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Q: connection string in with a local db c# windows forms I've created a database using Visual Studio 2015 by adding it locally as a mdf file
following this guide: Walkthrough creating a local database SQL Server Express
The exception that gets thrown is Format of the initialization string does not conform to specification starting at index 81.
I'm building a Windows Forms application.
My connection string looks like this string cs = @"Data Source=(LocalDB)\MSSQLLocalDB;AttachDbFilename=|DataDirectory|\TestDb.mdf;Integrated";
A: You can user Visual Studio's server explorer to connect to your DB and then you'll get the correct connection string:
http://www.c-sharpcorner.com/uploadfile/suthish_nair/how-to-generate-or-find-connection-string-from-visual-studio/
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{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 367
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\section{Introduction}
\noindent Silicon Photomultipliers (SiPMs) are multi-pixel APDs operated in Geiger mode. This solid-state photon detection technology provides good single photon detection capability and high photon detection efficiency. Further features are their compact size, insensitivity to magnetic fields and cost efficiency, which make them suitable for many research fields that require photon detection, such as particle physics, nuclear physics or medical imaging. \\
A position sensitive Cherenkov detector was built, consisting of an array of $8\times8$ SiPMs (Hamamatsu S10931-‐100P) with an active area of $3\times3$ mm$^2$ each and a pixel size of $100\times100$~$\mu$m$^2$. The signals are amplified with four 16 channel amplifiers that were built in-house and are based on Photonique amplifiers. In addition, a suitable light concentrator consisting of 64 pyramid-‐shaped funnels was developed. With an entrance surface of $7\times7$~mm$^2$ and an exit surface of $3\times3$ mm$^2$, this light concentrator, which is made out of brass and coated with aluminium, increases the detection area of the module, while providing sufficient position resolution, e. g. for the barrel DIRC detector~\cite{DIRC4PANDA} at the PANDA experiment at the FAIR facility in Darmstadt~\cite{PANDALOI}. Increasing the detection area of the detector by this method gives several advantages. One essential advantage is that the signal-to-noise ratio improves by increasing the sensitive area using light focusing and keeping the dark count rate constant~\cite{S2Nratio}. Another benefit is that the number of read-out channels can be kept low, thus the module can be built very compactly.\\
In previous work, simulations for the collection efficiency were performed~\cite{Ahmed} as well as a scan with a laser beam to measure the collection efficiency of the module. However, the beam spot diameter was as large as 1 mm and the step size was $250\;\mu$m~\cite{Gruber}. These two parameters have been improved significantly in the new tests, providing a more detailed picture of the characteristics of the SiPMs and the light concentrator. Also, a scan with a finite incident angle was performed. The new data allows to further optimize the light guide.
\section{Test Setup}
\label{sec:testsetup}
\noindent To test the position sensitive photon detector, the complete setup was put inside a dark box. The test setup consists of the detector module, a light source and two stepping motors which move the beam spot across the area of the scanned SiPMs. \\
The Hamamatsu 10931 $3\times3$ mm$^2$ SiPMs with a pixel size of $100\times100\;\mu$m$^2$ were chosen because they have the highest photon detection efficiency and an adequate dynamic range. The 10931 sensor series has the photon detection maximum at $\lambda=440$ nm. For the scan, a light source with a wavelength near that maximum looked reasonable and an LED with a wavelength range of $465$ nm $<\lambda<475$ nm was used. \\
The light source was set to emit pulses instead of a continuous wave in order not to saturate the sensor.
The pulse rate of the LED was about 900 kHz with a pulse width of about 6.5 ns. \\
To reduce the beam spot diameter from $1.3\pm0.1$ mm at the LED exit to $108\pm4\;\mu$m at the SiPM surface, an optical setup, including 3 biconvex lenses and a $10\;\mu$m pinhole were included into the test setup. This optical apparatus, which is presented in figure~\ref{fig:schematic_laser} was moved by the two stepping motors, which changed the beam spot position on the detector and the array by steps of 100 $\mu$m. This guaranteed that each pixel of the SiPM was triggered by the light beam.
\begin{figure}[hbt]
\centering
\includegraphics[width=\columnwidth,keepaspectratio]{Laser_schematic.pdf}
\caption{Schematic of optomecanical items and laser beam.}
\label{fig:schematic_laser}
\begin{center}
\begin{tabbing}[h]
\hspace*{0cm}\=\hspace*{0.85cm}\=\hspace*{0.55cm}\=\kill
ad fig.~\ref{fig:schematic_laser}.: \> \> \textsf{A:} \> LED beam exit\\
\> \> \textsf{B:} \> biconvex lens with f = 30 mm \\
\> \> \textsf{C:} \> 10 $\mu$m pinhole, serves as point-like light source \\
\> \> \textsf{D:} \> collimating biconvex lens with f = 100 mm \\
\> \> \textsf{E:} \> focusing biconvex lens with f = 200 mm \\
\end{tabbing}
\end{center}
\end{figure}
\newline
\noindent During the tests, the coordinate convention was defined as follows: The x- and z-axis build a plane parallel to the detector surface and the y-axis is parallel to the beam direction. Figure~\ref{fig:motor} shows a schematic of the optical setup and its mounting on the stepping motors. Furthermore, it gives an overview of the chosen coordinate convention.
\begin{figure}[hbt]
\centering
\includegraphics[width=\columnwidth,keepaspectratio]{Motor_setup.pdf}
\caption{Schematic of motor and optical setup, the coordinates x, y and z of movements are defined. }
\label{fig:motor}
\end{figure}
Due to the fact that the motors are high precision tools and that the weight had to be completely poised in order to keep the precision of the motors at its high level, some measures had to be taken. The beam spot could be moved in an x- and z-direction. In order to reduce the wiggling of the motor tips, cage plates were mounted to serve as stabilisers. The optical apparatus is fixed via fixation cage plates on the x-axis motor tip, the beam direction is parallel to the y-axis of this setup.
\noindent Figure~\ref{fig:darkbox} shows the opto-motoric setup together with the detector module inside the dark box.
\begin{figure}[hbt]
\centering
\includegraphics[width=\columnwidth,keepaspectratio]{Darkbox.pdf}
\caption{Test setup inside dark box. On the left side of the box the optical and motor setup is mounted. On the right side of the box sits the detector prototype. }
\label{fig:darkbox}
\end{figure}
\section{Scanned Channels and scanning mode}
\noindent Due to timing restraints not all 64 sensors could be scanned. Thus, three adjacent SiPMs were chosen randomly for the test. These sensors are referred to as F2, F3 and F4. Their position on the detector module surface can be seen in figure~\ref{fig:F2F3F4}.
\begin{figure}[hbt]
\centering
\includegraphics[width=0.8\columnwidth,keepaspectratio]{Channels_neu1.pdf}
\caption{Detector module with light concentrator. The scanned sensors are highlighted by the rectangular frame. }
\label{fig:F2F3F4}
\end{figure}
\noindent The sensors were scanned in three different ways. In the first two setups, all three sensors were scanned at once, with and without light concentrator. In order to test the behaviour of the collection efficiency in dependence of the incident beam angle, each sensor was scanned separately with light concentrator and an incident beam angle of about $15^\circ$.
\section{Data Acquisition}
\noindent For the data acquisition, a LeCroy 735Zi WavePro digital oscilloscope was used. Three channels were used to acquire the signal, while the fourth one was used as trigger input. \\
The scope of the experiment was to extract the pulse height from the signal of the respective SiPM. The amplitude of the signal was measured by acquiring the minimum of each waveform during the acquisition window of 200 ns. To achieve good statistics, 1000 samples were taken per position of the photon source for each of the three channels respectively. The oscilloscope calculated the mean and standard deviation of 1000 samples of the amplitude. The acquired data for each channel was background corrected and then added up. The data is referred to as $\langle a\rangle_{LC}$ and $\langle a\rangle_{noLC}$ for the mean amplitude with and without light concentrator respectively. \\%dieser Absatz its 1:1 aus der Diplomarbeit übernommen und sollte zitiert bzw. abgeändert werden!
These two data values (per channel) were saved into a text file, together with information about the coordinates of the beam position. \\
Taking into account the number of data points that need to be acquired during the scans, it is obvious that an automation routine is beneficial. Such a routine was created with LabVIEW and regulates the beam spot movement by the motors as well as the data acquisition by the oscilloscope and the saving of the data. \\
Figure~\ref{fig:DAQ} shows a snapshot of the data acquisition with the oscilloscope.
\begin{figure}[hbt]
\centering
\includegraphics[width=\columnwidth,keepaspectratio]{OsziLive1.pdf}
\caption{Data acquisition with trigger and SiPM signals. Due to the beam diameter of about $108\;\mu$m (FWHM), only one SiPM sends a signal at a time, represented here by sensor F3. }
\label{fig:DAQ}
\end{figure}
\section{Results}
\subsection{Qualitative analysis}
\noindent The data, acquired during the scans was transformed into two dimensional histograms, using routines based on C++ and ROOT. Figure~\ref{fig:TripleScan}
shows the two dimensional histograms from a top view. It is possible to clearly distinguish between the original sensitive area and the enhanced sensitive area when the light concentrator is applied. Also, the reduced collection efficiency due to an incident beam angle is evident in~\ref{fig:TopViewc}. \begin{figure}[hbt]
\begin{center}
\addtolength{\belowcaptionskip}{5pt}
\begin{subfigure}[b]{1.0\columnwidth}
\centering
\includegraphics[width=\columnwidth,keepaspectratio]{noLC.pdf}
\caption{Mean intensity without light concentrator}
\label{fig:TopViewa}
\end{subfigure}
\begin{subfigure}[b]{1.0\columnwidth}
\centering
\includegraphics[width=\columnwidth,keepaspectratio]{LC.pdf}
\caption{Mean intensity with light concentrator}
\label{fig:TopViewb}
\end{subfigure}
\begin{subfigure}[b]{1.0\columnwidth}
\centering
\includegraphics[width=\columnwidth,keepaspectratio]{AngleLC.pdf}
\caption{Mean intensity with light concentrator and beam angle of about $15^{\circ}$}
\label{fig:TopViewc}
\end{subfigure}
\addtolength{\abovecaptionskip}{-10pt}
\caption{Two-dimensional histogram of the scan data for the 3 sensors (a) without LC, (b) with LC and (c) with LC and an incident beam angle of about $15^{\circ}$. The colour scheme gives the mean intensity of signal height of the SiPMs in mV. }
\label{fig:TripleScan}
\end{center}
\end{figure}
\newline
\noindent As can be seen in figure~\ref{fig:TripleScan}, it can be distinguished between active areas and the areas where no photons get detected. One reason for the inactive area is the finite rim which separates the funnels from each other. At these areas, photons get reflected. Another reason is that the sensors were not soldered in perfect alignment, resulting in an offset between the exit area of the light concentrator and the active area of the SiPMs. Figure~\ref{fig:qualitative} shows a comparison between the two dimensional histograms and microscope photos of the respective channels, illustrating the offset of the sensors in relation to the light concentrator.
\begin{figure}[hbt]
\centering
\includegraphics[width=\columnwidth,keepaspectratio]{qualitativeAN.pdf}
\caption{Histogram of mean intensity and photo of the sensors with the light concentrator on top. The arrows indicate areas where no photons get detected as a result of imperfections of the alignment of the sensor array and the light concentrator. }
\label{fig:qualitative}
\end{figure}
\subsection{Collection efficiency}
\noindent The collection efficiency of the light concentrator can be calculated by comparing the data from the scans with light concentrator to the scans without the light concentrator. The collection efficiency $\epsilon_{col}$ of one funnel of the light concentrator is defined by
\begin{equation}
\epsilon_{col}=\frac{n_d}{\alpha \cdot n_{d0}},
\label{equ:CollEffTotal}
\end{equation}
with $n_d$ being the number of photons detected with light concentrator, $n_{d0}$ the number of detected photons without light concentrator and $\alpha=(\frac{7}{3})^2\cdot0.93$ an area factor~\cite{Gruber}. The 0.93 in the area factor $\alpha$ is the geometric fill factor and puts into account the fact that the edges are rounded. \\
The area factor $\alpha$ represents the enlargement of the detection area of a SiPM and is in this specific case $A_{entrance}/A_{exit}$, where $A_{entrance}$ and $A_{exit}$ represent the entrance and exit area respectively. The collection efficiency $\epsilon_{col}$ was calculated, using the following equation for a certain funnel:\\
\begin{equation}
\epsilon_{col}=\frac{\displaystyle\sum \langle a\rangle_{LC}}{\displaystyle\sum \langle a\rangle_{{noLC}} \cdot \alpha}\\
\label{equ:CollEffSingle}
\end{equation}
\noindent \\Table~\ref{tab:results} shows the results for the collection efficiency for each sensor with incident beam angles of $0^\circ$ and $15^\circ$ respectively. The mean collection efficiency is also given.
\begin{table}[htbp]
\begin{center}
\begin{tabular}{|l|c|c|}
\hline
\multicolumn{1}{|c}{Channel} &\multicolumn{1}{|c|}{Angle} &\multicolumn{1}{|c|}{Collection Efficiency {\boldmath$\epsilon_{col}$}} \\ \hline
F2 & $0^\circ$ & $88.6\;\% $ \\
F3 & $0^\circ$ & $83.4\;\%$ \\
F4 & $0^\circ$ & $86.0\;\%$ \\ \hline
Mean & $0^\circ$ & $86.0\;\%$ $(\sigma=2.6\;\%)$ \\ \hline
F2 & $15^\circ$ & $56.8\;\%$ \\
F3 & $15^\circ$ & $55.4\;\%$ \\
F4 & $15^\circ$ & $58.4\;\%$ \\ \hline
Mean & $15^\circ$ & $56.7\;\%$ $(\sigma=1.5\;\%)$\\ \hline
\end{tabular}
\end{center}
\caption{Collection efficiencies for the evaluated three channels at two different photon incident angles. Standard deviations of the collection efficiencies are also shown, indicating the fluctuations of the collection efficiency funnel by funnel. }
\label{tab:results}
\end{table}
\subsection{Comparison to simulations}
\noindent Comparing the measured mean values with simulations of the collection efficiency of the light concentrator shows that the results are in good agreement with the simulations. The simulated collection efficiency for a light concentrator with a funnel length of $4.5$ mm and an incident beam angle perpendicular to the detector surface is about $86\;\%$. The mean of the measured collection efficiency for the light concentrator with an incident beam angle of $0^\circ$ is also about $86\;\%$. Applying an incident beam angle of $15^\circ$ results in a mean collection efficiency of about $57\;\%$, compared to the simulation value of $61\;\%$. Figure~\ref{fig:simulation} shows the results of the simulation for the light concentrator, which was done previously by the authors~\cite{Ahmed}. The figure displays the collection efficiency for different funnel lengths. The simulated collection efficiencies are given in dependence of the incident beam angle.
\begin{figure}[hbt]
\centering
\includegraphics[width=\columnwidth,keepaspectratio]{SimulationAhmed.pdf}
\caption{Simulation of the collection efficiency in dependence of the incident beam angle and different funnel lengths~\cite{Ahmed}. }
\label{fig:simulation}
\end{figure}
\section{Conclusion and outlook}
\noindent A prototype of a position sensitive SiPM array with a light concentrator was tested in order to evaluate the collection efficiency by scanning with a narrowly-focused LED light. The scans were performed with a light source of a beam spot diameter of $108\pm4\;\mu$m and a stepping size of $100\;\mu$m. These parameters have been improved significantly to earlier tests, giving a more detailed picture of the collection efficiency and uniformity.
In addition, the performance of the light concentrator collection efficiency was tested for two different incident light beam angles, $0^\circ$ and $15^\circ$.
The simulation agrees well with the data and can be used to further optimise the geometry of the light concentrator.\\
Ideas to optimise the detector include better alignment of the sensors to the concentrator or a slightly narrower exit area in order to remove the gaps in-between and to develop a different kind of light concentrator with plexiglas cones instead of a metal grid.
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"redpajama_set_name": "RedPajamaArXiv"
}
| 2,997
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\section{Introduction}
We define finite {\it toric partial orders} or {\it toric posets}, which are cyclic analogues of partial orders, but differ from an established notion of {\it partial cyclic orders} already in the literature; see Remark~\ref{disambiguation-remark} below.
Toric posets can be defined in combinatorial geometric ways
that are analogous to partial orders or posets:
\begin{enumerate}
\item[$\bullet$] Posets on a finite set $V$ correspond to open polyhedral cones that arise as chambers in {\it graphic hyperplane arrangements} in $\mathbb{R}^V$;
toric posets correspond to chambers occurring within
{\it graphic toric hyperplane arrangements} in the quotient space $\mathbb{R}^V/\mathbb{Z}^V$.
\item[$\bullet$] Posets correspond to {\it transitive closures} of acyclic orientations of graphs; toric posets correspond to a
notion of {\it toric transitive closures} of acyclic orientations.
\item[$\bullet$] Both transitive closure and toric transitive closure
will turn out to be {\it convex closures}, so that there is a notion of {\it toric Hasse diagram}
for a toric poset, like the Hasse diagram of a poset.
\end{enumerate}
We next make this more precise,
indicating where the main results will be proven.
\subsection{Posets geometrically}
We first recall
(e.g. from Stanley \cite{Stanley:73},
Greene and Zaslavsky \cite[\S 7]{Greene:83},
Postnikov, Reiner and Williams \cite[\S\S 3.3-3.4]{Postnikov:08})
geometric features of posets, specifically their relations
to graphic hyperplane arrangements and acyclic orientations, emphasizing
notions with toric counterparts.
Let $V$ be a finite set of cardinality $|V|=n$; often we will choose $V=[n]:=\{1,2,\ldots,n\}$. One can think of a {\it partially ordered
set} or {\it poset} $P$ on $V$ as a binary relation $i <_P j$ which is
\begin{itemize}
\item {\it irreflexive}: $ i\not <_P i$,
\item {\it antisymmetric}: $i <_P j$ implies $ j \not <_P i $, and
\item {\it transitive}: $i <_P j$ and $j <_P k$ implies $ i <_P k$.
\end{itemize}
However, one can also identify $P$ with a certain {\it open polyhedral cone} in $\mathbb{R}^V$
\begin{equation}
\label{cone-of-a-poset-defn}
c=c(P):=\{x \in \mathbb{R}^V: x_i < x_j \text{ if }i <_P j\}.
\end{equation}
Note that the cone $c$ determines the poset $P=P(c)$ as follows: $i <_P j$ if and only $x_i < x_j$ for all $x$ in $c$.
Each such cone $c$ also arises as a connected component in the complement
of at least one {\it graphic hyperplane arrangement} for a graph $G$, and
often arises in several such arrangements, as explained below.
Given a simple graph $G=(V,E)$,
the {\it graphic arrangement} $\mathcal{A}(G)$
is the union of all hyperplanes in $\mathbb{R}^V$ of the form
$x_i=x_j$ where $\{i,j\}$ is in $E$. Each point $x=(x_1,\ldots,x_n)$ in the {\it complement}
$\mathbb{R}^V {-} \mathcal{A}(G)$ determines an {\it acyclic orientation} $\omega(x)$
of the edge set $E$: for an edge $\{i,j\}$ in $E$, since
$x_i \neq x_j$, either
\begin{itemize}
\item $x_i < x_j$ and $\omega(x)$ directs $i \rightarrow j$, or
\item $x_j < x_i$ and $\omega(x)$ directs $j \rightarrow i$.
\end{itemize}
It is easily seen that the fibers of this map $\alpha_G: x \longmapsto \omega(x)$
are the connected components of the complement $\mathbb{R}^V {-} \mathcal{A}(G)$,
which are open polyhedral cones called {\it chambers}. Thus the map $\alpha_G$
induces a bijection between the set $\Acyc(G)$
of all acyclic orientations $\omega$
of $G$ and the set $\Chambers \mathcal{A}(G)$ of chambers $c$ of $\mathcal{A}(G)$:
\begin{equation}
\label{alpha-diagram}
\xymatrix{
\mathbb{R}^V {-} \mathcal{A}(G) \ar@{>>}[dr] \ar[rr]^{\alpha_G} & & \Acyc(G) \\
& \Chambers \mathcal{A}(G) \ar@{.>}[ur] &\\
}
\end{equation}
These two sets are well-known \cite[Theorem 7.1]{Greene:83}, \cite{Stanley:73}
to have cardinality
$$
|\Acyc(G)|=|\Chambers \mathcal{A}(G)|=T_G(2,0)
$$
where $T_G(x,y)$ is the {\it Tutte polynomial} of $G$ \cite{Tutte:54}.
Posets are also determined by their extensions to
{\it total orders} $w_1 < \cdots < w_n$, which are indexed by
permutations $w=(w_1,\ldots,w_n)$ of $V$. The total orders index the chambers
$$
c_w :=\{x \in \mathbb{R}^V: x_{w_1} < x_{w_2} < \cdots < x_{w_n}\}
$$
in the complement of the {\it complete graphic arrangement $\mathcal{A}(K_V)$},
also known as the {\it reflection arrangement of type $A_{n-1}$} or {\it braid arrangement}.
Given a poset $P$, its set $\mathcal{L}(P)$ of all {\it linear extensions} or
{\it extensions to a total order}
has the property that
$$
\overline{c(P)} = \bigcup_{w \in \mathcal{L}(P)} \overline{c}_w
$$
where $\overline{(\cdot)}$ denotes topological closure. Thus when one
{\it fixes} the graph $G$, chambers $c$ (or posets $P(c)$)
arising as $\alpha_G^{-1}(\omega)$ for various $\omega$ in $\Acyc(G)$
are determined by their sets $\mathcal{L}(P(c))$ of linear extensions.
The same poset $P$ or chamber $c=c(P)$ generally
arises in many graphic arrangements $\mathcal{A}(G)$,
as one varies the graph $G$, leading to
ambiguity in its labeling by a pair $(G,\omega)$ with
$\omega$ in $\Acyc(G)$. Nevertheless, this ambiguity
is well-controlled, in that there are two canonical choices
$
(\bar{G}(P),\bar{\omega}(P))
$
and
$
(\hat{G}^{\Hasse}(P),\omega^{\Hasse}(P))
$
with the following properties.
\begin{enumerate}
\item[$\bullet$] A graph $G$ has $c(P)$ occurring
in $\Chambers \mathcal{A}(G)$
if and only if
$
\hat{G}^{\Hasse}(P) \subseteq G \subseteq \bar{G}(P)
$
where $\subseteq$ is inclusion of edge sets.
In this case, $\alpha_G(c(P))=\omega$ where
$\omega$ is the restriction $\bar{\omega}(P)|_G$.
\item[$\bullet$] The map which sends
$(G,\omega) \longmapsto (\bar{G}(P),\bar{\omega}(P))$ is {\it transitive closure}.
It adds into $G$ all edges $\{i,j\}$ which lie on some {\it chain}
(= {\it totally ordered subset}) $C$ of $P$,
and directs $i \rightarrow j$ if $i <_C j$.
Alternatively phrased, transitive closure adds the
directed edge $i \rightarrow j$ to $(G,\omega)$ whenever there is
a directed path from $i$ to $j$ in $(G,\omega)$.
\end{enumerate}
The existence of a unique
{\it inclusion-minimal} choice $(\hat{G}^{\Hasse}(P),\omega^{\Hasse}(P))$,
called the {\it Hasse diagram} for $P$, follows
from this well-known fact \cite{Edelman:82, Edelman:02}:
the {\it transitive closure} $A \longmapsto \bar{A}$
on subsets $A$ of all possible oriented edges
$
\overleftrightarrow{K}_V=\{ (i,j) \in V \times V : i \neq j \},
$
is a {\it convex closure}, meaning that
\begin{equation}
\label{convex-closure-definition}
\text{ for } a \neq b \text{ with }
a,b \not\in \bar{A}\text{ and }
a \in \overline{A \cup \{b\}},
\text{ one has }b \notin \overline{A \cup \{a\}}.
\end{equation}
\subsection{Toric posets}
We do not initially define a toric poset $P$ on the finite set $V$
via some binary (or ternary) relation. Rather we
define it in terms of chambers in a {\it toric graphic arrangement}
$\mathcal{A}_{\tor}(G)=\pi(\mathcal{A}(G))$,
the image of the graphic arrangement $\mathcal{A}(G)$ under the
quotient map $\mathbb{R}^V \overset{\pi}{\rightarrow} \mathbb{R}^V/\mathbb{Z}^V$. These are
important examples of {\it unimodular toric arrangements} discussed by
Novik, Postnikov and Sturmfels in \cite[\S\S 4-5]{Novik:02}; see
also Ehrenborg, Readdy and Slone \cite{Ehrenborg:09}.
\begin{defn}
\label{toric-poset-defn}
A connected component $c$ of the complement
$\mathbb{R}^V/\mathbb{Z}^V {-} \mathcal{A}_{\tor}(G)$ is called a {\it toric chamber} for $G$;
denote by $\Chambers \mathcal{A}_{\tor}(G)$
the set of all toric chambers of $\mathcal{A}_{\tor}(G)$.
A {\it toric poset} $P$ is a set $c$ that arises as a toric chamber for
at least one graph $G$. We will write $P=P(c)$ and $c=c(P)$, depending
upon the context.
\end{defn}
\begin{ex}
When $n=2$, so $V=\{1,2\}$,
there are only two simple graphs $G=(V,E)$,
a graph $G_0$ with no edges and the complete graph $K_2$ with
a single edge $\{1,2\}$.
For both such graphs, the torus $\mathbb{R}^2/\mathbb{Z}^2$ remains connected
after removing the arrangement $\mathcal{A}_{\tor}(G)$, and hence
they each have only one toric chamber; call these chambers $c_0(=\mathbb{R}^2/\mathbb{Z}^2)$
for the graph $G_0$, and $c(=\mathbb{R}^2/\mathbb{Z}^2{-} \{x_1=x_2\})$
for the graph $K_2$. They represent two different toric posets $P(c_0)$
and $P(c)$,
even though their topological closures $\bar{c}=\bar{c}_0(=c_0)=\mathbb{R}^2/\mathbb{Z}^2$
are the same.
\end{ex}
A point $x$ in $\mathbb{R}^V/\mathbb{Z}^V$ does not have uniquely defined coordinates
$(x_1,\ldots,x_n)$. However, it is well-defined to speak of the
{\it fractional part} $\modone{x_i}$,
that is, the unique representative of the class of $x_i$ in $\mathbb{R}/\mathbb{Z}$ that
lies in $[0,1)$. Therefore a point $x$ in $\mathbb{R}^V/\mathbb{Z}^V {-} \mathcal{A}_{\tor}(G)$,
still induces an acyclic orientation $\omega(x)$ of $G$, as follows:
for each edge $\{i,j\}$ in $E$, since
$x_i \neq x_j \bmod{\mathbb{Z}}$, either
\begin{itemize}
\item $\modone{x_i} < \modone{x_j}$, and $\omega(x)$ directs $i \rightarrow j$, or
\item $\modone{x_j} < \modone{x_i}$, and $\omega(x)$ directs $j \rightarrow i$.
\end{itemize}
Denote this map $x \mapsto \omega(x)$ by
$
\mathbb{R}^V/\mathbb{Z}^V {-} \mathcal{A}_{\tor}(G)
\overset{\bar{\alpha}_G}{\longrightarrow}
\Acyc(G).
$
Unfortunately, two points lying in the same toric chamber
$c$ in $\Chambers_{\tor} \mathcal{A}_{\tor}(G)$ need not
map to the same acyclic orientation under $\bar{\alpha}_G$.
This ambiguity leads one naturally to the
following equivalence relation on acyclic orientations.
\begin{defn}
When two acyclic orientations $\omega$ and $\omega'$ of $G$ differ
only by converting one source vertex of $\omega$ into a sink of $\omega'$,
say that they differ by a \emph{flip}.
The transitive closure of the flip operation generates an
equivalence relation on $\Acyc(G)$ denoted by $\equiv$.
\end{defn}
A thorough investigation of this source-to-sink flip operation and
equivalence relation was
undertaken by Pretzel in~\cite{Pretzel:86}, and
studied earlier by Mosesjan \cite{Mosesjan:72}.
It has also appeared at other times in various contexts\footnote{
Pretzel called the source-to-sink flip {\it pushing down maximal
vertices}; in~\cite{Macauley:11}, it was called a \emph{click}.
In the category of representations of a quiver, it is
related to Bernstein, Gelfand and Ponomarev's
{\it reflection functors} \cite{Bernstein:73}.}
in the literature~\cite{Chen:10,Eriksson:09,Macauley:11,Speyer:09}.
Its relation to geometry of toric chambers $c=c(P)$ or
toric posets $P=P(c)$ is our first main result,
proven in \S~\ref{geometry-section}.
\begin{thm}
\label{geometric-interpretation-theorem}
The map $\bar{\alpha}_G$ induces a bijection
between $\Chambers \mathcal{A}_{\tor}(G)$ and $\Acyc(G)/\!\!\equiv$ as follows:
\begin{equation}
\label{alpha-floor-diagram}
\xymatrix{
\mathbb{R}^V/\mathbb{Z}^V {-} \mathcal{A}_{\tor}(G)
\ar@{>>}[d] \ar[r]^{\bar{\alpha}_G} & \Acyc(G) \ar@{>>}[d] \\
\Chambers \mathcal{A}_{\tor}(G) \ar@{.>}[r]_{\bar{\alpha}_G} & \Acyc(G)/\!\!\equiv
}
\end{equation}
In other words, two points $x,x'$ in $\mathbb{R}^V/\mathbb{Z}^V {-} \mathcal{A}_{\tor}(G)$
have $\bar{\alpha}_G(x)\equiv\bar{\alpha}_G(x')$ if and only if
$x,x'$ lie in the same toric chamber $c$ in $\Chambers \mathcal{A}_{\tor}(G)$.
\end{thm}
\noindent
The two sets $\Chambers \mathcal{A}_{\tor}(G)$ and $\Acyc(G)/\!\!\equiv$ appearing
in the theorem are known to have cardinality
$$
|\Acyc(G) / \!\! \equiv |=|\Chambers \mathcal{A}_{\tor}(G)|=T_G(1,0)
$$
where $T_G(x,y)$ is the Tutte polynomial of $G$;
see \cite{Macauley:08} and \cite[Theorem 4.1]{Novik:02}.
\begin{ex}
A {\it tree} $G$ on $n$ vertices has Tutte polynomial
$T_G(x,y)=x^{n-1}$. It will have $T(2,0)=2^{n-1}$
acyclic orientations $\omega$ and induced partial orders,
but only $T(1,0)=1$ toric chamber or toric partial order:
any two acyclic orientations of a tree are equivalent by a
sequence of source-to-sink moves.
\end{ex}
\begin{ex}
\label{four-cycle-toric-posets-example}
As a less drastic example, consider
$V=\{1,2,3,4\}$ and $G=(V,E)$ this graph:
$$
\tiny{
\xymatrix{
1\ar@{-}[r] & 2\ar@{-}[d] \\
3\ar@{-}[u]& 4\ar@{-}[l]
}
}
$$
It has Tutte polynomial $T_G(x,y)=x^3+x^2+x+y$, and hence
has $T_G(2,0)=2^3+2^2+2+0=14$ acyclic orientations $\omega$.
These $\omega$ fall into $T_G(1,0)=1^3+1^2+1+0=3$ different $\equiv$-classes
$[\omega]$, having cardinalities $4,4,6$, respectively,
corresponding to three different toric posets $P_i$
or chambers $c_i$ in $\Chambers \mathcal{A}_{\tor}(G)$:
$$
\tiny{
\begin{array}{rcccc}
P_1:
&
\xymatrix{
4 & \\
& 3 \ar[ul] \\
& 2 \ar[u] \\
1\ar[ur] \ar[uuu]
}
&
\xymatrix{
1 & \\
& 4 \ar[ul] \\
& 3 \ar[u] \\
2\ar[ur] \ar[uuu]
}
&
\xymatrix{
2 & \\
& 1 \ar[ul] \\
& 4 \ar[u] \\
3\ar[ur] \ar[uuu]
}
&
\xymatrix{
3 & \\
& 2 \ar[ul] \\
& 1 \ar[u] \\
4\ar[ur] \ar[uuu]
}
\\
& & & & \\
& & & & \\
& & & & \\
P_2:
&
\xymatrix{
1 & \\
& 2 \ar[ul] \\
& 3 \ar[u] \\
4\ar[ur] \ar[uuu]
}
&
\xymatrix{
2 & \\
& 3 \ar[ul] \\
& 4 \ar[u] \\
1\ar[ur] \ar[uuu]
}
&
\xymatrix{
3 & \\
& 4 \ar[ul] \\
& 1 \ar[u] \\
2\ar[ur] \ar[uuu]
}
&
\xymatrix{
4 & \\
& 1 \ar[ul] \\
& 2 \ar[u] \\
3\ar[ur] \ar[uuu]
}
\\
& & & & \\
& & & & \\
& & & & \\
P_3:
&
\xymatrix{
& 1 & \\
2\ar[ur]& &4\ar[ul] \\
& 3\ar[ur]\ar[ul]&
}
&
\xymatrix{
2 & 4 \\
1\ar[u]\ar[ur] & 3\ar[u]\ar[ul]
}
&
\xymatrix{
& 2 & \\
1\ar[ur]& &3\ar[ul] \\
& 4\ar[ur]\ar[ul]&
}
& \\
& &
\xymatrix{
& 3 & \\
2\ar[ur]& &4\ar[ul] \\
& 1\ar[ur]\ar[ul]&
}
&
\xymatrix{
1 & 3 \\
2\ar[u]\ar[ur] & 4\ar[u]\ar[ul]
}
&
\xymatrix{
& 4 & \\
1\ar[ur]& &3\ar[ul] \\
& 2\ar[ur]\ar[ul]&
}
\end{array}
}
$$
\end{ex}
{\it Toric total orders} (see \S~\ref{total-orders-section})
are indexed by the $(n-1)!$ cyclic equivalence classes of permutations
\begin{equation}
\begin{array}{rcl}
\label{typical-cyclic-permutation}
[w]:=[(w_1,w_2,\ldots,w_n)]=\left\{\right. &(w_1,w_2,\ldots,w_{n-1},w_n), & \\
&(w_2,\ldots,w_{n-1},w_n,w_1), &\\
&\vdots&\\
&(w_n,w_1,w_2,\ldots,w_{n-1}) &\left.\right\}
\end{array}
\end{equation}
and correspond to the toric chambers $c_{[w]}$
in the complement of the {\it toric complete graphic arrangement $\mathcal{A}_{\tor}(K_V)$}.
For a particular toric poset $P=P(c)$,
one says that $[w]$ is a {\it toric total
extension} of $P$ if $c_{[w]} \subseteq c$. Denote by $\mathcal{L}_{\tor}(P)$ the
set of all such toric total extensions $[w]$ of $P$.
Although it is possible (see Example~\ref{lack-of-determination-by-extensions-example} below)
for two different toric posets $P$ to have the same set $\mathcal{L}_{\tor}(P)$,
the following assertion (combining
Proposition~\ref{edge-subgraph-chamber-refinement} and
Corollary~\ref{determination-by-total-cyclic-extensions} below) still holds.
\begin{prop}
When one {\it fixes} the graph $G$, the toric chamber $c$
(or its poset $P=P(c)$) for which $\bar{\alpha}_G(c)=[\omega]$
is completely determined by its topological
closure $\overline{c}$.
Furthermore one has
$
\overline{c} = \bigcup_{w \in \mathcal{L}_{\tor}(P)} \overline{c}_{[w]}.
$
so that this closure depends only on the set of toric total extensions
$\mathcal{L}_{\tor}(P)$.
\end{prop}
\begin{ex}
The graph $G$ from Example~\ref{four-cycle-toric-posets-example}
and its three toric posets $P_1,P_2,P_3$ partition
the $(4-1)!=6$ different toric total orders on $V=\{1,2,3,4\}$
into their sets of toric total extensions $\mathcal{L}_{\tor}(P_i)$ as follows:
$$
\begin{aligned}
\mathcal{L}_{\tor}(P_1)
& = \{ [(1,2,3,4)] \},\\
\mathcal{L}_{\tor}(P_2)
& = \{ [(1,4,3,2)] \},\\
\mathcal{L}_{\tor}(P_3)
& = \{ [(1,2,4,3)],[(1,3,2,4)], [(1,3,4,2)],[(1,4,2,3)]\}.
\end{aligned}
$$
\end{ex}
As with posets, the same toric poset $P=P(c)$ arises as a chamber $c$
in {\it many} toric graphic arrangements $\mathcal{A}_{\tor}(G)$.
However, as with posets, this ambiguity is well-controlled,
in that there are two canonical choices of equivalence classes
$
(\bar{G}^{\tor}(P), [\bar{\omega}^{\tor}(P)])
$
and
$
(\hat{G}^{\torHasse}(P), [\omega^{\torHasse}(P)])
$
with the following properties.
\begin{enumerate}
\item[$\bullet$] A graph $G$ has $c(P)$ occurring in
$\Chambers \mathcal{A}_{\tor}(G)$ if and only if
$$
\hat{G}^{\torHasse}(P) \subseteq G \subseteq \bar{G}^{\tor}(P)
$$
where $\subseteq$ is inclusion of edges. In this case,
if $\bar{\alpha}_G(c(P))=[\omega]$, then
$\omega$ can be taken to be the
restriction to $G$ of a particular orientation
in the class $[\bar{\omega}^{\tor}(P)]$.
\item[$\bullet$] The map which sends
$(G,\omega) \longmapsto (\bar{G}^{\tor}, \bar{\omega}^{\tor})$ may be described by what
will be called (in \S~\ref{transitivity-section}) {\it toric transitive closure}:
one adds into $G$ all edges $\{i,j\}$ which lie on some {\it toric chain} $C$ in $P$.
Here a toric chain (see \S~\ref{chains-section}) is a subset $C \subset V$
which is totally ordered in {\it every} poset associated with
an orientation in the class $[\omega]$.
One directs $i \rightarrow j$ if there is a
{\it toric directed path} from $i$ to $j$ in $(G,\omega)$, as defined
in \S~\ref{directed-paths-section} below.
Alternatively phrased, toric transitive closure will add the
directed edge $i \rightarrow j$ to $(G,\omega)$ whenever there is
a toric directed path from $i$ to $j$ in $(G,\omega)$.
\end{enumerate}
The existence of the unique
{\it inclusion-minimal} choice $(\hat{G}^{\torHasse}(P),[\omega^{\torHasse}(P)])$,
which we will call the {\it toric Hasse diagram} of $P$, follows
from our second main result, proven in \S~\ref{convex-closure-section}.
\begin{thm}
\label{toric-convex-closure-theorem}
Considered as a {\it closure operation} $A \longmapsto \bar{A}^{\tor}$
on subsets $A$ of all possible oriented edges
$
\overleftrightarrow{K}_V=\{ (i,j) \in V \times V : i \neq j \},
$
toric transitive closure is a {\it convex closure}, that is,
it satisfies \eqref{convex-closure-definition} above.
\end{thm}
\begin{ex}
The toric poset $P_1=P(c_1)$ from Example~\ref{four-cycle-toric-posets-example}
appears as a chamber $c_1$ in $\Chambers \mathcal{A}_{\tor}(G_i)$ for
exactly four graphs $G_1,G_2,G_3,G_4$, each shown below with an
orientation $\omega_i$ such that $\bar{\alpha}_{G_i}(c_1)=[\omega_i]$.
$$
\tiny{
\xymatrix{
4 & \\
& 3 \ar[ul] \\
& 2 \ar[u] \\
1\ar[ur] \ar[uuu]
}
\qquad
\xymatrix{
4 & \\
& 3 \ar[ul] \\
& 2 \ar[u]\ar[uul] \\
1\ar[ur] \ar[uuu]
}
\qquad
\xymatrix{
4 & \\
& 3 \ar[ul] \\
& 2 \ar[u] \\
1\ar[ur] \ar[uuu] \ar[uur]
}
\qquad
\xymatrix{
4 & \\
& 3 \ar[ul] \\
& 2 \ar[u] \ar[uul]\\
1\ar[ur] \ar[uuu] \ar[uur]
}
}
$$
For any of these four pairs $(G_i,\omega_i)$ with $i=1,2,3,4$, one has that the
leftmost pair is its Hasse diagram $(\hat{G_i}^{\torHasse},\omega_i^{\torHasse})$,
and the rightmost pair is its toric transitive closure
$(\bar{G}^{\tor}_i,\bar{\omega}_i^{\tor})$.
\end{ex}
We close this Introduction with two remarks, one on terminology,
the other giving further motivation.
\begin{rem}
\label{disambiguation-remark}
Aside from the connection to toric hyperplane arrangements,
we have chosen the name ``toric partial order'',
as opposed to the arguably more natural term
``cyclic partial order'', because the latter is
easily confused with {\it partial cyclic orders}, the following
pre-existing concept in the literature, going
back at least as far as Megiddo~\cite{Megiddo:76}.
\begin{defn}
A \emph{partial cyclic order} on $V$ is a ternary relation $T \subseteq V\times V \times V$
that is
\begin{itemize}
\item \emph{antisymmetric}: If $(i,j,k)\in T$ then $(k,j,i)\not\in T$;
\item \emph{transitive}: If $(i,j,k)\in T$ and $(i,k,\ell)\in T$, then
$(i,j,\ell)\in T$;
\item \emph{cyclic}: If $(i,j,k)\in T$, then $(j,k,i)\in T$.
\end{itemize}
\end{defn}
\begin{defn}
\label{total-cyclic-order-defn}
When a partial cyclic order on $V$ is \emph{complete} in the sense that
for every triple $\{i,j,k\} \subseteq V$ of distinct elements,
$T$ contains some permutation of $(i,j,k)$, then $T$ is called a \emph{total cyclic order}. A total cyclic order on $V$ is easily seen to be the same a toric total order: specify a cyclic equivalence class $[w]$ as
in \eqref{typical-cyclic-permutation}, and then check that $[w]$ is
determined by knowing its restrictions $[w|_{\{i,j,k\}}]$
for all triples $\{i,j,k\}$.
\end{defn}
Partial cyclic orders have been widely studied, and
have some interesting features not shared by ordinary partial
orders. For example, every partial order can be extended to a total
order, but not every partial cyclic order can be extended
to a total cyclic order; an example of this on $13$ vertices
is given in \cite{Megiddo:76}.
\end{rem}
\begin{rem}
We mention a further analogy between posets and toric posets,
related to Coxeter groups, that was
one of our motivations for formalizing this concept.
Recall \cite{Bourbaki:02} that a {\it Coxeter system} $(W,S)$ is a group $W$
with generating set $S=\{s_1,\ldots,s_n\}$ having presentation
$
W= \langle S : (s_i s_j)^{m_{i,j}}=e \rangle
$
for some $m_{i,j}$ in $\{1,2,3,\ldots\} \cup \{\infty\}$,
where $m_{i,i}=1$ for all $i$ and $m_{i,j} \geq 2$ for $i \neq j$.
Associated to $(W,S)$ is the {\it Coxeter graph} on vertex set $S$
with an edge $\{s_i,s_j\}$ labeled by $m_{i,j}$ whenever $m_{i,j} > 2$, so that
$s_i,s_j$ do not commute; ignoring the edge labels, we will call this
the unlabeled Coxeter graph.
A {\it Coxeter element} for $(W,S)$ is an element of
the form $s_{w_1} s_{w_2} \cdots s_{w_n}$ for some choice of a total
order $w$ on $S$.
\begin{thm} Fix a Coxeter system $(W,S)$ with unlabeled Coxeter
graph $G$, and consider the map sending an
acyclic orientation $\omega$ in $\Acyc(G)$ having poset $P=\alpha_G(\omega)$
to the Coxeter element $s_{w_1} s_{w_2} \cdots s_{w_n}$ for any choice
of a linear extension $w$ in $\mathcal{L}(P)$.
\begin{enumerate}
\item[(i)]
This map is well-defined, and induces a bijection
(see \cite[\S V.6]{Bourbaki:02} and \cite{Cartier:69})
$$
\Acyc(G)
\longleftrightarrow
\{ \text{ Coxeter elements for }(W,S) \,\, \}.
$$
\item[(ii)]
It also induces a well-defined map on the toric equivalence classes
$[\omega]$ to the {\bf $W$-conjugacy classes} of all Coxeter elements,
and gives a bijection (see \cite{Eriksson:09, Macauley:08, Macauley:11, Shi:97}
and \cite[Remark 5.5]{Novik:02})
$$
\Acyc(G)/\!\!\equiv
\quad \longleftrightarrow \quad
\{ W\text{-conjugacy classes of Coxeter elements for }(W,S) \}.
$$
\end{enumerate}
\end{thm}
\noindent
We believe toric partial orders will play a key role in resolving
more questions about $W$-conjugacy classes.
\end{rem}
\section{Toric arrangements and
proof of Theorem~\ref{geometric-interpretation-theorem}}
\label{geometry-section}
Recall the statement of the theorem.
\vskip.1in
\noindent
{\bf Theorem~\ref{geometric-interpretation-theorem}.}
{\it
The map $\bar{\alpha}_G$ induces a bijection
between $\Chambers \mathcal{A}_{\tor}(G)$ and $\Acyc(G)/\!\!\equiv$ as follows:
$$
\xymatrix{
\mathbb{R}^V/\mathbb{Z}^V {-} \mathcal{A}_{\tor}(G)
\ar@{>>}[d] \ar[r]^{\bar{\alpha}_G} & \Acyc(G) \ar@{>>}[d] \\
\Chambers \mathcal{A}_{\tor}(G) \ar@{.>}[r]_{\bar{\alpha}_G} & \Acyc(G)/\!\!\equiv
}
$$
In other words, two points $x,x'$ in $\mathbb{R}^V/\mathbb{Z}^V {-} \mathcal{A}_{\tor}(G)$
have $\bar{\alpha}_G(x)\equiv\bar{\alpha}_G(x')$ if and only if
$x,x'$ lie in the same toric chamber $c$ in $\Chambers \mathcal{A}_{\tor}(G)$.
}
\vskip.1in
\noindent
Before embarking on the proof, we introduce one further geometric object
intimately connected with
\begin{itemize}
\item
the graphic arrangement $\mathcal{A}(G)=\bigcup_{\{i,j\} \in E} \{x \in \mathbb{R}^V: x_i = x_j\} \subset \mathbb{R}^V$, and
\item
the toric graphic arrangement
$\mathcal{A}_{\tor}(G)=\pi(\mathcal{A}(G))$, its image
under $\mathbb{R}^V \overset{\pi}{\rightarrow} \mathbb{R}^V/\mathbb{Z}^V$.
\end{itemize}
\begin{defn}
Define the \emph{affine graphic arrangement} in $\mathbb{R}^V$ by
\begin{equation}
\label{affine-arrangement-defn}
\mathcal{A}_{\aff}(G) := \pi^{-1}(\mathcal{A}_{\tor}(G)) = \pi^{-1}(\pi(\mathcal{A}(G)))
=\bigcup_{\substack{ \{i,j\} \in E\\ k \in \mathbb{Z}}} \{x \in \mathbb{R}^V: x_i = x_j+k \}.
\end{equation}
Call the connected components
$\hat{c}$ of the complement $\mathbb{R}^V {-}\mathcal{A}_{\aff}(G)$ {\it affine
chambers}, and denote the set of all such chambers $\Chambers \mathcal{A}_{\aff}(G)$.
\end{defn}
The reason for introducing $\mathcal{A}_{\aff}(G)$ and $\Chambers \mathcal{A}_{\aff}(G)$
is the following immediate consequence of the path-lifting
property for $\mathbb{R}^V \overset{\pi}{\rightarrow} \mathbb{R}^V/\mathbb{Z}^V$ as
a (universal) covering map (see e.g. \cite[Chap. 13]{Munkres:75}), along
with the definition \eqref{affine-arrangement-defn} of $\mathcal{A}_{\aff}(G)$
as the full inverse image under $\pi$ of $\mathcal{A}_{\tor}(G)$.
\begin{prop}
\label{path-lifting-prop}
Two points $x,y$ in $\mathbb{R}^V/\mathbb{Z}^V {-} \mathcal{A}_{\tor}(G)$
lie in the same chamber $c$ in $\Chambers\mathcal{A}_{\tor}(G)$ if
and only if they have two lifts $\hat{x}, \hat{y}$ lying in the
same affine chamber $\hat{c}$ in $\Chambers \mathcal{A}_{\aff}(G)$.
\end{prop}
\noindent
The point will be that, since affine chambers $\hat{c}$ are (open) convex
polyhedral regions in $\mathbb{R}^V$, it is sometimes easier to argue about
lifted points $\hat{x}$ rather than $x$ itself.
Our proof of
Theorem~\ref{geometric-interpretation-theorem}
proceeds by showing the map
$
\mathbb{R}^V/\mathbb{Z}^V {-} \mathcal{A}_{\tor}(G)
\overset{\bar{\alpha}_G}{\longrightarrow} \Acyc(G)
$
descends to
\begin{itemize}
\item a well-defined map
$
\Chambers \mathcal{A}_{\tor}(G)
\overset{\bar{\alpha}_G}{\longrightarrow} \Acyc(G)/\!\!\equiv,
$
\item which is surjective,
\item and injective.
\end{itemize}
\subsection{Well-definition}
We must show that when $x, y$ lie in the same toric chamber
$c$ in $\Chambers\mathcal{A}_{\tor}(G)$, then
$\bar{\alpha}_G(x) \equiv \bar{\alpha}_G(y)$.
As in Proposition~\ref{path-lifting-prop}, pick lifts
$\hat{x},\hat{y}$ in $\mathbb{R}^V$ and a path $\hat{\gamma}$ between
them in some affine chamber $\hat{c}$. Because these chambers are
open, one can assume without loss of generality that $\hat{\gamma}$
takes steps in coordinate directions only, and therefore that
$\hat{x}, \hat{y}$ differ in only a single coordinate: say
$\hat{x}_i \neq \hat{y}_i$, but $\hat{x}_j=\hat{y}_j$ for all $j \neq i$.
Furthermore, as $\bar{\alpha}_G(x)$ changes only when a coordinate
of $\hat{x}$ passes through an integer, without loss of generality, one may assume
$$
\begin{aligned}
\modone{\hat{x}_i}&=1-\varepsilon,\\
\modone{\hat{y}_i}&=\varepsilon
\end{aligned}
$$
for some arbitrarily small $\varepsilon > 0$.
Since the points on $\hat{\gamma}$ all avoid $\mathcal{A}_{\aff}(G)$,
and the $i^{th}$ coordinate will pass through $0$ at some point
on the path $\hat{\gamma}$, each of the
coordinates $\hat{x}_j(=\hat{y}_j)$ for indices $j$ with
$\{i,j\}$ in $E$ must have $0 < \modone{\hat{x}_j} < 1$.
Hence one can choose $\varepsilon$
small enough that all $j$ for which $\{i,j\}$ in $E$ satisfy
$$
\left( \modone{\hat{y}_i} = \right) \varepsilon < \modone{\hat{x}_j}
< 1 - \varepsilon \left( = \modone{\hat{x}_i} \right).
$$
One finds that $\bar{\alpha}_G(\hat{x})$ and $\bar{\alpha}_G(\hat{y})$
differ by changing $i$ from sink to a source, so
$\bar{\alpha}_G(\hat{x}) \equiv \bar{\alpha}_G(\hat{y})$,
as desired.
\subsection{Surjectivity}
It suffices to check that the map
$
\mathbb{R}^V/\mathbb{Z}^V {-} \mathcal{A}_{\tor}(G)
\overset{\bar{\alpha}_G}{\longrightarrow} \Acyc(G)
$
is surjective. Given an acyclic orientation $\omega$
of $G$, pick any linear extension $w_1 < \cdots < w_n$ of its associated
partial order $\alpha_G^{-1}(\omega)$ on $V$. Then choose real numbers
$0 < x_{w_1} < \cdots < x_{w_n} < 1$, so that
$$
x=(x_1,\ldots,x_n)=(\modone{x_1},\ldots,\modone{x_n})
$$
and hence $\bar{\alpha}_G(x)=\omega$.
\subsection{Injectivity}
The key to injectivity is the following lemma.
\begin{lem}\label{lem:key}
Suppose $x$ lies in a toric chamber $c$ in $\Chambers \mathcal{A}_{\tor}(G)$,
and $\bar{\alpha}_G(x)=\omega$.
Then for any $\omega' \equiv \omega$, there exists
some $x'$ in the same toric chamber $c$ having $\bar{\alpha}_G(x')=\omega'$.
\end{lem}
\begin{proof}
It suffices to check this when $\omega'$ is obtained from $\omega$ by
changing a source vertex $i$ in $\omega$ to a sink in $\omega'$.
Since $\bar{\alpha}_G(x)=\omega$, one must have for each $j$ with
$\{i,j\}$ in $E$ that
$$
(0 \leq ) \modone{x_i} < \modone{x_j} (< 1).
$$
Lift $x$ to $\hat{x}=(\modone{x_1},\ldots,\modone{x_n})$, and
choose $\varepsilon$ small enough so that each $j$ with $\{i,j\}$ in $E$ has
$
\modone{x_j} < 1-\varepsilon.
$
Define $\hat{y}$ to have all the same coordinates as $\hat{x}$ except for
$\hat{y}_i=-\varepsilon$, so that $\modone{\hat{y}_i}=1-\varepsilon$, and
hence $y:=\pi(\hat{y})$ has $\bar{\alpha}_G(y)=\omega'$ by construction.
Note that the straight-line path $\hat{\gamma}$
from $\hat{x}$ to $\hat{y}$ changes only the $i^{th}$ coordinate,
decreasing it from $\hat{x}_i$ to $\hat{y}_i=-\varepsilon$,
and hence never crosses any of the affine
hyperplanes in $\mathcal{A}_{\aff}(G)$. Therefore $\hat{x},\hat{y}$ lie
in the same affine chamber, and $x,y$ lie in the same toric
chamber $c$.
\end{proof}
Now suppose that points $x,x'$ in two toric chambers $c,c'$
have $\bar{\alpha}_G(x) \equiv \bar{\alpha}_G(x')$, and we must show that
$c=c'$. By Lemma~\ref{lem:key}, without loss of generality one
has $\bar{\alpha}_G(x) = \omega = \bar{\alpha}_G(x')$. Thus one can
lift $x,x'$ to $\hat{x},\hat{x}'$ having $\hat{x}_i,\hat{x}'_i$ in $[0,1)$
for all $i$, and hence
$\alpha_G(\hat{x})=\omega=\alpha_G(\hat{x}')$.
For each edge $\{i,j\}$ in $E$, say directed $i\to j$ in $\omega$,
one has both
$$
\begin{aligned}
0 \leq \hat{x}_i &< \hat{x}_j <1,\\
0 \leq \hat{x}'_i &< \hat{x}'_j <1.
\end{aligned}
$$
Thus every point $\hat{y}$ on the straight-line path
$\hat{\gamma}$ between $\hat{x}$ and $\hat{x}'$ also satisfies
$0 \leq \hat{y}_i < \hat{y}_j < 1$, avoiding all affine
hyperplanes in $\mathcal{A}_{\aff}(G)$. Thus $\hat{x},\hat{x}'$ lie
in the same affine chamber $\hat{c}$, so that $x,x'$ lie
in the same toric chamber, as desired. This completes the proof
of injectivity, and hence the proof of Theorem~\ref{geometric-interpretation-theorem}.$\qed$
\vskip.2in
One corollary to Theorem~\ref{geometric-interpretation-theorem} is a (slightly)
more concrete description of a toric chamber $c$.
\begin{cor}
\label{toric-chamber-as-union-cor}
For a graph $G=(V,E)$ and toric chamber $c$ in $\Chambers \mathcal{A}_{\tor}(G)$
with $\bar{\alpha}_G(c)=[\omega]$, one has
$$
c= \bigcup_{ \omega' \in [\omega] }
\bar{\alpha}_G^{-1}(\omega')
= \bigcup_{ \omega' \in [\omega] }
\{ x \in \mathbb{R}^V/\mathbb{Z}^V: \modone{x_i} < \modone{x_j} \text{ if }\omega'\text{ directs }i \rightarrow j\}.
$$
\end{cor}
\section{Toric extensions}
\label{extensions-section}
Recall that for two (ordinary) posets $P, P'$ on a set $V$, one
says that {\it $P'$ is an extension of $P$} when $i <_P j$ implies
$i <_{P'} j$. It is easily seen how to reformulate this geometrically:
$P'$ is an extension of $P$ if and only one has
an inclusion of their open polyhedral cones
$c(P') \subseteq c(P)$, as defined in \eqref{cone-of-a-poset-defn}.
This motivates the following definition.
\begin{defn}
Given two toric posets $P, P'$ say that {\it $P'$ is a toric
extension of $P$} if one has an inclusion of their open chambers
$c(P') \subseteq c(P)$ within $\mathbb{R}^V/\mathbb{Z}^V$.
\end{defn}
An obvious situation where this can occur is when
one has $G=(V,E)$ and $G'=(V,E')$ two graphs on the same vertex set $V$,
with $G$ an {\it edge-subgraph} of $G'$ in the sense that $E \subseteq E'$,
\begin{prop}
\label{edge-subgraph-chamber-refinement}
Fix $G=(V,E)$ a simple graph.
\begin{enumerate}
\item[(i)] Toric chambers in $\Chambers \mathcal{A}_{\tor}(G)$
are determined by their topological closures:
for any pair of chambers $c_1,c_2$
in $\Chambers \mathcal{A}_{\tor}(G)$, if
$\bar{c}_1=\bar{c}_2$ then $c_1=c_2$.
\item[(ii)]
If $G$ is an edge-subgraph of $G'$, then
$
\bar{c} = \bigcup_{c'} \bar{c}',
$
where the union runs over all toric chambers
$c'$ in $\Chambers \mathcal{A}_{\tor}(G')$ for which
$P(c')$ is a toric extension of $P(c)$.
\end{enumerate}
\end{prop}
\begin{proof}
For (i), first note that any toric chamber $c$ in $\Chambers \mathcal{A}_{\tor}(G)$
has boundary $\bar{c} {-} c$ contained in $\mathcal{A}_{\tor}(G)$.
Now assume two toric chambers $c_1, c_2$ in $\Chambers \mathcal{A}_{\tor}(G)$ have
$\bar{c}_1 = \bar{c}_2$, and we wish to show $c_1=c_2$.
Any point $x$ of $c_1$ has
$x \in c_1 \subseteq \bar{c}_1=\bar{c}_2$.
However, $x$ cannot lie in $\mathcal{A}_{\tor}(G)$ since
$c_1$ is disjoint from $\mathcal{A}_{\tor}(G)$, so $x$ does not lie
in $\bar{c}_2 {-} c_2 \subset \mathcal{A}_{\tor}(G)$ by our first
observation. Hence $x$ lies in $c_2$.
But then $c_1, c_2$ are connected components of
$\mathbb{R}^V / \mathbb{Z}^V {-} \Chambers \mathcal{A}_{\tor}(G)$, sharing the point $x$,
so $c_1=c_2$.
For (ii), we first argue that
\begin{equation}
\label{closure-described-via-lift}
\bar{c}=\pi\left( \overline{\pi^{-1}(c)} \right)
\end{equation}
using the fact that the covering map $\mathbb{R}^V \overset{\pi}{\rightarrow} \mathbb{R}^V/\mathbb{Z}^V$
is locally a homeomorphism. For any point $x$ in $\mathbb{R}^V/\mathbb{Z}^V$ there is
an open neighborhood $U$ which lifts to an open neighborhood $\hat{U}$,
mapping homeomorphically under $\pi$ to $U$. Hence $x$ is the
limit of a sequence $\{x_i\}_{i=1}^\infty$ of points in $c$
if and only if its lift $\hat{x}=\pi|_{\hat{U}}^{-1}(x)$ is a limit of
the sequence of points $\{\pi|_{\hat{U}}^{-1}(x_i)\}_{i=1}^\infty$ in
$\pi^{-1}(c)$. This shows \eqref{closure-described-via-lift}.
Since a toric chamber $c$ has $\pi^{-1}(c)$ given by a union of
affine chambers $\hat{c}$ in $\Chambers \mathcal{A}_{\aff}(G)$, in light of
\eqref{closure-described-via-lift},
it suffices to show that any affine chamber $\hat{c}$ in $\Chambers \mathcal{A}_{\aff}(G)$ has closure $\overline{\hat{c}}$
given by the union of the closures
$\overline{\hat{c}'}$ taken over all affine chambers $\hat{c}'$ in $\Chambers \mathcal{A}_{\aff}(G')$ that satisfy $\hat{c}' \subseteq \hat{c}$.
However, this
is clear since $\hat{c}$ is a polyhedron bounded by hyperplanes taken from $\mathcal{A}_{\aff}(G)$, while $\mathcal{A}_{\aff}(G')$ simply refines this decomposition with more hyperplanes.
\end{proof}
\section{Toric directed paths}
\label{directed-paths-section}
A particular special case of Proposition~\ref{edge-subgraph-chamber-refinement} is worth noting:
every graph $G=(V,E)$ is an edge-subgraph of the {\it complete graph $K_V$}.
As noted in the Introduction, acyclic orientations $\omega$ of $K_V$ correspond to total orders $w_1 < \cdots < w_n$, indexed by permutations $w=(w_1,\ldots,w_n)$ of $V=[n]:=\{1,2,\ldots,n\}$. It is easy to characterize the equivalence relation $\equiv$ on these total orders, and hence the toric chambers $\Chambers \mathcal{A}_{\tor}(K_V)$, in terms of cyclic shifts of these linear orders. However,
it is worthwhile to define this concept is a bit more generally-- it turns out to be crucial in Section~\ref{chains-section}.
\begin{defn}
\label{toric-directed-path-defn}
Given a simple graph $G=(V,E)$ and an acyclic orientation $\omega$ of
$G$, say that a sequence $(i_1,i_2,\ldots,i_m)$ of elements of $V$ forms
a {\it toric directed path in $\omega$} if $(G,\omega)$ contains all of these
edges:
\begin{equation}
\label{toric-directed-path-figure}
\xymatrix{
i_m & \\
&i_{m-1} \ar[ul] \\
&\vdots \ar[u]\\
&i_2 \ar[u] \\
i_1 \ar[uuuu] \ar[ur]&
}
\end{equation}
\end{defn}
\noindent
In particular, for small values of $m$, a toric
directed path in $\omega$
\begin{enumerate}
\item[$\bullet$]
of size $m=2$ is a directed edge $(i_1,i_2)$,
\item[$\bullet$]
of size $m=1$ is a degenerate path $(i_1)$ for any $i_1$ in $V$, and
\item[$\bullet$]
of size $m=0$ is the empty subset $\varnothing \subset V$.
\end{enumerate}
\begin{prop}
\label{flip-preserves-toric-directed-path}
An acyclic orientation $\omega$ of $G$
contains a toric directed path $(i_1,i_2,\ldots,i_m)$ if and only if every
acyclic orientation $\omega'$ in its $\equiv$-equivalence class contains a (unique) toric directed path
$$
(i_\ell,i_{\ell+1},\ldots,i_m,i_1,i_2,\ldots,i_{\ell-1})
$$
which is one of its cyclic shifts, that is, it lies in the cyclic
equivalence class $[(i_1,\ldots,i_m)]$.
\end{prop}
\begin{proof}
A toric directed path $(i_1,i_2,\ldots,i_m)$ has only one source, namely $i_1$,
and only one sink, namely $i_m$. The assertion follows by
checking that the effect of a source-to-sink flip at $i_1$ (resp. $i_m$)
is a cyclic shift to the toric directed path $(i_2,\ldots,i_m,i_1)$ (resp.
$(i_m,i_1,i_2,\ldots,i_{m-1})$).
\end{proof}
\begin{rem}
\label{Coleman-remark}
We point out a reformulation of the sink-to-source equivalence
relation $\equiv$ on $\Acyc(G)$, due to Pretzel \cite{Pretzel:86},
leading to a reformulation of toric directed paths, useful
in Section~\ref{Antichain-section} on toric antichains.
Given a simple graph $G=(V,E)$, say that a cyclic equivalence
class $I=[(i_1,\ldots,i_m)]$ of ordered vertices is a {\it directed
cycle} of $G$ if $m\geq 3$ and $G$ contains all of the (undirected)
edges $\{i_j,i_{j+1}\}_{j=1,2\ldots,m}$,
with subscripts taken modulo $m$. Given such a directed cycle $I$
define \emph{Coleman's $\nu$-function}~\cite{Coleman:89}
$$
\Acyc(G) \overset{\nu_I}{\longrightarrow} \mathbb{Z}
$$
where $\nu_I(\omega)$ for an acyclic orientation $\omega$ of $G$ is
defined to be the number of edges $\{i_j,i_{j+1}\}$ in $I$
which $\omega$ orients $i_j \rightarrow i_{j+1}$
minus the number of edges $\{i_j,i_{j+1}\}$
which $\omega$ orients $i_{j+1} \rightarrow i_j$.
It is easy to see that $\nu_I$ is preserved
by flips, and thus extends in a well-defined manner
to $\equiv$-classes $[\omega]$. In fact, Pretzel~\cite{Pretzel:86}
showed that this is a complete $\equiv$-invariant:
\begin{prop}\label{prop-pretzel}
Fixing the graph $G=(V,E)$, two acyclic orientations $\omega, \omega'$ in $\Acyc(G)$
have $\omega \equiv \omega'$ if and only if
$\nu_I(\omega)=\nu_I(\omega')$ for every directed cycle $I$ of $G$.
\end{prop}
Toric directed paths then have an obvious
characterization in terms of their $\nu_I$ function.
\begin{cor}
\label{cor:nu}
Given a directed cycle in $I=[(i_1,\ldots,i_m)]$ in $G$, an acyclic orientation $\omega$ in $\Acyc(G)$ contains a toric directed path lying in the
cyclic equivalence class $I$ if and only if $\nu_I(\omega)=m-1$.
\end{cor}
\end{rem}
\section{Toric total orders}
\label{total-orders-section}
An important special case of toric directed paths occurs when one
considers acyclic orientations of the complete graph $K_V$. Acyclic
orientations of $K_V$ correspond to permutations $w=(w_1,\ldots,w_n)$
of $V$ (or \emph{total orders}), and always form toric directed paths
in $w$. Hence their toric equivalence classes are the equivalence
classes $[w]$ of permutations up to cyclic shifts, or {\it toric total
orders}. This concept coincides with the pre-existing concept of
{\it total cyclic order} from
Definition~\ref{total-cyclic-order-defn}, even though toric {\it partial}
orders are not the same as {\it partial} cyclic orders. Therefore, we
can use these terms interchangeably.
By Theorem~\ref{geometric-interpretation-theorem}, these toric total
orders $[w]$ index the chambers $c_{[w]}$ in $\Chambers \mathcal{A}_{\tor}(K_V)$.
By Corollary~\ref{toric-chamber-as-union-cor}, one has this more concrete
description of such chambers:
\begin{equation}
\label{more-concrete-total-cyclic-chamber-description}
c_{[w]}= \bigcup_{ i=1 }^n
\{ x \in \mathbb{R}^V/\mathbb{Z}^V:
\modone{x_{w_i}} < \cdots < \modone{x_{w_n}}
< \modone{x_{w_1}} < \cdots < \modone{x_{w_{i-1}}} \}.
\end{equation}
\begin{defn}
\label{total-cyclic-extension-defn}
Given a toric poset $P=P(c)$ on $V$,
say that a toric total order $[w]$ on $V$ is a {\it toric total extension} of $P$ if the toric chamber $c_{[w]}$ of $\Chambers \mathcal{A}_{\tor}(K_V)$ is contained in $c$. Denote by $\mathcal{L}_{\tor}(P)$ the
set of all such toric total extensions $[w]$ of $P$.
\end{defn}
The following corollary is then a special case of Proposition~\ref{edge-subgraph-chamber-refinement}.
\begin{cor}
\label{determination-by-total-cyclic-extensions}
Fix a simple graph $G=(V,E)$.
Then any toric chamber/poset $c=c(P)$ in $\Chambers \mathcal{A}_{\tor}(G)$
has topological closure
$$
\bar{c} = \bigcup_{[w] \in \mathcal{L}_{\tor}(P)} \bar{c}_{[w]}.
$$
and is completely determined by its set $\mathcal{L}_{\tor}(P)$
of toric total extensions:
if $c_1,c_2$ in $\Chambers \mathcal{A}_{\tor}(G)$ have
$\mathcal{L}_{\tor}(P(c_1))=\mathcal{L}_{\tor}(P(c_2))$, then $c_1=c_2$.
\end{cor}
\begin{ex}
\label{lack-of-determination-by-extensions-example}
Corollary~\ref{determination-by-total-cyclic-extensions}
fails when one does {\it not} fix the graph $G$. For example,
when $V=\{1,2,3\}$, all $7$ of the {\it non-complete} graphs
$G \neq K_V=K_3$
share the property that $\Chambers \mathcal{A}_{\tor}(G)$ has only one
chamber $c=c(P)$ with $\mathcal{L}_{\tor}(P)=\{ [(1,2,3)], [(1,3,2)]\}$,
whose closure $\bar{c}$ is the entire torus $\mathbb{R}^3/\mathbb{Z}^3$.
However, the unique toric chambers $c$ for these $7$ graphs
are all different, when considered as {\it open} subsets of $\mathbb{R}^3/\mathbb{Z}^3$,
and therefore each represents a {\it different} toric poset $P=P(c)$.
On the other hand, the complete graph $V_V=K_3$ has $2$ different
toric equivalence classes of acyclic orientations, representing
two different chambers within the same toric arrangement $\mathcal{A}_{\tor}(K_3)$,
and two different toric posets: $P(c_{[(1,2,3)]})$ and $P(c_{[(1,3,2)]})$.
\end{ex}
\section{Toric chains}
\label{chains-section}
We introduce the toric analogue of a chain (= totally ordered subset)
in a poset, and explicate its relation to the toric directed paths
from Definition~\ref{toric-directed-path-defn} and the toric total
extensions from Definition~\ref{total-cyclic-extension-defn} (or
equivalently, total cyclic extensions).
As motivation, note that in an ordinary poset $P(c)$, a
chain $C=\{i_1,\ldots,i_m\} \subseteq V$ has the following geometric
description: there is a total ordering $(i_1,\ldots,i_m)$ of $C$ such
that every point $x$ in the open polyhedral cone $c=c(P)$ has
$x_{i_1}<x_{i_2}<\cdots<x_{i_m}$.
\begin{defn}\label{toric-chain-defn}
Fix a toric poset $P=P(c)$ on a finite set $V$.
Call a subset $C=\{i_1,\ldots,i_m\} \subseteq V$
a {\it toric chain} in $P$ if
there exists a cyclic equivalence class $[(i_1,\ldots,i_m)]$ of
linear orderings of $C$
with the following property:
for every $x$ in the open toric chamber $c=c(P)$ there exists some $(j_1,\dots,j_m)$ in $[(i_1,\ldots,i_m)]$ for which
\begin{equation}
\label{eq:toric-chain}
\modone{x_{j_1}}<\modone{x_{j_2}}<\cdots<\modone{x_{j_m}}.
\end{equation}
In this situation, we will say that $P|_C=[(i_1,\ldots,i_m)]$.
\end{defn}
\begin{rem}
\label{toric-chain-defn-remark}
Note that
\begin{enumerate}
\item[$\bullet$]
singleton sets $\{i\}$ and the empty subset $\varnothing \subset V$ are
always toric chains in $P$,
\item[$\bullet$]
subsets of toric chains are toric chains, and
\item[$\bullet$]
a pair $\{i,j\}$ is a toric chain in $P=P(c)$ if and only
if every point $x$ in the open toric chamber
$c$ has $\modone{x_i} \neq \modone{x_j}$;
in particular, this will be true whenever $c$ as appears as
toric chamber in $\Chambers \mathcal{A}_{\tor}(G)$ for a graph $G$ having
$\{i,j\}$ as an edge of $G$.
\end{enumerate}
\end{rem}
Though the definition of toric chain does not refer to a particular
graph $G$, there are several convenient characterizations that involve
a graph. In the following proposition, we list five equivalent
conditions.
The exception when $|C|\neq 2$ is needed because the last
condition is vacuously true whenever $|C|=2$; in this case, only the
first four are equivalent.
\begin{prop}\label{toric-chain-prop}
Fix a toric poset $P=P(c)$ on a finite set $V$, and
$C=\{i_1,\ldots,i_m\} \subseteq V$.
The first four of the following five conditions are equivalent,
and when $m=|C| \neq 2$, they are also equivalent to the fifth.
\begin{enumerate}
\item[(a)] $C$ is a toric chain in $P$, with $P|_C=[(i_1,\ldots,i_m)]$.
\item[(b)] For every graph $G=(V,E)$ and acyclic orientation $\omega$
of $G$ having $\bar\alpha_G(c)=[\omega]$, the subset $C$ is a chain
in the poset $P(G,\omega)$, ordered in
some cyclic shift of the order $(i_i,\dots,i_m)$.
\item[(c)] For every graph $G=(V,E)$ and acyclic orientation $\omega$
of $G$ having $\bar\alpha_G(c)=[\omega]$, the subset $C$ occurs as a
subsequence of a toric directed path in $\omega$, in some cyclic
shift of the order $(i_i,\dots,i_m)$.
\item[(d)] There exists a graph $G=(V,E)$ and acyclic orientation
$\omega$ of $G$ having $\bar\alpha_G(c)=[\omega]$ such that $C$ occurs as a
subsequence of a toric directed path in $\omega$, in some cyclic shift
of the order $(i_1,\dots,i_m)$.
\item[(e)] Every total cyclic extension $[w]$ in $\mathcal{L}_{\tor}(P(c))$ has the
same restriction $[w|_C]=[(i_1,\ldots,i_m)]$.
\end{enumerate}
\end{prop}
\noindent
The following easy and well-known lemma will be used in the proof.
\begin{lem}
\label{incomparability-lemma}
When two elements $i,j$ are incomparable in a finite poset $Q$ on $V$,
one can choose a linear extension $w=(w_1,\ldots,w_n)$
in $\mathcal{L}(Q)$ that has $i,j$ appearing consecutively,
say $(w_s,w_{s+1})=(i,j)$.
\end{lem}
\begin{proof}
Begin $w$ with any linear extension
$w_1,w_2,\ldots,w_{s-1}$
for the order ideal $Q_{<i} \cup Q_{<j}$,
followed by $w_s=i, w_{s+1}=j$, and
finish with any linear extension
$w_{s+2},w_{s+3},\ldots,w_n$
for $Q {-} \left(Q_{\leq i} \cup Q_{\leq j}\right)$.
\end{proof}
\begin{proof}[Proof of Proposition~\ref{toric-chain-prop}]
Note that if $|C|\leq 1$, all five conditions (a)-(e) are vacuously true,
so without loss of generality $|C| \geq 2$.
We will first show (a) implies (b) implies (c) implies (d) implies (e).
Then we will show that (e)
implies (a) when $|C|\geq 3$, and (d) implies (a) when $|C|=2$.
\vskip.1in
\noindent {\sf (a) implies (b).} Assume that $C$ is a toric chain of $P$,
with $P|_C=[(i_1,\ldots,i_m)]$,
and take a graph $G$ and orientation $\omega$ such that
$\bar{\alpha}_G(c)=[\omega]$.
We first show by contradiction that $C$ must be totally ordered in
$Q:=P(G,\omega)$. Assume not, and
say $i,j$ in $C$ are incomparable in $Q$. By Lemma~\ref{incomparability-lemma}
there is a linear extension $w=(w_1,\ldots,w_n)$ in $\mathcal{L}(Q)$
having $i,j$ appear consecutively, say $(w_s,w_{s+1})=(i,j)$.
Choose $x$ in $\mathbb{R}^n$ with
$0\leq x_{w_1}<\cdots<x_{w_n}<1$
and let $x'$ be obtained by $x$ by exchanging $x_i,x_j$,
that is $x'_i=x_j$ and $x'_j=x_i$.
Since $x=\modone{x}$ and $x'=\modone{x'}$,
one has $\bar{\alpha}_G(x)=\omega=\bar{\alpha}_G(x')$,
and hence $x, x'$ lie in $c=c(P)$.
The condition \eqref{eq:toric-chain} on $x, x'$ implies that
$[w|_C]=[w'|_C]$ should give the same cyclic order on $C$,
which forces $m=2$ and $C=\{i,j\}$. However, the average
$x''=\frac{x+x'}{2}$ gives a third point in $c$
having $\modone{x''_i}=x''_i=x''_j=\modone{x''_j}$,
contradicting \eqref{eq:toric-chain}.
Once one knows that $C$ is totally ordered in $Q$,
consideration of \eqref{eq:toric-chain} for the point $x$
chosen as above implies that $w|_C$ lies in $[(i_1,\ldots,i_m)]$,
and hence the same is true of $Q|_C$.
\vskip.1in
\noindent {\sf (b) implies (c).}
Assume for the toric poset $P=P(c)$,
every graph $G$ and orientation $\omega$ with
$\bar{\alpha}_G(c)=[\omega]$
has $C$ totally ordered in $P(G,\omega)$ by
a cyclic shift $(j_1,\ldots,j_m)$ in $[(i_1,\ldots,i_m)]$.
We will show that $C$ actually occurs in this order
as a subsequence of some toric directed path in $\omega$.
By Proposition~\ref{flip-preserves-toric-directed-path}, one is free to
alter $\omega$ within the class $[\omega]$.
So choose $\omega$ within $[\omega]$ among
all those for which $P(G,\omega)$ on $V$ totally orders $C$ as
$j_1 < \cdots < j_m$, but minimizing the cardinality $|Z|$
where
$$
Z:=\{z \in V: z \text{ there is a directed }\omega\text{ path from }j_m\text{ to }z\}
$$
Note that $Z$ is nonempty, since it contains $j_m$.
We claim that minimality forces $|Z|=1$, that is, $Z=\{j_m\}$.
To argue the claim by contradiction, assume $Z \neq \{j_m\}$.
Then one can find an $\omega$-sink $z \neq j_m$ in $Z$, as $V$ is finite,
and $\omega$ is acyclic.
Perform a sink-to-source flip at $z$ to create a new orientation
$\omega'$ in $[\omega]$. Then $\omega'$
still has $P(G,\omega')$ totally ordering $C$ as
$j_1 < \cdots < j_m$, but its set $Z'$ has $|Z'|<|Z|$ because
$Z' \subset Z {-} \{z\}$.
Now $Z=\{j_m\}$ means that $j_m$ is an $\omega$-sink. Create
$\omega'$ by flipping $j_m$ from sink to source.
Since $j_1$ is supposed to be comparable with
$j_m$ in $P(G,\omega')$, one must have
$j_m <_{P(G,\omega')} j_1$, that is, there is an $\omega'$-path
of the form
$j_m\rightarrow k \rightarrow \cdots \rightarrow j_1$;
possibly $k=j_1$ here.
But this means that prior to the sink-to-source flip of $j_m$,
one had a toric directed $\omega$-path
$k \rightarrow \cdots \rightarrow j_1 \rightarrow j_2 \rightarrow \cdots \rightarrow j_m$
that contained $C$, as desired.
\vskip.1in
\noindent {\sf (c) implies (d).}
Trivial.
\vskip.1in
\noindent
{\sf (d) implies (e).}
Assume the graph $G$ has $\bar{\alpha}_G(c)=[\omega]$ and
$C$ occurs in the order $(i_1,\ldots,i_m)$ as a
subsequence of a toric directed path in $\omega$.
We must show that every total cyclic extension $[w]$ of $P=P(c)$
has restriction $[w|_C] = [(i_1,\ldots,i_m)]$.
By Definition~\ref{total-cyclic-extension-defn}, one has $c_{[w]} \subseteq c$. By \eqref{more-concrete-total-cyclic-chamber-description},
one can pick a point $x$ in $c_{[w]}$, so that
$$
\modone{x_{w_1}} < \cdots < \modone{x_{w_n}}.
$$
Since $x$ also lies in $c$, one has $\bar{\alpha}_G(x) = \omega' \equiv \omega$.
Proposition~\ref{flip-preserves-toric-directed-path} implies
that $\omega'$ contains as a toric directed path
some cyclic shift $(j_1,\ldots,j_m)$ of $(i_1,\ldots,i_m)$.
Hence
$$
\modone{x_{j_1}} < \cdots < \modone{x_{j_m}},
$$
which forces $w|_C=(j_1,\ldots,j_m)$, as desired.
\vskip.1in
\noindent
{\sf (e) implies (a) when $|C|\geq 3$.}
Assume that every total cyclic extension $[w]$ of $P=P(c)$ has $w|_C$
lying in the same cyclic equivalence class $[(i_1,\ldots,i_m)]$.
We want to show that every point $x$ in $c$
satisfies \eqref{eq:toric-chain}.
Recall from Corollary~\ref{toric-chamber-as-union-cor}
that there is at least one graph $G$
and $\equiv$-class $[\omega]$ containing
$\bar{\alpha}_G(x)$, that is, $\bar{\alpha}_G(c)=[\omega]$.
It suffices to show that
the partial order $Q:=P(G,\omega)$ on $V$ induced by any orientation
$\omega$ in this $\equiv$-class has restriction $Q|_C$
to the subset $C$ giving a total order $(j_1,\ldots,j_m)$,
and this total order lies in $[(i_1,\ldots,i_m)]$.
Suppose that $Q|_C$ is {\it not} a total order;
say elements $i,j$ in $C$ are incomparable in $Q$. By Lemma~\ref{incomparability-lemma}, one can then
choose linear extensions $w,w'$ in
$\mathcal{L}(Q)$ that both have $i,j$ consecutive, and differ only
in swapping $i,j$, say $(w_s,w_{s+1})=(i,j)$ and $(w'_s,w'_{s+1})=(j,i)$.
Pick points $x,x'$ that satisfy
\begin{align*}
&0 \leq x_{w_1} < \cdots < x_{w_n} < 1\\
&0 \leq x'_{w'_1} < \cdots < x'_{w'_n} < 1.
\end{align*}
Since $x=\modone{x}, x'=\modone{x'}$,
one finds that $x,x'$ lie in $c_{[w]}, c_{[w']}$, respectively.
Also one has $\bar{\alpha}_G(x)=\omega=\bar{\alpha}_G(x')$ so that
both $x, x'$ lie in $c$. Hence $c_{[w]},c_{[w']} \subseteq c$, that
is, both $[w],[w']$ are total cyclic extensions in $\mathcal{L}_{\tor}(P(c))$.
However, since $|C| \geq 3$, there exists some third
element $k$ in $C {-} \{i, j\}$, and $[w],[w']$ differ
in their cyclic ordering of $\{i,j,k\}$. This contradicts
assumption (e), so $Q|_C$ is a total order.
Once one knows that $Q|_C$ is a total order $j_1 < \cdots <j_m$,
the above argument shows that $(j_1,\ldots,j_m)$ lies in the cyclic
equivalence class $[w|_C]$ for every $w$ in $\mathcal{L}_{\tor}(P)$, which
is $[(i_1,\ldots,i_m)]$ by assumption.
\vskip.1in
\noindent
{\sf (d) implies (a) when $|C|=2$.} Suppose
$\bar\alpha_G(c)=[\omega]$ and $C$ occurs as a subsequence of a toric
directed path in $\omega$, with $i_1<i_2$. By
Proposition~\ref{flip-preserves-toric-directed-path}, if
$\omega'\equiv\omega$, then $C$ occurs in a toric directed path in
$\omega'$. This means that for any $x$ with $\bar{\alpha}_G(x)=\omega'$,
we have $\modone{x_{i_1}}\neq\modone{x_{i_2}}$, and
so either $\modone{x_{i_1}}<\modone{x_{i_2}}$ or
$\modone{x_{i_2}}<\modone{x_{i_1}}$ must hold for every $x$ in
$c$. Thus $C$ is a toric chain of $P(c)$.
\end{proof}
\section{Toric transitivity}
\label{transitivity-section}
We next clarify the edges that are ``forced'' in a toric partial order,
an analogue of transitivity that we refer as \emph{toric transitivity}.
\begin{thm}
\label{thm:toric-transitivity}
Fix a toric poset $P=P(c)$ on $V$, and assume that $G=(V,E)$ has
$c$ appearing as a toric chamber in $\Chambers \mathcal{A}_{\tor}(G)$,
say $\bar{\alpha}_G(c)=[\omega]$. Then for any
non-edge pair $\{i,j\} \not\in E$, either
\begin{enumerate}
\item[(i)] $i,j$ lies on a toric chain in $P$, in which case
$c$ is also a toric chamber for $G^+=(V,E\cup \{i,j\})$,
and there is a unique extension $\omega^+$ of $\omega$
such that $\bar{\alpha}_{G^+}(c)=\omega^+$, or
\item[(ii)] $i,j$ lies on no toric chains in $P$, and then
the hyperplane $\modone{x_i = x_j}$ intersects the open toric chamber $c$.
\end{enumerate}
\end{thm}
\begin{proof}
Assertion (i) follows from Proposition~\ref{toric-chain-prop}: when $i,j$ lie
on a toric chain $C$ in $P$, assertion (b) of that proposition says that
they lie on a toric directed path in $\omega$
for every representative of the class $[\omega]$, and hence
the inequality $\modone{x_i} < \modone{x_j}$ (or its reverse inequality)
is already
implied by the other inequalities defining the points of
$\bar{\alpha}_G^{-1}(\omega)$ that come from the edges of $G$ induced by $C$.
For assertion (ii), note that whenever there exist no points $x$ of the open toric chamber $c$ having $\modone{x_i} = \modone{x_j}$, then every $x$ in $c$ has
either $\modone{x_i} < \modone{x_j}$ or
$\modone{x_j} < \modone{x_i}$. This shows that
$\{i,j\}$ is itself a toric chain in $P=P(c)$; see
Remark~\ref{toric-chain-defn-remark}.
\end{proof}
This suggests the following definition.
\begin{defn}
\label{toric-transitive-closure-defn}
Given a graph $G=(V,E)$ and $\omega$ in $\Acyc(G)$,
the {\it toric transitive closure} of the pair
$(G,\omega)$ is the pair
$
(\bar{G}^{\tor},\bar{\omega}^{\tor})
$
defined as follows.
The edges of $\bar{G}^{\tor}$ are obtained by
adding to the edges of $G$ all pairs
$\{i,j\}$ that are a subset of some toric directed path in $\omega$;
see the dotted edges in \eqref{toric-transitivity-figure} below.
The acyclic orientation
$\bar{\omega}^{\tor}$ orients the edge $i \rightarrow j$ if the
toric directed path contains a path from $i$ to $j$, rather than
from $j$ to $i$.\end{defn}
\begin{equation}
\label{toric-transitivity-figure}
\xymatrix{
i_m & \\
&i_{m-1} \ar[ul]\\
&i_{m-2} \ar[u]\ar@{-->}[uul]\\
&\vdots \ar[u]\\
&i_3 \ar[u]\ar@{-->}[uuuul]\\
&i_2 \ar[u]\ar@{-->}[uuuuul] \\
i_1 \ar[uuuuuu] \ar[ur]\ar@{-->}[uur]\ar@{-->}[uuuur]\ar@{-->}[uuuuur]&
}
\end{equation}
\begin{cor}
\label{toric-transitive-closure-independence}
The toric transitive closure
depends only upon the toric poset $P=P(c)$ which satisfies $\bar{\alpha}_G(c)=[\omega]$,
in the following sense:
given two graphs $G_i=(V,E_i)$ for $i=1,2$,
and $\omega_i$ in $\Acyc(G_i)$ with $\bar{\alpha}_{G_i}(c)=[\omega_i]$, then
\begin{enumerate}
\item[(i)]
$\bar{G}^{\tor}_1=\bar{G}^{\tor}_2$, and
\item[(ii)]
$\bar{\omega}^{\tor}_1 \equiv \bar{\omega}^{\tor}_2$.
\end{enumerate}
\end{cor}
\begin{proof}
Assertion (i) follows from the fact that $\{i,j\}$ appears as an edge
in $\bar{G}^{\tor}$ if and only if it is a subset of some toric chain of
$P$, and adding $\{i,j\}$ does not affect the toric poset $P=P(c)$,
according to Theorem~\ref{thm:toric-transitivity}(i). For assertion
(ii), note that iterating Theorem~\ref{thm:toric-transitivity}(i)
gives
$$
\bar{\alpha}_{\bar{G}^{\tor}}^{-1}(\bar{\omega}_1^{\tor})=
\bar{\alpha}_{G_1}^{-1}(\omega_1)=
c=\bar{\alpha}_{G_2}^{-1}(\omega_2)
=\bar{\alpha}_{\bar{G}^{\tor}}^{-1}(\bar{\omega}_2^{\tor}).
$$
Assertion (ii) then follows from Theorem~\ref{geometric-interpretation-theorem}.
\end{proof}
\begin{rem}
Note that the toric transitive closure of $\bar{A}^{\tor}$ is
always a subset of the ordinary transitive closure $\bar{A}$, since
any toric directed path that contains $(i,j)$ as a subsequence
also contains an ordinary directed path from $i$ to $j$.
\end{rem}
\section{Proof of Theorem~\ref{toric-convex-closure-theorem}}
\label{convex-closure-section}
Here we wish to regard a pair $(G,\omega)$ of a simple graph $G=(V,E)$
and acyclic orientation $\omega$ in $\Acyc(G)$ as a subset
$A \subset \overleftrightarrow{K}_V$ of the set
of all possible directed edges
$\overleftrightarrow{K}_V=\{ (i,j) \in V \times V : i \neq j \}$.
Then the toric transitive closure operation
$(G,\omega) \longmapsto (\bar{G}^{\tor},\bar{\omega}^{\tor})$
from Definition~\ref{toric-transitive-closure-defn}
may be regarded as a {\it closure operator} on $\overleftrightarrow{K}_V$,
that is, a map $A \longmapsto \bar{A}^{\tor}$
from $2^{\overleftrightarrow{K}_V}$ to itself, satisfying
\begin{itemize}
\item $A \subseteq \bar{A}^{\tor}$,
\item $A \subseteq B$ implies $\bar{A}^{\tor} \subseteq \bar{B}^{\tor}$, and
\item $\bar{\bar{A}}^{\tor}=\bar{A}^{\tor}$.
\end{itemize}
\noindent
Recall the statement of Theorem~\ref{toric-convex-closure-theorem}:
\vskip.1in
\noindent
{\bf Theorem~\ref{toric-convex-closure-theorem}.}
{\it
The toric transitive closure operation $A \longmapsto \bar{A}^{\tor}$
is a {\it convex closure}, that is,
$$
\text{ for } a \neq b \text{ with }
a,b \not\in \bar{A}^{\tor} \text{ and }
a \in \overline{A \cup \{b\}}^{\tor},
\text{ one has }b \notin \overline{A \cup \{a\}}^{\tor}.
$$
}
For the purposes of the proof, introduce one further
bit of terminology.
\begin{defn}
For $\omega$ in $\Acyc(G)$ and a toric directed path $C=(i_1,\ldots,i_m)$ in
$\omega$ of size $m\geq 3$, as in \eqref{toric-directed-path-figure},
call $(i_1,i_m)$ the {\it long edge} of $C$, and
call the other edges $(i_1,i_2),(i_2,i_3),\ldots,(i_{m-1},i_m)$ the
{\it short edges} of $C$.
\end{defn}
\begin{proof}[Proof of Theorem~\ref{toric-convex-closure-theorem}.]
Proceed by contradiction:
suppose $(i,j) \neq (k,\ell)$ are {\it not} in
$\bar{A}^{\tor}$, but both
\begin{enumerate}
\item[$\bullet$]
$(k,\ell)$ lies in $\overline{A\cup (i,j)}^{\tor}$,
say because $(i,j)$ creates a toric directed path $C$
also containing $(k,\ell)$, which was
not already present in $\bar{A}^{\tor}$,
and
\item[$\bullet$]
$(i,j)$ lies in $\overline{A\cup (k,\ell)}^{\tor}$,
say because $(k,\ell)$ creates a toric directed path $D$
also containing $(i,j)$, which was
not already present in $\bar{A}^{\tor}$.
\end{enumerate}
Introduce the (ordinary) partial order $Q$ on $V$ which
is the (ordinary) transitive closure of
$\bar{A}^{\tor} \cup \{(i,j),(k,\ell)\}$.
We use this to argue a contradiction in various cases.
\vskip.1in
\noindent
{\sf Case 1. Either $(i,j)$ is the long edge of $C$, or
$(k,\ell)$ is the long edge of $D$.}
By relabeling, assume without loss of generality
that $(i,j)$ is the long edge of $C$.
Then in $Q$, one has
\begin{equation}
\label{first-transitivity-inequalities}
i \leq k < \ell \leq j
\end{equation}
with at least one of the two weak inequalities being strict.
\vskip.1in
\noindent
{\it Subcase 1a. $(k,\ell)$ is also the long edge of $D$.}
Then in $Q$ one also has $k \leq i < j \leq \ell$, which
with \eqref{first-transitivity-inequalities} gives
$$
k \leq i \leq k < \ell \leq j \leq \ell
$$
forcing the contradiction $(i,j)=(k,\ell)$.
\vskip.1in
\noindent
{\it Subcase 1b. $(k,\ell)$ is a short edge of $D$.}
Then since $C$ has $(i,j)$ as its long edge and gives a toric directed path
containing $(k,\ell)$ (while $\bar{A}^{\tor}$ had no such path),
$C$ must contain a directed path from $k$ to $\ell$ with at least two steps.
Combining this with $D {-} \{(k,\ell)\}$ gives
a toric directed path in $\bar{A}^{\tor}$ that contains $(i,j)$; contradiction.
\vskip.1in
\noindent
{\sf Case 2. Both $(i,j),(k,\ell)$ are
short edges of $C, D$, respectively.}
In this case, $\bar{A}^{\tor}$ cannot contain a path
from $i$ to $j$, else replacing $(i,j)$ in $C$ with this path
would give the contradiction that $(i,j)$ is in $\bar{A}^{\tor}$.
Similarly, $\bar{A}^{\tor}$ cannot contain a path
from $k$ to $\ell$. Also note that, since $C$ (or $D$) is a directed path
containing all four of $\{i,j,k,\ell\}$, the four of them are totally ordered in $Q$.
We now argue in subcases based on how $Q$ totally orders $\{i,j,k,\ell\}$.
\vskip.1in
\noindent
{\it Subcase 2a. Either $Q$ has $i < j \leq k < \ell$
or $k < \ell \leq i < j$.}
In this case, adding $(i,j)$ to $\bar{A}^{\tor}$ cannot help to create
a directed path from $k$ to $\ell$, contradicting the existence of $C$.
\vskip.1in
\noindent
{\it Subcase 2b. Either $Q$ has $i \leq k < \ell \leq j$,
with at least one of the weak inequalities strict,
or $k \leq i < j \leq \ell$,
with at least one of the weak inequalities strict.}
Assume without loss of generality, by relabeling, that
one is in the first case $i \leq k < \ell \leq j$. But
then adding $(i,j)$ to $\bar{A}^{\tor}$ again cannot help to create
a directed path from $k$ to $\ell$, contradicting the existence of $C$.
\vskip.1in
\noindent
{\it Subcase 2c. Either $Q$ has $i \leq k \leq j \leq \ell$,
with at least two consecutive strict inequalities,
or $k \leq i \leq \ell \leq j$,
with at least two consecutive strict inequalities.}
Assume without loss of generality, by relabeling, that
one is in the first case $i \leq k \leq j \leq \ell$.
But then the consecutive strict inequalities either imply
the existence within $\bar{A}^{\tor}$ of
a directed path from $i$ to $j$,
or one from $k$ to $\ell$; contradiction.
\end{proof}
\section{Toric Hasse diagrams}
\label{Hasse-diagram-section}
For convex closures $A \longmapsto \bar{A}$, it is
well-known that for any subset $A$, its {\it extreme
points}
$$
\extreme(A): = \{ a \in A: a \not\in \overline{A {-} \{a\}} \}
$$
gives the unique set which is minimal under inclusion among all subsets
having the same closure as $A$; see \cite{Edelman:85}.
For ordinary transitive closure of an acyclic orientation $(G,\omega)$
as a subset of $\overleftrightarrow{K_V}$, its extreme points
are exactly the subset of directed edges $(i,j)$ in the usual {\it Hasse diagram} for its associated partial order $P$. This suggests the following definition.
\begin{defn}
\label{toric-Hasse-diagram-defn}
Given a graph $G=(V,E)$ and $\omega$ in $\Acyc(G)$,
corresponding to a subset $A$ of $\overleftrightarrow{K_V}$,
its {\it toric Hasse diagram} is the pair
$(\hat{G}^{\torHasse},\omega^{\torHasse})$ corresponding
to its subset of extreme points $\extreme(A)$ with respect to
the toric transitive closure operation $A \longmapsto \bar{A}^{\tor}$. The toric Hasse diagram of a toric poset $P$ is $(G^{})$
\end{defn}
Definition~\ref{toric-transitive-closure-defn} allows one to rephrase this as
follows:
\begin{itemize}
\item $\hat{G}^{\torHasse}$ is obtained from $G$ by
removing all {\it chord edges} $\{i_j,i_k\}$ with $|j-k| \geq 2$
from all toric directed paths $C=\{i_1,\ldots,i_m\}$ in $\omega$
that have $m=|C| \geq 4$, and
\item $\omega^{\torHasse}$ is the restriction $\omega|_{\hat{G}^{\torHasse}}$.
\end{itemize}
One then has the following analogue of
Corollary~\ref{toric-transitive-closure-independence}.
\begin{cor}
\label{toric-Hasse-diagram-closure-independence}
The toric Hasse diagram
depends only on the toric poset $P=P(c)$ having $\bar{\alpha}_G(c)=[\omega]$,
in the following sense:
given two graphs $G_i=(V,E_i)$ for $i=1,2$,
and $\omega_i$ in $\Acyc(G_i)$ with $\bar{\alpha}_G(c)=[\omega_i]$, then
\begin{enumerate}
\item[(i)]
$\hat{G}_1^{\torHasse}=\hat{G}_2^{\torHasse}$, and
\item[(ii)]
$\omega_1^{\torHasse} \equiv \omega_2^{\torHasse}$.
\end{enumerate}
\end{cor}
\begin{proof}
Same as the proof of Corollary~\ref{toric-transitive-closure-independence}.
The key point is that the toric directed paths
$C=\{i_1,\ldots,i_m\}$ in $\omega$ are the toric chains in $P$,
and when $|C| \geq 4$, removing chords from $C$ still keeps it a toric chain.
\end{proof}
\section{Toric antichains}
\label{Antichain-section}
Since chains in posets have a good toric analogue, one might ask if the same is true for antichains.
Recall that an {\it antichain} of an
ordinary poset $P$ on $V$ is a subset $A=\{i_1,\dots,i_m\} \subseteq V$
characterized
\begin{enumerate}
\item[$\bullet$] {\it combinatorially} by the condition
that no pair $\{i,j\}\subset A$ with $i \neq j$ are comparable,
that is, they lie on no chain of $P$, or
\item[$\bullet$] {\it geometrically} by the equivalent condition
that the $(|V|-m+1)$-dimensional linear subspace
$
\{x\in\mathbb{R}^V: x_{i_1}=x_{i_2}=\cdots=x_{i_m}\}
$
intersects the open polyhedral cone/chamber $c(P)$ in $\mathbb{R}^V$.
\end{enumerate}
In the toric situation, these two conditions
lead to different notions of toric antichains.
\begin{defn}
\label{toric-antichain-defn}
Given a toric poset $P=P(c)$ on the finite set $V$,
say that $A=\{i_1,\dots,i_m\} \subseteq V$
is a
\begin{enumerate}
\item[$\bullet$] {\it combinatorial toric antichain} of $P$ if no
$\{i,j\}\subset A$ with $i \neq j$ lie on a common toric chain of $P$.
\item[$\bullet$] {\it geometric toric antichain} if the
subspace $\{x\in\mathbb{R}^V/\mathbb{Z}^V: x_{i_1}=x_{i_2}=\cdots=x_{i_m}\}$
intersects the open toric chamber $c=c(P)$.
\end{enumerate}
\noindent
By analogy to the notion of the width of a poset, which is the size of its
largest antichain, define the \emph{geometric (resp. combinatorial) toric width} of a toric poset to be the size of the largest geometric (resp. combinatorial) toric antichain.
\end{defn}
Given a toric poset $P=P(c)$ and a graph $G=(V,E)$ with
$\bar{\alpha}_G(c)=[\omega]$, the definition and
Corollary~\ref{toric-chamber-as-union-cor} imply
that $A\subseteq V$ is a geometric toric antichain of $P$
if and only if $A$ is an antichain of $P(G,\omega')$
for some $\omega'\equiv\omega$.
The following proposition should also be clear.
\begin{prop}
In a toric poset $P$, every geometric toric antichain is a
combinatorial toric antichain. Thus its geometric toric width is
bounded above by its combinatorial toric width.
\end{prop}
The next example shows that the inequality between
these two notions of toric width can be strict.
\begin{ex}
Consider the toric poset $P=P(c)$ whose toric Hasse diagram is the
circular graph $G=C_6$ and for which $\bar{\alpha}_G(c)$ contains
the following representatives $\omega_1$, $\omega_2$ and $\omega_3$
of $\Acyc(G)$:
\[
\tiny{
\xymatrix{
5 & \\
4 \ar[u] & \\
3 \ar[u] & \\
2 \ar[u] & 6 \ar[uuul] \\
1 \ar[u] \ar[ur] & }\hspace{1in}
\xymatrix{
& \\
4 & \\
3 \ar[u] & \\
2 \ar[u] & 6 \\
1 \ar[u] \ar[ur] & 5 \ar[u]\ar[uuul] }\hspace{1in}
\xymatrix{
& \\
& \\
3 & 6 \\
2 \ar[u] & 5 \ar[u] \\
1 \ar[u] \ar[uur] & 4 \ar[u]\ar[uul] }}
\]
All three of these orientations satisfy $\nu_I(\omega_i)=2$ for the
directed cycle $I=[(1,2,3,4,5,6)]$ of $G$, where $\nu_I$ is
Coleman's $\nu$-function from Remark~\ref{Coleman-remark}.
Moreover, Proposition~\ref{prop-pretzel}
says that $\nu_I(\omega)=2$ must hold for any other $\omega$ in
$[\omega_i]$. It is easy to check that for any such $\omega$, the
directed graph $(G,\omega)$ must be isomorphic to either
$(G,\omega_1)$, $(G,\omega_2)$, or $(G,\omega_3)$.
Consequently, $P$ has no toric chains except for those of cardinality $0,1,2$,
that is, the empty set $\varnothing$, the $6$ singletons
and the $6$ edge pairs in $G$. From this one can easily
check that the combinatorial toric antichains of $P$ are the
empty set $\varnothing$, the $6$ singletons, the pairs $\{i,j\}$
which do not form edges of $G$, and the two triples $\{1,3,5\}, \{2,4,6\}$.
In particular, $P$ has combinatorial toric width $3$.
However, we claim neither of these triples $\{1,3,5\},\{2,4,6\}$ can
be a geometric toric antichain, so that the geometric toric width of
$P$ is $2$. To argue that $\{1,3,5\}$ is not a geometric toric antichain,
consider three paths of length $2$ in $G$ between the
elements of $\{1,3,5\}$, that is, the paths
$$
\begin{aligned}&1-2-3\\&3-4-5\\&5-6-1\end{aligned}
$$
The only way one could avoid having an $\omega$-directed path between two
elements of $\{1,3,5\}$ would be if $\omega$ orients both edges in each
of the three paths listed above
in opposite directions. But this would lead to $\nu_I(\omega)=0$
which is impossible for $\omega$ in $[\omega_i]$. The argument for
$\{2,4,6\}$ is similar.
\end{ex}
Despite the difference in the two notions of toric width,
one might still hope that one of the notions gives
a toric analogue for one or both of these two
classic results on chains and antichains in ordinary posets.
\begin{thm}\label{thm:Dilworth}
For any (ordinary) finite poset $P$, one has:
\begin{enumerate}
\item[(i)] Dilworth's Theorem \cite{Dilworth:50}:
$$
\max \{ |A|: A \text{ an antichain in }P \}=
\min \{ \ell: V=\cup_{i=1}^\ell C_i, \text{ with }C_i
\text{ chains in }P \}
$$
\item[(ii)] Mirsky's Theorem \cite{Mirsky:71}:
$$
\max \{ |C|: C \text{ a chain in }P \}=
\min \{ \ell: V=\cup_{i=1}^\ell A_i, \text{ with }A_i
\text{ antichains in }P \}.
$$
\end{enumerate}
\end{thm}
\noindent
One at least has the following
inequalities, coming from the easy observation
that a toric chain and toric antichain (whether combinatorial or geometric)
can intersect in at most one element.
\begin{prop}\label{prop-toric-dilworth}
For a toric poset $P$, both versions (geometric or
combinatorial) of a toric antichain lead to the following inequalities
holds:
\[
\begin{array}{lcl}
\max \{ |A|: A \text{ a toric antichain in }P \}&\leq&
\min \{ \ell: V=\cup_{i=1}^\ell C_i, \text{ with }C_i
\text{ toric chains in }P \} \\
\max \{ |C|: C \text{ a toric chain in }P \}&\leq&
\min \{ \ell: V=\cup_{i=1}^\ell A_i, \text{ with }A_i
\text{ toric antichains in }P \}.
\end{array}
\]
\end{prop}
\noindent
However, the following example shows that both inequalities in
Proposition~\ref{prop-toric-dilworth} can be strict: neither of
our two notions of toric antichain leads to a version
of Dilworth's Theorem, nor of Mirsky's theorem.
\begin{ex}\label{ex:mirsky-counterexample}
Consider the toric poset $P=P(c)$ whose toric Hasse diagram is the
circular graph $G=C_5$ and for which $\bar{\alpha}_G(c)$ contains
the following representatives $\omega_1$ and $\omega_2$ of
$\Acyc(G)$:
\[
\tiny{
\xymatrix{
4 & \\
3 \ar[u] & \\
2 \ar[u] & 5 \ar[uul] \\
1 \ar[u] \ar[ur] & }\hspace{1in}
\xymatrix{
& \\
3 & \\
2 \ar[u] & 5 \\
1 \ar[u] \ar[ur] & 4 \ar[u]\ar[uul] }}
\]
Both orientations above satisfy $\nu_I(\omega_i)=1$ for the directed
cycle $I=[(1,2,3,4,5)]$ of $G$. Proposition~\ref{prop-pretzel} says
that $\nu_I(\omega)=1$ must hold for any other $\omega$ in
$[\omega_i]$, and so for such an $\omega$, the directed graph
$(G,\omega)$ must be isomorphic to either $(G,\omega_1)$ or
$(G,\omega_2)$.
Consequently, $P$ has no toric chains except for those of cardinality $0,1,2$,
that is, the empty set $\varnothing$, the $5$ singletons
and the $5$ edge pairs in $G$. In particular, the maximum size of
a toric chain is $2$. From this one can also easily
check that the combinatorial toric antichains of $P$ are the
empty set $\varnothing$, the $5$ singletons, and the $5$ pairs $\{i,j\}$
which do not form edges of $G$. In fact,
all of these are also geometric toric antichains, so in this
example the two notions coincide, and for either one the
toric width is $2$.
However, as $|V|=5$, there is no partition of $V$ into
two toric chains (the analogue of Dilworth's Theorem fails),
nor into two toric antichains (the analogue of Mirsky's Theorem
fails).
\end{ex}
\bibliographystyle{amsplain}
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\providecommand{\MR}{\relax\ifhmode\unskip\space\fi MR }
\providecommand{\MRhref}[2]{%
\href{http://www.ams.org/mathscinet-getitem?mr=#1}{#2}
}
\providecommand{\href}[2]{#2}
|
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| 2,213
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{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 2,423
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#import "TiProxy.h"
@class TiHost;
@interface Bridge : NSObject {
@private
id callback;
NSString *basename;
@protected
NSURL *url;
TiHost *host;
}
-(id)initWithHost:(TiHost*)host;
-(void)boot:(id)callback url:(NSURL*)url preload:(NSDictionary*)preload;
-(void)booted;
-(void)shutdown:(NSCondition*)condition;
-(void)gc;
-(TiHost*)host;
- (NSString*)basename;
@end
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 4,983
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The Kampa Travel Pod Cross Air XL is a freestanding awning that will allow you to drive-away whilst still reserving your pitch on site for when you return. The Cross has been updated for 2019, and its large square living area will now benefit from new steeper walls that offer improved headroom. An optional 2 berth air annexe can also be attached if a sleeping area is required. The Cross comes complete with a large pre-attached front canopy and a fully waterproof clip-in groundsheet, however many sites do not allow for groundsheets, so the groundsheet can be left out if necessary. The Travel Pod Cross comes with a Multi Point Inflation System and can be pitched in around 10-15 minutes.
|
{
"redpajama_set_name": "RedPajamaC4"
}
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\section{Introduction and Preliminaries}\label{se1}
Ageing is a common phenomenon experienced by both living organisms and mechanical systems. It largely describes how a system/living organism improves or deteriorates with age. The study of stochastic ageing has receieved considerable attention from researchers in the last few decades. In the literature, many different types of stochastic ageing concepts (e.g., increasing failure rate (IFR), increasing failure rate in average (IFRA), etc.) have been developed to describe different ageing characteristics of a system. There are three types of ageings, namely, positive ageing, negative ageing and no ageing. A brief discussion on different ageing concepts could be found in Barlow and Proschan~\cite{bp}, and Lai and Xie~\cite{lx0}. Similar to these ageing concepts, there is another useful notion of ageing, called relative ageing, which describes how a system ages relative to another one.
\\\hspace*{0.2 in}The proportional hazard (PH) rate model, commonly known as Cox's PH model (see Cox~\cite{cox}), is widely used to analyze the failure time data in reliability and survival analysis. Later, different other models were introduced, namely, proportional mean residual lifetime model, proportional reversed hazard rate model, proportional odds model, etc (see Marshall and Olkin~\cite{mo}, Lai and Xie~\cite{lx0}, and Finkelstein~\cite{f8}). In many real life scenarios, the phenomenon of crossing hazards/mean residual lives has been observed (see Pocock et al.~\cite{pgk}, Champlin et al.~\cite{cmeg}, and Mantel and Stablein~\cite{ms}). To handle this crossing hazard rates problem, Kalashnikov and Rachev~\cite{kr} introduced a stochastic order (called ageing faster order in the hazard rate) based on the concept of relative ageing. Indeed, this approach could be considered as a reasonable alternative to the PH model. A detailed study of this order is done by Sengupta and Deshpande~\cite{sd}. In addition, they have also introduced two other similar kinds of stochastic orders. Later, Finkelstein~\cite{f6} proposed a stochastic order, based on mean residual lifetime functions, that describes the relative ageings of two life distributions, whereas Rezaei~\cite{rgi} introduced a similar stochastic order in terms of the reversed hazard rate functions. Some generalized orderings in this direction were proposed by Hazra and Nanda~\cite{hn0}.
\\\hspace*{0.2 in}The basic structures of most of the real life systems match with the so called coherent system. A system is called coherent if its all components are relevant and its structure function (see Barlow and Proschan~\cite{bp} for the definition) is monotonically non-decreasing with respect to each argument (which means that an improvement in performance of a component cannot decrease the lifetime of the system). The well known $k$-out-of-$n$ system is a special case of coherent systems. A system of $n$ components is said to be $k$-out-of-$n$ system if it functions as long as at least $k$ of its $n$ components function. Two extreme cases of a $k$-out-of-$n$ system are $1$-out-of-$n$ system (called parallel system) and $n$-out-of-$n$ system (called series system). Further, there is an one-to-one correspondence between a $k$-out-of-$n$ system and an $(n-k+1)$-th order statistic (of lifetimes of $n$ components). Thus, the study of a $k$-out-of-$n$ system is essentially the same as the study of an order statistic.
\\\hspace*{0.2 in}Stochastic comparisons of coherent systems is considered as one of the important problems in reliability theory. The list of results, so far developed, on various stochastic comparisons of $k$-out-of-$n$ systems with independent components could be found in Pledger and Proschan~\cite{pp}, Proschan and Sethuraman~\cite{ps}, Balakrishnan and Zhao~\cite{bz}, Hazra et al.~\cite{hkfn}, and the references therein. Further, stochastic comparisons of general coherent systems were considered in Esary and Proschan~\cite{ep}, Nanda et al.~\cite{njs}, Kochar et al.~\cite{kms}, Belzunce et al.~\cite{bfrr}, Navarro and Rubio~\cite{nr}, Navarro et al.~\cite{nass1, nass3, npd}, Samaniego and Navarro~\cite{sn}, Amini-Seresht et al.~\cite{azb}, to name a few. Note that all these results are developed using different stochastic orders, namely, usual stochastic order, hazard rate order, likelihood ratio order, etc. However, the study of coherent systems using ageing faster orders are not substantially done yet. Misra and Francis~\cite{mf}, Li and Lu~\cite{ll6}, and Ding and Zhang~\cite{dz} developed some results for $k$-out-of-$n$ systems using ageing faster orders. Later, Ding et al.~\cite{dfz} have given some sufficient conditions in terms of signature to compare the lifetimes of two coherent systems (with independent components) with respect to ageing faster orders. However, there is no such result where the sufficient conditions are given in terms of reliability functions. Furthermore, the coherent systems with dependent components are also not considered yet. Thus, one of our major goals of this paper is to provide some sufficient conditions (in terms of reliability functions) under which one coherent system dominates another one with respect to ageing faster orders.
\\\hspace*{0.2 in}One of the effective ways to enhance the lifetime of a system is by incorporating spares (or redundant components) into the system. Then the key question is $-$ how to allocate spares into the system so that the system's lifetime will be optimum in some stochastic sense? In Barlow and Proschan~\cite{bp}, it is shown that the allocation of active redundancy at the component level (of a coherent system) is superior to that at the system level with respect to the usual stochastic order. Later, many other researchers have studied this problem in different directions (see Boland and El-Neweihi~\cite{be}, Misra et al.~\cite{mdg}, Nanda and Hazra~\cite{nh}, Hazra and Nanda~\cite{hn8}, Zhao et al.~\cite{zzl}, Da and Ding~\cite{dd}, Zhang et al.~\cite{zad}, and the references therein). However, to the best of our knowledge, this problem using ageing faster orders is not studied yet. Thus, another goal of this paper is to derive some necessary and sufficient conditions under which the lifetime of a coherent system with active redundancy at the component level is larger (smaller) than that at the system level with respect to ageing faster orders.
\\\hspace*{0.2 in}The real life systems are either formed by new components or by used components. Consider two coherent systems, namely, a used coherent system (i.e., a coherent system formed by a set of new components, and then the system has been used for some time $t>0$) and a coherent system of used components (i.e., a coherent system formed by a set of components which have already been used for time $t>0$). It is a fact that a coherent system of new components does not always have larger lifetime than a coherent system made out of used components (see Navarro et al.~\cite{nffa}). Similarly, a used coherent system may or may not perform better than a coherent system of used components. The stochastic comparisons between these two systems have been done in numerous papers, see, for example, Li and Lu~\cite{ll}, Gupta~\cite{g}, Gupta et al.~\cite{gmk}, Hazra and Nanda~\cite{hn}, to name a few. However, to the best our knowledge, the ageing faster orders have not yet been used, as a tool, to compare these two systems. Thus, the study of stochastic comparisons between a used coherent system and a coherent system of used components is another thrust area that is to be focused here.
\\\hspace*{0.2 in}In what follows, we introduce some notation that will be used throughout the paper. For a random variable $W$ (with absolutely continuous cumulative distribution function), we denote its probability density function (pdf) by $f_W(\cdot)$, the cumulative distribution function (cdf) by $F_W(\cdot)$, the hazard rate function by $r_W(\cdot)$, the reversed hazard rate function by $\tilde r_W(\cdot)$ and
the survival/reliability
function by $\bar F_W(\cdot)$; $\bar F_W(\cdot)=1-F_W(\cdot)$.
\\\hspace*{0.3 in}Let us consider a coherent system with lifetime $\tau\left(\mbox{\boldmath$X$}\right)$ formed by $n$ components having dependent and identically distributed (d.i.d.) lifetime vector $\mbox{\boldmath$X$}=(X_1,X_2,\dots,X_n)$, where $X_i$'s are identically distributed, say $X_i\stackrel{\text{d}}=X,$ $i=1,2,\dots,n$, for some non-negative random variable $X$; here $\stackrel{\text{d}}=$ means equality in distribution.
Then the joint reliability function of $\mbox{\boldmath$X$}$ is given by
\begin{eqnarray*}
\bar F_{\mathbf{X}}(x_1,x_2,\dots,x_n)&=&P\left(X_1>x_1,X_2>x_2,\dots,X_n>x_n\right)
\\&=&K\left(\bar F_{X}(x_1),\bar F_{X}(x_2),\dots,\bar F_{X}(x_n)\right),
\end{eqnarray*}
where $K(\cdot,\cdot,\dots,\cdot)$ is a survival copula describing the dependency structure among components of the system. Indeed, this representation is well known through Sklar's Theorem (see Nelsen~\cite{n}). In the literature, many different types of survival copulas have been studied in order to describe different dependency structures among components. Some of the widely used copulas are Farlie-Gumbel-Morgenstern (FGM) copula, Archimedean copula with different generators,
Clayton-Oakes (CO) copula, etc. We refer the reader to Nelsen~\cite{n} for a detailed discussion on the copula theory, and its various applications. In what follows, we give a lemma that describes a fundamental bridge between a system and its corresponding components through the domination function.
\begin{lem}[Navarro et al.~\cite{nass1}]\label{II}
Let $\tau\left(\mbox{\boldmath$X$}\right)$ be the lifetime of a coherent system formed by $n$ d.i.d. components with the lifetime vector $\mbox{\boldmath$X$}=(X_1,X_2,\dots,X_n)$. Then the reliability function of $\tau\left(\mbox{\boldmath$X$}\right)$ can be written as
$$\bar F_{\tau\left(\mbox{\boldmath$X$}\right)}(x)=h\left(\bar F_{X}(x)\right),$$
where $h(\cdot):[0,1]\rightarrow [0,1]$, called the domination (or dual distortion) function, depends on the structure function $\phi(\cdot)$ (see Barlow and Proschan~\cite{bp} for definition) and on the survival copula $K$ of $X_1,X_2,\dots,X_n$. Furthermore, $h(\cdot)$ is an increasing continuous function in $[0,1]$ such that $h(0)=0$ and $h(1)=1$. $\hfill\Box$
\end{lem}
\hspace*{0.2 in}Below we give an example (borrowed from Navarro et al.~\cite{nass1}) that illustrates the result given in the above lemma.
\begin{ex}
Let $\tau\left(\mbox{\boldmath$X$}\right)=\min\{X_1,\max\{X_2,X_3\}\}$, where $\mbox{\boldmath$X$}=(X_1,X_2,X_3)$ is described by the FGM Copula (see Nelsen~\cite{n})
$$K(p_1,p_2,p_3)=p_1p_2p_3(1+\theta(1-p_1)(1-p_2)(1-p_3)), \text{ for }p_i\in(0,1),\; i=1,2,3,\text{ and }\theta \in[-1,1].$$
Further, let $X_1,X_2$ and $X_3$ be identically distributed with a random variable $X$. Then the minimal path sets (see Barlow and Proschan~\cite{bp}) of $\tau\left(\mbox{\boldmath$X$}\right)$ are given by $\{1,2\}$ and $\{1,3\}$. Let $X_{\{1,2\}}$, $X_{\{1,3\}}$ and $X_{\{1,2,3\}}$ be the lifetimes of the path sets $\{1,2\}$, $\{1,3\}$ and $\{1,2,3\}$, respectively. Then the reliability function of $\tau\left(\mbox{\boldmath$X$}\right)$ can be written as
\begin{eqnarray*}
\bar F_{\tau\left(\mbox{\boldmath$X$}\right)}(x)&=&P\left(\{X_{\{1,2\}}>x\}\cup\{X_{\{1,3\}}>x\}\right)
\\&=&P\left(X_{\{1,2\}}>x)+P(X_{\{1,3\}}>x\right)-P(X_{\{1,2,3\}}>x)
\\&=&\bar F_{\mathbf{X}}(x,x,0)+\bar F_{\mathbf{X}}(x,0,x)-\bar F_{\mathbf{X}}(x,x,x)
\\&=&K\left(\bar F_{X}(x),\bar F_{X}(x),1\right)+K\left(\bar F_{X}(x),1,\bar F_{X}(x)\right)-K\left(\bar F_{X}(x),\bar F_{X}(x),\bar F_{X}(x)\right)
\\&=&h\left(\bar F_{X}(x)\right),
\end{eqnarray*}
where
\begin{eqnarray*}
h(p)&=&K(p,p,1)+K(p,1,p)-K(p,p,p)
\\&=&2p^2-p^3-\theta p^3(1-p)^3, \text{ for }p\in(0,1)\text{ and }\theta\in[-1,1].
\end{eqnarray*}
\end{ex}
\hspace*{0.2 in}Stochastic orders are commonly used to compare two random variables (or two sets of random variables), and have been extensively studied in the literature due to their various applications in different branches of science and engineering. An encyclopedic information on this topic is nicely encapsulated in the book written by Shaked and Shanthikumar~\cite{ss} (also see Belzunce et al.~\cite{bmr}). For the sake of completeness, we give the following definitions of the stochastic orders that are used in our paper.
\begin{d1}\label{de1}
Let $X$ and $Y$ be two absolutely continuous random variables with cumulative distribution functions $F_X(\cdot)$ and $F_Y(\cdot)$, respectively, supported on $[0,\infty)$. Then $X$ is said to be smaller than $Y$ in
\begin{enumerate}
\item [$(a)$] hazard rate (hr) order, denoted as $X\leq_{hr}Y$, if $${\bar F_Y(x)}/{\bar F_X(x)}\text{ is increasing in } x \in [0,\infty);$$
\item [$(b)$] reversed hazard rate (rhr) order, denoted as $X\leq_{rhr}Y$, if $$ {F_Y(x)}/{ F_X(x)}\text{ is increasing in } x \in [0,\infty).$$
\end{enumerate}
\end{d1}
\hspace*{0.2 in}Similar to the above discussed stochastic orders, there are two more sets of stochastic orders which are useful to describe the relative ageings of two systems. The first set of stochastic orders, known as transform orders (namely, convex transform order, quantile mean inactivity time order, star-shaped order, super-additive order, DMRL order, s-IFR order, etc.), describes whether a system is ageing faster than another one in terms of the increasing failure rate, the increasing failure rate on average, the new better than used, etc. A detailed discussion on these orders could be found in Barlow and Proschan~\cite{bp}, Bartoszewicz~\cite{b}, Deshpande and Kochar~\cite{dk}, Kochar and Wiens~\cite{kw}, Arriaza et al.~\cite{ass}, Nanda et al.~\cite{nhga}, and the refernces therein. The second set of stochastic orders, called ageing faster orders, is defined based on monotonocity of ratios of some reliability measures, namely, hazard rate function, reversed hazard rate function, mean residual lifetime function, etc. For motivation and usefulness of these orders, we refer the reader to Kalashnikov and Rachev~\cite{kr}, Sengupta and Deshpande~\cite{sd}, Di Crescenzo~\cite{d}, Finkelstein~\cite{f6}, Razaei et al.~\cite{rgi}, Hazra and Nanda~\cite{hn0}, Misra et al.~\cite{mfn}, Kayid et al.~\cite{kiz}, and Misra and Francis~\cite{mf2}. Below we give the definitions of the ageing faster orders that are used in our paper.
\begin{d1}
Let $X$ and $Y$ be two absolutely continuous random variables with failure rate functions $r_X(\cdot)$ and $r_Y(\cdot)$, respectively, and reversed failure rate functions $\tilde r_X(\cdot)$ and $\tilde r_Y(\cdot)$, respectively.
Then $X$ is said to be ageing faster than $Y$ in
\begin{enumerate}
\item [$(a)$] failure rate, denoted as $X\underset{ c}{\prec} Y$, if
$$r_X(x)/r_Y(x)\text{ is increasing in }x\in[0,\infty);$$
\item [$(b)$] reversed failure rate, denoted as $X\underset{ b}{\prec} Y$, if
$$\tilde r_X(x)/\tilde r_Y(x)\text{ is decreasing in }x\in[0,\infty).$$
\end{enumerate}
\end{d1}
\hspace*{0.2 in}The theory of totally positive functions has various applications in different areas of probability and statistics (see Karlin~\cite{k}). Below we give the definitions of TP$_2$ and RR$_2$ functions. Different properties of these functions are used in proving the main results of our paper.
\begin{d1}
Let $\mathcal{X}$ and $\mathcal{Y}$ be two linearly ordered sets. Then, a real-valued function $\kappa(\cdot,\cdot)$ defined on $\mathcal{X}\times\mathcal{Y}$,
is said to be TP$_2$ (resp. RR$_2$) if
$$\kappa(x_1,y_1)\kappa(x_2,y_2)\geq(\text{resp. }\leq)\;\kappa(x_1,y_2)\kappa(x_2,y_1),$$
for all $x_1<x_2$ and $y_1<y_2$.$\hfill \Box$
\end{d1}
\hspace*{0.2 in}Throughout the paper increasing and decreasing, as usual, mean non-decreasing and non-increasing, respectively. Similarly, positive and negative mean non-negative and non-positive, respectively. Assume that all random variables considered in this paper are absolutely continuous and non-negative (i.e., distributional support is $[0,\infty)$). By $a\stackrel{\text{sgn}}=b$, we mean that $a$ and $b$ have the same sign, whereas $a\stackrel{\text{def.}}=b$ means that $a$ is defined as $b$. Further, we use bold symbols to represent vectors, and the symbol $\mathbb{N}$ is used to represent the set of natural numbers. We write $\tau_{k|n}$ and $\tau_{l|m}$ to represent the lifetimes of a $k$-out-of-$n$ and a $l$-out-of-$m$ systems, respectively. We use the acronyms $i.i.d.$ and $d.i.d.$ for `independent and identically distributed' and `dependent and identically distributed', respectively.
\\\hspace*{0.2 in}The rest of the paper is organized as follows. In Section~\ref{se2}, we discuss some useful lemmas which are intensively used in the proofs of the main results. In Section~\ref{se3}, we provide some sufficient conditions under which the lifetime of one coherent system is larger than that of an another system with respect to ageing faster orders in terms of the hazard and the reversed hazard rates. In Section~\ref{se4}, we discuss a redundancy allocation problem in a coherent system. We derive some necessary and sufficient conditions under which the allocation of active redundancy at the component level (of a coherent system) is superior to that at the system level with respect to ageing faster orders. Stochastic comparisons between a used coherent system and a coherent system made by used components are discussed in Section~\ref{se5}. The concluding remarks are given in Section~\ref{se6}.
\\\hspace*{0.2 in}All proofs of lemmas and theorems, wherever given, are deferred to the Appendix.
\section{Useful Lemmas}\label{se2}
In this section we discuss some lemmas which will be used in proving the main results of this paper. In the first lemma we discuss the sign change property of the integral of a function. The following lemma is adopted from Karlin~(\cite{k}, Theorem 11.2, pp. 324-325), and Hazra and Nanda~(\cite{hnx}, Lemma 3.5).
\begin{lem}\label{l1}
Let $\kappa(x,y)>0$, defined on $\mathcal{X}\times \mathcal{Y}$, be RR$_2$ (resp. TP$_2$), where $\mathcal{X}$ and $\mathcal{Y}$ are subsets of the real line.
Assume that a function $f(\cdot,\cdot)$ defined on $\mathcal{X}\times\mathcal{Y}$ is such that
\begin{itemize}
\item [$(i)$] for each $x\in \mathcal{X}$, $f(x,y)$ changes sign at most once, and if the change of sign does occur, it is from positive to negative, as $y$ traverses $\mathcal{Y}$;
\item [$(ii)$] for each $y \in \mathcal{Y}$, $f(x,y)$ is increasing (resp. decreasing) in $x\in \mathcal{X}$;
\item [$(iii)$] $\omega(x)=\int\limits_{\mathcal{Y}}\kappa(x,y)f(x,y)d\mu(y)$ exists absolutely and defines a continuous function of $x$, where $\mu$ is a sigma-finite measure.
\end{itemize}
Then $\omega(x)$ changes sign at most once, and if the change of sign does occur, it is from negative (resp. positive) to positive (resp. negative), as $x$ traverses $\mathcal{X}$.$\hfill\Box$
\end{lem}
\hspace*{0.2 in}In the following lemma we state an equivalent condition of a monotonic function. The proof is straightforward, and hence omitted.
\begin{lem}\label{l2}
Let $f(\cdot)$ and $g(\cdot)$ be two non-negative real-valued functions defined on $(a,b)\subseteq (0,\infty)$. Then $f(x)/g(x)$ is increasing (resp. decreasing) in $x$, if and only if for any real number $c$, the difference $f(x)-c g(x)$ changes sign at most once, and if the change of sign does occur, it is from negative (resp. positive) to positive (resp. negative),
as $x$ traverses from $a$ to $b$.$\hfill\Box$
\end{lem}
\hspace*{0.2 in}Some properties of the reliability functions of a $k$-out-of-$n$ and a $l$-out-of-$m$ systems are discussed in the next two lemmas. Lemma~\ref{l3} ($i$) is obtained in Esary and Proschan~\cite{ep}, whereas Lemma~\ref{l4} ($i$) is obtained in Nanda et al.~\cite{njs}. The other proofs are deferred to the Appendix.
\begin{lem}\label{l3}
Let $h_{k|n}(\cdot)$ and $h_{l|m}(\cdot)$ be the reliability functions of the $k$-out-of-$n$ and the $l$-out-of-$m$ systems with i.i.d. components, respectively, where $1\leq k\leq n$ and $1\leq l\leq m$. Further, let $H_{k|n}(p)=ph_{k|n}'(p)/h_{k|n}(p)$ and $H_{l|m}(p)=ph_{l|m}'(p)/h_{l|m}(p)$ for all $p\in(0,1)$. Then the following results hold.
\begin{itemize}
\item [$(i)$] $H_{k|n}(p)$ is decreasing in $p\in(0,1);$
\item [$(ii)$] ${H_{k|n}(p)}/{H_{l|m}(p)}$ is decreasing in $p\in(0,1)$, for all $k\leq l$ and $m-l\leq n-k$;
\item [$(iii)$] $(1-p){H_{k|n}'(p)}/{H_{k|n}(p)}$ is decreasing in $p\in (0,1)$.
\end{itemize}
\end{lem}
\begin{lem}\label{l4}
Let $h_{k|n}(\cdot)$ and $h_{l|m}(\cdot)$ be the reliability functions of the $k$-out-of-$n$ and the $l$-out-of-$m$ systems with i.i.d. components, respectively, where $1\leq k\leq n$ and $1\leq l\leq m$. Further, let $R_{k|n}(p)=(1-p)h_{k|n}'(p)/(1-h_{k|n}(p))$ and $R_{l|m}(p)=(1-p)h_{l|m}'(p)/(1-h_{l|m}(p))$ for all $p\in(0,1)$. Then the following results hold.
\begin{itemize}
\item [$(i)$] $R_{k|n}(p)$ is increasing in $p\in(0,1);$
\item [$(ii)$] ${R_{k|n}(p)}/{R_{l|m}(p)}$ is increasing in $p\in(0,1)$, for all $l\leq k$ and $n-k\leq m-l$;
\item [$(iii)$] $p{R_{k|n}'(p)}/{R_{k|n}(p)}$ is decreasing in $p\in (0,1)$.
\end{itemize}
\end{lem}
\section{Stochastic comparisons of two coherent systems}\label{se3}
In this section we compare two coherent systems with respect to ageing faster orders in terms of the failure and the reversed failure rates. We show that the proposed results hold for the $k$-out-of-$n$ and the $l$-out-of-$m$ systems with i.i.d. components.
\\\hspace*{0.2 in}Let $\tau_1\left(\mbox{\boldmath$X$}\right)$ and $\tau_2\left(\mbox{\boldmath$Y$}\right)$ (resp. $\tau_{k|n}\left(\mbox{\boldmath$X$}\right)$ and $\tau_{l|m}\left(\mbox{\boldmath$Y$}\right)$) be the lifetimes of two coherent systems (resp. $k$-out-of-$n$ and $l$-out-of-$m$ systems) formed by two different sets of d.i.d. components with the lifetime vectors $\mbox{\boldmath$X$}=(X_1,X_2,\dots,X_{n})$ and $\mbox{\boldmath$Y$}=(Y_1,Y_2,\dots,Y_{m})$, respectively. For the sake of simplicity of notation, let us assume that all $X_i$'s are identically distributed with a non-negative random variable $X$, and all $Y_j$'s are identically distributed with a non-negaive random variable $Y$. Further, let $h_1(\cdot)$ and $h_2(\cdot)$ be the domination functions of $\tau_1\left(\mbox{\boldmath$X$}\right)$ and $\tau_2\left(\mbox{\boldmath$Y$}\right)$, respectively. In what follows, we use the following notation. For $p\in(0,1)$,
$$H_i(p)=\frac{ph_i'(p)}{h_i(p)}, \quad i=1,2,$$ and $$R_i(p)=\frac{(1-p)h_i'(p)}{1-h_i(p)}, \quad i=1,2.$$
\hspace*{0.2 in}In the following theorem we show that under a set of sufficient conditions $\tau_1\left(\mbox{\boldmath$X$}\right)$ is ageing faster than $\tau_2\left(\mbox{\boldmath$Y$}\right)$ in terms of the failure rate.
\begin{thm}\label{t1}
Suppose that the following conditions hold.
\begin{itemize}
\item [$(i)$] $H_1(p)$ and ${H_1(p)}/{H_2(p)}$ are decreasing in $p\in(0,1)$;
\item [$(ii)$] $(1-p){H_1'(p)}/{H_1(p)}$ or $(1-p){H_2'(p)}/{H_2(p)}$ is decreasing in $p\in (0,1)$;
\item [$(iii)$] $X\underset{ c}{\prec} Y$ and $Y\leq_{rh} X $.
\end{itemize}
Then $\tau_1\left(\mbox{\boldmath$X$}\right)\underset{ c}{\prec}\tau_2\left(\mbox{\boldmath$Y$}\right)$.$\hfill\Box$
\end{thm}
\hspace*{0.2 in} The following corollary follows from Theorem~\ref{t1} and Lemma~\ref{l3}. It is worthwile to mention here that Theorem~3.1 (a) of Misra and Francis~\cite{mf} is the particular case of this corollary ($k=l$ and $m=n$).
\begin{cor}\label{c1}
Suppose that the $X_i$'s are i.i.d., and that the $Y_j$'s are i.i.d.
If $X\underset{ c}{\prec} Y$ and $Y\leq_{rh} X $, then ${\tau_{k|n}\left(\mbox{\boldmath$X$}\right)}\underset{ c}{\prec}$ ${\tau_{l|m}\left(\mbox{\boldmath$Y$}\right)}$ for $k\leq l$ and $m-l\leq n-k$.
\end{cor}
\begin{rem}
Let the assumption of Corollary~\ref{c1} hold. Then from Corollary~\ref{c1} we have
\begin{itemize}
\item [$(i)$] ${\tau_{k|n}\left(\mbox{\boldmath$X$}\right)}\underset{ c}{\prec}$ ${\tau_{l|n}\left(\mbox{\boldmath$Y$}\right)}$ for $k\leq l$;
\item [$(ii)$] ${\tau_{k|n}\left(\mbox{\boldmath$X$}\right)}\underset{ c}{\prec}$ ${\tau_{k|m}\left(\mbox{\boldmath$Y$}\right)}$ for $m\leq n$;
\item [$(iii)$] ${\tau_{l-r|m-r}\left(\mbox{\boldmath$X$}\right)}\underset{ c}{\prec}$ ${\tau_{l|m}\left(\mbox{\boldmath$Y$}\right)}$ for $r\leq l$.$\hfill\Box$
\end{itemize}
\end{rem}
\hspace*{0.2 in}One natural question may arise, which is whether the result stated in Theorem~\ref{t1} holds without the condition $Y\leq_{rh} X$. Below we cite a counterexample which shows that this condition could not be relaxed.
\begin{ce}
Consider two coherent systems $\tau_{1}(\mbox{\boldmath$X$})=\max\{X_1,X_2,X_3\}$ and $\tau_{2}(\mbox{\boldmath$Y$})=\max\{Y_1,Y_2,Y_3\}$, where $X_i$'s are i.i.d. with the reliability function given by $\bar F_X(x)=\exp\{-2x^3\}$, $x>0$, and $Y_i$'s are i.i.d. with the reliability function given by $\bar F_Y(x)=\exp\{-0.1x^2\}$, $x>0$. Then it could easily be verified that $X\underset{ c}{\prec} Y$ but $Y\nleq_{rh} X $ (In fact $Y\nleq_{st} X $). Now, by writing $k(x)=r_{\tau_{1}\left(\mbox{\boldmath$X$}\right)}(x)/r_{\tau_{2}\left(\mbox{\boldmath$Y$}\right)}(x)$, we have
\begin{eqnarray*}
k(x)=30x e^{-(2x^3-0.1x^2)}\left[\frac{1-\left(1-e^{-0.1 x^2}\right)^3}{1-\left(1-e^{-2 x^3}\right)^3}\right]\left[\frac{\left(1-e^{-2 x^3}\right)^2}{\left(1-e^{-0.1 x^2}\right)^2}\right],\quad x>0,
\end{eqnarray*}
which is non-monotone over $x>0$, and hence ${\tau_{1}\left(\mbox{\boldmath$X$}\right)}\underset{ c}{\nprec}$ ${\tau_{2}\left(\mbox{\boldmath$Y$}\right)}$.$\hfill\Box$
\end{ce}
\hspace*{0.2 in}In the following proposition we give a necessary and sufficient condition for the case when the lifetimes of the components of both coherent systems are identically distributed. The proof follows in the same line as in Theorem~\ref{t1}, and hence omitted.
\begin{pro}\label{p1}
Let $X_i$'s be identically distributed. Then
$\tau_1\left(\mbox{\boldmath$X$}\right)\underset{ c}{\prec}(\text{resp. }\underset{ c}{\succ})~\tau_2\left(\mbox{\boldmath$X$}\right)$ if and only if
\begin{eqnarray*}
{H_1(p)}/{H_2(p)}\text{ is decreasing (resp. increasing) in }p\in(0,1).
\end{eqnarray*}
\end{pro}
\hspace*{0.2 in}The following corollary, which is obtained in Theorem~2.1 of Misra and Francis~\cite{mf}, follows from Proposition~\ref{p1} and Lemma~\ref{l3}.
\begin{cor}\label{c2}
Suppose that the $X_i$'s are i.i.d. Then ${\tau_{k|n}\left(\mbox{\boldmath$X$}\right)}\underset{ c}{\prec}$ ${\tau_{l|m}\left(\mbox{\boldmath$X$}\right)}$ for $k\leq l$ and $m-l\leq n-k$.
\end{cor}
\begin{rem}
Let the assumption of Corollary~\ref{c2} hold. Then from Corollary~\ref{c2} we have
\begin{itemize}
\item [$(i)$] ${\tau_{k|n}\left(\mbox{\boldmath$X$}\right)}\underset{ c}{\prec}$ ${\tau_{l|n}\left(\mbox{\boldmath$X$}\right)}$ for $k\leq l$;
\item [$(ii)$] ${\tau_{k|n}\left(\mbox{\boldmath$X$}\right)}\underset{ c}{\prec}$ ${\tau_{k|m}\left(\mbox{\boldmath$X$}\right)}$ for $m\leq n$;
\item [$(iii)$] ${\tau_{l-r|m-r}\left(\mbox{\boldmath$X$}\right)}\underset{ c}{\prec}$ ${\tau_{l|m}\left(\mbox{\boldmath$X$}\right)}$ for $r\leq l$.$\hfill\Box$
\end{itemize}
\end{rem}
\hspace*{0.2 in}The following corollary given in Ding and Zhang~\cite{dz} follows from Proposition~\ref{p1}. It shows that a series system ages faster (in terms of the hazard rate) as its number of components increases whereas the reverse scenario is observed for the parallel system.
\begin{cor}
Suppose that the $X_i$'s are d.i.d.components with the common Archimedean copula generated by $\phi(\cdot)$. If $x\ln'\left[-\phi'(x)/(1-\phi(x))\right]$ is decreasing in $x>0$, then
\begin{itemize}
\item [$(i)$] ${\tau_{1|n}\left(\mbox{\boldmath$X$}\right)}\underset{ c}{\prec}$ ${\tau_{1|m}\left(\mbox{\boldmath$X$}\right)}$ for $m\leq n$;
\item [$(ii)$] ${\tau_{n|n}\left(\mbox{\boldmath$X$}\right)}\underset{ c}{\prec}$ ${\tau_{m|m}\left(\mbox{\boldmath$X$}\right)}$ for $n\leq m$. $\hfill\Box$
\end{itemize}
\end{cor}
\hspace*{0.2 in}Below we give an example that illustrates the result given in Proposition~\ref{p1}.
\begin{ex}\label{exx1}
Consider two coherent systems ${\tau_{1}\left(\mbox{\boldmath$X$}\right)}=\min\{X_1,\max\{X_2,X_3\}\}$ and ${\tau_{2}\left(\mbox{\boldmath$X$}\right)}=\min\{X_1,X_2,X_3\}$ which are formed by three identical components with lifetimes $X_1,X_2$ and $X_3$. Further, let the joint distribution function of $(X_1,X_2,X_3)$ be described by the FGM copula
$$K(p_1,p_2,p_3)=p_1p_2p_3(1+\theta (1-p_1)(1-p_2)(1-p_3)),$$
where $p_i\in(0,1)$, $i=1,2,3$, and $\theta\in [-1,1]$.
Then the domination functions of ${\tau_{1}\left(\mbox{\boldmath$X$}\right)}$ and ${\tau_{2}\left(\mbox{\boldmath$X$}\right)}$ are, respectively, given by
\begin{eqnarray*}
h_1(p)=2p^2-p^3-\theta p^3 (1-p)^3, \quad 0<p<1
\end{eqnarray*}
and
\begin{eqnarray*}
h_2(p)=p^3+\theta p^3(1-p)^3, \quad 0<p<1.
\end{eqnarray*}
These give
\begin{eqnarray*}
H_1(p)=\frac{ph_1'(p)}{h_1(p)}=\frac{4p^2-3(1+\theta)p^3+12\theta p^4-15\theta p^5+6\theta p^6}{2p^2-(1+\theta)p^3+3\theta p^4-3\theta p^5+\theta p^6}, \quad 0<p<1
\end{eqnarray*}
and
$$H_2(p)=\frac{ph_2'(p)}{h_2(p)}=\frac{3(1+\theta)p^3-6\theta p^6-12\theta p^4+15 \theta p^5}{(1+\theta)p^3-\theta p^6-3\theta p^4+3\theta p^5},\quad 0<p<1.$$
Writing $s_\theta(p)=H_1(p)/H_2(p)$, we have
$$s_\theta(p)=\frac{(4p^2-3(1+\theta)p^3+12\theta p^4-15\theta p^5+6\theta p^6)((1+\theta)p^3-\theta p^6-3\theta p^4+3\theta p^5)}{(2p^2-(1+\theta)p^3+3\theta p^4-3\theta p^5+\theta p^6)(3(1+\theta)p^3-6\theta p^6-12\theta p^4+15 \theta p^5)},\quad 0<p<1.$$
Now, consider the following two cases.
\\Case-I: Let $\theta=-.9,-.8,\dots, .4,.5$. Then it can be verified that $s_\theta(p)=H_1(p)/H_2(p)$ is decreasing in $p\in(0,1)$, and hence $\tau_1\left(\mbox{\boldmath$X$}\right)\underset{ c}{\prec}\tau_2\left(\mbox{\boldmath$X$}\right)$ follows from Proposition~\ref{p1}.
\\Case-II: Let $\theta=.75,.80,\dots, .95,1$. Then it can be checked that $s_\theta(p)=H_1(p)/H_2(p)$ is non-monotone over $p\in(0,1)$. Hence, by Proposition~\ref{p1}, we get that neither $\tau_1\left(\mbox{\boldmath$X$}\right)\underset{ c}{\prec}\tau_2\left(\mbox{\boldmath$X$}\right)$ nor $\tau_1\left(\mbox{\boldmath$X$}\right)\underset{ c}{\succ}\tau_2\left(\mbox{\boldmath$X$}\right)$ holds.
$\hfill\Box$
\end{ex}
\hspace*{0.2 in}In the following theorem we compare $\tau_1\left(\mbox{\boldmath$X$}\right)$ and $\tau_2\left(\mbox{\boldmath$Y$}\right)$ with respect to the ageing faster order in the reversed hazard rate.
\begin{thm}\label{t2}
Suppose that the following conditions hold.
\begin{itemize}
\item [$(i)$] $R_1(p)$ and ${R_1(p)}/{R_2(p)}$ are increasing in $p\in(0,1)$;
\item [$(ii)$] $p{R_1'(p)}/{R_1(p)}$ or $p{R_2'(p)}/{R_2(p)}$ is decreasing in $p\in (0,1)$;
\item [$(iii)$] $X\underset{ b}{\prec} Y$ and $X\leq_{hr} Y $.
\end{itemize}
Then $\tau_1\left(\mbox{\boldmath$X$}\right)\underset{ b}{\prec}\tau_2\left(\mbox{\boldmath$Y$}\right)$.$\hfill\Box$
\end{thm}
\hspace*{0.2 in} The following corollary immediately follows from Theorem~\ref{t2} and Lemma~\ref{l4}. Note that Theorem 3.1(b) of Misra and Francis~\cite{mf} is a particular case of this corollary ($k=l$ and $m=n$).
\begin{cor}\label{c3}
Suppose that the $X_i$'s are i.i.d., and that the $Y_j$'s are i.i.d.
If $X\underset{ b}{\prec} Y$ and $X\leq_{hr} Y $, then ${\tau_{k|n}\left(\mbox{\boldmath$X$}\right)}\underset{ b}{\prec}$ ${\tau_{l|m}\left(\mbox{\boldmath$Y$}\right)}$ for $l\leq k$ and $ n-k\leq m-l$.
\end{cor}
\begin{rem}
Let the assumption of Corollary~\ref{c3} hold. Then from Corollary~\ref{c3} we have
\begin{itemize}
\item [$(i)$] ${\tau_{k|n}\left(\mbox{\boldmath$X$}\right)}\underset{ b}{\prec}$ ${\tau_{l|n}\left(\mbox{\boldmath$Y$}\right)}$ for $l\leq k$;
\item [$(ii)$] ${\tau_{k|n}\left(\mbox{\boldmath$X$}\right)}\underset{ b}{\prec}$ ${\tau_{k|m}\left(\mbox{\boldmath$Y$}\right)}$ for $n\leq m$;
\item [$(iii)$] ${\tau_{k|n}\left(\mbox{\boldmath$X$}\right)}\underset{ b}{\prec}$ ${\tau_{k-r|n-r}\left(\mbox{\boldmath$Y$}\right)}$ for $r\leq k$. $\hfill\Box$
\end{itemize}
\end{rem}
\hspace*{0.2 in}The following counterexample shows that the result given in Theorem~\ref{t2} may not hold without the condition $X\leq_{hr}Y.$
\begin{ce}
Consider the coherent systems $\tau_{1}(\mbox{\boldmath$X$})=\min\{X_1,X_2\}$ and $\tau_{2}(\mbox{\boldmath$Y$})=\min\{Y_1,Y_2\}$, where $X_i$'s are i.i.d. with the cumulative distribution function given by $ F_X(x)=\exp\{-(2.1/x)^7\}$, $x>0$, and $Y_i$'s are i.i.d. with the cumulative distribution function given by $ F_Y(x)=\exp\{-(2/x)^3\}$, $x>0$. Then it is easy to verify that $X\underset{ b}{\prec} Y$ but $X\nleq_{hr} Y $ (In fact $X\nleq_{st} Y $). Now, by writing $l(x)=\tilde r_{\tau_{2}\left(\mbox{\boldmath$Y$}\right)}(x)/\tilde r_{\tau_{1}\left(\mbox{\boldmath$X$}\right)}(x)$, we have
\begin{eqnarray*}
l(x)=\left[\frac{24 x^4 e^{-(2/x)^3}\left(1-e^{-(2/x)^3}\right)}{7\times 2.1^7e^{-(2.1/x)^7}\left(1-e^{-(2.1/x)^7}\right)}\right]\left[\frac{1-\left(1-e^{-(2.1/x)^7}\right)^2}{1-\left(1-e^{-(2/x)^3}\right)^2}\right],\quad x>0,
\end{eqnarray*}
which is non-monotone over $x>0$, and hence ${\tau_{2|2}\left(\mbox{\boldmath$X$}\right)}\underset{ b}{\nprec}$ ${\tau_{2|2}\left(\mbox{\boldmath$Y$}\right)}$.$\hfill\Box$
\end{ce}
\hspace*{0.2 in}In the following proposition we discuss an analog of Proposition~\ref{p1} for the ageing faster order in the reversed hazard rate.
\begin{pro}\label{p2}
Let $X_i$'s be identically distributed. Then
$\tau_1\left(\mbox{\boldmath$X$}\right)\underset{ b}{\prec}(\text{resp. }\underset{ b}{\succ})~\tau_2\left(\mbox{\boldmath$X$}\right)$ if and only if
$${R_1(p)}/{R_2(p)}\text{ is increasing (resp. decreasing) in }p\in(0,1).$$
\end{pro}
\hspace*{0.2 in}The following corollary given in Theorem 2.2 of Misra and Francis~\cite{mf} follows from Proposition~\ref{p2} and Lemma~\ref{l4}.
\begin{cor}\label{c4}
Suppose that the $X_i$'s are i.i.d. Then ${\tau_{k|n}\left(\mbox{\boldmath$X$}\right)}\underset{ b}{\prec}$ ${\tau_{l|m}\left(\mbox{\boldmath$X$}\right)}$ for $l\leq k$ and $ n-k\leq m-l$.
\end{cor}
\begin{rem}
Let the assumption of Corollary~\ref{c4} hold. Then from Corollary~\ref{c4} we have
\begin{itemize}
\item [$(i)$] ${\tau_{k|n}\left(\mbox{\boldmath$X$}\right)}\underset{ b}{\prec}$ ${\tau_{l|n}\left(\mbox{\boldmath$X$}\right)}$ for $l\leq k$;
\item [$(ii)$] ${\tau_{k|n}\left(\mbox{\boldmath$X$}\right)}\underset{ b}{\prec}$ ${\tau_{k|m}\left(\mbox{\boldmath$X$}\right)}$ for $n\leq m$;
\item [$(iii)$] ${\tau_{k|n}\left(\mbox{\boldmath$X$}\right)}\underset{ b}{\prec}$ ${\tau_{k-r|n-r}\left(\mbox{\boldmath$X$}\right)}$ for $r\leq k$.$\hfill\Box$
\end{itemize}
\end{rem}
\hspace*{0.2 in}The following corollary obtained in Ding and Zhang~\cite{dz} immediately follows from Proposition~\ref{p2}. It shows that a series system ages faster (in terms of the reversed hazard rate) as its number of components decreases whereas the reverse scenario is observed for the parallel system.
\begin{cor}
Suppose that the $X_i$'s are d.i.d. components with the common Archimedean copula generated by $\phi(\cdot)$. If $x\ln'\left[-\phi'(x)/\phi(x)\right]$ is decreasing (resp. increasing) in $x>0$, then
\begin{itemize}
\item [$(i)$] ${\tau_{1|n}\left(\mbox{\boldmath$X$}\right)}\underset{ b}{\succ}(\text{resp. }\underset{ b}{\prec})$ ${\tau_{1|m}\left(\mbox{\boldmath$X$}\right)}$ for $m\leq n$;
\item [$(ii)$] ${\tau_{n|n}\left(\mbox{\boldmath$X$}\right)}\underset{ b}{\succ}(\text{resp. }\underset{ b}{\prec})$ ${\tau_{m|m}\left(\mbox{\boldmath$X$}\right)}$ for $n\leq m$. $\hfill\Box$
\end{itemize}
\end{cor}
\hspace*{0.2 in}The result stated in Proposition~\ref{p2} is revealed through the following example.
\begin{ex}\label{exx2}
Consider two coherent systems which are discussed in Example~\ref{exx1}. Then
\begin{eqnarray*}
R_1(p)=\frac{(1-p)h_1'(p)}{1-h_1(p)}=\frac{4p-(7+3\theta)p^2+3(1+5\theta)p^3-27\theta p^4+21\theta p^5-6\theta p^6}{1-2p^2+(1+\theta)p^3-3\theta p^4+3\theta p^5-\theta p^6},\quad 0<p<1
\end{eqnarray*}
and
$$R_2(p)=\frac{(1-p)h_2'(p)}{1-h_2(p)}=\frac{3(1+\theta)p^2-3(1+5\theta)p^3+27\theta p^4-21\theta p^5+6\theta p^6}{1-(1+\theta)p^3+\theta p^6+3\theta p^4-3\theta p^5},\quad 0<p<1.$$
Writing $v_\theta(p)=R_2(p)/R_1(p)$, we have
\begin{eqnarray*}
v_\theta(p)=&&\frac{3(1+\theta)p^2-3(1+5\theta)p^3+27\theta p^4-21\theta p^5+6\theta p^6}{1-(1+\theta)p^3+\theta p^6+3\theta p^4-3\theta p^5}
\\&&\times \frac{1-2p^2+(1+\theta)p^3-3\theta p^4+3\theta p^5-\theta p^6}{4p-(7+3\theta)p^2+3(1+5\theta)p^3-27\theta p^4+21\theta p^5-6\theta p^6},\quad 0<p<1.
\end{eqnarray*}
For $\theta=-1,-.8,\dots, .8,1$, it can be verified that $v_\theta(p)$ is increasing in $p\in(0,1)$. Hence $\tau_1\left(\mbox{\boldmath$X$}\right)\underset{ b}{\succ}\tau_2\left(\mbox{\boldmath$X$}\right)$ follows from Proposition~\ref{p2}.
\end{ex}
\section{Stochastic comparisons of coherent systems with active redundancy at the component level versus the system level}\label{se4}
Let ${\mbox{\boldmath $X$}}=(X_1,X_2,\dots,X_n)$ be a vector of random variables representing the lifetimes of $n$ d.i.d. components. Further, let $\{{\mbox{\boldmath $Y$}}_1,{\mbox{\boldmath $Y$}}_2,\dots,{\mbox{\boldmath $Y$}}_m\}$ be a set of $m$ vectors representing the lifetimes of $mn$ spares (or redundant components), where ${\mbox{\boldmath $Y$}}_i=(Y_{i1},Y_{i2},\dots,Y_{in})$ is a vector of $n$ d.i.d. random variables, for $i=1,2,\dots,m$. Assume that all $X_j$'s and $Y_{ij}$'s are identically distributed with a non-negaive random variable $X$. We write $T_C=\tau\left({\mbox{\boldmath $X \vee Y_1\vee Y_2\vee \dots \vee Y_m$}}\right)$ to denote the lifetime of a coherent system with active redundancies at the component level, where the symbol ${\mbox{\boldmath $X \vee Y_1\vee Y_2\vee \dots \vee Y_m$}}$ stands for a $n$-tuple vector ${\mbox{\boldmath $Z$}}=(Z_1,Z_2,\dots,Z_n)$ such that $Z_j$ represents the lifetime of a parallel system formed by $(m+1)$ independent components $\{X_j,Y_{1j},\dots,Y_{mj}\}$, for $j=1,2,\dots,n$. Further, we write
$T_S=\tau({\mbox{\boldmath $X$}})\vee \tau({\mbox{\boldmath $Y$}}_1)\vee \tau({\mbox{\boldmath $Y$}}_2)\vee \dots \vee \tau({\mbox{\boldmath $Y$}}_m)$ to denote the lifetime of a coherent system with active redundancies at the system level, where the symbol $\vee$ stands for maximum. Furthermore, it is assumed that $\tau({\mbox{\boldmath $X$}})$ and $ \tau({\mbox{\boldmath $Y$}}_i)$'s are independent, and they have the same domination function as $\tau({\mbox{\boldmath $Z$}})$ has. We denote this domination function by $h(\cdot)$. In what follows, we use the notation $R(p)=(1-p)h'(p)/(1-h(p))$, $p\in(0,1)$.
\\\hspace*{0.2 in}In the following theorem, we provide an equivalent condition to hold that the allocation of redundancy at the component level is better/worse than that at the system level with respect to the ageing faster order in terms of the hazard rate.
\begin{thm}\label{t41}
For $m\in \mathbb{N}$,
$T_S \underset{ c}{\prec}(\text{resp. }\underset{ c}{\succ})~ T_C$ holds if and only if
\begin{eqnarray}\label{km0}
\left(\frac{\left(1-h(p)\right)^{m}h'\left(p\right)}{1-\left(1-h(p)\right)^{m+1}}\right)\left(\frac{h\left(1-(1-p)^{m+1}\right)}{\left(1-p\right)^{m}h'\left(1-(1-p)^{m+1}\right)}\right)
\end{eqnarray}
$\text{ is decreasing (resp. increasing) in }p\in(0,1).$ $\hfill\Box$
\end{thm}
\hspace*{0.2 in}The following corollary follows from Theorem~\ref{km0}.
\begin{cor}\label{co1}
If all $X_i$'s and $Y_j$'s are i.i.d., then $\tau_{n|n}\left(\mbox{\boldmath$X$}\right)\vee \tau_{n|n}\left(\mbox{\boldmath$Y_1$}\right)\underset{ c}{\succ}\tau_{n|n}\left(\mbox{\boldmath$X\vee Y_1$}\right).$ $\hfill\Box$
\end{cor}
\hspace*{0.2 in}In the next theorem we discuss an analog of Theorem~\ref{t41} under the ageing faster order in the reversed hazard rate.
\begin{thm}\label{t3}
For $m\in \mathbb{N}$,
$T_S \underset{ b}{\prec} T_C$ holds if and only if
\begin{eqnarray*}\label{cs1}
~~~~~~~~~~~~~~~~~~~~~~~~~
\frac{R(p)}{R(1-(1-p)^{m+1})}\text{ is increasing in }p\in(0,1).
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\hfill\Box
\end{eqnarray*}
\end{thm}
\hspace*{0.2 in}Since the condition given in Theorem~\ref{t3} is involved with $m$, it is practically not easy to verify. In the following proposition we discuss a sufficient condition that could be useful to show the result.
\begin{pro}\label{p41}
If $pR'(p)/R(p)$ is decreasing and positive for all $p\in(0,1)$, then $T_S \underset{ b}{\prec} T_C$.
\end{pro}
\hspace*{0.2 in}The following corollary follows from Proposition~\ref{p41} and Lemma~\ref{l4}.
\begin{cor}
Suppose that all $X_i$'s and $Y_j$'s are i.i.d. Then, for $1\leq k\leq n,$ $$\tau_{k|n}\left(\mbox{\boldmath$X$}\right)\vee \tau_{k|n}\left(\mbox{\boldmath$Y_1$}\right)\vee \dots \vee\tau_{k|n}\left(\mbox{\boldmath$Y_m$}\right)\underset{ b}{\prec}\tau_{k|n}\left(\mbox{\boldmath$X\vee Y_1 \vee\dots \vee Y_m$}\right).$$
\end{cor}
\hspace*{0.2 in}Below we provide an example that illustrates the result given in Proposition~\ref{p41}.
\begin{ex}\label{exx3}
Let $m=1$. Consider a coherent system ${\tau\left(\mbox{\boldmath$X$}\right)}=\min\{X_1,X_2,\dots,X_{n}\}$ formed by $n$ identical components with the lifetime vector $\mbox{\boldmath$X$}=(X_1,X_2,\dots,X_{n})$. Further, let $\{X_1,X_2,\dots,X_n\}$ have the Gumbel-Hougard copula given by
$$K(p_1,p_2,\dots,p_{n})=\exp\left\{-\left(\sum\limits_{i=1}^{n}(-\ln p_i)^\theta\right)^{1/\theta}\right\},$$
where $p_i\in(0,1)$, $i=1,2,\dots,n$, and $\theta\in [1,\infty)$.
Then the domination function of ${\tau\left(\mbox{\boldmath$X$}\right)}$ is given by
$h(p)=p^{a}$,
where $a=n^{1/\theta}$ ($\geq 1$). This gives
$$R(p)=\frac{(1-p)h'(p)}{(1-h(p))}=\frac{a\left(p^{a-1}-p^a\right)}{1-p^a},\quad 0<p<1$$
and
$$\frac{pR'(p)}{R(p)}=\frac{a-1-ap+p^a}{1-p-p^a+p^{a+1}},\quad 0<p<1.$$
Since $((1-p^a)/(1-p))\leq a$, for all $a\geq 1$ and $p\in(0,1)$, we have $pR'(p)/R(p)\geq 0$, for all $p\in(0,1)$. Further,
\begin{eqnarray*}
\left[\frac{pR'(p)}{R(p)}\right]'=\frac{\gamma_1(p)}{\left(1-p-p^a+p^{a+1}\right)^2}, \quad 0<p<1,
\end{eqnarray*}
where $$\gamma_1(p)=a^2p^{a-1}+2(1-a^2)p^a+a^2 p^{a+1}-p^{2a}-1,\quad 0<p<1.$$
\end{ex}
Now,
$${\gamma_1}'(p)=p^{a-2}\gamma_2(p),\quad 0<p<1,$$
where $$\gamma_2(p)=a^2(a-1)-2a(a^2-1)p+a^2(a+1)p^2-2ap^{a+1},\quad 0<p<1.$$
Differentiating $\gamma_2(p)$ twice, we get
\begin{eqnarray*}
&&\gamma_2'(p)=-2a(a^2-1)+2a^2(a+1)p-2a(a+1)p^a,\quad 0<p<1
\end{eqnarray*}
and
\begin{eqnarray*}
\gamma_2''(p)=2a^2(a+1)\left(1-p^{a-1}\right)\geq 0,\quad 0<p<1.
\end{eqnarray*}
Thus, we have $\gamma_2'(p)\leq \gamma_2'(1)=0$, for all $p\in(0,1)$, which implies $\gamma_2(p)\geq \gamma_2(1)=0$, for all $p\in(0,1)$. Further, this implies $\gamma_1'(p)\geq 0$, for all $p\in(0,1)$, which gives $\gamma_1(p)\leq \gamma_1(1)=0$, for all $p\in(0,1)$.
Hence $pR'(p)/R(p)$ is decreasing in $p\in(0,1)$.
Thus, $T_S \underset{ b}{\prec} T_C$ follows from Proposition~\ref{p41}.
\section{Stochastic comparisons of a used coherent system and a coherent system of used components}\label{se5}
Let $X$ be a random variable representing the lifetime of a component/system. Then its residual lifetime at a time instant $t$ ($>0$) is denoted by $X_t$ and is defined as
$$X_t=(X-t|X>t).$$
We call $X_t$ as a used component/system. Let $\mbox{\boldmath$X$}=(X_1,X_2,\dots,X_n)$ be a vector of random variables representing the lifetimes of $n$ d.i.d. components. Then we write
$$\mbox{\boldmath$X$}_t=\left((X_1)_t,(X_2)_t,\dots,(X_n)_t\right),\quad t>0,$$
to represent a vector of $n$ used components $\{(X_1)_t,(X_2)_t,\dots,(X_n)_t\}$, $t>0$. Consequently, we write $\tau\left(\mathbf{X}_{t}\right)$ to denote the lifetime of a coherent system made by a set of components with the lifetime vector $\mbox{\boldmath$X$}_t$. Further, by $\left(\tau\left(\mathbf{X}\right)\right)_{t}=(\tau\left(\mathbf{X}\right)-t|\tau\left(\mathbf{X}\right)>t)$, we mean the lifetime of a used coherent system formed by a set of components with the lifetime vector $\mbox{\boldmath$X$}$. For the sake of simplicity, we assume that all $X_i$'s are identically distributed with a non-negative random variable $X$. In what follows, we denote the reliability function of $\tau\left(\mathbf{X}\right)$ by $h(\cdot)$, and we write $H(p)=ph'(p)/h(p)$, $0<p<1$.
\\\hspace*{0.2 in}In the following theorem we derive the necessary and sufficient condition for a used coherent system to be ageing faster than a coherent system of used components in terms of the hazard rate.
\begin{thm}\label{th1}
For any fixed $t\geq0$, $\tau\left(\mathbf{X}_{t}\right)\underset{ c}{\prec}\left(\tau\left(\mathbf{X}\right)\right)_{t}$ holds if and only if
\begin{eqnarray}\label{rs2}
{pH'(p)}/{H(p)}\text{ is decreasing in }p\in(0,1).
\end{eqnarray}
\end{thm}
\hspace*{0.2 in}In the following proposition we discuss the same result as in Theorem~\ref{th1} under a different set of sufficient conditions which is sometimes easy to verify. The proof follows from Theorem~\ref{th1}. Hence we omit it.
\begin{pro}\label{pp1}
For any fixed $t\geq 0$,
$\tau\left(\mbox{\boldmath$X$}_t\right)\underset{ c}{\prec}\left(\tau\left(\mbox{\boldmath$X$}\right)\right)_t$ holds if
\begin{eqnarray*}
(1-p){H'(p)}/{H(p)} \text{ is decreasing and negative in }p\in(0,1).
\end{eqnarray*}
\end{pro}
\hspace*{0.2 in}The following corollary follows from Proposition~\ref{pp1} and Lemma~\ref{l3}.
\begin{cor}
If the $X_i$'s are i.i.d., then ${\tau_{k|n}\left(\mbox{\boldmath$X$}_t\right)}\underset{ c}{\prec}\left({\tau_{k|n}\left(\mbox{\boldmath$X$}\right)}\right)_t$, for any fixed $t\geq 0$, and $1\leq k\leq n$.$\hfill\Box$
\end{cor}
\hspace*{0.2 in}In the following theorem we show a similar result as in Theorem~\ref{th1} for the ageing faster order in the reversed hazard rate.
\begin{thm}\label{th2}
For any fixed $t\geq 0$,
$\tau\left(\mbox{\boldmath$X$}_t\right)\underset{ b}{\prec}(\text{resp. }\underset{ b}{\succ})\;\left(\tau\left(\mbox{\boldmath$X$}\right)\right)_t$ holds if and only if, for all $q\in(0,1)$,
\begin{eqnarray}\label{rs0}
\left[\frac{h'(p/q)}{h'(p)}\right]\left[\frac{h(q)-h(p)}{1-h(p/q)}\right]\text{ is increasing (resp. decreasing) in }p\in(0,q).
\end{eqnarray}
\end{thm}
\hspace*{0.2 in}As a consequence of Theorem~\ref{th2}, we have the following corollary.
\begin{cor}\label{c02}
If the $X_i$'s are i.i.d., then $\tau_{1|n}\left(\mbox{\boldmath$X$}_t\right)\underset{ b}{\prec}\;\left(\tau_{1|n}\left(\mbox{\boldmath$X$}\right)\right)_t$, for any fixed $t\geq 0$.
\end{cor}
\section{Concluding Remarks}\label{se6}
In this paper, we study ageing faster orders (in terms of the hazard and the reversed hazard rates) which are useful to compare the relative ageings of two systems. To be more specific, we provide sufficient conditions under which one coherent system is ageing more faster than another one with respect to the hazard and the reversed hazard rates. Further, we consider a problem of allocation of redundancies into a coherent system. We show that, under some necessary and sufficient conditions, the allocation of active redundancy at the component level is superior (inferior) to that at the system level with respect to ageing faster orders, for a coherent system. Furthermore, a used coherent system and a coherent system made out of used components are compared with respect to these ageing faster orders.
Apart from these, we also show that most of our developed results hold for the well known $k$-out-of-$n$ and the $l$-out-of-$m$ systems. Nevertheless, we provide a list of examples to illustrate our proposed results. Some counterexamples are also given wherever needed.
\\\hspace*{0.2 in}Even though a vast literature exists on the study of different stochastic orders, there are a few results developed for the ageing faster orders. Since the ageing faster orders compare the relative ageings of two systems and the ageing is a common phenomenon experienced by each and every system, the study of ageing faster orders should be paid more attention from the researchers across the world. We believe that our study not only enriches the literature on ageing faster orders but also may be useful in some practical scenarios.
\\\hspace*{0.2 in}Similar to the problems considered in this paper, the study of other stochastic orders (as discussed in the introduction section), which describe the relative ageings of two systems, is under investigation, and will be reported in future.
\subsection*{Acknowledgments}
\hspace*{0.2 in}The authors are thankful to the Editor-in-Chief, the Associate Editor and the anonymous Reviewers for their valuable constructive comments/suggestions which lead to an improved version of the manuscript. The first author sincerely acknowledges the financial support from the IIT Jodhpur, Karwar-$342037$, India.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
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\section{Introduction}\label{sect:intro}
In statistical learning theory, empirical risk minimization is a standard technique whereby classifiers are formed from empirical data~\cite{devroye-gyorfi-lugosi1996}.
The idea is simple enough: when the underlying probability distributions characterizing the data are unknown, classifiers are found by minimizing an empirical form of the risk (probability of error) over some specified class of classifiers.
The technique is well-understood and has been generalized to include various cost criteria and problem settings.
In its generalized form, empirical risk minimization is sometimes referred to as $M$-estimation (the $M$ standing for minimization or maximization)~\cite{vandegeer2007}.
Recently, Nguyen, Wainwright and Jordan~\cite{nguyen_wainwright_jordan2010} applied $M$-estimation to the estimation of $f$-divergences (the Kullback-Leibler (KL) divergence~\cite{kullback1959} in particular) and to bounded likelihood ratio functions.
In this paper, we build on their ideas and develop a method for computing empirical quantization rules by maximizing the KL divergence.
We call the method \emph{empirical divergence maximization} (EDM) in deference to its similarity to empirical risk minimization and because the name is simple and descriptive.
The proposed formulation leads to an entirely different algorithm for computing the estimators than that employed in~\cite{nguyen_wainwright_jordan2010}, and the convergence rates reported here incorporate a margin condition not included in~\cite{nguyen_wainwright_jordan2010} that shows when fast convergence is possible.
As the name suggests, the criterion used in EDM is the KL divergence, a well-known information theoretic quantity that has enjoyed a prominent and long-standing place in both theory and practice.
Applications are numerous and range from detection and estimation problems~\cite{poor_thomas1977,chamberland_veeravalli2003,lhler_etal2004} to texture retrieval in image databases~\cite{do_vetterli2002} and from the study of neural coding~\cite{johnson_gruner2001} to linguistic problems~\cite{cai_kulkarni_verdu}.
Roughly speaking, the KL divergence quantifies the dissimilarity of two probability density functions (pdfs) and is therefore often regarded as a ``distance'', although it is not a distance metric.
Stein's Lemma~\cite[p. 309] {cover_thomas1991} fundamentally links the divergence to hypothesis testing by relating it to the decay rate of different error probabilities.
In fact, the divergence equals the optimal asymptotic error decay rate of a Neyman-Pearson test.
Thus increasing the divergence between two statistical hypotheses generally increases their discriminability.
The use of the KL divergence in quantization problems dates back nearly four decades~\cite{poor_thomas1977}.
In that time, various problem settings have been investigated including scalar, vector, and distributed quantization~\cite{poor_thomas1977,kassam1988,longo_lookabaugh_gray1990,gupta_hero2003}.
Until recently, however, most results addressing this type of quantization assumed full knowledge of the probability distributions of interest and did not explicitly address empirical designs.
Moreover, those works in the quantization literature most closely related to the present paper~\cite{gupta_hero2003,lazebnik_raginsky2009} invoke a small-cell assumption that forces partitions, designed to maximize the divergence, to resemble nearest neighbor partitions even when such partitions cannot well-approximate theoretically optimal partitions (see Fig.~\ref{fig:lr}).
Because of its flexibility, however, the EDM approach overcomes this shortcoming.
In~\cite{lazebnik_raginsky2009}, Lazebnik and Raginsky study a conceptually similar quantization problem to the one considered here, but the differences between the approaches are substantial.
For example, their information loss criterion is a difference of mutual informations, and while related to the KL divergence, this criterion measures a different quantity than the divergence loss studied here.
Their work is also placed in a machine learning setting where the data and the quantization values (labels) are jointly distributed and both play integral roles in their information criterion.
In this paper, the quantization values play a secondary role in the computation of our estimator.
To formalize the problem, let $\probmeasureHzero\text{~and~}\probmeasureHone$ be two probability measures defined on the probability space $([0,1]^{\ddimen},\mathcal{B})$, where $\mathcal{B}$ denotes the usual Borel $\sigma$-algebra and $d\geq1$.
Let $\disHzero$ and $\disHone$ denote the density functions of $\probmeasureHzero\text{~and~}\probmeasureHone$ with respect to Lebesgue measure and assume $P$ and $Q$ are absolutely continuous with respect to one another.
Then any quantization rule $\mapc:\mathbb{R}^{\ddimen} \mapsto \{0,\dotsc,\asize-1\}$ that operates on a random vector $X$ (distributed according to $\probmeasureHzero$ or $\probmeasureHone$) induces the probability mass functions (pmfs),
$\disHzero(\mapc)=(\disHzero_{0}(\mapc),\dotsc,\disHzero_{\asize-1}(\mapc))$ and $\disHone(\mapc)=(\disHone_{0}(\mapout),\dotsc,\disHone_{\asize-1}(\mapc))$, where $\disHzero_{\aindex}(\mapc)=\probmeasureHzero(\mapc(X)=\aindex)$ and similarly for $\disHone_{\aindex}(\mapc)$.
In this context, the KL divergence is defined as
\begin{equation*}
\kl{\disHzero(\mapc)}{\disHone(\mapc)}:=\sum_{\aindex=0}^{\asize-1} -\disHzero_{\aindex}(\mapc)\log{\bigg(\frac{\disHone_{\aindex}(\mapc)}{\disHzero_{\aindex}(\mapc)}\bigg)}.
\end{equation*}
In EDM, we maximize an empirical form of the KL divergence over some given class of quantization rules.
We therefore analyze an estimator of the form
\begin{equation*}
\widehat{\gamma}_{n}=\underset{\mapc\in\Gamma}{\arg\max} ~\empirfdiv{\mapc},
\end{equation*}
where $\empirfdiv{\mapc}$ represents an empirical KL divergence that is defined in Section~\ref{sec:EDM} (the subscript $n$ signifies that it is an empirical quantity that is based on $n$ samples from both $P$ and $Q$) and where $\Gamma$ denotes some class of quantization rules.
By design EDM estimators constructed rules $\widehat{\gamma}_{n}$ that induce maximally divergent pmfs, thereby best preserving the discriminability of $P$ and $Q$.
In other words, EDM estimators maximize the performance (in terms of KL divergence) of any downstream detector or classifier that operates on the quantized data.
The EDM formulation has several advantages: (i) it readily permits the application of empirical process theory which in turn provides the tools to quantify the estimator's error decay rates; (ii) it naturally leads to the Flynn and Gray algorithm which efficiently computes the quantization rules; (iii) it provides a systematic derivation for the Flynn and Gray algorithm; and (iv) the flexibility in candidate function classes allows efficient representation of the quantization rules and overcomes the small-cell constraint.
\section{Empirical Divergence Maximization}\label{sec:EDM}
The form of $\empirfdiv{\mapc}$ is taken from recent work by Nguyen et al.~\cite{nguyen_wainwright_jordan2010} and relies on rewriting the convex function $-\log(\cdot)$ appearing in the definition of the KL divergence.
Throughout the paper, we hold the number of quantization levels $L$ fixed.
\subsection{Expressing divergence using convex conjugates}
The notion of a \emph{convex conjugate} is based on the observation that a curve can either be described by its graph or by an envelope of tangents~\cite{rockafellar1970}.
More concretely, a (closed) convex function $f:\mathbb{R}\mapsto\mathbb{R}$ can be described as the pointwise supremum of a collection of affine functions $h(t)=tt^*-\mu^*$ such that the set of all pairs $(t^*,\mu^*)$ lie within the epigraph of its convex conjugate $f^*(t^*)$, i.e.,
\begin{equation}\label{equ:convexdual}
f(t)= \sup_{t^*}~\{t^*t-f^*(t^*)\}
\end{equation}
where by duality the convex conjugate $f^*(t^*)$ of $f(t)$ is defined by
\begin{equation*}
f^*(t^*)= \sup_{t}~\{tt^*-f(t)\}.
\end{equation*}
Now suppose $\mapc$ is an arbitrary quantization rule defined on $[0,1]^d$,
\begin{equation}
\mapc(x) = \sum_{i=0}^{L-1} i \, \indicator{R_i}(x),~x\in[0,1]^d,
\end{equation}
where $\{\region_{\partindex}\}_{\partindex=0}^{\asize-1}$ is a collection of disjoint sets partitioning $[0,1]^d$ and $\indicator{R_i}(\cdot)$ denotes the indicator function.
Using~\eqref{equ:convexdual}, we can write the divergence between the pmfs induced by $\mapc$ as
\begin{subequations}
\begin{align}
\kl{\disHzero(\mapc)}{\disHone(\mapc)}&=\sum_{\aindex=0}^{\asize-1} \disHzero_{\aindex}(\mapc)f\bigg(\frac{\disHone_{\aindex}(\mapc)}{\disHzero_{\aindex}(\mapc)}\bigg) \label{equ:klconvexconj1} \\
&=\sum_{\aindex=0}^{\asize-1} \disHzero_{\aindex}(\mapc)\cdot
\sup_{t^*}\Bigl\{t^* \frac{\disHone_{\aindex}(\mapc)}{\disHzero_{\aindex}(\mapc)} - f^*(t^*)\Bigl\}, \label{equ:klconvexconj2}
\end{align}
\end{subequations}
where $f(t)=-\log(t)$ for $t>0$ and $+\infty$ otherwise.
Calculating the convex conjugate, one finds
\begin{equation*}
f^*(t^*)=\begin{cases}-1-\log(-t^*)~~\text{if~~} t^*<0 \\
+\infty ~~\text{if~~} t^*\geq 0.
\end{cases}
\end{equation*}
Substituting this expression into~\eqref{equ:klconvexconj2}, we have the following expressions for the KL divergence
\begin{align*}
&\kl{\disHzero(\mapc)}{\disHone(\mapc)} \\
&\qquad =\sum_{\aindex=0}^{\asize-1} \disHzero_{\aindex}(\mapc)
\sup_{t^*_{\partindex}\in\mathbb{R}^{-}}\Bigl\{t^*_{\partindex} \frac{\disHone_{\aindex}(\mapc)}{\disHzero_{\aindex}(\mapc)} +1 +\log(-t^*_{\partindex})\Bigl\} \nonumber\\
&\qquad = \sum_{\aindex=0}^{\asize-1} \disHzero_{\aindex}(\mapc)
\negmedspace \sup_{\coeff_{\region_{\partindex}}\in\mathbb{R}^{+}}\Bigl\{\log(\coeff_{\region_{\partindex}}) - \coeff_{\region_{\partindex}} \frac{\disHone_{\aindex}(\mapc)}{\disHzero_{\aindex}(\mapc)} + 1\Bigl\}\nonumber\\
&\qquad = \negmedspace 1\negthinspace+\negmedspace\sum_{\aindex=0}^{\asize-1}
\sup_{\coeff_{\region_{\partindex}}\in\mathbb{R}^{+}}\negmedspace\Bigl\{\negthinspace\probmeasureHzero(\region_{\partindex})\log(\coeff_{\region_{\partindex}}) - \coeff_{\region_{\partindex}}\, \probmeasureHone(\region_{\partindex})\negmedspace \Bigl\},
\end{align*}
where in the second step we let $\coeff_{\region_{\partindex}}=-t^*_{\partindex}$, and in the last step use the fact that
$\disHzero_{\partindex}(\mapc)=\probmeasureHzero(\region_{\partindex}), \region_{\partindex}=\{x:\mapc(x)=\partindex\}$.
The validity of last expression is easily verified by differentiating it with respect to $\coeff_{\region_{\partindex}}$ and solving for the maximizers.
By defining the piecewise constant function
\begin{equation}\label{equ:piecerule}
\mapout(x):=\sum_{\partindex=0}^{\asize-1}\coeff_{\region_{\partindex}} \indicator{\region_{\partindex}}(x),\quad \coeff_{\region_{\partindex}}\in\mathbb{R}^{+}, x\in[0,1]^d
\end{equation}
we can write $\kl{\disHzero(\mapc)}{\disHone(\mapc)}$ in integral form:
\begin{equation}\label{equ:convexconjkl}
1 + \sup_{\mapout} \left\{\int_{[0,1]^{\ddimen}} \log(\mapout)~d\probmeasureHzero - \int_{[0,1]^{\ddimen}} \mapout~d\probmeasureHone \right\},
\end{equation}
where the supremum is taken over all functions of the form~\eqref{equ:piecerule}.
Note that the $\phi$ which achieves the supremum depends on $P$, $Q$, and $\{R_i\}_{i=0}^{L-1}$.
Below, we restrict $\phi$ to lie within a (more) specific class of rules and define the proposed quantization rule estimator in terms of the empirical counterpart to~\eqref{equ:convexconjkl}.
In addition, note that unlike $\gamma$, $\phi$ does not map $[0,1]^{d}$ to a set of indices.
We nevertheless refer to both as quantization rules since $\phi$ only assumes $L$ real values.
Note also that in terms of KL divergence, $\phi$ determines $\gamma$, i.e., if $\phi$ is known, a quantization rule $\gamma: [0,1]^{d}\mapsto \{0,\dotsc,L-1\}$ can be defined that induces the same pmfs as $\phi$.
This fact becomes important for the algorithm described in Section~\ref{sect:algorithm}.
\subsection{Empirical estimator}\label{subsect:empest}
To define a function class for $\phi$, we first consider different ``labelings'' of a uniform partition of $[0,1]^d$.
For a given positive integer $J$, let $\pi_J$ denote a tesselation of $[0,1]^{d}$ by uniform hypercubes $S_k, k=0,\dotsc,2^{dJ}-1$.
To each cell $S_k$, we can associate one of $L$ labels $\{0,\dots,L-1\}$, and thus for each different labeling of $\pi_{J}$, we can define another partition, $\pi_{R}$, with cells $\{R_i\}_{i=0}^{L-1}$ described by
\begin{equation}
R_i=\bigcup_{k:\,\text{label}(S_k)=i} \negthickspace S_k, \quad i=0,\dotsc,L-1.
\end{equation}
Now, for a given partition $\pi_R$ and positive constants $\lbound>0$ and $\ubound<\infty$, denote by $\candclass_{\pi_{R}}(\asize,J,\lbound,\ubound)$ the set of all $L$-level piecewise constant functions defined on $\pi_{R}$ that are bounded and positive:
\begin{equation*}
\candclass_{\pi_{R}}(\asize,J,\lbound,\ubound)=\negthickspace \left\{\mapout(x)=\negthickspace \sum_{\partindex=0}^{\asize-1}\coeff_{\region_{\partindex}} \indicator{\region_{\partindex}}(x)\negthickspace : \lbound\leq\coeff_{\region_{\partindex}}\leq\ubound \negthinspace \right\}.
\end{equation*}
Letting $\Pi_{R}$ denote the set of all partitions $\pi_{R}$, or equivalently the set of all different labelings of $\pi_J$, we define the candidate class $\candclass(\asize, J,\lbound,\ubound)$ of our empirical quantizers as
\begin{equation}\label{equ:defcandclass2}
\candclass(\asize, J,\lbound,\ubound):= \bigcup_{\pi_R \in \Pi_R} \candclass_{\pi_{R}}(\asize,J,\lbound,\ubound).
\end{equation}
Letting $\{\inputrv_{\obsindex}^{\disHzero}\}_{\obsindex=1}^{\noobs}$ and $\{\inputrv_{\obsindex}^{\disHone}\}_{\obsindex=1}^{\noobs}$ be training data distributed according to $\disHzero$ and $\disHone$, respectively, we define the function $\empirfdiv{\mapout}$ as an empirical counterpart to~\eqref{equ:convexconjkl}
\begin{equation}\label{equ:empirKL}
\empirfdiv{\mapout}:=1+\frac{1}{\noobs}\sum_{\obsindex=1}^{\noobs} \log{\mapout(\inputrv_{\obsindex}^{\disHzero})}
- \frac{1}{\noobs}\sum_{\obsindex=1}^{\noobs} \mapout(\inputrv_{\obsindex}^{\disHone})
\end{equation}
and define the proposed empirical quantization rule estimator as
\begin{equation}\label{equ:empirest2}
\mapempest:=\underset{\mapout\in\candclass(\asize,J,\lbound,\ubound)}{\arg\max} ~\empirfdiv{\mapout}.
\end{equation}
$\mapempest$ is our \emph{empirical divergence maximization} (EDM) estimator.
Note $\empirfdiv{\mapout}$ is not in general a KL divergence; it can in fact be negative for some ${\mapout\in\candclass}$.
It is a consistent estimator, however, converging to the ``best in class'' estimator as $\noobs\rightarrow \infty$~\cite{nguyen_wainwright_jordan2010}.
\subsection{Best in class and optimal quantization rules}
The \emph{best in class estimate} $\mapsopt$ is that element in $\candclass$ that maximizes $\abbrefdiv{\mapout}$,
\begin{equation}\begin{split}\label{equ:knownpq_KLest}
&\mapsopt:=\underset{\mapout\in\candclass(\asize,J,\lbound,\ubound)}{\arg\max} ~\abbrefdiv{\mapout},\quad \text{where} \\
\abbrefdiv{\mapout}&:=1 + \int_{[0,1]^{\ddimen}} \log(\mapout)~d\probmeasureHzero - \int_{[0,1]^{\ddimen}} \mapout~d\probmeasureHone.
\end{split}
\end{equation}
Note that $\abbrefdiv{\mapout}$, as opposed to $\empirfdiv{\mapout}$, is not an empirical quantity; its definition requires full knowledge of the distributions $\probmeasureHzero$ and $\probmeasureHone$.
We take the theoretically optimal quantization rule $\mapcopt$ to be the rule that maximizes the divergence over a class of piecewise constant functions that has an assumed boundary regularity (the regularity conditions play a role in the convergence analysis in Section~\ref{subsect:approxerror}).
The class definition uses the notion of a locally constant function: a function $f:[0,1]^{\ddimen}\mapsto\mathbb{R}$ is \emph{locally constant} at a point $\inputs\in[0,1]^{\ddimen}$ if there exists $\epsilon>0$ such that for all $y\in[0,1]^{\ddimen}$, the condition $\norm{\inputs-y}<\epsilon$ implies $f(y)=f(\inputs)$.
\begin{defin}[PC class~\cite{ruicastro_phd_thesis2007}]
A function $f:[0,1]^{\ddimen}\mapsto\{\coeff_{\partindex}\}_{\partindex=0}^{\asize-1}, \coeff_{\partindex}\in\mathbb{R}^{+}$ is a positive-valued piecewise constant function with $\asize$ levels if it is locally constant at any point $x\in[0,1]^{\ddimen}\setminus B(f)$, where $B(f)\subset [0,1]^{\ddimen}$ is a boundary set satisfying $N(r)\leq \beta r^{-(\ddimen-1)}$ for all $r>0$.
Here, $\beta>0$ is a constant and $N(r)$ is the minimal number of balls of diameter $r$ that covers $B(f)$.
Furthermore, let $f$ be uniformly bounded on $[0,1]^{\ddimen}$, that is $\lbound\leq f(x)\leq \ubound$ for all $x\in[0,1]^{\ddimen}$, where $\lbound>0$ and $\ubound<\infty$.
The set of all piecewise constant functions $f$ satisfying the above conditions is denoted by $\PCclass$.
\end{defin}
\noindent In short, we consider $\PCclass$ to be a class of likelihood-ratio quantization rules that have well behaved boundaries.
The theoretically optimal quantization rule is thus defined to be
\begin{equation}\label{equ:mapcopt}
\mapcopt:=\underset{\psi\in\PCclass}{\arg\max} ~\abbrefdiv{\psi}.
\end{equation}
It is well-known that $\mapcopt$ can always be constructed by thresholding the likelihood ratio~\cite{tsitsiklis1993a}.
In other words, the optimal quantization rule $\mapcopt$ can always be chosen to be a piecewise constant function whose boundary sets are level sets of the likelihood ratio $\disHone(x)/\disHzero(x)$.
\section{Solving for the estimator}\label{sect:algorithm}
To find $\mapempest$ in~\eqref{equ:empirest2}, we employ a modified form of the Flynn and Gray algorithm~\cite{flynn_gray1987} that iteratively maximizes the divergence between two pmfs over a set of quantization rules.
The method directly follows from the EDM formulation (although it was not originally proposed in this context) and searches for an optimal cell labeling for a given partition where the number of cells is much larger than the number of quantization levels.
\subsection{The Flynn and Gray algorithm}
For independent and identically distributed random variables $\inputrv_{1},\dotsc,\inputrv_{\noobs}$, the \emph{empirical measure} of a set ${A\in[0,1]^{\ddimen}}$, denoted $\probmeasureHzero_{\noobs}(A)$, is the sample average
\begin{equation}\label{equ:sample_average}
\probmeasureHzero_{\noobs}(A)=\frac{1}{\noobs}\sum_{k=1}^{\noobs} \indicator{A}(\inputrv_{k}).
\end{equation}
The sample average of a function $g:[0,1]^{\ddimen}\mapsto \mathbb{R}$ can thus be written with respect to $\probmeasureHzero_{n}$ as an \emph{empirical expectation},
\begin{equation}
\probmeasureHzero_{\noobs}(g)=\frac{1}{\noobs}\sum_{k=1}^{\noobs} g(\inputrv_{k})=\int g~d\probmeasureHzero_{\noobs}.
\end{equation}
Using this notation, we rewrite~\eqref{equ:empirKL} as
\begin{equation}\label{equ:empirKL1}
\empirfdiv{\mapout}=1 + \int_{[0,1]^{\ddimen}} \log(\mapout)~d\probmeasureHzero_{\noobs} - \int_{[0,1]^{\ddimen}} \mapout~d\probmeasureHone_{\noobs},
\end{equation}
where $\mapout\in\candclass$.
For any fixed partition $\partition_{\region}\in\partclass$, $\empirfdiv{\mapout}$ is maximized by assigning $\phi(x)$ the values $\probmeasureHzero_{\noobs}(\region_{i})/\probmeasureHone_{\noobs}(\region_{i})$ for $x\in R_i$.
For this assignment choice, $\empirfdiv{\mapout}$ can be expressed as
\begin{equation}\label{equ:flynn_gray_setup}
\empirfdiv{\mapout}=1 + \sum_{i=0}^{L-1} \int_{\region_{i}} \log\biggl(\frac{\probmeasureHzero_{\noobs}(\region_{i})}{\probmeasureHone_{\noobs}(\region_{i})}\biggr)~d\probmeasureHzero_{\noobs} - \int_{\region_{i}} \frac{\probmeasureHzero_{\noobs}(\region_{i})}{\probmeasureHone_{\noobs}(\region_{i})}~d\probmeasureHone_{\noobs},
\end{equation}
and the estimator $\mapempest$ can now be found by searching over $\Pi_R$ for the partition that maximizes~\eqref{equ:flynn_gray_setup}.
The Flynn and Gray algorithm~\cite{flynn_gray1987} is a straightforward method which accomplishes this task.
To apply it, we rewrite~\eqref{equ:flynn_gray_setup} as
\begin{align}\begin{split}
\empirfdiv{\mapout}&=\sum_{i=0}^{L-1} \probmeasureHzero_{\noobs}(\region_{i}) \left[\log\biggl(\frac{\probmeasureHzero_{\noobs}(\region_{i})}{\probmeasureHone_{\noobs}(\region_{i})}\biggr)+1\right] \\ &\qquad + \probmeasureHone_{\noobs}(\region_{i}) \biggl(-\frac{\probmeasureHzero_{\noobs}(\region_{i})}{\probmeasureHone_{\noobs}(\region_{i})}\biggr)\end{split} \\
&=\sum_{i=0}^{L-1} \probmeasureHzero_{\noobs}(\region_{i}) a_{i} + \probmeasureHone_{\noobs}(\region_{i}) b_{i} \\
&=\sum_{i=0}^{L-1} \sum_{k\in I_i} \probmeasureHzero_{\noobs}(S_k) a_{i} + \probmeasureHone_{\noobs}(S_k) b_{i},\label{equ:flynn_gray_coreeq}
\end{align}
where $a_{i}=\log\bigl(\probmeasureHzero_{\noobs}(\region_{i})/\probmeasureHone_{\noobs}(\region_{i})\bigr)+1$, $b_{i}=-\probmeasureHzero_{\noobs}(\region_{i})/\probmeasureHone_{\noobs}(\region_{i})$, and $I_i$ is the index set ${\{k:\text{label}(S_k)=i\}}$.
The algorithm maximizes $\empirfdiv{\mapout}$ by iterating two steps: it first holds the set of weights $\{a_i\}$ and $\{b_i\}$ fixed and finds the labels for each cell $S_k\in \pi_J, k=0,\dotsc,2^{dJ}-1$ that maximizes~\eqref{equ:flynn_gray_coreeq}, and then holds the cell labels of $\pi_J$ fixed and updates the weights $\{a_i\}$ and $\{b_i\}$ using the probabilities $\probmeasureHzero_{\noobs}(\region_{i}), \probmeasureHone_{\noobs}(\region_{i})$, $i=0,\dotsc,L-1$ found from the first step.
Flynn and Gray showed these steps monotonically increase~\eqref{equ:flynn_gray_coreeq}, and since $\empirfdiv{\mapout}$ is upper bounded by $1-m+\log{M}$ (follows from the boundedness of $\phi$), the algorithm converges to a local maximum.
The algorithm returns a locally optimal labeling of $\pi_{J}$ and locally optimal weights from which $\mapempest$ can be determined:
\begin{equation}
\mapempest(x)=-b_i \quad \text{for~}x\in R_i.
\end{equation}
The algorithm is outlined in the panel entitled Algorithm~\ref{algor:flynn_gray}.
An advantage of the Flynn and Gray algorithm is that it avoids the exhaustive combinatoric search over all possible labelings by only needing to examine each cell $S_k$ once per iteration.
From experiments, it has been observed that for moderate sized partitions $\pi_J$ ($<2^{16}$ cells) and for $L<10$, the algorithm converges very quickly ($<30$ iterations).
EDM provides a new derivation for the Flynn and Gray algorithm; however, it is interesting to note that it can also be based on the fact that
\begin{equation}\label{graycomment}
\kl{p}{q}= \sup_{\gamma}~ \kl{\disHzero(\mapc)}{\disHone(\mapc)}
\end{equation}
where the supremum is over all measurable quantization rules with an arbitrary number of levels~\cite{gray1990,poor1988}.
Because $\kl{p}{q}\geq \kl{\disHzero(\mapc)}{\disHone(\mapc)}$ for any quantization rule, one could use~\eqref{graycomment} to justify an approach similar to EDM and maximize $\kl{\disHzero(\mapc)}{\disHone(\mapc)}$ over a set of quantization rules for a fixed quantization level.
While this approach leads to similar (if not identical) estimators, EDM has the advantage of making a clear connection with empirical process theory which provides the theoretical tools to analyze the error convergence rate.
Note also that the original Flynn and Gray algorithm does not explicitly constrain the values of $\phi$ to lie within the range $[m,M]$.
However, to avoid computing unbounded estimates at any given iteration, we employ the K-T technique~\cite{krichevsky-trofimov1981} when computing $P_n(S_k)$ and $Q_n(S_k)$, ${k=0,\dotsc,2^{dJ}-1}$.
This technique simply preloads each cell $S_k$ by one half before calculating the sample averages, thereby avoiding the possibility of computing zero probability estimates $P_n(R_i)$, $Q_n(R_i)$.
Thus instead of~\eqref{equ:sample_average}, one computes
\begin{equation}
P_n(S_k) = \frac{1}{2^{dJ-1}+n}\left(\frac{1}{2}+\sum_{k=1}^{\noobs} \indicator{S_k}(X_{k}^{p})\right)
\end{equation}
and likewise for $Q_n(S_k)$.
Here, the choice of $1/2$ is not arbitrary; it is based on theoretical considerations of what \emph{a priori} distribution of the probabilities $P(S_k)$, $Q(S_k)$ influences the sample averages $P_n(S_k)$, $Q_n(S_k)$ the least~\cite{krichevsky-trofimov1981}.
\begin{algorithm}
\caption{Modified Flynn and Gray algorithm}
\label{algor:flynn_gray}
\renewcommand{\baselinestretch}{1}\large\normalsize
\begin{algorithmic}[1]
\REQUIRE \parbox[t]{7cm}{L, J, n, empirical cell probabilities $P_n(S_k)$ and $Q_n(S_k)$, stopping threshold $\epsilon$ \vspace*{.2cm}}
\STATE Initialize iteration index $l=0$
\STATE Randomly label cells $S_k\in\pi_{J}$
\STATE Compute $P_n^{(l)}(R_i), Q_n^{(l)}(R_i),~i=0,\dotsc,L-1$
\STATE Initialize weights \\ $a^{(l)}_i=\log\bigl(P_n^{(l)}(R_i)/Q_n^{(l)}(R_i)\bigr)+1$ \\ $b^{(l)}_i=-P_n^{(l)}(R_i)/Q_n^{(l)}(R_i)$
\STATE Compute $D^{(l)}_n(\phi)$
\STATE Initialize intermediary divergence $\widetilde{D}_n(\phi)=0$
\WHILE{$(D^{(l)}_n(\phi)-\widetilde{D}_n(\phi))/D^{(l)}_n(\phi)>\epsilon$}
\STATE Find new label for each cell $S_k$ by computing \\
$i^{l+1}=\arg\max_{i\in\{0,\dotsc,L-1\}} P_n(S_k) a^{(l)}_{i} + Q_n(S_k) b^{(l)}_{i}$ \\ (hold weights fixed)
\STATE Update probabilities for $i=0,\dotsc,L-1$ \\
$P_n^{(l+1)}(R_i)=\sum_{k:\text{label}(S_k)=i} P_n(S_k)$ \\
$Q_n^{(l+1)}(R_i)=\sum_{k:\text{label}(S_k)=i} Q_n(S_k)$ \\
(includes K-T preloading if necessary)
\STATE Compute intermediary divergence \\
$\widetilde{D}_n(\phi)=\sum_{i=0}^{L-1} P_n^{(l+1)}(R_i) a^{(l)}_{i} + Q_n^{(l+1)}(R_i) b^{(l)}_{i}$
\STATE Update weights \\
$a^{(l+1)}_i=\log\bigl(P_n^{(l+1)}(R_i)/Q_n^{(l+1)}(R_i)\bigr)+1$ \\ $b^{(l+1)}_i=-P_n^{(l+1)}(R_i)/Q_n^{(l+1)}(R_i)$
\STATE Compute new divergence \\
$D^{(l+1)}_n(\phi)=\sum_{i=0}^{L-1} P_n^{(l+1)}(R_i) a^{(l+1)}_{i} + Q_n^{(l+1)}(R_i) b^{(l+1)}_{i}$
\STATE $l=l+1$
\ENDWHILE
\ENSURE \parbox[t]{7cm}{$\mapempest$ (locally optimal labels of $\pi_J$ and weights $\{a_i,b_i\}$), $D_n(\mapempest)$}
\end{algorithmic}
\end{algorithm}
In short, one solves for an EDM estimator based upon the training data $\{\inputrv_{\obsindex}^{\disHzero}\}, \{\inputrv_{\obsindex}^{\disHone}\}$ by first computing the sample averages $P_n(S_k), Q_n(S_k)$ for each cell $S_k\in \pi_{J}$ and then providing these probabilities as input to the Flynn and Gray algorithm.
The algorithm is applied to two numerical examples in Section~\ref{sect:app}.
\subsection{Recursive dyadic partitions}\label{subsect:RDP}
Because $\Phi$ is based on a uniform dyadic partition, any EDM estimate $\mapempest$ can be viewed as a piecewise constant function supported on a \emph{recursive dyadic partition} (RDP).
RDPs are a systematic class of partitions that have proven to be effective in function estimation and classification problems~\cite{ruicastro_phd_thesis2007,scott_nowak2006}.
Their usefulness stems from their ability to adapt to boundaries (including $\PCclass$), thus allowing efficient computation of estimators and concise encoding of estimates.
In the present context, RDPs are important because they allow efficient encoding of $\mapempest$, and their properties are key in the approximation error analysis presented in Section~\ref{subsect:approxerror}.
RDPs are partitions composed of quasi-disjoint sets%
\footnote{Two sets are quasi-disjoint if and only if their intersection has Lebesgue measure zero.}
whose union equals the entire space $[0,1]^{\ddimen}$.
A RDP is any partition that can be constructed using only the following rules~\cite{ruicastro_phd_thesis2007}:
\begin{enumerate}
\item $\{[0,1]^{\ddimen}\}$ is a RDP.
\item Let $\partition=\{\cell_{0},\ldots,\cell_{k-1}\}$ be a RDP, where $\cell_{\partindex}=[u_{\partindex 1},v_{\partindex 1}]\times\ldots\times [u_{\partindex \ddimen},v_{\partindex \ddimen}]$.
Then
\begin{equation*}
\partition'= \{\cell_{0},\dotsc, \cell_{\partindex-1}, \cell_{\partindex}^{0},\dotsc, \cell_{\partindex}^{(2^{\ddimen}-1)}, \cell_{\partindex+1},\ldots,\cell_{k-1}\}
\end{equation*}
is a RDP, where
$\{\cell_{\partindex}^{0},\dotsc,\cell_{\partindex}^{(2^{\ddimen}-1)}\}$ is obtained by dividing the hypercube $\cell_{\partindex}$ into $2^{\ddimen}$ quasi-disjoint hypercubes of equal size.
