text:
\section{Introduction and preliminaries}
Given a Banach algebra $A$, Kamyabi-Gol {\it et
	al.} in \cite{kam} defined a  new product on $A$ which is denoted by $\ast$. In fact $a\ast b=aeb,$ for each $a,b\in A,$ where $e$ is an element of the closed unit ball $\overline{B^{0}_{1}}$ of $A$. A Banach algebra $A$ equipped with $\ast$ as its product is denoted by $A_{e}$. They studied some properties like amenability and Arens regularity  of $A_{e}$. In \cite{khod} some homological properties of $A_{e}$ like biflatness, biprojectivity and $\phi-$amenability discussed.

New notions of $\phi-$amenability and approximate notions of homological Banach theory introduced and studied for Banach algebras see\cite{sam}, \cite{zhang} and \cite{kan}. In fact a Banach algebra
a Banach algebra $A$ is called approximate $\phi-$contractible if there exists a net ($m_{\alpha})$ in $A$ such that $am_{\alpha}-\phi(a)m_{\alpha}\rightarrow 0,$ and $\phi(m_{\alpha})=1,$ for every $a\in A,$ where $\phi$ is a multiplicative linear functional on $A$. For more information see \cite{agha}.  Also a Banach algebra $A$ is called
	approximate biprojective if there exists a net of
bounded linear maps from $A$ into $A\otimes_{p}A$, say
$(\rho_{\alpha})_{\alpha\in I}$, such that
\begin{enumerate}
	\item [(i)] $a\cdot \rho_{\alpha}(b)-\rho_{\alpha}(ab)\xrightarrow{||\cdot||} 0$,
	\item [(ii)] $\rho_{\alpha}(ba)-\rho_{\alpha}(b)\cdot a\xrightarrow{||\cdot||} 0$,
	\item [(iii)] $\pi_{A}\circ\rho_{\alpha}(a)-a\rightarrow
	0$,
\end{enumerate}
for every $a,b\in A$. In \cite{agha-aust} the structure of approximate biprojective Banach algebras and its nilpotent ideals and also the relation with other notions of amenability are discussed.

We present some standard notations and definitions that we shall need
in this paper. Let $A$ be a Banach algebra. Throughout this work,
the character space of $A$ is denoted by $\Delta(A)$, that is, all
non-zero multiplicative linear functionals on $A$.
 For each
$\phi\in\Delta(A)$ there exists a unique extension $\tilde{\phi}$ to
$A^{**}$ which is defined $\tilde{\phi}(F)=F(\phi)$. It is easy to
see that $\tilde{\phi}\in\Delta(A^{**})$.
 The projective
tensor product
$A\otimes_{p}A$ is a Banach $A$-bimodule via the following actions
$$a\cdot(b\otimes c)=ab\otimes c,~~~(b\otimes c)\cdot a=b\otimes
ca\hspace{.5cm}(a, b, c\in A).$$ The product morphism $\pi_{A}:A\otimes_{p}A\rightarrow A$ is given by $\pi_{A}(a\otimes b)=ab,$ for every $a,b\in A.$

Let $A$ and $B$ be Banach algebras.
We denote by
$\phi\otimes \psi$  a  map defined by $\phi\otimes \psi(a\otimes
b)=\phi(a)\psi(b)$ for all $a\in A$ and $b\in B.$ It is easy to see
that $\phi\otimes \psi\in\Delta(A\otimes_{p}B).$

 Let $X$ and $Y$ be Banach $A-$bimodules. The map $T:X\rightarrow Y$ is called $A-$bimodule morphism, if
$$T(a\cdot x)=a\cdot T(x),\quad T(x\cdot a)=T(x)\cdot a,\qquad (a\in A,x\in X).$$
Also a net of $(T_{\alpha})$ of maps from $X$ into $Y$ is called approximate $A-$bimodule morphism, if
$$T_{\alpha}(a\cdot x)-a\cdot T_{\alpha}(x)\rightarrow 0,\quad T_{\alpha}(x\cdot a)-T_{\alpha}(x)\cdot a\rightarrow 0,\qquad (a\in A,x\in X).$$

The content of the paper is as follows. In section 2 we study $\phi-$homological properties of $A_{e}$ like $\phi-$biflatness and $\phi-$biprojectivity.  Approximate biprojectivity and Johnson pseudo-contractibility are two important notions of Banach homology theory, which we discuss for $A_{e}$ in section 3. We give some examples of matrix algebras to illustrate  the paper.
\section{$\phi-$homological properties of certain Banach algebras}
This section is devoted to the concepts of Banach homology related to a charater $\phi.$
\begin{Proposition}\cite[Proposition 2.3]{kam}\label{exis unit}
Let A be a Banach algebra and $e\in \overline{B^{0}_{1}}$ . Then $A_{e}$ is unital if and only if A is unital and $e$ is invertible.
\end{Proposition}
\begin{Proposition}\cite[Proposition 2.4]{kam}
	Let A be a Banach algebra and $e\in \overline{B^{0}_{1}}$ .Then the followings hold:
	\begin{enumerate}
		\item If $\phi$ is a multiplicative linear functional on A, then  $\phi(e)\phi$ is a
		multiplicative linear functional on $A_{e}$.
		\item If $A_{e}$ is unital and  $\psi$ is a multiplicative linear functional on $A_{e}$, then
		$\phi(a)=\psi(e^{-1}a)$ is a multiplicative linear functional on A.
	\end{enumerate}
\end{Proposition}
\begin{Proposition}\cite[Proposition 2.3]{khod}\label{khod}
	Let A be a Banach algebra and  $e\in \overline{B^{0}_{1}}$. If $A_{e}$ is unital
	then $(A_{e})_{e^{-2}}$ = A, ( isometrically isomorphism ).
\end{Proposition}
\begin{Proposition}\label{app left}
Suppose that $A$ is a Banach algebra and also suppose that $e\in \overline{B^{0}_{1}}$ and $\phi\in\Delta(A)$. Then the followings hold:
\begin{enumerate}
\item If $A$ is approximate $\phi-$contractible and $\phi(e)\neq 0$, then $A_{e} $ is approximately  $\psi$-contractible, where $\psi=\phi(e)\phi$.
\item If $A_{e}$ is unital and approximate $\psi-$contractible, then $A$ is approximate $\phi$-contractible, where $\phi(a)=\psi(e^{-1}a)$ for each $a\in A$.
\end{enumerate}
\end{Proposition}
\begin{proof}
Suppose that $A$ is approximately $\phi-$contractible. So there is a net $(m_{\alpha})$ in $A$ such that $$am_{\alpha}-\phi(a)m_{\alpha}\rightarrow 0,\quad \phi(m_{\alpha})=1,\qquad (a\in A).$$ Define $n_{\alpha}=\frac{m_{\alpha}}{\phi(e)}$. Thus we have
\begin{equation*}
\begin{split}
a\ast n_{\alpha}-\psi(a)n_{\alpha}&=aen_{\alpha}-\psi(a)n_{\alpha}\\
&=ae\frac{m_{\alpha}}{\phi(e)}-\psi(a)\frac{m_{\alpha}}{\phi(e)}\\
&=ae\frac{m_{\alpha}}{\phi(e)}-\phi(ae)\frac{m_{\alpha}}{\phi(e)}+\phi(ae)\frac{m_{\alpha}}{\phi(e)}-\psi(a)\frac{m_{\alpha}}{\phi(e)}\rightarrow 0,\qquad (a\in A_{e}).
\end{split}
\end{equation*}
Also $$\psi(n_{\alpha})=\psi(\frac{m_{\alpha}}{\phi(e)})=\phi(e)\phi(\frac{m_{\alpha}}{\phi(e)})=\phi(m_{\alpha})=1.$$ It follows that $A_{e}$ is approximate  $\phi-$contractible.

Suppose that $\phi(a)=\psi(e^{-1}a)$ and also suppose that $A_{e}$ is unital and approximately left $\psi-$contractible. It is easy to see that $\psi(a)=\phi(ea)$.  Let $(m_{\alpha})$ be a net in $A_{e} $ such that
$$a\ast m_{\alpha}-\psi(a)m_{\alpha}\rightarrow 0,\quad \psi(m_{\alpha})=1,\qquad (a\in A_{e}).$$ Since
\begin{equation*}
\begin{split}
a\ast m_{\alpha}-\psi(a)m_{\alpha}&=aem_{\alpha}-\psi(a)m_{\alpha}\\
&=aem_{\alpha}-\phi(ea)m_{\alpha}\\
&=aem_{\alpha}-\phi(e)\phi(a)m_{\alpha}\\
&=aem_{\alpha}-\phi(a)\phi(e)m_{\alpha}\\
&=aem_{\alpha}-\phi(ae)m_{\alpha},
\end{split}
\end{equation*}
we have
$$a\ast m_{\alpha}-\psi(a)m_{\alpha}=aem_{\alpha}-\phi(ae)m_{\alpha}\rightarrow 0$$ for each $a\in A.$ Replacing $a$ with $ae^{-1}$ we have $am_{\alpha}-\phi(a)m_{\alpha}\rightarrow 0.$ Regarding  $$1=\psi(m_{\alpha})=\phi(em_{\alpha})=\phi(e)\phi(m_{\alpha}),$$
we may suppose that $\phi(m_{\alpha})\neq 0,$ for each $\alpha.$ Now define $n_{\alpha}=\frac{m_{\alpha}}{\phi(m_{\alpha})}$. Clearly $\phi(n_{\alpha})=1.$ Also $$an_{\alpha}-\phi(a)n_{\alpha}=a\frac{m_{\alpha}}{\phi(m_{\alpha})}-\phi(a)\frac{m_{\alpha}}{\phi(m_{\alpha})}\rightarrow 0.$$ It finishes the proof.
\end{proof}
\begin{Example}
	Let  $A=\{\left(\begin{array}{ccc} a_{11}&a_{12}&a_{13}\\
	0&a_{22}&a_{23}\\
	0&0&a_{33}
	\end{array}
	\right)| a_{ij}\in \mathbb{C}\}$ and suppose that $e=\left(\begin{array}{ccc} \frac{1}{6}&\frac{1}{6}&\frac{1}{6}\\
	0&\frac{1}{6}&\frac{1}{6}\\
	0&0&\frac{1}{6}
	\end{array}
	\right)$. Clearly $A$ with matrix operations and $\ell^1$-norm is a Banach algebra. We know that  $e$ is invertible and by Proposition \ref{exis unit}, $A_{e}$ is unital. Define $\phi:A\rightarrow \mathbb{C}$ by $$\phi(\left(\begin{array}{ccc} a_{11}&a_{12}&a_{13}\\
	0&a_{22}&a_{23}\\
	0&0&a_{33}
	\end{array}
	\right))=a_{33}.$$ Clearly $\phi$ is a character(multiplicative linear functional) and $\phi(e)\neq 0$. Suppose conversely that $A_e$ is approximate $\psi-$contractible. By previous Proposition(2), $A$ becomes approximate $\phi-$contractible. On the other hand by the same arguments as in  the proof of \cite[Theorem 5.1]{sah taki}
 $A$ is not approximate $\phi-$contractible, which is a contradiction.
\end{Example}
Let $A$ be a Banach algebra and $\phi\in \Delta(A)$. $A$ is called
$\phi$-biprojective, if there exists a bounded $A$-bimodule morphism
$\rho:A\rightarrow A\otimes_{p}A$ such that $\phi \circ\pi_{A}
\circ\rho=\phi$. Also $A$ is called $\phi$-biflat if there exists a
bounded $A$-bimodule morphism $\rho:A\rightarrow (A\otimes_{p}
A)^{**}$ such that $\tilde{\phi} \circ \pi_{A}^{**} \circ\rho=\phi$. For more information about $\phi-$biflatness and $\phi-$biprojectivity, the reader refers to \cite{sah phi-biflat} and \cite{sah col}.

