text:
Alberni Valley Community Forest
Forest Stewardship Plan
Forest Manager
Community Forest News
Open House & AGM
Active Logging
What is a Forest Stewardship Plan?
The Forest Stewardship Plan (FSP) is a requirement of the Forest Range and Practices Act (2004). The FSP is a landscape level plan, which is focused on establishing strategies and results for conserving and protecting timber and non-timber resource values for forest management activities over the life of the plan. The FSP states measurable, enforceable results and strategies that must be consistent with objectives set by government for a variety of forest values (e.g. fish, water, biodiversity cultural and heritage resources, etc). The FSP is a plan, which shows how the AVCFC will manage all stated government objectives from a site-specific level to the landscape level over the life of the plan. This results based plan is intended to allow tenure holders to be responsive to changing market conditions and to allow forest professionals to be innovative in applying management strategies to meet and exceed environmental standards in a manner specific to their operating area.
The FSP is intended to be a succinct document stating the management objectives for the resources present in the operating areas or Forest Development Units (FDU) as they are called in the FSP. The FSP is not intended to state methodology or give direction regarding how the management objectives are to be met; that is the role of the AVCFC and their forest professionals to decide. The FSP results and strategies must be consistent with government objectives for the management of all resources, and they must be measurable and verifiable.
How is a Forest Stewardship Plan created?
Through a competitive bid process, Econ Consulting was contracted to prepare the FSP for the AVCF. The first stage of FSP preparation involves pre-planning and information gathering. This stage includes a review of the AVCF goals and legislated direction for forest management in BC and includes consideration of forest legislation, higher-level plans and existing FSP that overlap the AVCF tenure area.
The FSP text was then drafted and sent to the AVCF Directors for review. Once the Directors were satisfied with the plan, the mandatory 60-day review period was initiated. At this stage, local stakeholders are invited to view the FSP and attend a public open house to discuss and provide comment on the plan's content. Stakeholders include first nations, existing tenure holders, community groups and the general public. The review period provides an opportunity for stakeholders to provide input into the AVCF's management decisions.
Once the review period is concluded, the FSP is submitted to the Ministry of Forests Lands and Natural Resource Operations (MFLNRO). The MFLNRO reviews the plan to ensure its content meets legal requirements. The plan is then submitted to the District Manager (DM) whom will either provide written approval or rejection of the plan. Once the DM approves the FSP, the AVCF can proceed with applications for operational permits to build roads and harvest timber.
Where in this process is the AVCF Forest Stewardship Plan?
The AVCF FSP public review period ended on March 18, 2011. The FSP has now been submitted to the Ministry of Forests Lands and Natural Resource Operations (MFLNRO).
Links to FSP Documents:
AVCFC Forest Stewardship Plan (May 2016) ~ 13 MB PDF file
AVCF Forest Stewardship Plan (Amendment May 29, 2014) ~ 8 MB PDF file
AVCF Forest Stewardship Plan Supporting Document (May 10, 2011) ~ 2.6 MB PDF file
AVCF Forest Stewardship Plan Stocking Standards, Specified Areas (Amendment May 29, 2014) ~ 46 KB PDF file
Overview Map ~ 4.2 MB PDF file
Map of Sproat FDU (A) ~ 5.5 MB PDF file
Map of Taylor FDU (B) ~ 6.7 MB PDF file
© 2023 Alberni Valley Community Forest | Developed by Alberni Design


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text:
\section{Introduction}
J. Cuntz constructed in his paper~\cite{Cu} a new $C^{\ast }$-algebra
naturally associated with the $ax+b$-semigroup over $\mathbb{N}$.
The $C^{\ast }$-algebra $\mathcal{Q}_{\mathbb{N}}$ is generated by a unitary $u$ and
isometries $\{s_{n}\}_{n\in \mathbb{N^{\ast }}}$ with range projections
$e_{n}=s_{n}s_{n}^{\ast }$ satisfying the relations
$$
s_{n}s_{m}=s_{nm}, ~s_{n}u=u^{n}s_{n}, ~\sum_{k=0}^{n-1}u^{k}e_{n}
u^{-k}=1
$$
for $n, m\in \mathbb{N}^{\ast }$.

Cuntz showed in~\cite{Cu} $\mathcal{Q}_{\mathbb{N}}$ is simple and purely infinite.
Furthermore he showed $\mathcal{Q}_{\mathbb{N}}$ contains Bunce-Deddens algebra
and is also generated by Bost-Connes algebra adding one unitary which
corresponds to addition.\\
 In this paper, motivated by the above
construction of the $C^{\ast }$-algebra, we introduce a $C^{\ast }$-algebra $B_{\sigma }$
associated with a primitive substitution $\sigma$.
The $C^{\ast }$-algebra $B_{\sigma }$ is generated by
the Cuntz-Krieger algebra \cite{Cu-K} $\mathcal{O}_{E_{\sigma }^{(2)}}=
C^{\ast }(\{s_{f}\}_{f\in E_{\sigma }^{(2)}})$,
 where $E_{\sigma }^{(2)}$ is the set of edges with partial order arising from $\sigma$,
and a unitary $u$ corresponding to addition (Vershik map)
satisfying the following relations:
$$
u(\sum_{\text{f is maximal in $E_{\sigma }^{(2)}$}}s_{f})
=(\sum_{\text{f is minimal in $E_{\sigma }^{(2)}$}}s_{f})u
$$
and for $f\in E_{\sigma }^{(2)}$ which is not maximal in $E_{\sigma }^{(2)}$,
$$
us_{f}=s_{g},~\text{where $g$ is the successor of $f$.}
$$

Let $f_{1},f_{2},\dots ,f_{k-1}$ be maximal elements
in $E_{\sigma }^{(2)}$ and $f_{k}$ be not a maximal
element in $E_{\sigma }^{(2)}$.
Let $g_{k}$ be the successor of $f_{k}$ and $(g_{1},g_{2},\dots ,g_{k})$ be the unique path
such that $g_{1},\dots ,g_{k-1}$ are minimal.
By using the relations above, we can calculate
\begin{align*}
us_{f_{1}}s_{f_{2}}\dots s_{f_{k}}
& = u(\sum_{\text{g is maximal in $E_{\sigma }^{(2)}$}}s_{g})s_{f_{2}}\dots s_{f_{k}}\\
& = (\sum_{\text{g is minimal in $E_{\sigma }^{(2)}$}}s_{g})us_{f_{2}}\dots s_{f_{k}}\\
& = \dots =
(\sum_{\text{g is minimal in $E_{\sigma }^{(2)}$}}s_{g})^{k-1}s_{g_{k}}
& = s_{g_{1}}s_{g_{2}} \dots s_{g_{k}}.
\end{align*}

We show that when $\sigma$ is proper, $B_{\sigma }$ is simple and purely infinite and contains the crossed product $C^{\ast }$-algebra of the corresponding Cantor minimal system, studied by Durand-Host-Skau\cite{Du}, Giordano-Putnam-Skau\cite{Gi} and Herman-Putnam-Skau\cite{He}.
Furthermore we show that a certain $B_{\sigma }$ has a unique KMS-state and determine the $K$-groups of $B_{\sigma }$.

\section{$C^{\ast }$-algebras associated with substitutions}
In this paper, we study $C^\ast$-algebras associated with primitive substitutions,
we introduce the  definition of stationary ordered Bratteli diagrams following after~\cite{Du}.

Let $A$ be an alphabet consisting of at least $2$ letters.
We denote by $A^{+}$ the set of finite words of $A$.
Then a map $\sigma :A\rightarrow A^{+}$ is said to be a substitution on $A$.
By concatenation, for each $n>0$, the $n$-th iteration $\sigma ^{n}:A\longrightarrow A^{+}$
is again a substitution. For any $a\in A$, let $\kappa (a,b)\in \mathbb{N}$ denote the number of occurrences of $b$ in $\sigma (a)$. A substitution $\sigma$ is said to be primitive if there exists $n >0$
such that for every $a$, $b\in A$, $b$ occurs in $\sigma ^{n}(a)$.

Let $\sigma$ be a primitive substitution
on an alphabet $A$ of $N$ elements.
Then we construct the stationary ordered Bratteli diagram ($V_{\sigma }$,$E_{\sigma }$,$\leq $ )  associated to a substitution $\sigma$ as follows:
The vertex set $V_{\sigma }$ is given as a disjoint union
$V_{\sigma }^{(0)}\cup V_{\sigma }^{(1)}\cup V_{\sigma }^{(2)}\cup \dots $
where
$V_{\sigma }^{(0)}$ is one point set $\{x_{0}\}$ and each $V_{\sigma }^{(n)}$ $(n\geq 1)$ consists of
$N$ points. We denote the element of $V_{\sigma }^{(n)}$ by $x(a,n)$ ($a\in A$).
i.e. $V_{\sigma }^{(n)}=\{x(a,n)| a\in A \}$ $(n\geq 1)$.
The vertex $x(a,n)$ is called the vertex of label $a$ at level $n$.

The edge set $E_{\sigma }$ is given as a disjoint union
$E_{\sigma }^{(1)}\cup E_{\sigma }^{(2)}\cup E_{\sigma }^{(3)}\cup \dots$
such that $s(E_{\sigma }^{(n)})\subset V_{\sigma }^{(n-1)}$
and $r(E_{\sigma }^{(n)})\subset V_{\sigma }^{(n)}$ where $r(e)$ denote the range of the edge $e$ and $s(e)$ the source of $e$.
For each $a\in A$, $E_{\sigma } ^{(1)}$ contains exactly one edge from $x_{0}$ to $x(a,1)$
for each $a\in A$.
Each $E_{\sigma }^{(n)}$ consists of edges from $V_{\sigma }^{(n-1)}$ to $V_{\sigma }^{(n)}$.
For an integer $n>1$ and $a$, $b$ in $A$, the number of edges from $x(b,n-1)$ to $x(a,n)$  is equal to $\kappa (a,b)$.

For the Bratteli diagram $(V_{\sigma }$,$E_{\sigma })$ constructed above,
we have a range map $r:E_{\sigma }\longrightarrow V_{\sigma }$ and a source map
$s:E_{\sigma }\longrightarrow V_{\sigma }$ so that
$r(E_{\sigma }^{(n)})\subset V_{\sigma }^{(n)}$ and
$s(E_{\sigma }^{(n)})\subset V_{\sigma }^{(n-1)}$.

For each $n \geq 1$, we define a partial order on $E_{\sigma }^{(n)}$ as follows:

Two edges $e_{1}, e_{2}\in E_{\sigma }$ are compatible if and only if
$r(e_{1})=r(e_{2})$.

For each $n\geq 2$ and $a\in V_{\sigma }^{(n)}$, we
put $E_{\sigma }^{(n)}(a)=\{e\in E_{\sigma }^{(n)}\mid $ the label of $r(e)$ is $a.\}$.
Let $\sigma (a)=a_{1}a_{2},\dots,a_{m}$, then
we denote the elements of $E_{\sigma }^{(n)}(a)$ by $e(a,k)$ $(1\leq k\leq m)$
 so that the label of $s(e(a,k))$ is $a_{k}$ and $r(e(a,k))=a$.

Let $e_{1}$ and $e_{2}$ be in $E_{\sigma }^{(n)} (n\geq 1)$ with $r(e_{1})=r(e_{2})$
and let $a$ be the label of $r(e_{1})$,
then $e_{1}$ and $e_{2}$ are in $E_{\sigma }^{(n)}(a)$. When we write $e_{1}$ and $e_{2}$
as $e(a,i)$ and $e(a,j)$ $(1\leq  i,j \leq m)$, we put $e(a,i)<e(a,j)$ if $i<j$.\\

We basically follow the definition and property of ordered Bratteli diagrams
in~\cite{Du}, \cite{Gi},~\cite{He} and~\cite{Ho}.
we introduce some necessary notations, definitions
of ordered Bratteli diagrams.
\begin{notation}
Suppose that $(V_{\sigma },E_{\sigma },\leq )$ is the ordered Bratteli diagram
associated to a primitive substitution $\sigma$.\\
(a)Let $x = x(a,n)$ be in $V_{\sigma }^{(n)}$. Then
$a$ is said to be the label of $x(a,n)$ and $n$ the level of $x(a,n)$.
Let $L(x)$ denote the label of $x$ and $\ell (x)$ the level of $x$.\\
(b)Let $x = x(a,n) \in V_{\sigma }^{(n)}$. For $k\in \mathbb{Z}$ with $-(n-1)\leq k $,
let $x_{+k}$ denote the vertex $x(a,n+k)\in V_{\sigma }^{(n+k)}$.
Let $f$ be in $E_{\sigma }^{(n)}$ $(n\geq 1)$, then for $k\in \mathbb{Z}$ with
$-(n-2)\leq k$, we denote by $f_{+k}$ the edge in $E_{\sigma }^{(n+k)}$ such that
$r(f_{+k}) = r(f)_{+k}$ and $s(f_{+k}) = s(f)_{+k}$.
\end{notation}

\begin{definition}
Let $(V_{\sigma },E_{\sigma },\leq )$ be the stationary ordered Bratteli diagram associated with a primitive substitution $\sigma$.
Let $\mathcal{P}_{\sigma }$ denote the infinite path space associated with $\sigma$.
i.e.
$$
\mathcal{P}_{\sigma }=\{(f^{(1)},f^{(2)},\dots)\mid f^{(k)}\in E_{\sigma }^{(k)},~ r(f^{(k)})=s(f^{(k+1)}) ~\text{for any} ~k\geq 1 \}.
$$

Let $\widetilde{V}_{\sigma }=V_{\sigma }\backslash (V_{\sigma }^{(0)}\cup V_{\sigma }^{(1)})$.
For each $x\in \widetilde{V}_{\sigma }$, let $\mathcal{P}_{\sigma }(x)$ denote
the set of finite pathes connecting between the top vertex $x_{0}$ and $x$. i.e.
\begin{align*}
\mathcal{P}_{\sigma }(x) =
& \{(f^{(1)},f^{(2)},\dots,f^{(m)})|f^{(k)}\in E_{\sigma }^{(k)}
 ~(1\leq k\leq m),\\
& r(f^{(k)}) = s(f^{(k+1)}) ~(1\leq k\leq m-1), r(f^{(m)}) = x \}.
\end{align*}

Let $\mathcal{P}_{\sigma }^{(n)}=\cup _{x\in V_{\sigma }^{(n)}}P_{\sigma }(x)~(n\geq 2)$
and $\mathcal{P}_{\sigma }^{\text{fin}}$ denote the set $\cup _{n\geq  2}\mathcal{P}_{\sigma }^{(n)}$.
$\mathcal{P}_{\sigma }^{\text{fin}}$ is called the finite path space arising from $\sigma$.
\end{definition}

\begin{notation}
Let $\sigma$ be a primitive substitution. \\
(a)We let $E_{\sigma }^{\text{max}}$ and $E_{\sigma }^{\text{min}}$ denote the maximal and minimal edges in $E_{\sigma }$. i.e.
$$
E_{\sigma }^{\text{max}}=\{f\in E_{\sigma }\mid f ~\text{is maximal in}~ r^{-1}(r(f)).\}.
$$

Let $\mathcal{P}_{\sigma }^{\text{max}}$ and $\mathcal{P}_{\sigma }^{\text{min}}$
denote the set of maximal and minimal
infinite pathes in $\mathcal{P}_{\sigma }$.
i.e.
$$
\mathcal{P}_{\sigma }^{\text{max}} = \{(f^{(1)},f^{(2)},\dots )\in \mathcal{P}_{\sigma }\mid f^{(k)}\in E_{\sigma }^{(k)}\cap E_{\sigma }^{\text{max}} , r(f^{(k)}) = s(f^{(k+1)})~ (1\leq k)\}.
$$

Let $\mathcal{P}_{\sigma }^{\text{fin,max}}$ and $\mathcal{P}_{\sigma }^{\text{fin,min}}$ denote the set of maximal and  minimal finite pathes. i.e.
$$
\mathcal{P}_{\sigma }^{\text{fin,max}}=
\{(f^{(1)},f^{(2)},\dots ,f^{(m)})\in \mathcal{P}_{\sigma }^{\text{fin}}\mid f^{(k)}\in E_{\sigma }^{(k)}\cap E_{\sigma }^{\text{max}}~(1\leq k\leq m)\}.
$$
(b)For a finite path $f=(f^{(1)}$,$f^{(2)}$,$\dots$,$f^{(n)})$
$\in \mathcal{P}_{\sigma }^{\text{fin}}$, let $s(f)$ and $r(f)$ denote the vertex
$r(f^{(1)})\in V_{\sigma }^{(2)}$ and $r(f^{(n)})$ respectively and let $\ell (f)$ denote  the length of finite path $f$. i.e.~$\ell (f)=n$.

