text:
\section{Introduction}\label{se:intro}

The first edition of the seminal book \emph{Geometrie der Lage} by Karl Georg
Christian von Staudt appeared in 1847; see \cite{hart-08a} for publication
details. Projectivities are defined there by the invariance of harmonic
quadruples \cite[p.~49]{stau-47}: \emph{''Zwei einf\"{o}rmige Grundgebilde heissen
zu einander projektivisch {\projective}, wenn sie so auf einander bezogen sind,
dass jedem harmonischen Gebilde in dem einen ein harmonisches Gebilde im andern
entspricht.''} Next, after defining perspectivities, the following theorem is
established: Any projectivity is a finite composition of perspectivities and
vice versa. (It was noticed later that there is a small gap in von Staudt's
reasoning. A detailed exposition can be found in \cite{voel-08a}.) Any result
in this spirit now is called a \emph{von Staudt's theorem}.
\par
In the present article we shall be concerned with \emph{projective lines over
rings} (associative with a unit element) and the algebraic description of their
\emph{harmonicity preservers}, i.~e., mappings which take all harmonic
quadruples of a first projective line to harmonic quadruples of a second one.
There is a widespread literature on this topic. The following short review is
rather sketchy, as it does not fully reflect the varying (often rather
technical) assumptions on the underlying rings. Part of the presented material
is related with mappings which reappear in a more general setting in the
surveys \cite{brehm-08a} and \cite{veld-95a}.
\par
All harmonicity preserving bijections of the projective line over any
\emph{commutative field} $F$ of characteristic $\neq 2$ onto itself were
determined by O.~Schreier and E.~Sperner \cite[p.~191]{schrei+s-35a}. In terms
of an underlying $F$-vector space $V$ these transformations comprise precisely
the \emph{projective semilinear group} $\mathrm{P}\Gamma\mathrm{L}(V)$. The
case of a (not necessarily commutative) \emph{field} of characteristic $\neq 2$
was settled in several steps by G.~Ancochea \cite{anco-41a}, \cite{anco-42a},
\cite{anco-47a} and L.-K.~Hua \cite{hua-49a} (see also \cite{hua-52a}). For a
proper skew field $F$ one has to include mappings which arise from
\emph{antiautomorphisms} of $F$ (provided that $F$ admits any
antiautomorphism). A.~J.~Hoffman \cite{hoff-51a} ($F$ commutative) and R.~Baer
\cite[p.~78]{baer-52a} ($F$ arbitrary) proved that similar results hold if the
invariance of harmonic quadruples is replaced by the invariance of an arbitrary
cross ratio $k\neq 0,1$ in the centre of $F$. In this way the case of
characteristic $2$ need no longer be excluded. A detailed account with
historical remarks is given in \cite[pp.~56--57]{karz+k-88a}.
\par
There are several outcomes for the projective line over a \emph{ring $R$ with
stable rank $2$}: Loosely speaking, in the case of a \emph{commutative ring}
$R$ the result of Schreier and Sperner remains unaltered provided that $R$
contains ``sufficiently many'' units, in particular $2$ has to be a unit in
$R$. Contributions (under varying additional assumptions) are due to W.~Benz
\cite{benz-64a}, \cite[pp.~173--183]{benz-73a}, B.~V.~Limaye and N.~B.~Limaye
\cite{lima+l-77b}, N.~B. Limaye \cite{lima-71a}, \cite{lima-72a},
B.~R.~McDonald \cite{mcdon-81a}, and H.~Schaef\-fer \cite{schae-74a}. Little
seems to be known for non-commutative rings: B.~V.~Limaye and N.~B.~Limaye
(\cite{lima+l-77a}, \cite{lima+l-77c}) treated the case of a (not necessarily
commutative) \emph{local ring} $R$. They determined all bijections of the
projective line over $R$ such that all quadruples with a given cross ratio $k$
go over to quadruples with a given cross ratio $k'$, where $k, k'$ are elements
in the centre of $R$ other that $0,1$. Here the algebraic description is more
involved, since one has to use \emph{Jordan automorphisms} (or, in a different
terminology, \emph{semiautomorphisms}) of $R$. More information can be
retrieved from the surveys in \cite{bart+b-85}, \cite{benz+l+s-72a}, and
\cite{benz+s+s-81}.
\par
F.~Buekenhout \cite{buek-65a}, St.~P.~Cojan \cite{cojan-85a}, D.~G.~James
\cite{james-82a}, and B.~Klotzek \cite{klotz-88a} characterised those (not
necessarily injective) mappings between projective lines over fields which
satisfy a much weaker form of cross ratio preservation than the one mentioned
in the preceding paragraph. The link with ring geometry is achieved via a
recoordinatisation of the domain projective line in terms of a \emph{valuation
ring} \cite{james-82a}.
\par
It was pointed out by C.~Bartolone and F.~Di~Franco \cite{bart+f-79} that an
algebraic description of all harmonicity preserving bijections of the
projective line over an arbitrary ring is out of reach, even in the commutative
case. They therefore initiated the study of mappings which preserve
\emph{generalised harmonic quadruples} and succeeded in describing all such
mappings for commutative rings; see also M.~Kulkarni \cite{kulk-80a}. However,
this goes beyond the scope of the present article. With regard to the
non-commutative case, we refer to the work of C.~Bartolone and F.~Bartolozzi
\cite{bart+b-85}, D.~Chkhatarashvili \cite{chkh-98a}, L.~Cirlincione and
M.~Enea \cite{cirl+e-90}, and A.~A.~Lashkhi \cite{lash-89a}, \cite{lash-90a},
\cite{lash-97a}. Take notice that some of the quoted papers are merely short
communications without any proof. For harmonicity preserving mappings of other
geometric structures see \cite{bert-03a}, \cite{blunck-91a}, \cite{ferr-81a},
and the references therein. It is also worth noting that the invariance of
harmonic quadruples appears together with other conditions in an early paper
\cite{hua-45a} of L.-K.~Hua on a characterisation of certain transformations of
matrix spaces. However, as Hua pointed out in a subsequent note \cite{hua-45b},
the condition about harmonic quadruples is superfluous in that context, and it
afterwards disappeared from the so-called \emph{geometry of matrices}; cf.\ the
monographs \cite{huanglp-06a} and \cite{wan-96a}. An analogous result for
projective lines over certain semisimple rings is due to A.~Blunck and the
author \cite{blunck+h-05b}.
\par
The present article is organised as follows: In Section \ref{se:basic} we
collect the relevant notions and we recall the definition of harmonicity
preservers which arise from Jordan homomorphisms. Our main result is
Theorem~\ref{thm:staudt} in Section~\ref{se:staudt}. It shows that under
certain conditions there are no other harmonicity preservers between projective
lines over rings, but those which arise from Jordan homomorphisms. A major tool
in our proof is a lemma from \cite{lima+l-77b} which characterises Jordan
homomorphisms.

\section{Basic notions and examples}\label{se:basic}

All our rings are associative with a unit element $1$ which is inherited by
subrings and acts unitally on modules. The trivial case $1=0$ is excluded. The
group of units (invertible elements) of a ring $R$, say, will be denoted by
$R^*$.
\par
Let $R$ be a ring and let $M$ be a free left $R$-module of rank $2$. We say
that $a\in M$ is \emph{admissible} if there exists $b\in M$ such that $(a,b)$
is a basis of $M$ (with two elements). As a matter of fact, we do not require
that all bases of $M$ have the same number of elements; cf.\
\cite[p.~3]{lam-99a}.
\par
The following exposition is mainly taken from \cite[p.~785]{herz-95a}; see also
\cite[pp.~15--16]{blunck+he-05a} or \cite[pp.~899--904]{havl-12b}: The
\emph{projective line} over $M$ is the set $\bP(M)$ of all cyclic submodules
$Ra$, where $a\in M$ is admissible. The elements of $\bP(M)$ are called
\emph{points}. At times it will be convenient to use coordinates with respect
to some basis $(e_0,e_1)$ of $M$. Given any pair $(a,b)\in M^2$ let $(x_0,x_1)$
and $(y_0,y_1)$ be the coordinates of $a$ and $b$, respectively. The matrix
\begin{equation}\label{eq:matrix}
    \begin{pmatrix}x_0&x_1\\y_0&y_1\end{pmatrix}
\end{equation}
will be called the \emph{matrix of $(a,b)$ w.~r.~t.\ the basis $(e_0,e_1)$}.
The pair $(a,b)$ is a basis of $M$ if, and only if, the matrix in
(\ref{eq:matrix}) is invertible. Thus $(x_0,x_1)\in R^2$ is admissible (or,
said differently, a coordinate pair of a point) precisely when it is the first
(or second) row of a matrix in $\GL_2(R)$. One particular case deserves
explicit mention, since it links the group $R^*$ with the group $\GL_2(R)$: For
all $x,y\in R$ holds
\begin{equation}\label{eq:GL-R*}
    \begin{pmatrix}x&1\\y&1\end{pmatrix}\in\GL_2(R)
    \quad\mbox{if, and only if,}\quad
    x-y\in R^* .
\end{equation}
This is immediate from
\begin{equation}
    \begin{pmatrix}1&1\\0&1\end{pmatrix}
    \begin{pmatrix}x-y&0\\0&1\end{pmatrix}
    \begin{pmatrix}1&0\\y&1\end{pmatrix} =
    \begin{pmatrix}x&1\\y&1\end{pmatrix}.
\end{equation}
\par
By definition, each point $p\in\bP(M)$ has an admissible generator, say $a$. If
there exist $x,y\in R$ with $xy=1$ and $yx\neq 1$ then $ya$ is a non-admissible
generator of $p$, whereas $xa$ is an admissible generator of a point other than
$p$ \cite[Prop.~2.1 and Prop.~2.2]{blunck+h-00b}. We adopt from now on the
following convention: \emph{We only use admissible generators of points.} Two
admissible elements of $M$ generate the same point precisely when they are
left-proportional by a unit in $R$.
\par
Two points $p$ and $q$ are called \emph{distant}, in symbols $p \dis q$, if
$M=p\oplus q$. For all $a,b\in M$ holds $Ra \dis Rb$ precisely when the
coordinate matrix of $(a,b)$ w.~r.~t.\ any basis $(e_0,e_1)$ of $M$ is
invertible. The graph of the relation $\dis$, i.~e. the pair
$\big(\bP(M),{\dis}\big)$, is called the \emph{distant graph\/} of $\bP(M)$. It
is an undirected graph without loops, and it need not be connected. In order to
describe the \emph{connected components} of the distant graph we need some
prerequisites.
\par
The \emph{elementary linear group\/} $\E_2(R)$ is generated by the set of all
matrices
\begin{equation*}\label{eq:E(t)}
  E(t):=\begin{pmatrix}
  \XX t&1\\-1&0
  \end{pmatrix}
  \quad\mbox{with}\quad  t\in R ;
\end{equation*}
see \cite[p.~5]{cohn-66a}. Let $\Seq(R)$ be the set of all finite sequences in
$R$ (including the empty sequence). We adopt the shorthand notation
\begin{equation*}\label{eq:E(T)}
    E(T):= E(t_1)\cdot E(t_2)\cdots E(t_n)
    \quad\mbox{where}\quad T=(t_1,t_2,\ldots,t_n)\in\Seq(R).
\end{equation*}
(Note that $n\geq 0$, the length of $T$, is arbitrary.) From
$E(t)^{-1}=E(0,-t,0)$ follows that all matrices $E(T)$ with $T\in\Seq(R)$
comprise the entire group $\E_2(R)$. The subgroup of $\GL_2(R)$, which is
generated by $\E_2(R)$ and the set of all invertible diagonal matrices, is
denoted by $\GE_2(R)$. By definition, a \emph{$\GE_2$-ring\/} $R$ is
characterised by $\GL_2(R)=\GE_2(R)$.
\par
If $(e_0,e_1)$ is a basis of $M$ then the connected component of the point
$Re_0\in\bP(M)$ is given by the set of all points $p=R(x_0e_0+x_1e_1)$, where
$(x_0,x_1)$ is the first row of some matrix $E(T)$ with $T\in\Seq(R)$ or, said
differently,
\begin{equation}\label{eq:component}
    (x_0,x_1)=(1,0)\cdot E(T) \quad\mbox{for some}\quad  T\in\Seq(R).
\end{equation}
Furthermore, the distant graph $\big(\bP(M),{\dis}\big)$ is connected precisely
when $R$ is a $\GE_2$-ring \cite[Thm.~3.2]{blunck+h-01a}.
\par
A quadruple $(p_0,p_1,p_2,p_3)\in\bP(M)^4$ is \emph{harmonic} if its cross
ratio \cite[p.~787]{herz-95a} equals $-1\in R$, i.~e., there exists a basis
$(g_0,g_1)$ of $M$ such that
\begin{equation}\label{eq:H.g}
    p_0=R g_0,\quad p_1=R g_1,\quad  p_2=R(g_0+g_1),\quad p_3=R(g_0-g_1).
\end{equation}
In this case we write $\Harm(p_0,p_1,p_2,p_3)$. In terms of coordinates
w.~r.~t.\ some basis $(e_0,e_1)$ of $M$ there is an alternative description:
$\Harm(p_0,p_1,p_2,p_3)$ holds if, and only if, there is a matrix
$G\in\GL_2(R)$ such that
\begin{equation}\label{eq:H.G}
    (1,0)\cdot G,\quad
    (1,0)\cdot E(0)\cdot G,\quad
    (1,0)\cdot E(1)\cdot G,\quad
    (1,0)\cdot E(-1)\cdot G
\end{equation}
are coordinates of the points $p_0$, $p_1$, $p_2$, $p_3$, respectively. Indeed,
if (\ref{eq:H.g}) holds for some basis $(g_0,g_1)$ we can take as $G$ the
coordinate matrix of $(g_0,g_1)$ w.~r.~t.\ $(e_0,e_1)$ in order to obtain
(\ref{eq:H.G}). Conversely, the rows of $G$ provide the coordinates w.~r.~t.\
$(e_0,e_1)$ of an appropriate basis of $M$ to guarantee
$\Harm(p_0,p_1,p_2,p_3)$.
\par
From $\Harm(p_0,p_1,p_2,p_3)$ follows $p_0\dis p_1$ and $p_i\dis p_j$ for all
$i\in\{0,1\}$ and all $j\in\{2,3\}$. \emph{Therefore all four points belong to
the same connected component of the distant graph $(\bP(M),\dis)$.} By virtue
of (\ref{eq:GL-R*}), we have
\begin{equation}\label{eq:2}
    p_2\dis p_3
    \quad\mbox{if, and only if,}\quad 2\in R^* .
\end{equation}
The inequality $p_2\neq p_3$ holds precisely when $-1\neq 1\in R$. (In
\cite[4.7]{blunck+h-03b} these two conditions erroneously got mixed up.)
\par
If $p_0,p_1,p_2$ are three mutually distant points of $\bP(M)$ then there is a
unique point of $\bP(M)$, say $p_3$ with $\Harm(p_0,p_1,p_2,p_3)$. This is the
well known \emph{uniqueness of the fourth harmonic point}. Since
$\Harm(p_0,p_1,p_2,p_3)$ is equivalent to $\Harm(p_0,p_1,p_3,p_2)$, there holds
as well the \emph{uniqueness of the third harmonic point}. The latter (less
prominent) property will be used when proving Lemma~\ref{lem:2}.
\par
Let $M'$ be a free left module of rank $2$ over a ring $R'$. A mapping
$\mu:\bP(M)\to\bP(M')$ will be called a \emph{harmonicity preserver} if it
takes all harmonic quadruples of $\bP(M)$ to harmonic quadruples of $\bP(M')$.
No further assumptions, like injectivity or surjectivity of $\mu$ are made
here. A simple, though important, property is that any harmonicity preserver
$\mu:\bP(M)\to\bP(M')$ is \emph{distant preserving}, i.~e.,
\begin{equation}\label{eq:dist-pr}
    p_0\dis p_1
    \quad\mbox{implies}\quad
    p_0^\mu \dis p_1^\mu
    \quad\mbox{for all}\quad
    p_0,p_1\in\bP(M).
\end{equation}
This follows readily from the existence of points $p_2$ and $p_3$ with
$\Harm(p_0,p_1,p_2,p_3)$.
\par
We close this section by quoting several examples of harmonicity preservers
$\bP(M)\to\bP(M')$.

