Thunderstorms likely. High 69F. Winds W at 5 to 10 mph. Chance of rain 80%..
Thunderstorms early, then mainly cloudy after midnight with light rain possible. Low 43F. Winds WSW at 5 to 10 mph. Chance of rain 90%.
Clay County's Connor Robinson and Kaylee Mathis both earned first-team all region honors from the media.
Summer Parker was named third team all-region. Shelby Phillips, Taylor Asher, Kim Mills, Jacob Curry, Evan Langdon, Connor Farmer and Raven Abner were named honorable mention.
This was voted on by the working media of the 13th region only.
Matthew Taylor, Corbin; Evan Langdon, Clay County; Brett Norvell, South Laurel; Zach Patterson, Knox Central; Dakota Lawson, Whitley County; London Stephney, Bell County; Logan Rose, Jackson County; Adam Sizemore, North Laurel; Sochi Onouha, Oneida Baptist; Conner Farmer, Clay County; Hunter Hollingsworth, Whitley County; Jacob Curry, Clay County; Landon Young, North Laurel; Shawn Vaughn, Barbourville; Marcus Hollingsworth, Lynn Camp; Alex Hostettler, South Laurel; Taylor Spurlock, Harlan County; Deagon Finley, Red Bird; Issac Collett, Bell County; JD McVey, Barbourville; Maison Prater, Lynn Camp; Devin Bowling, Whitley County; Max Bird, Lynn Camp; Dalton Ponder, Williamsburg; Isaiah Alexander, North Laurel; Zack Bingham, Jackson County; Tyler Partin, Bell County; Matthew Gray, Barbourville; Elijah Woods, Oneida Baptist; Raven Abner, Clay County; Cade Barnes, Harlan;Alex Byrley, Corbin; Ben Goodin, Pineville; Mykel Griffin, Middlesboro; Conner Bain, Barbourville; Jon Eldridge, Harlan; Ryan Lambdin, Bell County; Gabe Price, Harlan County; Trent Downs, Whitley County; Hunter Brown, Williamsburg; Tyler Cole, Harlan County; David Collett, Red Bird; Austin Napier; Max Kitoko, Red Bird.
Natalie King, Harlan; Bradi Coe, Lynn Camp; Payeton Charles, Harlan; Shelby Stewart, Corbin; Hailee Valentine, North Laurel; Isabel Gray, North Laurel; Abba Smith, Barbourville; Taya Davis, Jackson County; Raigan King, Pineville; Reanna Middleton, Harlan County; Jadyn Robertson, Corbin; Emily Davis, Knox Central; Baylee Woody, Middlesboro; Mikkah Siler, Whitley County; Kaelie Frazier, Whitley County; Shelby Phillips, Clay County; Caylan Mills, Knox Central; Reis Anderson, Whitley County; Emilie Field-Darragh, OBI; Jasmine Rogers, Lynn Camp; Sara Mulligan, Jackson County; Alyssa Gibson, Red Bird; Marissa Scott, Corbin; Jada Rogers, Lynn Camp; Lillie Abbott, Williamsburg; Breann Turner, Harlan County; Kailey Owens, Middlesboro; Taylor Asher, Clay County; Morgan Blakely, Harlan County; Sabrina Lewis, Whitley County; Eden Lakes, Jackson County; Ashtyn Myers, Bell County; Brianna Gallagher, Barbourville; Kimberly Mills, Clay County; Hannah Medlin, Corbin.


text:
Modernism/modernity
American Valkyries: Richard Wagner, D. W. Griffith, and the Birth of Classical Cinema
Matthew Wilson Smith
10.1353/mod.2008.0040
Focusing on the period from 1910 to 1915, this article argues that Richard Wagner's work was crucial to the development of American cinema. Critics and artists not only advocated Wagner's composition techniques for film accompaniment, but also turned Wagner into an emblem for far broader reforms. These reforms included a greater integration of music and film, a conception of film as a high art, and a conception of film as a medium for national purification and bourgeoisification. These reforms, and their connection to Wagner, helped set the stage for the use of "The Ride of the Valkyries" in D. W. Griffith's The Birth of a Nation (1915).

redpajama_set_name: RedPajamaCommonCrawl


text:
Review: Ann Hampton Callaway Hits Redial in 'On My Way to You' – The New York Times
Standard / by Bash Daily Group Archive Feed / June 19, 2015 / No Comments
http://www.nytimes.com/2015/06/20/arts/music/review-ann-hampton-callaway-hits-redial-in-on-my-way-to-you.html?emc=edit_tnt_20150619
Review: Ann Hampton Callaway Hits Redial in 'On My Way to You'
A self-proclaimed "hopeless romantic" surrounded by "material girls": That's how the singer Ann Hampton Callaway described her younger self on Thursday evening at 54 Below, where she opened her new show, "On My Way to You." That younger self was a demure piano bar entertainer whose rich, extravagantly gorgeous voice hovered on the brink of tears as she poured her heart out in classic ballads and lachrymose original songs.
That was the '80s. Since then, Ms. Callaway, 57, has acquired spectacular accouterments: most prominently a layer of jazz. She has devoted entire programs to celebrating Ella Fitzgerald and Sarah Vaughan. And on Thursday, she called the jazz singer Kurt Elling to the stage for an extended interlude of shared vocal improvisation.
The new show might be described as an official happy ending to an exhaustive romantic search. Last November, Ms. Callaway married her partner, Kari Strand, whom she called "the love of my life." She mused about marriage equality and her sense of responsibility to make the relationship work. How to make it last was the subject of her improvisation with Mr. Elling, who has been married to the same woman for nearly 20 years. He wittily parried her pointed questions.
This duet suggested a valiant attempt by two highly accomplished pop-jazz performers to emulate the ease and spontaneity of those greatest of all jazz playmates, Ella Fitzgerald and Louis Armstrong. But the disparity between Mr. Elling's gritty blues growl and Ms. Callaway's refined ornamentation kept them at a distance. Ms. Callaway delivers perfect vocal simulations of jazz instruments but is too demure to get down and dirty. That said, the duet was illuminating for the tensions it exposed.
For the rest of the show, Ms. Callaway's band — John DiMartino on piano, Martin Wind on bass and Tim Horner on drums — provided solid pop-jazz arrangements on a mixture of original songs and standards.
For all the paths Ms. Callaway explored, the starting point was the emotional territory she occupied when I first saw her 30 years ago. Ms. Callaway is the same hopeless romantic she was then. The purity of her sound may have diminished slightly, but at heart Ms. Callaway remains a true believer in happily ever after. A medley of "My Foolish Heart" and "My Romance," and two Michel Legrand ballads with lyrics by Alan and Marilyn Bergman — "How Do You Keep the Music Playing" and "On My Way to You" — got to the heart of the matter.
Next Harold Battiste, New Orleans saxophonist, composer and educator, dies at 83 | NOLA.com
Previous Jon Batiste Brings Music to 'Late Show' and Streets of NYC – WSJ
He has helped me with everything from advertising gigs, setting up ads, contact information, contacting labels and media, producing my own CDs, distribution, and other invaluable advice. His passion for what he does helps keep me on a positive attitude when the business side of things seems bleak at times.
I am quite sure I would not have been able to achieve the success that I have over the last 13 years if it wasn't for the expertise, hard work, and guidance of Jim Eigo and Jazz Promo Services.
Dave Stryker, Guitarist / Bandleader / Clinician


I'm a preschool teacher, who wants to be pet sitting or taking care of pets as my side job. I'm a major animal lover, and will probably will fall in love with your pet in a matter of minutes. It's to hard not to :) I've taken classes here and there over the years with has to do with pet training! I also have my first aid and cpr certification if needed.


Today we will explore the city on foot, exploring the major landmarks around the Plaza de Mayo, site of the famous mass demonstrations by supporters of Juan Domingo Peron in 1945, organised by his wife Eva. We'll also visit the elaborate mausoleums of the famous Recoleta Cemetery, the final resting place of some of Argentina's most revered figures, including Eva Peron herself. Our tour will also take us via Palermo Park and the tango-filled streets of La Caminito in the La Boca district of the city. Settled originally by Italian migrants, the area is famed for its colourful houses and lively street performers.
This morning we board our flight to Mendoza, the beautifully located city at the heart of Argentinian wine production, with its incredible backdrop of the mighty Andes mountain range. We spend the morning exploring the city on foot, including the sweeping views from the Hill of Glory (Cerro de la Gloria), which commemorates the Army of the Andes' successful defeat of the area's Spanish rulers in 1817. In the afternoon we have free time to spend exploring the city, maybe taking the opportunity to visit the wine museum in preparation for tomorrow's tasting. Alternatively, Mendoza is a great place to try some of the local delicacies - alfajores are an Argentinian staple, a cakey biscuit filled with 'dulce de leche', perfect with an afternoon tea or coffee. We may even try mate - a bitter green tea drink that takes centre stage in most Argentinian social gatherings. Throughout our trip we may often see people walking around the towns and cities, flask in hand, as they go about their daily business. If you are offered some, take note - mate is served in a communal cup, with a 'bombilla' (straw) to drink through, and is passed around the circle, each drinker enjoying the mate before passing it back to the server to be re-brewed. Be careful not to say thank you, as this signifies that you have had enough!
30 minutes outside of Mendoza city we find Lujan de Cuyo - a wine growing region where row upon row of vines grow in neatly ordered lines under the shadow of the Andes mountains. Drinking wine in this landscape is an unforgettable experience, and today we will be jumping on our bikes and enjoying a leisurely cycle between three of the region's wineries. Although this area is home to many varieties of white, it is mainly known for its reds, particularly the famous malbec. We will have the chance to taste the multiple varieties of grape found in the area throughout this roughly 4km cycle (all flat!), and in the last winery we enjoy lunch with especially paired wines before returning to Mendoza.
This morning we set off on our 8 hour drive to San Juan via Leoncito National Park. The national park is a protected area with very strict regulations on air pollution, and due to the quality of the atmosphere it is also an important centre of astronomy. In fact, the whole area around San Juan and this section of the Northwest is renowned for its clear skies, and is therefore a fantastic place to see a solar eclipse. The highlight of the park is the observatory, and there are also many trails around the rocky terrain and across the streams of the park, amid herds of guanacos (a smaller relative of the llama) with outstanding panoramas of the Andes mountains. Our Explore Leader will discuss the options on our arrival, and whether it's a visit to the observatory or a longer hike, this is a great landscape to explore. We arrive into the charming town of San Juan in the early evening, and no doubt Argentina's famous steak will be on the menu for many!
A day full of weird and wonderful lunar landscapes awaits us today as we drive to the magnificent Talampaya National Park. Once the kingdom of dinosaurs, this park, along with its neighbour, Ischigualasto, which we will be visiting tomorrow, contains an important number of fossils, providing a sequential record of life through the Triassic period. The scrub desert boasts fossilised remains of a wide range of indigenous animals and plants from roughly 250 to 200 million years ago. Even for those amongst us who are not scientifically-minded, the park is an impressive show of geological wonders. The main attraction is the canyon, an imposing red rock trench carving through the desert floor. From here we can observe the oddly shaped rock towers with evocative names such as the Chimney, the Monk, the Elevator and the Cathedral - more akin to a Martian journey than our expectations of Andean scenery. We head on to Villa Union, our base for this evening.
Another other-worldly landscape awaits us today as we drive to Ischigualasto National Park. Popularly known as 'Moon Valley', this park, together with its neighbour Talampaya, was granted UNESCO heritage some years ago. The hike to Cerro Morado Hill will take us to many balcony-style look out points with excellent vistas across the whole valley. The round trip hike takes approximately three hours, and there is a moderate 400m climb up to the top of the hill, with chances of spotting centuries old plant life, unique to the valley. We will also explore the unique rock features found in the park, by bus, before heading to our accommodation near San Juan in preparation for the eclipse tomorrow.
Despite being somewhat overshadowed by the fame of Mendoza, San Juan is the second largest wine-producing region in the country, and this morning we have the opportunity to visit one of the many vineyards on the 'Ruta del Vino' to try some of the first-class local specialties. We take a flight later this afternoon back to the hustle and bustle of Buenos Aires, which will be a huge contrast from the quiet and open landscapes of the Andean region. This evening is free to enjoy something of Buenos Aires' dazzling nightlife.
Today we enjoy a full free day in Buenos Aires to explore more of one of South America's most captivating cities. There is much to enjoy just from wandering around the districts like La Boca and San Telmo, and you could also taste some of Argentina's most mouth-watering traditional food at one of the city's many parrillas (grills). Choripan comes highly recommended - a large chorizo in a homemade crusty bread roll, covered with spicy chimichurri sauce. Delicious! Alternatively, you can escape the city with a trip to an estancia (ranch) or take a boat trip on the Tigre delta, a fascinating insight into the manmade communities and stilted homes set around this swampy river.
Our tour ends this morning in Buenos Aires.
In San Juan you will have 3 nights in a hotel made up of differing cabins and rooms. Here you may be in a studio apartment, a double room or a small multi-share cabin. This is the best available place to stay to see the eclipse near San Juan.


