text:
\section{Introduction}

Artin groups, which are also called Artin-Tits groups, have been studied widely
since Tits introduced such groups (\cite{Tits}).
In particular, Artin groups are one of most important examples in geometric group theory.
Indeed for various nonpositively-curved or negatively-curved properties on discrete groups,
Artin groups are interesting targets.
In this paper we treat acylindrical hyperbolicity for Artin groups.

Let $\Gamma$ be a finite simple graph
with the vertex set $V=V(\Gamma)$ and the edge set $E=E(\Gamma)$.
Each edge $e$ has two endvertices, which we denote by $s_e$ and $t_e$.
We suppose that any edge $e$ is labeled by an integer $\mu(e)\ge 2$.
The \textit{Artin group} $A_{\Gamma}$ associated to $\Gamma$ is defined by the following presentation:
	\begin{align}\label{pres_standard_eq}
	A_{\Gamma}=\langle V(\Gamma)\mid \underbrace{s_e t_e s_e t_e \cdots}_{\text{length $\mu(e)$}}
	=\underbrace{t_e s_e t_e s_e\cdots}_{\text{length $\mu(e)$}} \quad \text{for all $e\in E(\Gamma)$ } 	\rangle.
	\end{align}
Free abelian groups, free groups and braid groups are typical examples of Artin groups.
Adding relations $v^2 = 1$ for all $v\in V(\Gamma)$ on (\ref{pres_standard_eq}),
we get the associated \textit{Coxeter group} $W_\Gamma$.
In terms of the properties of $W_{\Gamma}$,
we can define important classes of Artin groups.
The Artin group $A_\Gamma$ is said to be \textit{of finite type}
if $W_\Gamma$ is finite. The others are said to be \textit{of infinite type}.
Also the Artin group $A_\Gamma$ is said to be \textit{irreducible}
if $W_\Gamma$ is \textit{irreducible}, that is,
the defining graph $\Gamma$ is not decomposed as a join of two subgraphs
such that all edges between them are labeled by $2$.
It is well-known that $W_\Gamma$ is irreducible if and only if
$W_\Gamma$ can not be directly decomposed into two subgroups.
On the other hand it is unclear
whether $A_\Gamma$ is irreducible only if $A_\Gamma$ can not be directly decomposed
into two subgroups.
In general, Coxeter groups are understood very well,
but many basic questions for Artin groups are still open
(refer to \cite{Charney2008PROBLEMSRT} and \cite{MR3203644}).

Now, we consider nonpositively-curved or negatively-curved properties on Artin groups.
The following (\cite[Problem 4]{Charney2008PROBLEMSRT}) is one of the biggest open problems.
\begin{problem}\label{CAT(0)_conj}
Which Artin groups are \textit{CAT(0) groups}, that is, groups acting geometrically on CAT(0) spaces?
\end{problem}
\noindent
Here, \textit{CAT(0) spaces} are geodesic spaces where every geodesic triangle is not fatter than
the comparison triangle in the Euclidean plane (see \cite{BH} for the precise definition).
A group action is said to be \textit{geometric} if the action is proper, cocompact and isometric.
In recent geometric group theory, various properties as well as the CAT(0) property,
for example systolic property and the Helly property, are actively studied (\cite{HO20}, \cite{HO21} and so on).

In this paper we consider the following (\cite[Conjecture B]{haettel2019xxl}).
\begin{problem}\label{acyl_conj}
Are irreducible Artin groups of infinite type acylindrically hyperbolic?
\end{problem}
\noindent
The definition of acylindrical hyperbolicity is recalled in Section \ref{recall}.
We have many applications of acylindrical hyperbolicity.
See \cite{MR3589159}, \cite{Osin} and \cite{Osin2} and so on.
\begin{remark}
Clearly reducible Artin groups can be directly decomposed into two infinite subgroups.
On the other hand acylindrical hyperbolic groups can not be directly decomposed into two infinite subgroups
(\cite[Corollary 7.3]{Osin}).
Hence such Artin groups can not be acylindrically hyperbolic.
Also it is known that Artin groups of finite type have infinite cyclic centers
(\cite{MR323910}, \cite{MR422673}).
Since acylindrical hyperbolic groups do not permit infinite centers (\cite[Corollary 7.3]{Osin}),
such Artin groups can not be acylindrically hyperbolic.
We remark that the central quotients for Artin groups of finite type are
acylindrically hyperbolic
(See \cite{MR1914565}, \cite{MR2390326}, \cite{MR2367021} for braid groups,
and \cite{MR3719080} in general).
\end{remark}

Many affirmative partial answers for Problem \ref{acyl_conj} are known.
Indeed the following irreducible Artin groups of infinite type are known to be
acylindrically hyperbolic:
Right-angled Artin groups (\cite{MR2827012}, \cite{MR3192368});
Two-dimensional Artin groups such that the associated Coxeter groups are hyperbolic (\cite{alex2019acylindrical});
Artin groups of XXL-type (\cite{haettel2019xxl});
Artin groups of type FC such that the defining graphs have diameter greater than $2$ (\cite{MR3966610});
Artin groups that are known to be  CAT(0) groups by a result of Brady and McCammond (\cite{MR1770639}, \cite{KO});
Euclidean Artin groups (\cite{Calvez}).
Actually, except for Euclidean Artin groups,
all of these Artin groups are regarded as special cases of
the following irreducible Artin groups of infinite type,
which are known to be acylindrically hyperbolic:
Artin groups associated to graphs that are not joins (\cite{Charney});
\textit{Two-dimensional} Artin groups, that is,
Artin groups such that every triangle with three vertices $v_1,v_2,v_3$
of the defining graphs satisfies $\frac{1}{\mu((v_1,v_2))} + \frac{1}{\mu((v_2,v_3))}  + \frac{1}{\mu((v_3,v_1))} \le 1$
(\cite{Vaskou}).

In \cite{Charney},
Charney and Morris-Wright showed acylindrical hyperbolicity for
Artin groups of infinite type associated to graphs that are not joins,
by studying clique-cube complexes,
which are CAT(0) cube complexes,
and isometric actions on them.
In fact they constructed a WPD contracting element of such an Artin group
with respect to the isometric action on the clique-cube complex.
In this paper we generalize their result
by developing their study with additional discussion.
Our main theorem is the following.
\begin{theorem}\label{main}
Let $A_\Gamma$ be an Artin group associated to
$\Gamma$ with at least three vertices.
Suppose that $\Gamma$ is not a cone.
Then the following are equivalent:
\begin{enumerate}
\item $A_\Gamma$ is irreducible, that is,
$\Gamma$ can not be decomposed as a join of two subgraphs such that all edges between them are labeled by $2$;
\item $A_\Gamma$ has a WPD contracting element with respect to the isometric action on the clique-cube complex;
\item $A_\Gamma$ is acylindrically hyperbolic;
\item $A_\Gamma$ is directly indecomposable, that is, it is not decomposed as a direct product
of two nontrivial subgroups.
\end{enumerate}
\end{theorem}

\begin{remark}
When an Artin group $A_\Gamma$ is irreducible
and the defining graph $\Gamma$
is not a cone,
the center $Z(A_\Gamma)$ is known to be trivial.
This fact is shown in \cite{Charney}. We can give
an alternative proof based on Theorem \ref{main} (See Remark \ref{mainremark}).
\end{remark}

By Theorem \ref{main}, we can newly see that
many irreducible Artin groups of infinite type, for example,
Artin groups associated to defining graphs in Figure \ref{new_ex_fig}
are acylindrically hyperbolic.
\begin{figure}
\begin{center}
\includegraphics[width=12cm,pagebox=cropbox,clip]{graph.pdf}
\caption{Defining graphs $\Gamma$ of new examples $A_{\Gamma}$.}\label{new_ex_fig}
\end{center}
\end{figure}

We give an outline of this paper.
Section \ref{recall} contains preliminaries on acylindrically hyperbolic groups, WPD contracting elements and
CAT(0)-cube complexes.
Also Section \ref{artin} contains preliminaries on defining graphs of Artin groups and
Joins of graphs.
In Section \ref{clique} we treat clique-cube complexes and the actions by Artin groups on them following
\cite{Charney}.
In Section \ref{local_geom}, we study local geometry of clique-cube complexes.
Section \ref{main_proof} gives Proof of Theorems \ref{main}.
Indeed our main task is to construct a candidate of a WPD contracting element
and to show that it is really a WPD contracting element.

\section{Acylindrical hyperbolicity, WPD contracting elements and CAT(0)-cube complexes}\label{recall}
In this section, we collect some definitions and properties related to acylindrical hyperbolicity,
WPD contracting elements and CAT(0)-cube complexes for our use.
See \cite{Genevois} and references therein for details.

First of all we recall the definition of acylindrical hyperbolicity in \cite{Osin}.
\begin{definition}\label{acylindrically hyperbolic}
A group $G$ is \textit{acylindrically hyperbolic} if it admits an isometric action
on a hyperbolic space $Y$ which is \textit{non-elementary}  (i.e., with an infinite limit set)
and \textit{acylindrical} (i.e., for every $D\ge 0$, there exist some $R, N \ge 0$ such that,
for all $y_1, y_2\in Y$, $d_Y(y_1, y_2)\ge R$ implies
$\#\{g\in G |\ d_Y(y_1, g(y_1)), d_Y(y_2, g(y_2)) \le D\}\le N$).
\end{definition}

Next we recall a definition of a WPD contracting element.
\begin{definition}\label{WPD}
Let a group $G$ isometrically act on a metric space $X$. For $\gamma\in G$,
we say that:
\begin{itemize}
\item $\gamma$ is \textit{WPD} if for every $D \ge 0$ and $x \in X$, there exists some $M\ge 1$
such that $\#\{g\in G |\ d_X(x, g(x)), d_X(\gamma^M(x), g\gamma^M(x)) \le D\}<\infty$;
\item $\gamma$ is \textit{contracting} if $\gamma$ is \textit{loxodromic}, that is,
there exists $x_0\in X$ such that $\mathbb Z \to X; n \mapsto \gamma^n(x_0)$ is a quasi-isometry onto the image
$\gamma^\mathbb{Z}x_0:=\{\gamma^n(x_0) | n \in \mathbb Z\}$,
and $\gamma^\mathbb{Z}x_0$ is \textit{contracting}, that is, there exists $B\ge 0$ such that
the diameter of the nearest-point projection of any ball disjoint from $\gamma^\mathbb{Z}x_0$ onto $\gamma^\mathbb{Z}x_0$ is bounded by $B$.
\end{itemize}
\end{definition}

The following is a consequence of \cite{BBF}.
\begin{theorem}\label{BBF}
Let a group $G$ isometrically act on a geodesic metric space $X$.
Suppose that $G$ is not virtually cyclic.
If there exists a WPD contracting element $\gamma\in G$, then $G$ is acylindrically hyperbolic.
\end{theorem}

CAT(0) cube complexes are considerd as generalized trees in higher dimensions.
The following is a precise definition. See \cite[p.111]{BH}.

\begin{definition}\label{CAT(0)}
A \textit{cube complex} is a $CW$ complex constructed by gluing together cubes of
arbitrary (finite) dimension by isometries along their faces.
Furthermore, it is \textit{nonpositively curved} if the link of any of its vertices is
a \textit{simplicial flag} complex (i.e., $n+1$ vertices span an $n$-simplex if and only if they are pairwise adjacent),
and \textit{CAT(0)} if it is nonpositively curved and simply-connected.
\end{definition}

\begin{definition}
Let $X$ be a CAT(0) cube complex.
We define an equivalence relation for edges of $X$ as
the transitive closure of the relation identifying two parallel edges of a square.
For an equivalence class, a \textit{hyperplane} is defined as
the union of the midcubes transverse to the edges belonging to the equivalence class.
Then for any edge belonging to the equivalence class,
the hyperplane is said to be \textit{dual to} the edge.

For a hyperplane $J$,
we denote by $N(J)$
the union of the cubes intersecting $J$, that is,
the smallest subcomplex of $X$ containing $J$.
Also we denote by $X\!\setminus\!\!\setminus J$
the union of the cubes not intersecting $J$, that is,
the largest subcomplex of $X$ not intersecting $J$.
\end{definition}

See \cite{Sageev} for the following.
\begin{theorem}\label{separate}
Let $X$ be a CAT(0) cube complex and $J$ is a hyperplane.
Then $X\!\setminus\!\!\setminus J$ has exactly two connected components.
\end{theorem}
\noindent
Two connected components of $X\!\setminus\!\!\setminus J$
are often denoted by $J^+$ and $J^-$, respectively.

For convenience in Proof of Theorem \ref{main}, we prepare the following.
\begin{definition}\label{separatingdef}
Let $X$ be a CAT(0) cube complex.
For two vertices $x$ and $x'$ in $X$, we call a sequence of hyperplanes
$P_1,\ldots,P_M$ a \textit{sequence of separating hyperplanes from $x$ to $x'$}
if the sequence satisfies
$$x \in P_1^-, P_1^+\supsetneq P_2^+\supsetneq\cdots
\supsetneq P_{M-1}^+\supsetneq P_M^+\ni x'$$
for some connected components $P_i^+$ of $X\!\setminus\!\!\setminus P_i$
for all $i\in \{1,\ldots,M\}$.

For two hyperplanes $J$ and $J'$ in $X$, we call a sequence of hyperplanes
$P_1,\ldots,P_M$ a \textit{sequence of separating hyperplanes from $J$ to $J'$}
if the sequence satisfies
$$J^+ \supsetneq P_1^+\supsetneq P_2^+\supsetneq\cdots
\supsetneq P_{M-1}^+\supsetneq P_M^+ \supsetneq J'^+.$$
for some connected components $J^+$ of $X\!\setminus\!\!\setminus J$,
$J'^+$ of $X\!\setminus\!\!\setminus J'$
and $P_i^+$ of $X\!\setminus\!\!\setminus P_i$
for all $i\in \{1,\ldots,M\}$.
\end{definition}
\begin{remark}\label{separatingrem}
When $P_1,\ldots,P_M$ is a sequence of separating hyperplanes from $J$ to $J'$,
for two vertices $x\in J^-$ and $x'\in J'^+$,
$P_1,\ldots,P_M$ is a sequence of separating hyperplanes from $x$ to $x'$.
\end{remark}

The following is a part of \cite[Theorem 3.3]{Genevois},
which is used in Proof of Theorem \ref{main}.
\begin{theorem}\label{criterion}
Let a group $G$ isometrically act on a CAT(0) cube complex $X$.
Then $\gamma \in G$ is a WPD contracting element if there exist
two hyperplanes $J$ and $J'$ satisfying the following:
\begin{enumerate}
\item[(i)] $J$ and $J'$ are strongly separated, that is,
any hyperplane can not intersect both $J$ and $J'$;
\item[(ii)] $\gamma$ skewers $J$ and $J'$, that is, we have connected components $J^+$ of $X\!\setminus\!\!\setminus J$ and
$J'^+$ of $X\!\setminus\!\!\setminus J'$ such that
$\gamma^n (J^+) \subsetneq J'^+ \subsetneq J^+$ for some $n\in\mathbb N$;
\item[(iii)] $\mathrm{stab}(J) \cap \mathrm{stab} (J')$ is finite, where
$\mathrm{stab}(J)=\{g\in G\  | \ g(J)=J\}$ and $\mathrm{stab} (J')=\{g\in G\  | \ g(J')=J'\}$.
\end{enumerate}
\end{theorem}

\section{Defining graphs of Artin groups and joins}\label{artin}
\subsection{Defining graphs of Artin groups}\label{defining_graph}
We precisely describe a definition of
the defining graph of an Artin group and introduce related graphs.

