May 29 2008 Bush: "the fruits of the new strategy" in Iraq.
A sad ode to 2 black men lynched in Indiana.
The fruits that the bushman speaks of are strange and just as sad, indeed.
Yes, violence in Iraq is down to the lowest point since March of 2004. Civilian deaths are down. Sectarian killings are down. Yes, all these wonderful things are true, but Bush's gloating is ill-placed, false-faced, and enraging a a growing number of Americans and Iraqis who are able to see the truth of the consequences of our actions.
The cost of this fruit has been immense, and it continues to rise.
Telling us violence is down now is like telling an amputee with a new limb that she is better off than she was right after her arm was chopped off. Sure, she has seen an improvement, but it would have been a lot less painful if she'd never lost her arm in the first place.
and so have our service men and women.
No amount of 'improvement' will erase the loss and trauma we have enabled in Iraq.
Peace Operation, Mother of Two Springs, Lion Roar, Charge of the Knights, Iron Harvest and Phantom Phoenix are different names of military operations carried out by Iraqi national forces around Iraq "without discrimination."
Get it? While the news in the U.S. persists in feeble attempts to make the American forces look as if they are successful--the real action is being carried out by Iraqi troops. Yes, Virginia, there is an Iraq--and you better not stand in its way as it recovers its country from the invaders; from _all_ the invaders.
Our media gives much of the credit for gains in security to U.S. forces. It's a lie. We accomplish very little, and in fact tend to kill an inordinate number of civilians as we "make progress." This kind of progress only enrages more and more Iraqis and leads them into the resistance with increasing fervor.
Check out AlSumaria , an Iraqi media outlet.
"Yesterday, my brother went to the center to add his new born daughter. They added her name to the list easily and everything was great. At nigh, my wife told me "laith, your brother just added his new daughter to their ration food card. It took him less than five minutes to finish the procedures". I reminded her that I went to the office the last month and I have been told to apply in June. She insisted that things had changed and I can add our son to our food card. As a role in my life, I don't argue women because she wouldn't ever change her mind no matter what I say. I went to the office and I applied to add my son to my family's food card. I was surprised when the employee said "you cant add your son to the list now. You have to wait for one month". I asked him about the reasons, he said "it's the roles of the ministry of trade". Again, I asked why the roles don't allow me to add my son any time after he passed his first year, doesn't he need food to grow or he is not any more a member from the family? The man couldn't say a word but "it's the roles of the ministry of trade".
I cant describe my shock, not because my son doesn't have his food substances. I was shocked because the roles of my country are so strange to the extent that no mind can find real explanations for them.
I went back home and told my wife about the whole thing. She didn't say a word but her look asked me thousands of questions..."
$200 will feed a widow and her 2 children for a month. This is emergency relief for those in dire need. We are asking that you pledge a steady amount each month, for one year. This will enable us to budget for the year and determine how many families we will be able to help survive these horrid days in Baghdad.
Hey! You don't want to commit to a year? OK. Just send whatever you can, as soon as you can. We'll get 90% of it into the hands of needy widows and their children.
We are not a charity--just a few people trying to make a difference, so your contribution is not tax-deductible.
How we can expect Americans to take responsibility for the horrible consequences of the occupation of Iraq? Why should they when they are fed the pap of such as the NY Times history of the conflict? What a crock. What a travesty. What a vulgar insult to the memory of MItch, who died on 9/11?
What does the NY Times give us to help our understanding of the occupation?
An honest admission of the errors made at the onset.
A long and wonderfully clear exposition of the invasion and occupation.
AND NOT A @#$%ING THING ABOUT THE SOLDIERS LOST!
AND NOT A @#$%ING THING ABOUT THE INNOCENT CIVILIANS CRUSHED BY THE OCCUPATION!
Once again the American public is shielded from the part of the truth that would open their hearts to the true consequences of our actions.
Are we not strong enough to take the truth?
It's Memorial Day. At least let us think about the people; about the people we have lost; about the people 'they' have lost; about the people Ours and theirs) who remain traumatized by the violence we enabled; about the truth.
My NYTimes gives me the news about Iraq. And so do the Iraqi media with web outlets. They don't seem to match.
We are fed stories of "success" (the calm after the mini-genocides) and more progress on the political front (Malaki's U.S.-led government deals with opposition leaders).
Sadr City - At least 11 people were killed when gunfire broke out after a roadside bombing in a Shiite militia stronghold, the scene of a major military clampdown.
Al Rustumiya - Two people were killed and five others wounded in an explosion targeting a minibus.
Zayyuna - a bomb explosion targeting a police patrol killed three people.
Mandali - a child was killed in a suicide bomb attack on the head of the local awakening council Matlab Al Naddawi, wounding him and two of his guards.
Baldroz - a mortar round hit an Iraqi market killing two people.
Al Duluiya - five members of Awakening councils including leader Qassem Al Dulaimi were killed in a drive by shooting.
My NYTimes wrote today of the success of Iraqi troops as they drove through Sadr city without opposition.
This is not a triumph of American forces. It is the result of M. Sadr's deal with Malaki. We had little to do with it.
We are not helping. We are only prolonging the agony of the occupation. Iraq will not start to heal itself until the occupation ends. Our troops are dying for nothing. Bring them home now.
A father sent his 8 year old son to the store for bread. It was only a few doors down the street. When the explosion rattled the house the dad ran into the street. Tears washed his face as he cried, "Omar! Omar!" Sand was raining and he could not see through the curtain of debris flying through the air. Mom came out, too.
We'll see that 90% of it gets put into the hands of needy Baghdad moms; widows who have lost their homes and temporarily rely on the goodness of others for basic material support.
May 19 2008 Truth in Occupation: Iraqis will own Iraq again.
This fiction may evoke memories in 3 or 4 people, but everything has become material for my writing, as most of you already know.
"Don't you forget. We have been invaded over and over and over and over we have always remained ourselves."
He came to the U.S. many years ago but his once- or twice yearly leisurely trips back home had fed his super-ex-patriotism. Of course he stopped going home "after the occupation. Like Palestinians after the 2nd Intifada. It was too dangerous--and if we had any contact with Americans..."
We drank bitter coffee. He talked in a steady stream of memories and realities with a passionately positive attitude that belied his quiet demeanor. His voice was soft with love for Iraq.
"See that man? He was the prime minister under xxxxxxx," he said, indicating the straight backed, well dressed, and mustachioed gentleman across the room. "Big in the military. Big. There are many like him left at home They are just waiting."
We talked of the how hard it is for almost all immigrants when they get here. Many have to wait a generation for a meaningfully satisfying life here, and much is lost in the transition. We talked of many things, but mostly of the simple fact that Iraq will be, once again, owned by Iraqis. After all, "who owns Vietnam today?"
Nothing will change but the devastation will have taken place and will continue until we get out of there. And we talked of the shame of Iraq having to go through the pains of this meaningless occupation.
And the shame of all the military who die for nothing.
And the shame of all the civilians who die for nothing.
And the trauma; the deep trauma of prolonged exposure to insanity.
We read a melancholy poem written by a recent immigrant.
But mostly we talked about the uselessness of feeling bad, and the inevitable return of Iraq to Iraqis.
At times a previous wave of immigration washed into our consciousness; the reggae dancehall from the street vendor's stand.
The rain stopped. We got up to go and said our goodbyes.
"Thank you for helping my family," he said as we stepped away. I stopped and looked him in the eye, not quite understanding.
He smiled softly. "All Iraqi are my family."
For years now we have been told there is no need for a draft of U.S. forces because our volunteer army is able to handle the job.
Recently it has been revealed that the standards for military service have been reduced drastically so that felons previously excluded are now "fit for duty." And the medical standards have also been reduced. And most recently we were told that some 40,000 soldiers once declared "medically unfit for duty" are now "fit for duty." You really have to keep you eye on the moving target.
Today it is revealed that Blackwater International, our premier mercenary supplier of battlefield personnel, has an amazingly free hand in Iraq.
You may recall the atrocities that Blackwater committed recently. The Iraqi government was so outraged by the free-wheeling killing of innocents by Blackwater that it demanded they all be removed from Iraq. After an investigation the suspected perpetrators of the atrocities were named. None were brought to trial.
It was then revealed that Blackwater does not function under U.S. law. It is not answerable to Iraqi law either. Blackwater is answerable to Blackwater only, and this is a huge problem. Thier freedom to kill without thought of prosecution has led to ourtageous behaviour on thier part. "Shoot first and ask questions later," is no way to win the hearts and minds of the Iraqi populous. thier blatant disregard for humanitarian concerns paints us, once again, as a lawless nation.
"If the contractors were removed, we would have to leave Iraq," stated Patrick F. Kennedy, the under secretary for management. So as long as we stay then Blackwater must stay also.
So our all-volunteer army is adequate? NOT! We have bankrupt our military to the point that unless we use lawless mercenaries to help us (AT TREMENDOUS COST) we will be unable to continue the occupaqtion of Iraq.
Our reputation is damaged every day we continue to extoll the virtues of mercenaries who are above the law. No one will believe us when we say we respect the law.
Contact your elected officials today, please. Let them know that your support a quick and orderly withdrawal from Iraq. Tell them that if we are to speak to the world about democratic law and order there will be no one to believe us as long as we support lawlessness.
From the White House to the inititail invasion, to the occupation, Guantanamo, Abu Ghraib, and the slaughter of innocents in Baghdad streets we have trashed the rule of law, and everyone knows it!
We are no longer the leaders of the free world. We have become the leader in exported violence. Our contribution to world violence is far beyond the combined violence of all the terorists on the face of the earth.
Contact your elected officials today, please. Let them know that your support a quick and orderly withdrawal from Iraq. Tell them that if we are to speak to the world about democratic law and order there will be no one to believe us as long as we support lawlessness. There. I said it twice.
May 4 2008 Bremmer: Iraq should pay us for the gift of the occupation.
Today L. Paul Bremmer reminds us that the U.S. has shed "enormous amounts of blood and treasure helping get democratic Iraq on its feet." It may be ludicrous to call the U.S. installed government of Iraq democratic, but it is certainly true that the occupation is costing us dearly.
Bremmer goes on to argue that it is "time for the Iraqis to cover much more of the costs for training their security forces and for reconstruction projects." And here he exposes an oozing flaw in the American character. Bremmer is not alone in feeling that the Iraqis should start to pay for the "gift" of the occupation. Top leaders of our government echo this call, and it popular with those running for office to offer this sick solution to our own economic problems.
As a nation the U.S. lacks the sense of responsibility and willingness to shoulder the burden of the consequences of our actions. What happened to "You broke it, you fix it!"?
We are beginning to blame the victim, and this will only lead us all to despair. Blaming the victim It avoids serious analysis of the problem, and it directs hatred toward innocent, powerless Iraqis. Think of how Iraqis feel about our wanting the to pay for the repair of the damage they know we have caused.
Remember: we are not a charity; just a small dedicated group of family and friends who are putting cash into the hands of needy women in a Baghdad neighborhood still struggling to keep its moral compass; where all are welcome to live in mutual support during these incredibly horrible time.
Tearful and feeling defeated, a priest approached Archbishop Desmond Tutu for solice. "Our work for peace and justice is ignored - even undermined, " the priest said. "What can we do?"
Archbishop Tutu gently took her face in his hands, brought it close to his. Eye to eye, he slowly said the three words, "Find your voice."


Give all your guests a unexpected scare this Halloween by displaying this Skeleton Restroom Door Cover on your bathroom door. He seems to be a little shocked that you aren't giving him any privacy. Then you can use decorate your fridge by using the Halloween Refrigerator Door Cover to keep the scares coming for your guest. The Skeleton Restroom Door Cover measures 30 inches wide by 5 feet tall and can easily be hung up with tape on any door you wish.
I was so impressed and happy with the speedy delivery and protective packaging of the door hangings.
Very neat. Husband saw it and grinned, it'll make a great party decoration.


