text:
\section{Introduction}

Recollements of triangulated categories, defined in \cite{BBD}, provide a tool of decomposing triangulated categories into smaller ones. They are analogues of short exact sequences of groups or modules.
In this paper we restrict our attention to recollements whose terms are all derived categories of algebras.
Derived simple algebras, analogous to simple groups or simple modules, are those algebras whose derived categories cannot be decomposed further via recollements. For example, local algebras are derived simple \cite{AKL2}. In analogy to composition series, a stratification of a derived category is a sequence of recollements which iteratively decomposes the given category into derived simple ones. It is natural to ask for a derived version of Jordan--H\"older property (DJHP for short), that is, the existence and uniqueness of finite stratifications of derived categories.  It was shown that hereditary Artin algebras \cite{AKL2}, finite dimensional piecewise hereditary algebras \cite{AKL3}, commutative Noetherian rings \cite{AKLY2} and derived discrete algebras \cite{Q} satisfy DJHP.

DJHP fails for general algebras by results of Chen and Xi \cite{CX12, CX15}. The first finite dimensional example was proposed also by Chen and Xi. Another finite dimensional example was given in \cite[Example 7.6]{AKLY2}.
The aim of this paper is to generalise this example to obtain more finite dimensional algebras which do not satisfy DJHP. More precisely,  given two finite dimensional elementary algebras $A$ and $B$, we construct a $2\times 2$-matrix algebra $\Lambda(A,B)$. Our main result is:
\begin{theorem*}[\ref{t:maintheorem}] DJHP fails for $\Lambda(A,B)$ in the following two cases:
\begin{itemize}
\item[(1)] both $A$ and $B$ are derived simple, but neither is simple;

\item[(2)] one of $A$ and $B$ is derived simple but nonsimple and the other one is of infinite global dimension.
\end{itemize}
\end{theorem*}
As consequences of this result, we have
\begin{itemize}
\item[(a)] There are finite dimensional algebras of finite global dimension which do not satisfy DJHP. Indeed, if $A$ and $B$ are finite dimensional derived simple elementary algebras of positive finite global dimension, \emph{e.g.} Fibonacci algebras \cite{H,Membrillo-Hernandez94} and their generalisations \cite{LY16}, then $\Lambda(A,B)$ has finite global dimension and does not satisfy DJHP (Corollary~\ref{cor:finite-global-dimension}).

\item[(b)] Fix a local nonsimple finite dimensional elementary algebra $A$. Then for any finite dimensional elementary algebra $B$ applying the construction $\Lambda(?):=\Lambda(A,?)$ twice yields a finite dimensional algebra $\Lambda^2(B)$ which does not satisfy DJHP  (Corollary~\ref{cor:cardinality}).
\end{itemize}

During the preparation of this paper, Martin Kalck informed us that he found a family of finite dimensional algebras of global dimension 2 (hence quasi-hereditary) for which DJHP fails (see \cite[Proposition 3.4]{K}); Changchang Xi informed us that \cite[Theorem 1.1]{CX} can be used to construct finite dimensional algebras which do not satisfy DJHP. For two finite dimensional elementary algebras with the same number of isomorphism classes of simple modules, Chen and Xi constructed in the end of \cite[Section 5]{CX} an upper triangular $2\times 2$-matrix algebra, for which results similar to Theorem~\ref{t:maintheorem}, (a) and (b) can be obtained.

The paper is organised as follows: In Section \ref{s:preliminaries} we collect necessary definitions about recollements, stratifications and derived Jordan--H\"older property for later use. Section \ref{s:main result} contains our main result. We give the construction of the matrix algebra $\Lambda(A,B)$ and study its derived Jordan--H\"older property using stratifying ideals. In Section \ref{s:quiver-relation} we provide the underlying quiver and relations of $\Lambda(A,B)$.

Throughout the paper $k$ is a field, all algebras are finite dimensional algebras over $k$,  and all tensor products are over $k$ unless otherwise stated. For an algebra $A$, we denote by $\mathcal{D}(A)$ the unbounded derived category of right $A$-modules. For a quiver $Q$, we denote by $Q_0$ and $Q_1$ the sets of vertices and the set of arrows respectively. For an arrow $\alpha\in Q_1$, we denote by $\mathrm{s}(\alpha)$ and $\mathrm{t}(\alpha)$ the source and target of $\alpha$ respectively.

\medskip
\noindent{\it Aknowledgement.} The first-named author acknowledges support by National Natural Science Foundation of China No. 11671207, and by RFDP of Higher Education of China 20133207120013.
The second-named author acknowledges support by National Natural Science Foundation of China No. 11671207.

\section{Preliminaries} \label{s:preliminaries}

In this section we recall the definition of recollements, stratifying ideals and stratifying recollements, stratifications, derived simplicity and derived Jordan--H\"older property of algebras.

\subsection{Recollements}
Let $A,B,C$ be algebras.
A \emph{recollement}~\cite{BBD} of
${\mathcal D}(A)$ by ${\mathcal D}(B)$ and ${\mathcal D}(C)$ is a diagram
\begin{eqnarray*}\xymatrix@!=6pc{\mathcal{D}(B)
\ar[r]|{i_*=i_!} &\mathcal{D}(A) \ar@<+2.5ex>[l]|{i^!}
\ar@<-2.5ex>[l]|{i^*} \ar[r]|{j^!=j^*} &
\mathcal{D}(C)\ar@<+2.5ex>[l]|{j_*}
\ar@<-2.5ex>[l]|{j_!}}
\end{eqnarray*}
of
$k$-linear triangle functors such that
\begin{enumerate}
\item $(i^\ast,i_\ast)$,\,$(i_!,i^!)$,\,$(j_!,j^!)$ ,\,$(j^\ast,j_\ast)$
are adjoint pairs;

\item  $i_\ast,\,j_\ast,\,j_!$  are full embeddings;

\item  $i^!\circ j_\ast=0$ (and thus also $j^!\circ i_!=0$ and
$i^\ast\circ j_!=0$);

\item  for each $M\in \mathcal{D}(A)$ there are triangles
\[
\xymatrix@R=0.3pc{
i_! i^!(M)\ar[r]& M\ar[r]& j_\ast j^\ast (M)\ar[r]& i_!i^!(M)[1]\\
j_! j^! (M)\ar[r]& M\ar[r]& i_\ast i^\ast(M)\ar[r]& j_!j^!(M)[1]
}
\]
where the first two maps in each triangle are given by adjunctions.
\end{enumerate}

We say the recollement is {\it trivial} if one of the algebras $B$ and $C$ is trivial, or equivalently, one
of the full embeddings $i_*$, $j_!$ and $j_*$ is a triangle
equivalence.

For an algebra $A$, we denote by $\mathrm{rank}(A)$ the number of nonisomorphic simple $A$-modules and call it the \emph{rank} of $A$.

\begin{proposition}\label{prop:invariants}
 Let $A,B,C$ be algebras and assume that there is a recollement of ${\mathcal D}(A)$ by ${\mathcal D}(B)$ and ${\mathcal D}(C)$. Then
 \begin{itemize}
 \item[(i)] {\rm (\cite[Proposition 2.14]{AKLY2})}
 $A$ is of finite global dimension if and only if so are $B$ and $C$;
 \item[(ii)] {\rm (\cite[Proposition 6.5]{AKLY2})}
 $\mathrm{rank}(A)=\mathrm{rank}(B)+\mathrm{rank}(C)$.
 \end{itemize}
\end{proposition}

\subsection{Stratifying ideals and stratifying recollements}

Let $A$ be an algebra and $e\in A$ an idempotent. Following \cite[Definition 2.1.1]{CPS} we call the idempotent ideal $AeA$ a {\it stratifying ideal} of $A$ if the natural morphism $Ae\overset{\boldmath{L}}{\ten}_{eAe}eA\to AeA$ is an isomorphism in ${\mathcal D}(A)$. If $AeA$ is stratifying, then there exists a recollement of ${\mathcal D} (A)$ by ${\mathcal D} (A/AeA)$ and ${\mathcal D}(eAe)$ (see for example \cite[Example 4.5]{AKL1})
\[ \label{d:recollement}
\xy (-46,0)*{\mathcal{D}(A/AeA)}; {\ar (-19,3)*{}; (-35,3)*{}_{i^\ast}};
{\ar (-35,0)*{}; (-19,0)*{}|{i_\ast=i_!}}; {\ar(-19,-3)*{}; (-35,-3)*{}^{i^!}};
(-12,0)*{{\mathcal D}(A)}; {\ar (11,3)*{}; (-5,3)*{}_{j_!}};
{\ar (-5,0)*{}; (11,0)*{}|{j^*=j^!}}; {\ar (11,-3)*{}; (-5,-3)^{j_*}};
(20,0)*{{\mathcal{D}(eAe)}};
\endxy
\]
where the six functors are derived functors given by
\[
\begin{array}{c}
i^*=-\overset{\boldmath{L}}{\ten}_A A/AeA,\ \ i^!=\mathbf{R}\mathrm{Hom}_A(A/AeA,-),\\
i_*=\mathbf{R}\mathrm{Hom}_{A/AeA}(A/AeA,-)=-\overset{\boldmath{L}}{\ten}_{A/AeA}A/AeA=i_!,\\
j_!=-\overset{\boldmath{L}}{\ten}_{eAe} eA,\ \  j_*=\mathbf{R}\mathrm{Hom}_{eAe}(Ae,-), \\
j^!=\mathbf{R}\mathrm{Hom}_A(eA,-)=-\overset{\boldmath{L}}{\ten}_A Ae=j^*.
\end{array}
\]
We refer to it as the {\it stratifying recollement} of $\mathcal{D}(A)$ induced by $e$. The following lemma is a left-module version of {\cite[Lemma 2.12]{AKLY2}}.

\begin{lemma}[]
Let $A$ be an algebra and $e\in A$ an idempotent. Assume that $A/AeA$, as a left $A$-module, admits a projective resolution with components in $\opname{add}\nolimits(Ae)$ except in degree $0$. Then $AeA$ is a stratifying ideal of $A$. \label{l:criterion for stratifying ideal}
\end{lemma}

\subsection{Derived simplicity of algebras}

An algebra $A$ is said to be {\em derived simple} if there is no nontrivial recollement of ${\mathcal D}(A)$ by derived categories of algebras.
Known examples include  local algebras \cite[Proposition 4.11]{AKL2}, Fibonacci algebras  \cite{H,Membrillo-Hernandez94} and their generalisations \cite{LY16}, blocks of group algebras of finite groups and indecomposable symmetric algebras of
finite representation type \cite{LY12}, and indecomposable self-injective algebras of rank 2 with nonpositive Cartan determinant \cite[Remark 5]{HanQin16}.

\subsection{Stratifications and the derived Jordan--H\"older property}

Roughly speaking, a
\emph{stratification} is a way of breaking up a given derived
category into simple pieces using recollements. More rigorously, for an algebra $A$ a
stratification of ${\mathcal D}(A)$ is a full rooted binary tree whose root is the given
derived category ${\mathcal D}(A)$, whose nodes are derived categories of algebras and whose
leaves are derived categories of derived simple algebras such that a node together with its two child
nodes form a nontrivial recollement unless it is a leaf.  The leaves are called the \emph{simple factors} of the
stratification. By abuse of language, we will also call the corresponding derived simple algebras the simple factors of the stratification. If we allow trivial recollements in the above definition, we call the binary tree a \emph{pseudostratification}. Given a pseudostratification, we obtain a stratification by deleting the subtrees with trivial nodes and contracting the edges which are equivalences.

\begin{lemma}\label{lem:splitting-stratification}
Let $A$ be an algebra with a decomposition $A=A_1\times\cdots\times A_s$ into the direct product of algebras $A_1,\ldots,A_s$. Then any stratification of ${\mathcal D}(A)$ is the direct sum of pseudostratifications of ${\mathcal D}(A_1),\ldots,{\mathcal D}(A_s)$.
\end{lemma}
\begin{proof}
Let $B$ and $C$ be algebras such that there is a recollement of ${\mathcal D}(A)$ by ${\mathcal D}(B)$ and ${\mathcal D}(C)$. It follows from \cite[Lemma 2.3]{AKLY2} that there are algebras $B_1,\ldots,B_s$, $C_1,\ldots,C_s$ (some of which may be trivial) such that $B=B_1\times\cdots\times B_s$ and $C=C_1\times\cdots\times C_s$ and there are recollements of ${\mathcal D}(A_i)$ by ${\mathcal D}(B_i)$ and ${\mathcal D}(C_i)$, $i=1,\ldots,s$, which sum up to the given recollement. The desired result follows by induction.
\end{proof}

Analogous to the classical Jordan--H\"older property of finite groups, the derived Jordan--H\"older property of an algebra is about the existence and uniqueness of  finite stratifications. Let $A$ be an algebra. According to \cite[Corollary 7.4]{AKLY2}, any stratification of the derived category ${\mathcal D}(A)$ is finite. If the simple factors in any stratification of ${\mathcal D}(A)$ are unique up to ordering and up to derived equivalence, we say that the algebra $A$ satisfies the \emph{derived Jordan--H\"older property} (DJHP for short).
As examples, piecewise hereditary algebras \cite{AKL2,AKL3} and derived discrete algebras \cite{Q} satisfy DJHP.

\begin{proposition}\label{prop:DJHP-for-2-vertex-alg-with-finite-global-dimension}
Assume that $k$ is algebraically closed. If $A$ is of rank $2$ and of finite global dimension, then $A$ satisfies DJHP.
\end{proposition}
\begin{proof}
Assume that $A$ is not derived simple. Then there are algebras $B$ and $C$ such that there is a nontrivial recollement of ${\mathcal D}(A)$ by ${\mathcal D}(B)$ and ${\mathcal D}(C)$. By Proposition~\ref{prop:invariants}, both $B$ and $C$ are of rank $1$ and of finite global dimension, so they are derived equivalent to $k$. Therefore, any stratification of ${\mathcal D}(A)$ is a recollement and has simple factors two copies of $k$.
\end{proof}

According to \cite[Theorem 3.7]{LY16}, an algebra over an algebraically closed field which is of rank $2$ and of finite global dimensional is either derived simple or derived equivalent to a quasi-hereditary algebra.

