1}{2}$ and $\beta(a)$ in front of
$[M^c]$ and $[M^d]$ are optimal in the sense
that the criterion is false if any of the coefficients are
reduced.
Furthermore, there exists no $\beta(-1)$ such that for $M$ with $\Delta M1_{(\Delta
M\neq0)}\ge-1$, integrability of $\exp(\frac{1}{2}[M^c]_\infty+\beta(-1)[M^d]_\infty)$ suffices to
ensure that $\mathcal{E}(M)$ is a uniformly integrable martingale.
\end{theorem}
\begin{proof}
\textit{Sufficiency.} We proceed in a manner closely related to the
proof of Theorem \ref{theorem:OptimalNovikov}. Defining $g$ by putting
$g(x)=\log(1+x)-x/(1+x)$, we find by Lemma \ref{lemma:alphaFun} that
for $-1<a\le x$, $\beta(a)\ge \beta(x)$, yielding $g(x)\le
\beta(a)x^2$. Letting $a>-1$ and letting $M$ be a locally square
integrable local martingale with initial value zero, $\Delta M1_{(\Delta M\neq0)}\ge a$ and $\exp(\frac{1}{2}\langle
M^c\rangle_\infty+\beta(a)\langle M^d\rangle_\infty)$ integrable, we
obtain for all $t\ge0$ that $\log(1+\Delta M_t)+\Delta
M_t/(1+\Delta M_t)\le \beta(a) (\Delta M_t)^2$, and so Theorem III.7
of \cite{LM} shows that $\mathcal{E}(M)$ is a uniformly integrable
martingale. Thus, the condition is sufficient.
As in Theorem \ref{theorem:OptimalNovikov}, optimality of the $\frac{1}{2}$ in front of $[M^c]$ follows from \cite{AAN},
so it suffices to consider the coefficient $\beta(a)$ in front of
$[M^d]$. Thus, for $a>-1$, we need to prove that
for each $\eps>0$, there exists a locally square
integrable local martingale with initial value zero and $\Delta
M1_{(\Delta M\neq0)}\ge a$ such that
$\exp(\frac{1}{2}[M^c]_\infty+(1-\eps)\beta(a)[M^d]_\infty)$ is
integrable, while $\mathcal{E}(M)$ is not a uniformly integrable martingale.
\textit{The case $a>0$.} Let $\eps,b>0$, put $T_b = \inf\{t\ge0\mid N_t -
(1+b)t=-1\}$ and define $M_t = a(N^{T_b}_t-t\land T_b)$. Noting that
$[M]_\infty = a^2N_{T_b}$, we may argue as in the proof of Theorem
\ref{theorem:OptimalNovikov} and obtain that it suffices to identify
$b>0$ such that
\begin{align}
E\exp(T_b((1+b)\log(1+a)-a))&< 1+a\textrm{ and }\label{eqn:SqPrimaryFirstReq}\\
E\exp(N_{T_b}a^2(1-\eps)\beta(a))&<\infty.\label{eqn:SqPrimarySecondReq}
\end{align}
Let $f_b$ be as in Lemma \ref{lemma:PoissonCompMart}. As in the proof of Theorem
\ref{theorem:OptimalNovikov}, we obtain that $E\exp(T_bh(b))$ is
finite, where $h(x)=(1+x)\log(1+x)-x$, and furthermore obtain that with $\lambda(b) = -\log((1+a)\frac{b}{a})$,
$E\exp(-T_bf_b(\lambda(b)))<1+a$ for $b<a$. As $N_{T_b}=(1+b)T_b-1$
almost surely and $g(b)=h(b)/(1+b)$, we
then also obtain that $E\exp(N_{T_b}g(b))$ is finite. Thus, if we can choose $b\in(0,a)$ such that
\begin{align}
(1+b)\log(1+a)-a&\le -f_b(\lambda(b))\textrm{ and }\label{eqn:SqFirstReq}\\
a^2(1-\eps)\beta(a)&\le g(b),\label{eqn:SqSecondReq}
\end{align}
we will obtain the desired result, as (\ref{eqn:SqFirstReq}) implies
(\ref{eqn:SqPrimaryFirstReq}) and (\ref{eqn:SqSecondReq}) implies
(\ref{eqn:SqPrimarySecondReq}). As earlier noted,
(\ref{eqn:SqFirstReq}) always holds for $0<b<a$. As for
(\ref{eqn:SqSecondReq}), this requirement is equivalent to having that
$(1-\eps)g(a)\le g(b)$ for some $b\in (0,a)$, which by continuity of
$g$ can be obtained by choosing $b$ close enough to $a$. Choosing $b$
in this manner, we obtain $M$ yielding an example proving that the
coefficient $\beta(a)$ is optimal. This concludes the proof of
optimality in the case $a>0$.
\textit{The case $a=0$.} This follows similarly to the corresponding
case in the proof of Theorem \ref{theorem:OptimalNovikov}.
\textit{The case $-1<a<0$.} Let $\eps>0$, let $-1<b<0$, let $c>0$ and
define a stopping time $T_{bc}$ by putting $T_{bc} = \inf\{t\ge0\mid N_t -
(1+b)t\ge c\}$. Also define $M$ by $M_t = a(N^{T_{bc}}_t-t\land
T_{bc})$. As in the proof of Theorem \ref{theorem:OptimalNovikov}, in
order to obtain the desired counterexample, it suffices to choose
$-1<b<0$ and $c>0$ such that
\begin{align}
E\exp(T_{bc}((1+b)\log(1+a)-a))&< (1+a)^{-c}\textrm{ and }\label{eqn:SqPrimaryFirstReqMinus}\\
E\exp(T_{bc}a^2(1-\eps)\beta(a))&<\infty.\label{eqn:SqPrimarySecondReqMinus}
\end{align}
With $f_b$ as in Lemma \ref{lemma:PoissonCompMart}, we find as in the
proof of Theorem \ref{theorem:OptimalNovikov} that $E\exp(T_{bc}h(b))$
is finite. Furthermore, defining
$\lambda(b,c)=(c+1)^{-1}\log((1+a)^{-c}\frac{b}{a})$, it holds for $b$
with $a<b\le (1+a)^ca$ that $\lambda(b,c)\ge0$ and
$E\exp(-T_{bc}f_b(\lambda(b,c)))<(1+a)^{-c}$. Also, as
$N_{T_{bc}}\le(1+b)T_{bc}+c+1$, $E\exp(N_{T_{bc}}(1+b)^{-1}h(b))$ and
thus $E\exp(N_{T_{bc}}g(b))$ is finite. Therefore, if we can choose
$b\in(a,0)$ and $c>0$ such that
\begin{align}
(1+b)\log(1+a)-a&\le -f_b(\lambda(b,c)) \textrm{ and }\label{eqn:SqFirstReqMinus}\\
a^2(1-\eps)\beta(a)&\le g(b),\label{eqn:SqSecondReqMinus}
\end{align}
we obtain the desired result. By arguments as in the proof of the corresponding case of Theorem
\ref{theorem:OptimalNovikov}, we find that by first picking $b$ close
enough to $a$ and then $c$ large enough, we can ensure that both (\ref{eqn:SqFirstReqMinus}) and
(\ref{eqn:SqSecondReqMinus}) hold, yielding optimality for this case.
\textit{The case $a=-1$.} For this case, we need to show that for any
$\gamma\ge0$, it does not hold that finiteness of $E\exp(\gamma[M^d]_\infty)$
implies that $\mathcal{E}(M)$ is a uniformly integrable martingale. Let
$\gamma\ge0$. By Lemma \ref{lemma:alphaFun}, $\beta(a)$ tends to infinity as
$a$ tends to $-1$. Therefore, we
may pick $a>-1$ so small that $\beta(a)\ge \gamma$. By what we
already have shown, there exists $M$ with initial value zero and
$\Delta M1_{(\Delta M\neq0)}\ge -1$ such that
$E\exp(\beta(a)[M^d]_\infty)$ and thus $E\exp(\gamma[M^d]_\infty)$ is finite,
while $\mathcal{E}(M)$ is not a uniformly integrable martingale.
\end{proof}
\begin{corollary}
\label{coro:NovikovExtension}
Let $M$ be a local martingale with initial value zero and $\Delta
M\ge0$. If $\exp(\frac{1}{2}[M]_\infty)$ is integrable or if $M$ is
locally square integrable and $\exp(\frac{1}{2}\langle
M\rangle_\infty)$ is integrable, then $\mathcal{E}(M)$ is a uniformly
integrable martingale. Furthermore, this criterion is optimal in the
sense that if either the constant $\frac{1}{2}$ is reduced, or the
requirement on the jumps is weakened to $\Delta M\ge-\eps$ for some
$\eps>0$, the criterion ceases to be sufficient.
\end{corollary}
\begin{proof}
That the constant $\frac{1}{2}$ cannot be reduced follows from
Theorem \ref{theorem:OptimalNovikov} and Theorem
\ref{theorem:OptimalNovikov2}. That the requirement on the jumps cannot
be reduced follows by combining Theorem \ref{theorem:OptimalNovikov} and Theorem
\ref{theorem:OptimalNovikov2} with the fact that $\alpha$ and $\beta$ both are strictly
decreasing by Lemma \ref{lemma:alphaFun}.
\end{proof}
\bibliographystyle{amsplain}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 5,798 |
#ifndef CPP_INCLUDE_RPC_SSL_OPTIONS_H_
#define CPP_INCLUDE_RPC_SSL_OPTIONS_H_
#ifdef HAS_OPENSSL
#include <boost/asio/ssl.hpp>
#endif // HAS_OPENSSL
#include <algorithm> // std::find
#include <string>
#include <vector>
namespace xtreemfs {
namespace rpc {
class SSLOptions {
#ifdef HAS_OPENSSL
public:
SSLOptions(const std::string ssl_pem_path,
const std::string ssl_pem_cert_path,
const std::string ssl_pem_key_pass,
const std::string ssl_pem_trusted_certs_path,
const std::string ssl_pkcs12_path,
const std::string ssl_pkcs12_pass,
const boost::asio::ssl::context::file_format format,
const bool use_grid_ssl,
const bool ssl_verify_certificates,
const std::vector<int> ssl_ignore_verify_errors,
const std::string ssl_method_string)
: pem_file_name_(ssl_pem_path),
pem_file_pass_(ssl_pem_key_pass),
pem_cert_name_(ssl_pem_cert_path),
pem_trusted_certs_file_name_(ssl_pem_trusted_certs_path),
pkcs12_file_name_(ssl_pkcs12_path),
pkcs12_file_pass_(ssl_pkcs12_pass),
cert_format_(format),
use_grid_ssl_(use_grid_ssl),
verify_certificates_(ssl_verify_certificates),
ignore_verify_errors_(ssl_ignore_verify_errors),
ssl_method_string_(ssl_method_string) {}
virtual ~SSLOptions() {
}
std::string pem_file_name() const {
return pem_file_name_;
}
std::string pem_cert_name() const {
return pem_cert_name_;
}
std::string pem_file_password() const {
return pem_file_pass_;
}
std::string pem_trusted_certs_file_name() const {
return pem_trusted_certs_file_name_;
}
std::string pkcs12_file_name() const {
return pkcs12_file_name_;
}
std::string pkcs12_file_password() const {
return pkcs12_file_pass_;
}
boost::asio::ssl::context::file_format cert_format() const {
return cert_format_;
}
bool use_grid_ssl() const {
return use_grid_ssl_;
}
bool verify_certificates() const {
return verify_certificates_;
}
bool ignore_verify_error(int verify_error) const {
return std::find(ignore_verify_errors_.begin(),
ignore_verify_errors_.end(),
verify_error) != ignore_verify_errors_.end();
}
std::string ssl_method_string() const {
return ssl_method_string_;
}
private:
std::string pem_file_name_;
std::string pem_file_pass_;
std::string pem_cert_name_;
std::string pem_trusted_certs_file_name_;
std::string pkcs12_file_name_;
std::string pkcs12_file_pass_;
boost::asio::ssl::context::file_format cert_format_;
bool use_grid_ssl_;
bool verify_certificates_;
std::vector<int> ignore_verify_errors_;
std::string ssl_method_string_;
#endif // HAS_OPENSSL
};
} // namespace rpc
} // namespace xtreemfs
#endif // CPP_INCLUDE_RPC_SSL_OPTIONS_H_
| {
"redpajama_set_name": "RedPajamaGithub"
} | 3,118 |
Thank Folk For That
Home New Releases New Release: Jack Johnson – All The Light Above It Too /...
New Release: Jack Johnson – All The Light Above It Too / My Mind Is For Sale
US singer-songwriter Jack Johnson releases his 7th studio album, All The Light Above It Too, on September 8th via Johnson's own Brushfire Records label
The album's first single, My Mind Is For Sale has also been unveiled, accompanied by a lyric video which can be seen below…
All The Light Above It Too was recorded over the past year at Jack's Hawaii based Mango Tree Studio. For the first time in years, Johnson handled most of the instrumentation himself, echoing the four-track recordings that launched his career over 17 years ago
"This album shares what has been on my mind during the past year or so," says Johnson. "A year in which I sailed through the North Atlantic Gyre for a documentary about plastic pollution in the ocean. A year in which Trump was elected as the President of the United States. A year in which I camped, surfed, got stitches, explored, dreamed, shared time and endless conversations with my family and friends…all of which inspired these songs. I usually make sketches of the songs first then set up a time to actually record the album. This time around the original sketches became the final versions. I didn't want to lose any of the spirit that a song has in its rawest form"
Full Track listing for All The Light Above It Too is as follows…
1. Subplots
2. You Can't Control It
3. Sunsets For Somebody Else
4. My Mind Is For Sale
5. Daybreaks
6. Big Sur
7. Love Song
8. One Moon
9. Gather
10. Fragments Of The Sea
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News: Bright Eyes Reform & Announce 2020 Tour
New Release: Bon Iver – Blood Bank (10th Anniversary Edition)
Festivals: End Of The Road Announce 2020 Acts inc. Big Thief, Angel Olsen and Bright Eyes
Search Thank Folk For That
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© 2019 - Thank Folk For That | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 9,077 |
Q: Find all nodes that DON'T have NoSchedule taint? What is kubectl command that lists all nodes that don't have NoSchedule taint(s)?
What is kubectl command that lists all nodes that do have NoSchedule taint(s)?
A: You could get nodes and taints with using jsonpath.
kubectl get node -o jsonpath='{range .items[*]}{@.metadata.name}{"\t"}{@.spec.taints[*].effect}{"\n"}{end}'
Have NoSchedule
kubectl get node -o jsonpath='{range .items[*]}{@.metadata.name}{"\t"}{@.spec.taints[*].effect}{"\n"}{end}' | grep NoSchedule
Don't Have NoSchedule
kubectl get node -o jsonpath='{range .items[*]}{@.metadata.name}{"\t"}{@.spec.taints[*].effect}{"\n"}{end}' | grep -v NoSchedule
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 4,715 |
\section{Introduction}
M dwarfs have become targets of choice for many exoplanet surveys. This
is because low-mass planets (i.e. Earth- to Neptune-size) are easier
to detect by the Doppler or transit techniques around stars of lower
mass. The transits of smaller planets are also easier to detect when
they occur in the also smaller M dwarfs. In addition, M dwarfs have
much lower luminosities than the Sun, and their ``habitable zones''
(HZ) are closer in, which makes transits more likely to occur and
radial velocity variations easier to detect for planets in their
HZ. Earth-like planets within the HZ of M dwarfs are thus eminently
more detectable with current observational techniques than earth-like
planets in the HZ of G dwarfs \citep{Tarter2007,Gaidos2007}. M dwarfs
are also the most plentiful class of stars, constituting the largest
fraction ($>70\%$) of main sequence objects in the Galaxy and in the
vicinity of the Sun \citep{Reid2002,Covey2008,Bochanski2010}.
However, even nearby M dwarfs are generaly faint at the visible
wavelengths where most planet searches are conducted, and most
exoplanet detection techniques \--- with the notable exception of
micro-lensing \citep{Dong_etal.2009} \--- are currently restricted
to relatively bright stars. This significantly limits the number of M
dwarfs that can be targeted in exoplanet surveys. Doppler searches in
particular are usually restricted to stars with visual magnitudes
$V<12$, and less than $\sim$10\% of late-K and early-M stars within
30~pc are currently being monitored by the large-scale Doppler surveys
\citep{Butler2006,Mayor2009}. However, new surveys are pushing this
limit to fainter magnitudes \citep{Apps2009}, and high-resolution
spectrographs suitable for Doppler observations at near-infrared
wavelengths, where M dwarfs are relatively brighter, are being
developed \citep{Terada2008,Bean2010,Quirrenbach2010,Wang2010}. In any
case, only a fraction of all catalogued, nearby M dwarfs are bright
enough to be included in radial-velocity monitoring programs.
Transit surveys, on the other hand, can include much fainter
stars \citep{Irwin2009}. However, because they have a much lower
detection efficiency due to orbital inclination constraints, they
require extensive lists (thousands) of targets in order to detect any
significant number of transit events. For transit surveys, the Solar
Neighborhood census and its estimated $\approx$5,000 M dwarfs is
therefore too small, and transit programs would greatly benefit from
extending their target lists to much larger distance limits.
A fundamental obstacle to progress has been the lack of a large,
complete, and uniform catalog of bright M dwarfs suitable as targets
for exoplanet programs. In particular, most catalogs and surveys of M
dwarfs have focused on identifying the nearest objects, which are not
necessarily the brightest. Whereas the Hipparcos catalog
\citep{vanLeeuwen2007} provides a near-complete census of solar-mass
stars within 100 parsecs of the Sun, the bright magnitude limit of the
catalog excludes all but the very nearest M dwarfs \--- although it
lists stars as faint as $V=12-13$, the Hipparcos catalog is complete
only to $V<8$.
The widely utilized {\it Third Catalog of Nearby Stars}, or CNS3
\citep{gj1991}, which lists $\approx$3,800 stars, though predating the
Hipparcos survey, has historically provided a more complete list of
M dwarf candidates in the Solar Neighborhood. Many of the fainter
stars in the CNS3 have (ground-based) parallax measurements from a
variety of sources \citep{VLH95}. However, the CNS3 was largely
compiled based on a photometric analysis the high proper motion stars
catalogued by \citet{lhs,nltt}, and in large part using photometric
data collected by \citet{Gliese_1982} and
\citet{Weis1984,Weis1986,Weis1987}. The CNS3 has been largely used in
recent years to select M dwarf targets for exoplanet surveys
\citep{Marcy2001,Naef2003,Butler2004,Rivera2005,Endl2008,Bailey2009,Anglada2012a,Anglada2012b}. Unfortunately,
the catalog suffers from various sources of incompleteness. These
mainly consist of: (1) limited availability of quality data for stars
in the Luyten catalogs, at the time the CNS3 was compiled, (2)
kinematic bias in the Luyten catalog due a relatively high
($\mu>0.18\arcsec$ yr$^{-1}$) proper motion limit at the low end, and
(3) incompleteness of the Luyten catalogs even for stars with proper
motions above the fiducial limit.
Motivated mainly by the need to complete the census of the Solar
Neighborhood, several surveys have since been conducted to identify
the low-mass stars (mostly M dwarfs) suspected to be missing from the
CNS3. These have included a re-analysis of the proper motion catalogs
of \citet{nltt,lhs} in light of high quality photometric data
provided by the 2MASS survey \citep{Cutri.etal.2003}. This has led to
the identification of hundreds of additional nearby star candidates
that had previously been overlooked
\citep{Reid_Cruz.2002,Reidetal.2004}. In addition, cross-matching
against 2MASS and examination of Digitized Sky Survey images has
uncovered significant ($>1\arcmin$) errors in many of the coordinates
quoted in the Luyten catalogs, which was hitherto preventing efficient
follow-up studies \citet{Bakos.2002,SalimGould.2003}.
Parallelling these efforts, new proper motion surveys have been
conducted, mainly to find the high proper motion stars missing from
the Luyten catalogs with a focus on completing the stellar census of
the Solar Neighborhood
\citep{Lepine2002,Lepine2003b,Deacon2005,Levine2005,Lepine2005,Subasavage2005a,Subasavage2005b,Lepine2008}. In
addition, some surveys have also been reaching to lower proper motion
limits \citep{Lepine2005,Reid2007,Boyd2011}, potentially extending the
census of M dwarfs to larger distances. Recently, we have analyzed
data from theSUPERBLINK proper motion survey, which has a proper
motion limit $\mu>0.04\arcsec$ yr$^{-1}$, with an emphasis on the
identification of {\em bright} M dwarfs, rather than just {\em nearby}
ones; our search has turned up 8,889 candidate M dwarfs with infrared
magnitude $J<10$ \citep{LepineGaidos.2011}. Of these, we found that
only 982 were previously listed in the Hipparcos catalog, and another
898 in the CNS3. Most of the other 7009 stars were not commonly known
objects, and were identified as probable nearby M dwarfs for the first
time. With its high estimated completeness, especially in the northern
sky, the \citet{LepineGaidos.2011} census provides a solid basis for
assembling an extensive and highly complete catalog of bright M
dwarfs, suitable for exoplanet search programs.
Not all M dwarfs, however, are equally suitable targets for planet
searches. Some M dwarfs have significant photometric variability
(flares, spots) which are affecting transit searches
\citep{Hartman2011}; some display chromospheric emission affecting
Doppler searches \citep{Isaacson2010}. Because M dwarfs are relatively
faint stars, they often require considerable investement of observing
time on large telescopes to achieve exoplanet detection, and there is
value in identifying subsets of M dwarfs that are intrinsically more
likely to host detectable planets \citet{Herrero2011}. In
particular, one might be interested in selecting stars of higher
metallicity which may harbor more massive planets
\citep{Sousa_etal.2010}, or young stars with relatively luminous
massive planets which would be easier to detect through direct imaging
\citep{Mugrauer_etal.2010}. In addition, one would like to avoid
possible contaminants (e.g. background giants) or problematic systems
(e.g. very active stars) in order to optimize exoplanet survey
efficiencies.
Determining physical properties of the M dwarfs is also important in
order to better characterize the local populations of low-mass
stars. This is especially true since proximity makes them brighter and
thus more efficient targets for follow-up observations and detailed
study. Some of the bright M dwarfs may be close enough
(d$\lesssim$20pc) to warrant inclusion in the parallax programs
devoted to completing the census of low-mass stars in the Solar
Neighborhood \citep{Henry_etal.2006}, in which case it is also
important that the candidates first be vetted through spectral
typing.
\begin{deluxetable*}{llrrrrrrrrrcrrr}
\tabletypesize{\scriptsize}
\tablecolumns{15}
\tablewidth{0pt}
\tablecaption{Survey stars: positions and photometry. \label{table_photo}}
\tablehead{
\colhead{Star name} &
\colhead{CNS3\tablenotemark{a}} &
\colhead{R.A.(ICRS)} &
\colhead{Decl.(ICRS)} &
\colhead{$\mu_{R.A.}$} &
\colhead{$\mu_{Decl.}$} &
\colhead{Xray\tablenotemark{b}} &
\colhead{hr1\tablenotemark{b}} &
\colhead{FUV\tablenotemark{c}} &
\colhead{NUV\tablenotemark{c}} &
\colhead{V} &
\colhead{V} &
\colhead{J\tablenotemark{d}} &
\colhead{H\tablenotemark{d}} &
\colhead{K$_s$\tablenotemark{d}} \\
\colhead{} &
\colhead{} &
\colhead{(ICRS)} &
\colhead{(ICRS)} &
\colhead{$\arcsec$ yr$^{-1}$} &
\colhead{$\arcsec$ yr$^{-1}$} &
\colhead{cnts/s} &
\colhead{} &
\colhead{mag} &
\colhead{mag} &
\colhead{mag} &
\colhead{flag} &
\colhead{mag} &
\colhead{mag} &
\colhead{mag}
}
\startdata
PM I00006+1829 & & 0.163528 & 18.488850 & 0.335 & 0.195& & & & 20.04& 11.28&T& 8.44& 7.79& 7.64\\
PM I00012+1358S & & 0.303578 & 13.972055 & 0.025 & 0.144& & & & 19.85& 11.12&T& 8.36& 7.71& 7.53\\
PM I00033+0441 & & 0.829182 & 4.686940 &-0.024 &-0.085& & & & 21.18& 12.04&T& 8.83& 8.18& 7.98\\
PM I00051+4547 & Gl 2 & 1.295195 & 45.786587 & 0.870 &-0.151& & & & & 9.95&T& 6.70& 6.10& 5.85\\
PM I00051+7406 & & 1.275512 & 74.105217 & 0.035 &-0.023& & & & & 10.63&T& 7.75& 7.15& 6.97\\
PM I00077+6022 & & 1.927582 & 60.381760 & 0.340 &-0.027& 0.1700& -0.41& & & 14.26&P& 8.91& 8.33& 8.05\\
PM I00078+6736 & & 1.961682 & 67.607124 &-0.045 &-0.091& & & & & 12.18&P& 8.35& 7.72& 7.51\\
PM I00081+4757 & & 2.026727 & 47.950695 &-0.119 & 0.003& 0.2190& -0.27& 19.68& 18.91& 12.70&P& 8.52& 8.00& 7.68\\
PM I00084+1725 & GJ 3008 & 2.113679 & 17.424309 &-0.093 &-0.064& & & & 19.24& 10.73&T& 7.81& 7.16& 6.98\\
PM I00088+2050 & GJ 3010 & 2.224675 & 20.840403 &-0.065 &-0.247& 0.0899& -0.28& 21.07& 16.71& 13.90&P& 8.87& 8.26& 8.01\\
PM I00110+0512 & & 2.769255 & 5.208822 & 0.241 & 0.061& & & 22.85& 20.58& 11.55&T& 8.53& 7.88& 7.69\\
PM I00113+5837 & & 2.841032 & 58.617561 & 0.056 & 0.029& & & & & 11.21&T& 8.02& 7.31& 7.13\\
PM I00118+2259 & & 2.970996 & 22.984573 & 0.142 &-0.221& 0.4110& 0.28& & 22.37& 13.09&P& 8.86& 8.31& 7.99\\
PM I00125+2142En& & 3.139604 & 21.713478 & 0.189 &-0.290& & & & & 11.67&T& 8.84& 8.28& 8.04\\
PM I00131+7023 & & 3.298130 & 70.398003 & 0.045 & 0.139& & & & 19.93& 11.37&T& 8.26& 7.59& 7.39
\enddata
\tablenotetext{a}{Designation in the Third Catalog of Nearby Stars \citep{gj1991}}
\tablenotetext{b}{X-ray flux and hardness ratio from the ROSAT all-sky
points source catalog \citep{Voges.etal.1999,Voges.etal.2000}.}
\tablenotetext{c}{Far-UV and near-UV magnitudes in the GALEX fifth data release.}
\tablenotetext{d}{Infrared magnitudes from the Two Micron All-Sky
Survey \citep{Cutri.etal.2003}.}
\end{deluxetable*}
Spectral classification and analysis for a significant fraction of the
low-mass stars in the CNS3 was performed as part of the Palomar-MSU
spectroscopic survey \citep{Reid1995,Hawley1996}, hereafter PMSU. The
survey has notably provided formal spectral classification for 1,971
of the fainter CNS3 stars, confirming 1,648 of them to be nearby M
dwarfs. More recent spectroscopic follow-up surveys have mainly
focused on candidate nearby stars missing from the CNS3. Very high
proper motion stars from the Luyten catalogs
\citep{GizisReid.1997,ReidGizis.2005}, or stars discovered in the more
recent proper motion surveys
\citet{Scholz2002,Lepine2003,Scholz2005,Reyle2006} have thus been
targeted. Most notably, the ``Meet the Cool Neighbours'' program
(hereafter MCN) has determined spectral subtypes for several hundred M
dwarfs identified from the Luyten catalogs but not listed in the CNS3
\citep{CruzReid.2002,Cruzetal.2003,Reidetal.2003,Reidetal.2004,Cruzetal.2007,Reid2007}. As
with the other more recent surveys, the MCN program placed an emphasis
on the identification and classification of nearby, very-cool M
dwarfs, most of which are however relatively faint and unsuitable for
exoplanet surveys. It should be noted that while large numbers of M
dwarfs have also been identified and classified as part of the
spectroscopic follow-up program of the Sloan Digital Sky Survey
\citep{Bochanski2005,Bochanski2010,West2011}, most of them are
relatively distant sources, and generally too faint for exoplanet
surveys.
This is why a significant fraction of the bright M dwarf candidates
published in \citet{LepineGaidos.2011} had no available spectroscopic
data at the time of release. In order to assemble a comprehensive
database of M dwarfs targets suitable for exoplanet survey programs,
we are now conducting a spectroscopic follow-up survey of the
brightest M dwarf candidates from \citep{LepineGaidos.2011}. Our goal
is to provide a uniform catalog of spectroscopic measurements to
confirm the M dwarf classification, and initiate detailed studies of
their physical properties, as well as the tayloring of exoplanet
searches. In this paper, we present the first results of our survey,
which provides data for the 1,564 brightest M dwarf candidates north
of the celestial equator, with apparent near-infrared magnitudes
J$<$9. Observations are described in \S2. Our spectral classification
techniques are described in \S3, and our model fitting and effective
temperature determinations are given in \S4. Metallicity measurements
are presented in \S5. Activity diagnostics are presented and analyzed
in \S6. A kinematic study informed by our metallicity and activity
measurements is presented in \S7, followed by discussion and
conclusions in \S8.
\section{Spectroscopic observations}
\subsection{Target selection}
Targets for the follow-up spectroscopic program were selected from the
catalog of 8,889 bright M dwarfs of \citet{LepineGaidos.2011}. All
stars are selected from the SUPERBLINK catalog of stars with proper
motions $\mu>40$ mas yr$^{-1}$. Stars are identified as probable M
dwarfs based on various color and reduced proper motions cuts; all
selected candidates have, e.g., optical-to-infrared colors
$V-J>2.7$. The low proper motion limit of the SUPERBLINK catalog
excludes nearly all background red giants. The low proper motion
limit however also results in a kinematic bias, whose effects are
discussed in \S2.3 below.
While some astrometric and photometric data have already been compiled
for all the stars, most lack formal spectral classification. Spectral
subtypes have been estimated in \citet{LepineGaidos.2011} only based on a
calibrated relationship between M subtype and $V-J$ color. However,
the $V$ magnitudes of many SUPERBLINK catalog stars are based on
photographic measurements (from POSS-II plates); the resulting $V-J$
colors have relatively low accuracy and are sometimes
unreliable. Besides from affecting spectral type estimates, unreliable
colors can cause contamination of our sample of M dwarf candidates by
bluer G and K dwarfs, which would have otherwise failed the color
cut. These are strong arguments for performing systematic spectroscopic
follow-up observations, to provide reliable spectral typing and filter
out G/K dwarfs (or any remaining M giant contaminants).
A subsample of the brightest of the M dwarfs candidates, with apparent
infrared magnitude J$<9$ was assembled for the first phase of this
survey. We also restricted the sample to stars north of the celestial
equator. This initial list contains a total of 1,564 candidates. All
stars were indiscriminately targeted for follow-up observations,
whether or not they already had well-documented spectra. This would
ensure completeness and uniformity, and allows comparison of our
sample with previous surveys. In particular, our target list
includes M dwarf classification standards from \citet{KHM91} which
provide a solid reference for our spectral classification. The list
includes 557 stars that were previously observed in the PMSU
spectroscopic survey, and 161 that were obseved and classified as part
of the MCN spectroscopic program (including 82 stars observed in both
the PMSU and MCN).
Of the 1,564 M dwarf candidates, we found that 286 had been observed
at the MDM observatory by one of us (SL) prior to November 2008, as
part of a separate spectroscopic follow-up survey of very nearby
(d$<$20pc) stars \citep{Alpert2011}. The remaining targets were
distributed between our observing teams at the MDM Observatory
(hereafter MDM) and University of Hawaii 2.2-meter Telescope
(hereafter UH), with the MDM team in charge of higher declination
targets ($\delta>30$) and the UH team in charge of the lower
declination range ($0<\delta<30$). In the end we obtained spectra for
all 1,564 stars from the initial target list.
To check for any possible systematic differences arising from using
different telescopes and instruments, we observed 146 stars at both
MDM and UH. We call this subset the ``inter-observatory
subset''. Observations were obtained at different times at the two
observatories. Data were processed in the same manner as the rest of
the sample.
\begin{figure}
\vspace{-0.4cm}
\hspace{-0.8cm}
\includegraphics[scale=0.47]{f1.eps}
\caption{Examples of spectra collected at the MDM Obsevatory with
either the MkIII or CCDS low-resolution spectrographs, and using the
McGraw-Hill 1.3-meter or Hiltner 2.4-meter telescope. All spectra
covered the 6,200\AA-8,700\AA\ wavelength range with a resolution of
1.8\AA-2.4\AA\ per pixel, and a signal-to-noise ratio
20$<$S/N$<$30.\label{spec_mdm}}
\end{figure}
\begin{figure}
\vspace{-0.4cm}
\hspace{-0.8cm}
\includegraphics[scale=0.47]{f2.eps}
\caption{Examples of spectra collected at the University of Hawaii
2.2-meter telescope with the SNIFs spectrograph. Observations
covered the 5,200\AA-9,800\AA\ wavelength regime with a spectral
resolution R$\simeq$2000, and a signal-to-noise ratio
40$<$S/N$<$50.\label{spec_uh}}
\end{figure}
The full list of observed stars is presented in
Table~\ref{table_photo}. We used the standard SUPERBLINK catalog name
as primary designation, however we also include the more widely used
designations (GJ, Gl, and Wo numbers) for the 682 stars listed in the
CNS3 \citep{gj1991}. The CNS3 stars are often well-studied objects,
with abundant data from the literature; a majority of them (580) were
classified as part of the PSMU survey or MCN spectroscopic
program. Another 56 stars on our table which are not in the CNS3 were
however classified as part of the PMSU survey or the MCN program. The
remaining 821 stars are not in the CNS3, and were also not classified
as part of the PMSU survey or MCN progra; these are new
identifications for the most part, and little data existed about them
until now.
Table~\ref{table_photo} lists coordinates and proper motion vectors
for all the stars, along with astrometric parallaxes whenever
available from the literature. The table also lists X-ray source
counts from the ROSAT all-sky point source catalogs
\citep{Voges.etal.1999,Voges.etal.2000}, far-UV ($FUV$) and near-UV
($NUV$) magnitudes from the GALEX fifth data release, optical $V$
magnitude from the SUPERBLINK catalog \citep{LepineShara.2005}, and
infrared $J$, $H$, and $K_s$ magnitudes from 2MASS
\citep{Cutri.etal.2003}. More details on how those data were compiled
can be found in \citet{LepineGaidos.2011}. The optical ($V$ band)
magnitudes listed in Table~\ref{table_photo} come from two sources
with different levels of accuracy and reliability. For 919 stars in
Table~\ref{table_photo}, generally the brightest ones, the $V$
magnitudes come from the Hipparcos and Tycho-2 catalogs. These are
generally more reliable with typical errors smaller than $\pm$0.1
magnitude; those stars are flagged ``T'' in
Table~\ref{table_photo}. The other 645 objects have their $V$
magnitudes estimated from POSS-I and/or POSS-II photographic
magnitudes as prescribed in \citet{Lepine2005}. Photographic
magnitudes of relatively bright stars often suffer from large errors
at the $\sim$0.5 magnitude level or more, in part due to photographic
saturation; those stars are labeled ``P'' in Table~\ref{table_photo}.
\begin{deluxetable}{ccccc}
\tabletypesize{\scriptsize}
\tablecolumns{6}
\tablewidth{0pt}
\tablecaption{Definition of Spectral Indices.\label{tab:si}}
\tablehead{
\colhead{Index} &
\colhead{Numerator} &
\colhead{Denominator} &
\colhead{Reference}
}
\startdata
CaH2 & 6814-6846 & 7042-7046 & \citet{Reid1995} \\
CaH3 & 6960-6990 & 7042-7046 & \citet{Reid1995} \\
TiO5 & 7126-7135 & 7042-7046 & \citet{Reid1995} \\
VO1 & 7430-7470 & 7550-7570 & \citet{Hawley2002} \\
TiO6 & 7550-7570 & 7745-7765 & \citep{Lepine2003} \\
VO2 & 7920-7960 & 8130-8150 & \citep{Lepine2003}
\enddata
\end{deluxetable}
\subsection{Observations}
Spectra were collected at the MDM observatory in a series of 22
observing runs scheduled between June, 2002 and April, 2012. Most of
the spectra were collected at the McGraw-Hill 1.3-meter telescope, but
a number were obtained at the neighboring Hiltner 2.4-meter
telescope. Two different spectrographs were used: the MkIII
spectrograph, and the CCDS spectrograph. Both are facility
instruments which provide low- to medium-resolution spectroscopy in
the optical regime. Their operation at either 1.3-meter or 2.4-meter
telescopes is identical. Data were collected in slit spectroscopy
mode, with an effective slit width of 1.0\arcsec to 1.5\arcsec. The
MkIII spectrograph was used with two different gratings: the 300 l/mm
grating blazed at 8000\AA, providing a spectral resolution
R$\simeq$2000, and the 600 l/mm grating blazed at 5800\AA, which
provides R$\simeq$4000. The two gratings were used with either one of
two thick-chip CCD cameras ({\it Wilbur} and {\it Nellie}) both having
negligible fringing in the red. Internal flats were used to calibrate
the CCD response. Arc lamp spectra of Ne, Ar, and Xe provided
wavelength calibration, and were obtained for every pointing of the
telescope to account for flexure in the system. The spectrophotometric
standard stars Feige 110, Feige 66, and Feige 34 \citep{Oke1991} were
observed on a regular basis to provide spectrophotometric
calibration. Integration times varied depending on seeing, telescope
aperture, and target brightess, but were typically in the 30 seconds
to 300 seconds range. Between 25 and 55 stars were observed on a
typical night. Spectra for a total of 901 bright M dwarf targets were
collected at MDM.
Additional spectra were obtained with the SuperNova Integral Field
Spectrograph \citep[SNIFS; ][]{Lantz2004} on the University of Hawaii
2.2~m telescope on Mauna Kea between February 2009 and November 2012.
SNIFS has separate but overlapping blue (3200-5600\AA) and red
(5200-10000\AA) spectrograph channels, along with an imaging channel,
mounted behind a common shutter. The spectral resolution is
$\sim$1000 in the blue channel, and $\sim$1300 in the red channel; the
spatial resolution of the 225-lenslet array is 0.4\arcsec. SNIFS
operates in a semi-automated mode, acquiring acquisition images to
center the target on the lenslet array, and bias images and
calibration lamp spectra before and after each target spectrum. Both
twilight and dome flats were also obtained every night. Integration
times depended on $J$ magnitude but were 54~s for the faintest ($J=9$)
targets. Up to 75 target spectra were obtained in one night. Spectra
for 655 bright M dwarf targets were collected at UH with SNIFS.
\begin{deluxetable*}{lcrrrrrrrrcrrr}
\tabletypesize{\scriptsize}
\tablecolumns{14}
\tablewidth{0pt}
\tablecaption{Survey stars: spectroscopic data \label{table_spectro}}
\tablehead{
\colhead{Star name} &
\colhead{Observatory} &
\colhead{Julian Date} &
\colhead{CaH2$_{c}$\tablenotemark{a}} &
\colhead{CaH3$_{c}$} &
\colhead{TiO5$_{c}$} &
\colhead{VO1$_{c}$} &
\colhead{TiO6$_{c}$} &
\colhead{VO2$_{c}$} &
\colhead{Sp.Ty.\tablenotemark{b}} &
\colhead{Sp.Ty.} &
\colhead{$\zeta$} &
\colhead{log g\tablenotemark{d}} &
\colhead{$T_{eff}$\tablenotemark{d}} \\
\colhead{} &
\colhead{} &
\colhead{2,450,000+} &
\colhead{} &
\colhead{} &
\colhead{} &
\colhead{} &
\colhead{} &
\colhead{} &
\colhead{index} &
\colhead{adopted} &
\colhead{} &
\colhead{} &
\colhead{K}
}
\startdata
PM I00006+1829 & MDM & 4791.75 &\nodata&\nodata&\nodata&\nodata&\nodata&\nodata&\nodata& G/K&\nodata&\nodata&\nodata\\
PM I00012+1358S & UH22 & 5791.05 & 0.706 & 0.864 & 0.788 & 0.967 & 0.944 & 0.970 & 0.14 & M0.0& 1.08 & 5.0 & 3790 \\
PM I00033+0441 & UH22 & 5791.05 & 0.580 & 0.797 & 0.679 & 0.959 & 0.888 & 0.939 & 1.38 & M1.5& 0.93 & 4.5 & 3510 \\
PM I00051+4547 & MDM & 5095.80 & 0.603 & 0.824 & 0.664 & 0.956 & 0.911 & 1.014 & 1.10 & M1.0& 1.10 & 4.5 & 3560 \\
PM I00051+7406 & MDM & 5812.87 &\nodata&\nodata&\nodata&\nodata&\nodata&\nodata&\nodata& G/K&\nodata&\nodata&\nodata\\
PM I00077+6022 & MDM & 5838.82 & 0.364 & 0.620 & 0.392 & 0.905 & 0.630 & 0.798 & 4.60 & M4.5& 0.90 & 5.0 & 3140 \\
PM I00078+6736 & MDM & 5099.83 & 0.534 & 0.789 & 0.623 & 0.905 & 0.806 & 0.929 & 2.03 & M2.0& 0.96 & 5.0 & 3500 \\
PM I00081+4757 & MDM & 5098.84 & 0.410 & 0.700 & 0.420 & 0.871 & 0.684 & 0.814 & 3.80 & M4.0& 1.01 & 5.0 & 3280 \\
PM I00084+1725 & UH22 & 5791.06 & 0.671 & 0.842 & 0.785 & 0.970 & 0.947 & 0.970 & 0.34 & M0.5& 0.91 & 4.5 & 3600 \\
PM I00088+2050 & MDM & 4413.72 & 0.372 & 0.646 & 0.356 & 0.893 & 0.602 & 0.793 & 4.58 & M4.5& 0.99 & 5.0 & 3130 \\
PM I00110+0512 & UH22 & 5050.03 & 0.653 & 0.839 & 0.706 & 0.962 & 0.893 & 0.925 & 0.86 & M1.0& 1.16 & 4.5 & 3660 \\
PM I00113+5837 & MDM & 5811.90 &\nodata&\nodata&\nodata&\nodata&\nodata&\nodata&\nodata& G/K&\nodata&\nodata&\nodata\\
PM I00118+2259 & UH22 & 5050.03 & 0.439 & 0.729 & 0.424 & 0.923 & 0.694 & 0.798 & 3.50 & M3.5& 1.10 & 4.5 & 3260 \\
PM I00125+2142En& UH22 & 5791.06 & 0.721 & 0.865 & 0.851 & 0.973 & 0.967 & 0.988 & -0.18 & M0.0& 0.80 & 4.5 & 3690 \\
PM I00131+7023 & MDM & 5812.89 & 0.643 & 0.816 & 0.754 & 0.973 & 0.883 & 1.038 & 0.93 & M1.0& 0.88 & 4.5 & 3570
\enddata
\tablenotetext{a}{All spectral indices are corrected for instrumental effects, see Section 3.2.}
\tablenotetext{b}{Numerical spectral subtype M evaluated from the corrected spectral band indices (not-rounded).}
\tablenotetext{c}{H$\alpha$ spectral index, comparable to equivalent width.}
\tablenotetext{d}{Gravity and effective temperature from PHOENIX model fits.}
\end{deluxetable*}
Spectroscopic data and results are summarized in
Table~\ref{table_spectro}. SUPERBLINK names are repeated in the first
column and provide a means to match with the entries in
Table~\ref{table_photo}. The second and third columns list the
observatory and Julian date of the observations. The various
spectroscopic measurements whose values are listed in the remaining
columns are described in detail in the sections below.
\subsection{Reduction}
\begin{figure}
\epsscale{1.15}
\plotone{f3.eps}
\caption{Histogram of the $V-J$ color distribution of survey stars
with spectral morphologies inconsistent with red dwarf of subtype K5
and later (in green). The distribution of $V-J$ colors from the full
sample is shown in red. Stars with spectra inconsistent with red
dwarfs are close to the blue edge of the survey, which indicates
that they are most probably contaminants of the color-magnitude
selection, and not the result of mis-acquisition at the
telescope. \label{contaminant}}
\end{figure}
\subsubsection{MDM data}
Spectral reduction of the MDM spectra was performed using the CCDPROC
and DOSLIT packages in IRAF. Reduction included bias and flatfield
correction, removal of the sky background, aperture extraction, and
wavelength calibration. The spectra were also extinction-corrected and
flux-calibrated based on the measurements obtained from the
spectrophotometric standards. We did not attempt to remove telluric
absorption lines from the spectra. Many spectra were
collected on humid nights or with light cirrus cover, which resulted
in variable telluric features. However, telluric features generally do
not affect standard spectral classification or the measurement of
spectral band indices, since all the spectral indices and primary
classification features avoid regions with telluric absorption.
A more common problem at the MDM observatory was slit loss
from atmospheric differential refraction. Although this problem could
have been avoided by the use of a wider slit, the concomitant loss of
spectral resolution was deemed more detrimental to our science
goals. Instead, stars were observed as close to the meridian as
observational constraints allowed. In some cases, however, stars were
observed up to $\pm$2 hours from the meridian, resulting in noticeable
slit losses. Fluctuations in the seeing, which often exceeded the slit
width, played a role as well. As a result, the spectrophotometric
calibration was not always reliable, since the standards were only
observed once per night to maximize survey efficiency. Flux
recalibration was
therefore performed using the following procedure. Spectral indices
were measured and the spectra were classified using the classification
method outlined below (\S 3.1). The spectra were then compared to
classification templates assembled from M dwarf spectra from the Sloan
Digital Sky Survey (SDSS) spectroscopic database. Each spectrum
was divided by the classification template of the same alleged
spectral subtype. In many cases, the quotients yielded a flat
function, indicating that the spectrophotometric calibration was
acceptable. Other quotients yielded residuals consistent with first or
second order polynomials spanning the entire wavelength range, and
indicating problems in the spectrophotometric calibration. A
second-order polynomial was fit through the quotient spectra, and the
original spectrum was normalized by this fit, correcting for
calibration errors. Spectral indices were then measured again, and the
spectra reclassified; this yielded changes by 0.5 to 1.0 subtypes for
$\approx20\%$ of the stars (no changes for the rest). The
re-normalization was then performed again using the revised spectral
subtypes, and the procedure repeated until convergence for all stars.
Finally, all the spectra were wavelength-shifted to the rest frames
of their emitting stars. This was done by cross-correlating each
spectrum with the SDSS template of the corresponding spectral
subtype. Spectral indices were again re-measured, and the stars
re-classified. Any change in the spectral subtype prompted a repeat of
the flux-recalibration procedure outlined above, and the
cross-correlation procedure was repeated using the revised spectral
template until converegence.
A sequence of MDM spectra is displayed in Figure~\ref{spec_mdm},
which illustrates the wavelength regime and typical quality. Note the
telluric absorption features near 6,850\AA, 7,600\AA, and
8,200\AA. Signal-to-noise ratio is generally in the 30$<$S/N$<$50
range near 7500\AA.
\begin{figure*}
\vspace{-6.5cm}
\hspace{0.3cm}
\includegraphics[scale=0.87]{f4.eps}
\caption{Comparison between our spectral band index measurements for a
subset of 484 stars observed at MDM and UH, and the band index
measurements for the same stars as reported in the Palomar-MSU
spectroscopic survey of \citet{Reid1995}. Small but systematic offsets
are observed, which are explained by differences in spectral
resolution and spectrophotometric calibration between the different
observatories. These offsets demonstrate that the spectral indices
are instrument-dependent, but that the measurements can be corrected
by observing large subsets of stars at the different
obervatories. The red segments show the fits to the offsets, which
are used to calibrate corrections for each observatory, here using
the Palomar-MSU measurements as the standard.\label{pmscomp}}
\end{figure*}
\subsubsection{SNIFS data}
SNIFS data processing was performed with the SNIFS data reduction
pipeline, which is described in detail in \citet{Bacon:2001} and
\citet{Aldering:2006}. After standard CCD preprocessing (dark, bias,
and flat-field corrections), data were assembled into red and blue 3D
data cubes. The data cubes were cleaned for cosmic rays and bad
pixels, wavelength-calibrated using arc lamp exposures acquired
immediately after the science exposures, and spectrospatially
flat-fielded, using continuum lamp exposures obtained during the same
night. The data cubes were then sky-subtracted, and the 1D spectra were
extracted using a semi-analytic PSF model. We applied corrections to
the 1D spectra for instrument response, airmass, and telluric lines
based on observations of the Feige 66, Feige 110, BD+284211, or
BD+174708 standard stars \citep{Oke1991}. Because the SNIFS spectra
are from an integral field spectrograph, operating without a slit,
their spectrophotometry is significantly more reliable than the
slit-spectra obtained at MDM. In fact, it is possible to perform
synthetic photometry by convolving with the proper filter response.