Let $V$ be a finite set.
Denote by $\mathrm{diag}(V\times V):=\{(v,w)\in V\times V |\ v=w\}$ the diagonal set.
We consider the involution on the off-diagonal set
$$\iota:V\times V\setminus\mathrm{diag}(V\times V)\ni (v,w)\mapsto (w,v)\in V\times V\setminus\mathrm{diag}(V\times V).$$
Any $e\in V\times V\setminus\mathrm{diag}(V\times V)$ is often presented as $(s_e,t_e)$.
Then for any $e\in V\times V\setminus\mathrm{diag}(V\times V)$,
we have $s_{\iota(e)}=t_e$ and $t_{\iota(e)}=s_e$.
We take a symmetric map
$$\tilde{\mu}:V\times V\setminus\mathrm{diag}(V\times V) \to \mathbb{Z}_{\ge 2}\cup \{\infty\}.$$
Here `\textit{symmetric}' means satisfying $\tilde{\mu}\circ\iota=\tilde{\mu}$.
Put $E_m:=\tilde{\mu}^{-1}(m)$ for any $m\in \mathbb{Z}_{\ge 2}\cup \{\infty\}$.
Then we have
$$V\times V\setminus\mathrm{diag}(V\times V)=\bigsqcup_{m\in \mathbb{Z}_{\ge 2}\cup \{\infty\}}E_m.$$
Now we have a finite simple labeled graph $\Gamma$
with the vertex set $V(\Gamma)=V$,
the edge set $E(\Gamma)=\bigsqcup_{m\in \mathbb{Z}_\ge 2}E_m$
and the labeling $\mu:=\tilde{\mu}|_{E(\Gamma)}$.
Then the Artin group $A_\Gamma$ associated to $\Gamma$
is defined by the presentation (\ref{pres_standard_eq})
and $\Gamma$ is called the \textit{defining graph} of $A_\Gamma$.

For convenience, we define
two other finite simple graphs $\Gamma^c$ and $\Gamma^t$ as follows.
We define $\Gamma^c$ as a finite simple graph
with the vertex set $V(\Gamma^c)=V$
and the edge set $E(\Gamma^c)=E_{\infty}$.
Then $\Gamma^c$ is the so-called \textit{complement graph} of $\Gamma$.
Also we define $\Gamma^t$ as a finite simple graph with the vertex set $V(\Gamma^t)=V$
and the edge set $E(\Gamma^t)=\bigsqcup_{m\in \mathbb{Z}_{\ge 3}\cup\{\infty\}}E_m$.
See Figures \ref{new_ex_fig} and \ref{traditional_fig}.

\begin{figure}
\begin{center}
\includegraphics[width=12cm,pagebox=cropbox,clip]{traditional.pdf}
\caption{$\Gamma^t$ and $\Gamma^c$ with respect to $\Gamma$ in Figure \ref{new_ex_fig}.}\label{traditional_fig}
\end{center}
\end{figure}

\begin{remark}
In researches related to Coxeter groups and some traditional treatments of Artin groups,
$\Gamma^t$ with the label $\tilde{\mu}|_{E(\Gamma^t)}$
may be standardly used.
On the other hand we mainly use $\Gamma$
following many recent researches of Artin groups.
We only use $\Gamma^t$ as an aid in this paper.
\end{remark}

\subsection{Joins}
\begin{definition}
Let $\Lambda\neq \emptyset$ be an index set.
The \textit{join} $\ast_{\alpha\in \Lambda}\Gamma_\alpha$
of simple graphs $\Gamma_\alpha$ ($\alpha\in \Lambda$) is defined as a simple graph with
the vertex set $$V(\ast_{\alpha\in \Lambda} \Gamma_\alpha):=\bigsqcup_{\alpha\in \Lambda}  V(\Gamma_\alpha)$$
and the edge set $$E(\ast_{\alpha\in \Lambda} \Gamma_\alpha):=\bigsqcup_{\alpha\in \Lambda} E(\Gamma_\alpha)\sqcup
\bigsqcup_{\alpha,\beta\in \Lambda, \alpha\neq \beta}\{(v_\alpha, v_\beta)\mid v_\alpha\in V(\Gamma_\alpha), v_\beta\in V(\Gamma_\beta)\}.$$

A simple graph $\Gamma$ is said to be \textit{decomposable} (as a join)
if there exist an index set $\Lambda$ with $\# \Lambda\ge 2$ and
subgraphs $\Gamma_\alpha$ ($\alpha\in \Lambda$) of $\Gamma$
such that $\Gamma=\ast_{\alpha \in \Lambda}\Gamma_\alpha$,
which is called a \textit{join decomposition} of $\Gamma$ into factors $\Gamma_\alpha$ ($\alpha\in \Lambda$).
Also $\Gamma$ is said to be \textit{indecomposable} (as a join) if it is not decomposable.

A simple decomposable graph $\Gamma$ is called a \textit{cone}
if $\Gamma$ has a join decomposition into a subgraph
consisting of only one vertex $v_0$ and a subgraph $\Gamma'$:
$$\Gamma=\{v_0\}\ast \Gamma'.$$
\end{definition}

\begin{remark}
For any simple graph $\Gamma$,
it is indecomposable as a join if and only if its complement graph $\Gamma^c$ is connected.
\end{remark}

The following is a well-known fact. See Figures \ref{new_ex_fig} and \ref{traditional_fig}.
\begin{lemma}\label{join-decomposition_lem}
Let $\Gamma$ be a simple graph. Suppose that $\Gamma$ is decomposable.
Then $\Gamma$ has a unique join decomposition into indecomposable factors:
$$\Gamma=\ast_{\alpha\in \Lambda}\Gamma_\alpha.$$
\end{lemma}
\begin{proof}
Consider the decomposition of $\Gamma^c$ into connected components:
$$\Gamma^c=\bigsqcup_{\alpha\in \Lambda}(\Gamma^c)_\alpha.$$
Put $V_\alpha:=V((\Gamma^c)_\alpha)$ and
define $\Gamma_\alpha$ as the subgraph of $\Gamma$ spanned by $V_\alpha$.
Then $(\Gamma_\alpha)^c=(\Gamma^c)_\alpha$.
Also we have a join decomposition $\Gamma=\ast_{\alpha \in \Lambda}\Gamma_\alpha$.
\end{proof}

\begin{remark}
For any decomposable graph,
it is not a cone if and only if its indecomposable factors have at least two vertices, respectively.
\end{remark}

\section{Clique-cube complexes and actions on them}\label{clique}
In this section we consider clique-cube complexes and actions on them
following \cite{Charney}.
Let $A_\Gamma$ be an Artin group associated to a defining graph $\Gamma$
as in Subsection \ref{defining_graph}.
By a theorem of van der Lek (see \cite{Paris}, \cite{van}), for any subset $U\subset V=V(\Gamma)$,
the subgroup of $A_\Gamma$ generated by $U$ is itself an Artin group associated to
the full subgraph of $\Gamma$ spanned by $U$.
We denote this subgroup by $A_U$.
When $U$ is empty, we define $A_\emptyset =\{1\}$.
We say that $U$ \textit{spans a clique} in $\Gamma$ if any two elements of $U$ are joined by an edge in $\Gamma$.

\begin{definition}[\cite{Charney} Definition 2.1]\label{clique cube}
Consider the set
$$\Delta_\Gamma = \{U \subset V |\ U\text{ spans a clique in }\Gamma,\text{ or }U = \emptyset\}.$$
The \textit{clique-cube complex}, denoted by $C_\Gamma$, is the cube complex
whose vertices (i.e., $0$-dimensional cubes) are cosets $gA_U$ ($g\in A_\Gamma, U\in \Delta_\Gamma$),
where two vertices $gA_U$ and $hA_{U'}$
are joined by an edge (i.e., a $1$-dimensional cube) in $C_\Gamma$
if and only if $gA_U\subset hA_{U'}$ and $U$ and $U'$ differ by a single generator.
Note that in this case,
we can always replace $h$ by $g$, that is, $hA_{U'}=gA_{U'}$.
More generally, two vertices $gA_U$ and $gA_{U'}$ with $gA_U\subset gA_{U'}$
spans a $\#(U'\setminus U)$-dimensional cube $[gA_U , gA_{U'}]$ in $C_\Gamma$.
\end{definition}

The group $A_\Gamma$ acts on the clique-cube complex $C_\Gamma$ by left multiplication,
$h\cdot gA_U = (hg)A_U$.
This action preserves the cubical structure and is isometric.
The action is also co-compact with a fundamental domain $\bigcup_{U\in \Delta_\Gamma}[A_\emptyset, A_U]$,
where $[A_\emptyset, A_U]$ is a $\# U$-dimensional cube spanned by two vertices $A_\emptyset$ and $A_U$ in $C_\Gamma$.
However, the action is not proper.
In fact the stabilizer of a vertex $gA_U$ is the conjugate subgroup
$gA_U g^{-1}$, so all vertices except translates of $A_\emptyset$ have infinite stabilizers.
We also note that $C_\Gamma$ is not a proper metric space since it contains infinite valence vertices.
Also $C_\Gamma$ has infinite diameter if and only if $\Gamma$ itself is a clique.

\begin{remark}[\cite{Charney} Section 2]\label{label}
Each edge in $C_\Gamma$ can be labeled with a generator in $V$.
For example, the edge between $gA_U$ and $gA_{U\sqcup \{v\}}$ is labeled by $v$.
Any two parallel edges in a cube have the same label,
so we can also label the hyperplane dual to such an edge by $v$
and say that such a hyperplane is \textit{of $v$-type}.
It is easy to see that every hyperplane of $v$-type is the translate of a hyperplane
dual to the edge between $A_\emptyset$ and $A_{\{v\}}$.
If a hyperplane of $v$-type crosses another hyperplane of $v'$-type, then $(v,v')\in E(\Gamma)$.
In particular two different hyperplanes of the same type do not cross each other.
\end{remark}

\begin{theorem}[\cite{Charney} Theorem 2.2]
The clique-cube complex $C_\Gamma$ is CAT(0) for any graph $\Gamma$.
\end{theorem}

\begin{lemma}[\cite{Charney} Lemma 2.3]\label{linkGamma}
In the clique-cube complex $C_\Gamma$, the link of the vertex
$A_\emptyset$ is isomorphic to
the flag simplicial complex whose $1$-skeleton is $\Gamma$.
\end{lemma}

\begin{lemma}[\cite{Charney} Lemma 2.4]
If the clique-cube complex
$C_\Gamma$ is reducible, that is,
decomposable as a product of two subcomplexes, then $\Gamma$ is decomposable (as a join).
In particular, if $\Gamma$ is indecomposable, then $C_\Gamma$ is irreducible.
\end{lemma}

More strongly we can show the following.
It is not directly used in Proof of Theorem \ref{main},
but it will be of independent interest.
\begin{proposition}\label{C_reducible_prop}
The following are equivalent:
\begin{itemize}
\item[(1)] $C_{\Gamma}$ is reducible;
\item[(2)] $A_\Gamma$ is reducible, that is, $\Gamma^t$ is connected.
In other words,
$\Gamma$ is decomposed as a join of two subgraphs such that all edges between them are labeled by $2$;
\item[(3)] In addition to (2), $C_{\Gamma}$ is a direct product of $C_{\Gamma'}$ and $C_{\Gamma''}$
when $\Gamma$ is decomposed as a join of two subgraphs $\Gamma'$ and $\Gamma''$
such that all edges between them are labeled by $2$.
\end{itemize}
\end{proposition}

\begin{proof}
$(3)\Rightarrow (1)$ is obvious. We show that $(1)\Rightarrow (2)\Rightarrow (3)$.

We show $(1)\Rightarrow (2)$. Suppose that $C_{\Gamma}$ is reducible.
First we show that $\Gamma$ is decomposed as a join of two subgraphs.
We fix two subcomplexes $C'$ and $C''$ of $C_{\Gamma}$ satisfying $C_{\Gamma}=C'\times C''$.
Then for any vertex $v=(v', v'')$ of $C_{\Gamma}=C'\times C''$,
$\mathrm{Lk}_{C_{\Gamma}}v$ is the join of $\mathrm{Lk}_{C'}v'$ and $\mathrm{Lk}_{C''}v''$.
Let $\Gamma_v$, $\Gamma'_v$ and $\Gamma''_v$ be $1$-skeletons of
$\mathrm{Lk}_{C}v$, $\mathrm{Lk}_{C'}v'$ and $\mathrm{Lk}_{C''}v''$, respectively.
Then $\Gamma_v$ is the join of $\Gamma'_v$ and $\Gamma''_v$.
In particular $\Gamma_{A_{\emptyset}}$ is
the join of $\Gamma'_{A_{\emptyset}}$ and $\Gamma''_{A_{\emptyset}}$.
We put $\Gamma':=\Gamma'_{A_{\emptyset}}$ and $\Gamma'':=\Gamma''_{A_{\emptyset}}$.
Since $\Gamma$ is isomorphic to $\Gamma_{A_{\emptyset}}$ by Lemma \ref{linkGamma},
$\Gamma$ is regarded as the join of $\Gamma'$ and $\Gamma''$.
So far, the argument is based on Proof of Lemma \ref{linkGamma} in \cite{Charney}.