text:
The West Kowloon station is a public facility that people can enter freely, the rally organiser says. Photo: Alvin Lum
Hong Kong / Politics
Organisers promise next major protest against Hong Kong's extradition bill will be 'peaceful, rational and graceful' with the aim of gaining support from mainland Chinese visitors
They dismiss concerns about the risk of violence and chaos seen at past protests, with at least 2,000 people expected to march on Sunday afternoon
But MTR has stopped selling tickets for high-speed rail trips across border as rally is due to head to West Kowloon station after leaving Tsim Sha Tsui
Topic | Hong Kong extradition bill
Alvin Lum , Christy Leung , Naomi Ng
Published: 12:13am, 7 Jul, 2019
Updated: 11:24am, 7 Jul, 2019
Alvin Lum
Alvin Lum is an award-winning political journalist specialising in Hong Kong politics and the city's justice system. Previously, he worked for the Hong Kong Economic Journal and Citizen News.
Christy Leung
Christy writes about crime and security-related stories for South China Morning Post's city desk.
Naomi Ng
Naomi is a city news reporter at the South China Morning Post. She joined the Post in 2015 as a cadet reporter and has previously written for CNN International in Hong Kong, and Mizzima news in Myanmar.


Lana Del Rey is a successful pop singer with a carefully crafted persona who sings about The Way We Live Now. For reasons no one is quite clear on, music writers find the idea of a pop singer with a carefully crafted persona novel and interesting and enjoy writing pieces that attempt to explain what her carefully crafted persona and songs about The Way We Live Now tell us about The Way We Live Now. We have ranked some of these pieces below by how much someone who is not a music writer might want to read them.
1) Flavorwire, "Flowchart: So You Wanna Write A Lana Del Rey Thinkpiece..."
Here's a handy flowchart to pair you up with the right angle for your inevitable Lana thinkpiece.
2) Stereogum, "Premature Evaluation: Lana Del Rey Ultraviolence"
Inauthenticity was a a massively important part of her entire project, one of the engines that gave her entire persona its force.... Lana Del Rey is a construction.
3) Gawker, "Lana Del Rey, That's No Way For a Pop Star To Behave"
Pop music is clichéd—inherently, even—but Lana Del Rey is clichéd in a new way... We're so familiar with ourselves that Del Rey's kind of portraits can breed contempt, especially since they're less than wholesome, and moreover especially because they're unwholesome portrayals of female behavior. And in a culture that endlessly gazes at itself, sometimes she feels more like a symptom than an auteur.
t-4) Fader, "Lana Del Rey Is Anyone She Wants To Be"
Some criticize the way she seems to idealize powerlessness and servitude, while others appreciate her fluid embodiment of different identities, as well as her candor about both her desire and her weakness.
The New York Times, "Finding Her Future Looking To The Past"
Her songs and video clips demurely step into cultural minefields, exploring eroticism, mortality, power, submission, glamour, faith, pop-culture iconography and the meaning(s) of the American dream.
The Guardian, "Lana Del Rey: 'I Wish I Was Dead Already'"
7) Pitchfork, "Pretty When You Cry"
8) Flavorwire, "Lana Del Rey Is The Urban Outfitters of Music"
Until she produces her own original images, Lana Del Rey will remain a pop star for the Tumblr generation, for the girls who — season in and season out — look to Urban Outfitters to tell them who to be.
9) Grantland, "The United States of Lana"
10) Vox, "Here's why Lana Del Rey is so controversial"


Sign in to Gmail, and start a chat session with one or more of your contacts. When a contact sends you a YouTube link, you can watch the video right in Gmail's chat window.
When users instant message using Gmail's browser-based chat feature, they can now have multi-user chats, add emoticons to their messages, and view media like YouTube videos right in the chat window.


text:
Панделі Евандьєлі (1859 −1939) — албанський політичний діяч, двічі прем'єр-міністр Албанії. Він був першим ортодоксальним християнином, що очолив албанський уряд.

Народився у Корчі у 1859 році.

Примітки

Прем'єр-міністри Албанії

redpajama_set_name: RedPajamaWikipedia


text:
Mustangs stomp out Storm
Cathie Lawrence
Los Medanos College's Emma Spears takes a knock to her head to stop the ball in the 2-0 win over Napa.
Jesus Cano, [email protected]
A two-goal second half paved the way for the Los Medanos College Women's soccer team, as they picked up their third win of the season by defeating Napa Valley College 2-0.
Plenty of scoring opportunities were presented in the first half Tuesday afternoon at Mustang Stadium, but the Mustangs did not capitalize until the 53rd minute in the second half. Brianna Olmedo was on the attack but was tripped up in the penalty box by a Napa player, right in front of the eyes of center referee Francisco Franco. A penalty kick was awarded where defender Emma Spears put the ball away in the bottom right corner to edge LMC to a 1-0 lead.
"It felt good because we had just talked at halftime," Spears said. "After we get one goal our momentum and intensity increases and we could put some more in to take the game."
The intensity did pick up, as 10 minutes later Maggie Alvarez headed in a one time shot that seemed as it might give LMC another goal, but was saved on the line by a Napa defender.
LMC was however able to get an insurance goal late in the game. Adrianna Urrutia dribbled the ball in the Napa box and delivered a pass right next to Autumn Kish who rifled the ball in the bottom right corner to give LMC a comfortable 2-0 lead with seven minutes left in the match.
"She (Urrutia) played the perfect ball to my feet," Kish said. "I just got really lucky there."
In the first half, the Mustangs had multiple chances to put the game away from the beginning, but simply could not place the ball in the back of the net. Kish had two shots on goal, but was not successful. Brianna Olmedo had a one-on-one encounter with the goalkeeper, but took one extra step too much and ended up losing possession of the ball.
Goalkeeper Norma Jara preserved her clean sheet early in the game on a free kick shot that took an awkward bounce over the heads of many of her teammates, but managed to tip the ball out of bounds.
While LMC missed a lot of scoring plays, offside infraction were also a big issue in the first half as assistant referee Jose Chavez called three offside on scoring chances for the Mustangs.
"We've had trouble with the timing of our runs," head coach Zach Sullivan said. "We were the better team but struggled to score points."
The team did show improvement, as no offside infractions were called in the second half, where they were able to net in two goals, which took a lot of weight of the shoulder of team according to Sullivan.
"It was more of a relief," Sullivan added. "The more the game goes on the scarier it gets when you have not scored."
Many of the team members agreed that Napa was not a competitive opponent, and believed they should have won by a significantly greater margin.
"We won that game 2-0 but we should've easily put away at least four," Spears said. "They weren't that challenging and I think we outplayed them completely. We were definitely the better team."
LMC will be home once again on Friday Oct. 6 at 3 pm against Merritt College from Oakland.
For more information on the women's soccer team, please visit http://www.losmedanos.edu/Soccer/
Jesus Cano, Sports Writer
"There's only two people better than me, and I'm both of them!"
- LaVar Ball
Cathie Lawrence, Sports Photographer/ Staff Writer
I'm very excited to be working on the staff of the Experience. I continue to enjoy my position as a Sports Photographer and Staff Writer...


text:
4 ARTISTS YOU NEED TO KNOW BEFORE YOU HEAR THE NEW JONAS BROTHERS ALBUM: LINES, VINES, AND TRYING TIMES
While we still lack definite confirmation that the title of the Jonas Brothers' next album will be Lines, Vines And Trying Times, we can tell you (fairly) confidently about where the Jonas Brothers are coming from on their new album.
In a recent Billboard interview, Nick Jonas revealed that the boys drew heavy inspiration from Elvis Costello, The Zutons, Kings Of Leon and the reigning king of all entertainment, Neil Diamond. And while we can't tell you exactly what that sounds like through a Jonas filter, we can tell you a little bit about those bands. Consider it a Lines, Vines, And Trying Times primer for the uninitiated.
ELVIS COSTELLO: An intellectual's rockstar, he exploded into the world with My Aim Is True, a peppy, quirky, ska-infused downer that's widely considered one of the best debuts of all time. That's right. Peppy downer. Think about it. Elvis Costello videos to know: "Radio Radio" and the very Jonasy "Veronica."
THE ZUTONS: Punky British indie rockers who were way too catchy to stay indie long. Their first single, "Pressure Point," was featured in a Levi's commercial and on every nerd's mixtape in 2004. Zutons video to know: "Pressure Point."
KINGS OF LEON: Studs. Strike me dead for saying so, but this Tennessee-bred band of brothers makes the JoBros look like the nerdy nice guys who finish last in all the movies. Their music comes from the South and from the guts. Visceral rock that even Miley Cyrus and Taylor Swift can't resist. Kings of Leon video to know:. Kings of Leon video to know: "The Bucket."
NEIL DIAMOND: Need I say more? If you don't know Neil Diamond, you're still just waiting to be born. Here's a classic. Play it loud, my baby.


text:
Home » Newswire » First Nations, developer call for return and protection of sacred burial site
First Nations, developer call for return and protection of sacred burial site
by ahnationtalk on June 18, 2019138 Views
Credits: The Canadian Press
Indigenous leaders from across B.C. and parts of the U.S. gatherered on Friday, June 14 at a sacred burial site in Abbotsford, B.C., to call for its return to the Sema:th First Nation. Cold Water Ranch Developments, which owns the land, says has been unable to reach a settlement with the provincial government.
Use our news video on your website or broadcast: http://www.thecanadianpress.com/multi…
The Canadian Press is Canada's most trusted news leader in providing real-time, bilingual multimedia content for online, mobile and emerging platforms.
Business, Indigenous
http://nationtalk.ca/story/first-nations-developer-call-for-return-and-protection-of-sacred-burial-site


Some of them probably are, just like some of the football players that they cheer on are. Northeastern male cheerleader Carlos Munoz had this perspective on the association cheerleaders get with being gay, "You can look at cheerleading in two ways: as 'OK, I wear a skirt and I jump around saying cheers' or 'I'm a big man that picks up girls with one hand,'" he said. "It depends which way you put it. You can make it look good or you can make it look bad." His point of view is encouraging however there are still the people who look down on men who take up unconventional sports.
In an interesting twist of irony, many football players are required by their coaches to take ballet, conjuring up images of hulking muscle bound men in tutus. However, there is no assumption of homosexuality with these athletes. Whether they enjoy the dancing or not they are socially cleared of any homosexuality on the assumption that since they play football they are not gay and that they would never take up dancing unless forced to do so. One conclusion that can be drawn from this is that society on a whole does not think that "real" men would ever dance or cheerlead unless there was a way to simultaneously reaffirm their masculinity, in this case better play on the football field.


"Everyone these days wants to know how to slow down, but they want to know how to slow down really quickly. That's kind of the word that we live in now, a world stuck in fast forward. We are hurrying trough our lives, instead of actually living them. We are living the fast life instead of the good life. ... "


text:
Author: Sumit Kolhe
Home > Sumit Kolhe
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With headquarters in New Delhi, Delhi NCT, India, a few years back Intex was India's second-largest mobile phone company by sales. Intex was founded by Narendra Bansal back in 1996. Intex is an Indian smartphone, consumer electronics, and accessories manufacturer company. But in the last…
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text:
Q: Installing Virtualbox on Linux Mint 17 I am trying to install virtualbox 5 on mint.
Here is the output
$ sudo aptitude install virtualbox-5.0
The following NEW packages will be installed:
  virtualbox-5.0{b}
0 packages upgraded, 1 newly installed, 0 to remove and 1 not upgraded.
Need to get 64.2 MB of archives. After unpacking 161 MB will be used.
The following packages have unmet dependencies:
 virtualbox-5.0 : Depends: libdevmapper1.02.1 (>= 2:1.02.99) but 2:1.02.77-6ubuntu2 is installed.
                  Depends: libstdc++6 (>= 5.2) but 4.8.4-2ubuntu1~14.04.1 is installed.
                  Depends: libvpx2 (>= 1.4.0) which is a virtual package.
The following actions will resolve these dependencies:

     Keep the following packages at their current version:
1)     virtualbox-5.0 [Not Installed]

Accept this solution? [Y/n/q/?] Y
No packages will be installed, upgraded, or removed.
0 packages upgraded, 0 newly installed, 0 to remove and 1 not upgraded.
Need to get 0 B of archives. After unpacking 0 B will be used.