As with the MDM data, SNIFS spectra were shifted to the rest frames of
their emitting stars by cross-correlation to SDSS templates
\citep{Bochanski2007} of the corresponding spectral subtype. Spectral
indices were re-measured and the stars re-classified. This process was
repeated if there was a change in the spectral subtype determination.
A sequence of UH SNIFS spectra are displayed in Figure~\ref{spec_uh},
which show the wavelength range and typical data
quality. Signal-to-noise ratio is generally S/N$\approx$100 near
7500\AA.
\section{Spectral classification}
\subsection{Visual identification of contaminants}
We first examined all the spectra by eye to confirm the presence of
morphological features consistent with red dwarfs of spectral subtype
K5 and later. The main diagnostic was the detection of broad CaH and
TiO moleculars band near 7000\AA. Of the 1564 stars observed, 1408 were
found to have clear evidence of CaH and TiO. The remaining 156 stars
did not show those molecular features clearly, within the noise
limit, and were therefore identified as probable contaminants in the
target selection algorithm.
Most of these contaminants appeared to be early to mid-type K dwarfs,
with a few looking like G dwarfs affected by interstellar reddening. A
number of stars also displayed carbon features consistent with low gravity
objects, most probably K giants. We suspect that many of the G and
K dwarfs have inaccurate V-band magnitudes, making them appear redder
than they really are, which would explain their inclusion in our
color-selected sample. Interstellar reddening would also explain the
inclusion of more distant G dwarfs in our sample, due to their redder
colors. An alternate explanation, however, might be that the targets
were mis-acquired in the course of the survey, and that the spectra
represent random field stars. Indeed the very large proper motion of
the sources sometimes makes them difficult to identify at the
telescope, as they often have moved significantly from their positions
on finder charts. Stars in crowded field are particularly susceptible
to this effect. To verify this hypothesis, we compared the $V-J$ color
distribution of the contaminants to the distribution of the full
survey sample (Figure~\ref{contaminant}, top panel); the fraction of
contaminant stars in each color bin is also shown (bottom panel). The
two distributions are significantly different, with the contaminants
being dominated by relatively blue stars, and their fraction quickly
drops as $V-J$ increases. We can only conclude that the contaminants
are not mis-acquired stars, otherwise one would expect the two
distributions to be statistically equivalent. Rather, the majority of
the contaminants must have been properly acquired and are simply
moderately red FGK stars that slipped into the sample in the
photometric/proper motion selection, as suggested first.
\begin{figure*}
\vspace{-6.5cm}
\hspace{0.3cm}
\includegraphics[scale=0.87]{f5.eps}
\caption{Comparison between the {\em corrected} spectral band index
measurements from MDM and the band index measurements of the same
stars observed from UH. Observatory-specific corrections to the
CaH2, CaH3, and TiO5 indices (see Figure~\ref{pmscomp}) bring the
MDM and UH values in close agreement. Small offsets are however
observed for the TiO6, VO1, and VO2 indices, which we could not
calibrate against Palomar-MSU survey stars. The horizontal red lines
show the adopted offsets which are used to correct the MDM values to
bring them in line with the UH ones. Offsets are again believed to
be due to inter-observatory differences in the spectral resolution
and spectrophotometric calibration.\label{idx_comp}}
\end{figure*}
We find an overall contamination rate of $\simeq$10\% in our survey,
although most of the contamination occurs among stars with relatively
blue colors. The contamination rate is $\simeq$26\% for red dwarf
candidates in the $2.7<V-J<3.0$ color range, but this rate drops to
$\simeq$8\% for candidates with $3.0<V-J<3.3$, and becomes negligible
($<1\%$) in the redder $$V-J>3.3$$ candidates. The 156 stars
identified as contaminants are included in Table~\ref{table_photo} and
Table~\ref{table_spectro} for completeness and future
verification. Spectroscopic measurements for these stars, such as band
indices, subtypes, and effective temperatures, are however left blank.
\subsection{Classification by spectral band indices}
\subsubsection{M dwarf classification from molecular bandstrengths}
The spectra of M dwarfs are dominated by molecular bands from metal
oxides (mainly TiO, VO), metal hydrides (CaH, CrH, FeH), and metal
hydroxides (CaOH). The most prominent of these in the optical-red
wavelength range (5000\AA-9,000\AA) are the bands from titanium oxide
(TiO) and calcium hydride (CaH). The resulting opacities from those
broad moleciular bands significantly affect the broadband colors and
spectral energy distribution of M dwarfs
\citep{Jones1968,Allard2000,Kraw2000}. Early atmospheric models of M
dwarfs showed that the strength of the TiO and CaH bands depends on
effective temperature, but also on surface gravity and metal
abundances \citep{Mould.1976}.
Molecular bands have historically been the defining diagnostic and
classification features of M dwarfs. For stars that have settled on
the main sequence, one can assume that the surface gravity is entirely
constrained by the mass and chemical composition. Leaving only the
effective temperature and chemical abundances as general parameters in
the classification and/or spectroscopic modeling. For local disk stars
of solar metallicity, a classification system representing an
effective temperature sequence can thus be established based on
molecular bandstrentghs.
The detection and measurement of TiO and CaH molecular bands thus
forms the basis for the M dwarf classification system
\citep{JoyAbt1974}. Molecular bands become detectable starting at
spectral subtype K5 and K7, the latest subtypes for K dwarfs (there
are no K6, K8, or K9 subtypes). The increasing strength of the
molecular bands then defines a sequence running from M0 to M9. The
strength of both TiO and CaH molecular bands reach a maximum around
$T_{eff}\simeq$2700K. There is a turnaround in the correlation below
this point, and molecular bands become progressively weaker at lower
temperatures until they vanish \citep{CruzReid.2002}. The reversal and
weakening is thought to be due to the condensation of molecules into
dust, and their settling below the photosphere \citep{JonesTsuji1997}.
\begin{deluxetable*}{lcccccc}
\tabletypesize{\scriptsize}
\tablecolumns{7}
\tablewidth{0pt}
\tablecaption{Coefficients of the spectral index corrections, by
observatory.\label{tab:sic}}
\tablehead{
\colhead{{\it IDX}} &
\colhead{$A_{IDX:{\rm PMSU}}$} &
\colhead{$B_{IDX:{\rm PMSU}}$} &
\colhead{$A_{IDX:{\rm MDM}}$} &
\colhead{$B_{IDX:{\rm MDM}}$} &
\colhead{$A_{IDX:{\rm UH}}$} &
\colhead{$B_{IDX:{\rm UH}}$}
}
\startdata
CaH2 & 1.00 & 0.00 & 0.95 & 0.011 & 0.92 & 0.004 \\
CaH3 & 1.00 & 0.00 & 0.90 & 0.070 & 1.00 & -0.028 \\
TiO5 & 1.00 & 0.00 & 1.00 & 0.000 & 1.06 & -0.063 \\
VO1 & \nodata& \nodata& 1.00 & 0.040 & 1.00 & 0.000 \\
TiO6 & \nodata& \nodata& 1.00 & -0.021 & 1.00 & 0.000 \\
VO2 & \nodata& \nodata& 1.00 & 0.005 & 1.00 & 0.000
\enddata
\end{deluxetable*}
Sequences of classification standards were compiled in \citep{KHM91},
which identified the main molecular bands in the yellow-red spectral
regime, where TiO and CaH bandheads are most prominent. To better
quantify the classification system, a number of ``band indices'' were
defined by \citet{Reid1995}, which measure the ratio between on-band and
off-band flux, for various molecular bandheads. Calibration of these
band indices against classification standards provide a means to
objectively assign subtypes based on spectroscopic
measurements. Originally, these band indices measured the strengths of
various features near $7,000\AA$, where the most prominent CaH and TiO
features are found in early-type M dwarfs. These bands, however,
become saturated in late-type M dwarfs, which makes their use
problematic in later dwarfs. There is a VO band near $7,000$\AA,
located just between the main CaH and TiO features, which becomes
prominent only at later subtypes; and band index measuring this
feature was introduced as a primary diagnosis for the so-called
``ultra-cool'' M dwarfs, and provided a classification scale for
subtypes M7-M9 \citep{KHS95}. Additional band indices associated with
TiO and VO bands in the 7500\AA-9000\AA\ range, where molecular
absorption develops at later subtypes, were introduced as secondary
classification features \citep{Lepine2003}. These form the basis of
current classification methods based on optical-red spectroscopy.
Nearby low-mass stars associated with the local halo population have
long been known to show peculiar banstrength ratios, and in particular
to have weak TiO compared to CaH
\citep{Mould.1976,Mould_McElroy.1978}. A system to identify and
classify the metal-poor M subdwarfs based on the strength and ratio of
CaH and TiO was introduced by \citet{Gizis1997}, and expanded by
\citet{Lepine2003b} and \citet{Lepine2007}, as the spectroscopic
census of M subdwarfs grew larger. In this system, the CaH
bandstrengths are used as a proxy of $T_eff$ and determine spectral
subtypes, while the TiO/CaH band ratio is used to evaluate
metallicity. For that purpose, the $\zeta_{TiO/CaH}$ parameter, which
is a function of the TiO/CaH band ratio, was introduced by
\citet{Lepine2007} as a possible proxy for metallicity, and a
tentative calibration with [Fe/H] was presented by \citet{Woolf2009}.
One of the main issues in the current M dwarf classification scheme,
is that both TiO and CaH bandstrengths are used to determine the
spectral subtype, whereas TiO is now believed to be quite sensitive to
metallicity. This means, e.g., that moderately metal-rich M dwarfs may
be assigned later subtypes than Solar-metallicity ones. In addition,
the classification of young field M dwarfs may be affected by their
lower surface gravities, which also tend to increase TiO
bandstrengths and would thus yield to marginally later subtype
assignments compared with older stars of the same $T_{eff}$. These
caveats must be considered when one uses M dwarf spectral subtypes as
a proxy for surface temperature.
\begin{figure*}
\vspace{-6.5cm}
\hspace{0.3cm}
\includegraphics[scale=0.87]{f6.eps}
\caption{Variation in the corrected spectral indices as a function
of adopted spectral subtypes. The thin black lines show the
adopted, revised calibrations for spectral
classification. Blue circles represent the subset of classification
standards from \citet{KHM91} which we observed; the X-axis values of
those data points are the formal spectral subtypes adopted in
\citet{KHM91}, while the Y-axis values are the spectral index
measurements from our survey. For our own spectral typing scheme, we
calculate the average of the subtype values for the ${\rm
CaH2}_{c}$, ${\rm CaH3}_{c}$, ${\rm TiO5}_{c}$, and ${\rm
TiO6}_{c}$ indices given by the adopted relationships. The two VO
indices have relatively weak leverage on early type stars due to the
shallow slope in the relationships, and are not used for assigning
spectral subtypes of our (mostly early-type) survey
stars.\label{idx_spty}}
\end{figure*}
\subsubsection{Definition and measurement of band indices}
The strength of the TiO, CaH, and VO molecular bands are measured
using spectral band indices. These spectral indices measure the ratio
between the flux in a section of the spectrum affected by molecular
opacity to the flux in a neighboring section of the spectrum minimally
affected by molecular opacity. The latter section defines a
pseudo-continuum of sorts, although M dwarf spectra do not have a
continuum in the classical sense, because their spectral energy
distribution strongly deviates from that of a blackbody, and is
essentially shaped by atomic and molecular line opacities.
We settle on a set of six spectral band indices: CaH2, CaH3, TiO5,
TiO6, TiO7, VO1, and VO2. These band indixes, which we previously used
in \citet{Lepine2003} to classify spectra collected at MDM, measure
the strength of the most prominent bands of CaH, TiO, and VO in the
$6000{\rm \AA}<\lambda<8500{\rm \AA}$ regime. The spectral indices
and are listed in Table~\ref{tab:si} along with their definition. The
CaH2, CaH3, and TiO5 indices are the same as those used in the
Palomar-MSU survey, and were first defined in \citep{Reid1995}. The TiO6,
VO1, and VO2 indices were introduced by \citet{Lepine2003} to better
classify late-type M dwarfs, whose CaH2 and TiO5 indices become
saturated at cooler tempratuers and are not as effective for accurate
spectral classification of late-type M dwarfs. Each spectral band
index is calculated as the ratio of the flux in the spectral region of
interest (numerator) to the flux in the reference region
(denominator), i.e.:
\begin{equation} IDX = \frac{\int_{num} I(\lambda) d\lambda}{\int_{denom}
I(\lambda) d\lambda}
\end{equation}
Because the wavelength range for some indices is relatively narrow
(especially the denominator for CaH2, CaH3, and TiO5) it is important
that the spectra in which they are measured have their wavelengths
calibrated in the rest frame of the star, which is why special care
was made to correct all spectra for any significant redshift/blueshift
(see above).
Because the measured molecular bandheads are relatively sharp, and
because the spectral indices measuring them are defined over
relatively narrow spectral ranges, the index values are potentially
dependent on the spectroscopic resolution, and may thus depend
on the specific instrumental setup used for the observations. In
addition, the index values may be affected by systematic errors in the
spectrophotometric flux calibration, which can also be dependent on
the instrument and/or observatory where the measurements were
made. One way to verify these effects is to compare spectral index
measurements of the same stars obtained at different
observatories. Because of the significant overlap with the Palomar-MSU
survey, we can use those stars as reference sample, and recalibrate
the spectral indices so that they are consistent to those reported in
\citet{Reid1995}.
Our census have 557 stars in common with the PMSU spectroscopic
survey; these stars are all identified with a flag (``P'') in the last
column of Table~\ref{table_spectro}. We identify 206 stars from the
list observed at MDM and 281 stars from the list observed at UH which
have spectral index measurements
reported in the PMSU survey. The differences between our measured
CaH2, CaH3, and TiO5 and those reported in the PSMU catalog are
plotted in Figure~\ref{pmscomp}. Trends and offsets confirm the
existence of systematic errors, possibly due to differences in
resolution and flux calibration. To verify the spectroscopic
resolution hypothesis, we convolved the MDM spectra with a box kernel
5-pixel wide; we found that indeed the MDM indices for the smoothed
spectra had their offsets reduced by 0.01-0.02 units, bringing them
more in line with the PMSU indices. We also observe that the MDM
measurements tend to have a larger scatter than the UH ones; we
suggest that this may be due to spectrophometric calibration issues
with some of the MDM spectra, as discussed in \S2.3.1.
To achieve consistency in the measurements obtained at different
observatories, we adopt the values from the PMSU survey as a standard
of reference, and calculate corrections to the measurements from MDM and
UH by fitting linear relationships to the residuals. A corrections to a
spectral band index is thus applied following the general function:
\begin{equation}
IDX_{c} = A_{IDX:OBS}\ IDX_{OBS} + B_{IDX:OBS} \\
\end{equation}
where $IDX_{OBS}$ represents the measured value of an index at the
observatory $OBS$, and ($A_{IDX:OBS}$,$B_{IDX:OBS}$) are the
coefficients of the transformation from the observed value to the
corrected one ($IDX_{c}$). Hence the corrected values of the indices
CaH2, CaH3 and TiO5 for measurements done at MDM are defined as:
\begin{eqnarray}
{\rm CaH2_{c} = A_{\rm CaH2:MDM} CaH2_{\rm MDM} + B_{\rm CaH2:MDM} } \\
{\rm CaH3_{c} = A_{\rm CaH3:MDM} CaH3_{\rm MDM} + B_{\rm CaH3:MDM} } \\
{\rm TiO5_{c} = A_{\rm TiO5:MDM} TiO5_{\rm MDM} + B_{\rm TiO5:MDM} }
\end{eqnarray}
The measurements from the PSMU survey are used as standards for these
three indices, and we thus have by definition: $A_{\rm
CaH2:PMSU}=A_{\rm CaH3:PMSU}=A_{\rm TiO5:PMSU}=1.0$, $B_{\rm
CaH2:PMSU}=B_{CaH3:PMSU}=B_{\rm TiO5:PMSU}=0.0$.
For OBS=MDM and OBS=UH, the adopted correction coefficients are listed
in Table~\ref{tab:sic}. The corresponding linear relationships are
shown as red segments in Figure~\ref{pmscomp}.
To verify the consistency of the corrected spectral band index values,
we compare the corrected values for the stars observed at both MDM and
UH (the inter-observatory subset). The differences are shown in
Figure~\ref{idx_comp}. We find the corrected values CaH2$_{c}$,
CaH3$_{c}$, and TiO5$_{c}$ to be in good agreement, with no
significant offsets beyond what is expected from measurement
errors. The corrected values of all three spectral indices are listed
in Table~\ref{table_spectro}.
The TiO6, VO1, and VO2 spectral index mesurements from MDM and UH are
also compared in Figure~\ref{idx_comp}. Those were not measured in the
PMSU survey, since the indices were introduced later
\citep{Lepine2003}, and thus are displayed here without any
correction. Small but significant offsets between the MDM and UH
values again suggest systematic errors due to differences in spectral
resolution and flux calibration. This time we adopt the UH
measurements as fiducials, and determine corrections to be applied to
the MDM data. The corrections are listed in Table~\ref{tab:sic} and
the corresponding linear relationships are displayed in
Figure~\ref{idx_comp} as red segments. The corrected values of the
three spectral indices are also listed in Table~\ref{table_spectro}.
The scatter between the MDM and UH values, after correction, as well
as the scatter between the UH/MDM and PMSU values, provide an
estimate of the measurment accuracy for these spectral
indices. Excluding a few outliers, the mean scatter is $\approx$0.02
units (1$\sigma$) for the CaH2, CaH3, TiO5, TiO6, and VO1 indices, and
$\approx$0.04 units for VO2. This assumes that M stars do not show any
significant changes in their spectral morphology over time, and that
the spectral indices should thus not be variable.
\subsubsection{Spectral subtype assignments for K/M dwarfs}
Because of the correlation between spectral subtype and the depth of
the molecular bands, it is possible to use the values of the spectral
band indices to estimate spectral subtypes. This only requires a
calibration of the relationship between spectral index values and the
spectral subtypes, in a set of stars which were classified by other
means, e.g., classification standards. The system adopted in this
paper uses the spectral indices listed in Table~\ref{tab:si}, and
follows the methodology outlined in \citep{Gizis1997} and
\citep{Lepine2003}. Relationships are calibrated for each spectral
index, and spectral subtypes are calculated from the mean values
obtained from all relevant/available spectral indices. The mean values
are then be rounded to the nearest half integer, to provide formal
subtyping with half-integer resolution. The system is extended to
late-K dwarfs as well: an ``M subtype'' with a value $<0.0$ signifies
that star is a late-K dwarf: the star is classified as K7 for an index
value $\approx-1.0$ and as K5 for an index value $\approx-2.0$ (note:
there is no K6 subtype for dwarf stars, and K7 is the subtype
immediately preceding M0).
The original spectral-index classification method for M
dwarfs/subdwarfs is based on a relationship between subtype and with
the CaH2 index, which measures one of the most prominent band at all
spectral subtypes, and notably displays the deepest bandhead in
metal-poor M subdwarfs \citet{Gizis1997,Lepine2003}. The original
relationship is: $\left[ SpTy \right]_{CaH2} = 10.71 - 20.63
\ {\rm CaH2} + 7.91 \ \left( {\rm CaH2} \right)^2$. To verify this
relationship, we estimated spectral subtypes from our corrected
indices CaH2$_{c}$ for 16 spectroscopic calibration standards from
\citet{KHM91}, which were observed as part of our survey, and span a
range of spectral subtypes from K7.0 to M6.0. We found small but
significant differences in our estimated spectral subtypes and the
values formally assigned by \citet{KHM91}; subtypes estimated from
the \citet{Gizis1997} relationship tend to systematically
underestimate the standard subtypes by $\approx0.5$ units for stars
later than M3. To improve on the index classification method, we
performed a $\chi^2$ polynomial fit to recalibrate the relationship,
obtaining:
\begin{equation}
\left[ \rm SpTy \right]_{\rm CaH2} = 11.50 - 21.71 \ {\rm CaH2}_{c} + 7.99
\ \left( {\rm CaH2} \right)^2
\end{equation}
which does correct for the observed offsets at later types. Using this
this relatioship as a starting point, and guided by the formal
spectral subtype from the classification standards, we performed
additional $\chi^2$ polynomial fits to calibrate an index-subtype
relationships for CaH3:
\begin{equation}
\left[ \rm SpTy \right]_{\rm CaH3} = 18.80 - 21.68 \ {\rm CaH3}_{c}
\end{equation}
Where the corrected values of the spectral bands indices (see Eqs.2-5)
are used. The relationships are slightly different from those quoted in
\citet{Gizis1997} and \citet{Lepine2003} but are internally
consistent to each other, whereas an application of the older
relationships to our corrected band index measurements would yield
internal inconsistencies, with subtype difference up to 1 spectral
subtype between the relationships.
The ratio of oxides (TiO, VO) to hydrides (CaH,CrH,FeH) in M dwarfs is
known to vary significantly with metallicity
\citep{Gizis1997,Lepine2007}. In the metal-poor M subdwarfs, it is the
oxides bands that appear to be weaker, while hydride bands remain
relatively strong (in the most metal-poor ultrasubdwarfs, or usdM,
the TiO bands are almost undetectable). Therefore it makes sense to
rely more on the CaH band as the primary subtype/temperature
calibrator. The same $[\rm SpTy]_{\rm CaH2}$ and $[\rm SpTy]_{\rm
CaH3}$ relationships should be used to determine spectral subtypes
at all metallicity classes (i.e. in M subdwarfs as well as in M
dwarfs).
Because the TiO and VO bands are also strong in the metal-rich M
dwarfs, it is still useful to include these bands as secondary
indicators, to refine the spectral classification. In the late-type M
dwarfs, in fact, the CaH bandheads are saturating, and one has to rely
on the TiO and VO bands. In fact, the VO bands were originally used to
diagnose and calibrate ultracool M dwarfs of subtypes M7-M9 \citep{KHS95}.
The main caveat in using the oxide bands for spectral classification
is that this can potentially introduce a metallicity dependence on the
estimated spectral subtype, with more metal-rich stars being assigned
later subtypes than what they would have based on the strength of
their CaH bands alone. In any case, because our sample appears to be
dominated by near-solar metallicity stars, we calibrate additional
relationships between subtype and the TiO5 and TiO6 bands indices. We
first recalculate the subtypes by averaging the values of $\left[ SpTy
\right]_{CaH2}$ and $\left[ SpTy \right]_{CaH2}$, and perform a
$\chi^2$ fit of the TiO5 and TiO6 indices to the mean subtypews
calculated from CaH2 and CaH3, finding:
\begin{equation}
\left[ \rm SpTy \right]_{\rm TiO5} = 7.83 - 9.55 \ {\rm TiO5}_{c}
\end{equation}
\begin{displaymath}
\left[ \rm SpTy \right]_{\rm TiO6} = 9.92 - 15.68 \ {\rm TiO6}_{c}
+21.23 \ \left( {\rm TiO6}_{c} \right)^2
\end{displaymath}
\begin{equation}
- 16.65 \ \left( {\rm TiO6}_{c} \right)^3
\end{equation}
where again the corrected band indices are used. The relatively sharp
non-linear deviation in the TiO6 distribution around M3 forces the use
of a third order polynomial in the fit.
We also determine the relationships for the VO1 and VO2
band indices. This this after recalculating the subtypes from the
average of $\left[ \rm SpTy \right]_{\rm CaH2}$ and $\left[ \rm SpTy
\right]_{\rm CaH2}$, $\left[ \rm SpTy \right]_{\rm TiO5}$, and
$\left[ \rm SpTy \right]_{\rm TiO6}$, a $\chi^2$ fit again yields:
\begin{equation}
\left[ \rm SpTy \right]_{\rm VO1} = 69.8 - 71.4 \ {\rm
\left[\rm VO1\right]_{c}}
\end{equation}
\begin{displaymath}
\left[ \rm SpTy \right]_{\rm VO2} = 9.56 - 12.47 \ {\rm
\left[\rm VO2\right]_{c}}
+ 22.33 \ \left( {\rm \left[\rm VO2\right]_{c}} \right)^2
\end{displaymath}
\begin{equation}
- 19.59 \ \left( {\rm \left[\rm VO2\right]_{c}} \right)^3.
\end{equation}
The VO indices however make relatively poor estimators of spectral
subtypes for our sample, mainly because the shallow slope at earlier
subtypes provides little leverage. The VO2 index also shows
unexpectedly large scatter in the MDM spectra, including in the
classification standard stars, which we suspect is due the fact that
the index is defined very close to the red edge of the MDM spectral
range and is thus more subject to statistical noise and flux
calibration errors. We therefore do not include $\left[ SpTy
\right]_{VO12}$ and $\left[ SpTy \right]_{VO2}$ in the final
determination of the spectral subtypes.
\begin{deluxetable}{cc}
\tabletypesize{\scriptsize}
\tablecolumns{2}
\tablewidth{160pt}
\tablecaption{Distribution by spectral subtype of the 1564 survey
stars\label{table_st}}
\tablehead{
\colhead{Spectral subtype} &
\colhead{N}
}
\startdata
G/K\tablenotemark{a} & 160 \\
K7.0 & 27 \\
K7.5 & 101 \\
M0.0 & 177 \\
M0.5 & 160 \\
M1.0 & 152 \\
M1.5 & 147 \\
M2.0 & 141 \\
M2.5 & 119 \\
M3.0 & 125 \\
M3.5 & 125 \\
M4.0 & 72 \\
M4.5 & 36 \\
M5.0 & 13 \\
M5.5 & 4 \\
M6.0 & 2 \\
M6.5 & 2 \\
M7.0 & 1
\enddata
\tablenotetext{a}{Stars identified as earlier than K7.0 and/or with no
detected molecular bands.}
\end{deluxetable}
Fig.~\ref{idx_spty} plots all the corrected spectral band indices as a
function of the adopted spectral subtype. The relatively
small scatter ($\approx0.02$) in the distribution of $\left[\rm
CaH2\right]_{c}$, $\left[\rm CaH3\right]_{c}$, $\left[\rm
TiO5\right]_{c}$ and $\left[\rm TiO6\right]_{c}$ demonstrate
the internal consistency of the spectral type calibration for the four
indices. All four relationships have an average slope $\approx10$,
which means that since those indices have an estimated measurement
accuracy of $\approx0.02$ units, the spectral subtypes calculated by
combining the four indices should be accurate to about $\pm0.10$
subtype assuming that the measurement errors in the four indices are
uncorrelated. While this would make it possible to classify the stars
to within a tenth of a subtype, we prefer to follow the general
convention and assign spectral subtypes to the nearest half
integer.
To verify the consistency of the spectral classification, we
compare the spectral types evaluated independently for the list of 141
stars observed at both MDM and UH. We find that 82\% of the stars end
up with the same spectral type assigments from both observatories,
i.e. they have spectra assigned to the same half-subtype. All the
other stars have classifications within 0.5 subtypes. This is
statistically consistent with a $1\sigma$ error of $\pm$0.18 on the
spectral type determination, slightly larger than the assumed $0.1$
subtype precision estimated above. This suggests a $3\sigma$ error of
about $\pm$0.5, which justifies the more conservative use of
half-subtypes as the smallest unit for our classification.
The resulting classifications based on the CaH and TiO band index
measurements are listed in Table~\ref{table_spectro}. The numerical
spectral subtype measured from the average of the band indices is
listed to 2 decimal figures. These values are rounded to the nearest
half integers to provide our more formal spectral classifications to a
half-subtype precision. The non-rounded values are however useful for
comparison with other physical parameters as they provide a continuous
range of fractional values; these fractional subtype values are used
in the analysis throughout the paper
A histogram of the distribution of spectral subtypes is shown in
Figure~\ref{sptype_hist}, with the final tally compiled in
Table~\ref{table_st}. Most of the stars in our survey have
subtypes in the M0.0-M3.0 range. The sharp drop for stars of subtypes
K7.5 and K7.0 is explained by the color selection used in the
\citet{LepineGaidos.2011} catalog ($V-J>2.7$) which was originally
intended to select only M dwarfs; note that stars with subtypes
earlier than K7.0 are also excluded from the graph, and are probably
contaminants of the color selection in any case. Our deficit of K7.0
and K7.5 stars however spectra demonstrates that the adopted selection
criterion is efficient in excluding K dwarfs from the catalog. The
distribution of spectral subtypes also shows a marked drop in numbers
for subtypes M4 and later. This is a consequence of the relatively
bright magnitude limit ($J<9$) of our subsample, combined with the low
absolute magnitudes of late-type M dwarfs, which excludes most
late-type stars from our survey, since these tend to be fainter than
our magnitude limit, even relatively nearby ones.
\begin{figure}
\vspace{-2cm}
\hspace{-0.1cm}
\includegraphics[scale=0.42]{f7.eps}
\caption{Distribution of spectral subtype M, for the stars in our
survey, with K5=-2 and K7=-1. Most stars are found to be early-M
objects. The sharp drop at subtypes K7.5 and earlier is a
consequence of our initial $V-J>2.7$ color cut, which was meant to
select only stars cool enough and red enough to be M dwarf. The
sharp drop for subtypes M4 and later is a consequence of the high
magnitude limit ($J<9$) of our survey, which restricts the distance
range over which late-M stars are selected. The magnitude limit also
explains the slow drop in numbers from subtypes M0 to M3, whereas
one would normally expect the lower-mass M3 stars to be more common
than earlier-type objects in a volume limited
sample.\label{sptype_hist}}
\end{figure}
\subsection{Semi-automated classification using THE HAMMER}
\begin{figure}
\plotone{f8.eps}
\caption{Comparison of spectral types assigned by the spectral index
method and those assigned by the Hammer code, which is used for
stellar classification in the Sloan Digital Sky Survey. Subtypes
generally agree to within the advertized precision of the Hammer ($\pm$1
subtype, illustrated by the slanted lines), although the Hammer
subtypes are marginally later than the spectral index subtypes, by
0.27 subtype on average.\label{spec_comp}}
\end{figure}
To verify the accuracy and consistency of spectral typing based the
spectral-index method described above, we performed independent spectral
classification using the Hammer code \citep{Covey2007}. The Hammer was
designed to classify stars in the Sloan Digital Sky Survey
Spectroscopic database, including M dwarfs \citep{West2011}. The code
works by calculating a variety of spectral-type sensitive band
indices, and uses a best-fit algorithm to identify the spectral
subtype providing the best match to those band indices. For late-K and
M dwarfs, spectral subtypes are determined to within integer value
(K5, K7, M0, M1, ..., M9).
However, to ensure that the automatically determined spectral types were
accurate we used the manual ``eye check'' mode of the Hammer (version
1.2.5). This mode is typically used to verify that there are no
incorrectly typed interlopers. The Hammer allows the user to compare
spectra to a suite of template spectra to determine the best
match. \citet{West2011} have found that for late-type M dwarfs, the
automatic classifications were systematically one subtype earlier than
those determined visually. Our analysis confirms this offset, and we
therefore disregarded the automatically determined Hammer values to
adopt the visually determined subtypes.
The resulting subtypes are listed in Table~\ref{table_spectro}.
Some 170 stars were not found to be good fits to any of the K5, K7, or
M type templates, and thus identified as early-K or G dwarfs. This
subset includes all of the 156 stars that were visually identified as
non-M dwarfs on first inspection (see \S3.1). The remaining 14 stars
were initially found to be consistent with late-K stars, and
classified as K7.0 and K7.5 objects using the spectral index method
described in \S3.2 above; we investigated further to determine why the
stars were classified as early K using the Hammer. On closer
inspection, we found that 3 of the stars are indeed more consistent
with mid-K dwarfs that K7.0 or K7.5, and we thus overran the spectral
index classification and reclassified them as ''G/K'' in
Table~\ref{table_spectro}. For the other 11 stars flagged as mid-K
type with the Hammer, we determined that the stars do show significant
evidence for TiO absorption, which warrants that the stars retain
their spectral-index classifiction of K7.0/K7.5.
In the end Table~\ref{table_spectro} lists 159 stars from our initial
sample that are identified and early-type G and K dwarf contaminants,
and most likely made our target list due to inaccurate or unreliable
$V-J$ colors. The remaining 1405 stars are formally classified as
late-K and M dwarfs.
A comparison of spectral subtypes determined from the spectral-index
and Hammer methods is shown in Figure~\ref{spec_comp}. Because the
Hammer yields only integer subtypes, we have added random values in
the $-0.4,0.4$ range to facilitate the comparion. Slanted lines in
Figure~\ref{spec_comp} show the range expected if the two
classification methods (spectral index, Hammer) agree to within 1.0
subtypes. There is however a mean offset of 0.26 subtypes between the
spectral index and Hammer classifications, with the Hammer subtypes
being on average slighlty later.
Figure~\ref{spec_comp} also reveals a number of outliers with large
differences in spectral subtypes between the two methods. We found 48
stars with differences in spectral subtyping larger than $\pm$1.5. The
spectra from these stars were examined by eye: except for one star, we
found the band-index classifications to agree much better with the
observed spectra than the Hammer-determined subtypes. The one
exception is the star PM I11055+4331 (Gl 412B) which the band index
measurements classify as M6.5; in this case the Hammer determined
subtype of M 5.0 appears to be more accurate. This exception likely
occurs because of the saturation of the CaH2 and CaH3 subtypes in the
late-type star, which make the band-index classification less
reliable.
\subsection{Comparison with the ``Meet the Cool Neighbors'' survey}
\begin{figure}
\plotone{f9.eps}
\caption{Comparison of spectral types assigned by our spectral index
method and those assigned in the "Meet the Cool Neighbors" (MCN)
spectroscopic follow-up program. This shows all 161 stars in common
between our survey and the MCN program. Spectral index subtypes are
show before rounding up to the nearest half-integer; the MCN
subtypes have random values of $\pm$0.2 to help in the
comparison. The classifications generally agree to within $\pm$0.5
subtypes (slanted lines). Our subtypes are however marginally later
(by 0.28 subtypes on average) than the MCN
subtypes.\label{mcn_comp}}
\end{figure}
After the PMSU survey, the largest spectroscopic survey of M dwarfs in
the northern sky was the one presented in the ``Meet the Cool
Neighbors'' (MCN) paper series
\citep{CruzReid.2002,Cruzetal.2003,Reidetal.2003,Reidetal.2004,Cruzetal.2007,Reid2007}. In
this section we compared the spectral clssification from the MCN
survey with our own spectral type assignments.
To identify the stars in common between the two surveys, we first
performed a cross-correlation of the celestial coordinates of the
stars listed in the MCN tables, to the coordinates listed in the
SUPERBLINK catalog. This was performed for the 1077 stars in MCN which
are north of the celestial equator. We found counterparts in the
SUPERBLINK catalog to within 1\arcsec for 860 of the MCN stars. Of the
217 MCN stars with no obvious SUPERBLINK counterparts, 148 are
classified as ultracool M dwarfs (M7-M9) or L dwarfs (L0-L7.5) which
means they are very likely missing from the SUPERBLINK catalog because
they are fainter than the V=19 completeness limit of the catalog. Of
the 71 remaining stars, close examination of Digitized Sky Survey
scans failed to identify the stars at the locations quoted in MCN. A
closer examination of the fields around those stars identified 49
cases where a high proper motion stars could be found within 3\arcmin
of the quoted MCN positions. These nearby high proper motion stars are
all listed in SUPERBLINK, and have colors consistent with M dwarfs; we
therefore assumed that the quoted MCN positions are in error, and
matched those 49 MCN entries with the close by high proper motions
stars from SUPERBLINK. Of the remaining 22 stars we found 6 that have
proper motions below the SUPERBLINK limit of $\mu>40$ mas yr$^{-1}$
and another 5 stars with proper motions within the
SUPERBLINK limit but that appear to have been missed by the SUPERBLINK
survey. Finally, there were 11 MCN stars that we could not identify at
all on the Digitized Sky Survey images, and we can only assume that
the positions quoted in MCN are too large for proper identification,
and that the stars should be considered ``lost''.
Of the 909 stars in the MCN program with SUPERBLINK counterparts, we
found only 219 which satisfy the magnitude limit ($J<9$) of our
present sample of very bright M dwarfs. Of those, 52 stars have colors
bluer ($V-J<2.7$) than our sample limit; 48 of them are classified
as F and G stars in MCN, consistent with their bluer colors. The other
4 stars are classified as M dwarfs, although they have $V-J<2.7$
according to \citet{LepineGaidos.2011}. We infer that our $V-J$ colors
for those stars are probably underestimated, which suggests that our
color selection may be overlooking a small fraction of nearby M
dwarfs. In addition, we found another 6 stars which have $V$
magnitudes and $V-J$ colors within our survey range, but were rejected
by the additional infrared ($J-H$,$H-K$) color-color cuts used in
\citet{LepineGaidos.2011} to filter out red giants. All 6 stars are
very bright in the infrared, and it appears that at least one of the
$H$ or $K$ magnitudes listed in the 2MASS catalog may be in error,
making the stars appear to have $J-H$ and/or $H-K$ colors more
consistent with giants. Four of the stars are classified as M dwarfs
in MCN, the other two are late-K dwarfs. Overall, this makes a total
of 8 M dwarfs from the MCN census that were overlooked in our
selection out of the MCN subset of $\approx$150 nearby M dwarfs. This
suggests that our color cuts, combined with magnitude measurement
errors, might be missing $\sim5\%$ of the very bright, nearby M dwarfs.
In the end, this leaves only 161 stars in common between the MCN
program and our own spectroscopic survey. The stars are all classified
as late-K and M dwarfs by MCN, with subtypes ranging from K5 to
M5.5. All 159 stars are identified with a flag (''M'') in the last
column of Table~\ref{table_spectro}; we note that 82 of these stars
were also observed as part of the PMSU survey. We compare the spectral
type assignments from both surveys in Figure~\ref{mcn_comp}, where the
M dwarf subtypes from MCN are plotted against the (non-rounded)
subtypes calculated from the spectral-band indices. To ease the
comparison, random values of $\pm$0.2 are added
to the MCN subtypes. Overall, our classifications agree to within
$\pm$0.5 subtypes with the MCN values. The MCN subtypes, however, tend
to be marginally earlier on average, by 0.28 subtypes; this is in
contrast with the Hammer classifications (see above) which tend to be
slightly later than our own. For the MCN subtypes, the effect is more
pronounced for the earlier M dwarfs ($<$M2.5), where the mean offset
is 0.43 subtypes, whereas the mean offset is only 0.09 for the later
stars.
To investigate the difference in spectral subtype assignments, we
compare the recorded values of the CaH2, CaH3, and TiO5 indices
between the MCN program and our own survey. After a search of the
various tables published in the MCN series of papers, we identified 54
stars in common between the two programs, and for which values of
CaH2, CaH3, and TiO5 were also recorded in both. The differences
between the spectral index values are shown in
Figure~\ref{mcn_idx}. For our own survey, the corrected values of
these indices are used, i.e. CaH2$_c$, CaH3$_c$, and TiO5$_c$ as
defined in \S3.2.2. We find that the CaH2 and TiO5 are estimated
marginally higher in the MCN program than they are in our survey, and
this very likely explains the difference in spectral typing: the
higher index values yield margnially earlier spectral subtypes. This
again emphasizes the variation in the spectral index measurements due
to spectral resolution and other instrumental setups, and the need to
apply systematic corrections between observatories to obtain a uniform
classification system.
\begin{figure}
\hspace{1.0cm}
\includegraphics[scale=0.9]{f10.eps}
\caption{Differences between the spectral index values recorded in the
``Meet the Cool Neighbors'' (MCN) program and those measured in the
present survey. The values are compared for a subset of 55 stars in
common between the two surveys and for which values of CaH2, CaH3,
and TiO5 were recorded. For the present survey, we use the corrected
values (CaH2$_c$, CaH3$_c$, and TiO$_c$) as defined in \S3.2.2. The
MCN values tend to be marginally higher on average, especially for
stars of earlier spectral subtypes. This explains the marginally
earlier spectral subtype assigments in MCN compared with the
ones presented in this paper (see
Figure~\ref{mcn_comp}).\label{mcn_idx}}
\end{figure}
We note that smaller subsets of stars in our census may also have
spectroscopic data published in the literature, from various other
sources. This is especially the case for the 102 stars from the CNS3
and stars with very large proper motions $mu>0.2\arcsec$ yr$^{-1}$
which have been more routinely targeted for follow-up spectroscopic
observations. Additional pectroscopic surveys of selected bright M
dwarfs include, \citet{Scholz2002}, \citet{Scholz2005},
\citet{Reyle2006}, and \citep{Riaz2006}, which all have a few stars in
common with our catalog. Other surveys of nearby M dwarfs have mainly
been targeting fainter stars
\citep{Bochanski2005,Bochanski2010,West2011}, and do not overlap with
our present census.
\subsection{Color/spectral-type relationships}
\begin{figure}
\vspace{-0.2cm}
\hspace{0.1cm}
\includegraphics[scale=0.8]{f11.eps}
\caption{Variation of the UV-to-optical $NUV-V$ color index and the
optical-to-IR $V-J$ color index as a function of spectral subtype
M. Stars with more reliable V magnitudes from the Tycho-2 catalog
are shown in blue, stars with V magnitudes derived from less
reliable photometric measurements are shown in red. The distribution
of ultra-violet to optical colors ($NUV-V$) with spectral subtype
shows two populations, one with a tight correlation, consistent with
blackbody distribution and labeled ``quiescent'', and a scattered
population of stars with clear $UV$ excess, labeled ``active''. The
distribution of $V-J$ with subtype closely follows the relationship
used by \citet{LepineGaidos.2011} to predict subtypes from $V-J$
colors (dashed line) except for stars of later subtypes which have
redder colors than predicted. A revised relationship (full line) is
fitted to the data. Outliers point to stars with bad $V$ magnitude
measuremens.\label{color_spty}}
\end{figure}
\begin{deluxetable*}{cccccccccc}
\tabletypesize{\scriptsize}
\tablecolumns{8}
\tablewidth{0pt}
\tablecaption{Colors and $T_{eff}$ for red dwarfs in our survey as a
function of spectral subtype.\label{color_subtype}}
\tablehead{
\colhead{subtype} &
\colhead{n$_{NUV-V}$\tablenotemark{1}} &
\colhead{$\overline{NUV-V}$\tablenotemark{1}} &
\colhead{$\sigma_{NUV-V}$\tablenotemark{1}} &
\colhead{n$_{V-J}$} &
\colhead{$\overline{V-J}$} &
\colhead{$\sigma_{V-J}$} &
\colhead{$\overline{T_{eff}}$} &
\colhead{$\sigma_{T_{eff}}$}
}
\startdata
K 7.0 & 11 & 8.17& 0.21& 25 & 2.90 & 0.31 & 4073 & 98\\
K 7.5 & 47 & 8.60& 0.31& 97 & 2.89 & 0.16 & 3883 & 82\\
M 0.0 & 87 & 8.66& 0.36& 175 & 2.94 & 0.21 & 3762 & 71\\
M 0.5 & 79 & 8.74& 0.30& 159 & 3.11 & 0.34 & 3646 & 48\\
M 1.0 & 76 & 8.89& 0.39& 151 & 3.19 & 0.18 & 3565 & 44\\
M 1.5 & 57 & 9.07& 0.38& 147 & 3.36 & 0.23 & 3564 & 39\\
M 2.0 & 57 & 9.25& 0.47& 139 & 3.52 & 0.34 & 3518 & 57\\
M 2.5 & 48 & 9.45& 0.50& 118 & 3.69 & 0.28 & 3500 & 61\\
M 3.0 & 34 & 9.61& 0.38& 125 & 3.91 & 0.28 & 3423 & 62\\
M 3.5 & 28 & 9.69& 0.33& 124 & 4.17 & 0.33 & 3320 & 66\\
M 4.0 & 5 & 9.72& 0.35& 71 & 4.45 & 0.41 & 3204 & 76\\
M 4.5 & 1 &\nodata&\nodata& 36 & 4.81 & 0.46 & 3119 & 43\\
M 5.0 & 0 &\nodata&\nodata& 12 & 5.23 & 0.50 & 3014 & 61
\enddata
\tablenotetext{1}{Non-active (``quiescent'') red dwarfs only.}
\end{deluxetable*}
Spectral subtypes were inititially estimated in
\citet{LepineGaidos.2011} based on $V-J$ colors alone. Here we verify
this assumption and re-evaluate the color-magnitude relationship for
bright M dwarfs. The $V-J$ color index combines estimated optical ($V$)
magnitudes from the SUPERBLINK catalog to the infrared $J$
magnitudes of their 2MASS counterparts. The SUPEBLINK $V$ magnitudes
are estimated either from the Tycho-2 catalog $V_T$ magnitudes, or
from a combination of the Palomar photographic $B_J$ (IIIaJ), $R_F$
(IIIaF), and $I_N$ (IVn) magnitudes, as described in
\citet{LepineShara.2005}. Values of $V$ are more accurate for the
former ($\approx$0.1mag) than for the latter ($\gtrsim$0.5mag);
Table~\ref{table_photo} indicates the source of the V magnitude.
Mean values and dispersion about the mean of the $V-J$ colors are
listed in Table~\ref{color_subtype}, for each bin of half-integer
subtype; the table also lists how many stars of each type are in each
bin. The $V-J$ colors of our stars are also plotted as a function of
spectral subtype in Figure~\ref{color_spty} (top panel). The adopted
color-subtype relationship from \citet{LepineGaidos.2011}
is shown as a thick dashed line in Figure~\ref{color_spty}. Stars with
the presumably more reliable Tycho-2 magnitudes are shown in blue,
while stars with photographic $V$ magnitudes are shown in red. Stars
with Tycho-2 $V$ magnitudes appear to have marginally bluer colors
at a given subtype; this however is an effect of the visual
magnitude limit of the Tycho-2 catalog, which includes only the
brightest stars in the $V$ band and is thus more likely to list bluer
objects. The \citet{LepineGaidos.2011} relationship generally follows
the distribution at all subtypes, but with mean offsets up to
$\pm$0.4mag in $V-J$, especially at earlier and later subtypes. We
perform a $\chi^2$ fit to determine the following, improved
relationship:
\begin{displaymath}
\left[ \rm SpTy \right]_{V-J} = {\rm -32.79 + 20.75 ( V - J ) - 4.04 ( V - J )^2}
\end{displaymath}
\begin{equation}
{\rm + 0.275 ( V - J )^3}
\end{equation}
after exclusion of 3-$\sigma$ outliers. The relationship is shown in
Figure~\ref{color_spty} (solid line). There is a scatter of 0.7
subtype between $\left[ SpTy \right]_{V-J}$ and the subtype determined
from spectral band indices. While the spectroscopic classification is
more accurate and reliable, photometrically determined spectral
subtypes using the equation above should still be accurate to $\pm$0.5
subtype about 80\% of the time, and to $\pm$1.0 subtype 95\% of the
time, which may be useful for a quick assessment of subtype when
spectroscopic data is unavailable.
We also compare the near-UV to optical magnitude color $NUV-V$ for the
714 stars in our sample which have counterparts in GALEX; the
distribution is shown in Figure~\ref{color_spty} (bottom panel).