\section{Main result} \label{s:main result}

In this section we construct a matrix algebra $\Lambda = \Lambda(A,B)$ from two elementary algebras $A$ and $B$. We show that it has two natural stratifying ideals. By analysing the factors in the induced stratifying recollements, we give two cases when DJHP fails for $\Lambda$ (see Theorem \ref{t:maintheorem}).

\medskip

An algebra is said to be \emph{elementary} if all its simple modules are 1-dimensional, equivalently, if it is isomorphic to the quotient of the path algebra of a finite quiver by an admissible ideal. For an elementary algebra $A$ let $\opname{rad}(A)$ be the radical of the algebra $A$ and $\bar{A}=A/\opname{rad}(A)$ the semisimple quotient, which is the direct product of finite copies of $k$. We fix a retraction $\bar{A} \hookrightarrow A$ such that $\bar{A}$ becomes a subalgebra of $A$ (if $A$ is given by a quiver with relations, we take $\bar{A}$ to be the subspace spanned by all trivial paths). Then $A$ is isomorphic to $\bar{A} \oplus \opname{rad}(A)$ as a $k$-vector space. For $a\in A$, we will write $a=a_1+a_2$ with $a_1\in \bar{A}$ and $a_2\in \opname{rad}(A)$.

\subsection{The trivially twisted tensor product}\label{ss:tensor-product}

Let $A$ and $B$ be two elementary algebras.
Define a $k$-linear map $\tau: B \otimes A \rightarrow A \otimes B$ as the composition
$$B \otimes A \twoheadrightarrow (B \otimes A)/ (\opname{rad}(B) \otimes \opname{rad}(A)) \xrightarrow{\text{flip}} (A \otimes B) / (\opname{rad}(A) \otimes \opname{rad}(B)) \hookrightarrow A \otimes B$$
where the $\text{flip}$ is defined by switching the two components in the tensor products,
$(A \otimes B) / (\opname{rad}(A) \otimes \opname{rad} (B))$ is identified with $\bar{A} \otimes \bar{B} \oplus \bar{A} \otimes \opname{rad}(B) \oplus \opname{rad}(A) \otimes \bar{B}$ and naturally embedded into $A\otimes B$.
Thus for $a\in A$ and $b\in B$ we have
$$\tau(b \otimes a) = a \otimes b_1 + a_1 \otimes b_2 = a_1 \otimes b_1 + a_2 \otimes b_1 + a_1 \otimes b_2=a\otimes b-a_2\otimes b_2.$$
In particular, $\tau(b \otimes a)=0$ if $a\in\opname{rad}(A)$ and $b\in\opname{rad}(B)$.

Denote by $\mu_A$ and $\mu_B$ the multiplications of the algebras $A$ and $B$, respectively. Define a $k$-linear map $\mu_\tau$ on the tensor product $A\otimes B$:
$$\mu_\tau: (A \otimes B) \otimes (A \otimes B) \xrightarrow{\mathrm{id}_A \otimes \tau \otimes \mathrm{id}_B} A \otimes A \otimes B \otimes B \xrightarrow{\mu_A \otimes \mu_B} A \otimes B.$$
More precisely, for $a,a' \in A$ and $b,b' \in B$,
we have
\begin{align*}
\mu_\tau((a \otimes b) \otimes (a' \otimes b')) & =  \mu_A \otimes \mu_B (a \otimes (a' \otimes b_1 + a'_1 \otimes b_2) \otimes b')\\
& =  aa' \otimes b_1b' + aa'_1 \otimes b_2 b'\\
& = aa'\otimes bb'-aa'_2\otimes b_2 b'.
\end{align*}

\begin{proposition}\label{prop:twisted-tensor-product}
The vector space $A \otimes B$ with multiplication $\mu_\tau$ is an algebra.
\end{proposition}

\begin{proof}
By \cite[Proposition/Definition 2.3 and Remark 2.4 (1)]{CSV}, it is enough to prove
\begin{align*}\tau \circ (\mathrm{id}_B \otimes \mu_A) &= (\mu_A \otimes \mathrm{id}_B) \circ (\mathrm{id}_A \otimes \tau) \circ (\tau \otimes \mathrm{id}_A),\\
\tau \circ (\mu_B \otimes \mathrm{id}_A) &= (\mathrm{id}_A \otimes \mu_B) \circ (\tau \otimes \mathrm{id}_B) \circ (\mathrm{id}_B \otimes \tau).
\end{align*}
We only proof the first equality and the second is similar.
Let $b\in B$ and $a,a'\in A$. Then
\begin{align*}
\tau \circ (\mathrm{id}_B \otimes \mu_A)(b \otimes a \otimes a') & =  \tau( b \otimes aa')\\
& =  \tau( (b_1 + b_2) \otimes (a_1a'_1 + (a_1a'_2 + a_2a'_1 + a_2a'_2)) )\\
& =  aa' \otimes b_1 + a_1a'_1 \otimes b_2.
\end{align*}
On the other hand
\begin{align*}
(\mu_A \otimes \mathrm{id}_B) \circ (\mathrm{id}_A \otimes \tau) \circ (\tau \otimes \mathrm{id}_A)(b \otimes a \otimes a') & =
       (\mu_A \otimes \mathrm{id}_B) \circ (\mathrm{id}_A \otimes \tau)((a \otimes b_1 + a_1 \otimes b_2) \otimes a')\\
& =  (\mu_A \otimes \mathrm{id}_B)(a \otimes a' \otimes b_1 + a_1 \otimes a'_1 \otimes b_2)\\
& =  aa' \otimes b_1 + a_1a'_1 \otimes b_2.
\end{align*}
\end{proof}

Following \cite{CX} we call the algebra in Proposition~\ref{prop:twisted-tensor-product} the {\it trivially twisted tensor product} of $A$ and $B$ (over $k$) and denote it by $A\otimes_0 B$. If $A$ or $B$ is semisimple, then $A\otimes_0 B$ is the usual tensor product $A\otimes B$ of algebras. Moreover, $\bar{A}\otimes B$ and $A\otimes \bar{B}$ are subalgebras of $A\otimes_0 B$.

\begin{lemma}\label{l:twisted tensor}
\begin{itemize}
\item[(i)] $A\otimes \opname{rad}(B)$ and $\opname{rad}(A) \otimes B$ are two-sided ideals of $A \otimes_0 B$ with  quotients $A \otimes \bar{B}$ and $\bar{A} \otimes B$, respectively.

\item[(ii)] The restriction of $\mu_\tau$ to $(A \otimes \opname{rad}(B)) \otimes (\opname{rad}(A) \otimes B)$ vanishes.
\end{itemize}
\end{lemma}

\begin{proof}

(i) We check that $A \otimes \opname{rad}(B)$ is closed when multiplying with an element in $A \otimes_0 B$. The proof for $\opname{rad}(A) \otimes B$ is similar. For $a,a'\in A$ and $b=b_2 \in \opname{rad}(B),b' \in B$ we have
\begin{align*}
\mu_\tau((a \otimes b_2) \otimes (a' \otimes b')) & = \mu_A \otimes \mu_B (a \otimes  a'_1 \otimes b_2 \otimes b')\\
& = aa'_1 \otimes b_2 b' \in A \otimes \opname{rad}(B),
\end{align*}
and
\begin{align*}
\mu_\tau((a' \otimes b') \otimes (a \otimes b_2)) & =  \mu_A \otimes \mu_B (a' \otimes  (a \otimes b'_1 + a_1 \otimes b'_2)  \otimes b_2)\\
& =  a'a \otimes b'_1 b_2 + a'a_1 \otimes b'_2 b_2 \in A \otimes \opname{rad}(B).
\end{align*}

(ii) This follows from the fact that the restriction of $\tau$ to $\opname{rad}(B) \otimes \opname{rad}(A)$ vanishes.
\end{proof}

\subsection{The matrix algebra: construction and properties}\label{ss:construction} \label{ss:property}

Let $A$ and $B$ be two elementary algebras. We have the following data:
\begin{itemize}
\item[(1)] an $(\bar{A} \otimes B)$-$(A \otimes_0 B)$-bimodule $\bar{A} \otimes B$: the left $(\bar{A}\otimes B)$-module structure is the free module of rank $1$; the
right $(A \otimes_0 B)$-module structure is induced from the quotient map $A \otimes_0 B \twoheadrightarrow \bar{A} \otimes B = (A\otimes_0 B)/ (\opname{rad}(A) \otimes B)$ in Lemma~\ref{l:twisted tensor} (i); explicitly, we have
\[
(a'\otimes b')(a\otimes b)(a''\otimes b'')=a'aa''_1\otimes b'bb''
\]
for $a\otimes b, a'\otimes b'\in \bar{A}\otimes B$ and $a''\otimes b''\in A\otimes_0 B$;

\item[(2)] an $(A \otimes_0 B)$-$(\bar{A} \otimes B)$-bimodule $A \otimes \opname{rad}(B)$: $A \otimes \opname{rad}(B)$ as a two-sided ideal of $A \otimes_0 B$ is naturally an $(A\otimes_0 B)$-$(A \otimes_0 B)$-bimodule; by Lemma \ref{l:twisted tensor} (ii), $\opname{rad}(A) \otimes B$ annihilates $A \otimes \opname{rad}(B)$ from the right hand side, hence the right $(A\otimes_0 B)$-module structure on $A \otimes \opname{rad}(B)$ factors through $\bar{A} \otimes B$ via the quotient map $A \otimes_0 B \twoheadrightarrow \bar{A} \otimes B = (A\otimes_0 B)/ (\opname{rad}(A) \otimes B)$ in Lemma~\ref{l:twisted tensor} (i); the right ${\bar A} \otimes B$-module structure on $A\otimes \opname{rad}(B)$ can also be obtained via the embedding $\bar{A} \otimes B\hookrightarrow A\otimes_0 B$ because the composition of this embedding with the quotient map $A \otimes_0 B \twoheadrightarrow \bar{A} \otimes B$ is the identity of $\bar{A}\otimes B$; explicitly, we have
\[
(a'\otimes b')(a\otimes b)(a''\otimes b'')=a'aa''\otimes b'_1bb''+a'a_1a''\otimes b'_2bb''
\]
for $a\otimes b\in A\otimes \opname{rad}(B)$, $a'\otimes b'\in A\otimes_0 B$ and $a''\otimes b''\in\bar{A}\otimes B$;

\item[(3)] an  $(A \otimes_0 B)$-$(A \otimes_0 B)$-bimodule homomorphism $$\varphi:(A \otimes \opname{rad}(B)) \otimes_{\bar{A} \otimes B} (\bar{A} \otimes B) \rightarrow A \otimes_0 B,~~(a\otimes b)\otimes (a'\otimes b')\mapsto aa'\otimes bb'$$ for $a\otimes b\in A\otimes \opname{rad}(B)$ and $a'\otimes b'\in\bar{A}\otimes B$; it is the composition of the canonical isomorphism $(A \otimes \opname{rad}(B)) \otimes_{\bar{A} \otimes B} (\bar{A} \otimes B)\stackrel{\cong}{\rightarrow} A\otimes \opname{rad}(B)$ and the embedding $A\otimes \opname{rad}(B)\hookrightarrow A\otimes_0 B$;

\item[(4)] an $(\bar{A} {\otimes} B)$-$(\bar{A} {\otimes} B)$-bimodule homomorphism
\begin{align*}
\psi:(\bar{A} \otimes B) \otimes_{{A} \otimes_0 B} ({A} \otimes \opname{rad}(B)) \rightarrow \bar{A} {\otimes } B,~~(a\otimes b)\otimes (a'\otimes b')\mapsto aa'_1\otimes bb'
\end{align*}
for $a\otimes b\in \bar{A}\otimes B$ and $a'\otimes b'\in A\otimes \opname{rad}(B)$; it is the composition of the embedding $(\bar{A} \otimes B) \otimes_{{A} \otimes_0 B} ({A} \otimes \opname{rad}(B)) \hookrightarrow (\bar{A} \otimes B) \otimes_{{A} \otimes_0 B} ({A} \otimes_0 B)$ with the canonical isomorphism $(\bar{A} \otimes B) \otimes_{{A} \otimes_0 B} ({A} \otimes_0 B)\stackrel{\cong}{\rightarrow} \bar{A}\otimes B$. \end{itemize}
Moreover, the following two diagrams are commutative
\begin{align*}
\xymatrix@C=1pc{
(\bar{A} \otimes B)\otimes_{A\otimes_0 B}(A \otimes \opname{rad}(B)) \otimes_{\bar{A} \otimes B} (\bar{A} \otimes B) \ar[rr]^(0.6){\mathrm{id}_{\bar{A}\otimes B}\otimes\varphi} \ar[d]^{\psi\otimes\mathrm{id}_{\bar{A}\otimes B}} & &(\bar{A} \otimes B)\otimes_{A\otimes_0 B} (A\otimes_0 B)\ar[d]^\cong\\
(\bar{A}\otimes B)\otimes_{\bar{A}\otimes B}(\bar{A}\otimes B)\ar[rr]^(0.55)\cong && \bar{A}\otimes B
}
\\
\xymatrix@C=1pc{
(A\otimes\opname{rad}(B))\otimes_{\bar{A}\otimes B}(\bar{A}\otimes B)\otimes_{A\otimes_0 B}(A\otimes\opname{rad}(B))\ar[rr]^(0.6){\mathrm{id}_{A\otimes\opname{rad}(B)}\otimes\psi} \ar[d]^{\varphi\otimes\mathrm{id}_{A\otimes \opname{rad}(B)}} && (A\otimes\opname{rad}(B))\otimes_{\bar{A}\otimes B} (\bar{A}\otimes B)\ar[d]^\cong\\
(A\otimes_0 B)\otimes_{A\otimes_0 B}(A\otimes \opname{rad}(B))\ar[rr]^(0.6)\cong && A\otimes\opname{rad}(B).
}
\end{align*}
We check the commutativity of the first diagram, and the commutativity of all other diagrams in the section can be checked in a similar way. For $a\otimes b\in\bar{A}\otimes B$, $a'\otimes b'\in A\otimes\opname{rad}(B)$ and $a''\otimes b''\in \bar{A}\otimes B$, we have
\begin{align*}
(a\otimes b)\varphi((a'\otimes b')\otimes (a''\otimes b''))&=(a\otimes b)(a'a''\otimes b'b'')=aa'_1a''\otimes bb'b'',\\
\psi((a\otimes b)\otimes(a'\otimes b'))(a''\otimes b'')&=(aa'_1\otimes bb')(a''\otimes b'')=aa'_1a''\otimes bb'b''.
\end{align*}
Therefore we obtain an algebra
$$\Lambda(A,B) := \begin{pmatrix} A \otimes_0 B & A \otimes \opname{rad}(B) \\ \bar{A} \otimes B & \bar{A} \otimes B \end{pmatrix},$$
whose addition is componentwise and whose multiplication is given by matrix multiplication together with the homomorphisms $\varphi$ and $\psi$. See for example \cite{Green82} for the study on matrix rings. In Section~\ref{s:quiver-relation} we will describe the quiver with relations of $\Lambda(A,B)$ in terms of those of $A$ and $B$.

