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\section{Introduction}
\label{P}
Stochastic perturbations in the equations of motions are commonly used to model small perturbations (numerical,
empirical, and physical uncertainties) or thermodynamic fluctuations present in
fluid flows. Moreover, it is used for a better understanding of turbulence.
As a consequence stochastic partial differential equations (SPDEs) such as the stochastic Navier--Stokes equations
are gaining more and more interest in fluid mechanical research. First result can be traced back to the pioneering work by Bensoussan end Teman \cite{BeTe} in 1973.
Today there exists an abundant amount of literature concerning the dynamics of incompressible fluids driven by stochastic forcing. We refer to the lecture notes by Flandoli \cite{Fl}, the monograph of Kuksin and
Shyrikian \cite{KukShi} as well as the references cited therein for a recent overview. Definitely much less is known if compressibility of the fluid is taken into account. Fundamental questions of well--posedness and even mere existence of solutions to problems dealing with stochastic perturbations of compressible fluids are, to the best of our knowledge, largely open, with only a few rigorous results available.\\
First existence results were based on a suitable transformation formula that allows to reduce the problem to a random system of PDEs: The stochastic integral does no longer appear and deterministic methods are applicable, see \cite{MR1760377} for the 1D case, \cite{MR1807944} for a rather special periodic 2D. The latter one is based on
the existence theory developed by Va{\u\i}gant and Kazhikhov in \cite{MR1375428}. Finally, the work by Feireisl, Maslowski, Novotn\'y \cite{MR2997374} deals with the 3D case.
The first ``truly'' stochastic existence result for the compressible Navier--Stokes system perturbed by a general nonlinear multiplicative noise was obtained by Breit, Hofmanov\'a \cite{BrHo}. The existence of the so-called finite energy weak martingale solutions in three space dimensions with periodic boundary conditions was established. Extension of this result to the zero Dirichlet boundary conditions then appeared in \cite{2015arXiv150400951S,MR3385137}. For completeness, let us also mention \cite{BrFeHo2015B} where a singular limit result was proved.
The next step towards a better understanding of stochastic compressible fluids is the so-called relative energy inequality derived in \cite{BrFeHo2015A}.
Among other possible applications, it allows to compare a weak solution to the compressible system with arbitrary (smooth) processes, in particular
with a strong solution of the same problem. This gives rise to the weak--strong uniqueness principle: A weak (in the PDE sense) solution satisfying the energy inequality necessarily coincides with a strong solution emanating from the same initial data, as long as the latter one exists.
In the light of this result,
a natural question to ask is whether or not a strong solution exists at least locally in time. Results concerning the existence of strong solutions in three dimensions, however, do not exists at all.
In the present paper, we fill this gap by showing existence of local-in-time strong solutions (up to a positive stopping time) of the stochastic
compressible Navier--Stokes system
enjoying the regularity properties required by the weak--strong uniqueness principle established in \cite{BrFeHo2015A}.
We consider a stochastic variant of the \emph{compressible barotropic
Navier-Stokes system} describing the time evolution of the mass density $\varrho$ and the bulk velocity $\vc{u}$ of a fluid driven by
a nonlinear multiplicative noise. The system of equations reads
\begin{equation} \label{P1}
{\rm d} \varrho + {\rm div}_x (\varrho \vc{u}) \ \,{\rm d} t = 0
\end{equation}
\begin{equation} \label{P2}
{\rm d} (\varrho \vc{u}) + \left[ {\rm div}_x (\varrho \vc{u} \otimes \vc{u}) + a \nabla_x \varrho^\gamma \right] \,{\rm d} t = {\rm div}_x \mathbb{S} (\nabla_x \vc{u}) \ \,{\rm d} t + \mathbb{G} (\varrho, \varrho \vc{u}) {\rm d} W,
\end{equation}
where $\mathbb{S}(\nabla_x \vc{u})$ is the standard Newtonian viscous stress tensor,
\begin{equation} \label{P3}
\mathbb{S}(\nabla_x \vc{u}) = \mu \left( \nabla_x \vc{u} + \nabla_x^t \vc{u} - \frac{2}{3} {\rm div}_x \vc{u} \mathbb{I} \right) + \lambda {\rm div}_x \vc{u} \mathbb{I}, \qquad \mu > 0, \ \lambda \geq 0.
\end{equation}
The driving process $W$ is a cylindrical Wiener process defined on some probability space $(\Omega,\mathfrak{F},\mathbb{P})$ and the coefficient $\mathbb{G}$ is generally nonlinear and satisfies suitable growth assumptions. The precise assumptions will be specified in Section \ref{E}.
We focus on the periodic boundary conditions, for which the underlying spatial domain $\mathcal O \subset \mathbb{R}^N$ may be identified with the flat torus
\[
\mathcal O = \mathbb{T}^N = \left( (-\pi, \pi)|_{\{ - \pi, \pi \}} \right)^N ,\ N=1,2,3.
\]
The initial conditions are random variables
\begin{equation} \label{P4}
\varrho(0,\cdot) = \varrho_0, \ \vc{u}(0, \cdot) = \vc{u}_0,
\end{equation}
with sufficient space regularity specified later.
We study the system \eqref{P1}--\eqref{P4} in the framework of solutions that are strong in both PDE and probabilistic sense.
More precisely, such solutions possess sufficient space regularity for \eqref{P1}--\eqref{P4} to be satisfied pointwise (not only in the sense of distributions) and they are defined on a given probability space.
We introduce the notion of \emph{local strong pathwise solutions} which only exists up to a suitable stopping time, see Definition \ref{def:strsol}.
Next, we consider \emph{maximal strong pathwise solutions} which live on a maximal (random) time interval determined by the hypothetical
blow-up of the $W^{2,\infty}$-norm of the velocity $\vc{u}$, see Definition \ref{def:maxsol}.
Our main result, Theorem \ref{thm:main}, then states the existence of a unique maximal strong pathwise solution to problem \eqref{P1}--\eqref{P4}.
The deterministic approach to the local existence problem
for the compressible Navier-Stokes system is usually based on energy estimates. These are derived first for the unknown functions $\varrho$, $\vc{u}$ and then,
repeatedly, for their time derivatives up to a sufficient order to guarantee the required smoothness, see the nowadays probably
optimal result by Cho, Choe and Kim \cite{ChoChoeKim}. However, for obvious reasons related to the irregularity of sample paths of the Brownian motion, this technique is not suitable in the stochastic setting. Instead, the required space regularity must be achieved by differentiating the equations only with respect
to the space variables - a typical approach applicable to purely hyperbolic systems. The related references include works on the incompressible stochastic Navier--Stokes system \cite{BeFr,BrzP}, the incompressible stochastic Euler equations \cite{GHVic}, and also quasilinear hyperbolic systems \cite{JUKim}.
Similarly to Kim \cite{JUKim} (see also \cite{GHVic}), we use suitable cut-off operators to render all non-linearities in the equations
globally Lipschitz. The resulting (stochastic) system may admit \emph{global-in-time solutions}. Still, the approach proposed in
\cite{JUKim} and later revisited in \cite{GHVic} cannot be applied in a direct fashion for the following reasons:
\begin{itemize}
\item The energy method is applicable to \emph{symmetric} hyperbolic systems and their viscous perturbations.
\item In order to symmetrize (\ref{P1}), (\ref{P2}), the density must be strictly positive - the system must be out of vacuum.
\item For the density to remain positive at least on a short time interval, the maximum principle must be applied to the transport
equation (\ref{P1}). Accordingly, equation (\ref{P1}) must be solved exactly and not by means of a finite-dimensional approximation.
\item To avoid technical problems with non-local operators in the transport equation, the cut-off must be applied only to the velocity field.
\end{itemize}
In view of these difficulties and anticipating strict positivity of the density, we transform
the problem to a symmetric hyperbolic system perturbed by partial viscosity and the stochastic driving term, see Subsection \ref{subsec:symsyst}. Then cut-off operators in the spirit of
\cite{JUKim} are applied to the velocity field and this system is then studied in detail in Section \ref{A}. We use this technique to cut the nonlinear parts as well as to guarantee the nondegeneracy of the density, which leads
to global in time strong martingale solutions to this approximate system. The main ideas of the proof are as follows. First, we adapt a hybrid method similar to the one proposed in \cite{BrHo}: The equation of continuity is solved directly, while
the momentum equation is approximated by a finite dimensional Galerkin scheme. On this level, we are able to gain higher order uniform energy estimates by differentiating
in space.
Then, using the stochastic compactness method, we prove the existence of a strong martingale solution.
In Subsection \ref{subsec:uniq} we establish pathwise uniqueness and then the method of Gy\"ongy--Krylov \cite{krylov} is applied to
recover the convergence of the approximate solutions on the original probability
space, see Subsection \ref{subsec:pathwiseexist}. The existence of a unique strong pathwise solution therefore follows.
Finally, in Section \ref{subsec:strong} we employ the results of the previous sections to prove our main result, Theorem \ref{thm:main}. This last step is in the spirit of the recent treatment of the incompressible Euler system by Glatt-Holtz and Vicol \cite{GHVic}. However, the analysis is more involved due to the complicated structure of \eqref{P1}--\eqref{P4}. We rely on a delicate combination of stopping time arguments that allow to use the equivalence of \eqref{P1}--\eqref{P4} with the system studied in Section \ref{A}. As a consequence, also the corresponding existence and uniqueness result may be applied. One of the difficulties originates in the fact that we no longer assume the initial condition to be integrable in $\omega$. Thus the a priori estimates from Section \ref{A} are no longer valid. We present the details of the proof of uniqueness in Subsection \ref{s:un}, the existence of a local strong pathwise solution in Subsections \ref{ex1} and \ref{subsec:ex2} and we conclude with the existence of a maximal strong pathwise solution in Subsection \ref{ex3}.
\section{Preliminaries and main result} \label{E}
We start by introducing the notation and some basic facts used in the text. To begin, we fix
an arbitrarily large time horizon $T>0$.
\subsection{Analytic framework}
The symbols $W^{s,p}(\mathbb{T}^N)$ denote the Sobolov spaces of functions having distributional derivatives up to order $s$ integrable in $L^p(\mathbb{T}^N)$ for $p\in[1,\infty]$. We will also use $W^{s,2}(\mathbb{T}^N)$ for $s \in \mathbb{R}$ to denote the space of distributions $v$ defined on $\mathbb{T}^N$ with the finite norm
\begin{equation}\label{trigo}
\left\| v \right\|^2_{W^{s,2}(\mathbb{T}^N)} = \sum_{k \in \mathbb{Z}^N} (1 + |k|^s)^{2} |c_k(v)|^2 < \infty,
\end{equation}
where $c_k(v)$ are the Fourier coefficients of $v$ with respect to the standard trigonometric basis $\{ \exp(ik\cdot x) \}_{k \in\mathbb{Z}^N}$.
The shorten notation we will write $\|\cdot\|_{s,p}$ for $\|\cdot\|_{W^{s,p}(\mathbb{T}^N)}$
and $\|\cdot\|_p$ for $\|\cdot\|_{L^p(\mathbb{T}^N)}$.
The following estimates are standard in the Moser-type calculus and can be found e.g. in Majda \cite[Proposition 2.1]{Majd}.
\begin{enumerate}
\item For $u,v \in W^{s,2} \cap L^\infty(\mathbb{T}^N)$ and $|\alpha| \leq s$
\begin{equation}\label{E5}
\left\| \partial^\alpha_x (u v) \right\|_{2} \leq c_s \left( \| u \|_{\infty} \| \nabla^s_x v \|_{2} +
\| v \|_{\infty} \| \nabla^s_x u \|_{2} \right).
\end{equation}
\item For $u \in W^{s,2}(\mathbb{T}^N)$, $\nabla_x u \in L^\infty(\mathbb{T}^N)$, $v \in W^{s-1,2} \cap L^\infty (\mathbb{T}^N)$ and $|\alpha| \leq s$
\begin{equation}\label{E6}
\left\| \partial^\alpha_x (uv) - u \partial^\alpha_x v \right\|_{2} \leq c_s
\left( \| \nabla_x u \|_{\infty} \| \nabla^{s-1}_x v \|_{2} +
\| v \|_{\infty} \| \nabla^s_x u \|_{2} \right).
\end{equation}
\item For $u \in W^{s,2} \cap C(\mathbb{T}^N)$, and $F$ $s$-times continuously differentiable function on an open neighborhood of the compact set $G =
{\rm range}[u]$, $|\alpha| \leq s$,
\begin{equation} \label{E7}
\left\| \partial^\alpha_x F(u) \right\|_{L^2(\mathbb{T}^N)} \leq c_s \| \partial_u F \|_{C^{s-1}(G)} \| u \|^{|\alpha| - 1}_{L^\infty(\mathbb{T}^N)} \|
\partial^\alpha_x u \|_{L^2(\mathbb{T}^N)}.
\end{equation}
\color{black}
\end{enumerate}
\subsection{Stochastic framework}
The driving process $W$ is a cylindrical Wiener process defined on some stochastic basis $\left(\Omega, \mathfrak{F},(\mathfrak{F}_t )_{t \geq 0}, \mathbb{P}\right)$ with a complete, right-continuous filtration, and taking values
in a separable Hilbert space $\mathfrak{U}$. More specifically, $W$ is given by a formal expansion
\[
W(t)=\sum_{k\geq 1} e_k \beta_k(t).
\]
Here $\{ \beta_k \}_{k \geq 1}$ is a family of mutually independent real-valued Brownian motions
with respect to $\left(\Omega, \mathfrak{F},(\mathfrak{F}_t )_{t \geq 0}, \mathbb{P}\right)$ and $\{e_k\}_{k\geq 1}$ is an orthonormal basis of $\mathfrak{U}$.
To give the precise definition of the diffusion coefficient $\mathbb{G}$, consider $\rho\in L^2(\mathbb{T}^N)$, $\rho\geq0$, $\mathbf{q}\in L^2(\mathbb{T}^N)$ and let {$\,\mathbb{G}(\rho,\mathbf q):\mathfrak{U}\rightarrow L^2(\mathbb{T}^N, \mathbb{R}^N)$} be defined as follows
$$\mathbb{G}(\rho,\mathbf q)e_k=\mathbf{G}_k(\cdot,\rho(\cdot),\mathbf q(\cdot)).$$
We suppose that
the coefficients $\mathbf{G}_{k}:\mathbb{T}^N\times [0,\infty) \times\mathbb{R}^N\rightarrow\mathbb{R}^N$ are $C^s$-functions that satisfy uniformly in $x\in\mathbb{T}^N$
\begin{equation}\label{FG1}
\vc{G}_k (\cdot, 0 , 0) = 0,
\end{equation}
\begin{equation}
|\nabla^l \vc{G}_k (\cdot, \cdot, \cdot) | \leq \alpha_k, \quad \sum_{k \geq 1} \alpha_k < \infty \quad \mbox{for all}\ l\in\{1,...,s\},
\label{FG2}
\end{equation}
with $s\in\mathbb{N}$ specified below.
A typical example we have in mind is
\begin{align}\label{eq:model}
\mathbf{G}_k(x,\rho,\mathbf q)=\mathbf{a}_k(x)\rho+\mathbb A_k(x) \mathbf{q},
\end{align}
where $\mathbf{a}_k:\mathbb{T}^N\rightarrow\mathbb{R}^N$ and $\mathbb A_k:\mathbb{T}^N\rightarrow\mathbb{R}^{N\times N}$ are smooth functions, however, our analysis applies to general nonlinear coefficients $\mathbf{G}_k$.
We also introduce a new variable $r$ related to $\varrho$ through formula
\[
\varrho = \varrho(r) =\left(\frac{\gamma-1}{2a\gamma}\right)^\frac{1}{\gamma-1}r^\frac{2}{\gamma-1},
\]
together with the associated family of diffusion coefficients
\[
\mathbf{F}_k(\cdot, r, \vc{u}) = \frac{1}{\varrho(r)} \mathbf{G}_k (\cdot, \varrho(r), \varrho(r) \vc{u} ).
\]
Note that for the model case \eqref{eq:model} this implies
\[
\mathbf{F}_k(x,r,\mathbf u)=\mathbf{a}_k(x)+\mathbb A_k(x)\mathbf{u}.
\]
\begin{Remark} \label{bR2}
As we are interested in \emph{strong} solutions for which both $\varrho$ and $\vc{u}$ are bounded and $\varrho$ is bounded below away from zero, the hypotheses
\eqref{FG2} implies the same property for $\mathbf{F}_k$ restricted to this range. In addition, we have
\begin{align*}
\sum_k|\vc{F}_k(\cdot,r,\mathbf{u})|\leq \,c\,(1+|\mathbf{u}|).
\end{align*}
Moreover, it is enough to assume that
\eqref{FG2} holds only locally, meaning on each compact subset of $\mathbb{T}^N \times (0, \infty) \times \mathbb{R}^N$.
\end{Remark}
\medskip
Observe that if $\varrho$, $\vc{q}$ are $(\mathfrak{F}_t)$-progressively measurable $L^2(\mathbb{T}^N)$-valued processes such that
\[
\varrho \in L^2 \Big( \Omega\times[0,T]; L^2(\mathbb{T}^N) \Big), \ \vc{q} \in L^2 \Big(\Omega\times[0,T]; L^2(\mathbb{T}^N; \mathbb{R}^N) \Big),
\]
and $\mathbb{G}$ satisfies (\ref{FG1}), (\ref{FG2}), then
the stochastic integral
\[
\int_0^t \mathbb{G}(\varrho, \varrho \vc{u}) \ {\rm d} W = \sum_{k \geq 1}\int_0^t \vc{G}_k (\cdot, \varrho, \varrho \vc{u}) \ {\rm d} W_k
\]
is a well-defined $(\mathfrak{F}_t)$-martingale ranging in $L^2(\mathbb{T}^N; \mathbb{R}^N)$.
Next, we report the following result
by Flandoli and Gatarek \cite[Lemma 2.1]{FlaGat} which allows to show fractional Sobolev regularity in time for a stochastic integral.
\begin{Lemma} \label{flan}
Let $p \geq 2$, $\alpha \in[0, \frac{1}{2})$ be given. Let $\mathbb{G} = \{ \vc{G}_k \}_{k=1}^\infty$ satisfy, for some $m\in\mathbb{R}$,
\[
\mathbb{E} \left[ \int_0^T \left( \sum_{k=1}^\infty \| \vc{G}_k \|_{W^{m,2}(\mathbb{T}^N, \mathbb{R}^N) }^2 \right)^{p/2} \,{\rm d} t \right] < \infty.
\]
Then
\[
t \mapsto \int_0^t \mathbb{G} \ {\rm d} W \in L^p\Big(\Omega; W^{\alpha,p} \Big(0,T;W^{m,2}(\mathbb{T}^N; \mathbb{R}^N) \Big)\Big),
\]
and there exists a constant $c = c(\alpha,p)$ such that
\[
\begin{split}
\mathbb{E} \left[ \left\| \int_0^t \mathbb{G}\ {\rm d} W \right\|_{W^{\alpha,p} \Big(0,T;W^{m,2}(\mathbb{T}^N; \mathbb{R}^N) \Big) }^p \right]
\leq c(\alpha,p) \mathbb{E} & \left[ \int_0^T \left( \sum_{k=1}^\infty \| \vc{G}_k \|_{W^{m,2}(\mathbb{T}^N, \mathbb{R}^N) }^2 \right)^{p/2} \,{\rm d} t \right].
\end{split}
\]
\end{Lemma}
\begin{Remark} \label{Rflan}
Note that the above result further implies H\"older continuity of the stochastic integral due to the embedding
\[
W^{\alpha,p} \Big(0,T;W^{m,2}(\mathbb{T}^N; \mathbb{R}^N) \Big) \hookrightarrow C^\beta \Big(0,T;W^{m,2}(\mathbb{T}^N; \mathbb{R}^N) \Big) \quad \mbox{if} \quad
\beta<\alpha-\frac{1}{p}.
\]
\end{Remark}
Combining Lemma \ref{flan}, the hypotheses \eqref{FG1}, \eqref{FG2}, the estimate \eqref{E7}, and the embedding
\[
W^{s,2} (\mathbb{T}^N) \hookrightarrow C(\mathbb{T}^N), \ s > \frac{N}{2},
\]
we get in addition the following estimate for the stochastic integral appearing in \eqref{P2}.
\begin{Corollary} \label{Cflan}
Let $\vc{G}_k = \vc{G}_k(\varrho, \vc{q})$ satisfy \eqref{FG1}, \eqref{FG2} for a nonnegative integer $s$.
Let $p\geq 2$, $\alpha \in[0, \frac{1}{2})$. Suppose that
\[
\varrho, \ \vc{q} \in L^{ \beta p}\Big( \Omega \times (0,T); W^{s,2} (\mathbb{T}^N) \Big), \ \beta = \max\{s, 1\}.
\]
Then the following holds:
{\bf (i)} If $s = 0$, then
\[
t \mapsto \int_0^t \mathbb{G}(\varrho, \vc{q} ) \ {\rm d} W \in L^p\Big(\Omega; W^{\alpha,p} \Big(0,T;L^{2}(\mathbb{T}^N; \mathbb{R}^N) \Big)\Big),
\]
and
\[
\begin{split}
\mathbb{E} \left[ \left\| \int_0^t \mathbb{G}(\varrho, \vc{q})\ {\rm d} W \right\|_{W^{\alpha,p} \Big(0,T;L^{2}(\mathbb{T}^N; \mathbb{R}^N) \Big) }^p \right]
\leq c(\alpha,p) \mathbb{E} & \left[ \int_0^T \| [\varrho, \vc{q}] \|_{L^{2}(\mathbb{T}^N, \mathbb{R}^N)}^p \ \,{\rm d} t \right].
\end{split}
\]
{\bf (ii)} If $s > \frac{N}{2}$,
then
\[
t \mapsto \int_0^t \mathbb{G}(\varrho, \vc{q} ) \ {\rm d} W \in L^p\Big(\Omega; W^{\alpha,p} \Big(0,T;W^{s,2}(\mathbb{T}^N; \mathbb{R}^N) \Big)\Big),
\]
and
\[
\begin{split}
\mathbb{E} \left[ \left\| \int_0^t \mathbb{G}(\varrho, \vc{q})\ {\rm d} W \right\|_{W^{\alpha,p} \Big(0,T;W^{s,2}(\mathbb{T}^N; \mathbb{R}^N) \Big) }^p \right]
\leq c(\alpha,p) \mathbb{E} & \left[ \int_0^T \| [\varrho, \vc{q}] \|_{W^{s, 2}(\mathbb{T}^N, \mathbb{R}^N)}^{s p} \ \,{\rm d} t \right].
\end{split}
\]
\end{Corollary}
Finally, we define an auxiliary space $\mathfrak{U}_0\supset\mathfrak{U}$ via
$$\mathfrak{U}_0=\bigg\{v=\sum_{k\geq1}\alpha_k e_k;\;\sum_{k\geq1}\frac{\alpha_k^2}{k^2}<\infty\bigg\},$$
endowed with the norm
$$\|v\|^2_{\mathfrak{U}_0}=\sum_{k\geq1}\frac{\alpha_k^2}{k^2},\quad v=\sum_{k\geq1}\alpha_k e_k.$$
Note that the embedding $\mathfrak{U}\hookrightarrow\mathfrak{U}_0$ is Hilbert-Schmidt. Moreover, trajectories of $W$ are $\mathbb{P}$-a.s. in $C([0,T];\mathfrak{U}_0)$.
\subsection{Main result}
Let us first introduce the notion of local strong pathwise solution. Such a solution is strong in both PDEs and probabilistic sense but possibly exists only locally in time. To be more precise, system \eqref{P1}--\eqref{P2} will be satisfied pointwise (not in the sense of distributions) on the given stochastic basis associated to the cylindrical Wiener process $W$.
\begin{Definition}[Local strong pathwise solution] \label{def:strsol}
Let $\left(\Omega, \mathfrak{F},(\mathfrak{F}_t )_{t \geq 0}, \mathbb{P}\right)$ be a stochastic basis with a complete right-continuous filtration and let ${W}$ be an $(\mathfrak{F}_t) $-cylindrical Wiener process.
Let
$(\varrho_0,\mathbf{u}_0)$ be a $W^{s,2}(\mathbb{T}^N)\times W^{s,2}(\mathbb{T}^N;\mathbb{R}^N)$-valued $\mathfrak{F}_0$-measurable random variable, and let $\mathbb{G}$ satisfy \eqref{FG1},
\eqref{FG2}.
A triplet
$(\varrho,\vc{u},\mathfrak{t})$ is called a local strong pathwise solution to system \eqref{P1}--\eqref{P4} provided
\begin{itemize}
\item $\mathfrak{t}$ is an a.s. strictly positive $(\mathfrak{F}_t)$-stopping time;
\item the density $\varrho$ is a $W^{s,2}(\mathbb{T}^N)$-valued $(\mathfrak{F}_t)$-progressively measurable process satisfying
$$\varrho(\cdot\wedge \mathfrak{t}) > 0,\ \varrho(\cdot\wedge \mathfrak{t}) \in C([0,T]; W^{s,2}(\mathbb{T}^N)) \quad \mathbb{P}\text{-a.s.};$$
\item the velocity $\vc{u}$ is a $W^{s,2}(\mathbb{T}^N)$-valued $(\mathfrak{F}_t)$-progressively measurable process satisfying
$$ \vc{u}(\cdot\wedge \mathfrak{t}) \in C([0,T]; W^{s,2}(\mathbb{T}^N; \mathbb{R}^N))\cap L^2(0,T;W^{s+1,2}(\mathbb{T}^N; \mathbb{R}^N))\quad \mathbb{P}\text{-a.s.};$$
\item there holds $\mathbb{P}$-a.s.
\[
\begin{split}
\varrho (t\wedge \mathfrak{t}) &= \varrho_0 - \int_0^{t \wedge \mathfrak{t}} {\rm div}_x(\varrho\vc{u} ) \ \mathrm{d} s, \\
(\varrho \vc{u}) (t \wedge \mathfrak{t}) &= \varrho_0 \vc{u}_0 - \int_0^{t \wedge \mathfrak{t}} {\rm div}_x (\varrho\vc{u} \otimes\vc{u} ) \ \mathrm{d} s \\
& \qquad+ \int_0^{t \wedge \mathfrak{t}} {\rm div}_x \tn{S}(\nabla_x \vc{u}) \ \mathrm{d} s
- \int_0^{t \wedge \mathfrak{t}}\nabla_x p(\varrho)\ \mathrm{d} s + \int_0^{t \wedge \mathfrak{t}} {\tn{G}}(\varrho,\varrho\vc{u} ) \ {\rm d} W,
\end{split}
\]
for all $t\in[0,T]$.
\end{itemize}
\end{Definition}
In the above definition, we have tacitly assumed that $s$ is large enough in order to provide sufficient regularity for the strong solutions.
Classical solutions require two spatial derivatives of $\vc{u}$ to be continuous $\mathbb{P}$-a.s. This motivates the following definition.
\begin{Definition}[Maximal strong pathwise solution]\label{def:maxsol}
Fix a stochastic basis with a cylindrical Wiener process and an initial condition exactly as in Definition \ref{def:strsol}. A quadruplet $$(\varrho,\vc{u},(\mathfrak{t}_R)_{R\in\mathbb{N}},\mathfrak{t})$$ is a maximal strong pathwise solution to system \eqref{P1}--\eqref{P4} provided
\begin{itemize}
\item $\mathfrak{t}$ is an a.s. strictly positive $(\mathfrak{F}_t)$-stopping time;
\item $(\mathfrak{t}_R)_{R\in\mathbb{N}}$ is an increasing sequence of $(\mathfrak{F}_t)$-stopping times such that
$\mathfrak{t}_R<\mathfrak{t}$ on the set $[\mathfrak{t}<T]$,
$\lim_{R\to\infty}\mathfrak{t}_R=\mathfrak t$ a.s. and
\begin{equation}\label{eq:blowup}
\sup_{t\in[0,\mathfrak{t}_R]}\|\vc{u}(t)\|_{2,\infty}\geq R\quad \text{on}\quad [\mathfrak{t}<T] ;
\end{equation}
\item each triplet $(\varrho,\vc{u},\mathfrak{t}_R)$, $R\in\mathbb{N}$, is a local strong pathwise solution in the sense of Definition \ref{def:strsol}.
\end{itemize}
\end{Definition}
The stopping times $\mathfrak{t}_R$ in Definition 2.6 announce the stopping time $\mathfrak{t}$ which is therefore predictable.
It denotes the maximal life span of the solution which is determined by the time of explosion of the $W^{2,\infty}$-norm of the velocity field. Indeed, it can be seen from \eqref{eq:blowup} that
$$
\sup_{t\in[0,\mathfrak{t})}\|\vc{u}(t)\|_{2,\infty}=\infty\quad \text{on}\quad [\mathfrak{t}<T].
$$
Note that the announcing sequence $(\mathfrak{t}_R)$ is not unique. Therefore, uniqueness for maximal strong solutions is understood in the sense that only the solution $(\varrho,\vc{u})$ and its blow up time $\mathfrak{t}$ are unique.
Let us also point out that, later on, we will choose $s$ in order to have the embedding $W^{s,2}\hookrightarrow W^{2,\infty}$, i.e. at least $s>\frac{N}{2}+2$. Even though one might expect that the $W^{s,2}$-norm blows up earlier than the $W^{2,\infty}$-norm, this is not true. Indeed, according to Definition \ref{def:strsol} and Definition \ref{def:maxsol}, a maximal strong pathwise solution satisfies
$$\vc{u}(\cdot \wedge \mathfrak{t}_R)\in C([0,T];W^{s,2}(\mathbb{T}^N,\mathbb{R}^N))\qquad\mathbb{P}\text{-a.s.}$$
and hence the velocity is continuous in $W^{s,2}(\mathbb{T}^N,\mathbb{R}^N)$ on $[0,\mathfrak{t})$. Consequently, the blow up of the $W^{s,2}$-norm coincides with the blow up of the $W^{2,\infty}$-norm at time $\mathfrak{t}$. This fact reflects the nature of our a priori estimates (see Subsection \ref{UNIF}): roughly speaking, control of the $W^{2,\infty}$-norm implies control of the $W^{s,2}$-norm and leads to continuity of trajectories in $W^{s,2}$.
Finally, we have all in hand to formulate our main result.
\begin{Theorem}\label{thm:main}
Let $s\in\mathbb{N}$ satisfy $s>\frac{N}{2} + 3$.
Let the coefficients
$\mathbf{G}_k$ satisfy hypotheses \eqref{FG1}, \eqref{FG2} and let $(\varrho_0,\mathbf{u}_0)$ be an $\mathfrak{F}_0$-measurable, $W^{s,2}(\mathbb{T}^N)\times W^{s,2}(\mathbb{T}^N,\mathbb{R}^N)$-valued random variable such that $\varrho_0>0$ $\mathbb{P}$-a.s.
Then
there exists a unique maximal strong pathwise solution $(\varrho,\vc{u},(\mathfrak{t}_R)_{R\in\mathbb{N}},\mathfrak{t})$ to problem \eqref{P1}--\eqref{P4} with the initial condition $(\varrho_0,\vc{u}_0)$.
\end{Theorem}
\begin{Remark} \label{mr+}
The required regularity $s > \frac{N}{2} + 3$ is definitely higher than $s > \frac{N}{2} + 2$
for the deterministic problem, see Matsumura and Nishida \cite{MANI}, \cite{MANI1}, Valli and Zajaczkowski \cite{VAZA}. This is due to the loss of regularity with respect to the time variable pertinent
to the stochastic problems. Possibly optimal results could be achieved by working in the framework of $L^p$-spaces as
Cho, Choe, and Kim \cite{ChoChoeKim} and to adapt this approach to the stochastic setting in the spirit of Glatt-Holtz and Vicol \cite{GHVic}.
\end{Remark}
\begin{Remark} \label{bR1}
The method used in the present paper can be easily adapted to handle the same problem on the
whole space $\mathcal O = \mathbb{R}^N$, with relevant far field conditions for $\varrho$, $\vc{u}$, say
\[
\varrho \to \Ov{\varrho} ,\ \vc{u} \to 0 \ \mbox{as}\ |x| \to \infty.
\]
On the other hand, the case when the fluid interacts with a physical boundary, for instance $\mathcal O$ a bounded domain with the no-slip boundary condition for
$\vc{u}$, would require a more elaborate treatment.
\end{Remark}
\begin{Remark}
Let us also point out that most of our analysis applies to the stochastic compressible Euler system as well. Indeed, the only point where we rely on the positive viscosity $\mu$ is the proof of continuity of trajectories of a solution in $W^{s,2}$, see Subsection \ref{subsec:ident}. It is based on the variational approach within a Gelfand triplet which gives a very elegant proof, especially in comparison to the Euler setting where one would need to find another reasoning, cf. \cite{GHVic}.
\end{Remark}
\subsection{Rewriting the equations as a symmetric hyperbolic-parabolic problem}
\label{subsec:symsyst}
It is well known in the context of compressible fluids that existence of strong solutions is intimately related to the strict positivity of the density, i.e. the non-appearance of vacuum states. Anticipating this property in the framework of strong solutions we may rewrite \eqref{P1}--\eqref{P2} as a
hyperbolic-parabolic system for unknowns $r,\mathbf{u}$ where $r$ is a function of $\varrho$.
To be more precise, as the time derivative of $\varrho$ satisfies the deterministic equation (\ref{P1}), we have
\[
{\rm d} (\varrho \vc{u}) = {\rm d} \varrho \ \vc{u} + \varrho\ {\rm d} \vc{u},
\]
where, in accordance with \eqref{P1}
\[
{\rm d} \varrho = - {\rm div}_x (\varrho \vc{u}) \ \,{\rm d} t .
\]
Consequently, the momentum equation \eqref{P2} reads
\[
\varrho {\rm d} \vc{u} + \left[ \varrho \vc{u} \cdot \nabla_x \vc{u} + a \nabla_x \varrho^\gamma \right] \,{\rm d} t = {\rm div}_x \mathbb{S} (\nabla_x \vc{u}) \ \,{\rm d} t + \tn{G} (\varrho, \varrho \vc{u}) {\rm d} W,
\]
or, anticipating strict positivity of the mass density,
\begin{equation*
{\rm d} \vc{u} + \left[ \vc{u} \cdot \nabla_x \vc{u} + a \frac{1}{\varrho} \nabla_x \varrho^\gamma \right] \,{\rm d} t = \frac{1}{\varrho} {\rm div}_x \mathbb{S} (\nabla_x \vc{u}) \ \,{\rm d} t + \frac{1}{\varrho} \tn{G} (\varrho, \varrho \vc{u}) {\rm d} W.
\end{equation*}
Next, we rewrite
\[
a \frac{1}{\varrho} \nabla_x \varrho^\gamma = \frac{a \gamma}{\gamma - 1} \nabla_x \varrho^{\gamma - 1} = \frac{2 a \gamma}{\gamma - 1}\varrho^{\frac{\gamma - 1}{2}} \nabla_x \varrho^{\frac{\gamma - 1}{2}},
\]
and evoking the renormalized variant of \eqref{P1} (cf. \cite{BrHo})
\[
{\rm d} \varrho^{\frac{\gamma - 1}{2}} + \vc{u} \cdot \nabla_x \varrho^{\frac{\gamma - 1}{2}}\,{\rm d} t + \frac{\gamma - 1}{2} \varrho^{\frac{\gamma - 1}{2}} {\rm div}_x \vc{u} \,{\rm d} t = 0.
\]
Thus, for a new variable
\[
r \equiv \sqrt{ \frac{2 a \gamma}{\gamma - 1} } \varrho^{\frac{\gamma - 1}{2}},
\]
system \eqref{P1}, \eqref{P2} takes the form
\begin{equation} \label{E3}
{\rm d} r + \vc{u} \cdot \nabla_x r \ \,{\rm d} t + \frac{\gamma - 1}{2} r {\rm div}_x \vc{u}\,{\rm d} t = 0,
\end{equation}
\begin{equation} \label{E4}
{\rm d} \vc{u} + \left[ \vc{u} \cdot \nabla_x \vc{u} + r \nabla_x r \right] \,{\rm d} t = D(r) {\rm div}_x \mathbb{S} (\nabla_x \vc{u}) \,{\rm d} t + \tn{F} (r, \vc{u}) {\rm d} W,
\end{equation}
where
$$D(r)=\frac{1}{\varrho}=\left(\frac{\gamma-1}{2a\gamma}\right)^{-\frac{1}{\gamma-1}}r^{-\frac{2}{\gamma-1}}, \quad \tn{F} (r, \vc{u}) = \frac{1}{\varrho(r)} \tn{G} (\varrho(r), \varrho(r) \vc{u}).$$
Observe that the left hand side corresponds to a symmetric hyperbolic system, cf. Majda \cite{Majd}, for which higher order energy estimates can be obtained by differentiating \eqref{E3}, \eqref{E4} in $x$ up to order $s$,
cf. Gallagher \cite{Gall2000}, Majda \cite{Majd}. Unlike the more elaborated treatment proposed by Cho, Choe, and Kim \cite{ChoChoeKim}
giving rise to the optimal regularity space for the deterministic
Navier-Stokes system, the energy approach avoids differentiating the equations in the time variable - a procedure that may be delicate in the stochastic setting.
\subsection{Outline of the proof of Theorem \ref{thm:main}}
\label{subsec:outline}
In the \emph{deterministic} setting, system \eqref{E3}--\eqref{E4} can be solved via an approximation procedure. The so-obtained local in time strong
solution exists on a maximal time interval, the length of which can be estimated in terms of the
size of the initial data. However, in the \emph{stochastic} setting it is more convenient to work with approximate solutions defined on the whole time interval $[0,T]$. To this end, we introduce suitable cut-off operators applied to the $W^{2,\infty}$-norm of the velocity field.
Specifically, we consider
the approximate system in the form
\begin{align} \label{E3'}
{\rm d} r + \varphi_R (\|\vc{u} \|_{2,\infty})\Big[\vc{u} \cdot \nabla_x r \ + \tfrac{\gamma - 1}{2}\, r\, {\rm div}_x \vc{u}\Big]\ \,{\rm d} t &= 0,\\
\label{E4'}
{\rm d} \vc{u} + \varphi_R(\| \vc{u} \|_{2,\infty})\left[ \vc{u} \cdot \nabla_x \vc{u} + r \nabla_x r \right] \,{\rm d} t & =\varphi_R(\| \vc{u} \|_{2
,\infty}) D(r) {\rm div}_x \mathbb{S} (\nabla_x \vc{u}) \ \,{\rm d} t \\& + \varphi_R(\| \vc{u} \|_{2,\infty})\tn{F} (r, \vc{u}) {\rm d} W\nonumber,\\
r(0)=r_{0} ,\quad \mathbf{u}(0)&=\mathbf{u}_0,\label{approx:initial}
\end{align}
where $\varphi_R:[0,\infty)\rightarrow[0,1]$ are smooth cut-off functions satisfying
\begin{align*}
\varphi_R(y)=\begin{cases}1,\quad &0\leq y\leq R,\\
0,\quad & R+1\leq y.
\end{cases}
\end{align*}
Our aim is to solve \eqref{E3'}--\eqref{approx:initial} via the stochastic compactness method: First, we construct solutions to certain approximated systems, establish tightness of their laws in suitable topologies and finally deduce the existence of a strong martingale solution to \eqref{E3}--\eqref{E4}. The necessary uniform bounds are obtained through
a purely hyperbolic approach by differentiating with respect to the space variable and testing the resulting expression with
suitable space derivative of the unknown functions.
For the above mentioned reasons, the approximated densities must be positive on time intervals of finite length. Therefore the approximation scheme
must be chosen to preserve the maximum principle for (\ref{E3'}). To this end,
the approximate solutions to \eqref{E3'}--\eqref{approx:initial} will be constructed by means of a hybrid method based on
\begin{itemize}
\item
solving the (deterministic) equation of continuity \eqref{E3'} for a given $\vc{u}$ obtaining $r = r[\vc{u}]$;
\item
plugging $r = r[\vc{u}]$ in \eqref{E4'} and using a fixed point argument to get local in time solutions of a Galerkin approximation of \eqref{E4'};
\item
extending the Galerkin solution to $[0,T]$ by means of {\it a priori} bounds.
\end{itemize}
Note that the transport equation \eqref{E3'} is solved exactly in terms of a given velocity field $\vc{u}$
as the cut-off operators apply to $\vc{u}$ only.
\section{The approximated system}
\label{A}
In this section we focus on the approximated system \eqref{E3'}--\eqref{E4'}. More precisely, our aim is twofold: First, we establish existence of a strong martingale solution for initial data in $L^p(\Omega;W^{s,2}(\mathbb{T}^N))$ for all $1\leq p<\infty$ and some $s>\frac{N}{2}+2$; second, we prove pathwise uniqueness provided $s>\frac{N}{2}+3$, which in turn implies existence of a (unique) strong pathwise solution.
To this end, let us introduce these two concepts of strong solution
for the approximate system \eqref{E3'}--\eqref{E4'}. A strong martingale solution is strong in the PDEs sense but only weak in the probabilistic sense. In other words, the stochastic basis as well as a cylindrical Wiener process cannot be given in advance and become a part of the solution. Accordingly, the initial condition is stated in the form of a initial law. On the other hand, a strong pathwise solution is strong in both PDEs and probabilistic sense, that is, the stochastic elements are given in advance.
\begin{Definition}[Strong martingale solution] \label{def:strsolmart}
Let $\Lambda$ be a Borel probability measure on
$$W^{s,2}(\mathbb{T}^N)\times W^{s,2}(\mathbb{T}^N,\mathbb{R}^N).$$
A multiplet
$$\left(\left(\Omega, \mathfrak{F},(\mathfrak{F}_t )_{t \geq 0}, \mathbb{P}\right),r,\vc{u},W\right)$$
is called a strong martingale solution to the approximate system \eqref{E3'}--\eqref{E4'} with the initial law $\Lambda$, provided
\begin{itemize}
\item $\left(\Omega, \mathfrak{F},(\mathfrak{F}_t )_{t \geq 0}, \mathbb{P}\right)$ is a stochastic basis with a complete right-continuous filtration;
\item ${W}$ is an $( \mathfrak{F}_t ) $-cylindrical Wiener process;
\item $r$ is a $W^{s,2}(\mathbb{T}^N)$-valued $(\mathfrak{F}_t)$-progressively measurable process satisfying
$$r \in L^2 \Big(\Omega; C([0,T]; W^{s,2}(\mathbb{T}^N)) \Big);$$
\item the velocity $\vc{u}$ is a $W^{s,2}(\mathbb{T}^N)$-valued $(\mathfrak{F}_t)$-progressively measurable process satisfying
$$ \vc{u} \in L^2\Big( \Omega; C([0,T]; W^{s,2}(\mathbb{T}^N))\cap L^2(0,T;W^{s+1,2}(\mathbb{T}^N)) \Big);$$
\item $\Lambda=\mathbb{P}\circ[(r(0),\vc{u}(0))]^{-1};$
\item there holds $\mathbb{P}$-a.s.
\[
\begin{split}
r(t) &= r({0}) - \int_0^{t} \varphi_R (\|\vc{u} \|_{2,\infty})\Big[\vc{u} \cdot \nabla_x r \ + \tfrac{\gamma - 1}{2}\, r\, {\rm div}_x \vc{u}\Big]\ {\rm d}s, \\
\vc{u} (t) &= \vc{u}(0) - \int_0^{t} \varphi_R(\| \vc{u} \|_{2,\infty})\left[ \vc{u} \cdot \nabla_x \vc{u} + r \nabla_x r \right] \mathrm{d} s \\
& \qquad+ \int_0^{t} \varphi_R(\| \vc{u} \|_{2
,\infty}) D(r) {\rm div}_x \mathbb{S} (\nabla_x \vc{u}) \mathrm{d} s
+ \int_0^{t} \varphi_R(\| \vc{u} \|_{2,\infty})\tn{F} (r, \vc{u}) \ {\rm d} W,
\end{split}
\]
for all $t\in[0,T]$.
\end{itemize}
\end{Definition}
\begin{Definition}[Strong pathwise solution] \label{def:strsolpath}
Let $\left(\Omega, \mathfrak{F},(\mathfrak{F}_t )_{t \geq 0}, \mathbb{P}\right)$ be a given stochastic basis with a complete right-continuous filtration and let ${W}$ be a given $( \mathfrak{F}_t ) $-cylindrical Wiener process.
Then $(r,\vc{u})$
is called a strong pathwise solution to the approximate system \eqref{E3'}--\eqref{E4'} with the initial condition $(r_0,\vc{u}_0)$ provided
\begin{itemize}
\item $r$ is a $W^{s,2}(\mathbb{T}^N)$-valued $(\mathfrak{F}_t)$-progressively measurable process satisfying
$$r \in L^2 \Big(\Omega; C([0,T]; W^{s,2}(\mathbb{T}^N)) \Big);$$
\item the velocity $\vc{u}$ is a $W^{s,2}(\mathbb{T}^N)$-valued $(\mathfrak{F}_t)$-progressively measurable process satisfying
$$ \vc{u} \in L^2\Big( \Omega; C([0,T]; W^{s,2}(\mathbb{T}^N))\cap L^2(0,T;W^{s+1,2}(\mathbb{T}^N)) \Big);$$
\item there holds $\mathbb{P}$-a.s.
\[
\begin{split}
r(t) &= r_{0} - \int_0^{t} \varphi_R (\|\vc{u} \|_{2,\infty})\Big[\vc{u} \cdot \nabla_x r \ + \tfrac{\gamma - 1}{2}\, r\, {\rm div}_x \vc{u}\Big]\ {\rm d}s, \\
\vc{u} (t) &= \vc{u}_0 - \int_0^{t} \varphi_R(\| \vc{u} \|_{2,\infty})\left[ \vc{u} \cdot \nabla_x \vc{u} + r \nabla_x r \right] \mathrm{d} s \\
& \qquad+ \int_0^{t} \varphi_R(\| \vc{u} \|_{2
,\infty}) D(r) {\rm div}_x \mathbb{S} (\nabla_x \vc{u}) \mathrm{d} s
+ \int_0^{t} \varphi_R(\| \vc{u} \|_{2,\infty})\tn{F} (r, \vc{u}) \ {\rm d} W,
\end{split}
\]
for all $t\in[0,T]$.
\end{itemize}
\end{Definition}
The main result of this section reads as follows.
\begin{Theorem}\label{thm:appr}
Let the coefficients
$\mathbf{G}_k$ satisfy hypotheses \eqref{FG1}, \eqref{FG2} and let
\[
(r_0,\mathbf{u}_0)\in L^p(\Omega,\mathfrak{F}_0,\mathbb{P};W^{s,2}(\mathbb{T}^N)\times W^{s,2}(\mathbb{T}^N))
\]
for all $1\leq p<\infty$ and some $s\in\mathbb{N}$ such that $s>\frac{N}{2} + 2$. In addition, suppose that
\begin{equation*}
\| r_0 \|_{W^{1,\infty}(\mathbb{T}^N)} < R, \ r_0 > \frac{1}{R} \ \mathbb{P}\mbox{-a.s.}
\end{equation*}
Then there exists a strong martingale solution
to problem \eqref{E3'}--\eqref{E4'} with the initial law $\Lambda=\mathbb{P}\circ[(r_{0},\vc{u}_0)]^{-1}$. Moreover, there exists a deterministic constant $\underline{r}_R>0$ such that
\[
r(t, \cdot) \geq \underline{r}_R > 0 \quad \mathbb{P}\mbox{-a.s.}\quad \mbox{for all}\ t \in [0,T]
\]
and
\begin{equation} \label{RE2}
\mathbb{E}\bigg[\sup_{t \in [0,T]} \| (r(t),\vc{u}(t)) \|_{s,2} +\int_0^T \| \vc{u} \|^2_{s+1,2} \ \,{\rm d} t \bigg]^p \leq
c(R,r_0, \vc{u}_0,p) < \infty \quad \mbox{for all}\quad 1 \leq p
< \infty.
\end{equation}
Finally, if $s > \frac{N}{2} + 3$, then pathwise uniqueness holds true. Specifically, if $(r^1, \vc{u}^1)$, $(r^2, \vc{u}^2)$ are two strong solutions to \eqref{E3'}--\eqref{E4'} defined on the same stochastic basis with the same Wiener process $W$ and
$$\mathbb{P} \left[ r^1_0 = r^2_0, \ \vc{u}^1_0 = \vc{u}^2_0 \right] = 1,$$
then
\[
\mathbb{P} \left[ r^1 (t) = r^2(t), \ \vc{u}^1(t) = \vc{u}^2(t), \ \mbox{for all}\ t \in [0,T]\right] = 1.
\]
Consequently, there exists a unique strong pathwise solution to \eqref{E3'}--\eqref{E4'}.
\end{Theorem}
The rest of this section is dedicated to the proof of Theorem \ref{thm:appr} which is divided into several parts. First, in Subsection \ref{subsec:galerkin} we construct the approximate solutions to \eqref{E3'}--\eqref{E4'} by employing the hybrid method delineated in Subsection \ref{subsec:outline}. Second, in Subsection \ref{UNIF} we derive higher order energy estimates which hold true uniformly in the approximation parameter $n$. Third, in Subsection \ref{subsec:comp} we perform the stochastic compactness method: we establish tightness of the laws of the approximated solutions and apply the Skorokhod representation theorem. This yields existence of a new probability space with a sequence of random variables converging a.s. Then in Subsection \ref{subsec:ident}, we identify the limit with a strong martingale solution to \eqref{E3'}--\eqref{E4'}. Finally, in Subsection \ref{subsec:uniq} we provide the proof of pathwise uniqueness under the additional assumption that $s>\frac{N}{2}+3$ and in Subsection \ref{subsec:pathwiseexist} we employ the Gy\"ongy-Krylov argument to deduce the existence of a strong pathwise solution.
\subsection{The Galerkin approximation}
\label{subsec:galerkin}
To begin with, observe that for any $\vc{u} \in C([0,T]; W^{2,\infty}(\mathbb{T}^N))$, the transport equation (\ref{E3'}) admits a classical solution
$r = r[\vc{u}]$, uniquely determined by the initial datum $r_{0}$. In addition, for a certain universal constant $c$ we have the estimates
\begin{equation} \label{est1}
\begin{split}
\frac{1}{R}
\exp \left( - cR t \right)&\leq
\exp \left( - cR t \right) \inf_{\mathbb{T}^N} r_0 \leq r(t, \cdot) \leq \exp \left( cR t \right) \sup_{\mathbb{T}^N} r_{0} \leq R \exp \left( cR t \right) \\
|\nabla_x r (t,\cdot) | & \leq \exp \left( cR t \right)|\nabla_x r_0 |\leq
R \exp \left( cR t \right) \ t \in [0,T].
\end{split}
\end{equation}
Next, we consider the orthonormal basis $\left\{ \boldsymbol{\psi}_m \right\}_{m=1}^\infty$ of the space $L^{2}(\mathbb{T}^N; \mathbb{R}^N)$ formed by trigonometric functions and set
\[
X_n = {\rm span} \left\{ \boldsymbol{\psi}_1, \dots, \boldsymbol{\psi}_n \right\}, \quad \mbox{with the associated projection}\ P_n : L^2 \to X_n.
\]
We look
for approximate solutions $\vc{u}_n$ of \eqref{E4'} belonging to {$L^2\Big( \Omega; C([0,T]; X_n) \Big)$},
satisfying
\begin{equation} \label{est2}
\begin{split}
{\rm d}\left< \vc{u}_n, \boldsymbol{\psi}_i \right> &+ \varphi_R(\| \vc{u}_n \|_{2,\infty}) \left< \Big[[\vc{u}_n \cdot \nabla_x \vc{u}_n + r[\vc{u}_n] \nabla_x r[\vc{u}_n, r_{0,R}] \Big]; \boldsymbol{\psi}_i \right> \,{\rm d} t
\\
&=\varphi_R(\| \vc{u}_n \|_{2,\infty}) \left< D(r[\vc{u}_n]) {\rm div}_x \mathbb{S} (\nabla_x \vc{u}_n); \boldsymbol{\psi}_i \right> \ \,{\rm d} t \\ &
+ \varphi_R(\| \vc{u}_n \|_{2,\infty}) \left< \tn{F} (r[\vc{u}_n] , \vc{u}_n); \boldsymbol{\psi}_i \right> {\rm d} W,\quad i = 1, \dots, n.\\
\vc{u}_n(0)&=P_n \mathbf{u}_0.
\end{split}
\end{equation}
As all norms on $X_n$ are equivalent, solutions of \eqref{E3'}, \eqref{est2} can be obtained in a standard way by means of the Banach fixed point argument.
Specifically, we have to show that the mapping
\[
\vc{u} \mapsto \mathscr{T} \vc{u} : X_n \to X_n,
\]
\begin{equation} \label{est3}
\begin{split}
\left< \mathscr{T}\vc{u}; \psi_i \right> =& \left< \vc{u}_0; \psi_i \right>- \int_0^\cdot\varphi_R(\| \vc{u} \|_{2,\infty}) \left< \Big[\vc{u}\cdot \nabla_x \vc{u} + r[\vc{u}] \nabla_x r[\vc{u}, r_{0,n} ] \Big]; \boldsymbol{\psi}_i \right> \,{\rm d} t
\\
&+\int_0^\cdot\varphi_R(\| \vc{u} \|_{2,\infty}) \left< D(r[\vc{u}]) {\rm div}_x \mathbb{S} (\nabla_x \vc{u}); \boldsymbol{\psi}_i \right> \ \,{\rm d} t \\ &
+ \int_0^\cdot\varphi_R(\| \vc{u} \|_{2,\infty}) \left< \tn{F} (r[\vc{u}] , \vc{u}); \boldsymbol{\psi}_i \right> {\rm d} W,\quad i = 1, \dots, n.
\end{split}
\end{equation}
is a contraction on $\mathcal B=L^2(\Omega;C([0,T^\ast]; X_n))$ for $T^\ast$ sufficiently small. The three components of $\mathscr{T}$ appearing on the right hand side of \eqref{est3}
will be denoted by $\mathscr T_{det}^1$, $\mathscr T_{det}^2$ and $\mathscr T_{sto}$, respectively.
\color{black}
For $r_1 = r[\vc{u}]$, $r_2 = r[\vc{v}]$, we get
\begin{equation*} \label{T8}
\begin{split}
{\rm d} (r_1 - r_2) &+ \vc{v}_1 \cdot \nabla_x (r_1 - r_2) \,{\rm d} t - \frac{\gamma - 1}{2} {\rm div}_x \vc{v}_1 (r_1 - r_2) \,{\rm d} t \\
&= - \nabla_x r_2 \cdot (\vc{v}_1 - \vc{v}_2) - \frac{\gamma - 1}{2} r_2 {\rm div}_x (\vc{v}_1 - \vc{v}_2) \,{\rm d} t ,
\end{split}
\end{equation*}
where we have set
\[
\vc{v}_1 = \varphi_R(\| \vc{u} \|_{2,\infty}) \vc{u} , \ \vc{v}_2 = \varphi_R(\| \vc{v} \|_{2,\infty})\vc{v}.
\]
Consequently, we easily deduce that
\begin{align}\label{eq:new}
\sup_{0\leq t\leq T^\ast}\big\|r[\mathbf{u}]-r[\mathbf{v}]\big\|^2_{L^2}\leq T^\ast C(n,R,T)\sup_{0\leq t\leq T^\ast} \big\|\mathbf{u}-\mathbf{v}\big\|_{X_n}^2
\end{align}
noting that $r_1$, $r_2$ coincide at $t=0$ and that $r_j$, $\nabla_x r_j$ are bounded by a deterministic constant depending on $R$.
As a consequence of \eqref{est1}, \eqref{eq:new} and the equivalence of norms on $X_n$ we can show that the mapping
$\mathscr T_{det}=\mathscr T_{det}^1+\mathscr T_{det}^2$ satisfies the estimate
\begin{align}\label{Tdet}
\|\mathscr{T}_{det}\mathbf{u}-\mathscr{T}_{det}\mathbf{v}\|_\mathcal{B}^2\leq T^{\ast}C(n,R,T)\|\mathbf{u}-\mathbf{v}\|_{\mathcal B}^2.
\end{align}
Finally, by Burgholder-Davis-Gundy inequality we have (setting $J_R(\mathbf{w})=\varphi_{R+1}(\|\mathbf{w}\|_{2,\infty})\mathbf{w} $)
\begin{equation*}
\begin{split}
\|\mathscr{T}_{sto}\mathbf{u}&-\mathscr{T}_{sto}\mathbf{v}\|_\mathcal{B}^2=\,\mathbb{E}\sup_{0\leq t\leq T_\ast}\bigg\|\int_0^t\Big(\varphi_R(\|\vc{u}\|_{2,\infty})\mathbb F\big(r[\vc{u}],\vc{u}\big)-\varphi(\|\vc{v}\|_{2,\infty})\mathbb F\big(r[\vc{v}],\vc{v}\big)\Big)\,\mathrm{d} W\bigg\|_{X_n}^2\\
&\leq C(n,R)\,\mathbb{E}\int_0^{T^{\ast}}\sum_{k\geq1}\Big\|\,\varphi_R(\|\vc{u}\|_{2,\infty})\mathbf{F}_k\big(r[\vc{u}],J_R(\vc{u})\big)-\varphi_R(\|\vc{v}\|_{2,\infty})\mathbb \mathbf{F}_k\big(r[\vc{v}],J_R(\vc{v})\big)\Big\|_{X_n}^2\mathrm{d} s\\
&\leq C(n,R)\,\mathbb{E}\int_0^{T^{\ast}}\big|\varphi_R(\|\vc{u}\|_{2,\infty})-\varphi_R(\|\vc{v}\|_{2,\infty})\big|^2\sum_{k\geq1}\Big\|\,\mathbf{F}_k\big(r[\vc{u}],J_R(\vc{u})\big)\Big\|_{X_n}^2\mathrm{d} s\\
&+ C(n,R)\,\mathbb{E}\int_0^{T^{\ast}}\varphi_R(\|\vc{v}\|_{2,\infty})^2\sum_{k\geq1}\Big\|\, \mathbf{F}_k\big(r[\vc{u}],J_R(\vc{u})\big)-\mathbb \mathbf{F}_k\big(r[\vc{v}],J_R(\vc{v})\big)\Big\|_{X_n}^2\mathrm{d} s.
\end{split}
\end{equation*}
Using the growth conditions for $\mathbf{F}_k$ (see \eqref{FG2} and Remark \ref{bR2})
we gain
\begin{align}
\|&\mathscr{T}_{sto}\mathbf{u}-\mathscr{T}_{sto}\mathbf{v}\|_\mathcal{B}^2\nonumber
\\&\leq T^\ast C(n,R)\bigg(\mathbb{E}\|\mathbf{u}-\mathbf{v}\|_{2,\infty}+\,\mathbb{E}\int_0^{T_{\ast}}\big\|\,r[\vc{u}]-r[\vc{v}]\big\|_{L^2}^2\mathrm{d} s+\,\mathbb{E}\int_0^{T_{\ast}}\big\|J_R(\vc{u})-J_R(\vc{v})\big\|_{L^2}^2\mathrm{d} s\bigg)\nonumber\\
&\leq T^{\ast}C(n,R)\|\mathbf{u}-\mathbf{v}\|_{\mathcal B}^2.\label{Tsto}
\end{align}
Note that the last step was a consequence of \eqref{eq:new} and the equivalence of norms. Combining \eqref{Tdet} and \eqref{Tsto} shows that
$\mathscr T$ is a contraction for {a deterministic (small) time $T^\ast> 0$}. A solution to \eqref{E3'}--\eqref{E4'} on the whole interval $[0,T]$ can be obtained by decomposing it into small subintervals gluing the corresponding solutions together.
\subsection{Uniform estimates}
\label{UNIF}
In this subsection, we derive estimates that hold uniformly for $n \to \infty$, which yields a basis for our compactness argument presented in Subsection \ref{subsec:comp}.
{At this stage, the approximate velocity field $\vc{u}_n$ is smooth in the $x$-variable; whence the corresponding solution
$r_n = r[\vc{u}_n, r_{0,n}]$ of the transport equation (\ref{E3'}) shares the same smoothness with the initial datum $r_0$.}
Let $\alpha$ be a multiindex such that $|\alpha|\leq s$. Differentiating \eqref{E3'} in the $x$-variable we obtain
\begin{align} \label{E8'}
\begin{aligned}
{\rm d} \partial^\alpha_x r_n &+ \varphi_R(\| \vc{u}_n \|_{2,\infty})\left[\vc{u}_n \cdot \nabla_x \partial^\alpha_x r_n \ + \tfrac{\gamma - 1}{2} \,r_n\, {\rm div}_x \partial^\alpha_x \vc{u}_n \right]\ \,{\rm d} t \\
&=\varphi_R(\| \vc{u}_n \|_{2,\infty})\big[ \vc{u}_n \cdot \partial^\alpha_x \nabla_x r_n - \partial^\alpha_x \left( \vc{u}_n \cdot \nabla_x r_n \right) \big] \,{\rm d} t \\
&+ \tfrac{\gamma - 1}{2}\varphi_R(\| \vc{u}_n \|_{2,\infty})
\left[r_n \partial^\alpha_x {\rm div}_x \vc{u}_n - \partial^\alpha_x \left( r_n {\rm div}_x \vc{u}_n \right) \right] \,{\rm d} t \\
&=:T_1^n \,{\rm d} t +T^n_2 \,{\rm d} t .
\end{aligned}
\end{align}
Similarly, we may use the fact that the spaces $X_n$ are invariant with respect to the spatial derivatives, in particular, we deduce
that
\begin{align} \label{E9'}
\begin{aligned}
{\rm d} \left< \partial^\alpha_x \vc{u}_n; \boldsymbol{\psi}_i \right> &+ \varphi_R(\| \vc{u}_n \|_{2, \infty})\ \left< \left[ \vc{u}_n \cdot \nabla_x \partial^\alpha_x \vc{u}_n + r_n \nabla_x \partial^\alpha_x r_n \right] ; \boldsymbol{\psi}_i \right> \,{\rm d} t \\
& - \varphi_R(\| \vc{u}_n \|_{2,\infty}) \left< D(r_n) {\rm div}_x \mathbb{S} (\nabla_x \partial^\alpha_x \vc{u}_n); \boldsymbol{\psi}_i \right> \ \,{\rm d} t \\
&= \varphi_R(\| \vc{u}_n \|_{2,\infty})\left< \left[ \vc{u}_n \cdot \partial^\alpha_x \nabla_x \vc{u}_n - \partial^\alpha_x \left( \vc{u}_n \cdot \nabla_x \vc{u}_n \right) \right];
\boldsymbol{\psi}_i \right> \,{\rm d} t \\
&+ \varphi_R(\| \vc{u}_n \|_{2,\infty}) \left< \left[ r_n \partial^\alpha_x \nabla_x r_n -
\partial^\alpha_x \left( r_n \nabla_x r_n \right) \right]; \boldsymbol{\psi}_i \right> \,{\rm d} t \\
&- \varphi_R(\| \vc{u}_n \|_{2, \infty}) \left< \left[ D(r_n) \partial^\alpha_x {\rm div}_x \mathbb{S} (\nabla_x \vc{u}_n)
- \partial^\alpha_x \left( D(r_n) {\rm div}_x \mathbb{S} (\nabla_x \vc{u}_n) \right)
\right]; \boldsymbol{\psi}_i \right> \,{\rm d} t \\& + \varphi_R(\| \vc{u}_n \|_{2,\infty}) \left< \partial^\alpha_x \tn{F} (r_n, \vc{u}_n); \boldsymbol{\psi}_i \right> {\rm d} W\\
&=:T_3^n\,{\rm d} t +T^n_4 \,{\rm d} t +T_5^n\,{\rm d} t + \varphi_R(\| \vc{u}_n \|_{2, \infty}) \left< \partial^\alpha_x \tn{F} (r_n, \vc{u}_n); \boldsymbol{\psi}_i \right> {\rm d} W,\quad i = 1,\dots,n.
\end{aligned}
\end{align}
It follows from \eqref{E6} that the ``error'' terms may be handled as
\begin{equation} \label{E10'}
\begin{split}
\left\| T_1^n\right\|_{2} & \lesssim \varphi_R(\| \vc{u}_n \|_{2,\infty}) \Big[
\| \nabla_x \vc{u}_n \|_{\infty} \| \nabla_x^s r_n \|_{2} + \left\| \nabla_x r_n \right\|_{\infty} \| \nabla_x^s \vc{u}_n \|_{2} \Big] \\
\left\| T_2^n \right\|_{2} & \lesssim \varphi_R(\| \vc{u}_n \|_{2,\infty}) \Big[
\| \nabla_x r_n \|_{\infty} \| \nabla_x^s \vc{u}_n \|_{2} + \left\| {\rm div}_x \vc{u}_n \right\|_{\infty} \| \nabla_x^s r_n \|_{2} \Big]
\\
\left\| T_3^n \right\|_{2} & \lesssim \varphi_R(\| \vc{u}_n \|_{2,\infty})
\| \nabla_x \vc{u}_n \|_{\infty} \| \nabla_x^s \vc{u}_n \|_{2} \\
\left\| T_4^n \right\|_{2} & \lesssim
\| \nabla_x r_n \|_{\infty} \| \nabla_x^s r _n\|_{2},
\end{split}
\end{equation}
and
\begin{equation} \label{E11'}
\begin{split}
\left\| T_5^n \right\|_{2} & \lesssim \varphi_R(\| \vc{u}_n \|_{2,\infty})
\left\| \nabla_x D(r_n) \right\|_{\infty} \left\| \nabla_x^s \mathbb{S} (\nabla_x \vc{u}_n) \right\|_{2} \\
&\quad +\varphi_R(\| \vc{u}_n \|_{2,\infty})
\left\| {\rm div}_x \mathbb{S} (\nabla_x \vc{u}_n) \right\|_{\infty} \left\| \nabla_x^s D(r_n) \right\|_{2} .
\end{split}
\end{equation}
Multiplying \eqref{E8'} by $\partial^\alpha_x r_n$ and integrating the resulting expression by parts, we observe
$$
\int_{\mathbb{T}^N}\vc{u}_n \cdot \nabla_x \partial^\alpha_x r \partial^\alpha_x r_n\,\,{\rm d} {x}=-\frac{1}{2}\int_{\mathbb{T}^N}{\rm div}_x\vc{u}_n|\partial^\alpha_x r_n|^2\,\,{\rm d} {x};
$$
whence
\begin{align} \label{E12'}
\begin{aligned}
&\left\| \partial^\alpha_x r_n (t) \right\|_{2}^2 + (\gamma - 1)\int_0^t\varphi_R(\| \vc{u}_n \|_{2,\infty}) \int_{\mathbb{T}^N} r_n {\rm div}_x \partial^\alpha_x \vc{u}_n \partial^\alpha_x r_n \,{\rm d} {x}\,\mathrm{d}\sigma \\
&\quad\lesssim \left\| \partial^\alpha_x r_0 \right\|_{2}^2 +\int_0^t\varphi_R(\| \vc{u}_n \|_{2, \infty})
\left( \| \vc{u}_n \|_{1, \infty} \| \varrho \|_{s,2} + \| r_n \|_{1,\infty} \| \vc{u} _n\|_{s,2} \right)
\|\partial^\alpha_x r_n\|_2\,\,\mathrm{d}\sigma\\
\end{aligned}
\end{align}
provided $|\alpha| \leq s$.
To apply the same treatment to \eqref{E9'}, we use It\^o's formula for the function $f(\mathbf{C}^n)=\int_{\mathbb{T}^N}|\partial^\alpha_x\mathbf{u}_n|^2\,{\rm d} {x}$. There holds
\begin{equation} \label{E13'}
\begin{split}
\left\|\partial^\alpha_x \vc{u}_n(t) \right\|^2_2 \,{\rm d} {x} &+2\int_0^t\varphi_R(\| \vc{u}_n \|_{2,\infty})\int_{\mathbb{T}^N} \left[ \vc{u}_n \cdot \nabla_x \partial^\alpha_x \vc{u}_n + r_n \nabla_x \partial^\alpha_x r_n \right] \cdot \partial^\alpha_x \vc{u}_n \,{\rm d} {x}\,\mathrm{d}\sigma \\
&-2\int_0^t\varphi_R(\| \vc{u}_n\|_{2, \infty})\int_{\mathbb{T}^N} D(r_n) {\rm div}_x \mathbb{S} (\nabla_x \partial^\alpha_x \vc{u}_n) \cdot \partial^\alpha_x \vc{u}_n \,{\rm d} {x}\,\mathrm{d}\sigma \\
&= \left\| \partial^\alpha_x P_n\vc{u}_0 \right\|^2 + 2\int_0^t\int_{\mathbb{T}^N}\left[ T_3^n+T_4^n+T_5^n\right] \cdot \partial^\alpha_x \vc{u}_n \,{\rm d} {x}\,\mathrm{d}\sigma \\
& + 2\int_0^t \varphi_R(\| \vc{u}_n \|_{2, \infty})\int_{\mathbb{T}^N} \partial^\alpha_x \tn{F} (r_n, \vc{u}_n) \cdot \partial^\alpha_x \vc{u}_n \ {\rm d} W\\
&+\sum_{k\geq1} \int_0^t \varphi_R(\| \vc{u}_n \|_{2,\infty})\int_{\mathbb{T}^N}| \partial^\alpha_x \vc{F}_k (r_n, \vc{u}_n)|^2 \,{\rm d} {x}\,\mathrm{d}\sigma.
\end{split}
\end{equation}
Integrating by parts yields
\[
\begin{split}
\int_{\mathbb{T}^N} &\big[ \vc{u}_n \cdot \nabla_x \partial^\alpha_x \vc{u}_n + r _n \nabla_x \partial^\alpha_x r _n \big] \cdot \partial^\alpha_x \vc{u}_n \,{\rm d} {x}\\&= - \frac{1}{2} \int_{\mathbb{T}^N} |\partial^\alpha_x \vc{u}_n |^2 {\rm div}_x \vc{u}_n \,{\rm d} {x}
- \int_{\mathbb{T}^N}r_n {\rm div}_x \partial^\alpha_x \vc{u}_n \partial^\alpha_x r_n \,{\rm d} {x} - \int_{\mathbb{T}^N} \nabla_x r_n \cdot \partial^\alpha \vc{u}_n \partial^\alpha_x r_n
\end{split}
\]
as well as
\[
\begin{split}
-\int_{\mathbb{T}^N} &\big[ D(r _n) {\rm div}_x \mathbb{S} (\nabla_x \partial^\alpha_x \vc{u}_n) \big] \cdot \partial^\alpha_x \vc{u}_n \,{\rm d} {x}\\&=
\int_{\mathbb{T}^N} \nabla_x D(r _n) \cdot \mathbb{S} (\nabla_x \partial^\alpha_x \vc{u}_n) \cdot \partial^\alpha_x \vc{u}_n \,{\rm d} {x}
+ \int_{\mathbb{T}^N} D(r_n) \mathbb{S} (\nabla_x \partial^\alpha_x \vc{u}_n ) : \nabla_x \partial^\alpha_x \vc{u}_n \,{\rm d} {x}
\end{split}
\]
Summing up \eqref{E12'}--\eqref{E13'} and using \eqref{E10'}--\eqref{E11'} we observe that the term containing $r_n\partial^\alpha_x r_n{\rm div}_x \partial^\alpha_x \vc{u}_n$ on the left hand side cancels out and we may infer that
\begin{align*}
&\left\| (r_n(t),\mathbf{u}_n(t)) \right\|^2_{s,2} + \sum_{|\alpha| \leq s } \int_0^t\int_{\mathbb{T}^N}\varphi_R(\| \vc{u}_n \|_{2,\infty}) D(r_n) \mathbb{S} (\nabla_x \partial^\alpha_x \vc{u}_n ) : \nabla_x \partial^\alpha_x \vc{u}_n \,{\rm d} {x} \,\mathrm{d}\sigma \\
&\quad \lesssim \left\| (r_0 ,\mathbf{u}_0) \right\|^2_{s,2}+
\int_0^t \left[ \varphi_R(\| \vc{u}_n \|_{2, \infty})
\| \vc{u}_n \|_{1, \infty} \Big( \| r_n \|^2_{s,2} + \| \vc{u} \|^2_{s,2} \Big) + \| r_n \|_{1,\infty} \| r_n \|_{s,2} \| \vc{u} _n\|_{s,2} \right] \,{\rm d} t
\\
&\quad + \int_0^t \left[ \varphi_R(\| \vc{u}_n \|_{2,\infty})\| {\rm div}_x \mathbb{S} (\nabla_x \vc{u}_n ) \|_{\infty} \left\| D(r_n) \right\|_{s,2} \| \vc{u}_n \|_{s,2}
+ \| \nabla_x D(r_n) \|_{\infty} \| \vc{u}_n \|^2_{s,2} \right] \,\mathrm{d}\sigma\\
&\quad + \int_0^t \varphi_R(\| \vc{u}_n\|_{2,\infty})\int_{\mathbb{T}^N} \partial^\alpha_x \tn{F} (r_n, \vc{u}_n) \cdot \partial^\alpha_x \vc{u}_n \,{\rm d} {x}\ {\rm d} W \\
&\quad +\sum_{k\geq1}\int_0^t \varphi_R(\| \vc{u}_n \|_{2,\infty})\int_{\mathbb{T}^N} | \partial^\alpha_x \vc{F}_k (r_n, \vc{u}_n)|^2 \,{\rm d} {x} \ \,\mathrm{d}\sigma
\end{align*}
as long as $s > \frac{N}{2} + 2$.
\begin{Remark} \label{estR1}
Note that the above estimate {depends on $R$ only through the cut-off function $\varphi_R$}. Moreover, in accordance with \eqref{est1},
\[
\begin{split}
\varphi_R(\| \vc{u}_n \|_{2, \infty})
\| \vc{u}_n \|_{1, \infty} + \varphi_R(\| \vc{u}_n \|_{2,\infty})\| {\rm div}_x \mathbb{S} (\nabla_x \vc{u}_n ) \|_{\infty} &\lesssim cR, \\
\ \| r_n \|_{1,\infty} + \| \nabla_x D(r_n) \|_{\infty} &\lesssim c \exp \left( cRT \right) \Big( \| r_{0} \|_{1,\infty} + \| \nabla_x r_{0} \|_{\infty} \Big) \\
& \lesssim c(R) \exp \left( cRT \right),
\end{split}
\]
and, in view of \eqref{E7}, \eqref{est1},
\[
\left\| D(r_n) \right\|_{s,2} \leq c(R,T) \| r_n \|_{s,2}.
\]
\end{Remark}
In contrast with the preceding part,
the following inequalities \emph{depend on $R$}. Using \eqref{FG2} as well as \eqref{E7} we have
\begin{align*}
\sum_{k\geq1}\int_0^t&\varphi_R(\| \mathbf{u}_n \|_{2,\infty})\int_{\mathbb{T}^N} |\partial^\alpha_x \vc{F}_k (r_n, \vc{u}_n)|^2 \,{\rm d} {x} \ \,\mathrm{d}\sigma\\
&\lesssim \int_0^t \varphi_R(\| \mathbf{u}_n \|_{2,\infty}) \int_{\mathbb{T}^N}\sum_{k\geq1} |\nabla^{s-1} \vc{F}_k|^2 \,\,{\rm d} {x} \,\|(r_n,\vc{u}_n)\|_{\infty}^{2(|\alpha|-1)}\|(r_n,\vc{u}_n)\|_{s,2}^2\,\,\mathrm{d}\sigma \\
&\lesssim c(R,T) \int_0^t \|(r_n,\vc{u}_n)\|_{s,2}^2 \,\,\mathrm{d}\sigma.
\end{align*}
as well as
\begin{align*}
\mathbb{E}\bigg[\sup_{t\in(0,T)}&\bigg|\int_0^t \varphi_R(\|\vc{u}_n \|_{2,\infty})\int_{\mathbb{T}^N} \partial^\alpha_x \tn{F} (r_n, \vc{u}_n) \cdot \partial^\alpha_x \vc{u}_n\,\,{\rm d} {x} \ {\rm d} W\bigg|\bigg]^p\\
&\lesssim
\mathbb{E}\bigg[\sum_{k\geq1}\int_0^T \varphi_R(\| \vc{u}_n \|_{2,\infty})^2\bigg(\int_{\mathbb{T}^N} \partial^\alpha_x \vc{F}_k (r_n, \vc{u}_n) \cdot \partial^\alpha_x \vc{u}_n \,{\rm d} {x}\bigg)^2\,{\rm d} {x}\,{\rm d} t \bigg]^{\frac{p}{2}}\\
&\lesssim
\mathbb{E}\bigg[\int_0^T \varphi_R(\| \vc{u}_n \|_{2,\infty})^2\bigg(\sum_{k\geq1}\|\vc{F}_k (r_n, \vc{u}_n)\|^2_{s,2}\bigg) \|\vc{u}_n\|^2_{s,2}\,{\rm d} t \bigg]^{\frac{p}{2}}\\
&\lesssim \mathbb{E}\bigg[\int_0^T \varphi_R(\|\vc{u}_n\|_{2,\infty})^2 \|(r_n,\vc{u}_n)\|_\infty^{2(s-1)}\|(r_n,\vc{u}_n)\|_{s,2}^4\,\,\mathrm{d}\sigma\bigg]^\frac{p}{2}\\
&\lesssim c(R,T) \mathbb{E}\bigg[\sup_{t\in(0,T)}\| \vc{u}_n\|_{s,2}^2\int_0^T\|(r_n,\vc{u}_n)\|_{s,2}^2\,\,\mathrm{d}\sigma\bigg]^\frac{p}{2}\\
&\lesssim c(R,T) \left( \kappa \mathbb{E} \left[ \sup_{t\in(0,T)}\| (r_n, \vc{u}_n) \|_{s,2}^{2p} \right] +c(\kappa)\mathbb{E}\bigg[\int_0^T\|(r_n ,\vc{u}_n)\|_{s,2}^2\,\,\mathrm{d}\sigma\bigg]^p \right)
\end{align*}
where we also took into account the Burgholder-Davis-Gundy and weighted Young inequalities.
Finally, we apply the Gronwall lemma to conclude
\begin{align} \label{E14''}
&\mathbb{E}\bigg[ \left( \sup_{(0,T)}\left\| (r_n,\mathbf{u}_n) \right\|^2_{s,2} + \int_0^T\int_{\mathbb{T}^N}|\nabla^{s+1}\vc{u}_n|^2 \,{\rm d} {x} \,{\rm d} t \right)^p \bigg]
\lesssim c(R,T, s) \mathbb{E}\bigg[\left\| (r_0,\mathbf{u}_0) \right\|^{2p}_{s,2} +1\bigg]
\end{align}
whenever $s > \frac{N}{2} + 2$.
\subsection{Compactness}
\label{subsec:comp}
Now we have all in hand to set up our compactness argument leading to the existence part of Theorem \ref{thm:appr}.
Let us define the path space $\mathcal{X}=\mathcal{X}_r\times\mathcal{X}_\mathbf u\times\mathcal X_W$,
\begin{align*}
\mathcal{X}_\mathbf u&= C([0,T]; W^{\beta,2}(\mathbb{T}^N; \mathbb{R}^N)) ,\quad
\mathcal{X}_r= C([0,T]; W^{\beta,2}(\mathbb{T}^N)) ,\quad\mathcal{X}_W=C([0,T];\mathfrak{U}_0),
\end{align*}
\color{black}
where $\beta < s$ (not necessarily integer) can be chosen arbitrarily close to $s$, in particular, $\beta > \frac{N}{2} + 2$ so that we have the embedding
\[
W^{\beta,2}(\mathbb{T}^N)\hookrightarrow W^{2,\infty}(\mathbb{T}^N)
\]
needed to pass to the limit in the cut-off operators.
We denote by $\mu_{r_n}$ and $\mu_{\mathbf u_n}$ the law of $r_n$ and $\mathbf u_n$
on the corresponding path space. By $\mu_W$ we denote the law of $W$ on $\mathcal{X}_W$ and their joint law on $\mathcal{X}$ is $\mu^n$.
To proceed, it is necessary to establish tightness of $\{\mu^n;\,n\in\mathbb N\}$.
\begin{Proposition}\label{prop:bfutightness}
The set $\{\mu_{\mathbf u_n};\,n\in\mathbb N\}$ is tight on $\mathcal{X}_\mathbf u$.
\end{Proposition}
\begin{proof}
We start with a compact embedding relation
\[
C([0,T]; W^{s,2}(\mathbb{T}^N)) \cap C^\gamma ([0,T]; L^2(\mathbb{T}^N)) \hookrightarrow\hookrightarrow C([0,T]; W^{\beta,2}(\mathbb{T}^N)),\ \gamma > 0, \ \beta < s,
\]
that follows directly from the abstract Arzel\` a-Ascoli theorem.
Due to \eqref{est3}, $\mathbf{u}_n$ satisfies
\begin{align*
\begin{aligned}
\vc{u}_n(t)&=P_n\vc{u}_0 - \int_0^t\varphi_R(\|\vc{u}_n\|_{2,\infty})P_n\big[ \vc{u}_n \cdot \nabla_x \vc{u}_n + r_n \nabla_x r_n \big] \,\mathrm{d}\sigma \\ &+\int_0^t \varphi_R(\|\vc{u}_n\|_{2,\infty}) P_n\big[D(r_n) {\rm div}_x \mathbb{S} (\nabla_x \vc{u}_n)\big] \,\mathrm{d}\sigma
+\int_0^t \varphi_R(\|\vc{u}_n\|_{2,\infty}) P_n\mathbb{F} (r_n, \vc{u}_n) {\rm d} W.
\end{aligned}
\end{align*}
Now we decompose $\mathbf{u}_n$ into two parts, namely, $\mathbf{u}_n=\mathbf{Y}_n+\mathbf{Z}_n$, where
\begin{equation*}
\begin{split}
\mathbf{Y}_n(t)&=P_n\vc{u}_0 - \int_0^t\varphi_R(\|\vc{u}_n\|_{2,\infty})P_n\big[ \vc{u}_n \cdot \nabla_x \vc{u}_n + r_n \nabla_x r_n \big] \,\mathrm{d}\sigma \\
&+\int_0^t \varphi_R(\|\vc{u}_n\|_{2,\infty})P_n\big[D(r_n) {\rm div}_x \mathbb{S} (\nabla_x \vc{u}_n)\big] \,\mathrm{d}\sigma,\\
\mathbf{Z}_n(t)&=\int_0^t \varphi_R(\|\vc{u}_n\|_{2,\infty})P_n\tn{F} (r_n, \vc{u}_n) {\rm d} W.
\end{split}
\end{equation*}
By \eqref{E14''} and the continuity of $P_n$ on $L^{2}$ we have for any $\kappa\in (0,1)$ that
\begin{equation*
\mathbb{E} \left[ \|\mathbf{Y}_n\|_{C^\kappa([0,T];L^2(\mathbb{T}^N))} \right]\leq c(R),
\end{equation*}
while \eqref{E14''} combined with Corollary \ref{Cflan} (for $s = 0$), Remark \ref{Rflan} yields
the same conclusion for $\vc{Z}_n$, with $0 < \kappa < 1/2$.
\end{proof}
\begin{Proposition}\label{proprtightness}
The set $\{\mu_{r_n};\,n\in\mathbb N\}$ is tight on $\mathcal{X}_r$.
\end{Proposition}
\begin{proof}
The proof is completely analogous to Proposition \ref{prop:bfutightness}
using the equation \eqref{E3'} for $r_n$ and the uniform estimate \eqref{E14''}.
\end{proof}
Since also the law $\mu_W$ is tight as being a Radon measure on the Polish space $\mathcal{X}_W$ we can finally deduce tightness of the joint laws $\mu^n$.
\begin{Corollary}\label{cor:tight}
The set $\{\mu^n;\,n\in\mathbb N\}$ is tight on $\mathcal{X}$.
\end{Corollary}
Since the path space $\mathcal{X}$ is a Polish space we may use the classical Skorokhod representation theorem. That is, passing to a weakly convergent subsequence $\mu^\varepsilon$ (and denoting by $\mu$ the limit law) we infer the following result.
\begin{Proposition}\label{prop:skorokhod1}
There exists a subsequence $\mu^n$, a probability space $(\tilde\Omega,\tilde{\mathfrak F},\tilde\mathbb{P})$ with $\mathcal{X}$-valued Borel measurable random variables $(\tilde r_n,\tilde\mathbf u_n,\tilde W_n)$, $N\in\mathbb{N}$, and $(\tilde r,\tilde\mathbf u,\tilde W)$ such that
\begin{enumerate}
\item the law of $(\tilde r_n,\tilde\mathbf u_n,\tilde W_n)$ is given by $\mu^n$, $n\in\mathbb N$,
\item the law of $(\tilde r,\tilde\mathbf u,\tilde W)$ is given by $\mu$,
\item $(\tilde r_n,\tilde\mathbf u_n,\tilde W_n)$ converges $\,\tilde{\mathbb{P}}$-a.s. to $(\tilde r,\tilde{\mathbf u},\tilde{W})$ in the topology of $\mathcal{X}$.
\end{enumerate}
\end{Proposition}
\subsection{Identification of the limit}
\label{subsec:ident}
As the next step, we will identify the limit obtained in Proposition \ref{prop:skorokhod1} with a strong martingale solution to \eqref{E3'}--\eqref{E4'}, completing the proof of Theorem \ref{thm:appr}.
Let us first fix some notation that will be used in the sequel. We denote by $\mathbf{r}_t$ the operator of restriction to the interval $[0,t]$ acting on various path spaces. In particular, if $X$ stands for one of the path spaces $\mathcal{X}_r,\,\mathcal{X}_{\mathbf{u}}$ or $\mathcal{X}_{W}$ and $t\in[0,T]$, we defin
\begin{align*
\mathbf{r}_t:X\rightarrow X|_{[0,t]},\quad f\mapsto f|_{[0,t]}.
\end{align*}
Clearly, $ \mathbf{r}_t$ is a continuous mapping.
Let $(\tilde{\mathfrak F}_t^n)$ and $(\tilde{\mathfrak F}_t)$, respectively, be the $\tilde{\mathbb{P}}$-augmented canonical filtration of the process $(\tilde r_n,\tilde{\mathbf u}_n,\tilde{W}_n)$ and $(\tilde r,\tilde{\mathbf u},\tilde{W})$, respectively, that is
\begin{equation*}
\begin{split}
\tilde{\mathfrak F}_t^n&=\sigma\big(\sigma\big(\mathbf{r}_t\tilde r_n,\,\mathbf{r}_t\tilde{\mathbf u}_n,\,\mathbf{r}_t \tilde{W}_n\big)\cup\big\{\mathscr M\in\tilde{\mathfrak F};\;\tilde{\mathbb{P}}(\mathscr M)=0\big\}\big),\quad t\in[0,T],\\
\tilde{\mathfrak F}_t&=\sigma\big(\sigma\big(\mathbf{r}_t\tilde{r},\,\mathbf{r}_t\tilde{\mathbf u},\,\mathbf{r}_t\tilde{W}\big)\cup\big\{\mathscr M\in\tilde{\mathfrak F};\;\tilde{\mathbb{P}}(\mathscr M)=0\big\}\big),\quad t\in[0,T].
\end{split}
\end{equation*}
We claim that $(\tilde r,\tilde\vc{u},\tilde W)$ is a strong martingale solution to \eqref{E3'}-\eqref{E4'}. Indeed, in order to identify \eqref{E3'}, let us define the following functional
\begin{align*}
(r,\vc{u})&\mapsto L(r,\mathbf{u})_t:={r}(t)-{r}(0) + \int_0^t\varphi_R(\|{\vc{u}}\|_{2,\infty})\Big[{\vc{u}} \cdot \nabla_x {r} \ -\tfrac{\gamma - 1}{2}\,{r}\, {\rm div}_x {\vc{u}}\Big]\,\mathrm{d}\sigma.
\end{align*}
Since the couple $(r_n,\vc{u}_n)$ solves \eqref{E3'} on the original probability space, it holds $L(r_n,\vc{u}_n)_t=0$, $t\in[0,T]$. Thus, due to equality of laws we get
$$\tilde\mathbb{E}\|L(\tilde r_n,\tilde\vc{u}_n)_t\|_{2}^2=\mathbb{E}\|L(r_n,\vc{u}_n)_t\|_2^2=0.$$
With Proposition \ref{prop:skorokhod1} and \eqref{E14''} at hand, we may pass to the limit on the left hand side and deduce that $(\tilde r,\tilde\vc{u})$ solves \eqref{E3'}.
In order to identify \eqref{E4'}, we first note that since $\tilde W_n$ has the same law as $W$, there exists a collection of mutually independent real-valued $(\tilde{\mathfrak F}_t)$-Wiener processes $(\tilde{\beta}^{n}_k)_{k\geq1}$ such that $\tilde{W}_n=\sum_{k\geq1}\tilde{\beta}^{n}_k e_k$ , i.e. there exists a collection of mutually independent real-valued $(\tilde{\mathfrak F}_t)$-Wiener processes $(\tilde{\beta}_k)_{k\geq1}$ such that $\tilde{W}=\sum_{k\geq1}\tilde{\beta}_k e_k$.
As the next step, let us fix times $s,t\in[0,T]$ such that $s<t$ and let
$$h:\mathcal{X}_r|_{[0,s]}\times\mathcal{X}_\mathbf{u}|_{[0,s]}\times\mathcal{X}_W|_{[0,s]}\rightarrow [0,1]$$
be a continuous function.
We define functionals
\begin{align*}
(r,\vc{u})&\mapsto M^n(r,\mathbf{u})_t:={\vc{u}}(t)-{\vc{u}}(0) + \int_0^t\varphi_R(\|{\vc{u}}\|_{2,\infty})P_n\big[ {\vc{u}}\cdot \nabla_x {\vc{u}} +{r} \nabla_x {r} \big] \,\mathrm{d}\sigma\\
&\qquad\qquad\qquad\qquad -\int_0^t \varphi_R(\|{\vc{u}}\|_{2,\infty})P_n\big[D({r}) {\rm div}_x \mathbb{S} (\nabla_x {\vc{u}})\big] \,\mathrm{d}\sigma\\
(r,\vc{u})&\mapsto M(r,\mathbf{u})_t:={\vc{u}}(t)-{\vc{u}}(0) + \int_0^t\varphi_R(\|{\vc{u}}\|_{2,\infty})\big[ {\vc{u}}\cdot \nabla_x {\vc{u}} +{r}\nabla_x {r} \big] \,\mathrm{d}\sigma\\
&\qquad\qquad\qquad\qquad -\int_0^t \varphi_R(\|{\vc{u}}\|_{2,\infty})\big[D({r}) {\rm div}_x \mathbb{S} (\nabla_x {\vc{u}})\big] \,\mathrm{d}\sigma.
\end{align*}
Since $(r_n,\vc{u}_n)$ satisfies \eqref{E4'} on the original probability space, we have that
$$M^n(r_n,u_n)_t=\int_0^t\varphi_R(\|\vc{u}_n\|_{2,\infty})P_n\mathbb F(r_n,u_n)\,\mathrm{d} W.$$
Hence $M^n(r_n,\vc{u}_n)$ is an $L^2(\mathbb{T}^N)$-valued martingale and if $(f_j)$ is an orthonormal basis of $L^2(\mathbb{T}^N)$ then for all $j\in\mathbb{N}$
$$\mathbb{E}\left[h(\mathbf{r}_s r_n,\mathbf{r}_s \vc{u}_n,\mathbf{r}_s W)\langle M^n(r_n,\vc{u}_n)_t-M^n(r_n,\vc{u}_n)_s,f_j\rangle\right]=0,$$
\begin{align*}
\mathbb{E}\bigg[h(\mathbf{r}_s r_n,\mathbf{r}_s \vc{u}_n,\mathbf{r}_s W)&\bigg(\langle M^n(r_n,\vc{u}_n)_t,f_j\rangle^2-\langle M^n(r_n,\vc{u}_n)_s,f_j\rangle^2\\
&\qquad-\int_s^t\varphi_R(\|\vc{u}_n\|_{2,\infty})\|(P_n\mathbb{F}(r_n,\vc{u}_n))^*f_j\|_\mathfrak{U}^2\,\mathrm{d} \sigma\bigg)\bigg]=0,
\end{align*}
\begin{align*}
\mathbb{E}\bigg[h(\mathbf{r}_s r_n,\mathbf{r}_s \vc{u}_n,\mathbf{r}_s W)&\bigg(\beta_k(t)\langle M^n(r_n,\vc{u}_n)_t,f_j\rangle-\beta_k(s)\langle M^n(r_n,\vc{u}_n)_s,f_j\rangle\\
&\qquad-\int_s^t\varphi_R(\|\vc{u}_n\|_{2,\infty})\langle e_k,(P_n\mathbb{F}(r_n,\vc{u}_n))^*f_j\rangle_\mathfrak{U}\,\mathrm{d} \sigma\bigg]=0.
\end{align*}
Equality of laws implies the corresponding three expressions for $(\tilde r_n,\tilde \vc{u}_n)$ and finally due to Proposition \ref{prop:skorokhod1} and the uniform moment estimates from \eqref{E14''} we may pass to the limit to deduce
$$\tilde\mathbb{E}\left[h(\mathbf{r}_s \tilde r,\mathbf{r}_s \tilde\vc{u},\mathbf{r}_s \tilde W)\langle M(\tilde r,\tilde\vc{u})_t-M^n(\tilde r,\tilde\vc{u})_s,f_j\rangle\right]=0,$$
\begin{align*}
\tilde\mathbb{E}\bigg[h(\mathbf{r}_s \tilde r,\mathbf{r}_s \tilde\vc{u},\mathbf{r}_s\tilde W)&\bigg(\langle M(\tilde r,\tilde\vc{u})_t,f_j\rangle^2-\langle M(\tilde r,\tilde\vc{u})_s,f_j\rangle^2\\
&\qquad-\int_s^t\varphi_R(\|\tilde u\|_{2,\infty})\|(\mathbb{F}(\tilde r,\tilde\vc{u}))^*f_j\|_\mathfrak{U}^2\,\mathrm{d} \sigma\bigg)\bigg]=0,
\end{align*}
\begin{align*}
\tilde\mathbb{E}\bigg[h(\mathbf{r}_s \tilde r,\mathbf{r}_s \tilde \vc{u},\mathbf{r}_s \tilde W)&\bigg(\tilde\beta_k(t)\langle M(\tilde r,\tilde\vc{u})_t,f_j\rangle-\tilde\beta_k(s)\langle M(\tilde r,\tilde\vc{u})_s,f_j\rangle\\
&\qquad-\int_s^t\varphi_R(\|\tilde u\|_{2,\infty})\langle e_k,(\mathbb{F}(\tilde r,\tilde\vc{u}))^*f_j\rangle_\mathfrak{U}\,\mathrm{d} \sigma\bigg]=0.
\end{align*}
According to \cite[Proposition A.1]{degen} this finally yields \eqref{E4'}
and completes { the existence part of the proof of Theorem \ref{thm:appr}. Note that the strong continuity of
$r$ and $\vc{u}$ in $W^{s,2}(\mathbb{T}^N)$ $\mathbb{P}\text{-a.s.}$ can be deduced directly from the equations.} Indeed, using the variational approach, the momentum equation \eqref{E4'} is solved in the Gelfand triplet
$$W^{s+1,2}(\mathbb{T}^N;\mathbb{R}^N)\hookrightarrow W^{s,2}(\mathbb{T}^N;\mathbb{R}^N)\hookrightarrow W^{s-1,2}(\mathbb{T}^N;\mathbb{R}^N),$$
the stochastic integral has continuous trajectories in $W^{s,2}(\mathbb{T}^N;\mathbb{R}^N)$ due to the uniform estimates, Corollary \ref{Cflan} (part (ii)) and Remark \ref{Rflan}, while
the coefficients of the deterministic parts in the momentum equation belong to the space $L^2(0,T; W^{s-1,2}(\mathbb{T}^N; \mathbb{R}^N))$. Hence \cite[Theorem 3.1]{KrRo} applies and yields the desired continuity of the velocity field $\vc{u}$.
The continuity of $r$ then follows from the equation of continuity.
\subsection{Pathwise uniqueness}
\label{subsec:uniq}
To show pathwise uniqueness, we mimick the approach of Subsection \ref{UNIF}. The difference of two solutions
$(r^j, \vc{u}^j)$, $j =1,2$, satisfies
\begin{equation}\label{UU1}
\begin{split}
{\rm d} \partial^\alpha_x (r^1 - r^2) &= - \varphi_R \left( \| \vc{u}^1 \|_{W^{2, \infty}} \right) \partial^\alpha_x \left( \vc{u}^1 \cdot \nabla_x r^1 + \frac{\gamma-1}{2} r^1 {\rm div}_x \vc{u}^1 \right) \,{\rm d} t
\\ &+ \varphi_R \left( \| \vc{u}^2 \|_{W^{2, \infty}} \right) \partial^\alpha_x \left( \vc{u}^2 \cdot \nabla_x r^2 + \frac{\gamma-1}{2} r^2 {\rm div}_x \vc{u}^2 \right) \,{\rm d} t ,
\end{split}
\end{equation}
and
\[
\begin{split}
{\rm d} \partial^\alpha_x (\vc{u}_1 - \vc{u}_2 ) &= - \varphi_R \left( \| \vc{u}_1 \|_{W^{2, \infty}} \right) \partial^\alpha_x \Big( \vc{u}_1 \cdot \nabla_x \vc{u}_1 + r_1 \nabla_x r_1 - D(r_1) {\rm div}_x \mathbb{S}(\nabla_x \vc{u}_1) \Big) \,{\rm d} t \\
&+\varphi_R \left( \| \vc{u}_2 \|_{W^{2, \infty}} \right) \partial^\alpha_x \Big( \vc{u}_2 \cdot \nabla_x \vc{u}_2 + r_2 \nabla_x r_2 - D(r_2) {\rm div}_x \mathbb{S}(\nabla_x \vc{u}_2) \Big) \,{\rm d} t \\
&+ \Big[ \varphi_R \left( \| \vc{u}_1 \|_{W^{2, \infty}} \right) \partial^\alpha_x \mathbb{F}(r_1, \vc{u}_1) - \varphi_R \left( \| \vc{u}_2 \|_{W^{2, \infty}} \right) \partial^\alpha_x \mathbb{F}(r_2, \vc{u}_2)
\Big] {\rm d}W
\end{split}
\]
for $|\alpha| \leq m$.
Multiplying (\ref{UU1}) on $\partial^\alpha_x (r^1 - r^2)$, we get
\begin{equation}\label{UU2}
\begin{split}
\frac{1}{2} {\rm d} \left| \partial^\alpha_x (r^1 - r^2) \right|^2 &= - \varphi_R \left( \| \vc{u}^1 \|_{W^{2, \infty}} \right) \partial^\alpha_x \left( \vc{u}^1 \cdot \nabla_x r^1 + \frac{\gamma-1}{2} r^1 {\rm div}_x \vc{u}^1 \right) \partial^\alpha_x (r^1 - r^2)\,{\rm d} t
\\ &+ \varphi_R \left( \| \vc{u}^2 \|_{W^{2, \infty}} \right) \partial^\alpha_x \left( \vc{u}^2 \cdot \nabla_x r^2 + \frac{\gamma-1}{2} r^2 {\rm div}_x \vc{u}^2 \right) \partial^\alpha_x (r^1 - r^2)\,{\rm d} t .
\end{split}
\end{equation}
Similarly, using It\^{o}'s product rule we obtain
\begin{equation} \label{UU3}
\begin{split}
\frac{1}{2} & {\rm d} \left| \partial^\alpha_x (\vc{u}^1 - \vc{u}^2 ) \right|^2 \\ &= - \varphi_R \left( \| \vc{u}^1 \|_{W^{2, \infty}} \right) \partial^\alpha_x \Big( \vc{u}^1 \cdot \nabla_x \vc{u}^1 + r^1 \nabla_x r^1 - D(r^1) {\rm div}_x \mathbb{S}(\nabla_x \vc{u}^1) \Big) \cdot \partial^\alpha_x (\vc{u}^1 - \vc{u}^2 ) \,{\rm d} t \\
&+\varphi_R \left( \| \vc{u}^2 \|_{W^{2, \infty}} \right) \partial^\alpha_x \Big( \vc{u}^2 \cdot \nabla_x \vc{u}^2 + r^2 \nabla_x r^2 - D(r^2) {\rm div}_x \mathbb{S}(\nabla_x \vc{u}^2) \Big) \cdot \partial^\alpha_x (\vc{u}^1 - \vc{u}^2 )\,{\rm d} t \\
&+ \Big[ \varphi_R \left( \| \vc{u}^1 \|_{W^{2, \infty}} \right) \partial^\alpha_x \mathbb{F}(r^1, \vc{u}^1) - \varphi_R \left( \| \vc{u}^2 \|_{W^{2, \infty}} \right) \partial^\alpha_x \mathbb{F}(r^2, \vc{u}^2)
\Big] \cdot \partial^\alpha_x (\vc{u}^1 - \vc{u}^2 ) {\rm d}W\\
&+ \frac{1}{2} \Big( \varphi_R \left( \| \vc{u}^1 \|_{W^{2, \infty}} \right) \partial^\alpha_x \mathbb{F}(r^1, \vc{u}^1) - \varphi_R \left( \| \vc{u}^2 \|_{W^{2, \infty}} \right) \partial^\alpha_x \mathbb{F}(r^2, \vc{u}^2)
\Big)^2 {\rm d}t
\end{split}
\end{equation}
Now observe, by virtue of the standard embedding relation,
\[
\left| \varphi_R \left( \| \vc{u}^1 \|_{W^{2, \infty}} \right)- \varphi_R \left( \| \vc{u}^2 \|_{W^{2, \infty}} \right)
\right| \leq c_1(R) \left\| \vc{u}^1 - \vc{u}^2 \right\|_{W^{2, \infty}} \leq c_2 (R) \left\| \vc{u}^1 - \vc{u}^2 \right\|_{W^{m, 2}}
\]
as soon as $m > \frac{N}{2} + 2$. Thus we sum \eqref{UU2}, \eqref{UU3}, integrate over the physical space, and perform the same estimates as in Section \ref{UNIF} noting that the highest order terms in \eqref{UU2} read
\[
\begin{split}
\varphi_R \left( \| \vc{u}^1 \|_{W^{2, \infty}} \right) &\int_{ \mathbb{T}^N}\left( \vc{u}^1 \cdot \nabla_x \partial^\alpha_x r^1 - \vc{u}^2 \cdot \nabla_x \partial^\alpha_x r^2 \right)
\partial^\alpha_x \left( r^1 - r^2 \right) \ \,{\rm d} {x}\\
+ \frac{\gamma - 1}{2} \varphi_R \left( \| \vc{u}^1 \|_{W^{2, \infty}} \right) &\int_{ \mathbb{T}^N} \left( r^1 {\rm div}_x \partial^\alpha_{x} \vc{u}^1 - r^2 {\rm div}_x \partial^\alpha_{x} \vc{u}^2 \right) \partial^\alpha_x \left( r^1 - r^2 \right) \,{\rm d} {x}\\
= \varphi_R \left( \| \vc{u}^1 \|_{W^{2, \infty}} \right) &\int_{ \mathbb{T}^N}\left( (\vc{u}^1 - \vc{u}^2) \cdot \nabla_x \partial^\alpha_x r^1
\right)\partial^\alpha_x \left( r^1 - r^2 \right)
+ \frac{1}{2} {\rm div}_x \vc{u}^2
\left| \partial^\alpha_x \left( r^1 - r^2 \right) \right|^2 \ \,{\rm d} {x}\\
+ \frac{\gamma - 1}{2} \varphi_R \left( \| \vc{u}^1 \|_{W^{2, \infty}} \right) &\int_{ \mathbb{T}^N} (r^1 - r^2) {\rm div}_x \partial^\alpha_{x} \vc{u}^2 \partial^\alpha_x \left( r^1 - r^2 \right) \,{\rm d} {x} \\
+ \frac{\gamma - 1}{2} \varphi_R \left( \| \vc{u}^1 \|_{W^{2, \infty}} \right) &\int_{ \mathbb{T}^N} r^1 {\rm div}_x \partial^\alpha_{x} (\vc{u}^1 - \vc{u}^2)
\partial^\alpha_x \left( r^1 - r^2 \right) \,{\rm d} {x}
\end{split}
\]
where the last integral
\[
\varphi_R \left( \| \vc{u}^1 \|_{W^{2, \infty}} \right) \int_{\mathbb{T}^N} r^1 {\rm div}_x \left( \partial^\alpha_x (\vc{u}^1 - \vc{u}^2) \right) \partial^\alpha_x
(r^1 - r^2)\ \,{\rm d} {x}
\]
cancels, after by parts integration, with its counterpart in \eqref{UU3}, namely
\[
\varphi_R \left( \| \vc{u}^1 \|_{W^{2, \infty}} \right) \int_{\mathbb{T}^N} r^1 \left( \partial^\alpha_x (\vc{u}^1 - \vc{u}^2) \right) \cdot \nabla_x \partial^\alpha_x
(r^1 - r^2)\ \,{\rm d} {x}.
\]
Thus we deduce, exactly as in Subsection \ref{UNIF},
\begin{equation*
\begin{split}
{\rm d} &\left( \left\| r^1 - r^2 \right\|^2_{W^{m,2}} + \left\| \vc{u}^1 - \vc{u}^2 \right\|^2_{W^{m,2}} \right)
\\
& \leq c(R) \left[ \left( 1 + \sum_{j=1}^2 \left( \| r^j \|_{W^{m+1,2}}^2 + \| \vc{u}^j \|_{W^{m+2,2}}^2 \right)\right)
\left( \left\| r^1 - r^2 \right\|^2_{W^{m,2}} + \left\| \vc{u}^1 - \vc{u}^2 \right\|^2_{W^{m,2}} \right) \right] \,{\rm d} t \\
& + \Big[ \varphi_R \left( \| \vc{u}^1 \|_{W^{2, \infty}} \right) \partial^\alpha_x \mathbb{F}(r^1, \vc{u}^1) - \varphi_R \left( \| \vc{u}^2 \|_{W^{2, \infty}} \right) \partial^\alpha_x \mathbb{F}(r^2, \vc{u}^2)
\Big] \cdot \partial^\alpha_x (\vc{u}^1 - \vc{u}^2 ) {\rm d}W,
\end{split}
\end{equation*}
where $m > \frac{N}{2} + 2$.
Let us now set
$$G(t)=c(R) \left( 1 + \sum_{j=1}^2 \left( \| r^j(t) \|_{W^{m+1,2}}^2 + \| \vc{u}^j (t)\|_{W^{m+2,2}}^2 \right)\right)$$
and observe that if $s\geq m+1$ then the a priori estimates from Subsection \ref{UNIF} imply in particular that $G\in L^1(0,T)$ a.s. Applying the It\^o formula to the product we therefore obtain
\begin{align*}
&{\rm d}\left[\mathrm{e}^{-\int_0^t G(\sigma){\rm d}\sigma}\Big(\|r^1-r^2\|_{m,2}^2+\|\vc{u}^1-\vc{u}^2\|_{m,2}^2\Big)\right]\\
&=-G(t)\mathrm{e}^{-\int_0^t G(\sigma){\rm d}\sigma}\Big(\|r^1-r^2\|_{m,2}^2+\|\vc{u}^1-\vc{u}^2\|_{m,2}^2\Big)\,{\rm d} t \\
&\qquad+\mathrm{e}^{-\int_0^t G(\sigma){\rm d}\sigma}{\rm d}\Big(\|r^1-r^2\|_{m,2}^2+\|\vc{u}^1-\vc{u}^2\|_{m,2}^2\Big)\\
&\leq \mathrm{e}^{-\int_0^t G(\sigma){\rm d}\sigma}\Big[ \varphi_R \left( \| \vc{u}^1 \|_{W^{2, \infty}} \right) \partial^\alpha_x \mathbb{F}(r^1, \vc{u}^1) - \varphi_R \left( \| \vc{u}^2 \|_{W^{2, \infty}} \right) \partial^\alpha_x \mathbb{F}(r^2, \vc{u}^2)
\Big] \cdot \partial^\alpha_x (\vc{u}^1 - \vc{u}^2 ) {\rm d}W(t).
\end{align*}
Integrating over $[0,t]$ and taking expectation we observe that the stochastic integral vanishes due to the assumptions on $r,\vc{u}$ in Definition \ref{def:strsolmart} and consequently we may infer that
\[
\mathbb{E} \left[\mathrm{e}^{-\int_0^t G(\sigma){\rm d}\sigma}\Big( \left\| r^1(t) - r^2 (t)\right\|^2_{W^{m,2}} + \left\| \vc{u}^1(t) - \vc{u}^2 (t)\right\|^2_{W^{m,2}} \Big)\right] = 0
\]
whenever
\[
\mathbb{E} \left[ \left\| r^1_0 - r^2_0 \right\|^2_{W^{m,2}} + \left\| \vc{u}^1_0 - \vc{u}^2_0 \right\|^2_{W^{m,2}} \right] = 0.
\]
Since
$$\mathrm{e}^{-\int_0^t G(\sigma){\rm d}\sigma}>0\qquad\mathbb{P}\text{-a.s.}$$
and the trajectories of $r^i,\vc{u}^i$, $i=1,2,$ are continuous in $W^{m,2}(\mathbb{T}^N)$,
the pathwise
uniqueness from Theorem \ref{thm:appr} follows.
\subsection{\textcolor{black}{Existence of a strong pathwise approximate solution}}
\label{subsec:pathwiseexist}
In order to complete the proof of Theorem \ref{thm:appr}, we make use of
the Gy\"{o}ngy--Krylov characterization of convergence in probability introduced in \cite[Lemma 1.1]{krylov}.
It applies to situations when pathwise uniqueness and existence of a martingale solution are valid and allows to establish existence of a pathwise solution.
\begin{Lemma} \label{GK}
Let $(\mathcal{X},\tau)$ be a Polish space and $\left\{ Y_n; n \in \mathbb{}{N} \right\}$ a family of random variables ranging in $\mathcal{X}$. Let
\[
\nu_{m,n} \equiv \mathbb{P} \left[ [Y_m, Y_n ] \in B \right], \ B \ \mbox{a Borel set in}\ \mathcal{X} \times \mathcal{X}
\]
be the collection of joint laws.
Then $Y_n$ converges in probability only if any subsequence of joint probability laws $\{ \nu_{m_k, n_k} \}_{k \geq 0}$
contains a weakly converging subsequence to a $\nu$ such that
\[
\nu \left[ (u,v) \in \mathcal{X} \times \mathcal{X}, u = v \right] = 1.
\]
\end{Lemma}
We start with a regular initial initial data corresponding to $s > \frac{N}{2} + 3$ required for pathwise uniqueness of
strong solutions to the approximate problem \eqref{E3'}, \eqref{E4'}.
Going back to the construction of approximate solution we
denote by $\mu_{m,n}$ the joint law of
$$(r_m,\mathbf{u}_m,r_n,\mathbf{u}_n)\quad\text{on the space}\quad \mathcal{X}_r\times \mathcal{X}_{\mathbf{u}}\times \mathcal{X}_r\times \mathcal{X}_{\mathbf{u}},$$
where $r_m$, $\vc{u}_n$, $r_n$, $\vc{u}_n$ are the Galerkin solutions.
In addition, denoting $\mu_W$ the law of $W$ on $\mathcal{X}_W$, we introduce
the extended path space
$$\mathcal{X}^J= \mathcal{X}_r\times \mathcal{X}_{\mathbf{u}}\times \mathcal{X}_r\times \mathcal{X}_{\mathbf{u}}\times\mathcal{X}_W$$
and denote by $\nu_{m,n}$ the joint law of
$$(r_m,\mathbf{u}_m,r_n,\mathbf{u}_n,W)\quad\text{on}\quad\mathcal{X}^J.$$
The following result follows easily from the arguments of Subsection \ref{subsec:comp}.
\begin{Proposition}
The collection $\{\nu_{m,n};\,m,n\in\mathbb{N}\}$ is tight on $\mathcal{X}^J$.
\end{Proposition}
Let us take any subsequence $\{\nu_{m_k,n_k};\,k\in\mathbb{N}\}$. By the Skorokhod representation theorem, we infer (for a further subsequence but without loss of generality we keep the same notation) the existence of a probability space $(\bar{\Omega},\bar{\mathscr{F}},\bar{\mathbb{P}})$ with a sequence of random variables
$$(\hat r_{n_k},\hat\mathbf{u}_{n_k},\check r_{m_k},\check\mathbf{u}_{m_k},\bar W_k),\quad k\in\mathbb{N},$$
conver\-ging almost surely in $\mathcal{X}^J$ to a random variable
$$(\hat r,\hat\mathbf{u},\check r,\check\mathbf{u},\bar W)$$
and
$$\bar{\mathbb{P}}\big((\hat r_{n_k},\hat\mathbf{u}_{n_k},\check r_{m_k},\check\mathbf{u}_{m_k},\bar W_k)\in\,\,\cdotp\big)=\nu^{n_k,m_k}(\cdot).$$
Observe that in particular, $\mu^{n_k,m_k}$ converges weakly to a measure $\mu$ defined by
$$\mu(\cdot)=\bar{\mathbb{P}}\big((\hat r,\hat\mathbf{u},\check r,\check\mathbf{u})\in\,\,\cdotp\big).$$
As the next step, we recall the technique established in Subsection \ref{subsec:ident}. Analogously, it can be applied to both
$$(\hat r_{n_k},\hat\mathbf{u}_{n_k},\bar W_k),\;(\hat r,\hat\mathbf{u},\bar W)$$
and
$$(\check r_{m_k},\check\mathbf{u}_{m_k},\bar W_k),\;(\check r,\check\mathbf{u},\bar W)$$
in order to show that $(\hat r, \hat\mathbf{u},\bar W)$ and $(\check r, \check\mathbf{u},\bar W)$ are strong martingale solutions to the approximate system \eqref{E3'}--\eqref{E4'}.
Finally, since $r_{n_k}(0)=r_{m_k}(0)=r_{0}$, it follows that
\begin{align*}
\bar\mathbb{P}(\hat r(0)=\check r(0))=1.
\end{align*}
Since $\vc{u}_{n_k}(0)=P_{n_k}\vc{u}_0$, $\vc{u}_{m_k}(0)=P_{m_k}\vc{u}_0$, we obtain for every $\ell\leq n_k\wedge m_k$
\begin{align*}
\bar\mathbb{P}(P_{\ell}\hat \vc{u}_{n_k}(0)=P_\ell\check \vc{u}_{m_k}(0))=\mathbb{P}(P_{\ell} \vc{u}_{n_k}(0)=P_\ell\vc{u}_{m_k}(0))=1
\end{align*}
which leads to
$$\bar\mathbb{P}(\hat \vc{u}(0)=\check\vc{u}(0))=1.$$
Hence, in accordance with the pathwise uniqueness established in Theorem \ref{thm:appr}, we get the desired conclusion
\begin{align*}
\mu&\Big((r_1,\mathbf{u}_1,r_2,\mathbf{u}_2);\;(r_1,\mathbf{u}_1)=(r_2,\mathbf{u}_2)\Big)=\bar{\mathbb{P}}\Big((\hat r,\hat\mathbf{u})=(\check r,\check\mathbf{u})\Big)=1.
\end{align*}
Thus, we have all in hand to apply Lemma \ref{GK}, which implies that the original sequence $(r_n,\mathbf{u}_n)$ defined on the initial probability space $(\Omega,\mathfrak F,\mathbb{P})$ converges in probability in the topology of $\mathcal{X}_r\times \mathcal{X}_{\mathbf{u}}$ to a random variable $(r,\mathbf{u})$. Without loss of generality, we assume that the convergence is almost sure and again by the method from Subsection \ref{subsec:ident} we finally deduce that
the limit
is the unique strong pathwise solution to the approximate problem \eqref{E3'}--\eqref{E4'}. Let us denote this solution by $(r_R, \vc{u}_R)$.
\section{\textcolor{black}{Proof of the main result, Theorem \ref{thm:main}}}
\label{subsec:strong}
Throughout the remainder of the paper, we go back to the original problem \eqref{P1}--\eqref{P4} and prove Theorem \ref{thm:main}. Our approach relies on the equivalence between \eqref{P1}--\eqref{P2} and \eqref{E3}--\eqref{E4} which is valid provided the density remains strictly positive, cf. Subsection \ref{subsec:symsyst}. In addition, introducing suitable stopping times allows us to work with \eqref{E3'}--\eqref{E4'} instead of \eqref{E3}--\eqref{E4} and therefore we may apply the results of the previous section, namely, Theorem \ref{thm:appr}.
Nevertheless, there is an additional difficulty that originates in the fact that the initial condition is not assumed to be integrable in $\omega$ and the initial density is not bounded from below by a positive constant. Consequently, the a priori estimates from Subsection \ref{UNIF} are no longer valid and the initial condition has to be truncated for Theorem \ref{thm:appr} to be applicable.
For this reason, the proof of uniqueness as well as existence of a local strong pathwise solution is divided into two steps. First, we consider an additional assumption the initial data so that Theorem \ref{thm:appr} applies. Second, we avoid this hypothesis.
\subsection{Uniqueness}
\label{s:un}
Let us first take an additional assumption that
\begin{equation} \label{DATA1}
\varrho_0 \in L^\infty(\Omega; \mathfrak{F}_0, \mathbb{P}, W^{s,2}(\mathbb{T}^N) ),\ \vc{u}_0 \in L^\infty(\Omega; \mathfrak{F}_0, \mathbb{P}, W^{s,2}(\mathbb{T}^N,\mathbb{R}^N) ),
\ \varrho_0 > \underline{\varrho} > 0
\end{equation}
for some deterministic constant $\underline{\varrho} > 0$.
In this case, the pathwise uniqueness of \eqref{P1}--\eqref{P4} is a simple consequence of the pathwise uniqueness for \eqref{E3'}--\eqref{E4'} proved in Theorem \ref{thm:appr}.
To be more precise, let $[\varrho^i,\vc{u}^i,(\mathfrak{t}_R^i),\mathfrak{t}^i]$, $i=1,2,$ be two maximal strong pathwise solutions to \eqref{P1}--\eqref{P4} starting from $[\varrho_0,\vc{u}_0]$ satisfying \eqref{DATA1}.
Then
$$\left[r^i:=\sqrt{\frac{2a\gamma}{\gamma-1}}{(\varrho^i)}^{\frac{\gamma-1}{2}},\vc{u}^i\right],\quad i=1,2,$$
both solve \eqref{E3'}--\eqref{E4'} up to the stopping time $\mathfrak{t}_R^1\wedge\mathfrak{t}_R^2$ and their initial conditions coincide. Besides, the a priori estimates from Subsection \ref{UNIF} as well as the pathwise uniqueness from Subsection \ref{subsec:uniq} apply up to the stopping time $\mathfrak{t}_R^1\wedge\mathfrak{t}_R^2$
and we deduce that
$$
\mathbb{P}\Big([\varrho^1,\vc{u}^1](t\wedge\mathfrak{t}^1_R\wedge\mathfrak{t}^2_R)=[\varrho^2,\vc{u}^2](t\wedge\mathfrak{t}^1_R\wedge\mathfrak{t}^2_R),\ \text{for all }t\in[0,T]\Big)=1.
$$
Sending $R\to\infty$ implies by dominated convergence
$$
\mathbb{P}\Big([\varrho^1,\vc{u}^1](t\wedge\mathfrak{t}^1\wedge\mathfrak{t}^2)= [\varrho^2,\vc{u}^2](t\wedge\mathfrak{t}^1\wedge\mathfrak{t}^2),\ \text{for all }t\in[0,T]\Big)=1.
$$
As a consequence, the two solutions coincide up to the stopping time $\mathfrak{t}^1\wedge\mathfrak{t}^2$ and due to maximality of $\mathfrak{t}^1$ as well as $\mathfrak{t}^2$, it necessarily follows that $\mathfrak{t}^1=\mathfrak{t}^2$ a.s. This completes the proof of uniqueness under the additional assumption \eqref{DATA1}.
Let $(\varrho_0,\vc{u}_0)$ satisfy the hypotheses of Theorem \ref{thm:main}, define the set
$$ \Omega_{K}=\left\{\omega\in\Omega \ \Big|\ \|\vc{u}_0(\omega)\|_{s,2}<K, \ \|r_0(\omega)\|_{s,2}<K,\ \inf_{\mathbb{T}^N} r_0(\omega)>\frac{1}{K}\right\}$$
and note that $\Omega=\cup_{K\in\mathbb{R}}\Omega_{K}$.
Therefore, since $\Omega_K$ is $\mathfrak{F}_0$-measurable, the a priori estimates from Subsection \ref{UNIF} can be employed on $\Omega_K$ to obtain
\begin{align} \label{E14''}
&\mathbb{E}\bigg[ \mathbf{1}_{\Omega_K}\left( \sup_{t\in[0,T\wedge\mathfrak{t}^i_R]}\left\| (r^i(t),\mathbf{u}^i(t)) \right\|^2_{s,2} + \int_0^{T\wedge\mathfrak{t}^i_R}\|\vc{u}^i(t)\|_{s+1,2}^2 \,{\rm d} t \right)^p \bigg]
\lesssim c(R,T, s,K).
\end{align}
Accordingly, the method of pathwise uniqueness from Subsection \ref{subsec:uniq} can be applied on $\Omega_K$ which yields
$$
\mathbb{P}\Big(\mathbf{1}_{\Omega_K}[\varrho^1,\vc{u}^1](t\wedge\mathfrak{t}^1_R\wedge\mathfrak{t}^2_R)=\mathbf{1}_{\Omega_K}[\varrho^2,\vc{u}^2](t\wedge\mathfrak{t}^1_R\wedge\mathfrak{t}^2_R),\ \text{for all }t\in[0,T]\Big)=1
$$
and since $\mathbf{1}_{\Omega_K}\to \mathbf{1}_\Omega$, $\mathfrak{t}^i_R\to\mathfrak{t}^i$, $i=1,2,$ a.s., we may send $R,K\to\infty$ and apply the dominated convergence theorem to deduce that
$$
\mathbb{P}\Big([\varrho^1,\vc{u}^1](t\wedge\mathfrak{t}^1\wedge\mathfrak{t}^2)= [\varrho^2,\vc{u}^2](t\wedge\mathfrak{t}^1\wedge\mathfrak{t}^2),\ \text{for all }t\in[0,T]\Big)=1.
$$
The uniqueness part of Theorem \ref{thm:main} is thus complete.
\subsection{\textcolor{black}{Existence of a local strong solution for bounded initial data}}
\label{ex1}
Finally, we have all in hand to go back to our original problem \eqref{P1}--\eqref{P4} and establish the existence of a local strong pathwise solution with an a.s. strictly positive stopping time.
Let us first take the additional assumption \eqref{DATA1}.
Having constructed strong solutions for the approximate problem \eqref{E3'}--\eqref{E4'} in Subsection \ref{subsec:pathwiseexist}, which we denoted by $(r_R, \vc{u}_R)$, we define
\[
\tau_R = \inf \left\{ t \in [0,T]\ \Big| \ \| \vc{u}_R (t) \|_{2,\infty} \geq R \right\}
\]
(with the convention $\inf \emptyset=T$). Since $\vc{u}_R$ has continuous trajectories in $W^{s,2}(\mathbb{T}^N,\mathbb{R}^N)$ which is embedded into $ W^{2,\infty}(\mathbb{T}^N,\mathbb{R}^N)$, $\tau_R$ is a well-defined stopping time. Moreover, due to \eqref{DATA1}, the stopping time $\tau_R$ is a.s. positive
provided $R$ is chosen large enough. Next, we recall that, as stated in Theorem \ref{thm:appr},
$$r_R\geq \underline{r}_R>0\quad\text{for a.e.}\quad(\omega,t,x),$$
for some deterministic constant $\underline{r}_R$. Consequently, the density given by
$${\varrho}:=\left(\frac{\gamma-1}{2a\gamma}\right)^{\frac{1}{\gamma-1}}r^{\frac{2}{\gamma-1}}_R,$$
remains uniformly positive as well. Therefore, the unique solution $(r_R,\vc{u}_R)$ of the approximated system \eqref{E3'}--\eqref{E4'} with the initial condition
$$
\left(r_0:=\sqrt{\frac{2a\gamma}{\gamma-1}}{\varrho_0}^{\frac{\gamma-1}{2}} , \vc{u}_0\right)
$$
generates a local strong
pathwise solution
\[
\left({\varrho}:=\left(\frac{\gamma-1}{2a\gamma}\right)^{\frac{1}{\gamma-1}}r^{\frac{2}{\gamma-1}}_R ,\ \vc{u}_R,\tau_R\right)
\]
of the original problem \eqref{P1}--\eqref{P4} with the initial condition $(\varrho_0,\vc{u}_0)$.
\subsection{\textcolor{black}{Existence of a local strong solution for general initial data}}
\label{subsec:ex2}
In order to relax the additional assumption upon the initial datum \eqref{DATA1}, consider again a solution $(r_R,\vc{u}_R)$ of the approximate problem \eqref{E3'}--\eqref{E4'}; now with a stopping time
\[
\tau_K = \tau^1_K \wedge \tau^2_K \wedge \tau^3_K,
\]
\[
\begin{split}
\tau^1_K &= \inf \left\{ t \in [0,T]\ \Big| \ \| \vc{u}_R (t) \|_{s,2} \geq K \right\} \\
\tau^2_K &= \inf \left\{ t \in [0,T]\ \Big| \ \| r_R (t) \|_{s,2} \geq K \right\} \\
\tau^3_K &= \inf \left\{ t \in [0,T]\ \Big| \ \inf_{\mathbb{T}^N} r_R (t) \leq \frac{1}{K} \right\}
\end{split}
\]
with $K=K(R)\to\infty$ as $R\to\infty$ and
\[
K(R) < R \min \left\{ 1 , \frac{1}{c_{1,\infty}}, \frac{1}{c_{2, \infty}} \right\},
\]
where $c_{1,\infty}$, $c_{2, \infty}$ are the constants in the embedding inequalities
\[
\| r \|_{1,\infty} \leq c_{1,\infty} \| r \|_{s,2}, \ \| \vc{u} \|_{2, \infty} \leq c_{2, \infty} \| \vc{u} \|_{s,2}.
\]
The stopping time $\tau_K$ is chosen in such a way that on $[0,\tau_K)$
$$
\sup_{t\in[0,\tau_K]}\|\vc{u}_R(t)\|_{2,\infty}<R,\quad\sup_{t\in[0,\tau_K]}\|r_R(t)\|_{1,\infty}< R,\quad \inf_{t\in[0,\tau_K]}\inf_{\mathbb{T}^N}r_R(t)>\frac{1}{R}\quad\mathbb{P}\text{-a.s.}
$$
Next we observe that Theorem \ref{thm:appr} can be used to construct solutions with the stopping time $\tau_{K}$ for general initial data
as in Theorem \ref{thm:main}. Indeed let $(r_0,\vc{u}_0)$ be an $\mathfrak{F}_0$-measurable random variable taking values in $W^{s,2}(\mathbb{T}^N)\times W^{s,2}(\mathbb{T}^N,\mathbb{R}^N)$ such that $r_0 > 0$ $\mathbb{P}$-a.s.
and define the set
\[
U_{K(R)} = \left\{ [r, \vc{u}] \in W^{s,2}(\mathbb{T}^N)\times W^{s,2}(\mathbb{T}^N,\mathbb{R}^N)\ \Big|\ \| r \|_{s,2} < K, \ \| \vc{u} \|_{s,2} < K, \ r > \frac{1}{K} \right\}.
\]
Theorem \ref{thm:appr} then provides a (unique) solution $[{r}_M, u_M]$ to \eqref{E3'}--\eqref{E4'} with $R=M$ and with the initial condition $[{r}_0, \vc{u}_0]\mathbf{1}_{ [r_0, \vc{u}_0] \in \left\{ U_{K(M)} \setminus \cup_{J=1}^{M-1} U_{K(J)} \right\}}$. It also solves the original system \eqref{E3}, \eqref{E4} up to the stopping time $\tau_{K(M)}$.
Next, we find that
\begin{equation}\label{eq:limsol}
[r, \vc{u}] = \sum_{M=1}^ \infty [r_M, \vc{u}_M] \mathbf{1}_{ [r_0, \vc{u}_0] \in \left\{ U_{K(M)} \setminus \cup_{J=1}^{M-1} U_{K(J)} \right\}},
\end{equation}
solves the same problem with the initial data $[r_0,\vc{u}_0]$ up to the a.s. strictly positive stopping time
\[
\tau = \sum_{M=1}^\infty \tau_{K(M)} \mathbf{1}_{ [r_0, \vc{u}_0] \in \left\{ U_{K(M)} \setminus \cup_{J=1}^{M-1} U_{K(J)} \right\}}.
\]
Note that in particular that $[r,\vc{u}]$ has a.s. continuous trajectories in $W^{s,2}(\mathbb{T}^N)\times W^{s,2}(\mathbb{T}^N,\mathbb{R}^N)$ and the velocity also belongs to
$ L^2(0,T;W^{s+1,2}(\mathbb{T}^N,\mathbb{R}^N))$ $\mathbb{P}$-a.s.
Indeed, there exists a disjoint collection of sets $\Omega_M\subset\Omega$, $M\in\mathbb{N}$, satisfying $\cup_M \Omega_M=\Omega$ such that $[r,\vc{u}](\omega)=[r_M,\vc{u}_M](\omega)$ for a.e. $\omega\in\Omega_M$. And due to Theorem \ref{thm:appr}, the trajectories of $[r_M,\vc{u}_M]$ are a.s. continuous in $W^{s,2}(\mathbb{T}^N)\times W^{s,2}(\mathbb{T}^N,\mathbb{R}^N)$. On the other hand, we loose the integrability in $\omega$ as the initial condition is only assumed to be in $W^{s,2}(\mathbb{T}^N)\times W^{s,2}(\mathbb{T}^N,\mathbb{R}^N)$ a.s. and no integrability in $\omega$ is assumed. In particular, the estimate \eqref{RE2} is no longer valid for the solution \eqref{eq:limsol}.
To conclude, after the straightforward transformation to the original variables $[\varrho,\vc{u}]$, we obtain the existence of a local strong pathwise solution to problem \eqref{P1}--\eqref{P4} with a strictly positive stopping time $\tau$.
\subsection{Existence of a maximal strong solution}
\label{ex3}
In order to extend the solution $(\varrho,\vc{u})$ to a maximal time of existence $\mathfrak{t}$, let $\mathcal T$ denote the set of all possible a.s. strictly positive stopping times corresponding to the solution starting from the initial datum $(\varrho_0,\vc{u}_0)$. According to the above proof, this set is nonempty. Moreover, it is closed with respect to finite minimum and finite maximum operations. More precisely,
$$\sigma_1,\sigma_2\in\mathcal{T} \Rightarrow \sigma_1\vee\sigma_2\in\mathcal{T},$$
and
$$\sigma_1,\sigma_2\in\mathcal{T} \Rightarrow \sigma_1\wedge\sigma_2\in\mathcal{T},$$
for any stopping time $\sigma_2$. Let $\mathfrak{t}=\sup_{\sigma\in\mathcal T} \sigma$.
Then
we may choose an increasing sequence $(\sigma_M)\subset\mathcal{T}$ such that $\lim_{M\to\infty}\sigma_M=\mathfrak{t}$ a.s. Let $[\varrho_M,\vc{u}_M]$ be the corresponding sequence of solutions on $[0,\sigma_M]$. Due to uniqueness, this sequence defines a solution $(\varrho,\vc{u})$ on $\cup_M[0,\sigma_M]$ by setting $(\varrho,\vc{u}):=(\varrho_M,\vc{u}_M)$ on $[0,\sigma_M]$.
For each $R\in\mathbb{N}$ we now define
\[
\tau_R=\mathfrak{t} \wedge \inf \left\{ t \in [0,T]\ \Big| \ \| \vc{u} (t) \|_{2,\infty} \geq R \right\}.
\]
Then $(\varrho,\vc{u})$ is a solution on $[0,\sigma_M\wedge\tau_R]$ and sending $M\to\infty$ we obtain that $(\varrho,\vc{u})$ is a solution on $[0,\tau_R]$. Note that $\tau_R$ is not a.s. strictly positive unless $\|\vc{u}_0\|_{2,\infty}<R.$ Nevertheless, since $\vc{u}_0\in W^{s,2}(\mathbb{T}^N,\mathbb{R}^N)$ a.s. we may deduce that for almost every $\omega$ there exists $R=R(\omega)$ such that $\mathfrak{t}_{R(\omega)}(\omega)>0$.
To guarantee the strict positivity, we combine the two sequences of stopping times $(\sigma_R)$ and $(\tau_R)$ and define $\mathfrak{t}_R=\sigma_R\vee\tau_R$. Then each triplet $(\varrho,\vc{u},\mathfrak{t}_R)$, $R\in\mathbb{N}$, is a local strong pathwise solution with an a.s. strictly positive stopping time. Next, we observe that, by repeating the construction of a local strong pathwise solution, a solution on $[0,\mathfrak{t}_R]$ can be extended to a solution on $[0,\mathfrak{t}_R+\sigma]$ for an a.s. strictly positive stopping time $\sigma$. Thus, in order to show that $\mathfrak{t}_R<\mathfrak{t}$ on $[\mathfrak{t}<T]$, assume for a contradiction that $\mathbb{P}(\mathfrak{t}_R=\mathfrak{t}<T)>0$. Then we have $\mathfrak{t}_R+\sigma\in\mathcal{T}$ and hence $\mathbb{P}(\mathfrak{t}<\mathfrak{t}_R+\sigma)>0$ which contradicts the maximality of $\mathfrak{t}$. Consequently, $(\mathfrak{t}_R)$ is an increasing sequence of stopping times converging to $\mathfrak{t}$.
Moreover, on the set $[\mathfrak{t}<T]$ we have that
$$\sup_{t\in[0,\mathfrak{t}_R]}\|\vc{u}(t)\|_{2,\infty}\geq R.$$
Thus, the existence part of Theorem \ref{thm:main} is complete.
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} | 5,137 |
{"url":"http:\/\/openstudy.com\/updates\/55910588e4b04283662708d1","text":"## anonymous one year ago What polynomial has a graph that passes through the given points? (-2, 2) (-1, -1) (1, 5) (3, 67)\n\n1. Michele_Laino\n\nA possible solution is a polynomial of fourth grade, namely: $\\Large y = a{x^4} + b{x^3} + c{x^2} + dx + e$ where, a,b , c, d, and e are real coeeficient. Now using your data we can write this algebraic system: $\\Large \\left\\{ \\begin{gathered} - 1 = a - b + c - d + e \\hfill \\\\ 5 = a + b + c + d + e \\hfill \\\\ 2 = 16a - 8b + 4c - 2d + e \\hfill \\\\ 67 = 81a + 27b + 9c + 3d + e \\hfill \\\\ \\end{gathered} \\right.$ or using matrices: $\\Large \\left( {\\begin{array}{*{20}{c}} 1&{ - 1}&1&{ - 1} \\\\ 1&1&1&1 \\\\ {16}&{ - 8}&4&{ - 2} \\\\ {81}&{27}&9&3 \\end{array}\\;\\;\\begin{array}{*{20}{c}} 1 \\\\ 1 \\\\ 1 \\\\ 1 \\end{array}} \\right)\\quad \\left( {\\begin{array}{*{20}{c}} a \\\\ b \\\\ c \\\\ \\begin{gathered} d \\hfill \\\\ e \\hfill \\\\ \\end{gathered} \\end{array}} \\right) = \\left( {\\begin{array}{*{20}{c}} { - 1} \\\\ 5 \\\\ 2 \\\\ {67} \\end{array}} \\right)$\n\n2. UsukiDoll\n\nwhat if OP doesn't know matrices? These days Linear Algebra is taught at either a very well known private high school or public university\n\n3. Michele_Laino\n\nI think that she knows algebra of matrices, since it is the only method that I know to solve her problem @UsukiDoll\n\n4. Michele_Laino\n\napplying the \"Gauss Elimination Method\" we get: the subsequent equvalent system: $\\Large \\left( {\\begin{array}{*{20}{c}} 1&{ - 1}&1&{ - 1} \\\\ 0&2&0&2 \\\\ 0&0&{ - 12}&{10} \\\\ 0&0&0&6 \\end{array}\\;\\;\\left. {\\begin{array}{*{20}{c}} 1 \\\\ 0 \\\\ { - 15} \\\\ {10} \\end{array}} \\right|\\;\\begin{array}{*{20}{c}} { - 1} \\\\ 6 \\\\ { - 6} \\\\ {50} \\end{array}} \\right)$\n\n5. Michele_Laino\n\nas we can see there are infinity solutions, which are depending on one real parameter\n\n6. Michele_Laino\n\nI think that I have made an error, since the reqiested polynoimial is a polynomial of third grade, like below: $\\Large y = a{x^3} + b{x^2} + cx + d$ In that case more simpler methods can be applied in order to solve the corresponding algebraic system\n\n7. Michele_Laino\n\nrequested*\n\n8. Michele_Laino\n\nIn that case we have a squre system, and if the matrix of coefficient is a nonsingular matrix, then we have one and only one solution for coefficients a, b, c, d\n\n9. Michele_Laino\n\noops.. square system...\n\n10. Michele_Laino\n\nusing the polynomial of third grade, I got this system: $\\Large \\left\\{ \\begin{gathered} - 1 = - a + b - c + d \\hfill \\\\ 5 = a + b + c + d \\hfill \\\\ 2 = - a + 4b - 2c + d \\hfill \\\\ 67 = 27a + 9b + 3c + d \\hfill \\\\ \\end{gathered} \\right.$\n\n11. Michele_Laino\n\nor, using matrices: $\\Large \\left( {\\begin{array}{*{20}{c}} { - 1}&1&{ - 1}&1 \\\\ 1&1&1&1 \\\\ { - 8}&4&{ - 2}&1 \\\\ {27}&9&3&1 \\end{array}\\left. {\\begin{array}{*{20}{c}} {} \\\\ {} \\\\ {} \\\\ {} \\end{array}} \\right|\\;\\begin{array}{*{20}{c}} { - 1} \\\\ 5 \\\\ 2 \\\\ {67} \\end{array}} \\right)$\n\n12. Michele_Laino\n\nagain, I apply the \"Gauss Elimination Method\" and I get: $\\Large \\left( {\\begin{array}{*{20}{c}} { - 1}&1&{ - 1}&1 \\\\ 0&2&0&2 \\\\ 0&0&6&{ - 3} \\\\ 0&0&0&{ - 20} \\end{array}\\left. {\\begin{array}{*{20}{c}} {} \\\\ {} \\\\ {} \\\\ {} \\end{array}} \\right|\\;\\begin{array}{*{20}{c}} { - 1} \\\\ 4 \\\\ 2 \\\\ { - 40} \\end{array}} \\right)$\n\n13. Michele_Laino\n\nnow being a square system, and being the matrix of coefficients of our system a nonsingular matrix, then we can state that there is one and only one solution","date":"2016-10-23 03:26:44","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\": 0.7820084095001221, \"perplexity\": 1046.1633631379925}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2016-44\/segments\/1476988719139.8\/warc\/CC-MAIN-20161020183839-00261-ip-10-171-6-4.ec2.internal.warc.gz\"}"} | null | null |
// Copyright (c) Corporation for National Research Initiatives
package org.python.core;
import java.lang.reflect.*;
import java.io.*;
import java.util.Hashtable;
import java.util.Properties;
import java.util.zip.*;
/**
* Utility functions for "import" support.
*/
public class imp
{
public static final int APIVersion = 12;
private imp() { ; }
public static PyModule addModule(String name) {
PyObject modules = Py.getSystemState().modules;
PyModule module = (PyModule)modules.__finditem__(name);
if (module != null) {
return module;
}
module = new PyModule(name, null);
modules.__setitem__(name, module);
return module;
}
private static byte[] readBytes(InputStream fp) {
try {
byte[] buf = FileUtil.readBytes( fp );
fp.close();
return buf;
} catch (IOException ioe) {
throw Py.IOError(ioe);
}
}
private static InputStream makeStream(File file) {
try {
return new FileInputStream(file);
} catch (IOException ioe) {
throw Py.IOError(ioe);
}
}
private static PyObject createFromPyClass(String name, InputStream fp,
boolean testing,
String fileName)
{
byte[] data = readBytes(fp);
int n = data.length;
int api = (data[n-4]<<24)+(data[n-3]<<16)+(data[n-2]<<8)+data[n-1];
if (api != APIVersion) {
if (testing) {
return null;
} else {
throw Py.ImportError("invalid api version("+api+" != "+
APIVersion+") in: "+name);
}
}
//System.err.println("APIVersion: "+api);
PyCode code;
try {
code = BytecodeLoader.makeCode(name+"$py", data);
} catch (Throwable t) {
if (testing)
return null;
else
throw Py.JavaError(t);
}
Py.writeComment("import", "'" + name + "' as " + fileName);
return createFromCode(name, code);
}
public static byte[] compileSource(String name, File file) {
return compileSource(name, file, null, null);
}
public static byte[] compileSource(String name, File file,
String filename, String outFilename)
{
if (filename == null) {
filename = file.toString();
}
if (outFilename == null) {
outFilename = filename.substring(0,filename.length()-3)+
"$py.class";
}
return compileSource(name, makeStream(file), filename, outFilename);
}
static byte[] compileSource(String name, InputStream fp,
String filename)
{
String outFilename = null;
if (filename != null) {
outFilename = filename.substring(0,filename.length()-3)+
"$py.class";
}
return compileSource(name, fp, filename, outFilename);
}
static byte[] compileSource(String name, InputStream fp, String filename,
String outFilename)
{
try {
ByteArrayOutputStream ofp = new ByteArrayOutputStream();
if (filename == null)
filename = "<unknown>";
org.python.parser.ast.modType node = null; //*Forte*
try {
node = parser.parse(fp, "exec", filename, null);
} finally {
fp.close();
}
org.python.compiler.Module.compile(node, ofp, name+"$py",
filename, true, false, true,
null);
if (outFilename != null) {
File classFile = new File(outFilename);
try {
FileOutputStream fop = new FileOutputStream(classFile);
ofp.writeTo(fop);
fop.close();
} catch (IOException exc) {
// If we can't write the cache file, just fail silently
}
}
return ofp.toByteArray();
} catch (Throwable t) {
throw parser.fixParseError(null, t, filename);
}
}
private static PyObject createFromSource(String name, InputStream fp,
String filename)
{
byte[] bytes = compileSource(name, fp, filename);
Py.writeComment("import", "'" + name + "' as " + filename);
PyCode code = BytecodeLoader.makeCode(name+"$py", bytes);
return createFromCode(name, code);
}
private static PyObject createFromSource(String name, InputStream fp,
String filename,
String outFilename)
{
byte[] bytes = compileSource(name, fp, filename, outFilename);
Py.writeComment("import", "'" + name + "' as " + filename);
PyCode code = BytecodeLoader.makeCode(name+"$py", bytes);
return createFromCode(name, code);
}
static PyObject createFromCode(String name, PyCode c) {
PyModule module = addModule(name);
PyTableCode code = null;
if (c instanceof PyTableCode)
code = (PyTableCode)c;
PyFrame f = new PyFrame(code, module.__dict__, module.__dict__, null);
code.call(f);
return module;
}
private static Object syspathJavaLoaderLock = new Object();
private static ClassLoader syspathJavaLoader = null;
public static ClassLoader getSyspathJavaLoader() {
synchronized (syspathJavaLoaderLock) {
if (syspathJavaLoader == null) {
syspathJavaLoader = new SyspathJavaLoader();
}
}
return syspathJavaLoader;
}
private static PyObject createFromClass(String name, Class c) {
//Two choices. c implements PyRunnable or c is Java package
//System.err.println("create from class: "+name+", "+c);
Class interfaces[] = c.getInterfaces();
for (int i=0; i<interfaces.length; i++) {
if (interfaces[i] == PyRunnable.class) {
//System.err.println("is runnable");
try {
PyObject o = createFromCode(
name, ((PyRunnable)c.newInstance()).getMain());
return o;
} catch (InstantiationException e) {
throw Py.JavaError(e);
} catch (IllegalAccessException e) {
throw Py.JavaError(e);
}
}
}
return PyJavaClass.lookup(c);
}
private static PyObject loadBuiltin(String name, PyList path) {
if (name == "sys") {
Py.writeComment("import", "'" + name + "' as sys in " +
"builtin modules");
return Py.java2py(Py.getSystemState());
}
String mod = PySystemState.getBuiltin(name);
if (mod != null) {
Class c = Py.findClassEx(mod, "builtin modules");
if (c != null) {
Py.writeComment("import", "'" + name + "' as " + mod +
" in builtin modules");
try {
return createFromClass(name, c);
}
catch (NoClassDefFoundError e) {
throw Py.ImportError("Cannot import " + name +
", missing class " +
c.getName());
}
}
}
return null;
}
private static Class findPyClass(String modName) {
if (Py.frozenPackage != null) {
modName = Py.frozenPackage+"."+modName;
}
return Py.findClassEx(modName+"$_PyInner", "precompiled");
}
private static PyObject loadPrecompiled(String name, String modName,
PyList path)
{
if (Py.frozenModules != null) {
//System.out.println("precomp: "+name+", "+modName);
Class c;
if (Py.frozenModules.get(modName+".__init__") != null) {
//System.err.println("trying: "+modName+".__init__$_PyInner");
c = findPyClass(modName+".__init__");
if (c == null) return null;
Py.writeComment("import", "'" + modName + "' as " +
"precompiled package");
//System.err.println("found: "+modName+".__init__$_PyInner");
PyModule m = addModule(modName);
m.__dict__.__setitem__("__path__", new PyList());
}
else if (Py.frozenModules.get(modName) != null) {
c = findPyClass(modName);
if (c == null) return null;
Py.writeComment("import", "'" + modName + "' as " +
"precompiled module");
}
else return null;
//System.err.println("creating: "+modName+", "+c);
return createFromClass(modName, c);
}
return null;
}
static PyObject loadFromZipFile(String name, String modName,
SyspathArchive zipArchive) {
PyObject o = null;
ZipEntry pkgEntry = null;
String entryName = name;
String pyName = entryName +".py";
String className = entryName +"$py.class";
try {
String sourceName = entryName + "/__init__.py";
String compledName = entryName + "/__init__$py.class";
ZipEntry sourceEntry = zipArchive.getEntry(sourceName);
ZipEntry compiledEntry = zipArchive.getEntry(compledName);
if (sourceEntry != null || compiledEntry != null) {
Py.writeDebug("import", "trying package: " + modName +
" in jar/zip file " + zipArchive);
PyModule m = addModule(modName);
SyspathArchive subArchive = zipArchive.makeSubfolder(name);
PyList zipPath = new PyList(new PyObject[] { subArchive });
m.__dict__.__setitem__("__path__", zipPath);
o = loadFromZipFile("__init__", modName, subArchive);
if (o != null) {
return m;
}
}
ZipEntry pyEntry = zipArchive.getEntry(pyName);
ZipEntry classEntry = zipArchive.getEntry(className);
if (pyEntry != null) {
Py.writeDebug("import", "trying source entry: " + pyName +
" from jar/zip file " + zipArchive);
if (classEntry != null) {
Py.writeDebug("import", "trying precompiled entry " +
className + " from jar/zip file " +
zipArchive);
long pyTime = pyEntry.getTime();
long classTime = classEntry.getTime();
if (classTime >= pyTime) {
InputStream is =
zipArchive.getInputStream(classEntry);
o = createFromPyClass(modName, is, true,
classEntry.getName());
if (o != null) {
return o;
}
}
}
InputStream is = zipArchive.getInputStream(pyEntry);
return createFromSource(modName, is, pyEntry.getName(), null);
}
} catch (Exception e) {
Py.writeDebug("import", "loadFromZipFile exception: " +
e.toString());
e.printStackTrace();
}
return null;
}
private static boolean isSyspathArchive(PyObject entry, boolean isDir) {
if (entry instanceof SyspathArchive)
return true;
if (isDir)
return false;
return SyspathArchive.getArchiveName(entry.toString()) != null;
}
static PyObject loadFromPath(String name, PyList path) {
return loadFromPath(name, name, path);
}
static PyObject loadFromPath(String name, String modName, PyList path) {
//System.err.println("load-from-path:"+name+" "+modName+" "+path);
PyObject o = loadPrecompiled(name, modName, path);
if (o != null) return o;
int nlen = name.length();
String pyName = name+".py";
String className = name+"$py.class";
int n = path.__len__();
for (int i=0; i<n; i++) {
PyObject entry = path.__getitem__(i);
String dirName = entry.toString();
// The empty string translates into the current working
// directory, which is usually provided on the system property
// "user.dir". Don't rely on File's constructor to provide
// this correctly.
if (dirName.length() == 0) {
dirName = null;
}
// First check for packages
File dir = new File(dirName, name);
boolean isDir = dir.isDirectory();
if (isSyspathArchive(entry, isDir)) {
Py.writeDebug("import", "trying " + modName +
" in jar/zip file " + dirName);
if (!(entry instanceof SyspathArchive)) {
try {
entry = new SyspathArchive(dirName);
path.__setitem__(i, entry);
} catch (IOException exc) {
// Silently ignore corrupt and missing .zip files.
continue;
}
}
PyObject ret = loadFromZipFile(name, modName,
(SyspathArchive)entry);
// Not found in zip/jar file check next item in path
if (ret != null) {
return ret;
}
}
if (isDir && caseok(dir, name, nlen) &&
(new File(dir, "__init__.py").isFile() ||
new File(dir, "__init__$py.class").isFile()))
{
PyList pkgPath = new PyList();
PyModule m = addModule(modName);
pkgPath.append(new PyString(dir.toString()));
m.__dict__.__setitem__("__path__", pkgPath);
o = loadFromPath("__init__", modName, pkgPath);
if (o == null)
continue;
return m;
}
// Now check for source
File pyFile = new File(dirName, pyName);
File classFile = new File(dirName, className);
Py.writeDebug("import", "trying source " + pyFile.getPath());
if (pyFile.isFile() && caseok(pyFile, pyName, nlen)) {
if (classFile.isFile() &&
caseok(classFile, className, nlen)) {
Py.writeDebug("import", "trying precompiled " +
classFile.getPath());
long pyTime = pyFile.lastModified();
long classTime = classFile.lastModified();
if (classTime >= pyTime) {
PyObject ret = createFromPyClass(
modName, makeStream(classFile), true,
classFile.getPath());
if (ret != null)
return ret;
}
}
return createFromSource(modName, makeStream(pyFile),
pyFile.getAbsolutePath());
}
// If no source, try loading precompiled
Py.writeDebug("import", "trying " + classFile.getPath());
if (classFile.isFile() && caseok(classFile, className, nlen)) {
return createFromPyClass(modName, makeStream(classFile),
false, classFile.getPath());
}
}
return null;
}
static boolean caseok(File file, String filename, int namelen) {
if (Options.caseok)
return true;
try {
File canFile = new File(file.getCanonicalPath());
return filename.regionMatches(0, canFile.getName(), 0, namelen);
} catch (IOException exc) {
return false;
}
}
static PyObject loadFromClassLoader(String name,
ClassLoader classLoader)
{
PyObject ret;
String path = name.replace('.', '/');
InputStream istream;
// First check to see if a package exists (look for name/__init__.py)
boolean loadCompiled = false;
boolean loadSource = false;
istream = classLoader.getResourceAsStream(path+"/__init__.py");
if (istream != null) {
PyModule m = addModule(name);
m.__dict__.__setitem__("__path__", Py.None);
return createFromSource(name, istream, null);
}
// Finally, try to load from source
istream = classLoader.getResourceAsStream(path+".py");
if (istream != null) return createFromSource(name, istream, null);
return null;
}
private static PyObject load(String name, PyList path) {
PyObject ret = loadBuiltin(name, path);
if (ret != null) return ret;
ret = loadFromPath(name, path);
if (ret != null) return ret;
Py.writeDebug("import", "trying " + name + " in packagemanager");
ret = PySystemState.packageManager.lookupName(name);
if (ret != null) {
Py.writeComment("import", "'" + name + "' as java package");
return ret;
}
Py.writeComment("import", "'" + name +
"' not found (=> ImportError)");
return null;
}
public static PyObject load(String name) {
return import_first(name,new StringBuffer(""));
}
private static String getParent(PyObject dict) {
PyObject tmp = dict.__finditem__("__name__");
if (tmp == null) return null;
String name = tmp.toString();
tmp = dict.__finditem__("__path__");
if (tmp != null && tmp instanceof PyList) {
return name.intern();
} else {
int dot = name.lastIndexOf('.');
if (dot == -1) return null;
return name.substring(0, dot).intern();
}
}
// can return null, None
private static PyObject import_next(PyObject mod,
StringBuffer parentNameBuffer,
String name)
{
if (parentNameBuffer.length()>0) parentNameBuffer.append('.');
parentNameBuffer.append(name);
String fullName = parentNameBuffer.toString().intern();
PyObject modules = Py.getSystemState().modules;
PyObject ret = modules.__finditem__(fullName);
if (ret != null) return ret;
if (mod == null) {
// ?? intern superfluous?
ret = load(name.intern(), Py.getSystemState().path);
} else {
ret = mod.impAttr(name.intern());
}
if (ret == null || ret == Py.None) return ret;
if (modules.__finditem__(fullName) == null)
modules.__setitem__(fullName, ret);
else
ret = modules.__finditem__(fullName);
return ret;
}
// never returns null or None
private static PyObject import_first(String name,
StringBuffer parentNameBuffer)
{
PyObject ret = import_next(null,parentNameBuffer,name);
if (ret == null || ret == Py.None)
throw Py.ImportError("no module named "+name);
return ret;
}
// Hierarchy-recursively search for dotted name in mod;
// never returns null or None
// ??pending: check if result is really a module/jpkg/jclass?
private static PyObject import_logic(PyObject mod,
StringBuffer parentNameBuffer,
String dottedName)
{
int dot = 0;
int last_dot= 0;
do {
String name;
dot = dottedName.indexOf('.', last_dot);
if (dot == -1) {
name = dottedName.substring(last_dot);
} else {
name = dottedName.substring(last_dot, dot);
}
mod = import_next(mod,parentNameBuffer,name);
if (mod == null || mod == Py.None)
throw Py.ImportError("No module named " + name);
last_dot = dot + 1;
} while (dot != -1);
return mod;
}
public static PyObject import_name(String name, boolean top,
PyObject modDict)
{
if (name.length() == 0)
throw Py.ValueError("Empty module name");
PyObject modules = Py.getSystemState().modules;
PyObject pkgMod = null;
String pkgName = null;
if (modDict != null) {
pkgName = getParent(modDict);
pkgMod = modules.__finditem__(pkgName);
if (pkgMod != null && !(pkgMod instanceof PyModule))
pkgMod = null;
}
int dot = name.indexOf('.');
String firstName;
if (dot == -1)
firstName = name;
else
firstName = name.substring(0,dot);
StringBuffer parentNameBuffer = new StringBuffer(
pkgMod != null ? pkgName : "");
PyObject topMod = import_next(pkgMod, parentNameBuffer, firstName);
if (topMod == Py.None || topMod == null) {
if (topMod == null) {
modules.__setitem__(parentNameBuffer.toString().intern(),
Py.None);
}
parentNameBuffer = new StringBuffer("");
// could throw ImportError
topMod = import_first(firstName,parentNameBuffer);
}
PyObject mod = topMod;
if (dot != -1) {
// could throw ImportError
mod = import_logic(topMod,parentNameBuffer,name.substring(dot+1));
}
if (top)
return topMod;
else
return mod;
}
public static PyObject importName(String name, boolean top) {
return import_name(name,top,null);
}
public synchronized static PyObject importName(String name, boolean top,
PyObject modDict) {
return import_name(name,top,modDict);
}
/**
* Called from jpython generated code when a statement like "import spam"
* is executed.
*/
public static PyObject importOne(String mod, PyFrame frame) {
//System.out.println("importOne(" + mod + ")");
PyObject module = __builtin__.__import__(mod,
frame.f_globals,
frame.getf_locals(),
Py.EmptyTuple);
/*int dot = mod.indexOf('.');
if (dot != -1) {
mod = mod.substring(0, dot).intern();
}*/
//System.err.println("mod: "+mod+", "+dot);
return module;
}
/**
* Called from jpython generated code when a statement like
* "import spam as foo" is executed.
*/
public static PyObject importOneAs(String mod, PyFrame frame) {
//System.out.println("importOne(" + mod + ")");
PyObject module = __builtin__.__import__(mod,
frame.f_globals,
frame.getf_locals(),
getStarArg());
// frame.setlocal(asname, module);
return module;
}
/**
* Called from jpython generated code when a stamenet like
* "from spam.eggs import foo, bar" is executed.
*/
public static PyObject[] importFrom(String mod, String[] names,
PyFrame frame)
{
return importFromAs(mod, names, null, frame);
}
/**
* Called from jpython generated code when a stamenet like
* "from spam.eggs import foo as spam" is executed.
*/
public static PyObject[] importFromAs(String mod, String[] names,
String[] asnames, PyFrame frame)
{
//StringBuffer sb = new StringBuffer();
//for(int i=0; i<names.length; i++)
// sb.append(names[i] + " ");
//System.out.println("importFrom(" + mod + ", [" + sb + "]");
PyObject[] pynames = new PyObject[names.length];
for (int i=0; i<names.length; i++)
pynames[i] = Py.newString(names[i]);
PyObject module = __builtin__.__import__(mod,
frame.f_globals,
frame.getf_locals(),
new PyTuple(pynames));
PyObject[] submods = new PyObject[names.length];
for (int i=0; i<names.length; i++) {
PyObject submod = module.__findattr__(names[i]);
if (submod == null)
throw Py.ImportError("cannot import name " + names[i]);
submods[i] = submod;
}
return submods;
}
private static PyTuple all = null;
private synchronized static PyTuple getStarArg() {
if (all == null)
all = new PyTuple(new PyString[] { Py.newString('*') });
return all;
}
/**
* Called from jpython generated code when a statement like
* "from spam.eggs import *" is executed.
*/
public static void importAll(String mod, PyFrame frame) {
//System.out.println("importAll(" + mod + ")");
PyObject module = __builtin__.__import__(mod,
frame.f_globals,
frame.getf_locals(),
getStarArg());
PyObject names;
boolean filter = true;
if (module instanceof PyJavaPackage)
names = ((PyJavaPackage)module).fillDir();
else {
PyObject __all__ = module.__findattr__("__all__");
if (__all__ != null) {
names = __all__;
filter = false;
} else names = module.__dir__();
}
loadNames(names, module, frame.getf_locals(), filter);
}
private static void loadNames(PyObject names, PyObject module,
PyObject locals, boolean filter)
{
PyObject iter = names.__iter__();
for (PyObject name; (name = iter.__iternext__()) != null; ) {
String sname = ((PyString)name).internedString();
if (filter && sname.startsWith("_")) {
continue;
} else {
try {
locals.__setitem__(sname, module.__getattr__(sname));
} catch (Exception exc) {
continue;
}
}
}
}
static PyObject reload(PyJavaClass c) {
// This is a dummy placeholder for the feature that allow
// reloading of java classes. But this feature does not yet
// work.
return c;
}
static PyObject reload(PyModule m) {
String name = m.__getattr__("__name__").toString().intern();
PyObject modules = Py.getSystemState().modules;
PyModule nm = (PyModule)modules.__finditem__(name);
if (nm == null || !nm.__getattr__("__name__").toString()
.equals(name)) {
throw Py.ImportError("reload(): module "+name+
" not in sys.modules");
}
PyList path = Py.getSystemState().path;
String modName = name;
int dot = name.lastIndexOf('.');
if (dot != -1) {
String iname = name.substring(0, dot).intern();
PyObject pkg = modules.__finditem__(iname);
if (pkg == null) {
throw Py.ImportError("reload(): parent not in sys.modules");
}
path = (PyList)pkg.__getattr__("__path__");
name = name.substring(dot+1, name.length()).intern();
}
// This should be better "protected"
//((PyStringMap)nm.__dict__).clear();
nm.__setattr__("__name__", new PyString(modName));
PyObject ret = loadFromPath(name, modName, path);
modules.__setitem__(modName, ret);
return ret;
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 8,108 |
Design Directory > Design > Awards > International Design Awards
The International Design Awards (IDA) exists to recognize, celebrate and promote legendary design visionaries and to uncover emerging talent in Architecture, Interior, Product, Graphic, and Fashion Design.
Website idesignawards.com
Category Design Awards
1318 E, 7th Street
012345 4.1 based on 7 votes
Last updated Sep 1, 2020
News About International Design Awards
International Design Awards 2020
The International Design Awards (IDA) is calling established & student designers from around the world to submit their most innovative designs to its 14th annual design competition.
IDA Announces Winners of COVID-19 Design Innovation Grant
The International Design Awards (IDA) and the European Product Design Awards (ePDA) have unveiled the winners of the 'COVID-19 Design Innovation Grant.'
IDA Launches COVID-19 Design Innovation Grant
International Design Awards (IDA), in collaboration with European Product Design Awards (ePDA), has launched 'COVID-19 Design Innovation Grant,' worth over $14,000.
International Design Awards Announces 2019 Winners
The International Design Awards (IDA) has announced the winners of its 12th edition.
International Design Awards Launches 12th Annual Global Design Competition
The International Design Awards (IDA) is calling designers and architects from around the world to submit their most innovative designs to its 12th annual design competition. The competition is open to architects, interior designers, product designers, graphic designers and fashion designers.
IDA International Design Awards 2018
The 11th annual International Design Awards (IDA) is now open for submissions. The competition is open to architects, interior designers, product designers, graphic designers and fashion designers.
The final deadline for the 2016 International Design Awards is April 30. Set to recognize outstanding talent, IDA's mission is to encourage creativity and innovation across all design fields, including fashion, graphic, interior design, architecture and product design.
IDA Poster of the Year Competition 09
The International Design Awards invite designers worldwide to showcase their creative talents.
IDA Announces 2008 Winners
The winners of the 2008 International Design Awards competition have been named. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 7,846 |
Park Shin Hye is in a photo shoot in the April issue for Vogue Korea.
i really like her hair style...really cute!!
Some of the brands used in this photo shoot includes, Miu Miu, Codes Combine, Paul & Alice, Bell & Nouveu, Comme des Garcons by 10 Corso Como and H&M.
Wow I love these pictures!!!
she's gorgeous! jealous. haha. love the photography... her facial expression made me smile.
Amazing editorial *.* loveee the black dress !
Love it. great outfit. In the last picture you remind me of madeline in Paris.
these looks are super cute ! | {
"redpajama_set_name": "RedPajamaC4"
} | 6,922 |
Q: How to convert time.strptime into an integer for datatime.date() Python 2 How would I convert the result from strptime into an integer value or a value that can be used by date.date()?
convertTOdate = time.strptime('2007-07-18 10:03:19', '%Y-%m-%d %H:%M:%S')
duedate = datetime.datetime(convertTOdate)
A Solution on stackoverflow was to do:
Use time.mktime() to convert the time tuple (in localtime) into seconds since the Epoch, then use datetime.fromtimestamp() to get the datetime object.
from time import mktime
from datetime import datetime
dt = datetime.fromtimestamp(mktime(struct))
I do not want to get the local time as it would not work with my function
I am using Python 2
Thank you
A: You can use the following approach.
from datetime import datetime
def time_in_seconds(dt):
epoch = datetime.utcfromtimestamp(0)
delta = dt - epoch
return delta.total_seconds()
convertTOdate = datetime.strptime('2007-07-18 10:03:19', '%Y-%m-%d %H:%M:%S')
duedate = time_in_seconds(convertTOdate)
returns 1184752999.0 which is equivalent to 2007-07-18 10:03:19
duedate = datetime.utcfromtimestamp(duedate)
print duedate
Just remember before using the following two:
fromtimestamp give you the date and time in local time and utcfromtimestamp gives you the date and time in UTC.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 2,075 |
It's tough flipping between ABC and NBC on Monday nights. Thankfully, Dancing with the Stars is down to an hour and a half vs. two hour format, so I can watch the second half of Heroes without interruption.
I did, however, miss Cheetah Girl Sabrina's jive last night. From the judges comments, I can assume it was her usual great performance. She got a 27 and was tied for the lead in judges scores with Mel B. and Jane Seymour. Personally, I thought Mel B's jive was technically good, but she doesn't do anything for me in the personality department. I doubt she'll make it to the finals. Jane Seymour's mother died last week and I'm sure there's a lot of sympathy for her in that regard. I felt her tango was elegant but cold, although the judges disagreed. Still, while Dr. Medicine Woman has a fairly strong fan base, I don't see her having the same passion and energy of Sabrina.
Cameron Mathison also danced the tango and scored a 23. I thought he looked good, but the choreography was a bit odd. He and Edyta danced to a cover of Sonny and Cher's "The Beat Goes On" and I have to say, it just wasn't good tango music. In fact, the musical selections all night were lame--with the exception of Helio and Julianne's jive which was danced to "Kids in America." Helio was suffering from a sprained ankle and looked a bit restrained, but still the man can dance! The judges nicked him this time around and he limped away with a 24.
They gave higher marks to Jennie Garth who rebounded after a mishap last week scoring a 26 for her tango. Marie Osmond also did the tango and looked a bit matronly in my opinion--although the judges praised her. But she was far better than Wayne Newton's tango which was stiff and stodgy. He was wearing Britney Spears-type bad hair extensions going for an "Errol Flynn" look. Judge Len Goodman said he looked like the cat from Shrek. Meow! Cheryl Burke looked and danced great and the pair ended up with an 18.
Mark Cuban is improving, but I don't think he has the fan base to continue much further in this competition. His jive was a bit lacking in energy and he only netted a 20 from the judges. Floyd Mayweather gave one of his better performances--the jive suits his style--but he still looks like he's dancing around a boxing ring instead of a dance floor.
Results show is tonight--I'm thinking it will be Mark Cuban and Wayne Newton in the bottom two with Mark departing.
Meanwhile, over on NBC the Heroes plots move forward slooowly. My friend Peter and I agree that we're not digging the new Maya and Alejandro characters or the Hiro in 17th century Japan storyline. I almost cheered when Hiro was about to return to the present after getting Takezo Kensei on the right path and uniting him with his Japanese lady love. But Hiro also has feelings for samurai's love interest and decided to stay in 1671. He has managed to send notes on mini scrolls back to Ando in the present via his sword.
Peter is still missing his memory--which makes the appearance of his powers unsettling and using them a bit haphazard. He, too, is avoiding "rejoining the present"as, when presented with the box that holds the key to his identity, he decides not to open it. Micah and Niki are back--minus D.L. who got shot by Linderman at the end of last season. Someone explain to me why he didn't survive one gunshot wound, but Matt Parkman looks none the worse for wear after 3-4 bullets to the chest? D.L. had one of the coolest powers, too...Anyway, not too much going on with Niki and Micah, although at the end of the episode she'd turned herself into "The Company" asking to be "cured." If she's so down on having superpowers, why not bring Micah (who it appears was dropped off at his paternal grandmother's) to get cured as well?
The fly boy at Claire's new school lets her know that he's onto her. Claire runs away upset, afraid of being tagged as a "freak"--but the fly boy whisks her off to the beach revealing his own abilities. Meanwhile, Mr. Bennet has Mohinder rummage through Isaac's paintings looking for clues to the future. Mohinder transmits a disturbing image via his cameraphone to Bennet which shows him dead--and Claire and the fly boy kissing in the background. | {
"redpajama_set_name": "RedPajamaC4"
} | 1,851 |
{"url":"http:\/\/aliceinfo.cern.ch\/ArtSubmission\/node\/3796","text":"# Figure 29\n\n TRD tracklet to TPC track residuals in $y$ as a function of the $z$~coordinate of the TPC track ($z_{\\rm track}$)for supermodules 2~(left) and 6~(right). The colour code is linear in the numberof tracks. The upper and lower panels show the situation with the survey alignmentand with in addition the external alignment, respectively. The data are from a2012 run of \\pp collisions with B~=~$-$\\SI{0.5}{\\tesla}. The alignment set used for the lowerplots was deduced from the same run. The internal alignment is applied in allfour cases.","date":"2018-03-20 21:36:15","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\": 0.6334002017974854, \"perplexity\": 3064.345333327599}, \"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\/1521257647545.54\/warc\/CC-MAIN-20180320205242-20180320225242-00483.warc.gz\"}"} | null | null |
You are here: Home / About / Ground Source Heat Pump FAQs
What is ground source?
Ground source systems use the ground as a means of absorbing and releasing heat. The heat pump system is extremely efficient and saves energy because the ground is at a constant temperature rather than the ambient which varies. The system is used for heating in cold weather and cooling in warm weather.
What is ground source cooling?
Ground source cooling uses the ground as a means of rejecting heat given off by the process. This Is then re-used for the heating cycle and thereby improving seasonal efficiency.
What is ground source heating?
Ground source heating uses the ground as a means of rejecting the cooling process which is given off from the system. The heat is later recycled in cooler weather in order to improve the season efficiency.
What are ground source heat pumps?
They are the integral component of the ground source system. Ground source heat pumps (GSHPs) are used to generate the cooling or heating medium. They are extremely efficient. GSHPs can give COP's of between 4.5 and 6. Therefore for every one Kilowatt of energy used to generate heating or cooling they give in return 4.5-6kw of heating or cooling as a useable load. On the other hand a gas boiler only returns 0.9 kilowatts of useable load for every one kilowatt of energy used.
What are bore holes?
These are the means of rejecting heat or cooling from the heat pumps. They are drilled to depths of between 70m and 300m in order to obtain the load required for the specific building. The bore holes are drilled using large rotary rigs to ensure that the stability of the bore hole is maintained until the ground loop is installed.
What are irrigation bore holes?
These are used to abstract up to 20 cubic meters of water per day and do not necessitate obtaining a licence or permission for the work involved. Potable water is used as an alternative to or alongside rain water harvesting systems in order to improve water quantity and quality. The water can be used when filtered to flush toilets and also to provide water for showers. The quality of the water can be as good as drinking water in some locations depending on the geology of the area being drilled.
What are geothermal systems?
Also known as ground source systems. However, geothermal systems are drilled to a much greater depth in order to capture high temperature water. Ground source systems tend to limit the drilling to 300m due to the operational pressure drop on the system.
What is drilling?
Is the process by which rotary drilling rigs are used to drill the bore holes down to the required depth for the designed system? Specialist rigs using mud puppies are used to ensure that a quick and clean drilling operation is achieved.
What are HDPE loops for ground source?
Once the bore hole has been drilled HDPE pipe work is installed as a means of transferring the rejected heat or cooling into the ground. The diameter varies to suit each application 32mm, 40mm or 50mm. The HDPE ground loops are very robust and have a 25 year life expectancy. They are configured in a flow and return system leading up to a chamber where they are isolated on one side and a flow measurement fitted to the other. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 3,795 |
\section{Introduction}
\begin{figure}
\includegraphics[width=0.5\textwidth]{biz.png}
\caption{illustration of the Alipay marketing campaign.}
\label{biz}
\end{figure}
Digital platforms nowadays serve various demands of societies, for example, e-commerce, ride sharing, and personal finance. For-profit platforms have strong motivation to grow the sizes of their active user bases, because larger active user bases introduce more network effect, more advertisement audience, more cash deposit, etc.
To convert inactive users to active users, one established way is to use personalized recommendation systems \cite{youtubeRecsys}\cite{netflixRecsys}. A platform can infer users' personal interests from profile information and behavior data, and recommend content accordingly. Recommendation systems rely on good product design, "big data", efficient machine learning algorithms\cite{itemCF}, and high-quality engineering systems \cite{amatriain_basilico_2016}.
Recently, the approach of using promotion incentive, such as coupon, to convert users has become popular \cite{CHRISTINO201978}\cite{Li2019LatentDA}. To enjoy the incentive, users are required to finish specified tasks, for example subscribing to a service, purchasing a product, sharing promotional information on a social network, etc. The key decision to make in such promotion campaigns is how much incentive to give to each user.
Our work is in the hypothetical context of Alipay offline payment marketing campaign, though it can be easily generalized to other incentive promotion campaigns. Gaining offline payment market share is a main business objective of Alipay. Ren-chuan-ren-hong-bao (social network based red packet) is the largest marketing campaign hosted by Alipay to achieve this goal. In this campaign, Alipay granted coupons to customers to incentivize them to make mobile payments with the Alipay mobile app. Given its marketing campaign budget, the company needed to determine the value of the coupon given to each user to maximize overall user adoption. We illustrate the marketing campaign in \autoref{biz}.
We propose a two-stage framework for solving the personalized incentive decision problem. In the first stage, we model users' personal promotion-response curves with machine learning algorithms.
In the second stage, we formulate the problem as a linear programming (LP) problem and solve it by established LP algorithms.
In practice, modeling promotion-response curves is challenging due to data sparsity and noise. Real-world promotion-response datasets usually lack samples for certain incentive values, even though the total amount of samples is large. Such sparsity combined with noise causes inaccuracy in response curve modeling and sub-optimal decision in incentives. We introduce a novel isotonic neural network architecture, the deep-isotonic-promotion-network (DIPN), to alleviate this problem. DIPN incorporates our prior knowledge of the response curve shape by enforcing isotonicity and regularizing for smoothness. It out-performed regular DNN and other state-of-the-art shape-constrained models in our experiments. \autoref{demo} illustrates such an example.
Another well known challenge for response curve modeling is treatment bias in historical data. If data samples are not collected through randomized trials, naively fitting the relationship between incentive and response cannot capture the true causal effect of incentive. \autoref{demo2} illustrates an example on how a biased dataset causes sub-optimal incentive decision. In real-world marketing campaigns, collecting fully randomized incentive samples is cost-ineffective, because it means randomly giving a large amount of users random amount of incentives. We use the inverse propensity score (IPS) \cite{austin2011introduction} technique to correct treatment bias.
\begin{figure}
\includegraphics[width=0.5\textwidth]{demo.png}
\caption{An illustration of how noise in data causes a sub-optimal incentive decision, and DIPN is more immune to noise compared to DNN. Rounds and diamonds represent average response rates at different promotion incentive (cost) levels in training and testing data, respectively. The amount of training data at low cost levels is small and hence the average response rate is noisy. The estimation by DNN based on the training data indicates that a low incentive should be applied (marked as "wrong best cost"). When this decision is made, a large amount of users receive incentive at low levels, and a large amount of "testing data" become available, reducing the noise and revealing the suboptimality of the original decision. DIPN is shape-constrained and more immune to the low training data quality. Based on DIPN, correct decision can still be made.}
\label{demo}
\end{figure}
\begin{figure}
\includegraphics[width=0.5\textwidth]{demo2.png}
\caption{Illustration of how treatment bias in training dataset leads to wrong incentive decisions. There are three types of users: active (red), ordinary (blue), and inactive (green). Each type has higher response rate to higher incentive (cost). However, in collected data, user activity is negatively correlated with incentive. A DNN model without knowledge of this bias will follow the dotted line, which does not reflect the causal effect of incentive. Based on this DNN model, a decision engine will never use any incentive larger than the "wrong best cost".}
\label{demo2}
\end{figure}
\section{Related work}
\subsection{Large-scale or personalized optimization problems based on prediction}
\label{sec:combining-prediction-and-optimization}
Many companies, e.g. Linkedin, Pinterest, Yahoo!, have developed solutions for large-scale optimization based on predictive models. \cite{Xu:2015:SPE:2783258.2788615} focused on online advertisement pacing. The authors developed user response prediction models and optimized for performance goals under budget constraints.
\cite{agarwal2012personalized} focused on online content recommendation. The authors developed personalized click shaping plans by solving a LP problem in its dual space. \cite{Gupta:2016:MCI:2946645.3007062}\cite{Gupta:2017:OEV:3132847.3132849}\cite{Zhao:2018:NVC:3219819.3219906} study the problem of email volume control. The predicted responses at different levels of email volume served as coefficients of engagement maximization problems. To our knowledge, there are few published studies on large-scale personalized incentive optimization.
\subsection{Causal inference and counterfactual prediction}
Incentive response modeling should establish causal relationship between incentive and user response. The model is used in a counterfactual way to answer the question "what if users are given incentive levels different from that is observed". There is a large volume of literature on counterfactual modeling, e.g. \cite{bonner2018causal}\cite{hartford2016counterfactual}\cite{swaminathan2015counterfactual}. We use inverse propensity score (IPS) \cite{austin2011introduction}\cite{rosenbaum1983central} to weight sample data when the data collection process is not randomized. This is assumed in our framework unless otherwise stated.
\subsection{Shape-constrained models}
Shape constraints include constraints on monotonicity and convexity (concavity). They are a form of regularization, introducing bias and prior knowledge. Shape constraints are helpful in below scenarios.
\begin{itemize}
\item Monotonicity or convexity (concavity) is desired for interpretability. For example, in economy theory, marginal gain on increasing promotion incentive should be positive but diminishing \cite{kahneman2013prospect}. In our experience, promotion response is non-decreasing, but the marginal gain does not necessarily decrease. We thus propose to apply just monotonicity constraint to promotion response models.
\item Prior knowledge on function shapes exists, but training data is too sparse to guarantee such shapes without regularization. In our experience, this is usually true for promotion response modeling, because a reasonable model is required as early as possible after a promotion campaign kicks off, allowing very limited time for training data collection. At the same time, we do know that response rate should monotonically increase with incentive.
\end{itemize}
The related works section of \cite{gupta2018diminishing} summarized four categories of shape-constrained models:
\begin{itemize}
\item General additive models (GAMs). A GAM is a summation of multiple one dimensional functions. Each of the 1-d function takes one input feature, and is responsible for enforcing the desired shape for that input.
\item Max-affine functions, which express piece-wise linear convex functions as the max of a set of affine functions. If the derivative with respect to an input is restricted to be positive/negative, monotonicity can also be enforced.
\item Monotonic neural networks. Neural networks can be viewed as recursive functions, with the output of one layer serving as the next layer's input. For a recursive function to be convex increasing, it is sufficient if its input function and the function itself are convex increasing. For a recursive function to be convex, it is sufficient if its input function is convex and the recursion is convex increasing.
\item Lattice networks \cite{You:2017:DLN:3294996.3295058}. The simplest form of lattice network is linear interpolation built on a grid of input space. Monotonicity and convexity are enforced as linear constraints on first and second order derivatives when solving for the model. The grid dimension grows exponentially with the input dimension, so the authors ensembled multiple low-dimension lattice networks built on different subsets of inputs to handle high dimensional data.
\end{itemize}
\cite{gupta2018diminishing} showed that lattice network has more expressive power than monotonic neural network since the former allows convex and concave inputs to coexist, and that lattice network is at least as good as monotonic neural network in accuracy.
Our work proposes the DIPN architecture (discussed in \autoref{isotonic}) that constrains NN to be monotonic for one input i.e. the incentive level. It does not aim to compete with state-of-the-art shape constrained models in terms of expressiveness, but instead aim to provide high accuracy and interpretability for promotion response modeling.
\section{The personalized promotion framework}
\label{sec_pnp}
The two steps of our framework for making personalized incentive decisions are (1) incentive-response modeling and (2) user response maximization under incentive budget. As the second step is the decision making step, and the first step prepares coefficients for it, we start \autoref{sec_pnp} by describing the optimization problem in the second step assuming a incentive-response model is already available, and dive deep into our approach for the incentive-response model in \autoref{isotonic}.
\autoref{tab:math_symbol_meaning} summarizes mathematical notations necessary for this section.
\begin{table}[h!]
\caption{Notations}
\label{tab:math_symbol_meaning}
\begin{center}
\begin{tabular}{c|p{0.8\columnwidth}}
symbol&meaning\\
\hline
$x_i$& feature vector of user i\\
$c_i$& incentive for user i \\
$y_i$& response label for user i \\
$f(x, c)$& user response prediction function with two inputs: x is the user feature vector, and c is the incentive \\
$g_k(x, c)$& user cost prediction function for the k-th resource \\
$d_j$& the j-th incentive level bin after discretizing the incentive space,$d_j < d_{j+1}, \forall j$ \\
$D$& total number of incentive level bins after discretizing the incentive space\\
$z_{ij}$ & decision variable representing the probability of choosing $d_j$ for the i-th user \\
$B$& total campaign budget \\
$b$& budget per capita\\
\end{tabular}
\end{center}
\end{table}
The objective of our formulation is to maximize total future user responses, and the constraint is the limited budget. Here future response can an arbitrarily defined business metric, e.g. click-through rate or long-term user value.
\begin{equation} \label{nonlinear}
\begin{split}
&\max_{c_i}\sum_{i}f(x_i, c_i) \\
s.t. &\sum_ic_i\leq B
\end{split}
\end{equation}
\subsection{Solving the optimization problem}
Because the number of users is large, and $f(x_i, c_i)$ is usually nonlinear and non-convex, \autoref{nonlinear} can be difficult to solve. We can restrict promotion incentive $c_i$ to a fixed number of $D$ levels: $\{{d_j}| j=0,1, ... D-1\}$, and use assignment probability variables $z_{ij} \in [0,1]$ to turn \autoref{nonlinear} into an LP.
\begin{equation} \label{mip_lp}
\begin{split}
&\max_{z_{ij}}\sum_i\sum_jf(x_i, d_j)z_{ij} \\
s.t. &\sum_{i}\sum_{j}d_jz_{ij} \leq B \\
&\sum_{j}z_{ij}=1, \forall i \\
& z_{ij} \in [0, 1], \forall i, j
\end{split}
\end{equation}
Solution $z_{ij}$ of \autoref{mip_lp} should be on one of the vertices of its feasible polytope, hence most of the values of $z_{ij}$ should be $0$ or $1$. For fractional $z_{ij}$ in solutions, we treat $z_{ij}, \forall i$ as a probability simplex and sample from it. This is usually good enough for production usage.
\autoref{mip_lp} can be solved by commercial solvers implementing off-the-shelf algorithms such as primal-dual or simplex. If the number of users is too large, \autoref{mip_lp} can be solved by dual ascent \cite{boyd2011distributed}. Note the dual problem of the LP can be decomposed into many user-level optimization problems and solved in parallel. A few specialized large-scale algorithms can also be applied to problems with the structure of \autoref{mip_lp}, e.g. \cite{zhong2015stock}\cite{zhang2020solving}.
One advantage of \autoref{mip_lp} is that $f(x_i, d_j)$ is computed before solving the optimization problem. Hence the specific choice of the functional form of $f(x_i, d_j)$ does not affect the difficulty of the optimization problem. Whether $f(x_i, d_j)$ is a logistic regression model or a DNN model is transparent to \autoref{mip_lp}.
Sometimes promotion campaigns have more than one constraints. A commonly seen example is that overlapped subgroups of users are subject to separate budget constraints. In general, we can model any constraints by predictive modeling, and then formulate a multiple-constraint problem. Suppose for each user $i$ there are $K$ kinds of costs modeled by:
\begin{equation} \label{g_k}
\begin{split}
g_k(x_i, c_i), k=0,1,.. K-1
\end{split}
\end{equation}
The multiple-constraint optimization problem is:
\begin{equation} \label{multi-lp}
\begin{split}
&\max_{z_{ij}}\sum_{i}\sum_{j}f(x_i, d_j)z_{ij} \\
s.t. &\sum_{i}\sum_{j}g_k(x_i, d_j)z_{ij} \leq B_k, k=0,1,... K - 1 \\
&\sum_{j}z_{ij}=1, \forall i \\
& z_{ij} \in [0, 1], \forall i, j
\end{split}
\end{equation}
A frequently seen alternative to constrain the total budget $B$ is to constrain budget per capita $b$. The corresponding formulation is as below.
\begin{equation} \label{avg_lp}
\begin{split}
&\max_{z_{ij}}\sum_{i}\sum_{j}f(x_i, d_j)z_{ij} \\
s.t. &\sum_{i}\sum_{j}(g_k(x_i, d_j)-b_k)z_{ij}\leq 0, \forall k \\
&\sum_{j}z_{ij}=1, \forall i \\
& z_{ij} \in \{0, 1\}, \forall i, j
\end{split}
\end{equation}
\autoref{multi-lp} and \autoref{avg_lp} are both LPs since $f(x_i, d_j)$ and $g_k(x_i, d_j)$ are pre-computed coefficients. They can be solved by the same solvers used for \autoref{mip_lp}.