Formally, let $q\in\{0,\dotsc,2^{\ddimen-1}\}$ and $q=q_1q_2\dotsc q_{\ddimen}$ by the binary representation of $q$.
Then
\end{enumerate}
\begin{align*}
\cell_{\partindex}^{(q)}=&\Biggl[u_{\partindex 1}+\frac{v_{\partindex 1}-u_{\partindex 1}}{2}q_{1},
v_{\partindex 1}+\frac{u_{\partindex 1}-v_{\partindex 1}}{2}(1-q_{1})\Biggr] \times \\
&\ldots \times \Biggl[u_{\partindex \ddimen}+\frac{v_{\partindex \ddimen}-u_{\partindex \ddimen}}{2}q_{\ddimen},
v_{\partindex \ddimen}+\frac{u_{\partindex \ddimen}-v_{\partindex \ddimen}}{2}(1-q_{\ddimen})\Biggr].
\end{align*}
We say a RDP has maximal depth $J$ if the side length of its smallest hypercube equals $2^{-J}$.
Fig.~\ref{fig:RDP} illustrates a RDP approximating an elliptical boundary.
\begin{figure}
\centerline{\includegraphics[width=5cm]{example_RDP.eps}}
\caption{\small An example two-dimensional RDP ($\RDPdepth=3$).}
\label{fig:RDP}
\end{figure}
It should be clear RDPs describe tree structures where the root node is the entire space $[0,1]^d$ and the leaf nodes represent the different cells comprising the RDP.
Each branch can have different depths and thus the cells can have different sizes.
This property allows a RDP to have larger cells in locations where the function value is constant and smaller cells where the values change (around boundaries).
The combination of the systematic tree structure and the partition's adaptivity allow the estimator to be efficiently encoded, that is, the number of bits necessary to map an observation to its quantized value can be done efficiently~\cite{scott_nowak2006}.
For a fixed estimator $\mapempest$ (or a fixed labeling of $\pi_J$), a RDP can be easily constructed by repeating step 2 above (starting with the whole space), but only producing a split if the cells $S_k \in \pi_J$ on either side of the split, but within the hypercube of interest, have different values (labels).
\section{Error Decay Rates}\label{sec:error_rates}
We gauge the quality of $\mapempest$ by characterizing the decay rate of the estimation and approximation errors.
\emph{Estimation error} is defined as the difference $\abbrefdiv{\mapsopt}-\abbrefdiv{\mapempest}$ and quantifies the error caused by computing $\mapempest$ without knowledge of $\disHzero$ and $\disHone$.
As the number of samples $n$ increases, the estimation error decreases at a rate (exponent of $n$) that depends on the complexity of $\Phi$ and on the properties of $p$ and $q$.
\emph{Approximation error} is defined as $\abbrefdiv{\mapcopt}-\abbrefdiv{\mapsopt}$ and arises in cases where $\mapcopt\notin\candclass$.
To quantify its decay, we think of the candidates rules $\phi\in\Phi$ as being supported on RDPs and the rate of decay in terms of the depth parameter $J$.
We begin in a standard fashion with two basic inequalities that follow from the definitions of $\mapsopt$ and $\mapempest$: ${\abbrefdiv{\mapsopt}-\abbrefdiv{\mapempest}\geq 0}$ and ${\empirfdiv{\mapsopt}-\empirfdiv{\mapempest}\leq 0}$.
They imply that the estimation error is upper bounded by a difference of empirical processes
\begin{align*}
0&\leq \abbrefdiv{\mapsopt}-\abbrefdiv{\mapempest}\\
&\leq-[(\empirfdiv{\mapsopt}-\abbrefdiv{\mapsopt})-(\empirfdiv{\mapempest}-\abbrefdiv{\mapempest})] \\
&=-(\genempir{\mapsopt}-\genempir{\mapempest})/\sqrt{\noobs},
\end{align*}
where the second inequality results from adding and subtracting $\empirfdiv{\mapsopt}$ and $\empirfdiv{\mapempest}$, and where $\genempir{\mapc}=\sqrt{\noobs}(\empirfdiv{\mapc}-\abbrefdiv{\mapc})$.
Adding the approximation error to both sides of the inequality bounds the total error by the two component errors.
\begin{align}\label{equ:fundaineqal}
\begin{split}
0&\leq \underset{\text{total error}}{\underbrace{\abbrefdiv{\mapcopt}-\abbrefdiv{\mapempest}}} \\
&\leq -\underset{\text{upper bound on est. error}}{\underbrace{(\genempir{\mapsopt}-\genempir{\mapempest})/\sqrt{\noobs}}} + \underset{\text{approx. error}}{\underbrace{\abbrefdiv{\mapcopt}-\abbrefdiv{\mapsopt}}}
\end{split}
\end{align}
We use this equation and examine the estimation and the approximation errors separately, giving the final rate result for the expected total error.
\subsection{Estimation Error}\label{subsect:esterror}
Let $\expectv_p$ and $\expectv_q$ denote expectation operator with respect to $P$ and $Q$.
Then, by writing $\rvert\genempir{\mapempest}-\genempir{\mapsopt}\lvert$ as
\begin{equation*}
\begin{split}
&\Bigg\vert \frac{1}{\noobs}\sum_{\obsindex=1}^{\noobs} (\log{\mapout(\inputrv_{\obsindex}^{\disHzero})}-\log{\mapsopt(\inputrv_{\obsindex}^{\disHzero})})- \expectv_{\disHzero} \big[\log{\mapout(\inputrv)}\\
&-\log{\mapsopt(\inputrv)}\big] + (\mapout(\inputrv_{\obsindex}^{\disHone})-\mapsopt(\inputrv_{\obsindex}^{\disHone})) - \expectv_{\disHone} \big[\mapout(\inputrv)-\mapsopt(\inputrv)\big] \Bigg\vert,
\end{split}
\end{equation*}
it is clear that for a given quantization rule $\mapout$, the empirical averages above converge almost surely to their respective values by the strong law of large numbers.
But because $\mapempest$ can potentially be any element in $\candclass$, any characterization of the convergence rate must hold uniformly over $\candclass$.
It is well-known that uniform rates of convergence depend on the complexity of the function class from which the empirical estimators are drawn~\cite{vandegeer2000}.
Here we use the notion of \emph{bracketing entropy} to characterize complexity of $\candclass$.
Roughly speaking, the bracketing entropy of a function class $\mathcal{G}$ equals the logarithm of the minimum number of function pairs that upper and lower bound (bracket) all the members in $\mathcal{G}$ to within some tolerance $\delta$ and with respect to some norm (a precise definition can be found in~\cite[p. 16]{vandegeer2000}).
We denote the bracketing entropy by $H_B(\delta,\mathcal{G},L_2(\probmeasureHzero))$ and say $\mathcal{G}$ has \emph{bracketing complexity} $\covcomp>0$ if $H_B(\delta,\mathcal{G},L_2(\probmeasureHzero))\leq \compconst \delta^{-\covcomp}$ for all $\delta>0$ and for some constant $A>0$.
Because the members of $\Phi$ are uniformly bounded, it can be shown that $\Phi$ has bracketing complexity $\alpha=1$~\cite{michaellexa_phdthesis2008}.
This fact is incorporated into Theorem~\ref{thm:esterrorconverge} below; however, the proof of the theorem given in Appendix~\ref{app:esterror} assumes the bracketing complexity lies between zero and two.
The proof therefore yields a slightly more general result than that stated.
We now introduce two conditions on $p$ and $q$.
The first simply states that $p$ and $q$ are uniformly bounded.
\noindent\textbf{Condition 1.}
Assume $\denlbound\leq\disHzero(x),\disHone(x)\leq \denubound \text{~for all~} x\in[0,1]^{\ddimen}$, $\denlbound>0$, $\denubound<\infty$.
The second is a condition introduced by Mammen and Tsybakov~\cite{mammen_tsybakov1999,tsybakov2004} and involves a key parameter $\kappa$ that provides insight into when fast convergence rates are possible (i.e., rates faster than $\noobs^{-1/2}$).
The condition arises in a slightly different form in function estimation and Bayesian classification problems, and within these contexts, it can be related to the behavior of $p$ and $q$ near a boundary of interest%
\footnote{In Bayesian classification and function estimation, the condition is known as the \emph{margin condition}}.
For example, in Bayesian classification, small $\kappa$ implies a ``steep'' regression function at the Bayes decision boundary and thus easier classification; large $\kappa$ implies a ``flat'' transition and harder classification.
Van de Geer~\cite{vandegeer2007} describes $\kappa$ as an ``identifiability'' parameter in the sense that it characterizes how well $\phi\in\candclass$ can be distinguished from $\mapcopt$.
Because $\mapcopt$ is determined by $P$ and $Q$, this condition is ultimately a condition on these underlying distributions.
\noindent\textbf{Condition 2.}
There exists constants $\marconst>0$ and $\marp\geq 1$ such that for all $\mapout \in\candclass$,
\begin{equation}\label{equ:margincond}
\abbrefdiv{\mapcopt}-\abbrefdiv{\mapout}\geq \norm{\mapcopt-\mapout}_{L_2}^{\marp}/\marconst.
\end{equation}
If $\kappa$ is small (close to 1) for a given $P$ and $Q$, the difference $\abbrefdiv{\mapcopt}-\abbrefdiv{\mapout}$ is larger for $\phi\in\candclass$ close to $\mapcopt$ (those $\phi$ such that $\norm{\mapcopt-\mapout}_{L_2}\leq1$) compared to those distributions having larger $\kappa$ values.
Intuitively, this means that such $\phi$ are more distinguishable from $\mapcopt$ for those distributions satisfying Condition 2 with small $\kappa$ compared to those distributions satisfying Condition 2 with larger $\kappa$ values, where distinguishability is measured in terms of divergence loss $\abbrefdiv{\mapcopt}-\abbrefdiv{\mapout}$%
\footnote{Note that the distinguishability in terms of divergence loss is intimately connected with how $\mapempest$ is computed: since $\empirfdiv{\mapout}$ is a surrogate of $\abbrefdiv{\mapout}$, maximizing $\empirfdiv{\mapout}$ over $\mapout\in\candclass$ is a surrogate for maximizing $\abbrefdiv{\mapout}$ over $\mapout\in\candclass$, or equivalently minimizing $\abbrefdiv{\mapcopt}-\abbrefdiv{\mapout}$.}.
The following result shows that $\kappa$ effectively characterizes this aspect of the problem, and like for Bayesian classification and estimation, is a key parameter for the error convergence rate.
\begin{theorem}[Estimation error]\label{thm:esterrorconverge}
Let $\mapempest$, $\mapsopt$, and $\mapcopt$ be as defined in~\eqref{equ:empirest2},~\eqref{equ:knownpq_KLest} and~\eqref{equ:mapcopt} respectively. Suppose Conditions 1 and 2 are met for some constants $\denlbound, \denubound, \marconst,\text{~and~} \marp$.
Then for any $0<\epsilon<1$ we have
\begin{align*}\label{equ:vandegeerlem}
\abbrefdiv{\mapcopt}&-\expectv\abbrefdiv{\mapempest} \leq \Bigl(\frac{1+\epsilon}{1-\epsilon}\Bigr) \\
& \biggl[\textnormal{const}(\denlbound,\denubound,\marconst,\marp)~ \noobs^{-\frac{\marp}{2\marp-1}}+ \abbrefdiv{\mapcopt}-\abbrefdiv{\mapsopt}\biggr],
\end{align*}
for sufficiently large $\noobs$ where $\textnormal{const}(\denlbound,\denubound,\marconst,\marp)$ is a decreasing function of $\epsilon$.
\end{theorem}
The proof is provided in Appendix~\ref{app:esterror} (see also~\cite{mlexa2010a}) and directly follows from results of van de Geer\cite[pp. 206-207]{vandegeer2007}~\cite{vandegeer2000} and Mammen and Tsybakov~\cite{mammen_tsybakov1999}.
Depending on $P$ and $Q$, the decay rate of the estimation error can be as fast as $\noobs^{-1}$ ($\kappa=1$) and no worse than $\noobs^{-1/2}$ ($\kappa=\infty$).
In particular, if for a given $P$ and $Q$, the approximation error is nonzero (which is commonly the case in quantization problems), we have
\begin{align*}
\abbrefdiv{\mapcopt}-\abbrefdiv{\mapout}&\geq \abbrefdiv{\mapcopt}-\abbrefdiv{\mapsopt} \\
&\geq \text{const} \cdot \frac{\norm{\mapcopt-\mapout}_{L_2}}{M}
\end{align*}
where the first inequality follows from the definition of $\mapsopt$ and the second from the fact that $0\leq \tfrac{\norm{\mapcopt-\mapout}_{L_2}}{M} \leq 1$ for all $\phi\in\candclass$.
Thus with a nonzero approximation error, Condition 2 can be met with $\kappa=1$ and the rate $n^{-1}$ is achievable.
This situation is common in quantization problems because it is unusual in practice for the level sets of a likelihood ratio function to coincide with a RDP for a fixed depth $J$.
For example, consider the simple scenario where $p(x)$ is a zero-mean unit variance Gaussian density function and $q(x)$ is a Laplace density function (also zero-mean and unit variance), $x\in\mathbb{R}$.
With $L=8$ and $J=6$, the approximation error is $0.002$ and hence we expect a rate of $\mathcal{O}(n^{-1})$.
Fig.~\ref{fig:1D_gauss-laplace_rates} confirms this result experimentally by plotting $\abbrefdiv{\mapcopt}-\expectv\abbrefdiv{\mapempest}$ as a function of $n$.
(For each value of $n$, Gaussian and Laplacian data were generated and $\mapempest$ was computed using the Flynn and Gray algorithm.)
The black dashed curve on the left hand plot is shown for reference and equals $40n^{-1}+0.002$.
\begin{figure*}
\begin{minipage}{.49\linewidth}
\centerline{\includegraphics[width=7cm]{1D_gaussian_laplace_Dphi.eps}}
\end{minipage}
\begin{minipage}{.49\linewidth}
\centerline{\includegraphics[width=7cm]{1D_gaussian_laplace_densities.eps}}
\end{minipage}
\caption{Left: Total expected divergence loss $\abbrefdiv{\mapcopt}-\expectv\abbrefdiv{\mapempest}$ plotted as a function of the number of training samples $n$ (solid curve) for the case $L=8$, $J=6$ where $p(x)$ is a zero-mean unit variance Gaussian density and $q(x)$ is a zero-mean unit variance Laplace density. The dashed (black) curve is $\mathcal{O}(n^{-1})$, thus for this example, we have a fast rate of decay. Right: Probability density functions.}
\label{fig:1D_gauss-laplace_rates}
\end{figure*}
In contrast, if $P$ and $Q$ are such that $\mapcopt\in\candclass$, then Condition 2 is only met with $\kappa=2$, and therefore the resulting decay rate is $\mathcal{O}(n^{-2/3})$.
To derive this result, consider the subset of quantization rules $\mapout\in\candclass$ that share the same partition associated with $\mapcopt$.
Recalling~\eqref{equ:piecerule}, we can in this case write $\mapcopt$ as
\begin{equation*}
\mapcopt(\inputsv)=\sum_{i=0}^{\asize-1} \mapcopt_{i}\indicator{R_{i}}(\inputsv),
\end{equation*}
where $\{\mapcopt_{i}\}$ are the levels of $\mapcopt$.
Then by~\eqref{equ:knownpq_KLest} and the fact that $\mapcopt_i=P(R_i)/Q(R_i)$, we have
\begin{align*}
\abbrefdiv{\mapcopt}&=1+\int \log{(\mapcopt)}~d\probmeasureHzero-\int\mapcopt~d\probmeasureHone \\
&=\int\log{\mapcopt}~d\probmeasureHzero \\
&=\int\mapcopt \log{(\mapcopt)}~d\probmeasureHone \\
&= \sum_{i=0}^{\asize-1} \int_{\region_i} \mapcopt_{i} \log{(\mapcopt_{i})}~d\probmeasureHone.
\end{align*}
Now, because $(\cdot)\log{(\cdot)}$ is differentiable and continuous on the range of the positive real numbers, we can by Taylor's Theorem~\cite{fulks1978} expand $\mapcopt_{i}\log{\mapcopt_i}$ on $R_{i}$ around $c_{R_i}$ for each $i$ to obtain
\begin{equation}\label{equ:taylor1}
\abbrefdiv{\mapcopt}=\sum_{i} \int_{R_i} \mapcopt_{i} \log{(c_{R_i})} +\mapcopt_{i} - c_{R_i} + \mathcal{R}_{i}~d\probmeasureHone,
\end{equation}
where $\mathcal{R}_{i}=\dfrac{1}{2\delta_{i}}(\mapcopt_{i}-c_{R_i})^{2}, i=0,\dotsc,\asize-1$, are the Taylor remainders of the expansions with $\delta_{i}$ lying in between $\mapcopt_{i}$ and $c_{R_i}$.
By adding and subtracting $ \int \log{(\mapout)}~dP$, ~\eqref{equ:taylor1} can be rearranged to yield
\begin{align}
\abbrefdiv{\mapcopt}-\abbrefdiv{\mapout}&=\sum_{i}\int_{R_i}\mathcal{R}_{i}~dQ \nonumber \\
&\geq \dfrac{c}{2\ubound} \sum_{i}\int_{R_i} (\mapcopt_{i}-c_{R_i})^{2}~d\inputsv \\
&= \dfrac{c}{2\ubound} \norm{\mapcopt-\mapout}_{L_2}^{2}
\end{align}
where the inequality follows from replacing $\delta_i$ with $\ubound$ in the remainder term and using the fact that $q$ is lower bounded by $c$ (Condition 1).
Thus, when there is no approximation for the given distributions $P$ and $Q$ (and for given values of $J$, $L$, $m$, and $M$) the best guaranteed convergence rate is $\mathcal{O}(n^{-2/3})$.
Intuitively, this is reasonable since among those $\mapout$ that share the same partition as $\mapcopt$, it is harder to distinguish $\mapcopt$ compared to the case where there is a nonzero approximation error.
Nguyen, Wainwright and Jordan reported a similar result to Theorem~\ref{thm:esterrorconverge} in~\cite{nguyen_wainwright_jordan2010} .
In their investigation, they used an empirical estimator of the same form as~\eqref{equ:empirest2}, but did not consider quantization, nor did they incorporate a margin condition like Condition 2 into their formulation.
They considered a class of (inverse) likelihood ratio functions $\mathcal{F}$ that satisfies a complexity condition like Condition 1 and found that the difference ${\abbrefdiv{f^{*}}-\empirfdiv{\hat{f}_{\noobs}}}$ decays as $\mathcal{O}(\noobs^{-1/(2+\alpha)})$, where $D_n(\cdot)$ and $D(\cdot)$ are as defined in~\eqref{equ:empirKL} and~\eqref{equ:knownpq_KLest}, $f^*\in\mathcal{F}$ is the best in class likelihood ratio function, and $\hat{f}_n$ is an empirical estimator similar to~\eqref{equ:empirest2}.
Note that this rate is strictly less than the rate in Theorem~\ref{thm:esterrorconverge} even if $\marp$ is eliminated from the formulation (take $\marp\rightarrow\infty$).
\subsection{Approximation Error}\label{subsect:approxerror}
The approximation error analysis also requires that we now think of $\Phi$ as a class of piecewise constant functions (quantization rules) supported on RDPs.
As discussed in Section~\ref{subsect:RDP}, this is fully consistent with the definition given in~\eqref{equ:defcandclass2}.
With this in mind, we have the result:
\begin{theorem}[Approximation error]\label{thm:approxerrorconverge}
Let $\candclass(\asize,J,\lbound,\ubound)$, $\mapsopt$, and $\mapcopt$ be as defined in~\eqref{equ:defcandclass2},~\eqref{equ:knownpq_KLest}, and~\eqref{equ:mapcopt} respectively.
Suppose that Condition 1 is met for some constants $c$ and $C$.
Then the approximation error is bounded as
\begin{equation}\label{equ:vandegeerlem}
\abbrefdiv{\mapcopt}-\abbrefdiv{\mapsopt} \leq \textnormal{const}(\beta,\denlbound,\denubound,\lbound,\ubound,\asize)~2^{-\RDPdepth}.
\end{equation}
\end{theorem}
The proof of this result is given in Appendix~\ref{pf:approxerror} (see also~\cite{lexa_2011a}).
It follows a related, function estimation result in~\cite{ruicastro_phd_thesis2007} with one important exception: the KL divergence is not additive, thus unlike a mean squared error metric, the approximation error $\abbrefdiv{\mapcopt}-\abbrefdiv{\mapsopt}$ cannot be quantified cell by cell.
Details are provided in the proof.
The combination of Theorem~\ref{thm:esterrorconverge} and Theorem~\ref{thm:approxerrorconverge} gives the decay rate of the total expected error in terms of the number of training samples $n$ and the depth $J$ of the uniform dyadic partition $\pi_{J}$.
To balance the errors and obtain a rate only in terms of $n$, one can express $J$ as a function of $n$.
Setting $J=\lceil \kappa \ln{n}/(2\kappa-1)\ln{2} \rceil$ yields the final result
\begin{equation}
\abbrefdiv{\mapcopt}-\expectv\abbrefdiv{\mapempest} \leq \textnormal{const}\cdot \noobs^{-\frac{\marp}{2\marp-1}},
\end{equation}
for sufficiently large $\noobs$.
\section{Application: Quantization under communication constraints}\label{sect:app}
When signals are measured and digitized at one location but processed at another, communication of the data is necessary.
Because of ever present power, computing, and rate constraints, the raw data cannot be transmitted in full fidelity; instead a summary of the data is sent.
When the ultimate goal is classification or detection, one strategy to maximize performance and minimize communication costs is to heavily quantize the data such that the KL divergence is maximized.
This is perhaps the simplest strategy and hence attractive when communications are severely constrained.
Optimal likelihood-ratio partitions can be very different from typical nearest neighbor (Voronoi) partitions that are associated with quantizers designed to minimize mean squared error (see Figs.~\ref{fig:lr} and~\ref{fig:classification_maps}).
Nevertheless, past work in quantization for classification has forced a small-cell property in the design strategy resulting in partitions resembling nearest neighbor partitions~\cite{gupta_hero2003}.
Consequently, optimal partitions with disjoint regions, for example, cannot be well-approximated by these methods.
The EDM quantization method overcomes this shortcoming.
\begin{figure}
\centerline{\includegraphics[width=7cm]{gaussian-laplace_lr.eps}}
\caption{Plot of a likelihood-ratio function of a zero-mean bivariate Gaussian probability density function and a zero-mean bivariate Laplace probability density function.
The level sets of this function, which are concentric circles centered in the quadrants, form the boundaries of an optimal likelihood-ratio partition.
Such partitions are not well-approximated by optimal nearest neighbor partitions.}
\label{fig:lr}
\end{figure}
As an illustration, we consider $P$ to be a zero-mean bivariate Gaussian distribution and $Q$ to be a zero-mean bivariate Laplace distribution, both with identity correlation matrices; $P$ and $Q$ thus differ only in their basic shapes.
The plot of the likelihood ratio in Fig.~\ref{fig:lr} shows that the boundaries of the optimal likelihood-ratio partition are concentric circles in each quadrant.
Fig.~\ref{subfig:ex1a} depicts the best in class quantization rule along with its associated RDP in Fig.~\ref{subfig:ex1b}.
The result was generated with the Flynn and Gray algorithm but with $P_n(S_k)$ and $Q_n(S_k)$ in Algorithm~\ref{algor:flynn_gray} replaced by $P(S_k)$ and $Q(S_k)$.
(Data points lying outside of $[-5,5]^2$ were simply ignored.)
Convergence occurred in 8 iterations.
Fig.~\ref{subfig:ex1c} shows the empirical estimator generated from training sets each of size of two million samples.
In this case, the Flynn and Gray algorithm converged in 11 iterations.
\begin{figure*}
\centerline{\subfigure[Best-in-class rule]{\includegraphics[width=6cm]{best-in-class_rule_g-l.eps}\label{subfig:ex1a}}
\hfil
\subfigure[Associated best-in-class RDP]{\includegraphics[width=6cm]{best-in-class_partition_g-l.eps}\label{subfig:ex1b}}}
\centerline{\subfigure[EDM estimator]{\includegraphics[width=6cm]{estimator_rule_g-l.eps}\label{subfig:ex1c}}
\hfil
\subfigure[Associated estimator RDP]{\includegraphics[width=6cm]{estimator_partition_g-l.eps}\label{subfig:ex1d}}}
\caption{Best-in-class and EDM quantization rules, and their associated recursive dyadic partitions when $P$ is bivariate Gaussian and $Q$ is bivariate Laplace, $L=4$, $J=6$. Note that the different cell labelings (colors) are inconsequential in terms of the divergence.}\label{fig:classification_maps}
\end{figure*}
In comparison to the best in class quantization rule, Fig.~\ref{fig:classification_maps} shows the effect of the trying to estimate $P$ and $Q$ on $\pi_{J}$ for low probability regions (corner regions).
In other words, the lack of data within these regions makes approximating $P$ and $Q$ on $\pi_{J}$ difficult, especially by empirical averages.
More sophisticated density estimation methods would improve this aspect of the estimator, such as kernal based methods.
The estimator might also be improved if one approximates $P$ and $Q$ on a (data-dependent) RDP instead of on $\pi_J$ (see e.g.,~\cite{devroye-gyorfi-lugosi1996}).
\section{Conclusion}\label{sect:conclusion}
In summary, EDM quantization provides a means of finding quantization rules, or more generally, low dimensional transformations that best preserve the divergence between two hypothesized distributions.
EDM estimators can be computed using the Flynn and Gray algorithm, and they can exhibit fast error convergence rates as a function of the number of training samples.
The EDM formulation benefits from its connection to empirical process theory and possesses the flexibility to overcome the necessity of a small-cell constraint and allow efficient encoding.
\section*{Acknowledgments}
I thank John Thompson and Mike Davies for their financial support in presenting this work at conferences and for allowing me access to the computational resources of The University of Edinburgh. I also thank Don Johnson for many helpful discussions during the early stages of this work.
|
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| 6,388
|
{"url":"https:\/\/www.repository.cam.ac.uk\/handle\/1810\/213747","text":"### Recent Submissions\n\n\u2022 #### Elliptic curves over Q$_{\u221e}$ are modular \ufeff\n\nWe show that if $\\textit{p}$ is a prime, then all elliptic curves de ned over the cyclotomic $\\mathbb{Z}$$_{p}$-extension of Q are modular.\n\u2022 #### Connective constants and height functions for Cayley graphs \ufeff\n\nThe connective constant $\\mu$($\\textit{G}$) of an infinite transitive graph $\\textit{G}$ is the exponential growth rate of the number of self-avoiding walks from a given origin. In earlier work of Grimmett and Li, a locality ...\n\u2022 #### On the GL$_{n}$-eigenvariety and a conjecture of Venkatesh \ufeff\n\nLet \u03c0 be a cuspidal, cohomological automorphic representation of GL$_{n}$(A). Venkatesh has suggested that there should exist a natural action of the exterior algebra of a certain motivic cohomology group on the \u03c0-part of ...\n\u2022 #### Infinitely many monotone Lagrangian tori in del Pezzo surfaces \ufeff\n\nWe construct almost toric fibrations (ATFs) on all del Pezzo surfaces, endowed with a monotone symplectic form. Except for CP$^{2}$#CP$^{2}$, CP$^{2}$#2CP$^{2}$, we are able to get almost toric base diagrams (ATBDs) of ...","date":"2017-03-23 16:23:15","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\": 0.8305956721305847, \"perplexity\": 2404.606442490307}, \"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\/1490218187144.60\/warc\/CC-MAIN-20170322212947-00545-ip-10-233-31-227.ec2.internal.warc.gz\"}"}
| null | null |
Midland Community Stadium (MCS) is located in Midland, Michigan and is owned by Midland Public Schools. The stadium is notable because the fan seating is built into man-made earthen hills, and does not have scaffolding type bleachers that are typical for high school stadiums. There is aluminum bench seating for approximately 7,500 (4,000 on the home side; 3,500 on the visitor side), but for the yearly football grudge match between the two cross-town rivals, it is not uncommon for over 10,000 fans to attend.
History
The stadium was built adjacent to Midland High School during the summer of 1958 and dedicated as "Midland Stadium" on Oct. 17 of that year during a win over the Alpena Wildcats. The stadium, high school and Parkdale Elementary were built on the old Midland Airfield, replaced by Barstow Airport.
After the city's second high school, Herbert Henry Dow High School opened in 1968, the name was changed to Midland Community Stadium because the facility was used by both schools as their home field for football & soccer games and track meets.
The stadium was briefly home to the Tri-City Apollos of the Continental Football League for one season in 1969. The team was so bad that they were known locally as the Tri-City Apollo-gies. The league folded that same year.
Nearby Northwood University has played at MCS, when ticket demand greatly exceeds the 3,000 seat capacity of their Hantz Stadium.
Improvements
The field was upgraded in 2004 with installation of an artificial surface, AstroTurf XPe, the same product used at many professional and college sports facilities. Very few high school fields use artificial turf due to the high cost. Additional improvements that same year included paving & surfacing, fencing, a new synthetic track surface and audio & video cable installation to simplify setup for broadcast and/or taping of stadium events.
In 2021, a $1 million project was completed which introduced all aluminum stadium seating, additional handrails, and wheelchair-accessible seating with partner benches.
References
External links
Midland Public Schools website
Midland High School website
H. H. Dow High School website
Midland, Michigan
Sports venues in Michigan
Buildings and structures in Midland County, Michigan
Music venues completed in 1955
1955 establishments in Michigan
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"redpajama_set_name": "RedPajamaWikipedia"
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Q: Why does Q_OBJECT break QDoc? Problem
Since the upgrade from Qt 5.10 to Qt 5.11 I have started having problems to generate a documentation with QDoc for my existing projects.
One of the many issues are missing functions in the documentation, although the corresponding comments exist in the source code.
Research
I have managed to narrow the issue down to the inclusion of the Q_OBJECT macro, as shown by the provided code example (see below).
This is indeed mentioned in the Qt documentation:
If not specified by the Cpp.ignoretokens or Cpp.ignoredirectives variables, non-standard constructs (typically macros) can result in erroneous documentation.
Q_OBJECT is not supposed to cause problems though, because just a little bit further it is written:
The Q_OBJECT macro, however, is an exception: QDoc recognizes this particular non-standard construct, so there is no need specifying it using the Cpp.ignoredirectives variable.
In any case I do include qt-cpp-defines.qdocconf in my qdocconf file.
I have also tried to manually add Q_OBJECT to the ignore list
Cpp.ignoredirectives += Q_OBJECT
but the result is the same.
I experience the described issue under Windows 10 and Ubuntu 17. Under Windows 7 I cannot execute qdoc.exe at all.
What is the correct configuration of qdocconf to overcome this issue?
Minimal example
For a quick reproduction (in the real situation the declarations and the implementations are split and proper comments are added), please consider the following setup:
Foo.h
#include <QObject>
class Foo : public QObject
{
// Q_OBJECT // <-- uncomment this to break QDoc
public:
Foo() {}
void boo() {}
protected:
void moo() {}
};
Foo.cpp
#include "Foo.h"
/*!
\class Foo
*/
test.qdocconf
include($QT_INSTALL_DOCS/global/compat.qdocconf)
include($QT_INSTALL_DOCS/global/fileextensions.qdocconf)
include($QT_INSTALL_DOCS/global/qt-cpp-defines.qdocconf)
include($QT_INSTALL_DOCS/global/macros.qdocconf)
# Uncoment this for a test
# Cpp.ignoredirectives += Q_OBJECT
outputdir = html
headerdirs = .
sourcedirs = .
exampledirs = .
imagedirs = ./images
Results
*
*Good result (without Q_OBJECT)
Executing qdoc.exe test.qdocconf I get more or less the following:
*
*Foo
Contents
*
*Public Functions
*Protected Functions
*Detailed Description
Foo Class
*
*List of all members, including inherited members
Public Functions
Foo()
void boo()
Protected Functions
void moo()
Detailed Description
Member Function Documentation
Foo::Foo()
Default constructs an instance of Foo.
void Foo::boo()
[protected] void Foo::moo()
*
*Bad result (with Q_OBJECT)
Uncommenting the Q_OBJECT macro and running qdoc.exe again yelds the following result:
*
*Foo
Contents
*
*Detailed Description
Foo Class
Detailed Description
IMPORTANT: Foo, moo and boo are gone.
A: I know this question is a few years old already but I wanted to post an answer for future searchers that find this. I had this issue for both Q_OBJECT and Q_INVOKABLE macros in my .cpp file.
The solution is either to use an undocumented command in your .qdocconf file, includepaths, or to pass -I parameters to your command when you run qdoc.
I will only show how I got it working with my config.qdocconf file
...
# undocumented feature that simulates passing -I parameters to the command line
includepaths = . \
.. \
$QT_INSTALL_HEADERS \
$QT_INSTALL_HEADERS/QtCore \
$QT_INSTALL_HEADERS/QtGui \
$QT_INSTALL_HEADERS/QtQuick \
$QT_INSTALL_DOCS
...
You can also use absolute paths instead of $QT_INSTALL_HEADERS if needed.
An easy way to see where those special variables point to is to run qmake -query (use an absolute path to your qt install bin if needed for your qmake command)
Edit: For me, the $QT_INSTALL_HEADERS = C:/Qt/5.12.9/msvc2017_64/include
Edit 2: make sure you have clang installed on your system (via chocolately, homebrew, apt, or others) and if on windows that you run set LLVM_INSTALL_DIR=C:\Program Files\LLVM before you run qdoc - Instructions here: Installing Clang for QDoc
A: The only solution I came up with is to add the following preprocessor directives to the Q_OBJECT macro:
#ifndef Q_QDOC
Q_OBJECT
#endif //Q_QDOC
Q_QDOC is defined in the included qt-cpp-defines.qdocconf, so QDoc skips the macro, but it is not defined within the build system and the code compiles as usual.
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"redpajama_set_name": "RedPajamaStackExchange"
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St Leonard's Tower may refer to a number of buildings.
St Leonard's Tower, Newton Abbot, Devon.
St Leonard's Tower, West Malling, Kent
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Villa Frondega is a unique, charming property. It was refurbished from a traditional Spanish Finca and is still full of the character and charm of its past. The alluring property is located a few minutes' drive from the town of Pollensa. The owners have dedicated a long time in decorating and restoring the house to give it a modern touch in a totally rustic and traditional atmosphere. The villa has been lovingly restored to give it a totally rustic and traditional atmosphere, combined with modern day comforts to form a wonderful holiday destination.
From the main terrace, a wooden door gives access to the interior of the house. The spacious living room consists of wooden beams on the ceiling, ceramic floors and traditional furniture, all with the aim of making the stay more relaxing. The lounge area is attractive and is fully equipped with sofas, armchairs, TV and DVD, and a fireplace. In the back of the lounge you can find another rest area and a toilet. The kitchen is very well equipped and has an outside exit towards a covered terrace and pool area. All of the three bedrooms are located on the top floor, with a master double bedroom en-suite and two further twin bedrooms that share a bathroom. All of the rooms have air conditioning for your comfort.
If you're a fan of outdoor cooking then the lovely barbecue kitchen area, and large terraces will be perfect for you! With plenty of room for al-fresco dining, you can try your skills at some Spanish classics and enjoy a glass or two of the delicious variety of Balearic Island wines. The private swimming pool is Roman in style and is surrounded by a well-kept lawned garden from which you can admire the different fruit and floral trees around.
Villa Frondega is ideal for families who want to spend a pleasant holiday in an authentic Mallorcan finca totally restored.
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Q: _main() unresolved I wrote program in C using visual studio 2013 , but i got this error:
MSVCRTD.lib(crtexe.obj) : error LNK2019: unresolved external symbol _main referenced in function ___tmainCRTStartup
I don't know what is it mean and how to fix it.this is the code :
#include<stdio.h>
int getArr(int arr[]){
int n;
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d",&arr[i]);
return n;
}
void putArr(int arr[],int n)
{
for(int i=0;i<n;i++)
printf("%d\t",arr[i]);
printf("\n");
}
void sort(int Arr[],int nArr)
{
for(int i=0;i<nArr-1;i++)
{
int minIndex=i;
for(int j=i+1;j<nArr;j++)
if(Arr[j]<Arr[minIndex])
minIndex=j;
int t=Arr[i];
Arr[i]=Arr[minIndex];
Arr[minIndex]=t;
}
}
int main()
{
int arr[100];
int nArr;
nArr=getArr(arr);
sort(arr,nArr);
putArr(arr,nArr);
return 0;
}
A: Your code as written builds fine with VS 2012, VS 2013, or VS 2015 using the command-line tools (via the "Developer Command Prompt for VS xxxx" window).
Windows has three types of main for C/C++ console apps:
main: This is the traditional ANSI main which takes command-line parameters as char*
wmain: This is the Unicode main which takes command-line parameters as wchar_t*
_tmain: This is the _TCHAR version which can build as either ANSI or Unicode. This is what the default template uses and is set to build as Unicode.
If you changed your void main() to void wmain() it would build as well since the default template project settings is set to "Use Unicode Character Set" which on the command-line adds /D_UNICODE /DUNICODE
If you go to Project -> Properties -> General and set Character Set to "Use Multi-Byte Character Set" for All Configurations and All Platforms, then your use of void main() will successfully link. This uses /D_MBCS for the command-line instead of /D_UNICODE /DUNICODE
That said, with VS 2013's default project your code builds fine replacing the existing _tmain with your main even without changing the character set setting.
A: Visual Studio dont want that user declare variables in case, or for instructions
use a declaration a the start of function corpus. Good Luck
#include<stdio.h>
int getArr(int arr[]){
int n,i;
scanf("%d",&n);
for(i=0;i<n;i++)
scanf("%d",&arr[i]);
return n;
}
void putArr(int arr[],int n)
{
int i=0;
for(i=0;i<n;i++)
printf("%d\t",arr[i]);
printf("\n");
}
void sort(int Arr[],int nArr)
{
int j, i=0;
int t;
for(i=0;i<nArr-1;i++)
{
int minIndex=i;
for( j=i+1;j<nArr;j++)
if(Arr[j]<Arr[minIndex])
minIndex=j;
t=Arr[i];
Arr[i]=Arr[minIndex];
Arr[minIndex]=t;
}
}
int main()
{
int arr[100];
int nArr;
nArr=getArr(arr);
sort(arr,nArr);
putArr(arr,nArr);
return 0;
}
A: You created a windows application project you should select console Application when creating a new project or if you want to learn win32 programming you can use
below links:
https://msdn.microsoft.com/en-us/library/windows/desktop/ff381398(v=vs.85).aspx
http://www.catch22.net/
http://pravin.paratey.com/win32/
http://www.win32developer.com/tutorial/windows/windows_tutorial_1.shtm
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"redpajama_set_name": "RedPajamaStackExchange"
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Jessica Louise Nelson (ur. 14 czerwca 1991) – angielska piosenkarka i była członkini brytyjskiej grupy Little Mix. Grupa powstała w ósmej serii The X Factor w 2011 roku i jako pierwszy zespół wygrała konkurs. Od debiutu dziewczęcy zespół sprzedał ponad 50 milionów płyt na całym świecie, co plasuje je w pierwszej piątce najlepiej sprzedających się girlsbandów wszech czasów. W 2019 roku Nelson zaprezentowała swój film dokumentalny powstały we współpracy z BBC Three, w którym opowiedziała o swoich doświadczeniach z obrazem ciała i zastraszaniem w Internecie, zatytułowany Jesy Nelson: Odd One Out, który zdobył nagrodę Factual Entertainment Award na 25. National Television Awards. 14 grudnia 2020 roku zdecydowała się opuścić grupę z powodów związanych ze zdrowiem psychicznym.
Wczesne życie
Jessica Louise Nelson urodziła się 14 czerwca 1991 r. Wychowała się w Romford, we wschodnim Londynie. Jej rodzicami są John Nelson, biznesmen i Janice White, funkcjonariuszka policji ds. społeczności. Jej rodzice rozstali się, gdy miała pięć lat. Jest drugą najmłodszą z czworga dzieci. Ma starszą siostrę Jade, starszego brata Jonathana i młodszego brata Josepha.
Nelson uczęszczała do Jo Richardson Community School i Abbs Cross Academy and Arts College w Hornchurch w Londynie. Uczęszczała również do szkół teatralnych Sylvii Young i Yvonne Rhodes. Jedną z jej koleżanek ze szkoły była Rita Ora. Przed przesłuchaniem do The X Factor Nelson pracowała jako barmanka w Dagenham. W 2020 roku Nelson powiedziała, że jako dziecko miała małe role jako statystka w Był sobie chłopiec (2002) i Harry Potter i Czara Ognia (2005).
Kariera
Jej utworem z pierwszego przesłuchania był "Bust Your Windows" Jazmine Sullivan. Odkąd dołączyła do Little Mix, zmagała się z cyberprzemocą i walczyła z nią podczas swojego udziału w The X Factor. Nelson i Perrie Edwards zostały umieszczone w grupie o nazwie "Faux Pas", a Jade Thirlwall i Leigh-Anne Pinnock w zespole o nazwie "Orion". Jednak obu grupom nie udało się zrobić postępów. Później podjęto decyzję o sprowadzeniu czwórki wokalistek z powrotem i utworzeniu z nich czteroosobowej grupy Rhythmix, która miała kontynuować swoją przygodę w programie wchodząc w etap domów sędziowskich. W końcu dotarły do momentu show na żywo pod opieką Tulisy Contostavlos. 28 października 2011 roku ogłoszono, że grupa będzie miała nową nazwę Little Mix. 11 grudnia 2011 r. ogłoszono Little Mix zwyciężczyniami oraz pierwszą grupą, która wygrała program. Nelson wydała z grupą sześć albumów: DNA (2012), Salute (2013), Get Weird (2015), Glory Days (2016), LM5 (2018) i Confetti (2020). W grudniu 2020 roku ogłosiła odejście z grupy z powodu przedłużających się problemów ze zdrowiem psychicznym. Powiedziała: "Uważam, że ciągła presja bycia w zespole i spełniania oczekiwań jest bardzo trudna".
Życie prywatne
Nelson powiedziała, że prześladowanie w szkole mogło przyczynić się do tego, że jako nastolatka cierpiała na łysienie wywołane stresem. W swoim dokumencie BBC Odd One Out, Nelson głośno opowiadała o swojej walce z wizerunkiem ciała. Powiedziała, że stosowała głodówkę przed występami telewizyjnymi lub nagraniami wideo, a później objadała się. Powiedziała, że nadużycia ze strony internetowych trolli na Twitterze doprowadziły ją do próby samobójczej w 2013 roku, stwierdzając: "Czułam, że fizycznie nie mogę już dłużej tolerować bólu".
Nelson była w 10-miesięcznym związku z członkiem Diversity Jordanem Banjo, który zakończył się w 2013 roku. W 2014 roku Nelson zaczęła spotykać się z wokalistą Rixton, Jakiem Rochem. Para zaręczyła się 19 lipca 2015 r., ale później rozstała się w listopadzie 2016 roku. W 2017 r. była w krótkim związku z Chrisem Clarkiem, a następnie 16-miesięcznym związku z muzykiem Harrym Jamesem, który zakończył się w listopadzie 2018 roku. W styczniu 2019 r. Nelson rozpoczęła spotykać się z uczestnikiem Love Island, Chrisem Hughesem. W rocznicę ich związku w styczniu 2020 roku powiedziała: "Mogę szczerze powiedzieć, że nigdy w życiu nie byłam bardziej zakochana i szczęśliwa". W kwietniu 2020 roku Nelson i Hughes zerwali z powodów osobistych. Cztery miesiące później ogłosiła, że jest w związku z aktorem Seanem Sagarem.
Nelson ma 15 tatuaży, w tym cytat na prawym ramieniu: "Muzyka jest najsilniejszą formą magii". Jej szacowany majątek w 2020 roku wyniósł 5,8 miliona funtów.
W listopadzie 2020 roku publicysta Little Mix stwierdził, że Nelson zrobi dłuższą przerwę od popowej grupy z "prywatnych powodów medycznych". 14 grudnia 2020 r. Nelson ogłosiła odejście z Little Mix z powodu problemów ze zdrowiem psychicznym.
Dyskografia
Singles
Udział w pisaniu piosenek
Filmografia
Nagrody i nominacje
Przypisy
Urodzeni w 1991
Brytyjskie wokalistki popowe
Uczestnicy programów typu talent show
|
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{"url":"http:\/\/ehecregnet.compbio.sdu.dk\/1\/clustering_quality_measures\/9","text":"ClustEval clustering evaluation framework\nThe FDR relates the number of false positives to the total number of positively predicted elements. This estimates a likelihood of an element being negative, if it is predicted positive. This measure is based on the pairwise approach to calculate TP,TN,FP and FN. $FDR = \\frac{FP}{FP+TP}$ In the binary classification background we have two classes that we want to distinguish: positive and negative. In this scenario there are four possible outcomes:","date":"2022-05-18 02:35:47","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\": 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\": 0.46235087513923645, \"perplexity\": 644.2022949691703}, \"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-2022-21\/segments\/1652662521041.0\/warc\/CC-MAIN-20220518021247-20220518051247-00162.warc.gz\"}"}
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{"url":"https:\/\/math.stackexchange.com\/questions\/2763156\/are-the-sets-mathbb-x-0-1-4-15-56-x-h-and-mathbb-y-0-2-12-70","text":"Are the sets $\\mathbb X=\\{0,1,4,15,56,\u2026,x_h,\u2026\\}$ and $\\mathbb Y=\\{0,2,12,70,408,\u2026,y_i,\u2026\\}$ (excepting the elements $x_0=y_0=0$) disjunct?\n\nIn trying to answer another question I came to the problem as written in the title:\n\nQ1: Are the sets $\\mathbb X=\\{0,1,4,15,56,...,x_h,...\\}$ and $\\mathbb Y=\\{0,2,12,70,408,...,y_i,...\\}$ (excepting the elements $x_0=y_0=0$) disjunct?\n\nand how to approach a proof.\n\nI have two ways to describe the sets as sequences with their elements depending on their indexes: $$x_0=0, x_1=1, x_{h+1}=4x_h-1x_{h-1}\\\\ y_0=0, y_1=2, y_{i+1}=6y_i-1y_{i-1}\\\\ \\tag 1$$ and using $p=2+\\sqrt3$ and $q=3+\\sqrt8$ $$x_h = f(h)= \\sinh (h \\cdot \\ln p)\/\\sqrt 3 \\\\ y_i =g(i) = \\sinh (i \\cdot \\ln q)\/\\sqrt 8 \\tag 2$$ (The latter is a compacted version of a Binet-like formula (as it is known for instance for the Fibonacci numbers))\n\nWhat I've tried was (\"ansatz 1\") to look at the composition of $i=g\u00b0^{-1}(f(h))$ and see, whether I can prove, that $i$ is never integer when $h$ is integer, but looking at heuristics and working with the results a bit I do not see any further forcing and\/or reliable way to proceed.\n\nNext (\"ansatz 2\") I've looked at the prime-factorizations of $f(h)$ and $g(i)$ in terms of $h$ resp. $i$. This gives at least insight to some basic facts, for instance that $h$ must be even, then divisible by $3$ and from this $i$ must be divisible by $3$ and so on, and at least leading to a proof for small $h$ and $i$. But this is so far only reasoning for finitely many classes of cases.\n\nQ2: Can my ansatz 1 or ansatz 2 be improved? Or not?\nQ3: Is there any differnt route to approach this?\n\n\u2022 I propose anyone who says \"the set $\\{0, 1,4,15, 56...\\}$ and the set $\\{0,2,12,70, 408,...\\}$\" and expects it to be understood by hung by their entrails. Anyway the answer is, no, they are not distinct; the 37th element of $X$ and then 408th element of $Y$ are both \"Barbar the Elephant in a green Tutu\". That's the only element they both have in common. \u2013\u00a0fleablood May 7 '18 at 21:57\n\u2022 @fleablood - if you have tried it numerically, I think your software made a truncation error. The second sequence ($y$) has a larger quotient than the first sequence, so it is impossible, that a number in $Y$ with an index larger than an index in $X$ can be equal. \u2013\u00a0Gottfried Helms May 8 '18 at 6:15\n\nHere is a positive answer to question Q1, and by the way to the question:\n\nLet $k$ be a postive integer number. Then $2k^2+1$ and $3k^2+1$ cannot both be square numbers.\n\nDefine the sequences $(x_n),(y_n),(z_n),(r_n),(s_n),(t_n)$ by the following second order linear recurrences: \\begin{align*} &x_0=0,\\ x_1=1 &x_{n+1}=4x_n-x_{n-1}\\\\ &y_0=0,\\ y_1=2 &y_{n+1}=6y_n-y_{n-1}\\\\ &z_0=0, \\ z_1=1 &z_{n+1}=6z_n-z_{n-1}\\\\ &r_0=1, \\ r_1=2 &r_{n+1}=4r_n-r_{n-1}\\\\ &s_0=1, \\ s_1=5 &s_{n+1}=4s_n-s_{n-1}\\\\ &t_0=1, \\ t_1=3 &t_{n+1}=4t_n-t_{n-1} \\end{align*}\n\nIt is clear that all the above sequences are monotonically increasing sequences of natural numbers. Also it is clear that $y_n=2z_n$. Now we have the following factorization:\n\n\\begin{align*} &x_{2n}=2r_nx_{n} \\end{align*}\n\nFor the proof for the above factorization, we first show that: \\begin{align} \\tag{1}\\label{1}s_{q+k}+t_{q+k}&=t_ks_q+s_kt_q \\\\ \\tag{2}\\label{2}2x_{k+1}&=s_k+t_k\\\\ \\tag{3}\\label{3}r_{k}&=\\frac{1}{2}(t_k+t_{k-1})\\\\ \\tag{4}\\label{4}3x_k&=\\frac{1}{2}(s_k+s_{k-1})\\\\ \\tag{5}\\label{5}r_{k}&=\\frac{1}{2}(s_k-s_{k-1})\\\\ \\tag{6}\\label{6}x_{k}&=\\frac{1}{2}(t_k-t_{k-1}) \\end{align} It is clear that the rhs and lhs in the above relationships are also integer sequences, say $X_k$, satisfying the second order linear recurrence $X_{k+1}=4X_k-X_{k-1}$. Then for proving lhs=rhs, it suffices that they coincide at the index $0$ and the index $1$, which is an easy check. (For $\\eqref{1}$, the two indices $k$ and $q$ are treated separately and consecutively). Then multiply equations $\\eqref{3}$ and $\\eqref{4}$, then multiply $\\eqref{5}$ and $\\eqref{6}$, then make the difference, reduce and account for $\\eqref{1}$ with $q=k-1$ and finally make use of $\\eqref{2}$.\n\nIf there exist non-zero natural numbers $m,m'$ such that $y_m=x_{m'}$, then $2z_m=x_{m'}$, then $x_{m'}$ is even, then ${m'}$ is even, say ${m'}=2n$. Then, answering yes to question Q1 is equivalent to disprove that there exist non-zero $n,m$ such that $$z_m = r_n x_n$$ We then show that there is no such $m$, because $z_m$ is a monotonically increasing function of $m$ and because we have:\n\nProposition: For $j>0$, we have: $\\ \\ \\ \\ z_{3j-2} < r_{2j-1} x_{2j-1} < z_{3j-1}< r_{2j} x_{2j} < z_{3j}$\n\nWe will use the following:\n\nLemma: For $j>3$, we have: $197z_{j-3}<z_{j}\\le204z_{j-3}$.\n\nProof of the Lemma. \\begin{align*} z_{j}=6z_{j-1}-z_{j-2}=35z_{j-2}-6z_{j-3}&=204z_{j-3}-35z_{j-4} \\le 204z_{j-3}\\\\ &=197z_{j-3}+7z_{j-3}-35z_{j-4}\\\\ &=197z_{j-3}+(42-35)z_{j-4}-7z_{j-5}\\\\ &=197z_{j-3}+7(z_{j-4}-z_{j-5})\\\\ &>197z_{j-3} \\end{align*} since $z_m$ is a positive and monotonically increasing function of $m$.\n\nProof of the Proposition.\n\nWe show $r_{2j} x_{2j} < z_{3j}$ by induction on $j$. It is true for $j=1$, since $r_{2} x_{2}=28 < 35 = z_{3}$. Suppose (induction hypothesis) that $r_{2(j-1)} x_{2(j-1)} < z_{3j-3}$, for some $j>1$. \\begin{align*} r_{2j}&=4r_{2j-1}-r_{2j-2}=16r_{2j-2}-4r_{2j-3}-r_{2j-2}=15r_{2j-2}-4r_{2j-3}\\\\ x_{2j}&=4x_{2j-1}-x_{2j-2}=16x_{2j-2}-4x_{2j-3}-x_{2j-2}=15x_{2j-2}-4x_{2j-3} \\end{align*} But $4r_{2j-3}=r_{2j-2}+r_{2j-4}$ and $4x_{2j-3}=x_{2j-2}+x_{2j-4}$, then \\begin{align*} r_{2j}&=14r_{2j-2}-r_{2j-4}<14r_{2j-2}\\\\ x_{2j}&=14x_{2j-2}-x_{2j-4}<14x_{2j-2} \\end{align*} then \\begin{align*} r_{2j}x_{2j}&<196r_{2j-2}x_{2j-2}\\\\ &<196z_{3j-3} \\text{ by the induction hypothesis.}\\\\ & < z_{3j} \\text{ by the Lemma.} \\end{align*}\n\nWe similarly show $r_{2j} x_{2j} > z_{3j-1}$ by induction on $j$. It is true for $j=1$, since $r_{2} x_{2}=28 > 6 = z_{2}$. Suppose (induction hypothesis) that $r_{2(j-1)} x_{2(j-1)} > z_{3j-4}$, for some $j>1$. \\begin{align*} r_{2j}&=15r_{2j-2}-4r_{2j-3}=19r_{2j-2}+4(r_{2j-2}-r_{2j-3})>19r_{2j-2}\\\\ x_{2j}&=15x_{2j-2}-4x_{2j-3}=19x_{2j-2}+4(x_{2j-2}-x_{2j-3})>19x_{2j-2} \\end{align*} then \\begin{align*} r_{2j}x_{2j}&>361r_{2j-2}x_{2j-2}\\\\ &>361z_{3j-4} \\text{ by the induction hypothesis.}\\\\ & > z_{3j-1} \\text{ by the Lemma.} \\end{align*}\n\nWe similarly show $r_{2j-1} x_{2j-1} < z_{3j-1}$ by induction on $j$. It is true for $j=1$, since $r_{1} x_{1}=2 < 6 = z_{2}$. Suppose (induction hypothesis) that $r_{2j-3} x_{2j-3} < z_{3j-4}$, for some $j>1$. \\begin{align*} r_{2j-1}&=4r_{2j-2}-r_{2j-3}=16r_{2j-3}-4r_{2j-4}-r_{2j-3}=15r_{2j-3}-4r_{2j-4}\\\\ x_{2j-1}&=4x_{2j-2}-x_{2j-3}=16x_{2j-3}-4x_{2j-4}-x_{2j-3}=15x_{2j-3}-4x_{2j-4} \\end{align*} But $4r_{2j-4}=r_{2j-3}+r_{2j-5}$ and $4x_{2j-4}=x_{2j-3}+x_{2j-5}$, then \\begin{align*} r_{2j-1}&=14r_{2j-3}-r_{2j-5}<14r_{2j-3}\\\\ x_{2j-1}&=14x_{2j-3}-x_{2j-5}<14x_{2j-3} \\end{align*} then \\begin{align*} r_{2j-1}x_{2j-1}&<196r_{2j-3}x_{2j-3}\\\\ &<196z_{3j-4} \\text{ by the induction hypothesis.}\\\\ & < z_{3j-1} \\text{ by the Lemma.} \\end{align*}\n\nWe similarly show $r_{2j-1} x_{2j-1} > z_{3j-2}$ by induction on $j$. It is true for $j=1$, since $r_{1} x_{1}=2 > 1 = z_{1}$. Suppose (induction hypothesis) that $r_{2j-3)} x_{2j-3} > z_{3j-5}$, for some $j>1$. \\begin{align*} r_{2j-1}&=15r_{2j-3}-4r_{2j-4}=19r_{2j-3}+4(r_{2j-3}-r_{2j-4})>19r_{2j-3}\\\\ x_{2j-1}&=15x_{2j-3}-4x_{2j-4}=19x_{2j-3}+4(x_{2j-3}-x_{2j-4})>19x_{2j-3} \\end{align*} then \\begin{align*} r_{2j-1}x_{2j-1}&>361r_{2j-3}x_{2j-3}\\\\ &>361z_{3j-5} \\text{ by the induction hypothesis.}\\\\ & > z_{3j-2} \\text{ by the Lemma.} \\end{align*}\n\n\u2022 Thank you, Ren\u00e9! The idea of taking some more sequences into the boat might be a more generally useful trick, I'll see... I can come back to this only later and shall then look what this gives me. (+1) for the effort so far anyway. \u2013\u00a0Gottfried Helms May 8 '18 at 6:18","date":"2020-09-27 03:55:38","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\": 1, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9874195456504822, \"perplexity\": 478.3769501151485}, \"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-2020-40\/segments\/1600400250241.72\/warc\/CC-MAIN-20200927023329-20200927053329-00052.warc.gz\"}"}
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Renault's managing director Cyril Abiteboul says capturing the signature of Daniel Ricciardo is evidence of his team's rapid rise in Formula 1 and has set high targets ahead of the arrival of the Australian driver.
Ricciardo shocked F1 by confirming he will leave Red Bull at the end of the this season to join Renault having signed a two-year deal with the French manufacturer.
The Renault team boss is naturally thrilled to secure the seven-time F1 race winner's services and feels he can help speed up the French manufacturer's planned progress climbing the F1 ranks.
Renault returned to F1 as a fully-fledged manufacturer in 2016 and set out a long-term goal of battling for world championships. Last year Renault climbed to sixth place in the F1 world constructors' standings and currently sits fourth in the teams' championship heading into the summer break.
Given the high-profile signing, Abiteboul has underlined the need to give Ricciardo a front-running package in 2019.
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{
"redpajama_set_name": "RedPajamaC4"
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{"url":"https:\/\/socratic.org\/questions\/if-the-sum-of-interior-angle-measures-of-a-polygon-is-900-how-many-sides-does-th","text":"# If the sum of interior angle measures of a polygon is 900, how many sides does the polygon have?\n\nThen teach the underlying concepts\nDon't copy without citing sources\npreview\n?\n\n#### Explanation\n\nExplain in detail...\n\n#### Explanation:\n\nI want someone to double check my answer\n\n30\nDec 19, 2015\n\nthe polygon is *heptagon (7-sided polygon)\n\n#### Explanation:\n\nSum of interior angles is $180 \\left(n - 2\\right)$\n\nSince sum is $= 900$,\n\n$900 = 180 \\left(n - 2\\right)$\n\n$\\frac{900}{180} = \\frac{180 \\left(n - 2\\right)}{180}$\n\n$\\frac{900}{180} = \\frac{\\cancel{180} \\left(n - 2\\right)}{\\cancel{180}}$\n\n$5 = n - 2$\n\n$5 + 2 = n$\n\n$7 = n$\n\n\u2022 27 minutes ago\n\u2022 28 minutes ago\n\u2022 28 minutes ago\n\u2022 33 minutes ago\n\u2022 4 minutes ago\n\u2022 6 minutes ago\n\u2022 8 minutes ago\n\u2022 10 minutes ago\n\u2022 14 minutes ago\n\u2022 20 minutes ago\n\u2022 27 minutes ago\n\u2022 28 minutes ago\n\u2022 28 minutes ago\n\u2022 33 minutes ago","date":"2018-03-21 10:32:27","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 8, \"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\": 0.5452731251716614, \"perplexity\": 9370.153388969851}, \"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-2018-13\/segments\/1521257647612.53\/warc\/CC-MAIN-20180321102234-20180321122234-00388.warc.gz\"}"}
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\section{Introduction}
\label{sec:introduction}
Understanding the nature of dark matter is one of the most pressing problems in cosmology and particle physics~\cite{Bertone:2010zza,Jungman:1995df,Bergstrom00,Bertone05}. A key prediction of the standard cosmological model, based on the assumption that dark matter is cold, i.e. non-relativistic at the epoch of structure formation, is that a large number of dark matter substructures should exist in Milky Way-like galaxies \cite{Diemand2008,Springel2008}.
Detecting these subhalos will not only be a strong indicator for the existence of dark matter but also will give valuable information about its particle nature. Depending on their couplings and production mechanism, in fact, dark matter particles can achieve non-negligible velocity dispersions -- thus act as warm dark matter (WDM) -- leading to a suppression of the primordial density fluctuations on small scales, and thus a cutoff at small halo masses, as already realised long ago \cite{1982ApJ...258..415P}.
We discuss below the case of a thermal WDM relic, for which the cutoff in the power spectrum depends only on the WDM particle mass, but the discussion can be generalised to sterile neutrinos \cite{Dodelson1994,Shi:1998km,Abazajian:2001nj,Asaka2005,Boyarsky2009,Adhikari:2016bei} by taking into account the lepton asymmetry parameter (see next section).
The subhalo mass function of WDM and cold dark matter (CDM) differs for masses smaller than those of dwarf galaxies. On these scales, such substructures would be completely dark matter dominated, and contain too few stars to be observed directly. A number of strategies have been proposed to indirectly probe low mass dark matter halos, including analyses of the impact on Ly$\alpha$ forest observations \cite{Narayanan:2000tp,Viel2005,Boyarsky:2008xj,Viel2013,Baur:2015jsy,Garzilli:2015iwa,Irsic:2017ixq} and perturbations of strong gravitational lenses \cite{Dalal2002,Vegetti:2008eg,Li:2015xpc,Penarrubia:2017nzw,Asadi:2017ddk,Mao:2017auo,Daylan:2017kfh,Minor:2016jou,Despali:2016meh}.
Here, we study the prospects for identifying the nature of dark matter particles by studying the perturbations induced by sub-dwarf galaxy clumps on cold stellar streams. A stellar stream, created as a result of tidal disruption of a globular cluster or dwarf galaxy by the Milky Way potential, has more or less uniform stellar density along its length. A flyby subhalo gravitationally perturbs the stars in the stream resulting in a region of low stellar density or gap whose size increases with time. There has been a lot of work recently \cite{Yoon2011,Carlberg2012,Carlberg2013,Erkal2015,Erkal2015a,Sanders2016} focusing on the study of gaps in stellar streams as a result of subhalo encounters. Specifically, ref.~\cite{Erkal2015a} showed that with accurate measurements of the density and phase-space structure of stellar streams, which is possible with the near future galaxy surveys like LSST and the Gaia mission currently taking data, we will be able to measure subhalo impacts with mass as low as $10^7 ~\rm{M}_{\odot}$. In ref.~\cite{Bovy2016a}, a novel framework was presented for inferring properties of the impacting subhalos by analyzing the power spectrum of the fluctuations in the density and mean track of the perturbed stream. This method was applied on Pal 5 density data and was shown to be sensitive to subhalos with mass as low as $10^{6.5} ~ \rm{M}_{\odot}$. The authors also predicted that with better data this method will be sensitive to even $10^5 ~\rm{M}_{\odot}$ subhalos. In order to discriminate between WDM and CDM, we will in particular exploit the difference in the density power spectrum arising from the different subhalo populations.
This paper is structured as follows. In section \ref{sec:WDM}, we describe the WDM model that we used to simulate the stream-subhalo encounters in the WDM scenario. In section \ref{sec:simstream}, we discuss our method of generating fast stream density simulations of the GD-1 stream and how we model subhalo impacts on it. In section \ref{sec:results}, we present our results of the stream density perturbations and their power spectra, and apply the Approximate Bayesian Computation (ABC) technique to constrain the mass of the dark matter particle from the stream power spectra. We show that for typical cases we are able to distinguish between WDM and CDM models. In section \ref{sec:outlier}, we demonstrate the risks of using only one stream to constrain the particle mass of dark matter. With these examples we make a case for the need of applying our method on multiple streams to make robust predictions on the mass of the dark matter particle. Finally we conclude in section \ref{sec:conclusions}. In appendix \ref{sec:scaleradius} we discuss how the scale radius of subhalos in the WDM scenario differ from the CDM scenario and why this variation has little effect on our method.
\section{Modeling warm dark matter}
\label{sec:WDM}
We discuss below a thermal WDM relic, for which the cutoff in the power spectrum depends only on the WDM particle mass, but the discussion can be generalised to sterile neutrinos by taking into account the lepton asymmetry parameter\footnote{The lepton asymmetry is defined as $L_6 \equiv 10^6 (n_{\nu_e} - n_{\bar{\nu_e}})/s$, where $n_{\nu_e}$ and $n_{\bar{\nu_e}}$ are the number densities in electron neutrinos and anti-neutrinos respectively, and $s$ is the entropy density of the Universe. For very high and low values of lepton asymmetry the power spectrum of sterile neutrinos can be well approximated by that of a thermal WDM.}. For example, the power spectrum of a 3.3 keV thermal WDM matches to a high degree with that of a 7 keV sterile neutrino with lepton asymmetry parameter equal to 8.66 (see figure 1 in ref.~\cite{Bose2015}), with both power spectra deviating from the CDM case at around $\log(k) \gtrsim 1.0 ~h~\rm{Mpc}^{-1}$, where $k$ is the comoving wave number and $h=H_0/100$~km$/$s$/$Mpc is the dimensionless Hubble parameter. The 7 keV sterile neutrino is especially interesting since its decay products have been attributed to be the source of the 3.5 keV X-ray line \cite{Bulbul2014,Boyarsky2014} detected in stacked spectrum of clusters as well as in the spectra of Andromeda and Perseus cluster. Notice that the case of the 7 keV sterile neutrino with lepton asymmetry of 8.66 corresponds to the coldest in the 7 keV sterile neutrino family, so any constraints on it can be extended to all the other members of the family.
Various studies have set a lower bound on the mass of thermal WDM based on the observed clumpiness in the Ly-$\alpha$ forest. Ref.~\cite{Viel2013} found this lower bound to be 3.3 keV ($2\sigma$), while ref.~\cite{Baur:2015jsy} found it to be 4.09 keV (95\% CL) from Ly-$\alpha$ data alone, and 2.96 keV (95\% CL) when combining Ly-$\alpha$ data with CMB data from Planck 2016. Very recently, ref.~\cite{Yeche:2017upn} found this lower limit to be 4.17 keV (95\% CL) using two different Ly-$\alpha$ measurements, and 4.65 keV (95\% CL) when adding a third set of data. Notice that the constraints on the thermal WDM mass depend on the thermal history assumed for the intergalactic medium~\cite{Garzilli:2015iwa}.
In this work we model WDM as a thermal relic and emphasize on the case of 3.3 keV particle mass. We follow the same framework that was adopted in WDM simulation works \cite{Lovell2013,Bose2015,Bose2016}. The WDM and CDM power spectra are related by a transfer function $T(k)$:
\begin{equation}
P_{\rm{WDM}} (k) = T^{2}(k)P_{\rm{CDM}}(k),
\end{equation}
where $T(k)$ is approximated by the fitting formula \cite{Bode2001}:
\begin{equation}
T(k) = (1+(\alpha k)^{2\nu})^{-5/\nu},
\end{equation}
with $\nu$ and $\alpha$ constants. Ref.~\cite{Viel2005} found that for $k < 5~ h~\rm{Mpc}^{-1}$, the best fit of the transfer function is obtained with $\nu = 1.12$. $\alpha$ is the cutoff scale as a result of free streaming in the WDM power spectrum relative to CDM, which following ref.~\cite{Viel2005}, is given by
\begin{equation}
\alpha = 0.047\left(\frac{m_{\rm{WDM}}}{\rm{keV}}\right)^{-1.11}\left(\frac{\Omega_{\rm{WDM}}}{0.2589}\right)^{0.11}\left(\frac{h}{0.6774}\right)^{1.22}h^{-1}~\rm{Mpc}.
\end{equation}
Here $m_{\rm{WDM}}$ is the WDM particle mass and $\Omega_{\rm{WDM}}$ is the WDM contribution to the density parameter.
The length scale at which the transfer function drops by a factor of 2 is the \textit{half mode} wavelength, $\lambda_{\rm{hm}}$, and the mean mass enclosed within a sphere of diameter equal to $\lambda_{\rm{hm}}$ is the \textit{half mode} mass, $M_{\rm{hm}}$. The \textit{half mode} mass quantifies the threshold mass below which the WDM subhalo mass function is strongly suppressed. For example, for a 3.3 keV thermal WDM, $M_{\rm{hm}} \sim 2 \times 10^{8} h^{-1} \rm{M}_{\odot}$ \cite{Colin2008,Angulo2013,Viel2013}.
Based on this WDM model, ref.~\cite{Schneider2012} performed a series of
high-resolution N-body simulations and obtained a functional fit for the differential mass function of WDM relative to CDM as
\begin{equation}
\left(\frac{dn}{dM}\right)_{\rm{WDM}} = \left(1+ \frac{M_{\rm{hm}}}{M}\right)^{-\beta} \left(\frac{dn}{dM}\right)_{\rm{CDM}},
\end{equation}
where $M$ is the subhalo mass and $\beta$ is a free parameter found to be equal to 1.16. Ref.~\cite{Lovell2013} studied the abundance and structure of WDM subhalos within a Milky Way-like host halo using high resolution N-body simulations based on the Aquarius Project \cite{Springel2008}. They found that the above functional fit is improved by introducing another parameter $\gamma$, such that
\begin{equation}
\left(\frac{dn}{dM}\right)_{\rm{WDM}} = \left(1+ \gamma\frac{M_{\rm{hm}}}{M}\right)^{-\beta} \left(\frac{dn}{dM}\right)_{\rm{CDM}},
\label{eq:wdmcdm}
\end{equation}
with $\gamma = 2.7$ and $\beta = 0.99$.
Since the number of subhalo encounters experienced by a particular stellar stream depends on the number of subhalos within the average radius of the stream from the galactic center, we need a model for the radial dependence of the number density of WDM subhalos. In ref.~\cite{Springel2008}, it was shown that for a Milky Way like host galaxy, the radial distribution of CDM subhalos in the mass range $[10^5 - 10^9]~ \rm{M}_{\odot} $ is described by an Einasto profile. Furthermore, the CDM subhalo mass function for those halos were shown to be well described by $dn/dM \propto M ^{-1.9}$. These results were combined in ref.~\cite{Erkal2016} to estimate the normalized subhalo profile at a given mass and at a certain galactrocentric radius:
\begin{equation}
\left(\frac{dn}{dM}\right)_{\rm{CDM}} = c_{0}\left(\frac{M}{m_{0}}\right)^{-1.9}\exp\left\{ - \frac{2}{\alpha}\left [\left(\frac{r}{r_{-2}}\right)^{\alpha} - 1 \right]\right\}
\label{eq:dndMc}
\end{equation}
with $c_{0}=2.02 \times 10^{-13}~\rm{M}_{\odot}^{-1}$~kpc$^{-3}$, $m_{0}= 2.52\times 10^{7}~\rm{M}_{\odot}$, $\alpha = 0.678$ and $r_{-2} = 162.4$~kpc.
Ref.~\cite{Lovell2013} found that the radial distributions of WDM subhalos of different particle masses are very similar to one another and to the CDM case. We can therefore adopt the Einasto profile for the radial distribution of WDM subhalos. We combine eqs.~\eqref{eq:wdmcdm} and \eqref{eq:dndMc} to obtain an expression for the WDM subhalo profile,
\begin{equation}
\left(\frac{dn}{dM}\right)_{\rm{WDM}} = c_{0}\left(\frac{M}{m_{0}}\right)^{-1.9}\exp\left \{ - \frac{2}{\alpha}\left[\left(\frac{r}{r_{-2}}\right)^{\alpha} - 1 \right]\right \}\left(1+ \gamma\frac{M_{\rm{hm}}}{M}\right)^{-\beta}.
\label{eq:dndMw}
\end{equation}
Notice that we have ignored the evolution of the subhalo number density over the age of the stream, since we do not expect it to cause any significant change in the results. This is because gaps fill up over time due to the internal velocity dispersion in the stream and therefore very old gaps will not be visible today.
Baryonic effects have been shown to tidally disrupt dark matter subhalos thereby reducing their number. Ref.~\cite{Sawala2016} used APOSTLE simulations in the $\Lambda$CDM framework to estimate that the substructure abundance in the mass range $[10^{6.5} - 10^{8.5}]~\rm{M}_{\odot}$ inside a Milky Way mass halo is greatly affected over a lookback time of up to 5 Gyr. Within the radius of the GD-1, they predicted a reduction of dark matter substructures by $\sim 45 - 50 \%$. Taking this into account, we reduced the number of substructures in each mass decade within the subhalo mass range $[10^6 - 10^9]~\rm{M}_{\odot} $ by 47\%. For WDM subhalos this factor may be even greater as they are more easily tidally disrupted owing to their lower concentration at the time of their infall. However, we have ignored this in the present work.
Figure \ref{fig:Nsub} shows the cumulative number of subhalos for different WDM particle masses obtained by integrating eq.~\eqref{eq:dndMw} up to a subhalo mass of $10^{12}~ \rm{M}_{\odot}$, shown by the solid lines. The dashed lines represent the cumulative number of subhalos taking the 47\% reduction into account.
\begin{figure}[t]
\centering
\includegraphics[scale=0.65]{Nsubhalo_0p53times.pdf}
\caption{Cumulative number of subhalos within a Galactocentric radius of 23 kpc of a Milky Way sized host halo for different WDM particle mass scenarios. Dashed lines indicate the cumulative numbers taking into account the 47\% reduction of the number of subhalos due to baryonic effects.}
\label{fig:Nsub}
\end{figure}
\section{Modeling the GD-1 stream and subhalo impacts}
\label{sec:simstream}
\subsection{Generating a smooth stream}
\label{subsec:smoothstream}
In order to analyze the gap power spectrum of a stellar stream due to impacts from thermal WDM subhalos, we need a mock stream that had minimal gap inducing perturbations from baryonic effects. In refs.~\cite{Erkal2017,Pearson2017}, it was shown that the Milky Way bar can induce gaps in stellar streams such as Pal 5 \cite{Odenkirchen2001}, which is in prograde motion with respect to the pattern speed of the bar. Such gaps are not induced in streams like the GD-1 \cite{Grillmair2006} which is in retrograde motion with respect to the bar \cite{Koposov2010, Bovy2016,Erkal2017}. Furthermore, ref.~\cite{Amorisco2016} studied the effects of giant molecular clouds on the GD-1 stream and found that because of the giant molecular clouds' larger pericentre and steeper mass function, and also due to GD-1's retrograde motion there is no strong density perturbations induced on the GD-1 stream. Taking all these points into consideration, the density gaps found in GD-1 stream are expected to be solely due to dark matter subhalo impacts. Therefore, we generate a mock GD-1 stream for our present analysis.
To generate a mock GD-1 stream and simulate the impacts due to dark matter subhalos, we make use of the simple model of stream evolution and subhalo impacts in the space of orbital frequency $\boldsymbol{\Omega}$, and orbital angle $\boldsymbol{\theta}$, that was developed in ref.~\citep{Bovy2014} and is included in the \texttt{galpy} code \cite{Bovy2015}. For a detailed step-by-step explanation of this entire method see refs.~\cite{Bovy2014,Bovy2016a}. We briefly summarize the method here. Given a model host potential, current progenitor phase-space information, velocity-dispersion parameter $\sigma_{v}$ of the progenitor, and a disruption time $t_d$ at which disruption started, a leading
or trailing tail model of the stream is generated in ($\boldsymbol{\Omega},\boldsymbol{\theta}$) space. In this work we use the well-tested three component Milky Way potential $\texttt{MWPotential2014}$ from ref.~\cite{Bovy2015} as the host potential. The GD-1 progenitor's phase space coordinate was taken from ref.~\cite{Bovy2016} : $(\phi_{1}, \phi_{2}, D, \mu_{\phi_{1}},\mu_{\phi_{2}},V_{\rm{los}}) = (0^{\circ},-0^{\circ}.82 \pm 0^{\circ}.08, 10.1 \pm 0.2 ~ \rm{kpc}, -8.5 \pm 0.3 ~\rm{mas ~yr^{-1}}, -2.15 \pm 0.10 ~ \rm{mas ~ yr^{-1}}, -257 \pm 5 ~\rm{km ~s^{-1}} )$, where $\phi_{1}$ and $\phi_{2}$ are custom sky coordinates as used in ref.~\cite{Koposov2010}, $D$ is the distance to the progenitor, $\mu_{\phi_{1}}$ and $\mu_{\phi_{2}}$ are the proper motion along $\phi_{1}$ and $\phi_{2}$, and $V_{\rm{los}}$ is the line of sight velocity. It should be noted that the progenitor's location $\phi_{1}$ is set to $0^{\circ}$, $\sigma_{v} = 0.365$ km s$^{-1}$, and we model the whole GD-1 stream as a leading arm. The true extent of the GD1 stream is limited by the edge of current surveys as well as by the galactic disk. Hence, constraining the disruption time $t_d$ is difficult with the present data. We therefore have assumed that the disruption occurred 9 Gyr ago which made our stream very old and long.
From the progenitor orbit, an approximate Gaussian action ($\textbf{J}$) distribution for the tidally stripped stars is constructed \cite{Eyre2011}, which is then transformed into frequency space using the Hessian matrix evaluated at the progenitor's actions $(\partial \boldsymbol{\Omega}/\partial \textbf{J})_{\mathbf{J_{P}}}$. The resulting variance tensor in frequency space has the principal eigenvector along the direction in frequency space in which the stream spreads, $\boldsymbol{\Omega}_{\parallel}$. The eigenvalues of the eigenvectors perpendicular to $\boldsymbol{\Omega}_{\parallel}$ are less than the largest eigenvalue by a factor of 30 or more \cite{Sanders2013a}. Tidally disrupted stars are generated with a frequency distribution that is modeled as a Gaussian. This Gaussian has a mean equal to the frequency offset from the progenitor and its variance tensor is obtained by transforming the Gaussian action distribution to frequency space and multiplying by the magnitude of the parallel frequency, which is done for the purpose of simplifying analytic calculations. The dispersion of the stellar debris in frequency space is scaled by the velocity dispersion, $\sigma_{v}$, and the relative eigenvalues. Once stripped, the gravitational effects on the stellar debris from its progenitor is neglected and they are evolved solely in their host potential. Their future locations are computed based on their linear evolution in angle space with their frequency remaining constant. For a given disruption time $t_d$, stellar debris is generated with a distribution of constant stripping time with a maximum time $t_d$. Following refs.~\cite{Bovy2014,Bovy2016}, we can transform the stellar debris from ($\boldsymbol{\Omega},\boldsymbol{\theta}$) space to configuration space using linearized transformation near the track of the stream.
Based on this approach, we generate a smooth GD-1 stream whose properties, namely path of the smooth stream in Galactic latitude and longitude, parallel angle variation $\Delta\theta_{\parallel}$ with respect to the Galactic longitude, and density variation are similar to those shown in figure 1 of ref.~\cite{Bovy2016a}. The perigalacticon of the stream is $\sim 14$ kpc, the apogalacticon is $\sim 30$ kpc, and the average Galactocentric radius is $\sim 23$ kpc. As shown in figure \ref{phi12}, in the custom sky coordinates $(\phi_{1},\phi_{2})$, the full stream is aligned along $\phi_{1}$ and stretches over $\sim 100^{\circ}$, while along $\phi_{2}$ the stream only extends over $\sim 5^{\circ}$. The gray points are samples drawn from the smooth stream model after applying the angular cut $-55^{\circ} < \phi_{1} < -5^{\circ}$ to match the observed data from ref.~\cite{Koposov2010} which are shown as red points. The data agree well with the mock stream's angular position in the sky. The minor offset between the data and the simulated track of the stream is because the phase space coordinate of the progenitor from ref.~\cite{Bovy2016} was obtained from a fit in which the Milky Way potential was left to vary freely, whereas we have evolved the stream in a fixed \texttt{MWPotential2014}. This offset however does not affect how well we can distinguish between WDM and CDM.
\begin{figure}
\centering
\includegraphics[scale=0.6]{GD1stream_phi1_phi2.pdf}
\caption{GD-1 stream model generated using the framework developed in ref.~\cite{Bovy2014} and using \texttt{galpy}'s \texttt{MWPotential2014} for a stream age of 9 Gyr, velocity-dispersion parameter $\sigma_{v} = 0.365$ km$s^{-1}$ and using the phase-space coordinates from ref.~\cite{Bovy2016}. The blue line is the mean stream track of the full stream in angular coordinates. The gray points show a sampling of mock stream data from the model after applying the angular cut $-55^{\circ} < \phi_{1} < -5^{\circ}$ in order to match the observed stream. The red points are the stream data positions from ref.~\cite{Koposov2010}.}
\label{phi12}
\end{figure}
\subsection{Modeling stream-subhalo impacts}
\label{subsec:subhaloimpacts}
A close encounter of a dark matter subhalo with a stellar stream imparts perturbations to the orbits of the stars in the stream which can be computed by the impulse approximation \cite{Yoon2011,Carlberg2013,Erkal2015,Sanders2016}. In this approximation, the subhalo-stream encounter is modeled as an instantenous velocity kick imparted to the stars in the stream at the point of closest approach. In ref.~\cite{Sanders2016} it was shown that subhalo-stream interactions can be efficiently modeled in frequency-angle space by transforming the velocity kicks, $\delta\vect{v}^{g}$, in frequency-angle space using the Jacobians $\partial\vect{\Omega}/\partial\vect{v}$ and $\partial\vect{\theta}/\partial\vect{v}$. The equations of motion for the stars in frequency-angle space before the subhalo impact are $\vect{\Omega} = \vect{\Omega}_{0} = $ constant and $\vect{\theta} = \vect{\Omega}_{0}t + \vect{\theta}_{0}$, where $(\vect{\Omega}_{0},\vect{\theta}_{0})$ is the frequency-angle coordinate when the star was stripped from its progenitor. If the subhalo impacts at $t^{g}$, then the equations of motion of the star becomes $\vect{\Omega} = \vect{\Omega}_{0} + \delta\vect{\Omega}^{g}= $ constant and $\vect{\theta} = \vect{\Omega}_{0}t + \delta\vect{\Omega}^{g}(t - t^{g}) + \delta\vect{\theta}^{g} + \vect{\theta}_{0}$, where $\delta\vect{\Omega}^{g}$ and $\delta\vect{\theta}^{g}$ are the frequency and angle kicks corresponding to $\delta\vect{v}^{g}$. Following refs.~\cite{Sanders2016,Bovy2016a}, instead of computing $\delta\vect{v}^{g}$ over the full 6-dimensional phase space volume, we only compute along the 1-dimensional mean track of the stream at the time of impact. Moreover, the angle kicks, $\delta\vect{\theta}^{g}$ were shown to be small compared to the frequency kicks, $\delta\vect{\Omega}^{g}$ after one orbital period \cite{Sanders2016}, therefore in the context of the GD-1 stream, which is much older than its orbital period, we can neglect the angle kicks. Extensive tests of these approximations were performed in ref.~\cite{Bovy2016} and they concluded that the approximations work well at least in the subhalo mass range of interest here.
In order to simulate the effects of multiple subhalo impacts of different masses, at different times in the orbit and locations along the stream, we need a prescription for sampling multiple subhalo impacts. For this we follow the sampling procedure described in detail in section 2.3 of ref.~\cite{Bovy2016a}. The only difference in the WDM case is the Poisson sampling of the number of impacts of different masses.
We consider the subhalos to follow a Hernquist profile, and use the relation for the scale radius, $r_{\rm{s}}$, from ref.~\cite{Erkal2016},
\begin{equation}
r_{\rm{s}} = 1.05 ~\rm{kpc} \left(\frac{M_{\rm{sub}}}{10^8 \rm{M}_{\odot}}\right)^{0.5},
\label{eq:rs}
\end{equation}
which was obtained for CDM subhalos with Hernquist profile by fitting the circular velocity - mass relation from the publicly available Via Lactea II catalogs \cite{Diemand2008}. We elaborate in appendix \ref{sec:scaleradius} on why we are justified in using the same $r_{\rm{s}}$ fitting formula for WDM subhalos. Less massive and smaller dark matter halos need to pass closer to a stellar stream compared to more massive and larger subhalos to result in an observable effect on the stream. To capture this effect in our simulations, we set the maximum impact parameter equal to five times the scale radius of the dark matter subhalo following ref.~\cite{Bovy2016a}. The maximum impact parameter is therefore smaller for lower mass subhalos and larger for more massive subhalos. To sample the velocity distribution of the DM subhalos, we use a Gaussian with a velocity dispersion of 120 km s$^{-1}$ which is the measured radial velocity dispersion within a galactocentric distance of 30 kpc for a collection of halo objects as found in ref.~\cite{Battaglia2005}.
For a particular mass of thermal WDM, one can find the number of subhalos in a specific mass range within a spherical radius by integrating eq.~\eqref{eq:dndMw} over the subhalo mass range and the chosen radial range.
For the GD-1 stream generated in this work, the mean spherical radius of the stream is $\sim 23$ kpc. If dark matter constitutes of a thermal WDM of particle mass 3.3 keV (or a 7 keV sterile neutrino with lepton asymmetry parameter equal to 8.66), we find 0.47 subhalos in the mass range $[10^{6} - 10^{9}]~ \rm{M}_{\odot}$ within 23 kpc from the galactic center if subhalo disruption due to baryonic effects are not taken into account. By taking the 47\% reduction of number of subhalos due to baryonic effects, this number is reduced to $\sim 0.25$. As a comparison, in case of a 1 GeV CDM candidate in the same mass range and within the same radius, we find $\sim 57.78$ subhalos when no baryonic effect is included, and $\sim 30.6$ when we include baryonic effects. For the rest of this work we include the 47\% reduction of number of subhalos due to the effect of baryons in all our results.
The expected number of impacts over the age of a stream depends on the number density of subhalos as well as on the impact parameter (see eq.~(1) in ref.~\cite{Bovy2016a}). In the WDM case, there are fewer subhalos compared to the CDM case, and as a result the probability of stream-subhalo encounter is much smaller if dark matter is warm. This is shown in figure \ref{fig:impact_pdf} for different cases of dark matter particle mass. Each impact PDF is constructed out of 2100 simulations of the GD-1 stream evolved in different dark matter particle mass scenarios.
\begin{figure}
\centering
\includegraphics[scale=0.5]{impact_number_pdf.pdf}
\caption{PDF of the number of impacts that a GD-1 like stream had over 9 Gyr in different cases of dark matter particle mass. Each PDF was constructed out of 2100 simulations.}
\label{fig:impact_pdf}
\end{figure}
To find the impact rates, we compute the number of subhalos within 23 kpc in each mass decade by integrating eqs.~\eqref{eq:dndMc} and \eqref{eq:dndMw}. We determine the number of impacts $N$ over the entire mass range of subhalos in the WDM scenario by sampling the number of impacts from a Poisson distribution for the expected number of impacts. We then sample $N$ masses from the impact-mass distribution. Note that this distribution differs from eq.~\eqref{eq:dndMw}, because it needs to account for the shrinking maximum impact parameter as the subhalo mass decreases and vice versa. All other parameters are then sampled exactly as in the CDM case considered in ref.~\cite{Bovy2016a}. The total rate of impact is 0.72 for a 3.3 keV thermal WDM, 1.95 for a 5 keV thermal WDM, and 24.8 for a 1 GeV CDM particle. These rates are calculated by assuming that the rate of impact for each mass is that corresponding to its mass decade and is computed at the center of the logarithmic mass bin. In the subhalo mass bins $[10^{6} - 10^{7}]~ \rm{M}_{\odot}$, $[10^{7} - 10^{8}]~ \rm{M}_{\odot}$, and $[10^{8} - 10^{9}]~ \rm{M}_{\odot}$, the impact rates for the 3.3 keV WDM scenario are $\sim 0.04$, $\sim 0.17$ and $\sim 0.49$, for the 5 keV WDM are $\sim 0.17$, $\sim 0.6$, $\sim 1.18$ and for the 1 GeV CDM scenario they are $\sim 15.94$, $\sim 6.34$, and $\sim 2.53$, respectively.
In order to efficiently calculate the stream-subhalo impacts, we use the \emph{line-of-parallel-angle} approach as explained in ref.~\cite{Bovy2016a}. In this approach, multiple impacts occurring at the same time do not increase the computational cost. Therefore, the impacts are allowed to happen at a set of equally spaced discrete times along the past orbit of the stream. In order for the structure of the stream not to be affected by the discrete time sampling, the sampling needs to be high enough. Furthermore, sampling impacts more frequently than the radial period of the stream allows us to explore stream-subhalo encounters at different epochs of the streams orbit. In Appendix C of ref.~\cite{Bovy2016a} it was shown that the statistical properties of a perturbed stream converge for greater than 16 different impact times. We considered 64 different impact times for the 9 Gyr stream. This would amount to a time interval of $\sim 140$ Myr which is shorter than the radial period of $\sim 400$ Myr for the GD-1 stream.
\section{Results}
\label{sec:results}
\subsection{Mock perturbed GD-1 stream}
Using the above mentioned machinery for generating perturbed stellar streams, we carry out 2100 simulations each for the 3.3 keV and 5 keV thermal WDM and 1 GeV CDM scenarios for a GD-1 like stream, considering subhalo impacts in the mass range of $[10^6 - 10^9]~\rm{M}_{\odot}$. For each simulation, we transform the density of the stream from frequency-angle space to configuration space $(\phi_{1},\phi_{2})$ and apply angular cuts to match the observed extent of the GD-1 stream. As evident from figure \ref{phi12}, the angular extent of the stream along $\phi_{2}$ is very small, therefore for the rest of this work, we will only analyze the stream as a function of $\phi_{1}$.
In figure \ref{fig:denscont} we show the density contrast (density of the perturbed stream divided by that of the unperturbed stream) of four different cases from the 2100 simulations for the 5 keV and 1 GeV scenarios. Black curves denote randomly chosen density contrasts, while green curves denote density contrasts for cases whose power is close to the median power of the 2100 simulations as shown in figure \ref{fig:Pk_disp}. We treat the cases shown in green as fiducial cases which are used to constrain the mass of dark matter in section \ref{sec:infer}. The red curve in the 5 keV WDM scenario in the left panel of figure \ref{fig:denscont} corresponds to a case whose power is close to the upper $2\sigma$ bound of the power spectrum dispersion, while the blue curves in both panels of the same figure correspond to cases whose power is close to the lower $2\sigma$ bound as shown in figure \ref{fig:Pk_disp}. For the 5 keV WDM, the case close to the lower $2\sigma$ bound occurs for a stream that had no subhalo impacts, and as a result the density contrast is equal to one. We do not consider a case in the 1 GeV CDM scenario whose power is close to the upper $2\sigma$ bound of the power spectrum dispersion because as we will discuss in section \ref{sec:infer}, our method can not distinguish between dark matter models if the mass of dark matter is greater than $\sim 15$~keV. In section \ref{sec:outlier}, we investigate these cases of extreme power and how they can bias our analysis.
In our stream simulation, the progenitor emits new stars constantly. In scenarios where the stream had many impacts, the bulk of the stream gets disrupted and stars are pushed towards the extremities. In such cases, the stars which are pushed towards the progenitor add up to the new stars which have recently been emitted from the progenitor, causing a large overdensity near the location of the progenitor ($\phi_{1}=0^{\circ}$). Such large overdensities are somewhat unphysical since in a real stream some stars pass the progenitor, moving from one arm of the stream to the other, and our simulations ignore this effect. In order to remove this potentially artificial peak, we cut the stream at $\phi_{1} > -5^{\circ}$. This choice of the $5^{\circ}$ cut removes the large peaks close to the progenitor, while leaving the smaller physical peaks which are expected due to the disruption of the subhalo.
For each case we binned the density in $ 0.1^{\circ}$ $\phi_{1}$ bins. If we assume that LSST can go below the main-sequence turn-off point and accurately resolve almost all the member stars of the GD-1 stream which has $\approx1,000\,\mathrm{stars\, deg}^{-1}$ \cite{Koposov2010}, then this would correspond to a shot noise of 10\% for a bin width equal to $0.1^{\circ}$. Therefore, we have considered a constant noise of 10\% of the density contrast in each $\phi_{1}$ bin.
\begin{figure}[t]
\centering
\begin{minipage}{0.5\textwidth}
\centering
\includegraphics[width=\linewidth]{GD1like_dencont_5keV.pdf}
\end{minipage}%
\begin{minipage}{0.5\textwidth}
\includegraphics[width=\linewidth]{GD1like_dencont_1GeV.pdf}
\end{minipage}
\caption{Density contrast of four different realizations of the perturbed GD-1 stream. The left panel shows the case in which dark matter is composed of thermal WDM with particle mass 5 keV and the right panel shows the case for a 1 GeV CDM candidate. The black curves show randomly chosen density contrasts. The green curves denote density contrast for the fiducial cases which we treat as mock observed data and are analyzed for inferring the mass of dark matter in section \ref{sec:infer}. The red curve indicates a realization whose power is close to the upper $2\sigma$ bound of the power spectrum dispersion, while the blue curves indicate the realizations that are close to the lower $2\sigma$ bound. In the 5 keV case, this happens when the stream had no subhalo impact. See section \ref{sec:outlier} for discussion on these extreme cases. }
\label{fig:denscont}
\end{figure}
\begin{figure}[t]
\centering
\includegraphics[scale=0.49]{powerdispersion_1GeVCDM_5keV_3p3keV_0p53rate_cropped.pdf}
\caption{Power spectrum of the density contrast of the mock GD-1 stream in the 3.3 keV WDM (left panel), 5 keV WDM (middle panel) and 1 GeV CDM (right panel) scenarios. The black solid line is the median power of the 2100 simulations. For the 3.3 keV case, the median power is close to 0 and hence not shown. The error bars represent the $2\sigma$ (95\%) spread of the power spectra of the 2100 stream density simulations. The green solid curves (wherever shown) represent the power spectrum of the fiducial cases whose density contrasts are shown in figure \ref{fig:denscont}. The red and blue curves (wherever shown) represent the power for cases close to the upper and lower $2\sigma$ bound of the dispersion in power spectrum, respectively. The dashed curves indicate the median power corresponding to 10\% noise in the density and their color represents the case they correspond to. The colored dots specify the angular scales at which the power is above the noise floor.}
\label{fig:Pk_disp}
\end{figure}
We compute the 1-dimensional power spectrum of the density contrast of the stream $P_{\delta\delta}(k_{\phi_1}) = \langle \delta(k_{\phi_1})^2 \rangle$ using the \texttt{csd} routine in \texttt{scipy} \cite{Jones2001c} and do not divide by the sampling frequency. In the power spectrum plots, we have plotted the square root of the power along $y$-axis and inverse of the frequency (1/$k_{\phi_1}$) along $x$-axis. In this form, the power spectrum plots represent the power at a particular angular separation which is indicative of the correlation of the density contrast at that angular scale. In figure \ref{fig:Pk_disp}, we show the power spectrum of the density contrast of the GD-1 stream for the 3.3 keV and 5 keV WDM, and the 1 GeV CDM scenarios. The black solid curves show the median power spectrum of the 2100 simulations. For the 3.3 keV case, owing to very few impacts in majority of the simulations the median power is close to 0. The error bars show their $2\sigma$ dispersion around the median. The green solid curves show the power spectrum of our fiducial cases whose density contrasts are shown in figure \ref{fig:denscont}. The red and blue solid curves show a case close to the upper and lower $2\sigma$ bound of the power spectrum dispersion, respectively. In both the 3.3 keV and 5 keV WDM scenarios the lower bounds are close to zero. This is again due to either no subhalo impacts or impacts that did not result in significant density fluctuations. The dashed lines show the 10\% noise power and their color represents the case they correspond to. The noise power spectrum is the median of 10,000 Gaussian noise realizations of the stream density and therefore independent of the density bin width. For the fiducial 1 GeV case, the noise power is much lower than the other cases since the bulk of the stream is pushed away due to subhalo encounters leaving the density along most of the stream close to 0 (top right panel figure \ref{fig:denscont}). The density noise being 10\% of the density has therefore very low power. The colored dots represent the angular scales at which the power is above the corresponding noise floor and hence are used in the ABC analysis to infer the dark matter particle mass (see section \ref{sec:infer}).
As evident from figure \ref{fig:Nsub}, there are many more subhalos in the mass range $[10^6 - 10^9]~\rm{M}_{\odot}$ in the 1 GeV case compared to the 5 keV and 3.3 keV cases, which leads to a higher chance of stream-subhalos encounters as suggested in figure \ref{fig:impact_pdf}, resulting in higher density fluctuations along the stream in the 1 GeV case compared to the 5 keV and 3.3 keV cases. This is easily seen from the density contrasts plotted in figure \ref{fig:denscont}.
The difference in the density fluctuations translates to the power spectrum and hence explains the more power at the largest scales in the 1 GeV case compared to the 5 keV case. The wide dispersion in the power spectra in either case is due to the range of possible ways the stream-subhalo impacts can occur resulting in different levels of density fluctuations along the stream.
Comparing the 5 keV and 1 GeV cases in figure \ref{fig:Pk_disp}, it is clear that there is more power at smaller scales in the 1 GeV case. This is because if dark matter is cold then there are more low mass subhalos whose impacts with the stream give rise to density fluctuations at smaller scales. On the contrary, if dark matter is warm then as discussed in section \ref{sec:WDM}, substructures less massive than the half-mode mass are strongly suppressed. As a result, majority of the impacts are by the more massive subhalos, giving rise to large scale density fluctuations. The noise power appears flat since we considered a constant noise level throughout the stream. Power at scales below the noise floor do not convey any useful information.
\subsection{Inferring the dark matter particle mass}
\label{sec:infer}
In this section we use the statistical properties of gaps in a stellar stream to distinguish between CDM and WDM. To achieve this, we use the power spectrum of the density contrast of the perturbed stream to constrain the mass of the dark matter particle. We employ the ABC method to construct an approximate posterior probability distribution function (PDF) of the mass of the dark matter particle. The ABC method is a likelihood-free approach of Bayesian parameter inference in which an approximate posterior PDF of the parameters in the problem is constructed using simulator outputs and by comparing the outcome with observed data. We run the simulations by randomly drawing the mass of the dark matter particle from a prior distribution of the mass, which we have assumed to be a uniform distribution between [0.1,16] keV. The upper limit of the prior was picked based on the result presented in figure \ref{fig:Pk_conv}, which shows that the median power spectrum for cases with mass of WDM greater than 15 keV converge and are indistinguishable from each other. This is consistent with figure \ref{fig:Nsub}, which shows that increasing the mass of the WDM particle shifts the subhalo mass function close to the CDM case in the subhalo mass range $[10^6 - 10^9]~\rm{M}_{\odot}$.
\begin{figure}[t]
\centering
\includegraphics[scale=0.5]{power_5-10-15-1GeV_compare_0p53rate.pdf}
\caption{Power spectrum of the median of 2100 simulations of the GD-1 stream in different WDM particle mass scenarios. The median power spectra for WDM particle mass larger than $\sim 10$~keV tend to overlap, and hence it is no longer possible to distinguish at high significance between WDM models that have particle mass greater than 10 keV. The power spectrum of cases with WDM particle mass greater than 15 keV is very close to the 1 GeV CDM case and therefore WDM models with particle mass greater than $\sim 15$~keV can not be distinguished from the CDM model.}
\label{fig:Pk_conv}
\end{figure}
The ABC works by accepting those simulations which are within some pre-defined tolerance of the data summaries. As our data summaries we have taken the power at the three largest observed scales in the 5 keV case and the five largest observed scales in the 1 GeV case, since those scales are above the noise floor of the fiducial cases (see figure \ref{fig:Pk_disp}). We do not consider the 3.3 keV case here, since as seen in figure \ref{fig:Pk_disp} the median power is close to zero for this case. Additionally, we do not consider power at scales below the noise floor because they are very noisy and thus are incapable of discriminating between different WDM models. The level of noise in the data is therefore crucial for our method. The tolerances are made as small as it is allowed by the noise in the data. In order to sample the entire range of the prior distribution properly, we ran $\sim 120,000$ simulations. Since the most time consuming part of the ABC approach is running these simulations, we follow the same strategy as adopted in ref.~\cite{Bovy2016a}, i.e.~for each WDM mass we produce 100 simulations by adding 100 realizations of the noise. Therefore, effectively, we ran $\sim 12,000,000$ simulations.
\begin{figure}[t]
\centering
\begin{minipage}{.5\textwidth}
\centering
\includegraphics[width=\linewidth]{posterior_top3scales_simind83_5keVwdm_phi1-5degcut_GD1stream_6-9_X5.pdf}
\end{minipage}%
\begin{minipage}{0.5\textwidth}
\centering
\includegraphics[width=\linewidth]{posterior_top5scales_simind44_1GeVcdm_phi1-5degcut_GD1stream_6-9_X5.pdf}
\end{minipage}
\caption{Posterior PDFs of the mass of dark matter obtained by running simulations of the GD-1 stream and using the ABC method to match the power at the largest scales above the noise floor (shown by red dots in figure \ref{fig:Pk_disp}) to the fiducial cases. The left panel shows the posterior PDF in the 5 keV WDM scenario and the right panel shows the posterior PDF in the 1 GeV CDM scenario. For these typical cases the ABC is able to distinguish between WDM and CDM and also constrain the mass of WDM. }
\label{fig:PDF}
\end{figure}
Figure \ref{fig:PDF} shows the posterior PDF of the dark matter particle mass obtained by running simulations of the GD-1 stream and using the ABC method as explained above in the 5 keV WDM and the 1 GeV CDM scenarios. For the 5 keV case, the PDF peaks at $5.9^{+5.9}_{-2.1}$ keV (68\% confidence) and the 95\% upper limit is 15.0 keV. In the 1 GeV CDM case, the PDF plateaus indicating the true dark matter mass is beyond the upper limit of the prior. The 95\% lower limit is 4.3 keV.
Next, we investigate the case in which the intrinsic power of the stream is below the noise floor. As shown in the power spectrum dispersion plots in figure \ref{fig:Pk_disp}, such cases may arise when dark matter is warm and the stream suffers few or no subhalo impacts (e.g., the density contrast for the 5 keV case corresponding to the lower $2\sigma$ bound).
In such cases the measured power is dominated by noise. The left panel of figure \ref{fig:noise_PDF} shows one randomly picked realization of the noise in the 5 keV case which we use as the measured power. We consider the power at the three largest scales to be data summaries. Since the power at all these three scales are below the median noise floor, the ABC accepts any simulation whose power at those scales are below the noise floor. The resulting posterior PDF of the dark matter mass is shown in the right panel of figure \ref{fig:noise_PDF}, which demonstrates that the 95\% upper limit on the mass of WDM is 5.3 keV. This is consistent with the fact that for a lower mass WDM, there are fewer subhalos and hence lower chances of stream-subhalo encounters, compared to the case of a higher mass WDM.
\begin{figure}[!htb]
\centering
\begin{minipage}{.5\textwidth}
\centering
\includegraphics[width=\linewidth]{power_5keV_lo_simind1537_noisedata_phi1cut_10percenterr.pdf}
\end{minipage}%
\begin{minipage}{0.5\textwidth}
\centering
\includegraphics[width=\linewidth]{posterior_top3scales_simind1537_lo_noise_data_5keVwdm_5degcut_stream_6-9_X5.pdf}
\end{minipage}
\caption{Figure demonstrating the case in which the intrinsic power of the stream density is below the noise floor, as a result of either no subhalo impact or impacts that did not contribute to any significant density fluctuations. Left panel: One realization of the noise power (solid black line) which we treat as our mock measurement of the stream density power spectrum, and the 10\% noise floor (dashed black line). The black dots represent the data summaries which are the power at the three largest scales. Right panel: Posterior PDF of the dark matter mass obtained using the ABC method.}
\label{fig:noise_PDF}
\end{figure}
\section{Risks of using only one stream}
\label{sec:outlier}
In this section we discuss the perils of using only one stream for constraining the mass of WDM using the stream density power spectrum method presented above. As evident from figure \ref{fig:Pk_disp}, a stream-subhalo interaction can take place in a wide range of possible ways giving rise to the dispersion in the power spectrum. Therefore, it is not unexpected that we could end up with a stream that had either too few or too many subhalo hits resulting in its power spectrum to be $> 2\sigma$ away from the median. Such cases would result in a posterior that would rule out the true dark matter particle mass by $> 2\sigma$. We demonstrate this with three examples here. In the 1 GeV CDM scenario we pick a density simulation whose power spectrum is close to the lower bound of the $2\sigma$ dispersion, and for each 3.3 keV and 5 keV WDM scenarios we pick one realization with a power close to the upper $2\sigma$ bound. These cases are shown by the blue and red curves in figure \ref{fig:Pk_disp}. Treating them as mock data and considering a 10\% density noise, we carry out the ABC steps with power at the two largest scales for the 1 GeV case, five largest scales in the 3.3 keV case and four largest scales in the 5 keV case as data summaries to construct the posterior PDF for the mass of dark matter. The results are shown in figure \ref{fig:extreme_cases}. The posterior PDF predicts, in the 3.3 keV case (left panel), a lower 95\% limit on the mass of dark matter to be $4.5$ keV, in the 5 keV case (middle panel), a lower 95\% limit on the WDM particle mass to be $6$ keV and in the 1 GeV CDM case (right panel), the PDF constrains the dark matter particle mass to $5.7^{+5.0}_{-2.4}$ keV at 68\% and sets a 95\% upper limit at $14.6$ keV.
\begin{figure}[t]
\centering
\begin{minipage}{.333\textwidth}
\centering
\includegraphics[width=\linewidth]{posterior_top5scales_simind827_hi_3p3keVwdm_phi1-5degcut_GD1stream_6-9_X5.pdf}
\end{minipage}%
\begin{minipage}{0.333\textwidth}
\centering
\includegraphics[width=\linewidth]{posterior_top4scales_simind1169_hi_5keVwdm_phi1-5degcut_GD1stream_6-9_X5.pdf}
\end{minipage}%
\begin{minipage}{0.333\textwidth}
\centering
\includegraphics[width=\linewidth]{posterior_top2scales_simind717_low_1GeVcdm_phi1-5degcut_GD1stream_6-9_X5.pdf}
\end{minipage}
\caption{Posterior PDFs for the extreme cases. Left panel: upper $2\sigma$ case in the 3.3 keV scenario; middle panel: upper $2\sigma$ case in the 5 keV scenario; right panel: lower $2\sigma$ case in the 1 GeV scenario.}
\label{fig:extreme_cases}
\end{figure}
These results suggest that we can not rule out CDM or WDM simply based on the $2\sigma$ results of one stream. This shortcoming can be improved by using multiple stellar streams to constrain the mass of dark matter. This will not only remove the effects of the outlier cases but will also make the constraints more robust. We leave this analysis for a future publication.
\section{Conclusions}
\label{sec:conclusions}
In this paper we presented a methodology of investigating the particle nature of dark matter by analyzing the statistical properties of density fluctuations, in the form of power spectrum, induced in a stellar stream as a result of its gravitational encounter with dark matter subhalos. We accomplished this by exploiting the fact that if dark matter is warm then there will be fewer subhalos in the Milky Way host halo compared to the CDM case. This difference will result in a different rate of stream-subhalo encounters and will be apparent in the intrinsic power spectrum of the stream density. By using fast simulations of stream density in frequency-angle space we generated mock GD-1 stream data and exposed it to subhalo encounters in a 3.3 keV WDM, 5 keV WDM and 1 GeV CDM scenario. Due to the higher rate of impacts in the CDM scenario, we found that the stream density power spectrum has more intrinsic power in the CDM scenario compared to the WDM case.
Assuming that LSST will be able to resolve most of the member stars of the GD-1 stream without contamination, we applied LSST-like noise to our mock data. With \textit{Gaia}'s magnitude threshold of 20, it will not provide much direct data on GD-1 for the purpose of our study. Nevertheless, Gaia will improve our knowledge of the Milky Way's gravitational potential in general and of the orbits of streams, which will be pivotal for our method.
We used an ABC technique to perform rigorous inference on the dark matter mass using mock streams for the 5 keV WDM and the 1 GeV CDM fiducial cases.
In the WDM case when the intrinsic power of the stream density is greater than the noise, we constrained the dark matter mass to $5.9^{+5.9}_{-2.1}$ keV (68\% confidence) and $< 15$ keV at 95\%. In the CDM scenario we found the lower limit on the mass of dark matter to be $4.3$ keV at 95\% confidence. These results indicate that with our methodology we can not only distinguish between CDM and WDM but also constrain the mass of WDM, if it's a few keV.
Because of the paucity of subhalos in the WDM scenario, it is possible that a stream doesn't accumulate enough density fluctuations in its lifetime so that its intrinsic power is less than the noise. In such cases, we obtained an upper limit on the mass of WDM to be 5.3 keV at 95\% confidence. We also explored three outlier cases of stream density perturbations in section \ref{sec:outlier} which can seriously limit our method's ability to distinguish between WDM and CDM. We stress that this situation can be ameliorated by applying our method on multiple streams.
\subsection*{Acknowledgments}
We thank Mark Lovell for many useful discussions and for providing us with the data from Aquarius warm dark matter simulations. We also thank Tameem Adel for discussions at an early stage of this project. We acknowledge the support of the D-ITP consortium, a programme of the Netherlands Organization for Scientific Research (NWO) that is funded by the Dutch Ministry of Education, Culture and Science (OCW). N.~Bozorgnia is grateful to the Institute for Research in Fundamental Sciences in Tehran for their hospitality during her visit. G.B. (PI) and N.~Bozorgnia acknowledge support from the European Research Council through the ERC starting grant WIMPs Kairos. J.~Bovy acknowledges the support of the Natural Sciences and Engineering Research Council of Canada (NSERC), funding reference number RGPIN-2015-05235, and from an Alfred P. Sloan Fellowship. N.~Bozorgnia has received support from the European Union's Horizon 2020 research and innovation programme under the Marie Sklodowska-Curie grant agreement No 690575.
\begin{appendices}
\section{Scale radius of warm dark matter subhalos}
\label{sec:scaleradius}
In this appendix we discuss the scale radius of WDM subhalos and how it compares to that of CDM subhalos. As it was explained in section~\ref{subsec:subhaloimpacts}, we use eq.~\eqref{eq:rs} for the scale radius of CDM subhalos, which is obtained assuming Hernquist density profiles for the subhalos in the Via Lactea II catalog.
In order to compare the properties of WDM subhalos to their CDM counterparts, we use the density profiles of subhalos available from the Virgo consortium Aquarius project~\cite{Lovell2013} where in addition to a CDM simulation, four haloes with WDM particle mass of 1.5, 1.6, 2.0, and 2.3 keV are simulated. These include subhalos which are not spurious (see ref.~\cite{Lovell2013} for a detailed discussion of removing spurious subhalos from their halo catalogs) and are within 2 Mpc of the host halo center. We fit a Hernquist profile to the density profiles of CDM and WDM subhalos and check how the best fit scale radii change as a function of subhalo mass.
To find the best fit Hernquist scale radius, for each individual simulated subhalo, we minimize the following $\chi^2$ function:
\begin{equation}
\chi^2(r_s) = \sum_{i}^{N} \frac{(\rho_i - \rho_{\rm Hern}(r_i, r_s))^2}{\sigma_i^2},
\end{equation}
where $\rho_i$ is the value of the DM density of the simulated subhalo at the radial bin $i$ with $r_i$ denoting the bin center, $\sigma_i$ is the corresponding $1\sigma$ Poisson error, $N$ is the number of radial bins we consider to perform the fit, and $\rho_{\rm Hern}(r_i, r_s)$ is the Hernquist density profile evaluated at radius $r_i$ as a function of the scale radius $r_s$.
The density profiles are computed in spherical shells spaced equally in $\log(r)$. We find the inner and outer radii for performing the fit by using the criteria discussed in Springel {\it et al.}~\cite{Springel2008}. Namely, the inner radius for the fit, $r_{\rm min}$, is set to the radius in which convergence is achieved according to the Power {\it et al.} criterion~\cite{Power:2002sw}. The outer radius, $r_{\rm max}$, is set to the largest radius where the density of bound mass is more than 80\% of the total mass density.
After specifying the radial range for the fit, we introduce the following two criteria to find \emph{good} subhalos for performing the fit: {\it (i)} $r_{\rm max} - r_{\rm min} > 1$~kpc, and {\it (ii)} the number of bins is greater or equal to 5. These additional criteria ensure that we are not fitting over a small radial range and we have enough bins for performing the fit. Finally, we need to consider the effect of tidal stripping on the goodness of fit. With the criteria mentioned, we don't obtain a good fit for some subhalos since we are probing a radial range where tidal stripping becomes important and there is a sharp decrease in the DM density from one bin to the next. To avoid this, we set the last bin we consider, $i_f$, as the bin where the DM density in the next bin decreases by more than an order of magnitude, i.e.~where $\rho_{i_f+1} < 0.1~\rho_{i_f}$. With these criteria we retain 4482 CDM subhalos over the mass range of $[1.5 \times 10^6, 10^{11}]~\rm{M}_{\odot}$. Notice that the lower boundary of the subhalo mass range is set by the requirement that the subhalo has at least 100 particles.
\begin{figure}[t]
\centering
\includegraphics[scale=0.7]{rsHernFit.pdf}
\caption{The mean values of the best fit Hernquist scale radii of subhalos in the CDM and WDM Aquarius simulations as a function of the subhalo mass. The halos in the WDM simulations have WDM particle masses in the range 1.5 -- 2.3 keV.}
\label{fig:WDMrs}
\end{figure}
The result is shown in figure~\ref{fig:WDMrs} where we have plotted the mean values of the best fit Hernquist scale radius at different masses, considering equal size bins in $\log_{10}(M_{\rm sub})$. As it can be seen in the figure, the overall deviation between the best fit scale radii of CDM halos and WDM halos with different WDM particle mass is small. Although for $M_{\rm sub} \gtrsim 10^{9}~\rm{M}_{\odot}$, the scale radii for WDM halos are in general larger than those of CDM halos, for $M_{\rm sub} \leq 10^{9}~\rm{M}_{\odot}$ which is the subhalo mass range probed in our study there is no clear trend. Moreover, because the impact parameters considered in this work are much larger than the scale radius of the interacting subhalos with the streams, we use the same scale radius relation to describe both WDM and CDM subhalos.
To capture the overall variation of the scale radius with the subhalo mass, we find the best fit $r_s(M_{\rm sub})$ relation from the mean scale radii of CDM subhalos in the Aquarius simulations shown in figure \ref{fig:WDMrs},
\begin{equation}
r_{\rm{s}}|_{\rm{fit}} = 1.24 ~\rm{kpc} \left(\frac{M_{\rm{sub}}}{10^8 \rm{M}_{\odot}}\right)^{0.39}.
\label{eq:rsfit}
\end{equation}
This equation is slightly different from eq.~\eqref{eq:rs} which is our fiducial scale radius relation. In figure \ref{fig:CDMrs_Pk} we show how this difference in the scale radius relation will affect our analysis. The black solid line is the median power of 2500 simulations with the fiducial scale radius relation given by eq.~\eqref{eq:rs}. The error bars represent the $1\sigma$ dispersion of these simulations. The red curve is the median power of the simulations with the new scale radius fit given by eq.~\eqref{eq:rsfit}. The green and blue curves are the median power spectrum with 0.4 and 2.5 times scale radius but keeping the maximum impact parameter $b_{\rm{max}}$ fixed for a particular mass subhalo. All the power spectra are within $1\sigma$ dispersion of the fiducial case, suggesting that varying the scale radius within what we discussed will not affect our analysis.
\begin{figure}[t]
\centering
\includegraphics[scale=0.55]{power_1GeVCDM_vary_rs_0p53rate.pdf}
\caption{Comparison of the median density power spectrum in the 1 GeV CDM scenario with different choices for the scale radius. The black curve is for the fiducial scale radius relation given by eq.~\eqref{eq:rs}, the red curve is with the scale radius relation found in eq.~\eqref{eq:rsfit} from a fit to the scale radii of subalos in the Aquarius simulations. In both of these cases the maximum impact parameter $b_{\rm{max}}$ is fixed to 5 times the scale radius for a particular mass subhalo. The error bars represent the $1\sigma$ dispersion of the power spectrum around the median for the fiducial case. The blue and green curve shows the cases when the fiducial scale radii of the subhalos are increased and decreased 2.5 times respectively while keeping the maximum impact parameter the same as the fiducial case. }
\label{fig:CDMrs_Pk}
\end{figure}
\end{appendices}
\FloatBarrier
\bibliographystyle{JHEP}
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{
"redpajama_set_name": "RedPajamaArXiv"
}
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N. Kovaithangam ( – 12 October 2022) was an Indian politician.
Born in India. He was elected to the Tamil Nadu legislative assembly as a Tamil Maanila Congress candidate from Valparai constituency in 2001, and 2006 elections. On 23 March he joined DMK in presence of party president M. K. Stalin.
References
1940s births
Year of birth missing
2022 deaths
People from Coimbatore district
Tamil Nadu MLAs 2001–2006
Tamil Nadu MLAs 2006–2011
Tamil Nadu politicians
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 2,407
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La Guardia Principal (en inglés: Main Guard), originalmente llamado Guardia della Piazza, es un edificio situado en La Valeta, Malta. Fue construido en 1603 por la Orden de San Juan como puesto de vigilancia. Se encuentra en el centro de la plaza de San Jorge (St. George's Square), frente al palacio del Gran Maestre. Actualmente alberga la oficina del Fiscal General.
Historia
El Edificio de la Guardia Principal fue construido en 1603 para albergar al Regimento di Guardia, la guardia personal del Gran Maestre de la Orden de San Juan. Se encuentra en la plaza frente al palacio del Gran Maestre.
En 1814, los británicos añadieron un pórtico neoclásico al edificio existente. Sobre él se añadió un bajorrelieve del escudo de armas británico, junto con la siguiente inscripción debajo:
Los británicos continuaron utilizando el edificio como ubicación de los guardaespaldas, esta vez de los gobernadores de Malta, que residían en el Palacio del Gran Maestre. El servicio de la Guardia Principal era ocioso, muchos soldados pintaban o arañaban insignias del regimiento u otras cosas en las paredes del edificio.
En los años 1980, el Centro se trasladó a otro lugar y el edificio de la Guardia Principal se convirtió en un anexo de la Oficina del Fiscal General. En la segunda década del siglo XXI, había planes para ceder el edificio al Ayuntamiento de La Valeta pero no se llevaron a cabo. Din l-Art Ħelwa se ofreció a renovar el edificio.
Características
La fachada del edificio es de una sola planta, pero la parte trasera, que da a la calle Strait, tiene tres alturas. Esto se debe a la diferencia en los niveles de las calles. El reloj del pórtico, ubicado debajo de la inscripción, es otro remanente de los británicos. Inmovilizado durante años, el 4 de octubre de 2009, después de cambiar piezas y limpiarlo, volvió a medir el tiempo.
Referencias
Bibliografía
Edificios y estructuras de La Valeta
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
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{"url":"https:\/\/control.com\/forums\/threads\/scada-alarm-messaging.20594\/","text":"C\n\n#### Chris Signorelli\n\nI need advice on how to send SMS messages to many mobile phones upon the occurance of alarm tags in a PLC. I've already tried using a GSM modem (Sony Ericsson GM29) with alarm monitoring software such as: GSM-Control, MovBridge and U.C.ME and have limited success. Most of the problems have been with the GSM modem. I would like to know if someone has had experience with this technology and how they achieved the task.\n\nThe requirements of my system are the following:\n- No phone line available. Hence the need for\na GSM modem.\n- No hardware alarm diallers (No more\navailable hardware outputs on PLC).\n- Alarm Monitoring Software.\n- Software that supports OPC communications.\n- Software must support multiple message\nrecipients.\n- Software must support multiple alarms.\n- Cost: around $1500 AUD. The PLC I'm using is an Allen Bradley ControlLogix L55 the software I'm using is RSView SE Studio, RSLinx and RSLogix 5000. One thing I've been thinking is that although most software claims support for standard GSM modems (maybe due to the apparent infancy of software in this area) not all GSM modems are really supported. I'm thinking of trying out the Siemens MC35 GSM modem. Has anyone successfully used this modem in this kind of application? Some feedback on a solution to this problem would be greatly appreciated. M #### Matt Try looking at using a Scripting Language such as perl, I have used it for various application such as controlling Sprinklers via a web page, I.E talking out a comm port. There is an OPC module and a GSM modem module, basicly you would plug both modules toghether in your program and away you go. http:\/\/www.devlearn.com\/perl\/index.html http:\/\/learn.perl.org\/library\/beginning_perl\/ The above links will get you started, just google and you will be suprised what you find.. Regards Matt L #### Lars-G\u00f6sta Johansson Here at MAX-lab National Laboratory, Sweden we have for some years use Products from Klinkmann Automation with success. Our application is almost identical to yours. We use hardware such as ControlLogix,SLC500 and Siemens S7. As a alarm OPC server we use RSView32 and as OPC alarm client we use Klinkmanns GSM-Control. Our GSM-modem is a Siemens TC35 Currently we have some 50 alarms and send them to 5 mobilephones. Klinkmann ( http:\/\/www.klinkmann.com ) sell a kit for about 1510 EUR (2417 AUD) and it contains: GSM-Control software with HASP dongle key Siemens TC35IT modem, antenna, power supply and modem cable. One thing I have notice is if you shut down RSView (for maintanence) you have to restart GSM-control (Connection with RSView OPC server is lost). Maybe there is a workaround for this described in the manual. Best regards Lars-G Johansson MAX-lab National laboratory Sweden R #### Raymond van der Tas To send SMS messages to mobile phones upon the occurance of alarm tags in a PLC you do the following: - buy a GSM modem and connect it to the Windows XP professional Serial\/USB port. - Install an OPC DA Server (OPC Data Access) - Install ICONICS AlarmWorX32 Multimedia software In case your OPC DA server also includes the OPC A\/E specification (alarms & events) you can directy have AlarmWorX32 Multimedia subscribe to the OPC A\/E information and the alarms\/events will be presented on the screen, logged to a database and directed to FAX, EMAIL, SMS, Text-To-Speech... In case your OPC DA server is not that modern yet, you can configure the ICONICS OPC Alarm Server to set the alarm limits (Hi\/Low, HiHi\/LoLo), digital alarms (on\/off), rate of change alarms and deviation alarms. AlarmWorX32Multimedia will then subscribe to this OPC A\/E server. The software will start at about US$2000. It's a bit more than your budget allows, but then also provides you with a complete scheduler to have multiple users with their own working schedule. Also escalation is provided to ensure that an alarm will be redirected in case action is not taken in time.\n\nWhen connecting a GSM Modem to your PC (instead of using a standard modem) it can receive the\nAlarm Acknowledgement via SMS.\n\nTo get the budget you explain the enduser that AlarmWorX32 is web-enabled. This allows to open up a web browser to monitor and acknowledge the alarms in the list. Certainly handy when the mobile phone runs on low battery on the recharger was left behind on site.\n\nGood luck!\n\nRaymond van der Tas\n\nFor more details:\nhttp:\/\/www.iconics.com\/products\/alarmworx32mmx.asp\n\nR\n\nM\n\n#### Michael Meirovitz\n\nDear Chris,\n\nThe problems you experienced with U.C.ME are due the fact that the Sony Ericsson modem is not supported. You may use the Siemens TC35i or MC35i modem that works with no problems.\n\nThe Sony-Ericsson will be supported in the near future.\n\nBest regards,\nMichael Meirovitz\nControl-See Software Solutions\n\nC\n\n#### chuck\n\nif you can program in VB, or some similar language, i'd recommend using:\n\nhttp:\/\/www.activexperts.com\/activsms\/\n\nby far the most flexible component for paging. you end up building your alarm\/pager management tool in an Access or SQL database, but you end up with a system flexible to your needs.\n\nchuck\n[email\u00a0protected]\n\nP\n\nM\n\n#### Michael Meirovitz\n\nPlease note that you can use U.C.ME-OPC to send SMS message using ANY GSM cellular modem, or, a standard analog modem using TAP protocol.\n\n1. Much cheaper than cellular modem.\n2. Cost of a phone call may be cheaper than SMS.\n3. No problems of cellular coverage and reception.\n\nPrices starts at 800 \\$ US.\n\nSee more at http:\/\/www.controlsee.com\n\nMichael Meirovitz\n\nC\n\n#### Cinzia La Morgia\n\nDid you try SMS Server? SMS Server is a add-on for CitectSCADA that lets you to notify via SMS and\/or email the alarms coming from the plant. It is equipped with an address book and a calendar by means of which it is possible to define the persons, or groups of persons, to which notify the alarms and time scheduling. SMS Server is strongly configurable and allows you to select single alarms or categories of alarms, or groups of alarms and groups of categories of alarms.\n\nThe current version 2.11 is on our web site:\nhttp:\/\/www.cpsistemi.it\n\nAfter December 17th, the new beta version will be available. It will have support for Citect server redundancy and remote connections.\n\nGoodbye,\nCinzia La Morgia","date":"2020-01-29 04:12:51","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\": 0.26040029525756836, \"perplexity\": 5116.59753525946}, \"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-2020-05\/segments\/1579251788528.85\/warc\/CC-MAIN-20200129041149-20200129071149-00070.warc.gz\"}"}
| null | null |
No one wants to sit in a humid home. With a pint (2 liters) capacity, this unit works best in areas up to 0square feet. Il est à la fois éconergétique et . With its Energy Star rating, paired with its R410a refrigerant, this unit is energy efficient and. The electronic controls will allow you to change .
Danby DDR60B3WP Pdf User Manuals. Shop with confidence on eBay! Are you looking for a quick and easy to setup dehumidifier for your wet room? For areas up to 0sq.
Environmentally friendly R410A refrigerant. Low temperature feature allows . Fan speeds (High – Low). With its energy star rating, paired with its r410a refrigerant, this unit is energy efficient and environmentally friendly.
Déshumidificateur de marque danby Premiere DDR60B3WP. Assez puissant pour déshumidifier un sous-sol. The unit is designed to take out pints of water (2 litres) every hours and depending on the specific conditions can work its magic in areas up to 600 . Striving and aches related to joints are not rare in old age. Barry Lutz you bask studying and writing about instinctive wellness and nutritional wellness supplements, with special c. Usage sized pliable couplings are preferred to match the danby premiere ddr60b3wp pint dehumidifier.
Some of the most popular window . Enter your comment here: Fill in your details . Hours Monday-Thursday : 8:30am to 6:00pm Eastern Standard Time Friday : 8:30am to 4:00pm Eastern Standard Time. Well, if yes, then here is our complete review can help you make the right choice. Use these helpful tips buying guide to help you choose the right . FREE shipping from Pure n Natural. Costco Item Various (See Table Below).
No person wants to sit down in a humid home. Características: -pinta de Estados Unidos (2litros) Capacidad por horas. R4A respetuoso con el medio ambiente.
Función de reinicio automático. Anticongelante automático previene la . Marketplace Research Stock is a single goal for all the diligence, firm and country reports. One point of interest to utilizing excited carbon?
The review of the danby premiere ddr60b3wp. Voici les points à vérifier, avant de courir au magasin. You can find relief with this high- end dehumidifier by. Browserul tau nu suporta HTML5.
Download baros – daca maine ft. This recall involves various dehumidifiers manufactured by Gree Electric Appliances, Inc. Recalled model numbers are listed below.
The brand name and the pint capacity are printed on the front of the dehumidifier.
|
{
"redpajama_set_name": "RedPajamaC4"
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| 2,119
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/**
* com.boneyard.ioc.engine.annotation.Action Tests
* @author Patricio Ferreira <3dimentionar@gmail.com>
**/
define(['ioc/engine/annotation/action',
'ioc/engine/helpers/dependency',
'specs/ioc.spec'], function(Action, Dependency, IocSpec) {
describe('com.boneyard.ioc.engine.annotation.Action', function() {
before(function() {
this.actionSimple = null;
this.actionSpy = sinon.spy();
this.injector = {
inject: function() {},
resolve: function() {}
};
this.injectorMock = sinon.mock(this.injector);
});
after(function() {
delete this.actionSpy;
this.injectorMock.restore();
delete this.injectorMock;
delete this.injector;
delete this.actionSimple;
});
describe('#new()', function() {
it('Should return an instance of Action Annotation', function() {
this.actionSimple = new Action(IocSpec.$actions[0]);
expect(this.actionSimple).to.be.ok();
});
});
describe('#getTarget()', function() {
it('Should retrieve Action target', function() {
expect(this.actionSimple.getTarget()).to.be.ok();
});
});
describe('#getContext()', function() {
it('Should retrieve Action context', function() {
expect(this.actionSimple.getContext()).not.to.be.ok();
});
});
describe('#resolve()', function() {
it('Should resolve action target bone method\'s reference to the operation', function() {
var getIdStub = sinon.stub(Action.prototype, 'getId').returns('$bone!simple.listenTo');
var getInjectorStub = sinon.stub(Action.prototype, 'getInjector').returns(this.injector);
this.actionSimple = new Action(IocSpec.$actions[0]);
var getCompoundStub = sinon.stub(this.actionSimple.getTarget(), 'getCompound')
.returns({ id: 'simple', method: 'listenTo' });
var engineMock = sinon.mock(this.actionSimple.getEngine());
var fakeBone = { bone: function() {} };
var boneMock = sinon.mock(fakeBone);
boneMock
.expects('bone')
.once()
.returns(fakeBone);
engineMock
.expects('bone')
.once()
.withArgs('simple')
.returns(fakeBone);
this.injectorMock
.expects('inject')
.once()
.withArgs(this.actionSimple.getTarget());
var result = this.actionSimple.resolve();
expect(result).to.be.ok();
expect(result.getTarget().expression).to.be(this.actionSimple.getTarget().getExpression());
boneMock.verify();
engineMock.verify();
this.injectorMock.verify();
boneMock.restore();
engineMock.restore();
Action.prototype.getId.restore();
Action.prototype.getInjector.restore();
this.actionSimple.getTarget().getCompound.restore();
});
});
describe('#parameters()', function() {
it('Should resolve all dependencies on Action\'s parameters', function() {
var dependencyA = {
expression: '$bone!model',
target: this.actionSimple,
property: '0',
bone: this.actionSimple,
resolve: function() {}
};
var dependencyMockA = sinon.mock(dependencyA);
var dependencyB = {
expression: '$bone!simple.update',
target: this.actionSimple,
property: '2',
bone: this.actionSimple,
resolve: function() {}
};
var dependencyMockB = sinon.mock(dependencyB);
var dependencies = [dependencyA, dependencyB];
var dependenciesInvokeStub =
sinon.stub(this.actionSimple.getInjector().getDependencies(), 'invoke',
_.bind(function(action, injector) { return _.invoke(dependencies, 'resolve', injector); }, this));
dependencyMockA
.expects('resolve')
.once()
.withArgs(this.actionSimple.getInjector())
.returns(dependencyA);
dependencyMockB
.expects('resolve')
.once()
.withArgs(this.actionSimple.getInjector())
.returns(dependencyB);
var result = this.actionSimple.parameters();
expect(result).to.be.an('array');
expect(result).to.have.length(3);
dependencyMockA.verify();
dependencyMockB.verify();
});
});
describe('#execute', function() {
it('Should execute the operation on a target that it has been resolve', function() {
var getContextStub = sinon.stub(this.actionSimple, 'getContext').returns({});
var getIdStub = sinon.stub(this.actionSimple, 'getId').returns(this.actionSpy)
var parametersStub = sinon.stub(this.actionSimple, 'parameters').returns(['value'])
this.actionSimple.execute();
expect(this.actionSpy.calledOnce);
this.actionSimple.getId.restore();
this.actionSimple.getContext.restore();
this.actionSimple.parameters.restore();
});
it('Should NOT execute the operation on a target that it has not been resolve yet', function() {
var getContextStub = sinon.stub(this.actionSimple, 'getContext').returns(null);
this.actionSimple.execute();
this.actionSimple.getContext.restore();
});
});
describe('static#only()', function() {
it('Should retrieves $actions annotations', function() {
var result = Action.only(IocSpec);
expect(result).to.be.ok();
expect(result.$id).to.be(undefined);
expect(result.$specs).to.be(undefined);
expect(result.$plugins).to.be(undefined);
expect(result.$actions).to.be.ok();
});
});
});
});
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 5,419
|
Certifikat kan syfta på:
Certifikat (finans) – ett finansiellt instrument.
Digitalt certifikat – en datafil som består av användaruppgifter och kryptonycklar
Certifiering – en standardiserad prövning
cs:Certifikát
hr:Certifikat
it:Certificato
ka:ატესტატი
ja:証書
pl:Certyfikat
sq:Certifikata
scn:Certificatu
simple:Certificate
sk:Certifikát
uk:Сертифікат
ur:سند
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 9,521
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NewsRemortgagingResidential Mortgages
Dudley launch a number of high LTV products
Jake Carter
Dudley Building Society has launched a number of products, which are available at a maximum 85% LTV.
Notable products include an 85% LTV 2-year fix at 3.79%, an 85% LTV 3-year fix at 3.84%, and an 85% LTV 5-year fix at 3.89%.
In addition, an 85% LTV 2-year discount product at 3.79% has been made available.
John Phillips: "2021 was anything but a traditional year"
An 85% LTV part discounted for term product at 3.94%, is now also accessible.
All of the products have a maximum loan size of £1m and are available for purchase and remortgage.
Sam Ward, commercial director of Dudley Building Society, said: "We have listened to our introducers whose customers are looking for higher LTVs and we have responded positively with these new products.
"Along with competitive fixed rates, we are launching a discounted 2-year deal and a part and part product, which will be of particular interest to customers with a larger mortgage wishing to keep monthly costs down.
"It all adds up to another comprehensive offering from the Dudley in support of the intermediary market."
Number of house hunters per branch reaches new high
Gatehouse Bank ups maximum FTV on HPPs
Rightmove: Price of homes coming to market up 0.3%
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 9,148
|
Джордж Га́рднер (; 1812? — 1849) — шотландский ботаник и исследователь.
Биография
Джордж Гарднер родился в мае 1812 года (согласно некоторым источникам, в мае 1809 года или в 1810 году) в Глазго (или в деревне Ардентинни) в семье садовника. Учился в Ардроссане, в 1829 году поступил в Андерсонский университет в Глазго (ныне Университет Стратклайда). В 1835 году окончил университет со степенью по медицине.
В 1836 году Гарднер при поддержке Уильяма Джексона Гукера отправился в Бразилию, в июле прибыл в Рио-де-Жанейро. На протяжении пяти лет он путешествовал по Бразилии, в 1841 году вернулся в Англию. В 1842 году Джордж Гарднер был избран членом Лондонского Линнеевского общества.
В 1844 году Гарднер был назначен главным садоводом Королевских ботанических садов в Перадении на Цейлоне. Он начал подготавливать монографию флоры острова, однако закончить её не успел. С 1846 по 1847 Гарднер был главным редактором журнала Calcutta Journal of Natural History.
10 марта 1849 года Джордж Гарднер скончался.
Некоторые научные работы
Gardner, G. (1846). Travels in the interior of Brazil. 562 p.
Роды растений, названные в честь Дж. Гарднера
Примечания
Литература
Boulger, G.S. Gardner, George // Dictionary of National Biography. — 1885—1900. — Vol. 20. — P. 431.
Ботаники Шотландии
Ботаники по алфавиту
Члены Лондонского Линнеевского общества
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 3,723
|
{"url":"https:\/\/www.physicsforums.com\/threads\/what-is-resistance.865148\/","text":"# B What is resistance?\n\nTags:\n1. Apr 2, 2016\n\n### Sho Kano\n\nMy professor says resistance depends on,\n$R\\quad =\\quad \\frac { \\rho L }{ A }$\n\nAnd is defined as,\n$R\\quad =\\quad \\frac { V }{ I }$\n\nWhat does she mean? What is the difference?\nA definition is denoted by $\\equiv$, what is the difference between that and an ordinary equal sign?\n\n2. Apr 2, 2016\n\n### jfizzix\n\nElectrical resistance is a physical concept like force, and not a mathematical one (so we wouldn't necessarily use the triple equals sign here).\nIn a related example, kinetic energy has been at times defined as half the mass times the square of the velocity, but this an equation that doesn't work when you're going near the speed of light. There are other definitions of kinetic energy that work well with special relativity, and reduce to the usual formula at low velocities.\nThe amount of resistance (as far as I know) is always equal to voltage over current, so you could define electric resistance by that ratio, but it's important to consider that that resistance is not just one constant for a given material. It depends on temperature, pressure, and applied voltage, among other things.\n\nLast edited: Apr 2, 2016\n3. Apr 2, 2016\n\n### David Lewis\n\nR=V\/I is not the definition of resistance. It's a formula to calculate its value.\n\nResistance is the property of a circuit or circuit element that impedes current by converting the current's kinetic energy to heat.\n\nLast edited: Apr 2, 2016\n4. Apr 2, 2016\n\n### Staff: Mentor\n\nI disagree. Resistance is defined by $R=V\/I$, and the KE is negligibe, it is the PE in the field which is converted to heat.\n\n5. Apr 2, 2016\n\n### Staff: Mentor\n\nA definition in physics serves the same purpose as in ordinary language, it establishes a convention for the meaning of a given word or symbol. So $R\\equiv V\/I$ says that whenever we say $R$ we mean $V\/I$.\n\nWe can then use other information and definitions to deduce the relationship between $R$ and other quantities. Those deduced relationships are identified using the usual equals sign.\n\n6. Apr 2, 2016\n\n### David Lewis\n\nDale wrote: \"Resistance is defined by R=V\/I.\"\n\nDavid Lewis wrote: You apply 1V to a capacitor, the initial current is 1A, so the resistance of the capacitor at that moment is 1 ohm?\n\n7. Apr 2, 2016\n\n8. Apr 3, 2016\n\n### davenn\n\nthe apparent resistance of the capacitor is 1 Ohm, Yes\nthe current into and out of the cap is initially 1A and falls rapidly as the capacitor is energised ( reaches equilibrium)\nNOTE: NO current flows through the capacitor tho.\n\nand another note ... please learn to use the reply button so as to quote people ...\ndoing the \"he wrote, she wrote\" etc thing can get very confusing to read\n\nRegards\nDave\n\n9. Apr 3, 2016\n\n### Staff: Mentor\n\nThat isn't the way that capacitors work, but in principle, yes. In a capacitor I is not proportional to V so it's resistance is not constant. Materials for which I is proportional to V are called \"Ohmic\".\n\n10. Apr 3, 2016\n\n### Staff: Mentor\n\nThanks, that is an excellent paper!\n\n11. Apr 3, 2016\n\n### anorlunda\n\nDale is right, resistance is merely the relationship between voltage V and current I. Components do not need to be linear, nor passive, so that many relationships are possible. The picture attached illustrates a few. (The one labeled battery represents a battery with internal resistance.).\n\nThe arbitrary curve is not impossible. R could be defined as the slope of a line connecting any two points on the arbitrary curve. The value could be positive, negative or zero.\n\nIt is only when we have approximately linear materials, that R becomes nearly constant (see the black line in the chart). That is what your professor had in mind, but it is not the only possibility.\n\nThe international standard defines the ohm, not resistance. ohm is to resistance as a meter is to distance.\n\n12. Apr 3, 2016\n\n### David Lewis\n\nIt would be more accurate to say the reactance of the capacitor is 1 ohm. Resistance is not the only physical quantity that can impede current or influence voltage\/current ratios.\n\n13. Apr 3, 2016\n\n### davenn\n\nONLY when dealing with AC, not DC\n\nCapacitive reactance XC and Inductive reactance XL\nare very different beasts to plain resistance. Yes they are both measured in Ohms\nbut that is where the similarity ends\n\nYou cannot use Ohms Law, R = V\/I, to determine resistance in an AC circuit\n\nD\n\n14. Apr 3, 2016\n\n### anorlunda\n\nThat's true davenn in that form, but Ohms Law comes in several forms. See the Insights Article, AC Power Analysis: Part 1, Basics for the complex form.\n\n15. Apr 3, 2016\n\n### davenn\n\nwhich is what I said\n\nif capacitive and inductive reactances are followed up with some research by David Lewis\nhe's going to come across the term impedance pretty quickly\n\n16. Apr 3, 2016\n\n### SteamKing\n\nStaff Emeritus\nThis is a graphical illustration of resistance:\n\nLast edited by a moderator: May 7, 2017\n17. Apr 3, 2016\n\n### David Lewis\n\nIt may be AC or DC. The only requirement is that the voltage is changing.\n\n18. Apr 3, 2016\n\n### Staff: Mentor\n\nIf it is DC then the voltage is not changing, by definition.\n\n19. Apr 3, 2016\n\n### davenn\n\nthis is incorrect ... so you just contradicted yourself\nsee Dale's response\n\n20. Apr 4, 2016\n\n### houlahound\n\nDC voltage can change as much as it wants it just can't change polarity from pos\/neg...or neg\/pos.\n\nLast edited: Apr 4, 2016","date":"2017-08-19 18:52:03","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\": 2, \"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\": 0.6621074676513672, \"perplexity\": 1901.9167219012666}, \"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-2017-34\/segments\/1502886105712.28\/warc\/CC-MAIN-20170819182059-20170819202059-00150.warc.gz\"}"}
| null | null |
{"url":"https:\/\/www.physicsforums.com\/threads\/a-50-omega-lossless-transmission-line-impedance-question.902172\/","text":"# Homework Help: A $50\\Omega$ lossless transmission line impedance question\n\n1. Jan 31, 2017\n\n### David J\n\n1. The problem statement, all variables and given\/known data\n\na $50\\Omega$ lossless transmission line of length $0.4\\lambda$ is terminated in a load of $(40+j30)\\Omega$. Determine, using $Zin=Z0\\frac{ZL cos \\beta l+j Z0 sin \\beta l}{Z0 cos \\beta l +j ZL sin \\beta l}$ the input impedance to the line.\n\n$Z0=50\\Omega$\n\n$ZL=(40+j30)\\Omega$\n\n$\\beta l =2.513$ radians\n\n$cos\\beta l =-0.809$ radians or $0.999^0$\n\n$sin\\beta l=0.588$ radians or $0.0438^0$\n\n2. Relevant equations\n\n$Zin=Z0\\frac{ZL cos \\beta l+j Z0 sin \\beta l}{Z0 cos \\beta l +j Zl sin \\beta l}$\n\n3. The attempt at a solution\n\n$Zin=50\\Omega\\frac{(40+j30)(-0.809) +j(50)(0.588)}{50(-0.809) +j(40+j30)(0.588)}$\n\nSo $50\\Omega\\frac{(-32.36-j24.27)+(j29.4)}{(-40.45+(-17.64+j23.52)}$\n\nSo $50\\Omega\\frac{(-32.36+j5.13)}{(-58.09+j23.52)}$\n\n$=50\\Omega(0.5093+j0.118)$\n\nSo $Zin=25.465+j5.90$ or $26.14\\angle13^0$ or $26.14\\angle0.23$ radians\n\nI have been over this 5 times now to check and I feel I am correct but I would like a second opinion please, if possible, to identify any errors etc.\n\nThanks\n\n2. Jan 31, 2017\n\n### Staff: Mentor\n\nThe only problem I can see is that the sines and cosines should be unitless. The sine or cosine (or result of any trig function) is a pure number that represents a ratio. Otherwise, well done!\n\n3. Jan 31, 2017\n\n### David J\n\nAppreciated. thanks again","date":"2018-07-17 00:59:13","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\": 0.6792241930961609, \"perplexity\": 1049.374311019624}, \"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-2018-30\/segments\/1531676589536.40\/warc\/CC-MAIN-20180716232549-20180717012549-00040.warc.gz\"}"}
| null | null |
Q: PIP installing module gives warning but installs module I just installed pip and a module. The module installed correctly but it gives the following error.
WARNING: Ignoring invalid distribution -ip (d:\python 3.8\lib\site-packages)
I mean to ask if this is bad or should I ignore it.
[1]: https://i.stack.imgur.com/a6v5Y.png
|
{
"redpajama_set_name": "RedPajamaStackExchange"
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| 8,143
|
Q: SQL queries on find Average for 2 column
I have a table with 3 columns being appName, Hours and Rating, I want to find the Average hours and average rating for All the apps and display it. How do i actually find the average in this situation. thank you!
A: If appname is unique then just use group by on appname as follows:
select lower(appname), avg(hour) as Avg_Hour, avg(rating) as Avg_Rating
from youTablename
group by lower(appname);
|
{
"redpajama_set_name": "RedPajamaStackExchange"
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| 4,704
|
Home / News / Swedish Prosecutor Requests Arrest Warrant for Julian Assange
Swedish Prosecutor Requests Arrest Warrant for Julian Assange
Stockholm, May 20 ( Prensa Latina) Swedish Public Prosecutor's Office filed an arrest warrant this Monday against Wikileaks founder Julian Assange for alleged sexual offences he denies.
'I have asked the court in Uppsala (north of Stockholm) to arrest Assange in absentia, suspected of rape to a lesser degree for probable cause. If the court decides to arrest him, I will send an European arrest warrant for him to be sent to Sweden', prosecutor Eva-Marie Persson said in a statement.
WikiLeaks editor-in-chief Kristinn Hrafnsson said Assange's case has been 'mishandled at all times', denouncing that Ecuador will hand over to the United States on Monday the belongings left by the Australian journalist at the South American country's embassy in London after his arrest last month.
There has been considerable political pressure on Sweden to reopen its investigation, but there has always been political pressure around this case,' Hrafnsson said.
Assange was arrested on April 11 by British police at the Ecuadorian embassy in the United Kingdom after the government of Ecuador ended his seven-year asylum.
The cyberactivist is in a 50-week jail sentence in the United Kingdom for violating the terms of a bail bond in 2012, and also faces an extradition order to the United States, which wants to try him for spreading confidential information about U.S. national security.
In the case of a conflict between a European arrest warrant and an extradition order from the United States, British authorities will have to decide the order of priority, said the Swedish prosecutor.
Sweden reopened the case on May 13, since the accusation, which involves Assange in a minor degree of rape (for having unprotected sexual relations with a young woman while sleeping), is still in force until 2020.
The Public Prosecutor's Office filed legal actions against the computer programmer in May 2017, given the impossibility of advancing the investigations.
Iranian Foreign Minister condemns statements by French counterpart
U.S. Capitol Police warned of riots on January 3rd
Activists demand justice for Indigenous leader Milagro Sala
World Health Organization warns worst is yet to come
Argentina detects first case of British coronavirus variant
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"redpajama_set_name": "RedPajamaCommonCrawl"
}
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The Information Technology & Innovation Foundation says that utilizing robots would improve productivity and growth of the U.S. economy.
There's no doubt that robot adoption is growing all over the world. As businesses seek new and cost-effective ways to manage themselves, robot adoption is one way they're doing it. But a new report claims the United States is behind on robot adoption. The Information Technology & Innovation Foundation says that utilizing robots would improve productivity and growth of the U.S. economy.
But in their latest report they find that the U.S. trails southeast Asia and parts of Eastern Europe in robot adoption. The ITIF report took a look at 27 nations and found that the United States ranked 16th overall in its share of robots.
Robots are key tools for boosting productivity and living standards. To date, most robot adoption has occurred in manufacturing, where there are robots designed to perform a wide variety of manual tasks more efficiently and consistently than humans. But with continued innovation, robot use is spreading to many other sectors, too, from agriculture to logistics to hospitality. As this trend continues—making robots increasingly important to productivity and competitiveness economy-wide—robot adoption will be a vital economic indicator for policymakers to monitor as a sign of growth and progress.
You can read the ITIF report on robot adoption at the link below and draw your own conclusions.
What do you think of the report and robot adoption in the United States? Let us know in the comments below or on Google+, Twitter, or Facebook.
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"redpajama_set_name": "RedPajamaC4"
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\section{Introduction}
\noindent A log Calabi-Yau pair $(Y,D)$ consists of a proper variety $Y$ and an effective $\mathbb{Q}$-divisor $D$ such that $(Y,D)$ is log canonical and $K_X+D$ is $\mathbb{Q}$-linearly equivalent to zero: see, for instance, \cite{kol}. A Calabi-Yau variety can be viewed as $(Y,0)$. If $Y$ is a Fano variety such that $D$ is $\mathbb{Q}$-linearly equivalent to the anticanonical divisor, then $(Y,D)$ is a log Calabi-Yau pair, provided it is log canonical.
\noindent Let us take into account three dimensional log Calabi-Yau pairs. As well known, there exist finitely many deformation types of Fano threefolds. As a result, there are finitely many possible values for their Euler characteristic. Conjecturally, this should be true for the collection of all Calabi-Yau threefolds too. Here by Calabi-Yau threefold we mean a complex K\"{a}hler compact manifold with trivial canonical bundle and no $p$-holomorphic forms for $p=1,2$. Since general log Calabi-Yau pairs interpolate between these two extremes, it is natural to wonder whether they are bounded or not. In this paper, we prove the following result.
\begin{thm}
\label{mainresult}
There exists an integer $N_0$ such that, for every $n\geq N_0$ there exists a log Calabi-Yau threefolds $(Y, D)$ with the Euler characteristic of $Y$ given by
$$
e(Y)=-48n-46.
$$
\noindent Moreover, $Y$ is smooth and its Kodaira dimension is negative. Additionally, we have $K_{Y}+D=0$, where $D$ is a divisor isomorphic to a $K3$ surface.
\end{thm}
\noindent Recently, Di Cerbo and Svaldi in \cite{DCS} prove that log Calabi-Yau pairs are bounded. One of their assumption is that the pair are klt. Notice that there is no contradiction between their result and ours; indeed, the example in Theorem \ref{mainresult} is not klt but log canonical.
\vspace{2mm}
\noindent The proof of Theorem \ref{mainresult} is constructive. More specifically, we describe a collection of log Calabi-Yau threefolds with the properties mentioned above. First, take into account the Segre-Hirzebruch surface $\mathbb{F}_n$ for any positive integer $n$. Next, fix a suitable decomposable vector bundle on each $\mathbb{F}_n$, namely
$$
\m{V}:=\mathcal{O}_{\mathbb{F}_n} \oplus \mathcal{O}_{\mathbb{F}_n}(2C_0-F),
$$
where $C_0$ is the unique effective divisor on $\mathbb{F}_n$ such that $C_0^2=-n$ and $F$ is the class of the fiber with respect to the $\mathbb{P}^1$-bundle structure on $\mathbb{F}_n$. For any $n$ denote by $X$ the scroll defined as $\mathbb{P}(\m{V})$, the projective bundle of hyperplanes in $\m{V}$. For futher information about these scrolls, see, for instance, \cite{fanflam}.
\vspace{2mm}
\noindent If the linear system $|-2K_X|$ had a smooth member, then the double covering of $X$ - branched along it - would be a smooth Calabi-Yau manifold. Unfortunately, this is not the case. The base locus of $|-2K_X|$ is given by a smooth rational curve. Luckily, the multiplicity of the generic section along the base locus is three. This requires a careful analysis of the cohomology group $H^0(X,-2K_X)$, which can be carried out more easily for $n$ big enough.
\vspace{2mm}
\noindent If we blow up $X$ along the smooth curve in the base locus of the bianticanonical system, we obtain a smooth threefold $X_1$. The linear system $|-2K_{X_1}|$ is not basepoint free. The base locus is given by a smooth rational curve $\gamma_1$. In order to resolve a generic section of the linear system $|-2K_X|$, we blow up $X_1$ along $\gamma_1$. We obtain a smooth threefold $X_2$. The degree two branched cover $Y_2$ along a smooth section of $-2K_X-2E_1-4E_2$ is not normal. Taking the normalization of it is equivalent to taking the branched covering of $X_2$ along a smooth member of the linear system $-2K_{X_2}- 2E_2=-2K_X-2E_1-4E_2$.
\vspace{2mm}
\noindent Finally, in order to calculate the Euler characteristic of $Y_2$ for $n$ big enough, it suffices to determine that of $X_2$ and that of a smooth surface in $|-2K_{X_2}-2E_2|$. The former follows from the cohomology of blow ups along a submanifold and the latter from the Chern classes of it: see, for instance, \cite{GH}.
\vspace{2mm}
\noindent Our construction relies on the choice of the vector bundle $\m{V}=\mathcal{O}_{\mathbb{F}_n} \oplus \mathcal{O}_{\mathbb{F}_n}(2C_0-F)$. It is important to stress that this is only one of the possible choices in order to arrive at an unbounded family of log Calabi-Yau pairs. To be more precise, we analysed all the cases as $\m{V}$ varies among the rank $2$ vector bundle on $\mathbb{F}_n$ that are decomposable. Our method yields a double cover, which is a smooth Calabi-Yau threefold, only for a finite number of cases. We expect that for the great majority of the other cases the situation is similar to that presented in this paper: one can mimic the construction and obtain a log Calabi-Yau pair.
\vspace{2mm}
\noindent The paper is organized as follows. In Section 2 we recall some preliminary results. Section 3 is devoted to describing the bianticanonical system of the scroll $X$, in particular a desingularitazion of a generic section of it. At last, Section 4 concludes the exposition with the computation of the Euler characteristic, thus showing that it is in fact unbounded!
\vskip 0.7cm
{\bf Acknowledgements.} The authors are both supported by INdAM - GNSAGA and by FIRB2012 "Moduli spaces and applications", which are granted by MIUR. We would like to thank Ciro Ciliberto and Claudio Fontanari for helpful remarks on the problem and the manuscript.
\section{Some Preliminary Results}
\noindent In this section we recall some basic facts and prove some results that will be applied in what follows. For further details, the reader is referred to \cite{Hag}, p. 369 ff.
\noindent Let $S$ be a smooth projective surface and denote by $X$ the projective bundle associated to a rank $2$ vector bundle $\m{V}$ on $S$. To avoid confusion we recall that $X$ is the projective bundle $\mathbb{P}(\m{V})$ over the base $S$, where $\mathbb{P}(\m{V})$ is the projective bundle of hyperplanes in $\m{V}$. In what follows we will set $\tau$ to be $c_1(\mathcal{O}_X(1))$.
\begin{lem}
\label{PROP:CHERNXANDD}
Denote by $\varphi:X\rightarrow S$ the fibration given by the projective bundle structure. Then the following identities hold:
$$c_1(X)=2\tau+\varphi^*(c_1(S)-c_1(\m{V}));$$
$$c_2(X)=\varphi^*(c_2(S)-c_1(\m{V})c_1(S))+2\varphi^*c_1(S)\tau;$$
$$c_3(X)=2\varphi^*(c_2(S))\tau.$$
\end{lem}
\begin{proof}
\noindent We have the exact sequences
\begin{equation}
\label{SES:Rel1}
0\rightarrow T_{X/S}\rightarrow T_X\rightarrow\varphi^*T_{S}\rightarrow 0,
\end{equation}
\begin{equation}
\label{SES:Rel2}
0\rightarrow \mathcal{O}_X\rightarrow (\varphi^*\m{V}^{\vee})\otimes \mathcal{O}_X(1)\rightarrow T_{X/{S}}\rightarrow 0.
\end{equation}
\vspace{2mm}
\noindent Recall also that $H^*(X)$ is generated as an $H^*(S)$-algebra by $\tau$ with the single relation
\begin{equation}
\label{SES:Relaz}
\tau^2-\varphi^*c_1(\m{V})\tau=0.
\end{equation}
We have
\begin{equation*}
c_1((\varphi^*\m{V}^{\vee})\otimes \mathcal{O}_X(1))=\varphi^*c_1(\m{V}^{\vee})+2\tau=-\varphi^*c_1(\m{V})+2\tau,
\end{equation*}
\begin{equation*}
c_2((\varphi^*\m{V}^{\vee})\otimes \mathcal{O}_X(1))=\varphi^*c_2(\m{V}^{\vee})+\varphi^*c_1(\m{V}^{\vee})\tau+\tau^2=\varphi^*c_2(\m{V})-\varphi^*c_1(\m{V})\tau+\tau^2.
\end{equation*}
By \eqref{SES:Relaz}, this yields
\begin{equation}
c((\pi^*\m{V}^{\vee})\otimes \mathcal{O}_X(1))=1+(2\tau-\varphi^*c_1(\m{V})).
\end{equation}
From the exact sequences \eqref{SES:Rel1} and \eqref{SES:Rel2}, we get
\begin{multline}
\label{EQ:CHERNX}
c(X)=c(T_{X/S})\varphi^*c(T_S)=c((\varphi^*\m{V}^{\vee})\otimes \mathcal{O}_X(1))\varphi^*c(T_S)=\\
=(1+(2\tau-\varphi^*c_1(\m{V})))\varphi^*c(S)=\\
=1+[2\tau-\varphi^*c_1(\m{V})+\varphi^*c_1(S)]+[\varphi^*c_2(S)+\varphi^*c_1(S)(2\tau-\varphi^*c_1(\m{V}))]+[2\varphi^*c_2(S)\tau]=\\
=1+[2\tau+\varphi^*(c_1(S)-c_1(\m{V}))]+[\varphi^*(c_2(S)-c_1(\m{V})c_1(S))+2\varphi^*c_1(S)\tau]+[2\varphi^*c_2(S)\tau]
\end{multline}
\end{proof}
\noindent In order to determine the cohomology of line bundles on $X$, we are going to apply the following result. We will recall it here for the sake of completeness: see, for instance, \cite{Hag}, pag. 253, Ex 8.4 (a).
\begin{lem}
\label{LEMMA:HAGPUSH}
Let $\m{V}$ be a vector bundle on a smooth surface $S$. Let $X=\mathbb{P}(\m{V})$ and define $\tau$ as before. Then
$$\varphi_*\mathcal{O}_X(a\tau)=0 \qquad \mbox{ if } a<0,$$
$$\varphi_*\mathcal{O}_X(a\tau)=S^a(\m{V}) \qquad \mbox{ if } a\geq0,$$
$$R^i\varphi_*\mathcal{O}_X(a\tau)=0 \qquad \forall a\in \mathbb{Z} \mbox{ if } 0<i<\Rk(\m{V})-1 \mbox{ or if } i\geq \Rk(\m{V}),$$
$$R^{\Rk(\m{V})-1}\varphi_*\mathcal{O}_X(a\tau)=0 \qquad \mbox{ if } a> -\Rk(\m{V}).$$
\end{lem}
\section{A Generic Member of the Bianticanonical Linear System}
\noindent From now onwards, $S$ will be the Segre-Hirzebruch surface $\mathbb{F}_n=\mathbb{P}(\mathcal{O}_{\mathbb{P}^1}\oplus \mathcal{O}_{\mathbb{P}^1}(-n))$, with $n$ positive. Recall that $\Pic(\mathbb{F}_n)$ is generated by $C_0$, the only effective divisor on $S$ such that $C_0^2=-n$, and $F$, the class of a fiber of the $\mathbb{P}^1$-bundle. Hence, without loss of generality, any decomposable vector bundle of rank $2$, up to tensor product with a line bundle, can be written as $\m{V}=\mathcal{O}_{\mathbb{F}_n}\oplus\mathcal{O}_{\mathbb{F}_n}(-A)$, where $A=xC_0+yF$ and $x$ is nonnegative. For the sake of convenience, we will denote by the same symbol a divisor on $S$ and its pullback on $X$. We will denote, as before, by $X$ the projective bundle associated to $\m{V}$
\begin{prop}
\label{PROP:FIXLOCUSP2}
Consider a divisor $D=a\tau+G$ on $X$ where $G=bC_0+cF$ is the pullback of a divisor on $S$. Then the following hold:
\begin{itemize}
\item [i)] If $A=xC_0+yF$ with $y\geq 0$ (i.e., if $A$ is effective), $D$ is effective if and only if $a,b,c\geq 0$.
\item [ii)] If $A=xC_0-yF$ with $y>0$, $D$ is effective if and only if $a\geq 0$ and
$$(b,c)\in \bigcup_{r=0}^a S_r\qquad \mbox{ with } S_r=\{(b,c)\,|\, b,c\geq 0\}+(rx,-ry).$$
\item [iii)] If $A=xC_0-yF$ with $y>0$, the only prime and rigid divisors on $X$ are $\tau, C_0$ and $\tau+A$.
\end{itemize}
\end{prop}
\begin{proof}
i) $D=a\tau+bC_0+cF$ is effective if and only if $a\geq 0$; else we have $$\varphi_*\mathcal{O}_X(a\tau+bC_0+cF)=\varphi_*\mathcal{O}_X(a\tau)\otimes \mathcal{O}_S(bC_0+cF)=0.$$
Hence, we can assume $a\geq0$. Doing so, we have
$$H^0(\mathcal{O}_X(D))=\bigoplus_{r=0}^a\H^0(\mathcal{O}_S(bC_0+cF-rA))\supset H^0(\mathcal{O}_S(bC_0+cF)).$$
If $b,c\geq 0$ the divisor is effective.
ii) Assume, now, $A=xC_0-yF$ with $y>0$. In this case
$$V_r=\H^0(\mathcal{O}_S(bC_0+cF-rA))=\H^0(\mathcal{O}_S((b-rx)C_0+(c+ry)F))$$ and $H^0(\mathcal{O}_X(D))$ is not zero exactly when at least one of these spaces is not zero. $V_r$ is not zero exactly when $b\geq rx$ and $c\geq -ry$, i.e., when $(b,c)\in S_r$, so the second claim is proved.
\noindent iii) Finally assume $A=xC_0-yF$ with $y>0$, and consider the effective divisor $D=a\tau+bC_0+cF$ with $a,b\geq 0$ and $-ay\leq c\leq 0$. If $D$ is rigid then $(b,c)\in S_r$ for exactly one value or $r$ (with $0\leq r\leq a$). If $(b,c)\in S_r$ we can write $D$ as $a\tau+rA+b'C_0+c'F$ with $0\leq c'<y$. If $r<a$ we can assume $0\leq b'<x$ whereas, if $(b,c)\in S_a$, we can assume $b'\geq 0$. In both cases, the divisor $D=a\tau+rA+b'C_0$ is effective; hence, if $c'>0$, we have $h^0(\mathcal{O}_X(D))\geq 2$. This shows that we have to look for rigid divisors among the ones of the form
$$D=a\tau+rA+b'C_0,$$
where $a\geq 0, 0\leq r< a$ and $0\leq b'<x$ or with $a\geq 0, r=a$ and $b'\geq 0$. It is not difficult to see that, in this case, $h^0(\mathcal{O}_X(D))=1$, so we always get a rigid divisor. It is also easy to see that every such divisor can be written as a sum
$$a_1\tau+a_2C_0+a_3(\tau+A),$$
which proves that $\tau,C_0$ and $\tau+A$ are the only rigid prime divisors on $X$ when $A=xC_0-yF$ and $y>0$.
\end{proof}
We will be interested in the case $\m{V}=\mathcal{O}_{\mathbb{F}_n}\oplus\mathcal{O}_{\mathbb{F}_n}(-A)$ with $A=2C_0-F$. Recall that, in this case,
$$
K_X=-2\tau-2C_0-(n+2)F-2C_0+F=-2\tau - 4C_0 - (n + 1)F,
$$
so we have
$$
-2K_X=4\tau+8C_0+(2n+2)F.
$$
\noindent As we will see, if $n$ is big enough, the linear system $|-2K_X|$ does not have smooth members. Thus, we need to describe more closely the base locus and the type of singularities.
\begin{prop} The base locus of the bianticanonical linear series is given by the complete intersection $\sigma$ of the rigid divisors with class $\tau+A$ and $C_0$.
\end{prop}
\begin{proof} Since $2C_0-F$ is not effective, by Proposition \ref{PROP:FIXLOCUSP2} there are three rigid prime divisors, namely $\tau, C_0$ and $\tau + A$. The intersections of these three divisors are
$$
\tau(\tau+A)=0, \qquad \tau C_0 := \gamma, \qquad (\tau+A)C_0= (\tau- (2n+1)F)_{|_{C_0}}:=\sigma.
$$
The unique surface $T$ with class given by $\tau$ is a Segre-Hirzebruch surface $\mathbb{F}_n$ with standard generators for $\Pic(\tau)$ given by
$$C_0|_T=\gamma_T, \qquad F|_T=f_T.$$
By standard generators, we mean a basis of effective prime divisors under which the intersection product has representative matrix
$$\begin{bmatrix}
-a & 1 \\
1 & 0
\end{bmatrix}$$ where $a$ is the (positive) index of the Segre-Hirzebruch surface. In particular, the class of the curve $\gamma$ seen in $T$ is given by $\gamma_T$.
\noindent Denote by $R$ the only surface whose class is $\tau+A$. One can easily see that $R$ is again a Segre-Hirzebruch surface $\mathbb{F}_n$ if one considers the vector bundle $\m{V}'=\m{V}\otimes \mathcal{O}_S(A)$ and uses the identification
$$X=\mathbb{P}(\m{V})=\mathbb{P}(\mathcal{O}_S\oplus\mathcal{O}_S(-A))=\mathbb{P}(\mathcal{O}_S\oplus\mathcal{O}_S(A))=\mathbb{P}(\m{V}').$$
Indeed the class of $c_1(\mathcal{O}_{\mathbb{P}(\m{V}')}(1))=\tau'$ is $\tau+A$ under this identification. The standard generators for $\Pic(R)$ are
$$C_0|_R=\gamma_R, \qquad F|_R=f_R.$$
\noindent The surface $U$, whose class is $C_0$, is also a Segre- Hirzebruch surface $\mathbb{F}_m$ with $m=2n+1$. The standard generators for the Picard lattice are
$$(\tau+A)|_U=\gamma_U, \qquad F|_U=f_U.$$
\noindent Notice that
$$-2K_X=4(\tau+A)+(2n+6)F,$$
so, an eventual base point of $|-2K_X|$ cannot lie outside the surface $R$. In fact, $(2n+6)F$ is globally generated. It is easy to prove that $\tau+A$ is not a component of $|-2K_X|$, so the base locus of the bianticanonical linear series is contained in $R$. In fact, let us restrict $-2K_X$ to $R$. This yields
$$
(4\tau+8C_0+(2n+2)F)|_R = 8\sigma+(2n+2)f_R,
$$
which shows that $\sigma$ is contained in the base locus of the bianticanonical linear series. Conversely, given a point in such a base locus, it must belong to $\sigma$ because it is in $R$ and nowhere else than in $\sigma$ because $|f_R|$ is globally generated in $R$. Therefore the claim is proved.
\end{proof}
{\bf Remark.} \noindent The curves $\sigma$ and $\gamma$ are the intersection of $C_0$ with $\tau$ and $\tau+A$, respectively. We can also see them inside these surfaces and the following table describes their classes (a "-" simply means that the curve cannot be seen in that particular surface).
\begin{center}
\begin{tabular}{c|c|c|c|c}
& T ($\tau$) & R ($\tau+A$) & U ($C_0$)\\ \hline
$T\cap U = \gamma$ & $\gamma_T$ & - & $\gamma_U+(2n+1)f_U$\\
$R\cap U = \sigma$ & - & $\gamma_R$ & $\gamma_U$
\end{tabular}
\end{center}
From this description (as well as from adjunction) one can see that both $\gamma$ and $\sigma$ are smooth curves of genus $0$. Moreover, $\sigma$ is rigid in both $R$ and $U$, whereas $\gamma$ is rigid only in $T$.
\begin{prop}
\label{PROP:MULT3}
The generic member of the bianticanonical system has multiplicity $3$ along the base locus.
\end{prop}
\begin{proof}
\noindent Define $t,u$ and $r$ to be the sections (uniquely determined up to scalar) such that $$H^0(\mathcal{O}_X(\tau))=\langle t\rangle \qquad H^0(\mathcal{O}_X(C_0))=\langle u\rangle \qquad H^0(\mathcal{O}_X(\tau+A))=\langle r\rangle$$
so the zero loci of $t,u$ and $r$ describe $T,U$ and $R$, respectively. Define $D_i$ to be $-2K_X-i(\tau+A)$. Therefore, there exists a positive integer $N_0$ big enough such that for $n \geq N_0$ the following hold:
\begin{center}
\begin{tabular}{c|c}
$h^0(D_0) = 14n+61 $ & -\\
$h^0(D_1) = 11n+52$ & $h^0(\mathcal{O}_X(D_0))-h^0(\mathcal{O}_X(D_1)) = 3n+9$\\
$h^0(D_2) = 8n+40$ & $h^0(\mathcal{O}_X(D_1))-h^0(\mathcal{O}_X(D_2)) = 3n+12$\\
$h^0(D_3) = 5n+25$ & $h^0(\mathcal{O}_X(D_2))-h^0(\mathcal{O}_X(D_3)) = 3n+15$\\
$h^0(D_4) = 2n+7$ & $h^0(\mathcal{O}_X(D_3))-h^0(\mathcal{O}_X(D_4)) = 3n+18$
\end{tabular}
\end{center}
Let us now describe the sections of $\mathcal{O}_{X}(-2K_X))=\mathcal{O}_X(4(\tau+A)+(2n+6)F)$.
\vspace{2mm}
\noindent We have the exact sequence
\begin{equation}
\label{EQ:EXS1}
\xymatrix@R=1pc@C=2pc{
0\ar@{->}[r] & H^0(\mathcal{O}_X(D_1))\ar@{^{(}->}[r]^-{-\otimes r} & H^0(\mathcal{O}_X(-2K_X)) \ar[r] & H^0(\mathcal{O}_R(-2K_X))
}
\end{equation}
Notice that $-2K_X|_R = (8C_0+(2n+2)F)|_R=8\gamma_R+(2n+2)f_R$ hence, as $R=\mathbb{F}_n$, we have
$$h^0(\mathcal{O}_{R}(-2K_X))=h^0(\mathcal{O}_{\mathbb{F}_n}(8\gamma_R+(2n+2)f_R))=3n+9=h^0(\mathcal{O}_X(D_0))-h^0(\mathcal{O}_X(D_1)).$$
Thus, the restriction map $H^0(\mathcal{O}_X(-2K_X))\rightarrow H^0(\mathcal{O}_R(-2K_X))$ in Equation \ref{EQ:EXS1} is indeed surjective. Denote by $V_0$ a subspace of $H^0(\mathcal{O}_X(-2K_X))$ such that
$$V_0\oplus (H^0(\mathcal{O}_X(D_1))\otimes \langle r\rangle) \simeq H^0(\mathcal{O}_{R}(-2K_X)).$$
If $s\in H^0(\mathcal{O}_X(-2K_X))$, we have a decomposition of $s$ as
$$s=r\alpha_0+\beta_0$$
with $\alpha_0\in H^0(\mathcal{O}_X(D_1))$ and $\beta_0\in V_0$. In particular, $\beta_0$ does not vanish identically on $R$ (it vanishes on $\gamma_R$ and some $\mathbb{P}^1$'s transversal to $\gamma_R$).
We can iterate this process by restricting $\alpha_0$ on $R$. As before, we have the following exact sequences, namely:
\begin{equation}
\label{EQ:EXS2}
\xymatrix@R=1pc@C=2pc{
0\ar@{->}[r] & H^0(\mathcal{O}_X(D_2))\ar@{^{(}->}[r]^-{-\otimes r} & H^0(\mathcal{O}_X(D_1)) \ar[r] & H^0(\mathcal{O}_R(D_1))\ar[r] & 0\\
0\ar@{->}[r] & H^0(\mathcal{O}_X(D_3))\ar@{^{(}->}[r]^-{-\otimes r} & H^0(\mathcal{O}_X(D_2)) \ar[r] & H^0(\mathcal{O}_R(D_2))\ar[r] & 0 \\
0\ar@{->}[r] & H^0(\mathcal{O}_X(D_4))\ar@{^{(}->}[r]^-{-\otimes r} & H^0(\mathcal{O}_X(D_3)) \ar[r] & H^0(\mathcal{O}_R(D_3))\ar[r] & 0,
}
\end{equation}
where the surjectivity follows as before by inspecting the dimension of
$$H^0(\mathcal{O}_R(D_i))=H^0(\mathcal{O}_{\mathbb{F}_n}((8-2i)\gamma_R+(2n+2+i)f_R)$$
and observing that it equals $h^0(\mathcal{O}_X(D_i))-h^0(\mathcal{O}_X(D_{i+1}))$.
Then we can create the vector spaces $V_i$ such that
$$V_i\oplus (H^0(\mathcal{O}_X(D_{i+1}))\otimes \langle r\rangle) \simeq H^0(\mathcal{O}_{R}(D_i))$$
and sections $\alpha_i\in H^0(\mathcal{O}_X(D_{i+1}))$, $\beta_i\in V_i$ such that
$$\alpha_i=r\alpha_{i+1}+\beta_{i+1}.$$
Finally, the section $s$ has the following form:
\begin{equation}
s=r^4\alpha_3+r^3\beta_3+r^2\beta_2+r\beta_1+\beta_0.
\end{equation}
Notice that $D_0|_R,D_1|_R$ and $D_2|_R$ are divisors with $\sigma_R$ as fixed components so $\beta_i$ for $i=0,1,2$ will vanish on it (with multiplicity greater than or equal to $4$). But the same is not true for $D_3|_R$, which is very ample. In particular, $\beta_3$ can be chosen such that $\beta_3|_R$ vanishes at exactly $5$ points of $\sigma_R$ (this is equal to $\sigma_R\cdot D_3|_R$) which are free on $\sigma_R$ and whose associated curve cut $\sigma_R$ transversely at such points.
In particular, the generic element of $|-2K_X|$ has $\sigma$ as base curve and the multiplicity of $\sigma$ along the generic bianticanonical divisor is $3$.
\end{proof}
\section{Blowing up the Projective Bundle}
\noindent In this section we will describe a resolution of a generic member of the linear system $|-2K_X|$.
\vspace{2mm}
\noindent Near a point $P$ of $\sigma$ we can choose local coordinates $(x,y,z)$ such that $x=y=0$ is the local equation of $\sigma$ near $P$, $x=0$ and $y=0$ are the local equations of $R$ and $U$ respectively and $z$ is a coordinate on $\sigma$. We can also use $(y,z)$ as local coordinates on $R$. We write, locally
$$s=x^3f+x^4g+x^2y^4f_1+xy^6f_2+y^8f_3,$$
where $f$ is the local expression for $\beta_3$ and $g$ is the local expression for $\alpha_3$.
We can blow up $\sigma$ in $X$ and take the strict transform $\tilde{D}$ of $D:=\{s=0\}$. Near $P$ the blow up $X_1$ looks like
$$\{(x,y,z)\times (l_0:l_1)\,|\, xl_1-yl_0=0 \}.$$
In the local chart $U_0=\{l_0\neq 0\}$ we have coordinates $(x,z,l_1)$ with $y=xl_1$ and the local equation for the exceptional divisor $E$ which is $x=0$. The total transform of $D$ has equation
$$x^3(f+xg+x^3l_1^4f_1+x^4l_1^6f_2+x^5l_1^8f_3),$$
so
$$\overline{s}=f+xg+x^3l_1^4f_1+x^4l_1^6f_2+x^5l_1^8f_3$$
is a local equazion for $\tilde{D}$. Notice that $f(0,y,z)$ is not identically zero because $f$ is the local expression of $\beta_3$. From the proof of Proposition \ref{PROP:MULT3} we have also that $\tilde{D}$ is smooth along $\sigma$ and hence everywhere (since it is the strict transform of something that has base locus $\sigma$). Unfortunately, $\tilde{D}$ is not a bianticanonical divisors on $X_1$: the bianticanonical class is indeed $\tilde{D}+E$ so we can take a bianticanonical divisor on $X_1$ to be the union of $\tilde{D}$ and $E$. This is reduced, reducible and singular exactly along the intersection $E_1\cdot \tilde{D}$.
\begin{lem} The divisor $E$ is a Segre-Hirzebruch variety $\mathbb{F}_{n+1}$.
\end{lem}
\begin{proof} The curve $\sigma$ is a complete intersection. More precisely, it is the intersection of the two rigid divisors $C_0$ and $\tau+A$. Thus, the normal bundle is $\mathcal{O}_{\sigma}(C_0) \oplus \mathcal{O}_{\sigma}(\tau +A)$. By direct computation, this is isomorphic to $\mathcal{O}_{\mathbb{P}^1}(-n) \oplus \mathcal{O}_{\mathbb{P}^1}(-2n-1)$, which proves the claim.
\end{proof}
The Picard group of $X_1$ is generated by $\tau$, $C_0$, $F$ and the exceptional divisor $E_1$. By construction, the restriction of $\tau$ to $E_1$ is zero. Moreover, the restriction of $C_0$ to the exceptional divisor is an integer multiple of $f_1=F|_{E_1}$, the class of a fiber of $E_1$ seen as Segre-Hirzebruch surface. This follows from the intersection numbers that are calculated in the next section. Therefore, the Picard group of the exceptional divisor is generated by the restriction of $E_1$ and $F$, respectively. It is not difficult to check that the unique divisor $\gamma_1$ on $E_1$ such that $\gamma_1^2=-n-1$ is given by
$$
\gamma_1=-{E_1}_{|{E_1}}-(2n+1)F_{|E_1}.
$$
The strict transform of the divisor $-2K_X$ is equal to $-2K_X-3E_1$. Its intersection with $E_1$ is given by $3\gamma_1 + 5f_1$. This is an effective divisor on $E_1$, which is made up of the unique curve of self-intersection $-n-1$ and $5$ disjoint fibers. Since we have
$$
-2K_{X_1}=-2K_X-2E_1= (-2K_X-3E_1)+E_1,
$$
the sections of the bianticanonical divisor $-2K_{X_1}$ pass through the curve $\gamma_1$, which is the complete intersection of $\tau+A-E_1$ (strict transform of the divisor $\tau+A$) and $E_1$.
\vspace{2mm}
\noindent Therefore, we blow up $X_1$ along the curve $\gamma_1$ and obtain a new variety $X_2$ with exceptional divisor $E_2$. To determine its structure, we compute the normal bundle of
$\gamma_1$ which is given by
$$
N_{\gamma_1/X_1}=\mathcal{O}_{\gamma_1}(E_1) \oplus \mathcal{O}_{\gamma_1}(\tau +A - E_1)\simeq \mathcal{O}_{\mathbb{P}^1}(-n-1) \oplus \mathcal{O}_{\mathbb{P}^1}(-n).
$$
Therefore, the exceptional divisor $E_2$ is isomorphic to $\mathbb{F}_1$. Let us denote by $\gamma_2$ and $f_2$ the generators of $E_2$ such that $\gamma_2^2=-1$, $\gamma_2f_2=1$ and $f_2^2=0$. As in the case of $E_1$, we can take $f_2$ to be the restriction of $F$ to $E_2$. As for the other divisor, it is easy to check that
\begin{equation}
\gamma_2= -{E_2}_{|E_2}-(n+1)F_{|E_2}.
\end{equation}
The bianticanonical divisor of $X_2$ is thus given by
$$
-2K_{X_2}=-2K_{X}-2E_1-2E_2= (-2K_X - 2E_1 - 4E_2)+2E_2.
$$
Let us compute the restriction of the divisor $(-2K_X - 2E_1 - 4E_2)$ to $E_2$. An easy calculation shows that it is equal to $4\rho_2+6f_2$, which corresponds to the class of a smooth irreducible curve on $E_2 \simeq \mathbb{F}_1$. Thus, there is a smooth member of the linear system
$$
2(-K_{X_2}-E_2)=-2K_X - 2E_1 - 4E_2.
$$
Being $2(-K_{X_2}-E_2)$ even, we can consider the cyclic covering $\beta:Y_2\rightarrow X_2$ of degree two with branch along a smooth member of $-2K_{X_2}-2E_2= 2K_X - 2E_1 - 4E_2$.
\begin{lem}
$Y_2$ is a smooth threefold and $\beta^*E_2$ is a $K3$ surface. Moreover
the pair $(Y_2,\beta^*E_2)$ is a log Calabi-Yau.
\end{lem}
\begin{proof}
$Y_2$ is clearly smooth as the branch divisor has been chosen to be smooth. Moreover, by \cite{BPHV} pag. 55, we have also
\begin{equation}
\label{EQ:KY2}
K_{Y_2}=\beta^*(K_{X_2}+B_2/2)=-\beta^*(E_2)
\end{equation}
so $(Y_2,\beta^*E_2)$ is a log Calabi-Yau. Notice that $\beta^*(E_2)$ is a degree two covering of the Segre-Hirzebruch surface $\mathbb{F}_1$ branched along the intersection of $E_2$ with the branch divisor of the covering $\beta:Y_2\rightarrow X_2$. We have already seen that this intersection can be written as the smooth curve $B_2=4\rho_2+6f_2$ on $E_2\simeq \mathbb{F}_1$, i.e. it is a smooth bianticanonical curve on $\mathbb{F}_1$. This is enough to conclude that the canonical divisor of $\beta^*E_2$ is trivial. The Euler characteristic is of $\beta^*E_2$ can be calculated as $2e(E_2) - e(R_2)$, where $R_2$ is the ramification divisor of the restriction of $\beta$ to $\beta^*(E_2)$. Since $\beta$ is a degree two covering, the divisor $R_2$ is isomorphic to the branch divisor $B_2$. This is a curve of genus $9$, so the Euler characteristic of $R_2$ is $-16$. We have hence $e(\beta^*E_2)=24$ so we can conclude that $\beta^*(E_2)$ is a $K3$ surface.
\end{proof}
In the next section, we are going to calculate the Euler characteristic of $Y_2$ for every $n \geq N_0$. To conclude this section, let us prove the following result.
\begin{thm}
Let $Y_2$ be as above. Then we have:
$$
h^{1,0}(Y_2)=0, \qquad h^{2,0}(Y_2)=0, \qquad h^{3,0}(Y_2)=0.
$$
Moreover, $Y_2$ has negative Kodaira dimension.
\end{thm}
\begin{proof}
We need to determine $h^{q,0}(Y_2)$ for $q\geq 1$. Recall that
$$\beta_*\mathcal{O}_{Y_2}\simeq \mathcal{O}_{X_2}\oplus\mathcal{O}_{X_2}(-B_2/2)=\mathcal{O}_{X_2}\oplus\mathcal{O}_{X_2}(K_{X_2}+E_2)$$
and that $R^q\beta_*\mathcal{F}=0$ for all $\mathcal{F}$ coherent on $Y_2$ and for all $q\geq 1$. Hence, by Leray spectral sequence, we have
$$
H^q(\mathcal{O}_{Y_2}) \simeq H^q(\mathcal{O}_{X_2}) \oplus H^q(\mathcal{O}_{X_2}(K_{X_2}+E_2)).
$$
$X_2$ is birational to $X$, which is a projective bundle over $\mathbb{F}_n$ so the Hodge numbers $h^{q,0}(X_2)=h^{q,0}(X)$ are zero for $q\geq 1$. Hence we need to prove that $h^q(\mathcal{O}_{X_2}(K_{X_2}+E_2))$ is zero for $q\geq 1$ in order to conclude the proof.
If $q=3$ this is straightforward: we have
\begin{equation}
\label{EQ:SOMECOH1}
h^3(\mathcal{O}_{X_2}(K_{X_2}+E_2))=h^0(\mathcal{O}_{X_2}(-E_2))=0
\end{equation}
because $E_2$ is effective. We have $h^p(\mathcal{O}_{X_2}(K_{X_2}))=h^{3-p}(\mathcal{O}_{X_2})=h^{3-p}(\mathcal{O}_{X})$ so,
\begin{equation}
\label{EQ:SOMECOH2}
h^1(\mathcal{O}_{X_2}(K_{X_2}))=h^2(\mathcal{O}_{X_2}(K_{X_2}))=0\qquad\mbox{ and }\qquad h^3(\mathcal{O}_{X_2}(K_{X_2}))=1.\end{equation}
To compute $H^q(K_{X_2}+E_2)$ for $q=1,2$, let us consider the exact sequence
$$
0 \rightarrow \mathcal{O}_{X_2}(K_{X_2}) \rightarrow \mathcal{O}_{X_2}(K_{X_2}+E_2) \rightarrow \mathcal{O}_{E_2}(K_{X_2}+E_2) \rightarrow 0,
$$
which yields, using also Equations \ref{EQ:SOMECOH1} and \ref{EQ:SOMECOH2}, the exact sequences
\begin{equation}
\label{EQ:EXSEQ1}
0 \rightarrow H^1(\mathcal{O}_{X_2}(K_{X_2}+E_2)) \rightarrow H^1(\mathcal{O}_{E_2}(K_{X_2}+E_2)) \rightarrow 0
\end{equation}\vspace{-12mm}
\begin{equation}
\label{EQ:EXSEQ2}
0 \rightarrow H^2(\mathcal{O}_{X_2}(K_{X_2}+E_2)) \rightarrow H^2(\mathcal{O}_{E_2}(K_{X_2}+E_2)) \rightarrow H^3(\mathcal{O}_{X_2}(K_{X_2}))\rightarrow 0
\end{equation}
By adjunction, $\mathcal{O}_{E_2}(K_{X_2}+E_2)$ is the canonical divisor of $K_{E_2}$ so $H^1(\mathcal{O}_{E_2}(K_{E_2}))=H^{1,2}(\mathbb{F}_1)=0$ (or, alternatively, by Lemma 2.9 of \cite{CMR}). Hence, from the exact sequence \ref{EQ:EXSEQ1}, also $H^1(X_2, \mathcal{O}_{X_2}(K_{X_2}+E_2))$ is zero.
\vspace{2mm}
\noindent Both the second and the third term of the exact sequence \ref{EQ:EXSEQ2} have dimension $1$ so $h^2(\mathcal{O}_{X_2}(K_{X_2}+E_2))=0$.
\vspace{2mm}
\noindent In order to see that the Kodaira dimension is $-\infty$, it is enough to observe that $-K_{Y_2}$ is effective and this follows from Equation \ref{EQ:KY2}.
\end{proof}
\section{The Euler Characteristic}
\noindent In this section, we will calculate the Chern numbers of $X_2$. Recall that $X=\mathbb{P}(\m{V})$ with $\m{V}=\mathcal{O}_S\oplus\mathcal{O}_S(-A)$ and $A=2C_0-F$. If $X_1=\Bl_{\sigma}X$, where $\sigma$ is the rational curve cut out by $R$ and $U$. If $E_1$ is the class of the exceptional divisor, we can consider the complete intersection curve cut out by the two divisors $\tau +A - E_1$ and $E_1$. As for the notation, denote by $E_2$ the exceptional divisor of the second blow up.
\vspace{2mm}
We will apply the following lemma:
\begin{lem}
\label{LEM:BLOW}
Let $Z$ be a smooth complex threefold and let
$$\xymatrix{
C \ar@{^{(}->}[r]^{j} & Z,
}$$
where $C$ is a smooth curve. If $Z'=\Bl_C(Z)$ with exceptional divisor $E$ and blow up map $\pi:Z'\rightarrow Z$. Then the following hold:
\begin{align}
\label{EQ:CHHBL}
c_1(Z')=&\pi^*c_1(Z)-E\\
c_2(Z')=&\pi^*(c_2(Z)-\eta_C)-\pi^*c_1(Z)E\\
H^*(Z')=&H^*(Z)\oplus H^*(E)/H^*(C),
\end{align}
where $\eta_C$ is the class of $C$ in $H^4(Z)$. Moreover, if $\alpha_p\in CH^p(Z)$ and $p+q=3$ with $q\geq 1$, then
\begin{equation}
\label{EQ:INBL}
E\cdot(\pi^*\alpha_2)=0\qquad E^2\cdot(\pi^*\alpha_1)=-j^*\alpha_1\qquad E^3=-c_1(N_{C/Z}).
\end{equation}
\end{lem}
\begin{proof}
The first two identities can be found in \cite{GH}, p. 609. Let $\alpha_p$ be a class in $CH^p(Z)$ and consider the following commutative diagram
$$\xymatrix{
E \ar@{^{(}->}[r]^{\iota}\ar[d]_\pi & Bl_C(Z) \ar[d]^{\pi} \\
C \ar@{^{(}->}[r]_{j}& Z
}$$
If we assume that $q\geq 1$ we can write $E^q=E^{q-1}\cdot E=E^{q-1}\iota_*(1)$ so that
\begin{multline*}
E^q\cdot (\pi^*\alpha_p)=(E^{q-1}\pi^*\alpha_p)\iota_*(1)=
\iota^*(E^{q-1}\pi^*\alpha_p)\cdot 1=\iota^*(E^{q-1})(\pi\circ \iota)^*\alpha_p=\\
=\iota^*(E^{q-1})(j\circ \pi)^*\alpha_p=\iota^*(E^{q-1})\pi^*(j^*\alpha_p)=\pi_*(\iota^*E)^{q-1}\cdot (j^*\alpha_p).
\end{multline*}
The restriction of the exceptional divisor to itself is the tautological class of $E$ when seen as the total space of the projective bundle $\mathbb{P}(N_{C/Z})\rightarrow C$. If we denote by $h=c_1(\mathcal{O}_{\mathbb{P}(N_{C/Z})}(1))$, we have $\iota^*(E)^{q-1}=(-1)^{q-1}h^{q-1}$. By definition we have also $\pi_*(h^{q-1})=s_{q-2}(N_{C/Z})$, where $s_{n}(N_{C/Z})$ is the Segre class of level $n$ of the vector bundle $N_{C/Z}$. To conclude, it is enough to observe that $s_1(N_{C/Z})=-c_1(N_{C/Z})$ and $s_0(N_{C/Z})=1$.
\end{proof}
\noindent Recall that
\begin{align}
c_1(X)=&2\tau + 4C_0 + (n + 1)F,\\
c_2(X)=&4\tau C_0 + (2n + 4)\tau F + (-2n + 6)C_0F,\\
c_3(X)=&8\tau C_0 F
\end{align}
and that $\sigma$, the center of the first blow up, is the complete intersection of $\tau+A$ and $C_0$. Hence
$$N_{\sigma/X}=\mathcal{O}_\sigma(\tau+A)\oplus\mathcal{O}_{\sigma}(C_0)$$
and the class $\eta_{\sigma}$ of $\sigma$ in $H^4(X)$ is simply the class of $(\tau+A)C_0$. In order to simplify notation, we will write $\alpha$ to indicate both a class in $X$ and its pullback to $X_1$ and $X_2$. The first Chern class of $X_1$ is simply given by $c_1(X)-E_1$ whereas
$$c_2(X_1)=c_2(X)-(\tau+A)C_0-c_1(X)E_1.$$
We are blowing up a smooth rational curve so $$E_1=\mathbb{P}(N_{\sigma/X})=\mathbb{P}(\mathcal{O}_\sigma(\tau+A)\oplus\mathcal{O}_{\sigma}(C_0))$$
is the Segre-Hirzebruch surface $\mathbb{F}_{n+1}$. By \eqref{EQ:CHHBL}, we obtain that the Hodge structure of $X_1$ is pure and $h^{1,1}(X_1)=4$. To recap, we have
\begin{align}
c_1(X_1)=&c_1(X)-E_1,\\
c_2(X_1)=&c_2(X)-(\tau+A)C_0-c_1(X)E_1,\\
c_3(X_1)=&10\tau C_0 F.
\end{align}
Moreover, the relations that characterize the intersection theory on $X_1$ are (here we don't report the ones coming from $X$) given by
$$E_1\tau =0\qquad E_1C_0F=0\qquad E_1^2C_0=n\qquad E_1^2F=-1\qquad E_1^3=3n+1.$$
The first relation follows simply by observing that $\tau$ and $\sigma$ are disjoint so $\tau$ and $E_1$ do not intersect. The others follow from Lemma \ref{LEM:BLOW} using
$$C_0 j^*\sigma=C_0^2(\tau+A)=-n\qquad F j^*\sigma=C_0(\tau+A)F=1.$$
and
$$c_1(N_{\sigma/X})=C_0^2(\tau+A)+C_0(\tau+A)^2=-(3n+1).$$
The curve $\gamma_1$ is smooth and rational. If we blow it up, we obtain an exceptional divisor $E_2$ that is isomorphic to $\mathbb{F}_1$. Indeed, the normal bundle of such a curve is isomorphic to $\mathcal{O}_{\mathbb{P}^1}(-n-1) \oplus \mathcal{O}_{\mathbb{P}^1}(-n)$. Using the same argument as before, we have
\begin{align}
c_1(X_2)=&c_1(X_1)-E_2,\\
c_2(X_2)=&c_2(X_1)-(\tau+A-E_1)E_1-c_1(X_1)E_2,\\
c_3(X_2)=&12\tau C_0 F.
\end{align}
Continuing as before we get
$$E_2\tau=0\qquad E_2C_0 F=0\qquad E_2E_1C_0=0\qquad E_2E_1 F=0\qquad E_2E_1^2=0$$
$$E_2^2C_0=n\qquad E_2^2F=-1\qquad E_2^2E_1 =n\qquad E_2^3=2n+1$$
This is all we need to prove the following theorem
\begin{thm}
For every positive integer $n$ big enough there exists a pair $(Y,D)$ such that
\begin{itemize}
\item $Y$ is a smooth threefold of negative Kodaira dimension with $$e(Y)-48n-46\qquad \mbox{ and }\qquad h^{q,0}(Y)=0 \ \mbox{ for } q \geq 1;$$
\item $D$ is a smooth K3 surface;
\item $(Y,D)$ is a log canonical log Calabi-Yau pair;
\end{itemize}
\end{thm}
\begin{proof}
Fix $n \geq N_0$ and consider the projective bundle $X=\mathbb{P}(\m{V})$ over $\mathbb{F}_n$, where $$\m{V} = \mathcal{O}_{\mathbb{F}_n} \oplus \mathcal{O}_{\mathbb{F}_n}(-2C_0+F).$$ First, blow up the projective bundle along the base locus of the bianticanonical divisor obtaining $X_1$. Next, blow up such a variety along the base locus of the bianticanonical divisor to obtain $X_2$. Take the degree two covering of $X_2$ with branch $B_2$ as described in the previous sections to finally obtain $Y_2$. Then one can take $Y=Y_2$ and $D=\beta^*E_2$. Everything, apart form the calculation for the Euler characteristics, have been done in the previous sections.
\vspace{2mm}
\noindent In order to compute the Euler characteristic, recall that if $D$ is a smooth irreducible divisor on $X_2$, we have
\begin{equation}
c_2(D)=c_2(X_2)-c_1(X_2)D+D^2
\end{equation}
so
\begin{equation}
\e(D)=(c_2(X_2)-c_1(X_2)D+D^2)N_{D/X_2}.
\end{equation}
Hence, being the branch locus $B_2$ a smooth element of $|-2K_{X_2}-2E_2|$, we have that
$$e(B_2)=48n + 70.$$
The Euler number of $X_2$ is given by $12$ so we have, finally, \begin{equation}
\e(Y_2)=2\e(X_2)-\e(B_2)=2\cdot 12-(48n + 70)=-48n - 46.
\end{equation}
Although feasible by hands, we have done the last computation using \verb|Magma|\footnote{http://magma.maths.usyd.edu.au/}.
\end{proof}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 6,373
|
This is a placeholder page for Nancy Binder, which means this person is not currently on this site. We do suggest using the tools below to find Nancy Binder.
You are visiting the placeholder page for Nancy Binder. This page is here because someone used our placeholder utility to look for Nancy Binder. We created this page automatically in hopes Nancy Binder would find it. If you are not Nancy Binder, but are an alumni of Quincy High School Quincy, IL, register on this site for free now.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 2,478
|
package skyhussars.terrained;
import org.slf4j.Logger;
import org.slf4j.LoggerFactory;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Component;
import skyhussars.engine.terrain.TheatreLoader;
import skyhussars.persistence.terrain.TerrainDescriptor;
/**
* State manager for the Theatre editor.
*
* This class is the main interface between the UI and the underlying functionalities.
*
*/
@Component
public class TerrainEdService {
private static final Logger LOGGER = LoggerFactory.getLogger(TerrainEdController.class);
private final TerrainProperties terrainProperties;
private final TheatreLoader loader;
@Autowired
public TerrainEdService(TerrainProperties terrainProperties,TheatreLoader loader){
this.terrainProperties = terrainProperties;
this.loader = loader;
}
public boolean saveToFile(){
boolean success = false;
if (canSave()) {
loader.saveTheatre(terrainProperties.asDescriptor());
success = true;
}
return success;
}
public boolean delete(){
boolean success = loader.deleteTheatre(terrainProperties.asDescriptor());
if(success) terrainProperties.clear();
return success;
}
private boolean canSave() {
return terrainProperties.name.get() != null
&& terrainProperties.size.get() >= 1
&& terrainProperties.location.get() != null;
}
public TerrainProperties getTerrainProperties() {
return terrainProperties;
}
public TerrainEdService newTheatre(String name){
TerrainDescriptor terrain = loader.createTheatre(name);
terrainProperties.from(terrain);
return this;
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 6,834
|
\section{Introduction}\label{section:intro}
\subsection{Background}
This paper concerns the analysis of singular solutions to semilinear elliptic systems with power-law nonlinearity of type
\begin{equation}\label{eq:main}
-\Delta {\mathbf u}=|{\mathbf u}|^{\alpha-1} {\mathbf u},
\end{equation}
where $1<\alpha\leq\frac{n+2}{n-2}$, and ${\mathbf u} = (u_1,\dots,u_m)$, $m\geq 1$, is a $C^2$ vector-valued function defined on a domain in ${\mathbb R}^n$, $n\geq 3$. Our primary interest is in the case when each component of ${\mathbf u}$ is nonnegative and the domain is of the form $B_R\setminus \{0\}$, with $B_R$ being the ball of radius $R$ centered at the origin. It is by now well known that in cylindrical coordinates $t= -\log |x| \in {\mathbb R}$ and $\theta = \frac{x}{|x|}\in{\mathbb S}^{n-1}$, the transformation
\begin{equation}\label{eq:cyl}
{\mathbf u}(x) = |x|^{-\frac{2}{\alpha-1}} {\mathbf v}\left(-\log |x|,\frac{x}{|x|}\right),
\end{equation}
yields the system
\begin{equation}\label{eq:main-cyl}
\partial_{tt}{\mathbf v} + \mu \partial_t{\mathbf v} + \Delta_\theta {\mathbf v} - \lambda {\mathbf v} + |{\mathbf v}|^{\alpha-1}{\mathbf v} = 0,
\end{equation}
in $(-\log R,\infty)\times{\mathbb S}^{n-1}$, and vice versa, where $\Delta_\theta$ is the Laplace-Beltrami operator on ${\mathbb S}^{n-1}$ and $\lambda$ and $\mu$ are the constants fixed throughout this paper by
\begin{equation}\label{eq:lam-mu}
\lambda = \frac{2}{\alpha-1} \left( n - 2 - \frac{2}{\alpha-1} \right),\quad \mu = \frac{4}{\alpha-1} - n + 2.
\end{equation}
The scalar case of this system was introduced by Lane \cite{Lane-1870} and later studied by Emden \cite{Emden-1907} for describing distribution of mass densities in spherical polytropic star in hydrostatic equilibrium. Since its birth, this equation has been used in many applications such as astrophysics, kinetic theory, and quantum mechanics (see \cite{Goenner-2000}). The Lane-Emden equation has thus been subject to intensive studies in the last few decades and nowadays there is a vast amount of literature treating many aspects of the solutions to this equation and its diverse varieties.
One of the central questions\footnote{To the best of our knowledge there are three central questions in this area. The other two questions refer to the structure of singular sets (see \cite{Pacard-1993}), and non-existence theory (see \cite{Grigoryan}, \cite{Souplet-2009}). } and a technically difficult problem for differential equations and systems is the study of the singular solutions, that is, solutions that develop singularities. In the scalar case, the classical and subsequent works have considered the asymptotic behavior of the solutions close to isolated singularities, with an accurate description of the asymptotic behavior of solutions around such singular points; see e.g., \cite{A2, BVV, CL2, CGS, KMPS,V} and the references therein.
The system \eqref{eq:main} can be considered as a generalization of the Lane-Emden equation, and can also be viewed as a strongly coupled system of nonlinear Schr\"odinger equations (or more precisely the limiting system of the associated blowup solutions). In the latter point of view, there has been some development regarding classification of the global solutions, and compactness of the blowup sequence; see for instance \cite{CL, DHV} and the references therein. In the former point of view, there are many other types of generalizations, among which the Lane-Emden-Fowler systems have received considerable attention. Among possible references, we refer to \cite{BVR, BG, BVG, BM, DFF, PQS, SZ1996} for the classification of global solutions, non-existence theory of singular, positive solutions and local estimates of solutions to the Lane-Emden-Fowler systems. We refer to \cite{RZ,Z} for more general cooperative elliptic systems, and the references therein. One may also consult to the monograph \cite{DF} for a general theory regarding semilinear elliptic systems. To the best of the authors' knowledge, this is the first paper that conducts a thorough analysis on the qualitative behavior of the system \eqref{eq:main}, particularly regarding the classification of the solutions on the punctured space ${\mathbb R}^n\setminus\{0\}$ with respect to the balanced-energy-type functionals (subcritical case $1<\alpha<\frac{n+2}{n-2}$) and the Pohozaev identities (critical case $\alpha = \frac{n+2}{n-2}$), as well as the asymptotic behavior of local solutions around the isolated singularities.
The key difference between the system \eqref{eq:main} and its scalar version is, of course, the multiplicity of the components. The major observation in this paper is that the system \eqref{eq:main} turns out to be very sensitive to the settimg of multiple components in the case of the upper critical exponent (that is, $\alpha = \frac{n+2}{n-2}$) and lower critical exponent (that is, $\alpha = \frac{n}{n-2}$). Specifically, in the upper critical case $\alpha = \frac{n+2}{n-2}$, we discover a new Pohozaev invariant other than the usual one. The lower critical case is rather technical and we shall present the discussion on this issue in Section \ref{subsection:asym0}.