We find that stars become progressively redder as spectral subtype
increases, from $NUV-V=8$ at M0 to $NUV-V=10$ at M4. There is however
a significant fraction of M dwarfs which display much bluer $NUV-V$
colors at any given subtype. The excess in $NUV$ flux is strongly
suggestive of chromospheric activity (see \S6 below for a more
detailed analysis). We separate the active stars from the more
quiescent objects with the following condition:
\begin{equation}
\left[ \rm NUV - V \right] > 7.7 + 0.35 \left[\rm Spty\right] \
\end{equation}
where $\left[\rm Spty \right]$ is the mean spectral subtype calculated from
equations 8-11. After excluding active stars, we calculate the mean
values and scatter about the mean of $NUV-V$ for each half-integer
spectral subtype. Again those are listed in Table~\ref{color_subtype}
for reference; the table also lists the number of non-active stars
used to calculate the mean. There is not a sufficient number of stars
to calculate mean values and scatter at M4.5 (1 star) and M5.0 (0
star).
\section{Survey completeness}
\begin{figure}
\epsscale{1.1}
\plotone{f12.eps}
\caption{Number of spectroscopically confirmed M dwarfs as a function
of $sin(b)$, where $b$ is the Galactic latitude. Numbers from our
census are shown as filled circles, with errorbars showing the
Poisson noise. Distributions predicted from models with either
uniform space density, or with decreases perpendicular to the
Galactic plane following scale-heights of 200pc and 400pc, are shown
for comparison. The observed distribution is largerly
consistent with a uniform density in the local $d<70$pc
volume, with no evidence for incompleteness at low Galactic
latitude.\label{compl}}
\end{figure}
Our 1,564 spectroscopically confirmed M dwarfs are drawn from a
catalog with proper motion limit $\mu<40$ mas yr$^{-1}$. The low
proper motion limit of the SUPERBLINK catalog catches most of the
nearby stars, but potentially overlooks nearby M dwarfs with small
components of motion in the plane of the sky \--- either due to low
space motion relative to the Sun or to projection effects. The catalog
may also be affacted by other sources of incompleteness (e.g. missed
detection, faulty magnitude estimate) which means that at least some
very bright, nearby M dwarfs must be missing from our survey due to
kinematics bias and other effects.
To evaluate the completeness of our census, we first consider the
primary source of incompleteness: the kinematics bias of the proper
motion catalog. As discussed in \citet{LepineGaidos.2011}, the
completeness depends on the local distribution of stellar motions, and
increases with distance from the Sun. To estimate the kinematics bias
in our sample, we built a model reproducing the local distribution and
kinematics of nearby M dwarfs. We first assumed the stars to have a
uniform spatial distribution in the solar vicinity, and generated a
random distribution of $10^{5}$ objects within a sphere of radius
$d=70$pc centered on the Sun. We assigned transverse motions to all the
stars, assuming a velocity-space distribution similar to that of the
nearby (d$<$100pc) G dwarfs in the Hipparcos catalog
\citep{vanLeeuwen2007}. Because the distribution of stellar velocities
is not isotropic, we assigned transverse motions for each simulated star
based on the statistical distribution of transverse motions for
Hipparcos stars with sky coordinates within $30^{deg}$ of the
simulated object. We also used the simplifying assumption that the
local M dwarf population has a uniform distribution of absolute
magnitudes over the range $5<M_J<15$, which is the approximate range
of absolute magnitudes reported in the literature for M dwarfs.
We then counted the total number of stars in the simulation with
apparent magnitudes $J<9$, and calculated the fraction of those stars
with proper motions $\mu>40$ mas yr$^{-1}$. We found that 93\% of
nearby M dwarfs with $J<9$, on average, have proper motions above the
SUPERBLINK limit. M dwarfs with $J<9$ extend to a maximum distance of
63 parsecs; most of the stars which fail the proper motion cut are in
the higher distance range (d$<$50pc), and are stars near the bright
end of the luminosity distribution $M_J\approx5-6$. For this reason,
if we assume a luminosity function which increases at fainter absolute
magnitudes, the fraction of $J<9$ stars which fall within the proper
motion cut is increased, because more of the $J<9$ stars in the local
population are now M dwarfs of lower luminosities and closer
distances, which are more likely to have high proper motions. The
observed field M dwarf luminosity function does indeed increase for
early-type dwarfs, to reach a peak at $M_J\simeq8.0$
\citep{Reid2002,Bochanski2010}, which means that the 93\% completeness
estimated above must be a lower limit. To verify this, we tested a
luminosity function where the number of stars increases linearly with
absolute magnitude and doubles from $M_J=5$ to $M_J=8$; with this
model, our simulations showed that 96\% of all $J<9$ M dwarfs, would
have $\mu>40$ mas yr$^{-1}$, and thus be within the detection limit of
SUPERBLINK. Overall, this suggests that only about $5\%$ of all $J<9$
M dwarfs on the sky will be overlooked in our census because of the
proper motion bias.
The SUPERBLINK surveys does however suffer from various other sources
of incompleteness, such as the inability to detect moving stars in
saturated regions of photographic plates (i.e. in the immediate
vicinity of very bright stars), or a difficulty in detecting stars in
very crowded field. Also, the SUPERBLINK code has trouble detecting
the motions of relatively bright $V<12$ stars because of the saturated
cores of their point spread functions. In practice, this is mitigated
by incorporating data from the Tycho-2 catalog, which provides very
accurate proper motion measurements for bright stars. However the
Tycho-2 catalog itself has some level of incompleteness in the
$8<V<12$ manitude range.
One way to test incompleteness due to crowding or saturation is to
examine the distribution of SUPERBLINK stars as a function of Galactic
latitude. Crowding and saturation effects should be more pronounced in
low Galactic-latitude fields, where the stellar density is high. In
comparison, stars in high Galactic latitude fields will be easier to
detect with the SUPERBLINK code. The same is true for the Tycho-2
catalog, which should be more complete at high Galactic latitudes.
We calculated the number of spectroscopically confirmed M dwarfs in our
sample as a function of $sin(b)$, where $b$ is the Galactic
latitude. The distribution is shown in Figure~\ref{compl}; the number
of stars in each bin is plotted as a filled circle, with errobars
showing the $1\sigma$ Poisson error. The distribution increases with
$sin(b)$ because our stars are all located north of the celestial
equator. For comparison, we plot the trend expected of a uniform
distribution of stars in the local volume (blue histogram), assuming
the same total number of stars as in our census. We find our data to
be largely consistent with the uniform distribution model, and see no
evidence of a significant dip at low Galactic latitude, which one
would expect if the SUPERBLINK survey is incomplete at low $b$. This
suggests that our sample does not suffer from significant sources of
incompleteness due to saturation/crowding.
One might however argue that the uniform density assumption is invalid
for a census extending to $\simeq$70pc of the Sun, and that a slight
overdensity of stars should be detected at low Galactic latitudes. The
fact that we see no evidence of such an overdensity in our data would
then be indirect evidence of incompleteness at low Galactic
latitude. To examine this possibility, we estimated the expected
distributions from simulations in which the stellar density
perpendicular to the plane (i.e. along $Z$) decreases with scale
heights of 400pc or 200pc. Kinematics and absolute magnitude
distributions were
also simulated as described above, and stars were selected based on
$J<9$ and $\mu>40$ mas yr$^{-1}$. The mean distributions from the
400pc and 200pc scale-height models are shown in Figure~\ref{compl},
and do indeed predict a slight overdensity of objects at low Galactic
latitudes. Assuming that all the stars at high Galactic latitude are
detected in all the models, we find a $\sim$5\% excess of stars in the
400pc scale-height model over the observed number, and a $\sim$10\%
excess of stars in the 200pc scale-height model. One might then argue
that the SUPEBLINK census potentially has an additional 5\%-10\%
incompleteness level, due to incompleteness in the plane of the Milky
way. This is most probably an overestimate, however, because there is
no evidence that the local stellar distribution shows any significant
decrease with $Z$. Overall, the SUPERBLINK census appears to be
essentially complete at low Galactic latitudes.
Proper-motion selection may introduce an additional bias against
metal-rich and/or young stars, which tend to have lower components of
motion in the vicinity of the Sun. However this effect is expected to
be small, e.g. a few percent against [Fe/H] = 0 relative to [Fe/H] =
-0.5 at 45~pc \citep{Gaidos2012}. Nearly all likely members of the
Hyades \citep{Perryman1998}, Ursa Majoris \citep{King2003}, and TW
Hydrae \citep{Reid2003} young nearby moving groups have proper motions
that exceed 40~mas yr$^{-1}$ and thus would not be selected against using
the current selection method.
Finally, some bright M dwarfs may be overlooked because of the imposed
color cuts, due to magnitude errors and uncertainties. As described in
\S3.4 above, a handful of previously known M dwarfs from the MCN
census were overlooked in our target selection for precisely those
reasons. As suggested in \S3.4, it is possible that we may be
overlooking 5\% of bright M dwarfs because of magnitude errors. This,
combined with the estimated 96\% completeness in the proper motion
selection estimated above, suggests that our list of 1405 M dwarfs
likely includes $\approx91\%$ of all existing M dwarfs with infrared
magnitude $J<9$ as seen from Earth. There may still be $\approx$140
bright M dwarfs to be identified, although the precise number can only
be determined after these ``missing'' stars are found.
\section{Phoenix Model fits and $T_{eff}$ estimates}
\begin{figure}
\epsscale{1.2}
\plotone{f13.eps}
\caption{Effective temperatures for the bright M dwarfs in our survey
determined by fits to the PHOENIX models. The gray points represent
individual objects, while the large black points are the median
values of the objects within each half subtype bin. The error bars
are the interquartile ranges. Blue diamonds and red triangles
show the T$_{eff}$ estimates for subsets of stars in our survey
whose tempreatures were estimated from photometry
\citet{Casagrande2008} and model fits to infrared spectra
\citep{Rojas-Ayala2012} respectively. While both spectroscopic
estimates show evidence of a plateau at M2-M3, the photometric
estimates do not concur. \label{spt_color}}
\end{figure}
We compared our spectra to a grid of 298 models of K- and M-dwarf
spectra generated by the BT-SETTL version of PHOENIX \citep{Allard2010}.
BT-SETTL includes updated opacities (i.e. of H$_2$O), revised solar
abundances \citep{Asplund2009}, a refractory cloud model, and
rotational hydrodynamic mixing. The models include effective
temperatures $T_{eff}$ of 3000-5000~K in steps of 100~K, $\log g$ values
of 4, 4.5, and 5, and metallicities of [M/H] = -1.5, -1, -0.5, 0,
+0.3, and +0.5. For each temperature, $\log g$ and metallicity value,
we selected the model with $\alpha$/Fe that was closest to solar.
\begin{figure*}
\plotone{f14.eps}
\caption{Distribution of corrected CaH2+CaH3 vs TiO5 band index
values for the stars in our survey (grey dots - with brightness
levels correlating with spectral subtype). There is a tight correlation
between the two indices, which are both also correlated
with spectral subtypes, with earlier stars on the upper right of the
diagram as shown. The distribution is used as a guide to calibrate
the value of $\zeta$, with $\zeta=1$ assumed to trace the
CaH2+CaH3/TiO5 relationship for stars with average Galactic disk
abundances (near-solar). The iso-$\zeta$ contours from the earlier
calibrations of \cite{Lepine2007} and \cite{Dhital2012} are shown as
dotted and dashed lines, respectively. When applied to our corrected
spectral index values, they both diverge from the observed
distribution at earlier subtypes, with the \cite{Lepine2007}
overestimating the $\zeta$ of late-K and early-M dwarfs, while the
\cite{Dhital2012} calibration yield underestimates. This emphasizes
again the need to use properly recalibrated and corrected spectral
index values (see Figures~\ref{pmscomp}-\ref{idx_comp}). Our
revised, dataset-specific calibration is shown with the continuous
lines. Known metal-rich amd metal-poor stars are denoted in red and
blue, respectively.\label{cah_tio}}
\end{figure*}
The spectral density of model calculations varies with wavelength but
is everywhere vastly greater than the resolution of our spectra.
Model spectra were thus convolved with a gaussian with FHWM of the same
resolution as the spectra, and a corrective shift (typically less than
a resolution element) was found by cross-correlating the observed and
model spectra. Normalized spectra were ratioed and $\chi^2$
calculated using the variance spectrum of the observations. We
restricted the spectral range over which $\chi^2$ is calculated to
5600-9000\AA\ and excluded the problematic region 6400-6600\AA\ which
contains poorly-modeled TiO absorption \citep{Reyle2011}. We also
excluded regions where the telluric correction is rapidly changing with
wavelength, i.e. the slope, smoothed over 4 resolution elements or
11.7\AA, exceeds 1.37 \AA$^{-1}$. The model with the smallest $\chi^2$
value was identified. For a more refined estimate of effective temperature,
we selected the 7 best-fit models and constructed 10,000 linear
combinations of them; the ``effective temperature'' of each is the
weighted sum of the temperatures of the components. We again found
the model with the minimum $\chi^2$. We calculated the standard
deviation of $T_{eff}$ among the combination models as a function of the
maximum allowed $\chi^2$. We reported the maximum standard deviation
as a conservative estimate of uncertainty. We also calculated formal
95\% confidence intervals for $T_{eff}$ based solely on the expected
distribution of $\chi^2$ for $N-3$ degrees of freedom, where $N \sim
1100$ is the number of resolution elements used in the fit. The
parameters of the best-fit model, and the refined $T_{eff}$, standard
deviation, and confidence intervals are reported in
Table~\ref{table_spectro}.
Values of $T_{eff}$ calculated for individual stars are plotted in
Figure~\ref{spt_color} as a function of their spectral subtype (grey
dots). Median values for stars within each half-subtype bin are
plotted in black, with error bars showing the interquartile
ranges. Our model-fit algorithm prefers values that match the $T_{eff}$
model grid, which have a 100K grid step (i.e., 3500K is preferred over
3510K).
Our results suggest the existence of a $T_{eff}$ plateau spanning
M1-M3. To investigate this further, we compare our values to the
effective temperatures reported in \citet{Casagrande2008} for 18 of
the stars in our sample, and in \citep{Rojas-Ayala2012} for another 49
stars; our own spectral type determinations are combined to the
T$_{eff}$ measured by the other authors. The values are compared in
Figure~\ref{spt_color}. We find that our mid-type plateau is
corroborated with the \citep{Rojas-Ayala2012} values, but not with
those from \citet{Casagrande2008}, whose values decrease more linearly
with spectral subtype. The effective temperatures in
\citet{Casagrande2008} are based on photometric measurements while
the \citep{Rojas-Ayala2012} values are estimated by PHOENIX model fits
to infrared spectra. It is interesting that the fits to the optical
and infrared spectra yield T$_{eff}$ values which are in
agreement. The disagreement with the photometric determinations
however suggest that atmospheric models for M dwarfs are still not
well understood, in particular in the M1-M4 spectral regimes.
\section{The $\zeta$-Parameter and Metallicity Estimates}
\begin{figure}
\epsscale{2.2}
\plotone{f15.eps}
\caption{The $\zeta$ parameter as a function of spectral subtype. Top:
difference in $\zeta$ for the same stars measured at the two
observatories (MDM and UH). The scatter provides an estimate of the
measurement error on $\zeta$, which is significantly larger at
earlier subtypes. Bottom: adopted values of $\zeta$ for all the
stars in the survey. The larger scatter at subtypes M2 and earlier
can be fully accounted by the measurement errors. The scatter at
subtypes M3 and later is larger than the measurement error, and is
thus probably intrinsic and is evidence for intrinsic metallicity
scatter in the solar neighborhood.\label{zeta_comp}}
\end{figure}
\subsection{Recalibration of the $\zeta$ parameter}
The $\zeta_{TiO/CaH}$ parameter (denoted $\zeta$ for short) is a
combination of the TiO5, CaH2, and CaH3 spectral indices which was
shown to be correlated with metallicity in metal-poor M subdwarfs
\citep{Woolf2009}. The index was first described in
\citet{Lepine2007}, and a revised calibration has recently been
proposed by \citet{Dhital2012}. The index measures the relative
strength of the TiO molecular band around 7,000\AA\ with respect to the
nearby CaH molecular band. In cool stars, in fact, the ratio between
TiO and CaH is a function of both gravity and metallicity. The CaH
band is noticeably stronger in giants \citep{Mann2012}, and this
effect can be used as affective means to separate out M giants from M
dwarfs using optical spectroscopy. In the higher gravity M
dwarfs/subwarfs however, the TiO to CaH ratio is however believed to
be mostly affected by metallicity, although young stars may show
gravity effects as well.
The $\zeta$ parameter was originally introduced to rank metal-poor,
main-sequence M stars into three metallicity classes
\citep{Lepine2007}; stars with $0.5<\zeta<0.825$ are formally
classified as subdwarfs (sdM), $0.2<\zeta<0.5$ defines extreme
subdwarfs (esdM), while a $\zeta<0.2$ identifies the star as an
ultrasubdwarf (usdM). However, it is conjectured that $\zeta$ could
be used to measure metallicity differences in disk M dwarfs, i.e. at
the metal-rich end. Disk M dwarfs are generally found to have
$0.9<\zeta<1.1$, though it is unclear if variations in $\zeta$
correlate with metallicity for values within that range. Measurement
of Fe lines in a subset of M dwarfs and subdwarfs does confirm that
the $\zeta$ parameter is correlated with metallicity
\citep{Woolf2009}, with $\zeta\simeq1.05$ presumably corresponding to
solar abundances. However there is a significant scatter in the
relationship which raises doubts about the accuracy of $\zeta$ as a
metallicity diagnostic tool.
A important caveat is that the TiO/CaH ratio is not sensitive to the
classical iron-to-hydrogen ratio Fe/H, but rather depends on the
relative abundance of $\alpha$-elements to hydrogen ($\alpha$/H)
because O, Ca, and Ti are all $\alpha$-elements. Variations in
$\alpha$/Fe would thus weaken the correlation between $\zeta$ and
Fe/H. The $\alpha$/Fe abundance ratio is however relatively small in
metal-rich stars of the thin disk ($\pm$0.05dex) disk stars, and are
found to be significant ($\pm$0.2dex) mostly in more metal-poor stars
associated with the thick disk and halo \citep{Navarro_etal.2011}. It
is thus unclear whether typical $\alpha$/Fe variations would affect
the $\zeta$ parameter significantly in our subset, which is dominated
by relatively metal-rich stars.
On the other hand, it is clear that the index has significantly more
leverage at later subtypes. This is because the strengths of both the
TiO and CaH bands are generally greater, and their ratio can thus be
measured with higher accuracy. The index is much leass reliable at
earlier M subtypes however, and is notably inefficient for late-K
stars.
A more important issue is the specific calibration adopted for the
$\zeta$ parameter, which is a complicated function of the TiO5, CaH2,
and CaH3 indices. The $\zeta$ parameter itself is defined as:
\begin{equation}
\zeta = \frac{1-{\rm TiO5}}{1-[{\rm TiO5}]_{Z_{\odot}}},
\end{equation}
which in turns depend on $[{\rm TiO5}]_{Z_{\odot}}$, itself a function
of CaH2+CaH3. The function $[{\rm TiO5}]_{Z_{\odot}}$ represents the
expected value of the TiO5 index in stars of solar metallicity, for a
given value of CaH2+CaH3. In \citep{Lepine2007}, ${\rm
TiO5}]_{Z_{\odot}}$ was defined as:
\begin{displaymath}
[{\rm TiO5}]_{Z_{\odot}} = -0.050 - 0.118 \ {\rm CaH} + 0.670 \ {\rm CaH}^2
\end{displaymath}
\begin{equation}
- 0.164 \ {\rm CaH}^3
\end{equation}
where ${\rm CaH}={\rm CaH2}+{\rm CaH3}$. The more recent calibration
of \citet{Dhital2012}, on the other hand, uses:
\begin{displaymath}
[{\rm TiO5}]_{Z_{\odot}} = -0.047 - 0.127 \ {\rm CaH} + 0.694 \ {\rm CaH}^2
\end{displaymath}
\begin{equation}
- 0.183 \ {\rm CaH}^3 - 0.005 \ {\rm CaH}^4 .
\end{equation}
The difference between the two calibrations is mainly in the treatment
of late-K and early-type M dwarfs, as illustrated in
Figure~\ref{cah_tio}. When overlaid on the distribution of CaH2+CaH3
and TiO5 values from our current survey, however, the two calibrations
fail to properly fit the distribution of data points at the earliest
subtypes (high values of CaH2+CaH3 and TiO5). This results in the
\citet{Lepine2007} overestimating the metallicity at earlier subtypes,
while the \citet{Dhital2012} calibration tends to underestimate
metallicity.
In any case, the evidence presented in \S3.3 and \S3.4 which shows
that differences in spectral resolution and flux calibration can yield
differences in the TiO5, CaH2, and CaH3 spectral indices of the same
stars, also suggests that a calibration of the $\zeta$ parameter may
only be valid for data from a particular observatory/instrument. A
general calibration of $\zeta$ may only be adopted after corrections
have been applied as described in Section 3.2. Because we do not have
any star in common with the \citet{Dhital2012} subsample, we cannot
verify the consistency of their $\zeta$ calibration to our data at
this time. In addition, because we have now applied a correction to
our MDM spectral index measurements, the \citet{Lepine2007}
calibration of $\zeta$ may now be off, and should not be used for our
sample.
Instead, we recalibrate the $\zeta$ parameter again, using our
corrected spectral index values. Our fit of $\left[\rm
TiO5\right]_{c}$ as a function of $\left[{\rm
CaH}\right]_{c}=\left[{\rm CaH2}\right]_{c}+\left[{\rm
CaH3}\right]_{c}$ yields:
\begin{displaymath}
[{\rm TiO5}]_{Z_{\odot}} = 0.622 - 1.906 \ (\left[{\rm CaH}\right]_{c}) -
2.211 \ (\left[{\rm CaH}\right]_{c})^2
\end{displaymath}
\begin{equation}
- 0.588 \ (\left[{\rm CaH}\right]_{c})^3.
\end{equation}
We calculate the new $\zeta$ values using the corrected values of the
TiO5 index, i.e.:
\begin{equation}
\zeta = \frac{1-{\rm \left[TiO5\right]_{c}}}{1-[{\rm TiO5}]_{Z_{\odot}}}.
\end{equation}
All our values of $\zeta$ are listed in Table~\ref{table_spectro}.
In order to evaluate the accuracy of the $\zeta$ measurements, we
compared the values of $\zeta$ independently measured at both MDM and
UH for the 146 stars in our inter-observatory subset. Values are
compared in Figure \ref{zeta_comp} (top panel) which shows $\Delta
\zeta=\zeta_{MDM}-\zeta_{UH}$ as a function of spectral subtype. We
find a mean offset $\bar{\Delta\zeta}=-0.01$ and a
dispersion $\sigma_{\Delta\zeta}=0.10$. The small offset indicates
that the $\zeta$ measurements are generally consistent between the two
observatories. The dispersion $\sigma_{\Delta\zeta}$, on the other
hand, provides an estimate of the measurement accuracy. Splitting the
stars in three groups, we find the mean offsets and dispersions
$(\bar{\Delta\zeta},\sigma_{\Delta\zeta})$ to be (-0.086,0.244) for
subtypes K7.0-M0.5, (-0.011,0.103) for subtypes M1.0-M2.5, and
(0.008,0.036) for subtypes M3.0-M5.5. Assuming that stars do not show
significant variability in those bands, we adopt the dispersions as
estimates of the measurement errors on $\zeta$ for that particular
subtype range. It is clear from Figure~\ref{cah_tio} that early type
stars should have larger uncertainties in $\zeta$ because of the
convergence of the iso-$\zeta$ lines. The best leverage for estimating
metallicities from the TiO and CaH bandheads is at later types when
the molecular bands and well developed.
The overall distribution of $\zeta$ values as a function of spectral
subtype also shows a decrease in the dispersion as a function of
spectral type (Figure~\ref{zeta_comp}, bottom panel). In early-type
dwarfs (K7.0-M0.5), the scatter in the $\zeta$ values is relatively
large, with $\sigma_{\zeta}\simeq0.174$. It then drops to
$\sigma_{\zeta}\simeq0.100$ for subtypes M1.0-M2.5, and to
$\sigma_{\zeta}\simeq0.059$ for subtypes M3.0-M5.5. Note that the
scatter in the M3.0-M5.5 bin is a factor 2 larger than the estimated
accuracy of the $\zeta$ for that range, as estimated above. We
suggests this to be evidence of an intrinsic scatter in the $\zeta$
values for the stars in our sample, which we allege to be the
signature of a metallicity scatter. If we subtract in quadrature the
$0.035$ measurement error on $\zeta$, we estimate the instrinsic
scatter to be $\approx$0.05 units in $\zeta$. This intrinsic scatter,
which presumably affects all subtypes equally, is unfortunately
drowned in the measurement error at earlier subtypes ($<$M3).
\subsection{Comparison with other metallicity diagnostics}
\begin{figure}
\vspace{-0.6cm}
\hspace{-0.5cm}
\includegraphics[scale=0.9]{f16.eps}
\caption{Comparison between the $\zeta$ parameter values and
independent metallicity measurements for subsets of M dwarfs in our
survey. Top panel compares our $\zeta$ to the Fe/H estimated from
the (V-K,M$_K$) calibration of \citet{Neves2012} for the same
stars. Bottom panel compares our $\zeta$ values to the Fe/H
estimated from the infrared K-band index by
\citet{Rojas-Ayala2012}. Both distribution show weak positive
correlations.\label{zeta_feh}}
\end{figure}
\begin{figure*}
\plotone{f17.eps}
\caption{SDSS photometry of M dwarf stars synthesized from SNIFS
spectra and transmission functions convolved with unit airmass.
Upper left: r-Ks (where Ks is from the 2MASS point source catalog)
against M spectral subtype (where K7=-1 and K5=-2). Points are
colored by the $\zeta$ parameter, which measures TiO/CaH ratio and
is a metallicity diagnostic in the optical. The $\zeta$ values are
undefined for late K stars, which are plotted as black points. The
black curve is a running median (N=81). Upper right: difference of
r-Ks with respect to the running median vs. spectral type, showing
no obvious correlation with zeta. Lower left: g-r vs. r-z, showing
an apparent correlation between these colors and $\zeta$. The
contours are the empirical function for $\zeta$ derived by
\citet{West2011}. Lower right: SDSS g-r vs. r-z colors generated by
the PHOENIX/BT-SETTL model \citep{Allard} for log g=5,
Teff=3500K-4200K, and five different values of the metallicity as
noted in the legend. The model predicts the more metal-rich M dwarfs
to have {\em bluer} g-r colors, while being redder in r-z. The color
dependence on metallicity in most pronounced in late-type stars, and
nearly vanishes at K7/M0.\label{model}}
\end{figure*}
To test our $\zeta$ as a tracer of metallicity for dM stars from M3 to
M6, we compared the values to two recent [Fe/H] calibration techniques
for M dwarfs with solar metallicities. First, we used the photometric
calibration of \citet{Neves2012}, which is based on
optical-to-infrared V-K color and absolute magnitude M$_{K}$. The
method is sensitive to small variations in V-K/M$_{K}$ and thus
requires an accurate, geometric parallax. A total of 143 stars in our sample have
parallaxes, and thus can have their metallicities estimated with the
method. Figure~\ref{zeta_feh} (top panel) plots the estimated $\rm
[Fe/H]$ as a function of $\zeta$ for those 143 stars. The distribution
shows significant scatter, but we find a weak correlation of $\rm
[Fe/H]$ with $\zeta$, which we fit with the relationship:
\begin{equation}
[Fe/H]_{\rm N12} = 0.750 \zeta - 0.743.
\end{equation}
Stars are scattered about this relationship with a 1-$\sigma$
dispersion of 0.383 dex. One drawback of the photometric metallicity
determination is that it assumes the star to be single. Unresolved
double stars appear overluminous at a given color, and will thus be
determined to be metal-rich. Also, young and active stars often appear
overluminous in the color-magnitude diagram \citep{Hawley2002}, and
their metalicities based on V-K/M$_{K}$ would also be
overestimated. Multiplicity and activity could therefore contribute in
the observed scatter. Stars with $[Fe/H]_{\rm N12}>0.4$, in
particular, could be overluminous in the V-K/M$_{K}$ diagram, as their
$\zeta$ does not suggest them to be metal-rich.
Next, we retrieve metallicity measurements from
\citet{Rojas-Ayala2012}, who estimated $[Fe/H]$ based on the
spectroscopic calibration from infrared K-band atomic features. Their
list has 37 stars in common with our survey. The $[Fe/H]$ values are
plotted as a function of our $\zeta$ values in the bottom panel of
Figure~\ref{zeta_feh}. Again there is significant scatter, but we also
find a weak correlation which we fit with the relationship:
\begin{equation}
[Fe/H]_{\rm RA12} = 1.071 \zeta - 1.096,
\end{equation}
about which there is a dispersion of 0.654 dex. The statistics are
relativey poor at this time, and more metallicity measurements in
the infrared bands would be useful.
The weak correlation found in both distribution is interesting in
itself. Using a sample of stars spanning a wide range of metallicities
and $\zeta$ values, including metal-poor M subdwarfs, extreme
subdwarfs (esdM), and extremely metal-poor ultrasubdwarfs
(usdM), \citet{Woolf2009} determined a relationship of the form
$[Fe/H] = -1.685 + 1.632 \zeta$, over the range $0.05<\zeta<1.10$. All
the stars in the two distributions from the present survey have
$\zeta$ values between $\sim$0.9 and $\sim$1.2, and thus represent the
metal-rich end of the distribution. The weaker slopes we find in our
correlations (0.75 and 1.07) may indicate that the relationship levels
off at high metallicity end, which would make $\zeta$ much less useful
as a metallicity diagnostic tool in Solar-metallicity and metal-rich M
dwarfs. The correlations are however weak, and more accurate
measurement of $\zeta$ and $Fe/H$ would be needed to verify this
conjecture.
\subsection{The one M subdwarf: PM I20050+5426 (V1513 Cyg)}
The primary purpose of the $\zeta$ parameter is the identification of
metal-poor M subdwarfs, for which it has already proven effective. By
definition, M subdwarfs are stars with $\zeta<$0.82
\citep{Lepine2007}. Though we have a few stars with values of $\zeta$
just marginally under 0.82, only one star clearly stands out as a
definite M subdwarf: the star PM I20050+5426 (= Gl 781) which boasts a
$\zeta=0.58$ well within the M subdwarf regime. The star also clearly
stands out in Figure~\ref{cah_tio} where it lies noticeably below the
main locus at TiO5$_{c}\simeq$0.75.
PM I20050+5426 is also known as V1513 Cyg, a star previously
identified as an M subdwarf by \citet{Gizis1997}, and one clearly
associated with the Galactic halo \citep{Fuchs1998}. The star is also
notorious for being a flare star, with chromospheric
activity due not to young age but to the presence of a low-mass
companion on a close orbit \citep{Gizis1998}. Our own spectrum indeed
shows a relatively strong line of H$\alpha$ in emission, which is
extremely unusual for an M subdwarf. It is an interesting coincidence
that the brightest M subdwarf in the northern sky should turn out to
be a peculiar object.
In any case, because the TiO molecular bands are weaker in M subdwarfs
than they are in M dwarfs, the use of TiO spectral indices for
spectral classification leads to underestimates of their spectral
subtype. The convention for M subdwarfs is rather to base the
classification on the strengths of the CaH bandheads
\citep{Gizis1997,Lepine2003,Lepine2007}. We adopt the same convention
here, and recalculate the subtype from the mean of Equations 6 and 7
only (CaH2 and CaH3 indices). We thus classify PM I20050+5426 as an
sdM2.0, which is one half-subtype later than the sdM1.5 classification
suggested by \citep{Gizis1997}.
\subsection{Photometric dependence on metallicity}
A prediction of current atmospheric models is that metallicity
variations in M dwarfs yield significant variations in optical
broadband colors \citep{Allard2000}. The metal-poor M subdwarfs have
in fact long been known to have bluer V-I colors than the more
metal-rich field M dwarfs of the same luminosity
\citep{Monet_etal.1992,Lepine2003}. The bluer colors are due to
reduced TiO opacities in the optical, which make the spectral energy
distribution of M subdwarfs closer to that of a blackbody, while it
makes the metal-rich M dwarfs display extreme red colors.
Interestingly, the SDSS $g-r$ color index shows the opposite trend,
and is bluer in the more metal-poor stars. This is because the TiO
bands very strongly depress the flux in the 6000\AA-7000\AA\ (r-band)
range, an effect which in fact makes the metal-rich M dwarfs
degenerate in $g-r$, as the increased TiO opacities in cooler stars
balance out the reduced flux in $g$ from lower T$_{eff}$. This effect
is much weaker in metal-poor stars due to the reduced TiO opacity,
which makes metal-poor stars go redder as they are cooler, as one
would normally expect. This has been observed in late-type M
subdwarfs, which have significantly redder color that field M dwarfs
\citep{LepineScholz2008}. The color dependence of M dwarfs/subdwarfs
on metallicity is also predicted by atmospheric
models. Figure~\ref{model} (bottom-right panel) shows the predicted
$g-r$ and $r-z$ colors from the PHOENIX/BT-SETTL model of
\citet{Allard}. The models corroborate observations and predict redder
$g-r$ colors in metal-poor stars.
Although $ugriz$ photometry is not available for our stars (all of them
are too bright and saturated in the Sloan Digital Sky Survey), it is
possible to use the well-calibrated SNIFS spectrophotometry to
calculate synthetic broadband $riz$ magnitudes for the subset of stars
observed at UH. We first examine any possible correlation between the
optical to infrared $r-K_S$ color (taken as a proxy for V-I) and the
$\zeta$ index. Figure~\ref{model} plots $r-K_s$ as a function of
spectral subtype, with the dots color-coded for the $\zeta$ values of
their associated M dwarf (top-left panel). We find a tight
relationship between $r-K_s$ and spectral subtype, which we fit using
a running median. The residuals are plotted in the top-right panel,
and show no evidence of a correlation with $\zeta$. There are a
significant number of outliers with redder $r-K_s$ colors than the
bulk of the M dwarfs: these likely indicate systematic errors in
estimating the synthetic $r$ band magnitudes. The absence of any clear
correlation suggests that an optical-to-infrared color such as $r-K_S$
is not sensitive enough to detect small metallicity variations, at
least at the metal-rich end.
The synthetic $g-r$ and $r-z$ colors are plotted in Figure~\ref{model}
(lower-left panel). The redder stars ($r-z$>1.2) show a wide scatter
in $g-r$, on the order of what is predicted for stars with a range of
metallicities $-0.5<[Fe/H]<0.5$. Though we do not find a clear trend
between the synthetic $g-r$ colors and the $\zeta$ values measured in
the same stars, the high-$\zeta$ stars (red and orange dots on the
plot) do seem to have lower values of $g-r$ on average than the
low-$\zeta$ ones (green dots). The trend is suggestive of a
metallicity link to both the $g-r$ colors and the $\zeta$ values, and
should be investigated further with data of higher precision.
\section{Chromospheric activity}
\begin{deluxetable*}{lrrrrrrcccc}
\tabletypesize{\scriptsize}
\tablecolumns{11}
\tablewidth{0pt}
\tablecaption{Survey stars: distances, kinematics, and activity.\label{table_distance}}
\tablehead{
\colhead{Star name} &
\colhead{$\pi_{trig}$} &
\colhead{$\pi_{phot}$} &
\colhead{$\pi_{spec}$} &
\colhead{U} &
\colhead{V} &
\colhead{W} &
\colhead{EWHA} &
\colhead{H$\alpha$} &
\colhead{Xray} &
\colhead{UV} \\
\colhead{} &
\colhead{$\arcsec$} &
\colhead{$\arcsec$} &
\colhead{$\arcsec$} &
\colhead{km s$^{-1}$} &
\colhead{km s$^{-1}$} &
\colhead{km s$^{-1}$} &
\colhead{\AA} &
\colhead{active} &
\colhead{active} &
\colhead{active}
}
\startdata
PM I00006+1829 & \nodata & \nodata & \nodata &\nodata&\nodata&\nodata&\nodata& -& -& -\\
PM I00012+1358S & \nodata & 0.030$\pm$ 0.008& 0.028$\pm$ 0.008& -13.9& 12.1&\nodata& 0.41& -& -& -\\
PM I00033+0441 & 0.0342$\pm$ 0.0032& 0.031$\pm$ 0.008& 0.030$\pm$ 0.009& 8.5& -6.8&\nodata& 0.39& -& -& -\\
PM I00051+4547 & 0.0889$\pm$ 0.0014& 0.078$\pm$ 0.021& 0.083$\pm$ 0.024& -38.2&\nodata& -15.9& 0.47& -& -& -\\
PM I00051+7406 & \nodata & \nodata & \nodata &\nodata&\nodata&\nodata&\nodata& -& -& -\\
PM I00077+6022 & \nodata & 0.078$\pm$ 0.031& 0.091$\pm$ 0.027& -15.1&\nodata& -4.4& -3.11& Y& -& -\\
PM I00078+6736 & \nodata & 0.046$\pm$ 0.012& 0.055$\pm$ 0.016& 5.1&\nodata& -8.5& 0.39& -& -& -\\
PM I00081+4757 & \nodata & 0.071$\pm$ 0.028& 0.061$\pm$ 0.018& 7.9&\nodata& 1.8& -2.93& Y& -& Y\\
PM I00084+1725 & 0.0460$\pm$ 0.0019& 0.040$\pm$ 0.011& 0.040$\pm$ 0.012& 11.2& 0.7&\nodata& 0.41& -& -& -\\
PM I00088+2050 & \nodata & 0.078$\pm$ 0.031& 0.080$\pm$ 0.024& 9.5&\nodata& -10.1& -4.64& Y& -& Y\\
PM I00110+0512 & 0.0233$\pm$ 0.0038& 0.032$\pm$ 0.009& 0.031$\pm$ 0.009& -48.5& -14.3&\nodata& 0.35& -& -& -\\
PM I00113+5837 & \nodata & \nodata & \nodata &\nodata&\nodata&\nodata&\nodata& -& -& -\\
PM I00118+2259 & \nodata & 0.055$\pm$ 0.022& 0.054$\pm$ 0.016& -3.0&\nodata& -16.7& 0.30& -& Y& -\\
PM I00125+2142En& 0.0358$\pm$ 0.0028& 0.023$\pm$ 0.006& 0.024$\pm$ 0.007& -5.9&\nodata& -32.5& 0.44& -& -& -\\
PM I00131+7023 & \nodata & 0.037$\pm$ 0.010& 0.037$\pm$ 0.011& -6.2&\nodata& 16.8& 0.37& -& -& -
\enddata
\end{deluxetable*}
\begin{figure}
\vspace{-0.3cm}
\hspace{-0.6cm}
\includegraphics[scale=0.60]{f18.eps}
\caption{Fraction of active stars as a function of the spectral
subtype M. The rise at later subtypes is consistent with earlier
studies of field M dwarf, which shows increased activity levels in
mid-type M dwarfs. The fraction level at later subtype is however
higher than that measured in the SDSS spectroscopic
catalog.\label{frac_active}}
\end{figure}
To evaluate the presence of H$\alpha$ in emission in our M dwarfs, we
used the H$\alpha$ equivalent width index EWHA defined as:
\begin{equation}
EWHA = 100{\rm \AA} \left[ 1 - \frac{ 14{\rm \AA}
\int_{6557.61}^{6571.61} S(\lambda)
d\lambda}{ 100{\rm \AA} \left( \int_{6500}^{6550} S(\lambda) d\lambda +
\int_{6575}^{6625} S(\lambda) d\lambda \right)} \right],
\end{equation}
where $S(\lambda)$ is the observed spectrum. The EWHA index measures
the flux in a region (6557.61\AA-6571.61\AA), which includes the
$H\alpha$ line, in relation to a pseudo-continuum region spanning
6500\AA-6550\AA\ and 6575\AA-6625\AA; the calculation provides a value
in units of wavelength (\AA) like the traditional equivalent
width. Note that for an $H\alpha$ line in emission, values of the EWHA
index are negative, following convention. Assuming the W1-W2 region to
measure the true spectral continuum, then the EWHA index would measure
the true equivalent width of $H\alpha$. As it turns out, the $W1-W2$
region often has a higher mean flux than the $W3-W4$ region without
the $H\alpha$ emission component, which means that the EWHA index
systematically underestimate the equivalent width of the $H\alpha$
line. The index is however reproducible and more convenient for
automated measurement than, e.g. manual evaluation of the equivalent
using interactive software such as IRAF.
The EWHA index was measured for all spectra in our sample, and used to
flag active stars. Following \citet{West2011}, we defined a
star to be chromospherically ``active'' if EWHA $<-0.75\AA$, which
usually corresponds to a clearly detectable $H\alpha$ line in
emission. Values of the EWHA index are listed in
Table~\ref{table_distance}. Under the above criterion, 171 M dwarfs in
our survey are considered active.
\citet{Hawley1996} found that active stars (by their criterion,
EW(H$\alpha$) $> 1 \AA$) have slightly redder $V-K$ colors for the
same value of TiO5 index, an effective temperature proxy. We
calculated a median ($N=30$) locus of $V-K$ vs. TiO5 index for our
entire sample and found that 13 active stars are bluer than this
locus, while 45 are redder, seemingly confirming their result. The
large scatter in $V-K$ colors however prevents us from quantifying
this offset more precisely.
The fraction of stars that are active at each spectral subtype is shown
in Figure~\ref{frac_active}, with error bars computed from the binomial
distribution. The increase in the active fraction with spectral type
is consistent with previous studies
\citep{JoyAbt1974,Hawley1996,West2004,West2008,Kruse_etal.2010,West2011}.
Our active fractions are higher at subtype M4-M6 than the
\citet{Hawley1996} results, even when using their criterion to define active
stars (see above). This may be a result of a slightly different
definition for EW, or a result of the relatively small number of
objects. Our active fractions are also higher than the \citet{West2011}
results at each subtype. This discrepancy is likely caused by the
magnitude limit imposed in our survey: our objects are all nearby, and
relatively close to the Galactic plane (see Section 8), which makes
them statistically younger, as also suggested \citet{West2011}. The
active fractions for our stars are closer to the active fractions for
the \citet{West2011} stars in bins of stars closest to the Galactic
plane.
Another chromospheric activity diagnostic in M dwarfs is the detection
of X-rays. M dwarfs that are X-ray bright are often young, and this has
been used to identify members of nearby young moving groups
\citep[e.g.,][]{Gaidos1998, Zuckerman2001, Torres2006}, and other
young stars in the Solar Neighborhood \citep{Riaz2006}. Most recently,
\citet{Shkolnik2009,Shkolnik2012} and \citet{Schlieder2012} have used
the ratio of {\it ROSAT} X-ray flux to 2MASS {\it J} or {\it K}-band
flux to identify candidate members of young moving groups. This
technique is particularly effective for objects $\lesssim70$pc away,
which includes all the stars in our sample. The
\citet{LepineGaidos.2011} catalog, from which our targets are drawn,
was already cross-matched to the {\it ROSAT} All-Sky Bright Source
Catalog \citep{Voges.etal.1999} and the {\it ROSAT} All-Sky Survey
Faint Source Catalog \citep{Voges.etal.2000}.
We have computed the X-ray flux for our survey stars from the measured
count rate and hardness ratio (HR1) using the prescription in
\citet{Schmitt1995}. Figure~\ref{act_xray} shows the distribution of
X-ray flux as a function of $V-K$ color, for the 290 M dwarfs with
{\it ROSAT} detections. Dots are color-coded according to the strength
of the H$\alpha$ emission, as measured by the EWHA index. As
expected, M dwarfs with strong H$\alpha$ emission also tend to be more
X-ray bright. Objects with $\log F_X \ F_K > -2.6$ (above the dashed
line in Figure~\ref{act_xray}) are considered bright enough in X-ray
to qualify as chromospherically active, following the definition of
\citet{Schlieder2012}. Some 154 of the X-ray sources are active
based on their EWHA values, and all of them also qualify as active
stars based on their X-ray fluxes. On the other hand, 53 M dwarfs
identified as active based on X-ray flux do not display significant
H$\alpha$ emission in our spectra; most of them tend to be earlier M
dwarfs, in which H$\alpha$ emission is not as easily detected as in
later type objects because of their higher continuum flux near
$\lambda6563\AA$. There are also 22 stars in our survey which are active
based on $H\alpha$ but are not detected by ROSAT. This suggests that
only two thirds of the ``active'' stars will be diagnosed as such from
both X-ray and H$\alpha$ emission, while the other third will show
only either. This could be due to source confusion in the ROSAT X-ray
survey, variability in either X-ray or H$\alpha$ emission, or, in the
case of the X-ray flux, non-uniform sky coverage by ROSAT. We
calculated the luminosity ratio index $L_X/L_{H\alpha}$ of active
stars, as defined by the criterion EW(H$\alpha$) $>1 \AA$ following
the procedure of \citet{Hawley1996}, and adopting the relation $V-R
\approx 0.7 + 0.06 {\rm SpTy}$ to estimate an $R$ magnitude and the
continuum flux at the H$\alpha$ line. We find that the ratio is
insensitive to bolometric magnitude and spectral type, and has a
median value of 0.85. This is higher than the \citet{Hawley1996}
average of $\sim 0.5$, and may in part be due to Malmquist bias in the
flux-limited ROSAT survey favoring the inclusion of the most X-ray
luminous stars, as well as greater variation in the ratio because of
the elapsed time (two decades) between the ROSAT survey and
our observations.
\begin{figure}
\vspace{-0.3cm}
\hspace{-0.6cm}
\includegraphics[scale=0.53]{f19.eps}
\caption{X-ray luminosity normalized by the flux in the infrared K$_s$
band, plotted as a function of the optical-to-infrared color $V-K$,
for stars in our sample which have counterparts in the ROSAT all-sky
points source catalog. The color scheme shows the H$\alpha$ equivalent
width; active stars are found to have large X-ray flux, as expected
from chromospheric activity.\label{act_xray}}
\end{figure}
\begin{figure}
\vspace{-0.3cm}
\hspace{-0.6cm}
\includegraphics[scale=0.53]{f20.eps}
\caption{Normalized near-UV flux as a function of the
optical-to-infrared $V-K_s$ color. The color scheme shows the
strength of the H$\alpha$ equivalent width. Closed circles show
stars identified as active based on X-ray emission, closed circles
show stars with low or no detection in ROSAT. \label{act_uv}}
\end{figure}
Active stars can also be identified from ultra-violet
excess, as suggested in Section 3.4. \citet{Shkolnik2011,Shkolnik2012}
showed that {\it GALEX} UV fluxes can identify young M dwarfs in
nearby moving groups, and can identify active stars to larger
distances. Figure~\ref{act_uv} shows the {\it GALEX} NUV to 2MASS
{\it J} flux ratio, for the objects with UV detections. The dashed
line shows the selection criteria of \citet{Shkolnik2011}. Dot colors
represent the EWHA index values for the stars, while filled circles
indicate objects with $\log F_X / F_K > -2.6$, i.e. stars whose X-ray
flux does not identify them as being active. Overall, there is a good
correpondence between the different activity diagnostics. However,
there are some stars identified as active based on UV flux that are
not identified as such from their X-ray and/or $H\alpha$
emission. Again this suggests that a complete identification of active
M dwarfs in the solar vicinity may require a combination of diagnostic
features.
In any case our survey, which combines X-ray, UV, and $H\alpha$
diagnostics, provides a valuable subset for identifying low-mass young
stars in the Solar Neighborhood, and may potentially yield new members
of young moving groups, or even the identification of new moving
groups. The last three columns in Table~\ref{table_distance} display
flags for stars found to be active from either H$\alpha$, X-ray, or UV
flux. The flag indicates activity by a ``Y''. Absence of a flag does not
necessarily indicate absence of activity: the GALEX survey does not
cover the entire sky, and the ROSAT X-ray survey is not uniform in
sensitivity, so a non-detection in either does not necessarily
indicate quiescence. Activity diagnostics could also be
time-variable. H$\alpha$ equivalent width is particular are know to be
variable on various timescales \citep{Bell2012}. In any case, there is
a good correlation between the different diagnostics. We flag 175
stars as active based on H$\alpha$, 42 based on X-ray emission, and
172 based on UV excess. Overall, 252 stars are assigned one or more
activity flags: 19 stars have all three flags on, 99 stars get two
flags, and 137 get only one.
