text:
Category: Trophy 5-a-side
Scope FC wins maiden edition of Trophy 5-A-Side
After 11 weeks of a keenly contested tournament amongst 1, 080 teams, Scope FC emerged winner at the maiden edition of Trophy 5-a-side grassroots tournament tagged, "Field of Honour".
The third place and finale matches took place on Saturday, May 11 at the Campos Mini-Stadium, Lagos Island where Amsterdam FC of Obalende, Lagos beat Rovers FC, Oyo 6-0 to emerge as bronze winners.
In an intensely contested final between Scope FC of Lagos and Ifayemi FC of Ile-Ife, Scope put up a good fight, getting a well deserved 2-0 win over Ifayemi FC in the second half after both teams failed to score in the first half.
In her remarks, Managing Director, International Breweries, Annabelle Degroot, appreciated the winning team saying: "International Breweries is very proud of Scope FC as they embodied the true Trophy values by tackling each challenge with panache, tenacity and zeal to win. With this tournament International Breweries has created an avenue for our consumers to talk about their favorite sport and be a part of it. With former Super Eagles Captain, Joseph Yobo as the tournament ambassador, Trophy continues to be associated with football, a sport most acknowledged for encouraging team spirit."
Scope FC, were presented their cash prize of N3 million. They will also be given the opportunity to meet Samuel Eto'o, renowned African football superstar and participate in the continental finals taking place in Tanzania later this year. Third placed team, Amsterdam FC and second place Ifayemi FC were also presented their cash prizes of N500,000 and N1,000,000 respectively.
Trophy Brand Ambassador, Femi Adebayo; Managing Director, International Breweries, Annabelle Degroot; Trophy Brand Ambassador, Joseph Yobo (behind); Captain of the winning team, Scope FC, Lagos, Folorunsho Sunday; Trophy Brand Ambassador, Folarin Falana (behind) and Marketing Director, International Breweries,
Posted in Field of Honour, Scope FC, Sports, Trophy 5-a-side


text:
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The Online Home of HBCU Alumni
Virginia State hall of famer Raymond "Coach" Crittenden passes away at 83
Raymond Crittenden Jr., one of Virginia State's most prolific players, died Monday of natural causes. He was 83.
Post author By aka Tito
No Comments on Virginia State hall of famer Raymond "Coach" Crittenden passes away at 83
Raymond Celester Crittenden Jr., one of Virginia State's most prolific players, died Monday of natural causes. He was 83.
At Virginia State College (now Virginia State University), Crittenden started every football and basketball game and lettered in football, basketball and track. He was also all-CIAA in football and basketball, and held a 10-year record for most points scored in a basketball game.
He was inducted into the Virginia State University Hall of Fame in 1988 and the Hampton Roads African-American Sports Hall of Fame in 2004.
After getting his bachelor's degree from Virginia State, Crittenden went on to a long career as a high school teacher, administrator, football coach and basketball coach from 1954-88.
The grandfather of former New England Patriots Raymond Crittenden IV, he was laid to rest Saturday in Newport News, Virginia.
(via Examiner.com)
Tags CIAA, Raymond Crittendes, Virginia State University
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text:
\section{Introduction}
Dans toute cette partie on fait les hypothèses suivantes: $(W,S)$ est un système de Coxeter satisfaisant aux conditions H(Cox) et $R:W\to GL(M)$ est une représentation de réflexion réductible. On pose $G:=Im R$. Le but est d'étudier la structure de $G$.
\subsection{Le théorème fondamental}
\begin{theorem}
Soit $(W,S)$ un système de Coxeter satisfaisant aux hypothèses H(Cox) et soit $R:W \to GL(M)$ une représentation de réflexion de $W$ obtenue par la construction fondamentale. On pose $G:=Im\, R$. Alors:
\begin{enumerate}
  \item $H:=\bigcap_{s\in S}H(s)$ est le plus grand sous-espace de $M$ stable par $G$.
  \item La suite $(\ast)$ de $KG$-modules, où $M':=M/\bigcap_{s\in S}H(s)$ est exacte et non scindée:
  \[
   (\ast) \qquad {0}\to  \bigcap_{s\in S}H(s) \to M \xrightarrow{\pi_{M}} M' \to {0}
  \]
  où $\pi_{M}$ est la projection canonique.
  \item La représentation $R$ est irréductible si et seulement si $H=\{0\}$, si et seulement si $\Delta(G)\neq\{0\}$.
  \item On suppose que $R$ est réductible. Alors $G$ opère sur $M'$ et l'on a la suite exacte $(\ast\ast)$:
  \[
 (\ast\ast) \qquad  \{1\}\to N(G)\to G \xrightarrow{\pi} G' \to \{1\}
  \]
  où $N(G):=\{g|\in G, g\; \text{opère trivialement sur}\, M'\}=\ker \pi$, $G':=G/N(G)$ et $\pi$, la projection canonique, est un bon morphisme.\\
  De plus $M'$ est un $KG$-module simple et un $KG'$-module simple. $G'$ opère comme un groupe de réflexion sur $M'$.
  \item $N(G)$ est un groupe commutatif sans torsion et tous ses éléments non triviaux sont des applications unipotentes.
\end{enumerate}
\end{theorem}
\begin{proof}
1) Soit $V$ un sous-espace de $M$ stable par $G$. Si $V \, \nsubseteq \bigcap_{s\in S}H(s)$ alors $\exists v \in V, \exists s \in S$ tels que $s(v) \in V$ et $s(v)\neq v$. Dans ces conditions $s(v)-v=\lambda a_{s}$ ($\lambda \in K^{*}$) est dans $V$ et $a_{s} \in V$. Comme $\Gamma(G)$ est connexe, en appliquant les différents éléments de $S$ à $a_{s}$, nous voyons que $\forall t \in S$, $a_{t}\in V$. Il en résulte que $V=M$. Il est clair que $H\subset C_{M}(G)$ et que $C_{M}(G)$ est stable par $G$, donc $H=C_{M}(G)$.\\
Les 2), 3)  sont clairs.\\
4) Nous montrons que $\pi$ est un bon morphisme. Soient $s$ et $t$ deux éléments de $S$. par hypothèse $st$ est d'ordre fini $n$ car $W$ est $2$-sphérique. Alors $\pi(st)$ est d'ordre $n'$ où $n'$ est un diviseur de $n$ et $\pi(st)^{n'}$ est d'ordre $1$ donc $(st)^{n'}\in N(G)$, mais tous les éléments non triviaux de $N(G)$ sont d'ordre infini et $(st)^{n'}$ est d'ordre fini, donc $(st)^{n'}=1$ et $n'=1$, d'où le résultat.\\
5) Comme les éléments de $N(G)$ opèrent trivialement sur $H$ et $M/H$, toutes leurs valeurs propres sont égales à $1$, donc ce sont des applications unipotentes. On voit aussi que $N(G)$ est commutatif et sans torsion.
\end{proof}
\begin{corollary}
Si $G'$ est un groupe de Coxeter $2$-sphérique alors la suite $(\star\star)$ est scindée.
\end{corollary}
\begin{proof}

\end{proof}
\begin{notation}
Avec les hypothèses du théorème précédent, si $R$ est réductible, on dit que $R$ est une représentation de réflexion \textbf{affine} de $W$, que $G$ est un groupe de réflexion \textbf{affine} et que $N(G)$ est son sous-groupe des \textbf{translations}.
\end{notation}
Le problème ici est que l'on ne sait pas si $N(G)$ est non trivial ou non.
\subsection{Des bases adaptées}
Dans toute cette section $(W,S)$ est un système de Coxeter qui satisfait aux conditions H(Cox). Soit $R: W\to GL(M)$ une représentation de réflexion \textbf{réductible} de $W$. On garde les notations précédentes et on utilise la construction fondamentale.\\
Pour pouvoir faire des calculs explicites, on va choisir une base de $H$ et uns base de $M$ qui ont de bonnes propriétés. On suppose que $n_{0}:= \dim H\geqslant 1$ et l'on pose $n_{1}:=n-n_{0}$.
\begin{remark}
On a toujours $n_{0}\leqslant n_{1}$.
\end{remark}
\begin{proof}
En effet si l'on avait $n_{1}<n_{0}$, le groupe $G'=\pi(G)$ n'opérerait pas irréductiblement sur $M'(=M/H)$.
\end{proof}
Il en résulte qu'il existe $S_{1} \subset S$, $|S_{1}|=n_{1}$ tel que $\Delta(<S_{1}>)\neq 0$.
\begin{notation}
On pose $S_{0}:=S-S_{1}$ et $G_{i}:=<S_{i}>(i=0,1)$. On pose $S_{1}:=\{s_{i_{1}},s_{i_{2}},\cdots,s_{i_{n_{1}}}\}$ et on suppose que $i_{1}<i_{2}<\cdots<i_{n_{1}}$.
\end{notation}
Il est clair que $\pi(G_{1})\simeq G_{1}$.\\
Soit $M_{1}:=<a_{i_{1}},a_{i_{2}},\cdots,a_{i_{n_{1}}}>$. On a $M=H\oplus M_{1}$. Soit $\pi_{1}:M\to M/M_{1}$ la projection canonique. On a $\pi_{1}(M)=\pi_{1}(H)$. \textbf{On choisit} pour $i\in\{1,2,\cdots,n\}-\{i_{1},i_{2},\cdots,i_{n_{1}}\}$, $b_{i}$ dans $H$ de telle sorte que l'on ait $\pi_{1}(b_{i})=\pi_{1}(a_{i})$. Alors\\ $(b_{j_{1}},b_{j_{2}},\cdots,b_{j_{n_{0}}})$ est une base de $H$ et $\mathcal{B}:=(b_{j_{1}},b_{j_{2}},\cdots,b_{j_{n_{0}}};a_{i_{1}},a_{i_{2}},\cdots,a_{i_{n_{1}}})$ est une base de $M$.\\
On a $\underline{n}_{0}:=\{j_{1},j_{2},\cdots,j_{n_{0}}\}$ et on suppose que $j_{1}<j_{2}<\cdots<j_{n_{0}}$. On pose $\underline{n}_{1}:=\{i_{1},i_{2},\cdots,i_{n_{1}}\}$ et $\underline{n}=\{1,2,\cdots,n\}$. Si $j\in \underline{n}_{0}$, on peut écrire:
\begin{equation}
b_{j}=a_{j}+\sum_{k\in \underline{n}_{1}}\rho_{j}^{k}a_{k}
\end{equation}
Le théorème suivant donne la valeur des $\rho_{j}^{k}$ en fonction des coefficients de Cartan $c_{i,j}$ et aussi les différentes relations existant entre eux.
\begin{theorem}\label{calcul des $c_{i,j}$}
Avec les hypothèses et notations précédentes, on a:\\
1)
\begin{equation}
  Car(G_{1})
  \begin{pmatrix}
\rho_{j_{1}}^{i_{1}} & \rho_{j_{1}}^{i_{2}} & \cdots & \rho_{j_{1}}^{i_{n_{1}}} \\
\rho_{j_{2}}^{i_{1}} & \rho_{j_{2}}^{i_{2}} & \cdots & \rho_{j_{2}}^{i_{n_{1}}} \\
\vdots & \vdots & & \vdots\\
\rho_{j_{n_{0}}}^{i_{1}} & \rho_{j_{n_{0}}}^{i_{2}} & \cdots & \rho_{j_{n_{0}}}^{i_{n_{1}}}
\end{pmatrix}
= -
\begin{pmatrix}
c_{i_{1},j_{1}} & c_{i_{1},j_{2}} & \cdots & c_{i_{1},j_{n_{0}}}\\
c_{i_{2},j_{1}} & c_{i_{2},j_{2}} & \cdots & c_{i_{2},j_{n_{0}}}\\
\vdots & \vdots & & \vdots\\
c_{i_{n_{1}},j_{1}} & c_{i_{n_{1}},j_{2}} & \cdots & c_{i_{n_{1}},j_{n_{0}}}
\end{pmatrix}
 \end{equation}
 2) Une condition nécéssaire et suffisante pour que $H$ soit de dimension $n_{0}$ est:
 \begin{equation}
\begin{pmatrix}
c_{j_{1},i_{1}} & c_{j_{1},i_{2}} & \cdots & c_{j_{1},i_{n_{1}}}\\
c_{j_{2},i_{1}} & c_{j_{2},i_{2}} & \cdots & c_{j_{2},i_{n_{1}}}\\
\vdots & \vdots & & \vdots\\
c_{j_{n_{0}},i_{1}} & c_{j_{n_{o}},i_{2}} & \cdots & c_{j_{n_{0}},i_{n_{1}}}
\end{pmatrix}
Car(G_{1})^{-1}
\begin{pmatrix}
c_{i_{1},j_{1}} & c_{i_{1},j_{2}} & \cdots & c_{i_{1},j_{n_{0}}}\\
c_{i_{2},j_{1}} & c_{i_{2},j_{2}} & \cdots & c_{i_{2},j_{n_{0}}}\\
\vdots & \vdots & & \vdots\\
c_{i_{n_{1}},j_{1}} & c_{i_{n_{1}},j_{2}} & \cdots & c_{i_{n_{1}},j_{n_{0}}}
\end{pmatrix}
=Car(G_{0}).
\end{equation}
\end{theorem}
\begin{proof}
Soient $l\in \underline{n}$ et $j\in \underline{n_{0}}$. On a
\[
s_{l}(b_{j})=b_{j}=a_{j}+\sum_{k\in \underline{n}_{1}}\rho_{j}^{k}a_{k}=(a_{j}-c_{l,j}a_{l})+\sum_{k\in \underline{n}_{1}}\rho_{j}^{k}(a_{k}-c_{l,k}a_{j}).
\]
Donc on a les relations:
\begin{equation}
\forall l \in \underline{n}, \forall j \in \underline{n_{0}}, c_{l,j}+\sum_{k\in \underline{n}_{1}}\rho_{j}^{k}c_{l,k}=0
\end{equation}
qui sont nécessaires et suffisantes pour que $H$ soit de dimension $n_{0}$. Les formules de l'énoncé sont simplement les transcriptions matricielles des relations (4).
\end{proof}
Avec les hypothèses et notations précédentes, on voit que $n_{1}$ est le rang de la matrice $Car(G)$. Si l'on prend n'importe quelle sous-matrice carrée $T$ de $Car(G)$ de dimension $n'$ avec $n_{1}<n'$, alors le rang de $T$ est encore $n_{1}$. Ces remarques démontrent la proposition suivante:
\begin{proposition}
Si l'on prend un sous-ensemble $S'$ de $S$ tel que $|S'|>n_{1}$ et si $G'=<S'>$ et $R'$ est la restriction de $R$ à $G'$ opérant sur $<a_{s}|s\in S'>$, alors $R'$ est une représentation de réflexion réductible de $G'$.
\end{proposition}
\begin{proof}
Car $Car(G')$ est une sous-matrice de $Car(G)$.
\end{proof}
Nous généralisons légèrement le problème. Nous travaillons dans la base $\mathcal{B}$. Soit $\mathcal{G}:=\{g|g\in GL(M),g\, \text{fixe} \,H\}$. La matrice de $g\in\mathcal{G}$ est:
\[
\begin{pmatrix}
I_{n_{0}} & A(g)\\
0 & P(g)
\end{pmatrix}
\]
où $I_{n_{0}}$ est la matrice identité d'ordre $n_{0}$, $A(g)\in \mathcal{M}(n_{0},n_{1})$ ($\mathcal{M}(n_{0},n_{1})$ étant l'espace vectoriel des matrices à $n_{0}$ lignes et $n_{1}$ colonnes à coefficients dans $K$) et $P(g)$ est un élément de $GL_{n_{1}}(K)$.\\
On pose
\begin{align}
\mathcal{Z}:=\{z\in \mathcal{G}|z \,\text{opère comme une opération scalaire sur}\, M'\},\\
\mathcal{N}:=\{n\in \mathcal{Z}| n \, \text{opère comme l'identité sur }\, M'\},\\
\mathcal{Z'}:= \{z\in \mathcal{Z}|z \, \text{opère comme $-1$ sur}\, M'\}.
\end{align}
Si $g$ et $g'$ sont dans $\mathcal{G}$, alors
\[
gg'=
\begin{pmatrix}
I_{n_{0}} & A(g')+A(g)P(g)\\
0 & P(g)p(g')
\end{pmatrix}
\]
donc $P: \mathcal{G}\to GL_{n_{1}}(K)$ est un morphisme surjectif de groupes dont le noyau est le groupe $\mathcal{N}$. On a aussi la formule $A(gg')=A(g')+A(g)P(g')$, de là on déduit que
\[
g^{-1}=
\begin{pmatrix}
I_{n_{0}} & -A(g)P(g)^{-1}\\
0 & P(g)^{-1}.
\end{pmatrix}
\]
Soient $h$ et $h'$ dans $\mathcal{N}$. Alors
\[
hh'=
\begin{pmatrix}
I_{n_{0}} & A(h)+A(h')\\
0 & I_{n_{1}}
\end{pmatrix}
\]
avec $A(h)\in \mathcal{M}(n_{0},n_{1})$ et $A: \mathcal{N}\to \mathcal{M}(n_{0},n_{1}):h \mapsto A(h)$ est un isomorphisme de groupes.\\
Notons le résultat suivant:
\begin{proposition}
Avec les hypothèses et notations du début, on a:\\
1) $\mathcal{G}$ opère à droite sur $\mathcal{M}(n_{0},n_{1})$ et sous l'action de $\mathcal{G}$ chaque sous-espace $\mathcal{M}_{i}$ de $\mathcal{M}(n_{0},n_{1})$ de la forme
\[
\begin{pmatrix}
 & & 0 & &\\
u_{1} & u_{2} & \cdots & u_{n_{1}-1} &u_{n_{1}}\\
 & & 0 & &
\end{pmatrix}
\]
où les éléments non nuls sont sur la i-ième  ligne, est stable par $\mathcal{G}$. On obtient ainsi une décomposition en somme directe de $K\mathcal{G}$-modules simples $\mathcal{M}_{i}$ $(1\leqslant i \leqslant n_{0})$ de $\mathcal{M}(n_{0},n_{1})$.\\
2) Si $\mathcal{N}_{i}=A^{-1}(\mathcal{M}_{i})$, alors chaque $\mathcal{N}_{i}$ est un $K\mathcal{G}$-module simple et $\mathcal{N}\simeq \oplus_{i=1}^{n_{0}}\mathcal{N}_{i}$.
\end{proposition}
\begin{proof}
Soient $g \in \mathcal{G}$ et $h \in \mathcal{N}$. Alors
\[
ghg^{-1}=
\begin{pmatrix}
I_{n_{0}} & A(h)P(g)^{-1}\\
0 & I_{n_{1}}
\end{pmatrix}
\]
donc $\mathcal{G}$ opère à droite sur $\mathcal{M}(n_{0},n_{1})$ et chaque $\mathcal{M}_{i}$ est stable sous l'action de $\mathcal{G}$. Tout le reste est clair.
\end{proof}