Next we show that all edges between $\Gamma'$ and $\Gamma''$ are labeled by $2$.
Assume that we have $s\in V(\Gamma')$ and $t\in V(\Gamma'')$
such that $e=(s,t)$ is an edge of $\Gamma$ with label $m$ greater than $2$.
Let us consider three squares
$[A_{\emptyset}, A_{\{s,t\}}]$, $[sA_{\emptyset}, A_{\{s,t\}}]$ and $[tA_{\emptyset}, A_{\{s,t\}}]$
around $A_{\{s,t\}}$. Then we have a $3$-line subgraph of $\Gamma_{A_{\{s,t\}}}$
corresponding to these three squares (see Figure \ref{3-line_fig}).
\begin{figure}
\begin{center}
\includegraphics[width=12cm,pagebox=cropbox,clip]{nosquare.pdf}
\caption{A $3$-line full subgraph of $\Gamma_{A_{\{s,t\}}}$.}\label{3-line_fig}
\end{center}
\end{figure}
The $3$-line subgraph is a full subgraph of $\Gamma_{A_{\{s,t\}}}$,
because it follows from Lemma \ref{toy}, shown in the next section, that
there exists no square in $C_{\Gamma}$
containing both edges $[A_{\{s,t\}},tA_{\{s\}}]$ and $[A_{\{s,t\}},sA_{\{t\}}]$.
Since $C_\Gamma$ is $C'\times C''$ and edges $[A_{\emptyset}, A_{\{s\}}]$ and $[A_{\emptyset}, A_{\{t\}}]$
of $C_{\Gamma}$ correspond to edges of $C'\times \{A_\emptyset\}$ and $\{A_\emptyset\}\times C''$ respectively,
edges $[A_{\{t\}}, A_{\{s,t\}}]$ and $[A_{\{s\}}, A_{\{s,t\}}]$ of $C_{\Gamma}$ must correspond to
edges of $C'\times \{A_{\{t\}}\}$ and $\{A_{\{s\}}\}\times C''$ respectively.
Thus two middle vertices of the $3$-line full subgraph of $\Gamma_{A_{\{s,t\}}}=\Gamma'_{A_{\{s,t\}}}\ast \Gamma''_{A_{\{s,t\}}}$
belong to $\Gamma'_{A_{\{s,t\}}}$ and $\Gamma''_{A_{\{s,t\}}}$ respectively.
This contradicts the fact that any $3$-line full subgraph of a join of two graphs is contained in either of the join factors.

We show $(2)\Rightarrow (3)$.
Suppose that $\Gamma$ is decomposed as a join of two subgraphs $\Gamma'$ and $\Gamma''$
such that all edges between them are labeled by $2$. Then we have a bijection
\begin{equation}\label{delta}
\Delta_{\Gamma'}\times \Delta_{\Gamma''}\to \Delta_\Gamma;(T',T'')\mapsto T'\sqcup T''.
\end{equation}
Also since $A_{\Gamma}$ is a direct product of subgroups $A_{\Gamma'}$ and $A_{\Gamma''}$,
we have a group isomorphism
\begin{equation}\label{group}
A_{\Gamma'}\times A_{\Gamma''}\to A_\Gamma; (g',g'')\mapsto g'g''.
\end{equation}
Clearly (\ref{delta}) and (\ref{group}) imply a bijection from
vertices of $C_{\Gamma'}\times C_{\Gamma''}$ to vertices of $C_\Gamma$
\begin{equation*}
\phi^0:C_{\Gamma'}^0\times C_{\Gamma''}^0\to C_{\Gamma}^0;(g'A_{T'},g''A_{T''})\mapsto g'g''A_{T'\sqcup T''}.
\end{equation*}
Then this extends to a cubical isomorphism
\begin{equation*}
\phi:C_{\Gamma'}\times C_{\Gamma''}\to C_{\Gamma}.
\end{equation*}
such that $\phi([g'A_{T'}, g'A_{U'}]\times [g''A_{T''}, g''A_{U''}])=[g'g''A_{T'\cup T''}, g'g''A_{U'\cup U''}]$
for any $(T',T'')\in \Delta_{\Gamma'}\times \Delta_{\Gamma''}$ and
$(g',g'')\in A_{\Gamma'}\times A_{\Gamma''}$.
\end{proof}

\begin{lemma}[\cite{Charney} Lemma 3.2]\label{CM Lemma 3.2}
Suppose that $\Gamma$ is not a cone.
Then the action of $A_\Gamma$ on $C_\Gamma$ is minimal.
That is, for any point $x\in C_\Gamma$, we have
$Hull(A_\Gamma x)=C_\Gamma$
(the convex hull of the orbit of $x$ is all of $C_\Gamma$).
\end{lemma}

\begin{proposition}\label{finite normal subgroup}
Suppose that $\Gamma$ is not a cone.
Then a finite normal subgroup of $A_\Gamma$ is trivial.
In particular a finite center of  $A_\Gamma$ is trivial.
Also if $A_\Gamma$ is isomorphic to a direct product $A_1\times A_2$ and
$A_1$ is finite, then $A_1$ is trivial.
\end{proposition}
\begin{proof}
Let $N$ be a finite normal subgroup of $A_\Gamma$.
Put $$Fix(N):=\{x\in C_\Gamma | \ nx=x\text{ for any }n\in N \}.$$
Since $N$ is finite and $C_\Gamma$ is a complete CAT(0) space, we have $Fix(N)\neq \emptyset$.
Take any $x\in Fix(N)$.
Then $A_\Gamma x\subset Fix(N)$.
Indeed normality of $N$ implies that for any $g\in A_\Gamma$ and $n\in N$, there exists $n'\in N$
such that $ng=gn'$ and therefore we have  $ngx=gn'x=gx$.
Since $C_\Gamma$ is CAT(0), $Fix(N)$ is convex. Hence we have $ Hull(A_\Gamma x)\subset Fix(N)$.
By Lemma \ref{CM Lemma 3.2}, we have $Hull(A_\Gamma x)=C_\Gamma$.
Hence $Fix(N)=C_\Gamma$. In particular $A_\emptyset \in Fix(N)$.
In general $A_\emptyset$ is not fixed by any nontrivial element of $A_\Gamma$.
Hence $N$ must be trivial.
\end{proof}

\section{Lemmas on local geometry of clique-cube complexes}\label{local_geom}
In this section we give two lemmas related to local geometry of clique-cube complexes.
First one is used in Proof of Proposition \ref{C_reducible_prop}.
Second one is used in Proof of Theorem \ref{main}.

We recall that the dihedral group for any $r\in \mathbb{N}$ is defined as
$$I_2(r):=\left\{
\begin{array}{ll}
\langle s,t \mid s^2=1, t^2=1, st\cdots s =ts\cdots t \ (\text{length } r)\rangle & \text{ if }r\text{ is odd},\\
\langle s,t \mid s^2=1, t^2=1, st\cdots t =ts\cdots s \ (\text{length } r)\rangle & \text{ if }r\text{ is even}. \\
\end{array}
\right.
$$
It is well-known that $\#I_2(r)=2r$.

Let $A_\Gamma$ be an Artin group associated to a defining graph $\Gamma$
as in Subsection \ref{defining_graph}.

\begin{lemma}\label{toy}
Let $e=(s,t)$ be an edge of $\Gamma$ with label $m$ greater than $2$.
Then there exists no square in $C_{\Gamma}$
containing both edges $[A_{\{s,t\}},tA_{\{s\}}]$ and $[A_{\{s,t\}},sA_{\{t\}}]$, that is,
there exist no $p,q\in \mathbb{Z}$ such that $ts^p=st^q$ in $A_\Gamma$.
\end{lemma}
\noindent See Figure \ref{3-line_fig}.
\begin{proof}
Assume that we have a square in $C_{\Gamma}$
containing both edges $[A_{\{s,t\}},tA_{\{s\}}]$ and $[A_{\{s,t\}},sA_{\{t\}}]$.
Then the square have four vertices $A_{\{s,t\}}$, $tA_{\{s\}}$, $sA_{\{t\}}$ and
$ts^pA_\emptyset=st^qA_\emptyset$
where $p,q$ are some integers.
Then we have $p=q$.
Indeed if we consider the projection $A_{\{s,t\}}\to A_{\{s\}}\cong\mathbb{Z}$
defined by $s\mapsto s$ and $t\mapsto s$,
then $ts^p=st^q$ in $A_{\{s,t\}}$ implies $s^{1+p}=s^{1+q}$ in $A_{\{s\}}$ and thus $p=q$.
Hence we have $ts^p=st^p$ in $A_{\{s,t\}}$.
By the natural projection $A_{\{s,t\}}\to W_{\{s,t\}}$,
$ts^p=st^p$ in $A_{\{s,t\}}$ implies
$ts=st$ in $W_{\{s,t\}}$ if $p$ is odd,
and also $t=s$ in $W_{\{s,t\}}$ if $p$ is even.
Both cases contradict the fact that
we have $ts\neq st$ and $s\neq t$ in $W_{\{s,t\}}$
in the case where $m>2$.
\end{proof}
\begin{lemma}\label{twist}
Let $e=(s,t)$ be an edge of $\Gamma$ with label $m$ greater than $2$.
Let $\tau$ be an alternating word
$$\tau:=st\cdots s\left\{
\begin{array}{l}
\text{with length }m \text{ if }m\text{ is odd}, \\
\text{with length }m+1 \text{ if }m\text{ is even}.\\
\end{array}
\right.
$$
Let $U\in \Delta_\Gamma$ with $s,t\in U$.
Then there exists no square in $C_{\Gamma}$
containing both edges $[A_U,A_{U\setminus\{s\}}]$ and $[A_U,\tau A_{U\setminus\{t\}}]$, that is,
there exists no $g\in A_{U\setminus\{s\}}$ such that $gA_{U\setminus\{t\}}=\tau A_{U\setminus\{t\}}$.
Also, there exists no square in $C_{\Gamma}$
containing both edges $[A_U,A_{U\setminus\{t\}}]$ and $[A_U,\tau A_{U\setminus\{s\}}]$, that is,
there exists no $g\in A_{U\setminus\{t\}}$ such that $gA_{U\setminus\{s\}}=\tau A_{U\setminus\{s\}}$.
\end{lemma}
\noindent See Figure \ref{square_fig}.
\begin{figure}
\begin{center}
\includegraphics[width=12cm,pagebox=cropbox,clip]{square.pdf}
\caption{Squares around $A_{U}$.}\label{square_fig}
\end{center}
\end{figure}
\begin{proof}
We show this by induction on $k=\#U$.

(1) We treat the base case where $k=2$.
In this case, $U=\{s,t\}$.
It is enough to show that for any $p\in \mathbb{Z}$,
$t^p A_{U\setminus\{t\}}\neq \tau A_{U\setminus\{t\}}$ and $s^p A_{U\setminus\{s\}}\neq \tau A_{U\setminus\{s\}}$.

(1-1) Suppose that $m$ is odd.
Note that $\tau=st\cdots s$ is equal to an alternating word $ts\cdots t$ with length $m$ in
$$A_{\{s,t\}}=\langle s,t \mid st\cdots s =ts\cdots t \ (\text{length } m)\rangle= A_U\subset  A_\Gamma.$$
We assume that there exists $p\in \mathbb{Z}$
such that $t^p A_{U\setminus\{t\}}=\tau A_{U\setminus\{t\}}$.
Then we have $t^{-p}\tau \in A_{U\setminus\{t\}}$.
On the other hand clearly we have  $t^{-p}\tau\in A_{\{s,t\}}$.
Since $U\setminus\{t\}=\{s\}$,
there exists $q\in \mathbb{Z}$ such that $t^ps^q=\tau$ in $A_{\{s,t\}}$.
Then we have $p+q=m$.
Indeed if we consider the projection $A_{\{s,t\}}\to A_{\{s\}}\cong\mathbb{Z}$
defined by $s\mapsto s$ and $t\mapsto s$,
then $t^ps^q=\tau$ in $A_{\{s,t\}}$ implies $s^{p+q}=s^m$ in $A_{\{s\}}$ and thus $p+q=m$.
Hence we have $t^ps^{m-p}=st\cdots s=ts\cdots t$ in $A_{\{s,t\}}$.

(1-1-1)
We treat the case where $p$ is odd.
By the natural projection $A_{\{s,t\}}\to W_{\{s,t\}}$,
$t^ps^{m-p}=ts\cdots t$ in $A_{\{s,t\}}$ and
$t^ps^{m-p}=t$ in $W_{\{s,t\}}$ imply
$t=ts\cdots t$ in $W_{\{s,t\}}$.
Thus we have $1=s\cdots t$ in $W_{\{s,t\}}$.
This means that we have a projection $I_2((m-1)/2)\to W_{\{s,t\}}\cong I_2(m)$
defined by $s\mapsto s$ and $t\mapsto t$.
Thus we have $m-1=\#I_2((m-1)/2)\ge \#I_2(m)=2m$. This contradicts $m> 2$.

(1-1-2) We treat the case where $p$ is even.
Then $t^ps^{m-p}=st\cdots s$ in $A_{\{s,t\}}$ and
$t^ps^{m-p}=s$ in $W_{\{s,t\}}$ imply
$s=st\cdots s$ in $W_{\{s,t\}}$.
Thus we have $1=s\cdots t$ in $W_{\{s,t\}}$.
This means that we have a projection $I_2((m-1)/2)\to W_{\{s,t\}}\cong I_2(m)$
defined by $s\mapsto s$ and $t\mapsto t$.
Thus we have $m-1=\#I_2((m-1)/2)\ge \#I_2(m)=2m$. This contradicts $m> 2$.

By (1-1-1) and (1-1-2), for any $p\in \mathbb{Z}$, we have $t^p A_{U\setminus\{t\}}\neq \tau A_{U\setminus\{t\}}$.
Also by the same argument, for any $p\in \mathbb{Z}$, we have
$s^p A_{U\setminus\{s\}}\neq \tau A_{U\setminus\{s\}}$.

(1-2) Suppose that $m$ is even.
Note that alternating words $st\cdots t$ and $ts\cdots s$ with length $m$ are equal in
$$A_{\{s,t\}}=\langle s,t \mid st\cdots t =ts\cdots s \ (\text{length } m)\rangle= A_U\subset  A_\Gamma.$$
We assume that there exists $p\in \mathbb{Z}$
such that $t^p A_{U\setminus\{t\}}=\tau A_{U\setminus\{t\}}$.
Then it follows from the same argument as in (1-1) that
there exists $q\in \mathbb{Z}$ such that $t^ps^q=\tau$ in $A_{\{s,t\}}$.
Then we have $p+q=m+1$.
Indeed if we consider the projection $A_{\{s,t\}}\to A_{\{s\}}\cong\mathbb{Z}$
defined by $s\mapsto s$ and $t\mapsto s$,
then $t^ps^q=\tau$ in $A_{\{s,t\}}$ implies $s^{p+q}=s^{m+1}$ in $A_{\{s\}}$ and thus $p+q=m+1$.
Hence we have $t^ps^{m+1-p}=st\cdots s$ in $A_{\{s,t\}}$.
This implies $t^ps^{m-p}=st\cdots t$ in $A_{\{s,t\}}$.
Thus we have $t^ps^{m-p}=st\cdots t=ts\cdots s$ in $A_{\{s,t\}}$.

