text:
Internship format allows companies to fill digital roles easier. Source: Shutterstock
Tech grads in Singapore have no trouble finding a job — Survey
By Tech Wire Asia | 1 March, 2019
IT and digital technology students who graduated in Singapore last year reported an extremely high full-time employment rate of 91.7 percent.
The category of graduates also commanded the highest median salary at SG$4,100 (US$3,000). The figure is significantly higher than the overall median of SG$3,500 (US$2,500) according to the survey.
SINGAPORE NEEDS MORE FINTECH TALENT SAYS MICHAEL PAGE
Tech Wire Asia | 21 February, 2019
The 2018 Joint Graduate Employment Survey covered 11,200 fresh graduates from four universities – Nanyang Technological University (NTU), the National University of Singapore (NUS), Singapore Management University (SMU), and Singapore University of Social Sciences (SUSS).
Universities credited internships and work attachments programmes as the principal reason for IT graduates securing employment quickly after graduation.
Internships are also a way for companies to recruit for short-term and pilot projects, and also allows companies to sift through profiles and work with students to really find the right fit for the long-term.
HERE'S WHY DEPLOYING AI MIGHT NOT LEAD TO JOB CUTS
James Sivalingam | 17 January, 2019
Further, the interim employment in an authentic work environment readies students better for the workforce.
NTU Associate Provost for undergraduate education Professor Tan Ooi Kiang claims that these experiences give students the upper hand.
"Employers expect more from job candidates, and internships can open the door to multiple job offers well before graduation," Tan said.
Tan further commented that students that complete the short-term programme are far more driven and prepared for the global workplace.
ONLINE HIRING GROWS RAPIDLY IN ASIA — NEW REPORT
Soumik Roy | 31 December, 2018
Internships or work attachments for NTU students range from 10 to 30 weeks either locally or overseas. More than one-third of Business & Computing graduates were hired by their internship companies.
SUSS stated in a press release that work attachments are compulsory for full-time undergraduates. The university pointed out that work attachments make up at least 24 weeks of the students' courses.
Some digital projects are only a matter of weeks or months. So, the format of such programme allows companies to fill the roles easily.
This will help companies ensure sufficient headcount for short-term digital projects as well as IT students in exploring different industries.
SINGAPORE LEADS APAC AUTONOMOUS VEHICLES ADOPTION
Michelle Chan | 18 February, 2019
SMU graduate Tay Jing Ying said that her internship experience helped her gain exposure and expand opportunities.
"Apart from the technical knowledge gained, the curriculum also built my analytical thinking, resilience and interpersonal skills, which are crucial and relevant in the real world," said Tay.
Moreover, companies could use the duration to assess the suitability of each candidate before converting them into a permanent headcount.
She is currently working as a technology analyst at Bank of America Merrill Lynch, one of the many companies she interned at.
NUS graduate Celeste Ang interned at artificial intelligence start-up AppZen in the Silicon Valley. She joined the programme to help "develop [her] soft skills and an entrepreneurial mindset."
Ang is currently working with Unilever as a data scientist.
Be it big data or quantum computing, the demand for digitization is soaring across all industries and markets. There are infinite possibilities for how technologies will shape tomorrow's businesses.
Ultimately, technological breakthroughs rely on what's being accomplished by talents in their respective fields.
Companies that open doors for more student internships and work attachments are not merely adding value to their own interests. Subsequently, the effort will drive the digital talent pool to be more innovative in keeping us all ahead with technology.


text:
Q: How to disable xscreensaver lock on suspend? I have recently installed Ubuntu 12.04 LTS and have installed xscreensaver.
As I am the only user on this computer and I am only using the computer for music, movies, and some web browsing  I have no need for the extra security of a lock screen.
Under user accounts, automatic login is turned on.
Under brightness and lock, the lock is turned off and requires my password when waking from suspend is unchecked.
Under xscreensaver preferences, the lock screen box is unchecked.
Every time the computer goes into suspend I get an xscreensaver lock screen upon waking it. Is there a setting I'm missing or something?
From what I understand removing xscreensaver will fix this, but I would consider that a last resort if I can not resolve this issue as I like to use my photos as a screensaver.
I have just begun using Linux and don't know what I'm doing so your patience and any help would be much appreciated.

A: Take a look at this AskUbuntu Question.
Basically, you should also disable screen locking in power manager settings
If you are not using power manager, you can try to comment out line LOCK_SCREEN=TRUE in /etc/default/acpi-support

A: I suggest that you have -- only one GUI. Which GUI generates its different configuration file and sometimes messes up with programs including screensaver settings.
I resolved the same problem as yours by  installing only one GUI -- cinnamon as an example, or you can try another interface and I highly recommend using Tweak Tool/Power Management.

redpajama_set_name: RedPajamaStackExchange


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2-D static airfoil simulated with the [turbinesFoam](https://github.com/turbinesFoam/turbinesFoam) actuator line model.

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redpajama_set_name: RedPajamaGithub


text:
https://www.myjoyonline.com/completion-of-tema-mpakadan-railway-project-to-open-businesses-around-volta-lake-akufo-addo/-------https://www.myjoyonline.com/completion-of-tema-mpakadan-railway-project-to-open-businesses-around-volta-lake-akufo-addo/
Completion of Tema-Mpakadan Railway Project to open businesses around Volta Lake – Akufo-Addo
Source: Ebenezer Sabutey
President Akufo-Addo has hinted that the quest to open up Ghana's Volta Lake to connect businesses along the country's border communities will become a reality soon.
This is coming after he inspected work on the railway bridge across the Volta River as part of the Tema-Mpakadan Railway project.
He believes this will be a major boost for both domestic tourism and trade along the eastern corridor route.
The bridge project which is part of the ongoing construction work on the Tema to Mpakadan Railway line is more than 98% completed.
President Akufo-Addo expressed confidence that carting of goods along the route will be easier.
The president earlier inspected the completed Afienya modern railway station which will be a starting point for the coaches.
After commissioning the road, President Akufo-Addo and the team inspected the rail bridge on the Volta Lake which is about 300 metres, and the first of its kind in the West African Sub region.
He also noted that the railway bridge over the Volta Lake will open up the area for business and international trade.
"When successfully completed, the project will link up the Tema port to the lake transport network especially the North-South movement over a distance of some 400 kilometres and open up the lake for the transportation of not only fuel and cement but containerized cargo and agricultural produce from the northern enclave and Afram Plains to the port of Tema and other destinations", he mentioned.
The Minister for Railways Development, John Peter Amewu, also hinted that the construction of a rail bridge over the Volta Lake which connects the Eastern region to the Volta region will boost local economic growth and position the country as having one of the best infrastructure projects in the West African sub region.
The over 97 kilometres line from Tema to Mpakadan is however on schedule and will be handed over to government by March 2023.
John Peter Amewu
President Akufo-Addo
Tema-Mpakadan railway project
Volta lake
I don't have confidence Akufo-Addo is committed to fighting corruption – Sulemana Braimah
RTI Commission says cost of Akufo-Addo's private jet rentals can't be disclosed
Police investigate drowning of eight children on Volta Lake
8 pupils drown in Volta Lake
Duncan Williams, Eastwood Anaba call for suspension of National Cathedral construction until independent audit
Volta Region to host 66th Independence Day celebration
GES recalls Chiana SHS students to school
Hanke pegs Ghana's inflation at 91%; describes inflation as "disaster"
Parents of embattled Chiana SHS girls express gratitude to Akufo-Addo
Those seeking flagbearership position can help NPP 'break the 8' – Majority Leader


text:
\section{Introduction}

Cluster algebras were introduced and studied by Fomin-Zelevinsky and
Berenstein-Fomin-Zelevinsky in a series of articles
\cite{FominZelevinsky02} \cite{FominZelevinsky03}
\cite{BerensteinFominZelevinsky05} \cite{FominZelevinsky07}. It was
the discovery of Marsh-Reineke-Zelevinsky
\cite{MarshReinekeZelevinsky03} that they are closely connected to
quiver representations.  This link is similar to
the one between quantum groups and quiver representations discovered
by Ringel \cite{Ringel90} and investigated by Kashiwara, Lusztig,
Nakajima and many
others.  The link between cluster algebras and quiver representations
becomes especially beautiful if, instead of categories of quiver
representations, one considers certain triangulated categories deduced
from them: the so-called {\em cluster categories}.
These were introduced in
\cite{BuanMarshReinekeReitenTodorov04} and, for Dynkin quivers
of type $A_n$, in \cite{CalderoChapotonSchiffler04}.
If $k$ is a field and $Q$
a quiver without oriented cycles, the associated cluster category ${\mathcal C}_Q$ is
the `largest' $2$-Calabi-Yau category under the derived category of
representations of $Q$ over $k$. It was shown
\cite{BuanMarshReiten06}
\cite{BuanMarshReitenTodorov05}
\cite{CalderoChapoton06}
\cite{CalderoKeller05a}
\cite{CalderoKeller05b}
that this category
fully determines the combinatorics of the cluster algebra
associated with $Q$ and carries considerably more information.
This was used to prove significant new results on cluster algebras,
{\em cf.}\  {\em e.g.}\
\cite{BuanReiten06}
\cite{CalderoReineke06}.
We refer to
\cite{FominZelevinsky03a} \cite{Zelevinsky04} for more background on
cluster algebras and to
\cite{AssemBruestleSchifflerTodorov05}
\cite{AssemBruestleSchiffler06}
\cite{BuanMarshReiten04b}
\cite{BuanReiten05b}
\cite{BuanReitenSeven06}
\cite{BaurMarsh06}
\cite{GeissLeclercSchroeer06}
\cite{GeissLeclercSchroeer05c}
\cite{Iyama05a}
\cite{IyamaReiten06}
\cite{Ringel07}
\cite{Thomas06}
\cite{Zhu06}
for recent developments in the study of their links with
representations of quivers and finite-dimensional algebras.

The question arises as to whether, for a given quiver $Q$ without
oriented cycles, the cluster category is the `unique model' of the
associated cluster algebra. In other words, if we view the
cluster algebra as a combinatorial invariant associated with
the cluster category, is the category determined by this invariant ?

In this paper, we show that surprisingly, this question has a positive
answer. Namely, we prove that if $k$ is an algebraically closed
field and ${\mathcal C}$ an
algebraic $2$-Calabi-Yau category containing a cluster tilting
object $T$ whose endomorphism algebra has a quiver $Q$ without
oriented cycles, then ${\mathcal C}$ is triangle equivalent to the cluster
category ${\mathcal C}_Q$. Notice that this result is `of Morita type',
but much stronger than typical Morita theorems,
since we only need to know the {\em quiver} of the endomorphism algebra,
not the algebra itself.

We give several applications: First, we show on an example that
cluster categories naturally appear as stable categories of
Cohen-Macaulay modules over certain singularities. This yields an
alternative proof of Iyama-Yoshino's \cite{IyamaYoshino06}
classification of rigid Cohen-Macaulay modules
over a certain isolated singularity.
More examples may be obtained from
\cite{IyamaReiten06} and \cite{IyamaYoshino06}, {\em cf.}\
\cite{BurbanIyamaKellerReiten07}.

Secondly, we show that the quivers associated in
\cite{GeissLeclercSchroeer05b} with representation-infinite
finite-dimensional preprojective algebras are not mutation-equivalent
to quivers without oriented cycles. This last result was obtained
independently by C.~Geiss \cite{Geiss06}. It has been used in
\cite{DerksenWeymanZelevinsky07}, Example~8.7, to show that the class
of rigid quivers with potential is strictly greater than the class of
quivers with potential mutation-equivalent to acyclic ones.  An
application to the realization of cluster categories as stable
categories of Frobenius categories with finite-dimensional morphism
spaces is given in \cite{BuanIyamaReitenScott07} and
\cite{GeissLeclercSchroeer06a}. In \cite{HolmJoergensen06} and
\cite{HolmJoergensen06a}, the authors use our results to determine
which stable categories of representation-finite selfinjective
algebras of type $A$ and $D$ are higher cluster categories.  More
generally, in \cite{Amiot07}, the author obtains
a classification of `most' triangulated categories with finitely many
indecomposables by methods similar to ours.