For an infinite path $l \in \mathcal{P}_{\sigma }$,
let $s(l)$, the source of $l$, denote the vertex $r(l^{(1)})\in V_{\sigma }^{(1)}$. \\
(c)For any $f=(f^{(1)},f^{(2)},\dots ,f^{(m)})\in \mathcal{P}_{\sigma }^{(m)}$ and
$g=(g^{(1)}$,$g^{(2)}$,$\dots$,$g^{(n)})\in \mathcal{P}_{\sigma }^{(n)}$ with  $L(r(f))=L(s(g))$, we define the finite path
$fg\in \mathcal{P}_{\sigma }^{(m+n-1)}$ by $(f^{(1)}$,$f^{(2)}$,$\dots$,$f^{(m)}$,$g_{+(m-1)}^{(2)}$,
$g_{+(m-1)}^{(3)}$,$\dots$,$g_{+(m-1)}^{(n)})$.\\
(d)For any $f=(f^{(1)}$,$f^{(2)}$,$\dots$,$f^{(m)})\in \mathcal{P}_{\sigma }^{(m)}$ and
$l=(l^{(1)}$,$l^{(2)}$,$\dots )
\in \mathcal{P}_{\sigma }$ with $L(r(f))=L(s(l))$,
we define the infinite path
$fl\in \mathcal{P}_{\sigma }$ by $(f^{(1)}$,$f^{(2)}$,$\dots$,
$f^{(m)}$,$l_{+(m-1)}^{(2)}$,
$l_{+(m-1)}^{(3)}$,$\dots)$.
\end{notation}

\begin{definition}
Let $\sigma$ be a primitive substitution.
Then $(V_{\sigma },E_{\sigma },\leq )$ is said to be properly ordered if
$\ ^\#\mathcal{P}_{\sigma }^{\text{min}}=\ ^\#\mathcal{P}_{\sigma }^{\text{max}}=1$.
Then we write $\mathcal{P}_{\sigma }^{\text{min}}=\{p_{min}\}$ and
$\mathcal{P}_{\sigma }^{\text{max}}=\{p_{max}\}$.
Now we topologize the infinite path space $\mathcal{P}_{\sigma }$ by the family of cylinder sets
$\{U(f)\}_{f\in \mathcal{P}_{\sigma }^{\text{fin}}}$ where
$$
U(f)=\{g\in \mathcal{P}_{\sigma }\mid g^{(k)}=f^{(k)}\in E_{\sigma }^{(k)},~
1\leq k\leq \ell (f)\}.
$$

Then each $U(f)$ is also closed and we note that
$\mathcal{P}_{\sigma }$ is a compact Hausdorff space with a countable basis of
clopen sets.
Since $\sigma$ is primitive, the cardinality of $\mathcal{P}_{\sigma }$ is infinity
and so we note that $\mathcal{P}_{\sigma }$ is a Cantor set.
\end{definition}

\begin{definition}
A primitive substitution $\sigma$ on an alphabet $A$ is proper if there exist
$ b,c\in A$ such that\\
(1)For each $a\in A$, $b$ is the first letter of $\sigma (a)$.\\
(2)For each $a\in A$, $c$ is the last letter of $\sigma (a)$.
\label{proper}
\end{definition}
It is easily seen that if a substitution $\sigma$ is proper, then
the stationary ordered Bratteli diagram $(V_{\sigma }, E_{\sigma }, \leq )$ is properly ordered.

We recall the definition of Vershik map in \cite{Ho}.
\begin{definition}
Let $(V_{\sigma },E_{\sigma },\geq )$ be an ordered Bratteli diagram associated with a  primitive substitution $\sigma$.
For each $f=(f^{(1)}$,$f^{(2)}$,$\dots$ $)\in \mathcal{P}_{\sigma } \backslash \mathcal{P}_{\sigma }^{\text{max}}$,
there exists $k\geq 1$ such that $f^{(1)}$,$\dots$,$f^{(k-1)}\in E_{\sigma }^{\text{max}}$
and $f^{(k)}\not\in E_{\sigma }^{\text{max}}$.
Let $g^{(k)}$ be the successor of $f^{(k)}$ in $r^{-1}(r(f^{(k)}))$, then we write
the minimal finite path from $x_{0}$ to $s(g^{(k)})$ as
$(g^{(1)}$,$g^{(2)}$,$\dots$,$g^{(k-1)})$.
Then the infinite path $(g^{(1)}$,$g^{(2)}$,$\dots$,
$g^{(k)}$,$f^{(k+1)}$,$f^{(k+2)}$,$\dots)$
is said to be the successor of $f$.
We let $\lambda _{\sigma }$ be a map defined by
$$
\lambda _{\sigma }:\mathcal{P}_{\sigma } \backslash \mathcal{P}_{\sigma }^{\text{max}} \rightarrow \mathcal{P}_{\sigma }\backslash \mathcal{P}_{\sigma }^{\text{min}},~~
\lambda _{\sigma }(f)=
\text{the successor of }f.
$$

The map $\lambda _{\sigma }$ is said to be the Vershik map arising from the ordered Bratteli diagram $(V_{\sigma },E_{\sigma },\leq  )$.

When substitution $\sigma$ is proper,
we extend the Vershik map $\lambda _{\sigma }$ to the whole infinite path space
$\mathcal{P}_{\sigma }$ by defining $\lambda _{\sigma }(p_{\text{max}})=p_{\text{min}}$.
Then $\lambda _{\sigma }:\mathcal{P}_{\sigma }\longrightarrow \mathcal{P}_{\sigma }$ is
a homeomorphism, thus $(\mathcal{P}_{\sigma }$,$\lambda _{\sigma })$ is a topological
dynamical system. This dynamical system is called the Bratteli-Vershik system
associated to the ordered Bratteli diagram. It is known in \cite{Ho}
that when $\sigma$ is proper,
$\lambda _{\sigma }$ is minimal.
\end{definition}

\begin{definition}
Let $\sigma$ be a primitive substitution.
For any $x\in \widetilde{V}_{\sigma }$,
let $p_{\text{min}}(x)$ $($resp. $p_{\text{max}}(x)$ $)$ denote the unique minimal $($resp. maximal$)$ path from the top vertex $x_{0}$ to $x$.
Let $x\in \widetilde{V}_{\sigma }$, then there is a map:

$$
\mathcal{P}_{\sigma }(x)\backslash \{p_{\text{max}}(x)\}\ni f\longrightarrow
(\text{the successor of}~f)\in \mathcal{P}_{\sigma }(x)\backslash \{p_{\text{min}}(x)\}.
$$

We let the map be expressed by the symbol $\widehat{\lambda} _{\sigma }$.
\end{definition}

Let $(V_{\sigma },E_{\sigma },\geq )$ be stationary ordered Bratteli diagram associated to a proper primitive substitution $\sigma$. Then we let $T$ be the unitary operator defined by
$$
T:\ell ^{2}(\mathcal{P}_{\sigma }) \longrightarrow \ell ^{2}(\mathcal{P}_{\sigma }),~~~
T(\xi _{\ell })=\xi _{\lambda _{\sigma }(\ell )}.
$$

For each $f\in E_{\sigma }^{(2)}$, we let $V_{f}$ be the operator defined by
$$
V_{f}: \ell ^{2}(\mathcal{P}_{\sigma }) \longrightarrow \ell ^{2}
(\mathcal{P}_{\sigma }),~~
V_{f}\xi _{l}=\begin{cases}\xi _{fl} & L(r(f))=L(s(l)))\\
0 & L(r(f))\not=L(s(l)))\end{cases}.
$$