Avec les notations précédentes, la formule $A(ghg^{-1})=A(h)P(g)^{-1}$ montre que $A$ est $\mathcal{G}$-équivariante.

 Nous donnons maintenant une condition suffisante qui assure que $N(G)\neq\{1\}$.
 \begin{proposition}
Avec les hypothèses et notations précédentes, si $C_{G}(G_{1})\neq\{1\}$, en particulier si $Z(G_{1})\neq\{1\}$, on a $N(G)\neq\{1\}$.
\end{proposition}
\begin{proof}
Si $z\in C_{G}(G_{1})-\{1\}$, alors $z$ opère comme une opération scalaire sur $M'$, donc $z\in \mathcal{Z}\cap G$. Il en résulte que $\forall g\in G$, $[z,g]\in \mathcal{N}\cap G$ et comme $z\not \in Z(G)$ (qui est trivial), il existe $g\in G$ tel que $[z,g]\neq 1$, donc $N(G)\neq\{1\}$.
\end{proof}
\begin{corollary}
Avec les hypothèses et notations précédentes, on suppose en plus que $n=3$. Alors $N(G)\neq\{1\}$ si $2|pqr$.
\end{corollary}
\begin{proof}
Si $n=3$, on a $n_{0}=1$ et $n_{1}=2$ donc ou bien $G_{1}$ est isomorphe à $C_{2}\times C_{2}$ auquel cas le résultat est vrai ou bien $G_{1}$ est un groupe diédral et ceci quelque soit le choix de $S_{1}$. Comme $2|pqr$, l'un de ces groupes diédraux est d'ordre multiple de $4$, donc son centre est non trivial et nous appliquons la proposition 2 pour avoir le résultat.
\end{proof}
\begin{proposition}
Soient $W$ un groupe de Coxeter satisfaisant aux conditions $H(Cox)$ et $R :\to GL(M)$ une représentation de réflexion affine. On pose $G:=Im R$ et on suppose que $N(G)\neq{1}$. Soient $z \in \mathcal{Z}'$ et $\Gamma:=<G,z>$. Alors:\\
1) $N(\Gamma)$ = $N(G)$.\\
2) $|\Gamma /G|\leqslant 2$.
\end{proposition}
\begin{proof}
Il est clair que $N(G)\lhd \Gamma$ car $G$ normalise $N(G)$ et $z$ opère comme $-1$ sur $N(G)$.
Soit $\pi:\Gamma \to \Gamma/N(G)$ la projection canonique. Supposons que $N(\Gamma)\neq N(G)$ alors nous obtenons une contradiction. En effet $\pi(N(\Gamma))$ est un sous-groupe normal de $SL_{2}(M')$ puisque $\pi(N(\Gamma))$ est formé d'éléments de déterminants $1$. Il est clair que $N(\Gamma)\neq\Gamma$. les seuls sous-groupes normaux non triviaux de $SL_{2}(M')$ sont contenus dans son centre et ils sont formés d'applications scalaires. Comme tous les éléments de $\pi(N(\Gamma))$ n'ont que $1$ comme valeur propre, ils sont triviaux et $\pi(N(\Gamma))$ = $1$: on a $N(\Gamma)$ = $N(G)$. Comme $\forall g \in G$, $[z,g] \in N(\Gamma)=N(G)$, nous voyons que $z$ normalise $G$ d'où le résultat $|\Gamma /G|\leqslant 2$.
\end{proof}
\begin{corollary}
Avec les hypothèses et notations précédentes on a:\\
1) $\mathcal{N}$ et $\mathcal{Z'}$ normalisent $G$.\\
2) On a l'une des possibilités suivantes: $\mathcal{Z'} \subset G$ ou $\mathcal{Z'}\cap G = \emptyset$.
\end{corollary}
\begin{proof}
1) Comme $z$ était quelconque dans la proposition on voit que $\mathcal{Z'}$ normalise $G$ et comme $\mathcal{N}\subset <\mathcal{Z'}>$, $\mathcal{N}$ normalise $G$.\\
2) Si $\exists z \in G\bigcap \mathcal{Z'}$, alors $\forall z' \in \mathcal{Z'}$, $zz' \in N(G)$, donc $z' \in G$ et $\mathcal{Z'}\subset G$.
\end{proof}
\subsection{Opérations du groupe $G$}
On garde les notations précédentes. Il est clair que $G$ est un sous-groupe de $\mathcal{G}$ et que $N(G)$ est un sous-groupe de $\mathcal{N}$. Le groupe $G$ opère sur $N(G)$ (et sur $\mathcal{N}$) par conjugaison et sur $\mathcal{M}(n_{0},n_{1})$ par multiplication à droite. On note $g.$ l'opération de $g$ sur $\mathcal{M}(n_{0},n_{1})$.\\
L'application $P:G\to GL_{n_{1}}(K)$ est un morphisme de groupes dont le noyau est $N(G)$ et l'image est $G'$.\\
Nous donnons maintenant les matrices des éléments de $S$ dans la base $\mathcal{B}$ de $M$.\\
-- Si $i \in \underline{n}_{1}$, on a
\[
s_{i}=
\begin{pmatrix}
I_{n_{0}} & 0 \\
0 & P(s_{i})
\end{pmatrix}
\]
Si $k \in \underline{n}_{1}$, on sait que $s_{i}(a_{k})=a_{k}-c_{ik}a_{i}$ donc $P(s_{i})=I_{n_{1}}+T(s_{i})$ où
\[
T(s_{i})=
\begin{pmatrix}
& & 0 & &\\
-c_{ii_{1}} & -c_{ii_{2}} & \cdots & -c_{ii_{n_{1}-1}} & -c_{ii_{n_{1}}}\\
& & 0 & &
\end{pmatrix}
\]
et les éléments non nuls de $T(s_{i})$ sont situés sur la i-ième ligne.\\
Si $\zeta \in \mathcal{M}(n_{0},n_{1})$ on a $s_{i}.\zeta =\zeta P(s_{i})= \zeta+\zeta T(s_{i})$ d'où
\[
[s_{i},\zeta]=s_{i}.\zeta-\zeta=\zeta T(s_{i}).
\]
-- Si $j \in \underline{n}_{0}$ et si $i \in \underline{n}_{1}$, on a $s_{j}(a_{i})=a_{i}-c_{ji}a_{j}$, mais $b_{j}=a_{j}+\sum_{k \in \underline{n}_{1}}\rho_{j}^{k}a_{k}$ et l'on obtient $s_{j}(a_{i})=a_{i}-c_{ji}b_{j}+c_{ji}\sum_{k \in \underline{n}_{1}}\rho_{j}^{k}a_{k}$. On en déduit que
\[
s_{j}=
\begin{pmatrix}
I_{n_{0}} & A(s_{j})\\
0 & P(s_{j})
\end{pmatrix}
\]
où $A(s_{j})\in \mathcal{M}(n_{0},n_{1})$ et toutes les lignes de $A(s_{j})$ sont formées de $0$ à l'exception de la j-ième qui est $(-c_{ji_{1}},-c_{ji_{2}},\cdots,-c_{ji_{n_{1}}})$.\\
En écrivant que $s_{j}^{2}=id_{M}$, on obtient
\[
s_{j}^{2}=
\begin{pmatrix}
I_{n_{0}} & 0\\
0 & I_{n_{1}}
\end{pmatrix}
=
\begin{pmatrix}
I_{n_{0}} & A(s_{j})+A(s_{j})P(s_{j})\\
0 & I_{n_{1}}
\end{pmatrix}
\]
d'où la relation: (1)  $A(s_{j})+A(s_{j})P(s_{j})=0$. On a $P(s_{j})=I_{n_{1}}+T(s_{j})$ où
\[
T(s_{j})=
\begin{pmatrix}
c_{ji_{1}}\rho_{j}^{i_{1}} & c_{ji_{2}}\rho_{j}^{i_{1}} & \cdots & c_{ji_{n_{1}}}\rho_{j}^{i_{1}}\\
c_{ji_{1}}\rho_{j}^{i_{2}} & c_{ji_{2}}\rho_{j}^{i_{2}} & \cdots & c_{ji_{n_{1}}}\rho_{j}^{i_{2}}\\
\vdots & \vdots & & \vdots\\
c_{ji_{1}}\rho_{j}^{i_{n_{1}}} & c_{ji_{2}}\rho_{j}^{i_{n_{1}}} & \cdots &c_{ji_{n_{1}}}\rho_{j}^{i_{n_{1}}}
\end{pmatrix}
.
\]
Si $\zeta \in \mathcal{M}(n_{0},n_{1})$, alors
\[
s_{j}.\zeta=
\begin{pmatrix}
I_{n_{0}} & A(s_{j})+\zeta P(s_{j})+A(s_{j})P(s_{j})\\
0 & I_{n_{1}}
\end{pmatrix}
\]
d'où d'après la relation (1) $s_{j}.\zeta =\zeta P(s_{j})=\zeta+\zeta T(s_{j})$ et
\[
[s_{j},\zeta]=\zeta T(s_{j}).
\]
\subsection{Le cas $n_{0}=1$}
Nous étudions en détail le cas $n_{0}=1$ puis nous passerons au cas $n_{0}>1$.\\
Comme $P(G)$ est un sous-groupe de $GL_{n_{1}}(K)$, $P(G)$ stabilise chacun des $\mathcal{M}_{j}$ $(j\in \underline{n}_{0})$ et pour voir comment $P(G)$ opère sur $\mathcal{M}_{j}$, nous allons d'abord supposer que $\underline{n}_{0}=\{1\}$ (donc $\underline{n}_{1}=\{2,3,\cdots,n\}$).\\
Par hypothèse $H$ est de dimension $1$ engendré par $b$ où, en posant pour simplifier les notations $\rho_{i}:=\rho_{1}^{i}\, (2\leqslant i \leqslant n)$,
\[
b=a_{1}+\sum_{i=2}^{n}\rho_{i}a_{i}
\]
et les $\rho_{i}$ sont donnés par (en utilisant la relation (2) du théorème  2):
\[
Car(G_{1})
\begin{pmatrix}
\rho_{2}\\
\vdots\\
\rho_{n}
\end{pmatrix}
=-
\begin{pmatrix}
c_{21}\\
\vdots\\
c_{n1}
\end{pmatrix}
\]
De plus on a les relations:
\[
\begin{pmatrix}
c_{12} & c_{13} & \cdots & c_{1n}
\end{pmatrix}
Car(G_{1})^{-1}
\begin{pmatrix}
c_{21}\\
c_{31}\\
\vdots\\
c_{n1}
\end{pmatrix}
=(2).
\]
-- Si $i\in \underline{n}_{1}$,
\[
T(s_{i})=
\begin{pmatrix}
& & 0 & &\\
-c_{i2} & -c_{i3} & \cdots & -c_{i(n-1)} & -c_{in}\\
& & 0 & &
\end{pmatrix}
\]
où les éléments non nuls sont sur la i-ième ligne.\\
Si $\zeta:=(\zeta_{2} \zeta_{3} \cdots \zeta_{n})\in \mathcal{M}(n_{0},n_{1})$ alors
\[
[s_{i},\zeta]=\zeta T(s_{i})=\zeta_{i}(-c_{i2}  -c_{i3}  \cdots  -c_{i(n-1)}  -c_{in}).
\]
-- \[
T(s_{1})=
\begin{pmatrix}
c_{12}\rho_{2} & c_{13}\rho_{2} & \cdots & c_{1n}\rho_{2}\\
c_{12}\rho_{3} & c_{13}\rho_{3} & \cdots & c_{1n}\rho_{3}\\
\vdots & \vdots & & \vdots\\
c_{12}\rho_{n} & c_{13}\rho_{n} & \cdots & c_{1n}\rho_{n}
\end{pmatrix}
.
\]
De plus:
\[
[s_{1},\zeta]=\omega(\zeta)(-c_{12}  -c_{13}  \cdots  -c_{1(n-1)}  -c_{1n})
\]
où $\omega(\zeta)=-\sum_{k=2}^{n}\zeta_{k}\rho_{k}$ et $\omega(c_{i})=-\sum_{k=2}^{n}c_{ik}\rho_{k}=c_{i1}\quad (i \in \underline{n}_{1})$.\\
Dans la suite , on pose $c_{i}:=(-c_{i2}  -c_{i3}  \cdots  -c_{i(n-1)}  -c_{in}) \, (1\leqslant i \leqslant n)$; on obtient ainsi les formules:
\[
[s_{i},c_{k}]=-c_{ki}c_{i} \quad (i \in \underline{n},k \in \underline{n}).
\]
\begin{theorem}
Avec les hypothèses et notations précédentes, on a:\\
1) $(c_{2},c_{3},\cdots,c_{n})$ est une base de $\mathcal{M}(n_{0},n_{1})(=\mathcal{M}(1,n-1))$.\\
2) On a $c_{1}=\sum_{i=2}^{n}\lambda_{i}c_{i}$ et les $\lambda_{i}$ sont donnés par
\[
\begin{pmatrix}
c_{12} & c_{13} & \cdots & c_{1n}
\end{pmatrix}
Car(G_{1})^{-1}=
\begin{pmatrix}
\lambda_{2} & \lambda_{3} & \cdots & \lambda_{n}
\end{pmatrix}
.
\]
3) $\mathcal{M}(n_{0},n_{1})$ est un $KG$-module simple isomorphe au $KG$-module $M'$. De plus $G$ opère comme un groupe de réflexion sur lui.
\end{theorem}
\begin{proof}
