text:
Louis van Gaal's Man United tenure a strangely intriguing wreck
By Joe Prince-WrightMay 24, 2016, 2:20 PM EDT
Louis Van Gaal has left the building.
The 64-year-old Dutchman is out as Manchester United's manager after the club confirmed the worst kept secret in sport on Monday with his three-year contract cut short after United's failure to qualify for the UEFA Champions League next season.
[ MORE: Mourinho in talks with United ]
Two years after arriving at Old Trafford, we are all still a little confused as to what actually happened as the allure of LVG had us bamboozled, disgusted and intrigued, while perhaps now even a little sad that it's over.
Van Gaal divided opinion among United's fans, former players, pundits and the media for most of his tenure at Old Trafford. Now the end has arrived we're not really sure if United are better off, worse off or just in a similar spot to before LVG arrived in 2014.
[ MORE: LVG releases statement after exit ]
Towards the end there's been an inevitability about his departure — despite his insistence he'd be in charge next season — but it's worth noting that when the history books look back at Van Gaal's time in charge of one of the world's biggest teams, history will likely be kind to him. Right now it seems like a failure but compared to David Moyes' time in charge, it's not. He delivered two top five finishes and an FA Cup trophy even if he couldn't get them back to the UCL for next season. From that point of view he failed to meet his targets but despite all of the unrest about the tactics, personnel choices and his character, he didn't miss his targets by much.
[ MORE: Premier League money table 2015-16 ]
Despite the tedious, glacial rate of play he fostered among his team, there were flashes of brilliance throughout his reign which kept you interested, kept you thinking that maybe, just maybe, LVG's boys would spark into life and deliver the goods. Anthony Martial's late goal against Liverpool this season. Marcus Rashford's double against Arsenal. The 4-2 home win against Manchester City and the win at City earlier this season. All of those moments gave you a notion that maybe the previous struggles were all part of Van Gaal's masterplan.
It has been said many times before in sport and I think United's fans can vouch for this after the past two years: it's not the losing, it's the hope that kills you.
[ MORE: Cantona wants Guardiola over Mourinho ]
He arrived after success with the Netherlands at the 2014 World Cup and a mystique surrounding his name after the decades he spent on the European continent. Under LVG, United were meant to return to a swashbuckling giant, a force to be reckoned with in England and Europe. Yet it became clear quite early in his reign that that wasn't going to be the case.
From his bizarre player selections to his penchant for slow, possession based play, Van Gaal quickly ate away at any excitement United's fans had for him. Every now and then he would deliver memorable moments like throwing himself to the ground in the technical area, an impromptu speech on the mic at the end of season awards dinner and blurting out something about sex masochism in the media to keep everyone interested.
[ MORE: Best moments of LVG's bizarre tenure ]
He told his players to be "horny" for the win, called out Sam Allardyce's long-ball accusations and had a go at journalists most weekends. The LVG sideshow delivered. The play on the pitch didn't and that's what ultimately cost him his job as his three-year project to return United to greatness ground to a halt. There was so much to dislike about Van Gaal's egotistic comments and brash demeanor but at the same time there were so many reasons to admire him.
(Photo by Dave Thompson/Getty Images)
Van Gaal is a man who led the club like Sir Alex Ferguson did, someone who knew what it meant to be at the helm of a global team and treat the fans with the respect they deserve and foster a family environment throughout the club. He put everything into it, even if he didn't show it as he sat in the dugout writing notes on his clipboard during games. That angered fans and pundits who wanted to see more aggression a la Jurgen Klopp. They wanted more excitement and LVG couldn't deliver it.
United's fans were simply bored of seeing their team play. They recorded their lowest ever goals tally in the PL era this season. They went 11 home games in a season without scoring a goal in the first half. They simply didn't deliver an exciting product on the pitch.
Van Gaal pointed to his philosophy and way of playing and he certainly stamped that on the team, successful or not. They led the PL in average possession this season and in backwards passes and although he managed to get through his ideas clearly to the players, it just didn't work in the PL.
All the time, though, it was interesting to see the situation unfold.
Many times United's fans didn't want to look, just like when you pass a car wreck on the road but are guilty of rubber-necking. Everybody, no matter what team you supported, wanted to drive by and have a long glance at United's issues and then have their say on the matter without being up close at the scene to deal with the injury crisis', plus having to blood youngsters (somewhat successfully) early, help under-performing players recover and then deal with the pressure which comes with managing such an illustrious club.
(Photo by Mike Hewitt/Getty Images)
Perhaps the most damning factor of his tenure was player recruitment. Van Gaal was given over $375 million to rebuild a team demoralized by Moyes and then decimated in the summer the Dutchman arrived. The only money he spent wisely was on Martial and Daley Blind plus helped the likes of Rashford, Jesse Lingard, Timothy Fosu-Mensah and Cameron Borthwick-Jackson break into the team.
Mercurial talents such as Angel di Maria, Memphis and Radamel Falcao just didn't fit into his system. Until the bitter end he spoke about the club's failure to get in his transfer targets, the players he wanted and tried to defend his teams lack of cutting edge in the final third. In the end, his excuses fell on deaf ears. He tried to explain himself too much over the two years and his relationship with the media soured over time.
One of the most alluring things about Van Gaal is his personality. He is brash, confident and outspoken because he's been there, seen it and done it. He's won trophies in Holland, Spain, Germany and now in England after his FA Cup success. He is an Amsterdammer who believes in himself and his ability as a manager. That provided countless moments in press conferences where awkward silences would be conjured by LVG, all the time glaring at a journalist who dared to question his authority, his team selections or the mentality of his team. He continued that until the end and went at the media time and time again.
His last words to the media as Manchester United'a manager were at Wembley Stadium on Saturday, just as news was breaking from people in that very room that Jose Mourinho was set to take his job from him.
One last time he was bullish, combative and forthright.
"Thank you for the congratulations!" Louis van Gaal says, sarcastically, as he picks up #FACup & strides out of the press conference. #MUFC
— Joe Prince-Wright (@JPW_NBCSports) May 21, 2016
Asking questions to Van Gaal over the past few years in crowded press conferences (they were never empty because, well, nobody knew what he would say next) myself and others always felt his eyes latch onto you as he sized you up and decided if you were worthy of an answer. He played the game. He always kept everyone guessing, just like the man he tried to replicate at United, Sir Alex Ferguson, did. His need for control seemed overbearing but he was old school, a former school teacher from Amsterdam who was the heir to Johan Cruyff at Ajax but never got the chance to break through and replace his idol.
If his playing career wasn't anything special, his career as a coach has been. He built a dynasty at Ajax, won major trophies with huge clubs and led the Netherlands to the World Cup semifinals in 2014 with a team that excited the fans. His time with Manchester United will live long in the memory despite two seasons of mediocrity for a club of their size. Fourth-place and fifth-place Premier League finishes will not be celebrated in 10 years to come by United's fans but they'll be sat having a pint in a pub saying "do you remember when Van Gaal…"
His legacy won't be glorious among United's fans but it will be memorable. Van Gaal's legend lives on as intrigue and mystique remain as he saunters out of the exit door at Old Trafford.
(Photo by Paul Gilham/Getty Images)


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There are more than 300 MMOs listed in the WarCry Codex, with more being added every day. All the relevant details about what makes an MMO unique can be found on WarCry, and you can check out screenshots and trailers to see if you like a game's visual style. WarCry will feature news and reviews of current MMOs, and we'll be heading out to press events and conventions to preview upcoming MMOs you're pumped about like Marvel Heroes as well as some you may not have heard about yet. I'm looking at you, Wildstar.
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I'm excited to bring you the new WarCry, and expand the audience for the kind of immersive worlds I love to play. We'll be rolling out previews and fresh MMO content daily, and there will always be new giveaways and offers from partners like Red 5 Studios, Perfect World and Blizzard coming to you on WarCry.com.
Come check out WarCry to find your game.