The main difficulty in the proof is the construction of a triangle
functor between the cluster category and the given Calabi-Yau
category.  Our construction is based on the description
\cite{Keller05} of the cluster category as a stable derived category of a
certain differential graded category. This approach leads to
interesting connections between Calabi-Yau categories of dimensions
$2$ and $3$, which have been further investigated in
\cite{Tabuada07}: It turns out that each algebraic
Calabi-Yau category of dimension $2$ containing a cluster-tilting
subcategory is equivalent to a stable derived category
of a differential graded category whose perfect derived
category is Calabi-Yau of dimension~$3$.

A more direct approach, based on the universal
property of the cluster category \cite{Keller05}, has been discovered
by Michel Van den Bergh, who has kindly accepted to include his proof
as an appendix to this article.

It turns out that the main theorem and its proofs can be generalized
almost without effort to Calabi-Yau categories of any dimension $d\geq
2$. However, one has to take into account that in the $d$-cluster
category, the selfextensions of the canonical cluster tilting object
vanish in degrees $-d+2, \ldots, -1$. This condition therefore has to
be added to the hypotheses of the generalized main theorem.

\subsection*{Acknowledgments} This research started during a
stay of the first-named author at the Norwegian University of Science
and Technology (NTNU). He thanks the second-named author
and the members of her group at the NTNU for their warm hospitality.
Both authors thank Michel Van den Bergh for pointing out gaps
and detours in the original proof and for agreeing to include his own
proof as an appendix to this article.
They are grateful to Carl Fredrik Berg for pointing
out reference \cite{Benson98}.

\section{The main theorem and two applications}

\subsection{Statement} \label{subsection:statement}
Let $k$ be a perfect field. Let ${\mathcal E}$ be a $k$-linear Frobenius
category with split idempotents. Suppose that its stable category
${\mathcal C}=\ul{{\mathcal E}}$ has finite-dimensional $\operatorname {Hom}$-spaces and is
Calabi-Yau of CY-dimension $2$, {\em i.e.}\  we have bifunctorial isomorphisms
\[
D{\mathcal C}(X,Y) \stackrel{_\sim}{\rightarrow} {\mathcal C}(Y,S^2X) \: , \; X,Y\in{\mathcal C}\: , \;
\]
where $D$ is the duality functor $\operatorname {Hom}_k(?,k)$ and $S$ the
suspension of ${\mathcal C}$.

Let ${\mathcal T}\subset{\mathcal C}$ be a cluster tilting subcategory. Recall
from \cite{KellerReiten07} that this means that ${\mathcal T}$ is a
$k$-linear subcategory which is functorially finite in ${\mathcal C}$
and such that an object $X$ of ${\mathcal C}$ belongs to ${\mathcal T}$ iff
we have $\operatorname {Ext}^1(T,X)=0$ for all objects $T$ of ${\mathcal T}$.
As shown in \cite{KellerReiten07}, the category $\operatorname{mod}{\mathcal T}$ of
finitely presented ${\mathcal T}$-modules is then abelian. If it is hereditary,
the cluster category ${\mathcal C}_{\mathcal T}$, as defined in
\cite{BuanMarshReinekeReitenTodorov04},
is the orbit category of the bounded derived category $\operatorname {cd}^b(\operatorname{mod} {\mathcal T})$
under the action of the autoequivalence $S^2 \circ \Sigma^{-1}$
where $S$ is the suspension and $\Sigma$ the Serre functor
of $\operatorname {cd}^b(\operatorname{mod} {\mathcal T})$.

\begin{theorem}
If $\operatorname{mod}{\mathcal T}$ is hereditary,
then ${\mathcal C}$ is triangle equivalent to the cluster category
${\mathcal C}_{\mathcal T}$.
\end{theorem}

We will prove the theorem in section~\ref{section:proofs} below.
Now assume that $k$ is algebraically closed.
Let ${\mathcal R}$ be the radical of ${\mathcal T}$, {\em i.e.}\  the ideal such that for two
indecomposables $X,Y$, the space ${\mathcal R}(X,Y)$ is formed by the non
isomorphisms from $X$ to $Y$. Let $Q$ be the quiver of ${\mathcal T}$: Its
vertices are the isomorphism classes of indecomposables of ${\mathcal T}$
and the number of arrows from the class of an indecomposable $X$
to an indecomposable $Y$ is the dimension of the vector space
${\mathcal R}(X,Y)/{\mathcal R}^2(X,Y)$.

\begin{corollary}
If $k$ is algebraically closed and
for each vertex $x$ of $Q$, only finitely many paths start
in $x$ and only finitely many paths end in $x$,
then ${\mathcal C}$ is triangle equivalent to the cluster category
${\mathcal C}_Q$.
\end{corollary}

Note that under the assumptions of the corollary, the projective (right)
$kQ$-module $kQ(?,x)$ and the injective $kQ$-module $D kQ(x,?)$
are of finite total dimension and that the category $\operatorname{mod} kQ$
of finitely presented $kQ$-modules coincides with the category
of modules of finite total dimension. We will prove the
corollary in section~\ref{section:proofs} below.

\subsection{Application: Cohen-Macaulay modules}
\label{subsection:CM-modules}
Suppose that $k$
is algebraically closed of characteristic $0$. Let the cyclic group $G=\mathbb{Z}/3\mathbb{Z}$
act on the power series ring $S=k[[X,Y,Z]]$ such that a generator
of $G$ multiplies each indeterminate by the same primitive
third root of unity. Then the fixed point ring $R=S^G$ is a Gorenstein ring,
{\em cf.}\  {\em e.g.}\  \cite{Watanabe74}, and an
isolated singularity of dimension $3$, {\em cf.}\  {\em e.g.}\  Corollary~8.2 of
\cite{IyamaYoshino06}.
The category $\opname{CM}(R)$ of maximal Cohen-Macaulay modules is
an exact Frobenius category. By Auslander's results
\cite{Auslander76}, {\em cf.}\  Lemma 3.10 of \cite{Yoshino90}, its stable
category ${\mathcal C}=\ul{\opname{CM}}(R)$
is $2$-Calabi Yau. By work of Iyama \cite{Iyama05}, the module
$T=S$ is a cluster-tilting object in ${\mathcal C}$. The endomorphism
ring of $T$ over $R$ is the skew group ring $S*G$. Under the
action of $G$, the module $T$ decomposes into three indecomposable direct
factors $T=T_1\oplus T_2\oplus S^G$ and we see that its
endomorphism ring $S*G$ is isomorphic to
the completed path algebra of the quiver
\[
\xymatrix{  & S^G \ar[rd] \ar@<1ex>[rd] \ar@<-1ex>[rd] &  \\
          T_1 \ar[ru] \ar@<1ex>[ru] \ar@<-1ex>[ru]  &
        & T_2 \ar[ll] \ar@<1ex>[ll] \ar@<2ex>[ll] }
\]
subject to all the `commutativity relations' obtained by labelling
the three arrows between any consecutive vertices by $X$, $Y$ and $Z$.
The stable endomorphism ring of $T$ is thus isomorphic to the
path algebra of the generalized Kronecker quiver
\[
Q : \xymatrix{ 1 \ar[r] \ar@<1ex>[r] \ar@<-1ex>[r] & 2}.
\]
The theorem now shows that the stable category of Cohen-Macaulay
modules $\ul{\opname{CM}}(R)$ is triangle equivalent
to the cluster category ${\mathcal C}_Q$.

    As a further application, we give an alternative proof of a
theorem from \cite{IyamaYoshino06}, stating that the indecomposable
nonprojective rigid modules in $\opname{CM}(R)$ are exactly the
modules of the form $\Omega^i(T_1)$ and $\Omega^i(T_2)$ for $i\in\mathbb{Z}$.
For this, note that the indecomposable rigid objects in ${\mathcal C}_Q$
are exactly the images of the indeomposable rigid $kQ$-modules and
the $SP$, for $P$ indecomposable projective $kQ$-module
\cite{BuanMarshReinekeReitenTodorov04}. So they
correspond to the vertices of the
component of the AR-quiver of ${\mathcal C}_Q$ containing the
indecomposable projective $kQ$-modules. The corresponding component of
the AR-quiver of $\ul{\opname{CM}}(R)$
is the one containing $T_1$ and $T_2$.
Hence the indecomposable rigid objects in $\ul{\opname{CM}}(R)$ are all
$\tau$-shifts of these.  Finally, we use that $\tau=\Omega^{-1}$ in this case
\cite{Auslander76}.

Note that this application does not need the full force of the
main theorem. For we only use that the AR-quivers of ${\mathcal C}_Q$ and
$\ul{\opname{CM}}(R)$ are isomorphic, with the component of the projective
$kQ$-modules for ${\mathcal C}_Q$ corresponding to the component of $T_1$ and $T_2$
for $\ul{\opname{CM}}(R)$, and it is easy to see that this follows from
proposition 2.1 c) and lemma 3.5 of \cite{KellerReiten07},
{\em cf.}\  also \cite{BuanMarshReiten04}.

\subsection{Application: Non acyclicity} Let $k$ be an algebraically closed
field and $\Lambda$ the preprojective algebra of a simply laced Dynkin
diagram $\Delta$.  Then $\Lambda$ is a finite-dimensional
selfinjective algebra and the stable category ${\mathcal C}$ of
finite-dimensional $\Lambda$-modules is $2$-Calabi-Yau,
{\em cf.}\  \cite{CrawleyBoevey00}, and admits a
canonical cluster-tilting subcategory ${\mathcal T}'$ with finitely many
indecomposables, {\em cf.}\  \cite{GeissLeclercSchroeer05b}. Let $Q'$ be its
quiver.  For example, by [loc. cit.], the quivers $Q'$ corresponding
to $\Delta=A_5$ and  $\Delta=D_4$ are respectively
\[
\begin{xy} 0;<0.7pt,0pt>:<0pt,-0.63pt>::
(75,0) *+{\circ} ="0",
(50,50) *+{\circ} ="1",
(100,50) *+{\circ} ="2",
(25,100) *+{\circ} ="3",
(75,100) *+{\circ} ="4",
(125,100) *+{\circ} ="5",
(0,150) *+{\circ} ="6",
(50,150) *+{\circ} ="7",
(100,150) *+{\circ} ="8",
(150,150) *+{\circ} ="9",
"1", {\ar"0"},
"0", {\ar"2"},
"2", {\ar"1"},
"3", {\ar"1"},
"1", {\ar"4"},
"4", {\ar"2"},
"2", {\ar"5"},
"4", {\ar"3"},
"6", {\ar"3"},
"3", {\ar"7"},
"5", {\ar"4"},
"7", {\ar"4"},
"4", {\ar"8"},
"8", {\ar"5"},
"5", {\ar"9"},
"7", {\ar"6"},
"8", {\ar"7"},
"9", {\ar"8"},
\end{xy}
\quad\quad
\begin{xy} 0;<0.7pt,0pt>:<0pt,-0.63pt>::
(0,75) *+{\circ} ="0",
(50,0) *+{\circ} ="1",
(50,100) *+{\circ} ="2",
(50,150) *+{\circ} ="3",
(100,75) *+{\circ} ="4",
(150,0) *+{\circ} ="5",
(150,100) *+{\circ} ="6",
(150,150) *+{\circ} ="7",
"0", {\ar"1"},
"0", {\ar"2"},
"0", {\ar"3"},
"4", {\ar"0"},
"1", {\ar"4"},
"5", {\ar"1"},
"2", {\ar"4"},
"6", {\ar"2"},
"3", {\ar"4"},
"7", {\ar"3"},
"4", {\ar"5"},
"4", {\ar"6"},
"4", {\ar"7"},
\end{xy}
\]
\bigskip
Part b) of the following proposition was obtained independently
by C.~Geiss \cite{Geiss06}.
\begin{proposition} Suppose that $\Lambda$ is representation-infinite.
\begin{itemize}
\item[a)] The stable category ${\mathcal C}=\ul{\operatorname{mod}} \Lambda$ is not equivalent to the cluster
category ${\mathcal C}_Q$ of a finite quiver $Q$ without oriented cycles.
\item[b)] The quiver $Q'$ of the canonical cluster-tilting subcategory
of \cite{GeissLeclercSchroeer05b} is not mutation-equivalent to a
quiver $Q$ without oriented cycles.
\end{itemize}
\end{proposition}