(1-2-1) We treat the case where $p$ is odd.
By the natural projection $A_{\{s,t\}}\to W_{\{s,t\}}$,
$t^ps^{m-p}=ts\cdots s$ in $A_{\{s,t\}}$ and
$t^ps^{m-p}=ts$ in $W_{\{s,t\}}$ imply
$ts=ts\cdots s$ in $W_{\{s,t\}}$.
Thus $1$ and an alternating word $s\cdots t$ with length $m-2$ are equal in $W_{\{s,t\}}$.
This means that we have a projection $I_2((m-2)/2)\to W_{\{s,t\}}\cong I_2(m)$
defined by $s\mapsto s$ and $t\mapsto t$.
Thus we have $m-2=\#I_2((m-2)/2)\ge \#I_2(m)=2m$. This contradicts $m> 2$.

(1-2-2)
We treat the case where $p$ is even.
Then $t^ps^{m-p}=st\cdots t$ in $A_{\{s,t\}}$ and
$t^ps^{m-p}=1$ in $W_{\{s,t\}}$ implies
$1=st\cdots t$ in $W_{\{s,t\}}$.
This means that we have a projection $I_2(m/2)\to W_{\{s,t\}}\cong I_2(m)$
defined by $s\mapsto s$ and $t\mapsto t$.
Thus we have $m=\#I_2(m/2)\ge \#I_2(m)=2m$. This contradicts $m> 2$.

By (1-2-1) and (1-2-2), for any $p\in \mathbb{Z}$, we have $t^p A_{U\setminus\{t\}}\neq \tau A_{U\setminus\{t\}}$.
Also by the same argument, for any $p\in \mathbb{Z}$, we have
$s^p A_{U\setminus\{s\}}\neq \tau A_{U\setminus\{s\}}$.

(2) Suppose that $k>2$ and the statement is true for $k-1$.
Let $U=\{u_1, \ldots, u_{k-2}, s, t\}$, where $\# U=k$.
Assume that there exists $g\in A_{U\setminus\{s\}}$ such that $gA_{U\setminus\{t\}}=\tau A_{U\setminus\{t\}}$.
Then we have a square $[gA_{U\setminus\{s,t\}}, A_U]$ in $C_{\Gamma}$, which vertices are
$A_U$, $\tau A_{U\setminus\{t\}}$, $gA_{U\setminus\{s,t\}}$ and $A_{U\setminus\{s\}}$.
Since $C_{\Gamma}$ is CAT(0),
this square spans a cube together with other two squares $[A_{U\setminus\{u_1, s\}}, A_{U}]$ and $[\tau A_{U\setminus\{u_1, t\}}, A_{U}]$.
See Figure \ref{local_fig}.

\begin{figure}
\begin{center}
\includegraphics[width=12cm,pagebox=cropbox,clip]{squarecube.pdf}
\caption{Three squares around $A_U$ spans a cube.}\label{local_fig}
\end{center}
\end{figure}

Then this cube contains another square $[g_1A_{U\setminus\{u_1, s, t\}}, A_{U\setminus\{u_1\}}]$ as a face
where $g_1\in A_{U\setminus\{u_1, s\}}=A_{(U\setminus\{u_1\})\setminus\{s\}}$.
Then we have $g_1 A_{(U\setminus\{u_1\})\setminus\{t\}}=\tau A_{(U\setminus\{u_1\})\setminus\{t\}}$.
Since $\#(U\setminus\{u_1\})=k-1$,
this contradicts the inductive assumption.
\end{proof}

\section{Proof of Theorem \ref{main}}\label{main_proof}
In this section let $A_\Gamma$ be an Artin group associated to
$\Gamma$ with at least three vertices, and suppose that $\Gamma$ is not a cone.

\subsection{Proof of $(2)\Rightarrow(3),(3)\Rightarrow(4),(4)\Rightarrow(1)$ in Theorem \ref{main}}\label{part}
We show that $(2)\Rightarrow(3),(3)\Rightarrow(4),(4)\Rightarrow(1)$ in Theorem \ref{main}.

First $(2)\Rightarrow(3)$ follows from Theorem \ref{BBF}.

Next we show $(3)\Rightarrow(4)$.
Let $A_\Gamma$ be acylindrically hyperbolic. If $A_\Gamma$ is isomorphic to a direct product $A_1\times A_2$, then
either of them is finite by acylindrical hyperbolicity (\cite[Corollary 7.3]{Osin}).
Proposition \ref{finite normal subgroup} implies that the finite one is trivial.
Hence $A_\Gamma$ is directly indecomposable.

Finally $(4)\Rightarrow(1)$ is clear.

\subsection{Proof of $(1)\Rightarrow(2)$ in Theorem \ref{main}}
In this subsection, we give a proof of $(1)\Rightarrow(2)$ in Theorem \ref{main}.
Suppose that $A_\Gamma$ is irreducible.
We will construct an element $\gamma\in A_\Gamma$ and
show that $\gamma$ is a WPD contracting element with respect to the action on the clique-cube complex.
When $\Gamma$ is indecomposable, that is, not a join, a WPD contracting element is already given in \cite[Remark 4.5]{Charney}.
Hence it is enough to treat the case where $\Gamma$ is decomposable.
From now on we suppose that $\Gamma$ is decomposable.

We take a unique join decomposition $\Gamma=\ast_{1\le i\le k}\Gamma_i$
into indecomposable factors (see Lemma \ref{join-decomposition_lem}).
We put $V_i=V(\Gamma_i)$ and $E_i=E(\Gamma_i)$ for each $i\in\{1,\ldots,k\}$.
Then for every $i\in\{1,\ldots,k\}$, the complement graph $(\Gamma_i)^c$ of $\Gamma_i$
is connected and the complement graph $\Gamma^c$ of $\Gamma$ is presented
as the disjoint union of connected components $\Gamma^c=\bigsqcup_{1\le i \le k}(\Gamma_i)^c$.

We define $Q(\Gamma)$ as a finite simple graph with
\begin{itemize}
\item the vertex set $V(Q(\Gamma))=\{V_i\}_{1\leq i \leq k}$, and
\item the edge set $E(Q(\Gamma))=\{(V_i,V_j), (V_j,V_i) \mid 1\le i<j\le k,(\Gamma_i\ast\Gamma_j)^t\text{ is connected}\}$
\end{itemize}
Note that $(\Gamma_i)^t$ is connected since $(\Gamma_i)^c$ is connected,
$V((\Gamma_i)^c)=V((\Gamma_i)^t)=V_i$ and $E((\Gamma_i)^c)\subset E((\Gamma_i)^t)$.
Also we remark that the following are equivalent for different $i, j\in \{1,\ldots,k\}$:
\begin{itemize}
\item $(\Gamma_i\ast\Gamma_j)^t$ is connected;
\item there exists an edge with label greater than $2$ between a vertex of $\Gamma_i$ and a vertex of $\Gamma_j$.
\end{itemize}

We confirm our setting:
\begin{enumerate}
\item $\Gamma$ is decomposable, that is, $\Gamma^c$ is not connected, that is, $k\ge 2$;
\item $\Gamma$ is not a cone, that is, every connected component of $\Gamma^c$ has at least two vertices,
that is, $\#\Gamma_i\ge 2$ for any $i\in\{1,\ldots,k\}$;
\item $A_\Gamma$ is irreducible, that is, $\Gamma^t$ is connected, that is, $Q(\Gamma)$ is connected.
\end{enumerate}

We take a spanning tree $T$ of $Q(\Gamma)$.
We regard $T$ as a rooted tree with the root $V_1$.
By trading indices of $V_2,\ldots,V_k$ if necessary,
we suppose that $i<j$ if $V_i$ is not farer than $V_j$ from $V_1$ in $T$.
For each $i,j$ with $1\le i,j\le k$ and $(V_i,V_j)\in E(T)$,
take an edge $e_{i,j}\in E(\Gamma)$ with label $m_{i,j}=\mu(e_{i,j})$ greater than $2$
with $s_{i,j}:=s_{e_{i,j}}\in V(\Gamma_i)$ and $t_{i,j}:=t_{e_{i,j}}\in V(\Gamma_j)$
and assume $e_{j,i}:=\iota(e_{i,j})$.
For any $i,j\in \{1,\ldots,k\}$ with $(V_i,V_j)\in E(T)$,
let $\tau_{i,j}$ be an alternating word
$$\tau_{i,j}=s_{i,j}t_{i,j}\cdots s_{i,j}$$ with length $m_{i,j}$ if $m_{i,j}$ is odd,
and let $\tau_{i,j}$ be an alternating word
$$\tau_{i,j}=s_{i,j}t_{i,j}\cdots s_{i,j}$$ with length $m_{i,j}+1$ if $m_{i,j}$ is even.

\begin{lemma}\label{closed}
We have $n\in \mathbb{N}$,
for each $i\in \{1,\ldots,k\}$,
a closed path
$(v_{i,1},\ldots,v_{i, n},v_{i, n+1})$ with $v_{i,1}=v_{i,n+1}$ on $(\Gamma_{i})^c$
passing through every vertex at least once,
and for any $i, j\in \{1,\ldots,k\}$ with $(V_i,V_j)\in E(T)$,
$l(i,j)\in\{1,\dots,n\}$ such that $(v_{i,l(i,j)},v_{j,l(i,j)})=(s_{i,j},t_{i,j})(=e_{i,j})$
and $l(j,i)=l(i,j)$.
\end{lemma}
\begin{proof}
For any $i\in \{1,\ldots,k\}$,
we consider the minimum $n_i$ of length of closed paths
on $(\Gamma_{i})^c$ passing through every vertex at least once.
Put $n=\prod_{1\le i\le k}n_i$.
Then for any $i\in \{1,\ldots,k\}$,
by concatenating $\frac{n}{n_i}$ copies
of a closed path with length $n_i$ on $(\Gamma_{i})^c$
passing through every vertex at least once,
we have a closed path $(v'_{i,1},\ldots,v'_{i,n},v'_{i, n+1})$
with $v'_{i,1}=v'_{i,n+1}$ on $(\Gamma_{i})^c$
passing through every vertex at least $\frac{n}{n_i}$ times (in particular at least once).

We put $(v_{1,1},\ldots,v_{1, n},v_{1, n+1}):=(v'_{1,1},\ldots,v'_{1, n},v'_{1, n+1})$.
For any $j\in \{2,\ldots,k\}$, we define inductively
$(v_{j,1},\ldots,v_{j, n},v_{j, n+1})$ as follows.
Take $j\in \{2,\ldots,k\}$.
Assume that $(v_{i,1},\ldots,v_{i, n},v_{i, n+1})$
is defined for $i\in \{1,\ldots, j-1\}$ with $(V_i,V_j)\in E(T)$.
Then we put by $l(i,j)$ the minimum $l$ such that $v_{i,l}=s_{i,j}$.
By a cyclic permutation of  $v'_{j,1},\ldots,v'_{j, n}$,
we have $v_{j,1},\ldots,v_{j, n}$ such that $v_{j,l(i,j)}=t_{i,j}$.
By putting $v_{j,n+1}:=v_{j,1}$, we have
$(v_{j,1},\ldots,v_{j, n},v_{j, n+1})$.
For any $i,j$ with $1\le i<j\le k$ and $(V_i,V_j)\in E(T)$, we put $l(j,i):=l(i,j)$.
\end{proof}

For $1\leq l\le n$, put
$$\lambda_l:=v_{1,l}v_{2,l}\cdots v_{k,l}.$$
For any $i, j\in \{1,\ldots,k\}$ with $(V_i,V_j)\in E(T)$, we define
$\lambda_l(i,j)$ as
$\lambda_l(i,j):=\lambda_l$ if $l\neq l(i,j)$. Also we define $\lambda_{l(i,j)}(i,j)$ as
$$\lambda_{l(i,j)}(i,j):=
\begin{array}{lll}
\tau_{i,j} v_{1,l(i,j)}v_{2,l(i,j)}\cdots v_{i-1,l(i,j)}v_{i+1,l(i,j)} & &\\
\ \ \cdots v_{j-1,l(i,j)}v_{j+1,l(i,j)}\cdots v_{k,l(i,j)}
\end{array}
$$
if $i<j$, and $\lambda_{l(i,j)}(i,j):=\lambda_{l(j,i)}(j,i)$ if $i>j$,
where we note that $l(j,i)=l(i,j)$.
Moreover we define
$\gamma(i,j)$ as
\begin{align*}
\gamma(i,j):&=\lambda_1(i,j)\lambda_2(i,j)\cdots \lambda_{n}(i,j) \\
&=\lambda_{1}\lambda_{2}\cdots \lambda_{l(i,j)-1}\lambda_{l(i,j)}(i,j) \lambda_{l(i,j)+1}\cdots \lambda_{n}.
\end{align*}
Then we have $\gamma(j,i)=\gamma(i,j)$.

Take a closed path
\begin{equation}\label{pathtree}
(V_{i_1},\dots,V_{i_r},V_{i_{r+1}})
\end{equation}
with $i_1=i_{r+1}=1$
on the spanning tree $T$ of $Q(\Gamma)$
passing through every vertex at least once.
We define $\gamma$ as
\begin{equation}\label{gammaelement}
\gamma:=\gamma(i_1,i_2)\gamma(i_2,i_3)\cdots\gamma(i_r,i_{r+1}).
\end{equation}
We put
$$\gamma(0):=1, \gamma(1):=\lambda_1(i_1,i_2),\gamma(2):=\lambda_1(i_1,i_2)\lambda_2(i_1,i_2),\ldots,\gamma(n):=\gamma(i_1,i_2)$$
and for $a\in\{2,\ldots r\}$ and $l\in\{1,\ldots,n\}$, we put
$$\gamma((a-1)n+l):=\gamma(i_1,i_2)\cdots\gamma(i_{a-1},i_a)(\lambda_1(i_a,i_{a+1})\cdots\lambda_l(i_a,i_{a+1})).$$
In particular we have $\gamma(rn)=\gamma$.