Then $\{V_{f}\}_{f\in E_{\sigma }^{(2)}}$ are partial isometries
 and the next lemma is known.
\begin{lemma}
Operators
$\{V_{f}\}_{f\in E_{\sigma }^{(2)}}$ satisfy the Cuntz-Krieger relations. i.e.
$$
\sum_{f\in E_{\sigma }^{(2)}} V_{f}V_{f}^{\ast }=1, ~~~~
$$
$$
V_{g}^{\ast }V_{g}=\sum_{f\in E_{\sigma }^{(2)},L(s(f))=L(r(g))}V_{f}V_{f}^{\ast }
 ~~~(g\in E_{\sigma }^{(2)}).
$$
\label{Cuntz}
\end{lemma}

\begin{lemma}
(a)For any $f\in E_{\sigma }^{(2)}\backslash E_{\sigma }^{\text{max}}$,
$V_{\widehat{\lambda }_{\sigma }(f)}=TV_{f}$.\\
(b)
$T(\sum_{f\in E_{\sigma }^{(2)}\cap E_{\sigma }^{\text{max}}}V_{f})=
(\sum_{f\in E_{\sigma }^{(2)}\cap E_{\sigma }^{\text{min}}}V_{f})T$.
\label{substitution}
\end{lemma}
\begin{proof}
(a) For any $l\in \mathcal{P}_{\sigma }$,
$$
V_{\widehat{\lambda }_{\sigma }(f)}\xi _{l}=
\begin{cases}
\xi _{\lambda _{\sigma }(fl)} & \text{if} ~L(r(f))=L(s(l)) \\ 0 & \text{otherwise}
\end{cases}
=TV_{f}\xi _{l}
$$
(b)For any $l\in \mathcal{P}_{\sigma }$,
\begin{align*}
T(\sum_{f\in E_{\sigma }^{(2)}\cap E_{\sigma }^{\text{max}}}V_{f})\xi _{l}
& =
\xi _{\lambda _{\sigma }(gl)} ~~(g\in
\mathcal{P}_{\sigma }^{(2)}\cap \mathcal{P}_{\sigma }^{\text{fin,max}} ~\text{with}
~L(r(g))=L(s(l)))\\
& = \xi _{h\lambda _{\sigma }(l)} ~~(h\in
\mathcal{P}_{\sigma }^{(2)}\cap \mathcal{P}_{\sigma }^{\text{fin,min}}
 ~\text{with}~
L(r(h))=L(s(\lambda _{\sigma }(l))))\\
& = (\sum_{f\in E_{\sigma }^{(2)}\cap E_{\sigma }^{\text{min}}}V_{f})T\xi _{l}.
\end{align*}
\end{proof}

According to the discussion above, we obtain the following definition.
\begin{definition}
Let $\sigma$ be a proper primitive substitution. Then we define $B_{\sigma }$ to be the
universal unital $C^{\ast }$-algebra generated by a unitary $u$ and partial
isometries $\{s_{f}\}_{f\in E_{\sigma }^{(2)}}$ such that\\
(a)(Cuntz-Krieger relations)
$$
\sum_{f\in E_{\sigma }^{(2)}}s_{f}s_{f}^{\ast }=1,
$$
$$
s_{g}^{\ast }s_{g}=\sum_{f\in E_{\sigma }^{(2)},L(s(f))=L(r(g))}s_{f}s_{f}^{\ast }
~~(f\in E_{\sigma }^{(2)}).
$$
(b)For any $f\in E_{\sigma }^{(2)}\backslash E_{\sigma }^{\text{max}}$,
$$
s_{\widehat{\lambda }_{\sigma }(f)}=us_{f}.
$$
(c)
$$
u(\sum_{f\in E_{\sigma }^{(2)}\cap E_{\sigma }^{\text{max}}}s_{f})
=(\sum_{f\in E_{\sigma }^{(2)}\cap E_{\sigma }^{\text{min}}}s_{f})u.
$$
\label{def:definition}
\end{definition}

\begin{example}
In \cite{Cu}, Cuntz defined a simple $C^{\ast }$-algebra $B_{2}$ generated by a unitary $u$ and an isometry $s_{2}$ satisfying the following relations.
$$
~s_{2}u=u^{2}s_{2}, ~~~\sum_{k=0}^{2^{n}-1}u^{k}s_{2}^{n}s_{2}^{\ast n}
u^{-k}=1
$$
for any $n\geq 2$.

The $C^\ast$-algebra $B_{2}$ is isomorphic to $B_{\sigma }$ for a certain substitution $\sigma$.
Let $\sigma$ be a substitution on an alphabet $A=\{a,b\}$ defined by
$\sigma (a)=ab$ and $\sigma (b)=ab$.
Then $E_{\sigma }^{(2)}=\{f_{1},\widehat{\lambda }_{\sigma }(f_{1}),f_{2},
\widehat{\lambda }_{\sigma }(f_{2})\}$ where
$f_{1},f_{2}\in E_{\sigma }^{(2)}\cap E_{\sigma }^{\text{min}}$
with $L(r(f_{1}))=$a and  $L(r(f_{2}))=$b.
We put $s_{f_{1}}=s_{2}^{2}s_{2}^{\ast }$, $s_{\widehat{\lambda }_{\sigma }(f_{1})}=us_{2}^{2}s_{2}^{\ast }$, $s_{f_{2}}=u^{2}s_{2}^{2}s_{2}^{\ast }u^{-1}$
and $s_{\widehat{\lambda }_{\sigma }(f_{2})}=u^{3}s_{2}^{2}s_{2}^{\ast }u^{-1}$.
Then $u$, $\{s_{f}\}_{f\in E_{\sigma }^{(2)}}$ satisfy the relations (a)-(c) in Definition
\ref{def:definition}.
According to the Theorem \ref{th:simple} in \S3,
$B_{\sigma }$ is simple. So $B_{\sigma }$ is isomorphic to $B_{2}$.
\end{example}
\begin{remark}
The $C^{\ast }$-algebra associated with substituion can be defined in general primitive substitution, but we restricted the case when $\sigma$ is proper in this paper.

Let $\sigma$ be a primitive substitution. Then we let $T$ be the operator defined by
$$
T:\ell ^{2}(\mathcal{P}_{\sigma }) \longrightarrow \ell ^{2}(\mathcal{P}_{\sigma }),~~~
T\xi _{l}=\begin{cases}\xi _{\lambda _{\sigma }(l)} & l\not\in \mathcal{P}_{\sigma }^{\text{max}} \\ 0 & l\in \mathcal{P}_{\sigma }^{\text{max}}
\end{cases}.
$$

Then the next equation clearly holds.
$$
T^\ast T+\chi _{\mathcal{P}_{\sigma }^{\text{max}}}=TT^{\ast }+\chi _{\mathcal{P}_{\sigma }^{\text{min}}}.
$$

Then $T$ and $\{V_{f}\}_{f\in E_{\sigma }^{(2)}}$ satisfy the
Lemma \ref{Cuntz} and Lemma \ref{substitution}.
It is known that $C^{\ast }(\{V_{f}\mid f\in E_{\sigma }^{(2)}\})$
contains the set of compact operators $K(\ell ^{2}(\mathcal{P}_{\sigma }))$.
Since $\sigma $ is a primitive substitution, the cardinality of $\mathcal{P}_{\sigma }$ is infinity.
Let $\pi$ be the natural map
$C^{\ast }(T, \{V_{f} \}_{f\in E_{\sigma }^{(2)}}) \longrightarrow C^{\ast }(T, \{V_{f}\}_{f\in E_{\sigma }^{(2)}})$
/$K(\ell ^{2}(\mathcal{P}_{\sigma }))$. Then $\pi (T)$ is a unitary and $\{\pi (V_{f})\}_{f\in E_{\sigma }^{(2)}}$
are partial isometries satisfying the equations of the
Lemma \ref{Cuntz} and
Lemma \ref{substitution}.
When $\sigma$ is primitive,
we can realize the $C^{\ast }$-algebra $B_{\sigma }$ generated by the quatient above.
\end{remark}
\section{simple and purely infiniteness for $B_{\sigma }$}
\begin{notation}
Let $\sigma$ be a proper primitive substitution on an alphabet $A$.\\
(a)For any $f=(f^{(1)},f^{(2)},\dots ,f^{(n)})\in \mathcal{P}_{\sigma }^{\text{fin}}$, we denote by $s_{f}$ the partial isometry
$s_{f^{(2)}}s_{f^{(3)}_{-1}}\dots s_{f^{(n)}_{-(n-1)}}$.\\
(b) For each $x\in \widetilde{V}_{\sigma }$,
let $\nu (x)\in \mathbb{N}$ denote the cardinality of the set $\mathcal{P}_{\sigma }(x)$.
\label{notation}
\end{notation}

\begin{lemma}
Let $\sigma$ be a proper primitive substitution.\\
(a) $s_{f}s_{g}^{\ast }=0$ for any $f, g\in \mathcal{P}_{\sigma }^{\text{fin}}$ with
$L(r(f))\not=L(r(g))$.\\
(b) $\sum_{f\in E_{\sigma }^{(2)}\cap E_{\sigma }^{\text{max}}}s_{f}s_{f}^{\ast }
=(\sum_{g\in E_{\sigma }^{(2)}\cap E_{\sigma }^{\text{max}}}s_{g})
(\sum_{h\in E_{\sigma }^{(2)}\cap E_{\sigma }^{\text{max}}}s_{h})^{\ast }.$\\
(c)For any $f\in \mathcal{P}_{\sigma }^{\text{fin,min}}$ and $0\leq k\leq \nu (r(f))-1$,
$$
u^{k}s_{f}=s_{\widehat{\lambda }_{\sigma }^{k}(f)}.
$$
(d)Let $f,g\in \mathcal{P}_{\sigma }^{\text{fin,min}}$. If
$-(\nu (r(f))-1)\leq k\leq \nu (r(g))-1$, then
$$
s_{f}^{\ast }u^{k}s_{g}\in \{s_{p}^{\ast }s_{q}\mid p,q\in \mathcal{P}_{\sigma }^{\text{fin}}~\text{with}~r(p)=r(f)~\text{and}~r(q)=r(g).\}.
$$
(e)We write $\mathcal{P}_{\sigma }^{\text{min}}=\{p_{\text{min}}\}$.
Then we let $\delta _{\sigma }$ be a map defined by
$$
\delta _{\sigma }:\mathcal{P}_{\sigma }^{\text{fin,min}}\longrightarrow \{0,1\}, ~
\delta _{\sigma }(f)=\begin{cases}1 &  \text{if}~ L(r(f))=L(s(p_{\text{min}})) \\
0 & \text{otherwise}
\end{cases}.
$$
For any $f\in \mathcal{P}_{\sigma }^{\text{fin,min}}$,
$$
s_{f}(\sum_{g\in E_{\sigma }^{(2)}\cap E_{\sigma }^{\text{min}}}s_{g})
=\delta _{\sigma }(f)\sum_{h\in \mathcal{P}_{\sigma }^{(\ell (f)+1)}\cap
\mathcal{P}_{\sigma }^{\text{fin,min}}}s_{h}.
$$
(f)For any $f ,g\in \mathcal{P}_{\sigma }^{\text{fin,min}}$ with $L(r(f))=L(r(g))$, we have
\begin{align*}
s_{f}s_{g}^{\ast } =
& \sum_{h\in \mathcal{P}_{\sigma }^{(2)}\backslash \mathcal{P}_{\sigma }^{\text{fin,max}}}s_{f}s_{\widehat{\lambda}_{\sigma } (h)}s_{\widehat{\lambda}_{\sigma } (h)}^{\ast }s_{g}^{\ast }\\
& + \delta _{\sigma }(f)u(\sum_{p\in \mathcal{P}_{\sigma }^{(\ell (f)+1)}
\cap \mathcal{P}_{\sigma }^{\text{fin,max}}}s_{p})
(\sum_{q\in \mathcal{P}_{\sigma }^{(\ell (g)+1)}\cap
\mathcal{P}_{\sigma }^{\text{{fin,max}}}}s_{q})^{\ast }u^{-1}.
\end{align*}
\label{1}
\end{lemma}
\begin{proof}
According to the Definition \ref{def:definition}, (a)-(e) clearly hold.\\
(f) Let $f,g\in \mathcal{P}_{\sigma }^{\text{fin,min}}$, then we have
$$
s_{f}s_{g}^{\ast }=s_{f}u(\sum_{h\in \mathcal{P}_{\sigma }^{(2)}}s_{h}s_{h}^{\ast })u^{-1}s_{g}^{\ast }
$$
$$
= \sum_{h\in \mathcal{P}_{\sigma }^{(2)}\backslash \mathcal{P}_{\sigma }^{\text{fin,max}}}s_{f}s_{\widehat{\lambda }_{\sigma }(h)}s_{\widehat{\lambda}_{\sigma } (h)}^{\ast }s_{g}^{\ast }
+ s_{f}u(\sum_{h\in \mathcal{P}_{\sigma }^{(2)}\cap \mathcal{P}_{\sigma }^{\text{fin,max}}}s_{h}s_{h}^{\ast })u^{-1}s_{g}^{\ast }.
$$

According to the relation (c) in Definition \ref{def:definition} and Lemma 3.2.(b),(e),  we have
\begin{align*}
& s_{f}u(\sum_{h\in \mathcal{P}_{\sigma }^{(2)}\cap \mathcal{P}_{\sigma }^{\text{fin,max}}}s_{h}
s_{h}^{\ast })u^{-1}s_{g}^{\ast }\\
& = s_{f}u(\sum_{h\in \mathcal{P}_{\sigma }^{(2)}\cap \mathcal{P}_{\sigma }^{\text{fin,max}}}s_{h})
(\sum_{i\in \mathcal{P}_{\sigma }^{(2)}\cap \mathcal{P}_{\sigma }^{\text{fin,max}}}s_{i})^\ast u^{-1}s_{g}^{\ast }\\
& = s_{f}(\sum_{h\in \mathcal{P}_{\sigma }^{(2)}\cap
\mathcal{P}_{\sigma }^{\text{fin,min}}}s_{h})uu^{-1}
(\sum_{i\in \mathcal{P}_{\sigma }^{(2)}\cap \mathcal{P}_{\sigma }^{\text{fin,min}}}s_{i})^\ast s_{g}^{\ast }\\
& = \delta _{\sigma }(f)(\sum_{h\in \mathcal{P}_{\sigma }^{(2)}\cap
\mathcal{P}_{\sigma }^{\text{fin,min}}}s_{h})^{\ell (f)}uu^{-1}
(\sum_{i\in \mathcal{P}_{\sigma }^{(2)}\cap \mathcal{P}_{\sigma }^{\text{fin,min}}}s_{i})^{\ast \ell (g)}\\
& = \delta _{\sigma }(f)u(\sum_{p\in \mathcal{P}_{\sigma }^{(\ell (f)+1)}\cap \mathcal{P}_{\sigma }^{\text{fin,max}}}s_{p})
(\sum_{q\in \mathcal{P}_{\sigma }^{(\ell (g)+1)}\cap \mathcal{P}_{\sigma }^{\text{fin,max}}}s_{q})u^{-1} &
\end{align*}
and we conclude the proof.
\end{proof}

For any $f\in \mathcal{P}_{\sigma }^{\text{fin}}$,
we let $\eta (f)$ denote the number $\text{min}\{\nu (z)\mid z\in V_{\sigma }^{(\ell (f)-1)}\}$.
The following lemma holds.
\begin{lemma}
For any $f,g\in \mathcal{P}_{\sigma }^{\text{fin}}$ and $a,b\geq 0$, there exists $N\geq \ell (f)$ such that for any $n\geq N$,
$u^{a}s_{f}s_{g}^{\ast }u^{-b}\in $ linear span $\{ u^{c}s_{p}s_{q}^{\ast }u^{-d} \mid
p\in \mathcal{P}_{\sigma }^{(n)}$, $q\in  \mathcal{P}_{\sigma }^{(n-\ell (f)+\ell (g))}$,
$0\leq c<\eta (p)$ and $0\leq d<\eta (q)\}$.
\label{2}
\end{lemma}

\begin{proof}
According to the relation $u^{a}s_{f}s_{g}^{\ast }u^{-b}=\sum_{h\in \mathcal{P}_{\sigma }^{(n)}}u^{a}s_{f}s_{h}s_{h}^{\ast }s_{g}^{\ast }u^{-b}$
$(a,b\geq 0$, $n\geq 2$ and $f,g\in \mathcal{P}_{\sigma }^{\text{fin}})$ and
$\lim_{n\rightarrow \infty }\text{min}\{\nu (z)\mid z\in V_{\sigma }^{(n)}\}=\infty $
 as $\sigma$ is primitive, the lemma holds.
\end{proof}

For a proper and primitive substitution $\sigma$, we let
$$
\mathcal{A}_{\sigma }=\text{linear span} \{ u^{a}s_{f}s_{g}^{\ast }u^{-b} \mid a,b\geq 0 ~\text{and}~f,g\in \mathcal{P}_{\sigma }^{\text{fin}}\}
$$
and
$$
\mathcal{L}_{\sigma }=\{ u^{a}s_{f}s_{g}^{\ast }u^{-b} \mid
f,g\in \mathcal{P}_{\sigma }^{\text{fin}},
0\leq a<\eta (f) ~\text{and}~ 0\leq b<\eta (g)\}.
$$

According to the Lemma \ref{2}, the following lemma holds.
\begin{lemma}
$$
\mathcal{A}_{\sigma }=\text{linear span} ~\mathcal{L}_{\sigma }.
$$
\end{lemma}

\begin{lemma}
Let $\sigma$ be a proper primitive substitution, then
$$
\text{Alg}_{\mathbb{C}}\{u,\{s_{f} \}_{f\in E_{\sigma }^{(2)}} \}=\mathcal{A}_{\sigma }
$$
\label{lem:3}
\end{lemma}
\begin{proof}
Since $u,\{s_{f}\}_{f\in E_{\sigma }^{(2)}}$ are in $\mathcal{A}_{\sigma }$,
to prove the lemma,
it suffices to show that $\mathcal{A}_{\sigma }$ is an involutive algebra.
We let $X=u^{a_{1}}s_{f_{1}}s_{g_{1}}^{\ast }u^{-b_{1}},
Y=u^{a_{2}}s_{f_{2}}s_{g_{2}}^{\ast }u^{-b_{2}}\in \mathcal{L}_{\sigma }$ with
$f_{1},f_{2},g_{1}, g_{2}\in \mathcal{P}_{\sigma }^{\text{fin,min}}$.
According to the Lemma \ref{2}, we may assume that $\ell (g_{1})=\ell (f_{2})$. We have
$$
XY=u^{a_{1}}s_{f_{1}}s_{g_{1}}^{\ast }u^{a_{2}-b_{1}}s_{f_{2}}s_{g_{2}}^{\ast }u^{-b_{2}}.
$$

We assume $a_{2}-b_{1}\geq 0$ without loss of generality, then
$0\leq a_{2}-b_{1}<\nu(r(f_{2}))+ \eta (f_{2})$.

If $a_{2}-b_{1}\leq \nu (r(f_{2}))-1$, then by using the Lemma \ref{1} (c),
$XY
=u^{a_{1}}s_{f_{1}}s_{g_{1}}^{\ast }s_{\widehat{\lambda }_{\sigma }^{a_{2}-b_{1}}(f_{2})}s_{g_{2}}^{\ast }u^{-b_{2}}\in \mathcal{A}_{\sigma }$.

If $\nu (r(f_{2}))\leq a_{2}-b_{1}< \nu(r(f_{2}))+ \eta (f_{2})$, then
we apply the Lemma \ref{1} (f) to $s_{f_{1}}s_{g_{1}}^{\ast }$,
we have
\begin{align*}
& XY=\sum_{h\in \mathcal{P}_{\sigma }^{(2)}\backslash \mathcal{P}_{\sigma }^{\text{fin,max}}}u^{a_{1}}s_{f_{1}}s_{\widehat{\lambda}_{\sigma } (h)}s_{\widehat{\lambda}_{\sigma } (h)}^{\ast }s_{g_{1}}^{\ast }u^{a_{2}-b_{1}}s_{f_{2}}s_{g_{2}}^{\ast }u^{-b_{2}}\\
& + \delta _{\sigma }(f_{1})u^{a_{1}+1}(\sum_{p\in \mathcal{P}_{\sigma }^{(\ell (f_{1})+1)}\cap \mathcal{P}_{\sigma }^{\text{fin,max}}}s_{p})
(\sum_{q\in \mathcal{P}_{\sigma }^{(\ell (g_{1})+1)}\cap \mathcal{P}_{\sigma }^{\text{fin,max}}}s_{q})^{\ast }u^{a_{2}-b_{1}-1}
s_{f_{2}}s_{g_{2}}^{\ast }u^{-b_{2}}.
\end{align*}

It follows from the Lemma \ref{1} (d) that
$$
s_{\widehat{\lambda}_{\sigma } (h)}^{\ast }s_{g_{1}}^{\ast }u^{a_{2}-b_{1}}s_{f_{2}},~
s_{q}^{\ast }u^{a_{2}-b_{1}-1}s_{f_{2}}
\in \{s_{l}^{\ast }s_{r}\mid
l,r\in \mathcal{P}_{\sigma }^{\text{fin}}\}
$$
for any $h\in \mathcal{P}_{\sigma }^{(2)}\backslash \mathcal{P}_{\sigma }^{\text{fin,max}}$ and
$q\in \mathcal{P}_{\sigma }^{(\ell (g_{1})+1)}\cap \mathcal{P}_{\sigma }^{\text{fin,max}}$ and we conclude the proof.
\end{proof}