1) On a $\det (c_{2},c_{3}, \cdots,c_{n})=\det(Car(G_{1}))=\Delta(G_{1})\neq 0$ par hypothèse, donc $(c_{2},c_{3}, \cdots,c_{n})$ est un système libre. Comme $\mathcal{M}(n_{0},n_{1})$ est de dimension $n-1$, $(c_{2},c_{3}, \cdots,c_{n})$ est une base de $\mathcal{M}(n_{0},n_{1})$.\\
2) Si $c_{1}=\sum_{i=2}^{n}\lambda_{i}c_{i}$, alors les $\lambda_{i}$ sont donnés par les formules de l'énoncé.\\
3) Soient $M'=M/H$ et $\pi_{M}:M\to M'$ la projection canonique. On pose $a'_{i}:=\pi_{M}(a_{i})$ $(i\in \underline{n}_{1})$. Alors $(a'_{i})_{i\in\underline{n}_{1}}$ est une base de $M'$. On définit $\varphi:M'\to \mathcal{M}(n_{0},n_{1})$ par $\varphi(a'_{i}):=c_{i}$ $(i\in \underline{n}_{1})$.\\
Si $k\in \underline{n}$ et si $i\in \underline{n}_{1}$, on a $s_{k}.(\pi_{M}(a_{i}))=a'_{i}-c_{ki}a'_{k}=\pi_{M}(s_{k}(a_{i}))$;\\
 $s_{k}.\varphi(a'_{i})=s_{k}.c_{i}=c_{i}-c_{ki}c_{k}=\varphi(a'_{i}-c_{ki}a'_{k})=\varphi(s_{k}.a'_{i})$, donc $\varphi$ est $G$-équivariante.\\
 On a le diagramme de $KG$-modules:
 \[
 \begin{diagram}
\node{M} \arrow[1]{e,t}{\varphi \circ \pi_{M}} \arrow{s,r,l}{\pi_{M}}
\node{\mathcal{M}(n_{0},n_{1})}\\
\node{M'} \arrow{ne,b}{\varphi}
\end{diagram}
 \]
 Comme $M'$ est un $KG$-module simple, on voit que $\mathcal{M}(n_{0},n_{1})$ est aussi un $KG$-module simple isomorphe à $M'$. dans ces conditions, il est clair que $G$ opère comme un groupe de réflexions sur $\mathcal{M}(n_{0},n_{1})$.
\end{proof}
Nous étudions maintenant le cas général.
\begin{theorem}
On suppose que $n_{0}\geqslant 1$. Alors\\
1) $\forall j\in \underline{n}_{0}$, $M_{j}$ est un $KG$-module de réflexion simple isomorphe à $M'$ (en tant que $KG$-module).\\
2) Si $N(G)\neq \{1\}$ on a
\begin{enumerate}
  \item $A(N(G))\cap M_{j} \neq \{1\}$ et $(A(N(G))\cap M_{j})\otimes_{\mathbb{Z}}K=M_{j}$;
  \item $A(N(G))\otimes_{\mathbb{Z}}K=\mathcal{M}(n_{0},n_{1})$.
  \end{enumerate}
\end{theorem}
\begin{proof}
1) Comme $P(G)$ est un sous-groupe de $GL_{n_{1}}(K)$, $P(G)$ stabilise chaque $M_{j}$ $(j\in \underline{n}_{0})$.\\
Soit $j\in \underline{n}_{0}$. On appelle $c_{i}(j)$ l'élément suivant de $M$, où $i\in \underline{n}_{1}$:
\[
c_{i}(j)=
\begin{pmatrix}
&& 0 &&\\
-c_{ii_{1}} & -c_{ii_{2}} & \cdots & -c_{ii_{(n_{1}-1)}} & -c_{ii_{n_{1}}}\\
&& 0 &&
\end{pmatrix}
.
\]
Soit \[\zeta :=
\begin{pmatrix}
&& 0 &&\\
\zeta_{i_{1}} & \zeta_{i_{2}} & \cdots &\zeta_{i_{(n_{1}-1)}} & \zeta_{i_{n_{1}}}\\
&& 0 &&
\end{pmatrix}\]
un élément de $M_{j}$. Si $i \in \underline{n}_{1}$, un calcul simple montre que $[s_{i},\zeta]=\zeta T(s_{i})=\zeta c_{i}(j)$. Nous avons les mêmes formules que dans le cas $n_{0}=1$, d'où le résultat.\\
2) Le groupe $G$ opère sur $N(G)$ par conjugaison et sur $\mathcal{M}(n_{0},n_{1})$ par translations à droite. De plus $A_{|N(G)}:N(G)\to A(N(G))$ est un isomorphisme de groupes, donc $A(N(G))$ se décompose en somme directe de sous-groupes sous l'action de $G$:
\[
A(N(G))=\oplus_{j\in \underline{n}_{0}}(A(N(G))\cap M_{j}).
\]
On pose $A(N(G))_{j}:=A(N(G))\cap M_{j}$ et $N(G)_{j}:=A^{-1}(A(N(G))_{j})$. Il en résulte que:
\[
N(G)=\oplus_{j\in \underline{n}_{0}}N(G)_{j}
\]
et chaque $N(G)_{j}$ est un sous-groupe normal de $G$ qui n'est pas central car $Z(G)=\{1\}$.\\
Chaque $M_{j}$ ($j\in \underline{n}_{0}$) est un espace vectoriel de dimension $n_{1}$ et chaque $s\in S$ opère linéairement sur lui; son polynôme caractéristique est $P_{s}(X)=(X+1)(X-1)^{n_{1}-1}$ donc chaque $s$ dans $S$ opère comme une réflexion sur $M_{j}$ et $[s,M_{j}]$ est de dimension $1$. Pour tout $s$ dans $S$ on donne un générateur de $[s,M_{j}]$.
\begin{notation}
Soit $j\in \underline{n}_{0}$,
si $i\in \underline{n}$, on pose
\[
c_{i}(j)=
\begin{pmatrix}
&& 0 &&\\
-c_{ii_{1}} & -c_{ii_{2}} & \cdots & -c_{ii_{n_{1}-1}} & -c_{ii_{n_{1}}}\\
&& 0 &&
\end{pmatrix}
\]
les éléments non nuls étant sur la j-ième ligne.
\end{notation}
 On montre maintenant que $\forall i\in \underline{n}$ on a $[s_{i},M_{j}]=<c_{i}(j)>$.\\
 - Soit $j\in \underline{n}_{0}$. Pour tout $h\in N(G)_{j}$, il existe $s$ dans $S$ tel que  $[s,h]\neq \{1\}$, donc $s[s,h]s^{-1}=[h,s]=[s,h]^{-1}$: $\forall s \in S,\exists h\in N(G)_{j}$ tel que $shs^{-1}=h^{-1}$ car $Z(G)=\{1\}$.\\
 Soit $s_{k}\in \underline{n}_{0}$. Comme on a la relation $A(s_{k})+A(s_{k})P(s_{k})=0$ on voit que $[s_{k},M_{j}]=<A(s_{k})>$ car $A(s_{k})\neq 0$ sinon $<G_{1},s_{k}>$ opérerait irréductiblement sur $M$ contrairement au choix de $G_{1}$. De plus, on voit que $A(s_{k})=c_{k}(j)$. On a le résultat dans ce cas.\\
 - Soit $i\in \underline{n}_{1}$. Alors $T(s_{i})=c_{i}(j)$ et si $h\in N(G)_{j}$ on a
 \[
 A(h)=
 \begin{pmatrix}
&& 0 &&\\
h_{1} & h_{2} & \cdots & h_{n_{1}-1} & h_{n_{1}}\\
&& 0 &&
\end{pmatrix}
 \]
 où les éléments non nuls sont sur la j-ième ligne.\\
 On a $A(h)T(s_{i})=h_{i}c_{i}(j)$ et comme on a la relation $A(h)(I_{n_{1}}+P(s_{i}))=0=A(h)(2I_{n_{1}}+T(s_{i}))$, on obtient $2A(h)=-h_{i}c_{i}(j)$ et $A(h)=-\frac{h_{i}}{2}c_{i}(j)$. Il en résulte aussitôt que $[s_{i},M_{j}]=<c_{i}(j)>$.\\
  Le reste est clair car $A$ est un isomorphisme $G$-équivariant de $K$-espaces vectoriels.
\end{proof}
\subsection{Exemples}
Dans ce paragraphe nous donnons des exemples en dimension principalement  $4$. Le cas $n=3$ sera étudié en détail dans la partie suivant. On garde les hypothèses du paragraphe 1.
\subsubsection{Le cas $n=4$}
On suppose $n=4$ et $G=<s_{1},s_{2},s_{3},s_{4}>$. Nous énumérons tous les graphes connexes $\Gamma$ de cardinal $4$; chaque arête est décorée d'un entier $p\geqslant3$. On se donne un arbre couvrant  $\textit{T}$, $s_{1}$ une racine de cet arbre (qui sera un cercle plein)  et enfin des arêtes en gras que l'on ajoute à l'arbre pour obtenir le graphe $\Gamma$ . Pour le groupe de réflexion $G$ ainsi obtenu nous donnons la base adaptée, la matrice de Cartan ainsi que le discriminant de cette représentation. Enfin nous donnons les conditions nécessaires et suffisantes pour qu'il existe des applications sesquilinéaires non nulles $G$-invariantes avec $\sigma \in Aut K$ tel que $\sigma^{2}=id_{K}$ et le sous corps des points fixes de $\sigma$ contient $K_{0}$. Les applications obtenues sont $\sigma$-hermitiennes. Si $\sigma$ est l'identité de $K$ on obtient des applications bilinéaires symétriques. Pour chaque $p_{i}$ on choisit $\alpha_{i}$ une racine de $v_{p_{i}}(X)$.\\
	(I) Un chemin
\[
\begin{picture}(150,88)
\put(-20,40){(I)}
\put(29,42){\circle*{7}}
\put(32,42){\line(1,0){30}}
\put(65,42){\circle{7}}
\put(68,42){\line(4,0){31}}
\put(103,42){\circle{7}}
\put(106,42){\line(7,0){32}}
\put(142,42){\circle{7}}
\put(24,52){$s_{1}$}
\put(61,52){$s_{2}$}
\put(99,52){$s_{3}$}
\put(136,52){$s_{4}$}
\put(44,47){$p_{1}$}
\put(80,47){$p_{2}$}
\put(116,47){$p_{3}$}
\end{picture}
\]
\[
Car(G)=
\begin{pmatrix}
2 & -\alpha_{1} & 0 & 0\\
-1 & 2 & -\alpha_{2} & 0\\
0 & -1 & 2 & \alpha_{3}\\
0 & 0 & -1 & 2
\end{pmatrix}
,
\]
\[
 (\varphi(a_{i},a_{j})_{1\leqslant i,j\leqslant4})=
\begin{pmatrix}
2 & -\alpha_{1} & 0 & 0\\
-\alpha_{1} & 2\alpha_{1} & -\alpha_{1}\alpha_{2} & 0\\
0 & -\alpha_{1}\alpha_{2} & 2\alpha_{1}\alpha_{2} & -\alpha_{1}\alpha_{2}\alpha_{3}\\
0 & 0 & -\alpha_{1}\alpha_{2}\alpha_{3} & 2\alpha_{1}\alpha_{2}\alpha_{3}
\end{pmatrix}
\]
On a $\Delta(G)=16-4\alpha_{1}-4\alpha_{2}-4\alpha_{3}+\alpha_{1}\alpha_{3}=(4-\alpha_{1})(4-\alpha_{3})-4\alpha_{2}$. \\
	(II) Une étoile
\[
\begin{picture}(150,88)
\put(6,40){(II)}
\put(29,42){\circle{7}}
\put(32,42){\line(1,0){30}}
\put(65,42){\circle*{7}}
\put(68,42){\line(4,0){31}}
\put(103,42){\circle{7}}
\put(65,8){\line(0,1){30}}
\put(65,5){\circle{7}}
\put(24,52){$s_{2}$}
\put(61,52){$s_{1}$}
\put(99,52){$s_{3}$}
\put(72,2){$s_{4}$}
\put(44,47){$p_{1}$}
\put(80,47){$p_{2}$}
\put(68,20){$p_{3}$}
\end{picture}
\]
\[
Car(G)=
\begin{pmatrix}
2 & -\alpha_{1} & -\alpha_{2} & -\alpha_{3}\\
-1 & 2 & 0 & 0\\
-1 & 0 & 2 & 0\\
-1 & 0 & 0 & 2
\end{pmatrix}
\]
\[
 (\varphi(a_{i},a_{j})_{1\leqslant i,j\leqslant4})=
 \begin{pmatrix}
 2 & -\alpha_{1} & -\alpha_{2} & -\alpha_{3}\\
 -\alpha_{1} & 2\alpha_{1} & 0 & 0\\
 -\alpha_{2} & 0 & 2\alpha_{2} & 0\\
 -\alpha_{3} & 0 & 0 & 2\alpha_{3}
\end{pmatrix}
\]
On a $\Delta(G) = 4(4-\sum_{1}^{4}\alpha_{i})$.\\
	(III) Un seul circuit qui n'est pas un carré
\[
\begin{picture}(150,88)
\put(29,42){\circle{7}}
\put(32,42){\line(1,0){30}}
\put(65,42){\circle*{7}}
\put(68,42){\line(3,2){31}}
\put(103,65){\circle{7}}
\linethickness{0.6mm}
\put(68,42){\line(3,-2){31}}
\put(24,52){$s_{4}$}
\put(61,52){$s_{1}$}
\put(98,74){$s_{2}$}
\put(101,8){$s_{3}$}