What is wrong with the dependencies?
Output of sudo apt-cache policy virtualbox virtualbox-5.0 libdevmapper1.02.1 libstdc++6 libvpx2:
sudo apt-cache policy virtualbox virtualbox-5.0 libdevmapper1.02.1 libstdc++6 libvpx2
virtualbox:
  Installed: (none)
  Candidate: 4.3.36-dfsg-1+deb8u1ubuntu1.14.04.1
  Version table:
     4.3.36-dfsg-1+deb8u1ubuntu1.14.04.1 0
        500 http://archive.ubuntu.com/ubuntu/ trusty-updates/multiverse amd64 Packages
        500 http://security.ubuntu.com/ubuntu/ trusty-security/multiverse amd64 Packages
     4.3.10-dfsg-1 0
        500 http://archive.ubuntu.com/ubuntu/ trusty/multiverse amd64 Packages
virtualbox-5.0:
  Installed: (none)
  Candidate: 5.0.16-105871~Ubuntu~wily
  Version table:
     5.0.16-105871~Ubuntu~wily 0
        700 http://download.virtualbox.org/virtualbox/debian/ wily/contrib amd64 Packages
     5.0.2-102096~Ubuntu~trusty 0
        700 http://extra.linuxmint.com/ qiana/main amd64 Packages
libdevmapper1.02.1:
  Installed: 2:1.02.77-6ubuntu2
  Candidate: 2:1.02.77-6ubuntu2
  Version table:
 *** 2:1.02.77-6ubuntu2 0
        500 http://archive.ubuntu.com/ubuntu/ trusty/main amd64 Packages
        100 /var/lib/dpkg/status
libstdc++6:
  Installed: 4.8.4-2ubuntu1~14.04.1
  Candidate: 4.8.4-2ubuntu1~14.04.1
  Version table:
 *** 4.8.4-2ubuntu1~14.04.1 0
        500 http://archive.ubuntu.com/ubuntu/ trusty-updates/main amd64 Packages
        100 /var/lib/dpkg/status
     4.8.2-19ubuntu1 0
        500 http://archive.ubuntu.com/ubuntu/ trusty/main amd64 Packages
libvpx2:
  Installed: (none)
  Candidate: (none)
  Version table:

Output of sudo apt-get install virtualbox-5:
sudo apt-get install virtualbox-5.0
Reading package lists... Done
Building dependency tree
Reading state information... Done
Some packages could not be installed. This may mean that you have
requested an impossible situation or if you are using the unstable
distribution that some required packages have not yet been created
or been moved out of Incoming.
The following information may help to resolve the situation:

The following packages have unmet dependencies:
 virtualbox-5.0 : Depends: libdevmapper1.02.1 (>= 2:1.02.99) but 2:1.02.77-6ubuntu2 is to be installed
                  Depends: libstdc++6 (>= 5.2) but 4.8.4-2ubuntu1~14.04.1 is to be installed
                  Depends: libvpx2 (>= 1.4.0) but it is not installable
E: Unable to correct problems, you have held broken packages.

A: To install virtualbox, follow these steps:

*

*sudo apt-get remove virtualbox virtualbox-4.* virtualbox-5.0

*Download virtualbox.

*sudo sh -c 'echo "deb http://download.virtualbox.org/virtualbox/debian [name of your distribution] contrib" >> /etc/apt/sources.list.d/virtualbox.list

*wget -q https://www.virtualbox.org/download/oracle_vbox.asc -O- | sudo apt-key add -
This answer was taken from here.
`

A: Try taking out the download.virtualbox.org sources. I'd stick with your distribution's packages unless you have compelling reasons to switch

A: Edit /etc/apt/sources.list file and add one of following line :
deb http://download.virtualbox.org/virtualbox/debian trusty contrib

Save and exit
update using :sudo apt-get update
According to virtualbox_wiki you need to Install dkms
sudo apt-get install dkms

Setup oracle public key:
wget http://download.virtualbox.org/virtualbox/debian/oracle_vbox.asc
sudo apt-key add oracle_vbox.asc

Install Oracle VirtualBox
sudo apt-get update
sudo apt-get install virtualbox-5.0

redpajama_set_name: RedPajamaStackExchange


text:
Football Narratives: Recovery & Mental Health
Brawn, P, COMBES, Helen and ELLIS, Naomi (2015) Football Narratives: Recovery & Mental Health. Journal of New Writing in Health and Social Care, 2 (1). pp. 30-46. ISSN 2057-2921
Text (Football Narratives: Recovery and Mental Health)
Journal_of_New_Writing_in_Health_and_Social_Care_(Vol_2_Issue_1).jsp
Official URL: http://www.staffs.ac.uk/research/centre_for_health...
Much has been written about the importance of physical activity on psychological wellbeing (Carless, 2008; Carless & Douglas, 2008; Carless and Douglas, 2010) however, little research has been published which explores the impact of participating in a competitive team sport such as football on mental health, well-being and recovery. This paper explores the stories of people who experience mental health difficulties who also compete in a mental health and well-being football league. Interviews were conducted with seven men between the ages of 25-63 each of whom were participating in a mental health and wellbeing football league. The interviews were subject to a narrative analysis. One major narrative theme and seven narrative subthemes were extrapolated. The major narrative was that participants rediscovered their sense of identity through their participation in the league. Participating in competition led to a greater sense of wellbeing amongst members. Within this major theme two subthemes emerged, which were connected to previous sporting histories and a sense of personal growth through connection and community. The narratives include stories describing processes, opportunities and outcomes through participation in the league occasionally connected to recovery. The football league and associated training can enable people to tell stories about themselves which develop and maintain a positive sense of self and identity within a united football community.
Previous Faculty of Health Sciences > Psychology, Sport and Exercise
Helen COMBES


text:
VK Cup 2017 - Wild Card Round 1
→ Custom contest rules
This round has an unusual addition to the rules. For each problem you must use distinct language. Thus, if you have a submission that has passed at least one test or is being tested at the moment, then for this task you can use only this language and can not use it for other tasks. More details can be read by the link. Different implementations of the same language are considered to be the same language (for example, in C++ you can only solve one problem, regardless of the exact chosen compiler).
Tutorial (ru)
D. Lie or Truth
Vasya has a sequence of cubes and exactly one integer is written on each cube. Vasya exhibited all his cubes in a row. So the sequence of numbers written on the cubes in the order from the left to the right equals to a1, a2, ..., an.
While Vasya was walking, his little brother Stepan played with Vasya's cubes and changed their order, so now the sequence of numbers written on the cubes became equal to b1, b2, ..., bn.
Stepan said that he swapped only cubes which where on the positions between l and r, inclusive, and did not remove or add any other cubes (i. e. he said that he reordered cubes between positions l and r, inclusive, in some way).
Your task is to determine if it is possible that Stepan said the truth, or it is guaranteed that Stepan deceived his brother.
The first line contains three integers n, l, r (1 ≤ n ≤ 105, 1 ≤ l ≤ r ≤ n) — the number of Vasya's cubes and the positions told by Stepan.
The second line contains the sequence a1, a2, ..., an (1 ≤ ai ≤ n) — the sequence of integers written on cubes in the Vasya's order.
The third line contains the sequence b1, b2, ..., bn (1 ≤ bi ≤ n) — the sequence of integers written on cubes after Stepan rearranged their order.
It is guaranteed that Stepan did not remove or add other cubes, he only rearranged Vasya's cubes.
Print "LIE" (without quotes) if it is guaranteed that Stepan deceived his brother. In the other case, print "TRUTH" (without quotes).
In the first example there is a situation when Stepan said the truth. Initially the sequence of integers on the cubes was equal to [3, 4, 2, 3, 1]. Stepan could at first swap cubes on positions 2 and 3 (after that the sequence of integers on cubes became equal to [3, 2, 4, 3, 1]), and then swap cubes in positions 3 and 4 (after that the sequence of integers on cubes became equal to [3, 2, 3, 4, 1]).
In the second example it is not possible that Stepan said truth because he said that he swapped cubes only between positions 1 and 2, but we can see that it is guaranteed that he changed the position of the cube which was on the position 3 at first. So it is guaranteed that Stepan deceived his brother.
In the third example for any values l and r there is a situation when Stepan said the truth.


text:
Three persons were killed after being hit by a moving train near a level crossing in Ganjam district
At least three persons were killed after being hit by a moving train near a level crossing in Ganjam district late on Sunday night.
The mishap occurred near Haldiapadar level crossing of Berhampur between 11.30 pm and 12 am last night. The identity of deceased was not ascertained immediately. According to some locals, around 7-8 people, all stated to be labourers, were crossing the railway track without aware of the incoming Visakhapatnam-Tatanagar Express on the route.
The speeding train ran over at least three of them while others miraculously escaped the mishap. Upon being informed by the locals, the Government Railway Police (GRP) rushed to the spot today morning and recovered the mutilated bodies from the railway tracks. A probe has been launched into the incident.


Been used a bit more in anger this week.
Was actually a bit of a life saver as we turned up to do a job that had been booked in a while, with parking permits sorted and everything, only to discover that the road was closed and the closest we could park was about 200m away!
Fortunately it wasn't a huge job, just a smallish cherry to fell and a fig to reduce, and with the aid of the arbor trolley we managed to get it all to the truck in 4 trips, 3 for brash and one for logs.


The design for the Carl H. Lindner College of Business is based on the ambition to create a learning environment which encourages interaction, accommodates future change, and enhances occupants' well-being. The design resonates with and reinforces the University of Cincinnati's masterplan, and aims to reaffirm the campus as an active pedestrian precinct.
The Carl H. Lindner College of Business is designed as a flexible building that can accommodate various pedagogies, increased occupancies, and transition from closed to open offices.
Classrooms will accommodate different kinds of instruction: generous floor areas, height-adjustable furnishings, and audio-visual capacity allow for traditional front-facing lectures as well as group work and activated learning. The auditoria on the ground floor similarly accommodate for different types of pedagogy and program: the flat floor in the lecture hall allows for multiple use and seating configurations, while the tiered platforms and furniture in the auditorium can facilitate traditional lectures and group work simultaneously.
Egress, toilet capacity, MEP systems, partition walls, and façade design are all designed to allow for transitioning much of the closed offices into a more open office environment in the future, which will allow for improved energy savings, increased capacity, and more synergy and interaction.
The design targets LEED v4 Gold Certification and is designed to reduce the energy consumption by 24% less than a similar standard building (ENERGY STAR Medium Property). The façade design supports the building's energy performance and allows sufficient daylight into both offices and classrooms, contributing to improved well-being and learning abilities.


text:
\section*{Introduction}

Drozd \cite{Drozd}
showed that finite dimensional algebras $A$ over an algebraically closed
field $K$ are either of finite representation type, or of tame representation type
or of wild representation type.
Here, an algebra is of finite representation type if there are only a finite number
of isomorphism classes of indecomposable modules. An algebra is of
wild representation type if all module categories are full subcategories
of the module category of the algebra. An algebra is of tame representation
type if for every positive integer $d$ there are $A-K[X]$-bimodules $M_i(d)$
for $i\in\{1,\dots,n_d\}$ so that $M_i(d)$ is free as $K[X]$-module and so that
all but finite number of isomorphism classes of indecomposable
$d$-dimensional $A$-modules are isomorphic to
$M_i(d)\otimes_{K[X]}K[X]/(X-\lambda)$ for $\lambda\in K$ and $i\in\{1,\dots,n_d\}$.
If $n_d$ is minimal with respect to $d$ satisfying this property,
$A$ is of polynomial growth if there is an integer $m$ so that
$\lim_{d\rightarrow\infty}\frac{n_d}{d^m}=0$, and $A$ is domestic if there is an integer
$m$ so that $n_d\leq d\cdot m$ for all $d$.