In particular, it follows that the two above quivers are not
mutation-equivalent to quivers without oriented cycles. In the proof
of the proposition, we use the main theorem. Let us stress that,
as at the end of \ref{subsection:CM-modules}, we do
not need its full force but only use the isomorphism of AR-quivers.
This is the variant of the proof also given by C.~Geiss \cite{Geiss06}.

\begin{proof} a) Recall first that the AR-translation $\tau$ is isomorphic
to the suspension in any $2$-Calabi-Yau category, so that it is
preserved under triangle equivalences.  We know from
\cite{AuslanderReiten96} that the AR-translation $\tau$ of ${\mathcal C}$ is
periodic of period dividing $6$.  In particular, we have
$\tau^6(X)\stackrel{_\sim}{\rightarrow} X$ for each indecomposable $X$ of ${\mathcal C}$. But in
${\mathcal C}_Q$, for each indecomposable $X$ which is the image of
a preprojective $kQ$-module, the iterated
translates $\tau^{-n}(X)$, $n\geq 0$, are all pairwise non isomorphic
since $Q$ is representation-infinite.

b) Suppose that $Q'$ is mutation-equivalent to a quiver $Q$.
By one of the main results of \cite{GeissLeclercSchroeer07}, it follows that
${\mathcal C}$ contains a cluster-tilting subcategory ${\mathcal T}$ whose quiver is $Q$.
If $Q$ does not have oriented cycles, it follows from the main theorem
that ${\mathcal C}$ is triangle equivalent to ${\mathcal C}_Q$ in contradiction to a).
\end{proof}

\section{Proofs} \label{section:proofs}

\subsection{Proof of the corollary}
First recall from \cite{KellerReiten07} that the
category $\operatorname{mod}{\mathcal T}$ of finitely presented ${\mathcal T}$-modules is abelian
and Gorenstein of dimension at most $1$. It follows from our
hypothesis that each object of $\operatorname{mod}{\mathcal T}$ has a finite composition
series all of whose subquotients are simple modules
\[
S_M = {\mathcal T}(?,M)/{\mathcal R}(?,M)
\]
associated with indecomposables $M$ of ${\mathcal T}$ and that each of these
simple modules is of finite projective dimension. Thus each object of
$\operatorname{mod}{\mathcal T}$ is of finite projective dimension so that $\operatorname{mod}{\mathcal T}$ has to
be hereditary. Since $k$ is algebraically closed, it follows that
$\operatorname{mod}{\mathcal T}$ is equivalent to $\operatorname{mod} kQ$ and the claim of the corollary
follows from the theorem.

\subsection{Plan of the proof of the theorem} \label{subsection:plan}
Our aim is to construct a triangle
equivalence ${\mathcal C} \to{\mathcal C}_{\mathcal T}$ such that the triangle
\[
\xymatrix{{\mathcal C} \ar[r] \ar[rd] & {\mathcal C}_{\mathcal T} \ar[d]\\
  & \operatorname{mod} {\mathcal T}}
\]
becomes commutative, where the diagonal functor takes $X$ to
${\mathcal C}(?,X)|{\mathcal T}$. To
construct the triangle equivalence ${\mathcal C} \to {\mathcal C}_{\mathcal T}$, we use the
construction of ${\mathcal C}_{\mathcal T}$ given in \cite{Keller05}, namely, the category
${\mathcal C}_{\mathcal T}$ is the stable derived category $\opname{stab}({\mathcal T}\oplus D{\mathcal T}[-3])$ of
the differential graded (=dg) category whose objects are the objects
of ${\mathcal T}$ and whose morphism complexes are given by the graded modules
\[
{\mathcal T}(x,y)\oplus (D{\mathcal T}(y,x))[-3]
\]
endowed with the vanishing differential (the construction of the
stable derived category is recalled in section~\ref{subsection:proof} below).
Thus, we have to construct an equivalence
\[
{\mathcal C} \to \opname{stab}({\mathcal T}\oplus (D{\mathcal T})[-3]).
\]
We proceed in three steps: 1) We construct a dg category ${\mathcal A}$ and
a triangle functor
\[
{\mathcal C} \to \opname{stab}({\mathcal A}).
\]
We show moreover that the subcategory of indecomposables of the
homology $H^*{\mathcal A}$ is isomorphic to ${\mathcal T}\oplus (D{\mathcal T})[-3]$.

2) Using the fact that $k$ is perfect we
show that the dg category ${\mathcal A}$ is formal, {\em i.e.}\  linked to its homology
by a chain of quasi-isomorphisms. This yields the required triangle functor
\[
{\mathcal C} \to\opname{stab}({\mathcal A}) \stackrel{_\sim}{\rightarrow} \opname{stab}(H^*{\mathcal A}) \stackrel{_\sim}{\rightarrow} \opname{stab}({\mathcal T}\oplus
(D{\mathcal T})[-3])= {\mathcal C}_{\mathcal T}.
\]

3) In a final step, we show that the composed functor
${\mathcal C}\to{\mathcal C}_{\mathcal T}$ is fully faithful and that its image generates
${\mathcal C}_{\mathcal T}$.

\subsection{The proof}\label{subsection:proof}
Let ${\mathcal M}\subset{\mathcal E}$ be the preimage of ${\mathcal T}$ under the projection
functor. In particular, ${\mathcal M}$ contains the subcategory ${\mathcal P}$ of
the projective-injective objects in ${\mathcal M}$. Note that ${\mathcal T}$ equals
the quotient $\ul{{\mathcal M}}$ of ${\mathcal M}$ by the ideal of morphisms
factoring through a projective-injective.
For each object $M$ of ${\mathcal T}$, choose an ${\mathcal E}$-acyclic complex
$A_M$ of the form
\[
0 \to M_1 \to M_0 \to P \to M \to 0 \: , \;
\]
where $P$ is ${\mathcal E}$-projective and $M_0, M_1$ are in ${\mathcal M}$,
{\em cf.}\  \cite{KellerReiten07}. Note that if $\Omega M$ denotes
the kernel of $P\to M$, the induced morphism $M_0 \to \Omega M$
is automatically a right ${\mathcal M}$-approximation of $\Omega M$.
Let ${\mathcal A}$ be the dg (=differential graded) subcategory of the dg
category ${\mathcal C}({\mathcal E})_{dg}$ of complexes over ${\mathcal E}$ whose objects are
these acyclic complexes. Thus, for two objects $A_L$ and $A_M$ of
${\mathcal A}$, we have
\[
H^{n} {\mathcal A}(A_L,A_M) = \operatorname {Hom}_{\mathop{\text{Ch}}{\mathcal E}}(A_L, A_M[n]) \: , \;
\]
where $\mathop{\text{Ch}}{\mathcal E}$ denotes the homotopy category of complexes
over ${\mathcal E}$. To compute this space, let $G_1$ be the functor
${\mathcal E}\to \operatorname{Mod}{\mathcal M}$ taking an object $X$ to ${\mathcal E}(?,X)|{\mathcal M}$. The
image of $A_M$ under $G_1$ is a projective resolution of
the ${\mathcal M}$-module $\ul{M}=\ul{{\mathcal E}}(?,M)$. Thus we have
\[
\operatorname {Hom}_{\mathop{\text{Ch}}{\mathcal E}}(A_L, A_M[n]) \stackrel{_\sim}{\rightarrow} (\operatorname {cd}{\mathcal M})(\ul{L}, \ul{M}[n]) =
\operatorname {Ext}^n_{\mathcal M}(\ul{L},\ul{M}) \: , \;
\]
where $\operatorname {cd}{\mathcal M}$ denotes the (unbounded) derived category of
$\operatorname{Mod}{\mathcal M}$. Notice that by the Yoneda lemma, for each object
$N$ of ${\mathcal M}$, we have a canonical isomorphism
\[
\operatorname {Hom}_{\mathcal M}(G_1 N, \ul{M}) \stackrel{_\sim}{\rightarrow} \ul{M}(N) = \ul{{\mathcal E}}(N,M)={\mathcal C}(N,M).
\]
Using this we see that the vector space $\operatorname {Ext}^n_{\mathcal M}(\ul{L},\ul{M})$
is the homology in degree $n$ of the complex
\[
0 \to {\mathcal C}(L,M)\to 0 \to {\mathcal C}(L_0,M) \to {\mathcal C}(L_1, M) \to 0.
\]
Clearly it is isomorphic to ${\mathcal C}(L,M)$ for $n=0$.
Using the triangle
\[
S^{-2} L \to L_1 \to L_0 \to S^{-1} L
\]
and the fact that ${\mathcal C}(S^{-1}L_0, M)=0$ and ${\mathcal C}(S^{-1}L,M)=0$,
we see that the homology is isomorphic to
${\mathcal C}(S^{-2}L, M)=D{\mathcal C}(M,L)$ for $n=3$ and vanishes
for all other $n\neq 0$. More precisely, we see that the map $M \mapsto A_M$
extends to an equivalence whose target is the (additive) graded category
$H^*{\mathcal A}$ and whose source is the graded category ${\mathcal T}\oplus
(D{\mathcal T})[-3]$ whose objects are those of ${\mathcal T}$ and whose morphisms
are given by
\[
{\mathcal T}(L,M) \oplus (D{\mathcal T}(M,L))[-3].
\]
In particular, we have a faithful functor ${\mathcal T} \to  H^*{\mathcal A}$ which
yields an equivalence from ${\mathcal T}$ to $H^0{\mathcal A}$. We denote by
$\operatorname {cd}^b({\mathcal A})$ the full subcategory of the derived category $\operatorname {cd}{\mathcal A}$
whose objects are the dg modules $X$ such that the restriction of
the sum of the $H^nX$, $n\in\mathbb{Z}$,
to ${\mathcal T}$ lies in the category $\operatorname{mod}{\mathcal T}$ of finitely
presented ${\mathcal T}$-modules (by Proposition~2.1 a) of \cite{KellerReiten07},
this category is abelian). In particular, each representable ${\mathcal A}$-module lies
in $\operatorname {cd}^b({\mathcal A})$ (by Proposition~2.1 b) of \cite{KellerReiten07})
and thus the perfect derived category $\opname{per}\nolimits({\mathcal A})$
is contained in $\operatorname {cd}^b({\mathcal A})$. We denote by $\opname{stab}({\mathcal A})$ the
triangle quotient $\operatorname {cd}^b({\mathcal A})/\opname{per}\nolimits({\mathcal A})$. Recall from
\cite{KellerVossieck87} \cite{Rickard89b} that we have a triangle
equivalence
\[
\ul{{\mathcal E}} \stackrel{_\sim}{\rightarrow} \operatorname {cd}^b({\mathcal E})/\operatorname {cd}^b({\mathcal P}).
\]
Let $G: \mathop{\text{Ch}}^b({\mathcal E}) \to \operatorname {cd} {\mathcal A}$ be the functor which takes
a bounded complex $X$ over ${\mathcal E}$ to the functor
\[
A_M \mapsto \opname{Hom^\bullet}_{\mathcal E}(A_M, X) \: , \;
\]
where $\opname{Hom^\bullet}_{\mathcal E}$ is the complex whose $n$th component is formed
by the morphisms of graded objects, homogeneous of degree $n$,
and the differential is the supercommutator with the differentials
of $A_M$ and $X$.
We will show that $G$ takes $\operatorname {cd}^b({\mathcal P})$ to zero, that it
maps $\mathop{\text{Ch}}^b({\mathcal E})$ to $\operatorname {cd}^b({\mathcal A})$ and the subcategory of acyclic
complexes to $\opname{per}\nolimits({\mathcal A})$. Thus it will induce a triangle functor
\[
\operatorname {cd}^b({\mathcal E})/\operatorname {cd}^b({\mathcal P}) \to \operatorname {cd}^b({\mathcal A})/\opname{per}\nolimits({\mathcal A})
\]
and we will obtain the required functor as the composition
\[
\ul{{\mathcal E}} \stackrel{_\sim}{\rightarrow} \operatorname {cd}^b({\mathcal E})/\operatorname {cd}^b({\mathcal P}) \to
\operatorname {cd}^b({\mathcal A})/\opname{per}\nolimits({\mathcal A})=\opname{stab}({\mathcal A}).
\]
First recall that if $A$ is an ayclic complex and $I$ a left
bounded complex of injectives, then each morphism from $A$ to $I$
is nullhomotopic. In particular, the complex $\opname{Hom^\bullet}_{\mathcal E}(A_M,P)$
is nullhomotopic for each $P$ in $\mathop{\text{Ch}}^b({\mathcal P})$. Thus $G$ takes
$\mathop{\text{Ch}}^b({\mathcal P})$ to zero. Now, we would like to show that $G$ takes values
in $\operatorname {cd}^b({\mathcal A})$ and that the image of each bounded acyclic complex
is in $\opname{per}\nolimits({\mathcal A})$. For this, we need to compute
\[
(GX)(A_L)=\opname{Hom^\bullet}_{{\mathcal E}}(A_L, X)
\]
for $L$ in ${\mathcal M}$ and $X$ in $\mathop{\text{Ch}}^b({\mathcal E})$.
To show that the restriction of the sum of the homologies of $GX$ lies in
$\operatorname{mod}{\mathcal T}$, it suffices to show that this holds if $X$ is concentrated in one
degree. Moreover, if we have a conflation
\[
0 \to M_1 \to M_0 \to X \to 0
\]
of ${\mathcal E}$ with $M_i$ in ${\mathcal M}$, it induces a short exact sequence
of complexes
\[
0 \to GM_1 \to GM_0 \to GX \to 0.
\]
So we may suppose that $X$ is an object of ${\mathcal M}$ considered as a
complex concentrated in degree $0$. Then one computes that the space
\[
\operatorname {Hom}_{\mathop{\text{Ch}}{\mathcal E}}(A_L, X[n])
\]
is isomorphic to the homology in degree $n$ of the complex
\[
0 \to {\mathcal C}(L,X) \to 0 \to {\mathcal C}(L_0,X) \to {\mathcal C}(L_1,X) \to 0 \: , \;
\]
where ${\mathcal C}(L,X)$ is in degree $0$. For $n=0$, we find that the
homology is ${\mathcal C}(L,X)$. Using the triangle
\[
S^{-2} L \to L_1 \to L_0 \to S^{-1}L
\]
and the vanishing of ${\mathcal C}(S^{-1}L,X)$ and ${\mathcal C}(S^{-1}L_0,X)$ we see
that the homology in degree $n$ is ${\mathcal C}(L,S^2 X)$ for $n=3$ and
vanishes for all other $n\neq 0$.
This shows that the restriction of the sum of the
homologies of $GX$ to ${\mathcal T}$ lies in $\operatorname{mod}{\mathcal T}$ since the restriction of
$\ul{{\mathcal E}}(?,Y)$ to ${\mathcal T}$ lies in $\operatorname{mod}{\mathcal T}$ for each $Y$ in
$\ul{{\mathcal E}}$.