\put(103,20){\circle{7}}
\put(103,23){\line(0,1){40}}
\put(44,48){$p_{4}$}
\put(80,60){$p_{1}$}
\put(80,20){$p_{2}$}
\put(108,42){$p_{3}$}
\end{picture}
\]
\[
Car(G)=
\begin{pmatrix}
2 & -\alpha_{1} & -\alpha_{2} & -\alpha_{4}\\
-1 & 2 & -l & 0\\
-1 & -m & 2 & 0\\
-1 & 0 & 0 & 2
\end{pmatrix}
\]
avec $lm=\alpha_{3}$.\\
Si $\varphi$ $\neq 0$ $\in \Phi$, on a $\sigma(l)\alpha_{1}=m\alpha_{2}$ et
\[
(\varphi(a_{i},a_{j})_{1\leqslant i,j\leqslant4})=
\begin{pmatrix}
2 & -\alpha_{1} & -\alpha_{2} & -\alpha_{4}\\
-\alpha_{1} & 2\alpha_{1} & -m\alpha_{2} & 0\\
-\alpha_{2} & -l\alpha_{1} & 2\alpha_{2} & 0\\
-\alpha_{4} & 0 & 0 & 2\alpha_{4}
\end{pmatrix}
\]
On a $\Delta(G) = 4(4-\sum_{1}^{4}\alpha_{i})+\alpha_{3}\alpha_{4}-2(\alpha_{1}l+\alpha_{2}m)$;\\
si $\varphi$ est bilinéaire alors $\Delta(G) = 4(4-\sum_{1}^{4}\alpha_{i})+\alpha_{3}\alpha_{4}-4\alpha_{1}l$\\
	(IV) Un carré
\[
\begin{picture}(150,88)
\put(65,7){\circle{7}}
\put(65,10){\line(0,1){30}}
\put(65,42){\circle*{7}}
\put(68,42){\line(4,0){31}}
\put(103,42){\circle{7}}
\linethickness{0.6mm}
\put(67,9){\line(1,0){32}}
\put(103,7){\circle{7}}
\thinlines
\put(103,12){\line(0,1){29}}
\put(61,52){$s_{1}$}
\put(97,52){$s_{4}$}
\put(61,-5){$s_{2}$}
\put(97,-5){$s_{3}$}
\put(80,48){$p_{4}$}
\put(80,1){$p_{2}$}
\put(51,24){$p_{1}$}
\put(108,24){$p_{3}$}
\end{picture}
\]
\[
Car(G)=
\begin{pmatrix}
2 & -\alpha_{1} & 0 & -\alpha_{4}\\
-1 & 2 & -l & 0\\
0 & -m & 2 & -1\\
-1 & 0 & -\alpha_{3} & 2
\end{pmatrix}
\]
avec $lm=\alpha_{2}$.\\
Si $\varphi$ $\neq 0$ $\in \Phi$, on a $\sigma(l)\alpha_{1}=m\alpha_{3}\alpha_{4}$ et
\[
(\varphi(a_{i},a_{j})_{1\leqslant i,j\leqslant4})=
\begin{pmatrix}
2 & -\alpha_{1} & 0 & -\alpha_{4}\\
-\alpha_{1} & 2\alpha_{1} & -m\alpha_{3}\alpha_{4} & 0\\
0 & -l\alpha_{1} & 2\alpha_{3}\alpha_{4} & -\alpha_{3}\alpha_{4}\\
-\alpha_{4} & 0 & -\alpha_{3}\alpha_{4} & 2\alpha_{4}
\end{pmatrix}
\]
$\Delta(G)=4(4-\sum_{1}^{4}\alpha_{i})+(\alpha_{1}\alpha_{3}+\alpha_{2}\alpha_{4})-(\alpha_{1}l+\alpha_{3}\alpha_{4}m)$;\\
si $\varphi$ est bilinéaire alors $\Delta(G) = 4(4-\sum_{1}^{4}\alpha_{i})+(\alpha_{1}\alpha_{3}+\alpha_{2}\alpha_{4})-2\alpha_{1}l$.\\
	(V) Un graphe avec deux circuits:
\[
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\put(29,42){\circle{7}}
\linethickness{0.6mm}
\put(32,42){\line(1,0){30}}
\put(65,42){\circle{7}}
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\put(65,9){\circle*{7}}
\thinlines
\put(65,11){\line(0,1){28}}
\put(31,40){\line(1,-1){30}}
\put(68,10){\line(1,1){30}}
\put(24,52){$s_{2}$}
\put(61,52){$s_{3}$}
\put(99,52){$s_{4}$}
\put(61,-3){$s_{1}$}
\put(44,48){$p_{4}$}
\put(80,48){$p_{5}$}
\put(30,25){$p_{1}$}
\put(90,25){$p_{3}$}
\put(67,25){$p_{2}$}
\end{picture}
\]
\[
Car(G)=
\begin{pmatrix}
2 & -\alpha_{1} & -\alpha_{2} & -\alpha_{3}\\
-1 & 2 & -l_{1} & 0\\
-1 & -m_{1} & 2 & -l_{2}\\
-1 & 0 & -m_{2} & 2
\end{pmatrix}
\]
avec $l_{1}m_{1}=\alpha_{4}$ et $l_{2}m_{2}=\alpha_{5}$.\\
Si $\varphi$ $\neq 0$ $\in \Phi$, on a $\sigma(l_{1})\alpha_{1}=\alpha_{2}m_{1}$ et $\sigma(l_{2})\alpha_{2}=\alpha_{3}m_{2}$ et
\[
(\varphi(a_{i},a_{j})_{1\leqslant i,j\leqslant4})=
\begin{pmatrix}
2 & -\alpha_{1} & -\alpha_{2} & \alpha_{3}\\
-\alpha_{1} & 2\alpha_{1} & -m_{1}\alpha_{2} & 0\\
-\alpha_{2} & -\alpha_{1}l_{1} & 2\alpha_{2} & -\alpha_{3}m_{2}\\
-\alpha_{3} & 0 & -\alpha_{2}l_{2} & 2\alpha_{3}
\end{pmatrix}
\]
\begin{multline*}
\Delta(G)=4(4-\sum_{i=1}^{5}\alpha_{i})+(\alpha_{1}\alpha_{5}+\alpha_{3}\alpha_{4})\\-(2\alpha_{1}l_{1}+2\alpha_{2}l_{2}+2\alpha_{2}m_{1}+2\alpha_{3}m_{2})-(\alpha_{1}l_{1}l_{2}+\alpha_{3}m_{1}m_{2})
\end{multline*}
si $\varphi$ est bilinéaire alors $\Delta(G) = 4(4-\sum_{1}^{4}\alpha_{i})+(\alpha_{1}\alpha_{5}+\alpha_{3}\alpha_{4})-4(\alpha_{1}l_{1}+\alpha_{2}l_{2})-2\alpha_{1}l_{1}l_{2}$.\\
\pagebreak
	(VI) Le graphe complet.
\[
\begin{picture}(150,-20)
\put(70,7){$s_{1}$}
\put(31,40){\line(1,-1){30}}
\put(24,52){$s_{2}$}
\put(66,-30){\line(0,1){34}}
\put(29,42){\circle{7}}
\linethickness{0.6mm}
\put(30,40){\line(1,-2){35}}
\put(67,-30){\line(1,2){35}}
\put(32,42){\line(1,0){60}}
\put(68,42){\line(4,0){31}}
\put(103,42){\circle{7}}
\put(65,9){\circle*{7}}
\put(66,-33){\circle{7}}
\put(64,-45){$s_{4}$}
\put(99,52){$s_{3}$}
\put(31,40){\line(1,-1){30}}
\put(68,10){\line(1,1){30}}
\put(62,48){$p_{4}$}
\put(90,0){$p_{5}$}
\put(34,0){$p_{6}$}
\put(50,25){$p_{1}$}
\put(70,25){$p_{2}$}
\put(69,-5){$p_{3}$}
\end{picture}
\]
\linebreak[4]
\linebreak[4]
\[
Car(G)
\begin{pmatrix}
2 & -\alpha_{1} & -\alpha_{2} & -\alpha_{3}\\
-1 & 2 & -l_{4} & -m_{6}\\
-1 & -m_{4} & 2 & -l_{5}\\
-1 & -l_{6} & -m_{5} & 2
\end{pmatrix}
\]
avec $l_{4}m_{4}=\alpha_{4}$, $l_{5}m_{5}=\alpha_{5}$ et $l_{6}m_{6}=\alpha_{6}$.\\
Si $\varphi$ $\neq 0$ $\in \Phi$, on a $\sigma(l_{4})\alpha_{1}=m_{4}\alpha_{2}$, $\sigma(l_{5})\alpha_{2}=m_{5}\alpha_{3}$ et $\sigma(l_{6})\alpha_{3}=m_{6}\alpha_{1}$ et
\[(\varphi(a_{i},a_{j})_{1\leqslant i,j\leqslant4})=
\begin{pmatrix}
2 & -\alpha_{1} & -\alpha_{2} & \alpha_{3}\\
-\alpha_{1} & 2\alpha_{1} & -m_{4}\alpha_{2} & -l_{6}\alpha_{3}\\
-\alpha_{2} & -\alpha_{1}l_{4} & 2\alpha_{2} & -\alpha_{3}m_{5}\\
-\alpha_{3} & -m_{6}\alpha_{1} & -\alpha_{2}l_{5} & 2\alpha_{3}
\end{pmatrix}
\]
\begin{multline*}
\Delta(G)=4(4-\sum_{i=1}^{5}\alpha_{i})+(\alpha_{1}\alpha_{5}+\alpha_{3}\alpha_{4}+\alpha_{2}\alpha_{6})-2(l_{4}l_{5}l_{6}+m_{4}m_{5}m_{6})\\
-\alpha_{1}(2l_{4}+2m_{6}+l_{4}l_{5}+m_{5}m_{6})-\alpha_{2}(2l_{5}+2m_{4}+l_{5}l_{6}+m_{6}m_{4})-\alpha_{3}(2l_{6}+2m_{5}+l_{6}l_{4}+m_{4}m_{5})
\end{multline*}
Si $\varphi \neq 0$ est bilinéaire, $\Delta(G)$ devient:
\begin{multline*}
\Delta(G)=4(4-\sum_{i=1}^{5}\alpha_{i})+(\alpha_{1}\alpha_{5}+\alpha_{3}\alpha_{4}+\alpha_{2}\alpha_{6})-4l_{4}l_{5}l_{6}\\
-4(\alpha_{1}l_{4}+\alpha_{2}l_{5}+\alpha_{3}l_{6})-2(\alpha_{1}l_{4}l_{5}+\alpha_{2}l_{5}l_{6}+\alpha_{3}l_{6}l_{4}).
\end{multline*}
\textbf{Dans les propositions qui suivent on suppose toujours que $\Delta(G)=0$.}
\begin{proposition}
Avec les hypothèses et notations précédentes.\\
1) Dans le cas (I) on a une seule possibilité: $n_{0}=1$ et $p_{1}=p_{3}=4$, $p_{2}=3$, $\alpha_{1}=\alpha_{3}=2$, $\alpha_{2}=1$. On a le graphe $\Gamma(G)\simeq \Gamma(\tilde{C}_{3})$ et $G\simeq W(\tilde{C}_{3})$ groupe de Weyl affine de type $\tilde{C}_{3}$. De plus $b=a_{1}+a_{2}+a_{3}+\frac{1}{2}a_{4}$.\\
2) Dans le cas (II) on a deux possibilités, $n_{0}=1$ et\\
	(i) $p_{1}=p_{2}=3$, $p_{3}=4$, $\alpha_{1}=\alpha_{2}=1$, $\alpha_{3}=2$. On a le graphe $\Gamma(G)\simeq \Gamma(\tilde{B}_{3})$ et $G\simeq W(\tilde{B}_{3})$ groupe de Weyl affine de type $\tilde{B}_{3}$. De plus $b=a_{1}+\frac{1}{2}(a_{2}+a_{3}+a_{4})$.\\
	(ii) $p_{1}=p_{2}=5$, $p_{3}=3$, $\alpha_{1}=\tau$, $\alpha_{2}=3-\tau$, $\alpha_{3}=1$. On a $G\simeq W(\tilde{H}_{3})$ groupe de réflexion affine de type $H_{3}$. De plus $b=a_{1}+\frac{1}{2}(a_{2}+a_{3}+a_{4})$. Ce groupe sera étudié plus loin.
\end{proposition}
\begin{proof}
(I) On a $\Delta(G)=16-4\alpha_{1}-4\alpha_{2}-4\alpha_{3}+\alpha_{1}\alpha_{3}=(4-\alpha_{1})(4-\alpha_{3})-4\alpha_{2}=0$. D'après la proposition 13 de la partie 2, ceci n'est possible que si $\alpha_{1}=\alpha_{3}=2$ et $\alpha_{2}=1$, c'est à dire $p_{1}=p_{3}=4$ et $p_{2}=3$. On a donc un unique exemple :$G\simeq W(\tilde{C}_{3})$. En particulier on ne peut pas avoir $n_{0}=2$.\\
(II) On a $\Delta(G)=4(4-\alpha_{1}-\alpha_{2}-\alpha_{3})=0$. Toujours d'après la proposition 13 de la partie 2, ceci n'est possible que dans les deux cas suivants:\\
(i) $\alpha_{1}=\alpha_{2}=1$, $\alpha_{3}=2$ (par exemple) d'où $p_{1}=p_{2}=3$ et $p_{3}=4$. On a donc $G\simeq W(\tilde{B}_{3})$.\\
(ii) $\alpha_{1}=\tau,\alpha_{2}=3-\tau$ et $\alpha_{3}=1$, d'où $p_{1}=p_{2}=5$ et $p_{3}=3$. On a le résultat de l'énoncé. \\
Dans les deux cas, on ne peut pas avoir  $n_{0}=2$.
\end{proof}
\begin{proposition}
Avec les hypothèses et notations précédentes, on a une seule possibilité dans le cas (III): $n_{0}=1$. Si $G_{1}=<s_{1},s_{2},s_{3}>$, $\Delta(G_{1})=\frac{1}{2}\alpha_{4}(4-\alpha_{3})=8-2\alpha_{1}-2\alpha_{2}-2\alpha_{3}-(\alpha_{1}l+\alpha_{2}m)\neq 0$. $\alpha_{1}l$ et $\alpha_{2}m$ sont racines d'un polynôme $Q(X)\in K_{0}[X]$ du second degré.
On a: $b=a_{1}+\frac{l+2}{4-\alpha_{3}}a_{2}+\frac{m+2}{4-\alpha_{3}}a_{3}+\frac{1}{2(4-\alpha_{3})}a_{4}$.
\end{proposition}
\begin{proof}