Have you ever briefed a graphic designer about a print or digital job, but been left wondering what language they were speaking? Did you just sit there nodding, pretending to understand, and then come away more confused than ever?
Like all industry sectors, graphic designers have their own lingo and jargon. Some of it dates back to the days of hand drawn artwork, whereas other terms have come about more recently as a result of the digital evolution.
Vector graphic – made up of points connected along a path (vector); they don't lose quality when resized, so they're best for logos and other graphics that will need to be produced on large items, e.g. flags, banners etc.
Both paper and card stock come in different finishes, which affect the appearance and feel of the printed item.
A typeface consists of a series of fonts and a full range of characters, such as numbers, letters and punctuations.
In Part 2, we thought we'd take a look at some specific web design terms. These will help you communicate more effectively with your designer and developer – whether you're having a new website built or updating an existing one. So watch this space!


text:
40 LISTSThe Best and Worst of the NHL
The Best Players All Time
Best Defensemen in NHL History
The Coolest Hockey Team Logos
The Greatest Teams All Time
The All-Time Most Feared Fighters
The Best Current Players in the NHL
The Best (and Worst) NHL Mascots
Best Goaltenders in NHL History
The Best NHL Players with Their Numbers Retired
The Very Best Active Goalies
The Greatest Black NHL Players
The Most Insanely Brutal Hockey Injuries
The Very Best NHL Arenas
Best Teams in the 2016-2017 Season
The All-Time Best Hockey Coaches
The Worst Coaches in NHL History
History's Most Important NHL Goals
The Most Violent NHL Thugs
History's Most Brutal Hockey Fights
Toothless Smiles
The Worst Hair in NHL History
The Greatest Ever Coach Meltdowns
Top NHL Video Games, Ranked
Should Be in the Hall of Fame
HoF Inductees Who Actually Suck
The Worst Trades Made In-Season
The Best Finnish NHL Players
The Best German NHL Players
The Best Latvian NHL Players
The Best Russian NHL Players
The Best Swedish NHL Players
Gross Misuses of the Stanley Cup
Best Current Coaches
2020 Stanley Cup Predictions
2019-20's Dirtiest Players
Every Injury in the 2019-20 Season
2020 Hart Trophy Predictions
Tallest NHL Players Today
Shortest NHL Players Today
Most Likable Players
The Greatest American Hockey Players of All Time
Originally by Ranker Hockey
Updated November 5, 2019 15.6k votes 2.3k voters 30k views
List RulesVote up the greatest US-born players ever to play in the NBA
This list of the greatest American hockey players includes some of the NHL's best players, including Mike Modano, Jeremy Roenick, Jonathan Quick, and Ed Olczyk. Many of these great US-born hockey players have also played for the U.S. Olympic hockey team. Some of them are goal-scoring legends, while others are some of the NHL's greatest goalies and defensemen. Many of these American-born hockey players are on a list of Stanley Cup champions and have seen extensive NHL Playoffs experience.
Vote for your favorite American hockey players below, downvote the hockey players you don't think are that great - and if we missed anyone from the list, feel free to add them so that other hockey fans can vote. You cna also re-rank this list to include your own list of favorite hockey players as well.
Belleville, Canada
Position: Right wing
Brett Andrew Hull (born August 9, 1964) is a Canadian-American former ice hockey player and general manager, and currently an executive vice president of the St. Louis Blues of the National Hockey ...more
Pat LaFontaine
Position: Centerman
Patrick Michael LaFontaine (born February 22, 1965) is an American former ice hockey center in the National Hockey League (NHL) who spent his entire career playing for the league's New York ...more
Brian Leetch
Position: Defenseman
Brian Joseph Leetch (born March 3, 1968) is an American former professional ice hockey defenseman who played 18 National Hockey League (NHL) seasons with the New York Rangers, Toronto Maple Leafs, ...more
Christos Kostas "Chris" Chelios (born January 25, 1962) is an American former professional ice hockey defenseman. He was one of the longest tenured players in the National Hockey League, and is a ...more
#8 of 116 The Best Defensemen in NHL History#75 of 85 Great Athletes Who Waited Too Long To Retire#24 of 111 The Greatest Montreal Canadiens of All Time
Jeremy Shaffer Roenick (; born January 17, 1970) is an American former professional ice hockey player who played the majority of his career in the National Hockey League (NHL). He played for the ...more
Mike Modano
Michael Thomas Modano Jr. (; born June 7, 1970) is an American former professional ice hockey player, who played primarily for the Minnesota/Dallas Stars franchise. He spent the final season of his ...more
#3 of 75 People Who Should Be in the Hockey Hall of Fame#3 of 21 20 Videos of Pro Athletes Crying at Press Conferences#42 of 144 The Best NHL Players To Have Their Numbers Retired
Position: Right wing, Centerman
Patrick Timothy Kane II (born November 19, 1988) is an American professional ice hockey right winger for the Chicago Blackhawks of the National Hockey League (NHL). The Blackhawks selected him with ...more
#73 of 80 The Most Controversial Athletes in Sports History#11 of 63 The Shortest Players In The NHL Today#25 of 25 The Dirtiest Players In The NHL Right Now
Keith Tkachuk
Melrose, Massachusetts
Position: Left Wing, Centerman, Forward
Keith Matthew Tkachuk (; born March 28, 1972) is an American former professional ice hockey player who played in the National Hockey League (NHL) in a 19-year career with the Winnipeg Jets, Phoenix ...more
Mike Richter
Abington Township, Pennsylvania
Position: Goaltender
Michael Thomas Richter (born September 22, 1966) is an American former professional ice hockey goaltender. He played his entire career with the New York Rangers organization, and led the team to the ...more
Phil Housley
Position: Defenseman, Forward
Phillip Francis Housley (born March 9, 1964) is an American professional ice hockey coach and former player. He is currently an assistant coach for the Arizona Coyotes of the National Hockey League ...more
Joe Pavelski
Plover, Wisconsin
Position: Right wing, Centerman, Forward
Joseph James Pavelski (born July 11, 1984) is an American professional ice hockey player for the Dallas Stars of the National Hockey League (NHL). He had previously played for the San Jose Sharks for ...more
#62 of 190 The Best Current NHL Players#36 of 49 The Most Likable Players In The NHL Today#3 of 207 The Greatest San Jose Sharks Of All Time
Robbinsdale, Minnesota
Blake James Wheeler (born August 31, 1986) is an American professional ice hockey right winger who serves as captain of the Winnipeg Jets of the National Hockey League (NHL). He was drafted by the ...more
Joe Mullen
Joseph Patrick Mullen (born February 26, 1957) is an American former professional ice hockey player. He played 16 seasons in the National Hockey League (NHL) with the St. Louis Blues, Calgary ...more
#158 of 502 The Best NHL Players of All Time#11 of 80 The Best Pittsburgh Penguins of All Time#40 of 89 The Greatest St. Louis Blues Players Of All Time
Ryan McDonagh
Ryan Patrick McDonagh (born June 13, 1989) is an American professional ice hockey defenseman, an alternate captain for the Tampa Bay Lightning of the National Hockey League (NHL). Drafted in the ...more
#19 of 120 The Greatest New York Rangers of All Time#9 of 51 The Best Tampa Bay Lightning Players of All Time
Ryan Suter
Ryan Suter (born January 21, 1985) is an American professional ice hockey defenseman and alternate captain with the Minnesota Wild of the National Hockey League (NHL). He has also played for the ...more
#10 of 47 The Greatest Nashville Predators of All Time#4 of 57 The Greatest Minnesota Wild Players of All Time#9 of 25 Celebrities Born On Martin Luther King Jr. Day
Jonathan Quick
Jonathan Douglas Quick (born January 21, 1986) is an American professional ice hockey goaltender for the Los Angeles Kings of the National Hockey League (NHL). Quick was selected in the third round, ...more
#34 of 99 The Top Hockey Goaltenders of All Time#14 of 16 The Most Clutch Athletes in Sports Today#18 of 37 Who's the Best Goalie in the NHL Right Now?
Neal Broten
Roseau, Minnesota
Neal LaMoy Broten (born November 29, 1959) is an American former professional ice hockey player. A member of the 1980 US Olympic hockey team that won the gold medal at Lake Placid in 1980, Broten was ...more
Tony Amonte
Hingham, Massachusetts
Anthony Lewis Amonte (born August 2, 1970) is a retired American professional ice hockey player. He played right wing over 17 seasons in the National Hockey League (NHL) for the New York Rangers, ...more
John LeClair
St. Albans, Vermont
Position: Left Wing
John Clark LeClair (born July 5, 1969) is an American former professional ice hockey player who played 16 seasons in the National Hockey League (NHL) with the Montreal Canadiens, Philadelphia Flyers, ...more
T. J. Oshie
Timothy Leif "T. J." Oshie (born December 23, 1986) is an American professional ice hockey right winger for the Washington Capitals of the National Hockey League (NHL). He was selected by the St. ...more
Gary Suter
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Mark Howe
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redpajama_set_name: RedPajamaCommonCrawl


text:
\section{Introduction}
The interplay between group theory and operator algebras dates back to the seminal papers by Murray and von Neumann \cite{rings-of-operators} and by choosing different completions of a discrete countable group $\Gamma$ one obtains interesting analytic objects; for instance the Banach algebra $\ell^1(\Gamma)$, the full and reduced $\Cstar$-algebras $\Cstar(\Gamma)$ and $\Cstarred(\Gamma)$, and the group von Neumann algebra $L\Gamma$. In general there are many norms on, say, $\ell^1(\Gamma)$ such that the completion with respect to this norm gives a $\Cstar$-algebra, and the question of when the $\Cstar$-completion is unique (in which case $\Gamma$ is said to be \emph{$\Cstar$-unique}) has been studied by various authors  \cite{Ng-Permanence-properties, Boidol, Barnes}.   A $\Cstar$-unique discrete group is evidently amenable and it is, to the best of the authors' knowledge, an open question whether the converse is true, although it is known to be false in the more general context of locally compact groups \cite{Ng-Permanence-properties}. More recently, the paper \cite{GMR} put emphasis on the question of when the complex group algebra ${\mathbb C}\Gamma$ has a unique $\Cstar$-completion. As is easily seen \cite[Proposition 6.7]{GMR}, if $\Gamma$ is locally finite (i.e.~if every finitely generated subgroup is finite) then ${\mathbb C}\Gamma$ has a unique $\Cstar$-completion, and \cite[Question 6.8]{GMR} asks if the converse is true. The present paper provides partial evidence towards a positive answer to this, in that we prove that for the following classes of non-locally finite groups have several $\Cstar$-completions.