We confirm the following for convenience:
\begin{enumerate}
\item $k$ is the number of indecomposable factors of a unique join decomposition of $\Gamma$;
\item $n$ is common length of the closed paths on $(\Gamma_i)^c$ for all $i\in\{1,\ldots,k\}$ taken in Lemma \ref{closed};
\item $r$ is length of the closed path (\ref{pathtree}) on $T$ taken above.
\end{enumerate}

For $1\leq l\le n$, put
$$U_l:=\{v_{1,l},v_{2,l},\ldots, v_{k,l}\}.$$
Then $U_l$ spans a clique in $\Gamma$, that is, $U_l\in \Delta_\Gamma$.
Hence we have a $k$-dimensional cube $[A_\emptyset, A_{U_l}]$ in $C_\Gamma$.
Hyperplanes dual to edges of $[A_\emptyset,A_{U_l}]$ are of $v_{i,l}$-type ($i\in \{1,\ldots,k\}$).
We denote by $H_{i,l}$ ($i\in \{1,\ldots,k\}$) such hyperplanes.

\begin{lemma}\label{key0}
For $i\in \{1,\ldots,k\}$, $l\in \{1,\ldots,n\}$ and $a\in\{1,\ldots,r\}$, we have the following:
\begin{enumerate}
\item $H_{i,l}\cap H_{i,l+1}=\emptyset$ where we put $H_{i,n+1}:=H_{i,1}$;
\item $H_{i,l}\cap \lambda_l H_{i,l}=\emptyset$;
\item $H_{i,l(i_{a},i_{a+1})}\cap \lambda_{l(i_{a},i_{a+1})}(i_{a},i_{a+1})H_{i,l(i_{a},i_{a+1})}=\emptyset$;
\item $[A_\emptyset,A_{U_l}]\cap [A_\emptyset,A_{U_{l+1}}]=\{A_\emptyset\}$ where we put $U_{n+1}:=U_{1}$;
\item $[A_\emptyset,A_{U_l}]\cap \lambda_l [A_\emptyset,A_{U_l}]=\{A_{U_l}\}$;
\item $[A_\emptyset,A_{U_{l(i_{a},i_{a+1})}}]\cap \lambda_{l(i_{a},i_{a+1})}(i_{a},i_{a+1})[A_\emptyset,A_{U_{l(i_{a},i_{a+1})}}]=\{A_{U_{l(i_{a},i_{a+1})}}\}$.
\end{enumerate}
\end{lemma}
\begin{proof}
(1) $H_{i,l}$ and $H_{i,l+1}$ are of $v_{i,l}$-type and of $v_{i,l+1}$-type, respectively.
Note that $v_{i,l}\neq v_{i,l+1}$ implies $H_{i,l}\neq H_{i,l+1}$.
Since  $(v_{i,l}, v_{i,l+1})\notin E(\Gamma)$, we have $\{v_{i,l}, v_{i,l+1}\}\notin \Delta_\Gamma$
and thus we see $H_{i,l}\cap H_{i,l+1}=\emptyset$.

(2) Since $[A_{U_l\setminus\{v_{i,l}\}},A_{U_l}]\subset N(H_{i,l})$,
we have $[\lambda_l A_{U_l\setminus\{v_{i,l}\}},\lambda_l A_{U_l}]\subset \lambda_l N(H_{i,l})$.
Note that $\lambda_l A_{U_l}=A_{U_l}$ and $\lambda_l A_{U_l\setminus\{v_{i,l}\}}=v_{1,l}\cdots v_{i,l} A_{U_l\setminus\{v_{i,l}\}}$.
Now assume that $H_{i,l}\cap \lambda_l H_{i,l}\neq \emptyset$.
Since $H_{i,l}$ and $\lambda_l H_{i,l}$ are of $v_{i,l}$-type, we see $H_{i,l}= \lambda_l H_{i,l}$ (see Remark \ref{label}).
Then we have
$$[\lambda_l A_{U_l\setminus\{v_{i,l}\}},\lambda_l A_{U_l}]=[v_{1,l}\cdots v_{i,l} A_{U_l\setminus\{v_{i,l}\}},A_{U_l}] \subset N(H_{i,l}).$$
Hence we have $A_{U_l\setminus\{v_{i,l}\}}=v_{1,l}\cdots v_{i,l} A_{U_l\setminus\{v_{i,l}\}}$.
This means $v_{i,l}\in A_{U_l\setminus\{v_{i,l}\}}$.
However it follows from \cite{van} that $\{v_{i,l}\}\cap (U_l\setminus\{v_{i,l}\})=\emptyset$ implies
$A_{\{v_{i,l}\}}\cap A_{U_l\setminus\{v_{i,l}\}}=A_\emptyset=\{1\}$.
This contradicts $v_{i,l}\neq 1$ in $A_\Gamma$.

(3) When $i\neq i_a, i_{a+1}$, by the same argument as in (2), we see that
$$H_{i,l(i_{a},i_{a+1})}\cap \lambda_{l(i_{a},i_{a+1})}(i_{a},i_{a+1})H_{i,l(i_{a},i_{a+1})}=\emptyset.$$
Now assume that $i=i_a$ or $i=i_{a+1}$ and
$$H_{i,l(i_a,i_{a+1})}\cap \lambda_{l(i_{a},i_{a+1})}(i_{a},i_{a+1})H_{i,l(i_{a},i_{a+1})}\neq \emptyset.$$
Note that
$\lambda_{l(i_{a},i_{a+1})}(i_{a},i_{a+1})A_{U_{l(i_{a},i_{a+1})}}=A_{U_{l(i_{a},i_{a+1})}}$,
and
$$\lambda_{l(i_{a},i_{a+1})}(i_{a},i_{a+1})A_{U_{l(i_{a},i_{a+1})}\setminus\{v_{i,l(i_{a},i_{a+1})}\}}=\tau_{i_a,i_{a+1}} A_{U_{l(i_{a},i_{a+1})}\setminus\{v_{i,l(i_{a},i_{a+1})}\}}.$$
Since $H_{i, l(i_{a},i_{a+1})}$ and $\lambda_{l(i_{a},i_{a+1})} H_{i,l(i_{a},i_{a+1})}$
are of $v_{i,l(i_{a},i_{a+1})}$-type, we see $H_{i,l(i_{a},i_{a+1})}= \lambda_{l(i_{a},i_{a+1})} H_{i,\lambda_{l(i_{a},i_{a+1})}}$.
Then we have
\begin{equation*}
\begin{split}
[\lambda_{l(i_{a},i_{a+1})} A_{U_{l(i_{a},i_{a+1})}\setminus\{v_{i,l(i_{a},i_{a+1})}\}},\lambda_{l(i_{a},i_{a+1})} A_{U_{l(i_{a},i_{a+1})}}]\\
=[\tau_{i_a,i_{a+1}} A_{U_{l(i_{a},i_{a+1})}\setminus\{v_{i,{l(i_{a},i_{a+1})}}\}},A_{U_{l(i_{a},i_{a+1})}}] \subset N(H_{i,{l(i_{a},i_{a+1})}}).
\end{split}
\end{equation*}
Hence we have
$$\tau_{i_a,i_{a+1}}A_{U_{l(i_a,i_{a+1})}\setminus\{v_{i,l(i_a,i_{a+1})}\}}=A_{U_{l(i_a,i_{a+1})}\setminus\{v_{i,l(i_a,i_{a+1})}\}}$$
This contradicts Lemma \ref{twist}.

(4), (5) and (6) follow from (1), (2) and (3), respectively.
\end{proof}

The following is a key lemma.
\begin{lemma}\label{key}
For $a\in\{1,\ldots,r\}$, we have
$$H_{i_a,l(i_a,i_{a+1})}\cap \lambda_{l(i_a,i_{a+1})}(i_a,i_{a+1})H_{i_{a+1},l(i_a,i_{a+1})}=\emptyset.$$
\end{lemma}
\begin{proof}
$H_{i_a,l(i_a,i_{a+1})}$ and $\lambda_{l(i_a,i_{a+1})}(i_a,i_{a+1})H_{i_{a+1},l(i_a,i_{a+1})}$
are of $v_{i_a,l(i_a,i_{a+1})}$-type and of $v_{i_{a+1},l(i_a,i_{a+1})}$-type, respectively.
Note that $v_{i_a,l(i_a,i_{a+1})}\neq v_{i_{a+1},l(i_a,i_{a+1})}$ implies
$H_{i_a,l(i_a,i_{a+1})}\neq \lambda_{l(i_a,i_{a+1})}(i_a,i_{a+1})H_{i_{a+1},l(i_a,i_{a+1})}$.
Note $$\lambda_{l(i_a,i_{a+1})}(i_a,i_{a+1})A_{U_{l(i_a,i_{a+1})}}=A_{U_{l(i_a,i_{a+1})}},$$
$$\lambda_{l(i_a,i_{a+1})}(i_a,i_{a+1})A_{U_{l(i_a,i_{a+1})}\setminus\{v_{i_{a+1},l(i_a,i_{a+1})}\}}
=\tau_{l(i_a,i_{a+1})}A_{U_{l(i_a,i_{a+1})}\setminus\{v_{i_{a+1},l(i_a,i_{a+1})}\}}.$$
Consider
$$[A_{U_{l(i_a,i_{a+1})}\setminus\{v_{i_a,l(i_a,i_{a+1})}\}}, A_{U_{l(i_a,i_{a+1})}}]
\subset N(H_{i_a,l(i_a,i_{a+1})}),$$
$$[\tau_{i_a,i_{a+1}}(i_a,i_{a+1})A_{U_{l(i_a,i_{a+1})}\setminus\{v_{i_{a+1},l(i_a,i_{a+1})}\}}, A_{U_{l(i_a,i_{a+1})}}]
\subset \lambda_{l(i_a,i_{a+1})}(i_a,i_{a+1})N(H_{i_{a+1},l(i_a,i_{a+1})}).$$
Now assume that
$$H_{i_a,l(i_a,i_{a+1})}\cap \lambda_{l(i_a,i_{a+1})}(i_a,i_{a+1})H_{i_{a+1},l(i_a,i_{a+1})}\neq \emptyset.$$
Since $$A_{U_{l(i_a,i_{a+1})}}\in N(H_{i_a,l(i_a,i_{a+1})})\cap \lambda_{l(i_a,i_{a+1})}(i_a,i_{a+1})N(H_{i_{a+1},l(i_a,i_{a+1})}),$$
two edges $$[A_{U_{l(i_a,i_{a+1})}\setminus\{v_{i_a,l(i_a,i_{a+1})}\}}, A_{U_{l(i_a,i_{a+1})}}] \text{ and}$$
$$[\tau_{i_a,i_{a+1}}(i_a,i_{a+1})A_{U_{l(i_a,i_{a+1})}\setminus\{v_{i_{a+1},l(i_a,i_{a+1})}\}}, A_{U_{l(i_a,i_{a+1})}}]$$
must span a square.
This contradicts Lemma \ref{twist}.
\end{proof}

Noting Lemma \ref{key0}, we define a sequence of hyperplanes for any $i\in\{1,\ldots,k\}$:
$$\ldots,J_{i,-1}, J_{i,0}, J_{i,1},\ldots,J_{i,2rn}, J_{i,2rn+1}, \ldots,$$
a sequence of $k$-dimensional cubes:
$$\ldots,K_{-1},K_0,K_1,\cdots,K_{2rn},K_{2rn+1},\ldots,$$
a sequence of vertices of $C_\Gamma$:
$$\ldots,w_{-1},w_0,w_1,\ldots,w_{2rn},w_{2rn+1},\ldots$$ as follows.
First we define for $a\in\{1,\ldots,r\}$ and $l\in\{1,\dots,n\}$,
\begin{align*}
J_{i,2((a-1)n+l)-1}&:=\gamma((a-1)n+l-1)H_{i,l},\\
J_{i,2((a-1)n+l)}&:=\gamma((a-1)n+l)H_{i,l},\\
K_{2((a-1)n+l)-1}&:=\gamma((a-1)n+l-1)[A_\emptyset,A_{U_l}],\\
K_{2((a-1)n+l)}&:=\gamma((a-1)n+l)[A_\emptyset,A_{U_l}],\\
w_{2((a-1)n+l)-1}&:=\gamma((a-1)n+l-1)A_{U_l},\\
w_{2((a-1)n+l)}&:=\gamma((a-1)n+l)A_\emptyset,\\
\end{align*}
Second we put for any $b\in\{1,\dots,2rn\}$ and $c\in \mathbb{Z}$,
$$J_{i,2rnc+b}:=\gamma^cJ_{i,b}, K_{2rnc+b}:=\gamma^cK_{b}, w_{2rnc+b}:=\gamma^cw_{b}.$$
Then we have $$K_{2rnc+b}=[w_{2rnc+b-1}, w_{2rnc+b}].$$
Also we have two connected components
$J_{i,2rnc+b}^-$ and $J_{i,2rnc+b}^+$ such that
$$C_\Gamma \!\setminus\!\!\setminus J_{i,2rnc+b}=J_{i,2rnc+b}^- \sqcup J_{i,2rnc+b}^+,$$
$$w_{2rnc+b-1}\in J_{i,2rnc+b}^-, w_{2rnc+b}\in J_{i,2rnc+b}^+.$$
Then Lemma \ref{key0} implies for any $i\in\{1,\ldots,k\}$,
$$\cdots\subsetneq J_{i,-1}^-\subsetneq  J_{i,0}^-\subsetneq J_{i,1}^-\subsetneq \cdots \subsetneq J_{i,2rn}^-\subsetneq J_{i,2rn+1}^-\subsetneq \cdots ,$$
$$\cdots\supsetneq J_{i,-1}^+\supsetneq J_{i,0}^+\supsetneq J_{i,1}^+\supsetneq \cdots \supsetneq J_{i,2rn}^+\supsetneq J_{i,2rn+1}^+\supsetneq \cdots.$$
Note that
$$J_{i,0}^-\not\ni w_0=A_{\emptyset}\in J_{i,1}^-,$$
$$J_{i,2rn}^+\ni w_{2rn}=\gamma A_\emptyset \not\in J_{i,2rn+1}^+.$$

Let $\ell$ be a path from  $w_0=A_\emptyset$ to $w_{2rn}=\gamma A_\emptyset$
which diagonally penetrates each of cubes $K_{1},\ldots,K_{2rn}$ in order.
Then the set of all hyperplanes intersecting with the path $\ell$ is
$\{J_{i,d}\}_{i\in\{1,\ldots,k\}, d\in\{1,\ldots,2rn\}}$.
Hence we have the following.
\begin{lemma}\label{key3}
The set
$\{J_{i,d}\}_{i\in\{1,\ldots,k\}, d\in\{1,\ldots,2rn\}}$
is the set of all hyperplanes separating $w_0=A_\emptyset$ and $w_{2rn}=\gamma A_\emptyset$.
\end{lemma}

Now we give the final lemma for Proof of Theorem \ref{main},
where recall Definition \ref{separatingdef} and Remark \ref{separatingrem}.
\begin{lemma}\label{key4}
\begin{enumerate}
\item For any $i\in\{1,\ldots,k\}$,
the sequence of hyperplanes $$J_{i,1},\ldots,J_{i,2rn}$$
is a sequence of separating hyperplanes
from $J_{i,0}$ to $J_{i,2rn+1}$, where $J_{i,0}$ and $J_{i,2rn+1}$
are of $v_{i,n}$-type and $v_{i,1}$-type, respectively
(in particular, from $w_0=A_\emptyset$ to $w_{2rn}=\gamma A_\emptyset$).
\item
The sequence of hyperplanes
\begin{align*}
&J_{i_1,1},J_{i_1,2},\ldots,J_{i_1,2l(i_1,i_{2})-1},\\
&J_{i_{2},2l(i_1,i_{2})},J_{i_{2},2l(i_1,i_{2})+1},
\ldots,J_{i_{2},2n},\\
&J_{i_{2},2n+1},J_{i_{2},2(n+1)},
\ldots,J_{i_{2},2(n+l(i_{2},i_{3}))-1},\\
&\cdots \\
&J_{i_{r},2((r-2)n+l(i_{r-1},i_{r}))},J_{i_{r},2((r-2)n+l(i_{r-1},i_{r}))+1},
\ldots,J_{i_{r},2(r-1)n},\\
&J_{i_{r},2(r-1)n+1},J_{i_{r},2((r-1)n+1)},
\ldots,J_{i_{r},2((r-1)n+l(i_r,i_{r+1}))-1},\\
&J_{i_{r+1},2((r-1)n+l(i_r,i_{r+1}))},J_{i_{r+1},2((r-1)n+l(i_r,i_{r+1}))+1},
\ldots,J_{i_{r+1},2rn}
\end{align*}
is a sequence of separating hyperplanes
from $J_{1,0}$ to $J_{1,2rn+1}$, where $J_{1,0}$ and $J_{1,2rn+1}$
are of $v_{1,n}$-type and $v_{1,1}$-type, respectively
(in particular, from $w_0=A_\emptyset$ to $w_{2rn}=\gamma A_\emptyset$).