Bocian, Holm and Skowro\'nski
\cite{BocianHolmSkowronski2004,BocianHolmSkowronski2007,HolmSkowronski2006}
classified all weakly symmetric
algebras of domestic type up to derived equivalence and Bia\l kowski,
Holm and Skowro\'{n}ski \cite{BialkowskiHolmSkowronski2003a,
BialkowskiHolmSkowronski2003b,HolmSkowronski2009}
did the same thing for all weakly symmetric
 non-domestic
algebras of polynomial growth. The result is a finite number of
families of algebras given by quivers and relations, depending on
certain parameters. In the case  of domestic algebras the
classification is complete, whereas in case of weakly symmetric
algebras of polynomial growth it is not known if certain choices
of parameters may lead to the same derived equivalence class.
This question if certain parameters lead to derived equivalent algebras
will be called the scalar problem in the sequel.

The main result of this paper is that the derived equivalence
classification of weakly symmetric algebras of polynomial growth
coincides (up to some scalar problems) with the
classification up to stable equivalences of
Morita type as given in
\cite{BocianHolmSkowronski2004,BocianHolmSkowronski2007,
HolmSkowronski2006} and \cite{BialkowskiHolmSkowronski2003a,
BialkowskiHolmSkowronski2003b,HolmSkowronski2009}. As a
consequence, the Auslander-Reiten conjecture holds for a stable
equivalence of Morita type between two weakly symmetric algebras of
polynomial growth.

For weakly symmetric algebras of domestic type we can show more. We prove that
the derived equivalence classification coincides with the classification up
to stable equivalences (no scalar problem occurs in this case) and that  the Auslander-Reiten conjecture holds
for stable equivalences between weakly symmetric algebras of domestic types.

As mentioned above the classification
of symmetric algebras of polynomial growth up to derived equivalence
leaves open if for certain choices of parameters the
algebras are derived equivalent or not.
The same questions concerning
the same choices of parameters remain open also for the
classification up to stable equivalences of Morita type.

Moreover, using our result \cite{ZhouZimmermannTameBlocks}, we show that
for tame symmetric algebras with periodic modules,
the derived equivalence classification coincides with the classification up
to stable equivalence of Morita type, up to some scalar problem, and that  the Auslander-Reiten conjecture holds
for stable equivalences between tame symmetric algebras with periodic modules.

The paper is organised as follows. In Section~\ref{stableinvsect}
we recall basic definitions and recall
the stable invariants we use in the sequel. In
Section~\ref{weaklysymmdomesticsection} we prove the
main result Theorem~\ref{domesticmaintheorem}
for weakly symmetric algebras of domestic representation type and in
Section~\ref{polysection} we show the main result
Theorem~\ref{NonDomesticClassification}
for stable equivalences of Morita type for weakly
symmetric algebras of polynomial
growth.

As soon as we use statements on representation type we shall always assume that
$K$ is an algebraically closed field.

\section{Stable categories; definitions, notations and invariants}

\label{stableinvsect}

\subsection{The definitions}

Let $K$ be a field and let $A$ be a finite dimensional
$K$-algebra. The stable category $A-\underline{mod}$ has
as objects all finite dimensional $A$-modules and  morphisms
from na $A$-module $X$ to another $A$-module $Y$ are the
equivalence classes
of $A$-linear homomorphisms $Hom_A(X,Y)$, where
two morphisms of $A$-modules are declared to be equivalent if
their difference factors through a projective $A$-module.

Two algebras are called {\em stably equivalent}
if $A-\underline{mod}\simeq B-\underline{mod}$ as additive categories.
Let $M$ be an $A-B$-bimodule and let $N$ be a
$B-A$-bimodule. Following Brou\'e the couple $(M,N)$
defines a {\em stable equivalence of Morita type} if
\begin{itemize}
\item
$M$ is projective as $A$-module and is projective as $B$-module
\item
$N$ is projective as $B$-module and is projective as $A$-module
\item
$M\otimes_BN\simeq A\oplus P$ as $A-A$-bimodules, where $P$ is a projective
$A-A$-bimodule.
\item
$N\otimes_AM\simeq B\oplus Q$ as $B-B$-bimodules, where $Q$ is a projective
$B-B$-bimodule.
\end{itemize}
Of course, if $(M,N)$ defines a stable equivalence of Morita type,
then $M\otimes_B-:B-\underline{mod}\simeq A-\underline{mod}$ is an equivalence. However,
it is known that there are algebras which are stably equivalent but for which there
is no stable equivalence of Morita type.

\subsection{Invariants under stable equivalences}

Auslander and Reiten conjectured that if $A$ and $B$ are stably equivalent,
then the number of isomorphism classes of
 non-projective simple $A$-modules is equal to the number of
isomorphism classes of  non-projective simple $B$-modules. The conjecture
is open in general, but we shall use in this article the following positive
result \cite{Pogorzaly} of
Pogorza\l y: If $A$ is stably equivalent to $B$ and if $A$ is selfinjective special biserial that is not Nakayama,
then $B$ is self-injective special biserial as well and the Auslander-Reiten
conjecture holds in this case.

Krause and Zwara \cite{KrauseZwara} showed that stable equivalences preserve the representation type.
Suppose $A$ and $B$ are two finite dimensional stably equivalent
$K$-algebras. If $A$ is of domestic representation
type (respectively of polynomial growth, respectively of tame
representation type), then so is $B$.

Reiten \cite{Reiten} show that being selfinjective is almost
invariant under stable equivalences. More precisely, let $A$ be a selfinjective
finite dimensional $K$-algebra which is stably equivalent
to a  finite dimensional indecomposable $K$-algebra $B$.
Then $B$ is either selfinjective or a radical squared zero Nakayama algebra.

Finally, if $A$ and $B$ are stably equivalent finite dimensional
algebras with Loewy length at least three,
then the stable Auslander-Reiten quivers are isomorphic as translation quivers.

\subsection{Invariants under stable equivalence of Morita type}

Much more is known for stable equivalences of Morita type. We shall cite only
those properties used in the sequel.

Xi has shown \cite{Xi2008} that if two finite dimensional $K$-algebras
$A$ and $B$ are stably
equivalent of Morita type, then the absolute value of the Cartan determinants
are equal. Even better, the elementary divisors, including their multiplies,
of the Cartan matrices different from $\pm 1$ coincide for $A$ and for $B$.

Liu showed \cite{Liu2008}
that if $A$ is an indecomposable finite dimensional $K$-algebra,
and $B$ is a finite dimensional $K$-algebra stably
equivalent of Morita type to $A$, then $B$ is indecomposable as well.

If $K$ is an algebraically closed field of characteristic $p>0$ and $A$ a finite dimensional
$K$-algebra. Define $[A,A]$ the $K$-space generated by all elements $ab-ba\in A$
for all $a,b\in A$. Then, $$\{x\in A\;|\;x^{p^n}\in[A,A]\}=:T_n(A)$$
is a $K$-subspace of $A$ containing $[A,A]$.
Liu and the authors have shown \cite{LZZ}
that
$$dim_K(T_n(A)/[A,A])=dim_K(T_n(B)/[B,B])$$
for all $n\in{\mathbb N}$. If $A$ and $B$ are symmetric and stably equivalent of
Morita type, then taking the orthogonal space with respect to the symmetrising
form, observing that $[A,A]^\perp=Z(A)$, Liu, K\"onig and the first author have
shown \cite{KLZ} that $$Z(A)/T_n(A)^\perp\simeq Z(B)/T_n(B)^\perp$$
as rings.

\section{Weakly symmetric  algebras of domestic representation type }

\label{weaklysymmdomesticsection}

Selfinjective algebras of domestic type split
into two subclasses: standard algebras and non-standard ones.

A selfinjective algebra of tame representation type is called {\em standard} if its
basic algebra admit simply connected Galois coverings. A
selfinjective algebra of tame representation type is called {\em nonstandard}
if it is not standard.

\subsection{Standard domestic weakly symmetric algebras}

\begin{Thm}\cite[Theorem 1]{BocianHolmSkowronski2004}
\label{DomesticStandardSingular}
For an algebra $A$ the following
statements are equivalent:
\begin{itemize}
\item[(1)] $A$ is representation-infinite domestic weakly
symmetric algebras having simply connected Galois coverings and
the Cartan matrix $C_A$ is singular.

\item[(2)] $A$ is derived equivalent to the trivial extension
$T(C)$ of a canonical algebra $C$ of Euclidean type.

\item[(3)] $A$ is stably equivalent to the trivial extension
$T(C)$ of a canonical algebra $C$ of Euclidean type.
\end{itemize}
Moreover, the trivial extensions $T(C)$ and $T(C')$ of two
canonical algebras $C$ and $C'$ of Euclidean type are derived
equivalent (respectively, stably equivalent) if and only if the
algebras $C$ and $C'$ are isomorphic.
\end{Thm}

From this theorem, we know that   a standard weakly symmetric
algebras of domestic representation type with {\em singular Cartan
matrice} is symmetric.

\begin{Cor} The class of indecomposable standard weakly
symmetric  algebras of domestic representation type with singular
Cartan matrices is closed under stable equivalences.

\end{Cor}

\begin{Proof}
 In fact, let $A$ be an indecomposable algebra stably
 equivalent to an indecomposable standard weakly
symmetric  algebras of domestic representation type.
Then by the preceding theorem, $A$ is stably equivalent to
the trivial extension
$T(C)$ of a canonical algebra $C$ of Euclidean type.
Again by the preceding theorem, $A$ is standard weakly
symmetric  algebras of domestic representation type.
\end{Proof}

For standard weakly symmetric algebras of domestic
representation type with {\em nonsingular Cartan matrices,}  we have the
following derived norm forms.
$$\unitlength0.6cm
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 \put(9.4,3){\circle{2.0}}
\put(6.8,3){\circle{2.0}} \put(7.85,2.9){\vector(0,1){0.3}}
\put(8,2.9){$\bullet$} \put(8.35,2.9){\vector(0,1){0.3}}
\put(4.8,2.9){$\beta$} \put(11,2.9){$\alpha$}

 \put(0,1.9){$A(\lambda)$}
\put(0,1){$\lambda\in K\backslash \{0\}$}

\put(4.8,0.5){$\alpha^2=0, \beta^2=0,
\alpha\beta=\lambda\beta\alpha$}
\end{picture}$$

$$\unitlength0.5cm
\begin{picture}(30,15)
 \put(18.1, 11.1){\vector(1,1){1.8}}

\put(16.5, 11.5){$\beta_1$}

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 \put(14.5, 13){$\beta_2$}

\put(16, 13){\vector(-2,1){1.8}}

\put(13, 13.2){$\beta_3$}

\put(14, 14){\vector(-1,0){1.8}}

\put(11, 13.0){$\beta_4$}

\put(12, 14){\vector(-2,-1){1.8}}

\put(10.7, 8.8){$\beta_{q-3}$}

\put(10, 9){\vector(2,-1){1.8}}

\put(12.2, 8.5){$\beta_{q-2}$}

\put(12, 8){\vector(1,0){1.8}}

\put(14, 8.8){$\beta_{q-1}$}

\put(14, 8){\vector(2,1){1.8}}

\put(16.4, 10.3){$\beta_{q}$}

\put(16, 9){\vector(1,1){1.8}}

\put(19.2, 11.5){$\alpha_{1}$}

\put(20, 9){\vector(-1,1){1.8}}

\put(21, 13){$\alpha_{2}$}

\put(22, 8){\vector(-2,1){1.8}}

\put(22.5, 13.5){$\alpha_{3}$}

\put(24, 8){\vector(-1,0){1.8}}

\put(24.5, 13){$\alpha_{4}$}

\put(26, 9){\vector(-2,-1){1.8}}

\put(18.8, 10.3){$\alpha_{p}$}

\put(20.7, 8.8){$\alpha_{p-1}$}

\put(22.4, 8.5){$\alpha_{p-2}$}

\put(24.1, 8.8){$\alpha_{p-3}$}

\put(20, 13){\vector(2,1){1.8}}

\put(22, 14){\vector(1,0){1.8}}

\put(24, 14){\vector(2,-1){1.8}}

\multiput(9, 11)(0.1,0.2){10}{\circle*{0.01}}

\multiput(9, 11)(0.1,-0.2){10}{\circle*{0.01}}

\multiput(27, 11)(-0.1,0.2){10}{\circle*{0.01}}

\multiput(27, 11)(-0.1,-0.2){10}{\circle*{0.01}}

 \put(1, 12){$A(p, q)$}

 \put(1, 10.5){$1\leq p\leq q$}
 \put(1, 9){$p+q\geq 3$}

 \put(9, 6){$\alpha_1\alpha_2\cdots \alpha_p\beta_1\beta_2\cdots
 \beta_q=\beta_1\beta_2\cdots \beta_q\alpha_1\alpha_2\cdots \alpha_p$}