Let us briefly illustrate how the new Pohozaev invariant comes into play in the analysis of the system \eqref{eq:main} in the upper critical case. For the sake of clarity, let us assume that the solution ${\mathbf u}$ is rotationally symmetric, so that the cylindrical transformation ${\mathbf v}$ is a function of $t$ only. After some manipulation, one can obtain the usual Pohozaev identity,
\begin{equation}\label{eq:brief}
\left| \frac{d{\mathbf v}}{dt} \right|^2 = \frac{(n-2)^2}{4} |{\mathbf v}|^2 -\frac{n-2}{n} |{\mathbf v}|^{\frac{2n}{n-2}} + \kappa,
\end{equation}
for the system \eqref{eq:main-cyl}, with a constant $\kappa$, also known as the usual Pohozaev invariant. Due to the presence of the multiple components, we have
\begin{equation}\label{eq:2nd-poho-1}
\left| \frac{d{\mathbf v}}{dt} \right|^2 - \left( \frac{d|{\mathbf v}|}{dt} \right)^2 = \frac{1}{|{\mathbf v}|^2} \sum_{1\leq i<j\leq m} \left( v_i\frac{dv_j}{dt} - v_j \frac{dv_i}{dt} \right)^2 \geq 0,
\end{equation}
and the equality on the rightmost side does not hold in general. This shows that $\kappa$ alone is insufficient to analyze the behavior of $|{\mathbf v}|$ completely, due to the discrepancy \eqref{eq:2nd-poho-1} between $|d{\mathbf v}/dt|$ and $|d|{\mathbf v}|/dt|$. In this paper, we find that there is another constant $\kappa_*$ such that
\begin{equation}\label{eq:|v|-poho}
\left( \frac{d |{\mathbf v}|}{dt} \right)^2 = \frac{(n-2)^2}{4} |{\mathbf v}|^2 -\frac{n-2}{n} |{\mathbf v}|^{\frac{2n}{n-2}} + \kappa + \frac{\kappa_*}{|{\mathbf v}|^2},
\end{equation}
and we shall call this constant the new Pohozaev invariant.
Thanks to an anonymous referee, we also observe a more precise characterization of the new invariant. Multiplying by $v_i$ and $-v_j$ in the $j$-th and respectively in the $i$-th component of the system \eqref{eq:main-cyl} (with $\alpha = \frac{n+2}{n-2}$), and then adding the resulting equations together side by side, we deduce that
\begin{equation*}
\frac{d }{dt} \left( v_i\frac{dv_j}{dt} - v_j \frac{dv_i}{dt} \right) = 0,\quad 1\leq i,j\leq m.
\end{equation*}
Thus, there exists a constant $k_{ij}$ such that for each $1\leq i,j\leq m$ we have
\begin{equation}\label{eq:2nd-poho-2}
v_i\frac{dv_j}{dt} - v_j \frac{dv_i}{dt} = k_{ij}.
\end{equation}
Inserting \eqref{eq:2nd-poho-2} into \eqref{eq:2nd-poho-1} and comparing it with \eqref{eq:|v|-poho}, we find that
\begin{equation}\label{eq:2nd-poho}
\kappa_* = - \sum_{1\leq i<j\leq m} k_{ij}^2.
\end{equation}
Due to such a precise characterization, we also find an explicit solution featuring $\kappa_* \neq 0$, at least in the two-particle systems (i.e., $m=2$); see Remark \ref{remark:Phi*-const3}. Without the radial symmetry, we obtain a more general formula \eqref{eq:kappa*} for the new Pohozaev invariant.
We point out that the analysis of the behavior of solutions to system \eqref{eq:main} involve both $\kappa$ and $\kappa_*$. As surprising as it may sound, one can construct radially symmetric solutions to the two-particle system (i.e., $m=2$) having non-removable singularity, even if the associated standard Pohozaev invariant $\kappa$ is zero, see Remark \ref{remark:Phi*-const3} for full details. This is a significant difference from the case of scalar equation, where $\kappa$ fully determines the behavior of the solution around the isolated singularity, and especially $\kappa = 0$ is a sufficient and necessary condition to have removable singularity.
Classification of solutions in higher dimensional systems (i.e., $m\geq 3$) is of independent interests and would lead to a more complete picture of this problem. Nonetheless, it is beyond the scope of this article, and we shall not push further towards this direction here.
On the technical level, the system \eqref{eq:main} exhibits some subtleties compared to the scalar case. One of the main tools we employ in the study of \eqref{eq:main} is the method of moving spheres, which has been considered in \cite{JLX, LZ} and then continuously developed especially in the frame of the fractional Laplace operator (see, e.g., \cite{JLX2, CJSX}). The use of such a method in the case of systems requires particular attention, since the procedure can be continued in some components but should stop in others.
Another technical tool is the balanced-energy-type monotonicity functional (see e.g., \eqref{eq:Phi} below), which yields the Pohozaev identity in the upper critical case $\alpha = \frac{n+2}{n-2}$, combined with the blowup analysis. Such energy functional has been a classical tool for the study of scalar case (see, e.g., \cite{BVV, A2, KMPS} and many others). We believe that the argument presented in this paper regarding the energy functional is more effective, due to an easy observation on the scaling relation \eqref{eq:Phi-scale} that is standard in the framework of free boundary problems.
\subsection{Main Results}\label{subsection:result}
The main results are as follows. First we classify the solutions on the entire space, via the method of moving spheres.
\begin{theorem}\label{theorem:main-g} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in ${\mathbb R}^n$ with $1<\alpha\leq\frac{n+2}{n-2}$.
\begin{enumerate}[(i)]
\item If $1<\alpha<\frac{n+2}{n-2}$, ${\mathbf u}$ is trivial.
\item If $\alpha = \frac{n+2}{n-2}$, then ${\mathbf u}$ is of the form
\begin{equation}\label{eq:u-g}
{\mathbf u} (x) = (n(n-2))^{\frac{n-2}{4}} \left( \frac{r}{r^2 + |x-z|^2}\right)^{\frac{n-2}{2}} {\mathbf e},
\end{equation}
for some $z\in{\mathbb R}^n$, $r\geq 0$ and a unit nonnegative vector ${\mathbf e}\in{\mathbb R}^m$.
\end{enumerate}
\end{theorem}
\begin{remark}\label{remark:main-g} Theorem \ref{theorem:main-g} (ii) was already proved by O. Druet, E. Hebey and J. V\'etois \cite [Proposition 1.1]{DHV} via the method of moving spheres. Here we contain the result and the proof for the reader's convenience.
\end{remark}
Next we classify the solutions in the punctured space, through the limiting energy levels or the Pohozaev invariants of the associated energy functional and the blowup analysis, which is standard in the framework of free boundary problems. For the upper critical case $\alpha = \frac{n+2}{n-2}$, we introduce a new Pohozaev invariant, which will play the central role.
\begin{theorem}\label{theorem:main-s} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in ${\mathbb R}^n\setminus\{0\}$ with $1< \alpha \leq\frac{n+2}{n-2}$, and let $\Phi(r,{\mathbf u})$ be as in \eqref{eq:Phi} for all $r>0$.
\begin{enumerate}[(i)]
\item If $1<\alpha\leq\frac{n}{n-2}$, then ${\mathbf u}$ is trivial.
\item If $\frac{n}{n-2} < \alpha < \frac{n+2}{n-2}$, then $\Phi(r,{\mathbf u})$ converges as $r\rightarrow 0$ and $r\rightarrow\infty$, and
\begin{equation}\label{eq:class-subcrit-Phi}
\{\Phi(0+,{\mathbf u}), \Phi(+\infty,{\mathbf u})\} \subset\left\{-\frac{\alpha-1}{\alpha+1}\lambda^{\frac{\alpha+1}{\alpha-1}},0\right\}.
\end{equation}
\begin{enumerate}[(a)]
\item $\Phi(0+,{\mathbf u}) = 0$, if and only if ${\mathbf u}$ is trivial.
\item $\Phi(+\infty,{\mathbf u}) = -\frac{\alpha-1}{\alpha+1}\lambda^{\frac{\alpha+1}{\alpha-1}}$, if and only if ${\mathbf u}$ is homogeneous of degree $-\frac{2}{\alpha-1}$, hence of the form
\begin{equation}\label{eq:u-s}
{\mathbf u}(x) = \lambda^{\frac{1}{\alpha-1}} |x|^{-\frac{2}{\alpha-1}} {\mathbf e},
\end{equation}
where $\lambda$ is given by \eqref{eq:lam-mu} and ${\mathbf e}\in{\mathbb R}^m$ is a unit nonnegative vector.
\end{enumerate}
\item If $\alpha = \frac{n+2}{n-2}$, then $\Phi_*(r,{\mathbf u})$ as in \eqref{eq:Phi*} is well-defined for all $r>0$, and there are constants $\kappa({\mathbf u})$ and $\kappa_*({\mathbf u})$ such that $\kappa({\mathbf u}) = \Phi(r,{\mathbf u})$ and $\kappa_*({\mathbf u}) = \Phi_*(r,{\mathbf u})$ for all $r>0$. Moreover,
\begin{equation}\label{eq:class-crit-kappa}
\kappa({\mathbf u}) \geq -\frac{2}{n} \left(\frac{n-2}{2}\right)^n,
\end{equation}
and
\begin{equation}\label{eq:class-crit-kappa*}
-\left(\frac{2}{n}\left(\frac{n-2}{2}\right)^n + \kappa({\mathbf u})\right) \left(\frac{n-2}{2}\right)^{n-2}\leq \kappa_*({\mathbf u}) \leq 0,
\end{equation}
where the equalities of the lower bounds of both $\kappa({\mathbf u})$ and $\kappa_*({\mathbf u})$ hold only simultaneously.
\begin{enumerate}[(a)]
\item $\kappa({\mathbf u}) = \kappa_*({\mathbf u}) = 0$ if and only if ${\mathbf u}$ has removable singularity at the origin, hence of the form \eqref{eq:u-g}.
\item If $\kappa({\mathbf u})^2 + \kappa_*({\mathbf u})^2 > 0$, then ${\mathbf u}$ has non-removable singularity at the origin, and is rotationally symmetric. Moreover, the cylindrical transformation ${\mathbf v}$ as in \eqref{eq:cyl} satisfies \eqref{eq:|v|-poho}.
\item $\kappa({\mathbf u})= -\frac{2}{n} (\frac{n-2}{2})^n$ and $\kappa_*({\mathbf u}) = 0$ if and only if ${\mathbf u}$ is homogeneous of degree $-\frac{n-2}{2}$, hence of the form
\begin{equation}\label{eq:u-s-2}
{\mathbf u}(x) = \left( \frac{n-2}{2} \right)^{\frac{n-2}{2}} |x|^{-\frac{n-2}{2}} {\mathbf e},
\end{equation}
where ${\mathbf e}$ is a unit nonnegative vector.
\end{enumerate}
\end{enumerate}
\end{theorem}
\begin{remark}\label{remark:Phi*-const3}
Due to the characterization \eqref{eq:2nd-poho} of the second Pohozaev invariant, one can find a rotationally symmetric solution to \eqref{eq:main-cyl} for which $\kappa_* \neq 0$, even for the case $m=2$. The following example was raised to us by an anonymous referee. Given a pair $(\kappa,\kappa_*)$ of admissible constants, satisfying the bounds \eqref{eq:class-crit-kappa} and \eqref{eq:class-crit-kappa*}, such that the equation
\begin{equation}\label{eq:Phi*-const3-1}
\frac{(n-2)^2}{4} \rho^2 - \frac{n-2}{n} \rho^{\frac{2n}{n-2}} + \kappa + \frac{\kappa_*}{\rho^2} = 0
\end{equation}
has two distinct positive roots. Following the classical work \cite{F} by R.H. Fowler, one obtains a positive, non-constant, periodic solution $\rho$ to
\begin{equation*}
\left(\frac{d\rho}{dt}\right)^2 = \frac{(n-2)^2}{4} \rho^2 - \frac{n-2}{n} \rho^{\frac{2n}{n-2}} + \kappa + \frac{\kappa_*}{\rho^2}.
\end{equation*}
Then one can easily verify that the mapping
\begin{equation}\label{eq:Phi*-const3-ex}
{\mathbf v} (t) = \rho(t) \left( \cos \left( \sqrt{-\kappa_*} \int_{t_0}^t \frac{ds}{\rho^2(s)}\right), \sin \left( \sqrt{-\kappa_*} \int_{t_0}^t \frac{ds}{\rho^2(s)}\right) \right)
\end{equation}
solves the system \eqref{eq:main-cyl} with $\alpha = \frac{n+2}{n-2}$, $n\geq 3$ and $m = 2$, having $\kappa$ and $\kappa_*$ as the first and respectively the second Pohozaev invariant.
Note that $\rho$ oscillates between two positive values, which implies that the corresponding solution ${\mathbf u}$ under the reverse cylindrical transformation \eqref{eq:cyl} has non-removable singularity at the origin. It should be stressted out that one can also have two distinct positive roots to \eqref{eq:Phi*-const3-1} even when $\kappa=0$, provided that $\kappa_*<0$ with $|\kappa_*|$ small. This shows the existence of singular solutions having trivial, standard Pohozaev invariant.
\end{remark}
The subsequent theorems are concerned with the local solutions in the punctured unit ball. First we deduce the asymptotic radial symmetry, by the combination of the method of moving spheres and moving plane; a similar argument appears in \cite[Theorem 1.2]{CJSX}. This result is particularly important to define the second Pohozaev invariant for local solutions.
\begin{theorem}\label{theorem:asym-rad} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in $B_1\setminus\{0\}$ with $1<\alpha\leq\frac{n+2}{n-2}$. Then
\begin{equation}\label{eq:asym-rad}
{\mathbf u}(x) = (1+O(|x|)) \bar{\mathbf u}(|x|)\quad\text{as }x\rightarrow 0,
\end{equation}
where $\bar{\mathbf u}(r)$ is the average of ${\mathbf u}$ over $\partial B_r$.
\end{theorem}
Utilizing the classification of solutions in the punctured space and the asymptotic radial symmetry, we obtain the exact asymptotic behavior of local solutions around the singularity.
\begin{theorem}\label{theorem:main} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in $B_1\setminus\{0\}$ with $1<\alpha\leq \frac{n+2}{n-2}$. Then either ${\mathbf u}$ has a removable singularity at the origin, or the following alternatives hold.
\begin{enumerate}[(i)]
\item If $\frac{n}{n-2}<\alpha <\frac{n+2}{n-2}$, then
\begin{equation}\label{eq:asym+sub}
|{\mathbf u}(x)| = (1 + o(1)) \lambda^{\frac{1}{\alpha-1}} |x|^{-\frac{2}{\alpha-1}} \quad\text{as }x\rightarrow 0,
\end{equation}
where $\lambda$ is given as in \eqref{eq:lam-mu}.
\item If $\alpha= \frac{n+2}{n-2}$, then there are $c,C>0$ such that
\begin{equation}\label{eq:asym+crit}
c|x|^{-\frac{n-2}{2}} \leq |{\mathbf u}(x)| \leq C|x|^{-\frac{n-2}{2}} \quad\text{as }x\rightarrow 0,
\end{equation}
where $c$ depends on ${\mathbf u}$ while $C$ is determined by $n$ and $m$ only.
\item If $1<\alpha<\frac{n}{n-2}$, then there are $c,C>0$ such that
\begin{equation}\label{eq:asym-}
c|x|^{2-n} \leq |{\mathbf u}(x)| \leq C |x|^{2-n}\quad\text{as } x\rightarrow 0,
\end{equation}
where both $c$ and $C$ depend on ${\mathbf u}$.
\item If $\alpha = \frac{n}{n-2}$, then
\begin{equation}\label{eq:asym0}
|{\mathbf u}(x)| = (1+o(1)) \left(\frac{(n-2)^2}{2|x|^2(-\log |x|)}\right)^{\frac{n-2}{2}}\quad\text{as }x\rightarrow 0.
\end{equation}
\end{enumerate}
\end{theorem}
The paper is organized as follows. In the next section, we present the balanced-energy-type monotonicity formula and introduce the second Pohozaev invariants for the upper critical case. In Section \ref{section:global}, we classify the solutions of \eqref{eq:main} on the whole space, proving Theorem \ref{theorem:main-g}. In Section \ref{section:sing}, we investigate the properties of the solutions on the punctured space, and present the proof of Theorem \ref{theorem:main-s}. Section \ref{section:upper} is devoted to the {\it a priori} estimates for the local solutions, which will play one of the key roles in the subsequent analysis, while we prove the asymptotic radial symmetry, Theorem \ref{theorem:asym-rad}, in Section \ref{section:asym-rad}. Finally, we derive the exact asymptotic behavior of the local solutions of \eqref{eq:main} for all $1<\alpha\leq\frac{n+2}{n-2}$, in Section \ref{section:asym}. The proof of parts (i)-(iv) in Theorem \ref{theorem:main} are presented in the end of Section \ref{subsection:asym+sub}-\ref{subsection:asym0}, respectively.
\subsection{Notation and Terminology}\label{subsection:notation}
We say that ${\mathbf u}$ has a removable singularity at the origin, provided that $|{\mathbf u}|$ is bounded in any neighborhood of origin. We say that ${\mathbf u}$ has a non-removable singularity at the origin, if ${\mathbf u}$ does not have a removable singularity at the origin, that is, $|{\mathbf u}|$ (but not necessarily all the components of ${\mathbf u}$) is unbounded in any neighborhood of the origin.
By $B_r(z)\subset{\mathbb R}^n$ ($n\geq 3$) we denote the ball of radius $r$ centered at $z$, and in case $z=0$, we shall simply write it by $B_r$. In addition, $\omega_n$ is the volume of the unit ball $B_1\subset{\mathbb R}^n$. Given an open set $\Omega\subset{\mathbb R}^n$, we shall denote by $\partial \Omega$ the topological boundary of $\Omega$. Moreover, when $\partial \Omega$ is $C^1$, $\nu$ denotes the unit normal on $\partial \Omega$ pointing towards the origin. $\nabla_\sigma$ will denote the tangential derivative on $\partial\Omega$.
${\mathbb S}^{n-1}$ is the unit sphere in ${\mathbb R}^n$, and is also identified with $\partial B_1$. Note that $n\omega_n$ is the area of ${\mathbb S}^{n-1}$. By $\nabla_\theta$ and $\Delta_\theta$ we shall write the derivative and, respectively, the Laplace-Beltrami operator on ${\mathbb S}^{n-1}$.
Any vector in the target space ${\mathbb R}^m$ ($m\geq 1$) is written in bold. Given a vector ${\mathbf a}\in{\mathbb R}^m$, we denote by $a_i$ the $i$-th component of ${\mathbf a}$. By $|{\mathbf a}|$ we denote its $l^2$-norm, i.e., $|{\mathbf a}| = (\sum_{i=1}^m a_i^2)^{1/2}$. By ${\mathbf a}\geq 0$ (resp., ${\mathbf a} \leq 0$) or by saying that ${\mathbf a}$ is nonnegative (resp., nonpositive) we indicate that $a_i \geq 0$ (resp., $a_i \leq 0$) for each $1\leq i\leq m$. For two vectors ${\mathbf a}$ and ${\mathbf b}$, ${\mathbf a}\cdot{\mathbf b} = \sum_{i=1}^m a_ib_i$. Also given two vectorial $C^1$-functions ${\mathbf f}$ and ${\mathbf g}$, $\nabla{\mathbf f}: \nabla{\mathbf g} = \sum_{i=1}^m (\nabla f_i)\cdot (\nabla g_i)$.
The constants $C,C_0,C_1,C_2,\cdots$ will always be positive, generic, determined by $n$, $m$ and $\alpha$ only, unless otherwise stated. We shall also call these constants universal. In addition, we shall fix $\lambda$, $\mu$ and $\bar\lambda$ throughout the paper as in \eqref{eq:lam-mu} and \begin{equation}\label{eq:lamb}
\bar\lambda = \frac{\alpha-1}{\alpha+1}\lambda^{\frac{\alpha+1}{\alpha-1}}.
\end{equation}
\section{Monotonicity Formula and Pohozaev Invariant}\label{section:energy}
We consider the balanced-energy-type functional
\begin{equation}\label{eq:Phi}
\begin{split}
\Phi(r,{\mathbf u}) &= \frac{r^{\mu+1}}{n\omega_n} \int_{\partial B_r} \left( \left| \frac{\partial {\mathbf u}}{\partial \nu} - \frac{2}{(\alpha-1)r} {\mathbf u} \right|^2 - |\nabla_\sigma {\mathbf u}|^2 \right) d\sigma \\
&\quad + \frac{2r^{\mu+1}}{(\alpha+1)n\omega_n}\int_{\partial B_r} |{\mathbf u}|^{\alpha+1} \,d\sigma - \frac{\lambda r^{\mu-1}}{n\omega_n} \int_{\partial B_r} |{\mathbf u}|^2 d\sigma,
\end{split}
\end{equation}
where $\lambda$ and $\mu$ are given as in \eqref{eq:lam-mu}. Note that $\lambda \geq 0$ if and only if $\alpha \geq \frac{n}{n-2}$, and $\mu \geq 0$ if and only if $1<\alpha \leq \frac{n+2}{n-2}$.
Let us introduce the scaling function
\begin{equation}\label{eq:ur}
{\mathbf u}_r(x) = r^{\frac{2}{\alpha-1}} {\mathbf u}(rx).
\end{equation}
Note that the problem \eqref{eq:main} is preserved under this scaling. That is, if ${\mathbf u}$ solves \eqref{eq:main} in $B_R\setminus\{0\}$ then ${\mathbf u}_r$ solves \eqref{eq:main} in $B_{R/r}\setminus\{0\}$. In terms of ${\mathbf u}_r$, one may easily observe that $\Phi$ satisfies the following scaling relation
\begin{equation}\label{eq:Phi-scale}
\Phi(rs,{\mathbf u}) = \Phi(s,{\mathbf u}_r),
\end{equation}
for any $r,s>0$.
Recall from \eqref{eq:cyl} the cylindrical transformation ${\mathbf v}$, in terms of which $\Phi$ can be represented as
\begin{equation}\label{eq:Phi-cyl}
\Phi(r,{\mathbf u}) = \Psi(-\log r,{\mathbf v}),
\end{equation}
where $\Psi(t,{\mathbf v})$ is given by
\begin{equation}\label{eq:Psi}
\Psi(t,{\mathbf v}) = \frac{1}{n\omega_n}\int_{{\mathbb S}^{n-1}} \left(|\partial_t{\mathbf v}|^2 - |\nabla_\theta {\mathbf v}|^2 - \lambda |{\mathbf v}|^2 + \frac{2}{\alpha+1} |{\mathbf v}|^{\alpha+1} \right) \,d\theta.
\end{equation}
\begin{proposition}\label{proposition:Phi-monot} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in $B_R\setminus\{0\}$ with $1<\alpha\leq\frac{n+2}{n-2}$, and let $\Phi(r,{\mathbf u})$ be as in \eqref{eq:Phi}. One has
\begin{equation}\label{eq:Phi'}
\frac{d}{dr}\Phi(r,{\mathbf u}) = \frac{2\mu r^\mu}{n\omega_n} \int_{\partial B_r} \left| \frac{\partial {\mathbf u}}{\partial \nu} - \frac{2}{(\alpha-1)r} {\mathbf u} \right|^2 \,d\sigma,
\end{equation}
where $\mu$ is given as in \eqref{eq:lam-mu}. In particular, the following are true.
\begin{enumerate}[(i)]
\item If $1<\alpha<\frac{n+2}{n-2}$, then $\Phi(r,{\mathbf u})$ is nondecreasing for $0<r<R$. Moreover, $\Phi(r,{\mathbf u})$ is constant for $r_1< r< r_2$, if and only if ${\mathbf u}$ is homogeneous of degree $-\frac{2}{\alpha-1}$ in $B_{r_2}\setminus \bar{B}_{r_1}$, i.e.,
\begin{equation}\label{eq:u-hom}
{\mathbf u}(x) = |x|^{-\frac{2}{\alpha-1}} {\mathbf u}\left( \frac{x}{|x|}\right)\quad\text{in }B_{r_2}\setminus \bar{B}_{r_1}.
\end{equation}
\item If $\alpha = \frac{n+2}{n-2}$, then $\Phi(r,{\mathbf u})$ is constant for $0<r<R$.
\end{enumerate}
\end{proposition}
\begin{proof} The computation is easy if one chooses the cylindrical coordinate. Since \eqref{eq:Phi-cyl} holds with $t=-\log r$, we have
\begin{equation*}
\begin{split}
r \dot\Phi(r,{\mathbf u}) &= - \Psi'(t,{\mathbf v}) \\
&= -\frac{2}{n\omega_n}\int_{{\mathbb S}^{n-1}} (( \partial_{tt}{\mathbf v} - \lambda{\mathbf v} + |{\mathbf v}|^{\alpha-1}{\mathbf v})\cdot \partial_t{\mathbf v} - \nabla _\theta {\mathbf v}: \nabla_\theta\partial_t{\mathbf v} )\,d\theta\\
&= -\frac{2}{n\omega_n}\int_{{\mathbb S}^{n-1}} (\partial_{tt}{\mathbf v} + \Delta_\theta{\mathbf v} -\lambda{\mathbf v} + |{\mathbf v}|^{\alpha-1}{\mathbf v})\cdot\partial_t{\mathbf v} \,d\theta\\
&= \frac{2\mu}{n\omega_n} \int_{{\mathbb S}^{n-1}} |\partial_t{\mathbf v}|^2\,d\theta,
\end{split}
\end{equation*}
where $\dot\Phi$ and $\Psi'$ denote $d\Phi/dr$ and respectively $d\Psi/dt$, and the right side is evaluated at $t=-\log r$. In addition, when deriving the last equality we used \eqref{eq:main-cyl}. Rephrasing the rightmost side in terms of ${\mathbf u}$, we arrive at \eqref{eq:Phi'}.
The assertion on the monotonicity of $\Phi$ is now clear from \eqref{eq:Phi'}. On the other hand, the assertion on the homogeneity can be shown as follows. We see that if $\alpha\neq \frac{n+2}{n-2}$, then one has $\mu\neq 0$. Hence, the assumption that $\Phi(r,{\mathbf u})$ being constant for $r_1<r<r_2$ along with \eqref{eq:Phi'} yields that for any $r_1<r<r_2$,
\begin{equation*}
\frac{\partial {\mathbf u}}{\partial \nu} = \frac{2}{(\alpha-1)r} {\mathbf u} \quad\text{on }\partial B_r,
\end{equation*}
where $\nu$ is the unit normal pointing towards the origin. Thus, ${\mathbf u}$ is homogeneous of degree $-\frac{2}{\alpha-1}$ in $B_{r_2}\setminus \bar{B}_{r_1}$.
\end{proof}
\begin{remark}\label{remark:Phi-monot} As a matter of fact, \eqref{eq:Phi'} holds for $\alpha>\frac{n+2}{n-2}$, and hence $\Phi(r,{\mathbf u})$ is nonincreasing in this case, since $\mu<0$ for $\alpha>\frac{n+2}{n-2}$.
\end{remark}
\begin{remark}\label{remark:Phi-monot1}
For the case $\alpha = \frac{n+2}{n-2}$, we obtain from Proposition \ref{proposition:Phi-monot} (ii) a constant $\kappa({\mathbf u})$ such that
\begin{equation}\label{eq:kappa}
\kappa({\mathbf u}) = \Phi(r,{\mathbf u}),
\end{equation}
for any $0<r<R$. Since there is a one-to-one correspondence between the nonnegative solutions ${\mathbf u}$ of \eqref{eq:main} and ${\mathbf v}$ of \eqref{eq:main-cyl} via the cylindrical transform \eqref{eq:cyl}, we shall write $\kappa({\mathbf u})$ by $\kappa({\mathbf v})$ as well. In view of \eqref{eq:Phi-cyl}, it is clear that
\begin{equation}\label{eq:kappa-cyl}
\kappa({\mathbf v}) = \Psi(t,{\mathbf v}),
\end{equation}
for any $t>-\log R$. We shall call $\kappa$ the first Pohozaev invariant.
\end{remark}
Let us construct the second Pohozaev invariant in a general setting, that is without rotational symmetry. For $\alpha = \frac{n+2}{n-2}$, let us define, formally for the moment, the quantity
\begin{equation}\label{eq:Phi*}
\begin{split}
\Phi_*(r,{\mathbf u}) &= \frac{1}{4}(r\dot{f}(r,{\mathbf u}))^2 - \frac{(n-2)^2}{4} f(r,{\mathbf u})^2 - \kappa({\mathbf u}) f(r,{\mathbf u}) \\
&\quad + \frac{n-2}{n}f(r,{\mathbf u})^{\frac{2n-2}{n-2}} - 2\int_0^r \left(\frac{\rho}{n\omega_n}\int_{\partial B_\rho} |\nabla_\sigma {\mathbf u}|^2 \,d\sigma\right)\dot{f}(\rho,{\mathbf u}) \,d\rho \\
&\quad + \frac{2n-2}{n}\int_0^r \left(\frac{\rho}{n\omega_n}\int_{\partial B_\rho} |{\mathbf u}|^{\frac{2n}{n-2}}\,d\sigma - f(\rho,{\mathbf u})^{\frac{n}{n-2}}\right) \dot{f}(\rho,{\mathbf u})\,d\rho,
\end{split}
\end{equation}
where $\dot{f}$ denotes $df/dr$, and
\begin{equation}\label{eq:f}
f(r,{\mathbf u}) = \frac{1}{n\omega_n r} \int_{\partial B_r} |{\mathbf u}|^2\,d\sigma.
\end{equation}
Notice that $\Phi_*(r,{\mathbf u})$ is well-defined only if the last two double integrals on the right side are finite. Moreover, once $\Phi_*(r,{\mathbf u})$ becomes well-defined, we may also deduce from
\begin{equation}\label{eq:rf'}
r \dot{f}(r,{\mathbf u}) = - \frac{2}{n\omega_n}\int_{\partial B_r} {\mathbf u}\cdot \left( \frac{\partial {\mathbf u}}{\partial \nu} - \frac{n-2}{2r} {\mathbf u}\right)\,d\sigma
\end{equation}
the following scaling relation of $\Phi_*$,
\begin{equation}\label{eq:Phi*-scale}
\Phi_*(rs,{\mathbf u}) = \Phi_*(s,{\mathbf u}_r),
\end{equation}
which holds for any $r,s>0$. On the other hand, in terms of the cylindrical transformation ${\mathbf v}$, one has
\begin{equation}\label{eq:Phi*-cyl}
\Phi_*(r,{\mathbf u}) = \Psi_*(-\log r,{\mathbf v}),
\end{equation}
where $\Psi_*(t,{\mathbf v})$ is given by
\begin{equation}\label{eq:Psi*}
\begin{split}
\Psi_*(t,{\mathbf v}) &= \frac{1}{4}(g'(t,{\mathbf v}))^2 - \frac{(n-2)^2}{4} g(t,{\mathbf v})^2 - \kappa({\mathbf v}) g(t,{\mathbf v}) \\
&\quad + \frac{n-2}{n} g(t,{\mathbf v})^{\frac{2n-2}{n-2}} + 2\int_t^\infty \left(\frac{1}{n\omega_n}\int_{{\mathbb S}^{n-1}} |\nabla_\theta{\mathbf v}|^2\,d\theta \right) g'(\tau,{\mathbf v})\,d\tau\\
&\quad - \frac{2n-2}{n}\int_t^\infty \left(\frac{1}{n\omega_n} \int_{{\mathbb S}^{n-1}} |{\mathbf v}|^{\frac{2n}{n-2}}\,d\theta - g(\tau,{\mathbf v})^{\frac{n}{n-2}} \right) g'(\tau,{\mathbf v}) \,d\tau,
\end{split}
\end{equation}
with $g'$ being $dg/dt$ and
\begin{equation}\label{eq:g}
g(t,{\mathbf v}) = \frac{1}{n\omega_n} \int_{{\mathbb S}^{n-1}} |{\mathbf v}|^2\,d\theta.
\end{equation}
\begin{proposition}\label{proposition:Phi*-const} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in $B_R\setminus\{0\}$ with $\alpha = \frac{n+2}{n-2}$, and let $\Phi_*(r,{\mathbf u})$ be as in \eqref{eq:Phi*}. Then $\Phi_*(r,{\mathbf u})$ is well-defined and is constant for $0<r<R$.
\end{proposition}
We shall postpone the proof to Section \ref{section:asym-rad}, since proving the well-definedness of $\Phi_*(r,{\mathbf u})$ essentially relies the asymptotic radial symmetry of local solutions to \eqref{eq:main} (see Theorem \ref{theorem:asym-rad}).
\begin{remark}\label{remark:Phi*-const2} Knowing that $\Phi_*(r,{\mathbf u})$ is constant, we obtain a constant $\kappa_*({\mathbf u})$ such that
\begin{equation}\label{eq:kappa*}
\kappa_*({\mathbf u}) = \Phi_*(r,{\mathbf u}),
\end{equation}
for any $0<r<R$. We shall call this constant the second Pohozaev invariant. As with the first Pohozaev invariant, we will also write it by $\kappa_*({\mathbf v})$ whenever ${\mathbf v}$ is the cylindrical transformation. Clearly,
\begin{equation}\label{eq:kappa*-cyl}
\begin{split}
\kappa_*({\mathbf v}) = \Psi_*(t,{\mathbf v}),
\end{split}
\end{equation}
for any $t>-\log R$. In Section \ref{section:sing} and Section \ref{subsection:asym+sub} we will observe that $\kappa_*({\mathbf v}) = 0$ if and only if ${\mathbf v}(t,\theta) = (1+o(1))|{\mathbf v}(t,\theta)|{\mathbf e}$ uniformly for $\theta\in{\mathbb S}^{n-1}$ as $t\rightarrow\infty$, with some unit nonnegative vector ${\mathbf e}\in{\mathbb R}^m$.
\end{remark}
\section{Solutions on the Whole Space}\label{section:global}
In this section we classify the smooth solutions of \eqref{eq:main} on the whole space ${\mathbb R}^n$. The analysis is based on the method of moving spheres along with the Kelvin transform, and we follow essentially the argument proposed by Li and Zhang \cite[Section 2]{LZ}, with only a minor modification. Nevertheless, we shall contain the full argument here for the reader's convenience.
Given $z\in{\mathbb R}^n$ and $r>0$, we shall write ${\mathbf u}_{z,r}^*$ by the Kelvin transform of ${\mathbf u}$ with respect to the sphere $B_r(z)$, that is,
\begin{equation}\label{eq:kelvin}
{\mathbf u}_{z,r}^*(y) = \left(\frac{r}{|y-z|}\right)^{n-2} {\mathbf u} \left( z + \frac{r^2}{|y-z|^2}(y-z) \right).
\end{equation}
Let us remark that if ${\mathbf u}$ is a solution of \eqref{eq:main} in ${\mathbb R}^n$, then
\begin{equation}\label{eq:kelvin-pde}
-\Delta {\mathbf u}_{z,r}^* = \left( \frac{r}{|y-z|} \right)^{(\alpha-1)\mu} |{\mathbf u}_{z,r}^*|^{\alpha-1} {\mathbf u}_{z,r}^*\quad\text{in }{\mathbb R}^n\setminus\{z\},
\end{equation}
where $\mu$ is given by \eqref{eq:lam-mu}. Note that $\mu\geq 0$ if and only if $1<\alpha\leq\frac{n+2}{n-2}$. The non-negativity of $\mu$ will play a key role when comparing ${\mathbf u}$ and ${\mathbf u}_{z,r}^*$.
We begin with a basic lemma that holds for any nonnegative, superharmonic function, as a starting point of the method of moving sphere.
\begin{lemma}[Lemma 2.1 in \cite{LZ}]\label{lemma:basic} Let $v\in C^2({\mathbb R}^n)$ be a super-harmonic and nonnegative function on ${\mathbb R}^n$. Then for each $z\in{\mathbb R}^n$, there exists $r_0>0$, which may depend on $v$ and $z$, such that for all $0<r<r_0$,
\begin{equation}\label{eq:basic}
v_{z,r}^* \leq v\quad\text{in }{\mathbb R}^n\setminus B_r(z).
\end{equation}
\end{lemma}
The next lemma is an analogue of \cite[Lemma 2.4]{CGS} which claims that either the inequality \eqref{eq:basic} must hold until the solution becomes symmetric (with respect to a sphere) or it must fail on a compact subset of ${\mathbb R}^n$. The proof is given in that of \cite[Lemma 2.2]{LZ}, and we shall not repeat it here.
\begin{lemma}\label{lemma:basic2} Let $v\in C^2({\mathbb R}^n)$, $z\in{\mathbb R}^n$ and $r_0>0$ be such that
\begin{equation}\label{eq:basic2-pde}
-\Delta (v - v_{z,r_0}^*) \geq 0 \quad\text{in }{\mathbb R}^n\setminus \bar{B}_{r_0}(z),
\end{equation}
and
\begin{equation}\label{eq:basic2-asmp}
v_{z,r_0}^* < v \quad\text{in }{\mathbb R}^n\setminus\bar{B}_{r_0}(z).
\end{equation}
Then there is a small $\epsilon>0$ such that for any $r_0<r<r_0 + \epsilon$,
\begin{equation}\label{eq:basic2}
v_{z,r}^* < v \quad\text{in }{\mathbb R}^n\setminus B_r(z).
\end{equation}
\end{lemma}
Now let us turn our interest to the nonnegative, smooth global solutions ${\mathbf u}$ of \eqref{eq:main}. Given $z\in{\mathbb R}^n$, let us define, for each $1\leq i\leq m$,
\begin{equation}\label{eq:u22}
r_i(z) = \sup\{r>0:\text{$(u_i)_\rho^*\leq u_i$ in ${\mathbb R}^n\setminus B_\rho(z)$ for any $0<\rho<r$}\}.
\end{equation}
Since each component $u_i$ of ${\mathbf u}$ is nonnegative and superharmonic, Lemma \ref{lemma:basic} applies to $u_i$. from which we know that $r_i(z) > 0$ for each $1\leq i\leq m$. Thus, we have
\begin{equation}\label{eq:rb}
\bar{r}(z) = \inf_{1\leq i\leq m} r_i(z)>0.
\end{equation}
Let us remark that we have defined $\bar{r}(z)$ by the infimum, instead of minimum, of finite set of indices $\{1,2,\cdots,m\}$, since $r_i(z)$ as a supremum could be infinite. Moreover, if $r_i(z) = \infty$ for all $1\leq i\leq m$, we shall say that $\bar{r}(z) = \infty$.
The following lemma takes care of the case when $\bar{r}(z)$ is either finite or infinite. The proof is essentially the same with \cite[Lemma 1.2, Lemma 1.3]{DHV}, which deals with the upper critical case $\alpha = \frac{n+2}{n-2}$ only, whence we shall skip the details.
\begin{lemma}\label{lemma:u2} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in ${\mathbb R}^n$ with $1<\alpha\leq\frac{n+2}{n-2}$, $z\in{\mathbb R}^n$ be arbitrary and $\bar{r}(z)$ be as in \eqref{eq:rb}. If $\bar{r}(z)$ is finite, then
\begin{equation}\label{eq:u2}
{\mathbf u}_{z,\bar{r}(z)}^* = {\mathbf u}\quad\text{in }{\mathbb R}^n\setminus\{z\}.
\end{equation}
If $\bar{r}(z_0) = \infty$ for some $z_0\in{\mathbb R}^n$, then $\bar{r}(z) = \infty$ for all $z\in{\mathbb R}^n$.
\end{lemma}
We are now ready to classify the smooth global solutions.
\begin{proof}[Proof of Theorem \ref{theorem:main-g}] In view of Lemma \ref{lemma:u2}, we observe that $\bar{r}(z)$ defined in \eqref{eq:rb} is either finite or infinite for all $z\in{\mathbb R}^n$. If $\bar{r}(z)$ is finite for all $z\in{\mathbb R}^n$, then we have \eqref{eq:u2} at every point $z\in{\mathbb R}^n$. In this case, we may apply \cite[Lemma 11.1]{LZ} that there are $a_i\geq 0$, $r_i>0$ and $z_i\in{\mathbb R}^m$ for $1\leq i\leq m$ such that
\begin{equation}\label{eq:u-g1}
u_i(x) = a_ir_i^{-\frac{n-2}{2}}\left( \frac{r_i}{r_i^2 + |x-z_i|^2} \right)^{\frac{n-2}{2}}.
\end{equation}
On the other hand, if $\bar{r}(z)$ is infinite for all $z\in{\mathbb R}^n$, we have \eqref{eq:u22} for all $r>0$ at any $z\in{\mathbb R}^n$. Due to \cite[Lemma 11.2]{LZ}, there are $b_i\geq 0$ for $1\leq i\leq m$ such that
\begin{equation}\label{eq:u-g2}
u_i(x) = b_i.
\end{equation}
Suppose that ${\mathbf u}$ satisfies \eqref{eq:u-g2}, that is, ${\mathbf u}$ is constant everywhere on ${\mathbb R}^n$. As ${\mathbf u}$ being a nonnegative solution of \eqref{eq:main} in ${\mathbb R}^n$, ${\mathbf u}$ must be zero everywhere. Hence, Theorem \ref{theorem:main-g} (i) and (ii) are all satisfied under this assumption.
Next, let us consider the case that $u_i$ satisfies \eqref{eq:u-g1} for all $1\leq i\leq m$. This part is the same with the proof of \cite[Proposition 1.1]{DHV}, so we omit the details.
\end{proof}
\section{Solutions in Punctured Space}\label{section:sing}
\subsection{Radial Symmetry of Singular Solutions}\label{subsection:sing}
This section is devoted to the radial symmetry of nonnegative, singular solutions of \eqref{eq:main}. To be more precise, ${\mathbf u}$ is a nonnegative solution of \eqref{eq:main} in the punctured space ${\mathbb R}^n\setminus\{0\}$ that has a non-removable singularity at the origin, i.e.,
\begin{equation}\label{eq:sing}
\limsup_{x\rightarrow 0} |{\mathbf u}(x)| = \infty.
\end{equation}
The proof relies again on the method of moving spheres used in the previous section. The proof for the case of a single equation has already been established by Jin, et al. \cite[Proposition 2.1]{JLX}. Nevertheless, the multiplicity in the components here makes the comparison argument more subtle, as observed in the previous section. Let us also address that the method of moving plane also works (c.f. \cite[Theorem 8.1]{CGS}) after a suitable modification.
\begin{lemma}\label{lemma:rad} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in ${\mathbb R}^n\setminus\{0\}$ with $1<\alpha\leq\frac{n+2}{n-2}$. If ${\mathbf u}$ satisfies \eqref{eq:sing}, then ${\mathbf u}$ is radially symmetric.
\end{lemma}
\begin{proof} Let $z\in{\mathbb R}^n\setminus\{0\}$ be arbitrary. Arguing similarly as with Lemma \ref{lemma:basic} (whose proof can be found in \cite[Lemma 2.1]{LZ}), there exists some $0<r_0<|z|$ such that for any $0<r\leq r_0$,
\begin{equation*}
(u_i)_{z,r}^* \leq u_i\quad\text{in ${\mathbb R}^n\setminus (B_r(z)\cup\{0\})$ for each $1\leq i\leq m$}.
\end{equation*}
Hence, one can define, as with \eqref{eq:u22} and \eqref{eq:rb},
\begin{equation*}
r_i(z) = \sup\{r>0: (u_i)_{z,\rho}^* \leq u_i\text{ in ${\mathbb R}^n\setminus (B_\rho(z)\cup\{0\})$ for any $0<\rho<r$}\},
\end{equation*}
and
\begin{equation*}
\bar{r}(z) = \inf_{1\leq i\leq m} r_i(z).
\end{equation*}
We first claim that
\begin{equation}\label{eq:rbz-rad}
0<\bar{r}(z)\leq |z|.
\end{equation}
The positivity of $\bar{r}(z)$ is clear.To prove the second inequality in \eqref{eq:rbz-rad}, let us first observe that by \eqref{eq:sing}, there exist some sequence $x_j\rightarrow 0$ and a component $u_i$ such that $u_i(x_j) \rightarrow \infty$. If $\bar{r}(z)>|z|$, then by its definition, there should exist $\rho>|z|$ such that
\begin{equation}\label{eq:rbz-rad-f}
(u_i)_{z,\rho}^* \leq u_i\quad\text{in }{\mathbb R}^n\setminus B_\rho(z).
\end{equation}
Now let $y_j$ be the reflection of $x_j$ with respect to $\partial B_\rho(z)$, i.e.,
\begin{equation*}
y_j = z + \left( \frac{\rho}{|x_j - z|}\right)^2 (x_j-z).
\end{equation*}
Since $x_j\rightarrow 0$, we have $y_j \in {\mathbb R}^n\setminus B_\rho(z)$ for all sufficiently large $j$, and moreover,
\begin{equation*}
y_j \rightarrow y_0 = \left(1- \left( \frac{\rho}{|z|} \right)^2 \right) z.
\end{equation*}
Thus, if we take $\rho$ close enough to $|z|$, we have $y_0\neq 0$, whence $u_i$ is smooth at $y_0$. However, \eqref{eq:rbz-rad-f} implies
\begin{equation*}
\begin{split}
u_i(y_0) &= \lim_{j\rightarrow\infty} u_i(y_j) \geq \lim_{j\rightarrow\infty} ((u_i)_{z,\rho}^* (y_j)) \geq \left(\frac{|z|}{\rho}\right)^{n-2} \lim_{j\rightarrow\infty} u_i(x_j) = \infty,
\end{split}
\end{equation*}
a contradiction.
From \eqref{eq:rbz-rad}, we can also claim that
\begin{equation*}
\bar{r}(z) = |z|.
\end{equation*}
The argument is based on the proof of \cite[Proposition 2.1]{JLX} with the corresponding modification shown in Lemma \ref{lemma:u2}, which amounts to the number of nontrivial components. The main idea is that if $\bar{r}(z) < |z|$, then \eqref{eq:sing} together with the maximum principle implies that
\begin{equation}\label{eq:rbz-rad2-f}
u_i > (u_i)_{z,\bar{r}(z)}^*\quad\text{in }{\mathbb R}^n\setminus(\bar{B}_{\bar{r}(z)}(z)\cup\{0\}),
\end{equation}
at least for one $1\leq i\leq m$. Then we must have $|{\mathbf u}| > |{\mathbf u}_{z,\bar{r}(z)}^*|$ in ${\mathbb R}^n\setminus (\bar{B}_{\bar{r}(z)}(z)\cup\{0\})$, and the strong maximum principle yields that the strict inequality in \eqref{eq:rbz-rad2-f} must hold for all nontrivial components. Hence, as with Lemma \ref{lemma:basic2}, we obtain some $\epsilon>0$ such that \eqref{eq:rbz-rad2-f} holds for all $1\leq i\leq m$ with $\bar{r}(z)$ replaced by some $\bar{r}(z) < r< \bar{r}(z) + \epsilon$, a contradiction to \eqref{eq:rbz-rad-f}. The details are omitted.
To this end, we have proved that for each $z\in{\mathbb R}^n\setminus\{0\}$ and for any $0<r<|z|$,
\begin{equation*}
(u_i)_{z,r}^* \leq u_i \quad\text{in ${\mathbb R}^n\setminus (B_r(z)\cup\{0\})$ for each $1\leq i\leq m$}.
\end{equation*}
Thus, one may deduce from \cite[Lemma 2.1]{JLX} that $u_i$ is radially symmetric for each $1\leq i\leq m$.
\end{proof}
\subsection{Limiting Energy Levels and Pohozaev Invariants}\label{subsection:kappa}
Knowing the radial symmetry of singular solutions, we may classify the nonnegative solutions on the punctured space, using the balanced-energy-limit. The idea is to consider both {\it blowups} and {\it shrink-downs} of ${\mathbf u}$ under the scaling \eqref{eq:ur}. Here by saying a blowup or a shrink-down under the scaling ${\mathbf u}_r$ we indicate a limit of ${\mathbf u}_r$ as $r = r_j\rightarrow 0+$, or respectively $r = r_j\rightarrow \infty$ in $C_{loc}^2({\mathbb R}^n\setminus\{0\};{\mathbb R}^m)$. The following lemma provides the compactness of the sequence ${\mathbf u}_r$ in order to have both the blowups and the shrink-downs.
\begin{lemma}\label{lemma:u-s} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in ${\mathbb R}^n\setminus\{0\}$ with $1<\alpha\leq\frac{n+2}{n-2}$. If ${\mathbf u}$ satisfies \eqref{eq:sing}, then for each $1\leq i\leq m$,
\begin{equation}\label{eq:u-sup-s}
u_i(x) \leq \left(\frac{\alpha-1}{2n}\right)^{-\frac{1}{\alpha-1}}|x|^{-\frac{2}{\alpha-1}} \quad\text{in }{\mathbb R}^n\setminus\{0\}.
\end{equation}
\end{lemma}
\begin{proof} Let $u_i$ be a positive component of ${\mathbf u}$. Then, since $u_i$ is superharmonic in ${\mathbb R}^n\setminus\{0\}$, it follows from the extended maximum principle \cite[Theorem 1]{GilSer} that
\begin{equation}\label{eq:u-liminf-s}
\liminf_{x\rightarrow 0} u_i(x) > 0.
\end{equation}
Now let $v = u_i^{1-\alpha}$. Then $v$ satisfies, in ${\mathbb R}^n\setminus\{0\}$,
\begin{equation*}
\Delta v \geq \frac{\alpha}{\alpha-1} \frac{|\nabla v|^2}{v} + \alpha - 1.
\end{equation*}
Hence, for each $r>0$, the auxiliary function
\begin{equation*}
w(x) = v(x) - \frac{\alpha-1}{2n} |x|^2
\end{equation*}
becomes subharmonic in $B_r\setminus\{0\}$. Then by \eqref{eq:u-liminf-s}, $w$ is bounded around the origin, and thus, it follows from the extended maximum principle \cite[Theorem 1]{GilSer} that
\begin{equation*}
0 \leq \limsup_{x\rightarrow 0} w(x) \leq \sup_{\partial B_r} w = \sup_{\partial B_r} v - \frac{\alpha-1}{2n} r^2.
\end{equation*}
In terms of $u_i$, we obtain
\begin{equation*}
\inf_{\partial B_r} u_i \leq \left(\frac{\alpha-1}{2n}\right)^{-\frac{1}{\alpha-1}}r^{-\frac{2}{\alpha-1}}.
\end{equation*}
Now the radial symmetry obtained in Lemma \ref{lemma:rad} yields \eqref{eq:u-sup-s}.
\end{proof}
The next lemma gives the compactness of the sequence ${\mathbf u}_r$, and hence the existence of both blowup and shrink-down of ${\mathbf u}$.
\begin{lemma}\label{lemma:u-cpt} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in ${\mathbb R}^n\setminus\{0\}$ with $1<\alpha\leq\frac{n+2}{n-2}$. Then there is some $0<\gamma<1$ such that ${\mathbf u}_r$ is uniformly bounded in $C^{2,\gamma}(K;{\mathbb R}^m)$ on each compact set $K\subset {\mathbb R}^n\setminus\{0\}$.
\end{lemma}
\begin{proof} If ${\mathbf u}$ does not satisfy \eqref{eq:sing}, then ${\mathbf u}$ is bounded around the origin, and the origin becomes a removable singularity. According to Theorem \ref{theorem:main-g}, if $1<\alpha<\frac{n+2}{n-2}$, ${\mathbf u}$ is trivial, while if $\alpha = \frac{n+2}{n-2}$, ${\mathbf u}$ is globally bounded and satisfies $|{\mathbf u}(x)| = O(|x|^{2-n})$ as $|x|\rightarrow\infty$. Hence, in any case, ${\mathbf u}_r$ is bounded uniformly for all $r>0$ on a fixed compact subset of ${\mathbb R}^n\setminus\{0\}$.
On the other hand, if ${\mathbf u}$ satisfies \eqref{eq:sing}, Lemma \ref{lemma:u-s} implies that ${\mathbf u}_r$ is globally bounded in ${\mathbb R}^n\setminus\{0\}$. Thus, regardless of the removability of the singularity at the origin, we know that ${\mathbf u}_r$ is uniformly bounded in each compact subset of ${\mathbb R}^n\setminus\{0\}$.
Now since ${\mathbf u}_r$ also solves \eqref{eq:main} in ${\mathbb R}^n\setminus\{0\}$, it follows from the interior regularity theory \cite[Theorem 6.2 and Theorem 6.19]{GT} that ${\mathbf u}_r$ is uniformly bounded in $C^{2,\gamma}(K;{\mathbb R}^m)$ on each compact set $K\subset {\mathbb R}^n\setminus\{0\}$, for some $0<\gamma<1$. This finishes the proof.
\end{proof}
Let $\Phi(r,{\mathbf u})$ be the balanced-energy-type functional defined by \eqref{eq:Phi}. Recall from Proposition \ref{proposition:Phi-monot} that $\Phi(r,{\mathbf u})$ is monotone increasing in $r>0$ for $1<\alpha<\frac{n+2}{n-2}$, while it is constant for $\alpha = \frac{n+2}{n-2}$.
\begin{lemma}\label{lemma:u-Phi} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in ${\mathbb R}^n\setminus\{0\}$ with $1<\alpha\leq\frac{n+2}{n-2}$, and let ${\mathbf u}_0$ (resp., ${\mathbf u}_\infty$) be a blowup (resp., a shrink-down) under the scaling ${\mathbf u}_r$. Then $\Phi(r,{\mathbf u}_0) = \Phi(0+,{\mathbf u})$ (resp., $\Phi(r,{\mathbf u}_\infty) = \Phi(\infty,{\mathbf u})$) for all $r>0$. In particular, both ${\mathbf u}_0$ and ${\mathbf u}_\infty$ are homogeneous of degree $-\frac{2}{\alpha-1}$, provided that $1<\alpha<\frac{n+2}{n-2}$.
\end{lemma}
\begin{proof} Since the argument for shrink-downs is the same, we shall only present it for blowups. Let ${\mathbf u}_0$ be a blowup with a sequence $r_j\rightarrow 0+$. Then due to the scaling relation \eqref{eq:Phi-scale}, we have, for any $r>0$,
\begin{equation*}
\Phi(r,{\mathbf u}_0) = \lim_{j\rightarrow\infty} \Phi(r,{\mathbf u}_{r_j}) = \lim_{j\rightarrow\infty} \Phi(rr_j,{\mathbf u}) = \Phi(0+,{\mathbf u}),
\end{equation*}
where the existence of $\Phi(0+,{\mathbf u})$ follows from the compactness of ${\mathbf u}_r$ (Lemma \ref{lemma:u-cpt}) and the monotonicity of $\Phi(r,{\mathbf u})$ (Proposition \ref{proposition:Phi-monot} (i)). This proves the first assertion of Lemma \ref{lemma:u-Phi}. The second assertion on the homogeneity follows again from Proposition \ref{proposition:Phi-monot} (i).
\end{proof}
\begin{lemma}\label{lemma:class-sub} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in ${\mathbb R}^n\setminus\{0\}$ with $1<\alpha \leq \frac{n+2}{n-2}$. Suppose further that ${\mathbf u}$ is homogeneous of degree $-\frac{2}{\alpha-1}$.
\begin{enumerate}[(i)]
\item If $1<\alpha\leq\frac{n}{n-2}$, then ${\mathbf u}$ is trivial.
\item If $\frac{n}{n-2} <\alpha \leq\frac{n +2}{n-2}$, then either ${\mathbf u}$ is trivial, or ${\mathbf u}$ is of the form \eqref{eq:u-s}.
\end{enumerate}
\end{lemma}
\begin{proof} Since ${\mathbf u}$ is homogeneous of degree $-\frac{2}{\alpha-1}$, the cylindrical transform ${\mathbf v}$ introduced in \eqref{eq:cyl} satisfies
\begin{equation}\label{eq:u0-pde-S}
\Delta_\theta {\mathbf v}- \lambda {\mathbf v} + |{\mathbf v}|^{\alpha-1}{\mathbf v} = 0\quad\text{on }{\mathbb S}^{n-1},
\end{equation}
where $\Delta_\theta$ is the Laplace-Beltrami operator, and $\lambda$ is given by \eqref{eq:lam-mu}.
\begin{case} $1<\alpha\leq\frac{n}{n-2}$.
\end{case}
In view of \eqref{eq:lam-mu}, we have $\lambda\leq 0$. As a nonnegative solution of \eqref{eq:u0-pde-S}, we see that each component $v_i$ satisfies $\Delta_\theta v_i \leq 0$ on ${\mathbb S}^{n-1}$. This implies that $v_i$ does not attain any strict local minimum on ${\mathbb S}^{n-1}$. As ${\mathbb S}^{n-1}$ being a compact manifold, $v_i$ must be a constant. This argument holds for all $1\leq i\leq m$, which makes ${\mathbf v}$ a nonnegative, constant vector on ${\mathbb S}^{n-1}$. However, a nonnegative constant solution of \eqref{eq:u0-pde-S} must be trivial because $\lambda\leq 0$. Returning back to ${\mathbf u}$, it indicates that ${\mathbf u}$ is trivial on $\partial B_1$. As each of its component being nonnegative and superharmonic, ${\mathbf u}$ must be trivial in the whole domain, which proves Lemma \ref{lemma:class-sub} (i).
\begin{case} $\frac{n}{n-2}<\alpha<\frac{n+2}{n-2}$.
\end{case}
Suppose that ${\mathbf u}$ is a nontrivial solution in the punctured space. Then by the non-negativity and the super-harmonicity of each component of ${\mathbf u}$, $|{\mathbf u}|$ is positive everywhere. As is homogeneous of degree $-\frac{2}{\alpha-1}$, ${\mathbf u}$ must have a non-removable singularity at the origin, i.e., \eqref{eq:sing} holds. By Lemma \ref{lemma:rad}, ${\mathbf u}$ is radially symmetric, whence ${\mathbf u}$ is a positive constant vector, ${\mathbf a}$, on $\partial B_1$.
By \eqref{eq:u0-pde-S} we have $|{\mathbf a}| = \lambda^{\frac{1}{\alpha-1}}$. By the homogeneity, we see that ${\mathbf u}$ is of the form $\lambda^{\frac{1}{\alpha-1}} |x|^{-\frac{2}{\alpha-1}} {\mathbf e}$ with some nonnegative unit ${\mathbf e}\in{\mathbb R}^m$, proving Lemma \ref{lemma:class-sub} (ii).
\end{proof}
We are in a position to prove Theorem \ref{theorem:main-s} (i) and (ii).
\begin{proof}[Proof of Theorem \ref{theorem:main-s} (i) and (ii)] Let ${\mathbf u}_0$ and ${\mathbf u}_\infty$ be a blowup and, respectively, a shrink-down of ${\mathbf u}$. According to Lemma \ref{lemma:u-Phi}, both ${\mathbf u}_0$ and ${\mathbf u}_\infty$ are homogeneous of degree $-\frac{2}{\alpha-1}$. Hence, it follows from Lemma \ref{lemma:class-sub} (i) that if $1<\alpha\leq\frac{n}{n-2}$, both ${\mathbf u}_0$ and ${\mathbf u}_\infty$ are trivial. This in turn yields by Lemma \ref{lemma:u-Phi} that $\Phi(0+,{\mathbf u}) = \Phi(\infty,{\mathbf u}) = 0$. Due to the monotonicity of $\Phi(r,{\mathbf u})$, $\Phi(r,{\mathbf u}) = 0$ for all $r>0$. Thus, by Proposition \ref{proposition:Phi-monot} (i), ${\mathbf u}$ is homogeneous of degree $-\frac{2}{\alpha-1}$. Theorem \ref{theorem:main-s} (i) is now an immediate consequence of Lemma \ref{lemma:class-sub} (i).
Now let us consider the case $\frac{n}{n-2}<\alpha<\frac{n+2}{n-2}$. By Lemma \ref{lemma:u-Phi} and Lemma \ref{lemma:class-sub} (ii), any blowup ${\mathbf u}_0$ is either trivial or of the form \eqref{eq:u-s}. If ${\mathbf u}_0$ is trivial, then clearly $\Phi(r,{\mathbf u}_0) = 0$ for all $r>0$, which along with Lemma \ref{lemma:u-Phi} implies that $\Phi(0+,{\mathbf u}) = 0$. On the other hand, if ${\mathbf u}_0$ is of the form \eqref{eq:u-s}, then a simple computation shows that $\Phi(r,{\mathbf u}_0) = -\bar\lambda$ for all $r>0$, with $\bar\lambda$ given as in \eqref{eq:lamb}. Thus, again from Lemma \ref{lemma:u-Phi} it follows that $\Phi(0+,{\mathbf u}) = - \bar\lambda$. The converse statement is obviously true, whence we have proved that $\Phi(0+,{\mathbf u}) \in \{-\bar\lambda,0\}$, and $\Phi(0+,{\mathbf u}) = 0$ if and only if all the blowups are trivial, while $\Phi(0+,{\mathbf u}) = -\bar\lambda$ if and only if all the blowups are of the form \eqref{eq:u-s}.
Further, the same assertion holds for any shrink-down ${\mathbf u}_\infty$, proving that $\Phi(\infty,{\mathbf u}) \in \{-\bar\lambda,0\}$, and $\Phi(\infty,{\mathbf u}) = 0$ if and only if all the shrink-downs are trivial, while $\Phi(\infty,{\mathbf u}) = -\bar\lambda$ if and only if all the shrink-downs are of the form \eqref{eq:u-s}.
Now if $\Phi(0+,{\mathbf u}) = 0$, then since $\Phi(r,{\mathbf u})$ is nondecreasing in $r$ and $\Phi(\infty,{\mathbf u})\in \{-\bar\lambda,0\}$, we must have $\Phi(r,{\mathbf u}) = 0$ for all $r>0$. Hence, by Lemma \ref{lemma:u-Phi} and Lemma \ref{lemma:class-sub} (ii), ${\mathbf u}$ is either trivial or of the form \eqref{eq:u-s}. However, the latter yields that $\Phi(0+,{\mathbf u}) = -\bar\lambda$, a contradiction. Thus, ${\mathbf u}$ must be trivial. Of course, the converse is also true.
Similarly, $\Phi(\infty,{\mathbf u}) = -\bar\lambda$ implies that ${\mathbf u}$ is of the form \eqref{eq:u-s}. This finishes the proof of Theorem \ref{theorem:main-s} (ii).
\end{proof}
The analysis on the case $\alpha = \frac{n+2}{n-2}$ is more subtle. Our approach relies on the Pohozaev invariants of which the first one $\kappa({\mathbf u})$ was introduced in \eqref{eq:kappa}. In the following we focus on the second Pohozaev invariant $\kappa_*({\mathbf u})$, which was briefly introduced in Remark \ref{remark:Phi*-const2}. More importantly, we shall observe that this second invariant appears solely due to the multiplicity of the components of \eqref{eq:main}.
\begin{lemma}\label{lemma:class-crit} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in ${\mathbb R}^n\setminus\{0\}$ with $\alpha = \frac{n+2}{n-2}$. Then $\Phi(r,{\mathbf u})$ and $\Phi_*(r,{\mathbf u})$ in \eqref{eq:Phi} and \eqref{eq:Phi*} are well-defined, and there are constants $\kappa({\mathbf u})$ and $\kappa_*({\mathbf u})$ satisfying \eqref{eq:kappa} and respectively \eqref{eq:kappa*}. Moreover, the inequalities \eqref{eq:class-crit-kappa} and \eqref{eq:class-crit-kappa*} holds and the equalities of the lower bounds only occur simultaneously.
\end{lemma}
\begin{proof}
The proof can be divided into two cases; first we consider the case where ${\mathbf u}$ is not rotationally symmetric, and then we treat the other case. We shall prove the equivalent statements for the cylindrical transformation ${\mathbf v}$. Since ${\mathbf v}$ will be fixed throughout the proof, we shall omit the dependence of $\Psi$, $\Psi_*$, $\kappa$ and $\kappa_*$ on ${\mathbf v}$ here.
Suppose that ${\mathbf u}$ is not rotationally symmetric. Due to Lemma \ref{lemma:rad}, ${\mathbf u}$ has a removable singularity at the origin. Thus, its cylindrical transformation ${\mathbf v}$, given as in \eqref{eq:cyl}, satisfies
\begin{equation}\label{eq:class-crit-1}
|{\mathbf v}(t,\theta)| + |\partial_t{\mathbf v}(t,\theta)| \leq Ce^{-\frac{n-2}{2}t}\quad\text{on }{\mathbb S}^{n-1},
\end{equation}
as $t\rightarrow\infty$, with some constant $C>0$ independent of $t$. This combined with \eqref{eq:kappa-cyl} implies that
\begin{equation}\label{eq:class-crit-2}
\kappa = \lim_{t\rightarrow\infty} \Psi(t) = 0.
\end{equation}
On the other hand, the estimate \eqref{eq:class-crit-1} also ensures the well-definedness of $\Psi_*(t)$ given by \eqref{eq:Psi*} for all $t\in{\mathbb R}$. To prove that $\Psi_*(t)$ is constant for any $t\in{\mathbb R}$, we need to compute the derivatives of $g$, given by \eqref{eq:g}. Utilizing \eqref{eq:main-cyl}, \eqref{eq:kappa-cyl} and \eqref{eq:class-crit-2} one can verify that
\begin{equation*}
g'' = \frac{2}{n\omega_n} \int_{{\mathbb S}^{n-1}} \left( \frac{(n-2)^2}{2} |{\mathbf v}|^2 + 2|\nabla_\theta{\mathbf v}|^2 - \frac{2n-2}{n} |{\mathbf v}|^{\frac{2n}{n-2}} \right) \,d\theta,
\end{equation*}
from which it follows that
\begin{equation}\label{eq:Psi*'}
\begin{split}
&\Psi_*'(t) \\
&= g' \left( \frac{g''}{2} - \frac{(n-2)^2}{2} g - \frac{2}{n\omega_n} \int_{{\mathbb S}^{n-1}} \left( |\nabla_\theta {\mathbf v}|^2 - \frac{n-1}{n} |{\mathbf v}|^{\frac{2n}{n-2}} \right)\,d\theta\right) \\
&= \frac{g'}{n\omega_n}\int_{{\mathbb S}^{n-1}} \left( |\partial_t {\mathbf v}|^2 -\frac{(n-2)^2}{4} |{\mathbf v}|^2 - |\nabla_\theta {\mathbf v}|^2 + \frac{n-2}{n} |{\mathbf v}|^{\frac{2n}{n-2}} \right) \,d\theta\\
&= 0.
\end{split}
\end{equation}
Thus, $\Psi_*(t)$ is constant for any $t\in{\mathbb R}$, and there must exist a constant $\kappa_*({\mathbf v})$ such that \eqref{eq:kappa*-cyl} holds for all $t$. Moreover, one can also verify from \eqref{eq:class-crit-1} that
\begin{equation*}
\kappa_* = \lim_{t\rightarrow\infty} \Psi_*(t) = 0.
\end{equation*}
This proves the lemma for the case where ${\mathbf u}$ is rotationally symmetric.
Next we consider the case ${\mathbf u}$ is rotationally symmetric, so that the cylindrical transformation ${\mathbf v}$ becomes a function of $t$ only. In this case, we have already observed that \eqref{eq:|v|-poho} holds with $\kappa_*$ given by \eqref{eq:2nd-poho}. Note that under the rotational symmetry of ${\mathbf v}$, $g$ as in \eqref{eq:g} is identical to $|{\mathbf v}|^2$. Hence, one can easily observe from \eqref{eq:Psi*} and \eqref{eq:|v|-poho} that
\begin{equation}\label{eq:kappa*-cyl-rad}
\Psi_*(t) = \frac{(g')^2}{4} - \frac{(n-2)^2}{4} g^2 - \kappa g + \frac{n-2}{n} g^{\frac{2n-2}{n-2}} = \kappa_*,
\end{equation}
as desired.
Let us now prove the bounds in \eqref{eq:class-crit-kappa} and \eqref{eq:class-crit-kappa*}. Since we have already verified above that $\kappa = \kappa_* = 0$ if ${\mathbf v}$ is not rotationally symmetric, it suffices to consider the situation where ${\mathbf v}$ is rotationally symmetric. Then one can follow the derivation of \eqref{eq:2nd-poho} and verify that $\kappa_* \leq 0$. Hence, we are only left with proving the lower bounds of $\kappa$ and $\kappa_*$.
Set
\begin{equation*}
f(s) = \frac{(n-2)^2}{4} s^2 - \frac{n-2}{n} s^{\frac{2n-2}{n-2}} + \kappa s,
\end{equation*}
and let us rephrase the second identity in \eqref{eq:kappa*-cyl-rad} as
\begin{equation}\label{eq:kappa*-cyl-rad-re}
\frac{(g')^2}{4} = f(g) + \kappa_*.
\end{equation}
Utilizing $\kappa_* \leq 0$ in the identity above, we see that $f(g) \geq 0$. Since either $g(t)=0$ and $g(t)>0$ for all $t$, and $g(t)=0$ yields $\kappa=0$, we can focus on the case $g(t)>0$ for all $t$. Then $\frac{1}{g} f(g) \geq 0$ as well, from which it follows that
\begin{equation*}
\kappa \geq - \frac{(n-2)^2}{4} g + \frac{n-2}{n} g^{\frac{n}{n-2}} \geq -\frac{2}{n} \left( \frac{n-2}{2} \right)^n.
\end{equation*}
This verifies the lower bound \eqref{eq:class-crit-kappa} of $\kappa$.
To verify the the lower bound \eqref{eq:class-crit-kappa*} of $\kappa_*$, let us remark that
\begin{equation*}
\left( \frac{2}{n} \left(\frac{n-2}{2}\right)^n + \kappa \right)\left(\frac{n-2}{2}\right)^{n-2} = f \left( \left(\frac{n-2}{2}\right)^{n-2} \right).
\end{equation*}
Now suppose towards a contradiction that there is a solution ${\mathbf v}$ having $\kappa_* < -f ( (\frac{n-2}{2})^{n-2})$. Then it follows from \eqref{eq:kappa*-cyl-rad-re} that $\min\{g (t) : t\in{\mathbb R}\} > (\frac{n-2}{2})^{n-2}$, or equivalently, $\min\{ |{\mathbf v}(t)| : t\in{\mathbb R}\} > (\frac{n-2}{2})^{\frac{n-2}{2}}$. In view of \eqref{eq:main-cyl}, this implies that
\begin{equation}\label{eq:class-crit-kappa*-1}
v_i'' = \frac{(n-2)^2}{4} v_i - |{\mathbf v}|^{\frac{4}{n-2}} v_i \leq - \delta v_i,
\end{equation}
for each $1\leq i\leq m$, where $\delta = \min\{|{\mathbf v}(t)| : t\in{\mathbb R}\} - (\frac{n-2}{2})^{\frac{n-2}{2}} > 0$. Hence, $v_i$ is a concave function. However, \eqref{eq:u-sup-s} shows that $v_i$ is uniformly bounded for all $t$, which indicates that $v_i(t) \rightarrow a_i$ and $v_i''(t) \rightarrow 0$ as $t\rightarrow\infty$ for some $a_i>0$. However, this is a contradiction against \eqref{eq:class-crit-kappa*-1}, which proves the lower bound \eqref{eq:class-crit-kappa*} of $\kappa_*$.
Finally, let us investigate the scenario when the equalities of the lower bounds in \eqref{eq:class-crit-kappa} and \eqref{eq:class-crit-kappa*} hold. Suppose that the equality of the lower bound in \eqref{eq:class-crit-kappa*} occur. That is,
\begin{equation}\label{eq:class-crit-kappa*-2}
\kappa + \left(\frac{n-2}{2}\right)^{2-n} \kappa_* = - \frac{2}{n} \left(\frac{n-2}{2}\right)^n.
\end{equation}
Arguing similarly as above, one can deduce that $\min\{|{\mathbf v}(t)|:t\in{\mathbb R}\} \geq (\frac{n-2}{2})^{\frac{n-2}{2}}$ and $v_i'' \leq 0$ in ${\mathbb R}$ for each $1\leq i\leq m$. Again $v_i$ is a concave function that is uniformly bounded in ${\mathbb R}$, so $v_i(t) \rightarrow a_i$, for some $a_i\in{\mathbb R}$, and $v_i''(t) \rightarrow 0$ as $t\rightarrow\infty$. Thus, $|{\mathbf v}(t)|\rightarrow |{\mathbf a}|$ with ${\mathbf a} = (a_1,\cdots,a_m)$, and it follows from $v_i''(t) \rightarrow 0$ and the first equality in \eqref{eq:class-crit-kappa*-1} that $|{\mathbf a}| = (\frac{n-2}{2})^{\frac{n-2}{2}}$. On the other hand, we also have $v_i'(t)\rightarrow 0$ as $t\rightarrow\infty$, so sending $t\rightarrow\infty$ in the second equality of \eqref{eq:kappa-cyl} yields that
\begin{equation*}
\kappa = \lim_{t\rightarrow\infty} \left( |{\mathbf v}'(t)|^2 - \frac{(n-2)^2}{4} |{\mathbf v}(t)|^2 + \frac{n-2}{n} |{\mathbf v}(t)|^{\frac{2n}{n-2}} \right) = -\frac{2}{n} \left(\frac{n-2}{2}\right)^n.
\end{equation*}
Thus, \eqref{eq:class-crit-kappa*-2} forces $\kappa_* = 0$, and the final assertion of the lemma is proved.
\end{proof}
Let us finish this section by proving Theorem \ref{theorem:main-s} (iii).
\begin{proof}[Proof of Theorem \ref{theorem:main-s} (iii)]
The well-definedness and the bounds of $\kappa$ and $\kappa_*$ are proved in Lemma \ref{lemma:class-crit}. The other assertions can be proved as follows.
First consider the assertion (iii)-(a). If ${\mathbf u}$ is not radially symmetric, then by Lemma \ref{lemma:rad}, ${\mathbf u}$ has a removable singularity at the origin, as desired. On the other hand, if ${\mathbf u}$ is radially symmetric, one can deduce from \eqref{eq:|v|-poho} that the cylindrical transformation ${\mathbf v}$, which is now a function of $t$ only, satisfies
\begin{equation}\label{eq:kappa-cyl-rad-reg}
\left(\frac{d|{\mathbf v}|}{dt} \right)^2 = \frac{(n-2)^2}{4} |{\mathbf v}|^2 - \frac{n-2}{n} |{\mathbf v}|^{\frac{2n}{n-2}}.
\end{equation}
Hence, the classical work such as \cite{F} or \cite{CGS} applies to $|{\mathbf v}|$, proving the `only if' part of the assertion (iii)-(a). The `if' part can be verified through a direct computation.
Let us move on to the case $\kappa^2 + \kappa_*^2 > 0$. From the assertion (iii)-(a), we see that ${\mathbf u}$ must have a non-removable singularity at the origin. According to Lemma \ref{lemma:rad}, ${\mathbf u}$ is radially symmetric, so one can follow the computation in Section \ref{section:intro} and deduce \eqref{eq:|v|-poho}.
Finally, assume that $\kappa = -\frac{2}{n} (\frac{n-2}{2})^n$ and $\kappa_* = 0$. It follows from \eqref{eq:|v|-poho} that
\begin{equation*}
\left(\frac{d|{\mathbf v}|}{dt}\right)^2 - \frac{(n-2)^2}{4}|{\mathbf v}|^2 + \frac{n-2}{n} |{\mathbf v}|^{\frac{2n}{n-2}} + \frac{2}{n} \left(\frac{n-2}{2}\right)^n = 0,
\end{equation*}
whence $|{\mathbf v}|$ has to be constant in ${\mathbb R}$, and the constant has to be $(\frac{n-2}{2})^{\frac{n-2}{2}}$. In terms of ${\mathbf u}$ this implies that ${\mathbf u}$ is homogeneous of degree $-\frac{n-2}{2}$ and is of the form \eqref{eq:u-s-2}. This constitutes the `only if' part of the assertion (iii)-(c). The `if' part follows easily from a direct computation.
\end{proof}
\section{A Priori Estimate and Harnack-Type Inequality for Local Solutions}\label{section:upper}
In this section, we prove {\it a priori} upper bounds for local solutions of \eqref{eq:main} in $B_1\setminus\{0\}$ with $1<\alpha \leq \frac{n+2}{n-2}$ which further allows us to derive related Harnack inequalities, interior gradient estimates and the compactness of scaling functions. Our analysis is divided into two cases, according to the subcritical range $1<\alpha<\frac{n+2}{n-2}$ and the critical range $\alpha = \frac{n+2}{n-2}$. The former is based on the non-existence of the smooth, positive, global solution in Theorem \ref{theorem:main-g} (i) along with a blowup argument. The latter uses the method of moving spheres presented in the previous section, essentially following the work of Li and Zhang \cite{LZ}.
\subsection{A Priori Bound for $1<\alpha<\frac{n+2}{n-2}$}\label{subsection:upper-sub}
We begin with the upper bound for the subcritical case, which is (much) simpler than the critical case.
\begin{proposition}\label{proposition:sup-sub} Let $1<\alpha<\frac{n+2}{n-2}$ and suppose that ${\mathbf v}\in C^2(B_1;{\mathbb R}^m)\cap C(\bar{B}_1;{\mathbb R}^m)$ is a nonnegative solution of
\begin{equation}\label{eq:main-reg}
-\Delta {\mathbf v} = |{\mathbf v}|^{\alpha-1}{\mathbf v} \quad\text{in }B_1.
\end{equation}
Then there exists $C>0$, depending only on $n$, $m$ and $\alpha$, such that
\begin{equation}\label{eq:sup}
|{\mathbf v}(x)| \leq C(1-|x|)^{-\frac{2}{\alpha-1}} \quad\text{in }B_1.
\end{equation}
\end{proposition}
\begin{proof}
Note that $w = v_1 + \cdots + v_m$ satisfies
\begin{equation*}
\frac{1}{c} w^\alpha \leq -\Delta w \leq c w^\alpha,
\end{equation*}
for some $c>1$, depending only on $m$ and $\alpha$. Thus, we can follow the proof of \cite[Theorem 2.1]{PQS} and obtain the desired inequality. We omit the details.
\end{proof}
\subsection{A Harnack-Type Inequality for $\alpha=\frac{n+2}{n-2}$}\label{subsection:upper-crit}
Our approach to achieve the Harnack-type inequality for $\alpha=\frac{n+2}{n-2}$ follows the line of the scalar case in Li and Zhang \cite[Lemma 5.1]{LZ}. In our system setting, the problem becomes very sensitive on the number of nonzero components, and we modify the proof of \cite[Lemma 5.1]{LZ} in this direction.
\begin{proposition}\label{proposition:har-crit} Let ${\mathbf v}\in C^2(B_2;{\mathbb R}^m)\cap C(\bar{B}_2;{\mathbb R}^m)$ be a nonnegative solution of
\begin{equation}\label{eq:main-crit}
-\Delta {\mathbf v} = |{\mathbf v}|^{\frac{4}{n-2}}{\mathbf v}\quad\text{in }B_2.
\end{equation}
Then, there exists $C>0$ depending only on $n$ and $m$, such that
\begin{equation}\label{eq:har-crit}
\left( \min_{i \in I_m} \inf_{\partial B_2} v_i\right) |{\mathbf v}(x)| \leq C (1-|x|)^{-\frac{n-2}{2}} \quad\text{in }B_1,
\end{equation}
where $I_m$ is the set of indices $1\leq i\leq m$ such that $v_i$ is nontrivial.
\end{proposition}
\begin{proof} If ${\mathbf v}$ is trivial, then $I_m = \emptyset$, whence there is nothing to prove. Thus, we shall assume that ${\mathbf v}$ is not trivial, so that $I_m \neq\emptyset$. Then for each $i\in I_m$, we know from the super-harmonicity and the non-negativity of $v_i$ that $\inf_{\partial B_2} v_i > 0$, whence $(\min_{i\in I_m} \inf_{\partial B_2} v_i)^{-1}$ is a positive, finite number.
If $|{\mathbf v}(x)| \leq C_1(1 - |x|)^{-\frac{n-2}{2}} $ in $B_1$ for some $C_1>0$ depending only on $n$ and $m$, then the claim \eqref{eq:har-crit} is true, since the maximum principle and the super-harmonicity of each component of ${\mathbf v}$ implies that $\inf_{\partial B_2} v_i \leq v_i(0)$. Thus, let us assume that for all $j\geq 1$ there are nonnegative solutions ${\mathbf v}_j$ of \eqref{eq:main-crit} and points $x_j\in \bar{B}_1$ such that
\begin{equation}\label{eq:sup-f}
M_j:= \sup_{|x|\leq 1} \left((1-|x|)^{\frac{n-2}{2}} |{\mathbf v}_j(x)|\right) =(1-|x_j|)^{\frac{n-2}{2}} |{\mathbf v}_j(x_j)| \to \infty.
\end{equation}
We know that $x_j\in B_1$ (instead of $\partial B_1$) since ${\mathbf v}_j$ is continuous on $\bar{B}_1$. Moreover, we shall set
\begin{align}\label{eq:rj-f}
r_j &= \frac{1}{2}(1-|x_j|)>0,\\
\label{eq:dj}
\delta_j &= |{\mathbf v}_j(x_j)|^{-\frac{\alpha-1}{2}} = 2r_j M_j^{-\frac{2}{n-2}} \rightarrow 0,\\
\label{eq:Rj}
R_j &= \frac{r_j}{\delta_j} = \frac{1}{2} M_j^{\frac{2}{n-2}} \rightarrow \infty.
\end{align}
It should be noted that due to \eqref{eq:sup-f}, we have
\begin{equation}\label{eq:uj-xj}
|{\mathbf v}_j(x)| \leq \left( \frac{1 - |x_j|}{1-|x|}\right)^{\frac{2}{\alpha-1}} |{\mathbf v}_j(x_j)| \leq 2^{\frac{2}{\alpha-1}} |{\mathbf v}_j(x_j)|\quad\text{in }B_{r_j}(x_j).
\end{equation}
In addition, inserting \eqref{eq:rj-f} into \eqref{eq:sup-f}, we obtain
\begin{equation}\label{eq:uj-xj2}
|{\mathbf v}_j(x_j)| = (2r_j)^{-\frac{2}{\alpha-1}} M_j.
\end{equation}
With \eqref{eq:uj-xj} and \eqref{eq:uj-xj2} at hand, one can following the proof of \cite[Lemma 5.1]{LZ} to deduce that the sequence of the scaled function,
\begin{equation*}
{\mathbf w}_j(x) = \delta_j^{\frac{n-2}{2}} {\mathbf v}_j( \delta_j x + x_j)\quad\text{in }B_{R_j},
\end{equation*}
converges to $w_0$ in $C_{loc}^2({\mathbb R}^n;{\mathbb R}^m)$ for certain ${\mathbf w}_0\in C^2({\mathbb R}^n;{\mathbb R}^m)$, which is a nonnegative solution of
\begin{equation}\label{eq:ut0-pde}
-\Delta{\mathbf w}_0 = |{\mathbf w}_0|^{\frac{4}{n-2}}{\mathbf w}_0\quad\text{in }{\mathbb R}^n,
\end{equation}
satisfying
\begin{equation}\label{eq:ut0-sup}
|{\mathbf w}_0(x)|\leq 2^{\frac{2}{\alpha-1}} \quad\text{in }{\mathbb R}^n,
\end{equation}
as well as
\begin{equation}\label{eq:ut0-0}
|{\mathbf w}_0(0)| = 1.
\end{equation}
We omit the details here.
With only a minor modification, one may apply Lemma \ref{lemma:basic} to each component $w_{i,j}$ of ${\mathbf w}_j$, with $i\in I_m$, and obtain a number $s_{i,j}(z)>0$, corresponding to each $z\in{\mathbb R}^n$, such that for all $0<r<s_{i,j}(z)$,
\begin{equation}\label{eq:wjk-basic}
(w_{i,j})_{z,r}^* \leq w_{i,j}\quad\text{in } B_{1/(2\delta_j)}(z) \setminus B_r(z).
\end{equation}
Here we choose $j$ large enough so that $B_{1/(2\delta_j)}(z)\subset B_{1/\delta_j}$, which is possible due to \eqref{eq:dj}. One may refer to the proof of \cite[Theorem 1.5]{LZ} for the details.
Let us now replace $s_{i,j}(z)$ by the supremum value of $r$ such that \eqref{eq:wjk-basic} holds, that is,
\begin{equation}\label{eq:sijz}
s_{i,j}(z) = \sup\{ r: (w_{i,j})_{z,\rho}^* \leq w_{i,j}\text{ in $B_{1/(2\delta_j)}(z) \setminus B_r(z)$ for any $0<\rho<r$}\}.
\end{equation}
Now with $s_{i,j}(z)$ defined as in \eqref{eq:sijz}, we shall set, analogously to \eqref{eq:rb},
\begin{equation}\label{eq:rbj}
\bar{s}_j(z) = \inf_{i \in I_m} s_{i,j}(z).
\end{equation}
Then we have
\begin{equation}
(w_{i,j})_{z,\bar{s}_j(z)}^* \leq w_{i,j}\quad\text{in $B_{1/(2\delta_j)}(z) \setminus B_{\bar{s}_j(z)}(z)$ for each $i \in I_m$},
\end{equation}
and respectively,
\begin{equation}\label{eq:wij-kelvin-pde}
-\Delta (w_{i,j} - (w_{i,j})_{z,\bar{s}_j(z)}^*) \geq 0 \quad\text{in }B_{1/(2\delta_j)}(z) \setminus \bar{B}_{\bar{s}_j(z)}(z).
\end{equation}
Now let us assume towards a contradiction that
\begin{equation}\label{eq:vj-f}
\min_{i\in I_m}\inf_{\partial B_2} v_{i,j} \geq j \left( \sup_{|x|\leq 1} (1-|x|)^{\frac{n-2}{2}} |{\mathbf v}_j(x)| \right)^{-1} = \frac{j}{M_j}.
\end{equation}
In terms of $w_{i,j}$, one may rewrite \eqref{eq:vj-f} as
\begin{equation}\label{eq:wj-f}
\begin{split}
\min_{i\in I_m} \inf_{\partial B_{1/\delta_j}} w_{i,j} &= \delta_j^{\frac{n-2}{2}} \min_{i\in I_m} \inf_{\partial B_1(x_j)} v_{i,j} \\
&\geq \delta_j^{\frac{n-2}{2}} \min_{i\in I_m} \inf_{\partial B_2} v_{i,j}\\
&\geq j \delta_j^{n-2},
\end{split}
\end{equation}
where in the derivation of the first inequality we used the super-harmonicity of $v_{i,j}$, the maximum principle and the fact that $B_1(x_j)\subset B_2$, while the second inequality follows from \eqref{eq:vj-f}, \eqref{eq:dj} and the fact that $2r_j = 1-|x_j| \leq 1$.
In view of \eqref{eq:wj-f}, one may easily deduce that for any $z\in{\mathbb R}^n$,
\begin{equation}\label{eq:rbj-f}
\lim_{j\rightarrow\infty} \bar{s}_j(z) = \infty.
\end{equation}
Suppose that \eqref{eq:rbj-f} is false, and there exists some $L>0$, independent of $j$, such that
\begin{equation}\label{eq:rbj-ff}
\bar{s}_j(z) \leq L.
\end{equation}
Then by the definition of the Kelvin transform (see \eqref{eq:kelvin}), we have, for any $i\in I_m$,
\begin{equation}\label{eq:wj-ff}
\begin{split}
\sup_{\partial B_{1/(4\delta_j)}(z)} (w_{i,j})_{z,\bar{s}_j(z)}^* &= (4\delta_j\bar{s}_j(z))^{n-2} \sup_{\partial B_{4\delta_j\bar{s}_j^2(z)}} w_{i,j} \\
&\leq (4\delta_j L)^{n-2} \delta_j^{\frac{n-2}{2}} \sup_{B_{4\delta_j^2L^2}} v_{i,j} \\
&\leq (8L)^{n-2} \delta_j^{n-2},
\end{split}
\end{equation}
where in deriving the first and the second inequality we used \eqref{eq:rbj-ff} and, respectively, \eqref{eq:uj-xj} with \eqref{eq:uj-xj2}. According to \eqref{eq:wj-f} and \eqref{eq:wj-ff}, for each $i\in I_m$,
\begin{equation}\label{eq:wj-fff}
\inf_{\partial B_{1/(4\delta_j)}(z)} (w_{i,j} - (w_{i,j})_{z,\bar{s}_j(z)}^*) \geq (j - (8L)^{n-2}) \delta_j^{n-2} > 0,
\end{equation}
for all sufficiently large $j$, where in the first inequality we used $w_{i,j}\geq \inf_{\partial B_{1/\delta_j}} w_{i,j}$ on $\partial B_{1/(4\delta_j)}(z)$, which follows from the maximum principle, the super-harmonicity of $w_{i,j}$ in $B_{1/\delta_j}$ and the fact that $B_{1/(4\delta_j)(z)} \subset B_{1/\delta_j}$. With \eqref{eq:wj-fff} at hand, we may apply the maximum principle to \eqref{eq:wij-kelvin-pde} and observe that for any $i\in I_m$,
\begin{equation}\label{eq:wj-f4}
(w_{i,j})_{z,\bar{s}_j(z)}^* < w_{i,j} \quad\text{in }B_{1/(2\delta_j)(z)}\setminus \bar{B}_{\bar{s}_j(z)}(z).
\end{equation}
Now that $w_{i,j}$ satisfies \eqref{eq:wij-kelvin-pde} and \eqref{eq:wj-f4} for each $i\in I_m$, we can follow a similar argument to that in the proof of \cite[Lemma 5.2]{LZ} and deduce that there exist $\bar{s}_{i,j}(z) > \bar{s}_j(z)$ and $0<\epsilon_{i,j} < \bar{s}_{i,j}(z) - \bar{s}_j(z)$ such that for any $\bar{s}_j(z)< r < \bar{s}_j(z) + \epsilon_{i,j}$,
\begin{equation}\label{eq:wj-f5}
(w_{i,j})_{z,\bar{s}_j(z)}^* < w_{i,j}\quad\text{in $B_{1/(2\delta_j)(z)}\setminus B_r(z)$ for each $i\in I_m$}.
\end{equation}
Clearly, \eqref{eq:wj-f5} violates the definition of $\bar{s}_j(z)$ in \eqref{eq:rbj}. Hence, the claim \eqref{eq:rbj-f} should be true, under the assumption \eqref{eq:vj-f}.
Knowing that \eqref{eq:wj-f} is true for all $z\in{\mathbb R}^n$ (under the assumption \eqref{eq:vj-f}), we have for any $z\in{\mathbb R}^n$ and $r>0$ that
\begin{equation}\label{eq:wij-fff}
(w_{i,j})_{z,r}^* \leq w_{i,j}\quad\text{in $B_{1/(2\delta_j)}(z)\setminus B_r(z)$ for any $i\in I_m$},
\end{equation}
for all sufficiently large $j$ such that $\bar{s}_j(z) > r$. On the other hand, recall from the beginning of this proof that ${\mathbf w}_j\rightarrow{\mathbf w}_0$ in $C_{loc}^2({\mathbb R}^n;{\mathbb R}^m)$ with some ${\mathbf w}_0\in C^2({\mathbb R}^n;{\mathbb R}^m)$ satisfying \eqref{eq:ut0-pde}, \eqref{eq:ut0-sup} and \eqref{eq:ut0-0} with $\alpha = \frac{n+2}{n-2}$. This implies $({\mathbf w}_j)_{z,r}^* \rightarrow ({\mathbf w}_0)_{z,r}^*$ in $C_{loc}^2({\mathbb R}^n\setminus \{z\};{\mathbb R}^m)$ for each $z\in{\mathbb R}^n$ and any $r>0$. Thus, we may pass to the limit with $j\rightarrow\infty$ (possibly along a subsequence) in \eqref{eq:wij-fff} in any compact domain of type $B_R(z)\setminus B_r(z)\subset {\mathbb R}^n\setminus\{z\}$, which gives
\begin{equation}\label{eq:wi0-fff}
(w_{i,0})_{z,r}^* \leq w_{i,0} \quad\text{in ${\mathbb R}^n\setminus B_r(z)$ for any $i\in I_m$}.
\end{equation}
As $z\in{\mathbb R}^n$ and $r>0$ in \eqref{eq:wi0-fff} being arbitrary, we conclude from \cite[Lemma 11.2]{LZ} that $w_{i,0}$ is constant for each $i\in I_m$. Then as $w_{i,0}$ being a nonnegative (global) solution of \eqref{eq:ut0-pde}, $w_{i,0}$ must be trivial for each $i\in I_m$. On the other hand, for any $i\not\in I_m$, $v_i$ is already trivial and so is the limit $w_{i,0}$. Consequently, ${\mathbf w}_0$ is a trivial solution, a contradiction against \eqref{eq:ut0-0}. Therefore, the assumption \eqref{eq:vj-f} must fail, which implies \eqref{eq:har-crit} with some constant $C>0$, depending only on $n$ and $m$. This finishes the proof.
\end{proof}
\subsection{Universal Upper Bounds for $1<\alpha\leq\frac{n+2}{n-2}$}\label{subsection:sup}
With Proposition \ref{proposition:sup-sub}, we obtain a universal upper estimate for (local) singular solutions for the subcritical case. Let us remark that this bound is not sharp for $1<\alpha\leq\frac{n}{n-2}$, although we obtain a universal constant as well as a universal neighborhood in the estimate. The sharp bounds for those cases will be given separately in Section \ref{subsection:asym-} and Section \ref{subsection:asym0}.
\begin{lemma}\label{lemma:u-sup} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in $B_1\setminus\{0\}$ with $1<\alpha<\frac{n+2}{n-2}$. Then there exists $C>0$, depending only on $n$, $m$ and $\alpha$, such that
\begin{equation}\label{eq:u-sup}
|{\mathbf u}(x)| \leq C|x|^{-\frac{2}{\alpha-1}} \quad\text{in }B_{1/2}\setminus\{0\}.
\end{equation}
\end{lemma}
\begin{proof} Let $x_0\in B_{1/2}\setminus\{0\}$ and set $r = \frac{1}{2}|x_0|$. Since $\bar{B}_r(x_0)\subset B_1\setminus\{0\}$, one can define
\begin{equation*}
{\mathbf v}(x) = r^{\frac{2}{\alpha-1}} {\mathbf u}(rx + x_0)\quad\text{in }\bar{B}_1.
\end{equation*}
As ${\mathbf u}$ being a nonnegative solution of \eqref{eq:main} in $B_1\setminus\{0\}$, we see that ${\mathbf v}$ is a nonnegative solution of \eqref{eq:main-reg}. Moreover, ${\mathbf v}$ is continuous up to the boundary of $B_1$. Hence, Proposition \ref{proposition:sup-sub} applies to ${\mathbf v}$ and taking $x=0$ in \eqref{eq:sup} we obtain \begin{equation*}
|{\mathbf v}(0)| \leq C,
\end{equation*}
which in terms of ${\mathbf u}$ can be rephrased as
\begin{equation*}
|{\mathbf u}(x_0)| \leq Cr^{-\frac{2}{\alpha-1}}.
\end{equation*}
Since $x_0\in B_{1/2}\setminus\{0\}$ was arbitrary and $r=\frac{1}{2}|x_0|$, the proof is finished.
\end{proof}
\begin{remark}\label{remark:u-sup}
For $1<\alpha<\frac{n+2}{n-2}$, one may take an alternative approach as follows. Let $w=u_1+u_2+\dots+u_m$. Then $w\geq 0$ and $\frac{1}{c_1} w\leq |{\mathbf u}|\leq c_1w$ in $B_1\setminus\{0\}$ with $c_1 = m^{\frac{1}{2}}$. Hence, $w$ satisfies $\frac{1}{c_2} w^{\alpha}\leq -\Delta w\leq c_2 w^\alpha$ in $B_1\setminus\{0\}$ with $c_2 = m^{\frac{\alpha-1}{2}}$. By \cite[Corollary IV]{SZ2002} it follows that $w\leq C|x|^{-\frac{2}{\alpha-1}}$ in $B_{1/2}\setminus\{0\}$, where $C$ depends only on $n$, $m$ and $\alpha$. This together with the inequality $|{\mathbf u}| \leq c_1w$ yields \eqref{eq:u-sup}.
\end{remark}
From the Harnack-type inequality in Proposition \ref{proposition:har-crit}, we obtain an upper estimate for the critical case $\alpha = \frac{n+2}{n-2}$.
\begin{lemma}\label{lemma:u-sup-crit} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in $B_1\setminus\{0\}$ with $\alpha = \frac{n+2}{n-2}$. Then there exists $C>0$, depending only on $n$ and $m$, such that
\begin{equation}\label{eq:u-sup-crit}
\left( \min_{i\in I_m} \inf_{\partial B_{3/4}} u_i\right) |{\mathbf u}(x)| \leq C |x|^{-\frac{n-2}{2}} \quad\text{in }B_{1/2}\setminus\{0\},
\end{equation}
where $I_m$ consists of all indices $1\leq i\leq m$ such that $u_i$ is nontrivial.
\end{lemma}
\begin{proof} If ${\mathbf u}$ has a removable singularity at the origin, then $-\Delta {\mathbf u} = |{\mathbf u}|^{\frac{4}{n-2}} {\mathbf u}$ in $B_1$ (instead of $B_1\setminus\{0\}$), whence one may apply Proposition \ref{proposition:har-crit} to ${\mathbf u}$ after scaling, and observe that
\begin{equation*}
\left( \min_{i\in I_m} \inf_{\partial B_{3/4}} u_i\right) |{\mathbf u}(x)| \leq C\left(\frac{3}{4}-|x|\right)^{-\frac{n-2}{2}} \leq C\left(\frac{3}{4}\right)^{-\frac{n-2}{2}} \quad\text{in }B_{1/2}\setminus\{0\},
\end{equation*}
which implies \eqref{eq:u-sup-crit}.
Henceforth, let us assume that ${\mathbf u}$ does not have a removable singularity at the origin. Clearly $I_m\neq \emptyset$, and by the super-harmonicity and the non-negativity of $u_i$ with $i\in I_m$, we have $u_i > 0$ in $B_1\setminus\{0\}$ for all $i\in I_m$.
Now let $x_0\in B_{1/2}\setminus\{0\}$ and $r = \frac{1}{8}|x_0|$. Since $\bar{B}_{2r}(x_0)\subset B_1\setminus\{0\}$, one can define
\begin{equation*}
{\mathbf v}(x) = r^{\frac{n-2}{2}} {\mathbf u}(rx + x_0)\quad\text{in }\bar{B}_2.
\end{equation*}
Obviously, $v_i$ is nontrivial if and only if $i\in I_m$. On the other hand, as ${\mathbf u}$ being a nonnegative solution of \eqref{eq:main} in $B_1\setminus\{0\}$, ${\mathbf v}$ becomes a nonnegative solution of \eqref{eq:main-crit}. Hence, it follows from \eqref{eq:har-crit} that
\begin{equation}\label{eq:v0-sup-crit}
|{\mathbf v}(0)| \leq C\left( \min_{i\in I_m} \inf_{\partial B_2} v_i\right)^{-1} = C \left( \min_{i \in I_m} \inf_{B_{2r}(x_0)} u_i \right)^{-1},
\end{equation}
where $C>0$ depends only on $n$ and $m$.
Now let $J_m\subset I_m$ consists of all components $u_i$ having non-removable singularity at the origin. Note that $J_m$ may not be equal to $I_m$. By the super-harmonicity and the positivity, the maximum principle implies that $\liminf_{x\rightarrow 0} u_i(x) = \infty$ for each $i\in J_m$. On the other hand, if $i\in I_m\setminus J_m$ (provided that $I_m\setminus J_m \neq \emptyset$), $u_i$ is bounded at the origin, and again by the maximum principle, one has $\liminf_{x\rightarrow 0}u_i(x) \geq \inf_{\partial B_{3/4}} u_i$. Hence, one should have $\inf_{\partial B_{2r}(x_0)} u_i \geq \inf_{\partial B_{3/4}} u_i$ for any $i\in I_m$. This along with \eqref{eq:v0-sup-crit} yields
\begin{equation*}
|{\mathbf u}(x_0)| \leq C \left( \min_{i\in I_m} \inf_{B_{3/4}} u_i \right)^{-1} r^{-\frac{n-2}{2}},
\end{equation*}
which proves the lemma.
\end{proof}
\begin{remark}\label{remark:u-sup-crit} We shall obtain later in Section \ref{subsection:asym+crit} without the term in the parenthesis, provided that ${\mathbf u}$ has a non-removable singularity at the origin.
\end{remark}
Due to Lemma \ref{lemma:u-sup} and Lemma \ref{lemma:u-sup-crit}, we obtain the standard Harnack inequality and interior gradient estimate.
\begin{lemma}\label{lemma:u-har-Du} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in $B_1\setminus\{0\}$ with $1<\alpha\leq\frac{n+2}{n-2}$. Then there exists $C>0$ such that for each $1\leq i\leq m$,
\begin{equation}\label{eq:u-har}
\sup_{B_r\setminus \bar{B}_{r/2}} u_i \leq C \inf_{B_r\setminus \bar{B}_{r/2}} u_i \quad\text{for any }0<r<\frac{1}{2},
\end{equation}
and
\begin{equation}\label{eq:Du-Linf}
|\nabla u_i(x)| \leq C\frac{u_i(x)}{|x|} \quad\text{in }B_{1/2}\setminus\{0\}.
\end{equation}
Moreover, the constant $C$ in \eqref{eq:u-har} depends only on $n$, $m$ and $\alpha$, provided that $1<\alpha<\frac{n+2}{n-2}$.
\end{lemma}
\begin{proof} After a scaling argument we may also say that \eqref{eq:u-sup} and \eqref{eq:u-sup-crit} hold in $B_{3/4}\setminus\{0\}$, instead of $B_{1/2}\setminus\{0\}$. Consider $u_i$, $1\leq i\leq m$, as a nonnegative solution of $-\Delta u_i = a(x)u_i$ in $B_1\setminus\{0\}$, where $a(x)=|{\mathbf u}|^{\alpha-1}$. Due to \eqref{eq:u-sup} if $1<\alpha<\frac{n+2}{n-2}$, and to \eqref{eq:u-sup-crit} if $\alpha=\frac{n+2}{n-2}$, we know that $0\leq a(x)\leq C|x|^{-2}$ in $B_{3/4}\setminus\{0\}$. Thus, \eqref{eq:u-har} follow easily from the classical Harnack inequality \cite[Corollary 9.25]{GT}. With \eqref{eq:u-har} at hand, one may also prove \eqref{eq:Du-Linf} by the classical gradient estimate \cite[Theorem 3.9]{GT}.
\end{proof}
\section{Asymptotic Radial Symmetry of Local Solutions}\label{section:asym-rad}
This section is devoted to the proof of Theorem \ref{theorem:asym-rad}. Let us address that a similar argument was also used in \cite[Theorem 1.2]{CJSX}, which is concerned with fractional Laplacian, scalar equations.
\begin{proof}[Proof of Theorem \ref{theorem:asym-rad}]
If the origin is a removable singularity, then the conclusion \eqref{eq:asym-rad} is clear. Hence, we shall assume that the origin is a non-removable singularity.
Recall from \eqref{eq:kelvin} that ${\mathbf u}_{z,r}^*$ is the Kelvin transform of ${\mathbf u}$ with respect to the sphere $\partial B_r(z)$. Since the origin is a non-removable singularity of ${\mathbf u}$, one may prove, with a minor modification of the proof of Lemma \ref{lemma:rad}, that there is some small $\epsilon>0$ such that for any $z\in B_{\epsilon/2}\setminus\{0\}$ and any $0<r\leq |z|$,
\begin{equation}\label{eq:asym-rad1}
(u_i)_{z,r}^* \leq u_i\quad\text{in $B_1\setminus (B_r(z)\cup\{0\})$ for each $1\leq i\leq m$}.
\end{equation}
The key observation here is that \eqref{eq:asym-rad1} implies, for any $a>\frac{1}{\epsilon}$ and $e\in\partial B_1$,
\begin{equation}\label{eq:asym-rad2}
u_i^*(y) \leq u_i^*(y_a)\quad\text{if $y\cdot e > a$ and $|y_a|>1$ for each $1\leq i\leq m$},
\end{equation}
where
\begin{equation*}
u_i^*(y) = (u_i)_{0,1}^*(y) = |y|^{2-n} u_i(|y|^{-2}y),\quad y_a = y + 2(a-y\cdot e)e,
\end{equation*}
and $H_a(e)$ is the half-space $\{x:x\cdot e > a\}$. Note that $y_a$ is the reflection point of $y$ with respect to the hyperplane $\partial H_a(e)$. To prove the claim \eqref{eq:asym-rad2}, let us note first that $y\in B_{1/\epsilon}$ if and only if $\frac{y}{|y|^2} \in B_\epsilon$. Now we shall choose some $z\in B_{\epsilon/2}\setminus\{0\}$ and some $0<r<|z|$ such that
\begin{equation}\label{eq:asym-rad3}
\frac{y_a}{|y_a|^2} - z = \left(\dfrac{r}{\left| \frac{y}{|y|^2} - z\right|}\right)^2 \left( \frac{y}{|y|^2} - z \right).
\end{equation}
In other words, $\frac{y_a}{|y_a|^2}$ is the reflection point of $\frac{y}{|y|^2}$ with respect to $\partial B_r(z)$. We shall ask in addition that
\begin{equation}\label{eq:asym-rad4}
\frac{|y_a|}{|y|} \leq \frac{1}{r}\left|\frac{y}{|y|^2} - z\right|.
\end{equation}
Before we actually find such $z$ and $r$, let us verify that along with \eqref{eq:asym-rad3} and \eqref{eq:asym-rad4}, \eqref{eq:asym-rad1} implies \eqref{eq:asym-rad2} as follows.
Given $y\in{\mathbb R}^n$ such that $y\cdot e> a$ and $|y_a|>1$, and $0<r<|z|<\frac{\epsilon}{2}$ such that \eqref{eq:asym-rad3} and \eqref{eq:asym-rad4} hold, let us write by $x$ and $x_{z,r}^*$ the points $\frac{y}{|y|^2}$ and respectively $\frac{y_a}{|y_a|^2}$. Then since $y\cdot e>a>\frac{1}{\epsilon}$ and $|y_a|>1$, we have $x\in B_r(z)$, and $x_{z,r}^*\in B_1\setminus B_r(z)$. Hence, one may proceed, using \eqref{eq:asym-rad1}, as
\begin{equation*}
\begin{split}
u_i^*(y) & = \frac{1}{|y|^{n-2}} \left(\frac{|x_{z,r}^* - z|}{r}\right)^{n-2} (u_i)_{z,r}^*(x_{z,r}^*) \\
& \leq \frac{1}{|y|^{n-2}} \left(\frac{|x_{z,r}^* - z|}{r}\right)^{n-2} u_i(x_{z,r}^*) \\
& \leq u_i^*(y_a),
\end{split}
\end{equation*}
proving \eqref{eq:asym-rad2}, where in deriving the first equality we used \eqref{eq:asym-rad3} while the last inequality follows from \eqref{eq:asym-rad4}. Thus, we only need to prove that there actually exist $0<r<|z|<\frac{\epsilon}{2}$ satisfying \eqref{eq:asym-rad3} and \eqref{eq:asym-rad4}. However, it only involves an elementary argument to verify \eqref{eq:asym-rad3} and \eqref{eq:asym-rad4} as well as $0<r\leq |z|<\frac{\epsilon}{2}$, by choosing $r=|z|$ and
\begin{equation*}
z = \frac{1}{|y|^2}y + \frac{|y_a|^2}{|y|^2 - |y_a|^2}\left( \frac{1}{|y|^2}y- \frac{1}{|y_a|^2}y_a \right) = \frac{1}{|y|^2 - |y_a|^2}(y-y_a).
\end{equation*}
With the claim \eqref{eq:asym-rad2} at hand, one may invoke \cite[Theorem 6.1 and Corollary 6.2]{CGS} to finish the proof. That is, from the former one obtains some $C>0$, independent of $\epsilon$, such that
\begin{equation*}
u_i^*(y) \leq u_i^*(x) \quad\text{if }|x|>1\text{ and }|y|\geq |x| + \frac{C}{\epsilon}\text{ for each $1\leq i\leq m$}.
\end{equation*}
As $u_i^*$ being a nonnegative superharmonic function, the latter implies
\begin{equation*}
u_i^* = \left( 1+ O\left( \frac{1}{R}\right) \right) \left(\inf_{\partial B_R} u_i^*\right)\quad\text{uniformly on $\partial B_R$ as $R\rightarrow\infty$},
\end{equation*}
which in terms of $u_i$ implies the asymptotic radial symmetry claimed as in \eqref{eq:asym-rad}. Hence, the proof is finished.
\end{proof}
With the asymptotic radial symmetry as well as the uniform estimate achieved in the previous section, we are ready to prove Proposition \ref{proposition:Phi*-const}, finally showing the existence of the second Pohozaev invariant (see \eqref{eq:kappa*}).
\begin{proof}[Proof of Proposition \ref{proposition:Phi*-const}] Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in $B_R\setminus\{0\}$ with $\alpha = \frac{n+2}{n-2}$, and let $\Phi_*(r,{\mathbf u})$ be as in \eqref{eq:Phi*}. To avoid the triviality, let us also assume that ${\mathbf u}$ is a nontrivial solution. Let us prove the well-definedness of $\Phi_*(r,{\mathbf u})$.
In the following, we shall denote by $C$ a positive generic constant independent of $r$. With $f(r,{\mathbf u})$ given as in \eqref{eq:f}, it follows immediately from \eqref{eq:u-sup-crit} and \eqref{eq:Du-Linf} that
\begin{equation}\label{eq:f-Lip}
f(r,{\mathbf u}) \leq C\quad\text{and}\quad r|\dot{f}(r,{\mathbf u})| \leq C f(r,{\mathbf u}) \quad\text{for any }0<r<\frac{R}{2}.
\end{equation}
On the other hand, by the asymptotic radial symmetry \eqref{eq:asym-rad}, we have
\begin{equation*}
|\Delta ({\mathbf u} - \bar{\mathbf u})| \leq C|x||\bar{\mathbf u}|^{\frac{n+2}{n-2}}\quad\text{ in } B_{2r}\setminus \bar{B}_r, \quad \text{as } r\rightarrow 0+,
\end{equation*}
where $\bar{\mathbf u}(r)$ is the average of ${\mathbf u}$ over the sphere $\partial B_r$. Hence, it follows from the interior gradient estimate \cite[Theorem 3.9]{GT} and the Harnack inequality \eqref{eq:u-har} that
\begin{equation*}
|\nabla ({\mathbf u}-\bar{\mathbf u})| \leq C|{\mathbf u}|\quad\text{on } \partial B_r,
\end{equation*}
and in particular,
\begin{equation}\label{eq:Dut-Linf}
|\nabla_\sigma {\mathbf u} | \leq C|{\mathbf u}|\quad\text{on }\partial B_r,
\end{equation}
where $\nabla_\sigma {\mathbf u}$ is the tangential derivative of ${\mathbf u}$ on $\partial B_r$.
By means of \eqref{eq:Dut-Linf} and \eqref{eq:f-Lip}, we deduce that
\begin{equation}\label{eq:Phi*-1}
\left|\int_0^r \left(\frac{\rho}{n\omega_n}\int_{\partial B_\rho} |\nabla_\sigma {\mathbf u}|^2\,d\sigma\right) \dot{f}(\rho,{\mathbf u})\,d\rho\right| \leq C\int_0^r \rho f(\rho,{\mathbf u})^2 \,d\rho,
\end{equation}
provided that $r>0$ is sufficiently small. Similarly, one may also prove from \eqref{eq:asym-rad} and \eqref{eq:f-Lip} that
\begin{equation}\label{eq:Phi*-2}
\left|\int_0^r \left(\frac{\rho}{n\omega_n}\int_{\partial B_\rho} |{\mathbf u}|^{\frac{2n}{n-2}}\,d\rho - f(\rho,{\mathbf u})^{\frac{n}{n-2}}\right)\dot{f}(\rho,{\mathbf u})\,d\rho\right| \leq C\int_0^r \rho f(\rho,{\mathbf u})^{\frac{2n-2}{n-2}} \,d\rho.
\end{equation}
By the first inequality in \eqref{eq:f-Lip}, we see that the right sides of both \eqref{eq:Phi*-1} and \eqref{eq:Phi*-2} are of order $r^2$, proving the well-definedness of $\Phi_*(r,{\mathbf u})$.
Proving that $\Phi_*(r,{\mathbf u})$ is indeed constant in $0<r<R$ is now easy by considering the cylindrical version $\Psi_*(t,{\mathbf v})$ defined as in \eqref{eq:Psi*}. Since the computation is very similar with \eqref{eq:Psi*'}, we omit the details.
\end{proof}
\section{Exact Asymptotic Behavior of Local Solutions}\label{section:asym}
With the {\it a priori} estimates and the classification of the solutions on the punctured space, we are now ready to investigate exact asymptotic behavior of local solutions near the isolated singularity at the origin. Before we begin our analysis, let us provide the basic integrability of the solution.
\begin{lemma}\label{lemma:dist} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in $B_1\setminus\{0\}$ with $\alpha>1$. One has ${\mathbf u} \in L^\alpha(B_1;{\mathbb R}^m)$. In particular, if $\alpha \geq \frac{n}{n-2}$, then ${\mathbf u}$ is a distribution solution of \eqref{eq:main} in $B_1\setminus\{0\}$ in $B_1$, that is,
\begin{equation*}
- \int_{B_1} {\mathbf u}\cdot \Delta {\mathbf v} \,dx= \int_{B_1} |{\mathbf u}|^{\alpha-1}{\mathbf u}\cdot{\mathbf v} \,dx \quad\text{for any }{\mathbf v} \in C_0^\infty(B_1;{\mathbb R}^m).
\end{equation*}
\end{lemma}
\begin{proof}
Recall from the proof of Proposition \ref{proposition:sup-sub} and Remark \ref{remark:u-sup} that $w = u_1 + \cdots + u_m$ satisfies $\frac{1}{c} w^\alpha \leq -\Delta w \leq c w^\alpha$, with some $c>1$ depending only on $m$ and $\alpha$. By \cite{BL}, $w\in L^\alpha(B_1)$ which implies that ${\mathbf u}\in L^\alpha(B_1;{\mathbb R}^m)$. The second assertion can be proved similarly as in \cite{CGS}, and we omit the details.
\end{proof}
\subsection{Case $\frac{n}{n-2}<\alpha<\frac{n+2}{n-2}$}\label{subsection:asym+sub}
The upper bound \eqref{eq:u-sup} and the classification of solutions on the punctured space allow us to capture the exact asymptotic behavior of local solutions to \eqref{eq:main}, by means of the blowup analysis. Let us recall from Section \ref{section:sing} that a blowup ${\mathbf u}_0$ is a limit of ${\mathbf u}_r$ along a sequence $r=r_j\rightarrow 0+$ in $C_{loc}^2({\mathbb R}^n\setminus\{0\};{\mathbb R}^m)$.
\begin{lemma}\label{lemma:hom} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in $B_1\setminus\{0\}$ with $\frac{n}{n-2}<\alpha<\frac{n+2}{n-2}$, and let $\Phi(r,{\mathbf u})$ be as in \eqref{eq:Phi}. Then $\Phi(0+,{\mathbf u})\in \{-\bar\lambda,0\}$, where $\bar\lambda$ is given by \eqref{eq:lamb}. Moreover, the following are true.
\begin{enumerate}[(i)]
\item $\Phi(0+,{\mathbf u}) = 0$ if and only if
\begin{equation}\label{eq:hom-reg}
|{\mathbf u}(x)| = o(|x|^{-\frac{2}{\alpha-1}})\quad\text{as }x\rightarrow 0.
\end{equation}
\item $\Phi(0+,{\mathbf u}) = -\bar\lambda$ if and only if
\begin{equation}\label{eq:hom-sing}
|{\mathbf u}(x)| = (1+ o(1)) \lambda^{\frac{1}{\alpha-1}} |x|^{-\frac{2}{\alpha-1}}\quad\text{as }x\rightarrow 0,
\end{equation}
where $\lambda$ is given by \eqref{eq:lam-mu}.
\end{enumerate}
\end{lemma}
\begin{proof} Due to the estimates \eqref{eq:u-sup} and \eqref{eq:Du-Linf}, we know that $\Phi(r,{\mathbf u})$ in \eqref{eq:Phi} is uniformly bounded for all $0<r<\frac{1}{2}$. This combined with the monotonicity (Proposition \ref{proposition:Phi-monot} (i)) implies that $\Phi(0+,{\mathbf u})$ exists. Hence, we may argue analogously as the proof of Lemma \ref{lemma:u-Phi} and observe that any blowup ${\mathbf u}_0$ of ${\mathbf u}$ satisfies $\Phi(r,{\mathbf u}_0) = \Phi(0+,{\mathbf u})$ for all $r>0$. As ${\mathbf u}_0$ being a nonnegative solution of \eqref{eq:main} in ${\mathbb R}^n\setminus\{0\}$, it follows from Lemma \ref{lemma:class-sub} (ii) that $\Phi(0+,{\mathbf u}) = 0$ if and only if any blowup ${\mathbf u}_0$ of ${\mathbf u}$ is trivial, while $\Phi(0+,{\mathbf u}) = -\bar\lambda$ if and only if any blowup of ${\mathbf u}_0$ is of the form $\lambda^{\frac{1}{\alpha-1}} |x|^{-\frac{2}{\alpha-1}}{\mathbf e}$ with some nonnegative unit vector ${\mathbf e}\in{\mathbb R}^m$. In other words, $\Phi(0+,{\mathbf u}) = 0$ if and only if $|{\mathbf u}_r| \rightarrow 0$ uniformly on $\partial B_1$, while $\Phi(0+,{\mathbf u}) = - \bar\lambda$ if and only if $|{\mathbf u}_r| \rightarrow \lambda^{\frac{1}{\alpha-1}}$ uniformly on $\partial B_1$, where ${\mathbf u}_r$ is the scaling function defined by \eqref{eq:ur}. This finishes the proof.
\end{proof}
The next lemma shows that \eqref{eq:hom-reg} is sufficient for the origin to be a removable singularity.
\begin{lemma}\label{lemma:remv+sub} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in $B_1\setminus\{0\}$ with $\frac{n}{n-2}<\alpha<\frac{n+2}{n-2}$. If ${\mathbf u}$ satisfies
\begin{equation}\label{eq:remv+sub}
|{\mathbf u}(x)| = o(|x|^{-\frac{2}{\alpha-1}} )\quad\text{as }x\rightarrow 0,
\end{equation}
then the origin is a removable singularity.
\end{lemma}
\begin{proof} Under the assumption \eqref{eq:remv+sub}, we claim that
\begin{equation}\label{eq:remv-re}
|{\mathbf u}(x)| \leq c|x|^{-\frac{2}{\alpha-1} + \delta}\quad\text{in }B_{r_0}\setminus\{0\},
\end{equation}
for some $\delta>0$, $r_0>0$ and $c>1$, where $c$ and $r_0$ may depend on ${\mathbf u}$.
Consider the auxiliary function
\begin{equation}\label{eq:vpe}
\varphi_\epsilon (x) = ( C_0 r_0^{-\delta} |x|^\delta + \epsilon )|x|^{-\frac{2}{\alpha-1}}\quad\text{in }{\mathbb R}^n\setminus\{0\},
\end{equation}
where $C_0>0$ is the (universal) constant from \eqref{eq:u-sup}, $r_0>0$ is a small radius to be determined later and $\epsilon>0$ is an arbitrary small number. By direct computation, we observe that
\begin{equation*}
\Delta \varphi_\epsilon = - ( C_0r_0^{-\delta} (\lambda + \mu\delta -\delta^2) |x|^\delta + \epsilon \lambda )|x|^{\frac{2\alpha}{1-\alpha}}\quad\text{in }{\mathbb R}^n\setminus\{0\},
\end{equation*}
with $\lambda$ and $\mu$ given by \eqref{eq:lam-mu}. Note that for $\alpha > \frac{n}{n-2} $, we have $\lambda > 0$. Thus, taking $\delta>0$ sufficiently small depending only on $\lambda$ and $|\mu|$, we obtain
\begin{equation}\label{eq:vpe-pde0}
\Delta \varphi_\epsilon \leq -\frac{\lambda}{2|x|^2} \varphi_\epsilon\quad\text{in }{\mathbb R}^n\setminus\{0\}.
\end{equation}
Let us fix $1\leq i\leq m$ and consider the $i$-th component $u_i$ of ${\mathbf u}$ as a solution of $\Delta u_i = -a(x) u_i$ in $B_1\setminus\{0\}$ with $a(x) = |{\mathbf u}|^{\alpha-1}$. Due to \eqref{eq:remv+sub}, there exists $r_0>0$ such that $0\leq a(x) \leq \frac{\lambda}{2|x|^2}$ in $B_{r_0}\setminus\{0\}$, and hence, it follows from \eqref{eq:vpe-pde0} that $\varphi_\epsilon$ is a supersolution of $\Delta u_i = -a(x) u_i$ in $B_{r_0}\setminus\{0\}$. That is,
\begin{equation}\label{eq:vpe-pde-re}
\Delta \varphi_\epsilon \leq - a(x) \varphi_\epsilon\quad\text{in }B_{r_0}\setminus\{0\}.
\end{equation}
On the other hand, choosing $C_0>0$ to be the constant for which $|{\mathbf u}|$ satisfies \eqref{eq:u-sup}, we have $u_i \leq C_0r_0^{-\frac{2}{\alpha-1}} \leq \varphi_\epsilon$ on $\partial B_{r_0}$. Utilizing the assumption \eqref{eq:remv+sub} again, one can find a sufficiently small $0<r<r_0$ such that $u_i \leq \epsilon |x|^{-\frac{2}{\alpha-1}} \leq \varphi_\epsilon$ in $B_r\setminus\{0\}$. Therefore,
\begin{equation}\label{eq:vpe-u-bdry}
u_i \leq \varphi_\epsilon\quad\text{on }(\partial B_{r_0}) \cup (B_r\setminus\{0\}).
\end{equation}
In view of \eqref{eq:vpe-pde-re} and \eqref{eq:vpe-u-bdry}, we may apply the maximum principle in $B_{r_0}\setminus B_r$ and obtain $u_i \leq \varphi$ in $B_{r_0}\setminus \bar{B}_r$. Combining this inequality with \eqref{eq:vpe-u-bdry}, we arrive at
\begin{equation}\label{eq:vpe-u-int}
u_i \leq \varphi_\epsilon\quad\text{in }B_{r_0}\setminus\{0\}.
\end{equation}
Since the parameters $C_0$, $r_0$ and $\delta$ in the definition \eqref{eq:vpe} of $\varphi_\epsilon$ are independent of $\epsilon$, we can take $\epsilon\rightarrow 0$ in \eqref{eq:vpe-u-int} and obtain
\begin{equation*}
u_i(x) \leq C_0r_0^{-\delta}|x|^{-\frac{2}{\alpha-1} + \delta}\quad\text{in }B_{r_0}\setminus\{0\}.
\end{equation*}
Now that this inequality holds for any $1\leq i\leq m$, we arrive at \eqref{eq:remv-re} with $c = C_0r_0^{-\delta}\sqrt{m}$.
Since $a(x) = |{\mathbf u}|^{\alpha-1}$, we have from \eqref{eq:remv-re} that $0\leq a(x) \leq c|x|^{-2+(\alpha-1)\delta}$ on $B_{r_0}\setminus\{0\}$, which certainly implies $a\in L^{\frac{n}{2-\eta}}(B_1)$ for some small $\eta>0$. According to Lemma \ref{lemma:dist}, $u_i$ satisfies $-\Delta u_i = a(x) u_i$ in $B_1$ in the distributional sense for each $1\leq i\leq m$, whence the classical result by Serrin \cite[Theorem 1]{S} yields that $u_i$ has a removable singularity at the origin. This proves the lemma.
\end{proof}
\begin{remark}\label{remark:remv+sub} One may have noticed that the proof of Lemma \ref{lemma:remv+sub} works for the upper critical case, $\alpha = \frac{n+2}{n-2}$, without any modification.
\end{remark}
We are ready to prove Theorem \ref{theorem:main} (i).
\begin{proof}[Proof of Theorem \ref{theorem:main}] Suppose that ${\mathbf u}$ has a non-removable singularity at the origin. Then by Lemma \ref{lemma:remv+sub}, ${\mathbf u}$ does not satisfy \eqref{eq:remv+sub}, whence it follows from Lemma \ref{lemma:hom} that ${\mathbf u}$ satisfies \eqref{eq:hom-sing}, which proves \eqref{eq:asym+sub}
\end{proof}
\subsection{Case $\alpha = \frac{n+2}{n-2}$}\label{subsection:asym+crit}
The asymptotic behavior for the case $\alpha = \frac{n+2}{n-2}$ becomes more subtle, due to the presence of the second Pohozaev invariant $\kappa_*$ given by \eqref{eq:kappa*}. The following lemma is the local version of Theorem \ref{theorem:main-s} (iii). Let us remark that the proof is similar to the classical argument (c.f. the proof of \cite[Theorem 1.2]{CGS}); however, the key difference is that we apply the radial symmetry to the second Pohozaev identity \eqref{eq:kappa*}, instead of the first identity \eqref{eq:kappa}.
\begin{lemma}\label{lemma:hom+} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in $B_1\setminus\{0\}$ with $\alpha=\frac{n+2}{n-2}$. Also set $\kappa({\mathbf u})$ and $\kappa_*({\mathbf u})$ as in \eqref{eq:kappa} and respectively \eqref{eq:kappa*}. Then $\kappa({\mathbf u})$ and $\kappa_*({\mathbf u})$ satisfy \eqref{eq:class-crit-kappa} and respectively \eqref{eq:class-crit-kappa*}. Moreover, the following are true.
\begin{enumerate}[(i)]
\item $\kappa({\mathbf u}) = \kappa_*({\mathbf u}) = 0$ if and only if
\begin{equation}\label{eq:hom+-reg}
|{\mathbf u}(x)| = o(|x|^{-\frac{n-2}{2}}) \quad\text{as }x\rightarrow 0.
\end{equation}
\item $\kappa({\mathbf u})^2 + \kappa_*({\mathbf u})^2 >0$ if and only if there are $c,C>0$ such that
\begin{equation}\label{eq:hom+-sing}
c|x|^{-\frac{n-2}{2}} \leq |{\mathbf u}(x)| \leq C|x|^{-\frac{n-2}{2}}\quad\text{as }x\rightarrow 0,
\end{equation}
where $c$ depends on ${\mathbf u}$ while $C$ is determined by $n$ and $m$ only.
\item $\kappa({\mathbf u}) = -\frac{2}{n} (\frac{n-2}{2})^n$ and $\kappa_*({\mathbf u}) = 0$ if and only if
\begin{equation}\label{eq:hom+-hom}
|{\mathbf u}(x)| = (1+ o(1)) \left(\frac{n-2}{2}\right)^{\frac{n-2}{2}} |x|^{-\frac{n-2}{2}}\quad\text{as }x\rightarrow 0.
\end{equation}
\end{enumerate}
\end{lemma}
\begin{proof} The existence of $\kappa({\mathbf u})$ and $\kappa_*({\mathbf u})$ are proved in Proposition \ref{proposition:Phi-monot} (ii) and respectively Proposition \ref{proposition:Phi*-const}. Now let ${\mathbf u}_0$ be any blowup of ${\mathbf u}$, and write $r_j\rightarrow 0+$ by the blowup sequence. By the scaling relation \eqref{eq:Phi-scale} of $\Phi(r,{\mathbf u})$, we see that
\begin{equation*}
\kappa({\mathbf u}_0) = \Phi(1,{\mathbf u}_0) = \lim_{j\rightarrow\infty} \Phi(1,{\mathbf u}_{r_j}) = \lim_{j\rightarrow\infty} \Phi(r_j,{\mathbf u}) = \kappa({\mathbf u}).
\end{equation*}
However, ${\mathbf u}_0$ is a nonnegative solution of \eqref{eq:main} (with $\alpha = \frac{n+2}{n-2}$) in ${\mathbb R}^n\setminus\{0\}$, whence Lemma \ref{lemma:class-crit} yields $\kappa({\mathbf u}_0)$ satisfies \eqref{eq:class-crit-kappa}, and so does $\kappa({\mathbf u})$. Similarly, one may deduce from the scaling relation \eqref{eq:Phi*-scale} of $\Phi_*(r,{\mathbf u})$ that $\kappa_*({\mathbf u}) = \kappa_*({\mathbf u}_0)$, and by Lemma \ref{lemma:class-crit}, $\kappa_*({\mathbf u})$ verifies \eqref{eq:class-crit-kappa*}.
Suppose that $\kappa({\mathbf u}) = \kappa_*({\mathbf u}) = 0$, and let ${\mathbf v}$ be the cylindrical transformation of ${\mathbf u}$ as in \eqref{eq:cyl}. Rephrasing the estimates \eqref{eq:Phi*-1} and \eqref{eq:Phi*-2} in terms of ${\mathbf v}$, the second Pohozaev identity \eqref{eq:kappa*-cyl} becomes (as $t\rightarrow\infty$),
\begin{equation}\label{eq:kappa*-cyl2}
(g')^2 = (n-2)^2 g^2 - \frac{4(n-2)}{n} g^{\frac{2n-2}{n-2}} + O \left( \int_t^\infty e^{-2\tau} g(\tau)^2\,d\tau\right),
\end{equation}
where $g$ is given by \eqref{eq:g} and $g'=dg/dt$. Since the term $O(\int_t^\infty e^{-2\tau} g(\tau)^2\,d\tau)$ decays exponentially, and is comparably smaller than $g(t)$, the behavior of $g'$ is determined by the nonnegative roots of
\begin{equation*}
(n-2)^2g^2 - \frac{4(n-2)}{n} g^{\frac{2n-2}{n-2}} = 0,
\end{equation*}
which are $0$ and $(\frac{n(n-2)}{4})^{\frac{n-2}{2}}$ respectively. In particular, $g(t)$ must be either non-increasing and converging to $0$, or nondecreasing and converging to $(\frac{n(n-2)}{4})^{\frac{n-2}{2}}$.
If $g(t)\rightarrow 0$ as $t\rightarrow\infty$, then by the asymptotic radial symmetry we have $|{\mathbf v}(t,\cdot)| \rightarrow 0$ uniformly on ${\mathbb S}^{n-1}$ as $t\rightarrow \infty$. After the inverse cylindrical transform via \eqref{eq:cyl}, we arrive at \eqref{eq:hom+-reg}, as desired.
Now let us show that the other alternative, i.e., $g(t)\rightarrow (\frac{n(n-2)}{4})^{\frac{n-2}{2}}$ as $t\rightarrow\infty$, cannot occur. Suppose that this is true. Then again from the asymptotic radial symmetry it follows that $|{\mathbf u}_r|\rightarrow (\frac{n(n-2)}{4})^{\frac{n-2}{2}}$ uniformly on $\partial B_1$ as $r\rightarrow 0+$. This implies that any blowup ${\mathbf u}_0$ of ${\mathbf u}$ must be of the form $(\frac{n(n-2)}{4})^{\frac{n-2}{2}}|x|^{-\frac{n-2}{2}}{\mathbf e}$ for some nonnegative unit vector ${\mathbf e}\in{\mathbb R}^m$. In particular, ${\mathbf u}_0$ has a non-removable singularity at the origin, and hence Theorem \ref{theorem:main-s} (iii) yields that $\kappa({\mathbf u}_0)$ or $\kappa_*({\mathbf u}_0)$ is non-zero, a contradiction to $\kappa({\mathbf u}) = \kappa({\mathbf u}_0) = 0$ or, respectively, $\kappa_*({\mathbf u}) = \kappa_*({\mathbf u}_0) = 0$. Hence, the assertion (i) is proved.
Now let us consider the case when $\kappa({\mathbf u})^2 + \kappa_*({\mathbf u})^2>0$. Let ${\mathbf u}_0$ be any blowup of ${\mathbf u}$. Then due to the asymptotic radial symmetry of ${\mathbf u}$, ${\mathbf u}_0$ is radially symmetric on the punctured space. Hence, by Lemma \ref{lemma:u-s}, we know that $|{\mathbf u}_0| \leq C|x|^{-\frac{n-2}{2}}$ where $C>0$ depends only on $n$ and $m$. Since ${\mathbf u}_0$ is an arbitrary blowup of ${\mathbf u}$, this proves the upper bound in \eqref{eq:hom+-sing}.
On the other hand, by Theorem \ref{theorem:main-s} (iii)-(b), the cylindrical transform ${\mathbf v}_0$ of ${\mathbf u}_0$ satisfies \eqref{eq:|v|-poho}. Due to R. H. Fowler \cite{F}, $|{\mathbf v}_0|$ has to be bounded uniformly away from zero, with the bound determined solely on the value of $n$, $\kappa(v_0) =\kappa(u_0) = \kappa(u)$ and $\kappa_*({\mathbf v}_0) = \kappa_*({\mathbf u})$. This proves that $|{\mathbf u}_0| \geq c|x|^{-\frac{n-2}{2}}$ for some $c>0$ depending only on $n$, $\kappa({\mathbf u})$ and $\kappa_*({\mathbf u})$. Since $c$ is independent of the blowup ${\mathbf u}_0$, the lower bound in \eqref{eq:hom+-sing} is proved. Thus, the assertion (ii) is proved.
The final assertion regarding \eqref{eq:hom+-hom} follows immediately from Theorem \ref{theorem:main-s} (iii)-(c), since the latter implies that the blowup of ${\mathbf u}$ is unique and is of the form \eqref{eq:u-s-2}, if and only if $\kappa({\mathbf u}) = -\frac{2}{n}(\frac{n-2}{2})^n$ and $\kappa_*({\mathbf u}) = 0$.
\end{proof}
As with Lemma \ref{lemma:remv+sub}, we observe that \eqref{eq:hom+-reg} is a sufficient condition to have a removable singularity.
\begin{lemma}\label{lemma:remv+crit} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in $B_1\setminus\{0\}$ with $\alpha=\frac{n+2}{n-2}$. If ${\mathbf u}$ satisfies
\begin{equation*}
|{\mathbf u}(x)| = o(|x|^{-\frac{n-2}{2}} )\quad\text{as }x\rightarrow 0,
\end{equation*}
then the origin is a removable singularity.
\end{lemma}
\begin{proof} As mentioned in Remark \ref{remark:remv+sub}, the same proof of Lemma \ref{lemma:remv+sub} works here as well, whence we leave out the details to the reader.
\end{proof}
\begin{proof}[Proof of Theorem \ref{theorem:main} (ii)]
Suppose that the origin is a non-removable singularity, and let us write by $\kappa$ and $\kappa_*$ the first and respectively the second Pohozaev invariant. As a contraposition to Lemma \ref{lemma:remv+crit}, \eqref{eq:remv+sub} fails. Thus, by Lemma \ref{lemma:hom+}, one has $\kappa^2 + \kappa_*^2 >0$. Then the asymptotic bounds in \eqref{eq:asym+crit} follows from the second alternative, \eqref{eq:hom+-sing}, of Lemma \ref{lemma:hom+}, and the proof is finished.
\end{proof}
\subsection{Case $1<\alpha<\frac{n}{n-2}$}\label{subsection:asym-}
The asymptotic analysis for the case $1<\alpha<\frac{n}{n-2}$ is very simple. It is noticeable that the monotonicity formula is not required here. We also mention that one can reduce our study to the scalar case by considering $w = u_1+ u_2 + \cdots + u_m \geq 0$, and directly apply the results in \cite{L}. Nevertheless, we shall give a more direct proof, for the sake of completeness.
We shall begin with the sharp upper estimate.
\begin{lemma}\label{lemma:sup-} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in $B_1\setminus\{0\}$ with $1<\alpha<\frac{n}{n-2}$. Then there is $C>0$, depending only $|{\mathbf u}|$, such that
\begin{equation}\label{eq:sup-}
|{\mathbf u}(x)| \leq C|x|^{2-n}\quad\text{as } x\rightarrow 0.
\end{equation}
\end{lemma}
\begin{proof} Lemma \ref{lemma:dist} asserts that ${\mathbf u}\in L^\alpha(B_1)$. Since $1<\alpha<\frac{n}{n-2}$ and ${\mathbf u}$ satisfies the Harnack inequality \eqref{eq:u-har}, it is easy to verify that
\begin{equation}\label{eq:u-sup-0}
|{\mathbf u}(x)| = o(|x|^{-\frac{2}{\alpha-1}})\quad\text{as }x\rightarrow 0.
\end{equation}
Utilizing \eqref{eq:u-sup-0}, and noting that $n-2<\frac{2}{\alpha-1}$, one may argue with a blowup argument to prove that for any $n-2<q<\frac{2}{\alpha-1}$, there is some $0<r_q<1$, depending only on $n$, $m$, $\alpha$ and $q$, such that
\begin{equation}\label{eq:u-sup-q}
|{\mathbf u}(x)| < |x|^{-q}\quad\text{in }B_{r_q}\setminus\{0\}.
\end{equation}
Now let $r_q$ be as in \eqref{eq:u-sup-q}. Due to Lemma \ref{lemma:dist} again, $\Delta {\mathbf u} = - |{\mathbf u}|^{\alpha-1}{\mathbf u}\in L^1(B_1)$, whence one can decompose ${\mathbf u}$, in $B_{r_q}\setminus\{0\}$, as
\begin{equation}\label{eq:u-exp}
{\mathbf u}(x) = |x|^{2-n}{\mathbf a} - \int_{B_{r_q}} |x-y|^{2-n} \Delta {\mathbf u}(y)\,dy + {\mathbf h}(x),
\end{equation}
where ${\mathbf a}$ is a nonnegative vector in ${\mathbb R}^m$ and ${\mathbf h}$ is a nonnegative and harmonic, vectorial function on $B_{r_q}$. However, owing to the estimate \eqref{eq:u-sup-q}, it is not hard to see from the equation $\Delta {\mathbf u} = -|{\mathbf u}|^{\alpha-1}{\mathbf u}$ that there is $C_q>0$, depending only on $n$, $m$, $\alpha$ and $q$, such that
\begin{equation}\label{eq:newton}
\left| \int_{B_{r_q}} |x-y|^{2-n} \Delta {\mathbf u}(y)\,dy \right| \leq \int_{B_{r_q}} |x-y|^{2-n} |y|^{-\alpha q}\,dy \leq C_q|x|^{2-n}.
\end{equation}
Thus, choosing $n-2<q<\frac{2}{\alpha-1}$ so as to depend only on $n$ and $\alpha$, and selecting $r_q$ and $C_q$ in \eqref{eq:newton} correspondingly, we derive the sharp estimate \eqref{eq:sup-} from \eqref{eq:u-exp}.
\end{proof}
Next we consider a sufficient condition to have a removable singularity.
\begin{lemma}\label{lemma:remv-} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in $B_1\setminus\{0\}$ with $1<\alpha<\frac{n}{n-2}$. If ${\mathbf u}$ satisfies
\begin{equation}\label{eq:remv-}
|{\mathbf u}(x)| = o( |x|^{2-n} ) \quad\text{as }x\rightarrow 0,
\end{equation}
then the origin is a removable singularity.
\end{lemma}
\begin{proof} Under the assumption \eqref{eq:remv-}, one has ${\mathbf u}\in L^q(B_1;{\mathbb R}^m)$ for any $1\leq q<\frac{n}{n-2}$. Since $1<\alpha<\frac{n}{n-2}$ and $|\Delta {\mathbf u}| \leq |{\mathbf u}|^\alpha$, we have $-\Delta {\mathbf u} \in L^{\frac{q}{\alpha}}(B_1;{\mathbb R}^m)$ for any $\alpha<q<\frac{n}{n-2}$. Thus, the $L^p$ theory \cite[Theorem 9.9]{GT} (applied to each component of ${\mathbf u}$) and a bootstrap argument based on the Sobolev inequality yields ${\mathbf u} \in W^{2,p} (B_1;{\mathbb R}^m)$ for any $1<p<\infty$. In particular, it follows from the Sobolev embedding that ${\mathbf u}\in C^{1,\gamma}(B_1;{\mathbb R}^m)$ for any $0<\gamma<1$, and thus ${\mathbf u}$ must have a removable singularity at the origin.
\end{proof}
We are in a position to prove Theorem \ref{theorem:main} (iii).
\begin{proof}[Proof of Theorem \ref{theorem:main} (iii)] Suppose that ${\mathbf u}$ has a non-removable singularity at the origin. By Lemma \ref{lemma:remv-}, we know that ${\mathbf u}$ does not satisfy \eqref{eq:remv-}, or equivalently, there is some $\delta>0$, a component, say $u_1$, and a sequence $r_j\rightarrow 0+$ such that
\begin{equation*}
\sup_{\partial B_{r_j}} u_1\geq \delta r_j^{2-n}.
\end{equation*}
By the Harnack inequality \eqref{eq:u-har}, we know that
\begin{equation*}
\inf_{\partial B_{r_j}} u_1 \geq c_0\delta r_j^{2-n},
\end{equation*}
where $c_0>0$ depends only on $n$, $m$ and $\alpha$. Taking $\delta>0$ smaller, if necessary, such that $c\delta \leq \inf_{\partial B_{1/2}} u_1$, it follows from the maximum principle that
\begin{equation*}
u_1(x) \geq c_0\delta |x|^{2-n}\quad\text{in }B_{1/2}\setminus\{0\},
\end{equation*}
proving the asymptotic lower bound in \eqref{eq:asym-}. The asymptotic upper bound in \eqref{eq:asym-} is established in Lemma \ref{lemma:sup-}. Hence, the theorem is proved.
\end{proof}
\begin{remark}\label{remark:asym-} As mentioned in the beginning of this section, the proof of Theorem \ref{theorem:main} (iii) can also be deduced by considering the function $w = u_1+u_2 + \cdots + u_m \geq 0$. Then $w$ satisfies $C_1w^\alpha \leq -\Delta w\leq C_2 w^\alpha$ in $B_1\setminus\{0\}$, where $C_1,C_2>0$ depend on $n$, $m$ and $\alpha$ only, and the claim in Theorem \ref{theorem:main} (iii) follows now from existent results in the literature, such as \cite[Theorem 2 and Remark 2]{L}.
\end{remark}
\subsection{Case $\alpha = \frac{n}{n-2}$}\label{subsection:asym0}
The analysis of the lower critical exponent, $\alpha = \frac{n}{n-2}$, exhibits its own subtlety, due to the multiplicity of components in \eqref{eq:main}, as with the upper critical case, $\alpha = \frac{n+2}{n-2}$. To briefly discuss this point, let us first give the asymptotic upper bound.
\begin{lemma}[Lemma 1 in \cite{A2}]\label{lemma:ub-sup-0} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} with $\alpha = \frac{n}{n-2}$ in $B_1\setminus\{0\}$. Then for each $1\leq i\leq m$,
\begin{equation}\label{eq:ub-sup-0}
\bar{u}_i(r) \leq \left(\frac{(n-2)^2}{2}\right)^{\frac{n-2}{2}} r^{2-n} (-\log r)^{\frac{2-n}{2}}\quad\text{as }r\rightarrow 0,
\end{equation}
where $\bar{u}_i$ is the average of $u_i$ over the sphere $\partial B_r$.
\end{lemma}
\begin{proof} Note that for each $1\leq i\leq m$, $\bar{u}_i$ satisfies, for $0<r<1$,
\begin{equation*}
\dot{\bar{u}}_i + \frac{n-1}{r} \dot{\bar{u}}_i + \bar{u}_i^{\frac{n}{n-2}} = 0,
\end{equation*}
whence the conclusion follows directly from \cite[Lemma 1]{A2}.
\end{proof}
Let us remark that the constant $(\frac{1}{2}(n-2)^2)^{\frac{n-2}{2}}$ in \eqref{eq:ub-sup-0} is exact in view of \eqref{eq:asym0}. Due to the fact that ${\mathbf u}$ consists of multiple components, there is not an easy way to prove that $|\bar{\mathbf u}|$ also satisfies \eqref{eq:ub-sup-0} with exactly the same constant. This prevents us from applying the argument in \cite[Section 2]{A2}, which deals with the scalar version of \eqref{eq:main} with $\alpha = \frac{n}{n-2}$. Instead, we mainly follow \cite[Section 3]{A2}, where a sign-changing problem is considered. The idea is to consider several refinements of the usual monotonicity formula $\Psi(t,{\mathbf v})$ introduced in \eqref{eq:Psi}.
Due to the refined upper bound \eqref{eq:ub-sup-0}, we shall consider a new cylindrical transformation ${\boldsymbol\phi}$ defined so as to satisfy
\begin{equation}\label{eq:cyl-n}
{\mathbf u}(x) = |x|^{2-n} (-\log |x|)^{\frac{2-n}{2}} {\boldsymbol\phi}\left(-\log |x|,\frac{x}{|x|}\right).
\end{equation}
Then the problem \eqref{eq:main} (with $\alpha = \frac{n}{n-2}$) can be reformulated in terms of ${\boldsymbol\phi}$ as
\begin{equation}\label{eq:main-cyl-n}
\partial_{tt} {\boldsymbol\phi} + (n-2) \left( 1 - \frac{1}{t} \right) \partial_t {\boldsymbol\phi} + \Delta_\theta {\boldsymbol\phi} = \frac{n-2}{2t} \left(n-2 -\frac{n}{2t}\right) {\boldsymbol\phi} - \frac{1}{t} |{\boldsymbol\phi}|^{\frac{2}{n-2}} {\boldsymbol\phi}.
\end{equation}
\begin{remark}\label{remark:cyl-n} Due to the asymptotic radial symmetry \eqref{eq:asym-rad} of ${\mathbf u}$, ${\boldsymbol\phi}$ satisfies $|{\boldsymbol\phi} - \bar{\boldsymbol\phi}| = O(e^{-\gamma t})$ as $t\rightarrow\infty$, for some $\gamma >0$, where $\bar{\boldsymbol\phi}(t)$ is the average of ${\boldsymbol\phi}(t,\theta)$ over $\theta\in{\mathbb S}^{n-1}$. In particular, one has (by arguing similarly as in the derivation of \eqref{eq:Dut-Linf})
\begin{equation}\label{eq:w-tan-Linf}
|\nabla_\theta {\boldsymbol\phi}(t,\theta)| \leq C e^{-\gamma t}\quad\text{in }(t_0,\infty)\times{\mathbb S}^{n-1},
\end{equation}
for some large $t_0$ and $C$ independent of $t$. Moreover, it follows from the sharp estimate \eqref{eq:ub-sup-0} and the gradient estimate \eqref{eq:Du-Linf} that
\begin{equation}\label{eq:w-dt-Linf}
|{\boldsymbol\phi}(t,\theta)| + |\partial_t {\boldsymbol\phi}(t,\theta)| \leq C\quad\text{in }(t_0,\infty)\times{\mathbb S}^{n-1}.
\end{equation}
\end{remark}
In comparison with \eqref{eq:main-cyl}, we obtain the first refinement of the monotonicity formula $\Psi(t,{\mathbf v})$, given as
\begin{equation}\label{eq:E}
\begin{split}
E(t,{\boldsymbol\phi}) &= \frac{1}{n\omega_n} \int_{{\mathbb S}^{n-1}} \left( t |\partial_t {\boldsymbol\phi}|^2 - t|\nabla_\theta {\boldsymbol\phi}|^2 + \frac{n-2}{n-1} |{\boldsymbol\phi}|^{\frac{2n-2}{n-2}} \right)\,d\theta \\
&\quad - \frac{n-2}{2n\omega_n} \left( n -2 - \frac{n}{2t} \right)\int_{{\mathbb S}^{n-1}} |{\boldsymbol\phi}|^2\,d\theta.
\end{split}
\end{equation}
Note that $E(t,{\boldsymbol\phi})$ is well-defined for any $t$ whenever ${\boldsymbol\phi}(t,\cdot)$ is defined on ${\mathbb S}^{n-1}$, due to the smoothness of ${\mathbf u}$.
The next lemma is concerned with the monotonicity of $E(t,{\boldsymbol\phi})$.
\begin{lemma}\label{lemma:E-monot} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in $B_1\setminus\{0\}$ with $\alpha = \frac{n}{n-2}$, and ${\boldsymbol\phi}$ be the cylindrical transformation as in \eqref{eq:cyl-n}. Then
\begin{equation}\label{eq:E'}
\begin{split}
E'(t,{\boldsymbol\phi}) &= - \frac{(2n-4)t-2n+3}{n\omega_n} \int_{{\mathbb S}^{n-1}} |\partial_t {\boldsymbol\phi}|^2 \,d\theta \\
&\quad - \frac{1}{n\omega_n}\int_{{\mathbb S}^{n-1}} \left( |\nabla_\theta {\boldsymbol\phi}|^2 \,d\theta + \frac{n(n-2)}{4t^2} |{\boldsymbol\phi}|^2 \right)\,d\theta.
\end{split}
\end{equation}
In particular, $E(t,{\boldsymbol\phi})$ is nonincreasing for $t>\frac{2n-3}{2n-4}$, and $E(\infty,{\boldsymbol\phi})$ exists.
\end{lemma}
\begin{proof} The proof of \eqref{eq:E'} follows easily from taking the inner product of \eqref{eq:main-cyl-n} with $t\partial_t{\boldsymbol\phi}$ and integrating the both sides over ${\mathbb S}^{n-1}$. We omit the details.
With \eqref{eq:E'} at hand, we know that $E(t,{\boldsymbol\phi})$ is nonincreasing for $t>\frac{2n-3}{2n-4}$. Thus, the existence of $E(\infty,{\boldsymbol\phi})$ follows immediately from that $E(t,{\boldsymbol\phi})$ is uniformly bounded from below as $t\rightarrow\infty$. However, \eqref{eq:w-tan-Linf} yields
\begin{equation*}
\lim_{t\rightarrow\infty}\int_{{\mathbb S}^{n-1}} t|\nabla_\theta{\boldsymbol\phi}|^2\,d\theta = 0,
\end{equation*}
which along with \eqref{eq:w-dt-Linf} ensures that
\begin{equation*}
\liminf_{t\rightarrow\infty} E(t,{\boldsymbol\phi}) > -\infty,
\end{equation*}
as desired.
\end{proof}
In order to have the full strength of the existence of $E(\infty,{\boldsymbol\phi})$, we shall prove the following, which is the system version of \cite[Lemma 3.2]{A2}. Although the proof is almost identical, we shall present the argument for the sake of completeness.
\begin{lemma}[Essentially due to \cite{A2}]\label{lemma:w-dt-int} Let ${\boldsymbol\phi}$ be as in Lemma \ref{lemma:E-monot}. Then
\begin{equation}\label{eq:w-dt-int}
\lim_{t\rightarrow\infty} \int_{{\mathbb S}^{n-1}} t|\partial_t {\boldsymbol\phi}|^2 \,d\theta = 0.
\end{equation}
\end{lemma}
\begin{proof} By \eqref{eq:w-tan-Linf} and \eqref{eq:w-dt-Linf}, one may integrate the both sides of \eqref{eq:E'} from $t_0 = \frac{2n-3}{2n-4}$ to $\infty$, and use the existence of $E(\infty,\phi)$ to deduce that
\begin{equation}\label{eq:w-dt-int1}
\int_{t_0}^\infty \int_{{\mathbb S}^{n-1}} \tau |\partial_\tau {\boldsymbol\phi}|^2 \,d\theta \,d\tau < \infty.
\end{equation}
Hence, it is sufficient to prove that $\int_{{\mathbb S}^{n-1}} t|\partial_t{\boldsymbol\phi}|^2\,d\theta$ is a Cauchy sequence in $t\rightarrow \infty$.
In order to do so, we differentiate \eqref{eq:main-cyl-n} in $t$ and find that ${\boldsymbol\psi} = \partial_t{\boldsymbol\phi}$ solves
\begin{equation}\label{eq:psi-pde}
\begin{split}
&\partial_{tt} {\boldsymbol\psi} + (n-2)\left(1-\frac{1}{t}\right) \partial_t{\boldsymbol\psi} - \frac{n-2}{2t} \left(n - 2 - \frac{n+4}{2t} \right) {\boldsymbol\psi} + \Delta_\theta {\boldsymbol\psi}\\
&= -\frac{n-2}{2t^2} \left( n -2 - \frac{n}{t}\right){\boldsymbol\phi} + \frac{1}{t} |{\boldsymbol\phi}|^{\frac{2}{n-2}}\left( \frac{1}{t} {\boldsymbol\phi} - \frac{2}{n-2} \frac{{\boldsymbol\phi}\cdot {\boldsymbol\psi}}{|{\boldsymbol\phi}|^2}{\boldsymbol\phi} - {\boldsymbol\psi} \right).
\end{split}
\end{equation}
Taking the inner product of \eqref{eq:psi-pde} with $t\partial_t{\boldsymbol\psi}$ and integrating over ${\mathbb S}^{n-1}$, one may verify after some computation that the functional
\begin{equation}\label{eq:J}
\begin{split}
J(t,{\boldsymbol\psi}) &= \frac{1}{n\omega_n}\int_{{\mathbb S}^{n-1}} \left(t|\partial_t{\boldsymbol\psi}|^2 - t|\nabla_\theta {\boldsymbol\psi}|^2 - \frac{n-2}{2} \left( n-2 - \frac{n+4}{2t}\right) |{\boldsymbol\psi}|^2\right) \, d\theta \\
&\quad - \frac{1}{n\omega_n}\int_t^\infty \int_{{\mathbb S}^{n-1}} \frac{n-2}{\tau}\left(n-2 - \frac{n}{\tau}\right) {\boldsymbol\phi}\cdot \partial_\tau{\boldsymbol\psi} \, d\theta\,d\tau \\
&\quad + \frac{1}{n\omega_n} \int_t^\infty \int_{{\mathbb S}^{n-1}} |{\boldsymbol\phi}|^{\frac{2}{n-2}} \left( \frac{1}{\tau}{\boldsymbol\phi}- \frac{2}{n-2} \frac{{\boldsymbol\phi}\cdot{\boldsymbol\psi}}{|{\boldsymbol\phi}|^2} {\boldsymbol\phi} - {\boldsymbol\psi} \right)\cdot \partial_\tau{\boldsymbol\psi}\,d\theta\,d\tau
\end{split}
\end{equation}
satisfies
\begin{equation}\label{eq:J'}
\begin{split}
J'(t,{\boldsymbol\psi}) &=-\frac{(2n-4)t - 2n+3}{n\omega_n} \int_{{\mathbb S}^{n-1}} |\partial_t{\boldsymbol\psi}|^2\,d\theta \\
&\quad - \frac{1}{n\omega_n} \int_{{\mathbb S}^{n-1}} \left( |\nabla_\theta {\boldsymbol\psi}|^2 + \frac{(n+4)(n-2)}{t^2} \int_{{\mathbb S}^{n-1}} |{\boldsymbol\psi}|^2 \right) \,d\theta,
\end{split}
\end{equation}
provided that the last two double integrals in \eqref{eq:J} are finite, i.e., $J(t,{\boldsymbol\psi})$ is well-defined for all $t$ large.
Assuming for the moment that $J(t,\psi)$ is well-defined for all $t$ large, one may proceed as in the proof of \cite[Lemma 3.2]{A2}. Note that \eqref{eq:J'} implies the monotonicity of $J(t,{\boldsymbol\psi})$ for $t\geq t_0 = \frac{2n-3}{2n-4}$. Analogously with Remark \ref{remark:cyl-n}, the asymptotic radial symmetry \eqref{eq:asym-rad} implies exponential decay of $|\nabla_\theta {\boldsymbol\psi}|$ as well as the uniform boundedness of $|{\boldsymbol\psi}|$ and $|\partial_t{\boldsymbol\psi}|$. Hence, one may deduce as in the proof of Lemma \ref{lemma:E-monot} that $J(t,{\boldsymbol\psi})$ is uniformly bounded from below as $t\rightarrow\infty$. As $J(t,{\boldsymbol\psi})$ being nonincreasing in $t\geq t_0$, $J(\infty,{\boldsymbol\psi})$ exists, and thus, integrating \eqref{eq:J} from $t_0$ to $\infty$ yields that
\begin{equation}\label{eq:w-dt-int2}
\int_{t_0}^\infty \int_{{\mathbb S}^{n-1}} \tau |\partial_\tau {\boldsymbol\psi}|^2 \,d\theta \,d\tau < \infty.
\end{equation}
Noting that
\begin{equation*}
\left|\frac{d}{dt} \left( t \int_{{\mathbb S}^{n-1}} |\partial_t{\boldsymbol\phi}|^2\,d\theta\right)\right| \leq \int_{{\mathbb S}^{n-1}} (|\partial_t{\boldsymbol\phi}|^2 + t |\partial_t {\boldsymbol\phi}|^2 + t|\partial_{tt}{\boldsymbol\phi}|^2) \,d\theta,
\end{equation*}
we conclude from \eqref{eq:w-dt-int1} and \eqref{eq:w-dt-int2} that $t \int_{{\mathbb S}^{n-1}} |\partial_t{\boldsymbol\phi}|^2\,d\theta$ is a Cauchy sequence in $t\rightarrow\infty$. Thus, \eqref{eq:w-dt-int} follows from \eqref{eq:w-dt-int1}.
To this end, we are only left with verifying the well-definedness of $J(t,{\boldsymbol\psi})$ for all $t\geq t_0$ with some $t_0$ large. As noted above, this boils down to proving that the last two double integrals in \eqref{eq:J} are finite. Due to the upper estimate \eqref{eq:ub-sup-0} and \eqref{eq:w-dt-int1}, it suffices to show that
\begin{equation}\label{eq:w-dt-int3}
\int_{t_0}^\infty \frac{1}{t} \int_{{\mathbb S}^{n-1}} (|{\boldsymbol\phi}| + |{\boldsymbol\psi}|) |\partial_t {\boldsymbol\psi}| \,d\theta\,dt < \infty.
\end{equation}
Owing to \eqref{eq:w-tan-Linf} and \eqref{eq:w-dt-Linf}, we have, in \eqref{eq:main-cyl-n} (recall that ${\boldsymbol\psi} = \partial_t{\boldsymbol\phi}$),
\begin{equation}\label{eq:main-cyl-n-re}
|\partial_t{\boldsymbol\psi}|= (n-2)|{\boldsymbol\psi}| + O\left(\frac{1}{t}\right),
\end{equation}
so multiplying \eqref{eq:main-cyl-n-re} by $\frac{1}{t} |{\boldsymbol\phi}|$ yields
\begin{equation}\label{eq:w-dt-int4}
\begin{split}
\int_{t_0}^\infty \frac{1}{t} \int_{{\mathbb S}^{n-1}} |{\boldsymbol\phi}||\partial_t {\boldsymbol\psi}|\,d\theta \,dt &\leq (n-2)\int_{t_0}^\infty \frac{1}{t} \int_{{\mathbb S}^{n-1}} |{\boldsymbol\phi}| |{\boldsymbol\psi}| \,d\theta \,dt + O(1) \\
& \leq \frac{n-2}{2} \int_{t_0}^\infty \int_{{\mathbb S}^{n-1}} |{\boldsymbol\psi}|^2\,d\theta\,dt + O(1) \\
& < \infty,
\end{split}
\end{equation}
where the second inequality follows from $|{\boldsymbol\phi}||{\boldsymbol\psi}| \leq \frac{1}{2t}|{\boldsymbol\phi}|^2 + \frac{t}{2} |{\boldsymbol\psi}|^2$, while the last inequality is derived from \eqref{eq:w-dt-int1}. On the other hand, multiplying \eqref{eq:main-cyl-n-re} by $\frac{1}{t} |{\boldsymbol\psi}|$, we deduce from \eqref{eq:w-dt-int1} that
\begin{equation}\label{eq:w-dt-int5}
\begin{split}
\int_{t_0}^\infty \frac{1}{t} \int_{{\mathbb S}^{n-1}} |{\boldsymbol\psi}| |\partial_t {\boldsymbol\psi}|\,d\theta\,dt \leq (n-2) \int_{t_0}^\infty \frac{1}{t} \int_{{\mathbb S}^{n-1}} |{\boldsymbol\psi}|^2\,d\theta\,dt < \infty.
\end{split}
\end{equation}
The claim \eqref{eq:w-dt-int3} follows readily from \eqref{eq:w-dt-int4} and \eqref{eq:w-dt-int5}. The proof is finished.
\end{proof}
Finally we have the classification of the blowup limit via the limiting energy levels $E(\infty,{\boldsymbol\phi})$.
\begin{lemma}\label{lemma:class0} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in $B_1\setminus\{0\}$ with $\alpha=\frac{n}{n-2}$, and ${\boldsymbol\phi}$ be its cylindrical transform as in \eqref{eq:cyl-n}. Also let $E(t,{\boldsymbol\phi})$ be as in \eqref{eq:E}. Then $E(\infty,{\boldsymbol\phi})\in \{-\frac{1}{n-1}(\frac{(n-2)^2}{2})^{n-1},0\}$. Moreover, the following are true.
\begin{enumerate}[(i)]
\item $E(\infty,{\boldsymbol\phi}) = 0$ if and only if
\begin{equation}\label{eq:class0-reg}
|{\mathbf u}(x)| = o\left(|x|^{2-n} (-\log |x|)^{\frac{2-n}{2}}\right)\quad\text{as }x\rightarrow 0.
\end{equation}
\item $E(\infty,{\boldsymbol\phi}) = -\frac{1}{n-1}(\frac{(n-2)^2}{2})^{n-1}$ if and only if
\begin{equation}\label{eq:class0-sing}
|{\mathbf u}(x)| = (1+o(1)) \left( \frac{(n-2)^2}{2} \right)^{\frac{n-2}{2}}|x|^{2-n} (-\log |x|)^{\frac{2-n}{2}}.
\end{equation}
\end{enumerate}
\end{lemma}
\begin{proof} Due to Lemma \ref{lemma:E-monot}, \eqref{eq:w-tan-Linf} and \eqref{eq:w-dt-int}, we have
\begin{equation}\label{eq:E-inf1}
E(\infty,{\boldsymbol\phi}) = \frac{1}{n\omega_n} \lim_{t\rightarrow\infty} \int_{{\mathbb S}^{n-1}} \left( \frac{n-2}{n-1} |{\boldsymbol\phi}|^{\frac{2n-2}{n-2}} - \frac{(n-2)^2}{2} |{\boldsymbol\phi}|^2\right) \,d\theta.
\end{equation}
In fact, \eqref{eq:w-tan-Linf} implies that whenever ${\boldsymbol\phi}(t_j,\theta)$ converges as $t_j\rightarrow\infty$, the limit is independent of $\theta\in{\mathbb S}^{n-1}$. Hence, along a convergent sequence ${\boldsymbol\phi}(t_j,\theta) \rightarrow {\mathbf a}$ (uniformly over $\theta\in{\mathbb S}^{n-1}$), we obtain from \eqref{eq:E-inf1} that
\begin{equation}\label{eq:E-inf2}
E(\infty,{\boldsymbol\phi}) = \frac{n-2}{n-1} |{\mathbf a}|^{\frac{2n-2}{n-2}} - \frac{(n-2)^2}{2} |{\mathbf a}|^2.
\end{equation}
Since the right hand side has at most three nonnegative roots, we conclude that the limit value $|{\mathbf a}|$ (under the uniform convergence of $|{\boldsymbol\phi}(t,\theta)|$ on ${\mathbb S}^{n-1}$ as $t\rightarrow\infty$) is unique.
To compute the limit value $|{\mathbf a}|$, let us take the inner product of \eqref{eq:main-cyl-n} with ${\boldsymbol\phi}$ and integrate the both sides over $(t_0,\infty)\times{\mathbb S}^{n-1}$ (with $t_0$ large). Then one may easily deduce from \eqref{eq:w-tan-Linf}, \eqref{eq:w-dt-Linf} and \eqref{eq:w-dt-int1} that
\begin{equation*}
\begin{split}
&\left| \int_{t_0}^\infty \frac{1}{n\omega_n \tau} \int_{{\mathbb S}^{n-1}} \left(\frac{(n-2)^2}{2} - |{\boldsymbol\phi}|^{\frac{2}{n-2}}\right)|{\boldsymbol\phi}|^2 \,d\theta\,dt \right| <\infty.
\end{split}
\end{equation*}
Now that $|{\boldsymbol\phi}|$ converges to $|{\mathbf a}|$ as $t\rightarrow\infty$ uniformly on ${\mathbb S}^{n-1}$, we must have either $|{\mathbf a}| = 0$ or $|{\mathbf a}| = (\frac{(n-2)^2}{2})^{\frac{n-2}{2}}$. Inserting this into \eqref{eq:E-inf2}, we deduce that either $E(\infty,{\boldsymbol\phi}) = 0$ if and only if $|{\mathbf a}| = 0$, or $E(\infty,{\boldsymbol\phi}) = -\frac{1}{n-1} (\frac{(n-2)^2}{2})^{n-1}$. Obviously, the assertions \eqref{eq:class0-reg} and \eqref{eq:class0-sing} follow immediately via inverse cylindrical transform \eqref{eq:cyl-n}.
\end{proof}
We are only left with proving that \eqref{eq:class0-reg} yields the removability of the singularity at the origin.
\begin{lemma}\label{lemma:remv0} Let ${\mathbf u}$ be a nonnegative solution of \eqref{eq:main} in $B_1\setminus\{0\}$ with $\alpha = \frac{n}{n-2}$. Suppose further that ${\mathbf u}$ satisfies
\begin{equation}\label{eq:remv0}
|{\mathbf u}(x)| = o(|x|^{n-2}(-\log |x|)^{\frac{n-2}{2}} )\quad\text{as }x\rightarrow 0.
\end{equation}
Then the origin is a removable singularity.
\end{lemma}
\begin{proof} Under the assumption \eqref{eq:remv0}, we claim that
\begin{equation}\label{eq:remv-re0}
|{\mathbf u}(x)| \leq c|x|^{2-n + \delta}\quad\text{in }B_{r_0}\setminus\{0\},
\end{equation}
for some small $\delta>0$, where $c>1$ and $r_0>0$ may depend on ${\mathbf u}$.
Consider the auxiliary function
\begin{equation*}
\varphi_\epsilon (x) = \left(C r_0^{-\delta} |x|^\delta + \epsilon (-\log |x|)^{\frac{2-n}{2}} \right)|x|^{2-n}\quad\text{in }B_{r_0}\setminus\{0\},
\end{equation*}
where $C_0>0$ is the (universal) constant chosen from \eqref{eq:u-sup-0}, $r_0>0$ is a small radius to be determined later and $\epsilon>0$ is an arbitrary small number. After some computations, one may verify that
\begin{equation*}
\Delta \varphi_\epsilon \leq \frac{C_1}{|x|^2\log |x|} \varphi_\epsilon\quad\text{in }B_{r_0}\setminus\{0\},
\end{equation*}
by choosing $\delta,r_0>0$ small, $C_1>0$ large. Here one may choose $\delta$ and $C_1$ to depend only on $n$.
Due to the assumption \eqref{eq:remv0}, we have $a(x) = |{\mathbf u}|^{\frac{2}{n-2}} = o (-|x|^2\log |x|)$, whence $\varphi_\epsilon$ becomes a supersolution of $\Delta u_i = -a(x) u_i$ in $B_{r_0}\setminus\{0\}$, by choosing $r_0>0$ sufficiently small, where $u_i$ is the $i$-th component of ${\mathbf u}$. The rest of the proof follows the same argument shown in the proof of Lemma \ref{lemma:remv+sub}, which eventually leads us to
$u_i \leq \varphi_\epsilon$ in $B_{r_0}\setminus\{0\}$.
Passing to the limit with $\epsilon\rightarrow 0$, we get
\begin{equation*}
u_i(x) \leq C_0 r_0^{-\delta} |x|^{2-n+\delta}\quad\text{in }B_{r_0}\setminus\{0\}.
\end{equation*}
Now that this inequality holds for any $1\leq i\leq m$, we arrive at \eqref{eq:remv-re0} with $c = C_0r_0^{-\delta}\sqrt{m}$.
Thus, it follows from \eqref{eq:remv-re} that $a(x) = |{\mathbf u}|^{\frac{2}{n-2}}\in L^{\frac{n}{2-\eta}}(B_1)$, for some $\eta>0$. We know from Lemma \ref{lemma:dist} that $u_i$ is a distribution solution of $-\Delta u_i = a(x) u_i$ in $B_1$ for each $1\leq i \leq m$. Hence, the classical result \cite[Theorem 1]{S} by Serrin implies that $u_i$ has a removable singularity at the origin, and the lemma is proved.
\end{proof}
Theorem \ref{theorem:main} (iv) is now merely a combination of Lemma \ref{lemma:class0} and Lemma \ref{lemma:remv0}.
\begin{proof}[Proof of Theorem \ref{theorem:main} (iv)]
If ${\mathbf u}$ has a non-removable singularity at the origin, then according to Lemma \ref{lemma:remv0}, ${\mathbf u}$ does not satisfy \eqref{eq:remv0}. By Lemma \ref{lemma:class0}, we have \eqref{eq:asym0}, proving the theorem.
\end{proof}
\noindent {\bf Acknowledgement.} S. Kim was supported by National Research Foundation of Korea (NRF) grant funded by the Korean government (NRF-2014-Fostering Core Leaders of the Future Basic Science Program). H. Shahgholian was supported in part by Swedish Research Council.
This work was partly conducted during the first and second author's visit to Royal Institute of Technology (KTH) in Stockholm. They wish to thank Henrik Shahgholian for the kind invitation and KTH for the hospitality.
The authors would like to thank the anonymous referees for their valuable comments. Especially, we are grateful for one of the referees who pointed out a precise characterization of the new Pohozaev invariant as well as the explicit solution featuring the nontrivial invariant in the two-particle system.
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{
"redpajama_set_name": "RedPajamaArXiv"
}
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Q: Return hgetall list from redis in nodejs I'm trying to return a json object so that I can pass it back before a page is rendered to populate a list. My problem is that I can't figure out how to pass the object data out from the hgetall callback function. Here is my example with comments on what I'm missing:
var redis = require("redis"),
client = redis.createClient();
function createMobs() {
var mobObject = {
name: "Globlin",
hp: 12,
level: 1
};
client.hmset("monsterlist", "mobs", JSON.stringify(mobObject));
var myMobs = function(object) {
return object;
};
var getMobs = function(callback) {
client.hgetall("monsterlist", function(err, object) {
callback(object);
});
};
// This is returning undefined instead of my mob
console.log("mobs: ", getMobs(myMobs));
// Goal is to return moblist
// return getMobs(myMobs);
}
exports.createMobs = createMobs;
A: The short answer is that you're not thinking asynchronously. Because you're using asynchronous functions in your function, your function must also be asynchronous.
Since you didn't post the rest of your code, here's the basic idea:
var client = require('redis').createClient();
function createMobs(callback) {
var mobObject = { name: 'Goblin' };
client.hmset('monsterlist', 'mobs', JSON.stringify(mobObject), function(err) {
// Now that we're in here, assuming no error, the set has went through.
client.hgetall('monsterlist', function(err, object) {
// We've got our object!
callback(object);
});
// There is no way to run code right here and have it access the object variable, as it would run right away, and redis hasn't had time to send you the data yet. Your myMobs function wouldn't work either, because it is returning a totally different function.
});
};
app.get('/create', function(req, res) {
createMobs(function(object) {
res.render('mobs.jade', {
mobs: object
});
});
});
Hopefully that helps clear things up.
A: The only way to deal with hgetall returning value are by Promises.
get all function:
async hashget(tag) {
return new Promise((resolve, reject) => {
redis.createClient({ port: portnumber, host: config.redis.host
}).hgetall(tag, (err, object) => {
if (err) {
reject(err);
} else {
resolve(Object.keys(object));
}
});
});
}
and handle it in the through the promise as
const result = await this.redis.hashget('asize').then((result) => {
return result;
});
return result;
A: I also encountered similar issue wherein I would need the result of hgetall back the function calling it. Here is how I got it working on my end.
export const test = async (hashId) => {
// reachedLimit will have value after the hgetall result on the `monitorLimit` function
const reachedLimit = await monitorLimit(hashId);
}
export const monitorLimit = (hashId) => {
let reachedLimit = false;
let attempts = 0;
return new Promise((resolve, reject) => redisDB.hgetall(hashId, (err, result) => {
if (err) {
// TODO reject logic here...
}
if (result) {
attempts = Number(result.tries);
if (attempts === 3) {
reachedLimit = true;
resolve({ reachedLimit });
}
}
attempts += 1;
redisDB.hmset(hashId, ['tries', attempts], (createErr) => {
if (createErr) {
// TODO reject logic here...
}
});
// expire in 3 minutes
redisDB.expire(hashId, 180);
resolve({ reachedLimit });
}));
};
A: my solution for this error
SyntaxError: Unexpected token o in JSON at position 1
res.send(Object.values(JSON.parse(JSON.stringify(object))))
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 5,816
|
the Asian Paints Colour of the Year 2023 Silver Escapade sets the tone
Asian Paints unveils the Asian Paints Colour of the Year 2023—Silver Escapade that sets the tone for the year to come
JAN 13, 2023 | By Tamanna Doctor
Mr. Amit Syngle, Managing Director and CEO of Asian Paints Ltd.; Photo courtesy Asian Paints
Asian Paints Colour of the Year 2023— 'Silver Escapade; Photo courtesy Asian Paints
The year is about to be silver-lickin' good! Because all you need is a sliver of silver to make the year brighter than ever.
On 13th January 2023, Asian Paints, in collaboration with St+art India Foundation unveils their latest project 'Asian Paints Art house' at the Mumbai Urban Art Festival (MUAF), providing an exclusive preview of 'Illusions', a part of the Sassoon Dock Art Project and announces the Asian Paints Colour of the Year 2023— 'Silver Escapade'.
The project 'Illusions' on the other hand reflects reality and is advocated by fluid and expansive views. The project commences from 13th January and is present till 22nd February.
The Asian Paints Art House project is an ode to the ocean. With an association to it, Sassoon Docks seemed like the prime location to host the project. With ample culture, art and people, the place was designed to entice and intrigue the attendees. It was created using waste material, debris, upcycled fabrics and other bits and pieces found around the dock. Considered to be a 'home', the art house inculcates the Asian Paints Colour of the Year— 'Silver Escapade' in a clever manner.
Unveiling the Asian Paints Colour of the Year 2023— 'Silver Escapade', the shade is characterised by time travel and technology, child craft, wellness, it also represents the shade of auspicious beginnings in our country. UK artist Steve Messam clutched onto the colour and created an installation that encapsulates it.
The Asian Paints Art house with St+art India Foundation also dropped 5 NFT's created by visual artist Amrit Pal Singh. The series labelled 'Toy Faces' is reminiscent of childhood experiences. Toys represent the part of our life which was devoid of cynicism. Symbolising fictional stories and characters predominant in that particular chuck of our life, Toy Faces aims to revisit the feeling of euphoria attached to them. For the MUAF, he extended his hand to five individual artists to celebrate various styles and mediums.
Mr. Amit Syngle, Managing Director and CEO of Asian Paints Ltd. stated- "We are proud to unveil the 'Asian Paints Art House' and the 2023 Colour of the Year at the Mumbai Urban Art Festival. We are thrilled for the public to see what has been created at Sassoon Dock, representing Mumbai city, its art and culture, in a way that Mumbai may have never witnessed or experienced before. Illusions, is another landmark project we've launched at Sassoon Dock, adding to our collective vision of celebrating Mumbai city and its people. I am also very excited for the NFTs that have been curated for the festival with a tinge of Colour of the Year – Silver Escapade. A call out to the St+Art India Foundation for co-curating the Mumbai Urban Art Festival and building onto our long-time partnership. A special mention to the Mumbai Port Authority for their support in making MUAF a possibility"
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 4,420
|
Senegalská reprezentace v malém fotbalu reprezentuje Senegal na mezinárodních akcích v malé kopané, jako je mistrovství světa nebo Africký pohár.
Historie
Již na své první akci, na mistrovství světa 2017, Senegal získal bronzové medaile, když porazil v boji o třetí místo Španělsko 5:0. Za zmínku také stojí, že v osmifinále Senegal vyřadil šestinásobné mistry Evropy Rumunsko 3:2 na penalty. V roce 2018 na Africkém poháru, se Senegal dostal do finále, kde podlehl Pobřeží slonoviny 4:3 na penalty. Na Africkém poháru v roce 2021 byl Senegal vyřazen kuriózním způsobem. Před čtvrtfinále s Pobřežím slonoviny oba týmy před zápasem přišly na hřiště v bílých dresech. Senegal měl jakožto domácí tým mít odlišnou barvu, ale dresy nedokázal sehnat. Původně se spekulovalo, že by se zápas odehrál následující den ráno. To byl ale problém, jelikož poté se mělo hrát semifinále a jeden z týmů by tak byl v nevýhodě. Zápas byl po domluvě Africké konfederace s nigerijskou asociací kontumován ve prospěch Pobřeží slonoviny. S českou reprezentací se Senegal zatím nikdy neutkal.
Výsledky
Mistrovství světa
Africký pohár
Reference
Reprezentace v malé kopané
Sport v Senegalu
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 2,515
|
\section{Introduction}
The importance of the issue with the Hubble constant as measured by
two different methods (the $H_0$ tension) can be appreciated from
recent comprehensive reviews on the subject
\cite{divalentino21,shah21} and by the fact
of special international conferences discussing exclusively this
particular issue \footnote{\texttt{https://www.eso.org/sci/meetings/2020/H0.html}}.
The {\it Planck} space observatory \cite{aghanim20} revealed a
statistically significant discrepancy between the
cosmological parameter $H_0$ as calculated
within the standard
cosmological model by using the Cosmic Microwave Background (CMB) power spectrum,
$H_0=(67.37\pm0.54)
\,km\,s$^{-1}$\,Mpc$^{-1}$,
and the values of this parameter obtained by using other methods --
mostly from direct local measurements \cite{riess20}.
One of these local measurements is based on optical and infrared (IR)
observations of variable Cepheid stars,
with the recent calculation of $H_0$ based on this method \cite{riess18} being
$H_0=(73.48\pm1.66)$\,km\,s$^{-1}$\,Mpc$^{-1}$.
Both local and {\it Planck}-derived estimates of $H_0$ have passed a number
of rigorous tests by considering many possible sources of systematic errors
\cite{efstathiou13,planck17,follin18},
but the discrepancy still remains.
Discussing the possible origin of this discrepancy, most authors and
reviewers focus primarily on
observational biases related to the method of standard-candles, Cepheids and type-Ia
supernovae (SN), and on proposals going far beyond the standard cosmological and
particle physics models. By contrast, possible biases intrinsic to the
CMB are passed by almost
without further thought.
\begin{figure}
\vspace{-0.5cm}
\hspace{-0.3cm}
\epsfig{figure=dust_t05k20210913z.eps,width=7.5cm}
\hspace{-0.1cm}
\epsfig{figure=h4.eps,width=8.5cm}
\vspace{-0.3cm}
\caption{{\small {\it Left:} Redshift dependences of the blackbody radiation energy
fraction $\eta(z)$ as observed in five {\it Planck}
frequency bands for $T_{\rm bb}=5$\,K; the constant shifts of the curves
with respect to each other have been normalised at $z=0$ by
subtracting from them their individual values $\eta(0)$;
{\it Right:}~CMB~power spectra (in the standard normalised presentation) generated by using
the code for anisotropies in the microwave background (CAMB tool)
for seven different values of $H_0$.}}
\label{fig:bb_slopes}
\end{figure}
\section{Distant foreground}
Various authors have reported that the contaminating emission from a medium
around distant extragalactic sources should affect correlations between CMB and
extragalactic cosmic structures traced by bright transient sources,
like supernovae (SNe) or gamma-ray bursts.
The signature of the distant foreground in the CMB,
based on the WMAP and {\it Planck}-mission results and traced by SNe was previously
reported by the author \cite{yershov12,yershov14},
who found a correlation
between the SN redshifts, $z_{\rm SN}$, and CMB temperature fluctuations
at the SNe locations, $T_{\rm SN}$.
By using different {\it Planck} frequency bands and computing
the fractions $\eta_\nu(z)$ of blackbody radiation energy as observed in each
frequency band $\nu$ for different redshifts $z$ one can estimate the regression
line slopes
for these bands \cite{yershov20}. The functions $\eta_\nu(z)$ are shown
in Fig.\,\ref{fig:bb_slopes}, as calculated for the blackbody temperature
$T_{\rm bb} = 5$\,K and five of the {\it Planck} frequency bands
$\nu= \{70, 100, 143, 217, 353\}$\,GHz. Note that the temperature of the
medium in thermal equilibrium with the CMB must exceed or be equal to 5\,K for the redshifts
$z > 0.835$, so for these redshifts the functions $\eta_\nu(z)$ appear
horizontal.
We can calculate slopes $\xi^c_\nu$ of these functions for different $T_{\rm bb}$
and compare them with the observed slopes $\xi^o_\nu$ corresponding to
different {\it Planck} frequency bands $\nu= \{70, 100, 143, 217, 353\}$\,GHz which,
according to our previous work \cite{yershov14}, are the following: $\xi^o_{70}=0.96\pm 0.63$, $\xi^o_{143}=1.09\pm 0.31$,
$\xi^o_{217}=0.61 \pm 0.36$ and $\xi^o_{353}= -0.99\pm 0.48$.
The slope for the 100\,GHz-band
was used as a reference for normalisation, so $\xi^o_{100}=1.00\pm 0.39$.
The calculated slopes $\xi^c_\nu$ of these functions averaged between the
temperatures $T_{\rm bb} = \{3, 4, 5, 6\}$\,K for each
of the {\it Planck} frequency bands $\nu= \{70, 143, 217, 353\}$\,GHz
are
$\xi^c_{70}=0.45\pm 0.03$, $\xi^c_{143}=1.63\pm 0.17$,
$\xi^c_{217}=0.73 \pm 0.74$ and $\xi^c_{353}= -1.48\pm 0.43$
(again, the slope for the 100\,GHz-band
was used as a reference).
They are within the 1-$\sigma$ tolerance interval with respect
to the above experimental values $\xi^o_\nu$, which indicates that the
temperature of the CMB-contaminating ingredient of the
intergalactic medium is very low, likely to be between 3\,K and 6\,K.
This can give clues as to the nature of the medium, which can be
coarse-grain (grey) dust, and which for some time has been suspected to
populate intergalactic space
\cite{eigenson49,zwicky57,gonzalez98,alton01}.
This ``grey'' dust
leaves little or no imprint on the spectral energy distribution of background sources.
However, it creates the long-known excess of radiation from some extragalactic objects in the
far IR at $\lambda \approx 500\,\mu$m, which extends up to centimetre wavelengths and
can interfere with the CMB radiation. Such a $500\,\mu$m radiation excess
was confirmed and measured by the
{\it Herschel} space observatory \cite{galliano11}.
In the 1990s, this excess was interpreted as an elevated spatial mass
density of cold dust \cite{reach95} with temperatures of 4 to 7\,K.
Here we confirm this interpretation from a completely different
point of view.
\section{CMB distrortion}
The angular sizes of the observed regions with the 500\,$\mu$m emission \cite{galliano11,lisenfeld02}
range from 0.02$^\circ$ to 0.5$^\circ$.
So this emission would effectively distort the CMB power spectrum at the
multipole moments $\ell \approx 360$ and higher, which would change
the estimated parameter $H_0$.
In order to quantify these changes we have used
the code for anisotropies in the microwave background \cite{lewis13} (CAMB)
which allows the extraction of different cosmological parameters
from theoretical CMB power spectra generated by the same code.
The calculated changes
are shown on the right panel of Fig.\,\ref{fig:bb_slopes}
for the first trough of the CMB power spectrum,
where its effect on the calculated parameter $H_0$
is quite strong.
In this code, the coefficients $C_\ell$ of the CMB power spectrum are calculated as
sums of the integrals $a_{\ell m}$, $|m| \le \ell$, which include
temperature fluctuations $\Delta T(x,\phi)$ over the celestial sphere, where
$x\in[-1,1]$ is the cosine of the latitude and $\phi\in[0,2\pi]$ is the longitude.
Conversely, the functions $\Delta T(x,\phi)$ are calculated by summing up the integrals
$a_{\ell m}$.
For a given CMB power spectrum, we have calculated a set of corresponding values of
$\Delta T(x,\phi)$
by using random $a_{\ell m}$ for
$\ell=0,1,\dots,\ell_{\tt max}$ with the
restriction $\ell_{\tt max}=500$.
We have taken five equal-spaced values of $H_0$, namely,
60, 65, 70, 75 and 80 [km s$^{-1}$ Mpc$^{-1}$]
plus the values 67.4 [km s$^{-1}$ Mpc$^{-1}$] (solid red curve on the right
panel of Fig.\,\ref{fig:bb_slopes})
and 73.5 [km s$^{-1}$ Mpc$^{-1}$]
(dashed red curve on the same panel)
corresponding, respectively,
to the {\it Planck} result and to the local measurements of $H_0$.
Additionally, for checking the consistency of our calculations we
have taken a few sets of normally distributed randomised values of
$a_{\ell m}^i$, $i=1,2,\dots,5$, so that for each of the selected
values of $H_0$, we have obtained five samples of $a_{\ell m}^{H_0,i}$ and,
correspondingly, five samples of values $\Delta T^{H_0,i}$.
For each of them, we have calculated the average of the CMB temperature
fluctuations $\overline{\Delta T}$ and its standard deviation $\sigma_{T}$.
Here we are mainly interested in the way the values $\sigma_T$ change
when the parameter $H_0$ is varied.
For each of these generated sequences, the trend of the
calculated values $\sigma_{T}$ was practically the same.
Namely, when the dispersion of the CMB temperature fluctuations increases, the
value of the estimated $H_0$ decreases, the difference between the two
discussed $H_0$ values 73.5 and 67.4 [km s$^{-1}$ Mpc$^{-1}$] being related to
$\Delta\sigma_{T}=-0.60\pm 0.04\,\mu$K.
This value is the measure of CMB contamination by
photons from the medium surrounding remote clumps of matter,
and it can thus be used for estimating the amount of cold
coarse-grain dust in the intergalactic medium.
\section{Discussion}
Between 2012 and 2019, new studies have appeared demonstrating that the dimming
of the type-Ia supernovae was different in different directions
\cite{cai12,cardenas13,bernal17,colin19,zhao19,rameez19,luongo21}.
Most of the authors of these
studies interpret their results in terms of anisotropic acceleration of the Universe.
However, anisotropy of acceleration violates the main cosmological principle. Therefore,
the grey-dust interpretation of the type-Ia supernovae anisotropic dimming becomes preferable:
it would be much more logical to assume non-uniform distribution of dust
rather than the Universe having different properties in different
directions.
What might be the origin of this coarse-grain dust?
Apart from the initial formation of dust particles within galaxies with
their further transport into the intergalactic space,
dust formation can also occur
directly in the intergalactic medium \cite{fabian94}.
The observational evidence for cold molecular clouds at the cooling flows
in galaxy clusters and the presence of dust in these regions is widespread
\cite{russell17}.
Besides this detectable dust clouds,
there exists yet another possibility of almost undetectable
coarse-grain dust existing in the intergalactic space which,
according to recent studies, should be seriously taken into
consideration. Namely, these macroscopic dust particles can be formed
by the process of hydrogen solidification under sufficiently low
temperatures, e.g., when the CMB temperature gets below 10\,K at $z<2$,
which was proposed in 1968 by F. Hoyle \cite{hoyle68} and
further discussed in the 1990s and early 2000s
\cite{pfenniger94,wardle99,pfenniger04}.
In the last decade, it was shown by experimental physicists that H$_2$ ice
became stable in vacuum and do not sublime if it
contains impurities \cite{schaefer07,walker13,kettwich15}
transported from galaxies into the intergalactic medium.
My conclusion is that the mechanism of contamination of CMB radiation
by some distant foreground emission from cold dust can explain the discrepancy
between the local measurements of $H_0$ and the {\it Planck}-derived value,
without invoking unnecessary assumptions that violate the basic cosmological
principles or break the standard cosmological and particle physics models.
\section*{References}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 5,138
|
import threading
import os
import time
from audio_pipeline.util import AcoustidLookup as lookup
from audio_pipeline.util import Process, Util, Exceptions
from audio_pipeline.util.MusicBrainz import PreferredRelease
from audio_pipeline import Constants
from ..util import Resources
from ...util import Utilities
from ...util.AudioFile import AudioFileFactory
MAX_RELEASES = 500
MATCHING_RATIO = .75
FEW_TRACKS = 3
WRONG_SINGLE = .4
class LoadReleases(threading.Thread):
fuzz = 3
def __init__(self, prev_buffer, next_buffer, current_release, starting_index=0, max_releases=MAX_RELEASES):
"""
Load next releases
"""
threading.Thread.__init__(self)
self.prev_buffer = prev_buffer
self.next_buffer = next_buffer
self.max_buffer = max_releases
self.starting_index = starting_index
self.current_release = current_release
self.loaded = 0
self.scanned = 0
self.processor = Process.Processor(Constants.processor, Constants.batch_constants.mb)
self.processor.populate()
self.tb_lookup = Constants.tb_lookup
self.acoustid_lookup = Constants.acoustid_lookup
def run(self):
time.sleep(.1)
num_dirs = len(self.current_release.directories)
if self.starting_index != 0:
self.load_release(self.next_buffer, self.starting_index)
self.loaded += 1
self.starting_index += 1
while self.current_release.current is not None and self.loaded < num_dirs and self.loaded < self.max_buffer:
# if buffers are full, wait for a signal from the model proper to load more releases,
# rather than constantly looping and wasting resources
self.current_release.cond.acquire()
if self.current_release.current is not None and \
len(self.next_buffer) + self.current_release.current[0] < num_dirs:
# haven't filled in next_buffer / have moved forward
# get next value
if len(self.next_buffer) > 0:
last_release = self.next_buffer[0]
else:
last_release = self.current_release.current
i = last_release[0] + 1
if i < num_dirs:
print("loading next " + str(i))
self.current_release.cond.release()
self.load_release(self.next_buffer, i)
self.current_release.cond.acquire()
self.loaded += 1
if self.current_release.current is not None and \
len(self.prev_buffer) <= self.current_release.current[0]:
# haven't filled in prev_buffer / have moved forward
# so get next buffer
if len(self.prev_buffer) > 0:
last_release = self.prev_buffer[0]
else:
last_release = self.current_release.current
i = last_release[0] - 1
if i >= 0:
print("loading prev " + str(i))
self.current_release.cond.release()
self.load_release(self.prev_buffer, i)
self.current_release.cond.acquire()
self.loaded += 1
self.current_release.cond.release()
print("Loaded: " + str(self.loaded))
print("All releases loaded")
def load_release(self, buffer, dir_index):
if len(buffer) >= self.max_buffer:
return None
directory = self.current_release.directories[dir_index]
files = os.listdir(directory)
indices = dict()
to_scan = set()
releases = list()
releases.append([])
i = 1
for f in files:
file = os.path.join(directory, f)
try:
file_data = AudioFileFactory.get(file)
except IOError:
continue
except Exceptions.UnsupportedFiletypeError:
print("Unsupported filetype")
continue
if not file_data.has_mbid() and Utilities.know_artist_name(str(file_data.album_artist)):
index = 0
else:
mbid_medium = (file_data.mbid.value, file_data.disc_num.value)
release_details = (file_data.album.value, file_data.album_artist.value, file_data.disc_num.value)
if mbid_medium in indices:
index = indices[mbid_medium]
elif release_details in indices:
index = indices[release_details]
else:
index = len(releases)
releases.append([])
if file_data.has_mbid():
indices[mbid_medium] = index
else:
indices[release_details] = index
to_scan.add(index)
releases[index].append(file_data)
# all releases are initially added to the to_scan pile; if they should not be
# (already scanned, already had metadata stuffed, have mbid & have more tracks
# than our 'confidence threshold') remove them
if ((file_data.has_mbid() and len(releases[index]) > FEW_TRACKS) or file_data.acoustid.value
or file_data.meta_stuffed.value) and index in to_scan:
to_scan.remove(index)
if self.acoustid_lookup:
for i in to_scan:
r = lookup.Release(releases[i])
print("Looking up " + ascii(releases[i]))
r.lookup()
if r.best_group:
p = PreferredRelease.BestRelease(r)
if len(releases[i]) <= FEW_TRACKS:
p.choose_release()
p.set_mbid(p.best_release)
self.scanned += 1
else:
match = p.mb_comparison(True)
if match is not None and (match > MATCHING_RATIO):
p.set_mbid(p.best_release)
self.scanned += 1
if len(releases[0]) <= 0:
releases.pop(0)
elif self.acoustid_lookup:
p = PreferredRelease.BestRelease(lookup.Release(releases[0]))
p.choose_release()
p.set_mbid(p.best_release)
self.scanned += 1
for release in releases:
release_track = release[0]
if (release_track.has_mbid() and (release_track.should_stuff_metadata() or
not release_track.has_minimum_metadata())) \
and (self.acoustid_lookup or self.tb_lookup):
print("Getting metadata")
release_meta = self.processor.get_release(release_track.mbid.value)
# stuff any additional MB metadata
for track in release:
release_meta.stuff_audiofile(track)
# if meta.disc_count is None:
# track.mbid.value = None
# pass
track.save()
release.sort(key=lambda x: x.track_num.value if x.track_num.value is not None else 0)
buffer.appendleft((dir_index, release))
class CurrentReleases:
starting_position = (-1, None)
def __init__(self, directories, timeout=.01):
self.directories = directories
self.__current = (-1, None)
self.__prev = (-1, None)
self.cond = threading.Condition()
def reset(self):
self.__current = self.starting_position
self.__prev = self.starting_position
@property
def current(self):
self.cond.acquire()
v = self.__current
self.cond.notify()
self.cond.release()
return v
@current.setter
def current(self, value):
self.cond.acquire()
self.__current = value
self.cond.notify()
self.cond.release()
@property
def prev(self):
self.cond.acquire()
v = self.__prev
self.cond.notify()
self.cond.release()
return v
@prev.setter
def prev(self, value):
self.cond.acquire()
self.prev = value
self.cond.notify()
self.cond.release()
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 5,215
|
external help file: JiraPS-help.xml
Module Name: JiraPS
online version: https://atlassianps.org/docs/JiraPS/commands/Get-JiraIssueCreateMetadata/
locale: en-US
schema: 2.0.0
layout: documentation
permalink: /docs/JiraPS/commands/Get-JiraIssueCreateMetadata/
---
# Get-JiraIssueCreateMetadata
## SYNOPSIS
Returns metadata required to create an issue in JIRA
## SYNTAX
```powershell
Get-JiraIssueCreateMetadata [-Project] <String> [-IssueType] <String> [[-Credential] <PSCredential>]
[<CommonParameters>]
```
## DESCRIPTION
This function returns metadata required to create an issue in JIRA - the fields that can be defined in the process of creating an issue.
This can be used to identify custom fields in order to pass them to `New-JiraIssue`.
This function is particularly useful when your JIRA instance includes custom fields that are marked as mandatory.
## EXAMPLES
### EXAMPLE 1
```powershell
Get-JiraIssueCreateMetadata -Project 'TEST' -IssueType 'Bug'
```
This example returns the fields available when creating an issue of type Bug under project TEST.
### EXAMPLE 2
```powershell
Get-JiraIssueCreateMetadata -Project 'JIRA' -IssueType 'Bug' | ? {$_.Required -eq $true}
```
This example returns fields available when creating an issue of type Bug under the project Jira.
It then uses `Where-Object` (aliased by the question mark) to filter only the fields that are required.
## PARAMETERS
### -Project
Project ID or key of the reference issue.
```yaml
Type: String
Parameter Sets: (All)
Aliases:
Required: True
Position: 1
Default value: None
Accept pipeline input: False
Accept wildcard characters: False
```
### -IssueType
Issue type ID or name.
```yaml
Type: String
Parameter Sets: (All)
Aliases:
Required: True
Position: 2
Default value: None
Accept pipeline input: False
Accept wildcard characters: False
```
### -Credential
Credentials to use to connect to JIRA.
If not specified, this function will use anonymous access.
```yaml
Type: PSCredential
Parameter Sets: (All)
Aliases:
Required: False
Position: 3
Default value: None
Accept pipeline input: False
Accept wildcard characters: False
```
### CommonParameters
This cmdlet supports the common parameters: -Debug, -ErrorAction, -ErrorVariable, -InformationAction, -InformationVariable, -OutVariable, -OutBuffer, -PipelineVariable, -Verbose, -WarningAction, and -WarningVariable.
For more information, see about_CommonParameters (http://go.microsoft.com/fwlink/?LinkID=113216).
## INPUTS
## OUTPUTS
### [JiraPS.Field]
## NOTES
This function requires either the `-Credential` parameter to be passed or a persistent JIRA session.
See `New-JiraSession` for more details.
If neither are supplied, this function will run with anonymous access to JIRA.