\smallskip
Put $\Lambda=\Lambda(A,B)$. By abuse of notation we write $1$ for the unit elements in all of the algebras $A$, $B$ and $\Lambda$. Consider the two natural idempotents in $\Lambda$
$${\footnotesize e_1=\begin{pmatrix} 1\otimes 1 & 0 \\ 0 & 0 \end{pmatrix},\hskip 20pt e_2 =\begin{pmatrix} 0 & 0 \\ 0 & 1\otimes 1 \end{pmatrix}.}$$
Then in $\Lambda$ it holds that $1= e_1 + e_2$ and $e_1e_2 = 0 = e_2e_1$. Since $A$ is elementary, the rank of $A$ equals $\mathrm{dim}\bar{A}$. The rank of a general algebra is smaller than or equal to the cardinality of a complete set of primitive orthogonal idempotents of $A$; the equality holds if it is elementary.

\begin{lemma} The following hold for $\Lambda$: \label{l:idempotents in Lambda}
\begin{itemize}
\item[(i)] $e_1 \Lambda e_1 = A \otimes_0 B$ and  $\Lambda / \Lambda e_1 \Lambda \cong \bar{A} \otimes \bar{B}$;

\item[(ii)]  $e_2 \Lambda e_2 = \bar{A} \otimes B$ and $\Lambda / \Lambda e_2 \Lambda \cong {A} \otimes \bar{B}$;

\item[(iii)] $\Lambda$ is elementary with $\mathrm{rank}(\Lambda) = 2\mathrm{rank}(A) \mathrm{rank}(B)$.

\end{itemize}
\end{lemma}

\begin{proof}
(i) It is clear that $e_1 \Lambda e_1 = A \otimes_0 B$. The ideal generated by $e_1$ is
\begin{align*}
\Lambda e_1 \Lambda & =  \begin{pmatrix} A \otimes_0 B & A \otimes \opname{rad}(B) \\ \bar{A} \otimes B & \bar{A} \otimes B \end{pmatrix} \begin{pmatrix} 1\otimes 1 & 0 \\ 0 & 0 \end{pmatrix}  \begin{pmatrix} A \otimes_0 B & A \otimes \opname{rad}(B) \\ \bar{A} \otimes B & \bar{A} \otimes B \end{pmatrix} \\
& =    \begin{pmatrix} A \otimes_0 B & 0 \\ \bar{A} \otimes B & 0 \end{pmatrix} \begin{pmatrix} A \otimes_0 B & A \otimes \opname{rad}(B) \\ \bar{A} \otimes B & \bar{A} \otimes B \end{pmatrix} \\
& =   \begin{pmatrix} A \otimes_0 B & A \otimes \opname{rad}(B) \\ \bar{A} \otimes B & \bar{A} \otimes \opname{rad}(B) \end{pmatrix}.
\end{align*}
Hence  $\Lambda / \Lambda e_1 \Lambda \cong \bar{A} \otimes B / (\bar{A} \otimes \opname{rad}(B)) \cong \bar{A} \otimes \bar{B}$.

(ii) Similar to (i).

(iii) Let $\{e_i\mid i\in I\}$ and $\{f_j\mid j\in J\}$ be complete sets of primitive orthogonal idempotents of $\bar{A}$ and $\bar{B}$, respectively. Then $e_i\otimes f_j$ ($i\in I, j\in J$) is a primitive idempotent of $A\otimes_0 B$ because $(e_i\otimes f_j )(A\otimes_0 B) (e_i\otimes f_j)=e_iAe_i\otimes_0 f_jBf_j$ is local. It follows that
\[
\{\begin{pmatrix}e_i\otimes f_j & 0\\ 0 &0\end{pmatrix},\begin{pmatrix}0&0\\ 0 & e_i\otimes f_j\end{pmatrix} \mid i\in I, j\in J\}
\]
is a complete set of primitive orthogonal idempotents of $\Lambda$. It follows that $\mathrm{rank}(\Lambda)\leq 2\mathrm{rank}(A)\mathrm{rank}(B)$.

On the other hand, the subspace
\begin{align*}
I&=\begin{pmatrix}
A\otimes\opname{rad}(B)+\opname{rad}(A)\otimes B & A\otimes \opname{rad}(B)\\
\bar{A}\otimes B & \bar{A}\otimes \opname{rad}(B)
\end{pmatrix}
\end{align*}
is an ideal of $\Lambda$ with semisimple quotient algebra
\[
\Lambda/I=\begin{pmatrix}
\bar{A}\otimes \bar{B} & 0\\
0 & \bar{A}\otimes \bar{B}
\end{pmatrix}.
\]
Therefore $\mathrm{rank}(\Lambda)\geq 2\mathrm{rank}(A)\mathrm{rank}(B)$. Altogether we have $\mathrm{rank}(\Lambda)=2\mathrm{rank}(A)\mathrm{rank}(B)$ and $\Lambda$ is elementary. Moreover,
$I=\opname{rad}(\Lambda)$ and $\bar{\Lambda}=\Lambda/I$.
\end{proof}

Sometimes we can recover $B$ from $\Lambda$.

\begin{lemma}\label{lem:recovering-B}
Assume that $A$ is local nonsimple and let $B'$ be another elementary algebra. If $\Lambda=\Lambda(A,B)$ and $\Lambda(A,B')$ are isomorphic as algebras, then $B$ and $B'$ are isomorphic as algebras.
\end{lemma}
\begin{proof} Put $\Lambda'=\Lambda(A,B')$ and suppose that $\phi:\Lambda\rightarrow\Lambda'$ is an algebra isomorphism.
By Lemma~\ref{l:idempotents in Lambda} (iii), $B$ and $B'$ have the same rank. Let $\{f_1,\ldots,f_m\}$ and $\{f'_1,\ldots,f'_m\}$ be complete sets of primitive orthogonal idempotents of $\bar{B}$ and $\bar{B}'$, respectively. Then
\[
\{\begin{pmatrix}1\otimes f_j & 0\\ 0 &0\end{pmatrix},\begin{pmatrix}0&0\\ 0 & 1\otimes f_j\end{pmatrix} \mid j=1,\ldots,m\}
\]
and
\[
\{\begin{pmatrix}1\otimes f'_j & 0\\ 0 &0\end{pmatrix},\begin{pmatrix}0&0\\ 0 & 1\otimes f'_j\end{pmatrix} \mid j=1,\ldots,m\}
\]
are complete sets of primitive orthogonal idempotents of $\Lambda$ and $\Lambda'$, respectively. Moreover, for $1\leq j\leq m$,
the ideal of $\Lambda$ generated by  $1-\begin{pmatrix} 1\otimes f_j & 0\\ 0 & 0\end{pmatrix}$ is
\begin{align*}
\Lambda (1-\begin{pmatrix} 1\otimes f_j & 0\\ 0 & 0\end{pmatrix})\Lambda=\begin{pmatrix}
A\otimes(\opname{rad}(B)+B(1-f_j)B) & A\otimes \opname{rad}(B)\\
\bar{A}\otimes B & \bar{A}\otimes B
\end{pmatrix},
\end{align*}
so
\begin{align*}
\Lambda/\Lambda(1-\begin{pmatrix} 1\otimes f_j & 0\\ 0 & 0\end{pmatrix})\Lambda&\cong A.
\end{align*}
Similarly, for $1\leq j\leq m$, we have
\begin{align*}
&\Lambda/\Lambda(1-\begin{pmatrix} 0 & 0\\ 0 & 1\otimes f_j\end{pmatrix})\Lambda\cong k,\\
&\Lambda'/\Lambda'(1-\begin{pmatrix} 1\otimes f'_{j} & 0\\ 0 & 0\end{pmatrix})\Lambda'\cong A,\\
&\Lambda'/\Lambda'(1-\begin{pmatrix} 0 & 0\\ 0 & 1\otimes f'_{j'} \end{pmatrix})\Lambda'\cong k.
\end{align*}
Therefore, it follows from \cite[Theorem 3.4.1]{DrozdKirichenko} that there exist an invertible element $\lambda\in\Lambda'$ and a permutation $\sigma$ on $\{1,\ldots,m\}$ such that $\phi\begin{pmatrix} 0 & 0\\ 0 & 1\otimes f_j\end{pmatrix}=\lambda\begin{pmatrix} 0 & 0\\ 0 & 1\otimes f'_{\sigma(j)}\end{pmatrix}\lambda^{-1}$  for any $1\leq j\leq m$. So $\phi\begin{pmatrix}0 & 0\\ 0 & 1\otimes 1\end{pmatrix}=\lambda\begin{pmatrix}0 & 0\\ 0 & 1\otimes 1\end{pmatrix}\lambda^{-1}$, and we obtain a chain of algebra isomorphisms
\begin{align*}
B&= \begin{pmatrix}0 & 0\\ 0 & 1\otimes 1\end{pmatrix}\Lambda\begin{pmatrix}0 & 0\\ 0 & 1\otimes 1\end{pmatrix}\\
&\cong \phi\begin{pmatrix}0 & 0\\ 0 & 1\otimes 1\end{pmatrix}\Lambda'\phi\begin{pmatrix}0 & 0\\ 0 & 1\otimes 1\end{pmatrix}\\
&\cong \begin{pmatrix}0 & 0\\ 0 & 1\otimes 1\end{pmatrix}\Lambda'\begin{pmatrix}0 & 0\\ 0 & 1\otimes 1\end{pmatrix}\\
&= B'.
\end{align*}
\end{proof}

\subsection{The matrix algebra: failure of DJHP}
Let $A$ and $B$ be two elementary algebras. In Section~\ref{ss:property} we defined a $2 \times 2$-matrix algebra $\Lambda = \Lambda(A,B)$ and two idempotents $e_1$ and $e_2$ of $\Lambda$.

\begin{proposition}\label{p:stratifying}
The idempotent ideals generated by $e_1$ and $e_2$ are both stratifying ideals of $\Lambda$. Consequently, ${\mathcal D}(\Lambda)$ has two stratifying recollements, induced by $e_1$ and $e_2$ and denoted by $(R1)$ and $(R2)$ respectively, such that
\begin{itemize}
\item[--]
$(R1)$ is a recollement of ${\mathcal D}(\Lambda)$ by ${\mathcal D}(\Lambda / \Lambda e_1 \Lambda)$ and ${\mathcal D}(e_1\Lambda e_1)$, where $\Lambda / \Lambda e_1 \Lambda\cong \bar{A} \otimes \bar{B}$, which is the direct product of $\mathrm{rank}(A)\mathrm{rank}(B)$ copies of $k$, and $e_1\Lambda e_1 = A\otimes_0 B$;
\item[--] $(R2)$ is a recollement of ${\mathcal D}(\Lambda)$ by ${\mathcal D}(\Lambda / \Lambda e_2 \Lambda)$ and ${\mathcal D}(e_2 \Lambda e_2)$, where $\Lambda / \Lambda e_2 \Lambda \cong A \otimes \bar{B}$, which is the direct product of $\mathrm{rank}(B)$ copies of $A$, and $e_2 \Lambda e_2 = \bar{A} \otimes B$, which is the direct product of $\mathrm{rank}(A)$ copies of $B$.
\end{itemize}
\end{proposition}

We need some preparation for the proof of Proposition~\ref{p:stratifying}.
Let $M$ be a left $B$-module with $B$-action given by $\rho_M:B\otimes M\rightarrow M$. By \cite[Sections 3.6, 3.7 and 3.8]{CSV}, the two equalities in the proof of Proposition~\ref{prop:twisted-tensor-product} shows that the map
\[
\xymatrix@R=0.5pc@C=1pc{
(A \otimes B) \otimes (A \otimes M)  \ar[rrr]^{\mathrm{id}_A\otimes \tau \otimes \mathrm{id}_M} &&& A \otimes A \otimes B\otimes M \ar[r]^(0.6){\mu_A\otimes \rho_M}  & A\otimes M\\
(a\otimes b)\otimes (a'\otimes m)\ar@{|->}[rrrr] &&&& aa'\otimes b_1m+aa'_1\otimes b_2m
}
\]
defines a left $(A\otimes_0 B)$-module structure on $A\otimes M$. We denote it by $A\otimes_0 M$.  Note that the ideal $A\otimes \opname{rad}(B)$ of $A\otimes_0 B$ as a left $(A\otimes_0 B)$-module is $A\otimes_0\opname{rad}(B)$.
For a left $B$-module homomorphism $f:M\to N$, the map
\[
\mathrm{id}_A\otimes f: A\otimes_0 M\longrightarrow A\otimes_0 N,~~a\otimes m\mapsto a\otimes f(m)
\]
is a left $A\otimes_0 B$-module homomorphism.