\subsection{Online decision making for new users}
Sometimes it is desired to be able to make incentive decisions for a stream of incoming users. It is not possible to put such users together and solve one optimization problem beforehand. However, if we can assume the dual variables is stable over a short period of time, such as one hour, we can solve the optimization problem with users in the previous hour, and reuse the optimal dual variables in the next hour. This assumption is usually not applicable to general optimization problems, but when the amount of users is large and user population is stable, it holds. This approach can also be viewed from the perspective of shadow price \cite{boyd2004convex}. The Lagrangian multiplier can be interpreted as the marginal objective gain when one more unit of budget is available. This marginal gain should not change rapidly for a stable user population.
We thus break down the optimization step into a dual variable solving step and a decision making step. The decision making step can make incentive decision for a single user. Consider the dual formulation of \autoref{mip_lp}, and let $\lambda$ be the Lagrangian multiplier for the budget constraint:
\begin{equation} \label{dual}
\begin{split}
&\min_{\lambda}\max_{z_{ij}}\sum_i\sum_jf(x_i, d_j)z_{ij} - \lambda(\sum_{i}\sum_{j}d_jz_{ij} - B) \\
s.t.
&\sum_{j}z_{ij}=1, \forall i \\
& z_{ij} \in [0, 1], \forall i, j \\
& \lambda > 0
\end{split}
\end{equation}
If $\lambda$ is given, \autoref{dual} can be decomposed into a per user optimization policy:
\begin{equation} \label{individual}
\begin{split}
&\max_{z_j}(f(x_i, d_j) - \lambda d_j)z_{ij}\\
s.t.
&\sum_{j}z_{ij}=1, \forall i \\
& z_{ij} \in [0, 1], \forall i, j \\
\end{split}
\end{equation}
\autoref{individual} is applicable to unseen users as long as $x_i$ is known.
\subsection{Challenges for two-stage framework}
"User's optimal incentive level" is the lowest level for a certain user to be converted. If we offer less incentive, we will lose a user conversion(situation A). If we offer more incentive, we waste a certain amount of marketing funds.
\begin{figure*}
\begin{minipage}{0.49\textwidth}
\includegraphics[width=0.9\textwidth]{SituA.png}
\label{situationA}
\end{minipage}
\hfill
\begin{minipage}{0.49\textwidth}
\includegraphics[width=0.9\textwidth]{SituB.png}
\label{situationB}
\end{minipage}
\caption{An illustration of two common bad cases in real world datasets, $d^*$ is the expected optimal incentive level, if $f(d_j)$ satisfy $f(d_j)-\lambda d_j < f(d^*)-\lambda d^*, \forall d_j \neq d^*$, our framework outputs the right answer $d^*$. But we often found some predicted score like $f(d_k), f(d_k) - \lambda d_k > f(d^*)-\lambda d^*$, become a wrong best incentive level.
}
\label{situation}
\end{figure*}
For illustration, we could relax constraint by supposing that optimal $z$ is on its $[0, 1]$ boundary, omit the notation of index $i$ and rewrite formula \ref{individual} as
\begin{equation} \label{individual_rewrite}
\begin{split}
&\max_{z_j}r(d_j, y_j)\\
s.t.
&y_j=f(x, d_j) \\
&\sum_{j}z_{j}=1 \\
& z_{j} \in [0, 1], \forall j \\
\end{split}
\end{equation}
where $r(d, y)=y - \lambda d$.
\autoref{demo} illustrates wrong prediction leads to wrong best incentive level. In this section, \autoref{situation} illustrates two situations of them in more details. Considering $d^*$ is the optimal incentive level for user $i$, according to \autoref{individual}, we hope our prediction function satisfy $f(d_j)-\lambda d_j < f(d^*)-\lambda d^*, \forall d_j \neq d^*$. If due to lack of data, $f(d_k)-\lambda d_k > f(d^*)-\lambda d^*$, incentive level $d_k \neq d^*$ will be chosen, and a wrong decision will be made.
\vspace{6pt}
\noindent\textbf{Situation A: non-monotonic}
the user response prediction function frequently gives a high prediction $f(d_k)$ at a lower incentive level $d_k$. And because $d_k < d^*$, user $i$ didn't get a satisfactory incentive, we will lose a customer.
In this situation, we introduce prior knowledge to constrain the response curve's shape: users' expected response monotonically increases with the promotion incentive. Therefore, $f(d_{k+1}) - f(d_k)$ must greater than zero.
\vspace{6pt}
\noindent\textbf{Situation B: non-smooth}
In other situations, a very high prediction $f(d_k)$ is given at a higher incentive level $d_k$. And because $d_k > d^*$, $d_k - d^*$ marketing funds is wasted.
In situation B, we introduce another prior knowledge to constrain the response curve's shape: the incentive-response curves are smooth. Therefore, $(f(d_{k}) - f(d_{k-1})) - (f(d_{k-1}) - f(d_{k-2}))$ should not be too large.
\vspace{10pt}
Based on the discussion of these two situations above, we introduce a novel deep-learning structure DIPN to avoid both situations as far as possible.
\section{Deep isotonic promotion network (DIPN)} \label{isotonic}
\begin{figure}
\centering
\includegraphics[width=0.5\textwidth]{dipn2_2.png}
\caption{The architecture of DIPN. DIPN composes of bias net and uplift net as shown on the left and right sides
. In both nets, sparse user features are one-hot encoded and mapped to their embedding, which are aggregated by summation. Then in bias net, concatenated sparse features are feed to one fully connected layer whose one dimension output activated by leaky ReLU is treated as logit of bias prediction. The bias prediction is inputted to uplift net. In uplift net, another fully connected layer takes concatenated sparse features as input and outputs $D$ ReLU activated positive values as uplift weights. The previous uplift weight will be inputted to the subsequent node for generating next uplift weight. The isotonic layer outputs the uplift weight representation $w(x)$.}
\label{dipnfig}
\end{figure}
In practice, promotion events often do not last for long, so there is business interest to serve promotion-response models as soon as possible after a campaign starts. Given limited time accumulating training data, incorporating prior knowledge to facilitate modeling is desirable. We choose to enforce the promotion response to be monotonically increasing with incentive level.
DIPN is a DNN model designed for learning user promotion-response curve. DIPN predicts the response value for a given discretized incentive level and a user's feature vector. We can get a user's response curve by enumerating all incentive levels. The response curve learned by DIPN satisfies both monotonicity and smoothness. DIPN achieves this using its isotonic layer (discussed later). Incentive level, which is a one-dimensional discrete scalar, and user features are all inputs to the isotonic layer. While incentive level is inputted to the isotonic layer directly, user features can be transformed by other layers. DIPN consists of bias net and uplift net. The term uplift refers to response increment due to incentive increment. While prediction result of the bias net gives the user's response estimate for minimum incentive, the uplift net learns the uplift response. The DIPN architecture is shown in \autoref{dipnfig}. In the remaining of \autoref{isotonic}, we focus on explaining the isotonic layer and learning process.
\subsection{Isotonic Embedding} \label{Isotonic embedding}
\begin{figure}
\centering
\includegraphics[width=0.48\textwidth]{isotonic_encoding2.png}
\caption{Logistic regression with isotonic embedding}
\label{ielrfig}
\end{figure}
Isotonic embedding transforms an incentive level to a vector of binary values. Each digit in the vector represents one level. All digits representing levels lower than or equal to the input level are ones. We use $e(c) \in \{0, 1\}^{D}$ to denote the $D$-digit isotonic embedding of incentive level $c$, and $d_j$ to denote the incentive level of the $j-th$ digit, thus
\begin{equation}
\begin{split}
& e_j(c)=
\begin{cases}
1,& \text{if } c\geq d_j\\
0,& \text{otherwise}
\label{ielr}
\end{cases}
\end{split}
\end{equation}
If we fit a logistic regression response curve with non-negative weights using isotonic embedding as input, the resulting curve will be monotonically increasing with incentive. Several examples are given in \autoref{ielrfig}.
\begin{equation}
\begin{split}
f(c) &= sigmoid(\sum_j w_je_j(c)+b) \\
&= {\frac{1}{(1 + exp(- \sum_j w_je_j(c)-b))}} \\
s.t. &w_j\geq 0, \forall j\\
\label{ielr}
\end{split}
\end{equation}
It is trivial to show the monotonicity of $f(c)$ since $sigmoid$ function is monotonic, and
\begin{equation}
\begin{split}
&(\sum_j w_je_j(c + \triangle c) + b) - (\sum_j w_je_j(c) + b) \\
&= \sum_{d_j=c}^{d_j=c+\triangle c}{w_j} \geq 0, \forall \triangle c \ge 0
\label{monotone-proving}
\end{split}
\end{equation}
\subsection{Uplift Weight Representation} \label{Uplift weight}
In DIPN, the non-negative weights are output of DNN layers. These weights are thus personalized. We name the personalized non-negative weight the uplift weight, because it is proportion to the uplift value, defined as the incremental response value corresponding to one unit increase in incentive value.
In a binary classification scenario, the prediction function of DIPN is:
\begin{equation}
\begin{split}
& f(x,c)=sigmoid(\sum_j w_j(x)e_j(c)+b)
\end{split}
\label{dipn-prediction-function}
\end{equation}
where $w_j(x)$ is the uplift weight representation, and $e_j(c)$ is the isotonic embedding. $w_j(x)$ is learned by a neural network using ReLU activation function in last layer so that $w_j(x) $ is non negative .
For two consecutive incentive levels $d_{j}$ and $d_{j+1}$, $w_{j+1}$ is a uplift measure for the incremental treatment effect of increasing incentive from $d_{j}$ to $d_{j+1}$. To see this, approximate $sigmoid$ function by its first order expansion around $d_j$:
\begin{equation}
\begin{split}
f(d_{j+1}) - f(d_{j}) &= g(z_{j+1}) - g(z_{j}) \\
&\simeq g'(z_{j})(z_{j+1} - z_{j}) \\
&=g'(z_{j})w_{j+1}
\label{firstorder}
\end{split}
\end{equation}
where $g$ is $sigmoid$ function and $g'$ is its first order derivative, $z_j = \sum_j w_je_j(d_j)+b$. Hence in a small region surrounding $d_{j}$, if $g'(z_{j})$ can be seen as a constant, $w_{j+1}$ is proportional to the uplift value at $f(d_j)$.
\subsection{Smoothness} \label{Monotonicity and smoothness}
Smoothness means that response value does not change much as the incentive level varies. Users' incentive-response curves are usually smooth when fitted with sufficient amount of unbiased data. We therefore add regularization in the DIPN loss function to enforce smoothness.
\begin{equation}
L=\frac{1}{M}\sum_i{log\_loss(f(x_i, c_i), y_i) +\alpha \cdot smoothness\_loss(w(x_i))}
\label{dipn-loss-function}
\end{equation}
where $M$ is the number of training data points, $log\_loss$ measures the degree of fitting to training data, $smoothness\_loss$ measures smoothness of predicted response curve, and $\alpha > 0$ balances the two losses.
The definition of $log\_loss$ and the definition of $smoothness\_loss$ are given in \autoref{log-loss-def} and \autoref{smoothness-loss-def} respectively. User index $i$ in \autoref{log-loss-def} and user feature vector $x_i$ in \autoref{smoothness-loss-def} are omitted for simplicity.
\begin{equation}
log\_loss(f(x, c), y) = -y log(f(x, c)) - (1 - y) log(1 - f(x, c))
\label{log-loss-def}
\end{equation}\begin{equation}
smoothness\_loss(w) = \frac{1}{D}\sum_{j}{(w_{j+1} - w_{j}) ^ 2 / (w_{j+1} w_{j})}
\label{smoothness-loss-def}
\end{equation}
One necessary and sufficient condition of smoothness of the predicted response curve is the uplift values of consecutive incentive levels are as close as enough. As we have proven in \autoref{Uplift weight} that the uplift value can be approximated by the uplift weight representation, we want the difference of the uplift weights $(w_{j+1} - w_{j})^2$ to be small enough. $(w_{j+1} w_{j})$ is added to denominator of $smoothness\_loss$ to normalize the differences over all incentive levels.
\subsection{Two Phases Learning} \label{Two Phases Learning}
For stability, the training process is split into two phases-bias net learning phase (BLP) and uplift net learning phase (ULP).
BLP learns the bias net while ULP learns the uplift net. In BLP, only samples with lowest incentive are used for training, and only bias net is activated which is easy to converge.
In ULP, all samples are used and variables in bias net are fixed. The fixed bias net sets up a robust initial boundary for uplift net learning. Bias net's prediction will be inputted to isotonic layer to enhance uplift net's capability based on the consumption that users with similar bias are more likely to have similar uplift responses.
Another difference in UWLP is smoothness loss's weight $\alpha$. $\alpha$ will be decaying gradually in UWLP as we observed that larger $\alpha$ helped model converging faster at the beginning of training. The logloss will dominates total loss eventually. $\alpha$'s updating formula is given as follow:
\begin{equation}
\begin{split}
&\alpha=max(\alpha^{l}, \alpha^{u}- \gamma \cdot global\_step)
\end{split}
\label{decayed-smoothness-loss-weight}
\end{equation}
where $\alpha^{u}$ is initial upper bound of $\alpha$, $\alpha^{l}$ is final lower bound of $\alpha$, $\gamma$ controls decaying speed.
\section{Experiment}
In experiments, we focus on comparing DIPN with other response models, solving the same optimization problems \autoref{mip_lp} with the same budget. Specifically, we compare (1) regular DNN, (2) ensemble Lattice network \cite{You:2017:DLN:3294996.3295058}, and DIPN. For each model, we search for a good architecture configuration and hyperparameter setting, and calculate below metrics:
\begin{itemize}
\item Logloss: Measures the likelihood of the fitted model to explain the observations.
\item AUC-ROC: Measures the correctness of the predicted probability order of being positive events.
\item Reverse pair rate (RPR): For each observation of user-incentive-response in the dataset, we make counterfactual predictions on the response of the same user given all possible incentives. For any pair of two incentives, if incentive $a$ is larger than $b$, and the predicted response of $a$ is smaller than that of $b$, we find one reversed pair. For $n$ incentives, there are $\frac{n(n-1)}{2}$ pairs. If there are $rp$ reversed pairs, the RPR is defined as $\frac{2rp}{n(n-1)}$. RPR can be viewed as the degree of violating the response model monotonicity. To obtain RPR for a population, we average all users' RPR values.
\item Equal pair rate (EPR): Similar to RPR, but instead of counting the pairs of reversed pairs, EPR counts the pairs having equal predicted response. We consider low EPR as a good indicator for monotonicity.
\item Max local slope standard deviation (MLSS): Similar to RPR, for each user, we make counterfactual prediction on response for each incentive. For every two consecutive incentives $c_1$ and $c_2$, assuming their predicted responses are $p_1$ and $p_2$, we can compute the local slope $s_1=\frac{p_1-p_2}{c_1-c_2}$. Consider a range on incentive $[c_i - r, c_i + r]$, we collect all local slopes inside this range, compute their standard deviation. Across all such incentive ranges, we use the maximum local slope standard deviation as MLSS. This metric reflects smoothness of the response curve. Average of all users MLSS is used as a model's MLSS.
\item Future response: Applying the strategy learned from training data to a new user population, following \autoref{individual}, the average response rates of this population. Higher response rate is better. With synthetic datasets, for which we know the ground truth data generation process, we can use the true response expectation on the promotion given by the strategy to evaluate the response outcome. With real-world datasets, we can search in the holdout dataset for the same type of user with the same promotion given by the strategy, and consider all such users will show the observed response on this promotion. This approach can be viewed as importance sampling.
\item Future cost error: Similar to future response, applying the strategy learned from the training data to a holdout population, the user-average cost may exceed the business constraint. We can follow the evaluation method for the future response metric, instead of compute the response rates, compute the response induced cost that exceeds budget.
\end{itemize}
\subsection{Synthetic and Production datasets}
We use synthetic dataset, for which we know the ground truth of data generation process, to evaluate our promotion solution. The feature space consists of three 1-in-n categorical variables, with $n_1$, $n_2$, and $n_3$ categories, respectively. The joint distribution of the three categories consists of $n_1n_2n_3$ different combinations. For each combination, we randomly generate four parameters $a \sim U(0, 1)$, $b = 1 - a$, $\mu \sim U(-50, 150)$, and $\delta \sim U(0, 50)$, where $U(l, u)$ is the uniform distribution bounded by $l$ and $u$. A curve $y=f(x)$ is then generated as follows:
\begin{flalign}
y = a + b\int_0^x \exp{-\frac{(t-\mu)^2}{2\delta^2}} dt, x\in[0, 100]
\label{gen}
\end{flalign}
The discrete version is:
\begin{flalign}
&y[i] = a + \frac{b}{100Z}\sum_{h=0}^i \exp{-\frac{(h-\mu)^2}{2\delta^2}}, i=0,1,..100 \\
&Z = \max_h\exp{-\frac{(h-\mu)^2}{2\delta^2}}, h=0,1,..100
\label{gen_discrete}
\end{flalign}
The curves of $y=f(x)$ is used as the ground truth of the expected response $y$ on different incentive $x$, for each joint category of users. Without loss of generality, we constrain the incentive range to be $[0, 100]$. It's easy to see that the curve is monotonically increasing, convex if $\mu < 0$, and concave if $\mu > 100$.
To generate pseudo dataset with noise and and sparsity, we first generate a random integer $z$ between 1 and 1000 with equal probability, for each feature combination. With $z$ being the number of data points for this combination, we randomly choose a promotion integer $p$ between 1 and 100 with equal probability for each data point. With promotion $p$ and expected response $y = f(p)$, a $0/1$ label is generated with probability $y$ of being 1.
On this dataset, we generated 20000 data points, 5000 training samples, 5000 validation samples and 10000 testing samples.
The evaluation metrics for the response models are shown below tables. \autoref{tab:table1} shows the simulation results on synthetic data with $n_1=2, n_2=2, n_3=2$. DIPN showed the highest future response rate and lowest future cost. \autoref{tab:table2} shows the simulation results on synthetic data with $n_3=2, n_2=5, n_3=7$. Again, DIPN showed the highest future response rate and lowest future cost. \autoref{tab:table3} shows the results on a real promotion campaign, DIPN showed the lowest future cost and the same future response rate as that of DNN. Overall DIPN consistently outperformed other models in our test.
\begin{table}[h!]
\begin{center}
\caption{Response model evaluation on synthetic data 1($n_1=2, n_2=2, n_3=2$)}
\label{tab:table1}
\begin{tabular}{c|c|c|c}
&\textbf{DNN} & \textbf{Lattice} & \textbf{DIPN}\\
\hline
LogLoss(lower is better) & 0.5713 & 0.5772 & 0.5770 \\
AUC-ROC & 0.6976 & 0.6931 & 0.6967 \\
RPR(lower is better) & 0.153 & 0 & 0\\
EPR(lower is better) & 0 & 0.048 & 0.001\\
MLSS(lower is better) & 0.003 & 0.007 & 0.006\\
Future response & 0.743 & 0.710 & 0.759\\
Future cost($constraint=11.0$) & 11.0 & 11.7$(*)$ & 11.0\\
\end{tabular}
\newline
\footnotesize{$(*)$optimization problem cannot be solved.}
\end{center}
\end{table}
\begin{table}[h!]
\begin{center}
\caption{Response model evaluation on synthetic data 2 ($n_1=3, n_2=5, n_3=7$)}
\label{tab:table2}
\begin{tabular}{c|c|c|c}
&\textbf{DNN} & \textbf{Lattice} & \textbf{DIPN}\\
\hline
LogLoss & 0.8189 & 0.6267 & 0.5822 \\
AUC-ROC & 0.7582 & 0.7579 & 0.7618 \\
RPR(lower is better) & 0.40 & 0.00 & 0.00\\
EPR(lower is better) & 0 & 0.05 & 0.00\\
MLSS(lower is better) & 0.33 & 0.01 & 0.01\\
Future response & 0.625 & 0.685 & 0.694\\
Future cost($constraint=11.0$) & 7.5$(*)$ & 11.0 & 10.9\\
\end{tabular}
\newline
\footnotesize{$(*)$optimization problem cannot be solved.}
\end{center}
\end{table}
\begin{table}[h!]
\begin{center}
\caption{Response model evaluation on Production dataset}
\label{tab:table3}
\begin{tabular}{c|c|c|c}
&\textbf{DNN} & \textbf{Lattice} & \textbf{DIPN}\\
\hline
LogLoss & 0.2240 & 0.2441 & 0.2189 \\
AUC-ROC & 0.7623 & 0.5000 & 0.7722 \\
RPR(lower is better) & 0.28 & 0.00 & 0.00\\
EPR(lower is better) & 0.00 & 1.00 & 0.08\\
MLSS(lower is better) & 0.06 & 0.00 & 0.01\\
Future response & 0.020 & 0.000 & 0.020\\
Future cost($constraint=0.05$) & 0.050 & 0.000$(*)$ & 0.048\\
\end{tabular}
\newline
\footnotesize{$(*)$optimization problem cannot be solved.}
\end{center}
\end{table}
\subsection{Online Results}
We deployed our solution at Alipay and evaluated it in multiple marketing campaigns using A/B tests. Our solution consistently showed better performance than baselines, and was eventually deployed to all users. We show A/B test results for three marketing campaigns for HuaBei. HuaBei is a online micro-loan service launched by Ant Financial. It has about 300 million users according to public data. The incentive is electronic cash voucher, with usability subject to different terms in different campaigns. These campaigns include:
\begin{itemize}
\item Preferred Payment: the voucher can be cashed if a user set Huabei as the default payment method in the mobile application of Alipay.
\item New User: the voucher can be cashed if a user activates the Huabei service.
\item User Churn Intervention: the voucher can be cashed when a user uses Huabei to make a purchase.
\end{itemize}
All these campaigns have their corresponding business targets strongly correlated with voucher usage, so we use the usage rates as well as the monetary costs as evaluation metrics. \autoref{tab:table4} shows the relative improvements of DIPN compared to DNN model baselines. DIPN models use $30\%$ traffic in each experiment. 95 percentile confidence intervals are shown in square brackets. In all of the experiments, costs were significantly reduced. In two experiments, average voucher use rates were significantly increased.
\begin{table}[h!]
\begin{center}
\caption{Online Results}
\label{tab:table4}
\begin{tabular}{p{0.08\textwidth}|p{0.16\textwidth}|p{0.15\textwidth}}
&\textbf{Cost (\%)} & \textbf{Usage Rate (\%)} \\
\hline
Payment Preferred & -6.05\%[-8.63\%, -3.47\%] & 1.86\%[-0.31\%, 4.04\%] \\
New User Gift & -8.58\%[-10.51\%, -6.66\%] & 5.20\%[3.16\%, 7.23\%]\\
User Churn Intervention & -9.42\%[-12.99\%, -5.84\%] & 8.45\%[4.53\%, 12.37\%]\\
\end{tabular}
\newline
\end{center}
\end{table}
\section{Conclusion}
We focus on the problem of massive-scale personalized promotion in the internet industry. Such promotions typically require maximizing total user responses with limited budget. The decision variables for solving such problems are the incentive associated with the promotion.
We propose a two-step framework for solving such personalized promotion problems. In the first step, each user's response to each incentive level is predicted, and stored as coefficients for the second step. Predicting all these coefficients for many users is a counterfactual prediction problem. We recommend using randomized promotion policy or appropriate causal inference techniques such as IPS to process training data. To deal with data sparsity and noise, we designed a neural network architecture that incorporate our prior knowledge: (1) users' expected response monotonically increases with promotion incentive, and (2) the incentive-response curves are smooth. The proposed neural network, DIPN, ensures monotonicity and regularizes the magnitude of the first order derivative change. In the second step, an LP optimization problem can be formulated with the coefficients computed in the first step. We discussed its variants including (1) optimizing with more than one constraints, (2) supporting the constraint on average value, and (3) making decision for unseen users.
In experiments on synthetic datasets, we compared three algorithms: DNN, Deep Lattice Network, and our proposed DIPN. We show that DIPN has better performance in terms of data fitting, constraint violation, and promotion decisions. We also conducted a online experiment in one of Alipay's promotion campaign, in which user engagement was the desired response. DIPN got 6.05\% budget saving without losing user engagement, compared to DNN.
There are many possible extensions to the proposed framework. For example, the promotion response modeling can adapt to any causal inference techniques besides IPS\cite{dudik2011doubly}, for example causal embedding \cite{bonner2018causal} and instrumental variable\cite{hartford2016counterfactual}. Also the optimization formulation can be changed according to business requirements, as long as the computational complexity can be handled.
\bibliographystyle{ACM-Reference-Format}
\section{Introduction}
ACM's consolidated article template, introduced in 2017, provides a consistent \LaTeX\ style for use across ACM publications, and incorporates accessibility and metadata-extraction functionality necessary for future Digital Library endeavors. Numerous ACM and SIG-specific \LaTeX\ templates have been examined, and their unique features incorporated into this single new template.
If you are new to publishing with ACM, this document is a valuable guide to the process of preparing your work for publication. If you have published with ACM before, this document provides insight and instruction into more recent changes to the article template.
The ``\verb|acmart|'' document class can be used to prepare articles for any ACM publication --- conference or journal, and for any stage of publication, from review to final ``camera-ready'' copy, to the author's own version, with {\it very} few changes to the source.
\section{Template Overview}
As noted in the introduction, the ``\verb|acmart|'' document class can be used to prepare many different kinds of documentation --- a double-blind initial submission of a full-length technical paper, a two-page SIGGRAPH Emerging Technologies abstract, a ``camera-ready'' journal article, a SIGCHI Extended Abstract, and more --- all by selecting the appropriate {\it template style} and {\it template parameters}.
This document will explain the major features of the document class. For further information, the {\it \LaTeX\ User's Guide} is available from \url{https://www.acm.org/publications/proceedings-template}.
\subsection{Template Styles}
The primary parameter given to the ``\verb|acmart|'' document class is the {\it template style} which corresponds to the kind of publication or SIG publishing the work. This parameter is enclosed in square brackets and is a part of the {\verb|documentclass|} command:
\begin{verbatim}
\documentclass[STYLE]{acmart}
\end{verbatim}
Journals use one of three template styles. All but three ACM journals use the {\verb|acmsmall|} template style:
\begin{itemize}
\item {\verb|acmsmall|}: The default journal template style.
\item {\verb|acmlarge|}: Used by JOCCH and TAP.
\item {\verb|acmtog|}: Used by TOG.
\end{itemize}
The majority of conference proceedings documentation will use the {\verb|acmconf|} template style.
\begin{itemize}
\item {\verb|acmconf|}: The default proceedings template style.
\item{\verb|sigchi|}: Used for SIGCHI conference articles.
\item{\verb|sigchi-a|}: Used for SIGCHI ``Extended Abstract'' articles.
\item{\verb|sigplan|}: Used for SIGPLAN conference articles.
\end{itemize}
\subsection{Template Parameters}
In addition to specifying the {\it template style} to be used in formatting your work, there are a number of {\it template parameters} which modify some part of the applied template style. A complete list of these parameters can be found in the {\it \LaTeX\ User's Guide.}
Frequently-used parameters, or combinations of parameters, include:
\begin{itemize}
\item {\verb|anonymous,review|}: Suitable for a ``double-blind'' conference submission. Anonymizes the work and includes line numbers. Use with the \verb|\acmSubmissionID| command to print the submission's unique ID on each page of the work.
\item{\verb|authorversion|}: Produces a version of the work suitable for posting by the author.
\item{\verb|screen|}: Produces colored hyperlinks.
\end{itemize}
This document uses the following string as the first command in the source file: \verb|\documentclass[sigconf,screen]{acmart}|.
\section{Modifications}
Modifying the template --- including but not limited to: adjusting margins, typeface sizes, line spacing, paragraph and list definitions, and the use of the \verb|\vspace| command to manually adjust the vertical spacing between elements of your work --- is not allowed.
{\bf Your document will be returned to you for revision if modifications are discovered.}
\section{Typefaces}
The ``\verb|acmart|'' document class requires the use of the ``Libertine'' typeface family. Your \TeX\ installation should include this set of packages. Please do not substitute other typefaces. The ``\verb|lmodern|'' and ``\verb|ltimes|'' packages should not be used, as they will override the built-in typeface families.
\section{Title Information}
The title of your work should use capital letters appropriately - \url{https://capitalizemytitle.com/} has useful rules for capitalization. Use the {\verb|title|} command to define the title of your work. If your work has a subtitle, define it with the {\verb|subtitle|} command.
Do not insert line breaks in your title.
If your title is lengthy, you must define a short version to be used in the page headers, to prevent overlapping text. The \verb|title| command has a ``short title'' parameter:
\begin{verbatim}
\title[short title]{full title}
\end{verbatim}
\section{Authors and Affiliations}
Each author must be defined separately for accurate metadata identification. Multiple authors may share one affiliation. Authors' names should not be abbreviated; use full first names wherever possible. Include authors' e-mail addresses whenever possible.
Grouping authors' names or e-mail addresses, or providing an ``e-mail alias,'' as shown below, is not acceptable:
\begin{verbatim}
\author{Brooke Aster, David Mehldau}
\email{dave,judy,steve@university.edu}
\email{firstname.lastname@phillips.org}
\end{verbatim}
The \verb|authornote| and \verb|authornotemark| commands allow a note to apply to multiple authors --- for example, if the first two authors of an article contributed equally to the work.
If your author list is lengthy, you must define a shortened version of the list of authors to be used in the page headers, to prevent overlapping text. The following command should be placed just after the last \verb|\author{}| definition:
\begin{verbatim}
\renewcommand{\shortauthors}{McCartney, et al.}
\end{verbatim}
Omitting this command will force the use of a concatenated list of all of the authors' names, which may result in overlapping text in the page headers.
The article template's documentation, available at \url{https://www.acm.org/publications/proceedings-template}, has a complete explanation of these commands and tips for their effective use.
\section{Rights Information}
Authors of any work published by ACM will need to complete a rights form. Depending on the kind of work, and the rights management choice made by the author, this may be copyright transfer, permission, license, or an OA (open access) agreement.
Regardless of the rights management choice, the author will receive a copy of the completed rights form once it has been submitted. This form contains \LaTeX\ commands that must be copied into the source document. When the document source is compiled, these commands and their parameters add formatted text to several areas of the final document:
\begin{itemize}
\item the ``ACM Reference Format'' text on the first page.
\item the ``rights management'' text on the first page.
\item the conference information in the page header(s).
\end{itemize}
Rights information is unique to the work; if you are preparing several works for an event, make sure to use the correct set of commands with each of the works.
\section{CCS Concepts and User-Defined Keywords}
Two elements of the ``acmart'' document class provide powerful taxonomic tools for you to help readers find your work in an online search.
The ACM Computing Classification System --- \url{https://www.acm.org/publications/class-2012} --- is a set of classifiers and concepts that describe the computing discipline. Authors can select entries from this classification system, via \url{https://dl.acm.org/ccs/ccs.cfm}, and generate the commands to be included in the \LaTeX\ source.
User-defined keywords are a comma-separated list of words and phrases of the authors' choosing, providing a more flexible way of describing the research being presented.
CCS concepts and user-defined keywords are required for all short- and full-length articles, and optional for two-page abstracts.
\section{Sectioning Commands}
Your work should use standard \LaTeX\ sectioning commands: \verb|section|, \verb|subsection|, \verb|subsubsection|, and \verb|paragraph|. They should be numbered; do not remove the numbering from the commands.
Simulating a sectioning command by setting the first word or words of a paragraph in boldface or italicized text is {\bf not allowed.}
\section{Tables}
The ``\verb|acmart|'' document class includes the ``\verb|booktabs|'' package --- \url{https://ctan.org/pkg/booktabs} --- for preparing high-quality tables.
Table captions are placed {\it above} the table.
Because tables cannot be split across pages, the best placement for them is typically the top of the page nearest their initial cite. To ensure this proper ``floating'' placement of tables, use the environment \textbf{table} to enclose the table's contents and the table caption. The contents of the table itself must go in the \textbf{tabular} environment, to be aligned properly in rows and columns, with the desired horizontal and vertical rules. Again, detailed instructions on \textbf{tabular} material are found in the \textit{\LaTeX\ User's Guide}.
Immediately following this sentence is the point at which Table~\ref{tab:freq} is included in the input file; compare the placement of the table here with the table in the printed output of this document.
\begin{table}
\caption{Frequency of Special Characters}
\label{tab:freq}
\begin{tabular}{ccl}
\toprule
Non-English or Math&Frequency&Comments\\
\midrule
\O & 1 in 1,000& For Swedish names\\
$\pi$ & 1 in 5& Common in math\\
\$ & 4 in 5 & Used in business\\
$\Psi^2_1$ & 1 in 40,000& Unexplained usage\\
\bottomrule
\end{tabular}
\end{table}
To set a wider table, which takes up the whole width of the page's live area, use the environment \textbf{table*} to enclose the table's contents and the table caption. As with a single-column table, this wide table will ``float'' to a location deemed more desirable. Immediately following this sentence is the point at which Table~\ref{tab:commands} is included in the input file; again, it is instructive to compare the placement of the table here with the table in the printed output of this document.
\begin{table*}
\caption{Some Typical Commands}
\label{tab:commands}
\begin{tabular}{ccl}
\toprule
Command &A Number & Comments\\
\midrule
\texttt{{\char'134}author} & 100& Author \\
\texttt{{\char'134}table}& 300 & For tables\\
\texttt{{\char'134}table*}& 400& For wider tables\\
\bottomrule
\end{tabular}
\end{table*}
\section{Math Equations}
You may want to display math equations in three distinct styles:
inline, numbered or non-numbered display. Each of
the three are discussed in the next sections.
\subsection{Inline (In-text) Equations}
A formula that appears in the running text is called an
inline or in-text formula. It is produced by the
\textbf{math} environment, which can be
invoked with the usual \texttt{{\char'134}begin\,\ldots{\char'134}end}
construction or with the short form \texttt{\$\,\ldots\$}. You
can use any of the symbols and structures,
from $\alpha$ to $\omega$, available in
\LaTeX~\cite{Lamport:LaTeX}; this section will simply show a
few examples of in-text equations in context. Notice how
this equation:
\begin{math}
\lim_{n\rightarrow \infty}x=0
\end{math},
set here in in-line math style, looks slightly different when
set in display style. (See next section).
\subsection{Display Equations}
A numbered display equation---one set off by vertical space from the
text and centered horizontally---is produced by the \textbf{equation}
environment. An unnumbered display equation is produced by the
\textbf{displaymath} environment.
Again, in either environment, you can use any of the symbols
and structures available in \LaTeX\@; this section will just
give a couple of examples of display equations in context.
First, consider the equation, shown as an inline equation above:
\begin{equation}
\lim_{n\rightarrow \infty}x=0
\end{equation}
Notice how it is formatted somewhat differently in
the \textbf{displaymath}
environment. Now, we'll enter an unnumbered equation:
\begin{displaymath}
\sum_{i=0}^{\infty} x + 1
\end{displaymath}
and follow it with another numbered equation:
\begin{equation}
\sum_{i=0}^{\infty}x_i=\int_{0}^{\pi+2} f
\end{equation}
just to demonstrate \LaTeX's able handling of numbering.
\section{Figures}
The ``\verb|figure|'' environment should be used for figures. One or more images can be placed within a figure. If your figure contains third-party material, you must clearly identify it as such, as shown in the example below.
\begin{figure}[h]
\centering
\includegraphics[width=\linewidth]{sample-franklin}
\caption{1907 Franklin Model D roadster. Photograph by Harris \& Ewing, Inc. [Public domain], via Wikimedia Commons. (\url{https://goo.gl/VLCRBB}).}
\Description{The 1907 Franklin Model D roadster.}
\end{figure}
Your figures should contain a caption which describes the figure to the reader. Figure captions go below the figure. Your figures should {\bf also} include a description suitable for screen readers, to assist the visually-challenged to better understand your work.
Figure captions are placed {\it below} the figure.
\subsection{The ``Teaser Figure''}
A ``teaser figure'' is an image, or set of images in one figure, that are placed after all author and affiliation information, and before the body of the article, spanning the page. If you wish to have such a figure in your article, place the command immediately before the \verb|\maketitle| command:
\begin{verbatim}
\begin{teaserfigure}
\includegraphics[width=\textwidth]{sampleteaser}
\caption{figure caption}
\Description{figure description}
\end{teaserfigure}
\end{verbatim}
\section{Citations and Bibliographies}
The use of {\rm B\kern-.05em{\sc i\kern-.025em b}\kern-.08emT\kern-.1667em\lower.7ex\hbox{E}\kern-.125emX}\ for the preparation and formatting of one's references is strongly recommended. Authors' names should be complete --- use full first names (``Donald E. Knuth'') not initials (``D. E. Knuth'') --- and the salient identifying features of a reference should be included: title, year, volume, number, pages, article DOI, etc.
The bibliography is included in your source document with these two commands, placed just before the \verb|\end{document}| command:
\begin{verbatim}
\bibliographystyle{ACM-Reference-Format}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 3,060 |
Q: getting null pointer exception because can't create bean
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:aop="http://www.springframework.org/schema/aop"
xmlns:tx="http://www.springframework.org/schema/tx"
default-autowire="byName"
xmlns:oxm="http://www.springframework.org/schema/oxm"
xmlns:mvc="http://www.springframework.org/schema/mvc"
xsi:schemaLocation="http://www.springframework.org/schema/oxm http://www.springframework.org/schema/oxm/spring-oxm-4.3.xsd
http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-4.3.xsd
http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-4.3.xsd
http://www.springframework.org/schema/aop http://www.springframework.org/schema/aop/spring-aop-4.3.xsd
http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx-4.3.xsd">
<bean id="userRoleDAOTarget"
class="com.revguru.crs.security.dao.UserRoleDAOImpl"
scope="prototype">
<property name="sessionFactory" ref="hsdSessionFactory" />
</bean>
<bean id="userAuthDAOTarget" class="com.revguru.crs.security.dao.AuthenticationDAOImpl" scope="prototype">
<property name="sessionFactory" ref="hsdSessionFactory" />
</bean>
<bean id="hotelUserDAOTarget" class="com.revguru.crs.security.dao.HotelUserDAOImpl" scope="prototype">
<property name="sessionFactory" ref="hsdSessionFactory" />
</bean>
<bean id="userRoleDataObjprocessor" class="com.revguru.crs.security.service.UserRoleProcessor" scope="prototype">
<property name="userRoleDAO" ref="hsdSecurityObjectDAO" />
</bean>
<bean id="userAuthDataObjprocessor" class="com.revguru.crs.security.service.AuthenticationProcessor" scope="prototype">
<property name="authenticationDAO" ref="hsdSecurityAuthObjectDAO" />
<property name="messages" ref="userSectionMessageSource"/>
</bean>
<bean id="hotelUserDataObjProcessor" class="com.revguru.crs.security.service.HotelUserProcessor" scope="prototype">
<property name="hotelUserDAO" ref="hsdHotelUserDAO" />
<property name="securitySectionLogger" ref="generalServiceLogger"/>
</bean>
<bean id="SECURITY_SERVICE" class="com.revguru.crs.core.service.sc.SecurityService" scope="prototype">
<property name="userRoleProcessor" ref="userRoleDataObjprocessor" />
<property name="userAuthProcessor" ref="userAuthDataObjprocessor" />
<property name="hotelUserProcessor" ref="hotelUserDataObjProcessor" />
</bean>
<bean id="hsdSecurityObjectDAO" class="org.springframework.aop.framework.ProxyFactoryBean">
<property name="proxyInterfaces">
<value>com.revguru.crs.security.dao.UserRoleDAO
</value>
</property>
<property name="interceptorNames">
<list>
<value>hibernateInterceptor</value>
<value>userRoleDAOTarget</value>
</list>
</property>
</bean>
<bean id="hsdSecurityAuthObjectDAO" class="org.springframework.aop.framework.ProxyFactoryBean">
<property name="proxyInterfaces">
<value>com.revguru.crs.security.dao.AuthenticationDAO
</value>
</property>
<property name="interceptorNames">
<list>
<value>hibernateInterceptor</value>
<value>userAuthDAOTarget</value>
</list>
</property>
</bean>
<bean id="hsdHotelUserDAO" class="org.springframework.aop.framework.ProxyFactoryBean">
<property name="proxyInterfaces">
<value>com.revguru.crs.security.dao.HotelUserDAO
</value>
</property>
<property name="interceptorNames">
<list>
<value>hibernateInterceptor</value>
<value>hotelUserDAOTarget</value>
</list>
</property>
</bean>
<bean id="userSectionMessageSource"
class="org.springframework.context.support.ResourceBundleMessageSource" scope="prototype">
<property name="alwaysUseMessageFormat" value="true"/>
<property name="basenames">
<list>
<value>userauthResources/hsd_user_msg</value>
</list>
</property>
</bean>
</beans>
this is my bean file and it is getting reference to hsdsessionFsctory wich is defined as below:
<bean id="dataSource" destroy-method="close" class="org.apache.commons.dbcp.BasicDataSource">
<property name="driverClassName" value="com.mysql.jdbc.Driver" />
<property name="url" value="jdbc:mysql://127.0.0.1:3306/revguru"/>
<property name="username" value="root" />
<property name="password" value="decoder" />
</bean>
<bean id="hsdSessionFactory" class="org.springframework.orm.hibernate5.LocalSessionFactoryBean">
<property name="dataSource" ref="dataSource" />
<property name="hibernateProperties">
and when i run my project i get null pointer exception
public class UserLoginController {
private FacesContext facesContext = FacesContext.getCurrentInstance();
HttpSession session = (HttpSession)FacesContext.getCurrentInstance().getExternalContext().getSession(true);
private SecurityService securityService;
public void setSecurityService(SecurityService securityService) {
this.securityService = securityService;
}
public SecurityService getSecurityService() {
return this.securityService;
}
/**
* Authenticate the User
* @return
*/
public String authenticate() {
if(session.getAttribute("menus") != null) {
session.removeAttribute("menus");
}
if (session.getAttribute("CommonGlobalBean") != null) {
session.removeAttribute("CommonGlobalBean");
}
CommonGlobalBean commonGlobalBean = new CommonGlobalBean();
commonGlobalBean.setTimeZone(TimeZone.getDefault().getID());
UserLogin userLogin =(UserLogin)session.getAttribute("UserLogin");
String username = userLogin.getUserName();
String password = userLogin.getPassword();
UserInformation userInformation = null;
System.out.println("security serv ice is ");
if(securityService==null){
System.out.println("security serv ice is null:");
securityService= new SecurityService();
}
if (userLogin != null) {
System.out.println("username is:"+username);
System.out.println("password is:"+password);
if (securityService.authenticate(username,password)) {
Util.log("Get User Object by Email");
UserDataObject userDataObject = securityService.getUserObjectByEmail(userLogin.getUserName());
Util.log("User name : "+userDataObject.getUserFirstName());
userInformation = (UserInformation)session.getAttribute("UserObject");
if (userInformation != null) {
session.removeAttribute("UserObject");
}
userInformation = getUserInformation(userDataObject);
session.setAttribute("UserObject", userInformation);
if(!checkUserAsHotelRole(userInformation)) {
Util.log("User does not have hotel role.");
Util.reportError(facesContext, "login_user_not_valid_password", null);
}
HotelDataObject hotelDataObject = getAssignedHotel(userInformation);
session.setAttribute("hotelDataObjectInSession", hotelDataObject);
List<Menu> menus = new ArrayList<Menu>();
for (FeatureDataObject feature : userInformation.getFeatureDataObjects()) {
if (feature.getParentFeatureId() == null && feature.getFeatureTypeId().getId().intValue() == HSDServiceConstants.MENU && feature.getShowFeature() == 1) {
List<MenuItem> menuItems = getMenuItems(feature.getId(), userInformation.getFeatureDataObjects());
menus.add(new Menu(feature.getId(), feature.getFeatureName(), "", menuItems));
}
}
String languageCode = userLogin.getLanguageCode();
if (languageCode == null) {
userLogin.setLanguageCode("en");
facesContext.getViewRoot().setLocale(Locale.ENGLISH);
} else {
if(languageCode.equals("fr")) {
facesContext.getViewRoot().setLocale(Locale.FRENCH);
} else {
facesContext.getViewRoot().setLocale(Locale.ENGLISH);
}
}
Util.log(":: languageCode :: " + userLogin.getLanguageCode());
session.setAttribute("menus", menus);
session.setAttribute("CommonGlobalBean", commonGlobalBean);
/**
* get hotel general info details
*/
GeneralHotelController generalHotelController = new GeneralHotelController();
generalHotelController.loadGeneralInformation();
/**
* return to hotel general info page
*/
return "generalInfo";
} else {
Util.log("Invalid User.");
Util.reportError(facesContext, "login_user_not_valid_password", null);
}
}
return "";
}
which call another method to class as follow
public boolean authenticate(String email, String password) {
log.info("Authenticating : " + email);
UserDataObject user = getUserObjectByEmail(email);
if (user == null) {
log.info("User does not exist for email id :" + email);
return false;
}
return checkPassword(password, user.getPassword());
}
when i see it in my spring explorer it shows me Bean ref is unknown
this is my stack trace
javax.servlet.ServletException: java.lang.NullPointerException
javax.faces.webapp.FacesServlet.service(FacesServlet.java:659)
io.undertow.servlet.handlers.ServletHandler.handleRequest(ServletHandler.java:85)
io.undertow.servlet.handlers.FilterHandler$FilterChainImpl.doFilter(FilterHandler.java:129)
org.apache.myfaces.webapp.filter.ExtensionsFilter.doFilter(ExtensionsFilter.java:357)
io.undertow.servlet.core.ManagedFilter.doFilter(ManagedFilter.java:61)
io.undertow.servlet.handlers.FilterHandler$FilterChainImpl.doFilter(FilterHandler.java:131)
io.undertow.servlet.handlers.FilterHandler.handleRequest(FilterHandler.java:84)
io.undertow.servlet.handlers.security.ServletSecurityRoleHandler.handleRequest(ServletSecurityRoleHandler.java:62)
io.undertow.servlet.handlers.ServletDispatchingHandler.handleRequest(ServletDispatchingHandler.java:36)
org.wildfly.extension.undertow.security.SecurityContextAssociationHandler.handleRequest(SecurityContextAssociationHandler.java:78)
io.undertow.server.handlers.PredicateHandler.handleRequest(PredicateHandler.java:43)
io.undertow.servlet.handlers.security.SSLInformationAssociationHandler.handleRequest(SSLInformationAssociationHandler.java:131)
io.undertow.servlet.handlers.security.ServletAuthenticationCallHandler.handleRequest(ServletAuthenticationCallHandler.java:57)
io.undertow.server.handlers.PredicateHandler.handleRequest(PredicateHandler.java:43)
io.undertow.security.handlers.AbstractConfidentialityHandler.handleRequest(AbstractConfidentialityHandler.java:46)
io.undertow.servlet.handlers.security.ServletConfidentialityConstraintHandler.handleRequest(ServletConfidentialityConstraintHandler.java:64)
io.undertow.security.handlers.AuthenticationMechanismsHandler.handleRequest(AuthenticationMechanismsHandler.java:60)
io.undertow.servlet.handlers.security.CachedAuthenticatedSessionHandler.handleRequest(CachedAuthenticatedSessionHandler.java:77)
io.undertow.security.handlers.NotificationReceiverHandler.handleRequest(NotificationReceiverHandler.java:50)
io.undertow.security.handlers.AbstractSecurityContextAssociationHandler.handleRequest(AbstractSecurityContextAssociationHandler.java:43)
io.undertow.server.handlers.PredicateHandler.handleRequest(PredicateHandler.java:43)
org.wildfly.extension.undertow.security.jacc.JACCContextIdHandler.handleRequest(JACCContextIdHandler.java:61)
io.undertow.server.handlers.PredicateHandler.handleRequest(PredicateHandler.java:43)
io.undertow.server.handlers.PredicateHandler.handleRequest(PredicateHandler.java:43)
io.undertow.servlet.handlers.ServletInitialHandler.handleFirstRequest(ServletInitialHandler.java:292)
io.undertow.servlet.handlers.ServletInitialHandler.access$100(ServletInitialHandler.java:81)
io.undertow.servlet.handlers.ServletInitialHandler$2.call(ServletInitialHandler.java:138)
io.undertow.servlet.handlers.ServletInitialHandler$2.call(ServletInitialHandler.java:135)
io.undertow.servlet.core.ServletRequestContextThreadSetupAction$1.call(ServletRequestContextThreadSetupAction.java:48)
io.undertow.servlet.core.ContextClassLoaderSetupAction$1.call(ContextClassLoaderSetupAction.java:43)
io.undertow.servlet.api.LegacyThreadSetupActionWrapper$1.call(LegacyThreadSetupActionWrapper.java:44)
io.undertow.servlet.api.LegacyThreadSetupActionWrapper$1.call(LegacyThreadSetupActionWrapper.java:44)
io.undertow.servlet.api.LegacyThreadSetupActionWrapper$1.call(LegacyThreadSetupActionWrapper.java:44)
io.undertow.servlet.api.LegacyThreadSetupActionWrapper$1.call(LegacyThreadSetupActionWrapper.java:44)
io.undertow.servlet.api.LegacyThreadSetupActionWrapper$1.call(LegacyThreadSetupActionWrapper.java:44)
io.undertow.servlet.handlers.ServletInitialHandler.dispatchRequest(ServletInitialHandler.java:272)
io.undertow.servlet.handlers.ServletInitialHandler.access$000(ServletInitialHandler.java:81)
io.undertow.servlet.handlers.ServletInitialHandler$1.handleRequest(ServletInitialHandler.java:104)
io.undertow.server.Connectors.executeRootHandler(Connectors.java:202)
io.undertow.server.HttpServerExchange$1.run(HttpServerExchange.java:805)
java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1142)
java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:617)
java.lang.Thread.run(Thread.java:745)
whats is red mark in my spring explorer
while others don't have it
A: UserLoginController is not configured as a bean in your XML (at least not in the XML you've posted), therefore Spring knows nothing about it and you get no injections.
Try adding this (with full package of course):
<bean id="loginController" class="com.revguru.crs.security.dao.AuthenticationDAOImpl" scope="prototype">
<property name="securityService" ref="SECURITY_SERVICE" />
</bean>
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 4,337 |
Констал Н је најстарији модел пољских трамваја серије Констал. Трамвај се производио у четрдесетим и педесетим годинама прошлог века у фабрикама Констал, Сточња Гдањска и Сановаг. Произведено је 516 трамваја за пољске јавне превознике. У производњи је замењен трамвајима типа Констал 4Н.
Конструкција
Н је двосмерни двоосовински трамвај који потиче од немачке концепције КСВ. Трамвај има врата на обе стране и возачке кабине на сваком крају возила. Трамвај може возити на 1435 мм колосеку, али и на 1000 мм. Струја са контактне мреже се узима пантографом. Трамваји се могу спајати и у типичне композиције (Н+НД+НД), а цела композиција управља се само са једним трамвајем.
Верзије трамваја и реконструкције
Н1 (производња: 1948-1956)
Н2 (производња: 1950-1952)
Н3 (производња: 1952-1956)
НД1 (производња: 1948-1950)
НД2 (производња: 1950-1953)
НД3 (производња: 1953-1956)
2Н (производња: 1950-1956; ширина колосека: 1000 мм)
2Н1 (производња: 1950-1951; ширина колосека: 1000 мм)
2НД (производња: 1955; ширина колосека: 1000 мм)
Галерија
Референце
N | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 952 |
I love Jon & Vinny's so much.
This joint fast became a favorite for Jon and me in 2015. Chris first took us in August – we fell in love – so I made sure we returned to 412 N Fairfax when we came back to LA in October. Here are some of the things I lurve about Jon & Vinny's… I love that it's owned by the same duo as Animal and Son of a Gun, Jon Shook & Vinny Dotolo. I could probably stop there and you would already check it out, huh?! I love their pint-sized dining space; it's minimal in design, think clean lines and light wood. I love how the aligned pizza boxes serve as the only colorful décor – and they actually say 'Pizza' on them. I love that they're an all day eating spot serving breakfast, brunch, lunch and dinner. I love how their pasta is made in house. I love that their pizza isn't massive and overwhelming. I love how the crust is a bit charred and crisp. I love that they have a dessert menu and a tempting pastry case. I love the hip-hop playlist. | {
"redpajama_set_name": "RedPajamaC4"
} | 265 |
"use strict";
const UrlMessage = require(__dirname + "/../../lib/message/url-message");
exports.testBuildUrlMessageSanity = test => {
const url = "http://viber.com";
const message = new UrlMessage(url);
const messageBody = { "type": "url", "media": url };
test.deepEqual(message.toJson(), messageBody);
test.ok(!message.keyboard);
test.done();
};
exports.testBuildUrlMessageTestEncoding = test => {
const url = "http://viber.com/bla?what is real";
const message = new UrlMessage(url);
const messageBody = { "type": "url", "media": encodeURI(url) };
test.deepEqual(message.toJson(), messageBody);
test.ok(!message.keyboard);
test.done();
};
exports.testBuildUrlMessageWithKeyboard = test => {
const keyboard = { foo: "bar" };
const message = new UrlMessage("http://viber.com", keyboard);
test.deepEqual(message.keyboard, keyboard);
test.done();
};
exports.testBuildNullUrlMessage = test => {
test.throws(() => new UrlMessage(null));
test.done();
}; | {
"redpajama_set_name": "RedPajamaGithub"
} | 1,516 |
Schoolchildren, residents and church representatives gathered at the Rosendale Road estate on Armistice Day, for an Act of Remembrance.
Year 5 pupils from Rosendale Primary school took part in a ceremony to commemorate the First World War centenary, and remember those who gave their lives in conflict.
Rosendale Road is home to two war memorials marking both military and civilian deaths in wartime on Peabody estates. This includes a Lych gate listing the names of 35 men from the estate who died on active service in WW1. There is also a memorial plaque in honour of three residents killed on the estate during World War Two.
David said "I am delighted that children from Rosendale School were able to join us, along with people who have lived in this community for many years. That we are such a diverse group is a wonderful sign of hope."
Following the Act of Remembrance, poppy wreaths were placed at the foot of the memorial. | {
"redpajama_set_name": "RedPajamaC4"
} | 8,689 |
Q: Java best way to parse MongoDB file So, I read my MongoDB this way:
mongo = new MongoClient("localhost", 27017);
// Accessing the database
MongoDatabase database = mongo.getDatabase("myDb");
MongoCollection<Document> collection = database.getCollection("searchresults");
// Getting the iterable object
FindIterable<Document> iterDoc = collection.find();
int i = 1;
// Getting the iterator
Iterator it = iterDoc.iterator();
while (it.hasNext()) {
System.out.println(it.next());
i++;
}
}
As you can see, each line has several columns: Title, etc. So, when I iterate over myDB, i want to parse each line by its value instead of get all in one line.
Any suggestions?
A: You can try reading into a Document structure, then run another loop across each of the entries. This will give each value on its own line.
FindIterable<Document> iterDoc = database.getCollection("").find();
for(Document doc : iterDoc) {
for(Map.Entry<String, Object> entry : doc.entrySet()) {
System.out.println("Key: " + entry.getKey() + " Value: " + entry.getValue());
}
}
If you only want certain keys, use a projection in your find query
A: This is not a fitting answer to your question, but i would look in to a concept thats called Object-Document-Mapping (ODM). It simplifies some boilerplate code that you have to care about. A common library for MongoDB-ODM is called Morphia :)
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 2,147 |
Q: Tracking info empty in shipment email in Magento 2 I have created a shipment by adding Tacking info. I am getting the shipment email with empty Tacking info.
Next time if i click on "send Tracking Information. Then I will get the tracking info in shipment email.
My Magento version is 2.3.5
here is the screenshot of shipment email
I need to get this info during shipment creation.
Hopefully its because i used this event to send shipment email automatically whenever the shipment is created.
Vendor/Module/etc/events.xml
<event name="sales_order_shipment_save_after">
<observer name="sales_order_shipment_after"
instance="Vendor\Module\Observer\SalesOrderShipmentAfter" />
</event>
Vendor/Module/Observer/SalesOrderShipmentAfter.php
use Magento\Framework\Event\ObserverInterface;
class SalesOrderShipmentAfter implements ObserverInterface
{
protected $shipmentNotifier;
protected $_logger;
public function __construct(
\Psr\Log\LoggerInterface $logger,
\Magento\Shipping\Model\ShipmentNotifier $shipmentNotifier
) {
$this->_logger = $logger;
$this->shipmentNotifier = $shipmentNotifier;
}
public function execute(\Magento\Framework\Event\Observer $observer)
{
try{
$this->_logger->info('inside shipment observer');
$shipment = $observer->getEvent()->getShipment();
$this->shipmentNotifier->notify($shipment);
//$shipment->save();
}catch (\Exception $e) {
$this->_logger->info('--shipment observer--'.$e->getMessage());
}
}
}
Can anyone please help me on this. Thanks
A: Step 1: First create event.xml
Path- app\code\Vendor\Extension\etc\events.xml
<?xml version="1.0"?>
<config xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:noNamespaceSchemaLocation="urn:magento:framework:Event/etc/events.xsd">
<event name="sales_order_shipment_save_after">
<observer name="track_shipment" instance="Vendor\Extension\Observer\Shipment" />
</event>
</config>
Step 2: Then enter the following code in Observer.
Path – app\code\Vendor\Extension\Observer\Shipment.php
<?php
namespace Vendor\Extension\Observer;
use Magento\Framework\Event\ObserverInterface;
class Shipment implements ObserverInterface
{
public function execute(\Magento\Framework\Event\Observer $observer)
{
try {
$shipment = $observer->getEvent()->getShipment();
$tracksCollection = $shipment->getTracksCollection();
foreach ($tracksCollection->getItems() as $track) {
$track_number = $track->getTrackNumber();
$carrier_name = $track->getTitle();
}
} catch (\Exception $e) {
}
}
}
?>
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 33 |
Californian Marshall Mullen has one heck of a back garden. It has taken 5 years to perfect and he knows how to ride seamless lines and build great trails.
Next is NYC-MTL Court métrage, it's not new but it may provide inspiration for any big cycling adventures you may have this year. What is it? Cyclists from Québec, France, England, Australia and Saskatchewan ride Road bikes and Fixed Gear bikes from New York to Montréal in 4 days covering 400 miles.
Bobsleigh is amazing and it's crazy. See the 1.6km run from the 1984 Sarajevo Olympic Games ridden. There's no ice and it could be a bit more exciting admittedly but what a great place to explore and ride about. Alway amazes me how big facilities like this a built then left to gather dust.
Brother Cycles: Do not adjust your set! Their frames rock and they ride a bit crazy in London Town. This was a film made to promote their Swift frame. | {
"redpajama_set_name": "RedPajamaC4"
} | 7,812 |
{"url":"https:\/\/groupprops.subwiki.org\/w\/index.php?title=Formula_automorphism&oldid=19194","text":"# Formula automorphism\n\n(diff) \u2190 Older revision | Latest revision (diff) | Newer revision \u2192 (diff)\n\n## Contents\n\nBEWARE! This term is nonstandard and is being used locally within the wiki. [SHOW MORE]\nThis article defines a property that can be evaluated for an automorphism of an algebra in a variety of algebras. The evaluation of that property depends on the ambient variety, and not just on the automorphism or the algebra.\nView all such properties\n\n## Definition\n\nLet $\\mathcal{V}$ be a variety of algebras and $A$ be an algebra in $\\mathcal{V}$. A formula automorphism is an automorphism of $A$ given by:\n\n$x \\mapsto f(x, x_2, x_3, \\dots, x_n)$\n\nwhere $f$ is a word (or expression) in the $x_i$s, using the operations of $\\mathcal{V}$, and the $x_i$ are elements of $A$.\n\nA strong formula automorphism is a formula automorphism whose inverse is also a formula automorphism.\n\n## Particular cases\n\n### For groups\n\nIn the case of groups, the formula automorphisms are called monomial automorphisms. An automorphism such that both that and its inverse are monomial is termed a strong monomial automorphism.","date":"2021-01-21 09:23:00","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\": 0, \"img_math\": 10, \"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\": 0.9252972602844238, \"perplexity\": 699.1003202088316}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-04\/segments\/1610703524270.28\/warc\/CC-MAIN-20210121070324-20210121100324-00060.warc.gz\"}"} | null | null |
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Europe hits Google with a record $5 billion fine Google's officials immediately responded, saying the company willappeal the European Union fine by the Margrethe Vestager's team. Google's actions reduce the incentives for manufacturers to install and for users to seek out competing apps, it said.
British PM wins key Brexit vote despite ongoing rebellion But former transport minister Mr Hammond insisted its provisions were "entirely in line" with Mrs May's Brexit White Paper. After a day of tense negotiations between whips and rebel Tory MPs, matters were looking grim.
'Possible serial killer': Parolee nabbed in string of murders Houston residents were on high alert over the last few days, sharing surveillance footage of Rodriguez on social media. Rodriguez, who has distinctive head and neck tattoos, was arrested just before 07:00 local time (12:00 GMT).
Brewers All-Star reliever Josh Hader apologizes for past racist, homophobic tweets Teammate Lorenzo Cain called Hader a "great guy", and acknowledged that everyone has "said some insane stuff when we're young". Some of Hader's family removed his replica jersey while at the All-Star Game, according to Jeff Passan of Yahoo Sports.
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Trump has since been criticised by USA government officials for not holding Russian Federation accountable for meddling in the United States election.
The final question of the press conference asked Vladimir Putin whether Russian Federation had compromising material on Donald Trump.
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"President Trump is flat out wrong to blame the United States and take the word of a foreign adversary over the findings of his own intelligence experts and the House Intelligence Committee".
President Trump had previously said that he was going into the meeting with "low expectations".
US President Donald Trump is in the Finnish capital Helsinki , ahead of his summit with his Russian President Vladimir Putin . Niinistö also called attention to the importance of establishing and fostering a dialogue between the two heads of state.
"No prior president has ever abased himself more abjectly before a tyrant".
Democratic California Representative Jimmy Gomez accused Trump of continuing to "sell out" his own country to Russian Federation.
Putin also pushed back against claims that his government interfered in the US election. "All Americans should be anxious".
"Our expectations are grounded in realism, but our hopes are grounded in America's desire for friendship, cooperation and peace", Trump said. Trump wrote on Twitter.
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James Clapper, Director of National Intelligence from 2010 to 2017, called Trump's performance "an incredible capitulation". | {
"redpajama_set_name": "RedPajamaC4"
} | 5,875 |
Q: Reading in a list of csv floats with a prefix in C++ I am trying to read in a list of floats formatted like the example below from a file and into a vector, and print them back out again. Is there a way to modify the stream or the iterator to drop the 'f' that indicates float and correctly parse the data with minimal changes?
1.5f, 2.0f, 4.0f, 1.0f, 1.0f, 2.0f, 4.0f, 2.0f, 1.0f, 0.0f, 0.0f, 1.0f, 9.0f
Code here
std::ifstream infile("matrices.txt");
std::string s;
std::vector<float> A;
std::getline(infile,s,'\n');
std::stringstream mss(s);
std::copy(std::istream_iterator<float>( mss ), std::istream_iterator<float>(),std::back_inserter(A));
std::copy(A.begin(), A.end(), std::ostream_iterator<float>(std::cout, ", "));
A: You can request stringstream to ignore 2 characters every time a float is extracted.
float x{};
while (!mss.eof())
{
mss >> x;
mss.ignore(2);
vfloats.push_back(x);
}
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 6,101 |
\section{Introduction}
The problem of critical phenomena description is known for a long time and was successfully solved in 70's mainly by Wilson \cite{bib:WilsonNL}. Wilson's theory is field-theoretical one and essentially use formal expansion in series in $\varepsilon$, so called $\varepsilon$-expansion. Here $\varepsilon$ is deviation of space dimensions from a case of 4 dimensions.
Besides Wilson's approach using $\varepsilon$-expansion yet another field-theoretical approach working directly in the real 3-dimensional space is also known. This approach proposed by Parisi \cite{bib:Parisi73} has been used in calculations of critical exponents of the scalar model up to five-loop approximation\cite{bib:BakerPRL,bib:BakerPR}. Similar approach was proposed by Ginzburg\cite{bib:Ginzburg}.
Up to now both these approaches have been developed in detail and now they form the standard basis of the critical phenomena theory (see\cite{bib:Pelissetto2002,bib:ZJ-98,bib:ZJ00} and references therein).
However yet another field-theoretical approach to critical phenomena is possible. This approach is described in this paper. As in Parisi's approach we use renormalization group (RG) in real 3-dimensional space. But in contrast to latter we use point of renormalization at arbitrary momentum. Such an arbitrary point lets us to describe a massless field as well. Besides, we treat the changes of $T-T_C$ as explicit perturbations by additional term in the Lagrangian. Although such details are not critical for physical meaning of a theory, these changes allow to simplify essentially the physical sense of a theory. Besides that, such an approach allows to describe very simply not only the critical domain but also the crossover from critical domain to the domain of Landau theory. It is worth to point out that the aim of this paper is not to calculate critical exponents with high accuracy but to draw simple and intuitively clear physical picture of a critical phenomena. This is why the consideration is restricted to scalar $\phi^4$ model in one-loop approximation and consideration is as simple as possible.
\section{Perturbation theory and diagrams}
In this section we describe very briefly perturbation theory to introduce the notations. The details can be found in any quantum fields textbook, say by Ramond\cite{bib:Ramond}.
Statistical physics of thermally fluctuating scalar field $\phi$ with interactions described by term $\phi^4$ is defined by partitional function which can be represented as path integral
\begin{equation}
\label{Z}
Z(J) = \int e^{-S(\phi , J)}{\cal D\phi} \, ,
\end{equation}
where "action" is
\begin{equation}
S(\phi , J)= S_0(\phi , J)+S_I(\phi)=\int {\cal L}_0 d^3{\bf x} + \int {\cal L}_I d^3{\bf x}\, ,
\end{equation}
and "Lagrangian" is
\begin{equation}
\label{L}
{\cal L}_0 =\frac{1}{2}(\nabla\phi)^2 + \frac{\tau_0}{2}\phi^2 - J\phi \, , \qquad
{\cal L}_I =\frac{g_0}{4!}\phi^4 \, .
\end{equation}
Generally speaking it should be additional $T^{-1}$ in the Gibbs exponent in (\ref{Z}) and some coefficient $f$ in $(\nabla\phi)^2 /2$ in (\ref{L}). But one can redefine field $\phi \to T^{-1/2}f^{1/2}\phi$ so that these coefficients disappear. Given such field redefinition $\tau_0\,$, $g_0$ and $J$ are also redefined:
\begin{equation}
\tau_0 \to \tau_0 f^{-1} \, ,
\end{equation}
\begin{equation}
\label{g0}
g_0 \to g_0 T f^{-2} \, ,
\end{equation}
\begin{equation}
J \to J T^{-1/2} f^{-1/2} \, .
\end{equation}
It is easy to see that variational derivatives of $Z(J)$ by source field $J({\bf x})$ generates non-normalized correlators $G=Z(0)\langle \phi({\bf x}_1)\dots \phi({\bf x}_n)\rangle$ which can be named Green's functions in analogy with quantum field theory. To make path integral calculable, the exponent with the interaction term is expanded in series:
\begin{equation}
e^{-S_I} = \sum_n \frac{1}{n!}(-S_I)^n \, .
\end{equation}
Such a representation yields the well known perturbation theory which can be represented by Feynman diagrams. These diagrams consist of solid lines and elementary vertexes. A solid line corresponds to the propagator (two-point Green's function) with Fourier transform $(k^2+\tau_0)^{-1}\,$. Elementary vertex is $-g_0/ (4!)\,$.
Further one can introduce connected Green's functions $G_C$ and so called one participle irreducible Green's functions $\Gamma$ in a common way. Self-energy function (mass operator) $\Sigma(k)$ in a diagrammatic form is presented as a sum of two-point diagram (without external tails) which can not be divided in two parts by removing of one line. With self-energy function one can easily get the exact perturbated propagator:
\begin{equation}
G_C({\bf k})=\frac{1}{k^2+\tau_0 - \Sigma(k)} \, .
\end{equation}
Other details see in the textbook cited above.
\section{Renormalization of interaction parameter}
Let us describe 4-point one-participle irreducible Green's function $\Gamma_4\,$. It is defined by diagram:
\begin{center}
\includegraphics[width=8cm,keepaspectratio]{fish1.eps}
\end{center}
\noindent
Analytical representation of this diagram is:
\begin{equation}
\label{G4-0}
-\Gamma_4({\bf p}_1, {\bf p}_2, {\bf p}_3, {\bf p}_4) = g_0 - g_0^2 \sum I({\bf q}) \, ,
\end{equation}
where $I({\bf q})$ are loop integrals different from each other by permutations of diagram's tails; ${\bf q}$ is the momentum which flows throw a loop, it can be found from tail momenta ${\bf p}_a$ by momentum conservation law.
After that we renormalize the perturbation theory. Usually renormalization is used when loop integrals diverge. In three dimensions $\phi^4$ is super-renormalizable theory, there is the only divergent diagram and it is different from the shown above. But anyway it is possible to renormalize the convergent diagrams also. Such a renormalization improves the perturbation theory.
To renormalize perturbation theory we divide parameters into couples of terms $\tau_0=\tau+\tau_{ct}\,$, $g_0=g+g_{ct}\,$ and treat the term of Lagrangian with $\tau_{ct}$ as perturbation. Besides that, $\phi^4$-terms with $g$ and $g_{ct}$ are treated as separate perturbations. Because the new perturbation series is series in $g\,$, counterterms $\tau_{ct}\sim g$ and $g_{ct} \sim g^2$ at least. Generally speaking, the gradient term $(\nabla\phi)^2 /2$ should be divided into a main term and a counterterm as well. But it is not necessary in one-loop approximation of $\phi^4$-theory and we do not describe it.
According to the above, renormalized equation (\ref{G4-0}) up to $g^2$ is:
\begin{equation}
\label{G4R}
-\Gamma_4({\bf p}_1, {\bf p}_2, {\bf p}_3, {\bf p}_4) = g - g^2 \sum I({\bf q}) + g_{ct} \, .
\end{equation}
Certainly there is a continuum of variants to divide bare interaction constant $g_0$ into renormalized interaction parameter $g$ and counterterm $g_{ct}\,$. Thus we should add some renormalization conditions to choose one of them.
Without perturbation theory it does not matter what variant to choose. But if one uses perturbations then some variants are much better then the other. Obviously perturbation expansion works well when third term in r.h.s. of (\ref{G4R}) approximately compensates the second term. Thus it is convenient to choose renormalization condition as a condition of full compensation of the second and the third term when the tail momenta are equal to the some four vectors, so called renormalization point vectors. Then the perturbation expansion will work well when the tail momenta are close to renormalization point vectors.
Now we should choose the normalization point. To parametrize four vectors by one parameter we introduce for unity vectors ${\bf n}_a\,$ and define normalization point as ${\bf p}_a=\mu {\bf n}_a\,$ where $\mu$ is renormalization parameter. It is convenient to choose vectors ${\bf n}_a$ being inscribed in cube symmetrically as shown on FIG.\ref{fig:symnvectors}. Obviously with such vectors ${\bf n}_a$ vectors ${\bf p}_a=p{\bf n}_a$ obey momentum conservation low.
\begin{figure}[h]
\begin{center}
\includegraphics[width=4cm,keepaspectratio]{n1-4.eps}
\end{center}
\caption{Symmetric set of vectors ${\bf n}_a\,$. }
\label{fig:symnvectors}
\end{figure}
Further we restrict the consideration of $\Gamma_4({\bf p}_1, {\bf p}_2, {\bf p}_3, {\bf p}_4 )$ to a case when ${\bf p}_a=p{\bf n}_a\,$. Such a $\Gamma_4$ will be denoted as $\Gamma_4(p)\,$. For such a case ${\bf q}$ is the same for all permutations of tails and we can write:
\begin{equation}
\label{Gp-ct}
-\Gamma_4( p) = g -
\frac{3}{2}g^2\frac{1}{(2\pi)^3}\int\frac{d^3{\bf k}}{(k^2+\tau)[({\bf k}+{\bf q})^2 +\tau ]} + g_{ct} \, .
\end{equation}
With elementary geometry one can find $q=(2p)/\sqrt{3}\,$. According to said above we define renormalized $g$ by condition:
\begin{equation}
\label{gdef}
\left . -\Gamma_4(p)\right |_{p=\mu} = g \, .
\end{equation}
Loop integrals similar to integral in (\ref{Gp-ct}) are usually evaluated by a well known Feynman trick. Specifically this integral also can be calculated by elementary methods in cylindrical coordinates with z-axis along vector ${\bf q}\,$. Anyway integration and fixing of $g_{ct}$ by (\ref{gdef}) yields:
\begin{equation}
\label{Gp}
-\Gamma_4(p)=g - g^2\frac{3\sqrt{3}}{16 \pi}
\left [ \frac{1}{p} \arctan\left(\frac{p}{\sqrt{3\tau}}\right) -
\frac{1}{\mu} \arctan \left(\frac{\mu}{\sqrt{3\tau}}\right) \right ]
\, .
\end{equation}
Actually equations (\ref{gdef}) and (\ref{Gp}) define how renormalized interaction parameter $g$ changes when the scale of renormalization point $\mu\,$ changes. One need only substitute the new scale $\mu'$ as $p$ in (\ref{Gp}). But it is worth to remember that (\ref{Gp}) works well only when $p$ is close to $\mu\,$. Thus it is much better to describe large change of $\mu\,$ as step-by-step small changes of $\mu\,$. If each step is infinitesimal then this yields a differential equation. This is the key idea of renormalization group method. Substituting $p=\mu+d\mu\,$ in (\ref{Gp}) with simple algebra we get:
\begin{equation}
\label{dgdmtau}
\frac{d g_{\mu}}{d \mu}=
g^2_{\mu}\frac{3\sqrt{3}}{16 \pi}\cdot\frac{1}{\mu^2}
\left[\arctan\left(\frac{\mu}{\sqrt{3\tau}}\right) -
\frac{\mu / \sqrt{3\tau} }{1 + ( \mu / \sqrt{3\tau} )^2}
\right] \, ,
\end{equation}
where the subscript $\mu$ stresses that $g$ is a function of $\mu\,$. This equation is one of the key equations of our description. When $\tau=0\,$ it becomes more simple:
\begin{equation}
\label{dgdm0}
\frac{d g_{\mu}}{d \mu}= g^2_{\mu}\frac{3\sqrt{3}}{32}\cdot\frac{1}{\mu^2} \, .
\end{equation}
\section{Renormalization of "mass"}
The parameter $\tau$ corresponds to the square of the participle mass in quantum field theory. This parameter also undergoes renormalization. In one-loop approximation this renormalization represented by single diagram for self-energy function:
\begin{center}
\includegraphics[width=1.5cm,keepaspectratio]{tadpole.eps}
\end{center}
It is essential that in one-loop approximation $\Sigma\,$ is momentum independent. In our theory $\tau\,$ is $\mu$-independent constant while one-loop approximation is used.
It is worth to note that in the common theories $\tau$ is $\mu$-dependent nevertheless. This is reached by very refined trick which we do not use. Instead, we use another more physically clear trick which is described below.
\section{"Running" interaction parameter}
In this section we solve equation (\ref{dgdmtau}) but before that it is useful to describe solution of (\ref{dgdm0}). Obviously, when $\mu>>\sqrt{\tau}\,$ the solution of (\ref{dgdmtau}) is close to the solution of (\ref{dgdm0}), while the latter is much more simple. Solution of (\ref{dgdm0}) is:
\begin{equation}
g = \frac{\mu}{A + C\mu} \, ,
\end{equation}
where $A=3\sqrt{3}/32\approx 0.1624\,$, and $C$ is an integration constant. One can see easily that if $\mu \to \infty\,$ then $g \to C^{-1}\,$. Thus at large momentum $g$ is constant. Let us denote this constant as $g_{\infty}\,$. Then
\begin{equation}
\label{gmut0}
g(\mu)=\frac{g_{\infty}}{1+Ag_{\infty}\mu^{-1}} \, .
\end{equation}
Otherwise if $\mu \to 0\,$ then
\begin{equation}
\label{gmto0}
g(\mu\to 0) =\frac{\mu}{A}\approx 6.158 \mu \, .
\end{equation}%
Equation (\ref{gmto0}) shows a very important fact: the behaviour of $g$ in IR limit is universal, it does not depend on $g_{\infty}\,$. This means that if $\phi\,$ describes some field in real system, say in crystal, then interactions of its long-wave thermal fluctuations does not depend on interactions at atomic scale. Now we see this fact for $\tau=0\,$ (i.e. for a system at just critical point), further we will see this for $g_{\infty}^2>>\tau\ne 0\,$ as well.
Exact solution of (\ref{dgdmtau}) yields not too complex result:
\begin{equation}
g=\frac{\mu}{[(3\sqrt{3})/(16\pi)]\arctan(\mu/\sqrt{3\tau}) + C\mu} \, ,
\end{equation}
where $C$ is an integration constant. We see again that $g \to {\rm const}=C^{-1}=g_{\infty}\,$ while
$\mu \to \infty\,$. So we can rewrite the equation in the form:
\begin{equation}
\label{gmgen}
g(\mu)=\frac{g_{\infty}}{1 + g_{\infty}\mu^{-1}[(3\sqrt{3})/(16\pi)]\arctan(\mu/\sqrt{3\tau})} \, .
\end{equation}
Some essential point should be discussed here. Describing solution of differential equation mathematically, one may think that the integration constant $C\,$ and the parameter $g_{\infty}\,$ can depend on $\tau\,$. But by physical reasons this parameters should be $\tau$-indepedent. Indeed, one can see from (\ref{Gp}) that if $\mu\to \infty,$ then $g_{ct}\to 0\,$. Thus $g_{\infty}\,$ actually is bar parameter $g_0\,$ which obviously does not depends on $\tau\,$.
Further, if $\mu = 0\,$ then:
\begin{equation}
g(\mu=0)=\frac{g_{\infty}}{1 + g_{\infty}[3/(16\pi)]\tau^{-1/2}} \, .
\end{equation}
One can see very easily that if $\tau<<g^2_{\infty}\,$ then the behaviour is universal again:
\begin{equation}
g(\mu=0) = \frac{16\pi}{3}\sqrt{\tau} \approx 16.755 \sqrt{\tau}\, .
\end{equation}
Besides if $g_{\infty}>>\mu >> \sqrt{\tau} \,$ then
\begin{equation}
g(\mu)= \frac{32}{3\sqrt{3}}\mu \approx 6.158 \mu \, .
\end{equation}
This equation is also universal and it is the same that was derived above for the massless ($\tau=0$) field.
\section{Perturbation of proximity to critical point}
Experimentally, the proximity of a system to the critical point is usually controlled by temperature. Thus a direct way to describe system reaction to the changes of proximity to critical point requires to find a reaction to $g_0$ changes (see equation (\ref{g0})~). But if $T-T_C<<T_C\,$ then it does not matter what to change, $T$ or $T_C\,$. This is why the proximity to critical point can be controlled by $\tau_0\,$ changes, this is more convenient technically.
We describe $T-T_C$ changes as an explicit perturbation. Thus we add a perturbation term $t \lambda_0 \phi^2/2 \,$ to the "Lagrangian". Here the parameter $t$ controls the changes of $T-T_C\,$. Why a parameter $\lambda_0$ is added will be clear further.
In reality, $t$ is spatially homogeneous. However, perturbation with spatially homogeneous $t$ can not be renormalized directly. This is why we generalize the theory to spatially inhomogeneous $t({\bf x})\,$ which will be changed to homogeneous $t$ at the end of calculations. Actually we add yet another source field interacting with $\phi\,$ non-linearly.
Perturbation $t({\bf x}) \lambda_0 \phi^2/2 \,$ generates additional 3-tail vertex ($-\lambda_0/2\,$ in the analytical representation). One of its tails is $t({\bf x})\,$ which will be depicted in the diagrams as a waving line. As any vertex, $\lambda$-vertex can be renormalized. To renormalize it we divide it into renormalized vertex $\lambda$ and counterterm $\lambda_{ct}$ where $\lambda_{ct} \sim g$ at least. Further descriptions are generally the same as the renormalization of $g\,$. One-loop correction to $\Gamma_3$ is represented by diagram:
\begin{center}
\includegraphics[width=5cm,keepaspectratio]{G3.eps}
\end{center}
which analytically yields:
\begin{equation}
\label{G3I}
-\Gamma_3(p)=\lambda - \frac{1}{2}g\lambda \frac{1}{(2\pi)^3}\int
\frac{d^3{\bf k}}{(k^2+\tau)[({\bf k}+{\bf p})^2+\tau]} + \lambda_{ct} \, ,
\end{equation}
where ${\bf p}\,$ is waving tail momentum. Renormalization conditions we choose similar to condition for $g\,$:
\begin{equation}
\left . -\Gamma_3(p) \right |_{p=\mu} = \lambda \, .
\end{equation}
Integral in (\ref{G3I}) is just the same that in (\ref{Gp-ct}). So with full analogy to above descriptions we derive:
\begin{equation}
-\Gamma_3(p) = \lambda - \lambda g \frac{1}{8\pi} \left[
\frac{1}{p}\arctan\left(p / \sqrt{4\tau}\right) - \frac{1}{\mu}\arctan\left(\mu / \sqrt{4\tau}\right)
\right] \, ,
\end{equation}
\begin{equation}
\label{dlgen}
\frac{d \lambda_{\mu}}{d \mu} = \lambda_{\mu} \frac{g_{\mu}}{8\pi}\cdot\frac{1}{\mu^2}\left[
\arctan\left(\mu / \sqrt{4\tau}\right) - \frac{\mu / \sqrt{4\tau}}{1 + \left(\mu / \sqrt{4\tau}\right)^2}
\right] \, .
\end{equation}
For $\tau=0\,$ or $\mu>>\sqrt{\tau}\,$ the last equation can be simplified:
\begin{equation}
\label{dlt0}
\frac{d \lambda_{\mu}}{d \mu} = \lambda_{\mu} \frac{g_{\mu}}{16}\cdot\frac{1}{\mu^2} \, .
\end{equation}
\section{"Running" $\lambda$}
$\lambda(\mu)\,$ can be found easily when $\tau=0\,$ or $\mu>>\sqrt{\tau}\,$. Using (\ref{gmut0}) and(\ref{dlt0}) we obtain:
\begin{equation}
\lambda(\mu) = C \left(\frac{\mu}{\mu + g_{\infty}\left(3\sqrt{3}/32\right)}\right)^{\alpha} \, ,
\end{equation}
where $\alpha=2/(3\sqrt{3})\approx 0.385\,$, $C\,$ is an integration constant. Obviously if $\mu\to\infty\,$ then $\lambda \to C={\rm const}\,$. Thus the equation can be rewritten as:
\begin{equation}
\label{lmt0}
\lambda(\mu) = \lambda_{\infty}
\left(\frac{\mu}{\mu + g_{\infty}\left(3\sqrt{3}/32\right)}\right)^{\alpha} \, .
\end{equation}
Here the parameter $\lambda_{\infty}\,$ is $\tau\,$-independent by the same reason as $g_{\infty}\,$.
When $\mu<< g_{\infty}\,$ the equation can be simplified:
\begin{equation}
\label{lmt0asim}
\lambda(\mu) = \frac{\lambda_{\infty} \mu^{\alpha}}
{\left[ g_{\infty}\left(3\sqrt{3}/32\right)\right]^{\alpha}}
\, .
\end{equation}
If $\mu \sim \sqrt{\tau\,}$ or $\mu <\sqrt{\tau}$ then the differential equation (\ref{dlgen}) apparently can be solved by numerical methods only. But at least in critical domain it is easy to get universal correlation between $\lambda(\mu=0)$ and $\lambda(\mu_*)\,$, where $\mu_* \sim \sqrt{\tau}\,$, say $\mu_*=10\sqrt{\tau}\,$. Further $\lambda(\mu_*)\,$ can be evaluated by (\ref{lmt0}) or (\ref{lmt0asim}).
If $\mu<\mu_*=10\sqrt{\tau}\,$ and $\mu_*<<g_{\infty}\,$ then (\ref{gmgen}) can by simplified as
\begin{equation}
\label{gmgensimp}
g(\mu)=\frac{16\pi\mu}{3\sqrt{3}\arctan(\mu/\sqrt{3\tau})} \, .
\end{equation}
With such $g(\mu)\,$ equation (\ref{dlgen}) yields:
\begin{equation}
\label{llaster}
\int\limits^{\lambda(\mu_*)}_{\lambda(0)} \frac{d\lambda}{\lambda} =
\alpha \int\limits_0^{10}\frac{\left[\arctan(x/2) - (x/2)/[1+(x/2)^2] \right]}
{x\arctan(x/\sqrt{3})} dx \, ,
\end{equation}
where substitution $x=\mu/\sqrt{\tau}\,$ used.
In (\ref{llaster}) r.h.s. is a universal constant which can be evaluated numerically. For our purpose however it does not matter what its value is. The only fact we need is that according to (\ref{llaster})
\begin{equation}
\lambda(0) \sim \lambda(\mu_*) \, .
\end{equation}
With (\ref{lmt0asim}) this correlation yields in critical domain:
\begin{equation}
\label{l0rdy}
\lambda(0) \sim \tau^{\alpha/2} \, .
\end{equation}
\section{Critical exponents}
The last equation of the previous section allows us to derive critical exponents very simply. If the system is off the critical point by some non-zero $\tau\,$, then, in order to bring it to the critical point, the perturbation of the self-energy function $\Sigma(k=0)\,$ should be equal $\tau\,$. On the other hand, in the first order of perturbation theory $\Sigma=-\lambda(0)\Delta T\,$, where $\Delta T\,$ is the temperature shift (negative to shift down to critical temperature). Thus we have a self-consistent equation:
\begin{equation}
\tau=-\lambda(0)\Delta T \, .
\end{equation}
Using (\ref{l0rdy}) we instantly obtain:
\begin{equation}
\tau \sim \tau^{\alpha/2} (T-T_C) \, .
\end{equation}
An obvious transformation yields:
\begin{equation}
\tau \sim (T-T_C)^{\gamma} \, ,
\end{equation}
where
\begin{equation}
\gamma = \frac{2}{2-\alpha} \, .
\end{equation}
There is no field renormalization in one-loop approximation. This is why the Fisher exponent $\eta\,$ is zero and the propagator is $G(k)=(k^2+\tau)^{-1}\,$. Thus the critical exponent for correlation length is
\begin{equation}
\nu=\frac{\gamma}{2} = \frac{1}{2-\alpha} \, .
\end{equation}
Numerical estimations yields values $\gamma=1.24\,$, $\nu=0.62\,$ which are close to the known precise values (see\cite{bib:Pelissetto2002}) .
Derivations in this section above may look doubtful because $\lambda(0)\,$ changes while $\tau\,$ changes. Using of the first order perturbation may also seem doubtful. But we can justify the result by consideration of infinitesimal change of $\tau\,$ with infinitesimal change of temperature. Such an approach yields:
\begin{equation}
d \tau = \lambda(0) d T \sim \tau^{\alpha/2} d T \, .
\end{equation}
Integration of this differential equation yields exactly the same result that we have obtained before.
\section{Crossover to Landau theory}
The approach considered here allows us to determine not only critical behaviour but also crossover to Landau theory behaviour when $\tau\,$ is large. To describe this crossover one should find $\lambda(0)\,$ as a function of $\tau\,$ without approximations. Using (\ref{gmgen}) and (\ref{dlgen}) we obtain
\begin{equation}
\lambda(0) = \lambda_{\infty} e^{-F(y)} \, ,
\end{equation}
where function $F(y)\,$ is defined as
\begin{equation}
F(y) = y\int\limits_0^{\infty}
\frac{\arctan(x/2)-(x/2)/[1+(x/2)^2] }
{x + y\alpha^{-1}\arctan(x/\sqrt{3})}
\cdot \frac{dx}{x} \, ,
\end{equation}
$y=g_{\infty}/(8\pi\sqrt{\tau})\,$, $x=\mu/\sqrt{\tau}\,$.
A dependence of $\lambda(0)/\lambda_{\infty}\,$ on $y=g_{\infty}/(8\pi\sqrt{\tau})\,$ can be calculated by numerical methods. The result is shown on FIG.\ref{fig:lont}.
\begin{figure}[h]
\begin{center}
\includegraphics[width=12cm,keepaspectratio]{lont.eps}
\end{center}
\caption{Dependence of $\lambda(0)/\lambda_{\infty}\,$ on $y=g_{\infty}/(8\pi\sqrt{\tau})\,$ calculated numerically. }
\label{fig:lont}
\end{figure}
From FIG.\ref{fig:lont} one can see that if $\sqrt{\tau} << g_{\infty}/(8\pi)\,$ i.e $y >> 1\,$ then $\lambda(0)\,$ depends on $\tau\,$ as described in previous sections. This behaviour corresponds to critical domain. Otherwise if $\sqrt{\tau} >> g_{\infty}/(8\pi)\,$ i.e. $y << 1\,$ then $\lambda(0)\approx \lambda_{\infty} = {\rm const}\,$. This means that in this domain $d\tau \sim dT\,$. Such a behaviour corresponds to Landau theory. One can see also that condition of crossover $\tau \sim g_{\infty}/(8\pi)\,$ i.e. $y \sim 1\,$ coincides, at least up to some coefficient, with well known Levanyuk criterion.
\section{Conclusion}
In this paper a very simple approach to critical behaviour description has been presented. Generally this approach is close to the common field-theoretical theories using renormalization group method. Nevertheless the presented approach has at least one advantage: it allows one to draw a simple, intuitively clear physical picture of a critical phenomena.
Physical reason why non-trivial exponents appear is that the system's sensitivity to $T-T_C\,$ variations, described by $\lambda$-vertex at zero momentum, changes as system goes toward to critical point. If one takes in account this fact, then reasonable values of non-trivial exponents can be obtained by very simple calculations. Moreover, far from critical point $\lambda$-vertex ceases to depend on correlation length (which is $\sim 1/\sqrt{\tau}\,$). In this way, far from critical point the behaviour of a system becomes identical to the one described by Landau theory. Such a crossover can also be described in a simple way.
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{"url":"https:\/\/www.semanticscholar.org\/paper\/Divergent-on-average-directions-of-Teichm%C3%BCller-flow-Apisa-Masur\/2564a849260fb33faa26ad61c318eb1c142744f9","text":"Divergent on average directions of Teichm\u00fcller geodesic flow\n\n@article{Apisa2021DivergentOA,\ntitle={Divergent on average directions of Teichm{\\\"u}ller geodesic flow},\nauthor={Paul Apisa and Howard A. Masur},\njournal={Journal of the European Mathematical Society},\nyear={2021}\n}\n\u2022 Published 28 February 2018\n\u2022 Mathematics\n\u2022 Journal of the European Mathematical Society\nThe set of directions from a quadratic differential that diverge on average under Teichmuller geodesic flow has Hausdorff dimension exactly equal to one-half.\n3 Citations\nResearch Statement\nMy research focuses on geometry and dynamics, often involving moduli spaces like Mg, the variety whose points correspond to the choice of a complex structure on a closed orientable genus g surface.Expand\nSingular Vectors on Fractals and Projections of Self-similar Measures\nSingular vectors are those for which the quality of rational approximations provided by Dirichlet\u2019s Theorem can be improved by arbitrarily small multiplicative constants. We provide an upper bound onExpand\nQuantitative weak mixing for interval exchange transformations\n\u2022 Mathematics\n\u2022 2021\nWe establish a dichotomy for the rate of the decay of the Ces\u00e0ro averages of correlations of sufficiently regular functions for typical interval exchange transformations (IET) which are not rigidExpand\n\nReferences\n\nSHOWING 1-10 OF 27 REFERENCES\nSum of Lyapunov exponents of the Hodge bundle with respect to the Teichm\u00fcller geodesic flow\n\u2022 Mathematics\n\u2022 2014\nWe compute the sum of the positive Lyapunov exponents of the Hodge bundle with respect to the Teichmuller geodesic flow. The computation is based on the analytic Riemann-Roch Theorem and uses aExpand\nPeriodic geodesics on generic translation surfaces\nThe objective of the paper is to study properties of periodic geodesics on translation surfaces that hold for generic elements of the moduli space of translation surfaces. 2000 Mathematics SubjectExpand\nOn cusp excursions of geodesics and Diophantine approximation\nIn this article we describe some new examples of correspondence between Diophantine approximation and homogeneous dynamics, by characterizing two kinds of exceptional orbits of geodesic flowExpand\nHomogeneous approximation for flows on translation surfaces\nWe consider how quickly a typical point returns to neighborhoods of itself under the flow in a typical direction on a translation surface.\nErgodicity of billiard flows and quadratic differentials\n\u2022 Mathematics\n\u2022 1986\nOn considere un systeme billard de 2 objets de masses m 1 et m 2 . On montre que pour un ensemble dense de paires (m 1 ,m 2 ) ce systeme est ergodique\nClosed trajectories for quadratic differentials with an application to billiards\nOn considere les orbites periodiques pour le systeme dynamique d'une table polygonale avec des angles multiples rationnels de \u03c0: le billard rationnel\nPeriodic geodesics on translation surfaces\nAs shown by Masur in 80s, for any translation surface there exists a periodic geodesic of bounded length, the directions of periodic geodesics are dense in the unit circle, and the number ofExpand\nTopological transitivity of billiards in polygons\n\u2022 Mathematics\n\u2022 1975\nConsider a billiard in a polygon Q\u2282R2 having all angles commensurate with \u03c0. For the majority of initial directions, density of every infinite semitrajectory in configuration space is proved. AlsoExpand\nExceptional directions for the Teichm\u00fcller geodesic flow and Hausdorff dimension\n\u2022 Mathematics\n\u2022 2017\nWe prove that for every flat surface $\\omega$, the Hausdorff dimension of the set of directions in which Teichmuller geodesics starting from $\\omega$ exhibit a definite amount of deviation from theExpand\nThe set of non-uniquely ergodic d-IETs has Hausdorff codimension 1\/2\n\u2022 Mathematics\n\u2022 2018\nWe show that the set of not uniquely ergodic d-IETs has Hausdorff dimension d-3\/2 (in the (d-1)-dimension space of d-IETs) for d>4. For d=4 this was shown by Athreya-Chaika and for d=2,3 the set isExpand","date":"2021-12-08 09:56:03","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 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\": 0.7926027178764343, \"perplexity\": 1808.4141592816154}, \"config\": {\"markdown_headings\": false, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-49\/segments\/1637964363465.47\/warc\/CC-MAIN-20211208083545-20211208113545-00419.warc.gz\"}"} | null | null |
Maraya (en castellano: Espejos) es una conocida serie de televisión satírica siria sobre política, muy popular, creada por el comediante Yasser al-Azma.
Dio comienzo en 1982, protagonizada por Yasser al-Azma, logrando un éxito rotundo; y expuesto en canales satelitales árabes, desde los años ochenta y aún hoy en día, la serie refleja la diversidad de la cultura local. El héroe de la serie refleja al gran artista sirio Yasser Al-Azma, así como a artistas y artistas involucrados en la encarnación de muchas personalidades diferentes.
Desde su creación en 1982, muchas historias han sido exhibidas en un estilo cómico, criticando a un bello cínico. Su denominador común es insuperable ya que está estrechamente relacionado con las preocupaciones de la sociedad y la vida cotidiana.
Han actuado Maria Gabr, Salim Klass, Maha Al Masri, Wafa Moussalli, Sabah Barakat, Radwan Aqili, Marwan Al-Masri, Fadah Khattab, Nadine, Abdel Moneim Amairi Marayas, siendo la serie a contribuir al descubrimiento de muchos talentos: en dirección, composición y actuación.
La serie en 2004
Para la serie «Espejos 2004», preparada, escrita y protagonizada por el artista Yasser Al-Azmah, que regresó ese año bajo el título «Nos han vivido», cumpliendo con el público en toques cómicos a través del cual muchos fenómenos sociales que impregnan nuestro ambiente se prsentaron. Su denominador común es insuperable ya que estuvo relacionado con las preocupaciones de la sociedad y la vida cotidiana. Trabajaron Yasser Al-Azmah, Sabah Barakat, Hassan, Abdul Hakim Kotaivan, Essam Gee, Mohamed Khair, Basil Khayat, Yassin Dina Haroun, Nizar Abu Hajar, Mohammed Khaonda, Maxim Khalil, Mohammed Hariri, Mohammad Haddaqa, Fawzi Bishara, Adham asesor y Adnan Abu Ahamat. Hasta Dostoivsky ingresó a nuestras salas a través de "Espejos" para contribuir al renacimiento de la nación árabe.
La serie en 2006
Para la serie Espejos, de 2006, se prepararon 30 episodios, compuesto y preparado por Yasser Al-Azmah, y Nouruddin Al Hashemi, con la dirección de Hisham Sharbatji. La producción fue de Countryside Art Productions. Aunque el programa de series 2006, difiere de sus predecesores en términos de enriquecer una nueva colección de temas más coloridos y atrayendo a un grupo de actores distinguidos, pero sin apartarse de la esencia de desentrañar las preocupaciones sociales y la vida cotidiana en un estilo crítico y cínico.
La serie en 2011
Yasser Al-Azmah regresó a la pequeña pantalla después de un receso de cinco años, tras el final de Mirrors 2006. Su regreso en 2011, con una nueva serie. Esta parte de Espejos muestra que la mayoría de los actores, ya habían participado en varias partes anteriores, como Salim Kalas, Abed Fahd, Merah Jabr, Issam Abe Jee y muchos otros.
La serie en 2013
Se mostró en Argelia, pero no logró una alta tasa de visualización como sus partes anteriores.
Reparto más importante
Yasser Al-Azmah (2013-2011)
Samia al-Jaza'iri (1982-1984, 1996-1999)
Salim Klass (1982-2000, 2011)
Hala Hosni (1984-2011)
Bashar Ismail (1996-2006)
Maha Masri (1988-2000)
Salma El Masri (1998-2011)
Merah Jabr (1996-2011)
Tawfiq al-Asha (1982-2005)
Hani Shahin (1986-2000)
Abed Fahad (1986-2000, 2005, 2011)
Saif al-Din Subaei (1995-2011)
Essam Abe Gee (1986-2011)
Nadine Khoury (1982-1991)
Jihad Abdo (2003-2011)
Fadia Khattab (1982-1995)
Mohamed Qnoua (1995-2011)
Hassan Dakk (1995-2011)
Wafa Moussalli (1998-2011)
Mañana Argelina (2003-2011)
Mohamed Jabouli (1984-1997)
Ejemplo de adaptación de la realidad siria por la serie
Es un hombre que sueña, donde lleva la delantera en una insurrección contra el régimen gobernante. Con sudor en la frente, se despierta a la mañana siguiente. ¿Cómo podría tener un pensamiento tan nefasto? se preguntó. Unos minutos después, policías con chaquetas de cuero negro lo golpean y arrestan. Él debe unirse, y ni siquiera sintió la necesidad de preguntar el motivo de su arresto. El espectador entendió por qué, pues el gobierno puede incluso leer tus sueños.
Referencias
Enlaces externos
La serie, en 1998, en Youtube
Otros capítulos de Maraya, en Youtube.
الريف للإنتاج الفني el sitio oficial de la producción.
Series de televisión de Siria
Medios de comunicación en árabe | {
"redpajama_set_name": "RedPajamaWikipedia"
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{"url":"https:\/\/www.hackmath.net\/en\/math-problem\/4566","text":"# The farmer\n\nThe farmer brought potatoes to the market. In the first hour he sold two-fifths of the potatoes brought in, in the second hour he sold five-sixths of the remaining potatoes, and in the third hour, he sold the last 40 kg of potatoes.\n\n1. Express a fraction of what portion left after the first hour of sale.\n2. Calculate how many kilograms of potatoes the farmer sold in the second hour.\n3. Calculate how many kilograms of potatoes the farmer brought to the market.\n\nCorrect result:\n\na = \u00a00.6\nb = \u00a0200 kg\nc = \u00a0400 kg\n\n#### Solution:\n\n$a=1 - 2\/5=\\dfrac{ 3 }{ 5 }=0.6$\n\n2\/5c + b + 40 = c\nb = 5\/6 (c-2\/5c)\n\n2\/5\u2022c + b + 40 = c\nb = 5\/6\u2022(c-2\/5\u2022c)\n\n5b-3c = -200\n30b-15c = 0\n\nb = 200\nc = 400\n\nCalculated by our linear equations calculator.\n\nOur examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!\n\nLeave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...):\n\nBe the first to comment!\n\nTips to related online calculators\nNeed help calculate sum, simplify or multiply fractions? Try our fraction calculator.\nDo you have a linear equation or system of equations and looking for its solution? Or do you have quadratic equation?\n\n## Next similar math problems:\n\n\u2022 The store\nThe store received the same number of cans of peas and corn. The first day sold 10 cans of peas and 166 cans ofcorn so that left 5 times more peas than corn cans. How many cans of each kind were in the store?\n\u2022 Sumo competition\nIn the competition for the heaviest man was 5 competitors. The first three together weighed 553 kg. What is the minimal weigt of winner, if a contestant weighed with an accuracy of one kilogram.\n\u2022 Pears\nThere were pears in the basket, I took two-fifths of them, and left six in the basket. How many pears did I take?\n\u2022 Unknown number\nDetermine the unknown number that is equal to a quarter of the fifth of number, which is by 152 more than unknown number.\n\u2022 New bridge\nThanks to the new bridge, the road between A and B has been cut to one third and is now 10km long. How much did the road between A and B measure before?\n\u2022 Pipe\nSteel pipe has a length 2.5 meters. About how many decimetres is 1\/3 less than 4\/8 of this steel pipe?\n\u2022 One third\nIf 3\/5 is 360, how much is 1\/3?\n\u2022 Simple equation\nSolve for x: 3(x + 2) = x - 18\n\u2022 Eq-frac\nSolve the following equation with fractions: h + 1\/3 =5\/3\n\u2022 Eqn\nSolve equation with fractions: 2x\/3-50=40+x\/4\n\u2022 Unknown number\nI think the number - its sixth is 3 smaller than its third.\n\u2022 Fraction + eq\nSolve following simple equation with fractions: -5\/6(8+5b) = 75 + 5\/3b\n\u2022 Guess a fraction\nTom was asked to guess a fraction. The sum of 1\/2 the numerator and 1\/3 of its denominator is 30. If Tom subtracts 36 from its denominator, the fraction becomes 1\/3. What is the fraction that Tom was asked to guess? (Leave your answer in simplest form)\n\u2022 Football season\nThroughout the football season gave Adam the 23 goals more than Brano and 16 goals less than Karol. Edo gave the 6 goals more Du\u0161an. Karol gave 5 more than Edo. Brano and Edo gave together 17 goals. How many goals gave each boy and how many goals gave the\n\u2022 Every day 7 pages\nAdelka reads the book every day 7 pages. When she reads one more page a day she will read it three days earlier. How long will Adelka read a book? How much does a book of pages have?\n\u2022 Tippler\nBottle with cork cost 8.8 Eur. The bottle is 0.8 euros more expensive than cork. How much is a bottle and the cork?","date":"2020-07-03 17:15:21","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 1, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5571823120117188, \"perplexity\": 2828.664164653184}, \"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-2020-29\/segments\/1593655882634.5\/warc\/CC-MAIN-20200703153451-20200703183451-00152.warc.gz\"}"} | null | null |
Q: Ubuntu 16.04 snapd problem I am having a problem with installing/uninstalling snapd and I cannot install any package.
And it shows the following error:
error: cannot communicate with server: Posthttp://localhost/v2/snaps/code: dial unix /run/snapd.socket: connect: no such file or directory
This is the version I have:
snap --version
snap 2.31.1~14.04
snapd unavailable
series -
| {
"redpajama_set_name": "RedPajamaStackExchange"
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{"url":"https:\/\/en.wikipedia.org\/wiki\/Arellano%E2%80%93Bond_estimator","text":"# Arellano\u2013Bond estimator\n\nIn econometrics, the Arellano\u2013Bond estimator is a generalized method of moments estimator used to estimate dynamic panel data models. It was first proposed by Manuel Arellano and Stephen Bond in 1991 to solve the endogeneity,[1] heteroscedasticity and serial correlation problems in static panel data problem. The GMM-SYS estimator is a system that contains both the levels and the first difference equations. It provides an alternative to the standard first difference GMM estimator.\n\n## Qualitative description\n\nUnlike static panel data models, dynamic panel data models include lagged levels of the dependent variable as regressors. Including a lagged dependent variable as a regressor violates strict exogeneity, because the lagged dependent variable is necessarily correlated with the idiosyncratic error.\n\nWhen the strict exogeneity assumption is violated, commonly used static panel data techniques such as fixed effects estimators are inconsistent, because these estimators require strict exogeneity.\n\nAnderson and Hsiao (1981) first proposed a solution by utilising instrumental variables (IV) estimation.[2] However, the Anderson\u2013Hsiao estimator is asymptotically inefficient, as its asymptotic variance is higher than the Arellano\u2013Bond estimator, which uses a similar set of instruments, but uses generalized method of moments estimation rather than instrumental variables estimation.\n\nIn the Arellano\u2013Bond method, first difference of the regression equation are taken to eliminate the fixed effects. Then, deeper lags of the dependent variable are used as instruments for differenced lags of the dependent variable (which are endogenous).\n\nIn traditional panel data techniques, adding deeper lags of the dependent variable reduces the number of observations available. For example, if observations are available at T time periods, then after first differencing, only T-1 lags are usable. Then, if K lags of the dependent variable are used as instruments, only T-K-1 observations are usable in the regression. This creates a trade-off: adding more lags provides more instruments, but reduces the sample size. The Arellano\u2013Bond method circumvents this problem.\n\n## Formal description\n\nConsider the static linear unobserved effects model for ${\\displaystyle N}$ observations and ${\\displaystyle T}$ time periods:\n\n${\\displaystyle y_{it}=X_{it}\\mathbf {\\beta } +\\alpha _{i}+u_{it}}$ for ${\\displaystyle t=1,\\ldots ,T}$ and ${\\displaystyle i=1,\\ldots ,N}$\n\nwhere ${\\displaystyle y_{it}}$ is the dependent variable observed for individual ${\\displaystyle i}$ at time ${\\displaystyle t,}$ ${\\displaystyle X_{it}}$ is the time-variant ${\\displaystyle 1\\times k}$ regressor matrix, ${\\displaystyle \\alpha _{i}}$ is the unobserved time-invariant individual effect and ${\\displaystyle u_{it}}$ is the error term. Unlike ${\\displaystyle X_{it}}$, ${\\displaystyle \\alpha _{i}}$ cannot be observed by the econometrician. Common examples for time-invariant effects ${\\displaystyle \\alpha _{i}}$ are innate ability for individuals or historical and institutional factors for countries.\n\nUnlike a static panel data model, a dynamic panel model also contains lags of the dependent variable as regressors, accounting for concepts such as momentum and inertia. In addition to the regressors outlined above, consider a case where one lag of the dependent variable is included as a regressor, ${\\displaystyle y_{it-1}}$.\n\n${\\displaystyle y_{it}=X_{it}\\mathbf {\\beta } +\\rho y_{it-1}+\\alpha _{i}+u_{it}{\\text{ for }}t=1,\\ldots ,T{\\text{ and }}i=1,\\ldots ,N}$\n\nTaking the first difference of this equation to eliminate the fixed effect,\n\n${\\displaystyle \\Delta y_{it}=y_{it}-y_{it-1}=\\Delta X_{it}\\beta +\\rho \\Delta \\,y_{it-1}+\\Delta u_{it}{\\text{ for }}t=1,\\ldots ,T{\\text{ and }}i=1,\\ldots ,N.}$\n\nThis equation can be re-written as,\n\n${\\displaystyle \\Delta y=\\Delta R\\pi +\\Delta u.}$\n\nApplying the formula for the Efficient Generalized Method of Moments Estimator, which is,\n\n${\\displaystyle \\pi _{\\text{EGMM}}=[\\Delta R'Z(Z'\\Omega Z)^{-1}Z'\\,\\Delta R]^{-1}\\,\\Delta R'Z(Z'\\Omega Z)^{-1}Z'\\Delta y}$\n\nwhere ${\\displaystyle Z}$ is the instrument matrix for ${\\displaystyle \\Delta R}$.\n\nThe matrix ${\\displaystyle \\Omega }$ can be calculated from the variance of the error terms, ${\\displaystyle u_{it}}$ for the one-step Arellano\u2013Bond estimator or using the residual vectors of the one-step Arellano\u2013Bond estimator for the two-step Arellano\u2013Bond estimator, which is consistent and asymptotically efficient in the presence of heteroskedasticity.\n\n## Instrument matrix\n\nThe original Anderson and Hsiao (1981) IV estimator uses the following moment conditions:\n\n${\\displaystyle E(y_{it-I}\\,\\Delta u_{it})=0{\\text{ with }}I\\geq 2{\\text{ for each }}t\\geq 3.}$\n\nUsing the single instrument ${\\displaystyle y_{it-2}}$, these moment conditions form the basis for the instrument matrix ${\\displaystyle Z_{di}}$:\n\n${\\displaystyle Z_{di}={\\begin{bmatrix}NA&(t=2)\\\\y_{i1}&(t=3)\\\\y_{i2}&(t=4)\\\\\\vdots &\\vdots \\\\y_{T-2}&(t=T)\\end{bmatrix}}}$\n\nNote: The first possible observation is t = 2 due to the first difference transformation\n\nThe instrument ${\\displaystyle y_{it-2}}$ enters as a single column. Since ${\\displaystyle y_{it-2}}$ is unavailable at ${\\displaystyle t=2}$, all observations from ${\\displaystyle t=2}$ must be dropped.\n\nUsing an additional instrument ${\\displaystyle y_{it-3}}$ would mean adding an additional column to ${\\displaystyle Z_{di}}$. Thus, all observations from ${\\displaystyle t=3}$ would have to be dropped.\n\nWhile adding additional instruments increases the efficiency of the IV estimator, the smaller sample size decreases efficiency. This is the efficiency - sample size trade-off.\n\nThe Arellano\u2013Bond estimator uses the following moment conditions\n\n${\\displaystyle E(y_{it-I}\\,\\Delta u_{it})=0{\\text{ for }}t\\geq 3,\\,I\\geq 2.}$\n\nUsing these moment conditions, the instrument matrix ${\\displaystyle Z_{di}}$ now becomes:\n\n${\\displaystyle Z_{di}={\\begin{bmatrix}y_{i1}&0&0&0&0&0&\\cdots \\\\0&y_{i2}&y_{i1}&0&0&0&\\cdots \\\\0&0&0&y_{i3}&y_{i2}&y_{i1}&\\cdots \\\\\\vdots &\\vdots &\\vdots &\\vdots &\\vdots &\\vdots &\\ddots \\end{bmatrix}}}$\n\nNote that the number of moments is increasing in the time period: this is how the efficiency - sample size tradeoff is avoided. Time periods further in the future have more lags available to use as instruments.\n\nThen if one defines:\n\n${\\displaystyle \\Delta u_{i}={\\begin{bmatrix}\\Delta u_{i3}\\\\\\Delta u_{i4}\\\\\\Delta u_{i5}\\\\\\vdots \\end{bmatrix}}}$\n\nThe moment conditions can be summarized as:\n\n${\\displaystyle E(Z_{di}^{T}\\,\\Delta u_{i})=0}$\n\nThese moment conditions are only valid when the error term ${\\displaystyle u_{it}}$ has no serial correlation. If serial correlation is present, then the Arellano\u2013Bond estimator can still be used under some circumstances, but deeper lags will be required. For example, if the error term ${\\displaystyle u_{it}}$ is correlated with all terms ${\\displaystyle u_{it-s}}$ for s${\\displaystyle \\leq }$S (as would be the case if ${\\displaystyle u_{it}}$ were a MA(S) process), it would be necessary to use only lags of ${\\displaystyle y_{it}}$ of depth S + 1 or greater as instruments.\n\n## Systems GMM\n\nWhen the variance of the fixed effect term across individual observations is high, or when the stochastic process ${\\displaystyle y_{it}}$ is close to being a random walk, then the Arellano\u2013Bond estimator may perform very poorly in finite samples. This is because the lagged dependent variables will be weak instruments in these circumstances.\n\nBlundell and Bond (1998) derived a condition under which it is possible to use an additional set of moment conditions.[3] These additional moment conditions can be used to improve the small sample performance of the Arellano\u2013Bond estimator. Specifically, they advocated using the moment conditions:\n\n${\\displaystyle \\operatorname {E} (\\Delta y_{it-1}(\\alpha _{i}+u_{it}))=0{\\text{ for }}t\\geq 3}$\n\nThese additional moment conditions are valid under conditions provided in their paper. In this case, the full set of moment conditions can be written:\n\n${\\displaystyle \\operatorname {E} (Z_{SYS,i}^{T}P_{i})=0}$\n\nwhere\n\n${\\displaystyle P_{i}={\\begin{pmatrix}\\Delta u_{i}\\\\u_{i3}\\\\u_{i4}\\\\u_{i5}\\\\\\vdots \\end{pmatrix}}}$\n\nand\n\n${\\displaystyle Z_{SYS,i}={\\begin{pmatrix}Z_{di}&0&0&0\\\\0&\\Delta y_{i2}&0&0\\\\0&0&\\Delta y_{i3}&0\\\\0&0&0&\\ddots \\end{pmatrix}}.}$\n\nThis method is known as systems GMM.\n\n## Implementations in statistics packages\n\n\u2022 R: the Arellano\u2013Bond estimator is available as part of the plm package.[4][5][6]\n\u2022 Stata: the commands xtabond and xtabond2 return Arellano\u2013Bond estimators.[7][8]","date":"2020-02-18 04:43:18","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 51, \"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\": 0.621512234210968, \"perplexity\": 1075.4087193759874}, \"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-2020-10\/segments\/1581875143505.60\/warc\/CC-MAIN-20200218025323-20200218055323-00098.warc.gz\"}"} | null | null |
The consequences of low morale include low productivity, co-worker sabotage, and the suffocation of a charitable mission. Of all the downside risks facing your nonprofit, this is one that demands immediate and thoughtful intervention. The first step is to stop blaming contextual issues over which you have little influence. Accept responsibility for making workplace morale the strongest it can be—your mission deserves no less.
Share information widely and generously. Err on the side of sharing information that employees may "want to know," rather than hoarding information because you can. Regardless of the size of the organization, it is crucial for every employee and volunteer to be on the same page. No one likes to be the odd man or woman out.
Is your supervisor generous with sharing information and insights you need to succeed in your job? If "no," provide an example of the types of information you want to know that hasn't been made available to you.
How would you rate the sense of mutual trust between you and your supervisor? Give your supervisor a "10" if you believe that he or she trusts you with important assignments directly related to our mission. Assign a lower number if mutual trust isn't what it could and should be.
Invite employees to openly share ideas for improving workload distribution, reducing the fear of job loss, and recognizing employee achievements. Don't solicit ideas unless you're willing to implement at least one-third or more of the recommendations presented. Ask those offering suggestions to help design implementation (and evaluation) plans and timetables and offer sincere thanks to all who participate.
Insist on candor in the workplace. In Beyond Management, author Mark Addleson discusses the need for candid, "adult conversations" between knowledge workers in a 21st century workplace. When there has been a miscommunication or misunderstanding between staff members in your nonprofit, insist on a face to face conversation. Model that approach every day, even when it is easier to dash off an email, leave a voice mail, or send a text.
Melanie Lockwood Herman is Executive Director of the Nonprofit Risk Management Center. She welcomes your ideas about human resources risk, and questions about NRMC resources at Melanie@nonprofitrisk.org or 703.777.3504. | {
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ACCEPTED
#### According to
The Catalogue of Life, 3rd January 2011
#### Published in
Lichenotheca Veneta no. 75 (1869)
#### Original name
Lichen retiger Bory
### Remarks
null | {
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using System;
using System.Collections.Generic;
using Microsoft.Xna.Framework;
using Microsoft.Xna.Framework.Audio;
using Microsoft.Xna.Framework.GamerServices;
using Microsoft.Xna.Framework.Graphics;
using Microsoft.Xna.Framework.Input;
using Microsoft.Xna.Framework.Storage;
using Microsoft.Xna.Framework.Content;
namespace XNA3
{
/// <summary>
/// Handle the audio in the game
/// </summary>
public class AudioComponent : Microsoft.Xna.Framework.GameComponent
{
private AudioEngine audioEngine;
private WaveBank waveBank;
public SoundBank soundBank;
public AudioComponent(Game game)
: base(game)
{
// TODO: Construct any child components here
}
/// <summary>
/// Allows the game component to perform any initialization it needs to before starting
/// to run. This is where it can query for any required services and load content.
/// </summary>
public override void Initialize()
{
// Initialize sound engine
audioEngine = new AudioEngine("Content\\audio.xgs");
waveBank = new WaveBank(audioEngine, "Content\\Wave Bank.xwb");
//if (waveBank != null)
//{
soundBank = new SoundBank(audioEngine, "Content\\Sound Bank.xsb");
//}
base.Initialize();
}
/// <summary>
/// Allows the game component to update itself.
/// </summary>
/// <param name="gameTime">Provides a snapshot of timing values.</param>
public override void Update(GameTime gameTime)
{
audioEngine.Update();
base.Update(gameTime);
}
/// <summary>
/// Play a cue
/// </summary>
/// <param name="cue">cue to be played</param>
public void PlayCue(string cue)
{
soundBank.PlayCue(cue);
}
}
} | {
"redpajama_set_name": "RedPajamaGithub"
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You Are Here: DFG - Graduiertenkolleg Anlaufmanagement
InnoMetrics
OIA 2020
Study in cooperation with the VDI
You Are Here:DFG - Graduiertenkolleg Anlaufmanagement
Lehrstuhl für Betriebswirtschaftslehre, insbesondere Technologie- und Innovationsmanagement
Funding Body:
Co-Investigator:
Partner Organisation:
WZLforum gGmbH; Karrierepool WZL Aachen GmbH and Center for Doctoral Studies (CDS) RWTH Aachen
Production ramp-up is the period between completion of the development process (including product approval) and full capacity utilization. The ramp-up of production for businesses in the automotive supplier industry takes on a key role in the life cycle of products: in this phase, the prototype of the design stage is further developed and prepared for serial production. Currently, the ramp-up process is not adequately regulated; the realization of products under development is fraught with formidable difficulties. Therefore, the ramp-up represents a deciding factor which determines the success or failure of a particular product. The decision-making skills of a manager are pivotal to a succesful ramp-up. The complexities of decision-making problems during production ramp-up involve: dynamics, iInterdependence of a great number of design factors and Interdisciplinarity. The problem of complexity in the decision making process cannot be adequately solved through a centralized approach: in an inductive and decentralized process, decisions on the various sub-processes are to be optimized while systems interdependencies are taken into account. In this way, the ramp-up process can be monitored, improved and led to success. It is a goal of the research school to train young researchers in dealing with complex decision-making situations during production ramp-up. Our graduate students shall be provided with a better understanding of the basic processes which makes it possible for them to develop solution strategies in both academic and industrial contexts. A decision model will be developed which takes into account the systems behavior of the ramp-up process in relation to all factors essential to decision making. Further, a set of criteria will be developed which will help Systematically to make single decisions which are part of the overall process. The decision model will allow managers to anticipate problems in the ramp-up process on a meta-level and enable them proactively to develop solution strategies. The graduate research school is characterized by its interdisciplinary approach to ramp-up management research. Special emphasis is placed on finding a suitable research methodology which as a standardized frame of reference must be considered a prerequisite for scientific progress. | {
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Warnau er en by og kommune i det nordlige Tyskland, beliggende i Amt Preetz-Land i den sydvestlige del af Kreis Plön. Kreis Plön ligger i den østlige/centrale del af delstaten Slesvig-Holsten.
Geografi
Warnau er beliggende ved Bundesstraße 404 mellem Kiel og Bad Segeberg. Byen ligger ca. 9 km øat for Bordesholm, omkring 17 km syd for Kiel. I kommunen ligger Hochfelder See som er en del af Naturschutzgebiet Lütjensee - Hochfelder See südöstlich Gut Bothkamp.
Fra 1911 til 1961 havde Warnau jernbanestation på Kleinbahn Kiel–Segeberg.
Eksterne kilder/henvisninger
Kommunens websted.
Statistikamt Nord – Bevölkerung der Gemeinden in Schleswig-Holstein 4. Quartal 2014 (XLSX-Data) (Fortschreibung auf Basis des Zensus 2011)
Kommuner i Slesvig-Holsten
Byer i Slesvig-Holsten
Byer i Kreis Plön | {
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} | 1,056 |
{"url":"http:\/\/www.mathnet.ru\/php\/archive.phtml?wshow=paper&jrnid=mzm&paperid=8073&option_lang=eng","text":"Matematicheskie Zametki\n RUS\u00a0 ENG JOURNALS \u00a0 PEOPLE \u00a0 ORGANISATIONS \u00a0 CONFERENCES \u00a0 SEMINARS \u00a0 VIDEO LIBRARY \u00a0 PACKAGE AMSBIB\n General information Latest issue Forthcoming papers Archive Impact factor Subscription Guidelines for authors License agreement Submit a manuscript Search papers Search references RSS Latest issue Current issues Archive issues What is RSS\n\n Mat. Zametki: Year: Volume: Issue: Page: Find\n\n Mat. Zametki, 1977, Volume\u00a022, Issue\u00a04, Pages\u00a0517\u2013523 (Mi mz8073)\n\nAbsence of localization of the Laplace series on the sphere for functions of the Nikol'skii class $H_1^1(S^2)$\n\nA.\u00a0K.\u00a0Pulatov\n\nM.\u00a0V.\u00a0Lomonosov Moscow State University\n\nAbstract: In this article a\u00a0function is constructed belonging to the class $H_1^1(S^2)$ and having a\u00a0singularity at a\u00a0definite point on the sphere, as a\u00a0consequence of which localization fails for the Laplace series of this function at the diametrically opposite point. The constructed example shows that the sufficient condition of localization in $H_p^a$ of the spectral expansions in the class of all elliptic differential operators on an $n$-dimensional paracompact manifold cannot be improved (see\u00a0[1]).\n\nFull text: PDF file (403\u00a0kB)\n\nEnglish version:\nMathematical Notes, 1977, 22:4, 779\u2013783\n\nBibliographic databases:\n\nUDC: 517.4\n\nCitation: A.\u00a0K.\u00a0Pulatov, \u201cAbsence of localization of the Laplace series on the sphere for functions of the Nikol'skii class $H_1^1(S^2)$\u201d, Mat. Zametki, 22:4 (1977), 517\u2013523; Math. Notes, 22:4 (1977), 779\u2013783\n\nCitation in format AMSBIB\n\\Bibitem{Pul77}\n\\by A.~K.~Pulatov\n\\paper Absence of localization of the Laplace series on the sphere for functions of the Nikol'skii class $H_1^1(S^2)$\n\\jour Mat. Zametki\n\\yr 1977\n\\vol 22\n\\issue 4\n\\pages 517--523\n\\mathnet{http:\/\/mi.mathnet.ru\/mz8073}\n\\mathscinet{http:\/\/www.ams.org\/mathscinet-getitem?mr=499935}\n\\zmath{https:\/\/zbmath.org\/?q=an:0363.42008}\n\\transl\n\\jour Math. Notes\n\\yr 1977\n\\vol 22\n\\issue 4\n\\pages 779--783\n\\crossref{https:\/\/doi.org\/10.1007\/BF01146423}","date":"2021-10-28 14:28:46","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\": 0.37715962529182434, \"perplexity\": 6252.9082756813405}, \"config\": {\"markdown_headings\": true, \"markdown_code\": false, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-43\/segments\/1634323588341.58\/warc\/CC-MAIN-20211028131628-20211028161628-00516.warc.gz\"}"} | null | null |
Manac est une entreprise canadienne concentrant ses activités dans la conception et fabrication de semi-remorques.
Manac est le plus important fabricant au Canada et l'un des plus importants fabricants de remorques spécialisées en Amérique du Nordet occupe le rang des fabricants de semi-remorques en Amérique du Nord.
Manac se distingue pour sa spécialisation de divers produits et en offre plus de 125 sous les marques de commerce Manac, CPS, Peerless, Darkwing, UltraPlate, Ultravan, Liddell Canada, Alutrec et Cobra. L'entreprise a été fondée en raison d'approvisionnement pour Canam Steel Works (maintenant Groupe Canam), une compagnie sous opération du fondateur, Marcel Dutil . Son siège social est situé à Saint-Georges, Québec dans la région de la Beauce, et possède des usines à Laurier-Station, Québec, Val-des-Sources, Québec, Penticton, Colombie Britannique et Oran, Missouri.
Historique
En 1966 la compagnie est fondée par Marcel Dutil. Il doit répondre aux demandes croissantes de remorques pour livrer les produits d'aciers de Canam Steel Works qui était situé à Saint-Gédéon-de-Beauce. Le délai de livraison pour les remorques en 1966 était de plus de 6 mois, alors il décida de fabriquer des remorques dans un garage attenant à l'usine de Saint-Gédéon et fera ensuite les touches finales des unités dans sa grange arrière familiale à Saint-Georges. Durant la production d'unités pour Canam plusieurs petits transporteurs de la région commencent à prendre note de la rapidité de livraisons des semi-remorques Manac et commencent à enregistrer des commandes. La production de Manac durant ces premières années fut seulement des remorques à plateau et fera que 11 unités durant la première année.
En 1972, Manac fait l'acquisition des Aciers Canam. Avec cette acquisition le Groupe Canam Manac voit le jour. Manac aura son tout premier bureau de ventes, pièces et service à Boucherville en 1974, à côté de l'usine de tablier métallique de Canam. Les premiers fourgons en plastiques renforcés de fibres de verre Manac voient le jour en 1977 et en 1982, le premier fourgon en acier galvanisé (galvanisation) est née dans l'entreprise.
Manac est derrière la concurrence et commence la production de ses premiers fourgons en aluminium en 1986. Trois ans plus tard, l'entreprise commence sa réputation spécialisé avec la lancée de deux produits innovateurs : des plates-formes à rideaux latéraux et des semi-remorques à copeaux avec murs en acier.
Les premières plates-formes en acier et aluminium (Combo) sont fabriquées chez Manac en 1991 et en 1994, Manac voit une occasion d'expansion à Orangeville, Ontario. Manac acquiert une usine de pieds carrés pour sa production de fourgons à la norme.
En 2000, Manac lance un nouveau produit: la plate-forme 100 % aluminium et acquiert la division canadienne de Kalyn-Siebert, qui inclut une usine de pieds carrés à Trois-Rivières, Québec ainsi que la marque de commerce Fabrex. Un nouveau bureau de vente et centre de service ouvre ses portes à Mississauga, Ontario (maintenant à Etobicoke, Ontario).
En 2002 Manac poursuit son expansion dans le marché américain avec l'achat des actifs de CPS trailers Inc. à Oran, Missouri.
Le Groupe Canam Manac se divise en 2004 en raison de la décision de Canam voulant se concentrer sur la fabrication de composantes. Manac voit le jour en tant qu'entreprise privée.
L'ouverture de sa deuxième usine américaine voit le jour à Kennett en 2007 et, en 2009, à la suite de la crise économique mondiale des années 2008 et suivantes Manac fait l'acquisition de la propriété intellectuelle et des actifs de Trailmobile Canada, puis en 2011 des fardiers Liddell Canada.
Manac est le seul fabricant de semi-remorques en Amérique du Nord qui réussit en 2012 tous les essais de collision à 50 km/h dans un pare-chocs réalisés par l'Insurance Institute for Highway Safety (IIHS) des États-Unis. En 2013,Manac devient une entreprise publique et ses actions se transigent à la Bourse de Toronto (TSX :MA).
Manac voit en 2014 une occasion dans l'ouest canadien et annonce l'acquisition de Peerless Limited de McCoy Corporation, un fabricant de semi-remorques spécialisées. En 2015, l'entreprise quitte la Bourse de Toronto et redevient privée. En 2016, elle fête son cinquantième anniversaire.
Le 31 août 2018, Manac fait l'acquisition d'Alutrec Inc., fabricant spécialisé en conception et production de semi-remorques en aluminium de Laurier-Station, Québec. Alutrec offre une gamme de produits dans le segment des plateformes en aluminium qui est complémentaire à celle qu'offre Manac.
En novembre 2019, Manac complète l'acquisition de Cobra Trailers, une filiale de Rush Truck Centres of Canada spécialisée dans la conception et la fabrication de bennes basculantes en aluminium..
Marques de produits
CPS
Les semi-remorques CPS fut acquis par Manac en 2002. La marque de commerce CPS Dump Trailers fut lancée dans les 1980 dans le but d'aider l'industrie de la construction et d'agriculture avec leur besoin de semi-remorques. Manac utilise la marque de commerce CPS seulement aux États-Unis et fabrique les produits dans l'usine d'Oran, Missouri. La marque CPS on comme produits les suivants:
Bennes basculantes de recyclage
Bennes basculantes en demi-cercle
Bennes basculantes légère
Remorques à déchargement Central
Remorques légères à déchargement central
Semi-remorques à trémie
Lidell Canada
En 2011 Manac fait l'acquisition de la propriété intellectuelle des fardiers Liddell Canada qui était propriété des Remorques Nordic. Liddell Canada est une division de fardiers produite dans l'usine de Saint-Georges, Québec. Liddell Canada se différencie des autres fardiers dans le cas que les produits Liddell sont de hautes gammes et visés pour les fardiers spécialisés.
Peerless
En 2014, Manac acquiert Peerless Limited de Penticton en Colombie-Britannique. Peerless concoit et fabrique des remorques et châssis hautement spécialisés destinés aux industries du pétrole, gaz, exploitation minière, forestière et de construction. La compagnie Peerless utilise la marque Scona aux États-Unis.
Trailmobile
En 2009, Manac fait l'acquisition de la propriété intellectuelle de Trailmobile Canada. Trailmobile Trailer LLC était originellement un géant Américain dans la production de semi-remorques pour plus de 100 ans qui on fait faillite en 2001 mais leur opérations Canadiennes se sont poursuivies durant 8 ans. Trailmobile était une compagnie américaine pionnière dans les semi-remorques modernes, étant responsable de plus de durant la Deuxième Guerre mondiale. Manac utilise la marque Trailmobile pour vendre ses fourgons dans le marché américain étant donné que la marque Trailmobile est un nom beaucoup plus reconnu aux États-Unis.
Notes et références
Entreprise de transport ayant son siège au Québec
Entreprise de fabrication ayant son siège au Québec
Constructeur de camions | {
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Q: Problem installing 12.04 to dual-boot with 13.10 I have a system currently with Ubuntu 13.10, setup to boot to commandline.
We want to also have Ubuntu 12.04 installed as dual-boot.
I downloaded the ISO, burned to USB and tried installing. But the installation GUI won't start, see picture attached. The system isn't hang, you can still type on the commandline but it doesn't do anything.
I tried using 13.10 ISO and followed the same steps and was able to get to the install-GUI without issue.
Any idea?
| {
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Chen Xiaohong (born 15 May 1963) is a Chinese management science and engineering management specialist, an academician of the Chinese Academy of Engineering, and currently party secretary of Hunan University of Technology and Commerce.
Biography
Chen was born in Changsha, Hunan, on 15 May 1963, while her ancestral home in Yongxin County, Jiangxi. She received his bachelor's degree and master's degree from Central South University of Technology (now Central South University) in 1983 and 1986, respectively. She received her doctor's degree from Tokyo Institute of Technology in 1999.
She joined the Communist Party in May 1983. Chen taught at Central South University of Technology since 1986, what she was promoted to associate professor in December 1991 and to full professor in September 1994. She also served as dean of Business Administration School from June 1999 to April 2002, and dean of Business School from April 2002 to January 2009. Chen became president of Hunan University of Technology and Commerce, in September 2014, and then party secretary, the top political position in the university, beginning in April 2021. In March 2018, she became a member of the 13th National Committee of the Chinese People's Political Consultative Conference.
Honours and awards
2005 State Science and Technology Progress Award (Second Class)
27 November 2017 Member of the Chinese Academy of Engineering (CAE)
References
1963 births
Living people
People from Changsha
Engineers from Hunan
Central South University alumni
Tokyo Institute of Technology alumni
Academic staff of the Central South University
Members of the Chinese Academy of Engineering
Members of the 13th Chinese People's Political Consultative Conference | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 2,935 |
John Taylor Babbitt Foundation Makes Generous Donation to the Sudden Cardiac Arrest Foundation
Submitted by SCAFoundation on Wed, 10/31/2018 - 12:00am
PITTSBURGH, PA--The John Taylor Babbitt Foundation donated $5,000 to the Sudden Cardiac Arrest Foundation at its annual Board of Directors meeting on Saturday.
The JTB Foundation was established in memory of John Babbitt, who was 16-years-old when he collapsed and died from an undiagnosed heart condition while playing basketball in his Youth Ministry league in 2006. The JTB Foundation mission is to increase awareness about the severity and prevalence of hypertrophic cardiomyopathy within the young adult community.
"The Sudden Cardiac Arrest Foundation is leading our nation in educating the public about sudden cardiac arrest," said JoAnne Taylor Babbitt in presenting the donation. "The John Taylor Babbitt Foundation is pleased to support their vision to increase awareness about the importance of immediate bystander intervention to reduce the number of sudden cardiac deaths."
"We are very grateful to the John Taylor Babbitt Foundation for their generosity," said Henry Jampel, MD, MHS, chairman of the Sudden Cardiac Arrest Foundation Board of Directors. "We appreciate their support for our mission to raise awareness about sudden cardiac arrest and help save lives."
Ms. Babbitt is vice-president of the JTB Foundation and a member of the SCA Foundation Board of Directors. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 6,516 |
\subsection{The Used Methods and Tools}
\subsubsection{The Event-B Method}
Within the Event-B framework, asynchronous systems may be developed and structured using \textit{abstract systems} \cite{Abr96a,AbrialMussat98}.
\textit{Abstract systems} are the basic structures of the so-called \textit{event-driven} B.
An \textit{abstract system} \cite{Abr96a,AbrialMussat98} describes a mathematical model of a system behaviour\footnote{A system behaviour is the set of its possible transitions from state to state beginning from an initial state}.
An abstract system is made mainly of a state description (constants, properties, variables and invariant) and several \textit{event} descriptions.
Abstract systems are comparable to Action Systems \cite{BaKu83}; they describe a nondeterministic evolution of a system through guarded actions.
Dynamic constraints can be expressed within abstract systems to specify various liveness properties \cite{AbrialMussat98,ZB02-Cansell}.
The state of an abstract system is described by variables and constants linked by an invariant. Abstract systems may be refined to concrete ones like abstract machines \cite{ZB02-Cansell,AbrialCansellMery03}.
An event of a B abstract system is considered as the observation of one transition of the system. Events are spontaneous and show the way a system evolves.
An event $e$ is modelled as a \textit{guarded substitution}: $e \defs eG \Longrightarrow eB$ where $eG$ is the event \textit{guard} and $eB$ the event \textit{body} or \textit{action}.
The symbol $\Longrightarrow$ denotes the guard.\\
An event may occur or may be observed only when its guard holds.
The action of an event describes, with generalised substitutions, how the system state evolves when this event occurs.
Several events may have their guards hold simultaneously; in this case, only one of them occurs. The system makes internally a nondeterministic choice. If no guard is true the abstract system is blocking (deadlock).
\begin{figure}[htp]
{\small
\begin{center}
\begin{multicols}{2}
\noindent
\begin{boxedminipage}{5.1cm}
\begin{tabbing}
\hspace{0.1cm}\=\hspace{0.4cm}\=\hspace{0.4cm}\=\hspace{0.4cm}\kill
\>\texttt{eventName} $\defs$ \textsf{~~~~~~~} \\
\>\>\textsc{select}~~~
$P_{(gcv)}$ \\
\>\>\textsc{then} ~~~~
$GS_{(gcv)}$\\
\>\>\textsc{end}
\end{tabbing}
\end{boxedminipage}
\centerline{(SELECT Form)}
\noindent
\begin{boxedminipage}{4.1cm}
\begin{tabbing}
\hspace{0.1cm}\=\hspace{0.4cm}\=\hspace{0.4cm}\=\hspace{0.4cm}\kill
\>\texttt{eventName} $\defs$ \textsf{~~~~~~ } \\
\>\>\textsc{any} $bv$ \textsc{where}
$P_{(bv, gcv)}$ \\
\>\>\textsc{then} ~~~~
$GS_{(bv,gcv)}$\\
\>\>\textsc{end}
\end{tabbing}
\end{boxedminipage}
\centerline{(ANY Form)}
\end{multicols}
\caption{General forms of B events}
\label{figure:eventshape}
\end{center}
}
\end{figure}
An event has one of the general forms (Fig. \ref{figure:eventshape}) where $gcv$ denotes the global constants and variables of the abstract system containing the event; $bv$ denotes the bound variables (variables bound to \textsc{any}). $P_{(bv, gcv)}$ denotes a predicate $P$ expressed with the variables $bv$ and $gcv$; in the same way $GS_{(bv,gcv)}$ is a generalised substitution $S$ which models the event action using the variables $bv$ and $gcv$.
The \textsc{select} form is a particular case of the \textsc{any} form. The guard of an event with the \textsc{select} form is $P_{(gcv)}$.
The guard of an event with the \textsc{any} form is $\exists(bv).P_{(bv,gcv)}$.
\subsubsection{The B4free and ProB Tools}
We use the theorem prover
\textsf{B4free}\footnote{{\small\url{www.B4free.fr}}}
and the \textsf{ProB}\footnote{{\small \url{www.stups.uni-duesseldorf.de/ProB/}}} model checker.
\subsubsection*{Overview of B4free}
\textsf{B4free} is one of the public domain theorem prover dedicated to Event-B. The prover originated from the industrial commercialised tool called AtelierB. It was developed together with an emacs front-end. It does not have the other modules such as code generator or document generator available in the AtelierB tool.
However, the \textsf{B4free} tool is free and is convenient for experimentations with the B method: parsing and proving the obligation proofs related to the B method. A new public domain framework for B is now available: Rodin\footnote{\url{www.event-b.org/platform.html}}.
\subsubsection*{Overview of ProB}
\label{subsection:ovvProB}
\input{ovv_ProB}
\subsection{Modelling and Analysing the System}
We consider the four steps of the proposed method.
\subsubsection{Step 1: Building an Abstract Reference Model}
The Manet system is made of a set of nodes that communicate; they form a range; a node is identified as a process type.
\subsubsection*{Specifying a Node Process}
Each node has
an identifier, a location, an IP address, a connection relation that
indicates its neighbours, etc. Accordingly we have the $S_i$ part of
the node as a set of typed variables that denote the features.
A set of events ($E_i$) with the associated behaviours ($Evt_i$) define the process behaviours which lead the evolution of the system. As far as its behaviour is concerned, any node
may initiate a message for a given destination, send a message, receive
a message, forward a message, leave a net (a transmission range).
The behaviour described by these events is observed only when a net exists; that means the net
structuring events are related to those needed for the routing.
Also we deal with the creation of a network by nodes
which have a given range, other nodes may join or leave this
range. Therefore, we link the range of a node with a given abstract
network.
The formal specification of a MANET is then
a set of sequences of configurations of the considered nodes; that is
their state variables, resulting from the fusion of the node state variables;
the evolution is modelled through the enabling of
events which possibly modify the state space.
Concerning the interaction within the MANET system, we consider the
events of the nodes and also the common events related to the entire system network (including the ranges).
\subsubsection{Step 2: Deriving an Event-B Specification}
The derivation of a Event-B model from the abstract event-based model is quite straightforward. The state variables form the B invariant; the abstract events are translated as B events.
\subsubsection*{Resulting B specification of the MANET}
The specification of the structure of a MANET is achieved using a set of state variables
and an invariant that describes the nodes and their current
configurations:
\begin{center}
{\small
\begin{boxedminipage}{6cm}
\begin{tabbing}
\hspace{0.4cm}\=\hspace{0.4cm}\=\hspace{0.4cm}\=\hspace{0.4cm}\kill
\textbf{\textsc{system}} ${MANET}$\\
\textbf{\textsc{sets}} NODE, RANGE, MSG /* abstract sets */\\
\textbf{\textsc{variables}}\\
\> $nodes, ranges, messages, $ /* state variables*/ \\
\> $rangNodes, reqMsg, inReqMsg, \cdots$\\
\textsc{invariant} ~~ /* state space predicate */\\
\> $nodes \subseteq NODE$ $\land$ $ranges \subseteq RANGE$\\
$\land$ \> $messages \subseteq MSG$ \\
$\land$ \> $rangNodes \in ranges \rel nodes$\\
$\land$ \> $reqMsg \in nodes \rel messages $\\
$\land$ \> $inReqMsg \in nodes \rel messages$\\
$\land$ \> $inRspMsg \in nodes \rel messages$\\
$\land$ \> $waitReqMsg \in nodes \rel messages$\\
$\land$ \> $\cdots$\\
\textbf{\textsc{initialisation}}\\
\> $nodes,ranges, messages,rangNodes := \emptyset,\emptyset,\emptyset,\emptyset $ \\
$\|$\> $\cdots$
\end{tabbing}
\end{boxedminipage}
}
\end{center}
Event-B uses the set notation. The standard operators are written as usually ($\in,~\cup,~\cap,~\{\cdots\}$). The symbol $A \rel B$ denotes a relation between two sets.
The
behaviour of the system depends on the set of events that define the
nodes and the specific system events: the observation of a net creation
(\textit{newRange}); an
existing net may disappear if there is no more connected nodes (\textit{rmvRange}).
Other events considered for the network are the following;
\textit{joinRange}: a node joins a range;
\textit{leaveRange}: a node leaves a net range;
\textit{newNode}: a new node appears;
\textit{newMsg}: a node initiates a message.
At this stage, the behaviour part (the set of events) is specified using Event-B as follows:\\
\begin{center}
{\small
\begin{boxedminipage}{3cm}
\begin{tabbing}
\hspace{0.4cm}\=\hspace{0.4cm}\=\hspace{0.4cm}\=\hspace{0.4cm}\kill
\textbf{\textsc{system}} ${MANET}$ (continued)\\
\>$\cdots$ \\
\textbf{\textsc{events}}\\
\>\texttt{newNODE} $\defs~~~\cdots$ \\
;\>\texttt{newRANGE} $\defs~~~\cdots$ \\
;\>\texttt{joinRange} $\defs~~~\cdots$ \\
;\>\texttt{leaveRange} $\defs~~~\cdots$ \\
;\>\texttt{newMsg } $\defs~~~\cdots$
\end{tabbing}
\vspace{-0.3cm}
\textbf{\textsc{end}}
\end{boxedminipage}
}
\end{center}
The specification of the event \textit{joinRange} is depicted in
Fig. \ref{figure:eventJoinRange}. The other events are specified in
quite the same way.
\begin{figure}[htp]
{\small
\begin{center}
\begin{boxedminipage}{8cm}
\begin{tabbing}
\hspace{0.0cm}\=\hspace{0.4cm}\=\hspace{0.4cm}\=\hspace{0.4cm}\=\hspace{0.9cm}\=\hspace{0.9cm}\kill
\>\texttt{joinRange} $\defs$ ~~ /* a node joins a range */ \\
\>\textsc{ANY} $nd, rg$ \>\>\>\> \textsc{WHERE}\\
\>\> $nd \in nodes \land rg \in ranges $\\
\>\> $\land rg \in \dom (rangNodes)
\land nd \notin rangNodes[\{rg\}]$ \\
\> \textsc{THEN} ~~\\
\>\> $rangNodes := rangNodes \cup \{rg \mapsto nd\}$ \\
\> \textsc{END}
\end{tabbing}
\end{boxedminipage}
\end{center}
}
\caption{{\small Specification of the \texttt{joinRange} event}}
\label{figure:eventJoinRange}
\end{figure}
In the B language, $\dom$ denotes the domain of a relation; if $r$ is a relation and $e$ an element of its domain, $r[\{e\}]$ denotes the images of $e$ with $r$. $a \mapsto b$ denotes the couple $(a, b)$.
As far as the routing aspect is concerned we consider one of the widely studied
routing protocols of MANET: \textit{Ad-hoc On demand Distance Vector}
(AODV) \cite{Chlamtac2003}.
A part of the behaviour of our B specification is related to
the structuring and another part is about the routing protocol. Therefore we
complete the previous specification of the MANET system with
the events related to the routing protocol.
Within the AODV protocol, each node acts as a router, contributes to
construct routes and to forward messages to other nodes.
There are two phases of the protocol: route discovery and route
maintenance. Route discovery is achieved by exchanging Route Request (RREQ) and Route Response (RREP) messages.
The algorithm of the nodes is as follows: when a node desires to set up
a route to a destination node, it broadcasts a RREQ message to its
neighbours (the nodes in its range). The RREQ/RREP messages have the following main parameters: the source node Id, the destination node Id, the
number of hops.
When a node $nd$ receives a RREQ message,
\textit{i)} either $nd$ is itself a destination and $nd$ responds with
a RREP
or $nd$ is an active route to the searched destination node then
$nd$ responds with a route information using the RREP message;
\textit{ii)} otherwise $nd$ broadcasts the RREQ further with the hop
count of RREQ increased by 1.
The routing of messages is symmetric when a node receives a RREP message.
The Event B specification is completed with all the events related to
the routing protocol described above.
We give in the following (see Fig. \ref{figure:eventRREQ}) the specification of the \texttt{sndRREQ} event to illustrate
the specification principle. Here, any node ($sn$) may send a message ($msg$) that it
has already prepared ($msg \in reqMsg[\{sn\}]$) to all the nodes in its
range ($otherNodesInRange$). Exchanged messages are modelled using
abstract channels ($inRepMsg$,$repMsg$).\\
\vspace{-0.2cm}
\begin{figure}[htp]
{\small
\begin{center}
\begin{boxedminipage}{6cm}
\begin{tabbing}
\hspace{0.0cm}\=\hspace{0.4cm}\=\hspace{0.4cm}\=\hspace{0.4cm}\=\hspace{0.9cm}\=\hspace{0.9cm}\kill
\>\texttt{sndRREQ} $\defs$ ~~ /* route request from sn to dn */ \\
\>\textsc{ANY} $sn, msg$ \>\>\>\> \textsc{WHERE}\\
\>\> $sn \in nodes$ /* source */ \\
\>\> $\land msg \in MSG \land msg \in messages$ \\
\>\> $\land msg \in reqMsg[\{sn\}]$ /* a msg initiated by nd */ \\
\> \textsc{THEN} \\
\>\> \textsc{LET} $otherNodesInRange$ \\
\>\> \textsc{BE} $otherNodesInRange = \{ndi | ndi \in nodes$ \\
\>\> $\land ndi \neq sn \land rangNodes^{-1}(sn) = rangNodes^{-1}(ndi) \}$ \\
\>\> \textsc{IN}
\>\>$inReqMsg := $\\
\>\>\>\>$~ ~ ~ ~ ~ inReqMsg \cup (otherNodesInRange *\{msg\})$ \\
\>\>\> $\|$ \>$reqMsg := reqMsg - \{(sn \mapsto msg)\}$ \\
\>\> \textsc{END} \\
\> \textsc{END}
\end{tabbing}
\end{boxedminipage}
\end{center}
}
\caption{{\small Specification of the \texttt{sndRREQ} event}}
\label{figure:eventRREQ}
\end{figure}
The expression $otherNodesInRange *\{msg\}$ denotes the Cartesian product of the elements in $otherNodesInRange$ with the singleton $\{msg\}$.
The dynamic aspect of the system architecture is based on the
fact that the event guards depend on the variables $nodes, messages,
\cdots$ which themselves depend on the current event. That means in an event guard, we can consider any event from $nodes$ or any messages from $messages$, etc.
This is illustrated by
the non-deterministic form of the event specifications:\\
\centerline{\texttt{event} $\defs$ \textsc{ANY} $sn$ \textsc{WHERE} $sn
\in nodes$ \textsc{THEN} ... \textsc{END}}
The Event-B specification which is the specific model for the study of
the MANET is then an abstract system equipped with
all the events described before: structuring and routing events.\\
\subsubsection{Step 3: Analysis of the Specific Model}
A multi-facet analysis of the specific Event-B model of
the MANET system is performed. For this purpose two different tools are used but they cover different facets of the analysis:
B4free and ProB.
\subsubsection*{Consistency and Refinement of the System}
The previously described B abstract system is proved consistent using the
B4free tool. Then it is refined; more details are added to the state
space and the event specifications; for instance we consider the management of
the IP addresses of the nodes and exchanged messages.
Unlike in the abstract system where a packet destination is
nondeterministically selected, in the refinement the nodes and the
messages have IP addresses, therefore, the receiver node is checked
against the destination IP address.
The resulting refined system is also proved correct with respect to consistency using
the B4free tool. However to accomplish the proofs, we combine the use of
B4free and ProB. That is, when a proof
obligation is not discharged by B4free, we model-check the
specification and discover possible errors by displaying and analysing
the displayed error
state. Accordingly the feedback is propagated in the reference model and we iterate.
\subsubsection*{Liveness Properties Analysis}
Many properties of the MANET routing protocol are well-expressed using LTL (Linear Temporal Logic) formula which
is not supported by the B4free tool.
We express these liveness properties with the ProB LTL formalism.
Then we extend
the Event-B abstract system with these LTL properties; for example
\noindent
{\small \textbf{P$_1$}. \textsf{A route request is always followed by a response:}}\\
\centerline{\small $G(e(sndRREQ) \implies F(e(sndRREP)))$}
The resulting specification is model-checked using ProB.
After that we come to the conclusion that our model extended with the stated properties, is correct with respect to these properties.
\subsubsection{Step 4: Feedback to the Reference Model}
Using the multi-facet approach, with B4free and ProB
helps us to perform a complete analysis.
For illustration, in experiments with ProB, when a deadlock is
detected after the exploration of nodes and transitions that cover all the operations (the B events), the state corresponding to the deadlock is carefully analysed. In one case we discover that it corresponds to a situation (net partitioning) where there are nodes with
some packets to be transmitted but no node in the current net range. This
corresponds to a real-life situation which is due to the dynamic
aspect of the MANET and the mobility of nodes.
A feedback is then propagated first in the Event-B specification.
The model is corrected by strengthening the guard of message initiation by the hypothesis of
non-emptiness of the net range. Thus the analysis of the model runs without
errors\footnote{the experiment result tables, not displayed here, show 0
deadlocked states for hundreds of explored states and transitions.}.
\endinput
\section{Introduction}
\label{section:intro}
\input{./intro.tex}
\section{Decentralised Dynamic System}
\label{section:manet}
\input{./manet.tex}
\section{The Proposed Method}
\label{section:method}
\input{theMethod.tex}
\section{Application to the MANET System}
\label{section:appli2manet}
\input{appliMethodManet.tex}
\section{Discussion and Conclusion}
\label{section:conclusion}
\input{./conclu.tex}
\vspace{-0.2cm}
\small
\bibliographystyle{abbrv}
\input{main_ma_adhoc.bbl}
\end{document}
\section{Introduction}
This demo file is intended to serve as a ``starter file''
for IEEE conference papers produced under \LaTeX\ using
IEEEtran.cls version 1.7 and later.
I wish you the best of success.
\hfill mds
\hfill January 11, 2007
\subsection{Subsection Heading Here}
Subsection text here.
\subsubsection{Subsubsection Heading Here}
Subsubsection text here.
\section{Conclusion}
The conclusion goes here. this is more of the conclusion
\section*{Acknowledgment}
The authors would like to thank...
more thanks here
\subsection{Overview of Mobile Ad-hoc Network}
\input{ovv_manet.tex}
\subsection{Overview of the ProB Tool}
The ProB tool \cite{LeuschelButler:FME03,LeuschelTurner:ZB05} is an animator and a model checker for B specifications.
It supports automated consistency checking of B specifications (an abstract machine or a refinement with its state space, its initialisation and its operations). The consistency checking is performed on all the reachable states of the machine. ProB also provides a constraint-based checking; with this approach ProB does not explore the state space from the initialisation, it checks whether applying one of the operation can result in an invariant violation independently from the initialisation.
ProB provides functionalities to show graphical views of automata.
The functionalities of ProB are organised within three categories: \textit{Animation}, \textit{Verification} and \textit{Analysis}.
ProB tool is used in our study to help in discharging consistency
proof obligations
(invariant violation) and to check liveness properties.
\endinput
We list a few of them which are used in the experiments.\\
In the \textit{Verification} category, the functionalities include: \\
\noindent
{\small \textsf{Temporal Model Checking}}: starting from a set of
initialisation states (initial nodes), it systematically explores the
state space of the current B specification. \\
\noindent
{\small \textsf{LTL Model Checking}}: this functionality enables one to
check the specification against a given LTL property.\\
In the \textit{Analysis} category we consider: \\
\noindent
{\small \textsf{Compute Coverage}}: the state space (the nodes) and the
transitions of the current specification are checked, some statistics
are given on deadlocked states, live states\footnote{the already computed states.}, covered and uncovered operations.
\subsection{Overview of Mobile Ad-hoc Network (MANET)}
A Mobile Ad-hoc Network (MANET) system is a typical example of a dynamic decentralised system.
A Mobile Ad-hoc Network \cite{Chlamtac2003} is a network formed with wireless mobile
nodes (called ad-hoc nodes) which are the user equipments or devices.
A MANET has no dedicated network infrastructure, but each node serves as a part of the
network and acts a \textit{router} to forward messages or packets since
there is no router dedicated to that task.
A mobile ad-hoc network is formed only when a group of users
put together their resources to enable and perform communications; hence
a mobile ad-hoc network is dynamically created and may also disappear quickly.
In a MANET, the nodes communicate either by exchanging
directly or via intermediate nodes. Technically they use ISM
band\footnote{they are radio system frequencies initially dedicated to
industrial, scientific and medical usage.} and
more generally Wireless LAN technologies. Each node is equipped with one
or more radio interfaces with specific transmission features. The
\textit{transmission range} of a node is the transmission area accessible from
this node. All the nodes in this range are accessible directly (one
hop); they are called the neighbours. To address a known node which is not in its transmission range,
the sender node sends its packet to one of the neighbour nodes which is closer
to the destination node (according to the transmission ranges).
Each node may communicate directly or indirectly using relay nodes
(multi-hop), with other nodes that are outside the sender range.
\medskip
\textit{Dynamic Aspect.} One of the main features of a MANET is its
dynamic aspect: the structure or topology of the network is frequently
changing. A node may join or leave the net at any time, changing the net
topology. The structure or topology of the net is then highly dynamic.
\medskip
\textit{Mobility Aspect.} The ad-hoc nodes may move at any time and very frequently due to their mobile nature;
consequently this impacts not only on the net topology but also on its
quality; there may be route changes, information loss, partitions of the network into different networks, etc.
As far as routing is concerned, in classical infrastructure-based network,
there are one or several nodes called routers that are in charge of
routing packets between nodes. For this purpose the routers and the nodes are equipped with a routing table
where there is the information about how to join a given destination
node or a network identified with an Internet Address (IP address).
In the scope of MANET, efficient routing protocols development is a challenging concern.
A message or packet sent to a node reaches it unless the net is partitioned.
Concerning the time, it is assumed to be discrete and divided into frames.
A node has a set of neighbour nodes during a frame.
During a frame a node may be idle, it also may send messages, receive
messages, forward the received messages.
Before sending a message to a destination, a source node $sn$ which does
not have the destination node address, sends a route request to get this
destination address. The request travels through the net possibly with multi-hop
and reaches the destination which sends back its address. When the address is received
by $sn$ the latter can send its message to the right destination
address. \\
The study of MANET is an
active and challenging field as this type of network is rapidly
growing and supporting small and medium size applications
such as mobile services sharing, wireless peer-to-peer systems, etc.
We chose the field of MANET for this work because it is a challenging
field shared by the fields of computer networks and software
engineering.
From the software system point of view, the MANET system is a
typical decentralised, asynchronous system with dynamically evolving architecture.
Moreover, its properties (dynamicity, mobility, correctness, etc) need a
combined use of several verification techniques (namely a multifacet analysis).
\endinput
\subsection{Overview of the Method}
\begin{description}
\item[Step 1.]~To build an abstract formal model from the system at hand and
to state the desired global properties according to this formal
model; it is the \textit{reference model}; an abstract reference model may be event-based, state-based, process-based, algebraic, etc.
\item[Step 2.]~To systematically derive or translate from this reference model,
other formal models which are specific to various analysis techniques;
\item[Step 3.]~To perform analysis (verification of properties) with the specific
models or with their extensions, by adding specific properties to
the global ones;
\item [Step 4.]~To ensure the consistency between the reference model and the
specific ones by propagating the feedback from the specific
models study on the reference model and by updating consequently
the other specific models. Then, the analysis of each facet via a
specific model participates in the global system analysis.
\end{description}
\medskip
The first step on building a reference model needs methods that are appropriate to the system at hand. In the
current case of the decentralised system, we have a \textit{multi-process} system. We detail this step.
\subsection{On Building an Event-based Reference Model}
An event-based model is suitable for dynamic system.
The approach \cite{coloss:CA_ICFEM06} provides rigorous guidelines to help in discovering and expressing the desired
behaviours of a multiprocess system with dynamic architecture.
Our approach to build the reference model combines a process-oriented view (at low level, for elementary identified processes) and
an event-based one (at global level, for composing processes).
The method used to build the reference model is summarised as follows.
\begin{enumerate}
\item Structuring aspects: From the requirements, elementary \textit{types of processes} are identified to describe behaviours.
Several processes may have the same type.
Each identified \textit{type of process} $P_i$ that participates in
the global system model is specified by considering its space state $S_i$
and the events $E_i$ with their description $Evt_i$ that lead its
behaviour.
\centerline{$P_i \defs \langle S_i, E_i, Evt_i\rangle$}
The constituents $S_i$, $E_i$ and $Evt_i$ will be detailed latter on. To handle the dynamic architecture of global system, we impose that for each type of process, the events to \textit{join} and \textit{leave} the system be defined. Some events may be common to several processes; they handle
interaction and state sharing aspects.
\item Interaction aspects: Interaction involves communication. As far as communication is concerned we use
guarded events,
message-passing and ordering of event occurrences. The processes synchronise and communicate through
the enabling/disabling of the guards of their events. Therefore, if an event
is used to model a process which is waiting for a data, it may be blocked until the availability of the
data (enabling the event guard), which is the effect produced by
another process event. Consider for
example the case of processes exchanging messages, one process
waits for the message, hence there should be an event with a non-enabled guard, and another process sends the message via a behaviour of an event which guard is enabled.
Communications are modelled with abstract channels.
An abstract channel modelled as a set, is used to wait for a message
or to deposit it. Hence the interaction between the processes is
handled using common abstract channels. Therefore, the communication is
achieved in a completely decoupled way to favour dynamic structuring.
\item Composition of the processes:
All the described processes are (hierarchically) combined by a fusion operation ($\biguplus$) that merges
state spaces and the events of the processes into a single global system $S$.
$$ S \defs \biguplus_{i} P_i ~~\equiv~~ \biguplus_{i}~\langle S_i, E_i, Evt_i \rangle$$
According to the fusion operator, when the processes are merged, a set is introduced to identify the merged processes. Each feature that is modelled with a variable, results in a function from the set of process identifiers to a set of values (of the feature). The events of the processes are now defined by considering the elements of the identifier set (if the set is empty there is no more process).
\end{enumerate}
In the following we illustrate the four steps of the method.
| {
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{"url":"http:\/\/quoscav.dblbld.ru\/a-note-on-down-dating-the-cholesky-factorization-105.htm","text":"# A note on down dating the cholesky factorization\n\nIn linear algebra, the Cholesky decomposition or Cholesky factorization is a decomposition of a Hermitian, positive-definite matrix into the product of a lower triangular matrix and its conjugate transpose, which is useful e.g.\n\nfor efficient numerical solutions and Monte Carlo simulations.\n\nIf A, C are fixed, and B is variable but nice (low-rank), then you want what is called \"Cholesky update\".Lets say we have a block matrix $M =\\left( \\begin A & B\\ B^ & C \\end \\right)$ where M is positive definite. The matrix $M = LU$ can be decomposed in an algebraic manner into $L = \\begin A^ & 0 \\ B^ A^ & Q^ \\end$ where $\\begin Q = C - B^ A^ B \\end$ $*$ indicates transpose in this case Now lets say we have already carried out the cholesky decomposition for A, and C.(A, and C are also pos def) There is a formula for carrying out block Cholesky decomposition. So we have already calculated $A^$, and $C^$ (It is therefore straightforward to calculate the inverses $A^$, and $C^$ using forward substitution). The problem is indeed technical in its origin , but I'd hoped (perhaps naively) that the problem would also be of interest to other mathematicians.The Money Market Hedge: How It Works | Investopedia explains how to hedge foreign exchange risk using the money market, the financial market in which highly liquid and short-term instruments like Treasury ... Category: Forex Beginner Tags: money market hedge asked June 22, 2012. [VIDEO]Successful Forex Hedge Strategy that Makes Money - \u2026 v=SH8_WI5y0gc BY TIMON WELLER39 MIN164K VIEWS Nov 28, 2012 \u00b7 Successful Forex Hedge Strategy that Makes Money ... START A FOREX CTA\/CPO HEDGE FUND | SCG FUND SERVICES https:\/\/scgfundservices.com\/hedge-fund\/forex-cta-fund Forex or FX or retail off-exchange foreign currency transactions all refer to the same thing \u2013 trading foreign currencies for gain, usually in the spot market.Money Market Hedge In Forex - trading commodities how to forexoptionsmeal.info\/money-market-hedge-in-forex money market hedge in forex As the market is going down at the moment it be a waste to open the Buy position at the same rate as the Sell position. Money Market Hedge In Forex - trading platforms with \u2026 Before trading with real money I highly recommend ... Successful Forex Hedge Strategy That Makes Money - intra ...The Cholesky factorization of a stiffness matrix can be updated after modifying a local stiffness matrix which can be written as a sum of a few rank-one matrices.","date":"2018-07-20 22:08:49","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\": 0.2657286822795868, \"perplexity\": 1455.6697901849539}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2018-30\/segments\/1531676591837.34\/warc\/CC-MAIN-20180720213434-20180720233434-00205.warc.gz\"}"} | null | null |
When it comes to Tevaughn Campbell, Canada men's rugby sevens head coach Damian McGrath is totally honest: he's never coached an athlete like this before.
Sure, he's had lightning quick players.
But he's never had an athlete arrive in his squad with zero rugby background.
The official rosters provided to World Rugby offer a good reminder of that fact: players usually have their home club listed and in Campbell's case, yes, he's down as coming from the Montreal Alouettes.
The thing is, a year ago, Campbell didn't even know what rugby sevens was.
Credit Robin MacDowell, former national team player, now rugby sevens coach, with making the score.
Campbell, a native of Scarborough, Ont., played university football at the University of Regina and also ran track.
Before being drafted by the Saskatchewan Roughriders in 2015, Campbell shattered the 40-yard sprint record at the CFL draft combine.
It was during a gym session at the University of Regina that MacDowell first spotted him.
That was little over a year ago. MacDowell cornered Campbell and immediately showed him clips of American speedsters Perry Baker and Carlin Isles, both of whom are football converts. Then the coach made a bold call: let's get him to Vancouver to take in the 2017 Canada Sevens.
Campbell flew out, and was impressed by what he saw. McGrath moved quickly to invite him along for the Hong Kong Sevens and then the Singapore Sevens, which Canada went on to win, their first World Rugby tournament title.
Campbell returned to the CFL after that — he was traded to the Alouettes during the summer — but sevens never left his mind.
And now Campbell has four tries to his name, including a pair in his team's final match on Sunday at the USA Sevens in Las Vegas, a heartbreaking final-play loss to France.
"People look at raw athletic ability and think it can just translate," McGrath said. The timing, the spatial awareness, the fitness, the differences between rugby and Canadian football are night and day.
In Las Vegas Baker played a starring role. That came as no surprise, given that Baker's the reigning World Rugby men's sevens player of the year.
He scored a dazzling try in the semi final, beating four Fijian defenders on a scoring run that started in the shadow of his own goal-line.
Baker's coach, Mike Friday, is cautious when looking to compare the two players.
The key to transitioning players from one sport to another is to "focus on what they can do rather than what they can't do … that's the public perspective as well as when coaching them," Friday added.
Campbell has already come a long way, but his coach would like to see what another 12 months of rugby-focused training might do.
"I think he could be the kind of player Canada has rarely seen in sevens." But the plan is for Campbell to return to his CFL life in a month or so. He still makes far more money playing for the Alouettes, there's just no getting around that, his coach said.
So for now, Campbell is learning as much as he can. "It's like he's studying for an exam," McGrath said.
Campbell said he watches a lot of video, both of himself and of other players. Given his football background, that makes a lot of sense.
The off-field experience, of going to a new city for a week, of taking in new cultures alongside training for the weekend's main event, and of course bonding with teammates, has been the biggest surprise.
Yes, he'll be going back to the CFL, but he's hoping he'll be back in sevens — and he's also going to carry a message with him. | {
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{"url":"https:\/\/zbmath.org\/?q=an%3A0876.65089","text":"# zbMATH \u2014 the first resource for mathematics\n\nContinuous Runge-Kutta methods for neutral Volterra integro-differential equations with delay. (English) Zbl\u00a00876.65089\nExplicit and implicit continuous Runge-Kutta methods (i.e., Runge-Kutta methods with dense output) are used for the numerical treatment of neutral Volterra integro-differential equations with delay. The order of convergence is investigated and it is shown that the arising nonlinear systems can be solved by fixed point iteration as long as the considered problem is non-stiff. A few numerical experiments are included.\nReviewer:\u00a0E.Hairer (Gen\u00e8ve)\n\n##### MSC:\n 65R20 Numerical methods for integral equations 45J05 Integro-ordinary differential equations\nRODAS\nFull Text:\n##### References:\n [1] Baker, C.T.H.; Paul, C.A.H., Parallel continuous Runge-Kutta methods and vanishing lag delay differential equations, Adv. comput. math., 1, 367-394, (1993) \u00b7 Zbl\u00a00824.65055 [2] Bellen, A.; Jackiewicz, Z.; Zennaro, M., Stability analysis of one-step methods for neutral delay-differential equations, Numer. math., 52, 605-619, (1988) \u00b7 Zbl\u00a00644.65049 [3] Brunner, H., The numerical solutions of neutral Volterra integro-differential equations with delay arguments, () \u00b7 Zbl\u00a00828.65146 [4] Brunner, H.; van der Houwen, P.J., The numerical solution of Volterra equations, () \u00b7 Zbl\u00a00611.65092 [5] Enright, W.H., The relative efficiency of alternative defect control schemes for high order continuous Runge-Kutta formulas, SIAM J. numer. anal., 30, 1419-1445, (1993) \u00b7 Zbl\u00a00787.65046 [6] W.H. Enright and H. Hayashi, Convergence analysis for the numerical solutions of retarded and neutral type delay differential equations by continuous numerical methods, SIAM J. Numer. Anal., to appear. \u00b7 Zbl\u00a00914.65084 [7] Hairer, E.; N\u00f8rsett, S.P.; Wanner, G., Solving ordinary differential equations I. nonstiff problems, (1993), Springer Berlin \u00b7 Zbl\u00a00789.65048 [8] Hairer, E.; Wanner, G., Solving ordinary differential equations II. stiff and differential-algebraic problems, (1991), Springer Berlin \u00b7 Zbl\u00a00729.65051 [9] Hayashi, H., Numerical solution of retarded and neutral delay differential equations using continuous Runge-Kutta methods, () [10] Jackiewicz, Z., One-step methods of any order for neutral functional differential equations, SIAM J. numer. anal., 21, 486-511, (1984) \u00b7 Zbl\u00a00562.65056 [11] Jackiewicz, Z.; Lo, E., The numerical integration of neutral functional-differential equations by fully implicit one-step methods, Z. angew. math. mech., 75, 207-221, (1995) \u00b7 Zbl\u00a00830.65079 [12] Kamont, Z.; Kwapisz, M., On the Cauchy problem for differential-delay equations in a Banach space, Math. nachr., 74, 173-190, (1976) \u00b7 Zbl\u00a00288.34069 [13] Vermiglio, R., Natural continuous extensions of Runge-Kutta methods for Volterra integro-differential equations, Numer. math., 53, 439-458, (1988) \u00b7 Zbl\u00a00629.65145\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":"2021-04-21 03:55:46","metadata":"{\"extraction_info\": {\"found_math\": false, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.832627534866333, \"perplexity\": 3783.701573716719}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-17\/segments\/1618039508673.81\/warc\/CC-MAIN-20210421035139-20210421065139-00191.warc.gz\"}"} | null | null |
Pornography, Psychedelics and Technology (Routledge Revivals)
Essays on the Limits to Freedom, 1st Edition
By E. J. Mishan
First published in 1980, Pornography, Psychedelics And Technology: Essays on the Limits to Freedom focuses on the crucial connections between technological growth and the more salient features of social malaise in the latter part of the twentieth century. Professor Mishan is one of the few economists absorbed by the larger social questions, and does not believe that the growth in state intervention and the decline of social liberty are simply the result of intellectual confusion and bureaucratic momentum. He sees them as unavoidable consequences of scientific and technical progress. While agreeing with many of his fellow economists in acknowledging the virtues of a competitive market economy, Professor Mishan is acutely aware of its limitations. Following the growth of self-styled liberation movements, seen as manifestations of a move towards a world of greater individual emancipation and fulfilment, the author nevertheless groups such movements together with the rising indices of violence, suicide, family breakdown and hooliganism, which have become indicative of a growing disorientation and social disintegration. These developments and the hazards they entail, however, are bound up with the rapid scientific and technological progress of the post-war world.
1. The Spillover Enemy 2. Making the World Safe for Pornography 3. The New Inflation 4. On the Road to Repression and Control 5. Why LSD Should Be Legalised 6. Economic Growth: An Alarmist View
Are there elusive titles that you need and have been trying to source for years but thought that you would never be able to find?
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Drawing from over 100 years of innovative, cutting-edge publishing, Routledge Revivals is an exciting programme whereby key titles from the distinguished and extensive backlist of the many acclaimed imprints associated with Routledge will be re-issued.
The programme draws upon the illustrious backlists of Kegan Paul, Trench & Trubner, Routledge & Kegan Paul, Methuen, Allen & Unwin and Routledge itself.
Routledge Revivals spans the whole of the Humanities and Social Sciences, and includes works by some of the world's greatest thinkers including Emile Durkheim, Max Weber, Simone Weil, Martin Buber, Karl Jaspers and Max Beloff.
If you are interested in Revivals in the Behavioral Sciences, please visit https://www.routledge.com/series/PSYREVIVALS
BUSINESS & ECONOMICS / General
BUSINESS & ECONOMICS / Economic Conditions
BUSINESS & ECONOMICS / Economics / Macroeconomics | {
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Régis Blachère, né à Montrouge le et mort à Paris le , est un orientaliste, islamologue et arabisant français.
Biographie
Agrégé d'arabe (1924), il est membre de l'Institut (1972), directeur d'études à l'Institut des hautes études marocaines de Rabat (1930-1935), professeur d'arabe à l'École nationale des langues orientales (1935-1950), professeur de littérature arabe du Moyen Âge à la Sorbonne (1950-1970), directeur d'études à l'École pratique des hautes études (1950-1968), directeur de l'Institut d'études islamiques de l'université de Paris (1956-1965), directeur du Centre de lexicographie arabe, associé au CNRS (1962-1971).
On lui doit une traduction critique du Coran (1947) avec un essai de reclassement des sourates dans l'ordre chronologique de leur révélation.
Publications
Analecta, Institut français de Damas, Damas, 1975.
Le Coran, Presses universitaires de France, (Que sais-je ?, ), 2002
Dans les pas de Mahomet, Hachette, Paris, 1956.
Dictionnaire arabe-français-anglais (Langue classique et moderne), G.-P. Maisonneuve, 1960.
Dictionnaire arabe-français-anglais Arabic/French/English Dictionary - Langue classique et moderne, Maisonneuve et Larose, Paris, 1967.
Éléments de l'arabe classique, Quatrième édition revue et corrigée, G.-P. Maisonneuve, 1958.
Exercices d'arabe classique, Adrien Maisonneuve, Paris, 1970.
Extraits des principaux géographes arabes du Moyen Âge.
Grammaire de l'arabe classique, Maisonneuve et Larose, Paris
Histoire de la littérature arabe : des origines à la fin du de J.-C., Volume 1, Adrien Maisonneuve, Paris, 1952,
Histoire de la littérature arabe : des origines à la fin du de J.-C., Volume 2, Adrien Maisonneuve, Paris, 1964, ,
Histoire de la littérature arabe : des origines à la fin du de J.-C., Volume 3, Adrien Maisonneuve, Paris, 1964,
Introduction au Coran, Maisonneuve et Larose,
Le Coran. Traduction selon un essai de reclassement des sourates, G.-P. Maisonneuve, Paris, 1949-1977.
Le problème de Mahomet - Essai de biographie critique du fondateur de l'Islam, un volume de 135 pages, Presses universitaires de France, Paris, 1952.
Notes et références
Liens externes
Naissance à Montrouge
Naissance dans le département de la Seine
Historien français du XXe siècle
Historien de l'Islam médiéval
Orientaliste français
Islamologue français
Agrégé d'arabe
Traducteur du Coran en français
Auteur publié par les éditions Maisonneuve et Larose
Auteur publié par les Presses universitaires de France
Auteur publié par les éditions Hachette
Enseignant à l'Institut national des langues et civilisations orientales
Académie des inscriptions et belles-lettres
Naissance en juin 1900
Décès en août 1973
Décès dans le 15e arrondissement de Paris
Décès à 73 ans | {
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There are many tourist Spots and interesting places around Kuching which promises memorable experience.
Kuching has a great cleat to offer for adventurous gourmet, ranging from local Malaysian delicacies to Western specialties. Visitors may be surprised to find that they needn't miss out on their fast food favorites such as McDonalds, Pizza Hut, KFC and A&W. Some other food delicacies can also be found such as Japanese, Korean and even Indonesian food.
Kuching has entertainment and night life to suit most tastes such as live music pubs, hotel bars, plush karaoke lounges and discos.
Kuching offers not only good places to shop clothing but also Borneo arts, crafts and curios. Other famous sought after items are authentic pottery, birdnests and fabric of variety designs.
For those seeking a more adventurous path, whether of nature trekking or simply enjoying the tropical atmosphere, make sure not to miss these most visited spot. There are Bako National Part, Kubah National Park, Gunung Gading National Park, Semengoh Orang Utan Rehabilitation, Jongs Crocodile Farm and Matang Family Park.
For a relaxing time, try these memorable places such as Sarawak Cultural Village, Holiday Inn Damai Beach Resort, Holiday Inn Damai Lagoon Resort, Damai Rainforest Resort, Santubong Kuching Resort and Hilton Batang Ai Longhouse Resort.
Other places of interests include Main Bazaar, The Astana, Fort Margherita, The Square Tower, The Sarawak Museum, Main Post Office, Tua Pek Kong Temple, The Kuching Mosque, The Sarawak Steamship Building, The Islamic Museum, Chinese History Museum, The Cat Museum and many more.
For the onward destinations outside Kuching, the famous parks include Mulu National Park, Niah Caves, Lambir Hills, Bario and Kelabit Highlands. | {
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Panicked packing. Because that's how I roll.
1. Pulled out clothes for The Rowdy Boys. Put on the floor until I fish out the suitcases.
2. Ate two pieces of pizza and chugged a diet Mountain Dew.
3. Decided on an Oreo cookie pop tart for breakfast tomorrow over a S'mores pop tart.
4. Pushed The Rowdy Boys clothes to the side so I could work out.
5. Sat down to write today's blog post.
Oh, I did some dishes too.
There's so much yet to do. Pack. Find that lip gloss I've been looking for. Actually wrap the bride's gift; yeah, it's a lot to squeeze into a small amount of time. No fear. Not yet, anyway.
And hey, I'm going to apologize right now. I've successfully blogged three solid weeks of the A to Z Challenge. I have extremely high hopes of blogging despite my bridesmaid duties. Will you simply get a photo of a slice of cake Saturday? Maybe.
Here's a shoddily taken picture of the shoes I'll be wearing. I won't be offended if you'd all like to start placing bets on how long it will take for me to fall on my face.
At least I'll have an interesting blog topic.
Previous PostOld Timey Ham Sandwiches: A recipe post. I know, I'm shocked too. Next PostRehearsal Reprimands: And I'm not surprised.
I hate packing! Do it too early, and you're constantly fishing things you need now out of your suitcase. Do it too late, and you're bound to forget something.
Hope you have a fun trip! Picture of cake sounds good!
I know the feeling. I'm right there with you. I have to leave in a few hours for the weekend and haven't done anything!
It worked out…but we ended up leaving 40 minutes late! | {
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\section{Introduction}
\label{sec:intro}
In recent years, increasingly large transformer-based models such as BERT~\cite{Devlin2019BERTPO}, RoBERTa~\cite{Liu2019RoBERTaAR} and GPT-3~\cite{Brown2020LanguageMA} have demonstrated remarkable state-of-the-art (SoTA) performance in many Natural Language Processing (NLP) and Computer Vision (CV) tasks and have become the de-facto standard.
However, those models are extremely inefficient; they require massive computational resources and large amounts of data as basic requirements for training and deploying. This severely hinders the scalability and deployment of AI-based systems across the industry.
One highly effective method for improving efficiency is knowledge distillation~\citep{Ba2014DoDN,Hinton2015DistillingTK}, in which the knowledge of a large model defined as the teacher is transferred into a smaller more efficient model defined as the student. TinyBERT~\cite{Jiao2020TinyBERTDB} and MiniLM~\cite{Wang2020MiniLMDS,Wang2021MiniLMv2MS} stand out with their superior accuracy-speed-size tradeoff, both introducing a novel distillation method specially designed for transformers. In TinyBERT, the knowledge residing in the hidden states and attention matrices of the teacher is transferred to the student, and in MiniLM the student deeply mimics the teacher's self-attention module by transferring the multi-head self-attention relations as computed by a scaled dot-product of pairs of queries, keys and values. Both methods consist of a two-stage learning framework which first captures the general-domain knowledge of the teacher and then captures the task-specific knowledge of the teacher using fine-tuning. MiniLM achieves higher accuracy than TinyBERT for most downstream tasks, with the same number of parameters, and is much simpler to fine-tune since it is directly fine-tuned on the downstream task, differently from TinyBERT which is fine-tuned using a task-specific distillation from a fine-tuned teacher for a large number of epochs. In addition, TinyBERT fine-tuning requires a preliminary data-augmentation stage.
Knowledge distillation has shown promising results for reducing the number of parameters, with, however, several caveats: First, a drop in accuracy and a still limited speed-up/latency gain, specifically in challenging NLP tasks such as QA; for example, DistilBERT~\cite{Sanh2019DistilBERTAD} produces a x1.7 speed-up albeit with a 3\% accuracy drop on SQuAD1.1~\cite{Rajpurkar2016SQuAD1Q}. Secondly, in many cases the target computational budget (HW type, memory size, latency constraints, etc.) is not given at the time of training. This implies that a separate student model must be trained for each applicable inference scenario and its distinct computational budget.
Recent studies have attempted to address these concerns by proposing dynamic transformers. Length-Adaptive Transformer (LAT)~\cite{kim-cho-2021-length} introduced the Drop-and-Restore method, which reduces the sequence length at each layer, finally bringing back the dropped tokens in the last layer to allow for a wide range of NLP tasks despite the dropping of tokens. In addition, LAT proposed a one-shot training method that enables the model to be used for any computational budget during inference time.
Dynamic-TinyBERT~\cite{Guskin2021DynamicTinyBERTBT} utilizes TinyBERT~\cite{Jiao2020TinyBERTDB} distillation and some LAT techniques (Drop-and-Restore inference and evolutionary-search) to train an efficient model that can be used for a wide range of NLP tasks with optimal performance per any computational budget. Dynamic-TinyBERT achieves up to x3 speedup with <1\% accuracy loss vs. BERT-base on the SQuAD1.1 benchmark.
In this work we expand Dynamic-TinyBERT to generate a much more highly efficient model. First, we use a much smaller MiniLM model which was distilled from a RoBERTa-Large teacher rather than BERT-base. Second, we apply the LAT method to make the model length-adaptive, and finally we further enhance the model's efficiency by applying 8-bit quantization~\cite{Zafrir2019Q8BERTQ8}. The resultant QuaLA-MiniLM (Quantized Length-Adaptive MiniLM) model outperforms BERT-base with only 30\% of parameters, and demonstrates an accuracy-speedup tradeoff that is superior to any other efficiency approach (up to x8.8 speedup with <1\% accuracy loss) on the challenging SQuAD1.1 benchmark. Following the concept presented by LAT, it provides a wide range of accuracy-efficiency tradeoff points while alleviating the need to retrain it for each point along the accuracy-efficiency curve.
\begin{figure}[t]
\centering
\includegraphics[width=13cm, height=3.6cm]{quala_minilm_train.pdf}
\caption{QuaLA-MiniLM training process. To run the model with the best accuracy-efficiency tradeoff per a specific computational budget, we set the length configuration to the best setting found by an evolutionary search to match our computational constraint.}
\label{fig:quala_minilm_training_fig}
\end{figure}
\section{Method}
QuaLA-MiniLM is generated by applying the following optimization techniques on top of each other: MiniLM distillation, Length Adaptive Transformer, and Quantization. Figure~\ref{fig:quala_minilm_training_fig} demonstrates the training pipeline.
\subsection{MiniLM Distillation}
\label{sec:distillation}
MiniLM uses deep self-attention distillation to generate a task-agnostic small Language Model (LM). The student model learns to deeply mimic the multi-head self-attention relations -- which are obtained by a scaled dot-product of pairs of queries, keys and values of multiple relation heads of a single teacher's layer. Training a task-specific MiniLM requires two steps: first, training a task-agnostic MiniLM using multi-head self-attention distillation on a general-domain's data; second, fine-tuning it to a specific downstream task using standard supervised fine-tuning on the task dataset.
\begin{table}[t]
\centering
\caption{Inference performance on the SQuAD1.1 evaluation dataset. For all the length-adaptive (LA) models we show the performance both of running the model without token-dropping, and of running the model in a token-dropping configuration according to the optimal length configuration found to meet our accuracy constraint.}
\label{tab:models-performance-table}
\resizebox{0.9\columnwidth}{!}{%
\begin{tabular}{@{}lllllll@{}}
\toprule
Model & Model size (Mb) & Tokens per layer & Accuracy (F1) & Latency (ms) & FLOPs & Speedup \\ \bottomrule
BERT-base & 415.4723 & (384,384,384,384,384,384) & 88.5831 & 56.5679 & 3.53E+10 & 1x \\ \bottomrule
TinyBERT-ours & 253.2077 & (384,384,384,384,384,384) & 88.3959 & 32.4038 & 1.77E+10 & 1.74x \\
QuaTinyBERT-ours & 132.0665 & (384,384,384,384,384,384) & 87.6755 & 15.5850 & 1.77E+10 & 3.63x \\ \bottomrule
MiniLMv2-ours & 115.0473 & (384,384,384,384,384,384) & 88.7016 & 18.2312 & 4.76E+09 & 3.10x \\
QuaMiniLMv2-ours & 84.8602 & (384,384,384,384,384,384) & 88.5463 & 9.1466 & 4.76E+09 & 6.18x \\ \bottomrule
LA-MiniLM & 115.0473 & (384,384,384,384,384,384) & 89.2811 & 16.9900 & 4.76E+09 & 3.33x \\
LA-MiniLM & 115.0473 & (269, 253, 252, 202, 104, 34) & 87.7637 & 11.4428 & 2.49E+09 & 4.94x \\ \bottomrule
QuaLA-MiniLM & 84.8596 &(384,384,384,384,384,384) & 88.8593 & 7.4443 & 4.76E+09 & 7.6x \\
QuaLA-MiniLM & 84.8596 & (315,251,242,159,142,33) & 87.6828 & 6.4146 & 2.547E+09 & \textbf{8.8x} \\ \bottomrule
\end{tabular}
}
\end{table}
\subsection{LAT (Length Adaptive Transformer)}
\label{sec:lat}
After fine-tuning a transformer to a specific task, the model is trained with LengthDrop and LayerDrop in a process called Drop-and-Restore, in which tokens are dropped at a random rate at each layer, and are brought back in the last hidden layer to enable a wider range of tasks and to make the model robust to the choice of length configuration at inference time. A length configuration
is a sequence of retention parameters (l\textsubscript{1}, · · · l\textsubscript{L}) each of which corresponds to the number of word vectors that are kept at each layer. Drop-and-Restore training is done using inplace distillation and sandwich rule methods, as follows: the full model without LengthDrop is fine-tuned for the downstream task as usual by minimizing the supervised loss function, while simultaneously, randomly-sampled sub-models with length reduction (sandwiches) learn to mimic the predictions of the full model using knowledge distillation~\cite{Hinton2015DistillingTK}.