\section{Distances and kinematics}
\subsection{Spectroscopic distances}
\label{sec:distances}
\begin{figure}
\vspace{0.0cm}
\hspace{-0.2cm}
\includegraphics[scale=0.85]{f21.eps}
\caption{Absolute visual ($M_V$) and infrared ($M_J$) magnitudes for
the 631 stars in our survey for which geometric parallax
measurements exist in the literature. Top panels: absolute
magnitudes against $V-J$ color, which follow the color-magnitude
relationship used in \citet{LepineGaidos.2011} to estimate
photometric distances. Bottom panel: absolute magnitudes as a
function of spectral subtype, based on the spectral-index
classification described in this paper. Active stars are plotted
in green, and are found to be overluminous at a given $V-J$ color
and given spectral subtype, compared with non-active stars. The
offset in notable for stars of earlier M subtypes (or bluer $V-J$
colors).\label{dist_cal}}
\end{figure}
Astrometric parallaxes are available for 631 of the stars in our
sample, spanning the full range of colors and spectral subtypes. We
combine these data with our spectroscopic measurements to re-evaluate
photometric and spectroscopic distances calibrations for M dwarfs in
our census. Absolute visual magnitudes $M_V$ are calculated and are
plotted against both $V-J$ color and spectral subtype in
Figure~\ref{dist_cal}. M dwarfs with signs of activity (H$\alpha$, UV,
X-ray) are plotted in green, other stars are plotted in
black. The solid red lines are the best-fit second order polynomials,
when both active and inactive stars are used, and after elimination of
3$\sigma$ outliers. The equations for the fits, where
$spT$ is the spectral type (K7 is -1 and M0=0, etc) are:
\begin{eqnarray}
M_J = 1.194 + 1.823 (V-J) - 0.079 (V-J)^2\\
M_J = 5.680 + 0.393 (SpT) + 0.040 (SpT)^2
\label{eqn:mj_vs_spt}
\end{eqnarray}
where SpT are the spectral types, and with the least-squares fit
performed after exclusion of 3-sigma outliers. The 1$\sigma$
dispersion about these relationships are $\pm0.61$ mag for
(M$_J$,V-J), and $\pm0.52$ mag for (M$_J$,SpT). The smaller scatter in
the spectroscopic relationship suggests that spectroscopic distances
may be marginally more accurate than the photometric ones. We suspect
that the larger uncertainty on the photographic $V$ magnitudes may be
the cause.
The most notable feature in the diagrams is that active stars appear
to be systematically more luminous than non-active stars. This
corroborates the observation made previously by \citet{Hawley2002}, as
part of the PMSU survey. \citet{Hawley2002} found that active stars
were more luminous by 0.48mag in a diagram of $M_K$ against TiO5
index, used as proxy for spectral subtype. For stars in our census, we
find that among stars of spectral subtype M2 and earlier, active stars
are on average 0.46mag more luminous at a given subtype than
non-active stars; in the color-magnitude diagram, bluer stars of
colors $V-J<4.0$ which are active are on average 0.47mag more luminous
than non-active stars of the same color. Both values agree well with
the values quoted by \citet{Hawley2002}. The systematic overluminosity
of active stars is also responsible for some of the scatter in the
color-magnitude and spectral-type magnitude relationships. If we
exclude active stars, the scatter about the color-magnitude
relationship fals marginally to $\pm0.58$, and the scatter in the
spectral-type magnitude relationship falls to $\pm0.49$.
The offset in absolute magnitude between active and non-active stars
suggests that spectroscopic and photometric distances would be more
accurate for active stars in our census if their estimated absolute
magnitudes were made 0.46mag brighter that suggested by
Equation~\ref{eqn:mj_vs_spt}. We therefore adopt the following
relationships to be applied only on active stars of subtype M2.5 and
earlier:
\begin{eqnarray}
{[M_J]}_{early-active} = 0.734 + 1.823 (V-J) - 0.079 (V-J)^2\\
{[M_J]}_{early-active} = 5.220 + 0.393 (SpT) + 0.040 (SpT)^2
\label{eqn:mj_vs_spt2}
\end{eqnarray}
Again we define as ``active'' any star which qualifies as such base on
any one of our criteria (H$\alpha$, UV, X-ray). There are several
reasons that would explain why active, early-type stars are more
luminous at a given subtype. Activity in an early-type M dwarf could
mean that the star is younger \citep{Delfosse_1998}; early-type M
dwarfs with ages $<100$Myr are known to be overluminous at a given
color, due to lower surface gravity \citep{Shkolnik2012}. Older stars
could remain active due to interaction with a close companion
\citep{Morgan_2012}, in which case the active stars would also appear
overluminous due to this unresolved companion. Late-type M dwarfs,
however, can remain active for long periods of time, and would thus
not require the star to be young or have a close companion.
Splitting the active and inactive stars results in lowering the
scatter of non-active stars in the subtype-magnitude
relationship (to $\pm$0.5mag). Overall, our spectroscopic distances
for non-active M dwarfs provide a $1\sigma$ uncertainty of $\pm26\%$
on the distance. For active stars, we find a scatter of $\pm$0.6mag,
which suggests distance uncertainties of $\pm32\%$. Our photometric
distances estimated from (V,V-J) have similar though perhaps slightly
larger uncertainties. Photometric and spectroscopic parallaxes,
estimated from the above relationships for active and non-active
stars, are listed in Table~\ref{table_distance}.
We estimated the effect of two relevant sampling biases on the
calibration between $M_J$ and $V-J$ color. In Eddington bias,
photometric errors scatter more numerous, bluer, and intrinsically
brighter stars to redder apparent colors than redder stars are
scattered to bluer apparent colors \citep{Eddington1913}. The net
effect is to make stars at a given apparent color appear more luminous
than they are. In Lutz-Kelker (LK) bias, a form of Malmquist bias,
errors in trigonometric prallax will scatter more numerous and more
distant stars with lower parallax to higher apparent parallax values,
making them appear less luminous than they are \citep{Lutz1973}. By
taking the derivative of $M_J$ with respect $V-J$ color, multiplying
by the derivative of the number of stars in our $J$-magnitude-limited
catalog with respect to $M_J$, assuming that the errors in $V-J$ are
gaussian-distributed with standard deviation $\sigma_{V-J}$, and
integrating over the distribution, we find the Eddington bias in $M_J$
to be;
\begin{equation}
\Delta_E = -\ln 10 \left[1.918 = 0.178\left(V-J\right)\right]^2 \left(0.6 -
0.4\gamma\right) \sigma_{V-J}^2,
\end{equation}
where $\gamma$ is the power-law index of a luminosity function for M
stars which we take to be 0.325. Performing a similar derivation for
the effect of L-K bias on $M_J$, we find:
\begin{equation}
\Delta_{LK} = \frac{15}{\ln 10} \sigma_{\pi}^2,
\end{equation}
where $\sigma_{\pi}$ is the fractional error in parallax. Using
published parallax errors for our {\it Hipparcos} stars and adopting a
conservative $\sigma_{V-J} = 0.05$, we find that L-K bias usually
dominates over Eddington bias and that 74\% of our stars have a total
bias of less than +0.2 magnitudes. A running median ($N = 100$)
vs. $V-J$ color is highest ($\sim 0.15$) for the bluest ($V-J =
2.7$) stars and falling to less than +0.05 magnitudes for $V-J > 3.3$.
An analogous analysis can be performed for the bias in $M_J$
vs. spectral type, with a similar result. To debias values of $M_J$,
these values should be {\it subtracted} from our calibration but we do
not perform that operation here because of the small magnitude of the
effect, which would overestimate distances by about 2\% on
average. The correction would also seem negligible compared with the
intrinsic scatter in our color-magnitude and subtype-magnitude
relationships are of order $\pm0.5-0.6$, much larger than the L-K
correction.
\begin{figure}
\vspace{-0.3cm}
\hspace{-0.2cm}
\includegraphics[scale=0.44]{f22.eps}
\caption{Top: distribution of spectroscopic distances for the stars in
our survey, shown for three ranges of spectral subtypes. Early-type
stars are clearly sampled over a larger volume, which explains why
they dominate our survey. Bottom: distribution of Galactic scales
heights for the same stars, assuming that the Sun is hovering 15~pc
sbove the Galactic midplane. As expected from our magnitude-limited
sample, stars of later spectral subtypes (and lower luminosity) are
found at shorter distances. Our survey samples a region well within
the Galactic thin disk.\label{disthist}}
\end{figure}
Figure \ref{disthist} shows the distribution of photometric distances
for our complete sample using Equation \ref{eqn:mj_vs_spt} and the
M$_J$=M$_J$(V-J) color-magnitude relationship. The spectral subtypes are
plotted in separate colors and demonstrates that the earlier-type
stars, which are intrinsically brighter, are sampled to significantly
larger distances compared with the later-type stars. In the 20pc
volume, the M3-M4 stars still appear to dominate. We also plot the
Galactic height of the stars in our sample, adopting a Galactic height
of 15 pc for the Sun \citep{Cohen1995,Ng1997,Binney1997}. It is clear
that our survey is largely contained within the Galactic thin disk,
and barely extends south of the midplane. This is consistent with the
relative absence of metal-poor stars associated with the thick disk
and halo.
\subsection{Kinematic analysis}
\begin{figure*}
\vspace{-0.3cm}
\hspace{0.5cm}
\includegraphics[scale=0.82]{f23.eps}
\caption{Velocity-space projections for the M dwarfs in our
survey. Velocies are calculated based on photometric distances and
proper motions alone (no radial velocities used). Each star in our
census is thus displayed in only one panel, which corresponds to the
projection in which the radial velocity of the star has the smallest
contribution. Stars with significant levels of H$\alpha$ emission,
i.e. chromospherically active M dwarfs, are plotted in
red. \label{uvw}}
\end{figure*}
Accurate radial velocities are not available for most of the stars in
our sample, which prevents us from calculating the full (U,V,W)
components of motion for each individual star. However, it is possible
to use the distance measurements (for stars with parallaxes) or
estimates (for stars with no parallax), and combine them to the proper
motions to evaluate with some accuracy at least two of these
components for each star. More specifically, we calculate the (U,V,W)
components by assuming that the radial velocities $R_V=0$. We then
consider the (X,Y,Z) positions of the stars in the Galactic reference
frame, from the distances and sky coordinates. For stars with the
largest component of position in +X or -X, the radial velocity mostly
contibute to U, and has minimal influence on the values of V and
W. Likewise stars with the largest component of position in +Y or -Y
(+Z or -Z) make good tracers of the velocity distribution in U and W
(U and V). We use this to assign any one of (U,V) or (U,W) or (V,W)
velocity component doublet to every M dwarf in our catalog.
Estimated values of the components of velocity are listed in
Table~\ref{table_distance}. For each star, one of the components is
missing, which is the component that would most depend on the radial
velocity component based on the coordinates of the star. Again, the
other two components are estimated only from proper motion and
distance. For the distance, we use the trigonometric parallaxes
whenever available; otherwise the spectroscopic distances as are used,
based on the raltionships described in the previous section.
The resulting velocity distributions are displayed in
Figure~\ref{uvw}. We measure mean values of the velocity components
using all allowable values and find:
\begin{displaymath}
<U> = -8.1~{\rm km s^{-1}}, \sigma_U = 32.8~{\rm km s^{-1}},
\end{displaymath}
\begin{displaymath}
<V> = -17.0~{\rm km s^{-1}}, \sigma_V = 22.8~{\rm km s^{-1}},
\end{displaymath}
\begin{displaymath}
<W> = -6.9~{\rm km s^{-1}}, \sigma_W = 19.3~{\rm km s^{-1}}.
\end{displaymath}
The values are also largely consistent with those found from the PMSU
survey and desribed in \citep{Hawley1996}. They are also remarkably
similar to the moments of the velocity components calculated by
\citet{Fuchs2009} for SDSS stars of the Galactic thin disk, and which
are $<U> = -8.6, \sigma_U = 32.4$, $<V> = -20.0, \sigma_V = 23.0$, and
$<W> = -7.1, \sigma_W = 18.1$. The agreement suggests that our
distance estimates are reasonably accurate, and it corroborates
earlier results about the kinematics of the local M dwarf population
which indicate a larger scatter of velocities in $U$. The mean values
of $<U>$ and $<V>$ are consistent with the offsets from the local
standard of rest as described in \citet{DehnenBinney_1998}
Active stars are found to have significantly smaller dispersions in
velocity space. All 252 stars with at least one activity flags
(i.e. stars found to be active either from H$\alpha$, X-ray flux, or
UV excess) are plotted in red in Figure~\ref{uvw}. Those active stars
have first and second moments:
\begin{displaymath}
<U> = -9.3~{\rm km s^{-1}}, \sigma_U = 25.2~{\rm km s^{-1}},
\end{displaymath}
\begin{displaymath}
<V> = -13.4~{\rm km s^{-1}}, \sigma_V = 16.8~{\rm km s^{-1}},
\end{displaymath}
\begin{displaymath}
<W> = -6.6~{\rm km s^{-1}}, \sigma_W = 15.3~{\rm km s^{-1}}.
\end{displaymath}
The values are consistent with \citep{Hawley1996}, who also reported
that active M dwarfs tend to have a smaller scatter compared with
inactive M dwarfs. The smaller dispersion values suggest that these
active stars may be significantly younger than the average star in the
Solar Neighborhood.
In any case, we also note that the velocioty space distribution is
non-uniform and shows evidence for substructure. Our M dwarf data
shows velocity-space substructure as that observed and described in
\citet{Nordstrom2004,Holberg2008} for solar-type stars in the vicinity
of the Sun. This substructure is sometimes referred to as ``streams''
or ``moving groups'', although an analysis by \citet{Bovy_Hog2010}
shows that these groups do not represent coeval populations arising
from star-formation episodes. The velocity-space substructure is more
likely transient and associated with gravitational perturbations which
are the signature of the Galactic spiral arms
\citet{QuillenMinchev2005} and the Galactic bar \citet{Minchev2012}. A
simple description of the velocity-space distribution in terms of
means values and dispersions, or as velocity ellipsoid, is therefore only a
crude approximation of a more complex and structured distribution.
Finally, we note that the stars of our catalog that were previously
part of the CNS3 and stars with measured trigonometric parallaxes
(e.g. from the Hipparcos catalog) tend to have larger velocity
dispersions, with $(\sigma_U,\sigma_V,\sigma_W)$=(38.4,25.9,23.1),
while the newer stars have
$(\sigma_U,\sigma_V,\sigma_W)$=(26.2,19.4,15.3). The difference could
be due to systematic underestimation of the photometric/spectroscopic
distances, but a more likely explanation is that the CNS3 and parallax
subsample suffers from proper motion selection. This is because most
of the CNS3 stars and M dwarfs monitored with Hipparcos were selected
from historic catalogs of high proper motion stars, which have a
higher limit than the SUPERBLINK proper motion catalog used in the
LG2011 selection. This kinematic bias means that the current subset of
M dwarfs monitored for exoplanet programs suffers from the same
kinematic bias, which could possibly introduce age and metallicity
selection effects.
\section{Conclusions}
We have now compiled spectroscopic data for a nearly complete list of
M dwarfs in the northern sky with apparent magnitudes $J<9$. Our
survey identifies a total of 1,403 very bright M dwarfs. Our new
catalog provides spectral subtypes and activity measurements
($H\alpha$ emission) for all stars, as well as a rough indicator of
metallicity in the guise of the $\zeta$ parameter, which measures the
ratio of TiO to CaH bandstrengths. Only one of the stars in the survey
is unambiguously identified as a metal-poor M subdwarf (PM I20050+5426
= V1513 Cyg).
Our target stars were identified from the all-sky catalog of bright M
dwarfs presented in \citet{LepineGaidos.2011}. As such, our
spectroscopic survey suffers from the same selection effects and
completeness issues. The completeness and bias of the SUPERBLINK
proper motion survey, from which these stars were selected is
discussed a length in \citet{LepineGaidos.2011}. In the northern
hemisphere, the SUPERBLINK catalog is complete for proper motions $\mu
>40$~mas$^{-1}$. We show in \S4 that there is a kinematic bias in the
catalog which excludes stars with very low transverse motions (in the
plane of the sky), but the low proper motion limit means that less
than 5\% of stars within 65 parsecs of the Sun are in fact excluded in
the selection. In addition, we estimate that $\approx$5\% of the nearby,
bright M dwarfs may have escaped our target selection scheme due to
faulty magnitudes. Therefore, we estimate that our census most likely
include $>90\%$ of all existing M dwarfs in the northern sky with
$J<9$. Early-type K7-M1 dwarfs have absolute magnitudes
$M_J\approx5.5$, and our $J<9$ sample thus identifies them well to a
distance of about 50pc, as confirmed in Figure~\ref{disthist}. On the
other hand, later type M3-M4 dwarfs have $M_J\approx8$ and thus only
those at very close distance range ($<$15~pc) will be included in the
catalog. Their completeness will however be very high because the
proper motion bias excludes less than $1\%$ of the stars within that
distance range. In any case, the different survey volumes for
early-type and late-type stars means that our survey favors the former
over the latter by a factor of about 35 to 1. It is thus no surprise
that our spectroscopic catalog is dominated by early-type M dwarfs.
An important result of our spectroscopic analysis is the
identification of systematic errors in the spectral indices, which
measure the strenghts of the CaH, TiO, and VO molecular
bands. Systematic offsets between data obtained at MDM Observatory and
at the University of Hawaii 2.2-meter telescopes, as well as offsets
between these and the values measured for the sames stars in the
Palomar-MSU survey of \citet{Reid1995}, indicate that these spectral
indices are susceptible to spectral resolution and spectrophotometric
calibration, such that using the raw measurements may result in
systematic errors in evaluating spectral subtypes and the metallicity
$\zeta$ parameter. In Section 3.2 we outline a procedure for
calculating corrected indices, based on a calibration of systematic
offsets between two observatories. A proper calibration requires that
large numbers of stars be re-observed every time a new observatory
and/or instrumental setup is used, in order to calibrate the offsets
and correct the spectral indices. Only the corrected spectral indices
can be used reliably in the spectral subtype and $\zeta$
relationships, which are calibrated with respect to the corrected
values. We adopt the Palomar-MSU measurement as our standard of
reference for the spectral indices, and correct our MDM and UH
values accordingly.
In the end, this catalog provides a useful list of targets for
exoplanet searches, especially those based on the radial velocity
variation method. Current methods and instruments require relatively bright
stars to be efficient, and the stars presented in our spectroscopic
catalog all constitute targets of choice, having been vetted for
background source contamination. Our accurate spectral types will be
useful to guide radial velocity surveys and selct stars of
comparatively lower masses.
We also provide diagnostics for chromospheric activity from H$\alpha$
emission, X-ray flux excess, and UV excess. Besides being useful to
identify more challenging sources for radial velocity surveys, they
also isolate the younger stars in the census. Follow-up radial
velocity observations could tie some of the stars to nearby moving
groups, and these objects would be prime targets for exoplanet
searches with direct imaging methods.
\acknowledgments
{\bf Acknowledgments}
This material is based upon work supported by the National Science
Foundation under Grants No. AST 06-07757, AST 09-08419, and AST
09-08406. We thank Greg Aldering for countless instances of
assistance with SNIFS, the telescope operators of the UH 2.2m
telescope, and Justin Troyer for observing assistance. We thank Bob
Barr and the staff at the MDM observatory for their always helpful
technical assistance.
| {
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Rolf Erik Johan Samuelsson Toloue, ursprungligen Samuelsson, född 17 oktober 1971 i Hejnums församling i Gotlands län, är en svensk musikalartist.
Biografi
År 1987 vann han länsfinalen av Musik Direkt med rockgruppen B.A.G. Han medverkade som solist i Talang 89 samt deltog i den lokala tävlingen Tiljans Talanger. Genom deltagandet i Musik Direkt med B.A.G. blev det en turné 1987–1988 i Frankrike, Belgien, Luxemburg och Kanada. Samuelsson utbildade sig i dans vid Visby dansskola samt vid Kulturamas tvååriga musikalartistutbildning i Stockholm.
Samuelsson har ägnat sig åt tävlingsdans och blev den första svenska Europamästaren i rock'n'roll 1984, då tillsammans med partnern Eva Johansson. År 1985 blev han svensk mästare i rock'n'roll. Han har vunnit Lag-SM där hans tävlingsform var jitterbug. Vid nordiska mästerskapen i disco freestyle i Norge 1989 kom han tvåa.
Den första stora musikalen han medverkade i var Sound of Music på Göta Lejon i Stockholm 1996, där han gjorde rollen som Rolf. Han har vidare medverkat i West Side Story, Grease, Cats, Elisabeth, Jesus Christ Superstar, Tartuffe, Djungelboken, Sugar, Dirty Dancing vid produktioner i Danmark, Italien, Sverige och Tyskland.
I Tyskland finns han samlings-CD:n Musical Stars 2, med låten "To where you are", skriven av Linda Thomson och Richard Marx.
År 1996 tilldelades han Sigge Fürsts minnespris.
Samuelsson har även medverkat i svensk TV, bland annat Razzel, Lilla sportspegeln och Hylands hörna.
Filmografi
Katsching - Lite pengar har ingen dött av
Sherlock Gnomes (röst)
Askungen (röst)
Musikteater
2020 - Grease
2019 - Annie
2018 - Sound Of music
2017 - Das Phantom Der Oper
2015-2016 - Das Phantom Der Oper
2011-2010 - Chicago
2009-2008 - Shadows of Motown
2007 - Mangrant
2007 - Karibiens pirater
2006 - West Side Story
2006 - West Side Story
2006 - Djungelboken
2005 - Bratpack show
2005-2004 - Sugar
2003 - Cats
2003-2002 - Elisabeth
2002-2001 - Jesus Christ Superstar
2000 - West Side Story
2000 - Tartuffe]
1999 - Dirty Dancing
1998 - West Side Story
1998 - West Side Story
1998-1997 - Grease
1997-1996 - Sound of Music
1996 - Guldhatten
1995 - Dreamscape
1994 - Grönsakernas Värld
1994 - Den heliga natten
Referenser
Noter
Webbkällor
Johan Samuelsson stagement.se
Östgötateaterns skrivna biografi
CD Musical Stars Volume 2
Tysk musikalsida
Födda 1971
Män
Svenska musikalartister
Levande personer
Personer från Gotland | {
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{"url":"https:\/\/simple.wikipedia.org\/wiki\/Baud","text":"Baud\n\nVarious modulation schemes invented in the 20th century such as Phase-shift keying can make bit rates much higher than signal rates. In another example, gigabit ethernet has a symbol rate of 125MBd. Gigabit ethernet uses pulse-amplitude modulation and can transmit two bits of payload data per symbol. Gigabit ethernet uses four wires for transmission. It can transmit ${\\displaystyle {\\rm {125\\;MBd\\cdot 2\\;{\\tfrac {bit}{symbol}}\\cdot 4=1000\\;{\\tfrac {Mbit}{s}}}}}$.","date":"2017-10-16 22:07:08","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 1, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.6254438757896423, \"perplexity\": 2046.9661500445682}, \"config\": {\"markdown_headings\": false, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2017-43\/segments\/1508187820466.2\/warc\/CC-MAIN-20171016214209-20171016234209-00207.warc.gz\"}"} | null | null |
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| {
"redpajama_set_name": "RedPajamaGithub"
} | 6,872 |
Al Duhail loan Ahmed Yasser to Vissel Kobe
QNB Stars League title holders Al Duhail agreed the loan deal of defender Ahmed Yasser to Japan's Vissel Kobe.
Ahmed was loaned out to Al Rayyan in last season's January transfer window after he completed another loan spell at Spanish club Cultural Leonesa.
The 24-year-old had displayed a high level in the AFC Champions League as well and that caught the attention of top J1 League club Vissel Kobe, who sought him on loan to assist the likes of Spanish star Andres Iniesta.
Meanwhile, Al Duhail forward Ismail Mohammed successfully underwent an operation after he sustained a double fracture on his left leg during their 3-0 QNB Stars League Week 1 win over promoted team Al Shahania on Sunday.
He will be out of action at least for six to seven months.
Ismail was fouled in the 52nd minute of their match by Al Shahania's Nadir Awadh. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 3,023 |
Home Full Mail Server Solution w/ Virtual Domains & Users (Debian Etch, Postfix, Mysql, Dovecot, DSpam, ClamAV, Postgrey, RBL) - Page 10 >
Full Mail Server Solution w/ Virtual Domains & Users (Debian Etch, Postfix, Mysql, Dovecot, DSpam, ClamAV, Postgrey, RBL) - Page 10
VII. Secure Email
A. SSL Certificates
1. Self-signed server certificate
2. CA-Signed Certificate
In an ideal world, our users would be able to send/receive email whenever they were on the net, from any place in the world. Unfortunately, that's incredibly insecure... passwords are being tossed back and forth in plain text via the SMTP and IMAP protocols, and that means anyone who wanted to could just 'snoop' the password.
If your users don't need direct access to the mail solution, then don't give it to them! There's no point in stressing over a secure email setup if all your users need is Webmail! Just make their connection to the webmail server secure, and make sure that the webmail server uses a secure network connection when talking to your mail servers. Problem solved! If, on the other hand, your users do require the ability to send/receive mail via the internet without using webmail, well then, that just makes things more difficult. Not impossible, just difficult.
So, here's your problem: SMTP and IMAP send passwords over clear text. You can have them send passwords using MD5, but basic MD5 can be hacked. You can have them send passwords using MD5CRYPT, but then you're dealing with multiple implementations (not to mention the fact that not all email clients support MD5 passwords). The solution? TLS (Transport Layer Security). We're going to set up our solution to support an encrypted connection over the internet. While we could modify some of our existing servers to handle this, there's no point in over-complicating their setups. We're just going to run a seperate server to handle all of this: secure-mail.example.com
NOTE: In the original scenario, the small business had multiple static IP addresses. Since this is the case, we were able to run SMTP+TLS on port 25, if you don't have multiple IP addresses, then that is not possible. The reason is simple enough: While IMAPS(secure IMAP) runs a different port (993) than standard IMAP (143), SMTP+TLS runs on the SAME port as SMTP (25). So, using a firewall to route based on ports allows you to run seperate IMAP and IMAPS servers, but no firewall in the world can route port 25 to two different machines. Even with all of that though, you could always just run SMTP+TLS on a non-standard port... heck, it would even be more secure.
So, with all of that in mind, we're going to setup a secure mail server, which uses SMTP+TLS for sending mail, and IMAPS for receiving it.
The simplest form of encryption is having a simple self-signed certificate on the server. This will generate a warning message when the clients first connect, but they should be able to save it for further use. It is not really secure, since anyone can execute a man-in-the-middle attack if you don't save the certificate.
The next level is using a server certificate signed by a Certificate Authority (CA), either a commercial one, or perhaps the company internal CA. This way, the server certificate will be trusted, and if you now receive a warning, there is potentially something bad going on.
Last but definitely not least is using client certificates for logging in to the server, and using a server certificate to authenticate the server to the clients. This is quite secure, but it is not supported in all mail clients. Thunderbird among others do have support for it.
First create the directories, create the private key, and lastly create the certificate.
# mkdir -p /etc/ssl/example.com/mailserver/
# cd /etc/ssl/example.com/mailserver/
# openssl genrsa 1024 > mail-key.pem
# chmod 400 mail-key.pem
# openssl req -new -x509 -nodes -sha1 -days 365 -key mail-key.pem > mail-cert.pem
Note that "Common Name (eg, YOUR name)" MUST match the name of the server, which in this case is secure-mail.example.com
Using a real CA-signed certificate is no different from using a self-signed one. It's just another step in the key-pair creation. If your company has its own CA, then they should issue a certificate for the mail server. A Google search for be your own ca will give you enough answers to create one yourself, if you have the need.
Full Mail Server Solution w/ Virtual Domains & Users (Debian Etch, Postfix, Mysql, Dovecot, DSpam, ClamAV, Postgrey, RBL) - Page 10 - Page 1
Full Mail Server Solution w/ Virtual Domains & Users (Debian Etch, Postfix, Mysql, Dovecot, DSpam, ClamAV, Postgrey, RBL) - Page 10 - Page 11
By: Vecter
postfix, mysql, security, antivirus, debian
How to Compile A Kernel - The Debian Way
Configuring Apache for Maximum Performance
How to Install Podman as Docker alternative on Debian 11
How to set up an NFS Mount on Rocky Linux 8
The Perfect Server - Ubuntu 18.04 (Nginx, MySQL, PHP, Postfix, BIND, Dovecot, Pure-FTPD and ISPConfig 3.1)
How to Install Redis on Ubuntu 20.04
The Perfect Server - Ubuntu 18.04 (Bionic Beaver) with Apache, PHP, MySQL, PureFTPD, BIND, Postfix, Dovecot and ISPConfig 3.1 | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 8,482 |
\@startsection{section}{1}{\z@{The MECS Point Spread Function}
The Medium Energy Concentrator Spectrometer (MECS; Boella et al. 1997/a) is one of the four
narrow
field instruments on board the Beppo-SAX satellite (Boella et al. 1997/b).
The MECS operates in the energy band 1.3-10 keV with a field of view of $28'$
radius.
The MECS consists of three units each composed of a grazing incidence Mirror
Unit (MU), and of a position sensitive Gas Scintillation Proportional Counter
(GSPC).\\
The Point Spread Function of the MECS (PSF$_{\rm MECS}$) is the convolution between the
MU PSF and the detector PSF.
The MU and detector PSFs are described, respectively, by a lorentzian function L(r) and
a gaussian function G(r).
Both the lorentzian and the gaussian functions are energy-dependent.
Typically the detector PSF improves with
increasing energy, whereas the MU PSF improves with decreasing energy.
The detector PSF dominates the core of the PSF$_{\rm MECS}$ ($r \lower.5ex\hbox{\ltsima} 2'$) whereas the MU PSF
dominates the wings of the PSF$_{\rm MECS}$ ($ \lower.5ex\hbox{\gtsima} 2'$).\\
The analytical expression for the on-axis PSF as given in Boella et al. 1997/a is:
$${\rm
PSF_{MECS}}(r,E)=\frac{1}{2\pi\Big[R(E)\sigma^{2}(E)+\frac{r^{2}_{l}(E)}
{2(m(E)-1)}\Big]}\cdot $$
\begin{equation}
\Bigg\{R(E)\exp\Bigg(-\frac{r^2}{2\sigma^2(E)}\Bigg)+\Bigg[1+\Bigg(\frac{r}
{r_l(E)}\Bigg)^2\Bigg]^{-m(E)}\Bigg\},
\end{equation}
where $R(E)$, $\sigma(E)$, $r_{l}(E)$ and $m(E)$ are algebraic functions of the energy E.\\
The integral of the PSF over the entire plane is normalized to unity:\\
$2\pi\int_{0}^{\infty}{\rm PSF_{\rm MECS}}(r,E)r\,dr\equiv1$.\\
We used eq. 1 to evaluate the $50\%$ and $80\%$ power radii
($r_{50}(E)$ and $r_{80}(E)$) as a function of E:\\
\vskip -0.5truecm
$$2\pi\int_{0}^{r_{50}(E)}{\rm PSF_{\rm MECS}}(r,E)r\,dr=0.5,$$
\begin{equation}
2\pi\int_{0}^{r_{80}(E)}{\rm PSF_{\rm MECS}}(r,E)r\,dr=0.8\,.
\end{equation}
\begin{figure}[htb]
\vspace{9pt}
\epsfig{figure=proc_1.ps, height=7.5cm, width=7.5cm, angle=-90}
\caption{ {\rm PSF}$_{\rm MECS}$ Power Radii vs. Energy. The solid line represents the 50 \% power
radius, r$_{50}$(E), the dashed line represents the 80 \% power radius, r$_{80}$(E) (see eq. 2).}
\end{figure}
As shown in fig. 1 $r_{50}(E)$ is always $<2'$ and decreases with increasing energy.
This is because at radii $<2'$ the PSF$_{\rm MECS}$ is dominated by the gaussian PSF of
the detector, that improves with increasing energy.
The power radius $r_{80}(E)$ does not vary strongly with energy (see fig. 1)
because of the combined effect of the improvement of the detector PSF and degradation of the MU PSF
with increasing energy.
\begin{figure}[htb]
\vspace{9pt}
\epsfig{figure=proc_2.ps, height=7.5cm, width=7.5cm, angle=-90}
\caption{ Ratios between convolved profiles vs. radius r. Solid line: Ratio R$_{1}$(r) between
the convolved profile $\tilde{I}$(r,3 keV) and the convolved profile $\tilde{I}$(r,6 keV).
Dashed line: Ratio R$_{2}$(r) between the convolved profile $\tilde{I}$(r,9 keV) and the convolved
profile $\tilde{I}$(r,6 keV)}
\end{figure}
\begin{figure}[htb]
\vspace{9pt}
\epsfig{figure=proc_3.ps, height=7.5cm, width=7.5cm, angle=-90}
\caption{ Correction vectors for the Virgo cluster observed with the MECS. Dot-dot-dot-dashed line:
correction vector V$_{1}$(E$_{i}$) for the region $0'-2'$. Dashed line: correction vector
$V_{2}$(E$_{i}$) for the region $2'-4'$. Dash-dotted line: correction vector $V_{3}$(E$_{i}$) for
the region $4'-6'$. Dotted line: correction vector $V_{4}$(E$_{i}$) for the region $6'-8'$.
Solid line: correction vector $V_{5}$(E$_{i}$) for the region $8'-10'$. }
\end{figure}
\@startsection{section}{1}{\z@{Convolution of the Source Radial Profile with the PSF$_{\rm MECS}$}
A proper analysis of extended sources, like clusters of galaxies, requires
that the blurring introduced by the limited spatial resolution of the observing
instruments be correctly taken into account.
In practice, this is done by evaluating the convolution of the source surface
brightness profile, $I$, with the PSF$_{\rm MECS}$. We have approximated $I$ with the profile
of the Virgo cluster as observed with the PSPC instrument on board the ROSAT satellite,
because of the considerably better spatial resolution of this instrument respect to the
MECS.\\
The ROSAT Virgo cluster profile, $I_{\rm ROSAT}(r_{0})$, convolved with the PSF$_{\rm MECS}$
at any point P($r$) in polar coordinates ($r_{0},\varphi$) is:\\
\begin{figure}[htb]
\vspace{9pt}
\epsfig{figure=proc_4.ps, height=7.5cm, width=7.5cm, angle=-90}
\caption{ Virgo temperature profiles. The solid line is the temperature profile $T_{obs}(r)$
obtained from the observed spectrum, $S_{obs}$, whereas the dash-dotted line is the temperature
profile $T_{corr}(r)$ obtained from the corrected spectrum, $S_{corr}$.}
\end{figure}
\noindent
$\tilde{I}(r,E)=\int_{0}^{\infty}dr_0\cdot$
\begin{equation}
\int_{0}^{2\pi}d\varphi\,I_{\rm ROSAT}(r_0)\,
{\rm PSF_{\rm MECS}}(r,r_{0},\varphi,E)\,r_0\,.
\end{equation}
We computed the convolved profiles $\tilde{I}(r,3{\rm keV})$,
$\tilde{I}(r,6{\rm keV})$ and $\tilde{I}(r,9{\rm keV})$.\\
In order to estimate the spectral distortions introduced by the energy-dependent
PSF$_{\rm MECS}$ we calculated the ratios $R_{1}(r)=\frac{\tilde{I}(r,3{\rm keV})}
{\tilde{I}(r,6{\rm keV})}$ and $R_{2}(r)=\frac{\tilde{I}(r,9{\rm keV})}
{\tilde{I}(r,6{\rm keV})}$.\\
As shown in fig. 2, $R_{1}$ and $R_{2}$ are contained
within 0.09 of unity for any radius and within 0.06 for radii $>1'$.
The obvious implication is that spectral distortions introduced by the PSF$_{\rm MECS}$
are quite modest.
\@startsection{section}{1}{\z@{Spectra Correction Method}
The spectra obtained from the MECS observations of clusters of galaxies are affected by
the blurring effects of the PSF$_{\rm MECS}$. To correct these effects we
computed correction vectors $V(E_{i})$ where i=1,..,256 are the energy channels of the MECS.
These vectors
$V(E_{i})$ are quantities that, multiplied by the spectrum of the cluster
observed by the MECS, $S_{obs}(E_{i})$, give us the corrected spectrum of the cluster,
$S_{corr}(E_{i})$:
\begin{equation}
S_{corr}(E_{i})=S_{obs}(E_{i})V(E_{i}).
\end{equation}
We computed correction vectors for annular regions centered on the emission peak of the
Virgo cluster. We considered 5 annular regions with inner and outer radii of $0'-2'$,
$2'-4'$, $4'-6'$, $6'-8'$ and $8'-10'$ respectively. The correction
vector for the $j_{th}$ region spectrum, $V_{j}(E_{i})$, is:
\begin{equation}
V_{j}(E_{i})=\frac{\int_{2(j-1)}^{2j}2\pi\,r
I_{\rm ROSAT}(r,E_{i})dr}
{\int_{2(j-1)}^{2j}2\pi\,r\,\tilde{I}(r,E_{i})dr}
\end{equation}
where $\tilde{I}(r,E)$ is defined by eq. (3).\\
All the correction vectors evaluated for the considered regions are shown in fig. 3.\\
We note that: 1) $V_{1}(E_{i})$ has a mean value of $\sim 1.4$, while all other
correction vectors are contained between 0.9 and 1.0. This is because the number of
photons revealed is lower than the number of the photons emitted by the source, while the viceversa
is true for the others four regions.
2) All correction vectors show small variations with the energy and therefore the spectral
distortions are modest.
\@startsection{section}{1}{\z@{Temperature Gradients Measures in Virgo/M87}
The observed, $S_{obs}$, and corrected, $S_{corr}$, spectra obtained from the annular
regions described in the previous section have been used to measure temperature
profiles.
This has been done by fitting each spectrum in the energy range 1.4-5 keV with thermal
emission model (MEKAL in XSPEC ver. 9.01). In fig. 4 we show the temperature profiles
$T_{obs}(r)$ and $T_{corr}(r)$ obtained from the observed, $S_{obs}$, and the corrected,
$S_{corr}$, spectra.
The difference in value between $T_{obs}(r)$ and $T_{corr}(r)$ are negligible.
This demonstrates that the effects introduced by the PSF$_{\rm MECS}$ are very small.
We compared our results with those of the ROSAT and ASCA satellites.
To make this possible we fitted the spectra with the same models used by Nulsen and
B$\ddot{\rm o}$hringer (1995) in the analysis of the ROSAT PSPC data and by Matsumoto
et al. (1996) in the analysis of the ASCA GIS data:
Nulsen and B$\ddot{\rm o}$hringer used a Raymond-Smith model (Raymond \& Smith 1977) in the energy
band 0.5-2.4 keV,
while Matsumoto et al. used a thermal bremsstrahlung model plus gaussian line in the energy band
3-10 keV.
We recall that for $r<6'$ Nulsen and B$\ddot{\rm o}$hringer found temperatures in the energy range
1.1-2.3 keV, while Matsumoto et al. found temperatures in the energy range 2.-2.3 keV.
Fitting the $S_{corr}$ spectra with a Raymond-Smith model in the energy range 1.4-2.4 keV we found,
basically, the same temperatures as in the ROSAT analysis.
Using a bremsstrahlung model plus a gaussian line in the energy range 3-10 keV we found temperatures
consistent with those found from the ASCA GIS analysis (see their table 1).
Details about these comparisons are in D'Acri et al. (1998).
\vskip -2truecm
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 3,963 |
'Untitled' was commissioned by Terra Incognita for 'Curio', a series of seven site specific public interventions in the East End of London. It was projected from the Kobi Nazrul Community Centre onto the old Truman Brewery building in Hanbury Street at a scale of over 20 feet.
'Untitled' shows a woman repeatedly brushing her hair. Her hair is thrown forward so that her face is concealed. Her identity remains unknown. It was produced following an enquiry into the representation of women's unbraided long hair in religious art and also references metonymy, the theory within semiotics in which the part stands for the whole. | {
"redpajama_set_name": "RedPajamaC4"
} | 7,864 |
Q: Cancel a function if button is clicked So I have made an image slideshow automatic, by auto clicking the next button <.slick-next> every 7 seconds using the code below. I want to know if it is possible to cancel the function if the user clicks either of the buttons <.slick-next><.slick-prev> so then they can scroll through the images themselves rather than it auto changing when the user changes it to an image they want to look at. Also note it is not a button it is a div and i get it by class name!
<script>
setInterval(function () {
$(".slick-next").click();
}, 7000);
</script>
.slick-prev
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 3,320 |
<?php
declare(strict_types=1);
namespace Windwalker\Utilities;
/**
* The StrNotmalise class.
*
* @since __DEPLOY_VERSION__
*/
class StrNormalise
{
public static function splitCamelCase(string $str): array
{
return preg_split('/(?<=[^A-Z_])(?=[A-Z])|(?<=[A-Z])(?=[A-Z][^A-Z_])/x', $str);
}
public static function splitBy(string $str, string $separator = '_'): array
{
return array_filter(explode($separator, $str), 'strlen');
}
public static function splitAll(string $str): array
{
$strs = preg_split('#[ \-_\.]+#', $str);
$strs = array_map([static::class, 'splitCamelCase'], $strs);
return array_values(Arr::flatten($strs));
}
/**
* Separate a string by custom separator.
*
* @param string $input The string input (ASCII only).
* @param string $separator The separator to want to separate it.
*
* @return string The string be converted.
*
* @since 2.1
*/
public static function separate(string $input, string $separator = '_'): string
{
return implode($separator, self::splitAll($input));
}
/**
* Method to convert a string into camel case.
*
* @param string $input The string input (ASCII only).
*
* @return string The camel case string.
*/
public static function toCamelCase(string $input): string
{
return lcfirst(static::toPascalCase($input));
}
/**
* Method to convert a string into pascal case.
*
* @param string $input The string input (ASCII only).
*
* @return string The camel case string.
*
* @since 2.0
*/
public static function toPascalCase(string $input): string
{
return implode('', array_map('ucfirst', self::splitAll($input)));
}
/**
* Method to convert a string into dash separated form.
*
* @param string $input The string input (ASCII only).
*
* @return string The dash separated string.
*
* @since 2.0
*/
public static function toDashSeparated(string $input): string
{
return self::separate($input, '-');
}
/**
* Dash separated alias.
*
* @param string $input
*
* @return string
*/
public static function toKebabCase(string $input): string
{
return static::toDashSeparated($input);
}
/**
* Method to convert a string into space separated form.
*
* @param string $input The string input (ASCII only).
*
* @return string The space separated string.
*
* @since 2.0
*/
public static function toSpaceSeparated(string $input): string
{
// Convert underscores and dashes to spaces.
return static::separate($input, ' ');
}
/**
* Method to convert a string into dot separated form.
*
* @param string $input The string input (ASCII only).
*
* @return string The dot separated string.
*
* @since 2.1
*/
public static function toDotSeparated(string $input): string
{
// Convert underscores and dashes to dots.
return static::separate($input, '.');
}
/**
* Method to convert a string into underscore separated form.
*
* @param string $input The string input (ASCII only).
*
* @return string The underscore separated string.
*
* @since 2.0
*/
public static function toSnakeCase(string $input): string
{
// Convert spaces and dashes to underscores.
return static::separate($input, '_');
}
/**
* Alias of toSnakeCase()
*
* @param string $input
*
* @return string
*/
public static function toUnderscoreSeparated(string $input): string
{
return static::toSnakeCase($input);
}
/**
* Method to convert a string into variable form.
*
* @param string $input The string input (ASCII only).
*
* @return string The variable string.
*
* @since 2.0
*/
public static function toVariable(string $input): string
{
// Remove dashes and underscores, then convert to camel case.
$input = self::toCamelCase($input);
// Remove leading digits.
$input = preg_replace('#^[0-9]+.*$#', '', $input);
// Lowercase the first character.
$input[0] = strtolower($input[0] ?? '');
return $input;
}
/**
* Method to convert a string into key form.
*
* @param string $input The string input (ASCII only).
*
* @return string The key string.
*
* @since 2.0
*/
public static function toKey(string $input): string
{
// Remove spaces and dashes, then convert to lower case.
$input = self::toUnderscoreSeparated($input);
$input = strtolower($input);
return $input;
}
/**
* Convert to standard PSR-0/PSR-4 class name.
*
* @param string $class The class name string.
*
* @return string Normalised class name.
*
* @since 2.0
*/
public static function toClassNamespace(string $class): string
{
$class = trim($class, '\\');
$class = (string) str_replace(['\\', '/'], ' ', $class);
$class = ucwords($class);
return str_replace(' ', '\\', $class);
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 5,660 |
\section{Introduction}
This paper addresses the problem of constructing hardware-efficient cryptographic primitives
suitable for assuring trustworthiness of resource-constrained devices used in services and applications provided by the next generation of wireless communication networks.
The importance of improving security of wireless networks for our society is hard to overestimate.
In 2014, the annual loss to the global economy from cybercrimes was more than \$400 billion~\cite{Cy14}.
This number can quickly grow larger with the rapid growth of {\em Internet-of-Things} (IoT) applications.
In coming years "things" such as household appliances, meters, sensors, vehicles, etc. are expected to be accessible and controlled via local networks or the Internet, opening an entirely new range of services appealing to users.
The ideas of self-driving cars, health-tracking wearables, and remote surgeries do not sound like science-fiction any longer. The number of wirelessly connected devices is expected
to grow to a few tens of billions in the next five years~\cite{Er12}.
Unfortunately, the new technologies are appealing to the attackers as well. As processing power and connectivity become
cheaper, the cost of performing a cyberattack drops, making it easier for adversaries of all types to penetrate networks. Attacks are becoming more frequent, more sophisticated, and more widespread.
A connected household appliance becomes a target for all hackers around the globe unless appropriate security mechanisms are implemented. Household appliances typically do not have the same level of protection as computer systems. A compromised device can potentially be used as an entry point for a cyberattack on other devices connected to the network. The first proven cyberattack involving "smart" household appliances has been already reported in~\cite{Pr14}. In this attack, more than 750.000 malicious emails targeting enterprises and individuals worldwide were sent from more than 100.000 consumer devices such as home-networking routers, multi-media centers, TVs, refrigerators, etc. No more than 10 emails were initiated from any single IP-address, making the attack difficult to block based on location. The attack surface of future IoT with billions of connected devices
will be enormous.
In addition to a larger attack surface, the return value for performing a cyberattack grows.
The assets accessible via tomorrow's networks (hardware, software, information, and revenue streams) are expected to be much greater than the ones available today,
increasing incentive for cyber criminals and underground economies~\cite{Er5G}. A growing black market for breached data serves as a further encouragement.
The damage caused by an individual actor may not be limited to a business or reputation, but could have a severe impact on public safety, national economy, and/or national security.
The tremendous growth in the number and type of wirelessly connected devices, as well as the increased incentive for performing attacks change the threat landscape and create new challenges for security.
The situation is complicated even further by the fact that many
end-point IoT devices require utmost efficiency in the use of communication, computing, storage and energy resources.
A typical IoT device spends most of its "life" in a sleep mode. It gets activated at periodic intervals, transmits a small amount of data and then shuts down again. To satisfy extreme limitations
of resource-constrained IoT devices, very efficient cryptographic primitives for implementing encryption, data integrity protection, authentication, etc. are required.
Invertible mappings are among the most frequently used primitives in cryptography~\cite{St06}.
For example, the round function of a block cipher~\cite{pres07,pri12} has to be invertible in order
to result in unique decryption. Stream ciphers~\cite{hell-grain,DuH14} and hash functions~\cite{Quark,DuNS15} use invertible
state mappings to prevent incremental reduction of the entropy of the state.