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Q: How to fetch Rundeck Job output through terminal? The purpose of my code is to download the output log generated by a specific rundeck job number, that is hosted on my local rundeck server.
So I tried using
curl -o output.txt -0 https://xx.xxx.xx.xxx:xxxx/project/Project-Folder/execution/show/25139#output

However, when I do this, the file output.txt has no info, and is basically empty.
Is there some other way? I can fetch the output of a rundeck job with it's exec number, through terminal?

A: So I got to a conclusion after a little looking around and fiddling the way to fetch the output would be to run this command as a local command:
cat /var/lib/rundeck/logs/rundeck/${job.project}/job/${job.id}/logs/${job.execid}.rdlog

This should output the log.

redpajama_set_name: RedPajamaStackExchange


These Theater Solutions Indoor Speakers are perfect for any occasion or application. They were designed with durability in mind, so there is no need to worry about wear-and-tear. They make great additions to any surround sound home theater, will fit onto any bookshelf, can be used with any stereo system and are great to use in bathrooms and kitchens. Each speaker can handle 200 watts of power, and has a full range sound and functionality. Included mounting brackets allows hassle-free installation in any location.


The Asian Pacific American (APA) Community has contributed in so many significant ways to the strength and exceptionalism of America. For example, the APA work ethic and entrepreneurial spirit have yielded tremendous business success. Whether in small shops, large boardrooms, restaurants or startup ventures, Asian Pacific Americans have generated great economic wealth and opportunity for all Americans.
In fields as diverse as academia, medicine, law, engineering and in so many other professional endeavors, APAs have overachieved. Their impact is abundantly evident in music and the arts, cuisine, culture, sports and so much else of what makes us uniquely American.
As an APA myself, and a second generation American, I honor my ancestors, my elders and the pioneers of the APA community for their sacrifice in coming to this country, serving and building-up these United States and creating more opportunities for me and the next generations to live the American dream.
However, too many stories of sacrifice and service by the APA community still remain unheard. Too few people know stories surrounding the Chinese Exclusion Act or forgotten Filipino War Veterans. Few understand the depths of humiliation and economic damage suffered by wrongfully incarcerated Japanese Americans during WWII.
Having done nothing wrong and loyal to our flag, many Japanese Americans were sent to internment camps and had their businesses and property confiscated by the federal government simply because of their ancestry; this while their sons fought in the US Military for the country they loved.
As we celebrate all that APAs have given and continue to contribute to America, I hope we take time to better educate each other about the reality of these injustices.
You can find more information on the Asian American and Pacific Islander Heritage Month here: https://asianpacificheritage.gov/.


The Bryton Rider 10 GPS system is a highly sophisticated GPS system that can track your speed, altitude%, cadence, and even your heart rate! This type of GPS from Bryton is designed specifically for the casual cyclist looking to increase their overall performance and awareness. Think it is going to be difficult to set up and figure out? Think again because with the user friendly navigation you will be able to slide through the 28 different functions with ease, and at fraction of the price! The 2 inch display is plenty big for you to stay aware of your heartrate, and the other 3 windows of date. They windows are fully customizable to display anything you wish!


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Home / Unlabelled / A feel good Independence story
A feel good Independence story
By SCOTT WAIDE
In 1975, Jerry Nalau, became one of the first Papua New Guineans to become a District Commissioner when powers were transferred from the Australian Colonial administration to the new Papua New Guinea government.
The young patrol officer who was then already a veteran in the new administration became one of the few Morobeans sent to Kunidiawa in the Simbu Province as part of the Somare's administration to help unite the new country.
"Somare said to me: 'I want you to go to the highlands.' And I said: 'Somare, I'm not the only one! What about the others?'"
By then, Jerry Nalau had already served several years in Bougainville and Rabaul during the turbulent but positive pre-independence period.
Nalau recalls that in typical Somare humor, the Chief Minister responded: "You go to the highlands because you Finchafens took the Word of God to the highlands calling God's name… Anutu… Anutu…Anutu… I think if you go, they will respect you."It was a 13 hour flight on a slow DC3 that eventually took Jerry Nalau from Bougainville to Minj.
"Back then, large planes didn't land at Kuniawa. We landed at Minj and travelled by road to Kundiawa."
Now in his mid-seventies, Sir Jerry Nalau, former patrol officer, former Morobe premier, says with hint of mischief, that Chief Minister, Michael Somare, had told him that he could have whatever he asked for, later, if he agreed to take the assignment to the highlands.
So when Independence came and the Charles, the Prince of Wales, representing Queen Elizabeth II, was due to arrive, Jerry Nalau called the soon to be Prime Minister Somare.
"According to the Prince's itinerary, he was to travel to Manus to Wewak then onwards to Goroka. Then drive through Simbu and then rest in Mt. Hagen."
"I called Somare and I said: 'Somare, do you remember what you said? You sent me from Bougainville to Simbu and you promised you would give me whatever I wanted later."
"So I have a small request: I want Prince Charles to sleep in Kundiawa instead of going straight to Hagen."
According to Sir Jerry, Somare returned his call 15 minutes later. The new Prime Minister granted his request after going through a lot of trouble. "He said later, 'Jerry, it was a problem to me but your request is granted.'" The Prince rested in Kundiawa.
A feel good Independence story Reviewed by PNGBlogger on September 16, 2014 Rating: 5

redpajama_set_name: RedPajamaCommonCrawl


This is the first of 2 posts about the early years audit. The Department for Education will be posting here in the next few days about how the audit worked from their point of view.
The audit is a list of all the relevant content with some options editors can select when reviewing the content.
We need the content to be in good shape so we can create a taxonomy and help users find what they're looking for.
Editors across a number of organisations audited 723 content items.
33 of the content items (4.6%) will be updated to help make them clearer, more findable or more useful.
79 of the content items (10.9%) will be retired: withdrawn or redirected to other, more useful content.
32 of the content items (7.2%) will be updated to be clearer, more findable or more useful.
79 of the content items (17.8%) will be retired: withdrawn or redirected to other, more useful content.
It wouldn't be agile without a retrospective. Editors from the organisations we worked with came together to talk about what they did and what they learned.
We've now made significant changes to the spreadsheets where audits are recorded. We've also rewritten the instructions. This should help make future audits easier and more focused on helping teams uncover and resolve problems with content that affect how easily it can be found.
After the audit, the information architects start to put together the taxonomy. We're conducting a number of waves of user research to accompany this.
We're working on education as a theme at the moment. We'll be in touch early if this affects your organisation.
Teams wanting to prepare can make sure their content is well organised. If you've got more pages than your users need, there's never been a better time to tidy up.
Particular thanks to our colleagues in the Department for Education, Ofsted, the Education Funding Agency and the National College for Teaching and Leadership for being the first to embark on this process.
Image by James Lumb. Used under Creative Commons.
Nice. Could you post the audit instructions and spreadsheet template? Would be helpful for other organisations who would benefit from simular audits.
Interesting question. Bear in mind that the reason we're auditing is geared towards findability, titling, organisation etc. The template and instructions could be useful but would need to be adapted to what each organisation wants to achieve. Also, we're making some changes to the instructions. I think it'd be best to share this later, perhaps in a separate post. Does that sound ok?


Established Café came to Talbot Street in Belfast with a bang in 2013, offering state of the art coffee and stylish surroundings in the Cathedral Quarter of the city. Its polished interior design and impressive range of hot beverages attracts people far and wide, but one of its main draws is its commitment to producing great food and unique baked goods.
Stephen Arnott has been part of the cooking and baking team for over three years, and says that coming up with innovative new baking products is a massive factor in the establishment's success.
"We've always been busy, but in the last few years the biggest trend has been our "Pie and Drip" days. Every Sunday we bring out a "flavour of the week" pie that we make fresh on site. There's a real buzz every week around it, with queues out the door at times, and it's become a real social event for families and friends. With our "Pie – To – Go" option consumers can take the pies home to their families for Sunday lunch, which is really popular too.
Established have an apt of taking a popular traditional baked good and altering it to create something fresh and ingenious. Over the festive period, they changed up the time-honoured mince pie and created the contemporary mince pocket, that proved very popular.
"Our mince pie pockets were made with a fresh puff pastry and filled with cider and brandy soaked mincemeat. We experimented with different flavours, both savoury and sweet. They were a huge hit and we couldn't keep them on the shelves.
"Our freshly made cookies are also really popular. We did a special Scandi Spice double chocolate one over Christmas , and are currently running a white chocolate, coconut and miso cookie that's also been selling really well.
Social media has proved to be a very helpful tool for baking sales in Established. New products are launched on the store's Instagram page so people can see what's coming next.


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Iowan Nick Ryan
I am excited about the opportunity to help Gov. Huckabee. He is a man of great integrity and it is a great opportunity to help enhance his vision for a stronger and more prosperous America, there are many good people considering running for President. What stands-out most about Gov. Huckabee is his energetic, positive, and warm outlook about the future of America.
– Iowan Nick Ryan
Found on FOX News 6 years ago Politics
Share your thoughts on Iowan Nick Ryan's quotes with the community:
"Iowan Nick Ryan Quotes." Quotes.net. STANDS4 LLC, 2022. Web. 20 Jan. 2022. <https://www.quotes.net/authors/Iowan+Nick+Ryan+Quotes>.
Missing a quote of Iowan Nick Ryan?
Know another good quote of Iowan Nick Ryan? Don't keep it to yourself!
Who said: "The art of leadership is saying no, not yes – it's very easy to say yes.'"?
A. Donald Trump
B. Franklin Delano Roosevelt
C. Mao Tse-Tung
D. Tony Blair


UK police have launched an appeal after two British men disappeared from the Costa del Sol this week.
Daniel and Liam Poole, from Burgess Hill, flew to Malaga on March 31 but have not been heard from since April 1.
Family of the father and son, aged 46 and 22 and who have established links on the costa, fear they may be in trouble.
Sussex police said: "Have you seen either of them in Sussex or had contact with either of them since this date?
Alan Fenn of Sussex Police added: "This is extremely unusual behaviour from Daniel and Liam. They have been on holiday together before but never have they lost contact with family members.


The indictment, filed in Virginia, said patients were pointed towards doctors who prescribed opioids in a "careless and clinically unwarranted manner" as part of an effort to boost sales.
The indictment contains 28 counts of felony including one count of conspiracy to commit mail, wire and health care fraud; one count of health care fraud; four counts of mail fraud; and twenty-two counts of wire fraud.
Shares in Indivior fell by as much as 70% on announcement of the news. Shares in the Reckitt Benckiser Group were down 6%.