\begin{theoremletter}[see Proposition \ref{thm:central-element-of-infinite-order} and Corollary \ref{cor:poly-growth}]\label{thm:A}
The class of countable groups $\Gamma$ for which ${\mathbb C}\Gamma$ does not have a unique $\Cstar$-norm includes the following:
\begin{itemize}
\item[(i)] Groups of polynomial growth.
\item[(ii)] Torsion free, elementary amenable groups with a non-trivial, finite conjugacy class.
\item[(iii)] Groups with a central element of infinite order.
\end{itemize}
\end{theoremletter}

The key to the proof of of (i) and (ii)  is the so-called {strong Atiyah conjecture} (see Section \ref{sec:atiyah})  which predicts a concrete restriction on the von Neumann dimension of kernels of elements in the complex group algebra under the left regular representation  --- notably these are predicted to be either zero or one if the group in question is torsion free.

\subsection*{Acknowledgements}
The authors would like to thank Mikael R{\o}rdam and Thomas Schick for helpful discussions revolving around the topics of the present paper.
DK gratefully acknowledge the financial support from  the Villum Foundation (grant no.~7423) and from the Independent Research Fund Denmark (grant no.~7014-00145B).

\section{Basic results on $\Cstar$-uniqueness}
In what follows, all discrete groups are implicitly assumed to be at most countable. We will use several operator algebras associated to a discrete group $\Gamma$: the maximal $\Cstar$-algebra $\Cstar(\Gamma)$, the reduced $\Cstar$-algebra $\Cstarred(\Gamma)$ and the von Neumann algebra $L\Gamma$. For more information on these, we refer to \cite[\S{}2.5]{brown-ozawa}. We recall that $L\Gamma = (\lambda({\mathbb C}\Gamma))'' \subset \mathbb B(\ell^2\Gamma)$ is generated by the left regular representation $\lambda\colon {\mathbb C}\Gamma\to {\mathbb B}(\ell^2\Gamma)$ and carries a canonical, faithful, normal trace given by $\tau(x) = \ip{x\delta_e}{\delta_e}$. We begin by formally introducing the notion of $\Cstar$-uniqueness. In order to avoid a notational conflict with the already existing notions studied in  \cite{Ng-Permanence-properties, Boidol}, we emphasize that we are investigating the uniqueness of $\Cstar$-norms on the complex group algebra in contrast to the $\ell^1$-algebra.

\begin{defi}
Let $\Gamma$ be a discrete group. ${\mathbb C}\Gamma$ is said to be:
\begin{itemize}
\item[(i)] $\Cstar$-unique if it carries a unique $\Cstar$-norm;
\item[(ii)] $\Cstarred$-unique if no $\Cstar$-norm on ${\mathbb C}\Gamma$ is properly majorised by the reduced $\Cstar$-norm.
\end{itemize}
$\Gamma$ is said to be algebraically $\Cstar$- (respectively $\Cstarred$-)unique if ${\mathbb C}\Gamma$ is $\Cstar$- (respectively $\Cstarred$-)unique.
\end{defi}

Amenable groups are characterized by the property that the maximal and reduced $\Cstar$-algebras coincide, and thus a nonamenable group is never algebraically $\Cstar$-unique; on the other hand, for amenable groups the above notions coincide.
Note also that the class of {$\Cstar$-simple} groups, which has recently received a lot of attention \cite{BKKO, Boudec}, falls within the class of algebraically $\Cstarred$-unique groups.
As already mentioned in the introduction, algebraic $\Cstar$-uniqueness appeared in the recent paper \cite{GMR} in which the authors observed that locally finite groups have this property and asked if this characterizes the class of locally finite groups. Below we prove a few basic permanence results regarding algebraic $\Cstar$-uniqueness, but before doing so we give an alternative characterization, which is straightforward algebraic adaptation of the similar result for $\ell^1$-algebras  \cite[Proposition 2.4]{Barnes}.

\begin{lem} Let $\Gamma$ be a discrete group. Then
${\mathbb C}\Gamma$ is $\Cstar$-unique (respectively $\Cstarred$-unique) if and only if every nontrivial closed, two-sided ideal in $\Cstar(\Gamma)$ (respectively $\Cstarred(\Gamma)$) intersects ${\mathbb C}\Gamma$ non-trivially.
\end{lem}
\begin{proof}
We give the proof for the statement about algebraic $\Cstar$-uniqueness; the other case is obtained by replacing $\Cstar(\Gamma)$ by $\Cstarred(\Gamma)$ throughout the proof.
Assume that there is a non-trivial ideal $J\trianglelefteqslant \Cstar(\Gamma)$ intersecting ${\mathbb C}\Gamma$ trivially
 and denote by $q\colon \Cstar(\Gamma) \to \Cstar(\Gamma)/J$ the quotient map. Composing $q$ with the inclusion ${\mathbb C}\Gamma\hookrightarrow \Cstar(\Gamma)$ yields a faithful representation of ${\mathbb C}\Gamma$
 and it defines a $\Cstar$-norm on it that is properly majorised by the maximal norm by non-triviality of $J$. Conversely, if there is a $\Cstar$-norm on ${\mathbb C}\Gamma$ which is properly majorised by the norm coming from $\Cstar(\Gamma)$, then $\Cstar(\Gamma)$ surjects onto the corresponding quotient, and the kernel of this surjection is a non-trivial ideal intersecting ${\mathbb C}\Gamma$ trivially.
\end{proof}

\begin{cor}\label{cor:direct-products}
If $\Gamma$ and $\Lambda$ are discrete groups and  ${\mathbb C}(\Gamma\times \Lambda)$ is $\Cstarred$-unique then so are ${\mathbb C}\Gamma$ and ${\mathbb C}\Lambda$.
\end{cor}
\begin{proof}
Let $J\trianglelefteqslant \Cstarred(\Gamma)$ be a non-trivial ideal intersecting ${\mathbb C}\Gamma$ trivially. Then $J\otimes_{\min} \Cstarred(\Lambda) \trianglelefteqslant \Cstarred(\Gamma)\otimes_{\min} \Cstarred(\Lambda) = \Cstarred(\Gamma\times\Lambda)$ is a non-trivial ideal intersecting ${\mathbb C} (\Gamma\times\Lambda) = {\mathbb C} \Gamma\otimes_{{\operatorname{alg}}} {\mathbb C}\Lambda$ trivially.
\end{proof}

\begin{prop}\label{thm:central-element-of-infinite-order}
If $\Gamma$ is a discrete group with a central element of infinite order then ${\mathbb C}\Gamma$ is not $\Cstarred$-unique.
\end{prop}
\begin{proof}
Denote by $Z$ the subgroup in $\Gamma$  generated by a central element of infinite order. Then $\Cstarred(Z)\cong C(S^1)$ and $LZ\cong L^\infty(S^1)$ via the Fourier transform and we denote by $p\in LZ$ the projection corresponding to the characteristic function of the upper half circle $\{e^{i\theta}\mid \theta\in [0,\pi]\}$. Define $\pi:=\lambda_\Gamma p$; i.e.~the left regular representation of $\Gamma$ restricted to the invariant subspace $p\ell^2(\Gamma)$. Choosing a non-zero function $f\in C(S^1)$ supported in the \emph{lower} half circle we obtain a non-zero element  $x\in \Cstarred(Z)\subset \Cstarred(\Gamma)$ with $xp=0$ and hence the norm on $\Cstarred(\Gamma)$ induced by $\pi$ is not the one induced by $\lambda_\Gamma$. We now only need to see that $\pi$ is faithful on ${\mathbb C}\Gamma$. To this end, consider the trace-preserving conditional expectation $E\colon L\Gamma\to LZ$ \cite[Lemma 1.5.11]{brown-ozawa} and assume that $a\in {\mathbb C}\Gamma$ is in the kernel of $\pi$. Then  $a^*a$ is also in the kernel af $\pi$ and since $E$ is an $LZ$-bimodule map \cite[Proposition 1.5.7]{brown-ozawa} we get
\[
0=E(\lambda_\Gamma(a^*a)p)=E(\lambda_\Gamma(a^*a))p.
\]
However, $E({\mathbb C}\Gamma)\subset {\mathbb C} Z \simeq {\operatorname{Pol}}(z,\bar{z})\subset C(S^1)$ and therefore $E(\lambda_\Gamma(a^*a))=0$ and since $E$ is trace-preserving and the trace on $L\Gamma$ is faithful we conclude that $a^*a$, and hence $a$, is zero.
\end{proof}