Moreover the sequence contains a hyperplane of $v_i$-type for any $i\in\{1,\ldots,k\}$ and $v_i\in V(\Gamma_i)$.
\end{enumerate}
\end{lemma}
\begin{proof}
(1) The assertion is clear by Lemma \ref{key0} (1) (2).

(2)
Lemma \ref{key0} (1) (2) (3) and Lemma \ref{key} imply that
the given sequence in the assertion is a sequence of separating hyperplanes
from $J_{i_1,0}$ to $J_{i_{r+1},2rn+1}$.
Note that $i_1=1$ and $i_{r+1}=1$ by definition.

We show that the sequence contains
a hyperplane of $v_i$-type for any $i\in\{1,\ldots,k\}$ and $v_i\in V(\Gamma_i)$.
Recall that $(V_{i_1},\dots,V_{i_r},V_{i_{r+1}})$ taken as (\ref{pathtree}) is a closed path on $T$, which is a tree,
Thus if an edge $(V_i,V_j)\in E(T)$ is contained in the closed path, then so is the inverse edge $(V_j,V_i)$.
Also since $T$ is a spanning tree of $Q(\Gamma)$ and
the closed path $(V_{i_1},\dots,V_{i_r},V_{i_{r+1}})$ passes through every vertex at least once,
any $i\in \{1,\ldots ,k\}$, $V_i$ is contained in the closed path as a vertex.
Also recall that $T$ is the rooted tree with the root $V_1$
and that $i<j$ if $V_i$ is not farer than $V_j$ from $V_1$ in $T$.

Now take any $i\in\{1,\ldots,k\}$.
We consider two cases where $i=1$ and $i\neq 1$.

First suppose that $i=1$.
Note that $i_1=1$ and put $j=i_2$.
Take $a\in \{1,2,\ldots r\}$ such that $a\neq 1$, $i_a=1$ and $i_{a-1}=j$.
Then the sequence in the assertion contains two subsequences
\begin{align*}
J_{i_1,1},J_{i_1,2},\ldots,J_{i_1,2l(i_1,i_{2})-1},
\end{align*}
which are of $v_{1,1}$-type, of $v_{1,1}$-type, $\ldots$,
of $v_{1,l(1,j)}$-type, and
\begin{align*}
J_{i_{a},2((a-2)n+l(i_{a-1},i_{a}))},J_{i_{a},2((a-2)n+l(i_{a-1},i_{a}))+1},
\ldots,J_{i_{a},2(a-1)n},
\end{align*}
which are of $v_{1,l(j,1)}$-type, of $v_{1,l(j,1)+1}$-type, $\ldots$,
of $v_{1,n}$-type.
Note that $l(j,1)=l(1,j)$ by Lemma \ref{closed}.
Take any vertex $v_1\in V(\Gamma_1)$.
Then there exists some $l\in\{1,\ldots,n\}$ such that $v_{1,l}=v_1$ by Lemma \ref{closed}.
Hence the sequence  in the assertion contains a hyperplane of $v_1$-type.

Next suppose that $i\neq 1$.
Take the smallest $a\in \{1,\ldots r\}$ such that $i_a=i$.
Since $i_1=1$ and $i\neq 1$, we have $a\neq 1$.
We put $j=i_{a-1}$.
Then we have $a'\in \{1,\ldots r\}$ such that
$a\le a'$, $i_{a'}=i$ and $i_{a'+1}=j$.
Then the sequence contains two subsequences
\begin{align*}
J_{i_{a},2((a-2)n+l(i_{a-1},i_{a}))},J_{i_{a},2((a-2)n+l(i_{a-1},i_{a}))+1},
\ldots,J_{i_{a},2(a-1)n},
\end{align*}
which are of $v_{i,l(j,i)}$-type, of $v_{i,l(j,i)+1}$-type, $\ldots$,
of $v_{i,n}$-type, and
\begin{align*}
J_{i_{a'},2(a'-1)n+1},J_{i_{a'},2((a'-1)n+1)},
\ldots,J_{i_{a'},2((a'-1)n+l(i_{a'},i_{a'+1}))-1},
\end{align*}
which are of $v_{i,1}$-type, of $v_{i,1}$-type, $\ldots$,
of $v_{i,l(j,i)}$-type.
Note that $l(j,i)=l(i,j)$ by Lemma \ref{closed}.
Take any vertex $v_i\in V(\Gamma_i)$.
Then there exists some $l\in\{1,\ldots,n\}$ such that $v_{i,l}=v_i$ by Lemma \ref{closed}.
Hence the sequence in the assertion contains a hyperplane of $v_i$-type.
\end{proof}

\begin{figure}
\begin{center}
\includegraphics[width=12cm,pagebox=cropbox,clip]{sequence.pdf}
\caption{A part of a sequence of hyperplanes
for the case where $\Gamma=\Gamma_1\ast\Gamma_2$.}\label{sequence_fig}
\end{center}
\end{figure}

Now we can complete Proof of Theorem \ref{main} by Subsection \ref{part} and the following.
\begin{proof}[Proof of $(1)\Rightarrow(2)$ in Theorem \ref{main}]
We show $(1)\Rightarrow(2)$ in Theorem \ref{main}.
We consider $\gamma\in A_\Gamma$ of (\ref{gammaelement})
and hyperplanes $J=J_{1,0}$ and $J'=J_{1,2rn+1}$,
which are of $v_{1,n}$-type and $v_{1,1}$-type respectively.
We will confirm three conditions (i), (ii) and (iii) in Theorem \ref{criterion}.

(i) $\gamma$ skewers $(J,J')$, indeed it is clear that
$$J^+\supsetneq (\gamma^{-1}(J'^+)\supsetneq\gamma(J^+)\supsetneq) J'^+\supsetneq \gamma^2(J^+).$$

(ii) We show that $J$ and $J'$ are strongly separated.
Take any hyperplane $H$ with $J\cap H\neq \emptyset$.
When $H$ is of $v_i$-type for some $i$ and $v_i\in V(\Gamma_i)$,
take a hyperplane $H'$ of $v_i$-type
separating $A_\emptyset$ and $\gamma A_\emptyset$
such that $J\cap H'=\emptyset$ by Lemma \ref{key4} (2).
Then $H\cap H'=\emptyset$ by Remark \ref{label}.
Since $J\supset H'^-, H'^+\supset J'$ and  $J\cap H\neq \emptyset$,
$J'\cap H=\emptyset$.

(iii) We show $Stab(J)\cap Stab(J')=\{1\}$.
Assume that $Stab(J)\cap Stab(J')\neq \{1\}$.
Take $g\in Stab(J)\cap Stab(J')$ with $g\neq 1$.
Then we have a hyperplane $H$ separating with
$A_\emptyset$ and $gA_\emptyset$.

Then this $H$ is not of $v_1$-type for any $v_1\in V(\Gamma_1)$.
Indeed assume that $H$ is of $v_1$-type for some $v_1\in V(\Gamma_1)$.
Since $H\cap J\neq \emptyset$,
it follows from Lemma \ref{key4} (1) for the case where $i=1$
that $H$ does not separate
$A_\emptyset$ and $\gamma A_\emptyset$.
Note that we have at least one sequence of separating hyperplanes
$P'_1,\ldots,P'_{M'}$ from $A_\emptyset$ to $\gamma A_\emptyset$
such that $P'_1$ is of $v_1$-type and $P'_{M'}$ is of $v_{1,n}$-type.
For example, we can take such a sequence by considering a subsequence of
the sequence in Lemma \ref{key4} (1) for the case where $i=1$.
Now we take a longest sequence of separating hyperplanes
$P_1,\ldots,P_{M}$ from $A_\emptyset$ to $\gamma A_\emptyset$
such that $P_1$ is of $v_1$-type and $P_{M}$ is of $v_{1,n}$-type.
Then $Q_1=H, Q_2=P_1,\ldots,Q_{M+1}=P_M$ is a sequence of separating hyperplanes
from $gA_\emptyset$ to $g\gamma A_\emptyset$
such that $Q_1$ is of $v_1$-type and $Q_M$ is of $v_{1,n}$-type.
Then $g^{-1}Q_1, g^{-1}Q_2,\ldots,g^{-1}Q_{M+1}$ is a sequence of separating hyperplanes
from $A_\emptyset$ to $\gamma A_\emptyset$
such that $g^{-1}Q_1$ is of $v_1$-type and $g^{-1}Q_M$ is of $v_{1,n}$-type,
which contradicts the fact that $P_1,\ldots,P_M$ is longest.

We can suppose that $H$ is of $v_i$-type for some $v_i\in V(\Gamma_i)$ where $i\neq 1$.
Note that $H$ does not intersect with $J'$.
We can suppose that $A_\emptyset\in H^-$
by denoting by $H^-$ an appropriate connected component of $C_\Gamma\!\setminus\!\!\setminus H$.
Then
$H$ does not intersect with $\ell$, i.e.,
$$A_\emptyset, \gamma A_\emptyset,g \gamma A_\emptyset \in H^-  \text{ and } gA_\emptyset \in H^+$$
or $H$ intersects with $\ell$, i.e.,
$$A_\emptyset\in H^- \text{ and } gA_\emptyset, \gamma A_\emptyset,g \gamma A_\emptyset \in H^+.$$
Here $\ell$ is the path given above Lemma \ref{key3}.

Suppose that
$$A_\emptyset, \gamma A_\emptyset,g \gamma A_\emptyset \in H^-  \text{ and } gA_\emptyset \in H^+ .$$
Note that we have at least one sequence of separating hyperplanes
$P'_1,\ldots,P'_{M'}$ from $A_\emptyset$ to $\gamma A_\emptyset$
such that $P'_1$ is of $v_i$-type and $P'_{M'}$ is of $v_{1,n}$-type.
For example, we can take such a sequence by considering a subsequence of
the sequence in Lemma \ref{key4} (2).
Now take a longest sequence of separating hyperplanes
$P_1,\ldots,P_M$ from $A_\emptyset$ to $\gamma A_\emptyset$
such that $P_1$ is of $v_i$-type and $P_M$ is of $v_{1,n}$-type.
Then $Q_1=H, Q_2=P_1,\ldots,Q_{M+1}=P_M$ is a sequence of separating hyperplanes
from $gA_\emptyset$ to $g\gamma A_\emptyset$
such that $Q_1$ is of $v_i$-type and $Q_M$ is of $v_{1,n}$-type.
Then $g^{-1}Q_1, g^{-1}Q_2,\ldots,g^{-1}Q_{M+1}$ is a sequence of separating hyperplanes
from $A_\emptyset$ to $\gamma A_\emptyset$
such that $g^{-1}Q_1$ is of $v_i$-type and $g^{-1}Q_M$ is of $v_{1,n}$-type,
which contradicts the fact that $P_1,\ldots,P_M$ is longest.

Next suppose that
$$A_\emptyset\in H^- \text{ and } gA_\emptyset, \gamma A_\emptyset,g \gamma A_\emptyset \in H^+.$$
Note that $gA_\emptyset\in J$ and $g\gamma A_\emptyset\in J'$ as well as
$A_\emptyset\in J$ and $\gamma A_\emptyset\in J'$ since $g(J)=J$ and $g(J')=J'$.
Then we take a longest sequence of separating hyperplanes
$P_1,\ldots,P_M$ from $g A_\emptyset$ to $g\gamma A_\emptyset$
such that $P_1$ is of $v_i$-type and $P_M$ is of $v_{1,n}$-type
by using Lemma \ref{key4} (2).
Then $Q_1=H, Q_2=P_1,\ldots,Q_{M+1}=P_M$ is a sequence of separating hyperplanes
from $A_\emptyset$ to $\gamma A_\emptyset$
such that $Q_1$ is of $v_i$-type and $Q_M$ is of $v_{1,n}$-type.
Then $gQ_1, gQ_2,\ldots,gQ_{M+1}$ is a sequence of separating hyperplanes
from $gA_\emptyset$ to $g\gamma A_\emptyset$
such that $gQ_1$ is of $v_i$-type and $gQ_M$ is of $v_{1,n}$-type,
which contradicts the fact that $P_1,\ldots,P_M$ is longest.