On a $\Delta(G)=-\alpha_{4}(4-\alpha_{3})+2(8-2\alpha_{1}-2\alpha_{2}-2\alpha_{3}-(\alpha_{1}l+\alpha_{2}m))=0$. Comme $\alpha_{4}(4-\alpha_{3})\neq 0$ par les hypothèses générales, si $G_{1}=<s_{1},s_{2},s_{3}>$, on obtient $\Delta(G_{1})\neq 0$ donc $n_{0}=1$. On  a $\Delta(G_{1})=\frac{1}{2}\alpha_{4}(4-\alpha_{3})=8-2\alpha_{1}-2\alpha_{2}-2\alpha_{3}-(\alpha_{1}l+\alpha_{2}m)$ et $\alpha_{1}l$ et $\alpha_{2}m$ sont racines d'un polynôme $Q(X)\in K_{0}[X]$ du second degré. On a par un calcul facile:
\[
b=a_{1}+\frac{l+2}{4-\alpha_{3}}a_{2}+\frac{m+2}{4-\alpha_{3}}a_{3}+\frac{1}{2(4-\alpha_{3})}a_{4}.
\]
\end{proof}
Dans les cas (IV) et (V) soient $i\in \underline{n}$, $S_{i}=S-\{i\}$ et $R_{i}$ la restriction de $R$ à $R_{i}$ opérant sur $<\mathcal{A}-\{a_{i}\}>$. Si $n_{0}=1$ on ne peut pas avoir $R_{i}$ réductible pour tout $i$, donc il existe $i\in \underline{n}$ tel que $R_{i}$ est irréductible.\\
On s'intéresse d'abord au cas (IV).
\[
\Delta(G)= 16-4\alpha_{1}-4\alpha_{2}-4\alpha_{3}-4\alpha_{4}+\alpha_{1}\alpha_{3}+\alpha_{2}\alpha_{4}-\alpha_{1}l-\alpha_{3}\alpha_{4}m=0.
\]
\begin{proposition}
Avec les hypothèses et notations précédentes, on a deux possibilités dans le cas (IV).\\
1) On suppose que $n_{0}=1$ et que $\underline{n}_{0}=\{1\}$. On a $G_{1}=<s_{2},s_{3},s_{4}>$ (on peut faire ce choix d'après la remarque précédente).\\
La formule donnant $\Delta(G)$ montre que $\alpha_{1}l$ et $\alpha_{3}\alpha_{4}m$ sont racines du polynôme $Q(X)\in K_{0}[X]$:
\[
Q(X)=X^{2}-(16-4\alpha_{1}-4\alpha_{2}-4\alpha_{3}-4\alpha_{4}+\alpha_{1}\alpha_{3}+\alpha_{2}\alpha_{4})X+\alpha_{1}\alpha_{2}\alpha_{3}\alpha_{4}.
\]
On voit ainsi que $[K:K_{0}]\leqslant 2$ et aussi que
\[
b=a_{1}+\frac{1}{\Delta(G_{1})}[(4-\alpha_{3}+l)a_{2}+(2m+2)a_{3}+(m\alpha_{3}+4-\alpha_{2})a_{4}].
\]
2) On suppose maintenant que $n_{0}=2$: $\underline{n}_{0}=\{1,3\}$, $\underline{n}_{1}=\{2,4\}$. On a:
\[
G_{0}=<s_{1},s_{3}>,\,Car(G_{0})=
\begin{pmatrix}
2 & 0\\
0 & 2
\end{pmatrix}
\]
\[
G_{1}=<s_{2},s_{4}>,\,Car(G_{1})=
\begin{pmatrix}
2 & 0\\
0 & 2
\end{pmatrix}
\]
On a $b_{1}=a_{1}+\frac{1}{2}(a_{2}+a_{4}),b_{3}=a_{3}+\frac{1}{2}(la_{2}+\alpha_{3})$.\\
On a $p_{2}=p'=p_{4}$ et $p_{1}=p_{3}=p$; $\alpha_{2}=\alpha_{4}=4-\alpha_{1}$ et $\alpha_{3}=\alpha_{1}$. De plus $m=-1$ et $l=-\alpha_{2}$.
\end{proposition}\begin{proof}
1) Il suffit de faire les calculs.
2) On utilise les résultats généraux pour trouver la valeur de $b_{1}$ et de $b_{3}$. Enfin on a la relation:
\[
\begin{pmatrix}
-\alpha_{1} & -\alpha_{4}\\
-m & -1
\end{pmatrix}
\frac{1}{2}
\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}
\begin{pmatrix}
-1 & -l\\
-1 & -\alpha_{3}
\end{pmatrix}
=\begin{pmatrix}
2 & 0\\
0 & 2
\end{pmatrix}
\]
d'où $\alpha_{1}+\alpha_{4}=4$, $m+1=0$, $l\alpha_{1}+\alpha_{3}\alpha_{4}=0$, $\alpha_{2}+\alpha_{3}=4$. On en déduit: $4-\alpha_{1}-\alpha_{4}=0$, $4-\alpha_{2}-\alpha_{3}=0$, $m=-1$, $l=-\alpha_{2}$ d'où $\alpha_{1}\alpha_{2}=\alpha_{3}\alpha_{4}=\alpha_{3}(4-\alpha_{1})=(4-\alpha_{2})(4-\alpha_{1})$ donc $16-4\alpha_{1}-4\alpha_{2}=0$ et $4-\alpha_{1}-\alpha_{2}=0$; on voit aussi que $4-\alpha_{3}-\alpha_{4}=0$. On a donc tous les résultats.
\end{proof}
On peut voir que le sous-groupe $<s_{1},s_{2},s_{3}>$ est un groupe diédral affine (voir la partie 3) et la structure de $G$ est maintenant immédiate.\\
Dans le cas (V), on a
\begin{multline*}
\Delta(G)=4(4-\sum_{i=1}^{5}\alpha_{i})+(\alpha_{1}\alpha_{5}+\alpha_{3}\alpha_{4})\\-(2\alpha_{1}l_{1}+2\alpha_{2}l_{2}+2\alpha_{2}m_{1}+2\alpha_{3}m_{2})-(\alpha_{1}l_{1}l_{2}+\alpha_{3}m_{1}m_{2})
\end{multline*}
-- Si $n_{0}=1$, alors $\Delta(G)=0$ et si l'on se donne $l_{1}$ quelconque (mais différent de $0$) alors $m_{1}$ est déterminé et $l_{2}$ et $m_{2}$ sont racines d'un polynôme $Q(X)$ du second degré à coefficients dans $K_{0}(l_{1})$. \textbf{Il y a ainsi une infinité de représentations de réflexion réductibles non équivalentes}.\\
-- On considère maintenant le cas $n_{0}=2$.
\begin{proposition}
On choisit $\underline{n}_{0}=\{1,3\}$ et $\underline{n}_{1}=\{2,4\}$. On a $b_{1}=a_{1}+\frac{1}{2}(a_{2}+a_{4})$ et $b_{3}=a_{3}+\frac{1}{2}(l_{1}a_{2}+m_{2}a_{4})$. On a $p_{3}=p_{1}'$, $p_{5}=p'_{4}$, $\alpha_{1}+\alpha_{3}=4$, $\alpha_{4}+\alpha_{5}=4$, $\alpha_{1}l_{1}+\alpha_{3}m_{2}+2\alpha_{2}=0$ et enfin $m_{1}+l_{2}+2=0$. Le $R$-sous-groupe $<s_{1},s_{2},s_{3}>$ (resp. $<s_{1},s_{3},s_{4}>$) est un  groupe de réflexion affine quelconque et on l'a plongé dans un groupe de réflexion affine de rang $4$ tel que $H$ soit de dimension $2$.
\end{proposition}
\begin{proof}
On a
\[
Car(G_{1})=2
\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}
, Car(G_{0})=
\begin{pmatrix}
2 & -\alpha_{2}\\
-1 & 2
\end{pmatrix}
\]
et nous obtenons:
\[
\begin{pmatrix}
\rho_{1}^{2} & \rho_{3}^{2}\\
\rho_{1}^{4} & \rho_{3}^{4}
\end{pmatrix}
=\frac{1}{2}
\begin{pmatrix}
1 & l_{1}\\
1 & m_{2}
\end{pmatrix}
\]
d'où les valeurs de $b_{1}$ et de $b_{3}$. De plus on a les relations de l'énoncé. Comme tous les mineurs de rang $3$ de $Car(G)$ sont nuls, on a aussi: $\Delta(<s_{1},s_{2},s_{3}>)=0$ et $\Delta(<s_{1},s_{3},s_{4}>)=0$ et ceci démontre la proposition.
\end{proof}
\subsubsection{Le cas $n=5$}
On ne s'intéresse ici uniquement au cas où $\Gamma(G)$ est un chemin:\\ $G=<s_{i}|1\leqslant i\leqslant 5>$ et
\[
\begin{picture}(150,88)
\put(-80,38){$\Gamma(G)$=}
\put(-11,42){\circle*{7}}
\put(-8,42){\line(1,0){33}}
\put(29,42){\circle{7}}
\put(32,42){\line(1,0){30}}
\put(65,42){\circle{7}}
\put(68,42){\line(1,0){31}}
\put(103,42){\circle{7}}
\put(107,42){\line(1,0){30}}
\put(141,42){\circle{7}}
\put(-15,52){$s_{1}$}
\put(24,52){$s_{2}$}
\put(61,52){$s_{3}$}
\put(99,52){$s_{4}$}
\put(135,52){$s_{5}$}
\put(4,48){$p_{1}$}
\put(44,48){$p_{2}$}
\put(80,48){$p_{3}$}
\put(116,48){$p_{4}$}
\end{picture}
\]
Pour $i\in \{1,2,3,4\}$ soit $\alpha_{i}$ une racine de $v_{p_{i}}(X)$. On a
\[
Car(G)=
\begin{pmatrix}
2 & -\alpha_{1} & 0 & 0 & 0\\
-1 & 2 & -\alpha_{2} & 0 & 0\\
0 & -1 & 2 & -\alpha_{3} & 0\\
0 & 0 & -1 & 2 & -\alpha_{4}\\
0 & 0 & 0 & -1 & 2
\end{pmatrix}
\]
et \[
\Delta(G)=2[(4-\alpha_{1})(4-\alpha_{4})-\alpha_{3}(4-\alpha_{1})-\alpha_{2}(4-\alpha_{4})].
\]
Je connais trois solutions à l'équation $\Delta(G)=0$ et je ne sais pas s'il y en a d'autres.\\
1) $p_{1}=p_{4}=4$, $p_{2}=p_{3}=3$ alors $\alpha_{1}=\alpha_{4}=2$ et $\alpha_{2}=\alpha_{3}=1$.
\[
\begin{picture}(150,88)
\put(-80,38){$\Gamma(G)$=}
\put(-11,42){\circle{7}}
\put(-8,42){\line(1,0){33}}
\put(29,42){\circle{7}}
\put(32,42){\line(1,0){30}}
\put(65,42){\circle{7}}
\put(68,42){\line(1,0){31}}
\put(103,42){\circle{7}}
\put(107,42){\line(1,0){30}}
\put(141,42){\circle{7}}
\put(-15,52){$s_{1}$}
\put(24,52){$s_{2}$}
\put(61,52){$s_{3}$}
\put(99,52){$s_{4}$}
\put(135,52){$s_{5}$}
\put(4,48){$4$}
\put(44,48){$3$}
\put(80,48){$3$}
\put(116,48){$4$}
\end{picture}
\]
On obtient le groupe de Weyl affine de type $\tilde{C}_{4}$:
$G\simeq W(\tilde{C}_{4})$.\\
2) $p_{1}=p_{4}=3$, $p_{2}=p_{3}=5$ alors $\alpha_{1}=\alpha_{4}=1$ et $\alpha_{2}=\tau$ et $\alpha_{3}=3-\tau$.
\[
\begin{picture}(150,88)
\put(-80,38){$\Gamma(G)$=}
\put(-11,42){\circle{7}}
\put(-8,42){\line(1,0){33}}
\put(29,42){\circle{7}}
\put(32,42){\line(1,0){30}}
\put(65,42){\circle{7}}
\put(68,42){\line(1,0){31}}
\put(103,42){\circle{7}}
\put(107,42){\line(1,0){30}}
\put(141,42){\circle{7}}
\put(-15,52){$s_{1}$}
\put(24,52){$s_{2}$}
\put(61,52){$s_{3}$}
\put(99,52){$s_{4}$}
\put(135,52){$s_{5}$}
\put(4,48){$3$}
\put(44,48){$5$}
\put(80,48){$5$}
\put(116,48){$3$}
\end{picture}
\]
Un calcul simple montre que $(s_{2}s_{4}^{s_{3}})^{3}=1$, donc d'après la proposition 16 2) de la section 4, $G_{1}=<s_{1},s_{2},s_{3},s_{4}>\simeq W(H_{4})$ et il existe $z\in Z(G_{1})-\{1\}$, d'où d'après la proposition 3, $(zs_{5})^{2}\in N(G)$. On a $G\simeq N(G)\ltimes W(H_{4})$ et $N(G)$ est un groupe commutatif libre de rang $8$.\\
3) $p_{1}=p_{4}=5$, $p_{2}=p_{3}=3$ alors $\alpha_{1}=\tau$, $\alpha_{4}=3-\tau$, $\alpha_{2}=\alpha_{3}=1$.
\[
\begin{picture}(150,88)
\put(-80,38){$\Gamma(G)$=}
\put(-11,42){\circle{7}}
\put(-8,42){\line(1,0){33}}
\put(29,42){\circle{7}}
\put(32,42){\line(1,0){30}}
\put(65,42){\circle{7}}
\put(68,42){\line(1,0){31}}
\put(103,42){\circle{7}}
\put(107,42){\line(1,0){30}}
\put(141,42){\circle{7}}
\put(-15,52){$s_{1}$}
\put(24,52){$s_{2}$}
\put(61,52){$s_{3}$}
\put(99,52){$s_{4}$}
\put(135,52){$s_{5}$}
\put(4,48){$5$}
\put(44,48){$3$}
\put(80,48){$3$}
\put(116,48){$5$}
\end{picture}
\]
Ici on voit que $G_{1}=<s_{1},s_{2},s_{3},s_{4}>\simeq W(H_{4})$ et on a les mêmes résultats que dans le cas précédent.