Big ticket: Antwan Odom, Tennessee; Justin Smith, Cincinnati.
Others worth looking at: Bobby McCray, Jacksonville; Travis LaBoy, Tennessee; John Engelberger and Ebenezer Ekuban, Denver; Josh Thomas, Indianapolis; Mike Rucker, Carolina.
Big ticket: T Flozell Adams, Dallas; G Alan Faneca, Pittsburgh.
Others worth looking at: G Jamar Nesbit, New Orleans; G Floyd Womack, Seattle; T L.J. Shelton, Miami; G Jake Scott, Indianapolis.
Bottom line: All the tackles worth signing were tagged, other than Adams. Past Faneca, there's not a high-level guard available. And they're lousy with centers, so that's not a need.
Best bet: Think the first round of a tackle-rich draft for more help here. If the right one's available, they could move Travelle Wharton to left guard and upgrade in two spots. That's probably what they're thinking, since both free agency and the draft is bad for interior linemen this year.
Big ticket: Michael Turner, San Diego.
Others worth looking at: Julius Jones, Dallas; Derrick Ward, N.Y. Giants; Michael Pittman, Tampa Bay; Chris Brown, Tennessee; Rock Cartwright, Washington; Musa Smith, Baltimore; DeShaun Foster, Carolina.
Bottom line: If they're serious about starting DeAngelo Williams, then there are a number of guys they can get to platoon with.
Best bet: A guy like Brown or Ward would be a cheap-enough add at a position where quantity is best. Turner would be good, but his free agency has created its own momentum, and somebody's going to back the truck up. That money's better spent elsewhere for the Panthers.
Others worth looking at: Erik Coleman, N.Y. Jets; Gibril Wilson, N.Y. Giants; Eugene Wilson, New England; Madieu Williams, Cincinnati.
Bottom line: As usual, there are a bunch of guys here. Most of them can play well enough. Unless you spend for a difference-maker, as they already have, the majority of them live in the great middle class. Once Dallas franchised Ken Hamlin, there wasn't really a big-name left out there.
Best bet: The typical Panthers strategy is to throw a bunch of names in a sack and pull one out. They've had mixed results with this approach, enjoying the hits (Marlon McCree, Chris Harris) and enduring the misses (Shaun Williams, the utterly functional Deke Cooper).
Big ticket: Randy Moss, New England; Donte Stallworth, New England; Bernard Berrian, Chicago.
Others worth looking at: Drew Carter, Carolina; Muhsin Muhammad, Chicago; Bryant Johnson, Arizona; Ernest Wilford, Jacksonville; Andre Davis, Houston; D.J. Hackett, Seattle; Marty Booker, Miami; Devery Henderson, New Orleans.
Bottom line: They've already got a Steve Smith, so don't even waste time thinking about Moss. There are enough guys here that will work.
Best bet: The Panthers hope their second receiver next year is Dwayne Jarrett, so don't expect them to overspend here. Muhammad makes a level of sense, otherwise they need to add guys in bulk. Davis is intriguing because he'd add some pop to the kick return game.


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Kenneth L. Gentry, Jr. (n. 3 mai 1950) este un teolog reformat și prezbiter hirotonit în . El este cunoscut în special pentru sprijinul său pentru și publicarea pe teme de preterismului ortodoxe și postmilenialismul în escatologia creștină, precum și pentru theonomy și crearea de șase zile. El susține că fiecare dintre aceste distinctivele teologice sunt extensii logice și teologice ale teologiei sale fundamentale, care este calvină și reformată.

Nașteri în 1950
Nașteri pe 3 mai
Prezbiterieni americani
Reconstrucționism creștin
Oameni în viață

redpajama_set_name: RedPajamaWikipedia


What areas do our Removalists Coolana cover?
We offer local, short distance removal services all over Coolana, please check out our list to know more about the areas that we cover.
MiniMovers Removalists Coolana will help you in all areas of your move with a comprehensive removal package including professional prepackers and unpackers who can also be hired at an hourly rate. Our MiniMovers Boxes will help prevent damage and speed up the moving process, thus dramatically decreasing the moving time and costs. Plus with our Buy Back system, we'll let you save even more. Storage is not a problem either – ask us for solutions and we'll help you with your self-storage needs. Book now and have the biggest, exceptionally professional local Removalists in Coolana to come and help you move!
Coolana is a small suburb located almost halfway between Gatton and Ipswich and if you are looking for a rural property or vacant land at a reasonable price, it may be the suburb for you. Coolana is part of the Somerset Regional Council which offers a large range of recreational facilities for all residents.
There haven't been too many property sales in Coolana in recent years, however taking into consideration sales since July 2012, the average property sale price is $342,000. Both vacant land and developed properties have been up for sale, however there have only been three properties sold since this date.
Coolana offers the perfect living for families with children and while it may feel like you are living in the middle of nowhere, there are plenty of primary schools around the area and a small number of high schools within relatively easy reach (although there are no schools in the suburb). For sporting options, Lowood is the closest main area (around 10km to the north) and offers sporting groups such as sailing, soccer, dance, golf, bowls, cricket, gymnastics, pony club and much more. If Lowood doesn't offer what you are after, there are a large range of options in both Ipswich and Gatton (both within a 35km drive).
The areas surrounding Coolana are very much agricultural with a mixture of maize, lucerne, hay, pumpkins, cattle and horses. Driving through the area you will also notice the buildings that date back from the development of the area in 1884 including the old train stations.
Just because you are living in a predominately rural area doesn't mean you can't enjoy a day on the water. Both Lake Atkinson (Atkinson Dam( and Lake Wivenhoe are just 25km away and both offer on and off water activities including boating, fishing, walking trails and picnic areas.
If you are looking for a more rural lifestyle, and are happy to work out of the city, Coolana may be the perfect offering for you and your family.


text:
AYMAN YOUSEF AHMAD ABU DAOUD
Shoemaker
Prison:
Kishon Prison
Click here for a PDF of Ayman's Profile.
Name: Ayman Yousef Ahmad Abu Daoud
Residence: Wadi Al Hirya/Hebron
Date of Re-arrest: 13 February 2012
Date of birth: 10 January 1982
Prison: Isolation in Jalameh (Kishon) Prison
Education: Student at Al Quds Open University
Profession: Shoemaker
UPDATE 28 May 2013 – Ayman ended his hunger strike on 23 May 2013 after 40 days. His lawyer struck a deal with the IPS for him to be deported to Gaza for ten years in exchange for an end to his hunger strike. The deportation will be enforced three months after the signing of the deal.
Release in Prisoner Exchange and Re-Arrest
Ayman Abu Daoud was released in the prisoners exchange deal on 18 October 2011 after serving 7 years out of a 35 year sentence. A condition of his release is that he visit the Israeli-Palestinian Authority military Coordination Office (DCO) in Hebron every two months and "sign-in" as proof of his residency in Hebron, where he resides with his wife and children.
On 13 February 2012, 4 months after their release, Ayman and several ex-prisoners presented at the DCO to sign their names. Ayman entered the DCO and never came out.
At 2:30 PM, within an hour of Ayman's fateful DCO visit, the Abu Daoud family and their neighbors were surprised to find a large convoy of the Israeli Occupying Forces' (IOF) soldiers entering the neighborhood by foot and jeep to raid Ayman's home. During the raid, the IOF blindfolded and shackled Ayman's brother Ahmad, and forced him against a wall. Then, Ayman's family and their visitors were corralled into one room as the soldiers searched and ransacked the home. The family later learned that they were searching for any cash stored in the home. During the raid, the IOF confiscated more than 20,000 NIS. The officer in charge of the raid then confirmed to the family that Ayman was arrested at the DCO during his routine two-month appointment, based on the claim that he violated the terms of his release.
The raid was conducted in broad daylight in front of a number of children, including Ayman's, who still remember the minute details to this moment.
Interrogation and Charges
After his arrest at the DCO, Ayman was transferred to Jalameh/Kishon Interrogation Center and was interrogated while being denied lawyer and family visits for the first ten days of his interrogation. After 15-17 days of interrogation, Ayman was formally charged with distributing financial allocations to persons affiliated with a political party. Due to his status as an ex-prisoner released in a prisoner exchange, Ayman is subject to Article 186 of Military Order 1651, which allows for the military prosecution at Ofer Military Court to demand that he stand in front of a special committee who determines if he should serve the remainder of his previous sentence. Due to this law, Ayman faces a 28 year sentence based on his old charges.
Ayman Abu Daoud launched a hunger strike on 14 April 2013 in protest of his arbitrary re-arrest especially as he did not violate the terms of the agreement as claimed by the military prosecution. Furthermore, he is on hunger strike in protest of the prosecution's attempt to sentence him to the remainder of his previous sentence instead of trying and sentencing him based on the new list of charges and not the old ones.
Upon launching his strike, Ayman was transferred from Megiddo Prison to isolation in Jalameh as a punishment. Ayman announced that he will continue his hunger strike until he is released.
While Ayman was detained in Megiddo prison, he was visited by his father and mother regularly. However, his wife has only been allowed two visits. She last tried to visit him in April 2012 and was later banned by the IOF for "security reasons." Ayman is no longer allowed family visits since he started his hunger strike and was transferred to Jalameh Prison and put in isolation.
Ayman lives in Hebron, near his father, mother and six brother and four sisters. Ayman is married with two children, Mohammad and Qutaiba. Ayman has not suffered from any previous health conditions, although it is expected that his health will deteriorate during his hunger strike.
He is the brother of Raed Abu Daoud, a martyr of the second Intifada. Ayman's older brother Mahmoud is also detained in Israeli jails. He has been arrested three times, most recently on 2 September 2012. Mahmoud is is detained in Nafha during his latest 34 month sentence.


The new audio conferencing changes are taking some time getting used to. There is one thing, however, that is bugging the heck out of our users here. We are a NC and NY based company. Now when you create a new phone conf from the web, the *4000 number inserts the label "Philadelphia, PA" instead of what it used to say, "Domestic (US)". Not a huge deal, but we are sensitive to this change because we have competitors in Phil and so we have our users going in after the fact and deleting this line. I know the local exchange is in Phil, PA but please change it to be a neutral sounding "Domestic US" or something along those lines. Thanks.


disclaimer: The author reserves the right not to be responsible and takes over however no adhesion or warranty for the topicality, correctness or quality ort he information provided. The content of this website is for personal use only and any downloading, printing or other copying or reproduction on any content of this site is only permitted for personal use, any other use of the content is expressly prohibited. Dennis Karsten is not responsible for any contents linked or referred to from these pages. The links which are provided on this website are not in the control of Dennis Karsten and no responsibility is accepted for the material or operation of such sites are free from malicious programmes, viruses, worms or any harmful component or that the use of the relevant site will be uninterrupted or free from error or defects. Dennis Karsten intended not to use any copyrighted material for the publication. The copyright for any material created by the author is reserved. All copyright matters protected by Dennis Karsten. If sections of individual terms of this disclaimer are not legal or correct, the content or validity of the other parts remain uninfluenced by this fact.


text:
'use strict';

var _createClass = function () { function defineProperties(target, props) { for (var i = 0; i < props.length; i++) { var descriptor = props[i]; descriptor.enumerable = descriptor.enumerable || false; descriptor.configurable = true; if ("value" in descriptor) descriptor.writable = true; Object.defineProperty(target, descriptor.key, descriptor); } } return function (Constructor, protoProps, staticProps) { if (protoProps) defineProperties(Constructor.prototype, protoProps); if (staticProps) defineProperties(Constructor, staticProps); return Constructor; }; }();

function _classCallCheck(instance, Constructor) { if (!(instance instanceof Constructor)) { throw new TypeError("Cannot call a class as a function"); } }

!function ($) {

  /**
   * ResponsiveAccordionTabs module.
   * @module foundation.responsiveAccordionTabs
   * @requires foundation.util.keyboard
   * @requires foundation.util.timerAndImageLoader
   * @requires foundation.util.motion
   * @requires foundation.accordion
   * @requires foundation.tabs
   */

  var ResponsiveAccordionTabs = function () {
    /**
     * Creates a new instance of a responsive accordion tabs.
     * @class
     * @fires ResponsiveAccordionTabs#init
     * @param {jQuery} element - jQuery object to make into a dropdown menu.
     * @param {Object} options - Overrides to the default plugin settings.
     */
    function ResponsiveAccordionTabs(element, options) {
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      this.$element = $(element);
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      this._init();
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      Foundation.registerPlugin(this, 'ResponsiveAccordionTabs');
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    /**
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     * @function
     * @private
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    _createClass(ResponsiveAccordionTabs, [{
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       * @private
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        } else {
          $panels = $liHeads.children('[data-tab-content]').removeClass('accordion-content');
        };

        $panels.css({ display: '', visibility: '' });
        $liHeads.css({ display: '', visibility: '' });
        if (toSet === 'accordion') {
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').remove();
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  ResponsiveAccordionTabs.defaults = {};

  // The plugin matches the plugin classes with these plugin instances.
  var MenuPlugins = {
    tabs: {
      cssClass: 'tabs',
      plugin: Foundation._plugins.tabs || null
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    accordion: {
      cssClass: 'accordion',
      plugin: Foundation._plugins.accordion || null
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  // Window exports
  Foundation.plugin(ResponsiveAccordionTabs, 'ResponsiveAccordionTabs');
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redpajama_set_name: RedPajamaGithub


*Potluck Lunch, Family Activity FUN! Arts, Crafts, Contests! A gift for Mom!
Schedule: 11:00 Devotion 12:00 *Potluck Lunch and Family Activity FUN!
*Potluck item is voluntary, your choice.
– Mothers Day Sunday, May 8, 2016 from 11 am to 2 pm or so.
Spirit of Peace is a missional community of believers. We believe that the community has life that is greater than its parts. Spirit celebrates its Moms on Mother's Day! When you come right down to it, where would we be without our Moms? May God Bless All Moms!
Stop by and visit us for a Mother's Day gift.