\put(9, 4.5){$ \alpha_p\alpha_1=0, \beta_q\beta_1=0$}

\put(9, 3){$\alpha_i\alpha_{i+1}\cdots \alpha_p\beta_1\cdots
\beta_q  \alpha_1 \cdots \alpha_{i-1}\alpha_i=0, 2\leq i\leq p$}

\put(9, 1.5){$\beta_j\beta_{j+1}\cdots \beta_q\alpha_1\cdots
\alpha_p \beta_1 \cdots \beta_{j-1}\beta_j=0, 2\leq j\leq q$}

\end{picture}$$

$$\unitlength0.5cm
\begin{picture}(30,11)

\multiput(9, 7)(0.1,0.2){10}{\circle*{0.01}}

\multiput(9, 7)(0.1,-0.2){10}{\circle*{0.01}}

\put(16.5, 7.5){$\beta_1$}

 \put(18, 7){\vector(-1,1){1.8}}

 \put(14.5, 9){$\beta_2$}

\put(16, 9){\vector(-2,1){1.8}}

\put(13, 9.2){$\beta_3$}

\put(14, 10){\vector(-1,0){1.8}}

\put(11, 9.0){$\beta_4$}

\put(12, 10){\vector(-2,-1){1.8}}

\put(10.7, 4.8){$\beta_{n-3}$}

\put(10, 5){\vector(2,-1){1.8}}

\put(12.2, 4.5){$\beta_{n-2}$}

\put(12, 4){\vector(1,0){1.8}}

\put(14, 4.8){$\beta_{n-1}$}

\put(14, 4){\vector(2,1){1.8}}

\put(16.4, 6.3){$\beta_{n}$}

\put(16, 5){\vector(1,1){1.8}}

\put(19.2, 7){\circle{2}}

\put(19, 8){\vector(-1,0){0.01}}

\put(20.5, 7){$\alpha$}

 \put(1, 8){$\Lambda(n)$}

 \put(1, 6.5){$n\geq 2$}

 \put(9, 2){$\alpha^2=(\beta_1\beta_2\cdots\beta_n)^2, \alpha\beta_1=0, \beta_n\alpha=0$}

\put(9, 0.5){$\beta_j\beta_{j+1}\cdots \beta_n\beta_1\cdots
\beta_n \beta_1 \cdots \beta_{j-1}\beta_j=0, 2\leq j\leq n$}

\end{picture}$$

$$\unitlength0.5cm
\begin{picture}(30,15)

\multiput(9, 11)(0.1,0.2){10}{\circle*{0.01}}

\multiput(9, 11)(0.1,-0.2){10}{\circle*{0.01}}

\put(16.5, 11.5){$\beta_1$}

 \put(18, 11){\vector(-1,1){1.8}}

 \put(14.5, 13){$\beta_2$}

\put(16, 13){\vector(-2,1){1.8}}

\put(13, 13.2){$\beta_3$}

\put(14, 14){\vector(-1,0){1.8}}

\put(11, 13.0){$\beta_4$}

\put(12, 14){\vector(-2,-1){1.8}}

\put(10.7, 8.8){$\beta_{n-3}$}

\put(10, 9){\vector(2,-1){1.8}}

\put(12.2, 8.5){$\beta_{n-2}$}

\put(12, 8){\vector(1,0){1.8}}

\put(14, 8.8){$\beta_{n-1}$}

\put(14, 8){\vector(2,1){1.8}}

\put(16.4, 10.3){$\beta_{n}$}

\put(16, 9){\vector(1,1){1.8}}

\put(18.2, 11.3){\vector(1,2){1.2}}

\put(19.8, 13.4){\vector(-1,-2){1.2}}

\put(18.4, 10.9){\vector(1,-2){1.2}}

\put(19.2, 8.4){\vector(-1,2){1.2}}

\put(19.4, 9.7){$\gamma_1$}

\put(19.4, 11.7){$\alpha_2$}

\put(18, 9){$\gamma_2$}

\put(18, 12.5){$\alpha_1$}

 \put(1, 12){$\Gamma(n)$}

 \put(1, 10.5){$n\geq 1$}

 \put(9, 6){$\alpha_1\alpha_2=(\beta_1\beta_2\cdots\beta_n)^2=\gamma_1\gamma_2,$}

\put(9, 4.5){$\alpha_2\beta_1=0, \gamma_2\beta_1=0,
\beta_n\alpha_1=0$}

\put(9, 3){$\beta_n\gamma_1=0, \alpha_2\gamma_1=0,
\gamma_2\alpha_1=0$}

\put(9, 1.5){$\beta_j\beta_{j+1}\cdots \beta_n\beta_1\cdots
\beta_n \beta_1 \cdots \beta_{j-1}\beta_j=0, 2\leq j\leq n$}

\end{picture}$$

\begin{Thm}\cite[Theorem
2]{BocianHolmSkowronski2004}\label{DomesticStandardNonSingular}
For a domestic standard self-injective algebra $A$,    the
following statements are equivalent:
\begin{itemize}
\item[(1)]$A$ is weakly symmetric and the Cartan matrix $C_A$ is
nonsingular.

\item[(2)] $A$ is derived equivalent to an algebra of the form
$A(\lambda), A(p, q), \Lambda(n),  \Gamma(n)$.

\item[(3)] $A$ is stably equivalent to an algebra of the form
$A(\lambda), A(p, q), \Lambda(n), \Gamma(n)$.

\end{itemize}

Moreover, two algebras of the forms $A(\lambda), A(p, q),
\Lambda(n)$ or $\Gamma(n)$ are derived equivalent (respectively,
stably equivalent) if and only if they are isomorphic.
\end{Thm}
All these algebras except
$$A(\lambda)=k\langle X, Y\rangle/(X^2, Y^2, XY-\lambda YX), \lambda\not\in\{ 0, 1\}$$
are symmetric. Remark that except $\Gamma(n)$, all algebras are
special biserial algebras.

Since the class of indecomposable standard weakly
symmetric  algebras of domestic representation type is closed under stable equivalences, we have the following

\begin{Cor} \label{standardweaklysymmetricstablyequiv}
Two  standard weakly symmetric algebras of
domestic representation type are stably equivalent if and only
if they are derived equivalent.
\end{Cor}

\subsection{Non standard domestic weakly symmetric algebras}

Nonstandard self-injective algebras of domestic
representation type are classified up to derived and stable
equivalences in \cite{BocianHolmSkowronski2007}.

\begin{Thm}\cite[Theorem 1]{BocianHolmSkowronski2007}\label{DomesticNonStandard}
Any nonstandard representation-infinite
selfinjective algebra of domestic type is derived equivalent
(resp. stably equivalent) to an algebra $\Omega(n)$ with $n\geq
1$. Moreover, two algebras $\Omega(n)$ and $\Omega(m)$ are derived
equivalent (respectively, stably equivalent) if and only if $m =
n$.\end{Thm} The quiver with relations of $\Omega(n)$ is as
follows.

$$\unitlength0.5cm
\begin{picture}(30,11)

\multiput(9, 7)(0.1,0.2){10}{\circle*{0.01}}

\multiput(9, 7)(0.1,-0.2){10}{\circle*{0.01}}

\put(16.5, 7.5){$\beta_1$}

 \put(18, 7){\vector(-1,1){1.8}}

 \put(14.5, 9){$\beta_2$}

\put(16, 9){\vector(-2,1){1.8}}

\put(13, 9.2){$\beta_3$}

\put(14, 10){\vector(-1,0){1.8}}

\put(11, 9.0){$\beta_4$}

\put(12, 10){\vector(-2,-1){1.8}}

\put(10.7, 4.8){$\beta_{n-3}$}

\put(10, 5){\vector(2,-1){1.8}}

\put(12.2, 4.5){$\beta_{n-2}$}

\put(12, 4){\vector(1,0){1.8}}

\put(14, 4.8){$\beta_{n-1}$}

\put(14, 4){\vector(2,1){1.8}}

\put(16.4, 6.3){$\beta_{n}$}

\put(16, 5){\vector(1,1){1.8}}

\put(19.2, 7){\circle{2}}

\put(19, 8){\vector(-1,0){0.01}}

\put(20.5, 7){$\alpha$}

 \put(1, 8){$\Omega(n)$}

 \put(1, 6.5){$n\geq 1$}

 \put(3, 2){$\alpha^2=\alpha\beta_1\beta_2\cdots\beta_n,
 \alpha\beta_1\beta_2\cdots\beta_n+\beta_1\beta_2\cdots\beta_n\alpha=0, $}

\put(3, 0.5){$\beta_n\beta_{1}=0, \beta_j\beta_{j+1}\cdots
\beta_n\beta_1\cdots \beta_n \beta_1 \cdots \beta_{j-1}\beta_j=0,
2\leq j\leq n$}

\end{picture}$$

Remark that the algebra $\Omega(n)$ is  always weakly symmetric,
but it is symmetric only when the characteristic of the base field
is $2$. Note that $\Omega(n)$ is not special biserial.

\subsection{Domestic standard type   versus domestic non standard type}

\begin{Thm}\cite[Theorem 1.1]{HolmSkowronski2006}\label{SymmetricDomesticStandardVsNonStandard} A
standard symmetric algebra of domestic representation type cannot
be derived equivalent to a non-standard symmetric one.

\end{Thm}
 In course of the proof of the above theorem,  one needs to compare the algebras
$\Omega(n)$ with $A(1, n)$ in case the characteristic of $K$ is
$2$. It is proved in \cite{HolmSkowronski2006} that the dimensions
of   the centre modulo the first K\"{u}lshammer ideal,  for
$\Omega(n)$ and  $A(1, n)$, are different, but this dimension is
in fact invariant under stable equivalences of Morita type (\cite{LZZ}). So  a
standard symmetric algebra of domestic representation type cannot
be stably equivalent of Morita type to a non-standard symmetric
one. In fact, we can even prove more.

\begin{Lem}\label{WeaklySymmetricDomesticStandardVsNonStandard}
 A standard weakly symmetric algebra of domestic type
cannot be stably equivalent to a nonstandard one.
\end{Lem}

\begin{Proof}
Stably equivalent algebras of Loewy length at least $3$ have
isomorphic stable Auslander-Reiten quivers (\cite[Corollary X.
1.9]{AuslanderReiten}). Notice that the stable Auslander-Reiten
quiver of $A(\lambda)$ for $\lambda\neq 0$ consists of an
Euclidean component of type $\mathbb{Z}\tilde{A}_{1}$ and a
$\mathbb{P}_1(K)$-family of homogenous tubes. Since all algebras
in the above list of Theorems~\ref{DomesticStandardSingular},
\ref{DomesticStandardNonSingular}, \ref{DomesticNonStandard} are
of Loewy length at least $3$,
 by comparing the shapes of the stable
Auslander-Reiten quivers  as in \cite[Section
4]{HolmSkowronski2006}, we know that a standard weakly symmetric
algebra of domestic type $A$ in the list of
Theorems~\ref{DomesticStandardSingular},
\ref{DomesticStandardNonSingular} is stably equivalent to a
nonstandard one $B=\Omega(n)$ only when $A=A(1, n)$ and
$B=\Omega(n)$ for $n\geq 1$. However, notice that $A(1, n)$ is
special biserial and $\Omega(n)$ is not. By the result of
Pogorza\l y \cite[Theorem 7.3]{Pogorzaly}, they can never be
stably equivalent.
\end{Proof}

\subsection{Stable equivalence classification of domestic weakly symmetric algebras}

 \begin{Prop} \label{ClassificationSymmetricDomestic}
 \begin{enumerate}
 \item
 Two symmetric  algebras of domestic representation type   are
derived equivalent
 if and only if they are stably equivalent of Morita type if and only
 if they are stably equivalent.