\begin{exa}\label{exa:jordan}
Let $\alpha:R\to R'$ be a Jordan homomorphism, i.~e. a mapping satisfying
\begin{equation*}\label{eq:jordan}
  (x+y)^\alpha = x^\alpha + y^\alpha,\quad
  1^\alpha = 1',\quad
  (xyx)^\alpha = x^\alpha y^\alpha x^\alpha\quad
  \mbox{for all}\quad x,y\in R.
\end{equation*}
See, among others, \cite[p.~832]{herz-95a} or \cite[p.~2]{jacob-68a}. Also, let
$C$ be any connected component of the distant graph $(\bP(M),\dis)$. We select
bases $(e_0,e_1)$ and $(e_0',e_1')$ of $M$ and $M'$, respectively, subject to
the condition $Re_0\in C$. According to a result of A.~Blunck and the author
\cite[Thm.~4.4]{blunck+h-03b} the following (rather cumbersome) construction
gives a well defined mapping
\begin{equation}\label{eq:mu}
    \mu : C \to \bP(M') : p \mapsto p^\mu.
\end{equation}
By (\ref{eq:component}), any point $p\in C$ can be written in the form
$p=R(x_0e_0+x_1e_1)$ with
\begin{equation*}\label{eq:(x0,x1)}
    (x_0,x_1)=(1,0)\cdot  E(T)
\end{equation*}
for some $T\in\Seq(R)$, say $T=(t_1,t_2,\ldots,t_n)$ with $n\geq 0$. We use the
shorthand $T^\alpha:=(t_1^\alpha,t_2^\alpha,\ldots,t_n^\alpha)\in\Seq(R')$ and
let
\begin{equation}\label{eq:(x0',x1')}
    (x_0',x_1') := (1,0)\cdot E(T^\alpha).
\end{equation}
The point $p^\mu$ is defined as $R'(x_0'e_0'+x_1'e_1')$. By
\cite[Prop.~4.8]{blunck+h-03b}, $\Harm(p_0,p_1,p_2,p_3)$ implies
$\Harm(p_0^\mu,p_1^\mu,p_2^\mu,p_3^\mu)$ for all $p_0,p_1,p_2,p_3\in C$.
\par
The previous construction can be repeated for all connected components of the
distant graph on $\bP(M)$. Thereby is not necessary to stick to a fixed Jordan
homomorphism. Altogether this gives a globally defined harmonicity preserver
$\bP(M)\to \bP(M')$.
\par
One particular case, due to C.~Bartolone \cite{bart-89a}, deserves special
mention: Let $R$ be a ring of \emph{stable rank\/ $2$}
\cite[p.~1039]{veld-95a}. Then $\big(\bP(M),{\dis}\big)$ has a single connected
component, each of its points can be described in terms of at least one finite
sequence $T=(t_1,t_2)\in R^2$, and $\mu$ can be rewritten as
\begin{equation*}\label{eq:bart}
    \mu: \bP(M)\to \bP(M'): R\big((t_1t_2-1)e_0 + t_1 e_1\big)\mapsto
    R'\big((t_1^\alpha t_2^\alpha-1)e_0' + t_1^\alpha e_1'\big) .
\end{equation*}
\end{exa}

\begin{exa}\label{exa:hom}
We adopt the settings of Example~\ref{exa:jordan}, but we make the extra
assumption that $\alpha$ is a homomorphism of rings. Then
\begin{equation*}\label{eq:semilin}
    \sigma: M\to M' : x_0e_0 +x_1e_1 \mapsto x_0^\alpha e_0' +x_1^\alpha e_1'
    \quad\mbox{for all}\quad x_0,x_1\in R
\end{equation*}
is an $\alpha$-semilinear mapping and
\begin{equation*}\label{eq:alpha*}
    \alpha_* : \GL_2(R)\to\GL_2(R') : X \mapsto X^\alpha,
\end{equation*}
i.~e., $\alpha$ is applied to each entry of $X$, is a homomorphism of groups.
Thus for any basis $(a,b)$ of $M$ the image $(a^\sigma,b^\sigma)$ is a basis of
$M'$. Consequently, the mapping
\begin{equation*}\label{eq:lambda}
     \lambda:\bP(M)\to \bP(M') : Ra \mapsto R'(a^\sigma)
     \quad\mbox{(with~} a\in M
     \mbox{~admissible)}
\end{equation*}
is well defined, and it preserves harmonicity. The mapping $\mu$ from
(\ref{eq:mu}) is the \emph{restriction\/} of $\lambda$ to $C$. The matrix
$E(T^\alpha)$ from (\ref{eq:(x0',x1')}) now can be expressed as
$E(T)^{\alpha_*}$, since $E(t)^{\alpha_*}=E(t^\alpha)$ for all $t\in R$.
\end{exa}

\begin{exa}\label{exa:anti}
We adopt the settings of Example~\ref{exa:jordan}, but we make the extra
assumption that $\alpha$ is an antihomomorphism of rings. We have the
homomorphism
\begin{equation}\label{eq:alpha**}
    \alpha_{**} : \GL_2(R)\to\GL_2(R') :
    X \mapsto E(0)^{-1}\cdot\big((X^{-1})^{\Trans}\big)^\alpha\cdot E(0),
\end{equation}
where $(X^{-1})^\Trans$ denotes the transpose of $X^{-1}$ and $\alpha$ is
applied entrywise. (We must not use $\alpha_*$ in (\ref{eq:alpha**}), since
$(X^{-1})^\Trans$ need not be invertible.) A straightforward calculation shows
$E(t)^{\alpha_{**}}= E(t^\alpha)$ for all $t\in R$. Hence the matrix
$E(T^\alpha)$ from (\ref{eq:(x0',x1')}) now can be expressed as
$E(T)^{\alpha_{**}}$. This leads us to the definition of a mapping
\begin{equation*}\label{eq:delta}
    \delta : \bP(M)\to\bP(M') : R(x_0e_0+x_1e_1) \mapsto R'(x_0'e_0'+x_1'e_1')
\end{equation*}
which runs as follows: $(x_0,x_1)$ is chosen as the first row of any matrix
$X\in\GL_2(R)$ and $(x_0',x_1')$ is defined as the first row of the matrix
$X^{\alpha_{**}}$. By \cite[Ex.~4.8]{blunck+h-03b} this mapping is well
defined. An equivalent (and more lucid) definition of $\delta$ in terms of the
dual module of $M$ can be read off from \cite[Rem.~5.4]{blunck+h-01b} or
\cite[Prop.~3.3]{herz-87a}. Formula (\ref{eq:H.G}) provides an easy direct
proof for $\delta$ being a harmonicity preserver. The mapping $\mu$ from
(\ref{eq:mu}) is the \emph{restriction\/} of $\delta$ to $C$.
\par
It may happen that $\alpha:R\to R'$ is a homomorphism and an antihomomorphism.
Then $R^\alpha$ is a commutative subring of $R'$ and we have $(\det
X^{\alpha_{*}}) X^{\alpha_{**}} = X^{\alpha_{*}}$ for all $X\in\GL_2(R)$. So in
this case the mappings $\lambda$ and $\delta$ coincide.
\end{exa}

\section{Von Staudt's theorem}\label{se:staudt}

We already noted in Section~\ref{se:basic} that the distant graph on $\bP(M)$
has a single connected component if, and only if, $R$ is a $\GL_2$-ring. In
this case the following version of von Staudt's theorem provides a unified
algebraic description of harmonicity preservers, otherwise it gives only a
description on an arbitrarily chosen connected component.
\begin{thm}\label{thm:staudt}
\renewcommand{\theenumi}{\roman{enumi}
\renewcommand{\labelenumi}{{\rm (\theenumi)}
Let $M$ and $M'$ be free modules of rank $2$ over rings $R$ and $R'$,
respectively. Furthermore, let $R$ satisfy the two conditions:
\begin{enumerate}
\item\label{cond:i} Given $x_1,x_2,\ldots,x_5\in R$ there exists $x\in R$
    such that $x-x_1,x-x_2,\ldots,x-x_5$ are units in $R$.
\item\label{cond:ii} $2$ is a unit in $R$.
\end{enumerate}
Let $\mu:\bP(M)\to \bP(M')$ be a harmonicity preserver. Choose any connected
component, say $C$, of the distant graph $\big(\bP(M),{\dis}\big)$. Then there
exist a basis $(a_{0},a_{1})$ of $M$, a basis $(a_{0}',a_{1}')$ of $M'$, and a
Jordan homomorphism $\alpha:R\to R'$ such that the restriction of $\mu$ to $C$
admits the following description:
\begin{equation*}\label{eq:mu|C}
    \mu | C : C\to\bP(M') :
    R(x_0 a_{0} + x_1 a_{1})\mapsto R(x_0' a_{0}' + x_1' a_{1}') ,
\end{equation*}
where
\begin{equation}\label{eq:(x0,x1)(x0',x1')}
    (x_0,  x_1) = (1,0)\cdot E(T),
    \quad
    (x_0',  x_1') = (1,0)\cdot E(T^\alpha),
\end{equation}
and $T$ is any finite sequence of elements in $R$.
\end{thm}