Now we have to show that $G$ takes acyclic bounded complexes to
perfect dg ${\mathcal A}$-modules. For this, we first observe that we have a
factorization of $G$ as the composition
\[
\xymatrix{\mathop{\text{Ch}}^b({\mathcal E}) \ar[r]^{G_1} & \operatorname {cd}{\mathcal M} \ar[r]^{G_2} & \operatorname {cd}{\mathcal A}}
\]
where $G_1$ sends $X$ to $\opname{Hom^\bullet}(?,X)|{\mathcal M}$ and $G_2$ sends $Y$
to the dg module
\[
A_L \mapsto \opname{Hom^\bullet}(G_1 A_L, Y).
\]
Clearly the
functor $G_1$ sends ${\mathcal E}$-acyclic bounded complexes to bounded
complexes whose homology modules are in $\operatorname{mod} \ul{{\mathcal M}}$. Since
$\operatorname{mod}\ul{{\mathcal M}}$ lies in $\opname{per}\nolimits{\mathcal M}$, it follows that $G_1$ sends
bounded acyclic complexes to objects of $\opname{per}\nolimits_{\ul{{\mathcal M}}}({\mathcal M})$.
Under the functor $G_2$, the module $\ul{{\mathcal M}}(?,L)$ is sent
to $A_L$ and $G_2$ restricted to the triangulated subcategory
generated by the $\ul{{\mathcal M}}(?,L)$ is fully faithful. We claim
that this subcategory equals $\opname{per}\nolimits_{\ul{{\mathcal M}}}({\mathcal M})$. Indeed,
each object in $\opname{per}\nolimits_{\ul{{\mathcal M}}}({\mathcal M})$ is an iterated
extension of its homology objects placed in their respective
degrees. So it suffices to show that each object concentrated in
degree $0$ is the cone over a morphism between objects
$\ul{{\mathcal M}}(?,L)$, $L\in{\mathcal M}$. But this is clear since $\operatorname{mod} \ul{{\mathcal M}}$ is
equivalent to $\operatorname{mod}{\mathcal T}$, which is hereditary. It follows
that $G_2$ induces an equivalence from $\opname{per}\nolimits_{\ul{{\mathcal M}}}({\mathcal M})$
to $\opname{per}\nolimits{\mathcal A}$ and thus $G=G_2 G_1$ sends bounded acyclic
complexes to $\opname{per}\nolimits{\mathcal A}$.
Thus, we obtain the required triangle functor $F: {\mathcal C} \to
\opname{stab}({\mathcal A})$.

In section~\ref{subsection:formality} below, we will show that
${\mathcal A}$ is formal. Thus we get an isomorphism
\[
{\mathcal T}\oplus (D{\mathcal T})[-3] \stackrel{_\sim}{\rightarrow} {\mathcal A}
\]
in the homotopy category of small dg categories. This yields an
equivalence
\[
{\mathcal C}_{\mathcal T} = \opname{stab}({\mathcal T}\oplus (D{\mathcal T})[-3]) \stackrel{_\sim}{\rightarrow} \opname{stab}({\mathcal A}).
\]
By construction, it takes each object $T$ of ${\mathcal T}$ to the module
$T^\wedge={\mathcal T}(?,T)$ in ${\mathcal C}_{\mathcal T}$. Since ${\mathcal T}$ generates ${\mathcal C}$ and
the $T^\wedge$, $T\in{\mathcal T}$, generate ${\mathcal C}_{\mathcal T}$, it is enough to show
that $F$ is fully faithful. We thank Michel Van den Bergh for simplifying
our original argument: For each object $X$ of ${\mathcal C}$, we
have a triangle
\[
T_1 \to T_0 \to X \to ST_1
\]
with $T_0, T_1$ in ${\mathcal T}$. Thus, to conclude that $F$ induces
a bijection
\[
{\mathcal C}(T,X) \to {\mathcal C}_{\mathcal T}(FT,FX)
\]
for each $T\in{\mathcal T}$,
it suffices to show that $F$ induces bijections
\[
{\mathcal C}(T, T'[i]) \stackrel{_\sim}{\rightarrow} {\mathcal C}_{\mathcal T}(FT, FT'[i])
\]
for $T,T'$ in ${\mathcal T}$ and $0\leq i\leq 1$. This is
clear for $i=1$ and not hard to see for $i=0$. We conclude that
for each $Y$ of ${\mathcal C}$, $F$ induces bijections
\[
{\mathcal C}(T',Y[i]) \stackrel{_\sim}{\rightarrow} {\mathcal C}_{\mathcal T}(FT', FY[i])
\]
for all $T'$ in ${\mathcal T}$ and all $Y$ in ${\mathcal C}$ and $i\in\mathbb{Z}$. By the
above triangle, it follows that $F$ induces bijections
\[
{\mathcal C}(X,Y) \to {\mathcal C}_{\mathcal T}(FX,FY)
\]
for all $X,Y$ in ${\mathcal C}$.

\subsection{Formality}\label{subsection:formality} For
categories ${\mathcal T}$ given by `small enough' quivers $Q$, one can use the
argument of Lemma~4.21 of Seidel-Thomas' \cite{SeidelThomas01} to show
that the category ${\mathcal T}\oplus (D{\mathcal T})[-3]$ is intrinsically formal and
thus ${\mathcal A}$ is formal. We thank Michel Van den Bergh for pointing out
that for general categories ${\mathcal T}$ with hereditary module categories,
Seidel-Thomas' argument cannot be adapted.  Instead, we show directly
that ${\mathcal A}$ is formal (we do not know if ${\mathcal T}\oplus (D{\mathcal T})[-3]$ is
intrinsically formal). Of course, it suffices to show that the full
subcategory ${\mathcal A}'$ whose objects are the $A_M$ with indecomposable $M$
is formal.

Since $k$ is perfect, the category of bimodules over
a semi-simple $k$-category is still semisimple. From this, one
deduces that the category ${\mathcal T}$ is
equivalent to the tensor category of a bimodule over the
semi-simplification of ${\mathcal T}$, {\em cf.}\  Proposition~4.2.5 in
\cite{Benson98}. Using this we can construct a lift of
the functor $\opname{ind}\nolimits{\mathcal T}\to H^0 ({\mathcal A}')$ to a functor $\opname{ind}\nolimits{\mathcal T} \to Z^0({\mathcal A}')$,
where $\opname{ind}\nolimits{\mathcal T}$ denotes the full subcategory of ${\mathcal T}$ formed
by a set of representatives of the isomorphism classes of
the indecomposables.
We define a ${\mathcal T}$-bimodule by
\[
X(L,M) = \opname{Hom^\bullet}_{\mathcal E}(A_L, M) \: , \; L,M\in {\mathcal M} \: , \;
\]
where we consider $M$ as a subcomplex of $A_M$.
Note that $X$ is a right ideal in the category ${\mathcal A}'$,
that it is a $kQ$-subbimodule of $(L,M) \mapsto {\mathcal A}'(A_L,A_M)$
and that we have $fg=0$ for all homogeneous elements $f$, $g$
of $X$ of degree $>0$. The computation made above in the proof
that $G$ takes $\mathop{\text{Ch}}^b({\mathcal E})$ to $\operatorname {cd}^b({\mathcal A})$ shows that $X$ has
homology only in degree $3$ and that we have a bimodule
isomorphism
\[
D{\mathcal T}(M,L) \stackrel{_\sim}{\rightarrow} H^3 X(L,M) \: , \; L,M\in {\mathcal M}.
\]
Thus we have an isomorphism
\[
D{\mathcal T}[-3] \stackrel{_\sim}{\rightarrow} X
\]
in the derived category of ${\mathcal T}$-bimodules. We choose a projective bimodule
resolution $P$ of $D{\mathcal T}[-3]$ whose non zero components are
concentrated in degrees $1$, $2$ and $3$ (note that this is possible
since the bimodule category is of global dimension $2$). We obtain
a morphism of complexes of bimodules
\[
P \to X
\]
inducing an isomorphism in homology. We compose it with the inclusion
$X \to {\mathcal A}'$. All products of elements in the image of $P$ vanish since
they all lie in components of degree $>0$ of $X$. Thus
we obtain a morphism of dg categories
\[
{\mathcal T} \oplus P \to {\mathcal A}'
\]
inducing an isomorphism in homology.
This clearly shows that ${\mathcal A}'$ is formal.