\begin{Theorem}
	Let $A$ be a Banach algebra and $\phi\in\Delta(A)$. Suppose that $e\in \overline{B^{0}_{1}}$ and $\phi(e)\neq 0.$
	If $A$ is $\phi$-biprojective, then $A_{e}$ is $\psi=\phi(e)\phi$-biprojective.
\end{Theorem}
\begin{proof}
Since 	$A$ is $\phi$-biprojective, there exists  a bounded $A-$bimodule morphism $\rho:A\rightarrow A\otimes_{p}A$ such that $\phi\circ\pi_{A}\circ\rho =\phi.$ Define $\tilde{\rho}=\frac{1}{\phi(e)}\rho$. We show that $\tilde{\rho}$ is a bounded $A_e$-bimodule morphism. To see this, consider
\begin{equation*}
\begin{split}
\tilde{\rho}(a\ast b)=\frac{1}{\phi(e)}\rho(a\ast b)=\frac{1}{\phi(e)}\rho(a e b)&=ae\frac{1}{\phi(e)}\rho( b)\\
&=a\ast \frac{1}{\phi(e)}\rho( b)\\
&=a\ast \tilde{\rho}( b),\qquad (a,b\in A_{e}).
\end{split}
\end{equation*}
Also
\begin{equation*}
\begin{split}
\tilde{\rho}(a\ast b)=\frac{1}{\phi(e)}\rho(a\ast b)=\frac{1}{\phi(e)}\rho(a e b)&=\frac{1}{\phi(e)}\rho( a)be\\
&= \frac{1}{\phi(e)}\rho( a)\ast b\\
&= \tilde{\rho}( a)\ast b,\qquad (a,b\in A_{e}).
\end{split}
\end{equation*}
On the other hand, since $$\psi\circ \pi_{A_{e}}\circ \tilde{\rho}=\phi(e)\phi\circ\pi_{A}\circ\rho,$$
we have
\begin{equation*}
\begin{split}
\psi\circ \pi_{A_{e}}\circ\tilde{\rho}(a)=\phi(e)\phi\circ\pi_{A}\circ\rho(a)=\phi(e)\phi(a)=\psi(a),\quad (a\in A_{e}).
\end{split}
\end{equation*}
So $A_{e}$ is $\psi-$biprojective.
	\end{proof}
Using the similar arguments as in the proof of the previous theorem, we have the following corollary:
\begin{cor}
	Let $A$ be a Banach algebra and $\phi\in\Delta(A)$. Suppose that $e\in \overline{B^{0}_{1}}$ and $\phi(e)\neq 0.$
	If $A$ is $\phi$-biflat, then $A_{e}$ is $\psi=\phi(e)\phi$-biflat.
\end{cor}
Let $A$ be a Banach algebra and $\phi\in \Delta(A)$.  $A$ is called $\phi-$amenable if there exists a bounded net $(m_{\alpha})$ in $A$ such that $am_{\alpha}-\phi(a)m_{\alpha}\rightarrow 0$ and $\phi(m_{\alpha})=1,$ for every $a\in A,$ see \cite{kan}.
\begin{cor}
	Let $A$ be a Banach algebra and $\phi\in\Delta(A)$. Suppose that $e\in \overline{B^{0}_{1}}$ and $\phi(e)\neq 0.$
	If $A$ is $\phi$-biflat and $A$ has a left approximate identity, then $A_{e}$ is approximate  $\psi=\phi(e)\phi$-contractible.
\end{cor}
\begin{proof}
Since  $A$ is $\phi$-biflat and $A$ has a left approximate identity, by similar arguments as in the proof of \cite[Theorem 2.2]{sah taki} $A$ is  $\phi$-amenable. It is easy to see that   $\phi$-amenability of $A$ implies that $A$ is approximate $\phi-$contractible. Applying Proposition \ref{app left}, $A_{e}$ becomes approximate $\psi-$contractible.
\end{proof}
Let $A$ b a Banach algebra and $\phi\in\Delta(A).$ Then
 $A$ is called
approximate left $\phi$-biprojective if there exists a net of
bounded linear maps from $A$ into $A\otimes_{p}A$, say
$(\rho_{\alpha})_{\alpha\in I}$, such that
\begin{enumerate}
	\item [(i)] $\rho_{\alpha}(ab)-\phi(a)\rho_{\alpha}(b)\xrightarrow{||\cdot||} 0$,
	\item [(ii)] $\rho_{\alpha}(ba)-\rho_{\alpha}(b)\cdot a\xrightarrow{||\cdot||} 0$,
	\item [(iii)] $\pi_{A}\circ\rho_{\alpha}(a)-a\rightarrow
	0$,
\end{enumerate}
for every $a,b\in A$, see \cite{sah glas}.
\begin{Theorem}\label{app phi proj}
	Let $A$ be a Banach algebra and $\phi\in\Delta(A)$. Suppose that $e\in \overline{B^{0}_{1}}$ and $\phi(e)\neq 0.$
	If $A$ is approximate left $\phi$-biprojective, then $A_{e}$ is approximate left $\psi=\phi(e)\phi$-biprojective.
\end{Theorem}
\begin{proof}
	Since 	$A$ is approximate left $\phi$-biprojective, there exists  a net of bounded linear maps $(\rho_{\alpha})$ from $A$ into $A\otimes_{p}A$ such that
	$$\rho_{\alpha}(ab)-\phi(a)\rho_{\alpha}(b)\rightarrow 0,\quad \rho_{\alpha}(ab)-\rho_{\alpha}(a)\cdot b\rightarrow 0,\quad \phi\circ\pi_{A}\circ\rho(a)-\phi(a)\rightarrow 0.$$ Define $\tilde{\rho}_{\alpha}=\frac{1}{\phi(e)}\rho_{\alpha}$. We show that there exists  a net of bounded linear maps $(\tilde{\rho}_{\alpha})$ from $A_{e}$ in to $A_{e}\otimes_{p}A_{e}$ such that
	$$\tilde{\rho}_{\alpha}(a\ast b)-\psi(a)\tilde{\rho}_{\alpha}(b)\rightarrow 0,\quad \tilde{\rho}_{\alpha}(a\ast b)-\tilde{\rho}_{\alpha}(a)\ast b\rightarrow 0,\quad \psi\circ\pi_{A}\circ\tilde{\rho}(a)-\psi(a)\rightarrow 0.$$To see this, consider
	\begin{equation*}
	\begin{split}
	\tilde{\rho}_{\alpha}(a\ast b)-\psi(a)\tilde{\rho}_{\alpha}(b)
	&=\tilde{\rho}_{\alpha}(ae b)-\phi(a)\phi(e)\tilde{\rho}_{\alpha}(b)\\
	&=\frac{1}{\phi(e)}(\rho_{\alpha}(ae b)-\phi(a)\phi(e)\rho_{\alpha}(b))\\
	&=\frac{1}{\phi(e)}(\rho_{\alpha}(ae b)-\phi(ae)\rho_{\alpha}( b)+\phi(ae)\rho_{\alpha}( b)-\phi(a)\phi(e)\rho_{\alpha}(b))\\
	&\rightarrow 0
	\end{split}
	\end{equation*}
	Also
	\begin{equation*}
	\begin{split}
	\tilde{\rho}_{\alpha}(a\ast b)-\tilde{\rho}_{\alpha}(a)\ast b=\frac{1}{\phi(e)}\rho_{\alpha}(ae b)-\frac{1}{\phi(e)}\rho_{\alpha}(a)eb\rightarrow 0
		\end{split}
	\end{equation*}
	On the other hand, since $$\psi\circ \pi_{A_{e}}\circ \tilde{\rho}_{\alpha}=\phi(e)\phi\circ\pi_{A}\circ\rho_{\alpha},$$
	we have
	\begin{equation*}
	\begin{split}
	\psi\circ \pi_{A_{e}}\circ\tilde{\rho}_{\alpha}(a)-\psi(a)=\phi(e)\phi\circ\pi_{A}\circ\rho_{\alpha}(a)-\phi(e)\phi(a)\rightarrow \phi(e)\phi(a)- \phi(e)\phi(a)=0,\quad (a\in A_{e}).
	\end{split}
	\end{equation*}
	So $A_{e}$ is approximate left $\psi-$biprojective.
\end{proof}
\begin{Remark}\label{remark1}
	Let $A$ and $B $ be  Banach algebras and $e_{A}\in  \overline{B^{0}_{1}}^{A}$ and $e_{B}\in  \overline{B^{0}_{1}}^{B}$. Then there exist two sequences $(x_{n})$ and $(y_{n})$ in the unit ball $A$ and the unit ball B such that $x_{n}\rightarrow e_{A} $ and $y_{n}\rightarrow e_{B},$ respectively. Since $$||x_{n}\otimes y_{n}-e_{A}\otimes e_{B}||\leq ||x_{n}\otimes y_{n}-e_{A}\otimes y_{n}||+||e_{A}\otimes y_{n}-e_{A}\otimes e_{B}||\rightarrow 0,$$
	we have $e_{A}\otimes e_{B}\in  \overline{B^{0}_{1}}^{A\otimes_{p}B}$.  Define $T:A_{e_{A}}\otimes_{p}B_{e_{B}}\rightarrow {A\otimes_{p}B}_{e_{A}\otimes e_{B}}$ by
	$T(a\otimes b)=a\otimes b$ for every $a\in A$ and $b\in B.$ It is easy to see that $T$ is an isometric algebra isomorphism. Also $T$ is a bounded ${A\otimes_{p}B}_{e_{A}\otimes e_{B}}-$bimodule morphism.
\end{Remark}
\begin{Proposition}
	Let $A$ and $B $ be  Banach algebras and $e_{A}\in  \overline{B^{0}_{1}}^{A}$ and $e_{B}\in  \overline{B^{0}_{1}}^{B}$. Suppose that $\phi_{A}\in\Delta(A)$ and $\phi_{B}\in \Delta(B)$ which $\phi_{A}(e_{A})\neq 0$ and $\phi_{B}(e_{B})\neq 0.$ If $A$ and $B$ are $\phi_{A}-$biprojective and $\phi_{B}-$biprojective, repectively, then 	${A\otimes_{p}B}_{e_{A}\otimes e_{B}}$ is $\phi_{A}(e_A)\phi_{A}\otimes \phi_{B}(e_B)\phi_{B}-$biprojective.
\end{Proposition}
\begin{proof}
Since $A$ and $B$ are $\phi_{A}-$biprojective and $\phi_{B}-$biprojective, repectively, then by Theorem \ref{app phi proj}, $A_{e}$ and $B_{e} $ are $\phi_{A}(e_A)\phi_{A}-$biprojective and $\phi_{B}(e_B)\phi_{B}-$biprojective, respectively. So there exist a $A_{e_{A}}-$bimodule morphism
$\rho_{0}:A_{e_{A}}\rightarrow A_{e_{A}}\otimes_{p}A_{e_{A}}$ and a $B_{e_{B}}$-bimodule morphism $\rho_{1}:B_{e_B}\rightarrow
B_{e_B}\otimes_{p}B_{e_B}$  such that $\phi_{A}(e_A)\phi_{A} \circ\pi_{A}\circ\rho_{0}=\phi_{A}(e_A)\phi_{A}$
and $\phi_{B}(e_B)\phi_{B} \circ\pi_{B} \circ\rho_{1}=\phi_{B}(e_B)\phi_{B}$.
Define
$\theta:(A_{e_{A}}\otimes_{p}A_{e_{A}})\otimes_{p}(B_{e_{B}}\otimes_{p}B_{e_{B}})\rightarrow(A_{e_{A}}\otimes_{p}B_{e_{B}})\otimes_{p}(A_{e_{A}}\otimes_{p}B_{e_{B}})
$ by $$(a_{1}\otimes a_{2})\otimes (b_{1}\otimes b_{2})\mapsto
(a_{1}\otimes b_{1})\otimes (a_{2}\otimes b_{2}),$$ where $a_{1},
a_{2}\in A$ and $b_{1}, b_{2}\in B$. Clearly $\theta $ is an isometric algebra isomorphism.  Set $\rho=(T\otimes T)\circ \theta
\circ(\rho_{0}\otimes\rho_{1})\circ T^{-1}$,  where $T$ is the map defined as in Remark \ref{remark1}. We know that  $\rho$ is a bounded linear map from $ A\otimes_{p}B_{e_{A}\otimes e_{B}}$ into $ (A\otimes_{p}B_{e_{A}\otimes e_{B}})\otimes_{p}(A\otimes_{p}B_{e_{A}\otimes e_{B}})$. Consider
$$\pi_{A\otimes_{p}B_{e_{A}\otimes e_{B}}}\circ\theta(a_{1}\otimes a_{2}\otimes b_{1}\otimes
b_{2})=\pi_{A\otimes_{p}B_{e_{A}\otimes e_{B}}}(a_{1}\otimes b_{1}\otimes a_{2}\otimes
b_{2})=\pi_{A_{e_{A}}}(a_{1}\otimes a_{2})\otimes\pi_{B_{e_{B}}}(b_{1}\otimes b_{2}),$$
then clearly one can show that
$\pi_{A\otimes_{p}B_{e_{A}\otimes e_{B}}}\circ\theta=\pi_{A_{e_{A}}}\otimes\pi_{B_{e_{B}}}$. Hence,
$$\pi_{A\otimes_{p}B_{e_{A}\otimes e_{B}}}\circ\theta(\rho_{0}(a)\otimes\rho_{1}(b))=\pi_{A_{e_{A}}}\circ\rho_{0}(a)\otimes\pi_{B_{e_{B}}}\circ\rho_{1}(b)$$
and it is easy to see that $$\phi_{A}(e_A)\phi_{A}\otimes \phi_{B}(e_B)\phi_{B}
\circ\pi_{A\otimes_{p}B}\circ\theta(\rho_{0}\otimes\rho_{1})(a\otimes
b)=\phi_{A}(e_A)\phi_{A}\otimes \phi_{B}(e_B)\phi_{B}(a\otimes b),$$  the proof is complete.
	\end{proof}
\section{Approximate homological properties of certain Banach algebras}
In this section we investigate approximate biprojectivity and Johnson pseudo-contractibility of $A_{e}$.
\begin{Theorem}\label{Approximate homological}
Suppose that $A$ is a Banach algebra and also suppose that $e\in \overline{B^{0}_{1}}$. Then the followings hold:
\begin{enumerate}
\item If $A$ is approximately biprojective and $A_{e}$ is unital then $A_{e}$ is approximately biprojective.
\item If $A_{e}$ is unital and approximately biprojective, then $A$ is approximately biprojective.
\end{enumerate}
\end{Theorem}
\begin{proof}
To show (1), suppose that $A$ is approximately biprojective and $A_{e}$ is unital. It follows that there is an approximately $A-$bimodule morphism $(\rho_{\alpha}) $ from $A$ into $A\otimes_{p}A$ such that $\pi_{A}\circ\rho_{\alpha}(a)-a\rightarrow 0$ for each $a\in A.$ Note that
\begin{equation*}
\begin{split}
\rho_{\alpha}(a\ast b)-a\ast \rho_{\alpha}(b)&=\rho_{\alpha}(aeb)-a\ast \rho_{\alpha}(b)\\
&=\rho_{\alpha}(aeb)-ae\rho_{\alpha}(b)+ae\rho_{\alpha}(b)-a\ast \rho_{\alpha}(b)\rightarrow 0,
\end{split}
\end{equation*}
and
\begin{equation*}
\begin{split}
\rho_{\alpha}(a\ast b)- \rho_{\alpha}(a)\ast b&=\rho_{\alpha}(aeb)-\rho_{\alpha}(a)\ast b\\
&=\rho_{\alpha}(aeb)-\rho_{\alpha}(a)eb+\rho_{\alpha}(a)eb- \rho_{\alpha}(a)\ast b\rightarrow 0,
\end{split}
\end{equation*}
for each $a\in A_{e}.$ It implies that $(\rho_{\alpha})$ from $A_{e}$ into $A_{e}\otimes_{p}A_{e}$ is an approximately $A_{e}$-bimodule morphism. Define $T:A_{e}\otimes_{p}A_{e}\rightarrow A_{e}\otimes_{p}A_{e}$ by $T(a\otimes b)=ae^{-1}\otimes b.$ Note that using Proposition \ref{exis unit},  the definition of $T$ makes sense. It is easy to see that $$T(a\ast (c\otimes d))=a\ast T(c\otimes d),\quad T((c\otimes d)\ast a)=T(c\otimes d)\ast a,\qquad (a,c,d\in A).$$ Set $\tilde{\rho}_{\alpha}=T\circ \rho_{\alpha}$. Using direct calculations we can see that  $$\pi_{A_{e}}\circ \tilde{\rho}_{\alpha}=\pi_{A}\circ\rho_{\alpha}$$
It follows that
$$\pi_{A_{e}}\circ \tilde{\rho}_{\alpha}-a=\pi_{A}\circ\rho_{\alpha}-a\rightarrow 0,\qquad (a\in A_{e}).$$
Thus $A_{e}$ is approximately biprojective.