\begin{cor}\label{cor:conjecture-for-abelian}
An abelian group is algebraically $\Cstar$-unique if and only if it is locally finite (i.e.~pure torsion).
\end{cor}
\begin{rem}
The result in Corollary \ref{cor:conjecture-for-abelian} was also observed, independently and with different proofs, by Rostislav Grigorchuk, Magdalena Musat and Mikael R{\o}rdam (unpublished).
\end{rem}

\begin{rem}
The class of locally finite groups  has many stability properties --- for instance it is closed under subgroups, quotients and extensions and, moreover, being virtually locally finite is the same as being locally finite. However, verifying these properties for the class of $\Cstar$-unique groups seems to be a much bigger challenge.
\end{rem}

\section{The strong Atiyah conjecture and $\Cstarred$-uniqueness}

\subsection{The strong Atiyah conjecture}\label{sec:atiyah}
The key to our main result is the so-called strong Atiyah conjecture which is briefly described in the following. A good general reference is \cite[Chapter 10]{Luck02} where all of the results below can be found, and to which we also refer for the original references.
 Let $\Gamma$ be a discrete group and denote by $\frac{1}{|\mathop\mathrm{FIN}(\Gamma)|}{\mathbb Z}$ the additive subgroup in ${\mathbb Q}$ generated by the set
\[
\left\{\frac{1}{|\Lambda|} \ \Big\vert \ \Lambda \leq \Gamma \text{ a finite subgroup} \right\}.
\]
Given a matrix $A\in {\mathbb M}_n({\mathbb C}\Gamma)$ we denote by $L_A\in L\Gamma\otimes {\mathbb M}_n({\mathbb C})\subset {\mathbb B}(\ell^2(\Gamma)^n)$ the bounded operator given by left multiplication with $A$ (via the left regular representation of $\Gamma$). The strong Atiyah conjecture then predicts that
\[
\dim_{L\Gamma} \ker(L_A) \coloneqq (\tau\otimes \operatorname{Tr})(P_{\ker L_A}) \in \frac{1}{|\mathop\mathrm{FIN}(\Gamma)|}{\mathbb Z}.
\]
Here $\dim_{L\Gamma}(-)$ denotes the von Neumann dimension of the (right) Hilbert $L\Gamma$-module $\ker(L_A)$ defined as the un-normalized trace of the kernel projection $P_{\ker L_A}$; see \cite{Luck02} for details on this. It should be noted that the strong Atiyah conjecture is  false in general \cite[Theorem 10.23]{Luck02}, but is known to hold for all groups which have a bound on the order of finite subgroups and belong to Linnell's class
${\mathscr C}$ \cite[Theorem 10.19]{Luck02}, the latter being the smallest class of groups which contain all free groups, is closed under directed unions and extensions by elementary amenable groups (i.e., if $\Lambda\trianglelefteqslant \Gamma$, $\Lambda \in {\mathscr C}$ and $\Gamma/\Lambda$ is elementary amenable, then $\Gamma\in {\mathscr C}$).
The above discussion motivates  the following notion.
\begin{defi} Let $\Gamma$ be a countable group. The \emph{torsion multiplier} of $\Gamma$ is defined as
\[
\theta(\Gamma) = \frac{1}{\mathop\mathrm{lcm}\{|H|\mid H\leqslant\Gamma\text{ finite\}}}\in [0,1].
\]
\end{defi}
In this definition, and in what follows, we use the convention that the least common multiple (lcm) of an infinite set of natural numbers is infinity and that $\frac{1}{\infty}=0$.
Note that if $\Gamma$ has an upper bound on the set of finite subgroups, then
\[
 \frac{1}{|\text{FIN(G)}|}{\mathbb Z}= \left\{n\theta(\Gamma) \mid n\in {\mathbb Z}   \right\},
\]
and $\frac{1}{|\text{FIN(G)}|}{\mathbb Z}$ has $0$ as an accumulation point otherwise. In view of this, the strong Atiyah conjecture for a group $\Gamma$ with $\theta(\Gamma) > 0$ implies that the possible kernel dimensions are properly quantized in the sense that they can only take values in the discrete set $\left\{ n\theta(\Gamma) \mid n\in {\mathbb N}   \right\}\subset {\mathbb R}$.
Theorem \ref{thm:A} (i) and (ii) will follow directly from our main technical result, Theorem \ref{thm:main} below. The key idea in the proof is to play the aforementioned ``quantization'' of the kernel dimensions against an abundance of central projections in $L\Gamma$ with small traces which provide representations of $\Cstarred(\Gamma)$ with non-trivial kernels. To quantify this, we need the following definition.

\begin{defi}
The \emph{central granularity} of $\Gamma$ is defined as
\[
\sigma(\Gamma) = \inf \{\tau(p)\mid p\in {\operatorname{Proj}}(Z(L\Gamma)),\,p\neq 0\}\in [0,1].
\]
\end{defi}
We note that $\sigma(\Gamma) < 1$ if and only if $Z(L\Gamma)$ is nontrivial which is equivalent $\Gamma$ not being icc\footnote{recall that a group is \emph{icc} if all non-trivial conjugacy classes are infinite}.
The next proposition allows to compute the central granularity of $\Gamma$ in group-theoretic terms for many relevant groups. Recall that the \emph{FC-centre} $\Gamma_{\mathrm{fc}}$ is the normal subgroup of $\Gamma$ consisting of all elements with finite conjugacy classes.
\begin{prop}\label{prop:granularity}
Let $\Gamma_{\mathrm{fc}} \trianglelefteqslant \Gamma$ be the FC-centre of $\Gamma$. Then
\begin{enumerate}
\item[(i)] if $\Gamma_{\mathrm{fc}}$ is finite then $\sigma(\Gamma) = \frac{1}{|\Gamma_{\mathrm{fc}}|}$,
\item[(ii)] if $\Gamma_{\mathrm{fc}}$ contains an infinite finitely generated subgroup then $\sigma(\Gamma) = 0$.
\end{enumerate}
\end{prop}
\begin{proof}
The centre of $L\Gamma$ consists of elements whose associated Fourier series in $\ell^2(\Gamma)=L^2(L\Gamma, \tau)$ are supported only on $\Gamma_{\mathrm{fc}}$ and are constant along conjugacy classes, and is therefore  contained in the centre of $L\Gamma_{\mathrm{fc}}$; hence  $\sigma(\Gamma) \geqslant \sigma(\Gamma_{\mathrm{fc}})$. If $\Gamma_{\mathrm{fc}}$ is finite, $L\Gamma_{\mathrm{fc}} = {\mathbb C} \Gamma_{\mathrm{fc}}$ is a direct sum of matrix algebras, and its centre is the space of class functions. The projection with a minimal trace is just the projection $p = \frac{1}{|\Gamma_{\mathrm{fc}}|}\sum_{g\in \Gamma_{\mathrm{fc}}} g$ with trace $\sigma(\Gamma_{\mathrm{fc}}) = \frac{1}{|\Gamma_{\mathrm{fc}}|}$. It is clearly also conjugation invariant under the action of $\Gamma$, whence $\sigma(\Gamma_{\mathrm{fc}}) = \sigma(\Gamma) = \frac{1}{|\Gamma_{\mathrm{fc}}|}$.  \\
To see (ii), let $\Lambda\subseteq \Gamma_{\mathrm{fc}}$ be an infinite finitely generated subgroup. As $\Lambda$ is generated by a finite number of elements with finite conjugacy classes, its centraliser $C_\Gamma(\Lambda)$ is of finite index in $\Gamma$. On the other hand, as $\Lambda$ is finitely generated and infinite, $L\Lambda$ has a diffuse von Neumann subalgebra and thus projections of arbitrarily small trace. This can be seen as follows: if $L\Lambda$ has a direct summand of type $\mathrm{II}_1$, it is clear because such von Neumann algebras are diffuse. Otherwise $L\Lambda$ is of type $\mathrm{I}$, but then $\Lambda$ is virtually abelian \cite[Lemma 3.3]{Luck97}, and hence, being infinite by assumption, contains a copy of ${\mathbb Z}$ which generates a diffuse von Neumann algebra $L{\mathbb Z}\cong L^\infty(S^1)$.
In view of the above, for an arbitrary $\eps > 0$ there is a projection $p \in L\Lambda\subset L\Gamma_{\mathrm{fc}}$ of trace $\tau(p) < \frac{\eps}{[\Gamma:C_\Gamma(\Lambda)]}$.  Now let
\[
q\coloneqq \bigvee_{g\in \Gamma} p^g,
\]
where $p^g:=gpg^{-1}$.  Then $q$ is a central projection in $L\Gamma$. Moreover, $p$ is invariant under the centraliser $C_\Gamma(\Lambda)$ of $\Lambda$ in $\Gamma$ and upon choosing coset representatives $g_1,\dots, g_{[\Gamma:C_\Gamma(\Lambda)]}$ for $\Gamma/C_\Gamma(\Lambda)$ we obtain that
 \[
q = \bigvee_{i=1}^{[\Gamma:C_\Gamma(\Lambda)]} p^{g_i}
\]
and hence $\tau(q) \leqslant [\Gamma:C_\Gamma(\Lambda)]\cdot \tau(p) < \eps$. Thus $\sigma(\Gamma) = 0$.
\end{proof}