## RELATED LINKS
[about_JiraPS_CreatingIssues](../../about/creating-issues.html)
[Get-JiraField](../Get-JiraField/)
[New-JiraIssue](../New-JiraIssue/)
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 349
|
define(function (require, exports, module) {var React = require("react");
var Bootstrap = require("react-bootstrap");
var mouseupTimeout;
var MOUSEUP_TIMEOUT = 50;
module.exports = React.createClass({displayName: "exports",
propTypes: {
movex: React.PropTypes.bool,
movey: React.PropTypes.bool,
minx: React.PropTypes.number,
maxx: React.PropTypes.number,
miny: React.PropTypes.number,
maxy: React.PropTypes.number,
onChange: React.PropTypes.func
},
getDefaultProps: function() {
return {
movex: true,
movey: true
};
},
getInitialState: function() {
return {
left: undefined,
top: undefined,
mousedown: false,
containerleft: undefined,
containertop: undefined,
mouseoffsetleft: undefined,
mouseoffsettop: undefined
}
},
render: function() {
var $__0= this.props,onChange=$__0.onChange,className=$__0.className,other=(function(source, exclusion) {var rest = {};var hasOwn = Object.prototype.hasOwnProperty;if (source == null) {throw new TypeError();}for (var key in source) {if (hasOwn.call(source, key) && !hasOwn.call(exclusion, key)) {rest[key] = source[key];}}return rest;})($__0,{onChange:1,className:1});
var style = {};
if (this.state.left) style.left = this.state.left + "px";
if (this.state.top) style.top = this.state.top + "px";
return (
React.createElement("div", React.__spread({ref: "element", className: this._getClassName() }, other , {onMouseDown: this._handleMouseDown, onMouseUp: this._handleMouseUp, onMouseMove: this._handleMouseMove, style: style }),
this.state.mousedown &&
React.createElement("div", {className: "fu-draggable-area", onMouseDown: this._handleMouseDown, onMouseUp: this._handleMouseUp, onMouseMove: this._handleInnerMouseMove})
)
);
},
_getClassName: function() {
var
cx = React.addons.classSet,
className = this.props.className,
classes = {
"fu-draggable": true,
"fu-draggable-x": this.props.movex && !this.props.movey,
"fu-draggable-y": this.props.movey && !this.props.movex,
"fu-draggable-xy": this.props.movex && this.props.movey
};
if (className) { classes[className] = true; }
return cx(classes);
},
_handleMouseDown: function(event) {
var
self = this,
containerRect = this.refs.element.getDOMNode().parentNode.getBoundingClientRect(),
resizeRect = this.refs.element.getDOMNode().getBoundingClientRect(),
clientX = event.clientX,
clientY = event.clientY;
this.setState({
mousedown: true,
containerleft: containerRect.left,
containertop: containerRect.top,
left: this.props.movex ? event.clientX - this.state.containerleft - this.state.mouseoffsetleft : undefined,
top: this.props.movey ? event.clientY - this.state.containertop - this.state.mouseoffsettop : undefined,
mouseoffsetleft: event.clientX - resizeRect.left,
mouseoffsettop: event.clientY - resizeRect.top });
},
_handleMouseMove: function(event) {
if (this.state.mousedown) {
var
self = this,
left = this.props.movex ? event.clientX - this.state.containerleft - this.state.mouseoffsetleft : undefined,
top = this.props.movey ? event.clientY - this.state.containertop - this.state.mouseoffsettop : undefined,
clientX = event.clientX,
clientY = event.clientY;
if (this.props.movex && this.props.minx && left < this.props.minx) left = this.props.minx;
if (this.props.movex && this.props.maxx && left > this.props.maxx) left = this.props.maxx;
if (this.props.movey && this.props.miny && top < this.props.miny) top = this.props.miny;
if (this.props.movey && this.props.maxy && top > this.props.maxy) top = this.props.maxy;
this.setState({ left: left, top: top });
}
},
_handleInnerMouseMove: function(event) {
if (mouseupTimeout) clearTimeout(mouseupTimeout);
this._handleMouseMove(event);
},
_handleMouseUp: function() {
if (this.state.mousedown) {
if (mouseupTimeout) clearTimeout(mouseupTimeout)
var self = this;
setTimeout(function() {
if (self.state.mousedown) {
var
x = self.state.left,
y = self.state.top;
self.setState(self.getInitialState(), function() {
if (self.props.onChange) self.props.onChange(x, y);
});
}
}, MOUSEUP_TIMEOUT);
}
}
})
});
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 7,443
|
{"url":"https:\/\/mathoverflow.net\/questions\/434087\/how-large-is-the-set-of-unimodular-lattices-whose-sucesssive-minima-cannot-be-at","text":"# How large is the set of unimodular lattices whose sucesssive minima cannot be attained by a basis of lattice?\n\nRecall that the $$i$$-th successive minimum of $$L\\in \\mathcal L$$ (space of full rank lattices in $$\\mathbb R^d$$), denoted $$\\lambda_i(L)$$ is the infimum of the radii of the balls containing $$i$$-linearly independent vectors in $$L$$. All norms are Euclidean 2-norm.\n\nLet $$\\Lambda$$ be a unimodular lattice in $$\\mathbb R^d$$. For $$d\\le 4$$, it is possible to construct a basis $$\\{v_1,\\dots, v_d\\}$$ of $$\\Lambda$$ such that $$\\lambda_i(\\Lambda)=\\|v_i\\|$$ for any $$i$$.\n\nIf $$d>4$$, then the example $$\\frac{1}{2^{1\/d}}L_0$$, where $$L_0:=Span_{\\mathbb Z} \\{e_1,\\dots, e_{d-1},\\frac{1}{2}(e_1+\\cdots+e_d) \\}$$ shows it may be impossible to find such a basis. The coefficients $$\\frac{1}{2^{1\/d}}$$ is used to scale $$L_0$$ to a unimodular lattice.\n\nIt seems to me that such examples are really rare, but how rare?\n\nConsider the set $$S$$ of unimodular lattices whose sucesssive minima cannot be attained by a basis of lattice. Recall the space of unimodular lattices are be identified with the homogeneous space $$SL(d,\\mathbb R)\/SL(d,\\mathbb Z)$$. Can we see that $$S$$ is contained in (countable union) of submanifold\/subvariety of dimension strictly lower than the full dimension? Or show it has Haar measure zero?\n\nIf it does not have measure zero, can we give an estimate of its measure?\n\n(1) To begin with can anyone construct any examples different from this $$\\frac{1}{2^{1\/d}}L_0$$? (As Alison commented below small perturbation of this $$\\frac{1}{2^{1\/d}}L_0$$ example may work)\n\n(2) Can any one construct any examples significantly different from this $$\\frac{1}{2^{1\/d}}L_0$$. For example, with $$\\lambda_1(\\Lambda)$$ close to zero? Such lattices are often called \"unbounded\".\n\n\u2022 I'd expect that small perturbations of your example of $2^{-1\/d}L_0$ also fail to be spanned by their shortest vectors. Nov 8, 2022 at 0:08\n\u2022 @AlisonMiller Right, that is also what I am thinking but it is hard to see the set of those perturbations could be of \"full dimension\" or \"positive measure\". Nov 8, 2022 at 0:19\n\u2022 It's an open neighborhood of a point in the homogeneous space, so I'd expect it to have positive measure! Nov 8, 2022 at 0:41\n\u2022 @AlisonMiller , obviously it cannot hold that both of us are right, as you will have two open sets. My idea is simply - given a short basis which spaces the lattice, small\u2019\u2019 unipotent elements (smaller than the minima, ratios, etc), should still gives a short basis. And enough short unipotents form an open set. For the `bad\u2019\u2019 lattice, how do these perturbation idea work for a small Cartan element? It will be a statement in diophantine approximation\u2026\n\u2013\u00a0Asaf\nNov 11, 2022 at 16:09\n\u2022 @Asaf I agree they can't both be open -- but I'm claiming a weaker statement: that if $e'_1, \\dotsc, e'_d$ is a small perturbation of the standard basis $e_1, \\dotsc, e_d$, then the lattice ${\\mathrm Span}_{\\mathbb Z} \\{e_1,\\dots, e_{d-1},\\frac{1}{2}(e_1+\\cdots+e_d)$.will still have successive minima given by $\\|e_1\\|' \\dotsc, \\|e_d\\|'$ in some order. I haven't thought too hard about the boundary of this set but I expect it to occur where there are multiple choices of sets of vectors representing the successive minima, where one choice gives a basis but the other doesn't. Nov 11, 2022 at 19:05\n\nEdit 2 - Definitely the set of lattices spanned by their shortest vectors is of positive measure. For each such lattice $$\\Lambda \\in SL_{n}(\\mathbb{R})\/SL_{n}(\\mathbb{Z})$$, there exists $$\\epsilon=\\epsilon(\\Lambda)>0$$ such that $$G_{\\epsilon}.\\Lambda$$ consists of lattices with such a short basis (where $$G_{\\epsilon}$$ is appropriate small neighborhood of unity). Hence it is of positive measure. This also shows that the technique described below is specialized to well-rounded lattices and cannot be improved.\n\nEdit - the answer below is about well-rounded lattices, which are particular kind of lattices spanned by their shortest vectors...\n\nIt is well know by McMullen's work that that any compact $$A$$ orbit (where $$A$$ is the split Cartan of $$SL_{n}(\\mathbb{R})$$) contains a well-rounded lattice. This has been generalized by Weiss and Shapira for every closed $$A$$ orbit.\n\nThe following holds for generic well-rounded lattices -\n\nFor the set of generic well-rounded lattices, one can show that for each such lattice, there exists an open neighborhood such that the intersection of the open neighborhood with the set of well-rounded lattices is a submanifold (actually subvarity) of positive co-dimension. Picking $$\\epsilon>0$$, throwing away the elements high up in the cusp by throwing out a whole cuspidal region (say of measure $$<\\epsilon\/2$$), you end up with a precompact subset of the generic well-rounded lattices. Using compactness and the theorem I just mentioned, this set is covered by a finite amount of submanifolds (actually, subvarities) of proper codimension. Hence of zero Haar measure. This shows that the set of generic well-rounded lattices is of zero measure as $$\\epsilon$$ was arbitrary.\n\nOne can probably push this strategy further (by taking a sequence of shrinking cuspidal regions) to show that one can cover the generic well-rounded lattices by a union of submanifolds of proper co-dimension, hence also the Hausdorff dimension of them is not full.\n\nI need to think about the answer regarding non-generic well rounded lattices, but they must be rather special...\n\n\u2022 It looks like you're answering a slightly different question from the one that was asked here: as I understand it, \"well-rounded\" lattices have a basis of vectors which all have length equal to the minimal length of the lattice, while the question is allowing all successive minima of the lattice (which need not be equal) as lengths. Nov 7, 2022 at 18:12\n\u2022 @AlisonMiller , you are right, thank you, I clarified that in an edit...\n\u2013\u00a0Asaf\nNov 7, 2022 at 18:50\n\nTo begin with can anyone construct any examples different from this $$(1\/\\sqrt[d]{2})L_0$$?\n\nMy (somewhat imperfect) undertsanding of this is that the answer to this question is no. Questions related to this are studied in various works by Martinet (and others), see for example this and this, though this second paper is more relevant.\n\nThe setup is as follows. One starts with a lattice $$\\Lambda$$. Then one considers its Minkowskian Sublattice $$\\Lambda'$$, which has as its basis some set of linearly independent representatives of length $$\\lambda_i(\\Lambda)$$. One can then consider the quotient $$\\Lambda \/ \\Lambda'$$, and try to classify possible pairs $$(\\Lambda, \\Lambda')$$. This is done by letting $$d\\mathbb{Z}$$ be the annihilator of $$\\Lambda\/\\Lambda'$$, and then writing\n\n$$\\Lambda = \\mathsf{span}_{\\mathbb{Z}}\\langle \\Lambda, f_1,\\dots, f_k\\rangle$$\n\nfor $$f_i = \\frac{\\sum_i \\alpha_i^{(j)} e_i}{d}$$, and $$\\alpha^{(j)} = (\\alpha_1^{(j)},\\dots, \\alpha_n^{(j)})\\in(\\mathbb{Z}\/d\\mathbb{Z})^n$$ certain words of a $$\\mathbb{Z}\/d\\mathbb{Z}$$-linear code. In this way, the example of $$L_0$$ corresponds to applying construction A to the (binary) repetition code $$[1,1,\\dots,1]$$. But one can obtain other pairs $$(\\Lambda, \\Lambda')$$ (starting in dimension 9) by instead lifting other codes. The second link classifies the codes that can appear (in small dimensions). Each code leads to a (somewhat) different analogue of $$L_0$$, though the linked work does not contain unbounded lattices. Instead it contains the comment\n\nIt should be noted that the two mentioned applications make use only of results for well rounded lattices, that is, for lattices with minimal vectors spanning $$E$$. In other words, these lattices have equal successive minima $$\\lambda_1(L),\\dots, \\lambda_n(L)$$. Indeed, a deformation argument (see [Mar01, Theorem 1.5]) shows that all codes can be realized using well rounded lattices.\n\nAs their primary goal is to classify the codes which can arise from the above construction, they reduce their analysis to the case of well-rounded lattices (which are not unbounded in your sense).\n\nNone of this is useful for your overall of quantifying how \"large\" this set of lattices is, but may be useful for additional explicit examples of analogues of $$L_0$$.","date":"2023-01-27 05:42:25","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\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 55, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.884981632232666, \"perplexity\": 343.587815599309}, \"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-2023-06\/segments\/1674764494936.89\/warc\/CC-MAIN-20230127033656-20230127063656-00163.warc.gz\"}"}
| null | null |
package org.h2.build.code;
import java.io.File;
import java.io.FileWriter;
import java.io.IOException;
import java.io.RandomAccessFile;
import java.util.ArrayList;
/**
* Switched source code to a specific Java version, automatically to the current
* version, or enable / disable other blocks of source code in Java files.
*/
public class SwitchSource {
private final ArrayList<String> enable = new ArrayList<String>();
private final ArrayList<String> disable = new ArrayList<String>();
/**
* This method is called when executing this application from the command
* line.
*
* @param args the command line parameters
*/
public static void main(String... args) throws IOException {
new SwitchSource().run(args);
}
private void run(String... args) throws IOException {
String dir = null;
String version = null;
for (int i = 0; i < args.length; i++) {
String a = args[i];
if ("-dir".equals(a)) {
dir = args[++i];
} else if ("-auto".equals(a)) {
enable.add("AWT");
version = System.getProperty("java.specification.version");
} else if ("-version".equals(a)) {
version = args[++i];
} else if (a.startsWith("-")) {
String x = a.substring(1);
disable.add(x);
enable.remove(x);
} else if (a.startsWith("+")) {
String x = a.substring(1);
enable.add(x);
disable.remove(x);
} else {
showUsage();
return;
}
}
if (version == null) {
// ok
} else if ("1.5".equals(version)) {
disable.add("Java 1.6");
disable.add("Java 1.7");
} else if ("1.6".equals(version)) {
enable.add("Java 1.6");
disable.add("Java 1.7");
} else if (version.compareTo("1.7") >= 0) {
enable.add("Java 1.6");
enable.add("Java 1.7");
} else {
throw new IllegalArgumentException("version: " + version);
}
if (dir == null) {
showUsage();
} else {
process(new File(dir));
}
}
private void showUsage() {
System.out.println("Switched source code to a specific Java version.");
System.out.println("java "+getClass().getName() + "\n" +
" -dir <dir> The target directory\n" +
" [-version] Use the specified Java version (1.4 or newer)\n" +
" [-auto] Auto-detect Java version (1.4 or newer)\n" +
" [+MODE] Enable code labeled MODE\n" +
" [-MODE] Disable code labeled MODE");
}
private void process(File f) throws IOException {
String name = f.getName();
if (name.startsWith(".svn")) {
return;
} else if (name.endsWith(".java")) {
processFile(f);
} else if (f.isDirectory()) {
for (File file : f.listFiles()) {
process(file);
}
}
}
private void processFile(File f) throws IOException {
RandomAccessFile read = new RandomAccessFile(f, "r");
byte[] buffer;
try {
long len = read.length();
if (len >= Integer.MAX_VALUE) {
throw new IOException("Files bigger than Integer.MAX_VALUE are not supported");
}
buffer = new byte[(int) len];
read.readFully(buffer);
} finally {
read.close();
}
boolean found = false;
// check for ## without creating a string
for (int i = 0; i < buffer.length - 1; i++) {
if (buffer[i] == '#' && buffer[i + 1] == '#') {
found = true;
break;
}
}
if (!found) {
return;
}
String source = new String(buffer);
String target = source;
for (String x : enable) {
target = replaceAll(target, "/*## " + x + " ##", "//## " + x + " ##");
}
for (String x : disable) {
target = replaceAll(target, "//## " + x + " ##", "/*## " + x + " ##");
}
if (!source.equals(target)) {
String name = f.getPath();
File fileNew = new File(name + ".new");
FileWriter write = new FileWriter(fileNew);
write.write(target);
write.close();
File fileBack = new File(name + ".bak");
fileBack.delete();
f.renameTo(fileBack);
File fileCopy = new File(name);
if (!fileNew.renameTo(fileCopy)) {
throw new IOException("Could not rename "
+ fileNew.getAbsolutePath() + " to " + name);
}
if (!fileBack.delete()) {
throw new IOException("Could not delete " + fileBack.getAbsolutePath());
}
// System.out.println(name);
}
}
private static String replaceAll(String s, String before, String after) {
int index = 0;
while (true) {
int next = s.indexOf(before, index);
if (next < 0) {
return s;
}
s = s.substring(0, next) + after + s.substring(next + before.length());
index = next + after.length();
}
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 4,382
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The Bat Association of MSU stands with Palestine
Like many other colonial projects, the establishment of the state of Israel started with the forced removal of the native population, an estimated 700,000 Palestinians in 1948. This was followed by an attempt to erase their history and culture, an event known to Palestinians as Al Nakba (The Catastrophe) [1]. Since then, the state of Israel has continued to defy international law both within its borders through the establishment of apartheid laws disadvantaging non-jewish citizens [2], and outside its border through the ongoing occupation and annexation of Palestinian land [3], an ongoing Nakba that continues to this day and is on full display in Sheikh Jarrah [4].
This ongoing violent campaign of dispossession and ethnic cleansing would not be possible without support from other colonial powers, chief among those, the United States who shares in both historical and contemporary injustices against our native population, and contributes billions of dollars annually in military aid to help support the occupation [5].
This support has been bipartisan and championed as supporting the Jewish community, which has historically been the target of massive injustice. Any resistance to this support has been smeared as anti-semetic but the truth is much more simple. In the words of our current commander in chief Joe Biden, "If there were not an Israel, we would have to invent one to make sure our interests were preserved." [6] The United States supports these atrocities, not because it cares about the Jewish community, but to preserve our interests in the region.
Recognizing the role that international solidarity played in toppling apartheid in South Africa, in 2005, Palestinian civil society organizations put out a call to the international community to support the Palestinian cause by boycotting, divesting from and sanctioning the state of Israel until it complies with international law. The demands include the dismantling of the apartheid wall which resides mostly within Palestine and isolates communities. Ending racist and discriminatory laws within the state of Israel which privilege the Jewish population and disadvantage the native Palestinians, and allowing those who were forcibly expelled from their land to return [7].
The occupation of Palestine is not complicated. It is not a conflict, it is not a war, it is not a religious dispute over territory. It is an ongoing settler colonialist project predicated on ethnic cleansing which has used the real, historical and ongoing oppression of the Jewish community as a shield against criticism [8]. As an organization who believes in basic human rights, we stand in full support of the Palestinian call for the boycott, divestment and sanction of Israel, and we find it infuriating that this is presented as a debate. The merit of ethnic cleansing is not up for debate, the merit of apartheid is not up for debate. We additionally support the recent call by MSU student organizations of conscience, demanding the Michigan State University administration take a stance on this issue[9].
We are all complicit and we cannot be silent.
In love and solidarity,
#SaveSheikhJarrah
#EndIsraeliApartheid
#FreePalestine
#BDS
[1] Pappé, Ilan. The Ethnic Cleansing of Palestine. Oxford: Oneworld, 2006. Print.
[2] Hesketh, K., Bishara, S., Zaher, S., & Rosenberg, R. (2010). Inequality report: The palestinian arab minority in Israel December 2010. In Arab World Geographer (Vol. 13, Issues 3–4).
[3] https://www.theguardian.com/commentisfree/2021/may/17/palestinians-sheikh-jarrah-jerusalem-city-identity
[4] https://www.btselem.org/publications/202103_this_is_ours_and_this_too
[5] https://www.reuters.com/article/us-usa-israel-statement/u-s-israel-sign-38-billion-military-aid-package-idUSKCN11K2CI
[6] https://www.middleeastmonitor.com/20210104-if-israel-wasnt-there-the-us-would-have-to-invent-it-to-protect-its-interests/
[7] https://bdsmovement.net/call
[8] https://jewishvoiceforpeace.org/first-ever-40-jewish-groups-worldwide-oppose-equating-antisemitism-with-criticism-of-israel/
← Bat Association stands with MSU APIDA community
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
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This Nov 24th, we want to share Small Business Saturday with you! It's a holiday shopping tradition, backed by American Express, that celebrates small businesses like ours. And it wouldn't be a celebration without customers like you joining us.
So mark your calendar for Nov 24th – the Saturday after Thanksgiving – and get ready to Shop Small with us. Grab a friend or family member and come by MAIN STREET BOOKS between 10AM and 5PM on the big day.
Thank you for all your support, and we hope to see you Saturday, Nov 24th!
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 4,193
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Historical Archaeology and Indigenous Collaboration:
Discovering Histories That Have Futures
D. Rae Gould, Holly Herbster, Heather Law Pezzarossi, and Stephen A. Mrozowski
Hardcover ISBN 13: 9780813066219 - Pub Date: 2/11/2020 Details: 224 pages, 6x9 Subject(s): Archaeology
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Available for pre-order. This book will be available February, 2020
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Sample Chapters & Awards
"Deftly weaves historic records and archaeological research through an Indigenous lens to create a well-crafted story of the Nipmuc of New England. Through this lens, the reader will better recognize the struggles Indigenous people faced in colonial America as well as the struggles they continue to face as they try to reestablish their sovereign relationships with the United States."—Joe Watkins, coeditor of Challenging the Dichotomy: The Licit and the Illicit in Archaeological and Heritage Discourses
"Demonstrates how genuinely inclusive archaeology can and should be. The complicated and long underappreciated histories of New England's Native Nipmuc people are brought to life through the wholly compelling narratives of Nipmuc individuals from the seventeenth century to the present carefully pieced together from traditional knowledge, fragments of pottery and stone, snippets of documents, and the physical traces of meaningful spaces and places."—Audrey Horning, coeditor of Becoming and Belonging in Ireland AD c. 1200–1600: Essays in Identity and Cultural Practice
Highlighting the strong relationship between New England's Nipmuc people and their land from the pre-contact period to the present day, this book helps demonstrate that the history of Native Americans did not end with the arrival of Europeans. This is the rich result of a twenty-year collaboration between indigenous and nonindigenous authors, who use their own example to argue that Native peoples need to be integral to any research project focused on indigenous history and culture.
The stories traced in this book center around three Nipmuc archaeological sites in Massachusetts—the seventeenth century town of Magunkaquog, the Sarah Boston Farmstead in Hassanamesit Woods, and the Cisco Homestead on the Hassanamisco Reservation. The authors bring together indigenous oral histories, historical documents, and archaeological evidence to show how the Nipmuc people outlasted armed conflict and Christianization efforts instigated by European colonists. Exploring key issues of continuity, authenticity, and identity, Historical Archaeology and Indigenous Collaboration provides a model for research projects that seek to incorporate indigenous knowledge and scholarship.
D. Rae Gould, a member of the Nipmuc Nation of Massachusetts, is associate director of the Native American and Indigenous Studies Initiative at Brown University. Holly Herbster is principal investigator and senior archaeologist at the Public Archaeology Laboratory. Heather Law Pezzarossi is a visiting scholar in the Department of Anthropology at Syracuse University. Stephen A. Mrozowski, professor of anthropology and director of the Andrew Fiske Memorial Center for Archaeological Research at the University of Massachusetts Boston, is the author of The Archaeology of Class in Urban America.
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Hardcover: $110.00
The Archaeology of Removal in North America
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Maya Salt Works
Iconography and Wetsite Archaeology of Florida's Watery Realms
Archaeologies of Listening
The Archaeology of American Childhood and Adolescence
An Archaeology of Abundance
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 7,566
|
\section{Introduction}
Studying the properties of molecular clouds is crucial to understand star formation \citep{kennicutt2012}. The dominant constituent of molecular clouds is molecular hydrogen, (H$_2$), which is a perfectly symmetric molecule, rendering it largely invisible at the typical temperatures of molecular clouds. While observable in ultraviolet absorption against background sources, it can only be detected via emission in environments where the gas is excited to temperatures above a few hundred Kelvin. The second dominant species is neutral helium which remains inert in molecular clouds. Therefore, observational studies of molecular clouds largely rely on tracer species, namely emission from dust and molecules. The most important of these tracers is carbon monoxide (CO) \citep{bolatto2013}. CO has a relatively high abundance, canonically [CO/H$_2$] $\approx$ 10$^{-4}$ \citep{hollenbach1999}, making it the most abundant molecule after H$_2$ . The small dipole moment allows its rotational transitions to be easily excited at the cold temperatures of molecular clouds. A crucial CO observable is the J = (1-0) rotational transition at a rest frequency of 115.27 GHz.
It is common for the emission of the lowest rotational transition of CO to be used to measure the total molecular gas \citep{bolatto2013}. This is encoded in the CO-to-H$_2$ conversion factor, $\ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi$, and the related quantity $\alpha_{\rm CO}$. $\ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi$ is defined as:
\begin{equation}
\label{eq:xco}
\ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi = \frac{\COL}{W_{\rm CO}(J = 1-0)},
\end{equation}
where $\COL$ is the H$_2$ column density in cm$^{-2}$ and $W_{\rm CO}(J = 1-0)$ is the CO flux in K km s$^{-1}$. The fiducial Milky Way (MW) value is $X_{\rm CO,MW} = 2 \times 10^{20}$ cm$^{-2}$ (K km s$^{-1}$)$^{-1}$ \citep{bolatto2013}. This conversion factor has been used to estimate gas mass in local, resolved studies of MCs and the molecular gas mass in high redshift galaxies \citep[e.g. the COLDz survey][]{riechers2019}. A significant number of studies, both observational and theoretical, have been devoted to measuring, modelling or applying \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi. Prior work shows it varies with density, metallicity \citep{bell2006, shetty2011, lagos2012, narayanan2013, glover2016}, cosmic ray (CR) ionization rate (CRIR) \citep{bell2006, wolfire2010, bisbas2015, clark2015, glover2016, remy2017, papadopoulos2018} and the radiation field \citep{bell2006, wolfire2010, shetty2011, lagos2012, narayanan2013, clark2015, glover2016, gaches2018a, gong2018}. Previously, \cite{gaches2018a} found that far ultraviolet radiation feedback from forming stars can reproduce the higher \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi values measured towards diffuse star-forming clouds in the outer galaxy.
Traditional one-dimensional photo-dissociation region (PDR) models have long predicted that neutral carbon will exist only in a thin transitional layer between ionized carbon and CO \citep{hollenbach1999}. However, observations show that forbidden line emission from neutral carbon covers similar spatial extents as CO \citep[e.g.][]{ikeda1999, kulesa2005, lo2014}. It is posited that forbidden line emission from neutral carbon is a good tracer of the gas mass \citep{papadopoulos2004, offner2014, glover2015, glover2016}. Synthetic observations of hydrodynamic simulations show that \ifmmode {X_{\rm CI}} \else X$_{\rm CI}$ \fi has a smaller dispersion than \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi within a molecular cloud and is a better tracer in low metallicity gas which tends to become CO-dark \citep{offner2014, glover2015, glover2016}. Observational studies using \ifmmode {X_{\rm CI}} \else X$_{\rm CI}$ \fi as a tracer of gas mass performs as well as \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi \citep{lo2014}. \ifmmode {X_{\rm CI}} \else X$_{\rm CI}$ \fi is defined analogously to \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi:
\begin{equation}
\label{eq:xci}
{\rm X_{\rm CI} = \frac{N(H_2)}{W(CI)_{609 \, \mu m}}} ~~ {\rm cm^{-2}\, (K \, km \, s^{-1})^{-1}}
\end{equation}
where $W({\rm CI})_{\rm 609 \, \mu m}$ is the integrated flux of the $^3P_1\rightarrow ^3\!P_0$ transition at 609~$\mu$m.
\cite{gaches2019} (hereafter Paper I) presented a modified astrochemical code which includes CR attenuation in-situ. \citetalias{gaches2019} included CRs accelerated by accreting, embedded protostars and CR attenuation in one-dimensional astrochemical models of molecular clouds. We used the code to study the impact of changing the CR spectrum due to differing galactic environments and the effects of embedded CR sources for a subset of species including CO, HCO$^+$ and N$_2$H$^+$ and tested various prescriptions for constraining the CRIR. We found that ions are enhanced and neutrals are depleted in dense gas due to embedded CRs. Carbon chemistry is substantially altered depending on the assumed CR model: CRs produced by embedded sources create a significant reservoir of atomic carbon, mostly neutral, in dense gas. Embedded CRs reduce the amount of CO in clouds and warm the gas to over 30 K. In this letter we investigate the impact of the above effects on \ifmmode {X_{\rm CI}} \else X$_{\rm CI}$ \fi and \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi. In Section \ref{sec:methods} we describe the methods used in this paper. In Section \ref{sec:res} we present the results and discuss the implications for observations.
\section{Methods}\label{sec:methods}
We use the same astrochemical models from \citetalias{gaches2019} and summarize the methodology here. See \citetalias{gaches2019} for further details.
We generate synthetic protoclusters assuming the Tapered Turbulent Core \citep{offner2011} accretion model following the method described in \citet{gaches2018a}. We directly sample from the bi-variate protostellar mass distribution using the method of conditional probabilities. Each molecular cloud is described by a gas surface density and number of constituent protostars, $\Sigma_{\rm cl}$ and $N_*$, respectively. We only consider models where the star formation efficiency, $\varepsilon_g \equiv M_*/M_{\rm gas} \leq 50\%$.
We calculate the accelerated proton spectrum due to accretion shocks for each star in the protocluster. CR protons are assumed to be accelerated via Diffuse Shock Acceleration (DSA) \citep[reviewed by][]{drury1983, melrose2009} near the surface of the protostar. DSA predicts a power law spectra in momentum space, $j(p)$, with an injection momentum, $p_{\rm inj}$, set by the shock gas temperature and a maximum energy constrained by collisional energy losses and upstream diffusion \citep{gaches2018b}. The CR flux spectrum is
\begin{equation}
j(p) = j_0 \left ( \frac{p}{p_{\rm inj}} \right )^{-a},
\end{equation}
where $j_0$ is the normalization constant calculated from the total shock energy and efficiency, and $a$ is set by the shock compression factor. We find that the maximum proton energy is typically between 1 - 10 GeV \citep{gaches2018b}. We attenuate the CRs by the gas surface density out of each protostellar core, $\Sigma_{\rm core} = 1.22 \Sigma_{\rm cl}$, following \citet{padovani2009}. We assume the CRs within the core free-stream outwards since shallower attenuation produces too much CR heating in the core (see \citealt{gaches2018b}). CRs may also be attenuated by the accretion flow (Offner et al. 2019, sub.), although we do not include this in the model. The total number of CRs escaping into a natal molecular cloud embedding a protocluster is the sum of the CRs accelerated by the individual protostars and then attenuated into the surrounding gas: $j_{\rm cluster}(E) = \sum_i^{N_*} j_i(E)$.
We embed the protoclusters in the center of one-dimensional molecular clouds with a density profile, $n(r) = n_0 (R/r)^{2}$. We set the outer density to $n_0 = 100$ cm$^{-3}$, and the radius, $R$, is determined by the total column density, $\mu m_H N(R) = \Sigma_{\rm cl}$. We utilize a modified version of the photo-dissociation region astrochemistry code {\sc 3d-pdr} \citep{bisbas2012} described in \citetalias{gaches2019}\footnote{The code is public at \url{https://uclchem.github.io/3dpdr.html}}, which includes CR attenuation in-situ. The astrochemistry code uses CR spectra at the surfaces of the gas model as inputs, rather than a global CRIR. It is not known exactly how CRs transport through molecular clouds. Therefore, we consider two different transport regimes: diffusive (1/r) and rectilinear (1/r$^2$). We use the two external CR spectra from \citet{ivlev2015}: a model that extrapolates the Voyager 1 data, $\mathcal{L}$, and one that attempts to account for modulation from interstellar gas, $\mathcal{H}$.
We also consider the impact of FUV radiation and we irradiate the external surface of the molecular cloud with the normalized interstellar radiation field described in \citet{draine1978}. We model the chemistry with the gas-phase {\sc Umist12} network \citep{mcelroy2013}, which includes 215 species and $\sim3000$ reactions. The network does not include gas-grain reactions, freeze-out or any desorption processes. We do not include the grain-assisted recombination proposed in the reduced network presented by \citet{gong2017}. We explored the impact of grain-assisted recombination for C$^+$ and He$^+$ on our results and found no significant changes in the CO- to CI-to-H$_2$ conversion factors. The inclusion of grain chemistry will be investigated in future studies. We include a model following the canonical setup: a low ionization rate with no attenuation, denoted as LNA. Models denoted with H or L utilize the $\mathcal{H}$ and $\mathcal{L}$ external spectra described above. Models without embedded sources are denoted with NI, while those including sources in the diffusive or rectiliniear regimes are denoted DI or RI, respectively. We consider the six different CR models listed in Table \ref{tab:models}: LNA, LNI, LRI, LDI, HNI, and HDI.
{\sc 3d-pdr} calculates the CO line-integrated emissivity, $\epsilon$, for the J-ladder from J=0 to J=41 and the CI 307 $\mu$m and 609 $\mu$m emissivities assuming non-local thermodynamic equilibrium and using an escape probability method to account for the line opactiy. We calculate the line flux from the emissivity:
\begin{equation}
I = \frac{1}{2\pi} \int_0^R \epsilon \, dz ~~~ {\rm (erg \, s^{-1} \, cm^{-2}\, sr^{-1})},
\end{equation}
with
\begin{equation}
W = \frac{1}{10^5}\frac{c^3}{2 k_b \nu^3}I ~~~ {\rm (K \, km \, s^{-1})},
\end{equation}
where $c$ is the speed of light, $k_b$ is Boltzmann's constant and $\nu$ is the line frequency. This definition of integrated flux assumes that the interstellar medium is entirely optically thin. We calculate the H$_2$ column density from the astrochemical models
\begin{equation}
N({\rm H_2}) = \int_0^R x({\rm H_2}) \, n_H \, dz,
\end{equation}
where $x({\rm H_2})$ is the abundance of H$_2$ and $n_H$ is the gas density. Finally, we compute \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi using Equation \ref{eq:xco}.
\begin{deluxetable}{c|cccc}
\tablecolumns{5}
\tablecaption{Models from \citetalias{gaches2019}\label{tab:models}. L/H denotes a Low/High external CR spectrum, NI denotes no internal sources of CRs, DI denotes internal sources with $a = 1$ (diffusive transport), RI denotes internal sources with $a = 2$ (rectilinear transport) and NA denotes no internal sources or CR attenuation.}
\tablehead{\colhead{Name} & \colhead{Source Transport} & \colhead{Internal} & \colhead{External Field} & \colhead{Attenuation} }
\startdata
LDI\label{model:fid} & $r^{-1}$ & $\checkmark$ & $\mathcal{L}$ & $\checkmark$ \\
LRI\label{model:rec} & $r^{-2}$ & $\checkmark$ & $\mathcal{L}$ & $\checkmark$ \\
LNI\label{model:ni} & ... & ... & $\mathcal{L}$ & $\checkmark$ \\
LNA\label{model:na} & ... & ... & $\mathcal{L}$ & ... \\
HNI\label{model:hni} & ... & ... & $\mathcal{H}$ & $\checkmark$ \\
HDI\label{model:hdi} & $r^{-1}$ & $\checkmark$ & $\mathcal{H}$ & $\checkmark$ \\
\enddata
\end{deluxetable}
\section{Results and Discussion}\label{sec:res}
We present the results from the astrochemical models on the CO-to-H$_2$ and CI-to-H$_2$ conversion factors here. A more general discussion on the astrochemical impact of CRs accelerated within protoclusters is presented in \citetalias{gaches2019}.
\subsection{Effect of Cosmic Rays on \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi}
Figure \ref{fig:fideff} shows the CO-to-H$_2$ conversion factor as a function of cloud surface density, $\Sigma_{\rm cl}$, and star formation efficiency, $\varepsilon_g$, for four of the CR models in
Table \ref{tab:models}. We plot \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi normalized to the fiducial MW value $X_{\rm MW} = 2 \times 10^{20}\,{\rm cm}^{-2}\,({\rm K}\,{\rm km}\,{\rm s}^{-1})^{-1}$ \citep{bolatto2013}.
The behavior of \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi changes significantly with the assumed CR model. \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi varies only as function of surface density for the models without internal sources, LNI and HNI. There is a 0.2 dex offset in \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi between models using the high and low external cosmic ray spectrum for $\Sigma_{\rm cl} < 3$ g cm$^{-2}$ owing to increased temperatures at low extinction in model HDI. The decline in \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi at higher surface densities is the result of a larger turbulent line width because of two cooperating effects. First, there is a higher temperature due to the increasing importance of turbulent heating. Second, the turbulent linewidth produces brighter, but still optically thick, CO emission.
In the models with CRs that attenuate diffusively, LDI and HDI, \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi becomes a sensitive function of the star formation efficiency, losing much of the dependence on surface density. \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi is reduced by up to 0.5 dex due to embedded sources with the lowest values occurring for the highest star formation efficiencies. It is important to emphasize that CRs from embedded sources do little to reduce the overall amount of H$_2$ \citepalias{gaches2019}. However, they cause two effects which act to decrease \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi. First, while they reduce the amount of CO in deeply embedded regions of the cloud, they cause an enhancement of CO in low extinction gas due to an increase of HCO$^+$ and, following the formation of OH through H$_3^+$, the OH formation pathway becoming important \citep{bisbas2017}. Second, the increased CRIR leads to higher kinetic temperatures making the CO emission brighter overall.
Some prior work has investigated the effect of star formation on \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi. CR and FUV feedback from star formation external to the molecular cloud can be modeled by scaling their intensity linearly with the star formation rate (SFR) \citep{papadopoulos2010}. This is motivated by the relationship between the supernova rate and the SFR and implicitly assumes that CRs are mainly accelerated in supernova shocks. \citet{clark2015} used these relations to model how the SFR affects \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi in simulated molecular clouds. They found that \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi increases with the SFR if the cloud properties remain fixed. The increase of \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi with SFR is very weak if the density of the cloud scales with the SFR. \citet{bisbas2015} modelled the effect of enhanced CRs on the ${\rm [CO/H_2]}$ ratio, comparing different environments. They show that ${\rm [CO/H_2]}$ decreases substantially with an increase in the CRIR. By construction, these models only account for variations in the external CR flux and neglect CRs accelerated within protoclusters due to accretion, jets or stellar winds.
\subsection{Effect of Cosmic Rays on \ifmmode {X_{\rm CI}} \else X$_{\rm CI}$ \fi}
Forbidden line emission from neutral carbon is a possible tracer for molecular gas, as discussed above. Figure \ref{fig:fidCIeff} shows \ifmmode {X_{\rm CI}} \else X$_{\rm CI}$ \fi as a function of surface density, $\Sigma_{\rm cl}$, and star formation efficiency, $\varepsilon_g$. \ifmmode {X_{\rm CI}} \else X$_{\rm CI}$ \fi shows the same qualitative trends as \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi, although it is more sensitive to the CRIR: a spread of 1.2 dex in \ifmmode {X_{\rm CI}} \else X$_{\rm CI}$ \fi and 0.5 dex in \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi for the LDI model. The canonical model, LNA, which has no attenuation, exhibits a maximum value of \ifmmode {X_{\rm CI}} \else X$_{\rm CI}$ \fi$ \geq 4 \times 10^{21}$ cm$^{-2}$ (K km s$^{-1}$)$^{-1}$. Models using the high, external CR spectrum, HNI and HDI, exhibit a 0.2-0.8 dex reduction in \ifmmode {X_{\rm CI}} \else X$_{\rm CI}$ \fi compared to the low spectrum models, LNI and LDI, respectively.
The increased CRIR throughout the cloud in the high models and those with internal sources causes atomic carbon to exist outside a thin transition layer. Atomic carbon is formed in the dense gas through the destruction of CO by He$^+$:
\begin{equation*}
{\rm He + CR \rightarrow He^+ + e^-}
\end{equation*}
\begin{equation*}
{\rm He^+ + CO \rightarrow He + O + C^+}
\end{equation*}
with neutral carbon forming from recombination of C$^+$. Neutral carbon is also the result of direct dissociation of neutral molecules, such as CO, by CR protons and CR-generated photons. This enhancement leads to a reduced \ifmmode {X_{\rm CI}} \else X$_{\rm CI}$ \fi. Embedded sources cause \ifmmode {X_{\rm CI}} \else X$_{\rm CI}$ \fi to decrease by over an order of magnitude across two orders of magnitude increase in the star formation efficiency.
Neutral carbon emission is easily observable at high redshifts due to the line shifting to millimeter wavelengths. Starburst galaxies have higher SFRs producing extreme environments and more CO-dark gas \citep{wolfire2010, glover2016}. Thus, at high redshifts and in galaxies undergoing starbursts, CI may become an optimal tracer of molecular gas.
\begin{figure*}[ht!]
\centering
\includegraphics[width=\textwidth]{all_XCO_eff.pdf}
\caption{\label{fig:fideff}
Color shows $\log X_{\rm CO}/X_{\rm MW}$ where $X_{\rm MW} = 2 \times 10^{20}$ cm$^{-2}$ (K km s$^{-1}$)$^{-1}$ as a function of gas surface density, $\Sigma_{\rm cl}$, and star formation efficiency, $\varepsilon_g$. White shaded cells show regions where X$_{\rm CO}$ is consistent with Milky Way observations, $-0.3 \geq \log \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi/X_{\rm MW} \leq 0.3$. The hatched regions indicate different cosmic-ray environments, where we define $\langle \zeta \rangle_x$, the spatially-averaged CRIR, $\langle \zeta \rangle_x < 10^{-16}$ as ``quiescent'', $10^{-16} < \langle \zeta \rangle_x < 10^{-15}$ as ``star forming'' and $\langle \zeta \rangle_x > 10^{-15}$ as ``extreme.''}
\end{figure*}
\begin{figure*}[ht!]
\centering
\includegraphics[width=\textwidth]{all_XCI_eff.pdf}
\caption{\label{fig:fidCIeff} Same as Figure \ref{fig:fideff} but for $\log X_{\rm CI}/10^{-21}$.}
\end{figure*}
\subsection{Statistical Trends}
Figure \ref{fig:violins} statistically summarizes the impact of the various CR models on \ifmmode {X_{\rm CI}} \else X$_{\rm CI}$ \fi and \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi. The violin plots show the distribution of the logarithmic difference between X$_{\rm i}$ as calculated with the canonical model, LNA, and each of the CR models in Table \ref{tab:models} using the clouds across the $\Sigma_{\rm cl}-\varepsilon_g$ space as samples. These distributions represent the impact on \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi when CR attenuation or embedded sources are neglected. We find very little deviation in \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi when attenuation is included in quiescent models without internal sources. Comparison to the star-forming and extreme CR model without internal sources, HNI, shows that \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi will be over-estimated by 0.15 dex in calculations using the often-assumed CRIR of $\zeta \approx 10^{-17}$ s$^{-1}$. CRs from embedded sources, which propagate via diffusion, decrease \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi for all clouds. Furthermore, there is a substantial spread due to variation with the number of protostars, $N_*$. The high model with internal sources, HDI, logarithmic difference with the canonical model exhibits a dispersion of 0.3 dex, similar to the spread derived from MW observations \citep{bolatto2013}. If CRs from embedded sources transport as r$^{-2}$ there is no impact on \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi because the CRIR is lower and dominated by the CRs originating from external sources rather than internal.
The \ifmmode {X_{\rm CI}} \else X$_{\rm CI}$ \fi distributions in the right panel of Figure \ref{fig:violins} show much greater sensitivity to the CR model assumptions. All models differ significantly from the often-assumed canonical model in \ifmmode {X_{\rm CI}} \else X$_{\rm CI}$ \fi. \ifmmode {X_{\rm CI}} \else X$_{\rm CI}$ \fi decreases by 0.5 dex for the high model with no internal sources, HNI, and massive and inefficient star forming regions. In the case of a ``Quiescient'' CR environment, CRs from embedded sources have a larger impact on \ifmmode {X_{\rm CI}} \else X$_{\rm CI}$ \fi. The inclusion of CRs from embedded sources in star-forming and extreme environments, represented by HDI, reduces \ifmmode {X_{\rm CI}} \else X$_{\rm CI}$ \fi by nearly a dex compared to the canonical model.
\begin{figure}
\centering
\includegraphics[trim=12 0 0 0, clip, width=\textwidth]{violins_xfacs.pdf}
\caption{Logarithm of the ratio of X$_{\rm CO}$ and X$_{\rm CI}$ for a given cosmic-ray model $i \subset $ (LNI, HNI, LRI, LDI, HDI) compared to model LNA (low, external spectrum and no attenuation). Black line indicates the mean. Magenta line indicates the median.}
\label{fig:violins}
\end{figure}
\subsection{Comparisons to Galactic-scale Observations}
The hatching in Figure \ref{fig:fideff} denotes different CR environments: ``Quiescent'' regions with $\langle \zeta \rangle_x < 10^{-16}$ s$^{-1}$, ``Star Forming'' regions with $10^{-16} < \langle \zeta \rangle_x < 10^{-15}$ s$^{-1}$ and ``Extreme'' regions with $\langle \zeta \rangle_x > 10^{-15}$ s$^{-1}$, where $\langle \zeta \rangle_x$ is the spatially-averaged CRIR. These labels are motivated by observational surveys which show the majority of pointings through diffuse gas have $10^{-16} < \zeta < 10^{-15}$ s$^{-1}$. Low A$_V$ observations where $\zeta > 10^{-15}$ s$^{-1}$ are primarily sight-lines towards the galactic center \citep{indriolo2012, indriolo2015}.
There have been numerous observational studies measuring \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi in different environments within the MW and other galaxies \citep[see][and citations within]{bolatto2013}. Remarkably, in the MW and many of the Local Group galaxies, \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi is relatively constant on kpc scales. The consistency of \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi in the MW and Local Group can be explained by similar molecular cloud properties due to star-formation feedback \citep{narayanan2013}. There is a general trend in star-forming galaxies of low values of \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi towards the center and larger values in the outer disk \citep{sandstrom2013}.
The white shading in Figure \ref{fig:fideff} shows where \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi is consistent with the MW average value and spread. Models without embedded sources, LNI and HNI, are only consistent with the MW value for $\Sigma_{\rm cl} < 0.2$ g cm$^{-2}$ and $\Sigma_{\rm cl} < 0.6$ g cm$^{-2}$, respectively. Models with high surface density and low star formation efficiency, similar to clouds in the Galactic Center, exhibit a decreased \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi compared to clouds with $\Sigma \approx 1$ g cm$^{-2}$. The introduction of embedded sources increases the agreement with the MW \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi. Clouds with star formation efficiencies greater than 2\% in the low model with internal sources, LDI, have $\ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi = (2.5 \pm 1)\times10^{20}$ cm$^{-2}$(K km s$^{-1}$)$^{-1}$, consistent with the MW value. High models with the internal sources show $\ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi = 3 \pm 1.5 \times 10^{20}$ cm$^{-2}$(K km s$^{-1}$)$^{-1}$ .{\it Thus, CRs accelerated during the star formation process act to regulate \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi and reduce variation}.
Starburst galaxies tend to have lower values of \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi \citep{downes1998, papadopoulos1999, papadopoulos2012, salak2014}. Our models show that \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi always decreases towards regions with more extreme CR environments. Environmental changes, which occur in higher redshift galaxies due to enhanced supernova rates, will also decrease \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi and \ifmmode {X_{\rm CI}} \else X$_{\rm CI}$ \fi. In starburst galaxies, which have high star formation rates, this decrease could be compounded by CRs produced during the star formation process.
\subsection{Summary}
We found in \citetalias{gaches2019} that the inclusion of CR sources, specifically accreting protostars, embedded within molecular clouds and CR attenuation make the CRIR vary throughout the cloud. In this paper, we investigate the impact of different external CR fluxes and the inclusion of embedded CR sources on the CO-to-H$_2$ and CI-to-H$_2$ conversion factors. We find that differences in the CR flux caused by changes in the external environment and embedded star formation alter \ifmmode {X_{\rm CI}} \else X$_{\rm CI}$ \fi significantly and \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi by factors of a few. However, external environmental changes alone reduce \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi only by 0.2 dex, within the measured spread of \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi in the MW \citep{bolatto2013}. The difference in \ifmmode {X_{\rm CI}} \else X$_{\rm CI}$ \fi is more pronounced: it declines by an order of magnitude for the lowest surface density environments. The inclusion of embedded CR sources removes the strong dependence of \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi and \ifmmode {X_{\rm CI}} \else X$_{\rm CI}$ \fi on surface density and reduces the conversion factors by 0.6 and 1.2 dex, respectively. Embedded sources act to regulate \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi and reduce variation as a function of gas surface density and star formation efficiency. Clouds in low model including embedded sources, LDI, with star formation efficiences greater than 2\% are consistent with the observed MW value and spread of $X_{\rm CO, MW} = 2 \times 10^{20} \pm 0.3$ dex cm$^{-2}$ (K km s$^{-1}$)$^{-1}$. Observations of the CRIR in diffuse gas in the MW show that the average CRIR, $\langle \zeta \rangle \approx 10^{-16}$, which is represented by our models with a high surface CR spectrum. Models with this CRIR and ongoing star formation, are consistent with the observed MW value for all regions with star formation efficiencies greater than 1\%. Our models reproduce the trends of a decreasing \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi towards more extreme CR environments, such as those observed in the Galactic Center, the high redshift universe and starburst galaxies. Our results motivate the inclusion of CR physics and the possibility of cosmic-ray feedback from internal sources when modeling \ifmmode {X_{\rm CO}} \else X$_{\rm CO}$ \fi and \ifmmode {X_{\rm CI}} \else X$_{\rm CI}$ \fi.
\software{ \begin{itemize} \item {\sc 3d-pdr} \citep{bisbas2012}, CR implementation \citepalias{gaches2019}
\item {\sc matplotlib} \citep{matplotlib2007}
\item {\sc NumPy} \citep{numpy2011}
\item {\sc SciPy} \citep{scipy2001}
\item {\sc JupyterLab} \end{itemize}}
\acknowledgments
SSRO acknowledges funding by the National Science Foundation (NSF) grants: NSF AAG grant AST-1510021 and NSF CAREER grant AST-1650486. TGB acknowledges funding by the German Science Foundation (DFG) via the Collaborative Research Center SFB 956 ``Conditions and impact of star formation''. The authors thank the anonymous referee for their useful comments. The calculations performed for this work were done on the Massachusetts Green High Performance Computing Center (MGHPCC) in Holyoke, Massachusetts supported by the University of Massachusetts, Boston University, Harvard University, MIT, Northeastern University and the Commonwealth of Massachusetts.
\iffalse
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{
"redpajama_set_name": "RedPajamaArXiv"
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\section{Introduction}
When working with tensorial expressions, one usually encounters difficulties handling index manipulations due to complicated symmetries.
Techniques including group theoretical calculations and Young tableaux have been introduced to try to tackle these problems. However, their complexity grows quickly with the size of the problem.
The purpose of this paper is to present a formalism based on 2-spinors that aims to simplify the situation by utilizing the symmetry properties of irreducible spinors.
Let $(\mathcal{M}, g_{ab})$ be a 4-dimensional manifold with metric $g_{ab}$ of Lorentzian signature and admitting a spin structure with spin metric $\epsilon_{AB}$. It is well known that any tensor field on $\mathcal{M}$ can be expressed in terms of 2-spinors, which in turn can be decomposed into irreducible symmetric spinors \cite[Prop 3.3.54]{PR1}. For instance a valence $(3,0)$ spinor can be decomposed as
\begin{align}
T_{ABC}={}&T_{(ABC)}
+ \tfrac{1}{3} T^{D}{}_{D(B}\epsilon_{C)A}
- \tfrac{1}{3} \epsilon_{A(B}T^{D}{}_{C)D}
- \tfrac{1}{2} T_{A}{}^{D}{}_{D} \epsilon_{BC}.
\end{align}
Therefore, it is sufficient to work with with symmetric spinors. To fully establish this perspective, a symmetric product for symmetric spinors with a number of contractions is needed. It is the intention of this work to introduce the corresponding algebra and to derive its basic properties.
In particular, with these operations we stay within the algebra of symmetric spinors. This offers great simplifications, and speeds up the calculations.
Furtheremore, no relevant information is left in the indices, and we therefore get an index-free compact formalism.
We have previously described the decomposition of the covariant derivative \cite{AndBaeBlu14a}, leading to four fundamental spinor operators, which can be viewed as a special case. Also, the symmetric product is a generalization of some special operators, like the $\mathcal{K}^i$ operators defined in \cite[Definition II.4]{2016arXiv160106084A}. Therefore, all properties of such operators can easily be derived from the corresponding properties of the symmetric product described in this paper.
The formalism has many potential applications, see \cite{CompComplex},\cite{AdjointOp}. As a simple example, consider a condition of the form
\begin{align}
0={}&K_{AB}{}^{FH} L_{F}{}^{C} \varphi_{HC}
+ M_{(A}{}^{C}\varphi_{B)C},
\end{align}
for symmetric spinors $K, L, M, \varphi$. For arbitrary $\varphi$ a systematic computation, using the techniques of this paper, shows that the conditions on $K,L, M$ are of the form
\begin{align}
K^{G}{}_{(ABC}L_{|G|F)}={}&0,&
M_{AB}={}&\tfrac{1}{2} K^{CF}{}_{AB} L_{CF},
\end{align}
see Section~\ref{sec:example} for details. The same techniques have been used in \cite{JacBac2022} to derive conditions on the spacetime for the existence of second order symmetry operators for the massive Dirac equation.
The formalism is implemented in the \emph{SymSpin} \cite{SymSpinWeb} package for \emph{xAct} \cite{xActWeb} for \emph{Mathematica}.
In Section~\ref{sec:SymSpinAlg} we introduce the symmetric product and state basic properties in Theorem~\ref{thm:SymProdProperties}. The expansion of a product into symmetric products is discussed in Lemma~\ref{lem:IrrDecProduct}. The irreducible parts of the Levi-Civita connection, its commutators, curvature and Leibniz rules are discussed in Section~\ref{sec:Derivatives}. A concise form the the dyad components of such symmetric spinors is given in Section~\ref{sec:GHP}. The computer algebra implementation is discussed in Section~\ref{sec:SymSpinPackage} and Section~\ref{sec:Conclusions} contains some conclusions.
\section{Symmetric spinor algebra} \label{sec:SymSpinAlg}
Let $\mathcal{S}_{k,l}$ be the space of symmetric valence $(k,l)$ spinors. In abstract index notation, elements are of the form $\phi_{A_1 \dots A_k A'_1 \dots A'_l} \in \mathcal{S}_{k,l}$. Sometimes it is convenient to suppress the valence and/or indices and we write e.g. $\phi \in \mathcal{S}$ or $\phi \in \mathcal{S}_{k,l}$.
\subsection{Symmetric product}
Given two symmetric spinors, we introduce a product which involves a given number of contractions and symmetrization afterwards.
\begin{definition} \label{def:SymProd}
Let $k,l,n,m,i,j$ be integers with $i \leq min(k,n)$ and $j \leq min(l,m)$. The symmetric product is a bilinear form
\begin{align}
\overset{i,j}{\odot}: \mathcal{S}_{k,l} \times \mathcal{S}_{n,m} \to{}& \mathcal{S}_{k+n-2i,l+m-2j}.
\end{align}
For $\phi \in \mathcal{S}_{k,l}, \psi \in \mathcal{S}_{n,m}$, it is given by
\begin{align}
\label{eq:SymMultDef}
(\phi\overset{i,j}{\odot}\psi)_{A_1 \dots A_{k+n-2i}}^{A'_1 \dots A'_{l+m-2j}}={}&
\phi_{(A_1 \dots A_{k-i-1}}^{(A'_1 \dots A'_{l-j-1}| B_1 \dots B_i B'_1 \dots B'_j|}
\psi^{A'_{l-j} \dots A'_{l+m-2j})}_{A_{k-i} \dots A_{k+n-2i})B_1 \dots B_i B'_1 \dots B'_j}
\end{align}
\end{definition}
For many commutator relations we will need the following coefficients.
\begin{definition}
Define the associativity coefficients
\begin{align}
F_{i,r,k}^{t,m,M}={}&\sum_{p=0}^M\sum_{q=0}^{M-p}
\frac{(-1)^{t-p+q} \binom{k - m}{p} \binom{m}{M - p - q} \binom{k - m - p}{q} \binom{r - m}{t - p} \binom{i - t}{M - p - q} \binom{t - p}{q}}{\binom{i + k - M - p + 1}{M - p} \binom{M - p}{q} \binom{k - 2 m + r}{t}}.
\end{align}
\end{definition}
Observe that the limits can be restricted to $\max(0, m - r + t) \leq p \leq \min(k - m, M, t)$ and $\max(0, M - m - p, M - i - p + t) \leq q \leq \min(k - m - p, M - p, t - p)$ because the terms are zero outside this range.
For multiple products we will use the convention $\omega\overset{m,n}{\odot}\varphi\overset{t,u}{\odot}\phi=\omega\overset{m,n}{\odot}(\varphi\overset{t,u}{\odot}\phi)$.
\begin{theorem} \label{thm:SymProdProperties}
Let $\phi \in \mathcal{S}_{i,j}, \omega \in \mathcal{S}_{r,s}, \varphi \in \mathcal{S}_{k,l}$. The symmetric product $\odot$ of Definition \ref{def:SymProd} has the following properties:
\begin{subequations}
\begin{enumerate}
\item \label{part:SymProdProp1} It is graded anti-commutative:
\begin{align}
\phi\overset{m,n}{\odot}\omega ={}& (-1)^{m+n}\omega\overset{m,n}{\odot}\phi \label{eq:Commutativity}
\end{align}
\item \label{part:SymProdProp2} It is non-associative:
\begin{align} \label{eq:NonAssociative}
(\omega\overset{m,n}{\odot}\varphi)\overset{t,u}{\odot}\phi
={}&\negmedspace\negmedspace\sum_{M=0}^{\min(i,k)}\sum_{N=0}^{\min(j,l)}\negmedspace (-1)^{t+u+M+N} F_{i,r,k}^{t,m,M}F_{j, s, l}^{u, n, N}
\omega\overset{t+m-M,u+n-N}{\odot}\varphi\overset{M,N}{\odot}\phi.
\end{align}
\item \label{part:SymProdProp3} It is Hermitian:
\begin{align}
\overline{\phi\overset{m,n}{\odot}\omega} = \overline\phi\overset{n,m}{\odot}\overline\omega
\end{align}
\end{enumerate}
\end{subequations}
\end{theorem}
Combining the first two points, we get the following useful relation.
\begin{corollary}
\begin{align}
\phi\overset{t,u}{\odot}\omega\overset{m,n}{\odot}\varphi
={}&\negmedspace\negmedspace\sum_{M=0}^{\min(i,k)}\sum_{N=0}^{\min(j,l)}\negmedspace F_{i,r,k}^{t,m,M}F_{j, s, l}^{u, n, N}
\omega\overset{t+m-M,u+n-N}{\odot}\phi\overset{M,N}{\odot}\varphi
\label{eq:SymMultCommutator1}
\end{align}
\end{corollary}
\subsection{Irreducible decomposition}
A key property of the symmetric product is that the product of two symmetric spinors can always be decomposed in terms of symmetric products and spin metrics.
\begin{definition}
We will use the following notation for for products of spin metrics.
\begin{subequations}
\begin{align}
\epsilon_{A_1\dots A_p}^{B_1\dots B_p}&=\epsilon_{A_1}{}^{B_1}\dots \epsilon_{A_p}{}^{B_p},\\
\bar\epsilon_{A_1'\dots A_q'}^{B_1'\dots B_q'}&=\bar\epsilon_{A_1'}{}^{B_1'}\dots \bar\epsilon_{A_q'}{}^{B_q'}.
\end{align}
\end{subequations}
\end{definition}
\begin{lemma}\label{lem:IrrDecProduct}
For $\phi \in \mathcal{S}_{i,j}, \varphi \in \mathcal{S}_{k,l}$ with $p$ unprimed and $q$ primed contractions, we have the irreducible decomposition
\begin{align}
&\hspace{-3ex}\phi^{C_{1}\dots C_pC_1'\dots C_q'}_{A_1\dots A_{i-p}A_1'\dots A_{j-q}'}\varphi_{C_1\dots C_pC_1'\dots C_q'}^{B_1\dots B_{k-p} B_1'\dots B_{l-q}'}\nonumber\\
={}&(-1)^{p+q}\negmedspace\sum_{m=p}^{\min(i,k)}\sum_{n=q}^{\min(j,l)}\Bigl(
\frac{(-1)^{m+n}\binom{i - p}{m - p} \binom{k - p}{m - p}\binom{j-q}{n-q} \binom{l-q}{n-q}}{\binom{i + k - m - p + 1}{m - p}\binom{j + l - n - q + 1}{n-q}} \nonumber\\
&\times \epsilon_{(A_1\dots A_{m-p}}^{(B_1\dots B_{m-p}}(\phi\overset{m,n}{\odot}\varphi){}_{A_{m-p+1}\dots A_{i-p})(A_{n-p+1}'\dots A_{j-q}'}^{B_{m-p+1}\dots B_{k-p})(B_{n-q+1}'\dots B_{l-q}'}\bar\epsilon_{A_1'\dots A_{n-q}')}^{B_1'\dots B_{n-q}')}
\Bigr).
\end{align}
\end{lemma}
\begin{proof}
Let $\phi$ and $\varphi$ be symmetric of valence $(i,0)$ and $(k,0)$ respectively. By \cite[Prop 3.3.54]{PR1} the irreducible decomposition of the product must have the following form
\begin{align}
\label{eq:IrrDecAnsatz}
\phi_{A_1\dots A_i}\varphi^{B_1\dots B_k}=\sum_{m=0}^{\min(i,k)} c_m \epsilon_{(A_1\dots A_m}^{(B_1\dots B_m}(\phi\overset{m,0}{\odot}\varphi){}_{A_{m+1}\dots A_i)}^{B_{m+1}\dots B_k)}
\end{align}
Taking a trace of the summand, we find by partial expansions of the symmetrizations that
\begin{align}
&\hspace{-3ex}\epsilon_{(A_1\dots A_m}^{(B_1\dots B_m}(\phi\overset{m,0}{\odot}\varphi){}_{A_{m+1}\dots A_i)}^{B_{m+1}\dots B_{k-1}A_i)}\nonumber\\
={}&\tfrac{m}{i}\epsilon_{A_i(A_1\dots A_{m-1}}^{(B_1\dots B_m}(\phi\overset{m,0}{\odot}\varphi){}_{A_m\dots A_{i-1})}^{B_{m+1}\dots B_{k-1}A_i)}
+\tfrac{i-m}{i}\epsilon_{(A_1\dots A_{m}}^{(B_1\dots B_m}(\phi\overset{m,0}{\odot}\varphi){}_{A_{m+1}\dots A_{i-1})A_i}^{B_{m+1}\dots B_{k-1}A_i)}\nonumber\\
={}&\tfrac{m}{ik}\epsilon_{A_i(A_1\dots A_{m-1}}^{A_i(B_1\dots B_{m-1}}(\phi\overset{m,0}{\odot}\varphi){}_{A_{m}\dots A_{i-1})}^{B_{m}\dots B_{k-1})}
+\tfrac{m(m-1)}{ik}\epsilon_{A_i(A_1\dots A_{m-1}}^{(B_1|A_i|\dots B_{m-1}}(\phi\overset{m,0}{\odot}\varphi){}_{A_{m}\dots A_{i+1})}^{B_{m}\dots B_{k-1})}\nonumber\\
&+\tfrac{m(k-m)}{ik}\epsilon_{A_i(A_1\dots A_{m-1}}^{(B_1\dots B_m}(\phi\overset{m,0}{\odot}\varphi){}_{A_{m}\dots A_{i-1})}^{B_{m+1}\dots B_{k-1})A_i}
+\tfrac{(i-m)m}{ik}\epsilon_{(A_1\dots A_{m}}^{A_i(B_1\dots B_{m-1}}(\phi\overset{m,0}{\odot}\varphi){}_{A_{m+1}\dots A_{i-1})A_i}^{B_{m}\dots B_{k-1})}\nonumber\\
={}&\tfrac{m(i+k-m+1)}{ik}
\epsilon_{(A_1\dots A_{m-1}}^{(B_1\dots B_{m-1}}(\phi\overset{m,0}{\odot}\varphi){}_{A_{m}\dots A_{i-1})}^{B_{m}\dots B_{k-1})}.
\end{align}
Recursively for $p\leq \min(i,k)$ traces we get
\begin{align}
&\hspace{-3ex}\epsilon_{(A_1\dots A_m}^{(B_1\dots B_m}(\phi\overset{m,0}{\odot}\varphi){}_{A_{m+1}\dots A_i)}^{B_{m+1}\dots B_{k-p}A_{i-p+1}\dots A_i)}\nonumber\\
={}&\tfrac{m(i+k-m+1)}{ik}
\epsilon_{(A_1\dots A_{m-1}}^{(B_1\dots B_{m-1}}(\phi\overset{m,0}{\odot}\varphi){}_{A_{m}\dots A_{i-1})}^{B_{m}\dots B_{k-p}A_{i-p+1}\dots A_{i-1})}\nonumber\\
={}&\tfrac{m(i+k-m+1)}{ik}\tfrac{(m-1)(i+k-m)}{(i-1)(k-1)}
\epsilon_{(A_1\dots A_{m-2}}^{(B_1 \dots B_{m-2}}(\phi\overset{m,0}{\odot}\varphi){}_{A_{m-1}\dots A_{i-2})}^{B_{m-1}\dots B_{k-p}A_{i-p+1}\dots A_{i-2})}\nonumber\\
={}&
\epsilon_{(A_1\dots A_{m-p}}^{(B_1\dots B_{m-p}}(\phi\overset{m,0}{\odot}\varphi){}_{A_{m-p+1}\dots A_{i-p})}^{B_{m-p+1}\dots B_{k-p})}\prod_{q=0}^{p-1}\tfrac{(m-q)(i+k-m+1-q)}{(i-q)(k-q)}.
\nonumber\\
={}&
\epsilon_{(A_1\dots A_{m-p}}^{(B_1\dots B_{m-p}}(\phi\overset{m,0}{\odot}\varphi){}_{A_{m-p+1}\dots A_{i-p})}^{B_{m-p+1}\dots B_{k-p})}
\frac{\binom{1 + i + k - m}{p} \binom{m}{p}}{\binom{i}{p} \binom{k}{p}}
\end{align}
Taking $p\leq \min(i,k)$ traces in \eqref{eq:IrrDecAnsatz} gives
\begin{align}
&\hspace{-3ex}\phi_{A_1\dots A_{i-p}}^{C_1\dots C_p}\varphi^{B_1\dots B_{k-p}}_{C_1\dots C_p}\nonumber\\
={}&(-1)^p\phi_{A_1\dots A_i}\varphi^{B_1\dots B_{k-p}A_{i-p+1}\dots A_i}\nonumber\\
={}&(-1)^p\sum_{m=0}^{\min(i,k)} c_m \epsilon_{(A_1\dots A_m}^{(B_1\dots B_m}(\phi\overset{m,0}{\odot}\varphi){}_{A_{m+1}\dots A_i)}^{B_{m+1}B_{k-p}A_{i-p+1}\dots A_i)}\nonumber\\
={}&(-1)^p\sum_{m=p}^{\min(i,k)} c_m\frac{\binom{i + k - m +1}{p} \binom{m}{p}}{\binom{i}{p} \binom{k}{p}}\epsilon_{(A_1\dots A_{m-p}}^{(B_1\dots B_{m-p}}(\phi\overset{m,0}{\odot}\varphi){}_{A_{m-p+1}\dots A_{i-p})}^{B_{m-p+1}\dots B_{k-p})}.
\end{align}
With $m<p$, we get at least one contraction of the symmetric spinor $(\phi\overset{m,0}{\odot}\varphi)$ and the term drops out.
If we symmetrize over all free indices, only the $m=p$ term survives, and we get
\begin{align}
c_m =\frac{(-1)^m\binom{i}{m} \binom{k}{m}}{\binom{i + k - m +1}{m}}.
\end{align}
Hence
\begin{align}
&\hspace{-3ex}\phi_{A_1\dots A_{i-p}}^{C_1\dots C_p}\varphi^{B_1\dots B_{k-p}}_{C_1\dots C_p}\nonumber\\
={}&(-1)^p\sum_{m=p}^i \frac{(-1)^m\binom{i}{m} \binom{k}{m}}{\binom{i + k - m + 1}{m}}\frac{\binom{i + k - m +1}{p} \binom{m}{p}}{\binom{i}{p} \binom{k}{p}}\epsilon_{(A_1\dots A_{m-p}}^{(B_1\dots B_{m-p}}(\phi\overset{m,0}{\odot}\varphi){}_{A_{m-p+1}\dots A_{i-p})}^{B_{m-p+1}\dots B_{k-p})}\nonumber\\
={}&(-1)^p\sum_{m=p}^{\min(i,k)} \frac{(-1)^m\binom{i - p}{m - p} \binom{k - p}{m - p}}{\binom{i + k - m - p +1}{m - p}}\epsilon_{(A_1\dots A_{m-p}}^{(B_1\dots B_{m-p}}(\phi\overset{m,0}{\odot}\varphi){}_{A_{m-p+1}\dots A_{i-p})}^{B_{m-p+1}\dots B_{k-p})}.
\end{align}
By complex conjugation we get the corresponding decomposition for the primed indices.
\end{proof}
\subsection{Proof of Theorem~\ref{thm:SymProdProperties}}
To proof the main theorem and in particular \eqref{eq:NonAssociative}, we need the following intermediate identities. We restrict to unprimed indices, as the effect of primed indices can be superimposed.
We begin with a partial expansion of symmetrization of $B$ indices.
\begin{proposition}\label{prop:PartialSymMultExpansion}
Let $\omega \in \mathcal{S}_{r,0}, \varphi \in \mathcal{S}_{k,0}$. We have the partial expansion
\begin{align}
\label{eq:PartialSymMultExpansion}
&\hspace{-3ex}(\omega\overset{m,0}{\odot}\varphi){}_{A_1\dots A_{k+r-2m-t}B_1\dots B_t}\nonumber\\
={}&\sum_{p=0}^{t}\frac{\binom{k - m}{p} \binom{r - m}{t-p}}{\binom{k - 2 m + r}{t}}
\omega_{B_{p+1}\cdots B_{t}(A_1\dots A_{r-m-t+p}}^{C_1\dots C_m}\varphi_{A_{r-m-t+p+1}\dots A_{k+r-2m-t})B_1\dots B_pC_1\dots C_m}.
\end{align}
The sum can be limited to the range $\max(0,t+m-r)\leq p\leq \min(t,k-m)$.
\end{proposition}
\begin{proof}
Partial expansion of the symmetry for the indices $B_t, B_{t-1}, \dots B_1$ gives
\begin{align}
&\hspace{-3ex}(\omega\overset{m,0}{\odot}\varphi){}_{A_1\dots A_{k+r-2m-t}B_1\dots B_t}\nonumber\\
={}&\tfrac{r-m}{k+r-2m}\omega_{B_t(A_1\dots A_{r-m-1}}^{C_1\dots C_m}\varphi_{A_{r-m}\dots A_{k+r-2m-t}B_1\dots B_{t-1})C_1\dots C_m}\nonumber\\
&+\tfrac{k-m}{k+r-2m}\omega_{(A_1\dots A_{r-m}}^{C_1\dots C_m}\varphi_{A_{r-m+1}\dots A_{k+r-2m-t}B_1\dots B_{t-1})B_tC_1\dots C_m}\nonumber\\
={}&\sum_{p=0}^{t}\Bigl(\binom{t}{p}\tfrac{(r-m)!(k-m)!(k+r-2m-t)!}{(r-m-t+p)!(k-m-p)!(k+r-2m)!}\nonumber\\
&\times \omega_{B_{p+1}\cdots B_{t}(A_1\dots A_{r-m-t+p}}^{C_1\dots C_m}\varphi_{A_{r-m-t+p+1}\dots A_{k+r-2m-t})B_1\dots B_pC_1\dots C_m}\Bigr),
\end{align}
which can be simplified to \eqref{eq:PartialSymMultExpansion}.
\end{proof}
We aso need to make an irreducible decomposition of a product of two spinors with some contractions and symmetrizations.
\begin{proposition}\label{prop:PropEight}
Let $\phi \in \mathcal{S}_{i,0}, \varphi \in \mathcal{S}_{k,0}$.
\begin{align}
&\hspace{-3ex}\phi^{C_{1}\dots C_p(B_1\dots B_{t-p}}_{(A_1\dots A_{i-t}}\varphi_{A_{i-t+1}\dots A_{i+k-t-m-p})C_1\dots C_p}^{B_{t-p+1}\dots B_{t-p+m})}\nonumber\\
={}&\negmedspace\sum_{M=0}^{\min(i,k)-p} \sum_{q=0}^M\frac{(-1)^{q+M}\binom{m}{M - q} \binom{k - m - p}{q} \binom{i - t}{M - q} \binom{t - p}{q}}{\binom{i + k - M - 2p + 1}{M}\binom{M}{q} }
\epsilon_{(A_1\dots A_{M}}^{(B_1\dots B_M}(\phi\overset{M+p,0}{\odot}\varphi){}_{A_{M+1}\dots A_{i+k-t-m-p})}^{B_{M+1}\dots B_{t-p+m})}
\label{eq:PropEight}
\end{align}
\end{proposition}
\begin{proof}
Let $\bumpeq$ mean equal after lowering the $A$ indices, raising the $B$ indices and symmetrizing over the $A$ and $B$ index sets separately.
Using Lemma~\ref{lem:IrrDecProduct}, performing a partial expansion of the symmetries and noticing that $\epsilon_{A_i}^{A_j}\bumpeq 0$ and $\epsilon_{B_i}^{B_j}\bumpeq 0$ if $i\neq j$, we get
\begin{align}
&\hspace{-3ex}\phi^{C_{1}\dots C_p}_{A_1\dots A_{i-t}B_1\dots B_{t-p}}\varphi_{C_1\dots C_p}^{A_{i-t+1}\dots A_{i+k-t-m-p}B_{t-p+1}\dots B_{t-p+m}}\nonumber\\
={}&(-1)^{p}\negmedspace\sum_{M=0}^{\min(i,k)-p}\Bigl(
\frac{(-1)^{M+p}\binom{i - p}{M} \binom{k - p}{M}}{\binom{i + k - M - 2p + 1}{M}} \nonumber\\
&\times \epsilon_{(A_1\dots A_{M}}^{(A_{i-t+1}\dots A_{i-t+M}}(\phi\overset{M+p,0}{\odot}\varphi){}_{A_{M+1}\dots A_{i-t}B_1\dots B_{t-p})}^{A_{i-t+M+1}\dots A_{i+k-t-m-p}B_{t-p+1}\dots B_{t-p+m})}
\Bigr)\nonumber\\
\bumpeq{}&\negmedspace\sum_{M=0}^{\min(i,k)-p}\frac{(-1)^{M}\binom{i - p}{M} \binom{k - p}{M}}{\binom{i + k - M - 2p + 1}{M}}\Bigl( \nonumber\\
&\tfrac{(t-p)(k-m-p)}{(i-p)(k-p)} \epsilon_{B_1(A_1\dots A_{M-1}}^{A_{i-t+1}(A_{i-t+2}\dots A_{i-t+M}}(\phi\overset{M+p,0}{\odot}\varphi){}_{A_{M}\dots A_{i-t}B_2\dots B_{t-p})}^{A_{i-t+M+1}\dots A_{i+k-t-m-p}B_{t-p+1}\dots B_{t-p+m})}\nonumber\\
&+\tfrac{(i-t)m}{(i-p)(k-p)} \epsilon_{A_1(A_2\dots A_{M}}^{B_{t-p+1}(A_{i-t+1}\dots A_{i-t+M-1}}(\phi\overset{M+p,0}{\odot}\varphi){}_{A_{M+1}\dots A_{i-t}B_1\dots B_{t-p})}^{A_{i-t+M}\dots A_{i+k-t-m-p}B_{t-p+2}\dots B_{t-p+m})}
\Bigr).
\end{align}
Repeatedly expanding, we find
\begin{align}
&\hspace{-3ex}\phi^{C_{1}\dots C_p}_{A_1\dots A_{i-t}B_1\dots B_{t-p}}\varphi_{C_1\dots C_p}^{A_{i-t+1}\dots A_{i+k-t-m-p}B_{t-p+1}\dots B_{t-p+m}}\nonumber\\
\bumpeq{}&\negmedspace\sum_{M=0}^{\min(i,k)-p}\frac{(-1)^{M}\binom{i - p}{M} \binom{k - p}{M}}{\binom{i + k - M - 2p + 1}{M}}\Bigl(\sum_{q=0}^M\binom{M}{q}\tfrac{(t-p)!(k-m-p)!(i-t)!m!(i-p-M)!(k-p-M)!}{(t-p-q)!(k-m-p-q)!(i-t-M+q)!(m-M+q)!(i-p)!(k-p)!}
\nonumber\\
&\epsilon_{B_1\dots B_qA_1\dots A_{M-q}}^{A_{i-t+1}\dots A_{i-t+q}B_{t-p+1}\dots B_{t-p+M-q}}(\phi\overset{M+p,0}{\odot}\varphi){}_{A_{M-q+1}\dots A_{i-t}B_{q+1}\dots B_{t-p}}^{A_{i-t+q+1}\dots A_{i+k-t-m-p}B_{t-p+M-q+1}\dots B_{t-p+m}}
\Bigr).
\end{align}
Moving the $A$ indices down and the $B$ indices up and writing out the symmetrizations, we get
\begin{align}
&\hspace{-3ex}\phi^{C_{1}\dots C_p(B_1\dots B_{t-p}}_{(A_1\dots A_{i-t}}\varphi_{A_{i-t+1}\dots A_{i+k-t-m-p})C_1\dots C_p}^{B_{t-p+1}\dots B_{t-p+m})}\nonumber\\
={}&\negmedspace\sum_{M=0}^{\min(i,k)-p}\frac{(-1)^{M}\binom{i - p}{M} \binom{k - p}{M}}{\binom{i + k - M - 2p + 1}{M}}\Bigl( \sum_{q=0}^M\frac{(-1)^q\binom{m}{M - q} \binom{k - m - p}{q} \binom{i - t}{M - q} \binom{t - p}{q}}{\binom{M}{q} \binom{i - p}{M} \binom{k - p}{M}}
\nonumber\\
&\epsilon_{(A_{i-t+1}\dots A_{i-t+q}A_1\dots A_{M-q}}^{(B_1\dots B_q B_{t-p+1}\dots B_{t-p+M-q}}(\phi\overset{M+p,0}{\odot}\varphi){}_{A_{M-q+1}\dots A_{i-t}A_{i-t+q+1}\dots A_{i+k-t-m-p})}^{B_{q+1}\dots B_{t-p}B_{t-p+M-q+1}\dots B_{t-p+m})}
\Bigr).
\end{align}
After rearranging the indices, and simplifying, we get \eqref{eq:PropEight}.
\end{proof}
\begin{proof}[Proof of Theorem \ref{thm:SymProdProperties}]
Part~\ref{part:SymProdProp1} follows from the zee-zaw rule on the $m+n$ contracted indices and part~\ref{part:SymProdProp3} follows from complex conjugation of \eqref{eq:SymMultDef}.
Part~\ref{part:SymProdProp2} follows from the following argument.
Proposition~\ref{prop:PartialSymMultExpansion}, a renaming of the contracted indices and using the zee-zaw rule gives
\begin{align}
&\hspace{-3ex}(\phi\overset{t,0}{\odot}\omega\overset{m,0}{\odot}\varphi){}_{A_1\dots A_{i+k+r-2m-2t}}\nonumber\\
={}&\sum_{p=0}^{t}\frac{\binom{k - m}{p} \binom{r - m}{t-p}}{\binom{k - 2 m + r}{t}}\nonumber\\
&\times \omega_{B_{p+1}\cdots B_{t}(A_1\dots A_{r-m-t+p}}^{C_1\dots C_m}\phi^{B_1\dots B_t}_{A_{k+r-2m-t+1}\dots A_{i+k+r-2m-2t}}\varphi_{A_{r-m-t+p+1}\dots A_{k+r-2m-t})B_1\dots B_pC_1\dots C_m} \nonumber\\
={}&\sum_{p=0}^{t}(-1)^{m}\frac{\binom{k - m}{p} \binom{r - m}{t-p}}{\binom{k - 2 m + r}{t}}\nonumber\\
&\times \omega_{B_1\dots B_{m+t-p}(A_{i-t+k-m-p+1}\dots A_{i+k+r-2m-2t}}\phi^{C_1\dots C_p B_{1} B_{t-p}}_{A_{1}\dots A_{i-t}}\varphi_{A_{i-t+1}\dots A_{i-t+k-m-p}) C_1\dots C_p}^{B_{t-p+1}\dots B_{t-p+m}}.
\end{align}
Using Proposition~\ref{prop:PropEight}, contracting the spin metrics, and using the zee-zaw rule, we get
\begin{align}
&\hspace{-3ex}(\phi\overset{t,0}{\odot}\omega\overset{m,0}{\odot}\varphi){}_{A_1\dots A_{i+k+r-2m-2t}}\nonumber\\
={}&\sum_{p=0}^{t}\sum_{M=0}^{\min(i,k)-p} \sum_{q=0}^M(-1)^{m}\frac{\binom{k - m}{p} \binom{r - m}{t-p}}{\binom{k - 2 m + r}{t}}\frac{(-1)^{q+M}\binom{m}{M - q} \binom{k - m - p}{q} \binom{i - t}{M - q} \binom{t - p}{q}}{\binom{i + k - M - 2p + 1}{M}\binom{M}{q} }\nonumber\\
&\times \omega_{B_1\dots B_{m+t-p}(A_{i-t+k-m-p+1}\dots A_{i+k+r-2m-2t}}
\epsilon_{A_1\dots A_{M}}^{B_1\dots B_M}(\phi\overset{M+p,0}{\odot}\varphi){}_{A_{M+1}\dots A_{i+k-t-m-p})}^{B_{M+1}\dots B_{t-p+m}}\nonumber\\
={}&\sum_{p=0}^{t}\sum_{M=0}^{\min(i,k)-p} \sum_{q=0}^M(-1)^{m+q+M}\frac{\binom{k - m}{p} \binom{r - m}{t-p}\binom{m}{M - q} \binom{k - m - p}{q} \binom{i - t}{M - q} \binom{t - p}{q}}{\binom{k - 2 m + r}{t}\binom{i + k - M - 2p + 1}{M}\binom{M}{q} }\nonumber\\
&\times \omega_{B_{M+1}\dots B_{m+t-p}(A_1\dots A_{r+M+p-m-t}}
(\phi\overset{M+p,0}{\odot}\varphi){}_{A_{r+M+p-m-t+1}\dots A_{i+k+r-2m-2t})}^{B_{M+1}\dots B_{t-p+m}}\nonumber\\
={}&\sum_{p=0}^{t}\sum_{M=0}^{\min(i,k)-p} \sum_{q=0}^M(-1)^{q+t-p}\frac{\binom{k - m}{p} \binom{r - m}{t-p}\binom{m}{M - q} \binom{k - m - p}{q} \binom{i - t}{M - q} \binom{t - p}{q}}{\binom{k - 2 m + r}{t}\binom{i + k - M - 2p + 1}{M}\binom{M}{q} }\nonumber\\
&\times \omega^{B_{M+1}\dots B_{m+t-p}}_{(A_1\dots A_{r+M+p-m-t}}
(\phi\overset{M+p,0}{\odot}\varphi){}_{A_{r+M+p-m-t+1}\dots A_{i+k+r-2m-2t})B_{M+1}\dots B_{t-p+m}}.
\end{align}
Hence
\begin{align}
&\hspace{-3ex}(\phi\overset{t,0}{\odot}\omega\overset{m,0}{\odot}\varphi)\nonumber\\
={}&\negmedspace\sum_{p=0}^{t}\negmedspace\sum_{M=0}^{\min(i,k)-p} \sum_{q=0}^M
\frac{(-1)^{t-p+q}\binom{k - m}{p} \binom{r - m}{t-p}\binom{m}{M - q} \binom{k - m - p}{q} \binom{i - t}{M - q} \binom{t - p}{q}}{\binom{k - 2 m + r}{t}\binom{i + k - M - 2p + 1}{M}\binom{M}{q}}
(\omega\overset{t+m-p-M,0}{\odot}\phi\overset{M+p,0}{\odot}\varphi)\nonumber\\
={}&\negmedspace\sum_{p=0}^{t}\negmedspace\sum_{M=p}^{\min(i,k)} \sum_{q=0}^{M-p}
\frac{(-1)^{t-p+q}\binom{k - m}{p} \binom{r - m}{t-p}\binom{m}{M - p - q} \binom{k - m - p}{q} \binom{i - t}{M - p - q} \binom{t - p}{q}}{\binom{k - 2 m + r}{t}\binom{i + k - M - p + 1}{M-p}\binom{M-p}{q}}
(\omega\overset{t+m-M,0}{\odot}\phi\overset{M,0}{\odot}\varphi)\nonumber\\
={}&\negmedspace\sum_{M=0}^{\min(i,k)}\negmedspace F_{i,r,k}^{t,m,M}
(\omega\overset{t+m-M,0}{\odot}\phi\overset{M,0}{\odot}\varphi),
\end{align}
where we have made the change $M\rightarrow M-p$ and re-ordered the sums.
The limits can be restricted to $\max(0, m - r + t) \leq p \leq \min(k - m, M, t)$ and $\max(0, M - m - p, M - i - p + t) \leq q \leq \min(k - m - p, M - p, t - p)$ because the terms are zero outside this range.
The treatment of the primed indices is completely analogous.
\end{proof}
\subsection{Derivatives} \label{sec:Derivatives}
In \cite{AndBaeBlu14a}, the irreducible decomposition of the covariant derivative of a symmetric spinor was done in terms of fundamental spinor operators.
By extending the symmetric product to the space of linear, symmetric differential operators of valence $(k,l)$, $\mathcal{O}_{k,l}$, we can express the fundamental spinor operators in a compact way.
\begin{remark}
For $\nabla \in \mathcal{O}_{1,1}$ we have the fundamental spinor operators \cite[Definition 13]{AndBaeBlu14a}
\begin{align} \label{eq:FundSpinOps}
\mathscr{D} \varphi ={}& \nabla \overset{1,1}{\odot}\varphi, &&&
\mathscr{C} \varphi ={}& \nabla \overset{0,1}{\odot}\varphi, &&&
\mathscr{C}^\dagger \varphi ={}& \nabla \overset{1,0}{\odot}\varphi, &&&
\mathscr{T} \varphi ={}& \nabla \overset{0,0}{\odot}\varphi.
\end{align}
\end{remark}
On $\varphi \in \mathcal{S}_{k,l}$ we have the irreducible decomposition of the covariant derivative into fundamental operators \cite[Lemma 15]{AndBaeBlu14a},
\begin{align}
\nabla_{A_1}{}^{A_1'}\varphi{}_{A_2\dots A_{k+1}}{}^{A_2'\dots A_{l+1}'}={}&
(\mathscr{T}\varphi){}_{A_1\dots A_{k+1}}{}^{A_1'\dots A_{l+1}'}\nonumber\\
&-\tfrac{l}{l+1}\bar\epsilon^{A_1'(A_2'}(\mathscr{C}\varphi){}_{A_1\dots A_{k+1}}{}^{A_3'\dots A_{l+1}')}\nonumber\\
&-\tfrac{k}{k+1}\epsilon_{A_1(A_2}(\mathscr{C}^\dagger\varphi){}_{A_3\dots A_{k+1})}{}^{A_1'\dots A_{l+1}'}\nonumber\\
&+\tfrac{kl}{(k+1)(l+1)}\epsilon_{A_1(A_2}\bar\epsilon^{A_1'(A_2'}(\mathscr{D}\varphi){}_{A_3\dots A_{k+1})}{}^{A_3'\dots A_{l+1}')}.\label{eq:IrrDecGeneralDer}
\end{align}
Next, we write the commutators in the new notation.
Define the operator
\begin{align}
\square &= -(\nabla \overset{0,1}{\odot} \nabla) \in \mathcal{O}_{2,0},
\end{align}
and its complex conjugate $\overline{\square} \in \mathcal{O}_{0,2}$.
In index notation, it reads
$\square_{AB} = \nabla_{(A|A'|}\nabla_{B)}{}^{A'}$. Acting on $\varphi \in \mathcal{S}_{k,l}$ it can be expressed in terms of curvature via
\begin{subequations}
\begin{align}
\square \overset{0,0}{\odot} \varphi ={}& - k \Psi \overset{1,0}{\odot} \varphi - l \Phi \overset{0,1}{\odot} \varphi,\\
\square \overset{1,0}{\odot} \varphi ={}& -(k-1) \Psi \overset{2,0}{\odot} \varphi - l \Phi \overset{1,1}{\odot} \varphi + (k+2) \Lambda \overset{0,0}{\odot} \varphi, \\
\square \overset{2,0}{\odot} \varphi ={}& - (k-2) \Psi \overset{3,0}{\odot} \varphi - l \Phi \overset{2,1}{\odot} \varphi.
\end{align}
\end{subequations}
\begin{lemma}{\cite[Lemma 18]{AndBaeBlu14a}}\label{lemma:commutators}
Let $\varphi \in \mathcal{S}_{k,l}$. The operators $\mathscr{D}$, $\mathscr{C}$, $\mathscr{C}^\dagger$ and $\mathscr{T}$ satisfy the commutator relations
\begin{subequations}
\begin{align}
\mathscr{D} \mathscr{C} \varphi
={}&\tfrac{k}{k+1} \mathscr{C} \mathscr{D} \varphi - \overline{\square} \overset{0,2}{\odot} \varphi,
& k\geq 0, l\geq 2, \label{eq:DivCurl}\\
\mathscr{D} \mathscr{C}^\dagger \varphi
={}&\tfrac{l}{l+1} \mathscr{C}^\dagger \mathscr{D} \varphi
- \square \overset{2,0}{\odot} \varphi,
& k\geq 2, l\geq 0,\label{eq:DivCurlDagger}\\
\mathscr{C} \mathscr{T} \varphi
={}&\tfrac{l}{l+1} \mathscr{T} \mathscr{C} \varphi
- \square \overset{0,0}{\odot} \varphi,
& k\geq 0, l\geq 0,\label{eq:CurlTwist}\\
\mathscr{C}^\dagger \mathscr{T} \varphi
={}&
\tfrac{k}{k+1} \mathscr{T} \mathscr{C}^\dagger \varphi
- \overline\square \overset{0,0}{\odot} \varphi,
& k\geq 0, l\geq 0,\label{eq:CurlDaggerTwist}\\
\mathscr{D} \mathscr{T} \varphi
={}&
-(\tfrac{1}{k+1}+\tfrac{1}{l+1}) \mathscr{C} \mathscr{C}^\dagger \varphi
+\tfrac{l(l+2)}{(l+1)^2} \mathscr{T} \mathscr{D} \varphi
-\tfrac{l+2}{l+1}\square \overset{1,0}{\odot} \varphi
-\tfrac{l}{l+1}\overline\square \overset{0,1}{\odot} \varphi,
& k\geq 1, l\geq 0,\label{eq:DivTwistCurlCurlDagger}\\
\mathscr{D} \mathscr{T} \varphi
={}&
-(\tfrac{1}{k+1}+\tfrac{1}{l+1})\mathscr{C}^\dagger \mathscr{C} \varphi
+\tfrac{k(k+2)}{(k+1)^2}\mathscr{T} \mathscr{D} \varphi
-\tfrac{k}{k+1}\square \overset{1,0}{\odot} \varphi
-\tfrac{k+2}{k+1}\overline\square \overset{0,1}{\odot} \varphi,
& k\geq 0, l\geq 1,\label{eq:DivTwistCurlDaggerCurl}\\
\mathscr{C} \mathscr{C}^\dagger \varphi
={}&
\mathscr{C}^\dagger \mathscr{C} \varphi
+(\tfrac{1}{k+1}-\tfrac{1}{l+1})\mathscr{T} \mathscr{D} \varphi
-\square \overset{1,0}{\odot} \varphi
+\overline\square \overset{0,1}{\odot} \varphi,
& k\geq 1, l\geq 1.\label{eq:CurlCurlDagger}
\end{align}
\end{subequations}
\end{lemma}
\begin{lemma}
For symmetric spinors $\phi \in \mathcal{S}_{i,j}, \varphi \in \mathcal{S}_{k,l}$ we have the following Leibniz rules.
\begin{subequations}
\label{eq:SymMultLeibniz}
\begin{align}
\mathscr{T} (\phi {\overset{m,n}{\odot}}\varphi)={}&(-1)^{m + n}\varphi {\overset{m,n}{\odot}}\mathscr{T} \phi
+ \tfrac{(-1)^{m + n} n}{j + 1}\varphi {\overset{m,n - 1}{\odot}}\mathscr{C} \phi
+ \tfrac{(-1)^{m + n} m}{i + 1}\varphi {\overset{m - 1,n}{\odot}}\mathscr{C}^\dagger \phi\nonumber\\
& + \tfrac{(-1)^{m + n} m n}{(i + 1) (j + 1)}\varphi {\overset{m - 1,n - 1}{\odot}}\mathscr{D} \phi
+ \phi {\overset{m,n}{\odot}}\mathscr{T} \varphi
+ \tfrac{n}{l + 1}\phi {\overset{m,n - 1}{\odot}}\mathscr{C} \varphi\nonumber\\
& + \tfrac{m}{k + 1}\phi {\overset{m - 1,n}{\odot}}\mathscr{C}^\dagger \varphi
+ \tfrac{m n}{k l + k + l + 1}\phi {\overset{m - 1,n - 1}{\odot}}\mathscr{D} \varphi
\label{TwistLeibniz},\\
\mathscr{C} (\phi {\overset{m,n}{\odot}}\varphi)={}&\tfrac{(-1)^{m + n + 1} (l - n)}{j + l - 2 n}\varphi {\overset{m,n + 1}{\odot}}\mathscr{T} \phi
+ \tfrac{(-1)^{m + n} (j - n) (j + l - n + 1)}{(j + 1) (j + l - 2 n)}\varphi {\overset{m,n}{\odot}}\mathscr{C} \phi\nonumber\\
& + \tfrac{(-1)^{m + n + 1} m (l - n)}{(i + 1) (j + l - 2 n)}\varphi {\overset{m - 1,n + 1}{\odot}}\mathscr{C}^\dagger \phi\nonumber\\
& + \tfrac{(-1)^{m + n} m (j - n) (j + l - n + 1)}{(i + 1) (j + 1) (j + l - 2 n)}\varphi {\overset{m - 1,n}{\odot}}\mathscr{D} \phi
- \tfrac{j - n}{j + l - 2 n}\phi {\overset{m,n + 1}{\odot}}\mathscr{T} \varphi\nonumber\\
& + \tfrac{(l - n) (j + l - n + 1)}{(l + 1) (j + l - 2 n)}\phi {\overset{m,n}{\odot}}\mathscr{C} \varphi
+ \tfrac{m (- j + n)}{(k + 1) (j + l - 2 n)}\phi {\overset{m - 1,n + 1}{\odot}}\mathscr{C}^\dagger \varphi\nonumber\\
& + \tfrac{m (l - n) (j + l - n + 1)}{(k + 1) (l + 1) (j + l - 2 n)}\phi {\overset{m - 1,n}{\odot}}\mathscr{D} \varphi
\label{CurlLeibniz},\\
\mathscr{C}^\dagger (\phi {\overset{m,n}{\odot}}\varphi)={}&\tfrac{(-1)^{m + n + 1} (k - m)}{i + k - 2 m}\varphi {\overset{m + 1,n}{\odot}}\mathscr{T} \phi
+ \tfrac{(-1)^{m + n + 1} n (k - m)}{(j + 1) (i + k - 2 m)}\varphi {\overset{m + 1,n - 1}{\odot}}\mathscr{C} \phi\nonumber\\
& + \tfrac{(-1)^{m + n} (i - m) (i + k - m + 1)}{(i + 1) (i + k - 2 m)}\varphi {\overset{m,n}{\odot}}\mathscr{C}^\dagger \phi\nonumber\\
& + \tfrac{(-1)^{m + n} n (i - m) (i + k - m + 1)}{(i + 1) (j + 1) (i + k - 2 m)}\varphi {\overset{m,n - 1}{\odot}}\mathscr{D} \phi
- \tfrac{i - m}{i + k - 2 m}\phi {\overset{m + 1,n}{\odot}}\mathscr{T} \varphi\nonumber\\
& + \tfrac{n (- i + m)}{(l + 1) (i + k - 2 m)}\phi {\overset{m + 1,n - 1}{\odot}}\mathscr{C} \varphi
+ \tfrac{(k - m) (i + k - m + 1)}{(k + 1) (i + k - 2 m)}\phi {\overset{m,n}{\odot}}\mathscr{C}^\dagger \varphi\nonumber\\
& + \tfrac{n (k - m) (i + k - m + 1)}{(k + 1) (l + 1) (i + k - 2 m)}\phi {\overset{m,n - 1}{\odot}}\mathscr{D} \varphi
\label{CurlDgLeibniz},\\
\mathscr{D} (\phi {\overset{m,n}{\odot}}\varphi)={}&\tfrac{(-1)^{m + n} (k - m) (l - n)}{(i + k - 2 m) (j + l - 2 n)}\varphi {\overset{m + 1,n + 1}{\odot}}\mathscr{T} \phi\nonumber\\
& + \tfrac{(-1)^{m + n + 1} (j - n) (k - m) (j + l - n + 1)}{(j + 1) (i + k - 2 m) (j + l - 2 n)}\varphi {\overset{m + 1,n}{\odot}}\mathscr{C} \phi\nonumber\\
& + \tfrac{(-1)^{m + n + 1} (i - m) (l - n) (i + k - m + 1)}{(i + 1) (i + k - 2 m) (j + l - 2 n)}\varphi {\overset{m,n + 1}{\odot}}\mathscr{C}^\dagger \phi\nonumber\\
& + \tfrac{(-1)^{m + n} (i - m) (j - n) (i + k - m + 1) (j + l - n + 1)}{(i + 1) (j + 1) (i + k - 2 m) (j + l - 2 n)}\varphi {\overset{m,n}{\odot}}\mathscr{D} \phi\nonumber\\
& + \tfrac{(i - m) (j - n)}{(i + k - 2 m) (j + l - 2 n)}\phi {\overset{m + 1,n + 1}{\odot}}\mathscr{T} \varphi\nonumber\\
& + \tfrac{(- i + m) (l - n) (j + l - n + 1)}{(l + 1) (i + k - 2 m) (j + l - 2 n)}\phi {\overset{m + 1,n}{\odot}}\mathscr{C} \varphi\nonumber\\
& + \tfrac{(j - n) (- k + m) (i + k - m + 1)}{(k + 1) (i + k - 2 m) (j + l - 2 n)}\phi {\overset{m,n + 1}{\odot}}\mathscr{C}^\dagger \varphi\nonumber\\
& + \tfrac{(k - m) (l - n) (i + k - m + 1) (j + l - n + 1)}{(k + 1) (l + 1) (i + k - 2 m) (j + l - 2 n)}\phi {\overset{m,n}{\odot}}\mathscr{D} \varphi
\label{DivLeibniz}.
\end{align}
\end{subequations}
\end{lemma}
\begin{proof}
Collectively, the left hand sides can be written as $\nabla\overset{t,u}{\odot}(\phi\overset{m,n}{\odot}\varphi)$ where $t, u\in \{0,1\}$.
Let $\nabla_\phi$ and $\nabla_\varphi$ be $\nabla$ only differentiating $\phi$ respectively $\varphi$. From the relations \eqref{eq:Commutativity} and \eqref{eq:SymMultCommutator1} we get
\begin{align}
\nabla\overset{t,u}{\odot}(\phi\overset{m,n}{\odot}\varphi)
={}&\nabla_\phi\overset{t,u}{\odot}(\phi\overset{m,n}{\odot}\varphi)+\nabla_\varphi\overset{t,u}{\odot}(\phi\overset{m,n}{\odot}\varphi)\nonumber\\
={}&(-1)^{m+n}\nabla_\phi\overset{t,u}{\odot}(\varphi\overset{m,n}{\odot}\phi)+\nabla_\varphi\overset{t,u}{\odot}(\phi\overset{m,n}{\odot}\varphi)\nonumber\\
={}&(-1)^{m+n}\sum_{M=0}^{1}\sum_{N=0}^{1} F_{1,k,i}^{t,m,M}F_{1, l, j}^{u, n, N}
\varphi\overset{t+m-M,u+n-N}{\odot}\nabla\overset{M,N}{\odot}\phi\nonumber\\
&+\sum_{M=0}^{1}\sum_{N=0}^{1} F_{1,i,k}^{t,m,M}F_{1, j, l}^{u, n, N}
\phi\overset{t+m-M,u+n-N}{\odot}\nabla\overset{M,N}{\odot}\varphi.
\end{align}
Explicit calculations of the $F_{1,i,k}^{t,m,M}$ coefficients gives the relations \eqref{eq:SymMultLeibniz}.
\end{proof}
\subsection{GHP expansion}\label{sec:GHP}
In this section we collect equations to efficiently expand symmetric spinorial equations into GHP components. Let us first briefly review the formalism, see \cite{GHP} for details. Introducing a normalized spinor dyad $(o_A, \iota_A), o_A \iota^A=1$, a two dimensional subgroup of the Lorentz group is given by
\begin{align} \label{GHPDyadTrafo}
o_A \to \lambda o_A, \qquad \iota_A \to \lambda^{-1} \iota_A,
\end{align}
with non-vanishing, complex scalar field $\lambda$. A field $\phi$ is said to be of GHP weight $\{p,q\}$ if it transforms via
\begin{align}
\phi \to \lambda^p \bar \lambda^q \phi
\end{align}
under \eqref{GHPDyadTrafo} and its complex conjugate. The Levi-Civita connection has a natural lift of the form
\begin{align} \label{ThetaDef}
\Theta_{AA'} = \nabla_{AA'} - p \omega_{AA'} - q \bar{\omega}_{AA'}, \qquad \text{with } \omega_{AA'} = \iota^B \nabla_{AA'} o_B,
\end{align}
and is of weight zero in the sense that it maps $\{p,q\}$ weighted fields to $\{p,q\}$ fields. The GHP operators are given by the dyad expansion of \eqref{ThetaDef},
\begin{align} \label{ThetaDyad}
\Theta_{AA'} ={}& \iota_A \bar \iota_{A'} \tho - \iota_A \bar o_{A'} \edt - o_A \bar \iota_{A'} \edt' + o_A \bar o _{A'} \tho'.
\end{align}
The connection coefficients are defined as follows,
\begin{subequations} \label{GammaDef}
\begin{align}
\Theta_{AA'} o_B ={}& \Gamma_{AA'} \iota_B, \qquad \text{where }
\Gamma_{AA'} = -\iota_A \bar \iota_{A'} \kappa + \iota_A \bar o_{A'} \sigma + o_A \bar \iota_{A'} \rho - o_A \bar o _{A'} \tau, \\
\Theta_{AA'} \iota_B ={}& \Gamma'_{AA'} o_B, \qquad \text{where }
\Gamma'_{AA'} = -\iota_A \bar \iota_{A'} \tau' + \iota_A \bar o_{A'} \rho' + o_A \bar \iota_{A'} \sigma' - o_A \bar o _{A'} \kappa'.
\end{align}
\end{subequations}
To express the dyad expansion of a general symmetric spinor, it is convenient to define a symmetric spinor basis $\SymDyadBasis{n}{k}{m}{l}$ of weight $\{2n-k,2m-l\}$ by
\begin{align}
\SymDyadBasis{n}{k}{m}{l}{}_{A_1\dots A_k}^{A'_1\dots A'_l} &= o_{(A_1} \dots o_{A_n} \iota_{A_{n+1}} \dots \iota_{A_k)} \bar o^{(A'_1} \dots \bar o^{A'_m} \bar \iota^{A'_{m+1}} \dots \bar \iota^{A'_l)}.
\end{align}
In particular this allows us to mostly avoid spinor indices for the rest of this section. For a full contraction of two basis elements we find
\begin{align} \label{BasisFullContraction}
\SymDyadBasis{n}{k}{m}{l} \overset{k,l}{\odot} \SymDyadBasis{i}{k}{j}{l} =
(-1)^{(n+m)} \binom{k}{n}^{-1} \binom{l}{m}^{-1} \delta^{n}_{k-i} \delta^m_{l-j},
\end{align}
where $\delta^a_b=1$ if $a=b$ and zero otherwise. Now any $\phi \in \mathcal{S}_{k,l}$ can be expanded into
\begin{align} \label{eq:phiklGHPExpansion}
\phi
={}& \sum_{i=0}^k \sum_{j=0}^l (-1)^{k-i+l-j} \binom{k}{i}\binom{l}{j} \phi_{ij'} \SymDyadBasis{i}{k}{j}{l},
\end{align}
where the scalar components of weight $\{k-2i,l-2j\}$ are defined by
\begin{align} \label{eq:phiklGHPComponents}
\phi_{ij'} ={}& \SymDyadBasis{k-i}{k}{l-j}{l} \overset{k,l}{\odot}\phi.
\end{align}
The following two lemmas yield component expressions for general symmetric products and derivatives of symmetric spinors. This allows to expand general symmetric spinor differential equations into dyad components, without expanding the symmetrizations.
\begin{lemma}
For $\phi \in \mathcal{S}_{i,j}, \varphi \in \mathcal{S}_{k,l}$ the symmetric product has components
\begin{align} \label{eq:SymSpinGHP}
(\phi\overset{m,n}{\odot}\varphi)_{st'}
={}&\sum_{p=0}^{k}\sum_{q=0}^{l}G_{i,k}^{m,p,s}G_{j,l}^{n,q,t}\phi_{(s+m-p)(t+n-q)'} \varphi_{pq'},
\end{align}
with coefficients given by
\begin{align}
G_{i,k}^{m,p,s}={}&\sum_{r=0}^m (-1)^{r}\frac{\binom{i}{s+m-p} \binom{i +p-s-m}{r} \binom{s+m-p}{m-r}\binom{k}{p} \binom{k -p}{m - r} \binom{p}{r}}{\binom{i}{m}\binom{k}{m}\binom{m}{r}\binom{i+k-2m}{s}}.
\end{align}
\end{lemma}
\begin{proof}
For ease of notation we assume $\phi \in \mathcal{S}_{i,0}, \varphi \in \mathcal{S}_{k,0}$.
Using the observation that $\SymDyadBasis{p}{k}{0}{0}=\SymDyadBasis{0}{k-p}{0}{0}\overset{0,0}{\odot}
\SymDyadBasis{p}{p}{0}{0}$, where $\SymDyadBasis{0}{k-p}{0}{0}$ is a symmetric product of $\iota_A$ and $\SymDyadBasis{p}{p}{0}{0}$ is a symmetric product of $o_A$, we can use \eqref{eq:PartialSymMultExpansion} to obtain
\begin{align}
\SymDyadBasis{p}{k}{0}{0}{}_{A_1\dots A_{k-m}}^{B_1\dots B_m}={}&(\SymDyadBasis{0}{k-p}{0}{0}\overset{0,0}{\odot}
\SymDyadBasis{p}{p}{0}{0}){}_{A_1\dots A_{k-m}}^{B_1\dots B_m}\nonumber\\
={}&\sum_{q=0}^{m}\frac{\binom{p}{q} \binom{k - p}{m-q}}{\binom{k}{m}}
\SymDyadBasis{0}{k-p}{0}{0}{}^{(B_{q+1}\dots B_{m}}_{(A_1\dots A_{k-p-m+q}}\SymDyadBasis{p}{p}{0}{0}{}_{A_{k-p-m+q+1}\dots A_{k-m})}^{B_1\dots B_q)}\nonumber\\
={}&\sum_{q=0}^{m}\frac{\binom{p}{q} \binom{k - p}{m-q}}{\binom{k}{m}}
\SymDyadBasis{p-q}{k-m}{0}{0}{}_{A_1\dots A_{k-m}}\SymDyadBasis{q}{m}{0}{0}{}^{B_1\dots B_{m}}
\end{align}
Using this in the expansion \eqref{eq:phiklGHPExpansion}, we find
\begin{subequations}
\begin{align}
\varphi_{A_1 \dots A_{k-m}B_1\dots B_m}
={}& \sum_{p=0}^k \sum_{q=0}^m (-1)^{k-p}\frac{\binom{k}{p} \binom{k - p}{m - q} \binom{p}{q}}{\binom{k}{m}} \varphi_{p0'}
\SymDyadBasis{p-q}{k-m}{0}{0}{}_{A_1\dots A_{k-m}}
\SymDyadBasis{q}{m}{0}{0}{}_{B_1\dots B_m},\\
\phi_{A_1 \dots A_{i-m}}^{B_1\dots B_m}
={}& \sum_{r=0}^i \sum_{q=0}^m (-1)^{i-r}\frac{\binom{i}{r} \binom{i - r}{q} \binom{r}{m-q}}{\binom{i}{m}} \phi_{r0'}
\SymDyadBasis{r-m+q}{i-m}{0}{0}{}_{A_1\dots A_{i-m}}
\SymDyadBasis{q}{m}{0}{0}{}^{B_1\dots B_{m}}.
\end{align}
\end{subequations}
Contracting the $B$ indices, symmetrizing and using \eqref{BasisFullContraction} yield
\begin{align}
&\hspace{-3ex}\phi_{(A_1 \dots A_{i-m}}^{B_1\dots B_m} \varphi_{A_{i-m+1} \dots A_{i+k-2m})B_1\dots B_m} \nonumber\\
={}& \sum_{r=0}^i\sum_{p=0}^k \sum_{q=0}^m (-1)^{k+i-r-p}\frac{\binom{i}{r} \binom{i - r}{q} \binom{r}{m-q}\binom{k}{p} \binom{k - p}{m - q} \binom{p}{q}}{\binom{i}{m}\binom{k}{m}}\phi_{r0'} \varphi_{p0'}
\SymDyadBasis{p+r-m}{i+k-2m}{0}{0}{}_{A_1\dots A_{i+k-2m}}
\nonumber\\
&\times\SymDyadBasis{q}{m}{0}{0}{}_{B_1\dots B_{m}}\SymDyadBasis{q}{m}{0}{0}{}^{B_1\dots B_{m}}\nonumber\\
={}& \sum_{r=0}^i\sum_{p=0}^k \sum_{q=0}^m (-1)^{k+i-r-p+m-q}\frac{\binom{i}{r} \binom{i - r}{q} \binom{r}{m-q}\binom{k}{p} \binom{k - p}{m - q} \binom{p}{q}}{\binom{i}{m}\binom{k}{m}\binom{m}{q}}\phi_{r0'} \varphi_{p0'}
\SymDyadBasis{p+r-m}{i+k-2m}{0}{0}{}_{A_1\dots A_{i+k-2m}}.
\end{align}
The relation \eqref{BasisFullContraction} then gives
\begin{align}
(\phi\overset{m,0}{\odot}\varphi)_{s0'} ={}& \SymDyadBasis{i+k-2m-s}{i+k-2m}{0}{0} \overset{i+k-2m,0}{\odot}(\phi\overset{m,0}{\odot}\varphi)\nonumber\\
={}&\sum_{r=0}^i\sum_{p=0}^k \sum_{q=0}^m (-1)^{-r-p+m-q-s}\frac{\binom{i}{r} \binom{i - r}{q} \binom{r}{m-q}\binom{k}{p} \binom{k - p}{m - q} \binom{p}{q}}{\binom{i}{m}\binom{k}{m}\binom{m}{q}\binom{i+k-2m}{i+k-2m-s}}\phi_{r0'} \varphi_{p0'}
\delta^{m+s-p}_{r}\nonumber\\
={}&\sum_{p=0}^k \sum_{q=0}^m (-1)^{q}\frac{\binom{i}{m+s-p} \binom{i + p - m - s}{q} \binom{m+s-p}{m-q}\binom{k}{p} \binom{k - p}{m - q} \binom{p}{q}}{\binom{i}{m}\binom{k}{m}\binom{m}{q}\binom{i+k-2m}{i+k-2m-s}}\phi_{(m+s-p)0'} \varphi_{p0'} \nonumber\\
={}&\sum_{p=0}^kG_{i,k}^{m,p,s}\phi_{(s+m-p)0'} \varphi_{p0'}.
\end{align}
The primed indices gives an analogous expansion and the combination yields \eqref{eq:SymSpinGHP}.
\end{proof}
\begin{lemma}
The GHP components of fundamental spinor operators \eqref{eq:FundSpinOps} on $\phi \in \mathcal{S}_{k,l}$ take the form
\begin{subequations}
\begin{align} \label{DivGHP}
(\mathscr{D}\phi)_{ij'} ={}&
(\tho - (k-i) \rho - (l-j) \bar\rho)\phi_{(i+1)(j+1)'}
+ (\tho' - (i+1) \rho' - (j+1) \bar\rho')\phi_{ij'} \nonumber \\
&- (\edt - (k-i)\tau - (j+1) \bar{\tau}')\phi_{(i+1)j'}
- (\edt' - (i+1) \tau' - (l-j)\bar{\tau})\phi_{i(j+1)'} \nonumber \\
& + (k-i-1) \kappa \phi_{(i+2)(j+1)'}
- (k-i-1) \sigma \phi_{(i+2)j'}
- i \sigma' \phi_{(i-1)(j+1)'} \nonumber \\
&+ i \kappa' \phi_{(i-1)j'}
+ (l-j-1) \bar\kappa \phi_{(i+1)(j+2)'}
- (l-j-1) \bar\sigma \phi_{i(j+2)'} \nonumber \\
&- j \bar\sigma' \phi_{(i+1)(j-1)'}
+ j \bar\kappa' \phi_{i(j-1)'}, \\
(\mathscr{C}\phi)_{ij'} ={}& \bigl(
-(k-i+1)(\tho + i \rho - (l-j) \bar\rho)\phi_{i(j+1)'}
+i(\tho' +(k-i+1)\rho' -(j+1) \bar\rho')\phi_{(i-1)j'} \nonumber \\
&\hspace{1ex}{} + (k-i+1)(\edt + i \tau - (j+1) \bar\tau')\phi_{ij'}
-i(\edt' + (k-i+1)\tau'-(l-j) \bar\tau)\phi_{(i-1)(j+1)'} \nonumber \\
&\hspace{1ex}{} - (k-i+1)(k-i) \kappa \phi_{(i+1)(j+1)'}
+(k-i+1)(k-i) \sigma \phi_{(i+1)j'}
- i(i-1) \sigma' \phi_{(i-2)(j+1)'}\nonumber \\
&\hspace{1ex}{}+ i(i-1) \kappa' \phi_{(i-2)j'}
-(k-i+1)(l-j-1) \bar\kappa \phi_{i(j+2)'}
-i(l-j-1) \bar\sigma \phi_{(i-1)(j+2)'}\nonumber \\
&\hspace{1ex}{}+ (k-i+1)j \bar\sigma' \phi_{i(j-1)'}
+ij \bar\kappa' \phi_{(i-1)(j-1)'} \bigr) / (k+1), \\
(\mathscr{C}^\dagger\phi)_{ij'} ={}& \bigl(
-(l-j+1)(\tho + j \bar\rho - (k-i) \rho)\phi_{(i+1)j'}
+j(\tho' +(l-j+1)\bar\rho' -(i+1) \rho')\phi_{i(j-1)'} \nonumber \\
&\hspace{1ex}{} + (l-j+1)(\edt' + j \bar\tau - (i+1) \tau')\phi_{ij'}
-j(\edt + (l-j+1)\bar\tau'-(k-i) \tau)\phi_{(i+1)(j-1)'} \nonumber \\
&\hspace{1ex}{}- (l-j+1)(l-j) \bar\kappa \phi_{(i+1)(j+1)'}
+(l-j+1)(l-j) \bar\sigma \phi_{i(j+1)'}
- j(j-1) \bar\sigma' \phi_{(i+1)(j-2)'}\nonumber \\
&\hspace{1ex}{}+ j(j-1) \bar\kappa' \phi_{i(j-2)'}
-(l-j+1)(k-i-1) \kappa \phi_{(i+2)j'}
-j(k-i-1) \sigma \phi_{(i+2)(j-1)'}\nonumber \\
&\hspace{1ex}{}+ (l-j+1)i \sigma' \phi_{(i-1)j'}
+ij \kappa' \phi_{(i-1)(j-1)'} \bigr) / (l+1), \\
(\mathscr{T}\phi)_{ij'} ={}& \bigl(
(k+1-i)(l+1-j)(\tho + i \rho + j\bar\rho)\phi_{ij'} \nonumber \\
&\hspace{1ex}{}+(k+1-i)j(\edt + i\tau + (l-j+1)\bar\tau')\phi_{i(j-1)'} \nonumber \\
&\hspace{1ex}{}+i(l+1-j)(\edt' + (k-i+1)\tau' + j\bar\tau)\phi_{(i-1)j'} \nonumber \\
&\hspace{1ex}{}+ij(\tho' +(k-i+1)\rho' +(l-j+1)\bar\rho')\phi_{(i-1)(j-1)'} \nonumber \\
&\hspace{1ex}{}+(k-i+1)(k-i)(l+1-j) \kappa \phi_{(i+1)j'}
+ (k-i+1)(k-i)j \sigma \phi_{(i+1)(j-1)'} \nonumber\\
&\hspace{1ex}{}+i(i-1)(l+1-j) \sigma' \phi_{(i-2)j'}
+i(i-1)j \kappa' \phi_{(i-2)(j-1)'} \nonumber\\
&\hspace{1ex}{}+(k+1-i)(l+1-j)(l-j) \bar\kappa \phi_{i(j+1)'}
+ i(l+1-j)(l-j) \bar\sigma \phi_{(i-1)(j+1)'}\nonumber\\
&\hspace{1ex}{}+(k+1-i)j(j-1) \bar\sigma' \phi_{i(j-2)'}
+ij(j-1) \bar\kappa' \phi_{(i-1)(j-2)'} \big) / \big((k+1)(l+1)\bigr)
\end{align}
\end{subequations}
\end{lemma}
\begin{proof}
To prove \eqref{DivGHP}, we start by expanding the argument of $\mathscr{D}\phi$ using \eqref{eq:phiklGHPExpansion} and contract with a symmetric basis as in \eqref{eq:phiklGHPComponents},
\begin{align} \label{DivphiStep1}
(\mathscr{D}\phi)_{ij'}
={}& \SymDyadBasis{k-1-i}{k-1}{l-1-j}{l-1} \overset{k,l}{\odot} (\mathscr{D}\phi) \nonumber \\
={}& \sum_{n=0}^k \sum_{m=0}^l (-1)^{k-n+l-m} \binom{k}{n}\binom{l}{m} \SymDyadBasis{k-1-i}{k-1}{l-1-j}{l-1} \overset{k,l}{\odot} (\mathscr{D}(\phi_{nm'} \SymDyadBasis{n}{k}{m}{l}) ).
\end{align}
Next, we use the Leibniz rule \eqref{DivLeibniz}, but switch to the GHP connection $\Theta_{AA'}$ (so the fundamental spinor operators are with respect to $\Theta_{AA'}$ instead of $\nabla_{AA'}$) as the GHP components and the basis elements are GHP weighted,
\begin{align}\label{DivphiBasis}
\mathscr{D}( \phi_{nm'}\underset{k,l}{\overset{0,0}{\odot}}\SymDyadBasis{n}{k}{m}{l})={}&\SymDyadBasis{n}{k}{m}{l} \overset{1,1}{\odot}\mathscr{T} \phi_{nm}
+ \phi_{nm}\overset{0,0}{\odot}\mathscr{D} \SymDyadBasis{n}{k}{m}{l}.
\end{align}
From \eqref{ThetaDyad} and \eqref{GammaDef} we have
\begin{align}\label{Twistphinm}
\mathscr{T}\phi_{nm'} ={}& (\tho \phi_{nm'}) \SymDyadBasis{0}{1}{0}{1} - (\edt \phi_{nm'}) \SymDyadBasis{0}{1}{1}{1} - (\edt' \phi_{nm'}) \SymDyadBasis{1}{1}{0}{1} + (\tho' \phi_{nm'}) \SymDyadBasis{1}{1}{1}{1},
\end{align}
and
\begin{align} \label{DivBasis}
\mathscr{D} \SymDyadBasis{n}{k}{m}{l}
={}& n \Gamma \overset{1,1}{\odot} \SymDyadBasis{n-1}{k}{m}{l}
+ (k-n) \Gamma' \overset{1,1}{\odot} \SymDyadBasis{n+1}{k}{m}{l} +m \overline{\Gamma} \overset{1,1}{\odot} \SymDyadBasis{n}{k}{m-1}{l}
+(l-m) \overline{\Gamma'} \overset{1,1}{\odot} \SymDyadBasis{n}{k}{m+1}{l}.
\end{align}
Inserting \eqref{Twistphinm}, \eqref{DivBasis} back into \eqref{DivphiBasis} and expanding $\Gamma, \Gamma'$ into the basis we can use the contraction rules
\begin{subequations} \label{Basis11vec}
\begin{align}
\SymDyadBasis{n}{k}{m}{l} \overset{1,1}{\odot}\SymDyadBasis{0}{1}{0}{1} ={}&\frac{m n}{k l}\SymDyadBasis{n-1}{k-1}{m-1}{l-1}, \\
\SymDyadBasis{n}{k}{m}{l} \overset{1,1}{\odot}\SymDyadBasis{1}{1}{0}{1}={}&- \frac{m (k - n)}{k l} \SymDyadBasis{n}{k-1}{m-1}{l-1}, \\
\SymDyadBasis{n}{k}{m}{l} \overset{1,1}{\odot}\SymDyadBasis{0}{1}{1}{1}={}&- \frac{n (l - m)}{k l} \SymDyadBasis{n-1}{k-1}{m}{l-1}, \\
\SymDyadBasis{n}{k}{m}{l} \overset{1,1}{\odot}\SymDyadBasis{1}{1}{1}{1}={}&\frac{(k - n) (l - m)}{k l}\SymDyadBasis{n}{k-1}{m}{l-1},
\end{align}
\end{subequations}
which are easily verified by expanding out the symmetries. The result can now be substituted into \eqref{DivphiStep1}. Each term has a full contraction of the form \eqref{BasisFullContraction} which cancels the double sum due to the $\delta$ factors. After some elementary algebra, the end result is given by \eqref{DivGHP}. The other expansions can be verified along the same lines, the only minor computation that needs to be done is the analog of \eqref{DivBasis} and \eqref{Basis11vec}.
\end{proof}
\section{SymSpin: A computer algebra implementation in xAct}\label{sec:SymSpinPackage}
The \emph{xAct} \cite{xActWeb} suite for \emph{Mathematica} is an open source project mainly devoted to symbolic computation in differential geometry and tensor algebra. In this section we introduce our contributed package \emph{SymSpin} \cite{SymSpinWeb} which contains the formalism of Section~\ref{sec:SymSpinAlg}. For syntax and more examples, see \emph{SymSpinDoc.nb} on that page.
\subsection{Loading the package and defining structures}
Load the package, define a four dimensional manifold M4, and Lorentzian metric with
\begin{mma}
\mmain{<$\mspace{0.5mu}$<xAct\textasciigrave SymSpin\textasciigrave\linebreak
\$DefInfoQ=False;\linebreak
DefManifold[M4,4,\{a,b,c,d\}]\linebreak
DefMetric[\{1,3,0\},g[-a,-b],CD]}
\end{mma}%
By default the valence numbers are displayed for each operator and complex conjugates are written with $\dagger$. To keep the notation the same as in the rest of the paper, we can change the display form with
\begin{mma}
\mmain{SetOptions[DefAbstractIndex,PrintAs->PrimeDagger];\linebreak
SetOptions[DefSpinor,~PrintDaggerAs->AddBar];\linebreak
SetOptions[DefFundSpinOperators,ShowValenceInfo->False];}
\end{mma}%
Define the spin structure, initialize \emph{SymSpin} and define the fundamental spinor operators with
\begin{mma}
\mmain{DefSpinStructure[g,Spin,\{A,B,C,F,G,H,P,Q,R\},$\epsilon $,$\sigma $,CDe,\{";","$\nabla$"\},\linebreak
SpinorPrefix->SP,SpinorMark->"S"]\linebreak
InitSymSpin[$\sigma$];\linebreak
DefFundSpinOperators[CDe];}
\end{mma}%
\subsection{Example: Coefficients} \label{sec:example}
Assume that $K$, $L$ and $M$ are symmetric spinor fields, and we want to find under which conditions of $K$, $L$ and $M$ the equation
\begin{align}
0={}&K_{AB}{}^{FH} L_{F}{}^{C} \varphi_{HC}
+ M_{(A}{}^{C}\varphi_{B)C}.
\end{align}
holds for all symmetric spinor fields $\varphi$.
The following calculation leads to the conditions
\begin{align}
K^{G}{}_{(ABC}L_{|G|F)}={}&0,&
M_{AB}={}&\tfrac{1}{2} K^{CF}{}_{AB} L_{CF}.
\end{align}
We first define the symmetric spinor fields. For clarity we have added the valence numbers to the names of the spinors, but not the display form.
\begin{mma}
\mmain{DefSymmetricSpinor[$\varphi $20,2,0,Spin,"$\varphi $"]\linebreak
DefSymmetricSpinor[K40,4,0,Spin,"K"]\linebreak
DefSymmetricSpinor[L20,2,0,Spin,"L"]\linebreak
DefSymmetricSpinor[M20,2,0,Spin,"M"]}
\end{mma}%
One can start with the indexed version of the spinor equation.
\begin{mma}
\mmain{OriginalEq=0==K40[-A,-B,F,H]L20[-F,C]$\varphi $20[-H,-C]\linebreak+ImposeSym[M20[-A,C]*$\varphi $20[-B,-C]]}\\
\mmaout{0~\texttt{==}~K_{AB}{}^{FH} L_{F}{}^{C} \varphi_{HC} + \underset{\scriptscriptstyle(13)}{Sym}[M\varphi ]_{A}{}^{C}{}_{BC}}
\end{mma}%
To convert this to the new formalism, we need the irreducible decomposition of the product of the $L$ and $\phi$ spinor.
\begin{mma}
\mmain{IrrDecomposeSymMult[L20,$\varphi $20,\{0,0\}]}\\
\mmaout{L_{AB} \varphi_{CF}~\texttt{==}~- \underset{\scriptscriptstyle(13)(24)}{Sym}[\epsilon (L{\overset{1,0}{\odot}}\varphi)]_{ACBF} + (L{\overset{0,0}{\odot}}\varphi)_{ABCF} + \tfrac{1}{3} \underset{\scriptscriptstyle(13)(24)}{Sym}[\epsilon \epsilon ]_{ACBF} (L{\overset{2,0}{\odot}}\varphi)}
\end{mma}%
It is convenient to work with the expanded and canonicalized version
\begin{mma}
\mmain{L20$\varphi $20IrrDecEq=ToCanonical@ExpandSym@\%}\\
\mmaout{L_{AB} \varphi_{CF}~\texttt{==}~(L{\overset{0,0}{\odot}}\varphi)_{ABCF} - \tfrac{1}{4} \epsilon_{BF} (L{\overset{1,0}{\odot}}\varphi)_{AC} - \tfrac{1}{4} \epsilon_{BC} (L{\overset{1,0}{\odot}}\varphi)_{AF} - \tfrac{1}{4} \epsilon_{AF} (L{\overset{1,0}{\odot}}\varphi)_{BC}}\\
\mmanoout{ - \tfrac{1}{4} \epsilon_{AC} (L{\overset{1,0}{\odot}}\varphi)_{BF} + \tfrac{1}{6} \epsilon_{AF} \epsilon_{BC} (L{\overset{2,0}{\odot}}\varphi) + \tfrac{1}{6} \epsilon_{AC} \epsilon_{BF} (L{\overset{2,0}{\odot}}\varphi)}
\end{mma}%
To work efficiently we turn the original equation into an index-free version. One could also use the index-free version as a starting point.
\begin{mma}
\mmain{IndexFreeEq=ToIndexFree[ToCanonical@ContractMetric[OriginalEq\linebreak
/.EqToRule@L20$\varphi $20IrrDecEq]//.SymHToSymMultRule]\linebreak
/.MultScalToSymMultRule[Spin]/.SortSymMult[Not@FreeQ[\#,$\varphi $20]\&]}\\
\mmaout{0~\texttt{==}~M{\overset{1,0}{\odot}}\varphi + K{\overset{2,0}{\odot}}L{\overset{1,0}{\odot}}\varphi }
\end{mma}%
We can turn the spinor valued equation into a scalar equation by contracting it with a dummy spinor $T$ to turn the free indices into contracted dummy indices. This dummy spinor is defined by
\begin{mma}
\mmain{DefSymmetricSpinor[T20,2,0,Spin,"T"]}
\end{mma}%
As the field $\varphi$ and the dummy spinor $T$ both should be arbitrary, we see that the irreducible components of their product can be treated as independent arbitrary fields. For convenience we make a list of them with
\begin{mma}
\mmain{IrrDecComps=SymMult[T20,\#,0,Spin][$\varphi $20]\&/@Range[0,2]}\\
\mmaout{\texttt{\{}(T{\overset{0,0}{\odot}}\varphi)\texttt{, }(T{\overset{1,0}{\odot}}\varphi)\texttt{, }(T{\overset{2,0}{\odot}}\varphi)\texttt{\}}}
\end{mma}%
We can now contract our index-free equation with $T$.
\begin{mma}
\mmain{SymMult[T20,2,0]/@IndexFreeEq}\\
\mmaout{0~\texttt{==}~T{\overset{2,0}{\odot}}M{\overset{1,0}{\odot}}\varphi + T{\overset{2,0}{\odot}}K{\overset{2,0}{\odot}}L{\overset{1,0}{\odot}}\varphi }
\end{mma}%
Commute $T$ inside, so that $T$ is directly contracted with the field $\varphi$, so we obtain the independent spinors in the list \texttt{IrrDecComps}.
\begin{mma}
\mmain{\%//.CommuteSymMultRuleIn[T20]}\\
\mmaout{0~\texttt{==}~- M{\overset{2,0}{\odot}}T{\overset{1,0}{\odot}}\varphi + K{\overset{4,0}{\odot}}L{\overset{1,0}{\odot}}T{\overset{0,0}{\odot}}\varphi + \tfrac{1}{2}K{\overset{4,0}{\odot}}L{\overset{0,0}{\odot}}T{\overset{1,0}{\odot}}\varphi }
\end{mma}%
Now, these independent spinors are moved out and to the left.
\begin{mma}
\mmain{\%/.SortSymMultReverse[MemberQ[IrrDecComps,\#]\&]\linebreak
//.Flatten[CommuteSymMultRuleOut/@IrrDecComps]}\\
\mmaout{0~\texttt{==}~- (T{\overset{1,0}{\odot}}\varphi){\overset{2,0}{\odot}}M + (T{\overset{0,0}{\odot}}\varphi){\overset{4,0}{\odot}}K{\overset{1,0}{\odot}}L + \tfrac{1}{2}(T{\overset{1,0}{\odot}}\varphi){\overset{2,0}{\odot}}K{\overset{2,0}{\odot}}L}
\end{mma}%
From this one can conclude that the coefficients of $(T\overset{1,0}{\odot}\varphi)$ and $(T\overset{0,0}{\odot}\varphi)$ both have to be zero.
As a convenience, we have implemented all of the steps from the index-free equation to the final list of equations in one function.
\begin{mma}
\mmain{ExtractCoeffsIndexFree[IndexFreeEq,$\varphi $20]}\\
\mmaout{\texttt{\{}0~\texttt{==}~(K{\overset{1,0}{\odot}}L)\texttt{, }0~\texttt{==}~- M + \tfrac{1}{2}K{\overset{2,0}{\odot}}L\texttt{\}}}
\end{mma}%
This can be translated back to the indexed form with
\begin{mma}
\mmain{ToIndexed/@\%}\\
\mmaout{\texttt{\{}0~\texttt{==}~\underset{\scriptscriptstyle(2346)}{Sym}[KL]^{G}{}_{ABCGF}\texttt{, }0~\texttt{==}~\tfrac{1}{2} K^{CF}{}_{AB} L_{CF} - M_{AB}\texttt{\}}}
\end{mma}%
Performing this kind of calculation in the indexed form would require expansions of symmetries and several steps of irreducible decompositions of different products. This new method was heavily used in \cite{JacBac2022}.
\subsection{Example: Derivatives}
To also demonstrate how to work with derivatives we use the previously defined field $\varphi$ and define a valence $(3,2)$ field $\psi$ via
\begin{mma}
\mmain{DefSymmetricSpinor[$\psi $32,3,2,Spin,"$\psi $"]}
\end{mma}%
The covariant derivative
\begin{mma}
\mmain{CDe[-A,-A$\dagger $]@$\psi $32[-B,-C,-F,-B$\dagger $,-C$\dagger $]}\\
\mmaout{\nabla_{AA'}\psi_{BCFB'C'}}
\end{mma}%
can be decomposed into the fundamental spinor operators with
\begin{mma}
\mmain{\%==ToFundSpinOp[\%]}\\
\mmaout{\nabla_{AA'}\psi_{BCFB'C'}~\texttt{==}~- \tfrac{1}{3} \bar{\epsilon}_{A'C'} (\mathscr{C} \psi)_{ABCFB'} - \tfrac{1}{3} \bar{\epsilon}_{A'B'} (\mathscr{C} \psi)_{ABCFC'} - \tfrac{1}{4} \epsilon_{AF} (\mathscr{C}^\dagger \psi)_{BCA'B'C'}}\\
\mmanoout{ - \tfrac{1}{4} \epsilon_{AC} (\mathscr{C}^\dagger \psi)_{BFA'B'C'} - \tfrac{1}{4} \epsilon_{AB} (\mathscr{C}^\dagger \psi)_{CFA'B'C'} + \tfrac{1}{12} \epsilon_{AF} \bar{\epsilon}_{A'C'} (\mathscr{D} \psi)_{BCB'}}\\
\mmanoout{ + \tfrac{1}{12} \epsilon_{AF} \bar{\epsilon}_{A'B'} (\mathscr{D} \psi)_{BCC'} + \tfrac{1}{12} \epsilon_{AC} \bar{\epsilon}_{A'C'} (\mathscr{D} \psi)_{BFB'} + \tfrac{1}{12} \epsilon_{AC} \bar{\epsilon}_{A'B'} (\mathscr{D} \psi)_{BFC'}}\\
\mmanoout{ + \tfrac{1}{12} \epsilon_{AB} \bar{\epsilon}_{A'C'} (\mathscr{D} \psi)_{CFB'} + \tfrac{1}{12} \epsilon_{AB} \bar{\epsilon}_{A'B'} (\mathscr{D} \psi)_{CFC'} + (\mathscr{T} \psi)_{ABCFA'B'C'}}
\end{mma}%
Commutators can be handled like
\begin{mma}
\mmain{DivCDe@CurlDgCDe@$\psi $32}\\
\mmaout{(\mathscr{D} \mathscr{C}^\dagger \psi)}
\end{mma}%
\begin{mma}
\mmain{\%==(\%/.CommuteOp[DivCDe,CurlDgCDe])}\\
\mmaout{(\mathscr{D} \mathscr{C}^\dagger \psi)~\texttt{==}~\tfrac{2}{3}\mathscr{C}^\dagger \mathscr{D} \psi + \Psi {\overset{3,0}{\odot}}\psi + 2\Phi {\overset{2,1}{\odot}}\psi }
\end{mma}%
Derivatives of products can also be handled efficiently
\begin{mma}
\mmain{CurlDgCDe@SymMult[$\varphi $20,1,0]@$\psi $32}\\
\mmaout{(\mathscr{C}^\dagger \varphi {\overset{1,0}{\odot}}\psi)}
\end{mma}%
\begin{mma}
\mmain{\%$== $(\%/.SymMultLeibnizRules[CDe])}\\
\mmaout{(\mathscr{C}^\dagger \varphi {\overset{1,0}{\odot}}\psi)~\texttt{==}~\tfrac{2}{3}\psi {\overset{2,0}{\odot}}\mathscr{T} \varphi - \tfrac{5}{9}\psi {\overset{1,0}{\odot}}\mathscr{C}^\dagger \varphi - \tfrac{1}{3}\varphi {\overset{2,0}{\odot}}\mathscr{T} \psi + \tfrac{5}{6}\varphi {\overset{1,0}{\odot}}\mathscr{C}^\dagger \psi }
\end{mma}%
\section{Conclusions and discussion}\label{sec:Conclusions}
In this work, we introduced an algebra on symmetric 2-spinors and the corresponding \emph{SymSpin} package for the \emph{Mathematica} suite \emph{xAct}. In various research projects of the authors this algebra turned out to be a very efficient way to perform calculations.
For example in \cite{JacBac2022} it is used to derive conditions on the spacetime for the existence of second order symmetry operators for the massive Dirac equation. This greatly simplified the calculations compared to the earlier approach \cite{AndBaeBlu14a}, where only parts of the formalism were used to investigate symmetry operators for the massless Dirac and the Maxwell equations.
The formalism is very efficient for cases where each spinor appears only once in each product. Choosing a preferred ordering of the factors in each product, one can use the relations in Theorem \ref{thm:SymProdProperties} to rewrite them in a canonical form.
However, if a spinor appears multiple times in a product the relations in Theorem \ref{thm:SymProdProperties} can give non-trivial equations where a term of the same form can appear both in the left and right hand sides as well as in several equations. Solving these equations, it should be possible to develop a method to write such products in a canonical form. We plan to continue the development of these tools for such cases in the future.
\subsection*{Acknowledgements}
The authors are grateful to Simon Jacobsson for testing of the \emph{xAct} implementation and to Teake Nutma for \LaTeX{} typesetting code of \emph{Mathematica} expressions.
|
{
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Бурганкент () — село в Табасаранском районе Дагестана. Административный центр сельского поселения «Сельсовет Бурганкентский».
География
Селение расположено в 23 км к юго-востоку от районного центра села Хучни.
История
С 1931 по 1935 года являлось административный центр Табасаранского района.
Население
Инфраструктура
Здравоохранение
Фельдшерский пункт.
Примечания
Населённые пункты Табасаранского района
Бывшие районные центры Дагестана
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{
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{"url":"https:\/\/www.gamedev.net\/articles\/programming\/graphics\/rendering-with-shaders-implementing-lighting-effects-in-c-game-development-r4892\/","text":"\u2022 # Rendering with Shaders: Implementing Lighting Effects in C++ Game Development\n\nGraphics and GPU Programming\n\nThis is an excerpt from the book, Mastering C++ Game Development written by Mickey Macdonald and published by Packt Publishing. With this book, learn high-end game development with advanced C++ 17 programming techniques.\n\nOne of the most common uses for shaders is creating lighting and reflection effects. Lighting effects achieved from the use of shaders help provide a level of polish and detail that every modern game strives for. In this post, we will look at some of the well-known models for creating different surface appearance effects, with examples of shaders you can implement to replicate the discussed lighting effect.\n\n## Per-vertex diffuse\n\nTo start with, we will look at one of the simpler lighting vertex shaders, the diffuse reflection shader. Diffuse is considered simple since we assume that the surface we are rendering appears to scatter the light in all directions equally. With this shader, the light makes contact with the surface and slightly penetrates before being cast back out in all directions. This means that some of the light's wavelength will be at least partially absorbed. A good example of what a diffuse shader looks like is to think of matte paint. The surface has a very dull look with no shine.\n\nLet's take a quick look at the mathematical model for a diffuse reflection. This reflection model takes two vectors. One is the direction of the surface contact point to the initial light source, and the second is the normal vector of that same surface contact point. This would look something like the following:\n\nIt's worth noting that the amount of light that strikes the surface is partially dependent on the surface in relation to the light source and that the amount of light that reaches a single point will be at its maximum along the normal vector, and its lowest when perpendicular to the normal vector. Dusting off our physics knowledge toolbox, we are able to express this relationship given the amount of light making contact with a point by calculating the dot product of the point normal vector and incoming light vector. This can be expressed by the following formula:\n\nLight Density(Source Vector) Normal Vector\n\n$$LightDensity = Source Vector * Normal Vector$$$$Light Density = SourceVector \\cdot NormalVector$$\n\nThe source and normal vector in this equation are assumed to be normalized.\n\nAs mentioned before, some of the light striking the surface will be absorbed before it is re-cast. To add this behavior to our mathematical model, we can add a reflection coefficient, also referred to as the diffuse reflectivity. This coefficient value becomes the scaling factor for the incoming light. Our new formula to specify the outgoing intensity of the light will now look like the following:\n\nOutgoing Light = (Diffuse Coefficient x Light Density x Source Vector) Normal Vector\n\n$$Outgoing Light = (Diffuse Coefficient x Light Density x Source Vector) \\cdot Normal Vector$$\n\nWith this new formula, we now have a lighting model that represents an omnidirectional, uniform scattering.\n\nOK, now that we know the theory, let's take a look at how we can implement this lighting model in a GLSL shader. The full source for this example can be found in the Chapter07 folder of the GitHub repository, starting with the Vertex Shader shown as follows:\n\n#version 410\nin vec3 vertexPosition_modelspace;\nin vec2 vertexUV;\nin vec3 vertexNormal;\nout vec2 UV;\nout vec3 LightIntensity;\nuniform vec4 LightPosition;\nuniform vec3 DiffuseCoefficient ;\nuniform vec3 LightSourceIntensity;\nuniform mat4 ModelViewProjection;\nuniform mat3 NormalMatrix;\nuniform mat4 ModelViewMatrix;\nuniform mat4 ProjectionMatrix;\n\nvoid main()\n{\nvec3 tnorm = normalize(NormalMatrix * vertexNormal);\nvec4 CameraCoords = ModelViewMatrix *\nvec4(vertexPosition_modelspace,1.0);\n\nvec3 IncomingLightDirection = normalize(vec3(LightPosition -\nCameraCoords));\nLightIntensity = LightSourceIntensity * DiffuseCoefficient *\nmax( dot( IncomingLightDirection, tnorm ), 0.0 );\n\ngl_Position = ModelViewProjection *\nvec4(vertexPosition_modelspace,1);\nUV = vertexUV;\n}\n\nWe'll go through this shader block by block. To start out, we have our attributes, vertexPosition_modelspace, vertexUV, and vertexNormal. These will be set by our game application, which we will look at after we go through the shader. Then we have our out variables, UV and LightIntensity. These values will be calculated in the shader itself. We then have our uniforms. These include the needed values for our reflection calculation, as we discussed. It also includes all the necessary matrices. Like the attributes, these uniform values will be set via our game.\n\nInside of the main function of this shader, our diffuse reflection is going to be calculated in the camera relative coordinates. To accomplish this, we first normalize the vertex normal by multiplying it by the normal matrix and storing the results in a vector 3 variable named tnorm.\n\nNext, we convert the vertex position that is currently in model space to camera coordinates by transforming it with the model view matrix. We then calculate the incoming light direction, normalized, by subtracting the vertex position in the camera coordinates from the light's position.\n\nNext, we calculate the outgoing light intensity by using the formula we went through earlier. A point to note here is the use of the max function. This is a situation when the light direction is greater than 90 degrees, as in the light is coming from inside the object. Since in our case we don't need to support this situation, we just use a value of 0.0 when this arises.\n\nTo close out the shader, we store the model view projection matrix, calculated in clip space, in the built-in outbound variable gl_position. We also pass along the UV of the texture, unchanged, which we are not actually using in this example.\n\nNow that we have the shader in place, we need to provide the values needed for the calculations. We do this by setting the attributes and uniforms. We built an abstraction layer to help with this process, so let's take a look at how we set these values in our game code. Inside the GamePlayScreen.cpp file, we are setting these values in the Draw() function. I should point out this is for the example, and in a production environment, you would only want to set the changing values in a loop for performance reasons. Since this is an example, I wanted to make it slightly easier to follow:\n\nGLint DiffuseCoefficient = shaderManager.GetUniformLocation(\"DiffuseCoefficient \");\nglUniform3f(DiffuseCoefficient, 0.9f, 0.5f, 0.3f);\nglUniform3f(LightSourceIntensity, 1.0f, 1.0f, 1.0f);\nglm::vec4 lightPos = m_camera.GetView() * glm::vec4(5.0f, 5.0f, 2.0f, 1.0f);\nglUniform4f(lightPosUniform, lightPos[0], lightPos[1], lightPos[2], lightPos[3]);\nglm::mat4 modelView = m_camera.GetView() * glm::mat4(1.0f);\n\nglUniformMatrix4fv(modelViewUniform, 1, GL_FALSE, &modelView[0][0]);\nglm::mat3 normalMatrix = glm::mat3(glm::vec3(modelView[0]),\nglm::vec3(modelView[1]),\nglm::vec3(modelView[2]));\n\nglUniformMatrix3fv(normalMatrixUniform, 1, GL_FALSE, &normalMatrix[0][0]);\nglUniformMatrix4fv(MatrixID, 1, GL_FALSE, &m_camera.GetMVPMatrix()[0][0]);\n\nI won't go through each line since I am sure you can see the pattern. We first use the shader manager's GetUniformLocation() method to return the location for the uniform. Next, we set the value for this uniform using the OpenGL glUniform*() method that matches the value type. We do this for all uniform values needed. We also have to set our attributes, and as discussed in the beginning of the chapter, we do this in between the compilation and linking processes. In this example case, we are setting these values in the OnEntry() method of the GamePlayScreen() class:\n\nshaderManager.AddAttribute(\"vertexPosition_modelspace\");\nshaderManager.AddAttribute(\"vertexNormal\");\n\nThat takes care of the vertex shader and passed in values needed, so next, let's look at the fragment shader for this example:\n\n#version 410\nin vec2 UV;\nin vec3 LightIntensity;\n\n\/\/ Ouput data\nout vec3 color;\n\n\/\/ Values that stay constant for the whole mesh.\nuniform sampler2D TextureSampler;\n\nvoid main()\n{\ncolor = vec3(LightIntensity);\n}\n\nFor this example, our fragment shader is extremely simple. To begin, we have the in values for our UV and LightIntensity, and we will only use the LightIntensity this time. We then declare our out color value, specified as a vector 3. Next, we have the sampler2D uniform that we use for texturing, but again we won't be using this value in the example. Finally, we have the main function. This is where we set the final output color by simply passing the LightIntensity through to the next stage in the pipeline.\n\nIf you run the example project, you will see the diffuse reflection in action. The output should look like the following screenshot. As you can see, this reflection model works well for surfaces that are very dull but has limited use in a practical environment. Next, we will look at a reflection model that will allow us to depict more surface types:\n\n## Per-vertex ambient, diffuse, and specular\n\nThe ambient, diffuse, and specular (ADS) reflection model, also commonly known as the Phong reflection model, provides a method of creating a reflective lighting shader. This technique models the interaction of light on a surface using a combination of three different components. The ambient component models the light that comes from the environment; this is intended to model what would happen if the light was reflected many times, where it appears as though it is emanating from everywhere. The diffuse component, which we modeled in our previous example, represents an omnidirectional reflection. The last component, the specular component, is meant to represent the reflection in a preferred direction, providing the appearance of a light glare or bright spot.\n\nThis combination of components can be visualized using the following diagram:\n\nSource: Wikipedia\n\nThis process can be broken down into separate components for discussion. First, we have the ambient component that represents the light that will illuminate all of the surfaces equally and reflect uniformly in all directions. This lighting effect does not depend on the incoming or the outgoing vectors of the light since it is uniformly distributed and can be expressed by simply multiplying the light source's intensity with the surface reflectivity. This is shown in the mathematical formula\n\nIa = LaKa\n\nThe next component is the diffuse component we discussed earlier. The diffuse component models a dull or rough surface that scatters light in all directions. Again, this can be expressed with the mathematical formula\n\nId = LdKd(sn)\n\nThe final component is the specular component, and it is used to model the shininess of the surface. This creates a glare or bright spot that is common on surfaces that exhibit glossy properties. We can visualize this reflection effect using the following diagram:\n\nFor the specular component, ideally, we would like the reflection to be at is most apparent when viewed aligned with the reflection vector, and then to fade off as the angle is increased or decreased from this alignment. We can model this effect using the cosine of the angle between our viewing vector and the reflection angle, which is then raised by some power, as shown in this equation: (r v) p. In this equation, p represents the specular highlight, the glare spot. The larger the value input for p, the smaller the spot will appear, and the shinier the surface will look. After adding the values to represent the reflectiveness of the surface and the specular light intensity, the formula for calculating the specular effect for the surface looks like so:\n\nIs = LsKs(r v) p\n\nSo, now, if we take all of our components and put them together in a formula, we come up with\n\nI = Ia + Id + Is\n\nor breaking it down more,\n\nI = LaKa + LdKd(sn) + LsKs(r v) p\n\nWith our theory in place, let's see how we can implement this in a per-vertex shader, beginning with our vertex shader as follows:\n\n#version 410\n\/\/ Input vertex data, different for all executions of this shader.\nin vec3 vertexPosition_modelspace;\nin vec2 vertexUV;\nin vec3 vertexNormal;\n\n\/\/ Output data ; will be interpolated for each fragment.\nout vec2 UV;\nout vec3 LightIntensity;\n\nstruct LightInfo {\nvec4 Position; \/\/ Light position in eye coords.\nvec3 La; \/\/ Ambient light intensity\nvec3 Ld; \/\/ Diffuse light intensity\nvec3 Ls; \/\/ Specular light intensity\n};\n\nuniform LightInfo Light;\nstruct MaterialInfo {\nvec3 Ka; \/\/ Ambient reflectivity\nvec3 Kd; \/\/ Diffuse reflectivity\nvec3 Ks; \/\/ Specular reflectivity\nfloat Shininess; \/\/ Specular shininess factor\n};\n\nuniform MaterialInfo Material;\nuniform mat4 ModelViewMatrix;\nuniform mat3 NormalMatrix;\nuniform mat4 ProjectionMatrix;\nuniform mat4 ModelViewProjection;\nvoid main()\n{\nvec3 tnorm = normalize( NormalMatrix * vertexNormal);\nvec4 CameraCoords = ModelViewMatrix *\u00a0vec4(vertexPosition_modelspace,1.0);\n\nvec3 s = normalize(vec3(Light.Position - CameraCoords));\nvec3 v = normalize(-CameraCoords.xyz);\nvec3 r = reflect( -s, tnorm );\n\nfloat sDotN = max( dot(s,tnorm), 0.0 );\n\nvec3 ambient = Light.La * Material.Ka;\nvec3 diffuse = Light.Ld * Material.Kd * sDotN;\nvec3 spec = vec3(0.0);\n\nif( sDotN > 0.0 )\nspec = Light.Ls * Material.Ks *\u00a0pow( max( dot(r,v), 0.0 ), Material.Shininess );\n\nLightIntensity = ambient + diffuse + spec;\ngl_Position = ModelViewProjection * vec4(vertexPosition_modelspace,1.0);\n}\n\nLet's take a look at what is different to start with. In this shader, we are introducing a new concept, the uniform struct. We are declaring two struct, one to describe the light, LightInfo, and one to describe the material, MaterialInfo. This is a very useful way of containing values that represent a portion in the formula as a collection. We will see how we can set the values of these struct elements from the game code shortly.\n\nMoving on to the main function of the function. First, we start as we did in the previous example. We calculate the tnorm, CameraCoords, and the light source vector(s). Next, we calculate the vector in the direction of the viewer\/camera (v), which is the negative of the normalized CameraCoords. We then calculate the direction of the pure reflection using the provided GLSL method, reflect. Then we move on to calculating the values of our three components.\n\nThe ambient is calculated by multiplying the light ambient intensity and the surface's ambient reflective value. The diffuse is calculated using the light intensity, the surface diffuse reflective value of the surface, and the result of the dot product of the light source vector and the tnorm, which we calculated just before the ambient value. Before computing the specular value, we check the value of sDotN. If sDotN is zero, then there is no light reaching the surface, so there is no point in computing the specular component. If sDotN is greater than zero, we compute the specular component.\n\nAs in the previous example, we use a GLSL method to limit the range of values of the dot product to between 1 and 0. The GLSL function pow raises the dot product to the power of the surface's shininess exponent, which we defined as p in our shader equation previously.\n\nFinally, we add all three of our component values together and pass their sum to the fragment shader in the form of the out variable, LightIntensity. We end by transforming the vertex position to clip space and passing it off to the next stage by assigning it to the gl_Position variable.\n\nFor the setting of the attributes and uniforms needed for our shader, we handle the process just as we did in the previous example. The main difference here is that we need to specify the elements of the struct we are assigning when getting the uniform location. An example would look similar to the following, and again you can see the full code in the example solution in the Chapter07 folder of the GitHub repository:\n\nGLint Kd = shaderManager.GetUniformLocation(\"Material.Kd\");\nglUniform3f(Kd, 0.9f, 0.5f, 0.3f);\n\nThe fragment shader used for this example is the same as the one we used for the diffuse example, so I won't cover it again here.\n\nWhen you run the ADS example from the Chapter07 code solution of the GitHub repository, you will see our newly created shader in effect, with an output looking similar to the following:\n\nIn this example, we calculated the shading equation within the vertex shader; this is referred to as a per-vertex shader. One issue that can arise from this approach is that our glare spots, the specular highlights, might appear to warp or disappear. This is caused by the shading being interpolated and not calculated for each point across the face. For example, a spot that was set near the middle of the face might not appear due to the fact that the equation was calculated at the vertices where the specular component was near to zero.\n\nYou enjoyed an excerpt from the book, Mastering C++ Game Development written by Mickey Macdonald and published by Packt Publishing.\n\nUse the code ORGDB10 at checkout to get recommended eBook retail price for \\$10 only until May 31, 2018.\n\nReport Article\n\n## User Feedback\n\nThere are no comments to display.\n\n## Create an account\n\nRegister a new account","date":"2018-06-22 00:36:05","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\": 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\": 0.45910152792930603, \"perplexity\": 1335.19093825258}, \"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-2018-26\/segments\/1529267864303.32\/warc\/CC-MAIN-20180621231116-20180622011116-00030.warc.gz\"}"}
| null | null |
Northwestern stumbles but rebounds in weekend split
By Callie Counsellor Feb 25, 2013, 12:41pm CST
Share All sharing options for: Northwestern stumbles but rebounds in weekend split
by Callie Counsellor (@CCounsellor)
Northwestern stumbled out of the gate against No. 5 North Carolina, losing 11-8 in a matchup of title contenders. However, the Wildcats got back on track just two days later, defeating Vanderbilt 15-8 in the first conference game of the season.
Apparently, even the defending national champions can't spot an opponent a six-goal lead and expect to come back. No. 1 Northwestern (3-1, 1-0 ALC) found that out the hard way against North Carolina (3-1) on Friday night, as the Tar Heels turned a 1-1 game into a 7-1 lead less than 15 minutes into the match.
NU did not help its own cause with nine turnovers in the first half.
"We just made some really poor plays in the first half," coach Kelly Amonte Hiller said. "Bad decisions, (we) weren't coming up with the fifty-fifty ball. That's just going to eat away at you when you play good teams."
The Cats did begin a comeback to close the first half, pulling within 7-4, courtesy of goals from senior Erin Fitzgerald, who was named to the Tewaaraton Award Watch List this week; senior Ali Cassera; and junior Kelly Rich.
Both teams started the second half with runs—North Carolina a two-goal streak to restore a five-goal lead, and NU a three-goal run to cut the lead to two with 20 minutes remaining in the game.
Junior Alyssa Leonard scored one of those three goals and added another with 14 minutes remaining to lead the Cats in scores. She also won 12 draw controls in the match to give her 37 for the young season, over half of the team's total draw control wins.
The Tar Heels were able to hold on, though, thanks to several saves by freshman goalkeeper Megan Ward in the final minutes.
"Luckily we have another great opportunity Sunday," senior Taylor Thornton said after Friday's game. "I just think we need to take this feeling and bottle it up and every time we play, you have to feel that and remember this and never want to feel it again."
NU looked like a team on a mission Sunday against Vanderbilt (1-4, 0-1) in its first conference game of the year, led by a quartet of upperclassmen. Senior Amanda Macaluso tallied career-bests with five points and three goals. Leonard also set a career-best with four goals and Thornton won five draw controls. Senior Gabriella Flibotte dominated the midfield with four ground balls, four draw controls and four caused turnovers, as well as a goal.
The Cats controlled much of the first half play to lead 7-3 at the break, but it was a 7-0 run at the beginning of the second half, including two from Macaluso, that put the game away for NU.
"We started to get some momentum," Amonte Hiller said. "We were coming up with draw controls and Taylor (Thornton) and Gabby (Flibotte) did a great job on the draw stick. I was really excited how explosive we were offensively."
The win elevated Amonte Hiller's all-time record to 199-31, meaning she will go for win No. 200 in the Cats' home-opener on Wednesday against No. 20 Boston College (2-1). The Eagles suffered their first loss of the season on Saturday at the hands of No. 13 Ohio State.
"We've been on the road a bit," Amonte Hiller said, "(so) it's nice to have weekend where we don't have to travel and really test ourselves against a good, young Boston College team."
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Q: Webapp shown twice once started I added a few webapps to my starter panel and if I start one of it the chromium icon appears below of it. Chromium itself is locked to the starter separately, that means that once I start the webapp I got 2x chromium icons in my starter (see screenshot)
Can someone tell me if it is possible to disable that the second chromium icon is shown once I've started a webapp ?
|
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| 4,087
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(function() {
define(function(require) {
require('fire_list');
require('fire_model');
require('fire_model_by_ref');
require('fire_value');
return require('fire_value_by_ref');
});
}).call(this);
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Q: SFML 2.4 Window doesn't display shapes I tried writing the program myself, copying it from the official site and watching tutorials. Nothing helped. The problem is that the window doesn't display anything. Here is an example:
#include "SFML/Graphics.hpp"
#include <iostream>
int main()
{
sf::RenderWindow window( sf::VideoMode( 600, 600 ), "SFML WORK!" );
sf::CircleShape circle(150);
circle.setRadius(200);
circle.setPointCount(300);
while ( window.isOpen( ) )
{
sf::Event event;
while ( window.pollEvent( event ) )
{
switch (event.type)
{
case sf::Event::Closed:
window.close();
break;
}
}
window.clear( );
window.draw(circle);
window.display( );
}
}
It should, and in the tutorial it does, show a circle. In my case it just shows a white screen. Any ideas how to fix this? The program build is successfull, no errors found.
|
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| 3,053
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Theodore "Ted" Butterman (* 26. Januar 1935; † 31. August 2022) war ein US-amerikanischer Jazzmusiker (Trompete, auch Kornett, Gitarre) des Dixieland.
Leben und Wirken
Butterman war mehr als ein halbes Jahrhundert in der Chicagoer Swing- und Dixieland-Szene aktiv. In den späten 1950er-Jahren hinterließ Ted einen starken Eindruck in der Jazz-Revival-Szene der San Francisco Bay Area, wo er u. a. mit Frank "Big Boy" Goudie, Pete Allen, Dick Oxtot und Don Marchant auftrat ("Bugle Call Rag"). 1958 entstanden erste Aufnahmen mit The River Boat Five für Mercury Records ("From Natchez to Mobile"/"Ma! They're Comin' Down the Street"). 1961 trat er mit Little Brother Montgomery auf (South Side Blues, Riverside); ab den frühen 1960er-Jahren leitete er die Ted Butterman Gold Coast Jazz Band. In dieser Zeit spielten Butterman und seine Band oft im inzwischen aufgelösten Folk-Musik-Club "Gate of Horn" in der Dearborn Street in Chicago und tourten auch mit der Komikerin Phyllis Diller.
Das von Butterman ab 1966 geleitete Neo-Passé Swing Quintett brachte ihn dazu, zur Gitarre zu wechseln und eine lange Residenz in The Village Tavern in der Long Grove Township (Illinois) anzunehmen. Ab 1982 war er vor allem als Leiter der Jazzband Chicago Cubs Dixieland bekannt, mit der er bei den Spielen des Baseball-Teams Chicago Cubs auftrat. Im Bereich des Jazz war er laut Tom Lord zwischen 1958 und 2001 an 13 Aufnahmesessions beteiligt.
Bis Mitte der 2010er-Jahre trat er mit seiner Butterman Cubs Dixieland Band auf, zuletzt mit Klarinettist Eric Schneider, Rob Curtis, Banjo, Tom Barrett, Posaune und Steve Hart an der Tuba.
Diskographische Hinweise
Jim Kweskin with the Neo-Passé Jazz Band: Jump for Joy (Vanguard), (1967), mit Frank Chace, Kim Cusack, Johnny Frigo, Marty Grosz, Truck Parham, Wayne Jones
Ted Butterman's Neo Passe Swing Quintette: Live at the Village Tavern (2001), mit Eric Schneider, Dave Elias, Scott Black, Charley Weeks
Weblinks
Anmerkungen und Einzelnachweise
Jazz-Gitarrist
Jazz-Trompeter
Musiker (Vereinigte Staaten)
US-Amerikaner
Geboren 1935
Gestorben 2022
Mann
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"redpajama_set_name": "RedPajamaWikipedia"
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\section{Algorithm for $L_{n}$: Grade And Classify (GAC)} \label{signatures}
In this section we show an algorithm that works on $L_{n}$ (note that $n$ is also the number of processes). The algorithm terminates in ${\cal O}(\log(n))$ rounds. We will then discuss in Section \ref{sec:general} how to use this algorithm so solve LA on arbitrary join semi-lattices, and how to adapt it to work in ${\cal O}(\log(f))$ rounds.
\paragraph{High Level Description.} Our algorithm is based on the algorithmic framework of Zheng et al.~\cite{Garg:2018} adapted to tolerate Byzantine failures. As in the original, the algorithm works by continuously partitioning processes in \emph{masters} and \emph{slaves} sets. Partitioning is recursively operated in successive epochs. Processes that have been assigned to the same partition in each epoch are a ``group''. We indicate a generic group $G$ at epoch $ep$ with the string $s \cdot \sigma$, where $\sigma$ is in $\{s,m\}^{ep-1}$. As an example, the string $ssm$ indicates the set of process that at the end of epochs $0$ and $1$ entered the group of slaves (string $ss$) and then, at the end of epoch $2$, entered in the masters group (string $ssm$). When we write $p \in G$, we indicate that process $p$ belongs to the group of processes identified by string $G$.
The algorithm then enforces some properties on the partitions generated at an epoch $ep$ on a Group $G$:
\begin{itemize}
\item each master process in $G\cdot m$ adjusts its proposal to be a superset of each possible decision of a slave process in $G \cdot s$.
\item the height of a sub-lattice in which processes in $G\cdot m$ (or $G\cdot s$) are allowed to decide halves at each epoch.
\end{itemize}
Thanks to the above properties each group becomes independent and has to solve the lattice agreement on a lattice that has half of the height of the original.
A key concept for our algorithm is the one of {\em ``admissible value"}, a value is admissible, for a certain epoch, if it is ensured that it will not conflict with decisions of processes that have been elected as masters in a previous epoch. This is done by showing a cryptographical proof that such a value can be accepted by a slave since it is in the proposal value of each master the slave could conflict with.
After ${\cal O}(\log(n))$ rounds the algorithm terminates (each group operates on a lattice
constituted by a single point).
\input{gradecast}
\subsection{Detailed Algorithm Description}
Processes communicate by provable gradecasts. The three rounds necessary to execute a gradecast instance form a single epoch. We assume that in each epoch there are $n$ concurrent instances of gradecast running, one for each possible sender. The pseudo-code is in Algorithm \ref{gac:algorithm}.
\begin{algorithm*}
\caption{{\em GAC} - Algorithm for process $p_i$} \label{gac:algorithm}
\footnotesize
\begin{algorithmic}[1]
\algrenewcommand\algorithmicprocedure{\textbf{function}}
\State $t_d=0,t_u=n, t_{m}=\lfloor \frac{t_u}{2} \rfloor$
\State $G=\epsilon$
\State $pro_i=\emptyset$ \Comment{proposed value.}
\State $Proofs=\{\}$
\smallskip
\Procedure{LA-Propose}{$pro_i$}
\State $pro_i=pro_i$
\State {\sc gradecast}$(M=(pro_i,G,\bot))$ \Comment{Epoch 0 - start}
\State $P_i=${\sc rcv}()
\State $G=G \cdot s$
\State $updateproofs(pro_i,P_i,0)$ \Comment{Epoch 0 - end}
\For{$ep \in [1,\ldots, \log(n))+1]$} \label{alg:forcycle}
\State {\sc gradecast}$(M=(pro_i,G,Proofs))$
\State $P_i=${\sc rcv}()
\State $V_i=${\sc filter}$(P_i)$ \label{alg:masterset}
\If{{\sc classify}$(V_i)=s$}
\State $G=G \cdot s$
\State $t_d=t_d, t_{u}=t_m, t_{m}=t_{d}+\lfloor \frac{t_u-t_d}{2} \rfloor$
\State $updateproofs(pro_i,P_i,ep)$
\Else
\State $G=G \cdot m$
\State $pro_i= V_i$
\State $t_d=t_m,t_{u}=t_u, t_{m}=t_{d}+\lfloor \frac{t_u-t_{d}}{2} \rfloor$
\EndIf
\EndFor
\State {\sc decide}$(\bigoplus pro_i)$ \Comment{The $\bigoplus$ is needed since $V_i$ is a set of sets}
\EndProcedure
\bigskip
\Procedure{classify}{$V$}
\If{ $|V| \leq t_m$}
\State {\bf return} $s$
\Else
\State {\bf return} $m$
\EndIf
\EndProcedure
\medskip
\Procedure{filter}{$P$}
\State $V = \{\forall v \in M | M \in P \land M$ rank is greater than $0\,\, \land$ {\sc admissible}$(v,M) \land \,\, M$ source is in $G \land \,\, v$ in $E\}$
\State {\bf return} $V$
\EndProcedure
\medskip
\Procedure{admissible}{$v,M$}
\If{$\forall t \in [0,..,|G|-1]$ such that $G[0:t]$ terminates with $s$, there exists an admissibility proof for $v$ in $M$}
\State {\bf return} True
\Else
\State {\bf return} False
\EndIf
\EndProcedure
\medskip
\Procedure{updateproofs}{$pro_i,P_i,ep$}
\ForAll{$v \in pro_i$}
\State $proofv=[\bot]^{ep+1}$
\If{$ep>0$}
\ForAll{$t \in [0,\ldots,ep-1]$}
\If{$G[t]=s$}
\State Let $p$ be a valid proof from $P_i$ for $G[0:t]$ (it must exists since $v$ is admissible).
\State $proofv[t]=p$
\EndIf
\EndFor
\EndIf
\State Construct a valid proof $p$ for $v$ and group $G$ using messages in $P_i$. \label{alg2:updateproof}
\State $proofv[ep]=p$
\State $Proofs=Proofs \cup \{proofv\}$
\EndFor
\EndProcedure
\end{algorithmic}
\end{algorithm*}
\subsubsection{Epoch $ep =0$}
Epoch $0$ has a special structure. Correct processes belong to a single group $G=\epsilon$. In this epoch all values in $E \subseteq \{\{1\},\ldots,\{n\}\}$ are admissible (i.e., they do not have to carry an admissibility proof), and no classification step is executed.
That is, at the end of the epoch each process goes to group $G=s$ and no correct $p_i$ updates its value $pro_i$. This phase is performed to force Byzantine processes to commit to a certain set of values that cannot be changed later; values that are not associated with a seen-all proof, showing that they have been gradecasted in epoch $0$, will be ignored in the next epochs.
\subsubsection{Epoch $ep \geq 1$}
Epochs $1$ and onward share the same structure.
At the beginning of an epoch, all correct processes belonging to a same group share the values of three thresholds. At the beginning of epoch $ep=1$, the group $G=s$ encloses all the processes and the thresholds are $t_d=0$, $t_m=\frac{n}{2}$ and $t_u=n$.
\paragraph{Value gradecast.} An epoch starts by making each correct process $p_j$ in a group $G$ gradecast a message $M_j$ containing, its proposal value $pro_i$, the group to which $p_j$ belongs, and an {\em admissibility proof} for each element in $pro_i$ (the structure and the precise purpose of this proof is defined later).
Each other correct process $p_i$ in a group $G$ receives, by mean of the gradecast, a set of tuples: $P_i:\{(p_0,M_0,c_0, S_{p_0,M_0}), (p_1,M_{1},c_1, S_{p_1,M_{1}}), \ldots \}$.
We define a special set of values $V_i$ that is a subset of values in messages contained in $P_i$.
Set $V_i$ contains all ``{\em admissible values}" in $P_i$ and such that: (1) the rank of the message carrying the value is at least $1$ and (2) the sender of the message is in $G$.
Value $v$ is admissible for process $p_i$ in epoch $ep$ if the message that carries $v$ contains also an ``{\em admissibility proof}" proving that $v$ has been seen by all correct processes in $SLV(G[0:1], G[0:2],\ldots,G[0:ep-1])$, where $SLV(Set)$ is a filter function that removes from the set of string $Set$ all the strings that end with letter $m$. As an example considering $G=ssmsm$ we have $SLV(s,ss,ssm,
ssms,ssmsm)=\{s,ss, ssms\}$. Essentially, there must be a proof showing that $v$ has been seen with rank 1 by all processes in the epochs in which processes in $G$ have been classified as slave. The actual structure of this proof is described later (Section \ref{sec:proof}).
Process $p_i$ becomes a group slave if $|V_i| \leq t_m$, it becomes a group master if $t_m < |V_i|$.
If $p_i$ becomes a slave, it enters the group $G \cdot s$. Otherwise, it becomes a master, and it enters $G \cdot m$.
\medskip
\paragraph{Slaves actions} If a process $p_i$ is a slave, it updates its set of thresholds as $t_d=t_d, t_{u}=t_m, t_{m}=t_{d}+\lfloor \frac{t_u-t_d}{2} \rfloor$. Finally, a slave does not update its proposed value $pro_i$, in the next epoch it will have again the exactly same value it had in the current epoch.
Regarding the admissibility proof, a correct slave has the duty to collect an admissibility proof for its value proposed $pro_i$ this is done by collecting the seen-all proof generated by gradecasting its $pro_i$ at the beginning of the epoch.
\smallskip
\paragraph{Masters actions} If process $p_i$ is a master, it updates its set of thresholds as $t_d=t_m,t_{u}=t_u, t_{m}=t_{d}+\lfloor \frac{t_u-t_{d}}{2} \rfloor$. Moreover, it updates its value $pro_i= V_i$.
Regarding the admissibility proof, a correct master has no duty in creating an admissibility proof for its new $pro_i$, but it has to collect proofs to show that any value inserted in $pro_i$ was admissible in $G[0:ep-1]$.
\subsubsection{Admissibility proof}\label{sec:proof}
A message $m$ containing a value $v$ carries an admissibility proof for $v$ and group $G$ if message $m$ contains for each, also non-proper, prefix $G[0:j]$ of $G$ terminating with character $s$ (for $j$ in $\{0,1,\ldots,|G|-1\}$), a seen-all proof for $m$ with a sender $p$ in $G[0:j-1]$. From Oservation \ref{obs:byz}, it is immediate to see that an admissibility proof for $G$ implies that all correct processes in $SLV(G[0:1], G[0:2],\ldots,G[0:|G|-1])$ received $v$ in the value gradecast phase and ranked the source of the gradecast at least $1$. See Figure \ref{apf} for a graphical representation of the usefulness of such a proof.
Assume there exists an admissibility proof for value $v$ and group $G=ssms$, then there is a seen-all proof for the epochs $0,1,3$ and groups in $G_{s}:\{s,ss,ssms\}$(marked as green in the figure).
This implies that, in each of these epochs, value $v$ has been seen by each correct process with rank at least 1.
In particular, a seen-all proof for value $v$ and group $ss$ implies that $v$ has been gradecasted by a process in group $s$, and it has been received by all correct with rank at least $1$. This implies that $v$ has been inserted in the proposal of all correct masters in group $sm$ (in the figure we represent with an orange border the processes that have $v$ in their proposal). This means that a master in $ssm$ can update its proposal inserting $v$, and it knows that it will still be comparable with the decision of any process in a group with prefix $sm$.
Iterating the reasoning, a chain of seen-all proofs, the first for group $ss$ and the second for group $ssms$, implies that $v$ is in the proposal of all correct processes in a group with prefix
$sm$ or $ssmm$, and thus a future master in $ssmmsm$ can safely include $v$ in its proposal.
The necessity of a seen-all proof for epoch $0$, and thus for group $s$ is needed to force Byzantine processes to commit to at most $f$ values. This is due to the fact that $f$ Byzantine processes are able to create admissibility proofs for at most $f$ distinct values in epoch $0$.
\begin{figure}
\centering
\includegraphics[width=0.4\textwidth]{admissibilityproof}
\caption{Graphical representation of the purpose of an admissibility proof. \label{apf} }
\end{figure}
\subsubsection{Termination}
A process $p_i$ terminates the above algorithm when the epoch is $\log(n)+1$. Upon termination it decides its value $pro_i$.
\section{An universal transformer from LA to Generalised LA }\label{bzgc:const}
\cite{Faleiro:2012} introduced the Generalised Lattice Agreement problem. Such problem is essentially a version of LA in which processes are allowed to propose and decide on a, possibly infinite, sequence of values. The equivalent problem for Byzantine failures, in asynchronous system, has been defined in \cite{ipdps}. In this section we show a transformer algorithm that builds upon a generic ``one-shot" LA algorithm to create a Byzantine tolerant Generalised LA algorithm.
\paragraph{Synchronous Generalised LA.} We consider the definition of \cite{ipdps} adapted for a synchronous system.
In the Synchronous Generalised LA, each correct process $p_i$ receives input values from an infinite
sequence $Pro_i=\langle pro_{0},pro_{1}, pro_{2},\ldots \rangle$ where each $pro_k$ is a value inside a set of admissible values $E$ (note that $E$ is not necessarily finite). Without loss of generality we imagine that at each round $r$, $p_i$ receives a value $pro_r \in Pro_i$ (note that this is not restrictive since we could modify the lattice to admit a neutral element, such as $\emptyset$). A correct process $p_i$ must output an infinite number of decision values $Dec_i = \langle dec_{0},dec_{1}, dec_{2},\ldots \rangle$.
The sequence of decisions has to satisfy the following properties:
\begin{itemize}
\item {\bf Liveness:} each correct process $p_i \in C$ performs an infinite sequence of decisions $Dec_{i}= \langle dec_{0},dec_{1}, dec_{2},\ldots \rangle$;
\item {\bf Local Stability:} For each $p_i \in C$ its sequence of decisions is non decreasing (i.e., $dec_{h} \subseteq dec_{h+1}$, for any $dec_{h} \in Dec_{i}$);
\item {\bf Comparability:} Any two decisions of correct processes are comparable, even when they happen on different processes;
\item {\bf Inclusivity:} Given any correct process $p_i \in C$, if $Pro_i$ contains a value $pro_k$, then $pro_k$ is eventually included in $dec_{h} \in Dec_{i}$;
\item {\bf Non-Triviality:} Given any correct process $p_i \in C$ if $p_i$ outputs some decision $dec_k$ at a round $r$, then
\\ $dec_k \leq \bigoplus (Prop[0:r] \cup B[0:g(r)])$, where, $Prop[0:r]$ is the union of the prefixes, until index $r$, of all sequences $Pro_i$ of correct processes; and, $B[0:g(r)]$ is the union of all prefixes, until index $r$, of
$f$ infinite sequences $B_i$, one for each Byzantine process. Function $g$ is $g: \mathbb{N} \rightarrow \mathbb{N}$. Each $B_i$ is a sequence of elements in $E$.
\end{itemize}
Intuitively, function $g$ upper bounds, to $f \cdot g(k)$ the number of values that Byzantine processes can insert up to decision $k$.
\subsection{Transformer}
We now explain the high level idea behind the transformer.
Let ${\cal LA}$ be a one shot synchronous lattice agreement algorithm that terminates in $\delta$ rounds.
We divide the time in terms, a term lasts for $\delta$ rounds and it allows to execute, from start to termination, an instance of ${\cal LA}$.
At the beginning of term $k$, correct processes start the $k$-th instance of ${\cal LA}$, we denote it as $k$-${\cal LA}$. Each correct process receives from upper layer a stream of elements in $E$, and it batches such elements until a new instance of ${\cal LA}$ starts. Let $C_{k}$ be the $k$-th batch, at the beginning of term $k$, process $p_i$ starts the instance $k$-${\cal LA}$ with input $(p_i, dec_{k-1} \oplus C_{k})$, where $dec_{k-1}$ is the output of the $(k-1)$-${\cal LA}$ instance.
There are few minor details to add to this description to get an actual algorithm (see pseudocode in Algorithm \ref{trans:algorithm}), the most important being the mechanism needed to bound the number of values that could be added by Byzantine processes (recall that we have to upper bound their values by a function $g$). The key idea is to start the instance $k$-${\cal LA}$ with a set of admissible values $ P_{T(k-1)+\delta}$, that is the set of all subsets of $E$ of size less or equal to $T(k-1)+\delta$. Function $T$ is a function mapping each index of the decision sequence of a correct process to an upper bound on the maximum size of the decision, where the size is counted as number of elements in the decision. We assume $T(-1)=0$; we have that $T(0)=\delta \cdot n$: each correct process that starts the first instance of LA proposes at most $\delta$ values; the closed form for $k>0$ of $T(k)$ is in the statement of Lemma \ref{lemma:transf}.
\paragraph{Correctness discussion.}
Assuming that each ${\cal LA}$ is an instance of a correct LA algorithm (according to definition in Section \ref{definition:la}), we argue that the Generalised LA algorithm obtained by the transformer described above is a correct algorithm according to definition in Section \ref{bzgc:const}.
The liveness property is satisfied by the liveness of each instance of LA. The local stability and the inclusivity derive directly from the fact that once a process $p$ outputs $dec_{k-1}$, the next instance of LA will have as input a value that contains $(p, dec_{k-1} \oplus C_{k})$. Therefore, by inclusivity of LA we have that the decision of $k$-${\cal LA}$ contains the pair $(p, dec_{k-1} \oplus C_{k})$, and this means that $ dec_{k-1} \oplus C_{k} \leq dec_{k}$. It remains to show the non-triviality:
\begin{Lemma}\label{lemma:transf}
Consider the sequence of decisions of a correct process $p$ executing the transformer in Algorithm \ref{trans:algorithm}. This sequence respects the Non-Triviality property for a function $g(r)< (T(k)=\frac{\delta \cdot n ((f+1)^{k+1}-1)}{f})$
\end{Lemma}
\begin{proof}
We have to upper bound the number of different values of $E$ that could be inserted by byzantines in the decision $dec_k$ of a correct process. Without loss of generality we suppose that $E$ is infinite.
We will upper bound the total number $T(k)$ of elements in $E$ that could be in a decision $dec_k$ ($dec_k$ is the $k$-th elements of the sequence of decisions $Dec_i = \langle dec_{0},dec_{1}, dec_{2},\ldots \rangle$). Clearly, this upper bound is an upper bound on $g(k)$.
When $k=0$ we have that each process is allowed to propose a $(id,Y)$ where $|Y|=\delta$, by the Non-Triviality of LA we have $|dec_{0}| \leq n\delta$. When $k > 0$ we have the following recurrence relation on $T$ that is $T(k) \leq f \cdot T(k-1) +T(k-1) + n \delta$, where: the term $(n -f)\delta$ accounts for the new $\delta$ elements that each correct process may add (the $C$ sets); the term $T(k-1)$ account for the fact that each correct process proposes also its previous decision (that has maximum size $T(k-1)$); and finally the term $f \cdot T(k-1)+ f \delta$ accounts for the fact that each byzantine could propose a set of new elements of size $T(k-1)+\delta$. Essentially, a Byzantine could fake its decision $dec_{k-1}$ to be any subset of $E$ of size at most $T(k-1)$.
\end{proof}
\begin{algorithm*}
\caption{{\em Transformer} - Algorithm for process $p_i$} \label{trans:algorithm}
\footnotesize
\begin{algorithmic}[1]
\algrenewcommand\algorithmicprocedure{\textbf{upon event}}
\State $C=\emptyset$ \Comment{Batch of values}
\State $dec=\emptyset$ \Comment{Decision}
\Procedure{propose}{$pro_r$}
\State $C=C \cup \{pro_r\}$
\EndProcedure
\Procedure{ $r = \delta \cdot (k+1)$ for some $k \in \mathbb{N}$}{}
\State Start the instance $k$ of $LA$ over lattice ${\cal L}_{k}$ with admissible values $E_{k}=\Pi \times P_{T(k-1)+\delta}$
\State {\sc $k$-LA-Propose}$((p_i, dec \oplus C))$
\State $C=\emptyset$
\EndProcedure
\Procedure{decision from the current instance of LA}{$dec'$}
\State $X:\{y | (x,y) \in dec'\}$
\State $dec=\bigoplus X$
\State Decision$_k(dec)$
\EndProcedure
\end{algorithmic}
\end{algorithm*}
\subsection{The Provable Gradecast Primitive}\label{pgc:sec}
The algorithm makes use of the gradecast primitive introduced by Ben-Or et al. in ~\cite{Ben-or:2010}. Such primitive is similar to a broadcast, we have a sender process $p_i$ that sends a message $m$, each other process $p_j$, after 3 rounds, outputs a tuple $(p_i,m_j,c_j)$ where $c_j \in \{0,1,2\}$ is a score
of the correctness of $p_i$. The gradecast ensures the following properties:
\begin{itemize}
\item for any two correct processes $p_j,p_\ell$ if $c_j >0$ and $c_\ell >0$ than $m_j=m_\ell$.
\item for any two correct processes $p_j,p_\ell$ we have $|c_j - c_\ell| \leq 1$
\item if the gradecast sender is correct, than for any correct process $p_j$ we have $c_j=2$.
\item for any correct process $p_j$ if $(p_i,m_j,0)$ then $m_j=\bot$.
\end{itemize}
Intuitively, if we let processes communicate by mean of the gradecast primitive we force Byzantines to send at most two different messages to the set $C$ of correct processes, and one of these messages has to be $\bot$.
We modify the original gradecast to make it ``provable". In our version of the gradecast each correct process outputs a tuple composed by 4 objects $(p_i,m_j,c_i, S_{i,m_j})$ where $S_{i,m_j}$ is a special object that can either be $\bot$ or a {\em seen-all} proof. In case $S_{i,m_j}$ is different from $\bot$, then it is a cryptographic proof that can be shown to other processes and it implies that, any correct process $p \in C$ has seen a rank, for the gradecast of message $m_j$ from process $p_i$, that is at least $1$. Moreover, we have that if $c_i=2$ then $S_{i,m_j} \neq \bot$.
Practically, the modification of the original gradecast are contained, and are limited to the second and third round of the algorithm.
The original gradecast, with source $p_s$ works as follows: in the first round $p_s$ broadcasts a message $m$ to all processes; in the second round each correct process relays the message received by $p_s$ (it ignores messages from other processes); at the end of the second round a correct process selects the most frequent message received, and if such message was received by at least a quorum of $n-f$ processes then it relays the message at the beginning of the third round; at the end of the third round each correct selects the most frequent message received and it ranks it $2$ and delivers it if the message was received by at least $n-f$ processes; if it was received by at least $f+1$ it delivers it and ranks it $1$, otherwise it delivers $\bot$ with rank $0$.
In our version, see Algorithm \ref{pgc:algorithm}, the relaying process signs the relay sent at the beginning of round 3 (see line \ref{pgc:sign})
and a process that sees a message with rank 2 collects the $n-f$ (i.e., $2f+1$ if $n=3f+1$) signed messages (see line \ref{pgc:proof}). These signed messages constitute a proof that the message has been seen by all: delivered by each correct with rank at least 1. Note that the algorithm is for a single instance and a single determined sender, however one can trivially run in parallel an instance for each possible sender in the system.
We do not prove the properties discussed above for our version of gradecast, they immediately derives from the correctness of the original algorithm \cite{Ben-or:2010}. We do show the property introduced by the seen-all proof:
\input{pcgAlg}
\begin{observation}\label{obs:byz}
Consider an instance of gradecast with source $p_s$.
If a process $p_i$, whether Byzantine or correct, can produce a {\em seen-all} proof for a message $m$, then each process $p \in C$ delivered message $m$ with rank at least $1$ at the end of the gradecast instance.
\end{observation}
\begin{proof}
Let us suppose by contradiction that there exists two processes $p_i$ and $p_j$ such that (i) $p_i$ is able to show a {\em seen-all} proof for a message $m$ and (ii) $p_j$ is correct and it never delivered $m$ with rank $1$ or more, during the execution of the instance of provable gradecast algorithm.
If $p_i$ is able to show a {\em seen-all} proof, it means that it collected at least $n-f$ signed copies of $m$ (see line \ref{pgc:proof}). Considering that in the system we have at most $f$ Byzantine processes that can generate fake signed copies of $m$, it follows that the remaining $n-2f$ copies arrive from correct processes.
Since $n \ge 3f +1$, it follows that at least $f+1$ signed copies of $m$ have been generated by correct processes and sent at the beginning of round $3$.
A correct process broadcasts the message to all other processes in the system. This implies that $p_i$ receives at least $f+1$ copies of $m$ at the beginning of round $3$. It remains to show that $m$ is the most frequent message $p_i$ receives.
Notice that a correct process sends a signed message $m$ at the beginning of round $2$ only upon receiving from a Byzantine quorum of processes (see Line \ref{pgc:sign}), this implies that no two correct processes relay two different messages at the beginning of round 2.
Therefore, $m$ is the only message that a correct can send to $p_i$, and the Byzantines cannot create more than $f$ messages. This implies that $m$ is the most frequent message that $p_i$ receives and it will be received with frequency at least $f+1$. This contradicts the fact that $p_i$ does not deliver $m$ with rank $1$. \end{proof}
\section{Authenticated Channels and No Signature - Necessity of $3f+1$ processes in synchronous systems} \label{authchannel:impossibility}
In the following we show that $3f+1$ processes are necessary when there are no signatures.
\begin{Lemma}\label{lemmabase}
It does not exist any algorithm solving Byzantine LA on arbitrary lattices in a synchronous system with $3$ processes when one is faulty. The impossibility holds even when relaxing the Non-Triviality allowing $|B| \leq k$ for any fixed $k$.
\end{Lemma}
\begin{proof}
We first discuss the case of $k=1$.
Let ${\cal A}$ be an algorithm solving LA with $3$ processes when one is faulty. Since ${\cal A}$ works on arbitrary lattices it should also work on the lattices induced by the union operation on the power set of the first $6$ natural numbers with $E=\{\{1\},\{2\}, \{3\}, \{4\}, \{5\}, \{6\}\}$.
Now let us consider the hexagonal system of Figure \ref{3fstart}. Such a system is constituted by $6$ processes $p_1,p_2,p_3,p_4,p_5,p_6$ with an edge between each $p_i,p_j$ such that $i=j \pm 1$ and one edge between $p_1$ and $p_6$.
Each of the six processes has as input an unique value in $[1,6]$, just for simplicity process $p_i$ has input $\{i\}$.
Note that even if ${\cal A}$ is an algorithm for three processes, it is possible to execute ${\cal A}$ on the hexagon, but its behaviour does not necessarily follows the LA specification.
On the right of the figure we have $6$ triangles, each triangle is related to a corresponding edge in the hexagon.
The relationship is such that the view of two neighbour processes in the hexagon is equal to the view of two processes in a triangle where the third process is a Byzantine simulating the behaviour of the other processes in the hexagon.
As example: the view of processes $p_1,p_2$ in the hexagon is the same that $p_1,p_2$ would have in triangle $t_{green}$, analogously the view of $p_6,p_1$ in the hexagon is the same view of $p_6,p_1$ in $t_{blue}$. Note that ${\cal A}$ once executed on any of the triangle in the figure has to follow the LA specification.
A run of ${\cal A}$ on the hexagon in principle has an undefined behaviour. However we observe that a run of ${\cal A}$ on the hexagon eventually terminates on each process, this is because each process has a local view that is consistent with a system of 3 processes one of which is a Byzantine. Recall, that the local view of each process $p_i$ in the hexagon is exactly the same view that the process has in the two triangles on the right, and ${\cal A}$ being a correct algorithm when $3$ processes are considered the
algorithm will correctly terminate in each triangle in the right.
Moreover, each process will output a decision value that must be the same that the process will output in the corresponding triangles.
Let $dec_1,dec_2,dec_3,dec_4,dec_5,dec_6$ be the decisions of processes dictated by ${\cal A}$ (naturally we have $dec_i$ decision of $p_i$). As reference see Figure \ref{3fend}.
The triangles on the right impose a certain number of comparability relationships among these decisions. Recall, that each decision is a subset, not necessarily proper, of $[1,6]$, and that
the comparability in this setting is the relationship of inclusion. An example is triangle $t_{green}$ that imposes the comparability between $dec_1$ and $dec_2$, that is either $dec_1 \subset dec_2$ or $dec_2 \subseteq dec_1$.
In the following we use $ dec_i \leftrightarrow dec_j$ to indicate that $dec_i$ must be comparable with $dec_j$.
Before continuing with the proof we give the following technical observation.
\begin{observation}
Consider a collection of $m$ sets $S_1,S_2,\ldots,S_m$ such that for each $S_i$ we have $i \in S_i$. If it holds that $S_j \leftrightarrow S_{i+1}$ for all $j \in [1,m-1]$ then there exists an $|S_k| \geq m$.
\end{observation}
Therefore, let us take w.l.o.g. $dec_1$, and let us walk in clockwise direction for 3 steps on the hexagon. On this walk we have:
$dec_1 \leftrightarrow dec_2 \leftrightarrow dec_3 \leftrightarrow dec_4$, by the inclusivity property we have for each $dec_i$ that $i \in dec_i$.
Therefore we can apply our technical lemma and state that one of the $dec_i$ has cardinality at least $4$ violating the non-triviality property of ${\cal A}$ on some triangle (recall that $k$ being equal to $1$ we have $|B| \leq 1$).
The generalised proof for an arbitrary $k$ follows the same reasoning using: a lattice on the power set of the first $3(k+1)$ natural numbers, $E=\{ \{1\},\ldots, \{3(k+1)\}\}$ and a $3(k+1)-$gon instead of an hexagon. We then walk on the $3(k+1)-$gon for $k+2$ steps instead of $3$.
We will have a chain $dec_1 \leftrightarrow dec_2 \leftrightarrow \ldots \leftrightarrow dec_{k+3}$ where by inclusivity of ${\cal A}$ on the corresponding triangle we have $i \in dec_i$, and where our
observation shows that one of the decisions contains at least $k+3$ distinct elements of $E$ violating the non-triviality on some triangle. \end{proof}
\begin{theorem}\label{th:impossible}
It does not exist any algorithm solving Byzantine LA on arbitrary lattices in a synchronous system with $3f$ processes and $f$ faulty if $|B| \leq f$. The same holds if $n < 3f$
\end{theorem}
\begin{proof}
Let ${\cal A}$ be an algorithm that solves $LA$ on a system of $3f$ processes.
We build an algorithm ${\cal A}_{sym}$ for three processes using ${\cal A}$. In algorithm ${\cal A}_{sym}$ each process simulates $f$ processes of ${\cal A}$. Each of these simulated processes starts with proposal of the real process.
Once a simulated process decides the corresponding real process decides the same set.
It is immediate that ${\cal A}_{sym}$ is a correct lattice agreement algorithm with a relaxed non-triviality assumptions that includes at most $f$ values from a Byzantine.
This violates Lemma \ref{lemmabase}.
When the number of failures is greater than $n/3$ the same argument holds.
\end{proof}
The proof of Theorem \ref{th:impossible} does not work in a system with authenticated messages (i.e., signatures), it is therefore unkown whether $3f+1$ processes are necessary also in this model.
Interestingly, in \cite{ipdps} it is shown that, when the system is asynchronous, $3f+1$ processes are necessary also when authenticated messages are available.
\begin{figure}
\centering
\includegraphics[width=0.8\textwidth]{impossibility3fstart.pdf}
\caption{Starting configuration of the Hexagon. For each edge there is a corresponding triangle that is a legal starting configuration of ${\cal A}$ with one Byzantine. As an example, take $p_1,p_2$ in the hexagon.
They have the exactly same view of $p_1,p_2$ in $t_{green}$ where the other node is $b_{green}$ a Byzantine that simulates the behaviour of $p_3,p_4,p_5,p_6$ in the hexagon. \label{3fstart} }
\end{figure}
\begin{figure}
\centering
\includegraphics[width=0.8\textwidth]{impossibility3fend.pdf}
\caption{Ending configuration of the Hexagon. \label{3fend} }
\end{figure}
\section{Introduction}
Fault-tolerance is a key research topic in distributed computing \cite{10.5555/2821576}: reliable algorithms have been deeply investigated in classic message passing \cite{doi:10.1002/0471478210} and in newer model of computations \cite{popprot,icdcn,michail,doty}. A special mention has to be given to algorithms for distributed agreement \cite{105555,6812814} . They represent a cornerstone of todays cloud-based services. In particular, practical and efficient implementations of distributed consensus, transformed in the last 10 years the Internet from a large computers network in a world-scale service platform. Despite its fundamental role, real implementations of distributed consensus are still confined to small scales or tightly controlled environments; the well known FLP result, in fact, makes distributed consensus impossible to solve deterministically in asynchronous settings, where communication latencies cannot be bounded. To cope with this limit, practical systems trade off consistency criteria (allowing weaker \emph{agreement} properties) with liveness (guaranteeing run termination only in long-enough grace periods where the systems ``behaves'' like a synchronous one).
For such a reason, agreement properties weaker than consensus proved to be extremely effective for the implementation of a broad family of distributed applications, since they can be used in systems where consensus cannot be solved, or they can be faster than consensus algorithms circumventing time-complexity lower bounds.
\noindent {\bf Lattice Agreement} \textemdash~ In this paper we investigate an agreement problem that is weaker than consensus: the Lattice Agreement (LA) problem. In LA, introduced by Attiya et al.~\cite{Attiya:1995}, each process $p_i$ has an input value $x_i$ drawn from the join semilattice and must decide an output value $y_i$, such that \emph{(i)} $y_i$ is the join of $x_i$ and some set of input values and \emph{(ii)} all output values are comparable to each other in the lattice, that is they are all located on a single chain in the lattice (see Figure \ref{slattice}). LA describes situations in which processes need to obtain some knowledge on the global execution of the system, for example a global photography of the system.
In particular Attiya et al.~\cite{Attiya:1995} have shown that in the asynchronous shared memory computational model, implementing a snapshot object is equivalent to solving the Lattice Agreement problem.
Faleiro et al.~\cite{Faleiro:2012} have shown that in a message passing system a majority of correct processes and reliable communication channels are sufficient to solve LA in asynchronous systems, proposing a Replicated State Machine with commutative updates built on top of a generalized variant of their LA algorithm. Generalized Lattice Agreement (GLA) is a version of LA where processes propose and decide on a, possibly infinite, sequence of values. A recent solution of Skrzypczak et al.~\cite{Skrzypczak} considerably improves Faleiro's construction in terms of memory consumption, at the expense of liveness.
The restriction of having only commutative updates is justified by the possibility of developing faster algorithms. It is well known \cite{ doi:10.1002/0471478210} that consensus cannot be solved in synchronous systems in less than ${\cal O}(f)$ rounds, even when only crash failures are considered, on the other hand it has been shown \cite{Garg:2018} that, when crash failures are considered, Lattice Agreement can be solved in ${\cal O}(\log f)$ rounds \footnote{Actually, it can be solved faster than $\log(f)$ on specific lattices, the ones having height less than $\log f$ \cite{Garg:2018}, however in this paper we are interested only in worst case performances.}, in ${\cal O}(\log f)$ message delays in asynchronous \cite{Garg:2018b}.
\noindent {\bf Byzantine Failures} \textemdash~ All these papers consider process failures, but assume that such failures are non-malicious. More recently, some works started proposing LA algorithms that tolerate Byzantine failures.
The first one has been by Nowak and Rybicki~\cite{Nowak:2019}: they introduced LA with Byzantine failures and proposed a definition of Byzantine LA in which decisions of correct processes are not allowed to contain values proposed by Byzantine processes. Using such definition, authors have shown that the number of correct processes needed to solve LA depends on the structure of the lattice, and they have shown a LA algorithm for specific kinds of lattices.
Successively, Di Luna et al. \cite{ipdps} proposed a less restrictive definitions of Byzantine LA, in which correct processes can decide also values proposed by Byzantine. Authors have shown that, when their definition is used, LA can be solved for any possible lattice when $f < \frac{n}{3}$: they proposed a solution for Byzantine LA in asynchronous systems that terminates in $\mathcal O(f)$ rounds; the same paper also proposed a Generalized version of the algorithm and built on top of it a Replicated State Machine that executes commutative operations. In this paper we adopt the Byzantine LA definition from \cite{ipdps}, since it allows to circumvent the impossibility of \cite{Nowak:2019} and it is usable in many practical scenario. The same definition has been also used by Zheng and Garg~\cite{Zheng:aa}, where they show that LA can be solved in synchronous systems with $\mathcal O(\sqrt f)$ rounds also in presence of Byzantine failures.
\begin{figure}
\centering
\includegraphics[width=0.5\textwidth]{lattice}
\caption{Hasse diagram of the semilattice induced over the power set of $\{1,2,3,4\}$ using the union operation as join. Given any two elements $e,e'$ of the lattice if $e < e'$, then $e$ appears lower in the diagram than $e'$ and there is an ``upward" path, going from lower points to upper points, connecting $e$ to $e'$ (e.g., $\{1\} \leq \{1,3,4\}$, but $\{2\} \not\leq \{3\}$). Any two elements $e,e'$ of the semilattice have a join $e \oplus e'= e \cup e'$ and $e \oplus e' \geq e,e'$ (e.g., $\{1\} \oplus \{2,3\}=\{1,2,3\}$ ). The red edges indicate a possible chain (i.e., sequence of increasing values) that contains the output values.
}\label{slattice}
\end{figure}
\noindent {\bf Contributions} \textemdash~In this paper we present new contributions for the Byzantine LA problem in synchronous settings.
Our first results is for systems with only authenticated channels (i.e., signatures are not available), in such systems we show that Byzantine LA on arbitrary lattices cannot be solved, in synchronous systems, with $f=\lceil n/3 \rceil$ or more faulty processes (Section \ref{authchannel:impossibility}).
Interestingly, such proof shows that the algorithm of Zheng and Garg~\cite{Zheng:aa} is tight in the number of tolerable failures.
On the positive side we show algorithms that solve LA and Generalized LA, with and without signatures, having better running time that the state-of-the-art.
Looking at the model with signatures, we show a novel algorithm for LA that works in a synchronous system model, tolerates up to $f$ byzantine failures (where $f<\lceil n/3 \rceil$) and that terminates in ${\cal O}(\log f)$ rounds. The algorithm improves over the $LA_\beta$ algorithm from Garg at al.~\cite{Garg:2018} by using a similar construction, but adding tolerance to Byzantine failures. We make use of a modified Gradecast algorithm that allows processes to prove that a message has been seen by all correct processes in the system. (Sections \ref{signatures}-\ref{sec:general})
We conclude our investigation on LA by briefly discussing how to remove signatures and make our construction work only with authenticated channels trading-off part of its resiliency: we are able to tolerate only $f<\lceil n/4 \rceil$ failures (Section \ref{sec:nosig}).
In the last part of the manuscript, we devote our attention to Generalized Lattice Agreement (Section \ref{bzgc:const}). Specifically, we show a transformer that using as building block a generic LA algorithm creates a Generalized Lattice Agreement algorithm. At the best of our knowledge this is the first time GLA is investigated in synchronous systems.
\subsection{Related Work}
\label{sec:related_work}
\renewcommand{\arraystretch}{1.2}
\begin{table*}[t]
\footnotesize
\caption{Comparison of algorithms for Byzantine Lattice Agreement when the lattice has an height that is greater than $f$. \label{table:comparison}}
\begin{tabularx}{\textwidth}{ | c | c | X | X | X | X | }
\hline
{\em Model} & {\em Assumption} & {\em Paper} & {\em Resiliency} & {\em Time} & {\em Messages} \\
\hline
\hline
\multirow{4}{*}{({\sf SYNC})} & \multirow{2}{*}{Auth. Links (No Sign.)} & This paper & $f<\lceil n/4 \rceil$ & ${\cal O}(\log(f))$ & ${\cal O}(n^2 \log(f))$ \\
\cline{3-6}
& & Zheng et al. \cite{Zheng:aa} & $f <\lceil n/3 \rceil$ &${\cal O}(\sqrt{f})$ & ${\cal O}(n^2 \sqrt{f})$ \\
\cline{2-6}
& \multirow{1}{*}{Auth. Messages (Signatures)} & This paper &$f< \lceil n/3 \rceil$ & ${\cal O}(\log(f))$ & ${\cal O}(n^2 \log(f))$ \\
\hline
\hline
\multirow{2}{*}{({\sf ASYNC})}& Auth. Links (No Sign.)& Di Luna et al. \cite{ipdps} & $f< \lceil n/3 \rceil$ & $ {\cal O}(f) $ & ${\cal O}(n^2)$ \\
\cline{2-6}
& Auth. Messages (Signatures) & Di Luna et al. \cite{ipdps} & $f< \lceil n/3 \rceil$ & $ {\cal O}(f) $ & $ {\cal O}(f \cdot n)$\\
\hline
\end{tabularx}
\end{table*}
A map of the relevant related work is illustrated in Table \ref{table:comparison}, where we compare Byzantine fault-tolerant algorithms using the definition of LA from \cite{ipdps}.
Zheng and Garg~\cite{Zheng:aa} showed that Byzantine LA can be solved in synchronous settings with $\mathcal{O}(\sqrt{f})$ rounds. The algorithm they propose makes use of a modified version of the Gradecast~\cite{Ben-or:2010} algorithm as a building block. Furthermore, correct processes are asked to keep track of a lattice of safe values among which final values will be decided. This approach guarantees that Byzantine processes cannot pollute correct process decisions with values that are not safe, i.e. values that are known by every correct process. When signatures are available, the algorithm we propose matches the same resiliency $f<\lceil n/3 \rceil$ (that we show being optimal) while having a faster running time. When we remove signatures our construction has a worse resiliency $f<\lceil n/4 \rceil$ but it keeps a faster running time.
Di Luna et al.~\cite{ipdps} proposed a solution for Byzantine LA in asynchronous settings, also providing an algorithms for its generalized variant. Their algorithm bases its correctness on an initial phase were values to be proposed are broadcasted to build a safe set of values in the lattice from which the final decided values will be chosen. Byzantine processes that try to propose arbitrary values that are not contained in the safe set will see their message simply ignored.
\section{System Model, Notation, and Preliminaries}
We use the usual message passing models with unique identifiers (IDs). There is a set $\Pi$ of $n$ processes with unique IDs in $\{1,\ldots,n\}$ connected by a complete communication graph. The system is synchronous, and the execution of the algorithm can be divided in discrete finite time units called rounds. In each round a process is able to send messages to its neighbours ({\em send phase}), and receive all messages sent to it at the beginning of the round ({\em receive phase}). Processes in $\Pi$ are partitioned in two sets $F$ and $C$. Processes in $C$ are correct, they faithfully follow the distributed protocol. Processes in $F$ are Byzantine, they arbitrarily deviate from the protocol. As usual when Byzantine failures are considered, we assume that the communication channels are authenticated by mean of Message Authentication Codes (MAC).
The {\em authenticated channels} are the only assumption used in Section~\ref{authchannel:impossibility}. In Section~\ref{signatures} we assume that there is a public key infrastructure that allows processes to cryptographically sign messages, that can be lately verified by other processes. This model has {\em authenticated messages}. Byzantine processes are polynomially bounded and cannot forge signatures of correct processes.
For an easier presentation we explain our algorithms for the case of $n=3f+1$, where $f=|F|$, however they can be easily adapted for any other $n> 3f+1$.
\paragraph{Notation.}
With $\epsilon$ we indicate the empty string. Given a string $G$, with $|G|$ we indicate the length of the string ($|\epsilon|=0$), with $G[j]$ and $0 \leq j < |G|$ we indicate the character of string $G$ in position $j$. With $G[k:l]$ (given $0 \leq k \leq l \leq |G|$), we indicate the substring of $G$ between position $k$ and $l$.
As an example given $G=ssms$, we have $G[0]=s$ and $G[0:1]=ss$.
Given two strings $a$ and $b$ with $a \cdot b$ we indicate the string obtained by concatenating $b$ after $a$.
\subsection{The Byzantine Lattice Agreement Problem} \label{definition:la}
Each process $p_i \in C$ starts with an initial input value $pro_i \in E$ with $E \subseteq V$ (set $E$ is a set of allowed proposal values).
Values in $V$ form a join semi-lattice $L=(V,\oplus)$ for some commutative join operation $\oplus$: for each $u,v \in V$ we have $u \leq v$ if and only if $v= u \oplus v$.
Given $V'=\{v_1,v_2,\ldots,v_k\} \subseteq V$ we have $\bigoplus V'=v_1 \oplus v_2 \oplus \ldots \oplus v_k$.