We postpone the proof until we have established four auxiliary results. In all
of them we tacitly adopt the assumptions of Theorem~\ref{thm:staudt}.
Lemma~\ref{lem:1} is self-explanatory. In Lemma~\ref{lem:2} we exhibit a
mapping $\beta:R\to R'$ which can be viewed as a ``local coordinate
representation of $\mu$''. Next, in Lemma~\ref{lem:3}, we establish that ``new
local coordinates'' (describing other parts of the given projective lines) can
be chosen in such a way that the ``new local coordinate representations'' of
$\mu$ coincides with the ``old'' one. This observation is the backbone of our
demonstration. Afterwards, in Lemma~\ref{lem:4}, the mapping $\beta$ is shown
to be a Jordan homomorphism. The actual proof Theorem~\ref{thm:staudt} amounts
then to verifying that the given mapping $\mu|C$ coincides with the harmonicity
preserver which arises from $\beta$ according to Example~\ref{exa:jordan}. It
goes without saying that part of our demonstration follows the same lines as
previous work by other authors. Condition~(\ref{cond:i}) is taken from
\cite{lima+l-77b}. It is equivalent to the following property of the projective
line $\bP(M)$:
\begin{enumerate}
\item[(i')]\label{cond:i'} Given points $p_1,p_2,\ldots,p_5\in \bP(M)$, all
    of which are distant to some point $p_0\in\bP(M)$, there exists $p\in
    \bP(M)$ which is distant to $p_0,p_1,\ldots,p_5$.
\end{enumerate}
The equivalence follows easily from (\ref{eq:GL-R*}) upon choosing a basis
$(e_0,e_1)$ of $M$ with $p_0=Re_0$. Then $p_i=R(x_i e_0+e_1)$ for
$i\in\{1,2,\ldots,5\}$ and $p=R(x e_0+e_1)$. Take notice that neither the
elements $x_1,x_2,\ldots,x_5$ nor the points $p_1,p_2,\ldots,p_5$ are assumed
to be distinct.

\begin{lem}\label{lem:1}
$2$ is a unit in $R'$.
\end{lem}

\begin{proof}
Since $M$ is free of rank $2$, there exists a harmonic quadruple
$(p_0,p_1,p_2,p_3)$ in $\bP(M)^4$. We read off $p_2\dis p_3$ from (\ref{eq:2})
and (\ref{cond:ii}). Application of $\mu$ yields $p_2^\mu\dis p_3^\mu$ by
virtue of (\ref{eq:dist-pr}). Now (\ref{eq:2}) in turn shows that $2$ is a unit
in $R'$.
\end{proof}

\begin{lem}\label{lem:2}
Given bases $(e_0,e_1)$ of $M$ and $(e_0',e_1')$ of $M'$ such that
\begin{equation}\label{eq:H.e}
    (Re_0)^\mu = R'e_0',\quad (Re_1)^\mu = R'e_1',\quad
    \big(R(e_0\pm e_1)\big)^\mu = R'(e_0'\pm e_1')
\end{equation}
there exists a unique mapping $\beta:R\to R'$ with the property
\begin{equation}\label{eq:beta.e}
    \big(R(x e_0+ e_1)\big)^\mu = R'(x^\beta e_0'+ e_1')
    \quad\mbox{for all}\quad x\in R.
\end{equation}
This $\beta$ is additive and satisfies $1^\beta=1$.
\end{lem}

\begin{proof}
For any $x\in R$ the point $p:=R(x e_0+ e_1)$ is distant from $Re_0$. From
(\ref{eq:dist-pr}) follows $p^\mu \dis R'e_0'$ so that the point $p^\mu$ has a
unique generator of the form $x'e_0'+e_1'$ with $x'\in R'$. We therefore can
define a unique mapping $\beta:R \to R'$ satisfying condition (\ref{eq:beta.e})
by $x^\beta:=x'$.
\par
By (\ref{eq:GL-R*}), for all $x,y\in R$ with $x-y\in R^*$ the points
\begin{equation*}\label{eq:mitte}
\begin{aligned}
    &q_0:=R (x e_0+e_1),&
    &q_1:=R(y e_0+e_1),\\
    &q_2:=R\big((x+y)e_0+2 e_1\big),\quad&
    &q_3:=R\big((x-y)e_0\big)=R e_0
\end{aligned}
\end{equation*}
satisfy $\Harm(q_0,q_1,q_2,q_3)$. From (\ref{eq:beta.e}),
condition~(\ref{cond:ii}),
 and (\ref{eq:H.e}) follows
\begin{equation}\label{eq:mitte-mu}
\begin{aligned}
    &q_0^\mu=R (x^\beta e_0'+e_1'),&
    &q_1^\mu=R(y^\beta e_0'+e_1'),    \\
    &q_2^\mu=R\left(\left(\frac{x+y}{2}\right)^\beta e_0'+ e_1'\right),\quad&
    &q_3^\mu=R e_0'.
\end{aligned}
\end{equation}
We infer from (\ref{eq:dist-pr}) that $q_0^\mu \dis q_1^\mu$, and so $(x^\beta
e_0'+e_1',y^\beta e_0'+e_1')$ is a basis of $M'$. Now (\ref{eq:GL-R*}) yields
that $x^\beta-y^\beta$ is a unit in $R'$, whence
$q_3^\mu=R'\big((x^\beta-y^\beta)e_0'\big)$. By defining
\begin{equation}\label{eq:mitte'}
    q_2':=R'\big((x^\beta+y^\beta)e_0'+2 e_1'\big)
\end{equation}
we obtain $\Harm(q_0^\mu,q_1^\mu,q_2',q_3^\mu)$. The uniqueness of the third
harmonic point (see Section~\ref{se:basic}) shows $q_2'=q_2^\mu$. Comparing
(\ref{eq:mitte-mu}) with (\ref{eq:mitte'}) and taking into account
Lemma~\ref{lem:1} gives
\begin{equation}\label{eq:1/2}
    \left( \frac{x+y}{2} \right)^\beta = \frac{x^\beta+y^\beta}{2}
    \quad\mbox{for all}\quad x,y\in R
    \quad\mbox{with}\quad x-y\in R^*.
\end{equation}
Also $(R e_1)^\mu = R' e_1'$ implies $0^\beta=0$.
\par
Due to the last observation, condition~(\ref{cond:i}), Lemma~\ref{lem:1}, and
(\ref{eq:1/2}), we can apply the first part of \cite[Lemma~1]{lima+l-77b}. This
establishes that $\beta$ is additive. Moreover, (\ref{eq:H.e}) implies $1^\beta
= 1$.
\end{proof}

\begin{lem}\label{lem:3}
Let $(e_0,e_1)$, $(e_0',e_1')$ and $\beta$ be given as in
Lemma~{\rm\ref{lem:2}}. Let $t\in R$ be fixed. Then
\begin{equation}\label{eq:basis.f}
    (f_0,f_1):=(t e_0+ e_1,-e_0) \quad\mbox{and}\quad
    (f_0',f_1'):=(t^\beta e_0'+ e_1',-e_0')
\end{equation}
are bases of $M$ and $M'$, respectively, and there holds
\begin{equation}\label{eq:beta.f}
    \big(R(x f_0+ f_1)\big)^\mu = R'(x^\beta f_0'+ f_1')
    \quad\mbox{for all}\quad x\in R.
\end{equation}
\end{lem}

\begin{proof}
(a) The matrix of $(f_0,f_1)$ w.~r.~t.\ $(e_0,e_1)$ is $E(t)\in\E_2(R)$. So
$(f_0,f_1)$ is a basis of $M$. Likewise, $E(t^\beta)\in\E_2(R')$ shows that
$(f_0',f_1')$ is a basis of $M'$. We deduce $(Rf_0)^\mu = R' f_0'$ from
(\ref{eq:beta.e}), whereas (\ref{eq:H.e}) yields $(Rf_1)^\mu = R' f_1'$. Now
the additivity of $\beta$ together with $1^\beta=1$ gives
\begin{equation}\label{}
    \big(R(f_0\pm f_1)\big)^\mu = \Big(R\big( (t\mp 1)e_0 + e_1 \big)\Big)^\mu
        = R'\big( (t^\beta \mp 1)e_0' + e_1' \big)
        = R' (f_0'\pm f_1') .
\end{equation}
Consequently, as in Lemma~\ref{lem:2}, there is a unique mapping $\gamma:R\to
R'$ such that
\begin{equation}\label{eq:gamma.f}
    \big(R(x f_0+ f_1)\big)^\mu = R'(x^\gamma f_0'+ f_1')
    \quad\mbox{for all}\quad x\in R.
\end{equation}
Also, as before, $\gamma$ turns out to be additive with $1^\gamma=1$.
\par
(b) Consider a fixed $x\in R$ such that $1+x$ and $1-x$ are units. We define
\begin{equation*}\label{eq:g0123}
\begin{split}
    &g_0:=    (t+1)e_0+e_1 = f_0-f_1,\\
    &g_1:=    (t-1)e_0+e_1 = f_0+f_1,\\
    &g_2:=    2\big((t+x)e_0+e_1\big) = 2(f_0-x f_1),\\
    &g_3:=    2\big((1+xt)e_0+x e_1\big)   = 2(x f_0-f_1) .
\end{split}
\end{equation*}
The matrix of $(g_0,g_1)$ w.~r.~t.\ $(f_0,f_1)$ is in $\GL_2(R)$ due to $2\in
R^*$ and (\ref{eq:GL-R*}), whence $(g_0,g_1)$ is a basis. The equations
$(1+x)g_0+(1-x)g_1=g_2$ and $(1+x)g_0-(1-x)g_1=g_3$ yield that the points
$p_i:=Rg_i$, $i\in\{0,1,2,3\}$, satisfy $\Harm(p_0,p_1,p_2,p_3)$. We define
\begin{equation*}\label{eq:g0123'}
\begin{split}
    &g_0':=     \big((t^\beta+1)e_0'+e_1'\big) , \\
    &g_1':=     \big((t^\beta-1)e_0'+e_1'\big) ,\\
    &g_2':=    2\big((t^\beta+x^\beta)e_0'+e_1'\big) ,\\
    &g_3':=    2\big((1+x^\beta t^\beta)e_0'+x^\beta e_1'\big) ,
\end{split}
\end{equation*}
whence Lemma~\ref{lem:2} gives
\begin{equation}\label{eq:bilder}
     p_0^\mu= R'g_0' ,\quad  p_1^\mu= R'g_1' , \quad   p_2^\mu= R'g_2' .
\end{equation}
Now $\Harm(p_0^\mu,p_1^\mu,p_2^\mu,p_3^\mu)$ implies $p_0^\mu \dis p_2^\mu \dis
p_1^\mu$ so that
\begin{equation*}\label{eq:1+-x}
    \begin{pmatrix}
    t^\beta\ \pm 1 & 1 \\t^\beta +x^\beta  & 1
    \end{pmatrix}
    \in \GL_2(R')
\end{equation*}
which in turn, by (\ref{eq:GL-R*}), gives that $1+x^\beta$ and $1-x^\beta$ are
units in $R'$. We therefore are in a position to proceed as above in order to
establish $\Harm(R'g_0',R'g_1',R'g_2',R'g_3')$. By (\ref{eq:bilder}) and the
uniqueness of the fourth harmonic point, we obtain
\begin{equation*}\label{eq:p3.mu}
    p_3^\mu = R'g_3' = R'\big((1+x^\beta t^\beta)e_0'+x^\beta e_1'\big)
    = R'(x^\beta f_0' -  f_1').
\end{equation*}
On the other hand, writing $p_3=R\big((-x)f_0+f_1\big)$ allows us to apply
(\ref{eq:gamma.f}) which gives $p_3^\mu = R'\big((-x)^\gamma f_0' + f_1'\big)$.
The additivity of $\gamma$ yields
\begin{equation}\label{eq:beta=gamma.prov}
    x^\beta = x^\gamma \quad\mbox{for all}\quad  x\in R
    \quad\mbox{with}\quad 1+x
    \quad\mbox{and}\quad 1-x
    \quad\mbox{units.}
\end{equation}
\par
(c) If $x$ is any element of $R$ then, by condition~(\ref{cond:i}), there
exists $y\in R$ with $1+y$, $1-y$, $1+(x+y)$, and $1-(x+y)$ units. We infer
$y^\beta =y^\gamma$ and $(x+y)^\beta = (x+y)^\gamma$ from
(\ref{eq:beta=gamma.prov}) whence, by the additivity of $\beta$ and $\gamma$,
we obtain
\begin{equation*}\label{eq:beta=gamma}
    x^\beta = x^\gamma \quad\mbox{for all}\quad  x\in R.
\end{equation*}
This completes the proof of (\ref{eq:beta.f}).
\end{proof}