\begin{lemma}\label{lemma:small-support}
Let $\Gamma$ be a  discrete non-icc group. For every $\eps > 0$ there exists a nonzero projection $p\in Z(L\Gamma)$ with $\tau(p)< \sigma(\Gamma) + \eps$ and a nonzero, central element $x\in \Cstarred(\Gamma)$ with $xp=0$.
\end{lemma}
\begin{proof}
Since $\Gamma$ is non-icc, $Z(L\Gamma)\neq \CC1$ so $\sigma(\Gamma)<1$. Let ${\varepsilon}>0$ be given and assume, without loss of generality, that $\sigma(\Gamma)+{\varepsilon}<1$.
One has $Z(L\Gamma)=Z(\Cstarred(\Gamma))''=Z({\mathbb C}\Gamma)''$, as can bee seen for instance by using Kaplansky's density theorem together with the center valued trace, and noting that $Z({\mathbb C}\Gamma)$ consists of the elements whose coefficients are constant along conjugacy classes.
By Gelfand duality, $Z(\Cstarred(\Gamma))$ is isomorphic to the $\Cstar$-algebra $C(Z)$ of continuous functions on its Gelfand spectrum $Z$, which is a compact Hausdorff space; it is metrisable because $\Cstarred(\Gamma)$ is separable. The canonical trace $\tau$ thus gives a regular Borel probability measure $\mu$ on $Z$ \cite[Theorem 2.14]{rudin-real-and-complex} and an isomorphism $Z(L\Gamma)=Z(\Cstarred(\Gamma))''\cong L^\infty(Z,\mu)$ compatible with the natural inclusions. Projections in $Z(L\Gamma)$ correspond via this isomorphism to measurable subsets of $Z$ (up to null sets), and we therefore obtain a measurable subset $A\subset Z$ such that $0 < \mu(A)<\sigma(\Gamma)+{\varepsilon}$.  By regularity of $\mu$, there exist $K\subset A$ compact such that
\[
0<\mu(K)\leq \mu(A) < \sigma(\Gamma)+ {\varepsilon}<1.
\]
Now, there is a non-zero function $f\in C(Z)$ vanishing on $K$ (for instance, the distance function to $K$); letting $p$ be the projection corresponding to $K$ finishes the proof.
\end{proof}

The following lemma gives a concrete description of the decomposition of the left regular representation of a discrete group $\Gamma$ over the cosets of a finite index normal subgroup $\Lambda$. There, and in what follows, ${\operatorname{tr}}$ will denote the normalized trace on ${\mathbb M}_n({\mathbb C})$ while $\operatorname{Tr}$ will denote the un-normalized trace.

\begin{lemma}\label{lemma:coset-decomposition}
Let $\Lambda\trianglelefteqslant \Gamma$ be a normal subgroup of finite index.
For every choice of coset representatives $g_1,\dots, g_{[\Gamma: \Lambda]}\in \Gamma$ there
exists a trace-preserving inclusion of von Neumann algebras $\pi\colon (L\Gamma,\tau)\hookrightarrow (\mathbb M_{[\Gamma : \Lambda]}(L\Lambda),\tau\otimes\mathrm{tr})$ which restricts to corresponding inclusions at the level of reduced $\Cstar$-algebras and complex group rings, and which for $x\in L\Lambda$ is given by
\begin{equation}\label{eq:diagonal-elements}
\pi(x) = \mathop\mathrm{diag}(x^{g_1},x^{g_2},\dots,x^{g_{[\Gamma:\Lambda]}}),
\end{equation}
where $x^g = gxg^{-1}$ is the conjugation action of $g\in\Gamma$ on $L\Lambda$.
\end{lemma}
\begin{proof}
Choose coset representatives $g_1,g_2,\dots,g_{[\Gamma:\Lambda]}$ of $\Gamma/\Lambda$ and consider the isomorphisms of Hilbert spaces
\[
\ell^2(\Gamma) \cong \bigoplus_{i=1}^{[\Gamma:\Lambda]} \ell^2(g_i^{-1}\Lambda) \cong \bigoplus_{i=1}^{[\Gamma:\Lambda]} \ell^2(\Lambda).
\]
These induce a $\ast$-isomorphism $\pi\colon \mathbb B(\ell^2\Gamma) \xrightarrow{\cong} \mathbb M_n({\mathbb B}(\ell^2(\Lambda)))$. It is routine to check that $\pi$ restricts to a trace-preserving inclusion of ${\mathbb C}\Gamma$ into ${\mathbb M}_{[\Gamma :\Lambda]}({\mathbb C}\Lambda)$ which automatically implies the corresponding results for the reduced $\Cstar$-algebras and von Neumann algebras. Finally, for $h\in \Lambda$ we have
\[
\pi(h) = \mathop\mathrm{diag}(\lambda(h), \dots, \lambda(h)) \in \bigoplus_{i=1}^{[\Gamma:\Lambda]} \mathbb B(\ell^2(g_i^{-1}\Lambda))
\]
and thus formula \eqref{eq:diagonal-elements} follows in view of the identity
\[
h g_i^{-1} h' = g_i^{-1} h^{g_i}h',\quad h,h'\in\Lambda.\qedhere
\]
\end{proof}

\begin{thm}\label{thm:main}
Let $\Lambda$ be a discrete group satisfying the strong Atiyah conjecture and let $\Lambda\trianglelefteqslant \Gamma$ be a finite index inclusion of $\Lambda$ into a group $\Gamma$ as a normal subgroup. If $[\Gamma:\Lambda]^2\cdot \sigma(\Lambda) < \theta(\Lambda)$ then ${\mathbb C}\Gamma$ is not $\Cstarred$-unique.
\end{thm}
\begin{proof}
The assumption $[\Gamma:\Lambda]^2\cdot \sigma(\Lambda) < \theta(\Lambda)$ forces $\Lambda$ to be non-icc and applying
Lemma \ref{lemma:small-support} we get a projection $p\in Z(L\Lambda)$ with $\tau(p) < \frac{\theta(\Lambda)}{[\Gamma:\Lambda]^2}$ and a non-zero central element $x\in \Cstarred(\Lambda)$ with  $xp=0$. We are going to construct a representation of $\Cstarred(\Gamma)$ which is injective on ${\mathbb C}\Gamma$ but with $x$ in the kernel.
To this end, consider a set of coset representatives $g_1,\dots, g_{[\Gamma:\Lambda]}$ for $\Gamma/\Lambda$ and the $*$-homomorphism $\pi\colon L\Gamma \to {\mathbb M}_{[\Gamma:\Lambda]}(L\Lambda)$ provided by Lemma \ref{lemma:coset-decomposition}.
From this we obtain a central projection $q := \bigvee_{i=1}^{[\Gamma:\Lambda]} p^{g_i}\in Z(L\Lambda)$,
and  cutting $\pi$ with the complement of $ \tilde{q}:= \mathop\mathrm{diag}(q, \dots, q)\in Z({\mathbb M}_{[\Gamma : \Lambda]}(L\Lambda))$ we get a representation
\[
\pi_q\colon \Cstarred(\Gamma) \to \mathbb B \left(\ell^2(\Lambda)^{[\Gamma:\Lambda]}\tilde{q}^\perp\right),
\]
\[
a\mapsto \pi(a) \tilde{q}^\perp.
\]
As $xp=0$, we have $x\in \ker\pi_q$ in view of \eqref{eq:diagonal-elements}. Let $a\in {\mathbb C}\Gamma\cap \ker \pi_q$. This means that $\pi(a)\tilde{q}^\perp = 0$, and thus the kernel projection $r$ of $\pi(a)$ satisfies  $r \geqslant (\tilde{q}^\perp)$. Therefore
\[
(\tau\otimes\mathrm{Tr})(r)\geqslant (\tau\otimes\mathrm{Tr})(\tilde{q}^\perp) \geq  [\Gamma:\Lambda](1 - [\Gamma:\Lambda]\tau(p)) > [\Gamma:\Lambda] - \theta(\Lambda).
\]
On the other hand, the assumption  $[\Gamma:\Lambda]^2\cdot \sigma(\Lambda) < \theta(\Lambda)$ forces an upper bound on the order of finite subgroups in $\Lambda$, i.e.~$\theta(\Lambda)>0$, and since $\Lambda$ is furthermore assumed to satisfy the strong Atiyah conjecture we obtain (using the notation of Section \ref{sec:atiyah}) that
\[
\dim_{L\Lambda}(\ker (L_A)) = (\tau\otimes\mathrm{Tr})(P_{\ker L_A}) \in \{n\theta(\Lambda) \mid n\in {\mathbb Z}\}
\]
for any matrix $A\in {\mathbb M}_{[\Gamma :\Lambda]}({\mathbb C}\Lambda)$. Thus $(\tau\otimes\mathrm{Tr})(r)\leqslant [\Gamma:\Lambda] - \theta(\Lambda)$ unless $\pi(a) = 0$. This proves that $\pi_q$ is injective on ${\mathbb C}\Gamma$ and hence completes the proof.
\end{proof}

