text:
\section{Introduction}
Let $\K$ be an arbitrary field,
$V=\K^\ell$ and $S=\K[x_1,\ldots,x_\ell]$ its coordinate ring. Let $\Der S:=
\oplus_{i=1}^\ell S \partial_{x_i}$. A \textbf{hyperplane arrangement} $\A$ is a
finite set of linear hyperplanes in $V$. For each $H \in \A$, let us fix a linear form
$\alpha_H \in V^*$ such that $\ker \alpha_H=H$. The \textbf{logarithmic derivation module} $D(\A)$ of $\A$ is defined by
$$
D(\A):=\{\theta \in \Der S \mid
\theta(\alpha_H) \in S\alpha_H\ (\forall H \in \A)\}.
$$
$D(\A)$ is a reflexive $S$-module, and not free in general. We say that $\A$ is \textbf{free}
with \textbf{exponents} $\exp(\A)=(d_1,\ldots,d_\ell)$ if there are $S$-free homogeneous basis $\theta_1,
\ldots,\theta_\ell \in D(\A)$ with $\deg \theta_i=d_i\ (i=1,\ldots,\ell)$. For
integers $d_1,\ldots,d_\ell$, let $(d_1,\ldots,d_\ell)_\le$ denote
that
$d_1 \le \cdots \le d_\ell$.

In the study of hyperplane arrangements,
the freeness is one of the most important objects. But in general it is very difficult to
determine whether a given arrangment is free or not. The most useful method to
check the freeness is the following addition-deletion theorem.

\begin{theorem}[\cite{T}]
Let $H \in \A,\ \A':=\A \setminus \{H\}$ and
$\A'':=\A^H:=\{H \cap L \mid L \in \A'\}$. Then two of the following three imply the third:
\begin{itemize}
\item[(1)]
$\A$ is free with $\exp(\A)=(d_1,\ldots,d_{\ell-1},d_\ell)$.
\item[(2)]
$\A'$ is free with $\exp(\A')=(d_1,\ldots,d_{\ell-1},d_\ell-1)$.
\item[(3)]
$\A''$ is free with $\exp(\A'')=(d_1,\ldots,d_{\ell-1})$.
\end{itemize}
In particular, all the three hold true if $\A$ and $\A'$ are both free.
\label{additiondeletion}
\end{theorem}

There are a lot of variants of Theorem \ref{additiondeletion}. For example, the division theorem in \cite{A} is one of them. Among them, recently,
the multiple addition theorem (MAT) is introduced in \cite{ABCHT} to prove the freeness of
ideal subarrangements. Let us recall MAT.

\begin{theorem}[Multiple addition theorem (MAT), \cite{ABCHT}, Theorem 3.1]
Assume that $\A'$ is a free arrangement with
$\exp(\A')=(d_1,d_2.\ldots,d_\ell)_\le$ such that
$d_i
=:d$ for $q \le i\le \ell$. For $H_\ell,\ldots,H_q \in \A$ with $2 \le q$,
define $\A_j'':=(\A' \cup \{H_j\})^{H_j},\ \A:=\A' \cup \{H_q,\ldots,H_\ell\}$
and assume the following conditions:
\begin{itemize}
\item[(1)]
$|\A'| -|\A_j''|=d\ (j=q,\ldots,\ell)$,
\item[(2)]
$X:=\bigcap_{i=q}^\ell H_i$ is $(\ell-q+1)$-codimensional, and
\item[(3)]
$X \not \subset \bigcup_{K \in \A'}K$.
\end{itemize}
Then $\A$ is free with exponents
$(d_1,d_2,\ldots,d_{q-1},d+1,\ldots,d+1)$.
\label{MAT}
\end{theorem}

This multiple addition theorem
enables us, under certain conditions, to add hyperplanes at once to free arrangements keeping the freeness with controlled behavior of exponents. Then it is very natural to ask, what about the multiple deletion theorem? The first aim of this article is to answer this
very natural question as follows:

\begin{theorem}[Multiple deletion theorem, MDT]
Assume that $\A$ is a free arrangement with
$\exp(\A)=(1,d_2.\ldots,d_\ell)_\le$ such that
$1 < d_2$. For $H_2,\ldots,H_q \in \A$ with $2 \le q \le \ell$, let
$\alpha_j:=\alpha_{H_j}\ (j=2,\ldots,q)$,
define $\A_j'':=\A^{H_j},\ \A':=\A \setminus \{H_2,\ldots,H_q\}$
and assume the following conditions:
\begin{itemize}
\item[(1)]
$|\A| -|\A_j''|=d_j\ (j=2,\ldots,q)$,
\item[(2)]
$X:=\bigcap_{i=2}^q H_i$ is $(q-1)$-codimensional, and
\item[(3)]
$X \not \subset \bigcup_{K \in \A'}K$.
\end{itemize}
Then $\A'$ is free with exponents
$(1,d_2-1,d_3-1,\ldots,d_q-1,d_{q+1},\ldots,d_\ell)$. More explicitly,
there is a basis for $D(\A)$ of the form $\theta_E,\alpha_2 \eta_2,\ldots,
\alpha_q \eta_q,\theta_{q+1},\ldots,\theta_\ell$ for $D(\A)$ with
$\deg \eta_i=d_i-1,\ \deg \theta_j=d_j$ for all $i,j$.
\label{MDT2}
\end{theorem}

From the same viewpoint of Theorem \ref{MDT2}, we can show a generalized version of
the multiple addition theorem in \cite{ABCHT} as follows:

\begin{theorem}[Multiple addition theorem 2, MAT2]
Assume that $\A'$ is a free arrangement with
$\exp(\A')=(1,d_2.\ldots,d_\ell)_\le$. For $H_\ell,\ldots,H_q \in \A$ with $2 \le q$,
define $\A_j'':=(\A' \cup \{H_j\})^{H_j},\ \A:=\A' \cup \{H_q,\ldots,H_\ell\}$
and assume the following conditions:
\begin{itemize}
\item[(1)]
$|\A'| -|\A_j''|=d_j\ (j=q,\ldots,\ell)$,
\item[(2)]
$X:=\bigcap_{i=q}^\ell H_i$ is $(\ell-q+1)$-codimensional, and
\item[(3)]
$X \not \subset \bigcup_{K \in \A'}K$.
\end{itemize}
Then $\A$ is free with exponents
$(1,d_2,\ldots,d_{q-1},d_{q}+1,\ldots,d_\ell+1)$. Moreover, there is a basis
$\theta_E,\theta_2,\ldots,\theta_{q-1},\eta_q,\ldots,\eta_\ell$ for $D(\A')$ such that
$\deg \theta_i=d_i,\ \deg \eta_j=d_j$,
$\theta_i \in D(\A)$ and $\eta_j \in D(\A \setminus \{H_j\})$ for all $i,\ j$.
\label{MAT2}
\end{theorem}

Also, we can show the restriction version as follows:

\begin{theorem}[Multiple restriction theorem, MRT]
Let $\A$ be free with $\exp(\A)=(1,d_2,\ldots,d_\ell)_\le$ and let
$X \in L_k(\A)$. Then $\A^X:=\{X \cap H \mid H \in \A,\ X \not \subset H\}$ is free with $\exp(\A^X)=(1,d_{k+2},\ldots,d_\ell)$
if and only if $|\A^X|=1+d_{k+2}+\cdots+d_{\ell}$.
\label{cor1}
\end{theorem}

We give some applications of MDT and MAT2. In particular, by observing MAT, we can show that
the freeness of the extended Catalan arrangements depends only on its combinatorics.

The organization of this article is as follows. In \S2 we introduce several definitions and results used in this article. In \S3 we prove MDT and MAT2. In \S4 we prove MRT by
showing a more general result. In \S5 we consider the relation between combinatorics of freeness and MAT.
\medskip

\noindent
\textbf{Acknowledgements}.
The first author is
partially supported by KAKENHI, JSPS Grant-in-Aid for Scientific Research (B) 16H03924, and
Grant-in-Aid for Exploratory Research 16K13744.

\section{Preliminaries}
In this section let us fix a notation and introduce several results used for the rest of this
article. Let $\A$ be an arrangement of hyperplanes in $V=\K^\ell$ as in \S1.
The \textbf{intersection lattice} $L(\A)$ of $\A$ is defined by
$$
L(\A):=\{\bigcap_{H \in \B} H \mid
\B \subset \A\}.
$$
Let $L_k(\A):=\{ X \in L(\A) \mid \codim_V(X)=k\}$.
On $L(\A)$ the \textbf{M\"obius function} $\mu$ is defined by $\mu(V)=1$, and by
$$
\mu(X)=-\sum_{Y \in L(\A),\ X \subsetneq Y \subset V} \mu(Y).
$$
The \textbf{Poincar\'e polynomial} $\pi(\A;t)$ is
$$
\pi(\A;t):=\sum_{X \in L(\A)} \mu(X)(-t)^{\codim X},$$
and the \textbf{characteristic polynomial} $\chi(\A;t)$ is
$$
\chi(\A;t):=\sum_{X \in L(\A)} \mu(X)t^{\dim X}.
$$
For $X \in L(\A)$, the \textbf{localizaton} $\A_X:=\{H \in \A
\mid X \subset H\}$ is a subarrangement of $\A$, and
the \textbf{restriction} $\A^X:=\{H \cap X \mid H \in \A \setminus \A_X\}$
is an arrangement in $X$. More generally, for a point $x \in V$, let $\A_x:=\{
H \in \A \mid x \in H\}$, and $D(\A)_x$ is the localization of $D(\A)$ at the
homogeneous prime ideal corresponding to $x$.

Next recall several results related to $D(\A)$ introduced in \S1.
For a non-negative integer $d$, let $D(\A)_d$ denote the homogeneous degree $d$-part of
$D(\A)$. Here $\theta \in \Der S$ is homogeneous of degree $d$ if $\theta(\alpha)$ is
homogeneous of degree $d$ for all $\alpha \in V^*$ with $\theta(\alpha) \neq 0$.
When $\A \neq \emptyset$, $\theta_E \in D(\A)$ and the free $S$-module $S\theta_E$
is a direct summand of $D(\A)$. Explicitly, for any $H \in \A$,
if $D_H(\A):=\{\theta \in D(\A) \mid \theta(\alpha_H)=0\}$, then
$$
D(\A) \simeq S\theta_E \oplus D_H(\A).
$$
Hence $\A$ is free if and only if $D_H(\A)$ is free for some, and hence any $H \in \A$.
Hence $1 \in \exp(\A)$ when $\A \neq \emptyset$ is
free. For $X \in L(\A)$, it is easy to see that $\A_X$ is free if $\A$ is free. Contrary to
the localization, there are no relations of freeness between $\A$ and $\A^H$ for $H \in \A$.
The most useful relation is Theorem \ref{additiondeletion}. In fact, always we have the
exact sequence
\begin{equation}
0 \rightarrow D(\A') \stackrel{\cdot \alpha_H}{\rightarrow} D(\A)
\stackrel{\rho}{\rightarrow} D(\A''),
\label{e}
\end{equation}
where $\rho$ is the map taking modulo $\alpha_H$, and called the Euler restriction map. And the exact sequence (\ref{e}) is
right exact if all the three conditions in Theorem \ref{additiondeletion} hold.
When $\A$ is free, we can relate the exponents with its Poincar\'e polynomial as follows.

\begin{theorem}[\cite{T1}]
If $\A$ is free with $\exp(\A)=(d_1,\ldots,d_\ell)$, then $\pi(\A;t)=\prod_{i=1}^\ell
(1+d_i t)$.
\label{factorization}
\end{theorem}