This paper presents a sufficient condition for invertibility of mappings $x \rightarrow f(x)$ of type $\{0,1,\ldots,2^n-1\} \rightarrow \{0,1,\ldots,2^n-1\}$. Such a single-variable $2^n$-valued mapping can be interpreted as an $n$-variable Boolean mapping $\{0,1\}^n \rightarrow \{0,1\}^n$ in which the Boolean variable $x_i \in \{0,1\}$ represents the bit number $i$ of the input $x$
and the Boolean function $f_i: \{0,1\}^n \rightarrow \{0,1\}$ represents the bit number $i$ of the output $f(x)$, i.e.:
\begin{equation} \label{map}
\left(
\begin{array}{c}
x_0\\
\ldots \\
x_{n-1}\\
\end{array}
\right)
\rightarrow
\left(
\begin{array}{cc}
f_0(x_0,\ldots,x_{n-1})\\
\ldots \\
f_{n-1}(x_0,\ldots,x_{n-1})\\
\end{array}
\right)
\end{equation}
for $i \in \{0,1,\ldots,n-1\}$.
For example, the 4-variable Boolean mapping
\[
\left(
\begin{array}{c}
x_0\\
x_1\\
x_2\\
x_3\\
\end{array}
\right)
\rightarrow
\left(
\begin{array}{c}
x_0 \oplus 1\\
x_1 \oplus x_0\\
x_2 \oplus x_0 x_1\\
x_3 \oplus x_0 x_1 x_2\\
\end{array}
\right)
\]
corresponds to the single-variable 16-valued mapping
\[
x \rightarrow x + 1 ~~\mbox{(mod 16)},
\]
where ``$+$'' is addition modulo 16.
Note that the corresponding single-variable $2^n$-valued mapping may not have a closed form.
In order to check if a mapping of type~(\ref{map}) is invertible or not, we generally need an exponential in $n$ number of steps. The condition derived in this paper can be checked in $O(n^2 N)$ time,
where $N$ is the size of representation of the largest Boolean function in the mapping.
So, it can be used for constructing secure invertible mappings
with a small hardware implementation cost. For example, we show how the presented results can be used in
stream cipher design.
The paper is organized as follows.
Section~\ref{not} summarizes basic notations used in the sequel.
In Section~\ref{prev}, we describe previous work and
show that previous methods cannot explain invertibility of some mappings which
can be handled by the presented approach.
Section~\ref{main} presents the main result. Section~\ref{compl}
estimates the complexity of checking the presented condition.
Section~\ref{appl} shows how the presented results can be used in
stream cipher design.
Section~\ref{conc} concludes the paper.
\section{Preliminaries} \label{not}
Throughout the paper, we use "$\oplus$" to
denote the Boolean XOR, "$\cdot$"
to denote the Boolean AND and $\overline{x}$ to denote the Boolean complement of $x$,
defined by $\overline{x} = 1 \oplus x$.
The Algebraic Normal Form (ANF)~\cite{CuS09} of a Boolean function $f: \{0,1\}^{n} \rightarrow \{0,1\}$
(also called positive polarity {\em Reed-Muller canonical form}~\cite{Gr91}) is a polynomial in the Galois Field of order 2, $GF(2)$, of type
\[
f(x_0, x_1,\ldots,x_{n-1}) = \sum_{i=0}^{2^n-1} c_i \cdot
x_0^{i_0} \cdot x_1^{i_1} \cdot \ldots \cdot x_{n-1}^{i_{n-1}},
\]
where $c_{i} \in \{0,1\}$ are constants, ``$\cdot$'' is the Boolean AND, ``$\sum$'' is the Boolean XOR, the vector $(i_{0} i_{1} \ldots i_{n-1})$ is the binary expansion of $i$, and $x^{i_{j}}_{j}$ denotes the $i_{j}$th power of $x_{j}$ defined by
$x^{i_{j}}_{j} = x_{j}$ for $i_{j} = 1$ and $x^{i_{j}}_{j} = 1$ otherwise, for $j \in \{0,1, \ldots, n-1\}$.
The {\em dependence set} (or {\em support set}~\cite{espr}) of a Boolean function $f: \{0,1\}^n \rightarrow \{0,1\}$
is defined by
\[
dep(f) = \{j \ | \ f|_{x_j=0} \not = f|_{x_j=1}\},
\]
where $f|_{x_j=k} = f(x_0, \ldots, x_{j-1}, k, x_{j+1}, \ldots, x_{n-1})$
for $k \in \{0,1\}$.
A mapping $x \rightarrow f(x)$ on a finite set is called {\em invertible} if $f(x) = f(y)$
if, and only if, $x = y$. Invertible mappings are also called {\em permutations}.
\section{Previous Work} \label{prev}
In this section, we describe previous work on invertible mappings and
show that previous methods cannot explain invertibility of some mappings which
can be handled by the presented approach.
Many methods for constructing different classes of invertible mappings are known.
The simplest one is to compose simple invertible operations. A Substitution-Permutation Network (SPN)
is a typical example. An SPN consists of S-boxes, which permute input bits locally,
and P-boxes, which diffuse input bits globally. This method of construction
cannot explain the invertibility of, for example, the following 4-variable
mapping
\begin{equation} \label{ex}
\left(
\begin{array}{c}
x_0\\
x_1\\
x_2\\
x_3\\
\end{array}
\right)
\rightarrow
\left(
\begin{array}{c}
x_1\\
x_2\\
x_3 \oplus x_1 x_2\\
x_0 \oplus x_3\\
\end{array}
\right)
\end{equation}
since a non-invertible operation Boolean AND is used.
Feistel~\cite{Fe73} proposed a powerful technique which makes possible constructing invertible mappings from non-invertible basic operations. It is used in many block ciphers, including DES~\cite{DES77}.
The basic Feistel construction maps two inputs, $l$ and $r$, into two outputs as follows:
\[
\left(
\begin{array}{c}
l\\
r\\
\end{array}
\right)
\rightarrow
\left(
\begin{array}{cc}
r\\
l \oplus f(r)\\
\end{array}
\right)
\]
where $f$ is any single-variable function. The full Feistel construction iterates the above mapping any number of times. This method was extended in several directions, including {\em unbalanced}, {\em homogeneous}, {\em heterogeneous}, {\em incomplete}, and {\em inconsistent} Feistel networks~\cite{ScK96}. However, the Feistel construction requires at least two variables. It cannot explain the invertibility of mappings $x \rightarrow f(x)$ of type $\{0,1,\ldots,2^n-1\} \rightarrow \{0,1,\ldots,2^n-1\}$.
The presented method can explain it by looking into the structure of
Boolean functions $f_i$ representing the bits of the output $f(x)$.
Shamir~\cite{Sh93} introduced an interesting construction based on the fact that the mapping of type~(\ref{map})
is invertible for almost all inputs if each $f_i$ has the form:
\[
f_i(x_0,\ldots,x_{n-1}) = h_i(x_0,\ldots,x_{i-1}) x_i + g_i(x_0,\ldots,x_{i-1}) ~~~ (\mbox{mod} ~~ N)
\]
where $g_i$ and $h_i$ are arbitrary non-zero $i$-variable polynomials modulo a large RSA modulus $N$.
The "triangular" nature of functions $f_i$ makes it possible to perform
the inversion process similarly to the way we do Gaussian elimination
to solve a system of linear equations.
This approach can also handle
$g_i$'s and $h_i$'s which mix arithmetic and Boolean operations~\cite{KlS02,Kl05}. The approach presented in this paper is similar to the approach of Shamir in that it also uses triangulation to prove invertibility. However, our functions $f_i$ are of the type
\begin{equation} \label{el_eq}
f_i(x_0,\ldots,x_{n-1}) = x_j \oplus g_i(x_0,\ldots,x_{n-1}),
\end{equation}
where $j \in \{0,1,\ldots,n-1\}$ and $g_i$ does not depend on $x_j$.
Thus, in our case, the $i$th output may depend on the input $j$ such that $j > i$.
For example, in the mapping defined by equations~(\ref{ex}),
$f_2$ depends on $x_1,x_2$ and $x_3$.
Shamir's construction was further extended by Klimov and Shamir~\cite{KlS02,Kl05} to a class on invertible mappings based on T-functions. A single-variable function $f(x)$ of type
$\{0,1,$ $\ldots,2^n-1\} \rightarrow \{0,1,\ldots,2^n-1\}$ is defined to be a T-function
if each of its output bits $f_{i-1}(x)$ depends only on
the first $i$ input bits $x_0, x_1, \ldots, x_{i-1}$:
\[
\left(
\begin{array}{l}
{x}_0 \\
{x}_1 \\
{x}_2 \\
\ldots \\
{x}_{n-1} \\
\end{array}
\right)
\rightarrow
\left(
\begin{array}{l}
f_0(x_0) \\
f_1(x_0, x_1) \\
f_2(x_0, x_1, x_2) \\
\ldots \\
f_{n-1}(x_0, x_1, \ldots, x_{n-1}) \\
\end{array}
\right).
\]
A T-function is invertible if, and only if, each output bit $f_i$
can be represented as
\[
f_i(x_0,\ldots,x_i) = x_i \oplus g_i(x_0,\ldots,x_{i-1}).
\]
This fundamental result inspired the construction which we present in this paper.
The reader will easily notice that, in our construction~(\ref{el_eq}),
functions $g_i$'s are T-functions while functions $f_i$'s are not if $j > i$.
Another difference is that Klimov and Shamir targeted software implementation
and therefore focused on mappings
whose expression can be evaluated by a program with the minimal number
of instructions. We target the hardware implementation.
Therefore, we are interested in minimizing the number of Boolean operations
in the expressions of Boolean functions representing output bits.
The construction method which we present can be used as a starting point for
constructing nonlinear invertible mappings which have an efficient hardware implementation.
Another group of construction methods consider {\em permutation polynomials} which
involve only the arithmetic operation of addition, subtraction, and multiplication.
Permutation polynomials are well-studied in mathematics. Hermite~\cite{LiH94} made a substantial progress in characterizing univariate permutation polynomials modulo a prime $p$.
Dickson~\cite{Di01} described all univariate polynomials with degrees smaller than 5. However, the problem remains unsolved for high degree polynomials modulo a large prime $p$.
The problem appears simpler for the ring of integers modulo $2^n$. Rivest~\cite{Ri99} provided a complete characterization of all univariate permutation polynomials modulo $2^n$. He proved that a polynomial
\[
p(x) = a_0 + a_1 x + \ldots + a_d x^d
\]
with integral coefficients is a permutation polynomial for $n > 2$ if, and only if, $a_1$ is odd, $(a_2 + a_4 + \ldots)$ is even and $(a_3 + a_5 + \ldots)$ is even.
His powerful algebraic proof technique was further generalized by Klimov and Shamir~\cite{KlS02,Kl05} to polynomials of type
\[
p(x) = a_0 \circ a_1 x \circ \ldots \circ a_d x^d,
\]
where $\circ \in \{+,-,\oplus\}$, which mix arithmetic and Boolean operations. However, since the resulting polynomials are T-functions, they do not cover the construction method presented in this paper for the case when $i$th output depends on the input $j$ such that $j > i$.
\begin{figure}[t!]
\begin{center}
\includegraphics[width=2.3in]{figure1.pdf}
\caption{The structure of an $n$-bit Non-Linear Feedback Shift Register.}\label{f1}
\end{center}
\end{figure}
Finally, we would like to discuss the relation between the mappings of type~(\ref{map}) and
the state mappings generated by {\em Non-Linear Feedback Shift Registers (NLFSRs)}~\cite{Ja89}.
An $n$-bit NLFSR consists of $n$ binary stages, each capable of storing 1 bit of information, a nonlinear Boolean function, called {\em feedback function}, and a clock (see Figure~\ref{f1}).
At each clock cycle, the stage $n-1$ is updated to the value computed by the feedback function. The rest of the stages shift the content of the previous stage. Thus, an $n$-bit NLFSR with the feedback functions $f$ implements the state mapping of type
\[
\left(
\begin{array}{c}
x_0\\
x_1\\
\ldots \\
x_{n-1}\\
\end{array}
\right)
\rightarrow
\left(
\begin{array}{cc}
x_1\\
x_2\\
\ldots \\
f(x_0,\ldots,x_{n-1})\\
\end{array}
\right).
\]
where the variable $x_i$ represents the value of the stage $i$, for $i \in \{0,1,\ldots,n-1\}$.
It is well-known~\cite{Golomb_book} that an $n$-bit NLFSR is invertible if, and only if, its feedback function is of type
\[
f(x_0,\ldots,x_{n-1}) = x_0 \oplus g(x_1,x_2,\ldots,x_{n-1}).
\]
The mappings considered in this paper can be implemented by a more general type of non-linear state machines, shown in Figure~\ref{f2}. Since the content of stages is not any longer shifted from one stage to the next, but rather it is updated by some arbitrary functions,
such registers are not called shift registers any longer. Instead, they are called
{\em binary machines}~\cite{Golomb_book} or
{\em registers with non-linear update}~\cite{LiD14}.
Binary machines
are typically smaller and
faster than NLFSRs generating the same sequence~\cite{Du10aj,Du11a}.
For example, the 4-bit NLFSR with the feedback function $f(x_0,x_1,x_2,x_3) = x_0 \oplus x_3 \oplus x_1 x_2 \oplus x_2 x_3$ generates the same set of sequences as the 4-bit binary machine implementing the 4-variable mapping defined by equations~(\ref{ex}). We can see that the binary machine uses 3 binary Boolean operations, while the NLFSR uses 5 binary Boolean operations. Furthermore, the depth of feedback functions of the binary machine is smaller that the depth of the feedback function of the NLFSR. Thus, the binary machine has a smaller propagation delay than the NLFSR.
\begin{figure}[t!]
\begin{center}
\includegraphics[width=3.4in]{figure2.pdf}
\caption{The structure of an $n$-bit binary machine.}\label{f2}
\end{center}
\end{figure}
\section{Conditions for Invertibility} \label{main}
As we mentioned in Section~\ref{prev},
it is well-known how to construct invertible NLFSRs~\cite{Golomb_book}.
An $n$-bit NLFSR is invertible if, and only if, its
feedback function is of type:
\begin{equation} \label{eq1}
f(x_0,x_1,\ldots,x_{n-1}) = x_0 \oplus g(x_1,x_2,\ldots,x_{n-1}).
\end{equation}
The proof is simple, because every two consecutive states
of an NLFSR overlap in $n-1$ positions.
This implies that each state can have only two possible
predecessors and two possible successors.
If $f$ is in the form (\ref{eq1}), then the NLFSR states which correspond to the binary $n$-tuples $x = (x_0, x_1, \ldots, x_{n-1})$
and $y = (\overline{x}_0, x_1, \ldots, x_{n-1})$
always have different successors. The values of $f(x)$ and $f(y)$
depend on the value of $g(x_1, \ldots, x_{n-1})$ and on the value of $x_0$.
The value of $g(x_1, \ldots, x_{n-1})$ is the same for $x$ and $y$.
The value of $x_0$ is different for $x$ and $y$.
Thus, $f(x) \not= f(y)$.
It is also easy to see that, if both $x$ and $y$ have the same successor, $f$
cannot have the form~(\ref{eq1}).
In the general case of mappings $\{0,1\}^n \rightarrow \{0,1\}^n$,
any binary $n$-tuple can have $2^n$ possible predecessors
and $2^n$ possible successors. Therefore, to guarantee that
a mapping $\{0,1\}^n \rightarrow \{0,1\}^n$ is
invertible, we have to check that, for all $x, y \in \{0,1\}^n$, $x \not= y$ implies that $f_i(x) \not= f_i(y)$,
for some
$i \in \{0,1,\ldots,n-1\}$. This clearly requires
the number of steps which is exponential in $n$.
The main contribution of this paper is a
more restricted sufficient condition
which can be checked in $O(n^2 N)$ steps.
To formulate this condition, we first introduce the notion of a free variable of a function.
\begin{definition}[Free variable]
A variable $x_i \in dep(f)$ is called a free variable if
$f$ can decomposed as
\[
f(x_0,x_1,\ldots,x_{n-1}) = x_i \oplus g(x_0,x_1,\ldots,x_{n-1}).
\]
where $x_i \not\in dep(g_i)$.
\end{definition}
\begin{definition}[Set of free variables]
The set of free variables of $f$, $\Phi(f)$, contains all free variables of $f$
\[
\Phi(f) = \{x_i ~|~ x_i ~\mbox{is a free variable of}~ f\}.
\]
\end{definition}
Now we are ready to formulate the following sufficient condition for invertibility.
\begin{theorem} \label{t2}
A mapping of type~(\ref{map}) is invertible if the following two conditions hold:
\begin{enumerate}
\item For each $i \in \{0,1,\ldots,n-1\}$, $f_i$ is of type
\begin{equation} \label{th_main}
f_i(x_0,x_1,\ldots,x_{n-1}) = x_{j_i} \oplus g_i(x_0,x_1,\ldots,x_{n-1}),
\end{equation}
where $x_{j_i} \in \Phi(f_i)$.
\item Functions $f_{0}, f_{1}, \ldots, f_{n-1}$ can be re-ordered as $f_{i_0}, f_{i_1}, \ldots, f_{i_{n-1}}$ to satisfy the property
\[
dep(g_{i_k}) \subseteq \left\{
\begin{array}{l}
\emptyset, ~ \mbox{for} ~ k = 0\\
\displaystyle\bigcup_{j=0}^{k-1} dep(f_{i_j}), ~ \mbox{for} ~ 1 \leq k \leq n-1 \\
\end{array}
\right.
\]
where ``$\cup$'' stands for the union.
\end{enumerate}
\end{theorem}
{\bf Proof:} By contradiction.
Suppose that there exist a non-invertible mapping of type~(\ref{map}) for which the conditions of the theorem hold. Then, this mapping is of type
\begin{equation} \label{pr}
\left(
\begin{array}{l}
{x}_{i_0} \\
{x}_{i_1} \\
{x}_{i_2} \\
\ldots \\
{x}_{i_{n-1}} \\
\end{array}
\right)
\rightarrow
\left(
\begin{array}{l}
x_{j_0} \oplus g_{i_0} \\
x_{j_1} \oplus g_{i_1}(x_{j_0}) \\
x_{j_2} \oplus g_{i_2}(x_{j_0},x_{j_1}) \\
\ldots \\
x_{j_{n-1}} \oplus g_{i_{n-1}}(x_{j_0},x_{j_1}\ldots,x_{j_{n-2}}) \\
\end{array}
\right)
\end{equation}
where $x_{j_k} \in \Phi(f_{i_k})$, for $k \in \{0,1,\ldots,n-1\}$.
Since the mapping~(\ref{pr}) in non-invertible, there exist at least two input assignments $a = (a_{i_0},a_{i_1},\ldots,a_{i_{n-1}}) \in \{0,1\}^n$ and $a' = (a'_{i_0},a'_{i_1},\ldots,a'_{i_{n-1}}) \in \{0,1\}^n$, $a \not= a'$, which are mapped into the same output. We analyzing~(\ref{pr}) row by row, we can conclude that:
\begin{itemize}
\item ${a}_{i_1} \rightarrow a_{j_1} \oplus g_{i_0}$ and ${a'}_{i_0} \rightarrow a'_{j_0} \oplus g_{i_0}$ and $a_{j_0} \oplus g_{i_0} = a'_{j_0} \oplus g_{i_0}$ implies $a_{j_0} = a'_{j_0}$
\item $a_{i_1} \rightarrow a_{j_1} \oplus g_{i_1}(a_{j_0})$ and $a'_{i_1} \rightarrow a'_{j_1} \oplus g_{i_1}(a'_{j_0})$ and $a_{j_1} \oplus g_{i_1}(a_{j_0}) = a'_{j_1} \oplus g_{i_1}(a'_{j_0})$ and $a_{j_0} = a'_{j_0}$ implies $a_{j_1} = a'_{j_1}$
\item $\ldots$
\item
$a_{i_{n-1}} \rightarrow a_{j_{n-1}} \oplus g_{i_{n-1}}(a_{j_0},a_{j_1}\ldots,a_{j_{n-2}})$
and
$a'_{i_{n-1}} \rightarrow a'_{j_{n-1}} \oplus g_{i_{n-1}}(a'_{j_0},a'_{j_1}\ldots,a'_{j_{n-2}})$
and
$a_{j_{n-1}} \oplus g_{i_{n-1}}(a_{j_0},a_{j_1}\ldots,a_{j_{n-2}}) = a'_{j_{n-1}} \oplus g_{i_{n-1}}(a'_{j_0},a'_{j_1}\ldots,a'_{j_{n-2}})$
and
$a_{j_0} = a'_{j_0}$, $a_{j_1} = a'_{j_1}$, $\ldots$ $a_{j_{n-2}} = a'_{j_{n-2}}$
implies $a_{j_{n-1}} = a'_{j_{n-1}}$.
\end{itemize}
Therefore, $a = a'$. We reached a contradiction. Thus, the assumption that there exist a non-invertible mapping of type~(\ref{map}) for which the conditions of the theorem hold is not true.
\begin{flushright}
$\Box$
\end{flushright}
An example of mapping which satisfies the conditions of Theorem~\ref{t2} is:
\begin{equation} \label{cons}
\left(
\begin{array}{l}
{x}_0 \\
{x}_1 \\
{x}_2 \\
\ldots \\
{x}_{n-1} \\
\end{array}
\right)
\rightarrow
\left(
\begin{array}{l}
x_1 \oplus g_0 \\
x_2 \oplus g_1(x_1) \\
x_3 \oplus g_2(x_1,x_2) \\
\ldots \\
x_0 \oplus g_{n-1}(x_1,x_2\ldots,x_{n-1}) \\
\end{array}
\right)
\end{equation}
The reader may notice that mappings defined by Theorem~\ref{t2}
can be obtained by re-labeling variables in a T-function.
The re-labeling is given by the ordering of free variables.
So, the mapping~(\ref{th_main}) can be viewed
as a composition of a bit permutation and a T-function.
Since there are $n!$ bit permutations, the class of invertible mappings considered in this paper is
by a factor of $n!$ larger than the class of invertible mappings
based on T-functions.
Clearly, the re-labeling of variables does not change the implementation cost of a mapping.
However, it might drastically change the cycle structure of the underlying
state transition graph, as illustrated by the following example.
Consider the following 4-variable mapping based on a T-function:
\begin{equation} \label{m1}
\left(
\begin{array}{c}
x_0\\
x_1\\
x_2\\
x_3\\
\end{array}
\right)
\rightarrow
\left(
\begin{array}{c}
x_0\\
x_1\\
x_2 \oplus x_0\\
x_3 \oplus x_0 \oplus x_0 x_1\\
\end{array}
\right).
\end{equation}
This mapping has a quite uninteresting state transition graph shown in Figure~\ref{fm1}.
Let us re-label the variables as $(0,1,2,3) \rightarrow (1,2,3,0)$. We get the mapping:
\begin{equation} \label{m2}
\left(
\begin{array}{c}
x_0\\
x_1\\
x_2\\
x_3\\
\end{array}
\right)
\rightarrow
\left(
\begin{array}{c}
x_1\\
x_2\\
x_3 \oplus x_1\\
x_0 \oplus x_1 \oplus x_1 x_2\\
\end{array}
\right).
\end{equation}
which has the state transition graph shown in Figure~\ref{fm2}. All states, except the all-0 state, are included in one cycle. If we implement this mapping by the binary machine shown in Figure~\ref{f2} and initialize the binary machine to any non-zero state, then its output generates a pseudo-random sequence with period 15 which
satisfies the first two randomness postulates of Golomb~\cite{Golomb_book}, namely the balance property and the run property.
No sequence generated by a nonlinear Boolean function satisfy the third randomness postulate (two-level autocorrelation property).
Therefore, the mapping we obtained by re-labeling has much more interesting statistical properties than the original mapping.
\begin{figure}[t!]
\begin{center}
\includegraphics*[angle=270,width=1.6in]{figure3.pdf}
\caption{The state transition graph of the mapping~(\ref{m1}).}\label{fm1}
\end{center}
\end{figure}
\begin{figure}[t!]
\begin{center}
\includegraphics*[angle=270,width=2.7in]{figure4.pdf}
\caption{The state transition graph of the mapping~(\ref{m2}).}\label{fm2}
\end{center}
\end{figure}
The cycle structure of a mapping is important for some cryptographic applications, e.g. stream ciphers~\cite{robshaw94stream}. Obviously, if we iterate a mapping a large number of times, we do not want the sequence of generated states to be trapped in a short cycle.
It is worth noting
that formal verification tools based on reachability analysis can be adopted for finding short cycles in a state transition graph. There are dedicated tools to compute the number and length of cycles in state transition graphs of synchronous sequential networks, e.g. BNS~\cite{DuT11} which is based on SAT-based bounded model checking and BooleNet~\cite{DuTM05} which is based on Binary Decision Diagrams.
The mappings defined by Theorem~\ref{t2} are invertible for any choice of Boolean functions $g_i(x_0,x_2\ldots,x_{i-1})$.
The smaller is the number of Boolean operations in the expressions of $g_i$'s,
the smaller is its hardware implementation cost.
Note, that we are not restricted to represent $g_i$'s in ANF. Any Boolean expression combining AND, OR, NOT and XOR can be used.
Multi-level logic optimization tools, such as UC Berkeley tool ABC~\cite{abc}
can be applied to transform the ANF or other Boolean expression representing the function into
an optimized multi-level expression.
Clearly, the choice of $g_i$'s will be guided not only by the hardware cost,
but also by other criteria determining the cryptographic strength of the resulting function, e.g. nonlinearity, correlation immunity, algebraic degree, etc. (see~\cite{CuS09} for requirements on cryptographically strong functions).
Finally, we would like to point out that the conditions of Theorem~\ref{t2} are sufficient, but not necessary conditions for invertibility. For example, the following 4-variable mapping is invertible, but it does not satisfy them:
\[
\left(
\begin{array}{c}
x_0\\
x_1\\
x_2\\
x_3\\
\end{array}
\right)
\rightarrow
\left(
\begin{array}{c}
x_1 \oplus x_0 \oplus x_0 x_2\\
x_2 \oplus x_1 \oplus x_3\\
x_3 \oplus x_0 x_2\\
x_0\\
\end{array}
\right).
\]
\section{Condition Checking} \label{compl}
Next, let us estimate the number of steps required to check the conditions of Theorem~\ref{t2}.
Suppose that all Boolean functions of the mapping are represented in ANF.
Let $N$ be the ANF size of the largest function in the mapping.
The {\em size} of an ANF is defined as the total number of variables appearing in the ANF.
For example, the ANF $x_1 \oplus x_1 x_2$ is of size 3.
\begin{algorithm}[t!]
\caption{\footnotesize{Checks conditions of Theorem~\ref{t2} for a mapping of type~(\ref{map})
in which all Boolean functions are represented in ANF}} \label{alg}
\begin{algorithmic}[1]
\STATE $\Phi = \emptyset$;
\STATE flag = 0;
\FOR{every $i$ from 0 to $n-1$}
\IF{$f_i = x_j$ or $f_i = x_j \oplus 1$, for some $j \in \{0,1,\ldots,n-1\}$}
\STATE add $x_j$ to $\Phi$
\STATE mark $f_i$
\STATE flag = 1;
\ENDIF
\ENDFOR
\IF{flag = 0}
\STATE Return "Conditions are not satisfied"
\ENDIF
\STATE {\bf loop:}
\FOR{every $i$ from 0 to $n-1$}
\IF{$f_i$ is not marked}
\IF{$f_i$ is of type~(\ref{th_main}) and $x_k \in \Phi~ \forall x_k \in dep(g_i)$}
\STATE add $x_j$ to $\Phi$
\STATE mark $f_i$
\STATE go to {\bf loop}
\ENDIF
\ENDIF
\ENDFOR
\FOR{every $i$ from 0 to $n-1$}
\IF{$f_i$ is marked}
\STATE continue
\ELSE
\STATE Return "Conditions are not satisfied"
\ENDIF
\ENDFOR
\STATE Return "Mapping is invertible"
\end{algorithmic}
\end{algorithm}
Consider the pseudocode shown as Algorithm~\ref{alg}.
$\Phi$ is a set which keeps track of variables which are identified as free for some $f_i$. If this set is implemented as a hash table of size $n$, then adding a variable to the set or checking if a variable belongs to the set can be done in constant time.
In the first {\bf for}-loop, we check if each $f_i$ is of type $f_i = x_j$
or $f_i = x_j \oplus 1$, for some $j \in \{0,1,\ldots,n-1\}$. If yes, we
add $x_j$ to the set $\Phi$ and mark $f_i$. Since the steps 4, 5, 6 and 7 can be done
in $O(1)$ time, the complexity of the first {\bf for}-loop is $O(n)$.
If none of the functions $f_i$ are marked during the first {\bf for}-loop, the algorithm terminates with the conclusion that the conditions of Theorem~\ref{t2} are not satisfied.
In the second {\bf for}-loop, for each non-marked $f_i$, we check if it is of type~(\ref{th_main})
and if every $x_k \in dep(g_i)$ belongs to $\Phi$.
If yes, add $x_j$ to $\Phi$, mark $f_i$, and return to step 13.
The steps 15, 17 and 18 can be done in $O(1)$ time.
The step 16 requires $O(N)$ time.
Thus, the complexity of the second {\bf for}-loop is $O(n N)$.
Since we return to the step 13 at most $n$ times,
the overall complexity of steps 13-22 is $O(n^2 N)$.
In the third {\bf for}-loop, we check if all $f_i$ are marked.
If yes, the mapping is invertible. Otherwise, the algorithm
returns "conditions are not satisfied". Since the
conditions of Theorem~\ref{t2} are sufficient, but not necessary conditions for invertibility, in this case the mapping may or may not be invertible. The complexity of the third {\bf for}-loop is $O(n)$.
We can conclude that overall complexity of Algorithm~\ref{alg} is $O(n^2 N)$.
\section{Applications} \label{appl}
In this section we show how the presented results can be used in stream cipher design.
A possible way to construct a key stream generator for a stream cipher is
to run several FSRs in parallel and to combine their outputs with a nonlinear Boolean function~\cite{CuS09}.
The resulting structure is called a {\em combination generator}.
If the periods of FSRs are pairwise co-prime, then the period of the resulting key stream generator
is equal to the product of periods of FRSs~\cite{DuT05}. Examples of stream ciphers based on combination generators are VEST~\cite{cryptoeprint:2005:415}, Achterbahn-128/80~\cite{GaGK07}, and the cipher~\cite{GaGK06}. VEST uses 16 10-bit and 16 11-bit NLFSRs. Achterbahn-128/80 uses 13 NLFSRs of size from 21 to 33 bits. The cipher from~\cite{GaGK06} uses 10 NLFSRs of size 22-29, 31 and 32 bits.
At present it is not known how to construct large NLFSRs with a guaranteed long period.
Existing algorithms cover special cases only, e.g.~\cite{Du12a,Du13a}.
Small NLFSRs which are used in combination generators are computed by a random search.
Lists of $n$-bit NLFSRs of size $< 25$ bits with the period $2^n-1$ whose feedback functions contain up to 6 binary Boolean operations in their ANFs are available in~\cite{Du12l}.
It is known that, for example, there are no 20-bit NLFSRs with the period $2^{20}-1$ whose feedback function contains only
four binary Boolean operations in its ANF (i.e. implementable with four 2-input gates).
Using Algorithm~\ref{alg} to bound random search, in 12 hours we were able to find 63 20-variable nonlinear invertible mappings with the period $2^{20}-1$ whose functions $f_i$ contain no more than 4 binary Boolean operations in their ANFs in total. Two representatives are:
\begin{enumerate}
\item $f_{16} = x_{17} \oplus x_4 x_{11}$, $f_{13} = x_{14} \oplus x_{13}$, $f_4 = x_5 \oplus x_1$. \vspace{1mm}
\item $f_{19} = x_{0} \oplus x_{16}$, $f_{15} = x_{16} \oplus x_{3} \oplus x_1 x_{15}$.
\end{enumerate}
The omitted functions are of type $f_i = x_{(i+1)~mod~20}$, for $i \in \{0,1,\ldots,19\}$.
This shows that Algorithm~\ref{alg} is useful for finding nonlinear invertible
mappings with a small hardware implementation cost.
Other important properties, such as nonlinearity, correlation immunity, algebraic degree, etc., can then be used to further guide the search.
\section{Conclusion} \label{conc}
We derived a sufficient condition for invertibility of mappings of type $\{0,1,\ldots,2^n-1\} \rightarrow \{0,1,\ldots,2^n-1\}$ which can be checked in $O(n^2 N)$ steps, where $N$
is the ANF size of the largest Boolean function in the mapping.
The presented method can be used as a starting point for
constructing nonlinear invertible mappings which can be efficiently implemented in hardware.
Future work remains on constructing new cryptographic primitives based on the presented class of invertible mappings and evaluating their security and hardware cost. We also plan to investigate the usability our results in reversible computing.
\section{Acknowledgements}
This work was supported in part by the research grant No SM14-0016 from the Swedish Foundation for Strategic Research. The author would like to thank the anonymous reviewers for their valuable comments and suggestions to improve the
quality of the paper.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,566 |
Trust the Dice
"We're not critics. We're professional fan-girls." --- This blog is dedicated to movies and the entertainment industry. We use random selection to bring into light the best and worst of streaming films and entertainment news.
Number Rolled: 90
Movie Name/Year: Running Scared (2006)
Tagline: Every bullet leaves a trail.
Production Companies: New Line Cinema, Media 8 Entertainment, True Grit Productions, VIP 1 Medienfonds, VIP 2 Medienfonds, MDP Filmproduktion, International Production Company, Pierce/Williams Entertainment, Zero Gravity Management
Producer: Tony Grazia, Andreas Grosch, Stewart Hall, Sammy Lee, Matt Luber, Andrew Pfeffer, Michae A. Pierce, Brett Retner, Andreas Schmid, Kevan Van Thompson, Jeff G. Waxman
Director: Wayne Kramer
Writer: Wayne Kramer
Actors: Paul Walker, Cameron Bright, Vera Farmiga, Chazz Palminteri, Karel Roden, Johnny Messner, Ivana Milicevic, Alex Neuberger, Michael Cudlitz, Bruce Altman, Elizabeth Mitchell, Arthur J. Nascarella, John Noble, Idalis DeLeon, David Warshofsky, Jim Tooey, Thomas Rosales Jr., Jan Kohout
Blurb from Netflix: Low-level mobster Joey Gazelle is tasked with disposing of a gun. But when it's stolen and used in another crime, he must reclaim the gun quickly.
Selina's Point of View:
It's always a little sad to watch a movie starring an actor that's died since filming it. It makes it a little easier when you don't have to say anything bad about the performance.
Paul Walker (Furious 7, Brick Mansions, Pawn Shop Chronicles) was a decent actor, but in Running Scared he was out of this world. I cared about his version of Joey Gazelle. He was also blessed with a great supporting cast, even the kids. Cameron Bright (Motive, The Twilight Saga: Breaking Dawn – Part 2, Walled In) didn't have as many lines as you would think, so he had to get his character's point and motivation across with physical cues. That's a difficult thing for an adult actor, but he was only approximately 13 at the time the film came out. I'm super impressed that Bright was able to keep up with the rest of the cast.
Wayne Kramer (Mindhunters, This Film is Not Yet Rated, Blazeland) really nailed the script and the directing. I'm not at all familiar with the rest of his work, but now I want to be.
At the start of the film, I really thought I was just watching a generic, somewhat artistically filmed, action flick. I wasn't in the mood for an action movie, and I didn't expect a damn thing from it. By the end of the film, I was entranced by my TV and sitting on the edge of my seat.
There was some artistic camera work that I felt actually took away from the story, but for the most part that artistic touch tended to add to the general feel of the movie. Also, in some movies, starting at the end and then showing how they got there isn't a good idea. I feel this movie was one of those. The suspense factor would have been through the roof if I hadn't seen that end scene in the beginning of the film.
Below, we have the Rotten Tomatoes scores listed, like we do in all our reviews. I would like to note the significant difference between the opinions of the critics and the opinions of the audience. There's nearly a 30% difference, which is huge. I was curious about the discrepancy, so I actually read through a few pages of reviews by the critics.
The consensus seems to be that the movie was violent. Yeah. That's not a typo. The majority of the "rotten" labeled reviews from critics state that the critic didn't like the movie because it was violent.
Let me take a moment here to explain why that pisses me off.
This is a rated R action film. In none of the advertisements does it claim to be anything different.
In fact, there are three different movie posters for this film. In one, Paul Walker is holding a gun, while people cower in terror behind him. Another one shows Walker holding a gun, but looks like it's for a Fast and Furious (2001-) movie. In that one, beneath him are images of a man screaming, and the same two people cowering in terror. In the last one, Walker is holding a gun while people are cowering and there's another man holding a bigger gun behind him.
What the fuck did the critics expect that the violence level threw them off? This is not a rhetorical question. If you've got an idea, let me know, because I am fucking baffled.
So, here's the deal. If you're looking to watch something about tiny blue creatures dancing to a magical tune, this is not the movie you're looking for. If you're looking for a violent action film, this is one of the best you can find. Try not to expect fluff.
Cat's Point of View:
If I were asked to sum up this movie in one word, it would be 'intense.'
The movie opens with a sense of 'what the hell did I just get myself into?!' and then just keeps charging from there.
It's relentless, gritty, super violent, and just a hair away from being Rated NC-17. It was miraculous that it managed the R rating. Apparently, even writer-director Wayne Kramer (Crossing Over, The Cooler, Pawn Shop Chronicles) is said to have been shocked that it managed the lesser.
I saw so many familiar faces from both the large and small screens. I also have a fresh temptation to prod the showrunners of The Walking Dead (2010-), or its after-show, to find a situation where Abraham's Michael Cudlitz (Tenure, Stolen, Inside Out) finds a stash of gummy bears in honor of his role in this movie.
Cameron Bright (Ultraviolet, Little Glory, Final Girl) was a great choice for the creepy-quiet Oleg. He was excellent with his kicked-puppy look. It really sold his story, even though he had little dialogue. I also loved Vera Farmiga (The Departed, Goats, The Judge) in this. She was a real badass mom.
I have to say that Elizabeth Mitchell (Lost, V, Once Upon A Time) takes the cake for the creep-factor in this one.
It all boils down to one person, though. Of course, I'm referring to Paul Walker (Flags of Our Fathers, Takers, Hours). I've long been a fan of his, may he rest in peace. I don't even care that he tended to gravitate to the same sorts of roles over and over – he was excellent at what he did.
Watching his films, now, is a bit like having a visit with an old friend. For this role, he was a loud and foul-mouthed gangster of a friend; but who's counting?
I'd forgotten that I'd seen this one before – though, that was nearly a decade ago when the movie was released to DVD in 2006. It wasn't by any means a reflection on the film, itself. I think my only problem with it would be the occasional patches of shaky-cam. Otherwise, I really enjoyed the whole visceral ride.
Rotten Tomatoes Critic Score – 40%
Rotten Tomatoes Audience Score – 79%
Netflix's Prediction for Selina – 5/5
Selina's Trust-the-Dice Score – 4.5/5
Netflix's Prediction for Cat – 3.5/5
Cat's Trust-the-Dice Score – 4/5
P.S. Comic Book-type drawings during the beginning of the credits.
Movie Trailer:
Posted by Selina Tropiano and Cat Vaughn at 6:00 PM
Labels: 120 minutes, 2006, Action, Dark, Mob, Movie, Netflix, New Line Cinema, Nudity, Rated R, Review, Trigger
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The Heroes of Evil (2015) - Foreign Film Friday
Top 20 Movies to Look Out For In January (2021)
Invisible Sister (2015) | {
"redpajama_set_name": "RedPajamaCommonCrawl"
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30 pages, 5 1/2" x 8 1/2"
Explores the teacher-facilitator's preparation and how a teaching circle can be created in the classroom. This booklet complements Six Lessons with Delphi and Teaching Delphi, and is a helpful resource for parents and teacher-partners.
The Circle Sanctuary and Its Presence in the Room - The Heart-Mind - Preparation Before Entering the Classroom - The Sanctuary Approach and the Pristine Moment - Having Fun in School - Leaving Convention Behind - Interfacing the Academic Studies with Our Partners in Nature -What the Children Can Tell Us and What They Already Know - What Makes a Teacher for the Sanctuary School?
A teaching circle is a creative expanse of energy that penetrates the classroom. It connects all life into a single unit where no barriers can exist and no static, rigid paradigm can remain. The fluidity of the circle moves energy around from one conception base to another. It passes thoughts into higher degrees of conscious awareness and allows revelation to occur within the class. It creates mobility of thought. | {
"redpajama_set_name": "RedPajamaC4"
} | 4,535 |
package org.hl7.types;
/**
* <p>
* A restriction of entity name that is effectively a simple string used for a
* simple name for things and places.
* </p>
* <p>
* <code>TN</code> is an <code>EN</code> that consists of only one name part
* without any name part type or qualifier. The <code>TN</code>, and its
* single name part are therefore equivalent to a simple character string. This
* equivalence is expressed by a defined demotion to {@link ST} and promotion
* from <code>ST</code>.
* </p>
*
* <pre>
* type TrivialName alias TN specializes EN {
* demotion ST;
* promotion TN (ST x);
* };
*
* invariant(TN x) where x.nonNull {
* x.head.nonNull;
* x.tail.isEmpty;
* x.formatted.equal(x.head);
* };
*
* invariant(ST x) {
* ((TN)x).head.equal(x);
* };
* </pre>
*
* <p>
* Trivial names are typically used for places and things, such as Lake Erie or
* Washington-Reagan National Airport:
* </p>
*
* Example 12.
*
* <pre>
* <name>Lake Erie</name>
* <name>Washington-Reagan National Airport</name>
* </pre>
*
* @see <a
* href="http://www.hl7.org/v3ballot2007sep/html/infrastructure/datatypes/datatypes.htm#dt-TN"
* target="_" title="HL7 Version 3 Standard">Trivial Name (TN)</a>
*/
public interface TN extends EN {
// simple specialization with the same behavior
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 3,392 |
Banjos? Because that sounds good to me.
Time for a favorite banjo thread maybe?
I was hoping I was having one of those flashbacks I payed for some years back.I mean where is it? (OK where are they) Was I ripped off? Was that it!? Pfft!..... I do miss the colors so.I want my money back! I wonder if that Hippie is still sittin' there at Haight & Fell?....Hmmm. Most likely,that was "prime-time corner" for good acid man.Still is,but it's watered down now.
Would I have even noticed? Probably not! That or the dirty rotten Hippie sold me bunk. Ha,ha! LOL.
Anytime I was in Tahoe or Reno,we always caught a show.I've seen Liberace,back when I was 10.
Edit: Why does your post look like I wrote a note to myself?
Rick Wakeman - Jon Lord - Chick Corea ....going by one's I've seen live.
+1 and still going strong with The GrandMothers Of Invention!
I like Manzarek but he makes a dreadful interviewee.
Matthew Shipp. Finally saw his trio this past year. Mindblowing to say the least. | {
"redpajama_set_name": "RedPajamaC4"
} | 3,777 |
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Home News Speech of Ms. Astrid Bant, UNFPA Representative at the Launching event on "Branding design competition for the National Action Month on GE and GBV"
Speech of Ms. Astrid Bant, UNFPA Representative at the Launching event on "Branding design competition for the National Action Month on GE and GBV"
Mr. Pham Ngoc Tien, Director of Gender Equality Department, Ministry of Labour, Invalids and Social Affairs;
Mr. Kim Jinoh, Director, Korea International Cooperation Agency (KOICA) Viet Nam;
Representatives from Government Organizations, CSOs, International development partners, UN Colleagues and media;
A very good morning to you all,
On behalf of the United Nations Population Fund in Viet Nam, I would like to thank the Ministry of Labour, Invalids and Social Affairs (MOLISA) for organizing this important event to launch the branding design competition for the National Action Month on Gender Equality and Prevention and Response to the Gender-Based Violence. I also would like to thank KOICA for contributing funding to the project "Building a Model to respond to Violence Against Women and Girls in Viet Nam". The launch event of today is one of activities conducted under the framework of this project.
Violence against women and girls (VAWG) remains the most pervasive and unaddressed human rights violation on earth. Situation in Viet Nam is no different. Despite the great efforts made by the Government, CSOs, International organizations and development partners in the last decade, VAWG is still a serious issue threatening the peace and security of women and girls.
Although there has been little research conducted on the problem, available data shows that nearly one in four women worldwide experience sexual violence by an intimate partner in their lifetime, and up to one-third of adolescent girls report their first sexual experience as being forced.
In Viet Nam, a MOLISA and Action Aid survey in 2016, which was conducted in five cities and provinces found that 51 per cent of women admitted that they had experienced sexual harassment at least once.
Recently, through media we have witnessed shocking attacks on women and girls such as the case of a young girl in Dien Bien who was kidnapped, raped and killed by a group of men during New Lunar Year Festival; sexual harassment of nine primary school students by a male teacher in Bac Giang province, etc…Sexual violence against women and girls is a global pandemic, most of victims are women, and the risk is higher for adolescents and young people.
From 2013 up to now, UNFPA has been supporting MOCST, and then MOLISA to lead and coordinate annual national communication campaigns in response to 16 days activism on ending VAWG, and national action month on gender equality and prevention of GBV since it has been recognized by the Government in 2016.
Communication package, including logo and IEC materials of the campaigns were developed and used for several years. Recently, the International Women's Day on 8 March has been celebrated with the theme: "Think equal, build smart, innovate for change.
In order to have one voice and image on any single activities towards the issue of GE and GBV, we need to renew the branding of the campaign in creative and innovative ways to raise public awareness and mobilize joint-actions in promotion of the fundamental right of every child and every woman to live a life free of violence.
In this context, I would like to highlight three key messages as follows:
Firstly, information of the competition should be widely disseminated to draw the attention of public and mobilize the participation of different target groups, including ethnic minorities, people with disabilities, and LGBTi through various communication channels: mass media and social media…
Secondly, the communication package of the campaign should be innovative for changing the mindsets and behaviors, particularly men towards gender equality; challenging traditional cultural norms and gender stereotypes, which reinforce dominant male attitudes and violent behaviors; creating an enabling and supportive environment for Vietnamese women to step up and speak out, to make perpetrators feel shame and discomfort when inflicting violence against their partners, co-workers, fellow students, women and girls in the street.
Thirdly, we encourage all participants of the competition to adopt green behaviours by reducing the amount of printing, recycling paper, plastic and glass and suggesting eco-friendly materials and products for the branding package.
I am so happy to see that many young people gather today. Let's collaborate with us to make a real contribution to ending violence against women and girls in Viet Nam.
Ending violence against women should be a priority for every man and woman.
Together, we can make Viet Nam safer and more equitable for women and girls.
I thank all the distinguished guests for your attention and participation. I wish you all good health, happiness and success.
Xin Cam On!
UN in Viet Nam rallies partners to take a stand against sexual violence on the International Day to End Violence against Women
Branding design competition
Op-Ed: From "City of Peace" to Safe Cities for Women and Girls
Code of conduct: To Prevent Harassment, Including Sexual Harassment, at UNFPA events
UNFPA is committed to enabling events at which everyone can participate in an...
UN system Model Code of Conduct
The organizations of the United Nations system are committed to enabling events at... | {
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} | 3,948 |
The core members of a group calling itself "Lizard Squad" — which took responsibility for attacking Sony's Playstation and Microsoft's Xbox networks and knocking them offline for Christmas Day — want very much to be recognized for their actions. So, here's a closer look at two young men who appear to be anxious to let the world know they are closely connected to the attacks.
Kim Dotcom offers Lizard Squad members vouchers to stop the attack.
The LizardSquad reportedly only called off their attacks after MegaUpload founder Kim Dotcom offered the group some 3,000 vouchers for his content hosting service. The vouchers sell for $99 apiece, meaning that Dotcom effectively offered the group the equivalent of $300,000 to stop their seige.
On Dec. 26, BBC Radio aired an interview with two young men who claimed to have been involved in the attacks. The two were referred to in the interview only as "Member 1" and "Member 2," but both have each given on-camera interviews previously (more on that in a bit).
Vinnie Omari, speaking to Sky News on Dec. 27.
Nolan: "So why did you take the vouchers, then?
Member2, the guy that does most of the talking in the BBC interview, appears to be a 22-year-old from the United Kingdom named Vinnie Omari. Sky News ran an on-camera interview with Omari on Dec. 27, quoting him as a "computer security analyst" as he talks about the attacks by LizardSquad and their supposed feud with a rival hacker gang.