\begin{lem}\label{lem:4}
The mapping $\beta$ from Lemma~{\rm\ref{lem:2}} is a Jordan homomorphism.
\end{lem}

\begin{proof}
We make use of Lemma~\ref{lem:3} in the special case $t=0$, i.~e.,
$(f_0,f_1)=(e_1,-e_0)$ and $(f_0',f_1')=(e_1',-e_0')$. Given any $x\in R^*$ we
calculate the image of $R(xe_0+e_1) = R(-x^{-1}f_0 + f_1)$ according to
(\ref{eq:beta.e}) and (\ref{eq:beta.f}). This gives
\begin{equation*}\label{eq:invert}
    R'(x^\beta e_0'+e_1') =
    R'\big((-x^{-1} )^\beta f_0' + f_1'\big) =
    R'\big((-x^{-1})^\beta e_1' - e_0'\big).
\end{equation*}
Since $x^\beta e_0'+e_1'$ and $(-x^{-1})^\beta e_1' - e_0'$ are admissible
generators of the same point, there exists a unit $u' \in R'$ with $u'(x^\beta
e_0'+e_1') = (-x^{-1})^\beta e_1' - e_0'$. Now $u'x^\beta=-1$ implies that
$x^\beta$ is a unit in $R'$ and, by the additivity of $\beta$, we obtain
\begin{equation}\label{eq:inv}
    (x^\beta)^{-1} = (x^{-1})^\beta \quad\mbox{for all}\quad  x\in R^*.
\end{equation}
Due to $1^\beta=1$ and (\ref{eq:inv}) we are in a position to apply also the
second part of \cite[Lemma~1]{lima+l-77b} which establishes that $\beta$
satisfies
\begin{equation}\label{eq:xy+yx}
    (xy+yx)^\beta = x^\beta y^\beta +y^\beta x^\beta
    \quad\mbox{for all}\quad  x,y\in R.
\end{equation}
Recall that $2$ is a unit in $R$ by condition~(\ref{cond:ii}), and also a unit
in $R'$ by Lemma~\ref{lem:1}. Moreover, from Lemma~\ref{lem:2}, $\beta$ is
additive and satisfies $1^\beta=1$. It is well known that under these
circumstances (\ref{eq:xy+yx}) characterises $\beta$ as being a Jordan
homomorphism; see, e.~g., \cite[p.~47]{herst-69a} or \cite[p.~320]{hua-52a}.
\end{proof}

\begin{proof}[Proof of Theorem {\rm\ref{thm:staudt}}]
Choose any point of the connected component $C$, say $p_0=R a_{0}$, and any
$a_{1}\in M$ such that $(a_{0},a_{1})$ is a basis of $M$. Let $p_1:=R a_{1}$,
$p_2:= R(a_{0}+a_{1})$, and $p_3:= R(a_{0}-a_{1})$. Then
$\Harm(p_0,p_1,p_2,p_3)$ implies $\Harm(p_0^\mu,p_1^\mu,p_2^\mu,p_3^\mu)$ so
that there exists a basis $(a_{0}',a_{1}')$ of $M'$ satisfying
\begin{equation}\label{eq:H.a}
    (Ra_{0})^\mu= R' a_{0}',\quad
    (Ra_{1})^\mu=R' a_{1}',\quad
    \big(R(a_{0}\pm a_{0})\big)^\mu=R' (a_{0}'\pm a_{1}') .
\end{equation}
We apply Lemma~\ref{lem:2} to the bases $(a_0,a_1)$ and $(a_0',a_1')$, but
relabel the mapping $\beta$ from there as $\alpha$. So, by (\ref{eq:beta.e})
and Lemma~\ref{lem:4}, there exists a Jordan homomorphism $\alpha:R\to R'$ with
\begin{equation}\label{eq:alpha.a}
    \big(R(x a_0+ a_1)\big)^\mu = R'(x^\alpha a_0'+ a_1')
    \quad\mbox{for all}\quad x\in R .
\end{equation}
By (\ref{eq:component}), a point $p\in \bP(M)$ belongs to $C$ precisely when
there is at least one sequence $T\in\Seq(R)$ such that $p=R(x_0 a_{0} + x_1
a_{1})$ with $(x_0, x_1)=(1,0)\cdot E(T)$. It therefore remains to verify that
for all finite sequences $T\in\Seq(R)$ the coordinate rows $(x_0,x_1)$ and
$(x_0',x_1')$ from (\ref{eq:(x0,x1)(x0',x1')}) define points which correspond
under $\mu$. We proceed by induction on the length of $T$ which will be denoted
by $n$.
\par
For $n=0$ the sequence $T$ is empty and $E()$ is the identity matrix. Now
(\ref{eq:(x0,x1)(x0',x1')}) reads $(x_0, x_1) = (1,0)$, $(x_0', x_1') = (1,0)$,
and indeed $(R a_{0})^\mu = R'a_{0}'$ according to (\ref{eq:H.a}).
\par
For $n=1$ we have $T=(t_1)$ with $t_1\in R$. The assertion follows from
(\ref{eq:alpha.a}), since (\ref{eq:(x0,x1)(x0',x1')}) now takes the form $(x_0,
x_1) = (t_1,1)$, $(x_0', x_1') = (t_1^\alpha,1)$.
\par
Let $n\geq 2$ and suppose $T=(t_1,t_2,\ldots,t_n)\in\Seq(R)$. There is a unique
basis of $M$, say $(e_0,e_1)$, with $E(t_3,\ldots,t_n)$ being its matrix
w.~r.~t.\ $(a_0,a_1)$. We proceed analogously in $M'$ and obtain a basis
$(e_0',e_1')$ with $E(t_3^\alpha,\ldots,t_n^\alpha)$ being its matrix w.~r.~t.\
$(a_0',a_1')$. The following table displays for all $x\in R$ the coordinates of
certain elements of $M$ and $M'$:
\begin{equation}\label{eq:tabelle.e}
    \begin{array}{|c|l|c|l|}
    \hline
    \multicolumn{2}{|c|}{\mbox{Coordinates w.~r.~t.~} (a_0,a_1)}&
    \multicolumn{2}{|c|}{\mbox{Coordinates w.~r.~t.~} (a_0',a_1')}\\
      \hline
        e_0 & (1,0)\cdot E(t_3,\ldots,t_n) &
        e_0' & (1,0)\cdot E(t_3^\alpha,\ldots,t_n^\alpha)\\
        x e_0+e_1 & (1,0)\cdot E(x,t_3,\ldots,t_n) & x^\alpha e_0'+e_1' &
        (1,0)\cdot E(x^\alpha,t_3^\alpha,\ldots,t_n^\alpha)\\
    \hline
    \end{array}
\end{equation}
Those elements of $M$ and $M'$ which appear in the same row of table
(\ref{eq:tabelle.e}) generate corresponding points under $\mu$ due to the
induction hypothesis. In particular, as $x$ ranges in $\{0,1,-1\}$, we get
\begin{equation*}\label{eq:H.e-neu}
    (Re_0)^\mu= R'e_0',\quad
    (Re_1)^\mu= R'e_1', \quad
    \big(R(\pm e_0+e_1)\big)^\mu= R'(\pm e_0'+e_1').
\end{equation*}
Hence the bases $(e_0,e_1)$ and $(e_0',e_1')$ satisfy (\ref{eq:H.e}) so that
Lemma~\ref{lem:2} can be applied to them (without any notational changes). We
claim that $\alpha$, as defined via (\ref{eq:alpha.a}), coincides with the
Jordan homomorphism $\beta$ appearing in Lemma~\ref{lem:2}: Indeed, $\alpha$
satisfies the defining equation (\ref{eq:beta.e}) according to the second row
of table (\ref{eq:tabelle.e}) in conjunction with the induction hypothesis. We
now introduce bases $(f_0,f_1)$ of $M$ and $(f_0',f_1')$ of $M'$ as in
Lemma~\ref{lem:3}, but replace the arbitrary $t\in R$ from there by the given
$t_2\in R$. This gives a second table of coordinates:
\begin{equation}\label{eq:tabelle.f}
    \begin{array}{|c|l|c|l|}
    \hline
    \multicolumn{2}{|c|}{\mbox{Coordinates w.~r.~t.~} (a_0,a_1)}&
    \multicolumn{2}{|c|}{\mbox{Coordinates w.~r.~t.~} (a_0',a_1')}\\
      \hline
      f_0 & (1,0)\cdot E(t_2,\ldots,t_n) &f_0' & (1,0)\cdot E(t_2^\alpha,\ldots,t_n^\alpha)\\
    t_1 f_0+f_1 & (1,0)\cdot E(t_1,t_2,\ldots,t_n) &    t_1^\alpha f_0'+f_1' & (1,0)\cdot E(t_1^\alpha,t_2^\alpha,\ldots,t_n^\alpha)\\
    \hline
    \end{array}
\end{equation}
Since $\alpha=\beta$, we can read off from (\ref{eq:beta.f}) that
$\big(R(t_1f_0+f_1)\big)^\mu = R'(t_1^\alpha f_0'+f_1')$. Hence the coordinates
from the last row of table (\ref{eq:tabelle.f}) describe points which
correspond under $\mu$.
\end{proof}

\paragraph{Acknowledgements.} The author expresses his warmest thanks to Mark Pankov (Olsztyn) for the translation
of several articles from Russian to English.

redpajama_set_name: RedPajamaArXiv


text:
Riyadh - Al Dawadimi
12:44 VBA still in the virtual lead
At km 299, three quarters into today's special, Cornejo is still in control by 1'18' over Kevin Benavides, 3'09' over Barreda and 5'37' over Santolino. Van Beveren is 9'03' down, but his direct competitors for the overall are far behind, with Walkner at 22'09' and Sunderland and 25'17'. VBA is still in the virtual lead. This could be the first major change of the second week.
12:21 Maciej Giemza and Jan Brabec in the thick of it
Maciej Giemza and Jan Brabec went through km 298, three quarters of the way through, in 12th and 13th place, beating a biker of the calibre of Quintanilla! Only Xavier de Soultrait, who used to be a factory rider, is ahead of these two among the pure private competitors who go head to head with the pros.
12:13 'Santo' and Sherco second in the virtual overall
VBA and Yamaha are not the only ones who stand to profit from stage 7. Sherco's Santolini moved up to second place in the virtual overall at km 259, 7′13″ behind the Frenchman. Walkner is trailing the leader by 8′02″, followed by Sunderland at 8′31″ and Benavides at 9′28″. We have a Frenchman in first place and a French bike in second!
12:04 Adrien Van Beveren in the virtual overall lead
Starting in 22nd place this morning, Cornejo has turned the pecking order on its head at the last three checkpoints. 259 km into the stage, he leads the special by 1'03' on Benavides, 2'55' on Barreda, 4'22' on Luciano Benavides and Xavier de Soultrait at 4'34', 5 seconds ahead of Toby Price. Santolino is within 5 minutes of the leader and Brabec is 5'52' down.
Meanwhile, Adrien Van Beveren is pulling off a coup in the general standings! He is 12th at 7'27', good enough to take 12'15' back on Walkner and 15'22' on Sunderland, putting the Frenchman in the virtual lead at this time check. It's almost 150 km from here to Ad-Dawadimi.