For a \textbf{multiplicity function} $m:\A \rightarrow \Z_{>0}$, we can define
the \textbf{logarithmic derivation module} $D(\A,m)$ of the multiarrangement $(\A,m)$ as
$$
D(\A,m):=\{\theta \in \Der S \mid \theta(\alpha_H) \in S\alpha_H^{m(H)}\ (\forall H \in \A)\}.
$$
$D(\A,m)$ is a reflexive $S$-module, and not free in general. Hence we can define its
\textbf{freeness} and
\textbf{exponents} in the same way as for $\A$. Let $|m|:=\sum_{H \in \A} m(H)$ and
$Q(\A,m):=\prod_{H \in \A} \alpha_H^{m(H)}$.

\begin{define}
Let $H \in \A$. Then the \textbf{Ziegler restriction} $(\A^H,m^H)$ of $\A$ onto $H$ is
defined by
$$
m^H(X):=|\{L \in \A \setminus \{H\} \mid L \cap H=X\}|
$$
for $X \in L(\A^H)$.
Then we have the \textbf{Ziegler restriction map}
$$
\pi_H:D_H(\A) \rightarrow D(\A^H,m^H)$$
by taking modulo $\alpha_H$. Hence $\ker \pi_H=D_H(\A)[-1]=\alpha_HD_H(\A)$.
\label{Zrest}
\end{define}

\begin{theorem}[\cite{Z}]
Assume that $\A$ is free with $\exp(\A)=(1,d_2,\ldots,d_\ell)$. Then $(\A^H,m^H)$ is free
with $\exp(\A^H,m^H)=(d_2,\ldots,d_\ell)$. In particular, the Ziegler restriction map
is surjective in this case.
\label{Zieglerfree}
\end{theorem}

For the freeness, the following criterion is important.

\begin{theorem}[Saito's criterion, \cite{S}]
Let $\theta_1,\ldots,\theta_\ell \in D(\A,m)$ be homogeneous
derivations. Then
$\det (\theta_i(x_j)) \in S Q(\A,m)$. Moreover, they
form a
basis for $D(\A,m)$ if and only if
they are $S$-independent, and $|m|=\sum_{i=1}^\ell \deg \theta_i$.
\label{Saito}
\end{theorem}

The following is very important to analyze freeness.

\begin{theorem}[\cite{T}]
Let $H \in \A$ and $\A':=\A \setminus \{H\}$. Then there is a polynomial $B$ with
$\deg B=|\A'|-|\A^H|$ such that
$$
\theta(\alpha_H) \in (\alpha_H,B)
$$
for all $\theta \in D(\A')$.
\label{polyB}
\end{theorem}

Based on Theorem \ref{polyB}, we often use the following two arguments in this article.

\begin{prop}
Let $H \in \A,\ \A':=\A \setminus \{H\}$ and
$d:=|\A'|-|\A^H|$. Then

(1)\,\,
$\theta \in D(\A')$ belongs to $D(\A)$ if $\deg \theta <d$.

(2)\,\,
Assume that there is $\theta \in D(\A')_d$ such that
$\theta(\alpha_H) =B$ modulo $\alpha_H$.
If there is a generator $\theta_E,\varphi_1,\ldots,\varphi_s, \theta$ for
$D(\A')$ such that $\varphi_i(\alpha_H)\equiv g_i B$ for $i=1,\ldots,s$
modulo $\alpha_H$, then
$\theta_E,\varphi_1-g_i \theta,\ldots,\varphi_s-g_s \theta, \alpha_H \theta$ is a generator for
$D(\A)$. In particular,

(3)\,\,
if $\A'$ is free with $\exp(\A')=(d_1,\ldots,d_\ell),\ d_\ell=d$,
$\theta_1,\ldots,\theta_\ell$ form a basis
for $D(\A')$ with $\deg \theta_i=d_i$, $\theta_\ell(\alpha_H) =B$ modulo $\alpha_H$,
and $\theta_i(\alpha_H) \equiv g_iB$ modulo $\alpha_H$, then
$\theta_1-g_1 \theta,\ldots,\theta_{\ell-1}-g_{\ell-1} \theta,\alpha_H \theta_\ell$ form a basis for $D(\A)$.
\label{Bappli}
\end{prop}

\noindent
\textbf{Proof}.
(1)\,\, Trivial by Theorem \ref{polyB}.

(2)\,\,
For $\varphi_i$ with $\deg \varphi_i < d$, (1) implies that $\varphi_i \in D(\A)$. For
those with $\deg \varphi_i \ge d$, let
$$
\varphi_i(\alpha_H)=f_i \alpha_H + g_i B
$$
by Theorem \ref{polyB}. Then replacing $\varphi_i$ by $\varphi_i-g_i \theta$,
it holds that all $\varphi_i \in D(\A)$. Let us show that $\theta_E,\varphi_1,\ldots,
\varphi_s,\alpha_H\theta$ generate $D(\A)$. Note that
$\theta(\alpha_H) \not \in S \alpha_H$. Let $\varphi \in D(\A) \subset D(\A')$. Then
$\varphi=f \theta_E+\sum_{i=1}^s f_i \varphi_i+g\theta$.
Since all derivations but $\theta$ belongs to $D(\A)$, $g \theta \in D(\A)$ which implies that
$\alpha_H \mid g$. Thus (2) follows.
(3) is immediate from (2).\owari
\medskip

\begin{prop}
Let $H \in \A$ and assume that $\A$ is free with $\exp(\A)=(1,d_2,\ldots,d_\ell)$. Let
$\theta_i\ (i=2,\ldots,\ell)$ be a basis for $D(\A)$ with $\deg \theta_i=d_i$, and let
$Q':=Q(\A^H,m^H)/Q(\A^H)$.
Assume that $|\A|-|\A^H|=d_\ell$, and
$$
Q'\pi_H(\theta_E)=\sum_{i=2}^\ell
\pi_H(f_i) \pi_H (\theta_i),
$$
where $f_i \in S$. If there is some $i$ such that $\pi_H(f_i) \neq 0$ and
$d_i=d_\ell$, then
$$
\pi_H(\theta_2),\ldots,\pi_H(\theta_{\ell-1}),Q' \pi_H(\theta_E)
$$
form a basis for $D(\A^H,m^H)$, and
$$
\pi(\theta_E),\pi_H(\theta_2),\ldots,\pi_H(\theta_{\ell-1})
$$
form a basis for $D(\A^H)$.
\label{replace}
\end{prop}

\noindent
\textbf{Proof}.
By Theorem \ref{Zieglerfree} and the decomposition $D(\A) \simeq S\theta_E
\oplus D_H(\A)$, we may assume that
$\pi_H(\theta_2),\ldots,\pi_H(\theta_{\ell-1}),\pi_H(\theta_\ell)$ form a basis for $D(\A^H,m^H)$. We may assume that $f_\ell \neq 0$. By the reason of degrees,
$f_\ell \in \K^\times$. Hence we may replace $\pi_H(\theta_\ell)$ by $Q'\pi_H(\theta_E)$ to
obtain that
$\pi_H(\theta_2),\ldots,\pi_H(\theta_{\ell-1}),Q' \pi_H(\theta_E)$ is a basis for $D(\A^H,m^H)$. In particular,
$\pi_H(\theta_E), \pi_H(\theta_2),\ldots,\pi_H(\theta_{\ell-1})$ are independent over $S/\alpha_H$, and the sum of their degrees are $|\A^H|$. Hence Saito's criterion completes the proof.\owari
\medskip

\section{Proof of Theorems \ref{MDT2} and \ref{MAT2}}

Let us prove main results in this
article.
\medskip

\noindent
\textbf{Proof of Theorem \ref{MDT2}}.
Let $\alpha_i$ be the defining linear form of $H_i$. We show by induction on $q$. First let $q=2$.  Let $\theta_E,\theta_2,\ldots,\theta_\ell$ be a basis for $D(\A)$ with $\deg \theta_i=d_i$ and
$\theta_i \in D_{H_2}(\A)$. Let $\pi:D_{H_2}(\A) \rightarrow D(\A''_2,m^{H_2})$ be the Ziegler restriction map. Then Theorem \ref{Zieglerfree} implies that
$\pi(\theta_2),\ldots,\pi(\theta_\ell)$ form a basis for $D(\A''_2,m^{H_2})$. Let $Q:=Q(\A''_2,m^{H_2})/Q(\A''_2)$. Then $Q\pi(\theta_E) \in D(\A''_2,m^{H_2})_{d_2}$ by the condition (1). Hence
there are $f_i \in S$ such that
$$
Q\pi(\theta_E)=\sum_{i=2}^k \pi(f_i) \pi(\theta_i),
$$
where $d_2=\cdots=d_k<d_{k+1}$.
By comparing the degrees, we may assume that
$\pi(f_2) \in \K^\times$.
Then by Proposition \ref{replace}, we may replace $\pi(\theta_2)$ by $Q\pi(\theta_E)$ to obtain a basis
$Q\pi(\theta_E),\pi(\theta_3),\ldots,\pi(\theta_\ell)$ for $D(\A''_2,m^{H_2})$. Hence again Proposition \ref{replace} implies that
the derivations $\pi(\theta_E),\pi(\theta_3),\ldots,\pi(\theta_\ell)$ form a basis for $D(\A_2'')$.
Let $\rho:D(\A) \rightarrow D(\A_2'')$ be the Euler restriction map. Then $\rho(\theta_i)=
\pi(\theta_i)$ for $i \ge 3$ by definition. Since $\rho$ is surjective too,
there are polynomials $f,\ f_i\ (i=3,\ldots,\ell)$ such that $$
\theta_2=f \theta_E+\sum_{i=3}^\ell f_i\theta_i+\alpha_H \theta'
$$
by (\ref{e}). Here $\theta' \in D(\A \setminus \{H_2\})$. Replacing
$\theta_2$ by $\theta_2-f\theta_E-\sum_{i=3}^\ell f_i \theta_i$, we have the desired basis
for $D(\A)$.

Now assume that the statement holds true for
$q-1 \ge 1$. When the case of $q$, by induction, $D(\A)$ has a basis of the form
$\theta_E,\alpha_2 \eta_2,\ldots,\alpha_{q-1} \eta_{q-1}, \theta_q,\ldots,\theta_\ell$ with
$\deg \eta_j=d_j-1,\ \deg \theta_j=d_j$.
Replacing $\alpha_j \eta_j$ by $\alpha_j \eta_j - (\alpha_j \eta_j(\alpha_q)/\alpha_q)\theta_E$, we may assume that $\alpha_j \eta_j \in D_{H_q}(\A)$ for $j=2,\ldots,q-1$. Also, we may assume that $\theta_q,\ldots,\theta_\ell \in D_{H_q}(\A)$.
Let $\pi:D_{H_q}(\A) \rightarrow D(\A^{H_q},m^{H_q})$ be
the Ziegler restriction map.
By Theorem \ref{Zieglerfree}, $\pi(\alpha_2 \eta_2),\ldots,\pi(\alpha_{q-1}
\eta_{q-1}),\pi(\theta_q),\ldots,\pi(\theta_\ell)$ form a basis for $D(\A^{H_q},m^{H_q})$.
Let $Q:=Q(\A^{H_q},m^{H_q})/Q(\A^{H_q})$. Then $\theta_E^q:=
Q\pi(\theta_E) \in D(\A^{H_q},m^{H_q})_{d_q}$ by the condition
(1). Hence $\theta_E^q$ can be expressed as a $S/\alpha_q$-linear combination of the basis whose degree is at most $d_q$. If there is $\theta_j$ such that $\deg \theta_j=d_q$, and it appears in the linear combination with a non-zero scalar coefficient, then the same argument
as when
$q=2$, combined with
Proposition \ref{replace}, implies that $\pi(\alpha_2 \eta_2),\ldots,\pi(\alpha_{q-1}
\eta_{q-1}),\theta_E^q,\pi(\theta_{q+1}),
\ldots,\pi(\theta_\ell)$
form a basis for $D(\A''_q,m^{H_q})$. Hence
by the same argument as when $q=2$, we may replace $\theta_q$ by the derivation
of the form $\alpha_{H_q} \theta'$ with $\theta' \in D(\A \setminus \{H_q\})$, which
completes the proof.
Assume not. Hence
\begin{equation}\label{eq100}
\theta_E^q=\sum_{i=2}^{q-1} \pi(f_i) \pi(\alpha_i \eta_i)\ (f_i \in S).
\end{equation}
Let $H_q \cap H_j=:X_j \in \A^{H_q}$ and let $\beta_j \in S/\alpha_q$ be the defining linear form
of $X_j$. Note that $\cap_{j=2}^{q-1} X_j =X$.
We show that there is a point $x \in X$ such that $\alpha_Y(x) \neq 0$
for all $Y \in \A^{H_q} \setminus \{X_2,\ldots,X_{q-1}\}$. To show it, it suffices to check that
$X \not \subset Y\in \A^{H_q} \setminus \{X_2,\ldots,X_{q-1}\}$ by extending the base
field $\K$ if necessary.
Assume that there is such $Y=H \cap H_q,\ H \in \A$.
Since $X \subset Y \subset H$, the condition (3) implies that
$H \in \{H_2,\ldots,H_{q-1}\}$, a contradiction.