Joe Wulff's debut LP and CD, "Ghost Under Water" portrays a heartfelt narrative of loss and love through six unforgettable songs. With his contemporary sensibility singer-songwriter guitarist Joe also pays homage to the sounds of the early seventies.
Each LP includes a one-time use download code managed by Optimal in Germany.


text:
\section{Introduction}

Cohomological Hall algebra of a quiver (with potential) was introduced by Kontesivech and Soibelman in \cite{KS}, which became a central tool in the study of quantized Donaldson-Thomas invariants. The underlying space of the cohomological Hall algebra of a finite quiver  is the equivariant cohomology group of representations of the quiver, while multiplication is  defined by using pull-back and push-forward operations. The discovery of cohomological Hall algebras opens new directions for motivic Donaldson-Thomas invariants, mathematical physics, representation theory and so on. Different versions of cohomological Hall algebras have been introduced and studied, see \cite{FR, SS, SV, YZ} for instance. To find a simple algebraic description of a cohomological Hall algebra is a hot topic (cf. \cite{FR19,SV}).

Let $d$ be a positive integer and $Q$ the $d$-loop quiver. The cohomological Hall algebra $\mathcal{H}^d$ of $Q$ admits an algebraic description. Namely, it is isomorphic to the space of symmetric polynomials in all possible numbers of variables endowed with a shuffle-type algebra structure. In order to develop an explicit and combinatorial setup for the study of the quantized Donaldson-Thomas invariants of $Q$,  Reineke \cite{R} introduced the quantized cohomological Hall algebra $\mathcal{H}^d_q$ and the degenerate cohomological Hall algebra $\mathcal{H}^d_0$ of $Q$. In particular, the cohomological Hall algebra $\mathcal{H}^d$ is the specialization of $\mathcal{H}^d_q$ at $q=1$, while the degenerate cohomological Hall algebra $\mathcal{H}^d_0$ is the specialization of $\mathcal{H}^d_q$ at $q=0$. Surprisingly, the degenerate cohomological Hall algebra $\mathcal{H}_0^d$ admits a combinatorial realization via partitions, which plays a key role in the study of quantized Donaldson-Thomas invariants for $Q$ (cf. \cite{R}).

The aim of this note is to pursue a similar combinatorial realization of $\mathcal{H}^d_q$ as $\mathcal{H}^d_0$ and to understand the associativity of the multiplication of $\mathcal{H}^d_q$. To do this, we introduce an algebra $\mathbb{A}^d_q(\Lambda)$ associated to partitions, which can be viewed as a quantization of the algebra $\mathbb{A}^d(\Lambda)$ introduced in \cite[Definition 5.2]{R}.  The associativity of the multiplication of $\mathbb{A}^d_q(\Lambda)$ comes from combinatorial properties of a class of polynomials, which have already appeared in the algebraic description of (quantized) cohomological Hall algebra $\mathcal{H}^d_q$.
We show that
$\mathbb{A}^d_q(\Lambda)$ is isomorphic to the quantized cohomological Hall algebra $\mathcal{H}^d_q$ and the isomorphism specializes to the isomorphism between $\mathbb{A}^d(\Lambda)$ and the degenerate cohomological Hall algebra $\mathcal{H}^d_0$.

Throughout this note, we fix a positive integer $d$.

\section{Quantized cohomological Hall algebras}

In this section, we recall the algebraic descriptions of the cohomological Hall algebra $\mathcal{H}^d$, the quantized cohomological Hall algebra $\mathcal{H}^d_q$ and the degenerate cohomological Hall algebra $\mathcal{H}^d_0$ of the $d$-loop quiver.

\subsection{$\mathcal{H}^d$ of the $d$-loop quiver}

For a non-negative integer $n$, denote by $S_n$ the symmetric group on $n$ letters. Let $\mathbb{Q}[x_1,\ldots, x_n]$ be the ring of polynomials in variables $x_1, \ldots, x_n$ with rational coefficients. Denote by $\mathbb{Q}[x_1,\ldots, x_n]^{S_n}$ the subspace of $\mathbb{Q}[x_1,\ldots, x_n]$ consisting of symmetric polynomials.

 Let
\[\mathcal{H}^d=\bigoplus_{n\geq 0}\mathbb{Q}[x_1, \ldots, x_n]^{S_n}.
\]
We endow $\mathcal{H}^d$ with a multiplication $\ast$ as follows.
Given $f_1\in \mathbb{Q}[x_1,\ldots, x_{n_1}]^{S_{n_1}}$, $f_2\in \mathbb{Q}[x_1,\ldots, x_{n_2}]^{S_{n_2}}$, let
\begin{align}
&f_1\ast f_2(x_1,\ldots,x_{n_1+n_2})\nonumber\\
&=\sum_{\{i_1,\ldots,i_{n_1}\}} f_1(x_{i_1},\ldots,x_{i_{n_1}})f_2(x_{j_1},\ldots,x_{j_{n_2}})
\prod_{l=1}^{n_2}\prod_{t=1}^{n_1}
(x_{j_l}-x_{i_t})^{d-1},\label{qe0}
\end{align}
where     the sum ranges over all the subsets $\{i_1,\ldots,i_{n_1}\}$    of $\{1,2,\ldots,n_1+n_2\}$ and $\{j_1,\ldots,j_{n_2}\}=\{1,2,\ldots,n_1+n_2\}\setminus
\{i_1,\ldots,i_{n_1}\}$. According to \cite[Theorem 2.2]{KS}, the \textit{cohomological Hall algebra} of the $d$-loop quiver is isomorphic to $(\mathcal{H}^d, \ast)$.

\subsection{$\mathcal{H}^d_q$ and $\mathcal{H}^d_0$ of $d$-loop quiver}

Let $q$ be an indeterminate and $\mathbb{Q}[q]$ the ring of polynomials in $q$ with rational coefficients.
The (naive) {\it quantized cohomological Hall algebra} $\mathcal{H}_q^d$ is the graded $\mathbb{Q}[q]$-module
\[\bigoplus_{n\geq 0}\mathbb{Q}[q][x_1,\ldots,x_n]^{S_n}
\]
endowed with the multiplication
\begin{align}
&f_1\ast f_2(x_1,\ldots,x_{n_1+n_2})\nonumber\\
&=\sum_{\{i_1,\ldots,i_{n_1}\}} f_1(x_{i_1},\ldots,x_{i_{n_1}})f_2(x_{j_1},\ldots,x_{j_{n_2}})
\prod_{l=1}^{n_2}\prod_{t=1}^{n_1}
(x_{j_l}-qx_{i_t})^{d-1},\label{qe1}
\end{align}
where $f_1\in \mathbb{Q}[q][x_1,\ldots, x_{n_1}]^{S_{n_1}}$, $f_2\in \mathbb{Q}[q][x_1,\ldots, x_{n_2}]^{S_{n_2}}$ and  the sum ranges over all the subsets $\{i_1,\ldots,i_{n_1}\}$    of $\{1,2,\ldots,n_1+n_2\}$ and $\{j_1,\ldots,j_{n_2}\}=\{1,2,\ldots,n_1+n_2\}\setminus
\{i_1,\ldots,i_{n_1}\}$, see \cite[Definition 5.3]{R}.

We can specialize the algebra $\mathcal{H}^d_q$ to any $q\in \mathbb{Q}$.  The {\it degenerate cohomological Hall algebra} $\mathcal{H}^d_0$ of the $d$-loop quiver is the specialization $\mathcal{H}^d_q|_{q=0}$ of $\mathcal{H}^d_q$ at $q=0$. In particular, $\mathcal{H}^d_0$ has the same underlying space as $\mathcal{H}^d$ and its multiplication  reduces to
\begin{align*}
&f_1\ast f_2(x_1,\ldots,x_{n_1+n_2})\nonumber\\
&=\sum_{\{i_1,\ldots,i_{n_1}\}} f_1(x_{i_1},\ldots,x_{i_{n_1}})f_2(x_{j_1},\ldots,x_{j_{n_2}})
\left(\prod_{l=1}^{n_2}
x_{j_l}\right)^{(d-1)n_1}.
\end{align*}

\subsection{A combinatorial realization of $\mathcal{H}^d_0$}

Let $\mathbb{N}$ be the set of non-negative integers.
Let $n$ be a non-negative integer. A partition of length $n$ is a vector $\lambda=(\lambda_1,\lambda_2,\ldots, \lambda_n)\in \mathbb{N}^n$ such that $\lambda_1\leq \lambda_2\leq \cdots\leq \lambda_n$.  For a partition $\lambda$, we denote by $|\lambda|$ the length of $\lambda$. For an integer vector $\mu\in \mathbb{N}^n$, denote by $\mu_{\leq}$ the partition obtained by rearranging the sequence of $\mu$ increasingly.

Let $\Lambda_n$ be the set of partitions  of length $n$. Denote by $\Lambda$ the disjoint union of all $\Lambda_n$ for $n\geq 0$. Reineke \cite[Definition 5.2]{R} introduced an associative algebra $\mathbb{A}^d(\Lambda)$, which is the $\mathbb{Q}$-vector space with basis elements $\lambda\in \Lambda$ endowed with a multiplication as follows
\begin{align}
\mu\ast \nu=(\mu_1,\ldots, \mu_m, \nu_1+m(d-1), \ldots, \nu_n+m(d-1))_{\leq},
\end{align}
where $|\mu|=m$ and $|\nu|=n$. The multiplication of $\mathbb{A}^d(\Lambda)$ is obviously associative. It has been proved by \cite[Proposition 5.4]{R} that $\mathbb{A}^d(\Lambda)$ is isomorphic to $\mathcal{H}^d_0$.