The task that processes in $C$ want to solve is the one of Lattice Agreement, and it is formalised by the following properties:
\begin{itemize}
\item {\bf Liveness:} Each process $p_i \in C$ eventually outputs a decision value $dec_i \in V$;
\item {\bf Stability:} Each process $p_i \in C$ outputs a unique decision value $dec_i \in V$;
\item {\bf Comparability:} Given any two pairs $p_i,p_j \in C$ we have that either $dec_i \leq dec_j$ or $dec_j \leq dec_i$;
\item {\bf Inclusivity:} Given any correct process $p_i \in C$ we have that $pro_i \leq dec_i $;
\item {\bf Non-Triviality:} Given any correct process $p_i \in C$ we have that $dec_i \leq \bigoplus (X \cup B)$, where $X$ is the set of proposed values of all correct processes ($X:\{pro_i | \text{ with } p_i \in C \}$), and $B \subseteq E$ is $|B| \leq f$.
\end{itemize}
{\bf Lattice definitions.} Given two distinct elements $u,v \in V$ ve say that there exists a path between $u$ and $v$ if they are comparable; a path of length $k$ between $u$ and $v$ is a sequence of $k+1$ distinct elements $(e_0,e_2,\ldots,e_k)$ of the lattice such that
$e_0=u \leq e_1 \leq e_2 \leq \ldots \leq e_{k-1} \leq e_k=v$. As an example the path between $\{1,2,3\}$ and $\{1\}$ in the lattice of Figure \ref{slattice} has length 2.
We say that a $v \in V$ is minimal if it does not exists $u \in V$, with $u \neq v$, such that $u \oplus v= v$ (i.e., it does not exists an $u \leq v$). As in \cite{Zheng:aa} we define the
height of an element $v$ in a lattice $(V, \oplus)$ has the length of the longest path from any minimal element to $v$ in the lattice (as an example the Lattice in Figure \ref{slattice} has height $4$). A sub-lattice of $(V,\oplus)$ is a subset $U$ of $V$ closed with respect the join operation, the definition of height for a sub-lattice does not change.
{\bf Preliminaries.} In the rest of the paper we will assume that $L$ is a semi-lattice over sets ($V$ is a set of sets) and $\oplus$ is the set union operation. This is not restrictive, it is well known~\cite{Nation} that any join semi-lattice is isomorphic to a semi-lattice of sets with set union as join operation. An important lattice is the one on the power set of the first $\{1,\ldots,n\}$ natural numbers with the union as join operation (see Figure \ref{slattice}), we will use as shorthand for such lattice the notation $L_{n}$. Such lattice is interesting for our purpose, we will show that an algorithm solving lattice agreement exclusively on such lattice (the GAC of Section \ref{signatures}) can be used as building block to solve lattice agreement on an arbitrary lattice (Section \ref{sec:general}).
When $L_{n}$ is considered, the height of an element $e$ is equal to its cardinality (i.e, $|e|$ cfr. Figure \ref{slattice}), given a sub lattice of $L_{n}$ its height is upper bounded by the difference between the minimum and maximum cardinality of its elements.
\subsection{Correctness of GAC}\label{gac:correctness}
\begin{definition}
Let $A(G)$ be the set of values admissible for correct processes in group $G$ during the gradecast of epoch $|G|$.
\end{definition}
Given a group $G$ the lemma below shows that the set of admissible values of any other group $G'=G \cdot \sigma$ will be a subset, not necessarily proper, of $A(G)$.
\begin{Lemma} \label{lemma:nonincreasing}
Let us consider the group $G$ it holds $A(G') \subseteq A(G)$ for any $G' = G \cdot y$.
\end{Lemma}
\begin{proof}
The proof is by induction on $y$:
\begin{itemize}
\item {\bf Base case.} $G'=G\cdot \epsilon$: It is immediate by observing that $G'=G$. Thus $A(G)=A(G')$.
\item {\bf Inductive case.} We assume the above is true up to $G'=\sigma$, in the inductive step we have to show that it holds for the two possible extensions of $\sigma$.
Case (1): $G'=\sigma \cdot m$, in such a case the set of admissible values does not change. Thus $A(\sigma)=A(G')$. Case (2): $G'=\sigma \cdot s$, suppose that a value $v$
is in $A(G')$ but not in $A(\sigma)$. In order to be admissible for $G'$ there must exist a proof for each prefix of $G'$ ending with an $s$, this is by construction also an admissibility proof for $\sigma$. Therefore $A(G') \subseteq A(\sigma)$.
\end{itemize}
\end{proof}
\begin{definition}
Given an epoch $ep$ we define as $W_{ep}$ the set of values that have been gradecasted by a process in epoch $ep$ and that have been seen by all correct processes with gradecast rank at least $1$.
With $W_{ep}(G)$ we indicate the subset of $W_{ep}$ that was sent by processes claiming to belong to group $G$.
\end{definition}
\begin{Lemma} \label{lemma:masterdominate}
Consider a correct master $p_i$ in the group $G=\sigma \cdot m$. If $p_i$ decides, its decision, let it be $dec_i$, is comparable with the one of any correct slave in $G'= \sigma \cdot s$.
\end{Lemma}
\begin{proof}
To prove the above it is sufficient to show that each value $v$ in $A(G')$ is included in $dec_i$. In order for a value to be in $A(G')$ it must exist an admissibility proof for group $G'$, the admissibility proof implies two things:
\begin{itemize}
\item (1) that $v$ is in $W_{|\sigma|}(\sigma)$ (see Observation \ref{obs:byz});
\item (2) that $v$ is in $A(\sigma)$.
\end{itemize}
From the above, and the code of a correct master $p_i$, $v$ will be in the set $V_i$ of Line \ref{alg:masterset} at epoch $|\sigma|$. Since a master never removes a value from its $pro_i$ its decision $dec_i$ must include $v$.
This complete the proof since each correct slave $p$ in $G'$ never put in its proposal a value that is not in $A(G')$: by Lemma \ref{lemma:nonincreasing} the set of values that a slave will consider in any possible future execution of Line \ref{alg:masterset} is
included in $A(G')$.
\end{proof}
\begin{Lemma}
Let $p$ be a correct process that decides $dec_i$. Its decision respects Inclusivity and Non-Triviality.
\end{Lemma}
\begin{proof}
The inclusivity is immediate from the fact that its proposal $pro_i$ will be admissible for any epoch, thus, in each epoch, it will always be included in its set $V_i$.
It remains to show that $dec_i \leq \bigoplus (X \cup B)$. The non-trivial part is to show the bound of $f$ on the set $B$.
We will show by induction on epoch number $ep$ that the sets $B_{ep}$ of admissible values issuable by Byzantines are such that $B_{ep+1} \subseteq B_{ep}$ and that $|B_{0}|=f$.
Note that, $B_{ep+1} \subseteq B_{ep}$ is implied by Lemma \ref{lemma:nonincreasing}.
The bound on $|B_{0}|=f$ derives directly from the gradecast property: a single Byzantine cannot gradecast two different elements of $E$ in epoch $ep=0$.
\end{proof}
\begin{Lemma} \label{lemma:onceisforever}
Let $p$ be a correct process that at some epoch $ep \leq \log(n)+1$ updates its proposal $pro_i$ with a new value $w$. For any possible $G$ such that $p \in G$ and $\log(n)+1 \geq |G| \geq ep$ we have $w \in A(G)$.
\end{Lemma}
\begin{proof}
The proof is by induction:
\begin{itemize}
\item Base case: $|G|=ep$: in this case $p$ puts $w$ in its proposal when it belongs to group $G$, since $p$ is correct it does so if and only if $w \in A(G)$.
\item Inductive step: Let us suppose that it holds for $G$, we will show that it holds also for the two possible extensions $G'=G \cdot s$ and $G''=G \cdot m$. For $G'$ we have that
process $p$ being correct it updates the admissibility proof for $w$ (line \ref{alg2:updateproof}). This implies that $w \in A(G')$. Considering $G''$ we have that, by construction of the admissibility proof, $A(G'')=A(G)$ and thus $w \in A(G'')$ (by inductive hypothesis $w \in A(G)$).
\end{itemize}
\end{proof}
\begin{Lemma} \label{lemma:expdecreasing}
For any correct process $p_i \in G$ with $G \neq \epsilon$ we have $|A(G)| -| pro_i| \leq n/2^{|G|-1}$. Moreover, given process $p_i \in G$ we have $|A(G)| \leq t_u$ and $pro_i \geq t_d$, where $t_u$ and $t_d$ are the thresholds of group $G$.
\end{Lemma}
\begin{proof}
We will show that for each $p_i \in G$ it holds that $
|A(G)| \leq t_u$, that $|pro_i| \geq t_d$, and, that $ t_u - t_d\leq n/2^{|G|-1}$. Recall that $t_u,t_d$ and $t_m$ depend on $G$. The proof is by induction on $G$.
\begin{itemize}
\item Base case: $G=s$: it derives immediately from the structure of the lattice and the fact that $t_d=0$ and $t_u=n$ for all processes in $G$.
\item Inductive case: By inductive hypothesis the claim holds for $\sigma$. We have $G=\sigma \cdot m$ or $G=\sigma \cdot s$. Let $t_u,t_m,t_d$ be the thresholds of group $\sigma$, by inductive hypothesis we have $
|A(\sigma)| \leq t_u$ that $|pro_i| \geq t_d$ and that $ t_u - t_d\leq n/2^{|G|-1}$.
\begin{itemize}
\item Case $G= \sigma \cdot m$. First notice that for master processes the set of admissible values does not change, that is $A(\sigma)=A(G)$ neither the threshold $t_u$. In such a case we will show that the lattice ``shrinks from below" in the sense
that by updating the lower bound on $pro_i$ our claim holds. Since $p_i$ is in $G$ we have that it updates $pro_i$ and the new $pro_i$ contains a number of elements that are at least $t_{d}+\lfloor \frac{t_{u}-t_{d}}{2} \rfloor+1$, thus the new $t'_{d}=t_d+ \lfloor \frac{t_{u}-t_{d}}{2} \rfloor+1$.
By immediate algebraic manipulations we have $t_u - t'_d \leq \frac{t_u-t_d}{2}$ that proves our claim.
\item Case $G= \sigma \cdot s$. In such a case we will show that the lattice ``shrinks from above". Consider the generic $p_i \in G$ this implies that at the end of epoch $|\sigma|$ the set $V_i$ contained at most $t_m$ elements. It is immediate that
since $p_i$ is correct each value in $A(\sigma)$ that is not in $V_i$ cannot be admissible in the extension $G$. Therefore we have $|A(G)| \leq t_m $, now it remains to show that $|pro_i| \geq t_d$ but this is immediate from inductive hypothesis.
Thus we have $t_u'=t_m$ and $t'_d=t_d$, that implies $t_u'-t_d' \leq t_m - t_d\leq \frac{t_u-t_d}{2}$.
\end{itemize}
\end{itemize}
\end{proof}
\begin{observation}
Any correct process decides at epoch $\log(n)+1$.
\end{observation}
\begin{Lemma}
Given any pair $p_i,p_j$ of processes in $C$ their decision $dec_i$ and $dec_j$ are comparable.
\end{Lemma}
\begin{proof}
Let $G_i$ be the group where $p_i$ belongs at the end of epoch $r=\log(n)+1$, and let $G_j$ the analogous for $p_j$.
If $G_i$ and $G_j$ share a common prefix $\sigma$ such that $\sigma \cdot m$ is a prefix of $G_i$ and $\sigma \cdot s$ is a prefix of $G_j$, then Lemma \ref{lemma:masterdominate} shows the comparability. \\
The only case when the above (or the symmetric of the above) does not hold is if $G=G_i=G_j$.
In this case, we will show that $dec_i=dec_j$. Suppose the contrary, then we have that there exists at least a value $v \in dec_i$ and such that $v \not\in dec_j$. By Lemma \ref{lemma:onceisforever} we have $v \in A(G)$, and by Lemma \ref{lemma:nonincreasing} $v$ is in each prefixes of $G$.
Suppose $p_i$ inserted $v$ in its proposal in an epoch $ep$ such that it has been master again in epoch $ep' > ep$, however this implies that also $p_j$ is master in $ep'$, and $p_i$ being correct in epoch $ep'$ it gradecasts $v$ that will be included in the proposal of $p_j$.
The above implies that $v$ has been received and inserted by $p_i$ in its proposal exactly in the last epoch in which $p_i$ became master.
Let $ep_{last}$ be such an epoch. Note that if $t_u-t_{d} \leq 2$ then $v$ is also in the proposal of $p_j$ (both processes enter in the master group in $ep_{last}$): in case $t_u-t_d=2$ to become master each process has to collect enough values to trespass the threshold $t_m$, recall that $t_m=t_d+1$ and thus it has to collect $t_u$ values, but those and are all the admissible values (by Lemma \ref{lemma:expdecreasing}). In the other case, when $t_u-t_d=1$, a process has also to collect all admissible values ($t_m=t_d$ and $t_u=t_m+1$) (by Lemma \ref{lemma:expdecreasing}).
Therefore, $t_u -t_{d} > 2$ in $ep_{last}$, however by the structure of the algorithm and the number of epochs being $\log(n)+1$ we eventually have an epoch $ep' > ep_{last}$ such that $t_u-t_{d} =1$ and $t_m=t_d$, when this happens process $p_j$ will become master upon receipt of $v$ from $p_i$ (by Lemma \ref{lemma:onceisforever} $v$ is admissible for $p_j$). This contradicts the fact that $G_i =G_j$, since $p_i$ is never again a master after epoch $ep_{last}$.
\end{proof}
\noindent
From previous lemmas we have:
\begin{theorem}
Given the lattice $L$ constituted by the power set of the first $n$ natural numbers with union as join operation, GAC is a correct LA algorithm on $L$, that terminates in ${\cal O}(\log(n))$ rounds and tolerates up to $n/3-1$ Byzantine processes.
\end{theorem}
\section{Adapting GAC to work on arbitrary semi-lattices in $\log(f)$ rounds}\label{sec:general}
We first explain how to adapt the GAC algorithm to work in $\log(f)$ on $L_{n}$ when each correct proposes a different unique value in $\{\{1\},\{2\},\ldots, \{n\}\}$. We call such an algorithm GAC$_{fast}$, we then discuss how to adapt GAC$_{fast}$ to work on a generic join semi-lattice dropping the assumption of different proposal values.
The main idea is to modify epoch $0$ to satisfy two needs: (1) to force Byzantines to commit to a certain value; (2) to make all processes collect at least $n-f$ different proposal values.
This allows the thresholds to be set to $t_d=n-f, t_u=n, t_m=\lfloor n-\frac{f}{2} \rfloor$ in all processes at the end of epoch $0$. The modified epoch $0$ is Algorithm \ref{alg:epoch0}.
The code follows the old one with the notable exceptions that: a correct process updates its proposal by including all values, in $E$, that have been seen with rank at least $2$, and it updates its thresholds accordingly.
\begin{algorithm*}
\caption{ {\bf GAC}$_{fast}$: Collect and Commit Epoch $0$ - Algorithm for process $p_i$} \label{alg:epoch0}
\footnotesize
\begin{algorithmic}[1]
\algrenewcommand\algorithmicprocedure{\textbf{function}}
\State $G=\epsilon$
\State $pro_i$ \Comment{proposed value.}
\State $Proofs=\{\}$
\smallskip
\Procedure{LA-Propose}{$pro_i$}
\State $pro_i=pro_i$
\State {\sc gradecast}$(M=(pro_i,G,\bot))$ \Comment{Epoch 0 - start}
\State $P_i=${\sc rcv}()
\State $V_i = \{\forall v \in M | M \in P_i \land M$ rank is equal 2 $\land v \in E\}$
\State $G=G \cdot s$
\State $pro_i= V_i$
\State $updateproofs(pro_i,P_i,0)$ \Comment{Epoch 0 - end}
\State $t_d=n-f,t_u=n, t_{m}=t_d+\lfloor \frac{t_u-t_d}{2} \rfloor$
\State $\ldots$ \Comment{Remain as Algorithm \ref{gac:algorithm} but for line \ref{alg:forcycle}}
\EndProcedure
\end{algorithmic}
\end{algorithm*}
The remaining of the algorithm is the same as Algorithm \ref{gac:algorithm} but for line \ref{alg:forcycle} where we have $ep \in [1 \ldots, \log(f)+1]$.
\paragraph{Correctness discussion}
The same lemmas and observations of Section \ref{gac:correctness} hold with the following exceptions:
\begin{Lemma} \label{lemma:expdecreasingf}
For any correct process $p_i \in G$ with $G \neq \epsilon$ we have $|A(G)| -| pro_i| \leq f/2^{|G|-1}$. Moreover, given process $p_i \in G$ we have $|A(G)| \leq t_u$ and $pro_i \geq t_d$, where $t_u$ and $t_d$ are the thresholds of group $G$.
\end{Lemma}
\begin{proof}
We will show that for each $p_i \in G$ it holds $
|A(G)| \leq t_u$ that $|pro_i| \geq t_d$ and that $ t_u - t_d\leq f/2^{|G|-1}$. Recall that $t_u,t_d$ and $t_m$ depends on $G$. The proof is by induction on $G$.
\begin{itemize}
\item Base case: $G=s$: it derives immediately from the structure of epoch $0$, the property of the gradecast, the fact that there are at least $n-f$ correct processes. At the end of epoch $0$ we have $t_d=n-f$ and $t_u=n$ for all processes in $G$.
\item Inductive case: By inductive hypothesis the claim holds for $\sigma$. We have $G=\sigma \cdot m/(\sigma \cdot s)$. Let $t_u,t_m,t_d$ be the thresholds of group $\sigma$, by ind. hypothesis we have $
|A(\sigma)| \leq t_u$ that $|pro_i| \geq t_d$ and that $ t_u - t_d\leq f/2^{|G|-1}$.
\begin{itemize}
\item Case $G= \sigma \cdot m$. First notice that for master processes the set of admissible values does not change, that is $A(\sigma)=A(G)$ neither the threshold $t_u$. In such a case we will show that the lattice ``shrinks from below" in the sense
that by updating the lower bound on $pro_i$ our claim holds. Since $p_i$ is in $G$ we have that it updates $pro_i$ and the new $pro_i$ contains a number of elements that are at least $t_{d}+\lfloor \frac{t_{u}-t_{d}}{2} \rfloor+1$, thus the new $t'_{d}=t_d+ \lfloor \frac{t_{u}-t_{d}}{2} \rfloor+1$.
By immediate algebraic manipulations we have $t_u - t'_d \leq \frac{t_u-t_d}{2}$ that proves our claim.
\item Case $G= \sigma \cdot s$. In such a case we will show that the lattice ``shrinks from above". Consider the generic $p_i \in G$ this implies that at the end of epoch $|\sigma|$ the set $V_i$ contained at most $t_m$ elements. It is immediate that
since $p_i$ is correct each value in $A(\sigma)$ that is not in $V_i$ cannot be admissible in the extension $G$. Therefore we have $|A(G)| \leq t_m $, now it remains to show that $|pro_i| \geq t_d$ but this is immediate from inductive hypothesis.
Thus we have $t_u'=t_m$ and $t'_d=t_d$, that implies $t_u'-t_d' \leq t_m - t_d\leq \frac{t_u-t_d}{2}$.
\end{itemize}
\end{itemize}
\end{proof}
\begin{observation}
Any correct process decides at epoch $\log(f)+1$.
\end{observation}
\begin{theorem}
Given the lattice $L$ constituted by the power set of the first $n$ natural numbers with union as join operation and where each correct process proposes a distinct element in $\{\{1\},\ldots, \{n\}\}$, GAC$_{fast}$ is a correct LA that terminates in ${\cal O}(\log(f))$ rounds and tolerates up to $n/3-1$ Byzantine processes.
\end{theorem}
\subsection{Arbitrary Semi-lattices $L_{A}$}
We adapt GAC$_{fast}$ to work on an arbitrary join semi-lattice $L_{A}=(V_{A},\oplus)$, an arbitrary set $E_{A} \subseteq V_{A}$ of allowed proposal values, and an arbitrary mapping of proposal values and correct processes (recall that in previous section we were assuming a different proposal value for each correct). The adaptation works by running GAC$_{fast}$ on an intermediate semi-lattice $L^{*}$. Lattice $L^{*}$ is the one induced by the union operation over the power set of $V^{*}=\Pi \times E_{A}$. The set $V^{*}$ is constituted by all possible pairs process ID and initial proposed value $pro_i$ (each $pro_i$ is in $E_{A}$). Each correct process $p_i$ starts GAC$_{fast}$ with input $(p_i, pro_i)$.
\begin{algorithm*}
\caption{ ${\cal O}(\log f)$ Lattice Agreement Algorithm for arbitrary join semi-lattice - Algorithm for process $p_i$} \label{alg:final}
\footnotesize
\begin{algorithmic}[1]
\algrenewcommand\algorithmicprocedure{\textbf{function}}
\medskip
\Procedure{\bf LA-Propose}{$pro_i$}
\State Trigger GAC$_{fast}$ with proposal value $(p_i,pro_i)$
\State wait until GAC$_{fast}$ terminates with output $dec_i$
\State $X:\{y | (x,y) \in dec_i \land y \in E_{A}\}$
\State $dec'_i=\bigoplus X$
\State {\bf return} $dec'_i$
\EndProcedure
\end{algorithmic}
\end{algorithm*}
Note that epoch $0$, with is commitment functionality, forces the algorithm to effectively decides on a lattice $L^{*}$ that is the power set of a subset $X$ of $V^{*}$ of cardinality at most $n$ and at least $n-f$.
Such lattice is isomorphic to the lattice on which the correctness of GAC$_{fast}$ has been shown in Section \ref{sec:general}.
Once GAC$_{fast}$ terminates each process $p_i \in C$ has a decision $dec_i$. This decision $dec_i$ is a set $\{ (p_i,val_{i}), \ldots, (p_x,val_{x}) \}$, from such a set the process $p_i$ obtains a decision $dec'_i$ on $L_{A}$ where $dec'_i$ is $dec'_i = \bigoplus D_i$ given $D_i: \{y | \forall (x,y) \in dec_i \land y \in E_{A}\}$. This strategy enforces that the decisions $dec'_i$ and $dec'_j$ of any two correct processes $p_i,p_j$ are comparable points on the semi-lattice $L_{A}$. It is also immediate that non-triviality and inclusivity hold. The formal pseudocode of the reduction is Algorithm \ref{alg:final}.
From the above we have:
\begin{theorem} Given $f$ Byzantine processes and $n$ processes in total,
if $n \geq 3f+1$, there exists a Byzantine lattice agreement algorithm terminating in ${\cal O}(\log f)$ rounds in the authenticated message model.
\end{theorem}
\subsubsection{Message Complexity}
The provable gradecast generates at most ${\cal O}(n^2)$ messages at each round. Each epoch is composed by $3$ rounds and in each epoch all correct processes do a gradecast, thus we have a total of ${\cal O}(n^3 \log f)$ messages. However, as noted also in \cite{Zheng:aa}, it is possible to use ${\cal O}(n^2)$ messages in total to run $n$ parallel instances of gradecast (each message will be structured with $n$ locations one for each possible gradecaster). Therefore, our algorithm can be implemented using ${\cal O}(n^2 \log f)$ messages.
\section{Trade-off Between Signatures and Number of Processes}\label{sec:nosig}
Signatures are used to implement the seen-all proof of our provable gradecast primitive (explained in Section \ref{pgc:sec}). By assuming $n \geq 4f+1$ processes we may implement an interactive version of the seen-all proof that does not use signatures. We explain the interactive provable gradecast algorithm when $n=4f+1$, the extension for $n > 4f+1$ is immediate. The modifications with respect to Section \ref{pgc:sec} are as follows:
\begin{itemize}
\item The threshold of line \ref{pgc:freqr1} remains $f+1$, the one of line \ref{pgc:freqr2} becomes $3f+1$, the threshold of line \ref{pgc:relay} is $3f+1$. Therefore, a message will have rank $1$ if seen with multiplicity at least $f+1$, a message has rank $2$ if seen with multiplicity at least $3f+1$.
\item The seen-all proof $S_{p_s,m_i}$ is simply a set of IDs. These IDs are the ones of processes from which $p_i$ receives $m_i$ in the receive phase of round 3.
\end{itemize}
The seen-all proof $S_{p_s,m_i}$ is checked in an interactive way by querying each process contained in the set $S_{p_s,m_i}$. The proof passes if at least $2f+1$ of such processes confirm to have relayed $m_i$ in the relevant gradecast instance. It is obvious that by increasing the two thresholds we do not affect the original properties of the gradecast (discussed in Section \ref{pgc:sec}).
\begin{observation}\label{obs:byznosig}
Consider an instance of interactive provable gradecast with source $p_s$.
If a process $p_i$, whether Byzantine or correct, can produce a passable {\em seen-all} proof for a message $m$, then each process $p \in C$ delivered message $m$ with rank at least $1$ at the end of the gradecast instance.
\end{observation}
\begin{proof} In the interactive provable gradecast if a proof produced by process $p_i$ is passable, then it implies that at least a set of $f+1$ correct processes have relayed message $m$ to $p_i$ at the end of round $3$ of the gradecast. Being these processes correct they will relay $m$ to all other processes in the system. Therefore, any other process has received $m$ with frequency at least $f+1$ at the end of round $3$. This implies that each correct process delivers $m$ with rank at least $1$. \end{proof}
Moreover, we have the following observation:
\begin{observation}\label{obs:byzcorrnosig}
Consider an instance of interactive provable gradecast with source $p_s$.
Given a correct process $p_i$, if $p_i$ receives a message $m$ with rank $2$ then the corresponding proof is passable.
\end{observation}
Using the interactive provable gradecast and the straightforward interactive variant of the admissibility proof we have:
\begin{theorem} Given $f$ Byzantine processes and $n$ processes in total,
if $n \geq 4f+1$, then there exists a byzantine lattice agreement algorithm terminating in ${\cal O}(\log f)$ rounds in the authenticated channel model.
\end{theorem}
\paragraph{Message Complexity.}
We argue that the interactive provable gradecast generates at most ${\cal O}(n^2)$ messages at each round. The additional cost introduced by the interactive proof is at most ${\cal O}(n)$ per round: this is the number of messages that a process has to send to verify the admissibility of all values inside a set (we have to send at most one query message to each other process independently of how many values we want to check). Thus the total asymptotic cost remains the same: ${\cal O}(n^2 \log f)$ messages.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 6,100
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{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 2,423
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{
"redpajama_set_name": "RedPajamaStackExchange"
}
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define('echarts/chart/tree', [
'require',
'./base',
'../util/shape/Icon',
'zrender/shape/Image',
'zrender/shape/Line',
'zrender/shape/BezierCurve',
'../layout/Tree',
'../data/Tree',
'../config',
'../util/ecData',
'zrender/config',
'zrender/tool/event',
'zrender/tool/util',
'../chart'
], function (require) {
var ChartBase = require('./base');
var GOLDEN_SECTION = 0.618;
var IconShape = require('../util/shape/Icon');
var ImageShape = require('zrender/shape/Image');
var LineShape = require('zrender/shape/Line');
var BezierCurveShape = require('zrender/shape/BezierCurve');
var TreeLayout = require('../layout/Tree');
var TreeData = require('../data/Tree');
var ecConfig = require('../config');
ecConfig.tree = {
zlevel: 1,
z: 2,
calculable: false,
clickable: true,
rootLocation: {},
orient: 'vertical',
symbol: 'circle',
symbolSize: 20,
nodePadding: 30,
layerPadding: 100,
itemStyle: {
normal: {
label: { show: true },
lineStyle: {
width: 1,
color: '#777',
type: 'curve'
}
},
emphasis: {}
}
};
var ecData = require('../util/ecData');
var zrConfig = require('zrender/config');
var zrEvent = require('zrender/tool/event');
var zrUtil = require('zrender/tool/util');
function Tree(ecTheme, messageCenter, zr, option, myChart) {
ChartBase.call(this, ecTheme, messageCenter, zr, option, myChart);
this.refresh(option);
}
Tree.prototype = {
type: ecConfig.CHART_TYPE_TREE,
_buildShape: function (series, seriesIndex) {
var data = series.data[0];
this.tree = TreeData.fromOptionData(data.name, data.children);
this.tree.root.data = data;
this._setTreeShape(series);
this.tree.traverse(function (treeNode) {
this._buildItem(treeNode, series, seriesIndex);
if (treeNode.children.length > 0) {
this._buildLink(treeNode, series);
}
}, this);
var panable = series.roam === true || series.roam === 'move';
var zoomable = series.roam === true || series.roam === 'scale';
this.zr.modLayer(this.getZlevelBase(), {
panable: panable,
zoomable: zoomable
});
if (this.query('markPoint.effect.show') || this.query('markLine.effect.show')) {
this.zr.modLayer(ecConfig.EFFECT_ZLEVEL, {
panable: panable,
zoomable: zoomable
});
}
this.addShapeList();
},
_buildItem: function (treeNode, serie, seriesIndex) {
var queryTarget = [
treeNode.data,
serie
];
var symbol = this.deepQuery(queryTarget, 'symbol');
var normal = this.deepMerge(queryTarget, 'itemStyle.normal') || {};
var emphasis = this.deepMerge(queryTarget, 'itemStyle.emphasis') || {};
var normalColor = normal.color || this.zr.getColor();
var emphasisColor = emphasis.color || this.zr.getColor();
var angle = -treeNode.layout.angle || 0;
if (treeNode.id === this.tree.root.id) {
angle = 0;
}
var textPosition = 'right';
if (Math.abs(angle) >= Math.PI / 2 && Math.abs(angle) < Math.PI * 3 / 2) {
angle += Math.PI;
textPosition = 'left';
}
var rotation = [
angle,
treeNode.layout.position[0],
treeNode.layout.position[1]
];
var shape = new IconShape({
zlevel: this.getZlevelBase(),
z: this.getZBase() + 1,
rotation: rotation,
style: {
x: treeNode.layout.position[0] - treeNode.layout.width * 0.5,
y: treeNode.layout.position[1] - treeNode.layout.height * 0.5,
width: treeNode.layout.width,
height: treeNode.layout.height,
iconType: symbol,
color: normalColor,
brushType: 'both',
lineWidth: normal.borderWidth,
strokeColor: normal.borderColor
},
highlightStyle: {
color: emphasisColor,
lineWidth: emphasis.borderWidth,
strokeColor: emphasis.borderColor
}
});
if (shape.style.iconType.match('image')) {
shape.style.image = shape.style.iconType.replace(new RegExp('^image:\\/\\/'), '');
shape = new ImageShape({
rotation: rotation,
style: shape.style,
highlightStyle: shape.highlightStyle,
clickable: shape.clickable,
zlevel: this.getZlevelBase(),
z: this.getZBase()
});
}
if (this.deepQuery(queryTarget, 'itemStyle.normal.label.show')) {
shape.style.text = treeNode.data.label == null ? treeNode.id : treeNode.data.label;
shape.style.textPosition = this.deepQuery(queryTarget, 'itemStyle.normal.label.position');
if (serie.orient === 'radial' && shape.style.textPosition !== 'inside') {
shape.style.textPosition = textPosition;
}
shape.style.textColor = this.deepQuery(queryTarget, 'itemStyle.normal.label.textStyle.color');
shape.style.textFont = this.getFont(this.deepQuery(queryTarget, 'itemStyle.normal.label.textStyle') || {});
}
if (this.deepQuery(queryTarget, 'itemStyle.emphasis.label.show')) {
shape.highlightStyle.textPosition = this.deepQuery(queryTarget, 'itemStyle.emphasis.label.position');
shape.highlightStyle.textColor = this.deepQuery(queryTarget, 'itemStyle.emphasis.label.textStyle.color');
shape.highlightStyle.textFont = this.getFont(this.deepQuery(queryTarget, 'itemStyle.emphasis.label.textStyle') || {});
}
ecData.pack(shape, serie, seriesIndex, treeNode.data, 0, treeNode.id);
this.shapeList.push(shape);
},
_buildLink: function (parentNode, serie) {
var lineStyle = serie.itemStyle.normal.lineStyle;
if (lineStyle.type === 'broken') {
this._buildBrokenLine(parentNode, lineStyle, serie);
return;
}
for (var i = 0; i < parentNode.children.length; i++) {
var xStart = parentNode.layout.position[0];
var yStart = parentNode.layout.position[1];
var xEnd = parentNode.children[i].layout.position[0];
var yEnd = parentNode.children[i].layout.position[1];
switch (lineStyle.type) {
case 'curve':
this._buildBezierCurve(parentNode, parentNode.children[i], lineStyle, serie);
break;
case 'broken':
break;
default:
var shape = this._getLine(xStart, yStart, xEnd, yEnd, lineStyle);
this.shapeList.push(shape);
}
}
},
_buildBrokenLine: function (parentNode, lineStyle, serie) {
var solidLineStyle = zrUtil.clone(lineStyle);
solidLineStyle.type = 'solid';
var shapes = [];
var xStart = parentNode.layout.position[0];
var yStart = parentNode.layout.position[1];
var orient = serie.orient;
var yEnd = parentNode.children[0].layout.position[1];
var xMiddle = xStart;
var yMiddle = yStart + (yEnd - yStart) * (1 - GOLDEN_SECTION);
var xMiddleStart = parentNode.children[0].layout.position[0];
var yMiddleStart = yMiddle;
var xMiddleEnd = parentNode.children[parentNode.children.length - 1].layout.position[0];
var yMiddleEnd = yMiddle;
if (orient === 'horizontal') {
var xEnd = parentNode.children[0].layout.position[0];
xMiddle = xStart + (xEnd - xStart) * (1 - GOLDEN_SECTION);
yMiddle = yStart;
xMiddleStart = xMiddle;
yMiddleStart = parentNode.children[0].layout.position[1];
xMiddleEnd = xMiddle;
yMiddleEnd = parentNode.children[parentNode.children.length - 1].layout.position[1];
}
shapes.push(this._getLine(xStart, yStart, xMiddle, yMiddle, solidLineStyle));
shapes.push(this._getLine(xMiddleStart, yMiddleStart, xMiddleEnd, yMiddleEnd, solidLineStyle));
for (var i = 0; i < parentNode.children.length; i++) {
xEnd = parentNode.children[i].layout.position[0];
yEnd = parentNode.children[i].layout.position[1];
if (orient === 'horizontal') {
yMiddleStart = yEnd;
} else {
xMiddleStart = xEnd;
}
shapes.push(this._getLine(xMiddleStart, yMiddleStart, xEnd, yEnd, solidLineStyle));
}
this.shapeList = this.shapeList.concat(shapes);
},
_getLine: function (xStart, yStart, xEnd, yEnd, lineStyle) {
if (xStart === xEnd) {
xStart = xEnd = this.subPixelOptimize(xStart, lineStyle.width);
}
if (yStart === yEnd) {
yStart = yEnd = this.subPixelOptimize(yStart, lineStyle.width);
}
return new LineShape({
zlevel: this.getZlevelBase(),
hoverable: false,
style: zrUtil.merge({
xStart: xStart,
yStart: yStart,
xEnd: xEnd,
yEnd: yEnd,
lineType: lineStyle.type,
strokeColor: lineStyle.color,
lineWidth: lineStyle.width
}, lineStyle, true)
});
},
_buildBezierCurve: function (parentNode, treeNode, lineStyle, serie) {
var offsetRatio = GOLDEN_SECTION;
var orient = serie.orient;
var xStart = parentNode.layout.position[0];
var yStart = parentNode.layout.position[1];
var xEnd = treeNode.layout.position[0];
var yEnd = treeNode.layout.position[1];
var cpX1 = xStart;
var cpY1 = (yEnd - yStart) * offsetRatio + yStart;
var cpX2 = xEnd;
var cpY2 = (yEnd - yStart) * (1 - offsetRatio) + yStart;
if (orient === 'horizontal') {
cpX1 = (xEnd - xStart) * offsetRatio + xStart;
cpY1 = yStart;
cpX2 = (xEnd - xStart) * (1 - offsetRatio) + xStart;
cpY2 = yEnd;
} else if (orient === 'radial') {
if (parentNode.id === this.tree.root.id) {
cpX1 = (xEnd - xStart) * offsetRatio + xStart;
cpY1 = (yEnd - yStart) * offsetRatio + yStart;
cpX2 = (xEnd - xStart) * (1 - offsetRatio) + xStart;
cpY2 = (yEnd - yStart) * (1 - offsetRatio) + yStart;
} else {
var xStartOrigin = parentNode.layout.originPosition[0];
var yStartOrigin = parentNode.layout.originPosition[1];
var xEndOrigin = treeNode.layout.originPosition[0];
var yEndOrigin = treeNode.layout.originPosition[1];
var rootX = this.tree.root.layout.position[0];
var rootY = this.tree.root.layout.position[1];
cpX1 = xStartOrigin;
cpY1 = (yEndOrigin - yStartOrigin) * offsetRatio + yStartOrigin;
cpX2 = xEndOrigin;
cpY2 = (yEndOrigin - yStartOrigin) * (1 - offsetRatio) + yStartOrigin;
var rad = (cpX1 - this.minX) / this.width * Math.PI * 2;
cpX1 = cpY1 * Math.cos(rad) + rootX;
cpY1 = cpY1 * Math.sin(rad) + rootY;
rad = (cpX2 - this.minX) / this.width * Math.PI * 2;
cpX2 = cpY2 * Math.cos(rad) + rootX;
cpY2 = cpY2 * Math.sin(rad) + rootY;
}
}
var shape = new BezierCurveShape({
zlevel: this.getZlevelBase(),
hoverable: false,
style: zrUtil.merge({
xStart: xStart,
yStart: yStart,
cpX1: cpX1,
cpY1: cpY1,
cpX2: cpX2,
cpY2: cpY2,
xEnd: xEnd,
yEnd: yEnd,
strokeColor: lineStyle.color,
lineWidth: lineStyle.width
}, lineStyle, true)
});
this.shapeList.push(shape);
},
_setTreeShape: function (serie) {
var treeLayout = new TreeLayout({
nodePadding: serie.nodePadding,
layerPadding: serie.layerPadding
});
this.tree.traverse(function (treeNode) {
var queryTarget = [
treeNode.data,
serie
];
var symbolSize = this.deepQuery(queryTarget, 'symbolSize');
if (typeof symbolSize === 'number') {
symbolSize = [
symbolSize,
symbolSize
];
}
treeNode.layout = {
width: symbolSize[0],
height: symbolSize[1]
};
}, this);
treeLayout.run(this.tree);
var orient = serie.orient;
var rootX = serie.rootLocation.x;
var rootY = serie.rootLocation.y;
var zrWidth = this.zr.getWidth();
var zrHeight = this.zr.getHeight();
if (rootX === 'center') {
rootX = zrWidth * 0.5;
} else {
rootX = this.parsePercent(rootX, zrWidth);
}
if (rootY === 'center') {
rootY = zrHeight * 0.5;
} else {
rootY = this.parsePercent(rootY, zrHeight);
}
rootY = this.parsePercent(rootY, zrHeight);
if (orient === 'horizontal') {
rootX = isNaN(rootX) ? 10 : rootX;
rootY = isNaN(rootY) ? zrHeight * 0.5 : rootY;
}
if (orient === 'radial') {
rootX = isNaN(rootX) ? zrWidth * 0.5 : rootX;
rootY = isNaN(rootY) ? zrHeight * 0.5 : rootY;
} else {
rootX = isNaN(rootX) ? zrWidth * 0.5 : rootX;
rootY = isNaN(rootY) ? 10 : rootY;
}
var originRootX = this.tree.root.layout.position[0];
if (orient === 'radial') {
var minX = Infinity;
var maxX = 0;
var maxWidth = 0;
this.tree.traverse(function (treeNode) {
maxX = Math.max(maxX, treeNode.layout.position[0]);
minX = Math.min(minX, treeNode.layout.position[0]);
maxWidth = Math.max(maxWidth, treeNode.layout.width);
});
this.width = maxX - minX + 2 * maxWidth;
this.minX = minX;
}
this.tree.traverse(function (treeNode) {
var x = treeNode.layout.position[0] - originRootX + rootX;
var y = treeNode.layout.position[1] + rootY;
if (orient === 'horizontal') {
y = treeNode.layout.position[0] - originRootX + rootY;
x = treeNode.layout.position[1] + rootX;
}
if (orient === 'radial') {
x = treeNode.layout.position[0];
y = treeNode.layout.position[1];
treeNode.layout.originPosition = [
x,
y
];
var r = y;
var angle = (x - minX) / this.width * Math.PI * 2;
x = r * Math.cos(angle) + rootX;
y = r * Math.sin(angle) + rootY;
treeNode.layout.angle = angle;
}
treeNode.layout.position[0] = x;
treeNode.layout.position[1] = y;
}, this);
},
refresh: function (newOption) {
this.clear();
if (newOption) {
this.option = newOption;
this.series = this.option.series;
}
var series = this.series;
var legend = this.component.legend;
for (var i = 0; i < series.length; i++) {
if (series[i].type === ecConfig.CHART_TYPE_TREE) {
series[i] = this.reformOption(series[i]);
var seriesName = series[i].name || '';
this.selectedMap[seriesName] = legend ? legend.isSelected(seriesName) : true;
if (!this.selectedMap[seriesName]) {
continue;
}
this._buildSeries(series[i], i);
}
}
},
_buildSeries: function (series, seriesIndex) {
this._buildShape(series, seriesIndex);
}
};
zrUtil.inherits(Tree, ChartBase);
require('../chart').define('tree', Tree);
return Tree;
});define('echarts/layout/Tree', [
'require',
'zrender/tool/vector'
], function (require) {
var vec2 = require('zrender/tool/vector');
function TreeLayout(opts) {
opts = opts || {};
this.nodePadding = opts.nodePadding || 30;
this.layerPadding = opts.layerPadding || 100;
this._layerOffsets = [];
this._layers = [];
}
TreeLayout.prototype.run = function (tree) {
this._layerOffsets.length = 0;
for (var i = 0; i < tree.root.height + 1; i++) {
this._layerOffsets[i] = 0;
this._layers[i] = [];
}
this._updateNodeXPosition(tree.root);
var root = tree.root;
this._updateNodeYPosition(root, 0, root.layout.height);
};
TreeLayout.prototype._updateNodeXPosition = function (node) {
var minX = Infinity;
var maxX = -Infinity;
node.layout.position = node.layout.position || vec2.create();
for (var i = 0; i < node.children.length; i++) {
var child = node.children[i];
this._updateNodeXPosition(child);
var x = child.layout.position[0];
if (x < minX) {
minX = x;
}
if (x > maxX) {
maxX = x;
}
}
if (node.children.length > 0) {
node.layout.position[0] = (minX + maxX) / 2;
} else {
node.layout.position[0] = 0;
}
var off = this._layerOffsets[node.depth] || 0;
if (off > node.layout.position[0]) {
var shift = off - node.layout.position[0];
this._shiftSubtree(node, shift);
for (var i = node.depth + 1; i < node.height + node.depth; i++) {
this._layerOffsets[i] += shift;
}
}
this._layerOffsets[node.depth] = node.layout.position[0] + node.layout.width + this.nodePadding;
this._layers[node.depth].push(node);
};
TreeLayout.prototype._shiftSubtree = function (root, offset) {
root.layout.position[0] += offset;
for (var i = 0; i < root.children.length; i++) {
this._shiftSubtree(root.children[i], offset);
}
};
TreeLayout.prototype._updateNodeYPosition = function (node, y, prevLayerHeight) {
node.layout.position[1] = y;
var layerHeight = 0;
for (var i = 0; i < node.children.length; i++) {
layerHeight = Math.max(node.children[i].layout.height, layerHeight);
}
var layerPadding = this.layerPadding;
if (typeof layerPadding === 'function') {
layerPadding = layerPadding(node.depth);
}
for (var i = 0; i < node.children.length; i++) {
this._updateNodeYPosition(node.children[i], y + layerPadding + prevLayerHeight, layerHeight);
}
};
return TreeLayout;
});define('echarts/data/Tree', [
'require',
'zrender/tool/util'
], function (require) {
var zrUtil = require('zrender/tool/util');
function TreeNode(id, data) {
this.id = id;
this.depth = 0;
this.height = 0;
this.children = [];
this.parent = null;
this.data = data || null;
}
TreeNode.prototype.add = function (child) {
var children = this.children;
if (child.parent === this) {
return;
}
children.push(child);
child.parent = this;
};
TreeNode.prototype.remove = function (child) {
var children = this.children;
var idx = zrUtil.indexOf(children, child);
if (idx >= 0) {
children.splice(idx, 1);
child.parent = null;
}
};
TreeNode.prototype.traverse = function (cb, context) {
cb.call(context, this);
for (var i = 0; i < this.children.length; i++) {
this.children[i].traverse(cb, context);
}
};
TreeNode.prototype.updateDepthAndHeight = function (depth) {
var height = 0;
this.depth = depth;
for (var i = 0; i < this.children.length; i++) {
var child = this.children[i];
child.updateDepthAndHeight(depth + 1);
if (child.height > height) {
height = child.height;
}
}
this.height = height + 1;
};
TreeNode.prototype.getNodeById = function (id) {
if (this.id === id) {
return this;
}
for (var i = 0; i < this.children.length; i++) {
var res = this.children[i].getNodeById(id);
if (res) {
return res;
}
}
};
function Tree(id) {
this.root = new TreeNode(id);
}
Tree.prototype.traverse = function (cb, context) {
this.root.traverse(cb, context);
};
Tree.prototype.getSubTree = function (id) {
var root = this.getNodeById(id);
if (root) {
var tree = new Tree(root.id);
tree.root = root;
return tree;
}
};
Tree.prototype.getNodeById = function (id) {
return this.root.getNodeById(id);
};
Tree.fromOptionData = function (id, data) {
var tree = new Tree(id);
var rootNode = tree.root;
rootNode.data = {
name: id,
children: data
};
function buildHierarchy(dataNode, parentNode) {
var node = new TreeNode(dataNode.name, dataNode);
parentNode.add(node);
var children = dataNode.children;
if (children) {
for (var i = 0; i < children.length; i++) {
buildHierarchy(children[i], node);
}
}
}
for (var i = 0; i < data.length; i++) {
buildHierarchy(data[i], rootNode);
}
tree.root.updateDepthAndHeight(0);
return tree;
};
Tree.fromGraph = function (graph) {
function buildHierarchy(root) {
var graphNode = graph.getNodeById(root.id);
for (var i = 0; i < graphNode.outEdges.length; i++) {
var edge = graphNode.outEdges[i];
var childTreeNode = treeNodesMap[edge.node2.id];
root.children.push(childTreeNode);
buildHierarchy(childTreeNode);
}
}
var treeMap = {};
var treeNodesMap = {};
for (var i = 0; i < graph.nodes.length; i++) {
var node = graph.nodes[i];
var treeNode;
if (node.inDegree() === 0) {
treeMap[node.id] = new Tree(node.id);
treeNode = treeMap[node.id].root;
} else {
treeNode = new TreeNode(node.id);
}
treeNode.data = node.data;
treeNodesMap[node.id] = treeNode;
}
var treeList = [];
for (var id in treeMap) {
buildHierarchy(treeMap[id].root);
treeMap[id].root.updateDepthAndHeight(0);
treeList.push(treeMap[id]);
}
return treeList;
};
return Tree;
});
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 4,304
|
Discovery – amerykański prom kosmiczny
Discovery – amerykański kosmiczny program badawczy
Discovery – nadawca telewizyjny
Muzyka
Discovery – album muzyczny Electric Light Orchestra z 1979
Discovery – album muzyczny Mike'a Oldfielda z 1984 roku
Discovery – album muzyczny Daft Punk z 2001
Discovery – box set Pink Floyd z 2011
Stacje telewizyjne i producenci telewizyjni
Discovery Polska – grupa stacji telewizyjnych
Discovery Channel – ogólno-dokumentalna
Animal Planet – przyrodnicza
Discovery Travel and Living – turystyczna
Discovery Science – naukowo-techniczna
Discovery HD – ogólno-dokumentalna w rozdzielczości HDTV
Discovery Historia – historyczna
Discovery World – historyczna (do 18 kwietnia 2008 nosił nazwę Discovery Civilisation)
Discovery Home & Health – rozrywkowa
Discovery Real Time – rozrywkowa
Discovery Real Time Extra – rozrywkowa
Discovery Wings – militarna
Discovery Kids – dla dzieci
Discovery Turbo – motoryzacyjna
Planet Green – rozrywkowo-ekologiczna
Investigation Discovery – śledcza
Military Channel – militarna
DMAX
FitTV – fitnessowa
HD Theater – ogólna w rozdzielczości HDTV
TLC
Animal Planet HD – przyrodnicza w rozdzielności HDTV
Science Channel HD – naukowa w HD
TLC HD – w rozdzielczości HDTV
Planet Green HD – rozrywkowo-ekologiczna w HDTV
Discovery en Español
Discovery Familia – familijna
Discovery Civilization Channel – historyczna
Discovery Knowledge – historyczno-społeczna
Discovery Channel Radio – radio FM
Discovery Channel Pro Cycling Team – amerykańska zawodowa grupa kolarska
Discovery Networks Europe – europejski przedstawiciel Discovery
Discovery Networks Asia – azjatycki przedstawiciel Discovery
Discovery Networks Canada – kanadyjski przedstawiciel Discovery
Discovery Networks Latin America – przedstawiciel Discovery w Ameryce Łacińskiej
Discovery Studios – wytwórnia filmowa
Discovery Digital Media – zbiór mobilnych mediów Discovery
Inne
Discovery One – fikcyjny amerykański statek kosmiczny, pojawiający się w filmach i powieściach z cyklu Odyseja kosmiczna Arthura Clarke'a
Land Rover Discovery – samochód terenowy
MV Discovery – statek
RRS Discovery – żaglowiec
Discovery Commerce – amerykańskie wydawnictwo
Discovery Education – zestaw pomocy dydaktycznych dla uczniów
Disc-Overy – album Tinie Tempah
Discovery – angielska odmiana jabłoni
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 6,487
|
Lette is een uit Ieper afkomstig geslacht.
Geschiedenis
De bewezen stamreeks begint met Pierre Lettin die rond 1450 werd geboren, woonachtig was te Ieper en in die laatste plaats in 1493 overleed. Zijn achterkleinzoon Jean Lettin (†1626) vestigde zich in Leiden, werd daar poorter in 1594 en was lakenhandelaar en wolkoper; hij nam de naam Lette aan. Nageslacht van hem oefende het beroep uit van medicus, predikant en bestuurder. De jongste tak verwierf in de 19e eeuw de heerlijkheden Oostvoorne, Rugge, Groot- en Klein-Oosterland.
In 1923 en 1965 werd het geslacht opgenomen in het genealogische naslagwerk Nederland's Patriciaat.
Enkele telgen
Johannes Arnoldus Lette (1745-1820), notaris
Mr. Nicolaas Johannes Cornelis Lette, ambachtsheer van Oostvoorne, Rugge, Groot- en Klein-Oosterland (1787-1851), burgemeester van Brielle
Marie Jeane Noldine Lette (1813-1852); trouwde in 1839 met Olke Arnoldus Uhlenbeck (1810-1888), vice-admiraal
Jeanne Charlotte Anne Lette (1817-1856); trouwde in 1839 met mr. Abraham Matthieu de Rouville (1812-1881), burgemeester van Brielle, gouverneur van Curaçao
Mr. Sebastiaan Hendrik Lette, ambachtsheer van Oostvoorne, Rugge, Groot- en Klein-Oosterland (1821-1890), burgemeester van Den Helder, gemeenteraadslid van Brielle
Mr. Nicolaas Johannes Cornelis Lette, ambachtsheer van Oostvoorne, Rugge, Groot- en Klein-Oosterland (1853-1915), ambtenaar Openbaar Ministerie
Antoinette Lette (1825-1909); trouwde in 1854 met haar zwager Olke Arnoldus Uhlenbeck (1810-1888)
George Frederik Lette (1827-1884), burgemeester van Brielle, Oostvoorne en Rockanje, lid van Provinciale Staten van Zuid-Holland; trouwde in 1860 met Maria Joanna Arnoldina Lette Anemaet (1836-1912)
Nicolaas Joannes Cornelis Sebastiaan Hendrik Lette, ambachtsheer van Oostvoorne (1861-1937), burgemeester van Brielle en Oostvoorne
George Frederik Lette (1887-1946)
Nicolaas Joannes Cornelis Lette, ambachtsheer van Oostvoorne sinds 1946 (1917), cargadoor; trouwde in 1945 met Mary Francesca Heineken (1922), dochter van dr. Henry Pierre Heineken (1886-1971), directeur van Heineken
Antoinette Henriëtte Lette, ambachtsvrouwe van Oostvoorne van 1937-1946 (1890-1970); trouwde in 1912 met mr. Arnold Nicolaas Fabius (1883-1947), lid van de familie Fabius, Officier van Justitie
Maria Johanna Arnoldina Lette (1789-1811); trouwde in 1808 met mr. Sebastiaan Hendrik Anemaet (1786-1863), burgemeester van Nieuwe Tonge en Tweede Kamerlid en uit wie een tak met de naam Lette Anemaet
Nederlands patriciërsgeslacht
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 4,412
|
\section{Introduction}
3D hand pose estimation from depth imaging has drawn lots of attention from researchers \cite{sinha2016deephand} \cite{ye2016spatial} \cite{wan2016hand} due to its important role in applications of augmented reality (AR) and human-computer interface (HCI) \cite{zhou2016novel}. It aims to predict the 3D accurate positions for hand joints \cite{supancic2015depth} with monocular depth images, which is critical for gesture recognition \cite{chen2016static}. Though has been studied for several years \cite{supancic2015depth}, it is still challenging owing to high joint flexibility, large view variance, poor depth quality, severe self occlusion, and similar part confusion.
Recently, deep convolutional networks (ConvNets) have exhibited state-of-the-art performance across several computer vision tasks such as object classification \cite{krizhevsky2012imagenet}, object detection \cite{girshick2016region}, and image segmentation \cite{chen2016deeplab}. ConvNets have also been employed to solve the problem of hand pose estimation, often with complicated structure design such as multi-branch inputs \cite{tompson2014real}\cite{oberweger2015hands} and multi-model regression \cite{oberweger2015hands} \cite{oberwegertraining} \cite{gerobust} \cite{zhang2016learning}. Thanks to the great modeling capacity and end-to-end feature learning, deep ConvNets have achieved competitive accuracy for hand pose estimation. However, ConvNets remain unable to obtain significant advantage over traditional random forest based methods \cite{sun2015cascaded} \cite{wan2016hand}, which may result from the relatively shallow ConvNet structure (often 3 - 5 convolution layers \cite{tompson2014real} \cite{oberwegertraining} \cite{zhang2016learning}) and high risk of overfitting with relative small datasets compared to image classification.
In this paper, we explore multiple good practices with hand pose estimation in single depth images. Most importantly, inspired by model ensemble and multi-view voting \cite{krizhevsky2012imagenet}, we present a single deep ConvNet architecture named \emph{Region Ensemble Net (REN)}\footnote{Codes and models are available at \url{https://github.com/guohengkai/region-ensemble-network}} (Fig.\ref{fig_overview}) to directly regress the 3D hand joint coordinates with end-to-end optimization and inference. We implement it by training individual fully-connected (FC) layers on multiple feature regions and combining them as ensembles. In addition, we adopt several approaches to enhance the performance including residual connection \cite{he2015deep}, data augmentation and smooth $L_1$ loss \cite{girshick2015fast}. As shown in our experiments, REN significantly promotes the performance of our ConvNet, which outperforms state-of-the-art methods on three challenging hand pose benchmarks \cite{tompson2014real} \cite{tang2014latent} \cite{sun2015cascaded}. Evaluated on fingertip \cite{tompson2014real} and human pose benchmarks \cite{haque2016towards}, our REN also achieves the best accuracy.
This paper builds on our preliminary publication \cite{guo2017region}. Compared with it, this paper describes more techinical details and discusses several important factors for good practice, leading to slightly better results than \cite{guo2017region} with different region settings. In addition, we add results for one extra hand pose dataset \cite{sun2015cascaded}, and further evaluate our REN for fingertip detection and human pose estimation with state-of-the-art performance.
\begin{figure}[htb]
\centering
{\includegraphics[width=0.48\textwidth]{overview}}
\caption{Region ensemble network (REN) with four regions: First deep ConvNet is used to extract features of depth image. The feature maps from ConvNet are then divided into regions. Each region is finally fed into fully-connected (FC) layers and then fused to predict the hand pose. The green rectangles represent the receptive field of the top-left region on the feature maps.}
\label{fig_overview}
\end{figure}
\section{Related work}
We briefly review relevant hand pose estimation methods with ConvNets for depth imaging, and examine methodologies related to the proposed algorithm, including ensemble methods and multi-view testing for ConvNets. Finally we also introduce works using ConvNets for RGB-D fingertip detection and human pose estimation, which will be compared in our experiments.
\subsection{RGB-D Hand pose estimation with ConvNets}
Recently deep ConvNets have been applied on hand pose estimation for depth imaging. Tompson et al. \cite{tompson2014real} first use ConvNets to produce 2D heat maps with multi-scale inputs and infer the 3D hand pose with inverse kinematics. Oberweger et al. \cite{oberweger2015hands} directly regress the 3D joint locations with multi-scale and multi-stage ConvNets using a linear layer as pose prior. In \cite{oberwegertraining}, a feedback loop is employed to iteratively correct the mistakes of inference, in which three ConvNets are used for pose initialization, image synthesis and pose updating. Ge et al. \cite{gerobust} employ three ConvNets from orthogonal views to separately regress 2D heat maps for each views with depth projections and fuse them to produce 3D hand pose. In \cite{zhou2016model}, physical joint constraints are incorporated into a forward kinematics based layer in ConvNet. Similarly, Zhang et al. \cite{zhang2016learning} embed skeletal manifold into ConvNets and train the model end-to-end to render sequential prediction.
\subsection{Multi-model ensemble methods for ConvNets}
Traditional ensemble learning means that training multiple individual models and combining their outputs via averaging or weighted fusions, which is widely adopted in recognition competitions \cite{krizhevsky2012imagenet}. In addition to bagging \cite{krizhevsky2012imagenet} \cite{sun2013deep}, boosting is also introduced for people counting \cite{walach2016learning}. However, using multiple ConvNets for both training and testing requires huge cost of memory and time, which is not practical for applications.
\subsection{Multi-branch ensemble methods for ConvNets}
Single ConvNet with the fusion of multiple branches can also be regarded as a generalized type of ensemble. One popular strategy is to fuse different scaling inputs \cite{tompson2014real} \cite{oberweger2015hands} or different image cues \cite{guo2016two} \cite{li2016deeptrack} \cite{chen2016accurate} with multi-input branches. Another approach is to employ multi-output branches with shared convolutional feature extractor, either training with different samples \cite{li2016convolutional} or learning to predict different categories \cite{ahmed2016network}. Compared with multi-input ensemble, multi-output methods cost less time because inference of FC layers is much faster than that of convolutional layers. Our method also falls into such category, but we apply ensemble on feature regions instead of inputs.
\subsection{Multi-view testing for ConvNets}
Multi-view testing is widely adopted to improve accuracy for object classification \cite{krizhevsky2012imagenet} \cite{sermanet2013overfeat} \cite{he2014spatial}. In \cite{krizhevsky2012imagenet}, predictions from 10-crop (four corners and one center with horizontal flip) are averaged on single ConvNet. In \cite{sermanet2013overfeat} \cite{he2014spatial}, fully-convolutional networks are employed in testing with multi-scale and multi-view inputs. Then spatially average pooling is applied on the class score map to obtain the final scores. To best of our knowledge, such strategy has not been applied on 3D pose regression yet.
\subsection{RGB-D fingertip detection and human pose estimation for ConvNets}
Fingertips play an important role in human-computer interaction among the hand joints. Wetzler et al. \cite{wetzler2015rule} employ ConvNet for in-plane derotation of hand depth image and then use random forests or ConvNets for fingertip coordinate regression. Guo et al. \cite{guo2016two} introduce a two-stream ConvNet to detect the 3D fingertips, which makes use of both depth information and edge information with slow fusion strategy.
Human pose estimation is also important for HCI applications such as action recognition \cite{zhang2016rgb} \cite{chen2016novel}. Though ConvNets are widely used in human pose estimation for RGB images \cite{carreira2016human} \cite{bulat2016human}, there are limited number of works using ConvNets for depth imaging \cite{shi2015high} \cite{wang2013depth} due to relatively small size of training datasets. Haque et al. \cite{haque2016towards} introduce a viewpoint invariant model using ConvNets and recurrent networks (RNNs) for human pose estimation. Local regions from depth images are transformed into a learned feature space via ConvNets and then RNNs are leveraged to predict the offsets of pose sequentially with multi-task setting.
\section{Region Ensemble Network}
As in Fig.\ref{fig_overview}, REN starts with a ConvNet for feature extraction. Then the features are divided into multiple grid regions. Each region is fed into FC layers and learnt to fuse for hand pose prediction. In this section we introduce the basic network architecture, region ensemble structure and implementation details.
\subsection{Network architecture with residual connection}
\label{residual}
The architecture of our ConvNet for feature extraction consists of six convolutional layers with $3\times3$ kernels (Fig.\ref{fig_baseline}) and three pooling layers with $2\times2$ kernels. Each convolutional layer is followed by a Rectified Linear Unit (ReLU) activation. The ConvNet accepts inputs of a $96\times96$ depth image and outputs the feature maps with dimension of $12\times12\times64$. To improve the learning ability, two residual connections are adopted between pooling layers with $1\times1$ convolution filters for dimension increase as in \cite{he2015deep}. So there are totally eight convolutional layers in our model, which is deeper than ConvNets in \cite{zhang2016learning} with five layers.
For regression, we use two 2048 dimension FC layers with dropout rate \cite{srivastava2014dropout} of 0.5 for each regressor to avoid overfitting. The output of regressor is a $3\times J$ vector representing the 3D world coordinates for hand joints, where $J$ is the number of joints.
\begin{figure}[htb]
\centering
{\includegraphics[width=0.48\textwidth]{baseline}}
\caption{Structure of basic ConvNet for feature extraction. The ConvNet consists of six convolutional layers and three pooling layers. The dotted arrows represent residual connections with dimension increase \cite{he2015deep}. The non-linear activation layers following each convolutional layers are not showed in the figure.}
\label{fig_baseline}
\end{figure}
\subsection{Region ensemble structure}
\label{ensemble}
Multi-view testing averages predictions from different crops of original input image, which reduces the variance for image classification \cite{krizhevsky2012imagenet}. Because image classification is invariant to translation and cropping, multi-view testing is easy to apply by directly cropping on the input image. When it comes to pose regression, each cropped parts will correspond to different hand pose configurations. So we should adapt the 3D coordinates of hand pose to the cropped view. Meanwhile, using multiple inputs to feed the ConvNet one-by-one is time-consuming.
Because each activation in the convolutional feature maps is contributed by a receptive field in the input image domain, we can project the multi-view inputs onto the regions of the feature maps. By using separate regions as features, we can train separate regressors instead of single regressor. So multi-view voting could be extended to regression task by utilizing each regions to separately predict the whole hand pose and then combining the results.
Based on this inspiration, we define a tree-structured network consisting of a single ConvNet trunk and several regression branches as shown in Fig.\ref{fig_overview}. We first divide the feature maps of ConvNet into several regions. For each region, we feed it into the FC layers respectively as branches. There are several ways to combine different branches. A simple strategy is bagging, which averages all outputs of branches using average pooling. In order to boost the predictions from all the regions, we employ region ensemble strategy instead of bagging: features from the last FC layers of all regions are concatenated and used to infer the coordinates with an extra regression layer. The whole network can be trained end-to-end by minimizing the regression loss.
For region setting, we use nine regions with size of $6\times6$ located at four conners (left part in Fig. \ref{fig_region}, which is also the whole setting in \cite{guo2017region}), four centers near the edges (middle part in Fig. \ref{fig_region}) and the center of the feature maps. The receptive fields of different regions within the $96\times96$ image bounding are shown in Fig. \ref{fig_receptive}, which is similar to the corner and center crop in \cite{krizhevsky2012imagenet}. We will discuss the effect of different region settings on accuracy in Section \ref{setting}.
\begin{figure}[htb]
\centering
{\includegraphics[width=0.48\textwidth]{region}}
\caption{Different region setting for feature maps: four conners \cite{guo2017region} (left), four centers in each edges (middle), and multi-scale regions with the same center (right). Proposed REN adopts nine regions with size of $6\times6$ including the center of feature maps and all the eight regions in left and middle figures. }
\label{fig_region}
\end{figure}
\begin{figure}[htb]
\centering
{\includegraphics[width=0.48\textwidth]{receptive_h}}
\caption{Receptive fields for different region positions: (a) $62\times62$ for conners, (b) $62\times76$ or $76\times62$ for centers in each edges, and (c) $76\times76$ for the center of feature maps.}
\label{fig_receptive}
\end{figure}
There are three main differences between proposed methods and multi-view voting: 1) To our knowledge, all multi-view testing methods before are designed for image classification while our region ensemble can be applied on both classification and regression. By applying fusion FC layer in REN, different views of inputs are trained to simultaneously predict the same pose. 2) We adopt end-to-end training for region ensemble instead of testing only, making the ConvNet adjust the contributions from each views. 3) We replace the average pooling with one FC layer on concatenated features to learn the fusion parameters, which increases the learning ability of the network. We will perform the comparison in Section \ref{basic}.
\subsection{Implementation details}
\label{detail}
We implement our REN with Caffe \cite{jia2014caffe} written in C++. We use stochastic gradient descent (SGD) with a mini-batch size of 128. The learning rate starts from 0.005 and is divided by 10 every 20 epochs, and the model is trained for total 80 epochs. In the meanwhile, we use a weight decay of 0.0005 and a momentum of 0.9. Our model is trained from scratch with random initialization \cite{he2015delving}. Moreover, there are three important strategies for training: patch cropping, data augmentation, and smooth $L_1$ loss. The details are described below. And we will show the incremental contributions of these strategies in later section.
\noindent\textbf{Patch cropping}\hspace{2mm} For ConvNet inputs, we extract a cube with fixed size of 150mm from the depth image centered in the hand region. Then the cube is resized into a $96\times96$ patch of depth values normalized to $[-1, 1]$ as input for ConvNet. The 3D coordinates are also normalized to $[-1, 1]$ according to the cube. To compute the center, we first segment the foreground with fixed thresholds and calculate the centroid of foreground.
\noindent\textbf{Data augmentation}\hspace{2mm} We apply data augmentation during training, including translation within $[-10, 10]$ pixel, scaling within $[0.9, 1.1]$ and rotation within $[-180, 180]$ degree. Random augmentation effectively increases the size of training dataset, so it can improve the generalization performance.
\noindent\textbf{Smooth $L_1$ loss}\hspace{2mm} To deal with noisy annotations, we adopt similar smooth $L_1$ loss in \cite{girshick2015fast}:
\begin{displaymath}
\textrm{smooth}_{L_1}(x)=\left\{ \begin{array}{ll}
0.5x^2 & \textrm{if} |x| < 0.01 \\
0.01(|x| - 0.005) & \textrm{otherwise}
\end{array} \right.
\end{displaymath}
Because it is less sensitive to outliers than the $L_2$ loss, it can benefit the training of ConvNet.
\section{Experiments}
In this section, we first introduce the evaluation datasets and metrics for our experiments. Then we compare our REN with several state-of-the-art methods on public hand pose datasets. Next we explore several good practices of training ConvNets for hand pose estimation, discuss different region settings and also compare with traditional ensembles and multi-view testing. Finally we apply our REN on fingertip detection and human pose estimation for public benchmarks.
\subsection{Experiment setup}
\subsubsection{Datasets}
We conduct our experiments on four publicly RGB-D datasets: ICVL hand pose dataset \cite{tang2014latent}, NYU hand pose dataset \cite{tompson2014real}, MSRA hand pose dataset \cite{sun2015cascaded}, and ITOP human pose dataset \cite{haque2016towards}. For self-comparison, ICVL dataset is mainly used. More details for datasets are as follows:
\noindent\textbf{ICVL dataset}\hspace{2mm} The training set of ICVL dataset contains 300K images with different rotations, and the testing set contains 1.6K images. All the depth images are captured by Intel RealSense. Totally 16 hand joints are initialized by the output of camera and manually refined.
\noindent\textbf{NYU dataset}\hspace{2mm} The NYU dataset has 72K images for training and 8K for testing with 36 3D annotated joints, collected from Microsoft Kinect camera. Following \cite{tompson2014real}, 14 hand joints with front-view image are used in experiments. And this dataset is also used to evaluate fingertip detection on the 5 fingertip joints in \cite{wetzler2015rule} \cite{guo2016two}.
\noindent\textbf{MSRA dataset}\hspace{2mm} The MSRA dataset contains 9 subjects with 17 gestures for each subject. 76K depth images with 21 annotated joints are collected with Intel's Creative Interactive Camera. For evaluation, each subject is alternatively used as testing data when other 8 subjects are used for training. This is repeated 9 times and the average metrics are reported.
\noindent\textbf{ITOP dataset}\hspace{2mm} The ITOP dataset consists 18K training images and 5K testing images for front view and top view acquired by two Kinect cameras. Each depth image is labelled with fifteen 3D joint locations of human body.
\subsubsection{Evaluation metrics}
We employ different metrics for hand pose estimation and human pose estimation following the literatures \cite{tang2014latent} \cite{tompson2014real} \cite{haque2016towards}. For hand pose, the performance is evaluated by two metrics: 1) \textbf{average 3D distance error} is computed as the average Euclidean distance for each joint (in millimeters). 2) \textbf{percentage of success frames} is defined as the rate of frames in which all Euclidean errors of joints are below a variant threshold \cite{oberweger2015hands}. In addition, mean precision (mP) with a threshold of 15mm as defined in \cite{wetzler2015rule} is calculated for fingertip detection.
For human pose, we compute the \textbf{mean average precision (mAP)} \cite{haque2016towards}, which is defined as the average detected rate for all human body joints. A joint is counted as detected when the Euclidean distance between predicted position and ground truth is below 10cm.
\subsection{Comparison with the state of the art}
We compare our methods against several state-of-the-art approaches on ICVL dataset \cite{tang2014latent} \cite{oberweger2015hands} \cite{sun2015cascaded} \cite{zhou2016model} \cite{wan2016hand} \cite{wan2017cross}, NYU dataset \cite{tompson2014real} \cite{oberweger2015hands} \cite{oberwegertraining} \cite{sinha2016deephand} \cite{gerobust} \cite{zhou2016model} \cite{zhang2016learning} \cite{wan2017cross}, and MSRA dataset \cite{sun2015cascaded} \cite{choi2015collaborative} \cite{gerobust} \cite{wan2016hand} \cite{wan2017cross}. Overall, Fig.\ref{fig_sota_icvl} - \ref{fig_sota_msra} show that proposed REN obtains the best accuracy among all the algorithms for hand pose estimation.
In details, on ICVL dataset our method surpasses other methods with a large margin. And the mean error $7.31mm$ obtains a $0.80mm$ decrease compared with LSN \cite{wan2016hand}, which is a $9.87\%$ relative improvement. Similarly on NYU dataset, our results are more accurate ($12.69mm$) than other approaches, and reduce the error of \cite{zhang2016learning} by $10.3\%$. For MSRA dataset, our algorithm also significantly outperform all state-of-the-art methods for nearly all thresholds, with an average error of $9.79mm$. Surprisely, it reduces the mean error of \cite{gerobust} by $25.7\%$. Note that either LSN or multi-view ConvNets \cite{gerobust} employ multiple models with complicated design, while our REN only uses single model without multi-stage regression, which indicates the power for proposed region ensemble strategy. Fig. \ref{fig_vis_all} shows some good cases and bad cases for all datasets. We can find that the failure cases are often caused by severe occlusion and bad depth images.
\begin{figure*}[htb]
\centering
\begin{minipage}[b]{0.49\textwidth}
\centering
\centerline{\includegraphics[width=0.95\textwidth]{icvl_error}}
\end{minipage}
\begin{minipage}[b]{0.49\textwidth}
\centering
\centerline{\includegraphics[width=0.95\textwidth]{icvl_map_new}}
\end{minipage}
\caption{Comparison with state-of-the-arts on ICVL \cite{tang2014latent} dataset: distance error (left) and percentage of success frames (right).}
\label{fig_sota_icvl}
\end{figure*}
\begin{figure*}[htb]
\begin{minipage}[b]{0.49\textwidth}
\centering
\centerline{\includegraphics[width=0.95\textwidth]{nyu_error}}
\end{minipage}
\begin{minipage}[b]{0.49\textwidth}
\centering
\centerline{\includegraphics[width=0.95\textwidth]{nyu_map}}
\end{minipage}
\caption{Comparison with state-of-the-arts on NYU \cite{tompson2014real} datasets: distance error (left) and percentage of success frames (right).}
\label{fig_sota_nyu}
\end{figure*}
\begin{figure*}[htb]
\begin{minipage}[b]{0.49\textwidth}
\centering
\centerline{\includegraphics[width=0.95\textwidth]{msra_error}}
\end{minipage}
\begin{minipage}[b]{0.49\textwidth}
\centering
\centerline{\includegraphics[width=0.95\textwidth]{msra_map_new}}
\end{minipage}
\caption{Comparison with state-of-the-arts on MSRA \cite{sun2015cascaded} datasets: distance error (left) and percentage of success frames (right).}
\label{fig_sota_msra}
\end{figure*}
\begin{figure*}[htb]
\centering
{\includegraphics[width=0.95\textwidth]{vis_all}}
\caption{Example results on ICVL \cite{tang2014latent}, NYU \cite{tompson2014real} and MSRA \cite{sun2015cascaded} datasets: ground truth (first row) and region ensemble network (second row) for each datasets.}
\label{fig_vis_all}
\end{figure*}
For MSRA dataset we also report the average joint errors distributed over all yaw and pitch viewpoint angles as in \cite{sun2015cascaded} and \cite{gerobust}, shown in Fig. \ref{fig_sota_msra_angle}. On all angles our method achieve the best accuracy with the smallest deviation, which indicates the robustness for viewpoint variance.
\begin{figure*}[htb]
\begin{minipage}[b]{0.49\textwidth}
\centering
\centerline{\includegraphics[width=0.95\textwidth]{msra_yaw_new}}
\end{minipage}
\begin{minipage}[b]{0.49\textwidth}
\centering
\centerline{\includegraphics[width=0.95\textwidth]{msra_pitch_new}}
\end{minipage}
\caption{The average joint errors distributed over all yaw/pitch viewpoint angles on MSRA \cite{sun2015cascaded} dataset. The standard deviations of the error distributions are shown in the legend titles.}
\label{fig_sota_msra_angle}
\end{figure*}
\subsection{Self-comparison}
We perform self comparison experiments for different strategies and setting of region ensemble network on ICVL dataset \cite{tang2014latent}.
\subsubsection{Exploration study}
In this section, we focus on the investigation of good practices. Specifically, we incrementally introduce five strategies on a basic shallow network in Fig. \ref{fig_shallow}: 1) adding one convolution layer after each convolution layer to increase the depth of ConvNet. 2) adding residual connection edges across pooling layers as described in Section \ref{residual}. 3) using smooth $L_1$ loss \cite{girshick2015fast} instead of Euclidean $L_2$ loss for regression optimization. 4) augmenting the input patches as described in Section \ref{detail}. 5) proposed region ensemble.
\begin{figure}[htb]
\centering
{\includegraphics[width=0.48\textwidth]{shallow}}
\caption{Structure of basic shallow ConvNet with three convolution layers and three pooling layers. The non-linear activation layers following each convolution layers are not showed in the figure.}
\label{fig_shallow}
\end{figure}
The experimental results are summarized in Table 1. Combining all the strategies reduces the errors by 3.17mm (relative $30.2\%$), which is a significant improvement of accuracy. Among them, $L_1$ loss and region ensemble are two most important factors for performance boosting, because $L_1$ loss is more suitable for labels with relative large noise and region ensemble can help improve the generalization for model.
Qualitative comparison on ICVL dataset are shown in Fig.\ref{fig_vis} for region ensemble (second row, corresponding to the sixth row in Table 1) and basic network (third row, corresponding to the fifth row in Table 1). The estimations are more accurate for region ensemble especially for fingers.
\begin{table}[htb]
\label{table_add}
\caption{Average 3D distance error (mm) of incremental strategies on ICVL dataset \cite{tang2014latent}. Lower is better.}
\centering
\begin{tabular}{|l|c|}
\hline
\multicolumn{1}{|c|}{Strategy} & Error(mm) \\
\hline
\multicolumn{1}{|c|}{Shallow} & 10.48 \\
+Deeper & 10.02 \\
+Residual Edge & 9.73 \\
+Smooth $L_1$ Loss & 8.59 \\
+Augmentation & 8.36 \\
+Region Ensemble & \textbf{7.31} \\
\hline
\end{tabular}
\end{table}
\begin{figure*}[htb]
\centering
{\includegraphics[width=0.95\textwidth]{vis_icvl_}}
\caption{Example results on ICVL \cite{tang2014latent} dataset: ground truth (first row), basic network (second row, corresponding to the fifth row in Table 1), and region ensemble network (third row, corresponding to the seventh row in Table 1).}
\label{fig_vis}
\end{figure*}
\subsubsection{Region setting}
\label{setting}
According to the analysis in Section \ref{ensemble}, different region partitions are equal to different patterns of multi-view inputs. Here we explore the effect of different settings of regions, including: 1) multi-scale regions with three regions of size $12\times12$, $8\times8$ and $4\times4$, which is similar to multi-scale inputs as in \cite{tompson2014real} \cite{oberweger2015hands}. 2) four regions of size $6\times6$ (left parts in Fig. \ref{fig_region}), which is the setting in \cite{guo2017region}. 3) nine regions of size $6\times6$ (four as left parts, four as middle parts in Fig. \ref{fig_region} and one in the center), which is the setting in this paper. 4) nine regions of size $4\times4$ with similar positions as (3). 5) nine regions of size $8\times8$ with similar positions as (3).
From Fig. \ref{fig_grid}, regions with same size are significantly more accurate than multi-scale regions due to the balance parameter number for FC layers of different regions. And more regions with moderate size (i.e. $9\times6\times6$) obtain slightly better performance. Too large or too small receptive field hurts the accuracy of hand pose estimation.
\begin{figure}[htb]
\centering
{\includegraphics[width=0.48\textwidth]{icvl_grid}}
\caption{Comparison of different region settings (region number $\times$ region width $\times$ region height) for percentage of success frames on ICVL dataset \cite{tang2014latent}. }
\label{fig_grid}
\end{figure}
\subsubsection{Comparison with ensembles and multi-view testing}
\label{basic}
We compare with traditional ensembles and multi-view testing in this section. In details, we implement three baselines: 1) \emph{Basic} network has the same convolution structure in Fig.\ref{fig_baseline} and single regressor on the full feature map with two FC layers of 2048 dimensions. 2) \emph{Basic Bagging} network has nine basic networks as (1) that trained independently on the same data with different random order and augmentation. The average predictions of all the networks form the final prediction. 3) \emph{Multi-view Testing} trains single basic network as (1) but averages the predicted 3D hand poses with nine multi-view inputs. The inputs are cropped as in Section \ref{detail}, but on different centers with bias of $-d$/$0$/$d$mm on their x and y coordinates relative to the centroid. We use $d=26.5625$ to approximately match the nine region positions in REN.
Results in Fig.\ref{fig_ensemble} shows that ensemble based methods (both basic bagging and region ensemble) are significantly more effective that baseline network. And the performance of our region ensemble is much better than traditional bagging. Because REN only employs multiple FC layers instead of multiple complete ConvNets, it also costs less time and memory than traditional bagging. Meanwhile, the improvement from multi-view testing is limited for hand pose estimation, because the model is more sensitive to translation in regression tasks than that in classification.
\begin{figure}[htb]
\centering
{\includegraphics[width=0.48\textwidth]{icvl_ensemble_view}}
\caption{Comparison of ensembles and mutli-view testing for percentage of success frames on ICVL dataset \cite{tang2014latent}. }
\label{fig_ensemble}
\end{figure}
\subsection{Evaluation on other RGB-D tasks}
Here we also test our REN on challenging benchmarks for fingertip detection and human pose estimation and compare with state-of-the-art methods.
\subsubsection{Fingertip detection}
We compare the fingertip detection results to several state-of-the-art algorithms \cite{wetzler2015rule} \cite{guo2016two} on NYU dataset without retraining our REN model. Table 3 illustrates that our REN achieves the best performance among all the methods, with an average error of 15.6mm.
\begin{table}[htb]
\label{table_finger}
\caption{Mean precision (mP) and average 3D distance error (mm) for fingertips of different methods on NYU dataset \cite{tompson2014real}. Higher is better for mP and lower is better for error.}
\centering
\begin{tabular}{|c|c|c|}
\hline
Methods & mP & Error(mm) \\
\hline
CNN-DeROT \cite{wetzler2015rule} & 0.63 & - \\
DeepPrior \cite{oberweger2015hands} & 0.43 & 26.4 \\
FeedLoop \cite{oberwegertraining} & 0.38 & 23.2 \\
TwoStream \cite{guo2016two} & 0.50 & 19.3 \\
Model \cite{zhou2016model} & 0.40 & 24.4 \\
REN & \textbf{0.66} & \textbf{15.6} \\
\hline
\end{tabular}
\end{table}
\subsubsection{Human pose estimation}
The results for human pose estimation are reported in Table 4, where we compare our method with \cite{yub2015random} \cite{carreira2016human} using mAP metric on ITOP dataet. For frontal view, proposed REN with 84.9 mAP significantly outperforms RTW and REF. And the accuracy for lower body is much higher. For top-down view, our method is better than RTW and shows comparable performance with REF, which contains deeper ConvNets with 16 convolution layers in their models. See Fig. \ref{fig_itop_vis} for some visualization results.
\begin{table*}[htb]
\label{table_human}
\caption{Mean average precision (mAP, unit: \%) of different methods on frontal view and top view of ITOP dataset \cite{haque2016towards} using a 10cm threshold. Higher is better.}
\centering
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
& \multicolumn{3}{|c|}{mAP (front-view)} & \multicolumn{3}{|c|}{mAP (top-view)} \\
\hline
Body Part & RTW \cite{yub2015random} & REF \cite{carreira2016human} & REN & RTW \cite{yub2015random} & REF \cite{carreira2016human} & REN \\
\hline
Head & 97.8 & 98.1 & 98.7 & 98.3 & 98.1 & 98.2 \\
Neck & 95.8 & 97.5 & 99.4 & 82.2 & 97.6 & 98.9 \\
Shoulders & 94.1 & 96.6 & 96.1 & 91.8 & 96.1 & 96.6 \\
Elbows & 77.9 & 73.3 & 74.7 & 80.1 & 86.2 & 74.4 \\
Hands & 70.5 & 68.6 & 55.2 & 76.9 & 85.5 & 50.7 \\
Torso & 93.8 & 85.6 & 98.7 & 68.1 & 72.9 & 98.1 \\
Hips & 80.3 & 72.0 & 91.8 & 55.7 & 61.1 & 85.5 \\
Knees & 68.8 & 69.0 & 89.0 & 53.9 & 51.6 & 70.0 \\
Feet & 68.4 & 60.8 & 81.1 & 28.6 & 51.5 & 41.6 \\
\hline
Mean & 80.5 & 77.2 & \textbf{84.9} & 68.5 & \textbf{75.5} & \textbf{75.5} \\
\hline
\end{tabular}
\end{table*}
\begin{figure*}[htb]
\centering
{\includegraphics[width=0.95\textwidth]{itop_vis}}
\caption{Example results on ITOP \cite{carreira2016human} dataset.}
\label{fig_itop_vis}
\end{figure*}
\noindent\textbf{Implementation details}\hspace{2mm} For human pose, a small ConvNet is trained to predict the torso position as center. The size of cube is $800\times1200\times800\textrm{mm}^3$ for front-view and $600\times600\times1000\textrm{mm}^3$ for top-view. For data augmentation, random flip of image with probability of 0.5 is also used.
\section{Conclusion}
To boost the performance of single ConvNet for 3D hand pose estimation, we exploit several good practices and present a simple but powerful region ensemble structure by dividing the feature maps into different regions and jointly training multiple regressors on all regions with fusion. Such strategies significantly improve the accuracy of ConvNet. The experimental results demonstrate that our method outperforms all the state-of-the-arts on three hand pose datasets and one human pose dataset. Since region ensemble is easy to be introduced into ConvNets, we believe that proposed structure could be applied on more computer vision tasks and achieve more promising results.
\noindent\textbf{Acknowledgments}\hspace{2mm} This work is supported by The National Science Foundation of China (No. 61271390 and No. 91648116), and State High-Tech Development Plan (No. 2015AA016304). Thanks Shulan Pan for paper edition.
{\small
\bibliographystyle{ieee}
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"redpajama_set_name": "RedPajamaArXiv"
}
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Daycare Newsletter Examples daycare newsletter examples childcare newsletter examples daycare newsletter template templates. daycare newsletter examples childcare newsletter examples free child care newsletter templates printable. daycare newsletter examples how to create a daycare newsletter great idea a monthly or semi templates.
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"redpajama_set_name": "RedPajamaC4"
}
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When I go to create a new web project with Twitter Bootstrap, the files do not get downloaded. I see the same behavior in PHPSTORM 6. It looks like something due to Twitter Bootstrap changing their URL's in GITHUB. Any idea how to fix this?
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"redpajama_set_name": "RedPajamaC4"
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Q: How to retrieve more than 50 records using Spotipy API I'm using the Spotipy API to retrieve song data from Spotify. Here's my code:
import pandas as pd
import spotipy
from spotipy.oauth2 import SpotifyClientCredentials
sp = spotipy.Spotify(auth_manager=SpotifyClientCredentials(client_id='<my_client_id>',
client_secret='<my_client_secret'))
results = sp.search(q="artist:guns n' roses", limit=50)
d = []
for idx, track in enumerate(results['tracks']['items']):
d.append (
{
'Track' : track['name'],
'Album' : track['album']['name'],
'Artist' : track['artists'][0]['name'],
'Release Date' : track['album']['release_date'],
'Track Number' : track['track_number'],
'Popularity' : track['popularity'],
'Track Number' : track['track_number'],
'Explicit' : track['explicit'],
'Duration' : track['duration_ms'],
'Audio Preview URL' : track['preview_url'],
'Album URL' : track['album']['external_urls']['spotify']
}
)
pd.DataFrame(d)
Per the docs, it appears that Spotify has a limit of 50 records.
Is it possible to retrieve all records for a given string search? (e.g. by chunking requests, etc.)
Thanks!
A: The Spotify Web API can return a maximum of 1000 items. (In this example, it found 390 tracks, so it got all of them.)
Here is the code to get them:
import pandas as pd
import spotipy
from spotipy.oauth2 import SpotifyClientCredentials
sp = spotipy.Spotify(auth_manager=SpotifyClientCredentials(client_id='<my_client_id>',
client_secret='<my_client_secret>'))
d = []
total = 1 # temporary variable
offset = 0
while offset < total:
results = sp.search(q="artist:guns n' roses", type='track', offset=offset, limit=50)
total = results['tracks']['total']
offset += 50 # increase the offset
for idx, track in enumerate(results['tracks']['items']):
d.append (
{
'Track' : track['name'],
'Album' : track['album']['name'],
'Artist' : track['artists'][0]['name'],
'Release Date' : track['album']['release_date'],
'Track Number' : track['track_number'],
'Popularity' : track['popularity'],
'Track Number' : track['track_number'],
'Explicit' : track['explicit'],
'Duration' : track['duration_ms'],
'Audio Preview URL' : track['preview_url'],
'Album URL' : track['album']['external_urls']['spotify']
}
)
pd.DataFrame(d)
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 9,168
|
Poli Genova (Sofija, 10. veljače 1987.) je bugarska pjevačica.
Počela je pjevati s četiri godine. Tijekom 2005. i 2006. sudjelovala je na bugarskom izboru za Pjesmu Eurovizije kao dio ženske grupe Melody. U to vrijeme je i diplomirala klarinet na Nacionalnoj glazbenoj školi u Sofiji.
Godine 2011., pobijedila je na bugarskom natjecanju za Pjesmu Eurovizije s pjesmom Na iInat, tako da je mogla predstavljati Bugarsku na 56. Eurosongu u Düsseldorfu, u Njemačkoj. Međutim, u poluzavršnici nije dobila dovoljno bodova za plasman u samu završnicu natjecanja.
Bila je voditeljica na Dječjoj Pjesmi Eurovizije, održanoj 21. studenog 2015. u Areni Armeec u Sofiji u Bugarskoj. Nekoliko mjeseci kasnije, izabrana je da predstavlja Bugarsku na natjecanju za Pjesmu Eurovizije 2016., ovaj put s pjesmom Ako je ljubav zločin. Pjesma je bila četvrta s 307 bodova, ostvarivši dotad najuspješniji nastup Bugarske na Pjesmi Eurovizije.
O svojoj eurovizijskoj uspješnici Genova je rekla:
Izvori
Vanjske poveznice
Poli Genova - službena Facebook stranica
Bugarski pjevači
Eurovizijski izvođači
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 1,864
|
Arctic SDI catalogue
Type of resources
service-view (4)
RI_622 (4)
service-other (1)
Available actions
Viewable (3)
Geoscientific information (4)
Norway Digital (5)
Marine activities (4)
Government information (4)
Mareano (3)
Sea regions (3)
Oceanographic geographical features (3)
Natural environment (2)
Shrub (1)
Ditch (1)
Wet meadow (1)
Contact for the resource
Institute of Marine Research (4)
Government and Municipalities of Québec; Government and Municipalities of Québec; City of Montreal (4)
Norwegian Meteorological Institute (1)
Norwegian Mapping Authority (1)
NSDI Norway (6)
NSDI Canada (4)
Representation types
Update frequencies
On going (4)
Historical archive (1)
Under development (1)
From 1 - 10 / 10
Circumpolar Arctic basic map cache
Circumpolar basic map with transparency, combining topographic details from the Norwegian Mapping Authority and the Norwegian Polar Institute south to latitude X (southern France). It also includes map data from the other Arctic countries that is published via the Arctic Spatial Data Infrastructure (Arctic SDI). This map does not include place names and road names – these can be added using the circumpolar place name cache service.
Ocean currents Barents sea Coastal water:
Ocean currents Barents sea Coastal water: Surface currents in the Barents sea : Coastal, Atlantic and Arctic water, delivered by the Institute of Marine Research.
Havdata fra ROMS Arctic 20km
Ecoterritories
This file contains polygons delineating ecoterritories. The City of Montreal has identified 10 sectors, called ecoterritories, where the protection and enhancement of the natural spaces in it were considered a priority. These are vast territories of more than 15 hectares, grouping existing protected areas (large parks, nature reserves, etc.), urbanized areas as well as natural spaces still to be protected and enhanced. Ecoterritories were established by the Policy for the Protection and Development of Natural Environments (Ville de Montréal, 2004), and these are recognized in the Montreal Planning Plan. This update of the ecoterritories follows the adoption of the new Plan d'aménagement et de développement de l'agglomération de Montréal and its entry into force on April 1, 2015. http://ville.montreal.qc.ca/schema **This third party metadata element was translated using an automated translation tool (Amazon Translate).**
Watercourse and ditch
This dataset contains the plots of streams (stream, river) and major ditches in the metropolitan area of Montreal. Watercourses and ditch are the result of an analysis of aerial photos and/or field visits and/or specific ecological studies. These are constantly updated according to the advancement of local knowledge. These data remain indicative, as some watercourses and ditch have not been validated in the field. **This third party metadata element was translated using an automated translation tool (Amazon Translate).**
Ocean currents Barents sea Polar front
Ocean currents Barents sea Polar front : Surface currents in the Barents sea : Polar front, Coastal, Atlantic and Arctic water, delivered by the Institute of Marine Research.
Ocean currents Barents sea Atlantic water
Ocean currents Barents sea Atlantic water: Surface currents in the Barents sea : Coastal, Atlantic and Arctic water, delivered by the Institute of Marine Research.
Natural wasteland
This dataset contains polygons delineating the natural wastelands of the Greater Montreal area. Brownfield boundaries and type are the result of an analysis of aerial photos and/or field visits and/or specific ecological studies. These are constantly updated according to the advancement of local knowledge. Some types and their limits may therefore be imprecise. **This third party metadata element was translated using an automated translation tool (Amazon Translate).**
This dataset contains polygons delineating wetlands in the metropolitan area of Montreal. Wetland boundaries and composition are the result of aerial photo analysis and/or field visits and/or specific ecological studies. These are continually updated according to the advancement of knowledge in the community. These data remain indicative as some wetlands have not been validated in the field. **This third party metadata element was translated using an automated translation tool (Amazon Translate).**
Ocean currents Barents sea Arctic water
Ocean currents Barents sea Arctic water: Surface currents in the Barents sea : Coastal, Atlantic and Arctic water, delivered by the Institute of Marine Research.
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 3,921
|
\section{Introduction}
Over the last six years, the \hbox{X-ray}\ surveys carried out by {\it Chandra\/}\ and {XMM-{\it Newton\/}}\
(e.g., \citep{gia02,ale03,fio03}; see \citep{bh05} for a review)
have provided remarkable results in resolving a significant fraction of the
cosmic \hbox{X-ray}\ background (XRB; \citep{com95,gil07}),
up to $\approx$~80\% in the 2--8~keV band (e.g., \citep{feb04,hm06}).
Despite the idea that a large fraction of the accretion-driven energy density in the Universe
resides in obscured \hbox{X-ray}\ sources has been widely supported and accepted
(e.g., \citep{bar05,hop06}), until recently only limited information was available to
properly characterize the broad-band emission of the counterparts of the \hbox{X-ray}\ obscured sources
and provide a reliable estimate of their bolometric output.
In this context, {\it Spitzer\/}\ data have provided a major step forward the understanding of the
broad-band properties of the \hbox{X-ray}\ source populations. If, on one hand, {\it Spitzer\/}\ data have
allowed to pursue the ``pioneering'' studies of \cite{elv94} on the spectral energy
distributions (SEDs) of broad-line (Type~1), unobscured quasars at higher redshifts
(e.g., \citep{ric06}), on the other hand they have produced significant results in the
definition of the SEDs of narrow-line (Type~2), obscured AGN (e.g., \citep{pol06}).
In this work we aim at providing a robust determination of the bolometric luminosity for hard
\hbox{X-ray}\ selected obscured AGN. This result can be achieved by effectively disentangling
the nuclear emission related to the active nucleus from the host galaxy starlight, which represents
the dominant component (at least for our obscured sources) at optical and near-infrared (near-IR)
wavelengths.
\section{Sample selection and $K_{\rm s}$-band properties}
The sources presented in this work were selected from the HELLAS2XMM survey (\citep{fio03})
which, at the 2--10~keV flux limit of $\approx$~10$^{-14}$~\cgs, covers $\approx$~1.4
square degrees of the sky using {XMM-{\it Newton\/}}\ archival pointings (\citep{bal02}).
Approximately 80\% of the HELLAS2XMM sources have a spectroscopic optical classification in the
final source catalog (see \citep{coc07} for details).
In particular, we selected eight sources from the original sample of \cite{mig04} which
are characterized by faint (23.7--25.1) $R$-band magnitudes and bright $K_{\rm s}$-band counterparts
($\approx$~17.6--19.1); all of our sources are therefore classified as extremely red objects
(EROs, $R-K_{\rm s}>5$ in Vega magnitudes).
From the good-quality $K_{\rm s}$-band images, \cite{mig04} were able to study the
surface brightness profiles of these sources, obtaining a morphological classification.
While two sources are associated with point-like objects, the remaining six sources are extended,
showing a profile typical of elliptical galaxies.
In this latter class of sources, the active nucleus, although evident in the \hbox{X-ray}\ band,
appears hidden or suppressed at optical and near-IR wavelengths,
where the observed emission is clearly dominated by the host galaxy starlight.
The relatively good constraints on the nuclear emission
in the near-IR represent a starting point for the analysis of the {\it Spitzer\/}\ IRAC and
MIPS data.
Due to the faint $R$-band magnitudes of our sources, optical spectroscopy was not feasible
even with the 8-m telescope facilities; however, the bright near-IR counterparts of our
sources allowed us to obtain spectroscopic redshifts in the $K_{\rm s}$\ band with {\sc ISAAC}
at {\sc VLT} for two sources:
one point-like AGN is classified as a Type~1.9 quasar at a redshift of 2.09, while one extended
source has line ratios typical of a LINER at $z$=1.35 (see \citep{mai06} for further
details on these classifications).
For the remaining sources, the redshift has been estimated using the optical and near-IR
magnitudes, along with the morphological information, as extensively described in $\S$5.1 of
\cite{mig04}; all of the redshifts are in the range $\approx$~0.9--2.1.
The large column densities [$\approx10^{22}$ -- a~few$\times10^{23}$~cm$^{-2}$] and the
\hbox{2--10~keV}
luminosities [$\approx(1-8)\times10^{44}$~\rm erg~s$^{-1}$, once corrected for the absorption] place our sources
in the class of the high-luminosity, obscured AGN, the so-called Type~2 quasars (see, e.g.,
\citep{vig06} and references therein).
\section{{\it Spitzer\/}\ data}
For our sample of eight sources, we obtained IRAC observations of 480~s integration time and MIPS
observations at 24~\hbox{$\mu$m}\ for a total integration time per position of $\approx$~1400~s.
All of the sources are detected in the four IRAC bands and in MIPS;
the faintest source in MIPS has a 24~\hbox{$\mu$m}\ flux density of $\approx$~150~\hbox{$\mu$Jy}\
($\approx$~5$\sigma$ detection; see \cite{poz07} for further details on data reduction and
cleaning procedures).
\section{Analysis of the Type~2 quasar spectral energy distributions}
A reliable determination of the bolometric output of our AGN sample requires that the nuclear
component, directly related to the accretion processes, is disentangled from the emission of the
host galaxy, which provides a dominant contribution in the optical and $K_{\rm s}$\ bands (in the
case of extended sources, see \citep{mig04}).
To achieve this goal, we constructed SEDs for all our sources over the optical, near- and
mid-IR range. At the same time, we used {\it Spitzer\/}\ data to improve our previous estimates
on the source redshift when possible.
In the following, we consider the sample of six extended sources and two point-like objects
separately, since a different approach has been adopted for the two sub-samples.
\subsection{Extended sources}
As already pointed out, from the $K_{\rm s}$-band morphological analysis carried out by
\cite{mig04}, we know that at least up to 2.2~\hbox{$\mu$m}\ (observed frame) the stellar contribution
is mostly responsible for the emission of these sources.
At longer wavelengths, the emission of the active nucleus is expected to arise as reprocessed
radiation of the primary emission, while the emission from the galaxy should drop significantly,
assuming reasonable elliptical templates.
Although many models have been developed in the past to deal with circum-nuclear dust emission
(including the effects of the torus geometry and opening angle, grain size distribution and
density), in our study we adopted a more phenomenological approach.
To reproduce the observed data, we used a combination of two components, one for the host galaxy
and another related to the reprocessing of the nuclear emission.
For the galaxy component, we adopted a set of early-type galaxy templates obtained from the
synthetic spectra of \cite{bc03}, assuming a simple stellar population spanning a large range of
ages (see \citep{poz07} for details). For the nuclear component, we adopted the templates of
\cite{sil04}, which are based on the interpolation of the observed nuclear IR data
(at least, up to $\approx$~20~\hbox{$\mu$m}) of a sample of local AGN
through the radiative transfer models of \cite{gd94}.
The strength of such an approach is that the nuclear templates depend upon two quantities,
the intrinsic 2--10~keV luminosity (which provides the normalization of the SED) and the column
density (responsible for the shape of the SED), and these are known directly from the \hbox{X-ray}\
spectra (\citep{per04}), once the redshift is known.
We also used all the available information, extended over the {\it Spitzer\/}\ wavelength range, to
place better constraints on the source redshift than those reported in \cite{mig04}.
Overall, we find a good agreement with the redshifts presented in \cite{mig04}, although {\it Spitzer\/}\
allows us to provide estimates with lower uncertainties;
only for one source the redshift is significantly lower ($z\approx$~1 instead of $\approx$~2)
and likely more reliable.
The data are well reproduced by the sum of the two components; the emission from the galaxy
progressively becomes less important at wavelengths above $\approx$~4~\hbox{$\mu$m}\ (in the source
rest frame), where the nuclear reprocessed emission starts emerging significantly
(see Fig.~\ref{seds}, left panel), being dominant in MIPS at 24~\hbox{$\mu$m}.
Furthermore, the latter is fully consistent with the upper limits provided in the
$K_{\rm s}$\ band by \cite{mig04}.
\begin{figure}
\includegraphics[height=0.28\textheight]{vignali_fig1a.ps}
\includegraphics[height=0.28\textheight]{vignali_fig1b.ps}
\caption{\label{seds} Rest-frame SEDs for two representative Type~2 quasars of the current
sample: an obscured AGN hosted by an elliptical galaxy (on the left) and a point-like AGN
(on the right).
{\it (Left)} The observed data (filled circles) are reproduced by summing up (solid line)
the contribution of an early-type galaxy template (dot-dashed line) to the reprocessed
nuclear component (dashed line).
The dotted line shows the nuclear component obtained from the templates of \cite{sil04},
normalized using the \hbox{X-ray}\ luminosity and column density (i.e., without fitting the data; see text
and \citep{poz07} for details).
The downward-pointing arrow indicates the constraint on the nuclear emission derived from the
$K_{\rm s}$-band data (\citep{mig04}). The combination of the two templates is also consistent with the
$R$-$K_{\rm s}$ color.
{\it (Right)} The observed data (filled circles) are well reproduced by the red quasar template
from \cite{pol06} (solid line).
}
\end{figure}
\subsection{Point-like sources}
For the two point-like sources, we adopted a different strategy, since their emission in
the near-IR is dominated by the unresolved AGN. To reproduce their observed SEDs,
we extincted a Type~1 quasar template from \cite{elv94}
with several extinction laws, but we were not able to find a satisfactory solution.
Then we used the recently published red quasar template from \cite{pol06}
and found good agreement with the data (Fig.~\ref{seds},
right panel), consistently with the results obtained for some obscured AGN in the ELAIS-S1 field
(\citep{gru07}).
As in the AGN sub-sample described above, most of the uncertainty lies in the far-IR, where
a proper study of the SEDs would require MIPS data at 70 and 160~\hbox{$\mu$m}.
\section{Bolometric corrections
The determination of the SEDs is meant to be the first step toward the estimate of the bolometric
luminosities (\hbox{$L_{\rm bol}$}) of obscured AGN. The bolometric luminosities can be estimated from the
luminosity in a given band by applying a suitable bolometric correction \hbox{$k_{\rm bol}$};
typically, to convert the 2--10~keV luminosity into \hbox{$L_{\rm bol}$}, \hbox{$k_{\rm bol}$}$\approx$~30 is assumed,
although this value was derived from the average of few dozens of bright, mostly low-redshift
Type~1 quasars (\citep{elv94}).
For obscured sources, only few estimates are present in literature (e.g., \citep{pol06}).
We derived \hbox{$k_{\rm bol}$}\ by integrating the quasar SEDs over the \hbox{X-ray}\
(0.5--500~keV) and IR (1--1000~\hbox{$\mu$m}) intervals;
in the \hbox{X-ray}\ band, we converted the \hbox{2--10~keV} luminosity assuming a power law with photon
index $\Gamma=1.9$ (typical for AGN emission) and the observed column density (\citep{per04}).
To derive the bolometric corrections, we accounted for both the covering factor of the
absorbing material (i.e., the opening angle of the torus) and the anisotropy of the IR emission.
According to unification models of AGN, the former effect should be directly related to the
observed fraction of Type~2/Type~1 AGN which, in the latest models of \cite{gil07},
is $\approx$~1.5 in the luminosity range of our sample.
Furthermore, the torus is likely to re-emit a fraction of the intercepted radiation in a direction
which does not lie along our line-of-sight; the correction for this anisotropy, according to
the templates of \cite{sil04}, is $\approx$~10--20\% (given the column densities of our sources).
Once these corrections are taken into account, we obtain $\langle$\hbox{$k_{\rm bol}$}$\rangle\approx35$
(median \hbox{$k_{\rm bol}$}$\approx$~26), similar to the average value of \cite{elv94}; the Type~1.9 quasar at
$z$=2.09 has the highest \hbox{$k_{\rm bol}$}\ ($\approx$~97); see \cite{poz07} for a discussion on the
uncertainties in these estimates.
\section{Black hole masses and Eddington ratios}
For the six AGN hosted by elliptical galaxies, we can derive both the galaxy and black hole masses.
Since the near-IR emission is dominated by the galaxy starlight, we computed
the rest-frame $L_{\rm K}$ assuming the appropriate SED templates and then the galaxy masses using
\hbox{$M_{\star}$}/$L_{\rm K}\approx0.5-0.9$ (\citep{bc03}); all of our AGN are hosted by massive galaxies
($\approx1-6\times10^{11}$~M$_{\odot}$).
To estimate the black hole masses, we used the local \hbox{$M_{\rm BH}$}--$L_{\rm K}$ (\citep{mh03}) which, along
with the \hbox{$M_{\star}$}/$L_{\rm K}$ values, provides a \hbox{$M_{\rm BH}$}--\hbox{$M_{\star}$}\ relation.
Despite several attempts in the recent literature to investigate
whether and how the black hole mass vs. stellar mass relation
evolves with cosmic time, there is no consensus yet. In this work, we assume the findings of
\cite{pen06}, who found that in the redshift range covered by our sources, the \hbox{$M_{\rm BH}$}-\hbox{$M_{\star}$}\ relation
evolves by a factor of $\approx$~2 with respect to the local value; see \cite{poz07} for an
extensive discussion. Under this hypothesis, we obtain black hole masses for the six obscured
quasars hosted by elliptical galaxies of $\approx2.0\times10^{8}-2.5\times10^{9}$~M$_{\odot}$;
these values are broadly consistent with the average black hole masses obtained by \cite{md04}
for the Sloan Digital Sky Survey (SDSS) Type~1 quasars (using optical and ultra-violet mass scaling
relationships) in our redshift range ($\approx3.5\times10^{8}-8.6\times10^{8}$~M$_{\odot}$).
As a final step, we derived the Eddington ratios, defined as \hbox{$L_{\rm bol}$}/\hbox{$L_{\rm Edd}$}, where \hbox{$L_{\rm Edd}$}\ is the
Eddington luminosity. We note that the uncertainties related to these estimates are clearly large,
due to the uncertainties of the approach adopted to derive the bolometric luminosities (through the
templates of \citep{sil04}) and the black hole masses (see above). The average Eddington ratio
is $\approx$~0.05, suggesting that our obscured quasars may have already passed their
rapidly accreting phase and are reaching their final masses at low Eddington rates.
The Eddington ratios of our sources are significantly lower than those derived for the SDSS
Type~1 quasars in the same redshift range ($\approx$~0.3--0.4, see \citep{md04}).
\section{Summary}
We used optical, near-IR, and {\it Spitzer\/}\ IRAC and MIPS (at 24~\hbox{$\mu$m}) data to unveil the
reprocessed nuclear emission of eight hard \hbox{X-ray}\ selected Type~2 quasars at
\hbox{$z\approx0.9-2.1$}.
From proper modelling of the nuclear SEDs, we derived a median (average) bolometric correction of
$\approx$~26 ($\approx$~35). For the six obscured sources dominated by the host galaxy starlight
up to near-IR wavelengths, we also derived black hole masses of the order of
\hbox{$2.0\times10^{8}-2.5\times10^{9}$~M$_{\odot}$} and relatively low Eddington ratios ($\approx$~0.05),
suggestive of a low-activity accretion phase.
\begin{theacknowledgments}
The authors acknowledge partial financial support by the Italian Space Agency under the
contract ASI--INAF I/023/05/0.
\end{theacknowledgments}
\IfFileExists{\jobname.bbl}{}
{\typeout{}
\typeout{******************************************}
\typeout{** Please run "bibtex \jobname" to optain}
\typeout{** the bibliography and then re-run LaTeX}
\typeout{** twice to fix the references!}
\typeout{******************************************}
\typeout{}
}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 5,505
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{"url":"http:\/\/stackoverflow.com\/questions\/7352493\/failed-to-load-the-jni-shared-library-jdk","text":"# Failed to load the JNI shared Library (JDK)\n\nWhen I try opening Eclipse, a pop-up dialog states:\n\nFailed to load the JNI shared library \"C:\/JDK\/bin\/client\/jvm.dll\".\n\nFollowing this, Eclipse force closes.\n\nHere's a few points I'd like to make:\n\n\u2022 I checked to see if anything exists at that path. It does exist.\n\u2022 My Eclipse and Java SE Development Kit are both 64-bit. I checked my system, and it can handle 64-bit.\n\u2022 I've searched for this problem on Google and on Stack\u00a0Overflow, and the only answer I found was to download the 32-bit versions of JDK and Eclipse.\n\nWhat would be suggested to solve this issue?\n\n-\nI had, to my surprise, 5 or so Java installed onto my computer. Not only that, but it turned out that I downloaded the 32-bit Eclipse by accident. So, I uninstalled every Java I had, deleted Eclipse (Eclipse doesn't have an \"uninstall\"), and downloaded Eclipse Classic 64-bit for my machine, along with a 64-bit Java (see SPP's answer for that link). \u2013\u00a0 Mike S. Jul 4 '12 at 10:39\nAlso: if anyone has trouble \"deleting\" their Eclipse (e.g. the folder won't go away no matter how many times you remove it manually or via command prompt), try deleting it using safe mode with command prompt.. I have no idea why I had to do this, but that ended up working. \u2013\u00a0 Mike S. Jul 4 '12 at 10:42\n\nYou need a 64-bit trio:\n\n\u2022 64-bit OS\n\u2022 64-bit Java\n\u2022 64-bit Eclipse\n-\nThat is what I have though. Yet, this error still pops up. Meanwhile, I just downloaded the 32-bit Eclipse and Troubleshoot it's compatibility every time I open it up (highly undesirable). \u2013\u00a0 Mike S. Sep 12 '11 at 14:22\ntry to copy the java64-jre to eclipse\/jre (so eclipse will find eclipse\/jre\/bin\/java.exe). Does this solve your problem? \u2013\u00a0 Peter Rader Sep 13 '11 at 7:12\ni have all 3 installed , but i got the error too. i've done what \"jayesh kavathiya\" wrote , and it works . \u2013\u00a0 android developer Mar 27 '12 at 20:56\nSee my answer below, it is more comprehensive. I also had no need to edit the eclipse.ini file like 'jayesh kavathiya' suggested. \u2013\u00a0 sjas Jun 12 '12 at 12:26\nI ended up accepting this answer at the time because it was one of a few only answers I had at the approximate time of asking the question. I do advise everyone look at all the other spectacular answers though, a lot of them are very insightful. \u2013\u00a0 Mike S. Jun 21 '12 at 15:11\n\n# Working pairings of OS, JDK and Eclipse:\n\n\u2022 32-bit OS - 32-bit JDK - 32-bit Eclipse (32-bit only)\n\u2022 64-bit OS - 32-bit JDK - 32-bit Eclipse\n\u2022 64-bit OS - 64-bit JDK - 64bit Eclipse (64-bit only)\n\nI had several JDKs and JREs installed.\n\nEach of them had their own entry in the PATH variable, all was working more or less.\n\nJudging from the PATH variables, some installations were completely useless, since they were never used. Of course, the \"inactive\" Javas could be referenced manually from within Eclipse if I needed, but I never did that, so I really did not need them. (At least I thought so at that time...)\n\nI cleaned up the mess, deinstalled all current Java's, installed only JDK + JRE 1.7 64-bit.\n\nOne of the Eclipse 'installations' failed afterwards with the Failed to Load the JNI shared Library and a given path relative to the fresh installed JDK where it thought the jvm.dll to be.\n\nThe failing Eclipse was the only one of all my IDEs that was still a 32-bit version on my otherwise all-64-bit setup.\n\nAdding VM arguments, like so often mentioned, in the eclipse.ini was no use in my case (because I had only the wrong JDK\/JRE to relate to.)\n\nI was also unable to find out how to check if this Eclipse was a 32-bit or 64-bit version (I could not look it up in the Task Manager, since this Eclipse 'installation' would not start up. And since it had been a while since I had set it up, I could not remember its version either.)\n\nIn case you use a newer JDK and a older JRE you might be in for trouble, too, but then it is more likely a java.lang.UnsupportedClassVersionError appears, IIRC.\n\n-\nIt should only use the first \\bin it comes to in the Path variable \u2013\u00a0 ekinnear Jul 20 '12 at 18:21\nto check which version 32-bit or 64-bit, open eclipse.ini and look for the equinox launcher used (e.g. org.eclipse.equinox.launcher.win32.win32.x86_64_1.1.100.v20110502) \u2013\u00a0 Steve Oh Jun 6 '13 at 9:16\nIt was the comment by @ekinnear that solved my issue. I am not sure why it was looking for the JVM.dll in my windows\/bin folder but putting the proper path in place seems to have resolved it. \u2013\u00a0 James Mar 31 at 18:58\n\nMake sure your eclipse.ini file includes the following lines.\n\n-vm\nC:\\path\\to\\64bit\\java\\bin\\javaw.exe\n\n\nMy eclipse.ini for example:\n\n-startup\nplugins\/org.eclipse.equinox.launcher_1.1.1.R36x_v20101122_1400.jar\n--launcher.library\nplugins\/org.eclipse.equinox.launcher.win32.win32.x86_64_1.1.2.R36x_v20101222\n-product\norg.eclipse.epp.package.java.product\n--launcher.defaultAction\nopenFile\n--launcher.XXMaxPermSize\n256M\n-showsplash\norg.eclipse.platform\n-vm\nC:\\Program Files\\Java\\jdk1.6.0_32\\bin\\javaw.exe\n--launcher.XXMaxPermSize\n256m\n--launcher.defaultAction\nopenFile\n-vmargs\n-Dosgi.requiredJavaVersion=1.5\n-Xms40m\n-Xmx512m\n\n\nUse OS and Eclipse both 64 bit or both 32 bit keep same and config eclipse.ini.\n\n-\nShould be javaw.exe not javaws.exe. That threw me at first \u2013\u00a0 Dave H Aug 14 '12 at 20:14\nThis solves the problem. I added the -vm C:\\Program Files\\Java\\jdk1.6.0_32\\bin\\javaw.exe line to my eclipse.ini file and it started up just fine. Thanks for the help +1. \u2013\u00a0 prolink007 Sep 10 '12 at 13:36\nYou cannot use a relative path (as I tried to do). See stackoverflow.com\/questions\/4945178\/\u2026 \u2013\u00a0 user77115 Oct 10 '12 at 3:30\nI did this with my Java 6 SDK, and it didn't work. I tried it again, pointing to my Java 7 SDK, and it worked. I'm on Windows 8 64-bit, Java 6 and 7 (both 64-bit) and Eclipse Helios (32-bit). I was ready to uninstall and re-install Eclipse 64-bit, but it seems the java 7 SDK includes some new intelligence to run 32 bit apps better. \u2013\u00a0 Ryan Shillington Jan 9 '13 at 15:40\nthis works. but need to put -vm in the top of the ini file; at least before -vmargs or --launcher.xxx \u2013\u00a0 pinkdawn Mar 27 '13 at 1:33\n\nI resolved it by installing 64 bit JVM from\n\n-\nThanks for posting this link. \u2013\u00a0 Dan Doyon May 10 '12 at 23:31\n\nI have multiple versions of Java installed, both Sun JDK & JRockit, both 32 bit and 64-bit, etc. and ran into this problem with a fresh install of 64-bit Eclipse for Java EE (JUNO).\n\n## What did NOT work:\n\n64-bit trio as suggested by Peter Rader:\n\nI'm using 64-bit Eclipse on 64-bit OS (Windows 7).\n\nI ensured Sun JDK 7 64-bit was the default java version. When I typed \"java -version\" from command line (cmd.exe), Sun JDK 7 64-bit was returned...\n\njava version \"1.7.0\"\nJava(TM) SE Runtime Environment (build 1.7.0-b147)\nJava HotSpot(TM) 64-Bit Server VM (build 21.0-b17, mixed mode)\n\n\nThis did not resolve the problem for me.\n\n## What DID work:\n\nAdding -vm option to eclipse.ini as suggested by Jayesh Kavathiya:\n\nI added the following to eclipse.ini:\n\n-vm\nC:\/apps\/java\/jdk7-64bit\/bin\/javaw.exe\n\n\n## Note:\n\nI did not have to uninstall any of the various versions of JDK or JRE I have on my machine.\n\n-\n\nAnother option is:\n\nCreate a shortcut to the Eclipse.exe. Open the shortcut and change the target to:\n\n\"C:\\Program Files\\eclipse\\eclipse.exe\" -vm \"c:\\Program Files\\Java\\jdk1.7.0_04\\bin\\javaw.exe\"\n\n\nFor your installation, make sure the locations point to the correct Eclipse installation directory and the correct javaw.exe installation directory.\n\n(The 64\/32 bit versions of Eclipse and Java need to be the same, of course.)\n\n-\n\nFor a missing jvm.dll file, we can provide the path of the dll file in eclipse.ini file as\n\n-vm\nC:\\Progra~1\\Java\\jdk1.6.0_38\\jre\\bin\\server\\jvm.dll\n\n\nHere it is important to remove any space in the path and the double quotes. It worked for me when i removed the quotes and space.\n\nI hope it helps someone.\n\n-\n\nI had a similar problem. It was solved doing the following.\n\n\u2022 Move Eclipse to Program Files (not to Program Files (x86)).\n\u2022 Remove the path to the 32-bit version of Java from the 'path' environment variable.\n\nI have both versions of Java installed, but Eclipse kept trying to use the 32-bit one.\n\n-\n\nAs many folks already alluded to, this is a 32 vs. 64 bit problem for both Eclipse and Java. You cannot mix up 32 and 64 bit. Since Eclipse doesn't use JAVA_HOME, you'll likely have to alter your PATH prior to launching Eclipse to ensure you are using not only the appropriate version of Java, but also if 32 or 64 bit (or modify the INI file as Jayath noted).\n\nIf you are installing Eclipse from a company-share, you should ensure you can tell which Eclipse version you are unzipping, and unzip to the appropriate Program Files directory to help keep track of which is which, then change the PATH (either permanently via (Windows) Control Panel -> System or set PATH=\/path\/to\/32 or 64bit\/java\/bin;%PATH% (maybe create a batch file if you don't want to set it in your system and\/or user environment variables). Remember, 32-bit is in Program files (x86).\n\nIf unsure, just launch Eclipse, if you get the error, change your PATH to the other 'bit' version of Java, and then try again. Then move the Eclipse directory to the appropriate Program Files directory.\n\n-\n\nSure, you need to have a compatible version of JDK and Eclipse, but you also need to add in the eclipse.ini file the below lines:\n\n-vm\nyourdrive\\java\\bin\n\n\nMake them the first two lines of your eclipse.ini file.\n\n-\n\nAlternatively, get the same \"bit\" version of JRE and Eclipse and then create a new shortcut with the below target (replace the installed JRE and Eclipse location\/path):\n\n\"C:\\studio\\eclipse.exe\" -vm \"C:\\Program Files\\Java\\jre7\\bin\\server\\jvm.dll\" eclipse.vm=\"C:\\Program Files\\Java\\jre7\\bin\\server\\jvm.dll\" java.home=\"C:\\Program Files\\Java\\jre7\" java.runtime.version=1.7.0\n\n\nThat should do the trick.\n\n-\n\nYes, just make sure your versions of Eclipse and JDK are both 64-bit. Just to make sure everything is correct uninstalled JDK and install it in Program Files and not in Program Files (x86). At least that resolved my problem.\n\n-\n\nYou can solve that problem as many other replicated. You need that Eclipse and the JDK be 32-bits or both on 64-bits. The architecture of the OS doesn't matter while the others remains on the same type of arquitecture.\n\n-\n\nThank you misterfrb, I realised that Eclipse was giving this error, because I had just installed Oracle 10g Developer suite, and it was looking for the jvm.dll file in the C:\\DevSuiteHome_1 folder (I must have opted to install JDK again along with developer suite).\n\nAfter removing the DevSuiteHome lines from the paths variable and adding the correction location for 64-bit jvm.dll (not sure if this was necessary, didn't try without), Eclipse worked again, and Developer suite still does too.\n\n-\n\nOne of the easy ways to resolve it is to copy the jre folder from installed the JDK into the Eclipse installation folder. Make sure that JDK you copy from is the same architecture as your Eclipse installation.\n\nI had to configure my machine that way, because I run both Eclipse and Appcelerator Titanium Studio on my machine. The Studio needs 32-bit Java, while Eclipse needs 64-bit.\n\n-\n\nYou should uninstall all old [JREs][1] and then install the newest one... I had the same problem and now I solve it. I've:\n\nBetter install Jre 6 32 bit. It really works.\n\n-\n\nDownloaded 64 bit JVM from site and installed it manually and updated the system path variable. That solved the issue.\n\n1. Default JVM is installed in my system was in \"C:\\Program Files\n(x86)\\Java\\jre7\"\n2. Manually installed JVM got installed in \"C:\\Program Files\\Java\\jre7\" and after updating this pate to system path variable it worked.\n-\n\nMake sure you are starting Eclipse with Administrator rights.\n\n-\nShould this be a comment? \u2013\u00a0 Austin Henley Nov 11 '12 at 1:35\n\nJust check the PATH environment variable. In My Computer - > Properties -> Advanced System settings -> Environment Variables -> (left upper window \"User Variables for \"some name of PC\"\" ) just check the PATH variable. If it doesn't exist create it with the following -- > C:\\Program Files (x86)\\Java\\jre7\\bin <--\n\nI was faced with the same problem after had updated my Eclipse. I've found that the path asked 64-bit version, but I had the 32-bit in dif path. It was helpful for me. P.S.: I have a 64-bit OS, 32-bit JRE and 32-bit Eclipse. All works fine :)\n\n-\n\nUse the Elipse site itself.\n\nThe above link had an older 3.6 Eclipse which then failed to update itself due to Eclipse Bug #317785.\n\nMy solution was to just install 32-bit Java alongside 64-bit - this allowed the Zend installer to work.\n\nIt's depressing that amount of Java \/ Eclipse cruft one has to go through to get a PHP IDE.\n\n-\n\nThe answers above me got me tempted so much, that I decided to dry run all the possible combinations with OS, Eclipse and JVM trio. Anyway, whoever is digging down and reading my post, check the following as a hot spot (I am Windows\u00a07 user).\n\n1. You understand Program Files and Program File (x86) are two different folders... x86 stands for the 32-bit version of programs and the former is the 64-bit version.\n\n2. If you have multiple versions of Java installed with different bitness and release versions, which is bound to happen with so many open source IDEs, managers, administrative consoles, the best option is to set the VM argument directly in the eclipse.ini file. If you don't, Eclipse will go crazy and try searching itself which is not good.\n\nIf you are interested in extensive debugging and various commands, I used to support my points. Please look in Eclipse failing to launch. Thanks for reading this loooong post....\n\n-\n\nSimple, I have a 64-bit OS, 32-bit Eclipse and both JDK 32 & 64 installed... I just uninstalled the 64-bit JDK and Eclipse is working fine..\n\n-\n\nIf you use whole 64-bit trio and it still doesn't work (I've come to this problem while launching Android Monitor in Intellij Idea), probably wrong jvm.dll is being used opposed to what your java expects. Just follow these steps:\n\n1. Find the jvm.dll in your JRE directory: C:\\Program Files\\Java\\jre7\\server\\bin\\jvm.dll\n\n2. Find the jvm.dll in your JDK directory: c:\\Program Files\\Java\\jdk1.7.0_xx\\jre\\bin\\server\\\n\n3. Copy the jvm.dll` from JRE drectory into your JDK directory and overwrite the jvm.dll in JDK.\n\nDon't forget to make a backup, just in case. No need to install or uninstall anything related to Java.\n\n-\n\nI had the same issue after upgrading from Java 6 to Java 7. After I removed Java 6 (64 bit) and reinstalled Java 7 (64 bit), Eclipse worked. :)\n\n-\n\nI'm not sure why but I had the jre installed into my c:\\windows directory and java.exe and javaw.exe inside my windows\\system32 directory.\n\nObviously these directories were getting priority even AFTER adding the -vm flag to my eclipse.ini file.\n\nDelete them from here fixed the issue for me.\n\n-\n\nOn the download page of Eclipse, it should be written \"JRE 32 bits\" or \"JRE 64 bits\" and not \"Windows 32 bits\" or \"Windows 64 bits\".\n\nBe sure to use the correct version compatible with your JDE, as answered previously.\n\n-\n\nYou can install the 32-bit version of JDK on a 64-bit machine. See JDK 7 downloads.\n\n-\nYour answer is not helpful. It is already known that this is possible. The OP wants to avoid it. \u2013\u00a0 nalply Oct 5 '12 at 18:51\n\nI have experienced all of the Eclipse errors and this is one of them. The problem is Eclipse 64-bit version. Download the 32-bit version and launch it.\n\n-\n\nThe same occurred to me. I had 64-bit Eclipse, but my JDK was 32-bit. So I installed the 64-bit version and it's OK right now.\n\n-\n\nIn my case, I tried to launch java from the command prompt and got this error\n\nError: could not open \"C:\\Windows\\jre\\lib\\amd64\\jvm.cfg\"\n\nIt meant \"java\" was looked for in the PATH starting at this wrong directory. Deleting the folder C:\\Windows\\jre\\ solved the issue\n\n-\n\n## protected by Community\u2666Dec 4 '12 at 12:08\n\nThank you for your interest in this question. 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\section{Introduction}
\label{sec:introduction}
\noindent
\PARstart{P}{olar} codes, invented by Ar\i kan~\cite{Arikan:09p},
achieve the capacity of arbitrary binary-input symmetric DMCs.
Moreover, they have low encoding and decoding complexity and
an~exp\-li\-cit construction.
Following Ar\i kan's seminal paper~\cite{Arikan:09p},
his results have been extended in a~variety of important ways.
In \cite{STA:09a}, polar codes have been generalized to symmetric
DMCs with \emph{non-binary} input alphabet.
In \cite{KSU:10p}, the polarization phenomenon has been
studied for \emph{arbitrary kernel matrices}, rather
than Ar\i kan's original $2 \times 2$ polarization kernel,
and error~exponents were derived for each such kernel.
It was shown~in~\cite{TV:11} that, under \emph{list-decoding},
polar codes can achieve remarkably good performance at short
code lengths.
In terms of applications, polar coding has been used with great
success in the context of
multiple-access channels \cite{STY:10a,AbbeTelatar:10a},
wiretap channels~\cite{MahdavifarVardy:11p},
data compression~\cite{Arikan:10a,Abbe:11a},
write-once channels~\cite{BurshteinStrugatski:12a},~and
channels with memory~\cite{Sasoglu:11c}.
In this paper, however, we will~restrict~our attention
to the original setting introduced by Ar\i kan\ in~\cite{Arikan:09p}.
Namely, we focus on binary-input, discrete, memoryless,
symmetric~channels,
with the standard $2 \,{\times}\, 2$ polarization~kernel~under
standard successive cancellation decoding.
Although the construction of polar codes is \emph{explicit}, there is
only one known instance --- namely, the binary erasure~channel (BEC) ---
where the
construction is also \emph{efficient}. A first attempt at an efficient
construction of polar codes in the general case was made by Mori and
Tanaka~\cite{Mori:10z,MoriTanaka:09c}.
Specifically, it is shown in~\cite{Mori:10z} that
a key step in the construction of polar bit-channels
can~be viewed as an instance of density
evolution~\cite{RichardsonUrbanke:08b}.
Based on this observation, Mori and Tanaka~\cite{MoriTanaka:09c}
proposed a construction algorithm utilizing convolutions,
and proved that the number of convolutions needed scales
linearly with the code length.
However, as indeed noted in \cite{Mori:10z}, it is
not clear how one would implement such convolutions to be
sufficiently precise on one hand while being tractable
on the other hand.
\looseness=-1
In this paper, we further extend the ideas of
\cite{Mori:10z,MoriTanaka:09c}.
An~exact implementation of the convolutions discussed
in \cite{Mori:10z,MoriTanaka:09c}\linebreak
implies an algorithm with memory requirements
that grow exponentially with the code length. It is thus impractical.
Alternatively, one could use quantization (binning) to try and reduce
the memory requirements. However, for such quantization scheme
to be of interest, it must satisfy two conditions.
First, it must be fast enough, which usually translates
into a rather small number of quantization levels (bins).
Second, after the calculations have been carried out, we must
be able to interpret them in a precise manner. That is, the
quantization operation introduces inherent inaccuracy into
the computation, which we should be able to account for
so as to ultimately make a precise statement.
Our aim in this paper is to provide a method by which~polar codes
can be efficiently constructed. Our main contribution~consists of
two approximation methods. In both methods, the memory limitations are
specified, and not exceeded. One method is used to get a lower bound
on the probability of error of each polar bit-channel while the other
is used to obtain an upper bound. The quantization used to derive a lower
bound on the probability of error is called a \emph{degrading quantization},
while the other is called an \emph{upgrading quantization}. Both
quantizations transform the ``current channel'' into a new one with a
smaller output alphabet. The degrading quantization results in a
channel degraded with respect to the original one, while the upgrading
quantization results in a channel such that the original channel is
degraded with respect to it.
The fidelity of both degrading and upgrading approximations
is a function of a parameter $\mu$,
which can be freely set to~an arbitrary integer value.
Generally speaking, the larger $\mu$ is the better the approximation.
The running time needed in order to approximate all $n$ polar bit-channels
is $O(n \cdot \mu^2 \log \mu)$.
Our results
relate to both theory and practice of polar codes.
In practice, it turns out that the degrading~and~upgrading approximations
are typically very close,
even for relatively small values of the fidelity parameter $\mu$.
This is illustrated in what follows with the help of
two examples.
\begin{figure*}
\centering
\includegraphics[width=122mm]{Figures/bw-zoom-adjusted.pdf}
\caption{%
\hspace*{-1ex}Upper and lower bounds on
the bit-channel probabilities of error
for a polar code of length $n = 1,048,576$ on BSC($0.11$),
computed using degrading and upgrading algorithms with $\mu = 256$.
Only those $132$ bit-channels for which the gap between
the upper and lower bounds crosses the $10^{-9}$ threshold
are shown.
}
\label{fig:BSCthreshold}
\end{figure*}
\begin{figure*}
\centering
$\,$\\[2.00ex]
\subfigure[Binary symmetric channel $\mathrm{BSC}(0.001)$]%
{\label{subfig:011}\includegraphics[width=60mm]{Figures/rate0001.pdf}}
\hspace*{9ex}
\subfigure[binary-input AWGN channel with
$E_s/N_0=5.00$\,dB]%
{\label{subfig:0001}\includegraphics[width=60mm]{Figures/AWGN_graph.pdf}}
\caption{%
\hspace*{-1ex}Upper and lower bounds on $P_{W_{\mathrm{base}},n}(k)$
as a function of rate $R=k/n$, for two underlying channels
and two code lengths $n = 2^{10}$ and $n = 2^{20}$.
The upper bound is dashed while the lower bound is solid.
For both channels, the difference between the bounds can only
be discerned in the plot corresponding to $n=2^{20}$.%
}
\label{fig:BSCplots}
\end{figure*}
\vspace{0.50ex}
\noindent
{\bf Example\,1.}
Consider a polar code of length $n = 2^{20}$ for the binary symmetric
channel (BSC) with crossover probability~$0.11$. Let
$\mathcal{W}_0,\mathcal{W}_1,\ldots,\mathcal{W}_{n-1}$ be the corresponding bit-channels
(see the next section for a rigorous definition of a bit-channel).
The basic task in the construction of polar codes is that of classifying
bit-channels into those that are ``good'' and those that are ``bad.''
Let $P_e(\mathcal{W}_i)$ denote the probability of error on the
$i$-th bit-chan-nel (see \Eq{eq:errorProbDef} for a precise definition
of this quantity)~for~$i = 0,1,\ldots,n\,{-}\,1$.
We arbitrarily choose a threshold of $10^{-9}$
and say that the $i$-th bit channel is good if
$P_e(\mathcal{W}_i) \leq 10^{-9}$ and bad otherwise.
How well do our algorithms perform~in~determining for
each of the $n$ bit-channels whether it is good or bad?
Let us set $\mu = 256$ and compute upper and lower bounds
on $P_e(\mathcal{W}_i)$ for all $i$, using the
degrading and upgrading quantizations, respectively.
The results of this computation are illustrated in
Figure\,\ref{fig:BSCthreshold}.
In $1,048,444$ out of the $1,048,576$ cases,
we~can provably
classify the bit-channels into good and bad.
Figure\,\ref{fig:BSCthreshold} depicts
the remaining $132$ bit-channels
for which the upper
bound is above the threshold whereas the lower bound
is below the threshold.
The horizontal axis in
Figure\,\ref{fig:BSCthreshold}
is the bit-channel index while the vertical
axis is the gap between the two bounds.
We see that the gap between the upper and lower bounds,
and thus the remaining uncertainty as to the true
value of $P_e(\mathcal{W}_i)$, is very small in all cases.
\hfill\raisebox{-0.50ex}{$\Box$}
\vspace{1.00ex}
\noindent
{\bf Example\,2.}
Now suppose we wish to construct a polar code~of a given length $n$
having the \emph{best possible rate} while guaranteeing a certain
block-error probability $P_{\rm block}$ under successive cancellation
decoding. Ar\i kan~\cite[Proposition 2]{Arikan:09p} provides\footnote
{%
In \cite{Arikan:09p}, Ar\i kan\ uses the Bhattacharyya parameter
$Z(\mathcal{W}_i)$ instead of the probability of error $P_e(\mathcal{W}_i)$.
As we shall see shortly, this is of no real importance.
}
a union bound on the block-error rate of polar codes:
\be{union}
P_{\rm block}
\ \le \
\sum_{i \in \cA} P_e(\mathcal{W}_i)
\ee
where $\cA$ is the \emph{information set} for the code
(the set of unfrozen bit-channels). The construction
problem for polar codes can be phrased (cf.~\cite[Section\,IX]{Arikan:09p})
as the problem
of choosing an information set $\cA$ of a given size
$|\cA| = k$ so as to minimize the right-hand side of~\Eq{union}.
Assuming the underlying channel $W_{\mathrm{base}}$ and
the code length $n$ are fixed, let
\be{PWnk-def}
P_{W_{\mathrm{base}},n}(k)
\ \ \mbox{$\stackrel{\rm def}{=}$} \ \
\min_{|\cA| = k} \, \sum_{i \in \cA} \!P_e(\mathcal{W}_i)
\ee
Using our degrading and upgrading algorithms, we
can~effici\-ent\-ly compute upper and lower bounds on $P_{W_{\mathrm{base}},n}(k)$.
These~are plotted in Figure\,\ref{fig:BSCplots} for two
underlying channels:
BSC~with cross\-over probability $0.001$
and the binary-input AWGN channel with a symbol SNR of $5.00$\,dB
(noise variance $\sigma^2=0.1581$).
In all\footnote{%
The initial degrading (upgrading) transformation of the
binary-input~con\-tinous-output AWGN channel to a binary-input
channel with a finite output alphabet was done according to the method
of~Section~\ref{sec:continuousChannels}. For that calculation,
we used a finer value of $\mu = 2000$. Note that the initial
degrading (upgrading) transformation is performed only once.
}
our calculations, the value of $\mu$ did not exceed $512$.
\looseness=-1
As can be seen from Figure\,\ref{fig:BSCplots},
the bounds effectively coincide.
As an example, consider polar codes of length $2^{20}$
and suppose we wish to guarantee
$P_{W_{\mathrm{base}},n}(k) \le 10^{-6}$.
What is the best possible rate of such a code?
According~to~Figure\,\ref{subfig:011}, we can
efficiently construct (specify the rows of
a~generator matrix) a polar
code of rate $R = 0.9732$.
On the other hand, we can also prove that there is
no choice of an information set $\cA$ in
\Eq{PWnk-def} that would possibly produce
a polar code of rate $R \:{\ge}\: 0.9737$.
According~to~Figure\,\ref{subfig:0001},
the corresp\-onding numbers for the binary-input
AWGN channel are $0.9580$ and $0.9587$.
In practice, such minute differences in
the code rate are
negligible.
\hfill\raisebox{-0.50ex}{$\Box$}
\vspace{1.00ex}
From a theoretical standpoint, one of our main contributions is the
following theorem. In essence, the theorem asserts that
capacity-achieving polar codes can be constructed in
time that is polynomial (in fact, linear) in their length $n$.\vspace{0.50ex}
\begin{theo}
\label{theo:polyConstruction}
Let $W_{\mathrm{base}}$ be a
binary-input, symmetric, discrete me\-moryless channel of
capacity $I(W_{\mathrm{base}})$. Fix arbitrary real
constants $\varepsilon > 0$ and $\beta < 1/2$. Then
there exists an even integer
\be{mu0-def}
\mu_0 \ = \ \mu_0(W_{\mathrm{base}},\varepsilon,\beta)\ ,
\ee
which does \emph{not} depend on the code length $n$,
such that the following holds.
For all even integers $\mu \geq \mu_0$ and
all sufficiently large code lengths $n=2^m$,
there is a construction algorithm with running
time $O(n \cdot \mu^2 \log \mu)$ that produces
a polar code for $W_{\mathrm{base}}$ of rate
$
R \ge I(W_{\mathrm{base}}) - \varepsilon
$\,
such that
$
P_{\rm block} \le \smash{2^{-n^\beta}}
$,
where $P_{\rm block}$ is the probability
of codeword error under successive~cancellation decoding.\vspace{1.00ex}
\end{theo}
We defer the proof of Theorem\,\ref{theo:polyConstruction} to
Section~\ref{sec:analysis}. Here, let us briefly discuss two
immediate consequences of this theorem. First, observe that
for a given channel $W_{\mathrm{base}}$ and any fixed $\varepsilon$
and $\beta$, the integer $\mu_0$ in \Eq{mu0-def} is a constant.
Setting our fidelity parameter
in Theorem\,\ref{theo:polyConstruction}
to $\mu = \mu_0$ thus yields a construction algorithm
with running time that is \emph{linear in $n$}.
Still, some might argue that the complexity of construction
in Theorem\,\ref{theo:polyConstruction} does depend on
a fidelity parameter $\mu$, and this is unsatisfactory.
The following corollary eliminates this dependence altogether,
at the expense of super-linear construction
complexity.\vspace{0.50ex}
\begin{coro}
Let $W_{\mathrm{base}}$ be a
binary-input, symmetric, discrete me\-moryless channel of
capacity $I(W_{\mathrm{base}})$. Fix arbitrary real
constants $\varepsilon > 0$ and $\beta < 1/2$. Then
there is a construction algorithm with running time
$O(n \log^2\!n \log\log n)$
that for all sufficiently large code lengths $n$,
produces a polar code for $W_{\mathrm{base}}$ of rate
$R \ge I(W_{\mathrm{base}}) - \varepsilon$\,
such that
$
P_{\rm block} \le \smash{2^{-n^\beta}}
$.
\end{coro}
\begin{proof}
Set $\mu=2\floor{\log_2 n}$ in Theorem~\ref{theo:polyConstruction}
(in fact, we could have used any function of $n$ that grows
without bound).\vspace{0.50ex}
\end{proof}
\looseness=-1
We would now like to draw the reader's attention to
what~The\-orem~\ref{theo:polyConstruction} \emph{does not} assert.
Namely, given $W_{\mathrm{base}}$, $\varepsilon$ and $\beta$, the theorem
does not tell us how large $n$ must be, only that some values~of $n$
are large enough.
In fact, given $W_{\mathrm{base}}$, $\varepsilon,\beta$,
how large does $n$ need to be in order to guarantee the
\emph{existence} of a polar code with
$R \ge I(W_{\mathrm{base}}) - \varepsilon$
and $P_{\rm block} \le \smash{2^{-n^\beta}}$,
let alone the complexity of its construction?
This is one of the central questions in the
theory of polar codes. Certain lower bounds
on this value of $n$ are given in
\cite{GoliHassaniUrbanke:12c}.
In the other direction, the exciting recent
result of Guruswami and Xia~\cite[Theorem\,1]{GuruswamiXia:13aa}
shows that for any fixed $W$ and $\beta \le 0.49$, this
value of $n$ grows as a polynomial in $1/\varepsilon$.
The work of \cite{GuruswamiXia:13aa} further shows
that, for any fixed $W$ and $\beta$, the
parameter $\mu_0$ in \Eq{mu0-def} can be also taken
as a polynomial in $1/\varepsilon$.
\looseness=-1
The rest of this paper is oragnized as follows. In
Section~\ref{sec:polarCodes}, we briefly review polar codes and
set up the necessary notation.
Section~\ref{sec:degradeUpgrade} is devoted to
channel degrading and upgrading relations,
that will be important for us later on.
In Section~\ref{sec:highLevelAlg}, we give a high
level description of our algorithms for approximating
polar bit-channels. The
missing details in Section~\ref{sec:highLevelAlg} are then fully
specified in Section~\ref{sec:mergingFunctions}. Namely, we show
how to reduce the output alphabet of a channel so as
to get either~a~de\-graded or an upgraded version
thereof.
In Section~\ref{sec:continuousChannels}, we show how to either
degrade or upgrade a channel with continuous~output into
a~channel with a finite output alphabet of specified~size. In
Section~\ref{sec:variationsOnTheme}, we discuss certain
improvements to our general algorithms
for a specialized case. The accuracy of the
(improved) algorithms is then analyzed in Section~\ref{sec:analysis}.
\vspace{2ex}
\section{Polar Codes}
\label{sec:polarCodes}
In this section we briefly review polar codes with the primary aim of
setting up the relevant notation. We also indicate where the difficulty
of constructing polar codes lies.
Let $W_{\mathrm{base}}$ be the underlying memoryless channel through
which we are to transmit information.
If the input alphabet of $W_{\mathrm{base}}$ is $\mathcal{X}$ and its
output alphabet is $\mathcal{Y}$, we write $W_{\mathrm{base}}\!: \mathcal{X} \to \mathcal{Y}$.
The probability of observing $y \in \mathcal{Y}$ given that $x \in\mathcal{X}$
was transmitted is denoted by $W_{\mathrm{base}}(y|x)$.
We assume throughout that $W_{\mathrm{base}}$ has binary
input and so $\mathcal{X}=\mysett{0,1}$. We also assume that
$W_{\mathrm{base}}$ is symmetric.
As noted in~\cite{Arikan:09p}, a binary-input channel~$W_{\mathrm{base}}$
is symmetric if and only if there exists a permutation
$\pi$ of $\mathcal{Y}$
such that $\pi^{-1} = \pi$ (that is, $\pi$ is an involution) and
$W_{\mathrm{base}}(y|1) = W_{\mathrm{base}}(\pi(y)|0)$
for all $y \in \mathcal{Y}$ (see \cite[p.\,94]{Gallager:68b}
for an equivalent definition).
When~the\linebreak
permutation is understood from the context, we abbreviate $\pi(y)$ as
$\bar{y}$, and say that $\bar{y}$ and $y$ are \emph{conjugates}. For now, we
will further assume that the output alphabet $\mathcal{Y}$ of $W_{\mathrm{base}}$
is finite. This assumption will be justified in
Section~\ref{sec:continuousChannels}, where
we show how to
deal with channels that have continuous output.
Denote the length of the codewords we will be transmitting over $W_{\mathrm{base}}$ by $n = 2^m$. Given $\yyy = (y_0,y_1,\ldots,y_{n-1}) \in \mathcal{Y}^n$
and $\uuu = (u_0,u_1,\ldots,u_{n-1}) \in \mathcal{X}^n$,
let
$$
W_{\mathrm{base}}^n(\yyy|\uuu)
\ \ \mbox{$\stackrel{\rm def}{=}$}\ \
\prod_{i=0}^{n-1} W_{\mathrm{base}}(y_i|u_i)\; .
$$
Thus $W_{\mathrm{base}}^n$ corresponds to $n$ independent uses of the
channel~$W_{\mathrm{base}}$. A key paradigm introduced in \cite{Arikan:09p}
is that of transforming $n$ identical copies (independent uses) of the
channel $W_{\mathrm{base}}$ into $n$ polar {bit-channels}, through a
successive application of Ar\i kan\ channel transforms, introduced
shortly. For $i = 0,1,\ldots,n-1$, the $i$-th
\emph{bit-channel} $\mathcal{W}_i$ has a
binary input alphabet $\mathcal{X}$, an output alphabet $\mathcal{Y}^n \times \mathcal{X}^i$,
and transition probabilities defined as follows.
Let $G$ be the polarization kernel matrix of~\cite{Arikan:09p},
given by
$$
G
\ = \
\left[
\begin{array}{@{\hspace{0.750ex}}c@{\hspace{1.25ex}}c@{\hspace{0.750ex}}}
1 & 0\\
1 & 1\\
\end{array}
\right] \; .
$$
Let $\Gm$ be the $m$-fold Kronecker product of $G$ and let $B_n$ be
the $n \times n$ bit-reversal premutation matrix defined in
\cite[Section\,VII-B]{Arikan:09p}.
Denote $\uuu_{i-1} = (u_0,u_1,\ldots,u_{i-1})$.
Then
\begin{multline}
\label{eq:bitChannelDef}
\mathcal{W}_i\bigl(\yyy, \uuu_{i-1}|u_i\bigr)\ \ \mbox{$\stackrel{\rm def}{=}$} \\[0.75ex]
\frac{1}{2^{n-1}}\hspace*{-1ex}
\sum_{\mathbfsl{v} \in \mysett{0,1}^{n-1-i}\hspace*{-3ex}}
\hspace*{-1.00ex}
W_{\mathrm{base}}^n\Bigl(\yyy\,|\,(\uuu_{i-1},u_i,\mathbfsl{v}) B_n \Gm\Bigr) \; .
\end{multline}
Given the bit-channel output $\yyy$ and $\uuu_{i-1}$, the optimal
(maxim\-um-likelihood) decision rule for estimating $u_i$ is
$$
\widehat{u}_i
\ = \
\argmax\bigl\{
\mathcal{W}_i\bigl(\yyy, \uuu_{i-1}|0\bigr),\mathcal{W}_i\bigl(\yyy, \uuu_{i-1}|1\bigr)
\bigr\}
$$
with ties broken arbitrarily. This is the decision rule used
in successive cancellation decoding~\cite{Arikan:09p}.
As before, we let $P_e(\mathcal{W}_i)$ denote the probability that
$\widehat{u}_i \ne u_i$
under this rule, assuming that the {a priori} distribution
of $u_i$ is $\mathrm{Bernoulli}(1/2)$.
In essence, constructing a polar code of dimension $k$ is equivalent
to finding the $k$ ``best'' bit-channels. In~\cite{Arikan:09p},
one is instructed to choose the $k$ bit-channels $\mathcal{W}_i$
with the lowest
Bhattacharyya bound $Z(\mathcal{W}_i)$
on the probability of decision error $P_e(\mathcal{W}_i)$.
We note that the
choice of ranking according to these Bhattacharyya bounds stems from the
relative technical ease of manipulating them. A more straightforward
criterion would have been to rank directly according to the probability
of error $P_e(\mathcal{W}_i)$,
and this~is the criterion we will follow here.
Since $\mathcal{W}_i$ is well defined through
(\ref{eq:bitChannelDef}), this task is indeed explicit, and thus so is
the construction of a polar code. However, note that the output
alphabet size of each bit-channel is exponential in $n$. Thus a
straightforward evaluation of the ranking criterion is intractable for
all but the shortest of codes. Our main objective will be to
circumvent this difficulty.
As a first step towards achieving our goal, we recall that the
bit-channels can be constructed recursively using the Ar\i kan\ channel
transformations $\Wbitb \Wbit$ and $\Wbita
\Wbit$, defined as follows. Let $\Wbit\! : \mathcal{X} \to \mathcal{Y}$ be
a binary-input, memoryless, symmet\-ric (BMS) channel.
Then the output alphabet of
$\Wbitb \Wbit$ is $\mathcal{Y}^2$,
the output alphabet of $\Wbita \Wbit$
is $\mathcal{Y}^2 \times \mathcal{X}$,
and their transition probabilities are given by
\begin{multline}
\label{eq:badSplit}
\bigl(\Wbit {b} \Wbit\bigr)(y_1,y_2 | u_1)\,\ \mbox{$\stackrel{\rm def}{=}$} \\
\frac{1}{2} \sum_{u_2 \in \mathcal{X}} \Wbit(y_1|u_1 \oplus u_2) \Wbit(y_2|u_2)
\end{multline}
and
\begin{multline}
\label{eq:goodSplit}
\bigl(\Wbit {a} \Wbit\bigr)(y_1,y_2,u_1|u_2)\,\ \mbox{$\stackrel{\rm def}{=}$} \\
\frac{1}{2} \Wbit(y_1|u_1 \oplus u_2) \Wbit(y_2|u_2)
\end{multline}
One consequence of this recursive construction
is that the explosion in the output alphabet size happens
gradually: each tran\-sform application roughly squares
the alphabet size. We will take advantage of this fact
in Section~\ref{sec:highLevelAlg}.
\section{Channel Degradation and Upgradation}
\label{sec:degradeUpgrade}
As previously outlined, our solution to the explosion in growth of the output alphabet of $\mathcal{W}_i$
is to replace the channel $\mathcal{W}_i$
by an approximation. In fact, we will have two approximations, one yielding a ``better'' channel and the other yielding a ``worse'' one. In this section, we formalize these notions.
We say that a channel $\mathcal{Q} : \mathcal{X} \to \calZ$ is (stochastically) \emph{degraded} with respect to $\Wbit : \mathcal{X} \to \mathcal{Y}$, if there exists a channel $\cP: \mathcal{Y} \to \calZ$ such that for all $z \in \calZ$ and $x \in \mathcal{X}$,
\begin{equation}
\label{eq:degraded}
\mathcal{Q}(z|x) = \sum_{y \in \mathcal{Y}} \Wbit(y|x) \cdot \cP(z|y) \; .
\end{equation}
For a graphical depiction, see Figure~\ref{subfig:degraded}. We write $\mathcal{Q} \preccurlyeq \Wbit$ to denote that $\mathcal{Q}$ is degraded with respect to $\Wbit$.
\begin{figure}
\centering
\subfigure[Degrading]{\label{subfig:degraded}\includegraphics[width=52mm]{Figures/channelDegradation.pdf}}
\\
\subfigure[Upgrading]{\label{subfig:upgraded}\includegraphics[width=52mm]{Figures/channelUpgradation.pdf}}
\caption{Degrading and upgrading a channel $\Wbit$}
\label{fig:degradedUpgraded}
\end{figure}
In the interest of brevity and clarity later on, we also define the inverse relation: we say that a channel $\mathcal{Q}' : \mathcal{X} \to \calZ'$ is \emph{upgraded} with respect to $\Wbit :\mathcal{X} \to \mathcal{Y}$ if there exists a channel $\cP: \calZ' \to \mathcal{Y}$ such that for all $z' \in \calZ'$ and $x \in \mathcal{X}$,
\begin{equation}
\label{eq:upgraded}
\Wbit(y|x) = \sum_{z' \in \calZ'} \mathcal{Q}'(z'|x) \cdot \cP(y|z')
\end{equation}
(see Figure~\ref{subfig:upgraded}). Namely, $\mathcal{Q}'$ can be degraded to $\Wbit$. Similarly, we write this as $\mathcal{Q}' \succcurlyeq \Wbit$.
By definition,
\begin{equation}
\label{eq:degradedUpgradedIff}
\Wbit \preccurlyeq \Wbit' \quad \mbox{if and only if} \quad \Wbit' \succcurlyeq \Wbit \; .
\end{equation}
Also, it is easily shown that ``degraded'' is a transitive relation:
\begin{equation}
\label{eq:degradedTransitive}
\mbox{If} \quad \Wbit \preccurlyeq \Wbit' \quad \mbox{and} \quad \Wbit' \preccurlyeq \Wbit'' \quad \mbox{then} \quad \Wbit \preccurlyeq \Wbit'' \; .
\end{equation}
Thus, the ``upgraded'' relation is transitive as well.
Lastly, since a channel is both degraded and upgraded with respect to itself (take the intermediate channel as the identity function), we have that both relations are reflexive:
\begin{equation}
\label{eq:upgradedDegradedReflexiv}
\Wbit \preccurlyeq \Wbit \quad \mbox{and} \quad \Wbit \succcurlyeq \Wbit \; .
\end{equation}
If a channel $\Wgeneric'$ is both degraded and upgraded with respect to $\Wbit$, then we say that $\Wbit$ and $\Wgeneric'$ are \emph{equivalent}, and denote this by $\Wbit \equiv \Wgeneric'$. Since ``degraded'' and ``upgraded'' are transitive relations, it follows
that the ``equivalent'' relation is transitive as well.
Also, by (\ref{eq:degradedUpgradedIff}), we have that ``equivalent'' is a symmetric relation:
\begin{equation}
\label{eq:equivalentSymmetric}
\Wbit \equiv \Wbit'
\quad \mbox{if and only if} \quad
\Wbit' \equiv \Wbit \; .
\end{equation}
Lastly, since a channel $\Wbit$ is both upgraded and degraded with respect to itself, we have by (\ref{eq:upgradedDegradedReflexiv}) that ``equivalent'' is a reflexive relation.
Thus, channel equivalence is indeed an equivalence relation.
Let $\Wbit:\mathcal{X} \to \mathcal{Y}$ be a given BMS channel. We now set the notation for three quantities of interest.
i) Denote by $P_e(\Wbit)$ the probability of error under maximum-likelihood decision, where ties are broken arbitrarily, and the input distribution is $\mathrm{Bernoulli}(1/2)$. That is,
\begin{equation}
\label{eq:errorProbDef}
P_e(\Wbit) =
\frac{1}{2} \sum_{y \in \mathcal{Y}} \min\{\Wbit(y|0),\Wbit(y|1)\} \; .
\end{equation}
ii) Denote by $Z(\Wbit)$ the Bhattacharyya parameter,
\begin{equation}
\label{eq:BhattacharyyaDef}
Z(\Wbit) = \sum_{y \in \mathcal{Y}} \sqrt{\Wbit(y|0)\Wbit(y|1)} \; .
\end{equation}
iii) Denote by $I(\Wbit)$ the capacity,
\[
I(\Wbit) = \sum_{y \in \mathcal{Y}} \sum_{x \in \mathcal{X}} \frac{1}{2} \Wbit(y|x) \log \frac{\Wbit(y|x)}{\frac{1}{2}\Wbit(y|0)+\frac{1}{2}\Wbit(y|1)} \; .
\]
The following lemma states that these three quantities behave as expected with respect to the degrading and upgrading relations. The equation most important to us will be (\ref{eq:errorProbDegrading}).
\begin{lemm}[{\cite[page 207]{RichardsonUrbanke:08b}}]
\label{lemm:degradingAndEZI}
Let $\Wbit:\mathcal{X} \to \mathcal{Y}$ be a BMS\ channel and let $\mathcal{Q} : \mathcal{X} \to \calZ$ be degraded with respect to $\Wbit$, that is, $\mathcal{Q} \preccurlyeq \Wbit$. Then,
\begin{align}
\label{eq:errorProbDegrading}
P_e(\mathcal{Q}) & \geq P_e(\Wbit) \; , \\
\label{eq:BhattacharyyaProbDegrading}
\quad Z(\mathcal{Q}) & \geq Z(\Wbit) \; , \quad \mbox{and} \\
\label{eq:capacityDegrading}
\quad I(\mathcal{Q}) & \leq I(\Wbit) \; .
\end{align}
Moreover, all of the above continues to hold if we replace ``degraded'' by ``upgraded'', $\preccurlyeq$ by $\succcurlyeq$, and reverse the inequalities. Specifically, if $\Wbit \equiv \mathcal{Q}$, then the weak inequalities are in fact equalities.
\end{lemm}
\begin{proof}
We consider only the first part, since the ``Moreover'' part follows easily from Equation~(\ref{eq:degradedUpgradedIff}).
For a simple proof of (\ref{eq:errorProbDegrading}), recall the definition of degradation (\ref{eq:degraded}), and note that
\begin{multline*}
P_e(\mathcal{Q}) = \frac{1}{2}\sum_{z \in \mathcal{Z}} \min\{\mathcal{Q}(\mathcal{Z}|0),\mathcal{Q}(\mathcal{Z}|1)\} = \\
\frac{1}{2}\sum_{z \in \mathcal{Z}} \min\left\{\sum_{y \in \mathcal{Y}} \Wbit(y|0) \cdot \cP(z|y),\sum_{y \in \mathcal{Y}} \Wbit(y|1) \cdot \cP(z|y)\right\} \\
\geq \frac{1}{2}\sum_{z \in \mathcal{Z}} \sum_{y \in \mathcal{Y}} \min\{\Wbit(y|0),\Wbit(y|1)\} \cdot \cP(z|y) = P_e(\Wbit)
\end{multline*}
Equation (\ref{eq:BhattacharyyaProbDegrading}) is concisely proved in \cite[Lemma 1.8]{Korada:09z}. Equation (\ref{eq:capacityDegrading}) is a simple consequence of the data-processing inequality \cite[Theorem 2.8.1]{CoverThomas:06b}.
\end{proof}
Note that it may be the case that $y$ is its own conjugate. That is, $y$ and $\bar{y}$ are the same symbol (an erasure). It would make our proofs simpler if this special case was assumed not to happen. We will indeed assume this later on, with the next lemma providing most of the justification.
\begin{lemm}
\label{lemm:allPairs}
Let $\Wbit : \mathcal{X} \to \mathcal{Y}$ be a BMS\ channel. There exists a BMS\ channel $\Wgeneric' : \mathcal{X} \to \mathcal{Z}$
such that i) $\Wgeneric'$ is equivalent to $\Wbit$, and ii) for all $z \in \mathcal{Z}$ we have that $z$ and $\bar{z}$ are distinct.
\end{lemm}
\begin{proof}
If $\Wbit$ is such that for all $y \in \mathcal{Y}$ we have that $y$ and $\bar{y}$ are distinct, then we are done, since we can take $\Wgeneric'$ equal to $\Wbit$.
Otherwise, let $y_{\scriptscriptstyle{?}} \in \mathcal{Y}$ be such that $y_{\scriptscriptstyle{?}}$ and $\bary_{\scriptscriptstyle{?}}$ are the same symbol. Let the alphabet $\mathcal{Z}$ be defined as follows:
\[
\mathcal{Z} = \left(\mathcal{Y} \setminus \mysett{y_{\scriptscriptstyle{?}}}\right) \cup \mysett{z_1,z_2} \; ,
\]
where $z_1$ and $z_2$ are new symbols, not already in $\mathcal{Y}$. Now, define the channel $\Wgeneric':\mathcal{X} \to \mathcal{Z}$ as follows. For all $z \in \mathcal{Z}$ and $x \in \mathcal{X}$,
\[
\Wgeneric'(z|x) =
\begin{cases}
\Wbit(z|x) & \mbox{if $z \in \mathcal{Y}$,} \\
\frac{1}{2} \Wbit(y_{\scriptscriptstyle{?}}|x) & \mbox{if $z = z_1$ or $z = z_2$.}
\end{cases}
\]
We first show that $\Wgeneric' \succcurlyeq \Wbit$. To see this, take the intermediate channel $\cP : \mathcal{Z} \to \mathcal{Y}$ as the channel that maps (with probability 1) $z_1$ and $z_2$ to $y_{\scriptscriptstyle{?}}$, and all other symbols to themselves. Next, we show that $\Wgeneric' \preccurlyeq \Wbit$. To see this, define the intermediate channel $\cP : \mathcal{Y} \to \mathcal{Z}$ as follows.
\[
\cP(z|y) =
\begin{cases}
1 & \mbox{if $z=y$,} \\
\frac{1}{2} & \mbox{if $y = y_{\scriptscriptstyle{?}}$ and $z \in \mysett{z_1,z_2}$,} \\
0 & \mbox{otherwise.}
\end{cases}
\]
To sum up, we have constructed a new channel $\Wgeneric'$ which is equivalent to $\Wbit$, and contains one less self-conjugate symbol ($y_{\scriptscriptstyle{?}}$ was replaced by the pair $z_1,z_2$). It is also easy to see that $\Wgeneric'$ is BMS. We can now apply this construction over and over, until the resulting channel has no self-conjugate symbols.
\end{proof}
Now that Lemma~\ref{lemm:allPairs} is proven, we will indeed assume from this point forward that all channels are BMS\ and have no output symbols $y$ such that $y$ and $\bar{y}$ are equal. As we will show later on, this assumption does not limit us. Moreover, given a generic BMS\ channel $\Wbit: \mathcal{X} \to \mathcal{Y}$, we will further assume that for all $y \in \mathcal{Y}$, at least one of the probabilities $\Wbit(y|0)$ and $\Wbit(\bar{y}|0)$ is positive (otherwise, we can remove the pair of symbols $y,\bar{y}$ from the alphabet, since they can never occur).
Given a channel $\Wbit : \mathcal{X} \to \mathcal{Y}$, we now define for each output symbol $y \in \mathcal{Y}$ an associated \emph{likelihood ratio}, denoted $\mathrm{LR}_\Wbit(y)$. Specifically,
\[
\mathrm{LR}_\Wbit(y) = \frac{\Wbit(y|0)}{\Wbit(y|1)} =
\frac{\Wbit(y|0)}{\Wbit(\bar{y}|0)}
\]
(if $\Wbit(\bar{y}|0) = 0$, then we must have by assumption that $\Wbit(y|0) > 0$, and we define $\mathrm{LR}_\Wbit(y) = \infty$). If the channel $\Wbit$ is understood from the context, we will abbreviate $\mathrm{LR}_\Wbit(y)$ to $\mathrm{LR}(y)$.
\section{High-Level Description of the Algorithms}
\label{sec:highLevelAlg}
\looseness=-1
In this section, we give a high level description of our algorithms for approximating a bit channel. We then show how these approximations can be used in order to construct a polar code.
In order to completely specify the approximating algorithms, one has to supply two \emph{merging functions}, a degrading merging function \mbox{\code degrading\_merge} and an upgraded merging function \mbox{\code upgrading\_merge}. We will now define the properties required of our merging functions, leaving the specification of the functions we have actually used to the next section. The next section will also make clear why we have chosen to call these functions ``merging''.
For a degrading merge function \mbox{\code degrading\_merge}, the following must hold. For a BMS\ channel $\Wbit$ and positive integer $\mu$, the output of $\mbox{\code degrading\_merge}(\Wbit,\mu)$ is a BMS\ channel $\mathcal{Q}$ such that i) $\mathcal{Q} \preccurlyeq \Wbit$ is degraded with respect to $\Wbit$, and ii) The size of the output alphabet of $\mathcal{Q}$ is at most $\mu$. We define the properties required of $\mbox{\code upgrading\_merge}$ similarly, but with ``degraded'' replaced by ``upgraded'' and $\preccurlyeq$ by $\succcurlyeq$.
Let $0 \leq i < n$ be an integer with binary representation $i=\binaryRep{b_1,b_2,\ldots,b_m}$, where $b_1$ is the most significant bit. Algorithms~\ref{alg:highLevelDegrade} and \ref{alg:highLevelUpgrade} contain our procedures for finding a degraded and upgraded approximation of the bit channel $\mathcal{W}_i^{(m)}$, respectively. In words, we employ the recursive constructions (\ref{eq:badSplit}) and (\ref{eq:goodSplit}), taking care to reduce the output alphabet size of each intermediate channel from at most $2 \mu^2$ (apart possibly from the underlying channel $W_{\mathrm{base}}$) to at most $\mu$.
\begin{algorithm}
\SetInd{0.49em}{0.49em}
\caption{Bit-channel degrading procedure}
\label{alg:highLevelDegrade}
\Input{An underlying BMS\ channel $W_{\mathrm{base}}$, a bound $\mu=2\nu$ on the output alphabet size, a code length $n=2^m$, an index $i$ with binary representation $i=\binaryRep{b_1,b_2,\ldots,b_m}$.}
\Output{A BMS\ channel that is degraded with respect to the bit channel $\mathcal{W}_i$.}
\label{algline:QinitDegrade} $\mathcal{Q} \leftarrow \mbox{\code degrading\_merge}(W_{\mathrm{base}},\mu)$\;
\For{$j = 1,2,\ldots, m$}
{
\eIf{$b_j=0$}
{
$\mathcal{W} \leftarrow \mathcal{Q} b \mathcal{Q}$
}
{
$\mathcal{W} \leftarrow \mathcal{Q} a \mathcal{Q}$
}
$\mathcal{Q} \leftarrow \mbox{\code degrading\_merge}(\mathcal{W},\mu)$\;
}
\Return{$\mathcal{Q}$}\;
\end{algorithm}
\begin{algorithm}
\SetInd{0.49em}{0.49em}
\caption{Bit-channel upgrading procedure}
\label{alg:highLevelUpgrade}
\Input{An underlying BMS\ channel $W_{\mathrm{base}}$, a bound $\mu=2\nu$ on the output alphabet size, a code length $n=2^m$, an index $i$ with binary representation $i=\binaryRep{b_1,b_2,\ldots,b_m}$.}
\Output{A BMS\ channel that is upgraded with respect to the bit channel $\mathcal{W}_i$.}
$\mathcal{Q}' \leftarrow \mbox{\code upgrading\_merge}(W_{\mathrm{base}},\mu)$\;
\For{$j = 1,2,\ldots, m$}
{
\eIf{$b_j=0$}
{
$\mathcal{W} \leftarrow \mathcal{Q}' b \mathcal{Q}'$
}
{
$\mathcal{W} \leftarrow \mathcal{Q}' a \mathcal{Q}'$
}
$\mathcal{Q}' \leftarrow \mbox{\code upgrading\_merge}(\mathcal{W},\mu)$\;
}
\Return{$\mathcal{Q}'$}\;
\end{algorithm}
The key to proving the correctness of Algorithms~\ref{alg:highLevelDegrade} and \ref{alg:highLevelUpgrade} is the following lemma. It is essentially a restatement of \cite[Lemma 4.7]{Korada:09z}. For completeness, we restate the proof as well.
\begin{lemm}
\label{lemm:degradedPreserved}
Fix a binary input channel $\Wbit:\mathcal{X} \to \mathcal{Y}$, and denote
\[
\Wgeneric_{\boxast} = \Wbit b \Wbit \; , \quad \Wgeneric_{\varoast} = \Wbit a \Wbit \; .
\]
Next, let $\mathcal{Q} \preccurlyeq \Wbit$ be a degraded with respect to $\Wbit$, and denote
\[
\Wdegraded_{\boxast} = \mathcal{Q} b \mathcal{Q} \; , \quad \Wdegraded_{\varoast} = \mathcal{Q} a \mathcal{Q} \; .
\]
Then,
\[
\Wdegraded_{\boxast} \preccurlyeq \Wgeneric_{\boxast} \quad \mbox{and} \quad \Wdegraded_{\varoast} \preccurlyeq \Wgeneric_{\varoast} \; .
\]
Namely, the degradation relation is preserved by the channel transformation operation.
Moreover, all of the above continues to hold if we replace ``degraded'' by ``upgraded'' and $\preccurlyeq$ by $\succcurlyeq$.
\end{lemm}
\begin{proof}
We will prove only the ``degraded'' part, since it implies the ``upgraded'' part (by interchanging the roles of $\Wbit$ and $\mathcal{Q}$).
Let $\cP : \mathcal{Y} \to \mathcal{Z}$ be the channel which degrades $\Wbit$ to $\mathcal{Q}$: for all $z \in \mathcal{Z}$ and $x \in \mathcal{X}$,
\begin{equation}
\label{eq:degradedWtoTildeW}
\mathcal{Q}(z|x) = \sum_{y \in \mathcal{Y}} \Wbit(y|x)\cP(z|y) \; .
\end{equation}
We first prove $\Wdegraded_{\boxast} \preccurlyeq \Wgeneric_{\boxast}$. By (\ref{eq:badSplit}) applied to $\mathcal{Q}$, we get that for all $(z_1,z_2) \in \mathcal{Z}^2$ and $u_1 \in \mathcal{X}$,
\[
\Wdegraded_{\boxast}((z_1,z_2)|u_1) = \sum_{u_2 \in \mathcal{X}} \frac{1}{2} \mathcal{Q}(z_1|u_1 \oplus u_2) \mathcal{Q}(z_2|u_2) \; .
\]
Next, we expand $\mathcal{Q}$ twice according to (\ref{eq:degradedWtoTildeW}), and get
\begin{multline*}
\Wdegraded_{\boxast}((z_1,z_2)|u_1) = \\
\sum_{(y_1,y_2) \in \mathcal{Y}^2} \sum_{u_2}
\frac{1}{2} W(y_1|u_1 \oplus u_2) W(y_2 | u_2) \cP(z_1|y_1) \cP(z_2|y_2) \; .
\end{multline*}
By (\ref{eq:badSplit}), this reduces to
\begin{multline}
\label{eq:WprimeHalfWay}
\Wdegraded_{\boxast}((z_1,z_2)|u_1) = \\
\sum_{(y_1,y_2) \in \mathcal{Y}^2} \Wgeneric_{\boxast}((y_1,y_2)|u_1) \cP(z_1|y_1) \cP(z_2|y_2) \; .
\end{multline}
Next, define the channel $\cP^*:\mathcal{Y}^2 \to \mathcal{Z}^2$ as follows. For all $(y_1,y_2) \in \mathcal{Y}^2$ and $(z_1,z_2) \in \mathcal{Z}^2$,
\[
\cP^*((z_1,z_2)|(y_1,y_2)) = \cP(z_1|y_1) \cP(z_2|y_2) \; .
\]
It is easy to prove that $\cP^*$ is indeed a channel (we get a probability distribution on $\mathcal{Z}^2$ for every fixed $(y_1,y_2) \in \mathcal{Y}^2$). Thus, (\ref{eq:WprimeHalfWay}) reduces to
\begin{multline*}
\Wdegraded_{\boxast}((z_1,z_2)|u_1) = \\
\sum_{(y_1,y_2) \in \mathcal{Y}^2} \Wgeneric_{\boxast}((y_1,y_2)|u_1) \cP^*((z_1,z_2)|(y_1,y_2)) \; ,
\end{multline*}
and we get by (\ref{eq:degraded}) that $\Wdegraded_{\boxast} \preccurlyeq \Wgeneric_{\boxast}$. The claim $\Wdegraded_{\varoast} \preccurlyeq \Wgeneric_{\varoast}$ is proved in much the same way.
\end{proof}
\begin{prop}
\label{prop:algsDegradingUpgrading}
The output of Algorithm~\ref{alg:highLevelDegrade} (Algorithm~\ref{alg:highLevelUpgrade}) is a BMS\ channel that is degraded (upgraded) with respect to $\mathcal{W}_i^{(m)}$.
\end{prop}
\begin{proof}
The proof follows easily from Lemma~\ref{lemm:degradedPreserved}, by induction on $j$.
\end{proof}
Recall that, ideally, a polar code is constructed as follows. We are given an underlying channel $W_{\mathrm{base}} :\mathcal{X} \to \mathcal{Y}$, a specified codeword length $n = 2^m$, and a target block error rate $e_\mathrm{Block}$. We choose the largest possible subset of bit-channels $\mathcal{W}_i$ such that the sum of their probabilities of error $P_e(\mathcal{W}_i)$ is not greater than $e_\mathrm{Block}$. The resulting code is spanned by the rows in $B_n \Gm$ corresponding to the subset of chosen bit-channels. Denote the rate of this code as $R_\mathrm{exact}$.
Since we have no computational handle on the bit channels $\mathcal{W}_i$, we must resort to approximations. Let $\mathcal{Q}_i$ be the result of running Algorithm~\ref{alg:highLevelDegrade} on $W_{\mathrm{base}}$ and $i$. Since $\mathcal{Q}_i \preccurlyeq \mathcal{W}_i$, we have by (\ref{eq:errorProbDegrading}) that $P_e(\mathcal{Q}_i) \geq P_e(\mathcal{W}_i)$. Note that since the output alphabet of $\mathcal{Q}_i$ is small (at most $\mu$), we can actually compute $P_e(\mathcal{Q}_i)$. We now mimic the ideal construction by choosing the largest possible subset of indices for which the sum of $P_e(\mathcal{Q}_i)$ is at most $e_\mathrm{Block}$. Note that for this subset we have that the sum of $P_e(\mathcal{W}_i)$ is at most $e_\mathrm{Block}$ as well. Thus, the code spanned by the corresponding rows of $B_n \Gm$ is assured to have block error probability of at most $e_\mathrm{Block}$.
Denote the rate of this code by $R_\mathrm{degraded}$. It is easy to see that $R_\mathrm{degraded} \leq R_\mathrm{exact}$. In order to gauge the difference between the two rates, we compute a third rate, $R_\mathrm{upgraded}$, such that $R_\mathrm{upgraded} \geq R_\mathrm{exact}$ and consider the difference $R_\mathrm{upgraded} - R_\mathrm{degraded}$. The rate $R_\mathrm{upgraded}$ is computed the same way that $R_\mathrm{degraded}$ is, but instead of using Algorithm~\ref{alg:highLevelDegrade} we use Algorithm~\ref{alg:highLevelUpgrade}. Recall from Figure~\ref{fig:BSCplots} that $R_\mathrm{degraded}$ and $R_\mathrm{upgraded}$ are typically very close.
We end this section by noting a point that will be needed in the proof of Theorem~\ref{theo:totalRunningTime} below. Consider the running time needed in order to approximate all $n$ bit channels. Assume that each invocation of either $\mbox{\code degrading\_merge}$ or $\mbox{\code upgrading\_merge}$ takes time $\tau=\tau(\mu)$. Thus, the time needed for approximating a single bit channel using either Algorithm~\ref{alg:highLevelDegrade} or Algorithm~\ref{alg:highLevelUpgrade} is $O(m \tau)$. A naive analysis suggests that the time needed in order to approximate all $n$ bit-channels is $O(n \cdot m \tau)$. However, significant savings can be gained by noticing that intermediate calculations can be shared between bit-channels. For example, in a naive implementation we would approximate $W_{\mathrm{base}} b W_{\mathrm{base}}$ over and over again, $n/2$ times instead of only once. A quick calculation shows that the number of \emph{distinct} channels one needs to approximate is $2n-1-1$. That is, following both branches of the ``if'' statement of (without loss of generality) Algorithm~\ref{alg:highLevelDegrade} would produce $2^j$ channels for each level $1 \leq j \leq m$. Thus, the total running time can be reduced to $O((2n-2)\cdot \tau)$, which is simply $O(n \cdot \tau)$.
\section{Merging Functions}
\label{sec:mergingFunctions}
In this section, we specify the degrading and upgrading functions used to reduce the output alphabet size. These functions are referred to as $\mbox{\code degrading\_merge}$ and $\mbox{\code upgrading\_merge}$ in Algorithms \ref{alg:highLevelDegrade} and \ref{alg:highLevelUpgrade}, respectively. For now, let us treat our functions as heuristic (delaying their analysis to Section~\ref{sec:analysis}).
\begin{figure*}
\hfill
\subfigure[]{\label{subfig:degradingWQ}\includegraphics[width=72mm]{{Figures/degradingWQ.pdf}}}
\hfill\hfill
\subfigure[]{\label{subfig:degradingP}\includegraphics[width=72mm]{{Figures/degradingP.pdf}}}
\hfill{}
\caption{Degrading $\Wbit$ to $\mathcal{Q}$. (a) The degrading merge operation: the entry in the first/second row of a channel is the probability of receiving the corresponding symbol, given that a $0$/$1$ was transmitted. (b) The intermediate channel $\cP$.}
\label{fig:degradingWQP}
\end{figure*}
\subsection{Degrading-merge function}
We first note that the problem of degrading a binary-input channel to a channel with a prescribed output alphabet size was independently considered by Kurkoski and Yagi \cite{KurkoskiYagi:11a}. The main result in \cite{KurkoskiYagi:11a} is an optimal degrading strategy, in the sense that the capacity of the resulting channel is the largest possible. In this respect, the method we now introduce is sub-optimal. However, as we will show, the complexity of our method is superior to that presented in \cite{KurkoskiYagi:11a}.
The next lemma shows how one can reduce the output alphabet size by 2, and get a degraded channel. It is our first step towards defining a valid $\mbox{\code degrading\_merge}$ function.
\begin{lemm}
\label{lemm:degradeStep}
Let $\Wbit:\mathcal{X} \to \mathcal{Y}$ be a BMS channel, and let $y_1$ and $y_2$ be symbols in the output alphabet $\mathcal{Y}$. Define the channel $\mathcal{Q} : \mathcal{X} \to \mathcal{Z}$ as follows (see Figure~\ref{subfig:degradingWQ}). The output alphabet $\mathcal{Z}$ is given by
\[
\mathcal{Z} = \mathcal{Y} \setminus \mysett{y_1,\bar{y}_1,y_2,\bar{y}_2} \cup \mysett{z_{1,2},\barz_{1,2}}\; .
\]
For all $x \in \mathcal{X}$ and $z \in \mathcal{Z}$, define
\[
\mathcal{Q}(z|x) =
\begin{cases}
\Wbit(z|x) & \mbox{if $z \not\in \mysett{z_{1,2},\barz_{1,2}}$,} \\
\Wbit(y_1|x) + \Wbit(y_2|x) & \mbox{if $z = z_{1,2}$,} \\
\Wbit(\bar{y}_1|x) + \Wbit(\bar{y}_2|x) & \mbox{if $z = \barz_{1,2}$.}
\end{cases}
\]
Then $\mathcal{Q} \preccurlyeq \Wbit$. That is, $\mathcal{Q}$ is degraded with respect to $\Wbit$.
\end{lemm}
\begin{proof}
Take the intermediate channel $\cP : \mathcal{Y} \to \mathcal{Z}$ as the channel that maps with probability 1 as follows (see Figure~\ref{subfig:degradingP}): both $y_1$ and $y_2$ map to $z_{1,2}$, both $\bar{y}_1$ and $\bar{y}_2$ map to $\barz_{1,2}$, other symbols map to themselves. Recall that we have assumed that $\Wbit$ does not contain an erasure symbol, and this continues to hold for $\mathcal{Q}$.
\end{proof}
We now define the $\mbox{\code degrading\_merge}$ function we have used. It gives good results in practice and is amenable to a fast implementation. Assume we are given a BMS channel $\Wbit: \mathcal{X} \to \mathcal{Y}$ with an alphabet size of $2L$ (recall our assumption of no self-conjugates), and wish to
transform $\Wbit$ into a degraded version of itself while
reducing its alphabet size to $\mu$. If $2L \leq \mu$, then we are done, since we can take the degraded version of $\Wbit$ to be $\Wbit$ itself. Otherwise, we do the following. Recall that for each $y$ we have that $\mathrm{LR}(y) = 1/\mathrm{LR}(\bar{y})$, where in this context $1/0=\infty$ and $1/\infty = 0$. Thus, our first step is to choose from each pair $(y,\bar{y})$ a representative such that $\mathrm{LR}(y) \geq 1$. Next, we order these $L$ representative such that
\begin{equation}
\label{eq:yOrdered}
1 \leq \mathrm{LR}(y_1) \leq \mathrm{LR}(y_2) \leq \cdots \leq \mathrm{LR}(y_L) \; .
\end{equation}
We now ask the followin
: for which index $1 \leq i \leq L-1$ does the channel resulting from the application of Lemma~\ref{lemm:degradeStep} to $\Wbit$, $y_i$, and $y_{i+1}$ result in a channel with largest capacity? Note that instead of considering $\binom{L}{2}$ merges, we consider only $L-1$. After finding the maximizing index $i$ we indeed apply Lemma~\ref{lemm:degradeStep} and get a degraded channel $\mathcal{Q}$ with an alphabet size smaller by $2$ than that of $\Wbit$. The same process is applied to $\mathcal{Q}$, until the output alphabet size is not more than $\mu$.
\begin{algorithm}
\SetInd{0.49em}{0.49em}
\caption{The \protect\mbox{\code degrading\_merge} function}
\label{alg:degradeChannel}
\Input{A BMS\ channel $\Wbit:\mathcal{X} \to \mathcal{Y}$ where $\sizee{\mathcal{Y}} = 2L$, a bound $\mu=2\nu$ on the output alphabet size.}
\Output{A degraded channel $\mathcal{Q} : \mathcal{X} \to \mathcal{Y}'$, where $\sizee{\mathcal{Y}'} \leq \mu$.}
\tcp{Assume $1 \leq \mathrm{LR}(y_1) \leq \mathrm{LR}(y_2) \leq \cdots \leq \mathrm{LR}(y_L)$}
\For{$i = 1,2,\ldots, L-1$}
{
$\datum \leftarrow$ new data element\;
$\datum.a \leftarrow \Wbit(y_i|0) \; , \quad \datum.b \leftarrow \Wbit(\bar{y}_i|0)$\;
$\datum.a' \leftarrow \Wbit(y_{i+1}|0) \; , \quad \datum.b' \leftarrow \Wbit(\bar{y}_{i+1}|0)$\;
$\datum.\mathrm{deltaI} \leftarrow \calcDeltaI(\datum.a,\datum.b,\datum.a',\datum.b')$\;
$\insertRightmost(\datum)$\;
}
$\ell = L$\;
\While{$\ell > \nu$}
{
$\datum \leftarrow \getMin()$\;
$a^+ = \datum.a+\datum.a' \; , \quad b^+ = \datum.b+\datum.b'$\;
$\datumLeft \leftarrow \datum.\leftAlg$\;
$\datumRight \leftarrow \datum.\rightAlg$\;
$\removeMin()$\;
$\ell \leftarrow \ell -1$\;
\If{$\datumLeft \neq \nullAlg$}
{
$\datumLeft.a' = a^+$\;
$\datumLeft.b' = b^+$\;
$\datumLeft.\mathrm{deltaI} \leftarrow \calcDeltaI(\datumLeft.a,\datumLeft.b,a^+,b^+)$\;
$\valueUpdated(\datumLeft)$\;
}
\If{$\datumRight \neq \nullAlg$}
{
$\datumRight.a = a^+$\;
$\datumRight.b = b^+$\;
$\datumRight.\mathrm{deltaI} \leftarrow \calcDeltaI(a^+,b^+,\datumRight.a',\datumRight.b')$\;
$\valueUpdated(\datumRight)$\;
}
}
Construct $\mathcal{Q}$ according to the probabilities in the data structure and return it.
\end{algorithm}
In light of Lemma~\ref{lemm:degradeStep} and (\ref{eq:yOrdered}), a simple yet important point to note is that if $y_i$ and $y_{i+1}$ are merged to $z$, then
\begin{equation}
\label{eq:LRMean}
\mathrm{LR}(y_i) \leq \mathrm{LR}(z) \leq \mathrm{LR}(y_{i+1}) \; .
\end{equation}
Namely, the original LR ordering is essentially preserved by the merging operation. Algorithm~\ref{alg:degradeChannel} contains an implementation of our merging procedure. It relies on the above observation in order to improve complexity and runs in $O(L \cdot \log L)$ time. Thus, assuming $L$ is at most $2\mu^2$, the running time of our algorithm is $O(\mu^2 \log \mu)$. In contrast, had we used the degrading method presented in \cite{KurkoskiYagi:11a}, the running time would have been $O(\mu^5)$.
Our implementation assumes an underlying data structure and data elements as follows. Our data structure stores data elements, where each data element corresponds to a pair of adjacent letters $y_i$ and $y_{i+1}$, in the sense of the ordering in (\ref{eq:yOrdered}). Each data element has the following fields:
\[
a\;,\quad
b\;,\quad
a'\;,\quad
b'\;,\quad
\mathrm{deltaI}\;,\quad
\datumLeft\;,\quad
\datumRight\;,\quad
h \; .
\]
The fields $a$, $b$, $a'$, and $b'$ store the probabilities $\Wbit(y_i|0)$, $\Wbit(\bar{y}_i|0)$,
$\Wbit(y_{i+1}|0)$, and $\Wbit(\bar{y}_{i+1}|0)$, respectively. The field $\mathrm{deltaI}$ contains the difference in capacity that would result from applying Lemma~\ref{lemm:degradeStep} to $y_{i}$ and $y_{i+1}$. Note that $\mathrm{deltaI}$ is only a function of the above four probabilities, and thus the function $\calcDeltaI$ used to initialize this field is given by
\[
\calcDeltaI(a,b,a',b')=C(a,b)+C(a',b')-C(a^+,b^+) \; ,
\]
where
\[
C(a,b) = -(a+b) \log_2((a+b)/2)+a \log_2(a) +b \log_2(b) \; ,
\]
we use the shorthand
\[
a^+ = a + a' \; , \quad b^+ = b + b' \; ,
\]
and $0 \log_2 0$ is defined as $0$. The field $\datumLeft$ is a pointer to the data element corresponding to the pair $y_{i-1}$ and $y_i$ (or ``null'', if $i=1$). Likewise, $\datumRight$ is a pointer to the element corresponding to the pair $y_{i+1}$ and $y_{i+2}$ (see Figure~\ref{fig:mergeBeforeAfter} for a graphical depiction). Apart from these, each data element contains an integer field $h$, which will be discussed shortly.
We now discuss the functions that are the interface to our data structure: $\insertRightmost$, $\getMin$, $\removeMin$, and $\valueUpdated$. Our data structure combines the attributes of a doubly-linked-list \cite[Section 10.2]{CLRS:01b} and a heap\footnote{In short, a heap is a data structure that supports four operations: ``insert'', ``getMin'', ``removeMin'', and ``valueUpdated''. In our implementation, the running time of ``getMin'' is constant, while the running time of the other operations is logarithmic in the heap size.} \cite[Chapter 6]{CLRS:01b}. The doubly-linked list is implemented through the $\datumLeft$ and $\datumRight$ fields of each data element, as well as a pointer to the rightmost element of the list.
Our heap will have the ``array'' implementation, as described in \cite[Section 6.1]{CLRS:01b}. Thus, each data element will have a corresponding index in the heap array, and this index is stored in the field $h$. The doubly-linked-list will be ordered according to the corresponding $\mathrm{LR}$ value, while the heap will be sorted according to the $\mathrm{deltaI}$ field.
\label{subsec:mergingFunctions:degrade}
\begin{figure}[t]
\fbox{
\parbox{\linewidth}{
Before the merge of $y_i$ and $y_{i+1}$:
\[
\ldots \leftrightarrow \overbrace{(y_{i-1},y_i)}^{\datumLeft} \leftrightarrow \underbrace{(y_i,y_{i+1})}_{\mbox{merged to $z$}} \leftrightarrow \overbrace{(y_{i+1},y_{i+2})}^{\datumRight} \leftrightarrow \ldots
\]
After the merge, a new symbol $z$:
\[
\ldots \leftrightarrow (y_{i-1},z) \leftrightarrow (z,y_{i+2}) \leftrightarrow \ldots
\]
}}
\caption{Graphical depiction of the doubly-linked-list before and after a merge.}
\label{fig:mergeBeforeAfter}
\end{figure}
The function $\insertRightmost$ inserts a data element as the rightmost element of the list and updates the heap accordingly. The function $\getMin$ returns the data element with smallest $\mathrm{deltaI}$. Namely, the data element corresponding to the pair of symbols we are about to merge.
The function $\removeMin$ removes the element returned by $\getMin$ from both the linked-list and the heap. The function $\valueUpdated$ updates the heap due to a change in $\mathrm{deltaI}$ resulting from a merge, but does not change the linked list in view of (\ref{eq:LRMean}).
The running time of $\getMin$ is $O(1)$, and this is obviously also the case for $\calcDeltaI$. Due to the need of updating the heap, the running time of $\removeMin$, $\valueUpdated$, and $\insertRightmost$ is $O(\log L)$. The time needed for the initial sort of the $\mathrm{LR}$ pairs is $O(L \cdot \log L)$. Hence, since the initializing for-loop in Algorithm~\ref{alg:degradeChannel} has $L$ iterations and the while-loop has $L-\nu$ iterations, the total running time of Algorithm~\ref{alg:degradeChannel} is $O(L \cdot \log L)$.
Note that at first sight, it may seem as though there might be an even better heuristic to employ. As before, assume that the $y_i$ are ordered according to their likelihood ratios, and all of these are at least $1$. Instead of limiting the application of Lemma~\ref{lemm:degradeStep} to $y_i$ and $y_{i+1}$, we can broaden our search and consider the penalty in capacity incurred by merging arbitrary $y_i$ and $y_j$, where $i \neq j$. Indeed, we could further consider merging arbitrary $y_i$ and $\bar{y}_j$, where $i \neq j$. Clearly, this broader search will result in worse complexity. However, as the next theorem shows, we will essentially gain nothing by it.
\begin{theo}
\label{theo:degradeOptimal}
Let $\Wbit:\mathcal{X} \to \mathcal{Y}$ be a BMS channel, with
\[
\mathcal{Y} = \mysett{y_1,y_2,\ldots,y_L,\bar{y}_1,\bar{y}_2,\ldots,\bar{y}_L} \; .
\]
Assume that
\[
1 \leq \mathrm{LR}(y_1) \leq \mathrm{LR}(y_2) \leq \cdots \leq \mathrm{LR}(y_L) \; .
\]
For symbols $w_1,w_2 \in \mathcal{Y}$, denote by $I(w_1,w_2)$ the capacity of the channel one gets by the application of Lemma~\ref{lemm:degradeStep} to $w_1$ and $w_2$. Then, for all distinct $1 \leq i \leq L$ and $1 \leq j \leq L$,
\begin{equation}
\label{eq:theo:degradeOptimal_LRgeq1}
I(\bar{y}_i,\bar{y}_j) = I(y_i,y_j) \geq I(y_i,\bar{y}_j) = I(\bar{y}_i,y_j)\; .
\end{equation}
Moreover, for all $1 \leq i < j < k \leq L$ we have that either
\[
I(y_i,y_j) \geq I(y_i,y_k)\; ,
\]
or
\[
I(y_j,y_k) \geq I(y_i,y_k)\; .
\]
\end{theo}
We note that Theorem~\ref{theo:degradeOptimal} seems very much related to \cite[Lemma 5]{KurkoskiYagi:11a}. However, one important difference is that Theorem~\ref{theo:degradeOptimal} deals with the case in which the degraded channel is constrained to be symmetric, while \cite[Lemma 5]{KurkoskiYagi:11a} does not. At any rate, for completeness, we will prove Theorem~\ref{theo:degradeOptimal} in Appendix~\ref{sec:proofDegradingOptimal}.
\subsection{Upgrading-merge functions}
The fact that one can merge symbol pairs and get a degraded version of the original channel should come as no surprise. However, it turns out that we can also merge symbol pairs and get an \emph{upgraded} version of the original channel. We first show a simple method of doing this. Later on, we will show a slightly more complex method, and compare between the two.
\begin{figure*}
\hfill
\subfigure[]{\label{subfig:doubleUpgradingWQ}\includegraphics[width=72mm]{{Figures/doubleUpgradingWQ.pdf}}}
\hfill\hfill
\subfigure[]{\label{subfig:doubleUpgradingP}\includegraphics[width=72mm]{{Figures/doubleUpgradingP.pdf}}}
\hfill{}
\caption{First method of Upgrading $\Wbit$ to $\mathcal{Q}'$. (a) The upgrading merge operation. (b) The intermediate channel $\cP$.}
\label{fig:doubleUpgradingWQP}
\end{figure*}
As in the degrading case, we show how to reduce the output alphabet size by $2$, and then apply this method repeatedly as much as needed. The following lemma shows how the core reduction can be carried out. The intuition behind it is simple. Namely, now we ``promote'' a pair of output symbols to have a higher LR value, and then merge with an existing pair having that LR.
\begin{lemm}
\label{lemm:upgradeStep0}
Let $\Wbit:\mathcal{X} \to \mathcal{Y}$ be a BMS channel, and let $y_2$ and $y_1$ be symbols in the output alphabet $\mathcal{Y}$. Denote $\lambda_2 = \mathrm{LR}(y_2)$ and $\lambda_1 = \mathrm{LR}(y_1)$. Assume that
\begin{equation}
\label{eq:upgradeStep0:lambdaOrder}
1 \leq \lambda_1 \leq \lambda_2 \; .
\end{equation}
Next, let $a_1 = \Wbit(y_1|0)$ and $b_1 = \Wbit(\bar{y}_1|0)$. Define $\alpha_2$ and $\beta_2$ as follows. If $\lambda_2 < \infty$
\begin{align}
\label{eq:alpha1beta1_step0}
\alpha_2 & = \lambda_2 \frac{a_1 + b_1}{\lambda_2 + 1} &
\beta_2 & = \frac{a_1 + b_1}{\lambda_2 + 1} \; .
\end{align}
Otherwise, we have $\lambda_2 = \infty$, and so define
\begin{align}
\alpha_2 & = a_1 + b_1 &
\beta_2 & = 0 \; .
\end{align}
We note that the subscript ``$2$'' in $\alpha_2$ and $\beta_2$ is meant to suggest a connection to $\lambda_2$, since $\alpha_2/\beta_2=\lambda_2$.
For real numbers $\alpha$, $\beta$, and $x \in \mathcal{X}$, define
\[
t(\alpha,\beta|x) =
\begin{cases}
\alpha & \mbox{if $x=0$,} \\
\beta & \mbox{if $x=1$.}
\end{cases}
\]
Define the channel $\mathcal{Q}' : \mathcal{X} \to \calZ'$ as follows (see Figure~\ref{subfig:doubleUpgradingWQ}). The output alphabet $\calZ'$ is given by
\[
\calZ' = \mathcal{Y} \setminus \mysett{y_2,\bar{y}_2,y_1,\bar{y}_1} \cup \mysett{z_2,\bar{z}_2} \; .
\]
For all $x \in \mathcal{X}$ and $z \in \calZ'$,
\[
\mathcal{Q}'(z|x) =
\begin{cases}
\Wbit(z|x) & \mbox{if $z \not\in \mysett{z_2,\bar{z}_2}$,} \\
\Wbit(y_2|x) + t(\alpha_2,\beta_2|x) & \mbox{if $z = z_2$,} \\
\Wbit(\bar{y}_2|x) + t(\beta_2,\alpha_2|x) & \mbox{if $z = \bar{z}_2$.} \\
\end{cases}
\]
Then $\mathcal{Q}' \succcurlyeq \Wbit$. That is, $\mathcal{Q}'$ is upgraded with respect to $\Wbit$.
\end{lemm}
\begin{proof}
Denote $a_2 = \Wbit(y_2|0)$ and $b_2 = \Wbit(\bar{y}_2|0)$. First, note that
\[
a_1+b_1 = \alpha_2 + \beta_2 \; .
\]
Next, let $\gamma$ be defined as follows. If $\lambda_2 > 1$, let
\[
\gamma = \frac{a_1-\beta_2}{\alpha_2-\beta_2} = \frac{b_1-\alpha_2}{\beta_2 - \alpha_2} \; ,
\]
and note that (\ref{eq:upgradeStep0:lambdaOrder}) implies that $0 \leq \gamma \leq 1$.
Otherwise ($\lambda_1 = \lambda_2 = 1$), let
\[
\gamma = 1 \; .
\]
Define the intermediate channel $\cP : \calZ' \to \mathcal{Y}$ as follows.
\[
\cP(y|z) =
\begin{cases}
1 & \mbox{if $z \not\in\mysett{z_2,\bar{z}_2}$ and $y=z$,}\\
\frac{\alpha_2 \gamma}{a_2+\alpha_2} & \mbox{if $(z,y) \in \mysett{(z_2,y_1),(\bar{z}_2,\bar{y}_1)}$,}\\
\frac{a_2}{a_2+\alpha_2} & \mbox{if $(z,y) \in \mysett{(z_2,y_2),(\bar{z}_2,\bar{y}_2)}$,}\\
\frac{\alpha_2 (1-\gamma)}{a_2+\alpha_2} & \mbox{if $(z,y) \in \mysett{(z_2,\bar{y}_1),(\bar{z}_2,y_1)}$,}\\
0 & \mbox{otherwise.}
\end{cases}
\]
Notice that when $\lambda_2 < \infty$, we have that
\[
\frac{a_2}{a_2+\alpha_2} = \frac{b_2}{b_2+\beta_2} \quad \mbox{and} \quad \frac{\alpha_2}{a_2+\alpha_2} = \frac{\beta_2}{b_2+\beta_2} \; .
\]
Some simple calculations finish the proof.
\end{proof}
The following corollary shows that we do not ``lose anything'' when applying Lemma~\ref{lemm:upgradeStep0} to symbols $y_1$ and $y_2$ such that $\mathrm{LR}(y_1) = \mathrm{LR}(y_2)$. Thus, intuitively, we do not expect to lose much when applying Lemma~\ref{lemm:upgradeStep0} to symbols with ``close'' LR values.
\begin{coro}
\label{coro:equalLRGivesEquivalent}
Let $\Wbit$, $\mathcal{Q}'$, $y_1$, and $y_2$ be as in Lemma~\ref{lemm:upgradeStep0}. If $\mathrm{LR}(y_1) = \mathrm{LR}(y_2)$, then $\mathcal{Q}' \equiv \Wbit$. That is, $\Wbit$ and $\mathcal{Q}'$ are equivalent. Moreover, all of the above holds if we replace ``Lemma~\ref{lemm:upgradeStep0}'' by ``Lemma~\ref{lemm:degradeStep}''.
\end{coro}
\begin{proof}
The proof follows by noticing that the channel $\mathcal{Q}'$ we get by applying Lemma~\ref{lemm:upgradeStep0} to $\Wbit$, $y_1$, and $y_2$, is exactly the same channel we get if we apply Lemma~\ref{lemm:degradeStep} instead. Thus, we have both $\mathcal{Q}' \succcurlyeq \Wbit$ and $\mathcal{Q}' \preccurlyeq \Wbit$.
\end{proof}
In Lemma~\ref{lemm:upgradeStep0}, we have essentially transferred the probability $\Wbit(y_1|0) + \Wbit(\bar{y}_1|0)$ onto a symbol pair with a higher LR value. We now show a different method of merging that involves dividing the probability $\Wbit(y_1|0) + \Wbit(\bar{y}_1|0)$ between a symbol pair with higher LR value and a symbol pair with lower LR value. As we will prove latter on, this new approach is generally preferable.
\begin{figure*}
\hfill
\subfigure[]{\label{subfig:tripleUpgradingWQ}\includegraphics[width=72mm]{{Figures/tripleUpgradingWQ.pdf}}}
\hfill\hfill
\subfigure[]{\label{subfig:tripleUpgradingP}\includegraphics[width=72mm]{{Figures/tripleUpgradingP.pdf}}}
\hfill{}
\caption{Second method of Upgrading $\Wbit$ to $\mathcal{Q}'$. (a) The upgrading merge operation. (b) The intermediate channel $\cP$.}
\label{fig:tripleUpgradingWQP}
\end{figure*}
\begin{lemm}
\label{lemm:upgradeStep1}
Let $\Wbit:\mathcal{X} \to \mathcal{Y}$ be a BMS channel, and let $y_1$, $y_2$, and $y_3$ be symbols in the output alphabet $\mathcal{Y}$. Denote $\lambda_1 = \mathrm{LR}(y_1)$, $\lambda_2 = \mathrm{LR}(y_2)$, and $\lambda_3 = \mathrm{LR}(y_3)$. Assume that
\[
1 \leq \lambda_1 < \lambda_2 < \lambda_3 \; .
\]
Next, let $a_2 = \Wbit(y_2|0)$ and $b_2 = \Wbit(\bar{y}_2|0)$. Define $\alpha_1$, $\beta_1$, $\alpha_3$, $\beta_3$ as follows. If $\lambda_3 < \infty$
\begin{align}
\label{eq:alpha1beta1}
\alpha_1 & = \lambda_1 \frac{\lambda_3 b_2 - a_2}{\lambda_3 - \lambda_1} &
\beta_1 & = \frac{\lambda_3 b_2 - a_2}{\lambda_3 - \lambda_1} \; , \\
\label{eq:alpha3beta3}
\alpha_3 & = \lambda_3 \frac{a_2 - \lambda_1 b_2}{\lambda_3 - \lambda_1} &
\beta_3 & = \frac{a_2 - \lambda_1 b_2}{\lambda_3 - \lambda_1} \; .
\end{align}
Otherwise, we have $\lambda_3 = \infty$, and so define
\begin{align}
\alpha_1 & = \lambda_1 b_2 &
\beta_1 & = b_2 \; , \\
\alpha_3 & = a_2 - \lambda_1 b_2 &
\beta_3 & = 0 \; .
\end{align}
Let $t(\alpha,\beta|x)$ be as in Lemma~\ref{lemm:upgradeStep0}, and
define the BMS\ channel $\mathcal{Q}' : \mathcal{X} \to \calZ'$ as follows (see Figure~\ref{subfig:tripleUpgradingWQ}). The output alphabet $\calZ'$ is given by
\[
\calZ' = \mathcal{Y} \setminus \mysett{y_1,\bar{y}_1,y_2,\bar{y}_2,y_3,\bar{y}_3} \cup \mysett{z_1,\bar{z}_1,z_3,\bar{z}_3} \; .
\]
For all $x \in \mathcal{X}$ and $z \in \calZ'$, define
\[
\mathcal{Q}'(z|x) =
\begin{cases}
\Wbit(z|x) & \mbox{if $z \not\in \mysett{z_1,\bar{z}_1,z_3,\bar{z}_3}$,} \\
\Wbit(y_1|x) + t(\alpha_1,\beta_1|x) & \mbox{if $z = z_1$,} \\
\Wbit(\bar{y}_1|x) + t(\beta_1,\alpha_1|x) & \mbox{if $z = \bar{z}_1$,} \\
\Wbit(y_3|x) + t(\alpha_3,\beta_3|x) & \mbox{if $z = z_3$,} \\
\Wbit(\bar{y}_3|x) + t(\beta_3,\alpha_3|x) & \mbox{if $z = \bar{z}_3$.}
\end{cases}
\]
Then $\mathcal{Q}' \succcurlyeq \Wbit$. That is, $\mathcal{Q}'$ is upgraded with respect to $\Wbit$.