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text:
Entered Court Orders
In Contested Wills, Family Law Matters, Legal Definitions and Terms
The Court does not have jurisdiction to re open entered court orders but may vary the order where has been a change of circumstance.
The court refused to re open or vary an entered court order in Sugrim v Sugrim 2016 BCSC 1644 when after entering a desk order under the Family Relations act under S 57, (which has the effect of crystalizing matrimonial assets and severing jointly owned titles), the husband became incapacitated and was awarded $1.8 million in damages for the injury.
With the consent of the committee adult child of the patient, the wife's application to set aside the entered court order was dismissed on the basis that inter alia there was a risk that the wife and her children were acting out of self interest and not in the best interest of the patient.
The Sugren case stated:
[22] The PGTBC submits and I agree that the court does not have jurisdiction to reopen and set aside an entered order. The court may, however, entertain a variation of an order on the basis of a change of circumstance.
[23] I raised a question as to the validity of a consent to a s. 57 declaration. However, having reviewed the circumstances of this case and the authorities, I conclude that the declaration was valid.
[24] This was not a situation where one party at a judicial case conference requested a s. 57 declaration and the other party opposed it, as was the case in Harrison v. Harrison, 2007 BCCA 120. InHarrison, Finch C.J. found the death of Mr. Harrison was a material new circumstance giving rise to a reconsideration of the previous order to avoid a miscarriage of justice. In that case, the order was not entered, so the court was not functus officio and there was evidence of a possible miscarriage of justice which does not exist in the present case.
[25] The only issue before me is whether or not a committee has the authority to consent to an application to set aside the s. 57 declaration based on a material change in circumstance.
[26] The claimant submits that s. 15 of the PPA provides that a committee of the patient has all the rights, privileges and powers with regard to the estate of the patient as the patient would have if of age of majority and of sound and disposing mind. Those rights include the ability to commence defend or otherwise conduct legal proceedings.
[27] The PGTBC submits that the committee's power is qualified in that she may only act in the best interests of the respondent.
[28] In Beadle v. Beadle, 1984 CanLII 806 (BC CA), the PGTBC continued divorce proceedings and set an application for an undefended divorce down for hearing on behalf of an incapacitated claimant, citing Re Swartz, [1947] 2 W.W.R. 979 at 980 (B.C.C.A.) where the court agreed that a committee has the power to bring a divorce action on behalf of the patient and there was nothing in the PPA which limited that power. In Beadle, at para. 8, MacFarlane J.A. held that:
… [the] appointment [of a committee] contemplates … the full management of the affairs of the patient who is incapable of doing so herself. That includes, in my opinion, the management or conduct of any litigation which the patient has undertaken, or which might need to be undertaken in the best interests of the patient. To hold otherwise would put a severe limitation upon the proper management of a patient's affairs while she is incompetent.
[29] In Beadle, the court made a distinction between dealing with property matters and dealing with questions of status. In response to this concern, the court said that there are controls in place to ensure that a committee exercises caution in bringing proceedings which involve the status of the patient. Those controls lie in the discretion of the PGTBC under the PPA and under the Supreme Court Civil Rules where the court has the discretion to deny an appointment of a committee as the guardian ad litem of the patient in a proceeding.
[30] A litigation guardian must declare that he or she does not have any interest in the proceeding that is adverse in interest to the patient (Rule 20-2 of the Supreme Court Civil Rules).
[31] Protection of the interests of the incompetent party is the primary consideration of the court.
[32] In a concurring judgment, Lambert J.A. added (at para. 26):
… that where any proceedings are instituted by a committee in which matters of status and morals are involved, there is an obligation on the committee to bring before the court evidence that will satisfy the court that the proceedings are in the best interests of the person who is being represented by the committee.
[33] In this case, the claimant says there has been a material change in circumstance which should give rise to a variation of the s. 57 declaration because the claimant asserts there is no longer a marital rift that will lead to a final divorce order.
[34] The claimant asserts that there is a reasonable prospect of reconciliation as one party wishes to reconcile and the other party consents by way of his committee. This submission focuses on the status of the parties and not on what is in the patient's best interests. The respondent's wish prior to his incapacity was to separate from the claimant. It is unclear to me how the committee could come to the conclusion that he would now want to reconcile. It is open to the committee to satisfy the court that this change in status would be in the respondent's best interests even though he cannot consent.
[35] In Anderson v. Anderson Estates, [1990] O.J. No. 1123 (H. Ct. J.), the court considered whether an attorney appointed by power of attorney had the authority to exercise the patient's right to elect or consent to receive an equalization payment in lieu of entitlement under the deceased spouse's will. The court concluded that the fact the patient was incapacitated from personally making the election should not diminish her right. The Powers of Attorney Act, R.S.O. 1980, c. 386, s. 5 contemplated an attorney continuing to manage the affairs of the donor after the donor was deemed incompetent.
[36] In Anderson, at para. 13, the court said:
… the right to elect is a very personal decision that should only be exercised by the surviving spouse and not left to a stranger to the marriage, lest the stranger interfere with the testator's intentions without knowing whether or not the surviving spouse has made the choice to disregard his or her spouse's last wishes. …, this concern can somewhat be alleviated by the fact that an attorney under power of attorney will always have a fiduciary duty to act in the best interest of the donor. It may also be possible for the donor to set out in the power of attorney his or her wishes with respect to such an election.
[37] In the case before me, the claimant also asserts that it is in the respondent's best interests to "have a home to return to should he ever be able to leave the care of the facility in which he currently resides."
[38] This assertion begs the question of whether the respondent is welcome in the home if the claimant is not successful with this application.
[39] The respondent's incapacity is a material change in circumstance which may give rise to grounds to vary the declaration.
[40] I agree with the PGTBC's submission that the entered order cannot be reopened or re-heard and the only way a court can reconsider the order is on a variation application.
[41] A variation application brought by the committee of the respondent will only be successful if the committee can satisfy the court that it is in the patient's best interests to vary the order.
[42] Having considered the reasons for the application, I conclude that while a variation of the s. 57 declaration may well be in the best interests of the claimant and her children, there is no evidence that satisfies me that a variation of the s. 57 declaration is in the best interests of the respondent. There is a risk that the claimant and her children are acting out of self-interest. We will never know what the respondent would want at this time, and his interests must be protected.
[43] As far as I know, the respondent is being well cared for in a long-term care facility receiving 24-hour a day nursing care. His family visit him on a regular basis, and ensure that the care is adequate. He has sufficient funds in trust to finance his care. The PGTBC monitors the spending of those funds to make sure they are used in only his best interests and not in the interest of anyone else. Upon his death, those funds will be left to the beneficiaries of his estate. There is nothing in the material that leads me to believe the current situation is not in the best interests of the respondent. Accordingly, I dismiss the application.
Examination For Discovery Limited to 7 Hours
Court Inherent Jurisdiction to Prevent Miscarriage of Justice
Guarantee Claims


text:
Math scores slip for U.S. high school seniors
Posted: 12:44 PM, Apr 27, 2016
It's not a promising picture for the nation's high school seniors — they are slipping in math, not making strides in reading and only about one-third are prepared for the academic challenges of entry-level college courses.
Scores released Wednesday from the so-called Nation's Report Card show one-quarter of 12th-graders taking the test performed proficiently or better in math. Only 37 percent of the students were proficient or above in reading.
The average math score on the test last year was 152, down from 153 in 2013, the last time the test was given. It marks the first drop in math in a decade. For reading, scores were flat over the same period of time, and down five points from more than two decades ago when the test was first given to students in 1992.
Education Secretary John B. King, Jr., says schools have undergone "some of the most significant changes in decades" as teachers retool their classroom practices to adapt to new and higher standards.
"We know the results of those changes will not be seen overnight, so we need to be patient — but not passive — in continuing to pursue the goal of preparing all students for success after high school," King said.
Since 2009, more than 40 states have adopted the Common Core learning standards, which outline skills students should learn and know in math and reading by the end of each grade. They emphasize critical thinking, with less of a focus on memorization.
Peggy Carr, acting commissioner of the National Center for Education Statistics, which administers the test, said the report suggests a pattern in scores for reading that needs a closer look. "There is a gap, a widening of a gap of higher and lower ability students, and I think that's something we need to think about," said Carr.
Reading scores increased for the highest-performing students last year, up two points for the highest group of performers on the test; and down six points from 2013 for the lowest-performing seniors. Math scores saw no significant difference over two years for the highest group of test-takers, but declined for the group of students at the bottom.
Bill Bushaw, executive director of the National Assessment Governing Board, said the scores are disappointing.
"We're not making the academic progress that we need to so that there's greater preparedness for post-secondary, for work, for military participation. These numbers aren't going the way we want," Bushaw said.
The report estimates about 37 percent of students, for both reading and math, scored well enough to be considered likely to possess the knowledge and skills to be academically prepared for college-level work. That is not much different than how well-prepared seniors were in 2013, the last time the tests were given.
—The average math score was 152, on a 300-point scale. The average score for reading was 287 on a 500-point scale.
—No significant change was seen from 2013 in the average math score for any racial and ethnic groups. And it was the same for reading, with no real change seen from 2013 for any groups.
—In math, the average score for English language learners was higher last year, up six points compared to 2013.
The National Assessment of Educational Progress (NAEP) is considered a national yardstick by which to measure student achievement. The math test was given last year to about 13,200 twelfth-graders in public and private schools. About 18,700 students took the reading test.


text:
Search Dover Air Force Base: Search
Air Refueling
CRT: Cheyenne Mountain
Military Working Dog
A C-17A Globemaster III taxis down the flight line prior to takeoff Jan. 9, 2019 at Dover Air Force Base, Del. Thirteen C-17As are assigned to Dover AFB. (U.S. Air Force photo by Roland Balik)
Tags: Dover AFB, 436th Airlift Wing, C-17A, 436 AW, DAFB, Del., 326th Airlift Squadron, DE, Delaware, 3d Airlift Squadron, 436th AW, 436th Maintenance Group, Eagle Wing, C-17A Globemaster III, 712th AMXS, 736th Aircraft Maintenance Squadron, 736th AMXS, 512th Maintenance Group, 3d AS, 736 AMXS, 2019, Department of Defense, U.S. Air Force, USAF, C-17, AMC, Air Mobility Command, DoD, United States Air Force, Dover Air Force Base, Team Dover, Air Force Reserve Command, Aircraft, Del, U.S.A.F., Liberty Wing, 512th Airlift Wing, 3d, 3 AS, 3rd AS, 18th Air Force, 326 AS, 326th AS, 712 AMXS, 712th Aircraft Maintenance Squadron
Photo by: GS-09 / Roland Balik | VIRIN: 190109-F-BO262-1004.JPG


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Homespace invites you to their Laughtober Fest Fundraiser.
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RSVP No later than September 22.