\section{The  algebra $\mathbb{A}^d_q(\Lambda)$}

In this section, we will introduce a  $\mathbb{Q}[q]$-algebra associated to partitions, which can be viewed as a quantization of $\mathbb{A}^d(\Lambda)$.
\subsection{The polynomial $g_{m,n}$}
Let $x_1,x_2,\ldots$ be infinitely many formal variables. For a non-negative integer vector ${\bf a}=(a_1,\ldots, a_n)\in \mathbb{N}^n$, we write
\[x^{\bf a}:=\prod_{i=1}^nx_{i}^{a_i}.
\]
For a pair of positive integers $(m,n)$, denote by  $g_{m,n}\in \mathbb{Q}[q][x_1,\ldots, x_{m+n}]$ the following polynomial
\begin{eqnarray}\label{gmn}
g_{m,n}:=
\prod_{s=m+1}^{m+n}\prod_{t=1}^m(x_s-qx_t)^{d-1}
=\sum_{\tiny \begin{array}{l}{\bf a}\in \mathbb{N}^m \\
{\bf b}\in \mathbb{N}^n\end{array}}c_{{m,n}}^{{\bf a},
{\bf b}}x^{({\bf a}, {\bf b})},
\end{eqnarray}
where $c_{m,n}^{{\bf a}, {\bf b}}\in \mathbb{Q}[q]$ is the coefficient of the monomial $x^{({\bf a},{\bf b})}:=x^{(a_1,\ldots,a_m,b_1,\ldots,b_n)}$ in the expansion of $g_{m,n}$. Let $l$ be a positive integer, we also denote
\[g_{m,n}[l]:=\prod_{s=l+m+1}^{l+m+n}\prod_{t=l+1}^{l+m}(x_s-qx_t)^{d-1}.
\]

By definition, it is easy to see that $g_{m,n}$ is symmetric with respect to
$x_1,\ldots, x_m$ and $x_{m+1}, \ldots, x_{m+n}$, respectively.
Consequently, for any ${\bf a}\in \mathbb{N}^m$,
${\bf b}\in \mathbb{N}^n$
and   any permutations $\tau\in S_m,\sigma\in S_n$,
\[c_{m,n}^{\tau({\bf a}), \sigma({\bf b})}=c_{m,n}^{{\bf a}, {\bf b}}.\]

\begin{lemma}\label{l:cofficient-equality}
Let $l,m,n$ be positive integers and ${\bf a}\in \mathbb{N}^l, {\bf b}\in \mathbb{N}^m,{\bf c}\in \mathbb{N}^n$. We have
\begin{eqnarray}\label{f:coefficient-equality}
\sum_{\tiny \begin{array}{l}{\bf a}_1+{\bf a}_2={\bf a}\\ {\bf b}_1+{\bf b}_2={\bf b}
\end{array}}c_{l,m}^{{\bf a}_1, {\bf b}_1}c_{l+m,n}^{({\bf a}_2,{\bf b}_2), {\bf c}}
=\sum_{\tiny \begin{array}{l}{\bf c}_1+{\bf c}_2={\bf c}\\ {\bf b}_1+{\bf b}_2={\bf b}
\end{array}}c_{l,m+n}^{{\bf a}, ({\bf b}_2, {\bf c}_2)}c_{m,n}^{{\bf b}_1, {\bf c}_1},
\end{eqnarray}
where the sums are taking over all non-negative integer vectors ${\bf a}_1$, ${\bf a}_2$, ${\bf b}_1$, ${\bf b}_2$, ${\bf c}_1$, ${\bf c}_2$.
\end{lemma}

\begin{proof}
By definition, we have
\begin{eqnarray*}
g_{l,m}g_{l+m,n}&=&\prod_{s=l+1}^{l+m}\prod_{t=1}^l(x_s-qx_t)^{d-1} \prod_{s=l+m+1}^{l+m+n}\prod_{t=1}^{l+m}(x_s-qx_t)^{d-1} \\[5pt]
&=&\left(\sum_{\tiny \begin{array}{l}{\bf a}_1\in\mathbb{N}^l \\
{\bf b}_1\in\mathbb{N}^m\end{array}}c_{{l,m}}^{{\bf a}_1,{\bf b}_1}
x^{({\bf a}_1, {\bf b}_1)}\right) \left(\sum_{\tiny \begin{array}{l}
{\bf a}_2\in \mathbb{N}^l \\ {\bf b}_2\in \mathbb{N}^m\\ {\bf c}
\in \mathbb{N}^n\end{array}}c_{{l+m,n}}^{({\bf a}_2,{\bf b}_2),
{\bf c}}x^{({\bf a}_2, {\bf b}_2, {\bf c})}\right)\\[5pt]
&=&\sum_{\tiny \begin{array}{l}{\bf a}_1,{\bf a}_2\in \mathbb{N}^l \\ {\bf b}_1, {\bf b}_2\in \mathbb{N}^m\\ {\bf c}\in \mathbb{N}^n\end{array}}c_{{l,m}}^{{\bf a}_1,{\bf b}_1}c_{{l+m,n}}^{({\bf a}_2,{\bf b}_2),{\bf c}}x^{({\bf a}_1+{\bf a}_2, {\bf b}_1+{\bf b}_2, {\bf c})}.
\end{eqnarray*}
Thus the left hand side of (\ref{f:coefficient-equality}) is  the coefficient of the monomial $x^{({\bf a}, {\bf b}, {\bf c})}$ in the expansion of $g_{l,m}g_{l+m,n}$.

On the other hand, consider the  following polynomial
\begin{eqnarray*}
g_{l,m+n}(g_{m,n}[l])&=&\prod_{s=l+1}^{l+m+n}\prod_{t=1}^l (x_s-qx_t)^{d-1} \prod_{s=l+m+1}^{l+m+n}\prod_{t=l+1}^{l+m}(x_s-qx_t)^{d-1}\\[5pt]
&=&\left(\sum_{\tiny \begin{array}{l}{\bf a}\in \mathbb{N}^l\\ {\bf b}_1\in \mathbb{N}^m\\ {\bf c}_1\in \mathbb{N}n\end{array}}c_{l,m+n}^{{\bf a}, ({\bf b}_1,{\bf c}_1)}x^{({\bf a}, {\bf b}_1,{\bf c}_1)}\right)\left(\sum_{\tiny \begin{array}{l}{\bf b}_2\in \mathbb{N}^m\\ {\bf c}_2\in \mathbb{N}n\end{array}}c_{m,n}^{{\bf b}_2,{\bf c}_2}x^{({\bf 0}, {\bf b}_2,{\bf c}_2)}\right)\\[5pt]
&=&\sum_{\tiny \begin{array}{l}{\bf a}\in \mathbb{N}^l\\ {\bf b}_1,{\bf b}_2\in \mathbb{N}^m\\ {\bf c}_1,{\bf c}_2\in \mathbb{N}n\end{array}}c_{l,m+n}^{{\bf a}, ({\bf b}_1,{\bf c}_1)}c_{m,n}^{{\bf b}_2,{\bf c}_2}x^{({\bf a}, {\bf b}_1+{\bf b}_2,{\bf c}_1+{\bf c}_2)}.
\end{eqnarray*}
Then the right hand side of (\ref{f:coefficient-equality}) is  the coefficient of the monomial $x^{({\bf a}, {\bf b}, {\bf c})}$ in the expansion of $g_{l,m+n}(g_{m,n}[l])$. The equality (\ref{f:coefficient-equality}) follows from the following obvious relation
\[g_{l,m}g_{l+m,n}=g_{l,m+n}(g_{m,n}[l]).\]
\end{proof}

\subsection{An auxiliary algebra $\mathbb{A}^d_q(\hat{\Lambda})$}
Denote by $\hat{\Lambda}$ the set of disjoint union of all $\mathbb{N}^n$ for $n\geq 0$. Let $\mathbb{A}^d_q(\hat{\Lambda})$ be the free $\mathbb{Q}[q]$-module with basis elements $\lambda\in \hat{\Lambda}$. For an element $\lambda\in \hat{\Lambda}$, we denote by $|\lambda|$ the length of $\lambda$, that is, $\lambda\in \mathbb{N}^{|\lambda|}$. Denote by $\emptyset$ the unique element of $\mathbb{N}^0$.
Let $\mu\in \mathbb{N}^m, \nu\in \mathbb{N}^n$ with $m,n\geq 1$, we define
$
\mu\ast \emptyset=\mu=\emptyset \ast \mu
$ and
\begin{align}\label{auxi}
\mu\ast\nu=\sum_{\tiny\begin{array}{l}{\bf a}\in \mathbb{N}^m\\ {\bf b}\in \mathbb{N}^n \end{array}}c_{m,n}^{{\bf a}, {\bf b}}(\mu+{\bf a}, \nu+{\bf b}),
\end{align}
where $c_{m,n}^{{\bf a}, {\bf b}}$ is as defined in \eqref{gmn}. Extend the multiplication defined in \eqref{auxi} $\mathbb{Q}[q]$-linearly to a multiplication of $\mathbb{A}^d_q(\hat{\Lambda})$.

\begin{lemma}
$(\mathbb{A}^d_q(\hat{\Lambda}), \ast)$ is an associative algebra with unit.
\end{lemma}
\begin{proof}
By definition $\emptyset$ is the unit. It remains to prove the associativity, that is, for $\mu,\nu,w\in  \hat{\Lambda}$, one has  $(\mu\ast \nu)\ast w=\mu\ast(\nu\ast w)$. There is nothing to prove if one of $\mu,\nu,w$ is $\emptyset$. Without loss of generality, we may assume that $|\mu|=l, |\nu|=m,|w|=n \geq 1$. By definition,
\begin{align*}
(\mu\ast \nu)\ast w&=\sum_{\tiny \begin{array}{l} {\bf a}_1\in \mathbb{N}^l\\ {\bf b}_1\in \mathbb{N}^m\end{array}}c_{l,m}^{{\bf a}_1, {\bf b}_1}(\mu+{\bf a}_1, \nu+{\bf b}_1)\ast w\\
&=\sum_{\tiny \begin{array}{l} {\bf a}_1\in \mathbb{N}^l\\ {\bf b}_1\in \mathbb{N}^m\end{array}}c_{l,m}^{{\bf a}_1, {\bf b}_1} \sum_{\tiny \begin{array}{l} {\bf a}_2\in \mathbb{N}^l\\ {\bf b}_2\in \mathbb{N}^m\\ {\bf c}\in \mathbb{N}^n\end{array}}c_{l+m,n}^{({\bf a}_2, {\bf b}_2), {\bf c}}(\mu+{\bf a}_1+{\bf a_2}, \nu+{\bf b}_1+{\bf b}_2, w+{\bf c})\\
&=\sum_{\tiny \begin{array}{l} {\bf a}_1, {\bf a}_2\in \mathbb{N}^l\\ {\bf b}_1,{\bf b}_2\in \mathbb{N}^m\\ {\bf c}\in \mathbb{N}^n\end{array}}c_{l,m}^{{\bf a}_1, {\bf b}_1}
c_{l+m,n}^{({\bf a}_2, {\bf b}_2), {\bf c}}
(\mu+{\bf a}_1+{\bf a_2}, \nu+{\bf b}_1+{\bf b}_2, w+{\bf c}).
\end{align*}
Therefore, the coefficient of $(\mu+{\bf a}, \nu+{\bf b}, w+{\bf c})$ in $(\mu\ast\nu)\ast w$ is
\[\sum\limits_{\tiny \begin{array}{l}{\bf a}_1+{\bf a}_2={\bf a}\\ {\bf b}_1+{\bf b}_2={\bf b}
\end{array}}c_{l,m}^{{\bf a}_1, {\bf b}_1}c_{l+m,n}^{({\bf a}_2,{\bf b}_2), {\bf c}}.\]