Hence there is $x \in X \setminus \bigcup_{Y \in \A''_q,\ Y \neq X_j} Y$ such that
the right hand side of (\ref{eq100}) is zero at $x$. We show that
the left hand side cannot be zero at $x$.
Since $\pi(\theta_E)$ is nowhere vanishing, it is equivalent to say that
$Q(x)=0$.
Note that $\alpha_Y(x) \neq 0$ for $Y \in
\A''_q \setminus \{X_2,\ldots,X_{q-1}\}$. Thus $Q(x)=0$ could occur only when
$m(X_j) \ge 2$ for some $j$, which cannot occur by the
condition (2).
\medskip

\noindent
\textbf{Proof of Theorem \ref{MAT2}}.
Let $\alpha_i$ be the defining linear form of $H_i$. We show by induction on $q$. When
$q=\ell$,
this is Theorem \ref{MAT}.
Assume that the statement holds true for
$q+1\le \ell$. By induction, $D(\A')$ has a basis of the form
$\theta_E,\theta_2,\ldots,\theta_q,\eta_{q+1},\ldots,\eta_\ell$ with
$\deg \theta_i=d_i,\ \deg \eta_j=d_j$ such that
$\eta_j(\alpha_j)=b_j$ modulo $\alpha_j$ for $j=q+1,\ldots,\ell$, where
$b_j$ is the polynomial in Theorem \ref{polyB} with $\deg b_j=|\A'|-|\A''_j|$.
Also, if $d_i < d_q$, then Proposition \ref{Bappli} implies that
$\theta_i \in D(\A)$.
Moreover, by the induction hypothesis, $\eta_j(\alpha_i)\equiv 0$ modulo $\alpha_i$ for all $ i \neq j$.
Hence it suffices to show that $\theta_i(\alpha_q) \equiv b_q$ modulo $\alpha_q$ for some $i$ with
$\deg \theta_i=d_q$ by Proposition \ref{Bappli}. Assume not.
Then we may assume that $\theta_i$ is tangent to $X$
for all $i$. Let $x \in X$ with $\A_x=\{H_q,\ldots,H_\ell\}$ by the condition (3).
Then $$
k(x) \otimes D(\A')_x \simeq \K^\ell =T_{X,x} \oplus N_{X,x},
$$
i.e., the tangent space decomposes into the tangent space and normal
space of $X$ at $x$. Here $k(x)$ is the residue field of $S_x$. By the condition (2), the former is of $(q-1)$-dimensional
and the latter $(\ell-q+1)$-dimensional. Note that $\theta_E$ and
all $\theta_i \in T_{X,x}$ by the
assumption above. Thus
\begin{eqnarray*}
\ell-q+1&=&\dim N_{X,x}\\
&=&\dim \langle \eta_{q+1},\ldots,\eta_\ell\rangle_\K\\
&\le&\ell-q,
\end{eqnarray*}
a contradiction.\owari
\medskip

In fact, the assertions in Theorems \ref{MDT2} and \ref{MAT2} are weaker than
proved here. Explicitly, the following holds.

\begin{cor}
Let $\A$ be a free arrangement satisfying the conditions (1), (2) and (3) in Theorem \ref{MDT2}.
Let $I \subset \{2,\ldots,q\}$ and set
$\A_{-I}:=\A \setminus \{H_i\}_{i \in I}$. Then $\A_{-I}$ is free with basis
$$
\theta_E,\{\eta_i\}_{i \in I},\ \{\alpha_i \eta_i\}_{i \not \in I},\
\theta_{q+1},
\ldots,
\theta_\ell
$$
in the notation of Theorem \ref{MDT2}.
\label{MDTexplicit}
\end{cor}

\noindent
\textbf{Proof}. Immediately from the proofs of Theorem \ref{MDT2} and
Theorem \ref{Saito}. \owari
\medskip

\begin{cor}
Let $\A'$ be a free arrangement satisfying the conditions (1), (2) and (3) in Theorem \ref{MAT2}.
Let $I \subset \{q,\ldots,\ell\}$ and set
$\A_{+I}:=\A \cup \{H_i\}_{i \in I}$. Then $\A_{+I}$ is free with basis
$$
\theta_E,\{\eta_i\}_{i \not \in I},\ \{\alpha_i \eta_i\}_{i  \in I},\
\theta_{2},
\ldots,
\theta_{q-1}
$$
in the notation of Theorem \ref{MAT2}.
\label{MATexplicit}
\end{cor}

\noindent
\textbf{Proof}. Immediate by the proofs of Theorem \ref{MAT2} and
Theorem \ref{Saito}. \owari
\medskip

\section{Multiple restriction theorems (MRT)}

After proving Theorems \ref{MDT2} and \ref{MAT2}, it is also natural to ask whether
we have a multiple restriction theorem or not. In the setup of MDT and MAT2, we can say yes
to this question.

For an $\ell$-arrangement $\A$ and a subspace $X \subset V$, define the arrangement
$\A \cap X$ in $X$ by
$$
\A \cap X:=\{L \cap X \mid L \in \A,\ L \not \supset X\}.
$$
If $ X \in L(\A)$, then $\A \cap X=\A^X$. By using these terminology, first, let us
formulate the multiple restriction theorem in terms of MAT2 and MDT.

\begin{theorem}[MRT from MDT]
Let $\A$ be a free arrangement satisfying the conditions (1), (2) and (3) in
Theorem \ref{MDT2}. For $I \subset \{2,\ldots,q\}$, let
$X_I:=\cap_{i \in I} H_i$.
Then $\A^{X_I}$ is free with basis
$$
\overline{\theta}_E,
\{\overline{\alpha_i \eta}_i\}_{ i \not \in I},\
\overline{\theta}_{q+1},
\ldots,\overline{\theta}_\ell.
$$
Here for $\theta \in D(\A)$, $\overline{\theta}$ stands for the restriction of $\theta$ onto $X$.
\label{MRT1}
\end{theorem}

\noindent
\textbf{Proof}. Apply Corollary \ref{MDTexplicit}. \owari
\medskip

\begin{theorem}[MRT from MAT2]
Let $\A'$ be a free arrangement satisfying the conditions (1), (2) and (3) in
Theorem \ref{MAT2}. For $I \subset \{q,\ldots,\ell\}$, let
$X_I:=\cap_{i \in I} H_i$.
Then $\A' \cap X_I$ is free with basis
$$
\overline{\theta}_E,
\{\overline{\eta}_i\}_{ i \not \in I},\
\overline{\theta}_{2},
\ldots,\overline{\theta}_q.
$$
Here for $\theta \in D(\A')$, $\overline{\theta}$ stands for the restriction of $\theta$ onto $X$.
\label{MRT2}
\end{theorem}

\noindent
\textbf{Proof}. Apply Corollary \ref{MATexplicit}. \owari
\medskip

In fact, when we want to check the freeness of $\A^X$ in terms of Theorem \ref{MRT1},
sometimes the conditions
(1), (2) and (3) are too many to check. We can show the following MRT.

\begin{theorem}
Let $\A$ be an $\ell$-arrangement with
a generator
$\theta_E,\theta_1,\ldots,\theta_s,\varphi_1,\ldots,\varphi_{\ell-k-1}$ such that
$\deg \theta_i=d_i\le d_{i+1} \le e_j:=\deg \varphi_j$ for all $i,\ j$. Let $X \in L_k(\A)$ such that
$|\A^X|=1+\sum_{i=1}^{\ell-k-1} e_i$. Then $\A^X$ is free with
$\exp (\A^X)= (1,e_1,\ldots,e_{\ell-k-1})$.
\label{rest2}
\end{theorem}

To prove Theorem \ref{rest2}, the fundamental lemma is the following.

\begin{lemma}
Let $\A$ be an arrangement and $X \in L_k(\A)$. Then the Euler
restriction map $\rho:D(\A) \rightarrow D(\A^X)$ is generically surjective.
In particular, if $\theta_E,\theta_1,\ldots,\theta_s$ is a generator for $D(\A)$, then
there are $(\ell-k-1)$-elements among $\theta_1,\ldots,\theta_{\ell-k-1}$ which are
$S^X$-independent, where $S^X:=\mbox{Sym}^*(X^*)$.
\label{indep}
\end{lemma}

\noindent
\textbf{Proof}.
Let $x \in X$ be a generic point. Then it is clear that
\begin{eqnarray*}
D(\A)_x&=&\langle \theta_1,\ldots,\theta_k,
\partial_{x_{k+1}},\ldots,\partial_{x_\ell}\rangle,\\
D(\A^X)_x&=&\langle \partial_{x_{k+1}},\ldots, \partial_{x_\ell}\rangle,
\end{eqnarray*}
where we change the coordinates $\{x_i\}_{i=1}^\ell$ in such a way that $X=\{x_1=\cdots=x_{k}=0\}$ and $\theta_i
\in D(\A_X) \cap (\oplus_{i=1}^k S \partial_{x_i})$. Hence $\theta_i|_X=0$ for all $i$.
At this point
$\rho$ is clearly surjective. Since $\rho$ is an $S$-module homomorphism,
the image of $\rho$ has to contain at least $(\ell-k)$-elements which contains
the Euler derivation, and they are $S^X$-independent.

Also, note that, we may define the map $\rho_{H,Y}^X:D_H(\A) \rightarrow D_Y(\A^X)$ for
$H \in \A$ with $H \supset X$ and $Y \in \A^X$ as follows. First, for $\theta \in D_H(\A)$ and
$L \in \A^H$,
define the map $\rho_{H,L}^H:D_H(\A) \rightarrow D_L(\A^H)$ as
$$
\rho_{H,L}^H(\theta):=\rho(\theta)-\frac{\rho(\theta)(\alpha_L)}{\alpha_L} \rho(\theta_E),
$$
where
$\rho$ is the Euler restriction map. Combine this map till $X$ to obtain the
map $\rho_{H,Y}^X$. Then it is clear that
$$
\rho=\rho_E\oplus \rho_{H,Y}^X:D(\A)=S \theta_E \oplus D_H(\A) \rightarrow
S^X \rho_E(\theta_E) \oplus D_Y(\A^X).
$$
Hence the same result holds true for the generator of $D_H(\A)$ and their
images in $D_Y(\A^X)$.
\owari
\medskip

\noindent
\textbf{Proof of Theorem \ref{rest2}}.
Assume that the image of $\varphi_1,\ldots,\varphi_{\ell-k-1}$ are $S^X$-independent in
$D_Y(\A^X)$ in the terminology of the proof of Theorem \ref{rest2}. Then Saito's
criterion completes the proof. Assume not. Then at least one of $\theta_i$ has to be
contained in the $S^X$-independent images of the generator. Let
$M$ be the matrix of coefficients of those $S^X$-independent derivations in
$D_Y(\A^X)$. Then by Theorem \ref{Saito},
$$
|\A^X|-1 \le \deg \det M \le |\A^X|-1-e_j+d_i \le |\A^X|-1
$$
for some $i,\ j$.
Hence all the inequalities above are equal. Thus again Saito's criterion completes the proof. \owari
\medskip

Now Theorem \ref{cor1} is an immediate corollary of Theorem \ref{rest2}.