As a corollary, we deduce that some important families of  groups are not $\Cstarred$-unique. In particular this includes the groups mentioned in Theorem \ref{thm:A} (i) and (ii),  and together with Proposition \ref{thm:central-element-of-infinite-order} this completes the proof of Theorem \ref{thm:A}.
\begin{cor}\label{cor:poly-growth}
All groups in following classes are not $\Cstarred$-unique:
\begin{itemize}
\item[(i)] Torsion free, non-icc groups satisfying the strong Atiyah  conjecture; in particular all elementary
amenable, non-icc, torsion free groups.

\item[(ii)] Virtually polycyclic groups with infinite FC-centre; in particular, all infinite groups of polynomial growth.
\end{itemize}
\end{cor}
\begin{proof}
To see (i), note that the existence of a non-trivial finite conjugacy class implies the existence of a non-trivial central element in ${\mathbb C}\Gamma$ (namely the sum of the elements in the finite conjugacy class) and hence a non-trivial projection in $Z(L\Gamma)$; thus $\sigma(\Gamma)<1$. Moreover, since $\Gamma$ is torsion free, $\theta(\Gamma)=1$ and since $\Gamma$ is assumed to satisfy the strong Atiyah conjecture it follows $\Cstarred$-unique by Theorem \ref{thm:main}. The last statement in (i) follows directly from this since the elementary amenable groups are contained in Linnell's class ${\mathscr C}$ (see Section \ref{sec:atiyah}) for which the strong Atiyah conjecture is known to hold in the presence of a bound on the order of finite subgroups \cite[Theorem 10.19]{Luck02}.\\
To see (ii), let $\Lambda\trianglelefteqslant \Gamma$ be a normal finite index polycyclic subgroup of $\Gamma$.
As $\Gamma$ has infinite FC-centre, so does $\Lambda$ and the FC-centre of $\Lambda$ is moreover finitely generated by polycyclicity.
A classical result by Hirsch \cite[Theorem 3.21]{hirsch-soluble-iii}
 implies that the orders of finite subgroups of $\Lambda$ are bounded; thus $\theta(\Lambda) > 0$. On the other hand, $\sigma(\Lambda) = 0$ by Proposition \ref{prop:granularity} (ii). Moreover, polycyclic groups, being elementary amenable, satisfy the strong Atiyah conjecture. Thus, ${\mathbb C}\Gamma$ follows non-$\Cstarred$ unique by Theorem \ref{thm:main}.

Finally, the claim about infinite groups of polynomial growth follows by first observing that by Gromov's theorem \cite{Gromov-poly-growth} these are exactly finitely generated virtually nilpotent groups. As finitely generated nilpotent groups are polycyclic, the claim follows once we argue that virtually nilpotent groups automatically have infinite FC-centre. To see this, recall that a finitely generated virtually nilpotent group $\Gamma$ contains a finite index torsion free nilpotent normal subgroup $\Lambda$ (by polycyclicity and \cite[Theorem 3.21]{hirsch-soluble-iii}). Now it follows that the centre of $\Lambda$ is infinite, and therefore so is the FC-centre $\Lambda_{\mathrm{fc}}$; but as $\Lambda\trianglelefteqslant \Gamma$ is a finite index inclusion, $\Lambda_{\mathrm{fc}}\subseteq \Gamma_{\mathrm{fc}}$. Thus, $\Gamma_{\mathrm{fc}}$ is infinite.
\end{proof}

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redpajama_set_name: RedPajamaArXiv


text:
{% extends "base.html" %}

{% block title %}Challenges{% endblock title %}

{% block main_classes %}
class="header_img"
{% endblock %}

{% block content %}

Challenges

The dynamic grid challenges page is the central location to engage teams in the needs of your mission. Each individual challenge page provides functionality for an image, hashtag, background content, challenge description, functional specifications, associated datasets and email service to contact the challenge owner. Each page is also set up for threaded comments via Disqus and an associated real-time wiki via Hackpad to enable collaborative engagement. Project (solution) pages can be initiated directly from challenge pages

Potential descriptive content here could also include:

How do I choose a challenge?

[Describe here how your participants will choose or be assigned to challenges]

How do I find a team?

[How will collaboration happen?]

How will I submit a solution?

[Describe your solution submission process.]

All

Software

Hardware

Citizen Science

Data Visualization

			{% for i in object_list %}

{{ i.title }}

						{{ i.subtitle }}

			{% endfor %}

		{% comment %}

		{% endcomment %}

{% endblock content %}

{% block js %}

{% endblock %}

redpajama_set_name: RedPajamaGithub


text:
Q: Erro exceptions.OptimisticLockException ao tentar realizar UPDATE com JPA Estou tendo problema quando tento realizar um update com JPA
com a instrução:
ObjetoDE = thisObjetoBF.update(ObjetoDE);

ERROR:
Exception [EclipseLink-5010] (Eclipse Persistence Services - 2.4.2.v20130514-5956486):
    org.eclipse.persistence.exceptions.OptimisticLockException
    Exception Description:
        The object [ObjetoDE{ChaveUidK=355}]
        cannot be merged because it has changed or been deleted since it was last read.
    Class> com.xxx.yyyy.domain.admin.entity.TabelaDE

Comentei todos os campos passando para o Update somente a chave do registro e mesmo assim persiste o erro, Teriam alguma idéia ?
Desde já agradeço.

A: Descobri que isso ocorre devido a versão que fica desatualiza e consegui resolver da seguinte forma:
ObjetoDE old = this.repository.findByID(de.getCampoChave());
        de.setUpdateS(old.getUpdateS());

O link https://www.mkyong.com/jpa/jpa-optimistic-lock-exception-in-java-development/ explica os detalhes desse processo.

redpajama_set_name: RedPajamaStackExchange


The Present has been in existence for 40 years, and is the place to buy, examine and buy all the things for houses, yards and gardens. Please replace your billing particulars here to continue enjoying your access to probably the most informative and considered journalism within the UK.
Previous Previous post: How Would You Style Your Perfect House And Exactly where Would It Be?