The same voice can be heard on this video from Vinnie's Youtube channel, in which he enthuses about hackforums[dot]net, a forum that is overrun with teenage wannabe hackers who spend most of their time trying to impress, attack or steal from one another.
In a thread on Hackforums that Omari began on Dec. 26 using the Hackforums username "Vinnie" Omari says he's been given vouchers from Kim Dotcom's Mega, and wonders if the Hackforums rules allow him to sell the vouchers on the forum.
Hackforums user "Vinnie" asks about selling MegaUpload vouchers. | {
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} | 7,642 |
\section{Section I: Effective model and edge states}
\par
Based on the DFT results, the bases describing the low energy bands along the $\Gamma - A$ path can be simplified as Eq.~1 in the main text.
So the full Hamiltonian with SOC takes the form $H(k) = I \otimes H_0(k) + H_{\text{SOC}}$
under the spinful bases $(|\uparrow\rangle,|\downarrow\rangle)\otimes(|\phi_{1}\rangle, |\phi_{2}\rangle, |\phi_{3}\rangle, |\phi_{4}\rangle)$.
The symmetry allowed $H_{\text{SOC}}$ is given explicitly by Eqs.~S1--S3:
\begin{align}
h_{\text{SOC}}(k) =
\left(
\begin{array}{cc}
\Lambda_{\uparrow\uparrow} & \Lambda_{\uparrow\downarrow}\\
\Lambda^\dag_{\uparrow\downarrow} & \Lambda_{\downarrow\downarrow}\\
\end{array}
\right),
\end{align}
with
\begin{align}
\Lambda_{\uparrow\uparrow} = -\Lambda_{\downarrow\downarrow} =
\left(
\begin{array}{cccc}
-\lambda_d & 0& 0& 0 \\
0 & \lambda_d & 0 & 0 \\
0 & 0 & - \lambda_p -\lambda_1 g(k_z) & 0 \\
0 & 0 & 0 & \lambda_p + \lambda_1 g(k_z) \\
\end{array}
\right),
\end{align}
and
\begin{align}
\Lambda_{\uparrow\downarrow} =
\left(
\begin{array}{cccc}
0 & 0& 0& 0 \\
0 & 0 & i\lambda_2 k_+ & 0 \\
0 & 0 & 0 & 0 \\
i\lambda_2 k_+ & 0 & 0 & 0 \\
\end{array}
\right).
\end{align}
$\lambda_{d,p}$ are the on-site SOC strengths for $d$ and $p$ orbitals, respectively.
$\lambda_1$ is the first order SOC induced by the $z$-direction hopping of the $p$ orbitals,
and $\lambda_2$ is induced by the in-plane hopping between the $p$ and $d$ orbitals with opposite spin.
$g(k_z)$ is defined in the Eq.~(4) of main text.
\par
The $H'(k)$ used in Eq.~(5) of the main text is the 8-band Hamiltonian under
new basis ordering ( $|\phi_{1},\uparrow\rangle, |\phi_{2},\uparrow\rangle, |\phi_{3},\uparrow\rangle, |\phi_{4},\uparrow\rangle,
|\phi_{2},\downarrow\rangle, |\phi_{1},\downarrow\rangle, |\phi_{4},\downarrow\rangle, |\phi_{3},\downarrow\rangle $ ),
which is obtained from $H(k)$ by a unitary transformation as following:
\begin{align}
H'(k)= UH(k)U^\dag,
\end{align}
where $U$ is a unitary matrx:
\begin{align}
U= \left(
\begin{array}{cccccccc}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\
\end{array}
\right).
\end{align}
\par
In order to check the topological electronic properties of our model in $1T$-TiTe$_2$,
we plot the real space charge density distribution of the surface state as illustrated by the blue square in Fig.~3(d) of the main text.
As shown in Fig.~S1, the charge density of the 2D Dirac cone are mostly accumulated at the system surface.
As a result, we have sufficient evidence to point out that the 2D Dirac cone is formed by topological surface states.
\begin{figure}[!t]
\includegraphics[width=0.5\columnwidth]{FigS1}
\caption{The charge distribution in real space for the 2D Dirac cone that illustrated by the blue square in Fig.3(d) of the main text.}
\label{fig:figs1}
\end{figure}
\section{Section II : TSC phase region and Zak phase}
\par
In Fig.~S2(a), we plot the energy spectrum at the $\Gamma$ point as a function of the chemical potential $\mu$ for different bulk
superconducting gap $\Delta_0$. Obviously, we find that the increase of $\Delta_0$ does not change the region of the TSC phase. In Fig.~S2(b),
we show the low energy dispersions of the BdG Hamiltonian with different $\Delta_0$ at $\mu = 280$~meV,
which demonstrates that the gap size on the vortex line is increased monotonously as $\Delta_0$ increases.
In Fig.~S2(c), we calculate the evolution of the Zak phase $\phi$ as a function of $\mu$ with $\Delta_0 = $ 50 meV
for researching the topological property of the vortex line,
which manifests that $\phi=0$ at $\mu < 267$~meV,
$\phi=\pi$ at $\mu \in $ [267 meV, 291 meV],
and $\phi=0$ at $\mu > 291$ meV, respectively.
This result confirm the topologically nontrivial region at $\mu \in $ [267 meV, 291 meV].
\begin{figure}[!b]
\includegraphics[width=\columnwidth]{FigS2}
\caption{
(a) The low-energy spectrum at the $\Gamma$ point with $\Delta_0$ =
1 (red), 10 (blue), 50 (violet) meV, respectively.
In particular, we note that the energy spectrum calculated by $\Delta_0$ = 50 meV is multiplied 1/5.
(b) The energy dispersions of the BdG Hamiltonian at $\mu = 280$~meV with $\Delta_0$ =
1 (red), 5 (green), 10 (blue) meV, respectively.
(c) The evolution of the Zak phase as a function of $\mu$ with $\Delta_0 = $ 50 meV,
which manifests that the Zak phase $\phi=\pi$ at $\mu \in $ [267 meV, 291 meV].
}
\label{fig:figs2}
\end{figure}
\section{Section III : Band structure and TSC in Ti(Se$_{0.5}$Te$_{0.5}$)$_2$ $ $ }
\par
In this section, we calculate the band structures of Ti(Se$_{0.5}$Te$_{0.5}$)$_2$ as shown in Fig.~S3(a),
which are very similar to the results of $1T$-TiTe$_2$. Then we use the effective model to fit the DFT calculated band structures of Ti(Se$_{0.5}$Te$_{0.5}$)$_2$,
and list the fitted parameters in Table~S1.
The fitted band structures (red) are plotted in Fig.~S3(b)--(d).
The Fig.~S3(b) shows that our model and parameters successfully capture the band dispersions and energy position of the topologically nontrivial gap in Ti(Se$_{0.5}$Te$_{0.5}$)$_2$ along the $\Gamma-A$ path.
Fig.~S3(c)-(d) show the in-plane band dispersions below 0.5~eV are reproduced reasonably well.
\begin{table}[!b]
\renewcommand*{\arraystretch}{1.5}
\caption{Parameters used for Ti(Se$_{0.5}$Te$_{0.5}$)$_2$.
}
\label{table:table1}
\begin{tabular}{c|c|c|c|c|c}
\hline\hline
$E_d$ (eV) & $E_p$ (eV) & $t_1$ (eV) & $t_2$ (eV$\cdot${\AA}) & $t_{d}^{\varparallel}$ (eV$\cdot${\AA}$^2$) & $t_p^{\varparallel}$ (eV$\cdot${\AA}$^2$) \\
\hline
0.418 & 0.917 & 0.038 & 3.0 & 12.0 & $-$10.0 \\
\hline
\hline
$t_d^z$(eV) & $t_{p}^z$ (eV) & $\lambda_d$ (eV) & $\lambda_p$ (eV) & $\lambda_1$ (eV) & $\lambda_2$(eV$\cdot${\AA}) \\
\hline
0.004 & $-$0.236 & 0.082 & 0.135 & 0.094 & 2.0 \\
\hline
\end{tabular}
\end{table}
\begin{figure}[!b]
\includegraphics[width=0.6\columnwidth]{FigS3}
\caption{
Band structures of Ti(Se$_{0.5}$Te$_{0.5}$)$_2$ $ $. (a) the HSE06 calculations.
The size of red and blue circles represent the weight projections for the $d$ orbitals of Ti atoms and the $p$ orbitals of chalcogen atoms, respectively.
(b) - (d) show the fitted band structures from the effective model (dashed red bands) with the DFT calculations (solid black and blue bands) along high-symmetry $k$-path.
The blue bands are mostly contributed by the $d_{z^2}$ orbital, which are not considered in our effective model.
}
\label{fig:figs3}
\end{figure}
\par
Finally, we calculate its vortex BdG model with $\Delta_0 = $ 0.8 meV and plot the spectrum in Fig.~S4.
The calculated results obviously manifest that the spectrum at the $A$ point is always gapped,
while the energy spectrum at the $\Gamma$ point closes its gap both at the critical chemical potentials $\mu_{c1} = 348.785$~meV and $\mu_{c2} = 397.6$~meV.
These results exhibit similar spectrum gap closing behavior as shown in 1$T$-TiTe$_2$, indicating that Te doped 1$T$-TiSe$_2$ is also a promising TSC candidate.
\begin{figure}[!t]
\includegraphics[width=0.9\columnwidth]{FigS4}
\caption{
(a) and (b) are the BdG spectrum of Ti(Se$_{0.5}$Te$_{0.5}$)$_2$ with respect to chemical potential $\mu$ at the $A$ and the $\Gamma$ point, respectively.
The spectrum is always gapped at the $A$ point. At the $\Gamma$ point, two critical chemical potentials at about $348.785$ and $397.6$~meV
are marked by red crossing. The inset panel shows the detailed spectrum around $348.785$~meV.
}
\label{fig:figs4}
\end{figure}
\end{document} | {
"redpajama_set_name": "RedPajamaArXiv"
} | 1,616 |
Elmis rioloides is een keversoort uit de familie beekkevers (Elmidae). De wetenschappelijke naam van de soort werd in 1870 gepubliceerd door Kuwert.
Beekkevers | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 3,080 |
from pyramid.view import view_config
from pyramid.httpexceptions import HTTPFound
@view_config(route_name="logout", renderer="logout.mako")
def response(request):
try:
del request.session["identity"]
except KeyError:
# User was not logged in when he tried to access the logout page
url = request.route_url("login")
return HTTPFound(location=url)
return {
"project": request.registry.settings.get("site_name"),
"title": "You have logged out"
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 7,807 |
Q: Exported functions from another package I am following instructions in https://golang.org/doc/code.html#Workspaces link and I build my first Go program.
So, I tried to make library with this instruction = https://golang.org/doc/code.html#Library
and everything is perfect until building hello.go, its gives me this error.
/hello.go:10:13: undefined: stringutil.Reverse
I've already rebuild my reverse.go.
Thats my code:
package main
import (
"fmt"
"github.com/d35k/stringutil"
)
func main() {
fmt.Printf(stringutil.Reverse("!oG ,olleH"))
}
that's my reverse.go (same as docs)
package stringutil
func reverse(s string) string {
r := []rune(s)
for i, j := 0, len(r)-1; i < len(r)/2; i, j = i+1, j-1 {
r[i], r[j] = r[j], r[i]
}
return string(r)
}
and my gopath variable
export GOPATH=$HOME/GoLang
and my files ar in
GoLang/src/github.com/mygithubusername/
A: Golang Tour specify exported name as
A name is exported if it begins with a capital letter. And When
importing a package, you can refer only to its exported names. Any
"unexported" names are not accessible from outside the package.
Change the name of reverse func to Reverse to make it exportable to main package. Like below
package stringutil
func Reverse(s string) string {
r := []rune(s)
for i, j := 0, len(r)-1; i < len(r)/2; i, j = i+1, j-1 {
r[i], r[j] = r[j], r[i]
}
return string(r)
}
A: A different problem with the same symptoms I ran into was that I was having two functions with the same name in Package B. Since two functions wih the same name but different types are no problem, I didn't think about it too much, however, VSCode didn't show the methods when I tried to use them in Package A, despite them being capitalized.
According to the question asked here
this is not supported by Go and requires you to change the function name or use introspection/an interface.
Thought I'd share it since googling brought me to this question here multiple times and it might be something other Go learners might run into as well...
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 8,215 |
{"url":"https:\/\/www.physicsforums.com\/threads\/finding-a-current-in-a-circuit-with-a-dependent-source.847597\/","text":"# Homework Help: Finding a current in a circuit with a dependent source\n\n1. Dec 10, 2015\n\n1. The problem statement, all variables and given\/known data\n\nMy problem is in the images attached. Essentially, I just want to find $I_s$.\n\n2. Relevant equations\n\n$V = IR$\n\n3. The attempt at a solution\n\nI already have the original solution using KCL, which tells me that $I_s = 4 A$ and that $I = 1 A$, but I was trying to do it using KVL instead. After trying (as shown in the image), I keep getting a different answer. Do you see any errors in my methods? Is there anything extra I have to do when dealing with dependent voltage sources like in this example?\n\n#### Attached Files:\n\nFile size:\n304 KB\nViews:\n95\n\u2022 ###### Screen Shot 2015-12-10 at 5.27.09 PM.png\nFile size:\n256.1 KB\nViews:\n98\n2. Dec 10, 2015\n\n### Staff: Mentor\n\nIs it possible that you've conflated the current $I$ used to control the dependent source with the loop current in the third loop?\n\nNote that the branch current through the 12 \u03a9 resistor is made up of the two loop current that pass through it.\n\n3. Dec 10, 2015\n\nOkay. Didn't I account for the fact that there are two currents through that branch by subtracting the mutual voltages in [3]? Is the 3rd loop's current not simply I in this case?\n\n4. Dec 10, 2015\n\n### Staff: Mentor\n\nIf the third loop mesh current is $I$, then $I_s = I$, but the current in the branch (that is the current through the 12 Ohm resistor) is $I - I_2$\n\n5. Dec 10, 2015\n\nOk, I'll try it again. Also, when dealing with circuits involving both current and voltage sources, either dependent or independent, how exactly can one apply mesh analysis? Since you end up having to ascribe a voltage to the current sources, but the resistance of one should be infinite.\n\n6. Dec 10, 2015\n\n### Staff: Mentor\n\nYou'll learn about something called a supermesh. Essentially you draw a loop surrounding the current source, not passing through it, and add a constraint equation that links the two \"merged\" mesh's currents with the current source that they pass through.\n\n7. Dec 12, 2015\n\nI'll certainly look into it. Is this supposed to work for dependent voltage sources, too?\n\n8. Dec 12, 2015\n\n### Staff: Mentor\n\nThe supermesh is generally invoked to deal with current sources (dependent or independent) that border loops. Voltage sources that border loops don't pose a problem since they provide voltage values for the KVL equations.\n\nA current source that does not border loops is trivial to deal with, since it effectively \"solves\" the mesh current for the loop it belongs to.","date":"2018-07-20 20:46:41","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5384072661399841, \"perplexity\": 1025.4695192465178}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2018-30\/segments\/1531676591831.57\/warc\/CC-MAIN-20180720193850-20180720213850-00545.warc.gz\"}"} | null | null |
Why Is The Higgs Boson A 'Big Whoop' For All Of Us?
The physicists who discovered the so-called 'God Particle' were awarded the Nobel Prize this year, but one writer says people still aren't paying enough attention. Scientist Ainissa Ramirez tells host Michel Martin why more people should care about the Higgs Boson, and why they probably won't.
Copyright © 2013 NPR. For personal, noncommercial use only. See Terms of Use. For other uses, prior permission required.
I'm Michel Martin and this is TELL ME MORE from NPR News. We've talked before on this program about why Latinos in the U.S. are more likely to tweet and use other social media than other Americans. Today, we're going to hear from a Latino tech leader who wants to boost the Latino presence in the science and business of technology. We'll talk about that in just a few minutes.
But, first, more good news about science and innovation. This week, the Nobel Prize for physics was announced, and it went to Francois Englert and Peter Higgs for their discovery, the Higgs boson, popularly known as the God particle. Our next guest is really excited about this and wants you to be, too. Ainissa Ramirez is a materials scientist. She wrote a guest post for Forbes called "The Higgs Boson: Why You Should care About The God Particle. And, Sadly, Why You Don't." And she's with us now. Welcome, thanks so much for joining us.
AINISSA RAMIREZ: Great to be here.
MARTIN: You said in your post that the Higgs boson is the biggest scientific discovery of the 21st century, period. Why?
RAMIREZ: Good question. The Higgs boson links us to the Big Bang. There was an equation that had lots of different parameters in it, and it worked, if we found the Higgs boson. And we continued to use this equation without finding it. And, in the last year or so, we actually found it. So it confirms what we believed. So it's wonderful to have a theory, but it's great to have proof. And so that's why the Higgs boson is so significant.
MARTIN: Why is it called the God particle?
RAMIREZ: The God particle, well, that's a misnomer. And I think that's a marketing term. I think there was a book that came out a couple of years ago, and that was the title of the book, "The God Particle." The scientists actually wanted to call it the God [bleep] particle because it was so hard to find. It's been something that scientists have been looking for for over 50 years. But the God particle is a - it's a term because it's an ethereal thing - it's very difficult to find. Sort of like, air is all around us, but we can't really see it. This particle, the Higgs boson, is all around but it's not something that we can see, and it was really hard for us to find how we can see it.
MARTIN: Well, is it in part, though, because it does verify the scientific understanding of the creation of the universe? Is that why?
RAMIREZ: Absolutely. It's the link. So 13.7 billion years ago, there was this big bang, and we went from nothing to something. There were all these particles that were flying around and they had no mass. And now we live in a universe where there's mass. How did they get the mass? Well, it was the Higgs boson, which creates the mass, and that was the missing link. So it connects the beginning part of the universe to now.
MARTIN: One of the things you also said in your post is that there's this disconnect between the scientific community - if I can use that term - and the rest of us. You said that a few thousand scientists are losing their minds, crying and hugging each other. The rest of society is trying to figure out why this is a big whoop. Why do you think that is? I mean, it used to be that people were very engaged in science. I mean, the moon launch, you know, for example, you know, people landing on the moon. The whole country kind of stopped what they were doing to watch this, right? So why do you think that is?
RAMIREZ: Everyone remembers where they were when the moon launch happened. I think I was being propped up by my family to watch the screen, although I didn't know exactly what I was looking at. What's so obvious is that there's a footprint on the surface of the moon - tells us that we've made an accomplishment. A couple of scratches on paper just doesn't have the same effect, but it's also a footprint. So I think that's one of the things that was a disconnect. It's so obvious when a person leaves our planet to go to the moon. The other thing is that I think we need to spend a little bit more time as scientists to explain what we're doing to the general public.
And that was what my call to action was in that blog. We have a small percentage of people who are losing their minds, having huge parties and the rest of the world doesn't really know what they're excited about. So if we spend a lot of money to put together these huge experiments - the Large Hadron Collider, where the Higgs Boson was found costs on the upwards of about $10 billion. Let's spend a little bit of that money just educating people. Now maybe scientists aren't the best people to do that. I mean, we're very, very consumed with what we're doing. We're very, very busy and we may not have that skill set to make things accessible to other people. So in that blog, I was saying, well, let's employ. Let's get some help. Let's consider a PR firm or let's make a videogame or let's get Peter Higgs - one of the folks who got the Nobel Prize - let's have him do a cameo on a television show, so that we can just make this a little bit more personal about what's going on.
MARTIN: You said, I think the nerds got it wrong by not inviting everybody to the party, which is kind of different from the way we usually think about it as, you know...
RAMIREZ: That's right.
MARTIN: The nerds not being the ones who get the invitation. But can you help us understand a bit more about why this - what will change in our lives because of this discovery? I mean, as you pointed out, you know, people going to the moon kind of opens up all kinds of possibilities about travel, about seeing other worlds. What are other things that might change as a result of this discovery, this understanding - or this verification of this theory?
RAMIREZ: Well, those are the things that are hard to tell at this point. When the electron was discovered back in the 1890s, it wasn't clear what we were going to do with that. But this conversation wouldn't be possible without electrons because we use it for communication and the like. But it's more than the endpoint of science, it's the journey that's actually important. So on our way to finding the Higgs, we created this little thing called the World Wide Web.
We needed a way to communicate with scientists that were in hundreds of countries, and thousands of scientists for them to interface and have ways to communicate. And, you know, the Postal Service was no way that they were going to be able to handle all that amount of data. So we found a way to communicate and that was the World Wide Web. And you and I can't live without that today. So it's more than the end goal, it's the pursuit that makes us better people. And that's what I think the folks at CERN should share. You know, there's a lot of inventions that occur in the pursuit of these different things.
MARTIN: You know, it's interesting. On the one hand, though, we live in a world in which science and technology is more powerful and present than ever. It is accessible. So many people have a smart phone. So many people throughout the developing world do their banking on their phone, right? And yet, on the one hand, you feel that there's still this - what do you think it is? A lack of understanding of how we get there? You see what I'm saying? What do you think the disconnect is? We don't really invest in it? We don't really care about how we get there? It's like a pot - we just want to cook in, we don't care how it's made?
RAMIREZ: Yeah, it's a good question. I think that there's an infinite amount of information. What we need to do is have individuals that help us understand it. You know, I love Wikipedia. I tell my students not to use it when they're writing their papers, but I do love Wikipedia. It's a first stop. But some Wikipedia entries are just so high-level. I don't understand what's going on. I'm saying that when you get that information, it's great to be accurate and to tell us things precisely, but make it understandable. Maybe we need a next level down. So there's not a dearth of information. I don't think that's the problem. I think it's the understanding that's missing.
MARTIN: OK. So now we're friends. What is a material scientist? Now that you've told me, I have permission to ask you and you're not going to make me feel stupid. What is a material scientist? What do you do when you're not trying to explain all this stuff to us?
RAMIREZ: A material scientist is a person who's interested in how stuff works - and stuff is materials. The reason why you're sitting in that chair and it's not falling apart, the reason why the sky is blue - these are things that chemists and a material scientist would be interested in. And the thing that blew my mind when I first took my material science course is that everything that we understand has to do with the interaction of atoms. Now atoms are about 1/100th the thickness of your hair.
So imagine pulling one of your hairs and whittling it 100,000 times. That would be the width of an atom. The way that these small particles interact is what prevents me from falling through the floor, for my pen working, all these things. So I'm interested in how - I guess I would call myself an atom whisperer or an atom hacker because I'm interested in how - understanding the behavior of atoms. And then, suggesting that they behave in new ways so that I can create new properties of the stuff that I'm using.
MARTIN: I feel so much better now.
RAMIREZ: Does that make any sense?
MARTIN: Yes.
RAMIREZ: OK, good.
MARTIN: Ainissa Ramirez is a material scientist. She is the author, most recently of "Newton's Football: The Science Behind America's Game." And she was kind enough to join us from the studios of Yale University. Ainissa Ramirez, thanks so much for speaking with us.
RAMIREZ: Oh, thank you for having me.
Copyright © 2013 NPR. All rights reserved. No quotes from the materials contained herein may be used in any media without attribution to NPR. This transcript is provided for personal, noncommercial use only, pursuant to our Terms of Use. Any other use requires NPR's prior permission. Visit our permissions page for further information.
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Source: http://www.npr.org/templates/story/story.php?storyId=231446083&ft=1&f=46
Category: Heartbreaker Justin Bieber notre dame Daft Punk Breaking Bad Season 5 Episode 10 mick jagger
Oct 11, 2013 12:02:12 AM
Why Do Hard To Smoking Cigarettes / Desidieter Type of my ideal sessions can become a electronic cigarettes prove to be a great conversation starter hypnosis session, because this kind of is fast, very easy and gives one particular client excellent excellent value. Here A will outline a small amount of of the devices and methods Naturally i...
Snowden: mass surveillance making us less safe MOSCOW (AP) — Former National Security Agency systems analyst Edward Snowden says surveillance programs used by the United States to tap into phone and Internet connections around the world are making people less safe. In short video clips posted by the WikiLeaks website on Friday, Snowden said the NSA mass... | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 2,638 |
Q: Text Widget - How to avoid text being parsed as special characters This may seem a weird question, but I really don't know how to search this one.
Imagine that I have this Widget:
Text("Career highlights (9)")
This is being shown as this:
How can I show a simple (9) insted of what is shown?
A: Try this
Text("Career highlights \(9\)")
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 7,912 |
Wee Beastie was a brand of alcopop, as it is commonly referred to. It is part of the RTD (ready to drink) market and is popular in Scotland. It was available in 275ml and 700ml bottles and was 5.4% ABV. The 700ml bottles are labelled "Big Beastie". Wee Beastie is owned and distributed by Inver House Distillers. The drink is described as a carbonated, premixed blend of vodka, taurine and caffeine, flavoured raspberry and blackcurrant. The taste is similar to Red Bull.
In March 2006 the company was told to change their packaging. They were asked to eliminate the cartoonish spider mascots, as they might encourage children to drink it. They also had to tone down the "For Adults Only" warning, as it might appeal to underage drinkers.
This drink has now been discontinued and is no longer being produced.
Premixed alcoholic drinks
Alcopops | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 6,369 |
\section{Introduction}
Astrophysical jets produced by supermassive black holes are
usually explained by the magnetic Blandford-Znajek process
\cite{Blandford:1977ds}. Due to its electromagnetic nature,
only charged particles can be ejected as a jet in this way. In
this work, we quantitatively analyze a gravitational
mechanism for jet production. Through the Penrose process
\cite{Penrose:1971uk}, particles which have fallen into a
supermassive black hole's ergosphere from the accretion disk
can extract energy from the black hole and be
ejected along the rotation axis \cite{Gariel:2010}. It should
be noted that while this is subdominant with respect to the Blandford-Znajek mechanism, it can affect Dark Matter (DM)
particles as well, due to its gravitational nature.
The main goal of this article is to confirm the presence of a
DM beam produced by the mechanism described above
quantitatively. It will be shown that the DM beam is highly
collimated. Furthermore, the dependence of the DM density on the
mass of the black hole and on the distance to it will be
analyzed as well.
The existence of such a DM beam renders this phenomenon
potentially interesting for future DM
detection. Indeed, a number of recent tests have yielded
promising results for indirect DM detection: records from the PAMELA satellite
\cite{Adriani:2008zr} suggest a surprisingly high fraction of
positrons in cosmic ray measurements over {$10\,$}GeV. Similar
results were obtained by the AMS collaboration
\cite{Aguilar:2013qda}, the DAMPE collaboration
\cite{Ambrosi:2017wek} reported a peak in the flux of electron
and positron cosmic rays at around $1.4\,$TeV, the
H.E.S.S. telescopes \cite{Aharonian:2004wa} detected a
point-like source of very high-energy $\gamma$ rays from
Sagittarius A*, and the FERMI/LAT data hinted at an
excess in {$\gamma$} rays at energies of {$130\,$}GeV
\cite{Weniger:2012tx}. Of course, it is unknown whether the
detected signals are truly the result of DM particle
annihilation, but it is an interesting possibility.
The usual WIMP cross section is too small to explain the
observed fluxes of DM-annihilation products, so a boost factor
$B$, which usually lies in the range between {$10^{2}$}
and {$10^4$}, must be artificially added to DM models in order
to explain the observed measurements,
\cite{Bergstrom:2005ss,Bergstrom:2004cy,Hooper:2008kv,Meade:2009iu,Bergstrom:2009fa}. Several
potential possibilities for the physical origin of this
overdensity have been discussed in the literature such as
density inhomogeneities at tiny scales \cite{Lavalle:2007apj}
and Sommerfeld enhanced annihilation cross sections
\cite{Hisano:2003ec}. When an incoming beam of DM particles
annihilates with some target DM particles in the proximity of
the Earth, the boost factor $B$ is defined as
\begin{equation}
B=\frac{\rho_{B}}{\rho_{0}}\frac{\rho_{T}}{\rho_{0}}
\end{equation}
where {$\rho_{B}$} and {$\rho_{T}$} are the densities of the
beam and the target, respectively, and {$\rho_{0} \approx
0.4\,$GeV/cm$^3$} is the macroscopic DM energy density in
the solar system, as defined in \cite{Bergstrom:2009ib} and
\cite{Bergstrom:2012fi}.
With this motivation in mind, we study whether such DM
overdensities in our vicinity could be caused by rotating
black holes via jet production. In this situation, the target
DM particles are those in our immediate vicinity, therefore
{$\rho_{T}/\rho_{0}=1$}. We will show that even though this
mechanism produces a DM beam, the corresponding DM densities
are not sufficient to obtain a significant boost factor.
This article is organized as follows: we describe geodesics in
a spinning black hole geometry in Sec.~\ref{sec:geometry} and
gather the relevant ones in Sec.~\ref{sec:boost} to determine
the overdensities in the black hole jet. We introduce
approximations in Sec.~\ref{sec:approximation}, which allow us
to numerically evaluate the DM density in the beam and show
the findings in Sec.~\ref{sec:applications}. We conclude in
Sec.~\ref{sec:conclusion}. Throughout this article, we use
units where $G=c=1$.
\section{Geodesics in a Kerr Geometry}
\label{sec:geometry}
The system under consideration is a galaxy with a rotating
Kerr black hole in the center. We concentrate on particles
which follow geodesics leading them from the accretion disk
into the ergosphere. There, they can scatter with other
particles or simply decay. As depicted in Fig.~\ref{boost},
some of the products of these interactions can then follow a
geodesic leading out of the ergosphere and moving parallel to
the rotation axis due to the Penrose process as shown in
Ref.~\cite{Gariel:2010}. Collecting all geodesics with similar
behaviour results in the black hole agglomerating particles
from the accretion disk and releasing them along its rotation
axis and thus forming a jet.
\begin{figure}[t]
\centering
\includegraphics[scale=1]{DMsystem.pdf}
\caption{Relevant geodesics for
jet formation in a Kerr spacetime. The solid line depicts a geodesic of
the particle A in the accretion disk
falling into the black hole where it scatters off
the particle B following the dashed line. After the
collision, particle A leaves the black hole and moves
parallel to the rotation axis and particle B falls
towards the black hole
singularity.}
\label{boost}
\end{figure}
We consider the usual Kerr metric of a black hole with mass $M$ and
angular momentum $M a$. The motion of a particle for a non-rotating
neutral black hole ($a=0$) is completely determined by the particle's
mass, energy and angular momentum. However, for the case of a Kerr
black hole ($a \not = 0$) a fourth constant is needed, the Carter
constant, which for a particle of unit mass reads
\begin{equation}
Q=\left(u_{\theta}\right)^{2}+a^{2}\cos^{2}\left(\theta\right)\left(1-E^{2}\right)+\frac{\cos^{2}\left(\theta\right)}{\sin^{2}\left(\theta\right)}L^{2}\;{}\label{carter-constant-mass}
\end{equation}
where $u_\theta=\mathrm{d}x_{\theta}/\mathrm{d}\tau{}$ is the {$\theta$}
component of the proper four-velocity, $E=-u_{t}$ is the
particle's energy and $L=u_{\phi}$ is its angular momentum. With the help of the constants of motion
together with the initial position and velocity of the DM particle in
the accretion disk we can determine whether it has a chance
of ending up in the DM
jet together with its final position and velocity for a given time. In
the next section we take all these geodesics into account to obtain the DM
density in the jet.
\section{The DM density of the
ejected particles}\label{sec:boost}
To investigate the jet formation we obtain the overall density of the outgoing particles by taking
into account the contributions due to all the possible outgoing
velocities,
\begin{equation}
\rho_{\rm{}out}\left(r_{\mathrm{out}},\theta\right)=\int{} \frac{\mathrm{d}^{3}\mathbf {v}_{\rm{}out}}{r_{\mathrm{out}}^{2}\sin\left(\theta\right)v_{r,\,\mathrm{out}}}{\frac{\mathrm{d}^{3}N_{\rm{}out}\left(r_{\mathrm{out},\theta,\mathbf{v}_{\mathrm{out}}}\right)}{\mathrm{d}{}v_{r,{\rm{}out}}\mathrm{d}{}v_{\phi,{\rm{}out}}\mathrm{d}{}v_{\theta,{\rm{}out}}}}\;{},\label{rho-out}
\end{equation}
where
{$N_{\rm{}out}\left(r_{\mathrm{out}},\theta{},\mathbf{v}_{\rm{}out}\right)$}
is the total number of particles in the beam at
{$\left(r_{\mathrm{out},\theta}\right)$} with velocity
{$\mathbf{v}_{\rm{}out}$}. The number of outgoing particles is
computed by taking the number of particles that fall into the
ergosphere and filtering out all of those that follow
trajectories that do not end up in the beam,
\begin{equation}
N_{\rm{}out}\left(\mathbf{x},\mathbf{v}_{\rm{}out}\right)=\eta P_\text{scat}\int{}\mathcal{D}^{3}\mathbf{v}_\text{in}\rho_{\rm{}in} V_\text{in}\left(\mathbf x, \mathbf v_{\rm{}in}, \mathbf v_\text{out}\right)\,{}.\label{nout}
\end{equation}
Here $V_\text{in}$ is the volume occupied by the infalling
particles with initial velocities
{$\mathbf{v}_\text{in}=\left(v_{r},v_{\phi},v_{\theta}\right)$}
which end up in the beam and reach $\mathbf x$ with final
velocity $\mathbf v_\text{out}$. {$\eta$} is the efficiency of
the Penrose process. $P_\text{scat}$ gives the probability of
an infalling particle scattering in the ergosphere and not
before. The density of the infalling particles is denoted by
{$\rho_{\rm{}in}$}. The integration runs over
{$\mathcal{D}^{3}\mathbf{v}=\mathrm
d^{3}\mathbf{v}f\left(\mathbf{v}\right)$}, where
{$f\left(\mathbf{v}\right)$} is the distribution function for
the velocities of the infalling particles. In the next
Section, we will introduce approximations to simplify
Eqs.~(\ref{rho-out}) and (\ref{nout}), and afterwards compute
{$\rho_{\mathrm{out}}$} numerically.
\section{Approximations and Assumptions}\label{sec:approximation}
Since the system under consideration has a high dimensional parameter
space and partially includes unknown initial conditions,
approximations and assumptions are needed to
numerically compute Eqs.~(\ref{rho-out}) and (\ref{nout}) for
any specific galaxy. This section
provides a discussion of all such approximations and assumptions used in this
article.
\textbf{Volume of infalling particles:}
The volume occupied by the incoming particles is for convenience approximated
to be a ring of radius {$r_{\rm{}in}$}, height
{$\Delta{}z=z_{\rm{}max}-z_{\rm{}min}$}, and width
{$v_{r}\Delta{}t$}, i.e. ${V_\text{in}\left(\mathbf x, \mathbf v_{\rm{}in}, \mathbf
v_{\rm{}out}\right) =
2\pi{}r_{\rm{}in}v_{\rm{}r}\Delta{}t\Delta{}z\left(\mathbf x,
\mathbf v_{\rm{}in}, \mathbf v_{\rm{}out}\right)}$. A more sophisticated procedure would be to
start from the detection of the particles with given
velocities in some time interval and trace them back to the
ergosphere and to the accretion disk. In this way one would
collect all relevant geodesics such as the one in
Fig.~\ref{boost} and could ask where the detected particles
originated in the accretion disk, which DM densities were
present at that time and thus how many particles are actually
taking this geodesic. By approximating this with an infalling
ring, there are different counteracting effects, e.g. a larger
radius accounts for a bigger volume but also a smaller
density, there is a region in the accretion disk in which the
most particles relevant for jet formation
originate. Therefore, a suitable choice of $r_\text{in}$ has
to lie within this region in order to gather the most relevant
geodesics and thus to be a reasonable approximation to the
above procedure. To obtain an upper bound, we choose
$r_\text{in}$ so that the maximum feasible
$\rho_{\mathrm{out}}$ is achieved. For example, for the
Andromeda galaxy {the correct choice is {$r_{\mathrm{in}}=$}}
$0.1\,$pc.
\textbf{Carter constant:}
We assume that the value of the Carter constant does not
change due to the scattering in the ergosphere, as was done in
Ref.~\cite{Gariel:2010}. Furthermore, we assume that
outgoing particles with the correct Carter constant
{$Q_{\rm{}out}$} do indeed end up in the beam and do not
follow some other geodesic with the same value of
{$Q_{\rm{}out}$}. This leads to an upper bound for the
maximal obtainable overdensity. Keeping all other
parameters fixed and solving {$Q_{\rm{}in}=Q_{\rm{}out}$}
for {$z\left(\mathbf x,
\mathbf{v}_{\rm{}in},\mathbf{v}_{\rm{}out}\right)$}
results in the position the particles initially must have
in order to end up in the beam at $\mathbf x$. Here,
{$Q_{\mathrm{in}}$} is the Carter constant for the
ingoing particles. If it has no solution, there is no
geodesic connecting the point in the beam with the point
in the accretion disk.
This
makes sure that all and only the particles that can end up in the beam
are considered for each set of parameters
{$\{\mathbf x, \mathbf v_{\rm{}in}, \mathbf v_{\rm{}out}\}$}.
\textbf{Distribution function:}
The velocity distributions of the infalling particles are assumed to be
Gaussian,
i.e.
\begin{equation}
f\left(\mathbf{v}_\text{in}\right)\equiv{}\frac{\exp\left\{-\frac{1}{2} (\mathbf v_\text{in} - \mathbf v_0)^T \Sigma^{-1} (\mathbf v_\text{in} - \mathbf v_0)\right\}}{\sqrt{(2\pi)^{3}|\text{det}(\Sigma)|}} \,.
\end{equation}
We assumed the covariance matrix to be diagonal and isotropic
such that it reads $\Sigma= \text{diag}(\sigma^2, \sigma^2,
\sigma^2)$ with standard deviation $\sigma$. The orbiting of
the black hole dominates the mean velocity of the DM particles
in the halo, and consequently {$\mathbf{v_0} =
(0,v_{\phi,0},0)$}. We take explicit values for $v_{\phi,
0}$ from data on the rotation curve of the Milky Way in
\cite{Boshkayev:2018sbj} and assume that other galaxies
have similar velocity distributions. In order to estimate
the value of {$\sigma$} we refer to Fig.~2 of
Ref.~\cite{Brandt:2010ts} to obtain the mass accretion rate
{$\mathrm dM/\mathrm dt$} of supermassive black holes. By
setting this equal to the infall rate of particles around the
black hole we obtain a value for {$\sigma$}. Explicitly, the
equation to be solved is
\begin{equation}
\frac{\mathrm dM}{\mathrm dt}=4\pi{}r_{\rm{}in}^{2}\rho_{\rm{}in} \int_{0}^{1} \mathrm dv_{r} \frac{1}{\sqrt{2\pi{}\sigma^{2}}}e^{-\frac{v_{r}^{2}}{2\sigma^{2}}}v_{r} \label{dm-dt}\,{}.
\end{equation}
We solve Eq.~(\ref{dm-dt}) for {$\sigma$} numerically with the
appropriate parameters depending on the specific case
considered. The velocity distribution must then be multiplied
with the density of DM at {$r_{\rm{}in}$} in order to obtain
the total number of particles with velocity
{$\mathbf{v}_{\rm{}in}$} per unit of volume. As an example, for the Andromeda galaxy with
{$r_{\rm{}in}=0.1\,{}$}pc and assuming a cored DM profile
\cite{Tulin:2017ara}, the DM density can be approximated by
{$\rho_{\rm{}in}\left(0.1\,{\rm{}pc}\right)=30\,\rho_0$}.
\textbf{Penrose efficiency:}
The efficiency of the Penrose process depends on the angular momentum
of the black hole and can reach for maximally rotating black holes a value of
$\eta = 0.29$ \cite{Dolan:2011xt}. Since the angular momentum of the black
holes under consideration is not known we assume a Penrose efficiency
$\eta = 0.01$. Since the DM density depends linearly on $\eta$, one can easily adjust it for other angular momenta.
\textbf{Mean free path:}
In order to describe the path of incoming particles solely with one geodesic, the probability for avoiding any scattering in the accretion disk has to be implemented. Taking the mean free path of the particles such that in average one scattering event occurs within the ergosphere and assuming no significant change in DM density along the geodesic we take $P_\text{scat} = {\lambda_\text{mfp}}/{r_\text{in}} \approx {2M}/{r_\text{in}}$.
\section{Results}\label{sec:applications}
With the approximations in place, we determine the DM density
in a black hole DM beam numerically, i.e.
$\rho_{\mathrm{out}}(r_\text{out}, \theta)$ in
Eq.~\eqref{rho-out}. To this end we integrate over the initial
velocities of the DM particles in Eq.~\eqref{nout} as a
Riemann sum. In order to reduce the computation time as much
as possible the integration limits were chosen such that the
neglected portion of the parameter space contributes at most
{$0.1\%$} to the final result. Furthermore, when deciding on
the step size for the integration, a compromise had to be
taken. However, while the final result can indeed change by
as much as two orders of magnitude when decreasing the size of
the integration steps, the qualitative analysis remains
unchanged.
In the code provided in Ref. \cite{code} we implemented
the steps just described. The integration limits and
step size are numerically adjusted such that we use the sweet
spot between accuracy and computation time required for each
specific black hole in question. Additionally, for the
outgoing radial velocity there is a lower bound since
particles arriving to us today had to be sent by the black
hole at the earliest during the black hole formation such that
$v_{r,\text{out}}>r_\text{out}/t_\text{age}$, where
$t_\text{age}$ is the age of the black hole. DM particles that
are slower than this bound might reach us in the future, but
cannot have reached us yet and thus do not contribute to the
boost factor.
In Fig.~\ref{boostAndr} we show the density profile of the DM
beam created by the Andromeda black hole. There are immediate
lessons to be learned from this result. First, the beam is
highly collimated with an opening angle of $2 \theta_B \approx
10^{-5}$ which for visibility is stretched. Second, the beam
is far ranging such that even for a distance of $1\,$Mpc the
beam can still be distinguished. Third, the DM density of the beam increases by
many orders of magnitudes if evaluated closer to the black
hole and its rotation axis but never reaches a significant
overdensity. With these observations we can conclude that the
Andromeda black hole is in principle capable to produce a
sharp, far ranging but faint DM beam. As a sanity check we
analyzed the qualitative behavior of the DM beam for different
values of the black hole's angular momentum $a$. For a
Schwarzschild black hole, i.e. $a=0$, without adjusting $\eta$, the DM density is 8
magnitudes smaller compared to the case with $a=M/2$
\cite{code}. This background contribution is due to geodesics
reaching the target location without using any
Kerr-black-hole-related effects such as the Penrose
process. This contribution is negligible compared to the DM
density in the beam and thus effects uniquely attributed to
the Kerr metric are causing the DM beam. Furthermore, the
larger $a$ is chosen, the more collimated is the DM beam as
expected \cite{code}.
It should be noted that the obtained densities are much smaller than
the local DM density. As an example, we present here the
explicit result for the black hole in the center of the Andromeda
galaxy, for which we assume that its rotation axis points at
us. Taking the age of the galaxy equal to the one of the black hole to
be $t_\text{age} = 10^{10}\,\mathrm{yrs}$, the DM density as in
Eq.~\eqref{rho-out} in our local neighborhood is
$\rho_{\mathrm{out}}\approx{}10^{-12}\rho_{0}$. Despite the Andromeda
black hole being the most promising candidate, since $B \ll 1$ the
overdensity is negligible and cannot explain boost functions of the
order $10^2$ required for explaining indirect DM measurements in the
solar system.
\begin{figure}[t]
\centering
\includegraphics[scale=1]{DMdRout.pdf}
\caption{DM density profile in the beam represented in units of $\rho_{0}$ for various distances to the black hole $r_\text{out}$ and the rotation axis. In addition, the opening angle of the beam $2\theta_B$ is schematically indicated. Due to the inapplicability of our approximations, for the white pixels there is no data.
}
\label{boostAndr}
\end{figure}
\begin{figure}[t]
\centering
\includegraphics[scale=1]{DMmassRout.pdf}
\caption{The DM
density in the beam at various distances from a black hole $r_\text{out}$ with mass
$M$ for $\theta=10^{-8}$. The mass range includes both smaller black
holes at galactic centers and the largest black
holes yet observed. The two examples presented in
the article, Andromeda and Sagittarius A*, are
marked by the symbols A and S. }
\label{boostMassDist}
\end{figure}
In order to evaluate a large class of astrophysical black
holes we show in Fig.~\ref{boostMassDist} the DM density
{$\rho_{\mathrm{out}}$} depending on the mass of the black
hole and {$r_{\mathrm{out}}$}. As a second specific example we
choose Sagittarius A* and evaluate the boost function at a
distance to the black hole of $10\,$pc as shown in
Fig.~\ref{boostMassDist}. Choosing the target location this
close to the black hole would open up new possibilities for
indirect DM detection close to black holes by observing
annihilation or decay products from this region. However, as
seen in Fig.~\ref{boostMassDist} the boost factor for
Sagittarius A* is negligible and therefore it is not
worthwhile to examine this possibility further. For other
black holes, as seen in Fig.~\ref{boostMassDist} there is an
optimal boost function for smaller distances and masses in the
range of $10^8\,M_\odot$ to $10^9\,M_\odot$. However, since
these DM overdensities are orders of magnitude too small to be
relevant for any indirect DM observations, we conclude that,
within the scope of our investigation, DM beams are generated
by spinning black holes but are so faint that they do not
significantly contribute to DM overdensities.
\section{Discussion and conclusion}\label{sec:conclusion}
The aim of this paper was to investigate whether beams of DM
particles can be formed along the rotation axis of
supermassive rotating black holes through the Penrose
process. We showed that this is indeed the case and further
analyzed the intensity and profile of such jets. In
particular, they are shown to be very collimated and
long-ranged. However, the density of the DM particles
inside the beams is very small and cannot possibly generate
overdensities near the Earth. Therefore, within the scope of our analysis, this mechanism
can be excluded as a possible source of a boost for the local
DM detection rate.
We showed that the density is larger closer to the black hole
and becomes smaller and smaller further away. Furthermore, we
showed that a black hole mass of approximately
{$10^{8}-10^{9}\,M_{\odot{}}$} leads to the densest beams. The
density falls off sharply further away from the rotational
axis, indicating that the DM jet is highly collimated. By
tuning appropriately all parameters, a maximal value of
approximately {$40\,\rho_{0}$} can be obtained. Furthermore,
increasing the black hole's angular momentum creates a more
collimated beam.
We hope that our study inspires future calculations. Moreover,
a closer analysis of the produced overdensites near the
rotating black hole at the center of the Milky Way might allow
to find a source of DM annihilation which is relevant for
local DM detection.
\begin{acknowledgments}
We appreciate financial support of our work by the DFG cluster of
excellence ``Origin and Structure of the Universe''. Furthermore, we
are thankful for helpful input from Alexis Kassiteridis and comments
from Cecilia Giavoni and Marc Schneider.
\end{acknowledgments}
\bibliographystyle{unsrt}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 1,453 |
Q: Hide some field in a standard page on insert and show on update In my contact standard page layout, I am trying to hide some field during insert and show it only on update. Is it possible ? What is the best way to do it ? The thing on my mind at the moment is using javascript or create a visual force page that will be put in the contact layout but I am not sure if that idea is correct. Please help. Thank you...
A: You will end up overriding either the insert page or the Edit page .The insert page override is by overriding the new Button and probably the easiest approach .
You can use fieldsets to keep it dynamic and flexibility to drag and drop the fields by admin if field changes .
The Javascript buttons will not be supported in lightning experience as well so its always good to design for future and avoid them
A: I think you can do that without need to a custom VisualForce page. You can use JQuery to scan the page for this field and remove it. JavaScript will be able to determine if this is insert or update record based on page parameters.
You can load the code in a {!REQUIRESCRIPT} on a custom link or button on the page.