 \item The class of    symmetric
algebras  of domestic representation type  is closed under stable
equivalences, hence is closed under stably equivalences of Morita type and
derived equivalences.
\end{enumerate}
\end{Prop}

\begin{Proof}
Corollary~\ref{standardweaklysymmetricstablyequiv} shows that for
standard weakly symmetric algebras the notions of derived
equivalence and of stable equivalence are the same.
Lemma~\ref{WeaklySymmetricDomesticStandardVsNonStandard} shows
that standard weakly symmetric algebras cannot be stably
equivalent to non standard weakly symmetric algebras.
Theorem~\ref{DomesticNonStandard} shows that for non standard
domestic weakly symmetric algebras derived and stable equivalence
are the same notion. This proves (1).

\medskip

Now let $A$ be an indecomposable algebra stably equivalent to an
indecomposable  symmetric algebra $B$ of domestic representation
type in the list of algebras in
Theorems~\ref{DomesticStandardSingular},
\ref{DomesticStandardNonSingular} and
 \ref{DomesticNonStandard}. Then by a result of Idun Reiten
\cite[Theorem 2.6]{Reiten} (see also
\cite[Theorem]{LiuPaquette}), $A$ itself is also self-injective or
$A$ is isomorphic to a radical square zero algebra  whose quiver
is of type $A$ with linear orientation. But the latter algebra has
finite representation type, and thus cannot be stably equivalent
to a representation-infinite algebra. We infer that $A$ is
self-injective and now  by \cite[Proposition
5.2]{BocianHolmSkowronski2004}, $A$ is weakly symmetric. By
\cite[Corollary 2 and the discussions afterwards]{KrauseZwara},
$A$ is of domestic type. We thus proved that $A$ is weakly
symmetric of domestic type.

To finish the proof of the second assertion, one needs to show that
$A$ is symmetric and thus to  exclude the cases where $A$ is
stably equivalent to $A(\lambda)$ with $\lambda\in K\backslash\{0,
1\}$ or to $\Omega(n)$ with $n\geq 1$ in case of $char K\neq 2$,
since these are the only cases where non symmetric algebras may
occur.

If $A$ is stably equivalent to $A(\lambda)$ with $\lambda\in
K\backslash\{0, 1\}$, then by  \cite[Theorem 7.3]{Pogorzaly}, $B$
is special biserial. Since the Auslander-Reiten conjecture is
proved for self-injective special biserial algebras (\cite[Theorem
0.1]{Pogorzaly}), $B$ is also local and is thus necessary
isomorphic to $A(1)$. However,  $A(1)$  is never stably
equivalent to $A(\lambda)$ with $\lambda\neq 1$ by an unpublished
result of Rickard.

If $char(K)\neq 2$ and   $A$ is stably equivalent to $A=\Omega(n)$
with $n\geq 1$,  then  by
Lemma~\ref{WeaklySymmetricDomesticStandardVsNonStandard}, $B$ is
necessarily nonstandard and $B=\Omega(n)$ for some $n$. But the
fact that $B$ is symmetric implies that  $char(K)=2$ which is a
contradiction.
\end{Proof}

In fact  we can extend (at least partially) the above result to
all weakly symmetric algebras of domestic representation type.

\begin{Thm}\label{domesticmaintheorem}
\begin{enumerate}
\item Two weakly symmetric  algebras of domestic
representation type are derived equivalent if and only if they are
stably equivalent of Morita type if and only if they are stably
equivalent.

\item  The class of   weakly symmetric
algebras  of standard domestic representation type  is closed
under stable equivalences, hence under stably equivalences of
Morita type, and derived equivalences.
\end{enumerate}
\end{Thm}

\begin{Proof} Since by
Lemma~\ref{WeaklySymmetricDomesticStandardVsNonStandard},
a standard weakly symmetric algebra of domestic type
cannot be stably equivalent to a nonstandard one,
Theorems~\ref{DomesticStandardSingular},
~\ref{DomesticStandardNonSingular} and ~\ref{DomesticNonStandard}
imply the first assertion.

For the second assertion, let $A$ be an indecomposable algebra
stably equivalent to a weakly symmetric standard algebra $B$ of
domestic representation type in the list of algebras in
Theorems~\ref{DomesticStandardSingular} and
~\ref{DomesticStandardNonSingular}. By \cite[Corollary 2 and the
discussions afterwards]{KrauseZwara}, $A$ is of domestic type.  As
in the proof of Proposition~\ref{ClassificationSymmetricDomestic},
$A$ is still self-injective. Since by
Proposition~\ref{ClassificationSymmetricDomestic}(2), the class of
symmetric algebras of domestic type is closed under stable
equivalences, one can assume that $B$ is not symmetric.  Now
$B=A(\lambda)$ with $\lambda\not\in\{ 0, 1\}$. Since
$B=A(\lambda)$ is a local special biserial algebra, so is $A$ by
\cite[Theorems 0.1 and 7.3]{Pogorzaly}. The algebra $A$ is thus
weakly symmetric.

 \end{Proof}

\begin{Rem}
\begin{enumerate}
\item If the class of  self-injective
nonstandard algebras of domestic type is closed under under stable
equivalences (hence under stably equivalences of Morita type  and
derived equivalences),  so is    the class of weakly symmetric
algebras of   domestic representation type.

\item The second assertion of the above theorem answers a question
of \cite[Page 47]{BocianHolmSkowronski2004}.

\item If we have a complete Morita equivalence classification of all
self-injective algebras of domestic type, the classification up to
stable equivalence (of Morita type) might be feasible as well.
\end{enumerate}
\end{Rem}

As a consequence,   the Auslander-Reiten conjecture holds for this
class of algebras.

\begin{Cor}
\begin{enumerate}
\item Let $A$ be an indecomposable algebra stably equivalent
to an indecomposable   symmetric algebra $B$ of
domestic type. Then $A$ has the same number of simple modules as
$B$.

\item Let $A$ be an indecomposable algebra stably equivalent
to an indecomposable weakly symmetric standard algebra $B$ of
domestic type. Then $A$ has the same number of simple modules as
$B$.
\end{enumerate}
\end{Cor}

\begin{Rem} \rm If the class of  self-injective
nonstandard algebras of domestic type is closed under under stable
equivalences, then Auslander-Reiten conjecture holds for a stable
equivalence between two weakly symmetric algebras of domestic
type.
\end{Rem}

\section{Non-domestic self-injective  algebras of polynomial growth}

\label{polysection}

The   derived equivalence classification of the standard (resp.
non-standard) non-domestic symmetric algebras of polynomial growth is
achieved in \cite[Page 653 Theorem]{BialkowskiHolmSkowronski2003a}
(resp.\cite[Theorem 3.1]{BialkowskiHolmSkowronski2003b}). They
give a complete list of representatives of derived equivalence
classes in terms of quiver with relations. We recall the derived
norm forms following \cite{HolmSkowronski2009}.

We have five quivers and various relations.

\unitlength1cm

\begin{picture}(15,6.5)

\put(1.2,6){$Q_2(1)$}
\put(1,5){$\bullet$}
\put(2,5){$\bullet$}
\put(1.1,5.3){\vector(1,0){.8}}
\put(1.9,4.95){\vector(-1,0){.8}}
\put(.4,5.1){\circle{1}}
\put(.88,5.1){\vector(0,-1){.1}}
\put(1.4,5.4){\scriptsize $\gamma$}
\put(1.4,4.7){\scriptsize $\beta$}
\put(0,5){\scriptsize $\alpha$}

\put(5.2,6){$Q_2(2)$}
\put(5,5){$\bullet$}
\put(6,5){$\bullet$}
\put(5.1,5.3){\vector(1,0){.8}}
\put(5.9,4.95){\vector(-1,0){.8}}
\put(4.4,5.1){\circle{1}}
\put(4.88,5.1){\vector(0,-1){.1}}
\put(6.7,5.1){\circle{1}}
\put(6.22,5.1){\vector(0,1){.1}}
\put(5.4,5.4){\scriptsize $\sigma$}
\put(5.4,4.7){\scriptsize $\gamma$}
\put(4,5){\scriptsize $\alpha$}
\put(6.99,5){\scriptsize $\beta$}

\put(9.6,2){$Q_3(3)$}
\put(9,1){$\bullet$}
\put(10,1){$\bullet$}
\put(11,1){$\bullet$}
\put(9.1,1.3){\vector(1,0){.8}}
\put(9.9,.95){\vector(-1,0){.8}}
\put(10.1,1.3){\vector(1,0){.8}}
\put(10.9,.95){\vector(-1,0){.8}}
\put(9.4,1.4){\scriptsize $\alpha$}
\put(9.4,.7){\scriptsize $\gamma$}
\put(10.4,1.4){\scriptsize $\sigma$}
\put(10.4,.7){\scriptsize $\beta$}

\put(.6,3){$Q_3(1)$}
\put(0,1){$\bullet$}
\put(1,1){$\bullet$}
\put(2,1){$\bullet$}
\put(.1,1.3){\vector(1,0){.8}}
\put(.9,.95){\vector(-1,0){.8}}
\put(1.1,1.3){\vector(1,0){.8}}
\put(1.9,.95){\vector(-1,0){.8}}
\put(1,1.9){\circle{.8}}
\put(1,1.48){\vector(1,0){.1}}
\put(.4,1.4){\scriptsize $\beta$}
\put(.4,.7){\scriptsize $\gamma$}
\put(1.4,1.4){\scriptsize $\delta$}
\put(1.4,.7){\scriptsize $\sigma$}
\put(1,2.1){\scriptsize $\alpha$}

\put(5.1,3){$Q_3(2)$}
\put(4.5,1){$\bullet$}
\put(5.5,1){$\bullet$}
\put(6.5,1){$\bullet$}
\put(4.6,1.3){\vector(1,0){.8}}
\put(5.4,.95){\vector(-1,0){.8}}
\put(5.6,1.3){\vector(1,0){.8}}
\put(6.4,.95){\vector(-1,0){.8}}
\put(5.5,2.2){$\bullet$}
\put(5.4,1.5){\vector(0,1){.7}}
\put(5.55,2.2){\vector(0,-1){.7}}
\put(4.9,1.4){\scriptsize $\alpha$}
\put(4.9,.7){\scriptsize $\beta$}
\put(5.9,1.4){\scriptsize $\epsilon$}
\put(5.9,.7){\scriptsize $\xi$}
\put(5.25,1.8){\scriptsize $\delta$}
\put(5.65,1.8){\scriptsize $\gamma$}

\end{picture}

Define for any $\lambda\in K\setminus\{0,1\}$ the algebras

\begin{eqnarray*}
\Lambda_2&:=&KQ_2(1)/(\alpha^2\gamma,\beta\alpha^2,\gamma\beta\gamma,
\beta\gamma\beta,\beta\gamma-\beta\alpha\gamma,\alpha^3-\gamma\beta)\\
\Lambda_2'&:=&KQ_2(1)/(\alpha^2\gamma,\beta\alpha^2,\beta\gamma,
\alpha^3-\gamma\beta)\\
\Lambda_3(\lambda)&:=&KQ_2(2)/(\alpha^4,\gamma\alpha^2,\alpha^2\sigma,
\alpha^2-\sigma\gamma-\alpha^3,\lambda\beta^2-\gamma\sigma,
\gamma\alpha-\beta\gamma,\sigma\beta-\alpha\sigma)\\
\Lambda_3'(\lambda)&:=&KQ_2(2)/(\alpha^2-\sigma\gamma,
\lambda\beta^2-\gamma\sigma,\gamma\alpha-\beta\gamma,
\sigma\beta-\alpha\sigma)\\
\\
\Lambda_5&:=&KQ_3(1)/(\alpha^2-\gamma\beta,\alpha^3-\delta\sigma,
\beta\delta,\sigma\gamma,\alpha\delta,\sigma\alpha,\gamma\beta\gamma,
\beta\gamma\beta,\beta\gamma-\beta\alpha\gamma)\\
\Lambda_5'&:=&KQ_3(1)/(\alpha^2-\gamma\beta,\alpha^3-\delta\sigma,
\beta\delta,\sigma\gamma,\alpha\delta,\sigma\alpha,\beta\gamma)\\
\Lambda_9&:=&KQ_3(2)/(\beta\alpha+\delta\gamma+\epsilon\xi,\gamma\delta,
\xi\epsilon,\alpha\beta\alpha,\beta\alpha\beta,
\alpha\beta-\alpha\delta\gamma\beta)\\
\Lambda_9'&:=&KQ_3(2)/(\beta\alpha+\delta\gamma+\epsilon\xi,\gamma\delta,
\xi\epsilon,\alpha\beta)\\
A_1(\lambda)&:=&KQ_3(3)/(\alpha\gamma\alpha-\alpha\sigma\beta,
\beta\gamma\alpha-\lambda\cdot\beta\sigma\beta,
\gamma\alpha\gamma-\sigma\beta\gamma,
\gamma\alpha\sigma-\lambda\sigma\beta\sigma)\\
A_4&:=&KQ_3(2)/(\beta\alpha+\delta\gamma+\epsilon\xi,
\alpha\beta,\gamma\epsilon,\xi\delta)
\end{eqnarray*}

Further, denote the trivial extensions of tubular canonical algebras as usual, in particular
$$\Lambda(2, 2, 2, 2, \lambda)=T(C(2,2,2,\lambda))\mbox{ for
}\lambda\in  K \setminus \{0, 1\}$$
$$\Lambda(3, 3, 3)=T(C(3,3,3))$$
$$\Lambda(2, 4, 4)=T(C(2,4,4))$$
$$\Lambda(2, 3, 6)=T(C(2,3,6)).$$

The precise quivers with relations of the trivial extensions
of tubular canonical algebras in question are displayed in
\cite{HolmSkowronski2009}.
We will refrain from presenting them here since we do not really need this
information in such details.