To show (2), suppose that $A_{e}$ is unital and approximately biprojective. By Proposition \ref{khod}, we know that $A=(A_{e})_{e^{-2}}$. Now applying (1) it is easy to see that $A$ is approximately biprojective.
\end{proof}
A Banach algebra $A$ is called biprojective if there exists a bounded $A-$bimodule morphism $\rho:A\rightarrow A\otimes_{p}A$ such that $\pi_{A}\circ\rho(a)=a$ for each $a\in A,$ see  \cite{run}.
\begin{Example}
	Let  $A=\{\left(\begin{array}{ccc} a_{11}&a_{12}&a_{13}\\
	a_{21}&a_{22}&a_{23}\\
	a_{31}&a_{32}&a_{33}
	\end{array}
	\right)| a_{ij}\in \mathbb{C}\}$. With the matrix operations  and $\ell^1$-norm, $A$ becomes a Banach algebra.  Suppose that $e=\left(\begin{array}{ccc} \frac{1}{4}&\frac{1}{4}&0\\
	0&\frac{1}{4}&0\\
	0&0&\frac{1}{4}
	\end{array}
	\right)$. Clearly $e$ is invertible and $A$ is unital. So by Proposition \ref{exis unit}, $A_{e}$ is unital. It is well-known that $A$ is biprojective, see \cite{run}. So $A$ is approximately biprojective. Applying previous theorem $A_{e}$ becomes approximately biprojective.
\end{Example}
\begin{Definition}
We say that a Banach algebra $A$ has approximate (F)-property(or $A$ is AFP) if  there is an approximate $A-$bimodule morphsim $(\rho_{\alpha})$ from $A$ into $(A\otimes_{p}A)^{**}$ such that $\pi^{**}_{A}\circ\rho_{\alpha}(a)-a\rightarrow 0,$ for each $a\in A.$
\end{Definition}
 For the motivation of this definition see \cite{askar}.
 \begin{Proposition}
 	If $A$ is AFP and $A_{e}$ is unital, then $A_{e}$ is approximately biprojective.
 \end{Proposition}
\begin{proof}
Since $A$ is AFP, there exists an approximate $A-$bimodule morphsim $(\rho_{\alpha})$ from $A$ into $(A\otimes_{p}A)^{**}$ such that $\pi^{**}_{A}\circ\rho_{\alpha}(a)-a\rightarrow 0,$ for each $a\in A.$ It is easy to see that $(\rho_{\alpha})$  is an approximate $A_{e}-$bimodule morphsim from $A_{e}$ into $(A_{e}\otimes_{p}A_{e})^{**}$ such that $\pi^{**}_{A_{e}}\circ\rho_{\alpha}(a)-a\rightarrow 0,$ for each $a\in A_{e}.$  Let $T:A_{e}\otimes_{p}A_{e}\rightarrow A_{e}\otimes_{p}A_{e}$ be the same map as in the proof of Theorem \ref{Approximate homological}. Clearly $T$ is $A_{e}$-module morphism, so is $T^{**}$. Similar to the proof of Theorem \ref{Approximate homological}, for the net $(T^{**}\circ \rho_{\alpha})$ is an approximate $A_{e}-$bimodule morphism from $A_{e}$ into $(A_{e}\otimes_{p}A_{e})^{**}$ such that
$$\pi^{**}_{A_{e}}\circ T^{**}\circ\rho_{\alpha}(a)-a=\pi^{**}_{A}\circ \rho_{\alpha}(a)-a\rightarrow 0,\quad (a\in A).$$ We denote the identity of $A_{e}$
with $a_{0}$ and define $m_{\alpha}=\rho_{\alpha}(a_{0})$. Clearly $(m_{\alpha})$ is a net in $(A_{e}\otimes_{p}A_{e})^{**}$ which satisfies $$a\ast m_{\alpha}-m_{\alpha}\ast a\rightarrow 0,\qquad \pi_{A_{e}}^{**}(m_{\alpha})a-a\rightarrow 0,\quad (a\in A_{e}).$$
 Take $\epsilon>0$ and arbitrary finite subsets $F\subseteq A_{e}$,
$\Lambda\subseteq (A_{e}\otimes_{p}A_{e})^{*}$ and $\Gamma \subseteq A_{e}^*$.
Then we have
$$||a\ast
m_{\alpha}-m_{\alpha}\ast a||<\epsilon,\quad
||\pi_{A_{e}}^{**}(m_{\alpha})a-a||<\epsilon,\quad (a\in
F).$$ It is well-known that for each $\alpha$, there exists a net
$(n^{\alpha}_{\beta})_{\beta}$ in $A_{e}\otimes_{p}A_{e}$ such that
$n^{\alpha}_{\beta}\xrightarrow{w^{*}}m_{\alpha}$. Since
$\pi^{**}_{A_{e}}$ is a $w^{*}$-continuous map, we have
$$\pi_{A_{e}}(n^{\alpha}_{\beta})=\pi_{A_{e}}^{**}(n^{\alpha}_{\beta})\xrightarrow{w^{*}}\pi_{A}^{**}(m_{\alpha}).$$
Thus we have $$|a\ast
n^{\alpha}_{\beta}(f)-a\ast m_{\alpha}(f)|<\frac{\epsilon}{K_{0}},\quad
|
n^{\alpha}_{\beta}\ast a(f)-m_{\alpha}\ast(f)|<\frac{\epsilon}{K_{0}}$$
and
$$|\pi_{A_{e}}(n^{\alpha}_{\beta})(g)-\pi_{A_{e}}^{**}(m_{\alpha})(g)|<\frac{\epsilon}{K_{1}},$$
for each $ a\in F$, $ f\in\Lambda$ and $g\in A^*$, where $K_{0}=\sup\{||f||:f\in
\Lambda\}$ and $K_{1}=\sup\{||g||:g\in
\Gamma\}$. Since $a\ast m_{\alpha}-m_{\alpha}\ast a\rightarrow 0$ and  $\pi_{A_{e}}^{**}(m_{\alpha})\ast a-a\rightarrow 0,$ we can find
$\beta=\beta(F,\Lambda,\Gamma,\epsilon)$ such that
$$|a\ast n^{\alpha}_{\beta}(f)-n^{\alpha}_{\beta}\ast a(f)|<c\frac{\epsilon}{K_{0}},\quad |\pi_{A_{e}}(n^{\alpha}_{\beta})\ast a(g)-a(g)|<\frac{\epsilon}{K_{1}},\quad (a\in F,f\in \Lambda,g\in \Gamma)$$
for some $c\in \mathbb{R^{+}}$. Using Mazur's lemma, we have a net
$(n_{(F,\Lambda,\Gamma,\epsilon)})$ in $A_{e}\otimes_{p}A_{e}$ such that $$||a\ast
n_{(F,\Lambda,\Gamma,\epsilon)}-n_{(F,\Lambda,\Gamma,\epsilon)}\ast a||\rightarrow
0,\quad || \pi_{A}(n_{(F,\Lambda,\Gamma,\epsilon)})\ast a-a||\rightarrow
0,\quad (a\in F).$$ Define $\rho_{(F,\Lambda,\Gamma,\epsilon)}:A_{e}\rightarrow
A_{e}\otimes_{p}A_{e}$ by $\rho_{(F,\Lambda,\Gamma,\epsilon)}(a)=a\ast
n_{(F,\Lambda,\Gamma,\epsilon)} $ for each $a\in A_{e}.$ It is clear that
$\rho_{(F,\Lambda,\Gamma,\epsilon)}(a\ast b)=a\ast
\rho_{(F,\Lambda,\Gamma,\epsilon)}(b)$ for each $a,b\in A.$ Also
\begin{equation}
\begin{split}
||\rho_{(F,\Lambda,\epsilon)}(a\ast b)-\rho_{(F,\Lambda,\Gamma,\epsilon)}(a)\ast b||&=||ab\ast
n_{(F,\Lambda,\Gamma,\epsilon)}-a\ast( n_{(F,\Lambda,\Gamma,\epsilon)}\ast b)||\\
&\leq ||a||||b\ast n_{(F,\Lambda,\Gamma,\epsilon)}-
n_{(F,\Lambda,\Gamma,\epsilon)}\ast b||\rightarrow 0,
\end{split}
\end{equation}
for each $a,b\in A_{e}.$ Also
\begin{equation}
\begin{split}
||\pi_{A_{e}}\circ\rho_{(F,\Lambda,\Gamma,\epsilon)}(a)-a||&=||\pi_{A_{e}}(a\ast
n_{(F,\Lambda,\Gamma,\epsilon)})-a||\\&=||\pi_{A_{e}}(a\ast
n_{(F,\Lambda,\Gamma,\epsilon)})-\pi_{A_{e}}(
n_{(F,\Lambda,\Gamma,\epsilon)}\ast a)+\pi_{A_{e}}(
n_{(F,\Lambda,\Gamma,\epsilon)}\ast a)-a||\\
&\leq ||\pi_{A_{e}}(a\ast
n_{(F,\Lambda,\Gamma,\epsilon)})-\pi_{A_{e}}(
n_{(F,\Lambda,\Gamma,\epsilon)}\ast a)||+||\pi_{A_{e}}(
n_{(F,\Lambda,\Gamma,\epsilon)})\ast a-a||\\
&\rightarrow 0,
\end{split}
\end{equation}
for each $a\in F$. Thus with respect to the net $(\rho_{(F,\Lambda,\Gamma,\epsilon)})_{(F,\Lambda,\Gamma,\epsilon)},$ $A_{e}$ becomes approximately biprojective.

	\end{proof}
A Banach algebra $A$ is called Johnson pseudo-contractible,  if
there exists a not necessarily bounded net $(m_{\alpha})$ in
$(A\otimes_{p}A)^{**}$ such that $a\cdot m_{\alpha}=m_{\alpha}\cdot
a$ and $\pi^{**}_{A}(m_{\alpha})a-a\rightarrow 0,$ for every $a\in
A,$ see \cite{sah new amen1} and \cite{sah1}.

A Banach algebra $A$ is called biflat,   if  there is a bounded $A-$bimodule morphsim $\rho$ from $A$ into $(A\otimes_{p}A)^{**}$ such that $\pi^{**}_{A}\circ\rho_{\alpha}(a)=a$, for each $a\in A$, see \cite{run}.

\begin{Proposition}
Let $A$ be a Banach algebra and $e\in \overline{B^{0}_{1}}$. Suppose that $A_{e}$ is unital.
Then  $A$ is Johnson pseudo-contractible if and only if  $A_{e}$ is Johnson pseudo-contractible.
\end{Proposition}
\begin{proof}
Since $A_{e} $ is unital, by Proposition \ref{exis unit} $A$ is unital.  So using \cite[Theorem 2.1]{askar}, Johnson pseudo-contractibility of $A$ implies that $A$ is amenable. Thus by \cite[Exercise 4.3.15]{run}, $A$ is biflat. Then by \cite[Theorem 2.4]{khod} $A_{e}$ is biflat. Since $A_{e}$ is unital, biflatness of $A_{e}$	 gives the amenability of $A_{e}$.