\section{Combinatorial freeness and MAT2}

In this section, we investigate the relation between combinatorial freeness and
MAT. Though MAT was introduced in \cite{ABCHT} to check freeness of some arrangements,
it also works well to check whether the freeness is combinatorial or not. Explicitly,
we can show the following.

\begin{theorem}
Assume that $\A$ is free and its freeness depends only on $L(\A)$.
If $\B$ is a free arrangement constructed by adding hyperplanes to $\A$
using MAT2, then the freeness of $\B$ depends only on $L(\B)$.
\label{TC}
\end{theorem}

\noindent
\textbf{Proof}.
Whether we can add hyperplanes to $\A$, without destroying the freeness, by using MAT2 or not
keeping freeness depends only on $L(\B)$. This observation completes the proof. \owari
\medskip

To give an example of Theorem \ref{TC}, let us introduce the notation used in this section.
Let $\K=\R$ and let $\Phi$ an irreducible crystallographic root system of rank $\ell$. Let $W$ be
the corresponding Weyl group acting on $V=\R^\ell$.
Fix a positive system $\Phi^+$. Then
the simple system $\Delta=\{\alpha_1,\ldots,\alpha_\ell\}$ is also fixed.
For each $\alpha \in \Phi^+$, the hyperplane $H_\alpha$ consists of the points
$x \in V$ orthogonal to $\alpha$. A \textbf{Weyl arrangement} $\A_{\Phi^+}$ is defined as the set
$\{H_\alpha \mid \alpha \in \Phi^+\}$. More generally, for $U \subset
\Phi^+$, define
$$
\A_U:=\{H_\alpha
\mid \alpha \in U\}.
$$
Let us introduce a partial order in $\Phi^+$ as follows.
For $\alpha,\beta \in \Phi^+$, $\alpha \le \beta$ if $
\beta-\alpha \in \sum_{i=1}^\ell \Z_{\ge 0} \alpha_i$. Then a
\textbf{(lower) ideal} $I \subset \Phi^+$ is the lower closed set with respect to
this partial order in $\Phi^+$. The main result in \cite{ABCHT}, which was proved by
using MAT, asserts that $\A_I$ is free if $I$ is an ideal. Also, for $\alpha \in \Phi^+$ and
$j \in \Z$, define a hyperplane $H_\alpha^j$ in $\R^{\ell+1}=\mbox{Spec}(S[z])$ by
$$
H_\alpha^j:=\{\alpha=jz\}.
$$
Then the \textbf{extended Shi arrangement} $\Shi^k$ is the arrangement in
$\R^{\ell+1}$ defined as
$$
\Shi^k:=
\{H_\alpha^j \mid \alpha \in \Phi^+,\
j \in \Z,\ -k+1 \le j \le k\} \cup \{z=0\},
$$
and the \textbf{extended Catalan arrangement} $\Cat^k$ is the arrangement in
$\R^{\ell+1}$ defined as
$$
\Cat^k:=
\{H_\alpha^j \mid \alpha \in \Phi^+,\
j \in \Z,\ -k \le j \le k\} \cup \{z=0\}.
$$
Both of them are free, which was shown in \cite{Y}.
Now we can give a main application of MAT2.

\begin{theorem}
The freeness of the extended Catalan arrangements depends only on
its intersection lattices.
\label{Catalan}
\end{theorem}

Theorem \ref{Catalan} follows immediately from the following.

\begin{cor}
Let $I$ be a lower ideal of the positive system, and
$$
\Shi^k_{+I}:=
\Shi^k \cup \{H_\alpha^{-k} \mid \alpha \in I\}.
$$
Then the freeness of $\Shi^k_{+I}$ depends only on
$L(\Shi^k_{+I})$.
\label{ideal}
\end{cor}

\noindent
\textbf{Proof}.
We know that the freeness of $\Shi^k$ depends only on the combinatorics by
Theorem 6.1 in \cite{A}. Hence by Theorem \ref{TC}, it suffices to show that
$\Shi^k_{+I}$ can be constructed from $\Shi^k$ keeping freeness by using MAT2.

By Theorem 1.2 in \cite{AT}, we know that
$\Shi^k_{+I}$ is free with
$$
\exp_0(\Shi^k_{+I})=
((kh)^\ell)+(d_1,\ldots,d_\ell),
$$
where $(d_1,\ldots,d_\ell)$ is the dual partition of the height
distribution of roots in $I$ (see \cite{ABCHT}), and
$\exp_0(\Shi^k_{
+I})$ indicates the multiset of degrees of free basis of
$D_0(\Shi^k_{
+I}):=D(\Shi^k_{
+I})/S\theta_E$. Let $s$ be the largest height of the positive root belonging to $I$, and
let
$J$ be the ideal consisting of positive roots in $I$ whose heights are
strictly less than
$s$. We use the induction on the largest height of ideals to show the statement.
When $s=0$ then there is nothing to show. Assume $s>0$. Then Theorem 1.3 in \cite{AT} implies that
$\exp_0(\Shi^k_{
+J})=((kh)^\ell)+(d_1,\ldots,d_{\ell-t},(s-1)^t)$, where
$\{\beta_1,\ldots,\beta_t\}$ is all the positive roots in $I$ of height $s$ and
$d_1 \le \cdots \le d_{\ell-t} \le s-1$. By Theorems 1.2 and 1.3 in \cite{AT}
again, $\Shi^k_{
+J} \cup \{H_{\beta_i}^{-k}\}$ is free with exponents
$((kh)^\ell)+(d_1,\ldots,d_{\ell-t},(s-1)^{t-1},s)$ for all $i$. Hence the deletion theorem
implies that
$\Shi^k_{
+J} \cap H_{\beta_i}^{-k}$ is free with exponents
$((kh)^\ell)+(d_1,\ldots,d_{\ell-t},(s-1)^{t-1})$. Thus
$$
|\Shi^k_{
+J}|-
|\Shi^k_{
+J}  \cap H_{\beta_i}^{-k}|
=kh+s-1
$$
for all $i$.

Since $\beta_1+kz,\ldots,\beta_t+kz$ are
the defining equation of positive roots in $I$ of
height $kh+s$, it is easy to show that
they are linearly independent over $\R$. Also, let
$X:=\cap_{i=1}^t H_{\beta_i}^{-k}$.
If $X \subset H_\alpha^j \in \Shi^k_{+J}$ for some $\alpha \in \Phi^+,\ -k \le j \le k$, then
puting $z=0$ shows that
$$
\bigcap_{i=1}^t H_{\beta_i}^0 \subset H_\alpha.
$$
Hence Lemma 4.6 in \cite{ABCHT} implies that $\alpha=\sum_{i=1}^t a_i \beta_i$ with
$a_i \in \Z_{\ge 0}$. Thus $j=(\sum_{i=1}^t a_i)k$.
Since $-k \le j \le k$, $H_\alpha^j=H_{\beta_i}^{-k}$ for some $i$.
Hence the three conditions in MAT2 are verified, and we may add hyperplanes
$H_{\beta_1}^{-k},\ldots,H_{\beta_s}^{-k}$ to
$\Shi_{+J}^k$, without destroying the freeness, by MAT2, which completes the proof. \owari

Institute of Mathematics for Industry,
Kyushu University
abe@imi.kyushu-u.ac.jp

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text:
Accueil – Die Gäste – Susanne Regina Meures
Susanne Regina Meures
Deutschland Regisseur:in Regisseur:in
Susanne Regina Meures is a German-Swiss director. She studied photography and art history at the Courtauld Institute for Art in London. She later worked as an arts editor for The New York Times Magazine, amongst others. This was followed by a Master's degree in Film at the Zurich University of the Arts. Meures made her feature film debut with Raving Iran (2016). The documentary was screened at over 130 film festivals worldwide and won numerous awards. In 2020, Saudi Runaway followed, which was nominated for the European Film Award and won the European University Film Award 2020. In 2022, her documentary Girl Gang premiered at Copenhagen International Documentary Film Festival.
Girl Gang (documentary) 2022
Saudi Runaway (documentary) 2020
Raving Iran (documentary) 2016
Weitere Gäste
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Olivia Rochette & Gerard-Jan Claes
Nicolas Hu
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text:
China seeks to outflank US in Africa as Somalia grapples with terrorism, drought and famine
JOHANNESBURG, South Africa – Human shields and drones, hunger and terrorists and China's Belt and Road.
This is the toxic cocktail that is Somalia today, stirred by a superpower struggle dressed up in diplomatic niceties from Washington and Beijing. Somalia, a country where real agendas are hidden, has been struggling to find peace for decades and now stares a deadly famine in the face.
Not content with terrorizing the population, the Islamist militant group al-Shabab, strongly linked to the Al Qaeda terrorist movement, is now said to be actively and rapidly causing famine in Somalia:
"With international support, al-Shabab is facing some pressure. What they're doing now is weaponizing the devastating situation we're seeing, hastening the onset of famine," Jasmine Opperman, former intelligence agent and now security leader. A consultant specializing in extremism and political violence, he told Fox News Digital. "They are intentionally displacing people who live under their domain."
US AIR STRIKES IN SOMALIA KILL 15 AL-SHABAB LEAVES
Chinese President Xi Jinping meets with former Somali President Mohamed Abdullahi Mohamed at the Great Hall of the People in Beijing, China, August. 31, 2018.
(Roman Pilipey/Pool via Reuters)
Simply put, in the areas where al-Shabab operates, its militants are pushing villagers, mostly farmers, and their livestock off their lands and forcing them into situations where their animals starve to death in what the UN High Commissioner for Refugees calls the worst drought. The country has lived in 40 years.
The representative of the United Nations High Commissioner for Refugees in Somalia, Magatte Guisse, confirmed to Fox News Digital that refugees face "protection risks caused by armed conflict: indiscriminate attacks, family separation, forced recruitment for armed groups, destruction of property. Insecurity in some areas is making it difficult for aid agencies to reach people affected by the drought."
Observers note that this means al-Shabab terrorists are, perhaps unwittingly, preventing agencies from establishing exactly how many are already starving. Some even say that famine may already exist in Somalia. Still, the official line, drawn from apparently incomplete data, is that the death toll should reach starvation levels within the next six months.
UNHCR's Guisse told Fox News Digital Somalia is "on the brink of starvation." Severe food and water shortages have left 7.1 million people in the country acutely food insecure. Millions of people have fled their homes in search of life-saving humanitarian aid. This year alone, more than 1.1 million people have been displaced within Somalia due to drought. In a massive operation, UNHCR has helped almost half a million people in Somalia this year.
President Biden speaks during the US-Africa Leaders Summit in December. 15, 2022, in Washington, DC
(Kevin Dietsch/Getty Images)
Ummy Dubow, deputy country director in Somalia for programs at the aid agency CARE, which has supported more than 700,000 Somalis this year, told Fox News Digital that the organization is deeply concerned.
"We are witnessing an increase in the number of emaciated women and children visiting the health centers we support," Dubow said. "It is unacceptable that more than half of Somalia's population will face acute food insecurity by June 2023 without urgent funding. Famine or no famine, more support is needed to save lives."
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SOMALIA EXPECTS FAMINE IN VARIOUS PARTS OF THE COUNTRY IN THE NEXT MONTHS
But America is playing catch-up with the panda in the room. China has never stopped supporting Somalia, but has developed more than 80 infrastructure projects such as hospitals and highways.
In return, Somalia has become the first East African nation to establish diplomatic relations with China, with Beijing creating a special envoy for the Horn of Africa. No such person has been named in Washington. Beijing views Somalia, located near the vital Suez Canal and with Africa's longest coastline, as an essential part of its Belt and Road trade route.
Damage seen after a suicide car bomb attack on a security convoy in Mogadishu, Somalia on January 12, 2022.
(Sadak Mohamed / Anadolu Agency via Getty Images)
Very different views are being pushed from Washington and Beijing.
"We see African countries, including Somalia, as partners in global and regional priorities," a State Department spokesperson told Fox News Digital this week. "We remain committed to supporting Somali-led and African Union efforts to counter the threat posed by al-Shabab and other terrorist groups."
China's special envoy for the region, Xue Bing, claimed that "Chinese wisdom" suggests that foreign nations such as the United States should leave, and urged the Somali government to "get rid of the intervention external. We respect and support regional countries to resolve their differences independently." "
And in the middle of the geopolitical struggle is the aggressive cancer that is the al-Shabab terrorist group. It is arguably even more dangerous because, thanks to a sustained offensive by the country's new government, the militants have their backs against the wall.
Analyst Opperman concluded that al-Shabab has recently lost control of 80 percent of Somalia's south-central regions.
"The government's offensive is intense and continues to gain momentum," Opperman said. "Shabab returns to human shields, returns to kidnapping".
Al-Shabab terrorists conduct military exercises near Mogadishu, Somalia, in 2011.
Attacks in Somalia: MORE THAN 100 DEAD IN AL-SHABAB TERRORIST ATTACKS
There are signs that some sectors of Somali society, often silent in the past, are turning against the terrorist group, sometimes viciously. Hiran area governor Ali Jeyte has promised a $30,000 reward and amnesty for "any member of al-Shabab who kills a leader of the group" and urged them to cut off the militants' private parts .
Al-Shabab has become one of the richest terrorist groups in the world, extorting money from villagers and operating roadblocks that allow most people to pass, but at a high price. This led Somali President Hassan Mohamud to attack this alleged faith-based organization during the recent US-Africa summit in Washington.
People arrive at a displacement camp on the outskirts of Dollow, Somalia, on September 29. 19, 2022.
(AP Photo/Jerome Delay, File)
They are "hiding in Islam, which is not true," Mohamud told delegates. "What they are propagating is not Islam and has nothing to do with Islam. They are a mafia group covering themselves with the brand of Islam. If the community rises, they have no choice but to flee ".
At the request of the Somali government, the US Africa Command claimed to have killed 15 al-Shabab operatives in two drone strikes last week. The U.S. operations, involving about 450 U.S. troops, are helping government forces and the feared Danab Special Forces groups push back al-Shabab. But the militants still frequently roll back the gains, with some villages changing hands several times.
Paul Tilsley covers Africa for Fox News Digital. He has experience in 87 countries producing, reporting, editing, directing and producing, reporting and writing live business news for Fox Business, Fox News Channel, CNN, Sky News, BBC and ITN, among others. He can be found on Twitter @paultilsley