Now we see $Stab(J)\cap Stab(J')=\{1\}$.
\end{proof}

\begin{remark}\label{mainremark}
In \cite[Theorem 3.3]{Charney},
it is shown that $A_\Gamma$ is centerless
under the setting in Theorem \ref{main} with (1),
This claim can be proved based on Theorem \ref{main}.
Indeed Theorem \ref{main} $(1)\Rightarrow(2)$ implies that $A_\Gamma$ is acylindrically hyperbolic.
Then the center of $A_\Gamma$ is finite by acylindrical hyperbolicity (\cite[Corollary 7.3]{Osin}).
Proposition \ref{finite normal subgroup} implies that it is trivial.
Hence $A_\Gamma$ is centerless.
\end{remark}

redpajama_set_name: RedPajamaArXiv


text:
NewsGermany Looks To Placate Brussels As Its Cannabis Plans...
FeaturedNewsRegulations
Germany Looks To Placate Brussels As Its Cannabis Plans Teeter
DELAYS in submitting its proposals to the European Commission and potential hold-ups once submitted could see Germany's adult-use cannabis plans pushed back into 2025.
Originally scheduled for submission before the end of 2022, Germany's draft cannabis law will not now be submitted for Europe's approval until the end of March.
With the European Commission (EC) still coming to terms with the rapid global changes in the approach to cannabis there are some concerns that Germany will need to work on a Plan B, or abandon the initiative, if its initial submission is rejected.
And, this is leading to calls for the decision-making process to include Europe's politicians in its Parliament and The Council of the European Union.
German Health Statement
In a statement to BusinessCann The German Federal Ministry of Health confirmed the draft law is currently 'being formulated within the federal government'.
It said it is also seeking 'expert opinion to shed light on the effects of a controlled sale of cannabis on health and youth protection and consumption', as well as undertaking 'a systematic review of the literature' pertaining to those countries who have some form of legalized recreational cannabis.
Going on to say that in the meantime it was continuing to talk to the EC with the above 'set to be completed by Spring 2023'.
Once submitted to the EC it will most probably be dealt with under an established protocol for enacting new member state legislation, known as the TRIS Notification Procedure.
This will see Germany's proposals subject to a three-month standstill period during which the initiating country cannot enact the domestic legislation until it has received feedback from the EC.
EC Power To Delay Plans
The EC, or another member state, can also submit a 'detailed opinion' which has the effect of extending the standstill period for a further three months. And, the EC can also block progress for a further 18 months.
In a statement to BusinessCann the Home Affairs Directorate of the European Commission reiterated that personal drug consumption was a matter for nation states.
It continued: "A formal notification has not yet been submitted by the German authorities. Therefore, since we have not yet received the formal German request for consultation, we cannot make further comments at this stage.
"The existing EU law lays down minimum criminal sanctions for illicit drug trafficking, and prohibits the cultivation of cannabis.
"We are aware and we are closely following these developments, notably to understand the impact changes in cannabis policies. This includes the impact on health, crime, environment or social aspects.
"Once the legislation is adopted, the Commission will examine its compliance with the EU acquis."
EC Cultivation Error
Keen-eyed readers will notice an obvious error in this statement which indicates some confusion within the EC on how to address the process.
This is the reference to the cultivation of cannabis being 'prohibited in EU law', which is plainly untrue with numerous medical cannabis cultivation facilities in situ across the continent.
Germany's outline plans were submitted to the EC in an Eckpunktepapier (cornerstone paper) in October last year, by the ruling Coalition, which came into power in late 2021.
However, the EC said it lacked detail and asked to see a draft of the law.
Germany's initial proposals recommend allowing those over the age of 18 to access cannabis through licensed stores with the aim of eliminating the illicit market and improving public health, in particular that of young people.
European cannabis cultivation
Niklas Kouparanis, CEO and Co-Founder of Bloomwell Group, one of Germany's largest cannabis businesses with 250 employees, acknowledges that the delays were due to the fact that the Eckpunktepapier was not sufficient for the EC.
The feedback from the EC prompted Germany to embark on two further exercises – the literature review and further expert input, which is being undertaken by a non-profit drug researchers ISD Hamburg – with the aim of 'demonstrating how prohibition has failed' and how this new law can 'protect public health', he said.
"Because of the notification by the EC, Germany will need to draft the whole law, as the 12 pages of Eckpunktepapier is not sufficient."
Battling Bureaucracy
Come this spring when the draft is submitted to the EC, what will possibly be the most important decision in the history of the European cannabis industry will rest with the EC officials in the Home Affairs Directorate headed by Monique Pariat.
Little is known of her views on cannabis and drugs, but with the politicians frozen out of the picture, for now, and Brussels bureaucrats tending to err to the status quo, it will irk many cannabis advocates if Germany's plans stumble at this hurdle.
There is also much domestic, anti-cannabis sentiment from various quarters such as the Police and Customs as well as the opposition Christian Democrats (CSU).
Bavaria's Health Minister Klaus Holetschek, of the CSU, met with Ms Pariat late last year to urge Brussels to say no.
And, Germany's Health Minister Karl Lauterbach, who is driving the reform, has made it clear that if the EC does say 'no', then that will more than likely be the end of its cannabis plans.
60/40 In The Betting
Mr Kouparanis believes Germany's plans will be approved by the EC, but there will be some to-ing and fro-ing between the two before a final draft is settled.
"There are two statements that the EC can provide to Germany one is that this cannot be implemented, and secondly these are our suggestions.
"I believe it will be the latter, but if there is a huge pushback from the EU, it could well delay the implementation of the law into 2025."
His best 'guesstimate' on the chances of success are 60/40 with the cannabis law coming into effect by Q1 or Q2 next year.
A third scenario is for the EC to reject the plans, the German coalition government to park the project and lose the 2025 election meaning the cannabis reform baton would be passed to fellow EU members, such as Czechia.
Finn Age Hänsel, Founder and Managing Director of leading Germany cannabis company Sanity Group says he would have 'loved things to go faster' but says it's 'better that it's done thoroughly and Germany has a good law'.
Adding: "Looking at the timeline now I would say Q4 2024. The notification process could take six months and then there is the German parliament approval and further details around licensing and cultivation which need to be ironed out."
Pathways To Progress
Many opponents have highlighted the obstacles to reform in the 2004 Schengen agreement, and European regulations around the freedom of movement of goods.
But, as BusinessCann has reported previously there are a number of pathways for Germany to introduce its legislation which allow it to comply with its international and European obligations.
As a signatory to the 1961 Single Convention on Narcotic Drugs (SCND), there are two possible compliant ways to proceed namely; under article 2 paragraph 9 which recognises the legality of using prohibited substances for industrial purposes, or, legalising as a scientific experiment.
The second, the scientific option, could see Germany introduce a trial for, say, a handful of years, with it being branded as an experiment, placing it out of the reaches of the EC or the UN.
Mr Kouparanis believes Germany needs to explore this option and is hopeful that Mr Lauterbach's air of resignation in relation to an EC 'No' is merely a guise. And, that his team is working up a boil-in-the-bag, Plan B, scientific experiment, which Germany could enact immediately.
He said: "As the biggest country in Europe Germany has to take the lead and over time – whether it's 2024 or not – we have to emphasise to fellow EU members that the number one priority is health and human protection. I believe this can only be done by drying up the illicit market.
"What we are doing could lead to a global change in cannabis perception with Europe saying we are now not working to the 1961 Convention anymore. This would also prompt change at the UN.
"Science is important. Switzerland and Holland are set to undertake trials which should provide the evidence required to get cannabis reform through on a political level.
Prohibition Does Not Work
"We have to convince the EC and member states that prohibition does not work and if the EC says no then we will have to take it to the political level."
Mr Hänsel agrees with his fellow countryman that the chances of success are currently around 60/40.
Sanity is well-connected in German political circles and from its discussions with the Government and politicians he believes Germany has a strong hand.
"The scientific study which is now being undertaken is an attempt to demonstrate comprehensively to Europe that for Germany, and other societies, it is better to have a legalised drug that is under control, rather than an illegal one which is not."
The Bloomwell team – Niklas Kouparanis, second left
He highlighted cannabis tourism as a main European concern. Many of Germany's neighbours such as Hungary and Poland, are expressing reservations.
One way to counter this is to permit sales solely to German residents, although this will also present the authorities with significant monitoring and enforcement challenges.
He added: "If Germany can get this right, and it gets the green light, we will see many other countries follow suit across Europe. So it's good that Germany is leading the way and tackling the challenges that face cannabis reform in Europe. This could eventually form the model for others to follow."
German cannabis lawyer Kai-Friedrich Niermann is following developments closely. He believes Germany will secure approval from the EC through the Notification process
He told BusinessCann: "No treaty under international law can ever oblige a country to prosecute its own citizens if it contradicts its own legal framework or constitutional principles.
"If the personal use and consumption of drugs does not violate the EU acquis, then logically the steps upstream for consumption, such as cultivation and trafficking, must also be possible under strictly controlled conditions.
"I assume that this understanding will prevail in the short to medium term in the official notification procedure."
Political Involvement
Last year witnessed the first co-ordinated pan-European efforts to present a united front on cannabis reform involving Germany, Czechia, Luxembourg, Malta and the Netherlands.
This led to the Council of the European Union adopting a new, human-rights based approach to drugs which has been branded a 'paradigm shift' which could smooth the pathway to continental cannabis reform.
This was driven by politicians so it's hard to imagine a scenario in which pro-cannabis political reformers in the EU will be totally frozen out of the decision-making process.
Mr Friedrich-Niermann agrees: "At the moment, it is difficult to foresee whether only the Commission will actually act on this matter.
"Parts of the European Parliament have already made it clear that they want to bring about a change in the legal framework for cannabis. The Council of the European Union could also become active here."
In 2021, Mr Niermann co-authored a paper on German cannabis reform which predicted legalisation by April 1, 2024 – as yet, it remains unclear whether that deadline will be met.
Nevertheless, with a fair wind there is a decent possibility that Europe will have spawned the world's largest, regulated cannabis market by the time the next Federal German elections take place in October 2025.
#Bloomwell Group
#European cannabis
#European Commission
#German cannabis
#Recreational cannabis
#Sanity Group
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text:
Weekend Box Office (March 9 - 11, 2007)
THIS WEEKEND The mighty Spartans won a glorious victory at North American theaters as the bloody war epic 300 exploded with a record-breaking opening and powered the overall marketplace to the biggest March weekend in box office history. Selling more tickets than all its enemies in the top ten combined, the ancient battle film exceeded even the loftiest of industry expectations conquering every multiplex it invaded. Despite the colossal strength of 300, holdovers performed well with most witnessing relatively small declines of 35% or less.
Capitalizing on intense pre-release anticipation, the Warner Bros. actioner 300 rallied to a staggering $70.9M opening weekend, according to final studio figures, ruling the box office with the greatest of ease. The violent and stylish R-rated tale played in only 3,103 theaters and averaged a sensational $22,844 per theater. The tally included a potent $3.4M from 62 higher-priced Imax venues ($54,839 average) marking a new opening weekend record for the large-screen format. Rival studios were scared away from the frame as no other major film dared to go head-to-head in wide release. The lack of competition helped to keep the focus of moviegoers on just one entertaining feature.
300 set a new March opening weekend record beating the $68M bow of Ice Age: The Meltdown from last year. That PG-rated toon played to a wider family audience and averaged a weaker $17,163 from nearly 4,000 theaters. The saga of King Leonidas and his battalion of brave Spartan warriors grossed a stunning $28.1M on Friday (including midnight shows from Thursday night), dropped an understandable 12% to $24.7M on Saturday, and slipped only 27% to $18.1M on Sunday.
300 also generated the third largest opening ever for an R-rated film trailing just The Matrix Reloaded ($91.8M) and The Passion of the Christ ($83.8M). And among non-sequels, it was the fifth biggest debut in history following Spider-Man ($114.8M), Harry Potter and the Sorcerer's Stone ($90.3M), Passion, and The Da Vinci Code ($77.1M). 300 also posted the sixth largest bow in studio history for Warner Bros. after the four Potter pics and the first Matrix sequel.
The sheer size of the audience was eye-popping for the stylish film which chronicles the Battle of Thermopylae between the warriors of Sparta and the mighty Persian army led by its ruler Xerxes in 480 B.C. Historical war epics like The Last Samurai and Troy made tons of money worldwide ($450-500M each) but after flops like Alexander and The Alamo, Hollywood ran the genre into the ground. Warner Bros. developed a new look that audiences would crave with 300 which is based on the graphic novel by Frank Miller. With digital effects, a stylized look, and brilliant marketing materials, the film began generating excitement last fall when the first trailers debuted. The studio should send a case of Cristal to the team that cut the trailers as they certainly ignited the spark leading to the fever-pitched anticipation.
With a reported budget of only $65M, 300 will easily become a major moneymaker for the studio especially since international theatrical and worldwide video revenue look to be explosive. The film had no pricey stars and featured epic battle scenes created by computers thereby eliminating the need to shoot on location with thousands of extras. In fact, only one scene in the enite film was shot outdoors. 300 debuted in only a handful of overseas markets this weekend but box office was impressive there as well. The film opened at number one in Greece, Singapore, Malaysia, Taiwan, and the Philippines with a combined gross of $6.5M from just 337 prints for a $19,288 per-print average which is phenomenal given the average ticket prices in those countries. The bloody actioner invades Korea later this week and attacks France, Italy, Spain, Mexico, and the U.K. on the following weekend.
For those in a less violent mood this weekend, Buena Vista's middle-aged motorcycle movie Wild Hogs was the ticket. The Tim Allen-John Travolta biker comedy dropped only 31% in its second weekend to $27.6M giving the studio a fantastic $77M in only ten days of release. Moviegoers are paying no attention to the universally poor reviews for Hogs which has now generated the second highest ten-day start of any film this year after Ghost Rider's $79M. The star-driven comedy could be on course to reach $150M or more domestically giving Disney a lucrative hit.
Ticket buyers were fixated on either 300 or Wild Hogs this weekend as the dynamic duo combined for a towering $98.5M in grosses accounting for a whopping 73% of all cash spent on the top ten films. Overall, the top ten posted its second best performance of 2007 with $135.7M narrowly trailing the $138.1M three-day tally from Presidents' Day weekend when Ghost Rider attacked. The North American box office is clearly alive and well.
The rest of the top five saw three films that were separated by less than $150,000. Third place went to Disney's Bridge to Terabithia which grossed $6.8M, down only 24%, for a $66.9M cume. Also in its fourth weekend, Sony's Ghost Rider fell 42% to $6.7M raising the total to $104M making the Nicolas Cage actioner the first film of 2007 to break the $100M mark. 300 and Wild Hogs could also join the century club as early as next weekend.
Fifth place went to the well-reviewed serial killer pic Zodiac which took in $6.6M, down a disturbing 50%, for a ten-day tally of $23.6M. Paramount's $65M production hoped to benefit from word-of-mouth, but instead suffered the worst drop by far of any film in the top ten thanks in part to competition from its R-rated foe 300. A disappointing final take of $34-37M seems likely making it director David Fincher's lowest grossing film ever.
A pair of funnymen followed. The Eddie Murphy comedy Norbit shed one third of its audience grossing $4.3M and lifted its total to $88.3M for Paramount. Jim Carrey's psychological thriller The Number 23 dipped 37% to $4.1M and upped its cume to a less impressive $30.2M for New Line.
Music and Lyrics, the romantic comedy starring Hugh Grant and Drew Barrymore, followed with $3.7M. Off only 24%, the Warner Bros. title has taken in $43.8M to date. Universal's thriller Breach collected $2.5M, down 31%, for a $29M sum. The slave trade drama Amazing Grace rounded out the top ten with $2.5M. Distributors Samuel Goldwyn and Roadside Attractions added over 200 theaters and enjoyed the smallest dip in the top ten sliding just 12%. Cume to date stands at $11.4M.
Fox Searchlight generated the biggest opening weekend average of the year with the launch of Mira Nair's The Namesake which bowed to $248,552 from only six locations for a muscular per-theater average of $41,425. Starring Kal Penn, the PG-13 film about an Indian family and their American-born children platformed in New York, Los Angeles, San Francisco, and Toronto and will expand on Friday into additional markets. Reviews were mostly positive.
Also opening in limited release was the Korean monster movie The Host with $314,488 from 71 theaters for a mild $4,429 average. The Magnolia release about a family that fights a mutated sea creature made its world premiere at the Cannes Film Festival last year and has already played in most of Asia. Critics were overwhelmingly giving praise.
Fox Faith, the new division of Fox dedicated to uplifting religious-themed pictures, opened its new film The Ultimate Gift over the weekend to $1.2M from 816 sites for a poor $1,520 average. The PG-rated film stars James Garner and Abigail Breslin and did not earn many positive reviews.
A pair of struggling films tumbled out of the top ten over the weekend suffering large declines. The Samuel L. Jackson-Christina Ricci pic Black Snake Moan fell 55% in its second weekend to $1.9M. The Paramount Vantage release has collected only $7.3M in its first ten days and should end with around $10M. Fox's comedy Reno 911!: Miami collapsed in its third weekend dropping 65% to $1.4M. With $19.1M in 17 days, look for a finish just north of $20M.
The top ten films grossed $135.7M (a new March record) which was up a stunning 48% from last year when Failure to Launch opened at number one with $24.4M; and up a solid 35% from 2005 when Robots debuted on top with $36M.
For NEW reviews of 300 and The Host visit The Chief Report.
Be sure to check back on Thursday for a complete summary, including projections, for next weekend when Premonition, I Think I Love My Wife, and Dead Silence all open.
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# Title Mar 9 - 11 Mar 2 - 4 % Chg. Theaters Weeks AVG Cumulative Distributor
1 300 $ 70,885,301 3,103 1 $ 22,844 $ 70,885,301 Warner Bros.
2 Wild Hogs 27,601,291 39,699,023 -30.5 3,296 2 8,374 77,016,616 Buena Vista
3 Bridge to Terabithia 6,779,315 8,904,512 -23.9 3,210 4 2,112 66,934,360 Buena Vista
4 Ghost Rider 6,670,463 11,553,469 -42.3 3,347 4 1,993 104,012,146 Sony
5 Zodiac 6,641,870 13,395,610 -50.4 2,379 2 2,792 23,595,973 Paramount
6 Norbit 4,278,099 6,398,696 -33.1 2,505 5 1,708 88,295,349 Paramount
7 The Number 23 4,082,292 6,486,304 -37.1 2,489 3 1,640 30,233,459 New Line
8 Music and Lyrics 3,744,084 4,888,337 -23.4 2,280 4 1,642 43,761,368 Warner Bros.
9 Breach 2,505,640 3,617,670 -30.7 1,505 4 1,665 29,021,080 Universal
10 Amazing Grace 2,478,036 2,826,987 -12.3 1,000 3 2,478 11,392,034 Goldwyn / Roadside
11 Black Snake Moan 1,884,690 4,143,199 -54.5 1,253 2 1,504 7,294,743 Par. Vantage
12 Reno 911!: Miami 1,400,035 3,942,926 -64.5 1,745 3 802 19,129,868 Fox
13 The Ultimate Gift 1,241,079 816 1 1,521 1,241,079 Fox Faith
14 Daddy's Little Girls 1,229,553 2,275,340 -46.0 798 4 1,541 30,121,004 Lionsgate
15 The Astronaut Farmer 1,015,921 2,212,155 -54.1 1,450 3 701 9,354,438 Warner Bros.
16 Night at the Museum 922,474 1,455,768 -36.6 739 12 1,248 244,766,562 Fox
17 The Lives of Others 852,558 781,095 9.1 163 5 5,230 3,460,248 Sony Classics
18 Pan's Labyrinth 780,957 1,194,796 -34.6 549 11 1,423 35,407,181 Picturehouse
19 Because I Said So 712,545 1,334,505 -46.6 709 7 1,005 41,615,740 Universal
20 The Last King of Scotland 656,921 961,122 -31.7 490 24 1,341 16,262,830 Fox Searchlight
Top 5 $ 118,578,240 $ 80,038,918 48.2
Top 20 vs. 2006 146,363,124 102,500,972 42.8
Last Updated : March 12, 2007 at 5:45PM ET
Watch Gitesh Pandya's weekly box office preview on CNN International airing live each Friday at 9:50am ET.