As proposed by LAT, we run a multi-objective evolutionary search~\cite{Cai2020OnceFA,Wang2020HATHT} to optimize the accuracy-efficiency trade-off per each computational budget without additional training. The algorithm finds the optimal length configurations per possible computational constraints by generating an initial population of length configurations, and evolving the population at each iteration by mutation and crossover to consist only of configurations that lie on a newly updated accuracy-efficiency Pareto frontier. This process repeats for many iterations until best tradeoff is found.
\subsection{Quantization}
\label{sec:quantization}
We use the traditional post-training quantization~\cite{jacob2018quantization,wang2019haq} to quantize the model. Quantization for neural network is the process of approximating used floating-point numbers \textit{r} by low bit width numbers \textit{q} by the following mapping: \textit{r=S(q-Z)} where \textit{S, Z} represents the scale-factor and the zero-point values. Post-training quantization requires calibration of samples from representative datasets and collection of tensor statistics such as min and max values to determine the scale-factor and zero point values. Typically, we expect an INT8 8-bit model to gain more instruction throughput over the FP32 model (e.g., x4 for Intel DL Boost) and to gain about x4 lower memory bandwidth over the FP32 model, and therefore deliver higher inference efficiency.
\section{Experiments Setup}
We first fine-tune a pre-trained general-MiniLM for 5 epochs on the SQuAD1.1 dataset. We use the publicly available pre-trained MiniLMv2 with 6-layers and a hidden size of 384 distilled from RoBERTa-large\footnote{\url{https://github.com/microsoft/unilm/tree/master/minilm}}. We train our task-specific MiniLM to be length adaptive by running another 10 epochs of fine-tuning with Drop-and-Restore. After training, we run an evolutionary search to identify the length configurations that maximize the performance per any computational budget, namely those that achieve the best tradeoff between high accuracy and low number of floating operations (F1 vs. FLOPs). We leverage an open-source quantization tool~\cite{ipex} to obtain a QuaLA-MiniLM. See appendix section for a full description of system configurations. Results may vary.
\section{Results}
\label{sec:results}
\begin{figure}[t]
\centerfloat
\resizebox{0.7\textwidth}{!}{
\begin{tikzpicture}
\definecolor{s1}{RGB}{197, 90, 17}
\definecolor{clr_orange}{RGB}{255, 127, 0}
\definecolor{clr_green}{RGB}{31, 182, 83}
\definecolor{clr_purple}{RGB}{182, 83, 204}
\definecolor{clr_fuchsia}{RGB}{145, 92, 130}
\begin{axis}[
width=\linewidth,
xlabel={FLOPs ratio [BERT-base FLOPs / model FLOPs]},
ylabel={F1},
xmin=1, xmax=16,
ymax=90.5,
ymajorgrids=false,
legend columns=1,
legend pos=north east,
legend style={at={(0.7,0.99)},anchor=north west},
xtick pos=left, ytick pos=left,
xtick={1, 2,3,4, 5, 6,7,8,9,10,11,12,13,14,15,16},
ytick={84, 84.5, 85,85.5,86,86.5,87,88, 88.5, 89, 89.5 ,90, 90.5, 91},
extra y ticks = {87.5},
extra y tick style = {brown},
yticklabels={84,84.5,85,85.5,86,86.5,87,88, {BERT 88.5}, 89, 89.5, 90, 90.5, 91},
xticklabels={1, 2,3,4,5, 6,7,8,9,10,11,12,13,14,15,16},
typeset ticklabels with strut,
enlarge x limits=false,
scale only axis]
\addlegendimage{blue}
\addlegendimage{green}
\addlegendimage{yellow}
\addplot[green, mark=*] table [x=speedup, y=f1, col sep=comma] {evo_minilm_ld.csv};
\addplot[blue, mark=*] table [x=speedup, y=f1, col sep=comma] {quala_minilm.csv};
\addplot[yellow, mark=*] table [x=speedup, y=f1, col sep=comma] {DynamicTinyBERT.csv};
\addplot[black ,mark=*, mark size=3pt, nodes near coords={TinyBERT}, every node near coord/.append style={xshift=20pt,yshift=3pt}] coordinates{(2, 87.5)};
\addplot[black ,mark=*, mark size=3pt, nodes near coords={DistilBERT}, every node near coord/.append style={xshift=25pt,yshift=6pt}] coordinates{(1.3, 85.8)};
\addplot[black ,mark=*, mark size=3pt, nodes near coords={MiniLM}, every node near coord/.append style={xshift=-25pt,yshift=-6pt}] coordinates{(7.42, 88.7)};
\addplot[black ,mark=*, mark size=3pt] coordinates{(1, 88.5)};
\addplot[mark=none, black] coordinates {(1,88.5) (16,88.5)};
\addplot[mark=none, brown] coordinates {(1,87.5) (16,87.5)};
\addlegendentry{QuaLA-MiniLM}
\addlegendentry{LA-MiniLM}
\addlegendentry{Dynamic-TinyBERT}
\end{axis}
\end{tikzpicture}
}
\hfill
\vspace{2mm}
\caption{We evaluate F1 vs. FLOP-ratio of LA-MiniLM and QuaLA-MiniLM models configured with each of the optimal length configurations found by the evolutionary search. Each length configuration determines a single performance result of the model. FLOP-ratio is computed by dividing the number of FLOPs used to run BERT-base by the number of FLOPs used to run the model with a specific length-configuration. Evaluation is done on the SQuAD1.1 evaluation set.}\label{fig:graph}
\end{figure}
\subsection{Accuracy-efficiency trade-off}
Figure~\ref{fig:graph} shows the significant boost to inference performance achieved by our approach. Both LA-MiniLM and QuaLA-MiniLM outperforms BERT-base, Dynamic-TinyBERT, TinyBERT, and DistilBERT models in terms of both accuracy and efficiency, with up to x14 reduction of FLOPs and less than 1\% accuracy drop vs. BERT-base. QuaLA-MiniLM exhibits lower accuracy than LA-MiniLM with the same FLOPs count (since compression of weights from 32-bit to 8-bit does not affect the number of floating/int operations), but runs x2 faster due to its x4 instruction throughput gain (see the next section for a speedup gain analysis).
\subsection{Inference speedup}
Inference performance in terms of model size and latency is presented in Table~\ref{tab:models-performance-table}. The QuaLA-MiniLM model is x5 smaller than BERT-base, x3 smaller than TinyBERT, and runs about x8.8 faster than BERT-base, x5 than TinyBERT and x2 than quantized-TinyBERT, while sacrificing less than 1\% accuracy vs. BERT-base. LA-MiniLM configured with the optimal length configuration achieves x1.5 inference speedup over the LA-MiniLM model without token-dropping.
\section{Conclusions and future work}
We propose QuaLA-MiniLM, which leverages sequence-length reduction and low-bit representation techniques to further enhance MiniLM inference performance, enabling adaptive sequence-length sizes to accommodate different computational budget requirements with an optimal accuracy-efficiency tradeoff. Experiments on the SQuAD1.1 benchmark dataset demonstrate the effectiveness of our method compared with previous work on BERT compression. In future work we intend to explore how leveraging Sparsity can further boost MiniLM performance.
\bibliographystyle{abbrvnat}
| {
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{"url":"https:\/\/plainmath.net\/38606\/graph-the-sets-of-points-whose-polar-coordinates-satisfy-the","text":"# Graph the sets of points whose polar coordinates satisfy the\n\nNompsypeFeplk 2021-11-19 Answered\nGraph the sets of points whose polar coordinates satisfy the given polar equation.\nThe given polar equation is written as follows:\n$$\\displaystyle-{\\frac{{\\pi}}{{{4}}}}\\le\\theta\\le{\\frac{{\\pi}}{{{4}}}},-{1}\\le{r}\\le{1}$$\n\n### Expert Community at Your Service\n\n\u2022 Live experts 24\/7\n\u2022 Questions are typically answered in as fast as 30 minutes\n\u2022 Personalized clear answers\n\n### Solve your problem for the price of one coffee\n\n\u2022 Available 24\/7\n\u2022 Math expert for every subject\n\u2022 Pay only if we can solve it\n\n## Expert Answer\n\nZachary Pickett\nAnswered 2021-11-20 Author has 8745 answers\n\nConsider the polar coordinates as $$\\displaystyle{P}{\\left({r},\\theta\\right)}.$$\nHere, r is the directed distance from origin O to point Pand $$\\theta$$ is the directed angle from initial ray to OP.\nFrom the given equation, the value of $$\\theta$$ varies from $$\\displaystyle-{\\frac{{\\pi}}{{{2}}}}$$ to $$\\displaystyle{\\frac{{\\pi}}{{{2}}}}$$ and r varies from 1 to 2.\nPolar point traces a circle when r is fixed at constant value and it traces a straight line when $$\\theta$$ is fixed at constant value.\nFrom the analysis, draw the sets of points whose polar coordinates satisfy the given polar equation as shown in\nFigure 1.\n\n### Expert Community at Your Service\n\n\u2022 Live experts 24\/7\n\u2022 Questions are typically answered in as fast as 30 minutes\n\u2022 Personalized clear answers\n\n### Solve your problem for the price of one coffee\n\n\u2022 Available 24\/7\n\u2022 Math expert for every subject\n\u2022 Pay only if we can solve it\n...","date":"2022-01-26 10:45:01","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 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\": 0.8881770372390747, \"perplexity\": 1699.4649267753757}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": false}, \"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-2022-05\/segments\/1642320304947.93\/warc\/CC-MAIN-20220126101419-20220126131419-00202.warc.gz\"}"} | null | null |
2. Kim, HJ, Kim, NC, Wang, YD., .. Li YR … Taylor, JP. (2013). Mutations in prion-like domains in hnRNPA2B1 and hnRNPA1 cause multisystem proteinopathy and ALS. Nature.
3. Li, YR, King, O, Shorter, J, & Gitler, AD (2013). Stress granules as crucibles of ALS pathogenesis. J. of Cell Biology, 201(3), 361–72.
4. Li YR., & Matsunami, H. (2013). Unfolding the mystery of olfactory receptor gene expression. Dev Cell, 27(2), 1–2.
5. Mainland, J. D., Keller, A., Li, Y. R… Matsunami, H. (2013). The missense of smell: functional variability in the human odorant receptor repertoire. Nat. neuroscience.
7. Aoki R.* and Li YR*#. alpha-synuclein promotes neuroprotection through NF-kappaB-mediated transcriptional regulation of protein kinase C-delta. Sci Signal 4, jc6 (2011).
8. Swerdlow DI, Holmes MV, … Li YR, …, Keating BJ, Sattar N.The interleukin-6 receptor as a target for prevention of coronary heart disease: a mendelian randomisation analysis. Lancet 379, 1214-1224 (2012).
9. Li, YR, Xian, RR, Ziober, A… June, CH, Zhang, PJ, and Tchou, J, "Mesothelin expression is associated with poor outcomes in breast cancer.," Breast Cancer Res. Treat., vol. 147, no. 3, pp. 675–84, Oct. 2014.
10. Jiang, Yue,* Li, YR*., Ma, Minghong and H. Matsunami. "Muscarinic acetylcholine receptor M3 modulates odorant receptor activity via inhibition of β-arrestin-2 recruitment." Nature Communications, vol. 6, p. 6448 (2015).
11. Li, YR…Keating, BJ, Hakonarson, H. The genomic landscape of pediatric autoimmune diseases. Nature Medicine. (2015).
12. Li, YR…Keating, BJ, Hakonarson, H. Heritability and genetic sharing of pediatric age-of-onset autoimmune diseases. Nat. Commun. 6, 8442 (2015).
13. van Ingen, G…. Li, YR… Hakonarson, H. et al. Genome-wide association study for acute otitis media in children identifies FNDC1 as disease contributing gene. Nat. Commun. 7, 12792 (2016).
14. Gershuni, V., Li, YR… Tchou, TC. Breast cancer subtype distribution is different in normal weight, overweight, and obese women. Breast Cancer Res. Treat. 1–7 (2017). | {
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Jacques Miller
Jacques Miller AC FRS FAA
Jacques Francis Albert Pierre Meunier
(1931-04-02) 2 April 1931 (age 90)
Discoveries of the function of the thymus and the T cell and B cell subsets of mammalian lymphocytes
Gairdner Foundation International Award (1966)
Scientific career
The Walter and Eliza Hall Institute of Medical Research, Melbourne
Ludwik Gross;
Sir Gustav Nossal
Jacques Francis Albert Pierre Miller AC FRS FAA (born 2 April 1931) is a French-Australian research scientist. He is known for having discovered the function of the thymus and for the identification, in mammalian species of the two major subsets of lymphocytes (T cells and B cells) and their function.
1 Early life and education
3 Awards and honours
7 Seminal publications
Early life and education[edit]
Miller was born on 2 April 1931 in Nice, France, as J.F.A.P. Meunier, and grew up in France, Switzerland and China, mostly in Shanghai. After the outbreak of World War II, in anticipation of Japan's entry into the war, his family moved in 1941 to Sydney, Australia, and changed their last name to "Miller". He was educated at St Aloysius' College in Sydney, where he met his future colleague, Sir Gustav Nossal.[1]
Miller studied medicine at the University of Sydney, and had his first experience of laboratory research in the laboratory of Professor Patrick de Burgh where he studied virus infection.[1]
In 1958, Miller travelled to the United Kingdom on a Gaggin Research Fellowship from the University of Queensland. He was accepted to the Chester Beatty Research Institute of Cancer Research (part of the Institute of Cancer Research, London) and as a PhD student at the University of London.[2] Miller chose to study the pathogenesis of lymphocytic leukemia in mice, expanding on the research of Ludwik Gross into murine leukemia virus. Miller showed that experimental animals without a thymus at birth were incapable of rejecting foreign tissues and resisting many infections, thus demonstrating that the thymus is vital for development and function of the adaptive immune system. Prior to this, the thymus was believed to be a vestigial organ with no function.[3] His discovery has led many to describe Miller as the "world's only living person who can claim to have been the first to have described the functions of a human organ".[4] In 1963, Miller continued his work into the function of the thymus at the National Institutes of Health.
In 1966, Miller returned to Australia to become a research group leader at the Walter and Eliza Hall Institute of Medical Research in Melbourne, at the invitation of its new director Sir Gustav Nossal, the successor of Sir Macfarlane Burnet. There, with student Graham Mitchell, he discovered that mammalian lymphocytes can be separated into what were later called T cells and B cells, and that these interact to allow normal antibody production (T cell help). Miller went on to show that the thymus produces the T cells, that it removes autoreactive T cells (central T cell tolerance) and several other landmark findings in immunology. These are considered crucial to understanding diseases such as cancer, autoimmunity and AIDS, as well as processes such as transplant rejection, allergy and antiviral immunity.[1] Miller was also the first to provide evidence that thymus-derived immune cells are important for the defense against certain tumors,[5] which forms the basis for modern cancer immunotherapy.
Semi-retired since 1996, Miller is still involved in immunological research.[4]
Miller has had a longstanding interest in art, and studied art in the 1980s. His art has been exhibited at venues in Melbourne.[6]
Awards and honours[edit]
1966 Gairdner Foundation International Award
1967 Scientific Medal of the Zoological Society of London[1]
1970 Elected a Fellow of the Royal Society, London
1971 Macfarlane Burnet Medal and Lecture of the Australian Academy of Science[7]
1974 Paul Ehrlich and Ludwig Darmstaedter Prize
1978 Rabbi Shai Shacknai Memorial Prize[1]
1981 Officer of the Order of Australia (AO)[8]
1982 Elected Foreign Associate for the United States National Academy of Science
1983 International St Vincent Prize; World Health Organization[1]
1990 Sandoz Prize for Immunology[1]
1990 Peter Medawar Prize for the Transplantation Society[9]
1992 Croonian Prize, Royal Society
1995 J. Allyn Taylor International Prize in Medicine[1]
2000 Florey Medal
2001 Royal Society of London Copley Medal
2001 Centenary Medal[10]
2003 Prime Minister's Prize for Science[11]
2003 Appointed a Companion of the Order of Australia (AC)[12]
2015 ANZAAS Medal
2018 Japan Prize for Medicine and Medicinal Science[13]
2019 Albert Lasker Award for Basic Medical Research[14]
French Australians
^ a b c d e f g h Miller, Jacques (2005). "A scientific odyssey: unravelling the secrets of the thymus". The Medical Journal of Australia. 183 (11/12): 582–584. doi:10.5694/j.1326-5377.2005.tb00041.x. S2CID 40679566.
^ Mellor, Lise (2008). "Miller, Jacques F A P". Faculty of Medicine Online Museum and Archive. The University of Sydney. Retrieved 12 February 2012.
^ Miller, Jacques F. A. P. (July 2011). "The golden anniversary of the thymus". Nature Reviews Immunology. 11 (7): 489–495. doi:10.1038/nri2993. PMID 21617694. S2CID 21191923(Full text requires subscriber login or payment) CS1 maint: postscript (link)
^ a b Miller, Jacques; Slattery, Robyn (19 September 2011). "Celebrating a scientific breakthrough". Health Report (Interview: audio). Interviewed by Dr Norman Swan. Australia: ABC Radio National. Retrieved 12 February 2012.
^ Miller, JF (1963). "Effect of thymectomy on the induction of skin tumours by 3,4-benzopyrene". Nature. 199 (4896): 920–2. Bibcode:1963Natur.199..920M. doi:10.1038/199920a0. PMID 14079914. S2CID 39669302.
^ "Jacques Miller: Authentic Australian Art". Jacques Miller. 2008. Archived from the original on 2 April 2012. Retrieved 12 February 2012.
^ "Macfarlane Burnet Medal and Lecture". Australian Academy of Science. Retrieved 22 February 2017.
^ Officer of the Order of Australia, 26 January 1981, It's an Honour
^ "The Medawar Prize: 1990". The Transplantation Society Awards. The Transplantation Society. Retrieved 12 February 2012.
^ Centenary Medal, 1 January 2001, It's an Honour
^ "Professor Jacques Miller: The Modern 'Father' of Immunology". 2003 Prime Minister's Prize for Science. Department of Industry, Innovation, Science, Research and Tertiary Education. Archived from the original on 29 February 2012. Retrieved 12 February 2012.
^ Companion of the Order of Australia (AC), 9 June 2003, It's an Honour
^ 2018 Japan Prize Honors Pioneers in Medical Science and Energy, 30 January 2018, Japan Prize Awards
^ Albert Lasker Award 2019
Miller, Jacques; Slattery, Robyn (19 September 2011). "Celebrating a scientific breakthrough". Health Report (Interview: audio). Interviewed by Dr Norman Swan. Australia: ABC Radio National. Retrieved 12 February 2012.
"Institute hosts quintuple anniversary to celebrate science legends" (Press release). Walter and Eliza Hall Institute. 2 June 2011. Archived from the original on 24 March 2012. Retrieved 12 February 2012.
Darmondy, Louise. "WEHI Revisited: Professor Jacques Miller". Walter and Eliza Hall Institute. Archived from the original (streaming audio) on 7 April 2011. Retrieved 21 September 2011.
Seminal publications[edit]
Miller J. F. (30 September 1961). "Immunological function of the thymus". Lancet. 2 (7205): 748–9. doi:10.1016/s0140-6736(61)90693-6. PMID 14474038.
Miller J. F. (26 June 1964). "The thymus and the development of immunologic responsiveness". Science. 144 (3626): 1544–51. Bibcode:1964Sci...144.1544M. doi:10.1126/science.144.3626.1544. PMID 14169340.
Miller J. F.; Mitchell, G. F. (18 November 1967). "The thymus and the precursors of antigen reactive cells". Nature. 216 (5116): 659–63. Bibcode:1967Natur.216..659M. doi:10.1038/216659a0. PMID 6082462. S2CID 4282516.
Miller J. F.; Sprent J. (1 July 1971). "Cell-to-cell interaction in the immune response. VI. Contribution of thymus-derived cells and antibody-forming cell precursors to immunological memory". J Exp Med. 134 (1): 66–82. doi:10.1084/jem.134.1.66. PMC 2139027. PMID 5105057.
Copley Medallists (2001–present)
Jacques Miller (2001)
John Pople (2002)
John Gurdon (2003)
Harry Kroto (2004)
Paul Nurse (2005)
Stephen Hawking (2006)
Robert May (2007)
Roger Penrose (2008)
Martin Evans (2009)
David Cox / Tomas Lindahl (2010)
Dan McKenzie (2011)
John E. Walker (2012)
Andre Geim (2013)
Alec Jeffreys (2014)
Peter Higgs (2015)
Richard Henderson (2016)
Andrew Wiles (2017)
Jeffrey I. Gordon (2018)
John B. Goodenough (2019)
Alan Fersht (2020)
Jocelyn Bell Burnell (2021)
Scopus author
Retrieved from "https://en.wikipedia.org/w/index.php?title=Jacques_Miller&oldid=1018496030"
Companions of the Order of Australia
Fellows of the Australian Academy of Science
Fellows of the Royal Society
French emigrants to Australia
Foreign associates of the National Academy of Sciences
Recipients of the Copley Medal
Sydney Medical School alumni
WEHI alumni
People educated at St Aloysius' College (Sydney)
Academics of the Institute of Cancer Research
CS1 maint: postscript
Use dmy dates from January 2021
Articles with CINII identifiers
Articles with Scopus identifiers
This page was last edited on 18 April 2021, at 10:52 (UTC). | {
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NewsNationalDemocracy 2018
30 years and $3 billion later, Olmsted Locks and Dam on Ohio River to open
By: Chris Welch
The Olmsted Locks and Dam project, located on the Ohio River between Illinois and Kentucky, has taken 30 years and $3 billion to complete. But after three decades, the project is slated to be operational in October.
The engineers and workers behind the development say the investment is worth it.
"The Ohio flows into the Mississippi 16 miles downriver from us, and then it's open passage all the way down to New Orleans and into the ocean," explains Mick Awbreyk, the Army Corps of Engineers deputy chief for the Olmsted Division. "There is no locks and dams from the Olmsted to the ocean."
He says the innovative details that went into the project are "phenomenal and truly world class."
Awbreyk will see the opening of the project, which began in 1988.
So, why has it taken so long to complete?
"The nature of the project; a lot of different things have led to the duration," explains Awbreyk of the timeline. "One: it hasn't been fully funded until calendar 2013, fiscal years 2014 and that's the nature of civil works projects."
The dam crosses one of the busiest shipping lanes in the United States. Awbreyk says about 91 million tons of goods on average come through the area per year.
"The equivalent of 25,000 semi-trucks worth of cargo passes through this site every year," says Awbreyk. "You can imagine the wear and tear on the interstates and the highways that would have an extra 25,000 vehicles that will be on there per day."
The project's price tag has grown to around $3 billion. Workers here say it's money Americans will see come back to them.
"It's widely important. The return on investment on this particular project is $640 million per year to the nation," says Awbreyk. "That's a net number, so ultimately it's a little under a $3 billion price tag; it will pay for itself in five years." | {
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From The Photographic News, March 17, 1882: "I should like to see your boxing pictures," said the Prince of Wales to Mr. Muybridge on Monday at the Royal Institution, when the galloping horse, the running deer, the trotting bull, the halting pig, and the racing dogs had successively crossed the screen in life-like measure.
"I shall be very happy to show them, your Royal Highness," responded the clever photographer; and promptly there was thrown up the screen two athletes, who pounded away at one another right merrily, to the infinite delight of the audience in general and the Prince of Wales in particular. Mr. Muybridge, in this case, had taken rapid successive pictures of a pair of boxers as they assume one fighting position after another, and then these photographs were rapidly thrown on the screen in the same order by means of his zoopracticoscope. This is a boxing-match reproduced in all its photographic reality. "I don't know that pictures teach us anything useful," said Mr. Muybridge, "but they are generally found amusing."
You Did it Then: Rediscover intermittency as the matrix of movies by constructing a two-second "movie" using nothing more than Muybridge's photos above. Those photos are just like the serial photographs Muybridge projected for the Prince of Wales. Click the button on the left for instructions and files to download. Find your inner Edweard Muybridge.
You Do it Now: Shoot an object in motion using the continuous shooting function (burst mode) of your digital camera. From your image sequence create a two-second "movie." Find the frames per second rate that works right for your movie. | {
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Q: Nest.JS TypeORM "SCRAM-SERVER-FIRST-MESSAGE: client password must be a string" I dont know why, but since I was cloning my working repository I am using on AWS to a local machine and try to run it, I am getting the following error:
"SCRAM-SERVER-FIRST-MESSAGE: client password must be a string"
Error: SASL: SCRAM-SERVER-FIRST-MESSAGE: client password must be a string
at Object.continueSession (C:\Users\thehe\Documents\workspace\work\nft-trading-server\node_modules\pg\lib\sasl.js:24:11)
at Client._handleAuthSASLContinue (C:\Users\thehe\Documents\workspace\work\nft-trading-server\node_modules\pg\lib\client.js:257:10)
at Connection.emit (node:events:390:28)
at C:\Users\thehe\Documents\workspace\work\nft-trading-server\node_modules\pg\lib\connection.js:114:12
at Parser.parse (C:\Users\thehe\Documents\workspace\work\nft-trading-server\node_modules\pg-protocol\src\parser.ts:104:9)
at Socket.<anonymous> (C:\Users\thehe\Documents\workspace\work\nft-trading-server\node_modules\pg-protocol\src\index.ts:7:48)
at Socket.emit (node:events:390:28)
at addChunk (node:internal/streams/readable:315:12)
at readableAddChunk (node:internal/streams/readable:289:9)
at Socket.Readable.push (node:internal/streams/readable:228:10)
at TCP.onStreamRead (node:internal/stream_base_commons:199:23)
POSTGRES_HOST=localhost
POSTGRES_PORT=5432
POSTGRES_USER=admin
POSTGRES_PASSWORD=admin
POSTGRES_DB=nftapi01
PORT=5000
Does anyone know where that is coming from and how to fix this? I am not sure why I get this locally. I can connect to the pg database with the credentials of the .dev.env file but the Nest app wont start.
A: Are you importing the "dotenv" package? you need it for access to environment variables.
On your server.js file put: require('dotenv').config();
You said that you cloned your repository... do you have a package.json file in your project? (inside this file you declare what is your main file: "main": "server.js")
Of course, you need access to read your .env file, check them!
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 5,640 |
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Front page - Fighting the U-boats - Allied Warship Commanders
Listing of officers
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Allied Warship Commanders
Ronald James Burch DSO, RN
Born 9 Jun 1907
Died 23 Jul 1940 (33) HMS Narwhal (N 45)
If you can help with photo or any information on this Officer please
use our comment form.
1 Nov 1928 S.Lt.
1 Feb 1930 Lt.
1 Feb 1938 Lt.Cdr.
28 Jun 1940 DSO
Warship Commands listed for Ronald James Burch, RN
Ship Rank Type From To
HMS Narwhal (N 45) Lt.Cdr. Submarine 26 Feb 1940 23 Jul 1940 (+)
We currently have no career / biographical information on this officer.
Events related to this officer
Submarine HMS Narwhal (N 45)
HMS Narwhal (Lt.Cdr. E.R.J. Oddie, RN) ended her 5th war patrol at Rosyth. (1)
HMS Narwhal (Lt.Cdr. R.J. Burch, RN) joined convoy ON 17 off Methil. The convoy was to proceed to Bergen, Norway where it arrived 7th.
On the 4th Narwhal left the convoy to proceed to Scapa Flow to participate in direction finding trials.
This was Narwhal's 6th war patrol. As no log is available for this period it is not possible to display a map of this patrol. (1)
HMS Narwhal (Lt.Cdr. E.R.J. Oddie, RN) arrived at Scapa Flow ending her 6th war patrol. At Scapa Flow Narwhal was to participate in direction finding trials. Narwhal was delayed for one day by very bad weather en-route. (1)
HMS Narwhal (Lt.Cdr. R.J. Burch, RN) conducted exercises off Scapa Flow. (2)
HMS Narwhal (Lt.Cdr. R.J. Burch, RN) departed Scapa Flow for Rosyth. She made the passage together with HMS Swordfish (Lt. P.J. Cowell, RN) that was en-route from Scapa Flow to Blyth. They were escorted by HMS Buttermere (Lt. E.G.P.B. Knapton, RN). (2)
HMS Narwhal (Lt.Cdr. R.J. Burch, RN) had barely arrived at Rosyth when she was ordered to sea as part of the deployment of submarines for operation R.3 (assistance to Finland by the occupation of Norwegian key points). This was her 7th war patrol. (As no log is available for this period it is not possible to display a map of this patrol.) Shortly after, it was learnt that Finland had signed an armistice but the submarines were recalled only on 16 March. It was an uneventful except when the submarine was forced to dive for a Do 17 aircraft on the morning of the 17th. (1)
HMS Narwhal (Lt.Cdr. R.J. Burch, RN) was undocked. The date she was docked is currently not known to us. (2)
HMS Narwhal (Lt.Cdr. R.J. Burch, RN) laid a dummy minefield south-east of the Firth of Forth, She then set course for Blyth. (2)
HMS Narwhal (Lt.Cdr. R.J. Burch, RN) encountered trawler Polar Prince (194 GRT, built 1915) fishing in the East Coast mine barrage. The ship was placed under arrest and was escorted by Narwhal to Methil. (2)
HMS Narwhal (Lt.Cdr. R.J. Burch, RN) arrived at Blyth. She departed later the same day for Immingham where she was to embark 50 mines. (2)
HMS Narwhal (Lt.Cdr. R.J. Burch, RN) arrived at Immingham. After embarking a full outfit of 50 mines she departed Immingham for her 8th war patrol. She was ordered to lay a minefield in the Heligoland Bight (minefield F.D. 1).
For the daily positions of HMS Narwhal during this patrol see the map below.
4 Apr 1940 (position 54.37, 6.35)
Between 0413 and 0422 hours, HMS Narwhal (Lt.Cdr. R.J. Burch, RN) laid minefield F.D. 1 (50 mines) in the North Sea in position 54°37'N, 06°35'E. (1)
HMS Narwhal (Lt.Cdr. R.J. Burch, RN) ended her 8th war patrol at Blyth. (1)
HMS Narwhal (Lt.Cdr. R.J. Burch, RN) was docked at Blyth. The date of undocking is currently not known to us. (2)
HMS Narwhal (Lt.Cdr. R.J. Burch, RN) departed Blyth for Immingham to embark a full outfit of 50 mines. (2)
HMS Narwhal (Lt.Cdr. R.J. Burch, RN) arrived at Immingham. After embarking mines she departed later the same day for her 9th war patrol. She was ordered to lay a minefield (F.D. 5) off Skagen.
As no log is available for this period it is not possible to display a map of this patrol. (1)
At 0425 hours HMS Narwhal (Lt.Cdr. R.J. Burch, RN was south of Kristiansand South and proceeding to her minelaying position when she was informed that heavy German naval units [Gneisenau, Scharnhorst and Hipper] were returning to Germany and this was initially believed to be through the Kattegat. She attempted to intercept but returned to her original course when she was told that the enemy force was returning home west of Jutland.
At 2145 hours, the submarine sighted three enemy destroyers of the Leberecht Maas class in 57°55'N, 09°27'E and made an enemy report. No German destroyer was actually operating in the area. (1)
13 Apr 1940 (position 57.26, 10.45)
Between 1628 and 1644 hours HMS Narwhal (Lt.Cdr. R.J. Burch, RN) laid minefield F.D. 5 (50 mines) north of Laeso Island in position 57°26'N, 10°45'E.
Later the same day Narwhal made a torpedo attack on a German convoy south of Skagen. No hits were obtained. The attack was unobserved and the intended victims remain unidentified.
(All times are zone -1)
2131 hours - Sighted dark objects later identified as a large and a small merchant ship escorted by two torpedo-boats or trawlers. Started attack.
2158 hours - Fired 6 torpedoes from 6000 yards. No hits were obtained. (1)
The German merchant Togo (5042 GRT, built 1938) is damaged when she hits a mine laid by the British submarine HMS Narwhal (Lt.Cdr. R.J. Burch, RN) on 13 April 1940 north of Laeso Island in position 57°26'N, 10°45'E
The German auxiliary minesweeper M 1302 / Schwaben (436 GRT, built 1937) is sunk when she hits a mine laid by the British submarine HMS Narwhal (Lt.Cdr. R.J. Burch, RN) on 13 April 1940 north of Laeso Island in position 57°28'N, 10°46'E. Minesweepers of her flotilla [the 13th] detonated 21 mines of the field.
HMS Narwhal (Lt.Cdr. R.J. Burch, RN) arrived at Immingham. After embarking mines she departed later the same day for her 10th war patrol. She was ordered to lay a minefield (F.D. 6) off Skagen.
1 May 1940 (position 57.30, 10.43)
Between 0727 and 0753 hours HMS Narwhal (Lt.Cdr. R.J. Burch, RN) laid minefield F.D. 6 (50 mines) in the Kattegat in position 57°30'N, 10°43'E.
Later the same day HMS Narwhal attacked a German convoy and torpedoed and sank the German troop transport Buenos Aires (6097 GRT, built 1912) and torpedoed and damaged the German troop transport Bahia Castillo (8580 GRT, built 1918) in the Kattegat in position 57°39'N, 11°03'E.
The Bahia Castillo reaches port but is declared a total loss.
1725 hours - Sighted 9 or 10 merchant vessels escorted by surface escorts and aircraft. Started attack in which 6 torpedoes were fired from about 1000 yards. All 6 were heard to explode. Several hits must have been obtained for sure.
Depth charging started after the 4th torpedo exploded. 75 Charges were dropped during the next 3 hours but they caused no damage to Narwhal. (1)
The German auxiliary minesweeper M 1102 / H.A.W. Müller is slightly damaged when ship hit a mine laid by the British submarine HMS Narwhal (Lt.Cdr. R.J. Burch, RN) on 1 May 1940 in the Skagerrak south of Cape Skagen in position 57°30'N, 10°43'E.
HMS Narwhal (Lt.Cdr. R.J. Burch, RN) ended her 10th war patrol at Immingham. (1)
HMS Narwhal (Lt.Cdr. R.J. Burch, RN) departed Immingham for her 11th war patrol. She was ordered to lay a minefield (F.D. 10) off Feistein Island, Norway (south-west of Stavanger).
HMS Narwhal (Lt.Cdr. R.J. Burch, RN) received a signal cancelling minefield F.D. 10. Minefield F.D. 12 had to be laid instead near Bud, Norway. (1)
11 May 1940 (position 62.58, 6.48)
Between 1701 and 1730 hours (zone -1) HMS Narwhal (Lt.Cdr. R.J. Burch, RN) laid minefield F.D. 12 (50 mines) off Bud, Norway in position 62°58'N, 06°48'E. (1)
HMS Narwhal (Lt.Cdr. R.J. Burch, RN) ended her 11th war patrol at Blyth. (1)
The German armed trawler V 1109 / Antares (291 GRT, built 1929) sank on a mine laid by the British submarine HMS Narwhal (Lt.Cdr. R.J. Burch, RN) on 11 May 1940 off Bud, Norway in position 62°58'N, 06°48'E. Eighteen men were rescued (including five wounded) but seventeen were missing, including the Norwegian pilot.
HMS Narwhal (Lt.Cdr. R.J. Burch, RN) departed Blyth for Immingham to embark a full outfit of mines. (2)
HMS Narwhal (Lt.Cdr. R.J. Burch, RN) arrived at Immingham. After embarking mines she departed later the same day for her 12th war patrol. She was ordered to lay a minefield (F.D. 16) off Jaerens, Norway (south of Stavanger).
3 Jun 1940 (position 58.46, 5.25)
Between 1937 and 2001 hours (zone -1) HMS Narwhal (Lt.Cdr. R.J. Burch, RN) laid minefield FD 16 (50 mines) off Jaerens Point, Norway in position 58°46'N, 05°25'E. (1)
The German merchant Palime (2863 GRT, built 1937) and the German minesweeper M 11 (874 tons, built 1939) both hits mines. The M 11 sank while the Palime is beached and later declared a total loss. The mines were laid by the British submarine HMS Narwhal (Lt.Cdr. R.J. Burch, RN) on 3 June 1940 off Jaerens Point, Norway in position 58°46'N, 05°25'E. It is possible that Clara M. Russ was damaged on 28 September 1940 (see below) and that UJ 175 was lost on this field on 7 January 1941. However flooders had been set for 19 July 1940 so there is a strong element of doubt.
HMS Narwhal (Lt.Cdr. R.J. Burch, RN) arrived at Immingham. After embarking mines she departed later the same day for her 13th war patrol. She was ordered to lay a minefield (F.D. 19) off Haugesund, Norway.
12 Jun 1940 (position 59.26, 5.10)
Between 2119 and 2217 hours (zone -1) HMS Narwhal (Lt.Cdr. R.J. Burch, RN) laid minefield F.D. 19 (50 mines) off Haugesund, Norway in position 59°26'N, 05°10'E.
Course was then set towards Utsire where Narwhal was to patrol during 13-16 June. (1)
HMS Narwhal (Lt.Cdr. R.J. Burch, DSO, RN) departed Blyth for Immingham to embark a full outfit of mines. (2)
HMS Narwhal (Lt.Cdr. R.J. Burch, DSO, RN) arrived at Immingham. After embarking mines she departed later the same day for her 14th war patrol. She was ordered to lay a minefield (F.D. 21) off Kristiansund, Norway.
4 Jul 1940 (position 63.15, 7.39)
Between 1025 and 1040 hours (zone -1) HMS Narwhal (Lt.Cdr. R.J. Burch, DSO, RN) laid minefield F.D. 21 (50 mines) north of Kristiansund, Norway in position 63°15'N, 07°39'E. (1)
At 0408 hours, near 60°08'N, 02°41.5'E, HMS Narwhal (Lt.Cdr. R.J. Burch, DSO, RN) sighted a flying boat and dived. Two bombs were heard but this was probably Hudson "O" of 269 Squadron which attacked HMS Sunfish in error.
Also on this day the German auxiliary submarine chaser UJ D / Treff VIII (356 GRT, built 1937) sank on a mine laid by the British submarine HMS Narwhal (Lt.Cdr. R.J. Burch, DSO, RN) on 4 July 1940 north of Kristiansund, Norway in position 63°15'N, 07°34'E. Although many documents show her as UJ D, she was apparently renamed UJ B on 1 June. Of a crew of 35, 13 were killed or missing. (1)
HMS Narwhal (Lt.Cdr. R.J. Burch, DSO, RN) ended her 14th war patrol at Blyth. (1)
HMS Narwhal (Lt.Cdr. R.J. Burch, DSO, RN) departed Blyth for Immingham to embark a full outfit of mines.
Narwhal arrived at Immingham later the same day. (2)
HMS Narwhal (Lt.Cdr. R.J. Burch, DSO, RN) departed Immingham for her 15th war patrol. She was ordered to lay a minefield (F.D. 22) north-west of Kristiansund, Norway near position 63º16'N, 07º13'E. (2)
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Boy George scratched
Viewers left asking how Boy George acquired facial bruising and scratches
On Sunday, wayward felon Boy George appeared on the BBC's The Voice with five deep scratches and a corker of a bruise on his face. One can only wonder how this self-loving sort acquired such.
A scratched and bruised Boy George on The Voice on Sunday 20th March 2016
The singer – born George O'Dowd – failed to respond to those who questioned him about the injuries on Twitter.
Martin Ellis Jones March 21, 2016 At 7:28 am
Bitch!
Dodge down unda March 21, 2016 At 8:43 am
I wish somebody would just drop this little pervert out of a low flying cargo plane over Iran.
At least we wouldn't have to keep seeing his stupid, grinning mush all over the media.
Maryam Hashemi March 22, 2016 At 12:46 am
He will fit in very well. I don't think you are aware of what iranian actually look like 🙂 do a google I age search and find out
Image search I meant 😉
HONEST GUV March 22, 2016 At 2:33 pm
I did try to find out what he would look like after the plane treatment and all I found was a picture of a pot of strawberry jam.
HONEST GUV March 21, 2016 At 9:04 am
Handcuffed to a wall and beaten with a metal chain perhaps?
Jackthelad. March 21, 2016 At 11:57 am
AAAH, POOR BOY……………………………………………………………………..(Just kidding folks !!!!!)
Martin Ellis Jones March 21, 2016 At 12:31 pm
Which bitch?
Ian Freeman March 21, 2016 At 2:48 pm
He's an excellent mentor on The Voice. A breath of fresh air and he knows the business inside out.
Matthew Steeples March 21, 2016 At 2:49 pm
A shame he has no remorse for the appalling abuse he subjected his victim to. The BBC should be ashamed of having hired this repugnant man in my personal opinion. Next they'll welcome back Stuart Hall and then Jonathan King.
Jerome Lyne March 21, 2016 At 5:12 pm
I think the bbc are doing a video link with rolf Harris on the next show asking him what he thinks of the young talent, only a rumour.
However much he knows about the popular music business, is he the sort of person we want to have around acting as role model for, and mentoring our children and young people? How about Harris or King on child care and raising teenagers. I would suggest not and if I wouldn't ask him into my house why do I have to have him grace a TV screen? Come the day soon when they scrap the licence fee completely and charge the public its true worth – nothing.
Henry Worsly is a role model we need in this day and age, and I hope they name the new polar research ship after him.
Well Ian, Why not let him mentor one of your siblings if he's such a breath of fresh air.
mel March 21, 2016 At 3:46 pm
I've nothing against him personally but he really seems to be a bit of a deviant.
Chaim Paddaman March 21, 2016 At 4:00 pm
It is evidence based that the BBC don't have an issue with celebrities breaking the law, will even defend them, but will smear the ordinary Joe Blogs that commits a crime or misdemeanour. A corporation that trade in double standards.
Buffoon Boris & Rotten Randy… Get A Room!
The blithering buffoon that is Boris Johnson and the randy rotter that is Prince Andrew should do the decent thing… Disappear from public view forever and get a room together with their weird wack job wives.
Friendless Fergie
Simpleton Sarah, Duchess of York "pauses" her unsurprisingly unpopular 'Fergie & Friends' YouTube channel as her increasingly friendless husband continues to face sexual assault allegations in New York.
Deviant Robert Durst Defeats Justice In Death
That billionaire murderer Robert Durst effectively becomes an innocent man in death is utterly outrageous; that he has defeated justice is ludicrous and the key question now is: "Who gets his fortune?" | {
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Home > Technology > Digital Ag
Precision Ag Means Different Things
Precision Ag is a term used to indicate the use of modern technologies in a growing operation. Global-Positioning Systems (GPS) are utilized to steer implements and machines automatically, Mobile Implement Control Systems (MICS) have been developed to run and record information from machines and implements, and Geographic Information Systems (GIS) are leveraged to gather and manage agricultural data collected through equipment computers and other hardware. A Precision Ag operation will typically include the use of external remote aerial, space-based, or ground-based sensors and other informative data sources. Typically, a Farm Management Information System (FMIS) will be employed to house all of the information a Precision Ag grower is capable of producing.
A grower that uses precision ag technology will do more than simply store this information. Using the systems described above, a grower that utilizes precision ag technology and data can leverage their operational data to produce site-specific agronomic decisions. This enables field prescriptions – be it seed, chemical, fertilizer, irrigation or otherwise – down to a site-specific point based on the varying levels of potential in a particular field. This technology and data allow for quick, in-field decisions and further yield optimization while preserving both natural and operational resources. There is also the added benefit of easing a grower's federal reporting requirements through streamlined data collection and delivery systems.
Common Precision Ag Solutions
There are many options available to handle this data. Here is a summary of the main types of systems in the agricultural industry today:
GPS (Global Positioning System)
Auto-Steer and GPS guidance uses GPS signals to automatically steer or guide a machine or implement. Various accuracies are available and can steer or locate machines and devices within a 1 inch to 2 foot repeatable position (Examples: RTK 1-2", OmniStar 2-5", WAAS 6-24", GLONASS 6-24").
MICS (Mobile Implement Control Systems)
Located in a cab, these GPS controlled systems control the machine/implement itself. These devices can also record and distribute data to other systems.
Planter/Seed Monitors use GPS position to monitor location of seed population, overlaps, and skips while recording and controlling how, where, and when seed is planted.
Yield Monitors use GPS position, sensors, and associated electronics mounted on a harvester to quantify the yield for the crop being harvested on an instantaneous and averaging basis. The Yield Monitor data can be combined with the Plant/Seed Monitor data for effective evaluation of farming practices by soil type, variable rate application, nutrient program, etc.
GIS (Geographic Information Systems)
GIS (Geographic Information Systems) programs use GPS positions and Geo-referenced maps to capture/store data, analyze, and manage different layers of data. FMIS (Farm Management Information Systems) is a system which visually coordinates and manipulates data mainly from agricultural practices as well as provides general record keeping management. Uses data from MICS controllers and Remote sensed data to create and manage valuable datasets to prescribe decisions and work orders to a field with a MICS controller.
The act of detection and/or identification of an object, series of objects, or landscape without having the sensor in direct contact with the object. Common examples include Satellite imagery taken from space (100-500 miles above the earth) to determine vegetative growth and patterns in a field or Aerial imagery taken within airspace to determine vegetative growth and patterns in a field. (Examples: Planes, Unmanned Aerial Systems (UAS/Drones)
Other Typical Sensor Types:
Ground: Used to measure temperature, moisture, EC (electro conductivity), PH, organic matter, etc.
Weather: Ground or satellite based equipment used to measure temperature, moisture, sun reflectiveness, wind speed, etc.
Mobile: A hand held device or mounted to an implement which records infrared, temp, nitrogen, PH, and other signatures from a plant or soil.
How could precision ag work in crop insurance?
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Lu Lan (chin. upr. 卢兰, chin. trad. 盧蘭, pinyin Lú Lán; ur. 2 maja 1987 w Changzhou) – chińska badmintonistka, dwukrotna medalistka mistrzostw świata oraz dwukrotnie srebrna medalistka mistrzostw Azji.
Największym sukcesem badmintonistki jest złoty i brązowy medal mistrzostw świata w grze pojedynczej. Tytuł najlepszej zawodniczki zdobyła podczas turnieju w 2009 roku w Hajdarabad. Brązowy medal zdobyła w 2007 roku w stolicy Malezji. Na mistrzostwach Azji zdobyła dwa srebrne medale: w 2007 i 2011 roku.
Zwyciężczyni Polish Open w 2004 roku w grze pojedynczej.
Linki zewnętrzne
Profil zawodniczki na stronie BWF
Chińskie badmintonistki
Ludzie urodzeni w Changzhou
Urodzeni w 1987 | {
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Mužská dvouhra bett1Hulks Indoors 2020 probíhala okolo poloviny října 2020. Do singlové soutěže tenisového turnaje v Kolíně nad Rýnem, hraného na tvrdém povrchu Lanxess Arény, nastoupilo dvacet osm hráčů. Jednalo se o premiérový ročník turnaje dodatečně zařazeného do kalendáře okruhu ve formě náhrady za zrušené události kvůli koronavirové pandemii.
Vítězem se stal nejvýše nasazený Němec Alexander Zverev, jenž ve finále zdolal kanadskou turnajovou trojku Félixe Augera-Aliassimeho po dvousetovém průběhu 6–3 a 6–3. V probíhající sezóně si tak připsal premiérové turnajové vítězství, které představovalo dvanáctý singlový titul na okruhu ATP Tour. O týden později triumfoval rovněž v Kolíně nad Rýnem na navazujcím bett1Hulks Championship a potřetí v kariéře ovládl dvě události v řadě.
Nasazení hráčů
Čtyři nejvýše nasazení hráči obdrželi volný los do druhého kola.
Pavouk
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Externí odkazy
Bett1Hulks Indoors
Tenis v Německu v roce 2020
ATP Tour 2020 | {
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Концертна зала імені Ватросла́ва Лиси́нського () — концертна зала та виставковий центр у Загребі. Розташована на площі Степана Радича, між загребським Головним вокзалом та Вулицею міста Вуковара в районі Трнє. Названа на честь Ватрослава Лисинського — хорватського композитора середини XIX століття. Будівля має у своєму складі велику залу на 1841 місце та малу залу — на 305. У великій залі встановлено орган. Фоє часто використовується для проведення виставок.
Історія
Рішення про спорудження в Загребі нової багатофункціональної концертної зали було прийнято в 1957 році. Проєкт майбутньої будівлі розроблявся групою архітекторів під керівництвом Маріяна Хаберле. Будівельні роботи розпочалися в 1961 році, проте повінь у Загребі 1964 року і спричинені нею фінансові труднощі відтермінували закінчення будівництва майже на десятиліття. Відкриття зали відбулося тільки 29 вересня 1973 року .
1990 року в залі пройшов конкурс «Євробачення». Для його успішного проведення виконано першу реконструкцію будівлі. У 1992 році було повністю замінено дах. Також роботи з часткової реконструкції та заміни інтер'єрів проводилися в 1999 та 2009 роках
Сучасність
У концертній залі проводяться видовищні заходи всіх видів: від оперних, балетних і театральних постановок до концертів популярних музикантів. Так, у залі імені Лисинського регулярно виступають Мирослав Шкоро та Марко Перкович Томпсон. У ній також проводяться багато міжнародних конгресів і конференцій.
У перші 30 років існування зали її відвідало близько 10 мільйонів осіб. У 2007 році на площах зали пройшли близько 450 заходів, на яких побувало 760 тисяч осіб.
Примітки
Посилання
Koncertna dvorana Vatroslava Lisinskog
Культура Загреба
Концертні зали
Місця проведення Євробачення | {
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function sdk {
COMMAND="$1"
QUALIFIER="$2"
case "$COMMAND" in
l)
COMMAND="list";;
ls)
COMMAND="list";;
h)
COMMAND="help";;
v)
COMMAND="version";;
u)
COMMAND="use";;
i)
COMMAND="install";;
rm)
COMMAND="uninstall";;
c)
COMMAND="current";;
o)
COMMAND="outdated";;
d)
COMMAND="default";;
b)
COMMAND="broadcast";;
esac
#
# Various sanity checks and default settings
#
mkdir -p "$SDKMAN_DIR"
# Always presume internet availability
SDKMAN_AVAILABLE="true"
if [ -z "$SDKMAN_OFFLINE_MODE" ]; then
SDKMAN_OFFLINE_MODE="false"
fi
# ...unless proven otherwise
__sdkman_update_broadcast_and_service_availability
# Load the sdkman config if it exists.
if [ -f "${SDKMAN_DIR}/etc/config" ]; then
source "${SDKMAN_DIR}/etc/config"
fi
# no command provided
if [[ -z "$COMMAND" ]]; then
__sdk_help
return 1
fi
# Check if it is a valid command
CMD_FOUND=""
CMD_TARGET="${SDKMAN_DIR}/src/sdkman-${COMMAND}.sh"
if [[ -f "$CMD_TARGET" ]]; then
CMD_FOUND="$CMD_TARGET"
fi
# Check if it is a sourced function
CMD_TARGET="${SDKMAN_DIR}/ext/sdkman-${COMMAND}.sh"
if [[ -f "$CMD_TARGET" ]]; then
CMD_FOUND="$CMD_TARGET"
fi
# couldn't find the command
if [[ -z "$CMD_FOUND" ]]; then
echo "Invalid command: $COMMAND"
__sdk_help
fi
# Check whether the candidate exists
local sdkman_valid_candidate=$(echo ${SDKMAN_CANDIDATES[@]} | grep -w "$QUALIFIER")
if [[ -n "$QUALIFIER" && "$COMMAND" != "offline" && "$COMMAND" != "flush" && "$COMMAND" != "selfupdate" && -z "$sdkman_valid_candidate" ]]; then
echo -e "\nStop! $QUALIFIER is not a valid candidate."
return 1
fi
# Validate offline qualifier
if [[ "$COMMAND" == "offline" && -n "$QUALIFIER" && -z $(echo "enable disable" | grep -w "$QUALIFIER") ]]; then
echo -e "\nStop! $QUALIFIER is not a valid offline mode."
fi
# Check whether the command exists as an internal function...
#
# NOTE Internal commands use underscores rather than hyphens,
# hence the name conversion as the first step here.
CONVERTED_CMD_NAME=$(echo "$COMMAND" | tr '-' '_')
# Execute the requested command
if [ -n "$CMD_FOUND" ]; then
# It's available as a shell function
__sdk_"$CONVERTED_CMD_NAME" "$QUALIFIER" "$3" "$4"
fi
# Attempt upgrade after all is done
if [[ "$COMMAND" != "selfupdate" ]]; then
__sdkman_auto_update "$SDKMAN_REMOTE_VERSION" "$SDKMAN_VERSION"
fi
}
| {
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<?xml version="1.0" encoding="utf-8"?>
<merge xmlns:android="http://schemas.android.com/apk/res/android"
android:layout_width="match_parent"
android:layout_height="match_parent" >
<com.huewu.pla.lib.MultiColumnListView
xmlns:pla="http://schemas.android.com/apk/res-auto"
android:id="@+id/list"
android:layout_width="match_parent"
android:layout_height="match_parent"
pla:plaColumnNumber="2"
pla:plaLandscapeColumnNumber="3"
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Mensur Bajramović je rođen 15. kolovoza 1965. godine u Zenici, gdje je i započeo svoju košarkašku karijeru. Prve nastupe u timu Čelika ubilježio je 1980. godine, a u periodu od 1982-85. nastupao i za sarajevsku Bosnu. Nakon toga se vratio u Čelik, odakle je kasnije otišao igrati u Tursku i Hrvatsku.
Po povratku u rodnu Zenicu nastupao je još neko vrijeme u ekipi Čelika, da bi 1997. godine prestao aktivno igrati košarku. Po završetku igračke karijere Mensur Bajramović je ušao u trenerske vode i na početku nove karijere vodio svoj dotadašnji klub Zenica Čelik. U međuvremenu je određeni vremenski period proveo i kao izbornik mlade muške košarkaške reprezentacije BIH u sezoni 1999./2000., ali i na poziciji pomoćnika izbornika najbolje selekcije BiH u mandatima Sabita Hadžića i Draška Prodanovića. U sezoni 2004./2005. Bajramović je bio trener košarkaškog kluba Bosna ASA iz Sarajeva, s kojima je osvojio prvo mjesto u Ligi 10 prvenstva BiH, a vodio ih je i kroz Ligu 6. Mensur Bajramović je trenutno izbornik muške košarkaške reprezentacije Bosne i Hercegovine.
Bosanskohercegovački košarkaši
Životopisi, Zenica | {
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Q: How to connect two monitors to my lenovo g50-30 laptop
I have lenovo(g50-30) laptop with one HDMI and one VGA port
I want to connect two monitors to my laptop.
1st one connect with VGA cable and 2nd connect with HDMI.
I tried hard but can'tconnect two monitors at same time which displays three screens(laptop + VAG + HDMI). I can see either screens, laptop+VGA or laptop+HDMI.
How can i connect all three screens?
A: You can't. That laptop doesn't support your desired configuration.
You can use a "USB to VGA Adapter" to add extra monitors.
| {
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{"url":"http:\/\/forum.symthic.com\/off-topic\/3466-computer-thread\/index385.html?s=83bb92f23a33d83f218c95c1f46c64da9f38d175","text":"Welcome to symthic forums! We would love if you'd register!\nYou don't have to be expert in bit baking, everyone is more than welcome to join our community.\n\nYou are not logged in.\n\nRezmer\n\nPosts: 4,259\n\nDate of registration\n: Apr 6th 2012\n\nPlatform: PC\n\nLocation: From the heart of Europe.\n\nBattlelog:\n\nReputation modifier: 17\n\nTuesday, October 14th 2014, 3:38am\n\nSUE THEM\n[Aristocrat's Shoes]\n\n### Quoted from \"Darktan13\"\n\nTLDR -\nTeamwork is where players function by themselves, but their effectiveness is multiplied when they work together. Not a checklist of \"did we bring a healer so we can start playing?\"\n\nCannot into Rankine\n\nPosts: 1,235\n\nDate of registration\n: Dec 19th 2013\n\nPlatform: PC\n\nLocation: 45.4000\u00b0 N, 75.6667\u00b0 W\n\nBattlelog:\n\nReputation modifier: 8\n\nTuesday, October 14th 2014, 4:13am\n\nuninstalled that shitty killer software and did this\nApparently we do have a facebook... You should go ahead and like it\n\nIf only people would talk back it wouldn't be so lonely.[In B4 I get overwhelmed]\n\n\"Talk sense to a fool and he calls you foolish.\" - Euripides, Bacchae\n\n\"You are not entitled to your opinion. You are entitled to your informed opinion. No one is entitled to be ignorant.\"- Harlan Ellison\n\n### Quoted from \"LeGarcon\"\n\nHardline is a fun and sometimes silly Cops and Robbers sorta thing and I think that's great. Or it would be if it didn't suck.\n\n### Quoted from \"Rezal\"\n\nintel best ethernet controllers\n\nammo regen pls\n\n### Quoted from \"tehmoriz\"\n\nTL;DR: all vehicles in BF4 are screwed up in some way or another.\n\n### Quoted from \"Aenonar\"\n\nI know I shouldn't be surprised that something like that even exists but....\n\n### Quoted from \"NoctyrneSAGA\"\n\nRezmer\n\nPosts: 4,259\n\nDate of registration\n: Apr 6th 2012\n\nPlatform: PC\n\nLocation: From the heart of Europe.\n\nBattlelog:\n\nReputation modifier: 17\n\nTuesday, October 14th 2014, 4:50am\n\nintel best ethernet controllers\n[Aristocrat's Shoes]\n\n### Quoted from \"Darktan13\"\n\nTLDR -\nTeamwork is where players function by themselves, but their effectiveness is multiplied when they work together. Not a checklist of \"did we bring a healer so we can start playing?\"\n\nCannot into Rankine\n\nPosts: 1,235\n\nDate of registration\n: Dec 19th 2013\n\nPlatform: PC\n\nLocation: 45.4000\u00b0 N, 75.6667\u00b0 W\n\nBattlelog:\n\nReputation modifier: 8\n\nTuesday, October 14th 2014, 6:26am\n\n### Quoted from \"Rezal\"\n\nintel best ethernet controllers\n\nAgreed it's a good thing I got an Intel nic on my mobo.\nApparently we do have a facebook... You should go ahead and like it\n\nIf only people would talk back it wouldn't be so lonely.[In B4 I get overwhelmed]\n\n\"Talk sense to a fool and he calls you foolish.\" - Euripides, Bacchae\n\n\"You are not entitled to your opinion. You are entitled to your informed opinion. No one is entitled to be ignorant.\"- Harlan Ellison\n\n### Quoted from \"LeGarcon\"\n\nHardline is a fun and sometimes silly Cops and Robbers sorta thing and I think that's great. Or it would be if it didn't suck.\n\n### Quoted from \"Rezal\"\n\nintel best ethernet controllers\n\nammo regen pls\n\n### Quoted from \"tehmoriz\"\n\nTL;DR: all vehicles in BF4 are screwed up in some way or another.\n\n### Quoted from \"Aenonar\"\n\nI know I shouldn't be surprised that something like that even exists but....\n\n### Quoted from \"NoctyrneSAGA\"\n\nBe Creative.\n\nPosts: 7,811\n\nDate of registration\n: Mar 9th 2012\n\nPlatform: PC\n\nLocation: Portugal\n\nReputation modifier: 19\n\nTuesday, October 14th 2014, 9:18am\n\n### Quoted from \"Pheo\"\n\nI love how most of DDR4 on Amazon is just a bunch of overpriced Corsair crap.\n\nYou kind of had to know that going with the 5820K.\n\nI turn down for no one\n\nPosts: 6,877\n\nDate of registration\n: May 3rd 2012\n\nPlatform: PC\n\nLocation: Boston\n\nReputation modifier: 14\n\nTuesday, October 14th 2014, 2:42pm\n\n### Quoted from \"Pheo\"\n\nI love how most of DDR4 on Amazon is just a bunch of overpriced Corsair crap.\n\nYou kind of had to know that going with the 5820K.\n\nI did. I was also expecting them to actually stock G.Skill Ripjaws 4 in their warehouse by now instead of having 3rd party sellers trying to sell them for a extra $60+ dollars. Moderator Posts: 1,491 Date of registration : Sep 1st 2012 Platform: PC Location: Pittsburgh, Pennsylvania Battlelog: Reputation modifier: 14 Tuesday, October 14th 2014, 2:51pm ### Quoted from \"R\" ### Quoted from \"Arekan\" How bad good is my stuff? Keep in mind the video card was changed to a GTX 760. Looks good! GTX 760 is a good card. You certainly don't need 750W btw. You could probably save a couple bucks by getting a 550W or 600W PSU. You might wanna get a better CPU, especially since the 8 core AMDs have been coming down in price recently. fx8320 is only 143.99, so not even 20$ more than fx6350\n\n@Sheepnub\nI have a VS229H-P and I would recommend it. Great monitor.\n\nFor as cheap as it is he should probably keep the 750W PSU. I have a 700 and my PC only uses 341 right now, but if I were to ever change the CPU that would go up, and i'm running an Fx-8120. Speaking of which, the prices on them have gone down, but there was never a huge difference between the 8-core's and the rest of AMD's processors. They're a little better at multitasking, which is what AMD has always been good at, but they run a lot hotter, mine averages about 59C under full load and safe max temp on that chip is rated at 61C. That is with a stock cooler of course, but they're still hot.\n\ntl;dr: If you want to save money cut the PSU but I wouldn't recommend it, the difference between AMD 8 and 6 cores is very small\n\nHow to Ignore users - Symthic Forum Rules\n\n### Collection of hilarious numbers.\n\nRocktopus\n\nPosts: 2,657\n\nDate of registration\n: Feb 16th 2013\n\nPlatform: PC\n\nReputation modifier: 9\n\nSunday, October 26th 2014, 6:27am\n\nWhat type of mouse pad do you guys prefer and why? I used to have a plastic one several years ago, but it got sort of scratched after a few years and I got rid of it. Now I have Steelseries cloth one, but it's like a magnet for dust, cat hair and whatever. It also feels like it slows hand movement a bit.\nSo what do you prefer?\n\nI turn down for no one\n\nPosts: 6,877\n\nDate of registration\n: May 3rd 2012\n\nPlatform: PC\n\nLocation: Boston\n\nReputation modifier: 14\n\nSunday, October 26th 2014, 6:40am\n\nCloth. I hate using hard ones. I want to try and get a Aritsan Hayate.\n\nThe only man who ever reported his own post!\n\nPosts: 2,980\n\nDate of registration\n: Jul 19th 2012\n\nPlatform: PC\n\nLocation: Grenoble, France\n\nBattlelog:\n\nReputation modifier: 14\n\nSunday, October 26th 2014, 11:42am\n\nI'm happy with Razer Goliathus Speed\nMy cat as well, given how he like to sleep on it while I play makes playing more challenging\n\nSent from phone.\nStats thingy\n\n### Quoted from \"Cheapnub\"\n\nI'm a battlefield player, good sir. I don't play metro.\n\n### Quoted from \"Suiizide\"\n\nPC is no longer PC master race. It's PC mustard race, because consoles need to ketchup :'D\n\n### Quoted from \"Riesig\"\n\nDICE gave so much into making commander better, but lemmings be lemmings I guess.\n\n### Quoted from \"MrT3a\"\n\nAs a good guy that don't want to use overly glitched weapons, I'll quit using the MTAR and switch to the ACWR until it's fixed\n\nThe world needs more people like you\n\n+1, I think we're all in agreement that more MrT3as would be an awesome thing\nAlthough if that was the case they'd use up so much of the world's awesome that there'd be none left for the rest of us!\n\n### Quoted from \"CobaltRose\"\n\nyes, I know, I'm a big-ass hypocrite\n\n11 guests","date":"2018-08-18 08:29:53","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\": 0.17282886803150177, \"perplexity\": 7394.414347476359}, \"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-2018-34\/segments\/1534221213508.60\/warc\/CC-MAIN-20180818075745-20180818095745-00134.warc.gz\"}"} | null | null |
Il Giro del Canavese era una corsa in linea maschile di ciclismo su strada riservata agli Under-23 che si svolgeva a Valperga, in Italia. Svoltosi dal 1992 al 2014, dal 2005 al 2008 fece parte del circuito UCI Europe Tour come gara di classe 1.2U.
Albo d'oro
Aggiornato all'edizione 2008.
Note
Competizioni ciclistiche italiane
Valperga
Sport nella città metropolitana di Torino
Canavese
Ciclismo in Piemonte | {
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{"url":"https:\/\/indico.cern.ch\/event\/848732\/contributions\/4529337\/","text":"# The 16th International Workshop on Tau Lepton Physics (TAU2021) (Virtual Edition)\n\nSeptember 27, 2021 to October 1, 2021\nIndiana University\nAmerica\/Indiana\/Indianapolis timezone\n\n## Exploring Neutrino Masses and Mixing in the Seesaw Model with $L_e-L_\\tau$ Gauged Symmetry\n\nOct 1, 2021, 10:50 AM\n2h\nVirtual (Indiana University)\n\n### Virtual\n\n#### Indiana University\n\nPoster contribution Tau2021 Abstracts\n\n### Description\n\nIn the poster, we have taken $L_e-L_\\tau$ gauge symmetry to study neutrino phenomenology in the framework of type-(I+II) seesaw mechanism. In the model, three heavy right-handed neutrinos, a scalar singlet, and one scalar triplet are added to the Standard Model. As a result, the active neutrino-mass matrix has a two-zero $A_1$ texture which helps explain neutrino oscillation parameters like $\\theta_{13}, \\theta_{23}. \\theta_{12},$ the sum of active neutrino masses etc. The model also explains neutrinoless double $\\beta$ decay and lepton flavor violation with reasonable accuracy. The branching ratio of $\\tau \\rightarrow e \\gamma$ and $\\tau \\rightarrow \\mu \\bar{\\mu} \\mu$ also stay well below the experimental upper bound.\n\nWhat is your topic? Neutrino Physics","date":"2022-05-28 07:47:37","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\": 0.7634655237197876, \"perplexity\": 2795.4385793467204}, \"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-2022-21\/segments\/1652663013003.96\/warc\/CC-MAIN-20220528062047-20220528092047-00202.warc.gz\"}"} | null | null |
{"url":"https:\/\/www.expatbridge.nl\/Feb-22+27081.html","text":"# calcium metal water reaction\n\nWater Hardness - Department of Chemistry\n\nThe water is treated with a coination of slaked lime, Ca(OH) 2, and soda ash, Na 2 CO 3. Calcium precipitates as CaCO 3, and magnesium precipitates as Mg(OH) 2. These solids can be collected, thus removing the scale-forming ions from the water supply. .\n\nReactions of metals - Metals - National 5 Chemistry \u2026\n\n17\/8\/2020\u00b7 When metals react with water, metal hydroxides and hydrogen gas are formed. This can be represented in different ways as shown: Word equation \\[Calcium + water \\to calcium\\,hydroxide + \u2026\n\nABC-R Lab #1 Test Calcium Reactions Name Reaction #1 Calcium metal reacts with the weak acid water.\n\nReaction #1 Calcium metal reacts with the weak acid water. Ca (s) + 2 H 2 O (l) Ca(OH) 2 (s) + H 2 (g) 1. Assign oxidation nuers. Write the nuer above the element in the equation\n\nHow to Mix Calcium Chloride and Water | Sciencing\n\nCalcium chloride is a chemical compound made up of calcium ions and chlorine ions. The ions are held together by an ionic, or weak salt bond. Mixing calcium chloride with water is an exothermic reaction, which means that the coination of the two substances\n\nMetal ions in aqueous solution - Wikipedia\n\nA metal ion in aqueous solution or aqua ion is a ion, dissolved in water, of chemical formula [M(H 2 O) n] z+.The solvation nuer, n, determined by a variety of experimental methods is 4 for Li + and Be 2+ and 6 for elements in periods 3 and 4 of the periodic table.\n\nCalcium Inhibition of Ribonuclease H1 Two-Metal Ion \u2026\n\n26\/2\/2014\u00b7 The metal ions facilitate an S N 2-type reaction, in which the nucleophilic \u201cattacking\u201d oxygen replaces the \u201cleaving\u201d oxygen in the broken P\u2013O bond of the phosphate on the opposite side. 4,5 Similar Mg 2+ ion selectivity has been found among many TMC enzymes 6\n\nCalcium - Periodic table\n\nCalcium is reactive and, for a metal, soft. With a bit of effort, it can be cut with a sharp knife. In contact with air, calcium develops a mixed oxide and nitride coating, which protects it from further corrosion. Calcium reacts easily with water and acids and the metal\n\nphysical appearance of calcium metal in cameroon\n\nSodium, Chemical Element - reaction, water, uses, \u2026 Physical properties Sodium is a silvery-white metal with a waxy appearance. It is soft enough to be cut with a knife. The surface is bright and shiny when first cut, but quickly becomes dull as sodium reacts with\n\nTransformation of meta-stable calcium silie hydrates \u2026\n\nAs a first step, we investigate reaction rates and mechanistic pathways for cement mineral growth in the absence of CO2 by coupling water chemistry with XRD and NMR spectroscopic data. We find that semi-crystalline calcium (alumino-)silie hydrate (Al-CSH) forms as a \u2026\n\nEquilibrium in the reaction of barium with calcium chloride\n\nThe reaction of calcium with strontium chloride has been studied by Ostertage (11). The experimental method which was used was similar to that used by Rinck (10) in studying the s odium-potassium; systems. Steel capsules were used to contain the lyzed.\n\nExplain why calcium metal after reacting with water \u2026\n\nWhen calcium metal is added to water, the gas evolved does not ch fire but the same gas evolved on adding potassiun metal to water ches fire. Explain why ? asked Oct 30, 2017 in Class X Science by priya12 ( -12,635 points)\n\nThe reaction of calcium oxide with water will yield _____. \u2026\n\nWhen calcium oxide (chemical formula: CaO) reacts with water (chemical formula: H2O), the following reaction takes place: CaO + H_2O -> Ca(OH)_2 The product of this reaction is calcium hydroxide\n\nMore sf in water, this time calcium metal. Not as \u2026\n\nNo memes, rage comics, image macros, reaction gifs, or other \"zero-content\" material. Ever. Likewise, simple pictures of uninteresting and garden variety chemistry-related things are not appreciated. If a caption or explanation is included this helps, but please\n\ncalcium metal burns in oxygen gas to form solid process\n\ndimers, Al2Cl6, in the solid state {due to Calcium burns in oxygen with a brick-red flamewith a green flame to form simple metal oxides Coustion 2014516- or gas, but is usually a gas (air) for Soot, for example, is a form of solid exhaust a fuel to be burned, a source of oxygen,\n\nCalcium + Calcium Nitrate \u2192 ? | Physics Forums\n\n23\/2\/2005\u00b7 I don''t believe that calcium hydroxide can be formed by simply dissolving calcium nitrate in water. And thus the point of dissolving calcium metal. You should recall the reaction where a metal dissolved in water will produce hydrogen gas and its corresponding basic oxide in this case $CaO_{(s)}$ which will react with water to form calcium hydroxide.\n\nMetals Non-metals Reaction of Metals with water class \u2026\n\nReaction of potassium metal with water: Potassium metal forms potassium hydroxide and liberates hydrogen gas along with lot of heat when it reacts with water. K + H 2 O KOH + H 2 Reaction of calcium metal with water: Calcium forms calcium hydroxide along with hydrogen gas and heat when it reacts with water\u2026\n\nWhat Is Calcium Carbonate''s Reaction With Hydrochloric \u2026\n\n2\/4\/2020\u00b7 Calcium carbonate reacts with hydrochloric acid to form calcium chloride, water and carbon dioxide. The reaction between these two compounds requires two parts hydrochloric acid to one part calcium chloride. This reaction is fairly rapid and energetic at high\n\nWhat Metals React With Water to Produce Hydrogen? | \u2026\n\nWhen sodium metal reacts with water, the resulting heat melts the metal almost immediately into a grey-silver ball. The hydrogen gas evolved during this reaction propels the ball rapidly across the surface of the water, leaving a white trail of sodium hydroxide that eventually dissolves into a clear solution.\n\nCARBONATE EQUILIBRIA - Department of Land, Air and Water \u2026\n\n1. The solution pH of water in equilibrium with carbon dioxide and essentially devoid of other controlling species. 2. The reaction of calcium carbonate saturated solutions with free access to carbon dioxide. In essence this is the equilibrium of lime with air or soil\n\nCalcium Oxide and its Exothermic Reactions\n\nI conducted tests to assess if the exothermic reaction of calcium oxide and water could ignite straw. A layer of clean, dry straw was placed on a metal tray. A pile of 500g of calcium oxide was placed on the straw, and then 500ml of water was added.\n\nCalcium - Compounds | Britannica\n\nCalcium - Calcium - Compounds: The most important calcium compound is calcium carbonate, CaCO3, the major constituent of limestone, marble, chalk, oyster shells, and corals. Calcium carbonate obtained from its natural sources is used as a filler in a variety of products, such as ceramics, glass, plastics, and paint, and as a starting material for the production of calcium oxide. Synthetic\n\nGCSE CHEMISTRY - How do the Alkaline Earth Metals \u2026\n\nThe Periodic Table The Alkaline Earth Metals - Reaction with Water. How does Magnesium React with Water? Magnesium will not react with cold water. Even finely powdered magnesium reacts only very slowly. Magnesium will react with gaseous water (steam) to form magnesium oxide and hydrogen gas.\n\nSolubility Product for Calcium Hydroxide\n\nA saturated solution of Ca(OH) 2 can be prepared by the reaction of calcium metal with water. Calcium is oxidized by water, yielding calcium hydroxide and hydrogen gas. Ca(s) + 2 H 2O(l) !Ca(OH) 2 (s) + H 2(g) (4) PROCEDURE 1 Check out any required\n\nThe reaction between calcium and water in more detail \u2026\n\nThe reaction between calcium and water in more detail Note: You will not need to know this information but it may help to provide you with a better understanding of what takes place during this reaction. As calcium is a group 2 metal, it has two electrons in its outer\n\nSolution: Calcium metal reacts with water | Clutch Prep\n\nProblem: Calcium metal reacts with water to form calcium hydroxide and hydrogen gas. How much hydrogen is formed when 0.50 g of calcium are added to water? 1. 0.050 g 2. 0.025 g 3. 0.50 g 4. 0.10 g \ud83e\udd13 Based on our data, we think this question is relevant for","date":"2021-04-11 09:21:48","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 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\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.6663050055503845, \"perplexity\": 5467.305397500004}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-17\/segments\/1618038061820.19\/warc\/CC-MAIN-20210411085610-20210411115610-00065.warc.gz\"}"} | null | null |
Brown Rice and Kerosine is the third album by Australian folk-rock group Redgum. The title is taken from the first track, and the album was released around the time Redgum changed from a part-time band to a full-time job for its members.
"100 Years On" was released as a single. As noted on a sticker on the cover, the song "Liberal Values" was to have been included on the album but was removed for legal reasons. It was a parody of Bacharach/David's "(The Man Who Shot) Liberty Valance." In April 2019, a recording of a live performance of "Liberal Values" from 1980 was uploaded to YouTube.
It was originally released as a record and was very briefly available on CD in the late 80s. Some tracks were included on the 2004 collection Against the Grain.
Track listing
Side A
"100 Years On" (J. Schumann)
"Lear Jets Over Kulgera" (M. Atkinson)
"Caught in the Act" (M. Atkinson/J. Schumann/V. Truman/C. Timms)
"Yarralumla Wine" (M. Atkinson)
"Where Ya Gonna Run to" (J. Schumann)
Side B
"Brown Rice and Kerosine" (M. Atkinson)
"The Federal Two-Ring Circus" (M. Atkinson)
"Your O.S. Trip" (M. Atkinson)
"The Last Frontier" (J. Schumann)
"Parramatta Gaol 1843" (M. Atkinson/V. Truman)
Charts
References
1981 albums
Redgum albums | {
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HomeNewsForty Years of Women at Stevens: First Female Students Pioneered the University's Future
9 Jan 2012 Campus & Community
Forty Years of Women at Stevens: First Female Students Pioneered the University's Future
When Martha Connolly '75 was mulling her many college acceptances in the spring of 1971, the prospect of achieving landmark status at Stevens Institute of Technology, which was admitting women as undergraduates for the first time 101 years after its founding, was certainly enticing.
"The opportunity to break ground – to be the first – was a draw," she recalls.
But what closed the deal for Connolly was something else entirely – the personal attention of Robert Seavy, Dean of Admissions, who took her on a private tour of the campus and astounded her by knowing the names and majors of every student he encountered.
"This was clearly a place that really focused on undergraduate education," said Connolly.
When she arrived at Stevens that fall, Connolly was one of 19 women to join the Class of 1975, and quickly learned that the campus had been forewarned. The Indicator had signaled the special status of these pioneering freshmen by publishing their photographs in that fall's edition of the magazine. In classes, she was often the only woman.
"I would be the first name the professor learned, and thus the first to be called to the blackboard. And I could never cut class. They'd know, of course," she said with a smile.
That first year, the women shared rooms in a relatively luxurious building set aside for married students and received invites galore for parties across campus, including on the decommissioned World War II transport ship, the SS Stevens, which was anchored in the Hudson River and functioned as a dormitory.
For full coverage of the celebration of the 40th anniversary of Stevens becoming fully coeducational, visit Women at Stevens.
"They really rolled out the red carpet for us," said Connolly, who earned a bachelor's and master's degree in four years at the university.
Lenore Schupak '74, who enrolled in Stevens with the first class of women but took so many classes that she earned her degree in only three years and became the university's first female graduate, believes her classmates were special for reasons besides their gender.
"I think Dean Seavy went through a very careful selection process," said Schupak. "He wanted to make sure the first women not only had the academic credentials to succeed, but were independent-minded and able to think on their feet."
The university found ways to engage its first female students in areas outside of academics. While there were no women's sports teams in 1975, the coaches created opportunities to involve the new freshmen.
"I was interested in racquet sports and was asked to join the squash and tennis teams as manager," said Schupak, who also joined the yacht club and was a photographer for The Stute. "I even got to practice with the boy's tennis team. It was a very well-rounded experience."
Women engineers were almost unheard of in the early 1970s. The year Stevens admitted them, a mere 361 women across the country had earned undergraduate degrees in engineering, and women accounted for less than one percent of the Ph.D. students to receive doctorates in engineering, according to data from the National Science Foundation.
Despite the odds, both Connolly and Schupak excelled at Stevens. Both women went on to pioneering careers in their respective fields – biomedical engineering and biosciences for Connolly and environmental engineering for Schupak. Connolly went on to become the first female graduate of Johns Hopkins University's biomedical engineering doctoral program, and she currently heads the Maryland Technology Enterprise Institute (Mtech) Maryland Industrial Partners Program at the University of Maryland, where she has spent her career fostering the state's bioscience industry. Schupak also continued to be a trailblazer, working at an early alternative energy start-up and eventually for General Motors, helping the company implement early environmental compliance rules.
Spring forward 40 years and Stevens is a vastly different place, its landscape altered in part by the many talented and ambitious women who followed in the footsteps of these pioneers. Women now make up 25 percent of undergraduates and occupy many leadership roles on campus.
This year, for example, the chair of the Honor Board, the head of Gear and Triangle, the honor society, and the editor of the yearbook are all women. And there are now a total of 13 women's athletic teams on campus, all quite renowned. The strength of the programs allows Stevens to recruit scholar-athletes from all over the country, including Laura Barito '11, a mechanical engineering major who was recently chosen as NCAA Woman of the Year.
"I felt like I fit in right away," Barito said. "The attitude toward men's and women's athletic teams was very equal. All told, this was one of the most balanced experiences you could find in engineering."
Senior Kendra Appleheimer, who served as vice president of the Student Government Association last year under a fellow female president and has held impressive internships at ITT, the Metropolitan Transportation Authority and Air Cruisers, agreed that women are highly influential on campus today.
"I do feel our impact is significant," she said. "Those trailblazing women who were the first engineering students here started this tradition."
Women are so well integrated on campus today that most of the time Appleheimer barely notices that they are still a minority. During her internship at Air Cruisers, she was the only woman on a large project team, but didn't realize it until someone pointed it out two weeks after she started.
"We're all working toward the same goal and the sense of community is so strong here that I don't even think about gender," she said.
If they are anything like their predecessors from 40 years ago, Appleheimer, Barito and their fellow female students can expect great success upon graduating from Stevens – by any standards. | {
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Nationwide Mutual Insurance v. Philadelphia Electric Co.
Temple Univ. Hosp., Inc. v. United States
The Hospital relies primarily on Nationwide Mut. Ins. Co. Philadelphia Elec. Co which did not arise under the…
Sanchez v. City of Espanola
The claim of the plaintiff was based on a tort and she was privileged to exact payment of the whole sum from…
Full title:NATIONWIDE MUTUAL INSURANCE COMPANY v. PHILADELPHIA ELECTRIC CO. and Carr…
Court:United States District Court, E.D. Pennsylvania
Date published: Nov 30, 1977
443 F. Supp. 1140 (E.D. Pa. 1977)
United States District Court, E.D. Pennsylvania
entering judgment in plaintiff insurer's favor and against joint tortfeasors on share of settlement including interest and costs, but not attorney's fees
Summary of this case from Temple Univ. Hosp., Inc. v. United States
Civ. A. No. 76-14.
Christopher C. Fallon, Jr., Philadelphia, Pa., for plaintiff.
Barry F. Greenberg, Philadelphia, Pa., for Phila. Electric.
David F. Campbell, Philadelphia, Pa., for Carr Duff.
Henry J. Lotto, Philadelphia, Pa., for Bellmont.
RAYMOND J. BRODERICK, District Judge.
In this action based on diversity of citizenship, the plaintiff, Nationwide Mutual Insurance Company (Nationwide) brought suit against Philadelphia Electric Company (PECO), Carr Duff Electrical Pole Line Construction, Inc. (Carr Duff), Bellmont Fund, Inc. (Bellmont), and Curtis Jones (Jones) to recover contribution pursuant to the Pennsylvania Uniform Contribution Among Joint Tortfeasors Act.
This action was tried before a jury which returned a verdict in favor of the plaintiff against Bellmont and Carr Duff. The Court granted PECO's motion for a directed verdict at the close of the plaintiff's case. At the close of all the evidence the Court granted Jones' motion for a directed verdict. Bellmont has moved this Court for a new trial and/or for a judgment n.o.v.; the plaintiff has moved for an award of interest; Carr Duff has moved for a new trial; and all parties have requested the Court to mold the verdict. After carefully considering the grounds urged in these motions, the Court has determined that it must deny Bellmont's motion for judgment n.o.v., grant plaintiff's motion for an award of interest, and deny Carr Duff's motion for a new trial and/or a judgment n.o.v.
The parties stipulated to the following facts: Sometime during the year 1966, Bellmont engaged Jones to construct an apartment building on a site located near the intersection of Red Lion and Clare Roads in Philadelphia. Bellmont was the owner of the project and Jones was the general contractor. In order to provide electric power for workers at the construction site, PECO was requested to install temporary electric service. In response to this request, on October 20, 1966 PECO installed a transformer on a pole on the north side of Red Lion Road which reduced the voltage normally carried on the overhead lines along Red Lion Road so that electricity could be provided at the construction site. After this service was provided, Bellmont and/or Jones complained to PECO that they were not able to secure sufficient electric power to operate their power tools, and asked PECO for an estimate to relocate the electric service to a point closer to the work being performed. At the request of Bellmont and/or Jones, Carr Duff, an electric pole and line contractor, installed three power poles at the site and strung wire between them. Thereafter, PECO connected the wires erected by Carr Duff with the power lines running along Red Lion Road and changed the location of the transformer.
It was further stipulated that on March 10, 1967, Altman Brothers, a carpentry subcontractor working on the construction site, engaged plaintiff Nationwide's insured, Walter Hinkle, to lift wood. Hinkle dispatched a hydraulic truck crane to the job site. At about 2:30 p.m., the cable on the crane became tangled. As George Baker and others worked to untangle the cable, the cable came in contact with overhead power lines previously installed by Carr Duff. George Baker was electrocuted and subsequently died. Baker's estate sued Walter Hinkle (insured by plaintiff), PECO, Carr Duff, Bellmont Fund and Curtis Jones. Shortly before the commencement of the trial of the tort action, Nationwide, acting on behalf of its insured, Walter Hinkle, settled the tort action with the estate of Baker for $75,000, an amount which all parties agreed was reasonable. Although Nationwide had obtained a release running in favor of all the defendants, none of the present defendants chose to voluntarily contribute to the settlement. (N.T. 1-47-51).
At trial before this Court, plaintiff contended that not only was its insured, Hinkle, negligent, but that the defendants were also negligent and that their negligence was a proximate cause of the accident. As a joint tortfeasor, plaintiff claimed the right to contribution from its fellow joint tortfeasors pursuant to the Uniform Contribution Among Joint Tortfeasors Act, 12 P.S. § 2082 et seq. This Act provides:
§ 2082.
For the purpose of this act, the term "joint tortfeasors" means two or more persons jointly or severally liable in tort for the same injury to persons or property, whether or not judgment has been recovered against all or some of them.
(1) The right of contribution exists among joint tortfeasors; (2) A joint tortfeasor is not entitled to a money judgment for contribution until he has by payment discharged the common liability or has paid more than his pro rata share thereof; (3) A joint tortfeasor who enters into a settlement with the injured person is not entitled to recover contribution from another joint tortfeasor whose liability to the injured person is not extinguished by the settlement.
As stated by Justice Eagen in Swartz v. Sunderlund, 403 Pa. 222, 225, 169 A.2d 289, 291 (1961):
. . . a reading of the act signifies that only two conditions must exist before the right of contribution arises, namely, (1) that one joint tort-feasor has discharged the common liability or paid more than his prorata share; (2) that the liability of the other joint tort-feasor to the injured persons has been extinguished by the settlement.
In addition, as stated by Judge Scalera in W.D. Rubright v. International Harvester Co., 358 F. Supp. 1388, 1392 (W.D.Pa. 1973):
When . . . settlement occurs before the injured plaintiff has proven his original case at trial, the settling tortfeasor cannot enforce his right to contribution unless in a separate proceeding he proves that:
(1) The settlement figure was reasonable.
(2) The parties from whom he seeks contribution were in fact joint tortfeasors. Swartz v. Sunderland, 403 Pa. 222, 169 A.2d 289 (1961), Restatement Restitution § 86(d).
At trial the defendants did not question the reasonableness of the settlement. It was stipulated that "reasonableness" had been agreed to at the time the settlement was made. It was also stipulated that the liability of all defendants had been extinguished by the plaintiff's payment at settlement. The sole question presented to the jury was whether the defendants were in fact joint tortfeasors, the parties having agreed that the Court should mold the verdict. The jury returned a verdict in favor of the plaintiff, Nationwide, against the defendants, Carr Duff, and the defendant, Bellmont.
Bellmont's Motion For A Judgment n.o.v.
In the instant case, plaintiff, Nationwide, seeks to enforce its right to contribution against both Bellmont and Carr Duff. Bellmont was the owner of property upon which an apartment building was to be constructed. Bellmont engaged Jones as general contractor for the construction of the apartment. Bellmont in turn engaged defendant Carr Duff, an electrical pole and line contractor, for the installation of three poles upon which Carr Duff strung electrical power lines. These lines were subsequently connected by PECO to its power lines running along a road adjacent to the construction site. It was these lines with which the deceased came in contact and was electrocuted. Evidence was presented to the jury that Carr Duff had failed to comply with the National Electrical Code by failing to insulate these wires.
Plaintiff Nationwide based its case against Bellmont solely upon Restatement (Second) of Torts, § 416 which provides:
One who employs an independent contractor to do work which the employer should recognize as likely to create during its progress a peculiar risk of physical harm to others unless special precautions are taken, is subject to liability for physical harm caused to them by the failure of the contractor to exercise reasonable care to take such precautions, even though the employer has provided for such precautions in the contract or otherwise.
The charge to the jury as to Bellmont's liability was based solely on § 416. And, as we have heretofore pointed out, the jury returned a verdict against Bellmont. Bellmont now seeks a judgment n.o.v. on the ground that § 416 was not applicable. We hold that § 416 is applicable and deny Bellmont's motion for judgment n.o.v.
Section 416 was adopted in Pennsylvania in Philadelphia Electric Co. v. Julian, 425 Pa. 217, 228 A.2d 669 (1967). This section provides an exception to the general rule that one who engages an independent contractor is not responsible for the negligent acts or omissions of the contractor or its employees. As stated by the Pennsylvania Superior Court in McDonough v. United States Steel, 228 Pa. Super. 268, 324 A.2d 542 (1974),
When an employer has exercised care in choosing a careful and competent contractor to do work on the employer's premises, and has entrusted the control and possession of the premises, and the performance of the task to that contractor, the employer is generally shielded from liability to third parties due to the negligence of the contractor. An employer, however, must use reasonable care to make the premises safe, or warn the contractor of any dangerous condition thereon. Grace v. Henry Disston Sons, Inc., 369 Pa. 265, 85 A.2d 118 (1952).
There are, however, exceptions to this general rule of non-liability. . . . The exception does not rest upon any personal negligence of the employer, but is a rule of vicarious liability. The rule is stated in Section 416 of the Restatement of Torts, 2d [supra].
324 A.2d at 545.
Judge Hoffman continued, stating the situations in which this exception would apply:
Section 416 is . . . applicable only to situations in which the negligence of the independent contractor consists of the failure to take the precautions necessary for the safe performance of a task. The risk of harm must arise from the peculiar or inherent nature of the task or the manner of performance, and not the ordinary negligence which might attend the performance of any task. "Liability does not ordinarily extend to so called `collateral' or `casual' negligence on the part of the contractor or his servants in the performance of the operative details of the work. The negligence for which the employer [of a general contractor] is liable . . . must be such as is intimately connected with the work authorized and such as is reasonably likely from its nature." Van Arsdale v. Hollinger, 68 Cal.2d 245, 252, 66 Cal.Rptr. 20, 24, 437 P.2d 508, 512 (1968) (quoting Harper, Law of Torts (1933) § 292); see Thorne v. United States, 479 F.2d 804 (9th Cir. 1973). Prosser has stated that the principle is applicable "to work in which there is a high degree of risk in relation to the particular surroundings, or some rather specific risks or set of risks to those in the vicinity . . . The emphasis is on the peculiar character of the risk, and the need for special care." Prosser, Law of Torts, 3rd Ed. (p. 486).
324 A.2d at 546. Likewise in Jacobini v. IBM, 57 Pa. D. C.2d 8 (1972), the court stated that § 416 is applicable only to injuries resulting from conditions bearing an unusual risk of physical injury.
As stated in McDonough, supra, and Moss v. Swann Oil Co., 423 F. Supp. 1280 (E.D.Pa. 1977), for this vicarious liability to apply, it is first necessary that the independent contractor commit an act of negligence which causes the plaintiff's injury. Here, the jury found that Carr Duff, an independent contractor, was negligent in failing to insulate the wires and that this negligence was the proximate cause of George Baker's death. In imposing § 416 liability it is also necessary that the work in question presents a peculiar risk of harm unless special precautions are taken. McDonough and Moss, supra, Restatement (Second) of Torts, § 416, Comment (d).
Restatement (Second) of Torts, § 416, Comment (d) provides in part:
In order for the rule stated in this Section to apply, it is not essential that the work which the contractor is employed to do be in itself an extra-hazardous or abnormally dangerous activity, or that it involve a very high degree of risk to those in the vicinity. It is sufficient that it is likely to involve a peculiar risk of physical harm unless special precautions are taken, even though the risk is not abnormally great. A "peculiar risk" is a risk differing from the common risks to which persons in general are commonly subjected by the ordinary forms of negligence which are usual in the community. It must involve some special hazard resulting from the nature of the work done, which calls for special precautions.
Based on these considerations, the evidence presented to the jury was more than sufficient to support the verdict against Bellmont.
In molding the verdict, however, we must next consider the question of whether Nationwide is entitled to contribution from both Bellmont and Carr Duff. Nationwide claims that the $75,000 it paid the estate of George Baker was in settlement of its own, Bellmont's and Carr Duff's liability and that each must contribute a one-third share. We disagree.
In this action for contribution pursuant to the Pennsylvania Uniform Contribution Among Joint Tortfeasors Act, supra, the parties agree that Pennsylvania law must be applied.
In Parker v. Rodgers, 125 Pa. Super. 48, 189 A. 693 (1937), the Pennsylvania Superior Court held in an analogous situation that where an employee is found negligent, the employer's share is identical to the employees in that the employer has not committed the tort but was liable solely on the basis of the respondeat superior doctrine.
Parker was a tort action to recover damages suffered in an automobile collision wherein the jury found that Mr. Rodgers, who owned and operated one of the automobiles, Glenzenger, the employee of a Mrs. Paugh, who was operating the second automobile, and Mrs. Paugh were, all three, liable. Rodgers' insurance carrier attempted to obtain contribution in the action on the theory that as among the three defendants, each was liable for one-third of the judgment. The court held that Mrs. Paugh and her driver were liable for only one-half of the judgment. In arriving at this conclusion the court reasoned that Mrs. Paugh was liable only for the negligence of her driver on the principle of respondeat superior. The court said:
The question now raised arises by reason of the fact that the three judgment debtors here involved were not in the same class. The rights of the plaintiff against the defendants and the rights of the defendants inter sese are not the same. The claim of the plaintiff was based on a tort and she was privileged to exact payment of the whole sum from any one or more of the defendants, while the claims of the defendants among themselves are founded on the application of equitable principles and do not sound in tort. . . .
This case, as we have said, is to be differentiated from others by the fact that the defendants are not all in the same class. . . .
Rodgers and Glenzinger were the persons who committed the actual tort or wrong as distinguished from a legal tort. Mrs. Paugh was not present when the accident occurred, but was only liable as the master of her servant and driver, Glenzinger. It was not alleged nor shown that she was personally and directly guilty of any trespass or was responsible by reason of anything which she personally did or omitted to do. Her responsibility even to the plaintiff did not arise from any act which was on her part morally wrong, but her liability was based on a legal principle that has become a part of the positive law of the Commonwealth that the negligence of a servant acting within the scope of his employment is imputed to the master. We agree with the learned judge of the court below that Mrs. Paugh's liability being purely derivative and because she and the driver are responsible to the plaintiff for one and the same act of negligence committed by the servant alone, both reason and justice require it to be held that Rodgers is not entitled to contribution since he has paid but one-half of the judgment.
In this connection it is to be observed that as between Mrs. Paugh and her driver she was entitled to recover from the driver the amount she was compelled to pay through his sole fault.
125 Pa. Super. 52, 53, 189 A. at 695-696.
In Russell v. United States, 113 F. Supp. 353 (M.D.Pa. 1953), plaintiff brought suit for damages done to his home as a result of a collision between a tractor-trailer and a United States Post Office bus. The United States was sued as the owner of the bus and as the employer of its driver. Also sued were the driver of the tractor-trailer, the owner of the tractor and the owner of the trailer. In a decision by Judge Follmer, the court found that Mr. Shank had operated the bus negligently and that Mr. Mitcheltree had negligently operated the tractor-trailer and that both were the proximate cause of the damage to the plaintiff's house when it was rammed by the tractor-trailer. The court also found that Mitcheltree was the employee of both Pitoniak, the owner of the tractor, and Mushroom, the owner of the trailer. The court, relying upon Parker, supra and Fisher v. Diehl, 156 Pa. Super. 476, 40 A.2d 912 (1945), resolved the issue of contribution among the joint tortfeasors as follows:
Since Mitcheltree, whose liability is primary, is the agent of the other third-party defendants, the right of contribution of the United States of America, which is governed by the law of Pennsylvania, is for one-half of the judgment.
The defendant, United States of America, is entitled to contribution from Theodore L. Mitcheltree, Cyril A. Pitoniak and the Mushroom Transportation Company, Inc., for one-half of the damage allotted to the plaintiffs in this case, to wit, $536.50.
113 F. Supp. at 356-357.
The liability of the defendant, Bellmont, is a vicarious liability based on § 416. In a contribution action such as this, Bellmont's share of the liability is therefore identical to the share of Carr Duff whom the jury found had committed the actual tort. Nationwide, Bellmont and Carr Duff were found to have been jointly and severally liable for the death of George Baker. Had a tort action been successfully brought by Baker's estate, it would have been privileged to exact payment of the entire amount of damages from any one or more of the defendants. However, this is a suit for contribution and it is not the total number of defendants involved who could have been liable to the deceased's estate but the number of directly and primarily liable parties which determines the number of pro-rata shares. Only two parties were directly and primarily liable, Carr Duff and Hinkle. Therefore liability for contribution is divisible in only two parts, one part being based upon the negligence of Carr Duff and the other part being based upon the negligence of Hinkle, the plaintiff's insured.
We therefore find that the total contribution to which Nationwide is entitled is $37,500, one-half the amount it paid in settlement.
In the event the judgment herein is paid by Bellmont, it is entitled to recover the amount so paid from Carr Duff. Restatement (Second) of Torts, § 416, Comment (c) states, ". . . the contractor who is employed to take the precautions is under a duty to indemnify his employer for any liability which the contractor's negligence in failure to take reasonably adequate precautions may bring upon him."
The rationale of that rule springs from the fact that one found liable under § 416 is deemed secondarily liable for the active negligence of the contractor who is considered primarily liable. As stated in Ohio Cas. Ins. Co. v. Bank Bldg. and Equipment Corp., 300 F. Supp. 632, 635-636 (W.D.Pa. 1968):
[under § 416] liability is imposed upon the owner, not because of his own actions, but by operation of law. As such, the liability of the owner of the land is secondary to that of the primary tort-feasor.
"But the important point to be noted in all the cases is that secondary as distinguished from primary liability rests upon a fault that is imputed or constructive only, being based on some legal relation between the parties, or arising from some positive rule of common or statutory law or because of a failure to discover or correct a defect or remedy a dangerous condition caused by the act of the one primarily liable." Builders Supply Co. v. McCabe, 366 Pa. 322, 327, 328, 77 A.2d 368, 371, 24 A.L.R.2d 319.
The right to indemnity as between a tortfeasor secondarily and one primarily liable has long been recognized in Pennsylvania. Builders Supply Co. v. McCabe, 366 Pa. 322, 77 A.2d 368 (1951). Globe Indemnity Co. v. Agway, Inc., 456 F.2d 472 (3d Cir. 1972). In McCabe, Justice Horace Stern, writing for the Pennsylvania Supreme Court, stated,
The right of indemnity rests upon a difference between the primary and the secondary liability of two persons each of whom is made responsible by the law to an injured party. It is a right which enures to a person who, without active fault on his own part, has been compelled, by reason of some legal obligation, to pay damages occasioned by the initial negligence of another, and for which he himself is only secondarily liable. The difference between primary and secondary liability is not based on a difference in degrees of negligence or on any doctrine of comparative negligence, — a doctrine which, indeed, is not recognized by the common law. . . . It depends on a difference in the character or kind of the wrongs which cause the injury and in the nature of the legal obligation owed by each of the wrongdoers to the injured person. Secondarily liability exists, for example, where there is a relation of employer and employee, or principal and agent; if a tort is committed by the employee or the agent recovery may be had against the employer or the principal on the theory of respondeat superior, but the person primarily liable is the employee or agent who committed the tort, and the employer or principal may recover indemnity from him for the damages which he has been obliged to pay.
77 A.2d at 370 (citations omitted) (emphasis in original).
In addition the Pennsylvania Uniform Contribution Among Joint Tortfeasors Act specifically states that it does not impair any rights to indemnity under existing law. 12 P.S. § 2087.
We find these principles applicable to the instant case. Bellmont, found liable pursuant to § 416, is secondarily liable for the death of George Baker. Carr Duff, on the other hand, was one of the two parties whose active negligence caused the death.
A final comment on this issue is necessary before moving on to the next. In denying Bellmont's motion for judgment n.o.v., we have been persuaded not only by the considerations heretofore discussed, but we have also considered what might result from the entry of a judgment n.o.v. in this situation. A determination that a defendant found vicariously liable in a proceeding brought pursuant to the Uniform Contribution Among Joint Tortfeasors Act should not have a judgment entered against it on the ground that in the same proceeding liability has been found against the party whose tortious acts gave rise to the vicarious liability, would produce a most inequitable result. Such a determination would leave the plaintiff with a judgment only against the actual tortfeasor and in many instances such a judgment might not be collectible. For instance in the situation wherein an employer is found to be vicariously liable for its employee, the plaintiff might be left with only an uncollectible judgment against the employee.
Plaintiff's Motion for an Award of Interest.
Plaintiff has moved this Court to award interest on the portion of the $75,000 settlement which represents the pro-rata shares of the other joint tortfeasors. Marrazzo v. Scranton-Nehi Bottling Co., 438 Pa. 72, 263 A.2d 336 (1970). We have molded the verdict to require Bellmont and Carr Duff to contribute to Nationwide one-half of the amount paid in settlement, or $37,500. In addition to this portion of the settlement, plaintiff seeks to recover interest on this amount from June 15, 1970, the date the settlement was paid to the estate of the deceased. Plaintiff contends that the jury's verdict establishes that Carr Duff and Bellmont should have contributed to the settlement, and that equity and fairness mandate that the plaintiff, who was deprived of the use of the funds representing defendant's pro-rata share, should recover interest from the defendants who benefited by not having to pay their share during this six year period.
The recovery of interest by a settling joint tortfeasor was considered in W.D. Rubright v. International Harvester Co., 358 F. Supp. 1388 (W.D.Pa. 1973). In Rubright, Judge Scalera, after a detailed analysis of the Pennsylvania case law, concluded that the plaintiff in the case before him was not entitled to interest.
As detailed in Rubright, recovery under the Pennsylvania Uniform Contribution Among Joint Tortfeasors Act has been characterized by the Pennsylvania Supreme Court as a recovery in assumpsit or contract rather than in tort. Harger v. Caputo, 420 Pa. 528, 218 A.2d 108 (1966).
Moreover, the right to recovery under the principles of contribution embodied in the Pennsylvania Uniform Contribution Among Joint Tortfeasors Act is a right which
. . . has its foundation in, and is controlled by, the principles of equity and natural justice. . . .
The doctrine of contribution rests upon the principle that when the parties stand in aequali jure the law requires equality, which is equity, and one of them is not to be obliged to bear a common burden in ease of the rest. Thus, the doctrine is founded on the broad equitable maxim that equality is equity, and that he who receives a benefit must incur the burden.
The right of contribution is also explained as a quasi contractual right which arises by reason of an implied engagement on the part of each obligor to help bear the common burden. P.L.E., Contribution § 2.
The contribution action is clearly distinct from the tort out of which the right of contribution arose. As pointed out in Rubright, the Pennsylvania courts have viewed the right to contribution provided by the Uniform Contribution Among Joint Tortfeasors Act as a dual one — a right sounding in both equity and quasi contract. 358 F. Supp. at 1392.
The substantive law controlling the allowance of additional damages in the nature of interest for a breach of a quasi contractual duty or for the failure to perform an equitable duty is set forth in the Restatement of Contracts, § 337 and the Restatement of Restitution, § 156, respectively.
Section 337 permits the recovery of interest from the time of performance where the defendant has committed a breach of contract by failing to perform at a required time, if the value of such performance is stated in the contract or is ascertainable by mathematical calculation.
Section 337 reads:
If the parties have not by contract determined otherwise, simple interest at the statutory legal rate is recoverable as damages for breach of contract as follows:
(a) Where the defendant commits a breach of a contract to pay a sum of money, or to render a performance the value of which in money is stated in the contract or is ascertainable by mathematical calculation from a standard fixed in the contract or from established market prices of the subject matter, interest is allowed on the amount of the debt or money value from the time performance was due, after making all the deductions to which the defendant may be entitled.
(b) Where the contract that is broken is of a kind not specified in Clause (a), interest may be allowed in the discretion of the Court, if justice requires it, on the amount that would have been just compensation if it had been paid when performance was due.
Section 156, upon which the court in Rubright relied and upon which this court shall rely, states
Subject to the rules stated in § 157, a person who has a duty to pay the value of a benefit which he has received, is also under a duty to pay interest upon such value from the time he committed a breach of duty in failing to make restitution if, and only if:
(a) the benefit consisted of a definite sum of money, or
(b) the value of the benefit can be ascertained by mathematical calculation from the terms of an agreement between the parties or by established market prices, or
(c) payment of interest is required to avoid injustice.
These principles are equally applicable to breaches of equitable as well as quasi contractual duties. Rubright at 1394.
As applied to the instant case, the above concepts lead to a conclusion that plaintiff Nationwide, under the circumstances of this case, is entitled to interest. In making this award this court does not disturb the settled law of Pennsylvania that interest as such is not allowed in tort actions on the theory that damages in tort are unliquidated. The action in this case is one sounding in quasi contract and equity, not tort. Here the amount recoverable was liquidated and agreed to have been reasonable.
McGonnell v. Pittsburgh Rys. Co., 234 Pa. 396, 83 A. 282 (1912).
It was stipulated by all the parties and the stipulation was read into the record at trial, that the plaintiff settled the tort case with the estate of the deceased for an amount which all parties agreed was reasonable at the time of settlement. The settlement of the tort action took place as that CASE neared trial. (N.T. 1-45-51). The only question for the jury in this case was, therefore, whether the defendants were joint tortfeasors, an issue which they resolved in the affirmative.
In Rubright, the stipulation as to reasonableness took place immediately prior to the trial for contribution.
The determination of at what point in time an agreement as to reasonableness should attach was discussed in Pennsylvania Railroad Co. v. Erie Avenue Warehouse Co., 206 F. Supp. 725, 726 (E.D.Pa. 1962). There, Judge Van Dusen stressed the tense of the verb used by the parties in their stipulation to reasonableness of an amount paid in settlement of a tort action. The court stated:
At a pre-trial conference (incident to the trial of this third-party action) held December 15, 1969, defendant stipulated that "the amount paid in settlement * * * to wit, $75,000., was fair and reasonable" (N.T. 42 of Document No. 65a). See Pre-Trial Order (Document No. 54) and Report of Pre-Trial Conference (Document No. 51). In view of this stipulation using the past tense of the verb, namely "was," defendant agreed that the $75,000 was the fair and reasonable settlement amount in the past, namely, when it was paid and approved by the court on August 6, 1958, and not at the time of the pre-trial conference of December 15, 1959, or any later date. (emphasis added).
Nationwide has, therefore, met the requirements of the Pennsylvania Uniform Contribution Among Joint Tortfeasors Act, supra, and may enforce its right to contribution. It should also be noted that it was agreed by the parties that this court would mold the verdict and enter judgment. We will include in our judgment interest at six per cent (6%) on one-half of the settlement from June 15, 1970, the date of payment, to the date of judgment.
The award of interest in this case falls squarely within the provisions of the Restatement of Restitution, § 156. Under subsection (a) interest can be granted only when the sum due is definite. As demonstrated above, the sum due the plaintiff from the defendant joint tortfeasor was not only definite but considered by both parties to be reasonable. As a result, the award of interest falls within the scope of § 156(a).
This award additionally comes within § 156(c) where payment of interest is required to avoid injustice. The plaintiff paid the sum of $75,000 to the estate of the deceased on June 15, 1970 in return for a release running in favor of all parties. Since that time the plaintiff has been without the benefit of those funds while defendants Bellmont and Carr Duff have benefited from the plaintiff's payment of their portion of the settlement. In Pennsylvania Railroad Co. v. Erie Avenue Warehouse Co., 206 F. Supp. 725, 726 (E.D.Pa. 1962), wherein the Court of Appeals had held that a defendant and a third-party defendant were equally liable and that the defendant who had settled in full was entitled to interest on one-half of the settlement, Judge Van Dusen, who was the trial judge, in making his finding that interest was due from the date of payment of settlement stated that, "Strong policy reasons favor the payment of interest to a party who has advanced money in order to settle the principal action on a fair and reasonable basis."
We shall, therefore, award plaintiff interest on defendant's pro-rata share of the settlement from the date the plaintiff paid the settlement, at which time plaintiff's right to contribution arose.
Carr Duff's Motion for a New Trial.
At the close of the plaintiff's case, PECO moved for a directed verdict in connection with plaintiff's claim that PECO was negligent and a joint tortfeasor. The plaintiff did not oppose this motion and the Court granted it. At the close of all the evidence PECO moved for a directed verdict in connection with the cross-claim for negligence filed against it by Carr Duff. The Court also granted this motion. Carr Duff now contends that the Court erred in making these findings.
The Court granted these motions after applying the standard set forth by the United States Supreme Court in Brady v. Southern Railroad, 320 U.S. 476, 479-80, 64 S.Ct. 232, 234, 88 L.Ed. 239 (1943):
When the evidence is such that without weighing the credibility of the witnesses there can be but one reasonable conclusion as to the verdict, the court should determine the proceeding by . . . directed verdict. . . .
In the instant case, without weighing the credibility of the witnesses, and viewing the evidence in the light most favorable to the plaintiff and Carr Duff, the only verdict which the jury could have reasonably found would have been for PECO.
Carr Duff contends that it should not be necessary to prove that PECO had notice and knowledge that a crane would be used underneath the power lines. We disagree.
The law in Pennsylvania regarding this issue is clear. In Stark v. Lehigh Foundries, 388 Pa. 1, 130 A.2d 123 (1957), the jury found the possessor of the land and the electric company liable for the injuries sustained by plaintiff when the crane on which he was working came in contact with a power line. While affirming the judgment as to the possessor, the Court entered a judgment n.o.v. for the electric company, holding that the plaintiff had not proved that the electric company had actual notice of the crane's use underneath the power lines, stating
Where a supplier of electricity has installed its high tension lines in a safe and proper manner on the land of another and has neither knowledge nor notice that the possessor of the land is conducting an activity thereon which makes the line dangerous to people working on the land, it is not subject to liability for the electrocution of a workman resulting from this activity. 388 Pa. at 12-13, 130 A.2d at 130-31.
In Dunnaway v. Duquesne Light Co., 423 F.2d 66 (3d Cir. 1970), the court held that an electric company's knowledge of the use of cranes and other equipment in the general construction area is insufficient to charge the electric company with notice that a crane would be used under its wires.
In Reed v. Duquesne Light Co., 1946, 354 Pa. 325, 47 A.2d 136, and Stark v. Lehigh Foundries, Inc., 1957, 388 Pa. 1, 130 A.2d 123, the Supreme Court of Pennsylvania held that such notice could not be imputed from an electric company's knowledge of the use of cranes in the general area of the high tension wires. In those cases as in the present case, "the position of the crane on the day of the accident immediately underneath the power line was not permanent." Stark v. Lehigh Foundries, Inc., 388 Pa. at 14, 130 A.2d at 131. A reasonable periodic inspection by the electric company would not have revealed any crane under or dangerously close to the lines unless the inspection happened to occur the day of the accident. 423 F.2d at 69-70.
There was no evidence in the instant case that PECO had any knowledge that a crane was being used under its wires. Although George Mittleman, the construction superintendent on the project, testified that, a crane was used on several occasions (N.T. 2-28) and that PECO was aware that an apartment building four stories high was to be constructed at the site (N.T. 1-63), there was no testimony that PECO had actual knowledge as to the position and use of the crane on the day of the accident.
Carr Duff argue that there was evidence that PECO was negligent in moving its transformer and energizing the wires, citing Meehan v. Philadelphia Electric Company, 424 Pa. 51, 225 A.2d 900 (1967). In Meehan, there was affirmative evidence by plaintiff's expert that the conduct of the electric company fell below the standard of care required under the circumstances. There is no such evidence in this case. Harold Carr, the president of Carr Duff, who testified as both an expert and a fact witness, testified that PECO did a "fine job", that it did the work in a proper fashion, and that it did the job and energized the system in conformity with the proper standards that it always used. (N.T. 4-154). Joseph Warren, an expert witness called by Carr Duff, testified that PECO did nothing wrong in energizing the electric line and that it was a perfectly proper thing to do. (N.T. 4-62). The Court did not err in granting PECO's motion for a directed verdict on plaintiff's claim and on the cross-claim of Carr Duff for the reasons heretofore stated.
While we have not discussed every contention that was raised, we have, however, reviewed the record and considered all of the grounds alleged by the parties and we hold that none of them, either singly or collectively, has sufficient substance to merit any further discussion.
Accordingly, the court will enter orders denying Bellmont's motion for judgment n.o.v., granting plaintiff Nationwide's motion for an award of interest, and denying Carr Duff's motion for a new trial and/or judgment n.o.v. In addition, the court will enter a judgment order in favor of plaintiff Nationwide and against defendants Bellmont and Carr Duff in the sum of $54,279.96 (which sum includes interest in the amount of $16,779.96) together with costs and such judgment order will provide that a right of indemnity exists in favor of defendant Bellmont against defendant Carr Duff.
explaining that one of the joint tortfeasors must have discharged the common liability through settlement, paid more than its pro rata share in the process, and extinguished the liability of the other tortfeasor
Summary of this case from Mendelson v. Delaware River Bay Authority
working in a crane on a cable which became entangled in overhead power line — peculiar risk when contractor failed to insulate in compliance with National Electric Code
Summary of this case from Sharkey v. Airco, Inc.
In Nationwide Mutual Insurance Co. v. Philadelphia Electric Co., 443 F. Supp. 1140 (E.D.Pa. 1977), the District Court, applying the Pennsylvania Uniform Contribution Among Tortfeasors Act, adopted the Superior Court's rationale in Parker, supra, and held it applicable in interpreting the statute.
Summary of this case from Jones v. Harrisburg Polyclinic Hospital | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 3,148 |
Robert Weld Mitchell MBE (10 August 1915 – 13 December 1994) was a Progressive Conservative party member of the House of Commons of Canada and a lawyer.
He was born in Toronto, the son of Percy Dawson Mitchell and Olive Weld, and was educated at Ridley College and Osgoode Hall. Mitchell was called to the Ontario bar in 1958. He was a director of William Weld Co. Ltd., served as president of the London Boy Scouts Association, was president of the London Chamber of Commerce from 1962 to 1963 and served as vice-president of Supertest Petroleum, later part of BP Canada.
Between 1940 and 1945, Mitchell served in the Canadian Army with The Perth Regiment.
He was first elected at the London riding in the 1953 general election. After serving only one federal term, the 22nd Canadian Parliament, Mitchell left federal politics and did not seek re-election. He died in 1994.
References
1915 births
1994 deaths
Members of the House of Commons of Canada from Ontario
Lawyers in Ontario
Politicians from Toronto
Progressive Conservative Party of Canada MPs | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 6,978 |
\section{Introduction}
\label{intro}
The sequence OEIS A281505 concerns odd legs in right triangles with integer side lengths and prime hypotenuse. By the parametrisation of Pythagorean triples, these are positive integers of the form $x^2 - y^2$, where $x,y \in \mathbb N$ and $x^2 + y^2$ is prime. Even legs are those of the form $2xy$, where $x, y \in \mathbb N$ and $x^2 + y^2$ is an odd prime.
Let $\mathcal A$ be the set of odd legs, and $\mathcal B$ the set of even legs that occur in such triangles. Consider the quantities
\[
\mathcal A(N) = \{ n \in \mathcal A: n \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N \}, \qquad \mathcal B(N) = \{ n \in \mathcal B: n \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N \}
\]
as $N \to \infty$.
Let $\mathcal P$ denote the set of primes. By a change of variables, observe that
\[
\mathcal A(N) = \# \{ ab \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N: \frac12 (a^2+b^2) \in \mathcal P \}.
\]
Additionally, note that
\[
\mathcal B(2N) = \mathcal C(N),
\]
where
\[
\mathcal C(N) = \# \{1< ab \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N: a^2+b^2 \in \mathcal P \}.
\]
We estimate $\mathcal C(N)$, which is equivalent to estimating $\mathcal B(N)$ and similar to estimating $\mathcal A(N)$.
Let
\[
\eta = 1 - \frac{1 + \log \log 2}{\log 2} \approx 0.086
\]
be the Erd\H{o}s--Ford--Tenenbaum constant. This constant is related to the number of distinct products in the multiplication table, and also arises in other contexts, for example, see \cite{For2008}, \cite{FLP}, and \cite{MPP}.
\begin{thm} \label{UpperBound}
We have
\[
\mathcal C(N) \ll \frac{N}{(\log N)^\eta} (\log \log N)^{O(1)}.
\]
\end{thm}
Since every prime $p\equiv1\pmod4$ is representable as $a^2+b^2$ with $a,b$ integral,
we have $\mathcal C(N)$ unbounded. In fact, using the maximal order of the divisor function, we have
$\mathcal C(N) \ge N^{1-o(1)}$ as $N\to\infty$.
We obtain a strengthening of this lower bound.
\begin{thm} \label{LowerBound}
We have, as $N\to\infty$,
\[
\mathcal C(N) \ge\frac{N}{(\log N)^{\log4-1+o(1)}}.
\]
\end{thm}
Note that $\log4-1\approx 0.386$. Since $\mathcal B(2N) = \mathcal C(N)$, we obtain the same bounds for $\mathcal B(N)$. By essentially the same proofs, one can also deduce these bounds for $\mathcal A(N)$.
To motivate the outcome, consider the following heuristic. There are typically $\approx(\log n)^{\log 2}$ divisors of $n$, which follows from the normal number of prime factors of $n$, a result of Hardy and Ramanujan \cite{HR}.
Moreover, given a factorisation $n=ab$, the ``probability"
of $a^2+b^2$ being prime is roughly $(\log n)^{-1}$. Since $\log 2 < 1$, we expect the proportion $\mathcal C(N)/N$ to decay logarithmically. In the presence of biases and competing heuristics, this \emph{prima facie} prediction should be taken with a few grains of salt. We use Brun's sieve and the Hardy--Ramanujan inequality to formally establish our bounds. In addition, for Theorem \ref{LowerBound} we use a result of Harman and Lewis \cite{HL2001}
on the distribution of Gaussian primes in narrow sectors of the complex plane.
We write $\mathcal P$ for the set of primes. We use Vinogradov and Landau notation. As usual, we write $\omega(n)$ for the number of distinct prime divisors of $n$, and $\Omega(n)$ for the number of prime divisors of $n$ counted with multiplicity. The symbols $p$ and $\ell$ are reserved for primes, and $N$ denotes a large positive real number.
\section*{Acknowledgments and a dedication}
The authors were supported by the National Science Foundation under Grant No. DMS-1440140 while in residence at the Mathematical Sciences Research Institute in Berkeley, California, during the Spring 2017 semester. The authors thank John Friedlander and Roger Heath-Brown for helpful comments and
Tomasz Ordowski for suggesting the problem.
This year (2017) is the 100th anniversary of the publication of the
paper {\it On the normal number of prime factors of
a number $n$}, by Hardy and Ramanujan, see \cite{HR}.
Though not presented in such terms, their paper ushered in the subject of probabilistic number theory.
Simpler proofs have been found, but the original
proof contains a very useful inequality, one which we are happy to use yet again. We dedicate
this note to this seminal paper.
\section{An upper bound}
\label{UpperSection}
In this section, we establish Theorem \ref{UpperBound}. The Hardy--Ramanujan inequality \cite{HR}
states that there exists a positive constant $c_0$ such that uniformly for $i \in \mathbb N$ and $N\ge3$ we have
\[
\# \{n \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N: \omega(n) = i \} \ll \frac N {\log N}
\frac{(\log \log N + c_0)^{i-1}}{(i-1)!}.
\]
By Mertens's theorem and the fact that the sum of the reciprocals of prime powers higher than the
first power converges, there is a positive constant $c_1$ such that
\begin{equation}
\label{eq:c1}
\sum_{p^\nu\leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N}p^{-\nu}\leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec\log\log N+c_1 \qquad (N \ge 3).
\end{equation}
Let ${\alpha}} \def\bfalp{{\boldsymbol \alpha}$ be a parameter in the range $1 < {\alpha}} \def\bfalp{{\boldsymbol \alpha} < 2$, to be specified in due course. We begin by bounding the size of the exceptional set
\[
\mathcal E_1 := \{ n \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N: \omega(n) > L \},
\]
where
\begin{equation} \label{Ldef}
L = \lfloor {\alpha}} \def\bfalp{{\boldsymbol \alpha} \log \log N \rfloor.
\end{equation}
By the Hardy--Ramanujan inequality, we have
\[
\# \mathcal E_1
\ll \frac{N}{\log N} \sum_{i > L} \frac{(k+c_0)^{i-1}}{(i-1)!}= \frac{N}{\log N}\sum_{j\ge L}\frac{(k+c_0)^j}{j!},
\]
where $k= \log \log N$, and therefore
\[
\frac{\log N}N \# \mathcal E_1 \ll \frac{(k+c_0)^{L}} {L!}
<\left(\frac{(k+c_0)e}{L}\right)^L=\left(\frac e{\alpha}} \def\bfalp{{\boldsymbol \alpha}+O\left(\frac1k\right)\right)^L.
\]
Note that we have used here the elementary inequality $1/L!<(e/L)^L$, which holds for all positive
integers $L$ and follows instantly from the Taylor series for $e^L$.
Thus,
\begin{equation} \label{FirstTerm}
\# \mathcal E_1 \ll \frac{N }{(\log N)^{1-{\alpha}} \def\bfalp{{\boldsymbol \alpha} + {\alpha}} \def\bfalp{{\boldsymbol \alpha} \log {\alpha}} \def\bfalp{{\boldsymbol \alpha}}}.
\end{equation}
For an integer $n\ge2$, write $P(n)$ for the largest prime factor of $n$, and let $P(1)=1$. By de Bruijn \cite[Eq. (1.6)]{deB} we may bound the size of the exceptional set
\[
\mathcal E_2 := \{ n \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N: P(n) \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N^{1/\log \log N}\}
\]
by $N/(\log N)^2$ for all sufficiently large numbers $N$. (Actually, the denominator may be taken as any fixed
power of $\log N$.)
Next, we estimate
\[
\mathcal C^*(N):= \# \{ ab \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N: ab \notin (\mathcal E_1 \cup \mathcal E_2),~ a^2+b^2 \in \mathcal P \}.
\]
For $n$ counted by $\mathcal C^*(N)$, we see by symmetry that we have $n = ab_0 \ell$ for some $a,b_0, \ell \in \mathbb N$ with $\ell > N^{1/\log\log N}$ prime and $a^2 + b_0^2 \ell^2$ prime. Thus
\begin{equation} \label{pause}
\mathcal C^*(N) \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec 2 \sum_{\substack{{ab_0 \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N^{1- 1/\log \log N} }\\ \omega(ab_0) \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec L}} S(a, b_0),
\end{equation}
where
\[
S(a,b_0) = \sum_{\substack{ N^{1/\log \log N} < \ell \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec \frac N{ab_0} \\ \ell,\;a^2 + b_0^2 \ell^2 \in \mathcal P}}1.
\]
We turn our attention to $S(a,b_0)$. We may assume that $ab_0$ is even and $\gcd(a,b_0) = 1$, for otherwise $S(a,b_0)= 0$. Observe that
\[
S(a, b_0) \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec \# \{ m \in (z,X]: \gcd(m(a^2+b_0^2 m^2), P(z)) = 1 \},
\]
where
\[
z = N^{(\log \log N)^{-3}}, \quad P(z) = \prod_{p < z} p, \quad X = \frac N{ab_0}.
\]
To bound this from above, we apply Brun's sieve \cite[Corollary 6.2]{FI2010} with
\[
\mathcal A = \Bigl \{ m(a^2+b_0^2 m^2): 1 \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec m \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec X \Bigr \},
\]
and with the completely multiplicative density function $g$ defined by
\[
g(p) = \begin{cases}
1/p, & \text{if } p \mid ab_0 \text{ or } p \not \equiv 1 \mmod 4 \\
3/p, & \text{if } p \nmid a b_0, ~p \equiv 1 \mmod 4.
\end{cases}
\]
For this to be valid, we need to check that
\begin{equation} \label{BrunHyp}
|r_d(\mathcal A)| \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec g(d) d \qquad (d \mid P(z)),
\end{equation}
where
\[
r_d(\mathcal A) = |\mathcal A_d| - Xg(d), \quad \mathcal A_d = \{ n \in \mathcal A: n \equiv 0 \mmod d \}
\]
and $P(z) = \prod_{p < z} p$. We begin by noting that if $p \in \mathcal P$ then the congruence
\[
m (a^2 + b_0^2 m^2) \equiv 0 \mmod p
\]
has $g(p)p$ solutions $m \mmod p$. Observe that any divisor $d$ of $P(z)$ must be squarefree; thus, by the Chinese remainder theorem, the congruence
\[
m (a^2 + b_0^2 m^2) \equiv 0 \mmod d
\]
has $g(d)d$ solutions $m \mmod d$. By periodicity, we now have
\[
r_d(\mathcal A) = \# \{ m \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec M: m(a^2 + b_0^2 m^2) \equiv 0 \mmod d \} - Mg(d),
\]
where $M = X - d \lfloor X/d \rfloor$. This confirms \eqref{BrunHyp}, since $0 \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec M < d$ and $0 < g(d) \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec 1$.
We also need to check that
\[
\log z \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec \frac{\log X}{c \log ( V(z)^{-1} \log X)},
\]
where $V(z) = \prod_{p < z} (1- g(p))$, and where
\[
(c/e)^c = e, \qquad c \approx 3.59.
\]
This follows from the inequalities
\[
X \ge N^{1/\log \log N}, \qquad V(z) \gg (\log z)^{-2}.
\]
Now \cite[Corollary 6.2]{FI2010} tells us that
\[
S(a,b_0) \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec X^{3/4} + 2XV(z) \ll \frac{N(\log \log N)^{O(1)}}{(\log N)^2 ab_0}.
\]
(Note that we might equally have used the version of Brun's sieve in \cite[p. 68]{HRi}, which
is less precise, but somewhat easier to utilise.)
Substituting this into \eqref{pause} yields
\begin{equation} \label{subI}
\mathcal C^*(N) \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec \frac{N(\log \log N)^{O(1)}}{(\log N)^2} I,
\end{equation}
where
\[
I = \sum_{j+k \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec L} \sum_{\substack{a \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N \\ \omega(a)=j}} a^{-1} \sum_{\substack{b_0 \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N \\ \omega(b_0)=k}} b_0^{-1}.
\]
It follows from the multinomial theorem that
\begin{align*}
I &\leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec \sum_{j+k \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec L} j!^{-1} \Bigl(\sum_{p^v \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N} p^{-v}\Bigr)^j
k!^{-1} \Bigl(\sum_{p^v \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N} p^{-v}\Bigr)^k \\
&=
\sum_{j+k \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec L} (j+k)!^{-1} {j+k \choose j} \Bigl( \sum_{p^v \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N} p^{-v} \Bigr)^{j+k}.
\end{align*}
Letting $m=j+k$, the binomial theorem now gives
\[
I \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec \sum_{m \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec L} m!^{-1} \Bigl( 2 \sum_{p^v \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N} p^{-v} \Bigr)^m \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec \sum_{m \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec L} \frac{(2 \log \log N + 2c_1)^m}{m!},
\]
where $c_1$ is as in \eqref{eq:c1}. In view of \eqref{Ldef}, we now have
\begin{align*}
I& \ll L!^{-1} (2 \log \log N + 2c_1)^L<\left(\frac{2e\log\log N+2ec_1}{L}\right)^L\\
&=\left(\frac{2e}{{\alpha}} \def\bfalp{{\boldsymbol \alpha}}+O\left(\frac1L\right)\right)^L\ll(\log N)^{{\alpha}} \def\bfalp{{\boldsymbol \alpha}(1+\log2- \log {\alpha}} \def\bfalp{{\boldsymbol \alpha})}.
\end{align*}
Substituting this into \eqref{subI} yields
\begin{equation} \label{CstarBound}
\mathcal C^*(N) \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N(\log \log N)^{O(1)} (\log N)^{{\alpha}} \def\bfalp{{\boldsymbol \alpha}(1+\log 2 - \log {\alpha}} \def\bfalp{{\boldsymbol \alpha}) - 2}.
\end{equation}
By \eqref{FirstTerm}, our estimate for $\#\mathcal E_2$, and \eqref{CstarBound}, we have
\[
\mathcal C(N) \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec \mathcal C^*(N) + \# \mathcal E_1 + \# \mathcal E_2 \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N (\log \log N)^{O(1)} (\log N)^{-\mathcal M},
\]
where
\[
\mathcal M = \min \{ 1 - {\alpha}} \def\bfalp{{\boldsymbol \alpha} + {\alpha}} \def\bfalp{{\boldsymbol \alpha} \log {\alpha}} \def\bfalp{{\boldsymbol \alpha}, ~ 2 ,~ 2 - {\alpha}} \def\bfalp{{\boldsymbol \alpha} - {\alpha}} \def\bfalp{{\boldsymbol \alpha} \log 2 + {\alpha}} \def\bfalp{{\boldsymbol \alpha} \log {\alpha}} \def\bfalp{{\boldsymbol \alpha} \}.
\]
We now choose $1 < {\alpha}} \def\bfalp{{\boldsymbol \alpha} < 2$ so as to maximise $\mathcal M$.
One might guess that this ${\alpha}} \def\bfalp{{\boldsymbol \alpha}$ solves
\[
1- {\alpha}} \def\bfalp{{\boldsymbol \alpha} + {\alpha}} \def\bfalp{{\boldsymbol \alpha} \log {\alpha}} \def\bfalp{{\boldsymbol \alpha} = 2 - {\alpha}} \def\bfalp{{\boldsymbol \alpha} - {\alpha}} \def\bfalp{{\boldsymbol \alpha} \log 2 + {\alpha}} \def\bfalp{{\boldsymbol \alpha} \log {\alpha}} \def\bfalp{{\boldsymbol \alpha},
\]
and indeed ${\alpha}} \def\bfalp{{\boldsymbol \alpha} = (\log 2)^{-1}$ does maximise $\mathcal M$ on the interval $(1,2)$.
With this choice of ${\alpha}} \def\bfalp{{\boldsymbol \alpha}$, we have
\[
\mathcal M = 1 - \frac{1 + \log \log 2}{\log 2} = \eta,
\]
completing the proof of Theorem \ref{UpperBound}.
\section{A lower bound}
\label{LowerSection}
In this section, we establish Theorem \ref{LowerBound}. Let
\[
\mathcal L_0 = \{ (a,b) \in \mathbb N^2:1< ab \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N,~ a^2 + b^2 \in \mathcal P \}.
\]
Writing $P(n)$ for the largest prime factor of $n>1$, and $P(1) = 1$, put
\[
\mathcal L_1 = \{(a,b) \in \mathcal L_0: P(ab) \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N^{1/\log \log N} \}.
\]
Let $\varepsilon$ be a small positive real number, and let
\begin{align*}
\mathcal L_2 &= \{(a,b) \in \mathcal L_0 \setminus \mathcal L_1: \omega(a) > (1+\varepsilon) \log \log N \}, \\
\mathcal L_3 &= \{(a,b) \in \mathcal L_0 \setminus \mathcal L_1: \omega(b) > (1+\varepsilon) \log \log N \}.
\end{align*}
Finally, write
\[
\mathcal L = \mathcal L_0 \setminus (\mathcal L_1 \cup \mathcal L_2 \cup \mathcal L_3).
\]
As we seek a lower bound, we are free to discard some inconvenient elements of $\mathcal C(N)$. Thus, by the Cauchy--Schwarz inequality, we have
\begin{equation} \label{Cauchy}
\mathcal C(N) \ge (\#\mathcal L)^2 / \mathcal S(N),
\end{equation}
where $\mathcal S(N)$ is the number of quadruples $(a,b,c,d) \in \mathbb N^4$ such that
\[
ab=cd\hbox{ and }(a,b),(c,d)\in\mathcal L.
\]
We first show that
\begin{equation} \label{L0bound}
\# \mathcal L_0 \gg N.
\end{equation}
For this, we use existing work counting Gaussian primes in narrow sectors. For convenience, we state the relevant result \cite[Theorem 2]{HL2001}.
\begin{thm}[Harman--Lewis]
\label{HL}
Let $X$ be a large positive real number, and let ${\beta}} \def\bfbet{{\boldsymbol \beta}, {\gamma}} \def\Gam{{\Gamma}$ be real numbers in the ranges
\[
0 \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec {\beta}} \def\bfbet{{\boldsymbol \beta} \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec \pi/2, \qquad X^{-0.381} \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec {\gamma}} \def\Gam{{\Gamma} \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec \pi/2.
\]
Then
\[
\# \{ (a,b) \in \mathbb N^2: a^2 + b^2 \in \mathcal P \cap [0,X], ~\arctan(b/a) \in [\beta, \beta + {\gamma}} \def\Gam{{\Gamma}) \} \gg \frac{{\gamma}} \def\Gam{{\Gamma} X}{\log X}.
\]
The implied constant is absolute.
\end{thm}
For positive integers $i \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec \frac{ \log N}{10 \log 2}$, we apply this with
\[
{\beta}} \def\bfbet{{\boldsymbol \beta} = {\gamma}} \def\Gam{{\Gamma} = \frac \pi {2^{i+1}}, \qquad X = 2^{i-2}N.
\]
By Jordan's inequality
\[
\frac 2\pi x \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec \sin x \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec x \qquad (0 \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec x \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec \pi/2),
\]
observe that if $a,b \in \mathbb N$, $a^2 + b^2 \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec X$ and ${\theta}} \def\bftet{{\boldsymbol \theta}} \def\Tet{{\Theta} = \arctan(b/a) \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec \pi 2^{-i}$ then
\[
ab \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec X \sin {\theta}} \def\bftet{{\boldsymbol \theta}} \def\Tet{{\Theta} \cos {\theta}} \def\bftet{{\boldsymbol \theta}} \def\Tet{{\Theta} = \frac12 X \sin (2 {\theta}} \def\bftet{{\boldsymbol \theta}} \def\Tet{{\Theta}) \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec X {\theta}} \def\bftet{{\boldsymbol \theta}} \def\Tet{{\Theta} \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N2^{i-2} \cdot \frac \pi {2^i} \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N.
\]
Thus
\[
\# \mathcal L_0 \gg \sum_{i \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec \frac{\log N}{10 \log 2}} \frac N{\log N} \gg N,
\]
confirming \eqref{L0bound}.
Next, we show that $\# \mathcal L_j = o(N)$ ($j=1,2,3$).
\begin{lemma} \label{discard0}
We have $\# \mathcal L_1 = o(N)$.
\end{lemma}
\begin{proof}
By de Bruijn \cite[Eq. (1.6)]{deB}, we have
\[
\sum_{a \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec \sqrt N} \sum_{\substack{b \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N/a \\ P(b) \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N^{1/\log \log N}}} \ll \sum_{a \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec \sqrt N} \frac N{a (\log N)^2} \ll \frac N{\log N}.
\]
Thus, by symmetry, we have $\# \mathcal L_1 \ll \frac N{\log N}$.
\end{proof}
\begin{lemma} \label{discard} We have
\[
\# \mathcal L_j = o(N) \qquad (j = 2,3).
\]
\end{lemma}
\begin{proof}
As $\# \mathcal L_2 = \# \mathcal L_3$, we need only show this for $j=2$. Taking out a prime factor $\ell > N^{1/\log \log N}$ of $ab$, we have
\[
\# \mathcal L_2 \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec 2\sum_{\substack{a \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N^{1-1/\log \log N} \\ \omega(a) > (1+\varepsilon) \log \log N } } \sum_{b \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec a^{-1} N^{1- 1/\log \log N}} S_{a,b},
\]
where
\[
S_{a,b} = \sum_{\substack{N^{1/\log \log N} < \ell \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec \frac N{ab} \\ \ell,\: a^2 + b^2 \ell^2 \in \mathcal P}}1.
\]
As in the last section, Brun's sieve implies that
\[
S_{a,b}
\ll \frac{N(\log \log N)^{O(1)}}{ab (\log N)^2}.
\]
Therefore
\begin{equation}
\label{lem3.3eq}
\# \mathcal L_2 \ll \frac {N(\log \log N)^{O(1)}} {\log N} \sum_{\substack{a \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N^{1-1/\log \log N}\\\omega(a) \ge T} }a^{-1},
\end{equation}
where
\begin{equation}
\label{Tdef}
T = \lfloor (1+\varepsilon) \log \log N \rfloor.
\end{equation}
As in the prior section, the multinomial theorem implies that
\begin{align*}
\sum_{\substack{a\leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N^{1-1/\log\log N}\\\omega(a)\ge T}}\frac1a&
\leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec\sum_{j\ge T}\frac1{j!}\left(\log\log N+c_1\right)^j
\ll_\varepsilon \frac1{T!}(\log\log N+c_1)^T\\
&\leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec \left(\frac{e\log\log N+ec_1}{T}\right)^T\ll(\log N)^{(1+\varepsilon)(1-\log(1+\varepsilon))}.
\end{align*}
Since $(1+\varepsilon)(1-\log(1+\varepsilon))<1$,
using this estimate in \eqref{lem3.3eq} completes the proof of the lemma.
\end{proof}
Combining \eqref{L0bound} with Lemmas \ref{discard0} and \ref{discard} gives
\begin{equation} \label{Lbound}
\# \mathcal L \gg N.
\end{equation}
\begin{lemma} \label{L2}
If $c' > \log 4 -1$ then
\[
\mathcal S(N) \ll_{c'} N (\log N)^{c'}.
\]
\end{lemma}
\begin{proof}
One component of the count is when $(a,b)=(c,d)$. This is the diagonal case, and it is
easily estimated. By the sieve, the number of pairs $(a,b)\in\mathcal L$ with $a\leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec b$ is at most
\[
\sum_{a\leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec\sqrt{N}}\sum_{b\leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N^{1-1/\log\log N}/a}\sum_{\substack{\ell\leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N/ab\\a^2+\ell^2b^2\in\mathcal P}}1
\leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec \frac{N(\log\log N)^{O(1)}}{(\log N)^2}\sum_{a,b}\frac1{ab}\leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N(\log\log N)^{O(1)},
\]
which is negligible. (Note that this estimate shows that \eqref{Lbound} is essentially tight.)
For the nondiagonal case we imitate
\S \ref{UpperSection}. If $(a,b,c,d)$ is counted by $\mathcal S(N)$, put
\[
g =\gcd(a,c), \quad a = gu, \quad c= gv,
\]
so that
\[
ub = vd, \quad d = uw, \quad b = vw.
\]
Recall \eqref{Tdef}, and let $\mathcal G$ be the set of $(g,u,v,w_0) \in \mathbb N^4$ such that
\[
guvw_0 \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N^{1-1/ \log \log N}, \qquad \omega(gu), \omega(vw_0), \omega(gv), \omega(uw_0) \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec T,\qquad u\ne v.
\]
As $P(ab) > N^{1/ \log \log N}$, we see by symmetry that
\begin{equation} \label{cS1}
\mathcal S(N) \ll N(\log\log N)^{O(1)} + \sum_{(g,u,v,w_0) \in \mathcal G} S(g,u,v,w_0),
\end{equation}
where
\[
S(g, u ,v, w_0) = \sum_{\substack{\ell\in\mathcal P,\,N^{1/\log \log N} < \ell \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec \frac N{guvw_0} \\ (gu)^2 + (vw_0)^2 \ell^2, \;(gv)^2 + (uw_0)^2 \ell^2 \in \mathcal P}} 1.
\]
The fact that $u \ne v$ ensures that there are three primality conditions defining $S(g,u,v,w_0)$. To bound $S(g,u,v,w_0)$ from above, we may assume without loss that $guvw_0$ is even, and that the variables $g,u,v,w_0$ are pairwise coprime, for otherwise $S(g,u,v,w_0) = 0$.
Paralleling \S \ref{UpperSection}, an application of Brun's sieve reveals that
\begin{equation} \label{FinalBrun}
S(g, u ,v, w_0) \ll \frac{ N (\log \log N)^{O(1)}} {guvw_0 (\log N)^3}.
\end{equation}
Substituting \eqref{FinalBrun} into \eqref{cS1} yields
\begin{equation} \label{cS2}
\mathcal S(N) \ll N(\log\log N)^{O(1)} + \frac{N (\log \log N)^{O(1)}}{(\log N)^3} \mathcal I,
\end{equation}
where
\[
\mathcal I = \sum_{k_1 + \cdots + k_4 \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec 2T} \prod_{i =1}^4 \Bigl(\sum_{n \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N:\, \omega(n) = k_i} n^{-1} \Bigr)
\]
and $T$ is as in \eqref{Tdef}.
With $U = 2T$, it follows from the multinomial theorem that
\begin{align*}
\mathcal I &\leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec \sum_{k_1 + \cdots + k_4 \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec U} \prod_i k_i!^{-1} \Bigl(\sum_{p^v \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N} p^{-v}\Bigr)^{k_i}
\\
&= \sum_{m \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec U} m!^{-1} \sum_{k_1 + \cdots + k_4 = m} { m \choose k_1, k_2, k_3, k_4 }
\Bigl( \sum_{p^v \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N} p^{-v} \Bigr)^m,
\end{align*}
and a further application of the multinomial theorem gives
\[
\mathcal I \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec \sum_{m \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec U} m!^{-1} \Bigl(4 \sum_{p^v \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N} p^{-v} \Bigr)^m \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec \sum_{m \leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec U} \frac{(4 \log \log N + 4c_1)^m}{m!}.
\]
As $U = 2(1+\varepsilon)\log \log N+O(1)$, we now have
\begin{align*}
\mathcal I &\ll \frac{ (4 \log \log N +4c_1)^U}{U!}<\left(\frac{4e\log\log N+4ec_1}{U}\right)^U\\
&=\left(\frac{4e}{2+2\varepsilon}+O\left(\frac1U\right)\right)^U\ll(\log N)^{2(1+\varepsilon)(1+\log2-\log(1+\varepsilon))}.
\end{align*}
Substituting this into \eqref{cS2} yields
\begin{align*}
\mathcal S(N)& \ll N(\log \log N)^{O(1)} (\log N)^{2(1+\varepsilon)(1+\log 2 - \log (1+\varepsilon)) - 3}\nonumber\\
&\leqslant} \def\ge{\geqslant} \def\pleq{\preccurlyeq} \def\ple{\prec N(\log\log N)^{O(1)}(\log N)^{\log4-1+2\varepsilon(1+\log2)}.
\end{align*}
As $c' > \log 4 -1$, we may choose $\varepsilon > 0$ to give
$\mathcal S(N) \ll_{c'} N (\log N)^{c'}$.
\end{proof}
Combining \eqref{Cauchy} and \eqref{Lbound} with Lemma \ref{L2} establishes Theorem \ref{LowerBound}.
\section{A final comment}
We conjecture that Theorem \ref{UpperBound} holds with equality. For a lower bound, one might
restrict attention to those pairs $(a,b)$ with $\omega(a)\approx\omega(b)\approx\frac1{2\log2}\log\log N$.
The upper bound for the second moment is analysed as in the paper, getting $N/(\log N)^{\eta+o(1)}$; we expect that a more refined analysis would give
\[
\frac {N (\log \log N)^{O(1)}} {(\log N)^\eta}
\]
here. The difficulty is in obtaining this same estimate as a lower bound for the first moment.
This would follow if we had an analogue of Theorem \ref{HL} in which $a,b$ have a restricted number of prime
factors. Such a result holds for the general distribution of Gaussian primes, at least if one restricts
only one of $a,b$, see \cite{FI}.
\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 8,892 |
Richard Lester, född 19 januari 1932 i Philadelphia, Pennsylvania, är en amerikansk filmregissör, verksam i Storbritannien. Han är främst känd för sitt arbete med The Beatles-filmer under 1960-talet och med Stålmannen-filmer under 1980-talet.
Richard Lester flyttade 1953 till London och började arbeta inom TV. Han gjorde bland annat den Oscarsnominerade kortfilmskomedin The Running Jumping & Standing Still Film (1960) med Spike Milligan och Peter Sellers. Han valdes sedan att regissera The Beatles i filmen A Hard Day's Night (1964), vilken året därpå följdes av Hjälp!. Därefter regisserade han filmer som Greppet (1965, originaltitel The Knack ... and How to Get It), för vilken han belönades med Guldpalmen på filmfestivalen i Cannes, En kul grej hände på väg till Forum (1966) och Petulia (1968).
Under 1970-talet gjorde han bland annat De tre musketörerna (1973) och De fyra musketörerna (1974), vilka båda vann Evening Standard British Film Award i kategorin Bästa komedi, samt actionfilmen Juggernaut (1974) och komedin Royal Flash (1975), med flera. Under 1980-talet regisserade han Superman II (1980) och Stålmannen går på en krypto-nit (1983), samt Jakten på miljonerna och The Return of the Musketeers (1989). Därefter har han dragit sig tillbaka från regissörsarbetet. 1991 regisserade han dock en konsertfilm för Paul McCartney.
Filmografi i urval
Referenser
Externa länkar
Amerikanska filmregissörer
Amerikanska TV-regissörer
Män
Födda 1932
The Beatles
Levande personer
Personer från Philadelphia | {
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} | 1,155 |
Bug #71168 fixed one Unix page but there are many others that need to be looked at. Here are a few I saw.
Are these servers still supported?
but it doesn't like there are official packages for 7 yet.
it still needs more careful review).
> Are these servers still supported? | {
"redpajama_set_name": "RedPajamaC4"
} | 3,118 |
Q: Marking nested makes terms query and aggregation not working to achieve proper query on attributes array , i declared the field as nested. Now the terms query nor the normal match query is not giving any results. Is there something i should do specific about this type ?
A: To make this working , you need to use
*
*Nested query for querying nested fields
*Nested filter for filtering
*Nested aggregation for aggregation
This is due to the change that the for each element in the nested array , a new nested lucene document is made and you need separate set of query/agg to access them. Else it looks in the root document , where it is not present.
| {
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TO THE LIGHTHOUSEjudging from her own experience, would all thetime be feeling, This is not what we want; thereis nothing more tedious, puerile, and inhumanethan this; yet it is also beautiful and necessary.Well then, well then? she asked, somehow expect-ing the others to go on with the argument, asif in an argument like this one threw one's ownlittle bolt which fell short obviously and left theothers to carry it on. So she listened again towhat they were saying in case they should throwany light upon the question of love. | {
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Liu Xin (; * 10. November 1975 in Zhenjiang, Jiangsu, Volksrepublik China) ist eine chinesische Fernsehmoderatorin und Journalistin.
Leben
Liu Xin wurde in Zhenjiang in der Provinz Jiangsu der Volksrepublik China geboren. Von 1993 bis 1997 studierte sie Englische Sprache und Literatur an der Universität Nanjing und beendete ihr Studium mit dem Bachelor of Arts. Im Mai 1995 gewann Liu einen internationalen Wettbewerb für öffentliche Reden, der von der English-Speaking Union (ESU) in London ausgetragen wurde. Nach Abschluss ihres Studiums im Jahr 1997 begann Liu für den Fernsehsender CCTV News, dem Vorläufer von CGTN und ein Kanal von China Central Television (CCTV), zu arbeiten. 2011 war sie Mitgründerin eines Nachrichtenbüros der CCTV in Genf und hielt für fast sechs Jahre die Position des Bureau Chief.
Im Jahr 2017 startete Liu das Programm The Point with Liu Xin, das zum Hauptabendprogramm von CGTN wurde.
Privates
Liu ist mit einem Deutschen türkischer Herkunft verheiratet. Sie haben zusammen zwei Kinder.
Einzelnachweise
Fernsehmoderator
Journalist (China)
Chinese
Geboren 1975
Frau | {
"redpajama_set_name": "RedPajamaWikipedia"
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The brooch « Rosie » is adorned with a big rose with garnet purple and black colors surrounded by her olive green leaves.
These SILK roses and LEAVES are completely handcrafted by the designer Manuela Biocca thanks to traditional tools for flower making art (paint, shape, assembly).
Pined to your top or even your handbag, it will not miss to brighten up your outfit giving it a romantic and colorful touch. | {
"redpajama_set_name": "RedPajamaC4"
} | 4,091 |
Climate Change Educational Materials Released
Need to engage teachers in learning about the carbon cycle? Trying to explain the evidence for climate change? Want to help forest landowners think about growing resilient forests, sequestering carbon, or understanding why using wood products removes carbon from the atmosphere?
You can find engaging activities and teaching resources for these topics and more in a new educational module produced by the University of Florida environmental education team with Project Learning Tree (PLT). The module, Southeastern Forests and Climate Change was produced for PINEMAP, the USDA/NIFA-funded integrated research-education-Extension project focusing on loblolly pine and climate change. This secondary module is designed for educators to use in middle and high school classrooms and is framed on the research activities associated with PINEMAP, but explains these concepts in the broader context of southern forests. Adaptations to select activities are being developed by Extension educators in Kentucky to feature Appalachian hardwoods, and this information will be available to others as well.
The website makes these activities and supplemental resources (videos, slide presentations, answer keys, etc.) available to interested educators: http://sfrc.ufl.edu/extension/ee/climate. Feel free to share this resource with your networks.
The collection of 14 activities helps biology, agriculture, and environmental science teachers focus on the interactions between climate and forests and the ways we can manage forests to adapt to and mitigate future change. It has also been used with 4-H youth and forest landowners in workshops and presentations. It is a great regional complement to PLT's other secondary modules including Focus on Forests and Forests of the World. This module retains the tried and true features of PLT's materials: engaging activities, teacher background, step-by-step instructions, interdisciplinary focus on a controversial issue, science-based perspectives, critical thinking skills, data analysis, modifications and enrichment suggestions for adaptations, and correlations to national science standards. The module also offers some new additions to PLT materials: a systems thinking connection in each activity and supplemental activities on the website, a research connection and videos with researchers explaining their work, and quotes from the pilot test teachers that provide encouraging words of wisdom on each activity.
This material can be defined in terms of several current buzzwords--STEM education or Education for Sustainability, for example. It engages learners in understanding science, using math skills, applying technology, integrating economics and justice, building skills in systems and critical thinking, enhancing group process and communication, and considering how we can approach the challenges of the future together. It is just good environmental education!
Between now and December 2015 copies of books will be available at no charge through the PINEMAP grant and mini-grants available to southeastern PLT Coordinators who wish to distribute this module in workshops. Extension faculty can use the resource and/or assist their state PLT coordinator with workshops. PINEMAP researchers are also available to assist.
It's For the Birds
Bringing Nature Home
The Clearwater Conservancy is sponsoring a presentation by University of Delaware professor Dr. Doug Tallamy. Dr. Tallamy is the author of a relatively new book entitled "Bringing Nature Home." The presentation is entitled "It's for the Birds" and will take place at the Centre County Convention and Visitors Bureau on November 20 in State College, PA. Tallamy will discuss what we can do to give birds what they need from our landscapes and explore the consequences of failing to do so. This presentation is worth seeing. Dr. Tallamy presents some intriguing information that will open your eyes to a whole new way of looking at our landscapes.
Dr. Doug Tallamy
"It's for the Birds"
Many bird populations in the U.S. are in steep decline, in part because our gardens and managed landscapes occupy more space than natural areas and we have not designed them with birds in mind. To do that we can no longer view plants only as ornaments but must consider all of their roles when selecting them for our landscapes. Tallamy will discuss what birds need from our landscapes to breed successfully, the important roles native plants play in maintaining food webs vital to birds, emphasize the benefits of designing landscapes with these roles in mind, and explore the consequences of failing to do so. Landscaping in this crowded world carries both moral and ecological responsibilities that we can no longer ignore.
A book signing and sale will follow Doug's talk.
The cost is only $15 or $10 for students. Pre-registration is required online only at www.clearwaterconservancy.org
Below is a story about the author that appeared in the New York Times in March 2008 by Anne Raver.
To Feed the Birds First Feed the Bugs
DOUG TALLAMY and his wife, Cindy, built their house seven years ago in the middle of 10 acres of former hayfields. But they don't sit inside much. Most of their spare time is spent cutting Oriental bittersweet and Japanese honeysuckle out of cherry and oak trees. They saw down thickets of autumn olive and multiflora rose and paint the cut stems with an herbicide that goes down into the roots and kills them.
The land was so thick with multiflora rose that they couldn't walk, so Mr. Tallamy cut paths with hand loppers. They work with handsaws, not a chain saw. And they paint on the herbicide, rather than spraying it, because they don't want to damage the treasures below: under those thorny rose bushes might be seedlings of black oak, Florida dogwood, black gum or arrowwood viburnum, which, if protected from deer, could flourish in the cleared space.
To read the rest of the story click here.
Pennsylvania's Northern Bobwhite Quail
woodcrestpoint.com
Saw this headline today, "Bobwhite Quail Close to Extinction in PA and NJ." Having spent eight years of my career in eastern Virginia where I was stationed on a small state forest that had a quail habitat management area I saw the impacts habitat improvement can have on the population. Small things such as burning old fields, planting warm season grasses, and disking area on a 3 year rotation greatly improved the quail population.
We are celebrating the 3 year anniversary of the release of the Pennsylvania Game Commission's "Quail Management Plan." The mission of the Northern Bobwhite Quail Management Plan for Pennsylvania was to maintain and restore wild breeding populations of Northern Bobwhite Quail in suitable habitats.
"The northern bobwhite quail is one of the most popular game birds in North America. Its native range included most of the eastern United States north to southern Maine, southern New York, southern Ontario, central Wisconsin and south central Minnesota, west to very southeastern Wyoming, eastern Colorado, eastern New Mexico, and eastern Mexico south to Chiapas. Twenty-two subspecies have been recognized. Since the mid 1960's, the bobwhite's range and populations have declined dramatically. Northern bobwhites were relatively common across southern Pennsylvania farmland and brush lands until about 1945. Populations declined rapidly between 1945-1955, but made a recovery in the early 1960's. Since 1966, the range and populations of bobwhites have declined to the point that most counties in the commonwealth no longer have bobwhites as a breeding species." (From PA Game Commission, Quail Management Plan, 2011)
In doing a quick internet search to see if anything existed on the current status of the management plan and quail populations I came across a number of interesting news articles but nothing current to provide us with an indication of the success or failure of the plan being implemented here. Unfortunately, this leaves us wondering…What is the current status of quail in Pennsylvania? Has the management plan been working? Or, are the headlines correct and wild quail populations are close to extinction here in PA?
Bobwhite Quail Close to Extinction in Pa. and NJ
By Edward Colimore, The Philadelphia Inquirer, October 12, 2014
Bill Haines Jr. used to see wild quail on his family's farm all the time when he was growing up. He heard their distinctive "bobwhite" calls and thought nothing of it. Fifty years ago, the small chicken-like bird thrived across parts of the state. Coveys of them were common. Hunters flushed them out by the scores while walking through brushy fields.
Now, their singing has all but stopped. The number of wild bobwhite quail has fallen off so precipitously that — except for small pockets — they're close to extinction in New Jersey and Pennsylvania, and barely holding on in Delaware, wildlife ecologists say. Choked forests, paved roads, housing developments, herbicides, and pesticides have destroyed food sources and nesting grounds. The birds disappeared as their habitat disappeared.
Click here for the rest of the story.
These news releases may be of interest to you as well. They are from 2011 when the quail management plan was released.
Can the bobwhite quail make a comeback in Pennsylvania?
By Marcus Schneck, October 05, 2011
The northern bobwhite quail – a familiar and popular species that even the experts are not sure still exists in the wild, non-stocked state in Pennsylvania – now has a management plan in the Keystone State. The Pennsylvania Board of Game Commissioners on Tuesday approved an aggressive plan, which has a starting point of next July. Although quail can be found in many areas of Pennsylvania, no one seems certain if any of those birds are native to the state or the result of the estimated 60,000 to 70,000 pen-raised birds stocked by hunting dog enthusiasts across the state each year.
A life preserver for bobwhite quail?
By Ad Crable, September 12, 2011
The last time I heard the whistle of a bobwhite quail calling its own name in Lancaster County was at least 20 years ago. I was hiking through tall grass in a long-deserted farm that had become part of the Muddy Run pumped-storage reservoir project. I stopped dead and tried whistling back. Bob-bob-white. Memories of chasing coveys of quail over hill and dale with a pointer and shotgun on Illinois farmland as a young boy came rushing back. I've not heard one here since. That's because there are few wild reproducing quail left in Pennsylvania.
Tree Species for Planting in Asian Longhorned and Emerald Ash Borer Beetle Areas
The table of tree species that do well in urban environments and are NOT hosts for Asian longhorned beetle (ALB) or emerald ash borer (EAB) has been revised recently. The table provides a starting point for identifying potential trees species to plant in areas impacted by these two pests. Tree planting, whether a replacement or new, is critical in helping communities provide valuable tree cover for aesthetics, clean air, stormwater reductions, and wildlife habitat.
Asian longhorned beetle
ALB and EAB have had a significant impact on communities in the Northeastern and Midwestern US. Numerous requests have been received from communities for a list of trees that do well in urban environments and are not ALB or EAB host trees. To develop this list, the Forest Service combined all the recommended replacement trees for areas where ALB has been found and added information on EAB tree replacement from the Michigan State University website. Information about each tree species was compiled from numerous references which are listed in The Replacement Tree Table. Drafts of the table were sent to urban forestry professionals for review. The table was edited using their suggestions.
To promote species diversity, the Forest Service included many trees that do well in urban areas but realize the table does not include every non-host tree species. Users of The Replacement Tree Table are encouraged to use this as a starting reference for potential replacement trees. The list will then need to be narrowed by researching the species that best fit your specific site characteristics.