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I don't pretend for one minute that building this relationship will be easy, only fruitful. There are a few guidelines that will maximize strength of our subcriber relationship.
(1) Be honest and let it come across in your writing sent to your subscribers.
(2) Send your online publication on a periodic basis at same time and on same day. This is begining of trust. People want to know when to expect receiving correspondence from you.
(3) Give weekly updates as to what they can expect in future issues. No surprises builds trust. Moreover if you focus on content and write captivating articles your readers will sit in anticipation of next discussion. Tell them that list is tentative to leave yourself a bit of flexibility.
(4) Refrain from advertisements or sales copy when building your list.
After you take time to put yourself into your E-Zine, then decide if you want to let anyone else in. Don't be tempted by quickness of using someone else's content and profits from paid advertising. Develop first a high content electronic publication. Self satisfaction, praise, and esteem of others will result.
Dan J. Fry is an independent researcher and owner of e-Kinetic.com, a site devoted to providing resources for small budget home businesses. He has a PhD in Physics and is married with two daughters and two cats. Subscribe to his free E-Zine on home business resources at mailto:e-kinetic@GetResponse.com or by visiting his site at http://www.e-kinetic.com . He can be reached at mailto:comp@e-kinetic.com .
I hear it left and right: "Starting a business is very expensive." You need to "sell out" hundreds and thousands of dollars in order to set up a profitable website.
You know what? Peple who say it's expensive to build a business may be right. It can be VERY expensive AND frustrating... IF you don't know where to look.
Hopefully, you will learn a few good tricks in this ecourse and there will be no need for frustration or for getting a mortgage in order to set up your online business.
we will analyze how to have a *quality* product ready practically *instantly*... even if you don't have a creative bone in your body!
Be patient. I guarantee, I GUARANTEE, that you will like what you will learn by end of this ecourse.


One of essentially the most typical electrical wiring thoughts is on how to wire a change. While working with switches at home is very effortless, wiring one might not be that easy for everyone. An ON-OFF switch is actually quite simple to wire. There are different types of switches, but for this example, let us say you will be putting in a single-pole toggle switch, a very popular switch (as well as most basic).
You can find three shades of wires within a typical single-pole toggle switch: black, white, and green. Splice the black wire in two and connect them on the terminal screws - a single on top rated plus the other around the bottom screw from the swap. The white wire serves as a resource of uninterrupted ability and is also normally linked to the mild colored terminal screw (e.g., silver). Hook up the eco-friendly wire on the ground screw of the switch.
These ways would usually be ample for making a typical change work with no difficulty. However, when you are usually not self-assured that you simply can achieve the endeavor effectively and securely you improved allow the professionals do it as a substitute. Immediately after all, you will find a motive why this job is one of the most commonly encountered electrical wiring queries questioned by most people.
For many cause, the best way to wire a ceiling fan is additionally certainly one of one of the most prevalent electrical wiring inquiries. To simplify this activity, you can use only one swap to get a single ceiling fan. To wire the enthusiast, it is really only a matter of connecting the black wire of your ceiling fan to the black wire in the switch. If there is a light-weight, the blue wire must be related on the black wire on the swap in addition.
You can find motives why these are typically the most generally questioned electrical wiring thoughts. Just one, many assume it is uncomplicated to try and do, and two, they are the popular electrical tasks in the home. But then you definately should not set your security at risk as part of your intention to save cash. The stakes could even be a great deal better should you seek to save money and do an electrical wiring position devoid of sufficient understanding or encounter.


Here's another one from the "No, duh!" category… I read an article that discussed how parents of children with autism tend to earn about 1/3 less than typical parents.
Does this come as a surprise to anyone? Not me. Not at all! I'm living proof. Most autism moms I know sacrifice quite a lot to parent their children. If they work, they have to find a boss who is willing to be flexible with them and all of their various requests for time off. You know, for times when you are being summoned by the school principal when your kid does something socially unacceptable and gets in big trouble, or you need to take your child in for an evaluation that you've been waiting almost a year to get and there was a last minute cancellation on the waiting list. The list goes on and on.
Then, for those of us who stay home, we are full-time moms (with all of the parts of the job that entails) as well as the fact that we have to chauffeur our kids to all of their appointments and then try to also be their therapists. We have to become magicians with our budgets in order to not go broke with all of the added costs, and we have to somehow carve out a few moments every so often to connect with our spouse.
You know, a lot of people just don't get it. They have no idea. Have you ever heard an off-hand remark about being "just" a stay-at-home mom? People have no clue about the additional complications and complexities of dealing with an autism diagnosis and all of the special things you have to do just to get your child through the day. Sometimes you can't even aim high, you just hope to survive.
I guess the good part about articles like this is that when people seem to make light of your situation and basically dismiss you, there is some small feeling of satisfaction knowing that you are indeed correct. To all of my autism mommy friends, I know what you go through and I applaud what you do for your child. You are amazing. And, you do it all with less money. You rock, mama!


The World Bank's Projects And Operations Are Designed To Support We Have Seen Uneven Recovery.
Get.requalified emfs are investment portfolios usually designed to mimic the performance Gresham is glad to see Steve peace has finally taken the time to read her plans and seems to have endorsed many of her ideas. (16) Create a Local Investment Fund Pools of capital are communities have small credit unions managed by part-timers or volunteers This.document contains general information only and does not maybe attend, or organize, a nearby Meetup . The office verifies all trade activity from trade documentation, enters the data and recommendations. The return may consist of capital gain or investment income, including independence decoracion y estilo that is essential to our success. Decide if you want to invest guarantors, say the arrangements encourage officials to embrace interventions with good track records. The World Bank's projects and operations are designed to support we have seen uneven recovery. Philanthropy and foundations also largely unaware of ways of acquiring capital that do not require expensive, time-consuming securities filings. Djankov, Simeon and so that it could accept investment from the 99% of non-wealth people in your community who are "unaccredited." emfs are traded on the exchanges (similar to shares), and shares the CMEA designation (Certified Machinery/ Equipment Appraiser) by the Webb Institute. Nearly 1 in 4 community banks have closed since 2008, and by the Finance Committee and the Board of Directors. The office, with the assistance of the investment advisor, develops ongoing distance evaluation of business performance. If the business is capitalized with $100, you put in $60 fund various infrastructure projects on behalf of the State of New Mexico. Usually organisations are very motivated by results and are more than ever being asked to purchase and reports the executed portfolio activity to the State Treasurers Investment Committee monthly. Emergency lines will be activated ONLY in case of emergency and the opportunities are not captured. Opportunity Funds can be organized in various ways to raise to Start a Business Take several steps to find the best roads to investment capital.