\section{A generalization to higher dimensions}

\subsection{Negative extension groups} \label{subsection:negative}
Let $k$ be a field and  $H$ a finite-dimensional hereditary $k$-algebra.
We write $\nu$ for the Serre functor of the bounded derived
category $\operatorname {cd}=\operatorname {cd}^b(\operatorname{mod} H)$ and $S$ for its suspension functor.
Let $d\geq 2$ be an integer. Let ${\mathcal C}={\mathcal C}^{(d)}_H$ be the {\em $d$-cluster category},
{\em i.e.}\  the orbit category of $\operatorname {cd}$ under
the action of the automorphism $\nu^{-1} S^d$, and $\pi : \operatorname {cd} \to {\mathcal C}$ the
canonical projection functor. We know from \cite{Keller05} that ${\mathcal C}$
is canonically triangulated and $d$-Calabi Yau and that
$\pi$ is a triangle functor. Moreover, the image $\pi(H)$ of $H$
in ${\mathcal C}$ is a $d$-cluster tilting object, {\em cf.}\  {\em e.g.}\  \cite{KellerReiten07}.
The fact that the module $H$ is projective and concentrated in
degree $0$ yields vanishing properties for the negative selfextension groups
of $\pi(H)$ if $d\geq 3$:

\begin{lemma} We have
\[
\operatorname {Hom}(\pi(H), S^{-i}\pi(H))=0
\]
for $1 \leq i \leq d-2$.
\end{lemma}

\begin{proof} Put $T=H$. For $p\in\mathbb{Z}$,  let $\operatorname {cd}_{\leq p}$ and
$\operatorname {cd}_{\geq p}$ be the $(-p)$th suspensions of the canonical left,
respectively right, aisles of $\operatorname {cd}$, {\em cf.}\  \cite{KellerVossieck88}.
We have to show that the groups
\[
\operatorname {Hom}(T,  \nu^{-p} S^{pd-i} T)
\]
vanish for all $p\in\mathbb{Z}$ and all $1\leq i \leq d-2$. Suppose that $p=-q$ for
some $q\geq 0$. Then we have
\[
\operatorname {Hom}(T,  \nu^{-p} S^{pd-i} T)) = \operatorname {Hom}(T, \nu^{q} S^{-qd-i} T)
\]
and the last group vanishes since $T$ lies in $\operatorname {cd}_{\leq 0}$ and
$\nu^{q} S^{-qd-i}T$ lies in $\operatorname {cd}_{\geq q(d-1)+i}$ and we have
$q(d-1)+i>0$. Now suppose that $p\geq 1$. Then we have
\[
\operatorname {Hom}(T,  \nu^{-p} S^{pd-i} T) = \operatorname {Hom}(\nu^{p} T , S^{pd-i} T) = \operatorname {Hom}(\nu^{p-1}(\nu T), S^{pd-i} T)
\]
and this group vanishes since we have $\nu^{p-1}(\nu T) \in \operatorname {cd}_{\geq p-1}$ (because
$\nu T=\nu H$ is in $\operatorname{mod} H$) and $S^{pd-i}T\in \operatorname {cd}_{\leq -(pd-i)}$ and
\[
(pd-i)-(p-1) = p(d-1) -i+1 \geq d-1-i+1\geq d-i \geq 2.
\]
\end{proof}

\subsection{A characterization of higher cluster categories}
Let $d\geq 2$ be an integer, $k$ an algebraically closed
field and ${\mathcal C}$ a $\operatorname {Hom}$-finite
algebraic $d$-Calabi Yau category containing a $d$-cluster tilting object $T$.

\begin{theorem} Suppose that $\operatorname {Hom}(T, S^{-i}T)=0$ for $1\leq i\leq d-2$.
If $H=\operatorname {End}(T)$ is hereditary, then, with the notations of section~\ref{subsection:negative},
there is a triangle equivalence ${\mathcal C} \stackrel{_\sim}{\rightarrow} {\mathcal C}^{(d)}_H$ taking $T$ to $\pi(H)$.
\end{theorem}

Notice that by the lemma above, the assumption on the vanishing of
the negative extension groups is necessary. These assumptions imply
that the endomorphism algebra is Gorenstein of dimension $\leq d-1$,
as we show in lemma~\ref{lemma:Gorenstein} below. For $d\geq 3$,
this does not, of course, imply that the endomorphism
algebra is hereditary if its quiver does not have oriented
cycles, but it implies that the global dimension is
at most $d-1$.

We will prove the theorem below in section~\ref{subsection:proof2}.
In \cite{IyamaYoshino06}, Theorem~1.3, the reader will find an example
from the study of rigid Cohen-Macaulay modules which shows
that the vanishing of the negative extension groups does not follow from
the other hypotheses. The following simple example, based on an idea
of M.~Van den Bergh, is similar in spirit:

\subsection{Example}
Let $\tilde{H}$ be the path algebra of a quiver with underlying
graph $A_6$ and alternating orientation.
Put $\tilde{\operatorname {cd}}=\operatorname {cd}^b(\operatorname{mod}\tilde{H})$ and let ${\mathcal C}$ be the orbit category of
$\tilde{\operatorname {cd}}$ under the automorphism
$F=\tau^4$ (where $\tau=S^{-1}\nu$). Then ${\mathcal C}$ is $3$-Calabi Yau: Indeed, one
checks that $F^2=\tau^{-1} S^2$ in $\tilde{\operatorname {cd}}$, which clearly yields $\nu=S^3$ in
${\mathcal C}$. The following diagram shows a piece of the Auslander-Reiten
quiver of $\tilde{\operatorname {cd}}$ which is a `fundamental domain' for $F$.
To obtain the Auslander-Reiten quiver of ${\mathcal C}$, we identify the left and right
borders.
\[
\begin{xy} 0;<1pt,0pt>:<0pt,-1pt>::
(0,0) *+{P_1} ="0",
(22,22) *+{P_2} ="1",
(0,44) *+{P_3} ="2",
(22,67) *{\circ} ="3",
(0,89) *{\circ} ="4",
(22,111) *{\circ} ="5",
(45,0) *{\circ} ="6",
(67,22) *{\circ} ="7",
(45,44) *{\circ} ="8",
(67,67) *{\circ} ="9",
(45,89) *{\circ} ="10",
(67,111) *{\circ} ="11",
(89,0) *{\circ} ="12",
(112,22) *{\circ} ="13",
(89,44) *{\circ} ="14",
(112,67) *{\circ} ="15",
(89,89) *{\circ} ="16",
(112,111) *{\circ} ="17",
(134,0) *{\circ} ="18",
(156,22) *{\circ} ="19",
(134,44) *{\circ} ="20",
(156,67) *+{SP_3} ="21",
(134,89) *{\circ} ="22",
(156,111) *{\circ} ="23",
(179,0) *+{FP_1} ="24",
(201,22) *+{FP_2} ="25",
(179,44) *+{FP_3} ="26",
(201,67) *{\circ} ="27",
(179,89) *{\circ} ="28",
(201,111) *{\circ} ="29",
"0", {\ar"1"},
"2", {\ar"1"},
"1", {\ar"6"},
"1", {\ar"8"},
"2", {\ar"3"},
"4", {\ar"3"},
"3", {\ar"8"},
"3", {\ar"10"},
"4", {\ar"5"},
"5", {\ar"10"},
"6", {\ar"7"},
"8", {\ar"7"},
"7", {\ar"12"},
"7", {\ar"14"},
"8", {\ar"9"},
"10", {\ar"9"},
"9", {\ar"14"},
"9", {\ar"16"},
"10", {\ar"11"},
"11", {\ar"16"},
"12", {\ar"13"},
"14", {\ar"13"},
"13", {\ar"18"},
"13", {\ar"20"},
"14", {\ar"15"},
"16", {\ar"15"},
"15", {\ar"20"},
"15", {\ar"22"},
"16", {\ar"17"},
"17", {\ar"22"},
"18", {\ar"19"},
"20", {\ar"19"},
"19", {\ar"24"},
"19", {\ar"26"},
"20", {\ar"21"},
"22", {\ar"21"},
"21", {\ar"26"},
"21", {\ar"28"},
"22", {\ar"23"},
"23", {\ar"28"},
"24", {\ar"25"},
"26", {\ar"25"},
"26", {\ar"27"},
"28", {\ar"27"},
"28", {\ar"29"},
\end{xy}
\]
Using the mesh category of this quiver,
it is not hard to check that the sum of the images of the indecomposable
projectives $P_1, P_2, P_3$ in ${\mathcal C}$ is a $3$-cluster tilting
object whose endomorphism ring $H$ is the path algebra on the full subquiver
with the corresponding $3$ vertices. On the other hand,
the image of $P_3$ in ${\mathcal C}$ has a one-dimensional space of $(-1)$-extensions.
Note that ${\mathcal C}=\tilde{\operatorname {cd}}/F$ is nevertheless an orbit category and admits
the $3$-cluster category
\[
{\mathcal C}^{(3)}_{\tilde{H}}=\tilde{\operatorname {cd}}/F^2
\]
as a `2-sheeted covering'.

\subsection{Proof}\label{subsection:proof2}
The proof of the theorem follows the lines of the one in section~\ref{subsection:proof}:
Let ${\mathcal T}$ be the full subcategory of ${\mathcal C}$ whose objects are the
direct sums of direct factors of $T$. Let $M$ be an object of ${\mathcal T}$.
We construct an ${\mathcal E}$-acyclic complex $A_M$
\[
0 \to M_{d+1} \to M_d \to \ldots \to M_1 \to M_0 \to 0
\]
which yields a resolution of the ${\mathcal M}$-module
\[
\ul{{\mathcal E}}(?,M): {\mathcal M}^{op}\to \operatorname{Mod} k
\]
as in part b) of Theorem~5.4 in \cite{KellerReiten07}. Thus
we can take $M_0=M$ and the morphism $M_1 \to M_0$ is a deflation with projective
$M_1$. Each morphism
\[
M_i \to Z_{i-1}=\operatorname {ker}(M_{i-1} \to M_{i-2}) \: , \; i\geq 2\: , \;
\]
yields a ${\mathcal T}$-approximation in ${\mathcal C}$. Our vanishing assumption
then implies that $M_2$, \ldots , $M_{d-1}$ are projective.
As in section~\ref{subsection:proof}, we let
${\mathcal A}$ be the dg subcategory of the dg
category ${\mathcal C}({\mathcal E})_{dg}$ of complexes over ${\mathcal E}$ whose objects are
these acyclic complexes. Thus, for two objects $A_L$ and $A_M$ of
${\mathcal A}$, we have
\[
H^{n} {\mathcal A}(A_L,A_M) = \operatorname {Hom}_{\mathop{\text{Ch}}{\mathcal E}}(A_L, A_M[n]).
\]
One computes that this vector space is isomorphic to ${\mathcal C}(L,M)$
for $n=0$, to ${\mathcal C}(L,\Sigma M)=D{\mathcal C}(M,L)$ for $n=d+1$ and vanishes
for all other $n$. Here we use again our vanishing hypothesis.
We see that the map $M \mapsto A_M$
extends to an equivalence whose target is the (additive) graded category
$H^*{\mathcal A}$ and whose source is the graded category ${\mathcal T}\oplus
(D{\mathcal T})[-(d+1)]$ whose objects are those of ${\mathcal T}$ and whose morphisms
are given by
\[
{\mathcal T}(L,M) \oplus (D{\mathcal T}(M,L))[-(d+1)].
\]
Now the proof proceeds as in~\ref{subsection:proof} and we obtain
a triangle functor $F: {\mathcal C} \to {\mathcal C}_{\mathcal T}$ taking the subcategory
${\mathcal T}$ to $\operatorname {add} \pi(H)$ and whose restriction to ${\mathcal T}$ is an equivalence.
By lemma~\ref{lemma:equivalences} below, $F$ is an equivalence.