Tree Species for Planting in Asian Longhorned and ... | {
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{"url":"http:\/\/cicy.itp.tuwien.ac.at\/wiki\/index.php\/Documentation_on_nef.x","text":"# Documentation on nef.x\n\nnef.x is a package to analyze nef partitions of a reflexive polytope. Such nef partitions determine complete intersections of Calabi-Yau type in toric varieties of, in principle, arbitrary codimension. Given a reflexive polytope in terms of a combined weight system or a list of points the main objective of the program is to determine the nef partitions and the Hodge numbers of the corresponding Calabi-Yau varieties. Further features include the calculation of the corresponding reflexive Gorenstein cones as well as information about the fibration structure.\n\nThe corresponding routines are listed in the header file Nef.h.\n\n## Nef partitions and reflexive Gorenstein cones\n\nConsider a dual pair of d-dimensional reflexive polytopes $\\Delta\\subset M_\\mathbb{R},\\Delta^*\\subset N_\\mathbb{R}$. A partition $V=V_0\\cup\\dots\\cup V_{r-1}$ of the set of vertices of \u0394 * into disjoint subsets $V_0,\\dots,V_{r-1}$ is called a nef partition of length r if there exist r integral upper convex \u03a3(\u0394 * )-piecewise linear support functions $\\phi_l:N_\\mathbb{R}\\rightarrow \\mathbb{R}$, $l=0,\\dots,r-1$ such that\n\n$\\phi_l(v)= \\begin{cases} 1 & \\text{if } v \\in V_l, \\\\ 0 & \\text{otherwise.} \\end{cases}$\n\nEach \u03d5l corresponds to a divisor $D_{0,l}=\\sum_{v \\in V_l} D_{v}$ on the toric variety $\\mathbb{P}_{\\Delta^*}$ associated to \u0394 * , and the intersection of all these divisors\n\n$X=D_{0,0}\\cap\\dots\\cap D_{0,r-1}$\n\ndefines a family $X\\subset \\mathbb{P}_{\\Delta^*}$ of Calabi-Yau complete intersections of codimension r.\n\nMoreover, each \u03d5l corresponds to a lattice polytope \u0394l defined as\n\n$\\Delta_l= \\Big\\{ x\\in M_\\mathbb{R} \\,\\Big|\\, (x,y)\\geq -\\phi_l(y) \\quad \\forall y\\in N_\\mathbb{R} \\Big\\} .$\n\nThe sum of the functions \u03d5l is equal to the support function of the anticanonical divisor $K_{\\mathbb{P}_{\\Delta^*}}^{-1}$ and, therefore, the corresponding Minkowski sum is $\\Delta_0 + \\dots + \\Delta_{r-1} = \\Delta$. Moreover, the knowledge of the decomposition $V=V_0\\cup\\dots\\cup V_{r-1}$ is equivalent to that of the set of supporting polytopes $\\Pi(\\Delta)=\\{\\Delta_0,\\dots,\\Delta_{r-1}\\}$, and therefore this data is often also called a nef partition.\n\nFor a given nef partition \u03a0(\u0394) the polytopes (The brackets $\\langle \\cdots \\rangle$ denote the convex hull.)\n\n$\\nabla_{l'}= \\big\\langle \\{0\\}\\cup V_{l'} \\big\\rangle ~\\subset N_\\mathbb{R}$\n\ndefine again a nef partition $\\Pi^\\ast(\\nabla)=\\{\\nabla_0,\\dots,\\nabla_{r-1}\\}$ such that the Minkowski sum $\\nabla=\\nabla_0+\\dots+\\nabla_{r-1}$ is a reflexive polytope. Then, its dual $\\nabla^*$ is also reflexive, and $\\Pi^\\ast(\\nabla)$ is called the dual nef partition. This is the combinatorial manifestation of mirror symmetry in terms of dual pairs of nef partitions of \u0394 * and $\\nabla^*$, which we summarize in the following diagram\n\n$\\begin{array}{rcl} M_{\\mathbb{R}}\\qquad\\qquad && \\qquad\\qquad N_{\\mathbb{R}}\\\\ & & \\\\ \\Delta=\\Delta_0+\\ldots+\\Delta_{r-1} && \\Delta^*=\\langle\\nabla_0,\\ldots,\\nabla_{r-1}\\rangle\\\\ & & \\\\ &~~~~ (\\Delta_l,\\nabla_{l'})\\ge-\\delta_{l\\,l'} ~~~~&\\\\ & & \\\\ \\nabla^*=\\langle\\Delta_0,\\ldots,\\Delta_{r-1}\\rangle && \\nabla=\\nabla_0+\\ldots+\\nabla_{r-1} \\end{array}$\n\nIn the horizontal direction, we have the duality between the lattices M and N and mirror symmetry goes from the upper right to the lower left. The complete intersections $X\\subset \\mathbb{P}_{\\Delta^*}$ and $\\check{X} \\subset \\mathbb{P}_{\\nabla^*}$ associated to the dual nef partitions are then mirror Calabi-Yau varieties.\n\nThere are two constructions to build new nef partitions from old ones: projections and direct products. Given a nef partition $V=V_0\\cup\\dots\\cup V_{r-1}$ where one of the subsets Vl, say V0, consists of a single vertex v, the nef condition implies that the projection $\\Delta^*_v$ of \u0394 * along v is reflexive. Moreover,the Calabi-Yau complete intersection X is given by $D \\cap X'$ with $X' = \\bigcap_{l=1}^{r-1} D_{0,l}$. Since D can only intersect the toric divisors that correspond to points bounding the reflexive projection along v, the variety X is isomorphic to the variety $X' \\subset \\mathbb{P}_{\\Delta^*_v}$, where $\\mathbb{P}_{\\Delta^*_v}$ is obtained from the projection $\\Delta^*_v$. In hep-th\/0410018 such nef partitions were called trivial. In nef.x they are labeled by P for projection, see -P.\n\nSuppose we are given two lattices M(1),M(2) and two reflexive polytopes $\\Delta^{(1)}\\subset M^{(1)}_{\\mathbb{R}}$, $\\Delta^{(2)}\\subset M^{(2)}_{\\mathbb{R}}$ such that (1)) * and (2)) * admit nef partitions $V^{(1)} = \\bigcup_l V_l^{(1)}$ and $V^{(2)} = \\bigcup_l V_l^{(2)}$, respectively. Then $\\Delta= \\Delta^{(1)} \\times\\Delta^{(2)}$ is reflexive with respect to $M=M^{(1)}\\oplus M^{(2)}$ and dual to \u0394 * whose set of vertices V is $\\{ (v^{(1)},0) \\,|\\, v^{(1)}\\in V^{(1)}\\} \\cup \\{ (0,v^{(2)}) \\,|\\, v^{(2)} \\in V^{(2)}\\}$. V admits a nef partition induced from the nef partitions V(1) and V(2). Such a nef partition is called a direct product since the corresponding Calabi-Yau complete intersection X is a direct product $X=X^{(1)} \\times X^{(2)}$ in $\\mathbb{P}_{\\Delta^*} = \\mathbb{P}_{(\\Delta^{(1)})^*} \\times \\mathbb{P}_{(\\Delta^{(2)})^*}$.\n\nOne can reformulate the duality of nef partitions in terms of reflexive Gorenstein cones as follows. We extend the lattices M and N to $tilde M = \\mathbb{Z}^r \\oplus M$ and $\\tilde N = \\mathbb{Z}^r \\oplus N$ and set $\\tilde d = d+r$.\n\nA $\\tilde d$-dimensional rational polyhedral cone C in $\\tilde M_{\\mathbb{R}}$ is called Gorenstein if $C\\cap(-C)=\\{0\\}$, there exists an element $n_C \\in \\tilde N_{\\mathbb{R}}$ such that $\\langle x,n_C\\rangle > 0$ for any nonzero $x \\in C$, and all vertices of the $(\\tilde d -1)$-dimensional convex polytope\n\n$\\Delta(C)=\\{x\\in C \\,|\\, \\langle x,n_C\\rangle=1\\}$\n\nbelong to $\\tilde M$. The polytope \u0394(C) is called the support of C. Conversely, any $(\\tilde d -1)$-dimensional lattice polytope \u039b determines a $\\tilde d$-dimensional Gorenstein cone C(\u039b) as the cone over \u039b with apex at lattice distance 1 from the hyperplane carrying \u039b; obviously \u0394(C(\u039b)) = \u039b. For any $m \\in C \\cap \\tilde M$, we define the degree of m as $\\deg m = \\langle m, n_C \\rangle$.\n\nA Gorenstein cone C is called reflexive if the dual cone\n\n$\\check C = \\{y \\in \\tilde N_\\mathbb{R} \\,|\\, \\langle x,y \\rangle \\geq 0\\; \\forall\\, x\\in C\\}$\n\nis also Gorenstein, i.e., there exists $m_{\\check C} \\in \\tilde M$ such that $\\langle m_{\\check C},y\\rangle > 0$ for all $y \\in \\check C \\setminus \\{0\\}$, and all vertices of the support $\\Delta ( \\check C ) = \\{ y \\in \\check C \\,|\\, \\langle m_{\\check C} , y \\rangle = 1 \\}$ belong to $\\tilde N$. We will call the integer $r = \\langle m_{\\check C} , n_C\\rangle$ the index of C (or $\\check C$).\n\nAny nef partition $\\Pi(\\Delta)=\\{\\Delta_0,\\dots,\\Delta_{r-1}\\}$ of length r of a reflexive polytope \u0394 determines a $\\tilde d$-dimensional dual pair of reflexive Gorenstein cones $C = C(\\Delta_1,\\dots,\\Delta_r)\\subset\\tilde M_{\\mathbb{R}}$, $\\check C = \\check C(\\nabla_1,\\dots,\\nabla_r)\\subset\\tilde N_{\\mathbb{R}}$ of index r by\n\n$C = \\{ (\\lambda_1, \\dots ,\\lambda_r,\\lambda_1x_1 + \\dots + \\lambda_r x_r) \\in \\tilde M_{\\mathbb{R}} \\,|\\, \\lambda_i \\geq 0, x_i \\in \\Delta_i, i=1,\\dots, r\\},$ $\\check C = \\{ (\\mu_1, \\dots ,\\mu_r,\\mu_1x_1 + \\dots + \\mu_r x_r) \\in \\tilde N_{\\mathbb{R}} \\,|\\, \\mu_i \\geq 0, x_i \\in \\nabla_i, i=1,\\dots, r\\}.$\n\nThere are, however, reflexive Gorenstein cones that do not come from nef partitions.\n\nA reflexive Gorenstein cone admits a representation in terms of the points of the underlying reflexive polytope as follows. Given a point $p \\in \\nabla_l$, the corresponding point $\\tilde{p} \\in \\check C(\\nabla_1,\\dots,\\nabla_r)$ is given as\n\n$\\tilde p = (\\phi_0(p),\\dots,\\phi_{r-1}(p),p).$\n\nwhere \u03d5l is the support function defined above. To see that the two descriptions of $\\check C$ are equivalent, note that both correspond to a cone whose support has vertices\n\n$(e_{i(1)},v_1),\\ldots,(e_{i(n)},v_n),(e_1, 0_N),\\ldots,(e_r,0_N),$\n\nwhere {ei} is the standard basis of $\\mathbb{Z}^r$, i(k) is the number such that $v_k\\in V^{(i(k))}$ and 0N is the origin in the N-lattice.\n\nThe Hodge numbers of a Calabi-Yau manifold X defined by means of a nef partition depend only on the structure of the corresponding reflexive Gorenstein cone in a manner described in math\/0103214 or alg-geom\/9509009. The corresponding formulas rely heavily on the counting of lattice points. For any lattice polytope \u039b let us denote by $\\ell(\\Lambda)$ the number of lattice points of \u039b and by $\\ell^*(\\Lambda)$ the number of lattice points in the interior of \u039b. It can be shown that\n\n$S_\\Lambda(t) = (1-t)^{\\dim \\Lambda + 1}\\sum_{k \\geq 0} \\ell(k\\Lambda) t^k$\n\nis a polynomial of degree $d\\le \\dim \\Lambda + 1$; S\u039b(t) is called the Ehrhart polynomial of \u039b. Similarly one can define a polynomial\n\n$T_\\Lambda(t) = (1-t)^{\\dim \\Lambda + 1}\\sum_{k \\geq 0} \\ell^*(k\\Lambda) t^k.$\n\nIn terms of a Gorenstein cone C over \u039b, with underlying lattice MC, S and T can be written as\n\n$S(C,t) = (1-t)^{\\dim C}\\sum_{m \\in C \\cap M_C} t^{\\deg m},$\n\n$T(C,t) = (1-t)^{\\dim C}\\sum_{m \\in \\mathrm{int}(C) \\cap M_C} t^{\\deg m}.$\n\nThe two polynomials satisfy a relation which is a consequence of Serre duality,\n\n$S(C,t) = t^{\\dim C}\\,T(C,t^{-1}),$\n\nwhich provides a stringent test on any results involving lattice point counting. For the computation of Hodge numbers, the S- and T- polynomials for all the faces of C(\u0394) as well as a polynomial called B, which is related to the poset structure of C(\u0394), are required.\n\n## Standard output\n\nIn this subsection we will explain in detail how to interpret the output of nef.x when called without any options.\n\nThe standard output slightly depends on whether the reflexive polytope is input as a combined weight system or as a collection of points. If the polytope was entered as a collection of points, the first line of the output takes the following form:\n\nM:# # N:# # codim=# #part=#\n\n\nNote that the input polytope is interpreted as $\\Delta\\subset M_\\mathbb{R}$ unless the option -N is used, while any output of a polytope in matrix format refers to its dual $\\Delta^*\\subset N_\\mathbb{R}$ except for the option -y. If the input is a CWS, the line starts with the CWS repeated before the symbol M.\n\n# M:# # N:# # codim=# #part=#\n\n\nwhere the first # stands for the sequence of numbers describing the CWS. The two numbers # after M correspond to the numbers of lattice points and vertices of $\\Delta\\subset M_\\mathbb{R}$ and the numbers # after N correspond to the numbers of lattice points and vertices of $\\Delta^*\\subset N_\\mathbb{R}$, respectively. The number r in codim=r is the length of the nef partition, i.e. the codimension of the corresponding Calabi-Yau complete intersection. The default value is 2, otherwise it is specified by the option -c*. The number n in #part=n is the number of all the nef partitions that nef.x has found, up to symmetries of the underlying lattice. If the symmetries of the underlying lattice should not be taken into account, use the option -s.\n\nThe subsequent lines contain the information about the various nef partitions. Note that the standard output suppresses the output of nef partitions which are equivalent under symmetries of the CWS. If the codimension is 2 the output line containing the information on a particular nef partition takes the following form:\n\nH:# [#] P:# V:# # #sec #cpu\n\n\nThe numbers # after H: are the Hodge numbers $h^{1,i}(X), i=1,\\dots,d-1$, where d is the dimension of the Calabi-Yau manifold X.\n\nThe number $in the square brackets [#] is the Euler number of X. If $h^{0,i}(X) \\not= 0$ for some $i=1,\\dots,d-1$ the Calabi-Yau manifold factorizes. See the option -D for this case. In any case, the full Hodge diamond is displayed with the option -H. The number # after P: is a counter specifying the nef partition. It runs from 0 to n - 1. Note that nef partitions corresponding to direct products and projections to nef partitions of lower length are omitted by default. To display them use the options -D, -Q for direct products and -P for projections. The sequence of numbers # separated by a single space after V: corresponds to the vertices that belong to the first part V0 of the nef partition. Note that the vertices are counted starting from 0. These numbers only make sense if the options -n, -Lv or -Lp are used. The vertices that belong to the second part <maht>V_1[\/itex] of the nef partition are not displayed, since they are simply the remaining ones. If the polytope entered also has points that are not vertices and if the option -Lp is used, then the second sequence of numbers # that is separated from the first sequence by two spaces corresponds to the non-vertex points that belong to the first part V0. For representations of the nef partition in terms of the Gorenstein cone see the option -g*. The number # before sec indicates the time that was needed to compute this partition. The number # before cpu indicates the number of CPU seconds that were needed to compute the Hodge numbers. For determining the nef partitions without computing the Hodge numbers see the option -p. If the length r is bigger than 2 the lines containing the information about the various nef partitions take the following form: H:# [#] P:# V0:# # V1:# # ... V(r-2):# # #sec #cpu Now, there are r - 1 expressions of the form Vi:# #, where i runs from 0 to r - 2 each representing a part Vi of the nef partition. The points and vertices in each Vi are listed in the same order as in the codimension two case. The final line of the output always takes the following form: np=# d:# p:# #sec #cpu The numbers # after d:, p:, np= are the numbers of nef partitions which are direct products, projections, and neither of the two, respectively. The total of the three numbers adds up to n, the total number of nef partitions as indicated in the first line after #part=. The following example illustrates the standard output of nef.x. We consider complete intersections of codimension 2 in $\\mathbb{P}^2\\times\\mathbb{P}^1\\times\\mathbb{P}^2$ discussed in arXiv:0704.0449[hep-th ] . Let $e_1,\\dots,e_5$ be the standard basis of $\\mathbb{R}^5$. We define the polytope $\\Delta^* \\subset N$ by the vertices $v_0,\\dots,v_7$ given by v0 = e1,v1 = e2,v2 = \u2212 e1e2,v3 = e3,v4 = \u2212 e3,v5 = e4,v6 = e5,v7 = \u2212 e4e5. By elementary toric geometry, we see that $\\mathbb{P}_{\\Delta^*} = \\mathbb{P}^2\\times\\mathbb{P}^1\\times\\mathbb{P}^2$ and the combined weight system can be read off from the linear relations v0 + v1 + v2 = 0,v3 + v4 = 0,v5 + v6 + v7 = 0. First, we enter the polytope by giving this combined weight system palp$ nef.x\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #colums' (= PolyDim #Points' or #Points PolyDim'):\n3 1 1 1 0 0 0 0 0 2 0 0 0 1 1 0 0 0 3 0 0 0 0 0 1 1 1\n3 1 1 1 0 0 0 0 0 2 0 0 0 1 1 0 0 0 3 0 0 0 0 0 1 1 1\nM:300 18 N:9 8 codim=2 #part=15\nH:19 19 [0] P:0 V:2 4 6 7 1sec 0cpu\nH:9 27 [-36] P:2 V:3 4 6 7 1sec 0cpu\nH:3 51 [-96] P:3 V:3 5 6 7 1sec 1cpu\nH:3 75 [-144] P:4 V:3 6 7 1sec 0cpu\nH:3 51 [-96] P:6 V:4 5 6 7 2sec 1cpu\nH:3 51 [-96] P:7 V:4 5 6 1sec 1cpu\nH:6 51 [-90] P:8 V:4 6 7 1sec 1cpu\nH:3 75 [-144] P:9 V:4 6 1sec 1cpu\nH:3 60 [-114] P:10 V:5 6 7 2sec 1cpu\nH:3 69 [-132] P:11 V:5 6 1sec 1cpu\nH:3 75 [-144] P:12 V:6 7 1sec 0cpu\nnp=11 d:2 p:2 0sec 0cpu\n\n\nEquivalently, we can use the option -N and enter the points of the polytope \u0394 * of the normal fan of $\\mathbb{P}^2\\times\\mathbb{P}^1\\times\\mathbb{P}^2$:\n\npalp$nef.x -N Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #colums' (= PolyDim #Points' or #Points PolyDim'): 5 8 Type the 40 coordinates as dim=5 lines with #pts=8 colums: 1 0 -1 0 0 0 0 0 0 1 -1 0 0 0 0 0 0 0 0 1 -1 0 0 0 0 0 0 0 0 1 0 -1 0 0 0 0 0 0 1 -1 M:300 18 N:9 8 codim=2 #part=15 H:3 51 [-96] P:0 V:2 3 4 7 1sec 1cpu H:3 51 [-96] P:1 V:2 4 6 7 2sec 1cpu H:3 60 [-114] P:2 V:2 4 7 2sec 1cpu H:3 51 [-96] P:3 V:2 6 7 1sec 1cpu H:3 69 [-132] P:4 V:2 7 1sec 1cpu H:9 27 [-36] P:5 V:3 4 6 7 1sec 0cpu H:3 75 [-144] P:6 V:3 4 7 0sec 0cpu H:19 19 [0] P:8 V:4 5 6 7 1sec 0cpu H:6 51 [-90] P:9 V:4 6 7 1sec 1cpu H:3 75 [-144] P:10 V:4 7 1sec 0cpu H:3 75 [-144] P:13 V:6 7 1sec 1cpu np=11 d:2 p:2 0sec 0cpu Note that both the points and the nef partitions are given in different orders. The polytope $\\Delta^* \\subset N_{\\mathbb{R}}$ has 9 points, 8 vertices and the interior point, while the dual polytope $\\Delta \\subset M_{\\mathbb{R}}$ has 300 points, 18 of which are vertices. The codimension is 2 and there are 15 nef partitions. There are 11 nef partitions listed, furthermore there are 2 nef partitions which are direct products, and 2 which are projections. According to the output the nef partitions e.g. 0 and 8 are given as follows (with the Hodge numbers and the Euler number of the corresponding Calabi-Yau 3-fold X): $0:\\; V_0=\\langle v_2, v_3, v_4, v_7 \\rangle,\\quad V_1=\\langle v_0, v_1, v_5, v_6 \\rangle h^{1,1}(X) = 3,\\, h^{2,1}(X) = 51,\\, \\chi(X) = -96.$ $8:\\; V_0=\\langle v_4, v_5, v_6, v_7 \\rangle,\\quad V_1=\\langle v_0, v_1, v_2, v_3 \\rangle, h^{1,1}(X) = 19,\\, h^{2,1}(X) = 19,\\, \\chi(X) = 0.$ ## Global parameters and limitations If the dimension of the polytope or the codimension of the nef partition are large, certain global variables in the header files Global.h and Nef.h may need to be modified. This depends very much on the problem to be treated by nef.x as well as on the CPU and the operating system of the computer nef.x is running on. Here we give a particularly nasty example: palp$ nef.x -Lp -N -c6 -P\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n7 9\nPlease increase POLY_Dmax to at least 12 = 7 + 6 - 1\n(nef.x requires POLY_Dmax >= dim N + codim - 1)\n\n\nThis means that in Global.h we need to set POLY_Dmax to at least 12:\n\n#define POLY_Dmax 12 \/* max dim of polytope *\/\n\n\nAfter recompiling PALP we get further but not far enough:\n\npalp$.\/nef.x -Lp -N -c6 -P Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #columns' (= PolyDim #Points' or #Points PolyDim'): 7 9 Type the 63 coordinates as dim=7 lines with #pts=9 columns: 1 0 0 0 0 -1 0 0 -1 0 1 0 0 0 -1 0 0 -1 0 0 1 0 0 -1 0 0 -1 0 0 0 1 0 -1 0 0 0 0 0 0 0 1 -1 0 0 0 0 0 0 0 0 0 1 0 -1 0 0 0 0 0 0 0 1 -1 M:5214 12 N:10 9 codim=6 #part=1 7 10 Points of Poly in N-Lattice: 1 0 0 0 0 -1 0 0 -1 0 0 1 0 0 0 -1 0 0 -1 0 0 0 1 0 0 -1 0 0 -1 0 0 0 0 1 0 -1 0 0 0 0 0 0 0 0 1 -1 0 0 0 0 0 0 0 0 0 0 1 0 -1 0 0 0 0 0 0 0 0 1 -1 0 -------------------------------------------------- 1 1 1 1 1 1 0 0 0 d=6 codim=2 1 1 1 0 0 0 1 1 1 d=6 codim=2 nef.x: Vertex.c:613: Finish_Find_Equations: Assertion _V->nv<64' failed. Aborted This can be remedied by adjusting the global variable VERT_Nmax in Global.h as follows (it should not be too large): #define VERT_Nmax 128 \/* !! use optimal value !! *\/ After recompilation it works for a while. Then the following error occurs Unable to alloc space for _BL This means that the program has run out of memory. ## Help screen The help screen for nef.x is: palp$ .\/nef.x -h\n\nThis is '.\/nef.x': calculate Hodge numbers of nef-partitions\nUsage: .\/nef.x <Options>\n\nOptions: -h prints this information\n-f or - use as filter; otherwise parameters denote I\/O files\n-N input is in N-lattice (default is M)\n-H gives full list of Hodge numbers\n-Lv prints L vector of Vertices (in N-lattice)\n-Lp prints L vector of Points (in N-lattice)\n-p prints only partitions, no Hodge numbers\n-D calculates also direct products\n-P calculates also projections\n-t full time info\n-cCODIM codimension (default = 2)\n-Fcodim fibrations up to codim (default = 2)\n-y prints poly\/CWS in M lattice if it has nef-partitions\n-S information about #points calculated in S-Poly\n-T checks Serre-duality\n-s don't remove symmetric nef-partitions\n-n prints polytope only if it has nef-partitions\n-v prints vertices and #points of input polytope in one\nline; with -u, -l the output is limited by #points:\n-uPOINTS ... upper limit of #points (default = POINT_Nmax)\n-lPOINTS ... lower limit of #points (default = 0)\n-m starts with [d w1 w2 ... wk d=d_1 d_2 (Minkowski sum)\n-R prints vertices of input if not reflexive\n-V prints vertices of N-lattice polytope\n-Q only direct products (up to lattice Quotient)\n-gNUMBER prints points of Gorenstein polytope in N-lattice\n-dNUMBER prints points of Gorenstein polytope in M-lattice\nif NUMBER = 0 ... no 0\/1 info\nif NUMBER = 1 ... no redundant 0\/1 info (=default)\nif NUMBER = 2 ... full 0\/1 info\n-G Gorenstein cone: input <-> support polytope\n\n\n## The options in detail\n\nWe will explain all the options of nef.x in the order of their appearance in the help screen. Here is a rough guide in terms of specific topics:\n\n### -h\n\nThis option prints the help screen.\n\n### -f or -\n\nThis option switches off the prompt for the input data. This is useful for building pipelines.\n\n### -N\n\nThe option -N interprets the input polytope in the N-lattice instead of the M-lattice.\n\nThe following example of a complete intersection of degree (2,2) in $\\mathbb{P}^3$ illustrates the difference. In order to point out the difference we display the points in the two lattices with the option -Lv.\n\npalp$nef.x -Lv Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #colums' (= PolyDim #Points' or #Points PolyDim'): 3 4 Type the 12 coordinates as dim=3 lines with #pts=4 colums: -1 0 0 1 -1 0 1 0 -1 1 0 0 M:5 4 N:35 4 codim=2 #part=0 3 4 Vertices in N-lattice: -1 -1 -1 3 -1 -1 3 -1 -1 3 -1 -1 -------------------- 1 1 1 1 d=4 codim=0 np=0 d:0 p:0 0sec 0cpu Without the option -N, the output polytope with 35 points and no nef partition is the dual of the input polytope. palp$ nef.x -Lv -N\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #colums' (= PolyDim #Points' or #Points PolyDim'):\n3 4\nType the 12 coordinates as dim=3 lines with #pts=4 colums:\n-1 0 0 1\n-1 0 1 0\n-1 1 0 0\nM:35 4 N:5 4 codim=2 #part=2\n3 4 Vertices in N-lattice:\n-1 0 0 1\n-1 0 1 0\n-1 1 0 0\n--------------------\n1 1 1 1 d=4 codim=0\nH:[0] P:0 V:2 3 (2 2) 0sec 0cpu\nnp=1 d:0 p:1 0sec 0cpu\n\n\nWith the option -N, the output polytope is the same as input polytope with 4 points and the expected nef partition corresponding to the complete intersection of degree (2,2) in $\\mathbb{P}^3$. Note that the order of the points in the output is the same as in the input. This last feature is the main advantage of the option -N. The reason is that the basis chosen does not respect the order given by the combined weight system that was entered. This can be extremely inconvenient at times. The option -N provides a way to work around this issue: first use the option -Lv to obtain the vertices for a given CWS. Then reorder them so that the basis of linear relations complies with the input and enter the reshuffled vertices into nef.x using the option -N. This will force the linear relations chosen by nef.x to be the same as the CWS.\n\nTo see this we consider the example used in the description of the standard output, the complete intersections of codimension 2 in $\\mathbb{P}^2\\times\\mathbb{P}^1\\times\\mathbb{P}^2$ discussed in arXiv:0704.0449[hep-th].\n\npalp$nef.x -Lv Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #colums' (= PolyDim #Points' or #Points PolyDim'): 3 1 1 1 0 0 0 0 0 2 0 0 0 1 1 0 0 0 3 0 0 0 0 0 1 1 1 3 1 1 1 0 0 0 0 0 2 0 0 0 1 1 0 0 0 3 0 0 0 0 0 1 1 1 M:300 18 N:9 8 codim=2 #part=15 5 8 Vertices in N-lattice: 0 0 0 0 1 0 -1 0 0 0 1 0 0 0 -1 0 0 0 0 1 0 0 0 -1 -1 0 0 0 0 1 0 0 -1 1 0 0 0 0 0 0 ---------------------------------------- 1 1 0 0 0 1 0 0 d=3 codim=3 0 0 1 0 1 0 1 0 d=3 codim=3 0 0 0 1 0 0 0 1 d=2 codim=4 H:19 19 [0] P:0 V:2 4 6 7 (0 3) (3 0) (1 1) 1sec 0cpu H:9 27 [-36] P:2 V:3 4 6 7 (0 3) (2 1) (2 0) 0sec 0cpu H:3 51 [-96] P:3 V:3 5 6 7 (1 2) (1 2) (2 0) 1sec 0cpu H:3 75 [-144] P:4 V:3 6 7 (0 3) (1 2) (2 0) 0sec 0cpu H:3 51 [-96] P:6 V:4 5 6 7 (1 2) (2 1) (1 1) 2sec 1cpu H:3 51 [-96] P:7 V:4 5 6 (1 2) (2 1) (0 2) 1sec 0cpu H:6 51 [-90] P:8 V:4 6 7 (0 3) (2 1) (1 1) 1sec 0cpu H:3 75 [-144] P:9 V:4 6 (0 3) (2 1) (0 2) 0sec 0cpu H:3 60 [-114] P:10 V:5 6 7 (1 2) (1 2) (1 1) 2sec 1cpu H:3 69 [-132] P:11 V:5 6 (1 2) (1 2) (0 2) 1sec 0cpu H:3 75 [-144] P:12 V:6 7 (0 3) (1 2) (1 1) 0sec 0cpu np=11 d:2 p:2 0sec 0cpu Note that the basis chosen does not respect the order given by the combined weight system that was entered. E.g. the weight vector 2 0 0 0 1 1 0 0 0 has changed to the linear relation 0 0 0 1 0 0 0 1 d=2 where the 0's and 1's are in a different order. This can be overcome using the option -N. We choose a basis for the lattice such that the vertices of the polytope satisfy the desired combined weight system 3 1 1 1 0 0 0 0 0 2 0 0 0 1 1 0 0 0 3 0 0 0 0 0 1 1 1: palp$ nef.x -N -Lv\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #colums' (= PolyDim #Points' or #Points PolyDim'):\n5 8\nType the 40 coordinates as dim=5 lines with #pts=8 colums:\n1 0 -1 0 0 0 0 0\n0 1 -1 0 0 0 0 0\n0 0 0 1 -1 0 0 0\n0 0 0 0 0 1 0 -1\n0 0 0 0 0 0 1 -1\nM:300 18 N:9 8 codim=2 #part=15\n5 8 Vertices in N-lattice:\n1 0 -1 0 0 0 0 0\n0 1 -1 0 0 0 0 0\n0 0 0 1 -1 0 0 0\n0 0 0 0 0 1 0 -1\n0 0 0 0 0 0 1 -1\n----------------------------------------\n1 1 1 0 0 0 0 0 d=3 codim=3\n0 0 0 1 1 0 0 0 d=2 codim=4\n0 0 0 0 0 1 1 1 d=3 codim=3\nH:3 51 [-96] P:0 V:2 3 4 7 (1 2) (2 0) (1 2) 1sec 0cpu\nH:3 51 [-96] P:1 V:2 4 6 7 (1 2) (1 1) (2 1) 1sec 1cpu\nH:3 60 [-114] P:2 V:2 4 7 (1 2) (1 1) (1 2) 2sec 1cpu\nH:3 51 [-96] P:3 V:2 6 7 (1 2) (0 2) (2 1) 1sec 0cpu\nH:3 69 [-132] P:4 V:2 7 (1 2) (0 2) (1 2) 0sec 0cpu\nH:9 27 [-36] P:5 V:3 4 6 7 (0 3) (2 0) (2 1) 1sec 0cpu\nH:3 75 [-144] P:6 V:3 4 7 (0 3) (2 0) (1 2) 0sec 0cpu\nH:19 19 [0] P:8 V:4 5 6 7 (0 3) (1 1) (3 0) 1sec 0cpu\nH:6 51 [-90] P:9 V:4 6 7 (0 3) (1 1) (2 1) 1sec 0cpu\nH:3 75 [-144] P:10 V:4 7 (0 3) (1 1) (1 2) 0sec 0cpu\nH:3 75 [-144] P:13 V:6 7 (0 3) (0 2) (2 1) 1sec 0cpu\nnp=11 d:2 p:2 0sec 0cpu\n\n\nThe order of the vertices being unchanged, the linear relations agree with the desired combined weight system.\n\n### -H\n\nThe option -H replaces the output lines starting with H: with the full Hodge diamond of the corresponding partition. Note that the information about the nef partitions is omitted. The following example of codimension 2 complete intersections in $\\mathbb{P}^7$ illustrates this option:\n\npalp$nef.x -H Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #colums' (= PolyDim #Points' or #Points PolyDim'): 7 1 1 1 1 1 1 1 7 1 1 1 1 1 1 1 M:1716 7 N:8 7 codim=2 #part=3 h 0 0 h 1 0 h 0 1 h 2 0 h 1 1 h 0 2 h 3 0 h 2 1 h 1 2 h 0 3 h 4 0 h 3 1 h 2 2 h 1 3 h 0 4 h 4 1 h 3 2 h 2 3 h 1 4 h 4 2 h 3 3 h 2 4 h 4 3 h 3 4 h 4 4 1 0 0 0 1 0 0 0 0 0 1 237 996 237 1 0 0 0 0 0 1 0 0 0 1 16sec 15cpu h 0 0 h 1 0 h 0 1 h 2 0 h 1 1 h 0 2 h 3 0 h 2 1 h 1 2 h 0 3 h 4 0 h 3 1 h 2 2 h 1 3 h 0 4 h 4 1 h 3 2 h 2 3 h 1 4 h 4 2 h 3 3 h 2 4 h 4 3 h 3 4 h 4 4 1 0 0 0 1 0 0 0 0 0 1 356 1472 356 1 0 0 0 0 0 1 0 0 0 1 42sec 41cpu np=2 d:0 p:1 0sec 0cpu ### -Lv The option -Lv prints the vertices of $\\Delta^*\\subset N_\\mathbb{R}$ and the non-negative linear relations among them in addition to the standard output. If only the vertices should be printed use the option -V. The output takes the following form: The part before the dashed line reads: D n Vertices in N-lattice: # # ... # # . . ... . . . . ... . . # # ... # # The first line means that %the N-lattice polytope \u0394 * has dimension D and is given by n vertices which are the columns of the subsequent $D \\times n$ array of numbers #. Below the dashed line the non-negative linear relations among these vertices are indicated as follows: Let $v_0, \\dots, v_{n-1}$ denote the n vertices corresponding to the n columns above the dashed line. Any IP simplex with vertices in $\\{v_0, \\dots, v_{n-1}\\}$ determines a linear relation $\\sum_{i=0}^{n-1} l_i\\,v_i = 0$, with li that are positive for the vertices of the IP simplex and 0 otherwise. It results in an output line of the form l_0 l_1 ... l_{n-1} d=l codim=c where $l = \\sum_{i=0}^{n-1} l_i$ is the degree of the linear relation and c is the codimension of the IP simplex. In other words, these lines give the set of generators of the cone of non-negative linear relations within the (n-D)-dimensional vector space of linear relations among the vertices. This set is completely fixed by the order of the vertices, and the conditions that each vector, i.e. each linear relation, is positive and primitive. The information contained in these lines can be very useful in conjunction with the option -F*. To suppress them see the option -V. Besides the standard output the degrees of the nef partition with respect to the linear relations are inserted in the output lines containing the information about the nef partitions as follows. Consider a nef partition of length r defined by r collections of vertices $V_0,\\dots, V_{r-1}$ such that every vertex is a member of some collection Vj. The (multi)degree of the nef partition $\\{V_0,\\dots,V_{r-1}\\}$ with respect to the linear relation $\\sum_{i=0}^{n-1} l_i\\,v_i = 0$ is the vector $(d_0,\\dots,d_{r-1})$ where $d_j = \\sum_{i:v_i \\in V_j} l_i$. Note that $\\sum_{j=1}^r d_j = l$, the degree of the linear relation. The degrees $(d_0,\\dots,d_{r-1})$ are the degrees of the polynomials defining the complete intersection. If the codimension is 2 the output lines describing the nef partitions take the following form H:# [#] P:# V:# # (d10 d11) ... (dn0 dn1) #sec #cpu or if the codimension r is bigger than 2 H:# [#] P:# V0:# # V1:# # ... V(r-2):# # (d10 ... d1(r-1)) ... (dn0 ... dn(r-1)) #sec #cpu The additional data is (d10 d11) ... (dn0 dn1) and (d10... d1(r-1)) ... (dn0 ... dn(r-1)), respectively, where n is the number of linear relations. If $d_i=(d_{i0},\\dots,d_{i,r-1})$ are the degrees with respect to the i-th linear relation, then $di0=d_{i0},\\ldots, di(r-1)=d_{i,r-1}$. The following example of a codimension 2 complete intersection taken from arXiv:hep-th\/0410018v2 illustrates this option palp$ nef.x -Lv\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #colums' (= PolyDim #Points' or #Points PolyDim'):\n5 1 1 1 1 1 0 0 4 0 0 0 1 1 1 1\n5 1 1 1 1 1 0 0 4 0 0 0 1 1 1 1 M:378 12 N:8 7 codim=2 #part=8\n5 7 Vertices in N-lattice:\n0 -1 0 1 0 0 0\n0 -1 1 0 0 0 0\n-1 0 0 0 0 0 1\n-1 1 0 0 1 0 0\n-1 1 0 0 0 1 0\n-----------------------------------\n1 1 1 1 0 0 1 d=5 codim=1\n1 0 0 0 1 1 1 d=4 codim=2\nH:2 64 [-124] P:0 V:0 6 (2 3) (2 2) 1sec 0cpu\nH:2 64 [-124] P:1 V:0 1 6 (3 2) (2 2) 1sec 0cpu\nH:2 74 [-144] P:2 V:2 3 5 (2 3) (1 3) 1sec 0cpu\nH:2 64 [-124] P:3 V:3 5 6 (2 3) (2 2) 1sec 0cpu\nH:2 86 [-168] P:4 V:3 5 (1 4) (1 3) 1sec 1cpu\nH:2 74 [-144] P:5 V:3 6 (2 3) (1 3) 1sec 0cpu\nnp=6 d:0 p:2 0sec 0cpu\n\n\nFrom the output we deduce that the 7 vertices of the 5-dimensional polytope satisfy the following linear relations:\n\nv_0 + v_1 + v_2 + v_3 + v_6 = 0,\\quad v_0 + v_4 + v_5 + v_6 = 0.\n\n\nThe first of these linear relations has degree 5, the second has degree 4. The corresponding IP simplices have codimension 1 and 2, respectively.\n\n### -Lp\n\nThe option -Lp prints all the points of the N-lattice polytope and the linear relations among them, including those that are not vertices. The output has the same structure as for the option -Lv. The points are ordered such that first the vertices $\\{v_0,\\ldots,v_k\\}$ are listed, then the points $\\{p_{k+1},\\ldots,p_{N-2}\\}$ which are not vertices and finally the origin pN \u2212 1. Note that there will be additional linear relations including the points which are neither vertices nor the origin. The following example of a codimension 2 complete intersection taken from hep-th\/0410018 illustrates this option:\n\n$.\/nef.x -Lp Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #columns' (= PolyDim #Points' or #Points PolyDim'): 5 1 1 1 1 1 0 0 10 2 2 2 2 0 1 1 5 1 1 1 1 1 0 0 10 2 2 2 2 0 1 1 M:378 6 N:8 6 codim=2 #part=4 5 8 Points of Poly in N-Lattice: -1 0 0 0 1 0 0 0 -1 0 1 0 0 0 0 0 -1 0 0 1 0 0 0 0 -1 2 0 0 0 0 1 0 -1 1 0 0 0 1 1 0 ---------------------------------------- 2 1 2 2 2 1 0 d=10 codim=0 1 0 1 1 1 0 1 d=5 codim=1 H:2 86 [-168] P:0 V:1 5 6 (2 8) (1 4) 2sec 2cpu H:2 68 [-132] P:1 V:2 3 4 (6 4) (3 2) 1sec 0cpu H:2 68 [-132] P:2 V:3 4 (4 6) (2 3) 1sec 0cpu np=3 d:0 p:1 0sec 0cpu The last two points are not vertices. There is one more linear relation including the point p6. Further example: Complete intersection Calabi-Yau fourfold of codimension two discussed in arXiv:0912.3524. Important: in Global.h set POLY_Dmax=7 or higher and recompile! See Global parameters and limitations. palp$ nef.x -Lp\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n6 3 2 1 0 0 0 0 0 0 0 0 13 6 4 0 1 0 0 0 0 0 1 1 7 3 2 0 0 1 0 0 0 0 1 0 8 3 2 0 0 0 1 0 0 1 1 0 15 6 4 0 0 0 0 1 1 1 1 1\n6 3 2 1 0 0 0 0 0 0 0 0 13 6 4 0 1 0 0 0 0 0 1 1 7 3 2 0 0 1 0 0 0 0 1 0 8 3 2 0 0 0 1 0 0 1 1 0 15 6 4 0 0 0 0 1 1 1 1 1 M:4738 39 N:15 11 codim=2 #part=11\n6 15 Points of Poly in N-Lattice:\n0 0 -2 3 0 0 0 0 0 0 0 -1 2 1 0\n0 2 -1 1 0 0 0 1 1 1 2 0 1 1 0\n0 1 -1 1 0 0 1 1 1 1 2 0 1 1 0\n0 1 -1 1 1 0 0 1 0 1 2 0 1 1 0\n0 -1 0 0 0 1 0 -1 -1 0 -1 0 0 0 0\n1 -1 0 0 0 0 0 0 0 0 -1 0 0 0 0\n---------------------------------------------------------------------------\n1 1 6 4 1 1 1 0 0 0 0 0 0 0 d=15 codim=0\n1 0 6 4 0 1 0 0 0 0 1 0 0 0 d=13 codim=2\n0 0 3 2 1 1 0 0 1 0 0 0 0 0 d=8 codim=2\n0 0 3 2 0 1 0 1 0 0 0 0 0 0 d=7 codim=3\n0 0 3 2 0 0 0 0 0 1 0 0 0 0 d=6 codim=4\n0 0 1 1 0 0 0 0 0 0 0 1 0 0 d=3 codim=4\n0 0 2 1 0 0 0 0 0 0 0 0 0 1 d=4 codim=4\n0 0 1 0 0 0 0 0 0 0 0 0 1 0 d=2 codim=5\nH:8 0 1113 [6774] P:0 V:0 4 7 (2 13) (1 12) (1 7) (1 6) (0 6) (0 3) (0 4) (0 2) 282sec 281cpu\nH:5 0 1115 [6768] P:1 V:0 2 3 5 9 11 12 13 (12 3) (12 1) (6 2) (6 1) (6 0) (3 0) (4 0) (2 0) 162sec 162cpu\nH:5 0 1115 [6768] P:2 V:1 5 6 8 (3 12) (1 12) (2 6) (1 6) (0 6) (0 3) (0 4) (0 2) 159sec 158cpu\nH:8 0 1113 [6774] P:3 V:1 6 7 8 10 (2 13) (1 12) (1 7) (1 6) (0 6) (0 3) (0 4) (0 2) 228sec 216cpu\nH:8 0 1113 [6774] P:4 V:0 1 7 8 (2 13) (1 12) (1 7) (1 6) (0 6) (0 3) (0 4) (0 2) 236sec 234cpu\nH:5 0 1115 [6768] P:5 V:0 1 4 7 8 (3 12) (1 12) (2 6) (1 6) (0 6) (0 3) (0 4) (0 2) 183sec 182cpu\nH:5 0 1115 [6768] P:6 V:4 5 6 (3 12) (1 12) (2 6) (1 6) (0 6) (0 3) (0 4) (0 2) 221sec 220cpu\nH:8 0 1113 [6774] P:7 V:4 6 7 10 (2 13) (1 12) (1 7) (1 6) (0 6) (0 3) (0 4) (0 2) 271sec 265cpu\nH:8 0 1113 [6774] P:9 V:5 6 (2 13) (1 12) (1 7) (1 6) (0 6) (0 3) (0 4) (0 2) 282sec 281cpu\nH:7 0 958 [5838] P:10 V:6 8 (1 14) (0 13) (1 7) (0 7) (0 6) (0 3) (0 4) (0 2) 272sec -4023cpu\nnp=10 d:0 p:1 272sec -4023cpu\n\n\nThe last four points are not vertices. There are three more linear relations including those points. Compare this to the output of the option -Lv:\n\npalp$nef.x -Lv Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #columns' (= PolyDim #Points' or #Points PolyDim'): 6 3 2 1 0 0 0 0 0 0 0 0 13 6 4 0 1 0 0 0 0 0 1 1 7 3 2 0 0 1 0 0 0 0 1 0 8 3 2 0 0 0 1 0 0 1 1 0 15 6 4 0 0 0 0 1 1 1 1 1 6 3 2 1 0 0 0 0 0 0 0 0 13 6 4 0 1 0 0 0 0 0 1 1 7 3 2 0 0 1 0 0 0 0 1 0 8 3 2 0 0 0 1 0 0 1 1 0 15 6 4 0 0 0 0 1 1 1 1 1 M:4738 39 N:15 11 codim=2 #part=11 6 11 Vertices in N-lattice: 0 0 -2 3 0 0 0 0 0 0 0 0 2 -1 1 0 0 0 1 1 1 2 0 1 -1 1 0 0 1 1 1 1 2 0 1 -1 1 1 0 0 1 0 1 2 0 -1 0 0 0 1 0 -1 -1 0 -1 1 -1 0 0 0 0 0 0 0 0 -1 ------------------------------------------------------- 1 1 6 4 1 1 1 0 0 0 0 d=15 codim=0 1 0 6 4 0 1 0 0 0 0 1 d=13 codim=2 0 0 3 2 1 1 0 0 1 0 0 d=8 codim=2 0 0 3 2 0 1 0 1 0 0 0 d=7 codim=3 0 0 3 2 0 0 0 0 0 1 0 d=6 codim=4 H:8 0 1113 [6774] P:0 V:0 4 7 (2 13) (1 12) (1 7) (1 6) (0 6) 284sec 283cpu H:5 0 1115 [6768] P:1 V:0 2 3 5 9 11 12 13 (12 3) (12 1) (6 2) (6 1) (6 0) 163sec 163cpu H:5 0 1115 [6768] P:2 V:1 5 6 8 (3 12) (1 12) (2 6) (1 6) (0 6) 159sec 158cpu H:8 0 1113 [6774] P:3 V:1 6 7 8 10 (2 13) (1 12) (1 7) (1 6) (0 6) 211sec 210cpu H:8 0 1113 [6774] P:4 V:0 1 7 8 (2 13) (1 12) (1 7) (1 6) (0 6) 235sec 234cpu H:5 0 1115 [6768] P:5 V:0 1 4 7 8 (3 12) (1 12) (2 6) (1 6) (0 6) 181sec 180cpu H:5 0 1115 [6768] P:6 V:4 5 6 (3 12) (1 12) (2 6) (1 6) (0 6) 220sec 220cpu H:8 0 1113 [6774] P:7 V:4 6 7 10 (2 13) (1 12) (1 7) (1 6) (0 6) 258sec 257cpu H:8 0 1113 [6774] P:9 V:5 6 (2 13) (1 12) (1 7) (1 6) (0 6) 282sec 281cpu H:7 0 958 [5838] P:10 V:6 8 (1 14) (0 13) (1 7) (0 7) (0 6) 271sec -4024cpu np=10 d:0 p:1 271sec -4024cpu ### -p The option -p computes the nef partitions without the (time-consuming) calculation of Hodge numbers. As an example we consider the codimension 4 cf. -c* complete intersections in $\\mathbb{P}^7$. Note that one must set POLY_Dmax in Global.h to at least 10. palp$ nef.x -c4 -p\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n8 1 1 1 1 1 1 1 1\n8 1 1 1 1 1 1 1 1 M:6435 8 N:9 8 codim=4 #part=5\nP:0 V0:2 3 V1:4 5 V2:6 7 0sec 0cpu\nnp=1 d:0 p:4 0sec 0cpu\n\n\nThe Hodge data in the line containing the partition information is omitted, and the computation time is 0. Without the option -p this line would look like this:\n\nH:1 65 [-128] P:0 V0:2 3 V1:4 5 V2:6 7 13127sec 13120cpu\n\n\nNote the computation time.\n\nFurther example: Complete intersection Calabi-Yau fourfold of codimension two discussed in arXiv:0912.3524. Important: in Global.h set POLY_Dmax=7 or higher and recompile! See Global parameters and limitations.\n\nInput with -p:\n\npalp$nef.x -p Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #columns' (= PolyDim #Points' or #Points PolyDim'): 6 3 2 1 0 0 0 0 0 0 0 0 13 6 4 0 1 0 0 0 0 0 1 1 7 3 2 0 0 1 0 0 0 0 1 0 8 3 2 0 0 0 1 0 0 1 1 0 15 6 4 0 0 0 0 1 1 1 1 1 6 3 2 1 0 0 0 0 0 0 0 0 13 6 4 0 1 0 0 0 0 0 1 1 7 3 2 0 0 1 0 0 0 0 1 0 8 3 2 0 0 0 1 0 0 1 1 0 15 6 4 0 0 0 0 1 1 1 1 1 M:4738 39 N:15 11 codim=2 #part=11 P:0 V:0 4 7 0sec 0cpu P:1 V:0 2 3 5 9 11 12 13 0sec 0cpu P:2 V:1 5 6 8 0sec 0cpu P:3 V:1 6 7 8 10 0sec 0cpu P:4 V:0 1 7 8 0sec 0cpu P:5 V:0 1 4 7 8 0sec 0cpu P:6 V:4 5 6 0sec 0cpu P:7 V:4 6 7 10 0sec 0cpu P:9 V:5 6 0sec 0cpu P:10 V:6 8 0sec 0cpu np=10 d:0 p:1 0sec 0cpu Input without -p (note the calculation time! (32-bit system)): palp$ nef.x\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n6 3 2 1 0 0 0 0 0 0 0 0 13 6 4 0 1 0 0 0 0 0 1 1 7 3 2 0 0 1 0 0 0 0 1 0 8 3 2 0 0 0 1 0 0 1 1 0 15 6 4 0 0 0 0 1 1 1 1 1\n6 3 2 1 0 0 0 0 0 0 0 0 13 6 4 0 1 0 0 0 0 0 1 1 7 3 2 0 0 1 0 0 0 0 1 0 8 3 2 0 0 0 1 0 0 1 1 0 15 6 4 0 0 0 0 1 1 1 1 1 M:4738 39 N:15 11 codim=2 #part=11\nH:8 0 1113 [6774] P:0 V:0 4 7 247sec 246cpu\nH:5 0 1115 [6768] P:1 V:0 2 3 5 9 11 12 13 141sec 141cpu\nH:5 0 1115 [6768] P:2 V:1 5 6 8 136sec 136cpu\nH:8 0 1113 [6774] P:3 V:1 6 7 8 10 183sec 182cpu\nH:8 0 1113 [6774] P:4 V:0 1 7 8 203sec 202cpu\nH:5 0 1115 [6768] P:5 V:0 1 4 7 8 157sec 156cpu\nH:5 0 1115 [6768] P:6 V:4 5 6 190sec 189cpu\nH:8 0 1113 [6774] P:7 V:4 6 7 10 226sec 225cpu\nH:8 0 1113 [6774] P:9 V:5 6 246sec 246cpu\nH:7 0 958 [5838] P:10 V:6 8 236sec 234cpu\nnp=10 d:0 p:1 236sec 234cpu\n\n\n### -D\n\nThe option -D also prints those nef partitions which are direct products of lower-dimensional nef partitions. If only direct products are to be printed use the option -Q. As an example we consider a codimension 2 complete intersection in $\\mathbb{P}^2\\times \\mathbb{P}^2$:\n\npalp$nef.x -D 3 1 1 1 0 0 0 3 0 0 0 1 1 1 3 1 1 1 0 0 0 3 0 0 0 1 1 1 M:100 9 N:7 6 codim=2 #part=5 H:4 [0] h1=2 P:0 V:2 3 5 D 0sec 0cpu H:20 [24] P:1 V:3 4 5 0sec 0cpu H:20 [24] P:2 V:3 5 0sec 0cpu H:20 [24] P:3 V:4 5 0sec 0cpu np=3 d:1 p:1 0sec 0cpu The last three nef partitions each describe a K3 surface. The first one is a $T^4=T^2\\times T^2$. The extra output triggered by -D is: H:4 [0] h1=2 P:0 V:2 3 5 D 0sec 0cpu h1=2 indicates that the Hodge number h1,0(T4) = 2. Furthermore the letter D indicates that the nef partition is a direct product. Compare this to the output without the option -D where the first nef partition is not shown: palp$ nef.x\n3 1 1 1 0 0 0 3 0 0 0 1 1 1\n3 1 1 1 0 0 0 3 0 0 0 1 1 1 M:100 9 N:7 6 codim=2 #part=5\nH:20 [24] P:1 V:3 4 5 0sec 0cpu\nH:20 [24] P:2 V:3 5 0sec 0cpu\nH:20 [24] P:3 V:4 5 1sec 0cpu\nnp=3 d:1 p:1 0sec 0cpu\n\n\n### -P\n\nThe option -P also prints nef partitions corresponding to projections. Consider for example a complete intersection of codimension 2 in $\\mathbb{P}^3$:\n\npalp$nef.x -P 4 1 1 1 1 4 1 1 1 1 M:35 4 N:5 4 codim=2 #part=2 H:[0] P:0 V:2 3 0sec 0cpu H:[0] P:1 V:3 0sec 0cpu np=1 d:0 p:1 0sec 0cpu Compared to the output without -P there is one additional line: H:[0] P:1 V:3 0sec 0cpu Let $v_0,\\dots,v_3$ denote the vertices of the polytope. The nef partition P:0 is then as follows: $0:\\; V_0=\\langle v_3 \\rangle,\\quad V_1=\\langle v_0, v_1, v_2 \\rangle.$ The part V0 only contains the vertex v3. Therefore the equation of the corresponding divisor D0,0 reads x3 = 0. A projection \u03c0 of \u0394 * along v3 yields a reflexive polytope $\\Delta_{v_3}^* = \\langle \\pi v_0,\\pi v_1,\\pi v_2 \\rangle$. Thus, we are left with a hypersurface $X' = D_{0,1} \\subset \\mathbb{P}^2 = \\mathbb{P}^3 \\cap D_{0,0}$. If there is a nef partition such that the dual nef partition in the M-lattice has a summand with only one vertex, then DP is displayed in the output. Further example: A complete intersection of codimension 6 which is reduced to codimension 3 by projections. We use the option -c* to set the codimension and -p to suppress the calculation of the Hodge numbers. Furthermore we list the vertices using the option -Lv: palp$ nef.x -P -c6 -p -Lv\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n6 1 1 1 1 1 1 0 0 0 6 1 1 1 0 0 0 1 1 1\n6 1 1 1 1 1 1 0 0 0 6 1 1 1 0 0 0 1 1 1 M:5214 12 N:10 9 codim=6 #part=1\n7 9 Vertices in N-lattice:\n-1 0 0 1 0 0 0 0 0\n-1 0 1 0 0 0 0 0 0\n0 -1 0 0 0 1 0 0 0\n0 -1 0 0 1 0 0 0 0\n-1 1 0 0 0 0 0 0 1\n-1 1 0 0 0 0 1 0 0\n-1 1 0 0 0 0 0 1 0\n---------------------------------------------\n1 1 1 1 1 1 0 0 0 d=6 codim=2\n1 0 1 1 0 0 1 1 1 d=6 codim=2\nP:0 V0:0 V1:2 V2:3 V3:4 7 V4:5 8 (1 1 1 1 1 1) (1 1 1 1 1 1) 0sec 0cpu\nnp=0 d:0 p:1 0sec 0cpu\n\n\nThe output shows that three elements of the nef partition contain only one vertex:\n\n P:0 V0:0 V1:2 V2:3 V3:4 7 V4:5 8 0sec 0cpu\n\n\nTherefore the variables associated to the vertices labeled by 0,2 and 3 can be set to zero and we are left with a complete intersection of codimension 3 in $\\mathbb{P}^2\\times \\mathbb{P}^2$.\n\n### -t\n\nThe option -t gives detailed information about the calculation times of the Hodge numbers. The Hodge numbers of a Calabi-Yau complete intersection are generated by the so called stringy E-function introduced by Batyrev and Borisov in alg-geom\/9509009. The combinatorial construction of the E-function involves the construction of a B-polynomial and an S-polynomial defined in alg-geom\/9509009. The option -t returns the accumulated computing times of the respective polynomials. We illustrate this option with the example of complete intersections of codimension 4 in $\\mathbb{P}^7$.\n\npalp$.\/nef.x -t -c4 Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #columns' (= PolyDim #Points' or #Points PolyDim'): 8 1 1 1 1 1 1 1 1 8 1 1 1 1 1 1 1 1 M:6435 8 N:9 8 codim=4 #part=5 BEGIN S-Poly 0sec 0cpu BEGIN B-Poly 11564sec 11558cpu BEGIN E-Poly 13126sec 13119cpu H:1 65 [-128] P:0 V0:2 3 V1:4 5 V2:6 7 13126sec 13119cpu np=1 d:0 p:4 0sec 0cpu This option can be useful for finding at which point in the calculation of the Hodge numbers the program crashes. Further example: Complete intersection Calabi-Yau fourfold discussed in arXiv:0908.1784. palp$ nef.x -t\n10 3 2 0 1 1 1 1 1 6 3 2 1 0 0 0 0 0\n10 3 2 0 1 1 1 1 1 6 3 2 1 0 0 0 0 0 M:2302 15 N:12 8 codim=2 #part=4\nBEGIN S-Poly 0sec 0cpu\nBEGIN B-Poly 61sec 57cpu\nBEGIN E-Poly 66sec 61cpu\nH:2 30 308 [1728] P:0 V:4 5 6 7 66sec 61cpu\nBEGIN S-Poly 0sec 0cpu\nBEGIN B-Poly 92sec 83cpu\nBEGIN E-Poly 100sec 91cpu\nH:5 5 448 [2736] P:1 V:5 6 7 100sec 91cpu\nBEGIN S-Poly 0sec 0cpu\nBEGIN B-Poly 152sec 138cpu\nBEGIN E-Poly 160sec 146cpu\nH:5 0 567 [3480] P:2 V:6 7 160sec 146cpu\nnp=3 d:0 p:1 0sec 0cpu\n\n\n### -c*\n\nThe option -c* where * is a positive integer r allows to specify the length of the nef partition and hence the codimension of the Calabi-Yau complete intersection. The default value for the codimension is 2. Note that the computation time can take several hours for r=4 or even days for r>4 and PALP may crash because the limits such as the number of vertices etc. set in Global.h may be exceeded. We illustrate this option with complete intersections of codimension 3 in $\\mathbb{P}^2\\times \\mathbb{P}^2$:\n\npalp$nef.x -c3 Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #columns' (= PolyDim #Points' or #Points PolyDim'): 3 1 1 1 0 0 0 3 0 0 0 1 1 1 3 1 1 1 0 0 0 3 0 0 0 1 1 1 M:100 9 N:7 6 codim=3 #part=7 H:[0] P:0 V0:1 3 V1:4 5 1sec 1cpu H:[0] P:1 V0:2 3 V1:4 5 1sec 0cpu np=1 d:1 p:5 0sec 0cpu Also hypersurfaces can be analyzed with nef.x. As an example we consider the quintic hypersurface in $\\mathbb{P}^4$: palp$ nef.x -c1\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n5 1 1 1 1 1\n5 1 1 1 1 1 M:126 5 N:6 5 codim=1 #part=1\nH:1 101 [-200] P:0 math 0sec 0cpu\nnp=1 d:0 p:0 0sec 0cpu\n\n\nCompare this to the output of poly.x:\n\npalp$poly.x Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #columns' (= PolyDim #Points' or #Points PolyDim'): 5 1 1 1 1 1 5 1 1 1 1 1 M:126 5 N:6 5 H:1,101 [-200] ### -F* The option -F* yields information about possible toric fibrations of the toric variety associated to the given reflexive lattice polytope. The polytopes associated to the toric fibers are restricted to be reflexive. By considering nef partitions for the given lattice polytope this option also computes possible fibrations of the corresponding complete intersection Calabi-Yau manifolds by lower-dimensional complete intersection Calabi-Yau manifolds. For more details see math\/0001106, hep-th\/0410018. In practice one should always use the option -F* in conjunction with either -Lv or -Lp. Here * is a non-negative integer s that specifies the maximal codimension s of the fiber polytope. The default value for s is 2. Note that this codimension does not need to coincide with the codimension of the corresponding complete intersection Calabi-Yau fiber. Besides the standard output and the output from the options -Lv or -Lp, the full information about fibration structures is listed below a second dashed line. The output takes the following form: ----------------------------------------------- #fibrations=# _ _ v v ... p p p v cd=# m: # # n: # # . . . . ... . . . . . . . . . . . . . ... . . . . . . . . . . . . . ... . . . . . . . . . v p _ v ... v _ _ p cd=# m: # # n: # # The number # in #fibrations=# specifies the number of fibrations by reflexive polytopes up to symmetry that have been found. Then each of the subsequent lines corresponds to one of these fibrations. The points of the given polytope are labeled by either v, p or _. This label indicates whether the corresponding point is a vertex (v), a non-vertex point (p) or not a point at all (_) of the fiber polytope. The latter correspond to the directions of the toric base. The non-negative integer # in cd=# specifies the codimension of the fiber polytope. The two positive integers # # after m: specify the numbers of points and vertices of the dual of the fiber polytope, respectively. The two positive integers # # after n: specify the numbers of points and vertices of the fiber polytope, respectively. We illustrate this option with a complete intersection of codimension 2 with several fibrations. In order to find all fibrations the argument of -F must be set to 3. This is an example where the interpretation of the fibration information depends on the choice of the nef partition. palp$ echo \"12 4 2 2 2 1 1 0 8 4 0 0 0 1 1 2\" | nef.x -f -Lp -F3\n12 4 2 2 2 1 1 0 8 4 0 0 0 1 1 2 M:371 12 N:10 7 codim=2 #part=5\n5 10 Points of Poly in N-Lattice:\n0 0 0 1 0 -1 0 0 0 0\n0 0 1 0 0 -1 0 0 0 0\n-1 4 0 0 0 0 0 1 2 0\n0 -1 0 0 1 0 0 0 0 0\n-1 2 0 0 0 1 1 1 1 0\n--------------------------------------------------\n4 1 2 2 1 2 0 0 0 d=12 codim=0\n4 1 0 0 1 0 2 0 0 d=8 codim=2\n2 0 1 1 0 1 0 0 1 d=6 codim=1\n2 0 0 0 0 0 1 0 1 d=4 codim=3\n1 0 0 0 0 0 0 1 0 d=2 codim=4\n--------------------------------------------- #fibrations=3\nv v _ _ v _ v p p cd=2 m: 35 4 n: 7 4\nv _ v v _ v v p v cd=1 m:117 9 n: 8 6\nv _ _ _ _ _ v p v cd=3 m: 9 3 n: 5 3\nH:4 58 [-108] P:1 V:0 2 (6 6) (4 4) (3 3) (2 2) (1 1) 1sec 0cpu\nH:3 65 [-124] P:2 V:0 2 3 (8 4) (4 4) (4 2) (2 2) (1 1) 1sec 0cpu\nH:3 83 [-160] P:3 V:3 5 (4 8) (0 8) (2 4) (0 4) (0 2) 1sec 1cpu\nnp=3 d:0 p:2 0sec 0cpu\n\n\nThere are three fibrations. The fiber polytope of the second fibration is of codimension 1, hence has dimension 5-1=4. As usual, we label the vertices and points by $v_0,\\ldots,v_6,p_7,p_8,p_9$. The vertices labeled with _ are v1 and v4, which are all in V1 for all the three nef partitions. Since we are considering a complete intersection of codimension 2, the corresponding Calabi-Yau threefold admits a fibration by K3 surfaces since the fiber has dimension 4-2=2. The linear relation of codimension 1 and degree 6 does not involve v1 and v4, hence it describes the fiber polytope. The degrees of the nef partitions with respect to this linear relation are given in the third parentheses in the lines containing the information of the nef partitions. Hence, the K3 fibers are $\\mathbb{P}(2,1,1,1,1)[3,3]$, $\\mathbb{P}(2,1,1,1,1)[4,2]$, and $\\mathbb{P}(2,1,1,1,1)[2,4]$, respectively. Note that the second fibration is an instance of the situation that a non-vertex point of the polytope becomes a vertex of the fiber polytope. Here, this is the point p8.\n\nThe fiber polytope of the first fibration is of codimension 2, hence has dimension 5-2=3. Naively, one would expect that the corresponding Calabi-Yau threefolds admit elliptic fibrations. This is indeed true for the first two nef partitions since the vertices labeled with _ are v2,v3, and v5, which are all in V1. Repeating the steps of the second fibration above in this case yields the complete intersection $\\mathbb{P}(4,1,1,2)[4,4]$ for both nef partitions. After discarding the trivial projection to the first coordinate, they become the hypersurfaces $\\mathbb{P}(1,1,2)[4]$.\n\nFor the third nef partition, however, the vertices and points of the fiber polytope only lie in the part V1 of the nef partition. Hence, the part V0 reduces the dimension of the base. The fiber of the corresponding Calabi-Yau threefold is only of codimension 1 in the 3-dimensional toric fiber, i.e. it is a K3 surface. In fact, the linear relation of codimension 2 and degree 8 involves all points of V1, hence it describes the fiber polytope. The degrees of the third nef partition with respect to this linear relation are the second parentheses in the line with P:3. Hence, the K3 fiber is $\\mathbb{P}(4,1,1,2)[8]$. This phenomenon is further described in hep-th\/0410018.\n\nFinally, the fiber polytope of the third fibration is of codimension 3, and hence has dimension 5-3=2. Naively, one would expect that the corresponding Calabi-Yau threefolds do not admit any fibrations since the codimension is also 2 and hence the fibers would be points. This is indeed the case for the first two nef partitions. For the third nef partition, the fiber polytope consists of the points v0,v6,p7, and p8, all of which lie in V1. Hence, the fiber of the corresponding Calabi-Yau threefold is only of codimension 1 in the 2-dimensional toric fiber, i.e. it is an elliptic curve. The degrees of the third nef partition with respect to the linear relation of codimension 3 are the fourth parentheses in the line with P:3. Hence, the elliptic curve is $\\mathbb{P}(2,1,1)[4]$.\n\nFurther examples:\n\nAn example is the degree 18 hypersurface in a crepant resolution of the weighted projective space $\\mathbb{P}(1,1,1,6,9)$. Since it is a hypersurface, we need to set the codimension to 1 using the option -c*.\n\npalp$echo \"18 9 6 1 1 1\" | nef.x -f -Lp -c1 -F 18 9 6 1 1 1 M:376 5 N:10 5 codim=1 #part=1 4 10 Points of Poly in N-Lattice: 0 0 -2 3 0 2 1 -1 0 0 0 3 -1 1 0 1 1 0 1 0 0 -1 0 0 1 0 0 0 0 0 1 -1 0 0 0 0 0 0 0 0 -------------------------------------------------- 1 1 9 6 1 0 0 0 0 d=18 codim=0 0 0 2 1 0 0 1 0 0 d=4 codim=2 0 0 1 1 0 0 0 1 0 d=3 codim=2 0 0 3 2 0 0 0 0 1 d=6 codim=2 0 0 1 0 0 1 0 0 0 d=2 codim=3 --------------------------------------------- #fibrations=1 _ _ v v _ p p p v cd=2 m: 7 3 n: 7 3 H:2 272 [-540] P:0 (18) (4) (3) (6) (2) 0sec 0cpu np=1 d:0 p:0 0sec 0cpu There is only one fibration whose fiber polytope has codimension 2. Since the whole polytope has dimension 4, the fiber polytope therefore has dimension 4-2=2, and the dimension of the fiber of the associated toric variety is also 2. Since we are considering a hypersurface, i.e. a complete intersection of codimension 1, the corresponding Calabi-Yau manifold $X\\,$ has dimension 4-1=3 and admits a fibration by elliptic curves since the fiber has dimension 2-1=1. We can specify the fiber even more by looking at the entries v, p and _ and comparing them to the linear relations of the same codimension as the fiber polytope above the second dashed line. We observe that the relation 0 0 3 2 0 0 0 0 1 d=6 codim=2 has precisely a zero for each point labelled by a _. Hence the fiber of the toric variety is (a crepant resolution of) the weighted projective space $\\mathbb{P}(1,2,3)$, and the fiber of $X\\,$ is a degree 6 curve in this weighted projective space. The last example is again a hypersurface, the degree 24 hypersurface in the crepant resolution of the weighted projective space $\\mathbb{P}(1,1,2,8,12)$. palp$ echo \"24 12 8 2 1 1\" | nef.x -f -Lp -c1 -F\n24 12 8 2 1 1 M:335 5 N:11 5 codim=1 #part=1\n4 11 Points of Poly in N-Lattice:\n0 0 -2 3 0 1 2 0 -1 0 0\n2 0 -1 1 0 1 1 1 0 0 0\n1 2 -1 1 0 1 1 1 0 1 0\n-1 1 0 0 1 0 0 0 0 1 0\n-------------------------------------------------------\n2 1 12 8 1 0 0 0 0 0 d=24 codim=0\n1 0 6 4 0 0 0 0 0 1 d=12 codim=1\n0 0 2 1 0 1 0 0 0 0 d=4 codim=2\n0 0 3 2 0 0 0 1 0 0 d=6 codim=2\n0 0 1 1 0 0 0 0 1 0 d=3 codim=2\n0 0 1 0 0 0 1 0 0 0 d=2 codim=3\n-------------------------------------------------- #fibrations=2\nv _ v v _ p p p p v cd=1 m: 39 4 n: 9 4\n_ _ v v _ p p v p _ cd=2 m: 7 3 n: 7 3\nH:3 243 [-480] P:0 (24) (12) (4) (6) (3) (2) 0sec 0cpu\nnp=1 d:0 p:0 0sec 0cpu\n\n\nThere are two fibrations, one of codimension 1 and one of codimension 2.\n\n\u2022 The same considerations as in the example above show that the latter yields an elliptic fibration of the corresponding Calabi-Yau threefold $X\\,$ with the same elliptic fiber.\n\u2022 The fiber polytope of the first fibration has dimension 4-1=3 and the dimension of the fiber of the associated toric variety is also 3. Since we are considering a complete intersection of codimension 1, the corresponding Calabi-Yau threefold $X\\,$ admits a fibration by K3 surfaces since the fiber has dimension 3-1=2. By comparing the points with te labels _ and the linear relations of codimension 1 with a 0 at these points, we see that the fiber is a degree 12 hypersurface in (a crepant resolution of) the weighted projective space $\\mathbb{P}(1,1,4,6)$.\n\u2022 Note that the points labelled with _ of the first fibration form a subset of the points labelled with _ of the second fibration. This means that the fiber polytope of the first fibration admits itself a fibration by a reflexive lattice polytope, the fiber being the fiber polytope of the second fibration. Hence the fibrations of the corresponding Calabi-Yau threefold $X\\,$ are compatible in the sense that the elliptic fibration factors through the K3 fibration.\n\u2022 Note that if one had specified the option -F1 instead of -F or -F2, only the first fibration would have been listed.\n\n### -y\n\nDepending on the input the option -y returns the CWS or the vertices of the M-lattice polytope if there is at least one nef partition. In order to trigger the output this nef partition may also be a projection. If there is no nef partition there is no output. Depending on the input the following output is given:\n\n\u2022 if there is a nef partition:\n\n- If the input is a CWS, the CWS is returned along with the polytope data.\n\n- If the input is a polytope in the M-lattice or N-lattice (cf. option -N) the M-lattice polytope is returned.\n\n\u2022 if there is no nef partition\n\n- If the input is a CWS, the CWS is returned without further information about the polytope.\n\n- If the input is a polytope there is no output.\n\nAs an example consider the codimension 2 complete intersection in $\\mathbb{P}^3$. If we enter the N-lattice polytope we get the following output:\n\npalp$nef.x -y -N Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #columns' (= PolyDim #Points' or #Points PolyDim'): 3 4 Type the 12 coordinates as dim=3 lines with #pts=4 columns: -1 0 0 1 -1 0 1 0 -1 1 0 0 3 4 Vertices of Poly in M-lattice: M:35 4 N:5 4 codim=2 #part=2 -1 -1 -1 3 -1 -1 3 -1 -1 3 -1 -1 Further examples: Example: Codimension 2 complete intersection in $\\mathbb{P}^3$, input is theCWS: palp$ nef.x -y\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n4 1 1 1 1\n4 1 1 1 1 M:35 4 N:5 4 codim=2 #part=2\n\n\nExample: Codimension 2 complete intersection in $\\mathbb{P}^3$, input is the M-lattice polytope:\n\npalp$nef.x -y Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #columns' (= PolyDim #Points' or #Points PolyDim'): 3 4 Type the 12 coordinates as dim=3 lines with #pts=4 columns: -1 -1 -1 3 -1 -1 3 -1 -1 3 -1 -1 3 4 Vertices of Poly in M-lattice: M:35 4 N:5 4 codim=2 #part=2 -1 -1 -1 3 -1 -1 3 -1 -1 3 -1 -1 Example without a nef partition, input is the CWS: palp$ nef.x -y\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n6 3 2 1 0 0 6 3 0 0 2 1\n6 3 2 1 0 0 6 3 0 0 2 1\n\n\nExample without a nef partition, input is the N-lattice polytope:\n\npalp$nef.x -y -N Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #columns' (= PolyDim #Points' or #Points PolyDim'): 3 5 Type the 15 coordinates as dim=3 lines with #pts=5 columns: 0 0 -1 2 0 -2 3 3 0 0 -1 1 1 1 1 Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #columns' (= PolyDim #Points' or #Points PolyDim'): The same holds if an M-lattice polytope is entered. ### -S The option -S gives information about the number of points in the reflexive Gorenstein cone and its dual (cf. options -g* and -d*) for each nef partition which is not a direct product or a projection. It displays the numbers $\\ell$ of lattice points and $\\ell^*$ of interior lattice points in degrees $k \\le (\\tilde d + 1)\/2$, where $\\tilde d$ is the dimension of the Gorenstein cone C, and the analogous data for the dual cone $\\check C$. These data enter the calculation of the (stringy) Hodge numbers via the S-polynomial (hence the name -S) as described in Nef partitions and Gorenstein cones. The output takes the following form. After the first line of the standard output, there is a part referring to the polytope $\\Delta(\\check C)$: #points in largest cone: layer: 1 #p: l1 #ip: 0 ... . ... . ... . layer: . #p: . #ip: . ... . ... . ... . layer: k #p: lk #ip: l*k where $l1 = \\ell(\\Delta(\\check C) ), \\ldots, lk =\\ell(k\\Delta(\\check C) )$, $l*1= \\ell^*(\\Delta(\\check C) ),\\ldots, l*k=\\ell^*(k\\Delta(\\check C) )$. Subsequently there is a second part referring to the polytope \u0394(C). #points in largest cone: layer: 1 #p: l1 #ip: 0 ... . ... . ... . layer: . #p: . #ip: . ... . ... . ... . layer: k #p: lk #ip: l*k where $l1 = \\ell(\\Delta(C) ), \\ldots, lk =\\ell(\\Delta(C) )$, $l*1= \\ell^*(\\Delta(C) ),\\ldots, l*k=\\ell^*(k\\Delta(C) )$. Then the rest of the standard output concerning the nef partitions follows. The following example illustrates this option. We consider a complete intersection of codimension 2 in $\\mathbb{P}^4$: palp$ nef.x -S\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n4 1 1 1 1\n4 1 1 1 1 M:35 4 N:5 4 codim=2 #part=2\n\n#points in largest cone:\nlayer: 1 #p: 6 #ip: 0\nlayer: 2 #p: 21 #ip: 1\nlayer: 3 #p: 56 #ip: 6\n\n#points in largest cone:\nlayer: 1 #p: 20 #ip: 0\nlayer: 2 #p: 105 #ip: 1\nlayer: 3 #p: 336 #ip: 20\nH:[0] P:0 V:2 3 0sec 0cpu\nnp=1 d:0 p:1 0sec 0cpu\n\n\nOne of the two nef partitions is a projection and is not analyzed. The output for the remaining nef partition has two blocks: The first block counts the numbers of points (after #p:) and points in the relative interior (after #ip:) of the Gorenstein cone $\\check C \\subset \\tilde N$ at degrees k = 1,2,3. Hence\n\n$\\ell(\\Delta(\\check C) ) =6,\\quad \\ell(2\\Delta(\\check C) ) =21,\\quad \\ell(3\\Delta(\\check C) ) =56, \\ell^*(\\Delta(\\check C) ) =0,\\quad \\ell^*(2\\Delta(\\check C) ) =1, \\quad \\ell^*(3\\Delta(\\check C) ) =6.$\n\nOne can check that the number of points at degree k = 1 indeed coincides with the number of points in the output of the option -g2.\n\nThe second block gives the same information for the dual Gorenstein cone $C \\subset \\tilde M$. Hence\n\n$\\ell(\\Delta(C) ) =20,\\quad \\ell(2\\Delta(C) ) =105,\\quad \\ell(3\\Delta(C) ) =336 \\ell^*(\\Delta(C) ) =0,\\quad \\ell^*(2\\Delta(C) ) =1,\\quad \\ell^*(2\\Delta(C) ) =20.$\n\nThe output of the option -d2 coincides with the number of points at degree k = 1.\n\n### -T\n\nThe option -T turns on an explicit check of the Serre duality relation (cf. Nef partitions and Gorenstein cones relating the S-- and T--polynomials. Normally the program actually uses that relation to avoid point counting beyond degree $(\\tilde d + 1)\/2$, but with -T the counting goes up to degree $\\tilde d$ and an error message is given if Serre duality is violated. This can be useful if one suspects that the program gives wrong Hodge numbers, for example because of numerical overflows. If nothing goes wrong, the only effect is a significantly increased computation time. The best way to illustrate this option is by combining it with -S. We consider the same example as for -S.\n\npalp$nef.x -S -T Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #colums' (= PolyDim #Points' or #Points PolyDim'): 4 1 1 1 1 4 1 1 1 1 M:35 4 N:5 4 codim=2 #part=2 #points in largest cone: layer: 1 #p: 6 #ip: 0 layer: 2 #p: 21 #ip: 1 layer: 3 #p: 56 #ip: 6 layer: 4 #p: 125 #ip: 21 layer: 5 #p: 246 #ip: 56 #points in largest cone: layer: 1 #p: 20 #ip: 0 layer: 2 #p: 105 #ip: 1 layer: 3 #p: 336 #ip: 20 layer: 4 #p: 825 #ip: 105 layer: 5 #p: 1716 #ip: 336 H:[0] P:0 V:2 3 0sec 0cpu np=1 d:0 p:1 0sec 0cpu Note how now the point counting proceeds up to degree 5. With these data we can compute the Ehrhart polynomial $S_{\\Delta(\\check C)}(t) = (1 - t)^5 (1 + 6t + 21 t^2 + 56 t^3 + 125 t^4 + 246 t^5 + \\dots)$ Since it has degree at most $\\tilde d = 5$, we find $S_{\\Delta(\\check C)} = 1 + t + t^2 + t^3.$ Similarly $T_{\\Delta(\\check C)}(t) = (1 - t)^5 (t^2 + 6t^3 + 21 t^4 + 56 t^5 + \\dots) = t^2 + t^3 + t^4 + t^5,$ and it is clear that the Serre duality relation is satisfied. A similar check can be performed for $C \\cap \\tilde M$ with the data from the second block. ### -s The option -s includes all nef partitions in the output, not just one representative for each class of nef partitions that are equivalent under symmetries of the CWS. Note that this option does not print all possible nef partitions as those corresponding to projections (cf. option -P) or direct products (cf. option -D) are omitted. The example we consider is a complete intersection of codimension 2 in $\\mathbb{P}^2\\times\\mathbb{P}^2$. We add the option -Lv in order to print the vertices and the CWS. palp$ nef.x -s -Lv\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n3 1 1 1 0 0 0 3 0 0 0 1 1 1\n3 1 1 1 0 0 0 3 0 0 0 1 1 1 M:100 9 N:7 6 codim=2 #part=31\n4 6 Vertices in N-lattice:\n0 0 0 1 0 -1\n0 0 1 0 0 -1\n-1 0 0 0 1 0\n-1 1 0 0 0 0\n------------------------------\n1 1 0 0 1 0 d=3 codim=2\n0 0 1 1 0 1 d=3 codim=2\nH:20 [24] P:2 V:4 5 (1 2) (1 2) 0sec 0cpu\nH:20 [24] P:4 V:0 5 (1 2) (1 2) 0sec 0cpu\nH:20 [24] P:5 V:0 4 (2 1) (0 3) 0sec 0cpu\nH:20 [24] P:6 V:0 4 5 (2 1) (1 2) 0sec 0cpu\nH:20 [24] P:8 V:1 5 (1 2) (1 2) 1sec 0cpu\nH:20 [24] P:9 V:1 4 (2 1) (0 3) 0sec 0cpu\nH:20 [24] P:10 V:1 4 5 (2 1) (1 2) 0sec 0cpu\nH:20 [24] P:11 V:0 1 (2 1) (0 3) 0sec 0cpu\nH:20 [24] P:12 V:0 1 5 (2 1) (1 2) 0sec 0cpu\nH:20 [24] P:14 V:2 3 (0 3) (2 1) 0sec 0cpu\nH:20 [24] P:16 V:2 5 (0 3) (2 1) 0sec 0cpu\nH:20 [24] P:17 V:2 4 (1 2) (1 2) 0sec 0cpu\nH:20 [24] P:18 V:2 4 5 (1 2) (2 1) 0sec 0cpu\nH:20 [24] P:19 V:0 2 (1 2) (1 2) 0sec 0cpu\nH:20 [24] P:20 V:0 2 5 (1 2) (2 1) 1sec 0cpu\nH:20 [24] P:21 V:0 2 4 (2 1) (1 2) 0sec 0cpu\nH:20 [24] P:22 V:1 3 (1 2) (1 2) 0sec 0cpu\nH:20 [24] P:23 V:1 2 (1 2) (1 2) 0sec 0cpu\nH:20 [24] P:24 V:1 2 5 (1 2) (2 1) 0sec 0cpu\nH:20 [24] P:25 V:1 2 4 (2 1) (1 2) 0sec 0cpu\nH:20 [24] P:26 V:0 3 (1 2) (1 2) 0sec 0cpu\nH:20 [24] P:27 V:0 1 2 (2 1) (1 2) 0sec 0cpu\nH:20 [24] P:28 V:3 4 (1 2) (1 2) 1sec 0cpu\nH:20 [24] P:29 V:3 5 (0 3) (2 1) 0sec 0cpu\nnp=24 d:1 p:6 0sec 0cpu\n\n\nNote that the CWS is symmetric under permutations of the vertices labeled by 0,1,4 and those labeled by 2,3,5. Furthermore there only exist three pairs of degrees of the complete intersection (up to exchange within a pair): {(1,2),(1,2)},{(0,3),(2,1)},{(1,2),(2,1)}. Therefore we conclude that there are only three inequivalent nef partitions. This is indeed confirmed by calling nef.x without the option -s.\n\npalp$nef.x -Lv Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #columns' (= PolyDim #Points' or #Points PolyDim'): 3 1 1 1 0 0 0 3 0 0 0 1 1 1 3 1 1 1 0 0 0 3 0 0 0 1 1 1 M:100 9 N:7 6 codim=2 #part=5 4 6 Vertices in N-lattice: 0 0 0 1 0 -1 0 0 1 0 0 -1 -1 0 0 0 1 0 -1 1 0 0 0 0 ------------------------------ 1 1 0 0 1 0 d=3 codim=2 0 0 1 1 0 1 d=3 codim=2 H:20 [24] P:1 V:3 4 5 (1 2) (2 1) 0sec 0cpu H:20 [24] P:2 V:3 5 (0 3) (2 1) 0sec 0cpu H:20 [24] P:3 V:4 5 (1 2) (1 2) 0sec 0cpu np=3 d:1 p:1 0sec 0cpu ### -n The option -n prints the points of the polytope in the N-lattice only if there is at least one nef partition which does not correspond to a projection or a direct product. In addition, the first line of the standard output is printed while the other lines are suppressed. palp$ nef.x -n\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n4 1 1 1 1\n4 1 1 1 1 M:35 4 N:5 4 codim=2 #part=2\n3 5 Points of Poly in N-Lattice:\n-1 0 0 1 0\n-1 0 1 0 0\n-1 1 0 0 0\n\n\nFurther examples:\n\nExample without a nef partition:\n\npalp$nef.x -n Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #columns' (= PolyDim #Points' or #Points PolyDim'): 6 3 2 1 0 0 6 3 0 0 2 1 6 3 2 1 0 0 6 3 0 0 2 1 M:21 5 N:12 5 codim=2 #part=0 Here the N-lattice polytope is not printed. Example: no output of the polytope if there is only a nef partition corresponding to a projection: Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #columns' (= PolyDim #Points' or #Points PolyDim'): 6 3 2 1 0 0 3 1 0 0 1 1 6 3 2 1 0 0 3 1 0 0 1 1 M:24 6 N:9 5 codim=2 #part=1 We can use the option -P to check that the nef partition corresponding a projection: palp$ nef.x -P\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n6 3 2 1 0 0 3 1 0 0 1 1\n6 3 2 1 0 0 3 1 0 0 1 1 M:24 6 N:9 5 codim=2 #part=1\nH:[0] P:0 V:4 DP 0sec 0cpu\nnp=0 d:0 p:1 0sec 0cpu\n\n\n### -v\n\nThe option -v prints the size of the matrix of vertices, the number of points and the vertices of the polytope that has been entered (M-lattice or N-lattice, depending on the input!). If the input is the weight matrix the M-lattice polytope is analyzed. The output is printed in a single line with the character E as seperator. Furthermore one can limit the output to polytopes whose number of points is limited to a lower and an upper bound:\n\n\u2022 -v -u#, where # is an integer $\\geq 0$, only gives output if the polytope has at most # points. The default value is the parameter POINT_Nmax which fixes the maximal number of points of a polytope at compilation.\n\u2022 -v -l#, where # is an integer $\\geq 0$, only gives output if the polytope has at least # points.The default value is 0.\n\nAfter closing the program a summary is printed. It contains information on how many of the polytopes examined satisfy the bounds and how many polytopes with # of points have been found.\n\nAs an example we consider the complete intersection of codimesnsion 2 in $\\mathbb{P}^3$ and $\\mathbb{P}^2\\times\\mathbb{P}^2$ with the weight matrices as input and without bounds.\n\npalp$nef.x -v Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #columns' (= PolyDim #Points' or #Points PolyDim'): 4 1 1 1 1 3 4 P:35 E -1 3 -1 -1E -1 -1 3 -1E -1 -1 -1 3 Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #columns' (= PolyDim #Points' or #Points PolyDim'): 3 1 1 1 0 0 0 3 0 0 0 1 1 1 4 9 P:100 E -1 2 -1 -1 2 -1 -1 2 -1E -1 -1 2 -1 -1 2 -1 -1 2E -1 -1 -1 2 2 2 -1 -1 -1E -1 -1 -1 -1 -1 -1 2 2 2 Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #columns' (= PolyDim #Points' or #Points PolyDim'): 2 of 2 35# 1 100# 1 Since we have entered a weight matrix the M-lattice polytope is analyzed. Let us discuss the first line of output: 3 4 P:35 E -1 3 -1 -1E -1 -1 3 -1E -1 -1 -1 3 The first two numbers indicate the number of rows and columns of the matrix of vertices in the M-lattice polytope. P:35 indicates that the M-lattice polytope has 35 points. The vertices of the M-lattice polytope are then written in one line with the separator E. The output of the second example is analogous. After we quit PALP by hitting enter without input the following output is given: 2 of 2 35# 1 100# 1 This means that 2 out of the 2 polytopes analyzed satisfy the bounds and that there is one polytope with 35 points and one with 100. Next we consider the same example as above but with the upper bound for the number of points set to 50: palp$ nef.x -v -u50\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n4 1 1 1 1\n3 4 P:35 E -1 3 -1 -1E -1 -1 3 -1E -1 -1 -1 3\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n3 1 1 1 0 0 0 3 0 0 0 1 1 1\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n\n1 of 2\n\n35# 1\n\n\nNow the second polytope exceeds the upper bound for the points as it has 100 points (cf. previous example). There is no output for the second polytope and the summary indicates that only one of the two polytopes analyzed satisfies the bounds.\n\nFurther examples:\n\nExample: same example as above but now we enter the N-lattice polytope and search for polytopes which have at least 7 points in the N-lattice.\n\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n3 4\nType the 12 coordinates as dim=3 lines with #pts=4 columns:\n-1 0 0 1\n-1 0 1 0\n-1 1 0 0\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n4 6\nType the 24 coordinates as dim=4 lines with #pts=6 columns:\n0 0 0 1 0 -1\n0 0 1 0 0 -1\n-1 0 0 0 1 0\n-1 1 0 0 0 0\n4 6 P:7 E 0 0 0 1 0 -1E 0 0 1 0 0 -1E -1 0 0 0 1 0E -1 1 0 0 0 0\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n\n1 of 2\n\n7# 1\n\n\nThere is no output for the first polytope because it has less than 7 vertices. As one can check using option -Lp the N-lattice polytope has only 5 points. There is output for the second polytope since it has seven points and therefore satisfies the bound.\n\n### -m\n\nThe option -m returns a nef partition of length 2 resulting from a partition d = d1 + d2 of the degree of a weight system. More precisely, the input data is a single weight system w and two positive integers d1,d2 such that\n\n \u2211 wi = d1 + d2 i\n\n. The input format is # d= # #, where the first # is the usual CWS, while the # after d= refer to d1 and d2 , respectively.\n\nAs always, w specifies $\\Delta^{(d)} \\subset M$ as the Newton polytope of degree d. Furthermore, the degrees d1,d2 specify Newton polytopes $\\Delta^{(d_1)}, \\Delta^{(d_2)}$ from which one obtains the Minkowski sum $\\Delta^{(d_1,d_2)}=\\Delta^{(d_1)}+\\Delta^{(d_2)} \\subseteq \\Delta^{(d)}$. If $\\Delta_1=\\Delta^{(d_1)}, \\Delta_2=\\Delta^{(d_2)}$ define a nef partition $(\\nabla_1, \\nabla_2)$ of the vertices of $(\\Delta^{(d_1,d_2)})^*$, then the data of this nef partition are given in the standard output.\n\nThe following example taken from hep-th\/0410018 illustrates this option. We consider the weighted projective space $\\mathbb{P}(1,1,1,1,4,6)$ specified by the weight vector $14\\; 1\\; 1\\; 1\\; 1\\; 4\\; 6$ of degree d=14. The polytope \u0394 = \u0394(14) is the Newton polytope of degree 14 monomials in this space. We first analyze the toric variety determined by \u0394(14):\n\npalp$nef.x -Lv Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #colums' (= PolyDim #Points' or #Points PolyDim'): 14 1 1 1 1 4 6 14 1 1 1 1 4 6 M:1271 13 N:10 8 codim=2 #part=2 5 8 Vertices in N-lattice: 0 -1 0 0 0 1 0 0 0 -1 1 0 0 0 0 0 0 -1 0 1 0 0 0 0 0 -4 0 0 1 0 -1 -1 1 -6 0 0 0 0 -1 -2 ---------------------------------------- 6 1 1 1 4 1 0 0 d=14 codim=0 1 0 0 0 1 0 1 0 d=3 codim=3 2 0 0 0 1 0 0 1 d=4 codim=3 H:1 149 [-296] P:1 V:3 4 5 7 8 (6 8) (1 2) (2 2) 2sec 1cpu np=1 d:0 p:1 2sec 1cpu So \u0394(14) has 1271 lattice points and 13 vertices, and (14)) * is the convex hull of the eight vertices shown in the output. By considering the weight systems below the dashed line one sees that $\\mathbb{P}_{(\\Delta^{(14)})^*}$ is the blowup of $\\mathbb{P}(1,1,1,1,4,6)$ along the divisors corresponding to the last two vertices of (14)) * . Now we want to use the option -m to see whether the partition 14 = 2 + 12 determines a nef partition via the Minkowski sum \u0394(2,12) = \u0394(2) + \u0394(12). palp$ nef.x -Lv -m\ntype degrees and weights [d w1 w2 ... wk d=d_1 d_2]:\n14 1 1 1 1 4 6 d=2 12\n14 1 1 1 1 4 6 d=2 12 M:1270 12 N:11 7 codim=2 #part=2\n5 7 Vertices in N-lattice:\n0 -1 0 0 0 1 0\n0 -1 1 0 0 0 0\n0 -1 0 1 0 0 0\n0 -4 0 0 1 0 -2\n1 -6 0 0 0 0 -3\n-----------------------------------\n6 1 1 1 4 1 0 d=14 codim=0\n3 0 0 0 2 0 1 d=6 codim=3\nd=12 2H:3 243 [-480] P:0 V:3 5 (2 12) (0 6) 7sec 6cpu\nnp=1 d:0 p:1 0sec 0cpu\n\n\nThe output indeed yields such a nef partition. Since not every monomial of degree 14 is a product of monomials of degree 2 and 12, the polytope \u0394(2,12) is only a proper subpolytope of \u0394(14). Consequently $\\mathbb{P}_{(\\Delta^{(2,12)})^*}$ is obtained from $\\mathbb{P}_{(\\Delta^{(14)})^*}$ by a further blowup along the vertex (0,0,0, \u2212 2, \u2212 3)T.\n\nBy using the option -m in the same way one can find that \u0394(6,8) = \u0394(14) and that 14 = 3 + 11 does not give rise to a nef partition.\n\n### -R\n\nThe option -R prints the vertices of the input polytope if it is not reflexive. To illustrate this we enter the CWS of a polytope which is not reflexive:\n\npalp$nef.x -R Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #columns' (= PolyDim #Points' or #Points PolyDim'): 6 3 2 1 0 0 5 0 0 1 1 3 3 4 Vertices of P: -1 1 0 0 0 1 -1 1 -1 0 0 0 The same output is given if we enter the N-lattice polytope itself. Without the option -R there is not output if the polytope is not reflexive: palp$ nef.x\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n6 3 2 1 0 0 5 0 0 1 1 3\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n\n\n### -V\n\nThe option -V prints the vertices of the polytope in the N-lattice together with the standard output. In contrast to the option -Lv the information about the linear relations is not given. Furthermore, in the lines containing the nef partitions the additional information about the degrees is left out. The option -V also works for non-reflexive polytopes.\n\nAs an example we consider a complete intersection of codimension 2 in $\\mathbb{P}^3$:\n\npalp$nef.x -V Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #columns' (= PolyDim #Points' or #Points PolyDim'): 4 1 1 1 1 4 1 1 1 1 M:35 4 N:5 4 codim=2 #part=2 3 4 Vertices of P: -1 0 0 1 -1 0 1 0 -1 1 0 0 H:[0] P:0 V:2 3 0sec 0cpu np=1 d:0 p:1 0sec 0cpu We can also enter the M-lattice polytope to get the same result: palp$ nef.x -V\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n3 4\nType the 12 coordinates as dim=3 lines with #pts=4 columns:\n-1 3 -1 -1\n-1 -1 3 -1\n-1 -1 -1 3\nM:35 4 N:5 4 codim=2 #part=2\n3 4 Vertices of P:\n-1 0 0 1\n-1 0 1 0\n-1 1 0 0\nH:[0] P:0 V:2 3 0sec 0cpu\nnp=1 d:0 p:1 0sec 0cpu\n\n\nExample: If the polytope is non-reflexive the output is the same as for the option -R:\n\npalp$nef.x -V Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #columns' (= PolyDim #Points' or #Points PolyDim'): 6 3 2 1 0 0 5 0 0 1 1 3 3 4 Vertices of P: -1 1 0 0 0 1 -1 1 -1 0 0 0 ### -Q The option -Q prints the information about the nef partitions and the Hodge numbers only if the corresponding complete intersection is a direct product (cf. option -D) up to lattice quotients. If none of the nef partitions is a direct product only the numbers of points and vertices in the M- and N-lattice, together with the codimension and the number of nef partitions is given. Example: Complete intersection of codimension 2 in $\\mathbb{P}^2\\times\\mathbb{P}^2$. As one can check using the option -D one of the nef partitions corresponds to a direct product: palp$ nef.x -Q\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n3 1 1 1 0 0 0 3 0 0 0 1 1 1\n3 1 1 1 0 0 0 3 0 0 0 1 1 1 M:100 9 N:7 6 codim=2 #part=5\nH:4 [0] h1=2 P:0 V:2 3 5 D 0sec 0cpu\nnp=4 d:1 p:0 0sec 0cpu\n\n\nExample: Complete intersection of codimension 2 in $\\mathbb{P}^3$. This example has no nef partition corresponding to a direct product. Then the output looks as follows:\n\npalp$nef.x -Q Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #columns' (= PolyDim #Points' or #Points PolyDim'): 4 1 1 1 1 4 1 1 1 1 M:35 4 N:5 4 codim=2 #part=2 np=2 d:0 p:0 0sec 0cpu The N-lattice polytope of $\\mathbb{P}^3$ has no nef partition corresponding to a direct product. Then the output looks as follows: \\begin{verbatim} palp$ nef.x -Q Degrees and weights d1 w11 w12 ... d2 w21 w22 ...'\n\n or #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n\n\n4 1 1 1 1 4 1 1 1 1 M:35 4 N:5 4 codim=2 #part=2 np=2 d:0 p:0 0sec 0cpu \\end{verbatim}\n\nExample: Complete intersection of codimension 2 in $\\mathbb{P}^2\\times\\mathbb{P}^2$ quotiented by $\\mathbb{Z}_3\\times\\mathbb{Z}_3$:\n\npalp$nef.x -Q Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #columns' (= PolyDim #Points' or #Points PolyDim'): 3 1 1 1 0 0 0 3 0 0 0 1 1 1\/Z3: 0 1 2 0 0 0\/Z3: 0 0 0 1 2 0 3 1 1 1 0 0 0 3 0 0 0 1 1 1 \/Z3: 0 1 2 0 0 0 \/Z3: 0 0 0 1 2 0 M:16 9 N:19 6 codim=2 #part=1 H:4 [0] h1=2 P:0 V:2 3 5 7 9 10 11 15 16 D 0sec 0cpu np=0 d:1 p:0 0sec 0cpu Note that the nef partition now contains more points due to the lattice refinement induced by the quotient. ???: The comment \"up to lattice quotient\" in the help screen may indicate that the scope of this option is larger than described here. ### -g* The option -g*, where * is an integer m=0,1,2, returns the points of the supports $\\Delta(\\check C)$ of the Gorenstein cones $\\check C \\subset \\tilde N_\\mathbb{R}$ associated to the nef partitions of length r of the input polytope $\\Delta^* \\subset N_\\mathbb{R}$. For the notation on Gorenstein cones see Nef partitions and Gorenstein cones. The default value is m=1. The standard output is changed as follows. The lines containing the information about the nef partition including the Hodge numbers, the parts of the nef partition etc. are suppressed. Instead, for each nef partition the points of $\\Delta(\\check C)$ are printed in the following form: D n Points of PG: (nv=#) # # ... # # . . ... . . . . ... . . # # ... # # The interpretation depends on the integer m. For m=2 the output is the list of points $\\tilde p \\in \\check C$. Note that since the origin 0N belongs to every part of the nef partition, it appears r times, each time another of the r support functions being equal to 1. For m=1 the redundant coordinate \u03d50(p) is omitted in $\\tilde p$ and we obtain vectors $\\tilde p'$. For m=0 all \u03d5i(p) are omitted and the resulting r-fold occurrence of 0N is reduced to just a single occurrence; information on the nef partition is lost and the output becomes just the list of lattice points of \u0394 * . The values of D, n and the # columns are summarized in the following table: $\\begin{array}{c|c|c|c} m & 0 & 1 & 2\\\\ \\hline D & d & \\tilde d-1 & \\tilde d\\\\ n & n & n+r-1 & n+r-1\\\\ \\# column & p & \\tilde p' & \\tilde p \\end{array}$ where n is the number of lattice points in \u0394 * and $d, r, \\tilde d$ are as in Nef partitions and Gorenstein cones. The number # in nv=# denotes the number of vertices of the cone $\\check C$. The order of the points is first the vertices, then the non-vertex points with the origin at the end. The following example illustrates this option. We consider complete intersections of codimension 2 in $\\mathbb{P}^2\\times\\mathbb{P}^1\\times\\mathbb{P}^2$ discussed in arXiv:0704.0449[hep-th]. The nef partitions for this example were discussed in the section on the standard output. With the choice of m=2 we obtain the information about the partition in terms of the Gorenstein cone. Let $v_0,\\dots,v_7$ denote the vertices of the polytope in the N-lattice. palp$ nef.x -N -g2\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #colums' (= PolyDim #Points' or #Points PolyDim'):\n5 8\nType the 40 coordinates as dim=5 lines with #pts=8 colums:\n1 0 -1 0 0 0 0 0\n0 1 -1 0 0 0 0 0\n0 0 0 1 -1 0 0 0\n0 0 0 0 0 1 0 -1\n0 0 0 0 0 0 1 -1\nM:300 18 N:9 8 codim=2 #part=15\n7 10 Points of PG: (nv=8)\n1 1 0 0 0 1 1 0 0 1\n0 0 1 1 1 0 0 1 1 0\n1 0 -1 0 0 0 0 0 0 0\n0 1 -1 0 0 0 0 0 0 0\n0 0 0 1 -1 0 0 0 0 0\n0 0 0 0 0 1 0 -1 0 0\n0 0 0 0 0 0 1 -1 0 0\n7 10 Points of PG: (nv=8)\n1 1 0 1 0 1 0 0 0 1\n0 0 1 0 1 0 1 1 1 0\n1 0 -1 0 0 0 0 0 0 0\n0 1 -1 0 0 0 0 0 0 0\n0 0 0 1 -1 0 0 0 0 0\n0 0 0 0 0 1 0 -1 0 0\n0 0 0 0 0 0 1 -1 0 0\n[output of further nef partitions]\n\n\nLet us consider for instance the nef partition P:8 as produced for by the option -Lp:\n\nH:19 19 [0] P:8 V:4 5 6 7 (0 3) (1 1) (3 0) 0sec 0cpu\n\n\nThis example was the focus of arXiv:0704.0449[hep-th]. The output of -g2 for this nef partition is:\n\n7 10 Points of PG: (nv=8)\n1 1 1 1 0 0 0 0 0 1\n0 0 0 0 1 1 1 1 1 0\n1 0 -1 0 0 0 0 0 0 0\n0 1 -1 0 0 0 0 0 0 0\n0 0 0 1 -1 0 0 0 0 0\n0 0 0 0 0 1 0 -1 0 0\n0 0 0 0 0 0 1 -1 0 0\n\n\nSince the first four vertices $v_0, v_1, v_2\\,$ and $v_3\\,$ are in $V_0\\,$, we have \u03d50(vi) = 1 and \u03d51(vi) = 0, hence the corresponding points of the Gorenstein cone take the form (1,0,vi) for $i=0,\\dots,3$. The next four vertices $v_4, v_5, v_6\\,$ and $v_7\\,$ are in $V_1\\,$, we have $\\phi_0(v_i) = 0\\,$ and $\\phi_1(v_i) = 1\\,$, hence the corresponding points of the Gorenstein cone take the form $(0,1,v_i)\\,$ for $i=4,\\dots,7$. Finally, the origin always belongs to every part of the nef partition, hence it appears as often as the codimension which here is $r=2\\,$. So $p_8 = 0\\,$ and $p_9 = 0\\,$. Once with $\\phi_0(p_8) = 0\\,$ and $\\phi_1(p_8) = 1\\,$ and once with $\\phi_0(p_9) = 1\\,$ and $\\phi_1(p_9) = 0\\,$.\n\n### -d*\n\nThe option -d*, where * is an integer m=0,1,2, returns the points of the Gorenstein cones $C \\subset \\tilde M_\\mathbb{R}$ associated to the nef partitions of length r of the polytope $\\nabla^* \\subset M_\\mathbb{R}$. For the notation on Gorenstein cones see the section on nef partitions and Gorenstein cones. This option can be used to determine the polytope $\\nabla^*$ for each of the nef partitions of the given polytope \u0394 * . The polytope $\\nabla^*$ can then be further analyzed with poly.x.\n\nThe integer m triggers the same output format as for the option -g*. The default value is m=1. The option -d2 automatically sets the flag -p.\n\nThe following example illustrates this option. We consider complete intersections of codimension 2 in $\\mathbb{P}^2\\times\\mathbb{P}^1\\times\\mathbb{P}^2$ discussed in arxiv:0704.0449[hep-th]. For more details on the nef partitions see the example in the section on the standard output and on -g*.\n\npalp$nef.x -N -d2 Degrees and weights d1 w11 w12 ... d2 w21 w22 ...' or #lines #colums' (= PolyDim #Points' or #Points PolyDim'): 5 8 Type the 40 coordinates as dim=5 lines with #pts=8 colums: 1 0 -1 0 0 0 0 0 0 1 -1 0 0 0 0 0 0 0 0 1 -1 0 0 0 0 0 0 0 0 1 0 -1 0 0 0 0 0 0 1 -1 M:300 18 N:9 8 codim=2 #part=15 7 63 Points of dual PG: (nv=27) 1 0 1 0 0 1 1 1 ... 0 1 0 1 1 0 0 0 ... -1 0 1 1 0 -1 -1 -1 ... -1 1 -1 0 1 1 1 -1 ... [63-8=55 more points] 0 1 0 1 -1 0 0 0 ... -1 0 -1 0 0 1 -1 -1 ... -1 1 1 1 1 -1 -1 1 ... [...] For each of the 11 nef partitions of the input polytope \u0394 * we get a 7-dimensional dual Gorenstein cone C from which the points of the polytope $\\nabla^*$ can be read off by omitting the first two entries of each column. The numbers of points and vertices of $\\nabla^*$ depend on which of the nef partitions is considered. The nef partition of interest in arxiv:0704.0449[hep-th] was P:8. The corresponding output of -d2 is 7 40 Points of dual PG: (nv=12) 0 0 1 1 1 0 0 0 ... 1 1 0 0 0 1 1 1 ... 0 0 -1 2 -1 0 0 0 ... 0 0 2 -1 -1 0 0 0 ... [40-8=32 more points] 0 1 0 0 0 1 1 0 ... -1 -1 0 0 0 2 -1 -1 ... -1 2 0 0 0 -1 -1 2 ... We see that the polytope $\\nabla^*$ has 39 points (the interior point appears twice) and 12 vertices. Let $e_1,\\dots,e_5$ be the standard basis of $\\mathbb{R}^5$. Let $\\check{v}_0,\\dots,\\check{v}_{11}$ denote the vertices of the polytope $\\nabla^*$. with $\\check{v}_0 = -e_4-e_5, \\check{v}_1 = e_3-e_4+2e_5, \\check{v}_2= -e_1+2e_2, \\check{v}_3 = 2e_1 -e_2, \\check{v}_4 = -e_1-e_2, \\check{v}_5 = e_3+2e_4-e_5, \\check{v}_6= e_3-e_4-e_5,$ $\\check{v}_7 = -e_4+2e_5, \\check{v}_8= -e_1+2e_2-e_3, \\check{v}_9 = 2e_1 - e_2 -e_3, \\check{v}_{10}=-e_1-e_2-e_3, \\check{v}_{11}=2e_4-e_5.$ From the above output we can read off the nef partition $\\check V = \\check V_0 \\cup \\check V_1$ of $\\nabla^*$: $\\check{V}_0 = \\langle \\check{v}_2, \\check{v}_3, \\check{v}_4, \\check{v}_8, \\check{v}_9, \\check{v}_{10} \\rangle, \\quad \\check{V}_1 = \\langle \\check{v}_0, \\check{v}_1, \\check{v}_5, \\check{v}_6, \\check{v}_7, \\check{v}_{11} \\rangle.$ We can check this by feeding the vertices back into nef.x with the options -N and -Lv. palp$ nef.x -N -Lv\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #colums' (= PolyDim #Points' or #Points PolyDim'):\n5 12\nType the 60 coordinates as dim=5 lines with #pts=12 colums:\n0 0 -1 2 -1 0 0 0 -1 2 -1 0\n0 0 2 -1 -1 0 0 0 2 -1 -1 0\n0 1 0 0 0 1 1 0 -1 -1 -1 0\n-1 -1 0 0 0 2 -1 -1 0 0 0 2\n-1 2 0 0 0 -1 -1 2 0 0 0 -1\nM:24 15 N:39 12 codim=2 #part=2\n5 12 Vertices in N-lattice:\n0 0 -1 2 -1 0 0 0 -1 2 -1 0\n0 0 2 -1 -1 0 0 0 2 -1 -1 0\n0 1 0 0 0 1 1 0 -1 -1 -1 0\n-1 -1 0 0 0 2 -1 -1 0 0 0 2\n-1 2 0 0 0 -1 -1 2 0 0 0 -1\n------------------------------------------------------------\n[linear relations]\nH:19 19 [0] P:0 V:1 2 3 4 5 6 13 15 17 ... [degrees]\nH:19 19 [0] P:1 V:2 3 4 8 9 10 16 17 18 ... [degrees]\nnp=2 d:0 p:0 0sec 0cpu\n\n\nWe see that the nef partition P:1 agrees with $\\check V = \\check V_0 \\cup \\check V_1$.\n\n### -G\n\nThe option -G works directly with Gorenstein cones which need not correspond to nef partitions. This is in particular useful for analyzing generalized Calabi-Yau manifolds as defined in hep-th\/9304045 and alg-geom\/9402002. The input polytope is interpreted as the support polytope \u0394(C) of a reflexive Gorenstein cone C, cf. the section on nef partitions and Gorenstein cones. The index r of the cone is 2 by default and can be set to different values with the -c*. The standard output contains information on the support polytopes of the cone and the dual cone and the string-theoretic Hodge numbers hij, $0\\le i, j\\le dim C- 2r$, see alg-geom\/9509009. If the input does not correspond to a reflexive Gorenstein cone of index r, no Hodge numbers and no N-lattice data can be computed; as usual, the number of facets is displayed instead. If the input corresponds to a reflexive Gorenstein cone of an index different from r, this is treated like a non-reflexive case but with a warning message.\n\npalp\\$ nef.x -G\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n4 2\nType the 8 coordinates as #pts=4 lines with dim=2 columns:\n0 0\n0 1\n1 0\n1 1\nM:4 4 N:4 4 H:[0] h0=0\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n3 1 1 1 1 1 1\n3 1 1 1 1 1 1 M:56 6 N:6 6 H:20 [24]\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n7 1 1 1 2 3 3 3\n7 1 1 1 2 3 3 3 M:154 18 F:9\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n7 1 1 2 2 2 3 3\n7 1 1 2 2 2 3 3 M:116 18 N:9 9 H:2 70 [-136]\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n3 2\nType the 6 coordinates as #pts=3 lines with dim=2 columns:\n0 0\n1 0\n0 1\nWarning: Input has index 3, should be 2! M:3 3 F:3\nDegrees and weights d1 w11 w12 ... d2 w21 w22 ...'\nor #lines #columns' (= PolyDim #Points' or #Points PolyDim'):\n1 1 1 0 0 0 0 2 0 0 1 1 1 1\n1 1 1 0 0 0 0 2 0 0 1 1 1 1 M:20 8 N:6 6 H:[0]\n\n\nAs the examples show, weight input d w1 ... wn requires $w_1+\\ldots w_n = rd$; in other words the weights qi = wi \/ d add up to r rather than to 1 as in the standard case. See arXiv:1204.1181[hep-th] for more information on how weight systems determine Gorenstein cones.\n\n-G cannot be combined with all of the other options. Nevertheless -N swaps the lattices M and N as usual; -H, -S, -T work as expected; -t works (without it no time information is given); -c* determines the index; -R displays the vertices of the input polytope if the cone is not reflexive of index r; -V displays the vertices of the support polytope of the dual (N lattice) cone; -g*, -d* display the full sets of points of the support polytopes in the lattices N or M, respectively (here no numbers can be specified with these options).\n\n## Strategy for future versions\n\nShould we keep both help screens, or try to get everything into 'nef.x -h'?\n\nIt would be good to have a documentation of the header file Nef.h: just type whatever information you can supply into the following listing, in the standard ('\/* ... *\/') C comment format.\n\n#define Nef_Max \t500000\n#define NP_Max 500000\n#define W_Nmax (POLY_Dmax+1)\n#define MAXSTRING\t100\n\n#undef\tWRITE_CWS\n#define\tWRITE_CWS\n\n#define Pos_Max (POLY_Dmax + 2)\n#define FIB_Nmax 10*EQUA_Nmax\n#define FIB_POINT_Nmax VERT_Nmax\n\ntypedef struct {\nLong W[FIB_Nmax][FIB_POINT_Nmax];\nLong VM[FIB_POINT_Nmax][POLY_Dmax];\nint nw;\nint nv;\nint d;\nint Wmax;\n} LInfo;\n\nstruct Poset_Element {\nint num, dim;\n};\n\nstruct Interval {\nint min, max;\n};\n\ntypedef struct Interval Interval;\n\ntypedef struct {\nstruct Interval *L;\nint n;\n} Interval_List;\n\ntypedef struct Poset_Element Poset_Element;\n\ntypedef struct {\nstruct Poset_Element x, y;\n} Poset;\n\ntypedef struct {\nstruct Poset_Element *L;\nint n;\n} Poset_Element_List;\n\ntypedef struct {\nint nface[Pos_Max];\nint dim;\nINCI edge[Pos_Max][FACE_Nmax];\n} Cone;\n\ntypedef struct {\nLong S[2*Pos_Max];\n} SPoly;\n\ntypedef struct {\nLong B[Pos_Max][Pos_Max];\n} BPoly;\n\ntypedef struct {\nint E[4*(Pos_Max)][4*(Pos_Max)];\n} EPoly;\n\ntypedef struct {\nLong x[POINT_Nmax][W_Nmax];\nint N, np;\n} AmbiPointList;\n\ntypedef struct {\nint n;\nint nv;\nint codim;\nint S[Nef_Max][VERT_Nmax];\nint DirProduct[Nef_Max];\nint Proj[Nef_Max];\nint DProj[Nef_Max];\n} PartList;\n\ntypedef struct {\nint n;\nint nv;\nint S[Nef_Max][VERT_Nmax];\n} Part;\n\ntypedef struct {\nint n, y, w, p, t, S, Lv, Lp, N, u, d, g, VP, B, T, H, dd, gd,\nnoconvex, Msum, Sym, V, Rv, Test, Sort, Dir, Proj, f;\n} Flags;\n\ntypedef struct {\nint noconvex, Sym, Test, Sort;\n} NEF_Flags;\n\nstruct Vector {\nLong x[POLY_Dmax];\n};\n\ntypedef struct Vector Vector\u00a0;\n\ntypedef struct {\nstruct Vector *L;\nint n;\nLong np, NP_max;\t\t\t\t\t\t} DYN_PPL;\n\nvoid part_nef(PolyPointList *, VertexNumList *, EqList *, PartList *,\nint *, NEF_Flags *);\n\nvoid Make_E_Poly(FILE *, CWS *, PolyPointList *, VertexNumList *, EqList *,\nint *, Flags *, int *);\n\nvoid Mink_WPCICY(AmbiPointList * _AP_1, AmbiPointList * _AP_2,\nAmbiPointList * _AP);\n\nint IsDigit(char);\n\nint IntSqrt(int q);\n\nvoid Die(char *);\n\nvoid Print_CWS_Zinfo(CWS *);\n`","date":"2017-09-20 02:02:16","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\": 252, \"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\": 0.5585659146308899, \"perplexity\": 1209.4606543393213}, \"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-2017-39\/segments\/1505818686117.24\/warc\/CC-MAIN-20170920014637-20170920034637-00376.warc.gz\"}"} | null | null |
\section{Introduction}
Topological insulators represent a new quantum state of matter for which gapless conducting states can develop at the surface of materials that are insulating in the bulk~\cite{Hasan2010RMP,Qi2011RMP,Haldane2017RMP}. These surface states disperse linearly around so-called Dirac points where they are protected by time-reversal symmetry and thus robust against perturbations (scattering). In the presence of an additional magnetic order, which breaks this time-reversal symmetry, a gap is opened at this Dirac point~\cite{Tokura2019NRP}. In such magnetic topological materials, the combination of non-trivial band topology and magnetic order may give rise to the emergence of a variety of novel quantum phenomena~\cite{Wan2011PRB,Yu2010SC,Chang2013SC,Chang2015NM,Qi2008PRB,Essin2009PRL,Mong2010PRB,Xiao2018PRL,Varnava2018PRB,He2017SC,Belopolski2019SC}. In recent years, an increasing number of magnetic materials, which have an intrinsic magnetic order and topological electronic states in the stoichiometric compositions, has been theoretically predicted as magnetic topological insulators~\cite{Chen2014PRL,Otrokov2019PRL,Zhang2019PRL,Li2019SA}, magnetic Dirac semimetals~\cite{Tang2016NP,Hua2018PRB}, and magnetic Weyl semimetals~\cite{Wang2016PRL,Xu2018PRB}. Such materials would not only provide a clean platform to realize the exotic topological phenomena under time-reversal symmetry breaking, but also show great potential for applications in quantum technology.
Recently, a strong experimental focus has been on MnBi$_2$Te$_4$~\cite{Otrokov2019Nat,Cui2019PRB,Gong2019CPL,Yan2019PRM,Yan2019PRB,Deng2020SC,Liu2020NM,Ge2020NSR,Hu2020NC,Lee2019PRR,Zeugner2019CM,Vidal2019PRB,Chen2019PRX,Swatek2020PRB,Hao2019PRX,Li2019PRX,Nevola2020,Xu2021PRB}, which has been predicted to be an intrinsic antiferromagnetic (AFM) topological insulator for which different topological states can be realized in bulk crystals as well as in thin films~\cite{Otrokov2019PRL,Li2019SA,Zhang2019PRL}. EuIn$_2$As$_{2}$ is another promising candidate for an intrinsic magnetic topological material~\cite{Xu2019PRL}. Different from the layered van der Waals-type MnBi$_2$Te$_4$, EuIn$_2$As$_{2}$ has a three-dimensional structure and crystallizes in the hexagonal $P6_3/mmc$ (No. 194) space group, with alternating stacking of Eu$^{2+}$ and [In$_2$As$_2$]$^{2-}$ layers along the $c$-axis~\cite{Xu2019PRL}. The Eu spins exhibit an A-type antiferromagnetic order below $T_N \simeq$ 18~K where they are parallel oriented within each layer and an antiparallel between neighbouring layers (along the $c$-axis)~\cite{Goforth2008IC,Zhang2020PRB}. It has been predicted that the AFM order in EuIn$_2$As$_{2}$ gives rise to an axion insulator with non-trivial topological states that are strongly influenced by the orientation of the magnetic moments. A topological crystalline insulator phase with gapless surface states on the (100), (010), and (001) surfaces is expected for in-plane oriented magnetic moments, whereas a higher-order topological insulator phase with chiral hinge states is predicted if the magnetic moments are out-of-plane oriented~\cite{Xu2019PRL}. From an experimental perspective, angle resolved photoemission spectroscopy (ARPES) studies have confirmed that EuIn$_2$As$_{2}$ has hole-type Fermi pockets around the bulk Brillouin zone center~\cite{Regmi2020PRB,Zhang2020PRB,Sato2020PRR}, together with a heavily hole-doped surface state and an inversion of the bulk band in the AFM state~\cite{Sato2020PRR} that is consistent with the theoretical prediction~\cite{Xu2019PRL}. A negative magneto-resistance seen in magneto-transport measurements has provided evidence for a rather strong spin scattering of the carriers by the localized magnetic moments~\cite{Goforth2008IC,Zhang2020PRB} that may affect the above described topological states. Electron spin resonance measurements have revealed that the spin dynamics in the vicinity of $T_N$ is governed by short-range AFM correlations of the Eu spins~\cite{Rosa2012PRB}. An optical spectroscopy study, which can directly probe the dynamics of the charge carriers and provide information about their interplay with the spin and lattice degrees of freedom, has not been reported to date (to our best knowledge).
Here, we present an infrared spectroscopy study of EuIn$_2$As$_2$ which reveals a strong interaction of the charge carriers with the Eu spins in terms of a cusp-like maximum of the free carrier scattering rate around $T_N$ and a sizeable exchange splitting of the bulk valence bands below $T_N$. Moreover, we observe corresponding anomalies of an infrared-active phonon mode around 185\icm\ which provide evidence for a sizeable spin-lattice coupling. These findings highlight a strong mutual interaction between the charge, spin and lattice degrees of EuIn$_2$As$_2$ that may also affect the predicted topological states and thus should be considered in the theoretical predictions and interpretation of experimental data.
\section{Experimental methods}
High-quality single crystals of EuIn$_2$As$_{2}$ with plate-like shapes have been synthesized with a self-flux method~\cite{Zhang2020PRB}. The in-plane resistivity exhibits a metallic temperature dependence, albeit with a cusp-like maximum around $T_N \simeq$ 18 K [see Fig.~\ref{Fig3}(b)]. The in-plane reflectivity $R(\omega)$ spectra were measured at a near-normal angle of incidence using a Bruker VERTEX 70v FTIR spectrometer with an \emph{in situ} gold overfilling technique~\cite{Homes1993}. Data from 30 to 20\,000\icm\ were collected at different temperatures with the sample mounted in an ARS-Helitran crysostat. The room temperature optical response function in the near-infrared to ultraviolet range (4\,000 -- 50\,000\icm) was measured with a commercial ellipsometer (Woollam VASE). The optical conductivity and related response functions and constants were obtained by performing a Kramers-Kronig analysis of $R(\omega)$~\cite{Dressel2002}. For the low frequency extrapolation below 30\icm, we used a Hagen-Rubens function ($R = 1 - A\sqrt{\omega}$). On the high frequency side, the extrapolation was anchored by the room temperature ellipsometry data.
\section{Results and discussions}
\subsection{Plasma edge and screened plasma frequency}
\begin{figure}[tb]
\includegraphics[width=\columnwidth]{Fig1}
\caption{ (color online) (a) Temperature dependent in-plane reflectivity spectra of EuIn$_2$As$_2$. The left inset shows the temperature-dependence of the reflectivity at 300\icm. The right inset displays the change of the reflectivity below $T_N \simeq$ 18~K in terms of the ratio of the spectra at 10 and 20~K. (b) Temperature dependence of the real part of the dielectric function $\varepsilon_1(\omega)$. Inset: magnified view of the $\varepsilon_1(\omega)$ spectra in the vicinity of the zero crossing. (c) Temperature evolution of the screened plasma frequency, $\omega^{\rm scr}_{p}$ as deduced from the zero crossing of $\varepsilon_1(\omega)$.}
\label{Fig1}
\end{figure}
Figure~\ref{Fig1}(a) shows the temperature-dependent spectra of the in-plane reflectivity, $R(\omega)$, of EuIn$_2$As$_2$. In the far-infrared range they show a typical metal-like response with a sharp plasma edge, below which the reflectivity increases rapidly and approaches unity toward the origin. The left inset details the temperature dependence of the low-frequency value $R(\omega = 300\icm)$ which exhibits a pronounced anomaly around $T_N \simeq$ 18~K. The small value of the plasma edge ($\sim 650$~\icm) suggests a rather low carrier density, consistent with the small hole pocket that has been observed in ARPES measurements~\cite{Regmi2020PRB,Zhang2020PRB,Sato2020PRR}. The slight temperature dependent upward shift of the plasma edge from around 650\icm\ at 300~K to 680\icm\ at 10~K indicates a corresponding weak increase of the plasma frequency with cooling. Note that the weak feature around 750\icm\ (marked by a star) is due to a plasmonic effect that will be discussed elsewhere. The far-infrared spectra also show a pronounced infrared-active phonon mode around 185\icm\ and there seem to be two additional, very weak modes around 80\icm\ and 215\icm. Toward higher frequency, starting from the mid-infrared range, the spectra reveal several interband transitions from the occupied states in the valence bands to the empty states in the conduction bands, that show up as kinks or peaks that are centered around 5\,000, 12\,000, 20\,000, 27\,500 and 32\,000\icm. The inset on the right-hand side of Fig.~\ref{Fig1}(a) shows the anomalous change of the reflectivity spectra in the AFM state, in terms of the ratio of the spectra at 10 and 20~K. It reveals characteristic peak-dip-peak features around 5\,000 and 12\,000\icm\ that arise from the magnetic splitting of the bulk valence bands and will be further discussed below.
\begin{figure*}[tb]
\includegraphics[width=\textwidth]{Fig2}
\caption{ (color online) (a) Optical conductivity of EuIn$_2$As$_{2}$ at different temperatures; the dashed line show the extrapolation of the Drude fits; Inset: Optical conductivity at room temperature in the full measured range up to 50\,000\icm. (b) Schematic of the band structure of EuIn$_2$As$_{2}$. The dashed lines denote the splitting of the valence bands below $T_N$. (c) Difference plot of $\sigma_1(\omega)$ across (10~K $-$ 20~K) and above (40~K $-$ 50~K) the AFM transition.}
\label{Fig2}
\end{figure*}
Figure~\ref{Fig1}(b) shows the corresponding temperature dependent the spectra of the real part of the dielectric function $\varepsilon_1(\omega)$. At low frequencies, $\varepsilon_1(\omega)$ is negative (a defining property of a metal) and can be well described with a Drude model, $\varepsilon(\omega) = \varepsilon_{\infty} - \frac{\omega_p^2}{\omega^2 + i\omega/\tau}$, where $\varepsilon_{\infty}$ is the high-frequency dielectric constant, $\omega_p = \sqrt{ne^2/\epsilon_0 m^\ast}$ is the plasma frequency, that is a measure of the ratio of the carrier density $n$ and the effective mass $m^\ast$ of the free carriers, and $1/\tau$ is their scattering rate. The zero crossing of $\varepsilon_1(\omega)$ (indicated by the horizontal dashed line in the inset) marks the screened plasma frequency $\omega_p^{\rm scr}$ of the free carriers, which is related to the plasma frequency through $\omega_p^{\rm scr} = \omega_p/\sqrt{\varepsilon_{\infty}}$. Fig.~\ref{Fig1}(c) shows that, similar to the low-frequency reflectivity in the left inset of Fig.~\ref{Fig1}(a), $\omega_p^{\rm scr}$ increases weakly toward low temperature, from about 580\icm\ at 300~K to 620\icm\ at 10~K, and exhibits an anomalous suppression around $T_N$.
\subsection{Optical conductivity and band splitting in the AFM state}
Fig.~\ref{Fig2}(a) displays the temperature dependence of the real part of the optical conductivity $\sigma_1(\omega)$ in the far-infrared to near-infrared range. The inset shows the room temperature spectrum of the optical conductivity over the full measured range up to 50\,000\icm. Below 500\icm\, the main features are a weak Drude peak centered at zero frequency and a sharp infrared-active phonon around 185\icm. The onset of the interband transitions occurs around 800\icm\ and is superimposed on the tail of the Drude peak. Towards higher frequency, there is a series of weak interband transitions up to 15\,000\icm\ that is followed by three much stronger interband transitions around 20\,000, 27\,500 and 32\,000\icm.
Fig.~\ref{Fig2}(b) shows a schematic of the band structure that is motivated by the reported band structure calculations~\cite{Xu2019PRL}. The assignment of the low-energy interband transition is indicated with colored arrows. In addition to a pair of conduction and valence bands (CB$_1$ and VB$_1$), for which the spin-orbit-coupling gives rise to an inverted band gap, there are two more valence bands (VB$_2$ and VB$_3$) that are degenerate around the $\Gamma$ point in the paramagnetic state (solid lines) and are expected to exhibit a magnetic splitting in the AFM state (dashed gray lines). This band assignment accounts for the quasi-linear conductivity in the frequency range 800 -- 3\,000\icm\ in terms of the interband transitions near the spin-orbit inverted gap (black arrow). The approximately linear frequency-dependent increase of the conductivity between 800 and 3\,000\icm\ agrees well with the presence of 3D linear bands near the Fermi level, that are also apparent from recent ARPES studies~\cite{Regmi2020PRB,Zhang2020PRB,Sato2020PRR}. The onset frequency of 800\icm\ provides an estimate of the spin-orbit gap of about 100~meV. A somewhat lower gap value will be obtained if the so-called Burstein-Moss shift, due to the Pauli-blocking effect by the holes in the valence band, is included. However, since the Drude peak is very weak, the latter effect should be rather small and thus has not been further considered. The absorption peak around 5\,000\icm\ is assigned to the interband transitions from VB$_2$ to the empty states in CB$_1$, as illustrated by the blue arrow in Fig.~\ref{Fig2}(c). The rapid increase of $\sigma_1(\omega)$ around 12\,000\icm\ can be understood in terms of the transitions from VB$_3$ to CB$_1$, as indicated by the orange arrow. Finally, the much stronger and sharp peaks around 20\,000 and 30\,000\icm\ are assigned to transitions from deeper valence bands that are also predicted by the band calculations~\cite{Xu2019PRL}.
Next we focus on the changes of the optical response across the AFM transition of EuIn$_2$As$_{2}$. Fig.~\ref{Fig2}(c) shows the $\sigma_1(\omega)$ difference spectrum between 10 and 20~K across $T_N \simeq 18$~K. It reveals two sets of peak-dip structures that are centered around the interband transitions at 5\,000 and 12\,000\icm, respectively, and are absent in the corresponding difference spectrum between 40 and 50~K. These characteristic peak-dip structures are interpreted in terms of a magnetic band splitting of VB$_2$ and VB$_3$ which is indicated by the dashed lines in Fig.~\ref{Fig2}(b). For the corresponding interband transitions, this magnetic splitting gives rise to a doublet of sub-bands that are located below and above the frequency of the interband transition in the paramagnetic state, respectively, e.g., VB$_{2a}$ $\rightarrow$ CB$_1$, VB$_{2b}$ $\rightarrow$ CB$_1$, VB$_{3a}$ $\rightarrow$ CB$_1$, and VB$_{3b}$ $\rightarrow$ CB$_1$. As shown in Fig.~\ref{Fig2}(c), from this peak-dip structures in the difference spectrum of $\sigma_1(\omega)$ we can estimate the magnitude of the magnetic band splitting, which amounts to $\sim$ 0.4~eV for VB$_2$ and $\sim$ 0.8~eV for VB$_3$. Note that a recent ARPES study found a similar magnetic splitting of VB$_2$ and a reconstruction of VB$_1$ by the emergence of an ``M''-shaped band due to the bulk-band inversion in the AFM state~\cite{Sato2020PRR}. The latter effect is likely weaker and thus not identified in the difference spectrum of the interband $\sigma_1(\omega)$ across $T_N$. The small anomaly of the free carrier plasma frequency around $T_N$, on the other hand, may be an indication of this band inversion.
\subsection{Drude response and spin fluctuations}
\begin{figure}[tb]
\includegraphics[width=\columnwidth]{Fig3}
\caption{ (color online) (a) Drude fit (red line) to the low-frequency optical conductivity of EuIn$_2$As$_{2}$ at 10 K (blue line). Note that the phonon mode at 185\icm\ is not included in this Drude fit. (b) Comparison of the temperature dependence of the dc resistivity, $\rho_{ab}$ (solid line), with the one of the zero-frequency value obtained from the extrapolation of the Drude fit to the optical conductivity, $1/\sigma_1(\omega = 0)$ (open circles). (c) Temperature dependence of the free carrier plasma frequency $\omega_p$ and the scattering rate $1/\tau$, as obtained from the Drude fit.}
\label{Fig3}
\end{figure}
Next we discuss the quantitative analysis of the temperature dependence of the free carrier response that has been obtained from a Drude fit to the low-frequency ${\sigma}_1(\omega)$ spectra. Fig.~\ref{Fig3}(a) shows as an example of the Drude fit to the data at 10~K from which we derive a (bare) plasma frequency of $\omega_p =$ 2056\icm\ and a scattering rate of $1/\tau =$ 50\icm. Corresponding Drude fits of the $\sigma_1(\omega)$ curves have been performed at all measured temperatures, as shown by the dashed lines in Fig.~\ref{Fig2}(a). Fig.~\ref{Fig3}(b) compares the temperature dependence of the dc resistivity $\rho \equiv 1/\sigma_1(\omega = 0)$ from electric transport measurements (black solid line), with the one of the inverse conductivity at zero frequency obtained from the Drude fit (open red circles). They both agree reasonably well concerning the temperature dependence, with a characteristic cusp-like maximum around $T_N$, and even the absolute values. This agreement confirms that the modelling of the optical data is meaningful and reliable. The corresponding plasma frequency $\omega_p$ (orange symbols) and the scattering rate $1/\tau$ (green symbols) are displayed in Fig.~\ref{Fig3}(c). Similar to the screened plasma frequency in Fig.~\ref{Fig1}(c), the bare plasma frequency shows a small increase toward low temperature in the paramagnetic state and only a weak anomaly around $T_N$. The electronic scattering rate, on the other hand, exhibits a pronounced, cusp-like maximum around $T_N$ that is similar to the one of the resistivity $\rho$ in Fig.~\ref{Fig3}(b). In return, this suggests that the cusp-like maximum of the resistivity $\rho = m^{\ast}/e^2 \tau n$ around $T_N$ arises from a corresponding increase of the free carrier scattering rate that is caused by the critical fluctuations of the Eu spins in the vicinity of $T_N$. Our optical data thus provide evidence for a rather strong interaction of the charge carriers with the (slow) spin fluctuations of the Eu moments in EuIn$_2$As$_2$.
To study the nature of the dominant scattering mechanism, we recall the Suezaki-Mori model that was proposed for AFM or order-disorder systems~\cite{Suezaki1968,Suezaki1969,Thomas1973}, where the scattering rate is given by
\begin{equation}
\label{scatteringrate_model}
\frac{1}{\tau(T)}= A + BT +C(1-D|\epsilon|^{2\beta}).
\end{equation}
The first and second terms represent the contributions of impurities or vacancies and of the phonons. The third term $1/\tau_C \propto 1-D|\epsilon|^{2\beta}$ accounts for the critical contribution, which is written in terms of the reduced temperature $\epsilon = (T - T_0)/T_0$ and the critical exponent $2\beta$. Using scaling estimates for an Ising-type model, the exponent should be $2\beta = 0.625$~\cite{KADANOFF1967,Thomas1973}. With this model, we can well reproduce the temperature dependence of scattering rate [red solid line in Fig.~\ref{Fig3}(c)], in particular, the upward-pointing cusp in the scattering rate that is due to the critical fluctuations. The fitted value of the critical exponent of $2\beta = 0.66(4)$ also agree with the prediction. This confirms that the scattering of the free carriers is strongly enhanced by the critical fluctuations of the Eu spins in the vicinity of $T_N$.
\subsection{Spin-lattice coupling}
\begin{figure}[tb]
\includegraphics[width=\columnwidth]{Fig4}
\caption{ (color online) (a) Line shape of the strong infrared-active phonon mode at temperature from 300 to 8 K in EuIn$_2$As$_2$. The black solid lines through the data denote the Fano fits. (b)--(e) Temperature dependence of the resonance frequency $\omega_0$, the linewidth $\gamma$, the oscillator strength $\Omega^2$, and the Fano parameter $1/q$ of the phonon. The vertical dashed line denotes the antiferromagnetic transition temperature $T_N$ at which all the parameters show anomalies.}
\label{Fig4}
\end{figure}
Finally, we turn to the infrared-active phonon mode at 185\icm. Its temperature dependence is detailed in Fig.~\ref{Fig4}(a) which shows the phonon line shapes after the electronic background (determined with a Drude fit) has been subtracted. With decreasing temperature, this mode shifts to higher frequency and becomes sharper. Meanwhile, the line shape becomes more strongly asymmetric. Such an asymmetric line shape is a characteristic signature of the coupling of the phonon to a much broader background of electronic and/or spin excitations.
The asymmetric phonon mode has been fitted with a Fano-type line shape~\cite{Fano1961} according to the following expression
\begin{equation}
\label{Fano}
\sigma_{1}(\omega)=\frac{2\pi}{Z_0} \frac{\Omega^2}{\gamma}
\frac{q^2 +\frac{4q (\omega - \omega_0)}{\gamma} -1}{q^2 (1 + \frac{4(\omega - \omega_0)^2}{\gamma^2})},
\end{equation}
where $\omega_{0}$, $\gamma$ and $\Omega$ are the resonance frequency, line width, and the strength of the phonon, respectively. The dimensionless parameter $q$ describes the asymmetry of the Fano profile. The parameter $1/q^2$ is a measure of the strength of the coupling between phonon and electron or spin. At $1/q^2 = 0$ the line shape is still symmetric and Lorentzian, whereas with increasing $1/q^2$ the line shape becomes more asymmetric. The best fits using this Fano model describe the phonon mode reasonably well, as shown by the black solid lines in Fig.~\ref{Fig4}(a). The temperature dependence of the fit parameters for $\omega_{0}$, $\gamma$, $\Omega^2$ and $1/q$ are summarized in Figs.~\ref{Fig4}(b) to \ref{Fig4}(e), respectively. They reveal some weak, but clearly resolved anomalies in the vicinity of $T_N \simeq$ 18~K which suggest that the spin-phonon coupling is also not negligible.
\section{Conclusions}
To summarize, the optical conductivity of the antiferromagnetic axion topological insulator candidate EuIn$_2$As$_2$ has been measured over a wide frequency range and at a variety of temperatures. In the far-infrared range, we observed a weak Drude response and a sharp infrared-active phonon mode. Towards higher frequency, there is a series of interband transitions that starts with a weak absorption edge around 800\icm\ and is followed by weak bands at 5\,000 and 12\,000\icm\ and strong bands at 20\,000, 27\,500 and 32\,000\icm. Based on reported band structure calculations, we assign the weak bands to transitions from the low-lying valence bands (VB$_2$ and VB$_3$) to an empty conduction band (CB$_1$). The gap magnitude is estimated to be 0.1~eV. Below $T_N \simeq 18$~K, we observed clear signs of a magnetic splitting of the valence bands VB$_2$ and VB$_3$ which amounts to about 0.4 and 0.8~eV, respectively. The free carrier response is also strongly affected by the antiferromagnetic transition. In particular, the scattering rate (width of the Drude peak) shows a pronounced, cusp-like maximum at $T_N$ that arises from critical antiferromagnetic fluctuations of the Eu spins that are strongly interacting with the charge carriers. An anomalous $T$-dependence in the vicinity of $T_N \simeq 18$~K is also seen for the infrared-active phonon mode around 185\icm\ which suggests that the spin-phonon coupling is also sizeable. Our study highlights that EuIn$_2$As$_2$ is of interest not only for its magnetic and topological properties, but also for the rather strong interaction amongst its charge, spin and lattice degrees of freedom. The latter may also affect the various topological surface states and need to be considered in comparing theory with experiments.
\begin{acknowledgments}
Work at the University of Fribourg was supported by the Schweizerische Nationalfonds (SNF) by Grant No. 200020-172611. B.S. acknowledges the support of the Fundamental Research Funds for the Central Universities, Grant No. 19lgpy260.
\end{acknowledgments}
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} | 9,822 |
Terrorist attacks are a constant threat for the events sector
A new wave of terror attacks is changing the events security landscape as fast-moving, co-ordinated threats become an increasing risk.
Magda Ibrahim
The devastating Paris attacks, California shootings and constant terror threats have placed security top of the agenda for event organisers.
But according to a 2016 EventHuddle poll, 93 per cent do not feel adequately trained for an incident.
"Marauding terrorist firearms threats are a new dimension," says Steve Cooper, former head of security for the Olympic Delivery Authority during the London 2012 games.
People laying flowers outside the Bataclan
concert hall in December 2015, a month after the Paris terror attacks
The almost simultaneous November 13 Paris attacks by gunmen and suicide bombers on the Bataclan concert hall, Stade de France, restaurants and bars left 130 people dead and hundreds injured.
Just weeks later, 14 people were killed and 22 seriously injured in a shooting at a conference centre in San Bernardino, California, while Brussels cancelled New Year celebrations after an apparent terror plot was uncovered.
With the Home Office assessing the UK's current threat from international terrorism as "severe", meaning an attack is thought highly likely, event organisers are naturally aware of their vulnerability.
Defence planning is key
Effective planning is the first line of defence, according to Mr Cooper and his colleague Chris Scott, who worked as head of emergency preparedness, testing and exercising for London 2012.
"Security must be properly integrated, but proportionality is key otherwise you will burn budgets and not be a lot more effective," says Mr Cooper.
Indeed, 48 per cent of events professionals expect costs to rise because of the need for greater security, according to MPI Meetings Outlook.
Planning and risk assessment for a large-scale event includes looking at the general political climate, previous threats and hoaxes, as well as collaboration with police, other emergency services and public authorities
"What the sad events of Paris have done is made people dust off their security plans," adds Mr Scott. "In a Paris scenario, there's no time, so rigorous testing means you can react quickly and there's a good chance of making the best decisions."
Planning and risk assessment for a large-scale event includes looking at the general political climate, previous threats and hoaxes, as well as collaboration with police, other emergency services and public authorities.
In-depth site visits are a chance to test venues and evacuation routes, whether there is a distinct alarm bell for a terror threat, as well as checking out whether mobile devices will work properly.
Even for smaller events, best-practice elements, such as a laminated contact list of useful names and numbers, can be simple but save valuable seconds, while close co-operation between the event owner, organiser and venue, as well as other suppliers, is critical.
Ian Cummings, a regional director at global agency CWT Meetings & Events, points out that while security has "shot to the top of the priority list for our clients, it's not a one size fits all". When it's an overseas event, having a team based in the destination providing insider information is valuable for extra insight, he adds.
Analysing the repurcussions
Clear communication channels proved crucial for Elliott Grant, business director of the Black Tomato Agency, when his team of nine were leading an 84-strong corporate incentive trip in Paris as the November attacks unfolded.
Having a dedicated point of contact as the team worked through its contingency plan meant instructions could be relayed speedily and accurately as attendees were taken to safety.
Grant explains that as he was in the UK, he could debunk any "misinformation they were hearing on Facebook and Twitter", and set up a text alert with verified news, while a crisis management team reorganised travel arrangements.
Despite the terrifying attacks, and those that took place at the Charlie Hebdo magazine offices in January 2015, it's pretty much business as usual for Paris and the rest of Europe.
Nearly half of corporate travel buyers report no change to their company's policy while, according to a Global Business Travel Association poll, 1 per cent temporarily suspended European travel in the days after the November incidents.
Meanwhile, MKG hospitality industry data shows the hotel sector wasn't hit as badly as might have been expected, and the Euro 2016 football tournament being held in France in June and large exhibitions should buoy hotel demand this year.
Events agency DRP's group chief executive Dale Parmenter says: "Paris was dreadful, but I don't think it will have the massive negative effect that occurred after 9/11."
What Paris did show is that unpredictability is the new normal.
Event venues across the UK responded accordingly, with companies such as Live Nation, AEG, which operates Wembley Stadium where England played football against France days after the Paris attacks, and major convention centre ExCeL London ramping up security.
At the Scottish Exhibition and Conference Centre in Glasgow, director of live entertainment John Langford explains: "Security measures on our campus are reviewed constantly and scaled to each event on a case-by-case basis."
Being heavy handed with security can also be a negative in events, where look and feel is crucial. "Consider what attendees will find appropriate," says Mr Scott.
Looking out for cyber criminals
The technology age has also changed the threat with cyber security a growing concern. Cyber security expert Jessica Barker says the major worry for events is that the industry lags behind sectors like finance, pharmaceutical or retail.
"Criminals often go for the low-hanging fruit, those least prepared," she warns. Dangers could include a cyber criminal accessing attendees' personal or financial details, or even hacking into venue or hotel systems to override electronic gate or door-locking.
"Anything that runs on the internet can be hacked and potentially used as an attack that can impact on physical safety," says Dr Barker. "That can extend as far as transport disruption and emergency services being able to respond to a physical attack."
However, technology can mean a greater ability to screen attendees and on-site staff as CCTV, biometrics and radio-frequency identification embedded in tickets allow risks to be spotted earlier.
Ultimately, "even in the most effectively policed countries, terrorist incidents cannot be completely mitigated", says David Burrill, director of security agency Burrill Green and former chief of staff for the intelligence and security centre of UK Armed Forces.
"Be aware of your vulnerabilities, be intelligence-led and be driven by a robust risk assessment," Mr Cooper concludes. "And remember, you're running an event, not a security event."
Should firms be cutting sick pay for unvaccinated employees?
The people vs AI: can a machine own intellectual property?
AI inventors: can AI own intellectual property rights? | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 5,763 |
module.exports = function(){
var routes = {
root : '/'
,connectionTest : '/testConnection'
,authentication : '/authentication'
,classes : '/classes/professor/:id'
,students : '/students/:idlist'
,presenceControl : '/presenceControl'
,studentAuthentication : '/studentAuthentication'
};
return routes;
}; | {
"redpajama_set_name": "RedPajamaGithub"
} | 5,487 |
Scrollside Ltd (t/a Zed FM) v Broadcasting Commission of Ireland
Judgment Irish Reports Cited authorities 9 Cited in 3 Precedent Map Related
Court Supreme Court
Judge Denham J.,Mr. Justice Kearns
Neutral Citation [2006] IESC 24
Docket Number [SC No.
[2006] IESC 24
Denham J.
McCracken J.
Kearns J.
[S.C. No: 405 of 2005]
SCROLLSIDE LTD v BROADCASTING COMMISSION OF IRELAND
Scrollside Limited Trading as "Zed FM"
Applicant/Appellant
Broadcasting Commission of Ireland
ADIO & TELEVISION ACT 1988 S6
RADIO & TELEVISION ACT 1988 S6(2)(a)
RADIO & TELEVISION ACT 1988 S6(2)(b)
SPIN COMMUNICATIONS LTD T/A STORM FM v INDEPENDENT RADIO & TELEVISION COMMISSION (IRTC) & MAYPRIL LTD 2001 4 IR 411 2002 1 ILRM 98
RADIO & TELEVISION ACT 1988 S6(1)
BROADCASTING ACT 2001 S60
OXFORD ENGLISH DICTIONARY "CHARACTER"
PLATO FILMS LTD v SPEIDEL 1961 AC 1090
WADE & FORSYTH ADMINISTRATIVE LAW 9ED 380
WHITE v DUBLIN CITY COUNCIL & ORS 2004 1 IR 545 2004 2 ILRM 509
SECRETARY OF STATE FOR EMPLOYMENT & SCIENCE v TAMESIDE BOROUGH COUNCIL 1977 AC 1014
ward of radio licence - Review of specialist decision maker - Curial deference - Licence awarded to former pirate broadcaster - Character of successful applicant - Whether unreasonable to consider experience obtained while broadcasting illegally - Policy to encourage cessation of illegal broadcasting - Spin Communications Ltd v IRTC [2001] 4 IR 411 and White v Dublin City Council [2004] IESC 35 [2004] 1 IR 545 followed; Secretary of State for Education and Science v Tameside MBC [1977] AC 1014 approved - Radio and Television Act 1988 (No 20), ss 6(2)(a) - Broadcasting Act 2001 (No 4), s 60 - Applicant's appeal dismissed (405/2006 - SC - 6/4/2006) [2006] IESC 24
Judgment delivered the 6th day of April 2006 by Denham J.
1. In this appeal Scrollside Limited, trading as "Zed FM", the applicant/appellant, hereinafter referred to as "the applicant", seeks an order determining that the High Court erred in refusing to quash the decision of the Broadcasting Commission of Ireland, the respondent, hereinafter referred to as "the respondent", granting a sound broadcasting contract for the operation of an alternative rock music broadcasting service, on the FM band in Dublin, to Dublin Rock Radio Limited trading as "Phantom Rock", and referred to hereinafter as "Dublin Rock".
2. The applicant submits that the decision of the respondent is flawed on a number of grounds, especially in relation to the illegal broadcasting operations carried out by various members of the Dublin Rock consortium under the name "Phantom Rock". The applicant submits that the respondent failed to give proper consideration to the character of the members of the Dublin Rock Consortium as required by s.6 of the Radio and Television Act, 1988. The applicant submits that the decision was unreasonable in that the respondent failed to give proper consideration to the character of Dublin Rock, in that it erred in conferring a benefit on Dublin Rock as a result of the illegal broadcasting. Further, that the respondent erred in conferring a benefit on Dublin Rock as a result of illegal broadcasting by a former pirate radio station "Phantom FM", that it prejudged the issue of the award of the sound broadcasting contract in favour of the Dublin Rock Consortium, and that it erred in having regard to the previous provision and operation by two distinct entities, Wireless Media Limited and Coxstone Limited, of services under temporary sound broadcasting contracts under the name "Phantom FM". It is obvious that these issues are all interrelated and that in essence they refer to previous pirate operations.
3. The respondent, on the otherhand, submits that the applicant has entirely failed to show that the decision of the respondent is null and void or vitiated by bias, as alleged, or at all.
4. On the 1st November, 2005, the High Court (O'Sullivan J.) gave a reserved judgment in this matter and ordered that the applicant's application be refused.
The applicant had sought:
(i) an order of certiorari by way of an application for judicial review to quash the respondent's decision of the 8th November, 2004, to award a sound broadcasting contract for the operation of an alternative rock music sound broadcasting service on the FM band in Dublin City and Country to Dublin Rock Radio Limited trading as "Phantom FM"; and,
(ii) A declaration that the said decision was ultra vires and void.
The decision of the respondent was challenged by the applicant on four grounds, namely:
(a) that the respondent failed to consider the "character" of Dublin Rock as required to do so by statute;
(b) that the respondent erred in law in conferring a benefit on Dublin Rock as a result of the illegal broadcasting by a number of individuals who formed part of the consortium comprising Dublin Rock;
(c) that the respondent prejudged the issue of the award of the contract and at all times intended to grant the contract to any applicant including "Phantom FM"; and,
(d) that the respondent had regard to broadcasting services under the name of "Phantom FM" provided by two distinct entities.
The High Court found against the applicant on each of the grounds advanced.
2 5.1 The learned trial judge pointed out that underlying the four grounds, advanced on behalf of the applicant, was the applicant's disquiet at the fact that 25% (or more) of the membership comprising Dublin Rock had been engaged in pirate broadcasting for a number of years up to eighteen months prior to the award of the contract and that the respondents had given credit for the experience and expertise gained during the illegal broadcasting, and erroneously took into account their track record under two separate temporary broadcasting contracts.
3 5.2 The respondent was established under the Radio and Television Act, 1988 to, amongst other activities, enter into contracts with providers of sound broadcasting services. The procedure is that the respondent first invites expressions of interest in the proposed contract, and in this case it received five, which were summarised by the executive of the respondent, and considered by it at a meeting where two of the bidders were short listed. Then these two were invited to an oral hearing at which each of them made a twenty minute submission followed by a twenty five minute question and answer session. After that questionnaires were sent out and the answers and all the material were considered at a final meeting where the decision was made. Subsequently the unsuccessful bidders were given feed-back reports explaining the decision.
6. High Court judgment
These judicial review proceedings were grounded on affidavits, principally the affidavit of Dermot O'Hanrahan for the applicant and Michael O'Keeffe on behalf of the respondent. Michael O'Keeffe was cross-examined at length.
2 6.1 The High Court held:
"The applicant asserts that the respondent had made its mind up effectively from the beginning to award a license to Dublin Rock because they were associated with Phantom FM. In this context it is instructive to consider, in part, the minutes of the initial meeting of the respondent on 6th September, 2004, when they were considering the five applicants for the contract. In relation to Zed FM (the applicant) this, in part, is what the minutes record:-"
"In general terms, the contents of the executive evaluation were endorsed. Zed FM was considered to have a credible and relevant mix in the ownership and control structures of the company. Specifically, the mix included well established and leading rock and radio industry figures in the Dublin market. In addition to contributing funds, the shareholders could also significantly contribute to the programme content.
The aim of the station to target a community of interest of alternative rock listeners, beyond the narrow focus of other applicant groups also appeals to a number of the members, including the proposal to target women. While some of the members welcomed the wider target group identified for this service, others said it was unclear as to how this might be achieved. This would require further clarification with the group.
The station's proposed low cost base was considered prudent, in view of the niche nature of the audience to be served and the need to establish a revenue base for this type of service. However, some members were concerned that this might impact on the quality of the service proposed and were of the view that this would require some further clarification with the group in the context of an oral hearing.
The applicant group's research proposals were considered to be well presented with an adequate sample size. In general terms, the applicant group's programming proposals were considered to be satisfactory and reflected the knowledge and commitment of members of the applicant group to serving the target audience. Some questions were raised regarding the staffing levels for the service and whether or not the service could be successfully operated on the levels proposed. This would require further consideration…the proposed management team for the service was considered to be appropriate and very experienced…in summary, this application was considered to be of a high standard and the programming ethos of the proposed service was considered to be relevant and appropriate to the Dublin market and the target audience in particular."
The applicant, as already stated, was short listed. At the final meeting on 8th November, 2004, the following appears in the minutes:-
"Zed FM's ownership and control proposals were considered in general terms to be satisfactory by the Commission. The company was considered to have a good...
Revenue Commissioners v Wen-Plast (Research and Developmenr) Ltd
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Ryanair Ltd v Commission for Aviation Regulaton
...1998 SI 505/1998 REG 14(2) EC DIR 96/67/EC ART 16(3) SCROLLSIDE LTD T/A ZED FM v BROADCASTING COMMISSION OF IRELAND UNREP SUPREME 6.4.2006 2006 IESC 24 COMMISSION v ITALY C-460/02 2004 ECR I-11547 Abstract: Judicial Review - Certiorari - Aer Rianta - Groundhandling charges - Onus of proof -...... | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 7,578 |
To my husband, Cenneth Niklasson—beloved best friend
## Contents
Sunday, November 11
Monday, November 12
Sunday, November 18
Monday, November 19
Tuesday, November 20
Wednesday, November 21
Thursday, November 22
Friday, November 23
Saturday, November 24
Sunday, November 25
Monday, November 26
Tuesday, November 27
Wednesday, November 28
Thursday, November 29
Saturday, December 1
Sunday, December 2
Monday, December 3
Tuesday, December 4
Wednesday, December 5
Thursday, December 13
Friday, December 14
Monday, December 17
Thursday, December 20
Friday, December 21
Saturday, December 22
Sunday, December 23
Wednesday, December 26
Epilogue
Author's Acknowledgments
## Sunday, November 11
For the first time in a week the sky cleared. The wan rays of November sunshine found their way through the clouds, and the spectators at the Visby trotting track turned their faces with yearning up toward the sun. It was the last race of the season, and there was a sense of anticipation in the air, mixed with a touch of melancholy. A chilly but enthusiastic crowd had gathered in the grandstands. They were drinking beer and hot coffee from plastic cups, eating hot dogs, and making notes in their track programs.
Henry "Flash" Dahlström got out his hip flask and took a good swig of his home-brewed liquor. It made him grimace, but it also warmed him nicely. With him in the stands sat the whole gang: Bengan, Gunsan, Monica, and Kjelle. All of them were rapidly advancing toward various states of intoxication.
The procession had just started. The snorting standardbreds, glossy with sweat, were lined up and prancing forward as the music blared from the loudspeakers. The drivers, with their legs wide apart, were firmly seated in their lightweight sulkies.
The odds were posted on a black tote board out near the track, with the numbers ticking past.
Henry leafed through the racing program. He ought to place a bet on Ginger Star, running in race number seven. No one else seemed to believe in her. She was only a three-year-old. He had followed the horse during the summer races, and even though she had a tendency to break into a gallop, she kept on getting better.
"Hey, Flash, take a look at Pita Queen. She's a beauty, don't you think?" Bengan slurred his words as he reached for the hip flask.
Henry had been given the nickname Flash because he had worked as a photographer for Gotlands Tidningar for many years before alcohol took over his life full-time.
"You're damn right. With that trainer..." he replied and then stood up to take his racing card to the window.
There was a line of betting windows, all with open wooden hatches. Wallets were eagerly pulled out, banknotes changed hands, and cards were handed in. One flight up was the track restaurant, where invited guests ate steak and drank strongbeer. Honored big-time players puffed on cigars, discussing the current condition of the horses and the technique of the drivers.
The race was about to begin. The first driver politely saluted the judges by giving a brief nod toward the judging tower. Over the loudspeakers the announcer called for the horses to take their places.
After four races Henry had an equal number of wins on his card. If luck was with him, he could win the jackpot with five in a row. Since he had also bet on the long shot Ginger Star in the last race, the winnings ought to be significant. If only she came up to his expectations.
The race began and he followed the sulkies on the track as closely as he could after consuming eight strongbeers and a countless number of shots. When the bell for the final lap rang, his pulse quickened. Ginger Star was running well, damned well, as a matter of fact. With each stride she closed in on the two favorites in the lead, and he seemed to be seeing her more clearly. The powerful neck, the snorting nostrils, and the ears pointing straight forward. She could do it.
Don't start galloping now, do not gallop. He was muttering this plea to himself like a mantra. His eyes were fixed on the young filly, who with furious energy was closing in on the leaders. Now she passed one of her rivals. Suddenly he became aware of the weight of the camera around his neck, and he was reminded that he had planned to take pictures. He snapped several photos, his hands relatively steady.
The red sand of the trotting track spurted up around the hooves that were pounding forward at breakneck speed. The drivers were using their whips on the horses, and the excitement rose among the spectators. Many in the stands were on their feet, some of them clapping, others shouting.
Ginger Star pulled forward on the outside and was now even with the horse in the lead. Then her driver used his whip for the first time. Dahlström stood up as he followed the horse through the lens of his camera.
When Ginger Star crossed the finish line ahead of the big favorite by a nose, a sigh of disappointment passed through the crowd. Dahlström was aware of scattered comments: "What the hell?" "It can't be true!" "Unbelievable!" "Damn it!"
But he dropped down onto the bench.
He had won all five races in a row.
The only audible sound was the sweep of the broom across the stable floor and the grinding jaws of the horses as they chewed their evening oats. Calm had settled in after the hectic race day. Fanny Jansson was sweeping with brisk, rhythmic strokes. Her body ached after all the hard work, and when she was done, she sank down onto a feed box outside Regina's stall. The horse peered out, and Fanny stuck her hand through the bars to stroke the horse's nose.
The slender, dark-skinned girl was alone in the stable. She had declined an invitation to join the others at a local restaurant to celebrate the end of the season. She could just imagine how rowdy it was bound to get. Worse than usual. She had been there several times before but didn't enjoy it. The horse owners would drink too much and try to hit on her. They called her "princess," pulled her onto their laps, and pinched her on the rear.
Some got bolder the more they drank. They would make comments about her body, both verbally and with their eyes. They were a pack of dirty old men.
She yawned, but she had no desire to bike home, either. Not really. Her mother had the day off from her job, and there was a good chance that she was drunk. If she was alone she would be sitting on the sofa with her mouth turned down in a sullen frown, with a bottle of wine in front of her. As usual, Fanny would feel guilty because she hadn't spent the day with her mother instead of with the horses. Her mother couldn't care less that it was a race day with tons of work to do. Nor did she understand that Fanny needed to get away from home. The stable was her lifeline. If she didn't have the horses, she didn't know what she would do.
Uneasiness seized her as she imagined an even worse scenario: that her mother might not be alone. If her so-called boyfriend, Jack, was there, they would get even drunker, and Fanny would have a hard time sleeping.
Tomorrow she had to be at school early, and she needed to get some sleep. Ninth grade was a torment that she wanted to get through as fast as possible. Fanny had tried to do her best when the term started, but things just kept getting worse. She was having a hard time concentrating, and she had started cutting classes fairly often. She just couldn't face it.
She had enough troubles outside of school.
## Monday, November 12
A bubble of saliva had formed at the corner of his mouth. With each exhalation it grew bigger until it burst and dribbled down his chin and onto the pillowcase.
The room was very bright. The blinds were rolled up and the dirty streaks on the windowpane were clearly visible. On the windowsill stood a solitary pot with an African violet that had long since perished.
Henry Dahlström slowly regained consciousness as the urgent ringing of the phone cut through the thick silence. The sound echoed off the walls in the shabby two-room apartment, persisting until it finally won out over sleep. Disconnected thoughts popped up in his mind, relentlessly bringing him back to reality. He had a vague feeling of happiness but couldn't remember what it was from.
The headache started the minute he swung his legs over the side of the bed. Cautiously he sat up. His vision blurred the pattern of the bedspread. Thirst made him get to his feet and stagger out to the kitchen. The floor swayed beneath him. He leaned against the door frame and looked at the chaos.
The kitchen cupboards stood wide open and the counter was covered with dirty glasses and plates, as well as scraps of food. There was burned coffee in the glass pot of the coffeemaker. Someone had dropped a plate on the floor. He could make out the remnants of fried herring and mashed potatoes among the pieces of china. On the kitchen table, beer cans were crowded together with liquor bottles, an overflowing ashtray, and a stack of racing cards.
Suddenly he remembered why he should be happy. He had brought home the jackpot as the sole winner of all five races. The sum was breathtaking, at least for him. Over eighty thousand kronor had been paid out to him, in cash, and gone straight into his pockets. He had never before had so much money in his possession.
His eyes flicked anxiously up and down over the half-empty cupboards. Surely he'd had enough sense to hide the money. If only none of the others...No, he refused to believe that. Although when it came to liquor or money, you never could tell.
He pushed aside the thought and tried to recall what he had done when he arrived home from the track the previous evening. Where the hell?
Oh, that's right. The broom closet. With trembling fingers he pulled out the package of vacuum cleaner bags. When he touched the bundle of banknotes, he breathed a sigh of relief. He sank down onto the floor, cradling the package in his hands as if it were a valuable porcelain vase. At the same time, thoughts about what he was going to do with the money flickered past. Fly to the Canary Islands and order drinks with little umbrellas. Maybe invite Monica or Bengan to come, too—or why not both of them?
An image of his daughter appeared. He really ought to send some of the money to her. She was grown up now and lived in Malmö. Contact between the two of them had been broken off long ago.
Henry stuffed the package back in the closet and stood up. Thousands of stars danced before his eyes.
The need for a drink became more urgent. The beer cans were empty, as were the liquor bottles. He lit one of the longer cigarette butts from the ashtray, swearing as he burned his finger.
Then he discovered a bottle of vodka under the table, and it turned out to have a decent slug left in the bottom. He greedily gulped it down, and the merry-go-round in his head eased up a bit. He went out to the patio and breathed in the cold, raw November air.
On the lawn lay an unopened can of strongbeer, of all things. He picked it up and definitely started feeling better. In the fridge he found a piece of sausage and a saucepan of dried mashed potatoes.
It was Monday evening. It was past six o'clock, and the state liquor store was closed. He had to go out and find some booze.
Henry took the bus downtown. The driver was nice enough to let him ride free, even though he could now afford to pay the fare. By the time he got out at Östercentrum, he was the only passenger. Rain was in the air, and it was dark and desolate on the streets. Most of the stores were closed at this time of night.
On one of the benches near the Allis hot dog stand sat Bengan with that new guy Örjan from the mainland. An unpleasant type, pale with dark, slicked-back hair and a sharp look in his eye. The muscles of his arms testified to how he had spent his time in the slammer, from which he had recently been released. He had apparently been sent up for aggravated assault and battery. Tattoos covered his arms and chest; part of one was visible inside the dirty collar of his shirt. Henry felt anything but comfortable with him, and things were made only worse by the fact that he always had that growling attack dog in tow. The animal was white with red eyes and a square snout. Ugly as sin. The guy bragged that his dog had bitten a toy poodle to death in Östermalm in the middle of downtown Stockholm. The fucking upper-class dame who owned the poodle went nuts and starting hitting Örjan with her umbrella until the police showed up and took charge. He had gotten off with a warning to buy a stronger leash. The incident was even reported on TV.
As Henry approached, a muted rumble issued from the dog's throat; the animal was lying at Örjan's feet. Bengan greeted him with a wobbly wave of his hand. It was apparent from far away that Henry's friend was quite inebriated.
"Hi, how are things? Congratulations again. It's so fucking great." Bengan gave his friend a befuddled look.
"Thanks."
Örjan pulled out a plastic bottle containing a colorless, unidentified liquid.
"Want some?"
"Sure."
The liquor had a pungent smell. After several sizable gulps, Henry's hands stopped shaking.
"That went down nice, didn't it?" Örjan asked the question without smiling.
"Absolutely," said Henry, and he sat down on the bench next to the other two men.
"How's it going for you?"
"Well, I've got my head up and my feet down."
Bengan leaned closer to Henry and breathed loudly in his ear.
"Shit, what about all that dough?" he muttered. "It's amazing. What are you thinking of doing with it?"
Henry cast a quick glance over at Örjan, who had lit a cigarette. He was staring out toward Östergravar and seemed to have stopped listening.
"We'll talk about it later," whispered Henry. "I want you to keep your mouth shut about the money. Don't tell anyone else about it. Okay?"
"Sure, no problem," promised Bengan. "Of course, buddy." He patted Henry on the shoulder and turned back to Örjan. "Give me a swig." He grabbed the bottle.
"Take it easy, damn it. Pianissimo."
Typical Örjan, thought Henry. He always has to sound so odd. Pianissimo—what the hell is that? The dog bared his teeth.
All Henry wanted right now was to buy some booze and get out of there.
"Have you got anything to sell?"
Örjan dug through a worn bag made of imitation leather. He pulled out a plastic bottle containing home-brewed liquor.
"Fifty kronor. But maybe you can afford to cough up more than that?"
"Naw. I've only got a fifty."
Henry handed over the banknote and reached for the bottle. Örjan kept his grip on it.
"Are you sure?"
"Yup."
"What if I don't believe you? What if I think that you've got more and you just don't want to pay more than that?"
"What the hell—let go!"
He yanked the bottle away from Örjan. At the same time he stood up. Örjan laughed and jeered, "Can't you take a joke?"
"I've got to go. See you. I'll be in touch."
He headed for the bus stop without looking back. He could feel Örjan's eyes fixed on his back like needles.
He was sitting in the living room, comfortably leaning back in the only armchair. On his way home he had passed a kiosk that was open at night, and he had bought some Grape Tonic, which he mixed with the booze to make himself a nice, tasty highball. He studied the glow from his cigarette in the dim light of the room, enjoying his solitude.
It didn't bother him that the apartment was still a mess from the party the night before.
He put an old Johnny Cash record on the stereo. The neighbor woman protested by pounding on the wall, presumably because the music was interfering with the Swedish soap opera on TV. He pretended not to notice because he despised everything that had to do with normal Swedish life.
During his professional days he had also avoided routines. As the foremost photographer at Gotlands Tidningar he'd had plenty of opportunities to plan his own work hours. When he eventually started his own business, of course, he did precisely as he pleased.
In moments of clarity he surmised that it was this freedom that had spelled the beginning of the end. It created space for his drinking, which slowly but surely nibbled away at his work, his family life, his free time, and finally took precedence over everything else. His marriage fell apart, his clients disappeared, and contact with his daughter became increasingly sporadic and then ceased altogether after a few years. In the end he had neither money nor a job. The only friends who remained were his drinking buddies.
He was roused from his reflections by a clattering sound on the patio. He stopped with the glass halfway to his lips.
Was it one of those damn kids in the area who was going around stealing bicycles and then painting and selling them? His own bicycle stood outside unlocked. They had tried to swipe it before.
Another clatter. He looked at his watch. Ten forty-five. Someone was out there—there was no doubt about that.
Might be an animal, of course, maybe a cat.
He opened the patio door and peered into the darkness. The little patch of grass that belonged to his corner property was lit up in the cold glow of the streetlight. Over by the pathway a shadow disappeared among the trees. Presumably just somebody out walking his dog. Henry pulled the door shut and locked it, just to be safe.
The interruption annoyed him. He switched on the ceiling light and looked around the apartment with distaste. He couldn't stand seeing all the clutter, so he stuck his feet into a pair of slippers and went down to his darkroom in the basement to check on the pictures he had taken during his evening at the harness races. He had taken a whole roll of Ginger Star, and a couple of shots just as she crossed the finish line. Her head thrust forward, her mane flying, and her nose ahead of all the others. What a feeling.
The building superintendent had been kind enough to let him use an old bicycle storage room. He had furnished it with an enlarger, trays for developer and fixer, and a rack for drying the pictures. The basement window was covered with pieces of black cardboard to keep out the daylight.
The only light source was a red bulb on the wall. In the faint glow of this lamp the work could be done without difficulty. He enjoyed spending time in his darkroom. Focusing one hundred percent of his attention on a task in silence and darkness. He had experienced this same feeling of calm only once before, during his honeymoon to Israel. One day he and Ann-Sofie had gone on a snorkeling expedition. As they moved below the surface of the silent sea, it was like being in another dimension. Undisturbed, untouched by the constant noise of the rest of the world. That was the only time he had gone snorkeling, but the experience had stayed with him as a pleasant memory.
He had been working for quite a while when he was interrupted by a discreet knock on the door. Instinctively he froze and listened carefully. Who could it be? It had to be close to midnight.
The knocking came again, slower and longer. He lifted the photo he was working on from the rinse bath and hung it up to dry as thoughts whirled through his mind.
Should he open the door? Common sense told him that it would be best not to. This might have something to do with his winnings. Someone who wanted the money. The news about his win certainly must have spread by now. The sound coming from the other side of the door signified danger. His mouth went dry. Although it could just as well be Bengan.
"Who is it?" he shouted.
The question hung in the darkness. No reply, utter silence. He sank down onto a stool, fumbled for the liquor bottle, and took several quick swigs. A few minutes passed and nothing happened. He sat totally still and waited, though he didn't know what he was waiting for.
Suddenly someone began pounding hard from a different direction, on the windowpane. He gave such a start that he nearly dropped the bottle on the floor. The last of his drunkenness vanished, and he stared up at the cardboard covering the window, hardly daring to breathe.
Then it came again. Hard, loud. As if the person out there wasn't using his knuckles but some sort of tool. The ceiling and walls closed in. Terror seized him by the throat. Here he sat, trapped like a rat, while someone out there was toying with him. Sweat broke out on his forehead, and his guts turned over. He needed to go to the toilet.
The pounding changed into a rhythmic thudding, a monotonous banging against the basement window. No one in the building would hear his cries for help. Not in the middle of the night on an ordinary weekday. Could the person or persons out there break the window? It would still be impossible to get in because the window was much too small. He had locked the door—he was sure of that.
All of a sudden there was silence. Every muscle in his body was on edge. He listened for sounds that weren't there.
For almost an hour he sat in the same frozen position before he dared to stand up. The hasty movement made him dizzy, and he staggered and saw flashing white stars in the dark. He had to go to the bathroom right now; he couldn't hold it any longer. His legs could barely support him.
When he opened the door he realized instantly that he had made a mistake.
Fanny studied herself in the mirror as she ran a comb through her shiny hair. Her eyes were dark brown, and her complexion was also dark. A Swedish mother and West Indian father. Mulatto, without having a trace of typical African features. Her nose was small and straight and her lips narrow. Raven-black hair that reached all the way to her waist. Some people took her to be Indian or North African, while others guessed that she came from Morocco or Algeria.
She had just stepped out of the shower and put on underpants and a big T-shirt. Freshly scrubbed with the stiff brushes that she bought at Åhléns department store. They tore at her body and made her skin tender. Her mother had asked her what she needed brushes like that for.
"For scrubbing myself. They make you a lot cleaner. And it's good for the skin," she replied. She explained that the smell from the horses clung to her. The shower had become her best friend.
She turned sideways and studied her thin body in profile. Her shoulders drooped. If she straightened her back, her breasts stuck out and seemed even bigger. That's why she always walked slightly bent over. She had developed early. By the seventh grade, she already had breasts. At first she had done everything she could to hide them. Big, baggy shirts helped.
The worst was in gym class. Even though she wore a sports bra that flattened out her breasts, they still were visible when she jumped or ran. The changes in her body made her feel sick. Why did everyone get so disgusting when they grew up? She shaved under her arms as soon as the slightest sign of hair appeared. Not to mention her crotch. And the blood that appeared every month, staining her panties and sheet when she bled through during the night. She despised her body.
The fact that she had dark skin didn't make things any better. She wanted to look like all the others. In her class there were only two others who were dark. They were twins, so at least they had each other. Two boys who had been adopted from Brazil. They were the school's best soccer players, and they were very popular because they looked like Roberto Carlos, the famous Brazilian wingback. For them the color of their skin was an advantage. But not for her. She didn't want to stand out.
She longed to have friends, to have her very own best friend. Someone to confide in, to share her worries. In school no one paid any attention to her anymore. Both there and at home, she was always alone. At the same time she was fully aware that this was her own fault. When she started in the ninth grade, kids would sometimes ask her to join them after doing their homework. She always said no. Not because she didn't want to, but because she had to rush home to walk Spot and take care of everything else that had to be done. Inviting a friend home was out of the question. The risk was too great that they would find a messy apartment reeking of smoke, with the blinds down and breakfast dishes still on the table. A depressed mother with a cigarette drooping from the corner of her mouth and a wineglass in her hand. No thanks, that wasn't something she wanted to put herself through, or a friend, either. It would just make everyone talk. How embarrassing that would be. The last thing she needed was more problems.
That was why Fanny was alone. The other kids got tired of asking her, and finally no one even bothered to talk to her. It was as though she didn't even exist.
## Sunday, November 18
The hail that was ricocheting off the galvanized roof woke Detective Superintendent Anders Knutas at his home, which was a stone's throw outside the ring wall in Visby.
He climbed out of bed and shivered as his feet touched the cold floor. He fumbled wearily for his bathrobe and pulled up the blinds. He peered out in surprise. Hail in November was unusual. The garden looked like something straight out of an old black-and-white Bergman movie. The trees mournfully stretched out their bare branches toward the steel gray sky. The asphalt on the residential street was wet and cold. Off in the distance a woman in a dark blue coat was struggling to cross the street with a baby buggy. Her shoulders were hunched against the wind and the sharp beads of ice that were peppering the ground. Two rumpled-looking sparrows huddled together under the currant bushes, although the sparse branches offered little protection.
Why should I get up at all? he thought as he crawled back under the warm covers. Lina had her back turned to him and seemed to be still asleep. He cuddled up next to her and kissed the back of her neck.
The thought of Sunday breakfast with warm scones and coffee finally convinced them to get out of bed. The local radio station was playing oldies requests, and the cat was sitting in the window trying to catch the drops of water on the other side of the pane. It didn't take long before the children came sauntering into the kitchen, still in their pajamas and nightgown. Nils and Petra were twins and had just turned twelve. They had Lina's freckles and curly red hair but their father's lanky build. They looked alike but were otherwise complete opposites. Petra had inherited her father's calm disposition, and she loved fishing, golf, and spending time outdoors. Nils was hot-tempered with a bellowing laugh and a talent for mimicry. He was also crazy about movies and music, just like Lina.
Knutas checked the thermometer outside the window. Thirty-six degrees. With a certain gloominess he noted that the crimson days of October were now gone. It was his favorite month: the crisp air, the leaves of the trees blazing with color ranging from ocher to purple, and the strong scent of earth and apples. Glittering bright red rowan-berries, and the woods filled with autumn chanterelles. Blue sky. Not too hot and not too cold.
Now October had been replaced with a dirty-gray November, which could hardly please anyone. The sun came up just after seven and went down before four. The days were going to get shorter and darker all the way until Christmas.
No wonder so many people got depressed at this time of year. Anyone who had to be outdoors was in a hurry to go back inside as fast as possible. People hunched their shoulders beneath the wind and rain, not even bothering to glance up at each other. We ought to hibernate, like bears, thought Knutas. This month is just a transitional period and nothing more.
The summer seemed long gone. Back then the island had looked entirely different. Each summer Gotland was invaded by hundreds of thousands of visitors who came to enjoy the unique nature, the sand beaches, and the medieval city of Visby. Of course the island needed tourists, but the visitors also meant a great deal of work for the police. Hordes of teenagers came to Visby to party at the numerous pubs. Problems with alcohol and drugs increased dramatically.
But this past summer all of that had been overshadowed by a serial killer who had ravaged the island, terrifying both tourists and Gotlanders alike. The police had worked under great pressure, and the enormous scrutiny from the media hadn't made their job any easier.
Afterward Knutas was unhappy about the way things had turned out. He brooded over the fact that the police hadn't seen the connection between the victims earlier and prevented the lives of more young women from being sacrificed.
He and his family had taken a five-week vacation, but when he went back to work, he felt anything but rested.
So far the fall had been uneventful, and that was exactly what he needed.
He had been standing outside the door, ringing the bell for almost five minutes. Surely Flash couldn't be such a sound sleeper? Now he kept his finger pressed on the shiny button, but no one responded inside the apartment.
He leaned down with some difficulty and shouted through the mail slot, "Flash! Flash! Open up, damn it!"
With a sigh he leaned against the door and lit a cigarette, even though he knew that the neighbor lady would complain if she happened to come past.
It was almost a week since he and Flash had met at Östercentrum; he hadn't seen him since. That wasn't like him. They should have at least run into each other at the bus station or at the Domus mall, if nowhere else.
He took one last drag on his cigarette and rang the neighbor's bell.
"Who is it?" squeaked a feeble voice.
"I'm pals with Flash...Henry Dahlström next door. I want to ask you a question."
The door opened a crack and an old woman peered at him from behind a thick safety chain.
"What's it about?"
"Have you seen Henry lately?"
"Has something happened?" An inquisitive glint appeared in her eyes.
"No, no, I don't think so. I'm just wondering where he is."
"I haven't heard a sound since all that racket last weekend. There was a terrible uproar. I suppose it was a drunken party, as usual," she snapped, giving him an accusatory look.
"Do you know if anyone else has a key to his apartment?"
"The building superintendents have keys to all the apartments. One of them lives in the building across the way. You can go over there and ask him. His name is Andersson."
When the building superintendent let him into the apartment, they found a chaos of pulled-out drawers, cupboards that had been emptied of their contents, and overturned furniture. Papers, books, clothes, and other junk had been scattered everywhere. In the kitchen the floor was littered with leftover food, cigarette butts, liquor bottles, and other garbage. The room smelled of old beer, cigarette smoke, and fried fish. Someone had also tossed the sofa cushions and bed linens around.
Both men stood in the middle of the living room, their mouths agape. Words came haltingly from the lips of Andersson.
"What the hell happened here?"
He opened the patio door and looked out.
"Nobody out there, either. There's only one other place to look."
They went downstairs to the basement. Along one side of the deserted corridor was a row of doors labeled with various signs: "Laundry Room," "Baby Buggies," "Bicycles." In the middle were the usual basement storerooms with chicken-wire doors. At the far end was an unmarked door.
From the darkroom issued a rotten odor that made their stomachs turn over. The stench just about knocked them to the floor. Andersson switched on the light, and the sight was appalling. On the floor lay Henry Dahlström, drenched in his own blood. He was lying on his stomach, face to the floor. The back of his head was smashed in, with an open wound as big as a fist. Blood had spattered the walls and even the ceiling. His outstretched arms were covered with small, brown blisters. His jeans had dark patches on the seat where he had shit.
Andersson backed out to the corridor.
"Have to call the police," he whimpered. "Do you have a cell phone? I left mine upstairs."
The other man replied only by shaking his head.
"Wait here. Don't let anyone in." The super turned on his heel and ran up the stairs.
When he came back, Flash's buddy was gone.
The gray concrete building made a dreary impression in the November darkness. Anders Knutas and his closest colleague, Detective Inspector Karin Jacobsson, climbed out of their car on Jungmansgatan in the Gråbo district.
An icy wind from the north made them hurry their steps toward Henry Dahlström's front door. A crowd of people had gathered outside the building. Some were talking to the police. The process of knocking on doors had begun, and the building superintendent had been taken in for questioning.
The apartment building seemed shabby. The outside light was broken, and in the stairwell the paint was peeling off the walls.
They greeted a male colleague, who showed them to the darkroom. When he opened the door to the basement, an unbearable stench enveloped them. The stale, nauseating, cadaverous odor told them that the body had already started to decompose. Jacobsson could feel how perilously close she was to vomiting. She had thrown up plenty of times at murder scenes, but she would prefer not to do so now. She pulled out a handkerchief and pressed it over her mouth.
Crime scene tech Erik Sohlman appeared in the doorway to the darkroom.
"Hi. The victim's name is Henry Dahlström. You probably know him—Flash, the old alcoholic who was a photographer? This was his darkroom. He was apparently still using it."
He tilted his head back in the direction of the basement room.
"His head has been bashed in, and it wasn't just a few blows, either. There's blood everywhere. I just wanted to warn you that it's not a pretty sight."
They paused in the doorway and looked down at the body.
"When did he die?" asked Knutas.
"He's probably been lying here close to a week, I would think. The body has started to rot, not too badly yet because it's reasonably cold down here. If he'd been here another day, the whole stairwell would have stunk."
Sohlman pushed a lock of hair back from his forehead and sighed.
"I've got to keep working. It will be a while before you can come in."
"How long?"
"A matter of hours. Actually I'd be happy if you could wait until tomorrow. We have a lot to do here. It's the same thing with his apartment."
"Okay."
Knutas studied the cramped room. Every inch of space had been put to use. Plastic trays were crowded next to jugs containing chemicals; there were scissors, clothespins, stacks of photographs, boxes and crates. In one corner was the enlarger.
A tray had been knocked over and the chemicals mixed with the blood.
When they exited through the front door, Knutas inhaled the fresh evening air deep into his lungs. It was eight fifteen. The rain pouring down from the dark sky was turning into wet snow.
## Monday, November 19
The next morning the investigative team gathered at police headquarters on Norra Hansegatan. An expensive remodeling had just been completed, and the criminal division had been assigned new offices. The meeting room was bright with a high ceiling, and it was twice as large as the old one.
Most of the decor was of simple Scandinavian design in gray and white, with birch furniture. In the middle of the room stood a long, wide table with room for ten on each side. At one end was a big white-board and a projector screen. Everything smelled new. The light-colored paint on the walls was barely dry.
Both sides of the room were lined with windows. One row of windows looked out on the street, the parking lot at Obs supermarket, and the eastern side of the ring wall. Beyond the wall the sea was visible. The other windows faced the corridor so that it was possible to see who was walking past. The thin cotton curtains could be closed for more privacy—the old yellow curtains had been replaced with white ones in a discreet pattern.
For once Knutas was several minutes late for the morning meeting. An amicable murmuring was going on as he stepped into the room with a coffee mug in one hand and a folder of papers in the other. It was past eight o'clock, and everyone was present. He removed his jacket, hung it over the back of his chair, and took his usual place at one end of the table. Taking a gulp of the bitter coffee from the office coffee machine, he studied his colleagues as they chatted with each other.
On his right sat Karin Jacobsson: thirty-seven years old, petite, with dark hair and brown eyes. On the job she was persistent and fearless, and she could be as irascible as a terrier. She was open and outgoing, but he knew very little about her personal life, even though they had been working together for fifteen years. She lived alone and had no children. Knutas didn't know whether she had a boyfriend or not.
He had spent all autumn without her working beside him, and he had missed her terribly. In connection with the homicides of the past summer, Karin Jacobsson had become the subject of an internal investigation regarding possible misconduct. The investigation was dropped, but the whole thing had taken its toll on her. She had been placed on leave while the investigation was ongoing, and then she had taken a vacation right afterward. He had no idea what she had done while she was away.
Right now she was immersed in a quiet conversation with Detective Inspector Thomas Wittberg. He looked more like a surfer than a police officer, with his thick blond hair and trim body. He was a twenty-seven-year-old playboy who constantly had new girlfriends, but his attention to his job was irreproachable. His talent for making contact with people had been of great use—as the head of an interrogation he was unbeatable.
Lars Norrby, on the other side of the table, was the direct opposite of Wittberg: tall, dark, and meticulous to the point of being long-winded. He could drive Knutas crazy with his fussing over details. At work they knew each other's habits inside out. They had joined the police force at the same time, and for a period they had patrolled together. Now they were both approaching fifty and were as familiar with the criminals on Gotland as they were with each other.
Detective Inspector Norrby was the police spokesman, as well as the assistant head of the criminal investigation unit—a situation that did not always please Knutas.
The technician of the group, Erik Sohlman, was intense, temperamental, and as zealous as a bloodhound; at the same time, he was incredibly methodical.
Birger Smittenberg, the chief prosecutor, was also sitting at the table. He was originally from Stockholm, but he had married a woman from Gotland. Knutas valued his knowledge and his strong sense of involvement.
Knutas began the meeting.
"The victim is Henry 'Flash' Dahlström, born in 1943. He was found dead just after six P.M. yesterday, in a basement room that he used as a darkroom. If you haven't all heard it already, he's the alcoholic who was once a photographer. He used to hang out down on Öster, and the most distinctive thing about him was the camera that he always wore around his neck."
No one at the table said a word. Everyone was listening intently.
"Dahlström was found with extensive contusions on the back of his head. There's no doubt that he was murdered. His body will be transported to the forensic medicine lab in Solna sometime today."
"Did you find the murder weapon?" asked Norrby.
"Not yet. We've searched both the darkroom and his apartment. Those are the only areas that we've cordoned off. Anything else would be pointless since the body has been lying there for a week, and Lord knows how many people have gone up and down the stairs during that time. Dahlström lived on the ground floor in a corner apartment. Right outside is the public passageway to Terra Nova. The whole area has been searched. The dark made our work more difficult, but the search was resumed as soon as it was daylight. Which was just a short time ago."
He looked at his watch.
"Who called it in?" asked the prosecutor.
"The body was discovered by one of the building superintendents. Apparently there are four of them. This one lives in the building across the way. His name is Ove Andersson. He said that a man claiming to be a good friend of the victim rang his doorbell around six P.M. yesterday. The man said that he hadn't seen Dahlström for several days and he wondered where he might be. They found him in the basement, but when the superintendent went up to his place to call the police, the friend took the opportunity to disappear."
"It seems fishy that he ran off. Maybe he was the murderer," Wittberg suggested.
"But if so, why would he contact the super?" objected Norrby.
"Maybe he wanted to get back inside the apartment to get something that he left behind, but he didn't dare break in again," Jacobsson piped in.
"Well, we can't rule that out, even though it doesn't sound very plausible," countered Norrby. "But why would he wait a whole week? There was always a risk that the body would be discovered."
Knutas frowned. "One alternative is that he disappeared because he was afraid of being a suspect. Maybe he was at the party, because it's obvious that a party took place in that apartment. No matter what, we need to get hold of him as soon as possible."
"Have we got a description?" asked Wittberg.
Knutas looked down at his papers. "Middle-aged, about fifty, according to the super. Tall and heavy. He has a mustache, and dark hair pulled back in a ponytail. Dark shirt, dark pants. He didn't notice the man's shoes. I think it sounds like Bengt Johnsson. He's probably the only one of the local winos who fits the description."
"It's got to be Bengan. Those two were always hanging out together," said Wittberg.
Knutas turned to the crime tech. "Erik, you can give us the technical details now."
Sohlman nodded. "We've gone over the apartment and darkroom, but we're far from done. If we start with the victim and his wounds, we need to look at the photos. I should warn you that they're rather nasty."
Sohlman switched off the lights and, using a computer, clicked the digital pictures onto the screen at the front of the room.
"Henry Dahlström was lying prone on the floor with extensive contusions to the back of his head. The perpetrator used a blunt instrument of some kind. My guess is a hammer, but the ME will eventually be able to tell us more. Dahlström was struck repeatedly on the head. The large amount of blood spatter resulted because the perpetrator first knocked a hole in the victim's skull and then continued to strike the bloody surface. Each time he delivered another blow, blood sprayed all over."
Sohlman used a pointer to show spatter that was visible on the floor, the walls, and the ceiling.
"The killer probably knocked Dahlström to the ground and then stood over him and kept striking as he lay there. As far as determining the time of death, I would estimate that the murder took place five or six days ago."
The victim's face was a blotchy yellowish gray shifting to green. His eyes had a dark, brownish-red color, and his lips were black and parched.
"The process of decomposition had begun," Sohlman went on impassively. "You can see the little brown blisters on the body and the corpse fluid that has started to seep out. The same substance is coming out of his mouth and nostrils."
His colleagues around the table grimaced. Jacobsson wondered how Sohlman could always manage to talk about bloody victims, rigor mortis, and decomposing bodies as if he were discussing the weather or his annual income tax returns.
"Everything in the place had been tossed, and the cupboards and boxes containing photos had been searched. The murderer was apparently looking for something. The victim also has defensive wounds on his arms. Here we can see bruises and scratches. So he attempted to resist. The bruise on his collarbone may have been made by a blow that missed its mark. We've taken blood samples, of course. We also found a cigarette butt in the basement hallway, and hairs that don't seem to have come from the victim. Everything has been sent to SCL but, as you know, it will take a while before we get any answers."
He took a sip of coffee and sighed. The response from SCL, the Swedish Crime Laboratory in Linköping, usually took at least a week, more often three.
Sohlman went on. "As far as evidence goes, we've found footprints in the flower bed outside the basement window. Unfortunately, the rain made them impossible to identify. On the other hand, we did get some footprints in the hallway outside the darkroom, and in the best-case scenario they should tell us something. The same footprints were in the apartment—which, by the way, was filled with bottles, ashtrays, beer cans, and a lot of other junk. We've secured quite a few fingerprints, as well as the footprints of four or five different individuals. We also searched the apartment."
The photos of the mess in Dahlström's place sent a clear message: The apartment had been completely turned upside down.
"Dahlström must have had something valuable at home, but I wonder what it might be," said Knutas. "An alcoholic living on welfare doesn't usually have assets of any great value. Did you find his camera?"
"No."
Sohlman cast another glance at his watch. He seemed eager to get away.
"You said that you found a cigarette butt in the basement. Could the murderer have waited outside the darkroom for Dahlström to come out?" asked Jacobsson.
"Quite possibly."
Sohlman then excused himself and left the room.
"In that case, the perp knew that Dahlström was inside the darkroom," Jacobsson went on. "He may have stood in the entryway for hours. What do the neighbors say?"
Knutas leafed through the investigative report.
"We kept knocking on doors until late last night. We haven't got all the reports in yet, but the neighbors in that stairwell confirm, as I mentioned, that there was a party at the apartment last Sunday. A bunch of people came staggering through the front door around nine P.M. A neighbor who encountered them in the entryway guessed that they had been to the racetrack because he heard some remarks about various horses."
"Oh, that's right, Sunday was the last race day of the season," Jacobsson reminded herself.
Knutas looked up from his papers. "Is that right? Well, the track isn't very far away, so they could have easily walked or bicycled home afterward. At any rate, there was a big racket in the apartment, according to the neighbors. A lot of noise and partying, with both male and female voices.
"The woman next door reported that the man who is probably Bengt Johnsson rang her doorbell first, to ask her whether she had seen Dahlström. She referred him to the building superintendent."
"Does her description of him match what the super told us?" asked Norrby.
"Yes, for the most part. An overweight man, younger than Dahlström, about fifty, she thought. Mustache and dark hair pulled back in a ponytail—a biker-type hairstyle, as she expressed it. Wearing shabby clothes, she also said."
Knutas gave a little smile.
"He had on dirty, loose-fitting jeans, with his stomach hanging out. A blue flannel shirt, and he was smoking. She recognized the man because she had seen him with Dahlström several times."
"Everybody knows who Henry Dahlström is, but what do we actually know about him?" asked Wittberg.
"He's been an alcoholoic for years," replied Jacobsson. "He usually hung out at Östercentrum or at the bus station with his buddies. Or at Östergravar in the summer, of course. Divorced, unemployed. He had been living on a disability pension for over fifteen years even though he didn't seem completely destitute. He paid his rent and bills on time, and he kept mostly to himself, according to the neighbors, aside from the occasional party. His friends say that he was utterly harmless, never got into fights or committed any sort of crime. He apparently kept up his interest in photography. This summer I ran into him one day as I was biking to work. He was in the process of photographing a flower near Gutavallen."
"What else do we know about his background?" Wittberg cast a glance at Jacobsson's papers lying on the table.
"He was born in 1943 in Visby Hospital," Jacobsson continued. "Grew up in Visby. In 1965 he married a woman from Visby, Ann-Sofie Nilsson. They had a child in 1967, a girl named Pia. Divorced in 1986."
"Okay, we'll find out more about him today," said Knutas. "And we've got to locate Bengt Johnsson."
He looked out the window.
"Since it's raining, the winos are probably sitting outside the Domus department store, in the mall. That would be the best place to start. Wittberg?"
"Karin and I can go."
Knutas nodded.
"I've started to collate the interviews with his neighbors, and I'd like to keep working on that," said Norrby. "And there are a couple of people I'd like to talk to again."
"That sounds fine," said Knutas, and then he turned to the prosecutor. "Birger, do you have anything to add?"
"No. Just keep me informed and I'll be happy."
"Okay. We'll stop here. But we'll meet again this afternoon. Shall we say three o'clock?"
After the meeting Knutas retreated to his office. His new office was twice as big as the old one. Embarrassingly big, he might say. The walls were painted a light color that reminded him of the sand at Tofta beach on a sunny day in July.
The view was the same as from their conference room next door: the parking lot at Östercentrum and in the distance, the ring wall and the sea.
On the windowsill stood a healthy-looking white geranium that had only recently stopped blooming in anticipation of winter. Jacobsson had given it to him for his birthday several years earlier. He had brought the potted plant with him from his old office, along with his beloved old desk chair made of oak with a soft leather seat. It spun nicely, and he often made use of that feature.
He filled his pipe, taking great care. His thoughts were on Henry Dahlström's darkroom and what he had seen there. When he thought about the man's crushed skull, he shuddered.
Everything pointed to a drunken brawl that got out of hand and came to an unusually brutal end. Dahlström had presumably taken a buddy down to the basement to show him some photographs, and they started arguing about something. Most cases of assault and battery started out that way, and every year some drunk or addict on Gotland was murdered.
In his mind he thought back, trying to summon up a picture of Henry Dahlström.
When Knutas had joined the police force twenty-five years earlier, Dahlström was a respected photographer. He worked for the newspaper Gotlands Tidningar and was one of the most prominent photographers on the island. At the time Knutas was a cop on the beat, patrolling the streets. Whenever big news events occurred, Dahlström was always the first on the scene with his camera. If Knutas met him at private functions, they would usually have a chat. Dahlström was a pleasant man with a good sense of humor, although he had a tendency to drink too much. Knutas would sometimes meet him heading home from a pub, drunk as a skunk. Occasionally he would give him a ride because the man was too drunk to get home on his own. Back then Dahlström was married. Later on he quit his job with the newspaper and started his own company. At the same time, his alcohol consumption seemed to increase.
Dahlström was once found passed out inside the thirteenth-century ruin of Saint Karin's church in the middle of Stora Torget, the central marketplace in Visby. He was lying on a narrow stairway, asleep, when he was discovered by a startled guide and his group of American tourists.
Another time he walked boldly into the Lindgård restaurant on Strandgatan and ordered a real feast consisting of five courses with wine, strongbeer, aquavit, and cognac. Afterward he asked for a cigar imported directly from Havana, which he puffed on as he enjoyed yet another liqueur. When the bill was presented, he openly admitted that, unfortunately, he was unable to pay due to a shortage of funds. The police were called. They took the sated and tipsy man down to the police station, but he was released a few hours later. Dahlström probably thought all the trouble was worth it.
Knutas hadn't seen Dahlström's wife in years. She had been notified about the death of her ex-husband. Knutas hadn't yet spoken to her, but she was scheduled to be interviewed later in the day.
He sucked on his unlit pipe and leafed through Dahlström's file. A few minor misdemeanors, but nothing serious. His friend Bengt Johnsson, on the other hand, had been convicted twenty or more times, mostly on burglary and minor assault charges.
It was strange that they hadn't heard from him.
Emma Winarve sat down on the worn sofa in the teachers' lounge. She was holding her mug of coffee in both hands to warm them. It was drafty in the old wooden building housing the Kyrk School in Roma. Her mug was inscribed with the words: "World's Best Mom." How ridiculous. A mother who had cheated on her husband and who, for the past six months, had also neglected her children because her mind was always on something else. She was fast approaching forty, and also fast losing all control.
The clock on the wall told her that it was nine thirty in the morning. Her colleagues were already crowding around the table, chatting congenially. The smell of coffee had permanently seeped into the curtains, books, papers, file folders, and the dirty-yellow wallpaper. Emma didn't feel like taking part in the conversation. Instead, she looked out the window. The leaves on the oak trees hadn't yet fallen. They were in constant motion, sensitive to the slightest gust of wind. In the yard next to the school, shaggy gray sheep stood huddled together, grazing. Their jaws were grinding as they ceaselessly chewed their cud. Roma's stone church with eight hundred years of history behind it stood there as steadfast as always.
Everything was going on as usual, no matter what storms might be raging inside of someone. It was incomprehensible that she could sit here, seemingly unperturbed, sipping endless cups of coffee, and no one even noticed a thing. Such as the fact that her mind was in the grip of a psychological battle. Or that her whole life was in the process of going to hell. But her colleagues merely sat around her, carrying on subdued conversations. As if nothing were happening.
In her mind's eye, video clips were playing in rapid succession: her daughter Sara's birthday when all Emma could do was cry; she and Johan rolling around in a hotel bed; her mother-in-law's searching eyes; Filip's cello concert, which had totally slipped her mind; her husband Olle's face when she once again rejected him.
She had gotten herself into an impossible situation.
Six months earlier she had met a man who had ended up changing everything. They got to know each other in connection with last summer's police hunt, when Emma's best friend became one of the killer's victims, and she herself came very close to meeting the same fate.
Johan had stepped into her path, and she couldn't just walk by him. He was so unlike everyone else she had ever known; so alive and intense about everything he did. She had never laughed so much with anyone else or felt so calm, almost spiritual. He made her discover sides of herself that she didn't even know existed.
She quickly fell madly in love with him, and before she knew it, he had totally invaded her life. When they made love she was filled with a sensuality that she had never experienced before. He made her relax. For the first time she didn't give a thought to how she looked or how he might judge her expertise in bed.
To be one hundred percent in the moment was something that she had known only from giving birth to her children.
Yet eventually she chose to break it off with him. For the children's sake, she decided to stay with Olle. When the drama of the serial murders was over and she woke up in the hospital with her family around her, she realized that she lacked the will to go through with a divorce, even though she felt that Johan was the great love of her life. Security counted more, at least at the time. With much anguish she put an end to their affair.
The whole family went to Greece on vacation because she needed to get away and have some distance from everything. But it hadn't turned out to be that simple.
When they were back home, Johan had written to her. At first she considered throwing out the letter, unread. But her curiosity got the better of her. Afterward she regretted it.
It would have been best for all parties concerned if she hadn't read even one line of that letter.
Karin Jacobsson and Thomas Wittberg walked down to Östercentrum as soon as the investigative meeting was over. The pedestrian street between the shops was almost deserted. The wind and rain were having their effect. They hurried into the mall at Obs supermarket and shook off the worst of the rain as they stood inside the glass doors.
The shopping center was quite modest: H&M, Guldfynd, a couple of beauty parlors, a health food store, a bulletin board. Obs with its rows of cashiers, then the bakery and pastry shop, the customer service counter, the Tips & Tobak betting parlor and tobacco shop. Restrooms in the back, a recycling station for bottles, and the exit leading to the parking lot. Along with weary retirees and the parents of small children, needing to rest their feet, drunks occupied the benches in the mall whenever the weather was bad.
Most of them kept a hip flask in a bag or pocket, but as long as they didn't do any drinking inside, the security guards left them in peace.
Jacobsson recognized two local winos sitting on the bench nearest the exit. They were filthy and unshaven, dressed in worn-out clothes. The younger man was leaning his head against the wall behind him and staring indifferently at the people walking past. He wore a black leather jacket and tattered running shoes. The older man had on a blue down jacket and knit cap. He was leaning forward with his head in his hands. Greasy locks of hair had crept out from under his cap.
Jacobsson introduced herself and Wittberg, even though she was fully aware that the two men knew who they were.
"We haven't done anything. We're just sitting here."
The man in the cap glanced up, his eyes crossed. And it's not even eleven o'clock in the morning, thought Jacobsson.
"Take it easy," Wittberg told them. "We just want to ask you a few questions."
He pulled a photo out of his pocket.
"Do you recognize this man?"
The younger drunk kept on staring straight ahead. He refused to give either of the police officers even a glance. The other man stared at the picture.
"Hell yes. That's Flash, of course."
"How well do you know him?"
"He's one of the gang, you know. Usually hangs out around here, or at the bus station. He's been doing that for twenty years. Of course I know Flash, everybody does. Hey, Jonas, you know who Flash is, don't you?"
He poked his pal in the side and handed him the photo.
"What a fucking stupid question. Everybody knows him."
The man named Jonas had pupils the size of peppercorns. Jacobsson wondered what he was high on.
"When did you last see him?" asked Wittberg.
"What did he do?"
"Nothing. We just want to know when you last saw him."
"Hmm, when the hell was it? What day is today? Monday?"
Jacobsson nodded. The man stroked his chin with fingers that had been stained dirty yellow from nicotine.
"I haven't seen him in several days, but sometimes he just takes off, you know."
Jacobsson turned to the other man.
"What about you?"
He was still staring straight ahead. His face is actually quite handsome, underneath all the dirt and stubble, she thought. His expression was defiant, showing a strong unwillingness to cooperate. She restrained a desire to stand right in front of him and wave her arms to force him to react.
"Can't remember."
Wittberg was starting to get annoyed.
"What did you say?"
"Why do you want to know? What did he do?" asked the older man in the cap.
"He's dead. Someone killed him."
"What the hell? Is that true?"
Now both men looked up.
"Yes, I'm afraid so. He was found dead last night."
"Are you fucking kidding me?"
"What we need to do now is try to find the person who did it."
"Sure, that's obvious. Come to think of it, I think the last time I saw him was at the bus station about a week ago."
"Was he alone?"
"He was there with his buddies—Kjelle and Bengan, I think."
"How did he seem?"
"What do you mean by 'seem'?"
"How did he act? Did he seem sick, or was he nervous in any way?"
"No, he was the same as usual. He never really says much. He was a little drunk, of course."
"Do you remember what day that was?"
"It was probably Saturday because there were a lot of people downtown. I think it was Saturday."
"A week ago?"
"That's right. But I haven't seen him since then."
Jacobsson turned to the other man.
"What about you? Have you seen him since then?"
"Nope."
Jacobsson suppressed the annoyed feeling that had begun to prickle at her throat.
"Okay. Do either of you know whether he'd spent time with any strangers lately?"
"No idea."
"Is there anyone who might want to harm him?"
"Not Flash, no. He never got into fights with anybody. He kept a low profile, if you know what I mean."
"Sure, I understand," said Jacobsson. "So do you happen to know where his pal Bengan might be? Bengt Johnsson?"
"Is he the one who did it?"
Behind the alcoholic fog, the older man looked genuinely surprised.
"No, no. We just want to talk to him."
"Haven't seen him in a while, have you?"
"Nope," said Jonas.
He was chewing gum so hard that his jaws made a cracking sound.
"The last time I saw him he was with that new guy from the mainland," the older man said. "The guy named Örjan."
"What's his last name?"
"I don't know because he hasn't lived here on Gotland for very long. He was in the slammer on the mainland."
"Do you know where we can find Bengt Johnsson?"
"He lives on Stenkumlaväg with his mother. Maybe that's where he is."
"Do you know the address?"
"Nope."
"All right then. Thanks for your help. If you see or hear anything that has to do with Flash, you should contact the police immediately."
"Sure," said the man with the cap, and then he, too, leaned back against the wall.
Johan Berg opened the morning paper as he sat at the kitchen table in his apartment on Heleneborgsgatan in Stockholm. The apartment was on the ground floor facing the courtyard, but that didn't bother him. The Södermalm district was the very heart of the city, and in his eyes there was no better place to live. One side of the building faced the waters of Riddarfjärden and the old prison island of Långholm with its bathing rocks and wooded walking paths. On the other side the shops, pubs, cafés, and subway were all within easy reach. The red subway line went directly to Karlaplan, and from there it was only a five-minute walk to the editorial headquarters of Swedish TV.
He subscribed to several daily newspapers: Dagens Nyheter, Svenska Dagbladet, and Dagens Industri. Currently Gotlands Tidningar was also in the stack that he plowed through each morning. After the events of the summer, his interest in Gotland had been given a boost. For more reasons than one.
He scanned the headlines: "Crisis in Housing for Elderly," "Police on Gotland Earn Less Than Officers on the Mainland," "Farmer Risks Losing EU Subsidies."
Then he noticed a news item: "Man Found Dead in Gråbo. Police Suspect Foul Play."
As he cleared away the breakfast dishes he thought about the article. Of course it sounded like an ordinary drunken fight, but his curiosity was aroused. He took a quick look at himself in the mirror and put a little gel on his dark curly hair. He was actually in need of a shave, but there was no time for that. His dark stubble would just have to grow out a bit. He was thirty-seven but looked younger. Tall and well built, with regular features and brown eyes. Women were always falling for him—and he'd taken advantage of that fact many times in the past. But not anymore. Ever since six months ago, only one woman existed for him: Emma Winarve of Roma on the island of Gotland. They had met when he was covering the hunt for a serial killer last summer.
She had turned his life upside down. He had never met a woman who moved him so deeply; she challenged him and made him think along whole new lines. He liked himself better when he was with her. When his friends asked him what was so special about Emma, he had a hard time explaining. Everything was just so obvious. And he knew that his feelings were reciprocated.
Things had gone so far that he thought she was actually considering leaving her husband, that it was just a matter of time. He had started fantasizing about moving to Gotland and working for one of the newspapers or for the local radio station. They would move in together, and he would be a stepfather for her two children.
Instead, just the opposite had happened. After the murderer was caught and the case was closed, she called it off. He was completely taken by surprise. His life fell apart. He was forced to take sick leave for several weeks, and when he recovered enough that he could take a vacation, she never left his thoughts for a moment.
When he came home he wrote her a letter. Quite unexpectedly, she answered, and then they started seeing each other again. They mostly met whenever Johan went to Gotland on a story. Occasionally she managed to get away to meet him in Stockholm. But he could tell that she wasn't comfortable with all the lying and that she was struggling with terrible feelings of guilt. Finally she asked for a two-month break. October and November. She explained that she needed some distance and time to think.
Suddenly they had no communication at all. No text messages, no e-mails, no phone calls.
But she had relented once. He was on Gotland on assignment and called her up. She happened to be feeling unhappy just then, and weak, so they met. A quick meeting that merely confirmed how strong their feelings were for each other, at least that's what he thought.
After that, nothing. He had made a couple of awkward attempts, but in vain. She was intractable.
At the same time, he understood. It was difficult for her, since she was married and had young children.
But weeks of restless nights, chain-smoking, and a constant, overwhelming longing for her had taken their toll on him, to put it mildly.
On his way to the subway station, he called Anders Knutas in Visby.
The police superintendent answered at once.
"Knutas."
"Hi. Johan Berg from Regional News here. How are things?"
"Fine, thanks. And you? It's been a while."
"Things are good. I saw an article in the paper about a possible homicide in Gråbo. Is it true?"
"We don't know much at this point."
"What happened?"
A brief pause. Johan could picture Knutas leaning back in his desk chair, filling his pipe. They'd had a great deal to do with each other when Johan reported on the murders from Gotland and then took an active role in solving the case.
"Last night a man was found dead in a basement on Jungmansgatan, in Gråbo."
"Of course."
"His injuries were such that we suspect he was murdered."
"How old was he?"
"Born in 1943."
"Known by the police?"
"Yes, but not because he had committed any crimes to speak of, although he was quite an inveterate alcoholic. He used to hang out downtown, drinking. A so-called local wino."
"Does it have to do with a drunken brawl?"
"It seems so."
"How was he killed?"
"I can't discuss that."
"When was the murder committed?"
"He'd been dead for several days. Maybe as long as a week."
"How could he be dead for so long if he was found in a basement?"
"He was inside a locked room."
"A basement storage room?"
"You could say that."
"Who found him?"
"The building superintendent."
"Had anyone reported him missing?"
"No, but a friend of his contacted the superintendent."
Knutas was starting to sound impatient.
"I see. Who was it?"
"Listen, I can't tell you that. I have to go now. You'll have to make do with what I've said, for the time being."
"Okay. When do you think you might have more to tell me?"
"I have no idea. Bye."
Johan switched off his cell phone, thinking that the murder didn't sound like something that Regional News would report on. Probably just an ordinary drunken fight that got out of hand. The story would be relegated to a few lines.
The Stockholm subway system on a Monday morning in November must be one of the most depressing places in the world, thought Johan as he leaned against the window with the black wall of the tunnel whizzing past an arm's length away.
The car was filled with sallow-faced people, weighed down by worries and the daily grind. No one was talking; the only sounds were the usual clanking and rattling of the subway. A few coughs and some sleepy rustling of giveaway newspapers. People stared at the ceiling, at the ad placards, at the floor, out the window, or at some indefinite point in midair. Everywhere but at each other.
The smell of wet clothing was mixed with perfume, sweat, and the dust burning on the heaters. Jackets were pressed next to coats, scarves next to caps, bodies against bodies, shoes against shoes, faces close to other faces, but without any sort of contact.
How can so many people be gathered in one place without making a sound? thought Johan. There's something sick about the whole thing.
It was mornings like this that could really make him long to get away.
When he emerged from the subway at Karlaplan he felt liberated. At least here he could breathe. The people around him were marching like tin soldiers toward buses, offices, schools, shops, the welfare center, a lawyer's office, or wherever they happened to be going.
He set off across the park near the church, Gustav Adolfskyrkan. The kids in the day-care center were outside, playing on the swing set in the biting wind. Their cheeks were as bright as ripe apples.
The huge edifice of TV headquarters loomed in the November fog. He waved hello to the statue of TV star Lennart Hyland before he stepped through the front door.
Up in the newsroom everyone was bustling around. The national morning news program was under way. At the elevators guests were hurrying past, along with anchormen, meteorologists, makeup artists, reporters, and editors—exiting the studios, or going to the bathrooms, or heading for the breakfast table. The row of picture windows offered a view of Gärdet, the big park in Östermalm, swathed in gray fog and swarming with lively dogs from the doggy day care on Grev Magnigatan. Brown, black, and spotted canines galloped around, playing on the big field and unaffected by the fact that it was a dreary Monday in November.