\subsection{Weakly symmetric standard non-domestic polynomial growth algebras}

\begin{Thm}
\cite[Page 653 Theorem]{BialkowskiHolmSkowronski2003a} Let $A$ be an
indecomposable standard non-domestic weakly symmetric algebra of
polynomial growth. Then $A$ is derived equivalent to one of the
following algebras:
\begin{itemize} \item  two
simple modules: $\Lambda_2'$ and  $\Lambda_3'(\lambda), \lambda
\in K \backslash\{0, 1\}$;

\item  three simple modules: $\Lambda_5'$ and $A_1(\lambda),
\lambda \in  K \backslash \{0, 1\}$;

\item four simple modules: $\Lambda_9'$  and $A_4$;

\item six simple modules: $\Lambda(2, 2, 2, 2, \lambda), \lambda
\in  K \backslash \{0, 1\}$;

\item eight simple modules: $\Lambda(3, 3, 3)$;

\item nine simple modules: $\Lambda(2, 4, 4)$;

\item ten simple modules: $\Lambda(2, 3, 6)$.
\end{itemize}
\end{Thm}

The above classification is complete up to the scalar problems in
$\Lambda_3'(\lambda)$ and in
$A_1(\lambda)$.
Remark that except $\Lambda_9'$ in case the characteristic
of the base field $K$ is different from $2$, all
algebras are symmetric.

We shall prove

\begin{Prop}\label{NonDomesticStandard}
The classification of  indecomposable standard non-domestic
weakly symmetric algebras of polynomial growth up to stable
equivalences of Morita type coincide with the derived equivalence
classification, modulo the scalar problems in
$\Lambda_3'(\lambda), \lambda \in K \backslash\{0, 1\}$ and in
$A_1(\lambda), \lambda \in  K \backslash \{0, 1\}$.
\end{Prop}

\begin{Proof}
Since a derived equivalence between self-injective algebras
induces a stable equivalence of Morita type (\cite{KellerVossieck},
\cite{Rickard1989}), we only need to show that two algebras from
the above list which are not derived equivalent are not stably
equivalent of Morita type, either.

Since the property of being symmetric is invariant under a stable
equivalence of Morita type (\cite[Corollary 2.4]{Liu2008}), in
case the characteristic of the base field is different from $2$,
$\Lambda_9'$ cannot be stably equivalent of Morita
type to any of the other algebras. We can concentrate on
the remaining symmetric algebras in the list.

Now the algebras of at least six simple modules are trivial
extensions of canonical algebras of tubular type
$(2, 2, 2, 2)$,  $ (3, 3, 3)$, $(2, 4, 4)$,  $(2, 3, 6)$
respectively. They have singular Cartan matrices, while all
other algebras have non-singular Cartan matrix. Since, by
a result of Xi \cite[Proposition 5.1]{Xi2008} the
absolute value of the Cartan determinant  is preserved by a
stable equivalence of Morita type, we can consider separately
those algebras of at most four simple modules and those of at
least six simple modules.

For trivial extension cases, by \cite[Propositions 5.1 and
5.2]{BialkowskiHolmSkowronski2003a}, two algebras of this type are
stably equivalent if and only if they are derived equivalent. So
the number of simple modules, which is an invariant under derived
equivalence, distinguishes them.

Now suppose two indecomposable standard non-domestic
weakly symmetric algebras of polynomial growth are stably
equivalent of Morita type and have non-singular
Cartan matrices. Then the algebras are algebras in the above list.
The following table
gives some stable invariants of Morita type which distinguish two algebras
in the above list.
$$\begin{array}{c||c|c|c|c|c|c}
Algebra\  A & \Lambda_2' &\Lambda_3'(\lambda)& \Lambda_5'&
A_1(\lambda)& \Lambda_9'& A_4\\
\hline
det\  C_A& 6 & 12 & 6 & 16 & 4 & 12 \\
dim  (Z(A)/R(A))& 3 & 4 & 2 &2 & 1 & 2
\end{array}$$

\end{Proof}

\subsection{Selfinjective non-standard non-domestic polynomial growth algebras}

Now we turn to  non-domestic non-standard selfinjective algebras
of polynomial growth.  Recall the following

\begin{Thm} (Bia\l kowski, Holm, Skowro\'nski
\cite[Theorem 3.1]{BialkowskiHolmSkowronski2003b})
Let $A$ be an indecomposable non-standard non-domestic
self-injective  algebra of polynomial growth. Then we get the following.
\begin{itemize}
\item If the base field is of characteristic $3$, then $A$ is derived equivalent to $\Lambda_2$.
\item else the base field is of characteristic $2$ and then
\begin{itemize}
\item if $A$ has
two simple modules, $A$ is derived equivalent to $\Lambda_3(\lambda),
\lambda \in K \setminus\{0, 1\}$
\item  if $A$ has three simple modules, $A$ is derived equivalent to $\Lambda_5$
\item if $A$ has four simple modules, $A$ is derived equivalent to $\Lambda_9$

\item else $A$ has five simple modules and then $A$ is derived equivalent to $\Lambda_{10}$
\end{itemize}
\end{itemize}
\end{Thm}

If the base field is of characteristic $2$ and $\lambda\neq\lambda'$,
we do not know if $\Lambda_3(\lambda)$ is
derived equivalent to $\Lambda_3(\lambda')$ or not.
We call this question the scalar problem.

The above classification is complete up to the scalar problem
in case the base field is of characteristic $2$ for the algebra
$\Lambda_3(\lambda), \lambda \in K \backslash\{0, 1\}$.

Remark that
the algebra $\Lambda_{10}$ is not   symmetric (even not weakly
symmetric), so it is not stably equivalent of Morita type to any
other algebra in the above list.

We can now prove
\begin{Prop}\label{NonDomesticNonStandard}The classification  of  indecomposable
non-standard non-domestic self-injective algebras of polynomial
growth up to stable equivalences of Morita type coincides with the
derived equivalence classification, modulo the scalar problem in
$\Lambda_3(\lambda)$.

\end{Prop}

\begin{Proof} We only need to consider the case of characteristic
two, since if the characteristic of $K$ is $3$, there is only one algebra. But notice
$det\  C_{\Lambda_3(\lambda)}=12$,  $det\  C_{\Lambda_5}=6$ and
$det\  C_{\Lambda_9}=4$.
\end{Proof}

\subsection{Standard algebras versus non-standard algebras}

Now one can prove that a non-standard algebra cannot be stably
equivalent to a standard one. The case of a derived equivalence is
proved by Holm-Skowro\'nski in \cite{HolmSkowronski2009}.

\begin{Thm}\cite[Main Theorem]{HolmSkowronski2009}
Let $A$ be a standard self-injective algebra of polynomial growth
and let $\Lambda$  be an
indecomposable,  nonstandard, non-domestic, symmetric algebra of
polynomial growth. Then $A$ and $\Lambda$ are not derived
equivalent.
\end{Thm}

\begin{Prop}\label{NonDomesticStandardVsNonStandard}
Let $A$ be an indecomposable standard weakly symmetric algebra and
$\Lambda$ be an  indecomposable  nonstandard  non-domestic
self-injective algebra of polynomial growth. Then $A$ and
$\Lambda$ are not stably equivalent of Morita type.
\end{Prop}

\begin{Proof} Suppose that $A$ and $\Lambda$ are stably equivalent
of Morita type. By Krause \cite[Page 605 Corollary]{Krause}, $A$ is also
non-domestic of polynomial growth.

We only need to consider
symmetric algebras. In fact, the only non-symmetric nonstandard,
non-domestic   self-injective algebra of polynomial growth
$\Lambda_{10}$ is only defined when the characteristic of the base field is
$2$, but in case of
characteristic $2$, all standard non-domestic weakly symmetric
algebras of polynomial growth are symmetric, as $\Lambda_9'$ is
non-symmetric if and only if characteristic of the base field is different from
$2$.

Now we consider symmetric algebras. In fact, the method of the
proof of \cite[Main Theorem]{HolmSkowronski2009} works. Since two
algebras which are  stably equivalent of Morita type have
isomorphic stable Auslander-Reiten quivers, by the shape of the
stable Auslander-Reiten quivers (see the second paragraph of
Section 4 of Holm-Skowro\'nski \cite{HolmSkowronski2009}),
we proceed by remarking the following facts.
\begin{enumerate}
\item If $K$ is of characteristic $3$, the algebras $\Lambda_2$ and
$\Lambda_2'$ are not stably equivalent of Morita type.  Indeed
Holm-Skowro\'nski \cite{HolmSkowronski2009} have shown that
$$
dim\ Z(\Lambda_2')/T_1(\Lambda_2')^{\perp}=3\neq 2
=dim\ Z(\Lambda_2)/T_1(\Lambda_2)^{\perp}.
$$

\item If $K$ is of characteristic $2$ and
$\lambda,\mu \in  K \setminus \{0, 1\}$, the algebras $\Lambda_3(\lambda)$ and
$\Lambda_3(\mu)'$  are not stably equivalent of Morita type. Indeed
Holm-Skowro\'nski \cite{HolmSkowronski2009} have shown that
$$
dim\ Z(\Lambda_3'(\lambda))/T_1(\Lambda_3'(\lambda))^{\perp}=3\neq
2=dim\ Z(\Lambda_3(\mu))/T_1(\Lambda_3(\mu))^{\perp}.
$$

\item If $K$ is of characteristic $2$, the algebras $\Lambda_5$ and
$\Lambda_5'$ are not stably equivalent of Morita type. Indeed
Holm-Skowro\'nski \cite{HolmSkowronski2009} have shown that
$$dim
Z(\Lambda_5')/T_1(\Lambda_3')^{\perp}=0\neq dim
Z(\Lambda_5)/T_1(\Lambda_3)^{\perp}.$$

\item If $K$ is of characteristic $2$, the algebras $\Lambda_9$ and
$\Lambda_9'$ are not stably equivalent of Morita type.  Indeed
Holm-Skowro\'nski \cite{HolmSkowronski2009} have shown that
$$dim HH^2(\Lambda_9)=4\neq 3=dim HH^2(\Lambda_9').$$
\end{enumerate}

Since $Z(A)/T_1(A)^{\perp}$ and $HH^2(A)$ are invariants under
stable equivalences of Morita type, this completes the proof.
\end{Proof}

\subsection{Stable equivalence classification of weakly
symmetric non-domestic polynomial growth algebras}

Combining Propositions~\ref{NonDomesticStandard},
\ref{NonDomesticNonStandard} and
\ref{NonDomesticStandardVsNonStandard}, we have

\begin{Thm} \label{NonDomesticClassification}
The classification  of  indecomposable
non-domestic weakly symmetric algebras of polynomial growth
up to stable equivalences of Morita type coincides with the derived
equivalence classification, up to the above mentioned scalar problems.
\end{Thm}

As a consequence, we can prove a special case of Auslander-Reiten
conjecture.

\begin{Thm} \label{ARNonDomestic}
\begin{enumerate}
\item Let $A$ be an indecomposable
algebra which is stably equivalent of Morita type to
   an indecomposable
non-domestic  symmetric algebra  $\Lambda$   of polynomial growth.
Then $A$ and $\Lambda$ have the same number of simple modules.