For converse, suppose that $A_{e}$ is Johnson pseudo-contractible. Since $A_{e}$ is unital by \cite[Theorem 2.1]{askar} $A_{e}$ is amenable, so is biflat. Applying \cite[Theorem 2.4]{khod}  follows that $A$ is biflat. Using Proposition \ref{exis unit}, $A$ is unital, thus by \cite[Exercise 4.3.15]{run} $A$ is amenable. So \cite[Lemma 2.1]{sah new amen1} implies that $A$ is Johnson pseudo-contractible.

	\end{proof}
\begin{Example}
Let  $A=\{\left(\begin{array}{ccc} a_{11}&a_{12}&a_{13}\\
	0&a_{22}&a_{23}\\
	0&0&a_{33}
	\end{array}
	\right)| a_{ij}\in \mathbb{C}\}$ and suppose that $e=\left(\begin{array}{ccc} \frac{1}{4}&\frac{1}{4}&0\\
	0&\frac{1}{4}&0\\
	0&0&\frac{1}{4}
	\end{array}
	\right)$. Clearly $e$ is invertible and $A$ is unital. So by Proposition \ref{exis unit} $A_{e}$ is unital. Using \cite[Theorem 2.5]{sah new amen1} we know that $A$ is not Johnson pseudo-contractible. So by previous proposition
	$A_{e}$ is not Johnson pseudo-contractible.
	\end{Example}
\begin{small}

redpajama_set_name: RedPajamaArXiv


British Columbia is an oasis for camping and boasts a plethora of sites for those who fancy the outdoors to get their fix. There are options for both short overnight trips and even long weekend getaways which make British Columbia the ultimate spot for the best camping experiences. In this article, we will look at some of the best camping sites in British Columbia. Given the variety offered by the province, whether you are into tent camping or you are more of a "Glamper," there will surely be a spot for you.
Haida Gwaii is one of the best places to visit in the province of British Columbia. It is amongst the Agate campground which is located inside Naikoon National park. They have over 32 tent sites that sit right on the edge of the ocean, and with such a view, you will want to go back every year.
If you are interested in taking on some weekend water sports, Shuswap Lake is one of the best places in British Columbia to do so. Located about four and a half hours east of Vancouver, visitors can rent houseboats or Sea-doo's and enjoy their weekend while lounging on the beach. The Swan Lake Campground also offers tent and RV sites, but you will want to call ahead of time and reserve a spot because this spot can get booked very fast.
Allouette Lake is located in Golden Ears Park which is a short one hour drive from Downtown Vancouver. It is a great spot to camp especially if you only have a short window of time. Golden Ears Parks includes a plethora of campsites that visitors can choose from. However, similar to Shuswap Lake, it is best to call ahead of time to reserve a site because they can book up very fast, especially on the weekends during the summertime.
If you are someone who isn't a fan of tent camping and want more of a luxury camping experience, then you might want to rent an RV and take a nice journey along the Great Northern Circle Route in Northern British Columbia. Along the way, the route has a bevy of campgrounds where you can stop for the night in the event you need a break for the night.


text:
Józef Albin Herbaczewski, Juozapas Albinas Herbačiauskas (ur. 20 października 1876 w Łankieliszkach, zm. 3 grudnia 1944 w Krakowie) – poeta litewski piszący po litewsku i po polsku, twórca litewskiego ekspresjonizmu, a także dramaturg, krytyk literacki, publicysta i tłumacz.

Życiorys
Pochodził z Suwalszczyzny. W latach 1901-1906 studiował na Uniwersytecie Jagiellońskim. Razem z Adomasem Varnasem założył w 1904 w Krakowie litewskie stowarzyszenie społeczno-kulturalne "Ruta". Związany był z kabaretem Zielony Balonik, w którym recytował swoje wiersze (Tadeusz Boy-Żeleński wspomina o nim w rozdziale Sukurmukur książki Znaszli ten kraj?... i inne wspomnienia). Publikował także w czasopiśmie "Museion" artykuły o tematyce litewskiej. Do roku 1924 był w Krakowie lektorem języka litewskiego, a w latach 1924-1935 pracował na Uniwersytecie w Kownie jako lektor języka polskiego i wykładowca literatury polskiej. Od 1933 mieszkał w Warszawie. Mimo że był zwolennikiem niepodległości Litwy, popierał zbliżenie polsko-litewskie. Zmarł w Krakowie ewakuowany po powstaniu warszawskim.

Twórczość
Do jego najważniejszych utworów należy groteskowy, ekspresjonistyczny dramat Potępienie (1906) i zbiory esejów – I nie wódź nas na pokuszenie... Szkicowane wizerunki dusz współcześnie wybitnych na tle myśli dziejowej (1911) i Amen. Ironiczna nauka dla umysłowo dojrzałych dzieci (1913). W utworach tych przeciwstawiał myśl chrześcijańską modnemu w okresie modernizmu okultyzmowi i satanizmowi, w tendencjach kulturowych epoki dostrzegając "obłęd antyreligijny". Na podobnym tle krytykował ostro twórczość Wyspiańskiego, Micińskiego, Żeromskiego i Brzozowskiego. Bardzo podkreślał także wychowawcze funkcje sztuki. Jako krytyk literatury odznaczał się ostrością sądów i pamfletowym ujmowaniem zagadnień. W publicystyce podejmował tematy pojednania i zbliżenia polsko-litewskiego, m.in. w książkach Odrodzenie Litwy wobec idei polskiej (1905), Głos bólu. Sprawa odrodzenia narodowego Litwy w związku ze sprawą wyzwolenia narodowego Polski (1912).

Przypisy

Bibliografia

Linki zewnętrzne
 Dodatkowe informacje o Herbaczewskim i polonistyce w Kownie w czasopiśmie "Nasz Czas"
 Barbara Stoczewska O Leonie Wasilewskim – z informacjami na temat działalności politycznej Herbaczewskiego
 Dzieła Józefa Albina Herbaczewskiego w bibliotece Polona

Absolwenci i studenci Uniwersytetu Jagiellońskiego (okres zaborów)
Litewscy pisarze
Litewscy pisarze polskojęzyczni
Litewscy tłumacze
Pisarze Młodej Polski
Pisarze związani z Kownem
Pisarze związani z Krakowem
Pisarze związani z Warszawą
Pochowani na Cmentarzu Rakowickim w Krakowie
Polscy Litwini
Polscy poeci XX wieku
Urodzeni w 1876
Wykładowcy Uniwersytetu Witolda Wielkiego
Zmarli w 1944

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text:
Free Books / Business / Business Law / Handbook Of The Law Of Contracts /
Evidence Of Custom And Usage. Part 3
This section is from the book "Handbook Of The Law Of Contracts", by Wm. L. Clark, Jr.. Also available from Amazon: Handbook of the law of contracts.
In Equity
In the application of equitable remedies, the granting or refusal of specific performance, the reformation of documents, or their rescission and cancellation, extrinsic evidence is much more freely admitted than at law. For instance, though, as we have seen, a man is ordinarily bound by the terms of an offer unequivocally expressed, and accepted in good faith, evidence has been admitted to show that the offer was made by inadvertence. Thus, where a person, immediately after dispatching an offer to sell several plots of land for a round sum, discovered that by a mistake in adding up the prices of the plots he had offered them for less than he intended, and informed the other party of the mistake without delay, but not before the latter had concluded the contract by acceptance, the court allowed the mistake to be shown, and refused specific performance, leaving the person to whom the offer was made to such remedy by way of damages as he could obtain in the common-law courts.60
Again, where a parol contract has been reduced to writing, or where a contract for a sale or lease of lands has been performed by the execution of a lease or conveyance, evidence may be admitted to show that a term of the contract is not the real agreement of the parties; and this is done for two purposes, and under two sets of circumstances.
Where a contract has been reduced to writing, or a deed executed, in pursuance of a previous agreement, and the writing or deed, owing to mutual mistake, fails to express the intention of the parties, a court of equity will rectify or reform the written instrument in accordance with their true intent; and this may be done even though the parties cannot be placed in the position they occupied when the contract was made.*1 In such cases, extrinsic, and, if necessary, parol evidence will be admited to show the true intent of the parties. There must have been a genuine agreement;02 its terms must have been expressed under mutual mistake;63 and the evidence must be clear and convincing.
59 Blackett v. Assurance Co., 2 Cromp. & J. 244; Brown v. Foster, 113 Mass. 13G, 18 Am. Rep. 463; Randolph v. Halden, 44 Iowa, 327; Greenstine v. Borchard, 50 Mich. 434, 15 N. W. 540, 45 Am. Rep. 51; Seavey v. Shurick, 110 Ind. 494, 11 N. E. 597; Wolff v. Campbell, 110 Mo. .114, 19 S. W. 622; O'Donohue v. Leggett, 134 N. Y. 40, 31 N. E. 269; Baltimore Baseball Club & Exhibition Co. v. Pickett, 78 Md. 375, 28 Atl. 279, 22 L. R. A. 690, 44 Am. St. Rep. 304; Holloway v. McNear, 81 Cal. 154, 22 Pac. 514; Gilbert v. Mc-Ginnis, 114 111. 28, 28 N. E. 382; Barnard v. Kellogg, 10 Wall. 383, 19 L. Ed. 987; Partridge v. Insurance Co., 15 Wall. 573, 21 L. Ed. 229; Globe Milling Co. v. Elevator Co., 44 Minn. 153, 46 N. W. 306; Menage v. Rosenthal, 175 Mass. 358, 56 N. E. 579. See "Customs and Usages," Dec. Dig. (Key-No.) §§ 14, 17; Cent. Dig. §§ 29, 34
60 Webster v. Cecil, 30 Beav. 62. See McCusker v. Spier, 72 Conn. 628, 45 Atl. 1011. See "Specific Performance," Dec. Dig. (Key-No.) § 52; Cent. Dig. §§ 155-159.
Where the mistake was not mutual, extrinsic evidence is only admitted in certain cases which appear to be regarded as having something of the character of fraud,64 and is admitted for the purpose of offering to the party seeking to profit by the mistake an option of abiding by a corrected contract, or having the contract annulled. Instances of such cases are where the mistake of one party was caused by the other, though not with any fraudulent intent, and was known to him before his position had been affected by. the con-'tract.65 In these cases it is probable that the court will not reform or correct the instrument unless the parties can be placed in statu quo.
prev: Evidence Of Custom And Usage. Part 2
next: Evidence Of Custom And Usage. Part 4

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text:
Boss: "It Seems Bury Happy To Keep Lowe"
Steve Davis says he had to explore other options after move for experienced striker went cold.
Steve Davis admits he made a move to sign the highly-rated Ryan Seager on loan from Southampton after his continued approaches to Bury over the availability over Ryan Lowe drew a blank.
The Crewe Alexandra boss made re-signing the vastly experienced Lowe his number one priority but due to their extended involvement in the FA Cup, David Flitcroft elected to keep the striker.
Steve told crewealex.net: "It went a bit quiet on Ryan (Lowe) so I had to consider other options and I was pleased to get Ryan Seager in on loan.
"It went cold over Lowey and that was more from Bury's side than ours because we were very keen on getting him back here.
"He is a Bury player and they it looks as though they are keen to keep him until the end of the season.
"We moved quickly to look at other options and had to go with a younger player, who is keen to get competitive football as part of his development."