Almost everyone was present for the morning meeting of Regional News: several cameramen, an early-morning editor, reporters, producers, and program planners. It was crowded in the lounge area of the newsroom. After they had discussed the latest broadcast, criticizing some parts and praising others, the editor Max Grenfors presented the day's roster of news stories. The assignments might very well change during the course of the meeting. Some reporter might have his own idea, or the objections to a story proposal might be so strong that it ended up in the wastebasket, or the discussion might take a new direction and lead to a reworking of all their plans. That's exactly the way things needed to function in a newsroom, thought Johan, who enjoyed the morning gatherings.
He briefly recounted to the others what he knew about the murder on Gotland. Everyone agreed that it sounded like a drunken fight. Johan was assigned to keep an eye on the situation since he was going to Gotland the next day anyway, to do a report on the controversy regarding a campground that was threatened with closure.
The Regional News editorial offices operated under high-pressure deadlines. Each day they produced a twenty-minute program, basically from scratch. A story that aired for two minutes usually took several hours to film and another two hours to edit. Johan was always nagging his bosses about giving the reporters more time.
He was not in favor of the changes that had been implemented since he had started out as a TV reporter ten years earlier. Nowadays the reporters hardly had time to look over their material before they had to submit it to the editor. This had a disastrous effect on quality. Good images that the cameraman had taken a lot of trouble to capture risked being lost because no one discovered them in all the rush. The cameramen were often disappointed when they saw the final story. As soon as management started taking shortcuts in the use of visual images, which were the real strength of TV, then things were really going downhill. Johan refused to write up his reports or do any editing until he had gone through all the material himself.
Of course there were exceptions. When time was tight and the story was thrown into editing twenty minutes before the broadcast, they still succeeded in putting it together.
Unpredictability was the real draw in terms of working in a news-room. In the morning he never knew how the day was going to go. Johan worked mostly as a crime reporter, and the contacts that he had established over the years were invaluable for the newsroom. He also had primary responsibility for covering Gotland, which had been placed under the domain of Regional News a little over a year ago. Swedish TV's large deficit meant that they had closed the local office on Gotland and moved the crew from Norrköping back to Stockholm. Johan was happy to take on Gotland, a place that had delighted him since he was a child. And now it was no longer just the island that attracted him.
Spot tugged at his leash. To think he's never learned to heel, thought Fanny angrily, but she didn't feel like yelling at him. The streets were deserted in the residential neighborhood where she was walking. A dark mist had settled over Visby, and the asphalt was shiny from the gentle rain. An inviting glow came from the curtain-framed windows of all the houses. How orderly everything was. Flowers on the windowsills, gleaming cars in the driveways, and charming mailboxes. Here and there a well-tended compost pile.
She had a good view inside the homes at this time of the evening, after dark. In one, copper utensils hung on the wall in the kitchen; another had a brightly painted, rustic grandfather clock. In a living room a little girl was jumping up and down on the sofa, talking to someone that Fanny couldn't see. Over there was a man holding a dustpan in one hand. A few crumbs must have landed on the rug, she thought and pressed her lips together. A man and woman were standing in another kitchen window; they seemed to be cooking together.
Suddenly the door to a big house opened. An elderly couple came out and went over to a waiting taxi as they chatted merrily. They were well dressed, and Fanny smelled the strong scent of the woman's perfume as they passed quite close to her. They didn't notice that she had stopped to watch them.
She was freezing in her thin jacket. Back home her mother was waiting in the silent and dark apartment. She worked the night shift at Flextronics. Fanny had met her father only a few times in her life, the last time when she was five years old. His band had been playing a gig in Visby, and he dropped by for a brief visit. The only thing she remembered about him was his big, dry hand holding hers, and his brown eyes. Her father was as black as night. He was a Rastafarian and came from Jamaica. In the photos she had seen, he had long tangled locks of hair. They call them dreadlocks, her mother had told her.
He lived in Stockholm, where he played drums in a band, and he had a wife and three kids in Farsta. That was all she knew.
She never heard from him, not even on her birthday. Sometimes she tried to imagine what it would be like if he and her mother had lived together. Maybe her mother wouldn't drink as much. Maybe she would be happier. Maybe Fanny wouldn't have to take care of everything: the cooking, cleaning, and laundry, taking Spot for a walk and doing the grocery shopping. Maybe she wouldn't have a guilty conscience about going out to the stables if her father was around. She wondered what he would say if he knew how things were for her. But he probably didn't care; she meant nothing to him.
She was simply the product of his love affair with her mother.
The first thing Jacobsson and Wittberg noticed was the group of sculptures. Almost two meters tall, made of concrete, and gathered in one place on the property. One depicted a rearing horse that was desperately whinnying at the clouds, another looked like a deer, a third was a moose with a disproportionately large head. Grotesque and phantomlike, they stood there in the pouring rain on the flat expanse of lawn.
They dashed from the car to the house, whose roof extended over the simple porch, offering some protection. A typical one-story building from the fifties with a basement and dirty gray stucco facade. The steps were rotting, and there seemed to be an imminent risk that they might put a foot right through them. The doorbell was almost inaudible. After a minute a tall, stout woman in her seventies opened the door. She was wearing a cardigan and a floral-patterned dress. Her hair was thick and white.
"We're from the police," Wittberg explained. "We want to ask you some questions. Are you Doris Johnsson, the mother of Bengt Johnsson?"
"That's right. Has he gotten mixed up in something again? Come in. You're getting soaked."
They sat down on the leather sofa in the living room. The room was cluttered with things. In addition to the sofa group, there were three armchairs, a rustic chiffonier, a TV, pedestals for flowers, and a bookshelf. The windowsills were crowded with potted plants, and every available space in the room held glass figurines in various designs. They all had one thing in common: they depicted animals. Dogs, cats, hedgehogs, squirrels, cows, horses, pigs, camels, and birds. In various sizes, colors, and poses, they were enthroned on tables and benches, in windows, and on shelves.
"You collect these things?" asked Jacobsson, rather foolishly.
The woman's lined face brightened. "Yes, I've been doing it for years. I have six hundred and twenty-seven pieces," she told them proudly. "So what was it you wanted?"
"Well, I'm sorry to say that we've brought some bad news," said Wittberg, leaning forward. "One of your son's friends has been found dead, and we suspect that someone killed him. His name is Henry Dahlström."
"Good gracious! Henry?" Her face turned pale. "He was murdered?"
"Unfortunately, that's probably what happened. We haven't caught the perpetrator, and that's why we're interested in talking to anyone who knew Henry. Do you know where Bengt is?"
"No, he didn't sleep here last night."
"Where was he?"
"I don't know."
"When did you last see him?" asked Jacobsson.
"Yesterday evening. He dropped by for only a minute. I was down in the basement, hanging up the laundry, so I didn't actually see him. He just called down the stairs to me. This morning he phoned to say that he was going to stay with a friend for a few days."
"I see. Who's the friend?"
"He didn't say."
"Did he give you a phone number?"
"No. He's a grown man, you know. I had the impression that he was staying with a woman."
"Why is that?"
"Because he was so secretive. Otherwise he usually tells me where he is."
"Did he call you on your home phone or on a cell?"
"The home phone."
"Do you have caller ID?"
"As a matter of fact, I do."
She got up and went out to the hall. After a minute she came back.
"No, it doesn't show anything. It must have been an unlisted number."
"Does he have a cell phone?"
Doris Johnsson stood in the doorway and gave the officers sitting on the sofa a defiant look.
"Before I answer any more questions, I want to know what happened. I knew Henry, too. You'll have to tell me what this is all about."
"Yes, of course," muttered Wittberg, who seemed to be quite affected by the domineering tone of the stout woman. Jacobsson noted that he used the formal means of address with her.
"Last night Henry was found by Bengt and the building superintendent. He was in his darkroom in the basement of the building where he lives. Someone had murdered him, but I can't go into the details. When the superintendent left to call the police, Bengt took off, and no one has heard from him since. It's urgent that we get in touch with him as soon as we can."
"He got scared, of course."
"That's very possible. But if we're going to catch the perpetrator, we need to talk to everyone who might have seen anything or who can tell us about Henry's actions during the days before the murder. Do you have any idea where Bengt might be, Mrs. Johnsson?"
"Hmm...He knows so many people. I suppose I could call around and ask."
"When did you last hear from Bengt, or rather when did you actually see him last?" Jacobsson interjected.
"Now let me see...Aside from yesterday evening...It must have been yesterday morning. He slept late, as usual. Didn't get up until eleven and then had his breakfast while I was eating lunch. Then he went out. He didn't say where he was going."
"How did he seem?"
"The same as always. He wasn't acting strange or anything like that."
"Do you know if anything unusual had happened lately?"
Doris Johnsson plucked at her clothing.
"No..." she said hesitantly.
Suddenly she threw out her hands.
"Well, yes. Henry won at the harness-racing track. He won the five-race jackpot, and he was the only winner, so it was a lot of money. Eighty thousand kronor, I think. Bengt told me about it the other day."
Jacobsson and Wittberg looked at her in astonishment.
"When did this happen?"
"It wasn't this past Sunday, so it must have been the previous Sunday. Yes, that's when it was, because they were at the track."
"And Henry won eighty thousand kronor? Do you know what he did with the money?"
"Bought booze, I assume. Part of it went straight to alcohol. As soon as they have a little cash, they start buying rounds for everybody."
"Who else belongs to his circle of friends?"
"There's a man named Kjell that he hangs out with a lot, along with a couple of girls. Monica and Gunsan. Though I suppose her real name is Gun."
"Last names?"
She shook her head.
"Where do they live?"
"I don't know that, either, but somewhere here in town. Also a man named Örjan, by the way. I think he just moved here recently. Bengt has been talking about him lately. I think he lives on Styrmansgatan."
They said good-bye to Doris Johnsson, who promised to call as soon as she heard anything from her son.
With the information about the track winnings, they now had a clear motive for the murder.
Knutas had brought along a packet of Danish open-faced sandwiches for lunch. His father-in-law had recently paid them a visit and delighted the whole family with the delicacies he had brought from Denmark. The three slices of dark rye bread each had a different kind of lunch meat: liver sausage topped with a piece of pickled squash; sliced meatballs with pickled beets; and his favorite, Danish sausage roll. And an ice-cold beer to go with this glorious repast.
He was interrupted by a knock on the door. Norrby stuck his head inside.
"Do you have a minute?"
"Of course."
Norrby folded his nearly six-foot-two frame into one of the visitor chairs in Knutas's office.
"I've been talking to one of the neighbors, who had something interesting to say."
"Let's hear it."
"Anna Larsson is an elderly woman who lives in the apartment above Dahlström's. On Monday night around ten thirty she heard Flash go out. He was wearing his old slippers, which made a special sound when he walked."
Knutas frowned. "How could she hear that from inside her apartment?"
"I know, that's something you might well ask, but it so happened that her cat was suffering from diarrhea."
"So?"
"Anna Larsson lives alone, and she doesn't have a balcony. She was just about to go to bed when her cat shit on the floor. It smelled so bad that she didn't want to have the garbage bag containing the shit in her apartment. She had already put on her nightgown and didn't want to go downstairs to the trash cans, for fear of running into one of her neighbors. So she put the bag on the landing outside her door for the time being. She thought that nobody would notice if she tossed it out first thing in the morning."
"Get to the point," said Knutas impatiently. Norrby's tendency to present too many details was sometimes annoying.
"Well, at the very moment that she opens her door, she hears Dahlström coming out wearing his slippers. He locks his door and goes downstairs to the basement."
"Okay," said Knutas, tapping his pipe on the table.
"Mrs. Larsson doesn't think any more about it. She goes to bed and falls asleep. In the middle of the night she's awakened by her cat me-owing. This time the cat has made a mess on the floor of her bedroom. That animal had a really bad stomachache."
"Hmm."
"She gets out of bed and cleans up everything. She now has another bag of cat shit that has to be put outside on the landing. When she opens the door, someone comes in the entrance one floor down and stops at Dahlström's door. But this time she doesn't hear Dahlström's shuffling slippers; this person is wearing real shoes. She's curious, so she stands there listening. The stranger doesn't ring the doorbell but the door opens and the person goes inside, and she doesn't hear any voices."
Now Knutas's interest was aroused. His pipe froze in midair.
"Then what happened?"
"Then everything was quiet. Not a sound."
"Did she have the impression that someone had opened Dahlström's door from the inside? Or did the person outside open it?"
"She thinks that the person outside opened it."
"Why didn't she tell us about this earlier?"
"She was interviewed on the evening when Dahlström's body was found. She says that she felt stressed and upset, so she mentioned only that she had heard him go down to the basement. Afterward I got to wondering how she could be so sure about it. That's why I went back to talk to her again."
"Good job," Knutas said. "It might have been the killer that she heard, but it could just as well have been Dahlström coming in from somewhere. This was several hours later, wasn't it?"
"Definitely, but it seems quite unlikely that he would have gone out, don't you think?"
"Maybe. Did the woman notice anything else after the person went inside?"
"No, she went back to bed and fell asleep."
"Okay. The question is whether the person had a key—assuming that it wasn't Dahlström, that is."
"There's no sign that the lock was forced."
"Maybe it was someone he knew."
"That seems most plausible."
When the investigative team met again that afternoon, Jacobsson and Wittberg started off by reporting on their encounter with Doris Johnsson and what she had told them about the winnings at the racetrack.
"Now at least we have a motive," said Jacobsson, concluding her report.
"That explains why the apartment was ransacked," said Knutas. "The murderer apparently knew that Dahlström had won big at the track."
"The money still hasn't turned up," added Sohlman, "so presumably the perpetrator found it."
"Bengt Johnsson comes immediately to mind," said Jacobsson. "I think we need to put out an APB on him."
"Considering that this involves a homicide, I have to agree." Knutas turned to Norrby. "We've obtained some new information from a witness."
His colleague told everyone about Anna Larsson and her sick cat in the apartment above.
"Damn," said Wittberg. "That indicates that the perp had a key. Which reinforces our suspicions about Johnsson."
"Why is that?" Jacobsson objected. "The perp could just as easily have killed Dahlström, then stolen his keys and gone up to his apartment."
"Or he might have just picked the lock," Sohlman interjected. "Dahlström had a regular cylinder lock on his door. A skilled burglar could have gotten it open without leaving any sign of forced entry. We didn't find any damage on first examination, but we'll take another look at the lock."
"I agree with Wittberg," said Norrby. "I think it was Bengt Johnsson. He was Dahlström's closest friend and it's likely that he had a spare key. Unless it was Dahlström himself who had decided to go out again in the middle of the night. Wearing real shoes this time."
"Sure, that's possible. But if it was Bengan, why would he then contact the super?" said Jacobsson, sounding skeptical.
"To divert suspicion from himself, of course," snapped Norrby.
"If the neighbor woman's testimony is accurate, then Dahlström was alive twenty-four hours after he went to the racetrack and had a party in his apartment," said Knutas. "That means he wasn't killed in connection with the party. The murder most likely took place late on Monday night or in the early hours of Tuesday morning. We'll soon have a more precise determination of the time from the medical examiner."
"By the way, we received another interesting piece of information from a witness," Norrby went on. "I was out there today, talking with all the neighbors for a second time. One of them who wasn't home gave me a call later on."
"Yes?"
Knutas leaned his head on his hands, preparing for another lengthy report.
"It's a girl who goes to Säve High School. She also heard someone in the stairwell late Monday night. She said it was Arne Haukas, the man who lives across from her on the floor below, meaning the same floor where Dahlström lived. Haukas is a PE teacher, and he usually goes out jogging in the evening. Normally he goes out around eight, but on Monday she heard him leave his apartment around eleven P.M. She also saw him from her window."
"Is that so? How can she be so sure of the time and day?"
"Her older sister from Alva was visiting. They were up late, talking, and they both saw him. This girl has been keeping an eye on him ever since she discovered that he's a bit of a Peeping Tom. He always looks in her window whenever he runs past. She thinks he goes jogging in the evening as a pretext for peering in people's windows."
"Does she have any proof for her allegations?"
"No. She actually sounded a little doubtful herself. She said that she wasn't sure about it, that it was just a feeling she had."
"Is this Haukas married?"
"No, he lives alone. And there could be some basis for the girl's uneasiness. I've only managed to make one phone call about the man so far, and that was to Solberga School, where he works. The principal, whom I happen to know personally, told me that several years ago Arne Haukas was accused of spying on the girls when they changed their clothes. The students claimed that he would barge into the locker room to tell them about something trivial. Four of them thought it was so unpleasant that they filed a complaint with the principal."
"What happened?"
"The principal had a talk with Haukas, who denied the allegations, and that was the end of the matter. It apparently never happened again. No other students have complained."
"There seem to be a lot of sleazy individuals living in that building," Wittberg interjected. "Alcoholics, sick cats, Peeping Toms...It makes you wonder what kind of madhouse that place is."
His comments prompted some merriment around the table. Knutas raised his hand in admonishment.
"In any event, we're not looking for a sex offender; we're looking for a murderer. But this PE teacher might have seen something since he was out running on the night of the murder. Has he been interviewed?"
"No, apparently not," replied Norrby.
"Then we need to do that today."
He turned to Jacobsson. "Anything new on Dahlström?"
"We know that he was employed as a photographer at Gotlands Tidningar. He worked there until 1980, when he resigned and started his own company, called Master Pictures. The business did well for the first few years, but in 1987 it went into bankruptcy, with major debts. After that, there's no information that Dahlström had any sort of job. He lived on welfare until he started receiving a disability pension in 1990."
"Where are his wife and daughter now?" asked Knutas.
"His ex-wife still lives in their old apartment on Signalgatan. His daughter lives in Malmö. Single, with no children. Or at least she's the only person listed at that address. Ann-Sofie Dahlström, his ex-wife, was on the mainland, but she'll be back home later this afternoon. She promised to come straight here from the airport."
"That's good," said Knutas. "We need to get in touch with the daughter, too. I want to put out an internal APB on Bengt Johnsson immediately. We need to ask everyone in his circle of acquaintances where they think he might be staying. Sohlman, you're in charge of examining the apartment door lock one more time. The question is: How many people knew about the money Dahlström won at the track? Everyone who was at the track with him that evening has to be interviewed. But did anyone else know?"
"In those kinds of circles, news like that probably spreads like wildfire," said Wittberg. "No one that we've talked to in town has said a word about the money, but they may have their reasons for not talking."
"You'll have to interview them again, along with all the others," said Knutas. "The money throws a whole new light on the case."
If there was one thing that Emma detested, it was sewing machines.
To think that anyone should have to bother with this kind of shit work, she thought, her mouth full of pins. Her sense of irritation was fast becoming a headache. She swore silently. Why should it be so damned difficult to make a pair of pants? When other people sewed in a zipper, they made it look ridiculously easy.
She was really trying her best, and she had armed herself with tons of patience before she started, promising herself that this time she wouldn't give up. She would not surrender to the slightest obstacle, although she had a tendency to do just that. She was certainly well aware of her own weaknesses.
She had been struggling with this sewing project for an hour, and she had already smoked three cigarettes to calm her nerves. Sweat broke out on her forehead as she tried to straighten out the denim fabric under the presser foot. Twice she had been forced to undo the seam when the zipper ended up buckling.
In school she had always hated sewing class. The silence, the sternness of the teacher. The fact that everything had to be so finicky—the seam allowance, the fitting of the pattern, the wrong and right side of the fabric. The only bad grade that she'd ever received on her report card in grade school was in sewing. It was a permanent reminder of her failure to make anything from pot holders to knitted caps.
The ring of her cell phone came like the arrival of a much anticipated guest. When she heard Johan's voice, fire raced through her breast.
"Hi, it's me. Am I interrupting anything?"
"No, but you know you're not supposed to call me."
"I couldn't help it. Is he home?"
"No, he plays floorball on Monday nights."
"Please don't be mad."
A brief silence. Then his voice again, low and gentle. Like a caress on her brow.
"How are you?"
"Fine, thanks. But I was just about to have a hysterical fit and throw my sewing machine out the window."
His soft laugh made her stomach lurch.
"You're trying to sew something? What happened to that vow you made?"
She was reminded of the time last summer when she had tried to mend a hole in his shirt with a needle and thread from his hotel. Afterward she had vowed never to try sewing anything again.
"It went to hell, just like everything else," she said without thinking.
"What? What do you mean?"
He was trying to sound neutral, but she could hear the hope in his voice.
"Oh, nothing. What do you want? You know you're not supposed to call," she repeated.
"I couldn't help myself."
"But if you don't leave me in peace, I won't be able to think," she said gently.
He tried to persuade her to meet him when he arrived in Gotland on the following day.
She refused, even though her body was screaming for him. It was a battle between reason and emotion.
"Don't keep doing this. It's hard enough as it is."
"But what are your feelings for me, Emma? Tell me honestly. I need to know."
"I think about you, too. All the time. I'm so confused. I don't know what I should do."
"Do you sleep with him?"
"You'd better hang up now," she said, annoyed.
He heard her light a cigarette.
"Come on, tell me. Do you? I want to know if you do."
She sighed deeply.
"No, I don't. I don't have the slightest desire to sleep with him. Are you satisfied?"
"But how long can you keep that up? You're going to have to make up your mind, Emma. Hasn't he noticed anything? Is he that insensitive? Doesn't he wonder why you're acting this way?"
"Of course he does, but he thinks it's a reaction to what happened this summer."
"You still haven't answered my question."
"What question?"
"What are your feelings for me?"
Another deep sigh.
"I love you, Johan," she said quietly. "That's what makes everything so difficult."
"But what the hell, Emma. We can't keep going on like this for much longer. Wouldn't it be better to make a clean break and tell him how things stand?"
"What the hell do you mean by 'how things stand'?" she roared. "You have no idea how things stand!"
"Yes, but—"
"But what?"
Her voice was angry now, and she was on the verge of tears.
"You have no fucking idea what it's like to be responsible for two young children! I can't sit on the sofa and cry all weekend because I miss you. Or decide to be with you just because I want to. Or need to. Or have to, in order to survive. Because surely you know that my whole life revolves around you, Johan. You're the first thing I think about when I wake up and the last thing I see in my mind's eye before I fall asleep. But I can't let this take over everything. I have to keep functioning. Take care of the house, my job, my family. Above all, I have to think of my children. What would happen to them if I left Olle? You go around over there in Stockholm with only yourself to think about. A good job, your own nice apartment in the center of town, and lots to do. If your longing for me starts to get difficult, there are plenty of things to divert your attention. You can go out to pubs, meet with friends, go to the movies. And if you're feeling sad and want to cry over me, you can do that, too. But where the hell can I go? Maybe I can sneak into the laundry room and cry. But I can't just go into town if I'm feeling unhappy and find something else to do. Or meet some new people who are fun? Not likely. Sure, there are plenty of people like that out here!"
She slammed down the phone just as she heard the front door open.
Olle was home.
Ann-Sofie Dahlström had the driest hands that Knutas had ever seen. And she kept rubbing them together so that flakes of skin came off and fell onto her lap. She wore her brown hair pulled back and fastened with a plastic barrette at the nape of her neck. Her face was pale and without a trace of makeup. Knutas began by expressing his condolences over the death of her ex-husband.
"We haven't had any contact for a long time. It's been years since we last talked...." Her voice trailed away.
"What was Henry like when you were married?"
"He was almost always working. There were plenty of late nights and working weekends. We didn't have much of a family life. I was the one who mostly took care of our daughter, Pia. Maybe it was partly my fault that things turned out the way they did. I probably shut him out. He started drinking more and more. Finally it got to be intolerable."
How typical for a woman, thought Knutas. An expert at taking the blame for her husband's bad habits.
"In what way was it intolerable?"
"He was almost always drunk and started neglecting his work. As long as he had a full-time job at Gotlands Tidningar, he managed well enough. The problems began when he started his own company and didn't have anyone looking over his shoulder. He started drinking in the middle of the week, didn't come home at night, and lost customers because he either failed to show up or didn't bother to deliver the photographs he had promised. I finally had to file for divorce."
As she talked, her hands continued their bizarre massage, making a faint scraping sound. She noticed Knutas's glance.
"My hands get like this in the winter, and no lotion does any good. It's the cold. There's nothing I can do about it," she added with a certain sharpness to her voice.
"No, of course not. Forgive me," Knutas apologized. He took out his pipe in order to focus on something else.
"How did his drinking affect Pia?"
"She became withdrawn and uncommunicative. She spent more and more time away from home. Told me that she was studying with friends, but her grades kept getting worse. She started skipping classes and then developed an eating problem. It took a long time for me to realize that it was serious. During the fall semester of her second year, the teachers concluded that she was suffering from anorexia, and she didn't get over it until she finished high school."
"But she stayed in school, in spite of her illness?"
"Yes. I don't think it was the most severe form of the disease, but there's no question that she had an eating disorder."
"What sort of help did you receive?"
"As luck would have it, I knew a doctor at the hospital who had worked at a clinic on the mainland—a clinic for patients with eating disorders. He helped me. I managed to persuade Pia to go over there with me. At the time she weighed only ninety-seven pounds, even though she's five foot nine inches tall."
"How did your husband react?"
"He didn't want to know anything about it. This was toward the end of our marriage."
"What does your daughter do now?"
"She lives in Malmö. She's a librarian at the municipal library."
"Is she married?"
"No."
"Children?"
"No."
"So how do you think she's doing?"
"What do you mean?"
"How is she?"
The woman sitting across from Knutas looked him in the eye without saying a word. Her right eyebrow was twitching. The silence was palpable. Finally it got so oppressive that he was forced to break it.
"How would you describe your contact with each other?"
"Regular."
"And what form does it take?"
"She calls me once a week. Always on Friday."
"How often do you see each other?"
"She usually comes to Gotland for a couple of weeks every summer. But she stays with friends."
"But you see each other?"
"Yes, naturally, we see each other. Of course."
The APB that was issued for Bengt Johnsson on the police-band radio brought results after a couple of hours. Jacobsson took the call from the local police in Slite. A boy who claimed to have seen Johnsson had come into the station. Jacobsson asked to speak to him.
"I think I know where the man is that you're looking for," said a young boy's voice on the phone.
"Really? Where is he?"
"In Åminne, in a cabin. It's an area near here, for summer houses."
"Have you seen him?"
"Yes, he was unloading things from a car outside one of the cabins."
"When was this?"
"Yesterday."
"Why did you happen to contact the police?"
"My best friend's father is on the police force in Slite. I told my friend that I'd seen a suspicious-looking guy out by the summer houses, and he told his father."
"Why did you think the man was suspicious?"
"He was dirty and had on ragged clothes. He seemed nervous and kept looking around, as if he didn't want anyone to notice him."
"Did he see you?"
"No, I don't think so. I was standing behind a tree. I waited to ride my bike past until he went inside the cabin."
"Was he alone?"
"I think so."
"Can you tell me anything else about how he looked?"
"Pretty old, maybe fifty or sixty. Very fat."
"Anything else? What about his hair?"
"He had dark hair, in a ponytail."
Jacobsson felt a vague lurch in her stomach.
"What was he unloading?"
"I couldn't tell."
"How did you happen to catch sight of him?"
"We live right next to the summer-house area. I was on my way home from visiting a friend."
"Could you point out the cabin?"
"Sure."
"Could I talk to one of your parents?"
"They're not home right now."
"Okay. Stay in the house. We'll be there in half an hour. Where do you live?"
Five minutes later Jacobsson and Knutas were in a car, heading east toward Åminne, a popular seaside vacation spot in the summer, located on the northeast side of the island. The local police were going out to the boy's home to await their colleagues.
Outside the car windows, the winter darkness was nearly impenetrable. There were no streetlights, and their only guides were the headlights of cars, as well as reflector posts that appeared at regular intervals. They passed an occasional house, a warm glow coming from its windows. A reminder that people lived out here in the countryside.
When they reached the boy's house, a Slite police car was in the driveway. The boy's name was Jon, and he looked to be about fifteen. Accompanied by his father, he led the way to the summer-house area. It was hard to see the houses in the dark. Without flashlights they would have been fumbling blindly. When they aimed the beams at the cabins, they saw that all of them were a dark Falun red with white trim. Each of them had a yard surrounded by a decorative fence. On this November evening the deserted area seemed almost ghostly. Jacobsson shivered and zipped up her jacket.
Suddenly they saw a light in one of the cabins at the very edge of the woods. It occurred to Knutas that they should have called for backup. Or dogs. Johnsson might not be alone. Knutas put his hand in the inside pocket of his coat, feeling for his service revolver.
Jacobsson was the only one who didn't have a weapon, since the investigation into her potential misconduct during the summer's serial killer case was ongoing, so she had to wait a short distance away. They sent the boy and his father home. The officers stopped before reaching the house and turned off their flashlights so they could discuss how to proceed.
An old Volvo Amazon was parked outside the fence. Knutas crouched down and crept forward, with the other two officers close behind. He paused under a window while the others took up position on either side of the front door.
Not a sound could be heard from inside. Cautiously Knutas stood up enough to peer through the window. In a matter of seconds his brain registered a complete picture of the room: the fireplace with a rocking chair in front of it, the table with four chairs, and an antique lamp hanging from the ceiling. All very cozy. On the table stood several bottles of beer. He signaled to his colleagues. No one there.
At that instant all three of them gave a start as someone moved inside the cabin. Knutas ducked down. The sound of someone clattering and rummaging around penetrated the walls. They waited. Knutas's legs were aching and his fingers were stiff from the cold. Again silence settled over the cabin. Knutas peeked inside and saw the back of a large man now seated in the rocking chair. The ponytail indicated that it was Bengt Johnsson. He had put more wood on the fire, and the flames were dangerously high. He had also moved the table over next to him. On the table stood a whiskey bottle, which looked as if it had been newly opened. Next to the bottle was a glass and an ashtray. The man was smoking as he stared into the fire. Then he leaned forward to take a gulp from the bottle. It was Johnsson, no doubt about it.
Visible to the right of the room was a hallway and part of the kitchen. Knutas had the feeling that Johnsson was alone, but he couldn't be absolutely sure. One of the police officers shifted his feet uneasily. It was freezing cold, and none of them was dressed for standing outdoors for any length of time.
Suddenly Johnsson stood up and looked right at the window. Knutas ducked so quickly that he fell over. Whether Johnsson had seen him or not, it was impossible to tell, but it was now or never.
Knutas took up position in front of the door with his weapon drawn and, after a nod of agreement from the other two officers, he kicked in the door with all his might.
They were greeted by Bengt Johnsson's look of bewilderment. He was obviously drunk, and he was once again sitting in the rocking chair with the glass in his hand.
"What the hell?" was all he managed to say when the three officers stormed in with their guns drawn.
The fire in the fireplace crackled pleasantly, and the kerosene lamps gave off a gentle glow. And there the man sat, peaceful as could be.
The situation was so absurd that Knutas felt an urge to laugh. He lowered his gun and said, "How are things going, Bengt?"
"Fine, thanks," slurred the man sitting next to the fire. "Nice of you to drop by."
Several Months Earlier
He made her unsure of herself. Fanny didn't know how she was supposed to act. He was probably twice her age. She really ought to think of him as a nice old man and nothing more. But there was something about the way he treated her that changed everything. In the beginning, he would grab a lock of her hair and cautiously tug at it, which was both playful and annoying at the same time. She would blush, finding the whole thing embarrassing because she sensed that it meant something more. Sometimes when she met his gaze, he would turn serious, and it felt as if his eyes were stripping her naked. She didn't find it entirely unpleasant. Sometimes she even thought him attractive when she studied him surreptitiously. He was muscular. He had thick, shiny hair with just a hint of gray at the temples. The wrinkles around his eyes and mouth revealed his age. His teeth were slightly yellow and crooked, with multiple fillings.
How could he look at her the way he did when he was so old? she had wondered. If was as if his eyes made her older than she was. Although he didn't always pay attention to her; sometimes he ignored her completely. Then, to her surprise, she would feel disappointed, as if she actually wanted him to notice her.
One time he had asked her if she wanted a lift. She said yes because it was windy and below freezing. He had a big car, and she got in. He put on some music—Joe Cocker. That was his favorite, he told her with a smile. She had never heard of Joe Cocker. He asked her what she liked to listen to. When she couldn't think of anything, he just laughed. It was great to sit there in his warm car and listen to his gentle laughter. It felt somehow safe.
The mere fact that she was sitting in that big fancy car made her feel more important.
## Tuesday, November 20
The morning dawned with a pale white sun that barely managed to rise above the horizon. The sea was still relatively warm, and from the surface a mist slowly lifted upward. The water merged with the sky, and in the haze it was impossible to distinguish one from the other. A seagull shrieked between Visby's medieval merchant buildings on Strandgatan. The rugged ring wall from the thirteenth century that surrounded the town was the best preserved in all of Europe.
From the harbor came the sound of a small fishing boat chugging its way into port with its nighttime catch of cod.
Knutas had just dropped off Lina at the hospital where she worked as a midwife. She started work at seven thirty in the morning, which suited him fine. He could drive her there and still arrive in time for the morning meeting.
They had been married for fourteen years, and he didn't regret a single day of it. They met when he was attending a police conference in Copenhagen. One evening he went to a restaurant on Gråbrødretorv with a colleague. Lina was working there as a waitress while she was studying. It was a warm summer evening, and she had on a short-sleeved blouse and black skirt. She had tried to bring some order to her unruly red hair by fastening it with a barrette, but stray locks kept on escaping and falling into her eyes. She had more freckles than anyone Knutas had ever seen. The tiny spots reached all the way to the tips of her milky-white fingers. She smelled of almonds, and when she leaned over the table, her arm brushed against his.
The next evening they had dinner together, and that was the beginning of a love the likes of which he had never even come close to before. The year that followed was filled with passionate encounters, exhausting good-byes, long nightly phone conversations, an aching sense of longing, and an ever-growing mutual feeling that they had found their partner in life. Lina finished her training, and without further ado she agreed to marry him and move to Gotland. He had just been promoted to head of the criminal division, and that was why they had decided to try living on Gotland.
It had turned out to be a good decision. Lina had no trouble adapting. With her open and cheerful manner she quickly made new friends and created her own life for herself. After only a couple of months she had found a temporary position at Visby hospital. They bought a house, and then it wasn't long before the twins were on their way. Knutas was thirty-five when they met, and he'd had a couple of previous long-term relationships, but he had never known how natural everything could feel. With Lina at his side, he was prepared for anything.
Of course they'd had their crises and arguments, just like everyone else. Lina had a quick temper, and when she started yelling in a strong Danish accent, he had a hard time understanding what she meant. He often couldn't help laughing, which only made her more furious. Even so, their arguments usually ended amicably. There was no sense of competition between the two of them.
Now her birthday was coming up, and he was feeling stressed. She was going to be forty-seven next Saturday, but this year he had no clue what to buy her.
And right now, he had other things on his mind. He was looking forward to the interview with Bengt Johnsson. The man had been drunk out of his mind when they took him in, so the interview had been postponed.
Smittenberg had decided to arrest him, having good reason to suspect him of murder, or at least manslaughter. That was the lowest degree of suspicion, and the evidence against Johnsson would have to be stronger for him actually to be arraigned. The prosecutor had three days to do that. He based the arrest on the argument that there was a risk Johnsson might obstruct the investigation if he was released. He had no alibi for the night of the murder, and he also had a great deal of money in his possession, although he couldn't explain where it had come from. Ten thousand kronor—money they assumed was part of Dahlström's winnings at the track. The fingerprints on the bills were being examined by the Fingerprint Center in Stockholm, and they expected to have an answer by morning. If it turned out that Dahlström's prints were on the bills, then things didn't look good for Johnsson.
Emma pedaled toward Roma, cursing herself for deciding to ride her bike to work. It was crazy how the cold and wind had picked up as she left the schoolyard and made her way out to the main road. The Kyrk School was located some distance from town. She started biking faster to get warm. On Tuesdays she finished teaching by twelve fifteen. She usually stayed at school to put in a couple more hours of work, but today she was planning to visit a friend. Then she was going to take the children into town to go shopping and stop at the pastry shop, as she had promised. They were in desperate need of new wardrobes.
The main road was quiet and deserted, with very little traffic at this time of year. She passed the lane that led to the cloister ruins where plays by Shakespeare were performed every summer. Then past Roma School and the public baths. Farther along, on the other side of the road, were the ramshackle buildings from the Roma sugar mill, which had been shut down. The windows in the yellow brick buildings gaped darkly at her. The sugar mill had been founded more than a hundred years earlier, but it was closed when profits began to plummet. The now deserted mill stood there as a sad reminder of how times had changed.
She lifted her face to the sky, closed her eyes, and inhaled the air deep into her lungs. Emma belonged to those who appreciated November. It was an in-between month without demands, unlike the summer, with its expectations of barbecue evenings, swimming excursions, and all the visits of friends and relatives. And God have mercy on anyone who wasn't outdoors when the sun was shining.
When the autumn darkness descended, she could retreat inside without a guilty conscience, and watch TV in the middle of the day if she felt like it, or read a good book. She could forget about putting on makeup and shuffle around wearing an old, nubby bathrobe.
In December, new demands appeared as Advent was celebrated, and preparations had to be made for Saint Lucia and Christmas Eve, with all the cooking, baking, buying Christmas presents, and putting up decorations.
For thirty-five years she had outwardly lived a good life. She was married and had two children; she had a teaching job and a great house in the middle of Roma. She had lots of friends and a good relationship with her parents and parents-in-law. Outwardly everything seemed fine, but her emotional life was in chaos. She would never have imagined how much her longing for Johan could hurt. It made her anxious, and it kept her awake at night. She had thought that her feelings for him would diminish with time. Oh, how she had deceived herself. They had seen each other only once in almost two months, and they had known each other for barely six months. By all rights their love ought to be dead. From a logical point of view, at least. But emotions and logic had nothing to do with each other.
She had tried to forget and to move on. She could see an uneasiness in the eyes of her children. Sara was only eight, and Filip was a year younger. Yet sometimes she imagined that they knew what was going on. More than Olle did. He carried on as usual. He seemed to think that they could go on forever, side by side, without touching each other. They were now like a couple of old friends. He seemed to have come to terms with the situation. Once she asked him how he could seem so content, in spite of everything. He replied that he wanted to give her time. Time after the trauma of Helena's death and everything else that followed. Olle was still under the illusion that all this had to do with the aftermath of the events of the past summer. And it was true that she thought often about Helena's horrible death. And she missed her terribly.
At first she had thought that the whole drama was the reason why she had fallen in love with Johan. That she had gone through some sort of emotional shock. But she couldn't stop thinking about him.
She seemed to see his face everywhere she turned—at the Konsum grocery store, in the schoolyard, when she went into town.
Her guilty conscience tormented her. To think she was capable of betraying Olle in such a dreadful way. The phone conversation with Johan had made her even more confused. Of course she wanted nothing more than to see him. But the consequences of such a meeting scared her to death.
When she looked at Olle she tried to conjure up the image of the man who had once sparked her love. The man to whom she had said yes in front of the altar. He was still the same person, after all. The same now as back then. They were supposed to grow old together, damn it. That's what they had decided long ago.
The throbbing above his temples started as soon as Johan disembarked from the plane. Shit. The last thing he needed right now was a headache. Accompanied by his colleague, cameraman Peter Bylund, he rented a car at the airport and drove straight to the old TV news-room that was still at their disposal. It was next to the Radio Gotland building, in the middle of Visby.
It smelled musty. Dust bunnies as big as balls of yarn lay in all the corners, and the computers were also covered with a fine layer of dust. It had been a while since anyone had been inside.
The story that was their priority for the day had to do with the future of the Björkhaga campground. It was a classic camping area from the late forties, idyllically located near a sandy beach on the west side of the island. During the summer months it was filled with tourists and Gotlanders alike. Many were regular visitors, who came back year after year because they appreciated a quieter campground, without all the facilities. Now the municipal grounds had been leased to a private individual. The plan was to transform the Björkhaga campground into a modern resort area. Protests from campers and the local inhabitants came quickly.
The story had all the makings of a good TV report: photos from the deserted campground that had given so many families and their children great pleasure over the years, and a fierce conflict in the form of outraged local residents versus a business-minded entrepreneur who had the municipal bigwigs behind him.
An easy job. From Stockholm, Johan had already scheduled the interviews, so it was just a matter of getting started. The biggest challenge for him was to keep away from Emma. Right now there were only a few miles between them.
The interrogation room was sparsely furnished with a table and four chairs. The tape recorder was as new as the furniture. This was the first time it would be put to use.
Bengt Johnsson didn't look as relaxed as he had the night before. Dressed in blue prison garb, he sat hunched on a chair, glaring at Jacobsson and Knutas, who were seated across from him. His dark hair was pulled back into a skimpy ponytail, and his mustache drooped, as did the corners of his mouth.
After the preliminary formalities were taken care of, Knutas leaned back and studied the man who was suspected of killing Henry Dahlström. Every interview had great significance for the investigative work. Establishing trust between the suspect and the interrogator was of the utmost importance. That was why Knutas took pains to proceed cautiously.
"How are you feeling?" he began. "Would you like something to drink?"
"Yes, damn it. A beer would taste good right now."
"Unfortunately, that's not something we can offer you." Knutas gave him a little smile. "How about a soda or some coffee?"
"I'll have a Coke."
Knutas rang for a soda.
"Am I allowed to smoke?"
"Sure."
"Great."
Johnsson shook a cigarette out of a crumpled pack of John Silvers and lit it with a slight tremor in his hand.
"Can you tell us when you last saw Henry?"
"It was the day after he won at the track. Or rather, the evening after. I was in town with a pal and Flash came over to see us. I was drunk, so I don't really remember much."
He was interrupted by the door opening. A police officer came in with the soda.
"What happened?"
"We just talked for a while."
"Who was your friend?"
"His name is Örjan. Örjan Broström."
"What did you do then?"
"Flash didn't stay long."
"Was he on foot when he left?"
"He went to catch a bus."
"And you didn't see him after that?"
"Nope."
"And this was on Monday, November twelfth, the day after you were at the track?"
"Yup."
"What time?"
"I'm not really sure, but most of the stores were closed and it was dark. There were hardly any people around, so I think it was pretty late."
"What do you mean by that? Ten or eleven at night?"
"No, no, damn it. It wasn't that late. Maybe seven or eight."
"And you didn't see Henry again after that night?"
"No, not until we found him in the darkroom, that is."
"The building superintendent says that you rang his doorbell. Is that right?"
"Yes."
"Why did you want to talk to him?"
"I hadn't seen Flash for a while. I get a little worried when a buddy suddenly isn't around."
"Why did you take off after you found him?"
Johnsson was silent for a moment before he resumed talking.
"Well, you see...I'd done something really stupid, something damn stupid."
"Okay," said Knutas. "What was it?"
"The whole gang was at the racetrack on Sunday, the last race day of the season, so it was extra festive. I was there with Flash and Kjelle, and two broads: Gunsan and Monica. We went over to Flash's place beforehand to have a bite to eat. And then when he won, he wanted to celebrate and we did, too. So we went back to his apartment afterward. We had a party there that night."
He fell silent. Knutas clearly sensed that this was a turning point in the interrogation. Now it was starting to get interesting.
"Well, Flash had won all this money at the track, eighty thousand big ones, in thousand-kronor bills. He showed me where he hid the money, in a box in the broom closet. Later, when the others were all in the living room, I just couldn't resist. I thought he wouldn't notice if I took a few thousand. I've been going through a real cash crunch, and Flash seemed to be really flush lately, so I thought that...well."
He paused and gave the officers a pleading look.
"But damn it, I didn't kill him. No, I didn't. I could never do anything like that. But I did take some of his money."
"How much?"
"I guess about twenty thousand," said Johnsson quietly.
"You only had ten thousand in the cabin. What happened to the rest of it?"
"I spent it. On a lot of booze. This thing with Flash really upset me."
"But why did you run away from the darkroom?" Knutas asked again.
"I was scared that you'd think I killed Flash because I stole his money."
"What were you doing on the evening of November twelfth?"
"What day was that?"
"Last Monday, when you saw Henry at the bus station."
"Like I told you, we were there until maybe eight or nine o'clock. Then I went home with Örjan. We spent the night drinking until I passed out on his sofa."
"What time was it then?"
"Don't know."
"Where does he live?"
"On Styrmansgatan, number fourteen."
"Okay. Then he should be able to back up your story."
"Sure, although we were both pretty far gone."
They were interrupted by a knock on the door. It was about the results from the Fingerprint Center. They took a short break and the officers left the room. Johnsson wanted to use the toilet.
Dahlström's fingerprints had been found on the bills. This finding was of little consequence if the police chose to believe Johnsson's story. Many other prints were also found, but none that matched any in police records.
"What do we do now?" asked Jacobsson as they got coffee from the office coffee machine.
"I don't know. Do you believe him?"
"Yes, actually, I do," she said, looking up at Knutas. "I think he sounds very convincing."
"I do, too. If only there was someone who could corroborate his story, we could release him right away. I think we can disregard the theft of the money for the time being."
"His pal, this Örjan, seems to keep popping up. We need to get hold of him," said Jacobsson.
"I'll talk to Birger about whether we should hold Bengt Johnsson any longer or not. I think we'll stop the interview here. Would you like some lunch?"
The choice of lunch restaurants in Visby during the wintertime was limited. Most of the pubs were open only in the evening, and so they usually ended up at the same place if they wanted a change from the meager offerings in the police department's cafeteria. Of course the lunch was more expensive, but it was worth every öre. The Cloister was furnished in classic inn style and had a well-respected chef. The owner, Leif Almlöv, was one of Knutas's best friends. When Knutas and Jacobsson stepped through the door, they were met by a great bustle and clatter and plenty of hurrying waitresses. All the tables were taken.
Leif caught sight of them and waved.
"Hi, how are things going?"
He gave Jacobsson a hug and shook hands with Knutas as he kept an eye on everything going on around them.
"Good. It's sure crowded in here today," said Knutas.
"There's a convention in town. It was like this yesterday, too. Total hysteria. What would you like to eat?"
"Looks like we're going to have to settle for hot dogs instead."
"No, no, don't even think of it. Of course I'll get you a table. Just wait here. Have a seat at the bar for the time being."
He called to the bartender to give them something to drink, on the house. As they sat down with glasses of light beer in front of them, Jacobsson lit a cigarette.
"Have you started smoking?" exclaimed Knutas in surprise.
"No, not at all. I only smoke when I go to a party or if I'm having problems."
"I see, and what would you call this?"
"The latter. I'm having some personal difficulties."
"Is it something you'd like to talk about?"
"No. Leif is waving to us—we have a table."
Sometimes Jacobsson could really drive Knutas crazy. She was overly secretive about her private life. She might tell him something about her travels, her relatives, or some social event that she had gone to, but he seldom found out anything important.
They didn't meet socially, except infrequently at a party. He had been to her place only a few times. She lived on Mellangatan, in a big three-room apartment with a view of the sea. The only male companion she ever talked about at any length was her large cockatoo named Vincent, who was the center of attention in his cage in the living room. The stories about him were legion: for one thing, he was a whiz at playing Ping-Pong with his beak, and he could scare off unwelcome visitors by growling like a dog.
Knutas didn't actually know very much about Karin Jacobsson except that she was interested in sports. She played soccer in Division Three and was by all accounts very good at it. She could always talk about soccer. She was a midfielder on the Visby P18 team that played in the mainland league, which meant that she often played matches off the island. Knutas imagined that if she operated on the same level as she did on the job, she was undoubtedly a tough player to tackle, in spite of her small size. She shared her interest in sports with Erik Sohlman. They were always talking about soccer.
Jacobsson was from Tingstäde parish in the north of the island. Her parents still lived in the same house on the edge of Tingstäde swamp, practically right across from the church. Knutas knew that she had a younger brother, but she never talked about him or her parents.
Many times he had wondered why she still lived alone. Karin was both charming and nice, and when she first started working with the Visby police, he had been slightly attracted to her. But that was just when he happened to meet Lina, so he had never fully examined his feelings. He didn't dare ask Karin about her love life; her sense of privacy blocked all attempts of that sort. Yet Knutas never held back from telling her about his own problems. She knew just about everything about him, and he considered her to be his best female friend.
Their food arrived, and they hungrily focused their attention on eating as they discussed the investigation. They both agreed that they believed Bengt Johnsson's story.
"Maybe the murder has nothing to do with the money Dahlström won at the track," said Jacobsson. "The perp could have stolen the cash as a diversion. He wants us to think that the murder was the result of a burglary. But then the question is: What was the real motive?"
"Do you know whether he was seeing anyone?"
"Well, that Monica who was at the track with him told us that they sometimes slept together, but it was nothing serious."
"What about in the past? Maybe there's a story farther back and none of his current friends knows anything about it."
"That's conceivable," said Jacobsson, drinking the last of the light beer she was having with her fish. "Do you think it might be about an ex-girlfriend who wanted revenge, or a jealous husband whose wife was sleeping with Dahlström, or some neighbor who got tired of all the coming and going in the stairwell?"
"I think the explanation could be even simpler than that. The most obvious motive is the track money—someone killed Dahlström for the money, plain and simple."
"Maybe." Jacobsson stood up. "I've got to run. We're going to track down Örjan Broström—Bengt's buddy."
"Okay. Good luck."
Most of the lunch guests had left the restaurant, and Leif sat down on the chair that Jacobsson had vacated.
He opened a frosty bottle of beer and took several long gulps.
"What an ordeal. Practically every customer wanted to order à la carte instead of choosing the daily special. The kitchen was an inferno, and the chef has been yelling at everyone. I had to console one of the waitresses who started sobbing."
"You poor guy," said Knutas with a laugh. "Is she cute?"
Leif made a wry face.
"Not much fun when you have to play nanny to every single person. Sometimes this place seems just like a day-care center. But never mind that, a lot of people means money in the bank, and that's what we need during the long, cold winter. How are things with you?"
"Lots of work—just like you. The difference is that the profits are scanty."
"How's the investigation going?"
"We've got someone under arrest, although between you and me, I doubt he's the guy. But I'm sure we'll solve this case, too."
"Wasn't it one of his drinking buddies who did it?"
"That seems the most likely, but we'll have to wait and see," said Knutas.
Even though he and Leif were close friends, he didn't like to discuss an investigation when he was in the middle of it. Leif was fully aware of this and respected his reticence.
"How are Ingrid and the kids?" asked Knutas.
"They're all fine. This morning I went out and bought tickets to Paris. I'm thinking of surprising Ingrid with a week of romance right after New Year's. We're celebrating our fifteenth wedding anniversary."
"Has it been that long?"
"Incredible but true."
"You always manage to come up with such good ideas. I can't think of what to buy Lina for her birthday. Do you have any suggestions?"
"No, you're going to have to think of something yourself. I've filled my quota when it comes to your wife's birthdays. At least until it's time for her fiftieth."
Knutas smiled with embarrassment. When Lina had turned forty they were going through a rough period financially. So the Almlövs had provided the place and the wait staff for the big celebration. Leif also happened to know the members of a band, and they had agreed to play for free. Leif was truly a thoughtful and generous friend. The entire Knutas family had been invited to the Almlöv mountain cabin and to their time-share apartment on the Costa del Sol in Spain.
The two families belonged to completely different economic brackets. This had bothered Knutas at first, but over time he had accepted this difference. Leif and Ingrid had a relaxed attitude toward their wealth, and they never talked about it.
Knutas asked for the bill, but Leif refused to let his friend pay for lunch. Every time Knutas came to the restaurant they had the same argument.
Johan was standing in front of the ATM on Adelsgatan when he noticed her. She came walking from Söderport, holding the hand of a child on either side. She was talking to them and laughing. Tall and slender, with her sand-colored hair hanging straight down to her shoulders. He saw the contours of her high cheekbones as she turned her head. She was wearing jeans and a short, lion-yellow quilted jacket. A striped scarf was wrapped around her neck. And she had on mocha-colored boots with fringe.
His mouth went dry and he turned his back to peer down at the ATM. "Receipt requested?" Should he turn around and say hello? Last night's conversation complicated matters. He didn't know whether she was still angry.
He had never met the children, just seen them from a distance. Would she notice him, or would she just walk past? There was hardly anyone on the street, which meant that she was bound to see him. He felt a slight panic and turned around.
She had stopped to look in a window a short distance away. He gathered his courage.
"Hi!" He looked right into her shining eyes.
"Hi, Johan."
The children looked up at him inquisitively, their cheeks red under brightly colored caps. One of them was slightly taller than the other.
"You must be Sara and Filip," he said, holding out his hand. "I'm Johan."
"How do you know our names?" asked the girl in her lilting Gotland accent.
She bore a striking resemblance to her mother. A mini-version of Emma.
"Your mother told me."
Emma's presence made him feel weak in the knees.
"Johan is sort of a friend of mine," Emma told the children. "He's a TV journalist and lives in Stockholm."
"Do you work for a TV station?" asked the girl, wide-eyed.
"I've seen you on TV," said the boy, who was smaller and blonder.
Johan was used to having children claim they had seen him, even though he knew it was very unlikely. He made an appearance only on those rare occasions when he did a stand-up, when reporters explain something with live video for the viewers.
But he didn't let on.
"Is that right?"
"Yes," said the boy solemnly.
"Next time don't forget to wave, okay?"
The boy nodded.
"How are things going?" Emma's question sounded rather indifferent.
"Fine, thanks. I'm here with Peter. We're doing a story on the Björkhaga campground."
"I see," she said without interest.
"What about you?"
"I'm good. Fine. Just fine."
She glanced quickly around, as if she were afraid that someone might notice them.
"I'm teaching, as usual. I've been really busy."
Johan felt a growing sense of irritation.
"How long are you staying?" she asked.
"I'm going home tomorrow or Thursday. It hasn't been decided yet. It depends."
"Uh-huh."
Silence settled between them.
"Come on, Mamma."
Filip was tugging at her arm.
"Okay, sweetie, I'm coming."
"Could we meet?"
He was forced to ask the question, even though she had already said no.
"No, I can't."
Her gaze shifted away from him. He tried to catch her eye.
The children were tugging at her. They didn't care about him anymore. They wanted to move on.
"Mamma," they both called.
Suddenly she looked him straight in the eye. And deep inside. For a second he felt everything stand still. Then she said exactly what he was hoping to hear.
"Call me."
Örjan Broström's apartment was on the fourth floor with windows facing Styrmansgatan. When they rang the doorbell, a dog started barking wildly. The barking was interspersed with a deep growl. They automatically took a step back.
"Who is it?" a man's voice said from the other side of the door.
"The police. Open up," ordered Wittberg.
"Just a minute," the voice said.
It turned out that Broström was not alone. Two beefy men with shaved heads were sitting in the kitchen playing cards, drinking beer, and smoking. They spoke an Eastern European language. Estonian, guessed Jacobsson.
"Who are your friends?" she asked as they sat down in the living room.
"Some of my buddies from Stockholm."
"From Stockholm?"
"That's right."
Broström gave her a sullen look. He was wearing a black vest that accentuated both his muscular arms and his chalk white skin. Not to mention all the tattoos. To her horror, Jacobsson noted that he had something resembling a swastika tattooed on his shoulder. He had greasy dark hair and a hard expression on his face. He kept one hand on the collar of the snarling attack dog as he lit a cigarette. In silence he peered at them through the smoke. An old trick among criminals was to let the cops speak first.
"Do you know Henry Dahlström?"
"I can't say that I really knew him. But I knew who he was."
"So you know what happened to him?"
"I know that he's dead."
"When did you last see him?"
"Don't remember."
"Think about it. We can always take you down to the station if that might help your memory," Wittberg suggested.
"Hell, that doesn't really seem necessary."
He made a face that might have been intended as a smile.
"Then you'd better start cooperating. You can begin by trying to recall when you last saw him."
"It must have been in town. That's the only place I ever saw him. We weren't really pals."
"Why not?"
"With that guy? An old drunk? Why would I want to hang out with him?"
"I have no idea, do you?"
Wittberg turned to Jacobsson, who shook her head. She was having a hard time relaxing in the cramped apartment with the dog on the other side of the table. The animal kept staring at her. The fact that he growled every once in a while didn't make things any better, nor did the hair standing up on his back or his stiff tail. She felt a strong urge to light a cigarette herself.
"Could you get rid of the dog?" she asked.
"What? Hugo?"
"Is that his name? It sounds a little too sweet for a dog like that."
"He has a sister named Josephine," muttered Örjan as he took the dog out to the men in the kitchen.
They heard the men exchange a few words and then burst out in raucous laughter. The kitchen door closed. Örjan came back, casting an amused glance in Jacobsson's direction. That's the first real sign of life in his eyes, she thought.
"When did you last see him?" Wittberg asked again.
"I guess it was one night a week ago when Bengan and I were at the bus station. Flash came over to talk to us."
"Then what did you do?"
"We just sat and drank."
"For how long?"
"Don't know. Maybe half an hour."
"What time was it?"
"Around eight, I think."
"Can you possibly remember what day that was?"
"It must have been last Monday, because on Tuesday I was busy with something else."
"What?"
"It's private."
Neither of the police officers felt like asking any more questions about that matter.
"Have you ever been to Henry Dahlström's apartment?" asked Jacobsson.
"No."
"How about his darkroom?"
Örjan shook his head.
"But he and Bengan were good pals, and you hang out with Bengan. How come you never went to his place?"
"It just never happened. I just moved here, damn it. I've only lived here for three months."
"Okay. So what did you do after that on Monday night, after Dahlström went home?"
"Bengan and I sat there for a while longer, even though it was fucking cold out, and then we came back here to my place."
"What did you do here?"
"We just sat and talked, watched TV, and drank a lot."
"Were the two of you here alone?"
"Yes."
"Then what happened?"
"I think we both crashed on the sofa. In the middle of the night I woke up and got into bed."
"Is there anyone who can confirm that what you're saying is true?"
"Don't think so, no."
"Did anyone call you during that time?"
"No."
"Was Bengan with you all night?"
"Yes."
"Are you sure about that? You were asleep, weren't you?"
"He passed out before I did."
"So what did you do?"
"Flipped through the TV channels."
"What did you watch?"
"Can't remember."
They were interrupted by one of the skinheads.
"Hey, Örjan, Hugo is getting restless. We're going to take him out for a walk."
Örjan looked at his watch.
"Good, he probably needs to go out. His leash is hanging on a hook in the hallway. And make sure he doesn't eat any leaves—they're not good for his stomach."
Amazing, thought Jacobsson. How considerate.
They left Örjan Broström without making any further progress. He was not someone they looked forward to meeting again.
When Knutas was back in his office after lunch, someone knocked on the door. Norrby's demeanor, which he normally kept under tight control, had now been shattered by an excitement that Knutas hadn't seen in his colleague for a long time.
"You won't believe this," Norrby gasped as he waved a sheaf of papers.
He dropped into one of the visitor's chairs.
"These are printouts from the bank, from Henry Dahlström's bank account. For years he had only one account, and that's where his disability pension was always deposited. See here?" said Norrby, pointing to the numbers on the page.
"Four months ago he opened a new account. Two deposits were made, both of them for the same large amount. The first was made on July twentieth, when the sum of twenty-five thousand kronor was deposited. The second was as late as October thirtieth, and for the same amount of twenty-five thousand."
"Where did the money come from?"
"It's a mystery to me."
Norrby leaned back in his chair and threw out his hands in a dramatic gesture.
"We now have a new lead!"
"So Dahlström was mixed up in some kind of monkey business. I've always had the feeling that this wasn't an ordinary robbery homicide. We need to call everyone in for another meeting."
Knutas looked at his watch.
"It's one forty-five. Shall we say two thirty? Will you tell the others?"
"Sure."
"In the meantime I'll call the prosecutor. Birger should be here, too."
When the investigative team had gathered, Norrby began by telling them about the deposits made to Dahlström's account.
The sense of focus in the room sharpened tangibly. Everyone automatically leaned forward, and Wittberg gave a long whistle.
"Jesus. Can we find out where the money came from?"
"Whoever made the deposit used an ordinary deposit slip. It doesn't give any information about the person. On the other hand, we do have the date of the deposit."
"What about the bank surveillance cameras?" Jacobsson suggested.
"We've already thought of that. The bank saves the tapes from the cameras for a month. The first bank tape from July is gone, but we have the one from October. If we're in luck, we can use it to trace the individual who made the deposits. We're picking it up right now."
"I've talked with the Swedish Forensic Lab. They're working hard on the evidence taken from the darkroom and apartment, and if we're lucky we'll have answers by the end of the week," Sohlman informed the others. "There are also palm prints and fingerprints from the basement window that we checked against the criminal records. We didn't come up with a match, so if they belong to the perp, he doesn't have a police record."
"What about the murder weapon?" asked Wittberg.
Sohlman shook his head.
"So far we haven't found it, but all indications are that it was a hammer, the ordinary kind that you can buy in any hardware store."
"All right. We need to proceed with the investigation as usual, but let's concentrate on finding out what Dahlström was up to. Who else among his acquaintances might know something? What about the building superintendent? Or the daughter? We still haven't had a proper interview with her. We're going to expand the interview process to include anyone who had contact with Dahlström or who may have seen him on the night of the murder—the bus driver, employees in kiosks and stores, more neighbors in the area."
"And the racetrack," Jacobsson interjected. "We should contact people at the track."
"But it's closed for the season," objected Wittberg.
"All the stables are still in operation. The horses have to be exercised, the stable personnel are working, and the drivers are there. It was at the track that he won all that money, after all."
"Absolutely," said Knutas. "All suggestions are welcome. One more thing before we adjourn—this has to do with how we're going to handle the media. So far, thank God, no journalist has published any details—as you know, we never allow that when it's a matter of a drunken brawl. But their interest in the case is going to grow if the news about the money gets out. So let's keep it under wraps; don't say a word to anyone. You know how easily word can spread. If any reporter starts asking you questions about the investigation, refer them to me or to Lars. I also think it's time for us to call in the National Criminal Police. I've asked for their assistance. Two officers will be arriving tomorrow."
"I hope Martin is one of them," said Jacobsson. "That would be great."
A murmur of agreement was heard.
Knutas shared their positive view of Martin Kihlgård, who had helped them with the investigation in the summer, but his relationship with the man did have its complications. Kihlgård was a cheerful and congenial person who was quite domineering and had an opinion about almost everything. Deep inside, Knutas was aware that his touchiness when it came to Kihlgård might have to do with a little-brother complex in relation to the gentleman from National. The fact that Karin Jacobsson had such an openly high opinion of his colleague didn't make the situation any better.
With a whir and a click the tape slipped into the VCR. Knutas and Jacobsson were alone in Knutas's office. A few seconds of grainy gray flickering, and then the inside of the bank appeared in black and white. They had to fast-forward a bit before they reached the time in question.
The clock in the upper-right-hand corner showed 12:23, and the date was October 30. Almost five minutes passed before anyone made the deposit in Dahlström's account. The bank was quite crowded because it was the lunch hour. This particular branch was centrally located in Östercentrum, and many people liked to take care of their banking at lunchtime. Two windows were open, with a female and a male teller behind the glass. On chairs near the window facing the street sat four people: an elderly man with a cane, a girl with long blond hair, a fat middle-aged woman, and a young man wearing a suit.
Knutas thought to himself that right now he might be looking at the very person who had murdered Henry Dahlström.
The door opened and two more people came into the bank. They didn't seem to be together. First a man who appeared to be in his fifties. He was wearing a gray jacket and checked cap with dark slacks and shoes. He walked forward without hesitation and took a number.
Behind him came another man, very tall and of slight build. He stooped a bit. He apparently already had a number, and he went to stand in front of the teller's window, as if he were next in line.
When he turned and glanced around the bank, Knutas saw that he had a camera hanging around his neck.
They recognized him at once. The man was Henry Dahlström.
"Damn it," groaned Knutas. "He deposited the money himself."
"There goes that possibility. How typical. It was too easy."
Jacobsson turned on the ceiling light.
"He got the money and then put it in the bank himself," she said. "Impossible to trace, in other words."
"Damned rotten luck. But why didn't the person just transfer the money directly into Dahlström's account? If he was so afraid of being discovered, it must have been an even bigger risk for him to meet Dahlström to give him the money than if he had transferred the sum directly."
"It certainly seems strange," Jacobsson agreed. "I wonder what the money was for. I'm convinced the story about the racetrack is true. Dahlström gambled regularly, and the track has always attracted a shady clientele. Something underhanded could have been going on there, maybe a dispute between two criminal elements. Maybe Dahlström was hired to spy for someone and take pictures, so that the person could keep tabs on his rivals."
"You've been watching too many movies," said Knutas.
"Shit," cried Jacobsson as she glanced at her watch. "Speaking of movies, I've got to get going."
"What are you going to see?"
"We're going to the Roxy to see a Turkish black comedy. It's a special showing."
"Who are you going with?"
"You'd really like to know, wouldn't you?"
She gave Knutas an annoying wink and disappeared into the hallway.
"Why are you always so secretive?" he shouted after her.
Several Months Earlier
Fanny had come home from school to an empty apartment.
Her feeling of relief was mixed with a dose of guilt. The less she saw of her mother lately, the better she felt. At the same time, she didn't think it was right to feel this way. You were supposed to like your mother. And besides, she was Fanny's only parent.
She opened the refrigerator and her mood sank. Her mother hadn't gone grocery shopping today, either.
Never mind. Right now she was going to do her homework. She was worried about Thursday's math test; math had never been her strong suit. She had just taken out her books and sharpened her pencils when the phone rang. The sound gave her a start. The phone hardly ever rang in their apartment.
To her astonishment it was him, and he wanted to invite her to dinner. She was both surprised and uncertain. She didn't know what to say.
"Hello, are you still there?" His smooth voice in the receiver.
"Yes," she managed to say, feeling her cheeks grow hot.
"Can you? Do you want to?"
"I've got homework to do. We're having a test."
"But you still have to eat, don't you?"
"Sure, of course I do," she said hesitantly.
"Is your mother home?"
"No, I'm here alone."
He sounded even more determined.
"Well then, it should be fine. If you study for the test now like a good girl, I can pick you up around seven. Then we'll have dinner together and I'll drive you straight home afterward. Surely there can't be any harm in that. And you'll have time to study, too."
He sounded so anxious that she felt compelled to say yes. But what were they going to talk about? At the same time, the invitation to go out to a restaurant was tempting. She could count on one hand the number of times she had gone out to eat. The last time was during a disastrous vacation the previous summer. Her mother had rented a car for a week and they took the boat to Oskarshamn so they could drive around Skåne and stay in youth hostels. It poured the whole time, and her mother drank every single day. On the last evening they went to a Chinese restaurant, and her mother got to talking to a group of Danish tourists. They drank a lot and started making a ruckus. Her mother got so drunk that she fell off her chair and pulled the whole tablecloth down with her. Fanny wanted to sink right through the floor.
She sat down at the kitchen table with her math books, wondering which restaurant they would go to. As long as it wasn't too fancy. What was she going to wear? Now she really couldn't concentrate on her math homework. Why had she said yes? Why was he inviting her out? Even though these thoughts were whirling around in her mind, she couldn't help feeling flattered.
Suddenly she heard keys rattling in the lock and then her mother's voice in the entryway.
"All right, Spot. Good dog. What dirty paws you have! Where's the towel?"
Fanny stayed where she was at the table without saying a word. She counted off the seconds: 1, 2, 3, 4...
Then it came. Four seconds this time.
"Fanny. Fanny!"
Slowly she stood up.
"What is it?" she called.
"Could you come and help me, please? My back hurts. Could you rinse off Spot? He's so filthy."
Fanny took the dog by the scruff of his neck and led him to the bathroom.
Her mother kept on chattering. She was clearly having one of her "up" days.
"We walked all the way out to Strandgärdet. I met a nice lady with a poodle. They just moved in. The dog's name is Salomon—can you imagine that? Spot really liked him. We took off their leashes, and they both went into the water, even though it's so cold. That's why he's so filthy, from rolling in the dirt afterward. God, I'm hungry. Did you go grocery shopping?"
"No, Mamma. I just got home from school. We have a math test, and I need to study."
As usual, her mother wasn't listening. Fanny heard her opening and closing cupboards in the kitchen.
"Don't we have anything in the freezer? Oh, look, this is great: fish casserole. I need to eat. How long does it have to be in the oven? Forty minutes. Good God, I'll starve to death. Oh, I really have to pee. Oooh."
She came rushing into the bathroom and sat down to pee while Fanny resolutely rinsed off the dog's dirty paws. Why did her mother always have to announce all her needs loud and clear so that everyone would know how she felt at every second? Her head was pounding with irritation.
"Make sure you dry him off properly so he won't catch cold," said her mother as she wiped off her crotch.
"Yes, Mamma."
How wonderful it would be if her mother showed the same concern for her daughter once in a while.
When Fanny came out of the bathroom, her mother was lying on the sofa with her eyes closed.
"Are you tired?"
"Yes, I need to rest for a while before going to work. Could you put the casserole in the oven when it's preheated?"
"Okay."
She sat down in the kitchen. Her mother seemed to have fallen asleep. She acts like a big baby, thought Fanny as she set the table. It was four o'clock. She now had three hours left. Two to study, she hoped, and one to get ready.
"What are you going to eat?" asked her mother when Fanny put the casserole on the table.
"Nothing. I'm not hungry yet. I'll fix something later."
"All right," said her mother, who already seemed to be thinking about something else.
Fanny was on the verge of telling her about the fun theater performance they had seen at school, but she could see that her mother wouldn't be able to concentrate enough to listen. Just as well to keep quiet.
His disappointment over the tape was still bothering Knutas as he drove the short distance home in the evening.
He shivered in the ice-cold car. Lina was always complaining about the fact that he stubbornly insisted on keeping the old Benz, even though they could afford a new car. So far he had managed to fend off her ideas about buying a new one. It was too expensive and too much trouble to have two cars, and besides, there wasn't room for more than one outside their house. And he would have a hard time giving up his Mercedes—there were too many memories and experiences attached to these comfortable old seats. It was as if he and the car felt a mutual affection for each other.
When he parked outside their house, he saw lights on in all the windows. A good sign; it meant that everyone was home. He was looking forward to a peaceful evening at home, but he found anything but an idyllic family scene when he opened the front door.
"Like hell I will! I don't give a shit about what she says!"
Nils pounded up the stairs and slammed his door. Petra was sitting at the kitchen table. Lina was standing at the stove with her back turned, clattering the pots and pans. He could see from the way she stood that she was angry.
"What's going on?"
Knutas asked the question even before he took off his coat.
His wife turned around. Her throat was flushed, and her hair was sticking out in all directions.
"Don't talk to me. It's been a hell of a day."
"So what have you two been up to?" asked Knutas, patting his daughter on the head. She instantly leapt up from her chair.
"What have the two of us been up to?" she shot back at him. "You should be asking what he's been up to. My so-called brother!"
And then she also pounded up the stairs.
"I had an awful day at work, and this is more than I can stand," said Lina. "You're going to have to deal with it."
"Did something bad happen?"
"We'll talk about it later."
He hung up his coat, took off his shoes, and then took the stairs in a couple of bounds. He summoned both children to the bedroom and sat down on the bed with them.
"Okay, tell me what's going on."
"Well, we were supposed to help set the table, but first we had to empty the dishwasher while Mamma cooked," said Nils. "I took out the silverware basket and started emptying it. But then Petra came and said that she wanted to do it."
"That's not what happened!"
"Quiet! I'm talking right now. That is what happened. You yanked it out of my hands even though I had already started."
Petra began to cry.
"Is that true?" asked Knutas patiently, turning to his daughter.
"Yes, but he always gets to do the silverware basket, just because it's easier. I thought it was my turn. I wanted to trade jobs, but he wouldn't. Then Mamma got mad and said that we should stop fighting and then Nils said that I was stupid."
Nils's face flushed with indignation.
"Yes, but I'd already started! You can't just come and yank it away from me! And then Mamma started yelling at me that it was all my fault!"
Knutas turned to his daughter.
"I agree that you can't just come and take away the silverware basket if Nils has already started to empty it. At the same time, Nils, you need to take turns when you empty the dishwasher from now on. And keep in mind that your mother is tired, and it's not much fun for her to listen to you fighting when she's trying to cook. And don't call your sister stupid, Nils."
"Okay, I'm sorry," he said sullenly.
Knutas put his arms around both children and gave them a hug. Petra relented, but Nils was still mad and pulled away.
"Come on, it wasn't that bad."
"Leave me alone," snapped Nils, giving his father an angry glare.
Knutas took Nils aside, and after some persuasion, his son reluctantly agreed to come downstairs for dinner.
Lina looked tired and worn out.
"So what happened?" asked Knutas when peace had once again settled over the household.
"Oh, we had a problem at work. I'll tell you later."
"But we want to hear about it, too," objected Petra.
"I don't know.... It's such an awful story," cautioned Lina.
"Please, Mamma. Tell us."
"Well, okay. A woman who was supposed to give birth to her first child came in this morning with labor pains. Everything looked fine, but when she started to push, we couldn't get the baby out. Anita thought we should give the mother an epidural to ease the contractions, but I wanted to wait."
Tears welled up in her eyes as she talked. Knutas reached for her hand under the table.
"Then the baby's heartbeat suddenly got fainter, so we had to do an emergency cesarean. But it was too late. The baby died. I feel like it was my fault."
"Of course it wasn't your fault. You did the best you could," Knutas assured her.
"Oh, that's so sad. Poor Mamma," said Petra, trying to console her.
"I'm not the one you should feel sorry for. I'm going upstairs to lie down for a while." Lina gave a big sigh and got up from the table.
"Shall I come with you?" asked Knutas.
"No, I'd rather be alone."
Usually her work was a source of great joy for Lina, but when things went wrong, she was very hard on herself. She would go over and over everything that had happened, brooding about what they could have done differently, whether they could have done this instead of that.
It wasn't really so strange, thought Knutas. She had to deal with life and death all day long. Just as he did.
## Wednesday, November 21
Pia Dahlström was a tall, dark, and very beautiful woman. Completely unlike her parents, both in appearance and demeanor. She was wearing a jacket, black pants, and high heels. Her hair was pinned up in a knot. She had arrived early because she had to leave that same morning. It was only 7:00 A.M., and police headquarters was still deserted.
Knutas had offered her coffee, which he had taken the trouble to make himself. It was rare that anyone bothered to make real coffee, even though the coffeemaker stood right next to the dreary office coffee machine. They chatted while they waited for the coffee to brew. She reminded him of Audrey Hepburn in the old movies from the fifties. Her big, dark eyes were rimmed with dark eyeliner, just like the movie star's eyes.
When the coffee was done brewing, she sat down on his visitor's sofa.
"Could you describe your relationship with your father?" Knutas asked, thinking that he sounded like a psychiatrist.
"We weren't close at all. His alcoholism prevented that. He started drinking more and more the older I got, or maybe I just noticed it more as I grew up."
She gave her beautiful head a slight shake. Not a strand of hair was out of place.
"He didn't care about me," she went on. "He never came to watch any of my riding lessons or gymnastics routines. Mamma was always the one who went to the PTA meetings and the quarterly teacher conferences. I can't remember him ever making a single sacrifice or doing anything for my sake. No, I really couldn't care less about him."
"I can understand that," said Knutas.
"You speak Gotland Swedish, but you sound like a Dane," she said with a smile.
"I'm married to a Dane, so I guess some of it has rubbed off. How did you react when you heard about your father's death?"
"I just felt empty inside. If he hadn't been murdered, he probably would have ended up drinking himself to death. When I was younger I was angry at him, but that feeling is long gone. He chose the life he was living. He used to have everything: a stimulating job, a family, and a house. But he chose booze over me and my mother."
"When did you last have contact with him?"
"The same day I passed my school exams," she said without changing expression.
"But that must be more than fifteen years ago," exclaimed Knutas in surprise.
"Seventeen, to be exact."
"How could it be that the two of you haven't had any contact since then?"
"It's very simple. He never called, and I never did, either."
"And you didn't have any contact with him after the divorce?"
"Sometimes I would spend the weekends with him, but it wasn't much fun. The fact that I was there didn't stop him from drinking. He never had any ideas about what we should do except stay in his apartment, and then his buddies would come over. They'd drink without paying any attention to me. Watch the races and soccer games on TV, or sometimes they'd sit there and look at girlie magazines. It was disgusting. Usually I'd end up going back home after an hour. Then I stopped going there at all."
"What about your relationship with your mother?"
"It's fine. I suppose it could be better, but I think it's at an acceptable level," she said, sounding as if she were talking about stocks and bonds.
She scratched her collarbone and her bra strap was visible for a moment. It was a glossy gold with a nice embroidered edge.
She's undoubtedly just as perfect underneath, thought Knutas, and then he was annoyed with himself for letting her femininity affect him.
"So how are you doing now?" he asked, to change the subject.
"Fine, thanks. I work at the municipal library in Malmö, and I like my job. I have lots of friends, both in Malmö and in Copenhagen."
"Do you live alone?"
"Yes."
"Do you know if your father had any enemies? You haven't had contact with him in so many years, but something from the distant past might also be important."
A frown appeared on her face. "Not that I can think of."
Not much more came out of the conversation. When Pia Dahlström left, a faint trace of her perfume lingered.
Several Months Earlier
"Are we going to eat here?"
She couldn't hide her disappointment. She had thought that they were going to a restaurant.
"That's right. I borrowed an apartment from a friend. The food is ready upstairs. Come on."
He led the way through the front entrance. The building was located on one of those posh streets near Södertorg, inside the ring wall. There was no elevator, so they had to trudge their way up to the fifth floor. When they reached the top landing, she was out of breath and had a growing sense of uneasiness in her chest. She looked at his trousers with the sharp creases. He suddenly seemed so old. What did he want with her here, anyway?
She had an urge to turn around and run back down the stairs, but then he took her hand.
"Wait till you see how nice it is."
He fumbled with the keys.
The apartment was the biggest one she had ever seen. It was on the top floor, with thick beams in the ceiling and a view of the sea. The living room was enormous, with a polished hardwood floor and big, colorful paintings on the walls. In one corner stood a table that was set with plates and glasses. He hurried over to light the candles in a candelabra.
"Come on," he said eagerly. "Come over here and have a look."
They went out on the balcony, which had a panoramic view. She could see the water and part of the harbor, the town, with its labyrinth of buildings, and the tower of the cathedral.
"Let's have some champagne."
He made it sound so natural that she felt very grown up. He came back with a bottle and two glasses. He eagerly filled them.
"Cheers."
She didn't dare refuse. Cautiously she took a sip. It tickled her nose but didn't taste very good. She hadn't tried much alcohol before. Just a couple of times when her mother had urged her to have some wine on a Saturday evening so that she wouldn't have to drink alone. Red wine tasted horrible. This was better. She took another sip.
"So, what do you think? Isn't this grand?" he said, putting his arm around her as if it were the most natural thing in the world. It made her feel uncomfortable. She didn't know how to react.
He drank another toast with her.
"Drink up, little lady. Then we'll go in and eat."
For dinner they first had toast with some kind of topping. She ate carefully, watching him to see what he did. He poured the rest of the champagne and clinked glasses with her again and again. She took small sips but soon began to feel dizzy. The conversation kept stalling. He asked her a number of questions but mostly talked about himself. Boasted about all the amazing trips he had made to exotic places in the world. As if he wanted to impress her.
She listened but said very little. Reluctantly she began to relax. It was wonderful to be sitting in such a beautiful room, feeling the warmth from the candles. To be eating such an elegant dinner with muted music in the background. The main course was pork tender-loin with saffron rice. And red wine with the food, much better than the wine she'd tasted at home. She drank the whole glass. He kept on talking as Fanny devoted herself to studying the movements of his lips. She started getting the giggles.
"Did you enjoy the food?" he asked as he stood up and started clearing away the plates.
"Yes, thank you. It was great." She snickered.
"That's good."
He looked so satisfied that she started laughing even more. To think he could be so pleased just because she was happy.
"Would you like some coffee? Or maybe you don't drink coffee?"
She shook her head.
"Where's the bathroom?"
"It's out in the hall, to the right. It says 'WC' on the door."
He pointed, eager to show her the way. She was in such a hurry to pee that she felt as if she would burst.
The bathroom was just as elegant as the rest of the apartment. It had a light dimmer. She played with the light switch, moving it back and forth. The bathroom was sparkling clean, and it smelled nice. Everything looked new and unused. The toilet paper had a pattern, and it was softer than what she was used to. She smiled at herself in the mirror, then giggled. To think she was allowed to enjoy all this luxury.
When she went back, he had dimmed the lights and was sitting on the sofa. On the low coffee table stood two glasses of wine and a tray with candles of varying sizes.
"Come here," he said softly.
She felt wary, didn't really know what he wanted. She sat down cautiously, some distance away from him.
"You're so beautiful. Do you know that?" he said gently.
He moved closer. Took her hand and played with her fingers. She hardly dared look at him. He put one hand on her leg. It felt warm and heavy through her jeans.
He left it there, not moving.
"You're so beautiful," he whispered.
Cautiously he tugged at a strand of her hair.
"And you have such lovely hair, black and shiny and thick."
He leaned back and stared at her.
"Your body...it's perfect. Do you know how sexy you are?"
She felt anxious and uncomfortable but couldn't utter a sound. No one had ever said anything like that to her before.
Suddenly he pulled her close and kissed her. She didn't know what to do, just sat there, motionless. Her head was spinning from the wine. His mouth pressed harder against hers, and he tried to open her lips with his tongue. She let him do it. His hands began groping under her shirt, sliding up toward her breasts. She felt his weight as he bent over her. Then his hand reached one of her breasts. She was frightened by his reaction. He moaned and whimpered. Started getting rough, tugging and pulling at her bra. His tongue whisked around in her mouth. Suddenly her thoughts were crystal clear. The only thing she knew was that she had to get away.
"Wait," she said. "Wait."
He didn't seem to hear but just kept tearing at her clothes.
"Wait a minute. I have to go to the bathroom," she added to make him stop.
"But I just want to touch you a little," he cajoled.
"Please, wait."
He put his hands on her back. They were sweaty now, he was sweaty all over. They sat motionless for a moment, and she listened to him breathing hard.
Then he loosened his grip. It seemed as if he were giving up.
He held her away from him and fixed his eyes on her breasts.
"Do you know how beautiful you are?" he whispered. "What are you doing to me?"
He began groping her again. Even rougher than before.
"No," she said. "I don't want to."
"Just a little. You can just give me a little."
He pushed her down onto the sofa, pulled down her zipper, and took a firm grip on her jeans, pulling them off with a jerk. They were so tight that her panties came off with them. She was completely exposed and realized that she didn't have a chance. She stopped struggling and lay still. He pushed her thighs apart.
Then she started to cry.
"I don't want to," she screamed. "Stop it! Stop it!"
All of a sudden he seemed to come to his senses. He let go of her.
When he drove her home, he didn't say a word the whole way. She didn't, either.
Against all odds, Emma had agreed to meet him for lunch. Johan had finished the interview with the county governor, which meant that he was free for the rest of the day. He was supposed to fly home in the morning.
They had agreed to meet at his hotel room. She didn't dare go anywhere else.
Grenfors had called to talk about the story Johan had been assigned to do back in Stockholm; it sounded totally uninteresting.
After the phone conversation, he sat in an armchair and looked at his watch. He had twenty minutes until Emma arrived. Should he order lunch now, to get that out of the way? It was probably a good idea. If the food was delivered faster, they would have more time to themselves. He grabbed the menu and scanned the selections: toast, Caesar salad, sole on a bed of spinach for two hundred and forty kronor—scandalous. Hamburgers with pommes frites—couldn't they just write French fries for once?
What would Emma like? What did she eat? Shrimp, shellfish—no, not shrimp soup. Pasta Bolognese—a fancy way of saying ordinary spaghetti with meat sauce. Something light, but not too light. But maybe she was super-hungry. How about an omelet?
He started to sweat. He would have to take a shower. Without making up his mind, he punched the number for room service. What did they recommend? What's fast, good, not too heavy, and not too expensive? Meatballs with cream sauce and lingonberries—sure, maybe not very elegant, but what the hell.
He ordered two portions and then tore off his clothes. Fifteen minutes left. Would the food come on time, or would they be interrupted in the midst of this longed-for rendezvous? At least he had been longing for it—as for her, he had no idea. What if she had agreed to meet him just to tell him that it was over?
As he got out of the shower, there was a knock on the door. No, it couldn't be...He needed to get dressed, comb his hair, and put on some aftershave. He stopped. Or was it their food? He crept over to the door with water dripping all over.
"Yes?"
"Room service," said a voice on the other side of the door. Relief flooded over him. Why did everything feel as if it were a matter of life and death?
The waiter started setting the table. No, no, that wasn't necessary, thanks. He couldn't offer him a tip, standing there like that in his underwear with a meager towel held up in front of him as a shield. Two minutes left. He threw on some pants and a clean shirt. Then it was twelve ten and she hadn't arrived. Time for a panic attack. What if she didn't come? Had he missed a text message on his cell phone? It was on the table. No, no messages. She had to come, damn it. He looked at himself in the mirror—pale, helpless, at the mercy of his stormy emotions and the despair that would inevitably flood over him if it turned out that she had changed her mind.
There was a knock on the door. He took such a deep breath that he saw stars. He shook his head. To think he couldn't take control of his own life.
It was unreal seeing her standing there in the corridor. With her dark eyes and rosy cheeks, she looked shamelessly perky and healthy. She smiled at him, and that was enough to make the floor disappear from under his feet.
"Mmmm...that smells good. Meatballs," she said without much enthusiasm.
How could he be so hopelessly stupid? Offering a teacher meatballs. That's what they probably had every day at school. What an idiot. They sat down at the table.
"Would you like a beer?"
"Sure, thanks."
What an absurd situation. Here they sat, each of them with a plate of food on the table, in a hotel room with cloudy skies outside, and it was the first time they had seen each other in almost a month. She had put on a little weight, he noticed. It suited her.
"How are you?"
The question sounded as artificial as the flowers on the table.
"Fine, thanks," she replied without looking up from the food. "What about you?"
"Not too bad."
The meatballs felt like cardboard in his mouth.
Silence.
They looked up from their plates at the same time and finished chewing with their eyes fixed on each other.
"Actually, I feel like hell," said Johan.
"Me, too."
"Miserable, in fact. I feel sick all the time."
"Same here. I keep feeling as if I'm going to throw up."
"The whole situation is rotten."
"Rotten to the core," she said, and her eyes danced.
They burst out laughing, but stopped abruptly. She took another bite of her food.
Johan leaned toward her, earnest now.
"I feel as if I'm only half alive. You know—I do all the usual things that I'm supposed to do. Get out of bed in the morning, have breakfast, go to work, but nothing is real. Everything seems to be happening somewhere else. I keep thinking that it's going to get better, but that'll never happen."
She carefully wiped her mouth with the napkin and got up from the table. She had a solemn look on her face. The only thing he could do was sit still. Quietly she pulled him up from his chair. They were almost the same height. She put her arms around him, kissed him on the neck. He felt her warm breath in his ear.
Her strong, hard body against his. They tumbled onto the bed, and she pressed herself against him, their legs intertwined, their arms wrapped tightly around each other. Her lips were soft and warm, her hair smelled like apple. He felt tears stinging his eyelids. Embracing her was like coming home.
He didn't really know what he did, or what she did; he knew only that he didn't want it to end.
It turned out that Martin Kihlgård from the national police did come after all. He was accompanied by Hans Hansson, who was a gaunt and unobtrusive man, compared with his boisterous colleague. Everyone in the criminal division welcomed Kihlgård with open arms. He was a big man whose clothes were always in disarray, but he was a respected and capable detective. There was much backslapping and handshaking all around. Karin Jacobsson gave him such a long hug that Knutas felt a pang of the same irritation he had felt last summer. Those two had gotten along so well that Knutas was jealous, even though he would never admit it out loud. Kihlgård was a big lug, but it was obvious that Jacobsson appreciated his outgoing personality.
When he caught sight of Knutas, Kihlgård's jovial smile got even bigger.
"Well, hello, Knutie," he shouted heartily, slapping him on the back. "How's it going, old boy?"
He sounds like Captain Haddock in the Tintin comics, thought Knutas as he returned the smile. He found it very annoying that Kihlgård had suddenly decided to call him Knutie.
They sat down in Knutas's office and started reviewing the case. No more than ten minutes passed before Kihlgård began grumbling about food.
"Aren't we going to have lunch?"
"Of course, it's almost time for it," said Jacobsson promptly. "Why don't we go to the Cloister? Anders's friend owns the place, and they have great food," she explained, turning to both officers from National.
"That sounds excellent," growled Kihlgård. "You get us a good table, okay, Knutie?"
Lunch was pleasant, in any event. Leif gave them a window table with a view of Saint Per's Ruin. Hans Hansson had never been to Gotland before, and he was impressed.
"It's even more beautiful than in the pictures we see. You live in a regular fairy tale city over here. I hope you appreciate it."
"Normally we don't think much about it," said Jacobsson with a smile. "But a trip to the mainland is always a good reminder. Then I realize how beautiful it is when I come back home."
"Same here," Knutas agreed. "I'd have a hard time living anywhere else."
They ate the grilled lamb and root-vegetable casserole with gusto. Kihlgård had no time to talk while he was eating, except once when he asked for more bread. Knutas was reminded that his colleague apparently had an insatiable appetite. The man was always eating, at all times of the day and night.
The restaurant was furnished in an old-fashioned style, with lighted candles and linen tablecloths on all the tables. The cozy atmosphere was particularly welcome now that it was overcast and cold outdoors. Leif surprised them with the restaurant's specialty, a homemade chocolate cake, with their coffee. Then he sat down to join them for a moment.
"How nice to have new lunch customers. Are you staying for a while?"
"We'll have to wait and see," said Kihlgård. "This is an amazingly delicious cake."
"Please come again, anytime. We're always happy to see all of our customers."
"I suppose it must be difficult in the wintertime."
"Yes, it's tough running a restaurant here that's open all year round. But we've managed to do all right, at least so far. Well, don't let me disturb you anymore."
Leif stood up and left.
"We've gone over the details of Dahlström's life, but what's the situation with alcoholics here on the island, in general?" asked Kihlgård. "For instance, how many are there?"
"I would estimate there are about thirty or so truly hard-core alcoholics, meaning individuals who drink all the time and have no job," replied Jacobsson.
"So they're homeless?"
"We actually don't have any homeless here, like you do in the city. Most of them have their own apartment or else they live in municipal housing for addicts scattered here and there."
"What about violent crime among this sort of people?"
"Occasionally they kill each other when they're drunk. We have a couple of murders a year, on average, that are drug or alcohol related. But usually that happens among the drug addicts. The alcoholics are generally harmless."
It was about time to go back to the office. Knutas waved to Leif to get the bill. The wonderful chocolate cake was on the house.
After seeing Emma again, Johan had a longing for fresh air. He took a walk to distract his thoughts.
Almedalen Park was quiet and deserted. The wet asphalt of the public footpath through the grass glittered in the glow of the streetlights, and he could hear the low quacking of the ducks in the pond, even though they were barely visible in the dark. He turned onto the shoreline pathway that ran from Visby all the way out to Snäckgärds Beach, two miles north. Here the wind picked up, and he turned up the collar of his jacket against the chill. Not a soul was in sight. The waves rolled in to shore, and seagulls shrieked. A large passenger ferry with its navigation lights shining through the darkness was approaching Visby Harbor.
He thought about Emma and couldn't comprehend how he had managed without her for so long. All his feelings had now been reawakened, and he realized that it would be rough to go on waiting. Even though their relationship had now entered a new phase. The anxious waiting was over, and he knew how she felt about him. And knowing this made him feel both calm and strong.
What he needed to do now was to come up with some good story ideas so that he could come back to the island as soon as possible. It was harder for Emma to find an excuse to go to Stockholm.
He passed the Maiden Tower, one of the ring wall's many defensive structures. There was an old legend about this particular tower. In the fourteenth century, King Valdemar Atterdag of Denmark was attempting to capture Visby and strip the city of its riches. A young woman helped him to gain access through one of the gates in the ring wall. The woman had fallen in love with the king, and he had promised to marry her and take her back to Denmark if she opened the gate for him and his men. She did as he asked, and the Danes then plundered Visby. But the king broke his promise and left the young woman to her fate after she had done what he asked. When her role was discovered, the townspeople punished her by walling her alive in the Maiden Tower. According to legend, her cries for help can still be heard. As Johan walked past in the dark, he could easily imagine her inside. The wind was howling, and perhaps it was her desperate cries that he heard in the wind. Even though he was freezing, he was enjoying the weather.
As he passed the Botanical Gardens, the rocks of Strandgärdet appeared, and in the distance shone the lights from the hospital.
Suddenly he heard a shout. A very real shout.
He stepped forward into the darkness and discovered an elderly woman lying on an embankment with a yapping terrier at her side.
"What happened?"
"I fell down and can't get up," complained the woman, her voice quavering. "My foot hurts terribly."
"Wait, let me help you," Johan reassured her, taking a firm grip on her arm. "Careful now, stand up slowly."
"Thank you so much. That was awful," moaned the woman as she got to her feet.
"Are you hurt? Can you put any weight on your foot?"
"Yes, it'll be fine. You're not the kind of man who mugs old women, are you?"
Johan couldn't help smiling. He wondered how he must look, in his black jacket, unshaven, and with his hair disheveled.
"You don't have to worry. My name is Johan Berg," he said.
"Thank goodness. I've had enough drama for one day. My name is Astrid Persson. Do you think you could walk me home? I live over on Backgatan, up there across from the hospital."
She pointed with a gloved finger.
"Of course," said Johan, taking her by the arm. In his other hand he held the little terrier's leash, and together they set off toward Backgatan.
Astrid Persson absolutely insisted on inviting him in for a cup of cocoa. Her husband, Bertil, had started to get worried, and he thanked Johan warmly for his help.
"You're not from Gotland, are you?"
"No, I'm here on an assignment. I'm a journalist for Swedish TV in Stockholm."
"Is that right? Are you here to report on the murder?"
"You mean the murder of Henry Dahlström?"
"Yes, exactly. Do you know anything about who did it?"
"No, we hardly know anything at all about the case. The police aren't saying much. At least so far."
"Ah, so that's how it is."
Bertil slurped his cocoa.
"He was a nice guy, that Dahlström."
"Did you know him?"
"Sure, of course I did. He helped me with some carpentry. He built our carport, and he did a really good job."
"He also did some work on the dormer window," his wife added. "He worked as a carpenter in his younger days, you know. Before he became a photographer."
"Is that right? And he managed to do carpentry work, in spite of his alcohol problem?"
"Oh yes, he did fine. It was as if he pulled himself together while he was working. I did notice that he smelled of liquor one time, but it didn't affect his work. He did the job he was supposed to do, showed up when he promised he would, and so on. Yes, he did an excellent job. And he was pleasant, too, not much of a talker but nice."
Astrid nodded in agreement. Her husband had carefully taped up her foot, which she was now resting on a stool.
"How long ago was this?" asked Johan.
"Well, let's see. We had the carport built several years ago. When was it?"
He looked at his wife.
"Four, maybe five years ago? And the dormer window was done last year."
"Did he help other people with this sort of work?"
"Sure he did. I heard about him from a friend in the local folklore society."
"Have you told the police about this?"
Bertil Persson looked embarrassed. He set his cocoa cup on the table.
"No, why should we? What does it matter that he was here and did a bit of carpentry work? Why would the police care about that?"
He leaned toward Johan and lowered his voice to speak confidentially.
"And besides, we paid him under the table. He was living on welfare and that's how he wanted it. You won't say anything, will you?"
"I hardly think the police would care about how he was paid, given the situation. They're conducting a murder investigation, and this would be important information for them to have. I can't keep it to myself."
Bertil raised his eyebrows.
"Really? But then we risk getting caught for hiring an illegal worker."
He looked upset. Astrid Persson put her hand on his arm.
"As I said, I don't think the police will take that very seriously," said Johan.
He stood up. He wanted to get out of there as quickly as possible.
"But I told you this in confidence," exclaimed Bertil Persson, looking as if he thought his days were numbered.
"I'm sorry, but there's nothing I can do."
The man grabbed Johan's arm, and his voice took on an ingratiating tone.
"But it can't be that important, can it? My wife and I are members of the church—it would be embarrassing if this got out. Can't we forget about the whole thing?"
"I'm sorry, but no," snapped Johan, pulling his arm away, a bit more brusquely than he intended.
He hurried out of the building after saying a rather strained goodbye.
Knutas sank onto his desk chair, holding what he hoped was his last cup of coffee for the day—at least if his stomach had anything to say about the matter. The preliminary autopsy report from the ME showed, exactly as expected, that Henry Dahlström had died as the result of contusions to the back of his head caused by a hammer. The perpetrator had delivered a series of blows, using both the blunt and claw end of the hammer.
The time of death was probably late on Monday night, November 12, or possibly early Tuesday morning. This coincided well with the circumstances known to the police. All indications were that the murder had occurred after 10:30 P.M., when Dahlström's neighbor heard him go down to the basement.
Knutas started meticulously filling his pipe as he continued studying the photos and reading the description of the victim's wounds.
Solving a homicide was like solving a crossword puzzle. Rarely was the solution discovered through direct means. Instead, it required leaving certain details alone for a day while concentrating on others. When he later returned to what he had set aside, new patterns would often emerge. And the same thing happened when he did crossword puzzles: He frequently found it very surprising that a particular problem had caused him so much trouble. When he looked at it again, the solution seemed crystal clear.
Knutas went over to the window, opened it slightly, and lit his pipe.
Then there were the witnesses. Dahlström's friends had nothing of any direct value to report. They had largely just confirmed what the police already knew. Nor had anything new emerged that might reinforce their suspicions about Johnsson, so the prosecutor had decided to release the man. He was still going to be charged with theft, but there was no reason to keep him in custody.
Knutas had practically ruled out the idea that Johnsson was the guilty party. On the other hand, he was giving a good deal of thought to the man named Örjan. An unpleasant type. He'd been in jail for aggravated assault and battery. And he seemed capable of murder.
When Örjan was interviewed he had denied it, of course, claiming that he hardly knew Dahlström. And this had been confirmed by others in their circle. But that didn't preclude the possibility that he might have killed Dahlström.
Arne Haukas, the PE teacher who lived in the same section of the building as Dahlström, had been questioned about his whereabouts on the night of the murder. He claimed that he had simply gone out jogging, as usual. He explained the late hour by saying that he'd been watching a movie on TV, and so he had postponed his run. There was a lighted ski trail nearby, so there was no problem with running at night. He hadn't seen or heard anything unusual.
Knutas's ruminations were interrupted by the phone ringing. It was Johan Berg, who told him about the carpentry work that Dahlström had done for Bertil and Astrid Persson on Backgatan. Knutas was surprised.
"Strange that we didn't hear about this before. Do you have the names of anyone else he did work for?"
"No, the old man wasn't happy when I said that I'd have to tell the police. But you could check with the local folklore society—that's where he heard about Dahlström."
"All right. Anything else?"
"No."
"Thanks for calling."
"You're welcome."
Knutas put down the phone, thinking about what he'd learned. So Dahlström had done work for people in their homes. The information opened a whole new avenue. In his mind he sent Johan words of gratitude.
Fanny went straight home from school. At the door she met her mother's boyfriend, Jack. He glanced at her but didn't even bother to say hi. He just hurried past. The door to the apartment wasn't locked, and Fanny realized at once that something was wrong. She peeked in the kitchen, but it was empty.
She found her mother stretched out on the sofa under a blanket. The blanket had slipped to one side, revealing that she was naked. On the table stood empty beer and wine bottles next to an ashtray filled with cigarette butts.
"Mamma," said Fanny, shaking her by the shoulder. "Wake up!"
Not a hint of life.
"Mamma," Fanny repeated with a sob rising in her throat. She shook her harder. "Mamma, please wake up."
Finally her mother opened her eyes and said in a slurred voice, "I have to throw up. Get me a bucket."
"Which one?"
"Bring the one under the kitchen counter. The red one."
Fanny dashed out to the kitchen to get the bucket, but she didn't find it in time. Her mother threw up all over the rug.
She helped her mother into the bedroom, pulled the covers over her, and set the bucket next to the bed. Spot had started licking up the vomit. She chased him away and then used some paper towels to wipe up the worst of it. But she could see that the rug would have to be washed. She ran hot water in the bathtub, poured in some laundry soap, and then lowered the rug into the water. She left it to soak in the tub while she cleaned up the living room, collecting all the bottles, emptying the ashtray, and airing out the place. When she was finished, she sank onto the sofa.
Spot whimpered. The poor thing needed to go out. She seriously considered calling her mother's sister to tell her that she couldn't handle things anymore. But she decided that she didn't dare; her mother would be furious. Yet what would happen if she kept on drinking like this? She risked losing her job, and then what?
Fanny didn't have the energy to think about that. Soon she wouldn't have the energy for anything at all.
## Thursday, November 22
The aroma of freshly brewed coffee and warm cinnamon rolls swept over Knutas as he stepped into the conference room the next morning. Someone had gone to a lot of trouble. He glanced at Kihlgård. It must have been him, of course. Everyone sitting at the table was in a lively mood. Jacobsson was joking with Wittberg, who had evidently been out partying last night. Knutas surmised that he was entertaining Jacobsson with a story about one of his girlfriends. He had a bottle of Coca-Cola in front of him, which was a clear sign that he had a hangover.
Kihlgård and Smittenberg were both leaning over a newspaper. The prosecutor was holding a pen in his hand, while Kihlgård was holding a roll, naturally. Good Lord, they were working on a crossword puzzle! Norrby and Sohlman were standing at the window, looking out at the rain mixed with hail and discussing the weather.
It was a virtual cocktail party. Incredible what effect fresh-baked goods could have.
Knutas took his usual place at one end of the table and loudly cleared his throat, but no one took any notice.
"All right, everyone," he ventured. "Shall we start?"
No reaction.
He gave Kihlgård a surly look. This was so typical of that darn fellow. To come here, all sweet and nice, bringing rolls and causing a disruption. Knutas had nothing against people enjoying themselves at work, but there was a time and a place. Besides, he was in a foul mood after having a big fight with Lina that morning.
It started with her complaining that clothes were scattered on the floor, that the cat hadn't been fed, and that he hadn't run the dishwasher last night, even though it was full and he was the last one to go to bed. Then she found out that, in spite of a solemn promise, he had forgotten to buy a new floorball stick for Nils, who had broken his old one, and he had a game to play tonight. That turned out to be the last straw. She blew up.
The noise in the conference room forced Knutas to get up from his chair and clap his hands.
"All right, could I have your attention?" he shouted. "Shall we get to work? Or maybe you've decided to devote the day to social activities?"
"An excellent idea," exclaimed Kihlgård. "Why don't we stay in, rent a good video, and make some popcorn? It's such awful weather today—I'm freeeeeezing."
His voice rose to a falsetto. He bent his forearms up and shook his palms at the same time as he wiggled his hips. Given the impressive bulk of his body, the dance was extremely funny. What a clown. Even Knutas couldn't help smiling a bit.
He started by telling them about the work Dahlström had done for payment under the table.
"How did we find this out?" asked Kihlgård.
"Actually it was that TV reporter, Johan Berg, who told me. The couple that lives on Backgatan didn't want to go to the police since it was a question of unreported payments."
"It's just amazing how people with money behave," exclaimed Jacobsson, whose expression had darkened as Knutas talked. "It's so damned wrong. People with high incomes who use illegal workers even though they could afford to pay them legitimately. And then when someone is murdered, they won't even go to the police because they're afraid of getting in trouble! That's about as low as it gets."
Her eyes were blazing as she glanced from one colleague to another.
"They can afford a lovely house and expensive vacations, but they won't pay their cleaning woman legally so that she could get insurance and retirement points and everything else that she's entitled to. They refuse to pay for that. They'll do everything to avoid paying taxes, without giving a thought to whether it's actually a crime. At the same time they expect free day-care centers to be provided for their children and a doctor to be available when they're sick, and they want the schools to offer good food. It's as if they can't see the connection between one thing and the other. It's so hopelessly stupid!"
Everyone at the table was looking at her in surprise. Even Kihlgård, who usually had some witty remark, didn't say a word. But maybe this was because his mouth was full of cinnamon roll, probably his third one.
"Take it easy, Karin," Knutas warned. "Spare us your diatribe."
"What do you mean? Don't you agree that it's damned wrong?"
Jacobsson glanced around the room, looking for sympathy.
"Do you have to turn everything into a political issue?" asked Knutas, sounding annoyed. "We're in the middle of a murder investigation here."
He deliberately turned away from her and looked at his other colleagues.
"So maybe we could go on now?"
Jacobsson didn't say another word, just sighed and shook her head.
"How did this couple get in touch with Dahlström?" asked Wittberg.
"Through friends of theirs who belong to the local folklore society. Apparently a number of people made use of his services."
"Maybe someone was unhappy with their garden shed," said Kihlgård with a snicker.
Knutas ignored his attempt at a joke and turned to Norrby.
"How's it going with the bank? Have you tracked down where the deposits came from?"
"Well, we've come to a dead end there. It's impossible to trace the money. Of course every bill has a serial number, but who keeps records of that? It's also impossible to find out who gave him the money since he made the deposit himself."
"Okay, then right now the important thing is to find out who hired Dahlström illegally. He could have been doing that kind of work for years. Strange that nobody he knew has said anything about it."
As Knutas left the meeting he had the distinct feeling that the issues associated with the murder were going to get much more complicated.
Johan's next meeting with Emma was about to occur much sooner than he had dared hope. The very next morning she called him at the hotel.
"I'm going to Stockholm tomorrow for a one-day conference, connected with my work."
"Are you kidding? Are we on the same flight?"
"No, I'm taking the boat. It was planned long ago."
"Does that mean that I can see you?"
"Yes. I wasn't thinking of staying overnight, but I can if I want to because there's a banquet in the evening. Teachers from all over the country are invited. I was planning to skip the banquet, but I can say that I've changed my mind and book a hotel room. That doesn't mean that I actually have to sleep there...."
He couldn't believe his ears.
"Are you serious?"
She laughed.
"Would you like to have dinner together tomorrow night? Or are you busy?" she asked.
He thought for a moment.
"Let me see...Tomorrow night I was planning to stay home alone in front of the TV and eat chips, so I guess I won't be able to meet you. Unfortunately."
His heart was singing.
"But seriously—we could go to a fantastic new place in Söder. It's small and noisy, but the food is superb."
"That sounds great."
He put down the phone and clenched his hand in a fist, in a gesture of victory. Could it be that she had finally given in?
From the beginning Grenfors had doubted that Regional News should do a story about the murder of Henry Dahlström. In his view, it had just been a drunken fight. He was not alone. Many of his colleagues shared this opinion, and consequently they had settled for only a brief mention of the case so far.
If the editors decided not to report a story in the beginning, it was difficult to sell the idea later on. News stories were perishable goods. A story that was super-hot one day might seem musty the next. Four days had passed since Dahlström's body had been found, and that was an eternity in the news business. Grenfors didn't sound especially interested when Johan called him after lunch.
"So what's new about it?"
"Dahlström was doing odd jobs for people in their homes. Carpentry work and things like that. Getting paid under the table, of course."
"You don't say." Grenfors yawned audibly.
Johan could picture the editor checking the TT wire service on his computer screen as they talked.
"Someone deposited money into his bank account. Twice. Twenty-five thousand kronor each time."
"So they might have been payments that he was getting for work done illegally."
"Maybe. But there's a lot to report about the case, and we haven't done a single story on it yet," countered Johan. "Good Lord, a man literally had his head bashed in with a hammer in his darkroom. And this happened on little Gotland—don't forget that. All the other stations have reported it, but we've hardly said a word. Now it turns out that the victim was working illegally for people, and on top of that, mysterious deposits were made to his bank account. And we're the only ones who know about it. All indications are that this was not your ordinary drunken fight. It's in our territory, for God's sake, and we do such a shitty job of reporting on Gotland."
"Have the police confirmed the information?"
"Not the bank deposits," Johan admitted. "We found that out from a bank teller. The police refused to confirm whether it was true, but I'm convinced that it is. I know how Knutas reacts in this type of situation. But he did confirm the part about Dahlström working illegally."
"That might be enough. But today we're reporting on the gang rape prosecution in Botkyrka and the trial of the cop killer in Märsta. That's a hell of a lot of crime stories for one broadcast."
Johan lost his temper.
"I don't think we can wait on this. We've been dragging our feet on this story, and now we're the only ones who have the new information. The newspapers might have the story by tomorrow!"
"That's the chance we'll have to take. It's not really that interesting. Finish up your assignment today, and then I need you back here in the newsroom tomorrow. But we won't run the story tonight. It fits in better with the Friday broadcast. That's all the time I have right now. Bye."
Johan was fuming as he put down the phone. What a fucking attitude! Every other news program had the story about the trial and the gang rape, but they were the only ones with this news about the murder. Generally he respected Grenfors as an editor, in spite of his shortcomings. But sometimes it was impossible to understand the man. If only he were consistent in his journalistic approach, at least! But one day he could be so overzealous that he would hound the reporters relentlessly to get what he wanted for the broadcast. The next day he would be like this. And they would sit in endless meetings, discussing over and over how they could do a better job on their own news program.
Johan didn't mince words as he sat in the car on the way out to Gråbo, complaining about incompetent editors. His cameraman Peter was equally indignant. He was the one who had found out about the deposits to Dahlström's account. He had met a girl at a bar in Visby, and her sister was a teller in the bank where the deposits were made.
And now they ran the risk of being upstaged by the local press. Again.
Gråbo seemed dead and gloomy in the biting wind. The bleak weather didn't exactly invite outdoor activities. The cars in the parking lot bore witness to the fact that the people living there had limited incomes. Most of the Fords had at least ten years on them. An old Mazda hesitantly pulled out of its parking space and rattled off. At the recycling station, someone had toppled over a shopping cart from the ICA grocery store.
On their way to Dahlström's section of the building, they passed a low wooden structure that looked like a communal laundry room. One wall was plastered with wads of snuff, and the windows were covered with graffiti. The playground in front had a sandbox, swings, and worn-looking wooden benches. Not a kid was in sight.
They walked around to the back of the building, where Dahlström had lived. The blinds were closed, preventing any curiosity-seekers from looking inside. The surrounding property consisted mainly of an overgrown lawn, and the patio was nothing more than a piece of wooden fencing with worn patio furniture that had seen better days. There was a stack of used disposable grills. Leaning against a cinder-block wall was a rusty bicycle and an overflowing garbage bag that seemed to contain empty cans. A rickety fence with peeling paint faced the passageway that continued on toward the woods.
They decided to try talking to the neighbors.
At the fourth apartment they tried, someone finally answered the doorbell. A young guy wearing only boxer shorts peered at them, bleary-eyed with sleep. His hair was dyed black and stood straight up like a scrub brush. An earring gleamed from one ear.
"Hi, we're from Regional News in Stockholm. We'd like to know something about the man who lived downstairs, the one who was murdered."
"Come on in."
He showed them to the living room and motioned for them to have a seat on the couch, while he sat down on a Windsor chair.
"A horrible thing, that murder," he said.
"What was your opinion of Dahlström?" asked Johan.
"A decent old guy. Nothing wrong with him. It didn't bother me that he was an alcoholic, at least. Besides, he had periods when he didn't drink as much, and then he spent a lot of time working on his photos."
"Was that something everybody knew about? The fact that he took photographs?"
"Sure. He used that old bicycle storage room as his darkroom. He's had it for the six years that I've lived here."
The guy looked as if he had just graduated from high school. Johan asked him how old he was.
"Twenty-three," he replied. He had moved away from home when he turned seventeen.
"What kind of contact did you have with Dahlström?"
"We said hello to each other if we met, of course, and sometimes he'd knock and ask if I had anything to drink. That's about all."
"Have you noticed anyone new visiting Dahlström lately? Anyone who was different in some way?"
He gave them a wry smile.
"Are you kidding? Just about anyone who came to visit him was different. Recently I saw a chick peeing in the flower bed."
"Did any of the neighbors complain?"
"I don't think it ever got that bad. Most people probably thought he was a pretty decent guy. But in the summer some did complain when he had parties outside, in back of the building."
"What are people around here saying about the murder?"
"Everyone's saying that the killer must have been someone that Flash knew, someone who had a key to his apartment."
"Why is that?"
"Well, the old lady who lives above him heard a sound at his door one night, about a week before his body was found. Someone went inside without ringing the doorbell while Flash was downstairs in the basement."
"Couldn't it have been Dahlström?" asked Peter.
"No, she could tell that it wasn't him. She knows the sound of the slippers that Flash wore."
"Who do you think would have a key?"
"No idea. He had one buddy that he hung out with more than others. I think his name is Bengan."
"Do you know his last name?"
"No."
"It must be Bengt Johnsson. He was the one the police arrested, but then they let him go. Apparently he had an alibi. Is there anything else you can tell us about Dahlström?"
"There was one strange thing that happened this summer. Flash was talking to a guy down by the harbor. It was fucking early in the morning, not even five o'clock. I happened to notice because they were standing in an odd place, between two containers outside a warehouse. As if they were up to something."
"So they weren't just hanging out and drinking?"
"The other guy wasn't one of Dahlström's usual buddies. I could see that at once. He looked much too neat to be a wino."
"Really? In what way?"
"He was wearing clean slacks and a polo shirt, like an executive on summer vacation."
"What else can you tell us about his appearance?"
"I don't really remember. I think he was younger than Flash, and he was very dark."
"Dark-skinned?"
"No, just really suntanned."
"What were you doing there so early in the morning?"
The guy smiled, looking a bit embarrassed.
"I was with a girl. We'd been out partying at Skeppet. That's a pub down at the harbor. Do you know it?"
Johan grimaced. He had a bad memory from the summer when he had spent the miserably wet Midsummer's Eve at Skeppet, and he ended up bent over a toilet all night.
"She had to catch the seven o'clock boat in the morning, so I went with her down to the harbor. We were just messing around a little, as they say. Before she had to go home."
"I suppose the police know about this?" said Johan.
"Oh no, they don't."
"Why not?"
"I don't like the police. I wouldn't tell them squat."
"Would it be okay for us to do an on-camera interview with you?"
"Not on your life. Then the cops will show up. And you can't say a word to them about what I've told you, either. I know all about protected informants, because my sister is a journalist and she told me that you're not allowed to reveal your sources."
Johan raised his eyebrows in surprise. This was some guy.
"That's true. Of course we won't say anything about the fact that you're the one who told us this. What kind of work do you do, by the way?"
"I'm studying at the college. Archaeology."
Even though they didn't get to do an on-camera interview, Johan was more than satisfied with the encounter. He had to contact Knutas—of course without telling him about where he had obtained the information. Knutas was fully aware of the ethical rules under which journalists did their work, so he would understand.
They tried to talk to the rest of the neighbors, but no one answered the doorbell. Behind the building it was deserted. They took a walk along the pathway. Peter filmed the surrounding area and suddenly gave a shout.
A police car was parked on the public footpath that led to the next residential area. Three uniformed officers stood together, talking. Two others were holding on to the leashes of dogs that were tracking something with their noses to the ground. Police tape had been put up to cordon off a grove of trees and bushes.
To their surprise, they noticed Knutas a short distance away.
"Hi," said Johan in greeting. "It's been a while."
"Yes, it has."
Knutas was not happy, to say the least. These confounded journalists kept turning up at the most inopportune moments. So far the investigation had been mostly spared any media attention. Reporters from the local press had called him this morning to ask questions. He didn't like it, but unfortunately it had become a natural part of his workload lately. On the other hand, he was grateful that Johan had tipped him off about Dahlström's moonlighting. Journalists were good at digging up their own information, and they were also available to relay information to the public when the police occasionally needed help. An interdependent relationship existed between the police and the media. But that didn't mean that the relationship was always easy to handle.
"What's going on here?" asked Johan.
Peter had immediately started the camera rolling, as he always did. Knutas realized that he might as well tell them the truth.
"We've discovered what we think is Dahlström's camera."
"Where?"
Knutas pointed to the grove of trees.
"Someone tossed it over there, and a canine unit found it a short time ago."
"What makes you think that it was his?"
"It's the same type of camera that he always used."
Just as Knutas spoke, they heard a shout from some shrubbery outside the area that had been cordoned off.
"We've got something," called one of the dog handlers.
The German shepherd began barking nonstop. Peter instantly turned the camera in that direction and jogged over to the shrubbery. Johan was right behind him. On the ground lay a hammer with brown splotches on the handle, the head, and the claw. Johan held out the microphone, and Peter let the camera record the ensuing commotion. They recorded the comments of the police, the camera on the ground, the dogs, and the drama when everyone present realized that the murder weapon had probably been found.
Johan couldn't believe his luck. It was sheer coincidence that they happened to show up at the decisive moment in a murder investigation, and then managed to get pictures of the whole thing.
They got Knutas to agree to an interview in which he confirmed that a discovery had just been made that might prove to be of interest. He refused to comment further, but that didn't matter.
Johan did a stand-up at the site with all the activity going on around him and reported that it was most likely the murder weapon that had just been found.
Before he and Peter left, Johan told Knutas, without revealing his source, about Dahlström's meeting down at the harbor.
"Why didn't this person come to the police?" asked Knutas angrily.
"The individual doesn't like the police. Don't ask me why."
Back in the car and with a gleeful smile on his face, Johan called Grenfors's direct line at the newsroom in Stockholm.
Several Months Earlier
He had called her cell phone again and again, asking her to forgive him. He had sent sweet picture messages and even a real bouquet of flowers. Fortunately, her mother had left for work before the flowers arrived.
Fanny had decided never to meet him alone again, but now she was wavering. He called and kept saying that he needed to made amends with her. No dinner this time. Horseback riding. He knew that was something she liked. He had a friend who owned some horses in Gerum, and they could each borrow one and go riding for as long as they liked. The invitation was tempting. Her mother couldn't afford riding lessons, and it was rare that she was allowed to ride any of the horses at the stable.
He suggested going riding on Saturday. At first she declined, but he wouldn't take no for an answer. He said he would call again on Friday night, in case she changed her mind.
She had such mixed feelings. More than two weeks had passed since that evening, and it no longer seemed as dangerous as it had then. Deep inside he was probably a very nice man.
When she stepped through the stable door on Friday afternoon, the horses greeted her with a low whinny. She pulled on her rubber boots and started working. Got out the wheelbarrow, the shovel, and rake. She took Hector out first and fastened his halter to chains on either side of the passageway. He had to stand there while she mucked out his stall. The horses stood on a bed of sawdust and hay, so the piles of manure were easy to gather up with the rake. It was much harder to deal with the urine, which soaked into the sawdust and turned it into heavy piles. She took care of one stall after the other. Eight stalls and almost two hours later, she was completely worn out, and her back ached. Her cell phone was ringing. What if it was him? Instead she heard her mother's twittering voice.
"Sweetheart, it's Mamma. Something has come up. The thing is that I've been invited to Stockholm for the weekend. Berit was supposed to go to the theater with a girlfriend, but the friend got sick, so Berit asked me to go with her instead. She won a whole theater package tour from the Bingo Lotto, you know, and we're going to see Chess and have dinner at the Operakällaran and stay at the Grand Hotel. Can you believe it! It's going to be great! The plane leaves at six, so I've really got to start packing. Is that okay with you?"
"Sure, that's fine. When are you coming home?"
"Sunday evening. The whole thing works out perfect because I don't have to go to work until Monday night. Oh, this is going to be so much fun. I'll leave you some money. But I don't have time to take Spot out, so you'd better come home soon. He's getting restless."
"I suppose I'll have to," Fanny said with a sigh.
She was supposed to ride Maxwell, but now she wouldn't have time. She would have to change her plans and head home.
When she got to their apartment she found her mother on her way out, with newly applied lipstick and blow-dried hair, her suitcase and purse in her hand.
After her mother finally left, and after walking Spot, Fanny lay down on her bed and stared at the ceiling.
Alone again. No one cared about her. Why did she even exist? She had an alcoholic mother who thought only about herself. As if that weren't enough, recently she had started giving some serious thought to her mother's extreme mood swings. One day she was as happy as a lark and full of energy, only to change the next day into a limp dishrag. Depressed, listless, and filled with dark thoughts. Unfortunately the bad days were getting more frequent, and that was when she would turn to the bottle. Fanny didn't dare criticize, because it always ended with her mother having a fit and threatening to kill herself.
Fanny had no one to talk to about her problems. She didn't know where to turn.
Sometimes she dreamt about her father, imagining that one day he would suddenly appear in the door, saying that he had come to stay. In her daydream she saw him embracing her and her mother. They celebrated Christmas together and went on vacations. Her mother was rosy-cheeked and happy and no longer drank. In certain dreams they would be walking along a beach in the West Indies, where her father was born. The sand was chalk white, and the sea was turquoise, just like in the colorful travel magazines she had seen. They watched the sunset together, with her sitting between her parents. That was the sort of dream that she never wanted to end.
She gave a start when Spot jumped up on the bed and licked away her tears. She hadn't even noticed that she was crying. Here she lay, all alone, with only a dog for company, when other families were having a cozy time at home. Maybe her classmates were visiting each other, watching a video or TV, listening to music or playing computer games. But what kind of life did she have?
Only one person had shown the slightest interest in her. She might as well see him again. To hell with everything. She would sleep with him, too, if that's what he really wanted. There had to be a first time, after all. He had said that he would call her tonight. The invitation to go horseback riding still stood, and she decided to say yes.
She got up and dried her tears. Heated up a meat pie in the microwave and ate it without much enthusiasm. Turned on the TV. The phone was silent. Wasn't he going to call after all? Now that she had made up her mind? The hours passed. She took a can of Coke out of the fridge, opened a bag of chips, and sat down on the sofa. It was nine o'clock, and he still hadn't called. She felt like crying again, but couldn't squeeze out more than a few dry sobs. He had probably given up on her, too. She started watching an old movie as she ate the whole bag of chips. Finally she fell asleep on the sofa with the dog beside her.
The sound of the phone ringing woke her. At first she thought it was the landline, but when she picked up the receiver she realized it was her cell phone ringing. She got to her feet and hurried out to the entryway to rummage through her jacket pockets. The phone stopped ringing. Then it started again. It was him.
"I have to see you...I have to. Listen here, honey. Couldn't we meet?"
"Sure," she said without hesitation. "You can come over here. I'm home alone."
"I'll be right over."
She regretted it the moment she saw him. He reeked of liquor. Spot started barking but soon gave up. The dog wasn't the menacing type.
She stood awkwardly in the center of the living room, unsure what to do, as he threw himself onto the sofa. Now that she had invited him over, she couldn't very well ask him to leave, could she?
"Would you like anything?" she asked uncertainly.
"Come here and sit down," he said, patting the sofa cushion next to him.
From the clock on the wall she saw that it was two in the morning. This whole thing was crazy, but she did as he said.
It took only a second before he was on top of her. He was rough and determined.
When he forced himself into her, she bit herself on the arm to keep from screaming.
## Friday, November 23
At the next day's morning meeting everyone was talking about the discovery of the murder weapon. It was a breakthrough in the investigation, of course. By all accounts, the blotches on the hammer were blood. The hammer had been sent to the Swedish Crime Lab for DNA analysis. But there were no fingerprints.
Most of them had seen on the evening news how the hammer was discovered. Naturally Kihlgård made jokes about the police officers' comments that were caught on tape, and he drew a good deal of laughter from the others. Knutas was only moderately amused. He was annoyed by the extent of the information presented in the news story. At the same time, he understood that the reporter was just doing his job. It was so typical that Johan should end up right in the thick of things. He had an incredible talent for showing up exactly when things were happening. Everything had gone so fast out there that no one had thought of reining him in before it was too late. Yet, once again Johan had provided new facts that would benefit the investigation, even though the police didn't know the source of his report about the witness at the harbor. After the case with the serial killer that past summer, Knutas had learned to trust the persistent TV reporter, although Johan could drive him crazy with all the information he managed to dig up. How he did it was a mystery. If he hadn't become a journalist, he would have made an excellent police detective.
The news program had started off with a long segment about the murder, the latest developments in the investigation, the payments Dahlström had received under the table, and the witness who had seen Dahlström with an unidentified man down at the harbor.
"Why don't we start with the unreported carpentry work?" said Norrby. "We've interviewed four people who hired Dahlström in addition to Mr. and Mrs. Persson. Two of them are members of the same folklore society as the Perssons. They all said more or less the same thing. Dahlström did a number of minor jobs for them. They paid him for the work, and that was that. Evidently he conducted himself in an exemplary manner, showed up when he was supposed to, and so on. They knew, of course, that he was an alcoholic, but he had been referred to them by friends."
"So it was through a referral from others that they got in touch with him?" asked Wittberg.
"Yes, and none of them had any complaints about his work. We're going to keep questioning people."
"The murder weapon wasn't the only thing we found yesterday. We also found his camera. Sohlman?"
"It's a professional camera, a Hasselblad. Dahlström's fingerprints were found on it, so we're confident that it did in fact belong to him. There was no film in it, and the lens was broken, so someone had treated it rather roughly."
"Maybe the murderer took the film," Jacobsson put in. "The darkroom had been searched, which indicates that the murder possibly had something to do with Dahlström's photography."
"Possibly. At the same time, we've received reports from SCL on the samples that were taken from Dahlström's apartment and darkroom. SCL have really outdone themselves—we've never received such quick results before," Sohlman murmured to himself as he leafed through the documents. "All the prints from glasses, bottles, and other objects have been analyzed. Many are from Dahlström's buddies who visited his apartment. But there are also prints that can't be ascribed to any of them. They may be from the perpetrator."
"Okay," said Knutas. "At least we know that much. As if the information about Dahlström's unreported carpentry work wasn't enough, Johan Berg has also found a witness claiming to have seen Dahlström with a man down at the harbor last summer. Unfortunately, this witness does not want to talk to the police."
From his notes he rattled off the description of the man at the harbor.
"They were standing in a narrow passageway between two containers and talking, around five in the morning. The witness recognized Dahlström and knew that this was far away from the places where he usually hung out. What do you think?"
"If there's one witness, there could be more," said Wittberg. "When exactly did this happen?"
"We don't know. Only that it was supposedly in the middle of the summer."
"Why was the witness down at the harbor so early in the morning?" asked Kihlgård.
"He was there with a girl who was going to take the morning ferry to Nynäshamn."
"So we're talking about a younger man. It might be one of Dahlström's neighbors. Wasn't there a young guy living in the building?"
"You're right about that. I think he lives on the floor upstairs."
Knutas glanced down at his papers.
"His name is Niklas Appelqvist. A student."
"If the witness, whoever he is, could at least tell us the name of the girl, then we could find out what day she left by looking at the passenger lists of Destination Gotland," said Jacobsson. "I think they keep the lists for three months."
"But how are we going to proceed if the witness doesn't want to talk to the police?" asked Norrby.
"Maybe the reporter would have better luck getting the information out of him than we would," said Jacobsson. "I think we should first ask Johan Berg for help. Maybe the witness is one of those types who's extremely hostile toward the police. For some inexplicable reason, those sorts of people do exist," she added sarcastically.
She turned to Knutas, giving him a big smile.
"So we're going to have to suck up to the reporter," she said gleefully. "And you're so good at that kind of thing, Anders."
Jacobsson gave him a friendly poke in the side. Kihlgård looked equally amused.
Knutas was annoyed, but he had to admit that she was right. Legally, he couldn't investigate the young man, but there was nothing to prevent him from asking Johan to find out the name of the girl. So the police were at the mercy of the journalist's goodwill. And that was a pisser.
Just as Johan entered the editorial offices of Regional News, his cell phone rang. It was Knutas.
"I wonder if you'd be willing to help us with something."
"What is it?"
"Do you think the witness would remember the name of the girl he was with when he saw Dahlström and another man down at the harbor?"
"I don't know. It sounded as if she was someone he spent only that one evening with."
"Could you ask him?"
"Sure. But it'll have to wait awhile. I just arrived at the newsroom."
The police wanted his help. How nice. This was a switch from the normal situation when, as a journalist, he had to beg, plead, and cajole to get any information. He would keep Knutas waiting for just a bit.
A pleasantly drowsy Friday mood had settled over the newsroom. Fridays often had a slower pace than usual because half of the evening news program was devoted to a longer story.
Grenfors was sitting alone at the big table in the middle of the room, the so-called news desk. It was the workplace for editors, anchormen, and broadcast producers—all the key people whose job it was to put together the programs, make decisions, and assign the stories. At this time of day the anchormen and producers hadn't yet put in an appearance. Most of the reporters were sitting at their own desks with phones pressed to their ears. In the morning they did their research and made appointments for interviews. The day often started off at a leisurely pace, which then accelerated and finally reached a crescendo of stress right before the broadcast. That's when they had to deal with stories that weren't finished in time, something in a report that had to be changed at the last minute because the editor wasn't happy with it, computers that crashed, video-editing machines that broke so that certain images couldn't be transmitted, and all sorts of other problems. Time was short, and they always worked up until the very last second. Everyone was used to that; it was their normal work tempo.
"Hi, there," Grenfors greeted Johan. "That was a good report yesterday. Great that we've got the story now. It feels like it's going to get bigger. We'll have to wait and see how it develops. Meanwhile...something else has come up."
The editor shuffled through the documents and newspapers that were heaped in a big, messy pile on the table.
"The police seized a record amount of Rohypnol in Kapellskär this morning. Could you look into it?"
Oh, right, look into it, thought Johan. That sounded easy enough, but he knew what Grenfors expected. A substantial story that he could use at the top of the broadcast, containing information that was a Regional News exclusive. He had strong doubts that it was a record amount. He had lost count of all the drug busts that had been made over the past year.
"Isn't National News doing the story?" he asked wearily. He had been hoping to go home early.
"Sure, but you know how they are. They do their report and we do ours. Besides, you have better contacts than all their reporters put together."
"Okay."
Johan went back to his desk. Before he got started, he called Niklas Appelqvist in Gråbo.
He answered at once. Yes, he had kept in touch with the girl for a short time. He might still have her last name and phone number somewhere. He recalled only that her first name was Elin and she lived in Uppsala. He promised to call back as soon as possible. Before Johan could pick up the receiver to call the Customs Agency, the phone rang. He heard his mother's voice.
"Hi, my dear boy. How are you? How was it on Gotland?"
"It was fine."
"Did you see Emma?"
"Yes, as a matter of fact, I did."
He was close to his mother, and by this time she knew almost everything about his complicated relationship with Emma. She listened and offered advice without expecting that he would follow it. She never judged him, and he appreciated that.
Johan's relationship with his mother had deepened after his father died of cancer almost two years ago now. There were four brothers, but Johan was the oldest, and he was closest to his mother. They had a need for each other. During the past year his mother had needed him more, and they had spent a great deal of time together, talking about his father and how life had changed. Especially for her, of course. She now lived alone in the big house in the suburb of Bromma. He had tried to persuade her to move so that she wouldn't have to take care of all the practical matters by herself. Her sons did help out quite a bit, but they also had their own lives.
She had now recovered from the worst of her grief. She had even started seeing a man who belonged to the same bowling club. He was a widower, and she seemed to enjoy his company. Whether there was anything romantic going on between them she had never mentioned, and Johan didn't want to ask. The fact that his mother was seeing this man took a lot of the pressure off because he no longer had to worry as much about her being alone.
Fanny was sitting at the kitchen table, looking at the reflection of her face in the window. She was alone. Her mother was at work, as usual. The neighbors across the courtyard had hung up their Advent stars already. In another month it would be Christmas Eve. Yet another Christmas alone with her mother. Other people got together with family and friends to celebrate with Christmas trees and presents. The coziest thing of all must be to sit around a big table and eat Christmas dinner together. A warm apartment, candles, and good company. But she and her mother had only each other. And Spot, of course. They never went to visit relatives. Fanny had begun to realize why. The relatives were afraid that her mother would either get drunk or have one of her outbursts. She was so unpredictable that no one could ever relax when she was around. They never knew what might happen. If someone said or did something that her mother took as a criticism, the rest of the evening would be ruined. That's why she and her mother were always alone. Not even her maternal grandmother was around anymore; she was senile and lived in a retirement home.
They never bought a real tree for Christmas, either. They just set up a dreary-looking plastic tree on the table, as if they were a couple of old retired people. They usually ate Christmas dinner in front of the TV. Store-bought meatballs, beet salad, and ready-made Jansson's Temptation, the traditional casserole of herring, potatoes, and onions in a cream sauce. All they had to do was heat it up in the microwave. Her mother would drink aquavit and wine and get more and more tipsy as the evening wore on. There was always some movie on TV that she wanted to see, but before long she would fall asleep on the sofa. Fanny would have to take Spot out for his evening walk. She hated Christmas. The fact that it was also her birthday didn't make matters any better. She was going to turn fifteen—that meant she was practically grown up. She felt like a child in an adult's body. She didn't want to get any older; she had nothing to look forward to. She leaned her head on her hands, inhaling the scent of her newly washed hair. In some strange way she found that comforting. She looked down at the curve of her breasts. They had caused all the problems; her body had ruined everything. If she hadn't gotten older, this whole thing would never have happened. Her body was a weapon that could be used both against others and against herself.
And him. Now she mostly felt sick whenever she thought of him. His sweaty hands would paw at her, wanting to get under her clothes; he whimpered and whined like a baby. He wanted to do all sorts of strange things with her, and she didn't dare protest. She felt disgusted with herself, revolted. He told her that now they were both involved, and she had to keep quiet about what they did together. He talked as if they shared a secret agreement, a pact. But that's not how it was. Deep in her heart, she knew that. He said that he needed her, that she was important to him, and he gave her presents, which she had a hard time resisting. And that made her feel guilty. She was equally at fault, and she had only herself to blame. But now she didn't want to go on. She wanted to get away from him, but for the life of her she couldn't imagine how to do that. In her daydreams she wished that someone would come around the corner and rescue her from everything. But no one ever showed up. She wondered what her father would say if he knew.
She went into the bathroom and opened the medicine cabinet. Spot followed and looked up at her with his sweet eyes. She took out the green box of razor blades and sat down on the toilet seat. Carefully she took out a blade and held it between her fingers. Tears welled up, hot and salty, and rolled down her cheeks to land on her lap. She held out one hand and studied her fingers. What use was this hand? The blue veins ran from her wrist and into her palm filled with her blood, which pumped through her body. How meaningless. Why was she born? To take care of her mother? So that some disgusting old man could paw at her?
She looked at Spot, and that was enough to make him wag his tail hesitantly. You're the only one who likes me, she thought. But I can't keep on living just for the sake of my dog.
She took a firm grip on the razor blade and pressed it against her leg, almost level with her kneecap. She wanted to watch it pierce her skin. She pressed harder and harder. It hurt. At the same time, it felt good, almost liberating. All her fear and pain collected there, in her leg instead of in her whole body. In one place. Finally the blood began to flow, running down her leg and onto the floor.
He saw Emma at once, as soon as she came through the door. He watched her for several seconds while she looked around. The restaurant was small, intimate, and very crowded. He was sitting in a corner at the back, and it was hard to see him from the entrance. Then she noticed him, and her face lit up. To think it was possible to be so beautiful. She was wearing a moss green jacket, and her hair was wet from the rain. It was unusual to see her in a restaurant in Stockholm, and he liked it.
They kissed. Her lips tasted of salty licorice, and she laughed into his mouth.
"What a day! I couldn't concentrate on anything. I didn't hear a thing they said. All I wanted was to get out of there. The course I was taking had absolutely nothing to say to me."
"Were the speakers boring?"
He could feel that his whole face was smiling.
She threw out her hands. "I'm sure they were brilliant, inspiring, and super-charismatic. Everybody else was very pleased. But for me, none of that mattered. I just sat there thinking about you and longing to get away."
Their hands met across the table, and Johan couldn't get his fill of looking at her.
This is how it should always be, he thought. On the ring finger of her left hand her wedding band gleamed, a reminder that he only had her on loan. Just as their food arrived, her cell phone rang. Johan could tell at once that it was her husband, Olle, calling.
"It was good," she said. "Interesting speakers. Mmm. I'm sitting here having a glass of wine with Viveka. Mmm. We're leaving soon. The banquet doesn't start until eight."
She glanced at Johan. Then she got a worried look on her face.
"What? He does? That's too bad. When did it start? Hmm. How high is his temperature? Oh no. Try to get him to drink some fluids...Is he throwing up, too? How typical that he should get sick when I'm not home. Aren't you supposed to play a match early tomorrow morning? Uh-huh. Okay. You and Sara aren't sick, too, are you? If he keeps on like that, you should probably give him some fluid-replacement mixture. Do we have any in the house? Hmm. I hope you get some sleep tonight."
"That was Olle," she explained unnecessarily. "Filip has the stomach flu. He's been throwing up all afternoon."
She took a sip of her wine and looked out the window. Just a quick glance, but enough for Johan to realize that everything was much more complicated than he wanted to believe. She had children that she shared with her husband and she always would. He had watched her as she talked on the phone, and he understood how much of an outsider he was. What did he know about childhood illnesses? He didn't even know Emma's children. They had no relationship to him.
After dinner he wanted to show her around. It had stopped raining, and they strolled down to Hornstull beach, past Reimersholme, and out to Långholmen. Even though it was dark, they walked across the Bridge of Sighs, along the path past the old Mälarvarvet, and over to the other side. The lights from Gamla Stan, the city hall, and Norr Mälarstrand were reflected in the water.
They sat down on a bench.
"Stockholm is so damn beautiful," said Emma with a sigh. "The water makes it seem like it's not a big city, even though there are so many people. I could see myself living here."
"You could?"
"Yes. I'm always so jealous when you tell me about everything going on here. All the people, the theater, the cultural events. It makes me really think about what I'm missing when I'm on Gotland. It's nice there, but nothing ever happens. And just the idea that I could be anonymous. I could sit here in a café and no one would recognize me. Just blend in with everyone else. Watch people and be entertained. And I don't really think the traffic is so bad. It must be the water," she said, looking out across the dark mirror of Riddarfjärden.
"Yes, I love this city. I always will."
"And yet you would be willing to move to Gotland?" she said, looking at him.
"For your sake, I would do anything. Anything at all."
When they went back to his apartment and got into bed like an ordinary married couple, Johan was struck by a feeling of unreality mixed with joy. They should be able to go to bed like this every night.
## Saturday, November 24
Saturday started out with snow mixed with rain, a strong wind, and the temperature hovering just above freezing. Knutas and his children had made breakfast and put a bouquet of flowers on the table next to Lina's place. Each of them was holding one of her birthday presents, and they had cleared their throats to make sure that their creaky morning voices would be able to handle the birthday song. On their way upstairs they started singing "Happy Birthday," each of them in a different key.
Lina sat up in bed, still dazed with sleep, her red hair in a cloud around her head. She gave them a big smile and looked with delight at the presents. She was childishly excited about receiving gifts and started with the ones from Petra and Nils: a book, nail polish, and a calendar with cute firefighters holding kittens. Lina had been in love once before, with a firefighter. The children liked to tease her about her weakness for men in uniforms. She saved the present from her husband for last. Knutas watched his wife with anticipation. He'd had trouble coming up with something, but then a brilliant idea had occurred to him. There was one thing that he knew she really wanted. In spite of countless diets and halfhearted attempts to start exercising, she hadn't managed to lose any weight. Consequently, he had filled a box with everything that might help her out: a year's membership to Gym 1 in Visby, a jump rope and weights for exercising at home, and an introductory package to Weight Watchers.
When Lina realized what his present was, her expression darkened and red blotches appeared on her throat. Slowly she raised her head and met her husband's eyes.
"What's all this supposed to mean?" Her eyes narrowed.
"What do you mean?" he stammered uncertainly and then began listing all the advantages of his gift. "You wanted to slim down, so here's everything you need. If you don't have time to go to the gym, you can work out at home, and Weight Watchers has a meeting for new members on Tuesday at Säve School. Plus you get a personal trainer for the first five times at the gym, so you'll learn how to use the machines correctly."
Knutas pointed eagerly at the brochure that was attached to the gift card.
"So you think I'm fat? That I'm not attractive anymore? Is that why you're giving me all these things? Because you want me to be more buff?"
Lina sat bolt upright in bed, and her voice rose to a falsetto. Startled, the children looked from one parent to the other.
"But you're always talking about wanting to lose weight. I just wanted to help you out."
"And you think this is the sort of thing that I'd want for my birthday? To be reminded how fat I am? Can't you at least let me enjoy my day?"
Now she was shouting and she had tears in her eyes. The children decided to leave the room.
Knutas lost his temper.
"What the hell is this? First you go on and on about weighing too much, and then when I give you things to help you lose a few pounds, you get mad. What the hell is that all about?"
He stomped downstairs and started banging the breakfast dishes around. Then he shouted to Lina, "Just ignore the whole thing. I'll take everything back. Forget all about it!"
He called to the children, "Here's breakfast, for anyone who wants it!"
"And what about you? Have you ever taken a look at yourself?" Lina yelled from upstairs. "I could buy you an arm exerciser for Christmas. And maybe some Viagra—that wouldn't hurt!"
Knutas didn't bother to reply. He could hear Lina still muttering angrily to herself upstairs. Sometimes he got really fed up with her hot temper.
The children came downstairs and ate their cornflakes in silence. Knutas spilled coffee on the tablecloth, but he didn't care. He looked at Petra and Nils. All three of them shook their heads in agreement. None of them could understand Lina's reaction.
"Go upstairs and talk to Mamma," said Petra after a while. "This is her birthday, after all."
Knutas sighed but followed his daughter's advice. Fifteen minutes later he had persuaded his wife that she wasn't at all fat, that he loved her just the way she was, and that she wasn't the slightest bit over-weight. No, she wasn't.
She was afraid of him. It started when he discovered the cuts.
They had done it again, in their secret place. The sexual act was a torment for her. Pain and disgust in a violent combination. It was as if she took pleasure in punishing herself. When he was done and lay next to her, gasping, he took hold of her wrist.
"What's this?" he said, sitting up on the sofa.
"Nothing."
She pulled her hand away.
He grabbed both of her hands and held them out.
"Were you trying to kill yourself?"
"No," she said, ashamed. "I just cut myself a little."
"What the hell for? Are you crazy?"
"No, it's nothing."
She tried to pull her hands away, but she couldn't.
"Did you cut yourself just for fun?"
"No, it's just something that I do. I've done it for years. I can't stop."
"Are you out of your mind!"
"Maybe I am."
She tried to laugh it off, but the laugh got caught in her throat. Fear was blocking the way.
"You can't keep doing this—you know that, don't you? What if someone finds out? Your mother or a teacher at school or someone else? Then they'll start asking a lot of questions. And you might not be able to keep quiet about us. They can manipulate you and coax you into talking. They might call in a bunch of psychologists and shit!"
His voice had gotten so loud that he was shouting. Saliva flew from his lips. He suddenly seemed dangerous, unpredictable. She drew the blanket tightly around her and watched him anxiously.
"No one is going to notice," she objected quietly.
"That's what you think. It's just a matter of time before someone sees those cuts. I forbid you to do it again. Do you hear me?"
He fixed his eyes on her. They were dark with anger.
"Okay, I promise. I'll stop."
He shook his head and went into the bathroom. She stayed on the sofa, unable to move as her panic grew. When he came back he had calmed down. He sat down next to her and stroked her arm.
"You can't keep doing this," he said in a gentle voice. "You might really hurt yourself. I'm worried about you. Don't you realize that?"
"Yes," she said. Tears were stinging her eyes.
"Now, now, honey," he consoled her. "I didn't mean to be so harsh. I was shocked when I saw those cuts, and I'm afraid of losing you. So I don't want to see any more of this, okay?"
He put his hand under her chin and looked her deep in the eyes.
"Promise me, my little princess."
She shuddered inside but nodded obediently.
In the car on the way back, she was convinced that she would never agree to see him again. In her mind she went over and over how she would phrase the words. She practiced the lines like a broken record.
He stopped a block away from her building, and turned off the engine. He wanted her to come and sit in the front seat for a last embrace before they parted. Lately he had made her sit in the back because he was afraid they might be seen.
When he had his nose pressed between her breasts, she gathered her courage.
"I think it's best if we don't see each other anymore."
Slowly he raised his head.
"What did you say?"
"I think it's best if we don't see each other anymore. We have to stop this."
His eyes grew dark and his voice turned icy.
"Why are you saying this?"
"Because I don't want to see you anymore," she stammered. "I just don't want to."
"What the hell are you saying?" he snarled. "Don't want to! What are you talking about? What do you mean by 'don't want to'? It's you and me!"
"But I don't want to meet anymore. I can't do this anymore."
Now she just wanted to get out of the car. His aggressive tone scared her. She tried to open the car door.
"You little bitch. Who the hell do you think you are?"
He threw himself at her and grabbed her hard by the arms. With his lips pressed close to her ear, he snarled, "Do you think you can just stop seeing me? You better be damn careful, because you're treading on thin ice. Don't think you can just start setting the terms. I'll fix things so that you never set foot in that stable again—do you understand? One word from me, and you won't be able to show your face there ever again. Is that what you want?"
She tried to pull herself out of his grip.
"Let me make one thing damn clear—our relationship is over when I say it's over. And not a word about this to a single person, or you can say good-bye to the stable forever. Just keep that in mind, you little slut!"
He pushed her away from him. Sobbing, she finally managed to open the door and stumble out of the car.
In the next instant he was gone. The last thing she heard was the tires screeching as he turned the corner.
Emma looked at her husband over the rim of her wineglass. They were still sitting at the table, talking after dinner as they usually did on the weekend. The children were watching Little Stars on TV, quite happy with bottles of Coke and a big bowl of popcorn. Olle seemed content. Was it really possible that he didn't suspect a thing?
He refilled her glass. How absurd, she thought. Yesterday I was sitting just like this with Johan.
"That was certainly delicious," he said.
She had served lamb burgers with yogurt sauce and homemade baba ghanoush. There was now a Lebanese restaurant in Visby, and they had tried it out on one of the rare occasions when they went out to dinner. The chef had given her the recipe when she asked him for it.
Yet another dinner in the long series of meals that they had shared. Olle asked her to tell him about the course she had taken in Stockholm, and so she did. They'd hardly had any time to talk since she had come home.
"How long did you stay at the banquet?"
"Oh, not very long," she replied evasively. "I don't know what time it was. Maybe one."
"Did you leave with Viveka?"
"Yes," she lied.
"Huh. I called your hotel this morning, but you weren't there. And your cell phone was off."
She felt a burning sensation shoot through her body. Now she was going to have to tell another lie.
"I must have been eating breakfast. What time did you call?"
"Eight thirty. I couldn't find Sara's sneakers."
He kept his eyes fixed on her. Emma took another sip of her wine to gain some time.
"That's when I was in the breakfast room. The battery on my cell had run out, so I left it in my room to recharge."
"Oh, so that's what happened," he said, sounding satisfied.
A perfectly natural explanation. Of course that was what happened. His trust in her had been built up over many years. Why should he doubt what she told him? She had never given him any cause to do that.
The lies burned inside her, and for her the relaxed mood was now gone. She started to clear the table.
"Hey, sit down," he objected. "That can wait."
Their conversation moved on to other topics, and her feeling of uneasiness soon disappeared. They put the children to bed and watched an exciting thriller on TV. She curled up on the sofa with Olle's arm around her, the same as always. And yet it wasn't.
## Sunday, November 25
Things finally fell apart the following morning. Emma's cell rang while she was in the shower, and Olle checked to see what the message was.
It said: "How are you? Longing for you. Kisses, Johan."
When she came out of the shower, Olle was sitting at the kitchen table. His face was white with fury, and he was holding her cell phone in his hand.
The floor gave way beneath her. She realized at once that he knew. Through the window she saw the children playing outside in the rain.
"What is it?" she asked in a feeble voice.
"What the hell is going on?" he said, his voice thick with anger.
"What do you mean?"
She could feel her lower lip quivering.
"You got a message," he shouted. "On this!" He waved her cell in the air. "From some Johan who is longing for you and sending you kisses. Who the hell is Johan?"
"Just wait and I'll explain," she pleaded as she cautiously sat down on the very edge of a chair across from him.
At that moment she heard the front door open.
"Mamma, Mamma, my mittens are wet," cried Sara. "Can I have another pair?"
"I'm coming," she called. She went out to the entryway and found another pair. Her hands were shaking.
"Here, sweetheart. Now go back out and play with Filip. Mamma and Pappa need to be alone to talk. So why don't you and your brother stay outside for a while. I'll call you when we're done."
She gave her daughter a kiss on the cheek and then went back to her husband in the kitchen.
"I've wanted to tell you, but it's been so difficult," she said, giving him an entreating look. "I've been seeing somebody for a while, but I'm so confused. I don't really know what I feel."
"What the hell are you talking about?"
His words cut right through her. She could hear how Olle was trying to control his anger by clenching his teeth. She didn't dare look at him.
"It can't be true! This is too fucking unbelievable!" he said.
He got up from the table and came to stand in front of her, still holding her cell phone in his hand.
"What the hell is going on here? Who is he?"
"He's the journalist who interviewed me after Helena was killed. The journalist from TV. Johan Berg," she said quietly.
Olle flung the cell phone to the floor with all his might. With a bang it was transformed into a pile of plastic and metal splinters. Then he turned to her.
"Have you been seeing him ever since then? Behind my back? For all these months?"
His face contorted with anger as he leaned toward her.
"Yes," she said weakly. "But you have to let me explain. We haven't been seeing each other the whole time."
"Explain!" he shouted. "You can explain to your lawyer. Get out! I want you out of here!"
He grabbed her hard by the arm and yanked her out of the chair.
"Get out! You don't belong here anymore. Leave right now, so I don't have to look at you. Go to hell! I never want to see you again! Do you hear me? Never!"
The children had heard the ruckus, and they now appeared in the doorway. At first they looked bewildered, then they both started to cry. That didn't stop Olle. He shoved Emma out onto the porch in her stocking feet and threw her jacket and boots after her.
"Here!" he yelled. "But you're not taking the car!" And he snatched away her car keys.
Then he slammed the door shut.
Emma put on her jacket and boots. The door opened again and her purse came flying out.
She was out in the cold. The street was deserted.
A Sunday morning in November, and it was over. She stared at the closed door. Her purse had fallen open and the contents were scattered all over the porch and front steps. Mechanically she gathered up everything, too numb to cry. She walked down to the gate and opened it, then turned right, although she didn't know why. She didn't notice the neighbor family a couple of houses away who were talking and laughing as they climbed into their car and drove off. The mother waved to Emma but got no response.
She felt empty inside, as if stunned. Her face felt rigid. What on earth had she done? Where should she go now? She couldn't go back to her own house.
The sports field next to the school was deserted. The wind was blowing from the north. She looked over at the main road where a few cars were driving past.
When did the buses go into town on Sundays? She had never needed to ask that question before.
## Monday, November 26
The temperature in the sauna was 176 degrees Fahrenheit. Knutas filled a wooden ladle and tossed more water on the glowing hot stones. The temperature rose even higher.
They had swum a mile and were more than satisfied. Once a week Anders Knutas and Leif Almlöv would go swimming together, at least in the wintertime. Knutas swam regularly at Solberga Baths during all seasons of the year. He actually preferred to swim alone. He always thought more clearly when he was in the water, swimming one lap after another. But this was a way for the two of them to meet. They had to put up with a good deal of joshing from their friends because they went to the swimming pool—something that was more typical for women. Men played tennis or golf together, or they went bowling.
In the sauna they would discuss all sorts of daily trivialities, or just sit in utter silence. That was the sign of a good friend, Knutas thought. He didn't care for loud people who insisted on jabbering incessantly, even when they had nothing sensible to say.
Knutas described Lina's birthday fiasco, which gave Almlöv a good laugh. They would never completely understand women—they could certainly agree on that.
They had sons the same age, and they talked about the problems of puberty that had started showing up. Their sons were classmates and friends. A week or so ago Almlöv had discovered them smoking in secret. It turned out that they had lit a couple of old cigarette butts. Almlöv's son, who wore his hair long—to the dismay of his parents—had managed to burn several locks on one side.
They talked about their surprise at getting older, about the anxiety of bulging stomachs and slack muscles, about getting gray hair on their chests. Knutas didn't think about old age and death very often, but sometimes he noticed how life seemed to be running away from him, and then he would wonder how much time he had left. He pictured himself getting older and older, with all the accompanying infirmities and immobility. How long would he be able to remain active? When he was thinking along those lines, he would start worrying about the fact that he smoked, although not much. Mostly he sucked on an unlit pipe, filling it and tending to it, but lighting it only a few times each day.
Almlöv was struggling with the same anxieties, even though he didn't smoke. He told Knutas that he had bought a home gym, and he was working out for an hour every morning. The results were quite evident, as Knutas noted with envy. He appreciated his friend's candor and the fact that he could confide in him. But when it came to Knutas's job, other rules applied. And Almlöv never asked him about his work. Even so, Knutas sometimes wished he could tell his friend about one thing or another. It was often good to talk to someone outside police headquarters, someone who had a different perspective. Lina was usually the one who served as his sounding board. She had helped him many times to think along new lines.
It was eleven o'clock by the time Knutas arrived at his office. On his desk was a handwritten note from Norrby along with the transcription of an interview from the Uppsala police. The young woman who was with the witness at the harbor had been tracked down to an address there. Only one passenger of the right age and from that city had taken the boat on the day in question. Her name was Elin Andersson. The Uppsala police had apparently agreed to assist the investigative team by interviewing her over the weekend. She had conceded that she knew Niklas Appelqvist and that they had been together at the harbor on the morning of July 20 before she left. But she had not noticed anyone in particular down at the harbor. So it was as they had guessed—Dahlström's young neighbor was the one who had provided Johan Berg with the information. Knutas was extremely annoyed that such an important witness refused to talk to the police. And it wasn't because he'd been in trouble with the police. A search of police records had come up negative.
When he entered the conference room half an hour later, Knutas noticed at once a sense of excitement in the air. Jacobsson and Kihlgård had gone through Dahlström's papers over the weekend, and from the look on their faces, it was clear that they had found something that they were dying to share with their colleagues. Kihlgård had two big cinnamon rolls on a plate in front of him next to a big mug of coffee. He ate as he fiddled with the papers. Crumbs fell on the table.
Knutas sighed. "Do the two of you have something to report?"
"You better believe it," said Kihlgård. "It turns out that Dahlström kept detailed records on his clients. We have a long list of names and dates, what he built, and how much he was paid."
"The work he did was much more extensive than we thought," added Jacobsson. "He had been doing carpentry jobs for over ten years. His first job was in 1990. Some of the people who made use of Dahlström's services are very well-known in Visby."
Everyone gave Jacobsson their full attention as she held up a list of names.
"Would you believe—wait till you hear this—city council chairman and Social Democrat Arne Magnusson?"
A gasp of surprise rippled through the room.
"Magnusson?" said Wittberg, laughing. "That can't be true! The guy who's always defending high taxes and talking about how great it is to pay them? That's too funny! He's the worst moralizer in all of Visby."
"Yes, he's always lobbying for the restaurants to close at one A.M. in the summertime and for smoking to be banned," snickered Sohlman.
"If this gets out...the journalists are going to have a field day." Norrby threw out his hands.
"A garden shed in 1997," read Jacobsson from the list. "Five thousand kronor, paid under the table, along with several bottles of liquor. Can you believe it?"
Knutas grew serious. "This is totally insane."
"Just wait. There are more surprises on the list," said Jacobsson. "Bernt Håkansson, chief surgeon at the hospital, and Leif Almlöv, restaurant owner and your good friend, Anders!"
"What the hell?" Knutas turned bright red in the face. "Is his name on there, too?"
"A sauna in the country for ten thousand—that was a tidy sum."
There was a glint of mischief in Jacobsson's eye. She was enjoying teasing him. Kihlgård looked equally pleased. They had certainly found something to entertain themselves. How nice for them.
"At least he's not alone. There are at least a dozen names."
"No one from here, I hope?" said Wittberg uneasily. "Don't tell me that, for God's sake."
"No, luckily there aren't any police officers on the list. On the other hand, there is someone with your last name. Roland Wittberg. Are you related to him?"
Wittberg shook his head.
"Let me see," said Knutas.
He recognized a number of the names.
"What are we going to do with this?"
"To start with, we'll check up on them and see if there are any other links to Dahlström," said Jacobsson, snatching the list back.
Knutas called Leif as soon as he was back in his office. He felt tremendously out of sorts.
"Why didn't you tell me that you had hired Dahlström?"
Silence.
"Are you there?"
"Yes."
Knutas heard a deep sigh on the phone.
"Why didn't you say anything about the sauna?" Knutas persisted.
"You know how it is with all the crooked dealings in the restaurant business. I thought that if it came out that I had hired an illegal worker for a private matter, then people would think I did the same thing at the restaurant. I would come under suspicion and the authorities would make life hell for me."
"Why didn't you think of that before you let him build that sauna?"
"I know it was really stupid. At the time things weren't going well with the restaurant, and Ingrid kept nagging me about the damn sauna. That's no excuse, but maybe it's an explanation of sorts. I hope that I haven't put you in an embarrassing situation."
"I'll manage. Besides, there are others who have reason to be uneasy. We have a list of plenty of people who did the same thing. You wouldn't believe your ears."
After hanging up the phone, Knutas leaned back in his chair and started filling his pipe. He was grateful that no police officers were on the list, and he accepted his friend's explanation. Good Lord, who hadn't done something stupid? Once, many years ago, he had actually swiped a package of underwear from a shop on Adelsgatan. He was standing in the store, holding the package, when he was suddenly seized with a wild impulse to find out what it felt like to steal something. He walked right out the door with the package under his arm. He was so nervous that he was shaking, but when he exited the store, a giddy feeling of joy came over him. A kind of invulnerability. It was as if the act made him untouchable. After he had gone far enough from the store to know that he had escaped undetected, he glanced at the package, only to find that he had taken the wrong size.
Knutas still felt ashamed every time he thought about what he had done. He turned his chair halfway around and looked out the window. Somewhere out there a murderer was walking around.
Nothing indicated that they would find him among Dahlström's circle of acquaintances. On the contrary. Dahlström was apparently mixed up in something, but they had no clue what it might be. Whatever it was, he had done a good job of hiding it. The question was how long it had been going on. Probably not much longer than the date of the first deposit in his bank account, Knutas guessed. July 20. The same day that Niklas Appelqvist saw Dahlström with an unidentified man down at the harbor. It seemed likely that on that occasion the man had handed the money over to Dahlström, who later in the day went to his bank to deposit it. Twenty-five thousand kronor. The next deposit was made in October, and for the same amount. Was it possible that the two deposits didn't actually have anything to do with each other? From the beginning Knutas had assumed that they were connected somehow, but now he was no longer certain. The explanation might be as simple as payments for various carpentry work. But why would someone who had hired Dahlström for something so trivial decide to meet him down at the harbor at five in the morning? It was obvious that the man didn't want to be recognized.
Fanny's muscles were pleasantly tired. Calypso had been wonderful. She had gone out riding, taking her favorite route through the woods even though it was a bit too long for the sensitive racehorse. But never mind. It was so seldom that she went out, and she just couldn't resist.
He was a gentle horse and responded to her prods without the least effort. He made her feel quite proficient. They galloped for long stretches along the soft forest path. Not a living creature as far as the eye could see. For the first time in a long while she had felt something that resembled joy. She felt a surge inside her chest as they raced forward. She stood up halfway in the saddle, urging the horse on. Tears rose in her eyes from the speed. Knowing that they were going faster than she could actually handle made the whole experience even more exhilarating. This was truly living: to see the horse's ears pointing forward, to hear his hooves pounding dully on the ground, to feel the animal's power and energy.
As they went back to the stable at a walk with the reins drooping, she felt so relaxed. She sensed a budding hope that everything was going to be all right. First and foremost, she had to break things off with him for good. He had called her cell about twenty times that day, but she had refused to answer it. He wanted to apologize. She had listened to his messages, and he sounded upset and remorseful. He tried to convince her that he didn't mean what he had said. This morning he had sent a picture message with hearts and flowers. None of that had any effect on her anymore.
It was over, no matter what he said. Nothing could make her change her mind. She had decided to ignore his threat about getting her thrown out of the stable. She had worked there for a year, and everyone knew her. They wouldn't pay any attention to him. And if he tried, she was thinking about revealing everything. By law it was a criminal offense for him to have sex with her; she was fully aware of that. She was no fool. And he was an old man. He might even end up in prison. It would serve him right. It would be so great to be rid of him, to have her body to herself, and to get out of doing all the shit he wanted her to do. She longed to have herself back. Her mother wasn't going to change, but Fanny would soon be fifteen, and she wouldn't have to live at home much longer. Maybe she could even move out next year when she started tenth grade. There were plenty of kids out in the country who did that. They lived in town during the week and went home on the weekends. Why couldn't she do that, too? All she had to do was tell the school counselor or nurse about her situation, and she was sure to get help.
When she gave Calypso a hug in his stall, she felt so grateful to the horse. It was as if he gave her strength and self-confidence. And the faith that everything was going to work out.
She had ridden her bike only three hundred yards when she saw the headlights. He came driving along the opposite side of the road, slowed down, and rolled down the window.
"Hi. Are you on your way home?"
"Yes," she responded, stopping.
"Wait there," he said. "I just have to drive a little farther and turn the car around. Wait right there."
"Okay."
Reluctantly she got off her bicycle and stood at the side of the road. She watched him drive off and had a strong urge to do the same. Just bike home as fast as she could to get away from him. The next second she changed her mind. She was going to tell him it was over. Once and for all.
When he returned, he wanted her to get in the car.
"But what should I do with my bike?" she asked, resigned.
"Leave it in the ditch. No one's going to take it. We can come back and get it later."
She didn't dare do anything but comply. Her legs were shaking as she got into the car.
"I have to go home. Mamma is at work, and I have to take Spot out for a walk."
"No problem. I just wanted to talk to you for a minute. Is that all right?"
He asked the question without looking at her.
"Okay," she said, glancing at him out of the corner of her eye.
His voice sounded strained, and he seemed tense. His jaw moved as if he were clenching his teeth.
She thought he was driving too fast but didn't dare object. It was dark out, with little traffic on the road. He headed south toward Klintehamn.
"Where are we going?"
"It's not far. You'll be home soon."
Fear began to creep into her veins. They were getting farther and farther away from town, and she now realized where they were going. She debated with herself and decided it wouldn't be a good idea to protest. The tense atmosphere in the car told her that it would be best not to.
When they reached the house he told her to take a shower.
"Why should I?" she asked.
"You reek of the stable."
She turned on the shower and the hot water struck her bare skin but she couldn't feel it. Mechanically she soaped up while thoughts zigzagged through her mind. Why was he acting so strange? She dried herself off with a bath towel, trying to rid herself of the uneasiness that crept over her. She told herself that he was just tense because of what had happened last time. For safety's sake she put all her clothes back on. In case she had to run away.
He was sitting in the kitchen reading a newspaper when she came downstairs. That made her feel calmer.
"You put your clothes back on?" he said, his voice stony. He gave her a distracted look—his glassy eyes were fixed on her, but it was as if he didn't really see her.
Her sense of relief vanished instantly. What was wrong with him? Was he on drugs? His question hung in the air.
"Yes," she said uncertainly. "I thought—"
"What exactly did you think, my dear?"
"I don't know. I have to go back...."
"Back? So you thought that we drove all the way out here just so you could take a shower?"
"No. I don't know."
"You don't know. Well, there's plenty that you don't know, sweetheart. But maybe it's just as well that you put your clothes back on. That might make it more interesting. We're going to play a little game, you see. Doesn't that sound like fun? You're so young that you like to play games, don't you?"
What had gotten into him? She tried to hold back the fear that shot up inside her, and she made an effort to act normal. It didn't help much. He grabbed her by the hair and forced her down on her knees.
"We're going to play dog and master. You're so fond of dogs, aren't you? You can be Spot. Is Spot hungry? Does Spot want a treat?"
As he talked he used his free hand to unbutton his pants, keeping a good grip on her hair with the other. She turned ice cold when she realized what he wanted. He pressed her hard against him. She felt nauseated but couldn't get away.
After a while he seemed to lose his concentration for a moment. He loosened his grip and then she saw her opportunity. She tore herself away and managed to pull free. Quickly she got to her feet and staggered out to the hall. She yanked open the door and dashed out. A fierce wind struck her. It was pitch dark and icy cold. The sea was roaring in the dark. She ran toward the road but he came after her. He knocked her down and slapped her in the face. He hit her so hard that she almost passed out.
"You damn little whore," he snarled. "Now I'm going to really let you have it."
Again he took her by the hair and then dragged her across the yard. The ground was soaking wet, and the water seeped through her clothes as she was pulled after him on all fours. Holes were torn in her pants, her hands were scraped badly, and blood ran from her nose. The sound of her sobs was drowned out by the howling wind.
Fumbling, he pulled out the key to the little building. The door opened with a screech. Abruptly he shoved her into the dark.
## Tuesday, November 27
When Majvor Jansson came home to her apartment after working the night shift, she discovered that the dog had peed on the hall rug. He jumped up and whimpered when she opened the front door. His water bowl was empty. She could tell at once that something was wrong. The door to Fanny's room stood wide open, and her bed had not been slept in. It was close to seven o'clock on this Tuesday morning, and it was clear that Fanny had not been in the apartment all night.
Majvor sat down on the sofa in the living room to think. Don't panic, she told herself. What was it that Fanny was going to do yesterday? Probably go out to the stable after school. She was always over there lately. They hadn't planned to see each other at home because Majvor had to go to work at five. That meant that Spot had been alone for fourteen hours! Anger bubbled up inside of her, but just as quickly it vanished. As she tried to gather her thoughts, a sense of uneasiness overtook her.
Fanny would never forget to come home if she knew that Spot was alone. Not of her own free will. Had she gone to a friend's house to spend the night? The likelihood of that happening was minimal, but she started looking through the apartment to see if her daughter had left a note. What about a message on her cell phone? She hurried out to the hall and pulled her cell out of her coat pocket. Nothing there, either. Spot had finished eating and was now whining loudly. He needed to go out.
As she walked between the apartment buildings, Majvor wondered what other possibilities there might be. Was Fanny mad at her? No, she didn't think so. They hadn't had a fight in a long time. In her heart she was aware that she might not be the sort of mother that Fanny really needed, but she couldn't help it. This was just how she was, and she didn't have the energy to make changes. It wasn't easy being a single mother.
Was this a sign of some kind of rebellion? Had Fanny run away with some friend that she didn't know about? Or a boy? Majvor hurried home with the dog, who now seemed much happier. She started making phone calls.
An hour later she was still at a loss. None of their relatives or acquaintances could tell her where Fanny had gone. She called the school. She learned that Fanny wasn't there, either. Anxiety was making her throat dry. She got out a bottle of wine and a glass. Something must have happened. What about the stable? Did she even have the phone number? A note with the number was stuck to the refrigerator. Fanny was always so organized. Majvor clutched the receiver tightly as she waited for someone to answer the phone.
"Hello," said a gruff male voice after the tenth ring.
She introduced herself. "Yes, hi, this is Majvor Jansson, Fanny's mother. Is Fanny there?"
As she spoke she realized that she didn't know who she was talking to, or even what the place looked like. Fanny had been working at the stable for over a year, but Majvor had never set foot in the place. Why hadn't she ever gone to visit? Now she cursed herself as it suddenly became crystal clear to her how little interest she had shown in her daughter. When was the last time she had helped Fanny with her homework? She didn't dare even think about that.
"No, she's not here," replied the man, sounding friendly. "She was here yesterday afternoon. But shouldn't she be in school right now?"
"She's not there, and she didn't come home last night, either."
Now the man on the phone sounded uneasy. "That's odd. Wait a minute," he said, and she heard him put down the receiver, then the sound of voices in the background as he shouted to someone. After a moment he was back.
"No, unfortunately, no one has seen her. I'm sorry."
A call to the hospital proved equally fruitless.
What about her room? Normally Majvor didn't go in there, since she and Fanny had a mutual understanding that the room was her private space.
At first glance everything looked the same as usual. The bed was neatly made and a book lay on the nightstand, next to the alarm clock. On the desk was a jumble of pens, several schoolbooks, elastic bands for her hair, scraps of paper, and newspapers. Majvor rummaged among Fanny's things, pulled out all the dresser drawers, then searched the bookshelf and the closet. She turned everything in the room upside down without finding any note with a message, or an address book or a phone number that might give her a clue about where Fanny had gone.
But hidden under several decorative pillows at the head of the bed she found what were obviously spots of blood on the reverse side of the bedspread. She tore off all the bedclothes. No blood on the sheet or the blanket, but under the bed she found a towel with more traces of blood. She was shaking all over as she punched in the phone number for the police.
As soon as he stepped inside, Knutas felt weighted down. He was glad that Sohlman had come with him. The whole apartment was depressing, with its cramped rooms and dreary colors. It was in a four-story building on Mästergatan in the Höken district, in the northeastern section of Visby and about half a mile outside the ring wall.
Majvor Jansson's eyes were red from crying when she opened the door. Since Fanny was not with her father, either, the police were taking the report of her disappearance very seriously. The bloodstains on the bedspread meant that there was reason to suspect an assault or a rape. That's why the police had decided to do a proper crime scene investigation of the girl's room. Sohlman immediately got to work.
Knutas noticed a faint smell of liquor on Majvor Jansson's breath.
"When did you last see Fanny?" he asked when they were sitting at the kitchen table.
"Yesterday morning. We had breakfast together before she left for school. I didn't have to go to work until five, but she usually goes to the stable after school, so we rarely see each other in the afternoon."
"How did she seem?"
"Tired. She's always tired, especially lately. That's probably because she doesn't eat properly. She's awfully thin."
"What did you talk about?"
"Nothing special. There's not much to talk about in the morning. She ate toast for breakfast, as usual. Then she left."
"What was the mood like between the two of you?"
"Same as always," replied Majvor flatly. At the same time she cast a pleading glance at Knutas, as if he might be able to tell her where her daughter was.
"What did she say when she left?"
"She just said bye."
"Is anything missing from the apartment? Clothes, a toiletry case, money?"
"I don't think so."
"And Fanny didn't leave a message? You're sure about that?"
"Yes, I've looked high and low."
"Tell me about Fanny. What's she like?"
"Well, what can I tell you? What are most kids like at that age? She doesn't say much, but I don't think she likes school. She's started cutting classes a lot. Maybe she's lonely. I don't know. She never brings any friends home."
"Why is that?"
"I have no idea. Maybe she's shy."
"Do you ever talk about these problems with your daughter?"
Majvor Jansson seemed disconcerted, as if it had never occurred to her that she was the one responsible for her daughter, and not the other way around.
"It's not easy to talk about things when you're a single mother and have to work all the time. I don't have a husband to support me. I have to do everything myself."
"I can understand that," said Knutas sympathetically.
Suddenly she fell apart and buried her face in her hands.
"Shall we take a break?" asked Knutas tactfully.
"No, we might as well get this over with so that you can start looking for her."
"Have you talked to anyone at her school about why she's been cutting classes?"
"Yes, a teacher called me. That was just a few days ago. He said that she hadn't been to his class for several weeks. We talked about the problem, but he seemed to think that she was just tired of school. I told Fanny that she had to go to school, and she promised to do better."
"Has Fanny mentioned anything new in her life? Someone new that she met?"
"No," replied Majvor, after giving it some thought. "I don't think so."
"Is there anyone in particular that she spends a lot of time with?"
"No, we don't have a big circle of friends, as they say."
"What about relatives?"
"My old mother lives at the Eken retirement home, but she's so out of it that it's almost impossible to talk to her. And I have a sister who lives in Vibble."
"Does your sister live alone?"
"No, she's married and has two children. Well, the son is her husband's, from a previous marriage."
"So they're the only cousins that Fanny has? How old are they?"
"Lena lives in Stockholm. I think she's thirty-two, and Stefan is forty. He lives here on Gotland, in Gerum. I was hoping that Fanny might have gone to stay with my sister."
Majvor started sobbing again. Knutas patted her arm.
"Now, now," he comforted her. "We're going to do everything we can to find her. I'm sure she'll turn up soon. Just you wait and see."
The message on his answering machine was a long one. Emma reported in a cracked, monotone voice that Olle now knew about everything. For the time being she was staying with her friend Viveka. She asked him not to try to contact her, and she promised to call when she could. Johan immediately tracked down Viveka's phone number, only to be told by Emma's friend that he needed to respect her wish to be left alone.
It was a form of psychological terror, and he had a hard time coping with it. He played a game of floorball but couldn't stop thinking about Emma. He went to a movie but left the theater without knowing what the film was about.
On Tuesday evening she called.
"Why won't you talk to me?" he asked.
"My whole life has fallen apart. Isn't that a good enough reason?" she said angrily.
"But I want to help you. I realize that this must be terribly hard for you. And I get so worried if we don't have any contact."
"Right now I can't be responsible for whether you're worried or not. I have enough to think about."
"How did he find out?"
"Your text message. You sent it while I was in the shower, and he checked my cell phone."
"I'm sorry, Emma. I'm really sorry. I shouldn't have texted you on a Sunday. That was stupid."
"The worst thing is that I still haven't had a chance to talk to the children. He won't answer the phone, and he's turned off the answering machine. I've gone over there, but nobody was home. And he took my keys, so I can't even get into the house." Her voice broke.
"Take it easy," he consoled her. "I assume he just has to let off some steam. He's had a shock. Isn't there anyone who could talk to him? What about your parents?"
"My parents! Not a chance. Do you know what he did? He called all of our friends and family and told them that I'd found someone else. He even called my grandmother in Lycksele! My parents are really upset with me. I've tried to talk to them, but they're siding with Olle. They can't understand how I could treat him so badly. And what about the children—why didn't I think about Sara and Filip? Everybody is against me. I don't know how I'm going to deal with this."
"Can't you come here? So you can get away from things?"
"No, I can't."
"Should I come over there?" he asked. "I could take some time off."
"What good would that do? Right now the most important thing is for me to have contact with my children. Do you have any idea what it's like not to be able to talk to your own children? I told you that I needed two months to think things through, but you refused to respect my wishes. You just couldn't let me have some time in peace and quiet. You kept calling and pressuring me, even though I told you not to. And now look what's happened! And it's all thanks to you, for God's sake!"
"So this is all my fault? What about you? Don't you think that you share some of the blame? I didn't force you into this, did I? You wanted to meet, too."
"All you can think about is yourself because you don't have to take anyone else into consideration. But I do. So leave me alone," she said, and slammed down the phone.
He noted that this was the second time she had done that recently.
That afternoon the real job got started of mapping out Fanny Jansson's activities during the past few days before she disappeared. The search was carried out on a wide front. The police interviewed everyone who worked at the stable, as well as the few relatives that she had. They visited her school to talk to her classmates and teachers. Their image of Fanny became clearer.
She appeared to be a solitary girl who would turn fifteen on Christmas Eve. Her classmates didn't think she was interested in being friends with any of them. When they started school together, some of them had tried to get her to join in various activities, but she always declined, and finally they gave up. She always seemed to be in a hurry to get home after school, until she started going to the stable, and then she was in a hurry to go there. No one really had much to say about her. They thought she was probably nice enough, but she never took the initiative to make contact with any of them, and that's why she had ended up alone. She only had herself to blame. She didn't seem to care, and that was also a bit irksome. Nothing seemed to bother her.
The teachers described her as quiet but smart—although lately something had changed. She seemed distracted for no obvious reason, and she had become even more withdrawn. At the same time, it wasn't easy to figure out kids her age. There were so many emotions at play; new patterns emerged, they started talking back, made friends and then dropped them; the boys started using snuff, the girls began wearing makeup and padding their bras, and the hormones practically gushed out of the kids. Irritability and aggression were common, and it wasn't always easy to keep up with all the mood swings or how a particular student was developing.
Her relatives didn't have much to say, either. They seldom saw Fanny. Her mother drank and had an unpredictable temperament, which prevented any sort of normal socializing. Of course they realized that it must be a difficult situation for Fanny, but that didn't mean that they wanted to get involved. They had enough problems of their own, they said dismissively.
Adult responsibility, thought Knutas. There is something called normal, decent adult responsibility. Isn't there any sort of collective feeling among people anymore? Nobody is prepared to deal with a child who goes astray, not even within their own family.
The neighbors all had the same impression of Fanny: a solitary, modest girl who seemed to carry a heavy responsibility at home. It was commonly known that her mother had a drinking problem.
The last person to see Fanny before she disappeared was a man at the stable. His name was Jan Olsson. According to him, she arrived at the stable around four, as usual, and worked with the horses. She was given permission to take one of them out for a ride. She was gone for about an hour and was elated when she returned. She didn't get to go out riding very often, so she was thrilled whenever she had the opportunity. Both she and the horse were sweaty, and Jan Olsson said that he suspected she had galloped harder than she really should have. But he didn't say anything because he felt sorry for the girl and thought she deserved to have a little fun.
When he was taking a cigarette break outside on the stable hill, he saw her pedaling off in the dark, heading toward home. After that there was no trace of the girl.
Knutas decided to go out to the racetrack to meet in person both the trainer who owned the stable and Jan Olsson. By now it was past seven o'clock, and when Knutas called the stable, everyone had left. He tried their home numbers, but no one answered. He would have to wait until first thing in the morning.
## Wednesday, November 28
The trotting track was located about half a mile from the center of town. When Knutas and Jacobsson drove up the stable hill, they came within a hair's breadth of colliding with a sulky. The huge gelding snorted and swerved to the side. The driver's admonishing words calmed the horse. Knutas got out of the car and inhaled the smell of horse and manure. He looked toward the racetrack, which was partially hidden in the cold and damp haze. The grandstands were barely visible through the mist.
On both sides of the stable hill stood rows of stables. A solitary horse was jogging around in an enclosure. A steel contraption of some kind was keeping the horse on the path and regulating its pace.
"It's called a horsewalker," said Jacobsson when she saw Knutas's look of puzzlement. "Horses that aren't going to be taken out riding can still get exercise. They may have an injury or be suffering from a cold or something else that means they shouldn't be ridden as hard as usual. Ingenious, isn't it?"
She led the way into the stable.
The horses had just been given their lunch feed, and the only sound was a pleasant munching along with an occasional stomping. Everything seemed very orderly. The floor was scrubbed clean, and the green-painted stalls were properly closed with locks. Halters hung on hooks outside each door. Shelves were filled with neat rows of supplies: bottles of liniment and baby oil, scissors, rolls of tape, hoof scrapers. Shin guards were stacked in baskets, along with rolls of binding tape, brushes, and other grooming tools. A barrel of sawdust stood in one corner. A black kitten lay on top of a feed box, sound asleep. In one window a radio was playing music at low volume.
They had made an appointment to meet with Sven Ekholm, who was both the trainer and the owner of the stable, but he was nowhere in sight. A stable girl appeared and took them over to a closed door that led to a coffee room.
Ekholm was sitting with his legs propped up on a round coffee table, talking on the phone. He motioned for them to sit down. Daylight was doing its best to penetrate through the dusty windowpanes. Spots of dried coffee marred the red plastic tablecloth. The table was covered with papers, stacks of racing newspapers, vitamin bottles, mugs, glasses, filthy riding shoes, rubber boots, and some three-ring binders. The ceiling was coated with spiderwebs. In one corner there was a kitchenette with a couple of hot plates, a dirty microwave, and a dusty coffeemaker. The walls were covered with finish-line photos of various horses, and a pile of dried roses lay on top of a cabinet. It wasn't hard to see what took priority in the world of these people.
Ekholm took his feet off the table and finished his phone conversation.
"Hello, and welcome. Would you like some coffee?"
They both said yes. Ekholm was a handsome man in his forties. He was muscular and moved with grace. His dark hair was tousled. He was wearing black pants and a gray turtleneck sweater. With some difficulty he managed to find clean cups, and after a moment they each had a cup of coffee; a plastic box of gingersnaps sat in front of them on the table.
"Can you tell us about Fanny Jansson?" Jacobsson began. "We understand that she spends a great deal of her free time here at the stable."
Sven Ekholm leaned back in his chair.
"She's a smart girl who works hard. Not very talkative, but she has a good way with the horses."
"How often is she here?" asked Knutas.
"How often is she here at the stable, you mean?" asked the trainer and then went on without waiting for a reply. "Probably four or five times a week, I would guess."
"When was she last here?"
"Yes, when was she last here?" Ekholm repeated. "I think the last time I saw her was a week ago, maybe on Thursday or Friday."
"How did she seem?"
"How did she seem?" Ekholm rubbed his chin. "I was busy driving, so I just said a quick hello. It might be better if you talked to the others in the stable—they spend more time with her than I do."
"Is Fanny paid for her work here?"
"Is she paid? No, that's how it is with stable girls, you know. They come here because they think it's fun to be around horses. To groom them and take care of them. That's how girls are at that age."
Sven Ekholm took a quick sip of coffee.
"How long has Fanny been coming to the stable?"
"How long has she been coming here? Hmm, maybe a year or so."
"Does she have a particularly good relationship with any of the employees?" asked Knutas, who was starting to get annoyed by the man's tendency to repeat every question.
"Any of the employees that she has a particularly good relationship with? Well, yes, that would be Janne. They seem to get along well. Otherwise she's quite shy, as I said."
"And how often are you here?" asked Jacobsson.
"Hm, what should I say? Twenty-five hours a day," he said with a grin. "Well, practically every day. I've been trying to take at least one day off every other weekend. I do have a wife and kids, too—I can't just live at the stable."
"How well do you know Fanny?"
"Not very well. She doesn't exactly welcome contact. I always have so much to do that I can't just sit around chatting with all the young girls who come here."
Why didn't Ekholm repeat the questions when Jacobsson asked them? Knutas found it enormously annoying.
"Where do you live?" Jacobsson went on.
"Right nearby. We've taken over my father's farm. Well, my father still lives there, in the guesthouse."
"Does your wife work at the stable, too?"
"Yes, she does. We have six full-time employees, and she's one of them."
"How is the work divided up?"
"We all help each other, training the horses and taking care of them, and lending a hand around the stable. It's a full-time job all year-round, even when the racing season is over."
"We'd like to talk to everyone. Can you arrange that?"
"Sure, no problem. Right now it's just me and Jan, I'm afraid. But later in the day, or tomorrow."
Knutas realized that he would have to ask one more question, just to see if the trainer had decided to stop repeating them.
"How many others work at the stable? Girls who work for free after school, and so on?"
"Girls who work for free after school, and so on? Well, we have quite a few of them. We used to have more, but it doesn't seem to be as popular as it once was. Or else maybe they have too much homework lately," said the trainer, giving Knutas a smile.
As they left the coffee room, Jacobsson noticed that her colleague's expression was as dark as a thundercloud.
The interview with the stable hand, Jan Olsson, went better.
The man was slightly older than the trainer, maybe forty-five, Knutas guessed. He was darker than most Swedes. Brown eyes that were almost black, distinct eyebrows that grew together, and a stubble that looked to be several days old. Wiry and muscular from years of working with horses. Not an ounce of fat on his body—that was evident from the shirt and dirty pants that he had on. He was not wearing a wedding ring. Knutas wondered if he lived with anyone but decided to wait to ask that question. Instead, he asked him to tell them once again what happened when Fanny left the stable. Olsson gave the same account as had been recorded in the previous report.
"Try to recall any details you can," said Knutas. "Anything that might seem insignificant could actually be important."
Jan Olsson ran his hand over the stubble on his face. He made a very frank and sympathetic impression.
"No, I really can't think of anything. She takes care of the horses and doesn't usually talk much. When she came back from her ride, she was happier than I've seen her in a long time. Her eyes were actually shining. After grooming Calypso and taking care of the harness, she said good-bye and left on her bike."
"What do you think might have happened to her?"
"I don't think she committed suicide, at any rate. She was much too happy and upbeat when she left here. I have a hard time imagining her going off to kill herself."
"How well do you know her?"
"Quite well, I think. She seems to like being here, but I understand that she doesn't have an easy home life. She's always in a hurry to rush home because she has to take the dog out. As I understand it, her mother is rather difficult, but I've never met her."
"Has Fanny ever talked about any friends or anyone she hung out with?"
"She doesn't seem to have any friends, since she spends all her time over here. Those of us who work in the stable are much older. Although she sometimes talks to Tom, who works in the next stable."
"Is that right?"
"I've seen them talking to each other on the stable hill once in a while. They seem to get along. Fanny isn't exactly the most open person, so I notice when she talks to anyone."
"Are they the same age?"
"God, no. He must be thirty, at least. He's American but I think he's lived in Sweden for a long time. You can tell because of the way he speaks Swedish."
"What's his last name?"
"Kingsley."
"And how long has he worked here?"
"At least a year, maybe more."
Tom Kingsley was busy wrapping the hind leg of a horse when they entered the adjoining stable. Knutas and Jacobsson kept back a safe distance.
"We've heard that you know Fanny Jansson, the girl who has disappeared. Is that right?" Knutas began.