\end{lemm}
\begin{proof}
Denote $a_1 = \Wbit(y_1|0)$, $b_1 = \Wbit(\bar{y}_1|0)$, $a_3 = \Wbit(y_3|0)$, and $b_3 = \Wbit(\bar{y}_3|0)$. Define the intermediate channel $\cP : \calZ' \to \mathcal{Y}$ as follows.
\[
\cP(y|z) =
\begin{cases}
1 & \mbox{if $z \not\in\mysett{z_3,\bar{z}_3,z_1,\bar{z}_1}$ and $y=z$,}\\
\frac{a_1}{a_1+\alpha_1} = \frac{b_1}{b_1+\beta_1} & \mbox{if $(z,y) \in \mysett{(z_1,y_1),(\bar{z}_1,\bar{y}_1)}$,}\\
\frac{\alpha_1}{a_1+\alpha_1} = \frac{\beta_1}{b_1+\beta_1} & \mbox{if $(z,y) \in \mysett{(z_1,y_2),(\bar{z}_1,\bar{y}_2)}$,}\\
\frac{a_3}{a_3+\alpha_3} & \mbox{if $(z,y) \in \mysett{(z_3,y_3),(\bar{z}_3,\bar{y}_3)}$,}\\
\frac{\alpha_3}{a_3+\alpha_3} & \mbox{if $(z,y) \in \mysett{(z_3,y_2),(\bar{z}_3,\bar{y}_2)}$,}\\
0 & \mbox{otherwise.}
\end{cases}
\]
Notice that when $\lambda_3 < \infty$, we have that
\[
\frac{a_3}{a_3+\alpha_3} = \frac{b_3}{b_3+\beta_3} \quad \mbox{and} \quad \frac{\alpha_3}{a_3+\alpha_3} = \frac{\beta_3}{b_3+\beta_3} \; .
\]
The proof follows by observing that, whatever the value of $\lambda_3$,
\[
\alpha_1 + \alpha_3 = a_2 \quad \mbox{and} \quad \beta_1+\beta_3 = b_2 \; .
\]
\end{proof}
The following lemma formalizes why Lemma~\ref{lemm:upgradeStep1} results in a merging operation that is better than that of Lemma~\ref{lemm:upgradeStep0}.
\begin{lemm}
Let $\Wbit$, $y_1$, $y_2$, and $y_3$ be as in Lemma~\ref{lemm:upgradeStep1}. Denote by $\Wupgraded_{123} : \mathcal{X} \to \calZ'_{123}$ the result of applying Lemma~\ref{lemm:upgradeStep1} to $\Wbit$, $y_1$, $y_2$, and $y_3$. Next, denote by $\Wupgraded_{23} : \mathcal{X} \to \calZ'_{23}$ the result of applying Lemma~\ref{lemm:upgradeStep0} to $\Wbit$, $y_2$, and $y_3$. Then $\Wupgraded_{23} \succcurlyeq \Wupgraded_{123} \succcurlyeq \Wbit$. Namely, in a sense, $\Wupgraded_{123}$ is a more faithful representation of $\Wbit$ than $\Wupgraded_{23}$ is.
\end{lemm}
\begin{proof}
Recall that the two alphabets $\calZ'_{123}$ and $\calZ'_{23}$ satisfy
\begin{align*}
\calZ'_{123} &= \mysett{z_1,\bar{z}_1,z_3,\bar{z}_3} \cup \mathcal{A} \; , \\
\calZ'_{23} &= \mysett{y_1,\bar{y}_1,z_3,\bar{z}_3} \cup \mathcal{A} \; ,
\end{align*}
where
\[
\mathcal{A} = \mathcal{Y} \setminus \mysett{y_1,\bar{y}_1,y_2,\bar{y}_2,y_3,\bar{y}_3}
\]
is the set of symbols not participating in either merge operation.
In order to prove that $\Wupgraded_{123}$ is degraded with respect to $\Wupgraded_{23}$, we must supply a corresponding intermediate channel $\cP : \calZ'_{23} \to \calZ'_{123}$. To this end, let
\[
\lambda_3 = \frac{\Wupgraded_{123}(z_3|0)}{\Wupgraded_{123}(z_3|1)} = \frac{\Wupgraded_{23}(z_3|0)}{\Wupgraded_{23}(z_3|1)} = \frac{\Wbit(y_3|0)}{\Wbit(y_3|1)}
\]
and
\[
\gamma = \frac{\Wupgraded_{123}(z_3|0)}{\Wupgraded_{23}(z_3|0)} = \frac{\Wupgraded_{123}(\bar{z}_3|1)}{\Wupgraded_{23}(\bar{z}_3|1)} \; .
\]
Note that in both Lemma~\ref{lemm:upgradeStep1} and \ref{lemm:upgradeStep0} we have that $\alpha_3/\beta_3=\lambda_3$. Next, we recall that in Lemma~\ref{lemm:upgradeStep0} we have that $\alpha_3+\beta_3 = a_2+b_2$ whereas in Lemma~\ref{lemm:upgradeStep1} we have $\alpha_3+\beta_3 = a_2+b_2 - \alpha_1 - \beta_1$. Thus, we conclude that $0 \leq \gamma \leq 1$. Moreover, since $0 \leq \gamma \leq 1$, we conclude that the following definition of an intermediate channel is indeed valid.
\begin{multline*}
\cP(z_{123}|z_{23}) = \\
\begin{cases}
1 & \mbox{if $z_{123} = z_{23}$ and $z_{123} \in \mathcal{A}$,}\\
1 & \mbox{if $(z_{23},z_{123}) \in \mysett{(y_1,z_1),(\bar{y}_1,\bar{z}_1)}$,}\\
\gamma & \mbox{if $(z_{23},z_{123}) \in \mysett{(z_3,z_3),(\bar{z}_3,\bar{z}_3)}$,}\\
\frac{(1-\gamma)\lambda_1}{\lambda_1+1} & \mbox{if $(z_{23},z_{123}) \in \mysett{(z_3,z_1),(\bar{z}_3,\bar{z}_1)}$,}\\
\frac{(1-\gamma)}{\lambda_1+1} & \mbox{if $(z_{23},z_{123}) \in \mysett{(z_3,\bar{z}_1),(\bar{z}_3,z_1)}$,}\\
0 & \mbox{otherwise.}
\end{cases}
\end{multline*}
A short calculation shows that $\cP$ is indeed an intermediate channel that degrades $\Wupgraded_{23}$ to $\Wupgraded_{123}$.
\end{proof}
At this point, the reader may be wondering why we have chosen to state Lemma~\ref{lemm:upgradeStep0} at all. Namely, it is clear what disadvantages it has with respect to Lemma~\ref{lemm:upgradeStep1}, but we have yet to indicate any advantages. Recalling the conditions of Lemma~\ref{lemm:upgradeStep1}, we see that it can not be employed when the set $\mysett{\lambda_1,\lambda_2,\lambda_3}$ contains non-unique elements. In fact, more is true. Ultimately, when one wants to implement the algorithms outlined in this paper, one will most probably use floating point numbers. Recall that a major source of numerical instability stems from subtracting two floating point numbers that are too close. By considering the denominator in (\ref{eq:alpha1beta1}) and (\ref{eq:alpha3beta3}) we see that $\lambda_1$ and $\lambda_3$ should not be too close. Moreover, by considering the numerators, we conclude that $\lambda_2$ should not be too close to both $\lambda_1$ and $\lambda_3$. So, when these cases do occur, our only option is Lemma~\ref{lemm:upgradeStep0}.
We now define the merge-upgrading procedure we have used. Apart from an initial step, it is very similar to the merge-degrading procedure we have previously outlined. Assume we are given a BMS channel $\Wbit: \mathcal{X} \to \mathcal{Y}$ with an alphabet size of $2L$ and wish to reduce its alphabet size to $\mu$, while transforming $\Wbit$ into a upgraded version of itself. If $2L \leq \mu$, then, as before, we are done. Otherwise, as in the merge-degrading procedure, we choose $L$ representatives $y_1,y_2,\ldots,y_L$, and order them according to their $\mathrm{LR}$ values, all of which are greater than or equal to $1$. We now specify the preliminary step: for some specified parameter epsilon (we have used $\epsilon = 10^{-3}$), we check if there exists an index $1 \leq i < L$ such that the ratio $\mathrm{LR}(y_{i+1}) / \mathrm{LR}(y_i)$ is less than $1+\epsilon$. If so, we apply Lemma~\ref{lemm:upgradeStep0} repeatedly, until no such index exists. Now comes the main step. We ask the following question: for which index $1 \leq i \leq L-1$ does the channel resulting from the application of Lemma~\ref{lemm:upgradeStep1} to $\Wbit$, $y_i$, $y_{i+1}$, and $y_{i+2}$ result in a channel with smallest capacity increase? After finding the minimizing index $i$, we indeed apply Lemma~\ref{lemm:upgradeStep1} and get an upgraded channel $\mathcal{Q}'$ with an alphabet size smaller by $2$ than that of $\Wbit$. The same process is applied to $\mathcal{Q}'$, until the output alphabet size is not more than $\mu$. As before, assuming the output alphabet size of $\Wbit$ is at most $2 \mu^2$, an implementation similar to that given in Algorithm~\ref{alg:degradeChannel} will run in $O(\mu^2 \log \mu)$ time.
As was the case for degrading, the following theorem proves that no generality is lost by only considering merging of consecutive triplets of the form $y_i$, $y_{i+1}$, $y_{i+2}$ in the main step. The proof is given in Appendix~\ref{sec:proofUpgradingOptimal}.
\begin{theo}
\label{theo:upgradeOptimal}
Let $\Wbit:\mathcal{X} \to \mathcal{Y}$ be a BMS channel. Denote by $I_\Wbit$ the capacity of $\Wbit$ and by $I(y_1,y_2,y_3)$ the capacity one gets by the application of Lemma~\ref{lemm:upgradeStep1} to $\Wbit$ and symbols $y_1,y_2,y_3 \in \mathcal{Y}$ such that
\[
1 \leq \mathrm{LR}(y_1) \leq \mathrm{LR}(y_2) \leq LR(y_3) \; .
\]
Let $\mathrm{LR}(y_1) = \lambda_1$, $\mathrm{LR}(y_2) = \lambda_2$, $\mathrm{LR}(y_3) = \lambda_3$, $\pi_2 = \Wbit(y_2|0) +\Wbit(y_2|1)$, and denote the difference in capacities as
\[
\Delta[\lambda_1;\lambda_2,\pi_2;\lambda_3] = I(y_1,y_2,y_3) - I_\Wbit \; .
\]
Then, for all $\lambda_1' \leq \lambda_1$ and $\lambda_3' \geq \lambda_3$,
\begin{equation}
\Delta[\lambda_1;\lambda_2,\pi_2;\lambda_3] \leq \Delta[\lambda_1';\lambda_2,\pi_2;\lambda_3'] \; .
\end{equation}
\end{theo}
We end this section by considering the running time of Algorithms~\ref{alg:highLevelDegrade} and \ref{alg:highLevelUpgrade}.
\begin{theo}
\label{theo:totalRunningTime}
Let an underlying BMS channel $W_{\mathrm{base}}$, a fidelity parameter $\mu$, and codelength $n=2^m$ be given. Assume that the output alphabet size of the underlying channel $W_{\mathrm{base}}$ is at most $\mu$. The running time of either Algorithm~\ref{alg:highLevelDegrade} or Algorithm~\ref{alg:highLevelUpgrade} is as follows. Approximating a single bit-channel takes $O(m \cdot \mu^2 \log \mu)$ time; approximating all $n$ bit-channels takes $O(n \cdot \mu^2 \log \mu)$ time.
\end{theo}
\begin{proof}
Without loss of generality, we consider Algorithm~\ref{alg:highLevelDegrade}. Recall that the output alphabet size of $W_{\mathrm{base}}$ is at most $\mu$. Thus, by induction, at the start of each loop the size of the output alphabet of $\mathcal{Q}$ is at most $\mu$. Therefore, at each iteration, calculating $\mathcal{W}$ from $\mathcal{Q}$ takes $O(\mu^2)$ time, since the output alphabet size of $\mathcal{W}$ is at most $2 \mu^2$. Next, we have seen that each invocation of $\mbox{\code degrading\_merge}$ takes $O(\mu^2 \log \mu)$ time. The number of times we loop in Algorithm~\ref{alg:highLevelDegrade} is $m$. Thus, for a single bit-channel, the total running time is $O(m \cdot \mu^2 \log \mu)$.
As was explained at the end of Section~\ref{sec:highLevelAlg}, when approximating all $n$ bit channels, the number of distinct channels that need to be approximated is $2n-2$. Thus, the total running time in this case is $O(n \cdot \mu^2 \log \mu)$.
\end{proof}
\section{Channels with Continuous Output Alphabet}
\label{sec:continuousChannels}
Recall that in order to apply either Algorithm~\ref{alg:highLevelDegrade} or \ref{alg:highLevelUpgrade} to an underlying BMS\ channel $W_{\mathrm{base}}$, we had to thus far assume that $W_{\mathrm{base}}$ has a finite output alphabet. In this section, we show two transforms (one degrading and the other upgrading) that transform a BMS\ channel with a continuous alphabet to a BMS\ channel with a specified finite output alphabet size. Thus, after applying the degrading (upgrading) transform we will shortly specify to $W_{\mathrm{base}}$, we will be in a position to apply Algorithm~\ref{alg:highLevelDegrade} (\ref{alg:highLevelUpgrade}) and get a degraded (upgraded) approximation of $\mathcal{W}_i$. Moreover, we prove that both degraded and upgraded versions of our original channels have a guaranteed closeness to the original channel, in terms of difference of capacity.
Let $W_{\mathrm{base}}$ be a given BMS\ channel with a continuous alphabet. We will make a few assumptions on $W_{\mathrm{base}}$. First, we assume that the output alphabet of $W_{\mathrm{base}}$ is the reals $\mathbb{R}$. Thus, for $y \in \mathbb{R}$, let $f(y|0)$ and $f(y|1)$ be the p.d.f.\ functions of the output given that the input was $0$ and $1$, respectively. Next, we require that the symmetry of $W_{\mathrm{base}}$ manifest itself as
\[
f(y|0) = f(-y|1) \; , \quad \mbox{for all $y \in \mathbb{R}$} \; .
\]
Also, for notational convenience, we require that
\begin{equation}
\label{eq:fsane}
f(y|0) \geq f(y|1) \; , \quad \mbox{for all $y \geq 0$} \; .
\end{equation}
Note that all of the above holds for the BAWGN channel (after renaming the input $0$ as $-1$).
We now introduce some notation. For $y \geq 0$, define the likelihood ratio of $y$ as
\begin{equation}
\label{eq:lambday}
\lambda(y) = \frac{f(y|0)}{f(y|1)} \; .
\end{equation}
As usual, if $f(y|1)$ is zero while $f(y|0)$ is not, we define $\lambda(y)=\infty$. Also, if both $f(y|0)$ and $f(y|1)$ are zero, then we arbitrarily define $\lambda(y) = 1$. Note that by (\ref{eq:fsane}), we have that $\lambda(y) \geq 1$.
Under these definitions, a short calculation shows that the capacity of $W_{\mathrm{base}}$ is
\[
I(W_{\mathrm{base}}) = \int_{0}^{\infty} \left( f(y|0) + f(y|1) \right) C[\lambda(y)] \; dy \; ,
\]
where for $1 \leq \lambda < \infty$
\[
C[\lambda] = 1 - \frac{\lambda}{\lambda+1} \log_2 \left( 1 + \frac{1}{\lambda} \right) - \frac{1}{\lambda+1} \log_2 \left( \lambda + 1 \right) \; ,
\]
and (for continuity) we define $C[\infty] = 1$.
Let $\mu = 2\nu$ be the specified size of the degraded/upgraded channel output alphabet. An important property of $C[\lambda]$ is that it is strictly increasing in $\lambda$ for $\lambda \geq 1$. This property is easily proved, and will now be used to show that the following sets form a partition of the non-negative reals. For $1 \leq i \leq \nu-1$, let
\begin{equation}
\label{eq:Aigeneral}
A_i = \myset{y \geq 0 : \frac{i-1}{\nu} \leq C[\lambda(y)] < \frac{i}{\nu} } \; .
\end{equation}
For $i = \nu$ we similarly define (changing the second inequality to a weak inequality)
\begin{equation}
\label{eq:Ailast}
A_\nu = \myset{y \geq 0 : \frac{\nu-1}{\nu} \leq C[\lambda(y)] \leq 1 } \; .
\end{equation}
As we will see later on, we must assume that the sets $A_i$ are sufficiently ``nice''. This will indeed be the case of for BAWGN channel.
\subsection{Degrading transform}
\label{subsec:continuousChannels:degrade}
Essentially, our degrading procedure will consist of $\nu$ applications of the continuous analog of Lemma~\ref{lemm:degradeStep}. Denote by $\mathcal{Q} : \mathcal{X} \to \mathcal{Z}$ the degraded approximation of $W_{\mathrm{base}}$ we are going to produce, where
\[
\mathcal{Z} = \myset{z_1, \bar{z}_1, z_2, \bar{z}_2, \ldots, z_\nu, \bar{z}_\nu} \; .
\]
We define $\mathcal{Q}$ as follows.
\begin{align}
\label{Wdegradedzi}
\mathcal{Q}(z_i|0) = \mathcal{Q}(\bar{z}_i|1) &= \int_{A_i} f(y|0) \, dy \; , \\
\label{Wdegradedbarzi}
\mathcal{Q}(\bar{z}_i|0) = \mathcal{Q}(z_i|1) &= \int_{A_i} f(-y|0) \, dy \; .
\end{align}
\begin{lemm}
The channel $\mathcal{Q} : \mathcal{X} \to \mathcal{Z}$ is a BMS\ channel such that $\mathcal{Q} \preccurlyeq W_{\mathrm{base}}$.
\end{lemm}
\begin{proof}
It is readily seen that $\mathcal{Q}$ is a BMS\ channel. To prove $\mathcal{Q} \preccurlyeq W_{\mathrm{base}}$, we now supply intermediate channel $\cP : \mathbb{R} \to \mathcal{Z}$.
\[
\cP(z | y) =
\begin{cases}
1 & \mbox{if $z = z_i$ and $y \in A_i$} \; ,\\
1 & \mbox{if $z = \bar{z}_i$ and $-y \in A_i$} \; ,\\
0 & \mbox{otherwise} \; .
\end{cases}
\]
\end{proof}
The following lemma bounds the loss in capacity incurred by the degrading operation.
\begin{lemm}
The difference in capacities of $\mathcal{Q}$ and $W_{\mathrm{base}}$ can be bounded as follows,
\begin{equation}
\label{eq:contDegradingCapDiff}
0 \leq I(W_{\mathrm{base}}) - I(\mathcal{Q}) \leq \frac{1}{\nu} = \frac{2}{\mu}\; .
\end{equation}
\end{lemm}
\begin{proof}
The first inequality in (\ref{eq:contDegradingCapDiff}) is a consequence of the degrading relation and (\ref{eq:capacityDegrading}). We now turn our attention to the second inequality.
Recall that since the $A_i$ partition the non-negative reals, the capacity of $W_{\mathrm{base}}$ equals
\begin{equation}
\label{eq:IWunderlyingAsSumOni}
I(W_{\mathrm{base}}) = \sum_{i=1}^\nu \int_{A_i} \left( f(y|0) + f(y|1) \right) C[\lambda(y)] dy \; .
\end{equation}
As for $\mathcal{Q}$, we start by defining for $1 \leq i \leq \nu$ the ratio
\[
\theta_i = \frac{\mathcal{Q}(z_i|0)}{\mathcal{Q}(z_i|1)} \; ,
\]
where the cases of the numerator and/or denominator equaling zero are as in the definition of $\lambda(y)$. By this definition, similarly to the continuous case, the capacity of $\mathcal{Q}$ is equal to
\begin{equation}
\label{eq:IWdegradedContDef}
I(\mathcal{Q}) = \sum_{i=1}^\nu \left( \mathcal{Q}(z_i|0) + \mathcal{Q}(z_i|1) \right) C[\theta_i] \; .
\end{equation}
Recall that by the definition of $A_i$ in (\ref{eq:Aigeneral}) and (\ref{eq:Ailast}), we have that for all $y \in A_i$,
\[
\frac{i-1}{\nu} \leq C[\lambda(y)] \leq \frac{i}{\nu} \; .
\]
Thus, by the definition of $\mathcal{Q}(z_i|0)$ and $\mathcal{Q}(z_i|1)$ in (\ref{Wdegradedzi}) and (\ref{Wdegradedbarzi}), respectively, we must have by the strict monotonicity of $C$ that
\[
\frac{i-1}{\nu} \leq C[\theta_i] \leq \frac{i}{\nu} \; , \quad \mbox{if $\mathcal{Q}(z_i|0)>0$} \; .
\]
Thus, for all $y \in A_i$,
\[
\size{C[\theta_i] - C[\lambda(y)]} \leq \frac{1}{\nu} \; , \quad \mbox{if $\mathcal{Q}(z_i|0)>0$} \; .
\]
Next, note that $\mathcal{Q}(z_i|0)>0$ implies $\mathcal{Q}(z_i|0) + \mathcal{Q}(z_i|1)>0$. Thus, we may bound $I(\mathcal{Q})$ as follows,
\begin{multline*}
I(\mathcal{Q}) = \sum_{i=1}^\nu \left( \mathcal{Q}(z_i|0) + \mathcal{Q}(z_i|1) \right) C[\theta_i] = \\
\sum_{i=1}^\nu \int_{A_i} \left( f(y|0) + f(y|1) \right) C[\theta_i] dy \geq \\
\sum_{i=1}^\nu \int_{A_i} \left( f(y|0) + f(y|1) \right) \left( C[\lambda(y)] - \frac{1}{\nu} \right) dy = \\
\left( \sum_{i=1}^\nu \int_{A_i} \left( f(y|0) + f(y|1) \right) C[\lambda(y)] dy \right) - \frac{1}{\nu} = I(W_{\mathrm{base}}) - \frac{1}{\nu} \; ,
\end{multline*}
which proves the second inequality.
\end{proof}
\subsection{Upgrading transform}
In parallel with the degrading case, our upgrading procedure will essentially consist of $\nu$ applications of the continuous analog of Lemma~\ref{lemm:upgradeStep0}. Denote by $\mathcal{Q}' : \mathcal{X} \to \calZ'$ the upgraded approximation of $W_{\mathrm{base}}$ we are going to produce, where
\[
\calZ' = \myset{z_1, \bar{z}_1, z_2, \bar{z}_2, \ldots, z_\nu, \bar{z}_\nu} \; .
\]
As before, we will show that the loss in capacity due to the upgrading operation is at most $1/\nu$.
Let us now redefine the ratio $\theta_i$. Recalling that the function $C[\lambda]$ is strictly increasing in $\lambda \geq 1$, we deduce that it has an inverse in that range. Thus, for $1 \leq i \leq \nu$, we define $\theta_i \geq 1$ as follows,
\begin{equation}
\label{eq:lambdaiUpgrading}
\theta_i = C^{-1}\left[\frac{i}{\nu}\right] \; .
\end{equation}
Note that for $i = \nu$, we have that $\theta_\nu = \infty$. Also, note that for $y \in A_i$ we have by (\ref{eq:Aigeneral}) and (\ref{eq:Ailast}) that
\begin{equation}
\label{eq:lambdayLambdaiupgrading}
1 \leq \lambda(y) \leq \theta_i \; .
\end{equation}
We now define $\mathcal{Q}'$. For $1 \leq i \leq \nu$, let,
\begin{equation}
\label{eq:piiDefinition}
\pi_i = \int_{A_i} \big( f(\alpha|0) + f(-\alpha|0) \big) \, d\alpha \; .
\end{equation}
Then,
\begin{equation}
\label{eq:WupgradedContDef}
\mathcal{Q}'(z|0) =
\begin{cases}
\frac{\theta_i \pi_i}{\theta_i+1} & \mbox{if $z=z_i$ and $\theta_i \neq \infty$} \; , \\
\frac{\pi_i}{\theta_i+1} & \mbox{if $z=\bar{z}_i$ and $\theta_i \neq \infty$} \; , \\
\pi_i & \mbox{if $z=z_i$ and $\theta_i = \infty$} \; , \\
0 & \mbox{if $z=\bar{z}_i$ and $\theta_i = \infty$} \; , \\
\end{cases}
\end{equation}
and
\begin{equation}
\mathcal{Q}'(z_i|1) = \mathcal{Q}'(\bar{z}_i|0) \; , \quad \mathcal{Q}'(\bar{z}_i|1) = \mathcal{Q}'(z_i|0) \; .
\end{equation}
\begin{lemm}
The channel $\mathcal{Q}' : \mathcal{X} \to \calZ'$ is a BMS\ channel such that $\mathcal{Q}' \succcurlyeq W_{\mathrm{base}}$.
\end{lemm}
\begin{proof}
As before, the proof that $\mathcal{Q}'$ is a BMS\ channel is easy. To show that $\mathcal{Q}' \succcurlyeq W_{\mathrm{base}}$, we must supply the intermediate channel $\cP$. The proof follows easily if we define $\cP : \calZ' \to \mathbb{R}$ as the cascade of two channels, $\cP_1 : \calZ' \to \mathbb{R}$ and $\cP_2 : \mathbb{R} \to \mathbb{R}$.
The channel $\cP_1 : \calZ' \to \mathbb{R}$ is essentially a renaming channel. Denote by $g(\alpha|z)$ the p.d.f.\ of the output of $\cP_1$ given that the input was $z$. Then, for $1 \leq i \leq \nu$,
\begin{multline}
g(\alpha|z) = \\
\begin{cases}
\frac{f(\alpha|0)+f(-\alpha|0)}{\pi_i} & \mbox{if $z=z_i$ and $\alpha \in A_i$} \; , \\
\frac{f(\alpha|0)+f(-\alpha|0)}{\pi_i} & \mbox{if $z=\bar{z}_i$ and $-\alpha \in A_i$} \; , \\
0 & \mbox{otherwise} \; .
\end{cases}
\end{multline}
Note that by (\ref{eq:piiDefinition}), the function $g(\alpha|z)$ is indeed a p.d.f.\ for every fixed value of $z \in \calZ'$.
Next, we turn to $\cP_2 : \mathbb{R} \to \mathbb{R}$, the LR reducing channel. Let $\alpha \in A_i$ and recall the definition of $\lambda(y)$ given in
(\ref{eq:lambday}). Define
the quantity $p_\alpha$ as follows,
\begin{equation}
p_\alpha =
\begin{cases}
\frac{\theta_i - \lambda(\alpha)}{(\lambda(\alpha) + 1)(\theta_i-1)} & \mbox{if $1 < \theta_i < \infty$} \; , \\
\frac{1}{2} & \mbox{if $\theta_i = 1$} \; , \\
\frac{1}{\lambda(\alpha)+1} & \mbox{if $\theta_i = \infty$ and $\lambda(\alpha) < \infty$} \; , \\
0 & \mbox{if $\lambda(\alpha) = \infty$} \; .
\end{cases}
\end{equation}
By (\ref{eq:lambdayLambdaiupgrading}) with $\alpha$ in place of $y$ we deduce that $0 \leq p_\alpha \leq 1/2$. We define the channel $\cP_2 : \mathbb{R} \to \mathbb{R}$ as follows. For $y \geq 0$,
\begin{equation}
\cP_2(y|\alpha) =
\begin{cases}
1-p_\alpha & \mbox{if $y = \alpha$} \; , \\
p_\alpha & \mbox{if $y = -\alpha$} \; ,
\end{cases}
\end{equation}
and
\[
\cP_2(-y|-\alpha) = \cP_2(y|\alpha) \; .
\]
Consider the random variable $Y$, which is defined as the output of the concatenation of channels $\mathcal{Q}'$, $\cP_1$, and $\cP_2$, given that the input to $\mathcal{Q}'$ was $0$. We must show that the p.d.f.\ of $Y$ is $f(y|0)$. To do this, we consider the limit
\[
\lim_{\epsilon \to 0} \frac{\mathrm{Prob}(y \leq Y \leq y+\epsilon)}{\epsilon} \; .
\]
Consider first a $y$ such that $y \in A_i$, and assume further that $\epsilon$ is small enough so that the whole interval between $y$ and $y+\epsilon$ is in $A_i$. In this case, the above can be expanded to
\begin{multline*}
\lim_{\epsilon \to 0} \frac{1}{\epsilon}\Big[ \mathcal{Q}'(z_i | 0) \cdot \int_{y}^{y+\epsilon}g(\alpha|z_i)(1-p_\alpha) \, d\alpha
\\
+ \mathcal{Q}'(\bar{z}_i | 0) \cdot \int_{-y-\epsilon}^{-y}g(\alpha|\bar{z}_i)p_{-\alpha} \, d\alpha \Big] \; .
\end{multline*}
Assuming that the two integrands are indeed integrable, this reduces to
\[
\mathcal{Q}'(z_i | 0) \cdot g(y|z_i)(1-p_y) + \mathcal{Q}'(\bar{z}_i | 0) \cdot g(-y|\bar{z}_i)p_y \; .
\]
From here, simple calculations indeed reduce the above to $f(y|0)$. The other cases are similar.
\end{proof}
As in the degrading case, we can bound the loss in capacity incurred by the upgrading operation.
\begin{lemm}
The difference in capacities of $\mathcal{Q}'$ and $W_{\mathrm{base}}$ can be bounded as follows,
\begin{equation}
\label{eq:contUpgradingCapDiff}
0 \leq I(\mathcal{Q}') - I(W_{\mathrm{base}}) \leq \frac{1}{\nu} = \frac{2}{\mu} \; .
\end{equation}
\end{lemm}
\begin{proof}
The first inequality in (\ref{eq:contUpgradingCapDiff}) is a consequence of the upgrading relation and (\ref{eq:capacityDegrading}). We now turn our attention to the second inequality.
For all $y \in A_i$, by (\ref{eq:Aigeneral}), (\ref{eq:Ailast}) and (\ref{eq:lambdaiUpgrading}), we have that
\[
C[\theta_i] - C[\lambda(y)] \leq \frac{1}{\nu} \; , \quad \mbox{if $\mathcal{Q}'(z_i|0)>0$} \; .
\]
Next, notice that by (\ref{eq:WupgradedContDef}),
\[
\theta_i = \frac{\mathcal{Q}'(z_i|0)}{\mathcal{Q}'(z_i|1)} \; , \quad \mbox{if $\mathcal{Q}'(z_i|0)>0$} \; .
\]
As in the degrading case, we have that $\mathcal{Q}'(z_i|0)>0$ implies $\mathcal{Q}'(z_i|0) + \mathcal{Q}'(z_i|1)>0$.
Thus, we may bound $I(\mathcal{Q}')$ as follows,
\begin{multline*}
I(\mathcal{Q}') = \sum_{i=1}^\nu \left( \mathcal{Q}'(z_i|0) + \mathcal{Q}'(z_i|1) \right) C[\theta_i] = \\
\sum_{i=1}^\nu \int_{A_i} \left( f(y|0) + f(y|1) \right) C[\theta_i] dy \leq \\
\sum_{i=1}^\nu \int_{A_i} \left( f(y|0) + f(y|1) \right) \left( C[\lambda(y)] + \frac{1}{\nu} \right) dy = \\
\left( \sum_{i=1}^\nu \int_{A_i} \left( f(y|0) + f(y|1) \right) C[\lambda(y)] dy \right) + \frac{1}{\nu} = I(W_{\mathrm{base}}) + \frac{1}{\nu} \; ,
\end{multline*}
which proves the second inequality.
\end{proof}
\section{Variations of Our Algorithms}
\label{sec:variationsOnTheme}
As one might expect, Algorithms~\ref{alg:highLevelDegrade} and \ref{alg:highLevelUpgrade} can be tweaked and modified. As an example, we now show an improvement to Algorithm~\ref{alg:highLevelDegrade} for a specific case. As we will see in Section~\ref{sec:analysis}, this improvement is key to proving Theorem~\ref{theo:polyConstruction}. Also, it turns out that Algorithm~\ref{alg:highLevelDegrade} is compatible with the result by Guruswami and Xia \cite{GuruswamiXia:13aa}, in the following sense: if we were to use algorithm Algorithm~\ref{alg:highLevelDegrade} with the same $n$ and $\mu$ dictated by \cite{GuruswamiXia:13aa}, then we would be guaranteed a resulting code with parameters as least as good as those promised by \cite{GuruswamiXia:13aa}.
Recall our description of how to construct a polar code given at the end of Section~\ref{sec:highLevelAlg}: obtain a degraded approximation of each bit channel through the use of Algorithm~\ref{alg:highLevelDegrade}, and then select the $k$ best channels when ordered according to the upper bound on the probability of error. Note that Algorithm~\ref{alg:highLevelDegrade} returns a channel, but in this case only one attribute of that channel interests us, namely, the probability of error. In this section, we show how to specialize Algorithm~\ref{alg:highLevelDegrade} accordingly and benefit.
The specialized algorithm is given as Algorithm~\ref{alg:specialized}. We note that the plots in this paper having to do with an upper bound on the probability of error were produced by running this algorithm.
The key observation follows from Equations~(26) and (27) in \cite{Arikan:09p}, which we now restate. Recall that $Z(\Wbit)$ is the Bhattacharyya parameter of the channel $\Wbit$. Then,
\begin{align}
Z(\Wbit b \Wbit ) &\leq 2Z(\Wbit)-Z(\Wbit)^2 \label{eq:ZATbad}\\
Z(\Wbit a \Wbit ) &= Z(\Wbit)^2 \label{eq:ZATgood}
\end{align}
\LinesNumbered
\begin{algorithm}
\SetInd{0.49em}{0.49em}
\caption{An upper bound on the error probability}
\label{alg:specialized}
\Input{An underlying BMS\ channel $W_{\mathrm{base}}$, a bound $\mu=2\nu$ on the output alphabet size, a code length $n=2^m$, an index $i$ with binary representation $i=\binaryRep{b_1,b_2,\ldots,b_m}$.}
\Output{An upper bound on $P_e(\mathcal{W}_i)$.}
$\ZAlg \leftarrow Z(W_{\mathrm{base}})$\;
$\mathcal{Q} \leftarrow \mbox{\code degrading\_merge}(W_{\mathrm{base}},\mu)$\; \label{alg:specialized:loopStarts}
\For{$j = 1,2,\ldots, m$}
{
\eIf{$b_j=0$}
{
$\mathcal{W} \leftarrow \mathcal{Q} b \mathcal{Q}$ \;
$\ZAlg \leftarrow \min\{Z(\mathcal{W}), 2\ZAlg-\ZAlg^2 \}$\; \label{alg:specialized:minBad}
}
{
$\mathcal{W} \leftarrow \mathcal{Q} a \mathcal{Q}$ \;
$\ZAlg \leftarrow \ZAlg^2$\; \label{alg:specialized:minGood}
}
$\mathcal{Q} \leftarrow \mbox{\code degrading\_merge}(\mathcal{W},\mu)$\; \label{alg:specialized:degrade}
}
\Return{$\min\{P_e(\mathcal{Q}), \ZAlg \}$}\; \label{alg:specialized:return}
\end{algorithm}
\begin{theo}
Let a codeword length $n=2^m$, an index $0 \leq i < n$, an underlying channel $W_{\mathrm{base}}$, and a fidelity parameter $\mu=2\nu$ be given. Denote by $\hat{p}_A$ and $\hat{p}_D$ the outputs of Algorithms~\ref{alg:highLevelDegrade} and \ref{alg:specialized}, respectively. Then,
\[
\hat{p}_A \geq \hat{p}_D \geq P_e(\mathcal{W}_i) \; .
\]
That is, the bound produced by Algorithm~\ref{alg:specialized} is always
as least as good as that produced by
Algorithm~\ref{alg:highLevelDegrade}.
\end{theo}
\begin{proof}
Denote by $\mathcal{W}^{(j)}$ the channel we are trying to approximate during iteration $j$. That is, we start with $\mathcal{W}^{(0)} = W_{\mathrm{base}}$. Then, iteratively $\mathcal{W}^{(j+1)}$ is gotten by transforming $\mathcal{W}^{(j)}$ using either $b$ or $a$, according to the value of $b_j$. Ultimately, we have $\mathcal{W}^{(m)}$, which is simply the bit-channel $\mathcal{W}_i$.
The heart of the proof is to show that after iteration $j$ has completed (just after line~\ref{alg:specialized:degrade} has executed), the variable $\ZAlg$ is such that
\[
Z(\mathcal{W}^{(j)}) \leq \ZAlg \leq 1 \; .
\]
The proof is by induction. For the basis, note that before the first iteration starts (just after line~\ref{alg:specialized:loopStarts} has executed), we have $\ZAlg = Z(\mathcal{W}^{(0)})$. For the induction step, first note that $2\ZAlg-\ZAlg^2$ is both an increasing function of $\ZAlg$ and is between $0$ and $1$, when $0 \leq \ZAlg \leq 1$. Obviously, this is also true for $\ZAlg^2$. Now, note that at the end of iteration $j$ we have that the variable $\mathcal{W}$ is degraded with respect to $\mathcal{W}^{(j)}$. Recalling (\ref{eq:BhattacharyyaProbDegrading}), (\ref{eq:ZATbad}) and (\ref{eq:ZATgood}), the induction step is proved.
\end{proof}
\begin{table}
\begin{tabular}{cccc}
& Algorithm~\ref{alg:highLevelDegrade} & Algorithm~\ref{alg:specialized} & Algorithm~\ref{alg:highLevelUpgrade} \\
$\mu=8$ & 5.096030e-03 & 1.139075e-04 & 1.601266e-11 \\
$\mu=16$ & 6.926762e-05 & 2.695836e-05 & 4.296030e-08 \\
$\mu=64$ & 1.808362e-06 & 1.801289e-06 & 7.362648e-07 \\
$\mu=128$ & 1.142843e-06 & 1.142151e-06 & 8.943154e-07 \\
$\mu=256$ & 1.023423e-06 & 1.023423e-06 & 9.382042e-07 \\
$\mu=512$ & & 9.999497e-07 & 9.417541e-07 \\
\end{tabular}
\caption{Upper and lower bounds on $P_{W_{\mathrm{base}},n}(k)}%{\hat{P}_\mathrm{FER}$ for $W_{\mathrm{base}}=\mathrm{BSC}(0.11)$, codeword length $n=2^{20}$, and rate $k/n = 445340/2^{20} = 0.42471$.}
\label{tbl:muandpe}
\end{table}
We end this section by referring to Table~\ref{tbl:muandpe}. In the table, we fix the underlying channel, the codeword length, and the code rate. Then, we compare upper and lower bounds on $P_{W_{\mathrm{base}},n}(k)}%{\hat{P}_\mathrm{FER}$, for various values of $\mu$. For a given $\mu$, the lower bound is gotten by running Algorithm~\ref{alg:highLevelUpgrade} while the two upper bounds are gotten by running Algorithms \ref{alg:highLevelDegrade} and \ref{alg:specialized}. As can be seen, the upper bound supplied by Algorithm~\ref{alg:specialized} is always superior.
\section{Analysis}
\label{sec:analysis}
As we've seen in previous sections, we can build polar codes by employing Algorithm~\ref{alg:specialized}, and gauge how far we are from the optimal construction by running Algorithm~\ref{alg:highLevelUpgrade}. As can be seen in Figure~\ref{fig:BSCplots}, our construction turns out to be essentially optimal, for moderate sizes of $\mu$. However, we are still to prove Theorem~\ref{theo:polyConstruction}, which gives analytic justification to our method of construction. We do so in this section.
As background to Theorem~\ref{theo:polyConstruction}, recall from \cite{ArikanTelatar:09c} that for a polar code of length $n = 2^m$, the fraction of bit channels with probability of error less than $2^{-n^\beta}$ tends to the capacity of the underlying channel as $n$ goes to infinity, for $\beta < 1/2$. Moreover, the constraint $\beta < 1/2$ is tight in that the fraction of such channels is strictly less than the capacity, for $\beta > 1/2$. Thus, in this context, the restriction on $\beta$ imposed by Theorem~\ref{theo:polyConstruction} cannot be eased.
In order to prove Theorem~\ref{theo:polyConstruction}, we make use of the results of Pedarsani, Hassani, Tal, and Telatar \cite{PHTT:11c}, in particular \cite[Theorem~1]{PHTT:11c} given below. We also point out that many ideas used in the proof of Theorem~\ref{theo:polyConstruction} appear --- in one form or another --- in \cite[Theorem~2]{PHTT:11c} and its proof.
\begin{theo}[Restatement of {\cite[Theorem~1]{PHTT:11c}}]
\label{theo:CSBound}
Let an underlying BMS channel $W_{\mathrm{base}}$ be given. Let $n=2^m$ be the code length, and denote by $\mathcal{W}_i^{(m)}$ the corresponding $i$th bit channel, where $0 \leq i < n$. Next, denote by $\mathcal{Q}_i^{(m)}(\nu)$ the degraded approximation of $\mathcal{W}_i^{(m)}$ returned by running Algorithm~\ref{alg:highLevelDegrade} with parameters $W_{\mathrm{base}}$, $\mu=2\nu$, $i$, and $m$. Then,
\[
\frac{\size{\myset{ i : I(\mathcal{W}_i^{(m)}) - I(\mathcal{Q}_i^{(m)}(\nu)) \geq \sqrt{\frac{m}{\nu}}}}}{n} \leq \sqrt{\frac{m}{\nu}} \; .
\]
\end{theo}
With respect to the above, we remark the following. Recall that in Subsection~\ref{subsec:continuousChannels:degrade} we introduced a method of degrading a continuous channel to a discrete one with at most $\mu=2\nu$ symbols. In fact, there is nothing special about the continuous case: a slight modification can be used to degrade an arbitrary discrete channel to a discrete channel with at most $\mu$ symbols. Thus, we have an alternative to the merge-degrading method introduced in Subsection~\ref{subsec:mergingFunctions:degrade}. Thus, it follows easily that Theorem~\ref{theo:CSBound} and thus Theorem~\ref{theo:polyConstruction} would still hold had we used that alternative.
We now break the proof of Theorem~\ref{theo:polyConstruction} into several lemmas. Put simply, the first lemma states that a laxer requirement than that in Theorem~\ref{theo:polyConstruction} on the probability of error can be met.
\begin{lemm}
\label{lemm:ZdeltaEpsilon}
Let $\mathcal{Q}_i^{(m)}(\nu)$ be as in Theorem~\ref{theo:CSBound}. Then, for every $\delta>0$ and $\epsilon>0$ there exists an $m_0$ and a large enough $\mu=2\nu$ such that
\begin{equation}
\label{eq:ZdeltaEpsilon}
\frac{\size{\myset{ i_0 : Z\left(\mathcal{Q}_{i_0}^{(m_0)}(\nu)\right) \leq \delta}}}{n_0} \geq I(W_{\mathrm{base}}) - \epsilon \; ,
\end{equation}
where
\[
n_0 = 2^{m_0} \quad \mbox{and} \quad 0 \leq i_0 < n_0 \; .
\]
\end{lemm}
We first note that Lemma~\ref{lemm:ZdeltaEpsilon} has a trivial proof: By \cite[Theorem~2]{Arikan:09p}, we know that there exists an $m_0$ for which (\ref{eq:ZdeltaEpsilon}) holds, if $\mathcal{Q}_{i_0}^{(m_0)}(\nu)$ is replaced by $\mathcal{W}_{i_0}^{(m_0)}$. Thus, we may take $\mu$ large enough so that the pair-merging operation defined in Lemma~\ref{lemm:degradeStep} is never executed, and so $\mathcal{Q}_{i_0}^{(m_0)}(\nu)$ is in fact equal to $\mathcal{W}_{i_0}^{(m_0)}$.
This proof --- although valid --- implies a value of $\mu$ which is
doubly exponential in $m_0$. We now give an alternative proof, which ---
as we have recently learned --- is a precursor to the result of
Guruswami and Xia~\cite{GuruswamiXia:13aa}. Namely, we state this
alternative proof since we have previously conjectured and now know by
\cite{GuruswamiXia:13aa} that it implies a value of $m_0$ which is not
too large.
\begin{proof}[proof of Lemma~\ref{lemm:ZdeltaEpsilon}]
For simplicity of notation, let us drop the subscript $0$ from $i_0$, $n_0$, and $m_0$. Recall that by \cite[Theorem~1]{Arikan:09p} we have that the capacity of bit channels polarizes. Specifically, for each $\epsilon_1 > 0$ and $\delta_1 >0$ there exists an $m$ such that
\begin{equation}
\label{eq:IpolarizationEpsilonDelta}
\frac{\size{\myset{ i : I\left(\mathcal{W}_{i}^{(m)}\right) \geq 1 - \delta_1}}}{n} \geq I(W_{\mathrm{base}}) - \epsilon_1 \; .
\end{equation}
We can now combine the above with Theorem~\ref{theo:CSBound} and deduce that
\begin{multline}
\label{eq:IpolarizationEpsilonDelta_DegradedComplicated}
\frac{\size{\myset{ i : I\left(\mathcal{Q}_{i}^{(m)}(\nu)\right) \geq 1 - \delta_1-\sqrt{\frac{m}{\nu}}}}}{n} \geq \\
I(W_{\mathrm{base}}) - \epsilon_1 - \sqrt{\frac{m}{\nu}} \; .
\end{multline}
Next, we claim that for each $\delta_2 > 0$ and $\epsilon_2 > 0$ there exist $m$ and $\mu = 2\nu$ such that
\begin{equation}
\label{eq:eq:IpolarizationEpsilonDelta_DegradedSimple}
\frac{\size{\myset{ i : I\left(\mathcal{Q}_{i}^{(m)}(\nu)\right) \geq 1 - \delta_2}}}{n} \geq I(W_{\mathrm{base}}) - \epsilon_2 \; .
\end{equation}
To see this, take $\epsilon_1 = \epsilon_2/2$, $\delta_1 = \delta_2/2$, and let $m$ be the guaranteed constant such that (\ref{eq:IpolarizationEpsilonDelta}) holds. Now, we can take $\nu$ big enough so that, in the context of (\ref{eq:IpolarizationEpsilonDelta_DegradedComplicated}), we have that both
\[
\delta_1+\sqrt{\frac{m}{\nu}} < \delta_2
\]
and
\[
\epsilon_1 + \sqrt{\frac{m}{\nu}} < \epsilon_2 \; .
\]
By \cite[Equation (2)]{Arikan:09p} we have that
\[
Z\left(\mathcal{Q}_{i}^{(m)}(\nu)\right) \leq \sqrt{1-I^2\left(\mathcal{Q}_{i}^{(m)}(\nu)\right)} \; .
\]
Thus, if (\ref{eq:eq:IpolarizationEpsilonDelta_DegradedSimple}) holds then
\[
\frac{\size{\myset{ i : Z\left(\mathcal{Q}_{i}^{(m)}(\nu)\right) \leq \sqrt{2\delta_2-\delta_2^2}}}}{n} \geq I(W_{\mathrm{base}}) - \epsilon_2 \; .
\]
So, as before, we deduce that for every $\delta_3 > 0$ and $\epsilon_3 > 0$ there exist $m$ and $\mu$ such that
\[
\frac{\size{\myset{ i : Z\left(\mathcal{Q}_{i}^{(m)}(\nu)\right) \leq \delta_3}}}{n} \geq I(W_{\mathrm{base}}) - \epsilon_3 \; .
\]
\end{proof}
The next lemma will be used later to bound the evolution of the variable $\ZAlg$ in Algorithm~\ref{alg:specialized}.
\begin{lemm}
\label{lemm:zetaProcess}
For every $m \geq 0$ and index $0 \leq i < 2^m$ let there be a corresponding real $0 \leq \zeta(i,m) \leq 1$. Denote the binary representation of $i$ by $i=\binaryRep{b_1,b_2,\ldots,b_m}$. Assume that the $\zeta(i,m)$ satisfy the following recursive relation. For $m > 0$ and $i' = \binaryRep{b_1,b_2,\ldots,b_{m-1}}$,
\begin{multline}
\label{eq:zetaCases}
\zeta(i,m) \leq \\
\begin{cases}
2\zeta(i',m-1)-\zeta^2(i',m-1) & \mbox{if $b_m = 0$} \; ,\\
\zeta^2(i',m-1) & \mbox{otherwise} \; .
\end{cases}
\end{multline}
Then, for every $\beta < 1/2$ we have that
\begin{equation}
\label{eq:zetaLimit}
\liminf_{m \to \infty} \frac{\size{\myset{i : \zeta(i,m) < 2^{-n^\beta} }}}{n} \geq 1 - \zeta(0,0) \; ,
\end{equation}
where $n=2^m$.
\end{lemm}
\begin{proof}
First, note that both $f_1(\zeta) = \zeta^2$ and $f_2(\zeta) = 2\zeta - \zeta^2$ strictly increase from $0$ to $1$ when $\zeta$ ranges from $0$ to $1$. Thus, it suffices to prove the claim for the worst case in which the inequality in (\ref{eq:zetaCases}) is replaced by an equality. Assume from now on that this is indeed the case.
Consider an underlying BEC with probability of erasure (as well as Bhattacharyya parameter) $\zeta(0,0)$. Next, note that the $i$th bit channel, for $0 \leq i < n = 2^m$, is also a BEC, with probability of erasure $\zeta(i,m)$. Since the capacity of the underlying BEC is $1-\zeta(0,0)$, we deduce (\ref{eq:zetaLimit}) by \cite[Theorem~2]{ArikanTelatar:09c}.
\end{proof}
We are now in a position to prove Theorem~\ref{theo:polyConstruction}.
\begin{proof}[Proof of Theorem~\ref{theo:polyConstruction}]
Let us first specify explicitly the code construction algorithm used, and then analyze it. As expected, we simply run Algorithm~\ref{alg:specialized} with parameters $W_{\mathrm{base}}$ and $n$ to produce upper bounds on the probability of error of all $n$ bit channels. Then, we sort the upper bounds in ascending order. Finally, we produce a generator matrix $G$, with $k$ rows. The rows of $G$ correspond to the first $k$ bit channels according to the sorted order, and $k$ is the largest integer such that the sum of upper bounds is strictly less than $2^{n^{-\beta}}$. By Theorem~\ref{theo:totalRunningTime}, the total running time is indeed $O(n \cdot \mu^2 \log \mu)$.
Recall our definition of $\mathcal{W}_i^{(m)}$ and $\mathcal{Q}_{i}^{(m)}$ from Theorem~\ref{theo:CSBound}. Denote the upper bound on the probability of error returned by Algorithm~\ref{alg:specialized} for bit channel $i$ by $\overline{P_e}(\mathcal{W}_i^{(m)}, \mu)$. The theorem will follow easily once we prove that for all $\epsilon>0$ and $0 < \beta < 1/2$ there exists an even $\mu_0$ such that for all $\mu = 2\nu \geq \mu_0$ we have
\begin{equation}
\label{eq:epsilonFromCapacity}
\liminf_{m \to \infty} \frac{\size{\myset{i : \overline{P_e}(\mathcal{W}_i^{(m)}, \mu) < 2^{-n^\beta} }}}{n} \geq I(W_{\mathrm{base}}) - \epsilon \; .
\end{equation}
By Lemma~\ref{lemm:ZdeltaEpsilon}, there exist constants $m_0$ and $\nu$ such that
\begin{equation}
\label{eq:ZkappaKappa}
\frac{\size{\myset{ i_0 : Z\left(\mathcal{Q}_{i_0}^{(m_0)}(\nu)\right) \leq \frac{\epsilon}{2}}}}{n_0} \geq I(W_{\mathrm{base}}) - \frac{\epsilon}{2} \; ,
\end{equation}
where
\[
n_0 = 2^{m_0} \quad \mbox{and} \quad 0 \leq i_0 < n_0 \; .
\]
Denote the codeword length as $n = 2^m$, where $m = m_0 + m_1$ and $m_1 > 0$. Consider an index $0 \leq i < n$ having binary representation
\[
i = \binaryRep{b_1,b_2,\ldots,b_{m_0},b_{m_0+1},\ldots,b_m} \; ,
\]
where $b_1$ is the most significant bit. We split the run of Algorithm~\ref{alg:specialized} on $i$ into two stages. The first stage will have $j$ going from $1$ to $m_0$, while the second stage will have $j$ going from $m_0+1$ to $m$.
We start by considering the end of the first stage. Namely, we are at iteration $j=m_0$ and line~\ref{alg:specialized:degrade} has just finished executing. Recall that we denote the value of the variable $\mathcal{Q}$ after the line has executed by $\mathcal{Q}_{i_0}^{(m_0)}(\nu)$, where
\[
i_0 = \binaryRep{b_1,b_2,\ldots,b_{m_0}} \; .
\]
Similarly, define $\ZAlg_{i_0}^{(m_0)}(\nu)$ as the value of the variable $\ZAlg$ at that point.
Since, by (\ref{eq:BhattacharyyaProbDegrading}), degrading increases the Bhattacharyya parameter, we have then that the Bhattacharyya parameter of the variable $\mathcal{W}$ is less than or equal to that of the variable $\mathcal{Q}$. So, by the minimization carried out in either line \ref{alg:specialized:minBad} or \ref{alg:specialized:minGood}, we conclude the following:
at the end of line~\ref{alg:specialized:degrade} of the algorithm, when $j = m_0$,
\[
\ZAlg = \ZAlg_{i_0}^{(m_0)}(\nu) \leq Z\left(\mathcal{Q}_{i_0}^{(m_0)}(\nu)\right) = Z(\mathcal{Q}) \; .
\]
We can combine this observation with (\ref{eq:ZkappaKappa}) to conclude that
\begin{equation}
\label{eq:ZAlgkappaKappa}
\frac{\size{\myset{ i_0 : \ZAlg_{i_0}^{(m_0)}(\nu) \leq \frac{\epsilon}{2}}}}{n_0} \geq I(W_{\mathrm{base}}) - \frac{\epsilon}{2} \; .
\end{equation}
We now move on to consider the second stage of the algorithm. Fix an index $i = \binaryRep{b_1,b_2,\ldots,b_{m_0},b_{m_0+1},\ldots,b_m}$. That is, let $i$ have $i_0$ as a binary prefix of length $m_0$. Denote by $\ZAlg[t]$ the value of $\ZAlg$ at the end of line~\ref{alg:specialized:degrade}, when $j=m_0+t$. By lines \ref{alg:specialized:minBad} and \ref{alg:specialized:minGood} of the algorithm we have, similarly to (\ref{eq:zetaCases}), that
\[
\ZAlg[t+1] \leq
\begin{cases}
2\ZAlg[t]-\ZAlg^2[t] & \mbox{if $b_{m_0+t+1} = 0$} \; ,\\
\ZAlg^2[t] & \mbox{otherwise} \; .
\end{cases}
\]
We now combine our observations about the two stages. Let $\gamma$ be a constant such that
\[
\beta < \gamma < \frac{1}{2} \; .
\]
Considering (\ref{eq:ZAlgkappaKappa}), we see that out of the $n_0=2^{m_0}$ possible prefixes of length $m_0$, the fraction for which
\begin{equation}
\label{eq:Z0good}
\ZAlg[0] \leq \frac{\epsilon}{2}
\end{equation}
is at least $I(W_{\mathrm{base}}) - \frac{\epsilon}{2}$. Next, by Lemma~\ref{lemm:zetaProcess}, we see that for each such prefix, the fraction of suffixes for which
\begin{equation}
\label{eq:Zm1}
\ZAlg[m_1] \leq 2^{-(n_1)^\gamma}
\end{equation}
is at least $1-\ZAlg[0]$, as $n_1 = 2^{m_1}$ tends to infinity.
Thus, for each such prefix, we get by (\ref{eq:Z0good}) that (in the limit) the fraction of such suffixes is at least $1 - \frac{\epsilon}{2}$. We can now put all our bounds together and claim that as $m_1$ tends to infinity, the fraction of indices $0 \leq i < 2^{m}$ for which (\ref{eq:Zm1}) holds is at least
\[
\left(I(W_{\mathrm{base}}) - \frac{\epsilon}{2}\right) \cdot \left( 1 - \frac{\epsilon}{2} \right) \geq I(W_{\mathrm{base}}) - \epsilon \; .
\]
By line (\ref{alg:specialized:return}) of Algorithm~\ref{alg:specialized}, we see that $\ZAlg[m_1]$ is an upper bound on the return value $\overline{P_e}(\mathcal{W}_i^{(m)})$. Thus, we conclude that
\[
\liminf_{m \to \infty} \frac{\size{\myset{i : \overline{P_e}(\mathcal{W}_i^{(m)}, \mu) < 2^{-(n_1)^\gamma} }}}{n} = I(W_{\mathrm{base}}) - \epsilon
\]
With the above at hand, the only thing left to do in order to prove (\ref{eq:epsilonFromCapacity}) is to show that for $m_1$ large enough we have that
\[
2^{-(n_1)^\gamma} \leq 2^{-n^\beta} \; ,
\]
which reduces to showing that
\[
(n_1)^{\gamma-\beta} \geq (n_0)^\beta \; .
\]
Since $\gamma > \beta$ and $n_0 = 2^{m_0}$ is constant, this is indeed the case.
\end{proof}
We end this section by pointing out a similarity between the analysis used here and the analysis carried out in \cite{HMTU:11a}. In both papers, there are two stages. The first stage (prefix of length $m_0$ in our paper) makes full use of the conditional probability distribution of the channel, while the second stage uses a simpler rule (evolving the bound on the Bhattacharyya parameter in our paper and using an RM rule in \cite{HMTU:11a}).
\appendices
\section{Proof of Theorem~\ref{theo:degradeOptimal}}
\label{sec:proofDegradingOptimal}
This appendix is devoted to the proof of Theorem~\ref{theo:degradeOptimal}. Although the initial lemmas needed for the proof are rather intuitive, the latter seem to be a lucky coincidence (probably due to a lack of a deeper understanding on the authors' part). The prime example seems to be Equation~(\ref{eq:deltaPrimeSumBingo}) in the proof of Lemma~\ref{lemm:bothDeltaPrimesSmaller}.
We start by defining some notation. Let $\Wbit:\mathcal{X} \to \mathcal{Y}$, $\nu$, $y_1,y_2,\ldots,y_\nu$ and $\bar{y}_1,\bar{y}_2,\bar{y}_\nu$ be as in Theorem~\ref{theo:degradeOptimal}.
Let $w \in \mathcal{Y}$ and $\bar{\w} \in \mathcal{Y}$ be a symbol pair, and denote by $(a,b)$ the corresponding probability pair, where
\[
a = p(w | 0 ) = p(\bar{\w} | 1) \; , \quad b = p(w | 1 ) = p(\bar{\w} | 0) \; .
\]
The contribution of this probability pair to the capacity of $\Wbit$ is denoted by
\begin{multline*}
C(a,b) = -(a+b) \log_2((a+b)/2)+a \log_2(a) +b \log_2(b) = \\
-(a+b) \log_2(a+b)+a \log_2(a) +b \log_2(b) + (a+b) \; ,
\end{multline*}
where $0 \log_2 0 = 0$.
Next, suppose we are given two probability pairs: $(a_1,b_1)$ and $(a_2,b_2)$ corresponding to the symbol pair $w_1,\bar{\w}_1$ and $w_2,\bar{\w}_2$, respectively. The capacity difference resulting from the application of Lemma~\ref{lemm:degradeStep} to $w_1$ and $w_2$ is denoted by
\[
\Delta(a_1,b_1;a_2,b_2) = C(a_1,b_1) + C(a_2,b_2) - C(a_1+a_2,b_1+b_2) \; .
\]
For reasons that will become apparent later on, we henceforth relax the definition of a probability pair to two non-negative numbers, \emph{the sum of which may be greater than $1$}. Note that $C(a,b)$ is still well defined with respect to this generalization, as is $\Delta(a_1,b_1;a_2,b_2)$. Furthermore, to exclude trivial cases, we require that a probability pair $(a,b)$ has at least one positive element.
The following lemma states that we lose capacity by performing a downgrading merge.
\begin{lemm}
\label{lemm:DeltaNonNegative}
Let $(a_1,b_1)$ and $(a_2,b_2)$ be two probability pairs. Then,
\[
\Delta(a_1,b_1;a_2,b_2) \geq 0 \;
\]
\end{lemm}
\begin{proof}
Assume first that $a_1,b_1,a_2,b_2$ are all positive. In this case, $\Delta(a_1,b_1;a_2,b_2)$ can be written as follows:
\begin{eqnarray*}
(a_1+a_2)\Bigg( &\frac{-a_1}{a_1+a_2} \log_2 \frac{(a_1+b_1)(a_1+a_2)}{a_1(a_1+b_1+a_2+b_2)} + \\
& \frac{-a_2}{a_1+a_2} \log_2 \frac{(a_2+b_2)(a_1+a_2)}{a_2(a_1+b_1+a_2+b_2)} \Bigg) + \\
(b_1+b_2)\Bigg( &\frac{-b_1}{b_1+b_2} \log_2 \frac{(a_1+b_1)(b_1+b_2)}{b_1(a_1+b_1+a_2+b_2)} + \\
& \frac{-b_2}{b_1+b_2} \log_2 \frac{(a_2+b_2)(b_1+b_2)}{b_2(a_1+b_1+a_2+b_2)} \Bigg)
\end{eqnarray*}
By Jensen's inequality, both the first two lines and the last two lines can be lower bounded be $0$. The proof for cases in which some of the variables equal zero is much the same.
\end{proof}
The intuition behind the following lemma is that the order of merging does matter in terms of total capacity lost.
\begin{lemm}
\label{lemm:DeltaDistributive}
Let $(a_1,b_1)$, $(a_2,b_2)$, and $(a_3,b_3)$ be three probability pairs. Then,
\begin{multline*}
\Delta(a_1,b_1;a_2,b_2) + \Delta(a_1+a_2,b_1+b_2;a_3,b_3) = \\
\Delta(a_2,b_2;a_3,b_3) + \Delta(a_1,b_1;a_2+a_3,b_2+b_3)\; .
\end{multline*}
\end{lemm}
\begin{proof}
Both sides of the equation equal
\[
C(a_1,b_1) + C(a_2,b_2) + C(a_3,b_3) - C(a_1+a_2+a_3,b_1+b_2+b_3) \; .
\]
\end{proof}
Instead of working with a probability pair $(a,b)$, we find it easier to work with a probability sum $\pi = a + b$ and likelihood ratio $\lambda = a/b$. Of course, we can go back to our previous representation as follows. If $\lambda = \infty$ then $a = \pi$ and $b=0$. Otherwise, $a = \frac{\lambda \cdot \pi}{\lambda+1}$ and $b = \frac{\pi}{\lambda+1}$. Recall that our relaxation of the term ``probability pair'' implies that $\pi$ is positive and it may be greater than $1$.
Abusing notation, we define the quantity $C$ through $\lambda$ and $\pi$ as well. For $\lambda = \infty$ we have $C[\infty,\pi] = \pi$. Otherwise,
\begin{multline*}
C[\lambda,\pi] =
\\
\pi \left(
- \frac{\lambda}{\lambda+1} \log_2 \left( 1+\frac{1}{\lambda} \right) - \frac{1}{\lambda+1} \log_2(1+ \lambda)
\right) + \pi \; .
\end{multline*}
Let us next consider merging operations. The merging of the symbol pair corresponding to $[\lambda_1,\pi_1]$ with that of $[\lambda_2,\pi_2]$ gives a symbol pair with $[\lambda_{1,2},\pi_{1,2}]$, where
\[
\pi_{1,2} = \pi_1 + \pi_2
\]
and
\begin{multline}
\label{eq:lambdaMean}
\lambda_{1,2} = \bar{\lambda}[\pi_1,\lambda_1;\pi_2, \lambda_2] = \\
\frac{ \lambda_1 \pi_1(\lambda_2+1)+ \lambda_2 \pi_2(\lambda_1+1)}{\pi_1(\lambda_2+1) + \pi_2(\lambda_1+1)}
\end{multline}
Abusing notation, define
\[
\Delta[\lambda_1,\pi_1;\lambda_2,\pi_2] = C[\lambda_1,\pi_1] + C[\lambda_2,\pi_2] - C[\lambda_{1,2},\pi_{1,2}] \; .
\]
Clearly, we have that the new definition of $\Delta$ is symmetric:
\begin{equation}
\label{eq:DeltaLRSymmetric}
\Delta[\lambda_1,\pi_1;\lambda_2,\pi_2] = \Delta[\lambda_2,\pi_2;\lambda_1,\pi_1] \; .
\end{equation}
\begin{lemm}
\label{lemm:DeltaMonotnicIncreasingInPi}
$\Delta[\lambda_1,\pi_1;\lambda_2,\pi_2]$ is monotonic increasing in both $\pi_1$ and $\pi_2$.
\end{lemm}
\begin{proof}
Recall from (\ref{eq:DeltaLRSymmetric}) that $\Delta$ is symmetric, and so it suffices to prove the claim for $\pi_1$. Thus, our goal is to prove the following for all $\rho > 0$,
\[
\Delta[\lambda_1,\pi_1 + \rho;\lambda_2,\pi_2] \geq \Delta[\lambda_1,\pi_1; \lambda_2,\pi_2] \; .
\]
At this point, we find it useful to convert back from the likelihood ratio/probability sum representation $[\lambda,\pi]$ to the probability pair representation $(a,b)$. Denote by $(a_1,b_1)$, $(a_2,b_2)$, and $(a',b')$ the probability pairs corresponding to $[\lambda_1,\pi_1]$, $[\lambda_2,\pi_2]$, and $[\lambda_1, \pi_1 + \rho]$, respectively. Let $a_3 = a' - a_1$ and $b_3 = b' - b_1$. Next, since both $(a_1,b_1)$ and $(a',b')$ have the same likelihood ratio, we deduce that both $a_3$ and $b_3$ are non-negative. Under our new notation, we must prove that
\[
\Delta(a_1+a_3,b_1+b_3;a_2,b_2) \geq \Delta(a_1,b_1; a_2,b_2) \; .
\]
Since both $(a_1,b_1)$ and $(a',b')$ have likelihood ratio $\lambda_1$, this is also the case for $(a_3,b_3)$. Thus, a simple calculation shows that
\[
\Delta(a_1,b_1;a_3,b_3) = 0 \; .
\]
Hence,
\begin{multline*}
\Delta(a_1+a_3,b_1+b_3;a_2,b_2) = \\
\Delta(a_1+a_3,b_1+b_3;a_2,b_2) + \Delta(a_1,b_1;a_3,b_3)
\end{multline*}
Next, by Lemma~\ref{lemm:DeltaDistributive},
\begin{multline*}
\Delta(a_1+a_3,b_1+b_3;a_2,b_2) + \Delta(a_1,b_1;a_3,b_3) = \\
\Delta(a_1,b_1;a_2,b_2) + \Delta(a_1+a_2,b_1+b_2;a_3,b_3) \; .
\end{multline*}
Since, by Lemma~\ref{lemm:DeltaNonNegative}, we have that $\Delta(a_1+a_2,b_1+b_2;a_3,b_3)$ is non-negative, we are done.
\end{proof}
We are now at the point in which our relaxation of the term ``probability pair'' can be put to good use. Namely, we will now see how to reduce the number of variables involved by one, by taking a certain probability sum to infinity.
\begin{lemm}
\label{lemm:DeltaLimit}
Let $\lambda_1$, $\pi_1$, and $\lambda_2$ be given. Assume that $0 < \lambda_2 < \infty$. Define
\[
\Delta[\lambda_1,\pi_1;\lambda_2,\infty] = \lim_{\pi_2 \to \infty} \Delta[\lambda_1,\pi_1;\lambda_2,\pi_2] \; .
\]
If $0 < \lambda_1 < \infty$, then
\begin{multline}
\label{eq:DeltaLimit}
\Delta[\lambda_1,\pi_1;\lambda_2,\infty] = \\
\pi_1 \left(
-\frac{\lambda_1}{\lambda_1+1} \log_2\left( \frac{1+\frac{1}{\lambda_1}}{1+\frac{1}{\lambda_2}} \right) - \frac{1}{\lambda_1+1}\log_2\left( \frac{\lambda_1+1}{\lambda_2+1}\right)
\right) \; .
\end{multline}
If $\lambda_1 = \infty$, then
\begin{equation}
\label{eq:DeltaLimitLambda1Infinity}
\Delta[\lambda_1,\pi_1;\lambda_2,\infty] = \pi_1 \left(
-\log_2\left( \frac{1}{1+\frac{1}{\lambda_2}} \right) \right) \; .
\end{equation}
If $\lambda_1 = 0$, then
\begin{equation}
\label{eq:DeltaLimitLambda1Zero}
\Delta[\lambda_1,\pi_1;\lambda_2,\infty] = \pi_1 \left(
- \log_2\left( \frac{1}{\lambda_2+1}\right)
\right) \; .
\end{equation}
\end{lemm}
\begin{proof}
Consider first the case $0 < \lambda_1 < \infty$. We write out $\Delta[\lambda_1,\pi_1;\lambda_2,\pi_2]$ in full and after rearrangement get
\begin{equation}
\label{eq:DeltaLimBreakupGeneralCase}
\begin{split}
&\pi_1 \left(
\frac{1}{\frac{1}{\lambda_{1,2}}+1} \log_2 \left( 1+ \frac{1}{\lambda_{1,2}} \right)
-\frac{1}{\frac{1}{\lambda_1}+1} \log_2 \left( 1+ \frac{1}{\lambda_1} \right)
\right) + \\
&\pi_1 \left(
\frac{1}{\lambda_{1,2}+1} \log_2 \left( 1+ \lambda_{1,2} \right)
-\frac{1}{\lambda_1+1} \log_2 \left( 1+ \lambda_1 \right)
\right) + \\
&\pi_2 \left(
\frac{1}{\frac{1}{\lambda_{1,2}}+1} \log_2 \left( 1+ \frac{1}{\lambda_{1,2}} \right)
-\frac{1}{\frac{1}{\lambda_2}+1} \log_2 \left( 1+ \frac{1}{\lambda_2} \right)
\right) + \\
&\pi_2 \left(
\frac{1}{\lambda_{1,2}+1} \log_2 \left( 1+ \lambda_{1,2} \right)
-\frac{1}{\lambda_2+1} \log_2 \left( 1+ \lambda_2 \right)
\right) \; ,
\end{split}
\end{equation}
where $\lambda_{1,2}$ is given in (\ref{eq:lambdaMean}). Next, note that
\[
\lim_{\pi_2 \to \infty} \lambda_{1,2} = \lambda_2 \; .
\]
Thus, applying $\lim_{\pi_2 \to \infty}$ to the first two lines of (\ref{eq:DeltaLimBreakupGeneralCase}) is straightforward. Next, consider the third line of (\ref{eq:DeltaLimBreakupGeneralCase}), and write its limit as
\[
\lim_{\pi_2 \to \infty} \frac{
\frac{1}{\frac{1}{\lambda_{1,2}}+1} \log_2 \left( 1+ \frac{1}{\lambda_{1,2}} \right)
-\frac{1}{\frac{1}{\lambda_2}+1} \log_2 \left( 1+ \frac{1}{\lambda_2} \right)}
{\frac{1}{\pi_2}}
\]
Since $\lim_{\pi_2 \to \infty} \lambda_{1,2} = \lambda_2$, we get that both numerator and denominator tend to $0$ as $\pi_2 \to \infty$. Thus, we apply l'H\^{o}pital's rule and get
\begin{multline*}
\lim_{\pi_2 \to \infty}
\frac{\frac{1}{(\lambda_{1,2}+1)^2} \left(\log_2 e - \log_2 \left( 1+ \frac{1}{\lambda_{1,2}} \right)\right)\frac{\partial \lambda_{1,2}}{\partial \pi_2}}
{\frac{1}{(\pi_2)^2}} =
\\
\lim_{\pi_2 \to \infty}
\frac{1}{(\lambda_{1,2}+1)^2} \left(\log_2 e - \log_2 \left( 1+ \frac{1}{\lambda_{1,2}} \right)\right) \cdot \\
\frac{\pi_1(\lambda_2+1)(\lambda_1+1)(\lambda_2 - \lambda_1)}{(\frac{\pi_1(\lambda_1+1) + \pi_2(\lambda_1+1)}{\pi_2})^2} =
\\
\frac{\pi_1(\lambda_2-\lambda_1)}{(\lambda_1+1)(\lambda_2+1)} \left(\log_2 e - \log_2 \left( 1+ \frac{1}{\lambda_2} \right)\right) \; ,
\end{multline*}
where $e=2.71828\ldots$ is Euler's number. Similarly, taking the $\lim_{\pi_2 \to \infty}$ of the fourth line of (\ref{eq:DeltaLimBreakupGeneralCase}) gives
\[
\frac{\pi_1(\lambda_2-\lambda_1)}{(\lambda_1+1)(\lambda_2+1)} \left(-\log_2 e - \log_2 \left( 1+ \lambda_2 \right)\right) \; .
\]
Thus, a short calculations finishes the proof for this case. The cases $\lambda_1 = \infty$ and $\lambda_1 = 0$ are handled much the same way.
\end{proof}
The utility of the next Lemma is that it asserts a stronger claim than the ``Moreover'' part of Theorem~\ref{theo:degradeOptimal}, for a specific value of $\lambda_2$.
\begin{lemm}
\label{lemm:bothDeltaPrimesSmaller}
Let probability pairs $(a_1,b_1)$ and $(a_3,b_3)$ have likelihood ratios $\lambda_1$ and $\lambda_3$, respectively. Assume $\lambda_1 \leq \lambda_3$. Denote $\pi_1 = a_1 + b_1$ and $\pi_3 = a_3 + b_3$. Let
\begin{equation}
\label{eq:lambda2_lemm:bothDeltaPrimesSmaller}
\lambda_2 = \lambda_{1,3} = \bar{\lambda}[\pi_1,\lambda_1;\pi_3, \lambda_3]\; ,
\end{equation}
as defined in (\ref{eq:lambdaMean}). Then,
\[
\Delta[\lambda_1,\pi_1;\lambda_2,\infty] \leq \Delta[\lambda_1,\pi_1;\lambda_3,\pi_3]
\]
and
\[
\Delta[\lambda_3,\pi_3;\lambda_2,\infty] \leq \Delta[\lambda_1,\pi_1;\lambda_3,\pi_3]
\]
\end{lemm}
\begin{proof}
We start by taking care of a trivial case. Note that if it is not the case that $0 < \lambda_2 < \infty$, then $\lambda_1 = \lambda_2 = \lambda_3$, and the proof follows easily.
So, we henceforth assume that $0 < \lambda_2 < \infty$, as was done in Lemma~\ref{lemm:DeltaLimit}. Let
\[
\Delta_{(1,3)} = \Delta[\lambda_1,\pi_1;\lambda_3,\pi_3] \; ,
\]
\[
\Delta'_{(1,2)} = \Delta[\lambda_1,\pi_1;\lambda_2,\infty] \; ,
\]
and
\[
\Delta'_{(2,3)} = \Delta[\lambda_3,\pi_3;\lambda_2,\infty] \; .
\]
Thus, we must prove that $\Delta'_{(1,2)} \leq \Delta_{(1,3)}$ and $\Delta'_{(2,3)} \leq \Delta_{(1,3)}$. Luckily, Lemma~\ref{lemm:DeltaLimit} and a bit of calculation yields that
\begin{equation}
\label{eq:deltaPrimeSumBingo}
\Delta'_{(1,2)} + \Delta'_{(2,3)} = \Delta_{(1,3)} \; .
\end{equation}
Recall that $\Delta'_{(1,2)}$ and $\Delta'_{(2,3)}$ must be non-negative by Lemmas~\ref{lemm:DeltaNonNegative} and \ref{lemm:DeltaMonotnicIncreasingInPi}. Thus, we are done.
\end{proof}
The next lemma shows how to discard the restraint put on $\lambda_2$ in Lemma~\ref{lemm:bothDeltaPrimesSmaller}.
\begin{lemm}
\label{lemm:oneDeltaPrimeSmaller}
Let the likelihood ratios $\lambda_1$, $\lambda_3$ and the probability sums $\pi_1$, $\pi_3$ be as in be as in Lemma~\ref{lemm:bothDeltaPrimesSmaller}. Fix
\begin{equation}
\label{eq:lambda2Range_lemm:bothDeltaPrimesSmaller}
\lambda_1 \leq \lambda_2 \leq \lambda_3 \; .
\end{equation}
Then either
\begin{equation}
\label{eq:leftMergeBetter_lemm:bothDeltaPrimesSmaller}
\Delta[\lambda_1,\pi_1;\lambda_2,\infty] \leq \Delta[\lambda_1,\pi_1;\lambda_3,\pi_3]
\end{equation}
or
\begin{equation}
\label{eq:rightMergeBetter_lemm:bothDeltaPrimesSmaller}
\Delta[\lambda_3,\pi_3;\lambda_2,\infty] \leq \Delta[\lambda_1,\pi_1;\lambda_3,\pi_3]
\end{equation}
\end{lemm}
\begin{proof}
Let $\lambda_{1,3}$ be as in (\ref{eq:lambda2_lemm:bothDeltaPrimesSmaller}), and note that
\[
\lambda_1 \leq \lambda_{1,3} \leq \lambda_3 \; .
\]
Assume w.l.o.g.\ that $\lambda_2$ is such that
\[
\lambda_1 \leq \lambda_2 \leq \lambda_{1,3} \; .
\]
From Lemma~\ref{lemm:bothDeltaPrimesSmaller} we have that
\[
\Delta[\lambda_1,\pi_1;\lambda_{1,3},\infty] \leq \Delta[\lambda_1,\pi_1;\lambda_3,\pi_3]
\]
Thus, we may assume that $\lambda_2 < \lambda_{1,3}$ and aim to prove that
\begin{equation}
\label{eq:lambda2lambda13SubGoal}
\Delta[\lambda_1,\pi_1;\lambda_2,\infty] \leq \Delta[\lambda_1,\pi_1;\lambda_{1,3},\infty] \; .
\end{equation}
Next, notice that
\begin{equation}
\label{eq:DeltaZeroLambda1Lambda2Equal}
\Delta[\lambda_1,\pi_1;\lambda_2,\infty] = 0 \; , \quad \mbox{if $\lambda_2 = \lambda_1$} \; .
\end{equation}
Thus, let us assume that
\[
\lambda_1 < \lambda_2 < \lambda_{1,3} \; .
\]
Specifically, it follows that
\[
0 < \lambda_2 < \infty
\]
and thus the assumption in Lemma~\ref{lemm:DeltaLimit} holds.
Define the function $f$ as follows
\[
f(\lambda_2') = \Delta[\lambda_1,\pi_1;\lambda_2',\infty] \; .
\]
Assume first that $\lambda_1=0$, and thus by (\ref{eq:DeltaLimitLambda1Zero}) we have that
\[
\frac{\partial f(\lambda_2')}{\partial \lambda_2'} \geq 0 \; .
\]
On the other hand, if $\lambda_1 \neq 0$ we must have that $0 \leq \lambda_1 < \infty$. Thus, by (\ref{eq:DeltaLimit})
we have that
\[
\frac{\partial f(\lambda_2')}{\partial \lambda_2'} = \frac{\pi_1}{(\lambda_1 + 1)(\lambda_2'+1)}\left( 1 - \frac{\lambda_1}{\lambda_2'} \right) \; ,
\]
which is also non-negative for $\lambda_2' \geq \lambda_1$. Thus, we have proved that the derivative is non-negative in both cases, and this together with (\ref{eq:DeltaZeroLambda1Lambda2Equal}) proves (\ref{eq:lambda2lambda13SubGoal}).
\end{proof}
We are now in a position to prove Theorem~\ref{theo:degradeOptimal}.
\begin{proof}[Proof of Theorem \ref{theo:degradeOptimal}]
We first consider the ``Moreover'' part of the theorem. Let $[\lambda_1,\pi_1]$, $[\lambda_2,\pi_2]$, and $[\lambda_1,\pi_1]$ correspond to $y_i$, $y_j$, and $y_k$, respectively. From Lemma~\ref{lemm:oneDeltaPrimeSmaller} we have that either (\ref{eq:leftMergeBetter_lemm:bothDeltaPrimesSmaller}) or (\ref{eq:rightMergeBetter_lemm:bothDeltaPrimesSmaller}) holds. Assume w.l.o.g.\ that (\ref{eq:leftMergeBetter_lemm:bothDeltaPrimesSmaller}) holds. By Lemma~\ref{lemm:DeltaMonotnicIncreasingInPi} we have that
\[
\Delta[\lambda_1,\pi_1; \lambda_2,\pi_2] \leq \Delta[\lambda_1,\pi_1;\lambda_2,\infty] \; .
\]
Thus,
\[
\Delta[\lambda_1,\pi_1; \lambda_2,\pi_2] \leq \Delta[\lambda_1,\pi_1;\lambda_3,\pi_3] \; ,
\]
which is equivalent to
\[
I(y_j,y_k) \geq I(y_i,y_k) \; .
\]
Having finished the ``Moreover'' part, we now turn our attention to the proof of (\ref{eq:theo:degradeOptimal_LRgeq1}). The two equalities in (\ref{eq:theo:degradeOptimal_LRgeq1}) are straightforward, so we are left with proving the inequality. For $\lambda \geq 0$ and $\pi > 0$, the following are easily verified:
\begin{equation}
\label{eq:ClambdaOneOverLambda}
C[\lambda,\pi] = C[1/\lambda,\pi] \; ,
\end{equation}
and
\begin{equation}
\label{eq:CincreasesWithLambda}
\mbox{$C[\lambda,\pi]$ increases with $\lambda \geq 1$} \; .
\end{equation}
Also, for $\bar{\lambda}$ as given in (\ref{eq:lambdaMean}), $\lambda_1,\lambda_2 \geq 0$, and $\pi_1 > 0, \pi_2 > 0$, it is easy to show that
\begin{equation}
\label{eq:lambdaBarIncreasesWithLambdas}
\mbox{$\bar{\lambda}[\lambda_1,\pi_1, \lambda_2,\pi_2]$ increases with both $\lambda_1$ and $\lambda_2$} \; .
\end{equation}
Let $[\lambda_1,\pi_1]$ and $[\lambda_2,\pi_2]$ correspond to $y_i$ and $y_j$, respectively. Denote
\[
\gamma = \bar{\lambda}[\lambda_1,\pi_1;\lambda_2,\pi_2]
\]
and
\[
\delta = \bar{\lambda}[1/\lambda_1,\pi_1;\lambda_2,\pi_2]
\]
Hence, our task reduces to showing that
\begin{equation}
\label{eq:LR1geq1Reduction}
C[\gamma,\pi_1+\pi_2] \geq C[\delta,\pi_1+\pi_2] \; .
\end{equation}
Assume first that $\delta \geq 1$. Recall that both $\lambda_1 \geq 1$ and $\lambda_2 \geq 1$. Thus, by (\ref{eq:lambdaBarIncreasesWithLambdas}) we conclude that $\gamma \geq \delta \geq 1$. This, together with (\ref{eq:CincreasesWithLambda}) finishes the proof.
Conversely, assume that $\delta \leq 1$. Since
\[
\bar{\lambda}[\lambda_1,\pi_1;\lambda_2,\pi_2] = \bar{\lambda}[1/\lambda_1,\pi_1;1/\lambda_2,\pi_2]
\]
we now get from (\ref{eq:lambdaBarIncreasesWithLambdas}) that $\gamma \leq \delta \leq 1$. This, together with (\ref{eq:ClambdaOneOverLambda}) and (\ref{eq:CincreasesWithLambda}) finishes the proof.
\end{proof}
\section{Proof of Theorem~\ref{theo:upgradeOptimal}}
\label{sec:proofUpgradingOptimal}
As a preliminary step toward the proof of Theorem~\ref{theo:upgradeOptimal}, we convince ourselves that the notation $\Delta[\lambda_1;\lambda_2,\pi_2;\lambda_3]$ used in the theorem is indeed valid. Specifically, the next lemma shows that knowledge of the arguments of $\Delta$ indeed suffices to calculate the difference in capacity. The proof is straightforward.
\begin{lemm}
For $i=1,2,3$, let $y_i$ and $\lambda_i$, as well as $\pi_2$ be as in Theorem~\ref{theo:upgradeOptimal}. If $\lambda_3 < \infty$, then
\begin{multline}
\Delta[\lambda_1;\lambda_2,\pi_2;\lambda_3] = \frac{\pi_2}{(\lambda_2+1)(\lambda_1-\lambda_3)} \Biggl[\\
\shoveright{(\lambda_3-\lambda_2)\left( \lambda_1 \log_2\left( 1 + \frac{1}{\lambda_1} \right) + \log_2(1+\lambda_1) \right) +} \\
\shoveright{(\lambda_2-\lambda_1)\left( \lambda_3 \log_2\left( 1 + \frac{1}{\lambda_3} \right) + \log_2(1+\lambda_3) \right) +} \\
(\lambda_1-\lambda_3)\left( \lambda_2 \log_2\left( 1 + \frac{1}{\lambda_2} \right) + \log_2(1+\lambda_2) \right)
\Biggr] \; .
\end{multline}
Otherwise, $\lambda_3 = \infty$ and
\begin{multline}
\Delta[\lambda_1;\lambda_2,\pi_2;\lambda_3 = \infty] = \\
\shoveright{\frac{\pi_2}{\lambda_2+1} \Biggl[
-\lambda_1 \log_2\left( 1+ \frac{1}{\lambda_1}\right) - \log_2(1+\lambda_1)} \phantom{\Biggr] \; .}\\
+\lambda_2 \log_2\left( 1+ \frac{1}{\lambda_2}\right) + \log_2(1+\lambda_2) \Biggr] \; .
\end{multline}
\end{lemm}
Having the above calculations at hand, we are in a position to prove Theorem~\ref{theo:upgradeOptimal}.
\begin{proof}[Proof of Theorem~\ref{theo:upgradeOptimal}]
First, let consider the case $\lambda_3 < \infty$. Since our claim does not involve changing the values of $\lambda_2$ and $\pi_2$, let us fix them and denote
\[
f(\lambda_1, \lambda_3) = \Delta[\lambda_1;\lambda_2,\pi_2;\lambda_3] \; .
\]
Under this notation, it suffices to prove that $f(\lambda_1, \lambda_3)$ is decreasing in $\lambda_1$ and increasing in $\lambda_3$, where $\lambda_1 < \lambda_2 < \lambda_3$. A simple calculation shows that
\begin{multline}
\frac{\partial f(\lambda_1, \lambda_3)}{\partial \lambda_1} = \frac{-\pi_2(\lambda_3-\lambda_2)}{(1+\lambda_2)(\lambda_3-\lambda_1)^2}
\Biggl[ \\
\lambda_3 \log \left( \frac{1+\frac{1}{\lambda_1}}{1+\frac{1}{\lambda_3}}\right) +
\log\left(\frac{1+\lambda_1}{1+\lambda_3} \right)
\Biggr] \; .
\end{multline}
So, in order to show that $f(\lambda_1, \lambda_3)$ is decreasing in $\lambda_1$, it suffices to show that the term inside the square brackets is positive for all $\lambda_1 < \lambda_3$. Indeed, if we denote
\[
g(\lambda_1,\lambda_3) = \lambda_3 \log \left( \frac{1+\frac{1}{\lambda_1}}{1+\frac{1}{\lambda_3}}\right) +
\log\left(\frac{1+\lambda_1}{1+\lambda_3} \right) \; ,
\]
then is readily checked that
\[
g(\lambda_1,\lambda_1) = 0 \; ,
\]
while
\[
\frac{\partial g(\lambda_1,\lambda_3)}{\partial \lambda_1} = \frac{\lambda_3-\lambda_1}{\lambda_1(\lambda_1+1)}
\]
is positive for $\lambda_3 > \lambda_1$. The proof of $f(\lambda_1, \lambda_3)$ increasing in $\lambda_3$ is exactly the same, up to a change of variable names.
Let us now consider the second case, $\lambda_3 = \infty$. Similarly to what was done before, let us fix $\lambda_2$ and $\pi_2$, and consider $\Delta[\lambda_1;\lambda_2,\pi_2;\lambda_3 = \infty]$ as a function of $\lambda_1$. Denote
\[
h(\lambda_1) = \Delta[\lambda_1;\lambda_2,\pi_2;\lambda_3 = \infty] \; .
\]
Under this notation, our aim is to prove that $h(\lambda_1)$ is decreasing in $\lambda_1$. Indeed,
\[
\frac{\partial h(\lambda_1)}{\partial \lambda_1} = \frac{-\pi_2 \log_2\left( 1 + \frac{1}{\lambda_1} \right)}{\lambda_2+1}
\]
is easily seen to be negative.
\end{proof}
\section*{Acknowledgments}
We thank
Emmanuel Abbe,
Erdal Ar\i kan,
Hamed Hassani,
Ramtin Pedarsani,
Uzi Pereg,
Eren \c{S}a\c{s}o\u{g}lu,
Artyom Sharov,
Emre Telatar,
and R\"{u}diger Urbanke
for helpful discussions.
We are especially grateful to Uzi Pereg and
Artyom~Sharov~for going over many incarnations of this paper and
offering valuable comments.
\twobibs{
\bibliographystyle{IEEEtran}
|
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| 5,144
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{"url":"https:\/\/rodneydyer.com\/post\/fixing-imports-to-hugo\/","text":"# Fixing Imports to Hugo\n\nThe solution to this, is to recognize that a lot of the content I make (in fact most of it) will be presented as output from R Markdown. The tutorials I create, the analyses I do, etc. As such, why not start with an R Markdown file in the first place and go from there? Enter blogdown\n\ndevtools::install_github(\"rstudio\/blogdown\")\n\nI then had to install python3 and pip to get the academic Hugo theme going.\n\nbrew update\nbrew install python3\npip3 install academic\n\nI then went and grabbed a BibTex file of my publications from Web Of Science. Not all of them are there but it gets things going. Then I had the hugo academic import them into the site.\n\nacademic import --bibtex ~\/Downloads\/savedrecs.bib \n\nI quickly noticed a flaw in this approach as Web of Sciences labels each of the entries with the ISI number, which academic uses for naming within Hugo. This means absolutely nothing to me (e.g., instead of DyerNason2004 that manuscript is noted as ISI:000221933000001\u2014totally useless). So, I went back to the file and manually changed the entry titles. Not a big deal.\n\nI then imported them using the code above and it was good. I got a folder\n\nNow, to go fix a few other things.\n\n## Fixing tags\n\nBy default, the import mechanism uses a semicolon separated list of tags from the keywords for each publication. Hugo assumes everything is just one long set of tag identifiers so tags like, gene flow, popgraph, and genetic structure are imported into Hugo as gene-flow-popgraph-genetic-structure, which is not the right thing. So I wrote the following quick script to loop though all the files just created make the right changes.\n\nfiles = list.files(\".\/content\/publications\/\", pattern=\"index.md\",full.names = TRUE,recursive = TRUE)\nfiles\nfor( file in files ) {\nprint(file)\n}","date":"2021-04-11 13:56:30","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\": 0.40430206060409546, \"perplexity\": 3014.136946329884}, \"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-2021-17\/segments\/1618038062492.5\/warc\/CC-MAIN-20210411115126-20210411145126-00071.warc.gz\"}"}
| null | null |
Q: Centering picture above button to different webpage I want to center a logo above a button that leads to a different page with js. I got that working but I can't get the image above the button.
https://jsfiddle.net/0bzagL6x/2/ I am not good with html and css. It is my first project.
<a class="btn" onclick="WebsiteRedirectForum()">
<img class="forum" src="img/forum.png" />
Forum
</a>
Thanks a lot!
Tom
A: Assuming that you want the images in each block to render above the "Forum" text, in this case, I would suggest usage of line breaks (<br>)
For example:
<a class="btn" onclick="WebsiteRedirectForum()">
<img class="forum" src="img/forum.png" /><br>
Forum
</a>
I hope this helps. Please comment if this does not answer your question.
A: There are many ways you could achieve this button layout using HTML and CSS.
The first thing I would do update the HTML markup and add a <span> tag around the button text - in this case, the word "Forum". This will allow us to select the text in CSS.
Heres a codepen with two solutions
The first is a simple solution using display: block and text-align: center.
The second uses flexbox and is slightly more complicated (and probably overkill).
If you're starting out I would recommend Code Academy as a good place to start learning HTML and CSS.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
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| 6,779
|
using System;
using System.Reflection;
using System.Resources;
using System.Runtime.InteropServices;
[assembly: AssemblyTitle("OvermanGroup.NuGet.Packager.Test")]
[assembly: AssemblyProduct("OvermanGroup.NuGet.Packager.Test")]
[assembly: AssemblyDescription("Unit Tests for OvermanGroup.NuGet.Packager")]
[assembly: AssemblyCompany("Overman Group, LLC.")]
[assembly: AssemblyCopyright("Copyright © 2014 Overman Group, LLC.")]
[assembly: AssemblyTrademark("")]
[assembly: AssemblyCulture("")]
[assembly: NeutralResourcesLanguage("en-us")]
#if DEBUG
[assembly: AssemblyConfiguration("Debug")]
#else
[assembly: AssemblyConfiguration("Release")]
#endif
[assembly: ComVisible(false)]
[assembly: CLSCompliant(true)]
[assembly: Guid("6ae2b50f-1473-408f-98b1-a9b9dee62342")]
[assembly: AssemblyVersion("1.1.0.0")]
[assembly: AssemblyFileVersion("1.1.0")]
[assembly: AssemblyInformationalVersion("1.1.0")]
|
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"redpajama_set_name": "RedPajamaGithub"
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| 8,191
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Scott Glynn Farlow is an Australian politician. He has been a Member of the New South Wales Legislative Council since the 2015 NSW state election, representing the Liberal Party.
Farlow previously served as Mayor of the Municipality of Strathfield, elected to the position at just 23 years of age.
Farlow is married to Penny George-Farlow who together have a son Christian and daughter Colette.
Early life
Farlow grew up in Sydney's Inner Western suburbs. He was the first member of his family to study at university, graduating with degrees in Law and Commerce from the University of Sydney. He later graduated from the College of Law with a Diploma of Legal Practice.
Farlow got his start in politics by joining the Sydney University Liberal Club, becoming a member of the Liberal Party soon after. He became active in the Young Liberal Movement of NSW and in 2008 was elected president of the organisation, going on to become its longest-serving president.
In 2004 Farlow was elected to Strathfield City Council and became mayor in 2007. At the age of 23, Farlow was the youngest mayor in Australia at that time.
Farlow worked as an advisor to a number of federal and state MPs during his career.
Farlow has also worked as an analyst and lobbyist for firms such as Deloitte and Hill+Knowlton.
Political career
Farlow was preselected by the Liberal Party to run as a candidate for the NSW Legislative Council at the March 2015 Election. He was subsequently elected and is serving an eight-year term set to expire in 2023.
In Farlow's inaugural speech to the NSW Parliament he highlighted increasing efficiency and transparency in government expenditure as one of his priority policy areas. He also spoke on the need to restore state rights in order to counter the issue of vertical fiscal imbalance and to restore competitive federalism.
Farlow has been appointed and chaired a number parliamentary committee's since entering the Upper House in 2015. In May 2015 he was appointed chair of the Legislative Council Standing Committee on Law and Justice. The same year Farlow was appointed a member of the Committee on the Ombudsman, the Police Integrity Commission and the Crime Commission. The following year Farlow was appointed to the Standing Committee on Social Issues, which he chaired whilst concurrently being a member of the General Purpose Standing Committee No. 6 as well as chairing the Standing Committee on Law and Justice. Farlow was appointed chair of the Regulation Committee as of December 2017.
In February 2017 Farlow was promoted to the position of Parliamentary Secretary to the Premier and Leader of the House in the Legislative Council; a position he held until April 2019. Following the 2019 New South Wales state election, in April 2019 Farlow was appointed as Parliamentary Secretary to the Treasurer, serving to 20 December 2021.
References
Living people
Members of the New South Wales Legislative Council
Liberal Party of Australia members of the Parliament of New South Wales
Year of birth missing (living people)
21st-century Australian politicians
|
{
"redpajama_set_name": "RedPajamaWikipedia"
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| 4,845
|
\section{Introduction}
Let $G$ be a
connected complex reductive group of dimension $n$, and let
$\pi:G\to GL(V)$ be a faithful representation of $G$. A generic {\em hyperplane section} $H_\pi$ corresponding
to $\pi$ is the preimage $\pi^{-1}(H)$ of the intersection of $\pi(G)$ with a generic affine hyperplane
$H\subset{\rm End}(V)$. There is a nice explicit formula for the self-intersection index of $H_\pi$ in $G$, and more generally,
for the intersection index of $n$ generic hyperplane sections corresponding to different representations
(see Theorem \ref{t.degree} below) in terms of the weight polytopes of the representations.
In this paper, I give a similar formula for the intersection indices of the {\em Chern classes}
of $G$ (defined in \cite{VK}) with generic hyperplane sections (see Theorem \ref{t.chern}).
The {\em Chern classes} of $G$ can be defined as the Chern classes of the
{\em logarithmic} tangent bundle over a {\em regular}
compactification of $G$ (see Section \ref{s.chern} for a precise definition).
Denote by $k$ the rank of $G$, i.e. the dimension of a maximal torus in $G$. Only the first $(n-k)$
Chern classes are not trivial \cite{VK}. These Chern classes are elements
of the {\em ring of conditions} of $G$, which was introduced by C.De Concini and C.Procesi (see \cite{CP2}).
They can be represented by subvarieties $S_1,\ldots, S_{n-k}\subset G$, where $S_i$ has codimension $i$. All enumerative
problems for $G$, such as the computation of the intersection index $S_iH_\pi^{n-i}$, make sense in the ring
of conditions.
First, I recall the usual Brion--Kazarnovskii formula for the
intersection indices of hyperplane sections. Choose a maximal torus $T\subset G$,
and denote by $L_T$ its character lattice. Choose also a Weyl chamber ${\cal D}\subset
L_T\otimes\mathbb{R}$.
Denote by $R^+$ the set of all positive roots of $G$ and denote by $\rho$
the half of the sum of all positive roots of $G$. The inner product
$(\cdot,\cdot)$ on $L_T\otimes\mathbb{R}$ is given by a nondegenerate
symmetric bilinear form on the Lie algebra of $G$ that is invariant under the adjoint action of $G$ (such a form exists
since $G$ is reductive).
\begin{thm} {\em \cite{Brion, Kaz}} \label{t.degree} If $H_\pi$ is a
hyperplane section corresponding to a
representation $\pi$ with the
weight polytope $P_\pi\subset L_T\otimes\mathbb{R}$ ,
then the self-intersection index of $H_\pi$ in the ring of conditions is equal to
$$n!\int\limits_{P_\pi\cap{\cal D}}\prod_{\alpha\in R^+}\frac{(x,\alpha)^2}{(\rho,\alpha)^2}dx.$$
The measure $dx$ on $L_T\otimes\mathbb{R}$ is normalized so that the covolume of $L_T$ is $1$.
\end{thm}
This theorem was first proved by B.Kazarnovskii \cite{Kaz}. Later, M.Brion proved
an analogous formula for arbitrary spherical varieties using a different method \cite{Brion}.
The integrand in this formula has the following interpretation.
The direct sum $L_T\oplus L_T$ can be identified with the Picard
group of the product $G/B\times G/B$ of two flag varieties. Here
$B$ is a Borel subgroup of $G$. Hence, to each lattice point
$(\lambda_1,\lambda_2)\in L_T\oplus L_T$ one can assign the
self-intersection index of the corresponding divisor in $G/B\times
G/B$. The resulting function extends to the polynomial
function $(n-k)!F$ on $(L_T\oplus L_T)\otimes\mathbb{R}$, where
$$F(x,y)=\prod_{\alpha\in R^+}\frac{(x,\alpha)(y,\alpha)}{(\rho,\alpha)^2}.$$
Note that the integrand is
the restriction of $F$ onto the diagonal $\{(\lambda,\lambda):\lambda\in L_T\otimes\mathbb{R}\}$.
This interpretation leads to another proof of the
Brion--Kazarnovskii formula (different from those of Kazarnovskii and Brion). Namely, take any regular
compactification $X$ of $G$ that lies over the compactification
$X_\pi$ corresponding to the representation $\pi$ (see Subsection \ref{ss.comp}
). Then reduce
the computation of $H_\pi^n$ to the computation of the
intersection indices of divisors in the closed orbits of $X$ (see Section \ref{s.proof}). All
closed orbits are isomorphic to the product of two flag varieties.
The precise algorithm for doing this was given by De Concini and
Procesi \cite{CP} in the case, where $X$ is a wonderful compactification of
a symmetric space. Then E.Bifet extended this algorithm to all regular compactifications of symmetric
spaces \cite{Bifet}. I will show that in the case, where a symmetric space is a reductive group, this
algorithm actually produces the Brion--Kazarnovskii formula if one uses the weight
polytope of $\pi$ to keep track of all transformations.
Moreover, the De Concini--Procesi algorithm
works not only for divisors. It can also be carried over to
the Chern classes
of $G$ (which are, in general, not linear combinations of complete intersections). In particular,
there is the following explicit formula for the intersection
indices of the Chern classes of $G$ with hyperplane sections. Assign to
each lattice point $(\lambda_1,\lambda_2)\in L_T\oplus L_T$ the
intersection index of the $i$-th Chern class of the tangent bundle over $G/B\times G/B$
with the divisor $D(\lambda_1,\lambda_2)$ corresponding to
$(\lambda_1,\lambda_2)$, that is the number $c_i(G/B\times
G/B)D^{n-k-i}(\lambda_1,\lambda_2)$. Extend this function to the polynomial
function on $(L_T\oplus L_T)\otimes\mathbb{R}$.
Since the Chern classes of $G/B$ are known the resulting function can be easily computed (see Section \ref{s.proof}).
The final formula is as follows.
Let $\mathbb{D}$ be the
differential operator (on functions on $(L_T\oplus L_T)\otimes\mathbb{R}$) given by the formula
$$\mathbb{D}=\prod_{\alpha\in R^+}(1+\partial_\alpha)(1+\widetilde\partial_\alpha),$$
where $\partial_\alpha$ and $\widetilde\partial_\alpha$ are directional derivatives along the
vectors $(\alpha,0)$ and $(0,\alpha)$, respectively. Denote by $[\mathbb{D}]_i$
the $i$-th degree term in $\mathbb{D}$.
\begin{thm} \label{t.chern} If $H_\pi$ is a generic hyperplane section corresponding to a
representation $\pi$ with the weight polytope $P_\pi\subset
L_T\otimes\mathbb{R}$, then the intersection index of $H_\pi^{n-i}$ with
the $i$-th Chern class of $G$ in the ring of conditions is equal
to
$$(n-i)!\int\limits_{P_\pi\cap{\cal D}}[\mathbb{D}]_iF(x,x)dx. \eqno (1)$$
The measure $dx$ on $L_T\otimes\mathbb{R}$ is normalized so that the covolume of $L_T$ is $1$.
\end{thm}
Since in general, the Chern classes of $G$ are not complete
intersections, this extends computation of the
intersection indices to a bigger part of the ring of conditions of $G$.
Theorem \ref{t.chern} also completes some results of \cite{VK}. Namely, the Chern classes $S_1,\ldots, S_{n-k}$
were used there as the main ingredients in an adjunction formula for the topological Euler characteristic of
complete intersections of hyperplane sections in $G$ (see Theorem 1.1 in \cite{VK}). Theorem \ref{t.chern}
in the present paper allows to make this formula explicit. E.g. if a complete intersection is just one hyperplane
section $H_\pi$, then
$$\chi(H_\pi)=
(-1)^{n-1}\int\limits_{P_\pi\cap{\cal D}}\left(n!-(n-1)![\mathbb{D}]_1+(n-2)![\mathbb{D}_2]-\ldots+k![\mathbb{D}]_{n-k}\right)F(x,x)dx.$$
There is also a formula for the Chern classes $c_i(X)$
of the tangent bundle over any regular compactification $X$ of $G$ in terms of $S_1$,
\ldots, $S_{n-k}$ (see Corollary 4.4 in \cite{VK}).
Theorem \ref{t.chern} allows to compute explicitly the intersection index of $c_i(X)$ with
a complete intersection of complementary dimension in $X$.
I am grateful to M.Brion, K.Kaveh, B.Kazarnovskii and A.Khovanskii for useful discussions.
\section{Preliminaries}
In this section, I recall some well-known facts which are used in the proof of Theorem \ref{t.chern}.
In Subsection \ref{ss.comp}, I define the
regular compactification $X$ of $G$ associated with a representation $\pi$ and describe the
orbit structure of $X$ in terms of
the weight polytope of the representation. In Subsection \ref{ss.picard}, I will relate the Picard group of $X$
to the space of virtual polytopes
analogous to the weight polytope of $\pi$. The notion of analogous polytopes is discussed in Subsection
\ref{ss.polytopes}.
Then I recall a formula for the integral of a polynomial function
over a simplex (Subsection \ref{ss.integral}), which is used to interpret the computation of intersection
indices in terms of integrals over the weight polytope.
\subsection{Polytopes}\label{ss.polytopes}
Let $P\subset\mathbb{R}^k$ be a convex polytope. Define the {\em normal fan}
$P^*$ of $P$. This is a fan in the dual space $(\mathbb{R}^k)^*$. To each face $F^i\subset P$ of dimension $i$
there corresponds a cone $F^*_i$ of dimension $(n-i)$ in $P^*$ defined as follows. The cone $F^*_i$
consists of all linear functionals in $(\mathbb{R}^k)^*$ whose maximum value on $P$ is attained on the interior of
the face $F^i$.
In particular, to each facet of $P$ there corresponds a one-dimensional cone, i.e. a ray, in $P^*$.
If the dual space $(\mathbb{R}^k)^*$ is identified with $\mathbb{R}^k$ by means of the Euclidean inner product, the ray
corresponding to a facet is spanned by a normal vector to the facet.
Two convex polytopes are called {\em analogous} if they have the same normal fan. All polytopes analogous to a given
polytope $P$ form
a semigroup $S_P$ with respect to Minkowski sum. This semigroup is also endowed with the action of the multiplicative
group $\mathbb{R}^{>0}$ (polytopes can be dilated). Hence, $S_P$ can be regarded as a cone in the vector space $V_P$, where
$V_P$ is the minimal group containing $S_P$ (i.e. the Grothendieck group of $S_P$). The elements of
$V_P$ are called {\em virtual} polytopes analogous to $P$.
We now introduce special coordinates in the vector space $V_P$.
Let $\Gamma_1$, \ldots, $\Gamma_l$ be the facets of $P$, and let $\Gamma_1^*$, \ldots, $\Gamma_l^*$ be the corresponding
rays in $P^*$. Choose a non-zero functional $h_i\in\Gamma_i^*$ in each ray. Call $h_i$ a {\em support function}
corresponding to the facet $\Gamma_i$.
For any polytope $Q$ analogous to $P$, denote by $h_i(Q)$ the maximal value
of $h_i$ on the polytope $Q$. For instance, if $h_i$ is normalized so that its value on the external unit normal to the
facet $\Gamma_i$ is 1, then $h_i(P)$ is up to a sign the distance from the origin to the hyperplane that contains the facet
$\Gamma_i$ (the sign
is positive if the origin and the polytope $P$ are to the same side of this hyperplane, and negative otherwise). The numbers
$h_1(Q),\ldots,h_l(Q)$ are called the {\em support numbers} of $Q$.
Clearly, the polytope $Q$ is uniquely defined by its support numbers. The coordinates
$h_1(Q),\ldots,h_l(Q)$ can be extended to the space $V_P$,
providing the isomorphism between $V_P$ and the coordinate space $\mathbb{R}^l$.
In what follows, we will deal with {\em integer} polytopes, i.e. polytopes whose vertices belong to a given lattice
$\mathbb{Z}^k\subset \mathbb{R}^k$. For such polytopes, the natural way to normalize the support functions is to require that $h_i(P)$
be equal to the {\em integral} distance from the origin to the hyperplane that contains the facet $\Gamma_i$. Suppose that
a hyperplane $H$ not passing through the origin is spanned by lattice vectors. Then the
{\em integral} distance from the origin to the hyperplane $H$ is the index in $\mathbb{Z}^k$ of the subgroup spanned by
$H\cap\mathbb{Z}^k$. To compute the integral distance one can apply a unimodular (with respect to the lattice $\mathbb{Z}^k$)
linear transformation of $\mathbb{R}^{k}$ so
that $H$ becomes parallel to a coordinate hyperplane. Then the integral distance is the usual Euclidean distance from
the origin to this coordinate hyperplane.
We will also use the notion of {\em simple} polytopes. A polytope in $\mathbb{R}^k$ is called {\em simple} if it is generic
with respect to parallel translations of its facets. Namely, exactly $k$ facets must meet at each vertex.
This implies that any other face is also the transverse intersection of those facets that contain it.
\subsection{Regular compactifications of reductive groups}\label{ss.comp}
With any representation $\pi:G\to GL(V)$ one can associate
the following compactification of $\pi(G)$. Take the projectivization $\mathbb{P}(\pi(G))$ of $\pi(G)$ (i.e. the set
of all lines in ${\rm End}(V)$ passing through a point of $\pi(G)$ and the origin),
and then take its closure in $\mathbb{P}({\rm End}(V))$.
We obtain a projective variety $X_\pi\subset\mathbb{P}({\rm End}(V))$
with a natural action of $G\times G$ coming from the left and right action of
$\pi(G)\times\pi(G)$ on ${\rm End}(V)$. E.g. when $G=(\mathbb{C}^*)^n$ is a complex torus,
all projective toric varieties can be constructed in this way.
Assume that $\mathbb{P}(\pi(G))$ is isomorphic to $G$.
Consider all weights of the representation $\pi$, i.e. all characters
of the maximal torus $T$ occurring in $\pi$. Take their convex hull $P_\pi$ in $L_T\otimes\mathbb{R}$.
Then it is easy to see that $P_\pi$ is a polytope
invariant under the action of the Weyl group of $G$. It is called the {\em weight polytope}
of the representation $\pi$.
The polytope $P_\pi$ contains
information about the compactification $X_\pi$.
\begin{thm}\label{t.equiv}
1) {\em (\cite{Tim}, Proposition 8)} The subvariety $X_\pi$ consists of a finite number of
$G\times G$-orbits. These orbits are in one-to-one correspondence with
the orbits of the Weyl group acting on the faces of the polytope
$P_\pi$. This correspondence preserves incidence relations. I.e.
if $F_1$, $F_2$ are faces such that $F_1\subset F_2$, then the orbit corresponding to $F_1$ is contained
in the closure of the orbit corresponding to $F_2$.
2) Let $\sigma$ be another representation of $G$.
The normalizations of subvarieties $X_\pi$ and $X_\sigma$ are isomorphic if and only if
the normal fans corresponding to the polytopes $X_\pi$ and $X_\sigma$
coincide. If the first fan is a subdivision of the second, then there
exists a $G\times G$--equivariant map from the normalization of $X_\pi$ to $X_\sigma$, and vice versa.
\end{thm}
The second part of Theorem \ref{t.equiv} follows from the general theory of spherical varieties
(see \cite{Knop}, Theorem 5.1) combined with the description of compactifications $X_\pi$
via colored fans (see \cite{Tim}, Sections 7, 8).
In what follows, we will only consider {\em regular} compactifications of $G$. The simplest example of a regular
compactification is the {\em wonderful compactification} constructed by De Concini and Procesi.
Suppose that the group $G$ is of adjoint type, i.e. the center of $G$ is trivial.
Take any irreducible representation $\pi$ with a strictly dominant highest weight.
It is proved in \cite{CP} that the corresponding compactification $X_\pi$ of the group $G$
is always smooth and, hence, does not depend on the choice of a highest
weight. Indeed, the normal fan of
the weight polytope $P_\pi$ coincides with the fan of the Weyl chambers and their faces, so the second
part of Theorem \ref{t.equiv}
applies.
This compactification is
called the {\em wonderful compactification} and is denoted by $X_{can}$.
Other {\em regular} compactifications of $G$ can be characterized as follows.
The normalization $X$ of $X_\pi$ is {\em regular} if first, it is smooth, and second, there is
a $(G\times G)$--equivariant map from $X$ to $X_{can}$.
These two conditions can be reformulated in terms of the weight polytope $P_\pi$.
Namely, the first condition implies that $P_\pi$ is {\em integrally} simple (see \cite{Tim} Theorem 9), i.e. it is simple
and the edges meeting at each vertex form a basis of $L_T$. The second condition
implies that none of the vertices of $P_\pi$ lies on the
walls of the Weyl chambers, i.e. the normal fan of $P_\pi$ subdivides the fan of the Weyl chambers
and their faces.
A regular compactification $X$ has the following nice properties (see \cite{Brion2} for details), which we will use
in the sequel.
The boundary divisor $X\setminus G$ is a divisor with normal crossings.
The $G\times G$--orbits of codimension $s$ correspond to the faces of $P_\pi$ of codimension $s$ and have rank $(k-s)$.
Recall that each face $F\subset P_\pi$ is the transverse intersection of several facets of $P_\pi$ (since $P_\pi$
is simple). Then the closure of the orbit corresponding to $F$ is the transverse intersection of the closures
of the codimension one orbits that correspond to these facets. Each closed orbit of $X$ (such orbits correspond to the
vertices of $P_\pi$) is isomorphic to the product of
two flag varieties $G/B\times G/B$.
\subsection{Picard groups of compactifications}\label{ss.picard}
Let $X$ be the normalization of the compactification $X_\pi$ of $G$.
We assume that $X$ is regular, and hence smooth. Then the second cohomology group $H^2(X)$ is isomorphic to the Picard group of $X$
(see \cite{Bifet}).
Denote by $V(\pi)$
the group of all integer
virtual polytopes analogous to the weight polytope $P_\pi$ and invariant under the action of the Weyl group.
There is a description of the Picard group of a regular complete symmetric space due to Bifet
(see \cite{Bifet}, Theorem 2.4, see also \cite{Brion}, Proposition 3.2).
In our case, this description can be reformulated as follows (such a reformulation is well-known in the toric case, and
in the reductive case it was suggested by K.Kiumars).
The Picard group ${\rm Pic}(X)$ of $X$ is canonically isomorphic to the quotient group of $V(\pi)$ modulo
parallel translations.
In particular, if $G$ is semisimple, then ${\rm Pic}(X)=V(\pi)$ (the only parallel translation
taking a $W$--invariant polytope to a $W$--invariant polytope is the trivial one).
The isomorphism takes the hyperplane section corresponding to a representation $\sigma$ to the weight polytope
of $\sigma$ and extends to the other divisors by linearity. Let us identify divisors in $X$ with the corresponding
polytopes using this isomorphism.
The variety $X$ has $l$ distinguished {\em boundary divisors} $\overline{\cal O}_1$, \ldots, $\overline{\cal O}_l$, which are the closures of
codimension one orbits. Let us describe the corresponding virtual polytopes.
Choose $l$ facets $\Gamma_1$, \ldots,
$\Gamma_l$ of $P_\pi$ so that each orbit of the Weyl group acting on the facets of $P_\pi$
contains exactly one $\Gamma_i$. E.g. take all facets that intersect the fundamental Weyl chamber.
Choose the support functions $h_1$, \ldots, $h_l$ corresponding to these facets
so that $h_i(P_\pi)$ is equal to the integral distance (with respect to the weight lattice $L_T$) from
the origin to the facet $\Gamma_i$.
\begin{lemma} \label{l.orbits} The closure $\overline{\cal O}_i$ of codimension one orbit corresponds to the virtual polytope
whose $i$-th support number is $1$ and the other support numbers are $0$.
\end{lemma}
\begin{proof}
Let $\sigma$ be any representation of $G$ whose weight polytope $P$ is analogous to $P_\pi$. Then $X$ is isomorphic to
the normalization of
the compactification $X_\sigma$. Thus a generic linear functional $f$ on $X_\sigma$ can
also be regarded as a rational function on $X$. Let us find the zero and the pole divisors of $f$.
The zero divisor is the divisor corresponding to the weight polytope of $\sigma$. The pole divisor
is a linear
combination of the divisors $\overline{\cal O}_1$, \ldots, $\overline{\cal O}_l$. It is not hard to show that the coefficients are
the support numbers $h_1(P_\sigma)$, \ldots, $h_l(P_\sigma)$, i.e. the integral distances from the origin to
the facets of $P_\sigma$
corresponding to $\Gamma_1$, \ldots, $\Gamma_l$. Indeed, for toric varieties, this statement is well-known
(see \cite{Fulton}, Section 3.4). In particular, this holds
for the closure $\o T$ in $X$ of the maximal torus $T\subset G$. Hence,
this also holds for $X$, and the divisor $D$ of a hyperplane section corresponding to $\sigma$
can be written as
$$D=h_1(P_\sigma)\overline{\cal O}_1+\ldots+h_l(P_\sigma)\overline{\cal O}_l.$$
It follows that $h_i(\overline{\cal O}_j)=0$, unless $i=j$.
\end{proof}
Another useful collection of divisors consists of the closures in $X$
of codimension one Bruhat cells in $G$. Denote these divisors by $D_1$, \ldots, $D_k$.
They can also be described as hyperplane sections corresponding to the irreducible representations of $G$
with fundamental highest weights $\omega_1$, \ldots, $\omega_k$, respectively.
Then to each dominant
weight $\lambda=m_1\omega_1+\ldots+m_k\omega_k$ there corresponds the {\em weight divisor} $D(\lambda)=m_1D_1+\ldots+m_kD_k$.
The polytope of this divisor is the weight polytope $P_{\lambda}$ of the irreducible representation with the
highest weight $\lambda$. Note that $\lambda$ is the only vertex of $P_{\lambda}$ inside the fundamental Weyl chamber.
Hence, it belongs to all facets of $P_{\lambda}$ corresponding to $\Gamma_1$,\ldots, $\Gamma_l$
(e.g. some of the facets might degenerate to the vertex $\lambda$). This implies the following lemma.
\begin{lemma} \label{l.cells} Let $D(\lambda)$ be the weight divisor corresponding to a
weight $\lambda\in L_T$ and let $P_\lambda$ be its polytope.
Then $h_i(P_\lambda)=h_i(\lambda)$ for any $i=1,\ldots,l$.
\end{lemma}
Combination of these two lemmas leads to the following result.
\begin{cor}\label{c.divisor}
Let $D$ be the divisor on $X$ corresponding to a polytope $P$. We assume that $P$ is analogous to $P_\pi$ and identify
the respective facets.
Then for any face
$F\subset P$ of codimension $s$ that intersects the fundamental Weyl chamber ${\cal D}$
and for any point $\lambda\in F\cap{\cal D}$, the divisor $D$ can be written uniquely as a linear combination of $D(\lambda)$
and of boundary divisors $\overline{\cal O}_i$ such that the corresponding facets $\Gamma_i$ do not contain $F$. Namely, if
$F=\Gamma_{i_1}\cap\ldots\cap\Gamma_{i_s}$, then
$$D=D(\lambda)+\sum_{j\in\{1,\ldots,l\}\setminus\{i_1,\ldots,i_s\}}[h_j(P)-h_j(\lambda)]\overline{\cal O}_j.$$
\end{cor}
\subsection{Integration of polynomials} \label{ss.integral}
Let $f(x_1,\ldots,x_k)$ be a homogeneous polynomial function of
degree $d$ defined on a real affine space $\mathbb{R}^k$ with coordinates
$(x_1,\ldots,x_k)$. Below I recall a useful formula expressing the
integral of $f$ over a simplex in $\mathbb{R}^k$ in terms of the
{\em polarization} of $f$. Recall that the {\em polarization} of $f$ is the unique
symmetric $d$-linear form $f_{pol}$ on $\mathbb{R}^k$ such that the restriction of $f_{pol}$
to the diagonal coincides with $f$. One can define $f_{pol}$ explicitly as follows:
$$f_{pol}(v_1,\ldots,v_d)=\frac{1}{d!}\frac{\partial^d}{\partial_{v_1}\ldots\partial_{v_d}}f,$$
where $\partial_{v_i}$ is the directional derivative along the vector $v_i$.
Let $\Delta\subset\mathbb{R}^k$ be a $k$-dimensional simplex with
vertices $a_0,\ldots,a_k$ and let $dx=dx_1\wedge
dx_2\wedge\ldots\wedge dx_k$ be the standard measure on $\mathbb{R}^k$.
\begin{prop}{\em \cite{Brion}}\label{p.int}
Let $f_{pol}$ be the polarization of $f$. It can be regarded as a
linear function on the $d$-th symmetric power of $V$. Then the
average value of $f$ on the simplex $\Delta$ coincides with the
average value of $f_{pol}$ on all symmetric products of $d$
vectors from the set \{$a_0$, \ldots, $a_k$\}:
$$\frac1{{\rm Vol}(\Delta)}{\int\limits_\Delta f(x)dx}=
\frac1{\binom{d+k}{k}}
\sum_{i_0+\ldots+i_k=d}
f_{pol}(\underbrace{a_0,\ldots,a_0}_{i_0},\ldots,\underbrace{a_k,\ldots,a_k}_{i_k}).$$
\end{prop}
\section{Chern classes}\label{s.chern}
In this section, I recall the definition of the Chern classes of spherical homogeneous spaces (see \cite{VK}
for more details). In the sequel, only Chern classes of $G\times G$--orbits in regular compactifications of $G$
will be used.
For these Chern classes, I prove a vanishing result for their intersection indices with
certain weight divisors in regular compactifications. This result will be important in Section \ref{s.proof} when
applying the De Concini--Procesi algorithm to the Chern classes of $G$.
Let $G/H$ be a spherical homogeneous space under $G$. Denote by $\mathfrak{g}$ and $\mathfrak{h}$ the Lie algebras of $G$ and $H$,
respectively, and denote by $m$ the dimension of $\mathfrak{h}$.
Define the {\em Demazure map} $p$ from $G/H$ to the Grassmannian ${\rm G}(m,\mathfrak{g})$ of $m$-dimensional subspaces in $\mathfrak{g}$
as follows:
$$p:G/H\to{\rm G}(m,\mathfrak{g}); \quad p: gH\to g\mathfrak{h} g^{-1}.$$
Let $C_i\subset{\rm G}(m,\mathfrak{g})$ be the Schubert cycle corresponding to a generic subspace $\Lambda_i\subset\mathfrak{g}$
of codimension $m+i-1$, i.e. $C_i=\{\Lambda\in{\rm G}(m,\mathfrak{g}): {\rm dim}(\Lambda\cap\Lambda_i)\ge1\}$. Then the $i$-th {\em Chern class}
$S_i(G/H)$ of $G/H$ is the preimage of $C_i$ under the map $p$:
$$S_i(G/H)=p^{-1}(C_i).$$
The class of $S_i(G/H)$ in the ring of conditions of $G/H$ is the same for all generic $C_i$ \cite{VK}.
It is related to the Chern classes of the tangent bundles over regular compactifications of $G$
\cite{Brion3,VK}. Namely, if $X$ is a regular compactification of $G/H$, then the closure of $S_i(G/H)$
in $X$ is the $i$-th Chern class of the {\em logarithmic} tangent bundle over $X$ that corresponds
to the divisor $X\setminus (G/H)$. This vector bundle is generated by all vector fields on $X$ that are tangent
to $G$--orbits in $X$. In what follows, this bundle will be called the {\em Demazure bundle} of $X$.
Let $X=G/H$ and $Y=G/P$ be two spherical homogeneous
spaces under $G$. Suppose that $H$ is a subgroup of $P$. Consider the $G$--equivariant map
$$p:X\to Y; \quad p: gH\mapsto gP.$$
Let $S_i(X)$ the $i$-th Chern class of $X$.
In general, it is not true that $S_i(X)$ is the inverse image under the map $p$
of a subset in $Y$. However, the intersection of $S_i(X)$ (when it is nonempty) with a fiber of $p$ has
dimension at least ${\rm rk}(P)-{\rm rk}(H)$.
{\begin{example}\em In what follows, we will mostly deal with the case, where $X$ and $Y$ are spherical
homogeneous spaces
under the doubled group $G\times G$. Namely, $X$ is a $G\times G$--orbit of a regular compactification
of the group $G$
and $Y$ is a partial flag variety constructed as
follows. Let $H\subset G\times G$ be the stabilizer of a
point in ${\cal O}$. Take the minimal parabolic subgroup $P\subset G\times G$ that contains $H$ and set $F=(G\times G)/P$.
\end{example}}
\begin{lemma}\label{l.fibers}
For a generic $S_i(X)$, there exists an open dense subset of $S_i(X)$ such that for any element $x$ of this subset
the intersection of the fiber $xP$ with $S_i(X)$ has dimension
greater than or equal to the ${\rm rk}(P)-{\rm rk}(H)$. In particular, the dimension of $p(S_i(X))$ satisfies
the inequality
$${\rm dim}~p(S_i(X))\le{\rm dim}~S_i(X)-({\rm rk}(P)-{\rm rk}(H)).$$
\end{lemma}
\begin{proof}
Choose a generic vector space $\Lambda\subset\mathfrak{g}$ of codimension ${\rm dim}~H+i-1$.
Denote by $\mathfrak{h}$ and $\mathfrak{p}$ the Lie algebras of $H$ and $P$
respectively. Then by definition
$S_i(X)$ consists of all cosets $gH$ such that $g\mathfrak{h} g^{-1}$ has a nontrivial intersection with $\Lambda$,
or equivalently $\mathfrak{h}\cap g^{-1}\Lambda g$ is nontrivial.
Let $gH$ be any element of $S_i(X)$. Estimate the dimension of the intersection of $S_i(X)$ with
the fiber $gP$ of the map $p$.
Note that for all $g$ from a dense open subset of $S_i(X)$, the intersection
$\mathfrak{h}\cap g^{-1}\Lambda g$ contains an element $v$ that is regular in $\mathfrak{h}$.
Denote by $C$ the centralizer
in $P$ of $v\in\mathfrak{h}\subset\mathfrak{p}$. Then ${\rm dim}(C\cap H)={\rm rk}(H)$ while $C$ has dimension at least ${\rm rk}(P)$.
Note that for any $c\in C$ the coset $gcH$ still belongs to
$S_i(X)$ since $c^{-1}g^{-1}\Lambda gc$ contains $c^{-1}vc=v$.
Hence, $S_i(X)\cap gP$ contains a set $gCH$ of dimension
at least ${\rm rk}(P)-{\rm rk}(H)$.
\end{proof}
Lemma \ref{l.fibers} is crucial for proving the following two vanishing results,
which extend Proposition 9.1 from \cite{CP} and rely on the same ideas.
Let $X$ be a regular compactification of $G$, and let $p:X\to X_{can}$ be its equivariant projection to the wonderful
compactification. Denote by $c_1$, \ldots, $c_{n-k}$ the Chern classes of the Demazure vector bundle
over $X$.
\begin{lemma} \label{l.vanish}
Let ${\cal O}$ be a $G\times G$--orbit in $X$ of codimension $s<k$, and $\overline{\cal O}\subset X$ its closure.
Suppose that the image $p({\cal O})$ under the map $p:X\to X_{can}$
coincides with the closed orbit of $X_{can}$. In terms of polytopes, this means that the face corresponding to
${\cal O}$ does not intersect the walls
of the Weyl chambers.
Let $\lambda$ be any weight of $G$, and $D(\lambda)$ the corresponding weight divisor. Then the homology class
$c_iD^{n-i-s}(\lambda)$ vanishes on $\overline{\cal O}$, i.e. the following intersection index is zero:
$$c_iD(\lambda)^{n-i-s}\overline{\cal O}=0.$$
\end{lemma}
\begin{proof} First of all, the intersection product $c_i\cdot\overline{\cal O}$ is the $i$-th Chern class
of the Demazure bundle over $\overline{\cal O}$ (see \cite{Bien}, Proposition 2.4.2). Hence, it can be realized as the
closure in $\overline{\cal O}$ of the
$i$-th Chern class $S_i({\cal O})$ of the spherical homogeneous space ${\cal O}$.
The computation of the intersection index
$c_i\overline{\cal O} D(\lambda)^{n-i-s}$ in $X$ thus reduces to the computation of the intersection index
$\o{S_i({\cal O})}D(\lambda)^{n-i-s}$ in $\overline{\cal O}$. The latter is equal to the intersection index
$S_i({\cal O})D(\lambda)^{n-i-s}$ in the ring of conditions of ${\cal O}$ since $D(\lambda)$ and $\o{S_i({\cal O})}$
have proper intersections with the boundary $\overline{\cal O}\setminus {\cal O}$.
To compute $S_i({\cal O})D(\lambda)^{n-i-s}$ we use the restriction of the map $p:X\to X_{can}$
to $\overline{\cal O}$. By the hypothesis the image $p(\overline{\cal O})$ is the closed orbit $F$ in $X_{can}$, so it is isomorphic to
the product $G/B\times G/B$ of two flag varieties. Then the divisor $D(\lambda)$
restricted to $\overline{\cal O}$ is the inverse image under the map $p$ of the divisor $D(\lambda,\lambda)$ in $F$. Indeed,
$D(\lambda)=p^{-1}(\widetilde D(\lambda))$, where $\widetilde D(\lambda)$ is the weight divisor in $X_{can}$ corresponding to $\lambda$.
It is easy to check that $\widetilde D(\lambda)\cap F=D(\lambda,\lambda)$ (see Proposition 8.1 in \cite{CP}).
Hence, all the intersection points in
$S_i({\cal O})D(\lambda)^{n-i-s}$ are contained in the preimage of
$p(S_i({\cal O})) D(\lambda,\lambda)^{n-i-s}$. But the latter is empty.
Indeed, since ${\cal O}$ has positive rank and $F$ has zero rank,
Lemma \ref{l.fibers} implies that
$${\rm dim}~p(S_i({\cal O}))<{\rm dim}~S_i({\cal O})=n-i-s.$$
\end{proof}
It remains to deal with the orbits in $X$ whose image under the map $p$ is not the closed orbit in $X_{can}$.
In this case, the face corresponding to such an orbit intersects the walls of the Weyl chambers, and hence, it
is orthogonal to some of the fundamental weights $\omega_1$,\ldots, $\omega_k$. Note that the
codimension one orbits ${\cal O}_1$,\ldots, ${\cal O}_k$ in $X_{can}$ are in one-to-one correspondence with the
fundamental weights $\omega_1$, \ldots, $\omega_k$. Namely, the facet corresponding to ${\cal O}_i$ is orthogonal to $\omega_i$.
Let $\overline{\cal O}_1$,\ldots, $\overline{\cal O}_k$ be the closures in $X_{can}$ of ${\cal O}_1$,\ldots, ${\cal O}_k$, respectively.
\begin{lemma}
Let ${\cal O}$ be a $G\times G$--orbit in $X$ of codimension $s<k$.
Suppose that the image $p({\cal O})$ under the map $p:X\to X_{can}$
is not closed and lies in the intersection $\overline{\cal O}_{i_1}\cap\ldots\cap \overline{\cal O}_{i_s}$. In terms of polytopes, this means
that the face corresponding to ${\cal O}$ is orthogonal to the weights $\omega_{i_1}$,\ldots, $\omega_{i_s}$.
Let $\lambda$ be any linear combination of the weights $\omega_{i_1}$,\ldots, $\omega_{i_s}$. Then
$$c_iD(\lambda)^{n-i-s}\overline{\cal O}=0.$$
\end{lemma}
\begin{proof}
We use the $G\times G$--equivariant map $r$ from $\overline{\cal O}_{i_1}\cap\ldots\cap \overline{\cal O}_{i_s}$
to a partial flag variety $G/P\times G/P$ constructed in \cite{CP} (see \cite{CP} Lemma 5.1 for details).
Consider the compactification $X_{i_1,\ldots,i_s}$
of $G$ corresponding to the irreducible representation $\pi_{i_1,\ldots,i_s}$ whose highest weight lies
strictly inside the cone spanned by $\omega_{i_1}$,\ldots, $\omega_{i_s}$. This compactification has a unique closed
orbit $G/P\times G/P$, where $P\subset G$ is the stabilizer of the highest weight vector in the representation
$\pi_{i_1,\ldots,i_s}$.
Clearly, the fan of the Weyl chambers and their faces subdivides the normal fan of the weight polytope of
$\pi_{i_1,\ldots,i_s}$. Hence, by Theorem \ref{t.equiv} there is an equivariant map $r:X_{can}\to X_{i_1,\ldots,i_s}$.
This map takes $\overline{\cal O}_{i_1}\cap\ldots\cap \overline{\cal O}_{i_s}$ to the closed orbit $G/P\times G/P$.
The composition $rp$ maps the orbit ${\cal O}$ to the closed orbit $G/P\times G/P$ of $X_{i_1,\ldots,i_s}$.
It is easy to show that
the divisor $D(\lambda)$ restricted to ${\cal O}$ is the preimage of the divisor $D(\lambda,\lambda)\subset G/P\times G/P$ under this map
(see \cite{CP} Section 8.1).
Now repeat the arguments of the proof of Lemma \ref{l.vanish}.
\end{proof}
These two lemmas imply the following vanishing result.
\begin{cor}\label{c.vanish}
Let ${\cal O}$ be any $G\times G$--orbit in $X$, and let $F$ be the face of the polytope of $X$ that corresponds to ${\cal O}$.
The intersection index
$$c_i D(\lambda_1)\ldots D(\lambda_{n-i-s})\overline{\cal O}$$
vanishes in the cohomology ring of $X$ in the following two cases:
1) The face $F$ does not intersect the walls of the Weyl chambers. Then weights $\lambda_1$,\ldots, $\lambda_{n-i-s}$
are any weights of $G$.
2) The face $F$ intersects a wall of the Weyl chambers and weights $\lambda_1$,\ldots, $\lambda_{n-i-s}$ are
orthogonal to $F$ (with respect to the inner product $(\cdot,\cdot)$ on $L_T\otimes\mathbb{R}$ defined in the Introduction).
\end{cor}
\section{Proof of Theorem \ref{t.chern}} \label{s.proof}
We use notation of Subsections \ref{ss.comp} and \ref{ss.picard}.
Let $X$ be any
regular compactification lying over the compactification $X_\pi$. Then the closure $\o H_\pi$ of $H_\pi$ in $X$
has proper intersections with all $G\times G$--orbits in $X$, and thus $S_iH^{n-i}_\pi$
coincides with the intersection index $\o S_i\o H_\pi^{n-i}$ in the cohomology ring of $X$.
Assume that $X$ corresponds to a representation of $G$ with the weight polytope
$P_0$. Let us compute $\o S_iD^{n-i}$ for a divisor $D$ under the assumption that the polytope $P$
corresponding to $D$ is analogous to $P_0$. After we establish the formula of Theorem \ref{t.chern} for such divisors,
it will automatically extend to the other divisors (in particular, for $\o H_\pi$) since any virtual polytope analogous
to $P_0$ is a linear combination of polytopes analogous to $P_0$. Since $X$ is regular, $P_0$ and hence $P$ are simple.
All computations are carried in the cohomology ring of $X$. First, break
$D^{n-i}$ into monomials of the form
$\overline{\cal O}_{i_1}\ldots\overline{\cal O}_{i_k}D(\lambda_1)\ldots D(\lambda_{n-i-k})$, where
$i_1$,\ldots, $i_k$ are distinct integers from $1$ to $l$ and
$\lambda_1$,\ldots, $\lambda_{n-i-k}$ are weights. Then every such monomial can be computed explicitly,
since the intersection $\overline{\cal O}_{i_1}\cap\ldots\cap\overline{\cal O}_{i_k}$ is either empty or isomorphic to the
product of two flag varieties.
\begin{wrapfigure}{l}{6.5cm}
\includegraphics[scale=.4]{PSL3.eps}
\caption{}
\end{wrapfigure}
Since we are going to intersect $D^{n-i}$ with $\o S_i$
we can ignore all monomials that are annihilated by $\o S_i$. Recall that $\o S_i$ is the $i$-th Chern class of the
Demazure bundle over $X$. In particular,
Corollary \ref{c.vanish} implies that $\o S_i$ annihilates the ideal $I\subset H^*(X)$
generated by the monomials of the form $D(\lambda_1)\ldots D(\lambda_{n-i-s})\overline{\cal O}$ such that
either the face of $P$ corresponding to the codimension $s$ orbit ${\cal O}$ does not intersect the walls of the
Weyl chambers or, if it does, the weights $\lambda_1,\ldots,\lambda_{n-i-s}$ are orthogonal to this face.
To keep track of our calculations we use a subdivision of the
polytope $P\cap{\cal D}$ into simplices coming from the barycentric subdivision of $P$ described below.
For each face $F\subset P$ choose a point $\lambda_F\in F$ as follows. If $F$ does not intersect the
walls of the Weyl chamber ${\cal D}$, then $\lambda_F$ is any point in the interior of the face. Otherwise, choose $\lambda_F$
so that the corresponding vector is orthogonal to the face $F$ (in particular, $\lambda_F$ will belong to
the intersection of the face with a wall of ${\cal D}$.)
If $F=P$ take $\lambda_F=0$.
An {\em $s$--flag} ${\cal F}$ is the collection $\{F_1\supset\ldots\supset F_s\}$ of $s\le k$
nested faces of $P$ such that each of them intersects
${\cal D}$, and $F_i$ has codimension $i$ in $P$.
Denote by $\overline{\cal O}_{\cal F}$ the closure in $X$ of the orbit corresponding
to the last face $F_s$, and by $\Delta_{\cal F}$ the $s$--dimensional simplex with the vertices $0$, $\lambda_{F_1}$,
\ldots, $\lambda_{F_s}$. In particular, when $s=k$, the simplex $\Delta_{\cal F}$ has full dimension and the orbit $\overline{\cal O}_{\cal F}$
is closed. The polytope ${\cal D}\cap P$ is the union of simplices $\Delta_{\cal F}$ over all
possible $k$--flags ${\cal F}$.
{\bf Example.} Take $G=PSL_3(\mathbb{C})$, and let $X=X_{can}$ be
its wonderful compactification. Let divisor $D$ be a hyperplane section corresponding
to the irreducible representation with a strictly dominant highest weight $\lambda$.
In this case, $P$ is a hexagon symmetric under the action of the Weyl group with
two edges $\Gamma_1$ and $\Gamma_2$ intersecting ${\cal D}$.
Then $\lambda_i=\lambda_{\Gamma_i}\in\Gamma_i$ is orthogonal to $\Gamma_i$ for $i=1,2$ and $\Gamma_1\cap\Gamma_2=\lambda$.
The subdivision of $P\cap{\cal D}$ into simplices
consists of two triangles $\Delta_1$ and $\Delta_2$ with the vertices $0$, $\lambda_1$, $\lambda$ and $0$, $\lambda_2$, $\lambda$,
respectively (see Figure 1).
\begin{lemma} \label{l.equiv} Denote by $f_d(x_1,\ldots,x_k)$ the sum of all
monomials of degree $d$ in $k$ variables $x_1$,\ldots, $x_k$.
The following identity holds in the cohomology ring of $X$ modulo the ideal $I$:
$$D^{n-i}\equiv k!\sum_{{\cal F}}{\rm Vol}(\Delta_{\cal F})
f_{n-k-i}(D,D(\lambda_{F_1}),\ldots,D(\lambda_{F_{k-1}}))\overline{\cal O}_{{\cal F}}\pmod I,$$
where the sum is taken over all possible $k$--flags ${\cal F}=\{F_1\supset\ldots\supset F_k\}$.
The volume form ${\rm Vol}$ is normalized so that the covolume of $L_T$ is equal to $1$.
\end{lemma}
\begin{proof} We will prove the following more general statement for $s$-flags. Denote by
$f_{d,s}(x_1,\ldots,x_s)$ the sum of all monomials of degree $d$ in $s$ variables.
Recall that $\Gamma_1$,\ldots, $\Gamma_l$ denote the facets of $P$
that intersect the Weyl chamber ${\cal D}$. An $s$-flag can be alternatively described by an ordered collection of
facets $\Gamma_{i_1}$,\ldots, $\Gamma_{i_s}$ such that their intersection $\Gamma_{i_1}\cap\ldots\cap\Gamma_{i_s}$ has codimension
$s$. Then $F_j=\Gamma_{i_1}\cap\ldots\cap\Gamma_{i_j}$. This is a one-to-one correspondence, since the polytope $P$ is simple.
Assign to each $s$-flag ${\cal F}$ the following number
$$c_{\cal F}=h_{i_1}(P)[h_{i_2}(P)-h_{i_2}(\lambda_{F_1})]\ldots[h_{i_s}(P)-h_{i_s}(\lambda_{F_s})].$$
In particular, when $s=k$, i.e. $F_s$ is just a vertex, the number $c_{\cal F}$
coincides with the volume of $\Delta_{\cal F}$ times $k!$. Indeed, by a unimodular linear transformation of $L_T\otimes\mathbb{R}$
we can map the hyperplanes containing the facets $\Gamma_{i_1}$,\ldots, $\Gamma_{i_s}$ to the coordinate hyperplanes.
Then $[h_{i_j}(P)-h_{i_j}(\lambda_{F_{j-1}})]$ is just the Euclidean distance from the vertex $\lambda_{F_{j-1}}$ of $\Delta_{\cal F}$
to the hyperplane containing $\Gamma_{i_j}$. Note that to define volumes we do not use the inner product $(\cdot,\cdot)$ on
the lattice $L_T$ defined in the introduction. We only use the lattice itself.
Then for any integer $s$ such that $1\le s\le k$ the following is true:
$$D^{n-i}\equiv\sum_{{\cal F}}c_{\cal F}
f_{n-s-i,s}(D,D(\lambda_{F_1}),\ldots,D(\lambda_{F_{s-1}}))\overline{\cal O}_{{\cal F}}\pmod I, \eqno(1)$$
where the sum is taken over all $s$--flags.
Prove by induction on $s$. We use the notations of Subsection \ref{ss.picard}.
For $s=1$, the statement coincides with the decomposition
$D=h_1(P)\overline{\cal O}_1+\cdots+h_l(P)\overline{\cal O}_l$ from Lemma \ref{l.orbits}.
Assume that the formula is proved for some $s<k$. Prove it for $s+1$. We now deal separately with each term on
the right hand side of formula (1). First subtract from every term
$f_{n-s-i,s}(D,D(\lambda_{F_1}),\ldots,D(\lambda_{F_{s-1}}))\overline{\cal O}_{\cal F} $
the element $f_{n-s-i,s}(D(\lambda_{F_s}),D(\lambda_{{\cal F}_1}),\ldots,D(\lambda_{F_{s-1}}))\overline{\cal O}_{\cal F}$ of the ideal $I$.
This operation does not change the identity (1). A simple calculation shows that
$$f_{n-s-i,s}(x,x_1,\ldots,x_{s-1})-f_{n-s-i,s}(x_s,x_1,\ldots,x_{s-1})
=(x-x_s)f_{n-s-i-1,s+1}(x,x_1,\ldots,x_{s-1},x_s).$$
Hence, after subtraction we can rewrite the difference as
$$(D-D(\lambda_{F_s}))f_{n-s-i-1,s+1}(D,D(\lambda_{F_1}),\ldots,D(\lambda_{F_s}))\overline{\cal O}_{\cal F}.$$
Since $\lambda_s$ lies in the intersection of $s$ facets
$\Gamma_{i_1}$,\ldots, $\Gamma_{i_s}$, Corollary \ref{c.divisor} implies that
$$(D-D(\lambda_{F_s}))\overline{\cal O}_{\cal F}=\sum_{j\ne i_1,\ldots,i_k}[h_j(P)-h_j(\lambda_{F_s})]\overline{\cal O}_j\overline{\cal O}_{\cal F}.$$
Note that $\overline{\cal O}_j\overline{\cal O}_{\cal F}$ is empty if and only if the intersection of $\Gamma_j$ with $\Gamma_{i_1}\cap\ldots\cap\Gamma_{i_s}$ is
empty.
Hence,
$$(D-D(\lambda_{F_s}))\overline{\cal O}_{\cal F}=\sum_{{\cal F}'}[h_j(P)-h_j(\lambda_{F_s})]\overline{\cal O}_{{\cal F}'},$$
where the sum is taken over all $(s+1)$-flags ${\cal F}'$ that extend ${\cal F}$, i.e.
${\cal F}'=\{F_1\supset\ldots\supset F_s\supset F_s\cap\Gamma_{j}\}$.
\end{proof}
It remains to compute the term
$$\o S_i\cdot f_{n-k-i}(D,D(\lambda_{F_1}),\ldots,D(\lambda_{F_{k-1}}))\overline{\cal O}_{{\cal F}} \eqno(2)$$
for each $k$-flag ${\cal F}$. Suppose that the closed orbit $\overline{\cal O}_{\cal F}$ is the intersection of $k$ hypersurfaces
$\overline{\cal O}_{i_1}$,\ldots, $\overline{\cal O}_{i_k}$. Then for any other codimension 1 orbit ${\cal O}_j$ (such that $j\ne i_1,\ldots,i_k$),
the intersection $\overline{\cal O}_{\cal F}\cap\overline{\cal O}_j$ is empty. Hence, $D$ in (2) can be replaced by
$D(\lambda_{F_k})$ since
$$D=D(\lambda_{F_k})+\sum_{j\ne i_1,\ldots,i_k}(h_j(P)-h_j(\lambda_{F_k}))\overline{\cal O}_j.$$
Note also that the evaluation of (2) reduces to the computation of
intersection indices in $\overline{\cal O}_{\cal F}$, which is the product of two flag varieties. We have that
$\o S_i\cdot\overline{\cal O}_{\cal F}=c_i(\overline{\cal O}_{\cal F})$ and $D(\lambda)\cdot\overline{\cal O}_{\cal F}=D(\lambda,\lambda)$. Here $c_i(\overline{\cal O}_{\cal F})$ is the $i$-th Chern class
of the tangent bundle over $\overline{\cal O}_{\cal F}$. Hence,
$$\o S_if_{n-k-i}(D(\lambda_{F_k}),D(\lambda_{F_1}),\ldots,D(\lambda_{F_{k-1}}))\overline{\cal O}_{{\cal F}}=$$
$$=c_i(G/B\times G/B)f_{n-k-i}(D(\lambda_{F_1},\lambda_{F_1}),\ldots,D(\lambda_{F_k},\lambda_{F_k})). \eqno (3)$$
The intersection product in the right hand side of this formula is taken in $G/B\times G/B$.
The function $F_i(\lambda)=c_i(G/B\times G/B)D(\lambda,\lambda)^{n-k-i}$ can be expressed
explicitly in terms
of the function $F$ defined in the Introduction, since the $i$-th Chern class of
$G/B\times G/B$ is the term of degree $i$ in the intersection product
$$\prod_{\alpha\in R^+}(1+D(\alpha,0))(1+D(0,\alpha)).$$
One way to compute $F_i$ is as follows.
Let $\mathbb{D}$ and $[\mathbb{D}]_i$ be the
differential operators defined in the Introduction. Then
$$F_i(x)=(n-k-i)![\mathbb{D}]_iF(x,x).$$
This easily follows from the formula for the polarization mentioned in Subsection \ref{ss.integral}
and the fact that $D^{n-k}(\lambda,\lambda)=(n-k)!F(\lambda,\lambda)$.
We can now apply Proposition \ref{p.int} to convert the sum (3) into the integral over the simplex $\Delta_{\cal F}$.
Indeed, by definition of the function $f_{n-k-i}$ we have that (3) can be rewritten as
$$\sum_{i_1+\ldots+i_k=n-k-i}
{F_i}_{pol}(\underbrace{\lambda_{F_1},\ldots,\lambda_{F_1}}_{i_1},\ldots,\underbrace{\lambda_{F_k},\ldots,\lambda_{F_k}}_{i_k}).$$
This is equal to the integral
$$\binom{n-i}{k}{\int\limits_{\Delta_{\cal F}} F_i(x)dx}/{{\rm Vol}(\Delta_{\cal F})}$$
by Proposition \ref{p.int} applied to the simplex $\Delta_{\cal F}$ (with the vertices $0$, $\lambda_{F_1}$,\ldots, $\lambda_{F_k}$)
and to the function $F_i(x)$.
Combining this with Lemma \ref{l.equiv} we get
$$\o S_iD^{n-i}=\frac{(n-i)!}{(n-k-i)!}
\sum_{{\cal F}}{\int\limits_{\Delta_{\cal F}} F_i(x)dx}=(n-i)!\int\limits_{P\cap{\cal D}}[\mathbb{D}]_iF(x,x)dx.$$
Note that when $i=0$, we get the Brion--Kazarnovskii formula.
\footnotesize
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"redpajama_set_name": "RedPajamaArXiv"
}
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Q: Characterisation of Galois Group with the action of $\sigma \in S_n$ on the roots Let $f \in K[X]$ be irreducible and separable with roots $x_1,...,x_n$ in a splitting field $L$ of $f$ over $K$. We identify $\text{Gal}(L|K)$ with $\text{Gal}(L|K)\cong G\subset S_n$.
How can I see the equivalence of the following two statements? (which means a characterisation of the galois group with the action of a $\sigma \in G$ on the roots $x_1,...,x_n$)
$(1)$ $\sigma \in G$.
$(2)$ If $P \in K[X_1,...,X_n]$ with $P(x_1,...,x_n)=0$, then for $P(X_{\sigma(1)},...,X_{\sigma(n)})$ it follows that $P(X_{\sigma(1)},...,X_{\sigma(n)})(x_1,...,x_n)=0$.
A: I prefer using three different notations, say $\sigma$ for the permutation
of indices ,$\tau$ for the permutation of roots and $t$
for the field homomorphism extending $\tau$ (when it exists) : thus $\tau(x_i)=x_{\sigma(i)}$.
$(1) \Rightarrow (2) $ If $t$ exists, and $P(x_1,x_2,\ldots,x_n)=0$, we have
$$
P(X_{\sigma(1)},...,X_{\sigma(n)})(x_1,...,x_n)=
P(x_{\sigma(1)},...,x_{\sigma(n)})=
P(\tau(x_1),\tau(x_2),\ldots,\tau(x_n))=
t(P(x_1,x_2,\ldots,x_n))=t(0)=0.
$$
$(2) \Rightarrow (1) $ The permutation $\tau$ is defined on $\lbrace x_1,x_2,
\ldots ,x_n\rbrace$ and we sould like to extend it to the the whole of
$L=K[x_1,\ldots,x_n]$. The obvious definition which comes to our mind is
$$
t(A(x_1,x_2,\ldots,x_n))=A(x_{\sigma(1)},\ldots,x_{\sigma(n)}) \tag{1}
$$
for any $A\in K[X_1,\ldots,X_n]$. The problem with (1) is that it might be an
incorrect definition, with two different values set for the same argument.
However, if $A(x_1,x_2,\ldots,x_n)=B(x_1,x_2,\ldots,x_n)$ for two polynomials
$A,B$, then the polynomial $C=A-B$ satisfies $C(x_1,x_2,\ldots,x_n)=0$, so
$C(x_{\sigma(1)},...,x_{\sigma(n)})=0$ by (2), and (1) will therefore yield
the same value in both cases.
So $t$ is correctly defined, and it follows immediately from its definition
that it is a homomorphism.
Alternatively, you can define $t$ as a "quotient map".
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"redpajama_set_name": "RedPajamaStackExchange"
}
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Q: Cant see changes in .java file when running Junit with Eclipse before mvn install I have multi module maven project. The technologies I use is Spring and Struts. I created test case and run it as :
@RunWith(SpringJUnit4ClassRunner.class)
@ContextConfiguration(locations = { "classpath:applicationContextSimpleTest.xml" })
public class CurrencySpeakerTest
Everything is fine... until I make some changes in source... I cant see them until i run mvn install on the parent project. Whats wrong? I tried to look into the debug configuration but I cant find the place where its said to look on compile source instead of .java files... I think this is the reason that junit is running the .class files instead of real one... anybody has any idea about that?
I have 3 module app ... they are build with parent module. Parent module is also the imported project in eclipse.I do the changes in one of three module in the test folder. I hit refresh and I do clean/build eclipse project manually ( if i enable automatically build it wont change anything). I run the test as JUnit with eclipse dialog. Eclipse seems not to see the changes in edited test file (.java).
Maybe this has something to do with the one main project which includes the other modules?
What is strange is that... if I edit any other classes not in the test folder and run web app with tomcat plugin, Eclipse sees the changes there?
I use mvn clean and mvn install but this is not the issue.
I use Eclipse to run the test but the Eclipse debugger cant see the changes in the code... for eg if I add System.out.println() somewhere and put the breakpoint there I run the test and Eclipse doesnt stop there because it looks somehow on the old code (I assume the .class file)...or if I delete the line,edit some String...nothing... it seems to point on the .class...why is that? Refresh and clean/build eclipse project doesnt work.
A: When I delete the target from my module where i write the test I got exception below. Which makes my theory true that eclipse is looking into compiled code and cant see the source... Why is that?
Class not found pl.erif.utils.CurrencySpeakerTest
java.lang.ClassNotFoundException: pl.erif.utils.CurrencySpeakerTest
at java.net.URLClassLoader$1.run(URLClassLoader.java:202)
at java.security.AccessController.doPrivileged(Native Method)
at java.net.URLClassLoader.findClass(URLClassLoader.java:190)
at java.lang.ClassLoader.loadClass(ClassLoader.java:307)
at sun.misc.Launcher$AppClassLoader.loadClass(Launcher.java:301)
at java.lang.ClassLoader.loadClass(ClassLoader.java:248)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.loadClass(RemoteTestRunner.java:693)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.loadClasses(RemoteTestRunner.java:429)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.runTests(RemoteTestRunner.java:452)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.runTests(RemoteTestRunner.java:683)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.run(RemoteTestRunner.java:390)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.main(RemoteTestRunner.java:197)
A: you may need to run a clean first
mvn clean compile
Are you editing java files outside of eclipse? If yes, you may need to hit f5 to refresh it. (It normally should detect changes)
If not, are you using the build automatically option in eclipse?
A: I had the same issue. It was due to a defective .classpath. Removing this .classpath and re-importing the project, generates a correct .classpath and fixed the environment.
To find more details make sure that eclipse is not generating the class files of your testcase in a wrong directory.
Ref: Unit test class runs old version in eclipse
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\section{ Introduction.}
While other kinds of application can be envisaged, the main motivation for
most of the work on the development of a relativistic description of
electromagnetic self interaction in string models arises from Witten's
observation \cite{Witten85} that currents are likely to occur naturally in
many conceivable kinds of cosmic strings arising from vortex defects of the
vacuum in plausible field theories.
In the first years after Witten's epoch making observation, the importance of
electromagnetic effects was actually overestimated in comparison with that of
the purely mechanical effects of the current, which were commonly
neglected\cite{Spergel_et89}. The fact that the latter would typically be far
more important was not generally recognised until after it was shown by Davis
and Shellard that ``vortons'', meaning equilibrium states of string loops,
could easily be sustained by the purely mechanical centrifugal effect of a
circulating current\cite{DavisShellard89}, whereas the magnetostatic support
mechanism that had previously been envisaged\cite{Copeland_et88} would usually
be too weak. This realisation lead to the development of a more realistic
kind of simplification in which electromagnetic effects (other than those
arising from a strong but purely external background field) were neglected
while attention was concentrated on the suitably non-linear treatment of the
purely mechanical effects of the current\cite{Carter89}\cite{Carter90}.
With adequate means now available for treating the primary mechanical effects
of currents in cosmic strings, it is reasonable to reconsider the secondary,
but not always negligible, effects due to electromagnetic interaction. A first
step towards the inclusion of such effects in a mechanically realistic
framework was taken by Peter\cite{Peter93}, in a study of circular vorton states.
This work confirmed the relative negligibility of the purely magnetostatic
support mechanism -- due to a spacelike current -- that had been considered
previously, but on the other hand it also showed that a potentially much more
important kind of ``spring'' effect can be produced by the electrostatic
effect of a timelike current.
The purpose of the present work is to provide the machinery needed for
generalising Peter's work\cite{Peter93} from strictly stationary states to
dynamically perturbed configurations. It will be shown that the kind of ``core
and sheath'' model introduced in early work\cite{Copeland_et88} can be
obtained in a very different manner using the regularisation scheme developed
here, which places it on a firmer footing and shows how it can be adapted in
such a way as to be fully compatible with the more realistic mechanical models
that are now available\cite{CarterPeter95} or that may be introduced in the
future to deal for example with string phenomena in which several independent
currents may be involved\cite{Carter94}.
The present treatment deals only with the leading order in the relevant
field gradients. This means that it does not incorporate radiation backreaction,
whose inclusion would require extending the scheme to the next differential
order, which is left for future work.
\section{ Overview of the problem.}
The primary issue dealt with here is the evaluation of the right hand side of
the two dimensional analogue of the familiar relativistic generalisation of
Newtons ``second law'' of motion for a point particle of mass $\mas$ and charge
$\qq$, which is given in terms of the unit tangent vector $\ue^\mu$ of the
worldline by
\begin{equation}} \newcommand{\fe}{\end{equation}\mas \acc_\mu=\qq \bF_{\mu\nu}u^\nu\, ,\hskip 1 cm
\acc^\mu=\ue^\nu\nabl_\nu\ue^\mu \, , \label}\newcommand{\bel}{\begin{equation}\label{0.0}\fe
for an electromagnetic field $\bF_{\mu\nu}=2\nabl_{[\mu}\bA_{\nu]}$, where
the gravitational field is allowed for via the connection that specifies the
Riemannian covariant differentiation operator $\nabl$.
In the case of a string with worldsheet stress energy tensor $\bar T} \def\hovT{\widehat T} \def\toT{\widetilde T{^{\mu\nu}}$
and electric surface current $\oj{^\nu}$, the corresponding the general
purpose ``second law'' of extrinsic motion that governs the evolution of its
worldsheet is given\cite{Carter89}\cite{Carter95} by
\begin{equation}} \newcommand{\fe}{\end{equation}\bar T} \def\hovT{\widehat T} \def\toT{\widetilde T{^{\mu\nu}}\Ke_{\mu\nu}^{\,\ \ \rho}=\ag^{\!\rho\mu}
\bF_{\mu\nu}\oj{^\nu} \ , \label}\newcommand{\bel}{\begin{equation}\label{0.1}\fe
where $\Ke_{\mu\nu}^{\,\ \ \rho}$ is the second fundamental tensor of the
worldsheet, and $\ag^{\!\rho}_{\,\mu}$ is the tensor of orthogonal projection .
The latter is defined in terms of the complementary tensor $\eta} \def\ag{\perp\!} \def\nabl{\nabla\!} \def\onab{\ov\nabla\!^{\rho}_{\
\mu}$ of tangential projection, i.e. the first fundamental tensor, simply by
$\eta} \def\ag{\perp\!} \def\nabl{\nabla\!} \def\onab{\ov\nabla\!^{\rho}_{\ \mu} + \ag^{\!\rho}_{\,\mu} = \bg^\rho_{\ \mu}$,
where $\bg_{\mu\nu}$ is the background spacetime metric. The first
fundamental tensor itself is definable as the square $\eta} \def\ag{\perp\!} \def\nabl{\nabla\!} \def\onab{\ov\nabla\!^{\rho}_{\ \mu}=
{\cal E}} \def\psie{\psi^\rho_{\ \nu}{\cal E}} \def\psie{\psi^\nu_{\ \mu}$ of the antisymmetric alternating tensor
which is specifiable by the formula ${\cal E}} \def\psie{\psi^{\mu\nu}=
2\ue^{[\mu}\ve^{\nu]}$ in terms of (but -- except for its orientation --
independently of the choice of) an orthonormal worldsheet tangential diad
consisting of a timelike unit vector, $\ue^\mu$ say, and the dually related
spacelike vector unit vector $\ve^\mu$, as characterised
by $\ue^\nu \ue_\nu=-1$, $\ve^\nu\ve_\nu=1$, $\ue^\nu\ve_\nu=0$. The second
fundamental tensor is defined in terms of the first one by
\begin{equation}} \newcommand{\fe}{\end{equation}\Ke_{\mu\nu}^{\,\ \ \rho}=\eta} \def\ag{\perp\!} \def\nabl{\nabla\!} \def\onab{\ov\nabla\!^\sigma_{\ \nu}\onab_\mu\eta} \def\ag{\perp\!} \def\nabl{\nabla\!} \def\onab{\ov\nabla\!^\rho_{\ \sigma}
\ , \hskip 1 cm \onab_\mu=\eta} \def\ag{\perp\!} \def\nabl{\nabla\!} \def\onab{\ov\nabla\!^\nu_{\ \mu}\nabl_\nu\ .\label}\newcommand{\bel}{\begin{equation}\label{0.2}\fe
As the natural generalisation to two dimensions of the familiar ``second
law'' (\ref{0.0}) the extrinsic dynamical equation (\ref{0.1}) should in
principle be applicable -- in the thin limit when higher order
longitudinal derivatives are negligible -- for any kind of string
model, including even the relatively complicated kind that would be
needed to describe a warm, dissipatively conducting terrestrial power
cable. Like its 1-dimensional analogue (\ref{0.0}), the 2-dimensional
equation of motion (\ref{0.1}) is straightforwardly applicable so long
as the field $\bF_{\mu\nu}$ can be considered to be of purely external
origin. However, again like (\ref{0.0}), the equation (\ref{0.1}) ceases
to be so obviously meaningful when one notices that although the locally
source free external contribution, $\extF_{\mu\nu}$ say, will be well behaved,
the self induced contribution $\widehat F} \def\extF{\widetilde F_{\mu\nu}$ to the total $\bF_{\mu\nu}=
\widehat F} \def\extF{\widetilde F_{\mu\nu}+\extF_{\mu\nu}$ will be divergent in the zero thickness limit.
For a realistic treatment one must recognise that, in a high resolution
description, the underlying physical system (which for a cosmic string will be
some kind of vortex defect of the vacuum) will in fact have a finite
effective thickness. This will provide a natural ``ultraviolet'' cut off
radius, $\rast$ say, for a scheme whereby the regularised value of any field
is covariantly specified as the average of its values at a distance $\rast$
in directions orthogonal to the world sheet.
In the point particle case the regularised field tensor $\widehat F} \def\extF{\widetilde F_{\mu\nu}$ will
have the form
\begin{equation}} \newcommand{\fe}{\end{equation} \widehat F} \def\extF{\widetilde F_{\mu\nu}={\qq\over\rast} \acc_{[\mu}\ue_{\nu]}\ ,\label}\newcommand{\bel}{\begin{equation}\label{0.4}\fe
in which -- as throughout the present discussion --
we are neglecting the higher derivative contributions that would be
needed to take account of radiation backreaction. If so desired,
the self interaction force due to (\ref{0.4}) in (\ref{0.0}) can be
allowed for by transferring the relevant term from the right to the
left and then absorbing it into the first term by a mass
renormalisation of the usual kind $\mas \mapsto \tmas$ where
\begin{equation}} \newcommand{\fe}{\end{equation}\tmas=\mas + {\qq^2\over 2 \rast} \, .\label}\newcommand{\bel}{\begin{equation}\label{0.5}\fe
Our purpose here is to carry out the analogous regularisation procedure -- and
thus to provide the option of a corresponding renormalisation if so desired --
for the two dimensional case of a string. In this case the leading
``ultraviolet'' divergence is not linear but only logarithmic, so that as well
as the radius $\rast$ characterising the effective radius of the string (or to
be more specific, that of its current distribution) one also needs a
macroscopic ``infrared'' cut off length, $\Rast$ say. It fortunately turns out
that no anomally arises in the leading order ``ultraviolet'' divergence
considered here, despite the fact that there is not any way of actually
performing the ``infrared'' cut off in a strictly covariant manner.
(It will not be so easy to avoid trouble with Lorentz invariance at the higher
order that would be needed for treating radiation backreaction.)
\section{Overview of the solution.}
The elegantly covariant result one obtains at leading order is as follows.
To start with -- as was remarked at the outset by Witten\cite{Witten85}, and
will be made obvious below -- the leading contribution to the regularised self
field $\widehat A} \def\extA{\widetilde A_\mu$ on the string world sheet will be expressible (using an
appropriate gauge) in terms of of the surface current $\oj{^\mu}$ there by a
simple proportionality relation of the form
\begin{equation}} \newcommand{\fe}{\end{equation} \widehat A} \def\extA{\widetilde A_\mu =\widehat{ l}\ } \def\Lambde{\Lambda\oj_\mu\label}\newcommand{\bel}{\begin{equation}\label{0.6}\fe
in which $\widehat{ l}\ } \def\Lambde{\Lambda$ is a dimensionless constant of the familiar form
\begin{equation}} \newcommand{\fe}{\end{equation} \widehat{ l}\ } \def\Lambde{\Lambda=2\ln\big\{ {\Rast/\rast} \big\}\, .\label}\newcommand{\bel}{\begin{equation}\label{0.7}\fe
The mathematically non-trivial part of the problem is the derivation of
the corresponding regularised field $\widehat F} \def\extF{\widetilde F_{\mu\nu}$. The worldsheet
tangential part can be obtained in the usual way by derivation of $\widehat A} \def\extA{\widetilde A_\mu$,
but since the regularised value is defined only on the worldsheet itself, it
cannot be directly differentiated in non tangential directions:
$\onab_\nu\widehat A} \def\extA{\widetilde A_\mu$ is directly meaningful, but $\nabl_\nu\widehat A} \def\extA{\widetilde A_\mu$ is not.
One can however obtain the required result by going through the regularisation
procedure again at first differentiated order.
The final result is most conveniently expressible in terms of the self-dual
part
\begin{equation}} \newcommand{\fe}{\end{equation} \Ce_{\mu\nu}^{\,\ \ \rho}={_1\over^2}\big(
\Ke_{\mu\nu}^{\,\ \ \rho}+\tilK_{\mu\nu}^{\,\ \ \rho}\big) \fe
of the second fundamental tensor, where the dual is defined by
\begin{equation}} \newcommand{\fe}{\end{equation} \tilK_{\mu\nu}{^\rho}=
{\cal E}} \def\psie{\psi_{\mu\sigma}{\cal E}} \def\psie{\psi_{\nu\tau}\Ke^{\sigma\tau\rho}=\Ke_{\mu\nu}{^\rho}
-\eta} \def\ag{\perp\!} \def\nabl{\nabla\!} \def\onab{\ov\nabla\!_{\mu\nu}\Ke^\rho \, ,\fe
in which $\Ke^\rho$ is the trace vector,
\begin{equation}} \newcommand{\fe}{\end{equation}\Ke^\rho=\Ke_\mu^{\ \mu\rho}=\onab_\mu\eta} \def\ag{\perp\!} \def\nabl{\nabla\!} \def\onab{\ov\nabla\!^{\mu\rho} \label}\newcommand{\bel}{\begin{equation}\label{0.9} \fe
(whose vanishing, $K^\mu=0$, would express the dynamical equations of motion
for a string model of the simplest Goto-Nambu type in the absence of any
external force or self interaction). The tensor $\Ce_{\mu\nu}^{\,\
\ \rho}$ is identifiable as the conformally invariant {\it conformation
tensor} whose original definition\cite{Carter92},
$ \Ke_{\mu\nu}^{\,\ \ \rho}=\Ce_{\mu\nu}^{\,\ \ \rho}+{_1\over ^2}
\eta} \def\ag{\perp\!} \def\nabl{\nabla\!} \def\onab{\ov\nabla\!_{\mu\nu}K^\rho$,
shows that it is trace free, $\Ce_{\mu}^{\,\ \mu\rho}=0$.
Although $\Ke^\rho$ will be usually be non zero in electromagnetically
interacting string models, its contribution cancels out in the final formula
for the self field, which reduces, as shown below, simply to
\begin{equation}} \newcommand{\fe}{\end{equation} \widehat F} \def\extF{\widetilde F_{\mu\nu}=2\eta} \def\ag{\perp\!} \def\nabl{\nabla\!} \def\onab{\ov\nabla\!_{\,[\mu}{^\rho}\onab_{\nu]}\widehat A} \def\extA{\widetilde A_\rho
+2\widehat A} \def\extA{\widetilde A^\rho \Ce_{\rho[\mu\nu]} \, .\label}\newcommand{\bel}{\begin{equation}\label{0.10}\fe
\section{Retarded Green measure for a string.}
To carry out the first differential order treatment given here (not to mention
the second differential order treatment that will need to be carried out in
future work to allow for radiation backreaction) the essential tool is the
appropriate Green formula for the relevant retarded solution. Since we are
concerned here only with the short range ``ultraviolet'' divergence
contribution it will suffice to use the flat space Green formula.
(For our present purpose it would make no difference if we used
the advanced Green formula: it is only at the next higher differential order
that the distinction becomes important.)
For an an isolated particle with charge $\qq$ the field will be given
simply by
\begin{equation}} \newcommand{\fe}{\end{equation} \bA^\mu={\qq\ue^\mu\over\retr^\nu \ue_\nu}\, , \label}\newcommand{\bel}{\begin{equation}\label{1.4}\fe
where $\ue^\mu$ is the future directed unit tangent vector to the trajectory
at the (unique) point where it is intersected by the past null cone from the
point of interest, and $\retr^\mu$ is the past directed null vector
-- as characterised by $\retr^\nu\retr_\nu=0$, $\retr^\nu \ue_\nu>0$ --
from this ``observation position'' to the intersection point on the worldline.
What we need is the analogous formula for a string, as provided in the
appropriate limit by the well known Li\'enard-Wiechert formula which solves
the source equation in the hyperbolic version
\begin{equation}} \newcommand{\fe}{\end{equation}\nabl_\nu\nabl^{\,\nu} A^\mu=-4\pi\hj^\mu\, ,\label}\newcommand{\bel}{\begin{equation}\label{1.5}\fe
(obtained from the gauge invariant version $\nabl_\nu
F^{\mu\nu}=4\pi\hj^\mu$ by imposition of the Lorentz condition $\nabl_\nu
A^\nu=0$) for a smooth current distribution $\hj{^\mu}$, in a background with
Minkowski metric $ ds^2=$ $(dx^{_1})^2+(dx^{_2})^2+(dx^{_3})^2-(dx^{_0})^2$,
in the form
\begin{equation}} \newcommand{\fe}{\end{equation}\bA^\mu=\int\int\int \hj{^\mu}
\, { dx^{_1}\, dx^{_2}\, dx^{_3}\over \retr_{_0} }\, . \label}\newcommand{\bel}{\begin{equation}\label{1.6}\fe
(Though not manifestly covariant, this formula does in fact give a well
defined Lorentz covariant result for a suitably bounded source
distribution.)
For the limiting case of a distribution confined to a string, with surface
current density $\oj{^\nu}$ that is tangential to the worldsheet, the result
will evidently be obtainable as an integral along the one dimensional curve
where the relevant (past or future) null cone from the point under
consideration intersects the worldsheet. With respect to a suitably
normalised parameter ${\cal G}$ this curve will have a tangent vector given by
\begin{equation}} \newcommand{\fe}{\end{equation} d\bar x^\mu=\tildr^\mu\, d{\cal G}\, , \hskip 1 cm
\tildr^\mu={\cal E}} \def\psie{\psi^{\mu\nu}r_\nu\, ,\label}\newcommand{\bel}{\begin{equation}\label{1.10}\fe
where, as before, $\retr^\mu=\bar x^\mu-x^\mu$ is the {\it retarded
null vector} specifying the ray that goes from the external observation
position $x^\mu$ to the relevant point $\bar x^\mu$ on the worldsheet.
It turns out that the dimensionless parameter ${\cal G}$ introduced in this way
directly specifies the relevant {\it Green measure}, in terms of which
the required analogue of (\ref{1.4}) giving the contribution
from a finite segment with beginning and end labeled by $-$ and $+$
will be given simply by
\begin{equation}} \newcommand{\fe}{\end{equation}\bA^\mu=\int_-^+ \oj{^\mu}\, {d{\cal G}}\, .\label}\newcommand{\bel}{\begin{equation}\label{1.7}\fe
The Green measure given by (\ref{1.10}) can be seen to be related to the
ordinary positive indefinite {\it proper} length measure $ds$ along the
curve by
\begin{equation}} \newcommand{\fe}{\end{equation} d{\cal G}={ds\over r_{\!\!_\perp}\!} \, ,\hskip 1 cm r_{\!\!_\perp}\!
=\sqrt{\tildr^\mu\tildr_\mu} =\sqrt{\ag^{\!\mu\nu}\retr_\mu\retr_\nu}
\, .\label}\newcommand{\bel}{\begin{equation}\label{1.8}\fe
In terms of worldsheet coordinates $\sigma} \def\tor{\tau^\ii$ and of the corresponding
2-dimensional intrinsic metric given by $\hg_{\ii\ij}= g_{\mu\nu}\bar
x^\mu{_{,\ii}} \bar x^\nu{_{,\ij}}$ (using a comma to denote partial
differentiation) the fundamental tensor will be given simply by $\eta} \def\ag{\perp\!} \def\nabl{\nabla\!} \def\onab{\ov\nabla\!^{\mu\nu}
=\hg^{\ii\ij}\bar x^\mu{_{,\ii}} \bar x^\nu{_{,\ij}}$, so
this works out as $d{\cal G}=$ $\vert\hg^{\ii\ij}\bar
x^\nu{\!_{,\ii}} \bar x^\rho{\!_{,\ij}}\retr_\nu \retr_\rho\vert^{-1/2}
\sqrt{\hg_{\ih\ik}\, d\sigma} \def\tor{\tau^\ih\, d\sigma} \def\tor{\tau^\ik }$.
Unlike (\ref{1.10}), the explicit formula (\ref{1.8}) becomes
indeterminate wherever the tangent direction $\tildr^\mu$ is null so
that the orthogonally projected distance $ r_{\!\!_\perp}\!$ vanishes. This can be
overcome at the expense of manifest covariance by referring to a
worldsheet frame consisting of any future directed timelike unit
tangent vector $\ue^\mu$, and the dually related spacelike unit vector
$\ve^\mu= {\cal E}} \def\psie{\psi^{\mu\nu}\ue_\nu$, which gives the everywhere well
behaved expression $d{\cal G}= (u_\nu r^\nu)^{-1}\ve_\mu d\bar x^\mu$.
What is commonly quoted in the literature is a more specialised
version\cite{Copeland_et90} that is given by $d{\cal G}= (\dot x^\nu r_\nu)^{-1}
\, d\sigma} \def\tor{\tau$, but that is valid only for coordinates $\sigma} \def\tor{\tau$, $\tau$ of {\it
conformal} type (meaning that $\dot x^\nu\dot x_{\,\nu}
+x^{\prime\nu}x^\prime_{\,\nu}=0$, $\dot x^\nu x^\prime_{\,\nu}=0$, where
$\dot x^\nu=\partial \bar x^\nu/\partial\tor$, $x^{\prime\nu}=\partial \bar
x^\nu/\partial\sigma} \def\tor{\tau$).
\section{ Regularised gradient operator for self field on worldsheet.}
It is almost immediately evident that the leading order contribution
to the regularised effective gauge potential on the string will be given
by an expression of the form (\ref{0.6}). The non-trivial part of the
calculation is the evaluation the corresponding effective gradient components
as analogously regularised by averaging over an infinitesimal cirle on which
$ r_{\!\!_\perp}\!=\rast$. The terms that remain at leading order give a result of the
form
\begin{equation}} \newcommand{\fe}{\end{equation}\widehat{\nabl_\nu A}{^\mu}= \widehat{ l}\ } \def\Lambde{\Lambda\Big( \onab_\nu\oj{^\mu}
+{_1\over ^2}\Ke_\nu \oj{^\mu}\Big)\, ,\label}\newcommand{\bel}{\begin{equation}\label{2.6}\fe using the same
regularisation coefficient $\widehat{ l}\ } \def\Lambde{\Lambda$ as was introduced in (\ref{0.6}).
For the tangentially differentiated components in the first term, the
postulate that this coefficient should be constant ensures that the
result of the regularisation procedure will be consistent with what one
would get by direct tangential differentiation of the regularised
components in the world sheet. This means that (\ref{2.6}) will be
expressible by $\widehat{\nabl_\nu A{^\mu}}= \widehat\nabl_\nu
\widehat A} \def\extA{\widetilde A{^\mu}$ in terms of the {\it regularised gradient operator} given
by \begin{equation}} \newcommand{\fe}{\end{equation}\widehat\nabl_\nu= \onab_\nu+ {_1\over ^2}\Ke_\nu \, .\label}\newcommand{\bel}{\begin{equation}\label{2.14}\fe
With this notation the required self interaction field will be
given simply by
\begin{equation}} \newcommand{\fe}{\end{equation}\widehat F} \def\extF{\widetilde F_{\mu\nu}=2\widehat\nabl_{[\mu}\widehat A} \def\extA{\widetilde A_{\nu]} \ .\label}\newcommand{\bel}{\begin{equation}\label{2.15}\fe
Reorganising this regularised effective field in terms of its distinct purely
tangential and mixed components (it has no purely orthogonal one) using
the tangentiality property $\ag^{\!\mu}_{\, \nu}\widehat A} \def\extA{\widetilde A^\nu=0$ implied by
(\ref{0.6}), which entails the identity
$\ag^{\!\rho}_{\,\nu}\onab_\mu\widehat A} \def\extA{\widetilde A{^\nu}=$ $\widehat A} \def\extA{\widetilde A^\nu \Ke_{\mu\nu}^{\, \ \
\rho}$, one finally obtains the quoted formula (\ref{0.10}).
\section{ Core and sheath representation.}
Having obtained the main result (\ref{2.15}), or equivalently (\ref{0.10}),
that was required, one naturally hopes to find that, as in the point particle
case, the effect of the ensuing self force contribution will be absorbable
into the left hand side of the ``second law'' equation (\ref{0.1}) by the use
of an appropriate two dimensional analogue of the renormalisation (\ref{0.5}).
The concept of such an adjustment was already introduced in the early work of
Copeland, Haws, Hindmarsh, and Turok\cite{Copeland_et88} who developped a
treatment whereby the current carrying cosmic string of strictly confined
``local type'' was considered as forming the core of a composite string with
an outer sheath of the extended ``global type'' constituted by its own
electromagnetic field. In a compound ``core and sheath'' string model of this
kind, the outer electromagnetic sheath provides an extra contribution to the
effective tension and energy per unit length that will be representable by a
worldsheet stress energy tensor of the form
\begin{equation}} \newcommand{\fe}{\end{equation}\hovT{^{\mu\nu}}=\widehat{ l}\ } \def\Lambde{\Lambda\big(\oj{^\mu} \oj{^\nu}- {_1\over ^2}\oj{^\rho}
\oj_\rho\eta} \def\ag{\perp\!} \def\nabl{\nabla\!} \def\onab{\ov\nabla\!^{\mu\nu} \big) \ ,\label}\newcommand{\bel}{\begin{equation}\label{1.2}\fe
where $\widehat{ l}\ } \def\Lambde{\Lambda$ is a dimensionless regularisation coefficient of the same form
(\ref{0.7}) as the one introduced above. Despite their mathematical similarity
at the end, it is to be noticed that the physical origin of the coefficient in
(\ref{0.6}) was very different from that of the one in (\ref{1.2}), which came
from an integral over a section through the field {\it outside} the current
distribution, whereas the $\widehat{ l}\ } \def\Lambde{\Lambda$ in (\ref{0.6}) comes from an integral with
support confined to the current carrying core.
In view of this distinction, it is remarkable that the two approaches turn out
in the end to agree perfectly. The form of the self field (\ref{0.10}) is
easily be seen to be such that the corresponding self force contribution in
(\ref{0.1}) can indeed be taken over to the left hand side and absorbed into
the first term by an adjustment of the form $\bar T} \def\hovT{\widehat T} \def\toT{\widetilde T{^{\mu\nu}}
\mapsto\toT{^{\mu\nu}}$, in which the required two dimensional analogue of the
mass renormalisation (\ref{0.5}) turns out to have precisely the form that
describes the total in the compound core and sheath model, namely
\begin{equation}} \newcommand{\fe}{\end{equation}\toT{^{\mu\nu}}=\bar T} \def\hovT{\widehat T} \def\toT{\widetilde T{^{\mu\nu}}+\hovT{^{\mu\nu}} \, ,\label}\newcommand{\bel}{\begin{equation}\label{1.3}\fe
where the constant coefficient $\widehat{ l}\ } \def\Lambde{\Lambda$ in (\ref{1.2}) is to be identified
precisely with the constant coefficient $\widehat{ l}\ } \def\Lambde{\Lambda$ in (\ref{0.6}).
This demonstration that it agrees with the result of the more refined analysis
presented here puts the compound ``core and sheath'' type of model on a firmer
footing than hitherto. The agreement is not limited to the prediction of the
extrinsic equation of motion discussed above, but still holds just as well for
the internal dynamics of the two kinds of model. To see this, let us consider
the dynamical evolution of the loop in an electromagnetic background that may
include a locally source free external field contribution $\extF_{\mu\nu}=$
$\nabl_{[\mu}\extA_{\nu]}$ as well as the self field contribution
(\ref{0.10}), so that the relevant effective total to be substituted in
(\ref{0.1}) is
\begin{equation}} \newcommand{\fe}{\end{equation} \bF_{\mu\nu}=\widehat F} \def\extF{\widetilde F_{\mu\nu}+ \extF_{\mu\nu}\, .\label}\newcommand{\bel}{\begin{equation}\label{3.1}\fe
The extrinsic dynamical equation (\ref{0.1}) is not sufficient
by itself to determine the evolution of the string configuration:
it is also necessary to know the equations of motion of the surface current
$\oj{^\mu}$ and any other independent internal fields on which the worldsheet
stress energy tensor $\bar T} \def\hovT{\widehat T} \def\toT{\widetilde T{^{\mu\nu}}$ may depend. Whatever the
nature of the internal fields, the internal momentum-energy
transport equation
\begin{equation}} \newcommand{\fe}{\end{equation} \eta} \def\ag{\perp\!} \def\nabl{\nabla\!} \def\onab{\ov\nabla\!^\rho_{\ \mu}\nabl_\nu\bar T} \def\hovT{\widehat T} \def\toT{\widetilde T{^{\mu\nu}}=\eta} \def\ag{\perp\!} \def\nabl{\nabla\!} \def\onab{\ov\nabla\!^{\rho\mu}\bF_{\mu\nu}
\oj{^\nu} \label}\newcommand{\bel}{\begin{equation}\label{3.2}\fe
must be satisfied.
Writing the ``sheath'' contribution (\ref{1.2}) in the form
\begin{equation}} \newcommand{\fe}{\end{equation}\hovT{^\nu}_\mu=\oj{^\nu}\widehat A} \def\extA{\widetilde A_\mu- {_1\over ^2}
\oj{^\rho}\widehat A} \def\extA{\widetilde A_\rho \eta} \def\ag{\perp\!} \def\nabl{\nabla\!} \def\onab{\ov\nabla\!^{\nu}_{\ \mu} \, ,\label}\newcommand{\bel}{\begin{equation}\label{3.3}\fe
one sees that, subject to the surface current conservation condition
\begin{equation}} \newcommand{\fe}{\end{equation} \onab_\nu \oj{^\nu}=0 \, , \label}\newcommand{\bel}{\begin{equation}\label{3.4}\fe
the self force contribution on the right in (\ref{3.2}) can also be taken over
to the left and absorbed into the first term by the same adjustment
$\bar T} \def\hovT{\widehat T} \def\toT{\widetilde T{^{\mu\nu}} \mapsto \toT{^{\mu\nu}}$ as before, where $\toT{^{\mu\nu}}$
is the total for the compound core and sheath model as given by (\ref{1.3}).
After performing such a transfer both for the extrinsic dynamical
equation (\ref{0.1}) and the extrinsic dynamical equation (\ref{3.2})
the results can be combined in a total force law of the form
\begin{equation}} \newcommand{\fe}{\end{equation} \onab_\nu\toT{^\nu}_{\!\mu}= \extF_{\mu\nu}\oj{^\nu}\, ,\label}\newcommand{\bel}{\begin{equation}\label{3.5}\fe
in which only the locally source free purely external field contribution
$\extF_{\mu\nu}$ is involved on the right hand side.
\section{ Action formulation for simply conducting case.}
What has done so far is valid even for very general string models involving
multiple conductivity and finite resistivity as in terrestrial power cables.
Let us now restrict attention to the simple strictly conservative kind of
model that is appropriate for a Witten type superconducting string, in which
the only independent dynamical variable can be taken to be a particle current
vector $c} \def\chie{\chi} \def\phie{\varphi} \def\pe{p} \def\pee{\bar p^\mu$ say, in terms of which the electromagnetic current will be
specifiable by a proportionality relation of the form
\begin{equation}} \newcommand{\fe}{\end{equation} \oj{^\mu}=\qq\,c} \def\chie{\chi} \def\phie{\varphi} \def\pe{p} \def\pee{\bar p^\mu\, , \label}\newcommand{\bel}{\begin{equation}\label{2.1} \fe
where $\qq$ is the relevant mean charge per particle. Such a
model will be specified by a {\it master function}, $\Lambde$ say, with a
a -- typically non-linear\cite{CarterPeter95} -- dependence on the single scalar variable
$\chie=c} \def\chie{\chi} \def\phie{\varphi} \def\pe{p} \def\pee{\bar p^\nuc} \def\chie{\chi} \def\phie{\varphi} \def\pe{p} \def\pee{\bar p_\nu$, so that its variation determines a corresponding
momentum vector given by $\delta\Lambde=\pee_\nu\deltac} \def\chie{\chi} \def\phie{\varphi} \def\pe{p} \def\pee{\bar p^\nu$ with
$\pee_\mu={\cal K}} \def\tilK{\tilde{\cal K} c} \def\chie{\chi} \def\phie{\varphi} \def\pe{p} \def\pee{\bar p_\mu$ where ${\cal K}} \def\tilK{\tilde{\cal K}=2 d\Lambde/d\chie$.
When self interaction is neglible, the mechanics will be governed just by a
Lagrangian density scalar of the simple form
\begin{equation}} \newcommand{\fe}{\end{equation} \bar L} \def\toLag{\widetilde L =\Lambde + \qq\,c} \def\chie{\chi} \def\phie{\varphi} \def\pe{p} \def\pee{\bar p^\nu\bA_\nu \, ,\label}\newcommand{\bel}{\begin{equation}\label{3.6}\fe
whose variation determines a gauge dependent total momentum covector:
$\delta\bar L} \def\toLag{\widetilde L=\pie_\nu\delta c} \def\chie{\chi} \def\phie{\varphi} \def\pe{p} \def\pee{\bar p^\nu$ where
\begin{equation}} \newcommand{\fe}{\end{equation} \pie_\nu=\pee_\nu+\qq\bA_\nu \, .\label}\newcommand{\bel}{\begin{equation}\label{3.7}\fe
In the application of the variation principle to the corresponding action
$\int \bar L} \def\toLag{\widetilde L\sqrt{\Vert\hg\Vert} \, d\sigma} \def\tor{\tau\, d\tor$, the current is not
an entirely free variable, but must be constrained so as to be conserved,
e.g. by taking it to be dual to the gradient of a freely variable stream function $\psie$
say, i.e. $c} \def\chie{\chi} \def\phie{\varphi} \def\pe{p} \def\pee{\bar p^\mu={\cal E}} \def\psie{\psi^{\mu\nu}\onab_\nu\psie$. For such a
model, the worldsheet stress energy tensor reqired for application of the
extrinsic equation of motion (\ref{0.1}) takes the form
\begin{equation}} \newcommand{\fe}{\end{equation} \bar T} \def\hovT{\widehat T} \def\toT{\widetilde T{^\mu}_{\nu}= c} \def\chie{\chi} \def\phie{\varphi} \def\pe{p} \def\pee{\bar p^\mu\pee_\nu+(\Lambde-c} \def\chie{\chi} \def\phie{\varphi} \def\pe{p} \def\pee{\bar p^\rho\pee_\rho)\eta} \def\ag{\perp\!} \def\nabl{\nabla\!} \def\onab{\ov\nabla\!^\mu_{\
\nu}\, . \label}\newcommand{\bel}{\begin{equation}\label{3.8}\fe
It is easy to see that appart from the kinematic particle conservation
law $\onab_\nuc} \def\chie{\chi} \def\phie{\varphi} \def\pe{p} \def\pee{\bar p^\nu=0$, whose consequence (\ref{3.4}) ensures that the loop
will be characterised by a conserved total charge,
$ Q=\oint{\cal E}} \def\psie{\psi_{\mu\nu}\,\oj{^\nu}\, dx^\mu$,
the remaining internal dynamical equations for the foregoing model will
consist just of the surface integrability condition
\begin{equation}} \newcommand{\fe}{\end{equation} \eta} \def\ag{\perp\!} \def\nabl{\nabla\!} \def\onab{\ov\nabla\!_{\,[\mu}{^\rho} \onab_{\nu]} \pie_\rho=0 \, , \label}\newcommand{\bel}{\begin{equation}\label{3.9}\fe
which ensures that the the tangentially projected part of the momentum
covector (\ref{3.7}) is proportional to the surface gradient of a
scalar, i.e. $ \eta} \def\ag{\perp\!} \def\nabl{\nabla\!} \def\onab{\ov\nabla\!^{\mu\nu}\pie_\nu=\she\onab^{\,\mu}\phie$,
where $\she$ is a fixed proportionality factor. It follows that the string
loop will be characterised by a conserved circuit integral given by
$ 2\piN} \def\pie{\pi} \def\she{ \sqrt{\kappa_{_0}}\, =\oint d\phie$. The inclusion of the proportionality constant
$\she$ allows the scalar $\phie$ to be adjusted\cite{CarterPeter95} in
order to agree with the phase of a boson condensate in an underlying
microscopic model, so that it will be periodic with period $2\pi$, in
which case $N} \def\pie{\pi} \def\she{ \sqrt{\kappa_{_0}}\, $ will be quantised as an integral winding number.
It can be seen that when the self interaction is taken into account using the
regularisation scheme given above, the foregoing system of equations will
still apply provided the field $\bA_\mu$ in the specification (\ref{3.7}) of
the total momentum per particle is interpreted in accordance with (\ref{3.1})
as the combination $\bA_\mu=\widehat A} \def\extA{\widetilde A_\mu+\extA_\mu$, in which $\extA_\mu$ is the
well behaved source free external contribution and $\widehat A} \def\extA{\widetilde A_\mu$ is the
regularised self field as given by (\ref{0.6}).
The same equations of motion can be obtained by a purely variational approach
(something that would not be possible at the next differential order when
radiation backreaction will be involved) within the framework of the ``core
and sheath'' model, whose dynamics will be expressible in variational form
using an appropriately modified Lagrangian given by $\toLag= \widetilde\Lambde
+\qqc} \def\chie{\chi} \def\phie{\varphi} \def\pe{p} \def\pee{\bar p{^\nu}\extA_\nu$ in which only the external field contribution
$\extA_\mu$ is involved, the self interaction contribution having been
absorbed into the uncoupled term by an additive renormalisation $\Lambde
\mapsto \widetilde{\Lambde}$ with
\begin{equation}} \newcommand{\fe}{\end{equation}\widetilde{\Lambde}=\Lambde+\widehat{\Lambde}\, , \hskip 1 cm
\widehat{\Lambde}={_1\over ^2}\,\widehat{ l}\ } \def\Lambde{\Lambda\qq^2c} \def\chie{\chi} \def\phie{\varphi} \def\pe{p} \def\pee{\bar p^\nuc} \def\chie{\chi} \def\phie{\varphi} \def\pe{p} \def\pee{\bar p_\nu\, .\label}\newcommand{\bel}{\begin{equation}\label{3.14}\fe
The ``sheath'' contribution, $\widehat{\Lambde}=$ ${1\over2}\,\widehat{ l}\ } \def\Lambde{\Lambda\qq^2
\chie$, is interpretable as the effective action density of the self generated
electromagnetic field, as evaluated\cite{Copeland_et88} by integration across
an external section. The effective momentum per particle will undergo a
corresponding adjustment $\pee_\mu\mapsto \widetilde\pe_\mu$ with
$\widetilde\pe_\mu= \pee_\mu+\widehat\pe_\mu$ where the ``sheath''
contribution is $\widehat\pe_\mu=\widehat{ l}\ } \def\Lambde{\Lambda\qq^2c} \def\chie{\chi} \def\phie{\varphi} \def\pe{p} \def\pee{\bar p_\mu$. The corresponding gauge
dependent total, $\widetilde\pie_\mu=$ $\widetilde\pe_\mu+\qq\extA_\mu$, is
the same as was given by (\ref{3.7}), i.e. one obtains
$\widetilde\pie_\mu=\pie_\mu$, so no adjustment is needed for the phase scalar
$\phie$ or its conserved winding number $N} \def\pie{\pi} \def\she{ \sqrt{\kappa_{_0}}\, $.
I wish to thank P. Peter, M. Sakellariadou, and A. Gangui for illuminating
discussions.
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ACCEPTED
#### According to
The Catalogue of Life, 3rd January 2011
#### Published in
Austral. J. Bot. 13:115. 1965
#### Original name
Agati formosa F.Muell.
### Remarks
null
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"redpajama_set_name": "RedPajamaGithub"
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NEW YORK (AP) — Missy Elliott knew she was being honoured by Essence — but not who was giving her the accolade. So there was plenty of emotion when close friend Janet Jackson not only handed her her trophy but praised her talent.
"Some rhyme, some rap; some act, some choreograph; some write hit songs; some create whole new sounds. Some women are able to make her mark in some of these fields; but there's only one woman who has made her mark in all of these fields," Jackson told a packed crowd on Thursday night.
Elliott — whose hits include "Work It" "Get Ur Freak On" and "One Minute Man" — began to cry during Jackson's tribute. The two shared a long embrace when Elliott went on stage for her "Visionary" award from Essence at the magazine's annual pre-Grammy event celebrating black women artists.
"That was a surprise," an overwhelmed and teary Elliott said.
She noted that some in the industry didn't think she could be successful — at times because she was a woman and a producer, at others because she didn't look like what was seen as traditionally beautiful. She recounted that one time she was even replaced her with someone who they thought was.
Among the celebrities in attendance were Janelle Monae, T.I. and "Orange Is the New Black" actress Danielle Brooks who turned out in an Adidas blue track suit in honor of one of Elliott's many fashion looks over the years.
The Essence event is just one of dozens being held around New York City ahead of Sunday's 60th annual Grammy awards, to be held at Madison Square Garden.
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"redpajama_set_name": "RedPajamaC4"
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It is a pleasure to read your satisfaction about our location and our breakfast.
We would like to thank you for this comment on Tripadvisor and we hope to see you soon in our establishment.
We thank you very much for this comment on Tripadvisor. We are glad to know that you enjoyed your stay and I will share your satisfaction with my staff as soon as possible.
We hope to see you again at the Mercure Rouen Centre Cathedrale.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 7,701
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The year 1763 in science and technology involved some significant events.
Astronomy
Publication posthumously of Nicolas Louis de Lacaille's Coelum australe stelliferum, cataloguing all his data from the southern hemisphere and including about 10,000 stars and a number of brighter star clusters and nebulae.
Publication of Edward Stone's The whole doctrine of parallaxes explained and illustrated by an arithmetical and geometrical construction of the transit of Venus over the sun, June 6th, 1761. Enriched with a new and general method of determining the places where any transit of this planet, and especially that which will be June 3d, 1769, may be best observed.
Mathematics
December 23 – Thomas Bayes' solution to a problem of "inverse probability" is presented posthumously in his "Essay towards solving a Problem in the Doctrine of Chances" read by Richard Price to the Royal Society, containing a statement of a special case of Bayes' theorem.
Medicine
Edward Stone publishes his discovery of the medicinal properties of salicylic acid.
Awards
Copley Medal: Not awarded
Births
January 31 (bapt.) – John Brinkley, English astronomer (died 1835)
May 12 – John Bell, Scottish surgeon (died 1820)
May 16 – Louis Nicolas Vauquelin, French chemist (died 1829)
August 16 – Giovanni Battista Guglielmini, Bolognese physicist (died 1817)
October 27 – William Maclure, Scottish American geologist (died 1840)
December 25 – Claude Chappe, French engineer (died 1805)
William Higgins, Irish chemist (died 1825)
Deaths
March 5 – William Smellie, Scottish obstetrician (born 1697)
July 11 – Peter Forsskål, Swedish naturalist (born 1732)
References
18th century in science
1760s in science
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{
"redpajama_set_name": "RedPajamaWikipedia"
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\section{Acknowledgments}
The DELVE project is partially supported by Fermilab LDRD project L2019-011 and the NASA Fermi Guest Investigator Program Cycle 9 No. 91201.
This work is supported by the Visiting Scholars Award Program of the Universities Research Association.
ABP acknowledges support from NSF grant AST-1813881.
This research received support from the National Science Foundation (NSF) under grant no. NSF DGE-1656518 through the NSF Graduate Research Fellowship received by SM.
JLC acknowledges support from NSF grant AST-1816196.
JDS acknowledges support from NSF grant AST-1714873.
SRM acknowledges support from NSF grant AST-1909497.
DJS acknowledges support from NSF grants AST-1821967 and 1813708.
DMD acknowledges financial support from the State Agency for Research
of the Spanish MCIU through the ``Centre of Excellence Severo Ochoa''
award for the Instituto de Astrofísica de Andaluc\'ia (SEV-2017-0709).
CPMB and MRLC acknowledge support from the European Research Council (ERC) under the European Union's Horizon 2020 research and innovation programme (grant agreement no.\ 682115).
This project used data obtained with the Dark Energy Camera (DECam), which was constructed by the Dark Energy Survey (DES) collaboration.
Funding for the DES Projects has been provided by
the DOE and NSF (USA),
MISE (Spain),
STFC (UK),
HEFCE (UK),
NCSA (UIUC),
KICP (U. Chicago),
CCAPP (Ohio State),
MIFPA (Texas A\&M University),
CNPQ,
FAPERJ,
FINEP (Brazil),
MINECO (Spain),
DFG (Germany),
and the collaborating institutions in the Dark Energy Survey, which are
Argonne Lab,
UC Santa Cruz,
University of Cambridge,
CIEMAT-Madrid,
University of Chicago,
University College London,
DES-Brazil Consortium,
University of Edinburgh,
ETH Z{\"u}rich,
Fermilab,
University of Illinois,
ICE (IEEC-CSIC),
IFAE Barcelona,
Lawrence Berkeley Lab,
LMU M{\"u}nchen, and the associated Excellence Cluster Universe,
University of Michigan,
NSF's National Optical-Infrared Astronomy Research Laboratory,
University of Nottingham,
Ohio State University,
OzDES Membership Consortium
University of Pennsylvania,
University of Portsmouth,
SLAC National Lab,
Stanford University,
University of Sussex,
and Texas A\&M University.
This work has made use of data from the European Space Agency (ESA) mission {\it Gaia} (\url{https://www.cosmos.esa.int/gaia}), processed by the {\it Gaia} Data Processing and Analysis Consortium (DPAC, \url{https://www.cosmos.esa.int/web/gaia/dpac/consortium}).
Funding for the DPAC has been provided by national institutions, in particular the institutions participating in the {\it Gaia} Multilateral Agreement.
Based on observations at Cerro Tololo Inter-American Observatory, NSF's National Optical-Infrared Astronomy Research Laboratory (2019A-0305; PI: Drlica-Wagner), which is operated by the Association of Universities for Research in Astronomy (AURA) under a cooperative agreement with the National Science Foundation.
This manuscript has been authored by Fermi Research Alliance, LLC, under contract No.\ DE-AC02-07CH11359 with the US Department of Energy, Office of Science, Office of High Energy Physics. The United States Government retains and the publisher, by accepting the article for publication, acknowledges that the United States Government retains a non-exclusive, paid-up, irrevocable, worldwide license to publish or reproduce the published form of this manuscript, or allow others to do so, for United States Government purposes.
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\section{Introduction}
\label{sec:intro}
The standard model of cosmology (\ensuremath{\rm \Lambda CDM}\xspace) is strongly supported by observations at large spatial scales \citep[e.g.,][]{Planck:2018,DES:2018}.
However, this fundamental model for the growth and evolution of our Universe remains challenging to test on scales smaller than our Milky Way.
Starting with the Sloan Digital Sky Survey \citep[SDSS;][]{York:2000}, large digital sky surveys have revolutionized our understanding of galaxies with stellar masses $\lesssim 10^5 \unit{M_\odot}$ (see recent reviews by \citealt{McConnachie:2012} and \citealt{Simon:2019}).
We now know that our Milky Way is surrounded by scores (and likely hundreds) of faint galaxies, which span orders of magnitude in luminosity \citep[e.g.,][and references therein]{Drlica-Wagner:2020}.
Discoveries of faint satellites around our nearest galactic neighbors have begun to extend these studies beyond the Milky Way \citep[e.g.\xspace,][]{Martin:2013,Chiboucas:2013,Muller:2015,Carlin:2016,Smercina:2018,Crnojevic:2019}.
Furthermore, we observe the tidal remnants of faint satellite galaxies traversing our own Galactic halo \citep[e.g.,][]{Ibata:2001a,Belokurov:2006,Shipp:2018} and the halos of other nearby galaxies \citep[e.g.,][]{Malin:1997,Ibata:2001b,Martinez-Delgado:2008,Mouhcine:2010}.
The discovery of the faintest galaxies and their remnants represents an observational frontier for large digital sky surveys, while the study of these systems continues to improve our understanding of \ensuremath{\rm \Lambda CDM}\xspace at the smallest observable scales.