text:
A Brief History Of Federal Government Shutdowns
History tells us that if there is a shutdown tonight, it will most likely be a short one.
Doug Mataconis · Friday, April 8, 2011 · 12 comments
NBC's Pete Williams has a post today listing the government shutdowns of the past, unfortunately the list is largely unreadable thanks to two huge floating ads. So, here it is (all starting times are midnight on the date given):
September 30 to October 11, 1976 (10 days)
October 31 to November 9, 1977 (8 days)
November 30 to December 9, 1977 (8 days)
September 30 to October 18, 1978 (18 days
November 20 to November 23, 1981 (2 days)
September 30 to October 2, 1982 (1 day)
December 17 to December 21, 1982 (3 days)
September 30 to October 3, 1984 (2 days)
October 3 to October 5, 1984 (1 day)
October 16 to October 18, 1986 (1 day)
December 18 to December 20, 1987 (1 day)
October 5 to October 9, 1990 (3 days)
December 5, 1995 to January 6, 1996 (21 days)
One of the interesting things to note is that, after 1980, the length of shutdowns was usually very short, 2 or 3 days at most. The exceptions to that are the disastrous shutdowns at the end of 1995. The main reason for that is that there was a change in the way a law called the Anti-Deficiency Act was interpreted in this situations (PDF of Congressional Research Service report):
For years, many federal agencies continued to operate during a funding gap, while "minimizing all nonessential operations and obligations, believing that Congress did not intend that agencies close down" while waiting for the enactment of annual appropriations acts or continuing resolutions.11 In 1980 and 1981, however, Attorney General Benjamin R. Civiletti issued two opinions that more strictly interpreted the Antideficiency Act in the context of a funding gap, along with its exceptions.12 The opinions stated that, with some exceptions, the head of an agency could avoid violating the Antideficiency Act only by suspending the agency's operations until the enactment of an appropriation. In the absence of appropriations, exceptions would be allowed only when there is "some reasonable and articulable connection between the function to be
performed and the safety of human life or the protection of property."
Basically then, before Civiletti's opinion, government shutdowns were relatively painless affairs because large numbers of government offices continued to operate. This means there was less pressure on Congress to actually resolve the dispute. Once it became clear that shutting the government down would have real consequences, though, Congress and the White House usually had an incentive to resolve their disputes quickly. The one exception since 1980 involved two competing and contentious personalities.
The lesson for 2011 from all this ? Two things, I think.
First, if there is a shutdown it's likely to be a short one. Unlike 1995/96, there don't seem to be the same type of personality conflicts between Boehner/Cantor and Reid/Obama that there were between Gingrich and Clinton. I don't think Boehner really wants a shutdown, but he also has to proceed in these negotiations in a way that isn't going to totally piss off the base. If that means that it will take a one or two day shutdown followed by a deal, then that may be what it will take.
Second, if you want to make government shutdowns less likely (although they are clearly less common now than they were in the 70's and 80s), then the best way to do that is to make a shutdown more painful by restricting even further the government functions that could operation during a shutdown. If Congress and the White House know that they will cause real pain if they don't resolve their disputes, then history suggests that they will make the deal rather than risking the pain and the inevitable political fallout.
FILED UNDER: Congress, Deficit and Debt, US Politics
SJ Reidhead says:
Friday, April 8, 2011 at 14:58
A brief history: Republicans always lose, end of story. It doesn't matter how badly the Dems are acting, the press will spin it their way.
reid says:
"The Liberal Media" — the biggest myth there is.
ponce says:
"Republicans always lose,"
I think this shutdown is a win-win for Boner.
If America blames Obama, he wins.
If America blames the Republicans, it will be the end of the fringe right freaks of the Tea Party who have plagued his brief reign.
Either way, Boner comes out ahead.
Axel Edgren says:
A brief history: Democrats have went more than halfway to meet the teepers billions-wise, but said scum wants to destroy Planned Parenthood and EPA as well so there might be a shutdown.
So I think you need to illuminate yourself. I suggest launching yourself into the sun.
Which is odd because the Tea Party people have stridently claimed they are just fiscal conservatives, not social cons.
RonR says:
For gen. info to liberals; TEA Party stands for "taxed enough already". By definition they ARE fiscal conservatives. Liberals seem to believe that money has some other source than taxes, this is the key to the current dispute. I do, however, believe that Sen Boehner should be asking for bigger cuts.
Saturday, April 9, 2011 at 09:29
"Liberals seem to believe that money has some other source than taxes"
Yes, and conservatives think that their profits should come on the backs of the people who actually work for a living. Please, let's call a spade a spade. The republican constituency consists of those who can afford to be taxed at a higher rate but never need to worry about it as long as their political puppets are in control of the government. Take away the Bush tax cuts and the budget deficit could be paid down with an additional $680 BILLION in currently sheltered tax breaks over the next 10 years (http://www.nytimes.com/2010/08/23/opinion/23krugman.html?_r=1)
Enough of the corrupt political process. And CERTAINLY enough of protecting the wealthy from doing their fair share
Farnham says:
… and the extreme fiscal conservatives seem to believe that roads, education, and national defense can be paid for without taxes.
D.Vader says:
Wednesday, April 13, 2011 at 13:10
@SJ Reidhead "A brief history: Republicans always lose, end of story."
Like they did in November?
No the issue isn't how the press will spin it, we all know the press will turn the republicans into boogiemen. The point is do these freshmen have the balls to do the job the promised. So far I am not convinced, this $38 billion in cuts is LAUGHABLE. It may pay for the national debts interest for a few days.
@Farnham "… and the extreme fiscal conservatives seem to believe that roads, education, and national defense can be paid for without taxes."
Seriously? 😛 Name me ONE extreme fiscal conservative that is active in his political career that thinks that?
@Axel "A brief history: Democrats have went more than halfway to meet the teepers billions-wise, but said scum wants to destroy Planned Parenthood and EPA as well so there might be a shutdown."
You know with all the support for Planned Parenthood out there, why can't they get private funding?
The EPA is overly powerful as it is. I wouldn't shed a tar for it's loss. I don't vote for the EPA so why do they get a say in my life?
@Jesse
You certainly are blinded son. Has anyone even figured out where this Stimulus cash went? I know some of it went to some lawyers that helped Hilary with her campaign.
While you cry, piss and moan about tax cuts because you have no respect for the producers of this country at least what Bush did was legal and above the board. So far much of this stimulus money seems to have disappeared into a black hole of political paybacks (yes by your pure as the driven snow Obama).
So your solution is the same tried solution…
Lets not regulate our politicians spending, they should just keep blowing money like they always have. Lets soak the rich (aka producers and job creators). So we can pay for some more handouts for those who don't produce.
Liberals…. grow up and get your head out of your collective asses. The government is not an extension of Mommy and Daddy. If you have pet programs you want funded, by all means use the power of your freedom to establish a foundation. The government is not some nanny state intended to save everyone from all their ills. Considering the amount of red tape the Gov't is a piss poor means to distribute social aid and the fact that politics gets mixed into that is a sham. Just more BS for democrats to whine about and pull heart stings.
Cry me a river and change your diaper.
Congress Looking At Ways To Make Government Shutdowns Impossible
GAO: Obama Broke The Law In Bergdahl Release Deal
It's Time To Make Government Shutdowns Impossible
Shutdown's Impact Likely To Be Limited
Shutting Down The Government Won't Stop Obamacare


Lynn and Colin being keen golfers decided to have their pre wedding shoot at Greenmount Golf Club. We had a stunning blue sky day and the course looked stunning ( even for one who is not the slightest bit interested in golf!!). They were both a little nervous but looked so relaxed on the shots when we got going. As I am behind with my uploading I can say now I look forward to photographing their wedding this coming Saturday!!!


text:
100,000 newborn babies will have their DNA completely decoded
One hundred thousand newborn babies in England will have their genomes sequenced, in a £105m research program that could pave the way for a full-scale neonatal screening scheme to detect rare genetic conditions.
Genomics England, a government-owned company, aims to read all the DNA carried by a nationally representative sample of babies shortly after they are born. The two-year project, which will be carried out in collaboration with the NHS, will start late next year.
"Generating this evidence will allow policy-makers to make well-informed decisions about whether and how whole-genome sequencing can be rolled out as part of a future newborn screening programme," said Richard Scott, chief medical officer of Genomics England.
Currently, babies are given a "heel poke test" when they are four or five days old. Biochemical analysis of these tiny blood samples detects nine potentially serious conditions so that treatment can start as soon as possible.
Whole-genome sequencing – reading all 3 billion "letters" of the baby's genetic code – is expected to expand the number of treatable conditions being discovered to around 200, many of them extremely rare.
"We estimate that there are approximately 3,000 children born in the UK a year who could be helped by this approach if adopted nationally," Scott said. The number of tests requiring further investigation among the 100,000 babies is likely to be in the range of 1,500 to 2,000.
"The genetic variants we select are well-established causes of serious childhood conditions that benefit from early detection and treatment," said David Bick, lead clinician for the Newborn Genomes Programme.
One of the goals is to put an end to the agonizing "diagnostic odyssey" that thousands of parents are subjected to if their young children develop a mentally or physically debilitating disorder with an unknown cause. Without genomic analysis, this usually takes four to five years, Scott said.
NHS trusts across England will select participants, chosen to be as representative as possible of the population, with an emphasis on achieving the ethnic diversity that has often been lacking in medical research projects, Bick added. Scotland, Wales and Northern Ireland may join later.
Selected parents will be contacted mid-pregnancy about potential participation, so they are ready for their baby to undergo more than a standard heel prick test. The benefits and risks will be explained, including the wider implications of storing a baby's genome over a lifetime.
The Genomics England project is one of eight newborn sequencing studies being planned around the world, but most of the others are on a smaller scale.
In its previous 100,000 Genomes Project, which focused on patients with cancer and rare diseases, Genomics England used equipment from US-based Illumina to read volunteers' DNA. The sequencing technology to be used for the Newborn Genomes program had not yet been determined, Scott said.
Although the Newborn Genomes program was designed in consultation with medical ethicists, several geneticists expressed doubts whether the NHS has the resources to carry out comprehensive genomic screening on top of its existing work.
"Using whole genome sequencing to screen newborn babies is a step into the unknown," said Frances Flinter, a clinical geneticist at Guy's and St Thomas' NHS Foundation Trust. "We must not rush to use this technology until both the science and the ethics are clear."
Why Candice Warner turned down Cricket Australia's offer of security for the Test match in South Africa
Dogecoin traders may enter short positions if this scenario plays out
Satellite images show how skiers are faring in one of the worst years on record
Public support for nurses' strike puts pressure on Sunak and divides Tories | Labor struggle
IWG reports rise in UK workplace visits as hybrid working takes hold | Commercial property


text:
Emerson on Fox News: Our Shameful Refusal to Call the Paris Attacks Islamic Terrorism
Interview on Fox News
https://www.investigativeproject.org/4720/emerson-on-fox-news-our-shameful-refusal-to-call
Martha MacCallum: We are back and we are waiting for comments from Secretary of State John Kerry. He's expected to address the horrific attack that took place in the offices of a satirical newspaper in Paris this morning. Steve Emerson is the executive director of the Investigative Project on Terrorism and he joins me now. Steve, your reaction to this and your thoughts on how prepared we are for something like this here at home?
Steve Emerson: First of all I wish I was surprised, but in the last four weeks as you know, Martha, France has experienced three similar attacks. Unfortunately French government officials refuse to categorize them as Islamic terrorist attacks due to political correctness. Fortunately they did characterize this one today. But our government today in this morning's press reaction by Josh Earnest refused to call it an Islamic terrorist attack; [all they] did [was] call it [plain] "terrorism" [and] just called it "extremism." The bottom line here, these are Islamic terrorist attacks and need to be called as such. The second thing is that these attacks are induced not by ISIS alone, not by al Qaeda alone, but by Muslim Brotherhood groups acting in the West that always spread the ideological poison that there is a "war against Islam." These are very groups that are invited to do outreach by the White House or by Western governments, and …[their incendiary message is] the number one inducement to carry out such attacks.
So are we prepared? Well, bottom line is the FBI has been doing a great job in stopping 98% of the terrorist plots since 9/11. But you can't stop all of them. So if armed terrorists want to go into a building of journalists or commercial activities, they can carry out such an attack and kill scores of people. The question is whether we can deter it or whether we can use informants and get [them] in place in time. And I don't know whether we can do that.
Martha MacCallum: You've touched on something that I think is so central to this fight because no, you cannot protect every single building in every city, it simply can't be done. But we did hear quite a bit about all the people, hundreds, who have passports who went to Syria and went to the Middle East to train with some of these groups. I often asked myself what the follow-up has been on that and how good of a job we're doing at figuring out who these people are before they walk into these buildings. Steve?
Steve Emerson: Here's the question Martha. I don't think we are doing a good enough job of identifying these volunteer fighters for ISIS or for jihadis overseas before or after; before they go or after they return. That's number one. Certainly in Europe they have not identified them properly and that is why they had the attacks in France last year, or in Belgium. Number two, you don't need to go to France, to Syria or Iraq anymore to become a jihadi. You go online to learn how to build a bomb. You can acquire weapons here. And three, the motivation is here online or in the communities among radical mosques in the United States or in Paris. That is the bottom line here. You don't see denunciations of radical Islam by name by mainstream Islamic groups. What you hear is that this has nothing to do with Islam. Islam is not the definition of terrorism, far from it. The point is there is a wing within Islam, radical Islamists, who believe in this and they have to be condemned. And unless they are ostracized and delegitimized, these attackers and these attacks are going to continue. And ultimately you can't seal off and insulate all of society from attacks. They're going to succeed ultimately. So as a counter-terrorist specialist, what I say here is that the only way to defeat this type of Islamic terrorism is ultimately to drive them from the blessings of the community of those radical Islamists that live within us. Unless we do it, we are going to continue to see [more of] his. And unless we call it what it is, radical Islamic extremism, [repeat] radical Islamic terrorism, we are going to [be giving] them a free pass.
Martha MacCallum: They are encroaching on new places and new ground as they have seen this morning. Steve, thank you very much. Good to have you with us.
Related Topics: Steven Emerson, Terrorist Attack, France, Paris, Islamic Extremism, Josh Earnest, ISIS, al Qaeda, Muslim Brotherhood
The Reality of our Reaction to Terrorism
Submitted by Hugh Andrew, Jan 8, 2015 06:25
I encourage you to read the article by my favorite reporter published in the National Post, Canada http://news.nationalpost.com/2015/01/07/christie-blatchford-terrorists-cowed-us-all-into-ridiculous-self-censorship/
The Deliberate President
Submitted by Michael Fox, Jan 8, 2015 04:32
That comes down directly from President Obama's orders. He will use the word Islam in describing Islam as a peaceful , loving religion and he will use the term terrorism.... but he is very careful to never link together the term, "Islamic terrorism.. '
Often it's as much as what a person doesn't say as it is what he does say.
terror attacks
Submitted by gee-gee, Jan 7, 2015 21:16
These people are following the Qur'an and the Sunna,as ordered by Muhammed.Call it what you will.They are fulfilling their religious duty.
In America,Britain,Australia,the people who have gone over to fight, are returning.What ? So they are now hardened warriors, That will be great for our countries.
I would exile them.The enemy is within the gates.Our time is coming.Sooner than one would think.Wake up your country needs you.
why are we not waking up to the fact that imams are behind this
Submitted by staffsgt7, Jan 7, 2015 20:46
We mention that there are radical mosques but mosques are not people. It is the imam/clerics that are behind this. They are teaching hate and intolerance and it is in the majority of mosques that they are doing this. A few years ago an article came out that said it was in 84% of the mosques that this is being taught. Their daily 'prayers' curse us - and that is ALL moslems doing that in their salat. And their Friday 'prayers' are even worse. The imam over Seal Team 6 and who knows how many imams over our troops over the time we have been in the Middle East, curse our people. This IS a majority of moslems that are behind this and that is also due to the funding of violent jihads via their wealth and their moral support by moslems just outright lying about what islam is AND obfuscating, deflecint, excusing away and a whole bunch of other ways to veer off of Islam's violence (and some sexual perversions too).