redpajama_set_name: RedPajamaArXiv


text:
Bozeat  is a village and civil parish in the postal district of Wellingborough, North Northamptonshire, England, about  south of Wellingborough on the A509 road, near Wollaston. At the time of the 2011 census, Bozeat's population (including Easton Maudit) was 2,052.

Origin of the name
Rev. Joseph Horace Marlow gives two possible origins of Bozeat's name:
Bozeat probably existed in Saxon times – Saxon coins have been found – and an early spelling of Bozeat was Bosgate, suggesting Bozeat may have meant Bosa(s) gate. Bosa was a common Saxon name and a Saxon Earl Bosa held land near here. In a similar way Strixton is named after the Saxon thane Strix. In Old English geat/gaet, and in Middle English yatt and zett are all recognised as meaning gate, opening or entrance to woods or land. All the various spellings of Bozeat over the years show some link to both Bosa and gate.

One possible French influence is shown in the Domesday Book (1086) spelling Bosiete and it is possible that the Normans slightly altered the name to make it more French. There may even be a link with the French Bosquet (small wood) or Latin Boscus (wood). With all the vagaries of spelling and spoken English over the centuries it is unlikely that Bozeat is still pronounced in the same way it was originally.

A second explanation of the name Bozeat meaning beautiful spring, from the French Beau-jet, assumes that the pronunciation of Bozeat was the same centuries ago as it is now. So although this theory is very appealing it does not have the same historical credibility as the first – Bosa's Gate.

The Oxford Dictionary of English Place-names says that Bozeat is derived from the "Gate or gap of a man called Bosa."

History
A little north-west of Bozeat a 48 ft circular Roman building was excavated in 1964-65. Two Roman kilns and five other buildings were also found during work on a housing estate.

Before the Norman Conquest, the Saxon thane, Strix (of Strixton) held some of the land here under Earl Waltheof, a powerful Saxon Earl of Northumbria. At the Norman Conquest, William I gave most of the land locally to his niece Judith, who became the first Countess of Northampton. Judith married Earl Waltheof, so the Saxon Earl and the Norman Lady became joint owners.

The Norman parish church dedicated to St Mary, built about 1130, is older than both Easton Maudit and Wollaston churches, and survived the Great Fire of Bozeat in 1729. It was rebuilt in 1880-83.

In the 15th century there was a thriving weaving industry within the farming community, but by the beginning of the 20th century the population had risen to 1,200 and boasted a cottage lace-making industry, a shoe industry and a windmill, with many independent tradesmen, making a very self-reliant village. At this time Bozeat had at least 20 shops, a school, five pubs and four churches.

There are many wells in the village, one of the most important being the communal 'Town Well', which came from a jet of water coming out of the side of a bank in Dychurch Lane. A charity was registered to pay for its upkeep.

Bozeat's shoe trade dates back to the middle of the 17th century but until the middle of the 18th century, it was considered a cottage industry. Men made and mended shoes in small buildings near their homes which were called 'shops', but although some remain today, they now serve a different purpose. The population grew rapidly when large shoe factories were built, providing much-needed employment for both men and women, and these remained Bozeat's main trade until 1982 when the last shoe factory closed.

The sports brand Gola was based in the village and was then known as the Bozeat Boot Company.

Transport
In January 1989 a by-pass was built west of the village to take the increasing traffic to Milton Keynes. In spring 2001, with more traffic passing the village and a number of accidents, a new roundabout was constructed to replace the junction of the A509 Wollaston Road. It made access into and out of the village safer. The village is connected by the W8 Wellingborough–Bozeat bus route.

Amenities
As of July 2017 the village had a primary school, a NISA supermarket, NHS doctors surgery, a soft furnishing specialist and a hairdresser's salon. Most trades are represented within the village. There is the Red Lion pub, bus services to Wellingborough and Northampton as well as local taxis. There are many specialist home-based businesses inline with the digital age. The Bozeat playing fields are marked for two football pitches and host outdoor exercise equipment, a table tennis table, a tennis court, children's play park, skate park and a pavilion building which contains a recycling shop on a saturday morning. There is also a Scout Group, which includes Beavers and Cubs, and a Women's Institute.

Services
The 20th century brought the utilities to Bozeat. Mains electricity was installed in 1925, mains water in 1949, sewerage in 1950, and gas in 1990. Three private housing estates were built during the 1960s and 1970s. 1999 saw the demolition of the Gola factory and 30+ more families settled into village life.

The village sign was erected in the millennium year after some months of dedicated fundraising. It depicts St Mary's Church, Bozeat and a windmill, representing aspects of village life that have continued over the years.

References

Bozeat 2000: The story of a Northamptonshire village, compiled by Philip Bligh, giving a history of Bozeat.

External links

Bozeat Parish Council website

Villages in Northamptonshire
North Northamptonshire
Civil parishes in Northamptonshire

redpajama_set_name: RedPajamaWikipedia


text:
Hulu won the day, but Facebook may be the future, investor says
Published Wed, Aug 3 2016 1:56 PM EDT Updated Thu, Aug 11 2016 4:02 PM EDT
Anita Balakrishnan@MsABalakrishnan
AndreyPopov | Getty Images
Now that online binge-watchers can get more real-time content, both traditional media companies and streaming companies may have to compete with tech giants for viewers, according to technology investor Jason Calacanis.
Calacanis: Linear TV has won the day
Binge with Carl Quintanilla
"I wouldn't be surprised if in three, four, five years from now we'll be on this network talking about Facebook, Google, YouTube, even Twitter, not just having the live rights to clips, but having the rights to carry things like CNN or CNBC or even full NBA games, " Calacanis told CNBC's "Squawk Alley " on Wednesday. "That's where this is headed, because the monetization on those networks will be even better because they have all the data about their users."
Time Warner announced Wednesday a 10 percent stake in Hulu, which will allow online live, on-demand streaming of channels like TNT, TBS, CNN, Cartoon Network, Adult Swim, truTV, Boomerang and Turner Classic Movies. Time Warner's shares popped about 3 percent Wednesday, after it also reported earnings.
Calacanis said it's a "nice boost" for Hulu to have this over-the-top service. In the past, he's said he sees a lot of growth left for Hulu's rival, Netflix.
But Calacanis said social media platforms also offer an exciting alternative to traditional cable bundles.
Twitter recently announced a deal to live stream a free new weekly NBA pregame show, and also has a deal with the NFL.
Facebook, while not as focused on long-form content, signed a deal for a soccer match between English clubs Manchester United and Everton. And during a recent hackathon, Facebook employees devised a way to use professional broadcast equipment on Facebook Live. Instagram, too, is doubling down on real-time content in light of the Rio Olympics, Brandon Gayle, head of global sports partnerships at Instagram, told "Squawk Alley."
"There's three areas where live TV is super relevant: Obviously, news, here we are," Calacanis said. "Sports, obviously a huge category, and this real-time reality TV that unfolds weekly. Those are very strong categories, and live is pretty exciting on these new platforms like Facebook and Twitter."
Disclosure: Comcast, which owns CNBC parent NBCUniversal, is a co-owner of Hulu.
Disclosure: CNBC parent NBCUniversal owns NBC Sports and NBC Olympics. NBC Olympics is the U.S. broadcast rights holder to all Summer and Winter Games through the year 2032.


text:
James McMurtry: It Had to Happen -- by George Graham
(Sugar Hill 1058 As broadcast on WVIA-FM 7/2/97)
The art of the singer-songwriter has become timeless, persisting beyond the frequent comings and goings of pop music fads. For over three decades, a performer getting up and weaving music and poetry together has never lost its relevance. While it tends to go in and out of vogue on the pop charts -- mostly out -- as an art-form, it is seemingly inexhaustible, despite its simple concept. We're in a period in which good singer-songwriters seem not to count much as far as the commercial music industry is concerned, nevertheless the artists are proliferating, with the number of even the really worthy singer-songwriters plying their trade approaching the overwhelming.
This week we have the work of yet another, whose music has that timeless quality the best practitioners evoke. It's James McMurtry, whose new fourth album is entitled It Had to Happen. And like many a singer-songwriter of intelligence and integrity, he was dropped by a major record company and moved to the ranks of the independents, in this case, going from Columbia to the usually folk-and-bluegrass label Sugar Hill, which has lately been carving out a niche as a home of first-rate folkies.
Thirty-five year old James McMurtry was born in Fort Worth, Texas after the marital breakup of his mother, a University of Richmond English professor, and his father, novelist Larry McMurtry, author of The Lonesome Dove and The Last Picture Show. James stayed mainly with his father in Houston, took up guitar by age seven, and also plunged into acting while still in school, even landing a film role. He began playing his songs before audiences while in college at the University of Arizona at Tucson. After a detour to Alaska, where he did some performing, he ended up working "day jobs" such as housepainter and bartender in San Antonio. His career began to take off in 1987 when he entered and won the songwriters' contest at the Kerrville, Texas, folk festival.
By 1989, he had gained John Mellancamp as a fan, who produced his debut album Too Long in the Wasteland, and served as executive producer of its follow-up, Candyland, three years later. Both were marked by McMurtry's inherited storytelling gift, along with a kind of jaundiced, but not quite bitter, attitude about life, reflected in some of his characters, and his often-sardonic manner of delivery. Three years against lapsed before McMurtry's last album Where Did You Hide the Body, produced by Don Dixon, and featuring Dixon's slightly more pop-oriented sound.
Now, less than two years after its predecessor comes It Had to Happen, McMurtry's first album recorded in his home state of Texas, with steel guitar player and multi-instrumentalist Lloyd Maines serving as producer. And for me, it's McMurtry's best, with some of his finest songs, usually woven around a storyline or interesting characters, set in some very tasteful and understated production, despite the generally electric instrumentation.
Once again, McMurtry's songs tell stories of, or mull over, the hand that life has dealt -- aging is often a theme -- delivered with both an attitude and a vocal style that is somewhere between a smirk and an air or resignation. The backing consists of a rocking little band that can set up a great groove, but always remembers its role to accompany and supplement the songs. Ronnie Johnson plays bass, Chris Searles is on drums and Lisa Mednick handles the keyboards, including accordion. There are a few additional guests such as producer Mains, who plays slide guitar, backing vocalist Randy Garibray, Jr., and Austin guitar notable Charlie Sexton.
The album gets under way with a song reflecting an interesting attitude toward one's state of life, presumably about aging. In the joint McMurtry/Wally Wilson composition, Paris, the main character reaches a point that he realizes that those around him feel he poses "no danger." The band provides some tasteful electric backing. <<>>
A similar subject is eloquently expounded upon in the following piece Peter Pan, whose chorus gives the children's storyline a new, and world-weary spin. <<>>
Another wonderfully sardonic viewpoint comes on No More Buffalo, which considers the obstacles to fame and fortune in the music business. <<>>
On the same subject, McMurtry throws a few zingers in the direction of philistine audiences on Be with Me, on which he certainly sounds the part of the curmudgeon, while the band plays a rocky little dirge, complete with mandolin and accordion. <<>>
The most musically upbeat song on the album is For All I Know, whose somewhat opaque lyrics are set in an infectious Zydeco-influenced groove. <<>>
One of my favorite pieces on the album is a set of childhood memories set into a song called 12 O'Clock Whistle, done mainly with just McMurtry's guitar and harmonica and Ronnie Johnson's bass as the accompaniment. <<>>
There is one cover song on It Had to Happen, a composition by Kinky Friedman called Wild Man from Borneo, written from the standpoint of the circus attraction in question, and by extension life as an exhibition. It's the kind of song that fits McMurtry perfectly. <<>>
The album ends with another of its gems. With some of McMurtry's best wordcraft, Jaws of Life, explores some of life's possibilities, and the unfulfillment of them. <<>>
James McMurtry is carrying on the great singer-songwriter tradition on his new, fourth and finest album It Had to Happen. Now off the major label scene, McMurtry is free to pursue his songs in a more laid-back intimate setting. But the album also sees his songwriting in peak form. He tells stories, examines the characters in his world and casts his own appealingly sardonic light on life. Of his own craft, McMurtry writes: "If I wrote straight autobiographical stuff, it'd be so boring. That's the same reason I try not to give opinions in songs. As soon as that happens, you're preaching... It's better to just tell the story and let listeners make up their own minds."
Sonically, the album is well done with a nice clean mix, and a minimum of studio gloss. The sound is like rest of the album -- plain and unadorned -- to let the songs and their stories be told. Producer Lloyd Mains and engineer/mixer Fred Remmert had the right idea.
Storytelling singer-songwriters are not exactly an endangered species. There are a lot of very good ones around. But on It Had to Happen James McMurtry blossoms and stands out as real master of the craft.
(c) Copyright 1997 George D. Graham. All rights reseved.


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Die Strada Statale 212 della Val Fortore (Abkürzung: SS 212) ist eine italienische Staatsstraße, die von Benevento nach Ripabottoni führt. Sie verbindet die italienischen Provinzen Benevento in Kampanien und Campobasso in Molise und ist 88,7 km lang.