Every person always looks forward to the summer where they could openly get out of the house and go to some barbeque event and appreciate every bite of the yummy spit roast. Some would love it with beer while others want it with wine. Yet why would you wait for the summer season when you can have these mouthwatering spit roasts anytime of the year?
It is an undeniable fact though that spit cooking can be a tedious task. Although some appreciate doing it in the backyard as its wonderful odor goes with the summer season wind. Others want it served quickly without doing the effort. This is why they prefer to get it from spit roast catering services like us. Contact Us For A Free Quote.
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North Tivoli food caterer has their very own secrets on how to come up with the most enjoyable and tasty spit roast. With this, deciding on can be a difficult task for you. An easy recommendation is that you have to go for a food caterer that has substantial experience in spit roast food catering. They surely know a great deal of variations for this meal that would certainly fulfill your special taste requirement for spit roast.
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text:
Pichon Lalande 2000
France > Bordeaux > Left Bank > Pauillac
Cabernet Sauvignon/Merlot
View all vintages of this wine | View all wines by Château Pichon Longueville Comtesse de Lalande
Bordeaux 2000 Pichon Lalande BT 2 \ 0 24 3,100.00 12 96 RP 24 344.444444 96.00 96.00 Pichon Lalande 0.75 10 Pichon Longueville Comtesse de Lalande
Sitting next to my former colleague, Pierre Antoine Rovani, at one of the tastings, he commented that he didn't like the striking green note in the aromatics of this wine, which I didn't detect at all, and a subsequent bottle at another tasting did not reveal it either. I do think there is a hint of bay leaf and a meatiness to it. In short, I find this to be a spectacular Pichon Lalande. Dense purple in color, with loads of coffee, mocha, creme de cassis, and chocolate notes, this is a somewhat unusual blend of 50% Cabernet Sauvignon, 34% Merlot, and a whooping 10% Petit Verdot, with a little bit of Cabernet Franc. The Petit Verdot certainly gives the wine more of a tapenade, floral note, which I think can be interpreted by some as herbal. This is a rich, opulent, stunning Pichon Lalande that is beginning to drink beautifully, yet should continue to improve for at least another 10-15 years and last 30 or more years. Drink 2020-2050.
Robert Parker, Wine Advocate (189), June 2010
James Suckling, Wine Spectator (Mar 31, 03), March 2003
A blend of 59% Cabernet Sauvignon, 34% Merlot, 6% Cabernet Franc, and 1% Petit Verdot, the 2000 Pichon Lalande is deep garnet-brick in color. It rolls out of the glass with easy-breezy scents of potpourri, dried cherries, fruit cake, and star anise, with wafts of sandalwood and crushed rocks. The medium-bodied palate is in a sweet-spot with loads of intermingled dried berries, floral, and exotic spice notes framed by soft, powdery tannins and plenty of freshness, finishing long and fragrant.
Lisa Perrotti-Brown MW, The Wine Independent, November 2022
Very, very dark crimson. Scented. Lead pencil. Smooth - not a rough edge in sight. Very well done! Low key compared to Longueville Baron, but very good. Super-smooth and suave. Date tasted 20th Jan 05.
Jancis Robinson MW, JancisRobinson.com (31 Aug 05), August 2005
Farr Vintners, April 2001
The 2000 Pichon-Longueville Comtesse de Lalande is spectacular, and certainly a noteworthy rival to the 1996, 1995, and even the extraordinary 1982. A blend of 50% Cabernet Sauvignon, 34% Merlot, 10% Petit Verdot, and the rest Cabernet Franc, this effort has a dense saturated purple color, a singular/distinctive yet possibly controversial aromatic smorgasbord existing of creme de cassis, vanilla, and violets, but also tapenade and tree bark. No doubt the Petit Verdot has imparted a certain almost olive-like component to the wine's aromatic profile. In the mouth, it is a wine of extraordinary density, opulence, great presence, and richness. As the 2000 sits in the glass, notes of lavender and melted licorice emerge along with the distinctive tapenade aromas interwoven with creme de cassis, espresso roast, and cedar. The wine is full-bodied, with extraordinary purity and a tremendous texture. I suspect this wine will be relatively approachable young but evolve effortlessly. This is a dramatic, almost flamboyant style that is not classic Pauillac in one sense, but a classic wine. Anticipated maturity: 2007-2025.
A dramatic effort, the 2000 is a blend of 50% Cabernet Sauvignon, 34% Merlot, 6% Cabernet Franc, and a whopping 10% Petit Verdot. Only 40% of the harvest made it into the grand vin. Its dense purple color is followed by fabulous aromas of creme de cassis intermixed with roasted espresso, violets, and toast. It has a superb entry on the palate, full body, an opulent texture, silky tannin, and a finish that lasts for nearly a minute. This stunning wine will provide unreal drinking early on, but age for three decades. Bravo! Anticipated maturity: 2005-2025.
Robert Parker, Wine Advocate (139), February 2002
A formidable effort, Pichon Lalande is one of the superstars of the vintage. The blend of 50% Cabernet Sauvignon, 34% Merlot, 10% Petit Verdot (from vines planted in 1932), and 6% Cabernet Franc has resulted in a spectacularly complex, rich wine. The 2000 offers an intensely spicy, blackberry, cassis, and smoky-scented perfume with hints of espresso, barbecue spice, and cedar. Enormously rich on the attack, with some smokiness, superb purity, high but sweet, ripe tannin. Powerful for this estate, yet with perfect harmony and purity, it will be drinkable between 2004-2030. Bravo!
James Suckling, Wine Spectator Weekly (16 Jan 03), January 2003
James Suckling, WineSpectator.com, January 2003


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The Wellness & Resource Fair gives participants access to free health screenings, consultations, and encourage healthy living.
8-9:00 Early volunteers will be asked to help vendors sign in and find their table and set up.
Volunteers will help man the HOPE truck-(Catholic Charities Mobile Unit)- serving coffee, general hospitality and sharing the mission of HOPE: feeding hot meals to the homeless, helping with food insecurities by providing a Mobile Food Pantry, and the ability to respond to the needs of people in a disaster.
9-1:00 Volunteers will man the entry table and welcome guests.
9-1:00 Volunteers will visit vendors and learn what they will provide, so volunteers can be ambassadors to attendees.
Volunteers will help set up & take down tables & chairs if need be.


text:
Home » Entretainment » What to Watch This Weekend: Our Top Binge Picks for May 23-24
What to Watch This Weekend: Our Top Binge Picks for May 23-24
Saturday night's all right for streaming…and Sunday is, too.
Another weekend is here so you know what that means: another two days to spend catching up on TV shows, old and new. And we're here to help you fully make the most of your time, providing our top binge picks for May 23-24.
This weekend, Netflix is offering up a new heartwarming drama that will give you flashbacks to the days of small town stories on The WB, and Homecoming is back for its mindf–k of a second season, with another star stepping in for Julia Roberts.
Plus, one of 2019's most underrated reality series is back with new episodes, a gripping crime drama is wrapping up and a highly anticipated new rom-com is here to provide the perfect entertainment for a date night at home.
Plus, we're celebrating the 10th anniversary of one of TV's most polarizing series finales ever…
If You Need a Dose of Southern Charm: Y'all, Sweet Magnolias goes down real smooth…and features the acting comeback of Jamie Lynn Spears! It's hard to believe this feel-good seemingly-fluffier-than-a-marshmallow show isn't airing on Hallmark Channel, given the name, but there's a bit of edge to this one the Christmas-loving network would blush over; JoAnna Garcia's Maddie is dealing with her husband (Chris Klein) leaving her for his pregnant mistress (Spears) in Serenity, South Carolina. Lots of wine nights with supportive girlfriends, a new business venture a flirtation with her son's coach and more small town antics ensue. (Where to Watch: Netflix)
If You Want to Remember the Good Times: With Vanderpump Rules' most recent season coming to a dramatic close, with the end of multiple decades-long friendships, why not go back to the early days of the Bravo reality hit to see its humble beginnings, back when the SUR gang's biggest concern was getting their shirt off in time for a parking lot brawl and not waking up their sleeping friend while having sex with his girlfriend? Ah, such simpler times. (Where to Watch: Right Here)
If You Want a Quick Slow-Burn: If you watched Homecoming's first season, with Julia Roberts as its anchor, you know a quick slow-burn is actually possible. The psychological thriller based on the hit podcast is back for its sophomore outing, with Janelle Monáe taking over for Roberts in her first major TV role. But one of the biggest treats of the second season is an expanded role for Hong Chau, fresh off of her run on Watchmen, and the scenes between the two female leads. (Where to Watch: Amazon Prime Video)
If You Are in the Mood For a Date Night: We've been waiting for this rom-com-gone-wrong for a long time. Issa Rae and Kumail Nanjiani, he of the recently ripped superhero bod, team up as a couple on the verge of breaking up who find themselves wrapped up in a murder mystery. It's a wacky and hilarious cocktail that'll spice up your weekend. (Where to Watch: Netflix)
If You Are Still Listening to the Rocketman Soundtrack on Repeat: Good news, the Elton John biopic starring a delightful (and Golden Globe-winning) Taran Egerton as the icon is finally available. Saturday night's alright for streaming! (Where to Watch: Hulu and Amazon Prime Video)
If You Need Some House Porn with Housewife Accents: Anyone else sick of their own home? Yeah, us too. So why not escape for a bit with the new season of Selling Sunset, one of 2019's most underrated TV series. The real estate reality soap (say that three times fast) is back in action, with the agents at the Los Angeles firm The Oppenheim Group once again fighting for clients amid their relationship drama: Heather Young is now dating Flip or Flop's Tarek El Moussa (Talk about a crossover), Chrishelle Stause is going through a high-profile divorce from This Is Us' Justin Hartley, and Christine Quinn got secretly married. Hey, you can't spell commission without "sin." That really made no sense but we're going with it. (Where to Watch: Netflix)
If You Want to Celebrate and Debate One of TV's Biggest Shows: 10 years ago, Lost came to a polarizing end, becoming one of early TV Twitter's biggest targets with its series finale. What better way to honor the ABC drama's legacy on May 31 than by rewatching the conclusion of the passengers on Flight 815's journey and fighting over it with your friends all over again. (Where to Watch: Hulu and Amazon Prime Video)
If You Want to Know How It All Ends: Chris Evans' eight-episode long Law & Order episode comes to an end next week, with Defending Jacob airing its finale and bringing resolution to the case….or does it? We highly recommend binging all of the previous episodes if you haven't already to form your own opinion about whether or not Jacob murdered his classmate. Oh, and to truly appreciate how stellar Michelle Dockery's performance. (Where to Watch: Apple TV+)
(E! and Bravo are both part of the NBCUniversal family.)
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Beautiful sunshine this morning! Temperatures are mainly in the mid 50's.
We do see rain on the radar near Memphis and in Arkansas. This rain will move toward us throughout the day. Don't believe the intensity will reach severe limits. Remember, to issue a Severe Thunderstorm Warning, the NWS must believe that winds of greater than 58 mph and hail at least 3/4" in diameter could occur from a particular storm. Memphis Radar snap from 9:14 am.
The Spotter Training class was very interesting yesterday. I encourage all you weather buffs to attend!
Hey Mike! I was just letting you know that we have had good rains here this morning! My golf game didn't get through, we were on the 12th about to go to the 13th when the Thunder Rolled! My score was awful, but hey...It's all for fun! Have a great day!


text:
\section{Introduction} Networks of elastoplastic springs are increasingly used in the modeling of the distribution of stresses in elastopastic media \cite{buxton,chen}, swarming of mobile router networks \cite{robot2,robot1}, and  other physical phenomena. According to Moreau \cite{moreau}, the stresses of springs of such a network can be described by a differential inclusion
({\it Moreau sweeping process})
\begin{equation}
-y'(t)\in N_{C(t)}(y(t)),\quad y(t)\in\mathbb{R}^m,
\label{eq:MSP0}
\end{equation}
where $C(t)\subset\mathbb{R}^m$ is a closed polyhedron that plays the role of a constraint,
\begin{equation}\nonumber
  N_{{C}}(x)=\left\{\begin{array}{ll}\left\{\zeta \in \mathbb{R}^n:\langle\zeta,c-x\rangle\leqslant 0,\  c\in {{C}}\right\},& {\rm if}\ x\in {{C}},\\
   \emptyset,& {\rm if}\ x\not\in {{C}},
\end{array}\right.
\end{equation}
and the dimension $m$ equals or smaller than the number of springs in the network.

\vskip0.2cm

\noindent Periodicity of the constraint $C(t)$ corresponds to periodicity of the external loading applied to the given network of springs. The fundamental result by Krejci \cite[Theorem 3.14]{Krejci1996} says that for $C(t)$ of the form $C(t)=C+c(t)$, where $C$ is a convex closed bounded set and $t\mapsto c(t)$ is a $T$-periodic vector-function, any solution of sweeping process (\ref{eq:MSP0}) converges to some $T$-periodic regime. For a class of continuum elastoplastic media with $T$-periodic loading the uniqueness of $T$-periodic response is established in Frederick-Armstrong \cite[p.~159]{frederick}. Sufficient conditions for the uniqueness of the response in sweeping processes can be drawn based on Adly et al \cite{adly}.
The non-uniqueness of the response for sweeping processes can of course be easily designed, see Fig.~\ref{fig1}a, where one gets a family of periodic solutions by moving a rectangle normal to its sides back and worth. However, as shown at Fig.~\ref{fig1}b, small perturbation of such a rectangle destroys the attracting family of
orbits of Fig.~\ref{fig1}a leaving only a single attracting solution.
\begin{figure}[h]
\begin{center}
\includegraphics[scale=0.78]{fig1.pdf}
\end{center}\vskip0.0cm
\caption{Sample trajectories (solid curves) of Moreau sweeping process with a moving constraint (dashed rectangle) that moves back and forth. The bold points are the initial conditions of  sample trajectories. The figure illustrates the type of attractor (solid black curves) when (a) the moving constraint is just a rectangle, (b) the moving constraint is a pentagon with a corner that accumulates all the trajectories.}
\label{fig1}
\end{figure}

\noindent That is why a natural question arises:
\begin{itemize}
\item[] whether or not any network of elastoplastic springs can always be slightly perturbed in way that destroys any potential family of periodic orbits in the respective sweeping process (\ref{eq:MSP0})?
\end{itemize}
\begin{figure}[h]\center
\vskip-0.7cm
\includegraphics[scale=0.78]{example2-sping-Ivan.pdf}\vskip-0.0cm
\caption{\footnotesize A one-dimensional network of 5 springs on 5 nodes with one displacement-controlled loading. The circled digits stand for numbers of nodes. The regular digits are the numbers of springs. The thick bar is the displacement-controlled loading $l_1(t)$. The stress-controlled loadings $f_1(t),...,f_5(t)$ are applied at nodes.
} \label{ex2newfig}
\end{figure}

\noindent As uniqueness of the response lies in the core of reliability of modeling prediction (see e.g. \cite{bouby,brazil}), the above-stated question is not of merely academic value. We introduce a simple example that answers this question negatively. Specifically, we show that the  cyclically loaded network of elastoplastic springs of Fig.~\ref{ex2newfig}  leads to a sweeping process with a family of attracting periodic orbits.