"Well, I can't say that I really know her. I've talked to her once in a while."
He didn't look up, just went on with his work.
"We need to ask you a couple of questions."
"Sure, I just need to finish this. I'm working on the last leg right now."
In spite of a distinctly American accent, his Swedish was fluent. When he was done, he stood up with a grimace and stretched out his back.
"What do you want to know?"
"How well do you know Fanny Jansson?"
"Not very well. We talk occasionally."
"How did you happen to meet each other?"
"Good Lord, we both work here. Of course we would see each other around the stables. We're always running into each other."
"What do you talk about?"
"Mostly about the horses, of course. But other things, too. How she's doing in school and about her home, and things like that."
"How do you think she's doing?"
"Not great, actually."
"What do you mean by that?"
"She complains about her mother, says that things are tough at home."
"In what way?"
"She told me that her mother drinks too much."
"So she has actually confided in you a great deal?"
"I don't know about that."
"Have you seen each other outside the stable?"
"No, no. Just here."
"Do you know whether she has met anyone new lately? A boyfriend, maybe?"
"I have no idea."
"When did you last see her?"
"It was on Saturday."
"Where?"
"Here, outside." He nodded toward the stables.
"How did she seem?"
"The same as usual."
"Do you have any idea where she might be?"
"Not a clue."
There was no one else at the stable to question. They left Tom Kingsley and went back to their car.
"What do you think happened?" asked Knutas as they drove back to police headquarters.
"It's possible that she might have killed herself."
"I have a hard time imagining that. She's too young. Fourteen-year-old girls who commit suicide are rare. They're usually at least a couple of years older. Besides, she didn't seem particularly depressed, even though things might have been worse than they seemed on the outside. I think all three men at the stable seem credible, although the trainer was damned irritating."
"I agree," said Jacobsson. "I didn't get any weird vibes from any of them."
By the afternoon Fanny had still not turned up. Her mother called Knutas to hear how the search was going. She was distraught. Her sister in Vibble, south of Visby, had stepped in to look after her. Knutas decided to begin searching the areas surrounding Fanny's apartment, her school, and the stable. A bulletin was broadcast on the local radio station and immediately attracted the interest of the media. Radio Gotland and both of the local newspapers, Gotlands Tidningar and Gotlands Allehanda, wanted to interview him.
Knutas tried to be generous with the press and agreed to brief interviews.
He dealt with one journalist after the other, and they all asked basically the same questions. He kept the interviews short, telling them only when Fanny had disappeared, where she was last seen, and what she looked like. He asked the reporters to say that the police were appealing to the public for help.
The search brought results. Fanny's bicycle was found by a passerby. It had been tossed into a ditch less than a kilometer from the stable. It was immediately taken in so that the techs could examine it.
Johan Berg also called.
"Hi. Am I disturbing you?"
"I'm very busy at the moment."
"I'm calling about the girl who disappeared. It just came over the wire service. What happened?"
Knutas gave him the same information that he had given to the other journalists, but he also told Johan about the bicycle. He thought he owed him that much.
"Do you suspect foul play?"
"Not at the moment."
"Do you think she might have committed suicide?"
"We can't rule out that possibility, of course."
"What's her home life like?"
"She and her mother live alone in an apartment here in Visby."
"Is she an only child?"
"Yes."
"The description says that she has a dark complexion. Was she adopted, or is her mother from some other country?"
"Her father is from the West Indies."
"Where does he live?"
"In Stockholm, with his wife and kids. They don't have any contact with each other."
"Could she have gone there?"
"We've talked to the father, of course. And she's not there."
"She could still have gone to Stockholm," said Johan.
"Sure."
"Did she take along any money, or her passport?"
"There's nothing to indicate that. All her belongings are still at home," replied Knutas impatiently. Why couldn't Johan Berg ever be satisfied with the same information he gave to all the other journalists? He never gave up asking more questions.
"The fact that her bike was found tossed aside could mean that she got into a car. Was it found near a road?"
"That's right. I have to go now."
"I realize that you've got your hands full, what with the murder investigation, too. Is there anything to indicate she might have fallen into the hands of the same perpetrator as Dahlström?"
"Not at the moment."
Knutas shook his head as he put down the phone. What a stubborn man that journalist was.
The next second the phone rang again. The switchboard told him that a woman from the youth clinic in Visby wanted to talk to him. He told the operator to put her through.
"Hi, my name is Gunvor Andersson, and I'm a midwife. The girl that I think you're looking for was here recently."
"Is that right? How do you know it was her?"
"I recognized her from the description on the radio. She was here several months ago, asking for birth control pills."
"Did she say why?"
"She said that she had a steady boyfriend. I asked her whether she really felt old enough to have intercourse. I said that we usually don't recommend the Pill for such young girls. She said that they had already done it. I told her that since she's under fifteen, it's a crime to have sexual intercourse with her. On the other hand, we can't very well refuse to give the Pill to a girl who wants to protect herself. We usually require a parent's consent in the case of such young girls, but when I said that I would have to call her mother, she didn't want anything more to do with us. She just got up and left. I tried to stop her, to say that we could talk about it, but before I knew it, she had walked right out the door."
"Did you find out who her boyfriend was?"
"No, unfortunately. She refused to say anything about him."
After Knutas finished talking to the woman, he called Majvor Jansson.
"Did you know that Fanny has a boyfriend?"
"No, I'm sure she doesn't."
"She went to the youth clinic to ask for birth control pills."
"What?"
"Yes, I've just talked to someone over there. She went there several months ago to get a prescription for the Pill, but when they told her that they would have to contact you, she left. I need you to think about this some more. Was there anything to indicate that she had a boyfriend? Was she spending time with anyone?"
There was a long silence on the other end of the line.
"She never said anything about it. But it's hard to keep tabs on her, because I work nights and I'm a single mother. She could always meet someone in the evening, since that's when I'm at work."
Majvor Jansson was clearly about to start crying again.
"I was thinking of trying to get a different shift, now that she's getting older. But I didn't think there was any danger yet. She's only fourteen, after all."
In the meantime, the search continued. A hundred volunteers offered to help the search-and-rescue groups that had been organized at various sites. The sense of alarm about what had happened to Fanny was growing with every hour that passed.
At 8:00 P.M. the investigative team gathered for a meeting at police headquarters. The mood was tense. Knutas told them about his phone conversation with the woman from the youth clinic and Fanny's failed attempt to obtain birth control pills. Sohlman, who looked worn out, told them about the results of searching Fanny's room.
"We've found three packets of morning-after pills hidden among the clothes in Fanny's closet. Two were empty; one still had both pills. That proves that she has had intercourse with someone."
"It doesn't take much detective work to come to that conclusion," Jacobsson interjected acidly. "But morning-after pills? Aren't they supposed to be used in extreme emergencies? Surely they're not meant to be used for birth control?"
She glanced around the room. When she saw the blank expressions on the faces of her colleagues, she realized that she worked with a bunch of middle-aged men who had all been cast from the same mold and who probably knew nothing about how that sort of pill worked.
"How many pills did she take?" asked Jacobsson, turning to Sohlman.
"There are two in each package, and from what I understand, that counts as one dose. So she took four pills, or two doses."
"Where do you get them? In a drugstore? Can a fourteen-year-old go out and buy them? Don't you have to be at least fifteen?"
No one at the table could answer Jacobsson's questions.
"All right," she said with a sigh. "I'll call the youth clinic."
Her colleagues looked relieved to get out of hearing any more embarrassing questions that they couldn't answer.
Sohlman went on. "Bloodstains and hairs that are not hers were found on the bedspread. They are short, dark, coarse hairs. In her bed we also found sperm and pubic hair, but we can't say yet who they're from. Everything has been sent to SCL. We also sent over some things that her mother didn't recognize and couldn't explain where Fanny had gotten them."
He read from a list: "One bottle of perfume, one necklace, several rings, one sweater, one dress, and two pairs of underwear. Quite sophisticated underwear, I might add," he said, clearing his throat. "We haven't found anything of interest on her bike."
When Sohlman fell silent, a heavy mood settled over the room. Their apprehension that Fanny was in trouble had been significantly reinforced by his report.
Wittberg broke the silence. "What the hell should we do?" he said with a resigned sigh. "What do we have to go on?"
"There's plenty we can do," Knutas objected. "While we wait for the lab results, we need to expand the search area. Tips have been coming in from the public, and they have to be processed."
"How should we divide up the work between the Dahlström investigation and this case?" asked Norrby.
"We'll work on them in tandem. We've done that before. Don't forget that we don't know what's happened to Fanny Jansson. She might turn up tomorrow."
When Johan came home from work on Wednesday evening, he found to his surprise that Emma was sitting on the steps. She looked pale and hollow-eyed, wearing her yellow quilted jacket.
"Emma, what are you doing here?" he exclaimed.
"I'm sorry that I was so mad yesterday, Johan. I just don't know what to do."
"Come inside."
She followed him in and without a word sank down on the sofa.
"I'm about to lose my footing altogether. Olle still won't let me talk to the children. I was thinking of going over to their school yesterday, but the school counselor advised me not to. She thinks that I should wait. I've talked to their teachers, and the children seem to be doing all right. The only thing they seem to know is that we're going through a crisis, and that I've taken a leave of absence from my job."
She pushed back her bangs. "Is it okay if I smoke?"
"Sure, go ahead and smoke. Do you want something to drink?"
"Yes, please. A glass of wine or a beer, if you have any."
Johan took two beers out of the fridge and sat down next to her.
"What are you thinking of doing?"
"That's exactly what I don't know," she said, sounding annoyed.
He touched her cheek.
"Have you quit your job?"
"I called in sick. Without giving any explanation. My job feels like the least important thing at the moment."
"Olle will calm down. You'll see. Don't worry about that. After a while you'll be able to talk to each other again."
"I just don't understand why he reacted so strongly. He's shown so little interest in me and our relationship during the past few years. He really shouldn't be surprised. But to hell with him. The only thing I can think about is Sara and Filip. You have no idea how tough this is."
He reached out his hand and caressed her cheek.
She grabbed his hand, kissed it, and put it on her breast. When he kissed her, the response was fierce. It was as if she were hungering for him, for physical contact, for solace. He wanted to transmit his own strength to her, to give her the energy she obviously needed. There was something disconsolate and desperate about the way she made love to him that night.
Afterward she fell asleep, curled up in his arms like a child. For a long time Johan lay in the dark, looking at her profile and listening to her breathing.
## Thursday, November 29
The media's interest in the disappearance of Fanny Jansson continued to grow as the hours passed. More and more people became involved in the search groups, and the police were using helicopters and infrared cameras in the woods around Visby as they intensified their search. On Thursday morning both evening papers ran big articles about the missing girl. Her picture dominated the front pages.
When Johan came into the Regional News editorial offices, he was met by Grenfors waving several newspapers in his hand.
"What the hell is this?" he shouted. His face was bright red. "Both Aftonbladet and Expressen have big spreads about the missing girl. Weren't you supposed to keep on top of this story?"
"Could you let me take off my jacket first?" Johan snapped back. He had waited at the Hornstull subway station for twenty minutes for a train that never came. The red line was having problems again. And then Stockholm Local Traffic had the nerve to raise the price of a monthly pass.
Grenfors stubbornly followed him as he went to his desk.
"How come we didn't have anything to report?" he continued, standing behind Johan.
Since Johan was painfully aware that he had been concentrating too much on Emma and too little on his job lately, he had no good answer. She had flown home this morning, and it would probably be a while before they saw each other again.
"I'll make some calls and check things out," he said.
"Maybe there's a connection to the murder of that alcoholic. The killer is still on the loose, after all."
"Do you think I should go over there?" asked Johan hopefully.
"That depends on what you find out."
He got out the local papers from the stack of dailies and listened to the Radio Gotland morning news on the Internet. It was true that they were reporting that Fanny Jansson was still missing, but the police also seemed to be working with a number of new clues. It was the same story as in the newspapers, which had reported how the search was being conducted and the fact that the girl's bicycle had been found.
It was damn stupid that he had been so lax at keeping tabs on the investigation. Regional News was now way behind in reporting the story. It was a big disadvantage that he wasn't on site in Gotland and able to follow developments. The evening papers were both speculating, of course, whether the same person who had murdered the alcoholic might have struck again.
With a sigh he picked up the phone and punched in Knutas's number. No answer, and his cell was turned off. Damn it. He tried Karin Jacobsson. He had dealt with her quite a bit during the summer. She sounded stressed.
"Jacobsson here."
"Hi, this is Johan Berg from Regional News. I wonder how it's going with the search for Fanny Jansson."
The voice on the other end of the line softened. Johan realized that he was still in the good graces of the Visby police, at least for the moment.
"We're working on a wide front. The search is now under way in the area around her school, her apartment building, and the racetrack, which is where she was last seen. But so far the results have been meager. We've found her bicycle, but I'm sure you already know that."
"Yes. Are there any prints on it?"
"You'll have to take that up with Anders Knutas. He's the only one who can decide what we tell the media."
"I've been trying to reach him, but he doesn't answer his phone."
"No, he's in a meeting with the new officers from the National Criminal Police right now. It will probably go on for another hour."
"Have you brought in more personnel from the NCP? Why is that?"
"As I said, you'll have to talk to Knutas."
"Okay. Thank you anyway. Bye."
He leaned back in his chair. The fact that the police were receiving more help from the NCP meant that they were taking a serious view of the case. Something else must have come to light, indicating that a crime was involved. He got up and went over to the desk where Grenfors was sitting with a phone pressed to his ear, as usual.
Sometimes Johan wondered how much time he wasted waiting for people to finish talking on the phone. He noticed that Grenfors had dyed his hair again. The editor had recently turned fifty, and he was meticulous about his appearance. He was always dressed in a sporty and youthful manner. On principle, he never ate lunch with his colleagues; instead, he preferred to make use of his pass to the gym in the television building. He was tall, slim, and trim. He looked good for his age. Max Grenfors was married to an attractive woman who was fifteen years younger and an aerobics instructor.
When the editor finally put down the phone, Johan told him what Jacobsson had said.
"Let's wait and see what Knutas has to say. It's too late for you to go over there today, unless they have something really significant to report. From here you can put together some text for the anchorman, so that we can at least keep the pot boiling. You and Peter can fly over tomorrow if it seems worthwhile."
That evening Johan went out with his friend Andreas. They started at the Vampire Lounge on Östgötagatan, where the drinks were cheap and the atmosphere relaxed. The female bartender had short cropped hair and was dressed all in black and wore big earrings. When she turned around to rinse some glasses, a tattoo was visible at the small of her back. She mixed each of them a frozen margarita in a glass with a spiral stem. The bar was filled with a relatively young crowd, most of them with a pack of Marlboro Lights in front of them on the bar. In the restaurants at lunchtime hardly anyone ever smoked, but in the evenings nearly everyone had a cigarette hanging from their lips.
"You seem a little out of sorts," said Andreas after they had run through the usual chitchat about their work and various sports events.
"Not really, just a little tired," said Johan as he lit a cigarette, like everyone else in the place.
"How are things going with your Gotland girlfriend, Emma?"
"Good, but it's tough, too, you know. With her husband and kids and everything."
Andreas shook his head. "Why are you getting mixed up with a married woman who has little kids? And who lives on Gotland! Could you make your life any more complicated?"
"I know," said Johan with a sigh. "But you don't understand because you've never really been in love with anyone."
"What the hell to you mean? Of course I have. I was with Ellen for five years," Andreas protested.
"Sure, but what do you really know about love? You had your doubts the whole time. You were always grumbling about one thing or another. The fact that she was a vegetarian, that she was always late, that she was messy, and that she didn't seem to have any plan for her life. And the fact that she kept studying and studying, but it never led anywhere and she never had any money. Have you forgotten about all that?"
Andreas let out a roar of laughter.
"Of course not, but do you know what she ended up doing? I ran into her downtown a month ago. Newly married with a baby on the way. She lives in Saltsjöbaden and is head of a big advertising agency. And on top of that, she's damn cute!"
"You see? You never can tell about anyone!" said Johan, laughing.
They started talking to three cheerful girls from Västberga, and then they all continued on to Kvarnen, the legendary Södermalm pub. Johan ran into some of his journalist colleagues and got into such an intense discussion about worldwide current events that both Andreas and the girls got bored and left.
When Johan caught a cab home around three in the morning, Emma was once more on his mind. What was she doing right now? He wanted to send her a text message but restrained himself. They had agreed that it would be her turn to call next.
## Saturday, December 1
Olle had suddenly called and invited her home for dinner. Finally she would be able to see the children. It was not even a week since she had last seen them, but it felt like a month—at least. She had called the night before and had a chance to talk to them for the first time since she was thrown out of the house. Both Sara and Filip sounded happy and strangely unaffected, in spite of everything that had happened. She wondered what was going on inside their young heads.
During the week various scenarios had fluttered through her mind. One moment it seemed right to get a divorce; the next she wanted nothing more than to be a family again, and she wished that she had never met Johan.
In the middle of everything she became very aware of the fragility of life. She was surrounded by stage sets that were ostensibly solid, but they could crumble at any moment and completely change everything.
At the same time, she was struck by her own stupidity. What was she thinking? That she could have an affair on the side, just to satisfy her own need for validation? She hadn't realized that she was playing with fire.
Was she prepared to sacrifice everything for Johan? She should have asked herself that question after the first kiss.
Her husband had given her his love, he had taken responsibility, he had kept the promise he had made when they got married. But what about her?
When he had reacted by throwing her out, the ground opened up beneath her.
Right now she had no idea what to think. Except that she was eager for the meeting with Olle to go well. She was deathly afraid that he was going to do something final, such as handing her divorce papers. There was something in Olle's voice when he called, a different tone that indicated something had changed. And it made her nervous.
She felt like a stranger on the evening of her visit—a guest in her own house. Olle looked happy when he opened the door. He took her coat and hung it up as if this were the first time she had ever come to the house. The situation was absurd. Irritation was just seconds from becoming visible on her face. But then the children came running out to the hall.
She was showered with soft kisses and fierce hugs. She loved holding their warm bodies close and breathing in their scent. Both children were eager to show her the gingerbread house that they had made with Pappa.
"Oh, how lovely," she told the children as they pointed to the towers and pinnacles. "It looks like a real castle!"
"It's a gingerbread castle, Mamma," said Filip.
Olle stood in the doorway. He was wearing an apron, his hair was disheveled, and he looked like a marvelous father. She felt an instinctive urge to give him a hug, but she controlled herself.
"Dinner is ready. Come on, let's eat."
When they had finished eating and the children were sent out to watch cartoons on TV, Olle refilled their wineglasses.
"Well, I've been wanting to have a proper discussion with you. That's why I asked you to come here tonight. I didn't want to talk on the phone."
"Okay," she said cautiously.
"I've been thinking and thinking. At first I was so mad. I never thought you would do something like this to me. When I found that text message, it made me see red. I really felt as if I hated you, and I wanted to tell the whole world about what you had done. It was as if I'd been living a lie. How could I have been so fucking stupid and not suspected anything? It was all so damn crazy. Not to mention how I felt about that jerk from TV. So many times I've been on the verge of going to Stockholm to rip him to shreds."
He took a sip of wine.
"But in spite of everything I realized that there was nothing to gain by punching him in the mouth. Maybe an assault charge, but that would undoubtedly make him happier than it would me."
Emma couldn't help smiling.
"My anger faded after a couple of days, and then I started to think more clearly. I thought about us, how we are together. I've replayed our whole life in here."
He tapped two fingers against his temple.
"Everything we've done together and all my feelings for you. I've come to the conclusion that I don't want to. Get a divorce, that is. Even though you've hurt me terribly, because you really have. But no matter how bad it is, I realize that I'm also partly to blame for the whole thing. I haven't paid much attention to you, I haven't listened when you wanted to talk. Not that I'm excusing what you did, but that might have contributed to it. It will take a while before I can trust you again, but I'm prepared to try."
Emma was totally confused. This was not what she had expected.
"Olle, I don't know. It's all so sudden. I don't know what to say."
"You don't have to say anything. But at least now you know what I want," he said and got up to make coffee.
They drank their coffee in front of the TV with the children and then put them to bed. She left the house without voicing any decision, either to Olle or to herself.
## Sunday, December 2
Five days had passed since Fanny Jansson had disappeared, and there had been no progress. The girl was still missing. With each day that passed, the police became more and more convinced that there had been foul play. Knutas's frustration grew. Not only did his mood get worse, his sleep was also affected. It was Sunday and the first day of Advent, but he was already awake by six o'clock. He had slept badly, with a hodgepodge of dreams. The dream images had merged into one another: Henry Dahlström with his head bashed in, Fanny Jansson wandering through the Botanical Gardens, Martin Kihlgård from the NCP chewing on pork chops served by Prosecutor Birger Smittenberg. Everything became jumbled together in his groggy mind, and he awoke exhausted, not knowing where he was or what time it was. He found himself staring at his wife's ear and realized that the whole thing had been a dream. Maybe it was the wind that had made him uneasy. It was roaring and howling over the roof, whistling through the rain gutters.
The weather had turned in the night. The wind was now coming from the north, and the temperature had dropped several degrees. Outside it was pitch dark, and snow was whirling in the gusty wind. Lina stretched out in bed next to him.
"Are you awake?" she asked, sounding sleepy.
"Yes. I was having such strange dreams."
"About what?"
"I can't really remember. It was just a mishmash."
"My poor boy," she murmured, nuzzling the back of his neck. "It must be your work that's getting to you. And look at this weather. Are you hungry?"
She was mixing Danish words with Swedish. He liked to tease her by saying that she still sounded as if she had oatmeal stuck in her throat when she talked. But he had adopted quite a few Danish words and expressions himself, and the children spoke an odd blend of Gotland Swedish and Danish.
When they sat down at the breakfast table he clearly noticed the pain. An aching, throbbing pain on the insides of his elbows, around his wrists, and at the backs of his knees, which confirmed the change in weather. It was a pain that he had lived with for as far back as he could remember. After the new weather conditions had gone on for a few days, the pain would vanish as quickly as it had appeared. There was no explanation for it, and no one in his family had experienced anything similar. By now Knutas was so used to it that he didn't think much about it. It was worse when the weather changed from warm to cold, like today.
He poured himself another cup of coffee. The uncertainty about Fanny Jansson was still gnawing at him.
Some of his colleagues were guessing it was suicide. He didn't believe in that theory, but as a matter of routine he had checked out several commonly used spots. One of them was Högklint outside of Visby, a steep precipice with a sharp drop to the sea below. But their search had turned up nothing.
As for the murder of Dahlström, they had made no further progress. The investigation had come to a standstill, and the only positive thing was that the media's interest in the case had begun to wane.
The impasse meant that Knutas could afford to take a day off and spend it with his family. Christmas was right around the corner. It was Shop Window Sunday, and they had made plans to meet Leif and Ingrid Almlöv to take a stroll downtown.
Knutas had been looking forward to forgetting all about the investigation, but the Almlövs immediately started talking about it.
"It's so horrible, the story about that girl who's missing," Ingrid began as soon as they had said hello. "She works at the stable where my father has his horse, Big Boy. Actually, we own half the horse."
"We own it together, but your father is the only one who's really interested. He was the one who wanted to buy it," said Leif.
"Well, it's terrible, at any rate. What do you think has happened to her?" asked Ingrid, turning to Knutas.
"It could be anything. Maybe she was in an accident, maybe she killed herself, or maybe she ran away from home. It doesn't have to involve any sort of crime."
"But you think that it does?"
Knutas didn't reply. Lina jumped in and started talking about the Christmas decorations that had been put up all over town.
The stores had made a real effort to create a holiday atmosphere. The wind had now subsided and the falling snow made everything look magical. Garlands of evergreen boughs were hung overhead between the buildings, and lights attached to the branches cast a warm glow over the streets. At Stora Torget, booths had been specially set up for the day, selling Christmas candy and handicrafts. Hot glögg and gingersnaps were available. The loudspeakers were blaring Christmas carols, and later in the afternoon people would gather to dance around the big Christmas tree in the middle of the marketplace. A fat Santa was handing out sparklers to the children. Even the smallest shops were open on this Sunday, and they hadn't seen so many people crowding onto the biggest shopping street, Adelsgatan, since high season last summer.
No matter where they turned, they saw familiar faces. They stopped to talk with people on every street corner. All four of them were well-known in Visby—Knutas in his capacity as detective superintendent, Lina as a midwife, and the Almlövs as restaurant owners. They went into a café to have hot cocoa with whipped cream and saffron rolls.
Knutas's cell phone rang. It was Karin Jacobsson.
"We've heard from Agneta Stenberg. She's the girl who works at the same stable as Fanny Jansson, but she's been away on vacation. She came home today, and she says that Fanny has a relationship with that man Tom Kingsley."
"What sort of proof does she have?"
"I've asked her to come in and talk to us. I thought you might want to be here."
"Of course. I'll be there in ten minutes."
Agneta Stenberg sat down on the sofa in Knutas's office, across from Knutas and Jacobsson. Her dark suntan was accentuated by her white turtleneck sweater. How on earth has she managed to get such a tan in only one week? thought Jacobsson.
Agneta got right to the point.
"I think that they're more than friends. I've seen them hugging and carrying on several times."
"Are you sure?"
"Of course I am."
"What do you mean by 'carrying on'?" asked Jacobsson.
Agneta squirmed nervously. She looked embarrassed.
"It's the kind of things that you notice. They stood very close together. You could see him stroking her arm. Intimate gestures that only happen between people if something is going on. Do you know what I mean?"
"Yes, we do," said Knutas. "When did this start?"
"They met on the stable hill and they've been talking to each other for a long time. It might have been in October that I noticed them hugging for the first time. It was near one of the outdoor stalls, a short distance away from the stable. It made me really uncomfortable, to tell you the truth. I mean, he's at least twice her age."
"What makes you think that there was anything strange about it? They could just be friends, giving each other a hug."
"I don't think so. When they caught sight of me, they let go of each other. And after that I've seen them hugging at other times."
"Did they do anything else?"
"No, not that I saw."
"Have any of you talked about this at the stable?"
"I mentioned it to a couple of people, but they thought it was probably just friendly hugging, that they were just friends."
"Why do you think they thought that?"
"It's because she's so young. No one could imagine that a nice guy like Tom would be seeing her. Everybody thinks he's so great."
"But you don't?"
"Oh sure, there's nothing wrong with him, but that doesn't mean that he might not be taking advantage of Fanny. She looks older than she is."
"Have you ever asked Fanny about her relationship with Tom?"
"No."
"What about Tom?"
"No. But maybe I should have." She gave them a solemn look. "What do you think has happened to her?"
Knutas's expression was worried as he replied. "We don't know," he said. "We really don't know."
Knutas called Tom Kingsley and asked him to come down to the station. He seemed reluctant but promised to be there within the hour.
"Maybe Kingsley is the secret boyfriend," said Knutas to Jacobsson as they had coffee and sandwiches while waiting for him to show up.
"It's possible," said Jacobsson between bites. "But why didn't he say anything about being close to her when we talked to him at the stable?"
"Maybe he was ashamed. I would be if I was seeing a fourteen-year-old girl."
"If it's true that they have a relationship, that alone would make him a suspect. If you're thirty years old and you start having an affair with a fourteen-year-old, there's something seriously wrong. That much is clear."
Tom Kingsley seemed nervous and tense when he at last showed up almost two hours later. He was wearing his stable clothes, and Knutas was bothered by the horse smell.
"I'm sorry about my clothes, but I've come straight from work," said Kingsley, as if he had read their minds.
"That's okay," lied Knutas. "When we met you at the stable the other day, you described your relationship with Fanny as superficial. You said that you don't know each other very well. Do you stick by what you said?"
Kingsley hesitated.
"Yes...you might say that."
"But you don't seem quite so certain anymore."
"That depends on what you mean."
Knutas felt a growing irritation. He found people who lied to his face tremendously annoying.
"In what way?"
"What does it mean to know someone well? I'm not sure."
"You said that you usually just chat a little."
"That's right."
"So you don't have any sort of closer relationship?"
"Not really."
"We've been given information that indicates otherwise. We've been told that you've been seeing each other. That you have a romantic relationship."
Tom Kingsley's expression darkened.
"Who the hell has been spreading lies like that?"
"We can't tell you that. But is it true?"
"Who the hell would say such a thing? They're out of their minds!"
"Just answer the question. Do you or do you not have a relationship with Fanny Jansson?"
"That's sick." Kingsley shook his head. "She's just a child, for God's sake."
Knutas was about to lose his patience.
"Yes, that's exactly what we want to know, and we have our reasons," he snapped. "Answer the question."
"Of course I don't. Fanny and I are just good friends. Nothing more. Nobody should be spreading lies that we're seeing each other."
"Why didn't you say anything about this before, when we talked to you the first time? Why didn't you say that you're in the habit of hugging each other?"
"We're not in the habit of hugging each other, damn it."
"But have you ever hugged?"
"I may have given her a little hug, but it was just a way of comforting her. She needs support. The girl has a terrible home life. Her mother drinks, and she doesn't have a father or any brothers or sisters. She has no friends. She's all alone. Can you understand that? She's all alone!"
Tom Kingsley had grown very angry.
"So you deny having a relationship with Fanny. Is that correct?" asked Knutas.
Kingsley merely shook his head in reply.
"How do you explain that people think the two of you have been seeing each other?"
"It must be their sick imaginations. Can't a guy even show a girl a little kindness and concern? This is crazy, damn it! Is Agneta the one who told you this? Agneta Stenberg?"
Knutas and Jacobsson looked at each other in surprise.
"Why would you think that?" they said in unison.
"Because she's jealous, of course. She's been following me around for months, but I told her that I wasn't interested. We had a party for the stable employees a while back, and that's when she really put the moves on me. I finally had to tell her to get lost."
Knutas was amazed at Kingsley's verbal prowess. He spoke perfect colloquial Swedish. If it weren't for a slight accent, anyone would take him for a native speaker.
When the interview was over, Knutas felt disappointed. He had been counting on catching Kingsley off guard so that he would be at a loss to come up with an answer. But that hadn't happened.
## Monday, December 3
There was no new trip to Gotland for Johan. It's just as well, he thought grimly. He hadn't heard a word from Emma all weekend. And yet they had just had such a cozy time together. He couldn't figure her out. If only she hadn't started to waver again.
At the moment Gotland seemed far away, also in terms of his work. Just as Grenfors finally seemed to be paying attention to the Gotland murder case, the police had reached an impasse. And besides, an act of madness had occurred at Stockholm's Medborgarplatsen in Södermalm at the very same time. Late on Monday afternoon, the news-room learned that a madman had gone berserk with a crowbar, killing at least one person. Five others were injured, including an infant. Regional News was tipped off about the event practically as it was happening. Johan immediately took off with a camerawoman. In the car on the way there, he sat with his cell phone pressed to his ear, alternating between talking to the duty officer, emergency services, and the newsroom.
The camerawoman drove swiftly and expertly through the traffic, constantly changing lanes to gain time and occasionally making an illegal move, which was necessary for anyone who wanted to make good time. At Medborgarplatsen she brazenly parked the car right on the open square and instantly pulled out her camera.
Ambulances and police cars were on the scene. The police were starting to cordon off the area, and crowds of people watched in dismay as medics tended to the wounded.
Johan interviewed both the police and witnesses, who said that the man, without any sort of provocation, had started attacking anyone who happened to cross his path. Finally he threw down the iron bar and disappeared down the stairs to the subway station at Björn's Garden. All traffic had been halted, and the police were searching the subway cars and platforms, using dogs.
The newsroom was seething with activity when Johan returned. Grenfors was talking on two phones at once, the program producer was running between video-editing machines to make sure the reports were all ready on time while he also kept in contact with the national news program, which of course was also working intently on the drama in Södermalm.
The idea was for the news programs to collaborate; interviews were divided up among the reporters, clips were exchanged back and forth. The Regional News footage was much in demand, since their camera-woman had been first on the scene. The producer was busy lining up appropriate individuals to interview live in the studio. The county police commissioner was called in, along with the head of the homeless shelter, since many people had gotten the impression that the man who had gone amok was homeless. In the meantime, he was still at large.
Regional News sent a direct feed from Medborgarplatsen. People had started arriving there to light candles and torches and to leave flowers. The casualty count was now at two, since the infant had died from his injuries.
On his way home in the subway, Johan was again struck by the unusual working conditions of journalists. When the most horrible events occurred, they had to set aside their own feelings because their first priority was to report the story. Their professional role took over, but it had nothing to do with a hyena mentality, as some people scornfully implied when they poured out their venom at the media. Johan thought that, like himself, most journalists were driven by a desire to get the story—it was that simple. It was all a matter of reporting, as quickly and accurately as possible, what had happened. It was each reporter's responsibility to gather as much material as he could in order to present the best possible report.
Back in the newsroom, they went through all the material, discussing it with the editors. What was relevant to include in the broadcast and what was not? All close-ups of the wounded were omitted, interviews with people who were clearly in shock were rejected, and anything that was considered an invasion of privacy was cut.
Each day was a new day with more ethical discussions. And behind each news story there was careful deliberation, especially in the case of stories of a sensitive nature, such as this one. Of course there were occasional oversights when a name or a photo was broadcast that should not have been made public; the editors didn't always see the story before it was shown, since time was often tight. Yet for the most part, things went smoothly with regard to the ethical rules that applied to all journalists. Of course, there was always the occasional rotten egg who crossed the prescribed boundaries. Some TV stations and newspapers had stretched those boundaries rather far, but still, this applied to only a handful of reporters.
## Tuesday, December 4
The perpetrator from the Medborgarplatsen attack was caught the next day as he lay sleeping in the corner of a garage in Skärholmen. That gave the media reports about the incident a new impetus.
That's the way the newsroom operated—the hottest story came first, and everything else had to wait. Something that was of intense interest one day could be completely forgotten the next. The list of priorities was constantly being revised at the morning meetings, during the day, and at the onset of each new event. The content of the workday was continuously being changed, renewed, and reversed to take in new points of view. One thing was certain—the job was seldom monotonous.
For that reason, the entire day had passed before Johan could think about Emma. But when he reached home, she once again dominated his thoughts. He called her even though he wasn't supposed to. She sounded tired.
"How are things going?"
"Better. I picked up the kids from school today."
"That's great."
"Yes."
Silence. Johan felt uneasiness settling in his stomach.
"Have you talked to Olle?"
"I'm at the house right now. He's reading a story to the children."
"What are you doing there? Have you moved back in?"
"No, but we have to be able to spend time together. You do understand that, don't you?"
She sounded annoyed, and she was speaking in a low voice, as if afraid that someone might hear.
"So he's not mad anymore?"
"Of course he's mad, but he has calmed down enough that we can talk, which means a lot to me. But I don't want to risk causing any more trouble by talking to you right now. Bye!"
Johan stared at the phone in bewilderment. At the same time the freezing temperature outdoors swiftly moved inside and took up lodging in his guts. All of a sudden she was giving priority to Olle again. She sounded as if he didn't mean shit to her, and that threat sapped him of all energy. He couldn't bear to lose her again.
## Wednesday, December 5
Emma stared at the indicator in her hand. It just couldn't be true. Did two intersecting blue stripes forming a plus sign really mean that she was pregnant? It had been so long since she'd done this sort of test. With a pounding heart she got out the package. The directions couldn't be clearer. A blue line in the window meant not pregnant. Two blue lines intersecting meant pregnant. How could this be possible? She and Johan had slept together only once recently, two weeks ago. And she could hardly remember the last time she had slept with her husband. Frantically she searched her memory. When was the last time with Olle? It must have been last summer. She counted the months since then: August, September, October, November, December. Good Lord, that would make her five months pregnant, and she ought to be showing more than she was. But her period was only three weeks late, and she'd had regular periods all fall. She felt suddenly faint when she realized what that meant. It had to be Johan. That Friday in October. His work had brought him to Gotland, and he had called her up. She was feeling weak and had agreed to meet him at the newsroom before he went back home. They had made love on the sofa. Damn it. How could she have such incredible bad luck? The one time they had given in when they were supposed to be taking a break from each other, and she ended up pregnant. That kind of thing could only happen to her.
She felt tears filling her eyes. This was more than she could take.
She just about jumped out of her skin when someone knocked on the bathroom door. She heard Olle's voice saying, "Emma, are you almost ready?"
"Yes, just a minute."
She tossed the indicator and the empty package in the wastebasket. She couldn't say anything about this right now. She needed time to think. Quickly she washed her hands and opened the door.
"What's wrong? Why are you so pale?" Olle gave her a worried look. "Are you sick?"
"You might call it that. I'm pregnant."
## Thursday, December 13
Every seat was taken in Visby Cathedral on this morning to celebrate the Saint Lucia holiday. Knutas was sitting with Lina and Nils on the third pew to the right of the center aisle. The cross vaulting in the church ceiling high overhead and the magnificent arches cast long shadows, in contrast to the glow of hundreds of burning candles. The churchgoers were whispering quietly to each other as they waited with anticipation. Only an occasional cough or shuffling of feet from the pews broke the gentle murmuring.
The Lucia procession in the cathedral was one of the high points of the year. Petra was one of the bridesmaids. She sang in the youth choir, which was now participating in this year's Lucia procession, just as all the other choirs had done for as far back as anyone could remember. Knutas glanced through the brochure about the church as they waited for the event to begin. Construction of the St. Maria Cathedral was started in the twelfth century with funds collected from the German ships that docked at Visby. In the beginning it was meant to serve only German merchants, but later it became the church of the entire German congregation. After the Reformation, it was opened to everyone. No extensive changes to the church had been made since the Middle Ages, and that seemed to Knutas quite evident as he sat there admiring the high ceiling, the beautifully painted windows, and the pulpit, which had probably been imported from the German city of Lübeck in the seventeenth century.
Suddenly a faint singing could be heard in the church, and everyone turned their heads to look back toward the entrance. The tones of the traditional Lucia song grew louder, and the white-clad figure of Lucia appeared in the doorway. Slowly she walked forward, wearing a long white dress. On her head was a wreath with candles. Behind her walked the bridesmaids, two by two, with tinsel wrapped around their waists. They each held a lit candle. Behind them came the star boys, wearing paper cones on their heads.
The glow of the candles made it a magical spectacle, as the young people dressed in white walked forward, singing in their clear voices. A star boy who couldn't have been more than ten or eleven sang so beautifully in a loud and lovely voice that Knutas felt tears fill his eyes. In the middle of the solo, his cell phone began vibrating in the inside pocket of his jacket. Cautiously he pulled out the phone and held it up to his ear. It was hard to make out what Karin Jacobsson was saying on the other end. He managed to squeeze past the other people sitting on the pew and went out to the entryway.
"This better be important. I'm here watching my daughter in the Lucia celebration at the cathedral," he said.
"Fanny Jansson was found dead out on Lojsta Heath."
It took almost an hour to reach the site. Jacobsson and Knutas took the 142 down to Hejde and then headed out to Lojsta Heath. Old limestone farm buildings stood at the turnoff into the woods. A flock of black sheep with shaggy winter coats was crowded together at the fence, staring at them as they drove past.
A police car was waiting to show them the way. They bumped over the unpaved forest road, which was normally used only by tractors. The snow on the ground between the trees was untouched, and there was no wind. The low mixed forest had dense undergrowth, with withered ferns, heather, and lingonberry bushes. Here and there a few remaining berries shone bright red among the snow-covered hillocks. At the end of the road the forest opened into a clearing where another police car was parked. A short distance away, near an embankment, crime scene tape had been put up. The air was cold and fresh.
Fanny's body lay in a hollow beneath several thick spruce trees covered with heavy green moss.
The site was relatively protected. The girl was fully dressed in dark riding pants, a short quilted jacket that was unbuttoned, and a brown woolen sweater that was torn at the neck. Her face was dark against the snow. Her beautiful long hair, which was spread out on the ground, seemed strangely alive. Her eyes were wide open, staring up at the sky. When Knutas took a closer look, he noticed that there were red specks in the whites of her eyes. Dark bruises covered her throat.
Her body had been found by a woman who was out riding. She had fallen off her horse when it was startled by a fox. The horse had wandered off and led her to the clearing. The woman had hurt her back in the fall, and she was also in such a state of shock that she had been taken to the hospital in Visby.
On their way back to the city, Knutas's cell phone started ringing. The third call was from Johan.
"What happened?" said the familiar voice on the phone.
"Fanny Jansson has been found dead," said Knutas wearily.
Jacobsson was driving the car so he could devote all his attention to answering the journalist's questions.
"Where?"
"In a wooded area out on Lojsta Heath."
"When?"
"At eight thirty this morning."
"Who found her?"
"A woman who was out horseback riding."
"Was she murdered?"
"All indications are that she was, yes."
"How?"
"I can't go into that right now."
"How long has she been there?"
"That's something the ME will have to determine. I can't answer any more questions. We're going to hold a press conference later today."
"When?"
"Sometime this afternoon. You'll have time to get here."
Johan and Peter landed right after lunchtime at Visby airport. The cab ride into town didn't take long.
Police headquarters in Visby had changed radically since they were last there. The ice blue metallic facade had been replaced with a soft beige stucco. The rooms were now bright and airy, and they had been decorated in a typically Nordic style that was very tranquil, with natural materials and muted shades of white and blue.
The old and rather shabby room in which they had previously held press conferences was nothing but a memory. The journalists were now shown into a spacious room on the ground floor with rows of stainless steel chairs facing a podium. Thin curtains hung in front of the windows that faced the drab wall of another building. The press had already started setting up their microphones at the podium. Johan counted four reporters from competing TV networks.
He was grateful that he had been entrusted with the task of reporting for all the news programs on Swedish TV. There hadn't even been any discussion about it. After Johan's highly praised reporting on the homicides last summer, the national editors had no doubts: Johan Berg was the man for the job. He was pleased that his report would be aired on all the news programs that evening. He felt a great satisfaction knowing that he would be reaching so many viewers and have such an impact.
He took a seat in the front row while Peter set up his camera. His colleagues from the local media greeted him. He recognized some of them from press conferences that summer.
A moment later Anders Knutas, Karin Jacobsson, Martin Kihlgård, and Lars Norrby all took seats on the podium.
"Welcome," Knutas began. "For those of you who don't know me, I'm Detective Superintendent Anders Knutas, and I'm in charge of this investigation."
He introduced the others and then went on.
"As you already know, the body of Fanny Jansson was found in a remote wooded area on Lojsta Heath. Her body was found around eight thirty this morning by an individual who was out horseback riding. Fanny Jansson was murdered. The injuries that she sustained could not have been self-inflicted, so there is no question of suicide, as we had previously speculated."
"What were her injuries?" asked Johan.
"I'm not at liberty to discuss that," replied Knutas curtly.
He sighed a bit. In spite of the fact that he had barely started on what he had intended to say, the questions had already begun. A number of hands were waving in the air. He had a hard time dealing with the eternal impatience of journalists.
"We'll answer your questions in a moment," he said, "but first I want to present some of the facts."
He had no intention of allowing them to run the show. The reporters lowered their hands.
"The body had been lying at the site for some time. We don't yet know how long. Fanny Jansson was fully dressed when she was found, and there are no signs of sexual assault. The crime scene has been cordoned off, and the area is being searched by our technicians. An ME will be here tomorrow to examine the body. The area will be kept under guard until the body can be moved and the technical investigation has been completed. That is all I have to say at the moment. Do any of you want to add anything?"
He gave his colleagues an inquiring look, but they shook their heads.
"Then we'll take questions."
"How long has the body been there?"
"It could be a matter of weeks, meaning the entire time that Fanny has been missing. But we're not at all sure about that. We'll have to wait for the ME's report."
"Was any sort of weapon found?"
"I have no comment."
"Can you say anything about how she was killed?"
"No, I can't."
"Did the perpetrator leave any clues?"
"I can't divulge that, out of consideration for the ongoing investigation."
"Does Fanny Jansson have any connection to the place where she was found?" asked Johan.
"Not as far as we know."
"Was she murdered at the site, or was she moved there?"
"We have reason to believe that she was killed somewhere else and that her body was later moved to the wooded area."
"What makes you think that?"
"As I said earlier, I can't divulge anything about any evidence or other information found at the crime scene," said Knutas with forced composure.
"Who found the body?"
"A local woman. I don't want to give her name."
"Are there any witnesses?"
"It's possible. We haven't yet started interviewing anyone who lives in the area. But we're going to appeal to the public for information. We want to talk to anyone who may have seen or heard anything suspicious, especially during the last few weeks, in connection with the place where the body was found. No information will be considered too insignificant. Everything is of interest."
Knutas gave them the number for the police hotline and the press conference was over.
That evening Johan presented live reports on all the news broadcasts, giving the television viewers the latest update. He and Peter had a late supper at their hotel and then went to bed.
Again Emma didn't answer her phone when Johan tried to call her. It had now been more than a week since they had last talked to each other. Her friend Viveka had explained to him that Emma was ill and wanted to be left alone. He would just have to wait until she decided to call.
The ME was expected on Gotland the following day, but that evening Sohlman was able to present to the investigative team a preliminary report along with some visual images.
"It's difficult to say how long she has been lying there, but her body is quite well preserved, as you can see, as a result of the cold weather. The perpetrator also covered the body with moss, so no animal got to her. Fanny was fully dressed when she was found, but her sweater was torn at the neck. Her clothing will be examined more closely when the ME arrives, but we're leaving her body where it is until he gets here tomorrow. I can make an educated guess and say that she died from lack of oxygen. Do you see the red specks in the whites of her eyes and the bruises on her neck? Without going out on a limb, we can assume that she was strangled.
"She apparently offered some resistance, since her sweater was torn. I'm hoping that the perpetrator has left some evidence on her clothing—skin particles or saliva, for instance. The body was protected by the woods and the moss. It was also lying in a hollow, so we hope we can find some traces from the killer. We've taken scrapings from under her fingernails. There are skin particles that most likely came from him. Everything is being sent to SCL, as usual.
"When it comes to the location of the body, we can conclude that she was probably killed elsewhere and was then dumped in the woods. There are no traces of blood or anything else that might indicate the murder was committed at the site. We haven't yet been able to examine the body, but we did discover one thing. She has cuts on her wrists."
Sohlman clicked through the photographs until he found the pictures of Fanny Jansson's hands. Cuts were clearly visible on both of her wrists.
"Someone has cut her here. She probably did it herself."
"So she did try to kill herself, after all," exclaimed Norrby.
"Well, I'm not so sure about that," Sohlman objected. "I think it's more likely that she was one of those girls who cut themselves. It's not all that uncommon among teenage girls who are depressed. She had cut herself in other places as well, for instance behind her ears. The cuts are superficial, so there's no question of a real suicide attempt. It's possible that there are more cuts hidden under her clothing."
"Why would she do that?" asked Wittberg.
"Girls who cut themselves do it because they don't know how to handle their fears," Jacobsson explained. "When they cut themselves, all their anxiety collects in that one spot. It's also possible that they experience the pain and the blood as liberating. It's something concrete and controllable. The moment they cut themselves, all their other anxieties disappear; their fear becomes concentrated in the part of their body that is being subjected to pain."
"But why would she cut herself in such odd places?"
"Probably so that it wouldn't be visible."
Knutas switched on the lights and looked at his colleagues with a serious expression on his face.
"We now have two murders to investigate. The question is whether there is any connection between them. What does a fourteen-year-old schoolgirl have in common with an alcoholic man in his sixties?"
"As I see it, there are two obvious connections," said Kihlgård. "First, alcoholism. Fanny's mother drinks, and Dahlström was an alcoholic. Second is the racetrack. Dahlström bet on the horses, and Fanny worked at a stable at the trotting track."
"Those are two reasonable connections," said Knutas. "Is there anything else that might not be as obvious? Anyone?"
No one replied.
"All right," he said. "That's all for now. Both lines of inquiry need to be explored without bias."
## Friday, December 14
It felt as if the dawn would never come on that cold December morning. Knutas was having oatmeal with his wife and children in the kitchen. They had lit candles, which made their shared breakfast a bit more pleasant. Lina and the kids had baked saffron rolls while he was out at the site where Fanny was found. He was going to need them. Today he had to pick up the ME at the airport and then drive back out to the forest clearing. He put on a wool sweater and got out his warmest winter jacket. The frost of the past few weeks was holding on.
The children were upset and worried, and they wanted to talk about Fanny's murder. They had been greatly affected by the death, since Fanny wasn't much older than they were and they knew her by sight. Knutas ran the palm of his hand over their cheeks as they stood at the front door on their way to school.
In the car on his way to the airport, he felt a cold sweat come over him, and he was overcome by such nausea that he had to pull over and stop for a moment. Everything swam before his eyes, and he felt a tight pressure in his chest. Occasionally he suffered from panic attacks, a form of anxiety, but it had been a long time since the last one. He opened the car door and tried to calm his ragged breathing. The images of Fanny's body, combined with his worries about his own children, had apparently brought on this attack. With his type of work, it was impossible to protect his kids from all the shit he was forced to deal with: drunkenness, drugs, and violence. As his children were growing up, society seemed to be getting more and more brutal. It was probably worse in the big cities, but even here on Gotland the change was noticeable.
He tried not to say too many negative things about his job. At the same time, he could seldom come home and tell them that he'd had an uplifting sort of day. Of course he was always relieved when a case was solved, but it was hardly a matter of feeling elated. When an investigation was successfully completed, he just felt tired afterward. There was no sense of catharsis, as some people might think. Instead, he mostly had a feeling of emptiness, as if he were utterly deflated. Then all he wanted to do was go home and sleep.
After a few minutes he felt better. He rolled down the window and slowly continued driving to the airport.
The ME was waiting for him outside the terminal. His plane had landed earlier than anticipated. It was the same doctor that Knutas had worked with last summer, a lean man with thinning hair and a horselike face. His extensive experience lent him an air of gravity and authority. On their way out to the site where the body was found, Knutas told the doctor about everything they knew so far.
By the time they arrived, it was ten fifteen in the morning, and Fanny Jansson's eyes were still staring up at the gray December sky. Knutas grimaced with dread as he thought about what the beautiful girl lying on the ground might have gone through. Her body looked so small and thin under her clothing. Her cheeks were brown and smooth, her chin softly childish. Knutas was annoyed to feel tears welling up in his eyes.
He turned his back and gazed at the woods, which were dense and inaccessible. Over near the tractor road he could see that the forest thinned out a bit. Since he had previously studied the map of the area, he knew that some distance away there were open fields and pastures. A crow cawed from far off, otherwise everything was silent except for a quiet rustling from the dark green branches of the trees. The ME was now fully involved in his examination, and would be for the next several hours. Erik Sohlman and a couple of the other techs were assisting him with his work.
Knutas realized that his presence wasn't needed. Just as he got into the car to drive back to police headquarters, Karin Jacobsson called him.
"There's one person who has ties to both Dahlström and Fanny Jansson."
"Really? Who is it?"
"His name is Stefan Eriksson, and he's the stepson of Fanny's aunt in Vibble. She has a daughter of her own, but she divorced the father early on and married someone else, a man who had a son from a previous marriage. Fanny and this Stefan have seen each other for years at various family gatherings and the like. He's forty years old, married with two children, and he also happens to own a horse at the stable."
"I know that. We've been down the whole list," said Knutas impatiently. "What about him?"
"He was an intern under Dahlström when he was in high school. He worked at the newspaper for two weeks. After that he was a temp for Gotlands Tidningar and later he also worked for Dahlström when he started his own business. This Eriksson owns a café in town, the Café Cortado on Hästgatan, but his hobby is photography."
"Is that right?" exclaimed Knutas in surprise. This was new information to him.
"He and Dahlström may have kept in contact over all these years, even though Eriksson denied it when Wittberg and I interviewed him. A most unpleasant type of person. I could easily imagine him—"
"All right, but let's not jump to any conclusions," Knutas interrupted her. "What else?"
"I asked him if he spends any time at the stable, and he said that he's there now and then. The staff at the stable confirm this. He would also occasionally drive Fanny home."
"Does he have a police record?"
"No. On the other hand, there have been a number of complaints filed against him for suspected neglect. His family used to raise sheep, and the animals were evidently treated badly, according to the person who complained. Eriksson no longer has any sheep."
"I want to talk to him myself. Where is he?"
"I think he's at home. He lives in...oh my God!"
Jacobsson abruptly fell silent.
"What is it?"
"Stefan Eriksson lives in Gerum, which is only a couple of miles from the place where Fanny Jansson's body was found."
"I'm ten minutes from there. I'm on my way."
Gerum is not a real town. It's just a church with a few scattered farms right next to the large and inaccessible Lojsta Heath. The landscape is flat, but Stefan Eriksson's farm and surrounding property were the exception. It stood on a hill with a panoramic view of the area. The farm consisted of a stone farmhouse with two wings and a large barn. A late-model Jeep was parked outside along with a BMW.
When Knutas rang the bell, he heard dogs barking inside. No one came to the door.
He took a stroll around the farm and looked in the windows of the separate wings. One was apparently used as an artist's studio, and there were paintings leaning against the walls. A painting of a woman's face was set on an easel in the middle of the room. Crowded onto a table splotched with paint were cans and tubes of paint along with paint-brushes.
As he peered in the windows, Knutas was interrupted by the sound of someone clearing his throat behind him. The detective was so startled that he jumped and dropped his pipe on the ground. A man was standing right behind him.
"Can I help you with something?"
Stefan Eriksson was almost six foot six inches tall, by Knutas's estimate. He had on a blue down jacket and a black knit cap.
Knutas introduced himself. "Could we go inside to talk? It's starting to get cold."
"Of course, come with me."
The man led the way inside. Knutas was practically knocked down by two Dobermans, who seemed beside themselves with joy.
"So you're not afraid of dogs?" asked Eriksson without making any attempt to calm the animals.
They sat down in what must have been the good parlor. To think that people still have rooms like this, thought Knutas. A remnant of bygone times.
Stefan Eriksson was clearly fond of antiques. A mirror in an elaborate gold frame hung on the wall. Next to it stood a bureau with curved legs and lion's claw feet; along one wall stood a grand cabinet with rounded feet. The room smelled stuffy and dusty. Knutas felt as if he were sitting inside a museum.
He declined the offer of coffee. His stomach growled, reminding him that lunchtime was long past.
"Well, I don't really understand what you want. I've already talked to the police," said the tall man, who had sat down on a plush armchair. The dogs had settled at his feet, with their eyes fixed on their master.
"I need to ask you a few additional questions, but first I would like to express my condolences."
The man sitting across from him did not change expression.
"It's true that Fanny was my cousin, but we hardly knew each other. And we're not real cousins, anyway. My father—"
"I know about the family ties," Knutas interrupted him. "How often did you see each other?"
"Very rarely. Sometimes at someone's birthday celebration. There were problems with her mother, so they didn't always come. Majvor can't keep away from the bottle."
"How well did you know Fanny?"
"There was a big age difference between us, so we didn't really have anything in common. She was a little girl who sometimes came to visit with her mother. She never said anything. You'd be hard-pressed to find a more silent girl."
"You own a horse at the stable where Fanny worked. Didn't you ever see each other there?"
"That old nag is practically useless. It costs a lot more to keep her than she ever brings in from racing. But of course I do stop by the stable once in a while. Occasionally Fanny was there at the same time."
"Did you sometimes give her a lift home?"
"Not very often."
"Which car did you drive?"
Stefan Eriksson shifted uneasily in his chair. A frown appeared on his face.
"What are you getting at? Am I under suspicion?"
"Not at all," said Knutas dismissively. "I'm sorry if I seem pushy, but we have to talk to everyone who knew Fanny."
"I understand."
"So which car did you drive?"
"The BMW that's parked outside."
"You knew Henry Dahlström, too, didn't you?"
"Yes, I was an apprentice for him eons ago when I was still in school. After I graduated I sometimes filled in for him at GT, and I also worked as a temp at Master's. I mean, Master Pictures, his company."
"How did you happen to meet him?"
"I'm interested in photography, and he was teaching a course that I attended when I was in high school. And then, as I mentioned, I was an apprentice for him."
"Did you keep in contact over the years?"
"No. When the business folded, he went completely downhill."
"Do you still take photographs?"
"When I can. I'm married and have children and we moved out here. The café that I own in town also takes up a lot of my time. It's Café Cortado, on Hästgatan," he added.
Knutas detected a note of pride in the man's voice. Café Cortado was one of the most popular cafés in town.
Suddenly the dogs rushed for the door and began barking. Knutas gave a start. Eriksson's face lit up.
"That's my wife and kids. Just a minute."
He got up and went out to the entryway. The dogs were barking wildly and jumping around.
"Hi, sweetheart. Hi, kids. How are you?"
Eriksson's voice took on an entirely different tone. It was suddenly filled with love and warmth.
His wife and children had clearly been out celebrating Lucia. Maja Eriksson came in to say hello. She was dark and sweet and soft-spoken. Knutas noticed the tender way in which Eriksson looked at his wife.
No, he thought. It can't possibly be him.
He thanked the man for his time and left.
The discovery of Fanny's body caused a big stir in the media. The evening papers devoted a great deal of attention to the news, as did Regional News and the local media on Gotland. There was much heated debate about what could have happened to the girl. The newspapers printed maps that allowed their readers to locate exactly where Fanny was found. The farms that were closest to the site received visits from reporters and photographers. The newspapers were filled with speculations and hunches about what the motive behind the murder might be, and the TV and radio stations broadcast interviews with the stable staff as well as with the girl's neighbors and classmates.
Without talking to Johan, Max Grenfors had called Majvor Jansson and persuaded her to agree to an interview. Grenfors was very pleased with his success in getting the mother to tell her story as an exclusive on Regional News. But he encountered quite a different reaction from Johan, who refused to interview her, which prompted Grenfors to give him a real tongue-lashing.
"I've managed to get her to agree to an exclusive interview, so of course we're going to talk to her!"
Johan was standing out in a field near the place where the body had been found. He was with Peter and a farmer who thought he had seen car headlights in the area late one night a couple of weeks earlier.
"I'm not interviewing someone who's in a state of shock," said Johan firmly. "The woman doesn't know what she's doing. She can't see the consequences at the moment."
"But she wants to do it. I talked to her myself!"
"Exactly what do you want me to ask her, one day after her daughter was found murdered? How does it feel?"
"Damn it, Johan. She wants to talk. Maybe it's a way for her to work through the whole thing. It's her own decision. She's unhappy with the police work and wants to say something about it. She also wants to appeal to the public for help in finding the murderer."
"Fanny was found yesterday. That's less than twenty-four hours ago. I can think of better ways to work through things than by talking on TV. In all good conscience, I don't think we can do it."
"For God's sake, Johan. I told her that you'd be at her sister's house in Vibble at two o'clock."
"Max, you can't trample on my journalistic integrity. I won't do it. I simply won't have this on my conscience. The woman is in shock and should be in a hospital. She's extremely vulnerable right now, and I think it's rotten if we try to take advantage of her weakness. She doesn't realize the impact of TV. We have to make certain decisions for people if they're not capable of doing it for themselves."
Johan glanced at Peter, who was standing next to him and rolling his eyes. He told Johan to give Grenfors his greetings and say that he refused to film an interview with the girl's mother. At the same time Johan could hear Grenfors breathing harder on the other end of the line.
"Just do the interview and we'll make the ethical decisions back here in the newsroom," shouted Grenfors. "See to it that you go out there to meet her. I want it on tonight's program. I've already promised the interview to Aktuellt, Rapport, and 24."
"And all of them want it?" asked Johan dubiously.
"You bet they do. So get going. Otherwise she might change her mind and talk to somebody else!"
"Fine. Let TV3 interview her, or the newspapers if they want to. But I won't do it."
"So you refuse?" Grenfors went on.
"What do you mean by 'refuse'?"
"You won't carry out the assignment that I've asked you to do. That's a dereliction of duty, damn it!"
"Call it whatever you like. I'm not going to do it."
Johan closed his phone, bright red in the face. His breath was visible in big billowing puffs all around him. He turned to Peter and the farmer.
"What a fucking pig."
"To hell with him," said Peter, in an attempt to console him. "Let's get back to work. I'm freezing to death."
The farmer, who had listened in astonishment to the phone argument while he waited to be filmed, was now interviewed. He told them about the car that he had seen driving along the tractor path one evening two weeks earlier when he was out in the barn tending to the evening milking. As he was crossing the barnyard, he saw lights from the road. No one ever drove out here so late at night. He couldn't say what type of car it was. He had stopped and waited for a while, but when the car didn't reappear, he gave up and went inside his house.
Johan and Peter headed back to town. They were planning two reports, one that dealt with the police work and another that focused on the feeling of shock the day after among the schoolkids, the stable staff, the neighbors, and the ordinary citizens of Visby.
Many had still been hoping that Fanny would be found alive, even though hope had dwindled with each day that passed. Now there was a great sense of sorrow.
Back at the hotel that evening, Johan tried to get hold of Grenfors, but the editor refused to talk to him. He had found a trainee to do the interview with Fanny's mother, but after discussions with the producer and editors, the piece was never aired. No one else seemed to be interested in it, either. It's just a matter of prestige, thought Johan when a colleague later recounted on the phone the wrangling going on in the newsroom. Good Lord, sometimes his job was like kindergarten.
The important thing was never to forget your purpose and to keep asking yourself why you were doing a particular story and whether it had general interest. And then you had to weigh that against the harm that you might cause people. He was sure that he had made the right decision when he refused to contact Majvor Jansson. No one could make him interview people who were in shock.
That was one lesson he had learned after all his years working in TV. On a few occasions he had done what some overzealous editor wanted and interviewed people who had just lost a loved one or who had been involved in an accident. Just in order to be accommodating. Afterward he had realized that it was wrong. Even though at the time of the interviews the individuals had wanted to talk in order to share their grief or to draw attention to a problem, they were confused and unable to think clearly. To dump the responsibility on them was indefensible. Besides, they didn't comprehend the scope of their participation. The impact of TV was enormous. Images and interviews could be repeated in all kinds of contexts, without allowing the person involved any opportunity to stop them. And each time his or her grief would be torn wide open.
She felt as if she were in a soundproof glass bubble, cut off from the rest of the world. Someone had pulled the cord, stopped the noise, brought the merry-go-round to a halt.
Emma was lying on her back on the floor of Viveka's small living room. Her friend was away for the weekend, so she had plenty of peace and quiet to think things through.
It was very tranquil in the living room. She didn't want any disturbing sounds—no radio, no TV, no music. She wished she could sink deep into an undemanding darkness that would simply embrace her.
Another body was growing inside of her body. A tiny human being that was part of her and Johan. Half him and half her. She closed her eyes and ran her hand over her smooth abdomen. Nothing was visible on the outside yet, but her body was sending her signals. Her breasts were tender, she had started suffering from morning sickness, and her craving for oranges was just as strong as during her previous pregnancies. She wondered what kind of person was inside her. A girl or a boy? A little sister or a little brother?
She let the tips of her fingers move in circles under her shirt, sliding down to her crotch and then back up to her sore nipples. The baby was telling her that he or she was inside, already taking nourishment through the umbilical cord, and growing bigger every day. She had figured out that she was in her eighth week. How far had the fetus developed? She and Olle had followed closely the various stages of development when she was pregnant with Sara and Filip. He had read aloud to her from a book about what was happening each week. They had been so filled with anticipation.
Now everything was different. This weekend she would have to make a decision. To have the baby or not. She had made a promise to Olle. He had reacted with surprising composure to the news that she was pregnant, even though it was quite clear that he was not the father. With icy determination he had told her that if she decided to have the child, their divorce would be inevitable. He had no intention of taking care of Johan's kid and being saddled with her lover for the rest of his life. If they were going to continue as a family, there was only one choice—to get rid of it, as he said. Get rid of it. The words sounded absurd to her ears. As if it were merely a matter of picking off a scab. Just scrape it off and flush it down the toilet.
She wished that someone else could make this decision for her. No matter which option she chose, it was going to be trouble.
## Monday, December 17
On Monday morning the phone began ringing the minute Knutas stepped through the doors of police headquarters.
"Hi, this is Ove Andersson, the building superintendent at Jungmansgatan. We met in connection with the murder of Henry Dahlström."
"Yes, of course."
"Well, the thing is that we're cleaning out the darkroom that Dahlström was using here. It's going to be a storage room for bicycles again. I'm standing in the room right now."
"Yes?"
"We've found something odd, behind a vent. It's a plastic bag with a package inside. It's taped up and I didn't want to open it because I thought I might destroy some evidence."
"What does it look like?"
"It's a brown paper package with ordinary tape around it, very lightweight, about the same size as a stack of postcards."
Under Knutas's intense supervision, Sohlman opened the carefully wrapped package, which had been delivered to the crime tech division. It turned out to contain photographs. Rather blurry, but there was no doubt about the subject matter. Almost identical, they all seemed to have been taken from the same angle. In the pictures they could distinguish a man who was having intercourse with a young woman, or rather a girl. She seemed to be half the size of the man. Her face was hidden, partly by him and partly by her long black hair. Her arms were stretched up in an unnatural position, as if she had been tied to something. The man was bending over her, almost covering the girl's body with his, but one of her legs was visible. She had dark skin.
Sohlman and Knutas looked at each other.
"It must be Fanny Jansson," Knutas said at last. "But who's the man?"
"God only knows."
Sohlman ran a hand over his forehead. He took out a magnifying glass and began scrutinizing the photos.
"Look at this. There's a painting hanging on the wall behind them. You can see a bit of red and a...What's that? Maybe a dog?"
He handed the magnifying glass to Knutas. One corner of the painting was visible.
"It looks like a dog lying on something red. It could be a cushion or a sofa."
Sohlman eagerly looked through the other pictures, but none of them revealed anything more.
Both men sank down on their chairs. Knutas dug his pipe out of his pocket.
"Well, we now have the connection," muttered Knutas. "Dahlström took pictures of someone who had a sexual relationship with Fanny Jansson. He must have photographed them on the sly and then blackmailed the man for money. That's where the twenty-five thousand came from. That would explain everything: the man at the harbor, the money, Fanny..."
"That means that the man we're looking at in these pictures is the perpetrator," said Sohlman, tapping his gloved index finger on the man's pale back.
"Presumably. It's easy to figure out why he killed Dahlström. But why Fanny? If it is her, that is. We can't be completely positive."
Knutas picked up one of the photographs and held it out.
"Who the hell is he?"
Knutas summoned the investigative team to a meeting to discuss the surprising discovery. The mood was one of nervous elation—rumors about the contents of the package had quickly spread through the corridors. Sohlman had scanned the photos so that he could project them on the screen at the front of the room. Wittberg was the first to speak.
"Are we positive that the girl in the photos is Fanny Jansson?"
"Her mother was just here, and she identified her. You can see the girl's watch on her left wrist. Fanny got that watch as a birthday present last year."
"How did the mother react?" asked Jacobsson.
"She fell apart," said Knutas with a sigh. "And who wouldn't, seeing their child in that sort of situation?"
"What kind of damn pervert is this guy?" growled Norrby.
"The only thing we've been able to determine so far is that we're dealing with a grown man—definitely not a boy her own age."
"It looks like she's tied up," Kihlgård interjected. "Her arms are stretched above her head. She's tied to something."
"Look at this," said Sohlman, putting up the most detailed of the photos. "There seems to be a painting in the background. The only thing we can really make out is the image of a dog lying on a red sofa or something similar. Yellow-patterned wallpaper with a faint border is visible in the background, as well as a glimpse of the back of a chair. It looks like an antique chair with a high back and carved decorations. The photographer took all the pictures from the same angle. The fact that they're so blurry could be because they were taken from outside, through a window. The question is: Where were the photos taken? It has to be somewhere in town or nearby, at some easily accessible place. Otherwise how would Dahlström have discovered Fanny and the unidentified man?"
"Maybe it's a storeroom," suggested Norrby. "Or a meeting room. Or it could be in the home of somebody that Dahlström knew."
"The room looks brightly lit. Can you see how the daylight is coming through the window? I have the impression that it's a big room," Jacobsson said.
"I really wonder how the man met Fanny," said Wittberg. "Could he be a friend of her mother?"
"How disgusting, if that's the case. That would be horrible." Jacobsson grimaced.
"I think the pictures look pornographic," said Kihlgård, holding one up. "It might very well be a sex ring. Maybe there was a whole gang of guys who were exploiting Fanny, and this is just one of them. Maybe she got drawn into prostitution and was forced to sell her body to the neighborhood men."
"Up until now we've been lucky to be spared that type of activity here on Gotland. At least as far as we know," said Knutas with a sigh.
"Or pedophiles," murmured Jacobsson. "Fanny might have been one of many children being exploited. We might have a pedophile ring right around the corner, and we don't have the faintest idea about it."
"The Internet. We have to check the Internet. I have a friend who's working on a big pedophile investigation in Huddinge. I'll ask her whether there might be anyone in that ring who has connections to Gotland."
"Good idea," said Knutas approvingly. "This could be about almost anything at all."
He was interrupted by his cell phone ringing. The others listened in silence to his murmuring. When he was finished, he looked at his colleagues alertly.
"That was Nilsson at SCL. The samples taken from Fanny Jansson's bedroom have been examined. No match was found in the police records, but the blood and hairs that were taken from her bed have been compared with the evidence from Dahlström's place. There's no doubt whatsoever—they match."
Late that evening Knutas went back home and found his entire family gathered in front of the TV. They answered his greeting by saying, "Shh—this is so exciting!"
He sighed and went out to the kitchen, opened the refrigerator, and took out a plate of leftovers, which he heated up in the microwave. The only one who wanted to keep him company was the cat, who rubbed against his leg and then hopped onto his lap and curled up. The cat seemed completely unaware of the problem she caused him. It wasn't easy to lean forward to eat with a cat curled up on his lap.
The idea that a murderer and sexual predator was on the loose on Gotland gave him goose bumps. At first the perpetrator had given in to Dahlström's demands and made two payments, but after that it clearly got to be too much. But actually deciding to murder the man who was blackmailing him was a big step. Maybe the killer thought he could get away with it if he made it look like a drunken brawl. And then there was the money from the racetrack. Most likely he knew about it and made use of the fact. He probably stole the money to mislead the police. The fact that Dahlström's apartment had been searched must mean that he was looking for the photos. The same with the darkroom. But his search had been in vain. The package was hidden inside a vent, and no one had bothered to look there, neither the killer nor the police.
After the murder the perpetrator left the scene. He tossed the murder weapon and the camera into a grove of trees some distance away. He presumably had a car parked farther away, near the next apartment complex.
Knutas poked at his food: meatballs with reheated pasta. He poured on some more ketchup and aimlessly stirred it into his food. He took a gulp of milk. Not a sound from the living room. The movie must be very exciting.
And then Fanny was killed. Although maybe that's where they really ought to start, since it was where the whole thing began. The story of the fourteen-year-old girl. How had the man met her in the first place? He must somehow be part of her world.
Knutas put that question aside for the time being and continued his train of thought. The man was using her sexually; there was no doubt about that. It was anyone's guess how long it had been going on. No one seemed to know that she was seeing anybody. He doubted that this was a love relationship in the usual sense. The man might have threatened her, or else she was dependent on him in some way. But what had prompted him to kill her? He had already gotten rid of Dahlström, so he wasn't being blackmailed anymore.
He was taking a big risk by committing another murder. It might not have been planned, of course. Maybe it happened as a result of some sex game. Fanny appeared to be tied up in the photos. Maybe the killer had strangled her by mistake and then dumped her body in the woods.
There was another alternative. Maybe Fanny had become so difficult that he found it necessary to kill her. Maybe she was threatening to expose him, or simply wanted to end the relationship.
The strange thing was that no one had noticed anything—not a single person.
His heart lurched when he thought about the body in the woods. The faces of various people flashed through his mind. Fanny's mother. How was she responsible for what had happened? Why hadn't she paid more attention to her daughter? Fanny had been all alone with this problem. She had felt so bad that she had even tried to harm herself. She was only fourteen and still a child. Yet no grown-up had cared about her, not even her mother.
It was the same situation at school. Even though the teachers had noticed that something was wrong with Fanny, they did nothing. She was there, right in front of everyone's eyes, but no one saw her.
## Thursday, December 20
As Knutas sat in his office drinking coffee, someone knocked on the door. Karin Jacobsson stuck her head in.
"Good morning! It sure is interesting how people can forget all about something and then suddenly remember the most important details."
She dropped onto a chair across from him and rolled her eyes.
"That guy Jan Olsson from the stable called to say that Fanny had gone out to visit Tom Kingsley."
"Is that right?"
"One time last fall Jan Olsson had to go over to Tom's place to drop something off."
"What did he drop off?" queried Knutas.
"He didn't say," Jacobsson went on impatiently. "But listen to this. Fanny's bicycle was outside Kingsley's house, and Olsson noticed that her jacket was hanging in the hallway."
"Did he see her?"
"No. Tom didn't invite him in."
"Okay, that's enough to bring Kingsley in. I'll call Birger so we can get a search warrant for his house."
Knutas reached for the phone to call the prosecutor.
"Sure, but there's just one problem," said Jacobsson dryly.
"What's that?"
"Tom Kingsley has left. He's on vacation in the States."
"For how long?"
"He has to be back at work on Monday, according to the stable owner. But he booked an open-ended round-trip ticket and hasn't yet made his return reservation. So we don't know when he'll be flying home."
"We're going in anyway."
Tom Kingsley's house stood in a wooded glade, not far from the racetrack. It was actually a summer cottage that he had been renting ever since he came to Gotland.
The road up to the house was not much wider than a tractor path. The police cars jolted their way forward. Knutas and Jacobsson were in the first car, with Kihlgård and Wittberg following behind. Prosecutor Smittenberg had immediately given the go-ahead to search the premises. Ordinarily, Tom Kingsley should have been notified, but no one knew where he was.
All the windows were dark. When they got out of the cars, it looked as if no one had been to the house in a while. The snow cover was untouched.
They had obtained keys from the landlord, whom Jacobsson had managed to locate during the course of the morning.
The ground floor of the house consisted of a small entryway and a living room on the right, with access to a cramped kitchen. The house was furnished simply but nicely: a dining table next to the window, a fireplace, and against one wall an old-fashioned wooden sofa with seat cushions covered with striped fabric. Between the kitchen and the living room was a woodstove. The kitchen, with windows facing the woods, was sparsely furnished: low kitchen benches, a pantry, an old electric stove, and a small refrigerator.
A narrow staircase curved up to the second floor, which had two small bedrooms and a hallway. It was neat and clean. Knutas lifted up the bedspreads. The bed linen had been removed and the mattresses underneath were worn. The police officers began methodically going through all the drawers and cupboards. Kihlgård and Jacobsson took the second floor, Knutas and Wittberg the first floor.
It wasn't long before Wittberg shouted, "Come and look at this!"
With tweezers he was holding a piece of paper that looked like instructions of some kind.
"Do you know what this is?"
The others shook their heads.
"It's instructions for taking morning-after pills."
## Friday, December 21
The discovery of the pill instructions in the home of Tom Kingsley, combined with the fact that he had definitely denied having any sort of close relationship with Fanny, made the prosecutor decide to issue a warrant for Kingsley's arrest in absentia. The fact that Fanny's fingerprints were found on the instructions made the police even more convinced that Kingsley was the man they were looking for. After checking with the airlines, they determined that a week earlier he had flown to Chicago on SAS. The Stockholm police were informed, and employees at the SAS ticket offices were told to keep a lookout for Kingsley and to sound the alarm when he booked his return flight.
Knutas felt relieved, even though he didn't know where Kingsley was. Now it was just a matter of waiting for his return.
In the meantime, he could take a much-deserved weekend to relax. Away from any kind of police work. He and Leif were going out to the Almlöv family summer house in Gnisvärd, on the coast about fifteen miles south of Visby, as they always did right before Christmas. He hadn't been sure that he could actually get away this time because of the investigation. But a warrant had been issued for Kingsley's arrest, and they couldn't do anything else until he returned home. So Knutas had decided that it would be possible, after all. It was only a twenty-minute drive from Visby, and he could be reached by cell phone if anything happened.
As for preparing for Christmas, he had done everything that was expected of him—the traditional buying of the Christmas tree with the children, and all the grocery shopping and cleaning that he had done together with Lina. Late one evening he had made his own pickled herring in a sherry sauce, as he always did for the Christmas and Midsummer holidays. During his lunch hour he had run out to buy Christmas presents and had actually managed to buy everything, wrap all the gifts, and compose the traditional rhymes to go with them.
Now he was ready for his reward. Two days of solitude eating good food and doing some fishing—interests that he shared with Leif.
He hurried home after work on Friday afternoon and packed a bag with clothes and his fishing gear. It had been snowing all day. The snowplows had been working nonstop to make the roads passable. Knutas couldn't remember the last time it had snowed so much on Gotland. If only it would stick around until Christmas.
In the car on their way south he felt himself relaxing more with every mile they put behind them. They were playing Simon and Garfunkel full blast. The wintry landscape slid past outside the windows, with expanses of white fields and an occasional farm.
A beautiful layer of snow covered the yard when they arrived.
It's actually silly to call this place a summer cabin, thought Knutas. It's more like a manor house. The typical Gotland-style limestone house from the mid-nineteenth century was impressive with its whitewashed walls, pitched roof, and smooth gables. During that era bigger houses were being built on Gotland to keep pace with the increasing prosperity in the countryside. This house had no less than seven rooms and a kitchen, divided into two wings. The farm also had a boathouse that was used as a storeroom and food cellar. Next to the house stood a sauna, only a few yards from the dock where Leif's fishing boat bobbed up and down all year-round.
The place was rather isolated. The nearest neighbor lived a couple of hundred yards away.
"I can just imagine how cold it is in the house," Leif warned his friend when he opened the heavy, creaking front door.
"It doesn't feel so bad," said Knutas as they stepped inside. He carried the bags of food out to the kitchen and started putting away the provisions. "But I suppose it will seem worse if we sit still."
"I'll turn on the electric heat and make a fire in the fireplace, but it will take a while to get rid of the dampness."
Several hours later they sat at the table with plates of roast beef and potatoes au gratin that smelled of garlic, along with a bottle of a robust Rioja wine. It had been a long time since Anders Knutas had felt so good.
"How many times have we done this? Is this the fifth or the sixth year? This year it seems even more necessary than usual."
"Yes, I think we both needed to get away," Leif agreed. "There's been a hell of a lot to do at the restaurant. The worst is when there are problems with the staff. One of my best waitresses had a miscarriage and had to go to the hospital. Then another waitress had to go to Stockholm because her mother died. And to top it off, I caught one of the bartenders stealing from the till. All within a couple of weeks. And as usual, these kinds of situations always happen at the least convenient time. Right now we're up to our necks in Christmas reservations. It's lucky I have such a superb chef, otherwise I'd never be able to handle all this. He can fix pretty much anything. I actually offered to stay home this weekend, but he persuaded me to go. I was thinking we could postpone this to some later date," he added apologetically.
"I'm glad we didn't. Tell him thanks from me." Anders took a sip of his wine. "At least you should be happy that the restaurant is doing so well. It's always full of people, and it's been like that ever since you opened. I don't know how you do it."
"So what about you? How's it going with the investigation?"
"Good. We finally seem to be on the right track."
"What a nasty business."
"It's been damn tough. When we know that a murderer is on the loose and we're blindly fumbling around, not being able to make sense of things...it's frustrating."
"So you're not doing that anymore? Blindly fumbling around, I mean?"
"No, I'm convinced that we're close to solving the case. You know that I can't discuss any investigations with you, but this much I can tell you: I think we're very close now."
"Is it someone you've suspected for a long time?"
"No. Actually, someone completely unexpected has turned up."
"So why haven't you caught him?"
"Enough questions, Leif. You know I can't answer them."
Leif held up his hands. "Of course. Would you like some more wine?"
They spent the rest of the evening playing chess in front of the fireplace. And they opened another bottle of Rioja.
It turned out to be a late night. They didn't get to bed until after midnight. Anders was given one of the upstairs bedrooms. The room was simply but beautifully furnished. The limestone walls were rough and bare. The slate roof was supported by heavy timbers. A wide wooden bed with a white flowered bedspread stood along one wall, and next to it were three country chairs painted blue. A little window with a deep recess faced the sea. The rhythmic sound of the waves lapping against the shore lulled him to sleep.
When he awoke, he had no idea how long he had slept. It was pitch dark in the room. He couldn't understand what had awakened him. He lay still with his eyes open and unseeing in the night, listening for sounds that weren't there.
He reached out his hand and turned on the lamp on the nightstand next to the bed. It was three ten in the morning.
His mouth was dry, and he needed to go to the bathroom.
Afterward he paused next to the window. He could hear the sea, but it seemed very calm. There was a light on in the boathouse. Strange. Was Leif out there at this hour? Maybe he had forgotten to turn off the light.
The snow gleamed white in the darkness, and the glow from the outdoor lights cast long shadows. Nothing was going on, so he went back to bed.
It took a long time before he fell asleep.
The days had passed and Johan hadn't heard a word from Emma. He had been back in Stockholm almost a week, since nothing new had occurred on Gotland that would warrant a trip to the island. At least nothing that he knew about. The police were being very tight-lipped. He had tried to put the pressure on Knutas several times, but he hadn't gotten anything useful out of him. Experience told him that they were close to catching the perpetrator. The police always reacted the same way when an investigation entered a sensitive stage. They all just clammed up.
He was longing for Emma terribly, but she refused to talk to him. Maybe a solution was near on more than one front. Oh, just let it happen, he felt at the same time. Bring on the shit, so we can get this over with once and for all. He was tired of all the worrying and all the planning for a future with Emma. Wondering how he would manage on Gotland, as a stepfather, as a man with responsibilities. Cooking pasta and reading good night stories and blowing noses and balancing things among Emma, her ex-husband, the kids, the in-laws; birthday parties, deciding where to spend Christmas Eve, and feeling torn between Stockholm and Gotland. And, to be honest, how much fun would it be to take over a family that already existed? He was a romantic who dreamed of getting married and eventually becoming a father. For Emma, none of it would be for the first time.
Marrying again, having children again. Did she even want to have children with him? They had never talked about that. Why hadn't they?
It was probably just as well that they put an end to things, once and for all. He might meet some girl in Stockholm who didn't have a broken marriage and kids as baggage. Then it could be a magical experience for both of them. Everything would be so much simpler—just the fact that they could live in Stockholm, close to their families, their work, and their friends. The conditions for having a successful and good life together would be so much better. Why make life more difficult than necessary? It was hard enough to make a relationship work. Did he also have to hassle with other people's children and ex-husbands? No thanks.
But there was just one hitch. Emma was the one he wanted.
## Saturday, December 22
On Saturday morning Anders woke up when Leif knocked on the door and barged into the room.
"Wake up, sleepy-head! It's eight o'clock and breakfast is ready!"
Groggy with sleep, he sat up in bed. Leif was looking shamelessly frisky.
"I've already been out to chop wood. It's glorious weather. Just have a look outside," he said, nodding toward the window.
Anders turned his head. To his great surprise, he saw the sun gleaming over the expanse of the sea, which was blue and relatively calm.
He had almost forgotten how beautiful the view was. When they arrived the day before it was already dark.
"Incredible! I'll be right down."
He took a quick hot shower. What a luxurious summer house this is, he thought as he admired the lovely tiled wall.
Breakfast was on the table when he came downstairs to the kitchen: a real Gotland loaf of rye bread, butter, cheese, liver sausage, ham, salami, and vegetables. The aroma of strong coffee was spreading through the kitchen. A fire was crackling in the fireplace.
Anders appreciated Leif's sense for food, and he dug in with a good appetite.
"What service," he said, grinning at his friend, who was sitting on the other side of the table, studying a nautical chart.
"Tomorrow it's your turn to make breakfast. I was thinking that we should take the boat out, since it's such good weather. A light wind and forty-one degrees."
"It's great to see sunshine in the middle of December. That's a real treat."
"Did you sleep well?"
Anders hesitated for a second. "Like a rock. How about you?"
"Same here. I always sleep well in the country."
Anders cleared the breakfast dishes and packed up his gear. He was looking forward to the fishing trip.
Two days left before Christmas. Anticipation was shining in the children's eyes. At the same time, she found herself as far away from the idyllic family scene and the serenity of Christmas as she could get. She woke up in Viveka's guest room, feeling sick. The pregnancy probably wasn't the only reason. It had been a late night. She and Viveka had consumed a lot of wine and stayed up talking half the night.
She might as well go ahead and drink wine. She no longer needed to think about the well-being of the baby. She had made up her mind, but she wasn't able to get an appointment for an abortion until after Christmas. She was going to have to spend the entire holiday noticing the clear signs of her pregnancy. A constant reminder of the child growing inside of her.
She still hadn't dared talk to Johan. She didn't want him to influence her decision. Of course it was selfish, but she didn't see any other option. She had chosen to lock him out. She had distanced herself from him completely and refused to speak to him on the phone. She defended her actions by telling herself that it was sheer self-preservation. It was lucky that he had gone back to Stockholm. That made things a little easier. To see him now would be disastrous. And she had to think about the children she already had.
They had decided to celebrate a completely normal Christmas, with the whole family together. To visit relatives and friends and do everything they usually did. She would just have to suffer through the nausea as best she could. She had only herself to blame, and Olle didn't seem to be the least bit sorry for her. There wasn't a trace of the sympathy he had exhibited when she was pregnant with his own children.
When she saw Sara and Filip she was filled with tenderness. They had no inkling of the chaos that was raging inside their mother.
The doorbell rang. With a sigh she got out of bed and fumbled for her bathrobe. It wasn't even ten o'clock.
When she opened the door she found herself looking at the faces of her husband and children.
"Good morning!" they cried in unison.
"You have to get dressed," Sara told her eagerly. "Hurry up!"
"What's going on?"
Emma cast an inquiring glance at Olle, who was looking sly.
"You'll see. Go and get ready. We'll wait."
Viveka was now up, and she came out to the entryway.
"Hi. Has something happened?"
"No. We're just here to pick up Emma," said Olle cheerfully.
"Come into the kitchen and wait." She turned to the children. "Would you like some juice?"
"Yes!"
Fifteen minutes later Emma was ready, and they set off. Olle drove south, heading away from Visby. In Vibble he turned onto a road leading through the woods.
"Where are we going?" she asked.
"You'll see soon enough."
They parked outside a solitary house and rang the bell. Dogs could be heard barking inside. The children were jumping up and down with excitement.
"That's Lovis," shouted Filip. "She's so cute!"
A young woman of about twenty-five opened the door, holding a baby in her arms, and with a golden retriever circling her legs. The dog was overjoyed to see the visitors.
Emma had to wait in the hall while the others hurried out to the kitchen. She could hear them whispering. Then they came out to join her, first Olle with an adorable golden puppy in his arms, followed closely by the children.
"Merry Christmas!" said Olle, handing her the puppy, who wagged her tail and stretched out her snout to lick Emma's hands. "You've always wanted to have a dog. She's yours, if you want her."
Emma felt herself beaming as she took the puppy in her arms. The dog was small, soft, and plump, and she eagerly licked Emma's face. The children were looking up at her happily. A ribbon was tied around the puppy's neck with a card attached: "To Emma with all my love—your Olle."
She sank down onto the bench in the hall, with the puppy climbing all over her.
"See how much she likes you?" Sara chattered.
"She just wants to keep licking and licking," said Filip with delight as he tried to pet the puppy.
"Do you want to keep her?" asked Olle. "You don't have to. We can leave her here."
Emma looked up at him without saying a word. Everything that had happened flashed through her mind. His coldness had scared her, but it probably was because he felt hurt. And with good reason. Of course she understood. She saw hope in the faces of her children. For their sake she would have to try.
"Yes," she said. "I want to keep her."
The call came into police headquarters as Jacobsson and Kihlgård were sitting in the pizzeria on the corner. The Stockholm police reported that Tom Kingsley had booked his return flight for the following day. He was due to land at Arlanda Airport at 2:45 P.M. They assumed that he planned to continue on to Visby the same day. The next flight for Visby was scheduled to depart at 5:10 P.M. The police at Arlanda would apprehend him at the airport and then escort him to Visby. Wittberg called to convey the information, and Jacobsson sent a text message to Knutas to update him.
"That's great," said Jacobsson, breathing a sigh of relief. "Maybe we can finally put an end to this whole story so we can have some time off during Christmas."
"I certainly hope so. If he really is the killer."
"And why wouldn't he be?"
"You just never know. Surely he should realize that he's going to come under suspicion sooner or later. There's nothing keeping him here. If Kingsley really is the perpetrator, we have to ask ourselves why he doesn't stay in the States. Why would he come back here and risk getting caught?"
"Maybe he's convinced that he's not a suspect."
"Sure. But it wouldn't surprise me at all if the guy turns out to be innocent and we have to start from scratch."
Kihlgård stuffed the last bite of the aromatic calzone into his mouth and wiped his lips with the back of his hand.
Jacobsson gave him a dubious look. "Optimist," she muttered.
"I think it's strange that Knutas seems so certain that Kingsley is the perp. Just because the investigation has come to a dead end, that doesn't mean he has to grasp at straws."
"Then how do you explain the morning-after pill?" Jacobsson objected.
Kihlgård leaned back and lowered his voice. "It could be that Fanny trusted Kingsley enough that she asked his advice about those blasted pills, and then she left the instructions at his place. That's not inconceivable, is it?"
Jacobsson looked at him skeptically. "Is that what you really believe?"
"Why not? We shouldn't lock ourselves into Kingsley. That's crazy." Kihlgård ran his hand through his thick mane, which was sprinkled with gray.
"So what should we do?" asked Jacobsson.
"How about having some dessert?"
Anders steered the little fishing boat out to sea. It was always so peaceful standing at the helm. Leif was preparing the nets on deck. He came from a family of fishermen and was quite experienced. When he was ready, he came to stand next to Anders in the wheelhouse.
"There's not much salmon on this side of the island, so we'll have to fish for cod instead."
"That's too bad. It would have been great to have fresh salmon for dinner."
"We can always try, by trolling. I'll toss out the lines behind the boat and let them trail in our wake. Now that it's so cold, the fish are right below the surface. If we're in luck, we'll catch a salmon or a steelhead."
They passed Tofta Beach, and Anders was amazed at how deserted it looked. The emptiness of the rippling sand dunes was a huge change from the hordes of swimmers in the summertime. Tofta was by far the most popular beach on the island, especially among young people. In the summer the beach towels were spread out so close together that you could hardly see the sand.
Leif gazed across the sea.
"Can you see the two Karlsö islands over there? It's incredible how clear they are."
Both islands stuck up from the water, the big one behind the little one. Anders had been out there so many times. His whole family went out to Big Karlsö every May to see the colonies of guillemets. That's when the unusual auks hatched their young.
Glints of sunlight kept coming through the clouds, and even though the wind was picking up, they decided to stay out at sea while the nets were in the water. Leif unpacked some sandwiches and a thermos of hot chocolate, which they enjoyed on deck. It was hard to believe that Christmas Eve was just a couple of days away.
Anders was tired, so he went into the cabin to lie down for a while. He fell asleep to the sound of the waves lapping against the hull. After an hour he woke up to find Leif nudging him.
"We have to pull up the nets. It's getting windy."
Anders was surprised to see how quickly the weather had changed. Gusts of wind met them as they came up on deck, and the sky was now dark. The boat was pitching back and forth as they pulled up the nets. It was a nice haul—they counted nine cod. The trolling lines brought in two salmon. Not exactly spectacular specimens, but still not bad.
"Now we'd better see about getting back home as fast as possible," said Leif. "I was listening to the marine report while you slept. There's a storm on the way."
It would take them an hour to get back to Gnisvärd. Darkness fell, and as they passed Tofta, the first squall set in. The boat listed abruptly. Anders, who was on his way up the companionway to the wheelhouse, fell headlong through the door.
"Damn it!" he shouted as he hit his head on the table.
It wasn't far now to land, but the boat was being tossed right and left. The fish were in buckets on deck, and when the first wave struck, Leif yelled, "We need to bring in the fish or they'll all end up back in the sea. Be careful when you open the door."
Leif kept his eyes fixed on the black water, battling the swells as best he could. Anders reached for the door handle and pushed open the door. One bucket had turned over, and the fish lay scattered on deck. The next wave crashed over the gunwale and washed some of the catch overboard.
Anders gathered up the remaining fish and threw them back into the bucket. God, this is nuts, he thought. Here I am practically risking my life just to save a few lousy fish. He could see Leif's tense face through the window.
Anders stumbled his way into the wheelhouse. His clothes were soaked through.
"Fucking hell. How's it going?" he asked Leif.
"Okay. We're close to shore, so it's going to be all right. But this weather is damn awful."
Suddenly the lights of the Gnisvärd dock appeared in the dark. Anders breathed a sigh of relief. They were only a hundred yards away.
When they once again had solid ground under their feet, Knutas realized how scared he had actually been. His legs could barely hold him up. They secured the boat and hurried back to the house.
"What an ordeal," Anders gasped. "Right now all I want is to get out of these clothes and take a hot shower."
"You do that," said Leif. "I'll make a fire in the meantime."
Up in his room, Anders discovered that his cell phone was gone. Damn, it must have been washed overboard when he was out on deck. Now Jacobsson wouldn't be able to reach him. He would have to ask Leif if he could borrow his cell. He also wanted to call Lina and tell her about their dramatic adventure. There was no phone in the house, in spite of all the other modern conveniences.
They warmed themselves up with some Irish coffee as they made dinner.
Leif prepared the salmon with an expert hand. He started by slitting open the fish with a sharp knife. Then he removed the guts and pulled the backbone away from the filets. Anders felt his mouth watering as he watched Leif brush the filets with oil, sprinkle them with herbs, and place them on a bed of coarse salt. Then he put the fish in the oven to bake.
When it was ready they hungrily launched into the salmon, washing it down with strongbeer. They talked about the day's drama. What an adventure. It could just as easily have ended in disaster. Outside the window the wind was blowing harder, and more snow was on the way.
After a number of shots of whiskey with their coffee, they were both feeling fairly intoxicated. They listened to some music and talked about trivial things, and by the time Anders went upstairs to bed, it was two in the morning. Leif had passed out on the sofa.
Anders fell into bed and should have fallen asleep instantly. Instead he was wide awake, thinking about the investigation, about Kingsley. According to Jacobsson's text message, the suspect was supposed to return home later today. The case that had consumed all his thoughts, day and night, for the past month would most likely be solved just in time for Christmas Eve. He was looking forward to enjoying Christmas dinner with his family and relatives without having to think about the murder. He suddenly felt a great longing for Lina and the kids. He had an urge to get in the car and drive straight home.
He realized that he wasn't going to be able to sleep. It was fruitless to try, so he got dressed and tiptoed down the stairs. The sofa in the living room was empty, so Leif must have gone to bed even though he hadn't heard him.
Anders sat down on one of the leather armchairs and started to fill his pipe. He lit it and inhaled deeply. He liked smoking when he was alone. He seemed to enjoy it more.
A painting caught his attention. It was of a woman with a dog resting on her lap. The woman was young and slender and wore a sleeveless red dress. Her eyes were closed, and her head was tilted toward one shoulder, as if she were asleep. Her lips were the same color red as her dress. The dog was looking out at the viewer. It was a beautiful painting.
Knutas leaned forward to see who the artist was. He got up from the armchair and ran his finger along the gilded frame. Moved his eyes to the wallpaper, which was a pale yellow with a slightly brighter border. Next to the painting stood a chair with a high back, richly decorated, and two turned posts with knobs. The details were merging into a puzzle, and slowly he realized where he had seen this chair before. It was without a doubt the distinctive chair back that was visible in Dahlström's photographs. Norrby, who was interested in antiques, had explained that it was an English Baroque chair.
At first he was overcome with utter confusion. How could Dahlström have taken pictures of Fanny in Leif's house? Had he and some companion exploited her in the summer house without Leif's knowledge? Did it happen while Dahlström was building the sauna?
His thoughts moved on and in his mind everything began coalescing to create an appalling pattern. Leif owned a horse at the stable and he had hired Dahlström. His appearance matched the description. It could just as well have been Leif in the photos. His friend of twenty years. An electric shock wave of ice-cold instinct shot through his body, making its way into every nook and cranny. He lost his grip on his pipe, which fell to the floor, scattering bits of tobacco over the rug.
He took another look at the painting to make sure that he was right. No, no. He couldn't believe it, refused to believe it. The thought passed through his mind that he should just go to bed and pretend that he hadn't noticed anything. He should bury his head in the sand and go on as usual. Part of him wished that he'd never seen that painting.
He tried to convince himself that there must be some other answer. Suddenly it occurred to him that Leif had been out in the boathouse the night before. What was he doing out there?
He had to go and take a look. Quickly he put on his shoes and jacket and then opened the front door as quietly as he could. He crossed the dark yard as his thoughts whirled. A jumble of irreconcilable images appeared in his mind: Leif in the sauna, on the ski slope, as Santa Claus at their house, playing soccer on the beach, standing in Dahlström's darkroom with a hammer in his hand and acting with cold-blooded brutality, bending over Fanny Jansson's young body in the photographs. He went around the corner of the house, and it took a few seconds before he noticed the figure in front of him. Suddenly he was standing face-to-face with Leif, who was holding his hands at a strange angle behind his back, as if he were hiding something. But Knutas never managed to see what it was.
## Sunday, December 23
Lina sounded worried when she called Karin Jacobsson early in the morning.
"I haven't heard from Anders since yesterday morning. Have you?"
"No, his cell is turned off. I've tried to call him several times."
"Leif doesn't answer, either. I just talked to Ingrid. I'm starting to worry. They were going out in the boat yesterday, and since then a real storm has blown in. I hope nothing has happened."
"I'm sure they're fine," Karin reassured her. "Anders said that he'd be here this afternoon. His cell battery probably ran out. Don't they have a phone at the summer house?"
"No. I'm thinking of driving out there to see if everything's all right. This is making me nervous. It's so unlike Anders not to call."
Jacobsson checked her watch. Ten fifteen. Kingsley wasn't supposed to land until that afternoon.
"Listen, I'll go out there myself. I can get away at the moment."
"Are you sure?"
"Yes. I'll be there in half an hour. We'll call you as soon as I get there."
"Thank you."
Jacobsson had tried to call Knutas on his cell many times without getting through, and she had started feeling uneasy herself. On her way out to Gnisvärd, she called the Marine Rescue Service. No, nothing had happened, as far as they knew. She got the same answer from the Coast Guard.
The road was slick. The temperature had dropped overnight and the slush had frozen, transforming the road into a sheet of ice. Jacobsson kept a safe distance from the other cars and was grateful that there was very little traffic.
When she came to the sign for Gnisvärd, she turned off and continued along a smaller road toward the old fishing village. The Almlöv summer house was half a mile away, in a secluded spot near the water. She had been there once before, for a crayfish party. The house had a marvelous location with its own dock.
The car was parked in the yard, and the boat was tied up at the dock. So they had to be close by.
It was almost eleven thirty. The house seemed deserted. No smoke from the chimney, and the lights were turned off. Of course it was daylight, but the clouds made it seem quite dark outside.
She knocked on the door. No answer. Pounded harder. Still no reaction.
She saw no sign of human activity anywhere, except for the footprints in the snow leading back and forth between the house and the dock. Maybe they were out taking a walk.
Imagine having a place like this, she thought enviously. Such peace. She looked out at the sea and the boathouse made of limestone. Farther down toward the water, right next to the dock, stood the sauna. That was the one that Dahlström had built. He had been paid under the table for it. She started walking across the yard. She didn't notice the person who appeared right behind her.
She heard only a brief rushing sound before she fell to the ground.
On the day before Christmas Eve the call that he had been dreading came through. Her words were like a battle tank that mowed him down. Powerful and inexorable.
"It's not going to work anymore. I can't keep doing this. I have to make up my mind, once and for all. I really care a lot for you, Johan, but I'm not ready to split up my family."
"I see," he said tonelessly.
"You have to understand. I just can't," she said, sounding more insistent. "It's for the sake of my children, too. They're still so young. And Olle and I get along fine, actually. It's not exactly a passionate sort of love, but it works."
"How nice for you."
"No, don't do that, Johan. I realize you're upset. This is really hard for me, too. Don't make things worse than they already are."
"Right."
"Don't be like that," she cried, sounding annoyed. "Don't make me feel even more guilty than I already do!"
"So that's how it is. You just call me up and tell me it's over, after you've said a hundred times that you love me, and that you've 'never felt this way about anyone else,'" he said, doing a terrible impression of her by raising his voice to a falsetto.
"Then in less than a minute you tell me that I have to understand, that I shouldn't make things worse than they already are, and that I shouldn't make you feel guilty. Thanks a fucking lot. How considerate of you. But you think you can just crush me underfoot like a cockroach. No problem at all. First you throw yourself into my arms and tell me that I'm the best thing that's ever happened to you—well, except for your kids that you're always talking about—and then you think it's perfectly all right to just call me up and say it's over!"
"It's good that you brought up the part about my children," she said, her voice icy cold. "That just confirms what I've suspected all along! You think it's a nuisance that I have kids! Unfortunately, they're part of the package, you know."
"Don't go saying that Sara and Filip have been some sort of obstacle, damn it. As you know, I've been fully prepared to take care of both you and the children. I've been daydreaming about moving to Gotland and maybe getting a job at the radio station or at one of the newspapers. The children would live with us, and I've thought about what my relationship would be with them. I wouldn't force things. I'd take it easy. I would just be there for them and do the best I could. That's what I've been thinking. And that maybe they would eventually get to know me and want to be with me, that we would play soccer and build tree houses and things like that. I love you—don't you understand that? Maybe you don't realize what that means. It's so damn easy for you to bring up the whole issue with the children. You're using Sara and Filip as some kind of fucking shield so that you won't have to change your life!"
"Great," she said sarcastically. "You said their names. I think that's the first time I've ever heard you do that! So now you seem to think it's time to show some interest in them. Well, it's a little late for that."
Johan sighed in resignation.
"Think whatever you like," he said. "But I'm sure that's exactly how things stand. You simply don't dare break things off with Olle. You're too scared. You should at least acknowledge this to yourself and stop putting the blame on anything else."
"You think you know everything," she snapped, now sounding on the verge of tears. "Maybe a lot of things have been happening over here that you don't know anything about. Everything is so easy for you, but life can be very complicated. I hope you'll learn that someday. You don't know shit about what I've been going through."
"Well, tell me! You've shut me out for weeks now. I've called and called, and the closest I can get to you is by talking to Viveka. How can I do anything if I don't know what's going on? Tell me what it is, and I'll help you. I love you, Emma. Can't you get that into your head?"
"No, I can't. I can't tell you what it is," she said in a stifled voice.
"What do you mean? What can't you tell me?"
"Nothing, Johan. I have to go now. Merry Christmas, have a nice holiday, Happy New Year, and have a great life!"
She hung up.
Karin Jacobsson woke to find herself tied to a bed. A rope had been wound around her body, and she was completely immobilized, as if she were in a vise. Her arms and legs were numb, and her head hurt. She tried to get her bearings in the room as best she could from her immovable position. She was in a child's bedroom that she recognized from her previous visit. On the table was an old-fashioned Parcheesi game with different-colored wooden cones as markers. There were chairs with homemade cushions covered with a tiny flower pattern and a Strindberg lamp. A polished hardwood floor, white cotton curtains at the window. How idyllic and homey it all was.
The house was quiet. Who had hit her?
What had happened to Knutas and Leif?
She listened for any sounds but couldn't hear a thing.
How long had she been lying here? She had arrived at about eleven thirty. Through the window she saw that it was still overcast and impossible to figure out how high the sun was in the sky.
She tried to move her hands, which were tied to the sides of the bed. The rope cut into her wrists.
It wasn't any better with her legs. With an effort she managed to lift her head and look around. Her jacket was lying on a chair. She arched her body, straining against the rope the way she had seen escape artists do. Tense and release, tense and release. Stubbornly she kept on, varying it by twisting and turning her wrists as she tried to loosen the rope.
At the same time her concern about Anders and Leif grew.
It bothered her that it was so quiet in the house. If someone had tied her up like this, shouldn't that person be close by? Karin felt her anger growing. She had no intention of lying here like some sacrificial lamb, waiting for someone to take her to slaughter. She tensed her muscles and arched her body up toward the ceiling as hard as she could.
The rope loosened enough to give her new hope. She repeated the movement. Suddenly she felt the rope release. The next instant she was able to free one hand and her left arm.
In a matter of minutes she was free and off the bed. She stretched her body, waved her arms, and shook out her legs to get the blood circulating. She crept over to the window and looked out. She could see the water, which was motionless and gray, the boathouse and the sauna down by the shore. Not a soul in sight. She put on her jacket and put her hand in her pocket for her cell phone and car keys. They were gone.
The plane landed on schedule at Arlanda Airport. After Tom Kingsley came through passport control, the police were waiting for him.
The arrest was undramatic. Kingsley mostly looked surprised. The police explained to him that he was under arrest. Then he was cuffed and escorted by two plainclothes officers to the domestic terminal to wait for the plane to Gotland later that afternoon.
The news that he had been arrested was received with relief and joy at police headquarters in Visby. Kihlgård called Knutas but got no answer. He tried Jacobsson's cell, but again with negative results.
"Why the hell can't we get hold of the two top officers when something is finally happening?" he roared.
"Karin was driving out to Gnisvärd this morning," said Wittberg. "Anders has apparently not answered his cell phone all weekend. She was worried that something might have happened. Hell, I forgot all about that."
"What do you mean by 'something might have happened'?" growled Kihlgård.
"He and Leif were going out in the boat, and there were nearly gale force winds."
Kihlgård looked at his watch.
"Let's drive out there. We've got time."
A dull thudding sound was audible as Jacobsson came out into the yard. It sounded like pounding and it was coming from inside the boathouse.
She peered through the window but couldn't see anything unusual. Not a sound. She stood still and waited. She pressed her body against the locked door to hear better. Then the thudding resumed, at a slower beat. It sounded almost halfhearted now.
She needed something to break the window. Her car stood where she had left it, next to Leif's. In the trunk she found a tire iron. It was now or never. With a crash the glass shattered and fell in like confetti. Jacobsson whispered through the broken window, "Anders, are you there?"
The whimper that came in response indicated that he had been gagged. She leaned down and looked inside. There in the dark she could make out her boss lying on the floor, his hands and feet bound, a rag stuffed in his mouth.
She turned around and looked up at the house. Not a sign of life. She reached inside for the latch and opened the window, cutting her hand on the broken glass. Damn it. She was bleeding, but that didn't matter. She climbed in.
When she looked into Knutas's eyes, she had never seen him so helpless. Quickly she started untying the rope that held the gag in place. He gasped when at last he was free.
"Thanks. I'd almost given up hope. I thought I was going to rot in this damn place."
"Where's Leif?" asked Jacobsson as she wrestled with the knots that held Knutas's wrists behind his back.
"I don't know. How did you happen to come out here?"
"We started getting worried when we didn't hear from you. But when I got here someone hit me over the head and tied me to a bed inside the house. I managed to get free and came out here looking for you. I heard you thumping."
"It was Leif."
Jacobsson paused. "What?"
"I think it was Leif who murdered both Dahlström and Fanny."
"Are you out of your mind!"
"No, I mean it. I'll explain later."
Something in his voice made her realize that it was true.
"Is the car still here?" he asked.
"Yes, it's parked outside."
"What about the boat?"
"It's at the dock."
"We have to get out of here. We have to get help."
The door was locked from the outside, so they climbed out the window and ran across the yard to the road.
After they had gone a hundred yards from the house, they heard a deafening boom. They turned around to a sea of fire. The sauna down by the water had exploded into an inferno of flames, building materials, and smoke. They watched the macabre spectacle in silence.
"He blew the whole thing up," gasped Knutas.
They approached the burning building and saw the flames reflected in the water.
The only thing Knutas could think about was whether Leif was inside.
Neighbors who had heard the explosion came driving up. They had alerted both the police and the fire department. Knutas and Jacobsson were tended to by their colleagues. Knutas managed to convince the medics that he didn't need to go to the hospital. He at least needed to stay at the scene long enough to see how things developed. Jacobsson felt the same. Finally they agreed to sit inside an ambulance to watch everything going on around them. Uniformed and armed police officers went into the house while others searched the area with dogs. The firefighters fought the blaze down at the dock, and several officers crept inside the boathouse with their guns drawn. The whole scene is right out of a movie, thought Knutas.
Gradually the police offices regrouped in the front yard. The firefighters had the fire under control, and now it was just a matter of putting it out completely. They had not yet found Leif Almlöv.
## Wednesday, December 26
The residential street was quiet and deserted, but inside the houses the Christmas dinner celebrations were fully under way. In some driveways sparklers were burning in the winter darkness, and cars were parked outside the gates.
He paused outside the fence to look at the house. There were lights in all the windows. Advent stars made of straw and wood gave off a gentle glow. In the living room a tall Advent candle in a cast-iron holder was visible along with two big amaryllis plants. The red flowers bore witness to much careful tending. He saw the family moving about inside. Back and forth between the living room and the kitchen. He knew that they had a dining table in the living room.
He caught a glimpse of Filip playing with a puppy. Did they have a new dog? Not a good sign. Not at all.
He opened the gate. The gravel crunched under his feet. The snow had vanished again, melting away on Christmas Eve. Now a gray haze had settled over the idyllic residential neighborhood in Roma.
He went up to the front porch and saw through the window that Olle had noticed him. Now there was no turning back. He took a deep breath and rang the bell.
## Epilogue
The chapel stood in a secluded spot near the fishing village of Kovik on the west side of the island, about five miles south of Gnisvärd.
It was built from Gotland limestone, with a single window like a porthole facing the cow pastures, the windswept boathouses, and the sea. The chapel had been dedicated to the memory of the men who had drowned at sea.
Leif Almlöv came from a family of fishermen who for generations had fished the stormy Baltic along Gotland's coast. That was where he was to be buried, in accordance with his last wishes. Only his immediate family were present.
Knutas sat in the back row of folding chairs that had been set up in the small space. He fixed his eyes on the flower-bedecked coffin in the front of the chapel as he pondered who Leif really was. Or rather, had become.
Everything seemed to have started with Fanny Jansson. Of course Leif had visited the stables on numerous occasions. This was confirmed by his father-in-law, with whom he shared ownership of the horse. That was where he had met the girl.
Then Leif had hired Dahlström to build the sauna out in the country, but the carpenter had discovered what Leif was doing with Fanny. Maybe Dahlström had spent the night there while he was working on the sauna and then saw something that he wasn't supposed to see.
That was the beginning of the end for everyone involved.
No one had any doubts that Leif was the perpetrator. It was his fingerprints that had been found in Dahlström's darkroom, in his apartment, and on the murder weapon. His hair and saliva were on Dahlström's clothes, and on Fanny's.
Several weeks had now passed since that fateful day out at Gnisvärd, which had ended with Leif perishing in the flames. The reason for the powerful explosion was the cylinders of gas that were kept in the storeroom next to the sauna. They could have blown up the boathouse, too; only a few yards separated the two buildings. A nasty chill spread through Knutas's body as he thought that his friend of twenty years might have been planning to blow him up. And what about Karin? The thought was inconceivable, but it was just as unbelievable that Leif could have murdered two people.
Leif's remains had been found in the ashes under the burned-out sauna. Whether he had committed suicide, they would never know. Knutas's thoughts turned again to Ingrid and the children. What sort of life was in store for them after all this? Was it even possible for them to go on?
And Fanny—she was just a child. Knutas felt a deep sorrow when he thought about the fourteen-year-old girl. She hadn't even had a chance to begin her adult life. At the same time he was weighed down by feelings of guilt. He wondered how much his friendship with Leif had interfered, and to what extent it had blinded him. He was fully aware that in his position as head of the homicide team, he was ultimately responsible for the investigation.
Outside the chapel the local press had gathered along with a number of curiosity seekers. Knutas declined to answer any questions. He slipped away and stared out at the horizon.
Three seagulls were flying low over the surface of the water. The sea was unusually still, and the new year had begun.
## Also by Mari Jungstedt
Unseen
## Author's Acknowledgments
This story is entirely a work of fiction. Any similarity between the characters in the book and real individuals is unintentional. Occasionally I've taken the liberty of changing certain things to suit the narrative. This includes the TV news division of Swedish TV on Gotland, which is very much alive, although in the book it has been closed down so that the responsibility for Gotland was moved to Stockholm. I have done this simply in order to tell the story in my own way. I have the greatest admiration for the existing team in Visby and for Swedish TV's regional news program Östnytt, which is responsible for covering Gotland in real life.
Any errors that may have slipped into the book are solely my own.
First and foremost, I want to thank my husband, the journalist Cenneth Niklasson, who is my greatest inspiration, my biggest supporter, and my most persistent critic.
Many thanks also to:
Gösta Svensson, former detective superintendent of the Visby police for his invaluable assistance with the police work.
Johan Gardelius and Bo Ekedahl, crime techs, the Visby police.
Martin Csatlos, the forensic medicine laboratory in Solna.
Neng Wanlayaphol, trotting-horse trainer, Visby Racetrack.
Mats Wihlborg, district prosecutor, Visby.
Jenny Ingårda and Eva Waltré, BRIS—Children's Rights in Society.
Il-nam Kroon, social worker.
Mikaela Säfvenberg, archaeologist and authorized guide, Gotland.
My mother, Kerstin, and my sister Ewa Jungstedt for their assistance on research trips to Gotland.
Tove Wiklander—for her constant positive support during our speed walks.
I would also like to give a warm thank-you to my publisher, Jonas Axelsson, for his faith in me, and to my editor, Ulrika Åkerlund, for all her help with this book.
And to my advance readers for their valuable opinions:
Anna-Maja Persson, journalist, Swedish TV.
Lena Allerstam, journalist, Swedish TV.
Lilian Andersson, editor, Bonniers Educational Books.
Bosse Jungstedt, my brother, and Kerstin Jungstedt, my sister-in-law.
Last, but not least, thanks to my beloved children, Rebecka and Sebastian Jungstedt, for their good humor, love, encouragement, and genuine patience with their mother's writing.
Älta, July 2004
Mari Jungstedt
This is a work of fiction. All of the characters, organizations, and events portrayed in this novel are either products of the author's imagination or are used fictitiously.
UNSPOKEN. Copyright © 2004 by Mari Jungstedt. English translation © 2007 by Tiina Nunnally. All rights reserved. No part of this book may be used or reproduced in any manner whatsoever without written permission except in the case of brief quotations embodied in critical articles or reviews. For information, address St. Martin's Press, 175 Fifth Avenue,
New York, N.Y. 10010.
www.minotaurbooks.com
Library of Congress Cataloging-in-Publication Data
Jungstedt, Mari, 1962–
[In denna stilla natt. English]
Unspoken / Mari Jungstedt; [translated by Tiina Nunnally].—1st ed.
p. cm.
ISBN: 978-1-4668-0733-4
I. Nunnally, Tiina, 1952–II. Title.
PT9877.2.U64I613 2007
839.73'8—dc22
2007019345
First published in Sweden under the title In Denna Stilla Natt by
Albert Bonniers Förlag
| {
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Q: Underbrace under regular text, including spaces Solution:
\newcommand{\undertext}[2] {\[\underbrace{\text{#1}}_{\text{#2}}\]}
Usage:
\undertext{textAbove}{textBelow}
I would simply like to use an underbrace under a regular, written sentence of text.
A: There are numerous solutions. Perhaps easiest would be to simply use mathmode, wrapped inside a new command:
\newcommand{\undertext}[2] {$\underbrace{\textrm{#1}}_{\textrm{#2}}$}
Then you can simply \undertext{do}{this}.
| {
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} | 1,360 |
{"url":"https:\/\/www.physicsforums.com\/threads\/integral-of-speed-as-a-function-of-displacement.596957\/","text":"# Integral of speed as a function of displacement\n\n1. Apr 15, 2012\n\n### Paul Czerner\n\nHello,\n\nI'm trying to figure out a way to do an integral of a speed\/velocity function v(x) described as a function of displacement (the speed changes based on the position of the particle). I want to know the distance traveled after a specific period of time, and I can't figure out how to formulate the integral for that.\n\nDo path integrals or integration by parts play a role in this? Thanks.\n\n2. Apr 16, 2012\n\n### haruspex\n\nYou have dx\/dt = f(x)?\nTurn that into dt = dx\/f(x) and integrate, if you can:\n\nt = $\\int$(1\/f(x)).dx\n\nHowever, you mentioned path integrals, so maybe there's more than one space dimension here?\nPlease provide the actual equation.\n\n3. Apr 16, 2012\n\n### Paul Czerner\n\nI don't have a specific problem to show, just an idea I want to follow up on.\n\nIt may be true that there are no path integrals (as of yet) as you mention involved in the problem, just one dimension so far. But I mention it just in case the solution may require expansion of the problem into more dimensions to find the solution.\n\nI've tried the integrating for t that you show, but maybe I'm not stating the problem correctly. I want to find the solution after a specific time t1 of the distance traveled, given the speed (one dimension) as it changes based on position x, so the displacement x' depends on position x and time t.\n\nI guess I'm looking for something in the form:\n\nx' = f(v(x),t1),\n\nand I'm sure it takes at least some form of integral. The velocity is a function of time only so far as it changes as it crosses space, and the position in space changes the velocity. Would it make sense to have a function of space as it changes over time, maybe stretching the space variable while keeping velocity constant with respect to a stretching space? Maybe I have to generate acceleration functions before going back to derive displacement? Or is there a simple solution I haven't remembered from my college days?\n\nThere must be some answer to something like this already, though I have been searching online and haven't found it yet, unless it's more complex than I'm familiar with. I'm thinking of solutions to a particle moving through a space-dependent force field.\n\nThanks in advance for any headway help.\n\n4. Apr 16, 2012\n\n### haruspex\n\nAs I understand it, you have speed v = v(x), so\nt = \u222b(1\/v(x)).dx\nIf you want to know the position x1 reached at time t1 then you have to solve\nt1 = $\u222b^{x1}_{0}$(1\/v(x)).dx\n\nE.g. suppose v(x) = a.x + b\nt1 = [ln(x+b\/a)\/a$]^{x1}_{0}$ = (ln(x1+b\/a) - ln(b\/a)]\/a\na.x1 = b.exp(a.t1) - b\n\nIf that's not what you're trying to do, please pick a specific v=v(x) so that we an discuss it more clearly.\n\n5. Apr 16, 2012\n\n### Paul Czerner\n\nThank you, it's kind of what I was looking for; basically, as you confirmed, it's an inverse problem where the variable that is to be solved is the boundary of the integral. I'll work on this a bit and see what I come up with.","date":"2018-08-19 02:19:07","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\": 0.731319010257721, \"perplexity\": 483.9776291617939}, \"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-2018-34\/segments\/1534221214538.44\/warc\/CC-MAIN-20180819012213-20180819032213-00028.warc.gz\"}"} | null | null |
Caprimulgus madagascariensis é uma espécie de ave da família Caprimulgidae.
Pode ser encontrada nos seguintes países: Comores, Madagáscar, Mayotte e Seychelles.
Os seus habitats naturais são: florestas subtropicais ou tropicais húmidas de baixa altitude e regiões subtropicais ou tropicais húmidas de alta altitude.
Referências
Caprimulgus
Aves descritas em 1840 | {
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{"url":"http:\/\/bootmath.com\/how-can-i-prove-that-leftsum_i0r-1i-binomai-binomn-ar-iright-leq-binomnr.html","text":"# How can I prove that $\\left|\\sum_{i=0}^r (-1)^i \\binom{a}{i} \\binom{n-a}{r-i}\\right| \\leq \\binom{n}{r}$?\n\nThis is a conjecture:\n\nHow can I prove that\n\n\\left|\\sum_{i=0}^r (-1)^i \\binom{a}{i} \\binom{n-a}{r-i}\\right| \\leq \\binom{n}{r}\n\nfor $0\\leq a \\leq n$, $0\\leq r \\leq n$ and $n,r,a \\in \\mathbb{N}$ ?\n\n#### Solutions Collecting From Web of \"How can I prove that $\\left|\\sum_{i=0}^r (-1)^i \\binom{a}{i} \\binom{n-a}{r-i}\\right| \\leq \\binom{n}{r}$?\"\n\nWe have $$\\left|\\sum_{i=0}^{r}\\left(-1\\right)^{r}\\dbinom{a}{i}\\dbinom{n-a}{r-i}\\right|\\leq\\sum_{i=0}^{r}\\dbinom{a}{i}\\dbinom{n-a}{r-i}=\\dbinom{n}{r}$$ where the last identity follows from the Chu-Vandermonde identity.","date":"2018-07-22 08:51:40","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9963268041610718, \"perplexity\": 542.8315517845681}, \"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\/1531676593142.83\/warc\/CC-MAIN-20180722080925-20180722100925-00341.warc.gz\"}"} | null | null |
KM Category : Featured Articles - Below
Hindi version of Kumbh Mela book launched in Lucknow
The event featured Shri Akhilesh Yadav, Honorable Chief Minister of Uttar Pradesh, as well first-hand insights from Harvard scholars and Kumbh administrators who were on the ground during the festival in 2013.
"A chance of a lifetime"
At an event on May 19 at the Museum of Fine Arts in Boston, faculty leaders from the Kumbh Mela project shared why studying the world's largest gathering provided so many lessons for fields such as business, public health, and urban planning.
TedX: Lessons from Kumbh Mela – Worlds Largest Gathering of Humanity
Watch SAI Director Tarun Khanna give a TedX talk about what the world can learn from the Kumbh Mela.
Delhi Notebook: How the wedding syndrome could fix India
In a recent column for The Financial Times, 'How the wedding syndrome could fix India,' Victor Mallet writes about how the Indian government manages one-off events but not longer-term projects, and cites SAI's recently- published book on the Kumbh...
Studying the Kumbh Mela from many perspectives
On Monday, January 18, the Harvard South Asia Institute (SAI) launched the book and exhibition Kumbh Mela: Mapping the Ephemeral Megacity in Mumbai at the Chhatrapati Shivaji Maharaj Vastu Sangrahalaya.
Asia Society: Lessons from the Kumbh Mela
At the Asia Society in New York on November 6, faculty leaders discussed how the Kumbh Mela is an opportunity to learn about megacities, possibly illuminating solutions to natural disasters that require temporary housing.
Asia Society: The 'Sheer Spectacle' of Kumbh Mela
"They had six weeks to get a city up and running with infrastructure, water supply, and electricity. That was an amazing surprise. But equally surprising was how quickly they dismantled it," says Rahul Mehrotra, Graduate School of Design, one of the faculty members who will speak about the project at the Asia Society on Nov. 6
Faith in a Smart City
"India is no stranger to confusion and that is why the much-talked-of jugaad, muddling through or miraculously rescuing a situation at the last moment is so common. But the Kumbh Mela, according to the Harvard team, was a carefully planned and efficiently executed operation."
Tracking disease at the world's largest religious festival
The Jana Swasthya Project has introduced a new mobile health surveillance system to help keep the millions of visitors healthy while they're at the festival.
The Jana Swasthya Project at the 2015 Kumbh Mela: Ushering in India's Mobile Health Revolution
The cornerstone of the project is a unique interactive visual analytic tool ("dashboard") that provides critical disease surveillance data to health officials in real time.
Kumbh Mela book launch in Delhi
On Monday, August 17, SAI launched the Kumbh Mela: Mapping the Ephemeral Megacity book and exhibition in Delhi, India. Shri Akhilesh Yadav, Honorable Chief Minister of Uttar Pradesh, was on hand to launch the book with Harvard faculty, to a crowd of over 250 people.
A powerful convergence
Read an article from the Harvard Gazette about SAI's launch of "Kumbh Mela: Mapping the Ephemeral Megacity." | {
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\section*{Introduction}
Algebraic classification (up to isomorphism) of algebras of small dimension from a certain variety defined by a family of polynomial identities is a classic problem in the theory of non-associative algebras. There are many results related to algebraic classification of small dimensional algebras in varieties of Jordan, Lie, Leibniz, Zinbiel and other algebras.
Another interesting approach of studying algebras of a fixed dimension is to study them from a geometric point of view (that is, to study degenerations and deformations of these algebras). The results in which the complete information about degenerations of a certain variety is obtained are generally referred to as the geometric classification of the algebras of these variety. There are many results related to geometric classification of Jordan, Lie, Leibniz, Zinbiel and other algebras \cite{ack, kv17,ikv17, kv16}.
In 1972, Kantor introduced the notion a conservative algebra as a generalization of Jordan algebras \cite{Kantor72}. Unlike other classes of non-associative algebras, this class is not defined by a set of identities. To introduce the notion of a conservative algebra, we need some notation.
Let $\mathbb V$ be a vector space, let $A$ be a linear operator on $\mathbb V,$ and let $B$ and $C$ be bilinear operators on $\mathbb V.$ For all $x,y,z \in \mathbb V,$ put
\begin{gather*}
[B,x](y)=B(x,y),\\
[A,B](x,y)= A(B(x,y))- B(A(x),y)-B(x,A(y)),\\
[B,C](x,y,z)=B(C(x,y),z) + B(x,C(y,z))+ B(y,C(x,z))\\
-C(B(x,y),z)- C(x,B(y,z))-C(y,B(x,z)).
\end{gather*}
Consider an algebra as a vector space $\mathbb V$ over a field $\mathbb{C}$, together with an
element $\mu$ of $\operatorname{Hom}(\mathbb V \otimes \mathbb V, \mathbb V),$ so that $a \cdot b =\mu(a \otimes b).$
For an algebra $(\mathbb V, \mathcal P)$ with a multiplication $\mathcal P$ and $x\in \mathbb V$ we denote by $L_x^{\mathcal P}$ the operator of left multiplication by $x.$
Thus, Kantor defines conservative algebras as follows:
\begin{definition}
An algebra $(\mathbb V, \mathcal P),$ where $\mathbb V$ is the vector space and $\mathcal P$ is the multiplication, is called a conservative algebra if there is a new multiplication $\mathcal P^*: \mathbb V\times \mathbb V\rightarrow \mathbb V$ such that
\begin{equation}\label{uno}
[L_b^{\mathcal P},[L_a^{\mathcal P}, {\mathcal P}]]=-[L_{{\mathcal P^*}(a,b)}^{\mathcal P},{\mathcal P}]
\textrm{, for all $a, b \in \mathbb V.$}
\end{equation}
Simple calculations take us to the following identity with an additional multiplication $\mathcal P^*,$ which must hold for all $a, b, x, y\in \mathbb V$:
\begin{equation}\label{dos}
\begin{split}
b(a(xy)-(ax)y-x(ay)) - a((bx)y) + (a(bx))y+(bx)(ay) -a(x(by))+(ax)(by)+x(a(by))= \\
= - \mathcal P^*(a,b)(xy)+(\mathcal P^*(a,b)x)y + x(\mathcal P^*(a,b)y).
\end{split}
\end{equation}
\end{definition}
The class of conservative algebras is very vast. It includes
all associative algebras,
all quasi-associative algebras,
all Jordan algebras,
all Lie algebras,
all (left) Leibniz algebras,
all (left) Zinbiel algebras,
and many other classes of algebras.
On the other side, all conservative algebras are "rigid" (by Cantarini and Kac) algebras \cite{kacan}.
However, this class is very hard to study and for now even the basic general questions about it remain unanswered, so it is a good idea to study its subclasses which are sufficiently wide but easier to deal with. In \cite{Kantor89} Kantor, studying the generalized TKK functor, introduced the class of terminal algebras which is a subclass of the class of conservative algebras.
\begin{definition}
An algebra $(\mathbb V, \mathcal P),$ where $\mathbb V$ is a vector space and $\mathcal P$ is a multiplication, is called a terminal algebra if for all $a\in \mathbb V$ we have
\begin{equation}\label{tress}
[[[{\mathcal P},a],{\mathcal P}],{\mathcal P}]=0.
\end{equation}
\end{definition}
Note that we can expand the relation (\ref{tress}), obtaining an identity of degree 4. Therefore, the class of terminal algebras is a variety.
Also, about terminal and conservative algebras see, for example \cite{Kantor90, KPP18, cfk19}.
The following remark is obtained by straightforward calculations.
\begin{remark}
A commutative algebra satisfying (\ref{tress}) is a Jordan algebra.
\end{remark}
Aside from Jordan algebras, the class of terminal algebras includes all Lie algebras, all (left) Leibniz algebras and some other types of algebras.
The following characterization of terminal algebras, proved by Kantor \cite[Theorem 2]{Kantor89}, provides a description of this class as a subclass of the class of conservative algebras.
\begin{remark}
An algebra $(\mathbb{V}, {\bf P})$ is terminal if and only if it is conservative
and the multiplication in the associated superalgebra ${\bf P}^*$ can be defined by
\begin{equation}
\label{terminal_associated}
{\bf P}^*(x,y)=\frac{2}{3}{\bf P}(x,y)+\frac{1}{3}{\bf P}(y,x).
\end{equation}
\end{remark}
Our method of classification of nilpotent terminal algebras is based on the calculation of central extensions of smaller nilpotent algebras from the same variety. The algebraic study of central extensions of Lie and non-Lie algebras has a very big story \cite{omirov,hac16,ss78}.
Skjelbred and Sund used central extensions of Lie algebras for a classification of nilpotent Lie algebras \cite{ss78}. After that, using the method described by Skjelbred and Sund were described all non-Lie central extensions of all $4$-dimensional Malcev algebras \cite{hac16}, all non-associative central extensions of $3$-dimensional Jordan algebras, all anticommutative central extensions of $3$-dimensional anticommutative algebras,
all central extensions of $2$-dimensional algebras \cite{cfk18}.
The algebraic classification of nilpotent algebras will be achieved by the calculation of central extensions of algebras from the same variety which have a smaller dimension.
Central extensions of algebras from various varieties were studied, for example, in \cite{ss78,omirov}.
Skjelbred and Sund \cite{ss78} used central extensions of Lie algebras to classify nilpotent Lie algebras.
Using the same method,
all non-Lie central extensions of all $4$-dimensional Malcev algebras \cite{hac16},
all non-associative central extensions of all $3$-dimensional Jordan algebras,
all anticommutative central extensions of $3$-dimensional anticommutative algebras,
all central extensions of $2$-dimensional algebras \cite{cfk18}
and some others were described.
One can also look at the classification of
$3$-dimensional nilpotent algebras \cite{fkkv},
$4$-dimensional nilpotent associative algebras \cite{degr1},
$4$-dimensional nilpotent Novikov algebras,
$4$-dimensional nilpotent bicommutative algebras,
$4$-dimensional nilpotent commutative algebras in \cite{fkkv},
$5$-dimensional nilpotent restricted Lie agebras \cite{usefi1},
$5$-dimensional nilpotent Jordan algebras,
$5$-dimensional nilpotent anticommutative algebras \cite{fkkv},
$6$-dimensional nilpotent Lie algebras \cite{degr3, degr2},
$6$-dimensional nilpotent Malcev algebras \cite{hac18},
$6$-dimensional nilpotent Tortkara algebras,
$6$-dimensional nilpotent binary Lie algebras \cite{ack}.
In the present paper, we classify nilpotent terminal algebras of dimension less than or equal to 4, and obtain the complete description of degenerations of these algebras.
\section{The algebraic classification of nilpotent terminal algebras}
\subsection{Method of classification of nilpotent algebras}
Throughout this paper, we use the notation and methods described in \cite{hac16,cfk18}
and adapted for the terminal case with some modifications (see also \cite{Jac} for a discussion of extensions of algebras in an arbitrary nonassociative variety). Therefore, all statements in this subsection are given without proofs, which can be found in the papers cited above.
Let ${\bf A}$ be a terminal algebra over $\mathbb C$ and $\mathbb V$ a vector space of dimension $s$ over the same base field. Then the space ${\rm Z_T^2}(\bf A,\mathbb V )$ is defined as the set of all maps $\theta :{\bf A} \times {\bf A} \to {\mathbb V}$ such that
\[\theta(b,a(xy) - (ax)y - x(ay)) - \theta(a,(bx)y) + \theta(a(bx),y) + \theta(bx,ay)\]
\[- \theta(a,x(by)) + \theta(ax,(by)) + \theta(x,(ab)y) = -\theta({\bf P}^*(a,b),xy) + \theta({\bf P}^*(a,b)x,y) + \theta(x,{\bf P}^*(a,b)y),\]
where ${\bf P}^*$ is given by (\ref{terminal_associated}). Its elements will be called \textit{cocycles}. For a linear map $f: \bf A \to \mathbb V$ define $\delta f: {\bf A} \times
{\bf A} \to {\mathbb V}$ by $\delta f (x,y ) =f(xy).$ One can check that $\delta f\in {\rm Z_T^2}({\bf A},{\mathbb V} ).$ Therefore, ${\rm B^2}({\bf A},{\mathbb V}) =\left\{ \theta =\delta f\ :f\in \operatorname{Hom}({\bf A},{\mathbb V}) \right\}$ is a subspace of ${\rm Z_T^2}({\bf A},{\mathbb V})$ whose elements are called \textit{coboundaries}. We define the \textit{second cohomology space} ${\rm H_T^2}({\bf A},{\mathbb V})$ as the quotient space ${\rm Z_T^2}({\bf A},{\mathbb V}) \big/{\rm B^2}({\bf A},{\mathbb V}).$
The equivalence class of $\theta \in {\rm Z^2}({\bf A},{\mathbb V})$ in ${\rm H^2}({\bf A},{\mathbb V})$ will be denoted by $[\theta].$
\bigskip
For a bilinear map $\theta :{\bf A} \times {\bf A} \to {\mathbb V}$ define on the linear space ${\bf A}_{\theta }:={\bf A}\oplus {\mathbb V}$ a bilinear product $[-,-]_{{\theta}}$ by
\[[x+x^{\prime },y+y^{\prime }]_{{\theta}}= xy +\theta (x,y) \mbox{ for all } x,y\in {\bf A},x^{\prime },y^{\prime }\in {\mathbb V}.\]
Then ${\bf A}_{\theta }$ is an algebra called an $s${\it{-dimensional central extension}} of ${\bf A}$ by ${\mathbb V}.$ The following statement can be verified directly:
\begin{lemma}
The algebra ${\bf A_{\theta}}$ is terminal \textit{if and only if} $\theta \in {\rm Z_T^2}({\bf A}, {\mathbb V}).$
\end{lemma}
Recall that the {\it{annihilator}} of ${\bf A}$ is defined as the ideal $\operatorname{Ann}({\bf A} ) =\{ x\in {\bf A}: x{\bf A}+{\bf A}x=0\}.$ Given $\theta \in {\rm Z_T^2}({\bf A}, {\mathbb V}),$ we call the set $\operatorname{Ann}(\theta)=\left\{ x\in {\bf A}:\theta (x, {\bf A} )+\theta ({\bf A},x ) = 0 \right\}$ the {\it{annihilator}} of $\theta.$
\begin{lemma}
\label{ann_of_ext}
$\operatorname{Ann}({\bf A}_{\theta }) = (\operatorname{Ann}(\theta) \cap \operatorname{Ann}({\bf A}))
\oplus {\mathbb V}.$
\end{lemma}
\bigskip
Therefore, $0 \neq \mathbb{V} \subseteq \operatorname{Ann}({\bf A}_{\theta}).$ The following lemma shows that every algebra with a nonzero annihilator can be obtained in the way described above:
\begin{lemma}
Let ${\bf A}$ be an $n$-dimensional terminal algebra such that $\dim(\operatorname{Ann}({\bf A}))=m\neq0.$ Then there exists, up to isomorphism, a unique $(n-m)$-dimensional terminal algebra ${\bf A}^{\prime }$ and a bilinear map $\theta \in {\rm Z_T^2}({\bf A}, {\mathbb V})$ with $\operatorname{Ann}({\bf A}')\cap \operatorname{Ann}(\theta)=0,$ where $\mathbb V$ is a vector space of dimension $m,$ such that ${\bf A}\cong {\bf A}^{\prime }_{\theta}$ and ${\bf A}/\operatorname{Ann}({\bf A})\cong {\bf A}^{\prime }.$
\end{lemma}
In particular, any finite-dimensional nilpotent algebra is a central extension of another nilpotent algebra of strictly smaller dimension. Thus, to classify all nilpotent terminal algebras of a fixed dimension, we need to classify cocycles of nilpotent terminal algebras ${\bf A}'$ of smaller dimension (with an additional condition $\operatorname{Ann}({\bf A}')\cap \operatorname{Ann}(\theta)=0$) and central extensions that arise from them.
\bigskip
We can reduce the class of extensions that we need to consider.
\begin{definition}
Let ${\bf A}$ be an algebra and $I$ be a subspace of ${\rm Ann}({\bf A})$. If ${\bf A}={\bf A}_0 \oplus I$
then $I$ is called an {\it annihilator component} of ${\bf A}$.
A central extension of an algebra $\bf A$ without an annihilator component is called a non-split central extension.
\end{definition}
Clearly, we are only interested in non-split extensions (in the contrary case we can cut off annihilator components lying in $\mathbb{V}$ until we obtain a non-split extension).
Let us fix a basis $e_{1},\ldots,e_{s}$ of ${\mathbb V},$ and $\theta \in {\rm Z_T^2}({\bf A},{\mathbb V}).$ Then $\theta$ can be uniquely written as $\theta (x,y) =\theta (x,y) =\sum_{i=1}^s\theta_{i}(x,y) e_{i},$ where $\theta_{i}\in {\rm Z_T^2}({\bf A},\mathbb C).$ Moreover, $\operatorname{Ann}(\theta)=\operatorname{Ann}(\theta_{1})\cap \operatorname{Ann}(\theta_{2})\cap\cdots \cap \operatorname{Ann}(\theta_{s}).$ Further, $\theta \in {\rm B^2}({\bf A},{\mathbb V}) $\ if and only if all $\theta_{i}\in {\rm B^2}({\bf A},\mathbb C).$ Using this presentation, one can determine whether the extension corresponding to a cocycle $\theta$ is split:
\begin{lemma} \cite[Lemma 13]{hac16}
\label{split_lindep}
Let $\theta (x,y) =\sum_{i=1}^s\theta_{i}(x,y) e_{i}\in {\rm Z_T^2}({\bf A},{\mathbb V})$ be such that $\operatorname{Ann}(\theta)\cap \operatorname{Ann}({\bf A}) = 0.$ Then ${\bf A}_{\theta }$ has an annihilator component if and only if $[\theta_{1}],[\theta_{2}],\ldots,[\theta_{s}]$ are linearly dependent in ${\rm H_T^2}({\bf A},\mathbb C).$
\end{lemma}
\bigskip
Some cocycles give rise to isomorphic extensions:
\begin{lemma}
Let $\theta, \vartheta \in {\rm Z_T^2}({\bf A}, {\mathbb V})$ be such that $[\theta] = [\vartheta].$ Then ${\bf A}_\theta \cong {\bf A}_\vartheta.$
\end{lemma}
By above, the isomorphism classes of extensions correspond to certain equivalence classes on ${\rm H_T^2}({\bf A},{\mathbb V}).$ These classes can be given in terms of actions of certain groups on this space. In particular, let $\operatorname{Aut}({\bf A})$ be the automorphism group of ${\bf A},$ let $\phi \in \operatorname{Aut}({\bf A}),$ and let $\psi \in \operatorname{GL}(\mathbb V).$ For $\theta \in {\rm Z_T^2}({\bf A},{\mathbb V})$ define
\[\phi \theta(x,y) =\theta (\phi(x), \phi(y)), ~ \psi\theta (x,y) = \psi(\theta(x,y)).\]
Then $\phi\theta, \psi\theta \in {\rm Z_T^2}({\bf A},{\mathbb V}).$ Hence, $\operatorname{Aut}({\bf A})$ and $\operatorname{GL}(\mathbb V)$ act on ${\rm Z_T^2}({\bf A},{\mathbb V}).$ It is easy to verify that ${\rm B^2}({\bf A},{\mathbb V})$ is invariant under both actions. Therefore, we have induced actions on ${\rm H_T^2}({\bf A},{\mathbb V}).$
\begin{lemma}
Let $\theta, \vartheta \in {\rm Z_T^2}({\bf A},{\mathbb V})$ be such that $\operatorname{Ann}({\bf A}_\theta) = \operatorname{Ann}({\bf A}_\vartheta) = \mathbb{V}.$ Then ${\bf A}_\theta \cong {\bf A}_\vartheta$ if and only if there exist a $\phi \in \operatorname{Aut}({\bf A}), \psi \in \operatorname{GL}(\mathbb V)$ such that $[\phi\theta] = [\psi\vartheta].$
\end{lemma}
\bigskip
Now we rewrite the above lemma in a form more suitable for computations.
Let ${\mathbb U}$ be a finite-dimensional vector space over $\mathbb C.$ The {\it{Grassmannian}} $G_{k}({\mathbb U})$ is the set of all $k$-dimensional linear subspaces of ${\mathbb V}.$ Let $G_{s}({\rm H_T^2}({\bf A},\mathbb C) )$ be the Grassmannian of subspaces of dimension $s$ in ${\rm H_T^2}({\bf A},\mathbb C).$ There is a natural action of $\operatorname{Aut} ({\bf A})$ on $G_{s}({\rm H_T^2}({\bf A},\mathbb C)):$ for $\phi \in \operatorname{Aut} ({\bf A}), W=\langle [\theta_{1}],[\theta_{2}],\dots,[\theta_{s}] \rangle \in G_{s}({\rm H_T^2}({\bf A},\mathbb C)),$ define
\[\phi W=\langle [\phi \theta_{1}], [\phi \theta_{2}],\dots,[\phi \theta_{s}]\rangle.\]
Note that this action is compatible with the action of $\operatorname{Aut} ({\bf A})$ on ${\rm H_T^2}({\bf A},{\mathbb V})$ and the above presentation of a cocycle as a collection of $s$ elements of ${\rm H_T^2}({\bf A},\mathbb C).$ Denote the orbit of $W$ under the action of $\operatorname{Aut} ({\bf A})$ by $\mathrm{Orb}(W).$ It is easy to check that given two bases of a subspace
\begin{equation*}
W=\langle [\theta_{1}],[\theta_{2}],\dots,[\theta_{s}] \rangle =\langle [\vartheta_{1}],[\vartheta_{2}],\dots,[\vartheta_{s}]
\rangle \in G_{s}({\rm H^2}({\bf A},\mathbb C)),
\end{equation*}
we have $\cap_{i=1}^s\operatorname{Ann}(\theta_{i})\cap \operatorname{Ann}({\bf A}) =\cap_{i=1}^s\operatorname{Ann}(\vartheta_{i})\cap \operatorname{Ann}({\bf A}).$ Therefore, we can introduce the set
\begin{equation*}
T_{s}({\bf A}) =\left\{ W=\left\langle [\theta_{1}],%
[\theta_{2}],\dots,[\theta_{s}] \right\rangle \in
G_{s}({\rm H^2}({\bf A},\mathbb C) ) : \cap_{i=1}^s \operatorname{Ann}(\theta_{i})\cap \operatorname{Ann}({\bf A}) =0\right\},
\end{equation*}
which is stable under the action of $\operatorname{Aut}({\bf A}).$
\medskip
Let us denote by $E({\bf A},{\mathbb V})$ the set of all {\it non-split} central extensions of ${\bf A}$ by ${\mathbb V}.$ By Lemmas \ref{ann_of_ext} and \ref{split_lindep}, we can write
\begin{equation*}
E({\bf A},{\mathbb V}) =\left\{ {\bf A}_{\theta }:\theta (x,y) =\sum_{i=1}^s\theta_{i}(x,y) e_{i}\mbox{ and }\langle [\theta_{1}],[\theta_{2}],\dots,
[\theta_{s}] \rangle \in T_{s}({\bf A})\right\}.
\end{equation*}
Also, we have the next result, which can be proved as \cite[Lemma 17]{hac16}.
\begin{lemma}
Let $\theta(x,y) =\sum_{i=1}^s \theta_{i}(x,y) e_{i}$ and $\vartheta(x,y) = \sum_{i=1}^s\vartheta_{i}(x,y) e_{i}$ be such that ${\bf A}_{\theta},{\bf A}_{\vartheta }\in E({\bf A},{\mathbb V}).$ Then the algebras ${\bf A}_{\theta }$ and ${\bf A}_{\vartheta }$ are isomorphic if and only if
\[\mathrm{Orb}\langle [\theta_{1}], [\theta_{2}],\dots,[\theta_{s}] \rangle = \mathrm{Orb}\langle [\vartheta_{1}],[\vartheta
_{2}],\dots,[\vartheta_{s}] \rangle. \]
\end{lemma}
Thus, there exists a one-to-one correspondence between the set of $\operatorname{Aut}({\bf A}) $-orbits on $T_{s}({\bf A})$ and the set of isomorphism classes of $E({\bf A},{\mathbb V}).$ Consequently, we have a procedure that allows us, given a terminal algebra ${\bf A}^{\prime }$ of dimension $n,$ to construct all non-split central extensions of ${\bf A}^{\prime }.$ This procedure is as follows:
\medskip
{\centerline{\it Procedure}}
\begin{enumerate}
\item For a given (nilpotent) terminal algebra $\bf{A}^{\prime }$
of dimension $n-s,$ determine ${T}_{s}(\bf{A}^{\prime })$
and $\operatorname{Aut}(\bf{A}^{\prime }).$
\item Determine the set of $\operatorname{Aut}(\bf{A}^{\prime })$-orbits on $%
{T}_{s}(\bf{A}^{\prime }).$
\item For each orbit, construct the terminal algebra corresponding to one of its
representatives.
\end{enumerate}
The above described method gives all (Leibniz and non-Leibniz) terminal algebras. But we are interested in developing this method in such a way that it only gives non-Leibniz terminal algebras, because the classification of all Leibniz algebras is given in \cite{demir}. Clearly, any central extension of a non-Leibniz terminal algebra is non-Leibniz. But a Leibniz algebra may have extensions which are not Leibniz algebras. More precisely, let ${\bf L}$ be a Leibniz algebra and $\theta \in {\rm Z_T^2}({\bf L}, {\mathbb C}).$ Then ${\bf L}_{\theta }$ is a Leibniz algebra if and only if
\begin{equation*}
\theta ( x, yz ) = \theta (xy, z )+ \theta (y, xz )
\end{equation*}
for all $x,y,z\in {\bf L}.$ Define the subspace ${\rm Z_L^2}({\bf L},{\mathbb C})$ of ${\rm Z_T^2}({\bf L},{\mathbb C})$ by
\begin{equation*}
{\rm Z_L^2}({\bf L},{\mathbb C}) =\left\{\begin{array}{c} \theta \in {\rm Z_T^2}({\bf L},{\mathbb C}) : \theta ( x, yz ) = \theta (xy, z ) + \theta (y, xz ) \text{ for all } x, y,z\in {\bf L}\end{array}\right\}.
\end{equation*}
Observe that ${\rm B^2}({\bf L},{\mathbb C})\subseteq{\rm Z_L^2}({\bf L},{\mathbb C}).$
Let ${\rm H_L^2}({\bf L},{\mathbb C}) =%
{\rm Z_L^2}({\bf L},{\mathbb C}) \big/{\rm B^2}({\bf L},{\mathbb C}).$ Then ${\rm H_L^2}({\bf L},{\mathbb C})$ is a subspace of $%
{\rm H_T^2}({\bf L},{\mathbb C}).$ Define
\begin{eqnarray*}
{\bf R}_{s}({\bf L}) &=&\left\{ {\bf W}\in {T}_{s}({\bf L}) :{\bf W}\in G_{s}({\rm H_L^2}({\bf L},{\mathbb C}) ) \right\}, \\
{\bf U}_{s}({\bf L}) &=&\left\{ {\bf W}\in {T}_{s}({\bf L}) :{\bf W}\notin G_{s}({\rm H_L^2}({\bf L},{\mathbb C}) ) \right\}.
\end{eqnarray*}
Then ${T}_{s}({\bf L}) ={\bf R}_{s}(
{\bf L})$ $\mathbin{\mathaccent\cdot\cup}$ ${\bf U}_{s}(
{\bf L}).$ The sets ${\bf R}_{s}({\bf L}) $
and ${\bf U}_{s}({\bf L})$ are stable under the action
of $\operatorname{Aut}({\bf L}).$ Thus, the terminal algebras
corresponding to the representatives of $\operatorname{Aut}({\bf L}) $%
-orbits on ${\bf R}_{s}({\bf L})$ are Leibniz algebras,
while those corresponding to the representatives of $\operatorname{Aut}({\bf L}%
) $-orbits on ${\bf U}_{s}({\bf L})$ are not
Leibniz algebras. Hence, we may construct all non-split non-Leibniz terminal algebras $%
\bf{A}$ of dimension $n$ with $s$-dimensional annihilator
from a given terminal algebra $\bf{A}%
^{\prime }$ of dimension $n-s$ in the following way:
\begin{enumerate}
\item If $\bf{A}^{\prime }$ is non-Leibniz, then apply the Procedure.
\item Otherwise, do the following:
\begin{enumerate}
\item Determine ${\bf U}_{s}(\bf{A}^{\prime })$ and $%
\operatorname{Aut}(\bf{A}^{\prime }).$
\item Determine the set of $\operatorname{Aut}(\bf{A}^{\prime })$-orbits on ${\bf U%
}_{s}(\bf{A}^{\prime }).$
\item For each orbit, construct the terminal algebra corresponding to one of its
representatives.
\end{enumerate}
\end{enumerate}
\medskip
\subsection{Notations}
Let us introduce the following notations. Let ${\bf A}$ be a terminal algebra with
a basis $e_{1},e_{2}, \ldots, e_{n}.$ Then by $\Delta_{ij}$\ we will denote the
bilinear form
$\Delta_{ij}:{\bf A}\times {\bf A}\longrightarrow \mathbb C$
with $\Delta_{ij}(e_{l},e_{m}) = \delta_{il}\delta_{jm}.$
The set $\left\{ \Delta_{ij}:1\leq i, j\leq n\right\}$ is a basis for the linear space of
bilinear forms on ${\bf A},$ so every $\theta \in
{\rm Z^2}({\bf A},\bf \mathbb V )$ can be uniquely written as $%
\theta ={\sum }_{i,j=1}^nc_{ij}\Delta_{ij},$ where $%
c_{ij}\in \mathbb C.$
We also denote by
$$\begin{array}{lll}
\T {i*}{j}& \mbox{the }j\mbox{th }i\mbox{-dimensional nilpotent Leibniz algebra}, \\
\T i{j}& \mbox{the }j\mbox{th }i\mbox{-dimensional nilpotent non-Leibniz terminal algebra}, \\
{\mathfrak{N}}_i& \mbox{the }i\mbox{-dimensional algebra with zero product}, \\
({\bf A})_{i,j} & j\mbox{th }i\mbox{-dimensional central extension of }\bf A. \\
\end{array}$$
Also, it is easy to see that every central extension of $\mathfrak{N}_i$ is a Leibniz algebra.
\subsection{The algebraic classification of 3-dimensional nilpotent terminal algebras}
Observe that a 2-dimensional nilpotent algebra is either $\mathfrak{N}_2,$ or is isomorphic to
\begin{align*
\begin{array}{llllll}
\T {2*}{01} &:& e_1 e_1 = e_2.
\end{array}
\end{align*}
Both of these algebras are nilpotent of index less than 4, and hence are terminal. All central extensions of 2-dimensional nilpotent algebras were described in~\cite{cfk18}. By a direct verification, we have the following list of all 3-dimensional nilpotent terminal algebras with annihilator of codimension 1 or 2
\begin{align}\label{3-dim-term}
\begin{array}{lllllllll}
\T {3*}{01}&:&\T {2*}{01}\oplus\mathfrak{N}_1 &:& e_1 e_1 = e_2; \\
\T {3*}{02}&:&(\mathfrak{N}_2)_{3,1} &:& e_1 e_1 = e_3, & e_2 e_2=e_3; \\
\T {3*}{03}&:& (\mathfrak{N}_2)_{3,2} &:& e_1 e_2=e_3, & e_2 e_1=-e_3; \\
\T {3*}{04}(\lambda)&:&(\mathfrak{N}_2)_{3,3} &:& e_1 e_1 = \lambda e_3, & e_2 e_1=e_3, & e_2 e_2=e_3; \\
\T {3*}{05}&:&(\T {2*}{01})_{3,1} &:& e_1 e_1 = e_2, & e_1 e_2=e_3; \\
\T 3{01}(\lambda)&:&(\T {2*}{01})_{3,2} &:& e_1 e_1 = e_2, & e_1 e_2=\lambda e_3, & e_2 e_1=e_3.
\end{array}
\end{align}
Notice that only $\T 3{01}(\lambda)$ from list \cref{3-dim-term} is non-Leibniz.
\subsection{The algebraic classification of $4$-dimensional nilpotent terminal algebras}
Analyzing the list of all the 4-dimensional nilpotent algebras with annihilator of codimension 2, we have only one terminal non-Leibniz algebra:
$$
\begin{array}{lllllllll}
\T 4{02}&:&(\T {2*}{01} )_{4,1} &:& e_1 e_1 = e_2, & e_1 e_2=e_4, & e_2 e_1=e_3.
\end{array}
$$
To complete the algebraic classification of 4-dimensional nilpotent terminal algebras, we need to describe all the 1-dimensional terminal non-Leibniz extensions of the algebras from the table \cref{3-dim-term}. This is done in Subsections~\ref{aut-and-H^2}--\ref{ext-T_05^3*} and summarized in the main theorem of the first part of the paper.
\begin{theorem}\label{main-alg}
Let $\mathbf A$ be a 4-dimensional nilpotent non-Leibniz terminal algebra over $\mathbb C$. Then $\mathbf A$ is isomorphic to one of the algebras $\T 4{01}-\T 4{44}$ or $\D 4{01}-\D 4{40}$ found below.
\end{theorem}
\begin{Proof}
The proof is split into several steps presented in Subsections~\ref{aut-and-H^2}--\ref{ext-T_05^3*} below.
\end{Proof}
\subsubsection{Automorphism and cohomology groups of $3$-dimensional nilpotent terminal algebras}\label{aut-and-H^2}
\[
\tiny
\begin{tabular}{|c|c|c|c|}
\hline
$\mathbf{A}$ & $\aut\mathbf{A}$ & ${\rm Z_T^2}(\mathbf{A})$ & ${\rm H_T^2}(\mathbf{A})$\\
\hline
$\T {3*}{01}$
&
$\begin{pmatrix}
x & 0 & 0\\
y & x^2 & u\\
z & 0 & v
\end{pmatrix}$
&
$\Big\langle
\begin{array}{l}
\Dt 11, \Dt 12, \Dt 13, \Dt 21,\\
\Dt 23, \Dt 31, \Dt 32, \Dt 33
\end{array}
\Big\rangle$
&
$\Big\langle
\begin{array}{l}
\Dl 12, \Dl 13, \Dl 21, \Dl 23\\
\Dl 31, \Dl 32, \Dl 33
\end{array}
\Big\rangle$
\\
\hline
$\T {3*}{02}$
&
$\begin{pmatrix}
x & y & 0\\
(-1)^{n+1} y & (-1)^n x & 0\\
z & u & x^2+y^2
\end{pmatrix}$
&
$\Big\langle
\begin{array}{l}
\Dt 11, \Dt 12, \Dt 13, \Dt 21\\
\Dt 22, \Dt 23, \Dt 31, \Dt 32
\end{array}
\Big\rangle$
&
$\Big\langle
\begin{array}{l}
\Dl 11, \Dl 12, \Dl 13, \Dl 21\\
\Dl 23, \Dl 31, \Dl 32
\end{array}
\Big\rangle$
\\
\hline
$\T {3*}{03}$
&
$\begin{pmatrix}
x & y & 0\\
z & u & 0\\
v & w & xu-yz
\end{pmatrix}$
&
$\Big\langle
\begin{array}{l}
\Dt 11, \Dt 12, \Dt 13, \Dt 21\\
\Dt 22, \Dt 23, \Dt 31, \Dt 32
\end{array}
\Big\rangle$
&
$\Big\langle
\begin{array}{l}
\Dl 11, \Dl 12, \Dl 13, \Dl 22\\
\Dl 23, \Dl 31, \Dl 32
\end{array}
\Big\rangle$
\\
\hline
$\T {3*}{04}$
&
$\begin{pmatrix}
x & y & 0\\
-\lambda y & x-y & 0\\
z & u & x^2-xy+\lambda y^2
\end{pmatrix}$
&
$\Big\langle
\begin{array}{l}
\Dt 11, \Dt 12, \Dt 13, \Dt 21\\
\Dt 22, \Dt 23, \Dt 31, \Dt 32
\end{array}
\Big\rangle$
&
$\Big\langle
\begin{array}{l}
\Dl 11, \Dl 12, \Dl 13, \Dl 21\\
\Dl 23, \Dl 31, \Dl 32
\end{array}
\Big\rangle$
\\
\hline
$\T {3*}{05}$
&
$\begin{pmatrix}
x & 0 & 0\\
y & x^2 & 0\\
z & xy & x^3
\end{pmatrix}$
&
$\Big\langle
\begin{array}{l}
\Dt 11, \Dt 12, \Dt 13,\\
\Dt 21, \Dt 22 - 3\Dt 31
\end{array}
\Big\rangle$
&
$\Big\langle
\begin{array}{l}
\Dl 13, \Dl 21,\\
\Dl 22 - 3\Dl 31
\end{array}
\Big\rangle$
\\
\hline
$\T 3{01}$
&
$\begin{pmatrix}
x & 0 & 0\\
y & x^2 & 0\\
z & (\lambda+1)xy & x^3
\end{pmatrix}$
&
$\begin{cases}
\Big\langle
\begin{array}{l}
\Dt 11, \Dt 12, -\Dt 13 + 3\Dt 31,\\
\Dt 21, {\color{black}\Dt 13 + \Dt 22}, \Dt 23
\end{array}
\Big\rangle & \lambda=0\\
\Big\langle
\begin{array}{l}
\Dt 11, \Dt 12, (\lambda-1)\Dt 13 + 3\Dt 31,\\
\Dt 21, \color{black}{\Dt 13 + \Dt 22}
\end{array}
\Big\rangle & \lambda\ne 0
\end{cases}$
&
$\begin{cases}
\Big\langle
\begin{array}{l}
\Dl 12, -\Dl 13 + 3\Dl 31,\\
{\color{black}\Dl 13 + \Dl 22}, \Dl 23
\end{array}
\Big\rangle & \lambda=0\\
\Big\langle
\begin{array}{l}
\Dl 12, (\lambda-1)\Dl 13 + 3\Dl 31,\\
\color{black}{\Dl 13 + \Dl 22}
\end{array}
\Big\rangle & \lambda\ne 0
\end{cases}$
\\
\hline
\end{tabular}
\]
\vskip0.5cm
Since the algebras $\T {3*}{01}, \T {3*}{02},\T {3*}{03},\T {3*}{04}$ and $\T {3*}{05}$ are Leibniz, it is natural to find the relation between the Leibniz and terminal cohomologies of these algebras in order to exclude those cocycles which give Leibniz algebras.
\vskip0.5cm
$$\begin{tabular}{|c|c|c|}
\hline
$\mathbf{A}$ & ${\rm H_L^2}(\mathbf{A})$ & ${\rm H_T^2}(\mathbf{A})$\\
\hline
$\T {3*}{01}$
&
$\langle \Dl 12, \Dl 13, \Dl 31, \Dl 33 \rangle$
&
${\rm H_L^2}(\mathbf{A})\oplus \langle \Dl 21, \Dl 23, \Dl 32 \rangle$
\\
\hline
$\T {3*}{02}$
&
$\langle \Dl 11, \Dl 12, \Dl 21 \rangle$
&
${\rm H_L^2}(\mathbf{A})\oplus\langle \Dl 13, \Dl 23, \Dl 31, \Dl 32\rangle$
\\
\hline
$\T {3*}{03}$
&
$\Big\langle
\begin{array}{l}
\Dl 11, \Dl 12, \Dl 13 - \Dl 31,\\
\Dl 22, \Dl 23 - \Dl 32
\end{array}
\Big\rangle$
&
${\rm H_L^2}(\mathbf{A})\oplus\langle \Dl 31, \Dl 32 \rangle$
\\
\hline
$\T {3*}{04}$
&
$\begin{cases}
\langle \Dl 11, \Dl 12, \Dl 21 \rangle, & \lambda\ne 0\\
\langle\Dl 11, \Dl 12, \Dl 21, \Dl 23\rangle, & \lambda=0
\end{cases}$
&
$
\begin{cases}
{\rm H_L^2}(\mathbf{A})\oplus \langle \Dl 13, \Dl 23, \Dl 31, \Dl 32 \rangle, & \lambda\ne 0\\
{\rm H_L^2}(\mathbf{A})\oplus \langle \Dl 13, \Dl 31, \Dl 32 \rangle, & \lambda=0
\end{cases}$
\\
\hline
$\T {3*}{05}$
&
$\langle \Dl 13\rangle$
&
${\rm H_L^2}(\mathbf{A})\oplus \langle \Dl 21,\Dl 22 - 3\Dl 31 \rangle$
\\
\hline
\end{tabular}$$
\subsubsection{$1$-dimensional central extensions of $\T {3*}{01}$}
Let us use the following notations:
\begin{align*}
\nb 1 = \Dl 12, \nb 2 = \Dl 13, \nb 3 = \Dl 31, \nb 4 = \Dl 33, \nb 5 = \Dl 21, \nb 6 = \Dl 23, \nb 7 = \Dl 32.
\end{align*}
Take $\theta=\sum_{i=1}^7\alpha_i\nb i\in {\rm H_T^2}(\T {3*}{01}).$
If
$$
\phi=
\begin{pmatrix}
x & 0 & 0\\
y & x^2 & u\\
z & 0 & v
\end{pmatrix}\in\aut{\T {3*}{01}},
$$
then
$$
\phi^T\begin{pmatrix}
0& \alpha_1& \alpha_2\\
\alpha_5& 0& \alpha_6\\
\alpha_3& \alpha_7& \alpha_4
\end{pmatrix} \phi=
\begin{pmatrix}
\alpha^*& \alpha_1^*& \alpha_2^*\\
\alpha_5^*& \alpha^{**}& \alpha_6^*\\
\alpha_3^*& \alpha_7^*& \alpha_4^*
\end{pmatrix},
$$
where
\begin{align*}
\alpha_1^* &= x^2(\alpha_1x + \alpha_7z),\\
\alpha_2^* &= u(\alpha_1x + \alpha_7z) + v(\alpha_2x + \alpha_6y + \alpha_4z),\\
\alpha_3^* &= u(\alpha_5x + \alpha_6z) + v(\alpha_3x + \alpha_7y + \alpha_4z),\\
\alpha_4^* &= v((\alpha_6 + \alpha_7)u + \alpha_4v),\\
\alpha_5^* &= x^2(\alpha_5x + \alpha_6z),\\
\alpha_6^* &= \alpha_6x^2v,\\
\alpha_7^* &= \alpha_7x^2v.
\end{align*}
Hence, $\phi\langle\theta\rangle=\langle\theta^*\rangle,$ where $\theta^*=\sum\limits_{i=1}^7 \alpha_i^* \nb i.$
We are interested in $\theta$ with $(\alpha_5,\alpha_6,\alpha_7) \neq (0,0,0).$ Moreover, the condition $\theta \in \mathbf{T}_1 (\T{3*}{01})$ gives us $(\alpha_2, \alpha_3, \alpha_4, \alpha_6, \alpha_7) \neq (0,0,0,0,0)$ and
$(\alpha_1 + \gamma\alpha_2, \alpha_5 + \gamma\alpha_3, \alpha_6, \alpha_7, \alpha_6 + \gamma\alpha_4, \alpha_7 + \gamma\alpha_4) \neq (0,0,0,0,0,0)$ for all $\gamma \in \mathbb{C}.$
\begin{enumerate}
\item Let $\alpha_6 \neq 0, \alpha_7 \neq 0$ and $\alpha_1 \alpha_6 - \alpha_5 \alpha_7=0.$
Taking $v =\frac{1}{x^2 \alpha_7},$
$z=-\frac{x \alpha_1}{\alpha_7}$ and $y=\frac{x (\alpha_1 \alpha_4 - \alpha_3 \alpha_7)}{\alpha_7^2},$
we get the family of representatives
$$
\langle \alpha_2^\star \nabla_2 + \alpha_4^\star \nabla_4 + \alpha_6^\star \nabla_6+ \nabla_7\rangle,
$$
where
\begin{align*}
\as 2 &= \frac 1{\alpha_7^3x}(\alpha_1\alpha_4\alpha_6 - \alpha_1\alpha_4\alpha_7 - \alpha_3\alpha_6\alpha_7 + \alpha_2\alpha_7^2),\\
\as 4 &= \frac 1{\alpha_7^2x^4}(ux^2\alpha_7(\alpha_6 + \alpha_7) + \alpha_4),\\
\alpha_6^\star &= \frac{\alpha_6}{\alpha_7}.
\end{align*}
\begin{enumerate}
\item $\alpha_6\neq -\alpha_7$
and $\alpha_1\alpha_4\alpha_6 - \alpha_1\alpha_4\alpha_7 - \alpha_3\alpha_6\alpha_7 + \alpha_2\alpha_7^2\neq 0$.
Then choosing $u =-\frac{\alpha_4}{x^2\alpha_7(\alpha_6 + \alpha_7)}$, where $x$ is such that $\as 2=1$, we have the family of representatives
$ \langle \nabla_2 + \alpha \nabla_6+ \nabla_7\rangle_{\alpha \neq -1,0}.$
\item $\alpha_6\neq -\alpha_7$
and $\alpha_1\alpha_4\alpha_6 - \alpha_1\alpha_4\alpha_7 - \alpha_3\alpha_6\alpha_7 + \alpha_2\alpha_7^2 = 0$. Then
choosing $u = -\frac{\alpha_4}{x^2 \alpha_7 (\alpha_6 + \alpha_7)},$ we have the family of representatives
$ \langle \alpha \nabla_6+ \nabla_7\rangle_{\alpha \neq -1,0}.$
\item $\alpha_6=- \alpha_7,$ $\alpha_4=0$
and $\alpha_2 \neq - \alpha_3$. Then choosing $x = \frac{\alpha_2 + \alpha_3}{\alpha_7},$ we have the representative
$ \langle \nabla_2 - \nabla_6+ \nabla_7\rangle,$ which will be joined with the family $ \langle \nabla_2 + \alpha \nabla_6+ \nabla_7\rangle_{\alpha \neq -1,0}.$
\item $\alpha_6=- \alpha_7,$ $\alpha_4=0$
and $\alpha_2 = - \alpha_3$. Then
we have the representative
$ \langle - \nabla_6+ \nabla_7\rangle,$
which will be joined with the family $ \langle \alpha \nabla_6+ \nabla_7\rangle_{\alpha \neq -1,0}.$
\item $\alpha_6=- \alpha_7$ and $\alpha_4\neq 0$.
Then choosing $x = \sqrt[4]{\frac{\alpha_4}{\alpha_7^2}},$ we have the family of representatives
$ \langle \alpha \nabla_2 +\nabla_4 - \nabla_6+ \nabla_7\rangle.$
It gives only two distinct orbits with representatives
$ \langle \nabla_2 +\nabla_4 - \nabla_6+ \nabla_7\rangle$ and
$ \langle \nabla_4 - \nabla_6+ \nabla_7\rangle.$
\end{enumerate}
\item Let $\alpha_6 \neq 0, \alpha_7 \neq 0$ and $\alpha_1 \alpha_6 - \alpha_5 \alpha_7\ne 0.$
Taking
$z=-\frac{\alpha_5x}{\alpha_6},$
$v = \frac x{\alpha_6\alpha_7}(\alpha_1\alpha_6 - \alpha_5\alpha_7),$
$y = \frac x{\alpha_6\alpha_7}(\alpha_4\alpha_5 - \alpha_3\alpha_6),$
and
$u = \frac x{\alpha_6\alpha_7^2}(\alpha_3\alpha_6^2 - \alpha_2\alpha_6\alpha_7 + \alpha_4\alpha_5(\alpha_7 -\alpha_6)),$
we get the family of representatives
$$ \langle \nabla_1 + \alpha_4^\star \nabla_4 + \alpha_6^\star \nabla_6+ \nabla_7\rangle,$$
where
\begin{align*}
\alpha_4^\star &= \frac 1{\alpha_7^3x}(\alpha_3\alpha_6^2 + \alpha_1\alpha_4\alpha_7 + \alpha_3\alpha_6\alpha_7 - \alpha_2\alpha_6\alpha_7 - \alpha_2\alpha_7^2 - \alpha_4\alpha_5\alpha_6), \\
\alpha_6^\star &= \frac{\alpha_6}{\alpha_7}.
\end{align*}
It gives two families of representatives of distinct orbits
$ \langle \nabla_1 + \alpha \nabla_6+ \nabla_7\rangle_{\alpha\neq 0} $
and
$ \langle \nabla_1 + \nabla_4 + \alpha \nabla_6+ \nabla_7\rangle_{\alpha\neq 0}$
depending on whether $\alpha_3\alpha_6^2 + \alpha_1\alpha_4\alpha_7 + \alpha_3\alpha_6\alpha_7 - \alpha_2\alpha_6\alpha_7 - \alpha_2\alpha_7^2 - \alpha_4\alpha_5\alpha_6=0$ or not.
\item Let $\alpha_6 = 0, \alpha_7 \neq 0.$
Taking
$y = \frac x{\alpha_7^2} (\alpha_1 \alpha_4 + \alpha_4 \alpha_5 - \alpha_3 \alpha_7),$ $z = -\frac{x \alpha_1}{\alpha_7},$
and $u = -\frac{\alpha_4v}{\alpha_7} $
we get a family of representatives
$$ \langle \alpha_2^\star \nabla_2 + \alpha_5^\star \nabla_5 + \nabla_7\rangle,$$
where
\begin{align*}
\alpha_2^\star &= \frac 1{x \alpha_7^2}( \alpha_2 \alpha_7-\alpha_1 \alpha_4), \\
\alpha_5^\star &= \frac{\alpha_5x}{\alpha_7v}.
\end{align*}
\begin{enumerate}
\item $\alpha_5 = 0$ and $\alpha_1\alpha_4 - \alpha_2\alpha_7=0$. Then
we have the representative $\langle\nabla_7\rangle,$
which will be joined with the family $ \langle \alpha \nabla_6+ \nabla_7\rangle_{\alpha \neq 0}.$
\item $\alpha_5 = 0$ and $\alpha_1\alpha_4 - \alpha_2\alpha_7\ne 0$. Then
choosing $x= \frac 1{\alpha_7^2}(\alpha_2 \alpha_7-\alpha_1 \alpha_4),$
we have the representative $\langle\nabla_2 + \nabla_7\rangle.$
which will be joined with the family $ \langle \nabla_2 + \alpha \nabla_6+ \nabla_7\rangle_{\alpha \neq 0}$,
\item $\alpha_5 \neq 0$ and $\alpha_1\alpha_4 - \alpha_2\alpha_7\ne 0$. Then
choosing $x= \frac 1{\alpha_7^2}(\alpha_2 \alpha_7-\alpha_1 \alpha_4)$ and
$v=\frac{\alpha_5\alpha_7x^2}{\alpha_2\alpha_7-\alpha_1\alpha_4}$
we get the representative $\langle \nabla_2 + \nabla_5 + \nabla_7 \rangle.$
\item $\alpha_5 \neq 0$ and $\alpha_1\alpha_4 - \alpha_2\alpha_7=0$. Then
choosing
$v = \frac {\alpha_5x}{\alpha_7}$
we get the representative $\langle \nabla_5 + \nabla_7 \rangle.$
\end{enumerate}
\item $\alpha_6 \neq 0$ and $\alpha_7 = 0.$
Taking
$y = \frac x{\alpha_6^2} (\alpha_1 \alpha_4 + \alpha_4 \alpha_5 - \alpha_2 \alpha_6),$
$z = -\frac{x \alpha_5}{\alpha_6},$ and $u = -\frac{\alpha_4v}{\alpha_6}$
we get the family of representatives
$$ \langle \alpha_1^\star \nabla_1 + \alpha_3^\star \nabla_3 + \nabla_6\rangle,$$
where
\begin{align*}
\alpha_1^\star &= \frac{\alpha_1x}{\alpha_6v}, \\
\alpha_3^\star &= \frac 1{x \alpha_6^2}(\alpha_3 \alpha_6-\alpha_4 \alpha_5).
\end{align*}
\begin{enumerate}
\item $\alpha_1\neq 0$ and $\alpha_3 \alpha_6- \alpha_4 \alpha_5\ne 0$. Then
we have the representative
$ \langle \nabla_1 + \nabla_3 + \nabla_6\rangle.$
\item $\alpha_1 = 0$ and $\alpha_3 \alpha_6- \alpha_4 \alpha_5\ne 0$. Then
we have the representative $ \langle \nabla_3 + \nabla_6\rangle.$
\item $\alpha_1 \neq 0$ and $\alpha_3 \alpha_6 - \alpha_4 \alpha_5=0$. Then
we have the representative $ \langle \nabla_1 + \nabla_6\rangle.$
\item $\alpha_1 = 0$ and $\alpha_3 \alpha_6 - \alpha_4 \alpha_5=0$. Then
we have the representative $ \langle \nabla_6\rangle.$
\end{enumerate}
\item $\alpha_5 \neq 0$, $\alpha_6=0$ and $\alpha_7=0.$
Taking
$u =\frac{-v x \alpha_3 - v z \alpha_4}{x \alpha_5}$ and $x = \frac{1}{\sqrt[3]{\alpha_5}},$
we get a family of representatives
$$ \langle \alpha_1^\star \nabla_1 + \alpha_2^\star \nabla_2 + \alpha_4^\star \nabla_4 + \nabla_5\rangle,$$
where
\begin{align*}
\alpha_1^\star &= \frac{\alpha_1}{\alpha_5}, \\
\alpha_2^\star &= \frac v{\alpha_5^2x^3}((\alpha_2\alpha_5 -\alpha_1\alpha_3)x + \alpha_4(\alpha_5 - \alpha_1)z), \\
\alpha_4^\star &= \frac{\alpha_4v^2}{\alpha_5x^3}.
\end{align*}
\begin{enumerate}
\item $\alpha_4 \neq 0$ and $\alpha_1 - \alpha_5\ne 0$. Then
choosing $z= -\frac{(\alpha_2\alpha_5 -\alpha_1\alpha_3)x}{\alpha_4 (\alpha_1 -\alpha_5)}$ and $v = \sqrt{\frac{\alpha_5x^3}{\alpha_4}}$
we have the family of representatives of distinct orbits
$ \langle \alpha \nabla_1 + \nabla_4 + \nabla_5\rangle_{\alpha \neq 1}.$
\item $\alpha_4 = 0$, $\alpha_1 - \alpha_5\ne 0$ and $\alpha_2\alpha_5-\alpha_1\alpha_3=0$. Then
we have the family of representatives of distinct orbits
$ \langle \alpha \nabla_1 + \nabla_5\rangle_{\alpha \neq 1}$. The corresponding extensions are split.
\item $\alpha_4 = 0$, $\alpha_1 - \alpha_5\ne 0$ and $\alpha_2\alpha_5-\alpha_1\alpha_3=0$. Then
we have the family of representatives of distinct orbits
$ \langle \alpha \nabla_1 +\nabla_2 +\nabla_5\rangle_{\alpha \neq 1}.$
\item $\alpha_4 \neq 0$, $\alpha_1 - \alpha_5=0$ and $\alpha_2 - \alpha_3\ne 0$. Then
choosing $x = \frac 1{\alpha_4\alpha_5}(\alpha_2-\alpha_3)^2$,
$v=\frac 1{\alpha_4^2\alpha_5}(\alpha_2- \alpha_3)^3$,
$u = -\frac {\alpha_2 - \alpha_3}{\alpha_4^2\alpha_5^2}(\alpha_4^2\alpha_5z + (\alpha_2 - \alpha_3)^2\alpha_3)$
we have the representative
$ \langle \nabla_1 + \nabla_2 +\nabla_4+ \nabla_5\rangle.$
\item $\alpha_4 \neq 0$, $\alpha_1 - \alpha_5=0$ and $\alpha_2 - \alpha_3= 0$. Then
choosing $v=\sqrt {\frac{\alpha_5x^3}{\alpha_4}}$ and $u=-\frac v{\alpha_5x}(\alpha_3x + \alpha_4z)$
we have the representative
$ \langle \nabla_1 + \nabla_4+ \nabla_5\rangle,$
which will be joined with the family $\langle \alpha \nabla_1 + \nabla_4+ \nabla_5 \rangle_{\alpha \neq 1}$.
\item $\alpha_4 = 0$, $\alpha_1 - \alpha_5=0$ and $\alpha_2 - \alpha_3\ne 0$. Then
choosing $u=-\frac{\alpha_3v}{\alpha_5}$ and $v=\frac{\alpha_5x^2}{\alpha_2 - \alpha_3}$
we have the representative
$ \langle \nabla_1 + \nabla_2+ \nabla_5\rangle,$
which will be joined with the family $\langle \alpha \nabla_1 + \nabla_2+ \nabla_5 \rangle_{\alpha \neq 1}$.
\item $\alpha_4 = 0$, $\alpha_1 - \alpha_5=0$ and $\alpha_2 - \alpha_3= 0$. Then choosing $u=-\frac{\alpha_3v}{\alpha_5}$
we have the representative
$ \langle \nabla_1 + \nabla_5\rangle,$
which gives a split extension.
\end{enumerate}
\end{enumerate}
\
Summarizing, we have the following representatives of distinct orbits:
$
\begin{array}{l}
\langle \nabla_1 + \nabla_4 + \alpha \nabla_6+ \nabla_7\rangle_{\alpha\neq 0},\\
\langle \nabla_1 + \alpha \nabla_6+ \nabla_7\rangle_{\alpha\neq 0},\\
\langle \nabla_2 +\nabla_4 - \nabla_6+ \nabla_7\rangle,\\
\langle \nabla_2 + \nabla_5 + \nabla_7 \rangle,
\end{array}
\begin{array}{l}
\langle \nabla_2 + \alpha \nabla_6+ \nabla_7\rangle,\\
\langle \nabla_4 - \nabla_6+ \nabla_7\rangle,\\
\langle \nabla_5 + \nabla_7 \rangle,\\
\langle \alpha \nabla_6+ \nabla_7\rangle,
\end{array}
\begin{array}{l}
\langle \nabla_1 + \nabla_6\rangle,\\
\langle \nabla_1 + \nabla_3 + \nabla_6\rangle,\\
\langle \nabla_3 + \nabla_6\rangle,\\
\langle \nabla_6\rangle,
\end{array}
\begin{array}{l}
\langle \alpha \nabla_1 +\nabla_2 +\nabla_5\rangle,\\
\langle \nabla_1 + \nabla_2 +\nabla_4+ \nabla_5\rangle,\\
\langle \alpha \nabla_1 + \nabla_4 + \nabla_5\rangle.
\end{array}
$
The corresponding algebras are:
\[\begin{array}{lllllllllll}
\T {4}{03}(\alpha)_{\alpha\neq 0}&:& e_1e_1 = e_2,& e_1e_2=e_4,& e_2e_3=\alpha e_4,& e_3e_2=e_4,& e_3e_3=e_4; \\
\T {4}{04}(\alpha)_{\alpha\neq 0}&:& e_1e_1 = e_2,& e_1e_2=e_4,& e_2e_3=\alpha e_4,& e_3e_2=e_4; \\
\T {4}{05} &:& e_1e_1 = e_2,& e_1e_3=e_4,& e_2e_3=- e_4,& e_3e_2=e_4,& e_3e_3=e_4; \\
\T {4}{06} &:& e_1e_1 = e_2,& e_1e_3=e_4,& e_2e_1= e_4,& e_3e_2=e_4; \\
\T {4}{07}(\alpha) &:& e_1e_1 = e_2,& e_1e_3=e_4,& e_2e_3=\alpha e_4,& e_3e_2=e_4; \\
\T {4}{08} &:& e_1e_1 = e_2,& e_2e_3=- e_4,& e_3e_2=e_4,& e_3e_3=e_4; \\
\T {4}{09} &:& e_1e_1 = e_2,& e_2e_1= e_4,& e_3e_2=e_4; \\
\T {4}{10}(\alpha) &:& e_1e_1 = e_2,& e_2e_3=\alpha e_4,& e_3e_2=e_4; \\
\T {4}{11} &:& e_1e_1 = e_2,& e_1e_2=e_4,& e_2e_3= e_4; \\
\T {4}{12} &:& e_1e_1 = e_2,& e_1e_2=e_4,& e_2e_3= e_4,& e_3e_1=e_4; \\
\T {4}{13} &:& e_1e_1 = e_2,& e_2e_3= e_4,& e_3e_1=e_4; \\
\T {4}{14} &:& e_1e_1 = e_2,& e_2e_3= e_4; \\
\T {4}{15}(\alpha) &:& e_1e_1 = e_2,& e_1e_2= \alpha e_4,& e_1e_3= e_4,& e_2e_1=e_4; \\
\T {4}{16} &:& e_1e_1 = e_2,& e_1e_2= e_4,& e_1e_3= e_4,& e_2e_1=e_4,& e_3e_3=e_4; \\
\T {4}{17}(\alpha) &:& e_1e_1 = e_2,& e_1e_2= \alpha e_4,& e_2e_1=e_4,& e_3e_3= e_4.
\end{array}\]
\subsubsection{$1$-dimensional central extensions of $\T {3*}{02}$}
Let us use the following notations:
\begin{align*}
\nb 1 = \Dl 11, \nb 2 = \Dl 12, \nb 3 = \Dl 21, \nb 4 = \Dl 13, \nb 5 = \Dl 23, \nb 6 = \Dl 31, \nb 7 = \Dl 32.
\end{align*}
Take $\theta=\sum_{i=1}^7\alpha_i\nb i\in {\rm H_T^2}(\T {3*}{02}).$
If
$$
\phi=
\begin{pmatrix}
x & y& 0 \\
(-1)^{n+1}y& (-1)^nx& 0\\
z & u & x^2+y^2
\end{pmatrix}\in\aut{\T {3*}{02}},
$$
then
$$
\phi^T\begin{pmatrix}
\alpha_1& \alpha_2& \alpha_4\\
\alpha_3& 0& \alpha_5\\
\alpha_6& \alpha_7& 0
\end{pmatrix} \phi=
\begin{pmatrix}
\alpha_1^*+\alpha^*& \alpha_2^*& \alpha_4^*\\
\alpha_3^*& \alpha^*& \alpha_5^*\\
\alpha_6^*& \alpha_7^*& \alpha^{**}
\end{pmatrix},
$$
where
\begin{align*}
\alpha_1^* &= \alpha_1(x^2 - y^2) + 2(-1)^{n+1}(\alpha_2 + \alpha_3)xy+(\alpha_4x+(-1)^{n+1}\alpha_5y+\alpha_6x+ (-1)^{n+1}\alpha_7y)z\\
&\quad - (\alpha_4y +(-1)^n\alpha_5x+\alpha_6y+(-1)^n\alpha_7x)u,\\
\alpha_2^* &= (-1)^n\alpha_2x^2 + \alpha_1xy + (-1)^{n+1}\alpha_3y^2 + (\alpha_4x + (-1)^{n+1}\alpha_5y)u + (\alpha_6y + (-1)^n\alpha_7x)z,\\
\alpha_3^* &= (-1)^n\alpha_3x^2 + \alpha_1xy + (-1)^{n+1}\alpha_2y^2 + (\alpha_4y + (-1)^n\alpha_5x)z + (\alpha_6x + (-1)^{n+1}\alpha_7y)u,\\
\alpha_4^* &= (\alpha_4x+(-1)^{n+1}\alpha_5y)(x^2+y^2),\\
\alpha_5^* &= (\alpha_4y+(-1)^n\alpha_5x)(x^2 + y^2),\\
\alpha_6^* &= (\alpha_6x+(-1)^{n+1}\alpha_7y)(x^2+y^2),\\
\alpha_7^* &= (\alpha_6y + (-1)^n\alpha_7x)(x^2+y^2).