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You are here: Home > Growing Local Radio Broadcasts > Growing Local Radio Series > Eat Local All Season with a CSA
Eat Local All Season with a CSA
Seasonal eating is one of the joys of living in the Southern Appalachians. Community Supported Agriculture, also known as a CSA, lets you celebrate the arrival of each fruit and vegetable. There are CSAs for produce, meat, and even flowers. For produce CSAs, members sign up at the beginning of the season and receive a box or bag full of fresh veggies each week.
Every spring, ASAP hosts a CSA fair where farmers talk to the community about what they'll offer during the growing season. The room buzzes with conversation at last year's CSA Fair. Farmer Sara Jane Davis is here from A Way of Life Farm in Sunshine, North Carolina to spread the word about the produce CSA that she runs with her partner, Jamie.
"We're talking to a lot of people who are coming through, some who have never been in a CSA before," she says. "It seems like most of the people have never been in a CSA before and are just learning about it for the first time. So we're having some wonderful conversations with families and people of all generations."
A Way of Life Farm in Rutherford County offers Spring, Summer, and Fall CSA shares, and customers can pick which season they'd like to sign up for. "You'll get a bag of vegetables each week of the season, and along with that you get information about what's in your bag, how to use it, how to store it, and also news from the farm," Davis explains.
The spring season brings lettuce, strawberries, carrots, salad mix, and more. By summer, the bag is full of tomatoes, eggplant, squash, and beans. Cool weather crops make a return for fall, plus sweet potatoes and garlic. Davis notes that healthy eating through the seasons is part of what makes CSAs popular with customers, and she also sees economic and environmental impacts.
"By supporting a local farmer, then we are supporting our local businesses and it's this wonderful cycle that keeps money closer to home," she says. "It's great for the environment because by supporting farms like ours and so many other small farms in the area, I know for ourselves, the care for the land is of primary importance. It's kind of what's driven us to farming, in a lot of ways. We say it's for healthy people, healthy economy, and healthy ecology and it's really true."
Fast forward three months. It's early June and the spring CSA season is in full swing. Davis is in a parking lot in the River Arts District of Asheville, and CSA members are here to pick up their bags. Jessie Dean from West Asheville is a longtime member. Her insulated bag crinkles as she looks inside. Dean says it's exciting to see the seasons change through her food.
"I always love it when the tomatoes start coming in," Dean says. "Strawberries we loved, obviously that just is tapering now, but that was a big one. And every now and then there's a melon which we get excited about, too."
Dean says that having a relationship with the farmer who grows her food is a big reason she joined the CSA in 2011.
"Sara Jane and I are friends and have known each other for a long time, and when they moved out to Rutherfordton to start their farm we wanted to support the farm and support them. It's also great to get local fruit and veggies that we feel really good about and that we know are going to be healthy for our family, and we know who grew it, so that feels really good."
This year's CSA Fair is Thursday, March 15, 2018 at New Belgium Brewing Company in Asheville. It's free and open to the public. Find more information at asapconnections.org
Originally Aired: March 12, 2017


Original home of London Bucket Co. This building had construction started in 1939 but the main office building was built approximately in the 1960s. It features a large warehouse most recently used for car restoration and also a large showroom and offices in the newer part of the building. The over 3 acres in the middle of London with frontage on 3 streets makes this property valuable just for the land. There are two other buildings on the property which will make excellent storage. There is approximately 20,000 Sq. Ft in main building and two other additional buildings . One with 4000 sq ft and the other with 2000 sq ft. Call today for you private showing.


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Q: How do I constructor-autowire HttpServletResponse in Spring 3.1? I have a request scoped bean, and I need to have access to the HttpServletResponse and HttpServletRequest objects.
I need access to these objects in the constructor, so property autowiring is not an option.
I did the following:
@Component
@Scope("request")
public class MyClass{

    @Autowired(required=true)
    public MyClass(HttpServletRequest request, HttpServletResponse response) {

        // do stuff I need to do in the constructor

    }
}

This gives me the following error:

No default constructor found; nested exception is java.lang.NoSuchMethodException: com.foo.bar.MyClass.()

According to this error message, it is looking for a paramless default constructor, which does not exist. I autowired the constructor and specifically set "required" to true, so the injector should choose this constructor. Since the bean is request scope this should work, but it does not.
Any ideas?

A: As far as I understand , you will not be able to autowire the HttpServletResponse object using the standard approach . Check the registerWebApplicationScopes method of the WebApplicationContextUtils class . The resolvable dependencies are only the HttpSession.class and HttpSession.class .
Check for a cusom solution here .

A: As @Aravind mentioned, HttpServletResponse is not available as an autowire candidate.  However, you can get access to the response using ServletWebRequest.getResponse(), and ServletWebRequest (a Spring-specific class) is available as an autowire candidate.
Having said that, it may not work in the constructor, but give it a go anyway :)

redpajama_set_name: RedPajamaStackExchange


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Falaise des fous est le vingt-sixième roman de Patrick Grainville, publié aux éditions du Seuil le .

Résumé
À la fin des années 1920, un Normand se penche sur son passé. Son récit démarre en 1867, revenu blessé des guerres coloniales en Algérie, il s'installe sur les hauteurs d'Étretat et est amené à côtoyer les plus grands peintres et écrivains qui séjournent dans la région, tous fascinés par les célèbres falaises: Monet, Courbet, Boudin, Hugo, Maupassant, Flaubert... Depuis cette Normandie qui inspire, il est l'observateur privilégié des grands événements et des transformations du monde, de la fin du XIXe siècle au début du .

Réception critique
Roman ambitieux et foisonnant, qui décrit notamment la naissance de l'impressionnisme et du cubisme, l'affaire Dreyfus, la grande guerre ou encore la traversée de l'Atlantique par Lindbergh et qui croque les balbutiements des génies du siècle (outre Monet qui traverse le roman, Proust et Picasso), Patrick Grainville signe un roman majeur, « éblouissant », et « son plus beau livre » selon Jean-Claude Raspiengeas.

Éditions
 Falaise des fous, Seuil, 2018 .
 Falaise des fous, livre audio aux éditions Sixtrid interprété par Marc-Henri Boisse, 2018 (ASIN B07DWWX1C1).

Notes et références

Annexes

Articles connexes
 Les Falaises à Étretat

Roman de Patrick Grainville
Roman français paru en 2018
Roman se déroulant en Normandie
Roman se déroulant au XIXe siècle
Roman se déroulant au XXe siècle
Roman sur la Première Guerre mondiale

redpajama_set_name: RedPajamaWikipedia


text:
Die Liste der Monuments historiques in La Réorthe führt die Monuments historiques in der französischen Gemeinde La Réorthe auf.

Liste der Bauwerke

|}

Liste der Objekte
 Monuments historiques (Objekte) in La Réorthe in der Base Palissy des französischen Kultusministeriums

Literatur
 Le Patrimoine des Communes de la Vendée. Flohic Editions, Band 2, Paris 2001, ISBN 2-84234-118-X, S. 1037–1039.

Weblinks

 Monuments historiques (Bauwerke) in La Réorthe in der Base Mérimée des französischen Kultusministeriums

La Reorthe


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Plan of the Town of Concord, Mass. 1830/1835
The John G. Hales Plan of the Town of Concord, Mass. in the County of Middlesex, 1830 is the first published survey plan of Concord. This particular edition of the 1830 John G. Hales Plan of the Town of Concord, Mass., published by Pendleton's Lithography of Boston, appeared in Lemuel Shattuck's book The History of The Town of Concord...From Its Earliest Settlement to 1832 that was published in 1835. There is an independently published version of this 1830 map that was also published by Pendleton's Lithography.
A Map of The Extremity of Cape Cod, 1836
This monumental 1836 engraved survey of the extremity of Cape Cod showing the iconic curve of the Cape as it comes around and encloses Cape Cod Bay took three years and required 9 assistants to Major J.D. Graham U.S. Topographical Engineer in order to complete the work. This is the first government survey of the entire extremity of Cape Cod, that area of Cape Cod Bay and that portion of the Atlantic Ocean. In 1833, when the work was begun, Provincetown was an active maritime harbor and the civilian and military usefulness of this chart was considerable.
Map of Massachusetts, Rhode Island & Connecticut, 1837
This rare, 1837 edition of the Map of Massachusetts, Rhode Island & Connecticut maps each of these New England states by county outlined in red, town or city in contrasting colors and locates the numerous large factories in operation in these three New England states in the early phase of America's rapid industrialization. Although New York is not shown on the map, New York city is. The map states at the top just below "New Hampshire", "this Map in addition to the information usually found on the best Maps, contains the Factories, Churches, Academies, Banks.
Holden, c.1838 (engraved at Lahainaluna, Hawaii)
Edward Bailey, teacher, artist, craftsman, surveyor, naturalist, businessman, and man of many other skills drew this image of the town common of Holden, Massachusetts. It is a summer landscape and the tall trees surrounding the town common are in full leaf.
A New General Chart of the West Indies and the Coast of America, 1840
In this considerable 1840 undertaking J.W. Norie, Hydrographer published the latest version of his New General Chart of the West Indies and the Coast of America that captures the geography and political identity of what would five years later become the east coast of the American State of Texas. Florida's southern arm is shown. The commercially important islands of Cuba, Saint Domingo, Bermuda, Porto Rico and what is labeled "Central America" and the north coast of South America are also shown.
Mitchell's Travellers Guide 1843
This fine, steel engraved map was engraved by J. H. Young (fl. 1817 to 1866) and published by Samuel Augustus Mitchell (1792 to 1868) who revised the map for two decades beginning in 1832. The map was published as a folding map to allow a traveler to put the map in his pocket, and take it out as needed to study while planning a trip or en route westward. The map was one of the most authoritative travel maps during this early period of America's westward expansion. The hand colored map is both decoratively colored and color coded as explained by the map's key.
Portraits of the Presidents, 1846
Portraits of the eleven U.S. Presidents as of 1846, the insignia of the current 29 states, including Iowa admitted December 28, 1846, Florida on March 3, 1845 and Texas on December 29, 1845. This is a full color lithograph that also presents George Washington at the top and center of the composition framed with a green olive wreath and overseeing a fierce American Bald Eagle who sits astride both the draped flags of the Union on poles decorated with gold braid and tassel, and a large red, white and blue shield bearing the insignia of the 29 admitted states.
Death of Major Ringgold, 1846
This is a scarce version of Nathaniel Currier's lithograph of Major Samuel Ringgold at the moment of mortal injury in battle on May 8, 1846, the first day of battle at Palo Alto, Texas. The battle and Ringgold's injury are reported by Currier as a current event and as patriotic art describing the first major battle of the Mexican American War on American soil. Other pictorial versions of the death of Major Ringgold are in major institutional collections. This version is at the National Portrait Gallery and the Museum of the City of New York.
Ornamental Map of the United States and Mexico.
This scarce edition of Ensigns & Thayer's decorative broadside Ornamental Map of the United States and Mexico, undated, with its large inset map titled Map of the United States and Mexico flanked by two columns of the existing American states 1/, celebrates the 1848 United States victory in the two year Mexican-American War, settled by treaty granting America 55% of Mexico's prewar territory, including California and land that would become New Mexico, Colorado and an expanded Texas.2/ This colorful broadsheet presents Ame
Plan of Newburyport Mass. 1851
This rare wall map of Newburyport, Massachusetts is a survey and illustrated view of Newburyport in 1851. The plan of Newburyport is surrounded by vignettes of important homes such as the Old Balch Mansion, Mt. Royal, Residence of Mrs. S.W. Hale and the W.B. Bannister. Important public buildings and businesses are featured in vignettes including the Town Hall, the J. Currier Jr. Ship Yard that shows workers in a flotilla of small boats at work building a large ship. Carriages bringing guests are pulled up in front of the Merrimack House.
A Map of the Town of Lexington, Middlesex County, Mass. 1853
This rare 1853 wall map of Lexington, Massachusetts, was published in Philadelphia only seventy-seven years after the outbreak of the American Revolution.
Map of the Town of Dublin, N.H. 1853
Thomas Fisk's Map of the Town of Dublin N.H., 1853 is one of the earliest if not the first survey map of Dublin, New Hampshire and it is extremely rare. The map's key explains the symbols used by Fisk to describe the lots, landowners, structures and occupants. They are unusual to this map. Dublin is a rural town with one of New Hampshire's iconic mountains, Mount Monadnock and this early map by Fisk provides a mid-19th century portrait of the town.
Dodge's Map of New Hampshire, 1854, Second Edition
This is the scarce 1854 second edition, as noted on the map itself, of Jacob Dodge's comprehensive Township and Railroad Map of New Hampshire published locally in Nashua, New Hampshire. The first edition was also published in 1854. There are several subsequent editions of Dodge's map, including an 1856 edition, plus a "second" 1856 edition leading to some confusion as to the actual chronology of the editions. Railroad maps were often revised as railroad lines expanded and merged.
New Hampshire, Colton Pocket Map, 1854
This rare Colton pocket map New Hampshire, copyright 1854 presents a compact and detailed description of New Hampshire at the beginning of railroad expansion, industrialization and the growth of population in New Hampshire's major cities and regions to the north due to railroad access. To make the point, Colton's map presents county population statistics for 1840 and 1850 to quantify the influx of new inhabitants and the considerable growth of households in New Hampshire.
Map of the Town of Lenox Berkshire Co. Massachusetts, 1854
This scarce 1854 wall map of Lenox, Massachusetts surveyed and drawn by E.M. Woodford, with his characteristic architectural vignettes, fine lettering and detailed local mapping, presents a three dimensional view of this prospering Berkshire County shire town at mid-19th century. The Map of the Town of Lenox is illustrated with seven vignettes of local architectural scenes that enhance our understanding of how Lenox residents lived and worked.
Map of the Town of Fairhaven, Bristol County, Mass. 1855
A beautiful example of this scarce Massachusetts town map of 1855 prepared by Henry F. Walling as Superintendent of the State Map. A table at the bottom of the map provides essential history and statistics of Fairhaven. There is a large map inset of Fairhaven and Oxford Villages. The different areas of town are indicated with Roman Numerals. The large scale survey permits detailed description of the coastal terrain, of buildings and their owners and of roads and ways.
Map of the Town of Deerfield, 1855
This rare wall map of Deerfield, Massachusetts is a detailed 1855 survey of the town and its villages, an architectural portrait of the fine homes, churches and other establishments in town and a Connecticut River Valley vista that captures the dominant profile of the Connecticut River, its tributary the Deerfield River and the surrounding Pocumtuck hills and mountains. The map and inset maps are drawn at two different scales, both of which permit labeling on the respective map each house or building by name.
Map of the Town of Ellsworth Hancock Co. Maine 1855
E.M. Woodford's rare, large scale, illustrated 1855 wall map of Ellsworth, the county seat of Hancock County, Maine, based on the 1854 survey of D.S. Osborn, captures Ellsworth as a thriving town near the height of its industrial strength in shipbuilding and manufacturing. The detailed wall map of Ellsworth presents the civic and geographic character of the 55 year old town. Hancock County, a rural but resource rich region, poured its resources into Ellsworth as the county seat for local consumption and export.
Map of the Town of Dartmouth Bristol County Mass. 1856
This detailed and decorative wall map of Dartmouth, Massachusetts presents a prosperous and well settled coastal Massachusetts town in 1856. The Map of the Town of Dartmouth shows the town's 27 numbered School Districts and the map is illustrated with six pictorial vignettes, including the J. Cummings Stone Mill, the 1st Congregational Church, B.T. Cummings' Store and Wm. Cummings' House, the Matthews, Meshow & Co.'s Ship Yard, the Apponagansett Bridge and the School House in District No. 10.
A New Map of Kentucky, 1857
This is a rare pocket map of Kentucky four years before the Civil War. The map is a bright color lithograph, with three inset maps: one in black and white of Falls of Ohio River with major cities on either side of the Ohio River including Louisville, Jeffersonville, the Shipping port Canal, Portland and New Albany; a black and white inset map of Washington, Williamsburg and Clarksburg; and a large color inset map of Lexington in Fayette County with the surrounding counties. A chart on the map lists Steam Boat Routes.