We let $D_{\sigma }$ denote the $C^{\ast }$-algebra generated by
$\{u^{k}s_{f}s_{f}^{\ast }u^{-k}\mid f\in \mathcal{P}_{\sigma }^{\text{fin}}, 0\leq k\}.$

It is clear that
any two of the form $s_{f}s_{f}^{\ast }$ $(f\in \mathcal{P}_{\sigma }^{\text{fin}})$ commute.
For $N\geq 2$, we let $D_{\sigma }^{(N)}$ denote the subalgebra of $D_{\sigma }$ generated by $\{s_{f}s_{f}^{\ast }\mid f\in \mathcal{P}_{\sigma }^{(N)}\}$. It is easily seen that $D_{\sigma }^{(N+1)}$ contains $D_{\sigma }^{(N)}$.

\begin{notation}
Let $\sigma$ be a proper primitive substitution. For any $f\in \mathcal{P}_{\sigma }^{\text{fin}}$, we let $\widetilde{U}(f)$ denote the set defined by
$$
\widetilde{U}(f)=\{g\in \mathcal{P}_{\sigma }^{\text{fin}}\mid
g^{(1)}=f^{(1)},g^{(2)}=f^{(2)},\dots,g^{(\ell (f))}=f^{(\ell (f))}\}.
$$
\end{notation}

\begin{lemma}
For any $f\in \mathcal{P}_{\sigma }^{\text{fin}}$ and $a\geq 0$, there exists $N\geq \ell (f)$ such that for any $n\geq N$,
$$
u^{a}s_{f}s_{f}^{\ast }u^{-a}\in  \text{linear span} \{ s_{g}s_{g}^{\ast }\mid
g\in \mathcal{P}_{\sigma }^{(n)}\}.
$$
\label{5}
\end{lemma}
\begin{proof}
According to the Lemma \ref{1} (c),
it suffices to show that for any $f\in \mathcal{P}_{\sigma }^{\text{fin,max}}$,
there exists $n\in \mathbb{N}$ such that
$us_{f}s_{f}^{\ast }u^{-1}\in $~linear span $\{ s_{g}s_{g}^{\ast }\mid g\in \mathcal{P}_{\sigma }^{(n)}\}$.
Since $\sigma $ is proper, we write $\mathcal{P}_{\sigma }^{\text{max}}=\{p_{\text{max}}\}$.
If $f\in \mathcal{P}_{\sigma }^{\text{fin,max}}$ with $p_{\text{max}}\not\in U(f)$,
then we let $S(f)$ denote the set
$$
S(f)=\{g\in \widetilde{U}(f)\mid
g^{(1)}, g^{(2)},\dots,g^{(\ell (g)-1)}\in E_{\sigma }^{\text{max}}~ \text{and}~ g^{(\ell (g))}\not\in E_{\sigma }^{\text{max}}\}.
$$

Then  by using the Cuntz-Krieger relations, we have
\begin{align*}
us_{f}s_{f}^{\ast }u^{-1}
 & = \sum_{g\in S(f)}us_{g}s_{g}^{\ast }u^{-1}
= \sum_{g\in S(f)}s_{\widehat{\lambda} _{\sigma }(g)}s_{\widehat{\lambda} _{\sigma }(g)}^{\ast }\\
 & \in \text{linear span}\{ s_{h}s_{h}^{\ast }\mid
h\in \mathcal{P}_{\sigma }^{(N)}\} ~(\exists N\geq \ell (f)).
\end{align*}

If $f\in \mathcal{P}_{\sigma }^{\text{fin,max}}$ with $p_{\text{max}}\in U(f)$, then
\begin{align*}
us_{f}s_{f}^{\ast }u^{-1}
& = u(1-\sum_{g\in \mathcal{P}_{\sigma }^{(\ell (f))},f\not=g}s_{g}s_{g}^{\ast })u^{-1}\\
& = 1-\sum_{g\in \mathcal{P}_{\sigma }^{(\ell (f))},f\not=g}s_{\widehat{\lambda _{\sigma }}(g)}s_{\widehat{\lambda _{\sigma }}(g)}^{\ast }
\end{align*}
and we conclude the proof.
\end{proof}

According to the Lemma \ref{5}, the following lemma holds.
\begin{lemma}
When $\sigma$ is proper and primitive,
$D_{\sigma }$ is the inductive limit of the subalgebra $D_{\sigma }^{(N)}$.
\end{lemma}

\begin{lemma}
Let $\sigma$ be a proper primitive substitution. Then
$C^{\ast }$-algebra $D_{\sigma }$ is isomorphic to $C(\mathcal{P}_{\sigma })$.
\label{lem:path}
\end{lemma}

\begin{proof}
The fact that $\mathcal{P}_{\sigma }$ is totally disconnected implies that linear span
$\{ \chi _{f}\mid f\in \mathcal{P}_{\sigma }^{\text{fin}} \}$ is dense in
$C(\mathcal{P}_{\sigma })$.
It clearly holds that
$\text{linear~span}\{\chi _{f}\mid f\in \mathcal{P}_{\sigma }^{(n)}\}$ is isomorphic to $D_{\sigma }^{(n)}$.
Since $D_{\sigma }$ is the inductive limit of $D_{\sigma }^{(n)}$,
$D_{\sigma }$ is isomorphic to $C(\mathcal{P}_{\sigma })$.
\end{proof}

According to the Definition \ref{def:definition}, we have the following lemma.
\begin{lemma}
Let $\sigma$ be a proper primitive substitution.
In this lemma, we put $e_{f}=s_{f}s_{f}^{\ast }$, then
the unitary $u$ and projections $\{e_{f}\}_{f\in \mathcal{P}_{\sigma }^{\text{fin}}}$
satisfy the following relations.\\
(i) $\sum_{f\in E_{\sigma }^{(2)}}e_{f}=1.$

For any $g\in \mathcal{P}_{\sigma }^{(n)}$,
$e_{g}=\sum _{h\in E_{\sigma }^{(2)},~L(r(g))=L(s(h))}e_{gh}$.\\
(ii)
$$
e_{\widehat{\lambda }_{\sigma }(f)}=ue_{f}u^{-1}
$$
for any $n\geq 2$ and $f\in \mathcal{P}_{\sigma }^{(n)}\backslash
(\mathcal{P}_{\sigma }^{(n)}\cap \mathcal{P}_{\sigma }^{\text{fin,max}})$.\\
(iii)
$$
u(\sum_{f\in \mathcal{P}_{\sigma }^{(n)}\cap \mathcal{P}_{\sigma }^{\text{fin,max}}}e_{f})u^{-1}
=\sum_{f\in \mathcal{P}_{\sigma }^{(n)}\cap \mathcal{P}_{\sigma }^{\text{fin,min}}}e_{f}.
$$
for any $n\geq 2$.
\label{lem:4}
\end{lemma}

According to the Lemma \ref{lem:4}, we define the following algebra.
\begin{definition}
Let $\sigma$ be a proper primitive substitution.
Then we define $\mathcal{F}_{\sigma }$ to be the universal unital $C^{\ast }$-algebra
generated by a unitary $u$ and projections $\{e_{f}\}_{f\in \mathcal{P}_{\sigma }^{\text{fin}}}$
satisfying the relations (i), (ii) and (iii) in Lemma \ref{lem:4}.
\label{definition:def2}
\end{definition}

Let $(\mathcal{P}_{\sigma }, \lambda _{\sigma })$ be a Bratteli-Vershik system
associated with a proper primitive substitution $\sigma$.
Then this gives a $C^{\ast }$-dynamical system
$(C(\mathcal{P}_{\sigma })$, $\mathbb{Z}$, $\widetilde{\lambda} _{\sigma })$, where
$$
\widetilde{\lambda} _{\sigma }^{n}(f)=f\circ \lambda _{\sigma }^{-n} ~(n\in \mathbb{Z}).
$$
\begin{prop}
Let $\sigma$ be a proper primitive substitution. Then $\mathcal{F}_{\sigma }$ is
isomorphic to the crossed product
$C(\mathcal{P}_{\sigma })\rtimes _{\widetilde{\lambda} _{\sigma }}\mathbb{Z}$.
\end{prop}
\begin{proof}
The crossed product is generated by projections $\chi _{f}$ $(f\in \mathcal{P}_{\sigma }^{\text{fin}})$ and a unitary $U$ implementing the automorphism $\widetilde{\lambda} _{\sigma }$.
It follows from Lemma \ref{lem:path} that
there is a homomorphism of $C(\mathcal{P}_{\sigma })$ into $\mathcal{F}_{\sigma }$ such that $\phi (\chi _{f})=e_{f}$ for any $f\in \mathcal{P}_{\sigma }^{\text{fin}}$.
We check that
$$
\phi (\widetilde{\lambda} _{\sigma }(\chi _{f}))=ue_{f}u^{-1}
$$
for any $f\in \mathcal{P}_{\sigma }^{\text{fin}}$.
If $f\in \mathcal{P}_{\sigma }^{\text{fin}}\backslash
\mathcal{P}_{\sigma }^{\text{fin,max}}$, then $\phi (\widetilde{\lambda }_{\sigma }(\chi _{f}))
=e_{\widehat{\lambda }_{\sigma }(f)}=ue_{f}u^{-1}.$
We write $\mathcal{P}_{\sigma }^{\text{max}}=\{p_{\text{max}}\}$.
If $f\in \mathcal{P}_{\sigma }^{\text{fin,max}}$ is such that $p_{\text{max}}\not\in U(f)$,
then we take the set $S(f)\subset \mathcal{P}_{\sigma }^{\text{fin}}$ (The definition of $S(f)$ is in the proof of Lemma \ref{5}.).
Then $\chi _{f}=\sum_{g\in S(f)}\chi _{g}$. So
$$
\phi (\widetilde{\lambda }_{\sigma }(\chi _{f}))
 = \sum_{g\in S(f)}e_{\widehat{\lambda }_{\sigma }(g)}
 = u(\sum_{g\in S(f)}e_{g})u^{-1}=ue_{f}u^{-1}.
$$

If $f\in \mathcal{P}_{\sigma }^{\text{fin,max}}$ such that $p_{\text{max}}\in U(f)$, then $\chi _{f}=1-\sum_{g\in \mathcal{P}_{\sigma }^{(\ell (f))},f\not=g}\chi _{g}$. So
$$
\phi (\widetilde{\lambda} _{\sigma }(\chi _{f}))
 = \phi (\widetilde{\lambda} _{\sigma }(1-\sum_{g\in \mathcal{P}_{\sigma }^{(\ell (f))},f\not=g}\chi _{g}))
 =1-\sum_{g\in \mathcal{P}_{\sigma }^{(\ell (f))},f\not=g}ue_{g}u^{-1}= ue_{f}u^{-1}.
$$

It follows from the discussion above
that $\mathcal{F}_{\sigma }$ provides a covariant representation of $(C(\mathcal{P}_{\sigma }), \mathbb{Z}, \widetilde{\lambda} _{\sigma })$.
It follows from the universal property of the crossed product
there is a homomorphism of $C(\mathcal{P}_{\sigma })\rtimes _{\widetilde{\lambda} _{\sigma }}\mathbb{Z}$
onto $\mathcal{F}_{\sigma }$ taking $U$ to $u$ and $\chi _{f}$ to $e_{f}$. Since
$\lambda _{\sigma }$ is minimal,
$C(\mathcal{P}_{\sigma })\rtimes _{\widetilde{\lambda }_{\sigma }}\mathbb{Z}$ is simple and
we conclude the proof.
\end{proof}

We consider a unitary $u$ and partial isometries $\{s_{f}\}_{f\in E_{\sigma }^{(2)}}$
satisfying the relations (a)-(c) of
Definition \ref{def:definition} which generate $B_{\sigma }$.
By the universal property of $B_{\sigma }$, for any $t\in \mathbb{T}$,
there exists an automorphism $\alpha _{t}$ of $B_{\sigma }$ such that
$\alpha _{t}(u)=u$, $\alpha _{t}(s_{f})=ts_{f}$ for each
$f$ in $E_{\sigma }^{(2)}$.
 Put
$$
E(x)=\int _{\mathbb{T}}\alpha _{t}(x)dt,~~~~~~x\in B_{\sigma }
$$
where $dt$ is the normalized Haar measure on $\mathbb{T}$.
Then $E$ is a conditional expectation onto $\mathcal{F}_{\sigma }$.

We consider a unitary $u$ and partial isometries $\{e_{f}\}_{f\in E_{\sigma }^{(2)}}$
satisfying the relations (i)-(iii) of
Lemma \ref{lem:4} which generate $\mathcal{F}_{\sigma }$.
By the universal property of $B_{\sigma }$, for any $t\in \mathbb{T}$,
there exists an automorphism $\beta _{t}$ of $B_{\sigma }$ such that
$\beta _{t}(u)=e^{it}u$, $\beta _{t}(e_{f})=e_{f}$ for each
$f$ in $E_{\sigma }^{(2)}$. Put
$$
F(x)=\int_{\mathbb{T}}\beta _{t}(x)dt,~~~~~x\in \mathcal{F}_{\sigma }.
$$

Then $F$ is a conditional expectation onto $D_{\sigma }$.

We put $G=F\circ E$, then $G$ is a conditional expectation of $B_{\sigma }$ onto  $D_{\sigma }$.

\begin{thm} Let $\sigma$ be a proper primitive substitution. Then $C^{\ast }$-
algebra $B_{\sigma }$ associated with $\sigma$ is simple and purely infinite.
\label{th:simple}
\end{thm}
\begin{proof}
Let $\sigma$ be a proper primitive substitution on an alphabet $A$ with $^\# A=c$.
To prove the theorem, it suffices to show that for any non-zero positive element
$X\in B_{\sigma }$, there exists $Z\in B_{\sigma }$ such that
$Z^{\ast }XZ$ is invertible.
Since $G$ is faithful, $G(X)\not=0$. We may assume that $\| G(X)\| =1$.
For arbitrary $0<\varepsilon <\frac{1}{3c^{2}+1}$, there exists
$Y\in $
$\text{Alg}_{\mathbb{C}}$
$\{u,\{s_{f}\}_{f\in E_{\sigma }^{(2)}} \}$
such that $\|  X-Y \| <\varepsilon $. Then we have
$$
\| G(Y)\| =\| G(X)-G(X-Y)\| \geq \| G(X)\|-\| X-Y\|  >\frac{3c^{2}}{3c^{2}+1}.
$$

For any $N\geq 2$, we let
\begin{align*}
& \mathcal{L}_{\sigma }^{(N)}=\{u^{a}s_{f}s_{g}^{\ast }u^{-b}\mid f,g\in \mathcal{P}_{\sigma }^{\text{fin,min}},
\text{min}\{\ell (f), \ell (g)\}=N, \\
& \text{max}\{\ell (f),
\ell (g)\}\leq 2N  ~\text{and}~ \text{if}~ u^{a}s_{f}s_{g}^{\ast }u^{-b}\in D_{\sigma }, \text{then}~
 u^{a}s_{f}s_{g}^{\ast }u^{-b}\in D_{\sigma }^{(N)}.\}
\end{align*}
and $\mathcal{A}_{\sigma }^{(N)}=\text{linear span}~\mathcal{L}_{\sigma }^{(N)}$.
According to the Lemma \ref{2} and Lemma \ref{5},
there exists $N>0$ such that $Y\in \mathcal{A}_{\sigma }^{(N)}$.
Now we choose pathes $\{p(f,v)|f\in \mathcal{P}_{\sigma }^{(N)},
v\in V_{\sigma }^{(1)}\}$ $\subset \mathcal{P}_{\sigma }^{\text{fin}}$ as follows:
$$
p(f,v)=p_{1}(f,v)p_{2}(f,v)\in \mathcal{P}_{\sigma }^{\text{fin}}
$$
where
$p_{1}(f,v)\in \widetilde{U}(f)$ is such that
when we write $p_{1}(f,v)=\widehat{\lambda} _{\sigma }^{k}(g) ~(g\in \mathcal{P}_{\sigma }^{\text{fin,min}} ~\text{with}~r(g)=r(p_{1}(f,v))~\text{and}~
0\leq k  )$, $\text{max}\{\nu (z)\mid z\in V_{\sigma }^{(2N)}\}\leq k\}$.

$p_{2}(f,v)\in \mathcal{P}_{\sigma }^{\text{fin}}$ is a sufficiently large length path such that $L(s(p_{2}(f,v)))=L(r(p_{1}(f,v)))$,
$L(r(p_{2}(f,v)))=L(v)$ and
$s_{p_{2}(f,v)}^{\ast }s_{g}s_{p_{2}(f,v)}=0$
for any $g\in \cup _{m=2}^{N+1}\mathcal{P}_{\sigma }^{(m)}$.

We claim that if $x=u^{a}s_{g}s_{h}^{\ast }u^{-b}\in \mathcal{L}^{(N)}_{\sigma }$
$(g,h\in \mathcal{P}_{\sigma }^{\text{fin,min}})$
with
$G(x)=0$, then $e_{p(f,v)}xe_{p(f,v)}=0$ for any $f\in \mathcal{P}_{\sigma }^{(N)}$ and  $v\in V_{\sigma }^{(1)}$, because
\begin{align*}
& e_{p(f,v)}xe_{p(f,v)} = s_{p(f,v)}s_{\widehat{\lambda }_{\sigma }^{-a}(p(f,v))}^{\ast }s_{g}s_{h}^{\ast }
s_{\widehat{\lambda }_{\sigma }^{-b}(p(f,v))}s_{p(f,v)}^{\ast }\\
& = s_{p(f,v)}s_{p_{2}(f,v)}^{\ast }s_{\widehat{\lambda }_{\sigma }^{-a}(p_{1}(f,v))}^{\ast }s_{g}s_{h}^{\ast }
s_{\widehat{\lambda }_{\sigma }^{-b}(p_{1}(f,v))}s_{p_{2}(f,v)}s_{p(f,v)}^{\ast }=0.
\end{align*}

For any $v\in V_{\sigma }^{(1)}$,
we let $\varphi _{v}^{(N)}$: $D_{\sigma }\longrightarrow D_{\sigma }$ be a map defined by
\begin{equation*}
\varphi _{v}^{(N)}(x)=\sum_{f\in \mathcal{P}_{\sigma }^{(N)}}e_{p(f,v)}xe_{p(f,v)}.
\end{equation*}

Then the restlicted map
$$
\varphi _{v}^{(N)}|_{D_{\sigma }^{(N)}}:D_{\sigma }^{(N)}\longrightarrow
C^{\ast }(\{e_{p(f,v)}|f\in \mathcal{P}_{\sigma }^{(N)}\})
$$
is an isomorphism.
According to the discussion above and the definition of $\mathcal{L}_{\sigma }^{(N)}$, we have
$$
\varphi _{v}^{(N)}(G(Y))=\sum_{f\in \mathcal{P}_{\sigma }^{(N)}}e_{p(f,v)}Ye_{p(f,v)}.
$$

So there exists a path
$p_{v}\in \{p(f,v)|f\in \mathcal{P}_{\sigma }^{(N)}\}$
such that
\begin{equation*}
e_{p_{v}}Ye_{p_{v}}=\| G(Y)\|e_{p_{v}}
\end{equation*}
for each $v\in V_{\sigma }^{(1)}$.

Put $Z=\sum_{v\in V_{\sigma }^{(1)}}\| G(Y)\| ^{-\frac{1}{2}}$
$s_{p_{v}}$, then $\| Z\| \leq \| G(Y)\|^{-\frac{1}{2}}c$ and we have
$$
Z^{\ast }YZ=\sum_{v\in V_{\sigma }^{(1)}}s_{p_{{v}}}^{\ast}s_{p_{v}}=1.
$$

Then
$\| 1-Z^{\ast }XZ\| =\|Z^{\ast }YZ-Z^{\ast }XZ \| \leq \| Z\|^{2}\| Y-X\|$
$<\frac{1}{3}$.

Thus $Z^{\ast }XZ$ is invertible and we conclude the proof.
\end{proof}

\begin{remark}
Let $\sigma$ be a proper primitive substitution and $n\in \mathbb{N}$.
Then $\{s_{f}\}_{f\in \mathcal{P}_{\sigma }^{(n+1)}}\subset B_{\sigma }$ satisfy the  relations $(a)$-$(c)$ in Definition \ref{def:definition} of $B_{\sigma ^{n}}$.