The Great Swindle (in original French: 'Au revoir la-haut'), a novel by globally known French crime writer Pierre Lemaitre, 66, is rendered in Persian by translator Parviz Shahdi and released by the Tehran-based publishing house Beh-Sokhan.
It is the second Persian translation of the book released in Iran; the first was by Iranian poet and literary translator Mahasti Bahreyni, 79, which was released last year by Mahi Publication, Mehr News Agency reported.
Shahdi, 81, studied comparative literature at Sorbonne University, spent 10 years in France and returned to Iran in 1984. After retirement, he was free to start professional translation from literary masterpieces mostly in French.
From among over 40 translations, the Great Swindle is Shahdi's latest work. It is said to be a hallmark of WWI novels in which Germans are not the most immediate enemy. The story is about a Machiavellian struggle between a corrupt French officer class and the browbeaten infantrymen under their command.
The engrossing novel was the 2013 winner of French top literature award Prix Goncourt. It focuses on the immediate aftermath of the war.
Heavy on extravagant characterization, kaleidoscopic description, third-person omniscient analysis, occasional first-person authorial digression, moral accounting and dissections of motivations, thoughts and prejudices of its protagonists, the book is rich in detail immersing the reader in its elaborately bleak world, The New York Times said in a review of the book.
Most front-line soldiers long to go home, but some commanders seize a last chance to enhance their reputations by advancing a bit farther on the battlefield.
So Pradelle, as everyone calls the lieutenant, sends two of his men out toward enemy lines.
Albert Maillard, a bank teller turned soldier, sees that the men have been shot in the back and realizes that Pradelle killed them to goad the troops into battle.
Pradelle tries to bury his crime by pushing Albert into a deep pit; Albert nearly suffocates; but another soldier, Edouard Pericourt, rescues him and is hit in the face by a chunk of shrapnel. Then the war ends.
Now Edouard lacks a lower jaw, smokes cigarettes through his nostrils and survives on morphine and heroin. Albert, after being buried alive with a putrefying horse, copes with flashbacks by donning a mask of the horse.
Edouard, creator of the therapeutic horse's head, also designs masks for his own mutilated face. The purpose is not to hide his gaping wound, but to express an idealized self: a pretty pink mouth set in a slightly condescending sneer, with two faded autumn leaves glued high up on the cheeks that looked like tears.
Pradelle, as pompous and ambitious as he is evil, embarks on a cynical plan to marry Edouard's well-connected sister and carry out a business scheme involving wartime graves and the reburial of soldiers' remains.
"The whole country was gripped by a frenzied desire to commemorate those who had died that was directly proportional to its revulsion for those who survived," Lemaitre wrote about the book.


Schneider Electric, the global specialist in the digital transformation of energy management and automation, has won the 'Hyperconverged Innovation of the Year' Category at the Storage, Virtualization and Cloud (SVC) Awards 2017.
The award, which was announced on December 27, 2017, was decided by popular vote, demonstrates the company's commitment to innovating technologies that support the adoption of both Edge and Cloud Computing, as well as meeting the challenges of Big Data and the IoT.
Schneider Electric's Micro Data Centre solutions enable the on-premise element of hybrid IT and edge data centre architectures to be deployed securely and scalably in any environment.
They can be customised to fit specific space constraints, and also built to withstand the specific environmental considerations of individual end-user applications.
So what exactly is a Micro Data Centres?
Here is an example of what a Micro Data Centre looks like.
Micro Data Centres form a key part of Schneider Electric's recently announced EcoStruxure™ architecture; an open, interoperable, IoT-enabled system architecture delivering enhanced value around safety, reliability, efficiency, sustainability, and connectivity for customers.
Throughout 2017 Schneider Electric announced a number of key Alliance Partnerships with industry-leading vendors, which have become instrumental to their micro data centre and hyperconvergence capabilities.
Customers within this space will often determine which products comprise the stack and need to be reassured that the technology will work immediately as promised, without interruption to service or downtime.
This award serves to prove Schneider Electric commitment to providing reliable Micro Data Centres and all the essential infrastructure to ensure 24/7 business critical applications availability.


text:
Magnusson Architecture + Planning Unveils St. Augustine Terrance in The Bronx
Apr 8, 2019 Architecture & Design
PDF Release
Magnusson Architecture + Planning (MAP) will debut St. Augustine Terrace, a new mid-rise building that will provide 112 units of affordable housing for low-income families in the Bronx. The project was commissioned by The Catholic Charities of the Archdiocese of New York and built on the site of the former St. Augustine's Church in the heart of the Morrisania neighborhood.
The design was inspired by the notion that a house of worship is a beacon of light in its community. From that idea came 13 floors of large, fully glazed elevator lobbies which provide spectacular views by day and then become a glowing tower by night.
"The beacon of light idea is a visual representation of the client's mission," said architect Fernando Villa, AIA, a principal at Magnusson Architecture and Planning (MAP). "The building shows what affordable housing can be, both for the community of people who live there and those in the surrounding neighborhood."
The placement of this south facing element was the most inclusive way to share the city vista with all residents. At the same time it gives light to the adjacent park in the evening.
One third of the units (35 studio apartments) will be set aside for adults with mental illness and the Beacon of Hope Division of Catholic Charities Community Services will provide supportive services. The remaining 77 units will include 19 one-bedroom, 37 two-bedroom, and 21 three-bedroom apartments.
Neighborhood Integration
Rock formations uncovered during construction were incorporated into the landscape at the front of the property providing a visual association with the park and a connection to the area's natural environment. They also supply an organic complement to the straight lines and right angels of the building. Just beyond the entry, decking extends out over the rocks to a look-out point, a spot for residents and visitors to enjoy as the topography slopes downward toward the west revealing more views. The bell and stones from the original church which are placed throughout the landscaping, celebrate the site's history and create yet another link to the neighborhood.
Amenities for Residents
With offices for social services on-site, this project is a prime example of the alternative ways that religious institutions can serve their communities. A large portion of the ground floor fronting Fulton Avenue will include supportive housing offices, the community/multi-use room and laundry. The ground floor will receive direct daylight and have access to a landscaped front yard. Below grade space will include bike storage and mechanical rooms.
The building, which has achieved LEED Gold status, will also be enrolled in the NYSERDA Multifamily Performance Program (MPP). As a NYC HPD funded project, it will comply with the Enterprise Green Communities certification program as well.
The resiliency and sustainability measures, quality building materials, broadly accessible panoramas, evening light provided to the nearby park and street corners, and elements that tie to the history of the place, all exemplify the inclusivity and care that underlie this work.
ABOUT MAP
For over three decades, Magnusson Architecture and Planning (MAP) has worked with non-profit groups, municipalities, and for-profit developers to improve the lives of those who live and work in their communities. MAP's award-winning buildings and urban revitalization projects serve as the foundation for vibrant and sustainable neighborhoods. Since our founding in 1986, we have been at the forefront of quality housing, green building and urban design.
MAP offers comprehensive architectural and planning services that blend design excellence with technical and planning expertise. The firm specializes in mixed-use developments including: affordable, supportive, senior, mixed-income, and market-rate housing; schools; community facilities; urban design projects; and, community planning.
In 2015 MAP established a Volunteer Program. Every month a team of staff, friends, and family donate personal time to projects that serve residents of the buildings the firm has designed, improve features in those buildings or support the clients of MAP's nonprofit partners.
ABOUT CATHOLIC HOMES OF NEW YORK
More than 1,000 new affordable apartments are under construction or in various stages of planning. In addition, Catholic Homes New York has renovated more than 2,000 apartments in 7 different neighborhoods over the past 8 years. Catholic Homes New York is an affiliate of The Catholic Charities of the Archdiocese of New York.
Contact: Abby Addams
Rhode Island School of Design's Furniture Design Department presents RISDXMABEO 2019 in Milan
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Choppy Tidings: On Higher GST Inflows
by Sumit Vishva December 3, 2020 December 3, 2020 Editorial of the day
The govt. must do more for sectors that are employment-intensive before the next Budget
Revenues from the GST crossed the ₹1-lakh crore mark for the second month in a row in November. The healthiest indirect tax collection streak since the national lockdown to combat COVID-19 in March, it is an encouraging sign for the economy that contracted less-than-expected in the second quarter. GST revenues in a month are a reflection of transactions from the previous month. The uptick in October's GST collections can be partly explained by the speedier unlocking of the economy in September as public transport restrictions were lifted in most parts of India. Some compliance-related relaxations for GST-paying businesses also pushed up collections significantly. There is no such compliance caveat attached to November's GST kitty — even though it is about ₹200 crore lower than the ₹1,05,155 crore from October. This suggests that the first month of the third quarter did see a genuine improvement in activity, whether it was driven by pent-up or festive demand, or a bit of both. This also ties in with other indicators from October, be it auto sales or the Purchasing Managers' Index (PMI) for manufacturing, which hit a 12-year high that month. With Deepavali falling in November this year (it was in October last year), some festive spending effect may be expected to come through in December's GST collections as well. The question is whether the improved demand trend will hold after the festival exuberance fades.
There are some early signs that the fervour and pace of economic activity already began to moderate in November. For instance, e-way bills generated by GST-covered entities fell nearly 14% to 55.3 million in November from 64.1 million in October, while the latest PMI indicates slower growth in orders and a further decline in employment as businesses are wary of the pandemic's lingering uncertainties. While India's infections have declined since September, some parts, including Delhi, Rajasthan and Gujarat, are seeing sharper spikes, prompting fresh travel curbs from there to the commercial capital Mumbai. In November, only Andhra Pradesh, Gujarat and Tamil Nadu saw double-digit growth in GST collections among major States, while Delhi's inflows fell 15% and Maharashtra's by 6% year-on-year. Thus, economic activity that has resumed remains fragile while the recovery seen so far has been uneven across States, thanks to the virus's unpredictable spread. The government has been conservative so far in its stimulus and support measures, despite which its fiscal math for 2020-21 will go awry. More direct and specific measures to help employment-intensive sectors, particularly those in distress such as travel and tourism or retail, are needed to ensure that further job losses today do not translate into tomorrow's demand slump.
1.Compliance (N)-the act of obeying an order, rule, or request. अनुपालन
2.Pushed Up (Phrasal Verb)-to make something higher in level or amount.
3.Pent-Up (Adj)-(of feelings) not expressed or released.
4.Exuberance (N)-a great amount. अधिकता
5.Fervour (N)-intense and passionate feeling. जोश
6.Wary (Adj)-feeling or showing caution about possible dangers or problems. सावधान
7.Lingering (Adj)-Lasting for a long time or slow to end.
Source:- The Hindu Editorial
The hindu editorial
Daily C.A One Liners, 1- December Updates