Next we construct a left $\Lambda$-module from $M$. For knowledge on modules over matrix rings, we refer to \cite{Green82}. Now we have a left $(A\otimes_0 B)$-module $A\otimes_0 M$ and a left $(\bar{A}\otimes B)$-module $\bar{A}\otimes M$. Moreover, we have
\begin{itemize}
\item[--] a left $A\otimes_0 B$-module homomorphism
\[
\xi=\xi_M:(A\otimes \opname{rad}(B))\otimes_{\bar{A}\otimes B}(\bar{A}\otimes M)\to A\otimes_0 M,~~(a\otimes b)\otimes(a'\otimes m)\mapsto aa'\otimes bm,
\]
for $a\otimes b\in A\otimes\opname{rad}(B)$ and $a'\otimes b'\in\bar{A}\otimes M$,
\item[--] a left $\bar{A}\otimes B$-module homomorphism
\[
\eta=\eta_M:(\bar{A}\otimes B)\otimes_{A\otimes_0 B}(A\otimes_0 M)\to \bar{A}\otimes M,~~(a\otimes b)\otimes (a'\otimes m)\mapsto aa'_1\otimes bm,
\]
for $a\otimes b\in\bar{A}\otimes B$ and $a'\otimes m\in A\otimes_0 M$.
\end{itemize}
They satisfy the following commutative diagrams
\begin{align*}
\xymatrix@C=1pc{
(\bar{A} \otimes B)\otimes_{A\otimes_0 B}(A \otimes \opname{rad}(B)) \otimes_{\bar{A} \otimes B} (\bar{A} \otimes M) \ar[rr]^(0.6){\mathrm{id}_{\bar{A}\otimes B}\otimes\xi} \ar[d]^{\psi\otimes\mathrm{id}_{\bar{A}\otimes M}} & &(\bar{A} \otimes B)\otimes_{A\otimes_0 B} (A\otimes_0 M)\ar[d]^\eta\\
(\bar{A}\otimes B)\otimes_{\bar{A}\otimes B}(\bar{A}\otimes M)\ar[rr]^(0.55)\cong && \bar{A}\otimes M
}
\\
\xymatrix@C=1pc{
(A\otimes\opname{rad}(B))\otimes_{\bar{A}\otimes B}(\bar{A}\otimes B)\otimes_{A\otimes_0 B}(A\otimes_0 M)\ar[rr]^(0.6){\mathrm{id}_{A\otimes\opname{rad}(B)}\otimes\eta} \ar[d]^{\varphi\otimes\mathrm{id}_{A\otimes_0 M}} && (A\otimes\opname{rad}(B))\otimes_{\bar{A}\otimes B} (\bar{A}\otimes M)\ar[d]^\xi\\
(A\otimes_0 B)\otimes_{A\otimes_0 B}(A\otimes_0 M)\ar[rr]^(0.6)\cong && A\otimes_0 M.
}
\end{align*}
So $\xi$ and $\eta$ define a left $\Lambda$-module structure on
\[
\begin{pmatrix}
A\otimes_0 M\\ \bar{A}\otimes M
\end{pmatrix}.
\]
We denote this module by $\widetilde{M}$. Note that $\widetilde{B}$ is naturally isomorphic to $\Lambda e_1$ as a left $\Lambda$-module.

Let $f:M\to N$ be a left $B$-module homomorphism. Then there are commutative diagrams
\begin{align*}
\xymatrix{
(\bar{A}\otimes B)_{A\otimes_0 B}(A\otimes_0 M)\ar[r]^(0.7){\eta_M} \ar[d]^{\mathrm{id}_{\bar{A}\otimes B}\otimes (\mathrm{id}_A\otimes f)}& \bar{A}\otimes M\ar[d]^{\mathrm{id}_{\bar{A}}\otimes f}\\
(\bar{A}\otimes B)_{A\otimes_0 B}(A\otimes_0 N)\ar[r]^(0.7){\eta_N} & \bar{A}\otimes N
}
\xymatrix{
(A\otimes \opname{rad}(B))\otimes_{\bar{A}\otimes B} (\bar{A}\otimes M)\ar[r]^(0.7){\xi_M}\ar[d]^{(\mathrm{id}_{A\otimes \opname{rad}(B)})\otimes (\mathrm{id}_{\bar{A}}\otimes f)} & A\otimes_0 M\ar[d]^{\mathrm{id}_A\otimes f}\\
(A\otimes\opname{rad}(B))\otimes_{\bar{A}\otimes B} (\bar{A}\otimes N)\ar[r]^(0.7){\xi_N} & A\otimes_0 N.
}
\end{align*}
It follows that
\[
\widetilde{f}:=\begin{pmatrix}
\mathrm{id}_A\otimes f & \\
 & \mathrm{id}_{\bar{A}}\otimes f
\end{pmatrix}
:\widetilde{M}
\longrightarrow
\widetilde{N}
\]
is a left $\Lambda$-module homomorphism.
Therefore, for a complex
\[
M^\bullet=\ldots\rightarrow M^{i-1}\stackrel{d^{i-1}}{\rightarrow} M^i\stackrel{d^i}{\rightarrow} M^{i+1}\rightarrow\ldots
\]
of left $B$-modules,
the sequence
\[
\widetilde{M^\bullet}:=\ldots\rightarrow
\widetilde{M^{i-1}}
\stackrel{\widetilde{d^{i-1}}}{\longrightarrow} \widetilde{M^{i}}
\stackrel{\widetilde{d^{i}}}{\longrightarrow} \widetilde{M^{i+1}}\rightarrow\ldots
\]
is a complex of left $\Lambda$-modules. Moreover, $M^\bullet$ is acyclic if and only if so is $\widetilde{M^\bullet}$ because as a complex of vector spaces $\widetilde{M^\bullet}$ is just $(A\oplus \bar{A})\otimes M^\bullet$. Therefore we have

\begin{lemma}\label{lem:tensoring-projective-resolution-for-lambda}
Let $M$ be a left $B$-module and let $P^\bullet$ be a free resolution of $M$ over $B$. Then $\widetilde{P^\bullet}$ is a projective resolution of $\widetilde{M}$ over $\Lambda$ with all components in $\opname{add}\nolimits(\Lambda e_1)$.
\end{lemma}

\begin{proof}[Proof of Proposition~\ref{p:stratifying}] The second statement follows from the first statement together with Lemma~\ref{l:idempotents in Lambda}. To prove the first statement, we construct a projective resolution of the left $\Lambda$-module $\Lambda/ \Lambda e_i \Lambda$ ($i=1,2$) all of whose components but the one in degree $0$ belong to $\opname{add}\nolimits(\Lambda e_i)$. The statement follows then by Lemma \ref{l:criterion for stratifying ideal}.

Multiplying $\begin{pmatrix} 0 & 0\\ 1\otimes 1 & 0\end{pmatrix}$ on the right defines an injective $\Lambda$-module homomorphism
\[
\Lambda e_2\longrightarrow \Lambda e_1
\]
with quotient $\Lambda/\Lambda e_2\Lambda$. So there is a short exact sequence
\[
0\rightarrow \Lambda e_2\rightarrow \Lambda e_1\rightarrow \Lambda/\Lambda e_2\Lambda \rightarrow 0,
\]
which is a projective $\Lambda$-resolution of $\Lambda/\Lambda e_2\Lambda$ with the desired property.

Consider the projection from $\Lambda e_2$ to $\Lambda/\Lambda e_1\Lambda$. Its kernel is naturally isomorphic to
\[
\widetilde{\opname{rad}(B)}=\begin{pmatrix}A\otimes_0 \opname{rad}(B) \\ \bar{A}\otimes \opname{rad}(B)\end{pmatrix}.
\]
By Lemma~\ref{lem:tensoring-projective-resolution-for-lambda}, $\widetilde{\opname{rad}(B)}$ admits a projective $\Lambda$-resolution with all components in $\opname{add}\nolimits(\Lambda e_1)$. This together with the projection from $\Lambda e_2$ to $\Lambda / \Lambda e_1 \Lambda$ provides a projective $\Lambda$-resolution of $\Lambda / \Lambda e_1 \Lambda$ with the desired property.
\end{proof}

\begin{corollary} The following statements are equivalent: \label{c:finite global dimension}
\begin{itemize}
\item[(i)] both $A$ and $B$ have finite global dimension;

\item[(ii)] the matrix algebra $\Lambda$ has finite global dimension;

\item[(iii)] the trivially twisted tensor product $A \otimes_0 B$ has finite global dimension.
\end{itemize}
\end{corollary}

\begin{proof} Applying Proposition~\ref{prop:invariants} (i) to the stratifying recollements $(R1)$ and $(R2)$ in Proposition~\ref{p:stratifying} we obtain the equivalence between (i) and (ii)  and between (i) and (iii), respectively.
\end{proof}

The following theorem is the main result of this paper.

\begin{theorem}  In the following two cases the matrix algebra $\Lambda$ does not satisfy the derived Jordan--H\"older property:
\begin{itemize}
\item[(1)] both $A$ and $B$ are derived simple but nonsimple;

\item[(2)] one of $A$ and $B$ is derived simple but nonsimple, and the other one has infinite global dimension.
\end{itemize} \label{t:maintheorem}
\end{theorem}

\begin{proof} For a stratification ${\mathcal S}$ of ${\mathcal D}(A)$ let $s({\mathcal S})$ denote the multiplicity of $k$ as simple factors of ${\mathcal S}$.

Consider the two stratifying recollements $(R1)$ and $(R2)$ in Proposition~\ref{p:stratifying}. Recall that $\Lambda/\Lambda e_1\Lambda\cong \bar{A}\otimes \bar{B}$, $e_1\Lambda e_1= A\otimes_0 B$, $\Lambda/\Lambda e_2\Lambda \cong A\otimes \bar{B}$ and $e_2\Lambda e_2= \bar{A}\otimes B$. By \cite[Lemma 7.1]{AKLY2}, $s({\mathcal S}_{\Lambda/\Lambda e_1\Lambda})=\mathrm{rank}(A)\mathrm{rank}(B)$ for any stratification ${\mathcal S}_{\Lambda/\Lambda e_1\Lambda}$ of ${\mathcal D}(\Lambda/\Lambda e_1\Lambda)$ and hence $s({\mathcal S}_1)\geq \mathrm{rank}(A)\mathrm{rank}(B)$ for any stratification ${\mathcal S}_1$ of ${\mathcal D}(A)$ starting from $(R1)$.

We will show that in both cases (1) and (2), $s({\mathcal S}_2)<\mathrm{rank}(A)\mathrm{rank}(B)$ for any stratification ${\mathcal S}_2$ starting from the recollement $(R2)$. It follows that DJHP fails for $\Lambda$.
In both cases we may assume that $A$ is derived simple but nonsimple. Then it follows from \cite[Lemma 7.1]{AKLY2} that any stratification of ${\mathcal D}(\Lambda/\Lambda e_2\Lambda)$ has simple factors $\mathrm{rank}(B)$ copies of $A$. Therefore it suffices to show that $s({\mathcal S}_{e_2\Lambda e_2})<\mathrm{rank}(A)\mathrm{rank}(B)$ for any stratification ${\mathcal S}_{e_2\Lambda e_2}$ of ${\mathcal D}(e_2\Lambda e_2)$. In view of Lemma~\ref{lem:splitting-stratification}, this is equivalent to showing that $s({\mathcal S}_{B})<\mathrm{rank}(B)$ for any stratification ${\mathcal S}_{B}$ of ${\mathcal D}(B)$. This is true in case (1) because $B$ is derived simple nonsimple. In case (2) this follows from Proposition~\ref{prop:invariants} since $B$ has infinite global dimension.
\end{proof}

\begin{remark}
The \emph{length} of a stratification is the number of its simple factors. In case (1) of Theorem~\ref{t:maintheorem}, if $A$ and $B$ are not local, namely, the ranks of $A$ and $B$ are greater than $1$, then the length of any stratification starting from $(R1)$ is different from that of any stratification starting from $(R2)$. Indeed, it follows from the proof of Theroem~\ref{t:maintheorem} that the length of any stratification starting from $(R1)$ is greater than or equal to $1+\mathrm{rank}(A)\mathrm{rank}(B)$, while the length of any stratification starting from $(R2)$ is precisely $\mathrm{rank}(A)+\mathrm{rank}(B)$.
\end{remark}

\begin{remark}
If $A$ (respectively, $B$) does not satisfy DJHP, nor does $\Lambda$. Indeed, we can start with the recollement $(R2)$ and obtain stratifications with different simple factors. Assume that both $A$ and $B$ satisfy DJHP. Following the proof of Theorem~\ref{t:maintheorem} we see that $\Lambda$ does not satisfy DJHP if $s({\mathcal S}_A)\mathrm{rank}(B)+s({\mathcal S}_B)\mathrm{rank}(A)<\mathrm{rank}(A)\mathrm{rank}(B)$ for a/any stratification ${\mathcal S}_A$ of ${\mathcal D}(A)$ and a/any stratification ${\mathcal S}_B$ of ${\mathcal D}(B)$. For example, this is the case if a/any stratification of ${\mathcal D}(A)$ (respectively, ${\mathcal D}(B)$) has no simple factors $k$, and $B$ (respectively, $A$) has infinite global dimension or a/any stratification of ${\mathcal D}(B)$ (respectively, ${\mathcal D}(A)$) has length strictly smaller than $\mathrm{rank}(B)$.
\end{remark}

Proposition~\ref{prop:DJHP-for-2-vertex-alg-with-finite-global-dimension} states that over an algebraically closed field all rank $2$ algebras of finite global dimension satisfy DJHP. This is far from being true for rank $2$ algebras of infinite global dimension.

\begin{corollary}\label{cor:local} Assume that both $A$ and $B$ are local nonsimple. Then $\Lambda$ does not satisfy DJHP.
\end{corollary}
\begin{proof}
This follows from Theorem~\ref{t:maintheorem} because local algebras are derived simple by \cite[Proposition 4.11]{AKL2}.
\end{proof}

We will see in Example~\ref{ex:akly2-example} that $\Lambda(k[x]/(x^2),k[x]/(x^2))$ is the algebra in \cite[Example 7.6]{AKLY2} which does not satisfy DJHP.