\subsection{Equivalences between $d$-Calabi Yau categories}
Let $d\geq 2$ be an integer, $k$ a field and ${\mathcal C}$ and ${\mathcal C}'$
$k$-linear triangulated categories which are $d$-Calabi Yau.
Let ${\mathcal T}\subset {\mathcal C}$ and ${\mathcal T}'\subset {\mathcal C}'$
be $d$-cluster tilting subcategories. Suppose that $F:{\mathcal C}\to{\mathcal C}'$
is a triangle functor taking ${\mathcal T}$ to ${\mathcal T}'$.

\begin{lemma} \label{lemma:equivalences}
$F$ is an equivalence iff the restriction of $F$ to ${\mathcal T}$ is
an equivalence.
\end{lemma}

\begin{proof} It follows from Proposition~5.5 a) of \cite{KellerReiten07},
{\em cf.}\  also part (1) of Theorem~3.1 of \cite{IyamaYoshino06},
that ${\mathcal C}$ equals its subcategory
\[
{\mathcal T}*S{\mathcal T}* \ldots * S^{d-1}{\mathcal T}
\]
and similarly for ${\mathcal C}'$. Suppose that the restriction of $F$ to ${\mathcal T}$
is an equivalence. Let $T\in{\mathcal T}$. By induction, we see that for each $1\leq i\leq d-1$,
the map
\[
{\mathcal C}(T,Y) \to {\mathcal C}'(FT, FY)
\]
is bijective for each $Y\in {\mathcal T}*S{\mathcal T} *\ldots * S^i{\mathcal T}$. Thus the map
\[
{\mathcal C}(S^j T, Y) \to {\mathcal C}'(S^j FT, FY)
\]
is bijective for all $j\in\mathbb{Z}$ and $Y$ in ${\mathcal C}$. Then
it follows that the map
\[
{\mathcal C}(X,Y) \to {\mathcal C}'(FX,FY)
\]
is bijective for all $X,Y$ in ${\mathcal C}$. Thus $F$ is fully faithful.
Since ${\mathcal T}'$ generates ${\mathcal C}'$, the functor $F$ is an equivalence.
Conversely, if $F$ is an equivalence and takes ${\mathcal T}$ to ${\mathcal T}'$,
then the image of ${\mathcal T}$ has to be ${\mathcal T}'$ since $F{\mathcal T}$ is
maximal $(d-1)$-orthogonal in ${\mathcal C}'$.
\end{proof}

\subsection{The Gorenstein property for certain $d$-Calabi Yau categories}
\label{subsection:Gorenstein}
Let $d\geq 2$ be an integer, $k$ a field and ${\mathcal C}$
a $k$-linear triangulated category which is $d$-Calabi Yau.
Let ${\mathcal T}\subset {\mathcal C}$ be a $d$-cluster tilting subcategory such that
we have
\[
{\mathcal C}(T, S^{-i}T')=0
\]
for all $1\leq i \leq d-2$ and all $T$, $T'$ of ${\mathcal T}$.

\begin{lemma} \label{lemma:Gorenstein}
The category $\operatorname{mod}{\mathcal T}$ is Gorenstein of dimension less than or
equal to $d-1$.
\end{lemma}

\begin{proof} As in \cite{KellerReiten07}, one sees that the functor
$\operatorname {Hom}(T,?)$ induces an equivalence from the category $S^{d}{\mathcal T}$ to
the category of injectives of $\operatorname{mod}{\mathcal T}$. So we have to show that
the ${\mathcal T}$-module $\operatorname {Hom}(?,S^{d}T)$ is of projective dimension $\leq d-1$
for each $T$ in ${\mathcal T}$. Put $Y=S^{d}T$. We proceed as in section~5.5 of
\cite{KellerReiten07}: Let $T_0 \to Y$ be a right
${\mathcal T}$-approximation of $Y$. We define
an object $Z_0$ by the triangle
\[
Z_0 \to T_0 \to Y \to SZ_0.
\]
Now we choose a right ${\mathcal T}$-approximation $T_1 \to Z_0$ and define
$Z_1$ by the triangle
\[
Z_1 \to T_1 \to Z_0 \to SZ_1.
\]
We continue inductively constructing triangles
\[
Z_i \to T_i \to Z_{i-1} \to SZ_i
\]
for $1<i\leq d-2$. By proposition~5.5 of \cite{KellerReiten07}, the
object $Z_{d-2}$ belongs to ${\mathcal T}$.
We obtain a complex
\[
0 \to Z_{d-2} \to T_{d-2} \to \ldots \to T_1 \to T_0 \to Y \to 0.
\]
We claim that its image under the functor
$F:{\mathcal C}\to \operatorname{mod}{\mathcal T}$ taking an object $X$ of ${\mathcal C}$ to
${\mathcal C}(?,X)|{\mathcal T}$ is a projective resolution of $FY$.
Indeed, by induction one checks that the object
$Z_i$ belongs to
\[
S^{d-i-1}{\mathcal T} * S^{-i}{\mathcal T}*S^{-i+1}{\mathcal T} * \ldots * S^{-1}{\mathcal T} * {\mathcal T}.
\]
Thus, for each $T\in{\mathcal T}$, we have ${\mathcal C}(T, S^{-1} Z_i)=0$ by our vanishing assumptions.
Moreover, the maps $FT_{i+1}\to FZ_i$ are surjective, by construction.
Therefore, the triangle
\[
S^{-1} Z_{i-1} \to Z_i \to T_i \to Z_{i-1}
\]
induces a short exact sequence
\[
0 \to FZ_{i} \to FT_i \to FZ_i \to 0
\]
for each $0\leq i \leq d-2$, where $Z_{-1}=Y$. This implies
the assertion.

\end{proof}

redpajama_set_name: RedPajamaArXiv


Who do you think will win? I realize this might not interest too many of my fellow Americans, but there are plenty of Europeans on f.ds, so there. My money is on Bayern or Real. Discuss.
Well I'm American and I love champions league but yeah Bayern or real will win. also does anyone internationally care about the WBC (world baseball classic)?
I am European and don't even know what the UEFA is.
Added a poll to see who will advance to the final 8. Everyone can see votes, so you will be accountable!
I think all of them will advance to the final 8, so should I click everything?
It seems okay now, right?


The son of President Buhari, Yusuf Buhari, just bagged his Master's Degree from a university in the United Kingdom.
The good news was disclosed to us through a Family friend to the family, Anas Mu'azu, who took to his Facebook wall to congratulate Yusuf Ibrahim in Hausa.
Yusuf and his sister Zahra, who has since married into the billionaire Indimi family, graduated from the University of Surrey, UK in July 2016.


By becoming an NHS Foundation Trust we will be accountable to our public and the local communities we serve through membership. We will have greater freedom and flexibility to determine how to deliver services which meet local priorities and deliver our aims. To view a short film about membership click here (link to You Tube).
Through membership, we will listen to your views and opinions which will help influence decisions and the Trust's future. Members (aged 11 and over) will be able to vote for and elect a Council of Governors (members aged 18 and over). The Council of Governors will advise our Board on how we should provide and improve services. This will be achieved by ensuring the views of local people (members) are taken into consideration when decisions about healthcare are made.
However membership is not a 'passport' to faster or preferential treatment. Where possible, to minimise costs and ensure we spend the maximum amount on patient services, we will seek to communicate with members via e-mail and through our website.
But we recognise that members will want to choose the method of communication which suits them and other options, including by post and text, will be available.
Is the carer of someone who is currently receiving treatment provided by Isle of Wight NHS Trust.
We will use established mechanisms to ensure that the views of young people below the age of 11, prisoners and those who are under-represented are heard by the Council of Governors and the NHS Foundation Trust Board of Directors. All Isle of Wight NHS Trust staff and volunteers will automatically become members unless they choose to 'opt out'.
Click here to download a membership leaflet.