Thus $B_{\sigma ^{n}}$ is isomorphic to $B_{\sigma }$.
\label{remark}
\end{remark}
\section{actions for $B_{\sigma }$}
Let $\sigma$ be a primitive substitution on an alphabet $A$.
For $x\in A^{\mathbb{Z}}$, we denote by $x_{[j,~k]}$ the word $x_{j}x_{j+1}\dots x_{k}$.
We define
\begin{align*}
X_{\sigma } =
 &  \{(x_{i})\in A^{\mathbb{Z}}\mid \text{For any} ~j<k, ~\text{there exists}~ n\in \mathbb{N} ~\text{and} ~a\in A\\
& \text{such that}~ x_{[j,k]}~ \text{is a subword of}~ \sigma ^{n}(a).\}
\end{align*}
and let $T_{\sigma }:X_{\sigma }\longrightarrow X_{\sigma }$ be a map defined by
$T_{\sigma }((x_{k})_{k\in \mathbb{Z}})=(x_{k+1})_{k\in \mathbb{Z}}$.
It is known in \cite{Qu} that for any primitive substitution $\sigma $, the substitution dynamical system $(X_{\sigma },T_{\sigma })$ is minimal and uniquely ergodic.

A substitution $\sigma :A\longrightarrow \{\text{the set of words of}$ $A \}$ naturally extends to a map
$\sigma :A^{\mathbb{Z}}\longrightarrow A^{\mathbb{Z}}$:
$$
\sigma (...w_{-2}w_{-1}.w_{0}w_{1}...)=...\sigma (w_{-2})\sigma (w_{-1}).\sigma (w_{0})\sigma (w_{1})....
$$

Let $\sigma$ be a proper primitive substitution.
An infinite word $w\in A^{\mathbb{Z}}$ is  called a periodic point of $\sigma$ if $\sigma (w)=w$ and is $T_{\sigma }$-periodic if there exists a positive integer $k$ with
$T_{\sigma }^{k}(w)=w$.
A proper substitution $\sigma$ is said to be periodic when there exists a periodic point of $\sigma$ when is also $T_{\sigma }$-periodic. When $\sigma$ is primitive, $|X_{\sigma }|<\infty$ if and only if $\sigma$ is periodic.

If $\sigma$ is periodic proper and primitive,  then we denote by $p_{\sigma }$ the smallest period of its fixed point $w$ and $u=w_{[0,~p_{\sigma }-1]}$.
As $\sigma (w)=w$, there exists $d_{\sigma }\geq 1$ such that
$\sigma (u)=u\dots u$ $(d_{\sigma }~times)$.
We need the following fact due to \cite{Du}.~$(\text{Proposition 16})$.
\begin{prop}
Let $\sigma$ be a proper primitive substitution.\\
(a)If $\sigma$ is non-periodic, then the Bratteli-Vershik
system $(\mathcal{P}_{\sigma },\lambda _{\sigma }) $ is isomorphic to the system $(X_{\sigma },T_{\sigma })$.\\
(b)If $\sigma $ is periodic, the system $(\mathcal{P}_{\sigma },\lambda _{\sigma })$ is
isomorphic to an odometer with a stationary base $(p_{\sigma },d_{\sigma },d_{\sigma },\dots)$.
\end{prop}
The next lemma follows from \cite{Da}$(\text{Theorem~VIII.4.1})$.
\begin{lemma}
If $\sigma$ is a periodic, proper  and primitive substitution, then
$\mathcal{F}_{\sigma }$ is isomorphic to Bunce-Deddens algebra with supernatural number
$p_{\sigma }d_{\sigma }^{\infty }$.
\end{lemma}

\begin{definition}
Let $\sigma$ be a primitive substitution on an alphabet $A$.
For any $a, b\in A$, let $\kappa (a,b)\in \mathbb{N}$ denote the number of occurrences of $b$ in $\sigma (a)$.
Let $M_{\sigma }$ denote its matrix; $M_{\sigma }=\{\kappa (a,b)\}_{a,b\in A}$.
A primitive substitution $\sigma$ is para-periodic if there exists $p\in \mathbb{N}$ such that $M_{\sigma }^{2}=pM_{\sigma }$. Let $\sigma$ be a para-periodic substitution, then we define a number $N_{\sigma }\in \mathbb{N}$ by
$M_{\sigma }^{2}=N_{\sigma }M_{\sigma }.$
\end{definition}

Let $\sigma$ be a proper primitive substitution.
It is known in \cite{Da} $(\text{Corollary~VIII.3.8})$ that there is a unique faithful trace $\varphi _{\sigma }^{(0)}$
on $\mathcal{F}_{\sigma }$.
 We let $\varphi_{\sigma }$ be a state on $B_{\sigma }$ defined by
$\varphi _{\sigma }=\varphi _{\sigma }^{(0)}\circ E$.

\begin{lemma}
Let $\sigma$ be a proper primitive substitution on an alphabet $A$.\\
(a)For any $f,g,h\in \mathcal{P}_{\sigma }^{\text{fin}}$ and $0\not=k\in \mathbb{Z}$,
$$
\varphi _{\sigma }(u^{k}s_{f}s_{f}^{\ast })=
\varphi _{\sigma }(s_{g}s_{h}^{\ast }u^{k}s_{h}s_{g}^{\ast })=0.
$$
(b)If $\sigma$ is para-periodic, then
for any $f,g\in \mathcal{P}_{\sigma }^{\text{fin}}$ with $L(r(f))=L(r(g))$, we have

$$
\varphi _{\sigma }(s_{f}s_{f}^{\ast })=N_{\sigma }^{\ell (g)-\ell (f)}
\varphi _{\sigma }(s_{g}s_{g}^{\ast }).
$$
\label{7}
\end{lemma}

\begin{proof}
(a)Let $f,g,h\in \mathcal{P}_{\sigma }^{\text{fin}}$.
According to the relation
$$
u^{k}s_{f}s_{f}^{\ast }=\sum_{g\in \mathcal{P}_{\sigma }^{(n)}}
u^{k}s_{f}s_{g}s_{g}^{\ast }s_{f}^{\ast }
$$
for any $n\geq 2$,
we may first assume that $|k|<\nu (r(f))$ and $k\not=0$. Then $s_{f}^{\ast }u^{k}s_{f}=0$.
Thus
$$
\varphi _{\sigma }(u^{k}s_{f}s_{f}^{\ast })=
\varphi _{\sigma }(s_{f}s_{f}^{\ast }u^{k}s_{f}s_{f}^{\ast })=0.
$$
$$
\varphi _{\sigma }(s_{g}s_{h}^{\ast }u^{k}s_{h}s_{g}^{\ast })
=\sum_{p\in \mathcal{P}_{\sigma }^{(n)}}
\varphi _{\sigma }(s_{p}s_{p}^{\ast }s_{g}s_{h}^{\ast }u^{k}s_{h}s_{g}^{\ast }
s_{p}s_{p}^{\ast })=0~(n\geq 2).
$$
(b)
We let $p_{b}^{(n)}$ denote the path in $\mathcal{P}_{\sigma }^{(n)}\cap \mathcal{P}_{\sigma }^{\text{fin,min}}$ such that $L(r(p_{b}^{(n)}))=b$ for any $b\in A$
and $n\in \mathbb{N}$.
Let $f,g\in \mathcal{P}_{\sigma }^{\text{fin}}$ with $L(r(f))=L(r(g))$ and $\ell (f)\leq \ell (g)$, then
$$
\varphi _{\sigma }(s_{f}s_{f}^{\ast })=\sum_{h\in \mathcal{P}_{\sigma }^{(\ell (g)-\ell (f)+2)}}\varphi _{\sigma }(s_{f}s_{h}s_{h}^{\ast }s_{f}^{\ast }).
$$

Since
$\varphi _{\sigma }(s_{f}s_{f}^{\ast })=\varphi _{\sigma }(s_{f^{'}}s_{f^{'}}^{\ast })$
for any $f,f^{'}\in \mathcal{P}_{\sigma }^{\text{fin}}$ with $r(f)=r(f^{'})$ and
\begin{align*}
\ ^\#\{h\in \mathcal{P}^{(\ell (g)-\ell (f)+2)}\mid L(s(h))=L(r(f)) ~\text{and} ~L(r(h))=b\}=N_{\sigma }^{\ell (g)-\ell (f)}\kappa (b,L(r(f)))
\end{align*}
as $\sigma$ is para-periodic, we have
\begin{align*}
\sum_{h\in \mathcal{P}_{\sigma }^{(\ell (g)-\ell (f)+2)}}\varphi _{\sigma }(s_{f}s_{h}s_{h}^{\ast }s_{f}^{\ast })
& = N_{\sigma }^{\ell (g)-\ell (f)}
\sum_{b\in A}\kappa (b,L(r(f)))\varphi _{\sigma }(s_{p_{b}^{(\ell (g)+1)}}
s_{p_{b}^{(\ell (g)+1)}}^{\ast })\\
& = N_{\sigma }^{\ell (g)-\ell (f)}
\varphi _{\sigma }(s_{g}s_{g}^{\ast }).
\end{align*}
\end{proof}

\begin{lemma}
Let $\sigma$ be a proper primitive substitution.
Let $x=u^{a}s_{f}s_{g}^{\ast }u^{-b}\in \mathcal{L}_{\sigma }$, where $f, g\in \mathcal{P}_{\sigma }^{\text{fin}, \text{min}}$.
For any $n>\ell (f)$, $x$ is written as the form
$$
x=\sum_{\text{finite}}u^{a_{k}}s_{f_{k}}s_{g_{k}}^{\ast }u^{-b_{k}}
$$
where $u^{a_{k}}s_{f_{k}}s_{g_{k}}^{\ast }u^{-b_{k}}\in \mathcal{L}_{\sigma }$,
$f_{k}\in \mathcal{P}_{\sigma }^{(n)}\cap
\mathcal{P}_{\sigma }^{\text{fin,min}}$ $~\text{and}~
\eta (f_{k})\leq a_{k}< \nu (r(f_{k}))+\eta (f_{k})$.

\label{6}
\end{lemma}
\begin{proof}
Let $x=u^{a}s_{f}s_{g}^{\ast }u^{-b}\in \mathcal{L}_{\sigma }$ and $n>\ell (f)$.
According to the relation $\sum_{h\in \mathcal{P}_{\sigma }^{(m)}}s_{h}s_{h}^{\ast }=1$
 $(m\geq 1)$,
we may assume that $\ell (f)=n-1$.
If $a\geq \text{min}\{\nu (z)\mid \ell (z)=n-1\}$, then
$$
x=u^{a}s_{f}s_{g}^{\ast }u^{-b}=\sum_{h\in E_{\sigma }^{(2)}}u^{a}s_{f}s_{h}s_{h}^{\ast }s_{g}^{\ast }u^{-b}.
$$

If $a<\text{min}\{\nu (z)\mid \ell (z)=n-1\}$, according to the Lemma 3.2 (f), we have
\begin{align*}
x
& = \sum_{h\in E_{\sigma }^{(2)}\backslash E_{\sigma }^{\text{max}}}u^{a}s_{f}
s_{\widehat{\lambda}_{\sigma } (h)}s_{\widehat{\lambda}_{\sigma } (h)}^{\ast }s_{g}^{\ast }u^{-b}\\
& + u^{a+1}(\sum_{h\in \mathcal{P}_{\sigma }^{(\ell (f)+1)}\cap \mathcal{P}_{\sigma }^{\text{fin,max}}}s_{h})
(\sum_{i\in \mathcal{P}_{\sigma }^{(\ell (g)+1)}\cap \mathcal{P}_{\sigma }^{\text{fin,max}}}s_{i})^{\ast }u^{-(b+1)}.
\end{align*}

The above equations satisfy the condition in Lemma \ref{6}.
\end{proof}

\begin{definition} Let $\mathcal{A}$ be a $C^{\ast }$-algebra, $\theta $ an $\mathbb{R}$-action on $\mathcal{A}$, $\varphi$ a state on $\mathcal{A}$ and $\beta>0$.
Then $\varphi$ satisfies the $\beta$-KMS condition if
for each $x$, $y\in \mathcal{A}$,
there is a function $f$,
bounded and continuous on $\{z\in \mathbb{C}\mid 0\leq \text{Im}(z)\leq \beta \}$ and
analytic on $\{z\in \mathbb{C}\mid 0<\text{Im}(z)<\beta \}$  such that for $t\in \mathbb{R}$,
$$
f(t)=\varphi (\theta _{t}(x)y),~~ f(t+i\beta )=\varphi (y\theta _{t}(x)).
$$
\end{definition}

It is known that $\varphi$ satisfies the $\beta$-KMS condition if and only if
$\varphi (x\theta _{i\beta }(y))=\varphi (yx)$ for any $x\in \mathcal{A}$ and $y$,
analytic element for $(\theta _{t})_{t\in \mathbb{R}}$.

Let $\sigma$ be a para-periodic proper substitution. Then let $\theta  :\mathbb{R}\longrightarrow \text{Aut}(B_{\sigma })$ be a map defined by
$$
\theta _{t}(u)=u,~~
\theta _{t}(s_{f})=N_{\sigma }^{it}s_{f}~~~(f\in E_{\sigma }^{(2)}).
$$

Then $\theta $ is an $\mathbb{R}$-action on $B_{\sigma }$.
\begin{prop} Let $\sigma$ be a para-periodic proper substitution. Then the state $\varphi_{\sigma }$ is a $1$-KMS state for $(\theta _{t})_{t\in \mathbb{R}}$.
\end{prop}
\begin{proof}
To prove the proposition, by Lemma \ref{lem:3} and Lemma \ref{6}, it suffices to show that
$\varphi _{\sigma }(xy)=\varphi _{\sigma }(y\theta _{i}(x))$ for any
$x=u^{a_{1}}s_{f_{1}}s_{g_{1}}^{\ast }u^{-b_{1}}$,
$y=u^{a_{2}}s_{f_{2}}s_{g_{2}}^{\ast }u^{-b_{2}}\in \mathcal{L}_{\sigma }$ with
$f_{1}$,$f_{2}$,$g_{1}$,$g_{2}\in \mathcal{P}_{\sigma }^{\text{fin,min}}$, $\ell (g_{1})=\ell (f_{2})$, $\eta (g_{1})\leq b_{1}< \nu (r(g_{1}))+\eta (g_{1})$ and
$\eta (f_{2})\leq a_{2}< \nu (r(f_{2}))+\eta (f_{2})$.

If $\ell (f_{1})+\ell (f_{2})\not=\ell (g_{1})+\ell (g_{2})$, then
$xy,yx\not\in \mathcal{F}_{\sigma }$ and so $\varphi _{\sigma } (xy)=\varphi _{\sigma } (yx)=0$.

We let $\ell (f_{1})+\ell (f_{2})=\ell (g_{1})+\ell (g_{2})$, then
$\ell (f_{1})=\ell (g_{2})$.
If $b_{1}\not=a_{2}$ or $f_{2}\not=g_{1}$, then $s_{g_{1}}^{\ast }u^{a_{2}-b_{1}}s_{f_{2}}=0$. Thus $xy=0$ and
$\varphi _{\sigma } (yx)=\varphi _{\sigma } (s_{g_{1}}s_{g_{1}}^{\ast }u^{a_{2}-b_{1}}s_{f_{2}}s_{g_{2}}^{\ast }u^{a_{1}-b_{2}}s_{f_{1}}s_{g_{1}}^{\ast })=0$.

Next we let $b_{1}=a_{2}$ and $g_{1}=f_{2}$, then $f_{1}=g_{2}$.
If $a_{1}\not=b_{2}$, then by using the Lemma \ref{7} (a), $\varphi _{\sigma } (xy)=\varphi _{\sigma }(u^{a_{1}-b_{2}}s_{f_{1}}s_{f_{1}}^{\ast })=0$ and $\varphi _{\sigma }(yx)=
\varphi _{\sigma }(s_{f_{1}}s_{g_{1}}^{\ast }u^{a_{1}-b_{2}}s_{g_{1}}s_{f_{1}}^{\ast })=0$.

We let $a_{1}=b_{2}$. According to the Lemma \ref{7} (b), we have
\begin{align*}
\varphi _{\sigma }(xy)
& = \varphi _{\sigma }(s_{f_{1}}s_{f_{1}}^{\ast })\\
& = N_{\sigma }^{\ell (g_{1})-\ell (f_{1})}\varphi _{\sigma }(s_{g_{1}}s_{g_{1}}^{\ast })\\
& = \varphi _{\sigma }(y\theta _{i}(x))
\end{align*}
and we conclude the proof.
\end{proof}

\section{The $K$-groups of $B_{\sigma }$}
We follow the definition of semigroup crossed product $C^{\ast }$-algebras
in~\cite{R}.
\begin{definition}
Let $B$ be a unital $C^{\ast }$-algebra and $\rho$ be an endomorphism of $B$ such that
$\rho$ is an isomorphism from $B$ onto $\rho (1)B\rho (1)$.
We define $B\rtimes _{\rho }\mathbb{N}$ to be the universal unital
$C^{\ast }$-algebra generated by $B$ and an isometry $s$ such that $\rho (b)=sbs^{\ast }$
for any $b\in B$.
\end{definition}
We denote by $\mu _{1}$ an isometry $\sum_{f\in E_{\sigma }^{(2)}\cap E_{\sigma }^{\text{min}}}s_{f}$ and let $\rho$ be a map defined by
\begin{equation*}
\rho :\mathcal{F}_{\sigma }\longrightarrow \mathcal{F}_{\sigma },~~~
\rho (x)=\mu _{1}x\mu _{1}^{\ast }.
\end{equation*}
Then $\rho \in \text{End}(\mathcal{F}_{\sigma })$ and it is easily seen that
$B_{\sigma }$ is isomorphic to $\mathcal{F}_{\sigma }\rtimes _{\rho }\mathbb{N}$.

We shall recall some facts on this dimension group $K_{\sigma }$ of a substitution dynamical system based on \cite{Du}.

\begin{definition}
Let $\sigma$ be a primitive substitution on an alphabet $A$ with $^\#A=d$,  then we let\\ $G_{\sigma }=\{x\in \mathbb{Q}^{d}\mid \text{there exists}~n>0 ~\text{such that}
~M_{\sigma }^{n}x\in \mathbb{Z}^{d}.\}$.\\
$H_{\sigma }=\{x\in \mathbb{Q}^{d}\mid \text{there exists}~n>0 ~\text{such that}~
M_{\sigma }^{n}x=0. \}$.\\
$K_{\sigma }=G_{\sigma }/H_{\sigma }$.

The group $K_{\sigma }$ is called the dimension group of a substitution dynamical system.
\end{definition}

We let $\phi$ be an endomorphism on $\mathbb{Z}^{d}$ defined by $\phi (x)=M_{\sigma }x$,  then the direct limit of $(\mathbb{Z}^{d},\phi )$ is isomorphic to $K_{\sigma }$.

Let $\sigma$ be a proper primitive substitution on an alphabet $A=\{c_{1},c_{2} \dots c_{d}\}$, then we let $C(\mathcal{P}_{\sigma },\mathbb{Z})$ denote the set of continuous functions with values in $\mathbb{Z}$. It follows from the Pimsner-Voiculescu exact sequence that
$K_{0}(\mathcal{F}_{\sigma })=C(\mathcal{P}_{\sigma },\mathbb{Z})/\widetilde{\lambda} _{\sigma }C(\mathcal{P}_{\sigma },\mathbb{Z})$
where $\widetilde{\lambda} _{\sigma }(f)=f-f\circ \lambda _{\sigma }^{-1}$, for each $f\in C(\mathcal{P}_{\sigma },\mathbb{Z})$ and
$K_{1}(\mathcal{F}_{\sigma })=\mathbb{Z}$.

For $n\geq 2$, we denote by $C_{n}$ the subgroup of $C(\mathcal{P}_{\sigma },\mathbb{Z})$ and $H_{n}$ the subgroup of $C_{n}$ defined by
$$
C_{n}=\{\sum_{f\in \mathcal{P}_{\sigma }^{(n)}}a_{f}\chi _{f}\mid a_{f}\in \mathbb{Z}\},
$$
$$
H_{n}=\{\sum_{f\in \mathcal{P}_{\sigma }^{(n)}}a_{f}\chi _{f}\in C_{n}\mid