\vskip0.2cm

\noindent The paper is organized as follows. In the next section we define a network of elastoplastic springs formally. In section~\ref{eq:sweepProcess} we derive a sweeping process \eqref{eq:MSP0} that governs the quasi-static evolution of such a network.
Section~\ref{step-by-step} is based on Moreau \cite{moreau} and Gudoshnikov-Makarenkov \cite{G-M}. It compiles a guide for closed-form computation of the quantities required for construction of a sweeping process of a given network of elastoplastic springs. This guide is then used in Section~\ref{sect:structStab} to construct the sweeping process of the network of elastoplastic springs of Fig.~\ref{ex2newfig}. We rigorously proof (Proposition~\ref{prop1} and Corollary~\ref{corr1}) that such a sweeping process admits a family of periodic orbits that persists under perturbations of the mechanical parameters of the network.

\section{A concise definition of a general network of elastoplastic springs}\label{defined}

\noindent We consider a network of $m$ elastoplastic springs on $n$ nodes that are connected according to a directed graph given by the $n\times m$ incidence matrix $D^\top$. The Hooke's coefficients  $a_1, ..., a_m$ of the springs are arranged into an $m\times m$-matrix $A={\rm diag}\left\{a_1,...,a_m\right\}.$ The elastic limits $[c_i^-,c_i^+]$ of springs are used to introduce a parallelepiped $C\subset\mathbb{R}^m$ as $C=[c_1^-,c_1^+]\times...\times[c_m^-,c_m^+].$ In addition the network comes with a collection of stress-controlled and displacement-controlled loadings $\{f_i(t)\}_{i=1}^n$ and $\{l_i(t)\}_{i=1}^q$ respectively. The stress-controlled loadings are simply applied at the $n$ nodes of the network and are supposed to satisfies the equation of static balance
\begin{equation}\label{balance!}
   f_1(t)+...+f_n(t)=0.
\end{equation}
As for the displacement-controlled loading $l_k(t),$ $k\in\overline{1,q}$, we consider a chain of springs which connects the left node $I_k$ of the constraint $k$ with its right node $J_k.$
To each displacement-controlled loading $l_k(t)$ we, therefore, associate a so-called {\it incidence vector}  $R^k\in\mathbb{R}^m$ whose $i$-th component $R^k_i$ is $-1,$ $0,$ or $1$ according to whether the spring $i$ increases, not influences, or decreases the displacement when moving from node $I_k$ to $J_k$ along the chain selected, see Fig.~\ref{figR}.
\begin{figure}[h]\center
\vskip-0.3cm
\includegraphics{drawing1-Ivan1.pdf}\vskip-0.0cm
\caption{\footnotesize Illustration of the signs of the components of the incidence vector $R^k\in\mathbb{R}^m.$ The dotted contour stays for the chain of the springs associated with the vector $R^k$.
} \label{figR}
\end{figure}
We assume that the displacement-controlled loadings $\{l_i(t)\}_{i=1}^m$ are independent in the sense that
\begin{equation}\label{A2}
{\rm rank}\left(D^\top R\right)=q.
\end{equation}
Mechanically, condition (\ref{A2}) ensures that the displacement-controlled loadings don't contradict one another. For example, (\ref{A2}) rules out the situation where two different displacement-controlled loadings connect same pair of nodes.

\section{A concise formulation of the sweeping process of a general network of elastoplastic springs}
\label{eq:sweepProcess}

\noindent In this section we follow Moreau \cite{moreau} (see also  Gudoshnikov-Makarenkov \cite{G-M}). If condition (\ref{balance!}) holds, then there exists a function
 $\bar h:\mathbb{R}\to\mathbb{R}^m$, such that
\begin{equation}\label{fh}
   f(t)=-D^\top \bar h(t).
\end{equation}
Then, under condition (\ref{A2}), there exists an $n\times q-$matrix $L$, such that
\begin{equation}
   R^\top D L=I_{q\times q}.
   \label{barxi}
\end{equation}

\sloppy

\noindent Introducing
\begin{equation}\label{U}   U=\left\{x\in D\mathbb{R}^n:R^\top x=0\right\},\qquad  V=A^{-1}U^\perp,
\end{equation}
where $U^\perp=\left\{y\in\mathbb{R}^m:\left<x,y\right>=0,\ x\in U\right\},$
the space $V$
becomes an orthogonal complement of the space $U$ in the sense of the scalar product
\begin{equation}\label{product}
   (u,v)_A=\left<u,Av\right>.
\end{equation}
Therefore, any element $x\in\mathbb{R}^m$ can be uniquely decomposed
 as
$$
   x=P_U x+P_V x,
$$
where $P_U$ and $P_V$ are linear (orthogonal in sense of \eqref{product}) projection maps on $U$ and $V$ respectively.
Define
\begin{align}
  g(t)&=P_V DL l(t),\label{g(t)}\\
 h(t)&=P_U A^{-1}\bar h(t),\label{h}\\
  N_C^A(x)&=\\
  &\hskip-1cm =\left\{\begin{array}{ll}\left\{\xi\in\mathbb{R}^m:\left<\xi,A(c-x)\right>\le 0,\  c\in C\right\},&\quad {\rm if}\ x\in C,\\
   \emptyset,& \quad {\rm if}\ x\not\in C,
\end{array}\right.\nonumber\\
  \Pi(t)&=A^{-1}C+h(t)-g(t),\label{Pi(t)}
\end{align}
Assuming that both $f:\mathbb{R}\to\mathbb{R}^n$ and $l:\mathbb{R}\to\mathbb{R}^q$ are Lipschitz continuous, we get that $h(t)$ and $g(t)$ are Lipschitz continuous as well, so that the function
$$
  y(t)=A^{-1}s(t)+h(t)-g(t)
$$
is absolutely continuous for any absolutely continuous $t\mapsto s(t).$

\begin{theorem}\label{thm1} {\rm \cite{moreau} (see also  \cite{G-M})} Assume that the network of elastoplastic springs  $(D,A,C,R,f(t),l(t))$ of section~\ref{defined} satisfies  the conditions (\ref{balance!}) and (\ref{A2}). Assume  that
$h:\mathbb{R}\to\mathbb{R}^m$ and $g:\mathbb{R}\to\mathbb{R}^m$ given by
(\ref{g(t)})-(\ref{h}) are Lipschitz continuous. Assume that safe load condition
\begin{equation}\label{A4}
\left(C+Ah(t)\right)\cap U^\perp\not=\emptyset
\end{equation}
holds on some time interval $[0,T].$
Then,
the function $s(t)=(s_1(t),...,s_m(t))$ defines the evolution of stresses of
the network  $(D,A,C,R,f(t),l(t))$ for $t\in[0,T]$ if and only if the function
$$
  y(t)=A^{-1}s(t)+h(t)-g(t)
$$
satisfies the differential inclusion (called {\it sweeping process})
\begin{eqnarray}
   -\dot y&\in& N^A_{\Pi(t)\cap V}(y),\quad {\rm for\ a.a.\ }t\in[0,T], \label{sw1}\\
y(0)&\in & \Pi(0)\cap V.\label{eq:initialElastic}
\end{eqnarray}
\end{theorem}

\noindent It remains to note that,  for Lipschitz continuous
$h:\mathbb{R}\to\mathbb{R}^m$ and $g:\mathbb{R}\to\mathbb{R}^m$,
sweeping process (\ref{balance!})-(\ref{A2}) has a unique Lipschitz-continuous solution for any initial condition (see e.g. Kunze and Monteiro Marques \cite[sect. 3]{kunze}).

\section{The shakedown condition}
\label{sect:conditions}

\noindent {The following} conditions will rule out the existence of constant solutions.

\begin{proposition}\label{prop3} {\rm \cite[Proposition~3]{G-M}} Assume that conditions of Theorem~\ref{thm1} hold. If
\begin{equation}\label{formulaprop3}
\|A^{-1}c^--A^{-1}c^+\|_A<\left\|g(t_1)-g(t_2)\right\|_A,
\end{equation}
for some $0\le t_1< t_2$,
where
\begin{eqnarray*}
   \|x\|_A&=&\sqrt{\left<x,Ax\right>},\\
   c^-&=&(c_1^-,...,c_m^-)^\top,\\
   c^+&=&(c_1^+,...,c_m^+)^\top.
\end{eqnarray*}
then sweeping process (\ref{sw1}) doesn't have any solutions that are constant on $[t_1,t_2].$
\end{proposition}

\begin{remark}\label{remarkprop3} Note, the left-hand-side in the squared inequality \eqref{formulaprop3} from the statement of Proposition~\ref{prop3} can be computed as \newline
$\|A^{-1}c^--A^{-1}c^+\|_A^2=
 \left<c^--c^+,A^{-1}(c^--c^+)\right>.$
\end{remark}

\section{A step-by-step guide to compute the quantities of the sweeping process from a network of elastoplastic springs}\label{step-by-step}

\noindent In this section we again follow Moreau \cite{moreau}, but use the notations and additional properties established in Gudoshnikov-Makarenkov \cite{G-M}. In particular, \cite[Lemma~1]{G-M} and \cite[formula~(49)]{G-M} say that
\begin{eqnarray}
\dim U&=&n-q-1.\label{dimU}\\
\dim V&=&m-n+q+1,\label{dimV}
\end{eqnarray}
provided that (\ref{A2}) is satisfied.

\vskip0.2cm

\noindent {\boldmath{\bf Step 1. The matrix $M.$}} According to (\ref{dimU}), there should exist an $n\times (n-q-1)-$matrix $M$ such that
\begin{equation}\label{RTDM}
  R^\top DM=0 \quad {\rm and}\quad {\rm rank}(DM)=n-q-1
\end{equation}
which allows to introduce $U_{basis}$ as
\begin{equation}\label{Ubasis}
  U_{basis}=DM.
\end{equation}

\noindent {\boldmath{\bf Step 2. The matrix $V_{basis}.$}}
According to (\ref{U}), $V_{basis}$ is an arbitrary matrix of $m-n+q+1=\dim V$ linearly independent columns that solves
\begin{equation}\label{Vbasis}
  (U_{basis})^\top AV_{basis}=0.
\end{equation}

\noindent {\boldmath{\bf Step 3. The matrix $D^\perp.$}} Define $D^\perp$  to be an $m\times(m-n+1)-$matrix of full rank that solves the equation
\begin{equation}\label{Dperp}
  (D^\perp)^\top D={\color{black}0_{(m-n+1)\times(m-n+1)}}.
\end{equation}

\noindent {\bf Step 4. Other quantities.} Using Steps 2 and 3, we can compute an $(m-n+q+1)\times q$-matrix $\bar L$ as
\begin{equation}\label{barLnew}
   \bar L=\left(\left(\begin{array}{c}
      R^\top \\
      (D^\perp)^\top\end{array}\right)V_{basis}\right)^{-1}\left(\begin{array}{c}
      I_{q\times q}\\ 0_{(m-n+1)\times q}\end{array}\right).
\end{equation}
It turns out that formula (\ref{g(t)}) can now be rewritten in closed-form as
\begin{equation}\label{g(t)ex}
   g(t)=V_{basis}\bar L l(t).
\end{equation}
To account for all possible functions $h(t)$ from (\ref{h}) we will simply take $h(t)$ as
\begin{equation}\label{h(t)}
  h(t)=U_{basis}H(t),
\end{equation}
where $H(t)$ is an arbitrary Lipschitz continuous control input. It is possible to compute $H(t)$ in terms of $f(t)$, but it is not of added value here.

\vskip0.2cm

\noindent Finally, for $\Pi(t)\cap V$ we have
\begin{equation}\label{222}
   \Pi(t)\cap V=\bigcap_{i=1}^{m}V_i(t),
\end{equation}

\noindent where
\begin{equation}\label{normals}
\begin{array}{rcl}
V_i(t)&=&\left\{x\in V:c_i^-\hskip-0.05cm+\hskip-0.05cma_ih_i(t)\hskip-0.05cm\le\right.\\
&&\hskip1cm \left.\le\hskip-0.05cm\left<n_i,Ax\hskip-0.05cm+\hskip-0.05cmAg(t)\right>\hskip-0.05cm\le c_i^+\hskip-0.05cm+\hskip-0.05cma_ih_i(t)\right\},\\
 n_i&=&V_{basis}\bar n_i,\\
  n_i&=&\left(\left(\begin{array}{c} R^\top\\ (D^\perp)^\top\end{array}\right)V_{basis}\right)^{-1}\left(\begin{array}{c} R^\top\\ (D^\perp)^\top\end{array}\right)e_i,
\end{array}
\end{equation}
and $e_i\in\mathbb{R}^m$ is the vector with 1 in the $i$-th component and zeros elsewhere.