The Dark Energy Camera \citep[DECam;][]{Flaugher:2015}, mounted on the 4\,m Blanco Telescope at Cerro Tololo Inter-American Observatory in Chile, is an exceptional instrument for exploring the faintest stellar systems.
The large field of view ($3 \deg^2$) and fast readout time ($27\unit{s}$) of DECam allow it to quickly image large areas of the sky.
DECam has been used by several large survey programs, including the Dark Energy Survey \citep[DES;][]{DES:2005,DES:2016}, the Survey of the Magellanic Stellar History \citep[SMASH;][]{Nidever:2017,Nidever:2021}, and the DECam Legacy Surveys \citep[DECaLS;][]{Dey:2019}.
Furthermore, the astronomical community has used DECam for targeted observing programs that have covered much of the remaining sky.
Here we present the {\bf DE}Cam {\bf L}ocal {\bf V}olume {\bf E}xploration survey (DELVE),\footnote{\url{https://delve-survey.github.io}} which is in the process of combining 126 nights of new DECam observations with existing public archival DECam data to assemble contiguous multi-band coverage of the entire high Galactic latitude ($|b| > 10 \deg$) southern sky (\figref{delve}).
The DELVE program consists of three main components:
(1) DELVE-WIDE seeks to complete DECam coverage of $\ensuremath{ {\sim}\,} 15000\xspace \deg^2$ of the high-Galactic-latitude southern sky in $g,r,i,z$;
(2) DELVE-MC provides deeper contiguous coverage of $2200\xspace \deg^2$ in $g,r,i$ surrounding the Large and Small Magellanic Clouds (LMC and SMC); and
(3) DELVE-DEEP performs deep imaging of $135\xspace \deg^2$ in $g,i$ around four Magellanic analogs in the Local Volume.
Each survey component will combine new observations with archival data that have been self-consistently processed with state-of-the-art data management pipelines.
The wide-area DELVE data are processed with the DES Data Management pipeline \citep[DESDM;][]{Morganson:2018}, which includes point-spread function (PSF) fitting and source modeling.
The deeper DELVE data around the Magellanic Clouds and Magellanic analogs are processed with the multiepoch point-source fitting pipeline used by SMASH \citep{NideverDorta:2020} to enable deep, accurate photometry in these regions.
The DELVE data will be released on regular intervals with the first public data release (DELVE DR1) documented here.
This paper summarizes the DELVE science program, survey design, progress to date, and contents of the first data release.
We start in \secref{motivation} by describing the scientific motivation for DELVE.
In \secref{progress} we document the survey strategy and observational progress to date.
In \secref{release} we describe DELVE DR1, including the input data set, processing, validation, and data access tools.
We provide several scientific demonstrations of the DELVE data in \secref{examples}, and we conclude in \secref{conclusion}.
Throughout this paper, all magnitudes are referenced in the AB system \citep{Oke:1974}, and all astronomical coordinates are provided in the Gaia-CRF2 reference frame \citep{Gaia:2018b} unless explicitly noted otherwise.
\begin{figure*}[t!]
\centering
\includegraphics[width=0.9\textwidth]{delve_footprint_n1024.pdf}
\caption{\label{fig:delve}
DELVE combines 126 nights of allocated time with public archival data to cover the southern equatorial sky with DECam.
DELVE will provide contiguous multi-band imaging with a $5\sigma$ point-source depth of
$g,r,i,z \gtrsim 23.5\xspace$ mag over $\ensuremath{ {\sim}\,} 15000\xspace \deg^2$ (turquoise region).
In addition, a region of $2200\xspace \deg^2$ around the Magellanic Clouds will be imaged to a depth of $g,r,i \geq 24.5\xspace$ mag (light-blue region).
Deep fields around four Magellanic analogs (Sextans B, NGC\,300, NGC\,55, and IC\,5152) will be imaged to a depth of $g,i \geq 25.5\xspace$ mag (dark-blue region).
The Galactic plane is indicated with a thick black line (dashed lines denote $b = {\pm}10\deg$), and the DES footprint is outlined in black.
This figure uses an equal-area McBryde--Thomas flat polar quartic projection in celestial equatorial coordinates.
}
\end{figure*}
\section{Scientific Motivation}
\label{sec:motivation}
The \ensuremath{\rm \Lambda CDM}\xspace model predicts that galaxies like the Milky Way inhabit large dark matter halos that grow hierarchically by merging with and/or accreting smaller galaxies.
There is ample evidence for the \ensuremath{\rm \Lambda CDM}\xspace paradigm on large scales; however, small-scale tests are challenging due to the low luminosities of the faintest galaxies that inhabit low-mass dark matter halos.
In particular, ultra-faint galaxies with stellar masses $\lesssim 10^5 \unit{M_\odot}$ have only been identified out to distances of a few Mpc \citep[e.g.\xspace,][]{McConnachie:2012, Martin:2013, Muller:2015, Carlin:2016, Smercina:2018, Crnojevic:2019}, while the census of the faintest dwarf galaxies (sometimes called ``hyper-faint'' dwarf galaxies; \citealt{Hargis:2014}) is incomplete even within the Milky Way halo \citep[e.g.\xspace,][]{Tollerud:2008,Hargis:2014,Kim:2018,Simon:2019,Drlica-Wagner:2020}.
Despite these observational challenges, the faintest galaxies provide crucial information about the role of environment and feedback on galaxy formation \citep[e.g.\xspace,][]{Maschenko:2007,Wheeler:2015,Wheeler:2019,Munshi:2018,Agertz:2020,Karunakaran:2020}, reionization and the first stars \citep[e.g.\xspace,][]{Bullock:2000,Shaprio:2004,Weisz:2014a,Weisz:2014b,Boylan-Kolchin:2015,Ishiyama:2016,Weisz:2017,Tollerud:2018,Graus:2019,Katz:2019}, and the nature of dark matter \citep[e.g.,][]{Bergstrom:1998,Spekkens:2013,Malyshev:2014,Ackermann:2015,Geringer-Sameth:2015,Brandt:2016,Bullock:2017,Nadler:2019b,Nadler:2020b}.
Ultra-faint galaxies also provide a unique opportunity to study the creation of heavy elements in some of the earliest star-forming environments \citep[e.g.,][]{Frebel:2015,Ji:2016,Roederer:2016}.
DELVE seeks to improve our understanding of dark matter and galaxy formation by studying the faintest satellite galaxies and their tidally disrupted remnants in a range of environments.
To accomplish this, DELVE consists of three survey programs, each with a specific observational aim.
\subsection{DELVE-WIDE}
The DELVE-WIDE program will complete DECam coverage over the entire high Galactic latitude ($|b|>10$) southern sky to provide a deep and accurate census of ultra-faint satellite galaxies around the Milky Way.
DELVE will reach a photometric depth in $g,r,i,z$ that is comparable to that of the first two years of DES.
Early DELVE-WIDE data have already resulted in the discovery of an ultra-faint satellite galaxy, Centaurus I \citep{Mau:2020}.
The full DELVE-WIDE survey will enable the detection of ultra-faint satellites similar to Centaurus I ($M_V = -5$ mag and $\mu = 27 \magn \asec^{-2}$) out to the virial radius of the Milky Way ($\ensuremath{ {\sim}\,} 300 \unit{kpc}$) with $>90\%$ efficiency \citep{Drlica-Wagner:2020}.
The combined model of the Milky Way and LMC satellite galaxy populations from \citet{Nadler:2020} predicts that the DELVE footprint (including the area covered by DES) contains $48 \pm 8$ satellite galaxies that are detectable by DELVE.
Given the existing population of confirmed and candidate satellite galaxies \citep{Drlica-Wagner:2020}, this model predicts that DELVE could discover over a dozen ultra-faint satellite galaxies.
The DELVE-WIDE program also provides exceptional sensitivity to stellar streams, the remnants of tidally disrupted dwarf galaxies and globular clusters.
These resolved stellar structures provide insight into the formation and evolution of the Milky Way stellar halo \citep[e.g.,][]{Bullock:2005}.
The composition, morphology, and orbital properties of detected structures capture the recent accretion history of the Milky Way, including the masses, orbits, and metallicities of recently accreted satellites \citep[e.g.,][]{Bonaca:2020b}.
Stellar streams also probe both the large- and small-scale distribution of dark matter around the Milky Way: they trace the gravitational potential of the Milky Way over a large range of radii \citep[e.g.,][]{Johnston:1999,Bovy:2016,Erkal:2016,Bonaca:2018}, and they offer a promising way to test dark matter clustering below the threshold of galaxy formation \citep[e.g.,][]{Johnston:2002,Carlberg:2013,Erkal:2016b,Banik:2019}.
The detection of stellar streams relies on deep, uniform coverage due to their low surface brightnesses ($\roughly32 \magn \asec^{-2}$) and large extents on the sky (tens of degrees).
Recent studies of stellar streams have emphasized the synergy between deep photometry with DECam, proper-motion measurements from {\it Gaia}, and radial velocities from massively multiplexed spectroscopic instruments \citep[e.g.,][]{Balbinot:2016,Shipp:2018,Jethwa:2018b,Shipp:2019,Li:2019,Li:2020,Shipp:2020,Bonaca:2020}.
DELVE-WIDE will extend the study of stellar streams by providing contiguous coverage across the southern hemisphere.
DELVE-WIDE will also enable a broad range of extragalactic science due to its wide, multi-band coverage.
In particular, DELVE-WIDE will enable extended southern-sky targeting for the Satellites Around Galactic Analogs (SAGA) program, which seeks to study more massive and luminous companions of Milky Way-like galaxies within 20--40\unit{Mpc} \citep{Geha:2017,Mao:2020}.
In addition, DELVE-WIDE will enable the search for strong gravitational lens systems, which can be used to probe the Hubble constant, dark energy, and the small-scale structure of dark matter \citep[e.g.,][]{Koopmans:2005,Treu:2010,Oguri:2010,Vegetti:2012,Treu:2018,Gilman:2019,Wong:2020}.
Furthermore, the DELVE-WIDE data can be used to study galaxies and galaxy clusters in a range of environments.
Several examples of extragalactic science with DELVE-WIDE can be found in \secref{examples}.
\subsection{DELVE-MC}
While it has long been hypothesized that the Magellanic Clouds (MCs) arrived with their own population of dwarf companions \citep[e.g.\xspace,][]{Lynden-Bell:1976,DOnghia:2008}, the observational evidence for this model has been strengthened by the discovery of many ultra-faint satellites surrounding the MCs \citep[e.g.,][]{Bechtol:2015,Koposov:2015,Drlica-Wagner:2015,Drlica-Wagner:2016,Torrealba:2018,Koposov:2018,Cerny:2020}.
These discoveries have stimulated a flurry of interest in simulating and modeling the satellite populations of the MCs.
Simulations predict that up to a third of the satellites around the Milky Way originated with the MCs \citep[e.g.\xspace,][]{Deason:2015,Wetzel:2015,Jethwa:2016,Dooley:2017b,Jahn:2019,Nadler:2020}.
Some studies suggest that the MCs should host more luminous satellites than are observed \citep[e.g.,][]{Dooley:2017b}, while others suggest that the MCs should host more faint satellites than are observed \citep[e.g.,][]{Jahn:2019}.
Furthermore, there is significant observational uncertainty in associating known ultra-faint galaxies with the MCs \citep{Kallivayalil:2018,Pardy:2020,Erkal:2020,Patel:2020}.
One issue in determining the satellite luminosity function of the MCs comes from the fact that the region around the MCs has only been observed by relatively shallow contiguous surveys \citep[e.g.\xspace,][]{Drlica-Wagner:2016,Mackey:2018} and by deep surveys with low fill factors \citep[e.g.\xspace,][]{Nidever:2017}.
The DELVE-MC program will provide deep, contiguous imaging of the MCs and their surrounding environment to robustly measure the population of faint satellites around the MCs.
DELVE has already started to probe this region with inhomogeneous early data, leading to the discovery an ultra-faint star cluster (DELVE 2; $M_V = -2.1$) located 12\unit{kpc} from the SMC and 28\unit{kpc} from the LMC \citep{Cerny:2020}.
The model of \citet{Nadler:2020} predicts that $\ensuremath{ {\sim}\,} 30\%$ of the ultra-faint galaxies contained within the DELVE footprint are associated with the MCs.
The stellar masses, star formation histories, and interaction histories of the MCs are expected to influence the properties of their satellite populations \citep[e.g.,][]{Jethwa:2016,Dooley:2017b,Jahn:2019}.
SMASH has used DECam to study the main bodies and periphery of the MCs with a deep, partially filled survey \citep[][]{Nidever:2017}.
Photometric metallicities from SMASH suggest that the LMC periphery is not as metal-poor as would be expected in a ``classical'' halo produced by \ensuremath{\rm \Lambda CDM}\xspace-style hierarchical assembly.
This observation is consistent with the hypothesis that the stellar envelope of the LMC may be dominated by material from the outer LMC disk, likely stirred up through a recent interaction with the SMC \citep{Choi:2018a,Choi:2018b,Nidever:2019a}.
In parallel, recent observations of {\it Gaia}-selected red giant branch (RGB) stars suggest that even more structure exists in the periphery of the MCs \citep{Belokurov:2018,Gaia:2020b}.
The deep, contiguous imaging of DELVE-MC will extend below the oldest main-sequence turn off (MSTO) of the MCs and will be sensitive to faint substructures that can provide clues about their interaction history \citep{Massana:2020}.
Comparisons between the stellar populations in the bodies, peripheries, and satellites of the MCs will help reconstruct the evolution of the MCs and their satellite system as they are accreted onto the Milky Way.
DELVE-MC is also well suited to study the gravitational wake of the LMC.
As the LMC moves through the Milky Way stellar halo, it pulls stars towards itself, creating an overdensity of stars along its past orbital path \citep[e.g.,][]{Garavito-Camargo:2019, Erkal:2020b}.
Recently, it has been shown that the Pisces Overdensity \citep{Watkins:2009} matches the properties of the expected wake \citep{Belokurov:2019}.
The wide coverage of DELVE will give a more complete view of the Milky Way's stellar halo close to the LMC, allowing us to better map the wake of the LMC and test the effect of dynamical friction.
This, in turn, may constrain alternative dark matter models that modify dynamical friction \citep[e.g.\xspace,][]{Lancaster:2020}.
\begin{figure*}[t!]
\center
\includegraphics[width=0.65\textwidth, trim=2.0cm 1.5cm 3.5cm 3.0cm, clip]{lv_gxs_MB_vs_dist_lt6Mpc_DELVE4.pdf}
\caption{$B$-band absolute magnitude vs.\ distance for nearby galaxies \citep{Karachentsev:2013}.
We highlight systems targeted by DELVE: ultra-faint Milky Way satellites (turquoise circles), the LMC/SMC (blue diamonds), and MC analogs in the Local Volume (navy diamonds). Also highlighted as light-green diamonds are MC analogs with existing DECam data prior to DELVE. Little is known about the faint dwarf galaxy population ($M_{\rm B} \gtrsim -10$) beyond $\ensuremath{ {\sim}\,} 1 \unit{Mpc}$, with the exception of a handful of satellites around Milky Way-mass galaxies (vertical dashed lines). Dotted horizontal lines denote the luminosities of the Milky Way, LMC, and SMC.}
\label{fig:lumdist}
\end{figure*}
\subsection{DELVE-DEEP}
As our understanding of satellite galaxies and stellar substructures in the Local Group has improved, searches for faint stellar systems have extended to more distant galaxies.
Within $\ensuremath{ {\sim}\,} 4\unit{Mpc}$, systematic imaging searches for dwarf companions and tidal debris have been undertaken for Centaurus\,A \citep[e.g.\xspace,][]{Crnojevic:2016,Crnojevic:2019,Muller:2019}, M\,101 \citep[e.g.\xspace,][]{Merritt:2014, Danieli:2017, Bennet:2019, Bennet:2020}, M\,94 \citep{Smercina:2018}, M\,81 \citep{Chiboucas:2013, Okamoto:2015, Okamoto:2019, Smercina:2020}, NGC\,253 \citep[e.g.,][]{Sand:2014, Toloba:2016, Romanowsky:2016}, and 10 other (approximately) Milky Way-mass hosts within the Local Volume \citep{Carlsten:2020a, Carlsten:2020b}.
A complementary approach has been taken by the SAGA Survey, which identifies satellites of Milky Way-mass hosts located at 20--40\unit{Mpc} via spectroscopic verification of bright companions selected from SDSS and DECam imaging \citep{Geha:2017,Mao:2020}.
The \ensuremath{\rm \Lambda CDM}\xspace model predicts that the abundance of satellite galaxies primarily depends on host halo mass \citep[e.g.,][]{Behroozi:2013,Moster:2013}.
However, scatter in the stellar mass--halo mass relation \citep[e.g.,][]{Behroozi:2013,Garrison-Kimmel:2014,Garrison-Kimmel:2017,Munshi:2021}, or the effects of reionization, tides, ram pressure stripping, and host infall time, may be relatively more important for satellites of low-mass hosts than they are for satellites of more massive hosts \citep[e.g.,][]{Dooley:2017b}.
Indeed, recent studies have shown that environmental effects of MC-mass hosts on their dwarf satellites are stronger than expected \citep[e.g.,][]{Garling:2020, Carlin:2019, Carlin:2020}.
While the discovery of ultra-faint satellites associated with the MCs is broadly consistent with \ensuremath{\rm \Lambda CDM}\xspace, interpreting the population of MC satellites in a cosmological context is complicated by the current location of the MCs within the Milky Way's gravitational potential.
The DELVE-DEEP program targets nearby isolated galaxies with stellar masses similar to those of the MCs (``MC analogs'').
Observations of satellites around isolated MC analogs will allow studies of satellite/host demographics in environments outside the influence of a massive host.
DELVE-DEEP will complement existing DECam observations of the SMC-analog NGC\,3109 \citep{Sand:2015} and other similar surveys such as MADCASH \citep{Carlin:2016,Carlin:2019,Carlin:2020} and LBT-SONG \citep{Davis:2021}.
By combining results from the DELVE-DEEP and other similar observing programs, it will be possible to perform a statistically robust analysis of the satellite populations of MC analogs.
DELVE-DEEP targets four isolated dwarf galaxies at 1.4--2.1\unit{Mpc} (\figref{lumdist} and \tabref{deep_targets}).
These galaxies were chosen as MC analogs: NGC\,55 and NGC\,300 have stellar masses roughly comparable to the LMC, while Sextans\,B and IC\,5152 are within an order of magnitude of the SMC ($\ensuremath{ {\sim}\,} 0.1-0.8\,M_{\rm \star, SMC}$).
NGC 55 is an irregular, barred spiral galaxy \citep{deVaucouleurs:1991}; therefore, it closely resembles the LMC not only in stellar mass but also in morphology. Unlike the LMC, NGC 55 is almost edge-on, which makes it well suited for stellar content studies. NGC 300 is an almost pure disk galaxy and is similar to M33, an LMC-mass satellite of M31.
Both NGC 55 and NGC 300 belong to the nearby Sculptor Group, which is actually an unbound filamentary structure extended along our line of sight \citep{Karachentsev:2003}.
Both Sextans B and IC 5152 are irregular dwarf galaxies.
While Sextans B is a member of a very loose group of dwarf galaxies (the NGC 3109 association; \citealt{Tully:2006}), IC 5152 is an exceptionally isolated object, with the nearest neighbor being NGC 55 at a distance of 800 kpc \citep{Karachentsev:2002}.
While all of our targets have been studied with deep imaging observations \citep[e.g.,][]{Tosi:1991,Zijlstra:1999,Tanaka:2011,Bellazzini:2014,Hillis:2016,Rodriguez:2016,Jang:2020}, systematic searches for dwarf satellites have not been possible due to limited sky coverage.
DELVE-DEEP will provide sensitivity to resolved stars $\gtrsim 1.5$ mag below the tip of the RGB of each target, enabling the detection of resolved satellite galaxies with $M_V \lesssim -7$ (comparable to the brightest ultra-faint satellites of the Milky Way).
DELVE-DEEP imaging covers the halos of each target \citep[$r_{\rm vir} \sim 110\unit{kpc}$ for an SMC-mass galaxy; e.g.,][]{Dooley:2017b} to provide a complete census of faint satellites with a total area of $\ensuremath{ {\sim}\,} 135\xspace \deg^2$.
According to predictions based on \ensuremath{\rm \Lambda CDM}\xspace, we expect to discover 5--17 satellites with $M_V \lesssim -7$ around the four targets \citep{Dooley:2017a,Dooley:2017b}.
In addition to searches for faint satellites, the DELVE-DEEP data will enable detailed studies of the target galaxies, including searches for globular clusters and measurement of their stellar density profiles to large radii and low surface brightnesses \citep[e.g.,][]{Pucha:2019}.
\input{table_deep}
\section{Survey Design and Progress}
\label{sec:progress}
DELVE began observing in 2019 February and has collected $\ensuremath{ {\sim}\,} 12000$ new DECam exposures as of 2021 January.
The DELVE imaging data are released via the NOIRLab Astro Data Archive\footnote{\url{https://astroarchive.noirlab.edu/}} without any proprietary period.
Observations for the three DELVE programs are scheduled by an automated open-source tool that optimizes for field availability, air mass, sky brightness, and seeing.\footnote{\url{https://github.com/kadrlica/obztak}}
DELVE observing has been performed on site at CTIO and from remote stations in Tucson and at Fermilab.
Remote observing from home commenced in 2020 October due to the COVID-19 pandemic.
The observational strategy and status for each of the DELVE programs are described in more detail below.
\subsection{DELVE-WIDE}
\label{sec:survey_wide}
DELVE-WIDE seeks to assemble contiguous DECam coverage over the entire southern sky with $|b| > 10\deg$ (\figref{delve}).
Two-band photometry provides sufficient color--magnitude information to separate old, metal-poor halo populations from Milky Way foreground; however, additional color information is useful for a wide range of science topics.
Thus, DELVE-WIDE observes preferentially in the $g,i$ bands and coordinates with other DECam programs to process and assemble coverage in the $r,z$ bands (see \secref{release}).
DELVE-WIDE nominally performs $3 \times 90 \unit{s}$ dithered exposures in $g,i$ using the same icosahedral tiling scheme used by DECaLS \citep{Dey:2019},\footnote{Based on the scheme of Hardin, Sloane, and Smith (\url{http://neilsloane.com/icosahedral.codes})} but with larger dithered offsets (corresponding to roughly three times the DECam CCD dimensions).
The DELVE tiling scheme allows the entire $15000\xspace \deg^2$ DELVE-WIDE footprint to be covered with three tilings in $g,i$ by $\ensuremath{ {\sim}\,} 43000$ DECam exposures.
However, a large fraction of the sky has already been covered by DES, DECaLS, and other DECam programs.
By preferentially targeting regions that lack sufficiently deep DECam data, the DELVE-WIDE footprint can be contiguously covered with 3 tilings in $g,i$ by $\ensuremath{ {\sim}\,} 20000$ new exposures.
DELVE-WIDE includes a fourth tiling when necessitated by observing constraints and existing DECam coverage.
As of 2021 January, DELVE-WIDE has collected 9778\xspace exposures.
When scheduling DELVE-WIDE, we use public metadata to assess the coverage and depth of archival DECam imaging.
We calculate the effective exposure time for each archival exposure from the shutter-open time, $T_{\rm exp}$, and the effective exposure time scale factor, $t_{\rm eff}$, which combines the achieved seeing, sky brightness, and extinction due to clouds to assess the effective depth of an exposure \citep{Neilsen:2015}.
We create maps of the summed effective exposure time in each band as a function of sky position at a resolution of $\ensuremath{ {\sim}\,} 3\farcm4$ (\HEALPix $\code{nside}=1024$).
Each DELVE-WIDE target exposure is compared to the existing summed effective exposure time at the target location of that exposure and is considered to have been covered if $>67\%$ of the exposure area exceeds a minimum depth threshold.
The minimum depth of each DELVE-WIDE target exposure is calculated from the tiling number, $N_{\rm tile} \in \{ 1, 2, 3, 4 \}$, the DELVE-WIDE shutter open time, $T_{\rm exp,0} = 90 \unit{s}$, and the minimum effective depth scale factor, $t_{\rm eff,min}$.
A DELVE-WIDE exposure in tiling $N_{\rm tile}$ is considered covered if the summed effective exposure time in that region of the sky is
\begin{equation}
\sum_{j} t_{{\rm eff},j} \times T_{{\rm exp}, j} > N_{\rm tile} \times t_{\rm eff, min} \times T_{\rm exp,0}.
\label{eqn:coverage}
\end{equation}
The cumulative exposure time from archival data (the left-hand side of the above equation) is calculated from archival exposures with $t_{{\rm eff},j} > 0.2$.
Motivated by experience from DES and other DECam programs, we set the minimum effective exposure time threshold as $t_{{\rm eff, min},g} = 0.4$ and $t_{{\rm eff, min},i} = 0.5$.
The DECam coverage map is updated with new archival data each semester, while completed DELVE exposures are included in real time during observing.
The DELVE-WIDE data are processed with the DESDM pipeline \citep{Morganson:2018} and serve as the basis for DELVE DR1.
The DELVE-WIDE data processing is described in more detail in \secref{processing}.
\subsection{DELVE-MC}
\label{sec:survey_mc}
The DELVE-MC program seeks to map the periphery of the MCs by covering a contiguous region of $\ensuremath{ {\sim}\,} 2200\xspace \deg^2$ extending $25\deg$ around the LMC and $15\deg$ around the SMC to a depth comparable to that of SMASH, $g=24.8$, $r=24.5$, and $i=24.2$ mag \citep{Nidever:2017,Nidever:2019a}.
DELVE-MC observes in three bands to leverage $(g-r)$ and $(r-i)$ colors to help separate compact blue galaxies from Magellanic MSTO stars at $g \gtrsim 23$ mag \citep[][]{Nidever:2019a}.
Roughly half of the DELVE-MC footprint has already been covered to the desired depth by DES and SMASH, and DELVE is in the process of supplementing the remaining $\ensuremath{ {\sim}\,} 1100 \deg^2$ with $\ensuremath{ {\sim}\,} 2200$ new exposures.
DELVE-MC targets a total integrated exposure time of $800\unit{s}$ in $g,r$, and $1000\unit{s}$ in $i$ by using $3 \times 267\unit{s}$ dithered exposures in $g,r$ and $3 \times 333 \unit{s}$ dithered exposures in $i$.
The DELVE-MC tiling scheme is the same as the DELVE-WIDE survey, and regions of missing DECam coverage are determined following the procedure described in \secref{survey_wide}.
Due to the low elevation of the MCs as seen from CTIO, DECam observations of the MCs generally have larger-than-average PSF FWHM values and correspondingly lower-than-average $\ensuremath{t_{\rm eff}}\xspace$ values.
Based on SMASH observations, we set $t_{{\rm eff, min},g} = 0.3$, $t_{{\rm eff, min},r} = 0.3$, and $t_{{\rm eff, min},i} = 0.45$ when calculating the existing coverage.
DELVE-MC observations are scheduled when the PSF FWHM is $< 1\farcs1$ (as estimated for an $i$-band exposure taken at zenith) to help improve crowded-field photometry.
As of 2021 January, DELVE-MC has collected 1467\xspace exposures.
The high stellar density in the DELVE-MC region motivates us to process the DELVE-MC data with a modified version of the \code{PHOTRED} pipeline \citep{NideverDorta:2020}.
This pipeline, called ``\code{DELVERED}'', was created to better handle large dithers between overlapping DELVE-MC exposures.
\code{DELVERED} first performs PSF photometry for each exposure and then performs forced PSF photometry for overlapping exposures.
At the exposure level, each night is processed separately.
This processing includes WCS correction for each CCD using {\it Gaia}\xspace DR2, PSF photometry using DAOPHOT \citep{Stetson:1987,Stetson:1994}, photometric zeropoint determination, and aperture correction.
The photometric zeropoint of each exposure is estimated in the same manner as for the NOIRLab Source Catalog \citep[NSC;][]{Nidever:2018,Nidever:2020a,Nidever:2020b} using all-sky catalogs and ``model magnitudes''.
Forced PSF photometry is then performed on overlapping exposures that pass quality cuts on seeing and zeropoints.
\code{DAOPHOT}/\code{ALLFRAME} \citep{Stetson:1994} is run on all CCD images overlapping each $0.25 \deg \times 0.25 \deg$ patch of sky (referred to as a ``brick'' from the DECaLS tiling scheme) using a master source list generated from a multi-band stacked image and \code{SourceExtractor}.
The \code{ALLFRAME} results are then calibrated using the zeropoints and aperture corrections determined at the exposure level.
Finally, weighted mean magnitudes, coordinates, and photometric variability indices are determined for each object using the measurements from the multiple exposures.
\subsection{DELVE-DEEP}
\label{sec:survey_deep}
DELVE-DEEP performs deep DECam imaging around four isolated dwarf galaxies:
Sextans\,B, NGC\,55, NGC\,300, and IC\,5152.
DELVE-DEEP will achieve $5\sigma$ depths of $g = 26.0\xspace \unit{mag}$ and $i = 25.0\xspace \unit{mag}$ for each target with total integrated exposure times of $4500\unit{s}$ in $g$ and $3000\unit{s}$ in $i$.
The nominal depth can be achieved in $15 \times 300\unit{s}$ exposures in $g$ and $10 \times 300\unit{s}$ exposures in $i$.
However, three of the four DELVE-DEEP targets (NGC\,55, NGC\,300, and IC\,5152) reside in the DES footprint; thus, we decrease the required exposure time by $900\unit{s}$ to account for the existing DES data and target $12 \times 300 \unit{s}$ $g$-band exposures and $7 \times 300 \unit{s}$ $i$-band exposures for these targets.
The DELVE-DEEP fields are chosen to roughly cover the angular region corresponding to the virial radius of the SMC \citep[$r_{\rm vir} \sim 110\unit{kpc}$;][]{Dooley:2017b} at the distance of each target (\tabref{deep_targets}).
This radius corresponds to $\ensuremath{ {\sim}\,} 3 \deg$ for NGC\,55, NGC\,300, and IC\,5152, but it is significantly larger for Sextans\,B due to its proximity ($\ensuremath{ {\sim}\,} 4.4\deg$).
Covering this large of a region around Sextans\,B is prohibitive given the DELVE-DEEP time allocation, and the coverage around Sextans\,B was reduced to $\ensuremath{ {\sim}\,} 3 \deg$.
Observations of each target start with the central pointing and progress outward.
Small dithers of $\ensuremath{ {\sim}\,} 2 \arcmin$ are applied in a hexagonal pattern to each tiling to cover the gaps between CCDs.
Star--galaxy separation is critical to the DELVE-DEEP science, and observations are performed only when the $i$-band zenith PSF is estimated to have ${\rm FWHM} < 0\farcs9$.
DELVE imaging for Sextans B was completed in 2019A, and imaging of NGC\,55 is $\roughly78\%$ complete as of 2020B.
The DELVE-DEEP program has collected 419\xspace exposures as of 2021 January.
The DELVE-DEEP data are processed with the DESDM image co-addition pipeline in a manner similar to the DES deep fields \citep*{HartleyChoi:2020}. While this has allowed for early visual inspection and catalog analysis, the DESDM pipeline is not optimized for stellar photometry in the crowded fields near the DELVE-DEEP targets. We are exploring the use of \code{DELVERED} to perform crowded-field stellar photometry in the DELVE-DEEP fields.
\section{Data Release}
\label{sec:release}
The first DELVE data release, DELVE DR1, is based on new and archival DECam data collected as part of DELVE-WIDE.
DELVE DR1 consists of a catalog of unique astronomical objects covering $\ensuremath{ {\sim}\,} 5000 \deg^2$ in each of $g,r,i,z$ and $\ensuremath{ {\sim}\,} 4000 \deg^2$ in all of $g,r,i,z$ simultaneously.
This section describes the DELVE DR1 data selection, processing, characterization, and validation.
The DELVE DR1 catalog can be accessed through the NOIRLab Astro Data Lab.\footnote{\url{https://datalab.noirlab.edu/delve}}
\input{table_summary.tex}
\subsection{Data Set}
\label{sec:data}
DELVE DR1 consists of 29929\xspace DECam exposures assembled from a combination of DELVE observations and archival DECam data.
The largest contributors to these data are DELVE itself, the DECam eROSITA Survey (DeROSITAS; PI Zenteno),\footnote{\url{http://astro.userena.cl/derositas}} DECaLS \citep[PI Schlegel;][]{Dey:2019}, and the Blanco Imaging of the Southern Sky Survey \citep[BLISS: PI Soares-Santos;][]{Mau:2019}.
However, over half of the exposures in DELVE DR1 come from $>150$ DECam community programs (\appref{propid}).
These data were downloaded from the NOIRLab Astro Data Archive.
The nominal DELVE DR1 region was defined as exposures having centroids with ${\ensuremath{\rm Dec}}\xspace < 0\deg$ and $b > 10\deg$.
This region was extended to the Galactic plane in the region of $120\deg < {\ensuremath{\rm RA}}\xspace < 140\deg$ to enable an extended search for the Jet stream \citep{Jethwa:2018}.
Exposures were selected to have exposure time $30\unit{s} < \ensuremath{T_{\rm exp}}\xspace < 350\unit{s}$ and effective exposure time scale factor $\ensuremath{t_{\rm eff}}\xspace > 0.3$ (this is slightly higher than the $t_{{\rm eff},j} > 0.2$ requirement applied in \secref{survey_wide}).
While no explicit cut was placed on the PSF FWHM, the cut on \ensuremath{t_{\rm eff}}\xspace removes exposures with very poor seeing (\figref{fwhm}).
Furthermore, exposures were required to have good astrometric solutions when matched to {\it Gaia}\xspace DR2 by \code{SCAMP} \citep{Bertin:2006} including $>250$ astrometric matches, $\chi^2_{\rm astrom} < 500$, $\Delta({\ensuremath{\rm RA}}\xspace) < 150\unit{mas}$, and $\Delta({\ensuremath{\rm Dec}}\xspace) < 150\unit{mas}$.
The key properties of the DELVE DR1 are listed in \tabref{summary}.
\begin{figure}[t]
\centering
\includegraphics[width=\columnwidth]{delve_dr1_fwhm.pdf}
\caption{PSF FWHM distributions for DECam exposures included in DELVE DR1.}
\label{fig:fwhm}
\end{figure}
\subsection{Data Processing}
\label{sec:processing}
All exposures in DELVE DR1 were consistently processed with the DESDM ``Final Cut'' pipeline \citep{Morganson:2018}.
Identical processing was performed at Fermilab and NCSA; DELVE DR1 is derived from the Fermilab processing.
DELVE implements updates to the DESDM pipeline that were developed for DES DR2 (see Section 3 of \citealt{DES-DR2:2021}).
Data were reduced and detrended using custom, seasonally averaged bias and flat images, and full-exposure sky background subtraction was performed \citep{Bernstein:2018}.
\code{SourceExtractor} \citep{Bertin:1996} and \code{PSFEx} \citep{Bertin:2011} were used to automate source detection and photometric measurement.
Astrometric calibration was performed against {\it Gaia}\xspace DR2 using \code{SCAMP} \citep{Bertin:2006}.
Photometric zeropoints for each CCD were derived by performing a $1\arcsec$ match between the DELVE Final Cut \code{SourceExtractor} catalogs and the ATLAS Refcat2 catalog \citep{Tonry:2018}, which places measurements from Pan-STARRS1 (PS1) DR1 \citep{Chambers:2016}, SkyMapper DR1 \citep{Wolf:2018}, and several other surveys onto the PS1 $g,r,i,z$-bandpass system.
Transformation equations from the ATLAS Refcat2 system to the DECam system were derived by comparing calibrated stars from DES DR1 (see \appref{transforms} for details).
Zeropoints derived from the DELVE processing and photometric calibration pipeline were found to agree with those derived by DES DR1 with an rms scatter of $\ensuremath{ {\sim}\,} 0.01 \unit{mag}$.
We built a multi-band catalog of unique sources by combining the \code{SourceExtractor} catalogs from each individual CCD image following the procedure of \citet{Drlica-Wagner:2015}.
We took the set of \code{SourceExtractor} detections with $\var{flags} < 4$, which allows neighboring and deblended sources, and $(\var{imaflags\_iso}\,\&\,2047) = 0$, which removes objects containing bad pixels within their isophotal radii \citep{Morganson:2018}.
We further required that each detection have a measured automatic aperture flux, a measured PSF flux, and a PSF magnitude error of $< 0.5$ mag.
We sorted \code{SourceExtractor} detections into $\ensuremath{ {\sim}\,} 3 \deg^2$ \HEALPix pixels ($\code{nside}=32$), and within each \HEALPix pixel we grouped detections into clusters by associating all detections within a $1\arcsec$ radius.
Occasionally, two real astronomical objects are located within $1 \arcsec$ of each other and were grouped into the same cluster.
We identified these ``double'' objects if they were contemporaneously detected on a single image.
If a double object was detected in two or more images, we split its parent cluster into two distinct clusters.
We did not split clusters when more than two objects were detected within $1\arcsec$.
This splitting procedure means that the DELVE DR1 catalog should be treated with care when searching for multiple closely separated objects.
Each cluster of detections was associated with an object in the DELVE DR1 catalog.
The astrometric position of each object was calculated as the median of the individual single-epoch measurements of the object.
We track two sets of photometric quantities for each object: (1) measurements from the single exposure in each band that had the largest effective exposure time (i.e., the largest $\ensuremath{t_{\rm eff}}\xspace \times T_{\rm exp}$), and (2) the weighted average (\var{WAVG}) of the individual single-epoch measurements.
The weighted average and unbiased weighted standard deviation were calculated following the weighted sample prescriptions used by DES \citep{Gough:2009,DES-DR2:2021}.\footnote{\url{https://www.gnu.org/software/gsl/doc/html/statistics.html\#weighted-samples}}
In addition, we track cluster-level statistics such as the number of detections in each band.
We follow the procedure of DES DR1 \citep{DES-DR1:2018} to calculate the interstellar extinction from Milky Way foreground dust.
We calculate the value of $E(B-V)$ at the location of each catalog source by performing a bi-linear interpolation in $({\ensuremath{\rm RA}}\xspace,{\ensuremath{\rm Dec}}\xspace)$ to the maps of \citet{Schlegel:1998}.
The reddening correction for each source in each band, $A_b = R_b \times E(B-V)$, is calculated using the fiducial interstellar extinction coefficients from DES DR1: $R_g = 3.185$, $R_r = 2.140$, $R_i = 1.571$, and $R_z = 1.196$ \citep{DES-DR1:2018}.
Note that, following the procedure of DES DR1, the \citet{Schlafly:2011} calibration adjustment to the \citet{Schlegel:1998} maps is included in our fiducial reddening coefficients.
The $A_b$ values are included for each object in DELVE DR1 but are not applied to the magnitude columns by default.
The list of the photometric and astrometric properties provided in DELVE DR1 can be found in \appref{tables}.
\subsection{Sky Coverage}
\label{sec:coverage}
We quantify the area covered by DELVE DR1 in the form of \HEALPix maps with angular resolution of $\ensuremath{ {\sim}\,} 0\farcm86$ ($\code{nside} = 4096$).
These maps were created by pixelizing the geometry of each DECam CCD using the \code{decasu}\footnote{\url{https://github.com/erykoff/decasu}} package built on \code{healsparse}.\footnote{\url{https://healsparse.readthedocs.io}}
This package calculates the geometry of each CCD using higher-resolution nested \HEALPix maps ($\code{nside} = 16384$) and sums the resulting covered pixels to generate lower resolution maps containing the fraction of each pixel that is covered by the survey.
This bypasses the computationally intensive \code{mangle} processing done by DES \citep{Swanson:2008,Morganson:2018} while retaining the same accuracy at a resolution of $\code{nside}=4096$.
We quantitatively estimate the covered area as the sum of the coverage fraction maps in each band independently, as well as the intersection of the maps in all four bands.
These values are reported in \tabref{summary} and visualized in \appref{depth}.
\subsection{Astrometry}
\label{sec:astrometry}
We assess the internal astrometric accuracy by comparing the distributions of angular separations of individual detections of the same objects over multiple exposures.
The median global astrometric spread is $26 \unit{mas}$ across all bands.
We find that this spread is fairly consistent within each band, with median offsets in $g,r,i,z$ of $24, 26, 25, 28\unit{mas}$.
Furthermore, we compare the DELVE DR1 catalog-coadd object locations to the locations of matched sources in the {\it Gaia}\xspace DR2 catalog \citep{Gaia:2018}, and we find an overall astrometric agreement of $\astroabs \unit{mas}$.
This comparison is somewhat circular, since the {\it Gaia}\xspace DR2 catalog was used for the image-level astrometric calibration; however, the good agreement confirms that no significant astrometric offsets have been introduced by the catalog coaddition procedure.
\begin{figure*}
\centering
\includegraphics[width=0.98\textwidth]{photo_gaia_r_n128.pdf}
\caption{Median difference between the DELVE DR1 photometry transformed into the {\it Gaia}\xspace $G$-band, $G_{\rm DELVE}$, and the measured {\it Gaia}\xspace magnitude, $G_{\it Gaia}\xspace$. The spatial distribution of the median difference in each pixel is shown in the left panel, while the right panel shows a histogram of the pixel values. A shift in the zeropoint can be seen at ${\ensuremath{\rm Dec}}\xspace \sim -30\deg$, which corresponds to the boundary between the ATLAS Refcat2 use of PS1 and SkyMapper (\secref{photrel}).
This comparison is restricted to the area with overlapping DELVE DR1 coverage in all four bands ($g,r,i,z$).
}
\label{fig:gaia}
\end{figure*}
\subsection{Relative Photometric Calibration}
\label{sec:photrel}
We assess the photometric repeatability in each band from the root-mean-square (rms) scatter between independent PSF measurements of bright stars.
For each band, we select stars with $16 < \var{WAVG\_MAG\_PSF} < 18$ mag and calculate the median rms scatter in $\ensuremath{ {\sim}\,} 0.2 \deg^2$ \HEALPix pixels ($\code{nside}=128$).
We estimate the median of the rms scatter over the entire footprint in each band.
This quantity is found to be $\ensuremath{ {\sim}\,} 5 \unit{mmag}$ and is listed for each band in \tabref{summary}.
We validate the photometric uniformity of DELVE DR1 by comparing to space-based photometry from {\it Gaia}\xspace DR2 (\figref{gaia}).
We transform the $g,r,i,z$ photometry from DELVE to the {\it Gaia}\xspace $G$ band following a set of transformations derived for DES \citep{Sevilla-Noarbe:2020,DES-DR2:2021}.
We compare the {\it Gaia}\xspace DR2 $G$-band magnitude in the AB system to the $G$-band AB magnitude of stars in DELVE with $16 < r < 20 \unit{mag}$ and $0 < i-z < 1.5 \unit{mag}$.
We calculate the median difference, $G_{\rm DELVE} - G_{\it Gaia}\xspace$, within each $\code{nside}=128$ \HEALPix pixel (\figref{gaia}).
We find that the median offset between DELVE DR1 and {\it Gaia}\xspace DR2 is $-4.8 \unit{mmag}$.
We estimate the photometric uniformity across the DELVE DR1 data as the standard deviation of the median differences across pixels, which yields a value of 9.1 \unit{mmag} (\tabref{summary}).
Since the distribution of residuals is non-Gaussian, we also calculate the 68\% containment, which is $15.3 \unit{mmag}$.
Similar comparisons between DES and {\it Gaia}\xspace DR2 have demonstrated that the nonuniformity of {\it Gaia}\xspace observations can be the dominant contributor to photometric nonuniformity estimated using this technique \citep{Burke:2018,Sevilla-Noarbe:2020,DES-DR2:2021}.
However, it is clear from the spatial structure in \figref{gaia} that systematics in the DELVE DR1 calibration dominate the nonuniformity relative to {\it Gaia}\xspace.
Furthermore, we observe a systematic shift of $11.2 \unit{mmag}$ relative to {\it Gaia}\xspace at ${\ensuremath{\rm Dec}}\xspace = -30\deg$.
This is the declination at which ATLAS Refcat2 switches from using PS1 to SkyMapper, and a similar feature can be seen in the residuals of Figure 3 in \citet{Tonry:2018}.
This offset is the dominant contributor to the broadening of the residuals between DELVE DR1 and {\it Gaia}\xspace DR2 seen in the left panel of \figref{gaia}.
The relative photometric calibration of DELVE could be improved in the future by performing an internal relative calibration such as the ``ubercalibration'' of SDSS and PS1 \citep[][]{Padmanabhan:2008,Schlafly:2012}, or the forward global calibration of DES \citep[i.e.,][]{Burke:2018}.
\begin{figure*}[t]
\centering
\includegraphics[width=0.49\textwidth]{maglim_psf_5sig_hist.pdf}
\includegraphics[width=0.49\textwidth]{maglim_psf_5sig_area.pdf}
\caption{(Left) Distribution of PSF magnitude limits for point-like sources at S/N=5. (Right) DELVE DR1 survey area in each band as a function of the limiting PSF magnitude (S/N=5). These distributions look similar when calculated from the \magauto limiting magnitude for all sources but are shifted brighter by $\ensuremath{ {\sim}\,} 0.4 \unit{mag}$.
}
\label{fig:maglim}
\end{figure*}
\subsection{Absolute Photometric Calibration}
\label{sec:photabs}
We do not have a precise estimate of the absolute photometric calibration of DELVE DR1, but a rough estimate can be derived by comparing to other data sets.
The absolute photometry of DELVE DR1 is tied to the HST CalSpec standard star C26202 via DES DR2, which was used to adjust the zeropoints of the ATLAS Refcat2 transformation equations (\appref{transforms}).
Thus, DELVE DR1 cannot have a better absolute calibration than DES DR2, which sets a lower limit on the statistical uncertainty of $2.2 \unit{mmag}$ per band and a systematic uncertainty of $11$ to $12 \unit{mmag}$ per band (see Table 1 of \citealt{DES-DR2:2021}).
The global offset between DELVE DR1 and {\it Gaia}\xspace DR2 (which is not seen in DES DR2) suggests that the absolute calibration cannot be better than $4.8 \unit{mmag}$.
Combining the maximum systematic uncertainty on the absolute calibration from DES DR2 and the DELVE DR1 offset relative to {\it Gaia}\xspace, we conservatively estimate that the absolute photometric accuracy of DELVE DR1 is $\lesssim \photabs \unit{mmag}$.
This is comparable to the absolute photometric accuracy estimated for the DES first-year cosmology data set \citep{Drlica-Wagner:2018}.
In the future, the absolute photometric calibration of DELVE can be improved and verified directly following a similar procedure to DES DR2 using HST CalSpec standards and/or pure hydrogen atmosphere white dwarf stars \citep{DES-DR2:2021}.
\subsection{Photometric Depth}
\label{sec:depth}
The photometric depth of DELVE DR1 can be assessed in several ways.
One common metric is to determine the magnitude at which a fixed signal-to-noise ratio (S/N) is achieved \citep[e.g.,][]{Rykoff:2015}.
The statistical magnitude uncertainty is related to the S/N calculated from the flux, $F/\delta F$, via propagation of uncertainties and Pogson's law \citep{Pogson:1856},
\begin{equation}
\delta m = \frac{2.5}{\ln 10} \frac{\delta F}{F}.
\end{equation}
\noindent Using this equation, we estimate the magnitude at which DELVE DR1 achieves S/N=5 ($\delta m \approx 0.2171$) and S/N=10 ($\delta m \approx 0.1085$).
We calculate these magnitude limits for point-like sources using \magpsf and for all sources using \magauto.
For each magnitude and S/N combination, we select objects and interpolate the relationship between $m$ and ${\rm median}(\delta m)$ in $\ensuremath{ {\sim}\,} 12 \unit{arcmin}^2$ \HEALPix pixels ($\code{nside} = 1024$).
The resulting median magnitude limits over the DELVE DR1 footprint are shown in \tabref{depth}.
We show histograms of the \magpsf magnitude limit for point-like sources at S/N=5 in the left panel of \figref{maglim}.
In the right panel of \figref{maglim} we show the DELVE DR1 area as a function of depth in each band.
The magnitude limits as a function of location on the sky are shown in \appref{depth}.
The median S/N=10 point-source depth of DELVE DR1 is comparable to the point-source depth from the first two years of DES \citep{Drlica-Wagner:2015}, which is consistent with the goal of the DELVE-WIDE program.
\input{table_depth}
\subsection{Object Classification}
\label{sec:classification}
DELVE DR1 includes the \code{SourceExtractor} \spreadmodel parameter, which can be used to separate spatially extended galaxies from point-like stars and quasars \citep[e.g.\xspace,][]{Desai:2012}.
Following DES \citep[e.g.,][]{DES-DR1:2018,DES-DR2:2021}, we define \extclass parameters as a sum of several Boolean conditions,
{\par\footnotesize\begin{align}\begin{split}
\var{ext}&\var{ended\_class\_g} = \\
~&((\spreadmodel[g] + 3\, \spreaderrmodel[g]) > 0.005) \\
+&((\spreadmodel[g] + \spreaderrmodel[g]) > 0.003) \\
+&((\spreadmodel[g] - \spreaderrmodel[g]) > 0.003).
\end{split}\end{align}}
\noindent When true, each Boolean condition adds one unit to the classifier such that an \extclass value of 0 indicates high-confidence stars, 1 is likely stars, 2 is likely galaxies, and 3 is high-confidence galaxies.
Objects that lack coverage in a specific band or where the \spreadmodel fit failed are set to a sentinel value of $-9$.
We calculate \extclass values similarly for each band; however, we recommend the use of \extclass[G] since the $g$ band has the widest coverage and deepest limiting magnitude.
We evaluate the performance of \extclass[g] by matching DELVE DR1 objects to data from the W04 (WIDE12H+GAMA15H) equatorial field of the wide layer of HSC-SSP PDR2 \citep{HSC-DR2}.
The superior image quality ($i$-band PSF FWHM $\ensuremath{ {\sim}\,} 0\farcs6$) and depth ($i \sim 25.9 \unit{mag}$) of HSC-SSP PDR2 enable robust tests of star--galaxy separation in DELVE DR1.
The matched data set covers $\ensuremath{ {\sim}\,} 95 \deg^2$ and contains $\ensuremath{ {\sim}\,} 2.3$ million matched objects.
Following \citet{DES-DR1:2018}, we select point-like sources from HSC-SSP DR2 based on the difference between the $i$-band PSF and model magnitudes of sources,
{\par\footnotesize\begin{align}\begin{split}
\var{hsc}&\var{\_stars} = \\
& ((\var{i\_psfflux\_mag} - \var{i\_cmodel\_mag}) < 0.03) \\
& ||~ (~ ( (\var{i\_psfflux\_mag} - \var{i\_cmodel\_mag}) < 0.1) \\
& ~~~ \& ~ (\var{i\_psfflux\_mag < 22})~).
\end{split}\end{align}}
We require that the PSF and model magnitudes are very similar for fainter sources, while the agreement is relaxed for brighter sources.
This selection results in $\ensuremath{ {\sim}\,} 1.7$ million matched objects classified as galaxies and $\ensuremath{ {\sim}\,} 0.6$ million matched objects classified as stars.
We use these objects to evaluate the differential performance of DELVE DR1 \extclass[g] as a function of magnitude in \figref{stargal}.
We report the performance of nominal star ($0 \leq \extclass[g] \leq 1$) and galaxy ($2 \leq \extclass[g]$) classifications integrated over the magnitude range $19 \leq \magauto[g] \leq 22$ mag in \tabref{summary}.
Spatial maps of high-confidence stars and galaxies are shown in \figref{stargalmap}.
The stellar density map clearly shows increasing stellar density toward the Galactic plane, and the galaxy density map shows the large-scale clustering of galaxies.
Stellar contamination of the galaxy sample on the eastern edge of the footprint correlates with the Galactic bulge.
The underdense regions in the northeast and southwest of the galaxy map correlate with interstellar extinction, which has not been corrected for when creating this map.
Color--magnitude diagrams for looser selections of stars and galaxies are shown in \figref{cmd}.
\begin{figure}
\centering
\includegraphics[width=0.9\columnwidth]{delve_stargal_sep.pdf}
\caption{\emph{Top:} Stars and galaxies occupy distinct regions of \spreadmodel space.
At fainter magnitudes it becomes harder to distinguish between the two types of objects.
\emph{Middle:} DELVE DR1 stellar classification performance as estimated from matched HSC observations.
\emph{Bottom:} Similar to the middle panel, but showing DELVE DR1 galaxy classification performance.}
\label{fig:stargal}
\end{figure}
\begin{figure*}[t!]
\centering
\includegraphics[width=0.49\textwidth]{delve_stars_n512.pdf}
\includegraphics[width=0.49\textwidth]{delve_galaxies_n512.pdf}
\caption{\label{fig:stargalmap} \textit{Left}: Stellar density map created with the $\extclass[g] = 0$ selection described in \secref{classification}. \textit{Right}: Analogous galaxy counts map created with the $\extclass[G] = 3$ selection. The region of lower galaxy density toward the northeast of the footprint can be attributed to higher interstellar extinction, which is not corrected for in this map. Color range units are number of objects per \HEALPix $\code{nside} = 512$ pixel. Both maps use a magnitude threshold of $\magauto[g] < 22$.}
\end{figure*}
\begin{figure}[t!]
\centering
\includegraphics[width=\columnwidth]{delve_cmd_lin.pdf}
\caption{\label{fig:cmd} Color--magnitude diagrams for objects in a $\ensuremath{ {\sim}\,} 30\deg^2$ region centered on ${\ensuremath{\rm RA}}\xspace,{\ensuremath{\rm Dec}}\xspace = (140,-20)\deg$.
\textit{Left}: PSF magnitudes for objects classified as stars ($0 \leq \extclass[g] \leq 1$).
\textit{Right}: Automatic aperture magnitude for objects classified as galaxies ($\extclass[g] \geq 2$).}
\end{figure}
\subsection{Known Issues}
\label{sec:issues}
The DELVE DR1 detrending and image reduction pipeline used flat-field images, bias images, bad pixel masks, and other image calibration products (i.e., ``supercals'') assembled during the processing of DES Y4.
These are believed to be the best available instrument calibrations for exposures taken prior to 2017 September (DECam exposure numbers $< 674970$).
However, newer calibrations are available from DES Y5 and Y6.
Applying these calibrations to exposures taken after 2017 September leads to a $\ensuremath{ {\sim}\,} 10 \unit{mmag}$ shift in the photometric zeropoints.
While the DELVE DR1 calibration to ATLAS Refcat2 removes these shifts, we note that more recent processing of the DELVE data uses the new DES calibration products and thus the derived zeropoints will change slightly in future releases.
Deriving updated calibration products extending beyond the end of DES in 2019 February is a topic of ongoing work.
As noted in \secref{photrel}, there is a residual photometric offset of $11.2\unit{mmag}$ that is apparent when comparing DELVE DR1 to {\it Gaia}\xspace DR2 across the boundary of ${\ensuremath{\rm Dec}}\xspace = -30\deg$.
This likely arises because ATLAS Refcat2 switches from PS1 to SkyMapper at this declination.
Internal relative photometric calibration of DELVE is expected to reduce inhomogeneities in the photometric calibration in the future.
Regions around bright stars suffer from scattered and reflected light artifacts \citep[e.g.,][]{DES-DR1:2018}.
Unlike DES, no effort was made to identify and remove affected CCDs prior to DELVE processing.
This can lead to regions of higher spurious object rates, biased object colors, and incorrect object sizes.
These regions can often be identified at catalog level by examining object colors \citep[i.e., Section 7.4 of][]{Drlica-Wagner:2018}, and most affected exposures were removed during DELVE DR1 validation.
Several problematic regions can be seen as spurious overdensities in the galaxy map shown in \figref{stargalmap}.
A prominent overdensity at $({\ensuremath{\rm RA}}\xspace,{\ensuremath{\rm Dec}}\xspace) \sim (155.5,-8)\deg$ has been associated with a PSF fitting failure that leads to incorrect star/galaxy separation.
Another spurious overdensity at $({\ensuremath{\rm RA}}\xspace,{\ensuremath{\rm Dec}}\xspace) \sim (190.5,-45)\deg$ (\HEALPix 10491) corresponds to a partial failure in the object matching procedure.
Each of these issues has been corrected in subsequent processing of the DELVE data.
The DESDM pipeline used to produce DELVE DR1 is not optimized for crowded-field photometry. This leads to severe blending and incompleteness in regions of high stellar density (see Section 4.6 of \citealt{DES-DR1:2018}). Blending is likely responsible for some of the increased stellar contamination in the galaxy sample close to the Galactic bulge seen in \figref{stargalmap}.
In general, the catalog creation process merged object detections located $< 1\arcsec$ from each other on different exposures (\secref{processing}).
The process for splitting two physical objects located within $1\arcsec$ required that these objects were detected on more than one exposure.
This splitting process was found to be implemented suboptimally in DELVE DR1.
Furthermore, two objects were not split if they were only resolved in one exposure due to lack of observations or reduced image quality in other observations.
Finally, higher-multiplicity systems (i.e., three or more objects within 1\arcsec) were merged into a single object.
Therefore, the DELVE DR1 catalog should be used with caution when analyzing multiple sources separated by $<1\arcsec$.
The process of associating individual detections into a unique object catalog will be improved in future data releases.
\subsection{Data Access}
\label{sec:access}
Access to DELVE DR1 is provided through the Astro Data Lab \citep{Fitzpatrick:2016,Nikutta:2020},\footnote{\url{https://datalab.noirlab.edu}} part of the Community Science and Data Center (CSDC) hosted by NOIRLab.
The Astro Data Lab is a science platform that provides online access and analysis services for large survey data sets.
It provides database access to catalog data from both a Table Access Protocol (TAP)\footnote{\url{http://ivoa.net/documents/TAP}} service and from direct PostgreSQL queries via web-based, command-line, and programmatic query interfaces.
The Astro Data Lab also provides the ability to cross-match catalogs against other tables held by Astro Data Lab; a cross-match has already been performed between DELVE DR1 and {\it Gaia}\xspace EDR3 \citep[\appref{tables};][]{Gaia:2020a}.
The Astro Data Lab also provides an image cutout service, built on the Simple Image Access (SIA) protocol, that points at the petabyte-scale holdings of NOIRLab's Astro Data Archive, including the DELVE DR1 images.
Furthermore, the Astro Data Lab provides a JupyterHub-based notebook analysis environment capable of accessing all of its data sets and services, as well as remote storage space for user databases and files.
We used templates provided by the Astro Data Lab to build the DELVE DR1 web pages\footnote{\url{https://datalab.noirlab.edu/delve}} and Jupyter notebooks demonstrating DELVE DR1 data access.\footnote{\url{https://github.com/noaodatalab/notebooks-latest/blob/master/05_Contrib/Galactic/DELVE_DR1/DELVE_access_dr1.ipynb}}
More detailed information on using the data services for DELVE DR1 science can be found on the Astro Data Lab website.
\section{Science Examples}
\label{sec:examples}
Below we present a few examples of scientific investigations that are possible with the DELVE DR1 data. This list is not intended to be comprehensive, but instead provides a glimpse at the wide range of topics that can be studied with DELVE DR1.
\subsection{Faint Milky Way Satellites}
\begin{figure*}[t]
\centering
\includegraphics[width=0.98\textwidth]{sextans_diagnostic.pdf}
\caption{The Sextans dwarf spheroidal galaxy observed with DELVE DR1. \textit{Left:} Normalized spatial density of stellar sources ($0 \leq \extclass[g] \leq 1$ using the prescription from \secref{classification}) smoothed with a $1\arcmin$ Gaussian kernel. \textit{Middle:} Normalized spatial density of galaxies ($\extclass[g] \geq 2$) smoothed by the same kernel. \textit{Right:} Color--magnitude diagram of all stellar sources within $10\arcmin$ of the system centroid. }
\label{fig:sextans}
\end{figure*}
One of the primary science objectives of DELVE is the discovery of ultra-faint satellites in the halo of the Milky Way.
Early DELVE data have already resulted in the discovery of three systems: Centaurus I, DELVE 1, and DELVE 2 \citep{Mau:2020,Cerny:2020}.
Additional satellites are expected to be discovered as the coverage, quality, homogeneity, and depth of the DELVE data continue to improve.
DELVE DR1 includes several known stellar systems spanning a range of luminosities.
The DELVE DR1 footprint contains the Milky Way dwarf spheroidal satellite galaxies Centaurus I \citep{Mau:2020}, Hydra II \citep{Martin:2015}, Leo IV \citep{2007ApJ...654..897B}, and Sextans \citep{1990MNRAS.244P..16I}.
Furthermore, DELVE DR1 contains several faint outer halo star clusters including BLISS 1 \citep{Mau:2019}, DELVE 1 \citep{Mau:2020}, Kim 3 \citep{2016ApJ...820..119K}, and Laevens 1/Crater 1 \citep{Laevens:2014a,2014MNRAS.441.2124B}.
The DELVE DR1 data can be used to study the extended stellar populations of known dwarf galaxies and star clusters.
Such studies can detect signatures of tidal disturbance, which can help inform the evolutionary history of these systems \citep[e.g.,][]{2019ApJ...885...53M}.
In \figref{sextans}, we show an example of the DELVE DR1 data surrounding the Sextans dwarf spheroidal galaxy, which is located at a distance of $84.7 \pm 0.4$ \unit{kpc} \citep{Vivas:2019}.
In the left (middle) panel, we plot the smoothed spatial distribution of stars (galaxies) after applying a selection based on \extclass[g] as presented in \secref{classification}.
We note similar variations in the central stellar density of Sextans as were reported by \citet{Roderick:2016}.
In the right panel, we present a color--magnitude diagram for the stars located within $10\arcmin$ of Sextans, demonstrating that the DELVE DR1 $g$-band depth provides high-precision photometry down to the subgiant branch of this system.
\subsection{Stellar Streams}
\label{sec:streams}
\begin{figure}[t!]
\centering
\includegraphics[width=0.98\columnwidth]{pal5_nofan_simple_gcut_extended01.pdf}
\caption{Leading tail of the disrupting globular cluster Palomar 5. Stellar objects were selected from DELVE DR1 with $0 \leq \extclass[g] \leq 1$ and a color--magnitude filter consistent with the stellar population of the Palomar 5 cluster (\secref{streams}).
}
\label{fig:pal5}
\end{figure}
The contiguous, deep coverage of DELVE DR1 enables the discovery and study of stellar streams at large Galactocentric distances \citep[e.g.\xspace,][]{Odenkirchen:2001,Newberg:2002,Belokurov:2006,Grillmair:2006,Bonaca:2012,Koposov:2014,Bernard:2014,Bernard:2016,Grillmair:2017,Drlica-Wagner:2015,Balbinot:2016,Shipp:2018,Shipp:2020}.
Following on other recent analyses \citep[e.g.\xspace,][]{Shipp:2019, Shipp:2020, Bonaca:2020, Li:2020}, DELVE DR1 photometry can be used to characterize known streams in tandem with complementary proper-motion measurements from {\it Gaia}\xspace and targeted spectroscopic follow-up.
The phase-space distribution of the population of stellar streams can be leveraged to construct a global map of the Milky Way's gravitational field \citep[e.g.,][]{Bovy:2016,Erkal:2016,Bonaca:2018}, while their internal structure may hold clues to the particle nature of dark matter \citep{Carlberg:2013,Erkal:2016b,Banik:2019}.
In \figref{pal5}, we show an example of the tidal disruption of the Palomar 5 (Pal 5) globular cluster \citep{Odenkirchen:2001}.
We select likely stars ($0 \leq \extclass[G] \leq 1$; \secref{classification}) consistent with the main sequence of Pal 5 in a de-reddened color--magnitude diagram.
Specifically, we select stars consistent with an old ($11.5\unit{Gyr}$), metal-poor (${\rm [Fe/H]} = -1.3$) isochrone from \citet{Dotter:2008} shifted to a distance of $22.5\unit{kpc}$.
To mitigate effects of nonuniform depth, we only include stars that are brighter than $g = 23.1$ mag.
The leading arm of Pal 5 resides in the DELVE DR1 footprint and is prominently detected extending to ${\ensuremath{\rm Dec}}\xspace \lesssim -5 \deg$, confirming similar results from \citet{Bonaca:2020}.
\subsection{Strong Lensing}
\begin{figure}[t!]
\centering
\includegraphics[width=0.98\columnwidth]{delve_quads.png}
\caption{DELVE composite color images ($g, r, i$) of the quadruply lensed quasar systems J1131GL \citep[left;][]{Sluse:2003} and 2M J1134-2103 \citep[right;][]{Lucey:2018}.
\label{fig:delve_quads}}
\end{figure}
Strong gravitational lens systems are excellent probes of galaxy structure, dark matter, and dark energy \citep[e.g.,][]{Treu:2010}.
The total catalog of confirmed strong lens systems numbers only in the hundreds \citep[e.g.,][and references therein]{Jacobs:2019a},
but recent wide-area surveys have rapidly increased the number of candidate systems through both visual inspection of image data and the use of machine-learning techniques.
Thousands of new lens candidates have been identified in DES \citep[e.g.\xspace,][]{Nord:2016,Diehl:2017, Jacobs:2019b,Jacobs:2019a}, DECaLS \citep[e.g.\xspace,][]{Huang:2020, Huang:2021}, and PS1 \citep[e.g.\xspace,][]{Canameras:2020}.
DELVE is expected to contain several thousand galaxy--galaxy lenses and $\ensuremath{ {\sim}\,} 100$ lensed quasars brighter than $i \sim 18.5 \unit{mag}$ based on the estimates of \citet{Collett:2015} and \citet{Treu:2018}, respectively.
In \figref{delve_quads}, we show DELVE images of two quadruply lensed quasars: J1131GL \citep[left;][]{Sluse:2003} and 2M J1134-2103 \citep[right;][]{Lucey:2018}.
DELVE has partnered with the {\it STRong lensing Insights into the Dark Energy Survey} \citep[STRIDES;][]{Treu:2018} to initiate searches for previously undiscovered galaxy--galaxy lenses and lensed quasars using the DELVE data.
\subsection{Galaxies in Different Environments}
While DELVE primarily focuses on faint galaxies in the Local Volume, the DELVE DR1 data can also be used to study more luminous galaxies at larger distances.
The wide coverage and depth of DELVE enable the investigation of galaxy populations spanning a range of different environments (from isolated galaxies to galaxy clusters).
The study of galaxies in different environments can help reveal the mechanisms responsible for variations in the structure and physical properties of galaxies as they form and evolve over cosmic time \citep[e.g.,][]{Darvish:2018,Chartab:2020}.
As an example, early community data processed by DELVE have been used to perform environmental studies of local analogs to high-redshift galaxies \citep{Santana-Silva:2020}.
Tidal streams around massive galaxies are a ubiquitous aspect of galaxy formation that has not yet been fully explored due to the deep imaging required to detect these systems beyond the Local Group \citep{Bullock:2005, Cooper:2010, Martinez-Delgado:2010, Morales:2018}.
A systematic survey for galactic remnants in a large sample of nearby galaxies is needed to understand whether the recent merger histories of the Milky Way and Andromeda are ``typical.''
DELVE has partnered with the {\it Stellar Stream Legacy Survey} \citep{Martinez-Delgado:2021}, a systematic imaging survey of tidal features around nearby galaxies that has reached surface brightnesses as faint as $\ensuremath{ {\sim}\,} 29 \magn \asec^{-2}$ using public data from the DESI Legacy Imaging Surveys \citep{Dey:2019}.
The DELVE data will enable a search for extragalactic stellar streams around several thousand galaxies in the southern hemisphere.
A catalog of extragalactic stellar streams will probe the current mass assembly rate of nearby galaxies, the stellar content and orbits of satellites, and the formation of stellar halos.
\subsection{Galaxy Clusters}
\begin{figure*}[t!]
\centering
\includegraphics[width=0.35\textwidth, trim=0cm -1.5cm 0cm 0cm]{abell1689_small.png}
\includegraphics[width=0.55\textwidth, trim=0cm 0.5cm 0cm 0cm, clip]{A1689_color-magnitude.pdf}
\caption{DELVE DR1 data around the rich galaxy cluster Abell 1689 at a redshift of $z=0.18$. (Left) False-color image assembled from $g,r,z$ images covering a $5\farcm8 \times 5\farcm8$ region around Abell 1689. (Right) Color--magnitude diagram of galaxies in Abell 1689 after a statistical background subtraction based on the field galaxy population. The red solid line is a linear red sequence relationship derived from the DES Y1 redMaPPer cluster catalog \citep{McClintock2019}, and the red dashed lines represent the expected 95\% containment region for red sequence galaxies. The red markers are spectroscopically confirmed members of the system.
}
\label{fig:a1689}
\end{figure*}
Galaxy clusters are a powerful tool for studying cosmology and galaxy evolution in dense environments \citep[][]{Allen2011,Zenteno16,Hennig2017,Bocquet2018,DESY1ClusterCosmology}.
Assembling a large population of galaxy clusters is essential for improving measurements of cosmological parameters \citep[][]{kidsviking,DES:2018,McClintock2019}.
DELVE collaborates closely with DeROSITAS (PI: Zenteno) to assemble contiguous imaging in $g,r,i,z$ and enable wide-area optical studies of galaxy clusters.
The optical data can be combined with data from large surveys in the submillimeter \citep[e.g.\xspace,][]{Bleem:2020,Hilton:2020} and X-ray \citep[e.g.\xspace, eROSITA;][]{Predehl:2020} regimes.
Multiwavelength studies of galaxy clusters can help mitigate systemics in optical studies \citep{Grandis2021} and provide information about the dynamical state of galaxy clusters, which is important for cosmological measurements \citep[e.g.\xspace,][]{Andrade-Santos:2012,Andrade-Santos:2017,Lovisari:2020a,Lovisari:2020b} and studies of galaxy evolution \citep[e.g.\xspace,][]{Zenteno20}.
As a proof of concept, \figref{a1689} shows DELVE DR1 data associated with the rich galaxy cluster Abell 1689 \citep[e.g.\xspace,][]{Limousin2007}.
We use a kernel-density estimate to subtract likely field galaxies from the color--magnitude diagram \citep{Pimbblet2002}, and we recover a prominent red sequence without the need for spectroscopic membership information or photometric redshifts.
The colors and magnitudes of DELVE DR1 galaxies (black circles) follow the linear red sequence relation derived from the DES Y1 redMaPPer catalog \citep[red line;][]{McClintock2019} demonstrating consistency between these data sets in the galaxy cluster regime.
In the future, we plan to assemble a catalog of galaxy clusters by using a cluster finder algorithm such as redMaPPer \citep{2014ApJ...785..104R} or WAZP \citep{Aguena:2021}.
These algorithms have been validated on DES data and can be naturally adapted to DELVE data.
\section{Summary}
\label{sec:conclusion}
DELVE seeks to study the fundamental physics of dark matter and galaxy formation through rigorous systematic studies of dwarf galaxies and stellar substructures in the Local Volume.
To do so, DELVE aims to complete contiguous coverage of the southern, high Galactic latitude sky in $g,r,i,z$ to a depth comparable to two years of DES observations.
As of 2021 January, DELVE has completed slightly less than half of the 126 nights of scheduled DECam observing.
The first major public release of DELVE DR1 combines observations from the DELVE-WIDE program with archival DECam data to cover $\ensuremath{ {\sim}\,} 4000 \deg^2$ of the northern Galactic cap to a $5\sigma$ depth of $g=24.3, r=23.9, i=23.3, z=22.8 \unit{mag}$ (\tabref{summary}).
The DELVE DR1 catalog contains PSF and automatic aperture measurements for $\ensuremath{ {\sim}\,} 520$ million astronomical objects produced by the DESDM pipeline and accessible through the NOIRLab Astro Data Lab.
DELVE DR1 has utility for a broad range of scientific investigations in the Local Volume and beyond.
Future DELVE data releases will increase the coverage, uniformity, and depth of the DELVE catalogs.
Furthermore, these releases will include products from the DELVE-MC and DELVE-DEEP programs, which are processed with the multiepoch point-source pipeline from SMASH.
We anticipate that DELVE DR1 and future DELVE data releases will be a valuable resource for the community in advance of the Vera C.\ Rubin Observatory Legacy Survey of Space and Time.
\input{ack.tex}
\facilities{Blanco (DECam), Astro Data Lab, {\it Gaia}\xspace}
\software{
\code{astropy} \citep{astropy:2018},
\code{fitsio},\footnote{\url{https://github.com/esheldon/fitsio}}
\HEALPix \citep{Gorski:2005},\footnote{\url{http://healpix.sourceforge.net}}
\code{healpy} \citep{Zonca:2019},\footnote{\url{https://github.com/healpy/healpy}}
\code{healsparse},\footnote{\url{https://healsparse.readthedocs.io/en/latest/}}
\code{matplotlib} \citep{Hunter:2007},
\code{numpy} \citep{NumPy:2020},
\code{PSFEx} \citep{Bertin:2011},
\code{scipy} \citep{Scipy:2020},
\SCAMP \citep{Bertin:2006},
\code{skymap},\footnote{\url{https://github.com/kadrlica/skymap}}
\code{SourceExtractor} \citep{Bertin:1996},
\SWARP \citep{Bertin:2002, Bertin:2010}
}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 5,371
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Q: how to update useState() and push to first position? Using const [projects, setProjects] = useState([]) I can update projects, but how can I push to projects at the first position?
I am using
setProjects([...projects, newData])
but that adds the newData at the end. I want it to be at the first position.
A: setProjects([newData, ...projects ])
A: You're almost there -
setProjects([newData, ...projects])
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 9,852
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A billionaire's baby deal!
One look into his infant son's trusting blue gaze and Aiden Langford knows his wild, carefree days are over. If only he can get Sarah Daltrey, his son's temporary guardian, to give him daddy lessons... Certainly the soft-hearted entrepreneur will agree to his ten-day proposal to stay as the nanny. Aiden just needs to keep his mind on parenting and off Sarah's seductive curves...
Being in the handsome tycoon's arms puts up Sarah's emotional defenses even as her body begs her to let go. But being a babysitter with benefits isn't on her agenda—especially for a father and son who might steal her heart...
Stay with me.
Sarah wasn't sure she'd heard Aiden's words correctly. They were surprising. They were scary—driving her to a place where she surrendered to her deep longing for him.
He granted the smallest fragment of a smile, looking at her with his heartbreaking blue eyes. He tenderly tucked her hair behind her ear, drawing his finger along her jaw to her chin. "I don't know what force in the universe brought you to me, Sarah. I only know that right now I need you. I want you. And I'd like to think that you want me, too."
The air stood still, but Sarah swayed, light-headed from Aiden's words. Their one night together had been electric, filling her head with memories she'd never surrender, but judging by the deep timbre of Aiden's voice, they might shatter what happened in Miami. "I don't want to ruin our friendship." And no-strings-attached only breaks my heart.
"Is that why you shut things down after Miami?"
"Yes." It wasn't the whole truth, but it was enough. As much as sleeping with Aiden might be a mistake, she didn't want to deprive herself of him. Would one more time really hurt? "And I've spent the last two nights regretting it."
"Then I say we have no more regrets."
Before she knew what was happening, he scooped her up into his arms.
* * *
Ten-Day Baby Takeover is part of Harlequin Desire's #1 bestselling series, Billionaires and Babies: Powerful men...wrapped around their babies' little fingers.
Dear Reader,
Thank you for picking up The Ten-Day Baby Takeover! It's the third book centered on the Langford siblings, and I have loved chronicling their dysfunctional dynamic. This story starts five months after Pregnant by the Rival CEO and is about Aiden, the mysterious older brother who has come home to New York to make peace with his family. Little does he know he's about to get a family of his own!
As an author, some books come together easily and some are an endless source of frustration. I wouldn't say that The Ten-Day Baby Takeover came "easily," but Aiden certainly materialized before my eyes like magic. I've been thinking about him for two books, so I guess my brain was ready to put him center stage.
Writing Aiden's story was so satisfying. He's everything I love in a hero—strong, the lone wolf, the one who claims he's not capable of being hurt, or capable of loving anyone. I won't spoil the story, but the arrival of the son he didn't know he had and his temporary caregiver, outspoken Sarah, brings big parts of Aiden's past to the surface. Deeply painful Langford family secrets come out—things we've wanted to know about for three books! In the end, only loving Sarah and sweet baby Oliver can make his heart whole again. I really hope you enjoy it!
Last, for anyone wanting to know whether Anna Langford and her husband, Jacob, had a girl or a boy, you'll find that out, too!
Please drop me a line any time at karen@karenbooth.net. I love hearing from readers!
Karen
KAREN BOOTH
The Ten-Day Baby Takeover
Karen Booth is a Midwestern girl transplanted in the South, raised on '80s music, Judy Blume and the films of John Hughes. She writes sexy big-city love stories. When she takes a break from the art of romance, she's teaching her kids about good music, honing her Southern cooking skills or sweet-talking her husband into whipping up a batch of cocktails. Find out more about Karen at karenbooth.net.
Books by Karen Booth
Harlequin Desire
That Night with the CEO
Pregnant by the Rival CEO
The CEO Daddy Next Door
The Best Man's Baby
The Ten-Day Baby Takeover
Visit her Author Profile page at Harlequin.com, or karenbooth.net, for more titles.
Join Harlequin My Rewards today and earn a FREE ebook!
Click here to Join Harlequin My Rewards
http://www.harlequin.com/myrewards.html?mt=loyalty&cmpid=EBOOBPBPA201602010002
For my dear friend in the writing world and the real world, Margaret Ethridge. I will always want to stay up way past my bedtime, talking and giggling in the dark with you.
Contents
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
Excerpt from _Expecting the Billionaire's Baby_ by Andrea Laurence
One
The lobby of LangTel's Manhattan headquarters was practically a shrine to order and quiet restraint. It was not the place to bring a fussy baby. Sarah Daltrey had done precisely that. Marble floors, towering ceilings and huge expanses of windows facing the street made any sound, especially baby Oliver's errant cries, echo and reverberate like crazy.
Sarah kissed his forehead, bouncing him on her hip as she paced in the postage stamp waiting area. For such a massive building, taking up nearly an entire city block, LangTel had been distinctly stingy with the amenities for the uninvited. Two chairs and a ten-by-ten rug sat opposite a closely guarded bank of elevators. It was clear that no one occupying this space should stay for long.
Oliver whimpered and buried his head in her neck. Poor little guy—none of this was his fault. Oliver hadn't asked to take a four-hour train ride that morning. He certainly hadn't asked to come to an ice-cold office building in the middle of his nap time. More than anything, Oliver hadn't asked to lose his mother three weeks ago, nor had he asked to have a father who refused to acknowledge his existence.
Sarah took her cell phone and dialed the number she'd memorized but wasn't about to add to her contacts. As soon as she got Oliver's dad to accept his paternal responsibility, she'd force herself to forget the string of digits that led to an office somewhere in this building, most likely the top floor. There would be no maintaining ties with Aiden Langford. Their connection was temporary, albeit of paramount importance. She had his son and he was going to take custody, even if it killed her.
"Yes. Hello. It's Sarah Daltrey. I'm calling for Aiden Langford. Again."
One of the two security guards manning the lobby gave her the side-eye. Meanwhile, the woman on the other end of the phone line expressed equal disdain with her snippy tone. "Mr. Langford has told me a dozen times. He does not know you. Please stop calling."
"I can't stop calling until he finally talks to me."
"Perhaps I can help you."
"No. You can't. This is a personal matter and Mr. Langford should appreciate that I'm not sharing the details of this situation with his assistant. I outlined it all in the email I sent to him." More like seven emails, but who's counting? "If I can just have five minutes of his time, I can explain everything." Five minutes was a lie. She'd need at least an hour to walk Mr. Langford through Oliver's schedule, his likes and dislikes, and to make sure he was off to as good a start as possible.
"Mr. Langford is very busy. I can't put through the call of every person who claims to need his time."
"Look. I just spent four hours on a train from Boston to New York and I'm downstairs in the lobby, caring for a ten-month-old sorely in need of a nap. I'm not leaving until I speak to him. I'll sleep here if I have to."
"I can have security escort you from the building, Ms. Daltrey. Surely you don't want that."
"Does LangTel want the embarrassment of their security removing a kicking and screaming woman with a baby from their lobby?"
Mr. Langford's assistant said everything with her momentary silence. "Can you hold, please? I'll see if there's anything I can do."
Sarah had very little hope for this, but what other options did she have? "Sure. I'll hold."
Just then, a statuesque woman with glossy brown hair dressed in a tailored gray dress and black pumps came through the revolving door. Sarah might not have noticed her, but she had a baby bump that was impossible to miss. The security guard beelined to her, taking the stack of papers in her arms. "Good afternoon, Ms. Langford. I'll get the elevator."
Anna Langford. Sarah recognized her now, from the research she'd done on the Langford family while trying to find a way to get to Aiden. Anna was one of two LangTel CEOs, along with her brother Adam. She was also Aiden Langford's younger sister.
Oliver dropped his favorite toy, a stuffed turtle, and unleashed a piercing wail. Sarah cringed, crouching down, scooting across the carpet in her wedge sandals, scrambling for Oliver's toy while cradling the phone between her ear and shoulder. Anna came to a dead stop and turned her head, zeroing in on Sarah and Oliver.
Great. Now we really are going to get kicked out of the lobby.
Anna frowned and strode closer, but when she removed her sunglasses, there was only empathy in her eyes. "Oh no. Somebody's unhappy."
Certain that she'd been banished to the land of horrible hold music, Sarah ended her call and tucked her phone into the diaper bag. "Sorry about that. It's nap time. He's tired." When she straightened to face Anna, she felt as if she needed a step stool. Anna was tall and in heels, while Sarah was height challenged even in her strappy sandals.
Anna shook her head. "Please don't apologize. This is the highlight of my day. He's adorable." She reached for Oliver's pudgy hand and smiled. He responded by gripping her finger, his head resting on Sarah's shoulder. "I'm Anna Langford, by the way."
"I'm Sarah. Daltrey. This is Oliver." Sarah watched as Oliver smiled shyly at Anna. He was such a sweet and trusting boy. Saying goodbye to him was going to be heartbreaking, especially after three weeks of caring for him all on her own, but that was her charge and there was nothing to be done about that. She was done with being a nanny, and caring for a child that wasn't her own, regardless of the circumstances, felt far too much like her old life.
Anna's eyes didn't stray from Oliver. "Nice to meet you both. I'm due to have my own little one in about six weeks. Middle of June. I have baby fever right now, big time." She studied the baby's face. "Your son's eyes are incredible. Such a brilliant shade of blue."
And exactly like your brother's.
Sarah cleared her throat. "He's not mine, technically. I'm his legal guardian. I'm in the process of connecting him with his father. That's why I'm here."
Confusion crossed Anna's face. "At LangTel. The father works here?"
Sarah had committed herself to discretion for the sake of everyone, especially Oliver, but this might be her one real chance to get to Aiden. She was getting nowhere with his assistant. "I came to see Aiden Langford. He's your brother, right? I need to speak to him about Oliver, but he won't take my phone calls."
"Oh." A flicker of surprise crossed Anna's face as her eyes darted between Oliver and Sarah. "Oh. Wow." She kneaded her temple with the tips of her fingers. "The lobby doesn't seem like a good place to talk about this. Maybe you should come upstairs with me."
* * *
Aiden's assistant buzzed his extension. "Mr. Langford? Your sister is here to see you. She's brought a visitor."
Visitor? "Sure. Send them in." Aiden set aside the LangTel global marketing report he'd been skimming, easily the driest thing he'd ever read, which was saying a lot. With more than a dozen years in business under his belt, he'd digested his fair share of dull financial projections and legal briefs. He preferred to rely on his gut when making decisions. Billions later, the strategy had served him well.
In walked Anna with a blonde woman he didn't know. To say the stranger was eye catching would've been dismissive. With full pink lips and big blue eyes, wearing a black sundress, she was natural femininity embodied. Their gazes connected and he noticed the faintest of freckles dotting her cheeks. His tastes in women were wide and varied, but this woman ticked off more of his "yes" boxes than he cared to admit. Unfortunately, one thing about her made her absolutely not his type—the baby asleep in her arms. As a skilled avoider of emotional entanglements, moms were not on his list of women suitable for dating.
"Aiden, I want you to meet Sarah Daltrey," Anna said softly.
That name ended all thought of sexy sundresses and freckles. "You're the woman who keeps calling. You just called from the lobby. How in the world did you get to my sister?"
Anna shushed him. "The baby. He's sleeping."
The baby. His brain whirred into overdrive. He'd read Sarah's email. Well, one of them at least. That was enough to help him decide he shouldn't speak to her. He'd had false paternity accusations thrown at him before. When you have a vast fortune and come from a family well-known for success, you might as well have a target on your back. "This isn't right." His gut told him this was all wrong. "I don't know what Ms. Daltrey is after, but I'm calling security." He reached for the phone, but Anna clapped her hand over his.
"Aiden. Don't. Just listen. Please. It's important."
"I don't know what she's told you, but it's all lies." His pulse throbbed in his ears.
"Five minutes is all I ask, Mr. Langford." Sarah's voice suggested nothing less than calm professionalism. Not exactly the approach of someone unbalanced. But a baby? Oh, no. "If you don't believe me and what I came to tell you, you won't need to call security. I'll leave on my own."
Anna eyed her brother, asking his opinion with an arch of her eyebrows.
With pleas from two women who were obviously not going to give up, what choice did he have? "If it will put an end to this, then fine. Five minutes."
"I'll leave you two to it." Anna stopped at the door, turning to Sarah. "Stop by my office when you're done. I'd love to get the title of that book you mentioned about getting a baby to sleep through the night."
Sarah nodded and smiled as if she and Anna were best friends. What was he in for? "Yes, of course. Thanks so much for your help." The door clicked shut when Anna left, leaving behind a suffocating silence. Sarah cleared her throat and stepped closer, the baby's head still resting on her shoulder. "It would be great if I could sit. He's really heavy."
"Oh, sorry. Of course." Aiden offered a seat opposite his desk. He didn't know what he was supposed to do with himself—stand, sit, cross his arms. Nothing felt right, so he settled on his chair.
"I know this is strange," she started. "So I'll just get right to it. Oliver's mom was my best friend from high school. Her name was Gail Thompson. Does that ring a bell? She told me she met you at the Crowne Lotus Hotel in Bangkok."
Aiden's shoulders tightened. These tidbits of information hadn't been in Sarah's email. She'd only mentioned that she was guardian of his baby. To his knowledge, nobody knew about his brief affair with Gail. They'd met in the hotel bar and spent three days together before she went back to the US. That was the last he'd ever heard from her. "I do remember the name. Yes. But that doesn't mean anything." He shifted in his seat. He knew exactly where this was going. And that made his stomach lurch.
"Nine months after you and Gail had your little tryst in Thailand—" she fluttered her hand at him "—Oliver came along. Eight months after that, Gail called me and told me she had late stage cancer. I was the only person she could sign over guardianship to. She had no siblings—her parents died in a car accident when she was in college. She knew that I used to be a nanny and it just made sense. She said she tried to call you, but had even less luck than I did. It's hard to be persistent when you're dying."
Aiden swallowed hard. Sarah's email had mentioned that the baby's mom had fallen ill. He'd assumed that she was still alive and that this was a scam for money to pay medical bills. "She passed away?" An inexplicable tug came from the center of his chest as his vision drifted to the child. All alone in the world. He'd known that feeling well when he was young, and he despised the idea of any child growing up that way.
"Yes." Sarah pressed her lips together and nodded. She cupped the back of Oliver's head and kissed him softly on the cheek. "That left Oliver with no mom. I was left in charge of finding you so I can sign over guardianship. I think it'd be best for everyone if we kept this as simple as possible and try to wrap it up today."
Today? Did she say what I think she said? No. That was not happening. "You expect to waltz into my office, hand me a baby I've never seen in my life, and then what? You go back to wherever you came from and I'm expected to raise this child? I don't think so, Ms. Daltrey. You aren't going anywhere until I know for certain that the baby is mine. We need lawyers. Paternity tests. I'm not convinced this isn't a big fat hoax."
Her lips pressed into a thin line, but she otherwise seemed unfazed by his reaction. "First off, it's Sarah and his name is Oliver. And I understand you're shocked, but that's not my fault. If you'd taken my phone call, you could've been prepared for this."
"I seriously doubt I would've felt prepared. It's the middle of the workday. I'm a single man and an incredibly busy one at that. I am not prepared to care for a baby I didn't know about five minutes ago." Anger bubbled up inside him, but it was more than this inconceivable situation. He disliked his own dismissive tone. Considering the way his father had treated him, he didn't want to reject the little boy. No child deserved that. Especially one who didn't know who his father was.
"I understand you'll want a paternity test, but I think that the minute you see him awake, you'll realize he's yours. He looks just like you. Especially his eyes. Plus, he has the same birthmark you have on your upper thigh." A flush of pink colored her cheeks. She cast her eyes at her lap, seeming embarrassed. Despite the nature of their conversation, Aiden found it extremely charming. Sarah seemed to be the sort of person who wore her heart on her sleeve, a quality that made her incredibly sexy, too. "I mean, Gail told me you have one. And that's where Oliver gets it."
Sarah carefully hitched up the baby's pant leg. The child must've been incredibly tired—he hardly stirred when she revealed the mark. Aiden's breath caught in his throat. He rounded the desk, dropping down on one knee before them. He had to see it up close. He had to know this was real. The shape and size of the birthmark were indeed the same as his—an oval about the height of a dime, tilting to one side. The dark brown color was a match. Is this possible?
He reached out to touch the mark, but stopped himself. "I'm sorry. I'm a little taken aback."
"It's okay. He's your son." Sarah's voice was sweet and even. Given the impression he had of her from that first email, she was not at all the woman he'd envisioned.
The boy's skin was powdery soft and warm. Aiden gently tugged his pant leg back down, then studied his face. His eyelids were closed in complete relaxation, lined with dark lashes. His light brown hair had streaks of blond, admittedly much like Aiden's, although Oliver had baby-fine curls and Aiden's hair was straight and thick. Still, he knew from his own baby pictures that his hair had once been like Oliver's. Was this possible? Was this really happening? And what was he supposed to do about it? He had no idea how to care for a baby. This would change his entire life. Just when he was getting settled back in New York and trying to find a place for himself in his own family.
Oliver shifted in Sarah's arms, and for an instant, he opened his eyes and looked right at Aiden. The familiar flash of blue was a shot straight to Aiden's heart. It was like staring into a mirror. Oh my God. He's mine.
Two
Things weren't going terribly. Awkward, yes. Terrible, no.
It was really only awkward on Sarah's side of things. Aiden was still on bended knee watching Oliver sleep, and it was impossible not to stare at him. She tried to look elsewhere, to feign interest in the framed black-and-white photographs of exotic locales on his walls, or the view out his office window overlooking the Manhattan skyline, but she could only sustain it for a few moments. His blue eyes would draw her back in, so vivid and piercing she was sure he could hypnotize her if their gazes connected for more than a few heartbeats. They were topped by dark brows that suited his hard-nosed demeanor, accentuated by just a few tiny crinkles at the corners. The scruff on his face was a warm cinnamon brown, neatly tended, but gave him an edge that made her wonder what he was like when he wasn't so guarded. And there was something about the way he carried himself—more than self-assured, he came across as superhuman. Bulletproof. Sarah was certain Aiden Langford did precisely what he wanted to do, when and how he wanted to do it. He was not the sort of man who cared to be told what to do.
Too bad she had to do exactly that. The thought made her pulse race like an overcaffeinated jackrabbit. There was no telling how he would react, but judging by the look on his face, there was a chance it might go okay. However much of a handsome jerk he'd been when she walked in the door, his demeanor had softened in the last few minutes, ever since he'd taken a good look at Oliver. Surely he realized now, that even in the absence of hard evidence like the results of a paternity test, the baby was his.
"So," Sarah started, recalling the speech she'd practiced many times, words she dreaded saying because they would signal the end of her time with Oliver. "I was thinking that I'll leave Oliver with you now and I'll check into a hotel while we get this straightened out. A paternity test is a quick thing. We'll get your name on Oliver's birth certificate. I'll sign over the power of attorney and guardianship. All we need is a lawyer and a few days and then I can be out of your hair."
A crease formed in the center of Aiden's forehead as he stared at her. "Out of my hair?" It was just as tough to look into his eyes as she'd guessed it would be—they really were the spitting image of Oliver's. She'd fallen in love with that shade of blue over the last three weeks. "I already told you that you are not handing me a baby and walking away." He stood and straightened his charcoal suit jacket, which showed off his wide shoulders and broad frame. The way he loomed over her only accentuated his stature. There must've been something in the water in the Langford household—the two she'd met were ridiculously tall. "It seems to me that the more sensible course is for you to keep Oliver until this gets straightened out. You said it yourself—you used to be a nanny. You're used to caring for a child. I have zero experience in this area."
Of course, most single men, especially those who notoriously played the field, weren't in a position to drop everything and care for a baby. But Aiden Langford wasn't most men. Didn't he have a pile of money to throw at the problem? "I used to be a nanny. Past tense. That's no longer my vocation." She stopped short of admitting that she didn't have the stomach for it anymore. "You'll need to hire someone. I wrote down the number for the top nanny agency in the city for you. One phone call and they'll send someone over to help you."
"So I'm not only supposed to work with a complete stranger to take care of a baby, but the baby is supposed to accept that, too?"
He'd gone for the jugular with that one, although he seemed to be doing nothing more than making his case. The thought of anyone aside from his own father caring for Oliver made Sarah's chest, especially everything in the vicinity of her heart, seize up. "I'm a businesswoman, Mr. Langford. I need to return to Boston and my work."
"Business? What sort of business?" Although he was following the logical course of their conversation, Sarah couldn't help but bristle at his dismissive tone.
"I run a women's apparel company. It's really taking off. We can't even keep up with demand."
"Good problem to have. Until your vendors get tired of waiting and move on to something else."
Wasn't that the truth. Half of her day was spent reassuring boutique owners that their orders would be there soon. "That's exactly why I need to be back in Boston. And don't forget that I have been caring for your child full-time for nearly a month. It's time I go back to my life and let Oliver start his new one. With you." That last part had been particularly difficult to say, but the fact that her voice hadn't cracked only bolstered her confidence. She hadn't even shed a tear. It was a miracle.
Aiden sat on the edge of his desk and crossed his arms. His suit jacket sleeves drew taut across his muscles. How was she supposed to hold her own in an argument when he was distracting her with his physique? "So, I'll pay you for your time."
Ah, so he did know how to throw money at a problem. He was just lobbing it in the wrong direction. A breathy punch of a laugh left her lips. "I'm not for hire."
"I'll pay you double whatever your going rate used to be."
She huffed.
"Fine. Triple."
"You're a terrible negotiator."
He shrugged. "I do what's necessary to get what I want."
"That would make me the most expensive nanny in the history of child care. I was paid very well for my services. I was very good at my job."
"You're only making my argument for me. Money is no object, Ms. Daltrey. If Oliver really is my son, he deserves the best. Sounds to me like that's you."
She shook her head. "No way. Absolutely not." This was not the way this was supposed to go. She needed to put an end to Aiden Langford and his money-throwing, muscle-bulging ways.
Oliver fussed and rubbed his eyes, moving his head fitfully as he woke.
Sarah had spoken too loudly. Nap time was apparently now over. She stood and attempted to hand the baby to Aiden. "Here. Take your son. At least for a minute."
Oliver refused, clinging to Sarah.
"See? He clearly wants to be with you. I'm a stranger to him. Would you really leave a baby with a stranger?"
She pursed her lips, calculating her best response. Of course she wouldn't do that. But after the extensive research she'd done on Aiden, he didn't really seem like a stranger. That, however, was not information she cared to share. Which meant she was back at nothing.
"Even worse," he continued. "A stranger who doesn't know how to change a diaper, or what to feed him, or what to do if he starts to cry."
"No idea? I know you have two younger siblings. You never babysat?"
Aiden threaded his fingers through his hair, tousling it in the process. "No."
Well, shoot. She couldn't hand over Oliver to a man he didn't know, especially not one who might not be able to care for him, even if that had been her plan. Her horribly simple plan. "I don't think it's a good idea for me to take Oliver to a hotel, either. He needs to get used to being with you. And you're apparently going to need to learn how to take care of him."
"Excuse me if I haven't thought it out quite that far yet. This is still a new concept for me." He blew out a breath, seeming deep in thought. "I guess the thing that makes the most sense is for you both to stay with me. Until we get things straightened out. And I can hire a nanny. I guess I have to buy a crib, too? I mean, really, this is a lot to pile on a person in one day."
He wasn't wrong. Maybe it would be in Oliver's best interest if she stayed for a couple of days, even if it would make it exponentially more difficult to say goodbye to him. As for the to-do list to get Aiden up and running with the baby, it was a long one if she was going to be thorough. They would need time. With the bad hand Oliver had been dealt in life, she owed it to him to spend a few days in New York so he could be off to the best possible start with Aiden. That was exactly what she'd promised Gail. "Okay. We'll stay at your place."
"You'll have to tell me what you want to be paid. I have no earthly idea how much money a nanny makes. Or even what a nanny does, other than everything a parent would do if they were around."
She'd first said no to Aiden's money on principle, but if she was going to help him with Oliver, she could get something from him that was far more valuable than a paycheck. She knew from her online snooping that he was a whiz when it came to growing companies. It was in his blood—the Langfords were one of the most successful entrepreneurial families in US history. Maybe he could help her solve the countless problems she was facing with trying to take her business to the next level.
"I don't want your money. I want your expertise."
"I'm listening." He cocked an eyebrow at her, threatening to make her throat close up.
"Business expertise. I want you to help me with my company. Help me find investors. Help me figure out my manufacturing issues and widen my distribution."
He nodded, clearly calculating. "That's a tall order. Between that and me going through baby school, this is going to take more than a few days. We'll need at least a week. At least."
How long could she do this? Every minute with Oliver only made her love him more. She clutched him, kissed his head, taking in his sweet baby smell. We don't have to say goodbye today, buddy. I guess that much is good. "Today is Friday. I'll give you ten days. I teach you how to care for Oliver. You help me with my company."
"I think I'd be a fool to say no. You have me in a corner here."
"I mean it, though. Ten days and I'm out of here."
"Like I said. In a corner."
"Okay, then. I want to have a say in the nanny you hire, too. And I want to help outfit the nursery."
Aiden then did the last thing she ever expected. He smiled. Not a lot, just enough to create the tiniest crack in his facade. Sarah felt as if she'd had the wind knocked out of her. His face lit up, especially his eyes. "Anything else?"
"That's all for now."
"Just so you know, fashion is outside my realm of expertise. Women's clothing isn't really my world."
Ah, but he hadn't let her finish. Given Aiden Langford's reputation for being a ladies' man, she had no doubt that he was well-versed in her specialty. "Actually, it's women's sleepwear and lingerie. Something tells me you know at least a little something about that."
Three
Oliver in her arms, Sarah climbed out of Aiden's black SUV, squinting behind sunglasses at the apartment building before them. About a dozen stories high, it had an antique brick facade blanketed in tidy sections of ivy and dotted with tall leaded glass windows. This was not what she'd envisioned for Aiden Langford's abode. She'd assumed a high-rise overlooking Central Park. Wasn't that his birthright? Ritzy address and an equally swanky apartment? Instead, he resided on Fifth Avenue at Twenty-sixth Street, in the Flatiron District with a view of Madison Square Park. She had a sneaking suspicion that Aiden was full of surprises. And that this was the first of many.
"Is that one yours?" She pointed at the highest floor. "The one on top with the biggest terrace?"
Aiden wheeled Sarah's suitcase from the car, lugging the teddy bear that was easily twice Oliver's size, while Aiden's driver John unloaded the remaining bags of toys and baby clothes. "The top four floors are my apartment."
Sarah gulped, surveying the manicured spaces—a formal balcony with stone columns and wrought iron on the lowest level all the way up to one that looked like a park in its own right, each spanning the building. He'd still gone for swanky, merely in a different corner of the city. "That's a lot of room for a single guy."
"My third floor is empty. And the fourth floor is all outdoors. I need my space."
"I'm surprised you don't live up by Anna and her husband. She was telling me she lives only a few minutes from your mom."
Aiden cast his sights down at her, his sunglasses revealing nothing but her own reflection. The crinkles in his forehead and the way his brows drew together were enough indication that he didn't like the question. The driver slammed the car tailgate. Sarah jumped.
"Like I said, I need my space." Aiden's voice was stern, like a father telling his wayward teenage daughter that she'd better be home before eleven.
Okay, then. Dropping the subject.
Together, they entered the beautifully appointed lobby. Black-and-white-checkerboard marble floors and a chandelier dripping with crystals hinted at both wealth and good taste. Sarah pushed Oliver in the stroller while she tried to remember to take deep breaths. Everything about this made her heart beat an uneven rhythm—entering into an agreement with a man she hardly knew, staying in his home, handing over the little boy she'd already grown to love more than she'd thought possible. She did everything she could to ignore the feeling in the pit of her stomach, the one saying that each passing minute was another step away from what she was supposed to be doing—leaving nannying behind, once and for all.
Stop being negative. This is good for Oliver. She had to believe that. Really, it was the best scenario for him—a transition period where his new dad could become acquainted with parenting. They'd find a nanny, set up the nursery. In ten days, this sweet little boy would be given the best possible start at a new life. And she'd get back to hers in Boston, a simple and solitary existence with its own rewards, the most notable of which was the chance to pursue a career that didn't leave her so open to heartbreak.
They stepped onto the elevator and Sarah closed her eyes to ward off her claustrophobia. Plus, every time she looked at Aiden, he got to her with his all-knowing gaze. No wonder the man had such a reputation with the ladies. Most women were probably too mesmerized by his penetrating stare to entertain a single lucid thought beyond, Of course, Aiden. Whatever you want, Aiden.
The elevator dinged, and John, loaded down with the bulk of the baby supplies, held the door for Sarah as she wheeled Oliver off the elevator. They entered a stunning foyer with glossy wood floors, an exotic carved console table and several colorful abstract paintings. Aiden followed with his laptop bag, Sarah's suitcase and the teddy bear, which was a nice counterpoint to his tailored gray suit and midnight-blue tie.
"Where would you like these, Mr. Langford?" John asked.
"Just leave them here. I'm not entirely sure where everything is going yet."
John did as instructed, neatly placing the bags on the table.
"Thank you so much for the help. I really appreciate it," Sarah said to John.
He turned and looked at her as if she had a unicorn horn sprouting from her forehead. "It's my job, ma'am."
"Well, we came with a lot of stuff. I'm sure Mr. Langford doesn't normally make you lug stuffed animals and diaper bags."
"I'm happy to do it. But thank you. For saying thank you." He smiled warmly.
Aiden watched the back and forth. "That's it for now, John. I'll let you know if I need anything else."
"I'll be downstairs, Mr. Langford." John stepped onto the elevator and the doors slid closed.
"He's really nice," Sarah said. "We talked quite a bit while we were figuring how to get the car seat into the SUV. He told me all about his wife and kids. Good guy."
"Of course. A very good guy." Everything in Aiden's voice said that he didn't know the first thing about his driver, and that it quite possibly had never occurred to him to ask.
"Now what?" Sarah wanted Aiden to take the lead. His house. His baby.
"Tell me why a baby needs a stuffed animal this large."
Sarah shrugged, unsubtly peeking ahead at what she could see of the apartment, which seemed to stretch on for days. "Kids love to have things to snuggle with. And eventually, Oliver will be bigger than the bear."
"Ah. I see."
"You'll learn."
"I have a feeling I won't have a choice." Aiden leaned her small suitcase against the wall and propped the bear up on top of it. "And how did you get all of this onto a train, then off a train and into the city, all by yourself?"
"Let's just say that I relied on the kindness of strangers. And I'm a very good tipper. I managed."
"You're resourceful. I'll give you that much."
Sarah went to get Oliver out of his stroller, but decided it was time to start the learning process. "Aiden. Here. You unbuckle him and get him out."
"You sure? I don't have the first clue what I'm doing."
"You have to start somewhere."
Aiden crouched down and Oliver messed with his hair while Aiden tried to decipher the maze of straps and buckles. Sarah watched, not wanting to interfere. Oliver was doing enough on his own, tugging on Aiden's jacket and kicking him in the chest.
Aiden sat back on his haunches, raking his hair from his face. "Is he always like this? So full of energy and into everything?"
"Unless he's asleep, yes. Now pick him up."
Aiden threaded his massive hands under the baby's tiny arms, lifting him as if he might break him if he went too fast, then holding Oliver awkwardly against his torso.
"Bend your arm and let him sit in the crook of your elbow." Sarah shifted Oliver into position. She straightened Aiden's suit coat while she was at it. She stood back and admired the change. The strong, strapping man holding her favorite baby on the planet was awfully sexy. "See? That wasn't so bad."
Oliver leaned toward Sarah, holding out his arms for her.
"I think he wants to be with you."
Sarah had to be firm. "He'll be fine. He needs to be with you. Let's start the tour so we can start planning the nursery. He'll stay in your arms if we're busy and there are things to look at."
Aiden blew out a breath and they strolled into the modern, open apartment. The space had very high ceilings and was decorated almost exclusively in white, black and gray. Everything was meticulous and neat, just like Aiden's office at LangTel. He was in for a big wake-up call when Oliver took over and there were toys everywhere. Best not to mention that, though. He'd learn.
To her right was a massive gourmet kitchen with an eight-burner stove and seating for six at the center island. Beyond the kitchen, she could see a hint of a dining room tucked away, then a staircase, and beyond that a room with a sofa and the beautiful windows she'd noticed on the front of the building. As a nanny, Sarah had seen grand displays of money, but nothing that hinted at this level of affluence. Although she was no real estate agent, the house had to be at least five thousand square feet if the other floors were the same size. By comparison, her Boston apartment probably could've fit inside the kitchen. When Aiden had said he needed his space, he wasn't kidding.
"The living room is at the front of the building, overlooking the park."
"Beautiful. Absolutely stunning." Sarah followed as Aiden led them in the opposite direction.
"This is the library." He nodded to his right, where black, open-back bookcases delineated the room. The shelves were packed with books. "The room with the French doors at the back of the building is my home office."
Aiden did a one-eighty and Sarah trailed behind him, past the dining room and stairs, to the living room. It was a grand and comfortable space with charcoal-gray sectional couches, a flat-screen TV above a stacked stone fireplace and a massive glass coffee table. "Another beautiful room."
"Thank you." He shifted Oliver in his arms, seeming ever-so-slightly more comfortable with holding him.
"Unfortunately, we're going to need to babyproof in here like nobody's business."
"Why? What's wrong with it?"
Sarah didn't know where to start. "There are outlets everywhere. The coffee table is a disaster waiting to happen. I can just see Oliver bonking his head. You'll probably have to put up a gate to keep him away from the fireplace. As for the rest of the house, that's going to need an overhaul, too. Those stairs will need a gate, too."
"Isn't that how children learn? By making mistakes?" There was no misconstruing the annoyance in his voice.
"Not on my watch, they don't. At least not the kind of mistakes that put a child in the emergency room."
A low grumble left his throat. "Talk about turning my entire life upside down." He shook his head and took what seemed like his hundredth deep breath. "I'll need you to make a list. We'll tackle it that way."
"Not a normal nanny responsibility, but okay."
"I thought you weren't a nanny anymore."
"I'm not."
"Well then. This is part of our business arrangement. You need my expertise. I need yours."
"Fine." Sarah walked over to a long, dark wood console table against the wall, plopping her handbag down to dig out a piece of paper. A handful of framed photographs were directly above—one taken from the viewpoint of someone skydiving, one looking straight down the side of a cliff with a waterfall and jungle in the periphery, and another of a group of men and donkeys on a narrow path carved into a mountainside. Each looked like something out of a movie. "Nice pictures. Are these from National Geographic?"
"Remembrances of my adventures."
"Wait. What? These are yours?"
Aiden nodded, fighting a smile. He joined her, Oliver in tow. Aiden was doing well with the baby, and she was happy to see him master his first few moments of dad duty. "I enjoy pushing the limits," he said.
Goose bumps cropped up on Sarah's arms. A man with a dangerous side held mysterious appeal, probably because it was the opposite of her personality. She'd fallen for a few guys who liked to live on the edge over the years. None of them was good at flexing their bravado in the realm of relationships.
"You're going to have to set aside your daredevil escapades for a little while. Skydiving is not an approved activity for a toddler."
He scowled. "I'm not enjoying this part, in case you're wondering. The part where you tell me how I have to construct my life around someone else's needs."
She patted him on the shoulder. "Welcome to parenthood. It's good for you. It'll remind you that the world doesn't revolve around you."
"Jumping out of an airplane reminds me that I'm still alive," Aiden countered. "And that I'd better find a way to enjoy my time on this planet."
There was a somber hint to that last string of words, but she was still piecing together who and what Aiden Langford truly was. It struck her as sad that he lived all alone in this big house, however much it was a showplace. Despite his protestations, Sarah couldn't imagine Oliver as anything less than a blessing in Aiden's life, quite possibly his salvation.
Oliver reached for the pictures, pointing to the skydiving snapshot. Aiden stepped close enough for him to touch it.
"Pretty cool, huh? I took that picture. I jumped out of an airplane. Maybe you and I can do that someday. Someday when Sarah isn't around to tell us what to do."
Oliver turned to Aiden, concentrating hard on his face. He flattened his palm against Aiden's cheek. Aiden reached up and covered Oliver's hand with his, a fascinated smile crossing his face. A sweet and tender moment, it left Sarah on the verge of tears. For the first time since she'd gotten off the train that afternoon, she was less worried about Aiden accepting fatherhood. They weren't out of the woods, but he was already showing signs of folding Oliver into his life. Which meant one step closer to Sarah being out of it.
Oliver needs his father. His new family. "For now, I still get to tell you what to do, at least when it comes to Oliver. I say it's time to find him a bedroom in this massive house of yours."
* * *
Aiden walked Sarah and Oliver up to the second floor, holding the little boy. He was slowly growing comfortable with this tiny human clutching the lapel of his suit coat, keeping him warm and reacting to the world Aiden walked through every day without giving it a second thought. It all was new to Oliver—sights and sounds, people and places. He didn't play the role of stranger though; he played explorer, full of curiosity. Aiden had to admire that disposition. He was cut from the same cloth.
They reached the top of the stairs and the hall where all four bedrooms were. At the far end was his master suite. There was only one other room furnished, for guests. The other two remained unused and unoccupied. With most of his family in the city, visitors weren't common, nor would they likely ever be. His friends, small in number and much like him in that they preferred to roam the globe, were not prone to planning a visit. No, the apartment with arguably too much space for a confirmed bachelor had been purchased with one thing in mind—breathing room.
He fought the sense that Sarah and Oliver were encroaching on his refuge. He made accommodations for no one and doing so put him on edge, but it was about more than covering electrical outlets and putting up gates. He hadn't come close to wrapping his head around his newfound fatherhood, even if he did accept that with the arrival of Sarah Daltrey, everything had changed.
He was counting on the results of the paternity test to help it all sink in. He'd already made the call to his lawyer. It would mean a lot to know that Oliver was truly his. Aiden had lived much of his own life convinced that Roger and Evelyn Langford—the people he called his parents—had lied to him about who Aiden's father was. Roger Langford's death nearly a year ago had made the uncertainty even more painful and the truth that much more elusive. He wasn't about to badger his mother, a grieving widow, over his suspicions. But he would confront her, eventually. He couldn't mend fences with his family until that much was known, and there was a lot of mending to be done. Aiden had made his own mistakes, too. Big, vengeful mistakes.
"I was thinking we could put Oliver in here." Aiden showed one of the spare rooms to Sarah. "It's the biggest. I mean, he is going to get bigger, isn't he?" Talk about things he hadn't considered...life beyond today, when Oliver would be older...preschool, grade school and beyond. No matter what, Aiden didn't need to think about where Oliver would go to school. He would be wherever Aiden was. There would be no shipping him off as his parents had done to him.
"Is it the closest room to yours?" Sarah asked.
"No. The smallest is the closest."
"That's probably a better choice for now." Without invitation, she ventured farther down the hall. "In here?" Sarah strolled in and turned in the small, but bright space—not much more than four walls and a closet. "This is better. It'll make it easier on you. He still gets up in the middle of the night."
"And I'll need to get up with him." He stated it rather than framing it as a question. He was prepared to do anything to feel less out of his element, as if any of this were logical to him, which it wasn't.
Oliver fussed and kicked, wanting to get down.
"Let's let him crawl around," Sarah said.
Aiden gently placed the little boy on the floor. He took off like a bolt of lightning, scrambling all over the room on his hands and knees.
Sarah pulled a few toys out of her bag and offered them to Oliver. "Yes. You'll need to get up with him and comfort him, especially when he's teething like he is now."
Aiden leaned against the door frame, acting as a barrier in case Oliver decided to escape. "Is that why he drools so much?"
Sarah smiled and sat on the floor with Oliver, tucking her legs beneath her, her dress flounced around her. "My mother used to say that's not drool. It's the sugar melting."
Aiden wasn't prone to smiling, let alone laughing, at things that were quaint and homey. But he couldn't have stopped if he'd wanted to. He drank in the vision of Sarah. She was so different from every woman he'd ever known. She was beautiful, but not made up. Eloquent, but not pretentious. There was no hidden agenda, nor did she seem concerned with impressing him. She just came right out with it, but didn't mow people over with her ideas. She simply stated what she found to be best, in a manner that made it seem as if it were the only logical choice.
Sarah again looked around the room. "We should probably order a crib online and see how quickly we can have it delivered, along with some other necessities. He'll need a dresser, a changing table. You should probably invest in a rocking chair for this room." She began counting on her fingers. "Then there's clothes, diapers, formula, bottles, toys, bath supplies, baby laundry detergent."
"Special laundry detergent?"
Pressing her lips together, she nodded. "When he's crying in the middle of the night, you don't want to be wondering if it's because his skin is irritated. One less thing to worry about."
Just when he thought he was getting a handle on things, a new spate of information came down the pike. "Like I said before, it'd be great if you could make some lists. You can use the computer in my home office and get a lot of that ordered."
"We need to call the nanny agency, too. They probably don't take calls after five on a Friday. Sounds like we have a busy night ahead of us. Oliver's going to need a bath, too." Oliver crawled over to Sarah with a stuffed toy in his hand and showed it to her.
Aiden's cell phone rang with a call from his sister Anna. "Excuse me for a minute. I need to make sure this isn't anything important."
"Sure thing. I'll call the nanny agency and Oliver can play. Avoiding outlets, of course."
"Right. The outlets." Gotta deal with that, ASAP. He accepted the call and stepped out into the hall. "Anna, hi. Everything okay?"
"I was calling to ask you the same thing. Is everything going well with Sarah and Oliver? I can't believe it, Aiden. A baby. It's so amazing. Are you just bursting at the seams?"
Aiden wandered into his room and sat on the leather bench at the foot of the bed. "More like my brain is about to implode. I don't know what I'm supposed to feel. At least you've had time to get used to the idea of becoming a parent. It's only been a few hours for me."
"I'm sure it will take some time, but I'm so excited for you. You know, the minute I looked into Oliver's eyes, I knew he was yours. He looks just like you. It's going to blow Mom's mind when she sees him."
Oh no. The one thing he hadn't yet taken into account. "Please tell me you haven't said anything to Mom. Or Adam for that matter, but especially not Mom. I need to figure out how best to deal with this."
"I haven't said a peep."
He exhaled a little too loudly, if only to make the weight of dealing with his mother subside. "Good." His mind often raced at the mere mention of his mom, thoughts quickly mired in bad memories and sad stories. He couldn't fathom the moment when she'd meet the son he hadn't known he had. Would he feel better about his suspicions, a misgiving he'd shared with no one other than Anna? Or would he feel worse? Either way, his mother's reaction to Oliver would be telling. If she accepted him unconditionally, he'd always wonder why she hadn't treated him the same way. If she rejected him, he'd have a hard time not blowing up at her.
"When are you going to tell her?" Anna asked.
"Tomorrow. Or maybe Sunday. I need time to get us settled." He rested his elbows on his knees. "Sarah's calling the nanny agency, we have an entire nursery of furniture to order and I'm apparently in Daddy School after that. I have to learn how to change a diaper and give him a bath."
Anna tittered.
"What's so funny?"
"I like the image of you bathing a baby. It's sweet. And unlike anything I ever imagined you doing."
"You and me both. I never thought I'd have kids." Not after everything with Dad.
"Sometimes life gives us unexpected gifts. I felt like that when I got pregnant."
Anna was carrying a miracle baby. Her doctor had told her it would be nearly impossible to conceive and even more difficult to carry a pregnancy to term, but she was doing great. "I hear you. I'm still getting used to it."
"Well, promise me you won't keep Oliver to yourself. I want to see him, too. I could even come over and take care of him if you need help. I could bring Jacob. It would be great practice for us."
Anna always managed to take the edge off his greatest concerns. Even if he'd come back to New York and everything had been a total disaster with the rest of his family, he still would've forged a better relationship with Anna. "Thanks. I'll definitely need your help after Sarah leaves."
"She's staying for ten whole days? How'd you convince her to do that? She seemed hell-bent on only being in town through the weekend when we first talked."
"We made a deal. She gives me ten days and I help her with her business." Aiden then realized that his sister might be able to help. Before she'd taken a job at LangTel, she'd been CFO for a company that manufactured women's workout clothes. "Did Sarah happen to tell you what she does?"
"She did. And the idea of you helping her with it is almost as amusing as the image of you giving Oliver a bath."
"I'm glad you find my life changes so entertaining right now. Do you think you could help me out with some contacts in the garment industry? I haven't talked to Sarah about it that much, but I know she needs manufacturing and warehousing and distribution. Maybe you know someone I could call."
"Oh, absolutely. Let me think about it and I'll email you a list."
"Perfect." One thing he could check off his to-do list. "God only knows how I'll get any work done on Monday when I'm back at the office. I doubt I'll get much rest this weekend."
"Don't worry about that. Work can wait. You're a dad now. That's the most important thing."
Four
Sitting on the floor in Oliver's room, Sarah ended her call with the nanny agency. She leaned down and kissed the baby's head. He'd been playing quietly in her lap for a few minutes. "Guess what? Your daddy's going to hire someone very nice to take care of you. Won't that be great?"
Oliver gnawed on a plastic teething ring, not interested in much else.
Sarah swept his soft curls to the side. "She'll play with you and take you for walks in the park and sing songs to you. Just like I do." Her voice wobbled as Oliver peeked up at her with wide eyes. She wrinkled her nose and forced herself to smile, if only to stop tears from gathering. The thought of leaving Oliver was as unhappy as it was inevitable. Getting attached to children who weren't her own was no longer part of her self-destructive pattern. Nor was getting wrapped up in the life of a single dad. The sooner she left Oliver with Aiden, the better.
"It really is too bad that you can't just stay and be his nanny," Aiden said.
Sarah nearly fainted. First out of surprise at his voice, then from the view as she slanted her sights to him. Leaning against the door frame, he stood there like he could hold up the whole world that way. He'd changed clothes. In a long-sleeved black T-shirt and a pair of jeans with a dark wash, he was now at a level of casual she hadn't pondered, although he had to take off the suit at some point. That thought sent her brain skipping ahead, especially now that she could better see the contours of his shoulders and how well-defined his chest was. No doubt about it—Aiden Langford logged his fair share of time in the gym.
Cut it out. The things cycling through her mind were not good—thoughts of peeling away his T-shirt and smoothing her hands over his chest, kissing him. Her curiosity was getting the best of her, and his presence was making it worse. Unfortunately, his expression was just as irresistible as the rest of the package—a look that said he didn't care what anyone else thought about, well, anything. Sarah could hardly keep her jaw in a place that suggested some measure of decorum. Forget ladylike—right now she was going for not ogling him like a sex-starved loon.
"I adore Oliver, but I told you I'm no longer a nanny."
Aiden stepped into the room and once again, something about the way he moved left her pulse unsettled. He held up his hands in surrender. "Got it. No more nannying for you. But did you call the service?"
"I did. They'll send candidates over on Monday morning. We can sit down before then and go over your priorities. And mine, of course."
"We? You know, I'm more than capable of conducting an interview. And you aren't going to have to put up with this person. I am."
She narrowed her focus on him. "You asked for my help." She stood and gathered Oliver in her arms, settling him on her hip. "Some of these nannies will embellish on their experience just to get the primo jobs. I'll see past that."
"This is one of those primo jobs?"
"With this house? Yes. And you're going to need someone at your beck and call with your schedule. I told them you need live-in help." Sarah didn't like this idea, although she couldn't arrive at a sensible reason why. She only knew that the myth of the nanny falling for the father of her charge was very real. It happened all the time. It had happened to her. If Aiden were to be judged on his looks alone, she could see most women falling for him. Add in the money, power and semiarrogant veneer? Forget it. It was only a matter of time.
"Wait a minute. I'm not just getting one new member of the household, I'm getting two? Can't the nanny live at her house and come over when I go to work?"
"That might work if you had a backup, like a family member. Otherwise, I can't imagine you waiting for the nanny to show up so you can go to work. What about your mom?"
Lightning fast, Aiden plucked Oliver from her arms. "My mother will not be taking care of him."
Sarah grappled with his hyperprotective reaction. A few hours ago, he'd been ready to banish her and the baby from LangTel corporate headquarters forever. Now, there was something else to contend with, something that Sarah sensed went deep. "Why? Most people would do anything to have a grandparent around to care for their child."
"Not me."
"Technically, I'm Oliver's legal guardian. I have a right to know why." None of this added up. Aiden's sister Anna had spoken warmly of her mother. Sarah had read about Evelyn Langford when she was researching Aiden. She sat on countless charity boards and was known for her generosity with children's hospitals, cancer research and battered women's shelters. By all reports, her benevolence had grown in the wake of her husband's death.
"I'm not saying my mother would hurt him. Not that. It's..." He closed his eyes for a moment and Sarah's breath hitched in her throat. No air would go in, nor would it come out. She was too in awe of this glimpse of vulnerability. It was so incongruous with his personality. He was showing a different side of himself, a side Sarah wanted to know. A side Sarah wanted to comfort. "It's complicated. Let's just say that for now, it's best if you know that my mother can't be relied upon for anything."
There was a finality to his tone that said Sarah should leave it alone. "Okay."
"What's the schedule for the rest of the night? I have some work I need to tend to."
Sarah consulted her phone—nearly five o'clock. "Oliver eats at five thirty. Bath time at six o'clock, story time at six forty-five. Bedtime is at seven."
"Is that Oliver's schedule or yours?"
"It's everybody's schedule. That's how things work with a baby. It makes him feel secure. He knows what happens and when."
It was impossible to ignore Aiden's attitude. Once again, he seemed put out. "I see. I guess I still have a lot to learn. We can order some takeout to come for us around eight. I trust that will work?"
She nodded. "Yes. That will give us the perfect time to talk about my business." There had to be some payoff for allowing herself to get in deeper, when she'd told herself she'd never do that.
"I spoke to my sister Anna about it briefly. She may be able to help. I wasn't kidding when I told you that I don't have many connections in that business. I can't promise you the world."
But you can ask the world of me. She stopped before the words left her lips, but she was all too familiar with handsome, powerful men who expected everything for very little in return. "Well, if nothing else, I'm sure you can give me some good advice. That alone could end up being very helpful."
"Come on. Let's go down to my office and we'll get the nursery furniture ordered."
They headed downstairs and Aiden led them to the double French doors, into one of the coziest, most gorgeous rooms Sarah had ever seen. The office had a different feel to it than the rest of the house, warmer and more colorful. The walls were a deep navy, and an ornate Oriental carpet sat in the center of the room, topped with a pair of club chairs and a massive oak desk. Bookshelves lined two of the walls from floor to ceiling.
"More books? Even with the home library?"
Aiden shrugged and rounded to the chocolate-brown leather desk chair. "I like to read. It's a nice escape."
"Escape? From what?"
"Excuse me?"
"From where I sit, you have a pretty perfect life. You have this gorgeous home, a job that tons of people would kill for and you don't seem to be hurting from the financial end of things. More than anything, you don't seem to do anything you don't want to do. At all. Ever."
For a moment, he just glared, not saying a word. He wasn't angry, nor was he pleased. "You say whatever you want to say, don't you?"
"It's not that bizarre a question. I've seen the pictures. Skydiving. Hiking the Andes. I'm just wondering what you need to escape from."
"Stress," he answered flatly, methodically spinning a pen on a pad of paper. She hadn't noticed his hands much before now and she was kicking herself for not paying better attention. His fingers moved gracefully, demonstrating their ability to do things deftly, but they were manly, too—strong. Able.
"Stress." Her stupid brain leaped ahead to methods of reducing stress and none of it had to do with reading. Again she was knee-deep in thoughts of what he looked like under that T-shirt.
"Yes." He opened his laptop and placed his fingers on the keyboard, but stopped before typing. "I don't even know where to start. Do I just search for baby crib?"
"Here. Let me do it." She carried Oliver around behind Aiden's desk and handed him the baby. Oliver settled in on Aiden's lap, but reached for the pen.
"Can I let him have this?"
"No. He'll put it in his mouth. You can run upstairs and grab a toy out of his room."
Aiden raised an eyebrow as if she'd made the most ludicrous suggestion ever.
She shrugged and waved him off. "Gotta start being Daddy sometime. Now shoo. Let me see what I can find online."
Aiden trekked out of the office with Oliver. Sarah rested her chin on her hand, watching as they made their way down the straight shot of the house, past the library and the kitchen, until they disappeared up the stairs. Aiden was so big, Oliver so tiny in his arms. She hoped to hell they would be okay on Sunday, after she left. She couldn't bear the thought of anything else.
She pulled up a browser window and quickly found a furniture place offering next day delivery in Manhattan. That was the genius part of being in a big city. Virtually anything could be delivered at any time. Once she was done, a delivery truck would be set to arrive in front of Aiden's building tomorrow morning. And she'd be one step closer to removing herself from Aiden's and Oliver's life.
Five
Aiden had learned one thing already—fatherhood was no walk in the park. He'd struggled through his first attempt at feeding Oliver his dinner. With no high chair, they'd had to improvise by wheeling Oliver's stroller into the dining room. The baby rubbed his eyes and turned his head, refusing every spoonful Aiden offered. He had to hand it to Sarah, though—she only gave advice when asked. She'd otherwise sat by quietly and watched as a man capable of orchestrating billion-dollar deals and negotiating with cantankerous CEOs was unable to convince a fussy toddler to take a single bite of food. Frustrated, he'd finally asked her to do it. She took over, Oliver downing an entire jar of baby food with hardly a single complaint. Aiden walked away from the dinner table with a bruised ego. And baby food on his jeans.
He wasn't sure what to make of bath time, either. But this time, Sarah took charge.
"This is the only tub you have in the house?"
Aiden failed to understand the question. The tub was perfect, in that it fit two people. For him, seduction was the only reason to get in a bathtub. "Yes. What's wrong with it?"
"It's huge."
"Of course it is. It's a two-person soaking tub." He cleared his throat, waiting for her next comment.
"Well, you're going to have to get in there with him. I refuse to bathe a child in the kitchen sink. It's not sanitary."
He turned and dropped his head until his chin was nearly flat against his chest. He was at least a foot taller than her, maybe more, and they were nearly toe-to-toe. She was still wearing the sundress from earlier in the day. Had that really been today? So much had happened, it was hard to wrap his brain around it. "So you're going to see me naked before we've known each other for eight hours? You take things quickly."
"Very funny. No, Oliver gets to get naked. You're putting on swim trunks. If I had a bathing suit with me, I'd do it myself. But I don't, and you need to bond with him."
He raised an eyebrow. "This from the woman who swore I'd have no problem feeding him dinner."
She shrugged. "Babies are unpredictable. The sooner you learn that, the better. I promise you that physical contact will help you and Oliver to bond. It's a scientific fact. Now go change. I'll get the water running."
"I like it hot."
"You'll get lukewarm and like it."
He grumbled, but made his way into his walk-in closet, closing the door behind him. He took off his clothes and plucked a pair of board shorts from the bottom drawer of his bureau, slipping into them and tying the white string at the waist. He opened the door. "Ready."
Sarah turned, glancing at him over her shoulder. Every muscle in his body tightened from that single flash of her eyes and the immediate connection he felt. Good God she was gorgeous, all deep blue eyes and skin flushed with rosy pink. She shied away. "So I see."
He liked getting that reaction. He liked it a lot. "What now?"
"Get in. I'll hand him to you." She tended to Oliver, who was pulling himself to standing at the edge of the bathtub. He bounced up and down on his toes while Sarah took off his pants and diaper.
"He seems excited."
"Just you wait. He loves bath time. It's a good thing you're in your trunks. I'm going to get soaked."
Aiden climbed into the tub, wrestling with the idea of Sarah, soaked, and the white-hot image it conjured. Sure, they only had ten days together, but that was plenty of time for him. In fact, it was his preference—a strict, short timetable. But was that a good idea? From a physical standpoint, sure. From every other standpoint, he didn't know. There were repercussions and awkward conversations to worry about. Dammit.
Sarah handed him the baby and he let Oliver sit on his lap while he wrapped his hands around his waist. The baby wasted no time slapping the surface of the water and sending it flying.
Sarah laughed and dropped a few plastic toys into the bath. "Told you."
Splash splash splash. Oliver looked at Sarah, who beamed at him as if she couldn't be any more in love with someone if she tried. She rested her elbows on the edge of the tub and leaned closer, flicking at the water with the tips of her fingers. Oliver giggled, then mimicked her in a far less delicate way. Splash splash splash. He laughed so hard his entire body shook. It was impossible not to find the fun in their game, even with water being flung at his face and shoulders, not to mention all over the bathroom.
"Is bath time always this chaotic?"
"Basically. Anything you can do to get him clean. And it helps relax him."
Splash splash splash. Another peal of Oliver's sweet giggles rang out.
"It relaxes him?"
"Believe it or not, yes. He has a lot of energy. This helps to get it out." Sarah pulled out a toiletry bag and poured a dollop of golden shampoo into her hand. "Get his hair wet. We don't have a cup, so just use your hands."
Aiden scooped water with one hand, curling his arm around wiggly Oliver. He started tentatively, unsure if the baby would like it, but quickly learned that he took no issue with water running down his face. Aiden had a little fish on his hands. How amazing it would be to teach him to swim, then snorkel and surf, another of Aiden's favorite pastimes. Small waves at first. It's dangerous. He was still getting used to these parental thoughts, but he was amazed how quickly they had kicked in. Especially when the topic of his mother had come up. He hadn't meant to impulsively take Oliver out of Sarah's arms. He only knew that was his gut talking—and reacting. Oliver would know nothing but unconditional love from his family. He wasn't certain his mother could offer it, and until she'd demonstrated as much, she would be kept on a very short leash.
Sarah leaned over and shampooed Oliver's head, his blond curls becoming matted and soapy. A soft fragrance filled the air.
"It smells nice," he said.
"It smells like baby, and that's the most wonderful smell in the world. Well, most of the time. There are times when it gets stinky, too."
"I bet." Like most things, there would be both good and bad to parenting. Aiden was optimistic about more good, mostly because he and Oliver had a clean slate. Aiden would not do to Oliver what his parents had done to him. Oliver would never wonder whether his father loved him. For that matter, he would never have to wonder who his father was. Once the paternity test was done, Aiden would have that sewn up for them both.
"Turn him around, facing you. So I can rinse out his hair."
He carefully turned Oliver in his hands, but it wasn't easy—it was like holding on to a greased-up watermelon with moving arms and legs. "I'm trying to figure out how I'm supposed to do this by myself."
"I ordered a seat that goes in the tub. That will help immensely. And it won't be long before he can sit up reliably in the bath on his own."
Now that he and Oliver were facing each other, Aiden had a chance to really study him. Oliver returned the gaze, chewing on a rubbery red fish. His eyes were so sweet and innocent, full of wonder. Aiden saw only hope, remarkable considering what the little guy had been through. As Sarah rinsed his hair, Aiden was overcome with the most unusual feeling. It was stronger than his inclination to protect Oliver from big waves. It was a need to keep him from everything bad. He never wanted Oliver's eyes to reflect anything but happiness. Had his own father ever looked at him like this? He didn't enjoy the role of pessimist, but the idea was implausible.
Sarah rolled a small bar of soap in her delicate hands and washed Oliver's back, shoulders and stomach, while Aiden held on tight. Every gentle caress showed someone who genuinely cared about her charge. He'd never really seen this side of any woman aside from on TV or in movies, and it was breathtaking to watch. If he were honest, he'd never done so many things with a woman that gave him a taste of what being a couple was like. Wining, dining and seduction were not the same. This was different.
Sarah swept her hair to one side, displaying the stretch of her graceful neck, the contours of her collarbone. Her skin was so touchable, and the urge to do exactly that was strong with her mere inches away. His hands were practically twitching at the idea. He had to set his mind on another course.
"So. Tell me more about you," he said.
She smiled and sat back on her haunches. "Not much to tell. Born and raised in Ohio, oldest of five. Moved to Boston to study fashion design, stayed for the good nanny jobs."
"Why not go right into design?"
She plucked a washcloth from the bathroom vanity and wiped her hands. "Nannying was a detour. I grew up helping out with my siblings, so it was a natural thing to care for children. And Boston is not cheap. Nannying pays well. It just worked."
"If you liked it that much and it paid well, how does that stop working?"
She looked down at the floor, her golden hair falling down around her face. "I burned out. Badly. Let's put it that way."
That didn't make sense. She didn't seem at all burned out on caring for Oliver. If anything, she had superhuman stamina and patience when it came to it. "And the rest? Surely there's a special guy in your life."
"There is." Her face lit up so brightly that it was as if someone had sucker punched him. So much for seduction. There was another man.
"His name is Oliver," she continued. "He's so sweet. He doesn't talk much. Drools a fair amount. Still learning how to walk. Exactly like I like my men."
He laughed and shook his head. She was ridiculously charming and clever, probably why he had such a strong reaction to the idea of her with a boyfriend.
She flipped her hair back and grinned at Oliver. "But seriously, the right guy hasn't walked into my life and I'm not about to wait. I'm too busy trying to build my business to think about stuff like that. Romance is not on my radar right now."
No wonder he'd been feeling as though he and Sarah might be kindred spirits, even though they came from different worlds. She wasn't looking for love. And neither was he. And with only ten days together, that might be perfect.
* * *
Sarah was ready to claim victory over bath time—Oliver was clean and she hadn't been caught staring at Aiden. It was a miracle since she'd been doing exactly that, sneaking peeks at his chest, broad and firm with the most perfect patch of dark hair in the center. Then there were his glorious shoulders and his sculpted biceps. She'd also spent a fair amount of time studying the tattoo on the inside of his forearm—a dark and intriguing pattern, impossible to decipher.
She bopped Oliver on the nose with the tip of her finger. "Hey, mister. It's time for somebody to get out of the bath and get into pj's."
Aiden furrowed his brow. "Sarah's no fun," he said to Oliver. "I don't know about you, but I'm good for at least another fifteen minutes."
She smiled. "The water will be freezing by then. And don't forget the schedule."
"Ah, yes. The schedule."
Aiden lifted Oliver out of the bath and handed him to Sarah, who had a towel at the ready. She wrapped up the baby, holding him close, gently drying his hair with an extra washcloth. Her vision drifted to Aiden as he climbed out and planted one foot on the edge of the tub and bent over to scrub his leg with the towel. She nearly bit right through her lip. His back was long and lean, his posture flaunting the definition—a railroad of muscle running north to that thick, touchable head of hair and south to a pleasingly tight rear view.
He dropped his foot and turned. Either she hadn't had time to turn away or she hadn't had the will. A devilish half smile crossed his face—a grin that said he knew she'd just committed his backside to memory. Sarah was petrified. If she shied away, she'd look even more guilty. It'd be tantamount to blurting, I had to look. You're too hot not to look. But if she kept staring, it would be hard to stop and that would further chip away at her resolve. No falling for the impossibly handsome single dad with the adorable baby.
"You're wet." Aiden nodded in her direction, wrapping the towel around his waist.
Sarah shifted Oliver to her hip. Her dress was streaked with dark patches and clung to her thighs. "Oh, shoot. Yeah. I should probably get out of this thing."
"Might as well get comfortable since we're in for the night."
"Comfortable?" No, not comfortable. I need to get uncomfortable.
"Unless an evening gown is more appropriate for story time. I'm still learning here." He took the baby from her. They were ridiculously cute together—Aiden bare chested and wearing a towel, Oliver bundled up in his arms. "If Oliver gets to wear pajamas, that's what I'm wearing, too. You might as well join us."
"I didn't pack for a pajama party. All I have is one of my nightgowns."
"I haven't seen your work yet. If I'm going to help you with your business, I need to know what you're selling."
"I'll show you pictures."
"Why? Too sexy?"
"No," she blurted, not taking the time to think.
"Then what's the problem?" He cast her a look of admonishment that left her quaking. "If this is what you do, you have to own it. You have to live it or it'll never work."
"I do live it. I do own it."
"Then show me. I promise I'll contain myself."
She stifled her exasperation. "Fine. Everything you need to get Oliver dressed is on your bed. I'll be right back."
"Don't take too long. I'm still figuring out this whole diaper thing."
Sarah hustled down the hall in her bare feet, muttering to herself. "Great job, Sarah. First you get caught staring and now he talked you into half-naked story time."
How had she ended up in this situation? Aiden. He was everything she hadn't expected. Once she'd gotten past the get-out-of-my-office exterior and been invited into his inner sanctum, he'd shown her a different side, one that was unfairly appealing. He was nicer, he was more amenable, he was generous. And then there were the things his physical presence did to her, making her tingle in places that hadn't tingled in more than a year. Not since she'd discovered that her employer Jason had been taking her to bed when he was in town and doing the same with countless other women when he traveled for work. She'd allowed herself to get caught up in their lives, and crossed the line no nanny should, and she'd paid the price. Her heart had been trampled by Jason, and even worse—she'd had to say goodbye to Chloe, his sweet, adorable daughter. That had hurt like nothing else. She couldn't repeat that mistake.
She ducked into her room and closed the door, sucking in a deep breath to reclaim some semblance of control. She would not be her own worst enemy. Time to get her act together.
Her eyes darted to her suitcase, perched on the bench at the foot of her bed. Unless some different pajamas had magically made their way inside, she had exactly one of her designs with her—a midthigh bias-cut nightgown with thin straps. The black raw silk held a subtle shimmer, embroidered with delicate silver threads at the hem and demure neckline. It didn't scream sexpot, but it wasn't anywhere close to frumpy either. Just risqué enough to give her an anxiety attack. Her shoulders dropped in defeat.
"He said I have to own it. I just have to do that." There was no more time for thinking. Aiden was indeed still figuring out the whole diaper thing, and Oliver would invariably pee all over him if she took too long. She wrestled her way out of her dress and threaded the chemise over her head. The silk skimmed her skin, reminding her precisely why her customers couldn't get enough of her nightgowns—they made a woman feel sexy.
But she could take the edge off. She grabbed her black cardigan and put it on, buttoning it up. She'd bought herself a small measure of modesty, but as she stole a passing glance in the mirror, she saw that she was not owning it—she was borrowing it. Frustration bubbled up inside her, but she couldn't simply traipse into Aiden's bedroom dressed for seduction. This would have to do. If she had to walk a narrow tightrope, she would. Even if she'd be donning a bizarre ensemble while doing it.
With no more time for second-guessing, she hurried back down the hall. Aiden's door was open. He was hunched over the side of the bed, attempting to dress Oliver.
Sarah joined them, perching gingerly on the edge of the mattress. The bedding was so soft and silky she had to stifle a moan of approval. In dark gray pajama pants and a black T-shirt that showed off the straight line of his shoulders, Aiden was dressed to kill. Why did everything about him have to be so enticing? "Looks like you did well with the diaper. What about the rest?"
"I'm worried I'm going to bend him in the wrong direction."
"Just think about how you would get ready for bed. Do that."
He arched an untamed eyebrow at her. "Then he's ready. Because I don't wear much to bed."
Of course he just had to plant that mental image in her head. He had to. "Then pretend you're putting on a shirt for work." She crossed her legs, noticing how parts of her were again tingling and zipping with electricity.
Aiden got Oliver into the sleeves and the legs, but then he hit another trouble spot with the snaps. "These things don't match up."
"Start at the top."
He did as she'd instructed and picked up Oliver when he was done. "Good?"
"Fantastic."
He sat next to her on the bed, Oliver in his lap. "You know, I can hardly see what you're wearing with that big old sweater over the top of it."
She wrapped her arms around her waist. "I was cold," she lied. Being this close to Aiden, she was about to go up in flames. "And we need to get going with bedtime."
"Right. The schedule."
Nightgown crisis averted, it was now acceptable to exhale.
"Since the crib doesn't come until tomorrow, Oliver can sleep with you tonight. It will be a nice way for you two to bond."
"But what if I roll onto him? What if he falls out of the bed?"
A breathy laugh escaped her lips.
"Something funny?" he asked.
"Careful. You sound like a dad."
"They're valid questions."
"And I'm glad you're concerned. We can put some dining chairs next to the side of the bed."
Aiden scratched his head, looking around the room. "Get up. Hold Oliver."
Sarah stood and took the baby, watching as Aiden pushed aside the bench at the foot of his bed and began tugging on his mattress. It only took a few pulls before it landed with a thump on the floor.
"There. Then if he rolls out of bed, he won't go far. It's only for one night."
Sarah could hardly believe her eyes. "Talk about problem solving."
"It's partly selfish. I won't get any sleep if I'm paranoid all night about what's going to happen to him."
She didn't bother containing her smile, even though she sensed that with every sweet thing Aiden did or said, she was being pulled more forcefully into his orbit. "You're turning into a dad right before my eyes."
Aiden retrieved the pillows from their resting place on the box spring—they hadn't made the trip. "I have a job to do, I don't shy away from it."
"I know, but you were bitching about baby gates a few hours ago. Now you're camping out in your own bedroom."
Aiden stepped over to her and took Oliver. "It was the bath. I guess it started to sink in that he needs me. It feels nice. Nobody's ever needed me like he does."
There was an edge of sadness to Aiden's voice that tugged at Sarah's heart. She needed to make a graceful exit, now. "You know, I think I'm going to get Oliver a bottle and walk you through the bedtime routine. Something tells me you'll do just fine."
"Oh, okay. Did you want to have dinner after he goes to sleep?"
I do. I really do. But I don't. "No, thank you. I've had a long day. I'll just turn in."
Aiden seemed puzzled, but didn't argue. "Okay."
Sarah retrieved Oliver's bottle and left Aiden to his own devices after a brief overview of what to do. Since Oliver's nap had been cut short that day, she was sure he'd fall asleep quickly. Apparently, exactly that had happened, since she didn't hear another peep for hours. She stayed in her room, tucked under the covers, trying to banish thoughts of Aiden from her head.
Deep in the middle of the night, Sarah woke to the sound of Oliver's cries. They caused her physical pain, made worse by the fact that she couldn't go to him. Aiden had to learn how to deal with it. The baby let out another screech and Sarah rolled onto her side, squinting at the alarm clock on the bedside table. Two forty-three.
This was normal. No big emergency. Oliver cried again and her instinct told her to go to him, and if she were honest, there was sympathy for Aiden, too. He'd been through a lot today and had risen to the occasion. She sat up, dangling her feet off the edge of the bed, listening. There was quiet. She was just about to lie back down when another cry came.
As did a knock on the door. "Sarah? Are you up?"
Hearing Aiden's voice in the middle of the night did something funny to her. "Yes. Need some help?"
Oliver wailed again.
"Yeah. If that's okay."
"Two secs." Sarah climbed out of bed and opened the door. There stood a nearly naked Aiden, wearing only a pair of gray boxer briefs, and a red-eyed Oliver. The baby lunged for her and Sarah took him, bouncing him in her arms to comfort him. "What's wrong, buddy? Why are we giving Daddy such a hard time on his first night?"
Aiden walked in and sat down on the edge of her bed, running his hands through his hair. "I thought I was in good shape. He started to get fussy and then he started to cry, so I got up and changed his diaper. He was pretty wet."
Sarah nodded, pacing back and forth, partly to comfort Oliver, and partly to distract herself from the vision of Aiden. "Good."
"He didn't want a bottle."
"Did you try his pacifier?"
"I couldn't find that thing. I think it's somewhere in the bed."
"That might help. Let's go look."
She followed Aiden down the hall into his bedroom, where he flipped on one of the bedside lamps. His daddy instincts were already becoming attuned. Most first-time parents would've flipped on the overhead light. She bounced Oliver up and down while Aiden kneeled down on the bed and began rummaging through the sheets.
"Ha. Found it." Aiden stood, victorious, and brought his finding to Oliver. The baby grabbed the pacifier with his hand and plugged it right into his mouth. "So that's what he wanted."
"Apparently."
Aiden blew out a breath. "I have a lot to learn. But thanks for your help." He reached for Oliver, but the baby was having none of that, clinging to Sarah and whimpering. "He wants you. Maybe you should take him for the rest of the night."
"Oh no. You have a way bigger bed than I do. And you two are supposed to be bonding, anyway."
"So sleep in my bed with us. I'm too tired to argue."
"That hardly seems appropriate."
"Why? We're going to have a baby between us. You'll have to trust me when I say that nothing will happen, however tempting you might be, Sarah Daltrey."
Tempting? Yeah, right.
"And there'll be plenty of room. You're practically a miniature human being."
"I'm not miniature."
"Like I said, too tired to argue. Just get in the bed. Please."
"Fine. But tonight only."
"I won't need to have you in my bed tomorrow night. There'll be a crib."
Well, that certainly solved that, didn't it? She walked around to the other side of the bed, and climbed in under the covers. Even with the mattress directly on the floor, she'd never been on a more comfortable bed in her entire life. Oliver must've been really unhappy to have had a hard time sleeping.
Aiden turned out the light and joined them, lying on his side, facing her and Oliver. The baby relaxed and let go of his iron grip on her shoulder, settling in on his back between them. The quiet was thick and nearly unbearable. She was too keenly aware that she was in bed with Aiden and that Aiden was aware of her and that Aiden was still awake. It was going to take forever to fall asleep.
"Sarah," Aiden whispered.
"What?"
"You let me see the nightgown."
Sleep deprivation makes me dumb. "I guess I did."
"It's gorgeous. I'm not surprised you're having a hard time keeping inventory."
She smiled to herself, there in the darkness.
"You were wrong when you said it wasn't sexy."
Sarah's heart galloped at an unhealthy speed. Now she'd really never get to sleep.
"It shows off your legs. You have nice legs," he continued. "And, well, it compliments other parts of you as well."
If it were possible to die from flattery, Sarah was DOA. What was she supposed to say to that? Was she supposed to reciprocate? Those boxers you're wearing sure show off your three percent body fat and ridiculously alluring physique?
"Sarah? Are you asleep?"
Dang. She should've pretended to be snoring. "Not quite," she whispered.
"Did I say something wrong?"
No. You said everything right. "I think we should get some sleep. And I don't want to wake the baby."
Six
Aiden was rarely overwhelmed. He didn't believe in it. Why panic when there's a lot to do? Tackle it, and move on. But bleary-eyed, navigating the maze of boxes in the hall outside Oliver's nursery and operating on very little sleep, he was officially off his game.
"How does a person who is so small need so much stuff?"
"You asked me the same thing yesterday. And I don't know why, they just do." Sarah balanced Oliver on her hip while she peeked into his room, where two delivery people were assembling furniture. "They should have the crib done soon. Somebody needs to sleep in his own bed tonight." She cleared her throat and looked square at Aiden. "That goes for me, too."
"Of course." Yeah, he'd gotten the hint last night when he'd tried to say a few nice things and she'd hardly reacted at all. Although he'd caught her staring when he'd climbed out of the bath, and the look in her eyes said she approved, so which was it? It seemed like she was attracted to him, but maybe not.
It might have been a bad idea to invite Sarah into bed last night, but that was also his first dose of sleep deprivation at the hands of a crying baby. He could already see how a parent could end up giving in to any number of demands, just to have a respite.
Of course he hadn't slept soundly. He'd worried that he might crush little Oliver, so he'd made a point of not moving, which didn't lend itself to relaxation. He'd been intensely aware of every peep the baby made, hoping he'd sleep through the rest of the night. Sarah's presence hadn't helped. He'd ended up in bed with women in fewer than twenty-four hours before, but never like this, and never with a woman like her. As he'd studied her in the soft light that morning, he found himself not only admiring her uncommon beauty—the scattering of faint freckles across her cheeks, and lips that could make him lose all sense of direction if he thought about them too much—he had to extol the gumption contained in her small frame. She'd gotten through to him when that was the last thing he allowed.
"How are we supposed to get all of this put away with a toddler crawling all over the house?" he asked.
"Now you understand the challenge of caring for a child. It's a constant juggling act."
Oliver struggled and kicked to get down from Sarah's hip.
"So I'm learning." He yawned and took another sip of his coffee. This was going to be a long day. Not that he wasn't looking forward to it. As much as he'd never imagined spending his weekend this way, and as tired as he was, yesterday had been incredible.
"He'll go down for his morning nap soon and he'll have the longer one in the afternoon. That should give us some time. Of course, it'd be a lot easier if we brought in reinforcements. Maybe you could call your family?"
Not this again. "I already told you I'm not ready for Oliver to be around my mom at all, let alone have her come in and spend any time with him on her own."
"What about Anna? Didn't she offer to watch him? She could take him for a walk and some fresh air."
That could work. Anna was Aiden's strongest ally in his family. He and his brother Adam had their moments, but he also represented some of the most painful parts of Aiden's childhood—their father pitting the boys against each other, and deeming Adam heir apparent, even when logic said that Aiden, as the oldest child, should've eventually been handed the reins at LangTel.
Aiden's other family ally was Anna's husband, Jacob. He and Aiden were cut from a similar cloth—both dealing with the price a man must pay when he's had a strained relationship with his father.
"Anna would love it."
Sarah let Oliver down onto the floor. With the help of a cardboard box, he pulled himself to standing and began pounding on the top. "Yes. Please call your sister."
Anna and Jacob were over to the apartment in less than an hour. Oliver was still taking his nap when they arrived. Sarah and Anna had gone up to Aiden's room, so Anna could watch Oliver sleep. Talk about baby fever—Anna had it.
"Now that you've had twenty-four hours to come to terms with it, how's fatherhood?" Jacob asked, settling in on the living room couch.
"Surreal. That's the best way to describe it." He scratched his head and glanced out the window—the sky was crystal clear. Not a single cloud. "Honestly, it's the only way to describe it. It doesn't feel real."
"I take it you're going to have a paternity test?"
"We have to for my name to be added to Oliver's birth certificate. Then Sarah will sign over guardianship."
"Is there any chance he might not be yours?"
After his first look into Oliver's eyes yesterday, and especially after seeing his birthmark, Aiden had been viewing the paternity test as nothing more than a formality. There was no way that Sarah could've popped into his life with a baby that looked just like him. But the question brought up feelings he'd wrestled with for so long. Had his dad done a paternity test? Was that the moment when things went wrong? When Roger Langford decided he wanted nothing to do with him?
"I don't want to entertain the thought, to be honest. I look in his eyes and I know he's my son. It's the best feeling, even if it has been out of the blue."
Jacob smiled and stretched his arm across the back of the couch. "I can't believe Anna and I are so close. Only six weeks until her due date. Talk about surreal, try touching someone's stomach and feeling a kick and realizing there's a tiny person in there."
"Are your parents excited about becoming grandparents? That's a lot of pressure on you as an only child."
"My mom says she is, but we'll have to wait and see what happens. Your mom, on the other hand. All she talks about is the baby."
That gave Aiden a sliver of optimism.
"Look who's up from his nap." Anna waltzed into the room, talking in a happy singsong, holding Oliver and smiling warmly at him. It was funny the effect that Oliver had on people. Aiden felt as though he'd been given a ray of sunshine.
Jacob got up from the couch and went to Anna, wrapping his arm around her back. "Holding a baby suits you. You look perfect."
Anna's smile only grew. "Oliver is perfect. I have the perfect nephew. It's a fact."
Aiden soaked up the joy radiating from his sister. He needed to share Oliver with his family, which meant he needed to get his mother up to speed and invite her over to meet him. It was time to let her in. Aiden walked over to Anna as Oliver cuddled with her. He pressed a gentle kiss to his petal-soft cheek. This child will always know love. Oliver would never doubt that he was wanted and adored. Not for a minute.
"If you guys are going to take him on a walk, I'll show you how to use the stroller."
* * *
As she watched the elevator doors to Aiden's apartment draw closed, and Oliver, along with Anna and Jacob, disappeared from sight, Sarah was struck by one thought. We're alone.
She turned and walked square into Aiden's chest.
"Slow down, champ." He grasped her shoulders. "I know they won't be gone long, but we have time to get down to business."
Get down to business. Why did her brain have to translate everything he said into a rambling internal dialogue about S-E-X?
"No time like the present." She laughed nervously. He still hadn't let go, and his warmth poured into her like a waterfall into a thimble. She was sure he was trying to hypnotize her with his blue eyes.
A smile rolled across his lips and Sarah was now distracted by his mouth. It was so tempting, so kissable, and she was dying of curiosity. Which version of Aiden Langford would she get if she kissed him? The powerful, broad-shouldered businessman? Walking sex in a suit? He'd probably want to be in charge in the bedroom. Or would it be the effortlessly sexy tattooed guy in board shorts? The one who needed space and jumped out of planes? She could see that guy taking his time, tending to the small touches that send a woman over the edge. She shuddered at the realization—she wanted both.
"I just want to thank you." With no other sound in the apartment, the timbre of his husky voice echoed in her head.
"For what?"
"For bringing Oliver to me. I never imagined I would feel like this."
"That's a big turnaround from the guy who ignored my emails and phone calls for three weeks."
He nodded like a man accepting guilt, a storm of blues and grays swirling in his eyes. "I know. And I apologize. I didn't want to believe it was true. It's impossible to know how you'll feel about parenthood until you have a child. If you'd asked me two days ago if I wanted a baby, I would've said absolutely not. I don't feel like that anymore."
Sarah's ovaries were whispering to her, God, he's good. A ridiculously hot guy confessing his tender feelings for a baby? Forget about it. After the bath last night and later being in his bed, Sarah was tempting fate. She needed to keep things professional. "It's been nice to watch you and Oliver connect. That makes me happy. Now let's go upstairs and get his room squared away."
Aiden released her from his grasp, leaving shockwaves of heat. He drew in a deep breath through his nose, studying her face. "Okay. Whatever you say."
Sarah did an abrupt about-face to lead the way upstairs.
"Hold on a sec," Aiden said.
"What?"
"Since when do I have a cookie jar?"
"Since you left me alone with the internet and your credit card. You have a little boy in the house. You need a cookie jar."
"I thought nannies didn't allow children to have things like cookies."
Sarah shrugged. "It's nice to show someone how much you love them by giving them something sweet." She resumed her trek to the stairs, holding her finger up in the air. "Just not every day."
Upstairs, the dark wood crib was waiting in the corner of Oliver's room. There was a combination dresser and changing table, and a beautiful rocking chair as well. All it needed were finishing touches—artwork, more books, his most precious toys—the special things that would make Oliver feel at home.
Sarah had already washed the crib bedding. "Let's make up his bed. I'll show you how to lower the side of the crib."
Aiden stood by her side, again making her nervous, as she showed him how to lift up on the side rail of the crib before lowering it. "Seems simple enough."
"Be sure that the side always goes up. You don't want him escaping."
"Or staging a coup."
"Very funny. Since Oliver's pulling up on furniture, the mattress is on the lowest setting. It goes at the top for a newborn. To save your back."
"Unless there's something I don't know, I wouldn't ever need to put it up higher, would I?"
Sarah started to put on the waterproof mattress pad and Aiden helped at the other end. "Maybe you'll get married and have another baby." Why she'd chosen to go on a fishing expedition was beyond her.
"I'd have to keep a woman around for more than a few days for that to happen."
Leave it to Aiden to casually own up to his playboy ways. "I take it your very short relationship with Oliver's mom was the norm?" Gail had been spare with her account of Aiden, saying that he was charming and sexy and up-front about not wanting anything serious. Sarah couldn't blame her for a second for going for it. She would've done the same thing if she were brave enough to have a fling. She'd never had a talent for walking away from an amazing guy.
"Remember when I said that I need space? That includes my love life."
"Space. That's such a cop-out." Sarah might have subjected herself to horrible heartbreak, but at least she'd taken chances for love.
"Excuse me?"
"So, you'll jump out of an airplane, but getting serious with someone is off-limits? You meet a woman and you decide before the start that it's going nowhere."
"No, I decide precisely where it's going. I know how it ends. I know my limitations and I accept them."
Sarah draped two small blankets over the end of the crib. "If that's what makes you happy, that's great. I just don't think you're being honest with yourself. You said you need space, but it didn't take long for you to get comfortable with Oliver."
"That's not the same thing, at all. Oliver needs me. And what about you? You're the one who said you won't make time for a boyfriend."
"My situation is completely different."
"How?"
"Because I refuse to treat a man as temporary." Exactly the way they tend to treat me. "But you treat women that way. It's sad, really."
"I don't need you to feel sorry for me."
"Oh, I don't. So don't worry about it." She shook her head. She had to escape this line of conversation. She'd learned enough frustrating details for one day. Aiden was everything she'd first thought—the guy who does not commit. "Can we please talk about my business? I need you to hold up your end of the deal. I'm at the point where Kama could either take off or crash and burn."
"Kama? Is that what it's called?"
"Yes. It's the Hindu god of desire. Our fabrics all come from India, so I thought it was fitting."
He nodded and jutted out his lower lip. "I like it. Simple. Elegant. Plus, it makes me think of the Kama Sutra and you know what that makes me think of."
You walked right into that one. "Will you please take this seriously? The next six months are crucial and I don't know what I'm supposed to do. I'm terrified it will fail." That was an understatement. She couldn't imagine a future without Kama. She'd be left to start all over again, doing what, she had no idea.
"It really is important to you, isn't it? Your little business of making nightgowns."
"Don't be so dismissive. It's my livelihood. My career."
"And like I said last night, if it means that much to you you have to own it. You were hiding it from me last night. That's troublesome."
I was hiding me from you last night. Not the same thing.
"And whatever you say, you still have nannying," he continued. "There will always be children to care for and you're so good at it. Not everyone is an entrepreneur."
"I don't know how many times I need to tell you. I am not a nanny."
"Now who's living a contradictory life? I watch you with Oliver and you clearly adore him. So you love kids, but refuse to earn a living that way?"
"I want the challenge of making this work."
"That's it?"
"That's it." That's all I'm going to tell you.
"Okay, then. Walk me through the whole thing."
"Let me show you." She pulled up some photos on her phone. She'd taken them to show the bank when things started to take off and she'd tried to get a loan for expansion. "Flip through these. You'll see some of what I'm up against."
He swiped at the screen. "It's tiny. How can you get anything done in this space?"
"Honestly? I have no idea, other than I have some incredible employees who are willing to put up with a lot. I have six people doing assembly. If I were going to keep up with demand, I could easily have two dozen, but I'd have nowhere to put them. Moving means a huge lease, more equipment, health insurance and finding qualified people. It's a lot. I barely sleep as it is."
"So outsource. Let someone else do your manufacturing."
"I can't fire these people. They've been with me since the beginning, and they all do exceptional work. They have families to support."
"You're destroying your margins."
"Not if I'm in with the right retailers and can demand a better price point. Plus, our margins will improve once we've streamlined our manufacturing."
"Okay, then. Why don't I become an investor? I'll write a check and we can be done."
Here was Aiden's propensity for clean and simple, in sharp focus. It wasn't merely his attitude toward romance—he did this with everything that could get messy. No matter what, he was not going to become her investor. She needed to see out this ten days and get out of Dodge before his eyes made her do something she would regret.
"I need guidance. I need someone to give me advice and help me make the right connections, not throw money at me and hope I'll go away. That's your role in our arrangement and now you're trying to get out of it." She didn't want to speak to him in this manner, but she hated being blown off. It felt too much like Jason discounting everything about her—her dreams, her desires, and most important, the feelings she'd thought were between them, the ones he'd said were a figment of her overactive imagination.
Aiden handed her back her phone, appraising her. It was as if she could see the gears turning in his head, and she was more than a little nervous to hear what he was going to say. "You're right. I said I'd help you and I will. Not having looked at your financials, I'm thinking we need to find you an exclusive partnership. Become a subsidiary of a larger fashion corporation to scale your production, help with facility and warehousing issues, and most importantly, take over distribution so you can focus on what you're good at."
Finally, he got what she wanted. And he wasn't trying to back out. "Yes. Great. I can spend my time designing." And I can be back on track.
Seven
Sunday had brought Aiden's less sunny side. He'd finally called his mother about Oliver that morning, which had not gone as he'd hoped. He'd assumed, and so had Sarah, that she would be eager to come over right away. Instead, she'd said she was busy and would stop by Monday. That response had prompted Aiden to hunker down in his home gym for hours, lifting weights and running on the treadmill. As if the man needed to be in better shape.
Sarah had tried her best to go about her day, working on Oliver's room while he played, and during his nap, doing research on apparel companies she could partner with, per Aiden's suggestion. She'd also taken Oliver for a long walk, with a stop at a bookstore for some of her favorite children's reads. Considering how much Aiden loved books, she knew the gesture would be appreciated. Maybe someday those books would make him think of her—the woman who'd brought him Oliver out of the blue. And slipped away just as fast.
Now that it was late Monday morning, Sarah was still awaiting the return of pleasant Aiden. He'd been a real jerk during the nanny interviews, which was not the way it should've gone. The agency was the top in the city. Money was no object. All signs led to this being a short and simple process. But she hadn't counted on Aiden stonewalling.
"What is your problem?" Sarah asked as the elevator doors closed on the fourth and final candidate. "For now, the agency has no more nannies to send. The woman was practically Mary Poppins and you tell her that you don't think she's right for Oliver?"
Aiden shoved his hands into the pockets of his dark gray suit pants. He'd ditched the jacket and tie for the interview, but otherwise dressed handsomely, which was driving Sarah crazy. It took too much work to be mad at him when he looked so good.
"Did you see her face when she wasn't talking? It was so cold and stern. I want Oliver to be happy, not scared out of his mind."
"Are you saying she had resting bitch face? Is that really what this has come down to? Because you're being ridiculous."
"I didn't like her. End of story."
Sarah grumbled. Aiden might be right about the woman's austere facial expressions, but she was otherwise perfect. Plus, she was in her fifties and happily married and there was a very petty part of Sarah that wasn't about to leave Aiden with a perky twentysomething.
She flipped through the candidates' résumés. "What about Frances? She had a very sunny personality and came with impeccable references. She was a nanny for Senator Meyers, for God's sake. Do you think just anyone gets that job?"
"And why doesn't she have that job anymore? I'm not sure I buy her answer."
"She wanted to be in New York to help with her sick aunt. The Senator and his family are in Washington, DC. Seems reasonable to me."
"What if her aunt's illness takes over her life? I need someone who is solely focused on Oliver. That's what's best for him."
"You've spent all of three days with him. How can you say that you know what's best?"
Aiden shot her a look that said she'd taken it too far. He swallowed so hard that his Adam's apple bobbed. "I'm trusting my gut. That's the best thing I have to go on right now." He turned and walked out of the entryway and into the kitchen.
Sarah followed. They had to get the nanny situation resolved. "Don't forget that I'm the expert on this subject. I'm telling you right now that you're an idiot of the highest order for sending those four women away."
"We have to keep looking."
"I'm only here for a week, Aiden. It's Monday. Your ten days are up on Sunday and I'm gone. What are you going to do then?"
He grabbed an apple out of the bamboo bowl on the kitchen island. "Maybe I need you to stay longer."
So that's what he was doing—avoiding the potential mess of someone who might not be right by trying to keep the one thing he knew would work—her. "You're trying to force me to stay by sending away the other nannies?"
"Listen to what you just said. Other nannies."
"No way. I'm done with that."
"Honestly, I don't believe you're capable of walking away from Oliver on Sunday. You love him. I can see it."
Why did he have to make this so much worse? His words cut to her core. They were the truth and he knew it. "Of course I love Oliver. How could I not? But he's not my child and just like every other child I've cared for, I eventually have to leave him." Just saying the words brought up an unholy mess of things she dreaded and terrible memories. If she was bad at anything, it was goodbye.
"So you've got leaving down to a science. You can do it. No problem." Everything in his tone was biting, dripping with sarcasm.
"I'm not heartless."
"Which is why I don't buy it."
"Look. You need to focus on holding up your end of our deal. Part of that is hiring a nanny. I'm calling the agency to see if they have anyone else for us to interview." She slapped the résumés down on the kitchen counter. "In the meantime, I'd appreciate it if you would please look these over again and see if you're willing to reconsider any of these applicants." She turned on her heel and took extralong strides to get to the stairs.
"Sarah. Hold on."
She turned back, just in time to see him push aside the résumés. "What?"
He blew out a breath. "I'm sorry. I'm sorry if it seems like I have ridiculous standards, but my gut is telling me that those women were not right. Remember, this is all new to me. Almost too new. I'm doing my best. I swear."
She crossed her arms, hoping it would make it easier to buffer her attraction to him. It was hopeless when he was being sweet about the baby and talking in that tone that made her want to flatten him against the wall and climb him like a tree. "I know you're trying. I'm just antsy about time. We don't have much and I have to get back to Boston and Kama."
"I know you do, which is the other thing I need to say to you. Anna and I are working on getting us into a charity fashion show, organized by Fad Forward Magazine. Apparently it's a big deal."
Holy crap. Sarah clamped her hand over her mouth to keep a string of elated profanity from leaving her lips. "The Forward Style show? Where is it this year?"
"Miami. I thought I'd just buy tickets, but you have to be invited, which seems ludicrous since it's a charity event..."
Sarah couldn't breathe. She'd seen the pictures in Fad Forward Magazine every year since she was a teenager. Their annual charity fashion show was a chance for designers to bring out their most adventuresome work, and was attended by fashion legends, rock stars, Hollywood bigwigs and sometimes even royalty.
"Everybody who's anybody will be there. But I'm not sure it will help me."
"Our target is Sylvia Hodge. She's the honorary host this year. Anna and I dug up some info that she's acquiring new brands, but she's about to spend six months in Europe and Asia, looking for designers. If we want to meet face-to-face with her, going to Miami is the only way. And we might have to just walk up to her and start talking. I can't get her to take my call."
"But you're Aiden Langford. Isn't your last name enough?"
"Sylvia Hodge's admin didn't seem to care who I was."
"If we go, what are our chances?"
He shrugged. "No idea. Right now I'm waiting to see if they'll let me buy tickets."
"But the tickets. I can't afford that. They're tens of thousands of dollars."
"It's my treat."
"But it goes so far beyond our agreement."
"You brought me Oliver. It's the least I can do."
* * *
After having been away from LangTel for much of Friday and all of Monday so far, Aiden had a mountain of work, but he couldn't focus, not even with the relative quiet of working from home. The clock on the wall was taunting him. Three forty-five. Fifteen minutes until his mother was set to arrive to meet Oliver.
Sarah poked her head into his office, Oliver on her hip. The baby smiled at him, sweetly tilting his head to the side. This child would be the death of him, in a good way.
"If you're still working, I can hang out with Oliver until your mom arrives," she said. "Then I'll clear out so you three can have some time alone."
Aiden was trying to be optimistic, clinging to the idea that Oliver would bridge the chasm between him and his mother, but he had too many reasons to believe that would not be the case. "Where are you headed?"
"Out for a run. With all of the excitement of waiting to hear about Miami, I'm way too tense. Plus, I haven't worked out in days. I feel like a slob."
His vision drifted over her. She was wearing black leggings that showed off her fit and healthy curves, with a formfitting top that left her bare shoulders on display. Her hair was back in a high ponytail. He stepped out from behind his desk and took Oliver, unable to keep from admiring her. He wrestled with a deep desire to thread his hands into the back of her golden blond hair and pull out the rubber band, tilt her head back and give her the sort of kiss that makes a woman linger for a moment afterward with her eyes half-open.
"You are not a slob. You look incredible."
"You're just saying that because you feel bad about the nanny interviews."
"I'm saying it because you're a beautiful woman and I'd be an idiot if I didn't at least say it out loud."
A wash of pink crossed her cheeks and she fought a smile. If only she knew that it made him that much more attracted to her. If only she knew that she was making every inch of his body draw tight and burn hot.
"Thank you. I appreciate the compliment." She pressed her lips together and gazed up at him. "If you're going to take Oliver, I'll just head out. I should be done in about an hour. I don't know how long your mom is planning on staying, but I can grab a cup of coffee if you want more time."
The gears in Aiden's head whirred. He'd first thought it would be better if he and his mom were alone with Oliver. Keep things simple. But the truth was that he couldn't imagine Sarah not being there. It didn't make any sense, although he wanted to know why. Then he realized that no matter the situation, Sarah calmed him. She took the edge off. She made him believe things would work out. Aside from his sister, he didn't have anyone in his life who did that, but this was different. Sarah wasn't obligated to make him feel good.
"What if I said I wanted you to stay?"
She scrunched up her adorable nose. "What? Really?"
"I could use the moral support. I could use someone on my side. Things with my mother are not easy. I think you've gathered that much by now."
"I have, although you haven't told me the reason why."
Because I don't want to talk about it. "It's complicated. If you're here, it'll keep the conversation light and fun. I could use that right now."
She looked down at herself. "Oh God. I look terrible. I should go change. I don't want her to see me like this."
Before he could think about what he was doing, his fingers cupped her chin. He shouldn't have crossed that line, but he couldn't help himself. "I think you look perfect. Don't change."
She didn't move. He didn't either. Neither of them said a thing, but their eyes connected, as if they were each digging deeper, wanting more.
Sarah broke the spell with a shake of her head. "You're sweet, but there's no way I'm wearing this to meet your mom. And I have no makeup on." She turned and headed out of his office. "Back in five minutes."
Aiden watched her jog away, her leggings accentuating every move. A ripple of steamy thoughts ran through his head—everything he wanted to do with her. It had been a long time since he'd wanted a woman as badly as he wanted Sarah. The question was whether the opportunity would present itself. So far, there was always something in the way.
Aiden wandered into the kitchen, where his housekeeper had put on a pot of coffee. She'd also stocked Sarah's cookie jar with an assortment of biscotti, some of it plain. Oliver regularly chowed down on teething biscuits, and Aiden decided that this was basically the same thing.
"Do you want a cookie?" He offered it to Oliver.
The baby snatched it from his hand and it went right into his mouth, like most things. His eyes grew wide once he'd gotten a taste. Aiden leaned against the counter, enjoying the moment. Sarah was going to leave behind a lot more than a cookie jar on Sunday.
"Good, huh? Just wait until you get older and I can take you out for hot fudge sundaes or we can get a hot dog at a baseball game." The thought brought with it a peculiar mix of hope and melancholy. Dads did those things with their children. Aiden very much looked forward to having those experiences with Oliver, but they were things he'd missed out on entirely.
Sarah hurried down the stairs. "This is as good as it's going to get. I really wish you would've given me some advance notice. I could've taken a shower and done something with my hair." Her face was flush with color, probably from rushing around. She wore a full black skirt that skimmed her knees and a white top that hinted at the curves he'd been admiring for days now. She'd put on the sandals she'd been wearing the day he met her, which gave her a few more inches of height. He still towered over her, but he loved the way they made her legs look.
"Once again, you look perfect."
The buzzer for the elevator rang.
"You're sweet. And you need to answer the door."
Aiden's heart went from racing over Sarah to plummeting to his stomach. His mother had arrived. He didn't bother with the intercom, hitting the button to grant her access to his floor. "Here goes nothing," he said to Sarah. He filed into the entryway, Oliver in tow and making excellent progress on his cookie.
When the doors slid open, Aiden managed a smile. It was only half-forced. He still loved his mother, despite his immense frustrations with her.
She actually gasped when she saw Oliver, breezing off the elevator in her usual garb of all black with a colorful scarf tied at her neck. She took a direct route to her grandson, her mouth softening to a tiny O. "Aiden, he looks just like you." She held on to his hand and shook her head in disbelief, but not enough to muss her short, dark hair. "What an angel." A tear rolled down her cheek, but then a steady stream started. Of the many reactions he'd anticipated from his mother, full-on crying was not one of them. She smiled through the tears, her eyes crinkling at the corners. "Can I hold him?"
Aiden was stricken with conflicting emotions, ones that didn't belong in one person's head at the same time. He wanted to protect Oliver. But at the same time, there was a yearning—so deep he could feel it—for his mother to accept and love Oliver. He had to take this leap, however much she could end up hurting either of them. "Yes. Of course."
He handed over Oliver, who seemed perplexed. She bounced him up and down as she plopped her handbag on the entry table.
"Come in, Mom. I want you to meet Sarah."
Sarah was pouring herself a glass of water. "Mrs. Langford. It's so nice to meet you. I've..." She paused and looked right at Aiden. "I've heard so much about you."
"Please, call me Evelyn. And I wish I could say the same about you. My son has been remarkably quiet about everything." She turned and shot Aiden a disappointed look, a wordless reprimand. Did she have any idea how hard he was working to keep up his hopes? Intentional or not, she expertly knocked them back down. "Not that I'm surprised. He keeps things to himself. Always."
Eight
Stress radiated off Aiden like August heat off a tin roof—jaw tense, shoulders rigid. Was he always this away around his mother? He must be, because she didn't seem to notice. She was too preoccupied with Oliver, sitting on the floor in the library, offering him toys from a bag Sarah had given her.
"He's smart. I can tell," she said to no one in particular.
Aiden stood sentry, arms crossed squarely at his chest. This was not a bonding moment for him. He was observing, like a hawk.
Sarah walked up behind him and placed her hand on his shoulder. He flinched, then relaxed under her touch. She might have underestimated this burden, and her heart ached because of it. Whatever there was between him and his mother, it was not good. Sarah desperately wanted to know more. Even if it was painful, she wanted to know.
"Is there anything I can get for you?" she whispered to Aiden.
He looked over his shoulder, lowering his face closer to hers. A waft of his heavenly scent hit her nose—warm and masculine, like the sheets on his bed, just like his entire bedroom. "Now who's the sweet one?" he muttered.
You are. When you want to be. "I'm sensing you could use a drink."
"It's only a little past four."
"It's five o'clock in Nova Scotia. I've heard it's lovely there this time of year. Bourbon?"
"On the rocks." He cracked a smile and his shoulders visibly relaxed. She fought the urge to dig her fingers into them, help him unwind while she committed the contours of his broad frame to memory.
"Cocktail, Evelyn?" Sarah asked.
She shook her head and started peekaboo with Oliver. "I'm too in love with my grandson for a drink. But I'll take a diet soda."
"Got it." Sarah went to work, getting Aiden his drink first. He was in greater need. "Here you go." Their fingers brushed when she gave him his glass, sending a tingly recognition through her.
"Thank you. For everything." His voice was low and soft, just as luxurious as the bourbon in the glass. Why did the man she was supposed to stay away from have to be so undeniably sexy?
The elevator rang again.
"That's odd," Aiden said. "I don't know who could be here."
"I'll get it." Sarah hurried to the entryway and buzzed the intercom. "Hello?"
"It's Liam Hanson for Mr. Langford. I'm one of the admins at Barkforth and Sloan."
The paternity test. "I thought you were coming tomorrow morning." Surely this was not the sort of commotion Aiden wanted while his mother was there.
"They asked me to come this afternoon since you need this done right away. They should've called you."
"Come on up."
Aiden joined her. "Great. With my mom here." He must have overheard. He raked his hand through his hair, again showing her the strain she hated seeing on any face, especially on one as handsome as his.
The elevator doors slid open. "Mr. Langford. I'm so sorry if there was a miscommunication. I promise this will be quick. Five minutes and I'm out of your hair."
Aiden nodded. "No, it's fine. We have to get it done. What do you need?"
"It's a cheek swab from you and one from the baby. That's it."
"He's in the other room."
Sarah followed Liam and Aiden into the library.
"Mom, I need Oliver for a minute. The lawyer's office needs to do the cheek swabs for the paternity test so we can take care of the legal end of things."
Evelyn picked up the baby, handing him to Aiden. "This seems silly. One look at him and it's obvious he's your son."
Aiden snatched Oliver away, much as he had the day Sarah had mentioned the idea of Evelyn caring for him. "It's important for Oliver and me, too. That we know he's my son, for sure."
She took a sip of her soda. "Of course."
Liam took out two plastic tubes and asked Aiden to open his mouth. He swabbed the inside of his cheek, then did the same to Oliver. "All done. We'll have this expedited. I understand the paperwork needs to be taken care of this week."
Aiden nodded. "Yes."
"If it takes until next week, I could always come back down from Boston, I suppose." Had she just said that? She should not be straying from the timeline she'd established. With every passing day, they were more comfortable with each other, she more attracted to him, her resolve becoming flimsy. There were signs he might be feeling the same way—the moment when he'd slid his fingers under her chin? The comments about how good she looked?
"It's best if we wrap this up quickly," Aiden said. "Sarah needs to leave on Sunday and return to her business. Oliver and I need to start our life together, too."
Why was her body filled with such utter disappointment at his words? This was what she'd wanted, and it was no time to switch her priorities. Stay with the plan, Sarah. "Aiden's right. I really can't be running back to New York."
"Certainly." Liam packed up everything in a messenger bag, which he slung across his chest. "We'll take care of it, Mr. Langford. We should have the results by Friday."
Aiden walked Liam to the elevator and returned with Oliver. There was a shift in the mood, one impossible to ignore. Aiden had said he wanted to keep things short with his mother for this first visit. Evelyn showed no sign of leaving.
"Can I hold him again?" she asked, getting up from the sofa.
Aiden took in a deep breath. "For a few minutes, then Sarah and I have some things to do. Oliver needs a bath and quiet time before bed."
Even if it might have been an excuse, Aiden was clinging to the schedule, and it was adorable. He sat on the couch next to his mother and tried to hand over the baby, but Oliver wanted to stay with Daddy.
"Don't you want to spend time with Grandma?" Evelyn leaned into Aiden, trying to catch Oliver's attention.
"He's still getting used to you, Mom. It might take time."
"Maybe he's unhappy after having that strange man put a stick in his mouth. The whole thing really is silly. He looks just like you."
"We have to follow the proper channels. It's important." Aiden held Oliver close. "He lost his mother a month ago, and that will be something he'll always wonder about. A child needs to know where he came from. He needs to know where he belongs." Aiden's voice cracked, which Sarah had never heard before, not even in the tender moments with Oliver.
"I wasn't trying to upset you."
Aiden cleared his throat. "Mom. Listen to yourself. There's this dark cloud hanging over us and Oliver has given us the perfect chance to talk about it, but you won't let yourself go there, will you? You'd rather keep your secret."
"I have no earthly idea what you're talking about."
"Yes, you do. I've brought this up with you at least three times since I've been back in New York. It's the reason I've been unhappy for most of my life. It's the reason I tried to take over LangTel. And you're sitting here, talking about the paternity of my son like this hasn't been a question in your own life. It's so frustrating I want to scream." His voice didn't waver in the slightest now. It was sheer determination.
Oliver whimpered.
"You've upset the baby," she said.
Oh no.
Aiden pulled Oliver closer and rubbed his back, pecking him on the forehead. "I think you need to leave. I'm not going to pretend that my entire existence in this family hasn't been built on a lie and you won't own up to it. I know that your husband was not my father. I know it with every fiber of my being. And we all act as if that isn't the case."
Sarah gasped. She didn't mean to, but how could she not? Roger Langford wasn't Aiden's father? How could that be?
Evelyn reached out and set her hand on Aiden's forearm. "Aiden, darling. Haven't we all been through enough with losing your father? Don't tear us apart even more. I love you and you're my son. That's all that matters."
Sarah turned to sneak upstairs. She had no business being in the room for this.
"No, Sarah. Don't leave," Aiden said.
Did the man have eyes in the back of his head? Sarah looked down at her feet. Damn noisy shoes.
"So that's your response," he said to his mother. "And when you say we lost my father, do you mean my actual father, or Roger? Because I know they're not the same. There's no other explanation for you sending me off to boarding school. There's no other reason why Adam would be deemed heir apparent when I'm the oldest."
"Aiden, there's no good in dredging up the past. And I really don't think we should discuss this in front of a stranger."
"A stranger? You're calling Sarah a stranger? She's nothing of the sort." Aiden bolted from his seat. "It's time for you to go now. You can come back when you're ready to talk. Until then, I don't want to see you."
"You're going to keep me from my grandson?"
"That's all on you. I have nothing to do with it."
Evelyn blew out a breath, just as determined as Aiden. Family gatherings must be a real barrel of laughs with the Langfords. "You'll change your mind. A baby needs his grandmother." She leaned forward and kissed Oliver's forehead, but Aiden kept both arms firmly around him. "Bye-bye, sweet boy. Grandma will see you soon."
With that, Evelyn Langford traipsed out of the room. Aiden stood, facing the entryway, his back to Sarah. All Sarah could hear was her own pulse thumping in her ears. What she was supposed to say? What was she supposed to do? There was no mistaking the pain in his voice. True or not, he believed that he'd been lied to about his father.
"I'm sorry you had to see that," he said. "I'm sure that's the last thing you would've wanted to trade your run for."
She went to him, her heart heavy. "No, Aiden. I'm glad I was here. I mean, if that's what you wanted."
"It is what I wanted. Honestly, your presence probably prevented a bigger blowup. One that's been coming for thirty years."
She reached for his arm, wanting to comfort him even more than that. Here she was, stepping into emotional quicksand, the very last place she belonged if she was going to leave on Sunday with her heart in one piece. "I don't believe that you would've blown up in front of Oliver. I really don't."
Aiden managed a smile, but it was as if it had been broken and cobbled back together. There was some part of him inside that was fractured. She'd sensed that about him the day they met, but now she was beginning to understand what had caused it. Her own family was so important to her. They were always there for her. Always. She couldn't imagine growing up the way Aiden had. He might have had money and privilege, but that didn't replace love. That didn't replace knowing where you came from.
"Thank you for being here. That's really all I can say." He tugged her into an embrace with one arm while he held on to Oliver with the other.
She sank against his chest, so drawn to him, every inch of her wanting to make things better. Each second in his arms was another step into his world, but she would've needed a heart of stone to walk away. He needed her. And in that moment, him needing her was everything.
* * *
Oliver was sound asleep in his crib, but Aiden stayed, studying the steady rise and fall of his chest as he slept like a starfish—arms above his head, legs splayed, tiny rosebud mouth open. After the upset of his mother's visit, Oliver filled Aiden with contentment he'd never known. If he never had anything more in his life than Oliver, even if he never got the truth from his mother, he could be happy.
He flipped the baby monitor on, then crept out of the room, quietly closing the door. A few steps down the hall, a heavenly smell hit his nose. Sarah was cooking dinner and judging by that one whiff, it was going to be delicious. She was his savior today, but not because she was preparing a meal. She'd been there for him when his composure crumbled and anger threatened to consume him. She'd been his rock.
That left him in a peculiar spot. If he were smart, he needed to work very hard to keep Sarah as a friend, and as part of Oliver's life. He couldn't imagine her not being involved, even if it was only an occasional phone call or a visit on Oliver's birthday. Considering his zero percent success rate with keeping a woman around for more than a few days, logic said he shouldn't allow them to be anything more than friends. He shouldn't cross that line, however attracted he was to her, even when every inch of him wanted her. Between her beauty, her spark and the sweet things she did for him, he didn't see how he was supposed to stay away. He only knew he had to. Giving in to the temptation of Sarah—sweeping her up in his arms and finally tasting her lips, would likely end with her never speaking to him again. For the first time in his life, he didn't want that.
"Whatever you're cooking, sign me up." He strolled into the kitchen.
She had two glasses of red wine waiting. Was she the perfect woman? She was reading his mind. He wanted nothing more than to relax and put his afternoon behind him.
"Good. Because otherwise, it's toast or a protein bar." Her back was to him, and she was humming—something he'd noticed she did every time she was busy in the kitchen. He forced himself still, to keep from walking up behind her, wrapping his arms around her waist, leaning down and kissing the graceful slope of her neck. He wanted to bury his face in her hair, inhale her sweet scent, get lost in her.
But he had to be good. So he slugged back some wine. "What are we having?"
She turned and smiled sweetly. "Pasta. Almost ready."
"Perfect. Thanks for opening a bottle of wine. If you hadn't, I would have." What was he doing? He'd been reprimanding himself moments ago about how Sarah had to stay a friend, and yet he couldn't stop from talking as if he was in pursuit.
She dished the pasta into two bowls. "Sorry it's not fancy."
"If you want fancy, we could take this bottle of wine up to the rooftop when we're done eating. It's a beautiful night." Was this a good idea? No. Did it sound like fun? Yes. "We'll need to bring the baby monitor."
"Okay, Dad." She elbowed him in the ribs and flashed a flirtatious smile. That was it. She was going to kill him before the night was over.
They ate at the kitchen island, chatting about Oliver, squabbling about nanny candidates and whether or not there would be any more people to interview. It wasn't long before dinner was done, the dishes were in the dishwasher and Sarah suggested they open a second bottle of wine.
"The terrace?" Stop encouraging this.
"I should grab a cardigan first. In case I get cold."
I can keep you warm. He pressed his lips together to keep the words from escaping. "Okay."
He waited for Sarah outside her room, then led her to the second-floor stairs at the back of the house that took them up to the empty third floor and finally up to the terrace.
The darkening night sky was streaked with purple and midnight blue, the city lights casting a glow across Sarah's face. She rushed across the stone pavers, like a little kid who couldn't contain her excitement. "It's so beautiful up here. Like your own private park." Holding out her arms, she turned enough to make her dress swirl around her legs.
"It's nice, isn't it?" He smiled, admired her, wishing he could make everything in the world conform to his will—why couldn't he have a free pass for a night, kiss her and have everything return to normal tomorrow?
"After meeting your mom, I think I understand why you need space."
"Very perceptive. Although the physical space is nice for anyone, especially in the city."
He led her to an outdoor sectional couch and lit a kerosene heater. She plopped down, tucking her leg underneath herself. He set the wine bottle on a low table and joined her, keeping his distance, staying in check.
"Do you want to talk about today?" she asked. "Maybe you'll feel better about everything if you just get it out."
He wasn't much of a talker, especially when it came to things like this, but Sarah wasn't like anyone he'd ever considered confiding in. She had no agenda, nothing to gain. And she had to be wondering what was going on. "I've suspected since I was eight that the man I called Dad wasn't my biological father."
Sarah pressed her lips into a thin line. "That's what I thought. I didn't want to eavesdrop on what you were saying to your mom, but it was hard not to hear."
"I'm glad you were there. It made it far less uncomfortable."
"It seemed pretty uncomfortable."
He had to laugh. She didn't shy away from the truth. "Honestly, that was nothing."
"How does an eight-year-old arrive at that conclusion?"
"I was home from boarding school for Christmas break and I overheard them arguing about it."
"Did you ever ask them about it?"
He took a sip of his wine, fighting back memories of standing in the hall of the Langford family penthouse apartment, late at night. He'd been unable to sleep and wanted to ask his mom for a warm glass of milk, but he'd instead heard her say something terrible. If I could take it back, I would. But I can't change the fact that he's not your child. "I never said anything to anyone until I was much older. You have to understand, Roger Langford was an imposing man. And he was never very warm to me. He was to Adam and Anna, but not to me. I didn't want to give him another reason to push me out of the family."
"Push you out?"
"They sent me to boarding school after Adam was born. I was seven. I didn't understand why, but they said it was for my own good. Now I suspect it was because he couldn't stand the sight of me."
Sarah's eyes became impossibly sad. He hated seeing that look on her face, even when it was his pain she was reflecting. She grasped his forearm, scooting closer on the couch. The distance he'd left was gone, and he was so glad. He craved having her close. He wanted nothing more than to erase the space between them.
"I can't even imagine what that must've been like. I'm sure you were just the sweetest boy. I'm so sorry."
Aiden wasn't much for pity, but it was healing to have Sarah see how wrong it all was. "It got worse over the years. I got in trouble at school a lot, mostly for fighting or practical jokes. I got kicked out of a few. That was never fun. It embarrassed my dad and confirmed to him that I didn't belong at home. I guess I was self-destructive, but I was confused. I certainly didn't feel loved."
Sarah was now rubbing his arm softly with her thumb. "Of course you didn't. No child should feel that way." She looked down at her lap and fiddled with the hem of her sweater. She was so gorgeous in the moonlight—it was like watching a museum masterpiece come to life. "I know that my arrival with Oliver was a shock, but you have such a big opportunity with him."
"Opportunity?"
She nodded and sucked in her lower lip. "You can give Oliver the childhood that you didn't have."
For a moment, it felt as though the earth didn't move. He'd thought that on some level, but hearing her say it made it clear how right it was. "I can break the cycle."
"Yes. Although I don't think you'll have real closure until your mom finally tells you the truth."
He downed the last of his wine. "I'm starting to wonder if that will happen."
"Have you told anyone else in your family?"
"I confided in Anna, but she thinks I'm crazy. She knows that something wasn't right in our household, but I don't think she wants to believe it. My father has only been gone about a year. Everyone is still grieving. No one wants to think ill of him."
"I don't think you're crazy, Aiden. It makes perfect sense to me."
Sarah's lips were right there, waiting for him, telling him everything he'd ever wanted to hear. The validation of his pain, his fears, the vulnerability that he wished didn't exist at all, was so powerful it made his entire body feel lighter. He wanted to kiss her so badly, to express his gratitude for her in a way that would leave no doubt in her mind that he appreciated her.
But he couldn't do that. Not when he needed her in Oliver's life. Not when he was sure it would ruin everything.
His phone buzzed with a text. Normally, he wouldn't stop to read it, especially when he was alone with a beautiful woman, but it was taking extreme effort to keep from kissing this one and it was his intention to do exactly that. "I'm sorry. I should check this." The message was from Anna—good news. "You and I are going to Miami. We have our tickets."
Sarah popped up out of her seat. "For Forward Style?"
He laughed, watching her bounce on her toes. "Yes. We go the day after tomorrow."
"Wednesday? Oh my God. We have to book flights. Who's going to take care of Oliver?" She looked him squarely in the face. "What am I going to wear?"
"We'll ask Anna and Jacob to stay with Oliver for the night. You know they'd love to do it."
Sarah's shoulders dropped with relief. "True. She and Jacob are so good with him. What about the rest?"
"We'll take the corporate jet. No need to worry about flights."
"Are you sure? That seems extravagant."
He grinned so wide it made his cheeks ache. The joy of seeing her happy and excited was his reward after a roller coaster of a day. "Yes. I'm sure. I told you I'd help you, and I'm a man of my word. As for what you should wear, we're in New York. Go shopping."
She shook her head. "There's no time. We have more nannies to interview tomorrow, and I know you need to get some work done. Plus, if I'm going to walk up to Sylvia Hodge and try to impress her, I need to be wearing one of my own designs."
"You're going to wear a nightgown?"
She slapped his arm and grimaced. "No, silly. I design other things. And I have the perfect gown at home. It's gorgeous. Emerald green, dangerously low-cut. Sylvia will love it." Her eyes flashed with mischief. "I just need to get Tessa in my office to overnight it to me."
Aiden felt like he couldn't breathe. In a little more than twenty-four hours, he'd be alone in Miami with Sarah and her dangerous dress. How he loved peril, especially at the hand of a beautiful woman. "Have her send you a swimsuit, too. We can't go to Florida without some fun in the sun."
Nine
Between leaving Oliver overnight, and knowing that in twelve hours she'd have to dazzle Sylvia Hodge at a fashion event most people would kill to attend, Sarah was so worked up she thought she might be sick. "I hope I haven't forgotten anything." Yesterday had been such a whirlwind, it'd be a modern miracle if she hadn't messed up something. Oliver'd had a fussy day, which probably meant he had another tooth coming in. Aiden had been shuttered in his home office for hours, coming out long enough to say no to three more nanny candidates.
Sarah had dealt with a million other details beyond that, including having Tessa, her assistant, overnight her gown and a few more clothes to the hotel in Miami. That meant she was in the same black sundress she'd had on the day she met Aiden. She didn't feel confident at all, but she'd only packed a weekend's worth of clothes when she'd come to New York. Two days had always been her plan.
"If we have any questions, we'll call you." Anna eased herself into a chair at the kitchen table. "I just want you to go to Miami and kick some serious butt."
"Any last-minute advice before we head to the airport? You worked in the garment industry for years. I really wish you could be there to make me look less incompetent."
Anna sat up straighter and reached across the table, placing her hand on Sarah's. "From the moment I met you, you struck me as nothing less than cool determination. You will have no problem with Sylvia Hodge. She's drawn to people who have a vision. Show her what you see for your future and everything else will fall into place. I promise."
Sarah blew out a breath. What a coup it was to have Anna's help, and Aiden's for that matter. He was bankrolling this venture, after all. But knowing he was putting so much money into it only made the pressure that much more intense.
"Ready?" Aiden strode into the room. In dark gray dress pants and a white dress shirt, he looked so good he could've sold her a magic bag of beans.
Jacob followed, holding Oliver. He'd just had his first diaper-changing lesson, courtesy of Aiden. The former daddy-in-training was teaching the daddy-to-be.
"You look extra handsome holding a baby," Anna said to Jacob, slowly pushing herself up out of the chair.
Jacob flashed his dazzling smile. "It's the ultimate fashion accessory. Women go crazy for it."
Anna rolled her eyes and sidled up next to them. "Don't push it."
Aiden watched Anna and Jacob with Oliver. He was anxious—Sarah could see it. That made a small part of her melt on the inside. There was nothing sexier than a man who was on edge about leaving his child.
For Aiden's sake, Sarah started the goodbyes. "Okay, sweet boy. Anna and Jacob are going to take very good care of you. We'll see you tomorrow when we get back." She kissed his forehead. Emotion washed over her. In a few short days she'd be doing this for real.
"Goodbye, buddy." Aiden's voice wobbled as he cupped the back of Oliver's head and kissed his cheek.
Aiden's driver John was waiting outside in the black SUV, idling at the curb. The ride to the private terminal in Teterboro, New Jersey, took nearly an hour with traffic. Aiden made work calls, leaving no time for them to talk. The way he laid down the law with people was inspiring and intimidating. Would she ever be that in control? Could she grow her company and give herself security, command respect and just tell people what she wanted? She had a hard time imagining she could muster that much mojo.
The car went through a security gate and drove up alongside the sleek white jet, tastefully marked with the royal blue LangTel logo. The plane's boarding stairs had been lowered to the tarmac. Aiden's driver opened the door for Sarah, and she dug her fingernails into the tender heels of her hands, reminding herself that this wasn't a movie. This was really happening. "Thank you so much, John. I don't know how you navigate traffic in this giant car, but I'm glad you can."
"Thank you, Ms. Daltrey."
She gently swiped at his arm. "Please, call me Sarah. It's only fair since I call you John."
An amused smile crossed his lips. "You're very gracious, Sarah. Excuse me while I retrieve the bags. I'll see you on board."
"Oh. You're joining us?"
"I always drive Mr. Langford. Everywhere."
Aiden climbed out of the car and placed his hand on her lower back, only amping up her nervousness. That touch from any other man would've reminded her this was really happening. It would've taken her out of the dream. But Aiden? He put her that much further into it. He nodded toward the plane, his sunglasses glinting. "Shall we?"
They climbed the stairs and stepped into the luxurious cabin, piling one surreal moment on top of countless others. There were a dozen or so oversize cream-colored leather seats, mahogany and chrome accents. Everything gleamed—even the flight attendant's red lipstick and white smile as she said, "Welcome aboard."
"Anywhere special you want me?" Sarah asked Aiden.
He removed his sunglasses and cocked an eyebrow. "Wherever you'd like to be is fine with me."
Her face flushed with heat. Damn him and his comebacks. Damn her and her brain that just had to go there. She took the seat closest to her. Aiden took the one directly opposite, facing her.
"May I get you a drink before takeoff, Ms. Daltrey?"
Sarah hadn't had a chance to introduce herself. Nor did she have a chance to respond.
"We'll have the usual, Genevieve," Aiden answered.
"Yes, Mr. Langford."
"I trust that champagne is okay?" Aiden asked Sarah.
"It's nine in the morning."
Aiden ruffled the newspaper open. "You're on edge. I see it on your face."
My face is fine. John boarded, taking a seat in the front. Sarah glanced out the window as the plane began to taxi. She hated to fly and seeing outside after takeoff would only make it worse. She lowered the shade, doing her best to act as if this was exactly where she should be. If she seemed on edge, it was because she was as far out of her element as she could imagine, and this was only the start of her day on the brink. There was much more to deal with—Sylvia Hodge, the fashion show, the countless glamorous people who would be in attendance, who would undoubtedly be wondering how someone like Sarah got in. And then there was the not-small matter of spending twenty-four hours with Aiden, when she already didn't trust herself with him.
Own it, Sarah. Own it. "I'm not on edge. I'm just thinking over the things Anna and I talked about this morning. For the moment when I meet Sylvia Hodge." God help me. "She gave me some great pointers." Sarah sat up straight and crossed her legs. If only she was in her black pencil skirt, short peplum jacket and pumps, she'd be the epitome of put-together. Thankfully, Tessa had sent that power suit, along with the dress she'd designed and a bathing suit. The clothes would help her fake her way through today, and then she'd be golden.
In the interest of control and modesty, Sarah had been explicit with Tessa about the swimsuit, asking for the plain black one at the very bottom of her dresser. Plain black. Got it? Hopefully she could talk Aiden out of a trip to the pool and she wouldn't even need it. She'd seen him without his shirt and managed to keep her own clothes on. No point in pressing her luck.
The flight attendant brought two champagne flutes, filled with golden bubbles. Aiden folded his newspaper and reached out to clink his glass with hers. "To success."
She admired his optimism—success was not a familiar concept. "Yes. I'm hoping for success."
His vision narrowed on her, a crease forming between his eyes. "There is no hoping. You need to walk up to Sylvia Hodge tonight, tell her what you do and tell her what you want. That's how you make deals. By taking charge."
"And how, exactly, do I take charge with Sylvia Hodge? She's a legend. She's put more designers on the map than anyone, and she's probably destroyed more. Just saying her name scares me."
"I can tell. And it's not good. But don't worry. I have a solution."
A solution? "Please. Do tell. I'm all ears."
"How do you feel about heights?"
Uh-oh. Mr. Adventure-seeker was at play here. "Absolutely mortified. So whatever it is that you're planning, just forget it. I'm not climbing or jumping off anything."
"No climbing or jumping. Just fun. You don't have to do anything other than sit there."
"A roller coaster?"
"Parasailing."
"Over water? In the sky?" Sarah's brain sputtered. As if she wasn't already nervous enough. "No way. My hair looks amazing today. I'm not giving up a God-given good hair day."
He leaned closer and rested his elbows on his thighs. "Sarah. You made me step outside my comfort zone. It's time for me to do the same for you. Trust me. It'll be good."
"Maybe I forgot to ask my assistant to send a bathing suit. Oh well. Your plan won't work."
He shook his head. "I was in the room when you asked her. Unless she's terrible at her job, you should be all set. Stop making excuses."
Sarah blew out an exasperated breath, downed half of her champagne and slumped back in her seat. Great. Now I get to risk life and limb right before I put my entire career on the line.
Three hours later, they were on the ground in Miami. Sarah stepped off the plane, thick balmy air hitting her skin as she squinted into the bright Florida sun. At least summery weather made it feel like vacation. A black SUV like the one Aiden had in New York was waiting for them planeside. John had them off to the hotel in no time. As much as Sarah felt out of place, traveling with Aiden did have an upside. No waiting to check your bag or slogging your way through security. It was lovely.
She glanced over at him while the car sped along a causeway, palm trees fluttering in the breeze outside. He was so good-looking it sometimes hurt to set her sights on him for too long, as if her eyes grew weary of handsome. What would it be like to be Aiden Langford's female companion on a trip like this? Romantic female companion. She already knew the VIP treatment was wonderful, but sleeping in the same bed with six-plus feet of pure man? Kissing him, taking off his clothes...the thought of it made her squirm in her seat. She quickly turned away and stared out her own window. She had to stay focused on business, even if romantic fantasies about Aiden were a nice escape.
They arrived at The Miami Palm hotel, situated on a private key, a small island just off the coast of downtown, connected to land by a gated bridge. Inside, the hotel lobby had classic Miami opulence—art deco chandeliers, towering potted palms and a tropical color scheme of cream, coral and sea green. Aiden was apparently a frequent customer—every employee, especially every female employee, knew his name. They didn't even have to check in. The bellman brought them straight up to the top floor. And one room.
"One room?" Sarah said under her breath. "Isn't that a little presumptuous?" It was her duty to feign indignation, at least while her brain attempted to determine what exactly Aiden was up to.
The bellman opened the door and stepped aside. Sarah nearly gasped when she walked into the luxurious space. Heck, if this was where they were staying, Aiden could presume whatever he wanted to. A sprawling living area was before them, with two large sectional couches. A black baby grand piano was beyond that, flanked by linen-upholstered armchairs. A dining table for ten was on the other side of the living area, with a wet bar beyond that. Along the length of the room were a trio of sliding doors leading to the terrace, with palm trees, a cloudless sky and the ocean completing the view.
"The presidential suite, sir." The bellman wheeled their suitcases inside.
Aiden peered down at her. "See? Nothing presumptuous. Two bedrooms. Two master baths. Separated by this big room. And don't worry, your door has a lock."
Now Sarah felt stupid for saying anything. This was a business trip. She needed to start acting as such.
"It's wonderful. Thank you so much for arranging this. I really appreciate it."
"Holding up my end of the bargain."
And nothing more.
"Time to get settled," he added. "We leave in forty-five minutes."
Her shoulders dropped. "So you were serious about parasailing? Really?"