Lingua Latina Verbs iOS application focuses on Latin verbs: grammatical aspect as well as their English counterparts. It was designed following Communicative Language Teaching (CLT) Approach. Namely, the app was planned to be a part of Presentation, Practice and Production language teaching methodology.
Presentation – Practice – Production, or PPP, is a method for teaching structures (e.g. grammar or vocabulary). As its name suggests, PPP is divided into three phases, moving from tight teacher control towards greater learner freedom. Practically the method relies on the progression from presentation, through controlled practice, to free production. Having this in mind Lingua Latina Verbs App was given three main parts Learn, Practice and Quiz.
Presentation in Lingua Latina Verbs iOS App is given in the Word List on the Home screen of the App and in customizing Learn and Quiz Modes.
Word List offers a list of Latin words and their English translations. The list is easily searchable by different categories (sematic or grammatical).
Learn Mode in Lingua Latina Verbs iOS App can easily be transformed into a presentation part. To achieve this, choose 0 for Number of Extra Words in SETTINGS.
This means that you will see in the Learn Mode only the correct answer to the question. A student is given an option to learn either vocabulary or grammar (Four principal parts of the Latin verb, or Conjugation: I, II, III, IV, or "irregular") in Learn Mode.
Learn and Practice Modes in Lingua Latina Verbs App represent a range of highly controlled to less controlled activities.
A student is given an option to practice either vocabulary or grammar (Four principal parts of the Latin verb, or Conjugation: I, II, III, IV, or "irregular") in Learn Mode.
Practice Mode consists of a number of games. The gameplay of each game includes the use of Latin words, that is, Latin verbs. These games are much less controlled than the activity in Learn Mode. Upon tapping an incorrect Latin verb, or an incorrect spelling in Word Search, the games are set to indicate lack of success which may lead to pausing or stopping the game and inviting the student to try again.
Memoria Ludum is a classic memory game that expects students to match only the same Latin verbs. As all card matching games, this practice game is best played with a friend(s). With this game students become more exposed to Latin verbs. The repetition of the revealed cards reinforces acquisition of Latin words. As the position of the cards with verbs is important, a student has to memorize both the position and the verb itself (spelling included).
Pontem Transitu game reveals Roman soldiers in chariots. Each chariot carries a word. An English definition of the sought word is on the bridge. A student is expected to tap the word on the chariot that corresponds with the English definition/translation on the bridge. If the correct word is tapped, the Roman soldier and the chariot go under the bridge safely landing. If the incorrect word is tapped, the soldier and the chariot disappear into the clouds with a sad "No!"In SETTINGS, you may choose.
Tiberim Transitu game is a fun game in which a student is trying to help an animal (elephant, rhino, pig, monkey…) cross the river by jumping from one raft onto another. The animal has to jump onto the raft (or boat) with the word that corresponds to the English translation/definition offered on the top of the screen. The animal is enabled to jump by a gentle tap on its back. The rafts with the correct and incorrect Latin words are moving all the time from left to right, and from right to left. The trick is to tap the animal at the moment the raft with the correct Latin word is just in front of the animal.
Aenigma Verbi game is a classic word search game in which a student attempts to cross out letters that make a certain Latin word following Latin spelling. A list of words is given in a column on the right hand side of the screen. A student looks for each word in a square filled with letters.
Nere Rota is a follow-up activity designed as a fun roulette with Roman numerals written on it. The Roman numerals stand for Latin verb conjugation. The roulette spins and stops on a random Roman numeral. Students are asked to think about and say out loud/ or write down any words from the verb conjugation that is shown.
Another idea is to turn this into a vocabulary peer testing or even a competition. One group writes down several (3 to 5) verbs from the 1st conjugation, another group writes down 3 to 5 verbs from the 3rd conjugation. Then, group one reads their Latin verbs without offering translation. The second group is expected to translate those verbs. Next, the groups switch roles so the second group reads their Latin verbs and the first group translates those verbs.
Lingua Latina Verbs did not include the Production Phase of PPP in the design of the app. It is left to the teachers/caregivers to involve students in using their knowledge of Latin verbs by writing a simple narrative using some of the words from the app and showing they have mastered vocabulary and grammar. Students may be asked to invent a short poem (e.g. an acrostic poem in Latin or English but using a Latin verb for the beginning of lines).
You can choose whether you wish the students to be tested on grammar (Four principal parts of the Latin verb, or Conjugation: I, II, III, IV, or "irregular") of Latin verbs or vocabulary. This can be chosen on the home screen of the app. Furthermore, you can choose which conjugations should be included in the quiz. For this option go to SETTINGS, then Word Settings – Active conjugations, and finally tap the Roman numerals representing verb conjugation to choose the desired conjugation(s). Also, you can choose the maximum number of correct words to be used in the Quiz in SETTINGS of the app (go to Word Settings, next Tap to select active word classes, and then tap next to the desired rhyming letter cluster).
History (Test Scores) section of the app shows individual student's progress.


text:
Kevin Hart On Bill Maher, Kathy Griffin Controversies: Comics 'Always Take Risks'
Hart weighed in on the recent controversies surrounding both comedians in an exclusive clip from his Wednesday appearance on "The View."
By Oliver Gettell	· June 7, 2017 December 6, 2020
This article originally appeared on Entertainment Weekly.
Comedy has always been a risky business, but Kevin Hart says the stakes are even greater in the era of social media.
In an excerpt from his Wednesday appearance on The View, Hart weighs in on the recent controversies sparked by Bill Maher uttering the N-word on the air and Kathy Griffin posing in a photo depicting a decapitated Donald Trump. According to Hart, comedians have to be especially careful because negative sentiments proliferate so quickly these days.
"Times are different now, and as a comedian, you have to understand and respect that," he says. "If you put yourself in a position to be viewed in a negative way from the public, with social media being the way that it is, if it's negative it's going to spread. … I think you have to use better judgment."
Regarding Maher's remark and Griffin's photo (both of which they apologized for), Hart says, "You're just looking at comics being comics. We always take risks, but sometimes it can be distasteful."
He adds, "Kathy Griffin, I get that it was a joke — it wasn't received well, you apologized, you should step away from it. Whatever happens after that, it happens. Bill Maher, I don't think Bill Maher is a racist, but you know the consequences of using the word, you know how many people view your show. It was stupid."
Watch the clip above for more, and catch Hart's full appearance Wednesday on The View.
TOPICS: Kevin Hart
Angela Bassett on New Season of 9-1-1, COVID-19 and Chadwick...
Regina King's Most Underrated Roles
Essence Wellness House is Back!
'This Is Us' Birth Mother Roundtable
Meet the Girls! We are back and better than ever for Season ...
Regina King And Tank Of Tank And The Bangas Meet For The Fir...
Trump Could Be Impeached Twice


The blouse is fitted and has raglan sleeves.
The bow tie has a plastic push-button in the back.
The buttons have a mother-of-pearl glow.
and the tabs on the collar get a more rugged with increased washings.
The dark blue version is 100% denim, with the detailing in stretch sky.
will have a varied appearance.


A condition for entitlement to old-age pension is accumulation of the required period of insurance and attainment of the stipulated age (i.e., retirement age or the age derived therefrom or 65 years of age).
On 1 January 2010, Act No. 155/1995 Coll., on pension insurance (the "Pension Insurance Act"), was amended. A new retirement age and the new required period of insurance for entitlement to old-age pension were set; changes were also made to the pension amount. The age of retirement also increased, taking effect from 30 September 2011.
Section 29(4) of the Pension Insurance Act (old-age pension upon fulfilling the conditions for entitlement to a disability pension).
- the retirement age is 65 years old.
- the manner of determining the age for entering into old-age retirement remains unchanged. For these groups of insured, the age limit stipulated by regulations valid prior to 1 January 1996 remains valid.
To ascertain the required period of insurance, only the year when one reaches retirement age is decisive. Insured individuals who have not accumulated the required period of insurance on the day that they reach retirement age will not receive an old-age pension on such date. Only after they accumulate the required period of insurance will they be able to receive an old-age pension.
20 years and, after 2013, they are at least 5 years older than the retirement age stipulated under Section 32 of the Pension Insurance Act for men of the same date of birth.
The entitlement to old-age pension arises for insured individuals who attain retirement age after 2014 and accumulated at least 30 years of insurance, provided they did not fulfil the conditions of Section 29(1) of the Pension Insurance Act; in this case, notionally credited periods of insurance shall not be included in the insurance period.
Insured individuals who attain 65 years of age and failed to accumulate the required period of insurance to be entitled to an old-age pension under Section 29(1), (2) or (3) of the Pension Insurance Act shall also be entitled to an old-age pension if they became disabled (regardless of the degree of disability) and accumulated the necessary period of insurance for entitlement to a disability pension or became disabled due to a work-related injury.
Recognition of insurance under Section 29(4) of the Pension Insurance Act does not prevent establishment of the entitlement to an old-age pension under Section 29(1), (2), (3) of the Pension Insurance Act.
The manner of determination of the old-age pension amount remains the same for insured individuals who do not raise the percentage assessment of pension through gainful activity after becoming entitled to a pension. The amount of the basic assessment in 2019 amounts to CZK 3 270, the amount of the percentage assessment amounts to 1.5% of the calculation basis for each whole year of insurance, but at least CZK 770 per month.
In 2019, the amount of the basic assessment of this type of old-age pension amounts to CZK 3 270 per month; the amount of the percentage assessment may not be lower than CZK 770 per month.
0.5% of the calculation basis in the case of first degree disability.
0.75% of the calculation basis in the case of second degree disability.
1.5% of the calculation basis in the case of third degree disability.
Even if the degree of disability changes for the individuals receiving this pension, the old-age pension is not recalculated and will continue to be paid out in the same amount.
The mentioned type of old-age pension shall not increase over the period of gainful activity after the establishment of the entitlement to this pension.
As of 31 December 2009, insured individuals may, after being recognised an old-age pension under Section 29 of the Pension Insurance Act, work and collect their old-age pension in the full amount regardless of the type and scope of the gainful activity carried out. In addition to this possibility, individuals receiving an old-age pension have the possibility to request that only half of the old-age pension be paid out while carrying out gainful activity.
If the age of retirement of the insured individual is less than 63 years old, the entitlement to an early old-age pension arises no earlier than on the day from which no more than three years remain until retirement age is reached.
It continues to apply that this pension cannot be recognised earlier than as of the day of filing of the application for this pension, with the date of establishment of the entitlement being deemed the day from which it is recognised and, in the case of recognition of early retirement, the entitlement to old-age pension under Section 29 of the Pension Insurance Act does not arise.
If individuals receiving full disability or partial disability pension reach 65 years of age prior to 1 January 2010, they lose entitlement to full disability or partial disability pension on 1 January 2010 and, at the same time, they become entitled to old-age pension in the amount of the hitherto paid out full disability or partial disability pension.
On the day that individuals receiving the disability pension reach 65 years of age after 31 December 2009, they lose entitlement to disability pension and, at the same time, they become entitled to old-age pension in the amount of the hitherto paid out disability pension.
A decision has to be issued to those individuals receiving a disability pension that they will be entitled to an old-age pension instead of a disability pension of the first, second or third degree.
Decisive at all times is the amount of the full or partial disability insurance as of 31 December 2009 in the case of individuals who reached 65 years of age prior to 1 January 2010 (point 1), or the amount of the disability pension as at the date of reaching 65 years of age after 31 December 2009 (point 2); any change in the degree of disability no longer has any impact on the amount of the old-age pension paid out.
This old-age pension shall not increase while individuals are involved in gainful activity after their entitlement to this pension has been established. The entitlement to this old-age pension does not preclude the entitlement to old-age pension under Section 29 of the Pension Insurance Act.