This is similar to this previous question
| {
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} | 2,020 |
{"url":"https:\/\/www.physicsforums.com\/threads\/cartans-identity.148962\/","text":"# Cartan's Identity\n\n1. Dec 21, 2006\n\n### SeReNiTy\n\nHi can someone help me prove this identity? I'm having trouble understand the role of the interior product or more precisely how to calculate with it.\n\nMy professor uses the \"cut\" notation _| but i don't see this in any text books. Can someone give me some hints on how to prove the identity?\n\n2. Dec 22, 2006\n\n### coalquay404\n\nThere's nothing difficult about proving Cartan's formula, although I admit that it can become a nightmare keeping track of the indices when you deal with a general $p$-form. For those who aren't aware of what Cartan's formula states, it says that the Lie derivative of a $p$-form $\\alpha$ along the flow generated by a vector field $X$ is equal to the anticommutator of the interior product and the exterior derivative, i.e.,\n\n$$\\pounds_X\\alpha = \\{d,\\iota_X\\}\\alpha = d(\\iota_X\\alpha) + \\iota_X(d\\alpha)$$\n\nThere's really no mystery here: simply plug in the definitions of the exterior derivative, interior product, and Lie derivative, and show that the left-hand and right-hand sides are equal.\n\nIn case you're really having problems understanding the interior product, note that it's simply a map which takes a vector and a $p$-form as arguments and spits out a $(p-1)$-form as an output. That is, the interior product is a map\n\n$$\\iota:T_xM\\times\\Lambda^p_x(M)\\to\\Lambda^{(p-1)}(M),\\,\\,\\,\\iota:(X,\\alpha)\\mapsto\\iota_X\\alpha.$$\n\nIt is defined so that\n\n$$\\iota_X\\alpha(X_{(2)},\\ldots,X_{(p)}) \\equiv \\alpha(X,X_{(2)},\\ldots,X_{(p)}).[\/itex] If this doesn't help, could you be a bit more specific about precisely what it is you're having difficulty with? Last edited: Dec 22, 2006 3. Dec 22, 2006 ### SeReNiTy My problem was i was trying to define the lie derivative of vector fields in the direction of vector fields ie, [X,Y] = LxY to equal the anticommutator. 4. Dec 23, 2006 ### coalquay404 You can define the Lie derivative of one vector field along the flow generated by another to be equivalent to the Lie bracket. That is, if $X,Y\\in\\mathfrak{X}(\\mathcal{M})$ then [tex] \\pounds_XY = [X,Y].$$\n\nHowever, Cartan's identity doesn't deal with the behaviour of vectors at all - it deals with the behaviour of $p$-forms. Look, suppose that you have some $p$-form $\\alpha$ and $X\\in\\mathfrak{X}(\\mathcal{M})$. Then Cartan's identity states that\n\n$$\\pounds_X\\alpha = \\{d,\\iota_X\\}\\alpha.$$\n\nIn order to prove this identity you need to have explicit knowledge of three things: the Lie derivative, the interior product, and the exterior derivative. Suppose that we have a metric on our manifold with an associated covariant derivative operator $\\nabla$. Then recall that, in component notation, the Lie derivative of $\\alpha$ along the flow generated by $X$ is\n\n$$(\\pounds_X\\alpha)_{i_1i_2\\ldots i_p} = X^m\\nabla_m \\alpha_{i_1i_2\\ldots i_p} +\\sum_{\\mu=1}^p\\alpha_{i_1\\ldots i_{\\mu-1}mi_{\\mu+1}\\ldots i_p}\\nabla_{i_\\mu}X^m$$\n\n(This is a standard identity for the Lie derivative - you should memorise not only this but also the more general expressions for the Lie derivative of a tensor of type $(r,s)$ along a vector field, and a weight-$w$ tensor density of type $(r,s)$ along a vector field.) The above equation gives you the left-hand side of Cartan's identity. Now focus on the right-hand side. Recall that the interior product has components\n\n$$(\\iota_X\\alpha)_{i_2\\ldots i_p} = \\frac{1}{p}X^{i_1}\\alpha_{i_1i_2\\ldots i_p}.$$\n\nIf you take the exterior derivative of this you should then have an expression for the components of $d(\\iota_X\\alpha)$. Now reverse the process so that you obtain an expression for the components of $\\iota_X(d\\alpha)$. Add them together to show that\n\n$$(d(\\iota_X\\alpha))_{i_1\\ldots i_p} + (\\iota_X(d\\alpha))_{i_1\\ldots i_p} = (\\pounds_X\\alpha)_{i_1\\ldots i_p}.$$\n\nYou've then proved that Cartan's identity holds for a general $p$-form.","date":"2017-05-30 01:38:02","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8453125953674316, \"perplexity\": 171.07705555033687}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2017-22\/segments\/1495463613738.67\/warc\/CC-MAIN-20170530011338-20170530031338-00342.warc.gz\"}"} | null | null |
namespace BrowserAutomationStudioFramework
{
ResourceModelScript::ResourceModelScript(QObject *parent) :
ResourceModelAbstract(parent)
{
dont_give_up = false;
}
int ResourceModelScript::GetSuccessAttempts()
{
return success;
}
void ResourceModelScript::SetSuccessAttempts(int val)
{
success = val;
}
int ResourceModelScript::GetFailAttempts()
{
return fails;
}
void ResourceModelScript::SetFailAttempts(int val)
{
fails = val;
}
int ResourceModelScript::GetNumberSimultaneousUse()
{
return number_simultaneous_use;
}
void ResourceModelScript::SetNumberSimultaneousUse(int val)
{
number_simultaneous_use = val;
}
int ResourceModelScript::GetIntervalBetweenUsage()
{
return interval_between_usage;
}
void ResourceModelScript::SetIntervalBetweenUsage(int val)
{
interval_between_usage = val;
}
QString ResourceModelScript::GetTypeId()
{
return QString("LinesFromScript");
}
bool ResourceModelScript::GetGreedy()
{
return greedy;
}
void ResourceModelScript::SetGreedy(bool Greedy)
{
this->greedy = Greedy;
}
bool ResourceModelScript::GetDontGiveUp()
{
return dont_give_up;
}
void ResourceModelScript::SetDontGiveUp(bool dont_give_up)
{
this->dont_give_up = dont_give_up;
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 6,125 |
require.config({
//baseUrl: ''
paths: {
'knockout': '../bower_components/knockout.js/knockout',
'knockout.validation': '../bower_components/knockout-validation/dist/knockout.validation',
'respond': '../bower_components/respond/dest/respond.min',
'sammy': '../bower_components/sammy/lib/sammy',
'bootstrap': '../bower_components/bootstrap/dist/js/bootstrap',
'jquery': '../bower_components/jquery/jquery',
'jquery.validation': '../bower_components/jquery.validation/dist/jquery.validate',
'jquery.unobtrusive.validation': '../bower_components/Microsoft.jQuery.Unobtrusive.Validation/jquery.validate.unobtrusive',
'jquery.unobtrusive.ajax': '../bower_components/Microsoft.jQuery.Unobtrusive.Ajax/jquery.unobtrusive-ajax',
'requirejs': '../bower_components/requirejs/require'
},
shim: {
'bootstrap': ['jquery'],
'respond': ['bootstrap'],
'jquery.validation': [],
'jquery.unobtrusive.validation': ['jquery', 'jquery.validation'],
'jquery.unobtrusive.ajax': ['jquery'],
'knockout.validation': ['knockout']
}
}) | {
"redpajama_set_name": "RedPajamaGithub"
} | 9,489 |
Q: setting an array from simple_xml data I am trying to create an array of titles from an xml feed using this code:
$url = 'https://indiegamestand.com/store/salefeed.php';
$xml = simplexml_load_string(file_get_contents($url));
$on_sale = array();
foreach ($xml->channel->item as $game)
{
echo $game->{'title'} . "\n";
$on_sale[] = $game->{'title'};
}
print_r($on_sale);
The echo $game->{'title'} . "\n"; returns the correct title, but when setting the title to the array i get spammed with this:
Array
(
[0] => SimpleXMLElement Object
(
[0] => SimpleXMLElement Object
(
)
)
[1] => SimpleXMLElement Object
(
[0] => SimpleXMLElement Object
(
)
)
[2] => SimpleXMLElement Object
(
[0] => SimpleXMLElement Object
(
)
)
Am I missing something when setting this array?
A: Use this:
$on_sale[] = $game->{'title'}->__toString();
or even better (in my opinion):
$on_sale[] = (string) $game->{'title'};
PHP doesn't know that you want the string value when you add the object to the array, so __toString() doesn't get called automatically like it does in the echo call. When you cast the object to string, __toString() is called automatically.
FYI: You don't really need the curly braces either, this works fine for me:
$on_sale[] = (string) $game->title;
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 1,318 |
{"url":"https:\/\/api-project-1022638073839.appspot.com\/questions\/how-do-you-find-the-derivative-of-cos-2-x-3-1","text":"# How do you find the derivative of cos^2(x^3)?\n\n$- 2 \\cos \\left({x}^{3}\\right) \\cdot \\sin \\left({x}^{3}\\right) \\cdot 3 {x}^{2}$","date":"2021-04-11 07:09:48","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 1, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.505025327205658, \"perplexity\": 713.7329687097074}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-17\/segments\/1618038061562.11\/warc\/CC-MAIN-20210411055903-20210411085903-00546.warc.gz\"}"} | null | null |
{"url":"https:\/\/www.clutchprep.com\/physics\/practice-problems\/154234\/the-kinetic-energy-of-an-object-is-quadrupled-its-momentum-will-change-by-what-f","text":"Intro to Momentum Video Lessons\n\nConcept\n\n# Problem: The kinetic energy of an object is quadrupled. Its momentum will change by what factor?a) 0b) 2c) 8d) 4e) none of the above\n\n###### FREE Expert Solution\n\nKinetic energy:\n\n$\\overline{){\\mathbf{K}}{\\mathbf{=}}\\frac{\\mathbf{1}}{\\mathbf{2}}{{\\mathbf{mv}}}^{{\\mathbf{2}}}}$\n\nMomentum:\n\n$\\overline{){\\mathbit{p}}{\\mathbf{=}}{\\mathbit{m}}{\\mathbit{v}}}$\n\nK in terms of p:\n\n$\\overline{){\\mathbf{K}}{\\mathbf{=}}\\frac{{\\mathbf{p}}^{\\mathbf{2}}}{\\mathbf{2}\\mathbf{m}}}$\n\nWe can prove the expression for K in terms of p by substituting v = p\/m\n\n79% (165 ratings)\n###### Problem Details\n\nThe kinetic energy of an object is quadrupled. Its momentum will change by what factor?\n\na) 0\nb) 2\nc) 8\nd) 4\ne) none of the above","date":"2021-09-21 02:12:29","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 3, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.29820021986961365, \"perplexity\": 1593.635007046987}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.3, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-39\/segments\/1631780057131.88\/warc\/CC-MAIN-20210921011047-20210921041047-00437.warc.gz\"}"} | null | null |
Q: what jquery isArrayLike checks nodeType There is a function named 'isArrayLike' in jquery which used by many functions for example $.each
function isArraylike( obj ) {
var length = obj.length,
type = jQuery.type( obj );
if ( type === "function" || jQuery.isWindow( obj ) ) {
return false;
}
if ( obj.nodeType === 1 && length ) {
return true;
}
return type === "array" || length === 0 ||
typeof length === "number" && length > 0 && ( length - 1 ) in obj;
}
I know it is used to see whether it is like an array,but i do not know why the second if.
It checks nodeType to make sure it's an element and why length?Is an element have length property?
Thx
A: If you find undocumented code that doesn't make sense, one thing you can do is look at why someone wrote it in the first place. Here's the commit that adds the nodeType check:
https://github.com/jquery/jquery/commit/3c7f2af81d877b24a5e5b6488e78621fcf96b265
Judging by the test that they added with it, it's to support form elements, which it will treat as an array of controls.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 315 |
Wet Distillers Grains are processed corn mash that has been dried to approximately 50% moisture. They have a shelf life of approximately 10 days and are sold to nearby markets.
For more information on Wet Distillers Grains products, please contact our merchandising team today. | {
"redpajama_set_name": "RedPajamaC4"
} | 5,331 |
Q: How to pass an object from one case statement to another I'm trying to get the "player" object from case "A" to to case "G" but of course I can't do that because the object was created in case A and I have to keep it there. How can I move it down there while keeping the same instance? Beginner to java. I want to be able to edit a player in the array by pressing "G"
package hockeyplayer;
import java.util.Scanner;
public class HockeyMain {
private static String choice;
private static HockeyPlayer[] players = new HockeyPlayer[12];
private static final String MENU = "Hockey Tracker\n"+
"A-Add Player\n"+
"G-Add game details\n"+
"S-Show players\n"+
"X-Quit\n";
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
do{
System.out.println(MENU);
choice = input.nextLine();
switch(choice){
case "A":
HockeyPlayer player = new HockeyPlayer();
players[player.getPlayerNumber()-1] = player;
break;
case "G":
break;
case "S":
break;
case "X":
}
}while(!choice.equals("X"));
}
}
package hockeyplayer;
import java.util.Scanner;
public class HockeyPlayer {
private String[] opponent = new String[10];
private int[] goalsScored = new int[10];
private int[] gameNumber = new int[10];
private String name;
private int playerNumber;
public HockeyPlayer() {
Scanner input = new Scanner(System.in);
System.out.println("What is the name of the player?");
name = input.nextLine();
System.out.println("What is the player's number?");
playerNumber = input.nextInt();
input.nextLine();
}
public String[] getOpponent() {
return opponent;
}
public void setOpponent(String[] opponent) {
this.opponent = opponent;
}
public int[] getGoalsScored() {
return goalsScored;
}
public void setGoalsScored(int[] goalsScored) {
this.goalsScored = goalsScored;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public int getPlayerNumber() {
return playerNumber;
}
public void setPlayerNumber(int playerNumber) {
this.playerNumber = playerNumber;
}
public void addGameDetails(){
Scanner input = new Scanner(System.in);
System.out.println("What game number was it?");
gameNumber[0] = input.nextInt();
input.nextLine();
System.out.println("Who were the opponents?");
opponent[0] = input.nextLine();
System.out.println("How many goals did the player score?");
goalsScored[0] = input.nextInt();
}
}
A: declare a HockeyPlayer instance outside of switch statement like this:
HockeyPlayer player = null;
do{
System.out.println(MENU);
choice = input.nextLine();
switch(choice){
case "A":
player = new HockeyPlayer();
players[player.getPlayerNumber()-1] = player;
break;
case "G":
player.addGameDetails(); //invoke the method here for case G
break;
case "S":
break;
case "X":
}
}while(!choice.equals("X"));
}
The idea is that HockeyPlayer object should be accessible to all the switch statements so you need to declare it somewhere where it is accessible.
A: move the declaration of the player variable outside of the do-while, still instantiating in the case "A":
HockeyPlayer player = null;
do {
System.out.println(MENU);
choice = input.nextLine();
switch(choice){
case "A":
player = new HockeyPlayer();
players[player.getPlayerNumber()-1] = player;
break;
case "G":
// you can use player here (assuming
break;
case "S":
break;
case "X":
}
} while(!choice.equals("X"));
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 1,261 |
\section{Introduction}
The transit method of exoplanet discovery has produced a small, but well constrained, sample of exoplanets that are unambiguously solid in terms of interior bulk composition. We call solid planets the ones that possess no H and He envelopes and/or atmospheres, i.e. their bulk radius is determined by elements (and their mineral phases) heavier than H and He. Such solid exoplanets are Kepler-10b~\citep[][]{Batalha:2011}, CoRoT-7b~\citep[][]{Queloz:2009, Hatzes:2011}, Kepler-36b~\citep[][]{Carter:2012} as well as - most likely: Kepler-20b,e,f; Kepler-18b, and 55 Cnc e in which the solid material could include high-pressure water ice (see references to Fig.~\ref{MRplot} in~\ref{4.2}).
There is an increased interest to compare their observed parameters to current models of interior planetary structure. The models, and their use of approximations and EOS, have evolved since 2005~\citep[][]{Valencia:2006, Fortney:2007, Grasset:2009, Seager:2007}, mainly because the more massive solid exoplanets (called Super-Earths) have interior pressures that are far in excess of Earth's model, bringing about corresponding gaps in our knowledge of mineral phases and their EOS~\citep[see a recent review by][]{Seager:2010}. Here we compute a new grid of models in order to aid current comparisons to observed exoplanets on the mass-radius diagram. As in previous such grids, we assume the main constituents inside the planets to be differentiated and model them as layers in one dimension.
The first part of this paper aims to solve the $2$-layer exoplanet model. The $2$-layer model reveals the underlying physics of planetary interior more intuitively, for which we consider 3 scenarios: Fe-MgSiO${}_{3}$ or Fe-H${}_{2}$O or MgSiO${}_{3}$-H${}_{2}$O planet.
The current observations generally measure the radius of an exoplanet through transits and the mass through Doppler shift measurement of the host star. For each assumption of core and mantle compositions, given the mass and radius input, the $2$-layer exoplanet model can be solved uniquely. It is a unique solution of radial dependence of interior pressure and density. As a result, all the characteristic physical quantities, such as the pressure at the center (p0), the pressure at core-mantle boundary (p1), the core mass fraction (CMF), and the core radius fraction (CRF) naturally fall out from this model. These quantities can be quickly determined by invoking the mass-radius contours.
The next part of this paper (Fig.~\ref{MRplot}) compares some known exoplanets to the mass-radius curves of 6 different $2$-layer exoplanet models: pure-Fe, $50\%$-Fe \& $50\%$-MgSiO${}_{3}$, pure MgSiO${}_{3}$, $50\%$-MgSiO${}_{3}$ \& $50\%$-H${}_{2}$O, $25\%$-MgSiO${}_{3}$ \& $75\%$-H${}_{2}$O, and pure H${}_{2}$O. These percentages are in mass fractions. The data of these six curves are available in Table~\ref{Table1}.
Up to now, a standard assumption has been that the planet interior is fully differentiated into layers: all the Fe is in the core and all the MgSiO${}_{3}$ is in the mantle. In section~\ref{nondiff}, we will change this assumption and discuss how the presence of Fe in the mantle affects the mass-radius relation.
The final part of this paper calculates the $3$-layer differentiated exoplanet model. Given the mass and radius input, the solution for the $3$-layer model is non-unique (degenerate), thus a curve on the ternary diagram is needed to represent the set of all solutions (see Fig.~\ref{ternary}). This curve can be obtained by solving differential equations with iterative methods. The ensemble of solutions is tabulated (Table~\ref{Table3}), from which users may interpolate to determine planet composition in $3$-layer model. A dynamic and interactive tool to characterize and illustrate the interior structure of exoplanets built upon Table~\ref{Table3} and other models in this paper is available on the website \url{http://www.cfa.harvard.edu/~lzeng}.
The methods described in this paper can be used to fast characterize the interior structure of exoplanets.
\section{Method}
Spherical symmetry is assumed in all the models. The interior of a planet is assumed to be fully differentiated into layers in the first part of the paper.
The $2$-layer model consists of a core and a mantle. The $3$-layer model consists of a core, a mantle and another layer on top of the mantle. The interior structure is determined by solving the following two differential equations:
\begin{equation}
\frac{dr}{dm}=\frac{1}{4 \pi \rho r^2}
\label{dfeq1}
\end{equation}
\begin{equation}
\frac{dp}{dm}=-\frac{G m}{4 \pi r^4}
\label{dfeq2}
\end{equation}
The two equations are similar to the ones in~\cite{Zeng_Seager:2008}. However, contrary to the common choice of radius r as the independent variable, the interior mass m is chosen, which is the total mass inside shell radius r, as the independent variable. So the solution is given as r(m) (interior radius r as a dependent function of interior mass m), p(m) (pressure as a dependent function of interior mass m), and $\rho$(m) (density as a dependent function of interior mass m).
The two differential equations are solved with the EOS of the constituent materials as inputs:
\begin{equation}
\rho=\rho(p,T)
\label{EOS}
\end{equation}
The EOS is a material property, which describes the material's density as a function of pressure and temperature. The EOS could be obtained both theoretically and experimentally.
Theoretically, the EOS could be calculated by Quantum-Mechanical Molecular Dynamics Ab Initio Simulation such as the VASP (Vienna Ab initio Simulation Package)~\citep[][]{Kresse:1993, Kresse:1994, Kresse:1996, French:2009}.
Experimentally, the EOS could be determined by high-pressure compression experiment such as the DAC (Diamond Anvil Cell) experiment, or shock wave experiment like the implosion experiment by the Sandia Z-machine~\citep{Yu:2011}.
The temperature effect on density is secondary compared to the pressure effect~\citep{Valencia:2006, Valencia:2007a}.
Therefore, we can safely ignore the temperature dependence of those higher density materials (Fe and MgSiO${}_{3}$) for which the temperature effect is weaker, or we can implicitly include a pre-assumed pressure-temperature (p-T) relation (for H${}_{2}$O it is the melting curve) so as to reduce the EOS to a simpler single-variable form:
\begin{equation}
\rho=\rho(p)
\label{EOS2}
\end{equation}
To solve the set of equations mentioned above, appropriate boundaries conditions are given as:
\begin{itemize}
\item{p0:} the pressure at the center of the planet
\item{p1:} the pressure at the first layer interface (the core-mantle boundary)
\item {p2:} the pressure at the second layer interface (only needed for $3$-layer model)
\item {p${}_{surface}$}: the pressure at the surface of the planet (set to 1 bar ($10^5 Pa$))
\end{itemize}
\section{EOS of Fe, MgSiO${}_{3}$ and H${}_{2}$O}
The 3 layers that we consider for the planet interior are Fe, MgSiO${}_{3}$ and H${}_{2}$O.
Their detailed EOS are described as follows:
\subsection{Fe}
We model the core of a solid exoplanet after the Earth's iron core, except that in our model we ignore the presence of all other elements such as Nickel (Ni) and light elements such as Sulfur (S) and Oxygen (O) in the core.
As pointed out by~\citet{Valencia:2010}, above $100$ GPa, the iron is mostly in the hexagonal closed packed $\epsilon$ phase. Therefore, we use the Fe-EOS by~\citet{Seager:2007}. Below $2.09*10^{4}$ GPa, it is a Vinet~\citep{Vinet:1987, Vinet:1989} formula fit to the experimental data of $\epsilon$-iron at p$\leq330$ GPa by~\citet{Anderson:2001}. Above $2.09*10^{4}$ GPa, it makes smooth transition to the Thomas-Fermi-Dirac (TFD) EOS~\citep{Seager:2007}. A smooth transition is assumed because there is no experimental data available in this ultrahigh-pressure regime.
The central pressure could reach $250$ TPa (terapascal, i.e., $10^{12}$ Pa) in the most massive planet considered in this paper. However, the EOS of Fe above $400$ GPa is beyond the current reach of experiment and thus largely unknown. Therefore, our best approximation here is to extend the currently available $\epsilon$-iron EOS to higher pressures and connect it to the TFD EOS.
The EOS of Fe is shown in Fig.~\ref{eosplot} as the upper curve (red curve).
\begin{figure}[htbp]
\begin{center}
\includegraphics[scale=0.9]{f1.eps}
\end{center}
\caption{$p-\rho$ EOS of Fe ($\epsilon$-phase of iron, red curve), MgSiO${}_{3}$ (perovskite phase, post-perovskite phase and its high-pressure derivatives, orange curve), and H${}_{2}$O (Ice Ih, Ice III, Ice V, Ice VI, Ice VII, Ice X, and superionic phase along its melting curve (solid-liquid phase boundary))}
\label{eosplot}
\end{figure}
\subsection{MgSiO${}_{3}$}
We model the silicate layer of a solid exoplanet using the Earth's mantle as a proxy. The FeO-free Earth's mantle with Mg/Si=1.07 would consist of mainly enstatite (MgSiO${}_{3}$) or its high-pressure polymorphs and, depending upon pressure, small amounts of either forsterite and its high-pressure polymorphs (Mg${}_{2}$SiO${}_{4}$) or periclase (MgO)~\citep[e.g.][]{Bina:2003}.
The olivine polymorphs as well as lower-pressure enstatite and majorite (MgSiO${}_{3}$ with garnet structure), are not stable above 27 GPa. At higher pressures, the system would consist of MgSiO${}_{3}$-perovskite (pv) and periclase or their higher-pressure polymorphs~\citep{Stixrude:2011, Bina:2003}. Given the high pressures at the H${}_{2}$O-silicate boundary usually exceeding 27 GPa, we can safely ignore olivine and lower-pressure pyroxene polymorphs. For the sake of simplicity, we also ignore periclase, which would contribute only 7 at.$\%$ to the silicate mantle mineralogical composition. There are also small amounts of other elements such as Aluminum (Al), Calcium (Ca), and Sodium (Na) present in Earth's mantle~\citep{Sotin:2007}. For simplicity, we neglect them and thus the phases containing them are not included in our model. We also do not consider SiC because carbon-rich planets might form under very rare circumstances, and are probably not common.
Some fraction of Fe can be incorporated into the minerals of the silicate mantle which could then have the general formula as (Mg,Fe)SiO${}_{3}$. For now, we simply assume all the Fe is in the core and all the Mg is in the mantle in the form of MgSiO${}_{3}$-perovskite and/or its high-pressure polymorphs. So the planet is fully differentiated. In a later section, we will discuss how the addition of Fe to the mantle can affect mass-radius relation and compare the differences between differentiated and undifferentiated as well as reduced and oxidized planets in Section~\ref{nondiff}.
We first consider the perovskite (pv) and post-perovskite (ppv) phases of pure MgSiO${}_{3}$. MgSiO${}_{3}$ perovskite (pv) is believed to be the major constituent of the Earth mantle. It makes transition into the post-perovskite (ppv) phase at roughly $120$ GPa and 2500 K (corresponding to a depth of 2600 kilometers in Earth)~\citep{Hirose:2010}. The ppv phase was discovered experimentally in 2004~\citep[by][]{Murakami:2004} and was also theoretically predicted in the same year~\citep[by][]{Oganov:2004}. The ppv is about 1.5$\%$ denser than the pv phase~\citep{Caracas:2008, Hirose:2010}. This 1.5$\%$ density jump resulting from the pv-to-ppv phase transition can be clearly seen as the first density jump of the MgSiO${}_{3}$ EOS curve shown in Fig.~\ref{eosplot}. Both the MgSiO${}_{3}$ pv EOS and MgSiO${}_{3}$ ppv EOS are taken from~\citet{Caracas:2008}. The transition pressure is determined to be $122$ GPa for pure MgSiO${}_{3}$ according to~\citet{Spera:2006}.
Beyond $0.90$ TPa, MgSiO${}_{3}$ ppv undergoes a two-stage dissociation process predicted from the first-principle calculations by~\citet{Umemoto:2011}. MgSiO${}_{3}$ ppv first dissociates into CsCl-type MgO and P2${}_{1}$c-type MgSi${}_{2}$O${}_{5}$ at the pressure of $0.90$ TPa and later into CsCl-type MgO and Fe${}_{2}$P-type SiO${}_{2}$ at pressures higher than $2.10$ TPa. The EOS of CsCl-type MgO, P2${}_{1}$c-type MgSi${}_{2}$O${}_{5}$, and Fe${}_{2}$P-type SiO${}_{2}$ are adopted from~\citet{Umemoto:2011, Wu:2011}. Therefore, there are two density jumps at the dissociation pressures of $0.90$ TPa and $2.10$ TPa. The first one can be seen clearly in Fig.~\ref{eosplot}. The second one cannot be seen in Fig.~\ref{eosplot} since it is too small, but it surely exists.
Since~\citet{Umemoto:2011}'s EOS calculation is until $4.90$ TPa, beyond $4.90$ TPa, a modified version of the EOS by~\citet{Seager:2007} is used to smoothly connect to the TFD EOS. TFD EOS assumes electrons in a slowly varying potential with a density-dependent correlation energy term that describes the interactions among electrons. It is therefore insensitive to any crystal structure or arrangements of atoms and it becomes asymptotically more accurate at higher pressure. Thus, the TFD EOS of MgSiO${}_{3}$ would be no different from the TFD EOS of MgO plus SiO${}_{2}$ as long as the types and numbers of atoms in the calculation are the same. So it is safe to use the TFD EOS of MgSiO${}_{3}$ as an approximation of the EOS of MgO and SiO${}_{2}$ mixture beyond $4.90$ TPa here.
\citet{Seager:2007}'s EOS is calculated from the method of~\citet{Salpeter:1967} above $1.35*10^{4}$ GPa. Below $1.35*10^{4}$ GPa, it is a smooth connection to TFD EOS from the fourth-order Birch-Murbaghan Equation of State (BME)~\citep[see][]{Birch:1947, Poirier:2000} fit to the parameters of MgSiO${}_{3}$ pv obtained by \textit{Ab initio} lattice dynamics simulation of~\citet{Karki:2000}.
~\citeauthor{Seager:2007}'s EOS is slightly modified to avoid any artificial density jump when connected with~\citeauthor{Umemoto:2011}'s EOS at $4.90$ TPa. At $4.90$ TPa, the ratio of the density $\rho$ between~\citeauthor{Umemoto:2011}'s EOS and~\citeauthor{Seager:2007}'s EOS is $1.04437$. This ratio is multiplied to the original~\citeauthor{Seager:2007}'s EOS density $\rho$ to produce the actual EOS used in our calculation for p$>4.90$ TPa.
A smooth transition is assumed because no experimental data is available in this ultrahigh-pressure regime. This assumption does not affect our low or medium mass planet models, since only the most massive planets in our model could reach this ultrahigh pressure in their MgSiO${}_{3}$ part.
The EOS of MgSiO${}_{3}$ is shown in Fig.~\ref{eosplot} as the middle curve (orange curve).
\subsection{H${}_{2}$O}
The top layer of a planet could consist of various phases of H${}_{2}$O.
Since H${}_{2}$O has a complex phase diagram, and it also has a stronger temperature dependence, thus the temperature effect cannot be ignored. Instead, we follow the solid phases along the melting curve (solid-liquid phase boundary on the p-T plot by~\citet{Chaplin:2012}).
Along the melting curve, the H${}_{2}$O undergoes several phase transitions. Initially, it is Ice Ih at low pressure, then subsequently transforms into Ice III, Ice V, Ice VI, Ice VII, Ice X, and superionic phase~\citep{Chaplin:2012, Choukroun:2007, Dunaeva:2010, French:2009}.
\subsubsection{Chaplin's EOS}
The solid form of water has very complex phases in the low-pressure and low-temperature regime. These phases are well determined by experiments. Here we adopt the Chaplin's EOS for Ice Ih, Ice III, Ice V, and Ice VI below $2.216$ GPa~\citep[see][]{Chaplin:2012, Choukroun:2007, Dunaeva:2010}. Along the melting curve (the solid-liquid boundary on the p-T diagram), the solid form of water first exists as Ice Ih (Hexagonal Ice) from ambient pressure up to $209.5$ MPa~\citep{Choukroun:2007}. At the triple point of $209.5$ MPa and $251.15$ K~\citep{Choukroun:2007, Robinson:1996}, it transforms into Ice III (Ice-three), whose unit cell forms tetragonal crystals. Ice III exists up to $355.0$ MPa and transforms into a higher-pressure form Ice V (Ice-five) at the triple point of $355.0$ MPa and $256.43$ K~\citep{Choukroun:2007}. Ice V's unit cell forms monoclinic crystals. At the triple point of $618.4$ MPa and $272.73$ K~\citep{Choukroun:2007}, Ice V transforms into yet another higher-pressure form Ice VI (Ice-six). Ice VI's unit cell forms tetragonal crystals. A single molecule in Ice VI crystal is bonded to four other water molecules. Then at the triple point of $2.216$ GPa and $355$ K~\citep{IAPWS:2011}, Ice VI transforms into Ice VII (Ice-seven). Ice VII has a cubic crystal structure. Ice VII eventually transforms into Ice X (Ice-ten) at the triple point of $47$ GPa and $1000$ K~\citep{Goncharov:2005}. In Ice X, the protons are equally spaced and bonded between the oxygen atoms, where the oxygen atoms are in a body-centered cubic lattice~\citep{Hirsch:1984}. The EOS of Ice X and Ice VII are very similar. For Ice VII (above $2.216$ GPa), we switch to the~\citep*{Frank:2004}'s EOS (FFH2004).
\subsubsection{FFH2004's EOS}
We adopt FFH2004's EOS of Ice VII for $2.216$ GPa$\leq$p$\leq37.4$ GPa. This EOS is obtained using the Mao-Bell type diamond anvil cell with an external Mo-wire resistance heater. Gold and water are put into the sample chamber and compressed. The diffraction pattern of both H${}_{2}$O and gold are measured by the Energy-Dispersive X-ray Diffraction (EDXD) technique at the Brookhaven National Synchrotron Light Source. The gold here is used as an internal pressure calibrant. The disappearance of the diffraction pattern of the crystal Ice VII is used as the indicator for the solid-liquid boundary (melting curve). The melting curve for Ice VII is determined accurately from $3$ GPa to $60$ GPa and fit to a Simon equation. The molar density ($\rho$ in $mol/cm^3$) of Ice VII as a function of pressure (p in GPa) is given by the following formula in FFH2004:
\begin{equation}
\rho(mol/cm^3)=0.0805+0.0229*(1-exp^{-0.0743*p})+0.1573*(1-exp^{-0.0061*p})
\label{Frankeq}
\end{equation}
We use Eq.~\ref{Frankeq} to calculate $\rho$ from $2.216$ GPa up to $37.4$ GPa. The upper limit $37.4$ GPa is determined by the intersection between FFH2004's EOS and~\citep*{French:2009}'s EOS (FMNR2009).
\subsubsection{FMNR2009's EOS}
FMNR2009's EOS is used for Ice VII, Ice X and superionic phase of H${}_{2}$O for $37.4$ GPa$\leq$p$\leq8.893$ TPa. This EOS is determined by Quantum Molecular Dynamics Simulations using the Vienna $Ab~Initio$ Simulation Package (VASP). The simulation is based on finite temperature density-functional theory (DFT) for the electronic structure and treating the ions as classical particles. Most of~\citeauthor{French:2009}'s simulations consider $54$ H${}_{2}$O molecules in a canonical ensemble, with the standard VASP PAW potentials, the $900$ eV plane-wave cutoff, and the evaluation of the electronic states at the $\Gamma$ point considered, for the $3$ independent variables: temperature (T), volume (V), and particle number (N). The simulation results are the thermal EOS p(T,V,N), and the caloric EOS U(T,V,N). The data are tabulated in FMNR2009.
In order to approximate density $\rho$ along the melting curve, from Table V in FMNR2009, we take $1$ data point from Ice VII phase at $1000$ K and $2.5 g/cm^3$, $4$ data points from Ice X phase at $2000$ K and $3.25 g/cm^3$ to $4.00 g/cm^3$, and the rest of the data points from the superionic phase at $4000$ K and $5.00 g/cm^3$ up to $15 g/cm^3$. Since the temperature effect on density becomes smaller towards higher pressure, all the isothermal pressure-density curves converge on the isentropic pressure-density curves as well as the pressure-density curve along the melting curve.
The FMNR2009's EOS has been confirmed experimentally by Thomas Mattson et al. at the Sandia National Laboratories. At $8.893$ TPa, FMNR2009's EOS is switched to the TFD EOS in~\citealt*{Seager:2007} (SKHMM2007).
\subsubsection{SKHMM2007's EOS}
At ultrahigh pressure, the effect of electron-electron interaction can be safely ignored and electrons can be treated as a gas of non-interacting particles that obey the Pauli exclusion principle subject to the Coulomb field of the nuclei. Assuming the Coulomb potential is spatially slowly varying throughout the electron gas that the electronic wave functions can be approximated locally as plane waves, the so-called TFD solution could be derived so that the Pauli exclusion pressure balances out the Coulomb forces~\citep{Eliezer:2002,Macfarlane:1984}.
In SKHMM2007, a modified TFD by~\citet{Salpeter:1967} is used. It is modified in the sense that the authors have added in a density-dependent correlation energy term which characterizes electron interaction effects.
Here,~\citeauthor{Seager:2007}'s EOS is slightly modified to connect to the FMNR2009's EOS. At $8.893$ TPa, the ratio of the density $\rho$ between the FMNR2009's EOS and ~\citeauthor{Seager:2007}'s EOS is $1.04464$. This ratio is multiplied to the original ~\citeauthor{Seager:2007}'s EOS density $\rho$ to produce the actual EOS used in our calculation for p$>8.893$ TPa. Only the most massive planets in our model could reach this pressure in the H${}_{2}$O-layer, so this choice of EOS for p$>8.893$ TPa has small effect on the overall mass-radius relation to be discussed in the next section.
The EOS of H${}_{2}$O is shown in Fig.~\ref{eosplot} as the lower curve (blue curve).
\section{Result}
\subsection{Mass-Radius Contours}
Given mass and radius input, various sets of mass-radius contours can be used to quickly determine the characteristic interior structure quantities of a $2$-layer planet including its p0 (central pressure), p1/p0 (ratio of core-mantle boundary pressure over central pressure), CMF (core mass fraction), and CRF (core radius fraction).
The $2$-layer model is uniquely solved and represented as a point on the mass-radius diagram given any pair of two parameters from the following list: $M$ (mass), $R$ (radius), p0, p1/p0, CMF, CRF, etc. The contours of constant $M$ or $R$ are trivial on the mass-radius diagram, which are simply vertical or horizontal lines. The contours of constant p0, p1/p0, CMF, or CRF are more useful.
Within a pair of parameters, fixing one and continuously varying the other, the point on the mass-radius diagram moves to form a curve. Multiple values of the fixed parameter give multiple parallel curves forming a set of contours. The other set of contours can be obtained by exchanging the fixed parameter for the varying parameter.
The two sets of contours crisscross each other to form a mesh, which is a natural coordinate system (see Fig.~\ref{contourplots}) of this pair of parameters, superimposed onto the existing Cartesian $(M,R)$ coordinates of the mass-radius diagram. This mesh can be used to determine the two parameters given the mass and radius input or vice versa.
\subsubsection{Fe-MgSiO${}_{3}$ planet}
As an example, the mesh of p0 with p1/p0 for Fe-core MgSiO${}_{3}$-mantle planet is illustrated in the first subplot (upper left corner) of Fig.~\ref{contourplots}. The mesh is formed by p0-contours and p1/p0-contours crisscrossing each other. The more vertical set of curves represents the p0-contours. The ratio of adjacent p0-contours is $10^{0.1}$ (see Table~\ref{Table1}). The more horizontal set of curves represents the p1/p0-contours. From bottom up, the p1/p0 values vary from 0 to 1 with step size 0.1.
\begin{figure}[htbp]
\begin{center}
\includegraphics[scale=0.3]{f2.eps}
\end{center}
\caption{Mass-Radius contours of $2$-layer planet. 1st row: Fe-MgSiO${}_{3}$ planet. 2nd row: MgSiO${}_{3}$-H${}_{2}$O planet. 3rd row: Fe-H${}_{2}$O planet. 1st column: contour mesh of p1/p0 with p0. 2nd column: contour mesh of CMF with p0. 3rd column: contour mesh of CRF with p0. To find out what p0 value each p0-contour corresponds to, please refer to Table~\ref{Table1}.}
\label{contourplots}
\end{figure}
Given a pair of p0 and p1 input, users may interpolate from the mesh to find the mass and radius. On the other hand, given the mass and radius of a planet, users may also interpolate from the mesh to find the corresponding p0 and p1 of a 2-layer Fe core MgSiO${}_{3}$-mantle planet.
Similarly, the contour mesh of p0 with CMF for the Fe-MgSiO${}_{3}$ planet is shown as the second subplot from the left in the first row of Fig.~\ref{contourplots}. As a reference point, for a pure-Fe planet with $p0=10^{11}$ Pa, $M=0.1254 M_{\oplus}, R=0.417 R_{\oplus}$.
The contour mesh of p0 with CRF for the Fe-MgSiO${}_{3}$ planet is shown as the third subplot from the left of the first row of Fig.~\ref{contourplots}.
\subsubsection{MgSiO${}_{3}$-H${}_{2}$O planet}
For $2$-layer MgSiO${}_{3}$-H${}_{2}$O planet, the $3$ diagrams (p0 contours pair with p1/p0 contours, CMF contours, or CRF contours) are the subplots of the second row of Fig.~\ref{contourplots}.
As a reference point, for a pure-MgSiO${}_{3}$ planet with $p0=10^{10.5}$ Pa, $M=0.122 M_{\oplus}, R=0.5396 R_{\oplus}$.
\subsubsection{Fe-H${}_{2}$O planet}
For $2$-layer Fe-H${}_{2}$O planet, the $3$ diagrams (p0 contours pair with p1/p0 contours, CMF contours, or CRF contours) are the subplots of the third row of Fig.~\ref{contourplots}.
As a reference point, for a pure-Fe planet with $p0=10^{11}$ Pa, $M=0.1254 M_{\oplus}, R=0.417 R_{\oplus}$.
\subsection{Mass-Radius Curves}\label{4.2}
For observers' interest, $6$ characteristic mass-radius curves are plotted (Fig.~\ref{MRplot}) and tabulated (Table~\ref{Table1}), representing the pure-Fe planet, half-Fe half-MgSiO${}_{3}$ planet, pure MgSiO${}_{3}$ planet, half-MgSiO${}_{3}$ half-H${}_{2}$O planet, 75$\%$ H${}_{2}$O-25$\%$ MgSiO${}_{3}$ planet, and pure H${}_{2}$O planet. These fractions are mass fractions. Fig.~\ref{MRplot} also shows some recently discovered exoplanets within the relevant mass-radius regime for comparison. These planets include Kepler-10b~\citep[][]{Batalha:2011}, Kepler-11b~\citep[][]{Lissauer:2011}, Kepler-11f~\citep[][]{Lissauer:2011}, Kepler-18b~\citep[][]{Cochran:2011}, Kepler-36b~\citep[][]{Carter:2012}, and Kepler-20b,c,d~\citep[][]{Gautier:2012}. They also include Kepler-20e ($R=0.868_{-0.096}^{+0.074} R_{\oplus}$~\citep[][]{Fressin:2011}, the mass range is determined by pure-silicate mass-radius curve and the maximum collisional stripping curve~\citep[][]{Marcus:2010}), Kepler-20f ($R=1.034_{-0.127}^{+0.100} R_{\oplus}$~\citep[][]{Fressin:2011}, the mass range is determined by $75\%$ water-ice and $25\%$ silicate mass-radius curves and the maximum collisional stripping curve~\citep[][]{Marcus:2010}), Kepler-21b ($R=1.64\pm0.04 R_{\oplus}$~\citep[][]{Howell:2012}, The upper limit for mass is $10.4 M_{\oplus}$: the 2-$\sigma$ upper limit preferred in the paper. The lower limit is $4 M_{\oplus}$, which is in between the "Earth" and "50$\%$ H${}_{2}$O-50$\%$ MgSiO${}_{3}$" model curves - the planet is very hot and is unlikely to have much water content if any at all.), Kepler-22b ($R=2.38\pm0.13 R_{\oplus}$~\citep[][]{Borucki:2012}, The 1-$\sigma$ upper limit for mass is $36 M_{\oplus}$ for an eccentric orbit (or $27 M_{\oplus}$, for circular orbit)), CoRoT-7b ($M=7.42\pm1.21M_{\oplus}$, $R=1.58\pm0.1 R_{\oplus}$~\citep[][]{Hatzes:2011, Leger:2009, Queloz:2009}), 55 Cancri e ($M=8.63\pm0.35M_{\oplus}$, $R=2.00\pm0.14 R_{\oplus}$~\citep[][]{Winn:2011}), and GJ 1214b~\citep[][]{Charbonneau:2009}.