text:
package org.zaproxy.zap.eventBus;

/**
 * A class that can publish event on the event bus
 * @author simon
 *
 */
public interface EventPublisher {

	/**
	 * The name by which the publisher is know - typically the full class name of the publisher.
	 * Only one publisher can register with a given name.
	 * @return the name of the publisher, never {@code null}.
	 */
	String getPublisherName();
}

redpajama_set_name: RedPajamaGithub


Your dream home awaits. Located one hour north of Prince Albert about five minutes from the Prince Albert National Park, this townhome combines the serenity and beauty of the boreal forest setting with luxury appointments and finishes. This condo features 1140 square feet of open concept The main level includes the kitchen, dining room, living room, a full bathroom, laundry room and the master bedroom with en-suite. The en-suite boasts a jetted soaker tub. Gleaming hardwood floors, vaulted ceilings and cozy gas fireplace make this a great place to call home. The living area faces the forest and has lots of natural light with the many windows. The lower walk-out level boasts two guest bedrooms and a full bath, family room, office area, bar, built-in-storage, sauna and has heated floors. The FA Room sound system is hooked up to outdoor speakers. A must see home. New H20 in 2017. Elk Ridge resort is a 4 season resort that offers 27 holes of championship golf, clubhouse, pub, fine dining, an indoor salt water pool, nature trails, a zip line, curling, skating, snow shoeing, snowmobile and cross-country ski trails are available. Boating and fishing are just a few minutes away at Waskesiu or McPhee Lake. Call today for your personal showing.


Colouring Book Review: Percy and the Colouring Adventure. Artists Edition.
Colouring Book Enchanted Forest Walk through with tips and coloured pages.
The Art of Mindfullness Colouring book. Finished pages, tips and walk through.
The Magical City. Walk through of the colouring pages.
Colouring Book 'Animal Kingdom' Colouring Tips and Walk Through.
Color Me Calm, A walk through of the colouring pages.


text:
info@owaytours.com
Free Walking Tour Cordoba
Guided Tour Mosque-Cathedral
Tour Medina Azahara
Guided Tour Mosque and Jewish Quarter
Tour Alcazar of the Christian Monarchs
Guided Tour Flamenco & Passion
City Legends Guided Tour
Rooftops & Tapas Guided Tour
Cordoba Segway Tour
Tourist map cordoba
Leisure and Entertaiment
OFFICIAL MOSQUE-CATHEDRAL TOUR
GUIDED TOUR OF THE MOSQUE-CATHEDRAL
Visit the Mosque-Cathedral of Cordoba, one of the most visited monuments in Spain, conserving the style from an Islamic Mosque together with the Cathedral. Its construction belongs to the 8th century and it became the second biggest Mosque in the world. You will discover inside a forest full of columns and double arches, the Maqsura, the Mihrab and the impressive chorus located close to the Mayor Chapel.
Patio de los Naranjos of the Mosque, next to the Audioguide Office (find our blue umbrella)
Adult: 22€
Children (8 to 12 years old): 11€
Children under 8 years old:free
Priority access to monuments (without queue)
Official tour guide
Radio Guidance
1 hour guided tour inside of the Mosque-Cathedral
Afternoon: Monday to Saturday at 13.15 pm
Evening: Monday, Tuesday and Friday at 17.30 pm
Shedule
Mosque Cathedral (Monday to Sunday)
-Winter Schedule: From the 1st of November to 28th of February: 10:00 to 18:00 h. (Sundays from 10 am to 11.30 am and 15:00 to 18:00 h)
-Summer Schedule: From the 1st of March to 31st of October: 10:00 to 19:00 h. (Sundays from 10 am to 11.30 am and 15:00 to 19:00 h)
Sessions every 30 minutes in the followings schedules:
-From 09:30 to 13:30h.
-From 16:00 to 18:30h. (Last session during the Winter Schedule at 17:30 h)
Get your ticket in tickets offices inside the monument. It will be open until 30 minutes before the closing time.
Mosque Cathedral:
-10€ general ticket entrance.
-5€ Reduced Ticket entrance: children between 10 – 14 years old and disabled.
-Free Ticket entrance: Born and/or residents in Córdoba, children under 10 years old and cardholders "Andalucia Junta 65".
-Special conditions for school groups, disables and old age. (Previous authorization)
Bell Tower:
-2€ General Ticket Entrance -Free Ticket: Born and/or residents in Córdoba.
Night visit "El Alma de Córdoba"
-18€ General Ticket -9€ reduced ticket entrance: Pensioners, disables and students under 26 years old. -Free Ticket: Children under 7 years old (not included radio guide and they must be accompanied by an adult)
From the train/bus station:
-Bus number 3 to "Fuensanta" until the stop called "San Fernando".
-Bus number 5 in direction to "Ciudad Sanitaria" until the stop called "Glorieta de la Media Luna".
-Parking "Fleming": Avenida Dr. Fleming, 14
-Parking "Centro histórico": Avenida de los Custodios s/n
-Parking city center: C/ Sevilla, 5
Inside the monument does not exists detailed information about the building, it is recommended to book a guided tour in order to understand better every detail inside this building declared world heritage by the UNESCO from the year 1984. The Mosque – Cathedral of Córdoba is one of the most famous monuments over the world. The monument is one of the most important elements concerning to the Islamic architecture. TripAdvisor placed in May in 2017 this historical heritage as the most interesting place in Europe and the third one over the world. This Mosque, that had a capacity for 30.000 believers, it was also the second biggest mosque in the world behind the main Mosque in Meca. The visit starts in the Patio de los naranjos (main courtyard of the Mosque), where you can observe the great bell tower that is keeping inside the marvellous alminar built by Abd al-Rahman III. Once inside you will be really surprised along your walk between 1.300 columns that form the building. From the first building ordered build by Abd al-Rahman I in the year 786, the trip thought the time takes us until the first expansion, where took place the proclamation of Abd al- Rahman III as the first Caliph, starting the Independent Caliphate of Al- Andalus, the most splendour period of Qurtuba. Later, you will be amazed with the following expansion by Al-Hakam II, considered the most important due to is reflecting the power in splendour, and there you can also observe the amazing Mirhab. Crossing this expansion you will enjoy the last expansion, it was built by the vizier Almansur. The last part of the visit will invite you to discover the magnificent cruise in the Cathedral, that was built from the year 1523 and it is the heart of that Islamic building. In this same moment you will be fascinated observing the great renaissance bolt that covers the cruise, but even more, when you point your view to the choir in the cathedral, with a precious mahogany wood created by the sculptor from Seville, Pedro Duque Cornejo.
Guided Tour Mosque Cathedral
Complete 3 Hours Tour Cordoba
Guided Tour Flamenco & Passion Tour
Rooftop & Tapas Guided Tour
MAPA · OWAY TOURS · MAP
© 2019 OWAY Tours Córdoba	| Aviso Legal | LOPD y Política de cancelación


Watching video HOW WOULD "YOU" SAVE OVERWATCH?
So today, time to get serious. How would you save Overwatch? To fix competitive, adding crazy changes for just a single season might be the game's salvation. Waiting for hero 29 may not be enough unless the abilities change the current stun meta. Let's discuss the options, and let me know what you think in the comments.
OVERWATCH FORUM CRIERS: NO COMP UNTIL LEVEL 200!


text:
5 minutes with MBA Graduate Greg Franks
Following his graduation from the AIB MBA programme, we spoke to Greg Franks about his career progression since the MBA and how he found the AIB flexible course structure.
How did you like the MBA programme?
The way the AIB programme was structured was great. The method of learning with the ability to see the starting point and the finish point, and knowing it was achievable in that timeframe really suited my lifestyle. The continuous flow of the subjects meant I was able to go with the momentum – it was a nice pattern of work and study.
Tell us about the networking experiences throughout the degree.
The networking was fantastic! We had a diverse range of people in our study group which was really good. We had people from the public sector, people who were in business themselves, engineers and an accountant. We were able to have a lot of different discussions at the table from a whole variety of backgrounds.
How has the AIB MBA helped your career?
Professionally, I have certainly made a number of improvements. I needed some knowledge to move to another level in terms of being able to perform, and that's certainly been achieved. I wasn't in my current job when I started studying, and the MBA actually helped me to win the job, particularly considering my boss had also done the same course.
This interview has been constructed by Laura Hutton from an AIB video testimonial. The very best has been done to accurately reflect the thoughts and opinions of the participating student.
Everyday Heroism in Action: Discovering New Horizons with the MBA
The MBA Directors Forum: Insights for our Students and Alumni
Not-For-Profit HR Insights from Carlie Gross of The Royal Flying Doctor Service
Discovering the Power of a Practical MBA: Mark Nesci's Graduate Story
Changes to FEE-HELP: What You Need to Know


Keeper of the Lost Cities: Flashback is the seventh book in the series. This time around, Jason Chan has drawn Wylie and Tam, and Sophie throwing a goblin throwing star at us.
This time we win. It was Sophie's only thought, even as the world around her fell into shadow and chaos.
Even as the Neverseen revealed their next move—their next twist in this deadly game.
A game Sophie had lost over and over again.
But this time she was ready.
This time she'd trained—and so had her friends.
This time, they knew how to fight back.
So Sophie's hand didn't shake as she reached for one of the throwing stars she'd learned to carry.
And she knew exactly where to aim.
But . . . things weren't always as they appeared.
And as everything flickered and faded, as her friends shouted and struggled to understand . . .
Illusion was her enemy's new secret weapon.


Whether you cycle daily, sometimes, or hardly ever we want to hear from you!
What would it take to get you on your bike?
What would you do to create a more inviting bicycle network?
What bike routes are currently working?
What links in the network are missing?
Where would you like new or improved bike routes in Victoria?
What is your vision for cycling in Victoria?
Workshop for future cyclists: If you rarely bike or never bike, we want to hear from you!
Hop on your bike and join us at these Information Stations this April and May. Come and spin the bike trivia wheel, map out your favourite bike routes, show us where you'd like future biking infrastructure improvements and complete a survey. We hope to see you in your neighbourhood soon!
Just drop by to an Information Station. An RSVP is not required.
Please take a few minutes to complete the online survey.
Your feedback is very important as we plan for the future of biking in Victoria.
To read more about cycling in Victoria, current infrastructure, upcoming projects, guiding policy, funding and route information click here.


The crunch has passed. The peak of the Corvallis restaurant year, OSU Graduation. And we broke a 4-year sales record, both for the day of Graduation/Father's Day, and for the week! Furthermore, considering our lower ticket prices, this translates into the highest volume of guests served for the entire 10 years of the restaurant.
Stress alleviated, I escape to the coast for my mid-week 'weekend'. Exhausted, I haven't slept more that a couple hours a night since last Thursday. But propped up on iced Americano from the Blodgett Store and what must be the euphoric endorphins of relief, I ride my bike over the Otter Crest Loop, the 'Old 101'.
I take a long walk out to the Japanese Tsunami dock, where gawkers and tourists take pictures, posed touching this relic of disaster. A huge floating obelisk of concrete and steel, stripped of the invasive species gathered on its thousands-miles-long journey.
On the long walk back to my van, I catch some stunning pictures of the Solstice sunset at Agate Beach. Only a few fellow sunset junkies seem to notice. I need to pay more attention to the awesome fireworks in the sky.