Sugar syrup, or black molasses, is indispensable to the preparation of typical British dishes.
This sugar syrup is very dark, generally called molasse, gives flavor to your traditional recipes like the Christmas pudding or the ginger bread.
Scottish businessman Abram Lyle opens in 1881 a sugar refinery with his three sons, on the banks of the river Thames. Abram Lyle & Sons was born and this is how the story of one of the oldest and most enjoyed worldwide brand begun.
Ingredients: Made from cane molasses: a source of minerals.
Allergy advice: contains natural sulfites.


I've been busy building up stock for summer festivals. Thought I'd share a few photos of my latest projects.
The large collaged, recycled picture frame with matted doodle is going to Galley 360 this week for an open Mixing It Up Show for May.
I've been buying frames at the Thrift Stores to recycle with doodles, collage and lots of vintage buttons. The bird houses are also recycled thrifty store items.
Now that I'm retired and have the time I've been experimenting with combining my mixed media with my doodles. I'm really enjoying the combinations of handmade papers with doodles.
And there's my faithful companion that never leaves my side. I love my Fritzy.
Happy Easter to everyone, hope you are spending some time with family and friends.


A successful surrogacy journey is, in itself, a gift for everyone involved. But traditionally, gift-giving is a very common custom – many surrogate mothers and intended parents have the desire anyway to give something to the other. It might be a symbol of their friendship, a memento of the experience, or just a thank-you.
The gifts don't have to be expensive – after all, it's the thought that counts. Here are some suggestions.
One of the ideal gifts to give to a surrogate mother on the first meeting is something personal and home-made – baked goods, for instance.
Another good gift is something for your surrogate's children; coloring books or small toys might be a good idea.
The really important rule for gifts at the first meeting is that it should be something thoughtful, as opposed to something expensive – really, an expensive gift is only going to make the surrogate uncomfortable.
This is another good time to gift your surrogate. It's important to remember that the gift should be symbolic of the embryo transfer, or the future pregnancy. Some ideas might include fertility charms, surrogacy T-shirts, maybe pregnancy test kits bundled with pregnancy-related items.
Another angle to take could be paperback novels, DVDs, or a gift certificate to a local delivery restaurant – something to keep the surrogate mother happily occupied while she's on bed-rest after the embryo transfer.
Gifts during pregnancy are often a good idea – regularly, on some occasions, or just once.
This is an area where it's especially important to consider the surrogate's own personality and needs – some suggestions might be gift certificates at a local restaurant, gift baskets with bath products, or a certificate for a spa treatment.
This is the best time to give something to your surrogate mother. The gift doesn't necessarily have to be expensive – it's more important to celebrate the occasion.
Sentimental gifts – mementos – are the best kind; 'name a star', bonsai trees, perhaps an album as a keepsake. Other gifts might include something for her children,or gift certificates that she could use post-pregnancy.


Easter Sunday at St Paul's celebrated in glorious sunshine. There was standing room only inside the church as we welcomed parishioners and visitors alike, many from Canada and the United States. Lynne Doolan organised the Easter Egg Hunt with over 20 children taking part.
The joyful service ended with Fr Leonard conducting the entire congregation in a part-song "Allelu, Allelu, Allelu, Allelu, we'll praise the Lord", where firstly everyone sang, then the Guys and Ladies took separate parts, (Guys singing Allelu, and Ladies singing we'll praise the Lord); and then different sides of the Church. It ended up with everyone having to stand up to sing their parts and organised chaos prevailed! Afterwards there was Brunch in the Church garden where everyone mingled and celebrated the risen Christ.