212
Straße in Kampanien
Straße in Europa


Here is another interesting freebie from Themefisher created with Photoshop.
The template comes with clean content, well organized elements and editable content.
If you want to host this files in your website you are free to do that with full credit.


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News Students set up Trans Day of Remembrance memorial
Students set up Trans Day of Remembrance memorial
Issy Kenney-Herbert
Last Friday morning, a group of students woke up at 6am to set up a memorial in honour of Transgender Day of Remembrance. Elliot Brooke, one of the organisers of this memorial, said the intent "was to honour the trans people who passed away this year".
Attached to the railings of the Radcliffe Camera, a trans pride flag with the words 'We Will Not Be Erased' alongside two large assortments of posters featuring the names of the 242 victims of violence against trans people in 2020 so far. The remaining 108 known victims are unnamed.
Speaking to Cherwell, Elliot Brooke emphasised the significance of this memorial with regards to how the trans community is often neglected, stating: "There was something cathartic, healing and empowering about putting something together that was going to get people's attention for a community that is often ignored and forgotten."
For trans and gender-non conforming people, the last year was one of the deadliest on record and a 6% increase in reported murders from 2019. As the statement alongside this memorial noted: "Unfortunately, we must recognise that the true number is likely far greater than 350, as often the death of many transgender and gender-non-conforming individuals goes unreported. In addition, this number does not include trans individuals who may have taken their own lives as a result of the transphobia they face from a society in which prejudice against them is still very much ingrained."
Since January 1, 2008, the TMM project has monitored reports of transgender murders. Since then, the number of cases recorded reached a total of 3665 in 75 countries and territories worldwide. In the US, 89% of those killed were trans or gender-diverse people of colour.
All of the organisers expressed a sense of solemness and grief they had putting up the memorial. River Deng, another of the organisers behind this and Oxford's Queer Week 2020, had another message: "This memorial also represents that there are people who care." He went on to say that this week has highlighted that "trans love is the most important thing to me, out of all this grief and loss our community suffers and will suffer, no matter what, people will come together to love each other".
The statement at the memorial affirmed the need to "recognise that allowing transphobia to persist in any form, overt or otherwise, renders us complicit in the rising rates of violence against the global transgender community".
Pax Butchart, one of the organisers involved with this and Merton's Trans Awareness Week, said to Cherwell that "Oxford is a city that is quite divided on this issue" and went on to say that while "Oxford can be proud of itself this week, it shouldn't get complacent about how much work there is to be done".
Image credit: Zachary Elliott


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El Kromfohrländer es una raza de perro de compañía originaria de Alemania.

Descripción

Apariencia
El Kromfohrländer tiene un parecido a un cruce entre un perro cobrador y un beagle y puede encontrarse en una gran variedad de mantos: pelo corto, largo y rizado. Tienen marcas marrones y blancas desde oscuros a claros.

Tamaño
Es un perro de tamaño medio, entre 38 y 46 cm en la cruz, con un peso entre los 10 y 14 kg .

Temperamento
El "Krom" es un perro con una naturaleza muy agradable y un cazador relativamente bueno. Leal y cariñoso, se adapta bien a un piso siempre que pueda tener un mínimo de 30 o 40 minutos de ejercicio físico diario.

Historia
El Krom aparece en Alemania en la década de 1940, los primeros aparecieron como resultado de un cruce accidental entre dos perros mestizos (o un Fox Terrier y una Gran basset grifón vendeano). Los cachorros resultantes se criaron hasta ser reconocidos como raza en 1955.

Referencias

Véase también

Grupo 9 de la FCI
Perros de compañía
Razas de perro de Alemania


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As PML-N eyeing power, internal rift deepens
Shafaqna Pakistan: In late last year, a power broker had predicted, nay, shared some information about the likely changes to take place in early this year.
"There will be change at government level before the next budget," he said. What will be the changes? That he didn't explain. By then, petition against the extension granted to army chief was not filed.
However, as situation took a dramatic turn after the Supreme Court verdict on extension and the subsequent legislation, a common-sense guess is that plan of change might have been delayed, if not averted, given the Imran Khan government's vociferous defense on every frontier. But the major opposition parties are not behind on this count. The PML-N, for instance, was in the lead to offer "unconditional support" to amendment in Army Act as authorized by Mian Nawaz Sharif, to the shock of many takers of his "vote ko izzat du" slogan.
Now after the legislation has been approved, smaller parties in the ruling alliance have started posing threat to the government giving credence to the speculations of change. The MQM's only minister, Khalid Maqbool Siddiqui, has resigned. The PML-Q's only minister, Tariq Bashir Cheema, was conspicuous by absence from Tuesday's cabinet meeting without any prior notice. On Monday night, he had expressed reservations against the government in a talk show. Also, partnership between the two key stakeholders of the system is not at par excellence.
In the given scenario, PML-N is expecting some return of the "unconditional support" to amendment which it extended at the cost of die-hard supporters. While this gesture has appeased the power players, rift in the party deepens. Some of its leaders are in trouble for violating the oath they were administered in famous London meeting where the Nawaz Sharif's decision of unconditional support to the amendment was conveyed under oath and others are in hot waters for abstaining from the vote on the amendment.
Three-decade old friendship between Shahid Khaqan Abbasi and Khawaja Asif appears to be the biggest casualty of this crisis. Khawaja was the man who had proposed the name of Shahid for premiership after Nawaz Sharif was disqualified. Shahid was the person who had connected Khawaja with the present power players. Not only introduced, they dined together with the most powerful man a number of times. Their present status of relationship? "They are virtually not at speaking term," said one party leader.
As Shahid came to attend the National Assembly session Tuesday, Khawaja avoided him, let alone shaking hands, according to journalists present in Parliament. Rana Tanvir and Ayaz Sadiq were sitting in another room with Khawaja by the time Shahid was present at the office of the opposition leader, Shehbaz Sharif. As Rana and Ayaz went to that office, Shahid sent them back saying he was in an important meeting with Miftah Ismail and Musaddaq Malik. (Miftah told The News that since they were busy in discussing a personal matter, other leaders left the place after hand-shake. Khawaja Asif denied having coming face to face with Shahid and also dispelled the impression of any bitterness.)
Bad blood started creating over the issue of "unconditional support". Shahid was the first to allege Khawaja for not taking him in the loop. Khawaja, on defensive, first explained to him and then went public to declare that it was Nawaz Sharif's decision to support the amendment and that he was only a messenger. This unexpected exposure was much to the embarrassment of Nawaz who was apparently holding the mantle of "civilian supremacy." Reaction followed. The PML-N social media supporters close to Maryam Nawaz started openly attacking Khawaja for "ditching the leadership."
While he was in the line of fire for making inside story of London meeting public, the top leadership was also unhappy with Rana Tanvir and Ayaz Sadiq for tacitly endorsing Khawaja's disclosure. Ayaz, in a rare display of bluntness, said in a parliamentary meeting held after the passage of the bill that Maryam should have asked his father about the veracity of Khawaja's disclosure instead of condemning others for conveying this message to party lawmakers. He said this in response to MNA Javed Latif's criticism.
It has been learnt that Khawaja is now thinking of quitting the post of parliamentary leader as more challenges are ahead like the appointments of chief election commissioner and the members of Election Commission of Pakistan. He doesn't go to talk shows any more. After the passage of amendment, he only participated in a couple of TV programs and has been refusing appearance since then. Meanwhile, his role is also being reduced. For example, the decisions of top leadership to Parliamentary Advisory Group are conveyed through Attaullah Tarar, deputy secretary general, instead of having them routed through Khawaja.
What will PML-N get in the changing scenario remains to be seen. According to an informed official, a power-sharing formula is under consideration to form a kind of national government. However, PML-N is hesitant to become part of any such arrangement. It wants either fresh elections or change in house.
On the other hand, Imran Khan has reportedly threatened to dissolve assemblies if he's forced to quit through no-confidence. A minister close to the power brokers has claimed that Shehbaz is expected to return Pakistan in a couple of weeks. His younger son is also in contact with powerful circles along with his close associates through an influential relative of somebody in Lahore. Meanwhile, consensus is being forged on new chief election commissioner. The PML-N is considering the name of Sikandar Sultan Raja, retired federal secretary and son-in-law of Saeed Mehdi, former principal secretary to Mian Nawaz Sharif during his second term as PM, after it was floated by the government . It has also been conveyed to the Parliamentary Advisory Group.
pml(n)
Supreme Court. MQM
Previous articleA rare goodwill gesture from Indian government towards Pakistan
Next articleNew footage shows Iranian missiles hitting Ukraine plane
Questions over Nawaz Sharif's health
Cafe photo adds to PTI suspicion over Nawaz's health ß
Un-Islamic provisions of NAB law: Will convictions of Nawaz, others on presumption be affected?
Shehbaz expected to return in 2nd week of Jan: Sanaullah
Pakistan's prison politics: Who was in and who was out in 2019
Is Opposition's Silence The Quiet Before The Storm?


Diesel mechanic Tony Yahle was presumed to be dead in the hospital after his heart stopped. He continued to lay with no heart beat for 45 minutes while doctors prepared to have his family come and say their goodbyes. He was pronounced dead then and there, but what happens next you won't believe.
As Yahle's family was coming to see him to say goodbye, he began to show signs of life. This absolutely blew doctors away! Then, just five days later, he fully awoke.
Dr. Raja Nazir is the cardiologist at the hospital where Tony was admitted. The whole situation completely blew his mind – he couldn't stop thinking or talking about it. How could something like this happen!?
Crediting his strong faith, Tony Yahle was able to walk away from the hospital.
Were you amazed by this family's story? So share this!
If You Are on Any of These Medications NEVER Use Apple Cider Vinegar- Your Life It's At Risk!!!


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Home >> Investors >> FCC Commissioner Brendan Carr Meets with Executives from T-Mobile and RigNet
FCC Commissioner Brendan Carr Meets with Executives from T-Mobile and RigNet
Commissioner Carr tours Valaris' Gulf Copper facility as he learns more about the recent launch of RigNet and T-Mobile's new 4G LTE network in the Gulf of Mexico deployed with 5G-ready equipment.
HOUSTON, Oct. 02, 2019 (GLOBE NEWSWIRE) -- RigNet (NASDAQ: RNET, the Company), the leading provider of ultra-secure, intelligent networking solutions, had members of its executive team including CEO, Steven Pickett, CIO/CTO, Brendan Sullivan, and SVP, of Government Services, Jamie Barnett Rear Admiral USN (Retired) along with representatives from T-Mobile (NASDAQ: TMUS) met with Commissioner Carr during a tour of Valaris' (NYSE: VAL; formerly known as EnscoRowan plc) Gulf Copper facility, including touring a state-of-the-art rig in Galveston, TX.
RigNet's executive team along with representatives from T-Mobile met with Commissioner Carr during a tour of Valaris' Gulf Copper facility, including touring a state-of-the-art rig in Galveston, TX.
During the tour, the company representatives informed Commissioner Carr about the new 4G LTE and 5G-capable network launched across more than 60,000 square miles of the Gulf of Mexico (GoM). This network now connects offshore workers, vessels and critical infrastructure in an area larger than the state of Florida.
"We had the opportunity to discuss with Commissioner Carr the critical importance such a network has to help with the safety of the crews on the platforms and the integrity of drilling operations in the Gulf," said Steven Pickett, CEO and President of RigNet. "Thousands of oilfield workers live and work on these platforms each day, and there are thousands of sensors on each platform that provide critical visibility to the safety and efficiency of drilling operations to onshore personnel."
Utilizing RigNet's already deployed 6Ghz Microwave Network, T-Mobile and RigNet have installed state of the art base stations that operate on the 600- and 700-megahertz spectrum; which is able to penetrate offshore assets more effectively than higher frequency solutions and can be picked up by most consumer mobile devices. The oil and gas drilling platforms within the coverage area will gain access to a low latency and high speed mobile connection.
"We are very happy to have met with Commissioner Carr to discuss the benefits and value of our Gulf of Mexico Microwave and LTE network," continues Steven Pickett. "As the FCC considers the unlicensed use of 6 GHz spectrum, we have stressed that this would create network interference and could endanger offshore crews and other LTE users in the Gulf. We believe the Gulf should be excluded from the FCC's rulemaking. With this new LTE network, T-Mobile and RigNet customers' critical communications will not be interrupted, helping to safeguard safety and drilling operations."
About RigNet, Inc.:
RigNet (NASDAQ: RNET) delivers advanced software, optimized industry solutions, and communications infrastructure that allow our customers to realize the business benefits of digital transformation. With world-class, ultra-secure solutions spanning IP connectivity, bandwidth- optimized OTT applications, IIoT big data enablement, and industry-leading machine learning analytics, RigNet supports the full evolution of digital enablement, empowering businesses to respond faster to high priority issues, mitigate the risk of operational disruption, and maximize their overall financial performance. RigNet is headquartered in Houston, Texas with operations around the world. For more information, please visit http://www.rig.net.
This press release includes "forward-looking statements" within the meaning of the safe harbor provisions of the United States Private Securities Litigation Reform Act of 1995 – that is, statements related to the future, not past, events. Forward-looking statements are based on the current expectations and include any statement that does not directly relate to a current or historical fact. In this context, forward-looking statements often address our expected future business and financial performance, including the expected benefits of acquiring and integrating other businesses, and often contain words such as "anticipate," "believe," "intend,", "will", "expect," "plan" or other similar words. These forward-looking statements involve certain risks and uncertainties, including those risks set forth in Item 1A – Risk Factors of the Company's most recent 10-K filing, and ultimately may not prove to be accurate. Actual results and future events could differ materially from those anticipated in such statements. For further discussion of risks and uncertainties, individuals should refer to RigNet's SEC filings. RigNet undertakes no obligation and does not intend to update these forward-looking statements to reflect events or circumstances occurring after this press release. You are cautioned not to place undue reliance on these forward-looking statements, which speak only as of the date of this press release. All forward-looking statements are qualified in their entirety by this cautionary statement.
Lee M. Ahlstrom
Senior Vice President and Chief Financial Officer
RigNet, Inc.
A photo accompanying this announcement is available at https://www.globenewswire.com/NewsRoom/AttachmentNg/b9ae3566-e1fe-42c4-8913-363b7f6d89c9
Source: RigNet, Inc.
1 yr.
RigNet is a registered trademark of RigNet, Inc.