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How to Alternate Tylenol & Motrin for a Fever
What Is Diethylpropion 75 mg?
Side Effects of Crocin
Ibuprofen 800 Mg Vs. Diclofenac
How Does Mucinex Work?
What Is Tylex 750?
Written by Erin Watson-Price
Tylex is the Mexican brand name for paracetamol, an alternative name for the chemical acetaminophen, sold as the pain reliever Tylenol in the United States 34. Tylex is distributed by Janssen-Cilag, a division of the Johnson and Johnson Company 1. The number 750 indicates the dosage. One Tylex 750 pill contains 750 mg of acetaminophen.
Is This an Emergency?
Tylex 750 is distributed as a pill to be taken orally. The adult dose is one pill every 6 to 8 hours, as needed. Unless specifically instructed by your physician, do not split or crush Tylex 750 pills.
Like Tylenol, Tylex is prescribed for mild pain and as a fever reducer 4. Tylex is suitable for relief of backaches, toothaches, menstrual pain, osteoarthritis pain and general aches and pains from cold or flu. Unlike ibuprofen or aspirin, Tylex does not reduce swelling.
No general side effects have been reported. Tylex does not produce the stomach pain and ulcers typical associated with ibuprofen, naproxen and aspirin. In rare cases, patients have reported allergic reactions such as itching and swelling particularly of the face, tongue and throat. Additional symptoms include rash, dizziness and trouble breathing. If you experience any of these symptoms, contact your physician immediately.
No general side effects have been reported.
In rare cases, patients have reported allergic reactions such as itching and swelling particularly of the face, tongue and throat.
Tylex increases the chances of bleeding or bruising. It should not be taken in combination with blood thinners such as warfrin. Tylex should not be taken in combination with non-steroidal anti-inflammatory drugs (NSAIDs) 5. Read over-the-counter cold and flu medications carefully to avoid combining hidden NSAIDs. Persons with liver damage or impaired liver function should consult their physician before taking Tylex. Consult your doctor before taking Tylex while breastfeeding. Metabolites of Tylex can pass from mother to child through breast milk.
Tylex increases the chances of bleeding or bruising.
Metabolites of Tylex can pass from mother to child through breast milk.
Drugs Used in Nebulizers
Nuvigil for Weight Loss
What Are the Dangers of Taking Excedrin Every Day?
Side Effects of Tandrilax
Neo Citran Side Effects
Universidad Nacional Autonoma de Mexico: 750 Tylex
WebMD: Paracetamol
RxList: Tylenol
Drugs.com: Tylex
Transplanted Yankee Erin Watson-Price lives in Birmingham, Ala., and has been writing freelance articles since 1997. She worked as writer/co-editor for Coast to Coast Dachshund Rescue's newsletter, "The Long and the Short of It." In 2007 she obtained a certification as a copy editor. Watson-Price holds a Bachelor of Arts in creative writing from Southern Illinois University-Edwardsville.
Copyright © 2023 Leaf Group Ltd., all rights reserved. // Leaf Group Lifestyle


Thousands of years ago, a melting glacier deposited sand and rock debris south of Cape Cod. What was left? The boomerang-shaped island of Nantucket.
In spite of Nantucket's relatively small size—less than 50 square miles—its inhabitants and history offer a bounty of captivating stories that have been passed down through generations. Remnants of those tales still remain, and history hunters and story seekers will find them in Nantucket.
Boaters will find not only a full range of services at the Nantucket Boat Basin, but also easy access to transportation, accommodations, shopping, and dining. The 240-slip marina is owned by Nantucket Island Resorts, which also operates and provides transportation to several premier hotels and restaurants on other parts of the island.
If you prefer to stay close to your boat in the marina but sleep on land, consider The Cottages & Lofts at the Boat Basin. The one-, two and three-bedroom units are located on the marina wharves, and each cottage displays a creative moniker on a quarterboard at its front door. All 29 cottages have kitchens and air conditioning, and if you're traveling with Rover, reserve a pet-friendly unit well in advance.
For those who prefer higher-end accommodations, the White Elephant Hotel and White Elephant Village offer guest rooms and suites, garden cottages and one- to three-bedroom residences furnished with state-of-the-art appliances. Amenities available to Village guests also include a heated pool and cabanas, fitness center and complimentary bicycles.
No Nantucket visit would be complete without a visit to the Whaling Museum, where exhibits and artifacts explain the time and culture far beyond that of the whaling industry. In addition to this, you'll learn about the island's earliest settlers, including the Quakers, shipmates' crafts and activities, the fire that destroyed much of the island, and the resulting exodus of islanders who sought other sources of income like gold mining in California.
The museum opens daily at 10 AM, and a number of tours and ticket options are available, some of which combine visits to other historic sites. Within steps of the Boat Basin and the seaport, Nantucket proper comprises boutiques and shops, art galleries, restaurants, and pubs. Here, or anywhere on the island, you won't find a Starbucks, Red Lobster, Macy's, or any other chain store or restaurant. Each shop is unique, and each shop-keeper offers a particular brand of Nantucket experience.
Classic New England style shingled homes line the coast in Nantucket.
You'll be hard-pressed to visit all in a weekend, so a return trip will not only prove necessary but, anticipated. Each time you visit, invest not only in yourself, but in the island community with a remembrance of your on-island experience: trousers or a skirt in Nantucket red, a Lightship basket, whale memento, or artwork created by an islander.
If you carry bicycles on board your boat, you'll enjoy using them on Nantucket. If not, bicycles and mopeds are readily available for rental. Five paved and well-marked trails range from 2.5- to 8-miles long and lead to scenic areas around the island where you'll enjoy views of both the sparkling sea and of protected conservation landscapes. Public transportation is available during the summer season to Siasconset, Madaket and Surfside beaches.
With all the walking, bicycling and beaching, feel free to sate your appetite with fine food and drink. Topper's at The Wauwinet—another accommodation option that offers a respite from bustling seaport activity—is the recipient of the coveted Wine Spectator "Grand Award." Closer to town, Brant Point Grill at the White Elephant offers a complete land-and-sea menu overlooking a well-manicured lawn and the harbor.
Some sources suggest that the island is sinking into the ocean, a claim that is unfounded. However, the scientific community at large verifies that the polar caps are melting and as a result, Earth's seas are rising. Consequently, low islands and coastal areas are living on borrowed time. All would agree that Nantucket is much closer to the end of her life than the beginning. Go soon and to experience her tales and treasures before she is gone forever.


As I mentioned previously in my post introducing the great shop Wishstrings, I'm very exciting to be working with the lovely Kelly to do a couple of posts about her store.
Wishstrings is the perfect place to go to if you're looking for affordable but amazingly cute gifts and presents. The wish strings themselves are my particular favourites, they come in so many different designs and charms. I love mine with an anchor and a little bird. When you tie them on you make a wish and once it falls off it will come true - how cute!
With all these nice nights I've been doing a lot of walking (since I live in the uttermost country side) so we grabbed the camera and took these photos as we walked. Thanks to M M Hall Photography for lugging around his camera for four miles!


The Montale house gives you the opportunity to discover the most iconic fragrances with a sampling kit (5x2ml) you can customise at will.
Choose your 5 samples and add them to the cart.
To offer a gift card is to offer a thousand gift possibilities. Choose the amount and add a personal message to it.


text:
Journalists Demand Salaries from Parliamentarian
Kong Sothanarith
PHNOM PENH - A small group of journalists gathered in protest outside the offices of a Cambodian ruling party lawmaker and head of a media center in Kandal province Wednesday, demanding they be paid for work performed last year.
The journalists said they are among more than 50 who have not received payment for work from the parliamentarian Chea Chamroeun and were prepared to burn tires outside the media center, underscoring how closely tied journalists and politicians often are in Cambodia.
Outside the Khmer Science Media Center, journalists said they were owed between $300 and $500 for work performed last year for Chea Chamroeun, a lawmaker for the Cambodian People's Party, who is the head of the media center and the rector of Chamroeun Polytechnic University.
The journalists, 53 in all, say they were not paid for work they did for an online and print publication that closed in September 2012.
Chea Chamroeun began negotiating with them on Wednesday, before they ignited tires outside the media center but after they had already filed complaints with the Club of Cambodian Journalists, Prime Minister Hun Sen and the ministries of Information and Labor.
"He promised to give all of us all the wages owed by the end of January," said Yin Leang Kong, a representatives of the journalists, following an hourlong negotiation.
"It is settled," Chea Chamroeun told reporters afterward. "There was no need for you to come and put your nose in."
Sam Rainsy Will Remain Ineligible for Election
Housing Activists Sentenced to Three Years Each in Prison
Likely Innocent, Two Men Returned to Prison for 2004 Murder


text:
NEW YORK, NY – APRIL 01: Traders work on the floor of the New York Stock Exchange as the website hosting service GoDaddy makes its initial public offering (IPO) on April 1, 2015 in New York City. GoDaddy, which manages about a fifth of the world's Internet domains will list on the New York Stock Exchange under the ticker 'GDDY' and priced its IPO at $20 Tuesday night. (Photo by Spencer Platt/Getty Images)
With IPO Priced Above Range, Web Host GoDaddy Raises $460 Million
Hosting firm's shares rise 30 percent on NYSE-debut day (GDDY)
Cheryl Kemp | Apr 01, 2015
This article originally appeared at The WHIR
In its New York Stock Exchange debut Wednesday, GoDaddy's shares rose 31 percent to $26.15 at 10 am ET after the company sold 23 million shares for $20 each. The IPO has raised $460 million, which is significantly more than GoDaddy's original aim to raise around $100 million, stated in the company's S-1 filing in June 2014.
On Tuesday, GoDaddy was expected to set an Initial Public Offering price of $20 per share, slightly higher than the the $17 and $19 range stated in an SEC filing last month. This share price values the company at up to $5.48 billion, including debt, according to Reuters.
This year has seen only the launch of two other tech stocks, which is considered low because there were 14 tech IPOs in the same period last year. Cloud storage provider Box was one of the tech IPOs this year with shares opening at $14 and rising as much as 65 percent on the first day of trading and are now trading at around $19.50.
Online marketplace Etsy is also rumored to be going public soon with an expected IPO priced at $14-$16 per share for a $1.78 billion valuation.
GoDaddy, which has been in business since 1997 and currently manages around a fifth of all domains, is much more well-established than new technology upstarts.
In 2011, it was acquired by a private equity consortium led by KKR & Co LP and Silver Lake Partners for $2.25 billion. KKR and Silver Lake each own 27.9 percent stakes, Technology Crossover Ventures (TCV) owns 12.6 percent, and company founder Bob Parsons owns approximately 28 percent. Following the IPO, the stakes KKR, Silver Lake, and Parsons will have in the company will drop to 23.9 percent, and TCV will hold 10.7 percent.
In 2014, GoDaddy reported $1.39 billion in revenue, a 23 increase from 2013, as well as adding 1.1 million customers bringing its total clients to 12.7 million which bring in $114 per customer on average. Yet it also reported ending the year with $143.3 million in net losses, albeit less than the $200 million it lost the previous year.
GoDaddy will be trading under the symbol "GDDY."
This piece ran first at http://www.thewhir.com/web-hosting-news/ipo-priced-range-web-host-godaddy-raises-460-million
Microsoft-Backed Security Startups Outsmart Hackers with Hackers
Millions of Verizon Customer Records Exposed through Open Amazon S3 Bucket
Azure Stack to Ship in September; Hardware on Sale Starting Today
Global Ransomware Attack Cripples Networks – Again


text:
Q: jQuery limit the results/selection returned Is it possible to limited the results returned in jQuery

One

Two

Three

Four

Five

Six

This will output:

OneTwoThreeFourFiveSix

But I want:

OneTwoThreeFour

Demo Fiddle: http://jsfiddle.net/JLqrc/

A: You can use :lt()
$('#box').children(':first').children(':lt('+num+')');

or as MackieeE commented to shorten the code
$('#box li:lt('+ num +')').text()

DEMO
Documentation : http://api.jquery.com/lt-selector/

A: You could use .slice too,
$('#box ul li').slice(0,4).text();

This gives you more control over what 4 elements are returned - as you can change it to .slice(1,4) and it'll return TwoThreeFourFive etc
Here's an example: JSFiddle

A: $(function(){
      var num = 4;
    console.log($("#box ul li:lt("+num+")" ).text());

  })

redpajama_set_name: RedPajamaStackExchange