text:
The Cost of Reclaimed Flooring
As a manufacturer and distributor of reclaimed building materials, we recognize that there is a lot of misinformation and confusion associated with the cost of reclaimed products. Cost can first be better understood by understanding the difference between a raw, unfinished product and a well-manufactured, ready to install product. There are many people who advertise raw, reclaimed products on internet sites, but these aren't products that you would want in your home. These products can include raw beams, partially milled wood and reclaimed flooring not re-milled into new form. Before purchasing reclaimed materials online, it's a good idea to take time to make sure you are comparing apples to apples and are fully aware of the condition of the product. See more
Bringing Home the Barns
Over a hundred years ago, thousands of barns across America were built with such care and quality craftsmanship that they can still be seen today if you were to take a drive down a rural road in nearly any state. These beautiful barns which were once an integral part of American life, are unfortunately in many cases, no longer needed and are being torn down and discarded. Never one to let history make its way into a landfill, Southend Reclaimed has salvaged millions of board feet of reclaimed barnwood through the years that have ended up in homes, offices and retail spaces across the country. See more
The Journey Of Reclaimed Wood
Southend Reclaimed Introduction
Everything You Want to Know About Reclaimed Products
What is meant by reclaimed building materials?
How is the process of reclamation done?
What is reclaimed flooring?
What is Green flooring?
What is barn board?
What are reclaimed woods?
What are textile mills?
What is meant by raw materials?
What is meant by reclaimed timber?
What is meant by beams?
What is an obsolete structure?
What is antique heart pine?
What is reclaimed oak?
What is antique oak?
What is wormy chestnut?
What is meant by solid plank flooring?
What is pre-finished engineered flooring?
What is unfinished engineered flooring?
What is a wear-layer?
What is the substrate?
What are relief cuts?
What is de-nailing?
What is kiln drying?
What is a vertical grain pattern?
What does heart content mean?
What are grain patterns?
What is meant by rustic?
What is sapwood?
What are blue sap stains?
What does patina mean?
What are circle saw marks
What is meant by distressed?
What does hand-hewn mean?
What is original face?
What is meant by rough sawn?
What does re-surfaced on all four sides mean?
What is meant by turn-key installation?
What is meant by custom pre-finishing
What is meant by hand-scraping
Why are there nail holes in your flooring?
What does FSC-certified mean?
What are solid bricks?
What are core bricks?
What is meant by antique brick?
Do you ship nationwide?
What is your typical lead-time for flooring orders?
What are your payment terms?
What is meant by reclaimed building materials? Reclaimed materials are salvaged from old structures that do not serve a functional purpose anymore. These salvaged, or "reclaimed" materials, can then be used again today for new building projects. Reclaiming is a great way to preserve history and build Green. [back to top] How is the process of reclamation done? Reclaiming antique building materials involves a number of detailed and dedicated steps. First, we find obsolete structures that were built using historic woods such as antique heart pine, antique oak, and wormy chestnut, then carefully remove them - saving them from the landfill and extinction. As the timbers are removed we select the finest ones, de-nail them by hand, and send them to the sawmill to be converted into flooring. Our sawmill runs these historic timbers through a tedious and perfected process that includes kiln drying - to safe-guard against shrinking and swelling - and precision tongue and groove molding - to ensure maximum strength and support. The result is a floor with an unmatched beauty, history, and character. [back to top] What is reclaimed flooring? Southend Reclaimed salvages valuable, unique and formerly used wood resources from obsolete structures. The wood species we find range from antique heart pine to antique oak to wormy chestnut. Southend then mills this wood into both solid and engineered flooring that adds distinction and rarity to your home or next building project. [back to top] What is Green flooring? Green flooring refers to environmentally friendly flooring. [back to top] What is barn board? As the name implies, barn board is reclaimed wood from antique barns. [back to top] What are reclaimed woods? They are various wood species reclaimed from historic structures. [back to top] What are textile mills? Textile mills are manufacturing facilities that provided goods for the apparel and industrial industries. They sprung up throughout the United States during the Industrial boom in the late1800's to early 1900's. Starting in the late 1900's as the textile industry began to move overseas, many of these historic mills had to shut their doors. Southend specializes in finding and extracting the rare materials left inside of them. [back to top] What is meant by raw materials? Raw materials refer to material in its unaltered form that has been removed from an antique structure. Southend surface cleans and palletizes these raw materials and ships them to flooring manufacturers, wood workers, and brick dealers. [back to top] What is meant by reclaimed timber? Reclaimed timbers refer to any large piece of salvaged lumber. [back to top] What is meant by beams? Beams usually refer to large antique heart pine timbers that were used as structural support in textile mills, manufacturing facilities and warehouses. These beams typically stretch 25+ feet long, 13 inches wide, and 15 inches thick. It is because of these unusually large timbers that wide plank flooring and custom millwork is possible. [back to top] What is an obsolete structure? An obsolete structure no longer serves a functional purpose in today's world. Many obsolete structures are torn down so the land may be built on again. Southend takes great pride is saving these historic structures from the landfill. [back to top] What is antique heart pine? Antique heart pine has a history and beauty you will not find in any other wood. Before North America was settled, longleaf pine trees grew throughout the Southeastern coastal states -- from Florida to Texas and northward to Virginia. The trees took hundreds of years to mature and grew to upwards of 150 feet tall. At maturity, longleaf pines were composed almost entirely of "heart" wood, which is the best part of the tree for use in building. The "heart" of this majestic wood formed through centuries of slow and steady growth when its resin hardened and petrified, leaving unmatched strength and beauty. It is "heart content" that makes antique heart pine stand apart from other wood. Since it is formed over hundreds of years as the resin hardens, it infuses the trees with an uncommon amount of strength and stability. In addition, the heart content produces unique colors - rich amber tones with streaks of honey - which add to this majestic wood's beauty. [back to top] What is reclaimed oak? Southend Reclaimed offers a distinct line of reclaimed antique oak - perhaps the most well-known of all the American wood species- at an affordable price. Reclaimed from country-side barns and other historic structures evoking elements of Americana, reclaimed oak flooring has its own unique story to tell. Understated hues, random nail holes, and other marks from ages past give this floor a universal appeal. Our standard line is comprised of a beautiful blend of red and white oak, just as it was used in the historic barns it came from. [back to top] What is wormy chestnut? American chestnut trees have been virtually decimated, due to a blight caused by an Asian bark fungus in the early 1900's. Before the species was devastated it was one of the most important forest trees growing throughout eastern North America. Today there are few, if any, mature specimens of the tree remaining. During the slow death of these trees they were often infested with small borers, which chewed pin sized holes in the wood, thus giving it the name "wormy chestnut." Southend Reclaimed now reclaims this special wood from old barns throughout the eastern United States. With its subtle hues, worm holes, and random nail holes, wormy chestnut has its own distinctive appeal admired by many. [back to top] What is meant by solid plank flooring? Solid plank flooring refers to standard tounge and groove construction. "Plank" flooring indicates long lengths and wide widths. [back to top] What is pre-finished engineered flooring? Engineered hardwood flooring consists of layers of wood pressed together and glued. In our case the top layer is a veneer slice of a reclaimed species pressed onto a mixed hardwood backing. The benefits of pre-finished engineered flooring are: 1) it helps rare reclaimed species to last longer since you're using less of it 2) it can be used in applications where standard solid wood flooring is problematic – such as moisture prone areas, concrete slabs, remodeling jobs (saw dust free=less mess) 3) It allows for very wide widths without the necessity of face nailing. This is because engineered flooring provides extra stability. It arrives to the customer sanded, finished, and ready to install. Also worth noting, is pre-finished engineered wood flooring is estimated to make up half or more of the hardwood flooring industries sales, and it is the fastest growing product line in the industry. [back to top] What is unfinished engineered flooring? Engineered hardwood flooring consists of layers of wood pressed together and glued. In our case the top layer is a veneer slice of a reclaimed species pressed onto a mixed hardwood backing. The benefits of unfinished engineered flooring are: 1) it helps rare reclaimed species to last longer since you're using less of it. 2) it can be used in applications where standard solid wood flooring is problematic – such as moisture prone area, concrete slabs, remodeling jobs 3) It allows for very wide widths without the necessity of face nailing. This is because engineered flooring provides extra stability. 4) This option allows you to pick the stain of your choice. [back to top] What is a wear-layer? The wear-layer is the top layer of engineered flooring, i.e., the part that will be seen and walked on once it is installed. [back to top] What is the substrate? The substrate is the foundation for the wear layer. [back to top] What are relief cuts? Relief cuts are vertical grooves running the length of the bottom-side of flooring. They add extra stability to the overall structural integrity of the floor. [back to top] What is de-nailing? Since we reclaim this material from historic structures, there is a high potential for them to contain nails from previous use. We, of course, de-nail the material, but the nails leave their "mark" in the process. Without a doubt, if you are drawn to the uniqueness of reclaimed antique heart pine, you will love the random nail holes which act as "footprints" to help tell its story. [back to top] What is kiln drying? Kiln drying is a process of carefully heating wood to release excessive moisture within the wood. It brings the lumber down to uniform moisture content to ensure that the lumber and/or flooring remains stable for a lifetime. An added benefit is that it also ensures that any possible dormant insects remaining in the reclaimed wood are eradicated. [back to top] What is a vertical grain pattern? Vertical grain is a pinstriped pattern achieved along the full length of the board by using the quarter sawing process. [back to top] What does heart content mean? Heartwood is formed when sapwood becomes inactive and is infused with additional resin compounds. It develops slowly in the center of the tree as the tree matures. The older the tree, the higher the heart content. According to the U.S. Forest Service a 200-year-old longleaf pine averages only 65% heart content (all the 200-year-old trees are now protected and cannot be cut). Longleaf heartwood exhibits rich colors when exposed to light and oxygen. As heart content decreases, color tones can vary widely from pale red to yellow. [back to top] What are grain patterns? Each tree has its own grain pattern. This is the direction in which the wood cell fibers grow. There are three distinct grain patterns: plain sawn, vertical (quarter sawn), and curly (marbled). Plain sawn has an arching grain. Vertical has pinstripes with no growth rings over 45 degrees perpendicular to the face. Curly is the rarest. [back to top] What is meant by rustic? A rustic grade of lumber or flooring displays heavy character marks with an aged patina. The more rustic the grade, the more character-filled the wood. [back to top] What is sapwood? Sapwood (non-heart) is the lighter colored wood on the outer perimeter of the log. It does not deepen in color and is not as hard as the heartwood. The best grades do not contain any sapwood. Lesser grades can have up to 50% sapwood and may today still be called heart pine. [back to top] What are blue sap stains? Blue sap stains are sapwood with a blue tint. [back to top] What does patina mean? Patina refers to marks of distinction that can only come through decades of oxidation and other exposure to elements from time. [back to top] What are circle saw marks? Circle saw marks are marks from the original "circle saw" that was used to cut it. Circle saws were used before the modern machinery we see today. Circle saw marks are a great way to test the antique age of a wood. [back to top] What is meant by distressed? The term "distressed" refers to the original time-worn faces of our reclaimed timbers. A distressed wood is textured with natural patina, original circle saw marks, and other century-old characteristics. [back to top] What does hand-hewn mean? Hand Hewn timbers are typically cut or shaped with hard blows of a heavy cutting instrument like an ax or chisel. The result is a rougher and more rustic look. [back to top] What is meant by original Face? Original face refers to any reclaimed product that's surface has not been altered, cut or changed by modern technologies. The result is a rustic, time-worn appearance. [back to top] What is meant by rough sawn? Rough sawn refers to an original face reclaimed wood. It has a textured- as opposed to smooth-surface. [back to top] What does re-surfaced on all four sides mean? This is when the reclaimed wood has been sawn on all 4 sides revealing a fresh face (see original face for an alternative to re-surfaced). Also referred to as S4S. [back to top] What is meant by turn-key installation? Southend Reclaimed is a full service company. We not only can provide you with the product we can also assist you in installing it as well. [back to top] What is meant by custom pre-finishing? We are able to offer any of our solid plank or unfinished engineered flooring lines pre-finished to suit any décor or preferences. [back to top] What is meant by hand-scraping? As the name implies, this type of flooring is literally scraped by hand to give an old-world, vintage look. There are many different degrees and styles of hand scraping. Please ask your Southend representative for more details. [back to top] Why are there nail holes in your flooring? Since we reclaim this material from historic structures, there is a high potential for them to contain nails from previous use. We, of course, de-nail them, but they leave their "mark" in the process. Without a doubt, if you are drawn to the uniqueness of reclaimed antique heart pine, you will love the random nail holes which act as "footprints" to help tell its story. [back to top] What does FSC-certified mean? FSC stands for "Forest Stewardship Council." In many forests around the world, logging still contributes to habitat destruction, water pollution, displacement of indigenous peoples, and violence against people who work in the forest and the wildlife that dwells there. Many consumers of wood and paper, and many forest products companies believe that the link between logging and these negative impacts can be broken, and that forests can be managed and protected at the same time. Forest Stewardship Council certification is one way to improve the practice of forestry. [back to top] What are solid bricks? Reclaimed solid bricks are the oldest bricks available today. These antique bricks – made from the very soil that the old mills were built upon – comprised the walls of the original structure; therefore they often reflect the land they came from. As a result, reclaimed "solids" tend to be oversized and filled with color variations and other time-worn marks. [back to top] What are core bricks? When air-conditioning was invented in the 1930's, the textile industry decided to brick in the large windows that had provided a natural cooling-system to that point. By this time, brick manufacturing had evolved from the look and feel of the solid bricks that comprised the original walls of the structure. The result came to be known as "core" bricks which introduced the three-hole design we still see today. This allowed the mortar to conjoin the bricks in an even more stable manner as well as reduced weight for shipping purposes. [back to top] What is meant by antique bricks? Antique bricks are typically at least 50 years old and have characteristics that cannot be duplicated by new bricks. Note: Most of our antique bricks are over 100 years old. [back to top] Do you ship nationwide? Of course! We work with top-notch freight companies that offer us incredible rates due to the frequency with which we ship with them. We pass these savings directly on to you. You just let us know where the material needs to be and we will get it there. [back to top] What is your typical lead-time for flooring orders? The usual lead-time is four weeks, though we can often rush delivery if needed. [back to top] What are your payment terms? We require a 50% deposit to begin work and the final 50% when the order is ready to ship. [back to top]


A Logic and Computation foundational course at European Summer School on Logic, Language, and Information (ESSLLI) 2009 in Bordeaux.
In this course, we adopt Turtle as a single uniform syntax that can be used for almost all technologies that are involved here. Readers who feel more at home with description logic syntax when dealing with OWL can find materials with alternative introductions to OWL 2 on our slides page.
Source files (office formats) will be provided later.