On the other hand, we have
\begin{eqnarray*}
\mu\ast(\nu\ast w)&=&\mu\ast \sum_{\tiny \begin{array}{l}{\bf b}_2\in \mathbb{N}^n\\ {\bf c}_2\in \mathbb{N}^n\end{array}}c_{m,n}^{{\bf b}_2, {\bf c}_2}(\nu+{\bf b}_2, w+{\bf c}_2)\\
&=&\sum_{\tiny \begin{array}{l} {\bf a}\in \mathbb{N}^l\\ {\bf b}_1,{\bf b}_2\in \mathbb{N}^m\\ {\bf c}_1, {\bf c}_2\in \mathbb{N}^n\end{array}}c_{l,m+n}^{{\bf a}, ({\bf b}_2, {\bf c}_2)}
c_{m,n}^{{\bf b}_1, {\bf c}_1}
(\mu+{\bf a}, \nu+{\bf b}_1+{\bf b}_2, w+{\bf c}_1+{\bf c}_2).
\end{eqnarray*}
Thus the coefficient of $(\mu+{\bf a}, \nu+{\bf b}, w+{\bf c})$ in $\mu\ast(\nu\ast w)$ is
\[\sum\limits_{\tiny \begin{array}{l}{\bf c}_1+{\bf c}_2={\bf c}\\ {\bf b}_1+{\bf b}_2={\bf b}
\end{array}}c_{l,m+n}^{{\bf a}, ({\bf b}_2, {\bf c}_2)}c_{m,n}^{{\bf b}_1, {\bf c}_1}.\]
We conclude that $(\mu\ast\nu)\ast w=\mu\ast(\mu\ast w)$ by Lemma \ref{l:cofficient-equality}.
\end{proof}
\begin{remark}
In fact,  $\mathbb{A}^d_q(\hat{\Lambda})$ is an $\mathbb{N}$-graded algebra
\[\mathbb{A}^d_q(\hat{\Lambda})=\bigoplus_{n\in \mathbb{N}}\mathbb{A}^d_q(\hat{\Lambda})_n,
\]
whose $n$-th component is the free $\mathbb{Q}[q]$-module with basis $\mathbb{N}^n$.
\end{remark}

\subsection{The algebra $\mathbb{A}^d_q(\Lambda)$ associated to partitions}

Let $\mu,\nu\in \hat{\Lambda}$, we denote by $\mu\sim \nu$ if $\mu$ is a permutation of $\nu$. Clearly, $\sim$ is an equivalence relation of $\hat{\Lambda}$ and
we may identify $\Lambda$ with the quotient $\hat{\Lambda}/\sim$
 of $\hat{\Lambda}$ by $\sim$. Let $W$ be the
  $\mathbb{Q}[q]$-subspace of $\mathbb{A}^d_q(\hat{\Lambda})$ spanned by
\[
\{\mu\circleddash\nu~|~ \text{for all $\mu,\nu\in \hat{\Lambda}$
such that $\mu\sim \nu$}\}.
\]
Here, to
distinguish with the minus of vectors, we use $\circleddash$ to represent the minus
of $\mathbb{A}^d_q(\hat{\Lambda})$.

\begin{lemma}\label{l:homogeneous}
The two-side ideal $( W)$ of
 $\mathbb{A}^d_q(\hat{\Lambda})$ generated by $W$ is exactly $W$.
\end{lemma}
\begin{proof}
Let $\mu\in \hat{\Lambda}$ with $|\mu|=m$ and $\sigma\in S_m$. For any $w\in \hat{\Lambda}$ with $|w|=n$, we have
\begin{align*
&(\mu\circleddash\sigma(\mu))\ast w\\
&=
\sum_{\tiny\begin{array}{l}{\bf a}\in\mathbb{N}^m\\
{\bf b}\in\mathbb{N}^n\end{array}}c_{m,n}^{{\bf a}, {\bf b}}
(\mu+{\bf a}, w+{\bf b})\circleddash\sum_{\tiny \begin{array}{l}{\bf a}\in \mathbb{N}^m\\
{\bf b}\in \mathbb{N}^n\end{array}}c_{m,n}^{{\bf a}, {\bf b}}
(\sigma(\mu)+{\bf a}, w+{\bf b})\\
&=\sum_{\tiny\begin{array}{l}{\bf a}\in\mathbb{N}^m\\
{\bf b}\in\mathbb{N}^n\end{array}}c_{m,n}^{{\bf a}, {\bf b}}
(\mu+{\bf a}, w+{\bf b})\circleddash\sum_{\tiny \begin{array}{l}{\bf a}\in \mathbb{N}^m\\
{\bf b}\in \mathbb{N}^n\end{array}}c_{m,n}^{{\bf a}, {\bf b}}
(\sigma(\mu)+\sigma({\bf a}), w+{\bf b})\\
&=\sum_{\tiny\begin{array}{l}{\bf a}\in\mathbb{N}^m\\
{\bf b}\in\mathbb{N}^n\end{array}}c_{m,n}^{{\bf a}, {\bf b}}[(\mu+{\bf a},w+{\bf b})\circleddash (\sigma(\mu+{\bf a}), w+{\bf b})].
\end{align*}
As a consequence, $(\mu\circleddash\sigma(\mu))\ast w$ belongs to $W$. Similarly, one can show that $w\ast (\mu\circleddash\sigma(\mu))$ belongs to $W$ as well.
\end{proof}

\begin{definition}\label{d:quantized-alg}
The algebra $\mathbb{A}^d_q(\Lambda)$ associated to partitions is the quotient algebra $\mathbb{A}^d_q(\hat{\Lambda})/(W)$.
\end{definition}

The following result is a direct consequence of Lemma \ref{l:homogeneous}, Definition \ref{d:quantized-alg} and (\ref{auxi}).

\begin{proposition}
\begin{enumerate}
\item[(1)] $\mathbb{A}^d_q(\Lambda)$ is an $\mathbb{N}$-graded associative algebra with unit, whose $n$-th component  is a free $\mathbb{Q}[q]$-module with basis $\Lambda_n$;
\item[(2)]  Let $\mu\in \Lambda_m, \nu\in \Lambda_n$. The multiplication of $\mu$ with $\nu$ in  $\mathbb{A}^d_q(\Lambda)$ is as follows
\begin{eqnarray}\label{e:multiplication}
\mu\ast\nu=\sum_{\tiny \begin{array}{l}{\bf a}\in \mathbb{N}^m\\ {\bf b}\in \mathbb{N}^n\end{array}}c_{m,n}^{{\bf a}, {\bf b}}(\mu+{\bf a}, \nu+{\bf b})_{\leq},
\end{eqnarray}
where $c_{m,n}^{{\bf a}, {\bf b}}$ is as defined in \eqref{gmn}.
\end{enumerate}
\end{proposition}

Note that $c_{m,n}^{{\bf a},{\bf b}}$ is a polynomial in $q$. Hence we can specialize the value of $c_{m,n}^{{\bf a},{\bf b}}$ at $q=0$. The following result suggests that $\mathbb{A}^d_q(\Lambda)$ is a quantization of $\mathbb{A}^d(\Lambda)$.
\begin{corollary}
By specializing $q$ to $0$, we obtain
\[\mathbb{A}^d_q(\Lambda)|_{q=0}\cong \mathbb{A}^d(\Lambda).
\]
\end{corollary}
\begin{example} Assume that $d=4$.
 Let $\mu=(0,2),\nu=(1)$. Then
\[g_{2,1}=(x_3-qx_1)^3(x_3-qx_2)^3.\]
One can compute $\mu\ast\nu$ in $\mathbb{A}^4_q(\Lambda)$ as follows.
\begin{align*}
\mu\ast\nu&=(q^6+3q^4+9q^2)\cdot(1, 3, 5)+(3q^2-3q^5)\cdot(2, 2, 5)\\
&\quad+(9q^4-3q^5-10q^3)\cdot(2, 3, 4)
+3q^4\cdot(3, 3, 3)+(3q^2-q^3)\cdot(0, 4, 5)\\
&\quad -9q^3\cdot(1, 4, 4) -3q\cdot(0, 3, 6) -3q\cdot(1, 2, 6)+(0, 2, 7).
\end{align*}
\end{example}
Let $q^{-1}$ be the formal inverse of $q$ and
$\mathbb{Q}[q^{\pm}]$ the Laurent polynomials in $q$
with rational coefficients. Let us consider the tensor
 product $\mathbb{Q}[q^{\pm}]\otimes_{\mathbb{Q}[q]}
 \mathbb{A}^d_q(\Lambda)$. In particular,
 $\mathbb{A}^d_q(\Lambda)$ is a subalgebra of
  $\mathbb{Q}[q^{\pm}]\otimes_{\mathbb{Q}[q]}\mathbb{A}^d_q(\Lambda)$.
 The bar involution $\bar{ } : \mathbb{Q}[q^\pm]\to \mathbb{Q}[q^\pm]$ given by $q^i\mapsto q^{-i}$($i\in \mathbb{Z}$),  induces a bar involution of
\begin{eqnarray*}
\bar{}:&&\mathbb{Q}[q^{\pm}]\otimes_{\mathbb{Q}[q]}\mathbb{A}^d_q(\Lambda)\to \mathbb{Q}[q^{\pm}]\otimes_{\mathbb{Q}[q]}\mathbb{A}^d_q(\Lambda),\\
&&f=\sum a_\mu\mu\mapsto \bar{f}=\sum\bar{a}_\mu\mu,
\end{eqnarray*}
where $a_\mu\in \mathbb{Q}[q^\pm]$. The following is a direct consequence of the property of the polynomial $g_{m,n}$ and (\ref{e:multiplication}).
\begin{proposition}
Let $\mu\in \Lambda_m,\nu\in \Lambda_n$. We have
\[\mu\ast\nu=q^{(d-1)mn}\overline{\nu\ast\mu}.
\]
\end{proposition}

\section{Isomorphism from $\mathbb{A}^d_q(\Lambda)$ to $\mathcal{H}^d_q$}

For a non-negative integer sequence $\lambda=(\lambda_1,\ldots, \lambda_n)\in \mathbb{N}^n$, we denote by
\begin{align}\label{eq2}
M_{\lambda}(x_1,\ldots,x_n):=\sum_{\sigma\in S_n}x_{\sigma(1)}^{\lambda_1}\cdots x_{\sigma(n)}^{\lambda_n}\in \mathcal{H}^d_q.
\end{align}
Let \[c({\lambda})=|\{\sigma\in S_n~|~ \lambda=(\lambda_{\sigma(1)}, \dots, \lambda_{\sigma(n)})\}|.
\]
Denote by $S_\lambda$ the set of different permutations of the parts of $\lambda$, that is,
\[S_\lambda=\{(t_1,\dots, t_n)~|~(t_1,\dots, t_n)=(\lambda_{\sigma(1)},\dots, \lambda_{\sigma(n)}), \text{for some $\sigma\in S_n$}\}.
\]
Clearly,
\[c(\lambda)=\frac{n!}{|S_\lambda|}.
\]
Let $m_\lambda(x_1,\ldots,x_n)$ be the monomial symmetric function of $\lambda$, that is, the sum of all the monomials $x_1^{\pi(1)}\cdots x_n^{\pi(n)}$, where $\pi\in S_\lambda$.
It is obvious that
\[M_{\lambda}(x_1,\ldots,x_n)=c(\lambda)\cdot m_{\lambda}(x_1,\ldots,x_n).\]
It is also easy to see that $m_{\lambda}(x_1,\ldots,x_n)=m_{\lambda'}(x_1,\ldots,x_n)$ and $M_{\lambda}(x_1,\ldots,x_n)=M_{\lambda'}(x_1,\ldots,x_n)$  for any $\lambda'\in S_{\lambda}$.