 \sum_{f\in \mathcal{P}_{\sigma }^{(n)},~L(r(f))=c_{k}}a_{f}=0 ~\text{for any}~
1\leq k\leq d\}.
$$

Then it clearly holds that $C_{n}\subset C_{n+1}$, $H_{n}\subset H_{n+1}$,
$C(\mathcal{P}_{\sigma },\mathbb{Z})=\cup _{n=2}^{\infty }C_{n}~\text{and}$
$\widetilde{\lambda} _{\sigma }C(\mathcal{P}_{\sigma },\mathbb{Z})=\cup _{n=2}^{\infty }H_{n}$.
We let $K_{n}=C_{n}/H_{n}$ and
$\varphi _{n}$ be a map on $K_{n}$ defined by
\begin{align*}
\varphi _{n}:K_{n}\longrightarrow K_{n+1},~\varphi _{n}(x+H_{n})=x+H_{n+1}
\end{align*}
for $n\geq 2$.

The group $C_{n}/H_{n}$ is isomorphic to $\mathbb{Z}^{d}$ for any $n\geq 2$ with
isomorphism $\psi _{n}$ defined by
\begin{align*}
\psi _{n}(\sum_{f\in \mathcal{P}_{\sigma }^{(n)}}a_{f}\chi _{f}+H_{n})=
(\sum_{g\in \mathcal{P}_{\sigma }^{(n)},L(r(g))=c_{k} }a_{g})_{1\leq k\leq d}.
\end{align*}

Then we have
$\phi \circ \psi _{n}=\psi _{n+1} \circ \varphi _{n}$ on $K_{n}$~$(n\geq 2)$.
Since $C(\mathcal{P}_{\sigma },\mathbb{Z})/\widetilde{\lambda} _{\sigma }C(\mathcal{P}_{\sigma },\mathbb{Z})$
is the direct limit of $(C_{n}/H_{n},\phi _{n})$ and
$K_{\sigma }$ is the direct limit of $(\mathbb{Z}^{d},\phi )$
 respectively, $K_{0}(\mathcal{F}_{\sigma })$ is isomorphic to $K_{\sigma }$.

Let $\sigma$ be a proper primitive substitution. For any $x+H_{\sigma }\in K_{\sigma }$,
there exists
$y+H_{\sigma }\in K_{\sigma }$ such that $x-M_{\sigma }y\in H_{\sigma }$.
Then we let $\tau _{\sigma }:K_{\sigma }\longrightarrow K_{\sigma }$ be a map defined by
$\tau _{\sigma }(x+H_{\sigma })=y+H_{\sigma }$.

\begin{thm} Let $\sigma$ be a proper primitive substitution. Then the $K$-groups of $B_{\sigma }$ are given by
\begin{equation*}
K_{0}(B_{\sigma })\simeq \mathbb{Z}\oplus \text{Coker}(1-\tau _{\sigma }), ~~~
K_{1}(B_{\sigma })\simeq \mathbb{Z}\oplus \text{Ker}(1-\tau _{\sigma }).
\end{equation*}
\end{thm}
\begin{proof}
It follows from \cite{Cu2} and the Pimsner-Voiculescu sequence in \cite{Pi}
that there is a exact sequence
$$
\begin{CD}
    {K_{\sigma }}@>{i_*-\rho _{\ast }}>> K_{\sigma }
@>i_*>> K_{0}(B_{\sigma }) \\
   @A{}AA
    @.
     @VV{}V \\
   K_{1}(B_{\sigma }) @<<i_*< \mathbb{Z} @<<{i_*-\rho _{\ast }}< \mathbb{Z}
\end{CD}
$$
where $\rho _{\ast }$ is the induced map of $\rho$ on $K_{0}(\mathcal{F}_{\sigma })$ and $K_{1}(\mathcal{F}_{\sigma })$.
Now $\rho _{\ast }=\tau _{\sigma }$ on $K_{0}(\mathcal{F}_{\sigma })$ and
$\rho _{\ast }=1$ on $K_{1}(\mathcal{F}_{\sigma })$ and we conclude the proof.
\end{proof}

\begin{example}
Let $\sigma$ be a para-periodic proper substitution. Since the map $\tau _{\sigma }$
on $K_{\sigma }$ is given by $\tau _{\sigma } (x+H_{\sigma })=\frac{x}{N_{\sigma }}+H_{\sigma }$, $K_{0}(B_{\sigma })\simeq \mathbb{Z}$ and $K_{1}(B_{\sigma })\simeq \mathbb{Z}$.
\end{example}

redpajama_set_name: RedPajamaArXiv


text:
Q: SQL Server Amount Split I have below 2 tables in SQL Server database.
Customer Main Expense Table
ReportID CustomerID  TotalExpenseAmount
1000       1          200
1001       2          600

Attendee Table
ReportID AttendeeName
1000        Mark
1000        Sam
1000        Joe

There is no amount at attendee level. I have need to manually calculate individual attendee amount as mentioned below. (i.e split TotalExpenseAmount based on number of attendees and ensure individual split figures round to 2 decimals and sums up to the TotalExpenseAmount exactly)
The final report should look like:
ReportID CustID AttendeeName TotalAmount AttendeeAmount
1000       1      Mark          200         66.66
1000       1      Sam           200         66.66
1000       1      Joe           200         66.68

The final report will have about 1,50,000 records. If you notice the attendee amount I have rounded the last one in such a way that the totals match to 200. What is the best way to write an efficient SQL query in this scenario?

A: You can do this using window functions:
select ReportID, CustID, AttendeeName, TotalAmount,
       (case when seqnum = 1
             then TotalAmount - perAttendee * (cnt - 1)
             else perAttendee
        end) as AttendeeAmount
from (select a.ReportID, a.CustID, a.AttendeeName, e.TotalAmount,
             row_number() over (partition by reportId order by AttendeeName) as seqnum,
             count(*) over (partition by reportId) as cnt,
             cast(TotalAmount * 1.0 / count(*) over (partition by reportId) as decimal(10, 2)) as perAttendee
      from attendee a join
           expense e
           on a.ReportID = e.ReportID
     ) ae;

The perAttendee amount is calculated in the subquery.  This is rounded down by using cast() (only because floor() doesn't accept a decimal places argument).  For one of the rows, the amount is the total minus the sum of all the other attendees.

A: Doing something similar to @Gordon's answer but using a CTE instead.
with CTECount AS (
                  select a.ReportId, a.AttendeeName,
                        ROW_NUMBER() OVER (PARTITION BY A.ReportId ORDER BY A.AttendeeName) [RowNum],
                        COUNT(A.AttendeeName) OVER (PARTITION BY A.ReportId) [AttendeeCount],
                        CAST(c.TotalExpenseAmount / (COUNT(A.AttendeeName) OVER (PARTITION BY A.ReportId)) AS DECIMAL(10,2)) [PerAmount]
                  FROM @Customer C INNER JOIN @Attendee A ON A.ReportId = C.ReportID

                  )
SELECT CT.ReportID, CT.CustomerId, AT.AttendeeName,
       CASE WHEN CC.RowNum = 1 THEN  CT.TotalExpenseAmount - CC.PerAmount *              (CC.AttendeeCount - 1)
            ELSE CC.PerAmount END [AttendeeAmount]
FROM @Customer CT INNER JOIN @Attendee AT
        ON CT.ReportID = AT.ReportId
    INNER JOIN CTECount CC
        ON CC.ReportId = CT.ReportID AND CC.AttendeeName = AT.AttendeeName

I like the CTE because it allows me to separate the different aspects of the query.  The cool thing that @Gordon used was the Case statement and the inner calculation to have the lines total correctly.

redpajama_set_name: RedPajamaStackExchange


http://www.svtuition.org/ and how it is used.
Like many other Web sites, http://www.svtuition.org/ makes use of log files. The information inside the log files includes Internet protocol ( IP ) addresses, type of browser, Internet Service Provider ( ISP ), date/time stamp, referring/exit pages, and number of clicks to analyze trends, administer the site, track user's movement around the site, and gather demographic information. IP addresses and other such information are not linked to any information that is personally identifiable.
site. These third-party ad servers or ad networks use technology to the advertisements and links that appear on http://www.svtuition.org/ send directly to your browsers.


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Alex Guarnaschelli Swears By This Advice On Alex Vs America
Dave Kotinsky/Getty Images
By Adrianna MacPherson/Jan. 3, 2022 10:56 am EST
Fans of chef, restaurateur, and television personality Alex Guarnaschelli may have heard of her latest show, "Alex vs America." The new program premiered on January 2, 2022, and the general premise may seem somewhat familiar to fans of "Beat Bobby Flay" and other similar culinary shows, in which a primary chef squares off against various competitors.
As Food Network explains, the show features two rounds of cooking, featuring Guarnaschelli competing against three different chefs, with any of them (even the star chef) liable to being sent home that episode if their plate is deemed the least impressive by the judges during a blind taste test. Additionally, while Guarnaschelli seems to be able to whip up just about anything thanks to her long, diverse career in the culinary industry, the episodes will be themed around particular types of cuisine. And that will allow each episode's participating cooks to be fearsome competitors specializing in that area — Guarnaschelli certainly isn't taking the home court advantage from that perspective.
The first few episodes of the competition focus on chefs that specialize in shellfish, beef, and spicy foods, with the competitors coming from all over the country, bringing their knowledge to the table as they battle it out for a $15,000 prize and, as Food Network says, the "ultimate bragging rights." The competitive angle is certainly not anything unusual for Guarnaschelli, who has duked it out with plenty of worthy competitors on "Iron Chef America" — though she received some solid advice for heading into this show.
Guarnaschelli knows it must be a 'balancing act' in this new competition show
Alex Guarnaschelli regularly shares her expertise and insight on both Twitter and Instagram, and with the premiere of "Alex vs America," she took to Twitter to answer some questions — and share the advice she swears by as she gets ready to face off with new competitors.
Though Guarnaschelli has some serious culinary chops, she doesn't count on being entirely undefeated throughout the whole series. As she wrote on Twitter, "To be honest, the producer of #AlexvsAmerica told me when we started this show: 'You can't win them all but you also can't lose them all. If either of those extremes happen, there is no actual show'!!" The simple advice makes sense — viewers certainly don't want Guarnaschelli to school the competitors without even breaking a sweat herself, but they also don't want the heavy hitters with particular specialties dancing circles around the renowned chef. As with many things, it's a balancing act.
Based on the buzz on Twitter after the premiere January 2, fans seem to be enjoying "Alex vs America" thus far. One wrote, "Love the authenticity of it and how you truly give your competitors the leading edge. Excellent dynamics for a great show!" Another found Guarnaschelli's latest project a breath of fresh air, saying, "It has such a different energy than any other competition show that's out there."
Viewers will have to stay tuned to see what Guarnaschelli's win rate is, and how she stacks up against her competitors in the coming episodes. "Alex vs America" airs Sundays at 10 | 9 pm CT.

redpajama_set_name: RedPajamaCommonCrawl


Addicts and substance abusers who are able to leave drugs in the rear view mirror have scaled one of the most difficult challenges they will ever face in their lives. Their efforts are to be commended and they should be very proud of their achievement. The dangers, however, do not end after rehab. Everyone knows that there is the possibility of relapse, but fewer people know about a common issue known as addiction transfer.
Addiction is a disease of both the body and the mind and addicts are in recovery for the rest of their lives. There is no cure. When the brain is addicted to a substance, it receives a level of pleasure that it can't seem to find in "normal" activities. In other words, achievements and other generally positive aspects of a person's life cannot compare to the euphoria provided by the substance of abuse.
When an addict loses their source of euphoria, they are often left with a void in their lives…an emptiness that must be filled by something that can offer a similar feeling without abusing the same substance. Further, an addict's self-image and feeling of self-worth can be significantly altered. While they may not be chasing the high from drugs, they may have a distorted view of their lives and bodies.
Addicts and their families must be aware of activities that mimic addiction. These may include anorexia or overeating, over-exercising, performing dangerous activities or other behavioral anomalies. These may all be signs that the addicted individual is looking replace the void that withdrawal from substances of abuse has left behind.
Addiction transfer can be managed and prevented, but it has to be identified first. A course of therapy along with reinforcement from families and friends can begin to minimize and eventually stamp out the effects of addiction transfer.


text:
Bair Hugger Warming Blanket
Chlorpyrifos/Pesticides
Chrysler/Jeep/Dodge Airbag
Onglyza
Power Morcellator
Roundup – Monsanto
Zimmer Biomet Shoulder
What Are Mass Torts?
9051 Mira Mesa Blvd. #261176
Xarelto Lawsuits
Extensive research has shown that use of the popular anticoagulant drug Xarelto may be linked to an increased risk of potentially life-threatening bleeding events, and as a result of this alleged Xarelto bleeding risk, patients across the country who suffered internal bleeding complications while taking the anticoagulant medication are filing product liability lawsuits against the drug's manufacturers. If you believe you have been adversely affected by bleeding side effects of Xarelto, contact a qualified product liability lawyer to discuss your options for legal recourse. You may be entitled to financial compensation for your injuries and medical bills, which you can pursue by filing a Xarelto bleeding lawsuit against Bayer Healthcare and Janssen Pharmaceuticals.
How Does Xarelto Work?