Examples of derived simple algebras of positive finite global dimension are given in \cite{H,Membrillo-Hernandez94,LY16}. So the next corollary shows that there are algebras of finite global dimension which do not satisfy DJHP.
In \cite[Proposition 3.4]{K} Kalck provides a family of finite dimensional algebras of global dimension 2 (hence quasi-hereditary) for which DJHP fails.

\begin{corollary}\label{cor:finite-global-dimension} Assume that both $A$ and $B$ are derived simple of positive finite global dimension. Then $\Lambda$ has finite global dimension and does not satisfy DJHP.
\end{corollary}
\begin{proof}
By Theorem~\ref{t:maintheorem}, it suffices to show that $\Lambda$ has finite global dimension,  which follows from Corollary \ref{c:finite global dimension}.
\end{proof}

The next corollary together with Lemma~\ref{lem:recovering-B} shows that the cardinality of the elementary algebras which do not satisfy DJHP is the same as the cardinality of all elementary algebras.

\begin{corollary}\label{cor:cardinality} Assume that $A$ is local nonsimple. Then $\Lambda(A,\Lambda(A,B))$ does not satisfy DJHP.
\end{corollary}

\begin{proof}
By Theorem~\ref{t:maintheorem}, it suffices to show that $\Lambda(A,B)$ has infinite global dimension,  which follows from Corollary \ref{c:finite global dimension}.
\end{proof}

\section{Quiver and relations}\label{s:quiver-relation}

Let $A$ and $B$ be elementary algebras and let
$$\Lambda=\Lambda(A,B) := \begin{pmatrix} A \otimes_0 B & A \otimes \opname{rad}(B) \\ \bar{A} \otimes B & \bar{A} \otimes B \end{pmatrix},$$
be defined as in Section~\ref{ss:construction}. The aim of this section is to describe the quiver with relations of $\Lambda$ in terms of those of $A$ and $B$.

\begin{theorem}\label{thm:quiver-with-relations}
Let $A = kQ /(\mathcal{R})$ and $B = kP /(\mathcal{S})$, where $Q$ and $P$ are finite quivers, and $\mathcal{R}$ and ${\mathcal S}$ are sets of minimal relations of $A$ and $B$, respectively. Then as an algebra $\Lambda$ is isomorphic to $k\Gamma/({\mathcal I})$, where
\begin{itemize}
\item
the set $\Gamma_0$ of vertices  is $\{i^j,~j^i\mid i\in Q_0, j\in P_0\}$, namely, it is two copies of $Q_0\times P_0$;
\item the set $\Gamma_1$ consists of three types of arrows:
\begin{itemize}
\item $\alpha^j: \mathrm{s}(\alpha)^j \rightarrow \mathrm{t}(\alpha)^j$, for $\alpha\in Q_1$ and $j\in P_0$,

\item $c(i,j): i^j \rightarrow j^i$, for $i\in Q_0$ and $j\in P_0$,

\item $\beta^i: \mathrm{s}(\beta)^i \rightarrow i^{\mathrm{t}(\beta)}$, for $\beta \in P_1$ and $i\in Q_0$;
\end{itemize}
\item
the set $\mathcal{I}$ consists of three types of relations:
\begin{itemize}
\item $r^j$, for $r\in\mathcal{R}$ and $j\in P_0$,

\item $c(i,j) \alpha^j$, for $i\in Q_0$, $j\in P_0$ and $\alpha \in Q_1$ with $\mathrm{t}(\alpha)=i$,

\item $s^i$, for $s\in \mathcal{S}$ and $i\in Q_0$,
\end{itemize}
where
\begin{align*}r^j& =\sum \lambda_{i_1,\ldots, i_s} \alpha_{i_1}^j \cdots \alpha_{i_s}^j
\end{align*}
for $r = \sum \lambda_{i_1, \ldots, i_s} \alpha_{i_1} \cdots \alpha_{i_s} \in \mathcal{R}$ and $j\in P_0$, and
\begin{align*}s^i &= \sum \mu_{j_1,\ldots,j_t} \beta_{j_1}^i c(i,\mathrm{s}(\beta_{j_1})) \beta_{j_2}^i c(i,\mathrm{s}(\beta_{j_2}))\cdots \beta_{j_{t-1}}^i c(i,\mathrm{s}(\beta_{j_{t-1}})) \beta_{j_t}^i\\
&=\sum \mu_{j_1,\ldots,j_t} \beta_{j_1}^i c \beta_{j_2}^i c\cdots \beta_{j_{t-1}}^i c \beta_{j_t}^i
\end{align*} for $s = \sum \mu_{j_1, \ldots, j_t} \beta_{j_1} \cdots \beta_{j_t} \in \mathcal{S}$ and $i \in Q_0$, where $c=\sum_{i\in Q_0,j\in P_0}c(i,j)$.
\end{itemize}
\end{theorem}

Before proving Theorem~\ref{thm:quiver-with-relations}, we first give two examples illustrating the construction of the quiver with relations in the theorem.

\begin{example}
Let $Q$ and $P$ be the following quivers with relations
\begin{align*}\xymatrix{1\ar[r]^{\alpha_1} & 2\ar[r]^{\alpha_2} & 3\ar[r]^{\alpha_3} & 4} ,&~~\alpha_3\alpha_2\alpha_1\\
\xymatrix{5\ar[r]^{\beta_1} & 6\ar[r]^{\beta_2} & 7 },&~~\beta_2\beta_1.
\end{align*}
Then $\Gamma$ is given by
\[\xymatrix@C=0.7pc@R=0.5pc{
  & & && & & 1^5 \ar[rrrr]|{\alpha_1^5} \ar@{-->}'[dd]|(0.75){c(1,5)}[ddd]  &  & & & 2^5 \ar@{-->}'[dd]|(0.75){c(2,5)}[ddd] \ar[rrrr]|{\alpha_2^5} &  & & & 3^5\ar@{-->}'[dd]|(0.75){c(3,5)}[ddd] \ar[rrrr]|{\alpha_3^5} &  & && 4^5 \ar@{-->}[ddd]|(0.45){c(4,5)}      \\ \\
  & &&1^6 \ar[rrrr]|{\alpha_1^6}\ar@{-->}'[dd]|(0.75){c(1,6)}[ddd]      & & &  & 2^6 \ar@{-->}'[dd]|(0.75){c(2,6)}[ddd] \ar[rrrr]|{\alpha_2^6} & & & & 3^6 \ar@{-->}'[dd]|(0.75){c(3,6)}[ddd] \ar[rrrr]|{\alpha_3^6} & & & & 4^6 \ar@{-->}[ddd]|(0.47){c(4,6)} \\
  &  &&& && 5^1 \ar@{..>}[ulll]|{\beta_1^1}      &  & & & 5^2 \ar@{..>}[ulll]|{\beta_1^2}   &  & &&  5^3 \ar@{..>}[ulll]|{\beta_1^3}  & & & & 5^4 \ar@{..>}[ulll]|{\beta_1^4}           \\
  1^7 \ar[rrrr]|{\alpha_1^7}\ar@{-->}[ddd]|{c(1,7)} &  & &&  2^7 \ar[rrrr]|{\alpha_2^7} \ar@{-->}[ddd]|{c(2,7)} &  & & & 3^7 \ar[rrrr]|{\alpha_3^7} \ar@{-->}[ddd]|{c(3,7)} &  & && 4^7 \ar@{-->}[ddd]|{c(4,7)} \\
  & & & 6^1 \ar@{..>}[ulll]|{\beta_2^1}      &  & & & 6^2 \ar@{..>}[ulll]|{\beta_2^2}  &  & && 6^3 \ar@{..>}[ulll]|{\beta_2^3}  & & & & 6^4 \ar@{..>}[ulll]|{\beta_2^4}  \\  \\
  7^1 & &&& 7^2 &&& & 7^3 &&& & 7^4 }\]
and ${\mathcal I}$ consists of
\begin{itemize}
\item[--] $\alpha_3^j\alpha_2^j\alpha_1^j$, $j=5,6,7$;

\item[--] $c(i,j)\alpha_{i-1}^j$, $i=2,3,4$, $j=5,6,7$;

\item[--] $\beta_2^i c(i,6) \beta_1^i$, $i=1,2,3,4$.
\end{itemize}

\end{example}

\begin{example}\label{ex:akly2-example}
Let $A=k[x]/(x^2)$ and $B=k[y]/(y^2)$. The quiver of $A$ (respectively, $B$) has a unique vertex, denoted by $a$ (respectively, $b$), and one loop $\alpha$ (respectively, $\beta$). By Theorem~\ref{thm:quiver-with-relations}, the quiver with relations of $\Lambda$ is
\[
\xymatrix@!=3pc{a^b\ar@(ul,dl)_{\alpha^b}\ar@<.7ex>[r]^{c(a,b)} & b^a\ar@<.7ex>[l]^{\beta^a}},~~\beta^a c(a,b) \beta^a,(\alpha^b)^2,c(a,b)\alpha^b.
\]
As $A$ and $B$ are both local nonsimple, DJHP fails for $\Lambda$ by Corollary~\ref{cor:local}. Note that $\Lambda$ is of rank $2$, so a stratification of $\mathcal{D}(\Lambda)$ is just a recollement. The factors of $(R1)$ are $\bar{A}\otimes\bar{B}= k$ and $A\otimes_0 B\cong k\langle x,y\rangle/(x^2,y^2,yx)$, while the factors of $(R2)$ are  $A\otimes\bar{B}= k[x]/x^2$ and $\bar{A}\otimes B= k[y]/y^2$, as also shown in \cite[Example 7.6]{AKLY2}.
\end{example}

The rest of this section is devoted to the proof of Theorem~\ref{thm:quiver-with-relations}.

\begin{proof}[Proof of Theorem~\ref{thm:quiver-with-relations}]
Recall from the proof of Lemma~\ref{l:idempotents in Lambda} (iii) that
\begin{align*}
\opname{rad}(\Lambda)&=\begin{pmatrix}
A\otimes\opname{rad}(B)+\opname{rad}(A)\otimes B & A\otimes \opname{rad}(B)\\
\bar{A}\otimes B & \bar{A}\otimes \opname{rad}(B)
\end{pmatrix}\\
& = \begin{pmatrix} (\opname{rad} A \otimes \bar{B}) \oplus (\bar{A} \otimes \opname{rad} B) \oplus (\opname{rad} A \otimes_0 \opname{rad} B) & A \otimes \opname{rad} B \\ \bar{A} \otimes B & \bar{A} \otimes \opname{rad} B \end{pmatrix}
\end{align*}
and
\[
\bar{\Lambda}=\begin{pmatrix}
\bar{A}\otimes \bar{B} & 0\\
0 & \bar{A}\otimes \bar{B}
\end{pmatrix}.
\]

We write $e_i$ ($i\in Q_0$) and $f_j$ ($j\in P_0$) for the primitive idempotents of $A$ and $B$ respectively, and $e_{i^j}, e_{j^i}$ the primitive idempotents in $k\Gamma$ corresponding to $i^j$ and $j^i$. Then the map
\[
e_{i^j}\mapsto \begin{pmatrix} e_i \otimes f_j & 0 \\ 0 & 0 \end{pmatrix} , \quad e_{j^i}\mapsto\begin{pmatrix} 0 & 0 \\ 0 & e_i \otimes f_j \end{pmatrix},\quad (i\in Q_0,~j\in P_0)
\]
defines an algebra isomorphism from $k\Gamma_0$ to $\bar{\Lambda}$. We identify these two semisimple algebras via this isomorphism.