The following proposition is already  known, see, for example, Carvalho and  D'Agostino \cite{CD},  we give a short proof here for the convenience of the readers.

\begin{proposition}\label{pmp}
Let $\lambda,\mu$ be two partitions of length $n$. Then
\[M_{\lambda}(x_1,\ldots,x_n)m_{\mu}(x_1,\ldots,x_n)
=\sum_{w\in S_\mu}M_{\lambda+w}(x_1,\ldots,x_n).\]
\end{proposition}

\begin{proof} Fix $w\in S_{\mu}$, choose a permutation $\tau\in S_n$ (not necessarily unique) such that $w_i=\mu_{\tau(i)}.$ Then we have
\begin{eqnarray*}M_{\lambda+w}(x_1,\cdots,x_n)&=&\sum_{\pi\in S_{n}}x_{\pi(1)}^{\lambda_1+w_1}\cdots x_{\pi(n)}^{\lambda_n+w_n}\\
&=&\sum_{\pi\in S_{n}}x_{\pi(1)}^{\lambda_1+\mu_{\tau(1)}}\cdots x_{\pi(n)}^{\lambda_n+\mu_{\tau(n)}}.
\end{eqnarray*}
On the other hand,
\begin{align*}
M_{\lambda}(x_1,\cdots,x_n)M_{\mu}(x_1,\cdots,x_n)
&=\sum_{\pi\in S_n}x_{\pi(1)}^{\lambda_1}\cdots x_{\pi(n)}^{\lambda_n}\sum_{\beta\in S_n}x_{\beta(1)}^{\mu_1}\cdots x_{\beta(n)}^{\mu_n}\\
&=\sum_{\pi\in S_n}\left(\sum_{\beta\in S_n}x_{\pi(1)}^{\lambda_1+
\mu_{\beta^{-1}(\pi(1))}}\cdots x_{\pi(n)}^{\lambda_n+\mu_{\beta^{-1}(\pi(n))}}\right),
\end{align*}
For a given monomial $x_{\pi(1)}^{\lambda_1+\mu_{\tau(1)}}\cdots x_{\pi(n)}^{\lambda_n+\mu_{\tau(n)}}$ of $M_{\lambda+w}(x_1,\ldots,x_n)$,
there are $c(\mu)$  permutations $\beta\in S_n$  in the summand
\[\sum_{\beta\in S_n}x_{\pi(1)}^{\lambda_1+
\mu_{\beta^{-1}(\pi(1))}}\cdots x_{\pi(n)}^{\lambda_n+\mu_{\beta^{-1}(\pi(n))}}\]
such that $\mu_{\beta^{-1}(\pi(i))}=\mu_{\tau(i)}$ for $i=1,\ldots,n$. Moreover, since there are $|S_{\mu}|$ different $\lambda+w$, and each $\lambda+w$ corresponds to $c(\mu)$ different permutations $\beta\in S_n$, we see that
\begin{align}
M_{\lambda}(x_1,\ldots,x_n)M_{\mu}(x_1,\ldots,x_n)
=c(\mu)\sum_{w\in S_\mu}M_{\lambda+w}(x_1,\ldots,x_n).
\end{align}
Therefore
\[M_{\lambda}(x_1,\ldots,x_n)m_{\mu}(x_1,\ldots,x_n)
=\sum_{w\in S_\mu}M_{\lambda+w}(x_1,\ldots,x_n).\]
This completes the proof.
\end{proof}

\begin{example}
For example, let $\lambda=(0,1,2),\mu=(1,1,3)$. Then there are 3 different permutations of $\mu$, i.e., $S_{\mu}=\{(3,1,1),(1,3,1),(1,1,3)\}$.
Then we have
\[M_{(0,1,2)}m_{(1,1,3)}=M_{(3,2,3)}+M_{(1,4,3)}+M_{(1,2,5)}=M_{(2,3,3)}
+M_{(1,3,4)}+M_{(1,2,5)}.\]
\end{example}

\begin{corollary}\label{pm}
Let $\lambda,\mu$ be two partitions of length $n$. Then
\[m_{\lambda}(x_1,\ldots,x_n)m_{\mu}(x_1,\ldots,x_n)
=\sum_{w\in S_\mu}\frac{c(\lambda+w)}{c(\lambda)} m_{\lambda+w}(x_1,\ldots,x_n).\]
\end{corollary}

\begin{proposition}\label{2pp}
Let $\lambda=(\lambda_1,\ldots,\lambda_n)\in\Lambda_n$  and $\mu=(\mu_1,\ldots,\mu_k)\in\Lambda_k$. Then
\[
\sum_{\{i_1,\ldots,i_n\}} M_{\lambda}(x_{i_1},\ldots,x_{i_n})
M_{\mu}(x_{j_1},\ldots,x_{j_k})=M_{(\lambda,\mu)_{\le}}(x_1,\ldots,x_{n+k}),
\]
where the sum takes over all the subsets $\{i_1,\ldots,i_n\}$ of $\{1,\ldots,n+k\}$ and $\{j_1,\ldots,j_k\}=\{1,\ldots,n+k\}\setminus\{i_1,\ldots,i_n\}$.
\end{proposition}

\begin{proof} By definition \eqref{eq2},
\begin{align*}
&\sum_{\{i_1,\ldots,i_n\}}M_{\lambda}(x_{i_1},\ldots,x_{i_n})
M_{\mu}(x_{j_1},\ldots,x_{j_k})\\
&=\sum_{\{i_1,\ldots,i_n\}}\left(\sum_{\sigma\in S_n}x_{\sigma(i_1)}^{\lambda_1}\cdots x_{\sigma(i_n)}^{\lambda_n}\right)\left(\sum_{\pi\in S_k}x_{\pi(j_1)}^{\mu_1}\cdots x_{\pi(j_k)}^{\mu_k}\right)\\
&=\sum_{(\{i_1,\ldots,i_n\},\sigma,\pi)}x_{\sigma(i_1)}^{\lambda_1}\cdots x_{\sigma(i_n)}^{\lambda_n}x_{\pi(j_1)}^{\mu_1}\cdots x_{\pi(j_k)}^{\mu_k}.
\end{align*}
It is easy to see that the set of triples $\{(\{i_1,\ldots,i_n\},\sigma,\pi)\}$
is in bijection with the set of permutations $S_{n+k}$. More precisely, a triple $(\{i_1,\ldots,i_n\},\sigma,\pi)$ can be associated to the permutation \[\tau=(\tau(1),\ldots,\tau(n+k))
:=(\sigma(i_1),\ldots,\sigma(i_n),\pi(j_1),\ldots,\pi(j_k)).\]
 Thus we have
\begin{align*}
&\sum_{\{i_1,\ldots,i_n\}}M_{\lambda}(x_{i_1},\ldots,x_{i_n})
M_{\mu}(x_{j_1},\ldots,x_{j_k})\\
&=\sum_{\tau\in S_{n+k}}x_{\tau(1)}^{\lambda_1}\cdots x_{\tau(n)}^{\lambda_n}x_{\tau(n+1)}^{\mu_1}\cdots x_{\tau(n+k)}^{\mu_k}\\
&=M_{(\lambda,\mu)}
(x_1,\ldots,x_{n+k})\\
&=M_{(\lambda,\mu)_{\leq}}(x_1,\ldots,x_{n+k}).
\end{align*}
This completes the proof.
\end{proof}

\begin{theorem}\label{t:main-thm}
Let $\Phi:\mathbb{A}^d_q(\Lambda)\to \mathcal{H}^d_q$ be the $\mathbb{Q}[q]$-linear map determined by
\begin{align}\label{maph}
\Phi(\lambda)=M_\lambda(x_1,\ldots, x_n),
\end{align}
where $\lambda$ is a partition of length $n$.
Then $\Phi$ is an algebra isomorphism between $\mathbb{A}^d_q(\Lambda)$ and $\mathcal{H}^d_q$.
\end{theorem}

\begin{proof}
Since $M_{\lambda}=c(\lambda)\cdot m_{\lambda}$,  $\{M_{\lambda}|\lambda\in\Lambda\}$ forms a basis of $\mathcal{H}_q^d$. It is easy to see that $\Phi$ is a vector space isomorphism. To show that $\Phi$ is an algebra isomorphism, we need to show $\Phi(\mu\ast\nu)=\Phi(\mu)\ast \Phi(\nu)$, that is,
\begin{eqnarray}\label{pp}
M_{\mu}\ast M_{\nu}=\sum_{\tiny\begin{array}{l}{\bf a}\in \mathbb{N}^n\\ {\bf b}\in \mathbb{N}^k\end{array}}c_{n,k}^{{\bf a}, {\bf b}}\cdot M_{(\mu+{\bf a}, \nu+{\bf b})_{\leq }},
\end{eqnarray}
for any two partitions $\mu=(\mu_1,\ldots,\mu_n)$ and $\nu=(\nu_1,\ldots,\nu_k)$.

By definition  \eqref{qe1},
\begin{align*}
&M_{\mu}\ast M_{\nu}(x_1,\ldots,x_{n+k})\\
&=\sum_{\{i_1,\ldots,i_n\}} M_{\mu}(x_{i_1},\ldots,x_{i_{n}})M_{\nu}(x_{j_1},\ldots,x_{j_{k}})
\prod_{l=1}^{k}\prod_{t=1}^{n}
(x_{j_l}-qx_{i_t})^{d-1}.
\end{align*}
Denote by $I=\{i_1,\ldots, i_n\}\subset \{1,\ldots, n+k\}$ and $J=\{j_1,\ldots, j_k\}$ such that $I\cup J=\{1,\ldots, n+k\}$.
Let \[g_{n,k}(I,J):=\prod_{s=1}^k\prod_{t=1}^n(x_{j_s}-qx_{i_t})^{d-1}.\] Clearly, $g_{n,k}(I,J)$ is symmetric with respect to $x_{i_1}, \ldots, x_{i_n}$ and $x_{j_1}, \ldots, x_{j_k}$, respectively. Hence we have
\begin{align*}
g_{n,k}(I,J)&=\sum_{\tiny\begin{array}{l}{\bf a}\in \mathbb{N}^n\\ {\bf b}\in \mathbb{N}^k\end{array}}c_{n,k}^{{\bf a}, {\bf b}}x_{i_1}^{a_1}\cdots x_{i_n}^{a_n}x_{j_1}^{b_1}\cdots x_{j_k}^{b_k}\\
&=\sum_{\tiny\begin{array}{l}{\bf a}\in \Lambda_n\\ {\bf b}\in \Lambda_k\end{array}}c_{n,k}^{{\bf a}, {\bf b}}m_{\bf a}(x_{i_1}, \ldots, x_{i_n})m_{\bf b}(x_{j_1},\ldots, x_{j_k}).
\end{align*}
Therefore,
\begin{align*}
&M_{\mu}(x_{i_1},\ldots, x_{i_n})M_\nu(x_{j_1},\ldots, x_{j_k})g_{n,k}(I,J)\\[5pt]
&=\sum_{\tiny\begin{array}{l}{\bf a}\in \Lambda_n\\ {\bf b}\in \Lambda_k\end{array}}c_{n,k}^{{\bf a}, {\bf b}}M_\mu(x_{i_1}, \ldots, x_{i_n})m_{\bf a}(x_{i_1}, \ldots, x_{i_n})M_\nu(x_{j_1}, \ldots, x_{j_k})m_{\bf b}(x_{j_1},\ldots, x_{j_k})\\[5pt]
&=\sum_{\tiny\begin{array}{l}{\bf a}\in \Lambda_n\\ {\bf b}\in \Lambda_k\end{array}}c_{n,k}^{{\bf a}, {\bf b}}\sum_{\lambda\in S_{\bf a}}M_{(\mu+\lambda)_\leq}(x_{i_1}, \ldots, x_{i_n})\sum_{w\in S_{\bf b}}M_{(\nu+w)_\leq}(x_{j_1}, \ldots, x_{j_k}) \\[5pt]
&=\sum_{\tiny\begin{array}{l}{\bf a}\in \mathbb{N}^n\\ {\bf b}\in \mathbb{N}^k\end{array}}c_{n,k}^{{\bf a}, {\bf b}}M_{(\mu+{\bf a})_\leq}(x_{i_1}, \ldots, x_{i_n})M_{(\nu+{\bf b})_\leq}(x_{j_1}, \ldots, x_{j_k}),
\end{align*}
where the second equality follows from  Proposition \ref{pmp}.
Consequently,
\begin{align*}
&M_\mu\ast M_\nu(x_1,\ldots, x_{n+k})\\
&=\sum_{\{i_1,\ldots,i_n\}} M_{\mu}(x_{i_1},\ldots,x_{i_{n}})M_{\nu}(x_{j_1},\ldots,x_{j_{k}})
g_{n,k}(I,J)\\[5pt]
&=\sum_{\tiny\begin{array}{l}{\bf a}\in \mathbb{N}^n\\ {\bf b}\in \mathbb{N}^k\end{array}}c_{n,k}^{{\bf a}, {\bf b}}\sum_{\{i_1,\ldots, i_n\}}M_{(\mu+{\bf a})_\leq}(x_{i_1},\ldots, x_{i_n})M_{(\nu+{\bf b})_\leq}(x_{j_1}, \ldots, x_{j_k})\\[5pt]
&=\sum_{\tiny\begin{array}{l}{\bf a}\in \mathbb{N}^n\\ {\bf b}\in \mathbb{N}^k\end{array}}c_{n,k}^{{\bf a}, {\bf b}}M_{(\mu+{\bf a},\nu+{\bf b})_{\leq}},
\end{align*}
where the last equality follows from Proposition \ref{2pp}.
\end{proof}
To conclude, we remark that in the special case $d=2$,  the product (\ref{e:multiplication}) can be defined combinatorially.
Given  $\mu\in \Lambda_n$ and $\nu\in \Lambda_k$, let $G_{\mu,\nu}$ denote the set of  direct bipartite graphs between nodes set $U=\{u_1,\ldots,u_n\}$ and $V=\{v_1,\ldots,v_k\}$ with an orientation of each edge.  Clearly, $G_{\mu,\nu}$ has $2^{n\cdot k}$ such direct bipartite graphs.
Given a    bipartite graph $G\in G_{\mu,\nu}$,
for $1\le i\le n$ and $1\le j\le k$,
let
\begin{align*}
a_i&=\mu_i+ \#\{\text{directed edges pointed to the node $u_i$}\},\\
b_j&=\nu_j+ \#\{\text{directed edges pointed to the node $v_j$}\}.
\end{align*}
Construct a vector $h(G)$ of $\mathbb{A}^2_q(\Lambda)$ according to $G$ as follows. The partition of $h(G)$ is  $(a_1,\ldots,a_n,b_1,\ldots,b_k)_{\le}$. The coefficient of $h(G)$ is
$(-q)^{m(G)}$,
where
\[m(G)=\sum_{i=1}^n\#\{\text{directed edges pointed to the node $u_i$}\}.\]
Now we can define
\begin{align*}\label{mp}
\mu\ast\nu=\sum_{G\in G_{\mu,\nu}}h(G).
\end{align*}