"Dead serious. I don't get to do this sort of thing nearly enough. I'm in Miami, I'm going parasailing. And you're coming with me."
"You know, I'm really more of a lounge-on-the-beach-with-a-mojito sort of girl."
"Although I'm enjoying the vision, that's not the plan today."
"But..."
He shook his head. "No buts. If you get scared, the boat can bring us in. But you won't get scared."
"Fine." Dejected, Sarah sucked in a deep breath and ambled to her room.
Once inside, Sarah's eyes were immediately drawn to her gown, hanging neatly on a dressing rod next to the closet. Her suit coat and skirt were behind the dress. They must have been steamed by the hotel staff. They looked the picture of perfection. The emerald-green silk of the dress was just as exquisite as she'd remembered, the beading on the bodice and trailing down onto the skirt equally sublime. This was a good thing. She would be confident in this dress. It would be her superhero costume, the one in which she set aside her everyday persona and became an invincible woman.
Time to step outside her comfort zone.
She turned, her vision drifted to the bed and her inner peace sizzled away like a bead of water on a hot skillet. Good God. No. There sat her beach cover-up, along with her bathing suit. The aqua-blue caftan was great thinking on Tessa's part. Sarah hadn't thought to ask for it. Next to it was indeed her black bathing suit—a plain one at that, precisely what she'd asked Tessa to send. Only that it wasn't the one she wanted. One-piece, dammit. One-piece. Not the teeny tiny bikini.
Ten
If Sarah seemed apprehensive during the flight, now she was downright agitated—trudging out of her room dressed for the beach in a pretty blue cover-up paired with sandals and a scowl. The hair that had been perfectly in place on the airplane was in a ponytail. Her makeup had been removed. Sarah didn't wear a lot of it, but there was a difference and he liked the change. It harkened back to the only morning she'd been in his bed, and the way he'd pored over her as she slept, wondering if it was a good idea to pursue any of the ideas ruminating in his head—thoughts of kisses bestowed and returned, and every satisfying thing that it could lead to.
"Ready?" he asked.
"You're making me do something I am literally terrified to do. So no, I'm not ready." Clutching her handbag, she plodded toward him as if she'd been banished to the gallows.
He placed his arm around her shoulders and gave her a gentle squeeze while unsubtly ushering her to the door. "You'll feel better after this. I promise."
"What do I get if I don't feel better? What if I feel worse? I'm already so worked up about tonight that I feel like I'm going to lose my lunch. We haven't even had lunch."
"You're just going to have to trust me."
She glared up at him when he opened the door. "You realize it's not my inclination to trust you, at all."
Her freckles again teased him, toyed with him. Now that they were completely alone, he wanted nothing more than to bend down and kiss her. Just get it over with so he could stop thinking so damn much and let his instincts take over. There was nothing stopping them—nothing stopping him, except the entirely foreign worry that sex might ruin what was already between them.
"You don't trust me even a little bit?"
"This is a trick question. If I say I trust you, it'll make your argument for letting a thin parachute carry me thousands of feet into the air over the Atlantic Ocean."
"Biscayne Bay. And it's five hundred feet, and that's only if they let us go all the way up. Not much more than a football field."
"Oh."
"I'll be right next to you. You can hold my hand."
"Oh. Okay." The faintest of smiles crossed her lips. "I guess that makes it a little better."
"Good. Let's go before you change your mind."
They headed down to the lobby. John was waiting outside and swiftly had them on their way to their beach adventure. This excursion was about more than distracting Sarah from her worries. He wanted her to see this side of him. She'd remarked about the photos in his apartment, comments that made him think she didn't understand why he did risky things. He hadn't always been the guy who jumped out of airplanes, but once you've done something that you could die doing, it takes away fear.
"I want to say one thing. Part of being successful in business is learning to fake fearlessness."
Sarah removed her sunglasses and shot him a very hot look of admonition. "Fake it? I assumed you were actually fearless. I've heard you on the phone. You're incredibly intimidating."
"I don't have to fake it now, but that doesn't mean there wasn't a time when I was scared to forge my own way."
"But you come from such an influential business family. Surely your dad helped you, even if you didn't have the best relationship."
It was Aiden's natural inclination to steer away from this topic, but Sarah knew his history. "That's one thing my dad offered, but I didn't want his help. By the time I graduated from college, I was too bitter to take anything from him anyway. I wanted to prove to my parents that I didn't need them. Now, granted, I had a trust fund that got me started, that was no small matter, but I did everything else on my own."
"Refusing your dad's help couldn't have made things better with your family."
"It didn't. But I also didn't feel that it was my responsibility to make things better." If ever there had been an understatement, that was it. "Regardless, I was terrified. I didn't know how to find the right companies to invest in or how to influence people. When a friend invited me to go skydiving in Peru, that changed my mindset. I realized that I could do anything because I had nothing to lose."
"Except maybe your life."
He laughed quietly. "Maybe. But when you give up a little control, you find out what you're made of. It's not my intention to scare you. I want to show you that you can do anything. You have no reason to be intimidated by Sylvia Hodge."
"I can think of fifty reasons, easy."
"No. Listen. You have a vibrant concept, you've demonstrated there's a demand for your product and most importantly, you have you. There's no substitute for a smart, creative mind. That brain of yours is pure gold."
Sarah's eyes swirled with wonder and emotion. Exquisite and teary, they took Aiden's breath away. He'd played a role in her reaction and that made his heart thump wildly.
"You make it sound like I can't fail."
"I don't think you can." Aiden blanketed Sarah's hand with his, unable to keep from touching her. He really did believe in her. After all, she'd found a way to reach him when he'd been determined to keep her away.
She turned her hand, allowing their palms to touch, wrapping her fingers around his. Her grip said that she didn't want to let go. Neither did he. Her skin was too soft, too warm. He'd waited too long for this. He had to know where this single touch led.
"Mr. Langford, the boat you hired is waiting. Anything I can carry out to the dock for you?"
John's voice yanked Aiden out from under the spell of Sarah. They were already in a parking lot adjacent to the beach.
"We'll be just fine, John." Even with the disruption, Sarah hadn't let go, and neither had he. Was his heart about to leave his body via his throat? Sarah was giving him a glimmer of hope he wasn't sure he should cling to. Why he was in any way unsure of himself with Sarah was a mystery. With any other woman, he knew precisely where hand-holding led...into his bed. With Sarah? They might never share more than what they just had.
They hiked across the hot white sand, sidestepping people soaking up the midday rays. The crew was waiting for them on a shiny blue speedboat, bobbing in the water. The winch, which held the line for the sail, was all set up on the back. Aiden greeted the young man standing sentry on the dock, introduced Sarah and helped her aboard.
"Here are your life jackets." An older man handed a red one to Sarah and a larger blue one to Aiden.
Sarah placed her handbag on one of the benches lining the perimeter of the hull. Aiden removed his T-shirt and put on the life vest. His eyes connected with Sarah's—she'd been watching, again filling him with ill-advised hope. She turned her attention to getting her flowing cover-up sleeves through the armholes of the life jacket.
"You should get rid of the top layer," he said.
She blew out a breath. "Yeah. Okay."
She tossed the vest aside and turned her back to him, suggesting she wanted privacy. But this gave him the perfect chance to watch. The aqua fabric skimmed the backs of her toned legs, over her pleasantly round bottom, revealing the feminine curve of her waist, and the beautiful contours of her back and shoulders. With string ties at the back and at the hips, her bikini left little to the imagination, but his mind was racing to fill in the details. Blood rushed to the lower half of his body. Heat surged. Again.
She turned and sat on the bench as the boat puttered from the dock. Aiden wasn't sure he could sit alongside her and keep his hands to himself, so he kneeled on the bench, steadying himself with his hand. Ocean air rushed as the boat picked up speed, cooling his overheated skin.
The crew prepared the harnesses and called Aiden and Sarah over. Even with the boat jostling them as it bounced over the waves, it wasn't hard to see Sarah's nervousness. Her back and shoulders were stiff as a board as she stepped into the straps and they hooked her onto the winch. Aiden took his place next to her, the two of them sitting on the platform at the back of the boat. One of the men released the chute. The wind caught it, yanking on the line.
Sarah grabbed his thigh. "Oh, my God. I'm going to die."
No. But I might. He swallowed hard and took her hand. "We're in this together."
"What a comforting thought," she yelled, as the boat gained speed.
One of the crewmen leaned in closer, grasping the top bar carrying the harnesses. "We're sending you up now. Give us a signal if you decide to come back down. Have fun."
The boat engine revved. The winch creaked. The rope began to unwind and they were lifted to standing.
Sarah yelped and squeezed his hand even harder. "Don't let go," she screamed as their feet left the deck and they were carried up into the air.
* * *
The line unrolled, steadily carrying them up into the warm, cloudless sky above the crystalline sea. She'd never before wondered what it felt like to be on the end of a kite string, but this had to be what it was like, floating free while tethered to safety. As they reached a height that she'd been sure would terrify her, elation bubbled up from the depths of her stomach, giving way to breathy giggles.
Aiden laughed. "You okay?" he yelled, still holding her hand.
"Yeah," she shouted. She did have to work at focusing on the freeness of the moment, rather than the fact that she and her feet were dangling hundreds of feet above the bay. She'd never willfully experienced a height like this, outside of visiting the top of the Sears Tower in Chicago and pressing her forehead against the window for a second before she clamped her eyes shut. Somehow, the tautness of the rope and the tug of the chute made it feel as though they couldn't fall. Or perhaps it was Aiden. He did scary things all the time and he always lived to tell the tale.
Careful not to look straight down, she took in the view—high-rise hotels lining the beach, people dotting the sand and countless shades of beautiful blue composing the vista of sea. She sucked in salt air, relished the wind against her skin, and more than anything, the comfort of Aiden's hand.
If someone had asked her a month ago if she'd ever do this, she would have said no way. Now she had to wonder why she'd never allowed herself to try. Bungee jumping was definitely not on her list, nor was BASE jumping or skydiving—basically, jumping of any kind was out of the question. But this—flying in the air with a handsome, hunky guy holding on to her? This she could do.
Aiden was such a huge part of everything she was feeling right now. He'd been so sweet in the car, giving her the pep talk that helped her step back from the proverbial ledge. Even though they were only friends, they'd grown close, and she couldn't help but compare him to other men. Aiden, even when he made her question what she was doing, did not doubt her ability to rise to the occasion. So many men had dismissed her, especially Jason. Not Aiden. He thought of her as more. That made her see those things in herself.
She smiled. Bad memories would not dog her today. Today was full of possibility. Today was about taking risks. Aiden leaned into her, sending a zap of electricity through her. It made them pitch to one side, which made her heart race. Sarah angled toward Aiden and that became their game, back and forth, laughing, smiling at each other, shoulders touching, hand in hand. Her heart swelled with the way she felt right now—unhindered. She could do anything today.
It was hard to know how long they'd been up in the air, but all too soon the rope was pulling them back to earth. Without much trouble, their feet settled on the boat's landing pad, while the crew rushed to bring in the sail and help them out of their harnesses. Minutes later they were at the dock and trekking back across the sand to the spot where John was waiting.
Taking her seat in the car with sandy feet, windswept hair and the blood brought to the surface of her skin, Sarah's nervousness had been obliterated. Feeling invincible and exhilarated, she wanted to hold on to this moment forever.
Aiden handed her a cold bottle of water, which John had brought for them. "So?"
She smiled, turning away from him for a moment and watching South Beach whiz by as they headed back to the hotel. She knew the answer Aiden was waiting for. He wanted confirmation that he'd been right and she'd been wrong. Part of her didn't want to give it to the guy who always got whatever he wanted, but he'd earned it.
She turned back. "You were right. I loved it. It was scary at first, but I loved it."
"And how are you feeling about tonight?"
She took in a deep breath through her nose, waiting for the old nervousness to creep back in. She was on such a high that she couldn't fathom that old negativity. She could do what she set out to do. She was going to dazzle Sylvia Hodge tonight, impress her with her gown, the photos she had ready on her phone, and prove to her that her company was worth investing in. "Honestly? I feel great. Which is scary in its own way since it's not normally the way I feel, but definitely better than the other kind of scared."
"A little bit of scared keeps you on your toes, but you can get rid of the rest of it. It doesn't help you accomplish what you want." He took a swig of his water and screwed the cap back on. That little bit of time in the sun had darkened his skin, giving it a golden glow, making him that much more touchable.
"You tan quickly."
He removed his sunglasses and lifted his T-shirt sleeve, revealing the rounded curves of his muscular biceps. "I guess I did get some sun." He glanced over, setting his sights on her, making her feel exposed in a wonderful way. "You did, too. Your shoulder is pink."
Sarah turned her head. Her cover-up had slipped down and she did indeed look sunburned. "Oh, man. And I used SPF 700."
"They make such a thing?"
She shook her head. "I'm exaggerating." She examined her other shoulder, the one that had been closest to Aiden and shaded by the chute. "This side is fine."
"We'll take a look at it when we get back to the room."
We will?
"I don't want you uncomfortable tonight. I want you to walk into that room in that dress and slay everyone."
The thought of slaying a room full of people was laughable. "You do realize I'm short, right? Room slaying is more for a woman who's five foot ten."
"All that matters is that you have confidence. And I have complete confidence that you will look stunning."
Sarah swallowed, hard. The man was lethal. Her resolve was doing more than melting away—she was having a hard time remembering why she'd ever needed it in the first place. Would it hurt anyone if she and Aiden gave in to a night of passion? She'd promised herself she'd stay away from single dads if she were caring for their children, but maybe this was different. She and Aiden had already shared so much more than she ever had with Jason. And she wasn't really Oliver's caretaker. Not for long, at least.
They arrived at the hotel and rode to their floor, both quiet. Considering all of the very sexual thoughts running through her head, Sarah was terrified to open her mouth.
"We have about an hour until we need to leave. I'm going to go ahead and hop in the shower." Aiden took his shirt off right there.
Sarah nearly choked. "Okay." She couldn't have moved if she wanted to. Not when the world's most gorgeous display was right before her. Considering that she'd never touched it, she had an irrational attachment to his chest, longed to spread her fingers across it, feel his skin against her palms and soak up the glory of Aiden Langford.
"Before you get in the shower, let me check your back for sunburn. I'm worried about that shoulder of yours."
"Oh, okay."
She stepped closer so carefully you'd think she was about to feed a lion from the palm of her hand.
"Turn around."
Goose bumps dotted her skin. He swept her ponytail to the side and teased her beach cover-up from her shoulder with his finger.
"Anything?" Her voice squeaked.
"I can't see very well. Take off this thing."
No no no. No taking off of things. Oh, but she wanted to. She really, really wanted to. He was so close, radiating heat right into her back. He towered behind her, making her wonderfully aware of his size. She crossed her arms, curled her fingers under the hem of the caftan and lifted it over her head.
"Much better."
She swore his voice was tailor-made for the bedroom. Why is it hotter now that I'm wearing fewer clothes?
He placed one hand on her left shoulder while brushing the right with the tips of his fingers, leaving behind a trail of white heat. His touch was heavenly and perfect. She could have stood there forever and let him ever-so-slightly caress her shoulder. "You're a little pink, but I think you'll live."
That's rich since I feel like fainting. "Okay. Thanks."
Eleven
Sarah finished her makeup—a few dabs of powder and another swipe of deep red lipstick. "I can do this," she muttered. "I was born to do this." I think.
She took a final look-see in the full-length mirror. Was this dress the right call? It was certainly stunning. And if any garment, aside from a Kama nightgown, could tell the story of her design aesthetic, this was it. Still, with a neckline aimed straight for her navel, it was a bold statement. For Aiden, it was practically a lie detector test. If he showed no interest while she was wearing it? She'd know precisely how stupid her thoughts about him had been. Like most things she designed, the dress was meant to be left in a puddle on the floor at the end of the night. It was meant to leave a man with few defenses.
Not that she had a single guard against Aiden. If he made a move, it'd be painful to say no. It didn't help that her brain wouldn't stop obsessing on the blissful moment when he'd caressed her shoulder and only their bathing suits had kept them apart. What would he have done if she'd glanced over her shoulder and uttered the words she'd been dying to say? Kiss me.
Some part of her hoped that he wanted her the way she wanted him. Another part—the grumpy, sensible part—said that it didn't matter. Crossing that line would be foolish. It would never end well. He was worldlier and infinitely more powerful than any man she'd been with. How could she possibly make an impression on a man like Aiden? How could she not disappoint him with her relatively narrow frame of reference in the bedroom? And if it happened, and proved to be a misstep, there would be many awkward conversations to endure over the coming days. All while she was trying to save her business.
It was best not to push her luck. There were only so many things she could conquer in one day—Sylvia Hodge and building a fashion empire at the top of today's list. She needed to force herself to stop barking up the handsome billionaire.
Sarah grabbed her evening bag and marched into the living room. Aiden was standing near the door, talking on his cell. Even seeing him only in profile, he was too much to take in with a single look. So her vision landed on his black dress shoes, perfectly polished. Eyes traveling north, she savored every inch—his long legs in black tux pants, his heavenly torso in a crisp white shirt, topped off with an untied bow tie. However ludicrous the thought, she considered begging him to forget the Sylvia thing. Face time with a fashion icon? Who cared? She needed face time with him. Face-to-face. Lips to lips.
He turned, his vision unsubtly washing over her. "I need to go," he said to his phone, then tucked it into his pocket.
Sarah waited for his verdict, her pulse racing and her mouth dry.
"You look absolutely gorgeous," he said. "The dress is stunning."
The dress. The dress is stunning. She couldn't ignore his choice of words.
"You clean up pretty nice yourself." She stopped short of mentioning that his suit pants might look better draped over a chair, his shirt flung over a lamp.
He cocked an eyebrow and tied his tie without so much as looking in the mirror. Surely her heart was never meant to withstand these flirtatious blows. "Good?" he asked.
"Your tie?"
"Yes."
"It's a little crooked." He'd done a spot-on job, but this was too good an excuse to touch him. She set down her bag and straightened the tie, quickly learning that with her height disadvantage, she was giving him a bird's-eye view down the front of her dress. His warm smell teased her nose, making the proximity of his mouth impossible to ignore. Why was she torturing herself? She patted his shoulder and stepped back. "You're perfect." Too perfect.
"Good. Let's get out of here."
John quickly had them to the warehouse where Forward Style was being held. The show moved from city to city each year, and to make it that much more exclusive—and elusive—the exact location was never revealed to guests until hours before it started. Judging by the jam of limousines and expensive cars in front of the venue, along with the mass of people and photographers standing behind barricades, Sarah could only imagine the mayhem if the address were publicized ahead of time.
Sitting in the car, waiting for their turn at the red carpet, Sarah's earlier calm faded. The closer they crept, it got worse. Cameras flashed. Spotlights beamed into the night sky. Sarah's pulse acted like it was auditioning to join a Miami music rhythm section. This world they were about to step into was all kinds of intimidating, but she'd wanted this since she was a teenager. You've waited long enough. This is your future.
A valet opened the door and Aiden was out first. Sarah scooted across the seat, and just when she had another pang of doubt, he took her hand and gave it a squeeze. Regardless of what it meant, she was so glad to have her fingers safely tucked inside his grasp. He didn't have to give a speech to bolster her now. His touch was all she needed.
She'd worried that the photographers' camera flashes would stop when she stepped out of the car, but they kept coming. Of course, it was part of the excitement of the evening, or quite possibly the allure of Aiden, but she soaked up every second. She stood tall and smiled, hanging on to Aiden's hand just as she had in the air above Biscayne Bay.
Across the threshold, gorgeous women in supershort dresses offered glasses of sparkling wine. Aiden and Sarah filtered into the warehouse, which had been done up with glitzy lighting and cascades of white fabric hanging from the tall ceilings. A din of conversation and thumping dance music filled the room. Models, designers, rock stars and the Hollywood elite were decked out in a dizzying array of fashion choices—everything from priceless gowns to ripped jeans. They all seemed to know each other, embracing, laughing and chattering away.
A handsome young man in a tux offered to show them to their seats. As they walked up the aisle, Sarah couldn't believe it as they got closer and closer to the front. They stopped at the first row, taking their seats between one notorious magazine editor and renowned twin sister fashion mavens. "How'd you manage these?" she whispered in Aiden's ear.
He put his arm around her and nestled his nose in her hair. "I made an additional donation. I figured it would add to your mystery. Everyone will want to know who you are."
Sarah looked up at him, their lips only inches apart. With every crazy thing going on in this room, hundreds of endlessly wealthy and fabulous people milling about, she could only think about planting her mouth on his. Earlier, she'd wanted him just for being his sexy self. Now she wanted to kiss him for countless reasons. "I don't know what to say. Thank you."
He smiled. "Of course."
The music faded and there was a rush for people to take their seats. The song changed to a delicate instrumental and a hush fell over the crowd. Sarah's heart threatened to explode. This was really happening. Out strode a grinning Sylvia Hodge, lithe and graceful in a silvery-gray gown with a slit up one leg. Her black hair was pulled back in a high ponytail, her wrist weighed down with a stunning collection of diamond bangles. She carried herself as a woman who had the world at her feet. And she did, so it worked.
She raised a microphone to her ruby-red lips. "I want to thank everyone for joining us for this year's Forward Style. It's much more than a fashion event—it's a community coming together to support a worthy cause. This year, all proceeds benefit art programs in our public schools. That makes me incredibly proud. We must nurture creativity whenever possible. Speaking of which, I know you're all ready to see what our brilliant designers have in store for us this evening. Without further ado, on with the show." Sylvia worked it on her way backstage, hips swaying, hemline flapping around her ankles.
The music changed to another driving dance beat, and before Sarah could put a single thought into Sylvia, the show started. An endless line of models filed down the runway as the secret Forward Style collections were revealed. Sarah focused on design details, following along in the program until she noticed that none of the other VIPs were doing the same. She set aside her guide. An up-and-coming designer wouldn't be obsessing over who made what, she'd be enjoying the creativity of her peers. She sat straighter, watching the show, keenly aware of two people—Aiden right next to her and Sylvia, now seated in the front row on the opposite side of the catwalk. One person to be her rock, the other her greatest challenge.
After an unbelievable display of fashion, the show drew to a close with the designers' curtain call. The crowd gave a standing ovation, furiously clapping. Sarah kept her eyes trained on Sylvia. Once the final bows had been taken, it was time to act. Goodbye comfort zone.
"I'm going in," she blurted to Aiden.
Surprise crossed his face. "Yeah. Go."
Sarah pushed her way through the throng of people, easier said than done when you're height challenged, but she was not about to be thwarted. Just fake it. When she reached Sylvia, she didn't think. She acted. She touched her arm and started talking. "Ms. Hodge. I'm Sarah Daltrey and I need to tell you about my company, Kama. I know you're shopping for new brands and you need to see my designs. My company makes women's sleepwear and lingerie. There's a gap in your company's portfolio when it comes to that category."
Sylvia looked both astonished and amused. "Not many people have the nerve to be so direct with me. Can you show me your work?"
"I'm wearing one of my own designs tonight."
Sylvia quickly eyed the dress. "Okay. Show me more."
Not a ringing endorsement, but Sarah still couldn't get out her phone fast enough. She pulled up the gallery and handed it over to Sylvia, then began explaining each image. That was the easy part. And now that she'd gotten over her initial nervousness, every word out of her mouth became more natural.
Sylvia flicked back through the pictures a second time, nodding, as Sarah tried to interpret what each facial tic might mean. She returned Sarah's phone then gave her a business card. "Call my office tomorrow morning and ask for Katie. She'll walk you through what else we'll need from you before we can talk any further. I trust you have your financials in order? A website? Designs for next season?"
Sarah's mind whirred into gear. This was happening. "Yes. Of course. Katie. I'll call her."
"She gets in very early. I'd call before her day gets too busy. You have talent, but it won't do you any good if you don't find the right partner."
Sylvia Hodge, who Sarah now regarded as a powerful and intimidating fairy godmother, disappeared into the crowd.
"Well, that was intense," Aiden quipped. "You just did it. You didn't need my help at all."
Sarah grasped his elbow with one hand and his lapel with the other. She wasn't sure she should be so forward, but taking what she wanted and faking it when necessary had actually worked. "But I did need your help. Now let's get out of here. This music is making me crazy."
They made a quick escape into the Miami night, heading back to the hotel. Sarah settled into her seat—as much as a girl can when she's floating on air for the second time in one day.
Aiden turned to her. "If things don't work out with Sylvia, I want to invest."
"Fashion's not in your wheelhouse, remember?"
"Sarah, you could sell me the Brooklyn Bridge right now. That was a really tough thing you did. And you killed it."
Pride swelled inside her, as did her yearning for Aiden. However much she'd wished earlier that she could touch him, kiss him, run her fingertips over the intricate patterns of that tattoo on his arm, the desire was tenfold now. "You're so sweet."
"That's not what you'd say if you knew the thoughts going through my head when I look at you in that dress."
"Thoughts?"
Her face was on fire. Swallowing became an impossible task. Kiss him. Just get it over with and if he turns you down, you can tell him it had been a momentary blip of insanity. The car came to a stop. Why are we not moving? We need to get to the hotel. Now.
"We're here, Mr. Langford," John said.
"Good." Aiden gazed deeply into Sarah's eyes, taking away her ability to breathe, let alone think. "Time to retire for the evening."
Halfway through the lobby, Sarah was overcome with the feeling that everything was about to change. And not just with her career.
* * *
Aiden unknotted his tie on the elevator. He couldn't take it anymore. He'd started something in the car, something that was not smart, but he'd had enough. Being with Sarah all day had been too great a test. Every inch of her was temptation...he would've berated himself for suggesting parasailing and subjecting himself to hours with her in a bikini if he hadn't so greatly enjoyed it. He'd thought at least a dozen times about untying one of those bows on her bathing suit, caressing her soft skin, leaving her bare to him.
Next to her in the elevator, the view was incredible, but looking was no longer enough. He needed to touch her, without the dress, needed her in his bed, so he could finally get lost in her sweetness and beauty. He'd be kissing her right here and now if he thought for even a minute that he'd be able to stop.
Finally, mercifully, the elevator reached their floor.
Sarah stopped halfway down the hall, planted her hand against the wall and kicked off her shoes. "These things are killing me."
Aiden laughed quietly. Her career was built on aesthetics, but she wasn't afraid to be real with him. He opened their door, ready to sweep Sarah into his arms and finally just kiss her.
She had other plans. "Sorry, but I need to do one thing."
Before he could say a single word, she darted into her room and closed the door.
Well damn.
He stood there for a moment, unsure of what to do, which made no sense. He almost always knew what to do. Come on, Sarah. Hurry up. But she didn't appear. And there was no sign that she would return. What could she possibly be doing in there? How long could it take? So that was that. He'd crossed a line with that comment in the car, and it hadn't worked. Time to admit defeat.
He stalked to his room and closed his door, disappointment threatening to consume him. This was not the way things typically went for him. Hell, he couldn't think of any time this had happened. Then again, he'd never known a woman like Sarah. He stepped out of his pants, which oddly enough made his arousal a more pressing matter. It felt good to be less constrained, but now it was impossible to ignore how badly he wanted Sarah. Should I go to her? Tell her how badly I want to make love to her? Or will I seem like a jerk? They only had a few more days together.
A quiet knock came at his door.
Aiden looked around the room, unsure if it was a construct of his imagination. A second knock came. He lunged for the door and opened it, pure evidence of how little he was thinking. In his boxers, there was no hiding how thrilled he was to see her.
"There you are," she said, slightly breathless, in the tempting black-and-silver nightgown.
"Where else would I be?"
"I thought you would wait for me. I mean...after the car. It seemed like you wanted to, um, spend some time together." Her eyes flashed with something he'd never seen, which was saying a lot since she was such an animated woman.
"I do want to spend time together." Why couldn't he find something smooth to say? Oh right, everything below his waist was doing the thinking.
"What are you up to?" She looked everywhere—at the ceiling, the floor, the windows with the view of the water. Either she was trying to ignore the erection in the room, or she was waiting for him to make his move.
"I'm up to this." He leaned down and cupped the side of her cheek, placing a soft and tentative kiss on her lips. He choked back a sigh of relief. He'd wanted this more than he'd known.
She popped up on to her tiptoes and reached for his neck. Before he could think twice about what she was doing, she had him in the throes of the most enthusiastic kiss anyone had ever laid on him. He stooped to get closer, her lips wild and untamed, as if she'd been sent to consume him. For an instant he had to wonder if he'd fallen asleep and was stuck in a dream, but then she nipped his lower lip and the heavenly sting brought him to the present.
"Do you?" He wasn't sure what he was trying to prove by inquiring about her intentions. They were pretty clear.
She drew in a deep breath and nodded, never taking her eyes off him. "I do. I want you to take off my nightgown and kiss me again. I want you to touch me. I need to touch you. And I don't want to stop until we're both exhausted."
She gazed up at him, her eyes open and honest, while he teased one of the skinny straps from her shoulder. The corners of her sweet lips turned up when he reached for the hem of her gown and tugged it over her head, dropping it to the floor.
He couldn't have stifled the groan in his throat if he'd wanted to. Having her stand before him, completely naked, was overwhelming. He didn't know where to start—he wanted every part of her at once. Her breasts were even more beautiful than he'd imagined—full and luscious. The outline of her bikini showed from their time in the sun—stretches of creamy, touchable skin paired with the golden glow of her tan.
He snaked his arms around her waist, leaning down to kiss her, taking her velvety bottom into his hands. Was he the luckiest man ever right now? He'd been fantasizing about this five minutes ago and now it was actually happening.
He wanted to kiss her in a way that projected how badly he wanted her, but their height difference made it difficult. He turned and sat on the edge of the bed, pulling her along. She stepped between his knees, her body bracketed by his legs. In his eyes, she was the embodiment of luscious femininity. His arms reined her in, holding her close, as his lips traveled from her pouty mouth to her neck, down the center of her chest, and he took her pert nipple into his mouth as he plucked at the other with his fingers. He listened to the way her breath halted, took note of the subtle moves she made to get closer to him. He had so much to learn about what pleased her, and about what would make her unravel.
She dug her fingers into his hair, massaging his scalp, soft and sexy moans coming from her gorgeous mouth. She dropped to her knees, seeking another kiss from him as she pulled at his boxer briefs and forced him to pop his hips up from the bed, so she could slide them down his legs. Kneeling before him, she took his length in hand, stroking firmly, making contact with her riveting gaze. It was everything he could do to hold his head up, let alone keep his eyes open. Her touch was bringing every nerve ending in his body alive, much as she was changing his entire life, bringing positive energy that hadn't been there before. She lowered her head and wrapped her lips around his tip, taking him into her warm and welcoming mouth. Aiden had to recline back onto his elbows—the way the blood was rushing through his body right now was enough to make him pass out.
She took her time with him, sweet and sensual with every pass, placing one hand flat on his stomach and caressing tenderly. The tension was coiling inside him like a rattlesnake about to strike. He couldn't hold on much longer.
"Sarah, come here."
She made a gentle popping sound when she released him from her mouth, but she kept her hand firmly wrapped around him. "I thought you were enjoying yourself."
The sight of her full mouth, her fingers on his body, her luscious curves, made it nearly impossible to form a coherent thought. "I was. You have no idea." He sat up and cupped the side of her face. "But I want to make love to you." His lips trailed from her mouth to her cheek and again down the graceful sweep of her neck. "If that's what you want."
"Aiden, I've wanted you from the moment I laid eyes on you. So, yes. That's what I want."
Did she really mean that? Aside from a few moments of flirtation, and the time he'd caught her staring after the bath with Oliver, he'd assumed the attraction was somewhat one-sided. Most women were very up-front with him. Sarah had hidden it well. How she somehow managed to become even sexier to him was beyond him, he only knew that this particular revelation did exactly that. Not many people surprised him. Sarah did.
"I have a condom in my suitcase," he said.
She climbed up onto the bed and rolled to her side. "Hoping to get lucky in Miami?"
He smiled to himself and shook his head. "That wasn't what I was thinking. I just happen to have them with me when I travel."
"Best to be prepared."
"Yes." He presented her with the foil packet and cocked an eyebrow. "Care to do the honors?"
"I do." The flirtation in her voice was enough to send him sailing past his peak, until she touched him. Then it was as if he were clutching the crumbling edge of a cliff with his fingertips.
She lay back on the bed and he followed, settling between her legs, sinking into her. He bestowed kisses on her forehead, her cheek and then her lips as they rocked together. For someone so small, Sarah had a lot of power resonating from her hips. She was already gathering around him, which made thinking too great a demand, so he didn't bother. He brushed her hair back from her forehead, enjoyed the feeling of her heels on the backs of his thighs and her hands roving up and down the channel of his spine.
Their gazes connected in the soft moonlight filtering through the window as she reached her peak, her body tugging on him in pulses as she gasped for breath. He loved that blissful look on her face—so incomprehensibly beautiful. He gave in to the pleasure, steeped in the knowledge that he'd made her happy. He dropped to his side and she didn't hesitate to curl right into him. They fit so wonderfully together and he couldn't wait to have her again. He was damn lucky to be with her, even if they might never have more than one perfect night.
Twelve
Sarah wasn't sure she'd ever been so tired. Nor had she been so blissfully happy in her own skin. There in Aiden's bed, half-asleep in the early morning light, she replayed her favorite moments with him. The most captivating memory came when Sarah had gotten up in the middle of the night to use the bathroom, only to find him wide-awake when she returned. He hadn't let her get more than a few steps before she was in his arms, his hands roving everywhere as his mouth explored hers and he led her back into the bathroom.
The shower was magnificent—marble and glass, with enough room to play. Hot water covered them in a deluge but there were moments when she could've sworn the heat all came from Aiden's hands. He spread soapy suds across her breasts, and gave her kisses that hardly let her come up for air, all as steam swirled and for the third time in twenty-four hours, it felt as if she were floating.
The man was a magician, every move born of some innate ability she didn't understand. He was always a step ahead of her, anticipating. She never asked for a thing, but again and again he did exactly what she would've asked for if she'd had the guts to put it into words. He'd asked her to sit on the shower bench, then dropped to his knees before her. He hitched one of her legs over his shoulder, and brought her to heights she'd never seen. She was accustomed to being the giver, not the receiver. Sitting back and letting Aiden take control had been heavenly.
But of course she'd begged to return to the role she relished, the one in which she pleased, as it was his turn to sit on the bench and his fingers combed through her hair as she took his steely length into her mouth. Judging by the extended string of dirty talk, which ended only when his body froze and he reached his apex, she'd made him happy—very happy. That turned her on more than anything.
The mattress bounced when Aiden shifted his weight and rolled away from her. She fought disappointment that it hadn't been his body's inclination to seek hers while they were in bed, but perhaps that was for the best. Now that it was morning, and the high of yesterday had faded, reality was creeping back in. She was going home on Sunday. Aiden and Oliver would be starting their new lives then, too. That was as it should be, precisely what she'd come to New York for. She'd done her job; she and Aiden had had their fun. Aiden was not the settling down type. He needed his space—he'd said as much. Becoming a dad was already an awful lot of settling down for a man who needed his freedom. More than being used to those things, he needed them the way everyone else needed air and water.
She'd promised herself she wouldn't get attached or involved. That she wouldn't cross that line with Aiden, however badly she'd wanted him. So that was her one mistake. She'd given in to the way he drew her in. Now it was time to return to their old dynamic or tempt fate. One mistake was forgivable. Two would be idiotic.
She gently peeled back the sheet and sat up, glancing at the alarm clock. It was only a few minutes after five. Best to let Aiden sleep. She could shower and get dressed, pack her things and be ready to go whenever he was. Careful not to disturb the bed, she tiptoed to the other side to get her nightgown. But the sight of that black silk against the pristine white carpet was a snapshot from her painful past—nearly an exact replica. It sent an avalanche of hurtful memories crashing through her head—the morning after what became the final time with Jason, when she'd scrambled through the pile of clothes at the foot of the bed, desperate to find her nightgown and her dignity. That was the morning he'd scoffed at the notion that "sleeping with the nanny" meant anything. It was the morning he'd laughed when she'd said I love you.
She closed her eyes, willing the tears away. You're stronger than this. It's not the same. You can stop before you get in too deep. Walk away. That chapter was gone. She'd turned the page.
She picked up her nightgown and crept out of Aiden's room. What happened in Miami stayed in Miami. That was the only way this was going to work.
* * *
Aiden woke to an empty bed. He even rolled to his side to touch the spot where he was certain Sarah had been last night. The sheets were cold, as if she'd never been there. He propped himself up on one elbow and raked his fingers through his hair, scanning the room for evidence he wasn't dreaming. But last night had happened. He and Sarah had made love. More than once. It wasn't his mind weaving a fantasy. It had been real.
He sat up to see if his clothes were where he'd left them. They weren't. They were draped over the arm of a chair. Most notably, her nightgown was missing from the floor. Huh. He'd made graceful exits from trysts. It'd never happened to him, but Sarah had a habit of keeping him on his toes. Luckily, there were only so many places she could be. He pulled on a clean pair of boxer briefs and began his search. He didn't need to go far.
"Morning." She spoke from behind the newspaper, seated at the dining table, a cup of coffee next to her. "I hope it's okay I ordered breakfast. I was starving and I figured we should get on with our day."
He rubbed his eyes and wandered over to her, struggling to make sense of this version of Sarah. This wasn't at all how he'd expected their morning would go. "You're already dressed and everything."
She didn't look up, eyes trained on the paper. "Packed and ready to go."
Sure enough, her suitcase was parked next to the front door. "You realize we can leave whenever we want, right? The jet will be waiting whenever we get to the airport. There was no need to rush." I was hoping for some morning sex to start my day.
"We have a lot to do today. I already spoke to Katie in Sylvia's office. We have a call this afternoon at four. I also called the nanny agency and told them we need more candidates to interview today. It's Thursday, Aiden. There's only so much time until I go home."
Well then. Aiden was used to being the distant one the morning after, the one who made it clear all roads ended here. He respected the tack Sarah took, even if it didn't add up. He'd been sure she was the girl who liked morning cuddles and romantic remembrances of the night before. Apparently not.
There was only one conclusion—her business was her top priority. He'd be a hypocrite to let that bother him. She'd had her big break and it was of her making. She'd be a fool to lose focus, even if they'd shared what he believed to be a special night.
It still didn't sit well with him, although he couldn't discern why. What was this uneasy feeling in his stomach? The one that made him want to take her hand and say sweet things. What was the feeling that made him hope she'd say sexy things in return, flutter her lashes and deliver a proposition he couldn't refuse with her unforgettable lips? Should we go back to bed?
He pulled out a chair and took a seat at the table, pouring himself a cup of coffee. Caffeine might help him find clarity.
Sarah finally made eye contact. "No shirt at the breakfast table?"
Was he in some alternate universe? "You were fine with me not wearing a shirt last night."
She cleared her throat and folded up the newspaper, casting it aside. "That was last night. Today is today."
"Okay, well. I don't feel like getting up and getting a shirt. So you'll have to put up with my chest. Hopefully you can control yourself."
She rolled her eyes. "I'll manage."
This was so stupid. He didn't put up with crap like this. "Did I miss a memo or something? Did we not enjoy ourselves last night? Did I do or say something wrong?"
She downed the last of her orange juice and folded her napkin. "Of course we enjoyed ourselves. It was nice."
Nice?
"But it's time for me to get back to work." With that, she got up from the table.
Aiden grasped her arm. "Okay. I get it." Touching her was a bad idea. He was overcome with that unfamiliar feeling again. It made him want to say things he'd normally never say. Can we talk? What are you thinking? What was wrong with him? Too much sun? "You're right. We need to get back to New York to take care of the nanny situation."
"The clock is ticking."
"It is." He let go of her arm, now struck with the feeling that not touching her was the bad idea. He needed to get his head on straight. He was not himself this morning. "Give me a few minutes to scarf down some breakfast and read the sports page, then I'll hop in the shower and we can be out the door."
"Do you want me to call John and let him know we'll be ready in a half hour?"
Talk about being in a rush. "Tell him forty-five minutes." He watched her walk away. She wasn't wearing her usual sundress and sandals. Today, she was all business—a straight black skirt and tailored jacket with heels. She looked every bit the role of take-charge entrepreneur. And maybe that was all she wanted to be.
His interactions with Sarah didn't improve over the next several hours. Not in the car on the way to the airport, not on the plane, not in the car back to the apartment. Aiden wasn't sure what he wanted from her—something more than a minimal acknowledgment that they'd shared a fantastic night? He hated the thought of wanting that or needing it, but he did. He had this need for her approval that he'd never experienced before. He needed her to say that she'd enjoyed it—even though he was certain she had. More than a small part of him wished that she'd say that she wanted to do it again. He'd been right to worry that sex would ruin their friendship. And for now, he had to focus on salvaging it.
Of course, his deadline with Sarah loomed. Ten days had seemed like a long time the day she walked into his office, but it had gone so fast. There was so much left to do, especially after the paternity results were back tomorrow. First to tackle was the nanny situation, which Aiden wasn't looking forward to revisiting.
They arrived back at his building midafternoon.
"I just got a text from the nanny agency," Sarah said. "They're sending one more candidate over at three."
"One? That's it? You'd think that with the money I would be paying that there would be more options than one more."
"Between your standards and mine, the pool is limited. Plus, it takes time to find a good nanny. And we don't have any." Sarah put on her sunglasses and climbed out of the car.
Aiden grumbled and followed, not wanting to chase after her, but he had to—she was walking at a clip. "Do you mind telling me what's going on? You've been weird since last night and I don't like it. If we need to talk about something, then please let's do it so that the next few days can be tolerable."
"I don't want to talk about it on the sidewalk. Can we wait until we get upstairs?"
John was blazing his own trail up the walk with their suitcases.
"Yes. Of course." Aiden turned and stopped him. "You know, John. I've got this. Why don't you knock off for the rest of the day? I'm sure you'd like to spend some time with your family." Aiden took the luggage.
"Sir?"
"Is there a problem?"
"No, sir. None at all. I just...you've never sent me home early before."
"Some things are more important than sitting around waiting on me. I realize it's your job, but I also just took you away from your family for a night. If I need to go anywhere, I'll get a cab."
John shook Aiden's hand. "Thank you so much, sir. I appreciate it. I'll be here bright and early tomorrow morning."
"Great. I might go into the office for a few hours." It'll keep my mind off the paternity test.
Aiden bid John his farewell and caught up to Sarah in the lobby. They rode in the elevator in silence. He didn't want to launch into everything right now anyway. He was looking forward to seeing Oliver too much and he didn't want to be in a bad mood when that happened.
His normally quiet apartment was noisy when he and Sarah stepped off the elevator. Music was playing and there was laughter, too—Oliver and Anna both, from the sound of it. Aiden left the suitcases in the entry, in search of the fun. He found them in the library. Jacob was lying on the floor, holding Oliver by the waist above him, letting him drop a few inches, and quickly catching him. Oliver unleashed peals of giggles, as did Anna, who was sitting next to Jacob on the floor. They were oblivious to Aiden and Sarah, too stuck in their happy world.
Aiden was overcome with a feeling impossible to label—longing, regret, sadness and joy. He loved seeing Anna and Jacob like this. He loved hearing Oliver's laugh. He loved seeing what a family looked like against the backdrop of his own home. It gave him an entirely different lens through which to see his future, a view that filled him with optimism and yet there was a nagging sense that not all was right. There were pieces missing. Aiden not only didn't know how to find the pieces, he didn't know what to look for.
Anna turned and her face lit up. "Look who's home, Oliver. It's Daddy."
Tears welled in his eyes. Daddy. That's who I am now.
Jacob got up and handed over Oliver, who snuggled right into Aiden's chest. "We had a great time, Aiden. You have an awesome little boy here." He helped Anna get up from the floor.
"You had a good time?" Aiden asked Oliver, rubbing his back and breathing in that magical baby smell.
"He had the best time," Anna said. "He's such a good baby."
Oliver tugged the sunglasses threaded on Aiden's shirt, smearing them with his tiny fingers. Aiden didn't care that they were five-hundred-dollar shades. He merely took the chance to kiss Oliver's temple and hold him close.
Sarah peeked around Aiden's shoulder. "Hey, sweet pea," she said.
Oliver's face lit up, and an elated gurgle rose out of him when she went in for a kiss. They rubbed noses, mere inches from Aiden's face. Sarah's soft, musical laugh filled his ears. Something deep inside him wanted to hold on to that moment forever. Disappointment washed over him when it ended.
Sarah was a missing piece. And that piece was leaving on Sunday.
Thirteen
As every minute passed, it became more difficult to be with Aiden. Sarah had entered territory she'd vowed to avoid, and she'd been a fool to think she could step out of it by adopting a steely demeanor. She could convince her brain of a lot of things, but it didn't mean her body was going to be on board. Just sharing the same air made everything harder—it only made her want him more.
The apartment buzzer rang. "That must be the final nanny," Sarah said to Oliver, scooping him up.
Aiden emerged from his office, where he'd been working. "What's this one's name? Lucy?"
His voice dripped with doubt, saddling Sarah with the fear that he'd turn down their final option. If he did, he could deal with the repercussions. Oliver was his responsibility, not hers, and she wasn't going to stay because Aiden refused to make a decision.
"Her name is Lily. Her credentials are exceptional. I think she could be the one."
"I read her résumé, Sarah. You don't have to keep selling me on these people."
Sarah choked back a frustrated grumble. If he were going to sabotage this, he'd better be prepared for a lecture when Lily left.
The elevator doors slid open and Lily roved into the foyer. Her wavy auburn hair was past her shoulders, barely tamed. She wore a swishy orange skirt that grazed the floor and a white tank top—not the professional interview attire Sarah expected. Maybe Aiden wouldn't have to ruin this. Maybe Lily would. "Langford residence?"
"It is. I'm Aiden." He stepped forward to shake her hand. "This is Sarah. She's been filling in as nanny until we find a permanent replacement."
His choice of words stung, especially after what had happened last night. It was confirmation of the way he saw her—as a temporary fixture. "Hi," Sarah said. "This is Oliver."
Lily's eyes grew impossibly large and she tilted her head as she went to him, taking his tiny hand. "Hello, Oliver. Aren't you the cutest thing ever?" Her voice was pure fairy-tale princess—full of magic.
Oliver was immediately taken, going to her.
"If it's okay with you," Lily said, "I'd like to play with him while we chat. I'm not big on formality."
Sarah never would've deigned to dictate the course of an interview when she was nannying, but she couldn't argue with Oliver's reaction to Lily. He was infatuated, babbling away and tangling his fingers in her hair. Aiden gestured for them to go into the library, where many of Oliver's toys were spread out on the floor.
Lily plopped down on the rug and jumped in with playtime. "I assume you've seen my résumé."
Aiden sat while Sarah stood, observing. "It's impressive," he answered. "You've worked for some very high-profile families."
"I've been lucky to have had the chance. And every child I've ever cared for has been wonderful. It worked well with those families because they appreciated my approach to nannying."
"And how would you describe that?" Aiden asked.
"Well, of course I'm firm with the children. They need some boundaries. But otherwise, I believe in letting them take the lead. If we go to a museum, we do what the child wants to do. If we do an art project, we make a mess if that suits the child's disposition. If we go to the park and he wants to dig in the dirt rather than play on the swings, we do that. We'll sing songs at the top of our lungs and splash water in the bathtub. Kids need freedom and space."
Sarah was taken aback. She'd never managed to deliver a spiel on her former vocation so eloquently. If it was rehearsed, it didn't come off that way. Then there was her choice of words—freedom and space. Aiden would have to reach to turn down Lily.
He sat forward and rested his elbows on his knees. "What else can you tell me?"
Lily launched into more of her philosophy of child-rearing, walking him through her typical weekly schedule for a toddler. She talked of long walks and afternoons in the park, of play dates and visits to the library.
Sarah had to step away as visions of Lily's plans appeared in her head. Oliver would have a wonderful, idyllic life and he'd be well cared for. It was everything she'd come looking for. If Aiden hired Lily, it would mean that Sarah had succeeded—she'd honored Gail's wishes and found Oliver his forever home. So why did she feel so empty? Why did it have to feel as if she were looking out the rear window of a car as it sped away?
"Sarah, can I speak to you for a moment?" Aiden's voice worked its way into her psyche, her weakness for him harder to ignore with his presence.
She sucked in a deep breath and shoved aside her feelings. "Yes, of course. What's up?"
"Am I crazy or did I just find a nanny? Lily is perfect."
She smiled and nodded, fighting her irrational tears. "I agree. She's fantastic. You and Oliver will be very happy with her. I think you should offer her the job." It was best to keep pushing him away, or else her heart would be a pile of rubble by the end of the weekend.
"Okay, then. It's decided." Not wasting a second, he strode into the library while Sarah followed. "Lily, I'd like to offer you the job. Can you start first thing Monday morning?"
Lily smiled awkwardly. "Oh. The agency should've told you I have another offer on the table right now. A family that's moving to France. I haven't decided yet if I'm going to take it or not. I promised them I'd decide before the end of the weekend."
"Whatever they're paying you, I'll double it," Aiden said.
Aiden was clearly committed to moving forward. Sarah reminded herself this was supposed to make her happy.
Lily got up from the floor. "I appreciate that, but it's not the money. Honestly, it's the chance to travel."
"I love to travel. I'd love to take Oliver on adventures all over the world and you can come with us."
Sarah could imagine the three of them globetrotting together. Talk about feeling left out...
Lily cast her sights down at the baby. "He's so sweet and you make a compelling case. If it's okay with you, I'd still like to have the weekend to think about it. I know you need someone on Monday, but the other family wouldn't need me to start for another month, so I could at least take over from..." Lily glanced at Sarah and pressed her hand to her chest. "I'm so sorry. I've completely forgotten your name."
Sarah blanched. I'm so out of the picture I'm a ghost. "Oh, no worries. It's Sarah."
"I could take over from Sarah until you find a permanent replacement."
"I guess I can't ask for much more than that," Aiden said. "But please, think about the things I said. I'm sure you and Oliver are a great match."
Lily bid her farewells, which included several sloppy kisses from Oliver. Even the baby was practically ready to send Sarah on her way.
Aiden let out an exaggerated exhale as the elevator doors closed. "I'm so relieved that's worked out, at least for the next month. I didn't want you to think I was trying to hold you hostage. I just needed someone I felt good about."
Sarah nodded. "I'm relieved, too." And feel so much worse.
"Now you can go to Boston on Sunday." He strolled into the kitchen and poured himself a glass of water. "Actually, you could go home earlier if you wanted to. The paternity results are in tomorrow and they're sending over the legal team to take care of the paperwork. You could go home Saturday. I mean, if you're eager to go."
Seriously? Sarah felt as though her heart should just throw up its hands and stomp out of the room. It hadn't even been twenty-four hours since they'd slept together and he was shooing her out the door? Her instincts that morning had been 100 percent correct. She needed to get out, not get wrapped up in the guy who hated the idea of being tied down. "Okay. I'll leave as soon as it makes sense. Speaking of which, I have that conference call with Katie in a few minutes. I'll take it in my room." Tears threatened again, but she had to keep it together. She turned to the stairs, but Aiden stopped her with a hand on her forearm.
"Sarah, wait."
She froze, the warmth of Aiden's fingers searing her skin. What? Did you change your mind? Do you actually want me to stay through Sunday?
"Use my office for your call."
* * *
Aiden wasn't sure what had gotten into him when he'd told Sarah she should leave before Sunday if she wanted to. It had seemed like the generous thing to do, but now he was kicking himself, even though he had to let her go sometime. Her actions suggested she hadn't wanted more than one night with him, and from the very beginning, she'd been laser-focused on the deadline. She'd delivered everything she'd promised. He'd done the same. Once the paperwork was done, their relationship could come to its logical conclusion. Except that Aiden had that uneasy feeling in the center of his chest again, the one that said something was wrong. He couldn't shake it, no matter how hard he tried.
Aiden played with Oliver in the library, unable to ignore how badly Sarah's call was going. She hadn't closed the doors to his office, so he couldn't avoid hearing her say things like, "I don't know. I'll have to get that together for you." Sylvia Hodge and her cohorts would likely only sink money into someone who was flawlessly prepared. He had to force himself to not walk in and offer his help. He'd done his part. He'd put her in the room with Sylvia Hodge. It was up to Sarah to make this happen.
Aiden's worst suspicions were confirmed when Sarah drifted out of the office. All traces of the excitement she'd had yesterday were gone.
"Well?" he asked.
"I feel like I just got hit by a train." Her voice was weary, as if she couldn't take another step.
"Those calls can be like that. It's not always a bad thing." He didn't want to give her false hope, but he couldn't stand to see her like this. Her upbeat air was gone and he missed it.
She pursed her lips and shook her head. "No. This was bad. There were so many questions I couldn't answer."
"I thought your financials were in order."
"They are, but they asked me things like what percentage of the market I can corner and how quickly I can do it. I can't answer that. That's what I need them for."
Aiden's stomach sank. Should he have prepared her more? Had he dropped the ball? "It's okay to not know the answer to everything."
"Judging by what Katie said, it's not. She said they don't work with companies that aren't as up-to-speed on the business end as they are on the design end. That's me, Aiden. I know the design end. I stumble through the rest of it."
"But you've accomplished a lot. They'll see that. And there's the value of your concept and product. I'm sure that it's Katie's job to be a bulldog, so Sylvia can step in and be your savior."
"I don't know why I tried to do any of this. Sylvia probably only agreed to have her people talk to me because I had her cornered and she didn't want to make a scene."
"Don't be so defeatist. You haven't had a definitive answer yet. And if this doesn't work out, you'll move on to the next thing. I've done it many times."
Sarah's jaw tensed in a way Aiden had never seen, not even that first day in his office when she'd been so frustrated. "There is no next thing, Aiden. I'm not you. I don't have a million amazing possibilities to juggle at one time. This is it for me. My career, my life, my paycheck. There's nothing else for me but this. This is the one thing I'm good at."
"Besides nannying."
"I don't know how many times I have to tell you that I'm done with that. Forever."
Yes, Sarah had said these things to him before, but he still didn't understand it. "Did something bad happen? Is that why you're so adamant about not going back to nannying?"
"Yes, something bad happened. Why else does a person decide they can't do something anymore?"
"Why didn't you tell me?"
"I don't talk about it with anyone, especially not someone I've known for a week."
Her dismissiveness felt like someone choking his heart. A week. In some ways it felt as if he'd known Sarah his whole life. "I'm just trying to help."
"I can't tell you. I'm too ashamed."
Now he had to find out what had happened. Sarah, ashamed? He couldn't imagine her doing a single dishonorable thing. "I'm not going to judge you. But I'd like to know what's going on. I think I've earned an explanation."
She stared at the ceiling, blinking back tears. "I got fired from my last job. I've never been let go in my entire life, and this family meant the world to me. It destroyed me. I took one more nannying job after that, but I only lasted a day. I had to do something else. I couldn't go back."
"So you were really attached to the children?"
"Child. Singular. A little girl named Chloe. She was a few months older than Oliver." She cast her sights down at the baby and pressed her lips together solidly. "I can't talk about this, Aiden. I really can't."
He pulled her into his arms, breathing in her sweet scent, overcome with the memory of how good this had felt last night. They fit together well. "It's okay to tell me. Maybe you'll feel better if you get it out. That's what you told me the night we sat up on the terrace and I was still upset about my mom."
She settled her head against his chest, trembling. "I became romantically involved with my boss. I fell in love with him."
"Go on." He choked back his discomfort at the thought of her with another man.
"I knew it was wrong, but I was so drawn to him and I adored his daughter and it just happened. His wife had passed away before I was hired and he seemed to need me and care about me, but I read the whole thing wrong."
He caressed her arm, closing his eyes and drawing in a deep breath. Her anguish poured into him. He longed to take it away. He also needed to know more. What kind of monster was this man who'd captured her heart and thrown it away? "What happened?"
Sarah looked up at him, but he didn't let go. He wanted her to know that he was there for her. "I told him the truth. I told him that I loved him. He actually laughed at me. He thought I was kidding. He thought we were having a fling. He'd assumed that I'd done it before, but I hadn't. And when I told him that it wasn't a joke, he fired me. He didn't want me around his daughter. He said he couldn't trust me anymore. Do you have any idea how awful that felt?"
"He couldn't trust you because you loved him?"
"Yes."
"That's horrible."
"It's the worst thing that's ever happened to me. Unrequited love is one thing, but it's quite another to have your bond with a child stripped away. I've always cared deeply for the children in my charge. I never knew another way. Leaving was always the hardest part, but at least it'd always been on good terms. This was just solid rejection. I was hollowed out. That's why I don't nanny anymore."
Oliver crawled over to them and pulled himself to standing with the help of Aiden's pant leg.
Sarah picked him up, tears streaming down her face. She smoothed his hair back and kissed his cheek. "I'm sorry. I know this is way more than you ever wanted to know about me. But now at least you know why Kama means so much. I can't go back to my old life."
Aiden had started over many times. He knew the appeal of a new beginning. "I understand. Completely."
She sighed and managed a smile before she handed over Oliver. "Thank you. I appreciate that. Now I need to go upstairs and regroup and try to figure out how I salvage this Sylvia Hodge thing. Are you okay to do bedtime on your own?"
"Yes, of course. I'm sure you want some alone time anyway." He really hoped the answer was that she didn't want to be by herself, that she wanted to stay up and talk after Oliver went to bed.
"I do. I need some time to think. Plus, a few more nights and you won't have me around to help. You might as well start acting like I'm not even here."
Fourteen
Sarah was working feverishly on an email to Katie and Sylvia Hodge Friday morning, when Aiden strolled into his home office, phone in hand. He grinned like a man without a single worry.
"Probability of paternity is 99.9 percent. Oliver is mine."
Sarah jumped out of the chair and raced from behind the desk, throwing her arms around him. "I knew it. I just knew it."
Unfortunately, the instant she was pressed against him, her body wanted to stay, especially when he returned the embrace with a firm squeeze, rocking her back and forth. With the clock ticking, should she take these happy moments? Even when they'd haunt her later?
"I knew in my heart that he was my son, but I don't think I realized how much it would mean to have the confirmation. Considering my own history, this gives me peace. Oliver and I are a family. No one can take that away from us." Aiden released her from their hug. It was impossible to ignore how enticing he was when he was so relaxed. Good news suited him well. "The lawyers will be by in an hour to do the final paperwork. They'll have it before the judge this afternoon. Then we'll be done."
Done. She was so close to being done, it wasn't even funny. She'd worried about awkward conversations after sleeping together, but Aiden hadn't said a thing. She respected a man who followed her lead, but part of her really wished he'd fought her on it beyond his minor protestation in Miami. His ready acceptance was another reminder that in the end, she was just another woman. Nothing more. "All sewn up. No more loose ends." It's for the best. And you know it.
He cleared his throat and walked over to the bookcase, straightening a book. "Have you thought at all about when you'll want to go?"
If only he knew how much that question hurt. "I'd like a little more time with Oliver." And you, if I'm being honest.
"I'm asking because I was thinking about having my family over tonight. For a celebration. Officially welcome Oliver into the Langford family. I definitely want you here for that."
Sarah could breathe a little easier. Maybe she was at least a notch above the other women he'd been with. "I'd love to be there. I think it's great you're involving your family. It's important to mend fences with your mom."
"This gives me another perfect opportunity to push her on it, but tonight's probably not the night, huh? Not when everyone is here."
"Agreed. Tonight should be happy. Leave the tough conversation for another day."
He turned to her while a soft smile crossed his face. "It means a lot to have someone to talk to about this."
Sarah grinned, even when she was dying a little more on the inside. Their ability to discuss painful things was one of the best parts of their friendship. He'd been so sweet with her when she'd opened up about Jason. He hadn't judged her. Not at all. "Good. I'm glad."
He knocked the bookcase with his knuckle. "I'll leave you to whatever you were working on. I'm going to get a workout in before the lawyers come by."
The afternoon was a flurry of activity. Documents were signed. Calls were made to the market to have food and drinks delivered. Oliver got an early bath before his first Langford family gathering. Sarah relished the hustle and bustle. It kept her mind off the clock, a constant reminder that it would soon be time to not only say goodbye to two people she cared about deeply, but after that came do-or-die time with Kama. What if everything with Sylvia Hodge blew up? Because that's where it seemed to be headed. The email she'd sent that afternoon had been answered with yet more questions. More doubts. More reasons they might say no. Then where would she be? Back in Boston, alone, her future a big fat question mark.
Aiden's brother Adam and his wife, Melanie, were the first to arrive. Adam was incredibly charming—just as magnetic as Aiden, with a smile that was nearly identical. That gave Sarah pause. Maybe Evelyn Langford wasn't keeping a secret. Sarah didn't have much time to think about it though, quickly hitting it off with Melanie, who was both down-to-earth and talkative.
"I have to say, Aiden. Fatherhood really agrees with you," Adam said as Oliver sat happily in Aiden's arms.
"Thank you. I really appreciate that." Aiden's response was more than polite conversation. His brother's kind words had resonated.
"I'll have to get some pointers from you when the time comes. Mel and I are trying to get pregnant," Adam said.
Melanie's eyes flashed. She swatted Adam on the arm. "I thought we weren't telling anyone."
Adam put his arm around her and pulled her closer, kissing her on the cheek. "We're with family. There are no secrets."
If only that were true with the Langfords.
Jacob and Anna arrived, both ecstatic to see Oliver. The baby had apparently fallen in love with his uncle Jacob during the Miami trip, since he readily went to him. Everyone chatted in the kitchen, wine and cocktails flowing, music in the background. No problems. No controversy. But despite Aiden's pledge to keep things light, Sarah had a sinking feeling that might change with the arrival of the final guest.
Aiden grabbed a carrot stick from a platter of veggies and dip. "Leave it to my mother to be late for her grandson's first party."
"Maybe she's stuck in traffic." Sarah arranged crackers on a plate.
"You know she likes making an entrance," Adam said. "It's annoying, but true."
The apartment buzzer rang. Aiden took in a deep breath and adopted the most forced smile Sarah had ever seen. "Mom's here." He soon returned with Evelyn.
She greeted everyone sweetly, saving Oliver for last. "There's my handsome grandson."
Oliver was content to stay with Jacob, but he humored his grandmother, laughing when she made a silly face and holding her finger with his pudgy hand.
Aiden smiled, but Sarah could see once again that he was having to try. What it must be like to live with something so big hanging over your head—Sarah could only imagine. It burdened him, greatly, and how could it not? It had made him the person he was today. Nothing was safe from the influence of the secret he was convinced his mother was keeping.
For nearly two hours, Sarah dedicated herself to being a comfort to Aiden, bringing him a fresh drink when needed, offering a reassuring smile or moment of eye contact, especially when he sat on the living room sofa with his mom. Every time he acknowledged Sarah with a smile or a nod, it shored up their solidarity. The friendship they'd forged would be one of her greatest takeaways from their ten days together. It could comfort her when she found herself wondering what would have happened if she hadn't put up a stone wall after Miami.
"How are you holding up?" She crouched next to him at the end of the couch when his mother had gone to use the bathroom. Things were winding down, which was good since Aiden seemed to have reached his limit. His eyes were tired, his jaw tense, brows drawn tightly together.
"She's making me crazy. She's spent the whole night planning things for Oliver's birthday next month and talking about how she wants to spend Christmas morning with him. It rubs me the wrong way. I can't help it."
Probably because she never did those things for you. Sarah bit down on her lip to keep from saying what Aiden already knew. "Maybe she's trying to make up for the past."
A slight smile crossed his face and he clasped his hand over hers. "You're so sweet. I love your optimism. But I'm pretty sure this is just her way of sweeping the past under the rug."
He was probably right. She didn't know why she had the need to put a positive spin on his mother's insensitivity, she only knew that she did. "So let's just get everyone to clear out."
"Yes. Oliver needs to get to bed anyway."
* * *
By the time Aiden's mom returned from the bathroom, he'd had enough for one day. Sarah was right. Everyone needed to go home. He got up from the couch. "I don't want to spoil the party, but I need to get Oliver to bed."
His mother smiled and nodded. "Such a good dad." She popped up onto her tiptoes and kissed Aiden on the cheek. "It's been a wonderful night. I only wish your father could've lived to meet his grandson."
That image left Aiden frozen with the words he wanted to say. It would be easier on everyone if he let it go, but after years on the periphery of his family, doubt festering in his head and heart, he not only wanted the truth, it was the only thing he could speak. "I'm not sure he would've accepted Oliver." He sure as hell didn't accept me.
"Of course he would have."
Again, the need—the thirst—for the truth was desperate. The fire inside him, the pain he lived with every day, blazed. "But he didn't accept me."
His mother's eyes were horror stricken. "Your father loved you."
Aiden calmly confronted her by looking her square in the eye. "Just tell me. I'm tired of wondering. I don't want to have to think about it anymore."
"But..."
He clasped his hand firmly over hers. "Mom. I love you, but there is no but. If you want to be a part of Oliver's life, you'll tell me the truth about who my father is."
Adam approached. "Everything okay?"
Aiden refused to let his mother off the hook. "Mom was going to finally tell me the truth about who my dad is. Weren't you?"
"You'd keep my grandson from me?"
Aiden nodded. "If you love me, you'll tell me."
His mother's eyes misted. Her lower lip trembled. "I don't want to hurt you. I never did."
"Mom, it's too late. This is your chance to start making it better." Aiden braced for what was to come.
His mother perched on the edge of a chair. "We tried to make it work, but your father..." She cast her eyes up at Adam, then Aiden. Anna joined them and took Aiden's hand, squeezing it tightly. "Your father couldn't deal with it. He looked at Aiden and all he saw was what he perceived as betrayal. That's why you were sent off to school. And I agreed, because I couldn't watch him be cruel to you and loving with Adam and Anna. Your father had such a temper. I was worried about what might happen if he got truly angry at you. That's why you were sent away."
Aiden swallowed, dogged by lonely memories from his childhood—birthdays in boarding school, phone calls from his mother where she acted as if this was all normal and summers at home as the odd man out. He was an expert at filing those things away, but he had to face them now. This is it. The moment I've waited for. "Then who's my father?"
Jacob, holding Oliver, had joined Anna. Melanie had crept closer, too, standing with Adam. They all had each other. Who did Aiden have? He turned and his vision landed on her—Sarah, standing there with concern painted on her face. She was his one true ally in this room.
"Your uncle Charlie is your biological father. I dated him before Roger. It was short, but that's when you were conceived. I lied to Roger when we first started going out. I didn't want him to know that I'd been with his brother. Roger and I fell in love and when we learned I was pregnant, we got married." She peered up at Aiden. The corners of her mouth were drawn down, deep creases between her eyes. She might be hurt, but she'd left him with the same scars. "It didn't take long after you were born for Roger to put two and two together. You have the same birthmark on your leg that Charlie had."
"The same one Oliver has." Aiden was amazed he'd said anything calmly considering the speed at which his mind and heart were racing.
"Yes." She gathered her composure. "Things were okay for a while, but everything changed when Adam was born. He always compared you two. He and Charlie had such a contentious relationship, it was no big surprise. And then Charlie died in the motorcycle accident and Roger couldn't handle it. It was such a tangle of emotion and you were the one it all got directed at, Aiden. I had to get you somewhere safe, where I knew you'd be okay. That's why I agreed to send you off to school."
Aiden stood there, thinking. With everything that had been launched at him, his mind was remarkably clear. The truth had washed away the dirt on the windows. He could see. The anger hadn't left, but it made sense now. He turned again to Sarah, who was standing back from the group. Of all the people in that room, she was the one he wanted to talk to. She was the one he wanted to confide in. He wanted to be alone with her. He wanted to feel good again. As to whether she wanted the same from him, he had no idea. Her face showed only sweet empathy as he walked Oliver over to her.
"I'm going to send everybody packing," he said softly.
"Good," Sarah said.
Aiden turned to his family. "Thanks everybody for coming today. It's been good for Oliver. And for me."
"Say something," his mother pleaded. "Please tell me you forgive me."
He could've been so cruel, but it wasn't in his heart. However misguided she might've been, she'd thought she was doing the right thing. "I forgive you, Mom. That doesn't mean I'm over it. We're talking about a lifetime of lies. It will take time. Someday I want you to tell me more about my real dad. For tonight, I think it's best if everyone goes."
Adam reached out to shake Aiden's hand. It might've been the first time that Aiden felt zero ill will toward his brother. They'd both been caught in the same dysfunctional dynamic. Adam might've reaped some of the good things, but he'd had his own burdens.
"You're my brother, Aiden," Adam said. "And I love you. I'm around if you need to talk about this."
"I'd like that," Aiden replied.
"I'm not sure what to say," his mother said. "Other than goodbye."
"Why don't we plan on you coming over next week for some time with your grandson?" Aiden answered. It was time for the healing to begin.
"I would love it." With that, Adam walked his mom and Melanie to the door.
Anna grasped Aiden's arm, tears in her eyes. "You were right," she muttered. "I didn't want to believe it. I'm so sorry."
He hugged his sister. "Don't apologize. You've never been anything but loving and supportive. You know, you're going to make such an amazing mom. I can't wait for Oliver to have a cousin."
Anna smiled through her tears and pecked him on the cheek. "I love you."
"I love you, too."
He walked Anna and Jacob to the entryway and watched as they stepped onto the elevator with Adam, Melanie and Evelyn, the door sliding closed. There were footsteps behind him. He knew it was Sarah, and not because she was the only other adult in the house. In only a week, he'd learned the tempo of her gait.
"If you want, I can do bedtime tonight," she said. "I'm sure you're exhausted."
Her voice was salve to his soul. He turned and it felt as if the universe was presenting him with the cure to all that ailed him. Whatever the problem, she made things better.
"Let's both put him to bed. Together."
"Oh. Sure. I'd like that."
The party had taken it out of Oliver. Aiden put him in his pajamas and Sarah read him a story, sitting with him in the rocking chair. Aiden leaned against the door frame, studying them together. If only he could capture that moment in a bottle, save it for later, after she was gone. Sarah's absence would leave a void in his and Oliver's life that would be impossible to fill. But it was what she wanted. She'd made that clear.
After only a few minutes, Oliver sacked out in Sarah's arms. She gently set him in the crib, and she and Aiden tiptoed into the hall.
"I'm sorry about tonight." Aiden reached for her arm. "I know we said that I should wait for another time, but I had to say something. It was killing me."
"I'm proud of you for doing it. Even though it hurts right now, be patient with yourself. Give yourself some time to process it. And in the end, Oliver's love will heal you. I truly believe that."
He would have smiled if it weren't so hard to breathe. She was so determined to make everything better, and that made her even more beautiful, that much more impossible to resist.
"I think I'll head to bed," she said. "I'm sure you want time to think about everything."
Could he risk his pride for a second time tonight? He had to. Even if she might say she didn't want him the way he wanted her. He might not get another chance. "Don't go, Sarah. Stay with me."
Fifteen
Stay with me. Sarah wasn't sure she'd heard his words correctly. They were surprising. They were scary—driving her to a place where she surrendered to her deep longing for him. Did he want her? She wasn't about to make presumptions about what Aiden had said. Not now. Not when she'd be opening herself up to more hurt. "Did you want to talk?"
He granted the smallest fragment of a smile, looking at her with his heartbreaking blue eyes, his gaze saying he didn't need to talk. He tenderly tucked her hair behind her ear, drawing his finger along her jaw to her chin. "I don't know what force in the universe brought you to me, Sarah. I only know that right now, I need you. I want you. And I'd like to think that you want me, too."
The air stood still, but Sarah swayed, lightheaded from Aiden's words. Their one night together had been electric, filling her head with memories she'd never surrender, but judging by the deep timbre of Aiden's voice, they might shatter what happened in Miami. "I don't want to ruin our friendship." And no-strings-attached only breaks my heart.
"Is that why you shut things down after Miami?"
"Yes." It wasn't the whole truth, but it was enough. As much as sleeping with Aiden might be a mistake, she didn't want to deprive herself of him. Would one more time really hurt? "And I've spent every minute since then regretting it."
"Then I say we have no more regrets."
Before she knew what was happening, he scooped her up into his arms. She'd never had a man carry her anywhere. Her small stature had always made her wonder why—apparently she'd had to wait for Aiden. She wrapped her hands around his neck and leaned into him. He took the few steps necessary to cross the threshold into his room. He set her gently on the bed and stretched out next to her.
She cupped the side of his face, the stubble of his beard scratching her palm. Her heart beat a frantic rhythm as she waited. Then his lips were on hers, soft and sensuous, and that made her pulse race faster. She closed her eyes to immerse herself in the world of Aiden—the silky soft sheets and his masculine smell, his solid, muscular body. He palmed her thigh, his hand inching along under her skirt, sending ribbons of electricity through her.
He rolled to his back, taking her with him until she was straddling his hips. Her dress was now hitched up around her and Aiden explored beneath her skirt again, slipping his fingers into the back of her lace panties and cupping her bottom. She rested her arms on the bed above his shoulders and dug her fingers into his thick hair, rocking her pelvis against his, his rock-hard erection rubbing against her apex. The need for Aiden had been building for two days, and everything he did made it more pronounced—his tongue exploring her mouth, his white-hot touch.
She sat back and scrambled her way through the buttons of his shirt. He untucked it, then shifted and rolled his magnificent shoulders out of his sleeves. She sat there in awe, reaching out and skimming her fingers along the contours of his shoulders. He was so incredible, inside and out. He clutched her neck, and brought her mouth back down onto his as she again rolled her hips, grinding against his crotch, making everything between her legs blaze with licks of fire—each pass was a tiny measure of gratification, and a bigger dose of torment. She needed him now.
"Touch me, Aiden."
She reached behind her and unzipped her dress, then Aiden threaded his hands beneath the skirt again and pushed the garment up over her head. She planted her hands on the bed next to his shoulders, and he traced the edge of her bra cups with his finger, dipping below the fabric edge and rubbing her nipple. The skin contracted hard beneath his touch, and goose bumps followed. A deep moan left her lips, just to let off a bit of the pressure. He was torturing her, his gaze never leaving her, the need in his eyes fierce and undeniable. He slid his finger under the strap and nudged it off her shoulder, then did the same on the other side. She was about to beg him to take off her bra when he snapped the clasp and the garment fell away. He cupped her breasts in his strong hands—such blissful relief that you'd think she'd waited a lifetime for his touch. She arched her back and her eyes drifted shut as he raised his head and flicked his tongue against one nipple, need shuddering through her. His lips closed on the tight bud, while his hand trailed down her stomach and slipped down the front of her panties, finding her apex.
His fingers teased, touching lightly, drawing gasps from her lungs as he took full control. She settled her forehead on his shoulder as he masterfully brought her closer to climax with firm circles and a steady pace. The pleasure rose inside her, cresting. The tension would build, then ebb, then surge back until she was again at the very edge. When it finally became too much and she gave way, she smashed her mouth into his shoulder to quiet her cries of ecstasy.
But now she only needed him more. She climbed off him and watched as a sexy grin crossed his face when she unbuckled his belt and unzipped the zipper. Not having the fortitude to tease him, she slipped his pants and boxers down at the same time. He was so primed it nearly stopped her dead in her tracks. She appreciated the dark, lusty expression on his face as she wrapped her fingers around him and stroked firmly. He watched for only a moment before his eyes drifted closed and his shoulders let go of all tension. She leaned in and pressed her lips against his, loving the way the depth of his kiss told her how much he appreciated each pass of her hands. His skin was so warm and smooth, but the pressure beneath the surface was intense. Pleasing him like this was so gratifying, but she needed him fully. She needed him to make love to her. She wanted them joined that way again.
"Aiden, make love to me."
He rolled to his side and ran his fingers through her hair, covering her face with kisses. "I want to. Now." He sat up and opened a drawer in the bedside table, handing her the condom packet. She tore it open as he stretched out again. He drew in a sharp breath when she rolled it onto his waiting erection. Then he watched as she slipped off her panties and kicked them to the floor when they were to her ankles.
He positioned himself between her legs when she lay back on the bed. "This is virtually the only thing I've thought about since we got back from Miami."
"Really?"
"Really."
He guided himself inside as she pulled his hips down and he sank into her, her body molding perfectly around him. He lowered his head and they settled into a long and tangled kiss. They moved together in their perfect rhythm, rocking back and forth as the kisses became more frantic, less refined. His breaths were ragged and shallow and it was clear he was close to climax. He reached down between them and placed his thumb against her center until she came and he quickly followed.
He collapsed next to her, catching his breath.
She was floating back down to earth, her mind a whirl of wonderful things. "That was incredible." But I want more.
He smiled, his eyes half-open. "I hope you aren't too tired. I want to make the most of our time together."
Make the most of it. Her thoughts, exactly.
* * *
In the new light of morning, Sarah again watched Aiden sleep. She lay on her side, one arm tucked under her pillow, studying his face, notably calm after last night. Aiden had let his guard down. He'd let her in and it had all been his idea. She hadn't had to push for a thing. She felt like a new person, emerging from her dark cocoon in the nick of time. One more day and she'd have been gone. Now, leaving was unimaginable. She'd be crumpling her own heart into a tiny ball. Certainly Aiden wouldn't let her. Their connection was too strong.
The events of the last week had turned everything upside down, but that meant she was out from under the menacing cloud, the one that had followed her for more than a year. Even more remarkable, the saddest thing she could imagine, Oliver losing his mother and Sarah losing her friend, had brought happiness. She could see a life with Aiden. She could imagine becoming Oliver's mom if that was where she and Aiden chose to take things. They could be a family.
There were still obstacles to overcome and the most pressing was no small thing—saying those three little words. But after last night with Aiden, riding out the aftereffects of a secret he'd feared his entire life, they'd cemented their bond. So as frightening as it was, she would take the leap of telling him her true feelings. When she'd sworn to never say it first again, she'd had no way of knowing that a man as extraordinary as Aiden would come into her life. He was different. They had a foundation. Synergy. There would be no sad ending after I love you.
Aiden shifted in the bed, scrunching up his face and groaning quietly. He snaked his arm around Sarah's waist and pulled her against him. "You're so far away."
She smiled as her eyes drifted shut and she inhaled his heady smell. "I'm right here."
He smoothed his hand over her bare bottom, gently squeezing. "So you are. My mistake." He nuzzled his way into her neck and she granted him access, even though it usually brought a fit of squealing. He peppered her skin with kisses that started soft and tentative but were now deeper and longer as their bodies pressed together.
"I usually don't like it if someone kisses my neck. After last night, you can kiss me wherever you want."
"It wasn't just last night. Earlier this morning was noteworthy, too."
She laughed quietly, but arched her back and hitched her leg up over his hip. Just thinking about it made her want him again. "It was wonderful."
He clasped her face and planted a kiss on her lips. "Thank you for everything last night."
"It's a little weird to say thank you for sex."
He shook his head and nudged at her nose with his own—such a sweet and tender gesture, it left her breathless. "No. I mean everything before we ended up in bed. You're just..."
She didn't want to be holding her breath, but she couldn't help it. Was he about to confess his feelings? Would he take her worry away and impart those three little words first?
He scanned her face, his eyes searching for something. "You're a miracle. I don't know how I got so lucky to have you and Oliver walk into my life, but I'm thankful. You've been there for me and I'm so appreciative."
She smiled wide, even though he hadn't relieved her of her greatest fear. "I like being there for you."
"I mean it, Sarah. I don't even want to think about the dark places my mind could have gone last night after everything with my mom. I've wasted so much time dreading that moment, worrying about what the truth would mean, but your presence made it all okay. You're like a magician."
A magician. A miracle. Both wonderful things to be called, but not quite what she was hoping for. It was hard to blame him. He'd been through so much with his family. It was no wonder that he was closed off, that he'd shored up his defenses so solidly that no woman had managed to make her way inside. She had to appreciate that he'd come so far since she'd met him. Maybe he needed a nudge. Maybe he needed to know that she wouldn't hurt him, that she would give her heart to him just as freely as she'd given her body.
Just do it. Just say it and let it come out. Open your heart. "I love you, Aiden." A warm wave hit her—contentment, satisfaction, accomplishment all rolled into one. This time she'd finally gotten it. She smiled and gazed into his eyes, but it became clear—within a few heartbeats—that something was wrong. His eyes weren't indifferent or angry...they were hurt. It wasn't at all what she'd expected. Of the many things she could've seen, that was not on the list of possible reactions to I love you.
Sixteen
I love you? No. This isn't happening.
Aiden had never before wanted so badly to be able to rattle off a string of words, but he couldn't. I love you was forever, and he wasn't ready for that. He was ready to ask if he could see her after their ten days were up, but the words she'd just said had ruined that possibility. There was only one good response, and he couldn't go there. "I don't take love lightly." In truth, he didn't take—or give—romantic love at all. He'd never told a woman he loved her. He'd never felt it. His relationship with Sarah was different, but they'd been caught in extraordinary circumstances and his feelings for Oliver were intertwined with his perception of her. Could it be love? His gut wasn't answering.
"I don't take it lightly either," Sarah pleaded. "But I love you. I know we haven't known each other for long, but this is what's in my heart. I had to say it."
Frustration nipped at him like an angry dog. Why was she pushing this? Why did she have to take such a huge leap? He was racing to keep up, out of control, with no idea where or how this would end. "I have feelings for you, Sarah. And they're good feelings. I'm just not ready to go there yet. It's too soon." Did people fall in love in ten days? If they did, what happened to those people? Were they still in love a year later? What if everything between them faded and fizzled?
"It's not too soon for me. Some people fall in love in a minute. There is no timetable."
"But there is for us. You just spent the last ten days reminding me of a deadline. I don't like the idea of being forced into something." He hated his biting tone, but he saw her as his safe place, and she'd turned that inside out. She was sabotaging what was between them, just as she had in Miami. This time, she wasn't making a unilateral decision. She'd pulled him into this one and forced him to participate. Did she not see that he'd already taken big steps with her? He'd never spent more than three days with a woman.
"Forced? You made the first move last night." She sat up in bed and yanked the covers over her. "And you knew I was leaving tomorrow, but you took me to bed anyway, knowing that you didn't have an inkling of serious feelings for me?"
"Of course I knew you were leaving. You've spent every waking minute of our time together reminding me of it."
"And that made it easy to sleep with me. No pesky Sarah to worry about after tomorrow."
"That's not fair. I wanted you. I still want you." At least he could say that much without reservation.
Oliver yelped over the baby monitor. Aiden tossed back the comforter and pulled on his boxer shorts. "I'll get him. We'll have to finish talking about this later. I don't want to argue in front of the baby."
Sarah rolled away from him. "Honestly? I don't want to talk about it at all."
"Why not?"
"Because there's no coming back from what I just said to you."
She wasn't wrong about that.
Aiden stalked down the hall, his mind reeling. He'd been thinking he might invite Sarah to spend next weekend with them, and see how that went. He certainly hadn't been thinking about labels. Love hadn't crossed his mind. It wasn't even on the map.
He opened the door to Oliver's room. The little guy was standing, holding on to the top rail of his crib, unsteady on the mattress. He bounced up and down when he saw Aiden, squealing and grinning. He picked up Oliver and kissed him on the forehead, holding him close. Two labels he didn't have to question were that of father and son. What they shared was love. But he wasn't able to put a label on what he felt for Sarah. And if he told her what she wanted to hear, just to make her happy for now, and it later ended up hurting her, he'd never forgive himself. He might not be able to say I love you, too, but that was better than taking it back later.
He changed Oliver and brought him down to the kitchen, warming up a bottle and sitting with him on the sofa in the living room. He tried to read the rhythm of Sarah's footsteps upstairs—there was no telling what she was doing, but she was busy. Was she pacing the floor, angry with him because he'd let her down? Was she rethinking what she'd said? Was she doing the inevitable—packing up to leave? He wouldn't blame her if she were, no matter how much it might hurt. She was a vibrant and beautiful young woman. Any man in the world would be a fool to say no to her, making Aiden a class A idiot. Still, he couldn't lie to her. He couldn't say he loved her when he wasn't sure what it meant.
His loose plan of asking if she wanted to date, although tantamount to picking out china for him, would clearly not be enough for her. Not now.
I love you.
Yeah, I'm not sure. Can we just go out to dinner?
Starting on dramatically different pages wasn't a recipe for romantic success. It was a setup for disaster. She'd already been hurt by the guy she worked for. He wouldn't hurt her like that—he was different. So maybe he was back to where he'd thought he'd needed to be a few days ago—preserve the friendship and set aside romance.
* * *
Sarah was about to wear a rut in the hardwood floor of Aiden's guest room. How could I have been so stupid?
When it came right down to it, Aiden was a case of unrequited love. And although it stung like crazy, at least Sarah knew what it was. The heartache ahead had a name. A label. She could say with confidence, I left because it was unrequited love. He wouldn't say it back to me and I'd already said it to him, so I had to leave. How does a girl come back from that? Her friends would answer, You don't come back from that. You leave. With your head held high and your dignity in place. And Sarah could smile and nod, knowing she'd done the right thing. Even when the moments came when she was crumbling to dust on the inside, she would know she'd had no choice.
Oliver was another matter. She'd already been destroyed by the notion of leaving him, precisely her fear. His place in her heart would always be there. Their relationship was quite the opposite of unrequited. It was the purest love she'd ever known. Aside from her family's, Oliver's love was the only love she'd never doubted. She saw it on Oliver's face when she walked into his room in the morning or when he'd woken up from a nap. She felt it when he was upset and she held him close, the two of them clinging to each other. She lived and breathed their love when he laughed. Oliver's love had filled her heart for a month and its absence would leave an unimaginable void, and there wasn't anything to do about it. Oliver belonged with his father, and his father didn't love her.
She slumped down on the bench at the foot of the bed. "Now what?" she asked aloud. She couldn't go downstairs and talk about this more. It would only hurt. And she wasn't going to try to convince Aiden that he loved her. She wanted him to just love her. She didn't much like the idea of hiding out in her room until tomorrow. That left only one option, the one Aiden had so generously provided her with yesterday after deciding on the nanny—leave today.
She wasn't ready to say goodbye to Oliver, but the truth was that she'd never be ready. She could spend a lifetime preparing and it would never make it any easier.
Her phone beeped with a notification. She walked over to the bedside table and looked at the screen—it was an email, from Katie.
Sarah,
Despite the gaps in your financial forecast, Sylvia would still like to continue talks about acquiring Kama. Sylvia and I would like to come to Boston first thing Monday morning to tour your facility, look over designs for next year and discuss our options. Does 9 a.m. suit you? I know today's Saturday, but I need to know ASAP.
Best,
Katie
How many signs could Sarah get from the universe before she stopped fighting? Aiden hadn't returned I love you. Sylvia Hodge wanted her back in Boston, ready to talk business. And she'd set Oliver up for the life she wanted him to have. That meant Sarah needed to say goodbye, get on the next train and not look back.
She typed her reply.
Katie,
Thanks so much. Tell Sylvia I will see you both Monday morning. Looking forward to it.
Sarah
With no more time wasted on overthinking, she got out her suitcase and started packing. The sooner she got out, the better. Luckily, she didn't have much, so it only took a few minutes. She then hopped in the shower, cleaned up and dressed in the same old sundress she'd worn the day she met Aiden. That seemed like a lifetime ago.
As she took each step down the stairs, the tears threatened to take over. She imagined it was like trying to get out of the ocean when a storm has come up out of nowhere. The waves roll you back as you swim, the tide pulling just as hard, ocean spray in your face, but you keep going because you have to get to shore. You have to save yourself. For what, you aren't sure. You only know that it's your instinct to survive. You'll do anything to make it.
She and her suitcase reached the landing. She raised the handle, and rolled it toward the foyer.
Aiden's voice from the kitchen stopped her dead in her tracks. "You're leaving?"
Oliver was playing on the floor with some plastic bowls and wooden spoons.
She bit into her lower lip. You can do this. "Yes. I have to. I got an email from Sylvia Hodge's office. They need me in Boston ready to talk Monday morning. I need time to prepare. And you don't need me anymore, so I might as well get out of your hair and let you and Oliver enjoy your weekend."
"Sarah. We didn't even finish talking about everything." He came out from behind the kitchen island, but thankfully didn't touch her. He instead crossed his arms. "We're just going to leave it all unsaid?"
She forced a smile and an enthusiastic nod. She'd never felt less happy or eager to do anything. "I don't think we need to talk about it anymore. I get it, Aiden. I do. I'm not going to try to get you to say things that aren't in your heart. You didn't do anything wrong."
"I just wish you'd give me some time to wrap my head around it."
The thing was—she didn't need more time. She knew exactly how much she loved him. She felt it in the depths of her belly right now, a terrible burning. She knew exactly how bad it was going to hurt to step onto that elevator. She couldn't wait. She couldn't give him another chance. Aiden might never get to the place she needed him to get to. It wasn't his fault. He'd been deeply hurt by his past. And he'd always been very up-front—he needed space.
"It's okay. I shouldn't have said anything this morning. Just forget it." She rushed over to Oliver and crouched down, raising his face with the tip of her finger. "Goodbye, sweet..." her voice cracked into a million pieces. Her lip shook. Her chest convulsed. She couldn't say it. Her heart wouldn't let her. She leaned down and placed a single kiss on the top of his head, committing to memory his smell and the feel of his soft curls against her lips. She would miss that so much. Forever.
She straightened and turned away from Aiden. The tears were streaming down her face in a deluge and she couldn't let him know that he'd gotten to her like this. "I have to go. I'll miss my train."
"Are you sure about this?" he asked, doing the thing she'd dreaded—grasping her arm.
She didn't look back. She hid. "I'm sure."
"At least let me call down to John and have him take you to the station. Let me do that much. Just to say thank you for everything."
Don't fight him. Just go. Just walk out. Save yourself. She nodded. "Okay. Great. Thanks."
With that, she rushed to the elevator, jabbed the button and walked on board. She dropped her head as the door closed, her tears dotting the floor. She couldn't look up. She couldn't watch everything she'd ever wanted disappear.
Sarah went immediately into autopilot, putting on her sunglasses to hide her eyes and marching through the lobby outside. Luckily, John was always waiting for Aiden—this time it paid off for her.
"Ms. Daltrey. Penn Station?"
"Yes. John. Thanks." She climbed into the backseat, sucking in a trembling breath. Just get me to the train. Then I'll be okay.
Her phone beeped with a text from Aiden.
This is stupid. Come back. We should talk.
Words weren't enough this morning. Not sure what's different now.
I need time. I'm sorry.
It's okay.
She stopped herself from typing the words she wanted to. I still love you even though you don't love me.
"Ms. Daltrey?" John asked from the front seat. "I have a message from Mr. Langford. He's asking me to bring you back to the house."
She blew out a breath. It was just like Aiden to snap his fingers and expect the world to conform to his wishes. "No. Please don't do that. Just pull over and drop me off and I'll get in a cab."
"Ma'am? I don't want to leave you, either."
Every sad feeling she'd had a few minutes ago was turning to frustration. "I'll text Mr. Langford. Please just keep driving."
She tapped out a message to Aiden.
Please don't put John in the middle of this. Let me go.
Waiting for Aiden's response was agony. She didn't want to argue. But she wasn't ready for the end, again.
Ok.
She tucked her phone into her bag. "All straightened out, John. It was just a misunderstanding."
"Oh, good. Okay. I'll have you at the station in no time."
"Great. The sooner, the better."
Seventeen
Day ten arrived with sunshine streaming through the windows and a giddy Oliver, full of energy and ready to take on the day. Right after breakfast, they'd started doing laps in the house. From the kitchen to the library to his office and back, Oliver walked while Daddy followed, holding his little hands to steady him. Oliver had discovered this new routine while they'd been playing last night before bed. Judging by the way he took to it and the enthusiasm with which he cruised along furniture, he'd be walking and running in a matter of days.
Aiden, however happy he was to share this milestone with Oliver, was dragging—no sleep and a gaping hole in your heart will do that to a guy.
Sarah was gone. And her absence was much more noticeable than Aiden had expected. The house felt strange and incomplete. Had it felt like this before she came along? He couldn't recall, exactly. It was quite different with Oliver there, but still, it wasn't the same without Sarah.
He missed everything about her—the way she hummed when she puttered around in the kitchen and her sweet smell when she walked past him. The way her face lit up when she laughed and the way she wouldn't let him get away with anything when she was mad. Memories shuffled through his mind—the day she managed to talk her way into one of the most secure office buildings in the world. She'd made his entire life turn on a dime that day, and done it in unflappable fashion. That night in the bathtub, when he'd first bonded with Oliver and Sarah had made it happen. That was also the night he'd caught her staring at him, the night he'd foolishly thought that seducing her would be like taking any other woman to bed. He'd relied on their ten-day deadline then. It made it easier for him to get what he wanted, no strings attached. Little did he know that Sarah was capable of tying up his heart and his head with those strings...and tugging them all the way back to Boston.
But what was he supposed to do? They were operating at different speeds. She was comfortable with bold strides. He needed to ease into it. He knew no other way.
His phone rang from the kitchen counter. His pulse picked up. Was it Sarah? He steered Oliver over and consulted the screen. Anna. Not the call he wanted, but maybe she could tell him to stop being such a wimp.
"Hey," Aiden said. "This is a surprise. It's a little early isn't it?"
"I figured you were already up with Oliver and I wanted to check in on you after the other night with Mom. How are you holding up?"
Aiden dragged a barstool around the kitchen island so he had a good view of Oliver, and sat down. "I'm doing fine. I've had years to stew over it. It's more of a relief than anything. And at least we can all get together now without it being hopelessly uncomfortable."
Anna blew out a breath. "Good. I'm glad you feel that way because I have something else I need to talk to you about. Jacob told me I should probably just butt out, and we kind of had a big argument about it, but I don't want to butt out. I can't not say something."
"What in the world are you talking about?"
"Sarah, Aiden. Don't you dare let her go back to Boston today without you two making a plan to keep seeing each other. I know how you are and I'm telling you right now that she's not like other women. She's a keeper, Aiden. I don't want you to blow it just because you've convinced yourself it's easier to play the field."
Aiden could only imagine what his face looked like right now—pure shock. Astonishment. "First off, why don't you tell me how you really feel? And second, how do you know there was anything going on between us?"
Anna huffed at what she apparently saw as Aiden's absurdity. "I saw the way you two were looking at each other the other night. And the minute that all of that stuff went down with Mom, she was the one you turned to. Right away. You didn't even hesitate. It's so obvious to me that you two are in love."
"How can you tell that from a look?"
"Am I wrong? There are feelings between you two, aren't there?"
"Well, yes, there are feelings between us. But that doesn't mean it's love. And besides, it's too late. She's already gone."
"What?" Anna shouted so loudly, she nearly blew out Aiden's eardrum.
"Careful or you'll go into labor."
"You let her leave? Why did you do that? Why would you be so stupid?"
Because she said she loved me and I couldn't say it back. The realization hit him, and the repercussions came at him just as hard. "It was moving too fast for me."
"The man who jumps out of airplanes thought it was moving too fast? Sounds to me like you're confused."
"Yeah. I guess I am. I just don't want to make a mistake. She means a lot to me. But I can't tell her I love her if I'm not sure. I don't even know how I'm supposed to know if it's really love. People always say that you'll know when it happens. Well, I don't know."
"Let me ask you this. How do you feel now that she's gone?"
"Horrible. Like somebody ripped my heart out of my chest."
"And what's the house like without her there?"
"Terrible. I'm thinking Oliver and I might need to move."
"And if you could do anything at all right now, what would you do?"
"Go see her. Apologize." Oh God. I love her. Aiden cast his sights down at Oliver, who was hitting the floor with a wooden spoon. I really am an idiot. I'm a complete jerk. He'd said to himself many times over the past ten days that he would never let Oliver go without. But in letting Sarah leave, he was not only depriving Oliver of the perfect mother, he was keeping himself from the one person who understood him and loved him despite his faults. Oliver had shown him unconditional love. But so had Sarah.
"Do you enjoy feeling like this?" Anna asked. "Because you know you can fix it."
"I can't fix it. I ruined it. She told me she loved me and I didn't say it back."
Anna gasped on the other end of the line.
"That's pretty much the end, isn't it?" His conscience was impossibly heavy. He'd trampled all over the heart of the woman he loved. "I mean, how do I come back from that?"
"Groveling."
"Groveling?"
"It's the only thing that works. Flowers help. Jewelry. Chocolate. A gift certificate for a massage. But mostly groveling. You need to get your butt up to Boston and beg for her forgiveness. You need to tell her how you feel."
"You think it will work?"
"Not sure, but I think you'll regret it forever if you don't try. Jacob and I can be over in a half hour to watch the baby."
None of this will be right without Oliver. "No. It's okay. I'm taking him with me."
* * *
Sarah went into the Kama office Sunday morning. Although it was their headquarters, that word was generous—it was really just an old warehouse she'd been renting for the last year. Sleep last night had been pointless—too many painful things wreaking havoc in her head. Too many things running through her heart, like water through a sieve. She'd been so scared of what would happen if she got too close and now she knew how right she'd been to fear it. Losing Oliver and Aiden was the worst thing that had ever happened to her. No doubt about that.
She didn't bother flipping on the lights as she wound through the sewing room with its massive cutting tables, stacked high with boxes of inventory ready to ship. She went straight back to her office and got to work—the act of a woman invested in her own success, but it felt like an empty gesture. A show. More faking it. Her heart wasn't in it, as much as she might very well be standing on the precipice of great success. On the inside, she was as empty as she'd ever been, which was a devastating realization. Her hard work was finally paying off, and she felt horrible. She'd seen low moments, but not like this.
Not like last night, when she couldn't get a single minute of relief because her eyes were like a faucet. Her heart had stubbornly chosen to ache and throb in her chest and remind her with every pointless beat that the difference between the love a person gives freely and the love they receive in return is what ends up breaking us. This was the second time she'd had to learn the lesson of how it empties a person—giving and giving, never refilling the tank. And she was as done as done could be. The fate of her business felt as inconsequential as a speck of dust floating in air. It was nothing worth holding on to if she couldn't have what she'd truly invested in—Aiden and Oliver.
But Aiden hadn't been able to go there. He just couldn't say I love you. If only he knew—or cared—three little words and she would've figured out a way to stay. She would've told him that she'd meant it. She would've done everything she could to make them all whole again, to knit them into the family they could have been. But apparently, for a man wealthy beyond anything she ever imagined, three words was too high a price to pay.
She tidied her office—going through the mail she'd missed over the last week, filing away things, neatening stacks of paper. She made sure her computer screen was free of smudges, and watered the pink orchid on her desk. She did every mindless task she could come up with, all in the interest of staying busy. If she couldn't move forward, she could at least tread water. She could keep her head above the rising tide. She had to fight back her thoughts of her last night with Aiden, of the connection they'd shared. There was no doubt in her mind that it had been more than sex that night. And she knew, deep down, that Aiden knew it, too. He just couldn't admit it to himself. He was too wounded.
Tessa popped into view. "Morning," she said, stepping foot into the office.
Sarah jumped. "You scared me." She pressed her hand to her chest. Her heart was pounding. "What are you doing here? You didn't need to come in today. You should be at home relaxing. Tomorrow's a big day. I need you on top of your game."
A mischievous smile spread across her face. "I came by to let somebody in. He was pretty sure you weren't going to let him in on your own."
"What? Who?"
Just then, Aiden appeared in her doorway, Oliver in his arms. "I had to talk my way in. I needed to bring Oliver to you. He misses you. I miss you."
The grin on Tessa's face had only grown. "I'll leave you three alone. See you tomorrow."
Sarah walked out from behind her desk, in shock. Was this a dream? Were Aiden and Oliver a mirage? Surely a figment of her imagination couldn't have the pull on her that Aiden did right now. All she wanted to do was fling her arms around him and kiss him. Oliver reached for her. The minute she had him, Aiden's arms were around them both.
"Sarah, I'm here because I had to tell you in person that I love you."
"But..." Tears rolled down her face. How could she possibly cry more? "You don't have to do this. Don't feel like you have to say that to me. And you really shouldn't feel like you have to travel hours with a baby to say that to me in person."
"But I do have to do those things. I have to make it up to you. And I have to tell you the truth." He loosened his grip, to see her better. "I've been falling in love with you since the first night, when you put me in the bathtub with this little guy. It's grown so fast that I didn't know what it was. I couldn't see it. I don't know if I was afraid or confused or what, but the minute you left, I knew it wasn't right."
She nodded eagerly, feeling as though a weight had been lifted. Her hunch had been right. And it hadn't taken long for Aiden to see it, too. "I know it happened fast. I thought I was crazy to say that to you yesterday, but I had to. Especially after everything with your family." She studied his face, his blue eyes nearly taking her breath away. "I couldn't not tell you that I love you. You deserved to know."
He sighed and looped her hair behind her ear, caressing her cheek. "I've spent my whole life homesick for a home I never even knew. And you showed up out of nowhere, and made that home for me in ten days."
"Technically, it was nine."
A breathy laugh left his lips. "You showed me what love is. You opened up my closed-up heart. And that heart is going to shrivel up and die without you. The home you built isn't going to work without you."
"What are you saying, Aiden?"
"I'm saying that I love you and we have to find a way to make this work."
"But you're in New York. I'm here. How are we going to manage that? You don't even have a permanent nanny."
"I called Lily from the plane and convinced her to take the job. She's flying up here tomorrow morning to take care of Oliver while I go into the LangTel regional office downtown for a few hours."
Sarah wasn't sure she was hearing him correctly. "You're going to hang out in Boston? For how long?"
He shrugged. "Depends on what Sylvia Hodge tells you tomorrow. Then we'll figure it out. Anna said that if Sylvia acquires Kama, she'll probably ask you to work out of New York so you're available for meetings and are more plugged in to the industry."
Sarah hadn't considered that. It was all still so new. "So we wait and see what happens tomorrow?"
"I was hoping Oliver and I can move in with you for a few days. I figure we'll put Lily in a hotel."
"I don't know. I need my space."
Aiden laughed and kissed the top of her head. "Darling, as long as you come back to me, you can have all the space you need."
Eighteen
For the third weekend in a row, Sarah was back in New York with Aiden and Oliver. She looked forward to these days more than anything, even when the back and forth was tiresome. Only one more week and the Kama office would move to Sylvia Hodge's Manhattan headquarters. She'd be in the city full-time. She, Aiden and Oliver would be together. Even though she and Aiden hadn't discussed their future, Sarah was more than content. It hadn't seemed necessary and her forcing of I love you had flopped—at least at first. Plus, Aiden was a complicated guy. Commitment wasn't easy for him. Just knowing that he loved her and wanted to be with her was enough for now.
Everything else workwise was already in place—when Sylvia decided to acquire Kama, it came together very fast. Last week, they'd moved everything into a new manufacturing space outside Boston. It was ten times bigger than the original facility and the air-conditioning worked—no small matter now that it was the middle of June. They'd hired ten new assembly people, three more employees to manage the warehouse. Tessa was overseeing the production facility, and she'd received a big fat raise for taking on her new responsibilities. Sarah couldn't have been any happier about being able to reward her for a job well done. The change also left much more time for Sarah to spend on designing, selecting fabrics and planning out the next several seasons. It was hard to believe, but everything on the work front was really coming together.
Saturdays at Aiden's were pretty low-key. Today had been no different, although they were anxiously awaiting a call from Jacob since Anna had gone into labor that morning. To pass the time, Sarah and Aiden had taken Oliver for a long walk, then grabbed some lunch. The baby had his nap after that, and they turned their dinner into a picnic up on the rooftop terrace. Now that it was nearly eight, Sarah was getting a bit of work done in Aiden's home office. Oliver was already in bed and Aiden had camped out with a book in the library.
Aiden's phone, which he'd left on the desk, rang. She glanced at the screen and grabbed it. "Oh my God. Aiden!" she yelled out. "It's Jacob. Get in here!" She answered the call. "Jacob? Aiden's in the other room. I figured I should answer. Is there news?"
"It's a girl," he said triumphantly. "Eight pounds, seven ounces. Twenty-one inches long. Big head of thick, black hair. She's beautiful."
Sarah loved hearing the good news, but Aiden was missing it. She got up from the desk to search for him. "Congratulations. I'm so happy. How's Anna?" She reached the library. No Aiden. Weird.
"She's tired, but she's doing great. We're both relieved the baby's finally here and she's healthy. It's been such a rocky road."
"Oh, I know. She's your miracle baby. It's wonderful." Into the kitchen she traveled. Still no Aiden. "Do you have a name yet?"
"Grace. It's Anna's middle name."
Sarah glanced into the living room, which was also empty. "That's so beautiful. You must be so thrilled."
"I am, but I also have a million more phone calls to make. If you could tell Aiden, that'd be great. I'm sure Anna will want to speak to you both at some point."
Up the stairs Sarah went. "I don't know if you guys will be up for it, but we have Oliver's birthday party next Saturday."
"Oh, right. We'll have to see how it goes, I guess. We might need to call you for some baby advice."
"Absolutely no problem. Whatever you need. Love to all three of you. Can't wait to see her." She ended the call, but didn't dare yell now that she was in the hall. She'd wake Oliver. Where in the world was Aiden?
The door to his bedroom was closed. Now that they'd been cohabitating for nearly a month, she didn't hesitate to open it. But the knob wouldn't turn. It was locked.
She leaned against the door, but heard nothing. She rapped quietly and waited. She couldn't text him—she had his phone. She knocked again. Finally, he answered.
"Hey," he said, seeming flustered. He raked his hand through his hair, poking his head through the narrow opening he'd left.
"Your sister had a baby girl. Her name is Grace. That's what's up. Where have you been?"
His shoulders dropped. "Damn. I can't believe I missed that call. That's a bummer. Is everything okay?"
"Everything's great. I told them you send your best." She tried to peek into his room, but could see nothing. "What are you doing in there? Can I come in?" He was behaving so strangely.
"I'm working on something. A surprise. But it's not ready."
She laughed quietly, curious what he was up to. Her birthday wasn't until October. "Like that's not cryptic. Do you want me to go away?"
"No. No. It's okay. I was going to call you up in a minute anyway. Just close your eyes and I'll lead you to the bed."
"If that's where we're going, it's not a surprise. Not that I won't enjoy it immensely." She elbowed him in the stomach, but he didn't take the joke. He was dead serious, so she decided to follow orders.
"Just a minute," he said when she was seated on the bed. "Be right back."
With her eyes closed, she listened for clues. He started singing. She'd never heard Aiden sing. Not once.
"How's it going in there?" she asked.
"You're so impatient." His voice was close—as if he were right next to her. He took her hand and she opened her eyes. "Ready?" He had a huge grin on his face.
"Yes." She trailed behind him into the bathroom. The lights were off. The marble countertops were covered in an array of lit candles. "Ooh. Bath night. So romantic."
"I had to make it romantic. For you."
She wrapped her arms around his waist, and he planted a soft and sensuous kiss on her lips. "That's adorable. I love that you made an extra effort for our Saturday night together."
He kissed her again, on her cheek, beneath her ear, on that extrasensitive spot on her neck. "It's more special than that."
Again, with the clues and mysterious phrasings. His kisses, however amazing, weren't helping. They made it difficult to think straight. "What kind of special?"
He cupped the side of her face and caressed her cheek with his thumb. "You're the best thing that's ever happened to me. I don't want to let you go."
Let me go? Oh my God. She clapped her hand over her mouth. "Aiden. Are you?"
"Shh. Just let me ask." He plucked a washcloth from the counter. A blue Tiffany box was beneath it. He popped it open, smiling wide and presenting her with the most gorgeous diamond ring she'd ever seen. "I love you so much. I want you to be my wife and Oliver's mother. Will you marry me?"
She blinked away tears, her heart about to burst from pure joy. "Yes. Of course." He placed the ring on her finger, and she popped up onto her toes, kissing him hard before stealing the chance to admire the diamond. "It's so gorgeous. Did you tell the people at Tiffany you were going to ask me in the bathroom?"
He laughed. "No. Do you know why I chose this room?"
She shrugged. "Because you like taking a shower with me?"
"That's part of it, but not the real reason." He took her left hand, straightening the ring. "That first night we gave Oliver a bath was the beginning of our life together. As a family. I wanted to acknowledge the start before we step into our future."
Tears welled again. "That is the sweetest thing ever."
"That was also the first time I caught you staring at me. That was sort of a big deal."
She swatted his arm. "I can't help it. You're just way too hot." Whatever she'd done to be lucky enough to have Aiden, she was glad she'd done it.
He cranked the faucet on the tub and turned his attention to her top, lifting it over her head. His heavenly lips skimmed her shoulder. "I can't wait to get in this bathtub with you and make love to you all night and talk about our future together."
Their future together. "Now I don't have to worry about the perfect guy walking into my life. I found him."
"Actually, I'm pretty sure you walked into his."
* * * * *
If you liked this story of a wealthy CEO tamed by the love of the right woman, pick up these other novels from Karen Booth.
THAT NIGHT WITH THE CEO
PREGNANT BY THE RIVAL CEO
THE CEO DADDY NEXT DOOR
THE BEST MAN'S BABY
Available now from Harlequin Desire!
***
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***
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Expecting the Billionaire's Baby
by Andrea Laurence
One
"You can do this, Cecelia."
Cecelia Morgan attempted to encourage herself as she looked over her portfolio for the hundredth time. Tomorrow, she was presenting her design plans to the board of directors of the new Bellamy Hotel. This was a big step for her and her company, To the Moon. The company she started after college specialized in children's furniture, bedding and toys. From the beginning she had targeted a high-end market, catering to wealthy parents who were looking for luxury products for their children.
The company had been a success from the very start. What had begun as a small online boutique had exploded into a series of stores across the United States after a celebrity posted on social media about how much they loved one of TTM's nursery designs. Cecelia had been forced to open her own production facility and warehouse outside her hometown of Royal, Texas, to keep up with the demand.
The portfolio on the desk in front of her, however, could take To the Moon to the next level. Designing furniture, toys and accessories for pampered little ones had been her first love, but now Cecelia was ready for her business to mature along with her tastes. The Bellamy Hotel was her chance to make this a reality.
The Bellamy was a brand-new five-star resort opening right outside Royal. Owner Shane Delgado had contacted Cecelia about decorating and furnishing the hotel about a month ago, after a previous designer had been fired well into the process. This would be a big step for Cecelia. If she could secure the contract with The Bellamy, it would give her the footing she needed to branch out into the luxury adult furniture market.
As her daddy always said, if you're not moving forward, you might as well be moving backward. She was successful, but that wasn't enough for the Morgans. Her subsidiary of To the Moon—Luna Fine Furnishings—could change everything for her.
She was shocked that Shane had reached out to her, given he was pretty clear he'd dismissed her as part of the mean girls clique, along with her best friends Simone and Naomi. Admittedly, she wasn't very nice to his girlfriend Brandee and recent gossip had been less than flattering about Cecelia and her friends. Some even suspected them of being behind the recent blackmailings. Shane was taking a huge leap of faith inviting her to submit her ideas for this incredible opportunity; she wasn't about to screw this up.
Cecelia gathered up everything into her portfolio binder and slipped it into her leather briefcase. She'd probably gone over it a hundred times already. She needed to stop fiddling with it and just let it lie. It was perfect. Some of her best work yet. As usual, she was putting too much pressure on herself. Her parents certainly didn't help matters. They always held Cecelia, their only child, to very high standards and never accepted anything less than perfection.
She supposed that was why she was so successful. Brent and Tilly Morgan were practically Texas royalty and had raised their daughter to follow in their footsteps. She went to the best private schools, rode horses and competed in dressage in high school, and went on to graduate summa cum laude with a business degree from a prestigious Ivy League university. Anything less for the younger Morgan would've been unacceptable.
While her parents had been supportive both emotionally and financially when it came to her company, Cecelia always worried that their support came at a price. If Luna Fine Furnishings wasn't the success that she hoped for, she might never hear the end of it. The last thing she needed was for her father to pat her on the back and tell her that maybe she needed to just stick with the baby things. You know...woman stuff. Or worse yet, to hand the business over to someone else and focus on settling down with Chip Ashford to make actual babies instead of baby furniture.
She wasn't opposed to settling down with Chip—he was her fiancé after all—but she certainly didn't want to throw away everything that she'd worked for in the process. Chip was a Texas senator, and he had been very supportive of her business so far. But Cecelia got the feeling that once they got married, Chip might feel the same way as her parents did.
It wasn't that she didn't want kids. Cecelia wanted her own children more than anything. But she was confident that she could be both a mother and the CEO of her own company. She didn't intend to set one ambition aside for the other.
A chime sounded on Cecelia's phone. She reached for it and tapped the screen to open up the Snapchat notification she'd just received for a private message. It took her a moment to realize what she was actually looking at. The picture was of a document with small text, but the header at the top brought a sinking feeling to her stomach. It read "Certificate of Birth" with the seal of the state of Texas on the bottom corner. The message across the screen was far more worrisome.
Somebody has got a secret.
Cecelia looked once more at the photo before it disappeared. It was then that she realized that this wasn't just any birth certificate, it was her original birth certificate. The one issued before she was adopted by the Morgans.
For a moment, Cecelia almost couldn't breathe. Her adoption had always been kept a secret. Everyone, including members of her extended family, believed that Cecelia was Brent and Tilly's biological daughter. Even Cecelia had believed it until her thirteenth birthday. That night, they'd told her that she was adopted but that they had kept it a secret for her own protection. The unfortunate truth was that her birth mother had been a junkie, and child services had taken Cecelia away from her when she was only a few weeks old. Her mother had overdosed not long after that, and she was put up for adoption. The Morgans thought that it was best if Cecelia's birth mother and that dark past were kept secret.
But someone had found out.
Cecelia didn't know how—she hadn't even seen her original birth certificate before. A new one had been issued when her adoption was finalized, so someone had done some serious sleuthing to find it.
Another image popped up on her screen. This one was a message written in letters cut from magazines like some sort of ransom note. She supposed that in some way, it was a ransom note. It demanded that twenty-five thousand dollars be wired to an account within twenty-four hours or her secret would be exposed to the entire town. It was signed, Maverick.
Considering everything that had been happening in Royal, Texas, lately, she should've known she would be targeted eventually. Maverick had been wreaking havoc on the lives of Royal residents for the past few months. This anonymous blackmailer had been the talk of the town, and everyone at the Texas Cattleman's Club had suspicions about who it could be. The most recent suspects had been Cecelia herself, along with Naomi and Simone.
Cecelia was a busy woman. She ran her own business, served as arm candy for her fiancé's various political events, was busy keeping up appearances for her parents and for Chip... She hardly had time in her schedule to get a manicure, much less to research and dig up dirt on her fellow residents. Her busy schedule and high standards made her come off as a bit snobbish, and Cecelia supposed she was, but she was no blackmailer. Unfortunately, the only way to prove it was to let everyone know that she was Maverick's latest victim.
That certainly wasn't an option. She couldn't have the whole town knowing that her entire life was a lie.
Unfortunately, this wasn't just her secret. Her parents had built their lives around their perfect "biological" daughter. They'd lied to countless family members and friends to keep up the charade, but they'd only done it to protect her. Paying Maverick was probably the only way to shield Brent and Tilly from the fallout.
But hers wasn't the only family she had to worry about. The Ashfords would have a fit. Chip came from a certain kind of family, and he believed that Cecelia was cut from the same cloth. Would Chip call off the engagement if he found out the truth? Their relationship was more about appearances and family alliances than love, but she hoped that Chip cared enough about her not to throw everything away if her secret got out. As far as she was concerned, she was a Morgan, through and through.
And as a Morgan, it was her responsibility to safeguard her and her family's reputation, or tomorrow's presentation would go down in flames. Her reputation where Shane was concerned was hanging on by a thread as it was. Surely, he wouldn't want a scandal to interfere with his hotel's grand opening.
But when did it stop? Would Maverick be content with the first payment, or would he drag this out until Cecelia was broke and her business was bankrupted?
Cecelia clutched her head in her hands and fought off a pending migraine. She'd suddenly found herself stuck between a rock and a hard place, and there was no easy way out of this. She either paid Maverick, or the truth of her adoption would be spread all over town. The clock was ticking.
She wasn't sure what her path forward would be, but Cecelia knew what she was doing next. In her life whenever a crisis arose, Cecelia always called her daddy. This conversation, however, was one that needed to be had in person. She didn't know how Maverick had found out about her adoption, but if her phone lines were tapped or her computer was being monitored, she couldn't risk anything but face-to-face communication.
* * *
It took Cecelia over an hour for her to reach her parents' mansion outside Houston. It was nearly ten o'clock by the time she arrived, but her parents would still be awake. As expected, she found her father sitting in his library. He was reading a book and smoking one of his favorite cigars.
Brent Morgan looked up in surprise when he noticed his daughter standing in the doorway of his library. "What are you doing here, sweetheart? Your mother didn't tell me were stopping by tonight."
Cecelia took a few steps into her father's favorite room and took a seat in the leather chair across from him. "She doesn't know I'm here. I'm in trouble, Daddy."
Furrowing his brow, he set aside his book and stubbed out his cigar. "What is it? Are you and Chip having problems?"
"No, this isn't about Chip." With a sigh, Cecelia told her father about the message she had received. His expression had morphed from concerned, to angry, to anxious as she spoke. "I've got twenty-four hours to wire them twenty-five thousand dollars, or everyone is going to know the truth."
"Our family can't afford a scandal like this. And imagine the pain this would bring to the Ashfords. Surely this isn't what you want. You're just going to have to pay him," he said, matter-of-factly.
Cecelia hated being put in a position where she had no options, and being under Maverick's thumb was the last place she wanted to be. The only real way to combat blackmail was by exposing the truth before the attacker could. If they beat Maverick to the punch they could put their own spin on her adoption and why they'd lied about it.
"Are you sure, Daddy? I mean, I know you and Mother were trying to protect me, but I'm a grown woman now. I'd rather the story not get out. However, would it be the end of the world if people discovered I was adopted? Does it change anything, really?"
"It absolutely does!" her father said with his face flushing red, making his salt-and-pepper hair appear more starkly white against his skin. "We've lied to everyone we know for thirty years. This would ruin our reputation. And what would the Ashfords think? They wouldn't understand. Neither would my customers or my friends. I could lose business. Hell, you could get thrown out of the Texas Cattleman's Club. It's social suicide, and your mother's heart couldn't take the scandal. No," he insisted. "This stays a secret. Period. I will loan you the money if you need it to pay the blackmailer, but you will pay him."
Cecelia noted the finality in her father's tone. It had been the same when she was an unruly child, the same when she was a teenager testing her boundaries. She was an adult now, but Brent Morgan was still in charge. She didn't have the nerve to go against him then, and she certainly didn't have the nerve to do it now. She'd come here for his advice, and she'd be a fool not to take it.
"No, I have the money. I'll make the transfer in the morning. I just hope it is enough to put an end to all of this."
"It has to be," her father said. "I refuse to have our family turned into laughingstocks."
Cecelia sighed in resignation and got up from her seat. "I'll take care of it, Daddy."
* * *
Deacon Chase turned his restored 1965 Corvette Stingray down the main street of Royal, Texas. It'd been thirteen years since he'd looked at this town in his rearview mirror and swore he'd never set foot in this narrow-minded, Texas dust trap again. The whole flight over from France, he questioned why he was coming back. Yes, it was good business, and working with his old friend from high school, Shane Delgado, had always been a pleasant experience. But when Shane mentioned that he wanted to build a resort in their hometown of Royal, he should have passed.
Then again, when else would he get the chance to show the town and the people who rejected him that he was better than them? Sure, back then he'd just been a poor kid with few prospects. He was the son of a grocery store clerk and the local car mechanic. He'd gotten to go to private school with all the rich kids only because his parents had been adamant that Deacon make something of himself, and they'd put every dime they had toward his schooling. Even then he had worked in the cafeteria to bridge the gap in tuition. Nobody else had expected much out of him, and those were the people who even acknowledged he existed. As far as most the residents of Royal were concerned, Deacon had never fit in, never would fit in and needed to accept his station in life.
No one had expected him to take his hobby of restoring cars and parlay the skills and money into restoring houses. They certainly hadn't expected him to take the profit from those houses and put it into renovating hotels. Now the kid who worked in the cafeteria was a billionaire and the owner of the most glamorous resort in Cannes, France, the Hotel de Rêve, among others.
The only person in Royal who had ever believed in him was Cecelia. Back in high school, she'd pushed him to be the best person he could be. Considering that she'd held herself to such high standards, he'd been flattered that she saw so much potential in him when most of the people in high school either ignored him or taunted him. Cecelia had said he was a diamond in the rough. Her diamond in the rough.
It'd certainly blown the minds of all the boys at school that Cecelia had chosen Deacon instead of one of them. What could he offer her after all? A free carton of milk with her lunch? It turned out that he'd had plenty to offer her. He could still remember how many hours they'd spent lying in the back of his pickup truck talking. Kissing. Dreaming aloud about their future together. Deacon and Cecelia had had big plans for their lives after graduation.
Step one had been to get the hell out of Royal, Texas. Step two had been to live happily-ever-after.
As Deacon came to a stop at the traffic light at the intersection of Main Street and First Avenue, he shook his head in disgust. He had been a fool to think any of that would ever happen. He might have fancy hotels and expensive suits, sports cars and a forty-foot yacht docked in the French Riviera, but Deacon knew, and everybody else knew, that Cecelia was too good for him.
It hadn't taken long for Cecelia to figure that out, too.
The light turned green, and Deacon continued down the road to where his father's old garage used to be. When he'd made his first million, Deacon had moved his parents out of Royal and into a nice subdivision in central Florida. There, they could enjoy their early retirement without the meddling of the snooty residents of Royal. His father had sold the shop, and now a new shopping center was sitting where it used to be. A lot had changed in the last thirteen years.
Deacon couldn't help but wonder how much Cecelia had changed. He tried not to cyberstalk her, but from time to time he couldn't help looking over the Houston society pages to see what she was up to. The grainy black-and-white pictures hardly did her beauty justice, he was certain. The last time he'd seen her, she'd been a young woman, barely eighteen. Even then, Deacon had been certain that she was the most beautiful woman he would ever see in person. He would bet that time had been kind to his Cecelia.
Not that it mattered. The most recent article he'd stumbled across in the paper had included the announcement of her engagement to Chip Ashford. He remembered Chip from high school. He was a rich, entitled, first-class douche bag. Deacon was fairly certain that that hadn't changed, but if Cecelia was willing to marry him, she certainly wasn't the girl that he remembered. Back then, she'd hardly given Chip the time of day.
Mr. and Mrs. Morgan must be so proud of her now. She'd finally made a respectable choice in a man.
Turning off the main drag, Deacon headed down the narrow country road out of Royal that led to his latest real estate acquisition. The rustic yet luxurious lodge that was to serve as his home base in the area stood on three acres of wooded land several miles outside town. He'd bought the property sight unseen when he decided to take on The Bellamy project with Shane. He couldn't be happier with the place. It was very much his style, although it was a far cry from the elegant European architecture and design that he'd become accustomed to.
He hadn't really needed to buy the home. Deacon had no real intention of staying in Royal any longer than he had to. But the businessman in him had a hard time passing up a good deal, and it seemed a shame to throw money away on renting a place while they built the hotel. He had no regrets. It was his happy retreat, away from the society jungles of Royal.
When he pulled up in front of the lodge, he was surprised to find Shane Delgado's truck parked out front. Deacon parked the Corvette in his garage, then stepped out front to meet his friend and business partner.
Deacon hadn't had many friends back in school. Basically none. But his side business of buying and restoring cars had drawn Shane's attention. Shane had actually bought Deacon's very first restoration, a 1975 cherry-red Ford pickup truck with white leather seats. Deacon had been damn proud of that truck, especially when Shane had handed over the cash for it without questioning his asking price. They'd bonded then over a mutual love of cars and had continued to keep in touch over the years. When they both ended up in the real estate development business, it was natural for them to consider working together on a few projects.
"What's wrong now?" Deacon asked as he joined Shane at the bottom of his front steps.
While the construction of The Bellamy had gone relatively smoothly, Deacon was the silent partner. Shane bothered him with details only when something had gone awry. He joked with Shane once that he was getting to the point that he dreaded the sight of his friend's face.
"For once," Shane said with a smile, "I'm just here to hang out and have a drink with my friend. Everything at the hotel is going splendidly. Tomorrow, Cecelia Morgan will be presenting her designs to the board, based on your recommendation. Assuming we like what Cecelia did, and I hope I'm not going too far out on a limb here, we'll be moving forward and getting that much closer to opening the hotel."
Deacon slapped his friend on the back of the shoulder. "I wouldn't have brought her on board if I didn't think she was the best designer for the job. Come on in," he said as they started up the massive stone stairs to the front door. "Have you eaten?" he asked as they made their way into his office for a drink.
Shane nodded. "I have. Brandee is constantly feeding me. By the end of the year, I'm going to weigh three hundred pounds."
"You're a lucky man," Deacon said as he poured them both a couple of fingers of whiskey over ice. Shane had recently gotten involved with Brandee Lawless, the owner of the nearby Hope Springs Ranch. She was a tiny blonde spitfire, and one hell of a cook. "I'd be happy to have Brandee feeding me every night."
"I bet you would," Shane said. "But you need to just stick with your cultured European women."
Deacon chuckled at his friend's remark. He had certainly taken advantage of the local delicacies while he was in Europe. Even though it'd been years since he and Cecelia had broken up, it had soothed his injured pride to have a line of beautiful and exotic women waiting for their chance to be with him. He would never admit to anyone, especially Shane, that not a one of them held a candle to Cecelia in his mind.
Deacon and Shane sat there together, sipping their drinks and enjoying each other's company. They didn't get a lot of opportunities to just hang out anymore. Deacon's office, however, just begged for gentlemen to spend time in comfortable chairs and shoot the shit. The walls were lined with shelves containing leather-bound books that, frankly, came with the house and Deacon would never read. They did create a nice atmosphere, though, along with the oil paintings of landscapes and cattle that hung there. It was all very masculine Texas style.
"Can I ask you something?" Shane asked.
"Sure. What?"
"You do know that Cecelia's business specializes in children's furniture, right?"
Deacon tensed in his chair. Perhaps his office made Shane too comfortable, since he felt like prying into Deacon's motivations for wanting Cecelia for the job. "Yeah, I know. I also know that she's managed to turn her small company into a furniture and accessories juggernaut since she started it. She's always had a good eye for design."
"She does, I won't argue that. But hiring her to decorate The Bellamy is a huge risk. She and Brandee aren't exactly fans of each other. And what if she and her friends are actually behind the cyberattacks? That's not the kind of publicity we'd want for our hotel. I don't have to remind you how much we stand to lose if our gamble doesn't pay off."
"That's why we just asked her to submit a proposal along with the two other design firms. We haven't hired anybody yet. If she's out of her depth in this, or acts suspicious in any way, we thank her for her time and send her on her way. It's not ideal, but not the end of the world, either."
Shane narrowed his gaze at him. He obviously suspected that Deacon had ulterior motives in wanting Cecelia involved in the project. Deacon understood. He wasn't entirely sure that he didn't.
"I'm not sold on either of the other firm's designs. She's last to present, so if she flops tomorrow, it's going to set the project back weeks while we find yet another designer and they start from scratch. We have hotel bookings starting day one. Every delay costs us money."
Deacon just nodded. He was well aware that he was taking a risk. But for some reason, he had to do it. Perhaps he was a glutton for punishment. Perhaps he was looking for any excuse to see her again. He wasn't sure. The only thing he was sure of was that everything would turn out fine. "Relax, Shane. The project will finish on time and on budget with the amazing decor you're hoping for."
"And how do you know that?" Shane asked, sounding unconvinced.
"Because," Deacon said confidently, "Cecelia hasn't failed at anything in her entire life. She's not going to start now."
Copyright © 2017 by Harlequin Books S.A.
ISBN-13: 9781488011511
The Ten-Day Baby Takeover
Copyright © 2017 by Karen Booth
All rights reserved. By payment of the required fees, you have been granted the non-exclusive, non-transferable right to access and read the text of this e-book on-screen. No part of this text may be reproduced, transmitted, down-loaded, decompiled, reverse engineered, or stored in or introduced into any information storage and retrieval system, in any form or by any means, whether electronic or mechanical, now known or hereinafter invented, without the express written permission of publisher, Harlequin Enterprises Limited, 225 Duncan Mill Road, Don Mills, Ontario M3B 3K9, Canada.
This is a work of fiction. Names, characters, places and incidents are either the product of the author's imagination or are used fictitiously, and any resemblance to actual persons, living or dead, business establishments, events or locales is entirely coincidental. This edition published by arrangement with Harlequin Books S.A.
® and ™ are trademarks of the publisher. Trademarks indicated with ® are registered are registered in the United States Patent and Trademark Office, the Canadian Intellectual Property Office and in other countries.
www.Harlequin.com
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{
"redpajama_set_name": "RedPajamaBook"
}
| 274
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Q: Issue with Java and StringIndexOutOfBoundsException I am trying to do a school assignment and I am a bit stuck with my code, when I run the code I get this:
Exception in thread "main" java.lang.StringIndexOutOfBoundsException: begin 0, end -1, length 25
at java.base/java.lang.String.checkBoundsBeginEnd(String.java:3756)
at java.base/java.lang.String.substring(String.java:1902)
at Demo4/d4.StringBuilder.poistaEkaSana(StringBuilder.java:32)
at Demo4/d4.StringBuilder.main(StringBuilder.java:18)
The goal is to use StringBuilder to remove first name firstname and print characters after space.
Any help is appreciated.
import fi.jyu.mit.ohj2.*;
public class StringBuilder {
public static void main(String[] args) {
String nimi = Syotto.kysy("Give Lastname Firstname");
StringBuilder sb = new StringBuilder();
removeFirstWord(sb);
System.out.println("Your last name is " + sb);
String lastname = giveFirstWord(name);
System.out.println("Your last name is " + lastname);
}
public static String removeFirstWord(StringBuilder sb) {
String bs = sb.toString();
int remove= bs.indexOf(" ");
String second= bs.substring(0, remove);
return second;
}
public static String giveFirstWord(String fname) {
int s1= fname.indexOf(' ');
String d2= fname.substring(s1, 14);
return d2;
}
}
A: Your class name is confusing with java.lang.StringBuilder.
Rename your class name( or use FQCN), and
String nimi = Syotto.kysy("Give Lastname Firstname");
java.lang.StringBuilder sb = new java.lang.StringBuilder(nimi); // add argument
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"redpajama_set_name": "RedPajamaStackExchange"
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"redpajama_set_name": "RedPajamaC4"
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{
"redpajama_set_name": "RedPajamaC4"
}
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EAST ASIAN STUDIES
CHINA INSTITUTE
Semester In Taiwan
Environment and Environmentalism in East Asia (2016)
Taiwan in Dynamic Transition Conference (2013)
TAIWAN IN DYNAMIC TRANSITION CONFERENCE
University of Alberta, Edmonton, Alberta
The international conference "Taiwan in Dynamic Transition," hosted by the Department of East Asian Studies at the University of Alberta, will take place on May 25-26. At the conference, distinguished scholars from Australia, Canada, New Zealand, Taiwan, the United Kingdom, and the United States will present original research on some of the most salient aspects of Taiwan's social, political, and cultural transformation in the 20th and 21st centuries.
The experts participating in this interdisciplinary conference represent a range of academic disciplines in the social sciences and humanities. Their presentations pertain to such topics as the influence of political discourse and education on the imagination of Taiwanese history and identity; democratization and political reform at local and national levels; cultural, racial, and gender tensions as seen through the lens of popular literature; Taiwan as a provider of foreign aid and developmental expertise; the influence of global norms on Taiwanese society and political institutions.
9:00-9:30 Welcome and Opening Remarks
Ryan Dunch, Chair, Department of East Asian Studies
Lois Harder, Associate Dean, Faculty of Arts
Michael Y. K. Tseng, Director General, Taipei Economic and Cultural Office, Vancouver
"Why is the East China Sea Peace Initiative Important?"
9:30-11:00 History, Identity, and Political Discourse
Panel Chair: Ashley Esarey, Department of East Asian Studies, University of Alberta
Rwei-ren Wu, Institute of Taiwan History, Academia Sinica "Nation-state formation at the interface: The Case of Taiwan"
Jennifer Wei, English Department, Soochow University "Familial Metaphors in Taiwanese Campaign Rhetoric: a comparative study of Mme. CKS and Annette Lu, and of Tsai Ing-wen in contrast"
I-Hsin Hsiao, Department of Sociology, University of Essex "A New Form of National Imagination in the High School History Curriculum: On the Non-Linear Development of Taiwanese History Course during Lee Teng-Hui and Chen Shui-Bian Eras"
Discussant: Maukuei Chang, Institute of Sociology, Academia Sinica
11:00-12:30 The Political Dimension: Political and Legal Reform
Panel Chair: Gordon Houlden, China Institute, University of Alberta
Jiunn-rong Yeh, National Taiwan University Law School "The Legacy of Incremental Constitutional Reform in Taiwan"
Benjamin Read, Politics Department, University of California, Santa Cruz "Urban Taiwan's State-Structured Neighborhood Governance: Deepening Democracy, Partisan Civic Engagement, Inverted Class Bias"
Eric Setzekorn, Department of History, The George Washington University "Military Reform in Contemporary Taiwan: The Lafayette Scandal, National Defense Law and All-Volunteer Force"
Discussant: Bruce Jacobs, Monash University
14:00-16:00 Global and Local Interconnections
Panel Chair: Thomas Gold, Department of Sociology, University of California, Berkeley
Szu-Chien Hsu, Institute of Political Science, Academia Sinica "Constructing a "Common Field" across the Taiwan Strait via Environmentalism? Case Studies of Cross-Strait Environment NGO Exchanges"
Lee Chia-wen, National Cheng Kung University "Taiwan's Death Penalty in the Local-Global Dynamics"
Gerald Chan, Department of Political Studies, University of Auckland "Taiwan as an 'established' aid donor: creating space between traditional and emerging donors?"
Lee Hyunji, Department of Political Science, University of Calgary "An Uninvited Guest at the Policymaking Table: The Rise of Mass Opinion in Post-Developmental States"
Discussant: Thomas Gold, University of California, Berkeley
9:00-11:00 Social Change, Gender, and Popular Culture
Panel Chair: Lin Jenn-Shann, Department of East Asian Studies, University of Alberta
Shu-ning Sciban, Shu-ning Sciban, Department of Germanic, Slavic and East Asian Studies, University of Calgary "How to Do Things with Neologisms: A Study of Wang Wenxing's Language"
Isabella Cheng, School of Languages and Area Studies, University of Portsmouth "Bridging Across or Sandwiched Between? The In-Between Identity of Chinese Immigrant Women in Taiwan"
Lin Pei-Yin, School of Chinese, University of Hong Kong "Voice from the Margin: Gender and Ethnicity in the Works of Rimuy Aki"
Jérôme Soldani, Department of Sociology and Anthropology, University of Ottawa "'Aborigines people are Strong, Chinese are Smart:' Racial Stereotypes on Autochthonous players in Taiwanese Baseball"
Discussant: Daniel Fried, Department of East Asian Studies, University of Alberta
11:00-12:30 Plenary Roundtable Discussion: Next Steps
Many thanks to our sponsors:
Ministry of Education Republic of China (Taiwan)
University of Alberta China Institute
Ministry of Foreign Affairs Republic of China (Taiwan)
©2015-2020 Taiwan Studies All rights reserved Designed and engineered by
|
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{"url":"https:\/\/zbmath.org\/?q=an%3A0104.24904","text":"# zbMATH \u2014 the first resource for mathematics\n\nDie Nilpotenz der $$H_ p$$-Gruppen. (German) Zbl\u00a00104.24904\n\ngroup theory\nFull Text:\n##### References:\n [1] Baer, R.: Partitionen endlicher Gruppen. Math. Z.75, 333-372 (1961). \u00b7 Zbl\u00a00103.01404 [2] Burnside, W.: Theory of groups of finite order, 2nd edition. Cambridge 1911. \u00b7 JFM\u00a042.0151.02 [3] Hughes, D. R., andJ. G. Thompson: TheH p-problem and the structure ofH p-groups. Pac. J. Math.9, 1097-1102 (1959). \u00b7 Zbl\u00a00098.25201 [4] Thompson, J. G.: Finite groups with fix-point-free automorphisms of prime order. Proc. Nat. Acad. Sciences (USA)45, 578-581 (1959). \u00b7 Zbl\u00a00086.25101\nThis reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.","date":"2021-07-25 04:36:52","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\": 0.6268195509910583, \"perplexity\": 7541.6286190921655}, \"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-2021-31\/segments\/1627046151563.91\/warc\/CC-MAIN-20210725014052-20210725044052-00326.warc.gz\"}"}
| null | null |
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IS BEING EXHAUSTED AT WORK AS DANGEROUS AS BEING DRUNK AT WORK?
For some workers, going to work chronically fatigued might be more dangerous than being at work drunk, according to a sleep expert speaking at the IAM141 2017 Safety Conference this week.
Mike Harnett is the Vice President of Human Factors for SIX Safety Systems and specializes in fatigue management. She has over 25 years of experience with IAM-represented organizations like NASA, as well as airport, rail, trucking, nuclear, manufacturing, mining, energy and first responder industries.
Airline union activists, managers, and safety advocates learned a surprising fact this week. Working with chronic, accumulative fatigue is more dangerous than working while moderately intoxicated – a lot more dangerous.
Mike Harnett, a featured speaker at the Conference, knows just how harmful a lack of sleep can be. She is an expert on the subject of fatigue management and sleep deprivation, and her research on the topic has provided valuable guidance to groups and organizations such as NASA, airports, transportation companies and labor groups such as the IAM.
"BEING AWAKE IS NOT ENOUGH"
With news reports of baggage handlers falling asleep in the underbellies of the planes they are working, only to awaken after take-off, sleeplessness has already become a severe issue for air carriers. Airport workers need high levels of physical athleticism, mental focus, and situational awareness to do their jobs safely. And, when the safety of the flying public is also taken into account, the need to avoid the kinds of impairments that come with knocking back a few beers before work is a no-brainer.
Yet, unlike being a little tipsy at work, it can be hard for employers and even workers to wake up to the dangers of chronic sleep deprivation. Nevertheless, the problem should be taken seriously, according to Harnett. Especially in the case of airport workers, who are uniquely vulnerable to the damage that can be caused by the effects of fatigue.
"IMPAIRED IS IMPAIRED"
Meanwhile, the symptoms of sleep deprivation are almost identical to intoxication. Loss of situational awareness, underestimation of risk, hindered visual perception, and reduced reaction times are all symptomatic of both drunkenness and fatigue.
Peer-reviewed studies have consistently demonstrated that a person who has been awake for only 17 hours has the equivalent impairment of a person with a blood-alcohol level of .05%. (In many states, a blood-alcohol level of .08% is enough to result in a DUI arrest.) Those who have stayed awake from 5:00 AM to 2:00 in the morning without sleeping will have reached a .08% blood alcohol level of impairment, and those who have gone without rest for 24 hours can expect an equivalent impairment of .1%. ??"Impaired is impaired," Mike Harnett told the crowd of some 120 union activists and company managers. "If you are impaired because you're drunk, or if you have these same impairments due to fatigue, you are creating the same hazard."
Fatigue is one of the most common causes of airline accidents, with most airport injuries happening early in the morning and late at night when workers are the most tired.
BETTER UNDERSTANDING IS NEEDED
Fixing the problem will require more than a nap. Harnett says that those suffering from chronic fatigue cannot accurately determine if they have reached a dangerous level of sleep deprivation. Even worse, the problem is rampant, with strong majorities of Americans completely unaware of the danger of chronic fatigue. Airport workers who must work sleep-defying shifts late at night or very early in the morning, and who get hit with mandatory overtime and inconsistent days off on a regular basis may be even more at risk than the overall population.
Harnett suggests that the solution to accidents caused by chronic fatigue will require a long-term partnership between workers and companies. Companies will need to begin understanding that fatigue can be a real threat and not merely a discipline issue. Company managers often think of sleep as a personal issue that shouldn't be factored into the work environment. In many cases, travel times to and from work are not factored into the space between work shifts. At many airlines, workers are severely punished for napping at the workplace before or after shifts. These policies need to change, Harnett says.
But, a lot of the burden is going to fall on the shoulders of airline workers themselves. There are real dangers associated with chronic fatigue, and most of these problems can be solved with better sleep.
Occupational Safety and Health Administration (OSHA) 1 (800) 321-6742
OSHA Website
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It is no secret that pursuing a graduate degree is emotionally, psychologically, and physically exhausting. Graduate school can be the first time students truly experience the deep frustrations of a competitive learning environment. Graduate students can come face to face with a deep, pervasive anxiety that seeps into everyday life, a constant questioning of capability, intelligence, and whether or not one is cut out to be there.
As mentioned in our Adjusting to Graduate Study section of PhDStudent.com, most individuals who apply for grad school are often in the top of their undergraduate classes, however, because you will be in class with the top of the top as a graduate student, you might find it more challenging to stand out. Here are some more tips to help you stand out, even when the competition is tense.
Competition is the silent topic that graduate students hate to discuss. It creeps into every classroom, lecture hall, and presentation. Looking at your colleagues, there will be people publishing more, teaching more, more extracurricular activities, or and people with more funding than you—making it easy to think you don't measure up. This self-deprecatory thinking causes students to ignore strategies that can help make them successful and instead of fixating on what colleagues are doing. What are these strategies?
Turn envy into admiration. Sit down with a colleague whose path you would like to know more about. You'll learn the steps required to achieve that level of success.
Run your own race. When thinking about what peers are doing, remember this is your degree, not someone else's. Classmates may be pursuing teaching-track positions while your goal may be a tenure-track research position. But because of all the teaching work they're doing, you might feel the need to work on your teaching strategies, even though teaching doesn't match up with your career goal.
When graduate programs advertise an overly positive picture of cohort cohesion – students sometimes feel ashamed for experiencing feelings of competition, something that is a normal part of life. Competition sometimes resurrects insecurities among students—wondering if classmates are having greater accomplishments—leading to feelings of worthlessness, which can be exaggerated when graduate schools promote a "We are in this together" mindset.
Assistantships are another way of funding your graduate education and are available at many schools with graduate programs. Unlike scholarships, assistantships are more like a work-study program. Graduate assistantships are additional educational opportunities for graduate students while pursuing their graduate education. With assistantships, graduate students are offered free or reduced tuition in exchange for leading discussions in undergraduate classes, proctoring exams and grading, or assisting professors with important research.
As mentioned in my previous posts (part 1 and part 2 of my Free Money series), scholarships and fellowships are not the only ways of getting free money toward your higher education, my post today is going to cover all you need to know about assistantships.
Assistantships are available on a limited basis in most graduate programs at universities and colleges throughout the nation. These positions provide funds for many masters and doctoral students. Each department or program has its own requirements and expectations so prospective students should be sure to do their research before applying to such a position.
Assistantships are a form of financial aid given by the college or university to graduate students who engage in teaching and/or research and provide students with training and valuable professional experience in higher education work environments. It is important to note that assistantship duties should not interfere unduly with academic studies, but rather should contribute to students' intellectual growth and degree goals. If you fell as if expectations of your assistantship are interfering with or conflicting with your studies, speak with your advisor or a department faculty member about your options.
Welcome to the PhDStudent Blogosphere, Ryan!
My name is Ryan, and I am currently in my final year of graduate school in the Public Policy and Political Economy doctoral program at the University of Texas at Dallas. My policy focus is on international development with an area specialty in Latin America and the Caribbean issues. However, I do find myself looking to Africa to see what development trends seem to be working there. I am fascinated with the intersection of politics and economics and how they relate to the development trajectories of countries abroad. I am currently in All But Dissertation status and am in the final stages of writing my dissertation with hopes to defend by January 2016. My dissertation is about how to conceptualize and measure rule of law in a new way and to systematically test what factors predict this conception of rule of law in a global data set.
Graduate school can be overwhelming, as most of you already know. There are classes and seminars to attend, research to do, labs to complete, exams to study for, and comps to take. I'm sure if there was something that could make your routine easier, you'd be up for it, right? Well, here's your chance to simplify your life, if only just a little bit, by learning which method of note-taking will work best for you.
Feel like your advisor is working you into the ground? Perhaps you have come to the conclusion that graduate programs have complete disregard for their students' personal lives and that they intentionally and ruthlessly work their students like slaves. I won't attempt to deny these accusations, but I do believe in seeing the bright side to any unfortunate situation. Below I've outlined a little pep talk to encourage you on your quest for that coveted postgraduate degree.
It seems that technology is advancing with each passing day, especially when it comes to our mobile phones. Smartphones have become such a part of our normal routine that we tend to panic when we forget them at home, when they run out of batteries, or when we drop them and crack the screens.
I'm basing my next series on these gadgets and something that we like to download onto them: applications. Apps, much like our beloved smartphones, come in all shapes and sizes; some help you get organized, and others are more educational. I'll include these topics and more in the upcoming series.
Students in grad school tend to burn out at one point or another as a result of losing interest in their thesis/dissertation topics or even their diminishing mental stamina. The learning apps in this post will include information and subjects for users to learn something new apart from their academic subjects that they've been learning for years.
I've been using Evernote, Adobe Reader, and Kindle Reader in the proposal drafting process. Super useful.
We've all been there: sick in bed and unsure if it would be good to grin and bear it and carry on about our day, or call in sick to stay at home and rest. Many graduate students probably end up ignoring their sicknesses and acting like they're okay, but this isn't always the best decision. There are plenty of reasons why you might want to think twice about going to classes, work, lab hours, etc. while you're sick: getting others sick if you're contagious, potentially getting sicker, and not allowing your body to rest and recuperate properly.
I understand not wanting to get sick when you have a ton of things going on as a college student, but the answer to getting better will usually not consist of continuing to do everything on your schedule. Sometimes, we get sick because our body needs us to slow down, take a break, and take better care of ourselves. I'm not telling you to ignore your priorities as a student (and any other roles you play in your professional and personal life), but give yourself some time every month or so to relax and not worry about grad school stresses. One of our other bloggers, John, even wrote about this subject last March; take a look at what he had to say.
As the infographic above describes, there are plenty of ways to stay healthy and allow your immune system to do its job in protecting your body. However, things we do to prevent sickness might not work all the time. If you happen to get ill, think about taking a day or two to recuperate so you can get back to being 100% again. Be careful with ignoring your symptoms and going about your days because you could end up worse and have to spend even more time to heal; plus, you could get classmates and coworkers sick, which they would not appreciate.
I recently wrote that the major ways to handle schedule changes in grad school are to set your priorities, budget your time, and know your limits. I also discussed that managing your time, maintaining your relationships, and creating new relationships are ways you can adjust to social changes in grad school. But what if your new schedule and changed social life get to be too much?
An adjustment change that many graduate students have to go through is figuring out how to balance their social lives and academic lives. Undergraduates have plenty of opportunities to be social, but grad students have a much heavier work load, and therefore, less time. However, there are still a number of ways you can have a healthy social life while in grad school.
I hope you've all been enjoying this blog series about adjusting to graduate school. Today's post is about scheduling changes: what you should expect in grad school and how you can transition well. When you become a grad student, you might feel that you're not able to fit everything you need to do within your week, but here are a few ways that you can alleviate this feeling.
Adjusting to graduate school is challenging for many reasons, but many students would probably tell you that the change that has brought the most difficulties is money. We already know that it's difficult just to pay for grad school, as described in our articles about paying for graduate school. However, I want to focus on how to deal with the financial adjustments that must come while attending grad school.
How were your study habits during undergraduate school? It might surprise you to know that even if you earned straight As, you'll need to adjust your studying habits according to your grad program. For the next 2–5 years, you'll be faced with plenty of challenges, so it'll be best if you're prepared and already know about some differences between undergrad and grad school. Here are some helpful tips that you can use, whether you just started grad school, you're almost done with grad school, or if grad school is just a twinkle in your eye.
I hope you've been able to read through my current blog series about adjusting to life as a graduate student and getting used to the many changes that occur when you come from either undergraduate school or the work force and start your graduate career. As part of the transition, sleeping schedules tend to change and, most of the time, worsen, the more time you're in grad school. However, there are plenty of ways to adjust your sleeping patterns, no matter what your usual sleeping schedule looks like. If you didn't get a chance to read the first part of adjusting to graduate school by getting better sleeping habits, then take a look at it HERE. For this post, I want to give some healthy tips of how you can break those sleeping habits in healthy ways.
When you transition into graduate school from either your undergrad or life in the field, you will probably find that your sleeping habits change and become unhealthy because of late nights studying and early morning classes. I want to give you two different ways to better the way you sleep in grad school: how to make your sleeping habits more consistent and how to temporarily break your habits in a healthy way. In this post, I'm going to address how you can make better sleeping habits for yourself.
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\section{Introduction}
Let $M \subset \R^n$ be an open, bounded and connected set with a smooth boundary, and consider the wave equation
on $M$,
\begin{align}
\label{eq:wave_isotropic}
&\p_t^2 u(t,x) - c(x)^2 \Delta u(t,x) = 0,
\quad &(t,x) \in (0, \infty) \times M,
\\&u(0,x) = 0,\ \p_t u(0,x) = 0,
\quad &x \in M, \nonumber
\\&\p_\nu u(t,x) = f(t,x),
\quad &(t,x) \in (0, \infty) \times \p M, \nonumber
\end{align}
where $c$ is a smooth stricty positive function on $\bar M$, and $\p_\nu$ is the normal derivative on
the boundary $\p M$.
Denote the solution of (\ref{eq:wave_isotropic}) by $u^f(t,x) = u(t,x)$, let $T > 0$, and define the operator
\begin{equation}
\label{eq:dtn}
\Lambda_{2T} : f \mapsto u^f|_{(0,2T) \times \p M}.
\end{equation}
Operator $\Lambda_{2T}$ models boundary measurements and is called the Neumann-to-Dirichlet operator.
Let us assume that $c|_{\p M}$ is known but $c|_M$ is unknown.
The inverse problem for the wave equation is to reconstruct the wave speed $c(x)$, $x \in M$,
using the operator $\Lambda_{2T}$.
Let $\Gamma \subset \p M$ be open, $\tau \in C(\bar \Gamma)$, and
consider a wave source
$f$ in $L^2((0,\infty) \times \p M)$
satisfying the support condition
\begin{equation}
\label{eq:source_supp_condition}
\supp(f) \subset \{ (t, y) \in [0, T] \times \bar \Gamma;\ t \in [T - \tau(y), T] \}.
\end{equation}
By the finite speed of progation for the wave equation \cite{Ga, Mi},
the solution $u^f$ satisfies then the support condition,
\begin{equation}
\label{eq:sol_supp_condition}
\supp(u^f(T)) \subset \{x \in M;\ \text{there is $y \in \bar \Gamma$ such that $d(x,y) \le \tau(y)$}\},
\end{equation}
where $d(x,y)$ is the travel time between points $x$ and $y$, see (\ref{domain_of_influence}) below. Let us denote the set in (\ref{eq:sol_supp_condition}) by $M(\Gamma, \tau)$ and call it the domain of influence.
The contribution of this paper is twofold.
First, we present a method to compute the volume of $M(\Gamma, \min(\tau, T))$ using the operator $\Lambda_{2T}$.
The method works even when the wave speed is anisotropic, that is, when the wave speed is given by a Riemannian metric tensor $g(x) = (g_{jk}(x))_{j,k}^n$, $x \in \bar M$.
In the case of the isotropic wave equation (\ref{eq:wave_isotropic}) we have $g(x) = (c(x)^{-2} \delta_{jk})_{j,k}^n$.
Second, assuming that the Riemannian manifold $(\bar M, g)$ is simple, we show that the volumes of $M(\Gamma, \tau)$
for $\tau \in C(\p M)$ contain enough information to
determine the metric tensor $g$ up to a change of coordinates in $M$.
We recall the definition of a simple compact manifold below,
see Definition \ref{def:simple}.
In the case of the isotropic wave equation (\ref{eq:wave_isotropic}) we can
determine the wave speed $c$
in the Cartesian coordinates of $M$.
Our method to compute the volume of
the domain of influence
is a quadratic minimization scheme in $L^2((0, 2T) \times \p M)$ for the source
$f$ satisfying the support condition (\ref{eq:source_supp_condition}).
After a finite dimensional discretization,
an approximate minimizer
can be computed by solving
a positive definite system of linear equations.
We show that the system can be solved very efficiently if
we use an iterative method, such as the conjugate gradient method,
and intertwine measurements with computation.
In particular, instead of solving the equation (\ref{eq:wave_isotropic}) computationally in an iteration step, we measure $\Lambda_{2T} f$
for two sources $f$:
one is the approximate minimizer given by the previous iteration step and the other is related to the time-reversed version of the approximate minimizer, see (\ref{eq:cg_step_measurements}) below.
We believe that our intertwined algorithm is more robust against noise than an algorithm where noise is propagated by simulation of the wave equation.
Let us consider next the problem to determine the metric tensor $g$ given the volumes of $M(\p M, \tau)$ for all
$\tau \in C(\p M)$. Our approach exploits the fact that
$C(\p M)$ is a lattice with the natural partial order
\begin{equation}
\tau \le \sigma
\quad \text{if and only if} \quad
\tau(y) \le \sigma(y)\ \text{for all $y \in \p M$}.
\end{equation}
Let us define the greatest lower bound of $\tau$ and $\sigma$ in $C(M)$ as their pointwise minimum and
denote it by $\tau \wedge \sigma$.
We recall that a subset of $C(M)$ is a meet-semilattice if it is closed under the binary operation $\wedge$.
Let us define the {\em boundary distance functions},
\begin{equation}
\label{eq:boundary_distance_functions}
r_x : \p M \to [0, \infty), \quad r_x(y) := d(x,y),
\end{equation}
for $x \in \overline M$.
We show that the volumes of
$M(\p M, \tau)$, $\tau \in C(\p M)$, determine the meet-semilattice,
\begin{equation}
\label{eq:semilattice_QM_intro}
\overline{Q(M)} = \bigcup_{x \in \overline M}
\{ \tau \in C(M);\ \tau \le r_x \}.
\end{equation}
Moreover, we show that if $(\bar M, g)$ is simple
then the boundary distance functions are the maximal elements of $\overline{Q(M)}$.
The set of boundary distance functions
determines the Riemannian manifold $(M,g)$ \cite{Ku_proc, KKL}. Thus the volumes of $M(\p M, \tau)$, $\tau \in C(\p M)$, determine $(M,g)$ if it is simple.
Our results give a new uniqueness proof for the
inverse problem for the wave equation in the case of
a simple geometry. Belishev and Kurylev have proved the uniqueness even when the geometry is not simple \cite{BeKu}.
Their proof is based on the boundary control method
\cite{AKKLT, Be3, KK, KKLima, Pestov}, originally developed for the isotropic wave equation \cite{Be}.
Our uniqueness proof might be the first systematic use of lattice structures in the context of inverse boundary value problems.
In previous literature, $M(\Gamma, \tau)$ has been defined in the case of a constant function $\tau$, see e.g. \cite{KKL} and the references therein.
In this paper we establish some properties of $M(\Gamma, \tau)$ when $\Gamma \subset \p M$ is open and $\tau \in C(\bar \Gamma)$.
In particular, we show that its boundary is of measure zero.
This important detail seems to be neglected in previous literature also in the case of a constant function $\tau$.
Our method to compute the volume of a domain of influence is related to the iterative time-reversal control method
by Bingham, Kurylev, Lassas and Siltanen \cite{ITRC}.
Their method produces a certain kind of focused waves,
and they also prove uniqueness for the
inverse problem for the wave equation using these waves. Moreover, they give a review of
methods that use time-reversed measurements
\cite{Bardos, Bardos2, Papa1, CIL, FinkD, FinkMain, Kliba}.
A modification of the iterative time-reversal control method is presented in \cite{DKL}.
\section{Main results}
Let $(M, g)$ be a $C^\infty$-smooth,
compact and connected Riemannian manifold
of dimension $n \ge 2$ with nonempty boundary $\p M$.
We consider the wave equation
\begin{align}\label{eq:wave}
&\p_t^2 u(t,x) + a(x,D_x) u(t, x) = 0, \quad (t,x) \in (0,\infty) \times M,
\\\nonumber& u|_{t=0} = 0, \quad \p_t u|_{t=0}=0,
\\\nonumber& b(x, D_x) u(t,x) = f(t,x), \quad (t,x) \in (0,\infty) \times \p M,
\end{align}
where $a(x, D_x)$ is a weighted Laplace-Beltrami operator and $b(x, D_x)$ is the
corresponding normal derivative.
In coordinates, $(g^{jk}(x))_{j,k=1}^n$ denotes the inverse of $g(x)$ and $|g(x)|$ the determinant of $g(x)$.
Then
\begin{align*}
a(x,D_x) u
&:= -\sum_{j,k=1}^n \mu(x)^{-1}|g(x)|^{-\frac 12}\frac {\p}{\p x^j}
\ll( \mu(x)|g(x)|^{\frac 12}g^{jk}(x)\frac {\p u}{\p x^k} \rr),
\\b(x, D_x) u
&:= \sum_{j,k=1}^n \mu(x)g^{jk}(x) \nu_k(x) \frac{\p u}{\p x^j},
\end{align*}
where $\mu$ is a $C^\infty$-smooth strictly positive weight function and
$\nu = (\nu_1, \dots, \nu_n)$ is the exterior co-normal vector of $\p M$
normalized with respect to $g$, that is $\sum_{j,k=1}^m g^{jk}\nu_j\nu_k=1$.
The isotropic wave equation (\ref{eq:wave_isotropic}) is a special case of (\ref{eq:wave})
with $g(x) := (c(x)^{-2} \delta_{jk})_{j,k=1}^n$ and $\mu(x) = c(x)^{n-2}$.
We denote the indicator function of a set $A$ by $1_A$,
that is, $1_A(x) = 1$ if $x \in A$ and $1_A(x) = 0$
otherwise.
Moreover, we denote
\begin{equation} \label{eq:integration_triangle_L}
L := \{ (t,s) \in \R^2; t + s \le 2 T,\ s > t > 0 \},
\end{equation}
and define the operators
\begin{align*}
&J f(t) := \frac{1}{2} \int_0^{2 T} 1_L(t,s) f(s) ds, \quad
R f(t) := f(2 T - t), \quad
\\&K := J \Lambda_{2T} - R \Lambda_{2T} R J, \quad
I f (t) := 1_{(0,T)}(t) \int_0^t f(s) ds,
\end{align*}
where $\Lambda_{2T}$ is the operator defined by (\ref{eq:dtn}) $u^f$ being the solution of (\ref{eq:wave}).
We denote by $dS_g$ the Riemannian volume measure of the manifold $(\p M, g|_{\p M})$.
Furthermore, we denote by $(\cdot, \cdot)$ and $\norm{\cdot}$
the inner product and the norm of $L^2((0, 2T) \times \p M; dt \otimes dS_g)$.
We study the regularized minimization problem
\begin{equation}
\label{eq:minimization_regularized}
\argmin_{f \in S} \ll ((f, K f) - 2(I f, 1) + \alpha \norm{f}^2 \rr),
\end{equation}
where the regularization parameter $\alpha$ is strictly positive and $S$
is a closed subspace of $L^2((0, 2T) \times \p M)$.
Operator $a(x, D_x)$ with the domain $H^2(M) \cap H^1_0(M)$ is self-adjoint on
the space $L^2(M; dV_\mu)$, where $dV_\mu = \mu |g|^{1/2} dx$ in coordinates.
Thus we call $dV_\mu$ the natural measure corresponding to $a(x, D_x)$
and denote it also by $m$.
In \cite{ITRC} it is shown that
\begin{equation}
\label{eq:inner_products}
(u^f(T), u^h(T))_{L^2(M; dV_\mu)}
= (f, K h).
\end{equation}
This is a reformulation of the Blagovestchenskii identity \cite{Bl}.
In Lemma \ref{lem:cross_term} we show the following identity
\begin{equation}
\label{eq:inner_product_with_1}
(u^f(T), 1)_{L^2(M; dV_\mu)} = (I f, 1).
\end{equation}
This is well known at least in the isotropic case, see e.g. \cite{Be2}.
The equations (\ref{eq:inner_products}) and (\ref{eq:inner_product_with_1}) imply that
\begin{align}
\label{eq:minimization}
(f, K f) - 2(I f, 1)
&= \norm{u^{f}(T)}_{L^2(M; dV_\mu)}^2 - 2 (u^{f}(T), 1)_{L^2(M; dV_\mu)}
\\\nonumber&=
\norm{u^{f}(T) - 1}_{L^2(M; dV_\mu)}^2 + C,
\end{align}
where $C = -\norm{1}_{L^2(M; dV_\mu)}^2$ does not depend on the source $f$.
Thus the minimization problem (\ref{eq:minimization_regularized}) is equivalent with the
minimization problem
\begin{equation*}
\argmin_{f \in S} \ll( \norm{u^{f}(T) - 1}_{L^2(M; dV_\mu)}^2 + \alpha \norm{f}^2 \rr).
\end{equation*}
For $\Gamma \subset M$ and $\tau : \bar \Gamma \to \R$, we define {\em the domain of influence},
\begin{equation}
\label{domain_of_influence}
M(\Gamma, \tau) := \{x \in M;\ \text{there is $y \in \bar \Gamma$ such that $d(x,y) \le \tau(y)$}\},
\end{equation}
where $d$ is the distance on the Riemannian manifold $(M,g)$.
In Section \ref{sec:regularization} we show the following two theorems.
\begin{theorem}
\label{thm:minimization_on_subspace}
Let $\alpha > 0$ and let $S \subset L^2( (0,2T) \times \p M)$ be a closed subspace.
Denote by $P$ the orthogonal projection
\begin{equation*}
P : L^2( (0,2T) \times \p M) \to S.
\end{equation*}
Then the regularized minimization (\ref{eq:minimization_regularized}) has unique minimizer $f_\alpha \in S$,
and $f_\alpha$ is the unique $f \in S$ solving
\begin{equation}
\label{eq:normal}
(PKP + \alpha) f = P I^+ 1,
\end{equation}
where $I^+$ is the adjoint of $I$ in $L^2( (0,2T) \times \p M)$.
Moreover, $PKP + \alpha$ is positive definite on $S$.
\end{theorem}
\begin{theorem}
\label{thm:indicator_functions}
Let $\Gamma \subset \p M$ be open, $\tau \in C(\bar \Gamma)$ and define
\begin{equation*}
S = \{f \in L^2((0, 2T) \times \p M);\ \text{$\supp(f)$ satisfies (\ref{eq:source_supp_condition})}\}.
\end{equation*}
Let $f_\alpha$, $\alpha > 0$,
be the minimizer in Theorem \ref{thm:minimization_on_subspace}.
Then in $L^2(M)$
\begin{equation*}
\lim_{\alpha \to 0} u^{f_\alpha}(T)
= 1_{M(\Gamma, \tau \wedge T)}.
\end{equation*}
\end{theorem}
We denote $m_\tau := m(M(\p M, \tau))$, for $\tau \in C(\p M)$, and $m_\infty := m(M)$.
Moreover, we define
\begin{align}
Q(M) &:= \{ \tau \in C(\p M);\ m_\tau < m_\infty \}, \label{eq:meet_semilattice_QM}
\\R(M) &:= \{ r_x \in C(\p M);\ x \in M \}, \nonumber
\end{align}
where $r_x$ is the boundary distance function defined by
(\ref{eq:boundary_distance_functions}) $d$ being the distance on the Riemannian manifold $(M,g)$.
We denote by $\overline{Q(M)}$ the closure of $Q(M)$ in $C(M)$.
In Section \ref{sec:maximal_elements} we prove the
equation (\ref{eq:semilattice_QM_intro}) and show
the following theorem.
\begin{theorem}
\label{thm:maximal_elements}
If $(M,g)$ satisfies the condition
\begin{itemize}
\item[(G)] $x_1, x_2 \in M$ and $r_{x_1} \le r_{x_2}$ imply $x_1 = x_2$,
\end{itemize}
then $R(M)$ is the set of maximal elements of $\overline{Q(M)}$.
\end{theorem}
Let $T \ge \max\{ d(x, y) ;\ x \in M,\ y \in \p M \}$.
Then the set of volumes,
\begin{equation}
\label{eq:volumes_for_uniqueness}
\mathcal V := \{m_\tau;\ \tau \in C(\p M)\ \text{and $0 \le \tau \le T$} \},
\end{equation}
determines the set $Q(M)$. Note that $r_x(y) \le T$, for all $x \in M$ and all $y \in \p M$, and that $m_\infty = \max \mathcal V$.
Moreover, by Theorem \ref{thm:indicator_functions} and equation (\ref{eq:inner_products}) we can compute the volume $m_\tau \in \mathcal V$ as the limit
\begin{equation}
\label{eq:volumes_via_minimizers}
m_\tau = \lim_{\alpha \to 0} (f_\alpha, K f_\alpha).
\end{equation}
The set $R(M)$ determines the manifold $(M,g)$ up to an isometry \cite{Ku_proc, KKL}.
Hence the volumes (\ref{eq:volumes_for_uniqueness}) contain enough information to
determine the manifold $(M,g)$ in the class of manifolds satisfying (G).
In section \ref{sec:maximal_elements}, we show that simple manifolds satisfy (G).
\begin{definition}
\label{def:simple}
A compact Riemannian manifold $(M, g)$ with boundary is {\em simple}
if it is simply connected, any geodesic has no conjugate points and
$\p M$ is strictly convex with respect to the metric $g$.
\end{definition}
Let us discuss Theorem \ref{thm:minimization_on_subspace} from the point of view of practical computations.
When the subspace $S$ is finite-dimensional, the positive definite system of linear equations (\ref{eq:normal})
can be solved using the conjugate gradient method.
In each iteration step of the conjugate gradient method we must evaluate one matrix-vector product.
In our case, the product can be realized by two measurements
\begin{equation}
\label{eq:cg_step_measurements}
\Lambda_{2 T}f, \quad \Lambda_{2 T} R J f,
\end{equation}
where $f$ is the approximate solution given by the previous iteration step.
The remaining computational part of the iteration step consists of a few inexpensive vector-vector operations.
Thus if we intertwine computation of a conjugate gradient steps with measurements (\ref{eq:cg_step_measurements}),
the computational cost of our method is very low.
\section{The open and the closed domain of influence}
\label{sec:domains_of_influence}
Let us recall that the domain of influence $M(\Gamma, \tau)$
is defined in (\ref{domain_of_influence}) for $\Gamma \subset M$ and $\tau : \bar \Gamma \to \R$.
We call $M(\Gamma, \tau)$ also the {\em closed} domain of influence and
define the {\em open} domain of influence
\begin{equation*}
M^0(\Gamma, \tau) := \{ x \in M;\ \text{there is $y \in \Gamma$ s.t. $d(x,y) < \tau(y)$} \}.
\end{equation*}
Let us consider the closed domain of influence $M(\Gamma, \tau)$ when $\Gamma \subset \p M$ is open and $\tau$ is a constant.
Finite speed of propagation for the wave equation guarantees that the solution $u^f$ at time $T$
is supported on $M(\Gamma, \tau)$ whenever the source $f$ satisfies the support condition
(\ref{eq:source_supp_condition}).
Moreover, using Tataru's unique continuation result \cite{Ta1, Ta2}, it is possible to show that the set of functions,
\begin{equation*}
\{ u^f(T);\ \text{$f \in L^2((0,2T) \times \p M)$ and $\supp(f)$ satisfies (\ref{eq:source_supp_condition})}\},
\end{equation*}
is dense in $L^2(M^0(\Gamma, \tau))$, see e.g. the proof of Theorem 3.16 and the orthogonality argument of Theorem 3.10 in \cite{KKL}.
It is easy generalize this for $\tau$ of form
\begin{equation*}
\tau(y) = \sum_{j=1}^N T_j 1_{\Gamma_j}(y), \quad y \in \p M,
\end{equation*}
where $N \in \N$, $T_j \in \R$ and $\Gamma_j \subset \p M$ are open, see \cite{ITRC}.
However, the fact that $M(\Gamma, \tau) \setminus M^0(\Gamma, \tau)$ is of measure zero, seems to go unproven in the literature.
In this section we show that this is indeed the case even for $\tau \in C(\bar \Gamma)$.
To our knowledge, this can not be proven just by considering the boundaries of the balls $B(y, \tau(y))$,
$y \in \Gamma$.
In fact, we give below an example showing that the union of the boundaries $\p B(y, \tau(y))$, for $y \in \p \Gamma$,
can have positive measure.
\begin{example}
Let $\mathcal C$ be the fat Cantor set, $M \subset \R^2$ be open, $g$ be the Euclidean metric,
$(0,1) \times \{0\} \subset \p M$ and $\Gamma = \ll( (0,1) \setminus \mathcal C \rr) \times \{0\}$.
Then the union $B := \bigcup_{y \in \p \Gamma} \p B(y, 1)$ has positive measure.
\end{example}
\begin{proof}
The fat Cantor set $\mathcal C$ is an example of a closed subset of $[0,1]$ whose boundary has positive measure, see e.g. \cite{Pugh}.
The map
\begin{equation*}
\Phi : (s, \alpha) \mapsto (s + \cos \alpha, s + \sin \alpha)
\end{equation*}
is a diffeomorphism from $\R \times (0, \pi/2)$ onto its image in $\R^2$.
The image of $H := \p \mathcal C \times (0, \pi/2)$ under $\Phi$
lies in $B$.
As $H$ has positive measure so has $B$.
\end{proof}
\begin{lemma}
\label{lem:characterization_of_domi}
Let $\Gamma \subset \p M$ be open and let $\tau \in C(\bar \Gamma)$.
Then the function
\begin{equation*}
r_{\Gamma, \tau}(x) := \inf_{y \in \Gamma} (d(x, y) - \tau(y))
\end{equation*}
is Lipschitz continuous and
\begin{align}
M(\Gamma, \tau) &= \{ x \in M;\ r_{\Gamma, \tau}(x) \le 0 \}, \label{eq:closed_domi_r}
\\M^0(\Gamma, \tau) &= \{ x \in M;\ r_{\Gamma, \tau}(x) < 0 \}. \label{eq:open_domi_r}
\end{align}
In particular, $M(\Gamma, \tau)$ is closed and $M^0(\Gamma, \tau)$ is open.
\end{lemma}
\begin{proof}
Let us define
\begin{equation*}
r(x) := r_{\Gamma, \tau}(x), \quad \tilde r(x) := \min_{y \in \bar \Gamma} (d(x, y) - \tau(y)),
\end{equation*}
and show that $\tilde r = r$. Clearly $\tilde r\le r$.
Let $x \in M$. The minimum in the definition of $\tilde r(x)$ is attained at a point $y_0 \in \bar \Gamma$.
We may choose a sequence $(y_j)_{j=1}^\infty \subset \Gamma$ such that $y_j \to y_0$ as $j \to \infty$.
Then
\begin{equation*}
r(x) \le d(x, y_j) - \tau(y_j) \to \tilde r(x)
\quad \text{as $j \to \infty$}.
\end{equation*}
Hence $\tilde r = r$.
Let us show that $\tilde r$ is Lipschitz.
Let $x \in M$, and let $y_0$ be as before.
Let $x' \in M$. Then
\begin{equation*}
\tilde r(x') - \tilde r(x)
\le d(x', y_0) - \tau(y_0) - \ll( d(x, y_0) - \tau(y_0) \rr) \le d(x, x').
\end{equation*}
By symmetry with respect to $x'$ and $x$, $\tilde r$ is Lipschitz.
Let us show (\ref{eq:closed_domi_r}).
Clearly $r(x) \le 0$ for $x \in M(\Gamma, \tau)$.
Let $x \in M$ satisfy $r(x) \le 0$, and let $y_0$ be as before.
Then
\begin{equation*}
d(x, y_0) - \tau(y_0) = \tilde r(x) = r(x) \le 0,
\end{equation*}
and $x \in M(\Gamma, \tau)$. Hence (\ref{eq:closed_domi_r}) holds.
The equation (\ref{eq:open_domi_r}) can be proven in a similar way.
\end{proof}
If $\Gamma \subset \p M$ and $\tau$ is a constant function, then
\begin{align*}
M(\Gamma, \tau) &= \{ x \in M;\ r_{\Gamma, \tau}(x) \le 0 \}
= \{ x \in M;\ \inf_{y \in \Gamma} d(x,y) \le \tau \}
\\&= \{ x \in M;\ d(x,\Gamma) \le \tau \}.
\end{align*}
Thus for a constant $\tau$, our definition of $M(\Gamma, \tau)$ coincides with the definition
of the domain of influence in \cite{KKL}.
\begin{lemma}
\label{lem:level_set_is_null}
Let $A \subset M$ be compact and let $\tau : A \to \R$ be continuous.
We define
\begin{equation*}
r(x) := \inf_{y \in A} (d(x,y) - \tau(y)), \quad x \in M.
\end{equation*}
If $\tau$ is strictly positive on $A$ or $A$ is a null set, then $\{x \in M;\ r(x) = 0\}$
is a null set.
We mean by a null set a set of measure zero with respect to the Riemannian volume measure.
\end{lemma}
\begin{proof}
Denote by $V_g$ the volume measure of $M$ and define
\begin{equation*}
Z := \{p \in M;\ r(p) = 0\}.
\end{equation*}
Let us show that
\begin{equation} \label{eq:A_does_not_matter}
V_g(Z) = V_g(Z \setminus A).
\end{equation}
If $V_g(A) = 0$, then (\ref{eq:A_does_not_matter}) is immediate.
If $\tau > 0$, then
\begin{equation*}
r(q) \le d(q,q) - \tau(q) = -\tau(q) < 0, \quad q \in A.
\end{equation*}
Hence $Z \cap A = \emptyset$ and (\ref{eq:A_does_not_matter}) holds.
Let $p \in M^\text{int}$. There is a chart $(U, \phi)$ of $M^\text{int}$ such that
$\phi(p) = 0$ and that the closure of the open Euclidean unit ball $B$ of $\R^n$
is contained in $\phi(U)$.
We denote $U_p := \phi^{-1}(B)$.
The sets $U_p$, $p \in M^\text{int}$, form an open cover for $M^\text{int}$,
and as $M^\text{int}$ is second countable,
there is a countable cover $U_{p_j}$, $j = 1, 2, \dots$, of $M^\text{int}$.
Hence
\begin{equation*}
V_g(Z) = V_g( (Z \setminus A) \cap (\p M \cup \bigcup_{j=1}^\infty U_{p_j}))
\le \sum_{j=1}^\infty V_g( (Z \setminus A) \cap U_{p_j}).
\end{equation*}
It is enough to show that $\phi((Z \setminus A) \cap U_p)$
is a null set with respect to the Lebesgue measure on $B$.
We define for $v = (v^1, \dots, v^n) \in \R^n$ and $x \in B$,
\begin{equation*}
|v|_{g(x)}^2 := \sum_{j,k = 1}^n v^j g_{jk}(x) v^k,
\quad |v|^2 := \sum_{j = 1}^n (v^j)^2,
\end{equation*}
where $(g_{jk})_{j,k=1}^n$ is the metric $g$ in the local coordinates on $\phi(U)$.
As $\overline B$ is compact in $\phi(U)$, there is $c_p > 0$ such that
for all $v \in \R^n$ and $x \in B$
\begin{equation*}
c_p |v|_{g(x)} \le |v| \le \frac{1}{c_p} |v|_{g(x)}.
\end{equation*}
As in the proof of Lemma \ref{lem:characterization_of_domi}
we see that $r$ is Lipschitz continuous on $M$.
Thus by Rademacher's theorem there is a null set $N \subset B$ such that
$r$ is differentiable in the local coordinates in $B \setminus N$.
We denote $Z_p := \phi((Z \setminus A) \cap U_p) \setminus N$.
Let $x \in Z_p$ and denote $p_x := \phi^{-1}(x)$.
As $A$ is compact and $q \mapsto d(p_x, q) - \tau(q)$
is continuous, there is $q_x \in A$ such that
\begin{equation*}
d(p_x, q_x) - \tau(q_x) = r(p_x) = 0.
\end{equation*}
We denote $s := d(p_x, q_x) = \tau(q_x)$.
As $p_x \notin A$ and $q_x \in A$, we have that $0 < s$.
As $M$ is connected and complete as a metric space, Hopf-Rinow theorem gives
a shortest path $\gamma : [0, s] \to M$
parametrized by arclength and joining $q_x = \gamma(0)$ and $p_x = \gamma(s)$.
For a study of shortest paths on Riemannian manifolds with boundary
see \cite{Ax2}.
As $p_x \in U_p$, there is $a \in (0, s)$ such that $\gamma|_{[a , s]}$
is a unit speed geodesic of $U_p \subset M^\text{int}$.
As $\gamma$ is parametrized by arclength,
\begin{equation*}
r(\gamma(t)) \le d(\gamma(t), q_x) - \tau(q_x) = t - s, \quad t \in [a, s],
\end{equation*}
and as $r(\gamma(s)) = r(p_x) = 0$,
\begin{equation*}
\frac{r(\gamma(t)) - r(\gamma(s))}{t - s} \ge \frac{t - s - 0}{t - s} = 1, \quad t \in (a,s).
\end{equation*}
The function $r \circ \gamma$ is differentiable at $s$ by the chain rule, and
\begin{equation*}
\p_t (r \circ \gamma)(s)
= \lim_{t \to s^-} \frac{r(\gamma(t)) - r(\gamma(s))}{t - s} \ge 1.
\end{equation*}
As $\gamma$ is a unit speed geodesic near $s$,
\begin{equation*}
c_p = c_p |\p_t \gamma(s)|_{g(x)} \le |\p_t \gamma(s)|
\le \frac{1}{c_p}|\p_t \gamma(s)|_{g(x)} = \frac{1}{c_p}.
\end{equation*}
Hence in the local coordinates in $B$
\begin{equation*}
|D r(x)| \ge D r(x) \cdot \frac{\p_t \gamma(s)}{|\p_t \gamma(s)|}
\ge c_p \p_t (r \circ \gamma)(s) \ge c_p.
\end{equation*}
Let $\epsilon > 0$.
There is $\delta(x) > 0$ such that
\begin{equation}
\label{eq:r_derivative_approximation}
|r(y) - r(x) - D r(x) \cdot (y - x)| \le c_p \epsilon |y - x|, \quad y \in B(x, \delta(x)),
\end{equation}
and $B(x, \delta(x)) \subset B$.
Here $B(x, \delta)$ is the open Euclidean ball with center $x$ and radius $\delta$.
The sets $B(x, \delta(x)/5)$, $x \in Z_p$, form an open cover for $Z_p$,
and as $Z_p$ is second countable,
there is $(x_j)_{j=1}^\infty \subset Z_p$ such that the sets
\begin{equation*}
B_j' := B(x_j, \delta(x_j)/5)
\end{equation*}
form an open cover for $Z_p$.
By Vitali covering lemma there is an index set $J \subset \N$ such that
the sets $B_j'$, $j \in J$, are disjoint and the sets
\begin{equation*}
B_j := B(x_j, \delta(x_j)), \quad j \in J,
\end{equation*}
form an open cover for $Z_p$.
We denote
\begin{equation*}
v_j := \frac{D r(x_j)}{|D r(x_j)|}, \quad \delta_j := \delta(x_j).
\end{equation*}
If $y \in Z_p \cap B_j$, then $r(y) = 0 = r(x_j)$ and by (\ref{eq:r_derivative_approximation})
\begin{equation*}
|v_j \cdot (y - x_j)| \le \frac{1}{|D r(x_j)|} c_p \epsilon |y - x_j| \le \epsilon \delta_j.
\end{equation*}
We denote by $\alpha_m$ the volume of the open Euclidean unit ball in $\R^m$
and by $V$ the Lebesgue measure on $B$.
Let $j \in J$.
Using a translation and a rotation we get such
coordinates that $x_j = 0$ and $v_j = (1, 0, \dots, 0)$.
In these coordinates
\begin{equation*}
Z_p \cap B_j \subset \{ (y^1, y') \in \R \times \R^{n-1};\ |y^1| \le \epsilon \delta_j, |y'| \le \delta_j \}.
\end{equation*}
Hence $V(Z_p \cap B_j) \le 2 \epsilon \delta_j \alpha_{n-1} \delta_j^{n-1}$.
Particularly,
\begin{equation*}
V(Z_p \cap B_j)
\le \epsilon \frac{2 \alpha_{n-1}}{\alpha_{n}} V(B_j)
= \epsilon c_n V(B_j'),
\end{equation*}
where $c_n := 2 \cdot 5^n \alpha_{n-1} / \alpha_{n}$.
Then
\begin{align*}
V(\phi((Z \setminus A) \cap U_p))
&= V(Z_p) = V(Z_p \cap \bigcup_{j \in J} B_j)
\le \sum_{j \in J} V(Z_p \cap B_j)
\\&\le \epsilon c_n \sum_{j \in J} V(B_j')
= \epsilon c_n V(\bigcup_{j \in J} B_j')
\le \epsilon c_n V(B).
\end{align*}
As $\epsilon > 0$ is arbitrary, $V(\phi((Z \setminus A) \cap U_p)) = 0$ and the claim is proved.
\end{proof}
\section{Approximately constant wave fields on a domain of influence}
\label{sec:regularization}
\begin{lemma}
\label{lem:cross_term}
Let $f \in L^2((0, 2T) \times \p M)$. Then the equation (\ref{eq:inner_product_with_1}) holds.
\end{lemma}
\begin{proof}
The map $h \mapsto u^h(T)$ is bounded $L^2((0, 2T) \times \p M) \to L^2(M)$, see e.g. \cite{LaTr}.
Thus is it enough to prove the equation (\ref{eq:inner_product_with_1}) for $f \in C_c^\infty( (0, 2T) \times \p M)$.
Let us denote
\begin{equation*}
v(t) := (u^f(t), 1)_{L^2(M; dV_\mu)}.
\end{equation*}
As $a(x,D_x) 1 = 0$ and $b(x,D_x)1 = 0$,
we may integrate by parts
\begin{align*}
\p_t^2 v(t) &= - (a(x, D_x) u^f(t), 1)_{L^2(M; dV_\mu)}
\\&= - \ll( (a(x, D_x) u^f(t), 1)_{L^2(M; dV_\mu)} - (u^f(t), a(x, D_x) 1)_{L^2(M; dV_\mu)} \rr)
\\&= (b(x, D_x) u^f(t), 1)_{L^2(\p M; dS_g)} - (u^f(t), b(x, D_x) 1)_{L^2(\p M; dS_g)}
\\&= (f(t), 1)_{L^2(\p M; dS_g)}.
\end{align*}
As $\p_t^j v(0) = 0$ for $j=0,1$,
\begin{align*}
v(T) &= \int_0^T \int_0^t \int_{\p M} f(s,x) dS_g(x) ds dt
\\&= \int_0^{2T} \int_{\p M} 1_{(0,T)}(t) \int_0^t f(s,x) ds dS_g(x) dt.
\end{align*}
\end{proof}
The proof of Theorem \ref{thm:minimization_on_subspace}
is similar to the proof of the corresponding result in \cite{ITRC}.
We give the proof for the sake of completeness.
\begin{proof}[Proof of Theorem \ref{thm:minimization_on_subspace}.]
We define
\begin{equation*}
E(f) := (f, Kf) - 2 (If, 1) + \alpha \norm{f}^2.
\end{equation*}
Then
\begin{equation}
\label{eq:energy_functional}
E(f) = \norm{u^{f}(T) - 1}_{L^2(M; dV_\mu)}^2 - \norm{1}_{L^2(M; dV_\mu)}^2 + \alpha \norm{f}^2.
\end{equation}
Let $(f_j)_{j=1}^\infty \subset S$ be such that
\begin{equation*}
\lim_{j \to \infty} E(f_j) = \inf_{f \in S} E(f).
\end{equation*}
Then
\begin{equation*}
\alpha \norm{f_j} \le E(f_j) + \norm{1}_{L^2(M; dV_\mu)}^2,
\end{equation*}
and $(f_j)_{j=1}^\infty$ is bounded in $S$.
As $S$ is a Hilbert space,
there is a subsequence of $(f_j)_{j=1}^\infty$ converging weakly in $S$.
Let us denote the limit by $f_\infty \in S$ and the subsequence still by $(f_j)_{j=1}^\infty$.
The map $h \mapsto u^h(T)$ is bounded
\begin{equation*}
L( (0, 2T) \times \p M) \to H^{5/6 - \epsilon}(M)
\end{equation*}
for $\epsilon > 0$, see \cite{LaTr}.
Hence $h \mapsto u^h(T)$ is a compact operator
\begin{equation*}
L^2( (0, 2T) \times \p M) \to L^2(M),
\end{equation*}
and $u^{f_j}(T) \to u^{f_\infty}(T)$ in $L^2(M)$ as $j \to \infty$.
Moreover, the weak convergence implies
\begin{equation*}
\norm{f_\infty} \le \liminf_{j \to \infty} \norm{f_j}.
\end{equation*}
Hence
\begin{align*}
E(f_\infty) &= \lim_{j \to \infty} \norm{u^{f_j}(T) - 1}_{L^2(M; dV_\mu)}^2 - \norm{1}_{L^2(M; dV_\mu)}^2 + \alpha \norm{f_\infty}^2
\\&\le \lim_{j \to \infty} \norm{u^{f_j}(T) - 1}_{L^2(M; dV_\mu)}^2 - \norm{1}_{L^2(M; dV_\mu)}^2 + \alpha
\liminf_{j \to \infty} \norm{f_j}^2
\\&= \liminf_{j \to \infty} E(f_j) = \inf_{f \in S} E(f),
\end{align*}
and $f_\infty \in S$ is a minimizer.
Let $f_\alpha$ be a minimizer and $h \in S$.
By orthogonality of the projection $P$ and identity (\ref{eq:inner_products}),
it is clear that $PKP$ is self-adjoint and positive semidefinite.
Denote by $D_h$ the Fr\'echet derivative to direction $h$.
As $f_\alpha = Pf_\alpha$ and $h = Ph$
\begin{equation*}
0 = D_h E(f_\alpha) = 2 (h, PKP f_\alpha) - 2 (h, PI^+ 1) + 2 \alpha (h, f_\alpha).
\end{equation*}
Hence $f_\alpha$ satisfies (\ref{eq:normal}).
As $PKP$ is positive semidefinite, $PKP + \alpha$ is positive definite
and solution of (\ref{eq:normal}) is unique.
\end{proof}
\begin{lemma}
\label{lem:approximation_by_simple}
Let $\Gamma \subset \p M$ be open, $\tau \in C(\bar \Gamma)$ and $\epsilon > 0$.
Then there is a simple function
\begin{equation*}
\tau_\epsilon(y) = \sum_{j=1}^N T_j 1_{\Gamma_j}(y), \quad y \in \p M,
\end{equation*}
where $N \in \N$, $T_j \in \R$ and $\Gamma_j \subset \Gamma$ are open,
such that
\begin{align*}
\tau - \epsilon &< \tau_\epsilon \quad \text{almost everywhere on $\Gamma$ and}
\\\tau_\epsilon &< \tau \quad \text{on $\bar \Gamma$}.
\end{align*}
\end{lemma}
\begin{proof}
As $\p M$ is compact, there is a finite set of coordinate charts covering $\p M$.
Using partition of unity, we see that it is enough to prove the claim in the case
when $\Gamma \subset \R^{n-1}$ is an open set.
But then $\tau$ is a continuous function on a compact set $\overline \Gamma \subset \R^{n-1}$,
and it is clear that there is a simple function with the required properties.
\end{proof}
\begin{lemma}
\label{lem:measures_by_approximation}
Let $\Gamma \subset \p M$ be open, $\tau \in C(\bar \Gamma)$ and let $\tau_\epsilon$, $\epsilon > 0$, satisfy
\begin{align*}
\tau - \epsilon &< \tau_\epsilon \quad \text{almost everywhere on $\Gamma$ and}
\\\tau_\epsilon &< \tau \quad \text{on $\bar \Gamma$}.
\end{align*}
Then
\begin{equation*}
\lim_{\epsilon \to 0} m(M(\Gamma, \tau_\epsilon)) = m(M(\Gamma, \tau)).
\end{equation*}
\end{lemma}
\begin{proof}
Let $\epsilon > 0$ and denote by $N \subset \Gamma$
the set of measure zero where $\tau - \epsilon \ge \tau_\epsilon$ as functions on $\Gamma$.
Let us show that $M^0(\Gamma, \tau - \epsilon) \subset M(\Gamma, \tau_\epsilon)$.
Let $x \in M^0(\Gamma, \tau - \epsilon)$. Then there is $y_0 \in \Gamma$
such that
\begin{equation*}
d(x,y_0) < \tau(y_0) - \epsilon.
\end{equation*}
As $\tau$ and the function $y \mapsto d(x,y)$ are continuous
and $\Gamma \setminus N$ is dense in $\Gamma$, there is $y \in \Gamma \setminus N$ such that
\begin{equation*}
d(x,y) < \tau(y) - \epsilon < \tau_\epsilon(y).
\end{equation*}
Hence $x \in M(\Gamma, \tau_\epsilon)$.
A similar argument shows that $M(\Gamma, \tau_\epsilon) \subset M^0(\Gamma, \tau)$.
Clearly $M^0(\Gamma, \tau - \epsilon_1) \subset M^0(\Gamma, \tau - \epsilon_2)$ for $\epsilon_1 \ge \epsilon_2 > 0$,
and
\begin{equation*}
\bigcup_{\epsilon > 0} M^0(\Gamma, \tau - \epsilon) = M^0(\Gamma, \tau).
\end{equation*}
Hence $m(M^0(\Gamma, \tau - \epsilon)) \to m(M^0(\Gamma, \tau))$ as $\epsilon \to 0$, and
\begin{align*}
0 \le m(M^0(\Gamma, \tau)) - m(M(\Gamma, \tau_\epsilon)) &\le m(M^0(\Gamma, \tau)) - m(M^0(\Gamma, \tau - \epsilon))
\\&\to 0, \quad \text{as $\epsilon \to 0$}.
\end{align*}
Moreover, by Lemmas \ref{lem:characterization_of_domi} and \ref{lem:level_set_is_null}
\begin{equation*}
m(M(\Gamma, \tau)) = m(M^0(\Gamma, \tau)) = \lim_{\epsilon \to 0} m(M(\Gamma, \tau_\epsilon)).
\end{equation*}
\end{proof}
\begin{proof}[Proof of Theorem \ref{thm:indicator_functions}.]
We may assume without loss of generality that $\tau \le T$, as
we may replace $\tau$ by $\tau \wedge T$ in what follows.
Let us denote
\begin{equation*}
S(\Gamma, \tau) := \{ f \in L^2((0,2T) \times \p M);\ \text{$\supp(f)$ satisfies (\ref{eq:source_supp_condition})}\}.
\end{equation*}
By the finite speed of propagation for the wave equation,
we have that $\supp(u^f(T)) \subset M(\Gamma, \tau)$ whenever $f \in S(\Gamma, \tau)$.
Hence for $f \in S(\Gamma, \tau)$,
\begin{align}
\label{eq:splitting_the_difference_1}
\norm{u^f(T) - 1}_{L^2(M; dV_\mu)}^2
&= \int_{M(\Gamma, \tau)} (u^f(T) - 1)^2 dV_\mu + \int_{M \setminus M(\Gamma, \tau)} 1 dV_\mu
\\\nonumber&= \norm{u^f(T) - 1_{M(\Gamma, \tau)}}_{L^2(M; dV_\mu)}^2
+ m(M \setminus M(\Gamma, \tau)).
\end{align}
Let $\epsilon > 0$.
By Lemmas \ref{lem:approximation_by_simple} and \ref{lem:measures_by_approximation}
there is a simple function $\tau_\delta$ satisfying
\begin{equation*}
\tau_\delta < \tau, \quad m(M(\Gamma, \tau)) - m(M(\Gamma, \tau_\delta)) < \epsilon.
\end{equation*}
By the discussion in the beginning of Section \ref{sec:domains_of_influence}, the set
\begin{equation*}
\{ u^f(T) \in L^2(M(\Gamma, \tau_\delta));\ f \in S(\Gamma, \tau_\delta) \}
\end{equation*}
is dense in $L^2(M(\Gamma, \tau_\delta))$.
Thus there is $f \in S(\Gamma, \tau_\delta) \subset S(\Gamma, \tau)$ such that
\begin{equation*}
\norm{u^f(T) - 1_{M(\Gamma, \tau_\delta)}}_{L^2(M; dV_\mu)}^2 \le \epsilon.
\end{equation*}
Then
\begin{equation*}
\norm{u^f(T) - 1_{M(\Gamma, \tau)}}_{L^2(M; dV_\mu)}^2 \le
\epsilon + \norm{1_{M(\Gamma, \tau_\delta)} - 1_{M(\Gamma, \tau)}}_{L^2(M; dV_\mu)}^2
\le 2 \epsilon.
\end{equation*}
Moreover, $E(f_\alpha) \le E(f)$ and equations (\ref{eq:splitting_the_difference_1}) and (\ref{eq:energy_functional}) give
\begin{align*}
&\norm{u^{f_\alpha}(T) - 1_{M(\Gamma, \tau)}}_{L^2(M; dV_\mu)}^2
\\&\quad= \norm{u^{f_\alpha}(T) - 1}_{L^2(M; dV_\mu)}^2 - m(M \setminus M(\Gamma, \tau))
\\&\quad\le E(f_\alpha) + \norm{1}_{L^2(M; dV_\mu)}^2 - m(M \setminus M(\Gamma, \tau))
\\&\quad\le \norm{u^f(T) - 1}_{L^2(M; dV_\mu)}^2 - m(M \setminus M(\Gamma, \tau)) + \alpha \norm{f}^2
\\&\quad= \norm{u^f(T) - 1_{M(\Gamma, \tau)}}_{L^2(M; dV_\mu)}^2 + \alpha \norm{f}^2
\le 2 \epsilon + \alpha \norm{f}^2.
\end{align*}
We may choose first small $\epsilon > 0$ and then small $\alpha > 0$ to
get $u^{f_\alpha}(T)$ arbitrarily close to $1_{M(\Gamma, \tau)}$ in $L^2(M)$.
\end{proof}
\section{The boundary distance functions as maximal elements}
\label{sec:maximal_elements}
\def\tilde{\widetilde}
We denote $M^0(\tau) := M^0(\p M, \tau)$, for $\tau \in C(\p M)$, and define
\begin{equation*}
\tilde Q := \{ \tau \in C(M);\ M \setminus M^0(\tau) \ne \emptyset \}.
\end{equation*}
\begin{lemma}
\label{lem:maximal_elements}
If $\tau$ is a maximal element of $\tilde Q$, then $\tau = r_x$
for some $x \in M$.
Moreover, if the manifold $(M,g)$ satisfies (G),
then $R(M)$ is the set of the maximal elements of $\tilde Q$.
\end{lemma}
\begin{proof}
Let $x \in M$ and $\tau \in C(\p M)$. Then $\tau \le r_x$ if and only if $x \notin M^0(\tau)$.
In other words,
\begin{equation}
\label{eq:characterization_of_Q}
\tilde Q = \{ \tau \in C(\p M);\ \text{there is $x \in M$ such that $\tau \le r_x$} \}.
\end{equation}
Moreover, $r_x \in \tilde Q$ for all $x \in M$.
Indeed, $r_x$ is continuous and trivially $r_x \le r_x$.
Suppose that $\tau$ is a maximal element of $\tilde Q$.
By (\ref{eq:characterization_of_Q}) there is $x \in M$ such that $\tau \le r_x$,
but $r_x \in \tilde Q$ and maximality of $\tau$ yields $\tau = r_x$
Let us now suppose that $(M,g)$ satisfies (G) and show that $r_x$
is a maximal element of $\tilde Q$.
Suppose that $\tau \in \tilde Q$ satisfies $r_x \le \tau$.
By (\ref{eq:characterization_of_Q}) there is $x' \in M$ such that $\tau \le r_x'$.
Hence
\begin{equation*}
r_x \le \tau \le r_{x'},
\end{equation*}
and (G) yields that $x = x'$.
Thus $\tau = r_x$ and $r_x$ is a maximal element of $\tilde Q$.
\end{proof}
\begin{lemma}
\label{lem:closure}
The set $\tilde Q$ is the closure of $Q(M)$ in $C(M)$.
\end{lemma}
\begin{proof}
Let us first show that $\tilde Q$ is closed.
Let $(\tau_j)_{j=1}^\infty \subset \tilde Q$ satisfy $\tau_j \to \tau$ in $C(\p M)$ as $j \to \infty$.
By (\ref{eq:characterization_of_Q}) there is $(x_j)_{j=1}^\infty \subset M$ such that $\tau_j \le r_{x_j}$.
As $M$ is compact there is a converging subsequence $(x_{j_k})_{k=1}^\infty \subset (x_j)_{j=1}^\infty$.
Let us denote the limit by $x$, that is, $x_{j_k} \to x$ as $k \to \infty$.
By continuity of the distance function,
\begin{equation*}
\tau(y) = \lim_{k \to \infty} \tau_{j_k}(y) \le \lim_{k \to \infty} r_{x_{j_k}}(y) = r_x(y), \quad y \in \p M.
\end{equation*}
Hence $\tau \in \tilde Q$ and $\tilde Q$ is closed.
Clearly $Q(M) \subset \tilde Q$ and it is enough to show that $Q(M)$ is dense in $\tilde Q$.
Suppose that $\tau \in \tilde Q$. Then there is $x_0 \in M$ such that $\tau \le r_{x_0}$.
Let $\epsilon > 0$. As $M \times \p M$ is compact and the distance function is continuous,
there is $r > 0$ such that
\begin{equation*}
\sup_{y \in \p M} |d(x, y) - d(x_0, y)| < \epsilon, \quad \text{when $d(x, x_0) < r$ and $x \in M$}.
\end{equation*}
Hence $\tau(y) - \epsilon \le r_{x_0}(y) - \epsilon < r_x(y)$ for all $y \in \p M$ and all $x \in B(x_0, r)$.
In other words,
\begin{equation*}
B(x_0, r) \subset (M \setminus M(\tau - \epsilon)),
\end{equation*}
and this yields that $\tau - \epsilon \in Q(M)$.
Functions $\tau - \epsilon$ converge to $\tau$ in $C(\p M)$ as $\epsilon \to 0$. Thus $\tau$ is in the closure of $A$.
\end{proof}
Lemmas \ref{lem:maximal_elements} and \ref{lem:closure} together prove Theorem \ref{thm:maximal_elements}.
Moreover, (\ref{eq:characterization_of_Q})
yields the equation (\ref{eq:semilattice_QM_intro})
in the introduction.
\begin{lemma}
\label{lem:simple_manifold}
If $(M,g)$ is simple or the closed half sphere, then (G) holds.
\end{lemma}
\begin{proof}
Let $x_1, x_2 \in M$ satisfy $x_1 \ne x_2$, and let us show that $r_{x_1} \nleq r_{x_2}$.
First, if $x_2 \in \p M$ then $r_{x_1}(x_2) > 0 = r_{x_2}(x_2)$.
Second, if $x_2 \in M^{\text{int}}$, then there
is the unique unit speed geodesic $\gamma$ and the unique point $y \in \p M$ such that
$\gamma(0) = x_1$, $\gamma(s) = x_2$ and $\gamma(s') = y$,
where $0 < s < s'$.
As $\gamma$ is a shortest path
from $x_1$ to $y$ (the shortest path if $(M,g)$ is simple) and the shortest path from $x_2$ to $y$,
\begin{equation*}
r_{x_1}(y) = s' > s = r_{x_2}(y).
\end{equation*}
\end{proof}
The closed half sphere is not simple, since for a point on the boundary the corresponding
antipodal point is a conjugate point.
Hence the manifolds satisfying (G) form a strictly larger class than the simple manifolds.
\vspace{1cm}
{\em Acknowledgements.}
The author would like to thank Y. Kurylev for useful discussions.
The research was partly supported by Finnish Centre of Excellence in Inverse Problems Research,
Academy of Finland COE 213476,
and partly by Finnish Graduate School in Computational Sciences.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 2,758
|
using System;
using System.Collections.Generic;
using System.Text;
using System.Threading.Tasks;
using Proto.TestFixtures;
using Xunit;
namespace Proto.Tests
{
public class WatchTests
{
[Fact]
public async void CanWatchLocalActors()
{
var watchee = Actor.Spawn(Actor.FromProducer(() => new DoNothingActor())
.WithMailbox(() => new TestMailbox()));
var watcher = Actor.Spawn(Actor.FromProducer(() => new LocalActor(watchee))
.WithMailbox(() => new TestMailbox()));
watchee.Stop();
var terminatedMessageReceived = await watcher.RequestAsync<bool>("?", TimeSpan.FromSeconds(5));
Assert.True(terminatedMessageReceived);
}
public class LocalActor : IActor
{
private readonly PID _watchee;
private bool _terminateReceived;
public LocalActor(PID watchee)
{
_watchee = watchee;
}
public Task ReceiveAsync(IContext context)
{
switch (context.Message)
{
case Started _:
context.Watch(_watchee);
break;
case string msg when msg == "?":
context.Sender.Tell(_terminateReceived);
break;
case Terminated msg:
_terminateReceived = true;
break;
}
return Actor.Done;
}
}
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 360
|
<?xml version="1.0" encoding="UTF-8"?>
<gresources>
<gresource prefix="/org/gnome/gjs">
<file>modules/tweener/equations.js</file>
<file>modules/tweener/tweener.js</file>
<file>modules/tweener/tweenList.js</file>
<file>modules/overrides/GLib.js</file>
<file>modules/overrides/Gio.js</file>
<file>modules/overrides/GObject.js</file>
<file>modules/overrides/Gtk.js</file>
<file>modules/cairo.js</file>
<file>modules/coverage.js</file>
<file>modules/gettext.js</file>
<file>modules/lang.js</file>
<file>modules/_lie.js</file>
<file>modules/mainloop.js</file>
<file>modules/jsUnit.js</file>
<file>modules/signals.js</file>
<file>modules/format.js</file>
<file>modules/package.js</file>
</gresource>
</gresources>
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 9,733
|
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