Xarelto (rivaroxaban) is an anticoagulant medication sold by Bayer Healthcare and Johnson & Johnson's Janssen Pharmaceuticals subsidiary, and used to prevent blood clots from forming in patients with atrial fibrillation or those undergoing knee or hip replacement surgery. Approved by the U.S. Food and Drug Administration (FDA) in 2011, Xarelto has been marketed by its manufacturers as a safer and more convenient alternative to warfarin, a blood thinner that has been on the market for decades. According to Bayer and Janssen, Xarelto is easier to take than warfarin and other older anticoagulant drugs, because it doesn't require regular checkups or changes in diet and lifestyle during treatment.
Xarelto Case Review
All case evaluations are free and carry no obligation.
What are the Possible Side Effects of Xarelto
Although Xarelto may require less patient monitoring than warfarin, the fact that the new-generation anticoagulant drug does not have an approved antidote makes it considerably more dangerous. Unlike warfarin, whose blood thinning effects can be reversed with a dose of vitamin K, there is no reversal agent for Xarelto, which means doctors may be unable to control internal bleeding complications in patients taking Xarelto, thereby increasing the risk of hospitalization or death. Among the possible side effects associated with Xarelto treatment are:
Abdominal bleeding
Abnormal liver function
Studies Linking Xarelto to Complications
Although Xarelto is designed to prevent blood clots and blood clot-related injuries, adverse event reports submitted to the FDA in connection with Xarelto treatment indicate that patients taking the blood thinner may actually be at an increased risk for blood clots that may trigger heart attacks or strokes if they travel to the brain. In a Phase III study intended to analyze the safety of Xarelto, 73% of the 16,000 patients who took at least one dose of the blood thinner experienced serious side effects, including anemia and bleeding complications. The Xarelto drug label also currently carries a black box warning highlighting the risk of dangerous spinal bleeds called spinal hematomas, which can lead to permanent paralysis.
Researchers have been examining the possible safety risks of Xarelto for years. In 2011, the New England Journal of Medicine published the results of the ROCKET AF study, which compared Xarelto to warfarin in patients with atrial fibrillation. According to the FDA, the study failed to provide adequate evidence that Xarelto is a more effective treatment than warfarin, and the new-generation blood thinner is also known to cause more internal bleeding than its predecessor. In light of the potential for Xarelto cause serious bleeding complications, former Xarelto users across the country are pursuing legal claims against Bayer and Janssen, alleging that the drug makers manufactured a defective drug and failed to adequately warn consumers and doctors about its risks.
Contact our expert Xarelto legal team today for help
Alert: Due to statute of limitations, the time you have to pursue a Xarelto claim is limited. Call us right away at 877-459-8260, email us at intake@monroelawgroup.com, or complete the form on this page now.
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Copyright © 2016 Monroe Law Group. All rights reserved. Privacy / Terms / Disclaimer
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Privacy / Terms / Disclaimer


By now, most homeowners would have heard of HDB's decision for all new HDB BTO units to be built with open-concept kitchens. Now this move was made in response to more home buyers requesting to have the kitchen integrated into the living area, and why not, because an open kitchen not only makes the living room feel spacious and welcoming, it also creates a multi-functional space for your family to work and play in. Here are 6 ways you can enhance the look of your open-concept kitchen and transform it into a cool place to hangout and dine in.
The best part is, you can find all these solutions in one convenient location. A one-stop shop for all your home renovation and furnishing needs, IMM houses a variety of furniture shops and interior design firms to help you with your home renovation. Browse and visit design studios easily as they're all located on the 3rd floor of the mall, including Outlook (#03-02B), G'Plan (#03-28), Space n Living (#03-28A), A Win Design (#03-28C), Artis (#03-28E), Spacious Planners (#03-31), Edgeline Planners (#03-32) and many more.
Aside from breaking down the walls in a bid to create a more spacious kitchen, designers at Posh Living (#03-28B) extended the space by incorporating an island that doubles as a gathering and dining area in the home. This way, you can save on having to craft out another dining area in your home, especially if you have limited floor area to work with.
Aside from spatial planning, ensure the kitchen is easy to maintain too. Choose quartz for your countertop as they are durable, non-porous, stain- and crack-resistant. They also come in a variety of colours and styles including stone-look finish in the design of this luxurious open-concept kitchen by U-Home Interior Design (#03-02A).
The team at U-Home Interior Design also accentuated the ceiling design by outlining the kitchen perimeters with wood-look finishes in this Scandinavian-inspired apartment. This simple strategy helps to not only highlight, but also frames the space in the home.
Another way you can anchor the kitchen island space is by choosing a large, iconic pendant lamp to illuminate the area and create a focal point, such as in this chic white-and-wood-hued home by Ultimate Design Station (#03-28D).
Prefer more colours in your home to add a cheerful vibe? Take inspiration from this café-style kitchen done by the team at Weiken.com (#03-28F). With splashes of pastel shades in pink, purple, yellow, orange and green, it adds some fun to the home environment.
Of course, when it comes to the design of the kitchen, furnishing it right is also important since you'd want a space that meets both aesthetic and functional needs. This way, you can whip up delicious meals for your family and friends conveniently! Song Cho's (#03-09) SC-S2 cooker hood (pictured) comes with active carbon that keeps your kitchen fresh and free of odour. The active carbon, which is placed externally for easy replacement, is good for use for up to 150 hours. With an intelligent sensor control panel, it is also sleek and quiet during operation. Pop by stores like Hoe Kee (#03-08), Sim Siang Choon (#03-10), Grohe (#03-11) and Asia Excel (#03-12) for all your kitchenware and accessories to ensure the heart of your home is well equipped with everything you need!
Brought to you by IMM.


Fellowships are paid or unpaid practical work experiences whose duration lasts longer than a semester.
First Generation is a term used to describe a student whose parents did not attend or graduate from college.
Internships are short-term, paid or unpaid practical work experiences.
Merit-Based opportunities are given to applicants based on achievements.
Need-Based opportunities are given to applicants based on financial need, or if students must come from a low-income household.
Renewable scholarships are scholarships that are able to be renewed each year. Once you are awarded a renewable scholarship, you are usually required to provide information to the providing organization each year in order to receive your scholarship funds.
Scholarships are financial awards provided to students who are entering or currently enrolled in a community college, university, or a graduate degree program.
Pre-College Programs are no-cost or fee-based leadership or academic opportunities exclusively for high school students.


The safety and traffic-flow service qualities of a freeway segment are a function of its prevailing conditions. Good weather, adequate pavement conditions, and an incident-free environment are assumed. Unfortunately, there are many time periods of varying length in which these assumptions are incorrect, and the mobility and safety of a freeway segment are compromised. A more comprehensive knowledge of how poor weather, snow and ice events in particular, impacts the safety and service provided by a freeway will improve the decision-making capabilities of the traveler and those responsible for roadway operations and maintenance. This project will study the traffic characteristics and crash occurrences during winter storm events within the freeway environment. The ability to estimate the operation and safety characteristics of a roadway based on readily available environmental data should assist in the development of winter maintenance standards and policies based on roadway user impacts. In addition, more quantifiable and informed resource allocation and operational-safety response decisions can be made. Based on data from the past, the impacts of a winter storm on traffic volumes and crash occurrence rates will be evaluated and quantified.


How can you write about a product and give your opinion on it if you clearly have not even tested yet?i have watched all the reviews of the phone and everyone seems to love it . What got me is the packaging ,the attention to detail and above all else..no fckibg bloatware.The quality of the phone just seems far superior than the iphone and galaxy by lightyearz.Why did razer join the race in smartphones? Because they can ez as that.
The lack of information that is in this article is shocking, no mention of the quick charge 4.0+ or the fact there is a USB Type -C Audio jack converter. the fact that Razer , a gaming laptop and peripheral company made a phone that is a far-cry better than a lot of cellphone manufacturers in terms of quality and what they offer. You need to do more research into products before even trying to make an article because it doesnt look like any research on the phone.


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K. Michelle Unveils 'Rebellious Soul' Cover & Tracklisting
by Elle Breezy July 16, 2013
K. Michelle luxuriates on the cover of her debut album Rebellious Soul, standing in a room full of candles, roses, a plush, velvet sofa, and of course, a bottle of spirits (likely V.S.O.P). The LP, which features the first single "V.S.O.P," will arrive on August 13 with eleven tracks.
The "Love & Hip-Hop: Atlanta" starlet is ready to be known for her artistry, rather than the antics that ensue on the VH1 small screen. "I was determined with this project that people were gonna finally see me for more than just a crazy, ratchet reality star," she said.
Preorder Rebellious Soul here, and check out the list of tracks below.
Rebellious Soul Tracklisting
1. "My Life"
2. "Damn"
3. "I Don't Like Me"
4. "Can't Raise a Man"
5. "V.S.O.P."
6. "Pay My Bills"
7. "Sometimes"
8. "It's Too Late"
9. "Hate on Her"
10. "When I Get a Man"
11. "A Mother's Prayer"
Taggedk. michelle
Andre Mieux aka A.M.X. – Bedroom Lovers
Robin Thicke's 'Blurred Lines' Nears 1 Million In UK Sales, Reclaims No.1 On UK Chart
Elle Breezy
View all posts by Elle Breezy →
Beyonce on New Album 4:'It Was Not About Singles'
Rihanna and Chris Brown Triple Dip In MTV
Ray J Sets Premiere For Second Round of VH1 Love
Ameriie Ready For New Album After Tying The Knot, Talks Feeling Special
Janet Jackson Dances Atop Club Play Songs Chart
Kenny 'Babyface' Edmonds' 'Fire & Rain' Most Added At Radio


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St. John Frizell writes about the liquored life for Bon Appétit, Saveur, and other magazines such as this one, and runs a café-bar in Red Hook, Brooklyn, called Fort Defiance. When not writing or bartending, he haunts the pubs around Port Authority, carrying a shoebox.
Community, September/October 2012: Issue No. 25, Sister City, Travel
Portland Fever
While other great food cities might rely on their ethnic communities (like Miami's Calle Ocho) or age-old culinary traditions (New Orleans' creole cooking), Portland's status as a food capital is homegrown—it stems directly from its wealth of fresh, local ingredients.
September 4, 2012 | By St. John Frizell
Eat, Magazine, September/October 2011: Issue No. 19, Sourcery
Chef Kurt Gutenbrunner Takes Us on a Tour of Austria's Homeland Flavors
He is a man of voracious appetites and seemingly boundless energy — no wonder his little hometown couldn't hold him.
August 31, 2011 | By St. John Frizell
Edible Manhattan is published six times a year and available by subscription, for sale at selected retailers and at other distribution spots throughout Manhattan. Please visit our sister magazines, Edible Brooklyn, Edible East End, and the Edibles in New York State. And visit Edible Communities to find the publication nearest you.
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Join the Edible Manhattan community.
Subscribe to the Edible Manhattan newsletter to keep up with local food news, events and more!


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TFPDL Best Movie Download Direct Link Free Movies Full Software Full Games
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in Anime November 29, 2022