\end{align*}
Hence, $\phi\langle\theta\rangle=\langle\theta^*\rangle,$ where $\theta^*=\sum\limits_{i=1}^7 \alpha_i^* \nb i.$
We are interested in $\theta$ with $(\alpha_4,\alpha_5,\alpha_6,\alpha_7) \neq (0,0,0,0).$ The condition $\theta \in \mathbf{T}_1 (\T{3*}{02})$ does not give us any new restrictions on parameters.
\begin{enumerate}
\item $\alpha_6=0$ and $\alpha_7=0.$ Then $\alpha_4\ne 0$ or $\alpha_5\ne 0$. If $\alpha_4\ne 0$ and $\alpha_5=0$, then choosing $x=y=1$, we have $\alpha^*_4\ne 0$ and $\alpha^*_5\ne 0$. The same holds for $\alpha_4=0$ and $\alpha_5\ne 0$. Thus, we shall assume from the very beginning that $\alpha_4\ne 0$ and $\alpha_5 \neq 0$.
\begin{enumerate}
\item $\alpha^2_4+\alpha^2_5\ne 0$. Then choosing $x=(-1)^{n+1}\frac{\alpha_4y}{\alpha_5},
z = \frac{(-1)^ny}{\alpha_5(\alpha_4^2 + \alpha_5^2)}(\alpha_1\alpha_4^2 + 2(\alpha_2 + \alpha_3)\alpha_4\alpha_5 - \alpha_1\alpha_5^2)$ and $u = \frac y{\alpha_5(\alpha_4^2 + \alpha_5^2)}(\alpha_2\alpha_4^2 -\alpha_3\alpha_5^2 - \alpha_1\alpha_4\alpha_5)$ we obtain the representative $\langle\as 3\nb 3+\as 4\nb 4\rangle$, where
\begin{align*}
\as 3 &= \frac{(-1)^ny^2}{\alpha_5^2}(\alpha_3\alpha_4^2 - \alpha_1\alpha_4\alpha_5 - \alpha_2\alpha_5^2),\\
\as 4 &= \frac{(-1)^{n+1}y^3}{\alpha_5^3}(\alpha_4^2 + \alpha_5^2)^2.
\end{align*}
Thus, we have two representatives $\langle \nabla_4 \rangle $ and $\langle \nabla_3 + \nabla_4 \rangle $ depending on whether $\alpha_3\alpha_4^2 - \alpha_1\alpha_4\alpha_5 - \alpha_2\alpha_5^2=0$ or not.
\item $\alpha^2_4+\alpha^2_5=0$. If $\alpha_4=-i\alpha_5$, then choosing $x=0$ and $y=n=1$, we obtain $\alpha^*_4=i\alpha^*_5$, so we shall assume from the very beginning that $\alpha_4= i \alpha_5.$
Taking $n=0$,
$u=\frac{i(\alpha_2x^2 + \alpha_1xy - \alpha_3y^2)}{\alpha_5(x + iy)}$ and
$z=-\frac{\alpha_3x^2 + \alpha_1xy - \alpha_2y^2}{\alpha_5(x + iy)}$
we have the representative $\langle \alpha_1^{\star} \nabla_1 + i \nabla_4 + \nabla_5 \rangle$, where
\begin{align*}
\as 1 &= \frac{x - iy}{\alpha_5(x + iy)^2}(\alpha_1 - i\alpha_2 - i\alpha_3).
\end{align*}
Thus, we have two representatives $\langle i \nabla_4 + \nabla_5 \rangle$ and $\langle \nabla_1 + i \nabla_4 + \nabla_5 \rangle$ depending on whether $\alpha_1 - i\alpha_2 - i\alpha_3=0$ or not.
\end{enumerate}
\item $\alpha_6\neq 0$ or $\alpha_7\neq0$, and $\alpha_6^2+\alpha_7^2\ne 0$.
Then we may make $\alpha_6\ne 0$ and $\alpha_7\neq 0$. After that, if we choose $x=(-1)^{n+1}\frac{\alpha_6y}{\alpha_7}\ne 0,$ then $x^2+y^2=\frac {y^2}{\alpha^2_7}(\alpha_6^2+\alpha_7^2)\ne 0$, $\alpha^*_6=\frac{(-1)^{n+1}y^3}{\alpha_7^3}(\alpha_6^2 + \alpha_7^2)^2\ne 0$ and $\alpha_7^* = 0$. Thus, we shall assume that $\alpha_6\ne 0$ and $\alpha_7=0$ from the very beginning. Taking $y = 0,$ we get $\alpha^*_7=0$ and
\begin{align*}
\alpha_1^* &= x (x \alpha_1 + (-1)^{n + 1} u \alpha_5 + z (\alpha_4 + \alpha_6)),\\
\alpha_2^* &= x ((-1)^n x \alpha_2 + u \alpha_4),\\
\alpha_3^* &= x ((-1)^n x \alpha_3 + (-1)^n z \alpha_5 + u \alpha_6),\\
\alpha_4^* &= x^3 \alpha_4,\\
\alpha_5^* &= (-1)^n x^3 \alpha_5,\\
\alpha_6^* &= x^3 \alpha_6.
\end{align*}
Consider $\alpha_2^* = 0, \alpha_3^* = 0$ as a system of linear equations on $z$ and $u.$ Its determinant is $(-1)^{n+1}x^2\alpha_4\alpha_5,$ so we have the following cases (each of which defines an $\operatorname{Aut}{\T {3*}{02}}$-invariant subset):
\begin{enumerate}
\item $\alpha_4 \neq 0$ and $\alpha_5 \neq 0$. Taking $u = \frac{(-1)^{n+1} x \alpha_2}{\alpha_4}, z = \frac{x (\alpha_2 \alpha_6-\alpha_3 \alpha_4)}{\alpha_4 \alpha_5},$ we get $\alpha_2^* = 0, \alpha_3^* = 0$ and
\begin{align*}
\alpha_1^* &= \frac{x^2}{\alpha_4\alpha_5}(\alpha_2\alpha_5^2 - \alpha_3\alpha_4^2 + \alpha_1\alpha_4\alpha_5 + \alpha_2\alpha_4\alpha_6 - \alpha_3\alpha_4\alpha_6 +\alpha_2\alpha_6^2),\\
\alpha_4^* &= x^3 \alpha_4,\\
\alpha_5^* &= (-1)^n x^3 \alpha_5,\\
\alpha_6^* &= x^3 \alpha_6.
\end{align*}
Thus, we obtain two families of representatives of distinct orbits $\langle \nabla_1 +\alpha \nabla_4+\beta \nabla_5+\nabla_6 \rangle_{\alpha\beta\neq 0, \beta \in \mathbb C_{\geq 0}}$ and $\langle \alpha \nabla_4+\beta \nabla_5+\nabla_6 \rangle_{\alpha\beta\neq 0, \beta \in \mathbb C_{\geq 0}}. $
\item $\alpha_4 \neq 0$ and $\alpha_5 = 0$. Taking $u = \frac{(-1)^{n+1} x \alpha_2}{\alpha_4}$, we get $\alpha_2^* = 0$ and
\begin{align*}
\alpha_1^* &= x (x \alpha_1 + z (\alpha_4 + \alpha_6)),\\
\alpha_3^* &= \frac{(-1)^nx^2}{\alpha_4}(\alpha_3\alpha_4 - \alpha_2\alpha_6),\\
\alpha_4^* &= x^3 \alpha_4,\\
\alpha_6^* &= x^3 \alpha_6.
\end{align*}
\begin{enumerate}
\item $\alpha_4 + \alpha_6 \neq 0$. Taking $z = -\frac{x\alpha_1}{\alpha_4 + \alpha_6},$ we get two series of representatives of distinct orbits $\langle \nabla_3 +\alpha \nabla_4 +\nabla_6\rangle_{\alpha\neq -1,0}$ and $\langle \alpha \nabla_4 +\nabla_6\rangle_{\alpha\neq -1,0}.$
\item $\alpha_4 + \alpha_6 = 0.$ In this case we get the series of representatives of distinct orbits $\langle \nabla_1 +\alpha \nabla_3 - \nabla_4+\nabla_6 \rangle_{\alpha \in \mathbb C_{\geq 0}}$ and two distinct representatives $\langle \nabla_3 - \nabla_4+\nabla_6 \rangle$ and $\langle - \nabla_4+\nabla_6 \rangle$, which will be joined with the families $\langle \nabla_3 +\alpha \nabla_4 +\nabla_6\rangle_{\alpha\neq -1,0}$ and $\langle \alpha \nabla_4 +\nabla_6\rangle_{\alpha\neq -1,0}$ found above.
\end{enumerate}
\item $\alpha_4 = 0$ and $\alpha_5 \neq 0.$ Consider $\alpha_1^* = 0, \alpha_3^* = 0$ as a system of linear equations on $z,u.$ Its determinant is $x^2(\alpha_5^2 + \alpha_6^2).$
\begin{enumerate}
\item $\alpha_5^2 + \alpha_6^2 \neq 0.$ Taking $u = \frac{(-1)^n x (\alpha_1 \alpha_5 - \alpha_3 \alpha_6)}{ \alpha_5^2 + \alpha_6^2}, z = -\frac{x ( \alpha_3 \alpha_5 + \alpha_1 \alpha_6)}{ \alpha_5^2 + \alpha_6^2},$ we get $\alpha^*_1 = \alpha^*_3 = 0$ and
\begin{align*}
\alpha_2^* &= (-1)^n x^2 \alpha_2,\\
\alpha_5^* &= (-1)^n x^3 \alpha_5,\\
\alpha_6^* &= x^3 \alpha_6.
\end{align*}
Therefore, we have two series of representatives of distinct orbits $\langle \nabla_2 + \alpha \nabla_5+\nabla_6 \rangle_{\alpha \in \mathbb C_{\geq 0}, \alpha \neq 0, i}$ and $\langle \alpha \nabla_5+\nabla_6 \rangle_{\alpha \in \mathbb C_{\geq 0}, \alpha \neq 0, i}.$
\item $\alpha_5^2 + \alpha_6^2 = 0.$ We may assume that $\alpha_5 = i \alpha_6$ (if $\alpha_5=-i\alpha_6$, then choose $n=-1$). Taking $n=0$, $u = -\frac{x\alpha_3}{\alpha_6}$and $z = 0,$ we get $\alpha_3 = 0$ and
\begin{align*}
\alpha_1^* &= x^2 (\alpha_1+i\alpha_3),\\
\alpha_2^* &= x^2 \alpha_2,\\
\alpha_5^* &= ix^3 \alpha_6,\\
\alpha_6^* &= x^3 \alpha_6.
\end{align*}
Therefore, we get the family of representatives of distinct orbits $\langle \nabla_1 + \alpha \nabla_2 + i \nabla_5 + \nabla_6 \rangle$ and two representatives $\langle \nabla_2 + i \nabla_5 + \nabla_6 \rangle, \langle i \nabla_5 + \nabla_6 \rangle$, which will be joined with the families $\langle \nabla_2 + \alpha \nabla_5+\nabla_6 \rangle_{\alpha \in \mathbb C_{\geq 0}, \alpha \neq 0, i}$ and $\langle \alpha \nabla_5+\nabla_6 \rangle_{\alpha \in \mathbb C_{\geq 0}, \alpha \neq 0, i}$ found above.
\end{enumerate}
\item $\alpha_4 = 0$ and $\alpha_5 = 0.$ Taking $z = -\frac{x\alpha_1}{\alpha_6}$ and $u = \frac{(-1)^{n+1}x\alpha_3}{\alpha_6},$ we get $\alpha_1^* = \alpha_3^* = 0$ and
\begin{align*}
\alpha_2^* &= (-1)^nx^2 \alpha_2,\\
\alpha_6^* &= x^3 \alpha_6.
\end{align*}
Therefore, we get two representatives $\langle \nabla_2 + \nabla_6 \rangle$ and $\langle \nabla_6 \rangle$, which will be joined with the families $\langle \nabla_2 + \alpha \nabla_5+\nabla_6 \rangle_{\alpha \in \mathbb C_{\geq 0}, \alpha \neq 0, i}$ and $\langle \alpha \nabla_5+\nabla_6 \rangle_{\alpha \in \mathbb C_{\geq 0}, \alpha \neq 0, i}$ found above. Note that representatives $\langle \nabla_2 + \nabla_6 \rangle$ and $\langle \nabla_6 \rangle$ define the same orbit.
\end{enumerate}
\item $\alpha_6\neq 0$ or $\alpha_7\neq0$, and $\alpha_6^2+\alpha_7^2=0$. Then we may assume that $\alpha_6= i \alpha_7\ne 0.$
\begin{enumerate}
\item $\alpha_4 \neq \pm i \alpha_5.$ We may assume that $\alpha_4 \neq 0$ (otherwise $\alpha_5\ne 0$, so we may take $x=y=1$ to make $\alpha^*_4\ne 0$). Then choosing $n=0$ and $y=-\frac{x\alpha_5}{\alpha_4}x,$ we get $\alpha^*_4=\frac{x^3}{\alpha_4^3}(\alpha_4^2 + \alpha_5^2)^2\ne 0$ and $\alpha_5^* = 0$, so we shall assume that $\alpha_4\ne 0$ and $\alpha_5=0$ from the very beginning. Now, choosing $n=y=0$, $u = \frac{i x \alpha_3}{\alpha_7}, z = -\frac{x (i \alpha_3 \alpha_4 + \alpha_2 \alpha_7)}{\alpha_7^2},$ we get $\alpha_2^* = \alpha_3^* = \alpha^*_5=0$ and
\begin{align*}
\alpha_1^* &= \frac{x^2}{\alpha_7^2}(\alpha_3\alpha_4\alpha_7 + \alpha_1\alpha_7^2 - \alpha_2\alpha_4\alpha_7 -i\alpha_3\alpha_4^2 - i\alpha_2\alpha_7^2 -i\alpha_3\alpha_7^2),\\
\alpha_4^* &= x^3 \alpha_4,\\
\alpha_6^* &= ix^3 \alpha_7,\\
\alpha_7^* &= x^3 \alpha_7.
\end{align*}
Thus, we get two series of representatives of distinct orbits $\langle \nabla_1 + \alpha \nabla_4 + i \nabla_6 + \nabla_7 \rangle_{\alpha \neq 0}$ and $\langle \alpha \nabla_4 + i \nabla_6 + \nabla_7 \rangle_{\alpha \neq 0}.$
\item $\alpha_4= i \alpha_5$. Then choosing $n=0$ we have
\begin{align*}
\alpha_1^* &= x^2 \alpha_1 - x (2 y (\alpha_2 + \alpha_3) + (u - i z) (\alpha_5 + \alpha_7)) - y (y \alpha_1 + (i u + z) (\alpha_5 + \alpha_7)),\\
\alpha_2^* &= x^2 \alpha_2 - y (y \alpha_3 + u \alpha_5 -i z \alpha_7) + x (y \alpha_1 + i u \alpha_5 + z \alpha_7),\\
\alpha_3^* &= (x + i y) z \alpha_5 + x (y \alpha_1 + x \alpha_3 + i u \alpha_7) - y (y \alpha_2 + u \alpha_7),\\
\alpha_4^* &= i (x + i y) (x^2 + y^2) \alpha_5,\\
\alpha_5^* &= (x + i y) (x^2 + y^2) \alpha_5,\\
\alpha_6^* &= i (x + i y) (x^2 + y^2) \alpha_7,\\
\alpha_7^* &= (x + i y) (x^2 + y^2) \alpha_7.
\end{align*}
Consider $\alpha_1^* = 0, \alpha_3^* = 0$ as a system of linear equations in $u,z.$ Its determinant is $i(x+iy)^2(\alpha_7^2 - \alpha_5^2).$
\begin{enumerate}
\item $\alpha_5^2 - \alpha_7^2 \neq 0$. Then by choosing the appropriate values of $u$ and $z$ we get $\alpha_1^* = 0, \alpha_3^* = 0$, $\alpha_2^* = i(i\alpha_1 + \alpha_2 + \alpha_3)(x - iy)^2$ and $\alpha^*_4,\alpha^*_5,\alpha^*_6,\alpha^*_7$ as above. Therefore, we have two series of representatives of distinct orbits $\langle \nabla_2 + i \alpha \nabla_4 + \alpha \nabla_5 + i \nabla_6 + \nabla_7 \rangle_{\alpha \neq \pm 1}$ and $\langle i \alpha \nabla_4 + \alpha \nabla_5 + i \nabla_6 + \nabla_7 \rangle_{\alpha \neq \pm 1}$.
\item $\alpha_5 = \alpha_7$. Then taking $n=0$, $z = \frac{-x y \alpha_1 + y^2 \alpha_2 - x^2 \alpha_3 - i u x \alpha_7 + u y \alpha_7}{(x + i y)\alpha_7}$ we get $\alpha_3^* = 0$ and
\begin{align*}
\alpha_1^* &= (x - i y) (x (\alpha_1- 2i\alpha_3) - y (i \alpha_1 + 2 \alpha_2)),\\
\alpha_2^* &= (x^2 + y^2) (\alpha_2-\alpha_3),\\
\alpha_4^* &= i (x + i y) (x^2 + y^2) \alpha_7,\\
\alpha_5^* &= (x + i y) (x^2 + y^2) \alpha_7,\\
\alpha_6^* &= i (x + i y) (x^2 + y^2) \alpha_7,\\
\alpha_7^* &= (x + i y) (x^2 + y^2) \alpha_7.
\end{align*}
\begin{enumerate}
\item $i\alpha_1 + \alpha_2 + \alpha_3\ne 0$, $\alpha_2 - \alpha_3\ne 0$ and $i\alpha_1 + 2\alpha_2\ne 0$. Then taking $y = \frac{(\alpha_1 - 2i\alpha_3)x}{i\alpha_1 + 2\alpha_2}$ we get the representative $\langle\as 2\nb 2+i \nabla_4 + \nabla_5 + i \nabla_6 + \nabla_7\rangle$, where
\begin{align*}
\as 2 &= \frac{(i\alpha_1 + 2\alpha_2)(\alpha_2 - \alpha_3)}{2\alpha_7 x(i\alpha_1 + \alpha_2 + \alpha_3)}.
\end{align*}
So, choosing the appropriate value of $x$, we get the representative $\langle \nabla_2 + i \nabla_4 + \nabla_5 + i \nabla_6 + \nabla_7\rangle.$
\item $i\alpha_1 + \alpha_2 + \alpha_3 = 0$. Then we have
\begin{align*}
\alpha_1^* &= i(x^2+y^2)(\alpha_2-\alpha_3),\\
\alpha_2^* &= (x^2+y^2)(\alpha_2-\alpha_3),
\end{align*}
Therefore, we get two representatives $\langle i\nabla_1 + \nabla_2 + i \nabla_4 + \nabla_5 + i \nabla_6 + \nabla_7\rangle$ and $\langle i \nabla_4 + \nabla_5 + i \nabla_6 + \nabla_7\rangle.$
\item $\alpha_2-\alpha_3 = 0$. Then $\alpha_1^* = (x-iy)^2(\alpha_1 - 2i\alpha_3)$, and we have only one new representative $\langle \nabla_1 + i \nabla_4 + \nabla_5 + i \nabla_6 + \nabla_7\rangle.$
\item $i\alpha_1 + 2\alpha_2=0$. Then
$
\alpha_1^* = x(x - i y) (\alpha_1- 2i\alpha_3),
$
so choosing $x=0$ we have two representatives $\langle \nabla_2 + i \nabla_4 + \nabla_5 + i \nabla_6 + \nabla_7\rangle$ and $\langle i \nabla_4 + \nabla_5 + i \nabla_6 + \nabla_7\rangle$ found above.
\end{enumerate}
\item $\alpha_5 = -\alpha_7$. Then taking $n=0$ and $z = \frac{x y \alpha_1 - y^2 \alpha_2 + x^2 \alpha_3 + i u x \alpha_7 - u y \alpha_7}{(x + i y) \alpha_7}$ we get $\alpha_3^*=0$ and
\begin{align*}
\alpha_1^* &= (x^2 - y^2)\alpha_1 - 2 x y (\alpha_2+\alpha_3),\\
\alpha_2^* &= (x^2 - y^2)(\alpha_2+\alpha_3) +2 x y \alpha_1,\\
\alpha_4^* &= -i(x + i y) (x^2 + y^2) \alpha_7,\\
\alpha_5^* &= -(x + i y) (x^2 + y^2) \alpha_7,\\
\alpha_6^* &= i (x + i y) (x^2 + y^2) \alpha_7,\\
\alpha_7^* &= (x + i y) (x^2 + y^2) \alpha_7.
\end{align*}
\begin{enumerate}
\item $(\alpha_1,\alpha_2+\alpha_3) \neq (0,0)$ and $\alpha_1^2 + (\alpha_2+\alpha_3)^2 \neq 0$. Then we may assume that $\alpha_1\ne 0$ and $\alpha_2+\alpha_3\ne 0$. In this case the equality $\alpha^*_1=0$ has two distinct roots $y_1=\mu_1x$ and $y_2=\mu_2x$, where at least one of $\mu_1,\mu_2$ is different from $\pm i$ (otherwise $\alpha_2+\alpha_3=0$). Let $\mu_1\ne\pm i$. Then choosing $y = \mu_1x$ we get $\alpha_1^* = 0$. Observe that in this case $x^2 + y^2\ne 0$ and $\alpha_1 = \frac{2(\alpha_2 + \alpha_3)\mu_1}{1 - \mu_1^2}$. Substituting this into $\alpha^*_2$, we obtain $\alpha_2^* = \frac{x^2(1 + \mu_1^2)^2(\alpha_2 + \alpha_3)}{1-\mu_1^2}\ne 0.$ Therefore, we have the representative $\langle \nabla_2 - i \nabla_4 - \nabla_5 + i \nabla_6 + \nabla_7\rangle.$
\item $\alpha_2+\alpha_3 \neq 0$ and $\alpha_1 = \pm i (\alpha_2+\alpha_3)\ne 0$. Then we get
\begin{align*}
\alpha_1^* &= \pm i (x \pm i y)^2 (\alpha_2+\alpha_3),\\
\alpha_2^* &= (x \pm i y)^2 (\alpha_2+\alpha_3),
\end{align*}
so we obtain two representatives $\langle i\nabla_1 + \nabla_2 - i \nabla_4 - \nabla_5 + i \nabla_6 + \nabla_7\rangle$ and $\langle -i\nabla_1 + \nabla_2 - i \nabla_4 - \nabla_5 + i \nabla_6 + \nabla_7\rangle.$
\item $(\alpha_1,\alpha_2+\alpha_3) = (0,0)$. Then we get the representative $\langle - i \nabla_4 - \nabla_5 + i \nabla_6 + \nabla_7\rangle.$
\end{enumerate}
\end{enumerate}
\item $\alpha_4 = -i\alpha_5$. Choosing $n=0$, we have
\begin{align*}
\alpha_1^* &= x^2 \alpha_1 - x (2 y (\alpha_2 + \alpha_3) + i z (\alpha_5 - \alpha_7) + u (\alpha_5 + \alpha_7)) - y (y \alpha_1 - i u (\alpha_5 - \alpha_7) + z (\alpha_5 + \alpha_7)),\\
\alpha_2^* &= x^2 \alpha_2 - y (y \alpha_3 + u \alpha_5 - i z \alpha_7) + x (y \alpha_1 - i u \alpha_5 + z \alpha_7),\\
\alpha_3^* &= (x - i y) z \alpha_5 + x (y \alpha_1 + x \alpha_3 + i u \alpha_7) - y (y \alpha_2 + u \alpha_7),\\
\alpha_4^* &= -i(x - iy) (x^2 + y^2) \alpha_5,\\
\alpha_5^* &= (x - i y) (x^2 + y^2) \alpha_5,\\
\alpha_6^* &= i (x + i y) (x^2 + y^2) \alpha_7,\\
\alpha_7^* &= (x + i y) (x^2 + y^2) \alpha_7.
\end{align*}
Consider $\alpha_2^* = 0, \alpha_3^* = 0$ as a linear system in $u, z.$ Its determinant is $-i((x - i y)^2 \alpha_5^2 + (x + i y)^2 \alpha_7^2).$ So, we may choose $x,y$ such that $x^2 + y^2 \neq 0, -i((x - i y)^2 \alpha_5^2 + (x + i y)^2 \alpha_7^2) \neq 0$ and $u$ and $z$ to make $\alpha_2^* = \alpha_3^*=0$. Observe that this does not change the conditions on $\alpha_4,\alpha_5,\alpha_6,\alpha_7$. So, we may assume that $\alpha_2=\alpha_3=0$ from the very beginning.
We may also suppose that $\alpha_5 + \alpha_7 \neq 0.$
\begin{enumerate}
\item $\alpha_5 \neq 0$. Then taking $y = \frac{\alpha_5 - \alpha_7}{i(\alpha_5+\alpha_7)}x$ we may make $\alpha_5^* = \alpha_7^*$, so we shall assume $\alpha_5=\alpha_7$. Now, choosing $n=y = u=z=0,$ we get
\begin{align*}
\alpha_1^* &= x^2 \alpha_1,\\
\alpha_4^* &= -ix^3 \alpha_7,\\
\alpha_5^* &= x^3 \alpha_7,\\
\alpha_6^* &= i x^3 \alpha_7,\\
\alpha_7^* &= x^3 \alpha_7.
\end{align*}
Therefore, we get two representatives $\langle \nabla_1 - i \nabla_4 + \nabla_5 + i\nabla_6 + \nabla_7 \rangle$ and $\langle - i \nabla_4 + \nabla_5 + i\nabla_6 + \nabla_7 \rangle.$
\item $\alpha_5 = 0$. Then $\alpha_4 = 0$, so that $\alpha^*_4=\alpha^*_5=0$. Choosing $n=0$ and the appropriate values of $u$ and $z$ we have $\alpha^*_2=\alpha^*_3=0$ and
\begin{align*}
\alpha_1^* &= (x - i y)^2 \alpha_1,\\
\alpha_6^* &= i (x + i y) (x^2 + y^2) \alpha_7,\\
\alpha_7^* &= (x + i y) (x^2 + y^2) \alpha_7.
\end{align*}
Therefore, we get two representatives $\langle \nabla_1 + i\nabla_6 + \nabla_7 \rangle$ and $\langle i\nabla_6 + \nabla_7 \rangle,$ which will be joined with the series $\langle \nabla_1 + \alpha \nabla_4 + i \nabla_6 + \nabla_7 \rangle_{\alpha \neq 0}$ and $\langle \alpha \nabla_4 + i \nabla_6 + \nabla_7 \rangle_{\alpha \neq 0}.$ Note that by above representatives $\langle \nabla_1 + i\nabla_6 + \nabla_7 \rangle$ and $\langle \nabla_2 + i\nabla_6 + \nabla_7 \rangle$ define the same orbit.
\end{enumerate}
\end{enumerate}
\end{enumerate}
Summarizing,
we have the following distinct orbits
\[ \langle \nabla_4\rangle,
\langle \nabla_3+ \nabla_4\rangle,
\langle \nabla_1+i\nabla_4+\nabla_5\rangle,
\langle i\nabla_4 +\nabla_5\rangle,\]
\[ \langle \alpha \nabla_4+\beta\nabla_5+\nabla_6 \rangle_{ \beta \in \mathbb C_{\geq 0}}, \langle \nabla_1+ \alpha \nabla_4+\beta\nabla_5+\nabla_6 \rangle_{\alpha\beta \neq 0, \beta \in \mathbb C_{\geq 0}},\]
\[ \langle \nabla_3+\alpha\nabla_4+\nabla_6 \rangle_{\alpha \neq 0}, \langle \nabla_1 +\alpha\nabla_3 - \nabla_4 +\nabla_6 \rangle_{\alpha \in \mathbb C_{\geq 0}}, \langle \nabla_1 + \alpha \nabla_2 + i \nabla_5 + \nabla_6 \rangle, \langle \nabla_2 +\alpha\nabla_5 +\nabla_6 \rangle_{\alpha \in \mathbb C_{\geq 0}}\]
\[\langle \nabla_1+\alpha\nabla_4+i\nabla_6 + \nabla_7 \rangle, \langle \alpha\nabla_4+i\nabla_6 + \nabla_7 \rangle,\]
\[\langle \nabla_2+i\alpha\nabla_4 + \alpha\nabla_5+i\nabla_6 + \nabla_7 \rangle_{\alpha \neq 0}, \langle i\alpha\nabla_4 + \alpha\nabla_5+i\nabla_6 + \nabla_7 \rangle_{\alpha \neq 0},\]
\[\langle i \nabla_1 + \nabla_2 + i \nabla_4 + \nabla_5 + i\nabla_6 + \nabla_7\rangle, \langle \pm i \nabla_1 + \nabla_2 - i \nabla_4 - \nabla_5 + i\nabla_6 + \nabla_7\rangle\]
The corresponding algebras are
$$ \begin{array}{lllllllllll}
\T {4}{18}&:& e_1 e_1 = e_3, & e_1e_3=e_4, & e_2 e_2=e_3; \\
\T {4}{19}&:& e_1 e_1 = e_3, & e_1e_3=e_4,& e_2e_1=e_4, & e_2 e_2=e_3; \\
\T {4}{20}&:& e_1 e_1 = e_3+e_4, & e_1e_3=ie_4, & e_2 e_2=e_3, & e_2e_3=e_4; \\
\T {4}{21}&:& e_1 e_1 = e_3, & e_1e_3=ie_4, & e_2 e_2=e_3,& e_2e_3=e_4; \\
\T {4}{22}(\alpha,\beta)_{\beta\in \mathbb C_{\geq0}}&:& e_1 e_1 = e_3, & e_1e_3=\alpha e_4,& e_2 e_2=e_3, &e_2e_3=\beta e_4,& e_3e_1=e_4; \\
\T {4}{23}(\alpha,\beta)_
\beta \in \mathbb C_{\geq0}}&:& e_1 e_1 = e_3+e_4, & e_1e_3=\alpha e_4,& e_2 e_2=e_3, &e_2e_3=\beta e_4,& e_3e_1=e_4; \\
\T {4}{24}(\alpha
&:& e_1 e_1 = e_3,& e_1e_3 = \alpha e_4, & e_2e_1 = e_4, & e_2e_2 = e_3, & e_3e_1 = e_4; \\
\T {4}{25}(\alpha)_{\alpha \in \mathbb C_{\geq0}}&:& e_1 e_1 = e_3+e_4, & e_1e_3=- e_4,& e_2e_1 = \alpha e_4, & e_2e_2 = e_3, & e_3e_1 = e_4; \\
\T {4}{26}(\alpha)&:& e_1 e_1 = e_3+e_4, & e_1e_2=\alpha e_4, & e_2 e_2=e_3, &e_2e_3=i e_4,& e_3e_1=e_4; \\
\T {4}{27}(\alpha)_{\alpha \in \mathbb C_{\geq0}}&:& e_1 e_1 = e_3, & e_1e_2= e_4, & e_2 e_2=e_3,& e_2e_3=\alpha e_4,& e_3e_1=e_4; \\
\T {4}{28}(\alpha)&:& e_1 e_1 = e_3 + e_4, & e_1e_3=\alpha e_4, & e_2 e_2=e_3,& e_3e_1=ie_4,& e_3e_2=e_4; \\
\T {4}{29}(\alpha)&:& e_1 e_1 = e_3, & e_1e_3=\alpha e_4, & e_2 e_2=e_3,& e_3e_1=ie_4,& e_3e_2=e_4; \\
\T {4}{30}(\alpha
&:& e_1 e_1 = e_3, & e_1 e_2 = e_4, & e_1e_3=i\alpha e_4, & e_2 e_2=e_3,& e_2e_3=\alpha e_4, & e_3e_1=ie_4,& e_3e_2=e_4; \\
\T {4}{31}(\alpha
&:& e_1 e_1 = e_3, & e_1e_3=i\alpha e_4, & e_2 e_2=e_3,& e_2e_3=\alpha e_4, & e_3e_1=ie_4,& e_3e_2=e_4; \\
\T {4}{32}&:& e_1 e_1 = e_3 + ie_4, & e_1 e_2 = e_4, & e_1e_3=i e_4, & e_2 e_2=e_3,& e_2e_3= e_4, & e_3e_1=ie_4,& e_3e_2=e_4; \\
\T {4}{33}&:& e_1 e_1 = e_3 + ie_4, & e_1 e_2 = e_4, & e_1e_3= -ie_4, & e_2 e_2=e_3,& e_2e_3= -e_4, & e_3e_1=ie_4,& e_3e_2=e_4; \\
\T {4}{34}&:& e_1 e_1 = e_3 -ie_4, & e_1 e_2 = e_4, & e_1e_3= -ie_4, & e_2 e_2=e_3,& e_2e_3= -e_4, & e_3e_1=ie_4,& e_3e_2=e_4.
\end{array}$$
The algebras above are pairwise non-isomorphic, except $\T {4}{23}(\alpha,0) \cong \T {4}{22}(\alpha,0)$ for $\alpha \neq -1,$ $\T {4}{23}(-1,0) \cong \T {4}{25}(0),$ $\T {4}{23}(0,\beta) \cong \T {4}{22}(0,\beta)$ for $\beta \neq i,$ $\T {4}{23}(0,i) \cong \T {4}{26}(0), \T {4}{24}(0) \cong \T {4}{22}(0,0), \T {4}{30}(0) \cong \T {4}{28}(0), \T {4}{31}(0) \cong \T {4}{29}(0).$
\newpage
\subsubsection{$1$-dimensional central extensions of $\T {3*}{03}$}
Let us use the following notations
\begin{align*}
\nb 1 = \Dl 11, \nb 2 = \Dl 12, \nb 3 = \Dl 13 - \Dl 31, \nb 4 = \Dl 22, \nb 5 = \Dl 23 - \Dl 32, \nb 6 = \Dl 31, \nb 7 = \Dl 32.
\end{align*}
Take $\theta=\sum_{i=1}^7\alpha_i\nb i\in {\rm H_T^2}(\T {3*}{03}).$
If
$$
\phi=
\begin{pmatrix}
x & y & 0\\
z & u & 0\\
v & w & xu-yz
\end{pmatrix}\in\aut{\T {3*}{03}},
$$
then
$$
\phi^T\begin{pmatrix}
\alpha_1& \alpha_2& \alpha_3\\
0& \alpha_4& \alpha_5\\
\alpha_6-\alpha_3& \alpha_7-\alpha_5& 0
\end{pmatrix} \phi=
\begin{pmatrix}
\alpha_1^*& \alpha_2^*-\alpha^*& \alpha_3^*\\
\alpha^*& \alpha_4^*& \alpha_5^*\\
\alpha_6^*-\alpha_3^*& \alpha_7^*-\alpha_5^*& \alpha^{**}
\end{pmatrix},
$$
where
\begin{align*}
\alpha^*_1 &= \alpha_1x^2 + \alpha_2xz + \alpha_4z^2 + v(\alpha_6x + \alpha_7z),\\
\alpha^*_2 &= x(2\alpha_1y + \alpha_2u) + z(\alpha_2y + 2\alpha_4u) + w(\alpha_6x +\alpha_7z) + v(\alpha_6y + \alpha_7u),\\
\alpha^*_3 &= (\alpha_3x+\alpha_5z)(xu - yz),\\
\alpha^*_4 &= y(\alpha_1y + \alpha_2u) + \alpha_4u^2 + w(\alpha_6y + \alpha_7u),\\
\alpha^*_5 &= (\alpha_3y+\alpha_5u)(xu - yz),\\
\alpha^*_6 &= (\alpha_6x+\alpha_7z)(xu - yz),\\
\alpha^*_7 &= (\alpha_6y+\alpha_7u)(xu - yz).
\end{align*}
Hence, $\phi\langle\theta\rangle=\langle\theta^*\rangle,$ where $\theta^*=\sum\limits_{i=1}^7 \alpha_i^* \nb i.$
We are interested in $\theta$ with $(\alpha_6,\alpha_7) \neq (0,0).$ Moreover, the condition $\theta \in \mathbf{T}_1 (\T{3*}{03})$ gives us $(\alpha_3,\alpha_5, \alpha_6 -\alpha_3, \alpha_7 - \alpha_5) \neq (0,0,0,0).$
If $\alpha_7\neq 0$, then taking $u=-\frac{\alpha_6y}{\alpha_7}$ we have $\alpha^*_7=0$, so we shall assume that $\alpha^*_6\ne 0$ and $\alpha^*_7=0$ from the very beginning. Choosing $y=0$, $z=-\frac{\alpha_3x}{\alpha_5}$, $u= \frac{\alpha_6x}{\alpha_5}$, $v = \frac x{\alpha_5^2\alpha_6}(\alpha_2\alpha_3\alpha_5-\alpha_3^2\alpha_4 - \alpha_1\alpha_5^2)$, $w = \frac x{\alpha_5^2}(2\alpha_3\alpha_4 - \alpha_2\alpha_5)$, we get the family of representatives
$ \langle \frac{\alpha_4}{\alpha_5 x}\nabla_4 + \nabla_5+ \nabla_6\rangle$. It gives two distinct representatives $ \langle \nabla_5+ \nabla_6\rangle$ and $ \langle \nabla_4 + \nabla_5+ \nabla_6\rangle$ depending on whether $\alpha_4=0$ or not.
The algebras corresponding to $ \langle \nabla_4 + \nabla_5+ \nabla_6\rangle$ and $ \langle \nabla_5+ \nabla_6\rangle$ are:
$$
\begin{array}{lllllllll}
\T {4}{35} &:& e_1 e_2=e_3, & e_2 e_1=-e_3,& e_2e_2=e_4,& e_2e_3=e_4,& e_3e_1=e_4,& e_3e_2=-e_4;\\
\T {4}{36} &:& e_1 e_2=e_3, & e_2 e_1=-e_3,& e_2e_3=e_4,& e_3e_1=e_4,& e_3e_2=-e_4.
\end{array}
$$
\subsubsection{$1$-dimensional central extensions of $\T {3*}{04}$}
Let us use the following notations
\begin{align*}
\nb 1 = \Dl 11, \nb 2 = \Dl 12, \nb 3 = \Dl 13, \nb 4 = \Dl 21,
\nb 5 = \Dl 23, \nb 6 = \Dl 31, \nb 7 = \Dl 32.
\end{align*}
Take $\theta=\sum_{i=1}^7\alpha_i\nb i\in {\rm H_T^2}(\T {3*}{04}).$
If
$$
\phi=
\begin{pmatrix}
x & y & 0\\
-\lambda y & x-y & 0\\
z & u & x^2-xy+\lambda y^2
\end{pmatrix}\in\aut{\T 3{04}},
$$
then
$$
\phi^T\begin{pmatrix}
\alpha_1 & \alpha_2 & \alpha_3\\
\alpha_4 & 0 & \alpha_5\\
\alpha_6 & \alpha_7 & 0
\end{pmatrix} \phi=
\begin{pmatrix}
\alpha_1^*+\lambda\alpha^* & \alpha_2^* & \alpha_3^*\\
\alpha_4^*+\alpha^* & \alpha^* & \alpha_5^*\\
\alpha_6^* & \alpha_7^* & 0
\end{pmatrix},
$$
where
\begin{align*}
\alpha^*_1 &= \alpha_1x^2 + \lambda(-\alpha_1 + \alpha_2 + \alpha_4) y^2 -2\lambda(\alpha_2 + \alpha_4) xy + (\alpha_3 + \alpha_6)xz - (\alpha_5 + \alpha_7) yz\\
&\quad- \lambda(\alpha_5 + \alpha_7) ux + \lambda(- \alpha_3 + \alpha_5 - \alpha_6 + \alpha_7) uy,\\
\alpha^*_2 &= \alpha_2x^2 - \lambda\alpha_4 y^2 + (\alpha_1 - \alpha_2)xy + \alpha_7xz + (\alpha_6 - \alpha_7)yz -\alpha_5\lambda uy + \alpha_3ux,\\
\alpha^*_3 &= (x^2 - xy + \lambda y^2)(\alpha_3x -\lambda\alpha_5 y),\\
\alpha^*_4 &= \alpha_4x^2 + (- \alpha_1 + (1-\lambda)\alpha_2 + \alpha_4)y^2 + (\alpha_1 - \alpha_2 - 2\alpha_4)xy + (\alpha_3 - \alpha_5)yz + \alpha_5xz\\
&\quad - \alpha_7\lambda uy + (- \alpha_5 + \alpha_6 - \alpha_7)ux + (- \alpha_3 + \alpha_5 - \alpha_6 + \alpha_7)uy,\\
\alpha^*_5 &= (x^2 - xy + \lambda y^2)(\alpha_5x + (\alpha_3 - \alpha_5)y),\\
\alpha^*_6 &= (x^2 - xy + \lambda y^2)(\alpha_6x -\lambda\alpha_7 y),\\
\alpha^*_7 &= (x^2 - xy + \lambda y^2)(\alpha_7x + (\alpha_6-\alpha_7)y).
\end{align*}
Hence, $\phi\langle\theta\rangle=\langle\theta^*\rangle,$ where $\theta^*=\sum\limits_{i=1}^7 \alpha_i^* \nb i.$
We are interested in $\theta$ with $(\alpha_3,\alpha_5,\alpha_6,\alpha_7) \neq (0,0,0,0)$ (if $\lambda \neq 0$) and $(\alpha_3,\alpha_6,\alpha_7) \neq (0,0,0)$ (if $\lambda = 0$). Moreover, the condition $\theta \in \mathbf{T}_1 (\T{3*}{04})$ gives us $(\alpha_4,\alpha_5,\alpha_6,\alpha_7) \neq (0,0,0,0).$
\begin{enumerate}
\item $\alpha_7\ne 0$. We have the following subcases.
\begin{enumerate}
\item $\alpha_6^2 - \alpha_6\alpha_7 + \lambda\alpha_7^2\ne 0$. Choosing $x=-\frac{(\alpha_6-\alpha_7)y}{\alpha_7}$, we have $\alpha^*_7=0$. Since $(\alpha^*_6)^2 - \alpha^*_6\alpha^*_7 + \lambda(\alpha^*_7)^2=(\alpha_6^2 - \alpha_6\alpha_7 + \lambda\alpha_7^2)(x^2 - xy + \lambda y^2)^3$, the condition $\alpha_6^2 - \alpha_6\alpha_7 + \lambda\alpha_7^2\ne 0$ is invariant under the action of the automorphism group. Thus, we may assume that $\alpha_7=0$ and $\alpha_6\ne 0$ from the very beginning. Then choosing $y=0$ we obtain $\alpha^*_7=0$ and
\begin{align*}
\alpha^*_1 &= x(\alpha_1x + (\alpha_3 + \alpha_6)z - \lambda\alpha_5u),\\
\alpha^*_2 &= x(\alpha_2x + \alpha_3u),\\
\alpha^*_3 &= \alpha_3 x^3,\\
\alpha^*_4 &= x(\alpha_4x + \alpha_5z + (- \alpha_5 + \alpha_6)u),\\
\alpha^*_5 &= \alpha_5 x^3,\\
\alpha^*_6 &= \alpha_6 x^3.
\end{align*}
\begin{enumerate}
\item $\alpha_5\ne 0$ and $\alpha_3\ne 0$. Then we choose $u=-\frac{\alpha_2x}{\alpha_3}$ and $z=-\frac{\alpha_4x+(- \alpha_5 + \alpha_6)u}{\alpha_5}$ and obtain the family of representatives $\langle\alpha\nb 1+\beta\nb 3+\gamma\nb 5+\nb 6\rangle_{\beta,\gamma\ne 0}$. It gives two families of representatives of distinct orbits: $\langle\nb 1+\alpha\nb 3+\beta\nb 5+\nb 6\rangle_{\alpha,\beta\ne 0}$ and $\langle\alpha\nb 3+\beta\nb 5+\nb 6\rangle_{\alpha,\beta\ne 0}$.
\item $\alpha_5\ne 0$ and $\alpha_3=0$. Then $\alpha^*_3=0$ and $\alpha^*_4$, $\alpha^*_5$, $\alpha^*_6$ are as above and
\begin{align*}
\alpha^*_1 &= x(\alpha_1x + \alpha_6z - \lambda\alpha_5u),\\
\alpha^*_2 &= \alpha_2x^2.
\end{align*}
Choosing $z=\frac{\lambda\alpha_5u-\alpha_1x}{\alpha_6}$, we have $\alpha^*_1=0$. Now we have the following subcases:
\begin{enumerate}
\item $\alpha_6^2 - \alpha_5\alpha_6 + \lambda\alpha_5^2\ne 0$. Then choosing $u=\frac{(\alpha_1\alpha_5 - \alpha_4\alpha_6)x}{\alpha_6^2 - \alpha_5\alpha_6 + \lambda\alpha_5^2}$ we have $\alpha^*_4=0$, so we obtain the family of representatives $\langle\alpha\nb 2+\beta\nb 5+\nb 6\rangle_{1-\beta+\lambda\beta^2\ne 0}$. It determines two families of representatives of distinct orbits: $\langle\nb 2+\alpha\nb 5+\nb 6\rangle_{\alpha \neq 0, 1-\alpha+\lambda\alpha^2\ne 0}$ and $\langle\alpha\nb 5+\nb 6\rangle_{\alpha \neq 0, 1-\alpha+\lambda\alpha^2\ne 0}$.
\item $\alpha_6^2 - \alpha_5\alpha_6 + \lambda\alpha_5^2=0$.
If $\lambda\not\in\{0,\frac 14\}$, then we have two families of representatives of distinct orbits
$\Big\langle \lambda\nb 2+\lambda\alpha\nb 4+\frac{1+\sqrt{1-4\lambda}}{2}\nb 5+\lambda\nb 6\Big\rangle, \Big\langle\nb 2+\alpha\nb 4+\frac{2}{1+\sqrt{1-4\lambda}}\nb 5+\nb 6\Big\rangle$ and 4 separate representatives
$\Big\langle\lambda \nb 4+\frac{1+\sqrt{1-4\lambda}}{2}\nb 5+\lambda \nb 6\Big\rangle, \Big\langle\nb 4+\frac{2}{1+\sqrt{1-4\lambda}}\nb 5+\nb 6\Big\rangle,$ $\Big\langle\frac{1\pm\sqrt{1-4\lambda}}{2\lambda}\nb 5+\nb 6\Big\rangle,$ the last two belonging to the family above.
If $\lambda=\frac 14$, then we have the family of representatives of distinct orbits $\langle\nb 2+\alpha\nb 4+2\nb 5+\nb 6\rangle$ and two separate representatives $\langle\nb 4+2\nb 5+\nb 6\rangle$ and $\langle 2\nb 5+\nb 6\rangle$.
If $\lambda=0$, then we have $\alpha_5=\alpha_6$, so we obtain the family of representatives of distinct orbits $\langle\nb 2+\alpha\nb 4+\nb 5+\nb 6\rangle$ and two separate representatives $\langle\nb 4+\nb 5+\nb 6\rangle$ and $\langle\nb 5+\nb 6\rangle$.
\end{enumerate}
\item $\alpha_3\ne 0$ and $\alpha_5=0$. Then $\alpha^*_5=0$, $\alpha^*_2$, $\alpha^*_3$ and $\alpha^*_6$ are as above and
\begin{align*}
\alpha^*_1 &= x(\alpha_1x + (\alpha_3 + \alpha_6)z),\\
\alpha^*_4 &= x(\alpha_4x + \alpha_6u).
\end{align*}
Choosing $u=-\frac{\alpha_4x}{\alpha_6}$, we get $\alpha^*_4=0$. Now we have two subcases:
\begin{enumerate}
\item $\alpha_3 + \alpha_6\ne 0$. Then choosing $z=-\frac{\alpha_1x}{\alpha_3 + \alpha_6}$, we obtain two families of representatives of distinct orbits $\langle \nb 2+\alpha\nb 3+\nb 6\rangle_{\alpha\not\in \{0,-1\}}$ and $\langle \alpha\nb 3+\nb 6\rangle_{\alpha\not\in \{0,-1\}}$.
\item $\alpha_3 + \alpha_6=0$. Then choosing $z=0$, we have the family of representatives of distinct orbits $\langle\nb 1+\alpha\nb 2-\nb 3+\nb 6\rangle$ and two separate representatives $\langle\nb 2-\nb 3+\nb 6\rangle$ and $\langle-\nb 3+\nb 6\rangle$ which will be joined with the families $\langle \nb 2+\alpha\nb 3+\nb 6\rangle_{\alpha\not\in \{0,-1\}}$ and $\langle \alpha\nb 3+\nb 6\rangle_{\alpha\not\in \{0,-1\}}$ found above.
\end{enumerate}
\item $\alpha_3=\alpha_5=0$. Then $\alpha^*_3=\alpha^*_5=0$, $\alpha^*_6$ is as above and
\begin{align*}
\alpha^*_1 &= x(\alpha_1x + \alpha_6z),\\
\alpha^*_2 &= \alpha_2x^2,\\
\alpha^*_4 &= x(\alpha_4x + \alpha_6u).
\end{align*}
Thus, choosing $z=-\frac{\alpha_1x}{\alpha_6}$ and $u=-\frac{\alpha_4x}{\alpha_6}$, we have two representatives depending on whether $\alpha_2=0$ or not: $\langle\nb 2+\nb 6\rangle$ and $\langle\nb 6\rangle$. They will be joined with the families $\langle\nb 2+\alpha\nb 5+\nb 6\rangle_{\alpha \neq 0, 1-\alpha+\lambda\alpha^2\ne 0}$ and $\langle\alpha\nb 5+\nb 6\rangle_{\alpha \neq 0, 1-\alpha+\lambda\alpha^2\ne 0}$ found above.
\end{enumerate}
\item $\alpha_6^2 - \alpha_6\alpha_7 + \lambda\alpha_7^2=0$ and $\alpha_5\ne 0$ and $\alpha_3^2 - \alpha_3\alpha_5 + \lambda\alpha_5^2\ne 0$. Then choosing $x=-\frac{(\alpha_3 - \alpha_5)y}{\alpha_5}$, we get $\alpha^*_5=0$. Since $(\alpha^*_3)^2 - \alpha^*_3\alpha^*_5 + \lambda(\alpha^*_5)^2=(\alpha_3^2 - \alpha_3\alpha_5 + \lambda\alpha_5^2)(x^2-xy+\lambda y^2)^3\ne 0$, we may assume that $\alpha_5=0$ and $\alpha_3\ne 0$ from the very beginning. Then choosing $y=0$ and $z=-\frac{\alpha_2x+ \alpha_3u}{\alpha_7}$ we have $\alpha^*_2=\alpha^*_5=0$ and
\begin{align*}
\alpha^*_1 &= \frac x{\alpha_7}((\alpha_1\alpha_7- \alpha_2\alpha_3 - \alpha_2\alpha_6)x - (\alpha_3^2 + \alpha_3\alpha_6 - \alpha_6^2 + \alpha_6\alpha_7)u),\\
\alpha^*_3 &= \alpha_3x^3,\\
\alpha^*_4 &= x(\alpha_4x - (\alpha_6 - \alpha_7)u),\\
\alpha^*_6 &= \alpha_6x^3,\\
\alpha^*_7 &= \alpha_7x^3.
\end{align*}
\begin{enumerate}
\item $\alpha_6 - \alpha_7\ne 0$. Then choosing $u=\frac{\alpha_4x}{\alpha_6 - \alpha_7}$ we have $\alpha^*_4=0$.
If $\lambda\not\in\{0,\frac 14\}$, then we have four families of representatives of distinct orbits $\Big\langle \nb 1+\alpha\nb 3+ \frac{1\pm\sqrt{1-4\lambda}}{2}\nb 6+\nb 7\Big\rangle_{\alpha\ne 0}$, $\Big\langle \alpha\nb 3+ \frac{1\pm\sqrt{1-4\lambda}}{2}\nb 6+\nb 7\Big\rangle_{\alpha\ne 0}$.
If $\lambda=\frac 14$, then $\alpha_6^2 - \alpha_6\alpha_7 + \lambda\alpha_7^2=0$ implies $\alpha_6=\frac 12\alpha_7$, so $\alpha_6 - \alpha_7\ne 0$ is satisfied. Thus, we have two families of representatives of distinct orbits $\langle\nb 1+\alpha\nb 3+\frac 12\nb 6+\nb 7\rangle_{\alpha\ne 0}$ and $\langle\alpha\nb 3+\frac 12\nb 6+\nb 7\rangle_{\alpha\ne 0}$.
If $\lambda = 0$, then $\alpha_6^2 - \alpha_6\alpha_7 + \lambda\alpha_7^2=0$ and $\alpha_6 - \alpha_7\ne 0$ imply that $\alpha_6=0$. Hence, we have two families of representatives of distinct orbits $\langle\nb 1+\alpha\nb 3+\nb 7\rangle_{\alpha\ne 0}$ and $\langle\alpha\nb 3+\nb 7\rangle_{\alpha\ne 0}$.
\item $\alpha_6 - \alpha_7=0$. Then $\alpha_6^2 - \alpha_6\alpha_7 + \lambda\alpha_7^2=0$ and $\alpha_7\ne 0$ imply that $\lambda = 0$. Now,
\begin{align*}
\alpha^*_1 &= \frac x{\alpha_7}((\alpha_1\alpha_7- \alpha_2\alpha_3 - \alpha_2\alpha_7)x - \alpha_3(\alpha_3 + \alpha_7)u),\\
\alpha^*_3 &= \alpha_3x^3,\\
\alpha^*_4 &= \alpha_4x^2,\\
\alpha^*_6 &= \alpha_7x^3,\\
\alpha^*_7 &= \alpha_7x^3.
\end{align*}
\begin{enumerate}
\item $\alpha_3 + \alpha_7\ne 0$. Then choosing $u=\frac{(\alpha_1\alpha_7- \alpha_2\alpha_3 - \alpha_2\alpha_7)x}{\alpha_3(\alpha_3 + \alpha_7)}$, we have $\alpha^*_1=0$. Thus, we obtain two families of representatives of distinct orbits $\langle\alpha\nb 3+\nb 4+\nb 6+\nb 7\rangle_{\alpha\ne 0,-1}$ and $\langle\alpha\nb 3+\nb 6+\nb 7\rangle_{\alpha\ne 0,-1}$.
\item $\alpha_3 + \alpha_7=0$. Then we obtain the family of representatives of distinct orbits $\langle\nb 1-\nb 3+\alpha\nb 4+\nb 6+\nb 7\rangle$ and two separate representatives $\langle-\nb 3+\nb 4+\nb 6+\nb 7\rangle$ and $\langle-\nb 3+\nb 6+\nb 7\rangle,$ which will be joined with the families $\langle\alpha\nb 3+\nb 4+\nb 6+\nb 7\rangle_{\alpha\ne 0,-1}$ and $\langle\alpha\nb 3+\nb 6+\nb 7\rangle_{\alpha\ne 0,-1}$ found above.
\end{enumerate}
\end{enumerate}
\item $\alpha_6^2 - \alpha_6\alpha_7 + \lambda\alpha_7^2=0$ and $\alpha_5\ne 0$ and $\alpha_3^2 - \alpha_3\alpha_5 + \lambda\alpha_5^2=0$. Choosing $x,y$ such that $(x^2 - xy + \lambda y^2)(\alpha_7x + (\alpha_6-\alpha_7)y)=1$, we have $(\alpha_3 - \alpha_5)y + \alpha_5x\ne 0$, since otherwise $x^2 - xy + \lambda y^2=(\alpha_3^2 - \alpha_3\alpha_5 + \lambda\alpha_5^2)\frac{y^2}{\alpha_5^2}=0$. Now, the suitable value of $z$ gives $\alpha^*_4=0$, so we shall assume $\alpha_7=1$ and $\alpha_4=0$ from the very beginning. The equality $\alpha_6^2 - \alpha_6\alpha_7 + \lambda\alpha_7^2=0$ takes the form $\alpha_6^2 - \alpha_6 + \lambda=0$, whence $\lambda=\alpha_6-\alpha_6^2$. On the other hand, $\alpha_3^2 - \alpha_3\alpha_5 + \lambda\alpha_5^2=0$ implies $\lambda=\frac{\alpha_3\alpha_5-\alpha_3^2}{\alpha_5^2}$. Therefore, $\alpha_6-\alpha_6^2=\frac{\alpha_3\alpha_5-\alpha_3^2}{\alpha_5^2}$, i.e. $\alpha_3^2-\alpha_3\alpha_5+\alpha_5^2(\alpha_6-\alpha_6^2)=0$. This equation has two solutions in $\alpha_3$, namely, $\alpha_3=\alpha_5\alpha_6$ and $\alpha_3=\alpha_5(1-\alpha_6)$.
\begin{enumerate}
\item $\alpha_3=\alpha_5\alpha_6$. Then
\begin{align*}
\alpha^*_2 &= (x + (\alpha_6-1)y)z + \alpha_5\alpha_6(x + (\alpha_6-1)y)u + \alpha_2x^2 + \alpha_1xy - \alpha_2xy,\\
\alpha^*_4 &= \alpha_5(x + (\alpha_6-1)y)z + ((\alpha_6 - \alpha_5 - 1)x + (\alpha_6^2 - \alpha_5\alpha_6 + \alpha_5 - 2\alpha_6 + 1)y)u\\
&\quad + (\alpha_1-\alpha_2)xy + (\alpha_2 + \alpha_2\alpha_6^2 - \alpha_1 - \alpha_2\alpha_6)y^2.
\end{align*}
We consider $\alpha^*_2=\alpha^*_4=0$ as a system of linear equations in $z$ and $u$. Its determinant is $(\alpha_5 + 1)(\alpha_6 - \alpha_5\alpha_6 - 1)( x + (\alpha_6-1)y)^2$. So, we have the following cases:
\begin{enumerate}
\item $\alpha_5 + 1\ne 0$ and $\alpha_6 - \alpha_5\alpha_6 - 1\ne 0$. Then choosing $x,y$ such that $x + (\alpha_6-1)y\ne 0$, $x - \alpha_6y\ne 0$ and $z,u$ such that $\alpha^*_2=\alpha^*_4=0$ we obtain
\begin{align*}
\alpha^*_1 &= (\alpha_1 - \alpha_2\alpha_6)(x - \alpha_6y)^2,\\
\alpha^*_3 &= \alpha_5\alpha_6(x - \alpha_6y)(x + (\alpha_6-1)y)^2,\\
\alpha^*_5 &= \alpha_5(x - \alpha_6y)(x + (\alpha_6-1)y)^2,\\
\alpha^*_6 &= \alpha_6(x - \alpha_6y)(x + (\alpha_6-1)y)^2,\\
\alpha^*_7 &= (x - \alpha_6y)(x + (\alpha_6-1)y)^2.
\end{align*}
\begin{itemize}
\item Let $\alpha_1 - \alpha_2\alpha_6\ne 0$. Taking $y=0$ and $x=\alpha_1 - \alpha_2\alpha_6$ we obtain the family of representatives $\langle\nb 1+\alpha\beta\nb 3+\alpha\nb 5+\beta\nb 6+\nb 7\rangle$, where $\alpha\not\in \{0,-1\}$, $\beta-\alpha\beta-1\ne 0$ and $\beta^2 - \beta + \lambda=0$.
If $\lambda\not\in\{0,\frac 14\}$, then $\beta=\frac{1\pm\sqrt{1-4\lambda}}2$. Observe that $\beta\ne 0$, so the condition $\beta-\alpha\beta-1\ne 0$ is equivalent to $\alpha\ne 1-\frac 1\beta=\frac{\sqrt{1-4\lambda}\mp 1}{\sqrt{1-4\lambda}\pm 1}$. Thus, we obtain two families $\Big\langle\nb 1+\frac{1\pm\sqrt{1-4\lambda}}2\alpha\nb 3+\alpha\nb 5+\frac{1\pm\sqrt{1-4\lambda}}2\nb 6+\nb 7\Big\rangle_{\alpha\not\in\left\{0,-1,\frac{\sqrt{1-4\lambda}\mp 1}{\sqrt{1-4\lambda}\pm 1}\right\}}$ of representatives of distinct orbits.
If $\lambda=\frac 14$, then $\beta=\frac 12$, and $\alpha\ne 1-\frac 1\beta$ becomes $\alpha\ne -1$. Thus, we obtain the family $\langle\nb 1+\frac 12\alpha\nb 3+\alpha\nb 5+\frac 12\nb 6+\nb 7\rangle_{\alpha\not\in\{0,-1\}}$ of representatives of distinct orbits.
If $\lambda=0$, then $\beta=0$ or $\beta=1$. If $\beta=0$, then the condition $\beta-\alpha\beta-1\ne 0$ becomes $-1\ne 0$; and if $\beta=1$, then it is $\alpha\ne 0$. Thus, we obtain two families $\langle\nb 1+\alpha\nb 5+\nb 7\rangle_{\alpha\not\in\{0,-1\}}$ and $\langle\nb 1+\alpha\nb 3+\alpha\nb 5+\nb 6+\nb 7\rangle_{\alpha\not\in\{0,-1\}}$ of representatives of distinct orbits.
\item Let $\alpha_1 - \alpha_2\alpha_6=0$. Then we obtain the same families as in the previous case, but without $\nb 1$.
\end{itemize}
\item $\alpha_6 - \alpha_5\alpha_6 - 1=0$. Then clearly $\alpha_6\ne 0$ and $\alpha_5=1-\frac 1{\alpha_6}$, so $\alpha_3=\alpha_6-1$. Moreover, $\alpha_6\ne 1$ since $\alpha_5\ne 0$. In particular, $\lambda\ne 0$. Choosing $z$ such that $\alpha^*_4=0$, we obtain
\begin{align*}
\alpha^*_1 &= \frac{x - \alpha_6y}{\alpha_6 - 1}(\alpha_1(\alpha_6-1)x + \alpha_6(\alpha_2(2\alpha_6^2 + 1) - \alpha_6(\alpha_1 + 2\alpha_2))y),\\
\alpha^*_2 &= \frac{x - \alpha_6y}{\alpha_6 - 1}(\alpha_2(\alpha_6-1)x + (\alpha_2\alpha_6(\alpha_6-1) - \alpha_1 + \alpha_2)y),\\
\alpha^*_3 &= (\alpha_6 - 1)(x - \alpha_6y)(x + (\alpha_6-1)y)^2,\\
\alpha^*_5 &= \left(1 - \frac 1{\alpha_6}\right)(x - \alpha_6y)(x + (\alpha_6-1)y)^2,\\
\alpha^*_6 &= \alpha_6(x - \alpha_6y)(x + (\alpha_6-1)y)^2,\\
\alpha^*_7 &= (x - \alpha_6y)(x + (\alpha_6-1)y)^2.
\end{align*}
Let $(\alpha_1,\alpha_2) =(0,0)$. Then we get the representative $\Big\langle(\alpha-1)\nb 3+\left(1-\frac 1{\alpha}\right)\nb 5+\alpha\nb 6+\nb 7\Big\rangle$, where $\alpha^2 - \alpha + \lambda=0$.
If $\lambda\not\in\{0,\frac 14\}$, then we obtain two representatives $\Big\langle\frac{-1\pm\sqrt{1-4\lambda}}2\nb 3+\frac{\sqrt{1-4\lambda}\mp 1}{\sqrt{1-4\lambda}\pm 1}\nb 5+\frac{1\pm\sqrt{1-4\lambda}}2\nb 6+\nb 7 \Big\rangle$ which will be joined with the families $\Big\langle\frac{1\pm\sqrt{1-4\lambda}}2\alpha\nb 3+\alpha\nb 5+\frac{1\pm\sqrt{1-4\lambda}}2\nb 6+\nb 7\Big\rangle_{\alpha\not\in\left\{0,-1,\frac{\sqrt{1-4\lambda}\mp 1}{\sqrt{1-4\lambda}\pm 1}\right\}}$.
If $\lambda=\frac 14$, then we obtain the representative $\langle -\frac 12\nb 3-\nb 5+\frac 12\nb 6+\nb 7\rangle$ which will be joined with the family $\langle\frac 12\alpha\nb 3+\alpha\nb 5+\frac 12\nb 6+\nb 7\rangle_{\alpha\not\in\{0,-1\}}.$
Let $(\alpha_1,\alpha_2) \neq (0,0).$ If $\alpha_2 =0,$ then
\begin{align*}
\alpha^*_1 &= \frac{x - \alpha_6y}{\alpha_6 - 1}((\alpha_6-1)x - \alpha_6^2y)\alpha_1,\\
\alpha^*_2 &= -\frac{x - \alpha_6y}{\alpha_6 - 1}y\alpha_1,
\end{align*}
so choosing $y \neq 0$ we may suppose for the rest of the case that $\alpha_2 \neq 0.$
\begin{itemize}
\item Let $\alpha_1-\alpha_2\alpha_6\ne 0$ and $\alpha_1 - \alpha_2 - 2\alpha_2\alpha_6^2 + 2\alpha_2\alpha_6\ne 0$. Then choosing $z=\frac{\alpha_6(\alpha_2\alpha_6(\alpha_6-1) - \alpha_1 + \alpha_2)y}{\alpha_6 - 1}-(\alpha_6-1)u$ and $x=\frac{(\alpha_1 - \alpha_2 -\alpha_2\alpha_6(\alpha_6-1))y}{\alpha_2(\alpha_6 - 1)}$ we obtain $\alpha^*_2=\alpha^*_4=0$ and
\begin{align*}
\alpha^*_1 &= (\alpha_1 - \alpha_2 - 2\alpha_2\alpha_6^2 + 2\alpha_2\alpha_6)^2(\alpha_1 - \alpha_2\alpha_6)\frac{y^2}{\alpha_2^2(\alpha_6 - 1)^2},\\
\alpha^*_3 &= (\alpha_1 - \alpha_2 - 2\alpha_2\alpha_6^2 + 2\alpha_2\alpha_6)(\alpha_1 - \alpha_2\alpha_6)^2\frac{y^3}{\alpha_2^3(\alpha_6 - 1)^2},\\
\alpha^*_5 &= (\alpha_1 - \alpha_2 - 2\alpha_2\alpha_6^2 + 2\alpha_2\alpha_6)(\alpha_1 - \alpha_2\alpha_6)^2\frac{y^3}{\alpha_2^3\alpha_6(\alpha_6 - 1)^2},\\
\alpha^*_6 &= (\alpha_1 - \alpha_2 - 2\alpha_2\alpha_6^2 + 2\alpha_2\alpha_6)(\alpha_1 - \alpha_2\alpha_6)^2\frac{\alpha_6y^3}{\alpha_2^3(\alpha_6 - 1)^3},\\
\alpha^*_7 &= (\alpha_1 - \alpha_2 - 2\alpha_2\alpha_6^2 + 2\alpha_2\alpha_6)(\alpha_1 - \alpha_2\alpha_6)^2\frac{y^3}{\alpha_2^3(\alpha_6 - 1)^3}.
\end{align*}
Choosing $y=\frac{\alpha_2(\alpha_6 - 1)(\alpha_1 - \alpha_2 - 2\alpha_2\alpha_6^2 + 2\alpha_2\alpha_6)}{\alpha_1 - \alpha_2\alpha_6}$ we obtain the representative $\Big\langle\nb 1+(\alpha-1)\nb 3+\left(1-\frac 1{\alpha}\right)\nb 5+\alpha\nb 6+\nb 7\Big\rangle$, where $\alpha^2 - \alpha + \lambda=0$.
If $\lambda\not\in\{0,\frac 14\}$, then $\alpha=\frac{1\pm\sqrt{1-4\lambda}}2$, so $\alpha-1=\frac{-1\pm\sqrt{1-4\lambda}}2$ and $1-\frac 1{\alpha}=\frac{\sqrt{1-4\lambda}\mp 1}{\sqrt{1-4\lambda}\pm 1}$, and we obtain the following two representatives $\Big\langle \nb 1+\frac{-1\pm\sqrt{1-4\lambda}}2\nb 3+\frac{\sqrt{1-4\lambda}\mp 1}{\sqrt{1-4\lambda}\pm 1}\nb 5+\frac{1\pm\sqrt{1-4\lambda}}2\nb 6+\nb 7 \Big\rangle$ which will be joined with the families $\Big\langle\nb 1+\frac{1\pm\sqrt{1-4\lambda}}2\alpha\nb 3+\alpha\nb 5+\frac{1\pm\sqrt{1-4\lambda}}2\nb 6+\nb 7\Big\rangle_{\alpha\not\in\left\{0,-1,\frac{\sqrt{1-4\lambda}\mp 1}{\sqrt{1-4\lambda}\pm 1}\right\}}$ found above.
If $\lambda=\frac 14$, then $\alpha=\frac 12$, so $\alpha-1=-\frac 12$ and $1-\frac 1{\alpha}=-1$, and we obtain the representative $\langle\nb 1-\frac 12\nb 3-\nb 5+\frac 12\nb 6+\nb 7\rangle$ which will be joined with the family $\langle\nb 1+\frac 12\alpha\nb 3+\alpha\nb 5+\frac 12\nb 6+\nb 7\rangle_{\alpha\not\in\{0,-1\}}$ found above.
\item Let $\alpha_1 - \alpha_2 - 2\alpha_2\alpha_6^2 + 2\alpha_2\alpha_6=0$. Then $\alpha_1=(2\alpha_6^2 - 2\alpha_6 + 1)\alpha_2$.
In this case $\alpha_3^*, \alpha_5^*, \alpha_6^*, \alpha_7^*$ are as above and
\begin{align*}
\alpha^*_1 &= \alpha_2 (x - y \alpha_6)^2 (2\alpha_6^2 - 2\alpha_6 + 1)\\
\alpha^*_2 &=\alpha_2 (x - y \alpha_6)^2.
\end{align*}
Therefore, we obtain the representative $\Big\langle (2 \alpha^2-2\alpha + 1)\nabla_1+\nabla_2+ (\alpha-1)\nabla_3 + (1-\frac{1}{\alpha})\nabla_5 +\alpha\nabla_6+\nabla_7\Big\rangle$, where $\alpha^2 - \alpha + \lambda=0$. Note that $2 \alpha^2-2\alpha + 1 = 1- 2\lambda.$
If $\lambda\not\in\{0,\frac 14\},$ then we obtain two representatives $\Big\langle (1-2\lambda)\nabla_1 + \nabla_2 +\frac{-1+\sqrt{1-4\lambda}}2\nb 3+\frac{\sqrt{1-4\lambda}- 1}{\sqrt{1-4\lambda}+ 1}\nb 5+\frac{1+\sqrt{1-4\lambda}}2\nb 6+\nb 7 \Big\rangle, \Big\langle (1-2\lambda)\nabla_1 + \nabla_2 +\frac{-1-\sqrt{1-4\lambda}}2\nb 3+\frac{\sqrt{1-4\lambda}+ 1}{\sqrt{1-4\lambda}- 1}\nb 5+\frac{1-\sqrt{1-4\lambda}}2\nb 6+\nb 7 \Big\rangle = \Big\langle (1-2\lambda)\lambda\nabla_1 + \lambda\nabla_2 -\lambda\frac{1+\sqrt{1-4\lambda}}2\nb 3-\Big(\frac{\sqrt{1-4\lambda}+ 1}{2}\Big)^2\nb 5+\lambda\frac{1-\sqrt{1-4\lambda}}2\nb 6+\lambda\nb 7 \Big\rangle$.
If $\lambda=\frac 14$, then we obtain the representative $\langle \frac 12\nb 1 + \nabla_2 -\frac 12\nb 3-\nb 5+\frac 12\nb 6+\nb 7\rangle$.
\item Let $\alpha_1-\alpha_2\alpha_6=0$. We may assume that $\alpha_6\ne\frac 12$ (and hence $\lambda\ne\frac 14$), since otherwise $\alpha_1=(2\alpha_6^2 - 2\alpha_6 + 1)\alpha_2$, which has already been considered.
In this case $\alpha_3^*, \alpha_5^*, \alpha_6^*, \alpha_7^*$ are as above and
\begin{align*}
\alpha^*_1 &= \alpha_2\alpha_6 (x - y \alpha_6)(x+(\alpha_6 - 1)y)\\
\alpha^*_2 &=\alpha_2 (x - y \alpha_6)(x+(\alpha_6 - 1)y).
\end{align*}
Therefore, we obtain the representative $\Big\langle \alpha\nabla_1+\nabla_2+ (\alpha-1)\nabla_3 + (1-\frac{1}{\alpha})\nabla_5 +\alpha\nabla_6+\nabla_7\Big\rangle$, where $\alpha^2 - \alpha + \lambda=0$.
If $\lambda\not\in\{0,\frac 14\},$ then we obtain two representatives $\Big\langle \frac{1+\sqrt{1-4\lambda}}2\nabla_1 + \nabla_2 +\frac{-1+\sqrt{1-4\lambda}}2\nb 3+\frac{\sqrt{1-4\lambda}- 1}{\sqrt{1-4\lambda}+1}\nb 5+\frac{1+\sqrt{1-4\lambda}}2\nb 6+\nb 7 \Big\rangle, \Big\langle \frac{1-\sqrt{1-4\lambda}}2\nabla_1 + \nabla_2 -\frac{1+\sqrt{1-4\lambda}}2\nb 3+\frac{\sqrt{1-4\lambda}+ 1}{\sqrt{1-4\lambda}- 1}\nb 5+\frac{1-\sqrt{1-4\lambda}}2\nb 6+\nb 7 \Big\rangle = \Big\langle \lambda\frac{1-\sqrt{1-4\lambda}}2\nabla_1 + \lambda\nabla_2 -\lambda\frac{1+\sqrt{1-4\lambda}}2\nb 3-\Big(\frac{\sqrt{1-4\lambda}+ 1}{2}\Big)^2\nb 5+\lambda\frac{1-\sqrt{1-4\lambda}}2\nb 6+\lambda\nb 7 \Big\rangle$
\end{itemize}
\item $\alpha_5 + 1=0$. Then $\alpha_3=-\alpha_6$. Observe that we may assume that $\alpha_6\ne\frac 12$ (and hence $\lambda\ne\frac 14$), since otherwise $\alpha_6 - \alpha_5\alpha_6 - 1=0,$ which was considered above. Choosing $z$ such that $\alpha^*_4=0$, we have
\begin{align*}
\alpha^*_1 &= \alpha_1x^2 + 2\alpha_2\alpha_6(\alpha_6 - 1)xy + \alpha_6(\alpha_6-1)(\alpha_1 - \alpha_2)y^2,\\
\alpha^*_2 &= \alpha_2x^2 + 2(\alpha_1 - \alpha_2)xy + (\alpha_2\alpha_6(\alpha_6-1) - \alpha_1 + \alpha_2)y^2,\\
\alpha^*_3 &= -\alpha_6(x - \alpha_6y)(x + (\alpha_6-1)y)^2,\\
\alpha^*_5 &= -(x - \alpha_6y)(x + (\alpha_6-1)y)^2,\\
\alpha^*_6 &= \alpha_6(x - \alpha_6y)(x + (\alpha_6-1)y)^2,\\
\alpha^*_7 &= (x - \alpha_6y)(x + (\alpha_6-1)y)^2.
\end{align*}
If $(\alpha_1,\alpha_2)=(0,0)$, then we obtain the representative $\langle-\beta\nb 3-\nb 5+\beta\nb 6+\nb 7\rangle$, where $\beta^2 - \beta + \lambda=0$.
If $\lambda\not\in\{0,\frac 14\}$, then we have two representatives $\Big\langle\frac{-1\mp\sqrt{1-4\lambda}}2\nb 3-\nb 5+\frac{1\pm\sqrt{1-4\lambda}}2\nb 6+\nb 7\Big\rangle$ which will be joined with the families $\Big\langle\frac{1\pm\sqrt{1-4\lambda}}2\alpha\nb 3+\alpha\nb 5+\frac{1\pm\sqrt{1-4\lambda}}2\nb 6+\nb 7\Big\rangle_{\alpha\ne 0,-1}$ found above.
If $\lambda=0$, then we have two representatives $\langle-\nb 3-\nb 5+\nb 6+\nb 7\rangle$ and $\langle-\nb 5+\nb 7\rangle$ which will be joined the families $\langle\alpha\nb 3+\alpha\nb 5+\nb 6+\nb 7\rangle_{\alpha\ne 0,-1}$ and $\langle\alpha\nb 5+\nb 7\rangle_{\alpha\ne 0,-1}$ found above.
Let $(\alpha_1,\alpha_2) \neq (0,0).$ Suppose that $\alpha_2=0$. Then we have $\alpha^*_2 = \alpha_1 y (2x - y).$ So, whenever $(\alpha_1,\alpha_2) \neq (0,0)$, we may find $x,y$ such that $\alpha^*_2\neq 0$ and $\alpha^*_7=1$. In the rest of the case we will suppose that $\alpha_2 \neq 0.$
The determinant of the equation $\alpha^*_2=0$ is $4(\alpha_2\alpha_6 + \alpha_1 - \alpha_2)(\alpha_1-\alpha_2\alpha_6)$.
\begin{itemize}
\item Let $(\alpha_2\alpha_6 + \alpha_1 - \alpha_2)(\alpha_1-\alpha_2\alpha_6)\ne 0.$
Suppose that $\alpha_2 = 2\alpha_1.$ Observe that in this case $(\alpha_2\alpha_6 + \alpha_1 - \alpha_2)(\alpha_2\alpha_6 - \alpha_1)\ne 0$ is equivalent to $\alpha_1^2(2\alpha_6 - 1)^2\ne 0$. Then $\alpha_2^* - 2\alpha_1^* = -2\alpha_1(2x-y)y(2\alpha_6-1)^2,$ so choosing appropriate $x,y$ we may suppose that $\alpha_2\ne 2\alpha_1.$
The equation $\alpha^*_2=0$ has two solutions $x_1=\mu_1 y$ and $x_2=\mu_2 y$, where $\mu_1,\mu_2\in{\mathbb C}$, $\mu_1\ne\mu_2$. As in the case 1(a) if $\mu_1^2 - \mu_1 + \lambda=\mu_2^2 - \mu_2 + \lambda=0$, then $\mu_1+\mu_2=1$. But $\mu_1+\mu_2=-\frac{2(\alpha_1 - \alpha_2)}{\alpha_2}$, whence $2(\alpha_1 - \alpha_2)=-\alpha_2$, which contradicts the assumption that $\alpha_2\ne 2\alpha_1$. Thus, we may choose $x=\mu_iy$ to make $\alpha^*_2=0$. Since in this case the condition $(\alpha_1,\alpha_2,\alpha_4)\ne (0,0,0)$ is invariant under automorphisms, we have $\alpha^*_1\ne 0$ for such a choice of $x$. Thus, we obtain the family of representatives $\langle\alpha\nb 1-\beta\nb 3-\nb 5+\beta\nb 6+\nb 7\rangle$, where $\beta^2 - \beta + \lambda=0$ and $\alpha\ne 0$. Then taking $y=0$ and $x=\alpha$, we obtain the representative $\langle\nb 1-\beta\nb 3-\nb 5+\beta\nb 6+\nb 7\rangle.$ It will be joined with the family $\langle\nb 1+\alpha\beta\nb 3+\alpha\nb 5+\beta\nb 6+\nb 7\rangle_{\alpha\not\in \{0,-1\}}.$
\item Let $\alpha_1-\alpha_2\alpha_6=0$. Then
\begin{align*}
\alpha^*_1 =\alpha_2\alpha_6(x + (\alpha_6-1)y)^2,
\alpha^*_2 =\alpha_2(x + (\alpha_6-1)y)^2.
\end{align*}
So, taking $x=\alpha_6y+\alpha_2$, we obtain the representative $\langle\beta\nb 1+\nb 2-\beta\nb 3-\nb 5+\beta\nb 6+\nb 7\rangle$, where $\beta^2 - \beta + \lambda=0$. If $\lambda\not\in\{0,\frac 14\}$, then we have two representatives $\Big\langle\frac{1\pm\sqrt{1-4\lambda}}2\nb 1+\nb 2-\frac{1\pm\sqrt{1-4\lambda}}2\nb 3-\nb 5+\frac{1\pm\sqrt{1-4\lambda}}2\nb 6+\nb 7\Big\rangle$.
If $\lambda=0$, then we have two representatives $\langle\nb 1+\nb 2-\nb 3-\nb 5+\nb 6+\nb 7\rangle$ and $\langle\nb 2-\nb 5+\nb 7\rangle$.
\item Let $\alpha_2\alpha_6 + \alpha_1 - \alpha_2=0$. Then
\begin{align*}
\alpha^*_1 =\alpha_2(1-\alpha_6)(x - \alpha_6y)^2,
\alpha^*_2 =\alpha_2(x - \alpha_6y)^2.
\end{align*}
Taking $x,y$ such that $(x + (\alpha_6-1)y)^2=\alpha_2(x - \alpha_6y)\ne 0$, we obtain the representative $\langle(1-\beta)\nb 1+\nb 2-\beta\nb 3-\nb 5+\beta\nb 6+\nb 7\rangle$, where $\beta^2 - \beta + \lambda=0$. If $\lambda\not\in\{0,\frac 14\}$, then we have two representatives $\Big\langle\frac{1\mp\sqrt{1-4\lambda}}2\nb 1+\nb 2-\frac{1\pm\sqrt{1-4\lambda}}2\nb 3-\nb 5+\frac{1\pm\sqrt{1-4\lambda}}2\nb 6+\nb 7\Big\rangle$.
If $\lambda=0$, then we have two representatives $\langle\nb 2-\nb 3-\nb 5+\nb 6+\nb 7\rangle$ and $\langle\nb 1+\nb 2-\nb 5+\nb 7\rangle$.
\end{itemize}
\end{enumerate}
\item $\alpha_3=\alpha_5(1-\alpha_6)$. Observe that we may assume that $\alpha_6\ne\frac 12$ (and hence $\lambda\ne\frac 14$), since otherwise $\alpha_3=\alpha_5\alpha_6$, which has already been considered.
Consider $\alpha^*_2=\alpha^*_4=0$ as a system of linear equations in $z$ and $u$. Then its determinant is the following polynomial in $x$ and $y$:
\begin{align*}
D(x,y)&=(\alpha_5^2\alpha_6 - \alpha_5^2 - \alpha_5 + \alpha_6 - 1)x^2\\
&\quad-(2\alpha_5^2\alpha_6^2 - 2\alpha_5^2\alpha_6 - 2\alpha_6^2 - \alpha_5 + 4\alpha_6 - 2)xy\\
&\quad+(\alpha_5^2\alpha_6^2 + \alpha_5\alpha_6 + \alpha_6^2 - 2\alpha_6 + 1)(\alpha_6 - 1)y^2.
\end{align*}
Suppose that $\alpha_5^2\alpha_6 - \alpha_5^2 - \alpha_5 + \alpha_6 - 1 = 0.$ In this case $\alpha_5^2+1\ne 0$, since otherwise $(\alpha_5^2+1)\alpha_6 = \alpha_5^2 + \alpha_5 + 1$ would imply $\alpha_5=0$, whence $\alpha_5^2+1=1$, a contradiction. Thus, $\alpha_6=\frac{\alpha_5^2 + \alpha_5 + 1}{\alpha_5^2+1}$ and hence $\alpha_3=-\frac{\alpha_5^2}{\alpha_5^2+1}$. Then $D(x,y) = \frac{y \alpha_5 (1 + \alpha_5)^3 ((1-\alpha_5)x + \alpha_5y )}{(1 + \alpha_5^2)^2}.$ If $\alpha_5+1=0,$ then $\alpha_5=-1$, $\alpha_6=\frac 12$, $\alpha_3=-\frac 12=\alpha_5\alpha_6$. This case has been considered above. Therefore, we may suppose that $D(x,y)$ is not identically zero, and hence we may solve $\alpha^*_2=\alpha^*_4=0$ in $z$ and $u$. Choosing these values of $z$ and $u$, we may suppose that $\alpha_2 = \alpha_4 = 0$ from the very beginning.
Then taking the appropriate values of $z$ and $u$ we have
\begin{align*}
\alpha^*_1&=\frac{\alpha_1(\alpha_5^2\alpha_6 - \alpha_5^2 - \alpha_5 + \alpha_6 - 1)(x + y(\alpha_6-1))^2(x-y\alpha_6)^2}{D(x,y)},\\
\alpha^*_3&=\alpha_5 (1 - \alpha_6) (x - y \alpha_6)^2 (x + y(\alpha_6-1)),\\
\alpha^*_5&=\alpha_5 (x - y \alpha_6)^2 (x + y(\alpha_6-1)),\\
\alpha^*_6&=\alpha_6 (x - y \alpha_6) (x + y(\alpha_6-1))^2,\\
\alpha^*_7&=(x - y \alpha_6) (x + y(\alpha_6-1))^2.
\end{align*}
Observe that for $x,y$ such that $\alpha_7^* = 1$ and $D(x,y)\ne 0$ we have $(\alpha_5^*)^2\alpha_6^* - (\alpha_5^*)^2 - \alpha_5^* + \alpha_6^* - 1 = \frac{D(x,y)}{(x + y(\alpha_6-1))^2}\ne 0,$ so we may assume that $\alpha_5^2\alpha_6 - \alpha_5^2 - \alpha_5 + \alpha_6 - 1 \neq 0$ from the very beginning.
Let $\alpha_5 \neq -1$. Taking $x = \frac{y (\alpha_6(\alpha_5-1)+1)}{\alpha_5+1},$ we have $ D(x,y) = \frac{y^2 \alpha_5^2 (2 \alpha_6-1)^2}{(\alpha_5+1)^2}\ne 0$, $x^2 - xy + \lambda y^2 = -\frac{y^2 \alpha_5 (2 \alpha_6-1)^2}{(\alpha_5+1)^2}\ne 0$, $\alpha_5^* = -\alpha_7^*$, so we get the representative $\langle \alpha \nabla_1 + (\beta-1)\nabla_3 - \nabla_5 + \beta \nabla_6 + \nabla_7 \rangle_{\beta^2 - \beta + \lambda=0},$ where
\[
\alpha = -\frac{\alpha_1 (\alpha_5+1) (\alpha_5^2\alpha_6 - \alpha_5^2 - \alpha_5 + \alpha_6 - 1)}{y \alpha_5^2 (2 \alpha_6-1)^2}
\]
\begin{itemize}
\item If $\alpha_1 \neq 0,$ then we take $y = -\frac{\alpha_1 (\alpha_5+1) (\alpha_5^2\alpha_6 - \alpha_5^2 - \alpha_5 + \alpha_6 - 1)}{\alpha_5^2 (2 \alpha_6-1)^2} \neq 0,$ so that $x^2 - xy + \lambda y^2 \neq 0$, $D(x,y) \neq 0$, and we get the representative $\langle \nabla_1 + (\alpha-1)\nabla_3 - \nabla_5 + \alpha \nabla_6 + \nabla_7 \rangle_{\alpha^2 - \alpha + \lambda=0}.$
If $\lambda \neq 0$, then we get two distinct representatives $\Big\langle \nb 1 - \frac{1\mp\sqrt{1-4\lambda}}2\nb 3-\nb 5+\frac{1\pm\sqrt{1-4\lambda}}2\nb 6+\nb 7\Big\rangle.$
If $\lambda = 0$, then we get two distinct representatives $\Big\langle \nb 1 - \nb 3-\nb 5 +\nb 7\Big\rangle$ and $\Big\langle \nb 1 -\nb 5+ \nb 6+\nb 7\Big\rangle.$
\item If $\alpha_1 = 0,$ then we get the same representatives but without $\nabla_1.$
\end{itemize}
\end{enumerate}
\item $\alpha_6^2 - \alpha_6\alpha_7 + \lambda\alpha_7^2=0$ and $\alpha_5=0$. We may assume that $\alpha_3=0$, since the case $\alpha_5=0$ and $\alpha_3\ne 0$ was considered in 1(b). Then $\alpha^*_3=\alpha^*_5=0$. A suitable choice of $x,y$ and $z$ gives $\alpha^*_7=1$ and $\alpha^*_2=0$, so we shall assume that $\alpha_7=1$ and $\alpha_2=0$. As above, $\lambda=\alpha_6-\alpha_6^2$. Now, choosing $z=-\frac{\alpha_1xy + (\alpha_4\alpha_6^2 - \alpha_4\alpha_6)y^2}{x+(\alpha_6 - 1)y}$ we obtain $\alpha^*_2=0$.
\begin{enumerate}
\item $\alpha_6\ne 1$. Then choosing $u=-\frac{\alpha_4x^2 + (\alpha_1 - 2\alpha_4)xy - (\alpha_1 - \alpha_4)y^2}{(\alpha_6 - 1)(x + (\alpha_6 - 1)y)}$ we have $\alpha^*_4=0$ and
\begin{align*}
\alpha^*_1&=(\alpha_1-\alpha_4\alpha_6)(x-\alpha_6y)^2,\\
\alpha^*_6 &= \alpha_6(x - \alpha_6y)(x + (\alpha_6-1)y)^2,\\
\alpha^*_7 &= (x - \alpha_6y)(x + (\alpha_6-1)y)^2.
\end{align*}
\begin{enumerate}
\item $\alpha_1-\alpha_4\alpha_6\ne 0$. Then choosing $y=0$ and $x=\alpha_1-\alpha_4\alpha_6$ we obtain the representative $\langle\nb 1+\alpha\nb 6+\nb 7\rangle$, where $\alpha^2 - \alpha + \lambda=0$. If $\lambda\not\in\{0,\frac 14\}$, then we have two representatives $\Big\langle\nb 1+\frac{1\pm\sqrt{1-4\lambda}}2\nb 6+\nb 7\Big\rangle$. We will join them with the families $\Big\langle\nb 1+\frac{1\pm\sqrt{1-4\lambda}}2\alpha\nb 3+\alpha\nb 5+\frac{1\pm\sqrt{1-4\lambda}}2\nb 6+\nb 7\Big\rangle_{\alpha\ne 0}$found above. If $\lambda=\frac 14$, then we have the representative $\langle\nb 1+\frac 12\nb 6+\nb 7\rangle$, which will be joined with the family $\langle\nb 1+\frac 12\alpha\nb 3+\alpha\nb 5+\frac 12\nb 6+\nb 7\rangle_{\alpha\ne 0}$ found above. If $\lambda=0$, then $\alpha_6=0$, since $\alpha_6\ne 1$ by assumption. So, we have the representative $\langle\nb 1+\nb 7\rangle$. It will be joined with $\langle\nb 1+\alpha\nb 5+\nb 7\rangle_{\alpha\ne 0}$.
\item $\alpha_1-\alpha_4\alpha_6=0$. Then we have the same representatives as above, but without $\nb 1$. We join them with the families found above.
\end{enumerate}
\item $\alpha_6=1$, so that $\lambda =0$. Then
\begin{align*}
\alpha^*_1 &=\alpha_1x(x - y),\\
\alpha^*_4 &=(\alpha_4(x-y) + \alpha_1y)(x - y),\\
\alpha^*_6 &= x^2(x - y),\\
\alpha^*_7 &= x^2(x - y).
\end{align*}
\begin{enumerate}
\item $\alpha_1\ne 0$ and $\alpha_1-\alpha_4\ne 0$. Then choosing $x=\alpha_1$ and $y=-\frac{\alpha_1\alpha_4}{\alpha_1-\alpha_4}$ we obtain the representative $\langle\nb 1+\nb 6+\nb 7\rangle$ which will be joined with the family $\langle\nb 1+\alpha\nb 3+\alpha\nb 5+\nb 6+\nb 7\rangle_{\alpha\ne 0}$ found above.
\item $\alpha_1=0$. Then we have two representatives $\langle\nb 4+\nb 6+\nb 7\rangle$ and $\langle\nb 6+\nb 7\rangle$ depending on whether $\alpha_4=0$ or not. Both of them belong to the families found above.
\item $\alpha_1-\alpha_4=0$. Then
\begin{align*}
\alpha^*_1 &=\alpha_1x(x - y),\\
\alpha^*_4 &=\alpha_1x(x - y),\\
\alpha^*_6 &= x^2(x - y),\\
\alpha^*_7 &= x^2(x - y).
\end{align*}
Thus, we have two representatives $\langle\nb 1+\nb 4+\nb 6+\nb 7\rangle$ and $\langle\nb 6+\nb 7\rangle$. The second one was found above.
\end{enumerate}
\end{enumerate}
\end{enumerate}
\item $\alpha_7=0$. We may assume that $\alpha_6=0$, since the case $\alpha_7=0$ and $\alpha_6\ne 0$ was considered in 1(a).
\begin{enumerate}
\item Let $\alpha_3^2 - \alpha_3\alpha_5 + \lambda\alpha_5^2 \neq 0.$ If $\alpha_5 \neq 0,$ we may take $x = \frac{(\alpha_5 - \alpha_3)y}{\alpha_5}$ and get $\alpha^*_5=0$. Observe that $(\alpha^*_3)^2 - \alpha^*_3\alpha^*_5 + \lambda(\alpha^*_5)^2=(\alpha_3^2 - \alpha_3\alpha_5 + \lambda\alpha_5^2)(x^2 - xy + \lambda y^2)^3$, so the condition $\alpha_3^2 - \alpha_3\alpha_5 + \lambda\alpha_5^2\ne 0$ is invariant under the automorphisms. Thus, we shall assume that $\alpha_5=0$ and $\alpha_3\ne 0$ from the very beginning. Choosing $y=0$, we have
\begin{align*}
\alpha^*_1 &= x(\alpha_1x + \alpha_3z),\\
\alpha^*_2 &= x(\alpha_2x + \alpha_3u),\\
\alpha^*_3 &= \alpha_3x^3,\\
\alpha^*_4 &= \alpha_4x^2.
\end{align*}
Taking $z = -\frac{\alpha_1x}{\alpha_3}, u = -\frac{\alpha_2x}{\alpha_3},$ we get two representatives $\langle \nabla_3 + \nabla_4 \rangle$ and $\langle \nabla_3 \rangle$ depending on whether $\alpha_4=0$ or not.
\item Let $\alpha_3^2 - \alpha_3\alpha_5 + \lambda\alpha_5^2 = 0, \alpha_5 \neq 0.$ Taking $x,y$ such that $(x^2-xy+\lambda y^2)(\alpha_5x + (\alpha_3-\alpha_5)y) = 1,$ we may suppose that $\alpha_5^* = 1$ and $\lambda = \alpha_3 - \alpha_3^2.$ Consider $\alpha_2^* = 0, \alpha_4^* = 0$ as a linear system in $u, z.$ Its determinant is $\alpha_3(x + (\alpha_3-1)y)^2.$ If $\alpha_3 = 0,$ then $\lambda = 0$ and we get a Leibniz cocycle. Therefore, we may suppose that this determinant is nonzero and the system $\alpha_2^* = 0, \alpha_4^* = 0$ has a unique solution. We get the family of representatives $\langle \alpha_1\nabla_1 + \alpha_3\nabla_3 + \nabla_5 \rangle,$ where $\alpha_3^2 - \alpha_3 + \lambda = 0,$ except for $(\lambda,\alpha_3) = (0,0)$ which gives a Leibniz cocycle. Therefore, we may suppose that $\alpha_2 = \alpha_4 = 0$ from the very beginning and
\begin{align*}
\alpha^*_1 &= \alpha_1(x-\alpha_3y)^2,\\
\alpha^*_3 &= \alpha_3(x-\alpha_3y)(x+(\alpha_3-1)y)^2,\\
\alpha^*_5 &= (x-\alpha_3y)(x+(\alpha_3-1)y)^2.
\end{align*}
If $\lambda\not\in\{0,\frac 14\}$, then we have the following representatives: $\Big\langle \nabla_1 + \frac{1\pm\sqrt{1-4\lambda}}{2}\nabla_3 + \nabla_5 \Big\rangle$ and $\Big\langle \frac{1\pm\sqrt{1-4\lambda}}{2}\nabla_3 + \nabla_5 \Big\rangle$. If $\lambda=\frac 14$, then we have the representatives $\langle\nb 1+\frac 12\nb 3+\nb 5\rangle$ and $\langle\frac 12\nb 3+\nb 5\rangle$. If $\lambda=0$, then we have the representatives $\langle \nabla_1 + \nabla_3 + \nabla_5 \rangle$ and $\langle \nabla_3 + \nabla_5 \rangle$.
\end{enumerate}
\end{enumerate}
\begin{landscape}
\tiny{
\begin{tabular}{|c|c|c|}
\hline
$\lambda=0$ & $\lambda=\frac 14$ & $\lambda\not\in\{0,\frac 14\}$\\
\hline
$\begin{array}{l}
\langle\nb 1+\alpha\nb 3+\beta\nb 5+\nb 6\rangle_{\alpha,\beta\ne 0},\\
\langle\alpha\nb 3+\beta\nb 5+\nb 6\rangle_{\alpha,\beta\ne 0},\\
\langle\nb 2+\alpha\nb 5+\nb 6\rangle_{\alpha\ne 1},\\
\langle\alpha\nb 5+\nb 6\rangle,\\
\langle\nb 2+\alpha\nb 4+\nb 5+\nb 6\rangle,\\
\langle\nb 4+\nb 5+\nb 6\rangle,\\
\langle \nb 2+\alpha\nb 3+\nb 6\rangle_{\alpha\ne 0},\\
\langle \alpha\nb 3+\nb 6\rangle_{\alpha\ne 0},\\
\langle\nb 1+\alpha\nb 2-\nb 3+\nb 6\rangle,\\
\langle\nb 1+\alpha\nb 3+\nb 7\rangle_{\alpha\ne 0},\\
\langle\alpha\nb 3+\nb 7\rangle_{\alpha\ne 0},\\
\langle\alpha\nb 3+\nb 4+\nb 6+\nb 7\rangle,\\
\langle\alpha\nb 3+\nb 6+\nb 7\rangle,\\
\langle\nb 1-\nb 3+\alpha\nb 4+\nb 6+\nb 7\rangle,\\
\langle\nb 1+\alpha\nb 5+\nb 7\rangle,\\
\langle\nb 1+\alpha\nb 3+\alpha\nb 5+\nb 6+\nb 7\rangle,\\
\langle\alpha\nb 5+\nb 7\rangle,\\
\langle\alpha\nb 3+\alpha\nb 5+\nb 6+\nb 7\rangle_{\alpha\ne 0},\\
\langle\nb 1+\nb 2-\nb 3-\nb 5+\nb 6+\nb 7\rangle,\\
\langle\nb 2-\nb 5+\nb 7\rangle,\\
\langle\nb 2-\nb 3-\nb 5+\nb 6+\nb 7\rangle,\\
\langle\nb 1+\nb 2-\nb 5+\nb 7\rangle,\\
\langle\nb 1-\nb 3-\nb 5+\nb 7\rangle,\\
\langle\nb 1-\nb 5+\nb 6+\nb 7\rangle,\\
\langle-\nb 3-\nb 5+\nb 7\rangle,\\
\langle-\nb 5+\nb 6+\nb 7\rangle,\\
\langle\nb 1+\nb 4+\nb 6+\nb 7\rangle,\\
\langle \nb 3 + \nb 4 \rangle,\\
\langle \nb 3 \rangle,\\
\langle \nb 1 + \nb 3 + \nb 5 \rangle,\\
\langle \nb 3 + \nb 5 \rangle
\end{array}$
&
$\begin{array}{l}
\langle\nb 1+\alpha\nb 3+\beta\nb 5+\nb 6\rangle_{\alpha,\beta\ne 0},\\
\langle\alpha\nb 3+\beta\nb 5+\nb 6\rangle_{\alpha,\beta\ne 0},\\
\langle\nb 2+\alpha\nb 5+\nb 6\rangle_{\alpha\ne 2},\\
\langle\alpha\nb 5+\nb 6\rangle,\\
\langle\nb 2+\alpha\nb 4+2\nb 5+\nb 6\rangle,\\
\langle\nb 4+2\nb 5+\nb 6\rangle,\\
\langle \nb 2+\alpha\nb 3+\nb 6\rangle_{\alpha\ne 0},\\
\langle \alpha\nb 3+\nb 6\rangle_{\alpha\ne 0},\\
\langle\nb 1+\alpha\nb 2-\nb 3+\nb 6\rangle,\\
\langle\nb 1+\alpha\nb 3+\frac 12\nb 6+\nb 7\rangle_{\alpha\ne 0},\\
\langle\alpha\nb 3+\frac 12\nb 6+\nb 7\rangle_{\alpha\ne 0},\\
\langle\nb 1+\frac 12\alpha\nb 3+\alpha\nb 5+\frac 12\nb 6+\nb 7\rangle,\\
\langle\frac 12\alpha\nb 3+\alpha\nb 5+\frac 12\nb 6+\nb 7\rangle,\\
\langle \frac 12\nb 1+ \nb 2 -\frac 12\nb 3-\nb 5+\frac 12\nb 6+\nb 7\rangle,\\
\langle \nb 3 + \nb 4 \rangle,\\
\langle \nb 3 \rangle,\\
\langle\nb 1+\frac 12\nb 3+\nb 5\rangle,\\
\langle\frac 12\nb 3+\nb 5\rangle
\end{array}$
&
$\begin{array}{l}
\langle\nb 1+\alpha\nb 3+\beta\nb 5+\nb 6\rangle_{\alpha,\beta\ne 0},\\
\langle\alpha\nb 3+\beta\nb 5+\nb 6\rangle_{\alpha,\beta\ne 0},\\
\langle\nb 2+\alpha\nb 5+\nb 6\rangle_{\alpha\ne\frac{1\pm\sqrt{1-4\lambda}}{2\lambda}},\\
\langle\alpha\nb 5+\nb 6\rangle,\\
\Big\langle \lambda\nb 2+\lambda\alpha\nb 4+\frac{1+\sqrt{1-4\lambda}}{2}\nb 5+\lambda\nb 6\Big\rangle,\\
\Big\langle\nb 2+\alpha\nb 4+\frac{2}{1+\sqrt{1-4\lambda}}\nb 5+\nb 6\Big\rangle,\\
\Big\langle\lambda \nb 4+\frac{1+\sqrt{1-4\lambda}}{2}\nb 5+\lambda \nb 6\Big\rangle,\\
\Big\langle\nb 4+\frac{2}{1+\sqrt{1-4\lambda}}\nb 5+\nb 6\Big\rangle,\\
\langle \nb 2+\alpha\nb 3+\nb 6\rangle_{\alpha\ne 0},\\
\langle \alpha\nb 3+\nb 6\rangle_{\alpha\ne 0},\\
\langle\nb 1+\alpha\nb 2-\nb 3+\nb 6\rangle,\\
\Big\langle \nb 1+\alpha\nb 3+ \frac{1+\sqrt{1-4\lambda}}{2}\nb 6+\nb 7\Big\rangle_{\alpha\ne 0},\\
\Big\langle \nb 1+\alpha\nb 3+ \frac{1-\sqrt{1-4\lambda}}{2}\nb 6+\nb 7\Big\rangle_{\alpha\ne 0},\\
\Big\langle \alpha\nb 3+ \frac{1+\sqrt{1-4\lambda}}{2}\nb 6+\nb 7\Big\rangle_{\alpha\ne 0},\\
\Big\langle \alpha\nb 3+ \frac{1-\sqrt{1-4\lambda}}{2}\nb 6+\nb 7\Big\rangle_{\alpha\ne 0},\\
\Big\langle\nb 1+\frac{1+\sqrt{1-4\lambda}}2\alpha\nb 3+\alpha\nb 5+\frac{1+\sqrt{1-4\lambda}}2\nb 6+\nb 7\Big\rangle,\\
\Big\langle\nb 1+\frac{1-\sqrt{1-4\lambda}}2\alpha\nb 3+\alpha\nb 5+\frac{1-\sqrt{1-4\lambda}}2\nb 6+\nb 7\Big\rangle,\\
\Big\langle\frac{1+\sqrt{1-4\lambda}}2\alpha\nb 3+\alpha\nb 5+\frac{1+\sqrt{1-4\lambda}}2\nb 6+\nb 7\Big\rangle,\\
\Big\langle\frac{1-\sqrt{1-4\lambda}}2\alpha\nb 3+\alpha\nb 5+\frac{1-\sqrt{1-4\lambda}}2\nb 6+\nb 7\Big\rangle,\\
\Big\langle (1-2\lambda)\nb 1+ \nb 2 + \frac{-1+\sqrt{1-4\lambda}}2\nb 3+\frac{\sqrt{1-4\lambda}- 1}{\sqrt{1-4\lambda}+ 1}\nb 5+\frac{1+\sqrt{1-4\lambda}}2\nb 6+\nb 7 \Big\rangle,\\
\Big\langle (1-2\lambda)\lambda\nabla_1 + \lambda\nabla_2 -\lambda\frac{1+\sqrt{1-4\lambda}}2\nb 3-\Big(\frac{\sqrt{1-4\lambda}+ 1}{2}\Big)^2\nb 5+\lambda\frac{1-\sqrt{1-4\lambda}}2\nb 6+\lambda\nb 7 \Big\rangle\\
\Big\langle \frac{1+\sqrt{1-4\lambda}}2\nb 1+ \nb 2 + \frac{-1+\sqrt{1-4\lambda}}2\nb 3+\frac{\sqrt{1-4\lambda}- 1}{\sqrt{1-4\lambda}+ 1}\nb 5+\frac{1+\sqrt{1-4\lambda}}2\nb 6+\nb 7 \Big\rangle,\\
\Big\langle \lambda\frac{1-\sqrt{1-4\lambda}}2\nabla_1 + \lambda\nabla_2 -\lambda\frac{1+\sqrt{1-4\lambda}}2\nb 3-\Big(\frac{\sqrt{1-4\lambda}+ 1}{2}\Big)^2\nb 5+\lambda\frac{1-\sqrt{1-4\lambda}}2\nb 6+\lambda\nb 7 \Big\rangle\\
\Big\langle\frac{1+\sqrt{1-4\lambda}}2\nb 1+\nb 2-\frac{1+\sqrt{1-4\lambda}}2\nb 3-\nb 5+\frac{1+\sqrt{1-4\lambda}}2\nb 6+\nb 7\Big\rangle,\\
\Big\langle\frac{1-\sqrt{1-4\lambda}}2\nb 1+\nb 2-\frac{1-\sqrt{1-4\lambda}}2\nb 3-\nb 5+\frac{1-\sqrt{1-4\lambda}}2\nb 6+\nb 7\Big\rangle,\\
\Big\langle\frac{1-\sqrt{1-4\lambda}}2\nb 1+\nb 2-\frac{1+\sqrt{1-4\lambda}}2\nb 3-\nb 5+\frac{1+\sqrt{1-4\lambda}}2\nb 6+\nb 7\Big\rangle,\\
\Big\langle\frac{1+\sqrt{1-4\lambda}}2\nb 1+\nb 2-\frac{1-\sqrt{1-4\lambda}}2\nb 3-\nb 5+\frac{1-\sqrt{1-4\lambda}}2\nb 6+\nb 7\Big\rangle,\\
\Big\langle\nb 1-\frac{1-\sqrt{1-4\lambda}}2\nb 3-\nb 5+\frac{1+\sqrt{1-4\lambda}}2\nb 6+\nb 7\Big\rangle,\\
\Big\langle\nb 1-\frac{1+\sqrt{1-4\lambda}}2\nb 3-\nb 5+\frac{1-\sqrt{1-4\lambda}}2\nb 6+\nb 7\Big\rangle,\\
\Big\langle-\frac{1-\sqrt{1-4\lambda}}2\nb 3-\nb 5+\frac{1+\sqrt{1-4\lambda}}2\nb 6+\nb 7\Big\rangle,\\
\Big\langle-\frac{1+\sqrt{1-4\lambda}}2\nb 3-\nb 5+\frac{1-\sqrt{1-4\lambda}}2\nb 6+\nb 7\Big\rangle,\\
\langle \nb 3 + \nb 4 \rangle,\\
\langle \nb 3 \rangle,\\
\Big\langle \nb 1 + \frac{1+\sqrt{1-4\lambda}}{2}\nb 3 + \nb 5 \Big\rangle,\\
\Big\langle \nb 1 + \frac{1-\sqrt{1-4\lambda}}{2}\nb 3 + \nb 5 \Big\rangle,\\
\Big\langle \frac{1+\sqrt{1-4\lambda}}{2}\nb 3 + \nb 5 \Big\rangle,\\
\Big\langle \frac{1-\sqrt{1-4\lambda}}{2}\nb 3 + \nb 5 \Big\rangle.
\end{array}$
\\
\hline
\end{tabular}
}
\end{landscape}
\newpage
Denote $\Theta=\frac{1+\sqrt{1-4\lambda}}{2}.$ The orbits above correspond to the following algebras:
{\tiny
\[\begin{array}{lllllllllll}
\D{4}{01}(\lambda,\alpha,\beta)&:& e_1 e_1 = \lambda e_3 + e_4, & e_1 e_3 = \alpha e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = \beta e_4, & e_3e_1 = e_4; \\
\D{4}{02}(\lambda,\alpha,\beta)&:& e_1 e_1 = \lambda e_3, & e_1 e_3 = \alpha e_4, & e_2 e_1=e_3 & e_2 e_2 = e_3, & e_2 e_3 = \beta e_4, & e_3e_1 = e_4; \\
\D{4}{03}(\lambda,\alpha)&:& e_1 e_1 = \lambda e_3, & e_1 e_2 = e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = \alpha e_4, & e_3e_1 = e_4; \\
\D{4}{04}(\lambda,\alpha)&:& e_1 e_1 = \lambda e_3, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = \alpha e_4, & e_3e_1 = e_4; \\
\D{4}{05}(\lambda,\alpha)&:& e_1 e_1 = \lambda e_3, & e_1 e_2 = \lambda e_4, & e_2 e_1=e_3 + \lambda\alpha e_4, & e_2 e_2 = e_3, & e_2 e_3 = \Theta e_4, & e_3e_1 = \lambda e_4; \\
\D{4}{06}(\lambda,\alpha)&:& e_1 e_1 = \lambda e_3, & e_1 e_2 = e_4, & e_2 e_1=e_3 + \alpha e_4, & e_2 e_2 = e_3, & e_2 e_3 = \Theta^{-1} e_4, & e_3e_1 = e_4; \\
\D{4}{07}(\lambda)&:& e_1 e_1 = \lambda e_3, & e_2 e_1=e_3 + \lambda e_4, & e_2 e_2 = e_3, & e_2 e_3 = \Theta e_4, & e_3e_1 = \lambda e_4; \\
\D{4}{08}(\lambda)&:& e_1 e_1 = \lambda e_3, & e_2 e_1=e_3 + e_4, & e_2 e_2 = e_3, & e_2 e_3 = \Theta^{-1} e_4, & e_3e_1 = e_4; \\
\D{4}{09}(\lambda,\alpha)&:& e_1 e_1 = \lambda e_3, & e_1 e_2 = e_4,& e_1 e_3 = \alpha e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_3e_1 = e_4; \\
\D{4}{10}(\lambda,\alpha)&:& e_1 e_1 = \lambda e_3,& e_1 e_3 = \alpha e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_3e_1 = e_4; \\
\D{4}{11}(\lambda,\alpha)&:& e_1 e_1 = \lambda e_3 + e_4,& e_1e_2 = \alpha e_4, & e_1 e_3 = -e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_3e_1 = e_4; \\
\D{4}{12}(\lambda,\alpha)&:& e_1 e_1 = \lambda e_3 + e_4,& e_1e_3 = \alpha e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_3e_1 = \Theta e_4, & e_3e_2 = e_4; \\
\D{4}{13}(\lambda,\alpha)&:& e_1 e_1 = \lambda e_3 + e_4,& e_1e_3 = \alpha e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_3e_1 = (1-\Theta)e_4, & e_3e_2 = e_4; \\
\D{4}{14}(\lambda,\alpha)&:& e_1 e_1 = \lambda e_3,& e_1e_3 = \alpha e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_3e_1 = \Theta e_4, & e_3e_2 = e_4; \\
\D{4}{15}(\lambda,\alpha)&:& e_1 e_1 = \lambda e_3,& e_1e_3 = \alpha e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_3e_1 = (1-\Theta)e_4, & e_3e_2 = e_4; \\
\D{4}{16}(\alpha)&:& e_1e_3 = \alpha e_4, & e_2 e_1=e_3 + e_4, & e_2 e_2 = e_3, & e_3e_1 = e_4, & e_3e_2 = e_4; \\
\D{4}{17}(\alpha)&:& e_1 e_1 = e_4, & e_1e_3 = -e_4, & e_2 e_1=e_3 + \alpha e_4, & e_2 e_2 = e_3, & e_3e_1 = e_4, & e_3e_2 = e_4; \\
\D{4}{18}(\lambda,\alpha)&:& e_1 e_1 = \lambda e_3 + e_4,& e_1e_3 = \Theta\alpha e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = \alpha e_4, & e_3e_1 = \Theta e_4,\\
&& e_3e_2 = e_4; \\
\D{4}{19}(\lambda,\alpha)&:& e_1 e_1 = \lambda e_3 + e_4,& e_1e_3 = (1-\Theta)\alpha e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = \alpha e_4, & e_3e_1 = (1-\Theta)e_4,
\\&& e_3e_2 = e_4; \\
\D{4}{20}(\lambda,\alpha)&:& e_1 e_1 = \lambda e_3,& e_1e_3 = \Theta\alpha e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = \alpha e_4, & e_3e_1 = \Theta e_4,
\\&& e_3e_2 = e_4; \\
\D{4}{21}(\lambda,\alpha)&:& e_1 e_1 = \lambda e_3,& e_1e_3 = (1-\Theta)\alpha e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = \alpha e_4, & e_3e_1 = (1-\Theta)e_4, \\&& e_3e_2 = e_4; \\
\D{4}{22}(\lambda)&:& e_1 e_1 = \lambda e_3 + (1-2\lambda)e_4,& e_1 e_2 = e_4,& e_1e_3 = (\Theta - 1) e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = (1-\Theta^{-1}) e_4,
\\&& e_3e_1 = \Theta e_4, & e_3e_2 = e_4; \\
\D{4}{23}(\lambda)&:& e_1 e_1 = \lambda e_3 + \lambda(1-2\lambda) e_4& e_1 e_2 = \lambda e_4& e_1e_3 = -\lambda\Theta e_4 & e_2 e_1=e_3 & e_2 e_2 = e_3 & e_2 e_3 = -\Theta^2 e_4,
\\&& e_3e_1 = \lambda(1-\Theta)e_4 & e_3e_2 = \lambda e_4; \\
\D{4}{24}(\lambda)&:& e_1 e_1 = \lambda e_3 + \Theta e_4,& e_1 e_2 = e_4,& e_1e_3 = (\Theta - 1) e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = (1-\Theta^{-1}) e_4,
\\&& e_3e_1 = \Theta e_4, & e_3e_2 = e_4; \\
\D{4}{25}(\lambda)&:& e_1 e_1 = \lambda e_3 + \lambda(1-\Theta)e_4,& e_1 e_2 = \lambda e_4,& e_1e_3 = -\lambda\Theta e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = -\Theta^2 e_4,
\\&& e_3e_1 = \lambda (1-\Theta)e_4, & e_3e_2 = \lambda e_4; \\
\D{4}{26}(\lambda)&:& e_1 e_1 = \lambda e_3 + \Theta e_4,& e_1 e_2 = e_4,& e_1e_3 = -\Theta e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = -e_4,
\\&& e_3e_1 = \Theta e_4, & e_3e_2 = e_4; \\
\D{4}{27}(\lambda)&:& e_1 e_1 = \lambda e_3 + (1-\Theta)e_4,& e_1 e_2 = e_4,& e_1e_3 = (\Theta-1) e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = -e_4,
\\&& e_3e_1 = (1-\Theta)e_4, & e_3e_2 = e_4; \\
\D{4}{28}(\lambda)&:& e_1 e_1 = \lambda e_3 + (1-\Theta)e_4,& e_1 e_2 = e_4,& e_1e_3 = -\Theta e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = -e_4,
\\&& e_3e_1 = \Theta e_4, & e_3e_2 = e_4; \\
\D{4}{29}(\lambda)&:& e_1 e_1 = \lambda e_3 + \Theta e_4,& e_1 e_2 = e_4,& e_1e_3 = (\Theta-1) e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = -e_4,
\\&& e_3e_1 = (1-\Theta)e_4, & e_3e_2 = e_4; \\
\D{4}{30}(\lambda)&:& e_1 e_1 = \lambda e_3 + e_4, & e_1e_3 = (\Theta-1) e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = -e_4, & e_3e_1 = \Theta e_4,
\\&& e_3e_2 = e_4; \\
\D{4}{31}(\lambda)&:& e_1 e_1 = \lambda e_3 + e_4, & e_1e_3 = -\Theta e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = -e_4, & e_3e_1 = (1-\Theta)e_4,
\\&& e_3e_2 = e_4; \\
\D{4}{32}(\lambda)&:& e_1 e_1 = \lambda e_3, & e_1e_3 = (\Theta-1) e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = -e_4, & e_3e_1 = \Theta e_4,
\\&& e_3e_2 = e_4; \\
\D{4}{33}(\lambda)&:& e_1 e_1 = \lambda e_3, & e_1e_3 = -\Theta e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3, & e_2 e_3 = -e_4, & e_3e_1 = (1-\Theta)e_4,
\\&& e_3e_2 = e_4; \\
\D{4}{34}&:& e_1 e_1 = e_4, & e_2 e_1=e_3 + e_4, & e_2 e_2 = e_3, & e_3 e_1 = e_4,& e_3 e_2 = e_4; \\
\D{4}{35}(\lambda)&:& e_1 e_1 = \lambda e_3, & e_1e_3 = e_4, & e_2 e_1=e_3 + e_4, & e_2 e_2 = e_3; \\
\D{4}{36}(\lambda)&:& e_1 e_1 = \lambda e_3, & e_1e_3 = e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3; \\
\D{4}{37}(\lambda)&:& e_1 e_1 = \lambda e_3 + e_4, & e_1e_3 = \Theta e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3,& e_2 e_3 = e_4; \\
\D{4}{38}(\lambda)&:& e_1 e_1 = \lambda e_3 + e_4, & e_1e_3 = (1-\Theta)e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3,& e_2 e_3 = e_4; \\
\D{4}{39}(\lambda)&:& e_1 e_1 = \lambda e_3, & e_1e_3 = \Theta e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3,& e_2 e_3 = e_4; \\
\D{4}{40}(\lambda)&:& e_1 e_1 = \lambda e_3, & e_1e_3 = (1-\Theta)e_4, & e_2 e_1=e_3, & e_2 e_2 = e_3,& e_2 e_3 = e_4.
\end{array}\]}
The algebras above are pairwise non-isomorphic, except for {\tiny \begin{gather*}
\D{4}{01}(\lambda,0,\beta) \cong \D{4}{02}(\lambda,0,\beta) \cong \D{4}{04}(\lambda,\beta),\quad \D{4}{01}(\lambda,\alpha,0)_{\alpha \neq -1} \cong \D{4}{02}(\lambda,\alpha,0) \cong \D{4}{10}(\lambda,\alpha),\quad \D{4}{01}(\lambda,-1,0) \cong \D{4}{11}(\lambda,0),\quad\\
\D{4}{03}(\lambda,0) \cong \D{4}{09}(\lambda,0),\quad \D{4}{03}\left(\lambda,(1-\Theta)^{-1}\right)_{\lambda \neq 0} \cong \D{4}{05}(\lambda,0)_{\lambda \neq 0}, \D{4}{03}\left(\lambda,\Theta^{-1}\right)\cong \D{4}{06}(\lambda,0),\quad \D{4}{04}(\lambda,0) \cong \D{4}{10}(\lambda,0),\quad\\
\D{4}{05}(1/4,\alpha) \cong \D{4}{06}(1/4,\alpha),\quad \D{4}{07}(1/4) \cong \D{4}{08}(1/4),\quad\\
\D{4}{05}(0,\alpha) \cong \D{4}{07}(0) \cong \D{4}{23}(0) \cong \D{4}{25}(0) \cong \D{4}{40}(0),\quad\\
\D{4}{12}(\lambda,0) \cong \D{4}{18}(\lambda,0),\quad \D{4}{12}(1/4,\alpha) \cong \D{4}{13}(1/4,\alpha),\quad \D{4}{12}(0,\alpha)_{\alpha \neq -1} \cong \D{4}{14}(0,\alpha),\quad \D{4}{12}(0,-1) \cong \D{4}{17}(0),\quad\\
\D{4}{13}(\lambda,0) \cong \D{4}{19}(\lambda,0),\quad \D{4}{14}(\lambda,0) \cong \D{4}{20}(\lambda,0),\quad \D{4}{14}(1/4,\alpha) \cong \D{4}{15}(1/4,\alpha),\quad \D{4}{15}(\lambda,0) \cong \D{4}{21}(\lambda,0),\quad\\
\D{4}{18}(1/4,\alpha) \cong \D{4}{19}(1/4,\alpha),\quad \D{4}{18}(0,0) \cong \D{4}{22}(0) \cong \D{4}{24}(0),\quad \D{4}{18}(1/4,-1) \cong \D{4}{19}(1/4,-1) \cong \D{4}{30}(1/4) \cong \D{4}{31}(1/4),\quad\\
\D{4}{20}(1/4,\alpha) \cong \D{4}{21}(1/4,\alpha),\quad \D{4}{20}(1/4,-1) \cong \D{4}{21}(1/4,-1) \cong \D{4}{32}(1/4) \cong \D{4}{33}(1/4),\quad\\
\D{4}{22}(1/4) \cong \D{4}{23}(1/4) \cong \D{4}{24}(1/4) \cong \D{4}{25}(1/4) \cong \D{4}{26}(1/4) \cong \D{4}{27}(1/4) \cong \D{4}{28}(1/4) \cong \D{4}{29}(1/4),\quad\\
\D{4}{37}(1/4) \cong \D{4}{38}(1/4),\quad \D{4}{39}(1/4) \cong \D{4}{40}(1/4).
\end{gather*}}
Moreover, the algebras $\D{4}{05}(0,\alpha) \cong \D{4}{07}(0) \cong \D{4}{23}(0) \cong \D{4}{25}(0) \cong \D{4}{40}(0), \D{4}{38}(0)$ are Leibniz.
\newpage
\subsubsection{$1$-dimensional central extensions of $\T {3*}{05}$}\label{ext-T_05^3*}
Let us use the following notations
\begin{align*}
\nb 1 = \Dl 13, \nb 2 = \Dl 21, \nb 3 = \Dl 22 - 3\Dl 31.
\end{align*}
Take $\theta=\sum_{i=1}^3\alpha_i\nb i\in {\rm H_T^2}(\T {3*}{05}).$
If
$$
\phi=
\begin{pmatrix}
x & 0 & 0\\
y & x^2 & 0\\
z & xy & x^3
\end{pmatrix}\in\aut{\T {3*}{05}},
$$
then
$$
\phi^T\begin{pmatrix}
0 & 0 & \alpha_1\\
\alpha_2 & \alpha_3 & 0\\
-3\alpha_3& 0 & 0
\end{pmatrix} \phi=
\begin{pmatrix}
\alpha^* & \alpha^{**} & \alpha^*_1\\
\alpha^*_2 & \alpha^*_3 & 0\\
-3\alpha^*_3 & 0 & 0
\end{pmatrix},
$$
where
\begin{align*}
\alpha^*_1 &= \alpha_1x^4,\\
\alpha^*_2 &= x^2(\alpha_2x - 2\alpha_3y),\\
\alpha^*_3 &= \alpha_3x^4.
\end{align*}
Hence, $\phi\langle\theta\rangle=\langle\theta^*\rangle,$ where $\theta^*=\sum\limits_{i=1}^3 \alpha_i^* \nb i.$
We are interested in $\theta$ with $(\alpha_2,\alpha_3) \neq (0,0).$ Moreover, the condition $\theta \in \mathbf{T}_1 (\T{3}{05})$ gives us $(\alpha_1, \alpha_3) \neq (0,0).$
Let $\theta$ be as above. Consider two mutually exclusive cases:
\begin{enumerate}
\item $\alpha_3 = 0.$ The conditions above imply that $\alpha_1 \neq 0, \alpha_2 \neq 0.$
Then choosing $x=\frac{\alpha_2}{\alpha_1},$ we have the representative $\langle \nabla_1+\nabla_2 \rangle.$
\item $\alpha_3 \neq 0.$ Choosing
$x = \frac{1}{\sqrt[4]{\alpha_3}}$ and $y=\frac{\alpha_2}{ 2 \sqrt[4]{\alpha_3^5}}$ we have the family of representatives $\langle\alpha\nabla_1 + \nabla_3 \rangle.$
\end{enumerate}
Summarizing, we have the following distinct orbits:
\[ \langle \nabla_1+\nabla_2 \rangle, \ \langle\alpha\nabla_1 + \nabla_3 \rangle.\]
They correspond to the following algebras:
\[\begin{array}{lllllllllll}
\T{4}{37}&:& e_1e_1 = e_2,& e_1e_2 = e_3,& e_1e_3 = e_4,& e_2e_1 = e_4; \\
\T{4}{38}(\alpha)&:& e_1e_1 = e_2,& e_1e_2 = e_3,& e_1e_3 = \alpha e_4,& e_2e_2 = e_4,& e_3e_1 = -3e_4.
\end{array}\]
\subsubsection{$1$-dimensional central extensions of $\T {3}{01}(\lambda)$}
\begin{enumerate}
\item $\lambda \neq 0.$
Let us use the following notations
\begin{align*}
\nb 1 = \Dl 12, \nb 2 = (\lambda-1)\Dl 13 + 3 \Dl 31, \nb 3 = \color{black}{\Dl 13 + \Dl 22}.
\end{align*}
Take $\theta=\sum_{i=1}^3\alpha_i\nb i\in {\rm H_T^2}(\T {3}{01}).$
If
$$
\phi=
\begin{pmatrix}
x & 0 & 0\\
y & x^2 & 0\\
z & (\lambda+1)xy & x^3
\end{pmatrix}
\in\aut{\T {3}{01}},
$$
then
$$
\phi^T\begin{pmatrix}
0 & \alpha_1 & (\lambda-1)\alpha_2\color{black}{+\alpha_3}\\
0 & \color{black}{\alpha_3} & 0\\
\color{black}{3\alpha_2}& 0 & 0
\end{pmatrix} \phi=
\begin{pmatrix}
\alpha^* & \alpha_1^* + \lambda\alpha^{**} & (\lambda-1)\alpha_2^*\color{black}{+\alpha_3^*}\\
\alpha^{**} & \color{black}{\alpha_3^*} & 0\\
\color{black}{3\alpha_2^*}& 0 & 0
\end{pmatrix},
$$
where
\begin{align*}
\alpha^*_1 &= x^2(\alpha_1x+({\color{black} 2}\alpha_3 - (\lambda + 1)(2\lambda + 1)\alpha_2)y),\\
\alpha^*_2 &= \alpha_2x^4,\\
\alpha^*_3 &= \alpha_3x^4.
\end{align*}
Hence, $\phi\langle\theta\rangle=\langle\theta^*\rangle,$ where $\theta^*=\sum\limits_{i=1}^3 \alpha_i^* \nb i.$
The condition $\theta \in \mathbf{T}_1 (\T{3}{01}(\lambda))$ gives us ${\color{black}(\alpha_2, \alpha_3)} \neq (0,0).$
\begin{enumerate}
\item $(\lambda + 1)(2\lambda + 1)\ne 0$.
\begin{enumerate}
\item $\alpha_3\ne 0$.
\begin{enumerate}
\item $\alpha_2 \ne \frac{\color{black} 2}{(\lambda + 1)(2\lambda + 1)}\alpha_3$. Then choosing $x=\frac 1{\sqrt[4]{\alpha_3}}$ and $y=-\frac{\alpha_1x}{{\color{black} 2}\alpha_3 - (\lambda + 1)(2\lambda + 1)\alpha_2}$, we obtain the family of representatives of distinct orbits
$\langle\alpha\nb 2+\nb 3\rangle$, where $\alpha\ne \frac{\color{black} 2}{(\lambda + 1)(2\lambda + 1)}$.
\item $\alpha_2=\frac{\color{black} 2}{(\lambda + 1)(2\lambda + 1)}\alpha_3$. Then we have two representatives
$\Big\langle\frac{\color{black} 2}{(\lambda + 1)(2\lambda + 1)}\nb 2+\nb 3\Big\rangle$ and $\Big\langle\nb 1+\frac{\color{black} 2}{(\lambda + 1)(2\lambda + 1)}\nb 2+\nb 3\Big\rangle$ depending on whether $\alpha_1=0$ or not. For convenience, we shall join the representative $\Big\langle\frac{\color{black} 2}{(\lambda + 1)(2\lambda + 1)}\nb 2+\nb 3\Big\rangle$ with the family $\langle\alpha\nb 2+\nb 3\rangle_{\alpha\ne \frac{\color{black} 2}{(\lambda + 1)(2\lambda + 1)}}$ found above.
\end{enumerate}
\item $\alpha_3=0$. Then $\alpha_2\ne 0$. Choosing $y=\frac{\alpha_1x}{(\lambda + 1)(2\lambda + 1)\alpha_2}$, we obtain the representative $\langle\nb 2\rangle$.
\end{enumerate}
\item $(\lambda + 1)(2\lambda + 1)=0$.
\begin{enumerate}
\item $\alpha_3\ne 0$. Then choosing $x=\frac 1{\sqrt[4]{\alpha_3}}$ and $y=-\frac{\alpha_1x}{{\color{black} 2}\alpha_3}$, we obtain the family of representatives of distinct orbits $\langle\alpha\nb 2+\nb 3\rangle$.
\item $\alpha_3=0$. Then $\alpha_2\ne 0$. So, we obtain two representatives $\langle\nb 2\rangle$ and $\langle\nb 1+\nb 2\rangle$ depending on whether $\alpha_1=0$ or not.
\end{enumerate}
\end{enumerate}
Summarizing, in the case (1) we have the following distinct orbits:
$$
\langle\nb 1+\nb 2\rangle_{\lambda\in\{-1,-\frac 12\}},
\left\langle\nb 1+\frac{\color{black} 2}{(\lambda + 1)(2\lambda + 1)}\nb 2+\nb 3\right\rangle_{\lambda\not\in\{-1,-\frac 12,0\}},
\langle\nb 2\rangle_{\lambda\ne 0},
\langle\alpha\nb 2+\nb 3\rangle_{\lambda\ne 0}.
$$
\item $\lambda=0.$
Let us use the following notations
\begin{align*}
\nb 1 = \Dl 12, \nb 2 = -\Dl 13 + 3 \Dl 31, \nb 3 = {\color{black}\Dl 13 + \Dl 22}, \nb 4 = \Dl 23.
\end{align*}
Take $\theta=\sum_{i=1}^4\alpha_i\nb i\in {\rm H_T^2}(\T {3}{01}(0)).$
If
$$
\phi=
\begin{pmatrix}
x & 0 & 0\\
y & x^2 & 0\\
z & xy & x^3
\end{pmatrix}
\in\aut{\T {3}{01}(0)},
$$
then
$$
\phi^T\begin{pmatrix}
0 & \alpha_1 & -\alpha_2{\color{black}+\alpha_3}\\
0 & \color{black}{\alpha_3} & \alpha_4\\
{\color{black} 3\alpha_2}& 0 & 0
\end{pmatrix} \phi=
\begin{pmatrix}
\alpha^* & \alpha^*_1 & -\alpha^*_2{\color{black}+\alpha_3^*}\\
\alpha^{**} & {\color{black}\alpha^*_3} & \alpha^*_4\\
{\color{black} 3\alpha^*_2}& 0 & 0
\end{pmatrix},
$$
where
\begin{align*}
\alpha^*_1 &= x (x^2 \alpha_1 - x y (\alpha_2 {\color{black}-2}\alpha_3) + y^2 \alpha_4),\\
\alpha^*_2 &= {\color{black} \alpha_2 x^4},\\
\alpha^*_3 &= x^3 (x \alpha_3 {\color{black}+} y \alpha_4),\\
\alpha^*_4 &= \alpha_4x^5.
\end{align*}
Hence, $\phi\langle\theta\rangle=\langle\theta^*\rangle,$ where $\theta^*=\sum\limits_{i=1}^4 \alpha_i^* \nb i.$
The condition $\theta \in \mathbf{T}_1 (\T{3}{01}(0))$ gives us $(\alpha_2, {\color{black}\alpha_3},\alpha_4) \neq (0,0,0).$
\begin{enumerate}
\item $\alpha_4\neq0$. Then choosing $y={\color{black}-}\frac{\alpha_3x}{\alpha_4}$ we have the family of representatives
$\langle \alpha_1^{\star} \nabla_1 + {\color{black}\alpha_2^*} \nabla_2+ \alpha^*_4\nabla_4 \rangle$,
where
\begin{align*}
\alpha^\star_1=& \frac{x^3}{\alpha_4}(\alpha_1\alpha_4{\color{black}+}\alpha_2\alpha_3{\color{black}-\alpha_3^2}).
\end{align*}
\begin{enumerate}
\item ${\color{black}\alpha_2}\ne 0$. Then we have the family of representatives of distinct orbits $\langle \alpha \nabla_1+ \nabla_2+ \nabla_4 \rangle$.
\item ${\color{black}\alpha_2}=0$. Then we have two representatives $\langle \nabla_4 \rangle$ and $\langle \nabla_1+ \nabla_4 \rangle$ depending on whether $\alpha_1\alpha_4{\color{black}+}\alpha_2\alpha_3{\color{black}-\alpha_3^2}=0$ or not.
\end{enumerate}
\item $\alpha_4=0.$
\begin{enumerate}
\item $\alpha_2 {\color{black}-2}\alpha_3\ne 0$. Then choosing $y = \frac{\alpha_1x}{\alpha_2 {\color{black}-2}\alpha_3}$ we have the representative $\langle\nb 2\rangle$ and the family of representatives of distinct orbits $\langle\alpha\nb 2+\nb 3\rangle_{\alpha\ne{\color{black} 2}}$ depending on whether $\alpha_3=0$ or not.
\item $\alpha_2 {\color{black}-2}\alpha_3=0$. Then we have two representatives $\langle{\color{black} 2}\nb 2+\nb 3\rangle$ and $\langle\nb 1{\color{black} +2}\nb 2+\nb 3\rangle$ depending on whether $\alpha_1=0$ or not. The representative $\langle{\color{black} 2}\nb 2+\nb 3\rangle$ will be joined with the family $\langle\alpha\nb 2+\nb 3\rangle_{\alpha\ne{\color{black} 2}}$.
\end{enumerate}
\end{enumerate}
Summarizing, in the case (2) we have the following distinct orbits:
$$
\langle\nb 1{\color{black} +2}\nb 2+\nb 3\rangle_{\lambda=0}, \langle \alpha \nabla_1+ \nabla_2+ \nabla_4 \rangle_{\lambda=0}, \langle \nabla_1+ \nabla_4 \rangle_{\lambda=0}, \langle\nb 2\rangle_{\lambda=0}, \langle\alpha\nb 2+\nb 3\rangle_{\lambda=0}, \langle \nabla_4 \rangle_{\lambda=0}.
$$
\end{enumerate}
Now, taking into account the both cases (1) and (2), we have the following distinct orbits:
$
\begin{array}{l}
\langle\nb 1+\nb 2\rangle_{\lambda\in\{-1,-\frac 12\}},\\
\left\langle\nb 1+\frac{\color{black} 2}{(\lambda + 1)(2\lambda + 1)}\nb 2+\nb 3\right\rangle_{\lambda\not\in\{-1,-\frac 12\}},
\end{array}
\begin{array}{l}
\langle \alpha \nabla_1+ \nabla_2+ \nabla_4 \rangle_{\lambda=0},\\
\langle \nabla_1+ \nabla_4 \rangle_{\lambda=0},\\
\end{array}
\begin{array}{l}
\langle\nb 2\rangle,\\
\langle\alpha\nb 2+\nb 3\rangle,
\end{array}
\begin{array}{l}
\langle \nabla_4 \rangle_{\lambda=0}.
\end{array}
$
Observe that $\left\langle\nb 1+\frac{\color{black} 2}{(\lambda + 1)(2\lambda + 1)}\nb 2+\nb 3\right\rangle=\langle(\lambda + 1)(2\lambda + 1)\nb 1+{\color{black} 2}\nb 2+(\lambda + 1)(2\lambda + 1)\nb 3\rangle$, the latter being $\langle\nb 2\rangle$, when $\lambda\in\{-1,-\frac 12\}$. Now, if $\lambda\not\in\{-1,-\frac 12\}$, then $\mathrm{Orb}\langle\nb 2\rangle=\mathrm{Orb}\langle\nb 1+\nb 2\rangle$. Thus, we may reorganize our orbits as follows:
$
\begin{array}{l}
\langle\nb 1+\nb 2\rangle,\\
\langle(\lambda + 1)(2\lambda + 1)\nb 1+{\color{black} 2}\nb 2+(\lambda + 1)(2\lambda + 1)\nb 3\rangle,
\end{array}
\begin{array}{l}
\langle \alpha \nabla_1+ \nabla_2+ \nabla_4 \rangle_{\lambda=0},\\
\langle \nabla_1+ \nabla_4 \rangle_{\lambda=0},
\end{array}
\begin{array}{l}
\langle\alpha\nb 2+\nb 3\rangle,\\
\langle\nb 4 \rangle_{\lambda=0}.
\end{array}
$
The corresponding algebras are:
$$
\begin{array}{lllllllll}
\T 4{39}(\lambda) &:& e_1 e_1 = e_2, & e_1 e_2=\lambda e_3+e_4, & e_1e_3 = (\lambda-1)e_4,\\
&& e_2 e_1=e_3, & e_3e_1 = 3e_4;\\
\T 4{40}(\lambda) &:& e_1 e_1 = e_2, & e_1 e_2=\lambda e_3+(\lambda + 1)(2\lambda + 1)e_4, & e_1e_3 = {\color{black}(2\lambda^2 + 5\lambda - 1)}e_4,\\
&& e_2 e_1=e_3, & e_2e_2={\color{black}(\lambda + 1)(2\lambda + 1)}e_4, & e_3e_1 = {\color{black} 6}e_4;\\
\T 4{41}(\alpha) &:& e_1 e_1 = e_2, & e_1 e_2=\alpha e_4, & e_1e_3=-e_4,\\
&& e_2 e_1=e_3, & e_2e_3 = e_4, & e_3e_1=3e_4;\\
\T 4{42} &:& e_1 e_1 = e_2, & e_1 e_2=e_4, & e_2 e_1=e_3, & e_2e_3 = e_4;\\
\T 4{43}(\lambda,\alpha) &:& e_1 e_1 = e_2, & e_1 e_2=\lambda e_3, & e_1e_3 = (\alpha(\lambda-1){\color{black}+1})e_4,\\
&& e_2 e_1=e_3, & e_2e_2={\color{black} e_4}, & e_3e_1 = 3{\color{black}\alpha} e_4;\\
\T 4{44} &:& e_1 e_1 = e_2, & e_2 e_1=e_3, & e_2e_3 = e_4.
\end{array}
$$
\section{The geometric classification of nilpotent terminal algebras}
\subsection{Degenerations of algebras}
Given an $n$-dimensional vector space ${\bf V}$, the set ${\rm Hom}({\bf V} \otimes {\bf V},{\bf V}) \cong {\bf V}^* \otimes {\bf V}^* \otimes {\bf V}$
is a vector space of dimension $n^3$. This space inherits the structure of the affine variety $\mathbb{C}^{n^3}.$
Indeed, let us fix a basis $e_1,\dots,e_n$ of ${\bf V}$. Then any $\mu\in {\rm Hom}({\bf V} \otimes {\bf V},{\bf V})$ is determined by $n^3$ structure constants $c_{i,j}^k\in\mathbb{C}$ such that
$\mu(e_i\otimes e_j)=\sum_{k=1}^nc_{i,j}^ke_k$. A subset of ${\rm Hom}({\bf V} \otimes {\bf V},{\bf V})$ is {\it Zariski-closed} if it can be defined by a set of polynomial equations in the variables $c_{i,j}^k$ ($1\le i,j,k\le n$).
The general linear group ${\rm GL}({\bf V})$ acts by conjugation on the variety ${\rm Hom}({\bf V} \otimes {\bf V},{\bf V})$ of all algebra structures on ${\bf V}$:
$$ (g * \mu )(x\otimes y) = g\mu(g^{-1}x\otimes g^{-1}y),$$
for $x,y\in {\bf V}$, $\mu\in {\rm Hom}({\bf V} \otimes {\bf V},{\bf V})$ and $g\in {\rm GL}({\bf V})$. Clearly, the ${\rm GL}({\bf V})$-orbits correspond to the isomorphism classes of algebras structures on ${\bf V}$. Let $T$ be a set of polynomial identities which is invariant under isomorphism. Then the subset $\mathbb{L}(T)\subset {\rm Hom}({\bf V} \otimes {\bf V},{\bf V})$ of the algebra structures on ${\bf V}$ which satisfy the identities in $T$ is ${\rm GL}({\bf V})$-invariant and Zariski-closed. It follows that $\mathbb{L}(T)$ decomposes into ${\rm GL}({\bf V})$-orbits. The ${\rm GL}({\bf V})$-orbit of $\mu\in\mathbb{L}(T)$ is denoted by $O(\mu)$ and its Zariski closure by $\overline{O(\mu)}$.
Let ${\bf A}$ and ${\bf B}$ be two $n$-dimensional algebras satisfying the identities from $T$ and $\mu,\lambda \in \mathbb{L}(T)$ represent ${\bf A}$ and ${\bf B}$ respectively.
We say that ${\bf A}$ {\it degenerates} to ${\bf B}$ and write ${\bf A}\to {\bf B}$ if $\lambda\in\overline{O(\mu)}$.
Note that in this case we have $\overline{O(\lambda)}\subset\overline{O(\mu)}$. Hence, the definition of a degeneration does not depend on the choice of $\mu$ and $\lambda$. It is easy to see that any algebra degenerates to the algebra with zero multiplication. If ${\bf A}\to {\bf B}$ and ${\bf A}\not\cong {\bf B}$, then ${\bf A}\to {\bf B}$ is called a {\it proper degeneration}. We write ${\bf A}\not\to {\bf B}$ if $\lambda\not\in\overline{O(\mu)}$ and call this a {\it non-degeneration}. Observe that the dimension of the subvariety $\overline{O(\mu)}$ equals $n^2-\dim\mathfrak{Der}({\bf A})$. Thus if ${\bf A}\to {\bf B}$ is a proper degeneration, then we must have $\dim\mathfrak{Der}({\bf A})>\dim\mathfrak{Der}({\bf B})$.
Let ${\bf A}$ be represented by $\mu\in\mathbb{L}(T)$. Then ${\bf A}$ is {\it rigid} in $\mathbb{L}(T)$ if $O(\mu)$ is an open subset of $\mathbb{L}(T)$.
Recall that a subset of a variety is called {\it irreducible} if it cannot be represented as a union of two non-trivial closed subsets. A maximal irreducible closed subset of a variety is called an {\it irreducible component}.
It is well known that any affine variety can be represented as a finite union of its irreducible components in a unique way.
The algebra ${\bf A}$ is rigid in $\mathbb{L}(T)$ if and only if $\overline{O(\mu)}$ is an irreducible component of $\mathbb{L}(T)$.
In the present work we use the methods applied to Lie algebras in \cite{GRH,GRH2}.
To prove
degenerations, we will construct families of matrices parametrized by $t$. Namely, let ${\bf A}$ and ${\bf B}$ be two algebras represented by the structures $\mu$ and $\lambda$ from $\mathbb{L}(T)$, respectively. Let $e_1,\dots, e_n$ be a basis of ${\bf V}$ and $c_{i,j}^k$ ($1\le i,j,k\le n$) be the structure constants of $\lambda$ in this basis. If there exist $a_i^j(t)\in\mathbb{C}$ ($1\le i,j\le n$, $t\in\mathbb{C}^*$) such that the elements $E_i^t=\sum_{j=1}^na_i^j(t)e_j$ ($1\le i\le n$) form a basis of ${\bf V}$ for any $t\in\mathbb{C}^*$, and the structure constants $c_{i,j}^k(t)$ of $\mu$ in the basis $E_1^t,\dots, E_n^t$ satisfy $\lim\limits_{t\to 0}c_{i,j}^k(t)=c_{i,j}^k$, then ${\bf A}\to {\bf B}$. In this case $E_1^t,\dots, E_n^t$ is called a {\it parametric basis} for ${\bf A}\to {\bf B}$.
To prove a non-degeneration ${\bf A}\not\to {\bf B}$ we will use the following lemma (see \cite{GRH}).
\begin{lemma}\label{main}
Let $\mathcal{B}$ be a Borel subgroup of ${\rm GL}({\bf V})$ and $\mathcal{R}\subset \mathbb{L}(T)$ be a $\mathcal{B}$-stable closed subset.
If ${\bf A} \to {\bf B}$ and ${\bf A}$ can be represented by $\mu\in\mathcal{R}$ then there is $\lambda\in \mathcal{R}$ that represents ${\bf B}$.
\end{lemma}
In particular, it follows from Lemma \ref{main} that ${\bf A}\not\to {\bf B}$, whenever $\dim({\bf A}^2)<\dim({\bf B}^2)$.
When the number of orbits under the action of ${\rm GL}({\bf V})$ on $\mathbb{L}(T)$ is finite, the graph of primary degenerations gives the whole picture. In particular, the description of rigid algebras and irreducible components can be easily obtained. Since the variety of $4$-dimensional nilpotent terminal algebras contains infinitely many non-isomorphic algebras, we have to fulfill some additional work. Let ${\bf A}(*):=\{{\bf A}(\alpha)\}_{\alpha\in I}$ be a family of algebras and ${\bf B}$ be another algebra. Suppose that, for $\alpha\in I$, ${\bf A}(\alpha)$ is represented by a structure $\mu(\alpha)\in\mathbb{L}(T)$ and ${\bf B}$ is represented by a structure $\lambda\in\mathbb{L}(T)$. Then by ${\bf A}(*)\to {\bf B}$ we mean $\lambda\in\overline{\cup\{O(\mu(\alpha))\}_{\alpha\in I}}$, and by ${\bf A}(*)\not\to {\bf B}$ we mean $\lambda\not\in\overline{\cup\{O(\mu(\alpha))\}_{\alpha\in I}}$.
Let ${\bf A}(*)$, ${\bf B}$, $\mu(\alpha)$ ($\alpha\in I$) and $\lambda$ be as above. To prove ${\bf A}(*)\to {\bf B}$ it is enough to construct a family of pairs $(f(t), g(t))$ parametrized by $t\in\mathbb{C}^*$, where $f(t)\in I$ and $g(t)=\left(a_i^j(t)\right)_{i,j}\in {\rm GL}({\bf V})$. Namely, let $e_1,\dots, e_n$ be a basis of ${\bf V}$ and $c_{i,j}^k$ ($1\le i,j,k\le n$) be the structure constants of $\lambda$ in this basis. If we construct $a_i^j:\mathbb{C}^*\to \mathbb{C}$ ($1\le i,j\le n$) and $f: \mathbb{C}^* \to I$ such that $E_i^t=\sum_{j=1}^na_i^j(t)e_j$ ($1\le i\le n$) form a basis of ${\bf V}$ for any $t\in\mathbb{C}^*$, and the structure constants $c_{i,j}^k(t)$ of $\mu\big(f(t)\big)$ in the basis $E_1^t,\dots, E_n^t$ satisfy $\lim\limits_{t\to 0}c_{i,j}^k(t)=c_{i,j}^k$, then ${\bf A}(*)\to {\bf B}$. In this case, $E_1^t,\dots, E_n^t$ and $f(t)$ are called a {\it parametric basis} and a {\it parametric index} for ${\bf A}(*)\to {\bf B}$, respectively. In the construction of degenerations of this sort, we will write $\mu\big(f(t)\big)\to \lambda$, emphasizing that we are proving the assertion $\mu(*)\to\lambda$ using the parametric index $f(t)$.
Through a series of degenerations summarized in the table below by the corresponding parametric bases and indices, we obtain the main result of the second part of the paper.
\begin{theorem}\label{main-geo}
The variety of $4$-dimensional complex nilpotent terminal algebras has 3 irreducible components: one of dimension $17$ determined by the family of algebras $\D 4{01}(\lambda,\alpha,\beta)$ and two of dimension 15 determined by the families of algebras $\T 4{41}(\alpha)$ and $\T 4{43}(\lambda,\alpha)$.
\end{theorem}
\begin{proof}[{\bf Proof}]
Thanks to \cite{kppv} the algebras
$\mathfrak{L}_2,$ $\mathfrak{L}_5,$ $\mathfrak{L}_{11}$ and $\mathfrak{N}_3(\alpha)$
define irreducible components in the variety of $4$-dimensional nilpotent Leibniz algebras.
Note that in \cite{kppv} right Leibniz algebras were considered, and here we use their opposite versions which are left Leibniz algebras (and hence are terminal):
$$
\begin{array}{lllllllll}
\mathfrak{N}_3(\alpha) & :& e_1e_1 = e_4 & e_1e_2 = -\alpha e_4 & e_2e_1 = \alpha e_4 & e_2e_2 = e_4& e_3e_3 = e_4 \\
\mathfrak{L}_2 &: & e_1e_1 = e_2 & e_1e_2 = e_3 & e_1e_3 = e_4 \\
\mathfrak{L}_5 &:& e_1e_1 = e_3& e_1e_2 = e_3& e_2e_2 = e_4& e_1e_3=e_4 \\
\mathfrak{L}_{11} &:& e_1e_1 = e_4& e_1e_2 = e_3 & e_1e_3=e_4 & e_2e_1=-e_3 & e_2e_2=e_4 & e_3e_1=-e_4. \\
\end{array}
$$
The list of all $4$-dimensional nilpotent non-Leibniz terminal algebras was found in Theorem~\ref{main-alg}.
All these algebras degenerate from one of the families: $\D 4{01}(\lambda,\alpha,\beta)$, $\T 4{41}(\alpha)$ or $\T 4{43}(\lambda,\alpha)$, as it is shown in the table below. By a straightforward computation we have $\dim\mathfrak{Der}(\D 4{01}(\lambda,\alpha,\beta))=2$ for $\alpha,\beta\ne 0$, and $\dim\mathfrak{Der}(\D 4{01}(\lambda,\alpha,\beta))>2$ otherwise. Hence, the closure of the orbit of $\D 4{01}(\lambda,\alpha,\beta)$ has dimension $4^2-2+3=17$. A similar argument yields that the dimensions of the orbit closures of $\T 4{41}(\alpha)$ and $\T 4{43}(\lambda,\alpha)$ are both equal to $15<17$. In particular, this shows that $\D 4{01}(\lambda,\alpha,\beta)$ cannot degenerate from $\T 4{41}(\alpha)$ or $\T 4{43}(\lambda,\alpha)$. Moreover, $\T 4{41}(\alpha)$ and $\T 4{43}(\lambda,\alpha)$\ do not degenerate from $\D 4{01}(\lambda,\alpha,\beta)$, since the squares of $\T 4{41}(\alpha)$ and $\T 4{43}(\lambda,\alpha)$ have dimension 3, while the square of $\D 4{01}(\lambda,\alpha,\beta)$ has dimension $2<3$. This completes the proof of the theorem.
\end{proof}
{\tiny
\begin{longtable}{|lcl|ll|}
\multicolumn{5}{c}{ {\bf Table.} {\it Degenerations of $4$-dimensional nilpotent terminal algebras.}}\\
\hline
$\T {4}{03}\left(\frac{1}{1 - t}\right)$ &$\to$& $\mathfrak{N}_3(\alpha)$&
$E^t_1= \frac{2 \alpha t}{\sqrt{t-1}} e_1 - \alpha t \sqrt{t-1} e_2 + \alpha t \sqrt{t-1} e_3$ &
$E^t_3= t e_3 + \frac{t}{4} \left(1 + \frac{1 - t}{\alpha^2} - t\right) e_4$\\
&&&
$E^t_2= -t \sqrt{t-1} e_2 + t \sqrt{t-1} e_3$ &
$E^t_4= t^2 e_4$ \\
\hline
$\T {4}{38}\left(t^{-1}\right)$ &$\to$& $\mathfrak{L}_2$&
$E^t_1= e_1$ &
$E^t_3= e_3$ \\
&&&
$E^t_2= e_2$&
$E^t_4= t^{-1} e_4$\\
\hline
$\D {4}{01}\left( 2t^{-1}, t^{-1}, t \right)$ &$\to$& $\mathfrak{L}_5$&
$E^t_1= 2 e_1 - 2 e_2 - 2 e_3$ &
$E^t_3= 4 t e_3 - 4t^{-1} e_4$\\
&&&
$E^t_2= -2 t e_2 - 2 e_3$ &
$E^t_4= 8 e_4$\\
\hline
$\D {4}{01}\left( -\frac 12, 1, \frac{1 + t^2}{4}\right)$ &$\to$& $\mathfrak{L}_{11}$&
$E^t_1= 4t^{-1} e_1 - 2t^{-1} e_2$ &
$E^t_3= -8t^{-1} e_3 + 32t^{-3} e_4$\\
&&&
$E^t_2= 4 e_2 - 8t^{-2} e_3$ &
$E^t_4= 32t^{-2} e_4$\\
\hline
\color{black} $\T {4}{43} \left(0,0\right)$ &\color{black} $\to$& \color{black} $\T {4}{01}$ &
\color{black} $E^t_1= te_1$ &
\color{black} $E^t_3= e_3$ \\
&&&
\color{black} $E^t_2= t^2e_2$&
\color{black} $E^t_4= e_4$\\
\hline
\color{black}$\T {4}{43} \left(0, 0\right)$ &\color{black}$\to$& \color{black}$\T {4}{02}$ &
\color{black} $E^t_1= te_1 + \frac 12 t^{-2}e_2$ &
\color{black} $E^t_3= t^3e_3 + \frac 12 e_4$ \\
&&&
\color{black} $E^t_2= t^2e_2 + \frac 12 t^{-1}e_3 + \frac 14 t^{-4}e_4$ &
\color{black} $E^t_4= e_4$\\
\hline
$\D {4}{01} \left(0, t + \alpha^{-1},t\right)$ &$\to$& $\T {4}{03} (\alpha)$ &
$E^t_1= \alpha^2 t e_2$ &
$E^t_3= \alpha^3 t^2 (e_1 - e_2)$\\
&&&
$E^t_2= \alpha^4 t^2 e_3$ &
$E^t_4= \alpha^6 t^4 e_4$
\\
\hline
$\T {4}{03} \left(\alpha \right)$ &$\to$& $\T {4}{04} (\alpha)$ &
$E^t_1= t^{-1} e_1$ &
$E^t_3= t^{-1} e_3$\\
&&&
$E^t_2= t^{-2} e_2$ &
$E^t_4= t^{-3} e_4$
\\
\hline
$\T {4}{07} \left(\frac{1}{t-1} \right)$ &$\to$& $\T {4}{05}$ &
$E^t_1= \frac{(t-1)^2}{1 - 3 t + t^2} e_1 + \frac{(t-1)^3}{(1 - 3 t + t^2)^2} e_2+ t e_3$ &
$E^t_2= \frac{(t-1)^4}{ (1 - 3 t + t^2)^2} e_2 + \frac{(t-1)^4 t}{ (1 - 3 t + t^2)^2} e_4$ \\
&&&
$E^t_3= - \frac{(t-1)^4}{ t (1 - 3 t + t^2)^2} e_2+e_3$ &
$E^t_4= -\frac{(t-1)^3}{ (1 - 3 t + t^2)^2} e_4$
\\
\hline
$\T {4}{03} \left(t \right)$ &$\to$& $\T {4}{06}$ &
$E^t_1= \frac{t - 1}{1 + t^2} e_1 + \frac{t^2 - 1}{(1 + t^2)^2} e_2 + \frac{1 - t}{1 + t^2} e_3$ &
$E^t_3= \frac{(1 - t) t}{1 + t^2} e_3$ \\
&&&
$E^t_2= \frac{(t - 1)^2}{(1 + t^2)^2} e_2 + \frac{(1 - t)^3}{(1 + t^2)^3} e_4$ &
$E^t_4= \frac{(1 - t)^3 t}{(1 + t^2)^3} e_4$\\
\hline
$\T {4}{03} \left( \alpha \right)$ &$\to$& $\T {4}{07} (\alpha)$ &
$E^t_1= -\frac{1}{ 1 +\alpha} e_1 - \frac{1}{ (1 +\alpha)^2} e_2 + \frac{1}{ 1 +\alpha} e_3$ &
$E^t_2= \frac{1}{(1 +\alpha)^2} e_2 + \frac{1}{ (1 +\alpha)^3} e_4$\\
&&&
$E^t_3= t^{-1} (e_2-(1 +\alpha) e_3)$ &
$E^t_4= -\frac{1}{ t(1 +\alpha)} e_4$ \\
\hline
$\T {4}{03} \left(-1 \right)$ &$\to$& $\T {4}{08}$ &
$E^t_1= t^{-1} e_1$ & $E^t_3= t^{-2} e_3$ \\
&&&
$E^t_2= t^{-2} e_2$ &
$E^t_4= t^{-4} e_4$\\
\hline
$\T {4}{03} \left(t+t^4 \right)$ &$\to$& $\T {4}{09}$ &
$E^t_1= \frac{t-1}{t^2 + t^5} e_1 + \frac{t-1}{t^3 (1 + t^3)^2}e_2 + \frac{1 - t}{t^2 + t^5} e_3$ &
$E^t_3= -\frac{(t-1 ) (-1 + t - t^2 - t^3 + 2 t^4 + t^7)}{ t (1 + t^3)^2 (-1 + t + t^4)} e_3$ \\
&&&
$E^t_2= \frac{(t-1)^2}{t^4 (1 + t^3)^2} e_2 + \frac{t-1}{(1 + t^3)^2} e_3$&
$E^t_4= -\frac{(t-1)^2 (-1 + t - t^2 - t^3 + 2 t^4 + t^7)}{ t^5 (1 + t^3)^4} e_4$\\
\hline
$\T {4}{03} (\alpha )$ &$\to$& $\T {4}{10}(\alpha )$ &
$E^t_1= t^{-2} e_1$ & $E^t_3= t^{-3} e_3$\\
&&&
$E^t_2= t^{-4} e_2$ & $E^t_4= t^{-7} e_4$\\
\hline
$\T {4}{03} (1+t^{-2})$ &$\to$& $\T {4}{11}$ &
$E^t_1= t^3 e_1$ & $E^t_3= t^5 e_3$ \\
&&&
$E^t_2= t^6 e_2 - t^6 e_3$ &
$E^t_4= t^9 e_4$\\
\hline
$\T {4}{41} (t^{-1})$ &$\to$& $\T {4}{12}$ &
$E^t_1= 3 e_1 + 3 e_2$ &
$E^t_3= 3t^{-1} e_3 + 9t^{-1} e_4$ \\
&&&
$E^t_2= 9 e_2 + 9 e_3 + 9t^{-1} e_4$ &
$E^t_4= 27t^{-1} e_4$\\
\hline
$\T {4}{12}$ &$\to$& $\T {4}{13}$ &
$E^t_1= e_1$ & $E^t_3= t^{-1} e_3$ \\
&&& $E^t_2= e_2$ & $E^t_4= t^{-1} e_4$\\
\hline
$\T {4}{03} (1/t )$ &$\to$& $\T {4}{14}$ &
$E^t_1= t^{-1} e_1$ & $E^t_3= t^{-2} e_3$ \\
&&& $E^t_2= t^{-2} e_2$ & $E^t_4= t^{-5} e_4$\\
\hline
$\T {4}{03} (\frac{1 - t}{\alpha} )$ &$\to$& $\T {4}{15}(\alpha)$ &
$E^t_1= e_1 + \frac{\alpha}{ t ( \alpha -1+ t)} e_2 - \frac{\alpha}{t ( \alpha-1 + t)} e_3$ &
$E^t_3= \frac{1 - t}{\alpha-1 + t} e_3 + \frac{\alpha (\alpha t -1+ t^2)}{ t^3 ( \alpha-1 + t)^2} e_4$ \\
&&&
$E^t_2= e_2 + \frac{(t-1) t}{\alpha-1 + t} e_3$ &
$E^t_4= \frac{-1 + t}{t ( \alpha-1 + t)} e_4$\\
\hline
$\T {4}{03} (1 + t )$ &$\to$& $\T {4}{16}$ &
$E^t_1= \frac{t}{(1 + t) X^2} e_1 + \frac{1 - t}{(1 + t) X^3} e_2 + \frac{1}{(1 + t) X^2}e_3$ &
$E^t_3= \frac{t}{(1 + t) X^3} e_3$ \\
\multicolumn{3}{|l|}{$X=-1 + t + t^2$}&
$E^t_2= \frac{t^2}{(1 + t)^2 X^4} e_2 + \frac{ 1 + t - t^2}{(1 + t)^2 X^5} e_4$ &
$E^t_4= \frac{t^2}{(1 + t)^2 X^6} e_4$\\
\hline
$\T {4}{03} (\frac{1}{1 - t^2} )$ &$\to$& $\T {4}{17}(\alpha)$ &
$E^t_1= t^2 X e_1 + t^2(-1 + t^2) X^2 e_2 + (t^2 - t^4) X^2 e_3$ &
$E^t_3= t^3 X^2 e_3$ \\
\multicolumn{3}{|l|}{$X=(-1 + \alpha + t^2)^{-1}$}&
$E^t_2= t^4 X^2 e_2 + t^4 ( t^2-1)(\alpha + t^2) X^4 e_4$ &
$E^t_4= t^6 X^4 e_4$\\
\hline
$\T 4{19}$ &$\to$& $\T 4{18}$ &
$E_1^t = \sqrt{1-t^2}e_1 + te_2 - te_3$ & $E_3^t = e_3$\\
&&& $E_2^t = e_2$ & $E_4^t = \sqrt{1-t^2}e_4$\\
\hline
$\T 4{23}(t^{-1},-t-1)$ &$\to$& $\T 4{19}$ &
$E_1^t = t^2e_1 - \frac{t^3}{t + 1}e_3$ & $E_3^t = t^4e_3$ \\
&&& $E_2^t = t^2e_2$ & $E_4^t = t^5e_4$\\
\hline
$\T 4{23}(it^{-1},t^{-1})$ &$\to$& $\T 4{20}$ &
$E_1^t = te_1$ & $E_3^t = t^2e_3$ \\
&&& $E_2^t = te_2$ & $E_4^t = t^2e_4$\\
\hline
$\T 4{20}$ &$\to$& $\T 4{21}$ &
$E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$\\
&&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\
\hline
$\T 4{23}(\alpha,\beta)$ &$\to$& $\T 4{22}(\alpha,\beta)$ &
$E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$ \\
&&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\
\hline
$\D 4{01}\left(\beta^2t^{-2},t+\alpha,t\right)$ &$\to$& $\T 4{23}(\alpha,\beta)$ &
$E_1^t = t^2\beta^{-2}(e_1 - e_2)$ & $E_3^t = t^2\beta^{-2}e_3$ \\
&&& $E_2^t = t\beta^{-1}e_2$ & $E_4^t = t^4\beta^{-4}e_4$\\
\hline
$\T 4{23}(\alpha,-(\alpha + 1)t)$ &$\to$& $\T 4{24}(\alpha)$ &
$E_1^t = te_1 - t(\alpha + 1)^{-1}e_3$ & $E_3^t = t^2e_3$ \\
&&& $E_2^t = te_2$ & $E_4^t = t^3e_4$\\
\hline
$\T 4{23}(-1,t)$ &$\to$& $\T 4{25}(\alpha)$ &
$E_1^t = e_1 + \alpha t^{-1}e_3$ & $E_3^t = e_3$ \\
&&& $E_2^t = e_2$ & $E_4^t = e_4$\\
\hline
$\T 4{23}(t,i)$ &$\to$& $\T 4{26}(\alpha)$ &
$E_1^t = \frac i{\alpha + i}e_1 -\frac{\alpha i}{(\alpha + i)^2t}e_3$ & $E_3^t = -\frac 1{(\alpha + i)^2}e_3 + \frac{\alpha}{(\alpha + i)^3t}e_4$ \\
&&& $E_2^t = \frac i{\alpha + i}e_2 - \frac{\alpha}{(\alpha + i)^2t}e_3$ & $E_4^t = -\frac i{(\alpha + i)^3}e_4$\\
\hline
$\T 4{23}\left(\frac{(\alpha^2 + 1)t}{\alpha - t},\alpha\right)$ &$\to$& $\T 4{27}(\alpha)$ &
$E_1^t = te_1 +\frac{(t-\alpha)t}{\alpha(\alpha^2 + 1)}e_3$ & $E_3^t = t^2e_3 - \frac{\alpha(t-\alpha)t^2}{\alpha^2 + 1}e_4$ \\
&&& $E_2^t = te_2 - \frac{(t-\alpha)t}{\alpha^2 + 1}e_3$ & $E_4^t = t^3e_4$\\
\hline
$\T 4{23}\left(\frac{i\alpha(1-t)}t,\frac{\alpha\sqrt{1-t}}t\right)$ &$\to$& $\T 4{28}(\alpha)$ &
$E_1^t = \frac{\alpha^2(t - 1)}t\left(-i e_1 + \frac 1{\sqrt{1-t}} e_2 - (\alpha - i) e_3\right)$ & $E_3^t = -\frac{\alpha^4(t - 1)}t e_3$ \\
&&& $E_2^t = \frac{\alpha^2(t - 1)}t(-e_1 + i\sqrt{1-t} e_2 + e_3)$ & $E_4^t = \frac{\alpha^6(t - 1)^2}{t^2}e_4$\\
\hline
$\T 4{28}(\alpha)$ &$\to$& $\T 4{29}(\alpha)$ &
$E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$ \\
&&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\
\hline
$\T 4{23}\left(1,(1 - \alpha)\sqrt{t - 1}\right)$ &$\to$& $\T 4{30}(\alpha)$ &
\multicolumn{2}{l|}{$E_1^t = \frac{(\alpha - 1)t(t - 1)}D\left(-e_1+\frac 1{\sqrt{t - 1}} e_2-\frac{(\alpha - 2)t - \alpha + 1}D e_3\right)$}\\
\multicolumn{3}{|l|}{$D=\alpha^2t^2 - 2\alpha^2t - 2\alpha t^2 + \alpha^2 + 2\alpha t + t^2 + 2t - 1$}
& \multicolumn{2}{l|}{$E_2^t = \frac{i(\alpha - 1)t(t - 1)}D\left(e_1+\sqrt{t - 1}e_2-\frac{(\alpha-1)t^2 - (2\alpha-3)t + \alpha - 1}D e_3\right)$}\\
&&& \multicolumn{2}{l|}{$E_3^t = \frac{(\alpha - 1)^2(t - 1)t^3}{D^2}\left(e_3+\frac{((\alpha-1) t - \alpha)(\alpha - 1)(t - 1)}De_4\right)$} \\
&&& \multicolumn{2}{l|}{$E_4^t = \frac{i(\alpha - 1)^3(t - 1)^2t^4}{D^3}e_4$}\\
\hline
$\T 4{30}(\alpha)$ &$\to$& $\T 4{31}(\alpha)$ &
$E_1^t = -\frac{i(\alpha^2 - 1)}{\alpha t}e_1 - \frac i{\alpha t}e_3$ & $E_3^t = -\frac{(\alpha^2 - 1)^2}{\alpha^2t^2}e_3 - \frac{i(\alpha + 1)^2(\alpha - 1)}{\alpha t^2}e_4$\\
&&& $E_2^t = -\frac{i(\alpha^2 - 1)}{\alpha t}e_2 + \frac 1t e_3$ & $E_4^t = i\frac{(\alpha^2 - 1)^3}{\alpha^3 t^3}e_4$\\
\hline
$\T 4{23}(\sqrt{t-1} + 1,1)$ &$\to$& $\T 4{32}$ &
$E_1^t = \frac 1{(\sqrt{t-1} + 1)^2}\left(e_1 - \sqrt{t-1}e_2 -\frac{\sqrt{t-1}}{(\sqrt{t-1} + 1)^2}e_3\right)$ & $E_3^t = \frac t{(\sqrt{t-1} + 1)^4}e_3$ \\
&&& $E_2^t = \frac i{(\sqrt{t-1} + 1)^2}\left(-e_1 - \frac 1{\sqrt{t-1}}e_2 + \frac {\sqrt{t - 1}}{(\sqrt{t - 1} + 1)^2}e_3\right)$ & $E_4^t = -\frac {it}{(\sqrt{t - 1} + 1)^6}e_4$\\
\hline
$\T 4{23}(\sqrt{t-1} - 1,1)$ &$\to$& $\T 4{33}$ &
$E_1^t = \frac 1t\left(e_1 - \sqrt{t - 1}e_2 -\frac{\sqrt{t - 1}}t e_3\right)$ & $E_3^t = \frac 1t e_3$ \\
&&& $E_2^t = \frac it\left(-e_1 -\frac 1{\sqrt{t - 1}}e_2 + \frac{\sqrt{t - 1}}t e_3\right)$ & $E_4^t = -\frac i{t^2}e_4$\\
\hline
$\T 4{23}(\sqrt{t-1} - 1,1)$ &$\to$& $\T 4{34}$ &
\multicolumn{2}{l|}{$E_1^t = \frac{\sqrt{t-1} - 1}{t\sqrt{t-1} - t + 2}\left(e_1 -\sqrt{t-1}e_2 -\frac{\sqrt{t-1}(t\sqrt{t-1} - t + 4)}{t(t\sqrt{t-1} - t + 2)}e_3\right)$}\\
&&&
\multicolumn{2}{l|}{$E_2^t = \frac{i(\sqrt{t-1}-1)}{t\sqrt{t-1} - t + 2}\left(-e_1 -\frac 1{\sqrt{t-1}}e_2 +\frac{t^2 - t\sqrt{t-1} - 3t + 4\sqrt{t-1}}{t(t\sqrt{t-1} - t + 2)}e_3\right)$}\\
&&&
\multicolumn{2}{l|}{$E_3^t = \frac{(\sqrt{t-1} - 1)^2}{(t\sqrt{t-1} - t + 2)^2}\left(te_3 +\frac{2\sqrt{t-1}(t - \sqrt{t-1})}{t\sqrt{t-1} - t + 2}e_4\right)$}\\
&&&
\multicolumn{2}{l|}{$E_4^t = -\frac{it(\sqrt{t-1} - 1)^3}{(t\sqrt{t-1} - t + 2)^3}e_4$}\\
\hline
$\D 4{01}\left(\frac 14,\frac 1{2t},\frac 1t\right)$ &$\to$& $\T 4{35}$ &
$E_1^t = -t^{-1}(2e_1 - e_2 - 2e_3)$ & $E_3^t = -2t^{-1}e_3$ \\
&&& $E_2^t = t^{-1}(2e_1 - (2t + 1)e_2 - 4e_3)$ & $E_4^t = 4t^{-2}e_4$\\
\hline
$\T 4{35}$ &$\to$& $\T 4{36}$ &
$E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$ \\
&&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\
\hline
$\T 4{38}\left(t^{-1}\right)$ &$\to$& $\T 4{37}$ &
$E_1^t = e_1 - \frac 1{2t}e_2$ & $E_3^t = e_3 - \frac 1{2t}(1 + \frac 1t)e_4$ \\
&&& $E_2^t = e_2 - \frac 1{2t}e_3 + \frac 1{4t^2}e_4$ & $E_4^t = \frac 1te_4$\\
\hline
\color{black} $\T 4{43}\left((1-\alpha)(t + 1)t^{-1}, t(\alpha - 1)^{-1}\right)$ &\color{black} $\to$& \color{black} $\T 4{38}(\alpha)$ &
\color{black} $E^t_1= e_1$ &
\color{black} $E^t_3= (1-\alpha)t^{-1}e_3$ \\
&&&
\color{black} $E^t_2= e_2$ &
\color{black} $E^t_4= e_4$\\
\hline
\color{black} $\T 4{43}\left(\lambda+t, -t^{-1}\right)$ &\color{black} $\to$& \color{black} $\T 4{39}(\lambda)$ &
\multicolumn{2}{l|}{\color{black} $E_1^t = e_1 - De_2$}\\
\multicolumn{3}{|l|}{\color{black} $D=(2t^2 + (4\lambda +5)t + 2\lambda^2 + 3\lambda + 1)^{-1}$}
& \multicolumn{2}{l|}{\color{black} $E_2^t = e_2 - (t + \lambda + 1)De_3 + D^2 e_4$}\\
&&& \multicolumn{2}{l|}{\color{black} $E_3^t = e_3 + t^{-1}(2t + 3\lambda + 3)De_4$} \\
&&& \multicolumn{2}{l|}{\color{black} $E_4^t = -t^{-1}e_4$}\\
\hline
\color{black}
$\T 4{43}\left(\lambda + t, \frac 2{(\lambda + 1)(2\lambda + 1)}\right)$ &\color{black}$\to$& \color{black}$\T 4{40}(\lambda)$ &
\multicolumn{2}{l|}{\color{black} $E_1^t = e_1 - \frac 12(\lambda + 1)(2\lambda + 1)D e_2$}\\
\multicolumn{3}{|l|}{\color{black} $D=(t(2t + 4\lambda + 3))^{-1}$}
& \multicolumn{2}{l|}{\color{black} $E_2^t = e_2 - \frac 12(\lambda + 1)(2\lambda + 1)(t + \lambda + 1)De_3 + \frac 14(\lambda + 1)^2(2\lambda + 1)^2D^2e_4$}\\
&&& \multicolumn{2}{l|}{\color{black} $E_3^t = e_3 - \frac 12(6t + 2\lambda^2 + 9\lambda + 7)De_4$} \\
&&& \multicolumn{2}{l|}{\color{black} $E_4^t = \frac 1{(\lambda + 1)(2\lambda + 1)}e_4$}\\
\hline
$\T 4{41}(t^{-2})$ &$\to$& $\T 4{42}$ &
$E_1^t = t^{-1}e_1 + t^{-1}e_2$ & $E_3^t = t^{-3}e_3 + 3t^{-3}e_4$\\
&&& $E_2^t = t^{-2}e_2 + t^{-2}e_3 + t^{-4}e_4$ & $E_4^t = t^{-5}e_4$\\
\hline
$\T 4{41}(0)$ &$\to$& $\T 4{44}$ &
$E_1^t = t^{-1}e_1 + t^{-1}e_2$ & $E_3^t = t^{-3}e_3 + 3t^{-3}e_4$ \\
&&& $E_2^t = t^{-2}e_2 + t^{-2}e_3$ & $E_4^t = t^{-5}e_4$\\
\hline
$\D 4{01}\left(\lambda, \alpha,\beta\right)$ &$\to$& $\D 4{02}\left(\lambda, \alpha,\beta\right)$
& $E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$\\
&&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\
\hline
$\D 4{01}\left( \frac{\lambda - t}{(t-1)^2}, t - \alpha t ,\alpha - \alpha t \right)$ &$\to$& $\D 4{03}\left(\lambda, \alpha \right)$
&
\multicolumn{2}{l|}{$E^t_1= \frac{(1 - \alpha ) \alpha (-1 + t)^2 t}{1 + \alpha ^2 \lambda + t - \alpha (1 + t)} e_1 +
\frac{(-1 + \alpha ) \alpha (-1 + t) t^2}{1 + \alpha ^2 \lambda + t - \alpha (1 + t)} e_2 -
\frac{(-1 + \alpha )^2 \alpha (-1 + t)^2 t}{ (1 + \alpha ^2 \lambda + t - \alpha (1 + t))^2} e_3$} \\
&&& \multicolumn{2}{l|}{$E^t_2= \frac{(-1 + \alpha ) \alpha (-1 + t) t}{ 1 + \alpha ^2 \lambda + t - \alpha (1 + t)}e_2 -
\frac{(-1 + \alpha ) \alpha ^2 (-1 + t)^2 t}{ (1 + \alpha ^2 \lambda + t - \alpha (1 + t))^2} e_3$ }\\
&&&
\multicolumn{2}{l|}{$E^t_3= \frac{(-1 + \alpha )^2 \alpha ^2 (-1 + t)^2 t^2}{(1 + \alpha ^2 \lambda + t - \alpha (1 + t))^2} e_3 +
\frac{ (-1 + \alpha )^2 \alpha ^4 (-1 + t)^4 t^2}{ (1 + \alpha ^2 \lambda + t - \alpha (1 + t))^3} e_4$}\\
&&&\multicolumn{2}{l|}{$E^t_4= -\frac{ (-1 + \alpha )^3 \alpha ^3 (-1 + t)^4 t^3}{ (1 + \alpha ^2 \lambda + t - \alpha (1 + t))^3} e_4$ }\\
\hline
$\D 4{01}\left(\lambda, 0,\alpha \right)$ &$\to$& $\D 4{04}\left(\lambda, \alpha\right)$
& $E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$\\
&&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\
\hline
$\D 4{01}\left(\lambda, t , \frac{1 + \sqrt{1 - 4 \lambda}}{2 \lambda }\right)$ &$\to$& $\D 4{05}\left(\lambda, \alpha\right)$
&
$E^t_1= -\frac{1 + \sqrt{1 - 4 \lambda}}{D} e_1 +
\frac{(1 + \sqrt{1 - 4 \lambda}) (D-2 \lambda \alpha )}{ t D^2} e_3$ &
$E^t_3= \frac{(1 + \sqrt{1 - 4 \lambda})^2}{ D^2} e_3 - \frac{(1 + \sqrt{1 - 4 \lambda})^4}{ 2 \lambda t D^3} e_4$ \\
\multicolumn{3}{|l|}{$D=1 + \sqrt{1 - 4 \lambda} + 2 \lambda (-1 + \alpha + \alpha t)$}&
$E^t_2= -\frac{1 + \sqrt{1 - 4 \lambda}}{ D} e_2 + \frac{(1 + \sqrt{1 - 4 \lambda})^2}{ t D^2} e_3$ &
$E^t_4= -\frac{(1 + \sqrt{1 - 4 \lambda})^3}{ D^3} e_4$
\\
\hline
$\D 4{01}\left(\lambda, t , \frac{1 -\sqrt{1 - 4 \lambda}}{2 \lambda }\right)$ &$\to$& $\D 4{06}\left(\lambda, \alpha\right)$
&
$E^t_1= -\frac{1 - \sqrt{1 - 4 \lambda}}{D} e_1 +
\frac{(1 - \sqrt{1 - 4 \lambda}) (D-2 \lambda \alpha )}{ t D^2} e_3$ &
$E^t_3= \frac{(1 - \sqrt{1 - 4 \lambda})^2}{ D^2} e_3 - \frac{(1 - \sqrt{1 - 4 \lambda})^4}{ 2 \lambda t D^3} e_4$ \\
\multicolumn{3}{|l|}{$D=1 - \sqrt{1 - 4 \lambda} + 2 \lambda (-1 + \alpha + \alpha t)$}&
$E^t_2= -\frac{1 - \sqrt{1 - 4 \lambda}}{ D} e_2 + \frac{(1 - \sqrt{1 - 4 \lambda})^2}{ t D^2} e_3$ &
$E^t_4= -\frac{(1 - \sqrt{1 - 4 \lambda})^3}{ D^3} e_4$
\\
\hline
$\D 4{01}\left(\lambda, t , \frac{1 + \sqrt{1 - 4 \lambda}}{2 \lambda }\right)$ &$\to$& $\D 4{07}\left(\lambda\right)$
&
\multicolumn{2}{l|}{$E^t_1=-\frac{D - (1 + \sqrt{1 - 4 \lambda}) \lambda}{D \lambda (1 + t)} e_1 +
\frac{2 (D -(1 + \sqrt{1 - 4 \lambda}) \lambda) (D -2 (1 + \sqrt{1 - 4 \lambda})\lambda + 2 \lambda^2)}{D^3 \lambda (1 + t)^2} e_3$} \\
\multicolumn{3}{|l|}{$D=1 + \sqrt{1 - 4 \lambda} - 2 \lambda$}&
\multicolumn{2}{l|}{$E^t_2= -\frac{D - (1 + \sqrt{1 - 4 \lambda}) \lambda}{D \lambda (1 +t)}e_2$} \\
&&&
$E^t_3= \frac{(D - (1 + \sqrt{1 - 4\lambda}) \lambda)^2}{D^2 \lambda^2 (1 + t)^2} e_3$ &
$E^t_4= \frac{(D - (1 + \sqrt{1 - 4 \lambda}) \lambda)^3}{ D^3 \lambda^3 (1 + t)^3} e_4$\\
\hline
$\D 4{01}\left(\lambda, t , \frac{1 - \sqrt{1 - 4 \lambda}}{2 \lambda }\right)$ &$\to$& $\D 4{08}\left(\lambda\right)$
&
\multicolumn{2}{l|}{$E^t_1=-\frac{D - (1 - \sqrt{1 - 4\lambda}) \lambda}{D\lambda (1 + t)} e_1 +
\frac{2 (D - (1 - \sqrt{1 - 4\lambda}) \lambda) (D - 2 (1 - \sqrt{1 - 4\lambda}) \lambda + 2\lambda^2)}{D^3\lambda (1 + t)^2} e_3$} \\
\multicolumn{3}{|l|}{$D=1 - \sqrt{1 - 4\lambda} - 2\lambda$}&
\multicolumn{2}{l|}{$E^t_2= -\frac{D - (1- \sqrt{1 - 4\lambda})\lambda}{D\lambda (1 +t)}e_2$} \\
&&&
$E^t_3= \frac{(D - (1 - \sqrt{1 - 4\lambda})\lambda)^2}{D^2\lambda^2 (1 + t)^2} e_3$ &
$E^t_4=- \frac{(D - (1 - \sqrt{1 - 4\lambda})\lambda)^3}{ D^3\lambda^3 (1 + t)^3} e_4$\\
\hline
$\D 4{01}\left( \frac{\lambda - t}{(t-1)^2}, \alpha - t^2, t - t^2 \right)$ &$\to$& $\D 4{09}\left(\lambda, \alpha \right)$
&
\multicolumn{2}{l|}{$E^t_1=(1-t)^2 t ( t^2-\alpha) D e_1 + (1 -t) t^2 ( t^2-\alpha) D e_2 + ( t-1 )^3
t (\alpha - t^2) D^2 e_3$ } \\
\multicolumn{3}{|l|}{$D=(-1 + \alpha (-1 + t) + t - \lambda t^2)^{-1}$}&
\multicolumn{2}{l|}{$E^t_2= (1 - t) t (-\alpha + t^2) D e_2 - (-1 + t)^2 t^2 (-\alpha + t^2) D^2 e_3$ } \\
&&&
\multicolumn{2}{l|}{$E^t_3=(1 - t)^2 t^2 (\alpha - t^2)^2 D^2 e_3 + (1 - t)^4 t^4 (\alpha - t^2)^2 D^3 e_4$ } \\\
&&&
\multicolumn{2}{l|}{$E^t_4=( 1-t )^4 ( t^3-\alpha t)^3 D^3 e_4$}\\
\hline
$\D 4{09}\left(\lambda, \alpha \right)$ &$\to$& $\D 4{10}\left(\lambda, \alpha \right)$
& $E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$\\
&&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\
\hline
$\D 4{01}\left( \frac{\lambda - t}{(t-1)^2}, t^2-1, t^2 - t\right)$ &$\to$& $\D 4{11}\left(\lambda, \alpha \right)$
&
\multicolumn{2}{l|}{$E^t_1=(1 - t)^3 (1 + t) D e_1 + (1 - t)^2 t (1 + t) D e_2 - \alpha (1 - t)^3 (1 + t)^2 D^2t^{-1} e_3$ } \\
\multicolumn{3}{|l|}{$D=(1 + \alpha \lambda t - t^2)^{-1}$}&
\multicolumn{2}{l|}{$E^t_2= (1 - t)^2 (1 + t) D e_2 - \alpha (1 - t)^3 (1 + t) D^2 e_3$ } \\
&&&
\multicolumn{2}{l|}{$E^t_3= (1 - t)^4 (1 + t)^2 D^2 e_3 + \alpha (1 - t)^6 t (1 + t)^2 D^3 e_4 $} \\\
&&&
\multicolumn{2}{l|}{$E^t_4=(1 - t)^7 (1 + t)^3 D^3 e_4$}\\
\hline
$\D 4{01}\left(\lambda\left(1-\frac {\alpha t}{\Theta^2}\right), \frac 1\Theta\left(\alpha+\frac{\lambda}{t}\right), \frac 1t\right)$ &$\to$& $\D 4{12}\left(\lambda,\alpha\right)$
& $E_1^t = -\alpha\Theta t^{-1}e_1 + \alpha\lambda t^{-1}e_2 + \alpha(\Theta - \alpha)t^{-1}e_3$ & $E_3^t = -\alpha^3 t^{-1}e_3$\\
&&& $E_2^t = -\alpha t^{-1}e_1 + \alpha(\alpha t + \lambda)(\Theta t)^{-1}e_2 + \alpha t^{-1}e_3$ & $E_4^t = \alpha^4 t^{-2}e_4$\\
\hline
$\D 4{01}\left(\lambda\left(1-\frac {\alpha t}{\Psi^2}\right), \frac 1{\Psi}\left(\alpha+\frac{\lambda}{t}\right), \frac 1t\right)$ &$\to$& $\D 4{13}\left(\lambda,\alpha\right)$
& $E_1^t = -\alpha\Psi t^{-1}e_1 + \alpha\lambda t^{-1}e_2 + \alpha(\Psi - \alpha)t^{-1}e_3$ & $E_3^t = -\alpha^3 t^{-1}e_3$\\
\multicolumn{3}{|l|}{$\Psi=1-\Theta$}&
$E_2^t = -\alpha t^{-1}e_1 + \alpha(\alpha t + \lambda)(\Psi t)^{-1}e_2 + \alpha t^{-1}e_3$ & $E_4^t = \alpha^4 t^{-2}e_4$\\
\hline
$\D 4{12}\left(\lambda,\alpha\right)$ &$\to$& $\D 4{14}\left(\lambda,\alpha\right)$
& $E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$\\
&&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\
\hline
$\D 4{13}\left(\lambda,\alpha\right)$ &$\to$& $\D 4{15}\left(\lambda,\alpha\right)$
& $E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$\\
&&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\
\hline
$\D 4{01}\left(t^2, \alpha , t^{-1}\right)$ &$\to$& $\D 4{16}\left(\alpha\right)$
&
$E^t_1= -\alpha D e_1 + \alpha(\alpha t - 1) D^2 e_3$ &
$E^t_3= -\alpha^3 t D^2 e_3 -\alpha^3 D^3 e_4$ \\
\multicolumn{3}{|l|}{$D=((\alpha^2 + \alpha + 1)t - \alpha - 1)^{-1}$}&
$E^t_2= -\alpha D e_1 + \alpha^2t D e_2 + \alpha(\alpha t - 1) D^2 e_3$ &
$E^t_4= \alpha^4 t D^3 e_4$\\
\hline
$\D 4{01}\left(t^2, -1 , t^{-1}\right)$ &$\to$& $\D 4{17}\left(\alpha\right)$
&
$E^t_1= D e_1 + ((\alpha - 2) t + \alpha)D^2 e_3$ &
$E^t_3= tD^2 e_3 + \alpha D^3 e_4$ \\
\multicolumn{3}{|l|}{$D=((\alpha + 1)t)^{-1}$}&
$E^t_2= D e_1 + tD e_2 + ((\alpha - 1)t + \alpha )D^2 e_3$ &
$E^t_4= tD^3 e_4$\\
\hline
$\D 4{01}\left(\lambda\left(1+\frac t{\Theta^2}\right), \alpha + \Psi, 1\right)$ &$\to$& $\D 4{18}\left(\lambda,\alpha\right)$
&
\multicolumn{2}{l|}{$E^t_1= t\Theta^3D(\alpha + \Psi)\left( e_1 - \Psi e_2 - \Theta(\Theta^2\alpha + \alpha t + \lambda) D e_3\right)$}\\
\multicolumn{3}{|l|}{$D=(\alpha t^2+\Theta\alpha(\alpha + \Theta) t+\Theta^2(\alpha + 1)(\Theta\alpha + \Psi))^{-1}$}&
\multicolumn{2}{l|}{$E^t_2= t\Theta(\alpha + \Psi)D\left(\Theta e_1+(t-\lambda)e_2 - (t^2+(\alpha - \Psi)\Theta t+\Theta(\Theta\alpha + \Psi)\Theta^2) D e_3\right)$}\\
\multicolumn{3}{|l|}{$\Psi=1-\Theta$}&
\multicolumn{2}{l|}{$E^t_3= t^3\Theta^4(\alpha + \Psi)^2D^2\left(e_3 -\Theta(\Theta\alpha + t) D e_4\right)$} \\
&&&
\multicolumn{2}{l|}{$E^t_4= t^4\Theta^6(\alpha + \Psi)^3D^3e_4$}\\
\hline
$\D 4{01}\left(\lambda\left(1+\frac t{\Psi^2}\right), \alpha + \Theta, 1\right)$ &$\to$& $\D 4{19}\left(\lambda,\alpha\right)$
&
\multicolumn{2}{l|}{$E^t_1= t\Psi^3D(\alpha + \Theta)\left( e_1 - \Theta e_2 - \Psi(\Psi^2\alpha + \alpha t + \lambda) D e_3\right)$}\\
\multicolumn{3}{|l|}{$D=(\alpha t^2+\Psi\alpha(\alpha + \Psi) t+\Psi^2(\alpha + 1)(\Psi\alpha + \Theta))^{-1}$}&
\multicolumn{2}{l|}{$E^t_2= t\Psi(\alpha + \Theta)D\left(\Psi e_1+(t-\lambda)e_2 - (t^2+(\alpha - \Theta)\Psi t+\Psi(\Psi\alpha + \Theta)\Psi^2) D e_3\right)$}\\
\multicolumn{3}{|l|}{$\Psi=1-\Theta$}&
\multicolumn{2}{l|}{$E^t_3= t^3\Psi^4(\alpha + \Theta)^2D^2\left(e_3 -\Psi(\Psi\alpha + t) D e_4\right)$} \\
&&&
\multicolumn{2}{l|}{$E^t_4= t^4\Psi^6(\alpha + \Theta)^3D^3e_4$}\\
\hline
$\D 4{18}\left(\lambda,\alpha\right)$ &$\to$& $\D 4{20}\left(\lambda,\alpha\right)$
& $E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$\\
&&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\
\hline
$\D 4{19}\left(\lambda,\alpha\right)$ &$\to$& $\D 4{21}\left(\lambda,\alpha\right)$
& $E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$\\
&&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\
\hline
$\D 4{01}\left(\lambda\left(1+\frac t{\Theta^2}\right), -\frac{\lambda^2}{\Theta^3}, 1\right)$ &$\to$& $\D 4{22}\left(\lambda\right)$
&
\multicolumn{2}{l|}{$E^t_1= \Theta^2D(\Theta e_1 - \lambda e_2 - \Theta\lambda D e_3)$}\\
\multicolumn{3}{|l|}{$D=(\Psi(t-2\lambda+3\Theta-1))^{-1}$}&
\multicolumn{2}{l|}{$E^t_2= \Theta D\left(\Theta e_1 + (t-\lambda)e_2 - \Theta^3\lambda^{-1}((2\Theta - 1)t - \Theta (2\lambda + 1) + 1) D e_3\right)$}\\
\multicolumn{3}{|l|}{$\Psi=1-\Theta$}&
\multicolumn{2}{l|}{$E^t_3= \Theta^4D^2\left(t e_3 - \Theta^3\lambda^{-1}(2\Theta - 1)(t - \Psi) D e_4\right)$} \\
&&&
\multicolumn{2}{l|}{$E^t_4= t\Theta^6 D^3 e_4$}
\\
\hline
$\D 4{01}\left(\lambda\left(1+\frac t{\Psi^2}\right), -\frac{\lambda^2}{\Psi^3}, 1\right)$ &$\to$& $\D 4{23}\left(\lambda\right)$
&
\multicolumn{2}{l|}{$E^t_1= \Psi^2D(\Psi e_1 - \lambda e_2 - \Psi\lambda D e_3)$}\\
\multicolumn{3}{|l|}{$D=(\Theta(t-2\lambda+3\Psi-1))^{-1}$}&
\multicolumn{2}{l|}{$E^t_2= \Psi D\left(\Psi e_1 + (t-\lambda)e_2 - \Psi^3\lambda^{-1}((2\Psi - 1)t - \Psi (2\lambda + 1) + 1) D e_3\right)$}\\
\multicolumn{3}{|l|}{$\Psi=1-\Theta$}&
\multicolumn{2}{l|}{$E^t_3= \Psi^4D^2\left(t e_3 - \Psi^3\lambda^{-1}(2\Psi - 1)(t - \Theta) D e_4\right)$} \\
&&&
\multicolumn{2}{l|}{$E^t_4= t\Psi^6 \lambda^{-1} D^3 e_4$}
\\
\hline
$\D 4{01}\left(\lambda\left(1-\frac t{\Theta^2}\right), -\frac{\lambda^2}{\Theta^3}, 1\right)$ &$\to$& $\D 4{24}\left(\lambda\right)$
&
$E^t_1= -\Theta D\left(\Theta e_1 - \lambda e_2 + \Theta^2 D e_3\right)$ &
$E^t_3= -t\Theta^2 D^2 e_3$ \\
\multicolumn{3}{|l|}{$D=(t - 2\Theta + 1)^{-1}$}&
$E^t_2= -D\left(\Theta e_1 - (t+\lambda) e_2 + \Theta^2 D e_3\right)$ &
$E^t_4= t\Theta^3 D^3 e_4$ \\
\hline
$\D 4{01}\left(\lambda\left(1-\frac t{\Psi^2}\right), -\frac{\lambda^2}{\Psi^3}, 1\right)$ &$\to$& $\D 4{25}\left(\lambda\right)$
&
$E^t_1= -\Psi D\left(\Psi e_1 - \lambda e_2 + \Psi^2 D e_3\right)$ &
$E^t_3= -t\Psi^2 D^2 e_3$ \\
\multicolumn{3}{|l|}{$D=(t - 2\Psi + 1)^{-1},\ \Psi=1-\Theta$}&
$E^t_2= -D\left(\Psi e_1 - (t+\lambda) e_2 + \Psi^2 D e_3\right)$ &
$E^t_4= t\Psi^3\lambda^{-1} D^3 e_4$\\
\hline
$\D 4{01}\left(\lambda\left(1+\frac t{\Theta^2}\right), -\Theta, 1\right)$ &$\to$& $\D 4{26}\left(\lambda\right)$
& $E_1^t = \Theta t^{-1}\left(\Theta e_1 - \lambda e_2 - \Theta t^{-1} e_3\right)$ & $E_3^t = \Theta^2 t^{-1} e_3$\\
&&& $E_2^t = t^{-1}\left(\Theta e_1 + (t-\lambda) e_2 - \Theta^2 t^{-1} e_3\right)$ & $E_4^t = \Theta^3 t^{-2} e_4$\\
\hline
$\D 4{01}\left(\lambda\left(1+\frac t{\Psi^2}\right), -\Psi, 1\right)$ &$\to$& $\D 4{27}\left(\lambda\right)$
& $E_1^t = \Psi t^{-1}\left(\Psi e_1 - \lambda e_2 - \Psi t^{-1} e_3\right)$ & $E_3^t = \Psi^2 t^{-1} e_3$\\
&&& $E_2^t = t^{-1}\left(\Psi e_1 + (t-\lambda) e_2 - \Psi^2 t^{-1} e_3\right)$ & $E_4^t = \Psi^3 t^{-2} e_4$\\
\hline
$\D 4{01}\left(\lambda\left(1+\frac t{\Theta^2}\right), -\Theta, 1\right)$ &$\to$& $\D 4{28}\left(\lambda\right)$
&
\multicolumn{2}{l|}{$E^t_1= \Theta^3D\left(\Theta e_1 - \lambda e_2 - t^{-1}\Theta^2(\Psi^2t + 4\Theta\lambda - \Theta) D e_3\right)$}\\
\multicolumn{3}{|l|}{$D=(\Psi t-2\lambda+\Theta)^{-1},\ \Psi=1-\Theta$}&
\multicolumn{2}{l|}{$E^t_2= \Theta^2D\left(\Theta e_1 + (t-\lambda) e_2 + t^{-1}\Theta((1 - 2\Theta)t^2 + \Theta(3\lambda-2\Psi)t -\Theta^2(4\lambda-1)) D e_3\right)$}\\
&&&
\multicolumn{2}{l|}{$E^t_3= \Theta^6 D^2\left(t e_3 + (2\lambda-\Theta)(t - \Theta) D e_4\right)$} \\
&&&
\multicolumn{2}{l|}{$E^t_4= t\Theta^9 D^3 e_4$}\\
\hline
$\D 4{01}\left(\lambda\left(1+\frac t{\Psi^2}\right), -\Psi, 1\right)$ &$\to$& $\D 4{29}\left(\lambda\right)$
&
\multicolumn{2}{l|}{$E^t_1= \Psi^3D\left(\Psi e_1 - \lambda e_2 - t^{-1}\Psi^2(\Theta^2t + 4\Psi\lambda - \Psi) D e_3\right)$}\\
\multicolumn{3}{|l|}{$D=(\Theta t-2\lambda+\Psi)^{-1},\ \Psi=1-\Theta$}&
\multicolumn{2}{l|}{$E^t_2= \Psi^2D\left(\Psi e_1 + (t-\lambda) e_2 + t^{-1}\Psi((1 - 2\Psi)t^2 + \Psi(3\lambda-2\Theta)t -\Psi^2(4\lambda-1)) D e_3\right)$}\\
&&&
\multicolumn{2}{l|}{$E^t_3= \Psi^6 D^2\left(t e_3 + (2\lambda-\Psi)(t - \Psi) D e_4\right)$} \\
&&&
\multicolumn{2}{l|}{$E^t_4= t\Psi^9 D^3 e_4$}\\
\hline
$\D 4{01}\left(\lambda\left(1-\left(2+\frac{2\lambda-1}{\Theta^2}\right)t\right), \frac\lambda{\Theta^2}\left(1-\frac\Theta t\right), \frac 1t\right)$ &$\to$& $\D 4{30}\left(\lambda\right)$
&
\multicolumn{2}{l|}{$E^t_1= -D\left(\Theta e_1 - \lambda e_2 + t\Psi^{-1}D e_3\right)$}\\
\multicolumn{3}{|l|}{$D=\frac{\lambda(t-\Theta)}{t\left((1-2\Theta)t+\Theta^2\right)},\ \Psi=1-\Theta$}&
\multicolumn{2}{l|}{$E^t_2= -D\left(e_1 - \Theta^{-1}((2\Theta-1)t + \lambda) e_2 + t\Psi^{-1}(2t-\Theta)(t-\Theta)^{-1} D e_3\right)$}\\
&&&
\multicolumn{2}{l|}{$E^t_3= tD^2\left((1-2\Theta)e_3 + \Theta^2\lambda^{-1}(t-\Theta)^{-1}D e_4\right)$} \\
&&&
\multicolumn{2}{l|}{$E^t_4= t D^3(2\Theta - 1) e_4$}
\\
\hline
$\D 4{01}\left(\lambda\left(1-\left(2+\frac{2\lambda-1}{\Psi^2}\right)t\right), \frac\lambda{\Psi^2}\left(1-\frac\Psi t\right), \frac 1t\right)$ &$\to$& $\D 4{31}\left(\lambda\right)$
&
\multicolumn{2}{l|}{$E^t_1= -D\left(\Psi e_1 - \lambda e_2 + t\Theta^{-1}D e_3\right)$}\\
\multicolumn{3}{|l|}{$D=\frac{\lambda(t-\Psi)}{t\left((1-2\Psi)t+\Psi^2\right)},\ \Psi=1-\Theta$}&
\multicolumn{2}{l|}{$E^t_2= -D\left(e_1 - \Psi^{-1}((2\Psi-1)t + \lambda) e_2 + t\Theta^{-1}(2t-\Psi)(t-\Psi)^{-1} D e_3\right)$}\\
&&&
\multicolumn{2}{l|}{$E^t_3= tD^2\left((1-2\Psi)e_3 + \Psi^2\lambda^{-1}(t-\Psi)^{-1}D e_4\right)$} \\
&&&
\multicolumn{2}{l|}{$E^t_4= t D^3(2\Psi - 1) e_4$}
\\
\hline
$\D 4{30}\left(\lambda\right)$ &$\to$& $\D 4{32}\left(\lambda\right)$
& $E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$\\
&&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\
\hline
$\D 4{31}\left(\lambda\right)$ &$\to$& $\D 4{33}\left(\lambda\right)$
& $E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$\\
&&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\
\hline
$\D 4{01}\left(\frac t{(t + 1)2}, 1 , 1 + \frac 1t\right)$ &$\to$& $\D 4{34}$
&
\multicolumn{2}{l|}{$E^t_1= -(t + 1)^2 D e_1 + (t + 1)t D e_2 + (t + 1)^2(t - 1)D^2 e_3$}\\
\multicolumn{3}{|l|}{$D=(t^2 + t - 1)^{-1}$}&
\multicolumn{2}{l|}{$E^t_2= -(t + 1)^2 D e_1 + (t + 1)^2t D e_2 + (t + 1)^2 D e_3$}\\
&&&
\multicolumn{2}{l|}{$E^t_3= -(t + 1)^2t^2 D^2 e_3 - (t + 1)^4t D^3 e_4$} \\
&&&
\multicolumn{2}{l|}{$E^t_4= (t + 1)^4t^2 D^3 e_4$}
\\
\hline
$\D 4{01}\left(\lambda, t^{-1}, 1\right)$ &$\to$& $\D 4{35}\left(\lambda\right)$
& $E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3 + (\lambda t^2)^{-1}e_4$\\
&&& $E_2^t = t^{-1}e_2 + (\lambda t)^{-1}e_3$ & $E_4^t = t^{-4}e_4$\\
\hline
$\D 4{35}\left(\lambda\right)$ &$\to$& $\D 4{36}\left(\lambda\right)$
& $E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$\\
&&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\
\hline
$\D 4{01}\left(\lambda,\Theta t^{-1},t^{-1}\right)$ &$\to$& $\D 4{37}\left(\lambda\right)$
& $E_1^t = te_1$ & $E_3^t = t^2e_3$\\
&&& $E_2^t = te_2$ & $E_4^t = t^2e_4$\\
\hline
$\D 4{01}\left(\lambda,(1-\Theta) t^{-1},t^{-1}\right)$ &$\to$& $\D 4{38}\left(\lambda\right)$
& $E_1^t = te_1$ & $E_3^t = t^2e_3$\\
&&& $E_2^t = te_2$ & $E_4^t = t^2e_4$\\
\hline
$\D 4{37}\left(\lambda\right)$ &$\to$& $\D 4{39}\left(\lambda\right)$
& $E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$\\
&&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\
\hline
$\D 4{38}\left(\lambda\right)$ &$\to$& $\D 4{40}\left(\lambda\right)$
& $E_1^t = t^{-1}e_1$ & $E_3^t = t^{-2}e_3$\\
&&& $E_2^t = t^{-1}e_2$ & $E_4^t = t^{-3}e_4$\\
\hline
\end{longtable}
}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,580 |
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50 CFR 17
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Docket No. FWS-R2-ES-2009-0083
MO 92210-0-0009
RIN:
1018-AV84
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FWS-R2-ES-2009-0083
Docket Name:
Three Forks Springsnail and San Bernardino Springsnail
Docket RIN
Public Comments:
Supporting/Related Materials:
Literature Cited for Three Forks and San Bernardio...