\begin{figure}[htbp]
\begin{center}
\includegraphics[scale=0.75]{f3.eps}
\end{center}
\caption{Currently known transiting exoplanets are shown with their measured mass and radius with observation uncertainties. Earth and Venus are shown for comparison. The curves are calculated for planets composed of pure Fe, 50$\%$ Fe-50$\%$ MgSiO${}_{3}$, pure MgSiO${}_{3}$, 50$\%$ H${}_{2}$O-50$\%$ MgSiO${}_{3}$, 75$\%$ H${}_{2}$O-25$\%$ MgSiO${}_{3}$ and pure H${}_{2}$O. The red dashed curve is the maximum collisional stripping curve calculated by~\citet{Marcus:2010}.}
\label{MRplot}
\end{figure}
\begin{sidewaystable}[htbp]
\caption{Data for the 6 characteristic mass-radius curves\label{Table1}}
\centering
\scalebox{0.65}{
\begin{tabular}{c | c c | c c c c | c c | c c c c | c c c c | c c |}
\\
\hline\hline
&
\multicolumn{2}{c}{Fe} &
\multicolumn{4}{c}{$50\% Fe$-$50\% MgSiO_3$} &
\multicolumn{2}{c}{MgSiO${}_{3}$} &
\multicolumn{4}{c}{$50\% MgSiO_3$-$50\% H_2O$} &
\multicolumn{4}{c}{$25\% MgSiO_3$-$75\% H_2O$} &
\multicolumn{2}{c}{H${}_{2}$O} \\ \hline
$log_{10}(p0)$ & Mass\footnotemark[1] & Radius\footnotemark[2] & Mass & Radius & CRF\footnotemark[3] & p1/p0\footnotemark[4] & Mass & Radius & Mass & Radius & CRF & p1/p0 & Mass & Radius & CRF & p1/p0 & Mass & Radius\\ \hline
9.6 & 0.001496 & 0.09947 & 0.00177 & 0.121 & 0.6892 & 0.2931 & 0.00623 & 0.2029 & 0.008278 & 0.2963 & 0.5969 & 0.2374 & 0.01217 & 0.3616 & 0.4413 & 0.3782 & 0.04663 & 0.58 \\
9.7 & 0.002096 & 0.1112 & 0.002481 & 0.1354 & 0.6888 & 0.2925 & 0.008748 & 0.227 & 0.01156 & 0.3286 & 0.6013 & 0.2398 & 0.01699 & 0.4009 & 0.4444 & 0.3803 & 0.06174 & 0.63 \\
9.8 & 0.002931 & 0.1243 & 0.003472 & 0.1513 & 0.6882 & 0.2918 & 0.01227 & 0.254 & 0.01615 & 0.3647 & 0.6051 & 0.2413 & 0.02351 & 0.4432 & 0.4475 & 0.385 & 0.08208 & 0.6853 \\
9.9 & 0.00409 & 0.1387 & 0.004848 & 0.169 & 0.6876 & 0.2909 & 0.01717 & 0.2838 & 0.02255 & 0.4049 & 0.6085 & 0.242 & 0.03213 & 0.4869 & 0.4513 & 0.3944 & 0.1091 & 0.7455 \\
10 & 0.005694 & 0.1546 & 0.006752 & 0.1885 & 0.6868 & 0.2898 & 0.02399 & 0.317 & 0.0313 & 0.4484 & 0.6119 & 0.2446 & 0.04399 & 0.535 & 0.4554 & 0.4016 & 0.1445 & 0.8102 \\
10.1 & 0.007904 & 0.1722 & 0.00938 & 0.2101 & 0.6858 & 0.2886 & 0.03343 & 0.3536 & 0.04314 & 0.4944 & 0.6165 & 0.2495 & 0.06029 & 0.588 & 0.4592 & 0.4073 & 0.1904 & 0.879 \\
10.2 & 0.01094 & 0.1915 & 0.01299 & 0.2339 & 0.6847 & 0.2871 & 0.04644 & 0.3939 & 0.05943 & 0.5449 & 0.6209 & 0.253 & 0.08255 & 0.6461 & 0.4629 & 0.4119 & 0.2494 & 0.9515 \\
10.3 & 0.01507 & 0.2126 & 0.01792 & 0.26 & 0.6834 & 0.2855 & 0.0643 & 0.4381 & 0.0817 & 0.6003 & 0.6249 & 0.2554 & 0.1127 & 0.7093 & 0.4662 & 0.4159 & 0.3249 & 1.028 \\
10.4 & 0.0207 & 0.2356 & 0.02464 & 0.2886 & 0.6819 & 0.2836 & 0.08866 & 0.4865 & 0.112 & 0.6607 & 0.6285 & 0.2571 & 0.1533 & 0.7777 & 0.4693 & 0.4197 & 0.4206 & 1.107 \\
10.5 & 0.02829 & 0.2606 & 0.03372 & 0.3197 & 0.6801 & 0.2815 & 0.1217 & 0.5391 & 0.1528 & 0.7262 & 0.6318 & 0.2585 & 0.2074 & 0.8509 & 0.4721 & 0.4234 & 0.5416 & 1.19 \\
10.6 & 0.0385 & 0.2876 & 0.04595 & 0.3535 & 0.6782 & 0.2792 & 0.1661 & 0.596 & 0.2074 & 0.7966 & 0.6347 & 0.2595 & 0.2789 & 0.9289 & 0.4747 & 0.427 & 0.6938 & 1.276 \\
10.7 & 0.05212 & 0.3167 & 0.06231 & 0.3901 & 0.6759 & 0.2767 & 0.2255 & 0.6573 & 0.2799 & 0.8718 & 0.6373 & 0.2603 & 0.3726 & 1.011 & 0.477 & 0.4305 & 0.8866 & 1.366 \\
10.8 & 0.07021 & 0.348 & 0.08408 & 0.4296 & 0.6735 & 0.274 & 0.3042 & 0.7228 & 0.3754 & 0.9515 & 0.6396 & 0.261 & 0.4946 & 1.098 & 0.4792 & 0.4339 & 1.132 & 1.461 \\
10.9 & 0.09408 & 0.3814 & 0.1129 & 0.4721 & 0.6708 & 0.2711 & 0.4075 & 0.7925 & 0.4998 & 1.035 & 0.6416 & 0.2615 & 0.6517 & 1.188 & 0.4811 & 0.437 & 1.444 & 1.562 \\
11 & 0.1254 & 0.417 & 0.1508 & 0.5177 & 0.6679 & 0.2682 & 0.542 & 0.8661 & 0.6607 & 1.123 & 0.6434 & 0.2618 & 0.8529 & 1.282 & 0.4827 & 0.4397 & 1.841 & 1.669 \\
11.1 & 0.1663 & 0.4548 & 0.2003 & 0.5663 & 0.6648 & 0.2651 & 0.7143 & 0.9429 & 0.8653 & 1.214 & 0.6448 & 0.2615 & 1.107 & 1.38 & 0.4839 & 0.4411 & 2.346 & 1.782 \\
11.2 & 0.2193 & 0.4949 & 0.2647 & 0.618 & 0.6616 & 0.2621 & 0.927 & 1.02 & 1.117 & 1.305 & 0.6455 & 0.2598 & 1.42 & 1.477 & 0.4843 & 0.44 & 2.985 & 1.901 \\
11.3 & 0.2877 & 0.537 & 0.348 & 0.6728 & 0.6582 & 0.259 & 1.2 & 1.101 & 1.437 & 1.4 & 0.6458 & 0.2589 & 1.818 & 1.58 & 0.484 & 0.44 & 3.77 & 2.023 \\
11.4 & 0.3754 & 0.5814 & 0.455 & 0.7307 & 0.6547 & 0.2559 & 1.545 & 1.186 & 1.842 & 1.499 & 0.6458 & 0.2582 & 2.321 & 1.688 & 0.4834 & 0.4403 & 4.735 & 2.147 \\
11.5 & 0.4875 & 0.6279 & 0.5919 & 0.7916 & 0.6512 & 0.2529 & 1.981 & 1.274 & 2.351 & 1.602 & 0.6453 & 0.2574 & 2.957 & 1.802 & 0.4828 & 0.4399 & 5.909 & 2.274 \\
11.6 & 0.6298 & 0.6765 & 0.7659 & 0.8554 & 0.6476 & 0.25 & 2.525 & 1.365 & 2.988 & 1.708 & 0.6445 & 0.2565 & 3.752 & 1.92 & 0.4822 & 0.4391 & 7.325 & 2.401 \\
11.7 & 0.8096 & 0.727 & 0.986 & 0.9221 & 0.644 & 0.2473 & 3.203 & 1.458 & 3.78 & 1.818 & 0.6435 & 0.2552 & 4.732 & 2.041 & 0.4813 & 0.4383 & 9.038 & 2.529 \\
11.8 & 1.036 & 0.7796 & 1.261 & 0.9907 & 0.6407 & 0.2451 & 4.043 & 1.554 & 4.763 & 1.933 & 0.6423 & 0.2536 & 5.936 & 2.164 & 0.4803 & 0.4377 & 11.11 & 2.66 \\
11.9 & 1.319 & 0.834 & 1.606 & 1.062 & 0.6374 & 0.2429 & 5.077 & 1.653 & 5.972 & 2.05 & 0.641 & 0.252 & 7.407 & 2.289 & 0.4792 & 0.4373 & 13.55 & 2.789 \\
12 & 1.671 & 0.8902 & 2.036 & 1.136 & 0.6342 & 0.2407 & 6.297 & 1.749 & 7.392 & 2.165 & 0.6394 & 0.2488 & 9.127 & 2.411 & 0.4778 & 0.4341 & 16.42 & 2.915 \\
12.1 & 2.108 & 0.9481 & 2.568 & 1.211 & 0.631 & 0.2385 & 7.714 & 1.842 & 9.043 & 2.275 & 0.6372 & 0.2449 & 11.12 & 2.529 & 0.4755 & 0.4299 & 19.77 & 3.039 \\
12.2 & 2.648 & 1.007 & 3.226 & 1.289 & 0.6278 & 0.2363 & 9.423 & 1.935 & 11.03 & 2.386 & 0.6345 & 0.242 & 13.52 & 2.647 & 0.4727 & 0.4275 & 23.68 & 3.16 \\
12.3 & 3.31 & 1.068 & 4.032 & 1.369 & 0.6247 & 0.2342 & 11.47 & 2.029 & 13.4 & 2.498 & 0.6317 & 0.2396 & 16.37 & 2.765 & 0.4695 & 0.4265 & 28.21 & 3.278 \\
12.4 & 4.119 & 1.13 & 5.018 & 1.451 & 0.6216 & 0.2321 & 13.87 & 2.121 & 16.18 & 2.608 & 0.6285 & 0.2371 & 19.72 & 2.882 & 0.467 & 0.4248 & 33.49 & 3.393 \\
12.5 & 5.103 & 1.193 & 6.216 & 1.534 & 0.6184 & 0.23 & 16.73 & 2.213 & 19.48 & 2.717 & 0.6252 & 0.2353 & 23.68 & 2.998 & 0.4646 & 0.4237 & 39.62 & 3.506 \\
12.6 & 6.293 & 1.257 & 7.665 & 1.618 & 0.6153 & 0.228 & 20.1 & 2.304 & 23.36 & 2.825 & 0.6219 & 0.2339 & 28.31 & 3.111 & 0.4623 & 0.423 & 46.72 & 3.616 \\
12.7 & 7.727 & 1.321 & 9.39 & 1.7 & 0.6126 & 0.2267 & 24.07 & 2.394 & 27.94 & 2.933 & 0.619 & 0.2323 & 33.71 & 3.222 & 0.46 & 0.4225 & 54.92 & 3.724 \\
12.8 & 9.445 & 1.386 & 11.45 & 1.783 & 0.61 & 0.2254 & 28.68 & 2.483 & 33.24 & 3.037 & 0.6162 & 0.2306 & 39.97 & 3.33 & 0.4579 & 0.4214 & 64.22 & 3.826 \\
12.9 & 11.49 & 1.451 & 13.91 & 1.865 & 0.6074 & 0.2241 & 33.99 & 2.569 & 39.33 & 3.138 & 0.6134 & 0.2288 & 47.15 & 3.434 & 0.4556 & 0.4199 & 74.79 & 3.924 \\
13 & 13.92 & 1.515 & 16.81 & 1.947 & 0.6048 & 0.2227 & 40.05 & 2.65 & 46.26 & 3.234 & 0.6105 & 0.2266 & 55.31 & 3.534 & 0.4531 & 0.4177 & 86.85 & 4.017 \\
13.1 & 16.78 & 1.579 & 20.21 & 2.027 & 0.6024 & 0.2213 & 46.88 & 2.728 & 54.07 & 3.325 & 0.6075 & 0.2238 & 64.47 & 3.627 & 0.4503 & 0.4148 & 100.3 & 4.104 \\
13.2 & 20.14 & 1.642 & 24.2 & 2.106 & 0.5999 & 0.2198 & 54.49 & 2.799 & 62.77 & 3.409 & 0.6042 & 0.2207 & 74.65 & 3.713 & 0.4472 & 0.4114 & 115.3 & 4.183 \\
13.3 & 24.05 & 1.704 & 28.85 & 2.183 & 0.5973 & 0.2182 & 63.08 & 2.866 & 72.58 & 3.488 & 0.6006 & 0.2178 & 86.14 & 3.793 & 0.4438 & 0.4087 & 131.9 & 4.256 \\
13.4 & 28.6 & 1.765 & 34.23 & 2.258 & 0.5947 & 0.2166 & 72.75 & 2.928 & 83.62 & 3.563 & 0.5967 & 0.2152 & 99.08 & 3.87 & 0.4404 & 0.4064 & 150.3 & 4.322 \\
13.5 & 33.87 & 1.824 & 40.45 & 2.331 & 0.5921 & 0.2151 & 83.59 & 2.986 & 95.97 & 3.631 & 0.5928 & 0.2129 & 113.6 & 3.94 & 0.437 & 0.4043 & 170.8 & 4.382 \\
13.6 & 39.94 & 1.881 & 47.59 & 2.401 & 0.5895 & 0.2136 & 95.68 & 3.039 & 109.8 & 3.695 & 0.5888 & 0.2106 & 129.7 & 4.006 & 0.4338 & 0.4022 & 193.6 & 4.435 \\
13.7 & 46.92 & 1.937 & 55.76 & 2.468 & 0.5869 & 0.2123 & 109.1 & 3.088 & 125.1 & 3.754 & 0.5847 & 0.2083 & 147.6 & 4.064 & 0.4306 & 0.4001 & 218.7 & 4.483 \\
13.8 & 54.93 & 1.99 & 65.07 & 2.531 & 0.5844 & 0.2114 & 124 & 3.132 & 142 & 3.807 & 0.5807 & 0.2061 & 167.2 & 4.116 & 0.4274 & 0.3979 & 246.6 & 4.525 \\
13.9 & 64.08 & 2.042 & 75.65 & 2.59 & 0.582 & 0.2106 & 140.3 & 3.17 & 160.6 & 3.854 & 0.5768 & 0.2038 & 188.8 & 4.162 & 0.4242 & 0.3956 & 277.3 & 4.561 \\
14 & 74.51 & 2.091 & 87.65 & 2.645 & 0.5797 & 0.21 & 158.2 & 3.204 & 181 & 3.895 & 0.5728 & 0.2015 & 212.5 & 4.202 & 0.4209 & 0.3932 & 311.3 & 4.592 \\
14.1 & 86.37 & 2.138 & 101.2 & 2.697 & 0.5775 & 0.2094 & 177.8 & 3.232 & 203.3 & 3.93 & 0.5688 & 0.1992 & 238.4 & 4.236 & 0.4175 & 0.3908 & 348.7 & 4.618 \\
14.2 & 99.8 & 2.183 & 116.5 & 2.745 & 0.5755 & 0.2089 & 199.2 & 3.255 & 227.7 & 3.959 & 0.5648 & 0.1971 & 266.8 & 4.265 & 0.4142 & 0.3884 & 390.1 & 4.639 \\
14.3 & 115 & 2.226 & 133.8 & 2.789 & 0.5735 & 0.2085 & 222.5 & 3.274 & 254.2 & 3.983 & 0.5607 & 0.1951 & 297.9 & 4.288 & 0.4109 & 0.3861 & 435.9 & 4.656 \\
14.4 & 132.1 & 2.266 & 153.1 & 2.829 & 0.5716 & 0.2082 & 248.1 & 3.288 & 283.3 & 4.002 & 0.5567 & 0.1932 & 332.1 & 4.307 & 0.4076 & 0.3841 & 486.4 & 4.669 \\
\hline
\end{tabular}
}
\end{sidewaystable}
\footnotetext[1]{mass in Earth Mass ($M_{\oplus} = 5.9742\times10^{24} kg$)}
\footnotetext[2]{radius in Earth Radius ($R_{\oplus} = 6.371\times10^{6} m$)}
\footnotetext[3]{CRF stands for core radius fraction, the ratio of the radius of the core over the total radius of the planet, in the two-layer model}
\footnotetext[4]{p1/p0 stands for core-mantle-boundary pressure fraction, the ratio of the pressure at the core-mantle boundary (p1) over the pressure at the center of the planet (p0), in the two-layer model}
\subsection{Levels of planet differentiation: the effect of Fe partitioning between mantle and core\label{nondiff}}
All models of planets discussed so far assume that all Fe is in the core, while all Mg, Si and O are in the mantle, i.e., that a planet is fully differentiated. However, we know that in terrestrial planets some Fe is incorporated into the mantle. There are two separate processes which affect the Fe content of the mantle: (1) mechanical segregation of Fe-rich metal from the mantle to the core, and (2) different redox conditions resulting in a different Fe/Mg ratio within the mantle, which in turn affects the relative size of the core and mantle. In this section we show the effects of (1) undifferentiated versus fully-differentiated, and (2) reduced versus oxidized planetary structure on the mass-radius relation for a planet with the same Fe/Si and Mg/Si ratios.
For simplicity, here we ignore the H${}_{2}$O and gaseous content of the planet and only consider the planet made of Fe, Mg, Si, O. To facilitate comparison between different cases, we fix the global atomic ratios of Fe/Mg=1 and Mg/Si=1; these fit well within the range of local stellar abundances~\citep{Grasset:2009}.
In particular, we consider a double-layer planet with a core and a mantle in two scenarios.
One follows the incomplete mechanical separation of the Fe-rich metal during planet formation, and results in addition of Fe to the mantle as metal particles. It does not change EOS of the silicate components, but requires adding an Fe-EOS to the mantle mixture. Thus, the planet generally consists of a Fe metal core and a partially differentiated mantle consisting of the mixture of metallic Fe and MgSiO${}_{3}$ silicates. While the distribution of metallic Fe may have a radial gradient, for simplicity we assume that it is uniformly distributed in the silicate mantle. Within scenario 1, we calculate three cases to represent different levels of differentiation:
\begin{description}
\item[Case 1: complete differentiation] metallic Fe core and MgSiO${}_{3}$ silicate mantle. For Fe/Mg=1, CMF=0.3574.
\item[Case 2: partial differentiation] half the Fe forms a smaller metallic Fe core, with the other half of metal being mixed with MgSiO${}_{3}$ silicates in the mantle. CMF=0.1787.
\item[Case 3: no differentiation] All the metallic Fe is mixed with MgSiO${}_{3}$ in the mantle. CMF=0 (no core).
\end{description}
The other scenario assumes different redox conditions, resulting in different Fe/Mg ratios in mantle minerals, and therefore requiring different EOS for the Fe${}^{2+}$-bearing silicates and oxides. More oxidized mantle means adding more Fe in the form of FeO to the MgSiO${}_{3}$ silicates to form (Mg,Fe)SiO${}_{3}$ silicates and (Mg,Fe)O magnesiow\"{u}stite (mv), thus reducing the amount of Fe in the core. The exact amounts of (Mg,Fe)SiO${}_{3}$ and (Mg,Fe)O in the mantle are determined by the following mass balance equation:
\begin{equation}
x~FeO + MgSiO3~\xrightarrow~(Mg{}_{\frac{1}{1+x}},Fe{}_{\frac{x}{1+x}})SiO{}_3 + x~(Mg_{\frac{1}{1+x}},Fe_{\frac{x}{1+x}})O
\label{massbalance}
\end{equation}
$x$ denotes the relative amount of FeO added to the silicate mantle, $x$=0 being the most reduced state with no Fe in the mantle, and $x$=1 being the most oxidized state with all Fe existing as oxides in the mantle. This oxidization process conserves the global Fe/Mg and Mg/Si ratios, but increases O content and thus the O/Si ratio of the planet since Fe is added to the mantle in the form of FeO. Because in stellar environments O is excessively abundant relative to Mg, Si, and Fe (e.g., the solar elemental abundances~\citep{Asplund:2009}), it is not a limiting factor in our models of oxidized planets.
We calculate the following three cases to represent the full range of redox conditions:
\begin{description}
\item[Case 4: no oxidization of Fe] $x$=0. Metal Fe core and MgSiO${}_{3}$ silicate mantle. For Fe/Mg=1, O/Si=3, CMF=0.3574.
\item[Case 5: partial oxidization of Fe] $x$=0.5. Half the Fe forms smaller metal core, the other half is added as FeO to the mantle. O/Si=3.5, CMF=0.1700.
\item[Case 6: complete oxidization of Fe] $x$=1. All Fe is added as FeO to the mantle, resulting in no metal core at all. O/Si=4, CMF=0.
\end{description}
Notice that Case 4 looks identical to Case 1. However, the silicate EOS used to calculate Case 4 is different from Case 1 at ultrahigh pressures (beyond $0.90$ TPa). For Cases 4, 5, and 6, the (Mg,Fe)SiO${}_{3}$ EOS is adopted from~\citet{Caracas:2008} and~\citet{Spera:2006} which only consider perovskite (pv) and post-perovskite (ppv) phases without including further dissociation beyond $0.90$ TPa, since the Fe${}^{2+}$-bearing silicate EOS at ultrahigh pressures is hardly available. On the other hand, Comparison between Case 1 and Case 4 also shows the uncertainty on mass-radius relation resulting from the different choice of EOS (see Table~\ref{Table2}). Fe${}^{2+}$-bearing pv and ppv have the general formula: (Mg${}_{y}$,Fe${}_{1-y}$)SiO${}_{3}$, where $y$ denotes the relative atomic number fraction of Mg and Fe in the silicate mineral. The (Mg${}_{y}$,Fe${}_{1-y}$)SiO${}_{3}$ silicate is therefore a binary component equilibrium solid solution. It could either be pv or ppv or both depending on the pressure~\citep{Spera:2006}. We can safely approximate the narrow pressure region where pv and ppv co-exist as a single transition pressure from pv to ppv. This pressure is calculated as the arithmetic mean of the initial transition pressure of pv $\rightarrow$ pv+ppv mixture and the final transition pressure of pv+ppv mixture $\rightarrow$ ppv~\citep{Spera:2006}. The pv EOS and ppv EOS are connected at this transition pressure to form a consistent EOS for all pressures.
Addition of FeO to MgSiO${}_{3}$ results in the appearance of a second phase, magnesiow\"{u}stite, in the mantle according to Eq.~\ref{massbalance}. The (Mg,Fe)O EOS for Cases 4, 5, and 6 is adopted from~\citet{Fei:2007}, which includes the electronic spin transition of high-spin to low-spin in Fe${}^{2+}$. For simplicity, we assume that pv/ppv and mw have the same Mg/Fe ratio.
Fig.~\ref{f4} shows fractional differences ($\eta$) in radius of Cases 2 \& 3 compared to Case 1 as well as Cases 5 \& 6 compared to Case 4 ($r_{0}$ is radius of the reference case, which is that of Case 1 for Cases 2 \& 3 and is that of Case 4 for Case 5 \& 6):
\begin{equation}
\eta=\frac{r-r_{0}}{r_{0}}
\label{fractionaldiff}
\end{equation}
Oxidization of Fe (partitioning Fe as Fe-oxides from the core into the mantle) makes the planet appear larger. The complete oxidization of Fe makes the radius 3$\%$ larger for small planets around 1 $M_{\oplus}$, then the difference decreases with increasing mass within the mass range of 1 to 20 $M_{\oplus}$. Undifferentiated planets (partitioning of metallic Fe from the core into the mantle) appear smaller than fully differentiated planets. The completely undifferentiated planet is practically indistinguishable in radius for small planets around 1 $M_{\oplus}$, then the difference increases to 1$\%$-level around 20 $M_{\oplus}$. The mass, radius, CRF and p1/p0 data of Cases 1 through 6 are listed in Table~\ref{Table2}.
\begin{figure}[htbp]
\begin{center}
\includegraphics[scale=0.75]{f4.eps}
\end{center}
\caption{fractional differences in radius resulting from Fe partitioning between mantle and core.
black curve: Case 1 \& Case 4 (complete differentiation and no oxidization of Fe); solid brown curve: Case 2 (partial (50\%) differentiation: 50$\%$ metallic Fe mixed with the mantle); solid pink curve: Case 3 (no differentiation: all metallic Fe mixed with the mantle); dashed brown curve: Case 5 (partial (50\%) oxidization of Fe); dashed pink curve: Case 6 (complete oxidization of Fe)}
\label{f4}
\end{figure}
\begin{sidewaystable}[htbp]
\caption{Data of Cases 1 through 6\label{Table2}}
\centering
\scalebox{0.60}{
\begin{tabular}{c | c c c c | c c c c | c c | c c c c | c c c c | c c |}
\\
\hline\hline
&
\multicolumn{4}{c}{Case 1 (CMF=0.3574)} &
\multicolumn{4}{c}{Case 2 (CMF=0.1787)} &
\multicolumn{2}{c}{Case 3} &
\multicolumn{4}{c}{Case 4 (CMF=0.3574)} &
\multicolumn{4}{c}{Case 5 (CMF=0.1700)} &
\multicolumn{2}{c}{Case 6} \\ \hline
$log_{10}(p0)$ & Mass\footnotemark[1] & Radius\footnotemark[2] & CRF\footnotemark[3] & p1/p0\footnotemark[4] & Mass & Radius & CRF & p1/p0 & Mass & Radius & Mass & Radius & CRF & p1/p0 & Mass & Radius & CRF & p1/p0 & Mass & Radius\\ \hline
9.6 & 0.002009 & 0.1302 & 0.5972 & 0.3847 & 0.002395 & 0.1381 & 0.4736 & 0.5634 & 0.004165 & 0.1659 & 0.002009 & 0.1302 & 0.5972 & 0.3847 & 0.002526 & 0.1428 & 0.4584 & 0.5625 & 0.004921 & 0.1804 \\
9.7 & 0.002816 & 0.1456 & 0.5967 & 0.3839 & 0.003359 & 0.1545 & 0.4731 & 0.5626 & 0.005846 & 0.1856 & 0.002816 & 0.1456 & 0.5967 & 0.3839 & 0.003542 & 0.1597 & 0.4579 & 0.5617 & 0.006905 & 0.2018 \\
9.8 & 0.003941 & 0.1628 & 0.5961 & 0.383 & 0.004704 & 0.1727 & 0.4726 & 0.5616 & 0.008194 & 0.2076 & 0.003941 & 0.1628 & 0.5961 & 0.383 & 0.004959 & 0.1786 & 0.4574 & 0.5607 & 0.009673 & 0.2256 \\
9.9 & 0.005504 & 0.1819 & 0.5954 & 0.382 & 0.006573 & 0.1929 & 0.4719 & 0.5604 & 0.01146 & 0.232 & 0.005504 & 0.1819 & 0.5954 & 0.382 & 0.00693 & 0.1995 & 0.4567 & 0.5596 & 0.01352 & 0.252 \\
10 & 0.00767 & 0.203 & 0.5946 & 0.3807 & 0.009165 & 0.2154 & 0.4711 & 0.5591 & 0.016 & 0.2589 & 0.00767 & 0.203 & 0.5946 & 0.3807 & 0.009662 & 0.2227 & 0.456 & 0.5582 & 0.01887 & 0.2812 \\
10.1 & 0.01066 & 0.2263 & 0.5936 & 0.3792 & 0.01275 & 0.2401 & 0.4702 & 0.5575 & 0.02228 & 0.2887 & 0.01066 & 0.2263 & 0.5936 & 0.3792 & 0.01344 & 0.2484 & 0.455 & 0.5567 & 0.02625 & 0.3135 \\
10.2 & 0.01477 & 0.252 & 0.5924 & 0.3774 & 0.01767 & 0.2674 & 0.469 & 0.5557 & 0.03093 & 0.3215 & 0.01477 & 0.252 & 0.5924 & 0.3774 & 0.01863 & 0.2766 & 0.454 & 0.5548 & 0.03639 & 0.3489 \\
10.3 & 0.02039 & 0.2802 & 0.591 & 0.3754 & 0.02442 & 0.2974 & 0.4678 & 0.5536 & 0.04278 & 0.3575 & 0.02039 & 0.2802 & 0.591 & 0.3754 & 0.02574 & 0.3076 & 0.4527 & 0.5528 & 0.05027 & 0.3878 \\
10.4 & 0.02805 & 0.311 & 0.5894 & 0.3732 & 0.03362 & 0.3303 & 0.4663 & 0.5513 & 0.05893 & 0.3968 & 0.02805 & 0.311 & 0.5894 & 0.3732 & 0.03544 & 0.3416 & 0.4513 & 0.5504 & 0.06915 & 0.4301 \\
10.5 & 0.03842 & 0.3447 & 0.5876 & 0.3706 & 0.0461 & 0.3661 & 0.4647 & 0.5487 & 0.08081 & 0.4395 & 0.03842 & 0.3447 & 0.5876 & 0.3706 & 0.04859 & 0.3786 & 0.4497 & 0.5479 & 0.09465 & 0.476 \\
10.6 & 0.05239 & 0.3814 & 0.5856 & 0.3679 & 0.06292 & 0.405 & 0.4628 & 0.5459 & 0.1102 & 0.4858 & 0.05239 & 0.3814 & 0.5856 & 0.3679 & 0.06631 & 0.4189 & 0.448 & 0.5451 & 0.1289 & 0.5257 \\
10.7 & 0.07111 & 0.4211 & 0.5833 & 0.3649 & 0.08548 & 0.4472 & 0.4608 & 0.5429 & 0.1496 & 0.5356 & 0.07111 & 0.4211 & 0.5833 & 0.3649 & 0.09006 & 0.4624 & 0.446 & 0.5422 & 0.1744 & 0.5789 \\
10.8 & 0.09605 & 0.464 & 0.5809 & 0.3617 & 0.1155 & 0.4927 & 0.4587 & 0.5398 & 0.2017 & 0.5888 & 0.09605 & 0.464 & 0.5809 & 0.3617 & 0.1217 & 0.5094 & 0.4439 & 0.5392 & 0.2345 & 0.6358 \\
10.9 & 0.1291 & 0.5101 & 0.5782 & 0.3584 & 0.1553 & 0.5416 & 0.4564 & 0.5366 & 0.2702 & 0.6456 & 0.1291 & 0.5101 & 0.5782 & 0.3584 & 0.1636 & 0.5599 & 0.4417 & 0.5361 & 0.3113 & 0.6946 \\
11 & 0.1725 & 0.5595 & 0.5753 & 0.355 & 0.2077 & 0.5939 & 0.454 & 0.5334 & 0.3596 & 0.7056 & 0.1725 & 0.5595 & 0.5753 & 0.355 & 0.2187 & 0.6139 & 0.4394 & 0.533 & 0.408 & 0.7548 \\
11.1 & 0.2294 & 0.6123 & 0.5723 & 0.3515 & 0.2762 & 0.6495 & 0.4515 & 0.5302 & 0.4744 & 0.7684 & 0.2294 & 0.6123 & 0.5723 & 0.3515 & 0.2903 & 0.6709 & 0.4371 & 0.5302 & 0.5315 & 0.8177 \\
11.2 & 0.3033 & 0.6684 & 0.5692 & 0.348 & 0.3652 & 0.7085 & 0.4489 & 0.5271 & 0.6178 & 0.8323 & 0.3033 & 0.6684 & 0.5692 & 0.348 & 0.3833 & 0.7313 & 0.4347 & 0.5275 & 0.6905 & 0.884 \\
11.3 & 0.399 & 0.7278 & 0.566 & 0.3445 & 0.48 & 0.7708 & 0.4463 & 0.5241 & 0.8018 & 0.8996 & 0.399 & 0.7278 & 0.566 & 0.3445 & 0.5024 & 0.7945 & 0.4323 & 0.5255 & 0.8932 & 0.9536 \\
11.4 & 0.5219 & 0.7906 & 0.5627 & 0.3412 & 0.6266 & 0.8358 & 0.4437 & 0.5216 & 1.036 & 0.9699 & 0.5219 & 0.7906 & 0.5627 & 0.3412 & 0.6542 & 0.8604 & 0.43 & 0.5237 & 1.15 & 1.026 \\
11.5 & 0.679 & 0.8565 & 0.5594 & 0.3379 & 0.8118 & 0.9031 & 0.4412 & 0.5199 & 1.331 & 1.043 & 0.679 & 0.8565 & 0.5594 & 0.3379 & 0.8478 & 0.9294 & 0.4278 & 0.5219 & 1.472 & 1.101 \\
11.6 & 0.8779 & 0.9251 & 0.5561 & 0.3352 & 1.047 & 0.9733 & 0.4388 & 0.5181 & 1.702 & 1.119 & 0.8779 & 0.9251 & 0.5561 & 0.3352 & 1.093 & 1.001 & 0.4255 & 0.5201 & 1.874 & 1.178 \\
11.7 & 1.128 & 0.996 & 0.553 & 0.333 & 1.343 & 1.046 & 0.4365 & 0.5163 & 2.166 & 1.197 & 1.128 & 0.996 & 0.553 & 0.333 & 1.402 & 1.076 & 0.4233 & 0.5184 & 2.373 & 1.258 \\
11.8 & 1.443 & 1.07 & 0.55 & 0.3307 & 1.715 & 1.122 & 0.4341 & 0.5145 & 2.741 & 1.277 & 1.443 & 1.07 & 0.55 & 0.3307 & 1.789 & 1.154 & 0.4211 & 0.5167 & 2.99 & 1.34 \\
11.9 & 1.837 & 1.146 & 0.547 & 0.3284 & 2.179 & 1.2 & 0.4318 & 0.5127 & 3.453 & 1.36 & 1.837 & 1.146 & 0.547 & 0.3284 & 2.272 & 1.233 & 0.419 & 0.5152 & 3.75 & 1.423 \\
12 & 2.327 & 1.225 & 0.5441 & 0.3262 & 2.755 & 1.281 & 0.4295 & 0.511 & 4.303 & 1.442 & 2.327 & 1.225 & 0.5441 & 0.3262 & 2.871 & 1.316 & 0.4168 & 0.5137 & 4.68 & 1.509 \\
12.1 & 2.934 & 1.306 & 0.5412 & 0.324 & 3.468 & 1.364 & 0.4273 & 0.5094 & 5.306 & 1.522 & 2.934 & 1.306 & 0.5412 & 0.324 & 3.61 & 1.4 & 0.4147 & 0.5123 & 5.818 & 1.596 \\
12.2 & 3.682 & 1.39 & 0.5383 & 0.3219 & 4.344 & 1.448 & 0.425 & 0.5077 & 6.522 & 1.603 & 3.682 & 1.39 & 0.5383 & 0.3219 & 4.519 & 1.486 & 0.4126 & 0.5109 & 7.203 & 1.684 \\
12.3 & 4.6 & 1.475 & 0.5355 & 0.3197 & 5.403 & 1.533 & 0.4229 & 0.5068 & 7.986 & 1.685 & 4.6 & 1.475 & 0.5355 & 0.3197 & 5.63 & 1.574 & 0.4105 & 0.5095 & 8.886 & 1.775 \\
12.4 & 5.72 & 1.562 & 0.5327 & 0.3176 & 6.671 & 1.617 & 0.4208 & 0.5068 & 9.724 & 1.766 & 5.72 & 1.562 & 0.5327 & 0.3176 & 6.983 & 1.664 & 0.4085 & 0.5081 & 10.93 & 1.866 \\
12.5 & 7.071 & 1.649 & 0.53 & 0.3161 & 8.202 & 1.702 & 0.4189 & 0.5067 & 11.8 & 1.848 & 7.082 & 1.651 & 0.5299 & 0.3155 & 8.626 & 1.754 & 0.4064 & 0.5065 & 13.39 & 1.96 \\
12.6 & 8.686 & 1.736 & 0.5275 & 0.3152 & 10.04 & 1.787 & 0.417 & 0.5065 & 14.27 & 1.929 & 8.729 & 1.74 & 0.527 & 0.3133 & 10.61 & 1.847 & 0.4043 & 0.5049 & 16.37 & 2.054 \\
12.7 & 10.62 & 1.822 & 0.5251 & 0.3143 & 12.23 & 1.871 & 0.4152 & 0.5064 & 17.18 & 2.01 & 10.71 & 1.831 & 0.5242 & 0.311 & 13 & 1.94 & 0.4021 & 0.503 & 19.96 & 2.15 \\
12.8 & 12.93 & 1.908 & 0.5228 & 0.3134 & 14.83 & 1.955 & 0.4134 & 0.5063 & 20.6 & 2.09 & 13.09 & 1.923 & 0.5213 & 0.3086 & 15.86 & 2.034 & 0.3999 & 0.5009 & 24.29 & 2.248 \\
12.9 & 15.67 & 1.993 & 0.5206 & 0.3125 & 17.9 & 2.038 & 0.4116 & 0.5059 & 24.57 & 2.168 & 15.93 & 2.015 & 0.5183 & 0.3061 & 19.29 & 2.129 & 0.3976 & 0.4985 & 29.48 & 2.347 \\
13 & 18.9 & 2.078 & 0.5183 & 0.3116 & 21.53 & 2.12 & 0.4098 & 0.5054 & 29.15 & 2.244 & 19.3 & 2.108 & 0.5153 & 0.3033 & 23.36 & 2.224 & 0.3952 & 0.4958 & 35.73 & 2.448 \\
13.1 & 22.69 & 2.161 & 0.5161 & 0.3104 & 25.76 & 2.201 & 0.4079 & 0.5049 & 34.37 & 2.316 & 23.28 & 2.2 & 0.5121 & 0.3004 & 28.19 & 2.32 & 0.3927 & 0.4928 & 43.22 & 2.551 \\
13.2 & 27.12 & 2.242 & 0.5139 & 0.3092 & 30.67 & 2.279 & 0.406 & 0.5044 & 40.27 & 2.384 & 27.97 & 2.293 & 0.5089 & 0.2972 & 33.9 & 2.417 & 0.3901 & 0.4893 & 52.21 & 2.656 \\
13.3 & 32.27 & 2.322 & 0.5116 & 0.3078 & 36.33 & 2.354 & 0.404 & 0.5039 & 46.99 & 2.449 & 33.46 & 2.385 & 0.5054 & 0.2937 & 40.62 & 2.513 & 0.3873 & 0.4853 & 62.98 & 2.762 \\
13.4 & 38.2 & 2.399 & 0.5093 & 0.3066 & 42.8 & 2.426 & 0.402 & 0.5038 & 54.61 & 2.511 & 39.88 & 2.477 & 0.5018 & 0.29 & 48.52 & 2.61 & 0.3844 & 0.481 & 75.88 & 2.87 \\
13.5 & 45.05 & 2.472 & 0.5069 & 0.3056 & 50.18 & 2.494 & 0.4001 & 0.5042 & 63.25 & 2.569 & 47.37 & 2.569 & 0.4981 & 0.2862 & 57.81 & 2.708 & 0.3813 & 0.4765 & 91.32 & 2.981 \\
13.6 & 52.85 & 2.542 & 0.5046 & 0.3048 & 58.56 & 2.557 & 0.3982 & 0.5045 & 72.97 & 2.624 & 56.06 & 2.66 & 0.4942 & 0.2821 & 68.68 & 2.806 & 0.3782 & 0.4714 & 109.8 & 3.093 \\
13.7 & 61.72 & 2.608 & 0.5024 & 0.3044 & 68.09 & 2.618 & 0.3963 & 0.5049 & 83.87 & 2.675 & 66.14 & 2.752 & 0.4901 & 0.2779 & 81.38 & 2.905 & 0.375 & 0.466 & 131.9 & 3.208 \\
13.8 & 71.81 & 2.67 & 0.5002 & 0.3042 & 78.88 & 2.674 & 0.3945 & 0.5055 & 96.04 & 2.722 & 77.81 & 2.842 & 0.4859 & 0.2735 & 96.2 & 3.004 & 0.3716 & 0.4604 & 158.4 & 3.326 \\
13.9 & 83.26 & 2.728 & 0.4982 & 0.3041 & 91.07 & 2.727 & 0.3928 & 0.5062 & 109.6 & 2.765 & 91.29 & 2.933 & 0.4816 & 0.2689 & 113.5 & 3.104 & 0.3682 & 0.4544 & 190 & 3.446 \\
14 & 96.2 & 2.781 & 0.4962 & 0.3042 & 104.8 & 2.775 & 0.3911 & 0.507 & 124.6 & 2.804 & 106.8 & 3.024 & 0.4772 & 0.2642 & 133.6 & 3.204 & 0.3647 & 0.4481 & 227.8 & 3.568 \\
14.1 & 110.8 & 2.831 & 0.4944 & 0.3043 & 120.1 & 2.819 & 0.3895 & 0.5079 & 141.3 & 2.839 & 124.7 & 3.114 & 0.4727 & 0.2593 & 157 & 3.306 & 0.3612 & 0.4415 & 273 & 3.694 \\
14.2 & 127.2 & 2.876 & 0.4926 & 0.3045 & 137.3 & 2.859 & 0.388 & 0.509 & 159.6 & 2.87 & 145.2 & 3.205 & 0.468 & 0.2542 & 184.1 & 3.408 & 0.3575 & 0.4345 & 327.1 & 3.823 \\
14.3 & 145.5 & 2.917 & 0.491 & 0.3049 & 156.4 & 2.894 & 0.3865 & 0.5102 & 179.9 & 2.897 & 168.7 & 3.295 & 0.4633 & 0.2489 & 215.5 & 3.511 & 0.3538 & 0.4271 & 391.8 & 3.955 \\
14.4 & 166 & 2.954 & 0.4894 & 0.3054 & 177.5 & 2.925 & 0.3851 & 0.5115 & 202.3 & 2.92 & 195.5 & 3.386 & 0.4584 & 0.2435 & 251.8 & 3.615 & 0.3499 & 0.4195 & 469.1 & 4.091 \\
\hline
\end{tabular}
}
\end{sidewaystable}
\subsection{Tabulating the Ternary Diagram}
For the $3$-layer model of solid exoplanet, points of a curve segment on the ternary diagram represent all the solutions for a given mass-radius input. These ternary diagrams are tabulated (Table~\ref{Table3}) with the intent to make comparison to observations easier.
Usually, there are infinite combinations (solutions) of Fe, MgSiO${}_{3}$ and H${}_{2}$O mass fractions which can give the same mass-radius pair. All the combinations together form a curve segment on the ternary diagram of Fe, MgSiO${}_{3}$ and H${}_{2}$O mass fractions~\citep{Zeng_Seager:2008, Valencia:2007b}. This curve segment can be approximated by $3$ points on it: two endpoints where one or more out of the $3$ layers are absent and one point in between where all $3$ layers are present to give the same mass and radius. The two endpoints correspond to the minimum central pressure (p0${}_{min}$) and maximum central pressure (p0${}_{max}$) allowed for the given mass-radius pair. The middle point is chosen to have the central pressure p0${}_{mid}$=$\sqrt{\text{p0}{}_{max}*\text{p0}{}_{min}}$.
Table~\ref{Table3} contains 8 columns:
1st column: $Mass$. The masses range from 0.1 $M_{\oplus}$ to 100 $M_{\oplus}$ with 41 points in total. The range between 0.1 and 1 $M_{\oplus}$ is equally divided into 10 sections in logarithmic scale. The range between 1 and 10 $M_{\oplus}$ is equally divided into 20 sections in logarithmic scale. And the range between 10 and 100 $M_{\oplus}$ is equally divided into 10 sections in logarithmic scale.
2nd column: $Radius$. For each mass $M$ in Table~\ref{Table3} there are $12$ radius values, $11$ of which are equally spaced within the allowed range: $R_{Fe}(M)+(R_{H_2O}(M)-R_{Fe}(M))*i$, where $i=0, 0.1, 0.2,... , 0.9, 1.0$. The $12$-th radius value ($R_{MgSiO_3}(M)$) is inserted into the list corresponding to the pure-MgSiO${}_{3}$ planet radius (see Table~\ref{Table1}) for mass $M$. Here $R_{Fe}(M)$, $R_{MgSiO_3}(M)$, and $R_{H_2O}(M)$ are the radii for planets with mass $M$ composed of pure-Fe, pure-MgSiO${}_{3}$, and pure-H${}_{2}$O correspondingly.
Overall, there are $41*12=492$ different mass-radius pairs in Table~\ref{Table3}.
For each $(M,R)$, $3$ cases: p0${}_{min}$, p0${}_{mid}$, and p0${}_{max}$ are listed.
3rd column: central pressure p0 (Pascal) in logarithmic base-10 scale.
4th column: p1/p0, the ratio of p1 (the first boundary pressure, i.e., the pressure at the Fe-MgSiO${}_{3}$ boundary) over p0.
5th column: p2/p1, the ratio of p2 (the second boundary pressure, i.e., the pressure at the MgSiO${}_{3}$-H${}_{2}$O boundary) over p1.
6th column: Fe mass fraction (the ratio of the Fe-layer mass over the total mass of the planet).
7th column: MgSiO${}_{3}$ mass fraction (the ratio of the MgSiO${}_{3}$-layer mass over the total mass of the planet).
8th column: H${}_{2}$O mass fraction (the ratio of the H${}_{2}$O-layer mass of over the total mass of the planet).
6th, 7th and 8th columns always add up to one.
\\
\begin{center}
\begin{longtable}{|c|c|c|c|c|c|c|c|}
\caption{Table for Ternary Diagram} \label{Table3}\\
\hline \multicolumn{1}{|c|}{\textbf{M($M_{\oplus}$)}} &
\multicolumn{1}{c|}{\textbf{R($R_{\oplus}$)}} &
\multicolumn{1}{c|}{$\mathbf{log_{10}(p0)}$} &
\multicolumn{1}{c|}{\textbf{p1/p0}} &
\multicolumn{1}{c|}{\textbf{p2/p1}} &
\multicolumn{1}{c|}{$\mathbf{Fe}$} &
\multicolumn{1}{c|}{$\mathbf{MgSiO_3}$} &
\multicolumn{1}{c|}{$\mathbf{H_2O}$} \\ \hline
\endfirsthead
\multicolumn{8}{c}%
{{\bfseries \tablename\ \thetable{} -- continued from previous page}} \\
\hline \multicolumn{1}{|c|}{\textbf{M($M_{\oplus}$)}} &
\multicolumn{1}{c|}{\textbf{R($R_{\oplus}$)}} &
\multicolumn{1}{c|}{$\mathbf{log_{10}(p0)}$} &
\multicolumn{1}{c|}{\textbf{p1/p0}} &
\multicolumn{1}{c|}{\textbf{p2/p1}} &
\multicolumn{1}{c|}{$\mathbf{Fe}$} &
\multicolumn{1}{c|}{$\mathbf{MgSiO_3}$} &
\multicolumn{1}{c|}{$\mathbf{H_2O}$} \\ \hline
\endhead
\hline \multicolumn{8}{|r|}{{Continued on next page}} \\ \hline
\endfoot
\hline \hline
\endlastfoot
0.1 & 0.3888 & 10.9212 & 0 & 0 & 1 & 0 & 0 \\
0.1 & 0.3888 & 10.9212 & 0 & 0 & 1 & 0 & 0 \\
0.1 & 0.3888 & 10.9212 & 0 & 0 & 1 & 0 & 0 \\
0.1 & 0.4226 & 10.9066 & 0.133 & 0 & 0.759 & 0.241 & 0 \\
0.1 & 0.4226 & 10.9126 & 0.102 & 0.046 & 0.818 & 0.172 & 0.01 \\
0.1 & 0.4226 & 10.9186 & 0.024 & 1 & 0.953 & 0 & 0.047 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
100 & 4.102 & 13.0979 & 1 & 1 & 0 & 0 & 1 \\
\end{longtable}
\end{center}
A dynamic and interactive tool to characterize and illustrate the interior structure of exoplanets built upon Table~\ref{Table3} and other models in this paper is available on the website \url{http://www.cfa.harvard.edu/~lzeng}.
\subsection{Generate curve segment on ternary diagram using Table~\ref{Table3}}
One utility of Table~\ref{Table3} is to generate the curve segment on the $3$-layer ternary diagram for a given mass-radius pair. As an example, for M=$1M_{\oplus}$ and R=$1.0281R_{\oplus}$, the table provides $3$ p0's. For each p0, the mass fractions of Fe, MgSiO${}_{3}$, and H${}_{2}$O are given to determine a point on the ternary diagram. Then, a parabolic fit (see Fig.~\ref{ternary}) through the $3$ points is a good approximation to the actual curve segment. This parabola may intersect the maximum collisional stripping curve by~\citet{Marcus:2010}, indicating that the portion of parabola beneath the intersection point may be ruled out by planet formation theory.
\begin{figure}[htbp]
\begin{center}
\includegraphics[scale=0.6,angle=90]{f5.eps}
\end{center}
\caption{The red, orange and purple points correspond to $log_{10}(p0~(in~Pa))=11.4391$, $11.5863$, and $11.7336$. The mass fractions are (1) red point: FeMF=0.083, MgSiO${}_{3}$MF=0.917, H${}_{2}$OMF=0; (2) orange point: FeMF=0.244, MgSiO${}_{3}$MF=0.711, H${}_{2}$OMF=0.046; (3) purple point: FeMF=0.728, MgSiO${}_{3}$MF=0, H${}_{2}$OMF=0.272. MF here stands for mass fraction. The blue curve is the parabolic fit. The red dashed curve is the maximum collisional stripping curve by~\citet{Marcus:2010}.}
\label{ternary}
\end{figure}
\section{Conclusion}
The $2$-layer and $3$-layer models for solid exoplanets composed of Fe, MgSiO${}_{3}$, and H${}_{2}$O are the focus of this paper. The mass-radius contours (Fig.~\ref{contourplots}) are provided for the $2$-layer model, useful for readers to quickly calculate the interior structure of a solid exoplanet.
The $2$-parameter contour mesh may also help one build physical insights into the solid exoplanet interior structure.
The complete $3$-layer mass-fraction ternary diagram is tabulated (Table~\ref{Table3}), useful for readers to interpolate and calculate all solutions as the mass fractions of the $3$ layers for a given mass-radius input. The details of the EOS of Fe, MgSiO${}_{3}$, and H${}_{2}$O and how they are calculated and used in this paper are discussed in section~\ref{EOS} and shown in Fig.~\ref{eosplot}.
A dynamic and interactive tool to characterize and illustrate the interior structure of exoplanets built upon Table~\ref{Table3} and other models in this paper is available on the website \url{http://www.cfa.harvard.edu/~lzeng}.
The effect of Fe partitioning between mantle and core on mass-radius relation is explored in section~\ref{nondiff}, and the result is shown in Fig.~\ref{f4} and Table~\ref{Table2}.
With the ongoing Kepler Mission and many other exoplanet searching projects, we hope this paper could provide a handy tool for observers to fast characterize the interior structure of exoplanets already discovered or to be discovered, and further our understanding of those worlds beyond our own.
\section{Acknowledgements}
We acknowledge partial support for this work by NASA co-operative agreement NNX09AJ50A (Kepler mission science team).
We would like to thank Michail Petaev and Stein Jacobsen for their valuable comments and suggestions. This research is supported by the National Nuclear Security Administration under the High Energy Density Laboratory Plasmas through DOE grant $\#$ DE-FG52-09NA29549 to S. B. Jacobsen (PI) with Harvard University. This research is the authors' views and not those of the DOE.
Li Zeng would like to thank Professor Pingyuan Li, Li Zeng's grandfather, in the Department of Mathematics at Chongqing University, for giving Li Zeng important spiritual support and guidance on research. The guidance includes research strategy and approach, methods of solving differential equations and other numeric methods, etc.
Li Zeng would also like to give special thanks to Master Anlin Wang. Master Wang is a Traditional Chinese Kung Fu Master and World Champion. He is also a practitioner and realizer of Traditional Chinese Philosophy of Tao Te Ching, which is the ancient oriental wisdom to study the relation between the universe, nature and humanity. Valuable inspirations were obtained through discussion of Tao Te Ching with Master Wang as well as Qigong cultivation with him.
\clearpage
\bibliographystyle{bibstyle1}
| {
"redpajama_set_name": "RedPajamaArXiv"
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