\section{The sweeping process of the network of elastoplastic springs of Figure~\ref{ex2newfig}}
\label{sect:structStab}

The network of elastoplastic springs of Fig.~\ref{ex2newfig} is given by
\begin{equation}\label{Dex2new}
D\xi=
\left(\begin{array}{ccccc}
-1 &  1 &  0 &  0 & 0\\
 0 & -1 &  1 &  0 & 0\\
 0 &  0 & -1 &  1 & 0\\
 0 &  0 &  0 & -1 & 1\\
 0 & -1 &  0 &  1 & 0
\end{array}\right)
\left(\begin{array}{c}\xi_1\\
\xi_2\\
\xi_3\\
\xi_4\\
\xi_5
\end{array}\right), \quad R=\left(\begin{array}{l} 1 \\ 1 \\ 1 \\ 1 \\ 0\end{array}\right),
\end{equation}

\noindent some $5\times 5$ diagonal matrix $A$ of Hooke's coefficients and some intervals $[c_i^-,c_i^+]$, $i\in\overline{1,5}$, of elasticity bounds

\vskip0.2cm

\noindent Formula (\ref{dimU}) leads to
$$
  \dim U=5-1-1=3.
$$
The $5\times 3-$matrix $M$ that solves (\ref{RTDM}) and the respective $5\times 3-$matrix (\ref{Ubasis}) are found as
\begin{equation}\label{UbasisEx2new}
  M=\left(\begin{array}{ccc}
   0 & 0 & 0\\
   1 & 0 & 0\\
   0 & 1 & 0\\
   0 & 0 & 1\\
   0 & 0 & 0
   \end{array}\right),
   \quad U_{basis}=\left(\begin{array}{ccc}
    1 &  0 &  0\\
   -1 &  1 &  0\\
    0 & -1 &  1\\
    0 &  0 & -1\\
   -1 &  0 &  1
   \end{array}\right),
\end{equation}
and
 $H(t)$ in (\ref{h(t)}) is an arbitrary Lipschitz continuous function from $[0,T]$ to $\mathbb{R}^3.$

\vskip0.2cm

\noindent
According to (\ref{dimV}) and (\ref{Vbasis}), one gets
\begin{equation}\label{VbasisEx2}
\begin{array}{rcl}
   \dim V&=&5-5+1+1=2,\\
   V_{basis}&=&\left(\begin{array}{ccc}
1/a_1 & \ &0\\
1/a_2 &\ &-1/a_2\\
1/a_3 &\ &-1/a_3\\
1/a_4 &\ &     0\\
    0 &\ & 1/a_5
\end{array}\right)
\end{array}
\end{equation}

\noindent Following Step~3 of section~\ref{step-by-step}, we compute $\dim D^\perp=5-5+1=1$ and the $5\times 1$-dimensional solution of (\ref{Dperp}) is
{\color{black}\begin{equation}\label{0110-1}
   D^\perp=\left(\begin{array}{c}
   0\\ 1\\ 1\\ 0\\ -1
   \end{array}\right).
\end{equation}}
Therefore, according to formula (\ref{barLnew}),  the $2\times1-$matrix $\bar L$ computes as
\begin{equation}
  \bar L=\left(\left(\begin{array}{ccccc}
1 & 1 & 1 & 1 &  0\\
0 & 1 & 1 & 0 & -1\\
\end{array}\right)V_{basis}\right)^{-1}\left(\begin{array}{c}
1\\0\end{array}\right)
\label{eq:LBarEx2new}
\end{equation}
and by (\ref{g(t)ex}) we get
\begin{equation}\label{relation1}
   g(t)=V_{basis}\left(\left(\begin{array}{ccccc}
1 & 1 & 1 & 1 &  0\\
0 & 1 & 1 & 0 & -1\\
\end{array}\right)V_{basis}\right)^{-1}\left(\begin{array}{c}
1\\0\end{array}\right)l(t).
\end{equation}
On the other hand, formula (\ref{normals}) says that for each $i\in\overline{1,5}$, the normal vector $n_i$ is given by $n_i=$
\begin{equation}\label{relation2}
   V_{basis}\left(\left(\begin{array}{ccccc}
1 & 1 & 1 & 1 &  0\\
0 & 1 & 1 & 0 & -1\\
\end{array}\right)V_{basis}\right)^{-1}\hskip-0.2cm\left(\begin{array}{ccccc}
1 & 1 & 1 & 1 &  0\\
0 & 1 & 1 & 0 & -1\\
\end{array}\right)e_i.
\end{equation}
Note, formulas (\ref{relation1}) and (\ref{relation2}) hold for any $a_i,$ $i\in\overline{1,5}$ and any $c_i^-,c_i^+$, $i\in\overline{1,5}.$ Therefore, we see from formulas (\ref{relation1}) and (\ref{relation2}) that $n_1\parallel g(t)$ and $n_4\parallel g(t)$ for any values of the physical parameters of the network of Fig.~\ref{ex2newfig}. However, at this point we don't know whether or not the normals $n_1$ and $n_4$ have anything to do with the sides of the shape $\Pi(t)\cap V$ given by (\ref{222}), as it may happen that the constraints of \eqref{222} provided by $n_1$ and $n_4$ become redundant for a particular $h(t)$.
\begin{proposition}\label{prop1}  There is an open set of the parameters $a_i,$  $c_i^-,c_i^+$, $i\in\overline{1,5}$, and an open set of Lipschitz-continuous functions $H:[0,T]\mapsto\mathbb{R}^3,$ for which the vectors $n_1$ and $n_4$ are the normal vectors of the two opposite sides of the shape $\Pi(t)\cap V$. In particular, this open set of the parameters contains the point
\begin{equation}\label{para}
\begin{array}{l}
   c_i^-=-1,\ c_i^+=1,\ a_i=1,\\
    H(t)\equiv(-0.5,-0.8,-1)^\top.
\end{array}
\end{equation}
Here $[0,T]$ is an arbitrary chosen domain of the functions $t\mapsto H(t).$
\end{proposition}

\noindent {\bf Proof.} Without loss of generality we can consider $g(t)\equiv 0$. Indeed, since $g(t)$ acts along $V$, $g(t)$ simply translates $\Pi(t)\cap V$  within $V$, so that $g(t)$ doesn't change the shape of $\Pi(t)\cap V$.

\vskip0.2cm

\noindent Plugging (\ref{para}) into  (\ref{UbasisEx2new}) and using (\ref{h(t)}) we get
$$
  h(t)\equiv (-0.5,-0.3,-0.2,1,-0.5)^\top.
$$
Therefore, for the parameters (\ref{para}), formula (\ref{222}) says that $x\in\Pi(t)\cap V$ if and only if
\begin{equation}\label{system}
  \left\{\begin{array}{lllll}
     -1-0.5&\le&\left<n_1,x\right>&\le& 1-0.5,\\
    -1-0.3&\le&\left<n_2,x\right>&\le& 1-0.3,\\
    -1-0.2&\le&\left<n_3,x\right>&\le& 1-0.2,\\
    -1+1&\le&\left<n_4,x\right>&\le& 1+1,\\
    -1-0.5&\le&\left<n_5,x\right>&\le& 1-0.5,
  \end{array}\right.
\end{equation}
where
\begin{equation}\label{n1}  \renewcommand\arraystretch{2}
\begin{array}{l}n_1=n_4=\dfrac{1}{d}\left(-3,-1,-1,-3,-2\right)^\top,\\
n_2=n_3=\dfrac{1}{d}\left(-1,-3,-3,-1,2\right)^\top,\\
n_5=\dfrac{1}{d}\left(-2,2,2,-2,-4\right)^\top,\\
d=-8.
\end{array}
\end{equation}
Based on (\ref{relation2}), $n_1=n_4$ and $n_2=n_3$. Therefore, 1st and 4th lines of system (\ref{system}) as well as 2nd and 3rd lines combine, that reduces the number of double-sided inequalities to 3. Substituting the expressions (\ref{relation2}) with parameters (\ref{para}) into (\ref{system}) and plugging $x=V_{basis}v,$ where $v\in\mathbb{R}^2,$ system (\ref{system}) reduces to the following system
\begin{equation}\label{system2}
  \left.\begin{array}{lrlcll}
{\rm normals}\  n_1\ {\rm and}\  n_4: \quad&    0&\le&v_1&\le& 0.5,\\
{\rm normals}\  n_2\ {\rm and}\  n_3: \quad&     -1.2&\le&v_1-v_2&\le& 0.7,\\
{\rm normal}\  n_5: &     -1.5&\le&v_2&\le& 0.5.
  \end{array}\right.
\end{equation}
\begin{figure}[h]\center
\vskip-0.5cm
\includegraphics[scale=0.6]{shape.pdf}\vskip-0.2cm
\caption{\footnotesize The gray region stays for the set of $(v_1,v_2)$ given by inequalities (\ref{system2}). The dotted lines denote the sets of $(v_1,v_2)$ where equalities of (\ref{system2}) are attained (the dotted line $v_1=0$ coincides with the vertical axis and another dotted line $v_2=-1.5$ is not shown).
} \label{figshape}
\end{figure}

\noindent Fig.~\ref{figshape} illustrates that the two constraints  from \eqref{222} corresponding to normal vectors $n_1$ and $n_4$ constitute the opposite sides of the shape
 $\Pi(t)\cap V$. This properties persists under small perturbations of the parameters (\ref{para}). Indeed, formulas (\ref{222}) and (\ref{relation2}) imply that small perturbations of the parameters (\ref{para}) lead to small parallel displacements of the dotted lines of Fig.~\ref{figshape} (without rotations), so that the two opposite parallel sides will stay. The proof of the proposition is complete. \qed

\vskip0.2cm

\noindent In order to obtain the existence of a structurally stable family of non-stationary periodic solutions it is now remains to apply the displacement-controlled loading (\ref{relation1}) of sufficiently large amplitude. We will now use Proposition~\ref{prop3} to give an estimate for the required amplitude. In the case of a 5-spring network, formula (\ref{formulaprop3}) of Proposition~\ref{prop3}  follows from
\begin{equation}\label{fofo}
\sum_{i=1}^5\dfrac{1}{a_i}\left(c_i^+-c_i^-\right)^2<\left\|V_{basis}\bar L\right\|^2_A\cdot\left(l(t_1)-l(t_2)\right)^2.
\end{equation}
In the case of parameters (\ref{para}), formula (\ref{fofo}) reduces to
$$
\sum_{i=1}^5 2^2<\|n_1\|^2\cdot\left(l(t_1)-l(t_2)\right)^2,
$$
where $n_1$ is given by (\ref{n1}), or simply
$$
   \dfrac{160}{3}<\left(l(t_1)-l(t_2)\right)^2.
$$
Since $\frac{160}{3}\approx 53.3,$ we introduce $l(t)$ as follows
\begin{equation}\label{loadt}
   l(t)=\left\{\begin{array}{ll}
      t, & \ {\rm if}\ t\in[0,54],\\
      -t+54, & \ {\rm if}\ t\in[54,108],
      \end{array}\right.
\end{equation}
extended to $[0,\infty)$ by 108-periodicity.

\begin{corollary} \label{corr1} Consider the network of elastoplastic springs of Fig.~\ref{ex2newfig} with the parameters (\ref{para}).  Assume the displacement-controlled loading given by (\ref{loadt}), so that $T=108.$ Then, for any parameters $a_i,$  $c_i^-,c_i^+$, $i\in\overline{1,5}$, and any  Lipschitz-continuous functions $T$-periodic $H:[0,T]\mapsto\mathbb{R}^3,$ that are close to those in (\ref{para}), and for any Lipschitz-continuous $T$-periodic $l(t)$ close to (\ref{loadt}), the sweeping process
(\ref{sw1})-(\ref{eq:initialElastic}) admits a structurally stable family of non-stationary $T$-periodic solutions (swept by the opposite parallel sides of Fig.~\ref{figshape}). Accordingly, the mechanical model of Fig.~\ref{ex2newfig} admits an entire family of co-existing stress distributions that evolves $T$-periodically in time.
\end{corollary}

\section{Conclusions} In this paper we showed that sweeping processes of networks of elastoplastic springs (elastoplastic systems) inherit a designated structure that restrict possible dynamic transitions. Specifically, we gave an example of an elastoplastic system whose sweeping process admits a structurally stable family of non-stationary periodic solutions. Specifically, the structure given by the elastoplastic system locks the family of periodic solutions of the associated sweeping process, so that it persists under all such small perturbations of the sweeping process that come from small perturbations of the physical parameters of the elastoplastic system.

\sloppy

\section*{Compliance with Ethical Standards}

\noindent {\bf Conflict of Interest:} The authors have no conflict of interest.