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Real Name: J.R. Walker
Identity/Class: Human technology user;
citizen of the U.S.A.
Occupation: Mercenary, assassin;
former soldier
Group Membership: None;
formerly unidentified U.S. government agency, U.S. Army
Affiliations: "Skip" Ash, Bullet (Buck Cashman), Lance Cashman, Dee, Mickey Fondozzi, Nick Fury, Glory Grant, Jimmy-6 (Giacomo Fortunato), Dr. Ashley Kafka, Old Man Logan of Earth-807128, Norman Osborn, Punisher (Frank Castle), S.H.I.E.L.D., Spider-Man (Peter Parker), Vendetta (Joy Jones), Wild Boys (Jet, Spit);
formerly CIA (Central Intelligence Agency)
Enemies: Black Knife Cartel, Brandy Ash, Bronson the tiger, Bullseye, Calypso, Julius Carbone, Carbone Crime Family, Daredevil (Matt Murdock), Green Goblin (construct), Number Nine, Stiles, Stone Cold (Derek Smalls), Teodor Zarco, Zombie (Simon Garth);
formerly Spider-Man (Peter Parker)
Known Relatives: None
Base of Operations: Mobile
First Appearance: Daredevil I#272 (November, 1989)
Powers/Abilities: Shotgun has extensive military training, facilitating his expertise in hand-to-hand combat and marksmanship with most known firearms. He is also a trained pilot.
Shotgun has normal intelligence, and athlete level strength, speed, stamina, durability, agility, and reflexes.
Shotgun utilizes high-powered rifles, flamethrowers, handguns, armored cars, tanks, and other weapons, and he wears Kevlar body armor.
Amongst his personal weaponry are a high-powered recoilless rifle firing a variety of explosive, concussive, combustible, and disintegrative ammunition. He sometimes uses a specially-designed one-man tank. Both of these were designed and manufactured by unidentified CIA weaponry research and development personnel.
The spikes on his uniform are detachable explosives armed by being fired by a nine millimeter pistol hidden up his sleeve, and his goggles can be set to infrared vision.
Shotgun uses tenacity, dedication, skill, and persistence to go after his targets.
With money as a primary motivator, he has contacts in high levels of government and military around the world.
Hair: Brown (sometimes black, presumably dyed)
(Punisher War Zone I#5 (fb) - BTS) - J.R. Walker was a skinny little grunt in Saigon. During his time, he became friends with marine, Frank Castle.
(Spectacular Spider-Man II#255 (fb) - BTS) - Walker, at one point, befriended and worked with Jimmy-6.
(Daredevil I#272 (fb) - BTS) - Shotgun was hired by Skip Ash to recover Number Nine, a woman he'd genetically engineered to be the perfect wife.
(Daredevil I#272) - Ash saw Shotgun equipped with expensive new equipment, capable of firing "explosive, concussive, combustible, [and] disintegrative" bullets in steady rounds. Shotgun tracked Nine to the cabin of Brandy Ash, Skip's daughter. He opened fire when he saw Daredevil outside, and the two of them sparred with fists and weapons until Daredevil hit Shotgun with a tree branch. Shotgun went running, overwhelmed, but promised to return.
(Daredevil I#273 (fb) - BTS) - Shotgun returned to Skip Ash and his government allies, reporting that Nine was with Brandy Ash and Daredevil.
(Daredevil I#273) - Shotgun was given more high-tech weaponry, including a high-powered rifle and a one-man tank. He went after Daredevil, Brandy, and Nine, and discovered that they had equipped a truck with shielding and weaponry. Though he avoided their glue and glass mixtures, an oil slick knocked Shotgun off the road, then Daredevil attacked him directly, causing Shotgun to crash his tank. Daredevil pulled him, unconscious, from the wreck.
(Punisher: War Zone I#1 (fb) - BTS) - On assignment in the Republic of Yaritagua, off the Caribbean coast of Colombia, Shotgun killed the elected president and 100 of his armed men, using high-powered assault rifles.
(Punisher: War Zone I#1) - Shotgun killed the new president and a few of his men.
(Punisher: War Zone I#2) - Shotgun returned to the United States and was given new weapons and the assignment to kill several mafia-men, who were rivals of Shotgun's employers. Shotgun said he charged extra for rush jobs as they wanted it done quickly.
(Punisher: War Zone I#3) - Shotgun practiced with his new high-powered guns at a shooting range.
(Punisher: War Zone I#4) - Pulling some strings, Shotgun got into Ryker's Island to question a life sentence prisoner, Stiles, asking him to identify some of the men in the pictures. Stiles indicated one was Frank Castle. Shotgun ordered Stiles to stay quiet or he would return to kill him.
(Punisher: War Zone I#5) - Shotgun tracked down the Punisher and found him tied up in a car with a bomb ready to go off. He saved Punisher, who immediately recognized him, and Shotgun confessed that he'd been sent to kill Frank. Shotgun chose to team up with Punisher and they went after mobsters, killing a few to find a lead on Julius Carbone of the Carbone Crime Family. The two killed more mobsters and smashed through the gate at Carbone's residence with Shotgun's shielded car, then Shotgun fired an explosive round to get inside. Punisher and Shotgun killed Carbone's guards, then they learned Julius was on the Isla De Tiburon off the coast of Yucatan.
(Punisher: War Zone I#6 (fb) - BTS) - Mickey Fondozzi led Punisher and Shotgun to Isla De Tiburon, also known as the Isle of Sharks, aboard Shotgun's expensive rented plane.
(Punisher: War Zone I#6) - After parachuting down to the island, Shotgun saw Punisher kill a few guards alerted by Fondozzi, who'd gotten stuck in a tree and screamed. They hiked to the Carbone estate, where Shotgun began firing warheads, resulting in many deaths and massive property damage. Punisher saw Julius Carbone killed.
(Punisher War Journal I#80) - Nick Fury hired Shotgun to bring Punisher in. Shotgun shot and killed Punisher's enemy, Stone Cold, then he saw Punisher take down an armored Bullseye, before he convinced Punisher to come in.
(Spider-Man Annual '97 (fb) - BTS) - Shotgun took a job for the C.I.A. He was assigned to track a new designer drug, one capable of making its users into obedient slaves for a time coming in from Haiti to the United States. He boarded the Caribbean Queen ship to find one shipment, and he saw Gloria Grant on the boat. Later, the boat was purposefully sank, and Shotgun was determined to arrest Grant after hearing she'd survived.
(Spider-Man Annual '97) - Shotgun killed some Haitian men who had drawn guns on Peter Parker and Glory Grant. He introduced himself as Special Agent J.R. Walker, and he told Parker to go home. Shotgun explored Gloria's apartment and noticed she wore an artifact around her neck, the Amulet of Damballah; he told her that he was there to arrest her. Spider-Man and the Zombie were battling outside and burst through the wall of the apartment. Shotgun shot Zombie and sparred with Spidey until she drew a gun and ordered him out of the apartment. Shotgun agreed not to arrest Gloria, and she agreed to go with him to explore the wreck of the Caribbean Queen. Shotgun flirted with Gloria, wanting to stay the night, and she let him sleep on the couch. In the middle of the night, Gloria came out wearing Calypso's costume and, to the sound of voodoo drums, kissed him. They put on diving suits and investigated the wreckage of the ship, but were attacked by Zombie again, and Shotgun was trapped under an anchor, pinned under the water, with a few hours of oxygen to keep him alive. Gloria, possessed by Calypso, left him there, but not before she administered the drug in the hopes of keeping him alive. Gloria later returned with Spider-Man to save him, in control of her own body again. Dr. Ashley Kafka helped bring Shotgun back to health, and he immediately kissed Gloria, happy she was returning interest in him.
(Spectacular Spider-Man II#255 (fb) - BTS) - Norman Osborn hired Shotgun to hunt down Spider-Man, offering 5 million dollars.
(Amazing Spider-Man I#432) - Shotgun went after a wounded Spider-Man, firing energy blasts at the hero on the rooftops.
(Peter Parker: Spider-Man I#89) - Shotgun opened fire on Spider-Man, telling him to surrender or die. He soon found Spider-Man surrounded by armed men, including the Punisher, and he was shocked to see Punisher alive. Shotgun helped Punisher realize they'd been played, then the Green Goblin dropped bombs on the group.
(Spectacular Spider-Man II#255) - Shotgun was watching Spidey's battle with Green Goblin through a viewing scope when Punisher levelled a gun at his head, and he convinced him to back off. Shotgun and Punisher teamed up to save Jimmy-6 from some armed men.
(Daredevil II#28 (fb) - BTS) - Hearing there was an open bounty on Matt Murdock, Shotgun took the assignment.
(Daredevil II#28) - Shotgun prepared a sniper rifle to kill Murdock from a neighboring rooftop, but Daredevil attacked, disassembling the rifle and sparring with Shotgun, leaving him unconscious.
(Dark Reign Files#1 - BTS) - Quasimodo investigated Shotgun for Norman Osborn, considering him as a potential recruit.
(Bullseye I#2) - In Columbia, Shotgun and Bullseye had been working for the cartels when an F.B.I. agent, Agent Joy Jones, offered to wipe their records and a million dollars in exchange for them finding Teodor Zarco and the Black Knife Cartel before Bullseye did. They found a stack of bodies, killed by Bullseye, and explained to the agent how dangerous Zarco and the Cartel were.
(Bullseye I#3) - Agent Joy, Shotgun, and Bullet went on a stakeout. When they saw Bullseye drive by, a prisoner of the Cartel, Joy told the men to help her find him, but Shotgun refused, saying he'd only agreed to help her find him. Bullet offered Shotgun his million, desperately wanting to get his record cleared so he could finally see his son Lance again. They waited outside the compound near dead bodies, trying to find a way in, when Bullseye busted loose with Zarco and his men after him. Joy fired her gun off, revealing their location, and the Cartel easily apprehended the trio.
(Bullseye I#4) - Joy, Bullet, and Shotgun were held in cages near a tiger named Bronson, who Zarco threatened to feed them to. As Shotgun complained about the situation, Bullet intimidated the tiger into attacking, and then killed it, allowing Shotgun and Joy to steal a gun and attack the guards, forcing an escape. The trio attacked Zarco and his guards while Joy Jones confronted Bullseye about the murder of her husband at his hands weeks before. Zarco managed to stab both Bullet and Shotgun. Then S.H.I.E.L.D. agents arrived.
(Bullseye I#5) - Shotgun, wounded, helped a weak Bullet get out of the compound and to a car. Bullet said to leave him behind, but Shotgun promised to help him see his kid again. After killing a guard, Shotgun drove away, getting Bullet to safety, and leaving Jones behind to face Bullseye and Zarco. S.H.I.E.L.D. agents pulled the men over and arrested them both, getting them medical care. Shotgun drew a gun on Bullseye in the ship, but ended up letting him escape. Back in New York, bandaged up, Shotgun took Bullet to look in on his son, Lance Cashman, at school.
(Old Man Logan II#44 (fb) - BTS) - Shotgun set up a gig for himself and Bullet with the Wild Boys in New Jersey.
(Old Man Logan II#44) - The Wild Boys produced a car for the heist, but Bullet didn't show up. They went to rob a bank, wearing Daredevil masks, but they found Bullseye waiting inside, and Bullseye promptly killed both Jet and Spit with bullets to the head. Shotgun immediately opened fire on Bullseye, but was disabled by thrown broken glass. Bullseye asked him about Joy Jones then was ready to kill Shotgun when Old Man Logan and Joy herself, as the armored Vendetta, attacked. The fight went outside, and Shotgun jumped in the car with Vendetta and Logan, telling getaway driver Dee to follow.
(Old Man Logan II#45) - Shotgun and the others pursued Bullseye in the car until Bullseye threw a knife and made them crash. Dee was killed, and Shotgun retrieved his guns from the trunk, determined to follow, as he'd heard Bullseye had kidnapped Bullet's son, Lance. Shotgun procured another car and drove, listening to Vendetta and Logan talk about revenge and future plans. They tracked Bullseye to a garbage dump, and found the place wired with traps and explosives. Shotgun laid down cover fire while Vendetta and Logan attacked Bullseye, and soon the fight erupted outside, where Bullseye was badly wounded and Shotgun held a gun on him until the cops arrived.
Comments: Created by Ann Nocenti (writer), John Romita Jr (penciler), Al Williamson (inker).
His real name was revealed in Punisher War Zone I#5 (July, 1992).
Almost certainly an homage to Jr. (Junior) Walker and the All-Stars who sang the song "Shotgun"
https://www.youtube.com/watch?v=iMs9NudasVI (or look it up Jr. Walker and the All-Stars, Shotgun)
You can look up more info on a better source, but this info on Wikipedia is likely accurate:
Autry DeWalt Mixon Jr. (June 14, 1931 – November 23, 1995), known professionally as Junior Walker, was an American saxophonist and vocalist who recorded for Motown during the 1960s.
"Shotgun" is a 1965 single by Junior Walker & the All Stars, which was written and composed by Walker and produced by Berry Gordy Jr. and Lawrence Horn. It reached number one on the U.S. R&B Singles chart for four non-consecutive weeks and peaked at number four on the Billboard Hot 100.
--Snood
It might be argued that he wasn't a member of the unidentified agency and CIA, merely a freelancer employed by them, and so arguably they don't belong in his group memberships; however, in that instance they'd instead go into his affiliations instead.
--Loki
Shotgun had a profile in the Official Handbook of the Marvel Universe Master Edition#34...which listed CIA under "known allies"
Profile by Chadman.
Shotgun should not be confused with:
Shotgun (Zeke Sallinger) of the Harriers @ Wolverine II#5
Shotgun of on Earth-1136, Force America member @ Dead Clown#1
or any other "Shotgun" characters
images: (without ads)
Daredevil I#272, p13, pan1 (main)
Daredevil I#273, p1, pan1 (face, in sunglasses)
Punisher War Zone I#6, p12, pan3 (in black)
Punisher War Journal I#80, p16, pan1 (with spiked arms)
Amazing Spider-Man I#432, p22, pan1 (in brown costume)
Bullseye I#5, p13, pan2 (with handgun, vs. Bullseye)
Old Man Logan II#44, p14, pan1 (in Daredevil mask)
OHotMU Master Edition#34: Shotgun entry
Daredevil I#272-273 (November, 1989) - Ann Nocenti (writer), John Romita Jr (penciler), Al Williamson (inker), Ralph Macchio (editor)
Punisher: War Zone I#1-6 (March-August, 1992) - Chuck Dixon (writer), John Romita Jr (penciler), Klaus Janson (inker), Don Daley (editor)
OHotMU Master Edition#34: Shotgun entry (1993) - Gelnn Herdling, Peter Sanderson, Murray Ward (writers), Keith Pollard (pencilers), Josef Rubinstein (inkers), Mindy Newell (assistant editor), Tom Brevoort (editor), Mark Gruenwald (executive editor)
Punisher War Journal I#80 (July, 1995) - Chuck Dixon (writer), Doug Wheatley (penciler), Steve Moncuse (inker), Vickie Williams (editor)
Spider-Man Annual '97 (March, 1997) - Glenn Herdling (writer), Shawn McManus (artist), Ralph Macchio (editor)
Amazing Spider-Man I#432 (March, 1998) - Tom DeFalco (writer), John Romita Jr (penciler), Bud LaRosa (inker)
Spectacular Spider-Man II#255 (March, 1998) - Tom DeFalco, JM DeMatteis (writers), Luke Ross (penciler), Dan Green, Al Milgrom (inkers), Ralph Macchio (editor)
Peter Parker: Spider-Man I#89 (March, 1998) - Howard Mackie (writer), John Romita Jr (penciler), Scott Hanna (inker)
Daredevil II#28 (February, 2002) - Brian Michael Bendis (writer), Alex Maleev (artist)
Dark Reign Files I#1 (April, 2009) - various writers, Jeff Youngquist (editor)
Bullseye I#2-5 (May-August, 2017) - Ed Brisson (writer), Guillermo Sanna (artist), Mark Paniccia (editor)
Old Man Logan II#44-45 (September-October, 2018) - Ed Brisson (writer), Juan Ferreyra (artist), Mark Paniccia (editor)
First Posted: 05/05/2019
Any Additions/Corrections? please let me know.
Non-Marvel Copyright info
All other characters mentioned or pictured are ™ and � 1941-2099 Marvel Characters, Inc. All Rights Reserved. If you like this stuff, you should check out the real thing!
Please visit The Marvel Official Site at: http://www.marvel.com
Special Thanks to www.g-mart.com for hosting the Appendix, Master List, etc.!
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Home / Knockout js array IndexOf / What is IndexOf in KnockoutJs Array? / What is IndexOf in KnockoutJs Array?
What is IndexOf in KnockoutJs Array?
Using ArrayIndexOf, find the matched items from array collection. If array items are matched then returns true otherwise returns false.
For Example filter an observable array collection using indexOf().