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EASHL to return in NHL 16 – Beta starting soon
Will Star Wars take over the world again?
Destiny: A look at Xur's inventory since launch
Resident Evil: Revelations 2 Review
The Game Café: Lesser known games of 2017
James Cameron gives update on Avatar sequels
James Cameron, director of the 2009 blockbuster hit Avatar, recently gave an update on the movie's upcoming sequels. Speaking with Entertainment Weekly, Cameron touched upon a variety of topics ranging from the development process for the Avatar sequels to squashing concerns that the success of the original Avatar film was simply not a fluke. According to Cameron, "I'm in the process of doing another pass through all three scripts right now. Just refining. That's in parallel with the design process. The design process is very mature at this point. We've been designing for about a year and a half. All the characters, settings and creatures are all pretty much [set]."
He continues by noting "Obviously, expectations are going to be very high on these films, especially on Avatar 2, to make sure it wasn't just some big fluke the first time. So we've got to deliver. I've created a nice rod for my own back, so they say."
Avatar, released in 2009, broke international box office records through garnering more than $2.7 billion world wide. In doing so, Cameron eclipsed the success of his Titanic which previously held the record for box office sales on a global scale. Avatar also received critical praise eventually earning 9 Oscar nominations. The film won 3 of its 9 nominations taking home the Best Achievement in Cinematography, Best Achievement in Visual Effects, and Best Achievement in Art Directions Oscars.
Avatar 2 is set to hit theaters in 2017 with the final two entries of the Avatar series to arrive shortly thereafter.
What are your thoughts on the upcoming Avatar sequels? Will Cameron be able to match his success, both financially and critically, with the final three films in the Avatar series? Let your thoughts be known through commenting below!
avatarAvatar 2James Cameronsequel


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Junior Boks name new training squad
Curwin Bosch (Gallo)
Jaer ruled out as Kings make 4 changes
Marais: Bulls not gelling properly
Meyer: Super Rugby cull 'unfair'
Cape Town - Zain Davids and Stedman Gans, who missed the Junior Springboks' first two training camps because of Springbok Sevens commitments, have been included in a 35-man squad that will assemble in Stellenbosch on Tuesday for the team's third training camp, which concludes on Thursday, May 18.
Four players currently involved in Super Rugby - Curwin Bosch (Sharks), Cobus Wiese (Stormers), Madosh Tambwe (Lions) and Yaw Penxe (Southern Kings) are set to join the squad on Sunday, May 7, while Khuta Mchunu and Ernst van Rhyn will fall in with the squad in Stellenbosch as soon as they have sufficiently recovered from their various injuries.
The Junior Springboks will face the Varsity Cup Dream Team on Tuesday, April 25 in their fourth warm-up match, while another game against a local outfit is being planned for May 9.
"This camp will allow us to have a final look at the players with an eye on finalising our touring squad, and it will also serve as the last opportunity to get everything in place for the tournament," said coach Chean Roux.
"In line with this we opted to extend the camp for another week after our initial planning was to wrap it up on May 12.
"We have had very little time to work with some of the Vodacom Super Rugby players, while Zain, Stedman and Curwin Bosch, all three members of the Junior Bok squad last year, have not been with us at all up to now, so we will have an opportunity to see how they adapt to the structures.
"With a handful of players being in-and-out of the first two camps due to Vodacom Super Rugby, Blitzbok and Varsity Cup duty, it is important that we take the necessary time to name the best possible squad to travel to Georgia."
Roux said other vital elements of the camp would be to ensure that all the required structures are in place and that the team is ready for the international spectacle.
"We did most of the groundwork at the first two camps, which we will build on at this camp," said Roux.
"There were also a few things that we did not have sufficient time to work on at the last two camps, such as our set pieces, and this camp will afford us the necessary time to do so.
"Although I was pleased with the progress made by the squad since the first camp, we still have a long way to go to reach the standard of play which I would like us to display in Georgia. So we have hard work ahead in the next few weeks."
The SA Under-20s will conclude their preparations for the international spectacle at their holding camp in Stellenbosch from May 21-25. They will face France, Georgia and Argentina in the pool stages of the World Rugby Under-20 Championship in Tbilisi, Georgia, on May 31, June 4 and June 8, respectively.
Stedman Gans - Blue Bulls, Johan Grobbelaar - Blue Bulls, Wikus Groenewald - Western Province, Benhard Janse van Rensburg - Sharks, Daniel Jooste - Western Province, Rewan Kruger - Free State Cheetahs, Manie Libbok - Blue Bulls, Gianni Lombard - Golden Lions, Len Massyn - Golden Lions, Lee-Marvin Mazibuko - Western Province, Salmaan Moerat - Western Province, Reinhard Nothnagel - Golden Lions, Embrose Papier - Blue Bulls, Manuel Rass - Golden Lions, Carlu Sadie - Western Province, Wandisile Simelane - Golden Lions, Cornel Smit - Western Province, Hendre Stassen - Blue Bulls, Gerhard Steenekamp - Blue Bulls, Muller Uys - Western Province, Wayne van der Bank - Golden Lions, Ruben van Heerden - Blue Bulls, Boan Venter - Free State Cheetahs, Damian Willemse - Western Province, Jondre Williams - Western Province, Nama Xaba - Western Province
Read more on: junior springboks | junior world championship | cobus wiese | curwin bosch | cape town | rugby
Who should walk away the winner in the Team of the Year category at the 2020 Laureus World Sports Awards?
USA women's football team
Mercedes-AMG Petronas
Spain's men's basketball team
63% Springboks
24% Liverpool
3% USA women's football team
5% Mercedes-AMG Petronas
3% Toronto Raptors
1% Spain's men's basketball team


The limited edition guitars sometimes are very expensive. These very expensive guitars usually be consumed by guitar collectors.
Eddie Van Halen (EVH) collaborate with Fender to make the EVH brand guitars. This limited edition guitars are made only 300 instruments.
The result is this stunning "Black Beauty" with a slim-profile neck, Page BurstBucker pickups, gold Bigsby vibrato and an innovative six-position toggle that lets you isolate the neck and bridge pickups, or add the middle in any combination.
The Gibson JOHNNY WINTER FIREBIRD 6-String Electric Guitar. Gibson custom shop recreation of the slide guitarists 1963 Gibson Firebird V.


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Home » News » Husband's income Affidavit for Matrimonial Cases
Husband's income Affidavit for Matrimonial Cases
Reading Time - 1 minute
by Nikitha | Jan 7, 2020 | 0 comments
The Punjab and Haryana High Court, through its order dated 7 January directed the Family Courts across Punjab, Haryana, and Chandigarh to insist on an affidavit of assets, income and expenditure of the estranged husband before a wife seeks maintenance.
The bench headed by Justice Gurvinder Singh Gill issued nine commandments to be followed in all matrimonial cases, under the Hindu Marriage Act, Protection of Women from Domestic Violence Act, Criminal Procedure Code, Hindu Adoption and Maintenance Act, Special Marriage Act, the Indian Divorce Act, Guardian and Wards Act, and the Hindu Minority and Guardianship Act. The bench further added that the courts shall even appoint a local commissioner for visiting the place of residence or business of a party to assess the standard of living and social status. Further, the court granted the wife the right to approach revenue authorities for information on land held by the husband or seek information under the RTI.
Justice Gill clarified that not only a penal action will be taken against the party attempting to conceal vital information or mislead the court, but also it would be open to the court to consider drawing an "adverse inference" against such a party. The direction for filing an affidavit could be issued even in pending cases if the parties were not forthcoming about the information.
"In a large number of cases, it is noticed that the delay occurs as the hapless wife is left fending for herself, collecting documents and evidence as regards the financial status of her husband to seek appropriate maintenance," the bench stated.
He added that approaching authorities to procure income tax returns was a "Herculean task". Forcing a party to collect information in such manner resulted in delays against the spirit of the legislation, which provided for a 60-day timeframe for disposing of the pleas.
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Who would ever think you could visit your local hardware store to have a shirt embroidered? Think again, because Galveston's own Chalmers Hardware & Embroidery offers monogramming services in addition to supplying building materials to Island residents and businesses.
Need hats for the local baseball team? No problem. Chalmers offers embroidery and monogramming services for just about anything you can think of. Custom hats, shirts, and jackets, no job is too big or too small.
In fact, word is getting out about this unique store and their services. Recently, James Brockway of Brockway Realty, LLC, received compliments on his customized shirt while attending the Texas Realtors Conference in San Antonio and was happy to let other attendees in on the secret about Chalmers' professionalism and great service.
So, the next time you need a hammer and a monogrammed shirt, check out Chalmers Hardware & Embroidery in Galveston, on the corner of 20th and Broadway.


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Method | HTTP request | Description
------------- | ------------- | -------------
[**getAllMoviesRequest**](MoviesApi.md#getAllMoviesRequest) | **GET** /movies | Movie Scripts List
[**getMovieRequest**](MoviesApi.md#getMovieRequest) | **GET** /movies/{id} | All information regarding one movie

# **getAllMoviesRequest**
> [BasicMovie] getAllMoviesRequest()

Movie Scripts List

Returns a list of all available movie scripts.

### Example
```javascript
var InsightsServer = require('insights_server');

var apiInstance = new InsightsServer.MoviesApi();

var callback = function(error, data, response) {
  if (error) {
    console.error(error);
  } else {
    console.log('API called successfully. Returned data: ' + data);
  }
};
apiInstance.getAllMoviesRequest(callback);
```

### Parameters
This endpoint does not need any parameter.

### Return type

[**[BasicMovie]**](BasicMovie.md)

### Authorization

No authorization required

### HTTP request headers

 - **Content-Type**: application/json
 - **Accept**: application/json

# **getMovieRequest**
> DetailedMovie getMovieRequest(id)

All information regarding one movie

Returns all specific data available for one movie: meta information and all available characters.

### Example
```javascript
var InsightsServer = require('insights_server');

var apiInstance = new InsightsServer.MoviesApi();

var id = "id_example"; // String | The id of the movie.

var callback = function(error, data, response) {
  if (error) {
    console.error(error);
  } else {
    console.log('API called successfully. Returned data: ' + data);
  }
};
apiInstance.getMovieRequest(id, callback);
```

### Parameters

Name | Type | Description  | Notes
------------- | ------------- | ------------- | -------------
 **id** | **String**| The id of the movie. |

### Return type

[**DetailedMovie**](DetailedMovie.md)

### Authorization

No authorization required

### HTTP request headers

 - **Content-Type**: application/json
 - **Accept**: application/json

redpajama_set_name: RedPajamaGithub


text:
Q: How do I avoid repeated code when using useState hook in react? I have three places where I'm using the useState hook and I had to write it three times. I want to know if there is a way I could write once instead of having to repeat over again.
const Header = () => {
  const [btnOpen, setbtnOpen] = useState(false);
  const MobileNavigator = () => {
    setbtnOpen(!btnOpen);
  }
  const [dropDown, setdropDown] = useState(false);
  const dropdown = () => {
    setdropDown(!dropDown);
  }
  const [openSearchform, setSearchform] = useState(false);
  const formdown = () => {
    setSearchform(!openSearchform);
  }

A: you can use HOC(higher order component) for reusing component logic

*

*reference

*medium article with example

redpajama_set_name: RedPajamaStackExchange


So today, I thought I would finally indulge those of you who are in constant puzzlement over how to tell the difference between two of the more common "sand stars" found throughout the Indo-Pacific!
This is a surprisingly common question and its not just my colleagues who repeatedly ask me this -I frequently see these two species mistaken for one another in tropical aquarium and "shell collector" magazines, websites and so on..
Know Your Sand Stars! Convergence between Astropecten and Archaster?
So, first, we are talking about TWO different species from two different families in two entirely separate orders.
Astropecten in the Astropectinidae, Order Paxillosida (I will refer to the species Astropecten polyacanthus for simplicity's sake).
I should add the caveat that there's just over 100 species of Astropecten throughout the world and a fair number of them are present in the tropical Indo-Pacific.. So, there may be SOME variation if you are looking at different species of Astropecten wherever you may be reading this..
In spite of belonging to two completely different and distant groups of starfish, BOTH live in sandy bottoms with loose, unconsolidated sediment and possess the ability to dig and hide under that sand or sediment.
As a consequence these taxa display a number of superficial similarities that are adaptations for each respective species to live buried in a sandy, muddy or otherwise unconsolidated bottom environment.
In vertebrates, you would see this same kind of relationship-called convergent evolution between dolphins (which are mammals) and fish (which are not). Several consistent morphological features, which appear similar, but that arise in very separate evolutionary groups as adaptations to a particular way of life.
In the case between dolphins and fish-the fins and smooth lemon drop shape are adaptations to life swimming through the water.
In the case of these starfish-the location of the spines and the certain similarities in body shape are independent adaptations to digging through sediment.
So, How Do We Tell Them Apart?
So-if there's one immediate and obvious character to use-its the distinctive central row of plates on each arm.
Another immediately obvious difference between the two..
Archaster has actual suckered tube feet.... As seen here on this video..
This is as opposed to Astropecten which have pointed or knobby ends on their tube feet..
3. Body Shape and Spines on the Outline of the Body.
Archaster only has big flat spines on the lower side of its body that flank the oral surface.. If there are spines on the upper surface, they are small and tiny..
Arm is more triangular. And many species of Astropecten will have a very prominent perimeter of spines around the body's edge.
4. The Spines are Different.
Note that these spines are round and sharp rather than flat and blunt.
In Archaster, the spines are flattened and blunt!!
To the seasoned expert, there are actually MORE differences than this. But these were the characteristics that I thought were the easiest to tell apart for those who are simply observing the exterior.
Thank you Mah for simple ways to tell them apart.
Hi there! I'm currently *attempting* to write a paper on a specimen identical to the Archaster you have pictured multiple times here. Could you tell me the species name? I can't seem to find it.
The widespread species of Archaster is A. typicus. There's another species-Archaster angulatus which is endemic to Australia. Most likely you've got the former. Good luck!


My wife can't stop raving about how comfortable these boots are. When the boots finally arrived .she couldnt believe how warm and comfortable her feet were.
Love the look and feel of this boot. Comfortable from the beginning. They were out of my size 7.5 so I got an 8. With wool socks they were good. Went to play in the snow, no cold or wet toes! Wore them everyday.


Gamers Forum > Gamers-Forum Central > New Gamers > Wow, it really has been a long time.
View Full Version : Wow, it really has been a long time.
Well, some of you may have noticed that I haven't been active recently. (if you haven't, please rethink your life) Well for whatever reason the new servers don't connect with our computers.
They both run on the same wireless, and the only time I can manage to get on is like right now when our laptop loses connection to the right one and I have to try and find it again.
I have gone through surgery, had to have a bone removed on my wirst actually happened to be on Friday. I got to take off all the bandages yesterday, :-) and it still hurts to type though.
Hopefully this will not be good-bye forever.
Welcome Back, I never noticed you werent here because I've had trouble getting on recently.
Well if we stay at these servers, I should be ok!
Edit: And yes, I am not running on another network right now.
It was as if the server had said "Don't go" I had been seriously considering just resiging and hoping to come back later.