\bibliographystyle{plain}


text:
INLOCA
Ana Vallés
#Dramático publications are available before and after performances at Teatro María Guerrero and Teatro Valle-Inclánand at these bookstores
Price: 10 euros · No. pages: 93 · Collection: New Collection No. in the collection: 15 · Language: Spanish · Binding: Paperback ISBN: 978-84-9041-438-5 · Year of publication: March 2022 · Place of publication: Madrid, Spain
Dramaturgia y dirección Ana Vallés. Una creación de Matarile coproducida con el Centro Dramático Nacional con el apoyo de Agadic-Xunta de Galicia
21 ENE - 06 FEB | María Guerrero Theatre
AGOTADAS
Escucha la música que ha inspirado "INLOCA"
(A few subtitles, which can be chapters, which are motivations)
The devil on the beach. Lucifer, the light bearer. The most beautiful angel in heaven who is condemned for defying God's creative power. The fallen angel.
Deleuze's umbrella. The struggle against chaos, in the face of which science and philosophy create a protective umbrella, while art opens cracks and fissures through which chaos and light seep.
Tarkovsky's room. Light as space, creator of atmospheres, temperatures, appearances or distance. Light to inhabit or diffuse. Ghostly relations.
The womb of the old woman. The capacity to create. From a womb, from a head, from a sensitivity, from which nothing new is expected. A place from which one can ask questions and, perhaps, feel out some answers. The great gift, an epiphany.
Moving the corpse. Where to take it, finding space for our remains, the remains of old Europe which, as Agamben says in When the house burns down, we have patched up and painted over so many times that they seem intact.
The shape of the brain. Based on Europe has the shape of my brainby Cartarescu. Shared thinking. But also a general sense of ending, of the end of civilisation (the last of Steiner's axioms to define The idea of Europe).
Europe after the rain
Steiner's café, the strategists
The death of Mickey Mouse
Writer, theatre director and actress. Co-founder of the company Matarile in 1986. She promotes contemporary exhibition and dissemination spaces: Teatro Galán, the Festival Internacional de Danza En Pé de Pedra and saLa Montiel. These spaces have provided the backdrop for most of her artistic work.
She has directed over 30 productions under her company Matarile. In 2019, the Madrid Autumn Festival featured a retrospective exhibition of her work, including Daimon y la jodida lógica, Los limones, la nieve y todo lo demás and Teatro Invisible. In the same year, she was selected as a finalist for Best Director at the MAX Awards for Circo de Pulgas. In 2008, the Valladolid TAC Festival paid tribute to Ana in recognition of her artistic career.
Her shows have been performed in Boston, Belgrade, Paris, Novisad, Porto, Leipzig, Bologna, Dresden, Edinburgh, Almada, Bayonne, Bucharest, Rennes, Dinan, Douardenez, Lanester, Brasilia, Tavira, Warsaw, Viana do Castelo, Cluj-Napoca and Timisoara.
She has received numerous awards, including the Audience Award for Best Show at the 2009 Don Quixote International Festival in Paris for Animales Artificiales, the Most Innovative Show Award at the 11th Valladolid International Festival of Theatre and Street Arts for Historia Natural in 2005 and for Hombre Bisagra in 2015, the Audience Award for Best Show at the 2005 Feira do Teatro de Galicia for Historia Natural, the Compostela Award for Best Direction from Teatro Gallego and the Compostela Award for Best Lighting from Teatro Gallego for Café Acústico, and the María Casares Award for Best Direction from Teatro Gallego for The Queen is Dead.
In addition, Teatro Galán and the En Pé de Pedra Festival (theatre and festival managed by Matarile Teatro since 1993) received the 2005 Zapatilla Award from ARTEZ 05 (Performing Arts Magazine) for the programming team, the 2006 Galician Critics' Award for the international dance festival En Pé de Pedra and the 2010 Abrente Award at the Ribadavia International Theatre Festival for the theatrical work of Matarile Teatro. The companies were also nominated for the Huesca International Theatre and Dance Fair awards for the Best Dance Programme in Spain in 2005.
Ana has also worked as a stage manager for several venues, including Teatro de La Abadía, the Galician Centre for Dramatic Arts and the Auditorium of Galicia.
In 2016, Ediciones Invasoras published a collection of her texts (Cerrado por Aburrimiento – Staying Alive – Teatro Invisible – El Cuello de la Jirafa), and in 2021, she published the texts of El diablo en la playa in collaboration with Celeste González and Claudia Faci.


Student Virginia Education Association: NEA-SP Awards Packet Released!
Chapters! It's YOUR time to SHINE!
Please consider applying for one or more of these awards for the chance to be recognized at both SVEA Convention and at NEA Student Leadership Conference, if you win in Virginia. We want to show the wonderful things that Virginia Chapters are doing! The submission date is March 11th to Donna Hamilton (dhamilton@veanea.org).
The Awards Packet is available for download NOW under "2016 Award Information"! Good luck to all chapters!!


text:
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text:
MULTITALENTED BEAUTY
Meet Catherine Roque, Model, Writer, and Visual Artist
Catherine Roque, also known as Cathy, is a professional model from the Philippines under the international management of XLNT Worldwide Management Group (operating in Asia, the U.S., and Europe). She is the Program Manager for the Philippines for Inspitale, an online platform in which all kinds of writers share their real-life positive stories. She handles all aspects of operations, including promotions, recruitment, and addressing concerns of interested individuals in the Philippines. Previously, she wrote for the now-closed publication and platform The Start-Up Growth, in which she used to discuss leadership and life, based on her own personal experiences.
Catherine is the 2020 Ambassador for the Penmasters League Administration. She was once nominated and shortlisted as the Gawad Parangal sa Mundo ng Literatura Most Promising Writer of 2019. She once turned one of her personal sufferings into a piece she calls Revenge, by viewing it in a much different light, with a new point of view. Now, she is also writing for two U.S.- based magazines, Model Icon, and Tiara, which focus on modeling, pageants, and lifestyle.
She is Miss WOW London Season 3 Second Runner- Up and was chosen Miss Uniworld Philippines 2020, Executive Vice President of Mr. & Miss Nigeria Golden
Heart, and one of the key members of Miss Universal Queen. She was featured in World of Winners Pageant Magazine's December 2020 issue, specifically on page 11, representing the Philippines.
She earned her Bachelor's Degree in Communication in Media Production at Assumption College, San Lorenzo, Makati City, Philippines. She finished an online course, Moral Foundations of Politics at Yale University, and earned her certificate during the intense lockdown.


text:
Eat My Words
Bourdain and Brisket
By Layne Lynch
Thousands upon thousands migrated to Austin during SXSW to hear some tunes, stalk some celebs, and, of course, eat some darn barbecue. One of those thousands included the no-holds-barred No Reservations host – Anthony Bourdain. In a recent interview with the Huffington Post, Bourdain talked about pre-made egg salad sandwiches, Ruth Bourdain, and the best brisket he's ever had, which happens to be from Franklin Barbecue during SXSW. Here's what the famous foodie said of the moist-meat experience: "It is the best. It is the finest brisket I've ever had. I can't imagine anyone could surpass this. It's unearthly in its moistness, in its perfect balance. Listen — I waited in line an hour and a half to get in there… I stood on line. That's part of the experience, to stand in line. It's a whole social thing. But let me tell you, by the time I was, oh, 10 feet away from that brisket, looking at it, I understood why'd I'd waited on line with everybody else. I could tell just from the way it hit the cutting board that this was going to be just earth-shatteringly good. And it was indeed." To read the full interview, check it out here.
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text:
Alameda Covered a $1 Billions Loss for FTX in 2021 (Report)
FTX's sister trading desk, Alameda Research, bailed out the exchange before it suffered a possible $1 billion trade loss in early 2021.
This news comes amid claims from former FTX CEO Sam Bankman-Fried that FTX and Alameda were independently operated firms.
As reported by the Financial Times on Friday, people familiar with the matter said FTX suffered massive losses after a client's leveraged trade on an obscure crypto token went south. Meanwhile, the "buffers" designed to shield the exchange from losses on a bad trade failed to protect FTX.
A common risk-management procedure used by firms when lending money is to collect collateral from the borrower in advance. If the borrower's collateral value falls to a certain level, the lender clears the loan and sells the customers' assets on its own behalf to cover its costs.
The token in question, called MobileCoin, spiked from $6 to $70 in April 2021, only to crash just as fast shortly afterward. At the time, a trader borrowed against the coin with an unusually large position, which Alameda had to step in and help cover. The trading desk lost hundreds of millions of dollars in the process.
Additional blockchain evidence provided by Nansen also suggests that Alameda acted as FTX's lender of last resort when funds were short.
The event signals deep ties between Alameda and FTX, both companies founded by Sam Bankman Fried, despite the ex-CEO's claims that he had no knowledge of what was going on at Alameda.
Sam Bankman-Fried explained during an interview last month that Alameda itself had a leverage position worth billions open with FTX, prior to going bankrupt.
Though the position had been collateralized by FTT token, the FTT market was too illiquid and dropped too fast for FTX to liquidate the position, and thus stay solvent.
The post Alameda Covered a $1 Billions Loss for FTX in 2021 (Report) appeared first on CryptoPotato.
Bitcoin Bearish Signal: MPI Records Highest Value Since April 2022
Sales of Bitcoin Mining Rigs in Russia Surge in Q4: Report


I encountered Quinn's early in life because my fictional tastes ran heavily to South Seas literature, and Quinn's was a part of the standard background against which these stories moved. It was here, for example, that the drunken Scottish captain of the rusting copra freighter went noisily to hell after being betrayed by the Polynesian girl. It was to the florid literature of the area what Sidi-bel-Abbés was to the French Foreign Legion stories or what Utah was to Zane Grey. When a good man fell from grace, he fell in Quinn's.
On the first night of my arrival in Papeete a few weeks ago, I made my way promptly to Quinn's. It was easy to find, in the center of town and facing the long quay, and at first glance I felt a measure of disappointment. It was a flat, totally unpretentious building, open in front, with QUINN'S BAR spelled out in bamboo letters across the entrance. A mountain of beer crates stood on the sidewalk in front of the place, and across the street 1 could see the outlines of a cluster of rusty Korean fishing boats tied to the dock. I was standing beside the entrance, transfixed in the way a young priest might be when standing outside of Saint Peter's on his first trip to Rome, when there was a momentary explosion of shouts and curses in front of me and a dark body sailed through the air and landed with a brutal crash in the pile of beer crates. The body pulled itself up and shuffled out of sight across the street. I entered.
There is nothing faint-hearted about Quinn's rosy version of life. In the center, near the entrance, is a horseshoe-shaped bar, and behind the bartenders was as energetic and feverish a rock-'n'-roll combo as it has ever been my misfortune to encounter. The noise was painful, the air was compounded of smoke and liquor fumes, and the lights were mercifully low. Around the wall was a row of booths, many with curtains drawn to conceal whatever adventures the occupants did not care to share with the general public. In the first booth, as I entered, a mountainous woman in a flowered sarong smiled at me. She must have weighed 350 pounds, and the smile revealed that, although she had only one upper and one lower tooth, she somehow achieved a malocclusion. "Wanna live a little?" she whispered huskily. I shuddered, and took a seat at the bar.
I don't know how many fights occur during an average evening in Quinn's, but I've heard that many Tahitians will fight nowhere else. At one point, shortly after my drink arrived, two fights broke out simultaneously on the dance floor. One fight was of such high quality that the combatants in the lesser engagement gave up quickly and joined the spectators watching the main event. Bouncers moved in quickly and carried the fighters off the floor, though they all reappeared a few minutes later at the bar, dusting themselves of and buying drinks for their supporters.
I suppose this is as good a place as any to say a few words about the toilet at Quinn's Bar, since this seems to be the feature that is most earnestly discussed by American tourists. There is only one toilet, separated from the dance floor by a tattered cloth curtain, and this accommodates both men and women and, from what I could gather, accommodated them simultaneously. Whether the shrieks of outrage that I heard were good-natured or genuine, I have no way of knowing.
The bartender gave me a look of dark reproach when, after the fourth set of fighters had been carried out over the heads of the bouncers, I asked for my check. It was still early in the evening, but the noise and tension in the place created a prickly atmosphere. As I passed the fat lady in the front booth she suggested that I wouldn't be leaving so early if I had possessed a little sporting blood. l saw that she also had a rather handsome mustache.
I particularly was interested in your description of the toilet facilities. I visited Quinns in 1965 and found the facilities as described. It certainly came as a shock to a wandering Canadian.
I was in merchant in the 60s,i read about quinns being hardest pub in the world ,I went into shiping federation in Southampton they were looking for crew on the southern cross and one of the ports it called into was Tahiti signed on straight away weeks later I was standing in quinns wow 17yrs old. French leginers were there some of there noses were bigger than me.i rember reading about two of them fighting for two hrs there was one toilet, mesh over over tiles with water running underneath,its one pub I will never forget went there 2 more times really had good time.
Visited quinns 1966 I was steward on immigrant liner northern star.
I had the bounty of visiting Tahiti and its wonderful people in 1964 and 1961. In 1961 an Italian film maker was filming a documentary titled "Women of the World." They documented the goings on in Quinns. Google it for yourselves. I'm sure there was lots of action but after all these years I remember only the comradeship of my shipmates and the friendship and toleration of the Tahitian people towards we grossly misbehaving strangers. I wish the best of progress and prosperity for those noble islanders.
I was in Quinn's in early 1967. I was a e/r rating aboard the MV Rangitane, New Zealand Shipping Co. emigrant ship. I was with my mate Dave Chapman, also from Leeds. I ordered two beers and when I offered to pay the barman said they were paid for. This happened again so I asked the barman who was paying for them. He pointed to a little Tahitian with a flower behind his ear who gave me a suggestive smile. Dave & I moved to the other side of the bar where the episode was repeated. When we looked around half the guys in there had flowers behind their ears.
I visited Quinn's in August 1962 when the Coast Guard Cutter I was stationed on made a 5 day stop in Papeete. I remember that when Quinn's closed late at night that everone went to another bar that opened around midnight. Does anyone recall the name of that bar?
It just came back to me. The late night bar that opened after the other bars closed each night was called Lafayette's.
I have the dubious distinction of having visited the two most disreputable bars in the world, both (sob!), of happy memory. Quinn's of course, and on a later trip, the Banana Court in Rarotonga. They sure don't make 'em like that anymore. They were deliciously disreputable.
Was in Quinn's in 1970. The loo curtain was on the top part of the entrance, hid the face but not the bottom. Massive cockroaches on the wall in the booths. was a most amazing experience. Never to be forgotten.
Was there in 1972 and the Legionnaires were very much in evidence as were the very large bouncers. Came in on an oceanographic research vessel the RV Chain and stayed several weeks. At this time there was one notable addition to Quinn's. in the far right had corner, as you entered, there was a screened, (floor to ceiling), seating area which was reserved for the cruise ship tourists. Obviously so that they could vicariously experience Quinn s, but still remain "safe". They even had their own entrance just to separate them from the crowd. Glad I experienced it, but am sorry that it is gone like so many other unique places.