\item Let $A$ and  $\Lambda$  be two indecomposable algebras which
are both  non-domestic  weakly symmetric algebra of polynomial
growth or which are both non-standard self-injective algebras of
polynomial growth. If they are stably equivalent of Morita type,
then $A$ and $\Lambda$ have the same number of simple modules.
\end{enumerate}
\end{Thm}

\begin{Proof} (1). By Krause \cite[Page 605 Corollary]{Krause}
and Liu \cite[Corollary 2.4]{Liu2008},
$A$ is also symmetric, non-domestic, and of polynomial growth.
Then one can apply Theorem~\ref{NonDomesticClassification}, by
noticing that the existing scalar problems all occur in families
of algebras for which different scalars yield algebras having
the same number of simple modules.

(2) is a direct consequence of
Theorem~\ref{NonDomesticClassification} and
Proposition~\ref{NonDomesticNonStandard}.

\end{Proof}

\subsection{Periodic algebras}

Recently, Karin Erdmann and Andrzej Skowro\'nski
\cite{ErdmannSkowronski2010} have shown that
a non-simple indecomposable symmetric algebra $A$ is tame with
periodic modules if and only if $A$ belongs to one of the following
classes of algebras:
a representation-finite symmetric algebra, a non domestic symmetric
algebra of polynomial growth, or an algebra of quaternion type (in
the sense of \cite{ErdmannLNM}). Note that these three classes of
algebras are closed under stable equivalences of Morita type, by
Krause \cite[Page 605 Corollary]{Krause} and Liu \cite[Corollary
2.4]{Liu2008}.

In \cite{Asashiba} Asashiba, in combination with Holm-Skowro\'nski \cite{HolmSkowronski2006}, the statement displayed in
\cite[Theorem 3.4, Theorem 3.5, Theorem 3.6]{ErdmannSkowronskiperiodic}
proved that for
the class of representation-finite
selfinjective
indecomposable algebras are derived equivalent if and only if they are stable
equivalent.
In our  recent paper
\cite[Theorem 7.1]{ZhouZimmermannTameBlocks},
we proved that for the class of algebras of quaternion
type, derived equivalence classification coincide with the
classification up to stable equivalences of Morita type
(up to some scalar problems).
Therefore, combining the above Theorem~\ref{ARNonDomestic}, we
showed the following

\begin{Thm}
\begin{enumerate}
\item
    The class of  tame symmetric algebras  with
periodic modules is closed under stable equivalences of Morita
type, in particular under derived equivalences;

\item
For tame symmetric algebras  with periodic modules
derived equivalence classification coincide with the
classification up to stable equivalences of Morita type (up to
some scalar problems);

\item   The Auslander-Reiten conjecture holds for a stable
equivalence of Morita type between two tame symmetric algebra with
periodic modules.
\end{enumerate}
\end{Thm}

\section{Concluding remarks}

In Section~\ref{polysection}
we classified symmetric algebras of polynomial growth
only up to
stable equivalences of Morita type, whereas the results in Section~\ref{weaklysymmdomesticsection} concern stable equivalences in general.
It would be most interesting to get a stable equivalence classification
for general selfinjective algebras of polynomial growth,
in particular the validity of the Auslander-Reiten conjecture in general.
The crucial point for this more general statement would be
to first classify algebras which are stably equivalent (of Morita type) to
$\Lambda_9'$ in case the characteristic of $K$ is different from $2$
or to $\Lambda_{10}$, since these are the selfinjective polynomial growth
algebras which are  not symmetric. The second step is
to classify  weakly symmetric algebras of polynomial growth
up to stable equivalences.

If the for the class of selfinjective algebras
of polynomial growth the derived equivalence classification would coincide with the
classification up to stable equivalences (of Morita type) the Auslander-Reiten
conjecture for this class of algebras should follow.

Another point concerns the work of Martinez-Villa (\cite{Martinez-Villa}) of the
reduction of the Auslander-Reiten conjecture to selfinjective algebras.
If we can prove
that this reduction preserves representation type:
domestic type, polynomial growth etc,
then the conjectural result of the preceding paragraph would imply
the validity of the
Auslander-Reiten conjecture for algebras of polynomial growth.

The results of this paper present a second occurrence of a phenomenon of the same
kind: In our previous paper
\cite{ZhouZimmermannTameBlocks} we also obtained that the classification of a class of
tame symmetric algebras up to stable equivalence coincides with the derived equivalence
classification.
One might ask if two indecomposable tame symmetric algebras are stably equivalent of
Morita type if and only if they are derived equivalent.

redpajama_set_name: RedPajamaArXiv


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Breakfast is served in the sunny breakfast room overlooking the front terrace. breakfast includes traditional favourites the farmhouse is a classic english breakfast with grilled bacon, mushrooms, fried sausage, grilled tomato's and a choice of how you would like your egg cooked.
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Come and Celebrate Australia's National Day with celebrations in the Yass Valley at Coronation Park. The Rotary Club of Yass provide breakfast in the park from 8am. Face painting, jumping castle, playground to keep the children entertained! A lovely atmosphere so bring the whole family. Yass Valley Australia Day Citizen Awards are to recognise outstanding individuals and groups who make a real contribution and difference to their community. The Australia Day Ambassador will also address the audience. The winners of the Australia Day awards are announced in the park at 9.15am. Free entry so head along and enjoy a true Australia Day like no other in the Yass Valley.


Nanette Enriques is a 3rd Grade teacher in Davao Oriental, a region of Mindanao Island in the Philippines. When Typhoon Bopha hit on December 4, fifty families took shelter in her school.
During the storm the roof collapsed, injuring many inside and killing several. Enriques helped the wounded and described the terrifying night of the storm to LWR staff.
Is it okay for people who have experienced a disaster to be "happy"?


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The AGM of the Friends of St Anne's Church, Kew will be held on Saturday 8 June at 11:00 in the Parish Room (coffee served from 10:30). The papers for this meeting can be found here:
✠ Notice and Agenda
✠ Committee Member Nomination Form
✠ Chairman's Letter
Structure, Governance and Management
The Friends of St. Anne's Church, Kew, is a registered charity under the Charities Act constituted pursuant to a Constitution adopted on the 27th day of January 2001. It is managed by an Executive Committee composed of a Chairman, Treasurer and Secretary and two to four members (all elected at the charity's annual general meeting) plus the Vicar of St. Anne's Church and the two churchwardens.
The Charity's objects, as set out in the Constitution, are to aid in the restoration, preservation, repair, maintenance, improvement and beautification, for the benefit of the public, of St. Anne's Church and its churchyard and of the monuments, fittings, fixtures, stained glass, furniture, ornaments and chattels therein.
Having regard to the guidance issued by the Charity Commission on public benefit, the charity has undertaken to:
✠ Raise funds to help maintain the fabric of the church and churchyard
✠ Finance special projects to restore the churchyard
✠ Promote public awareness of the church and its place in the history of Kew.
Nigel J. Worn, President (Vicar of St Anne's Church, Kew)
Lorraine Neale, Chairman, elected
Cate Lyon, Churchwarden
Bill Cottle, Churchwarden
Anne Barker, Honorary Treasurer
Benjamin H Jones, Secretary, elected
Wendy Booth, Membership Secretary, elected
Paul Booth, IT Officer, elected
Gemma Roundell, elected
Linda Pethick, elected
Phyllis Cunningham, elected


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10.10.22 Комментарии Posted By W.T
Whether you are a man or woman, or anything in between, there are risks and consequences to using drugs. From procuring them to using them, there are risks at every step. But what are the differences?
Is the impact of substance abuse different between men and women?
Most notably:
How do Narcotics Impact a Menstrual Cycle
Some drugs can directly impact or stop the menstrual cycle and cause infertility, or the disordered cycle may increase pregnancy risk. If you are used to a disordered cycle, you may not even realise you're expecting.
Most of us know that if a woman uses drugs, these substances can be passed to the unborn baby, which then causes further complications.
Accessing support whilst pregnant may also be challenging – hard enough when you are looking for help for yourself, but when a child is involved, there is the fear of involvement of social services and the legal issues that would come with it.
How Cocaine Effects a Woman's Period
Cocaine is a vasoconstrictor - it affects and restricts the blood flow within the body. Regular cocaine use can interfere with hormone levels, disrupt the menstrual cycle and ultimately stop ovulation altogether. As a frequent cocaine user, you could go several months without having a period or not having one.
Another risk of regular cocaine use is that it could damage the fallopian tubes, making it more difficult to conceive a child in later life. Interestingly, a study conducted by the McLean Hospital in the USA by Marc Kaufman found that oestrogen counteracts the effect of cocaine. The study found that during the first half of their menstrual cycle, when oestrogen levels are high, the brain is protected from the damaging effects of cocaine use.
Other studies have suggested that women who are ovulating crave cocaine more than when they are not ovulating. This may make it more difficult if you are withdrawing or detoxing from cocaine and increases the risk of relapse. Though the research was conducted on rats, similar studies show human response is identical. Cocaine seeking was higher in female rats than male rats, "in female rats, the magnitude of cocaine craving was critically dependent on the phase of the oestrous cycle, demonstrating a novel role of ovarian hormones in the incubation of cocaine craving.
How Cocaine Effects Pregnancy
The primary function of the placenta is to transfer nutrients from the mother to the foetus. As with any nutrient or substance a pregnant woman takes, cocaine crosses the placenta barrier.
Cocaine use is a risk if pregnant or not, but a pregnant woman who uses cocaine dramatically increases her risk of high blood pressure. This, in turn, can cause fluid in the lungs, bleeding in the brain or heart failure. And that's only the start of it.
Cocaine use whilst pregnant can cause changes in the placenta and can cause death for both mother and baby. One study found that 38% of pregnant women who used cocaine lost their babies.
The placenta blood flow supplies the unborn baby with the oxygen it needs. Cocaine constricts the blood vessels, and after using cocaine, both the mother's and the unborn baby's heart rate and blood pressure increase. The constriction of blood vessels can affect the placenta and lead to a lack of oxygen supplied to the unborn baby. This can cause organ failure, placental abruption (separating the placenta from the uterus), premature birth and spontaneous abortion (miscarriage). One study has shown that pregnant cocaine users had a 1 in 3 chance of premature birth compared to a non-cocaine user with a 1 in 8 risk of premature birth.
Eclampsia (and pre-eclampsia) are severe complications of pregnancy that can result in maternal and foetal death if untreated. A little under 5% of pregnancies will be affected by pre-eclampsia, which increases significantly when the mother uses cocaine. Typical signs of eclampsia are headaches, seizures, blurred vision and seizures.
How Cocaine Effects New Borns
When conducting trials, they use medical-grade cocaine. They don't have to go to the neighbourhood dealer and get street cocaine, where the likelihood is that a bunch of cutting agents have been thrown in to make more money. The cocaine you're sniffing on your nights out is probably only 50-60% pure. So, if pure cocaine can do this to your unborn baby, imagine what the cutting and bulking agents can do to you and your unborn baby.
If you have been sniffing, smoking, or injecting cocaine throughout your pregnancy and are now about to deliver, you might ask, "How will it affect the baby?"
Unfortunately, your new baby will likely go through withdrawal. This is not a pleasant process in adults, and a new baby can include sleeplessness, tremors, muscle spasms and difficulty feeding.
Signs of prolonged cocaine use through pregnancy to your new baby can include a smaller head and an overall smaller baby. Your new baby's brain may be smaller at birth and will remain smaller than usual. Other issues can include brain, kidneys, genitals and heart defects.
Fear not, though! If you stop taking cocaine during pregnancy, you can dramatically reduce the risks. I always write about minimising harm and how to do drugs more safely, but there is no harm minimisation when it comes to pregnancy. Your cocaine use is not only putting you at risk.
Get help to stop. Get help to reduce the risk to you both. My only advice in minimising the risk is to stop! I know this is difficult and sometime may feel impossible, but reach out and ask for help.
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