The square of the radical of $\Lambda$ is
{\footnotesize
\begin{align*}
\opname{rad}^2 \Lambda & =  (\opname{rad} \Lambda)^2 \\
& =  \begin{pmatrix} \opname{rad}^2(A\otimes_0 B) \oplus (A \otimes \opname{rad} B) (\bar{A} \otimes B) & \opname{rad}(A\otimes_0 B) A \otimes \opname{rad} B + (A \otimes \opname{rad} B)(\bar{A} \otimes \opname{rad} B) \\ (\bar{A} \otimes B)\opname{rad}(A \otimes_0 B) + (\bar{A} \otimes \opname{rad} B)(\bar{A} \otimes B) & (\bar{A} \otimes B)(A \otimes \opname{rad} B) + \opname{rad}^2(\bar{A} \otimes B) \end{pmatrix}\\
& =  \begin{pmatrix} (\opname{rad}^2 A \otimes \bar{B}) \oplus (A \otimes \opname{rad} B) & (\opname{rad} A \otimes \opname{rad} B) \oplus (\bar{A} \otimes \opname{rad}^2 B) \\ \bar{A} \otimes \opname{rad} B  &  \bar{A} \otimes \opname{rad} B \end{pmatrix}.
\end{align*}}
Hence
\begin{align*}
\opname{rad} \Lambda / \opname{rad}^2 \Lambda  & =  \begin{pmatrix} (\opname{rad} A/ \opname{rad}^2 A) \otimes \bar{B} & \bar{A}  \otimes( \opname{rad} B / \opname{rad}^2 B) \\ \bar{A} \otimes \bar{B} & 0 \end{pmatrix}.
\end{align*}
It follows that the map
\[
\alpha^j \mapsto \begin{pmatrix} \alpha \otimes f_j & 0 \\ 0 & 0 \end{pmatrix}, \quad c(i,j)\mapsto\begin{pmatrix} 0 & 0 \\ e_i \otimes f_j & 0 \end{pmatrix}, \quad \beta^i \mapsto \begin{pmatrix} 0 & e_i \otimes \beta \\ 0 & 0\end{pmatrix} \quad (i\in Q_0,~j\in P_0)
\]
defines a $\bar{\Lambda}$-$\bar{\Lambda}$-bimodule isomorphism from the $k$-vector space spanned by the arrows of $\Gamma$ to $\opname{rad}(\Lambda)/\opname{rad}^2(\Lambda)$. This isomorphism then extends to a surjective algebra homomorphism
$$\tilde{\Phi}: k\Gamma \twoheadrightarrow \Lambda.$$

For any path $q =\alpha_1\alpha_2 \cdots \alpha_s$ of $Q$ and $j\in P_0$, $q^j:=\alpha_1^j \alpha_2^j \cdots \alpha_s^j$ is a path of $\Gamma$ and we have
$$\tilde\Phi(q^j) = \begin{pmatrix} q \otimes f_j & 0 \\ 0 & 0 \end{pmatrix}.$$
For any path $p =\beta_1 \beta_2 \cdots \beta_s$ of $P$ and $i \in Q_0$ we have
\begin{align*}
\tilde\Phi(c \beta_1^i c \beta_2^i c \cdots c \beta_s^i c ) & =  \begin{pmatrix} 0 & 0 \\ e_i \otimes \beta & 0  \end{pmatrix},\\
\tilde\Phi(c \beta_1^i c \beta_2^i c \cdots c \beta_s^i) & =  \begin{pmatrix} 0 & 0 \\ 0 & e_i \otimes \beta  \end{pmatrix},\\
\tilde\Phi(\beta_1^i c \beta_2^i c \cdots c \beta_s^i ) & =  \begin{pmatrix} 0 &  e_i \otimes \beta\\ 0 & 0  \end{pmatrix}.
\end{align*}
Therefore, the three types of relations in $\mathcal{I}$ belong to the kernel of $\tilde\Phi$: for $r\in\mathcal{R}$ and $j\in P_0$, $$\tilde{\Phi}(r^j) = \begin{pmatrix} r\otimes f_j & 0 \\ 0 & 0 \end{pmatrix} =0;$$
for $i\in Q_0$, $j\in P_0$ and $\alpha \in Q_1$ with $\mathrm{t}(a)=i$,
$$\tilde{\Phi}(c(i,j) \alpha^j) = \begin{pmatrix} 0 & 0 \\ e_i \otimes f_j & 0 \end{pmatrix} \begin{pmatrix} \alpha \otimes f_j & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ \alpha \otimes f_j & 0 \end{pmatrix} =0;$$
for $s = \sum \mu_{j_1, \ldots, j_t} \beta_{j_1} \cdots \beta_{j_t} \in \mathcal{S}$ and $i \in Q_0$,
\begin{align*}
\tilde{\Phi}(s^i) & =  \tilde{\Phi}(\sum \mu_{j_1,\ldots,j_t} \beta_{j_1}^i c\beta_{j_2}^i c \cdots \beta_{j_{t-1}}^i c \beta_{j_t}^i)\\
& =   \sum \mu_{j_1,\ldots,j_t} \begin{pmatrix} 0 & e_i \otimes \beta_{j_1}\beta_{j_2} \cdots \beta_{j_t} \\ 0 & 0 \end{pmatrix} \\
& =  \begin{pmatrix} 0 & e_i\otimes s \\ 0 & 0 \end{pmatrix} = 0.
\end{align*}
Hence $\tilde\Phi$ factors through a surjective algebra homomorphism
$$\Phi: \tilde{\Lambda} = k\Gamma /(\mathcal{I}) \twoheadrightarrow \Lambda.$$
We will show that $\Phi$ is bijective and thus an algebra isomorphism.

By abuse of notation we denote by $e_1$ the sum of all the idempotents $e_{i^j}$'s and $e_2$ the sum of all $e_{j^i}$'s in $\tilde\Lambda$. Then $\Phi(e_1) = e_1$ and $\Phi(e_2) = e_2$. Since $e_1$ and $e_2$ are orthogonal to each other, both algebras $\tilde\Lambda$ and $\Lambda$ decompose as $2\times 2$-matrix algebras and the decomposition is compatible with $\Phi$. We write $\Phi_{ij}$ for the induced map from $e_i \tilde\Lambda e_j$ to $e_i \Lambda e_j$ ($i,j = 1,2$). Then $\Phi = \bigoplus_{i,j=1}^{2} \Phi_{ij}$. The bijectivity of $\Phi$ is equivalent to the bijectivity of all $\Phi_{ij}$.

Bijectivity of $\Phi_{11}: e_1 \tilde\Lambda e_1 \rightarrow e_1 \Lambda e_1 = A \otimes_0 B$: The trivially twisted tensor product $A\otimes _0 B$ as  a $k$-algebra is generated by $e_i\otimes f_j$, $\alpha\otimes f_j$ and $e_i\otimes \beta$ for $i\in Q_0$, $j\in P_0$, $\alpha\in Q_1$ and $\beta\in P_1$, with generating relations $r\otimes f_j$ for $r\in\mathcal{R}$, $e_i\otimes s$ for $s\in \mathcal{S}$ and $(e_i \otimes \beta)(\alpha \otimes f_j)$. The preimages of $\alpha \otimes f_j$ and $e_i\otimes \beta$ in $e_1\tilde\Lambda e_1$ are $\alpha^j$ and $\beta^i c$ respectively. They also satisfy the generating relations of $A \otimes_0 B$. Indeed for $r=\sum \lambda_{i_1,\ldots, i_s} \alpha_{i_1} \cdots \alpha_{i_s} \in \mathcal{R}$, $r^j = \sum \lambda_{i_1,\ldots, i_s} \alpha_{i_1}^j \cdots \alpha_{i_s}^j=0$. For $s = \sum \mu_{j_1, \ldots, j_t} \beta_{j_1} \cdots \beta_{j_t} \in \mathcal{S}$, $ \sum \mu_{j_1,\ldots,j_t} \beta_{j_1}^i c \beta_{j_2}^i c \cdots \beta_{j_t}^i c =s^i c = 0$. For $i \in Q_0$, $j\in P_0$, $\alpha \in Q_1$ and $\beta \in P_1$, we have $\beta^i c \alpha^j =0$. Note that $e_1 \tilde\Lambda e_1$ as a $k$-algebra is generated by $e_{i^j}$, $\alpha^j$ and $\beta^ic$. Therefore $\Phi_{11}$ must be bijective.

Bijectivity of $\Phi_{22}: e_2 \tilde\Lambda e_2 \rightarrow e_2 \Lambda e_2 = \bar{A} \otimes B$. The tensor product $\bar{A} \otimes B$ as a $k$-algebra is generated by $e_i \otimes f_j$ and $e_i \otimes \beta$ for $i\in Q_0$ and $\beta \in P_1$ with generating relations $e_i \otimes s$ for $s\in \mathcal{S}$. The preimages of $e_i \otimes f_j$ and $e_i \otimes \beta$ in $e_2\tilde\Lambda e_2$ are $e_{j^i}$ and $c\beta^i$ respectively. They satisfy the generating relations of $\bar{A} \otimes B$. Indeed for $s = \sum \mu_{j_1, \ldots, j_t} \beta_{j_1} \cdots \beta_{j_t} \in \mathcal{S}$, $c s^i = \mu_{j_1,\ldots,j_t} c \beta_{j_1}^i c \beta_{j_2}^i c \cdots \beta_{j_t}^i = 0$. An a $k$-algebra $e_2\tilde\Lambda e_2$ is generated by $e_{j^i}$ and $c\beta^i$, so $\Phi_{12}$ must be bijective.

Bijectivity of $\Phi_{21}: e_2 \tilde\Lambda e_1 \rightarrow e_2 \Lambda e_1 = \bar{A} \otimes B$. For $i \in Q_0$, $j\in P_0$ and $\beta \in P_1$, the preimages of $e_i \otimes f_j$ and $e_i \otimes \beta$ are $c(i,j)$ and $c\beta^i c$ respectively. Consider the right multiplication map $r_c$ from $e_2\tilde\Lambda  e_2$ to $e_2\tilde\Lambda e_1$, sending an element $x$ to $xc$. This map is surjective according to the shape of $\Gamma$ and the description of ${\mathcal I}$. Hence $\dim_k e_2 \tilde\Lambda e_1 \leq \dim_k e_2 \tilde\Lambda e_2 = \dim_k e_2 \Lambda e_2  = \dim_k e_2 \Lambda e_1$. But the surjectivity of $\Phi_{21}$ implies $\dim_k e_2 \tilde\Lambda e_1 \geq \dim_k e_2  \Lambda e_1$. Hence $e_2 \tilde\Lambda e_1$ and  $e_2 \Lambda e_1$ have the same dimension and $\Phi_{21}$ is bijective.

Finally for the bijectivity of $\Phi_{12}: e_1 \tilde\Lambda e_2 \rightarrow e_1 \Lambda e_2 = A \otimes \opname{rad} B$, consider for each $c(i,j)$ the right multiplication map $r_{c(i,j)}: \tilde\Lambda e_{j^i} \rightarrow \tilde\Lambda e_{i^j}$, and the sum $r_c = \sum r_{c(i,j)}: \tilde\Lambda e_2 \rightarrow \tilde\Lambda e_1$. By Lemma \ref{l:injectivity} below, each $r_{c(i,j)}$ is injective, so is $r_c$. Let $r_c^1: e_1 \tilde\Lambda e_2 \rightarrow e_1 \tilde\Lambda e_1$ be the corresponding summand of $r_c$. It follows that $r_c^1$ is injective. The following commutative diagram
\[
\xymatrix@C=0.7pc{
  e_1\tilde\Lambda e_2 \ar[d]_{\Phi_{12}} \ar[rrrrrr]^{r_c^1} &&&&&& e_1\tilde\Lambda e_1 \ar[d]^{\Phi_{11}} \\
  e_1 \Lambda e_2 \ar@{=}[r] & A \otimes \opname{rad} B \ar[rrrr]^{\mathrm{inc}} &&&& A \otimes_0 B\ar@{=}[r] & e_1\Lambda e_1 }
\]
shows that $\Phi_{12}$ is injective, and hence bijective.
\end{proof}

\begin{lemma} Let $A=kQ/(\mathcal{R})$ where $Q$ is a finite quiver and $\mathcal{R}$ is a set of minimal relations. Suppose $c: \mathrm{s}(c) \rightarrow \mathrm{t}(c)$ is a path in $Q$ such that any path appearing in $\mathcal{R}$ is not of the form $pc$ for some path $p$ in $Q$. Then the right multiplication map $r_c: A e_{\mathrm{t}(c)} \rightarrow A e_{\mathrm{s}(c)},~~x\mapsto xc$, is injective. \label{l:injectivity}
\end{lemma}

\begin{proof}
Put $\tilde{A}=kQ$ and $I=(\mathcal{R})$. We have the following commutative diagram
\[\xymatrix{
  0 \ar[r] & I e_{\mathrm{t}(c)} \ar[d]_{\alpha} \ar[r] & \tilde{A} e_{\mathrm{t}(c)} \ar[d]_{\beta} \ar[r] &  A e_{\mathrm{t}(c)} \ar[d]_{r_c} \ar[r] & 0 \\
  0 \ar[r] & I e_{\mathrm{s}(c)} \ar[r] & \tilde{A} e_{\mathrm{s}(c)} \ar[r] & A e_{\mathrm{s}(c)} \ar[r] & 0   }\]
where the rows are exact and all the vertical maps are right multiplication with $c$. It is clear that $\alpha$ and $\beta$ are injective. Hence $r_c$ is injective if and only if $\Im(\beta) \cap I e_{\mathrm{s}(c)} = \Im(\alpha)$, {\em i.e.}\  $\tilde{A} c \cap I e_{\mathrm{s}(c)} = Ic$ in $\tilde {A}$. It suffices to show `$\subseteq$', {\em i.e.}\  if $x\in\tilde{A}$ satisfies $xc \in I e_{\mathrm{s}(c)}$, then $x\in I$. Write $xc = \sum_{r\in\mathcal{R}} a_1 r a_2$ with $a_1, a_2 \in \tilde{A}$, and write further $a_2 = a_2' + a_2''$ such that all paths appearing in $a_2'$ are of the form $pc$  for some path $p$ in $\tilde{A}$, and none of the paths appearing in $a_2''$ is of such form. Then $xc = \sum_{r\in \mathcal{R}} a_1 r a_2' + \sum_{r\in\mathcal{R}} a_1 r a_2''$ in $\tilde{A}$. The first summand is of the form $x'c$ for some $x' \in I$ and the second summand must be zero. Hence $xc = x'c$, namely $\beta(x) = \beta(x')$. It follows from the injectivity of $\beta$ that $x=x'\in I$.
\end{proof}

redpajama_set_name: RedPajamaArXiv


MD-IT is trying to take the pain out of data entry, by offering an electronic transcribing tool that can take the physical note-taking out of a doctor's day. Here Evan Frankel shows us the dictation-based EMR, which MD-IT uses to help physicians save time and be more efficient. This demo was given during the "The Doctor Will See You Now: Health 2.0 in the MD's Office (Part 2)" panel at the San Francisco Health 2.0 conference in 2011.