A plain white t-shirt. The familiar image calls to mind something nearly everyone can associate with. That was the idea behind the name for the Plain White T's, a pop-punk band whose hit single 'Hey There Delilah' reached number one on Billboard's Hot 100. This song is from Every Second Counts their first major-label album release (on Hollywood/Fearless Records), which came out in fall 2006.
Higgenson has been writing songs since the seventh grade. 'My parents weren't really great musicians,' he recalls. 'My mom played a little bit of guitar and my dad played a little bit of drums, but there was always music in the house. Whether it was Michael Jackson or Pat Benatar or Great White, it was just random stuff.
The band has played opening slots for Jimmy Eat World, AFI, and has performed on the Warped Tour three times. The PWT's album release, Every Second Counts on Hollywood/Fearless Records, is actually their fourth album. They self-released their first album, Come On Over, in 2001. Their second and third albums (Stop in 2002 and All That We Needed in 2005) were released by the indie punk-rock label, Fearless Records.
In addition to their musical success, the PWT's have added acting to their repertoire. Higgenson, Tirio and fellow bandmates Tim Lopez, Mike Retondo and De'Mar Hamilton play themselves in both speaking and performing roles in several guest appearances for the new ABC Family series Greek, which spotlights college life and the social minefield that comes with it.


text:
package mk.ukim.finki.wp.service.impl.db;

import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Service;

import mk.ukim.finki.wp.model.db.SpeachTrouble;
import mk.ukim.finki.wp.repository.db.SpeachTroubleRepository;
import mk.ukim.finki.wp.service.db.SpeachTroubleService;
import mk.ukim.finki.wp.service.impl.BaseEntityCrudServiceImpl;

@Service
public class SpeachTroubleServiceImpl extends
		BaseEntityCrudServiceImpl<SpeachTrouble, SpeachTroubleRepository> implements SpeachTroubleService {

	@Autowired
	private SpeachTroubleRepository repository;

	@Override
	protected SpeachTroubleRepository getRepository() {
		return repository;
	}

}


Guntersville, Ala. (June 23, 2018)—In a day filled with weather delays, and an evacuation of the entire venue, eventually the nearly 7,000 fans lined the shore to see four H1 Unlimited Hydroplane teams qualify on the 2-1/2 mile course on Lake Guntersville.
Andrew Tate driving the U-9 Delta/Realtrac was the top qualifier after two qualifying sessions with a speed of 157.866 mph. His speed established a new course record.
He was followed by Jimmy Shane (Maple Valley, Wash.) in the U-1 Miss HomeStreet with a speed of 154.425 mph. Third in qualifying was the U-11 Reliable Diamond Tool present J&D's driven by Tom Thompson (Cambridge, Md) at 148.306 mph. 2017 rookie-of-the-year Dustin Echols (Sultan, Wash.) was fourth with a speed of 140.265 mph in the U-440 Bucket List Racing.
The Alabama based U-27 Greater Gadsden presents Chase Building Group, U-21 Darrell Strong presents Payne West Insurance, and CARSTAR powers Miss Rock have yet to qualify for Sunday's race for the Southern Cup.
Runner-up Shane said, "We ripped off a couple 160's (mph), but received word that those speeds did not count because we had a fuel flow violation.
The fuel flow rule regulates teams to manage their fuel to under 4.2 gal per min. Shane said they were slightly over that number.
A high wind weather watch was issued in the morning and the event organizers were forced to evacuate the entire venue along Sunset Drive in Guntersville. This was a first in the sports 115-year history.
The orderly evacuation was announced one hour before the storm was schedule to hit the area. After about an hour, the storm passed through and the fans were allowed back. Soon after that, the course markers were reestablished and qualifying that originally was scheduled for Friday began.
The fans were on their feet cheering as the boats took to smooth water. Shane said the water "laid down and it was like glass" the remainder of the day.
Sunday's schedule includes testing from 9:00 am-10: 30 am (CDT), with preliminary Heats 1A and 1B at 11:30-noon. Preliminary Heats 2A and 2B run at 1:30-2:30 pm (CDT) and preliminary Heats 3A and 3B are scheduled between 3:00 pm-4:00 pm. The Southern Cup Final Heat is scheduled for 5:00 pm.


text:
Horimiya' Season 2 the Show Is Not Officially Canceled (Updated News)
by Farah Khan November 24, 2021, 11:36 PM 1.9k Views
Horimiya Season 2- Is It Going to Be Released?
All the anime fan lovers we are here with an amazing Japanese manga series. This series was written by Hiroki Adachi. Along with that, this series was further illustrated by Daisuke Hagiwara. This series is known as 'Horimiya'.
It is a romantic anime series that has been undoubtedly the favorite series of all anime lovers because of the chemistry between the two main lead characters of the show.
The lead characters Izumi and Kyoko have been hiding their real identity from their classmates. Kyoko has the characteristics of getting a lot aggressive and that completely depends on the circumstances.
Banished From the Hero's Party' Season 2 Release Date, Cast, Plot, Trailer
MeanwhileIzumi has characteristics of a light-mannered punk with various tattoos and piercings. Even though having so many differences, these 2 characters have a lot in common, especially with their secrets. they hold for each other and that makes them fall in love with each other.
From the date this series has been released it has become the most favorite anime of all the fans and critics. This series is a combination of beautiful animations and a great story in between the characters along with all other characters and the plot.
As season 1 of 'Horimiya' is coming to an end and lot of you might be thinking if there will be any season 2 or not. So here is everything you need to know about season 2 of Horimiya.
Horimiya Season 2 Release Date
Season 1 of 'Horimiya' season was released on 10th January 2021. It consists of 13 episodes and ended on 4th April 2021. This anime was produced by CloverWorks Studios and directed by Masashi Ishihama. Takao Yoshioka is the writer of this series.
Horimiya Season 2
In the second season of 'Horimiya, there is no official announcement from the producers yet. Just like other anime series 'Horimiya' is not much involved in spending time on revolving the plot instead it is direct and much interesting to watch. Izumi confessed his true feeling in episode 4 itself. After that Izumi and Kyoko start dating further in the fourth episode.
Mick Rock, Photographer Behind Iconic Images of David Bowie, Lou Reed, Dead at 72
This TV series of'Horimiya' has also caught up along with the manga series. The manga series has published chapter 122 recently in the month of March 2021. And the last episode of the TV series is an adaptation of this 122nd chapter. This episode will show the graduation of students from their high school.
Also as per the original manga series, a lot of the actual chapters have been skipped. A few of the episodes were covered in a montage at the end of the show during credits. And if we consider all these factors it is very unlikely for the second season to be out any soon.
As 'Horimiya' has a huge fanbase and we are expecting that this series might have a second season. However, we are not completely sure about it and for now, it's a no from the makers about the second part.
Other than this on 23rd November 2020, Horipro shared that they were making a live-action TV edition of the manga. Also, the first 3 episodes have also been released on 5th February 2021 and the full show was released on 17th February 2021.
This series had first made as a web manga with the name Hori-san to Miyamura-Kun. That web series has a total of 6 episodes adapted from the actual series. The first actual series was released on 5th October 2012 and after that, a total of 3 has been released to date.
Ginny & Georgia's Brianne Howey on 'Power-Hungry' Paul, 'Dark' Season 2, And More
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This system makes parking easier than ever. It uses front, side and rear sensors to assist you with parallel, angled and perpendicular parking . All you need to do is handle the brake, accelerator and gears (grade dependent).


text:
from django.db import migrations, models

class Migration(migrations.Migration):

    dependencies = [
        ('hs_core', '0040_auto_20190105_1844'),
    ]

    operations = [
        migrations.AddField(
            model_name='resourcefile',
            name='_size',
            field=models.BigIntegerField(default=-1),
        ),
    ]


text:
Belfast City Council is recruiting a new chief executive - and it's offering a big salary
Current boss Suzanne Wylie is leaving the top job later this month to take up a new post in Jersey
Belfast City Hall
Belfast City Council has advertised to recruit a new chief executive with a salary of almost £150,000.
Its current boss Suzanne Wylie is leaving the top job later this month to take up a new post in Jersey.
Her successor will receive an annual salary of between £130,859 and £147,217.
The job earns a higher wage than Stormont's first and deputy first ministers, who are each entitled to £123,000 a year.
The application process will run for three weeks until January 28 and interviews are expected to take place in mid-March.
Launching its job advertisement, the council said: "This is a rare and career-defining opportunity to serve as a chief executive of a capital city."
It said the successful candidate will act as an ambassador for the local authority on the national and international stage.
They will "drive inclusive growth", maximise the Belfast Region City Deal, support services through the Covid-19 pandemic and "deliver for Belfast and its residents", the council said.
A 16-page application form and 28-page candidate information pack, which includes a detailed job description, have been published on the council's website.
Ms Wylie has served as chief executive of Belfast City Council since 2014, becoming the first woman to hold the senior position.
Belfast City Council chief executive Suzanne Wylie (Image: Justin Kernoghan/Belfast Live)
Her salary in 2020-21 was between £140,000 and £145,000, according to the council's statement of accounts.
Last September it was confirmed Ms Wylie was leaving to take up a £250,000-a-year role as chief executive and head of the Public Service for the States of Jersey.
During the council's recruitment process, deputy chief executive Ronan Cregan will act up and take on overall responsibility for the local authority.
He will be supported by city solicitor John Walsh, who will take on responsibilities for political management issues.
Aspects of the job will also be allocated to other directors during the interim period.
The recruitment comes as Mid and East Antrim Borough Council seeks a new interim chief executive.
It follows the suspension of council boss Anne Donaghy during a probe into bullying and harassment allegations.
She has strongly denied any wrongdoing and is taking legal action against her employer for alleged discrimination.
Mid and East Antrim council has refused to make public a job description or salary for the new post, saying the recruitment process is "conducted confidentially".
Mid and East Antrim Borough Council 'confidentially' recruiting new interim chief executive
Belfast party bikes are a 'world of pain', says PSNI roads official


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text:
Matelea magnifolia är en oleanderväxtart som först beskrevs av Henri François Pittier, och fick sitt nu gällande namn av R. E. Woodson. Matelea magnifolia ingår i släktet Matelea och familjen oleanderväxter. Inga underarter finns listade i Catalogue of Life.

Källor

Oleanderväxter
magnifolia

redpajama_set_name: RedPajamaWikipedia


text:
STRATEGY & SECURITY
Japan's Marines have returned . . . to defend Senkakus, Okinawa from China
[FPI] Free Press International
Liberty Metron
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FPI / Geostrategy-Direct.com / December 6, 2019
Rising tension between Japan and China intensifying has led the Japanese Marines to re-form a fighting force that has not been seen since World War II.
Increased tensions between China and Japan over the Senkaku islands led Japan to re-form its amphibious Marines. / Video Image / YouTube
Concerns that Beijing might seize outlying islands led to the formation in 2018 of the 2,100-man Amphibious Rapid Response Brigade (ARDB), based at Sasebo on Kyushu—formed under the ground force's first unified command.
This year, ARDB soldiers with AAVs participated in the 2019 Iron Fist war games at Camp Pendleton, California in February, then in June took part in an amphibious landing exercise in Australia.
"Seventy-five years ago, the sight of 300 Japanese marines storming onto a Queensland beach in hulking tracked amphibious vehicles would have heralded a catastrophic setback to Australia's national security," Sebastien Roblin wrote in an analysis for National Interest. "But obviously, the world has changed quite a bit since World War II. The soldiers from Japan's Amphibious Rapid Deployment Brigade (ARB) were not invaders, but participants in the international 2019 Talisman Sabre exercise held biennially on Australian soil."
What is the operational concept behind the ARDB?
"Certainly, Japan shares with Australia, the Philippines and the United States a concern that China may seize key Pacific islands it could use to interdict maritime traffic. But Japan's constitution forbids its forces from coming to the aid of allies," Roblin wrote. "Thus, the ARDB's purpose remains specific: to rapidly recapture Japan's southwestern islands should they be occupied by Chinese forces. This remains in the United States and Australia's interests, as the Japanese island belt effectively constrains PLA Navy operations."
Disputes continue over the Senkaku/Diaoyu islands (the first name is Japanese, the latter Chinese), which are over 200 miles away from both mainland China and major Japanese islands.
Some Chinese scholars have argued that the more populated Nansei/Ryukyu island belt in southwestern Japan — which includes the Okinawa prefecture and is the site of a major U.S. military base — also rightfully belongs to China.
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