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Star Eagle Nic Naitanui aiming at being back for Western Derby
By Kim Hagdorn
Updated July 6, 2016 — 9.55am first published at 9.45am
Star West Coast ruckman Nic Naitanui looks certain to be back playing in time for a Western Derby showdown with Fremantle next month.
Naitanui revealed on Tuesday night that he was ahead of schedule in recovery from surgery to both his Achilles tendons last month, as the Fijian-born powerhouse celebrates the AFL's multicultural round this weekend.
Naitanui has targeted a derby return as expectations rise from around Fremantle that champion ruck force Aaron Sandilands is also back playing in time to take on arch enemy West Coast in round 20 on August 7.
Sandilands is recovering from three broken ribs and a punctured lung he copped from Naitanui in a big marking contest just minutes into the second term of their round three clash on April 9.
Nic Naitanui hopes to be back for the Western Derby. Credit:Getty Images
Naitanui plans to be playing late this month after having surgery just four days after West Coast dropped a home game to Adelaide on June 11.
He increased training on Tuesday with some light running, having last week week done exercises on a scooter and walked up steps in the terraces at his home ground in Perth.
Naitanui, 26, revealed on Fox Footy that he wants to play in time for the round 20 derby or even a week earlier with a road trip to Melbourne for an MCG engagement with Collingwood.
"I think I'm on track or even a little bit ahead with my schedules, so in my head it's round 19-20," he said.
"So I'm looking at the Collingwood game, I'm looking at the Freo game.
"Hopefully we're in a position to play finals and then I can have three or four games as a base, leading into that."
It has emerged that the highly influential big man had delicate key-hole surgery to release pressure and agonising pain in both Achilles.
The key-hole intrusion is expected assist a more rapid recovery than the original two months out of the game.
I went for my first run today, which is a bit weird considering a couple of weeks ago I was in a wheel chair,
Nic Naitanui
"I went for my first run today [Tuesday], which is a bit weird considering a couple of weeks ago I was in a wheel chair," Naitanui said.
"Fingers crossed I'm back within the next couple of weeks."
Naitanui disclosed that he had faced a real danger of missing the remainder of the season if his surgery had not been so successful.
"Yeh, there was," he said.
"I said I don't mind getting a few needles.
"I want to play and I don't want to miss the back end of the year. I want to get back."
He had numerous discussions with coach Adam Simpson and medical staff before being given undertakings that he would resume in time to play before the end of the home-and-away season.
Naitanui's availability is certain to be a significant factor in the Eagles' finals campaign, as well as a late season push for a finish inside the top-four.
"It's now about building back up and building some strength in my feet and my running and jumping again," Naitanui said.
"It was probably later into games and especially leading into having the surgery I was hindered late in games with jumping in the middle. But then also just getting around the ground, quick sharp movement.
"It would be the same thing with getting needles before the game and then being on crutches post-game.
"We sat down as a group and said 'is it really going to benefit me, am I going to get much out of the year or is it best to get the surgery done?'
"I think we made the right choice."
Kim Hagdorn is a veteran WA sports journalist, broadcasting on 6PR radio and contributing to WAtoday.


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Organic market steady as ample supplies met with good demand
Markets are quieter-than-normal as farmers continue to be busy with harvest operations. Photo: Thinkstock
By Phil Franz-Warkentin
Western Canadian organic markets were holding relatively steady in late September as harvest operations entered the final stages across the Prairies.
Variable crop conditions resulted in average crops overall, according to a number of organic buyers, with ample supplies and good demand countering each other to keep prices relatively steady.
Farmers are still busy in the field, resulting in a quieter-than-normal market environment, said Scott Shiels, of Grain Millers. "We expect that to pick up as we get into October and more of the crop is off," he added.
"The biggest factor is 'how was the harvest?'" according to Jason Freeman, general manager of Farmer Direct Organic Foods. "For most farmers, you take it all and put it together, it's an average year."
Shiels seconded the generally average production estimates. "The quality of the crop that's come in so far has been good," said Shiels, although he added that yields didn't quite live up to earlier expectations.
With the average harvest, Freeman didn't expect to see any supply shocks which would help keep prices relatively steady.
The U.S./Canadian exchange rate is also holding reasonably rangebound, and "as long as the Canadian dollar doesn't appreciate significantly, we expect specialty crop prices will remain relatively strong," said Freeman.
There's a lot of grain out there. – Jason Charles
However, he noted that the grains may not have as much support. "When farmers transition to organic, the first crops they grow are oats, wheat, and barley," said Freeman, noting that grain supplies are generally larger than the specialty crops.
"There's a lot of grain out there," said Jason Charles, director of organic marketing with Pipeline Foods. Some of that was due to an ample carryover for many crops, but good weather through the 2020 growing season and generally favourable yields were also factors.
While the big supplies are a bearish influence, "demand has been really good," said Charles adding, "we're buying a lot and looking to buy a bunch more."
Charles added that generally lower prices compared to the past ten years were encouraging some of that demand.
The good demand was tied in part to the COVID-19 pandemic and an increased interest in healthy eating, according to Freeman. "People are staying home and making hummus," he said, noting that social distancing measures have allowed more time for preparing food at home. While organic ingredients may be more expensive than their conventional counterparts, they are still cheap compared to prepared foods or restaurants.
barley, oats, wheat
Choosing the right cover crop to protect the soil


Steemit Rap Challenge Week #3-- Acid Droppin' N.M.C.
Week number 3 for this rap challenge! Shout out to @rondonson for puttin' this together, mad respect. Had to grab my recording equipment from the little bro to record this one, I hope ya'll like it! ENJOY!
Was I the only one who heard this?
Cmon folks, this guy is murdering people on Steemit, call the police!
Another week of straight flames, good stuff!
So dope! Keep it up!


Views from the corner garden: the voo doo lily revisited and . . .
Views from the corner garden: What is it? A voodoo lily, of course!
An acquaintance sent me this joke the other day. The picture in my head was so vivid I couldn't stop laughing. I laughed for a minute at least. I was on the subway, sitting on a newspaper, and a guy comes over and asks, "Are you reading that?" I didn't know what to say. So I said, "Yes," stood up, turned the page, and sat down again.


During the final week of the summer holidays children from Years 6-8 (plus some captains selected from Year 10) met Mr Suverkrop, Mrs Reynolds and Mrs Medcraft at Gatwick Airport to board a flight to Bordeaux in the South of France for a week of Watersports.
The trip was organised through a company offering water based activity holidays across three campsites in France. The children had the opportunity to try a full range of water sports, from kayaking to white water rafting and dart sailing to windsurfing. All of the activities were enjoyed immensely! You can now find some pictures in our galleries.
The tents provided by the camp came fully equipped with a bed frame and mattress and the camp had excellent toilet and showering facilities. The camp chef provides 3 healthy meals a day, including some themed evenings. French Night was a particular highlight with the children sampling snails and frogs legs. Within the wider camp resort children also had access to a small supermarket, waffle bar and take-away pizza shop.
Overall, the trip was yet again a success and many children are looking forward to coming again next summer! Mr Suverkrop will be holding assemblies in both the Junior and Senior schools in the coming weeks to inform pupils about the trip.
If your child is interested in coming on the Watersports Trip in 2018 please get in contact with Mr Suverkrop.


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