I'm falling in love with the Cool Treats set. I tried another card today with this set and the coordinating Tasty Treats Specialty Designer Series Paper, and I think it turned out to be just Sweet!
I liked the color combinations I started using yesterday, so I continued with the Peekaboo Peach and Calypso Coral combination for the two-step stamping technique on this card. The sentiment is from the Cool Treats set.
I used Peekaboo Peach for the basecard as well. I added the cone on a Stitched Shape oval with a scalloped matt in Peekaboo Peach. The little bow uses the 3/8" Peekaboo Peach Rached Ribbon, and the Pumpkin Pie enamels added interest to the card.
Here is a photos of yesterday's card and today's card together. You can use Designer Series paper or stamped images. Both create the cutest cards!
March 26 - We'll make cards using the Cool Treats stamp set and the Birthday Blast stamp sets in our March 26 calls. We'll use them to fill our free Sale-A-Bration Gold and White tins, which we'll decorate during our class. Stampin' Up! depleted its inventory of these tins, which were a free selection with a $50 order. However, I will have some of these tins for sale as part of this class, if you need one, when you register for the class. Watch for registration to open shortly.
Photos of our recent Spring-Easter class and our Dragonfly / Beautiful You class show you what our classes are like. Photos from the class are in a photo album on my Meetup site. While visiting the photo album, sign up for my Meetup site to receive notifications of upcoming classes.
Free selections, like the card tins, are already sold out, but other popular selections are still available. You can still choose the 6X6" Glimmer Paper Assortment Pack (143831). Just three weeks remain to make this one of your free selections with a $50 Order. Don't miss your favorite picks. Place your orders today.
Ending soon are the Sign-up bonuses for becoming a Stampin' Up! demonstrator also. Join Stampin' Up! as either a hobby demonstrator to make new friends and enjoy demonstrator product discounts of 20-25% on product purchases. Or you can turn your hobby into a business where you are your own boss, work your own hours, and do something you love!
Extra bonuses include two free stamp sets. Sign up as a demonstrator for $99 and get $125 in merchandise that YOU select from the current Stampin' Up! catalogs plus an additional 2 free stamp sets--any price!
Pick products that fit your stamping or scrapbooking style. Your kit also includes business supplies and catalogs. And your starter kit ships free!
Order total over $99? Signing up as a demonstrator is a great way to fill your wish list and get future purchases at demo discount prices.
Next year is Stampin' Up! 30th year in business and demonstrators are sure to see many 30-year Celebration specials.


Em is a Partner at Context Matters, Australia's leading Enterprise Agile consultancy. In 2012, she introduced the Scaled Agile Framework (SAFe) to Telstra, launching Australia's first Agile Release Train. This SAFe implementation is now one of the most widely referenced case studies in the world. More recently, in recognition of her experience with and contribution to SAFe, Em was invited to become a Scaled Agile Framework Program Consultant Trainer (SPCT). She is a recognised expert on scaling agile and considered a thought leader on scaling culture and creating agile tribes.
One of the challenges of DevOps is the clash of cultures between dev and ops. Teams of teams, or tribes, can help bust through these barriers.


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Home»Archives»Oman
Iran not talking to 'Great Satan'; Is talking to those who talk to U.S.
By World Tribune on July 28, 2019 Exclusives, Iran, Mideast, Oman, Top Story, World Links, World News, World Tribune
by WorldTribune Staff, July 28, 2019 Iranian officials over the weekend met with representatives of several nations who have the Trump administration's ear. The meetings followed comments on July 24 by Iranian President Hassan Rouhani who hinted that Iran was open to indirect talks with the United States over its nuclear program and U.S.-imposed sanctions […]
'Our diplomatic map has changed': Gulf Arab state allows Israeli flights
By World Tribune on December 10, 2018 Exclusives, Israel, Mideast, Oman, Rest of The Best, Top Stories, World Links, World Tribune
by WorldTribune Staff, December 10, 2018 Oman will allow Israeli passenger jets to use its airspace, Israeli Prime Minister Benjamin Netanyahu announced on Dec. 10. Speaking at an Israeli Foreign Ministry conference, Netanyahu revealed that Israeli passenger jets will now be permitted to fly over the Gulf Arab state on the southeastern end of the […]
Israel launches diplomatic offensive in Gulf Arab states
By World Tribune on October 31, 2018 Exclusives, Israel, Mideast, Oman, Rest of The Best, Top Stories, Top Story, UAE, World Links, World Tribune
by WorldTribune Staff, October 31, 2018 Israel has launched an unprecedented push for closer ties with Gulf Arab states. Communications Minister Ayoub Kara, at an Oct. 30 telecommunications conference in Dubai, called for "peace and security in every state." "Economic and scientific progress is what guarantees a future for the coming generations," Kara said. Israel […]
Oman defends Netanyahu's visit: 'Israel is a fact of life'
By World Tribune on October 28, 2018 Exclusives, Israel, Mideast, Oman, Palestinian Authority, Top Stories, Top Story, World Links, World Tribune
by WorldTribune Staff, October 28, 2018 Arab nations should consider Israel as an equal, Oman's foreign minister said after the nation was highly criticized in the Arab world for hosting Israeli Prime Minister Benjamin Netanyahu. "We should consider giving equal treatment to Israel among the other countries in the Middle East. Jews are part of […]
Who's who: Vote tally on UN resolution slamming U.S. on Jerusalem
By World Tribune on December 21, 2017 Afghanistan, Africa / Europe, Algeria, Americas, Argentina, Australia, Austria, Azerbaijan, Bahrain, Bangladesh, Bangladesh, Belarus, Belgium, Bolivia, Bosnia and Herzegovina, Brazil, Bulgaria, Burma, Cambodia, Canada, Chad, Chile, China, Colombia, Cote d'Ivoire, Croatia, Cuba, Cyprus, Czech Republic, Djibouti, Dominica, Ecuador, Egypt, Equatorial Guinea, Eritrea, Estonia, Ethiopia, EU, Exclusives, Finland, France, Georgia, Germany, Grace Vuoto, Great Britain, Haiti, Hamas, Hizbullah, Hong Kong, Hungary, Iceland, India, Indonesia, Iran, Iraq, Ireland, Israel, Italy, Japan, Jordan, Kazakhstan, Kenya, Kosovo, Kurdistan, Kuwait, Kyrgyz Republic, Laos, Latvia, Lebanon, Libya, Lithuania, Macedonia, Malaysia, Maldives, Mali, Mexico, Mideast, Mongolia, Montenegro, Morocco, NATO, NE Asia, Netherlands, New Zealand, Nicaragua, Niger, Nigeria, North Korea, Norway, Oman, Pakistan, Palestinian Authority, Panama, Peru, Philippines, Poland, Portugal, Puerto Rico, Qatar, Romania, Russia, Saudi Arabia, Serbia, Singapore, Slovakia, Somalia, South Africa, South Korea, South Sudan, Spain, Sri Lanka, Sudan, Sweden, Switzerland, Syria, Taiwan, Terrorism, Thailand, Top Stories, Top Story, Trinidad & Tobago, Tunisia, Turkey, U.S. Politics, Ukraine, United Nations, Uzbekistan, Venezuela, Vietnam, Washington, Yemen, Zimbabwe
Special to WorldTribune , December 21, 2017 Yes: 128 No: 9 Abstain: 35 Y AFGHANISTAN Y DOMINICA Y LITHUANIA SAO TOME AND PRINCIPE Y ALBANIA A DOMINICAN REPUBLIC Y LUXEMBOURG Y SAUDI ARABIA Y ALGERIA Y ECUADOR Y MADAGASCAR Y SENEGAL Y ANDORRA Y EGYPT A MALAWI Y SERBIA Y ANGOLA EL SALVADOR Y MALAYSIA […]
Oman, Iran's closest Gulf ally, signals change by joining Saudi alliance
By World Tribune on December 29, 2016 BB Middle East, Exclusives, Iran, Mideast, Oman, Rest of The Best, Saudi Arabia, Top Stories
Special to WorldTribune.com Radio Free Europe / Radio Liberty Oman, Iran's closest ally in the Persian Gulf region, has signaled that it may be changing course by joining a Saudi-led coalition of Muslim countries fighting terrorism. Arab news outlets said on December 28 that Oman's defense minister sent a letter to Saudi Arabia's Deputy Crown […]
Gulf States jointly condemn 'Iranian interference' in Saudi affairs
By World Tribune on January 10, 2016 Bahrain, BB Middle East, Exclusives, Iran, Kuwait, Mideast, Oman, Qatar, Saudi Arabia, Terrorism, Top Stories, UAE
Special to WorldTribune.com Radio Free Europe / Radio Liberty The Sunni Arab monarchies of the Persian Gulf have expressed support for Saudi Arabia in its diplomatic row with Iran, condemning what they described as Iranian interference in Riyadh's internal affairs. Meanwhile, Saudi Arabia accused Iran of sponsoring terrorism and warned that the kingdom was considering […]

redpajama_set_name: RedPajamaCommonCrawl


What should I complement my Green on Graphite Quinny Zapp Flex Stroller with?
The Quinny Zapp Cocoon & Hammock (Graphite) is the perfect accessory to complement the Quinny Zapp Flex Stroller. We would also recommend the Quinny Zapp Lux Carrycot (Grey on Graphite).


text:
Brands Category: Towns & Villages
Monton Village Community Association
Monton Village has a very active Community Association (MVCA) where residents and traders work together....
Salford More Info
Bolton is a town in Greater Manchester historically a former mill town,...
Bolton More Info
Walkden is a suburban town in the City of Salford,...
Walken More Info
Salford...
Coventry BID
Coventry is a city rich in history and culture....
Coventry More Info
Your Bromley
Your Bromley is the Business Improvement District (BID) for Bromley Town Centre. The BID promotes the town cen...
Bromley More Info
Worsley is a town in the metropolitan borough of the City of Salford...
Worsley More Info
Swinton is a lively town in Salford adjoining the towns of Pendlebury and Clifton...
Swinton More Info
Altrincham is a busy market town nestled between the urban sprawl of Manchester and rural tranquillity of Ches...
Altrincham More Info
Newchurch is a village on the Isle of Wight located between Sandown and Newport in the southeast of the island...
Sandown More Info
Monton is an urban village in Eccles, Salford and a upcoming destination to both live and visit...
Monton More Info
Chale is a village and civil parish on the Isle of Wight of England, in the United Kingdom. It is located thre...
Chale More Info


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Thomas Massie & Eddie Bernice Johnson
Compare the voting records of Thomas Massie and Eddie Bernice Johnson in 2017-18.
Thomas Massie
Represented Kentucky's 4th Congressional District. This is his 4th term in the House.
Eddie Bernice Johnson
Represented Texas's 30th Congressional District. This is her 13th term in the House.
Thomas Massie and Eddie Bernice Johnson are from different parties and disagreed on 74 percent of votes in the 115th Congress (2017-18).
Sept. 7, 2018 — Community Safety and Security Act
July 19, 2018 — Making appropriations for the Department of the Interior, environment, and related agencies for the fiscal year ending September 30, 2019, and for other purposes
June 21, 2018 — Agriculture and Nutrition Act of 2018
June 21, 2018 — Securing America's Future Act of 2018
May 9, 2018 — Citizens' Right to Know Act
Feb. 6, 2018 — Common Sense Nutrition Disclosure Act of 2017
Oct. 26, 2017 — Establishing the congressional budget for the United States Government for fiscal year 2018 and setting forth the appropriate budgetary levels for fiscal years 2019 through 2027
March 1, 2017 — SCRUB Act
Feb. 14, 2017 — Red River Gradient Boundary Survey Act
Sept. 6, 2018 — Providing for consideration of the bill (H.R. 6691) to amend title 18, United States Code, to clarify the definition of "crime of violence", and for other purposes, and providing for consideration of motions to suspend the rules
July 18, 2018 — O'Halleran of Arizona Amendment No. 27
July 18, 2018 — Grijalva of Arizona Amendment No. 25
June 28, 2018 — Clark of Massachusetts Amendment No. 15
June 21, 2018 — Providing for consideration of H.R. 4760, the Securing America's Future Act of 2018
April 27, 2018 — Duncan of Tennessee Part A Amendment No. 114
April 26, 2018 — DeFazio of Oregon Part A Amendment No. 42
April 12, 2018 — Proposing a balanced budget amendment to the Constitution of the United States
Feb. 6, 2018 — Providing for consideration of H.R. 772, the Common Sense Nutrition Disclosure Act of 2017; H.R. 1153, the Mortgage Choice Act of 2017; H.R. 4771, the Small Bank Holding Company Relief Act of 2018; and for other purposes
Dec. 21, 2017 — Providing for consideration of the Senate Amendment to H.R. 1370, the Department of Homeland Security Blue Campaign Authorization Act of 2017; and providing for consideration of H.R. 4667, making further supplemental appropriations for FY 2018, for disaster assistance for Hurricanes and Wildfires in calendar year 2017
Oct. 11, 2017 — Providing for consideration of S. 585, the Dr. Chris Kirkpatrick Whistleblower Protection Act of 2017; providing for proceedings during the period from October 16, 2017, through October 20, 2017; and providing for consideration of motions to suspend the rules
July 14, 2017 — Byrne of Alabama Amendment No. 17
June 26, 2017 — Veterans Expanded Trucking Opportunities Act of 2017


A bridle is a piece of equipment used to direct a horse. The "bridle" includes both the headstall that holds a bit that goes in the mouth of a horse, and the "reins" is a long, narrow strap attached at one end to the bit of a bridle. The rider controls a horse in riding by pulling the rein.
Leatherite offers quality Coated Webbing to the Equestrian Suppliers and Saddleries.