First comes marriage, then comes an earthbound love that is out of this world. From the day his parents named him, Yuzaki Nasa has been destined for outer space…or so he believes. But while trying to join the space program, he is thrown for a loop when an accident introduces him to Tsukasa, a mysterious girl with strange powers. Is she an alien, a moon goddess or something else? Since she insists on marrying him, Nasa has plenty of chances to find out.
Also Known As: TONIKAWA: Over the Moon for You
(Duration):…………..(30Min)
(Genres):…………….(Animation, Comedy, Romance)
(Ratings):……………(N/10)
:………………… IMDB INFO
(Language):…………..(Japanese | English Subtitle)
(File size):…………..(78MB)
Source:……………….1080p.WEBRip.x264
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Tonikaku Kawaii 2022-11-29
Tagged with: Tonikaku Kawaii
Categories Select Category 1080p Movies 1080p TV Series 2160p Movies 2160p TV Series 480p Movies 480p TV Series 720p Movies 720p TV Series Anime BluRay Complete Season Packs DVDRip DVDScr Ebooks Extra Games HDRip Jay Silverman Movies Movies Trailer Music Oscar Movies 2019 Oscar Movies 2020 PS4 Games Software Sports Top Feature Movies Tv Series WEB-DL WEBRip WEBSCR x265/HEVC
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Now that is what we're talking about Supernatural!
Fans cannot get over how perfect this episode was with developing the main characters and our new favourite, Cole. In the last review (click here), we said we wanted to know more about what Sam was going through, and this second episode followed through and then some. Demon-Dean's characterisation was also dealt with more than we could have hoped for, showing us how different Dean Winchester is as a black-eyed demon.
Sam has been fighting to find where his brother has run off to for the last four months, since the end of Season 9. Season 9 marked the end of an era when it comes to the Winchester brothers. The family business is no more, and that means, with Dean gone, that Sam has nothing to work for anymore except to save Dean. Until this episode, Sam did not know that Dean had been a demon since his disappearance so now, not only is he fighting to find his brother; he is also fighting for his brother's humanity. We got a chance in this episode to see how helpless Sam is but also how unwilling he is to give up on Dean. It is the perfect mirror of the beginning of Season 9 when Dean was willing to try by any means possible to save Sam, and now in Season 10, Sam is hell bent on saving Dean. I was delighted for this comparison. Seeing Sam slap the handcuffs on Demon-Dean was the perfect climax to the episode.
We keep talking about Demon-Dean in these reviews, but Jensen Ackles' performance is what is making the programme sing right now. After ten years, it is incredibly difficult to do something brand new to a show, and with Ackles delivering such a convincing Demon-Dean they have managed to pull it off. Dean in this episode is a bit of an enigma, his confrontation with Cole and Crowley displays his natural demon self which was unbelievably satisfying to watch, but something that many may not have picked up on in the episode was the piano scene when Sam and Dean first met. Dean was sitting at the piano playing a few bars (also what?? Dean can play the piano??) and drinking scotch. What was he playing though? Dean was playing 'Hey Jude', the lullaby his mother sang to him when he was a child. Why was he playing this? Could this mean he's still sentimental? Could this mean he's not fully demonised? Or maybe Dean is playing this as a mockery to his childhood, his innocence. We might never know, but it is definitely something to think about when it comes to Dean's character.
Last week's episode was a bit disappointing, but I would say many of our expectations were unrealistic for a forty minute show, but the second episode of the season made up for it in leaps and bounds. I have never been more excited for a Supernatural season. We are (once again) counting down the days until the next episode as if this were a brand new show!


This machine can print 12mugs at one time.
There some materials and tools coming with this package for you making some test. If want more materials, feel free to contact us.
Description: Qualified inkjet paper for transferring images onto hard surfaces such as coated mugs, coated plates, coated metal boards, etc.
Difference from other paper: It have a layer of coating on the print side, Printed color is bright, transfer greatly.
Note: This paper must be printing by sublimation ink. And this paper cannot use on the 100% pure cotton clothes.
This Art Design Software is a fantastic image and photo editing software for computers that run Windows. It features an intuitive and innovative user interface with support for layers, unlimited undo, special effects, and a wide variety of useful and powerful tools, a powerful yet simple image and photo editor tool. It has been compared to other digital photo editing software packages such as Adobe® Photoshop®, Corel® Paint Shop Pro®, Microsoft Photo Editor, and The GIMP. Now you can download it for free. Please ask seller about download.
Here we specially provide 700 templates for making Mug, Mouse pad, Tiles, anything you can imagine, you can easily add your image to the template and make personality mugs. Surprising!!!!


A San Luis Obispo native, Cal Poly graduate and 3rd generation Realtor, Stephanie Ronca has spent years understanding San Luis Obispo County, the local cities, communities, neighborhoods, school districts and market trends.
Today, as a Realtor at Wilson & Co. Sotheby's International Realty, Stephanie is passionate about helping her clients achieve their real estate goals. Stephanie works with both buyers and sellers and is able to pair her local knowledge and digital marketing background, to efficiently market listings and represent her clients extremely well.
Stephanie is on the Board of Directors for the Cal Poly Alumni Group's SLO Chapter, a member of the San Luis Obispo Association of Realtors and active with the SLO Young Professionals Networking Group. Stephanie currently resides in San Luis Obispo and enjoys hiking, golfing, playing tennis and visiting the local wineries with her friends.


text:
Full Prescribing Information
| Medication Guide |
Important Safety Information | Visit Patient Site
Visit Patient Site
Sleep and your patients
Sleep in the United States
Appropriate patients
Why SILENOR®
SILENOR® in elderly patients
SILENOR® savings
SILENOR® is indicated for the treatment of insomnia characterized by difficulty with sleep maintenance.
SILENOR® is contraindicated in individuals who have shown hypersensitivity to doxepin HCl, any of its inactive ingredients, or other dibenzoxepines. Serious side effects and even death have been reported following the concomitant use of certain drugs with MAO inhibitors (MAOIs). Do not administer SILENOR® if patient is currently on MAOIs or has used MAOIs within the past two weeks. The exact length of time may vary depending on the particular MAOI dosage and duration of treatment.
SILENOR® is contraindicated in individuals with untreated narrow angle glaucoma or severe urinary retention.
The failure of insomnia to remit after 7 to 10 days of treatment may indicate the presence of a primary psychiatric and/or medical illness that should be evaluated.
Complex behaviors such as "sleep-driving" (i.e., driving while not fully awake after ingestion of a hypnotic, with amnesia for the event) have been reported with hypnotics. These events can occur in hypnotic-naive as well as in hypnotic-experienced persons. Although behaviors such as "sleep-driving" may occur with hypnotics alone at therapeutic doses, the use of alcohol or other central nervous system depressants with hypnotics appears to increase the risk of such behaviors, as does the use of hypnotics at doses exceeding the maximum recommended dose. Due to the risk to the patient and the community, discontinuation of SILENOR® should be strongly considered for patients who report a "sleep-driving" episode. Other complex behaviors (i.e., preparing and eating food, making phone calls, or having sex) have been reported in patients who are not fully awake after taking a hypnotic. As with "sleep-driving", patients usually do not remember these events.
Amnesia, anxiety and other neuro-psychiatric symptoms may occur unpredictably.
Patients should not consume alcohol with SILENOR®. Patients should be cautioned about potential additive effects of SILENOR® used in combination with CNS depressants or sedating antihistamines.
In primarily depressed patients, worsening of depression, including suicidal thoughts and actions (including completed suicides), has been reported in association with the use of hypnotics. Doxepin, the active ingredient in SILENOR®, is an antidepressant at doses 10- to 100-fold higher than in SILENOR®. Antidepressants increased the risk compared to placebo of suicidal thinking and behavior in children, adolescents, and young adults in short-term studies of major depressive disorder (MDD) and other psychiatric disorders. Risk from the lower dose of doxepin in SILENOR® cannot be excluded.
Patients should not take SILENOR® unless they are prepared to get a full night's sleep. After taking SILENOR®, patients should confine their activities to those necessary to prepare for bed. Patients should avoid engaging in hazardous activities, such as operating a motor vehicle or heavy machinery, at night after taking SILENOR®, and should be cautioned about potential impairment in the performance of such activities that may occur the day following ingestion.
For faster onset and to minimize the potential for next day effects, SILENOR® should not be taken within 3 hours of a meal.
In clinical trials, the most common treatment-emergent adverse reaction was somnolence/sedation.
SILENOR® has not been studied in pregnant women. SILENOR® is excreted in human milk after oral administration. SILENOR® is not approved for use in children.
Please see full Prescribing Information before prescribing SILENOR®.
You are encouraged to report negative side effects of prescription drugs to the FDA. Visit www.fda.gov/medwatch or call 1-800-FDA-1088.
SILENOR® and A Good Day Starts at Night® are registered trademarks of Currax Pharmaceuticals LLC.
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Was Sam Pepper Really Funny?
If you haven't heard about the Sam Pepper incident until this very moment, allow me to guide you through it.
Sam Pepper is a YouTuber who mainly specializes in "prank" videos. I love a good prank because laughing is great, and I really think our society needs to learn how to laugh a little bit more. But what is the borderline between funny and mean? I think Sam took a big step over this line. Like, the line is in New York, and Sam is in Sydney.
Let me tell you that when I first watched the video, my reaction was, "What the hell?" I was confused at first because of the girls' behavior. Sure, they seemed embarrassed and confused, but no one fought back or punched him in the face. This fact did cause me to have a little bit of a freak-out, but then some wise comments on the video seriously made me think.
Would I be so brave if I were the victim of the prank? Because if I want to be totally honest right now, the answer to that is "no." I wouldn't be so brave. We all have different reactions to these kinds of situations because we haven't been taught how to react properly to harassment.
He is allegedly really happy that his video has started such a storm of indignation. He was "inspired" by a male friend of his who had experienced domestic violence from his partner.
Now, I do agree with some points made in Sam's "apology." Yes, we (as a society) totally need to react to the problem of street harassment, and, yes, we are often guilty of thinking that the victims of sexual harassment and domestic violence are just women.
But this does not excuse the fact that in Sam Pepper's original sexual harassment prank video, he mainly targeted women. In order to "raise awareness" about male domestic violence, do we honestly think that methods like the one Sam Pepper employed are the answer?
After all, I think that Sam's video was wildly inappropriate. Even if you are motivated by a good cause, this is not normal. This is clearly straight-up sexual harassment. Sure, do your research and speak up about an issue, but don't try to solve a problem by instigating it.
Do we need to talk about street harassment more? Yes. Do we need to teach women and men to stand up for themselves? Hell, yeah. But this is definitely NOT the way to do it.
Violet Baudelaire is 20 years old, bilingual (or maybe more), Christian, dreamy, bookworm. Studies in Hungary, lives in Ukraine, heart belongs to England. Wants to work with youth. Spends too much time on Internet. Loves rock music. A huge Sherlock fan. Thinks ordinary life is not cool at all. Wears checked shirts a lot. Loves sci fi. Especially Doctor Who. Obviously, Britain obsessed. Would love to travel around the world with a book and a camera in her hands.