Economic Analysis of the Critical Habitat Designation for the...
Questions and Answers: Proposed Rule to List the Three Forks...
News Bulletin - Re-opening of the Comment Period for the...
U.S. Forest Service Map
Literature Cited April 2011
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SUPPLEMENTARY INFORMATION:
Previous Federal Actions
Critical Habitat
New Information and Changes From the Previously Proposed Critical Habitat
Proposed Critical Habitat Designation
Boneyard Bog Springs Unit
Boneyard Creek Springs Unit
Consideration of Impacts Under Section 4(b)(2) of the Act
Draft Economic Analysis
Required Determinations—Amended
List of Subjects in 50 CFR Part 17
Proposed Regulation Promulgation
PART 17—ENDANGERED AND THREATENED WILDLIFE AND PLANTS
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Start Preamble Start Printed Page 71300
Fish and Wildlife Service, Interior.
Proposed rule; reopening of comment period.
We, the U.S. Fish and Wildlife Service (Service), announce the reopening of the public comment period on the April 12, 2011, proposed endangered status and designation of critical habitat for the Three Forks springsnail (Pyrgulopsis trivialis) and the San Bernardino springsnail (Pyrgulopsis bernardina) under the Endangered Species Act of 1973, as amended (Act). We are proposing to revise the previously proposed critical habitat for the Three Forks springsnail by increasing the size of the Boneyard Bog Springs Unit to 5.3 acres (2.1 hectares), and by adding an additional unit, the Boneyard Creek Springs Unit. In total, we are proposing to designate as critical habitat 17.1 acres (6.9 hectares) for the Three Forks springsnail. We also announce the availability of a draft economic analysis (DEA) of the proposed designation of critical habitat and an amended required determinations section of the proposal. We are reopening the comment period to allow all interested parties an opportunity to comment simultaneously on the revised proposed rule, the associated DEA, and the amended required determinations section. Comments previously submitted need not be resubmitted, as they will be fully considered in preparation of the final rule.
You may submit written comments by one of the following methods:
(1) Electronically: Go to the Federal eRulemaking Portal: http://www.regulations.gov. Search for Docket No. FWS-R2-ES-2009-0083, which is the docket number for this rulemaking.
(2) By hard copy: Submit by U.S. mail or hand-delivery to: Public Comments Processing, Attn: FWS-R2-ES-2009-0083; Division of Policy and Directives Management; U.S. Fish and Wildlife Service; 4401 N. Fairfax Drive, MS 2042-PDM; Arlington, VA 22203.
We will post all comments on http://www.regulations.gov. This generally means that we will post any personal information you provide us (see the Public Comments section below for more information).
Start Further Info
Steve Spangle, Field Supervisor, Arizona Ecological Services Field Office, 2321 West Royal Palm Road, Suite 103, Phoenix, AZ 85021; telephone (602) 242-0210; facsimile (602) 242-2513. Persons who use a telecommunications device for the deaf (TDD) may call the Federal Information Relay Service (FIRS) at (800) 877-8339.
End Further Info End Preamble Start Supplemental Information
We will accept written comments and information during this reopened comment period on our proposed listing and designation of critical habitat for the Three Forks springsnail and San Bernardino springsnail that published in the Federal Register on April 12, 2011 (76 FR 20464), revisions to the proposed critical habitat, our DEA of the proposed designation, and the amended required determinations provided in this document. We will consider information and recommendations from all interested parties. We are particularly interested in comments concerning:
(1) The reasons why we should or should not designate habitat as "critical habitat" under section 4 of the Act (16 U.S.C. 1531 et seq.), including whether there are threats to the species from human activity, the degree of which can be expected to increase due to the designation, and whether that increase in threat outweighs the benefit of designation such that the designation of critical habitat is not prudent.
(2) Specific information on:
(a) The distribution of the Three Forks springsnail and San Bernardino springsnail;
(b) The amount and distribution of the species' habitat;
(c) What areas occupied by the species at the time of listing that contain features essential for the conservation of the species we should include in the designation and why; and
(d) What areas not occupied at the time of listing are essential to the conservation of the species and why.
(3) Land use designations and current or planned activities in the subject areas and their possible impacts on proposed critical habitat.
(4) Any foreseeable economic, national security, or other relevant impacts, that may result from designating any area that may be included in the final designation. We are particularly interested in any impacts on small entities, and the benefits of including or excluding areas from the proposed designation that are subject to these impacts.
(5) Whether our approach to designating critical habitat could be improved or modified in any way to provide for greater public participation and understanding, or to assist us in accommodating public concerns and comments.
(6) Information on the extent to which the description of economic impacts in the DEA is complete and accurate.
(7) The likelihood of adverse social reactions to the designation of critical habitat and how the consequences of such reactions, if likely to occur, would relate to the conservation and regulatory benefits of the proposed critical habitat designation.
If you submitted comments or information on the proposed rule (76 FR 20464; April 12, 2011) during the initial comment period from April 12, 2011, to June 13, 2011, please do not resubmit them. We will incorporate them into the public record as part of this comment period, and we will fully consider them in the preparation of our final determination. Our final determination concerning critical habitat will take into consideration all written comments and any additional information we receive during both comment periods. On the basis of public comments, we may, during the development of our final determination, find that areas proposed are not essential, are appropriate for exclusion under section 4(b)(2) of the Act, or are not appropriate for exclusion.
You may submit your comments and materials concerning the proposed rule or DEA by one of the methods listed in the ADDRESSES section. We will not consider comments sent by email or fax or to an address not listed in the ADDRESSES section.
If you submit a comment via http://www.regulations.gov, your entire comment—including any personal identifying information—will be posted on the Web site. We will post all hardcopy comments on http://www.regulations.gov as well. If you Start Printed Page 71301submit a hardcopy comment that includes personal identifying information, you may request at the top of your document that we withhold this information from public review. However, we cannot guarantee that we will be able to do so.
Comments and materials we receive, as well as supporting documentation we used in preparing the proposed rule and DEA, will be available for public inspection on http://www.regulations.gov at Docket No. FWS-R2-ES-2009-0083, or by appointment, during normal business hours, at the U.S. Fish and Wildlife Service, Arizona Ecological Services Field Office (see FOR FURTHER INFORMATION CONTACT). You may obtain copies of the proposed rule and the DEA on the Internet at http://www.regulations.gov at Docket Number FWS-R2-ES-2009-0083, or by mail from the Arizona Ecological Services Field Office (see FOR FURTHER INFORMATION CONTACT).
It is our intent to discuss only those topics directly relevant to the proposed listing and designation of critical habitat for Three Forks springsnail and San Bernardino springsnail in this document. For more information on previous Federal actions concerning these species, refer to the proposed designation of critical habitat published in the Federal Register on April 12, 2011 (76 FR 20464), which is available online at http://www.regulations.gov (at Docket Number FWS-R2-ES-2009-0083) or from the Arizona Ecological Services Field Office (see FOR FURTHER INFORMATION CONTACT).
On April 12, 2011 (76 FR 20464), we published a proposed rule to list as endangered and designate critical habitat for the Three Forks springsnail and San Bernardino springsnail. We proposed to designate approximately 11.1 acres (ac) (4.5 hectares (ha)) in Arizona in two units located in Apache County as critical habitat for Three Forks springsnail and 2.013 ac (0.815 ha) in four units located in Cochise County as critical habitat for San Bernardino springsnail. That proposal had a 60-day comment period, ending June 13, 2011. We received no requests for a public hearing, and, therefore, no public hearing will take place.
Section 3 of the Act defines critical habitat as the specific areas within the geographical area occupied by a species, at the time it is listed in accordance with the Act, on which are found those physical or biological features essential to the conservation of the species and that may require special management considerations or protection, and specific areas outside the geographical area occupied by a species at the time it is listed, upon a determination that such areas are essential for the conservation of the species. If the proposed rule is made final, section 7 of the Act will prohibit destruction or adverse modification of critical habitat by any activity funded, authorized, or carried out by any Federal agency. Federal agencies proposing actions affecting critical habitat must consult with us on the effects of their proposed actions, under section 7(a)(2) of the Act.
In this document, we are notifying the public of changes to the proposed critical habitat rule. In the April 12, 2011, proposed rule (76 FR 20464), we mentioned that springsnails of the same genus as the Three Forks springsnail were recently found in a spring along Boneyard Creek between Three Forks Springs and Boneyard Bog Springs (Myers 2010, p. 1), but additional analysis was needed for a definitive determination of its taxonomy. Building on the field work of Myers (2010), Myers (2011, p. 5) found additional populations of Pyrgulopsis springsnails along Boneyard Creek. These additional populations are located in the same watershed and in between the two previously known locations, Three Forks Springs and Boneyard Bog Springs. The new populations found in Boneyard Creek are less than 1 mile (mi) (1.6 kilometer (km)) downstream from Boneyard Bog Springs and less than 2 mi (3.2 km) upstream of Three Forks Springs. Due to the proximity of these new populations in relation to Three Forks Springs and Boneyard Bog Springs, we believe that they are the same species. Two different species of springsnails occurring together in the same area is very rare (Liu et al. 2003, p. 2779). If there were different species of springsnails occurring together in this watershed, we can reasonably assume that other springsnail species would have been previously found in either the Three Forks Springs or Boneyard Bog Springs. Based on this information, we believe that the new populations of springsnails found in Boneyard Creek are Three Forks springsnails species.
Also, since publication of the April 12, 2011, proposed rule (76 FR 20464), we have new information regarding the taxonomy of springsnails in Sonora, Mexico. We mentioned in the proposed rule that a springsnail belonging to the same family as the San Bernardino springsnail occurs in two cienegas, or spring ecosystems, in Sonora, Mexico, about 0.25 miles (mi) (0.4 kilometers (km)) south of the San Bernardino National Wildlife Refuge, but additional research was needed to verify if they were the same species as San Bernardino springsnails. Since publication of the proposed rule, we have new information that verifies springsnails in the two cienegas (spring ecosystems in the desert Southwest) in Sonora, Mexico, are San Bernardino springsnails (Varela Romero and Myers 2010, p. 10). However, we will not designate critical habitat for the species in either of those cienegas, because we do not designate critical habitat outside the United States. As such, there are no changes to critical habitat as proposed on April 12, 2011, for the San Bernardino springsnail.
We are proposing to revise our proposed critical habitat designation for the Three Forks springsnail by increasing the size of the Boneyard Bog Springs Unit from 5.0 ac (2.0 ha) to 5.3 ac (2.1 ha) to capture an additional springhead that was discovered since the publication of the proposed rule. In addition, we are proposing a new unit, Boneyard Springs Creek Unit, which is approximately 5.8 ac (2.3 ha) in size, to encompass the newly discovered populations of Three Forks springsnails described above. In total, we are proposing to designate as critical habitat 17.1 ac (6.9 ha) for the Three Forks springsnail. For a full description of the previously proposed units for this species, please see the proposed critical habitat rule (76 FR 20464; April 12, 2011).
In the proposed listing and designation of critical habitat rule (76 FR 20464; April 12, 2011), we identified specific sites that were currently occupied by Three Forks and San Bernardino springsnails, which contained the physical and biological features that are essential to the conservation of the species, and which may require special management considerations or protection. Subsequent to the publication of the proposed listing and critical habitat rule, we discovered new populations of Three Forks springsnails in areas that contain the essential physical and biological features. Therefore, the purpose of this proposed revision to the proposed critical habitat is to include these new areas that are currently occupied by Three Forks springsnail, contain the physical or biological features essential to the conservation of Start Printed Page 71302the species, and meet the definition of critical habitat. We believe the additional unit included in the proposed designation would provide for the conservation of Three Forks springsnail by:
(1) Maintaining the physical and biological features essential to the conservation of the species where the species is known to occur, and
(2) Maintaining the current distribution, thus preserving genetic variation throughout the range of the species and minimizing the potential effects of local extirpation.
We are proposing to revise the previously proposed critical habitat for the Three Forks springsnail by increasing the size of the Boneyard Bog Springs Unit, and by adding an additional unit, the Boneyard Creek Springs Unit. The proposed critical habitat units constitute our current and best assessment of the areas that meet the definition of critical habitat for the species. Proposed critical habitat for the Three Forks Spring Unit for the Three Forks springsnail, and all previously proposed units for the San Bernardino springsnail, are unchanged from our descriptions in the April 12, 2011, proposed rule (76 FR 20464), and are not repeated in this document. We present below brief descriptions of the revised Boneyard Bog Springs Unit and the new Boneyard Creek Springs Unit, and reasons why they meet the definition of critical habitat for the Three Forks springsnail.
The proposed Boneyard Bog Springs Unit is a complex of springs, spring runs, spring seeps, and the segment of Boneyard Creek connecting them, and a small amount of upland area encircling them to make them a single unit of approximately 5.3 ac (2.1 ha), in the vicinity of UTM Zone 12 coordinate 659970, 3750730, in Apache County, Arizona. The entire unit is in Federal ownership and managed by the Apache-Sitgreaves National Forests of the U.S. Forest Service. The unit encompasses eight major springheads and spring runs, each of which flows several yards (meters) to Boneyard Creek, a tributary of the Black River. The spring complex contains spring seeps along the spring runs and the tributary. We are proposing to designate a single critical habitat unit that includes the springheads, spring runs, seeps, and that portion of Boneyard Creek that connects the spring runs. Boneyard Creek is occupied where spring seeps are present along it, and the proposed unit provides for springsnail movement among the occupied seeps, spring runs, and springs, and is essential for habitat connectivity. The area within the proposed unit contains approximately 3.3 feet (ft) (1.0 meter (m)) in width of upland area adjacent to the springheads, spring runs, spring seeps, and tributary segment. The moist soils and vegetation in the adjacent uplands are essential to the species because they produce food for the snails and protect the substrate.
Threats to the Three Forks springsnail in this unit that may require special management of the physical and biological features include wildfire, fire retardant used to fight wildfires, elk grazing, predation by nonnative crayfish, and potential competition from nonnative snails. Also, human-caused changes to the adjacent uplands, which may pose a threat to the aquatic habitats in this proposed unit, can be managed through conservation efforts by Arizona Game and Fish Department and through consultations between the U.S. Forest Service and the U.S. Fish and Wildlife Service under section 7 of the Act. This proposed unit contains all the primary constituent elements and supports all of the Three Forks springsnail's life processes.
The proposed Boneyard Creek Springs Unit is a complex of springs, spring runs, spring seeps, and the segment of Boneyard Creek connecting them, and a small amount of upland area encompassing them, in a single unit of approximately 5.8 ac (2.3 ha), in the vicinity of UTM Zone 12 coordinate 658300, 3749790, in Apache County, Arizona. The entire unit is in Federal ownership and managed by the Apache-Sitgreaves National Forests of the U.S. Forest Service. The unit encompasses at least 11 major springheads and spring runs, which each flow a distance of several yards (meters) to Boneyard Creek, a tributary of the Black River. The spring complex contains spring seeps along the spring runs and the tributary. We are proposing to designate a single critical habitat unit that includes the springheads, spring runs, seeps, and that portion of Boneyard Creek that connects the spring runs. Boneyard Creek is occupied where there are spring seeps along it and provides for springsnail movement among the occupied seeps, spring runs, and springs, and is essential for habitat connectivity. The area within the proposed unit contains approximately 3.3 ft (1.0 m) in width of upland area adjacent to the springheads, spring runs, spring seeps, and tributary segment. The moist soils and vegetation in the adjacent uplands are essential to the species, because they produce food for the snails and protect the substrate they use.
Threats to the Three Forks springsnail in this unit that may require special management of the physical and biological features include wildfire, fire retardant used to fight wildfires, elk grazing, predation by nonnative crayfish, and potential competition from nonnative snails. Also, human-caused changes to the adjacent uplands, which might pose a threat to the aquatic habitats, can be managed through conservation efforts by Arizona Game and Fish Department and through consultations between U.S. Forest Service and U.S. Fish and Wildlife Service under section 7 of the Act. This proposed unit contains all the primary constituent elements and supports all of the Three Forks springsnail's life processes.
Section 4(b)(2) of the Act requires that we designate or revise critical habitat based upon the best scientific data available, after taking into consideration the economic impact, impact on national security, or any other relevant impact of specifying any particular area as critical habitat. We may exclude an area from critical habitat if we determine that the benefits of excluding the area outweigh the benefits of including the area as critical habitat, provided such exclusion will not result in the extinction of the species.
When considering the benefits of inclusion for an area, we consider the additional regulatory benefits that area would receive from the protection from adverse modification or destruction as a result of actions with a Federal nexus (activities conducted, funded, permitted, or authorized by Federal agencies), the educational benefits of mapping areas containing essential features that aid in the recovery of the listed species, and any benefits that may result from designation due to State or Federal laws that may apply to critical habitat.
When considering the benefits of exclusion, we consider, among other things, whether exclusion of a specific area is likely to result in conservation; the continuation, strengthening, or encouragement of partnerships; or implementation of a management plan. In the case of Three Forks springsnail and San Bernardino springsnail, the benefits of critical habitat include public awareness of the presence of the species and the importance of habitat protection, and, where a Federal nexus Start Printed Page 71303exists, increased habitat protection for the species due to protection from adverse modification or destruction of critical habitat. In practice, situations with a Federal nexus exist primarily on Federal lands or for projects undertaken by Federal agencies.
We have not proposed to exclude any areas from critical habitat. However, the final decision on whether to exclude any areas will be based on the best scientific data available at the time of the final designation, including information obtained during the comment period and information about the economic impact of designation. Accordingly, we have prepared a draft economic analysis concerning the proposed critical habitat designation (DEA), which is available for review and comment (see ADDRESSES section).
The purpose of the DEA is to identify and analyze the potential economic impacts associated with the proposed critical habitat designation for the Three Forks springsnail and San Bernardino springsnail. The DEA describes the economic impacts of all potential conservation efforts for the Three Forks springsnail and San Bernardino springsnail; some of these costs will likely be incurred regardless of whether we designate critical habitat. The economic impact of the proposed critical habitat designation is analyzed by comparing scenarios both "with critical habitat" and "without critical habitat." The "without critical habitat" scenario represents the baseline for the analysis, considering protections already in place for the species (e.g., under the Federal listing and other Federal, State, and local regulations). The baseline, therefore, represents the costs incurred regardless of whether critical habitat is designated. The "with critical habitat" scenario describes the incremental impacts associated specifically with the designation of critical habitat for the species. The incremental conservation efforts and associated impacts are those not expected to occur absent the designation of critical habitat for the species. In other words, the incremental costs are those attributable solely to the designation of critical habitat, above and beyond the baseline costs; these are the costs we may consider in the final designation of critical habitat when evaluating the benefits of excluding particular areas under section 4(b)(2) of the Act. Thus, the analysis forecasts both baseline and incremental impacts likely to occur if we finalize the proposed listing and critical habitat designation. For a further description of the methodology of the analysis, see Chapter 2, "Framework for the Analysis," of the DEA.
The DEA provides estimated costs of the foreseeable potential economic impacts of the proposed critical habitat designation for the Three Forks springsnail and San Bernardino springsnail over the next 12 years, which was determined to be the appropriate period for analysis because limited planning information is available for most activities to forecast activity levels for projects beyond a 12-year timeframe. It identifies potential incremental costs as a result of the proposed critical habitat designation; these are those costs attributed to critical habitat over and above those baseline costs attributed to listing. The DEA quantifies economic impacts of Three Forks springsnail and San Bernardino springsnail conservation efforts associated with the following categories of activity: (1) Pesticide use, (2) groundwater pumping, (3) wildfire suppression, and (4) management of ungulate grazing. Additionally, the DEA quantifies economic impacts of additional administrative costs associated with the following categories of activity: (1) Additional effort to address adverse modification in a new consultation, and (2) incremental consultation resulting entirely from critical habitat designation. Total undiscounted costs are estimated at $70,700. The estimated costs are limited to administrative impacts that are likely to result from the designation of critical habitat.
As we stated earlier, we are soliciting data and comments from the public on the DEA, as well as all aspects of the proposed rule and our amended required determinations. We may revise the proposed rule or supporting documents to incorporate or address information we receive during the public comment period. In particular, we may exclude an area from critical habitat if we determine that the benefits of excluding the area outweigh the benefits of including the area, provided the exclusion will not result in the extinction of this species.
In our April 12, 2011, proposed rule (76 FR 20464), we indicated that we would defer our determination of compliance with several statutes and executive orders until the information concerning potential economic impacts of the designation and potential effects on landowners and stakeholders became available in the DEA. We have now made use of the DEA data to make these determinations. In this document, we affirm the information in our proposed rule concerning Executive Order (E.O.) 12866 (Regulatory Planning and Review), E.O. 12630 (Takings), E.O. 13132 (Federalism), E.O. 12988 (Civil Justice Reform), E.O. 13211 (Energy, Supply, Distribution, and Use), the Unfunded Mandates Reform Act (2 U.S.C. 1501 et seq.), the Paperwork Reduction Act of 1995 (44 U.S.C. 3501 et seq.), the National Environmental Policy Act (42 U.S.C. 4321 et seq.), and the President's memorandum of April 29, 1994, "Government-to-Government Relations with Native American Tribal Governments" (59 FR 22951). However, based on the DEA data, we are amending our required determination concerning the Regulatory Flexibility Act (5 U.S.C. 601 et seq.).
Under the Regulatory Flexibility Act, as amended by the Small Business Regulatory Enforcement Fairness Act (5 U.S.C. 802(2)), whenever an agency is required to publish a notice of rulemaking for any proposed or final rule, it must prepare and make available for public comment a regulatory flexibility analysis that describes the effect of the rule on small entities (i.e., small businesses, small organizations, and small government jurisdictions). However, no regulatory flexibility analysis is required if the head of an agency certifies the rule will not have a significant economic impact on a substantial number of small entities. Based on our DEA of the proposed designation, we provide our preliminary regulatory flexibility analysis. Based on comments we receive, we may revise this determination as part of our final rule.
According to the Small Business Administration, small entities include small organizations, such as independent nonprofit organizations; small governmental jurisdictions, including school boards and city and town governments that serve fewer than 50,000 residents; and small businesses (13 CFR 121.201). Small businesses include manufacturing and mining concerns with fewer than 500 employees, wholesale trade entities with fewer than 100 employees, retail and service businesses with less than $5 million in annual sales, general and heavy construction businesses with less than $27.5 million in annual business, special trade contractors doing less than $11.5 million in annual business, and agricultural businesses with annual sales less than $750,000. To determine if potential economic impacts to these small entities are significant, we considered the types of activities that Start Printed Page 71304might trigger regulatory impacts under this designation as well as types of project modifications that may result. In general, the term "significant economic impact" is meant to apply to a typical small business firm's business operations.
To determine if the proposed designation of critical habitat for the Three Forks springsnail and San Bernardino springsnail would affect a substantial number of small entities, we considered the number of small entities affected within particular types of economic activities, such as ranch operations. In order to determine whether it is appropriate for our agency to certify that this proposed rule would not have a significant economic impact on a substantial number of small entities, we considered each industry or category individually. In estimating the numbers of small entities potentially affected, we also considered whether their activities have any Federal involvement. Critical habitat designation will not affect activities that do not have any Federal involvement; designation of critical habitat only affects activities conducted, funded, permitted, or authorized by Federal agencies. In areas where the springsnails are present, once the species are listed, the Federal agencies are required to consult with us under section 7 of the Act on activities they fund, permit, or implement that may affect the species. If we finalize this proposed critical habitat designation, consultations to avoid the destruction or adverse modification of critical habitat would be incorporated into the existing consultation process.
In the DEA, we evaluated the potential economic effects on small entities resulting from implementation of conservation actions related to the proposed designation of critical habitat for the Three Forks springsnail and San Bernardino springsnail. Currently, livestock grazing is excluding from all units so no cattle operators will be impacted by the designation of critical habitat. The DEA does not anticipate impacts to small entities as a result of this designation, as all units are on State or federally owned land. Please refer to the DEA of the proposed critical habitat designation for a more detailed discussion of potential economic impacts.
In summary, we have considered whether the proposed designation would result in a significant economic impact on a substantial number of small entities. Information for this analysis was gathered from the Small Business Administration, stakeholders, and the Service. For the above reasons and based on currently available information, we certify that, if promulgated, the proposed critical habitat designation would not have a significant economic impact on small business entities. Therefore, an initial regulatory flexibility analysis is not required.
A complete list of all references cited in this proposed rule is available on the Internet at http://www.regulations.gov or upon request from the Field Supervisor, Arizona Ecological Services Field Office (see FOR FURTHER INFORMATION CONTACT).
The primary authors of this notice are the staff members of the Arizona Ecological Services Field Office, Southwest Region, U.S. Fish and Wildlife Service.
Start List of Subjects
Endangered and threatened species
Reporting and recordkeeping requirements
End List of Subjects
Accordingly, we propose to further amend part 17, subchapter B of chapter I, title 50 of the Code of Federal Regulations, which was proposed to be amended at 76 FR 20464, April 12, 2011, as follows:
Start Part
1. The authority citation for part 17 continues to read as follows:
Start Authority
Authority: 16 U.S.C. 1361-1407; 16 U.S.C. 1531-1544; 16 U.S.C. 4201-4245; Pub. L. 99-625, 100 Stat. 3500, unless otherwise noted.
End Authority
2. In § 17.95(f), amend the proposed entry for "Three Forks Springsnail (Pyrgulopsis trivialis)," which we proposed at 76 FR 20464 on April 12, 2011, by:
a. Revising proposed paragraph (f)(5);
b. Revising proposed paragraph (f)(7); and
c. Adding a new paragraph (f)(8), to read as set forth below.
§ 17.95
Critical habitat—fish and wildlife.
(f) Clams and Snails.
Three Forks Springsnail (Pyrgulopsis trivialis)
(5) Note: Index map of critical habitat for the Three Forks springsnail follows:
Start Printed Page 71305
(7) Boneyard Bog Springs Unit (2.1 ha; 5.3 ac). The Boneyard Bog Springs Unit consists of all areas within boundary points with the following coordinates in UTM Zone 12 with the units in meters using North American Datum of 1983 (NAD 83): 659968, 3750753; 659990, 3750731; 660021, 3750713; 660060, 3750717; 660070, 3750742; 660176, 3750787; 660190, 3750781; 660199, 3750758; 660208, 3750744; 660159, 3750685; 660125, 3750680; 660088, 3750684; 660081, 3750690; 660072, 3750691; 660072, 3750676; 660076, 3750675; 660076, 3750664; 660069, 3750664; 660067, 3750663; 660060, 3750654; 660052, 3750648; 660034, 3750649; 660029, 3750654; 660027, 3750663; 660008, 3750659; 659997, 3750649; 659997, 3750639; 659988, 3750639; 659982, 3750641; 659958, 3750660; 659954, 3750671; 659945, 3750675; 659942, 3750688; 659933, 3750685; 659904, 3750662; 659889, 3750669; 659885, 3750687; 659902, 3750702; 659919, 3750712; Thence returning to 659968, 3750753.
(8) Boneyard Creek Springs Unit (2.3 ha; 5.8 ac). The Boneyard Creek Springs Unit consists of all areas within boundary points with the following coordinates in UTM Zone 12 with the units in meters using North American Datum of 1983 (NAD 83): 658758, Start Printed Page 713063750008; 658765, 3749996; 658763, 3749984; 658732, 3749975; 658714, 3749981; 658698, 3749968; 658661, 3749971; 658655, 3749981; 658655, 3749998; 658642, 3750000; 658638, 3750024; 658623, 3750034; 658606, 3750036; 658580, 3750029; 658568, 3750020; 658553, 3750013; 658537, 3750005; 658519, 3749993; 658507, 3749985; 658492, 3749992; 658479, 3749976; 658469, 3749960; 658467, 3749945; 658460, 3749935; 658452, 3749913; 658405, 3749863; 658371, 3749841; 658343, 3749805; 658312, 3749789; 658273, 3749741; 658272, 3749733; 658268, 3749725; 658261, 3749722; 658254, 3749720; 658242, 3749699; 658211, 3749682; 658184, 3749655; 658140, 3749634; 658119, 3749610; 658074, 3749624; 658024, 3749603; 657999, 3749549; 657932, 3749492; 657916, 3749492; 657904, 3749509; 657912, 3749527; 657933, 3749545; 657982, 3749559; 658020, 3749623; 658072, 3749642; 658111, 3749632; 658129, 3749649; 658174, 3749667; 658201, 3749691; 658223, 3749705; 658246, 3749743; 658311, 3749811; 658336, 3749826; 658403, 3749893; 658410, 3749904; 658420, 3749908; 658434, 3749917; 658447, 3749962; 658473, 3749991; 658493, 3750013; 658509, 3750003; 658523, 3750019; 658528, 3750030; 658538, 3750043; 658564, 3750055; 658584, 3750053; 658598, 3750061; 658616, 3750068; 658657, 3750052; 658658, 3750032; 658656, 3750020; 658667, 3750002; 658666, 3749982; 658692, 3749984; 658712, 3749994; 658730, 3749994; Thence returning to 658758, 3750008.
Start Signature
Dated: November 8, 2011.
Rachel Jacobson,
Acting Assistant Secretary for Fish and Wildlife and Parks.
End Signature End Part End Supplemental Information
BILLING CODE 4310-55-P
BILLING CODE 4310-55-C
[FR Doc. 2011-29780 Filed 11-16-11; 8:45 am]
Continuity Information | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 6,922 |
"The New York City police chief said, 'We will never forget 9/11'" (joke)
Jokes about the September 11, 2001 terrorist attacks on New York City's World Trade Center are numerous, although not politically correct. English stand-up comedian Jimmy Carr told this joke:
"I saw the chief of the New York City police on the news. He said, 'We will never forget 9/11.' I thought, 'I should fucking hope not, it's your phone number.'"
The joke has been cited on Twitter since September 10, 2009.
Wikipedia: Jimmy Carr
James Anthony Patrick "Jimmy" Carr (born 15 September 1972) is an English stand-up comedian, television host and actor, known for his signature laugh, deadpan delivery, dark humour, and use of edgy one-liners. He is also a writer, actor, and presenter of radio and television. Carr moved to a career in comedy in 2000 and has become a successful comedian.[
Quotes / Jimmy Carr
"The American police have said they will never forget 9/11. Pretty hard too, I would think, considering it's your phone number."
Eigil Moe
@eimoe
RT @FunnyJoker The American police have said they will never forget 9/11. Pretty hard too I would think considering it's your phone number!
5:37 PM - 10 Sep 2009
Steve son of steve
@Stevebarrasford
Watching the news the other day. The chief of NY Police said "We'll never forget 9-11" Better not, it's your phone number
Jimmy Carr: Making People Laugh (2010) Movie Script
Tell you what we do,
we build a World Trade Centre.
You're sat there
with your arms crossed, thinking,
"That's gonna be a fucking big building."
We're going to have two of them.
I saw the chief of the New York City police
on the news.
He said, "We will never forget 9/11."
I thought, "I should fucking
hope not, it's your phone number."
Papa Shango™
@Papa_Shango
New York police officer recently.. "We will never forget 9/11"
Well i should hope not, it's your phone number!
10:16 AM - 13 May 2010
Om Robby
@RobbyIrawan
NYPD won't forget 9/11. it's their phone number.
I saw the chief of a New York City police on the news
submitted September 11, 2016 by EVOSexyBeast
I saw the chief of New York City police on the news, he said "We will never forget 9/11"
I said "Well I sure fuckin' hope not it's your phone number"
stabbymckiller
I don't know why I liked this so much lol
TittyTazed /
Because its Jimmy Carr's joke
New York City • Government/Law/Military/Religion /Health • Sunday, September 11, 2016 • Permalink | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 970 |
Peptostreptococcus canis is a bacterium from the family Peptostreptococcaceae.
References
Bacteria described in 2013
Peptostreptococcaceae | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 8,360 |
Now's the time to freshen up your knowledge and skills with a choice of lectures as well as hands-on demonstrations. This year's lecture programme includes topics identified by the GDC as "highly recommended to do as part of the minimum verifiable CPD requirement". Graham Ramsden, for example, a radiation protection adviser and medical physics expert at Public Health England, will be explaining the new regulations on radiation protection. Medical emergencies will be covered by Esther Stephenson, a SHO at the Oral and Maxillo-Facial Surgery department of St.John's Hospital, Livingston. Those requiring verifiable CPD on the early detection of oral cancer, should attend Dr Karishma Vijan's lecture on mastering the primary guidelines for prevention and detection.
New this year is the Future Experience where you'll be able to find out the scope of the latest digital technologies. You'll be able to peak into 'tomorrow's world' and review the next generation of dental equipment including 3D printers and 4D imaging. This event is going to be very 'hands on' and visitors are encouraged to take part in the scan challenge to be in with a chance of winning one of two 3D printers for the practice!
This year, for the first time, there will be a specific day dedicated to Foundation Dentists which is being sponsored by MyDentist. One of the keynote lecture theatres will be committed to the needs of trainee dentists. Topics will cover how to bridge the gap from DFT to Associate, organising your CPD, risk management, effective treatment planning and complaint handling. As well as formal lectures, there will be a specific Foundation Dentist Hub where you can get one-on-one career advice. The opportunities don't cease when the exhibition hall closes, for there will also be a Foundation Dentists' Ball on Thursday 4th October on the Sunborn super-yacht, which is conveniently moored alongside Excel London. Why not organise a reunion with your former students, where you can share your FD experiences? | {
"redpajama_set_name": "RedPajamaC4"
} | 9,587 |
> Aluminium Prints>Categories>Towns>Moscow
Aluminium Prints Moscow
St. Basil's Cathedral and the Spassky Tower of the Moscow Kremli
Moscow Kremlin and St Basil's Cathedral on the Red Square in Moscow
Moscow - Panoramic view of the Red Square with Moscow Kremlin and St Basil's Cathedral with rainbow
st. basil's cathedral and spassky tower on Red Square
Red Square, Moscow Kremlin at sunset. Moscow, Russia
Moscow,Russia,Red square,view of St. Basil's Cathedral in winter
Moscow Kremlin and St Basil's Cathedral on the Red Square in Moscow, Russia.
Moscow St. Basil's Cathedral Night Shot
View of Kremlin and Red Square in summer in Moscow, Russia.
Famous Moscow Fountain Friendship of Nations at late evening
Aerial View of Moscow City Skyline at Sunset, Moscow, Russia
Kremlin and St. Basil's Cathedral on the Red Square, Moscow, Russia
Red Square with Vasilevsky descent in Moscow
Moscow City skyline. Panorama
Aerial view of Moscow City with Moscow River, Russia, Moscow skyline with the historical architecture skyscraper and Moskva River.
winter panorama of the Moscow Kremlin, Russia
Moscow skyline at red evening light, Russian Federation
Sunrise over Moscow City district and Moscow river
Christmas Moscow. Russia. Spasskaya tower. St. Basil's Cathedral. Panorama of Moscow. Illumination against the backdrop of the Kremlin. Christmas decorations on the streets of Moscow. New Year
Beautiful panoramic view of the Moscow center
Moscow, St. Basil's Cathedral in Red square, Russia
Собор Василия Блаженного и никого S
Sunset over Moscow
Moscow,Russia,Red square,view of St. Basil's Cathedral
Sightseeing Of Moscow, Russia. Panoramic view of Moscow Kremlin and The Cathedral of Vasily the Blessed known as Saint Basil's Cathedral. Beautiful sunrise view of the russian capital city. Panorama
St. Basil's Cathedral
View of St. Basil's cathedral on the Red Square in summer in Moscow, Russia.
Cathedral of Christ the Saviour, Moscow
Interior of Komsomolskaya subway station in Moscow, Russia
Panorama of Red Square in Moscow, Russia
Panoramic view on Moscow Red Square, Kremlin towers, stars and Clock Kuranti, Saint Basil's Cathedral church Ivan bell tower. Panorama from hotel Russia. Moscow holidays vacation tours Putin residence
Moscow City skyline.
St. Basils cathedral on Red Square in Moscow, Russia
Russia. Sights of Moscow and its outskirts. view with altitude, 09/06/2019
Moscow State University Building, Russian Federation
Majestic view on Moscow-city - Sunset.Moscow riveк . Russian group of skyscrapers- very long built and not finished to now
St. Basil's Cathedral on Red Square in Moscow
St. Basil's Cathedral and Spassky Tower
Moscow Kremlin at night, Russia. Panoramic view of the Moscow city center in summer. Moscow cityscape with Moskva River in evening. Panorama of old Moscow Kremlin and Bolshoy Kamenny Bridge at dusk.
Basil's cathedral at Red square in Moscow
Moscow, Russia - Red square view of St. Basil's Cathedral at sunrise, nobody
12345Next page > | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 2,273 |
\subsection{Technical Lemmas}\label{ssec:technical}
\begin{lemma} \label{lem:TopKandIncreasingSequencegiveTopk-1}
Let $1\leq k\leq n$, let $R$ be a run,
let $x$ be a position of a $k$-stack of $R(|R|)$, and $y$ the
position of its topmost $(k-1)$-stack.
Let $a_i$ be the size of the $k$-stack of $R(i)$
at $\hist{\subrun{R}{i}{|R|}}{x}$ for each $0 \leq i \leq |R|$.
Assume that $\hist{R}{x}=\TOP k(R(0))$ and that $a_0=\min\{a_i:
0\leq i\leq |R|\}$.
Then $\hist{R}{y}=\TOP{k-1}(R(0))$.
\end{lemma}
Recall that the size of a $k$-stack is just the number of its $(k-1)$-stacks.
Before we prove the lemma, we state a proposition which is an immediate consequence of the definition of the history function
(recall that for every position $x_1\posNew{k}{x_2}$ we have $x_2\not=(0,\dots,0)$).
\begin{proposition}
Let $S$ be a run of length $1$, and $y$ a position of a $(k-1)$-stack in $S(1)$ for some $1\leq k\leq n$.
Assume that the last non-zero coordinate of $y$ and $\hist{S}{y}$ differ.
Then $\hist{S}{y}=\TOP{k-1}(S(0))$.
\end{proposition}
\begin{proof}[Lemma \ref{lem:TopKandIncreasingSequencegiveTopk-1}]
Set $m:=|R|$.
Let $b_i$ be the value of the last nonzero coordinate of
$\hist{\subrun{R}{i}{m}}{y}$ for each $0\leq i\leq m$
(notice that this is a level $k$ coordinate, as
$\hist{\subrun{R}{i}{m}}{y}$ always points to a $(k-1)$-stack).
We claim that
\begin{enumerate}
\item $b_i\geq a_0$ for each $0\leq i\leq m$ and
\item $z:=\hist{R}{y}$ is simple.
\end{enumerate}
Due to Proposition \ref{prop:HistPreservesPointerContainment} $z$
points into $\TOP k(R(0))$, and contains only links of level at most
$k$.
If the claims hold, we conclude that $b_0 = a_0$ whence
$z=\TOP{k-1}(R(0))$.
We prove the first claim by induction on $i$ (from $m$ to $0$).
For $i=m$ we have $b_m=a_m\geq a_0$.
Let $i<m$.
If $b_i=b_{i+1}$ we are done.
By the above proposition, $b_i\not=b_{i+1}$ implies that
$\hist{\subrun{R}{i}{m}}{y}=\TOP{k-1}(R(i))$.
Due to Corollary \ref{cor:positionContainment},
$\hist{\subrun{R}{i}{m}}{y}$ points into
$\hist{\subrun{R}{i}{m}}{x}$ whence
$a_i$ is the size of the topmost $k$-stack of $R(i)$.
Thus, $b_i=a_i\geq a_0$.
For the second claim we assume that
$z$ is not simple and derive a contradiction as follows.
Since $z$ points to a $(k-1)$-stack, the last link in $z$ is
of level (at least) $k$.
Recall the notation of the $\mathsf{pack}$ function from page
\pageref{def:pack}.
Coordinates of level greater than $k$ in
$\mathsf{pack}_{\NestingRank(z)}(z)$ are the same as in $\TOP
k(R(0))$,
the level $k$ coordinate is $b_0\geq a_0$,
and coordinates of level smaller than $k$ are zeroes.
So $\mathsf{pack}_{\NestingRank(z)}(z)\succeq\TOP{k-1}(R(0))$, and
points to a $(k-1)$-stack.
By Corollary \ref{cor:GoodStacksHaveSmallPositions}
$z=\TOP{k-1}(R(0))$ which contradicts our assumption that $z$ is not
simple. \qed
\end{proof}
\begin{lemma}\label{lem:ForNotEras}
Let $1\leq k\leq n$, let $R$ be a run with $m=\lvert R
\rvert$, $x$ a position of a $k$-stack of $R(|R|)$,
and $a_i$ the size of the $k$-stack of $R(i)$ at
$\hist{\subrun{R}{i}{|R|}}{x}$ for each $0 \leq i \leq |R|$.
Assume that $a_0<a_i$ for all $0<i<|R|$, and
$a_0\leq a_m$.
Then $R$ is a $\TOP0$-non-erasing run.
\end{lemma}
\begin{proof}
If $m = 0$ there is nothing to show. Otherwise,
$a_0<a_1$ whence the first operation of $R$ has to be $\push k$.
\begin{itemize}
\item First assume that $a_m>a_0$ whence
$a_1=\min\{a_i: 1\leq i\leq m\}$.
Let $y$ be the topmost $(k-1)$-stack of the $k$-stack of
$R(m)$ at $x$.
Application of Lemma
\ref{lem:TopKandIncreasingSequencegiveTopk-1} to
$\subrun{R}{1}{m}$, $x$ and $y$ implies that
$\hist{\subrun{R}{1}{m}}{y}=\TOP{k-1}(R(1))$.
Due to Proposition \ref{prop:history2topk-non-eras} applied
to $\subrun{R}{1}{m}$ and $k-1$ we obtain that
$\subrun{R}{1}{m}$ is a $\TOP{k-1}$-non-erasing run.
Since $\TOP0(R(0))\preceq\TOP{k-1}(R(1))$, $\TOP0(R(0))$
cannot be removed by $R$ if $\TOP{k-1}(R(1))$ is not removed
whence $R$ is a $\TOP0$-non-erasing run.
\item
Otherwise, $a_m=a_0$.
We apply the same argument as above, but to
$\subrun{R}{0}{m-1}$. We obtain that $\subrun{R}{0}{m-1}$ is a
$\TOP0$-non-erasing run.
Since $a_{m-1}>a_m=a_0$ and since only the topmost $k$-stack can
change its size, $x=\TOP k(R(|R|))$, and the operation between
$R(m-1)$
and $R(m)$ is $\pop k$ or $\col k$.
As the topmost $k$-stack of $R(m)$ has size $a_0$ and $\TOP
0(R(0))$ is present in $R(m-1)$, it is also present in $R(m)$.\qed
\end{itemize}
\end{proof}
Below we say that an $l$-stack $s$ \emph{occurs} in a $k$-stack $t$;
this includes occurring inside a link, and includes $s=t$.
\begin{lemma}\label{lem:ForClaim2}
Let $0\leq j<k\leq n$, let $R$ be some run, $x$ some position of a
$k$-stack in $R(|R|)$, and
$a_i$ the size of the $k$-stack at $\hist{\subrun{R}{i}{\lvert R
\rvert}}{x}$ in $R(i)$.
Assume that $a_i>a_{\lvert R \rvert}$ for all $0\leq i<\lvert R \rvert$.
Then every $j$-stack occurring in the $k$-stack at $x$ in $R(\lvert
R \rvert)$ occurs also in the $k$-stack at $\hist{R}{x}$ in $R(0)$.
\end{lemma}
\begin{proof}
It is enough to prove, for $1\leq b\leq a_{|R|}$ that the $b$-th
$(k-1)$-stack of the $k$-stack at $x$ in $R(\lvert R \rvert)$ is
equal to
the $b$-th $(k-1)$-stack of the $k$-stack at $\hist{R}{x}$ in $R(0)$.
We prove this by induction on the length of $R$.
For $|R|=0$ this is immediate.
Let $|R|\geq 1$.
In the light of the induction assumption,
it is enough to prove, for $1\leq b\leq a_{|R|}$, that the $b$-th
$(k-1)$-stack of the $k$-stack at $\hist{\subrun{R}{1}{|R|}}{x}$ in
$R(1)$ is equal to
the $b$-th $(k-1)$-stack of the $k$-stack at $\hist{R}{x}$ in $R(0)$.
\begin{itemize}
\item If the first operation in $R$ is of level below $k$, then
$\hist{\subrun{R}{1}{|R|}}{x}=\hist{R}{x}$ and only the topmost
$(k-1)$-stack is modified;
this is not one of the considered $(k-1)$-stacks, as
$a_0>a_{|R|}$.
\item If the first operation in $R$ is of level $k$, then
$\hist{\subrun{R}{1}{|R|}}{x}=\hist{R}{x}$ and some $(k-1)$-stacks
are removed or added, but none of the considered $(k-1)$-stacks,
as $a_0,a_1\geq a_{|R|}$
(this is also true for $\col k$, as performing $\col k$ is
equivalent to performing several $\pop k$).
\item If the first operation in $R$ is of level greater than $k$,
then the whole $k$-stacks at $\hist{\subrun{R}{1}{|R|}}{x}$ in
$R(1)$ and at $\hist{R}{x}$ in $R(0)$ are the same. \qed
\end{itemize}
\end{proof}
\begin{lemma}\label{lem:tech4}
Let $1\leq k\leq n$, $R$ a pumping run and $x$ a
position of a $k$-stack in $R(|R|)$.
Assume that the size of the $k$-stack at $x$ in $R(|R|)$ is
greater than that of the $k$-stack at
$\hist{\subrun{R}{i}{|R|}}{x}$ in $R(i)$ for some $i$.
Then $\TOP0(R(0))\neq\TOP0(R(|R|))$.
\end{lemma}
\begin{proof}
Let $m:=|R|$, and let $a_j$ be the size of the $k$-stack of $R(j)$
at $\hist{\subrun{R}{j}{|R|}}{x}$ for all $0\leq j\leq m$.
Take the maximal $b$ such that $a_b < a_m$ (note that $i\leq b$).
Since stack operations increase the number of stacks by at most one,
$a_{b+1}=a_b + 1$ whence maximality of $b$ implies
$a_{b+1}=\min\{a_j:b+1\leq j\leq m\}$.
Set $S:=\subrun{R}{b+1}{m}$.
Notice that $\hist{S}{x}=\TOP k(S(0))$ because only the topmost
$k$-stack can change its size.
Since $\hist{R}{\TOP0(R(m))}=\TOP 0(R(0))$, Proposition
\ref{prop:history2topk-non-eras} implies that $R$ is a
$\TOP0$-non-erasing run.
It means that $\TOP{k-1}(R(0))$ is present in $R(b)$.
Because the operation between $R(b)$ and $R(b+1)=S(0)$ is
necessarily $\push k$,
it implies that $\TOP0(R(0))\prec\TOP{k-1}(S(0))$.
Application of Lemma \ref{lem:TopKandIncreasingSequencegiveTopk-1} to
$S$ and $x$ shows that $\hist{S}{y}=\TOP{k-1}(S(0))$ for some
position $y$ in $R(m)$. Again using Proposition
\ref{prop:history2topk-non-eras}, we conclude that $S$ is a
$\TOP{k-1}$-non-erasing run and we obtain
\begin{equation*}
\TOP{0}(R(0)) \prec \TOP{k-1}(S(0)) \preceq \TOP{k-1}(R(m))
\prec \TOP{0}(R(m)). \tag*{\qed}
\end{equation*}
\end{proof}
\subsection{Main Technical Lemma}\label{ssec:pumping-tech}
Below we present our main technical lemma.
It shows how to find subruns of long runs which consist of a pumping
run followed by a $\TOP{0}$-non-erasing run.
Recall that the function $\mathsf{ctype}_\mathcal{X}$ maps
configurations to a finite set of types.
For each collapsible pushdown system $\mathcal{S}$,
let $\mathcal{T}_\mathcal{S}$ denote the image of $\mathsf{ctype}_\mathcal{X}$ with respect to
configurations of $\mathcal{S}$.
\begin{lemma}\label{lem:Pumpability}
Let $\mathcal{S}$ be an $n$-CPS,
$0\leq k\leq n$,
$R$ be a run of $\mathcal{S}$, and
\begin{align*}
G_k\subseteq \{i\ < \lvert R \rvert: \hist{\subrun{R}{i}{\lvert
R \rvert}}{\TOP{k}(R(\lvert R \rvert))}= \TOP{k}(R(i))\}.
\end{align*}
Furthermore, let $s^k$ be the $k$-stack of $R(0)$ to which
$\hist{R}{\TOP{k}(R(\lvert R \rvert))}$ points.
For $1\leq j\leq k$, let $r_j$ be the maximum of the sizes of $j$-stacks occurring in $s^k$.
Let
\begin{align*}
\hat N_0 := \lvert\mathcal{T}_\mathcal{S} \rvert +1 \text{ and } \hat N_j=r_j\cdot
2^{\hat N_{j-1}} \text{ for } 1\leq j\leq k.
\end{align*}
If $\lvert G_k\rvert \geq \hat N_k$,
then there are $0\leq x<y<z\leq \lvert R \rvert$ such that
\begin{enumerate}
\item \label{pumpability:1} $\mathsf{ctype}_\mathcal{X}(R(x))=\mathsf{ctype}_\mathcal{X}(R(y))$,
\item \label{pumpability:2}
$\subrun{R}{x}{y}$ is a pumping run,
\item \label{pumpability:3} $\hist{\subrun{R}{y}{\lvert R \rvert}}{\TOP{k}(R(\lvert R \rvert))}=\TOP{k}(R(y))$,
\item \label{pumpability:4} $\TOP0(R(x))\neq\TOP0(R(y))$ or
\begin{align*}
G_k\cap\{x,x+1,\dots,y-1\} \neq \emptyset,
\end{align*}
\item \label{pumpability:5} $z-1\in G_k$, and
\item \label{pumpability:6} $\subrun{R}{y}{z}$ is a $\TOP0$-non-erasing run.
\end{enumerate}
\end{lemma}
\begin{proof}
We prove the lemma by induction on $k$.
Consider the case that $k=0$.
By assumption $\lvert G_0 \rvert \geq \lvert \hat N_0 \rvert > \lvert
\mathcal{T}_\mathcal{S}\rvert$.
Thus, there are $x,y\in G_0$ with $x<y$
such that $\mathsf{ctype}_\mathcal{X}(R(x))=\mathsf{ctype}_\mathcal{X}(R(y))$.
Since $x,y\in G_0$,
\begin{align*}
&\hist{\subrun{R}{x}{\lvert
R \rvert}}{\TOP{0}(R(\lvert R \rvert))}= \TOP{0}(R(x)),\text{ and}\\
&\hist{\subrun{R}{y}{\lvert
R \rvert}}{\TOP{0}(R(\lvert R \rvert))}= \TOP{0}(R(y)).
\end{align*}
Due to Proposition \ref{prop:histTransitive}, we conclude
that
\begin{align*}
\hist{\subrun{R}{x}{y}}{\TOP{0}(R(y))}=\TOP{0}(R(x))
\end{align*}
which means that $\subrun{R}{x}{y}$ is a pumping run.
Since $x\in G_0$, we have $G_0\cap \{x, x+1, \dots,
y-1\}\neq\emptyset$.
By definition of $y$, $\subrun{R}{y}{\lvert R \rvert}$ is a pumping
run of length at least $1$. Due to the characterisation of pumping
runs (cf. Lemma \ref{lem:pumping-run}), this run starts with some
$\push{}$ operation.
Thus, for $z:=y+1$, we have $z-1\in G_0$ and $\subrun{R}{y}{z}$ is a $\TOP0$-non-erasing run.
Thus, $x$, $y$, and $z$ satisfy the claim of the lemma.
Now consider the case $k\geq 1$ and assume that the lemma holds for
all $k'<k$.
Let $a_i$ be the size of the $k$-stack of $R(i)$ at
position $\hist{\subrun{R}{i}{\lvert R \rvert}}{\TOP{k}(R(\lvert R
\rvert))}$ for $0\leq i\leq \lvert R \rvert$.
Due to
Proposition
\ref{prop:NonTopmostStacksDoNotChange} we know that
$a_i-a_{i-1}\leq 1$ for all $1\leq i \leq \lvert R \rvert$.
By definition $a_0\leq r_k$ whence
$\lvert G_k \rvert \geq \hat{N}_k\geq a_0\cdot
2^{\hat{N}_{k-1}}$.
Hence, we can apply Corollary \ref{cor:IncreasingNumberSequence} to
$(a_i)_{0\leq i \leq \lvert R \rvert}$ and obtain
indices $0\leq b < e \leq \lvert R \rvert$ such that
\begin{enumerate}
\item $e-1\in G_k$,
\item $a_b = \min\{a_i: b\leq i \leq e\}$,
\item $a_i>a_b$ for all $0\leq i < b$ and
\item $\lvert H_{b,e} \rvert \geq \hat N_{k-1}$ where
\begin{align*}
H_{b,e} =
\{ i: &\ b \leq i \leq e-1,\\
&a_i \leq a_j\text{ for all
} i\leq j \leq e
\text{, and }\\
&a_i < a_j\text{ for all } i<j \leq n_{G_k}(i)\}
\end{align*}
with $n_{G_k}(i):=\min\{ g\in G_k: g\geq i\}$.
\end{enumerate}
Set $R':=\subrun{R}{b}{e}$ and $G_{k-1}:=\{h-b:h\in H_{b,e}\}$.
Let us first assume that the following claims are true:
\begin{enumerate}[A)]
\item for each $h\in H_{b,e}$ we have $\hist{\subrun{R}{h}{e}}{\TOP{k-1}(R(e))}= \TOP{k-1}(R(h))$,
\item for all $i\leq e-1$,
$\hist{\subrun{R}{i}{e}}{\TOP{k}(R(e))}= \hist{\subrun{R}{i}{\lvert R \rvert}}{\TOP{k}(R(\lvert R \rvert))}$,
whence $a_i$ is the size of the $k$-stack in $R(i)$ at $\hist{\subrun{R}{i}{e}}{\TOP{k}(R(e))}$, and
\item if $t^{k-1}$ is the $(k-1)$-stack at
$\hist{R'}{\TOP{k-1}(R'(\lvert R' \rvert))}$, then
the size of every
$j$-stack occurring in $t^{k-1}$ for $j \leq k-1$ is bounded by $r_j$.
\end{enumerate}
We postpone the proof of these claims.
Claim A implies (by shifting from $R$ to $R'$) that for each $g\in
G_{k-1}$ we have $\hist{\subrun{R'}{g}{\lvert R'
\rvert}}{\TOP{k-1}(R'(\lvert R' \rvert))}= \TOP{k-1}(R'(g))$.
Together with Claim C this allows us to apply
the induction hypothesis to $k-1$, $R'$ and $G_{k-1}$.
We obtain three indices $0\leq x'<y'<z'\leq|R'|$; let
$x=x'+b$, $y=y'+b$, and
let $z$ be the smallest index such that $z\geq z'+b$ and $z-1\in G_k$
(it exists because $z'+b\leq e$ and $e-1\in G_k$).
Note that
\begin{enumerate}[1'.]
\item \label{pumpability:1Prime} $\mathsf{ctype}_\mathcal{X}(R(x))=\mathsf{ctype}_\mathcal{X}(R(y))$,
\item \label{pumpability:2Prime} $\subrun{R}{x}{y}$ is a pumping run,
\item \label{pumpability:3Prime} $\hist{\subrun{R}{y}{e}}{\TOP{k-1}(R(e))}=\TOP{k-1}(R(y))$,
\item \label{pumpability:4Prime} $\TOP0(R(x))\neq\TOP0(R(y))$ or
\begin{align*}
H_{b,e}\cap\{x,x+1,\dots,y-1\} \neq \emptyset,
\end{align*}
\item \label{pumpability:5Prime} $z'+b-1\in H_{b,e}$, and
\item \label{pumpability:6Prime} $\subrun{R}{y}{z'+b}$ is a $\TOP0$-non-erasing run.
\end{enumerate}
Note that items \ref{pumpability:1Prime}' and
\ref{pumpability:2Prime}' coincide with items \ref{pumpability:1}
and \ref{pumpability:2} of the lemma.
We now prove items \ref{pumpability:3} -- \ref{pumpability:6}.
\begin{enumerate}
\setcounter{enumi}{2}
\item Due to Corollary \ref{cor:positionContainment}, item
\ref{pumpability:3Prime}'
implies that
\begin{align*}
\hist{\subrun{R}{y}{e}}{\TOP{k}(R(e))}=\TOP{k}(R(y)).
\end{align*}
Together with Claim B this yields item \ref{pumpability:3}.
\item
Assume that $\TOP0(R(x))=\TOP0(R(y))$.
Note that this directly implies
\begin{align}\label{condition-k-1-equals}
\TOP{k}(R(x))=\TOP{k}(R(y) \text{ for all } 0\leq k \leq n.
\end{align}
Due to \ref{pumpability:4Prime}', there is some
\begin{align*}
h\in H_{b,e}\cap\{x,x+1,\dots,y-1\} \neq \emptyset.
\end{align*}
Items \ref{pumpability:2Prime}', \ref{pumpability:3Prime}', and
Claim A, after application of Corollary
\ref{cor:positionContainment}, imply
\begin{align*}
\hist{\subrun{R}{x}{y}}{\TOP{j}(R(y))}&=\TOP j(R(x)),\\
\hist{\subrun{R}{y}{e}}{\TOP{j}(R(e))}&=\TOP j(R(y)),\quad\mbox{and}\\
\hist{\subrun{R}{h}{e}}{\TOP{j}(R(e))}&=\TOP j(R(h))
\end{align*}
for all $j\geq k-1$.
Due to Proposition \ref{prop:histTransitive}, this implies that
\begin{align}\label{eq:hist-ok}
\hist{\subrun{R}{a}{b}}{\TOP{j}(R(b))}=\TOP j(R(a))
\end{align}
for each pair $a,b\in\{x,h,y,e\}$ with $a\leq b$.
With two applications of Lemma \ref{lem:PositionsDecrease} (to
$\subrun{R}{h}{y}$ and $\subrun{R}{x}{h}$) we obtain
that $\TOP{k-1}(R(h))$ is lexicographically bounded by
$\TOP{k-1}(R(x))=\TOP{k-1}(R(y))$ from below and from above
whence it is this position (the equality of the two positions
comes from equation \eqref{condition-k-1-equals}).
Claim B and equation \eqref{eq:hist-ok} (setting $j=k$) imply
that $a_x$, $a_h$ and $a_y$ are the sizes of the topmost
$k$-stacks of $x$, $h$ and $y$, respectively.
It follows that $a_x=a_h=a_y$.
Since $h\in H_{b,e}$,
there exists some $g\in G_k$ such that
$x\leq h\leq g$ and $a_j > a_h$ for all $h< j \leq g$.
As $a_y = a_h$, we conclude that $g < y$ whence
$G_k\cap \{x, x+1, \dots, y-1\} \neq \emptyset$.
\item $z-1\in G_k$ is satisfied by definition of $z$.
\item If $z=z'+b$, items \ref{pumpability:6} and
\ref{pumpability:6Prime}' coincide.
Assume that $z>z'+b$.
Because $z'+b-1\in H_{b,e}$, we know that $a_j > a_{z'+b-1}$ for
$z'+b\leq j\leq z-1$ because $z$ is minimal such that $z-1\geq z'+b-1$ and $z-1\in G_k$.
In particular, $z>z'+b$ implies $a_{z'+b} > a_{z'+b-1}$.
Recall that $z$ was chosed to satisfy $z\leq e$. This together
with $z'+b-1\in H_{b,e}$ implies that $a_z\geq a_{z'+b-1}$.
Thus, Lemma \ref{lem:ForNotEras} can be applied to $\subrun{R}{z'+b-1}{z}$.
It follows that $\subrun{R}{z'+b-1}{z}$ is a $\TOP0$-non-erasing run.
Since $\subrun{R}{y}{z'+b}$ is also a $\TOP0$-non-erasing run,
$\subrun{R}{y}{z}$ is one as well (cf.~Proposition
\ref{prop:topk-non-eras-composition}).
\end{enumerate}
Thus, $x,y$ and $z$ satisfy the lemma if Claims A -- C hold.
We continue with a simultaneous proof of Claims A and B.
We start with showing that
for each $h\in H_{b,e}$
\begin{align}\label{eq:k-dla-hR}
\hist{\subrun{R}{h}{\lvert R \rvert}}{\TOP{k}(R(\lvert R \rvert))}=\TOP{k}(R(h)).
\end{align}
Consider any $h\in H_{b,e}$.
If $h\in G_k$, the condition is satisfied by definition of $G_k$.
Otherwise, we conclude that $a_{h+1} > a_h$ by definition of
$H_{b,e}$ and the fact that $n_{G_k}(h)\geq h+1$.
But only the topmost $k$-stack can change its size whence equation
(\ref{eq:k-dla-hR}) holds.
Recall that $e-1\in G_k$, which implies that
\begin{align}\label{e-1isTOPk}
\hist{\subrun{R}{e-1}{\lvert R
\rvert}}{\TOP{k}(R(\lvert R \rvert))} = \TOP{k}(R(e-1)).
\end{align}
Together with equation (\ref{eq:k-dla-hR}) this implies
\begin{align*}
&\hist{\subrun{R}{h}{e-1}}{\TOP{k}(R(e-1))}=\\
&\hspace{0.5cm}=\hist{\subrun{R}{h}{ \lvert R \rvert}}{\TOP{k}(R(\lvert R \rvert))}=\TOP{k}(R(h))
\end{align*}
for each $h\in H_{b,e}$.
By definition of $H_{b,e}$, \mbox{$a_h=\min\{a_i: h\leq i \leq e\}$}.
Additionally, equation (\ref{e-1isTOPk}) implies that $a_i$ (for
$b\leq i\leq e-1$) is the size of the $k$-stack of $R(i)$ at
$\hist{\subrun{R}{i}{e-1}}{\TOP k(R(e-1))}$,
whence we may apply Lemma
\ref{lem:TopKandIncreasingSequencegiveTopk-1} to $x:=\TOP k(R(e-1))$
and to $\subrun{R}{h}{e-1}$.
This yields
\begin{align} \label{eqn:TOPK-1e-1}
\hist{\subrun{R}{h}{e-1}}{\TOP{k-1}(R(e-1))}=\TOP{k-1}(R(h))
\end{align}
for each $h\in H_{b,e}$.
We continue by case distinction on the operation between $e-1$
and $e$ in $R$.
\begin{enumerate}
\item Due to equation (\ref{e-1isTOPk}),
the operation at $e-1$ cannot be $\pop{k'}$ or $\col{k'}$ for
$k'>k$.
\item If the operation at $e-1$ is of level below $k$ or is a $\push{}$
operation, then
\begin{align*}
\hist{\subrun{R}{e-1}{e}}{\TOP{k-1}(R(e))}=\TOP{k-1}(R(e-1)).
\end{align*}
Due to equation (\ref{eqn:TOPK-1e-1}), this implies Claim A.
Together with (\ref{e-1isTOPk}) and Corollary
\ref{cor:positionContainment}, this implies
$$\hist{\subrun{R}{e-1}{e}}{\TOP
k(R(e))}=\hist{\subrun{R}{e-1}{|R|}}{\TOP{k}(R(|R|))}.$$
Using Proposition \ref{prop:histTransitive}, Claim B follows directly.
\item Assume that the operation at
$e-1$ is $\pop{k}$ or $\col{k}$.
We conclude immediately that
\begin{align*}
\hist{\subrun{R}{e}{\lvert R
\rvert}}{\TOP{k}(R(\lvert R \rvert))}=\TOP{k}(R(e)),
\end{align*}
because this is the only position $p$ of $R(e)$ that satisfies
$\hist{\subrun{R}{e-1}{e}}{p}=\TOP{k}(R(e-1))$ (and because $e-1\in
G_k$).
With Proposition \ref{prop:histTransitive}, Claim B follows directly.
Furthermore, $a_e$ is the size of the stack of $R(e)$ at $\TOP{k}(R(e))$.
By definition of $H_{b,e}$, we have $a_h=\min\{a_i:h\leq i\leq e\}$.
Application of Lemma \ref{lem:TopKandIncreasingSequencegiveTopk-1}
to $x:=\TOP k(R(e))$ and to $\subrun{R}{h}{e}$ for each $h\in H_{b,e}$ yields Claim A.
\end{enumerate}
For the proof of Claim C,
let $t^{k-1}$ be the $(k-1)$-stack of $R'(0)$ at the position
$\hist{R'}{\TOP{k-1}(R'(\lvert R' \rvert))}$
which is by definition the
$(k-1)$-stack of $R(b)$ at
$\hist{\subrun{R}{b}{e}}{\TOP{k-1}(R(e))}$.
Due to Corollary \ref{cor:positionContainment},
$\hist{\subrun{R}{b}{e}}{\TOP{k-1}(R(e))}$ points into
$\hist{\subrun{R}{b}{e}}{\TOP{k}(R(e))}$.
Hence, for $j\leq k-1$ every $j$-stack occurring in $t^{k-1}$
occurs also in the $k$-stack of $R(b)$ at
$\hist{\subrun{R}{b}{e}}{\TOP{k}(R(e))}$.
Due to Claim B,
$a_i$ is the number of $(k-1)$-stacks of the stack at
$\hist{\subrun{R}{i}{e}}{\TOP{k}(R(e))}$ for all $i\leq b$, and the
$k$-stack of $R(0)$ at $\hist{\subrun{R}{0}{e}}{\TOP k(R(e))}$ is
$s^k$.
We have $a_i>a_b$ for all $0\leq i<b$, so we can apply Lemma
\ref{lem:ForClaim2} to $\subrun{R}{0}{b}$ and
position $\hist{\subrun{R}{b}{e}}{\TOP{k}(R(e))}$.
We conclude that for $j\leq k-1$ every $j$-stack occurring in
$t^{k-1}$ occurs also in $s^k$. Thus, its size is bounded by $r_j$. \qed
\end{proof}
\subsection{Finitely Branching
Epsilon-Contractions}\label{ssec:finitely-branching}
The basic proof concept for the pumping lemma is as follows.
If we find a pumping run which starts and ends in configurations of the
same type, then we can apply Proposition \ref{prop:types1} to this run and
obtain arbitrarily long runs in the graph of the
CPS. But if we consider $\varepsilon$-contractions, all runs that we
construct may consist of $\varepsilon$-edges except for a bounded
number of transitions. In this case, the longer and longer runs would
perhaps always induce the same path in the $\varepsilon$-contraction.
In this section we show how to overcome this problem in the case of
finitely branching $\varepsilon$-contractions.
We first derive a technical condition that allows to
conclude that the $\varepsilon$-contraction of some collapsible
pushdown graph is infinitely branching.
This result basically uses the naive pumping approach described before
but we add some assumptions such that we really obtain larger and
larger runs that end in larger and larger stacks that belong to the
nodes of the $\varepsilon$-contraction.
Afterwards, we use this result
in order to define a bound on the difference of stack sizes between two
nodes of a finitely branching $\varepsilon$-contraction that are
connected by an edge. In the next section we use this fact in the
pumping construction in the following way: instead of talking about a
configuration being in some distance from the initial
configuration in the
$\varepsilon$-contraction, we talk about a configuration having stack
sizes bounded by some numbers.
Without loss of generality (by doubling the number of states of the system), we can assume that for each state $q$
transitions leading to state $q$ are all $\varepsilon$-transitions or are all non-$\varepsilon$-transitions.
\begin{proposition} \label{prop:inf-branching}
Let $\mathcal{S}$ be some CPS of level $n$ such that for each state $q$
transitions leading to state $q$ are all $\varepsilon$-transitions or are all non-$\varepsilon$-transitions.
Let $R$ be a run starting in a configuration of the
$\varepsilon$-contraction $\mathcal{G}$ of the graph of $\mathcal{S}$.
Then $\mathcal{G}$ is infinitely branching
if there are positions $0\leq x < y \leq \lvert R \rvert$ such that
\begin{enumerate}
\item $\mathsf{ctype}_\mathcal{X}(R(x))=\mathsf{ctype}_\mathcal{X}(R(y))$,
\item $\subrun{R}{x}{y}$ is a pumping run in
$\mathcal{P}_{>,\varepsilon}$, i.e., a pumping run such that
$\TOP0(R(x))\prec\TOP0(R(y))$ and all edges of
$\subrun{R}{x}{y}$ are labelled by $\varepsilon$, and
\item $\subrun{R}{y}{\lvert R \rvert}$ is a $\TOP{0}$-non-erasing
run ending with a non-$\varepsilon$-transition.
\end{enumerate}
\end{proposition}
\begin{proof}
Let $q$ be the state of $R(y)$.
We apply Proposition \ref{prop:types1} to $\subrun{R}{x}{y}$
and obtain infinitely many $\varepsilon$-labelled runs
$(R_i)_{i\in\ensuremath{\mathbb{N}}}$ from $R(x)$ to $c_i=(q,s_i)$ such that $\TOP0(s_i)\prec\TOP0(s_{i+1})$ for all $i\in\ensuremath{\mathbb{N}}$.
Now we apply Proposition \ref{prop:types2} to
$\subrun{R}{y}{\lvert R \rvert}$ and to $c_i$. We obtain a
$\TOP{0}$-non-erasing run
$S'_i$ from $c_i$.
It ends in the same state as $R$ whence it ends with a
non-$\varepsilon$-transition.
Let $S_i$ be the prefix of $S'_i$ which ends after the first
occurrence of a non-$\varepsilon$-transition.
Let $z\leq x$ be maximal such that $R(z)$ corresponds to a node of $\mathcal{G}$.
Then $U_i:=\subrun{R}{z}{y} \circ R_i \circ S_i$ connects $R(z)$
to one of its successors in $\mathcal{G}$ whose stack $t_i$ contains the
position $\TOP{0}(s_i)$.
Since $t_i$ contains only finitely many positions, and the
$(\TOP{0}(s_j))_{j\geq i}$ form an infinite sequence
of pairwise distinct positions, for each $i$ there is a $j\geq i$ such that
$\TOP{0}(s_j)$ is no position in $t_i$. This immediately implies
$t_i\neq t_j$. By induction, we conclude that the $U_i$ connect
$R(z)$ with infinitely many pairwise different successors in $\mathcal{G}$
whence $\mathcal{G}$ is infinitely branching at $R(z)$. \qed
\end{proof}
Now we are prepared to prove that in each finitely branching
$\varepsilon$-contraction of a collapsible pushdown system the stack
sizes grow only in a bounded manner from each node to its successors.
For the combinatorial part in the proof we use the
sequences from Definition \ref{def:SequencesMandN} without reference.
\begin{lemma}\label{lem:Bounded-Stack-Growth}
Let $\mathcal{S}$ be a CPS of level $n$ such that
the $\varepsilon$-contraction $\mathcal{G}$ of its configuration graph
is finitely branching and such that
transitions leading to some state $q$ are either all
$\varepsilon$-transitions or all
non-$\varepsilon$-transitions.
Set $c:=\lvert \mathcal{T}_\mathcal{S}\rvert+1$.
Let $R$ be a run starting in the initial configuration
whose last edge is not labelled by $\varepsilon$
and which corresponds to a path of length $m$ in $\mathcal{G}$.
The size of every $k$-stack of $R(\lvert R \rvert)$
is at most $M_k$ for all
$1\leq k\leq n$.
\end{lemma}
\begin{proof}
The proof is by induction on $m$. For $m=0$, the claim is trivial
(because $c\geq 2$ and the initial stack of any level has size
$1$).
Assume that we have proven the claim for all paths of length below
$m$ and assume that $R$ describes a path of length $m$ in $\mathcal{G}$.
Let $R(b)$ correspond to the $(m-1)$-st node of $\mathcal{G}$ on this path and
set $S:=\subrun{R}{b}{\lvert R\rvert}$.
Heading for a contradiction assume that $p$ is the position of a
$k$-stack in $R(\lvert R \rvert)$ such that the size of this stack
is greater than $M_k$.
For $0\leq i\leq \lvert S \rvert$, let $a_i$ be the size of the
$k$-stack of $S(i)$ at $\hist{\subrun{S}{i}{\lvert S \rvert}}{p}$.
By induction hypothesis, the size of any $k$-stack
of $S(0)$ is bounded by $M'_k$.
Thus, we have $a_{\lvert S \rvert}> M_k$ and $a_0\leq M_k'$.
Let $G\subseteq\{0, 1, \dots, \lvert S \rvert-1\}$ contain all
elements $i$ such that $a_i < a_j$ for all $i<j\leq \lvert S \rvert$.
Since $ a_i - a_{i-1} \leq 1$ for $1\leq i\leq \lvert S\rvert$,
for each $i$ in $\{M_k', M_k'+1, \dots, M_k\}$ we have an index $j$
such that $a_j=i$ and $a_j\in G$.
Using Lemma \ref{lem:MminusMPrimeGeqNPrime} we conclude that
$\lvert G \rvert \geq M_k - M_k' \geq N_{k-1}'$. Since $M_k'$ is a
bound on the sizes of $k$-stacks in $S(0)$, it follows that $G$ is
big enough in order to apply Lemma \ref{lem:Pumpability} for $k-1$.
We want to apply this lemma to the run
$T:=\subrun{S}{0}{\max(G)+1}$.
In order to satisfy the requirements of this lemma, we have to prove
that
$\hist{\subrun{T}{g}{\lvert T \rvert}}{\TOP{k-1}(T(\lvert T \rvert))}
= \TOP{k-1}(T(g))$
for all $g\in G$.
Choose $g\in G$ arbitrarily.
Since $a_{g+1}>a_g$, the $k$-stack at
$\hist{\subrun{S}{g}{\lvert S \rvert}}{p}$
is smaller than that at
$\hist{\subrun{S}{g+1}{\lvert S \rvert}}{p}$.
Due to Proposition \ref{prop:NonTopmostStacksDoNotChange},
this requires that
\begin{align*}
&\hist{\subrun{S}{g}{\lvert S \rvert}}{p}=\TOP{k}(S(g)),\quad \text{and}\\
&\hist{\subrun{S}{g+1}{\lvert S \rvert}}{p}=\TOP{k}(S(g+1)).
\end{align*}
Especially, $\hist{\subrun{S}{\lvert T \rvert}{\lvert S \rvert}}{p}=\TOP{k}(T(\lvert T \rvert))$ whence
$a_i$ for $i\leq \lvert T \rvert$ is the size of the stack at
$\hist{\subrun{T}{i}{\lvert T \rvert}}{\TOP{k}(T(\lvert T
\rvert))}$.
We conclude that for all $g\in G$ we have
\begin{align*}
\hist{\subrun{T}{g}{\lvert T \rvert}}{\TOP{k}(T(\lvert T \rvert))}=\TOP{k}(T(g)).
\end{align*}
Furthermore, for each $i> g$ we have $a_g < a_i$
whence we can apply
Lemma \ref{lem:TopKandIncreasingSequencegiveTopk-1} to $x:=\TOP
k(T(|T|))$ and to the run $\subrun{R}{g}{|T|}$ obtaining
that
\begin{align*}
\hist{\subrun{T}{g}{\lvert T \rvert}}{\TOP{k-1}(T(\lvert T
\rvert))}=\TOP{k-1}(T(g)).
\end{align*}
Application of Lemma \ref{lem:Pumpability} to $T$ and $k-1$
yields indices $0\leq x < y < z \leq \lvert T \rvert$ such that
\begin{enumerate}
\item $\mathsf{ctype}_\mathcal{X}(S(x))=\mathsf{ctype}_\mathcal{X}(S(y))$,
\item $\subrun{S}{x}{y}$ is a pumping run,
\setcounter{enumi}{3}
\item \label{prf:bounded-stack-growth-Enum-4}
$\TOP0(S(x))\neq\TOP0(S(y))$ or
\begin{align*}
G\cap\{x,x+1,\dots,y-1\} \neq \emptyset,
\end{align*}
\item $z-1\in G$, and
\item \label{prf:bounded-stack-growth-Enum-6}
$\subrun{S}{y}{z}$ is a $\TOP0$-non-erasing run.
\end{enumerate}
By definition of $G$, for $g\in G$ and $g<y$ we have $a_g<a_y$.
In other words the size of the $k$-stack in $S(y)$ at
$\hist{\subrun{S}{y}{|S|}}{p}$ is greater than that of the $k$-stack in
$S(g)$ at $\hist{\subrun{S}{g}{|S|}}{p}$.
Thus, if there is a $g\in G\cap\{x,x+1,\dots,y-1\}$,
application of Lemma \ref{lem:tech4} to $\subrun{S}{x}{y}$ shows that
$\TOP0(S(x))\neq\TOP0(S(y))$.
In the light of Property \ref{prf:bounded-stack-growth-Enum-4}), we
always have
$\TOP0(S(x))\neq\TOP0(S(y))$.
Since $z-1\in G$, we have $a_i>a_{z-1}$ for all $z\leq i\leq|S|$.
Application of Lemma \ref{lem:ForNotEras} to $\subrun{S}{z-1}{|S|}$
shows that $\subrun{S}{z-1}{|S|}$ is a $\TOP0$-non-erasing run.
Since Property \ref{prf:bounded-stack-growth-Enum-6}) implies that
$\subrun{S}{y}{z-1}$ is
$\TOP{0}$-non-erasing, we conclude using
Proposition
\ref{prop:topk-non-eras-composition}
that
$\subrun{S}{y}{\lvert S \rvert}$ is $\TOP0$-non-erasing run.
Recall that the last edge of $S$ is the only edge which is not
labelled $\varepsilon$.
Thus the assumptions of
Proposition \ref{prop:inf-branching} are satisfied by
$S$, $x$ and $y$ whence the lemma yields that
$\mathcal{G}$ is infinitely branching.
This contradicts our assumption.
Thus, we conclude that every $j$-stack in $R(\lvert R \rvert)$
has size bounded by $M_j$. \qed
\end{proof}
\subsection{Proof of the Pumping Lemma}\label{ssec:pumping-proof}
Having bounded the size of stacks in finitely branching
$\varepsilon$-contractions of pushdown graphs, we can prove the main
theorem.
Note that -- doubling the number of states of the system -- we can enforce
that for each $\varepsilon$-transition $\delta_1$ and each
non-$\varepsilon$-transition $\delta_2$, $\delta_1$ leads to a
different state than $\delta_2$.
Having obtained this condition the proof of the main theorem
follows from the following theorem.
\begin{theorem}
Let $\mathcal{S}$ be a CPS of level $n$ such that the $\varepsilon$-contraction $\mathcal{G}$ of its graph is finitely branching
and such that for each state $q$ transitions leading to state $q$ are all $\varepsilon$-transitions or are
all non-$\varepsilon$-transitions.
Let $c_m$ be some configuration of distance $m$ from the initial
configuration.
Let $S_1=(m+1)\cdot C_{\mathcal{S}}$ and $S_j=2^{S_{j-1}}$ for
$2\leq j\leq n$, where $C_{\mathcal{S}}=3 \cdot c \cdot 2^c$ with
$c=\lvert \mathcal{T}_\mathcal{S} \rvert +1$.
Assume also that in $\mathcal{G}$ there exists a path $p$ of length at
least $S_n$ which starts in $c_m$.
Then there are infinitely many paths in $\mathcal{G}$ which start in
$c_m$ and end in configurations having the same state as the last
configuration of $p$.
\end{theorem}
\begin{proof}
From Definition \ref{def:SequencesMandN} we obtain sequences
$M_i$ and $N_i$. Note that the
sequence $S_i$ defined in this lemma and the sequence $S_i$ defined
in that definition agree.
Due to the existence of $p$, there is a run
$R$ starting in $c_m$ such that $S_n$ transitions in $R$ are not
labelled by $\varepsilon$ and especially the last transition is not
labelled $\varepsilon$.
Let $G$ be the set of those $0\leq i < \lvert R \rvert$ such that
the transition between $R(i)$ and $R(i+1)$ is not labelled $\varepsilon$.
Since $\mathcal{S}$ is of level $n$, for any configuration $c'$ of $\mathcal{S}$ the only position of an $n$-stack in $c'$ is $\TOP{n}(c')=(0,0, \dots, 0)$.
Especially, every $g\in G$ satisfies
$\hist{\subrun{R}{g}{\lvert R\rvert}}{\TOP{n}(R(\lvert R\rvert))}=\TOP{n}(R(g))$.
Furthermore, we saw in
Lemma \ref{lem:Bounded-Stack-Growth} that $M_i$ is an upper bound
for the size of each
$i$-stack in $c_m$ for each $1\leq i\leq m$.
Thus, Lemma \ref{lem:CombinatoricSandN} implies that
$\lvert G \rvert = S_n\geq 3 N_n > \hat N_n$
and we can apply
Lemma \ref{lem:Pumpability} to $R$.
We obtain numbers $0\leq x<y<z\leq \lvert R \rvert$ such that
\begin{enumerate}
\item $\mathsf{ctype}_\mathcal{X}(R(x))=\mathsf{ctype}_\mathcal{X}(R(y))$,
\item $R_1:=\subrun{R}{x}{y}$ is a pumping run,
\setcounter{enumi}{3}
\item $\TOP0(R(x))\neq\TOP0(R(y))$ or
\begin{align*}
G\cap\{x,x+1,\dots,y-1\} \neq \emptyset,
\end{align*}
\item $z-1\in G$, and
\item $R_2:=\subrun{R}{y}{z}$ is a $\TOP{0}$-non-erasing run.
\end{enumerate}
$G\cap \{x, x+1, \dots, y_1\} =\emptyset$ is equivalent to saying
that all labels in $R_1$ are $\varepsilon$. Moreover, since $\TOP0(R(x))\neq\TOP0(R(y))$ in this case, we conclude that
$R_1\in \mathcal{P}_{>, \varepsilon}$.
As $z-1\in G$, $R_{y,z}$ ends by a non-$\varepsilon$-transition.
Thus, Proposition \ref{prop:inf-branching} implies that $\mathcal{G}$ is
infinitely branching which contradicts our assumptions.
Thus, $R_1$ contains at least one edge with a label different from
$\varepsilon$.
Due to Proposition \ref{prop:types1},
there are runs $(S_i)_{i\in\ensuremath{\mathbb{N}}}$
such that
\begin{itemize}
\item $S_i$ starts in $R(x)$,
\item contains at least $i$ transitions whose
label is not $\varepsilon$ and
\item $\mathsf{ctype}_\mathcal{X}(R(x)) \typOrd
\mathsf{ctype}_\mathcal{X}(S_i(|S_i|))$.
\end{itemize}
Let $T_i$ be the copy of $\subrun{R}{y}{\lvert R \rvert}$ obtained
by application of
Proposition \ref{prop:types3} starting in $S_i(|S_i|)$.
Then $U_i:=\subrun{R}{0}{x}\circ S_i \circ T_i$ is a run from
$c_m$ to $e_i:=T_i(\lvert T_i \rvert)$ that contains at least $i$
non-$\varepsilon$ labelled edges. Furthermore, the state of $e_i$
is the final state of $R$. Due to our assumption on the pushdown
system, this state determines whether the edge to $e_i$ is labelled
$\varepsilon$. Since the last edge of $R$ is not labelled
$\varepsilon$, the edge to $e_i$ is not labelled $\varepsilon$,
whence $e_i$ is a node in $\mathcal{G}$.
Thus, $U_i$ induces a path of length at least $i$ starting in
$c_m$ and ending in a configuration with the same state as the
final configuration of $p$. \qed
\end{proof}
\subsection{Positions and Histories of Stacks}
\label{sec:PosAndHist}
In this section we first introduce \emph{positions} of $i$-stacks in a
$k$-stack for $i\leq k$. These positions allow to access each substack
contained in a stack.
Afterwards we introduce the \emph{history function}.
Given a run $R$ and a position $x$ in the final stack of the run, this
function determines the origin of this position in the first stack of
$R$, i.e., it returns a position $y$ such that the stack at position
$x$ in the last stack of $R$ was created from the stack at position
$y$ in the first stack of $R$.
For a $k$-stack $s$ let us denote by $\lvert s \rvert$ its \emph{size}, i.e. the number of $(k-1)$-stacks $s$ consists of.
\begin{definition}
For each stack $s$ of level $k$ (where $1\leq k\leq n$)
we define the set of positions in $s$ as follows.
If $k=1$, a \emph{simple position} in $s$
is a number $x^1\in\ensuremath{\mathbb{N}}$ such that $x^1\leq \lvert s \rvert$.
If $k\geq 2$, a \emph{simple position} in $s$ is either a tuple $(0,\dots,0)$ of length $k$, or
a tuple $(x^k,\dots,x^1)$ where $1\leq x^k\leq \lvert s \rvert$
and $(x^{k-1},\dots,x^1)$ is a simple position in the $x^k$-th $(k-1)$-stack of $s$ (counted
bottom up).
We say that a simple position $x$ \emph{points to a $k$-stack} if
$k\in\ensuremath{\mathbb{N}}$ is maximal such
that $x$ ends in a sequence of $0$'s of length $k$.
A \emph{position} in $s$ is either a
simple position in $s$ or a sequence
$x:=x_0\posNew{k}{y}$ such that $x_0$ is a simple position pointing
to a $0$-stack $(a,k,t^k)$ in $s$ and $y$ is a position in $t^k$,
but $y \neq (0,\dots,0)$.\footnote{%
We forbid nonsimple positions ending in the simple position
$(0,0, \dots, 0)$ because of the following interpretation.
In a $0$-stack $s^0=(a,k,t^k)$ we actually do not consider $t^k$ to be
a $k$-stack but only the \emph{content} of a $k$-stack.
In this interpretation the application of $\col{k}$ when $s^0$ is
the topmost $0$-stack does not replace the topmost $k$-stack by
$t^k$ but the \emph{content} of the topmost $k$-stack by the
\emph{content} of $t^k$. This difference is only
of syntactical nature but it is useful to exclude such positions
when defining the history function.
}
A position $x$ \emph{points to a $k$-stack} if its rightmost simple
position points to a $k$-stack.
For $x,y$ positions in $s$ we say that $y$ \emph{points into the stack at}
$x$ (abbreviated $y$ \emph{points into} $x$) if
\begin{enumerate}
\item either $x$ points to a level $0$ stack and
$y=x \posNew{k}{z}$ or
\item $x$ points to a level $k\geq 1$ stack, $x$ and $y$ agree on all
entries where $x$ is nonzero, and $y\neq x$, i.e., $y$ extends the
position $x$ where $x$ starts to be constantly $0$.
\end{enumerate}
Let $s$ be some $n$-stack where $s^i$ denotes the topmost $i$-stack
of $s$.
The
\emph{position of the topmost $k$-stack of $s$} is
$\TOP{k}(s):= (\lvert s^n\rvert,\dots,\lvert s^{k+1}\rvert, 0, \dots, 0)$.
Finally, we define the nesting rank of a position. This rank counts the number
of simple positions involved in the position.
Let $\NestingRank(x):=0$ if $x$ is simple, and
$\NestingRank(x\posNew{k}{z}):=1+\NestingRank(x)+\NestingRank(z)$.
\end{definition}
\begin{remark}
We use the notation
$z \posNew{k}{z'}$ where $z$ is a non-simple position of a
$0$-stack that links to a $k$-stack and $z'$ points to some position
inside this linked stack.
\end{remark}
We next introduce the \emph{history function}. This function is
useful for giving semantical characterisations of the sets in the
family $\mathcal{X}$
defined by a grammar in Section \ref{sec:Runs}.
Our intuition of the history function is the following.
$\hist{R}{x}$ is the (unique) position of a $k$-stack in $s$ from
which $R$ created the $k$-stack at $x$ in $t$ in the sense that the
stack at $x$ in $t$ is a (possibly modified) copy of the stack at
$\hist{R}{x}$ in $s$ not only in terms of content but also in the
way it was produced by $R$. Here, a $\push{1}$ is understood as
copying the topmost $0$-stack and then completely replacing its
content.
\begin{definition}
Let $R$ be a run from stack $s$ to stack $t$ and
let $x$ be a position in $t$.
If $\lvert R\rvert=0$, then $\hist{R}{x}:=x$.
If $\lvert R\rvert=1$, we make a case distinction on the operation
performed by
$R$, and on the form of $x$.
\begin{itemize}
\item If $R$ performs a $\push{1}_{a,k}$ operation and
$ x = \TOP{0}(t) = (x^n, \dots, x^1)$,
then
\begin{align*}
\hist{R}{x}:=(x^n,
\dots, x^2,x^1-1).
\end{align*}
\item If $R$ performs a $\push{1}_{a,k}$ operation and $x$ is of the
form $\TOP{0}(t)\posNew{k}{(y^{k}_1,\dots,y^1_1)}$,
then we set
\begin{align*}
\hist{R}{x}:=
(\lvert u^n\rvert, \dots , \lvert u^{k+1} \rvert,
y^{k}_1,\dots, y^1_1)
\end{align*}
for $u^i$ the
topmost $i$-stack of $t$.
\item If $R$ performs a $\push{1}_{a,k}$ operation and $x$ is of the
form $\TOP{0}(t)\posNew{k}{(y^{k}_1,\dots,y^1_1)}\posNew{k'}{z}$,
then
\begin{align*}
\hist{R}{x}:=
(\lvert u^n\rvert, \dots , \lvert u^{k+1} \rvert,
y^{k}_1,\dots, y^1_1)\posNew{k'}{z}
\end{align*}
where $u^i$ is the
topmost $i$-stack of $t$.
\item If $R$ performs a $\push{i}$ operation for $2\leq i \leq n$ and
$x$ is of the form $(x^n, \dots, x^1)$
such that $(x^n, \dots, x^1)$
is $\TOP{i-1}(t)$ or points into $\TOP{i-1}(t)$,
then
\begin{align*}
\hist{R}{x}:=
(x^n, \dots, x^i, x^{i-1}-1, x^{i-2}, \dots, x^1)
\end{align*}
\item If $R$ performs a $\push{i}$ operation for $2\leq i \leq n$ and
$x$ is of the form
$(x^n, \dots, x^1)\posNew{k}{z}$
such that $x$ points into $\TOP{i-1}(t)$,
then
\begin{align*}
\hist{R}{x}:=
(x^n, \dots, x^i, x^{i-1}-1, x^{i-2}, \dots,
x^1)\posNew{k}{z}.
\end{align*}
\item If $R$ performs a $\col{k}$ operation and
$x=(x_0^n,\dots,x_0^1)$
points into\footnote{
Recall that $\TOP{k}(t)$ does \emph{not} point into
$\TOP{k}(t)$.}
$\TOP{k}(t)$,
then
\begin{align*}
\hist{R}{x} := \TOP{0}(s) \posNew{k}{(x_0^k, \dots, x_0^1)}.
\end{align*}
\item If $R$ performs a $\col{k}$ operation and
$x=(x_0^n, \dots, x_0^1)\posNew{k'}{y}$
points into $\TOP{k}(t)$,
then
\begin{align*}
\hist{R}{x} :=\TOP{0}(s)\posNew{k}{(x_0^k, \dots,
x_0^1)}\posNew{k'}{y}.
\end{align*}
\item In all other cases, we set $\hist{R}{x}:=x$.
\end{itemize}
If $\lvert R\rvert\geq 2$, we decompose $R= S\circ T$ where
$\lvert S\rvert = 1$, and we set
$\hist{R}{x}:=\hist{S}{\hist{T}{x}}$.
\end{definition}
Due to the inductive definition of the history function it is
compatible with decomposition of runs in the following sense.
\begin{proposition}\label{prop:histTransitive}
Let $R,S,T$ be runs such that $R=S\circ T$. If $x,y$ are positions
such that
$\hist{T}{x}=y$, then $\hist{R}{x}=z$ if and only if $\hist{S}{y}=z$.
\end{proposition}
\subsection{Related Work}
Hayashi \cite{hayashi-pumping} and Gilman \cite{gilman-pumping}
proved a pumping and a
shrinking lemma for indexed languages. It is shown in
\cite{AehligMO05} that
indexed
languages are exactly the string languages accepted by level $2$
collapsible pushdown systems. For higher levels, no shrinking
techniques are known so far.
Since our pumping lemma can be used only for finitely branching systems,
it cannot be used to show that certain string languages do not occur on certain
levels of the (collapsible) higher-order pushdown hierarchy.
Note that we do not know whether the string languages accepted by
nondeterministic level $n$ pushdown systems and by nondeterministic
level $n$ collapsible pushdown systems coincide for $n>2$.
Thus, it
is an interesting open question whether there is a stronger pumping lemma for runs of higher-order systems that could be used to separate
these classes of string languages.
\subsection{Characterisation of Returns and Colreturns}
We start with a definition of returns.
In Lemmas \ref{lem:return-wfrules} and \ref{lem:Change-level-Return}
we later see that the grammar from Section \ref{sec:Runs}
correctly describes the sets of returns.
\begin{definition}\label{def:return}
A run $R$ of length $m$ is called \emph{$k$-return} (where $1\leq
k\leq n$) if
\begin{itemize}
\item $\hist{R}{\TOP{k-1}(R(m))}$ points to the second topmost
$(k-1)$-stack\footnote{
Whenever we write ``the second topmost $(k-1)$-stack'' we
assume that it is in the same $k$-stack as the topmost
$(k-1)$-stack, i.e., we assume that the topmost $k$-stack has
size at least $2$.
} in the topmost $k$-stack of $R(0)$, and
\item $\hist{\subrun{R}{i}{m}}{\TOP{k-1}(R(m))}\neq\TOP{k-1}(R(i))$
for all $1\leq i \leq m-1$.
\end{itemize}
\end{definition}
The following propositions confirm our intuition about $k$-returns.
\begin{proposition}\label{prop:return-ends-k}
The last operation of a $k$-return $R$ is $\pop k$ or $\col k$.
\end{proposition}
\begin{proof}
Let $m:=\lvert R \rvert$.
Note that in order to satisfy
\begin{align*}
\hist{\subrun{R}{m-1}{m}}{\TOP{k-1}(R(m))}\neq\TOP{k-1}(R(m-1))
\end{align*}
the
last operation of $R$ is $\pop{j}$ or $\col{j}$ with $j\geq k$.
Heading for a contradiction assume that $j>k$.
It follows that
\begin{align*}
\hist{\subrun{R}{m-1}{m}}{\TOP{k}(R(m))}\neq \TOP{k}(R(m-1)).
\end{align*}
Let $i\leq m-2$ be maximal such that
\begin{align*}
x_{i}:=\hist{\subrun{R}{i}{m}}{\TOP k(R(m))}=\TOP k(R(i))
\end{align*}
Such $i$ exists because $i=0$ is of this form (cf.\ Corollary
\ref{cor:positionContainment}).
Due to Corollary \ref{cor:NonTopmostStacksDoNotChange}
(variant \ref{ntsdnc-varb}),
$\hist{\subrun{R}{i}{m}}{\TOP{k-1}(R(m))}$
points to the topmost $(k-1)$-stack of the $k$-stack to which
$x_{i}$ points. Thus, it points to the topmost $(k-1)$-stack of $R(i)$.
But this contradicts the definition of a return.\qed
\end{proof}
\begin{proposition}\label{prop:return-removes}
For every $k$-return $R$, the topmost $k$-stack of $R(0)$ after
removing its topmost $(k-1)$-stack is equal to the topmost $k$-stack
of $R(|R|)$.
If $x$ points into $\TOP{k}(R(\lvert R \rvert)$ then
$\hist{R}{x}$ points to the same position in the stack at
$\TOP{k}(R(0))$.
\end{proposition}
\begin{proof}
Let $m:=\lvert R \rvert$ and
let $l$ be the size of the topmost $k$-stack of $R(m)$. For
$1\leq i\leq l$,
let $t^{k-1}_i$ be the $i$-th $(k-1)$-stack (counting from the
bottom) of the topmost $k$-stack
of $R(m)$. Let $x_i$ be the position pointing to $t^{k-1}_i$.
By the above proposition, $R$ ends in $\pop{k}$ or $\col{k}$.
Note that the histories of $x_1, \dots, x_l$ with respect to
$\subrun{R}{m-1}{m}$ point to $(k-1)$-stacks, where the history of
$x_1$ points to the bottommost $(k-1)$-stack of some $k$-stack,
$x_{i}$ is directly below the history of $x_{i+1}$
for each $1\leq i< l$ and none of these histories point to the
topmost $(k-1)$-stack of $R(m-1)$.
Note that non-topmost $(k-1)$-stacks in the same $k$-stack are always
treated the same way by the history function. Thus,
a simple induction on the operations performed by
$\subrun{R}{0}{m-1}$ shows that this property is preserved by the
history function, i.e.,
$\hist{R}{x_1}, \hist{R}{x_2}, \dots, \hist{R}{x_l}$ point to the
first $l$ $(k-1)$-stacks of a $k$-stack. But by definition
$\hist{R}{x_l}=\hist{R}{\TOP{k-1}(R(m))}$ is the second topmost
$(k-1)$-stack of $\TOP{k}(R(0))$.
Application of Corollary \ref{cor:NonTopmostStacksDoNotChange} (variant \ref{ntsdnc-vara}) shows
that the $(k-1)$-stack at $x_i$ in $R(m)$ is
the same as the $(k-1)$-stack at $\hist{R}{x_i}$ in $R(0)$.
This proves the first part of this proposition.
Similarly, Corollary \ref{cor:NonTopmostStacksDoNotChange} implies the
preservation of pointers into
$\TOP{k}(R(m))$ which completes the proof.\qed
\end{proof}
\begin{corollary}\label{cor:push+return}
For every run $R$ which starts with a $\push k$ operation (including
arbitrary $\push 1_{a,l}$ for $k=1$), and continues with a
$k$-return, the topmost $k$-stacks of $R(0)$ and of $R(|R|)$ coincide.
Additionally, if $x$ points into $\TOP{k}(R(\lvert R \rvert)$ then
$\hist{R}{x}$ points to the same position in the stack at
$\TOP{k}(R(0))$.
\end{corollary}
Recall that wf-rules of the form $\Rule{X}{\delta YZ}$
must satisfy
the property that
whenever $R$ is a composition of a one-step run performing transition
$\delta$ with a run from $Y$, then the topmost $\lev(Y)$-stack of
$R(0)$ and $R(\lvert R \rvert)$ are the same.
Notice that in the grammars in Section \ref{sec:Runs} such rules
appear only when $\delta$ performs a $\push{}$ operation of some level
$k$, and $Y$ is a set of $k$-returns.
Since $\lev(X)=k$, the above corollary proves this property.
We now give a definition of $k$-colreturns.
Lemmas \ref{lem:colreturn-wfrules} and \ref{lem:Change-level-Return} show that, for such definitions, the grammar from Section \ref{sec:Runs}
correctly describes the sets of colreturns.
As already mentioned, the intuition of the definition is the following.
A $k$-colreturn is a run whose
last transition is $\col{k}$ from a stack where the topmost symbol is
a copy of the topmost symbol of the first stack.
\begin{definition}
A run $R$ of length $m$ is called \emph{$k$-colreturn} (where $1\leq
k\leq n$) if
\begin{itemize}
\item $\hist{R}{\TOP{k-1}(R(m))}$ is of the form
$\TOP{0}(R(0))\posNew{k}{x}$, where $x$ is simple, and
\item $\hist{\subrun{R}{i}{m}}{\TOP{k-1}(R(m))}$ is not simple for
all $0\leq i\leq m-1$.
\end{itemize}
\end{definition}
We first prove a decomposition result of $k$-returns and
$k$-colreturns into one transition followed by a sequence of shorter
returns or colreturns.
Later we deal with the change levels.
\begin{lemma}\label{lem:return-wfrules}
Let $R$ be some run.
Then $R$ is a $k$-return if and only if $R$ is of one of the
following forms.
\begin{enumerate}
\item $\lvert R\rvert=1$ and $R$ performs $\pop{k}$.
\item $R$ starts with an operation of level at most $k-1$, and
continues with a $k$-return.
\item $R$ starts with $\push1_{a,k}$ and continues with a $k$-colreturn.
\item $R$ starts with a $\push{j}$ for $j>k$ and continues with a $k$-return.
\item $R$ starts with a $\push{j}$ for $j\geq k$ (including $\push1_{a,l}$ for $j=k=1$) and decomposes as
$R=S \circ T \circ U$ where $S$ has length $1$, $T$ is a
$j$-return and $U$ is a $k$-return.
\end{enumerate}
\end{lemma}
\begin{lemma}\label{lem:colreturn-wfrules}
Let $R$ be some run.
Then $R$ is a $k$-colreturn if and only if $R$ is of one of the
following forms
\begin{enumerate}
\item $R$ has length $1$ and performs $\col{k}$.
\item $R$ starts with $\push{j}$ for some $j\geq 2$ and continues
with a $k$-colreturn.
\item $R$ starts with a $\push{j}$ (including $\push1_{a,l}$ for
$j=1$) and decomposes as
$R=S\circ T\circ U$ where $S$ has length $1$, $T$ is a
$j$-return and $U$ is a $k$-colreturn.
\end{enumerate}
\end{lemma}
Before we start the proof of these two lemmas, we state some auxiliary claims.
First, we observe that the history function either manipulates the simple
prefix of a position, or adds a simple prefix, or removes it.
This will be useful while analysing $k$-colreturns.
\begin{proposition} \label{prop:histPreservesLinkNesting}
Let $R$ be some run of length $m$.
Let $x \posNew{k}{y}$ be a position in $R(m)$
such that
\begin{align*}
\NestingRank(\hist{\subrun{R}{i}{m}}{x\posNew{k}{y}}) >
\NestingRank(\hist{R}{y})
\end{align*}
for all $i\leq m$.
Let $x'=\hist{R}{x}$.
Then $\hist{R}{x\posNew{k}{y}}=x'\posNew{k}{y}$ (neither $x$ nor $x'$
have to be simple).
Additionally, the $0$-stack of $R(m)$ at position $x$ is the same as
the $0$-stack of $R(0)$ at position $x'$.
\end{proposition}
\begin{proof}
Induction on $m$.
For $m=1$ we just analyse all cases.
For $m\geq 2$ we observe that the claim for any decomposition
$R= S \circ T$ follows from
the claim for $S$ and for $T$.\qed
\end{proof}
\begin{corollary} \label{cor:PushColStackdifference}
Let $R$ be a run of length $m \geq 2$, and $x$ a simple position in $R(m)$ such that $\hist{R}{x}$ is simple,
but $\hist{\subrun{R}{i}{m}}{x}$ is not simple for $1\leq i\leq m-1$.
Then, for some $k$, $R$ starts with $\push{1}_{a,k}$ and ends with $\col{k}$.
Additionally, $x=\TOP{k-1}(R(m))$ if and only if $\hist{R}{x}$ is the second topmost $(k-1)$-stack of $R(0)$.
\end{corollary}
\begin{proof}
The last operation of $R$ has to be a $\col k$ for some $k$, because
otherwise $\hist{\subrun{R}{m-1}{m}}{x}$ would be simple.
Then $\hist{\subrun{R}{m-1}{m}}{x}$ is of the form $\TOP
0(R(m-1))\posNew{k}{x'}$.
Proposition \ref{prop:histPreservesLinkNesting}, applied for
$\subrun{R}{1}{m-1}$,
shows that $\hist{\subrun{R}{1}{m}}{x}$ is of the form $z\posNew{k}{x'}$,
and that the topmost $0$-stack of $R(m-1)$, and the $0$-stack in
$R(1)$ to which $z$ points to are the same $k$-stack $u^k$, which is
in fact the topmost $k$-stack of $R(m)$.
Because $\hist{\subrun{R}{0}{1}}{z\posNew{k}{x'}}$ is simple,
necessarily $z=\TOP0(R(1))$, the first operation of $R(0)$ is
$\push1_{a,k}$, and
the topmost $k$-stack of $R(0)$ after removing its topmost $(k-1)$-stack is equal to $u^k$.
Additionally, $x$ points to the same position in the topmost
$k$-stack of $R(m)$, as $\hist{R}{x}$ in the topmost $k$-stack of
$R(0)$.
Thus $x=\TOP{k-1}(R(m))$ if and only if $\hist{R}{x}$ is the second topmost $k$-stack of $R(0)$.\qed
\end{proof}
The following lemma proves the intuition that $k$-colreturns make a
copy of the topmost stack symbol and finally use its collapse link of
level $k$ (the proof is almost the same as that of the previous corollary).
\begin{lemma}\label{lem:ColreturnDesreases}
Let $R$ be a colreturn.
Then the topmost $k$-stack of $R(|R|)$ is equal to the $k$-stack contained in the topmost $0$-stack of $R(0)$.
In particular its size is smaller than the size of the topmost $k$-stack of $R(0)$.
Additionally, the last operation of $R$ is $\col k$, and
$\hist{R}{\TOP k(R(|R|))}=\TOP k(R(0))$.
\end{lemma}
\begin{proof}
Let $m:=|R|$, and let $x:=\TOP{k-1}(R(m))$.
The last operation of $R$ has to be a $\col j$ for some $j$, because
otherwise $\hist{\subrun{R}{m-1}{m}}{x}$ would be simple.
Then $\hist{\subrun{R}{m-1}{m}}{x}$ is of the form $\TOP
0(R(m-1))\posNew{j}{x'}$.
Proposition \ref{prop:histPreservesLinkNesting}, applied for
$\subrun{R}{0}{m-1}$, implies that
$\hist{R}{x}=\hist{\subrun{R}{0}{m-1}}{\TOP0(R(m-1))}\posNew{j}{x'}$.
By definition of a $k$-colreturn it follows that $j=k$ and
$\hist{\subrun{R}{0}{m-1}}{\TOP0(R(m-1))}=\TOP0(R(0))$.
From this proposition we also conclude that the topmost $0$-stack of
$R(0)$ and the topmost $0$-stack of $R(m-1)$ store the same
$k$-stack $u^k$, which is in fact the topmost $k$-stack of $R(m)$.
Of course the size of $u^k$ is smaller than the size of the the
topmost $k$-stack of $R(0)$
because $\col k$ must decrease the size of the topmost $k$-stack
(see Remark \ref{rem:ColisPop}).
Corollary \ref{cor:positionContainment} implies that
$\hist{\subrun{R}{0}{m-1}}{\TOP k(R(m-1))}=\TOP k(R(0))$.
Since the last operation is $\col k$,
$\hist{R}{\TOP k(R(m-1))}=\TOP k(R(0))$. \qed
\end{proof}
The next two propositions describe which operations are allowed as the
first operation of a $k$-return and of a $k$-colreturn.
\begin{proposition}
\label{prop:ReturnsStartwithPushOrSmallLevel}
Let $R$ be a $k$-return.
The first operation of $R$ is neither $\col{j}$ for $j\geq k$ nor
$\pop{j}$ for $j>k$.
If the first operation of $R$ is $\pop{k}$, then $\lvert R \rvert=1$.
\end{proposition}
\begin{proof}
Let $m:=|R|$.
If the first operation of a run $R$ is $\col{j}$, $j\geq k$ then by
definition of the history function $\hist{\subrun{R}{0}{1}}{x}$ does
not point to any simple position inside $\TOP{j}(R(0))$ for all positions
$x$ in $R(1)$.
Thus, also $\hist{R}{x}$ does not point to any simple position inside
$\TOP{j}(R(0))$ for all positions $x$ in $R(m)$.
But if $R$ is a $k$-return, $\hist{R}{\TOP{k-1}(R(m))}$ is a simple position and points
into $\TOP{k}(R(0))$ whence it also points into $\TOP{j}(R(0))$.
Analogously, one shows that $R$ does not start with $\pop{j}$ for $j>k$.
If a $k$-return $R$ starts with $\pop{k}$, it follows that
$\hist{\subrun{R}{1}{m}}{\TOP{k-1}(R(m))}=\TOP{k-1}(R(1))$.
But this is not allowed if $1\leq m-1$. \qed
\end{proof}
\begin{proposition}
\label{prop:ColreturnsStartwithPush}
Let $R$ be a $k$-colreturn.
The first operation of $R$ is a $\push{}$ or $\col k$.
If the first operation of $R$ is $\col{k}$, then $\lvert R \rvert=1$.
\end{proposition}
\begin{proof}
Let $m:=|R|$.
If the first operation of a run $R$ is a $\pop{}$ then by
definition of the history function $\hist{\subrun{R}{0}{1}}{x}$ does
not point into $\TOP{0}(R(0))$ for all positions $x$ in $R(1)$.
Thus also $\hist{R}{x}$ does not point into $\TOP{0}(R(0))$ for all
positions $x$ in $R(m)$, in particular for
$x=\hist{R}{\TOP{k-1}(R(m))}$. This contradicts the definition of a
$k$-colreturn.
In $R(0)$ we have a position $\TOP{0}(R(0))\posNew{k}{x}$. Thus, the
only collapse operation which can be performed at $R(0)$ is a level
$k$ collapse, i.e., $\col k$.
If $R$ starts with $\col k$, then
$\hist{\subrun{R}{0}{1}}{y}=\TOP{0}(R(0))\posNew{k}{x}$ for some simple
$x$ only if $y$ is simple.
We conclude that $\hist{\subrun{R}{1}{m}}{\TOP{k-1}(R(m))}$ is simple
which implies $m=1$.\qed
\end{proof}
We state a last auxiliary lemma and then prove Lemmas
\ref{lem:return-wfrules} and \ref{lem:colreturn-wfrules}.
\begin{lemma} \label{lem:SecondReturnHistory}
Let $R$ be a $k$-return of length at least $2$.
Let $x$ be the position $\hist{\subrun{R}{1}{\lvert R
\rvert}}{\TOP{k-1}(R(\lvert R \rvert))}$. Then one of the
following holds.
\begin{enumerate}
\item $x$ points to the second topmost $(k-1)$-stack of $R(1)$ and
the first stack operation of $R$ is of level strictly below $k$ or
a $\push{j}$ for $j>k$.
\item $x$ points to the third topmost $(k-1)$-stack of $R(1)$ and
the first stack operation of $R$ is a push of level $k$.
\item There is a $j>k$ such that $x$ points to the second topmost
$(k-1)$-stack of the second
topmost $(j-1)$-stack of $R(1)$ and the first stack operation of
$R$ is $\push{j}$.
\item $x=\TOP0(R(1))\posNew{k}{\TOP{k-1}(u^k)}$, the first stack
operation of $R$ is $\push{1}_{a,k}$ and the topmost $0$-stack of
$R(1)$ is $(a,k,u^k)$.
\end{enumerate}
\end{lemma}
\begin{proof}
Since $R$ is a $k$-return, Proposition \ref{prop:histTransitive}
implies that
$\hist{\subrun{R}{0}{1}}{x}$ points to the second topmost
$(k-1)$-stack of $R(0)$.
We proceed by case distinction on the stack operation of
$S:=\subrun{R}{0}{1}$.
Due to Proposition \ref{prop:ReturnsStartwithPushOrSmallLevel} we
only have to consider the following cases.
\begin{itemize}
\item
Assume that $S$ performs a $\pop{j}$ operation for $j<k$, or a
$\col{j}$ operation for $j<k$, or a $\push{j}$ operation for
$2\leq j<k$, or a $\push1_{a,j}$ operation for $j\neq k>1$.
Then $x$ necessarily points to the second topmost $(k-1)$-stack of
$R(1)$ (because $S$ makes changes only inside the topmost
$(k-1)$-stack of $R(0)$).
\item
Assume that $S$ performs a $\push{k}$ for $k\geq 2$, or a
$\push1_{a,j}$ for $j\neq k=1$.
Then $x$ necessarily points to the third topmost $(k-1)$-stack of $R(1)$.
\item
Assume that $S$ performs a $\push{j}$ operation with $j > k$.
Then either $x$ points to the second topmost $(k-1)$-stack of
$R(1)$, or to the second topmost $(k-1)$-stack of the second
topmost $(j-1)$-stack of $R(1)$.
\item
Assume that $S$ performs $\push{1}_{a,k}$, and $k\geq 2$.
Then either $x$ points to the second topmost $(k-1)$-stack of $R(1)$,
or $x$ is of the form $\TOP0(R(1))\posNew{k}{\TOP{k-1}(u^k)}$
where $u^k$ is the $k$-stack stored in the topmost $0$-stack of
$R(1)$.
\item
Assume that $S$ performs $\push{1}_{a,k}$, and $k=1$.
Then either $x$ points to the third topmost $(k-1)$-stack of $R(1)$,
or $x$ is of the form $\TOP0(R(1))\posNew{k}{\TOP{k-1}(u^k)}$
where $u^k$ is the $k$-stack stored in the topmost $0$-stack of
$R(1)$. \qed
\end{itemize}
\end{proof}
Now we are prepared to prove Lemma \ref{lem:return-wfrules}, i.e., the
decomposition of returns into one transition followed by shorter
returns or colreturns.
\begin{proof}[of Lemma \ref{lem:return-wfrules}]
We first show that every return decomposes as required.
Let $R$ be a $k$-return of length $m$ (by definition $m\geq 1$). Set
$S:=\subrun{R}{0}{1}$.
When $S$ performs a $\pop{k}$ operation, and $m=1$, we immediately
get case one.
Otherwise,
we proceed by distinction of the cases of Lemma
\ref{lem:SecondReturnHistory} for the position
$x:=\hist{\subrun{R}{1}{|R|}}{\TOP{k-1}(R(m))}$.
\begin{itemize}
\item
Assume that $x$ points to the second topmost $(k-1)$-stack of $R(1)$.
Then $\subrun{R}{1}{m}$ is easily seen to be a $k$-return; we get
case 2 or case 4.
\item
Assume that $x$ points to the third topmost $(k-1)$-stack of $R(1)$.
Then the operation in $S$ was $\push{k}$ (or $\push{1}_{a,j}$ for $k=1$).
\begin{claim}
There is some $1<i<m$ such that
$\hist{\subrun{R}{i}{m}}{\TOP{k-1}(R(m))}$ points
to the second topmost $(k-1)$-stack of $R(i)$.
\end{claim}
Under the assumption that this claim holds,
choose the minimal such $i$.
From the choice of $i$ it follows immediately that
$U:=\subrun{R}{i}{m}$ is a $k$-return.
We show that $T:=\subrun{R}{1}{i}$ is also a $k$-return whence
$R$ decomposes as in case 5 of the lemma.
Indeed, $\hist{T}{\hist{U}{\TOP{k-1}(R(m))}}$ is the third
topmost $(k-1)$-stack of $R(1)$.
Since $\TOP{k-1}(R(i))$ is the $(k-1)$-stack directly on top of
$\hist{U}{\TOP{k-1}(R(i))}$,
and because $\hist{\subrun{R}{j}{m}}{\TOP{k-1}(R(m))}$ is not the
topmost $(k-1)$-stack of $R(j)$ for $0\leq j<m$ (definition of
$k$-return),
we can apply Proposition \ref{prop:Neihbour-position-lemma}
(variant \ref{npl-vara}) and
conclude that
$\hist{T}{\TOP{k-1}(R(i))}$ is the second topmost $(k-1)$-stack
of $R(1)$.
By the same proposition, if $\hist{T'}{\TOP{k-1}(R(i))}$ is the topmost
$(k-1)$-stack of $T'(0)$ for some proper suffix $T'$ of $T$, then
$\hist{T'\circ U}{\TOP{k-1}(R(m))}$ is the second topmost
$(k-1)$-stack of $T'(0)$ which contradicts the minimality of
$i$.
Thus, $T$ is a $k$-return and we showed that $R$ decomposes as
described in case $5$ of the lemma. We conclude this case by
proving the claim.
\begin{proof}[of Claim]
If the operation leading to $R(m)$ is $\pop k$, $i=m-1$ is a good
candidate.
Otherwise (Proposition \ref{prop:return-ends-k}), this operation
is $\col k$.
Then $\hist{\subrun{R}{m-1}{m}}{\TOP{k-1}(R(m))}$ is not simple.
Let $i<m-1$ be the last index for which
$\hist{\subrun{R}{i}{m}}{\TOP{k-1}(R(m))}$ is simple again (such
$i$ exists because $i=0$ is a good candidate).
From Corollary \ref{cor:PushColStackdifference}, applied to
$\subrun{R}{i}{m}$, we immediately obtain that
$\hist{\subrun{R}{i}{m}}{\TOP{k-1}(R(m))}$ is the second topmost
$(k-1)$-stack of $R(i)$,
so $i$ is a good candidate.
\end{proof}
\item
Assume that $S$ performs a $\push{j}$ operation with $j > k$, and
$x$ points to the second topmost $(k-1)$-stack of the second
topmost $(j-1)$-stack of $R(1)$.
Let $1< i \leq m$ be minimal such that
for $T:=\subrun{R}{1}{i}$ and $U:=\subrun{R}{i}{m}$
we have $\hist{U}{\TOP{j-1}(R(m))}=\TOP{j-1}(R(i))$.
We show that $T$ is a $j$-return and $U$ is a $k$-return whence we
are in case 5.
Due to the minimality of $i$ all proper suffixes $T'$ of $T$
satisfy the inequality
$\hist{T'}{\TOP{j-1}(T(\lvert T \rvert))}\neq \TOP{j-1}(T'(0))$.
Due to Corollary \ref{cor:positionContainment}, the
position $\hist{T \circ U}{\TOP{k-1}(R(m))}$ points
into
\begin{equation*}
\hist{T}{\TOP{j-1}(R(i))}=\hist{T\circ U}{\TOP{j-1}(R(m))}.
\end{equation*}
Thus, $\hist{T}{\TOP{j-1}(R(i))}$ points to the second topmost
$(j-1)$-stack of
$T(0)=R(1)$ and we conclude that $T$ is a $j$-return.
Due to Corollary \ref{cor:positionContainment}, we know that
$\hist{U}{\TOP{k-1}(R(m))}$ points into
\begin{equation*}
\hist{U}{\TOP{j-1}(R(m))}=\TOP{j-1}(R(i)).
\end{equation*}
On the other hand, by Corollary \ref{cor:push+return}, the only
position $x'$ in the topmost $(j-1)$-stack of $R(i)$ for which
$\hist{\subrun{R}{0}{i}}{x'}$ points to the second topmost
$(k-1)$-stack of $R(0)$
is $x$ pointing to the second topmost $(k-1)$-stack of $R(i)$.
By Proposition \ref{prop:histTransitive} we conclude that
$\hist{U}{\TOP{k-1}(R(m))}$ points to the second
topmost $(k-1)$-stack,
hence $U$ is a $k$-return.
\item
Finally, assume that $S$ performs $\push{1}_{a,k}$, and
$x$ is of the form $\TOP0(R(1))\posNew{k}{\TOP{k-1}(u^k)}$ where
$u^k$ is the $k$-stack stored in the topmost $0$-stack of
$R(1)$.
Let $i>1$ be minimal such that
$\hist{\subrun{R}{i}{m}}{\TOP{k-1}(R(m))}$ is simple.
Recall that $\hist{R}{\TOP{k-1}(R(m))}$ points to the second
topmost $(k-1)$-stack of $R(0)$.
From Corollary \ref{cor:PushColStackdifference}, applied to
$\subrun{R}{0}{i}$, we see that
$\hist{\subrun{R}{i}{m}}{\TOP{k-1}(R(m))}=\TOP{k-1}(R(i))$.
Since $R$ is a return, this implies $i=m$ and
due to the minimality of $i$, we conclude directly that
$\subrun{R}{1}{i}$ is a $k$-colreturn.
\end{itemize}
This concludes the proof that every $k$-return decomposes as required by
the lemma.
It is left to show that every run that decomposes as described by the
lemma is a $k$-return. Let $R$ be some run of length $m$. There are the
following cases.
\begin{enumerate}
\item If $\lvert R \rvert =1$ and it performs $\pop{k}$, the
definition of $\mathsf{hist}$ implies that $R$ is a $k$-return.
\item Assume that $R$ starts with an operation of level at most $k-1$, and
continues with a
$k$-return. Since history preserves positions of $(k-1)$-stacks
under operations of level at most $k-1$, and such operations also preserve the
existence of all $(k-1)$-stacks, the conditions for $R$ being a
return are trivially deduced from the fact that
$\subrun{R}{1}{m}$ is a return.
\item Assume that $R$ starts with $\push{1}_{a,k}$ and continues
with a $k$-colreturn. By definition of a colreturn, the position
$\hist{\subrun{R}{i}{m}}{\TOP{k-1}(R(m))}$ is not simple for all
$1\leq i< m$, whence this position is not $\TOP{k-1}(R(i))$.
Furthermore, $\hist{\subrun{R}{1}{m}}{\TOP{k-1}(R(m))}$ points into $\TOP0(R(1))$ and has nesting rank $1$, so $\hist{R}{\TOP{k-1}(R(m))}$ is simple.
By Corollary \ref{cor:PushColStackdifference} it follows that
$\hist{R}{\TOP{k-1}(R(m))}$ is the second topmost $(k-1)$-stack
of $R(0)$.
\item Assume that $R$ starts with a $\push{j}$ for $j>k$
and continues with a $k$-return. Then we conclude similar to the second case.
\item Assume that $R$ starts with a $\push{j}$ for $j\geq k$
(including
$\push1_{a,l}$ for $j=1$) and decomposes as
$R=S \circ T \circ U$ where $S$ has length $1$, $T$ is a
$j$-return and $U$ is a $k$-return.
We know that $x:=\hist{U}{\TOP{k-1}(R(m))}$ is the
second topmost $(k-1)$-stack of $U(0)$.
Corollary \ref{cor:push+return} applied for run $S\circ T$ implies
that $\hist{R}{\TOP{k-1}(R(m)),R)}$ ($=\hist{S\circ T}{x}$) is
the second topmost $(k-1)$-stack of $R(0)$.
For $j=k$, we know that
$\hist{T'}{\TOP{k-1}(U(0))}\neq \TOP{k-1}(T'(0))$ for every
suffix $T'$ of $T$ of positive length.
We apply Proposition \ref{prop:Neihbour-position-lemma} (variant \ref{npl-varb}) to
$\TOP{k-1}(U(0))$ and $x$ (the second topmost $(k-1)$-stack of
$U(0)$ and obtain
that $\hist{T'\circ
U}{\TOP{k-1}(R(m))}\neq\TOP{k-1}(T'(0))$ for every suffix
$T'$ of $T$.
From this we conclude directly that $R$ is a $k$-return.
For $j>k$, we also see that
$\hist{T'\circ U}{\TOP{k-1}(R(m))}\neq \TOP{k-1}(T'(0))$ for
every suffix
$T'$ of $T$ of positive length.
Indeed, if $\hist{T'\circ U}{\TOP{k-1}(R(m))}=\TOP{k-1}(T'(0))$,
then also $\hist{T'\circ U}{\TOP{j-1}(R(m))}=\TOP{j-1}(T'(0))$
(Corollary \ref{cor:positionContainment})
which is impossible because $T$ is a $j$-return.
Again, it is easy to conclude that $R$ is a $k$-return. \qed
\end{enumerate}
\end{proof}
Similarly, we now prove the decomposition of colreturns into one
transition followed by shorter returns or colreturns.
\begin{proof}[of Lemma \ref{lem:colreturn-wfrules}]
We first show that every $k$-colreturn $R$ decomposes as
described by the lemma.
Set $S:=\subrun{R}{0}{1}$ (the definition of a $k$-colreturn
requires $\lvert R\rvert \geq 1$).
If $R$ performs $\col k$ and $|R|=1$ we are in case 1 of Lemma
\ref{lem:colreturn-wfrules}.
Otherwise, due to Proposition \ref{prop:ColreturnsStartwithPush},
the operation in $S$ is $\push{}$.
As in the return case, we look at
$x:=\hist{\subrun{R}{1}{|R|}}{\TOP{k-1}(R(m))}$.
By definition of a $k$-colreturn,
$\hist{R}{\TOP{k-1}(R(m))}$ is of the form
$\TOP{0}(R(0))\posNew{k}{x'}$
for some simple position $x'$.
By Proposition \ref{prop:histTransitive}, we know that
$\hist{S}{x}=\hist{R}{\TOP{k-1}(R(m))} =
\TOP{0}(R(0))\posNew{k}{x'}$.
By case distinction on the possible $x$ satisfying this equation (in
dependence of the operation performed by $S$,
there are the following possibilities.
\begin{enumerate}
\item
$S$ performs a $\push{j}$ for $j\geq 2$ and
$x=\TOP{0}(R(1))\posNew{k}{x_1}$ for some simple $x_1$.
In this case, it is straightforward to see that $\subrun{R}{1}{m}$ is a
$k$-colreturn.
\item Otherwise,
$S$ performs a $\push{}$ operation of level $j$ and
$x=\TOP{0}(R(0))\posNew{k}{x_1}$ for some simple $x_1$.
Notice that $\TOP{0}(R(0))$ is the topmost $0$-stack of the second topmost $(j-1)$-stack of $R(1)$.
Recall that the last operation of $R$ is $\col k$ whence
$\hist{\subrun{R}{m-1}{m}}{\TOP{k-1}(R(m))}$ points into the
topmost $0$-stack of $R(m-1)$ and has nesting rank $1$.
Let $1<i<m$ be minimal such that
$\hist{\subrun{R}{i}{m}}{\TOP{k-1}(R(m))}$ has nesting rank $1$ and points into the topmost $(j-1)$-stack of $R(i)$.
Let $T:=\subrun{R}{1}{i}$, and let
$\hist{\subrun{R}{i}{m}}{\TOP{k-1}(R(m))}=y\posNew{k'}{y'}$.
Due to Proposition \ref{prop:histPreservesLinkNesting} (applied for run $T$),
we have $k'=k$, and $\hist{T}{y}=\TOP{0}(R(0))$.
Due to Corollary \ref{cor:positionContainment} and since $y$ points
into $\TOP{j-1}(R(i))$, $\hist{T}{\TOP{j-1}(R(i))}$ contains
$\TOP{0}(R(0))$
whence it points to the second topmost $(j-1)$-stack of $R(1)$.
The same corollary and the minimality of $i$ implies that for each
suffix $T'$ of $T$ of length at least $1$ we have
$\hist{T'}{\TOP{j-1}(R(i))}\neq \TOP{j-1}(T'(0))$.
Thus, $T$ is a $j$-return.
Let $U:=\subrun{R}{i}{m}$
We know that $\hist{S\circ T}{y}=\TOP{0}(R(0))$, and that $y$ is
in the topmost $(j-1)$-stack of $R(i)$.
By Corollary \ref{cor:push+return}, the only $y$ satisfying this
is $y=\TOP{0}(R(i))$.
It follows that $U$ is a $k$-colreturn.
\end{enumerate}
It is left to show that every run $R$ that decomposes as described by the
lemma is a $k$-colreturn.
In the first two cases we immediately see that $R$ is a $k$-colreturn.
So assume that $R=S\circ T\circ U$ where $S$ has length $1$ and performs a $\push{j}$ (including $\push1_{a,l}$ for $j=1$),
$T$ is a $j$-return and $U$ is a $k$-colreturn.
Let $m:=|R|$.
As $\hist{U}{\TOP{k-1}(R(m))}$ is of the form
$\TOP0(U(0))\posNew{k}x$ for simple $x$,
by Corollary \ref{cor:push+return} we immediately obtain that $\hist{R}{\TOP{k-1}(R(m))}=\TOP0(R(0))\posNew{k}x$.
In order to prove that $R$ is a $k$-colreturn, we still have to show
that the position
$\hist{\subrun{T}{i}{|T|}}{\TOP0(U(0))\posNew{k}x}$ is not simple for
all $0\leq i\leq |T|$.
Heading for a contradiction, assume that there is a greatest index
$i$ for which the position
$\hist{\subrun{T}{i}{|T|}}{\TOP0(U(0))\posNew{k}x}$ is simple.
Trivially $i<|T|$.
If $i=|T|-1$, the last operation of $T$ has to be $\push1_{a,k}$,
which is impossible in a $j$-return.
So $i\leq |T|-2$.
As $\hist{\subrun{T}{i+1}{|T|}}{\TOP0(U(0))\posNew{k}x}$ is not
simple (maximality of $i$), it has to be of the form
$\TOP0(T(i+1))\posNew{k'}x'$.
Proposition \ref{prop:histPreservesLinkNesting} applied to
$\subrun{T}{i+1}{|T|}$ implies that $k'=k$ and
$\hist{\subrun{T}{i+1}{|T|}}{\TOP0(U(0))}=\TOP0(T(i+1))$.
Due to Corollary \ref{cor:positionContainment}, also
$\hist{\subrun{T}{i+1}{|T|}}{\TOP{j-1}(U(0))}=\TOP{j-1}(T(i+1))$.
This is impossible because $T$ is a $j$-return.
\end{proof}
Up to now, we have only dealt with the general shape of $k$-returns
and $k$-colreturns. In Section \ref{sec:Runs} we divided these sets
further according to their \emph{change level}. We next formally
introduce this change level for every $k$-return and $k$-colreturn and
then complete the proof that the rules given in Section \ref{sec:Runs} correctly describe returns and colreturns.
The
\emph{change level} keeps track of the maximal level on
which the stack size was changed by the run $R$.
\begin{definition}
Let $R$ be a $k$-return or a $k$-colreturn.
For $1\leq i\leq n$, let $x^i$ be the size of the topmost $i$-stack
of $R(0)$; similarly $y^i$ for $R(|R|)$.
Then $\changelev(R):=\max\{ i: x^i\neq y^i\}$.
\end{definition}
\begin{remark}\label{rem:chl-big}
If $R$ is a $k$-return ($k$-colreturn), then Proposition
\ref{prop:return-removes} (Lemma \ref{lem:ColreturnDesreases},
respectively) implies that
the size of topmost $k$-stack of $R(0)$ and of $R(|R|)$ is
different, so $\changelev(R)\geq k$.
\end{remark}
We now give a characterisation of the change level of a $k$-return or
$k$-colreturn depending on the change level(s) of the subruns
occurring in its decomposition.
\begin{lemma} \label{lem:Change-level-Return}
Let $R$ be a $k$-return or a $k$-colreturn.
\begin{enumerate}
\item If $ \lvert R \rvert = 1$, then $\changelev(R)=k$,
\item If $R$ decomposes as $R=S \circ T$, where $S$ of length $1$
performs an operation of level $j$, and $T$ is a $k$-return or a
$k$-colreturn, then $\changelev(R)=\max\{j,\changelev(T)\})$.
\item If $R$ decomposes as $R= S\circ T\circ U$ where $S$ of length
$1$ performs a $\push{j}$ operation (including $\push1_{a,l}$ for
$j=k=1$), $T$ is a $j$-return, and
$U$ is a $k$-return or a $k$-colreturn, then
$$\changelev(R) =
\begin{cases}
\changelev(U) &\text{if } \changelev(T)=j\\
\max\{\changelev(T), \changelev(U)\} &\text{otherwise.}
\end{cases}
$$
\end{enumerate}
\end{lemma}
We begin the proof with an auxiliary proposition saying that
$k$-returns and $k$-colreturns cannot decrease the size
of the stacks of level greater than $k$.
\begin{proposition} \label{prop:Change-level-Behaviour}
Let $R$ be a $k$-return or $k$-colreturn such that $\changelev(R)>k$.
Then the size of the topmost $\changelev(R)$-stack of $R(0)$ is smaller than the size of the topmost $\changelev(R)$-stack of $R(|R|)$.
\end{proposition}
\begin{proof}
Let $m:=|R|$.
For a $k$-return, $\hist{R}{\TOP{k-1}(R(m))}$ points into in the
topmost $k$-stack of $R(0)$,
so by Corollary \ref{cor:positionContainment},
$\hist{R}{\TOP{k}(R(m))}=\TOP{k}(R(0))$.
For a $k$-colreturn we also have
$\hist{R}{\TOP{k}(R(m))}=\TOP{k}(R(0))$, due to Lemma
\ref{lem:ColreturnDesreases}.
In both cases, by Lemma \ref{lem:PositionsDecrease},
$\TOP k(R(0))\preceq \TOP k(R(m))$.
It follows that the size of the topmost $\changelev(R)$-stack
of $R(0)$ is smaller than the size of the topmost
$\changelev(R)$-stack of $R(m)$
(as for $i>\changelev(R)$, the size of the topmost $i$-stack
of $R(0)$ and of $R(m)$ is the same, and for
$i=\changelev(R)>k$ they differ). \qed
\end{proof}
Next we prove Lemma \ref{lem:Change-level-Return}.
\begin{proof}[Lemma \ref{lem:Change-level-Return}]
\begin{enumerate}
\item Case 1 is immediate.
\item
Assume we have case 2 of the lemma.
Notice that neither $S$ nor $T$ can change the size of the $i$-stack
for $i>\max\{j,\changelev(T)\})$.
If $j\neq \changelev(T)$, we see that one of the subruns changes the
size of the stack of level $\max\{j,\changelev(T)\})$, and the other
does not change it,
so we get $\changelev(R)=\max\{j,\changelev(T)\})$.
If $j=\changelev(T)$, $\changelev(T)\geq k$ (Remark
\ref{rem:chl-big}) implies that
the operation of the one-step run $S$ is necessarily $\push{}$
(cf.~Propositions
\ref{prop:ReturnsStartwithPushOrSmallLevel} and
\ref{prop:ColreturnsStartwithPush}).
Then the size of the stack of level $j$ is increased by $S$ and by
$T$ (cf.\ Proposition \ref{prop:Change-level-Behaviour}). Thus,
the claim follows immediately.
\item
Next, assume we have case 3 of the lemma.
None of the parts $S$, $T$, $U$ changes the size of the $i$-stack
for $i>\max\{\changelev(T),\changelev(U)\})$.
If $\changelev(T)=j$, Corollary \ref{cor:push+return}
implies that the topmost $j$-stack of $R(0)$ and of $U(0)$ is the same,
thus $\changelev(R)=\changelev(U)$.
So assume that $\changelev(T)>j$.
Then the size of the stack of level
$\max\{\changelev(T),\changelev(U)\})$ cannot be decreased by $S$ or
$T$ or $U$ (Proposition \ref{prop:Change-level-Behaviour}),
and at least one of $T$ and $U$ increases this value.
Thus, $\changelev(R)=\max\{\changelev(T),\changelev(U)\})$.\qed
\end{enumerate}
\end{proof}
\subsection{Non-Erasing Runs}
\label{sec:nonerasingRuns}
\begin{definition}
For $0\leq k \leq l$, let $\mathcal{N}_{k,\varepsilon}$ be the set of
$\TOP{k}$-non-erasing runs which is the set of runs
$R$ such that position $\TOP k(R(0))$ is present in every
configuration of $R$.
\end{definition}
Using $k$-returns we can characterise $\TOP{k}$-non-erasing runs in
the following way.
\begin{lemma}\label{lem:non-topk-erasing}
Let $R$ be some run and $0\leq k\leq n$.
$R$ is a $\TOP{k}$-non-erasing run if and only if $R$ has one
of the following forms.
\begin{enumerate}
\item \label{item:non-topk-erasingOne} $\lvert R \rvert=0$.
\item \label{item:non-topk-erasingTwo}
$R$ starts with an operation of level at most $k$, and
continues with a $\TOP{k}$-non-erasing run.
\item \label{item:non-topk-erasingThree}
$R$ starts with a $\push{j}$ (including arbitrary $\push 1_{a,l}$
for $j=1$) for $j\geq k+1$, and
continues with a $\TOP{j-1}$-non-erasing run.
\item \label{item:non-topk-erasingFour}
$R$ starts with a $\push{j}$ (including arbitrary $\push 1_{a,l}$
for $j=1$) and decomposes as $R=S\circ
T\circ U$, where $S$ has length $1$, $T$ is a $j$-return of change level $j$,
and $U$ is a $\TOP{k}$-non-erasing run.
\end{enumerate}
\end{lemma}
We start the proof by giving two propositions useful in the
right-to-left implication.
\begin{proposition}\label{prop:topk-non-eras-composition}
Let $R=S\circ T$ be a run such that $S$ and $T$ are $\TOP
k$-non-erasing runs for some $k$.
Then $R$ is a $\TOP k$-non-erasing run.
\end{proposition}
\begin{proof}
We claim the following.
Take some run such that $x$ and $y$ are simple positions in its
initial stack such that $x\preceq y$. If $y$ is present in all
configurations of the run, then $x$ is also present in all
configurations of the run.
Since $\TOP{k}(R(0))$ is present in all configurations of $S$, the
claim implies that all
$k$-stacks present in $R(0)=S(0)$ are also present in $S(\lvert S
\rvert)$. Thus, the topmost $k$-stack of $T(0)=S(\lvert S \rvert)$
is lexicographically greater or equal than $\TOP{k}(R(0))$.
Again using the claim, $\TOP{k}(R(0))$ is present in all
configurations of $T$ because $\TOP{k}(T(0))$ is present in all
configurations of $T$.
For the proof of the claim note that the statement of the claim is
preserved under composition of runs. Thus, we may consider a run $R$
of length $1$ such that $x\preceq y$ are positions in $R(0)$.
Since $\push{}$ operations do not delete positions in a stack, we
may assume that $R$ performs $\pop{j}$ or $\col{j}$.
Since an application of $\col{j}$ has the same effect as several
$\pop{j}$, it is sufficient to consider the $\pop{j}$ case (the
$\col{j}$-case then follows again by the composition closure
argument).
Assume that $R$ performs a $\pop{j}$ and $x$ is present in $R(0)$
but not in $R(1)$. Then $x$ points into or to the topmost
$(j-1)$-stack of $R(0)$. Since $x\preceq y$, $y$ must also point
into or to the topmost $(j-1)$-stack of $R(0)$. But then $y$ is not
present in $R(1)$. \qed
\end{proof}
\begin{proposition}\label{prop:history2topk-non-eras}
Let $0\leq k\leq n$, and let $R$ be a run such that
$\hist{R}{y}=\TOP k(R(0))$ for some position $y$ of $R(|R|)$.
Then $R$ is a $\TOP k$-non-erasing run.
\end{proposition}
\begin{proof}
Heading for a contradiction, assume that there is a minimal $i\leq
\lvert R \rvert$ such that
$x_0:=\TOP k(R(0))$ is not present in $R(i)$.
All simple positions in $R(i)$ are lexicographically smaller than $x_0$,
because $x_0$ was removed either by a $\mathsf{pop}$ operation, or
by a $\mathsf{col}$ operation (cf.\ Remark \ref{rem:ColisPop}).
Let $x_1$ be the simple prefix of $\hist{\subrun{R}{i}{\lvert R\rvert}}{y}$.
Due to Lemma \ref{lem:PositionsDecrease} applied to $\subrun{R}{0}{i}$, $x_0\preceq x_1$.
But this is a contradiction. \qed
\end{proof}
\begin{proof}[Lemma \ref{lem:non-topk-erasing}]
The proof of the right-to-left part is
by case distinction on the decomposition of $R$ according to the
four cases.
Case \ref{item:non-topk-erasingOne} is trivial and Cases
\ref{item:non-topk-erasingTwo} and
\ref{item:non-topk-erasingThree} follow directly from
Proposition \ref{prop:topk-non-eras-composition}.
We now investigate Case \ref{item:non-topk-erasingFour}.
Notice that $\hist{S\circ T}{\TOP{k}(T(\lvert T
\rvert))}=\TOP{k}(S(0))$: for $k<j$ it
follows from Proposition \ref{prop:return-removes}; for $k\geq j$ it
follows from Corollary \ref{cor:positionContainment}.
Thus, Proposition \ref{prop:history2topk-non-eras} applied to
$S\circ T$ and $y:=\TOP k(T(|T|))$ tells us that $S\circ T$ is a
$\TOP k$-non-erasing run.
Due to Proposition \ref{prop:topk-non-eras-composition}, also $R$ is
a $\TOP k$-non-erasing run.
Now concentrate on the left-to-right part.
Let $R$ be a run of length $m$ such that
$x:=\TOP{k}(R(0))$ is present in all configurations of $R$.
If $m = 0$, we are in case \ref{item:non-topk-erasingOne}.
Thus, assume that $m\geq 1$.
Note that the first operation cannot be $\col{j}$ or
$\pop{j}$ for
$j \geq k+1$ because this would delete position $x$ from
$R(1)$ (cf.\ Remark \ref{rem:ColisPop}).
Hence, one of the following cases applies.
\begin{itemize}
\item Assume that the first operation in $R$ is of level at most $k$. Then
$x=\TOP{k}(R(1))$ and $x$ is not removed during
$\subrun{R}{1}{m}$. Thus,
$R$ decomposes as in case \ref{item:non-topk-erasingTwo}.
\item Assume that the first operation in $R$ is $\push{j}$ for some
$j \geq k+1$
(in the rest of the proof, $\push{1}$ stands for arbitrary
$\push{1}_{a,k'}$).
Furthermore, assume that $y:=\TOP{j-1}(R(1))$ is present in all
configurations of
$T:=\subrun{R}{1}{m}$. Then $R$ decomposes as
as in case \ref{item:non-topk-erasingThree}.
\item Otherwise, the first operation is $\push{j}$ for some $j\geq k+1$
and there is a minimal $l\geq 1$ such that
$y:=\TOP{j-1}(R(1))$ is not present in $R(l)$.
We claim that $\subrun{R}{1}{l}$ is a $j$-return of change level
$j$ and that
the positions $\TOP{k}(R(l))$ and $\TOP{k}(R(0))$ agree.
Hence, $\subrun{R}{l}{m}$ is $\TOP{k}$-non-erasing and
$R$ decomposes as in case \ref{item:non-topk-erasingFour}.
Let us proof the claim.
Recall that
$x=\TOP{k}(R(0))$ is present in all configurations of
$\subrun{R}{0}{l}$.
Since $x$ points
into $x':=\TOP{j-1}(R(0))$ or $x'=x$,
\begin{equation}
\label{eq:test1}
x' \text{ is present in all configurations of }\subrun{R}{0}{l}.
\end{equation}
Hence we can apply
Lemma \ref{lem:TopPresentImpliesHistoryIsId} and conclude that
$\hist{\subrun{R}{0}{l}}{x'}=x'$ and
\begin{equation}
\label{eq:test2}
\text{if } \hist{\subrun{R}{l'}{l}}{x'}\text{ is
simple, it is equal to }x'.
\end{equation}
Thus, $\hist{\subrun{R}{1}{l}}{x'}=x'$ because no
non-simple
position $z$ in $R(1)$ satisfies
$x'= \hist{\subrun{R}{0}{1}}{z}$.
By definition of $\push{j}$,
$x'$ is the second topmost $(j-1)$-stack in
$R(1)$.
Since $y$ is directly above $x'$ and present in $R(l')$ for all
$1\leq l'<l$,
(\ref{eq:test1}) and (\ref{eq:test2}) imply that
$\hist{\subrun{R}{l'}{l}}{x'} \neq \TOP{j-1}(R(l'))$.
Finally, note that
from $R(l-1)$ to $R(l)$ the
$(j-1)$-stack above $x'$ (which is at $y$) is removed but $x'$ is
present in $R(l)$.
Using Remark \ref{rem:ColisPop}, the operation
is $\pop{j}$ or $\col{j}$ and $y$ points into the topmost
$j$-stack of $R(l-1)$ whence $x'$ points to the topmost
$(j-1)$-stack of $R(l)$.
In summary, $x'=\TOP{j-1}(R(l))$,
$\hist{\subrun{R}{1}{l}}{x'} = x'$ is the second topmost
$(j-1)$-stack of $R(1)$ and
$\hist{\subrun{R}{l'}{l}}{x'}$ is not the topmost $(j-1)$-stack of
$R(l')$ for
all $1\leq l' < l$.
Thus, $\subrun{R}{1}{l}$ is a $j$-return of change level $j$ and
the claim is proved.\qed
\end{itemize}
\end{proof}
\subsection{Pumping runs}
In this subsection we give a definition of pumping runs
and prove that the rules from Section \ref{sec:Runs} describe pumping
runs correctly.
\begin{definition}
For $x\in\{=,<\}$ and $y\in\{\varepsilon,{\,\not\!\varepsilon}\}$, let
$\mathcal{P}_{x,y}$ be the set of runs $R$ such that
\begin{itemize}
\item $\hist{R}{\TOP{0}(R(\lvert R \rvert))}=\TOP{0}(R(0))$, and
\item $\TOP{0}(R(\lvert R \rvert))=\TOP{0}(R(0))$ if and only if $x$
is $=$, and
\item $R$ uses only $\varepsilon$-transitions if and only if
$y=\varepsilon$.
\end{itemize}
A run $R$ is a \emph{pumping run} if it belongs to some $\mathcal{P}_{x,y}$.
\end{definition}
\begin{remark}
Lemma \ref{lem:PositionsDecrease} implies that for a pumping run
$R \in \mathcal{P}_{<, y}$, we have $\TOP{0}(R(0))\prec
\TOP{0}(R(\lvert R \rvert))$. In this sense the final stack of a
pumping run is is greater than its initial one.
\end{remark}
For the next proofs it is useful to distinguish all $k$-returns of
minimal change level (i.e., of change level $k$) from those of higher
change level.
\begin{definition}\label{def:ret-agree-not}
We set $\mathcal{R}_{k,=,x}:=\mathcal{R}_{k,k,x}$ and
$\mathcal{R}_{k,<,x} :=\bigcup_{i>k} \mathcal{R}_{k,i,x}$.
\end{definition}
\begin{remark}
A $k$-return $R$ in $\mathcal{R}_{k,=,x}$ satisfies
$\TOP{k}(R(0))=\TOP{k}(R(\lvert R \rvert))$.
Due to Proposition \ref{prop:Change-level-Behaviour},
a $k$-return $R$ in $\mathcal{R}_{k,<,x}$ satisfies
$\TOP{k}(R(0))\prec \TOP{k}(R(\lvert R \rvert))$.
\end{remark}
In the rest of this subsection we characterise pumping runs using wf-rules.
\begin{lemma}\label{lem:pumping-run}
Let $R$ be some run.
$R$ is a pumping run if and only if $R$ has one of the following forms.
\begin{enumerate}
\item $\lvert R \rvert = 0$.
\item \label{item:pumping-run-two}$R$ starts with a $\push k$
of any level (including arbitrary $\push 1_{a,l}$ for
$k=1$), and continues
with a pumping run.
\item \label{item:pumping-run-three}
$R$ starts with a $\push k$ of any level (including arbitrary
$\push 1_{a,l}$ for $k=1$), and decomposes
as $R=S\circ T\circ U$, where $S$ has length $1$, $T$ is a
$k$-return, and $U$ is a pumping run.
\end{enumerate}
Additionally, assuming that $R$ is a pumping run,
$\TOP{0}(R(0))=\TOP{0}(R(\lvert R \rvert))$
if and only if
\begin{itemize}
\item $R$ is of the first form, or
\item $R$ is of the last form, and $\TOP{k}(T(0))=\TOP{k}(T(\lvert T \rvert))$,
and $\TOP{0}(U(0))=\TOP{0}(U(\lvert U \rvert))$
\end{itemize}
\end{lemma}
The characterisation of pumping runs in
terms of the well-formed rules presented in Section \ref{sec:Runs}
follows immediately from the previous lemma.
\begin{remark}
Observe that $\hist{R}{x}=\TOP{0}(R(0))$ implies that the first
operation of $R$ is not $\pop{}$ or $\col{}$.
Indeed, after such an operation in $R(1)$ we have no position $y$ such
that $\hist{\subrun{R}{0}{1}}{y}=\TOP{0}(R(0))$ (which contradicts
with Proposition \ref{prop:histTransitive}).
\end{remark}
\begin{proof}[Lemma \ref{lem:pumping-run}]
The right-to-left direction of the first part is almost immediate.
In the third case we have to observe that $\hist{S\circ
T}{\TOP0(U(0))}=\TOP0(R(0))$; it follows from Corollary
\ref{cor:push+return}.
Now concentrate on the left-to-right direction of the first part of the lemma.
Let $R$ be a pumping run of length $m$.
If $m=0$, we are in case 1.
Thus, assume that $m\geq 1$.
Due to the above remark, $R$ starts with a $\push{}$ operation of
some level $k$.
Recall that $\hist{\subrun{R}{0}{1}}{x}=\TOP{0}(R(0))$ only if
$x=\TOP{0}(R(1))$ or if $x$ points to the topmost $0$-stack of
the second topmost $(k-1)$-stack.
By Proposition \ref{prop:histTransitive},
$\hist{\subrun{R}{1}{m}}{\TOP{0}(R(m))}$ is one of these positions
$x$.
Now there are two cases.
\begin{itemize}
\item If
$\hist{\subrun{R}{1}{m}}{\TOP{0}(R(m))}=\TOP{0}(R(1))$, then
$\subrun{R}{1}{m}$
is a pumping run and $R$ decomposes as in case \ref{item:pumping-run-two}.
\item Otherwise,
$\hist{\subrun{R}{1}{m}}{\TOP{0}(R(m))}$ points to the topmost
$0$-stack of the second topmost $(k-1)$-stack of
$R(1)$.
Due to Corollary \ref{cor:positionContainment}, we conclude
that
$\hist{\subrun{R}{1}{m}}{\TOP{k-1}(R(m)}$ points to the second
topmost $(k-1)$-stack of $R(1)$.
Let $2\leq i \leq m$ be minimal such that
$\hist{\subrun{R}{i}{m}}{\TOP{k-1}(R(m))}=\TOP{k-1}(R(i))$.
Notice that $T:=\subrun{R}{1}{i}$ satisfies all requirements of a $k$-return.
By Corollary \ref{cor:positionContainment} we know that
$\hist{\subrun{R}{i}{m}}{\TOP{0}(R(m))}$ points into
\begin{align*}
\hist{\subrun{R}{i}{m}}{\TOP{k-1}(R(m))}=\TOP{k-1}(R(i)).
\end{align*}
On the other hand, by Corollary \ref{cor:push+return}, the only
position $x$ in the topmost $(k-1)$-stack of $R(i)$ for which
$\hist{\subrun{R}{0}{i}}{x}=\TOP0(R(0))$ is $x=\TOP0(R(i))$.
By Proposition \ref{prop:histTransitive} we conclude that
$\hist{\subrun{R}{i}{m}}{\TOP{0}(R(m))}=\TOP0(R(i))$,
hence $U:=\subrun{R}{i}{m}$ is a pumping run.
\end{itemize}
Next we prove the last part of the lemma.
If $R$ is of length $0$ we immediately get $\TOP 0(R(0))=\TOP 0(R(|R|))$.
Let $R$ be a run satisfying item \ref{item:pumping-run-three} such that $\TOP{k}(T(0))=\TOP{k}(T(\lvert T \rvert))$ and $\TOP{0}(U(0))=\TOP{0}(U(\lvert U \rvert))$.
Because the operation in $S$ is $\push k$ we also have $\TOP{k}(R(0))=\TOP{k}(T(\lvert T \rvert))$.
By Corollary \ref{cor:push+return} we know that the topmost $k$-stack of $R(0)$ and of $T(|T|)$ are the same, so $\TOP{0}(R(0))=\TOP{0}(T(\lvert T \rvert))=\TOP{0}(R(|R|))$.
Finally assume that $R$ is a pumping run of length $m\geq 1$ such
that $\TOP{0}(R(m))=\TOP{0}(R(0))$.
Then $\hist{R}{\TOP{0}(R(m))} = \TOP{0}(R(m))$.
We already have observed that
$\hist{\subrun{R}{1}{m}}{\TOP{0}(R(m))}$ is simple.
Due to Lemma \ref{lem:PositionsDecrease} it is lexicographically bounded
from above by $\TOP{0}(R(m))$ and from below by
$\hist{R}{\TOP{0}(R(m))}=\TOP{0}(R(m))$. We conclude that
$\hist{\subrun{R}{1}{m}}{\TOP{0}(R(m))}=\TOP{0}(R(0))$.
From the analysis in the first part, we know that $R$ then satisfies
case \ref{item:pumping-run-three}, i.e., it decomposes as
$R=S\circ T \circ U$ where $S$ performs only one $\push{k}$, $T$ is
a $k$-return and $U$ is a pumping run. Using the same argument
again, we conclude that
$\TOP{0}(U(0))=\hist{U}{\TOP{0}(R(m))}=\TOP{0}(R(m))$.
Using Corollary \ref{cor:positionContainment} we also get that
$\TOP{k}(T(0))=\TOP{k}(R(0))=\TOP{k}(U(0))$. \qed
\end{proof}
In conclusion, Lemmas \ref{lem:pumping-run},
\ref{lem:non-topk-erasing}, \ref{lem:Change-level-Return},
\ref{lem:colreturn-wfrules} and \ref{lem:return-wfrules} show that the
Rules from Section \ref{sec:Runs} describe sets of runs that satisfy the
intended meaning described in that Section.
\subsubsection*{Types in previous papers.}
A similar concept of defining types were already present in \cite{parys2011} and \cite{parys-pumping}.
In both these papers types were used only for systems without collapse.
The types in \cite{parys2011} are defined completely semantically:
the definition is similar to our Lemma \ref{lem:run-equiv-type}.
Then it is necessary to prove that the type of $s^l$ does not depend on the choice of $s^n, s^{n-1},\dots,s^{l+1}$ present in the assumptions of the lemma.
We were unable to give a proof of the analogous fact for systems with collapse.
The types in \cite{parys-pumping} are much more similar to our types: they are also defined syntactically, i.e.~basing on possible transitions of the system.
But these types were defined only for one class of runs, namely for $k$-returns for each $k$.
The generalisation to an arbitrary family described by wf-rules required mainly the invention of a proper definition of these rules.
The generalisation to systems with collapse required mainly the invention of a proper definition of stacks (i.e.~that a $0$-stack should keep the copy of the linked stack, instead of just a link).
\section{Introduction}
\label{sec:intro}
\input{Chapter_Introduction}
\section{Collapsible Pushdown Graphs}
\label{sec:preliminaries}
\input{Chapter_Preliminaries}
\section{Proof Structure}
\label{sec:proofStructure}
\input{Chapter_Proof_Structure}
\section{Run Grammars}
\label{sec:grammars}
\input{Chapter_Grammars}
\section{A Family of Runs}
\label{sec:Runs}
\input{Chapter_Runs.tex}
\bibliographystyle{abbrv}
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Roger Federer, Rafael Nadal and Novak Djokovic are plus 30 years old and they still have success at the highest level. Dominic Thiem is already 24 years old and he may have a bigger success late in his career. But the former world No.
1 Boris Becker believes that Thiem should not think about the three legends. 'You should not make the mistake to be compared with Nadal or Federer', Becker told Kurier. 'These may be the best players of all time. Dominic Thiem is Dominic Thiem!
But I think he can be successful for a long time. Gunter Bresnik does not make him play every week, but coincides him breaks for the recovery. This is important to have a long career and a good year.' Becker won the Australian Open in 1991 and 1996.
What makes the first Grand Slam of the year more special? 'You are looking forward to come back under the sun after a long winter and play tennis outdoor. Yeah, the trip to DownUnder is hard and there is a ten-hour jet lag than Germany) - you have to do everything before.
But most of the players come two or three weeks before to play in Sydney, Brisbane or at the Hopman Cup in Perth. You only need time to suit to these conditions. The event is very well-organized and it is especially in favour of players.
The player's hotel is not far away from the club and you have good opportunities of training in Melbourne. So the conditions are ideal for players.' Are there special secrets to deal with the heat? 'Nothing special, except drinking enough and staying fit. | {
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You are here: Home » Regional news » Asia » The Halaal System of Pakistan (An Overview)
The Halaal System of Pakistan (An Overview)
By Mufti Yousuf 'Abdur-Razzaaq Khan
The history of Halaal Certification in Pakistan is not very old. However – Alhamdulillah! – Today, Pakistan is in a position to play a pivotal role in the world of Halaal.
In 1996, the first Halaal draft, consisting of only a few pages, was prepared. It was officially reviewed in 2010 and in the last 7 years, two very important standards were prepared and included in the Halaal System. In Pakistan, the standards are prepared by a National Standardization Body – the Pakistan Standard and Quality Control Authority (PSQCA) – which includes a dedicated committee tasked with preparing the Halaal standards. This committee has a team of near about 45 members, who comprise professionals and experts in every field of life. Since the basis of the Halaal Standards is the Shari'ah, there is a team of 8 to 10 dedicated Muftis who diligently perform their roles.
Another very special feature of this committee is that all the members pull their weight and put in their time and effort completely voluntarily (with no remuneration), with the sole intention of pleasing Allah d. Therefore, each member presents his view an opinion with complete personal freedom. An additional praiseworthy aspect of this organization is that it never attempts to force any policy upon its members. Instead, it merely plays the role of a guide and aide.
Alhamdulillaah! For the last 7 years, I have been a member of this committee and I am currently the vice chairmen of the Halaal Standardization Committee (NSC Halaal). To my knowledge, I can safely say that there may not be another committee like this anywhere else in the world, which boasts members who possess the kind of credentials that are required of them to serve the Halaal industry according to the Shari'ah.
The Current Halaal System is Divided into Two Sections:
Standard for Halaal System of Manufacturer.
Standard for Halaal Certification Body's.
The Halaal Certification Body applies the PS: 3733 to the manufacturing company while the State Organization, The Pakistan National Accreditation Council (PNAC) applies PS: 4992 to the Halaal Certification Body's. Since the private sector started offering this service in Pakistan before the government did, the government has allowed them to continue offering this service while keeping a check on them by means of a system called "accreditation". Remember, accreditation has not yet been declared compulsory in Pakistan and has been left optional. However, once the Pakistan Halaal Authority Act is ratified and passed and the moment this organization is operational, there is a possibility that accreditation will become compulsory.
An Introductory Review of PS: 3733[1]
This standard applies to any and every organization or company that plays any part in the production and preparation of foodstuffs and beverages, at whatever stage of the production process that may be. This standard is prepared with the support of the following international standards with regards to administrative matters:
ISO/IEC Guide 2, Standardization and Related Activities — General Vocabulary,
CODEX STAN l, General standard for the labelling of Prepacked foods,
CAC/RCP l, Recommended international code of practice, general principles of food hygiene,
CAC/RCP 58, Code of hygienic practice for meat,
ISO 22000, Food safety management systems – Requirements for any organization in the food chain,
ISO 22005, Traceability in the feed and food chain – General principles and basic requirements for system design and implementation,
ISO 9001, Quality management systems — Requirements.
With the guidance of our local Muftis, we also drew from the MS 1500 (Malaysian Standard) as well as the SMIIC Halaal Standards with regards to matters pertaining to the Shari'ah.
This Standard Comprises Two Main Sections:
The Shar'i Laws as a Standard
Administrative Standards
Clause 3 contains all the Shar'i terminologies such as the definitions of Halaal, Haraam, Najis, Makrooh-e-Tahreemi, Makrooh-e-Tanzeehi, Mashbooh, Haraam Animals, Haraam Birds, Carrion, Haraam beverages, Terrestrial and Aquatic Animals etc.
It also discusses the various types of slaughter, the method of slaughtering, animal rights, all the stages the animal passes through from the farm to the abattoir.
The fact That Mechanical Slaughter, all kinds of Stunning is prohibited, let it be clear that such meat may also not be imported into the country.
These definitions and regulations have been discussed in Clauses 3 to 9. Thereafter, from Clauses 9 to 13, administrative issues are discussed.
Clause 9 demands that each company should have its own Halaal Manual in which is clearly states its Halaal Policy. All the records should also be prepared and kept in accordance with the Halaal Standard to ensure that the Halaal System is implemented, which would then guarantee the Halaal status of whichever product is being manufactured.
Further on, it discusses the prerequisites which those individuals who will be implementing the Halal System in that particular company or organization need to meet. For example, the one in charge of the Halaal System may only be a Muslim and he needs to be educated in the basic aspects of the Deen and, together with that, he needs to go for training regarding this Halaal Standard etc.
All the stages of purchasing the raw materials and ingredients for the final preparation of the product, and right from the storage of the product to its transport should be scrutinized and wherever there is a possibility of the product becoming contaminated or Haraam, all those possibilities should be considered beforehand and a suitable solution should be sought. This is also a demand of this section which is referred to as "????? ???????????" in the Shari'ah.
It also calls for an internal audit so that the weaknesses and shortcomings of the organization or company are regularly brought to the fore and a report thereof is presented to top management who will, in turn, immediately take action so that the non-conformance or deficiency can be eliminated before the Halaal status of the product is jeopardized in any way. There is a famous quote of Hazrat 'Umar bin Khattaab h, who said: "Hold yourselves accountable before you are held accountable – the one who sets his record straight in this world, will have an easy reckoning on the Day of Judgement."
This is the purpose of internal audits – when the company keeps taking stock of itself, seeks out its own faults and deficiencies and takes the initiative to eradicate them, it will be very easy for them when another organization does an audit since the majority of the shortcomings would have already been addressed and resolved.
This system also demands that every possible precautionary measure be implemented, and the company should constantly check to see whether they are achieving the desired results.
Great emphasis is placed on cleanliness and hygiene at the facility at which the product is manufactured or prepared, which is also an important demand of the Shari'ah. It also mentions the way the product should be displayed in the market – it should not happen that a Halaal certified product is placed next to a Haraam or impure product. The system offers guidelines in this regard as well.
There are also guidelines regarding the packaging and labelling of the product once it is ready for the market. This is done so that the consumer can be fully aware of what he is purchasing, when it was produced, when it expires and, if he is purchasing meat, when it was slaughtered and packaged, what its sell-by date is, which animal it is, which Halaal Certification Body issued its certification etc. This section requires that all such information must be displayed on the packaging.
It is the responsibility and duty of the Halaal Certification Body to ensure that it does an audit and inspection of the company which it has certified and that it implements each aspect of this standard.
This standard does not allow to use the name, having the sensory profile of a Haraam product, like rum flavor, pork flavor, etc. and shall not be Halaal certified although the ingredients used are Halaal.
This standard stipulates principles and regulations for those organizations which, by assisting the government and utilizing its right to WILAAYAT-KHASS'AH (limited authority), issues Halaal certification. The following international standards were used while preparing this standard:
ISO 9000:200 5, Quality Management Systems – Fundamentals and Vocabulary
ISO/IEC 17000:2004, Conformity Assessment – Vocabulary and General Principles
ISO/IEC 17021:2011, Conformity Assessment – Requirements for bodies providing audit certification of management systems
ISO/IEC 17065.2012, Conformity Assessment – Requirements for bodies certifying products, processes
And services
ISO 19011, Guidelines /or quality and/or environmental management systems auditing,
ISO 22000, Food safety management systems – Requirements for any originations in the food chain,
ISO/TS 22003, Food safe4t management systems – Requirements for bodies providing audit and certification of food safety management systems.
In clause 3, all the terminologies related to Halaal certification are explained. This covers terms like Halaal certification, contract, auditor, experts of Shari'ah and science, Shari'ah, etc. Furthermore, it also explains the different roles of a Halal Certification Body.
Clause 4 deals with the limits and boundaries regarding matters like fairness, eligibility, transparency, confidentiality, the various responsibilities of the Halaal Certification Body etc. This clause also comprises a number of sub clauses which discuss Shar'i responsibilities as well as the status of the constitution.
Clause 6 covers conditions and prerequisites pertaining to the organization's structure. One such prerequisite is worth mentioning: The Halaal Certification Body shall be owned, managed and operated by the Muslims. The same applies to the administrative control as well as the over-all running of the organization, both of which may only be done by a Muslim.
The basis of this prerequisite is the command of the Shari'ah which I will briefly explain.
To Which Branch of Deen Does Halaal and Haraam Belong?
Matters of Halaal and Haraam are related to a branch of the Shari'ah called Diyaanaat. This has been mentioned in various books of jurisprudence written in Arabic.
The jurists explain that matters of Halaal and Haraam relate to the most delicate aspect of Islam, referred to as Deeniyaat (strictly religious matters) in the Urdu language or Diyaanaat in Arabic.
What is Meant by Diyaanaat?
Diyaanaat refers to the rights which are established between Allah d and his servant and which have a purely religious basis. One can also say that it refers to Huqooqullaah (the rights which Allah (has over his believing servants).
If we just cast a cursory glance at PS: 4992, we will see that this standard determines the entire structure of the organization. For example, it states that the organisation shall be a legal entity and that an individual's cannot offer this service, the organisation has to employ individuals who are specialists of both the Shari'ah as well as the technical aspects of the Halaal industry, it stipulates the education and training which the auditors need to have, it specifies the correct roles and responsibilities of the organisation and that each responsible person in each division needs to have the necessary skills and experience, it calls for regular training programs for the employees etc. It also demands that an impartial committee should be established which should get together once a year and do a completely impartial inspection of the organisation's annual activities and present a detailed report regarding their findings. This is done to ensure that the organisations activities remain transparent.
It also determines how an audit should be conducted, how many days the audit should take, which people are eligible to audit which industry, how an auditor's report should look, who will scrutinise the report and base a decision on it and what the conditions of eligibility of these individuals should be, which factors will lead to a decision being made in favour of the applicant and which factors will lead to a decision made against them, under which conditions the Halaal Certification Body will provide its logo (sign of Halaal testimony) in the event that the application of the applicant is approved, which rights will be reserved between the manufacturer and the Halaal Certification Body and what their legal status will be in the event of a contract being drawn up and signed between the two parties, how long the contract would last and that the certification body will do spot checks and numerous random and unannounced audits to determine whether the manufacturer is really implementing the Halaal system etc.
The Pakistan National Accreditation Council (PNAC) will determine whether the above-mentioned standards are being adhered to. In the event that they are being adhered to, it will issue an accreditation certificate in favour of the Halaal Certification Body, which will be internationally recognised, and which would serve as a sign of the eligibility and proficiency of this organisation, globally.
Our organisation has also passed the above-mentioned stages and attained its accreditation. In our experience, we can safely say that it is indeed an excellent system which has the potential of achieving its goals and aspirations.
At the moment, Pakistan is ready to enter the field of Halaal completely. If our standard does not surpass the international standards, it is in no way less. Once the Halaal Authority Act is passed in the Pakistani Parliament, Provincial Halaal Authorities will also be established in two provinces, Panjab and KPK. It looks like all these bodies will be up and running within the next 6 months, which will be of great assistance to the international Halaal industry together with playing a very vital role in the economy of our country
By Mufti Yousuf 'Abdur-Razzaaq Khan, CEO, SANHA HALAL ASSOCIATES PAKISTAN
Monday, 1 Rabee'ul Awwal, 1439
[1] This is the Pakistani Standard specification for Halaal Food Management System requirements for any organisation or company in the food chani (3rd Revision).
[3] PS:4992-2016(R), Pakistan Standard For Conformity Assessment – Requirements for Bodies Providing Halaal Certification (1st Revision).
NB: Other references in this document are written in Arabic and cannot be added.
Tags: Halaal accreditation, Halaal standards, Halal certification in Pakistan, Mufti Yousuf 'Abdur-Razzaaq Khan, PNAC, PSQCA
Category: Asia, Halal Integrity, Shariah Issues
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