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With only pride at stake, Jamshedpur host Bengaluru in their last fixture of the season
Written by: ISL-Indian Super League
The Men of Steel will be hoping to end their season on a good note.
Bengaluru FC will have a few things to put in place before they face the play-offs in the Hero Indian Super League.
The league toppers were the first team to qualify for the play-offs but the form of Carles Cuadrat's side since the international break has been a concern. They got their confidence back after a 3-0 win over FC Goa in the previous game and will now seek to finish the league stage on a winning note as they face Jamshedpur FC at the JRD Tata Sports Complex, Jamshedpur, on Wednesday.
Bengaluru FC have won only two of their last six matches.
Cesar Ferrando (Jamshedpur FC)
Jamshedpur have had a season full of mixed results and have played nine draws this season, the highest tally for a club in the league stage in ISL. After a shocking 1-4 defeat against Pune, they were held by Chennaiyin in their last match, a result that crushed their hopes of qualifying.
"Like the top four teams, we did not have a striker scoring consistently but most of our goals were shared between our attackers. The lack of finishing is perhaps why we have had so many draws. Of course, missing out on the top-four is a disappointment," said coach Cesar Ferrando.
With play-off spot no longer available, Jamshedpur will look to pocket a morale-boosting triumph.
"It's the last game in front of our home fans and we want to win it for them. I am delighted with the performance of the team and I wish we can finish the season on a high with a win against Bengaluru," said the coach.
Looking Forward: India earn confidence in stalemate against Syria
#INDSYR Ratings: Narender Gahlot & Rahul Bheke stand tallest as India hold Syria
India vs Syria Preview: Can the Blue Tigers eke out a win against the West Asian nation?
Carles Cuadrat (Bengaluru FC)
The Blues have faced Jamshedpur three times and both teams have a win each. Their last meeting ended in a thrilling 2-2 draw in October, with Sergio Cidoncha netting a late equaliser in added time.
"We have a lot of people who have stayed back in Bengaluru. We are coming here with a mix of players. We are going to start four players in the starting XI who have played most of the season with the second team (BFC B).
Cuadrat, who will be hoping to end Bengaluru's winless run away from home after three losses and a draw in their last four trips, said three second-team players will also start on the bench.
"We have the privilege because we are for sure going to finish first in the league. We are not fighting for second or third anymore. We have left some players to save them travel and avoid injuries. It's a chance for us to give the fringe players an opportunity," said Cuadrat.
Gurpreet Singh Sandhu and Amrinder Singh to battle it out for the ISL Golden Glove award
Former England head coach Sven-Goran Eriksson expresses interest for Indian national team job
Indian Player Ratings: Romeo Fernandes powers Delhi to a win against Pune with a goal and assist
The Men of Steel will be without Raju Gaikwad, who reportedly has viral fever. Tim Cahill is expected to make a cameo appearance.
Erik Paartalu is out for the season due to injury.
Probable Lineups
Subrata (GK), Tiri, Augustine, Gurung, Dhananchandra, Arques, Memo, Cidoncha, Morgado, Soosairaj, Passi.
Sandhu (GK), Serran, Juanan, Rino, Bheke, Delgado, Khabra, Udanta, Xisco, Chhetri, Miku.
Jamshedpur has again finished 5th just like last season.
Bengaluru will again top the table. They became the 1st side to secure a top of the table finish for 2 seasons in ISL history.
Bengaluru has scored most goals by any side via through balls (6).
The match will kick-off at 7:30 PM (27th February), at the JRD Tata Sports Complex. It will be broadcasted live via Star Sports, and streamed via Hotstar and Jio TV.
Published: Tue Feb 26, 2019 06:14 PM IST


text:
package uk.co.transputersystems.occam.il;

import java.util.Collections;

public class CompareLessThanOrEqual<TIdentifier> extends ILOp<TIdentifier> {
    public CompareLessThanOrEqual(TIdentifier id, String comment) {
        super(id, "CompareLessThanOrEqual", comment);
    }

    @Override
    public Integer getEncodedSize() {
        //TODO: Set encoded size properly
        return 1;
    }

    @Override
    public Iterable<String> getArgs() {
        return Collections.emptyList();
    }

    @Override
    public <TReturn, TContext> TReturn accept(ILOpVisitor<? extends TReturn, TIdentifier, TContext> visitor, TContext ctx) {
        return visitor.visitCompareLessThanOrEqual(this, ctx);
    }
}

redpajama_set_name: RedPajamaGithub


text:
America Meditating
America Meditating Radio
By Sister Jenna
Getting to Yes with Yourself with Negotiation Expert William Ury
William Ury, co-founder of Harvard's Program on Negotiation, is one of the world's best-known and most influential experts on negotiation. He is the co-author of Getting to Yes, the bestselling negotiation book in the world, and has taught negotiation to tens of thousands of people around the world. He has served as a mediator in conflicts ranging from boardroom battles to labor strikes and from family feuds to civil wars.
He has also served as a consultant to the Crisis Management Center at the White House; co-founded the Climate Parliament which Time magazine described as a "Google for global politics;" founded the Abraham Path Initiative, which seeks to build bridges between cultures and faiths and his work has been widely featured in the media from The New York Times to the Financial Times and from CNN to the BBC. He also has a popular TED talk, "The Walk from No to Yes." His newest book is entitled, "Getting to Yes with Yourself (and Other Worthy Opponents)."
Get the OFF TO WORK CD & Off the Grid Into the Heart CD by Sister Jenna. Like America Meditating on FB & follow us on Twitter.
Check Out Self Help Podcasts at Blog Talk Radio with America Meditating on BlogTalkRadio
No Replies to "America Meditating"


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Euro Truck Simulator 2 is headed to the Baltic states in next expansion
By Samuel Horti 03 March 2018
Glide through Latvia, Lithuania and bits of Russia, including Saint Petersburg.
Euro Truck Simulator 2's next expansion will let you ease around the bends of the Baltic states, developer SCS Software has revealed. The DLC will include the whole of Latvia, Lithuania, and Estonia, parts of southern Finland and a section of western Russia, including Saint Petersburg.
The team did not mention a release date for the expansion, provisionally named Beyond the Baltic Sea, so I wouldn't expect it imminently. They have "quite a lot of ground to cover" and the DLC will end up feeling "very different from the rest of the world we have so far", SCS said in a blog post.
You can see a map of the region that the expansion will add below—it's the purple bit.
Saint Petersburg is a particularly beautiful city, and I'm looking forward to seeing how the team handles its extravagant architecture. A lot of the expansion will be hours and hours of countryside, however, and you can see some screens of that below.
The game is slowly building up its reconstruction of Europe, having added the cities of northern Italy in December.
Samuel Horti
Samuel is a long-time PC Gamer freelancer who loves RPGs and making long lists of games he'll never have time to play. You can find him on the floor, struggling under the weight of his Steam backlog.
contribscontent


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100 NYC Title I Schools to Adopt Coding Curriculum
By Sri Ravipati
Source: Codesters.
A New York-based provider of classroom-friendly solutions for K–12 computer science instruction is bringing computer science to more than 5,000 students enrolled in 100 Title I schools in New York City.
Codesters teaches students Python through its platform that enables students to create interactive graphics, animation, elements and other projects. The company provides a pre-built curriculum with step-by-step lessons that give students instant feedback and assessments. The Codesters curriculum also includes lessons that integrate Common Core mathematics, so that students can "improve math outcomes while learning critical literacy of coding," according to the company's website. Additionally, teachers can utilize Codesters' learning management system to track student progress through a class dashboard.
"The environment provides a rich, differentiated experience for all students, which has allowed all my students to excel," a math teacher from a public school in Bronx, NY told Codesters. "I've watched students who continuously struggle in math come alive and make connections with the programming tools."
In addition to bringing computer science to 100 Title I schools, Codesters will provide content for the national CSforAll Consortium. Both announcements were made yesterday at the White House' Computer Science for All summit.
Further information is available on the Codesters site.
Sri Ravipati is Web producer for THE Journal and Campus Technology. She can be reached at [email protected].


Intrim have created Railing Revit files for their ConnectaRail, Hospital Handrail and Bump Rail Wall Protection to make drawing and specifying these systems easier.
Dry pressed reconstituted sandstone products, used in building design when trying to achieve a coloured cement or sandstone look opposed to a painted finish.
Classic Columns Australia have designed their architectural columns so they are quick and easy to install.
Rust Oleum Tub and Tile paint from Zinsser Asia Pacific is a product designed to rejuvenate bath tubs, sinks and tiled surfaces.
The Truebeam range from Hyne Timber is particularly suited for roof spans over large spaces and is treated for weather-exposed structural applications, where both strength and appearance are important design requirements.
Rust Oleum plastic paint from Zinsser Asia Pacific is a direct to plastic paint, which does not require primer. Rust Oleum plastic paint is suitable for revitalising indoor and outdoor plastics including outdoor furniture.
Gable vents are often seen as a cost effective and simple way to improve the exterior appearance of a home. They also play an important function in home ventilation and energy efficiency.


Posted on July 16, 2014, in Alchemical Transfiguration, Alchemy, Ancient Mysteries, Completion, Conscious Creation, Consciousness, Grace, Harmony, Heaven on Earth, Karmic Traces (memory), Living in Your Heart, Mastering Duality, Present Moment Awareness, Pure Being, Reality Shifts, Releasing Fear, Transcending Duality, Unconditional Love, Victoria Elohim, Victory, Your Reality, Your Souls Plan and tagged Alchemical Transfiguration, awakening, Central Sun, Divine Love, Eternal Love, Living in your Heart, Pleiadian Delegate, Releasing Fears, The Queen of Light, unconditional Love, Victoria Elohim, Victory. Bookmark the permalink.	3 Comments.
Thank you dear Laura for your great message, I understand that matter, but I know what it is and how must it be now. <3 I <3 Love <3 YOU <3 Love <3.


The purpose of the second visit was to see my relatives here. Visiting and hanging out with family is still the best part of every trip to Macau, hopefully every visit will have the same effect. Macau is an extremely unique place since it's culture, architecture, and language has Portuguese, Mandarin, and Cantonese influences. Hence, the visual, audio, and to the extent of a "spiritual" effect that Macau has on its visitors are one of a kind.
"The Ruins of St. Paul's refer to the façade of what was originally the Church of Mater Dei built in 1602-1640 and the ruins of St. Paul's College, which stood adjacent to the Church, both destroyed by fire in 1835. As a whole, the old Church of Mater Dei, St. Paul's College and Mount Fortress were all Jesuit constructions and formed what can be perceived as the Macao's "acropolis".
"The square is so named as since Ming Dynasty (1368-1644) it has been in front of the Leal Senado Building. During the era of Portuguese control this is where the local authorities used to review the troops on their inaugurations. In 1940, there was a statue of a Portuguese soldier named Mesquita in its centre. He was responsible for the deaths of many Chinese soldiers during the hostilities with the Qing Dynasty (1644-1911). Understandably, the statue has been destroyed by Chinese people and replaced by a fountain; this is why the square is also called "the Fountain".
Penha Hill is the third highest among the hills in Macau. Penha Hill is 62.7 meters above sea plane. Known as the Bishop´s Hill in the annals of colonial history, this leader amongst the popular destinations in Macau is identical with the city's colonial clout and authority. Forts, buildings, residences of the Penha hill are wonderful sights.
Macau's tall ranked officials, socialites and the famous ones reside at Bishop´s Hill. This makes Penha Hill become one of must-see sights in Macau.
There is a Church named The Penha Church on crest of the hill dates back to 1622. It was built by a group of Augustinian followers and dedicated to Our Lady of Penha. Existence of this monument in fact named Penha Hill as Bishop's Hill.
Whenever I get the chance to come back to Hong Kong, Macau will be on my list also.


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<!doctype html>

  <meta charset="utf-8">
  <meta name="viewport" content="width=device-width, initial-scale=1.0">
  <title>Gallery Image 1</title>
  <link rel="stylesheet" href="css/foundation.css">
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text:
package org.fmi.nlpir.annotators;

import java.util.Collection;
import java.util.List;

import edu.stanford.nlp.ie.machinereading.structure.EntityMention;
import edu.stanford.nlp.ie.machinereading.structure.MachineReadingAnnotations.RelationMentionsAnnotation;
import edu.stanford.nlp.ie.machinereading.structure.RelationMention;
import edu.stanford.nlp.ling.CoreAnnotations;
import edu.stanford.nlp.pipeline.Annotation;
import edu.stanford.nlp.util.CoreMap;

public class LocatedInAnnotator extends AbstractAnotator {

	@Override
	public String getAnnotated(Annotation annotation) {
		StringBuilder builder = new StringBuilder();
		for (CoreMap sentence : annotation.get(CoreAnnotations.SentencesAnnotation.class)) {
			Collection<RelationMention> triples = sentence.get(RelationMentionsAnnotation.class);

			for (RelationMention relationMention : triples) {
				List<EntityMention> entityMentionArgs = relationMention.getEntityMentionArgs();
				if (entityMentionArgs.get(0).getType().equalsIgnoreCase("LOCATION")
						&& entityMentionArgs.get(1).getType().equalsIgnoreCase("LOCATION") &&
						relationMention.getTypeProbabilities().getCount("Located_In") > 0.001) {
					builder.append(relationMention);
				}
			}
		}
		return builder.toString();
	}

}


We like to boast that we have the best game day experience in the nation in the Southeastern Conference, but the environment at K-State is as good as I have ever seen. In the days since our trip to Manhattan, many of our fans have told me it was the best they have ever been treated on a road trip. Considering how many road games some of our fans attend, that is saying a lot.
To reach Kellis Robinett at krobinett@kcstar.com. Follow him on Twitter: @kellisrobinett.


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Arsenal's Carabao cup semi-final tie with Liverpool postponed due to Covid outbreak
The English Football League (EFL) has postponed the Carabao Semi final tie between Arsenal and Liverpool that was to hold Thursday, 6th January.
This is coming after the authorities at the EFL accepted a request by Liverpool who are reportedly plagued with a of Covid outbreak.
The outbreak has left most of the team's coaching and playing staff with positive results, which has led to the shutdown of the club's training facility.
Consequently, the League Cup semi final tie will now be played on 13th January, first at Anfield, instead of the Emirates as it was initially booked. While the second leg will take place at Emirates Stadium on 20th January.
The 6 times UEFA Champions League winners have recently been suffering both on and off the pitch. While they have only picked up two points from a possible 9, they have lost the likes of Alisson and Firmino to Covid. It is so serious that they had to play Chelsea last time out without Manager, Jurgen Klopp who also tested positive for the virus.
Pepijn Lijnders, Liverpool's assistant manager who took charge of the side's 2:2 draw at Stamford Bridge has also been sidelined with the virus.
Categories: SPORTS NEWS
Tagged as: arsenal vs liverpool, Carabao cup semifinal, semi final
AFCON: Aim for Gold, Buhari urges Super Eagles
Selection Of Och'Idoma V: Show Restraint – Sen. Moro Calls For Calm