\begin{figure}[ht]
\begin{center}
\begin{tikzpicture}
\node at (0mm,0mm){$\bullet$};  \node [below] (0mm,0mm) {3};
\node at (10mm,0mm){$\bullet$}; \node at (10mm,-3mm) {2};
\node at (20mm,0mm){$\bullet$}; \node at (20mm,-3mm) {0};
\node at (5mm,10mm){$\bullet$}; \node at (5mm,13mm) {2};
\node at (15mm,10mm){$\bullet$};\node at (15mm,13mm) {1};

\draw[->](0mm,0mm)--(1mm,2mm);\draw(1mm,2mm)--(5mm,10mm);
\draw[->](0mm,0mm)--(3mm,2mm);\draw(3mm,2mm)--(15mm,10mm);

\draw[->](10mm,0mm)--(7.5mm,5mm);\draw(7.5mm,5mm)--(5mm,10mm);
\draw[->](15mm,10mm)--(12.5mm,5mm);\draw(12.5mm,5mm)--(10mm,0mm);

\draw[->](5mm,10mm)--(8mm,8mm);\draw(8mm,8mm)--(20mm,0mm);
\draw[->](20mm,0mm)--(16mm,8mm);\draw(16mm,8mm)--(15mm,10mm);

\end{tikzpicture}
\end{center}
\vspace{-.8cm}
\caption{A direct bipartite graph $G$ in $G_{\mu,\nu}$.}\label{PPP}
\end{figure}

For example, let $\mu=(1,2)$ and $\nu=(0,2,3)$. Then $G_{\mu,\nu}$ has $2^6$ direct bipartite graphs, we illustrate one of them as in Figure \ref{PPP}. Thus $a_1=\mu_1+2, a_2=\mu_2+2, b_1=\nu_1+1,b_2=\nu_2+1,b_3=\nu_3+0$ and $m(G)=4$. If $m=2$, then we have
$h(G)=(-q)^4\cdot(1,3,3,3,4).$

\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


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text:
Сёстры Папен, Кристина (, 8 марта 1905 — 18 мая 1937) и Ле́я (, 15 сентября 1911 — 24 июля 2001) — две сестры́-французские преступницы, 2 февраля 1933 года убившие в Ле Мане жену и дочь хозяина дома Ланселена, в котором они работали домашней прислугой (Кристина — кухаркой, Лея — горничной).

Дело сестёр Папен вызвало интерес всей Франции,
в том числе рабочего класса, литературных и интеллектуальных кругов,
подняв, среди прочих, вопросы эксплуатации рабочего класса.
Об этом деле писал, в частности французский психиатр и философ Жан Лакан, развивая тему параноидального психоза, как одной из причин параноидальных преступлений, приводя в пример сестёр Папен, вопреки заключению их судебно-психиатрической экспертизы.

По поводу дела сестёр французское общество разделилось на два лагеря:
одна часть — бо́льшая — считала сестёр виновными и требовала для них смертной казни;
другая — меньшая — говорила об эксплуатации трудящихся.

Жизнь до преступления

Детство
Сёстры росли в неблагополучной семье.
Во времена детства сестёр их отец пил, растлил их старшую сестру Эмилию, когда той было 11 лет.
Став взрослой, Эмилия постриглась в монахини.

Мать и отец сестёр во времена их детства развелись,
и отец уехал неизвестно куда.
Мать была гулящей, работала, где придётся.
Кристину взяла на воспитание её тётка,
но позже её у тётки отобрали и поместили в сиротский дом Бон-Пастер.
Лея воспитывалась в приюте Сен-Шарль.

Взрослая жизнь
Став взрослыми, сёстры стали работать домашней прислугой в частных домах.

Прежде чем трудоустроиться к Ланселенам,
Кристина сменила десять мест (домов) работы (прислугой).
Все предыдущие хозяева (наниматели) характеризовали её хорошо,
как хорошую работницу, трудолюбивую, чистоплотную и честную.
Однако Кристина всегда отказывалась трудоустраиваться в дом,
если вместе с ней на работу не брали её младшую сестру Лею.
Кристина — кухарка, Лея — горничная.

К моменту преступления Кристина проработала у Ланселенов 7 лет.

Сёстры зарабатывали у Ланселенов хорошую зарплату в 300 франков в месяц,
так что им удалось скопить 20 000 франков сбережений.

Преступление
В день своего преступления сёстры Папен до самого вечера оставались в доме одни,
в то время как мать и дочь Ланселен пошли по магазинам,
после чего мать с дочерью собирались ехать в гости (они были приглашены на обед),
так что в тот вечер мать и дочь Ланселен не собирались заходить домой.

Когда стемнело, Кристина и Лея закончили работы — из-за поломки отключилось электричество
(но вызванный электрик неисправностей не нашёл) —
поднялись к себе в комнату, разделись и легли в кровать.

Вскоре внизу послышался шум — это пришли мать и дочь Ланселен.
Кристина накинула на плечи блузку и спустилась вниз.
На площадке второго этажа она встретила хозяйку — Ланселен-мать.
На суде Кристина дала об этом такие показания:
«Госпожа обругала меня но не угрожала мне…
я схватила жестяной кувшин и изо всех сил ударила её по голове…».
Ланселен-дочь кинулась на помощь матери,
Кристина набросилась на неё
и пальцами вырвала ей один глаз (думала, что оба).

На шум прибежала Лея,
которая набросилась на Ланселен-мать,
пальцами вырвала ей глаза
и «ударяла её головой о кувшин».

О дальнейшем Кристина дала показания:
«Когда мы отправились на кухню за ножом и молотком, с ними [с матерью и дочерью Ланселен] уже было покончено».

После этого сёстры заперлись в своей комнате,
разделись и легли в постель, «крепко прижавшись друг к другу».

6 декабря 1932 года полицейский Делеа обнаружил на втором этаже особняка изувеченные трупы матери и дочери Ланселен.

Когда полиция выломала дверь в комнату сестёр,
они обе лежали голыми в своей постели
и Кристина сказала полицейским: «Мы ждали вас».

Суд

Судебный процесс по делу сестёр Папен состоялся 29 сентября 1933 года.
Кристине на тот момент было 28 лет, Лее — 22 года.

Сёстры так и не смогли объяснить, зачем они убили хозяек.
Кристина дала показания, что у Ланселенов им жилось неплохо
и претензий к хозяевам у неё нет.

Хозяин дома — поверенный Ланселен — даже после убийства сёстрами его жены и дочери,
характеризовал сестёр хорошо.
Будучи допрошен, он характеризовал Кристину, как честную, трудолюбивую женщину хорошего поведения.

На суде выражались подозрения,
что сёстры состояли друг с другом в лесбийской связи.
Основаниями для такого подозрения послужило следующее:
 сёстры не общались с мужчинами,
 были очень привязаны друг к другу,
 каждый выходной — даже в праздники — сёстры запирались в своей комнате и проводили в ней весь день;
 после их преступления полиция обнаружила их голыми в одной постели,
 в тюрьме сёстры умоляли надзирательниц позволить им встретиться, при этом Кристина кричала: «Верните мне мужа!», имея в виду Лею (!); когда Лею привели к ней в камеру, Кристина принялась обнимать и ласкать сестру. И надзирательницам пришлось их разлучить.
На суде сёстры — и Кристина, и Лея — отрицали лесбийскую связь между ними.

Психиатр-эксперт доктор Шварзиммер,
проводивший судебно-психиатрическую экспертизу сестёр,
дал на суде заключение (показания), что сёстры полностью вменяемы и ответственны за своё преступление,
в том числе у них нет плохой наследственности.
Адвокаты сестёр высказали сомнения в правильности его заключения
и вызвали дать показания на процессе других психиатров-экспертов.
Заключение экспертизы Шварзиммера было оспорено защитой сестёр на следующих основаниях:
 У сестёр плохая наследственность (вопреки тому, что говорил Шварзиммер (!)):
 дед — эпилептик,
 двоюродный брат умер в приюте для умалишённых,
 неврастеник дядя повесился у себя в комнате.
 Детство сестёр прошло в травмирующей психику обстановке:
 Гулящая мать плохо заботилась как об их отце (своём муже), так и о своих дочерях (сёстрах Папен), а однажды и вовсе исчезла;
 Отец был алкоголиком, и растлил их старшую сестру Эмилию, когда той было 11 (одиннадцать) лет;
 Развод родителей;
 После развода родителей Кристину вырвали из дома тётки, которая воспитывала её с любовью, и поместили в сиротский дом;
 У Леи тоже всё детство прошло в сиротском приюте, причём она попала туда, когда была ещё совсем маленькой.

На основании этих доводов защита сестёр на суде
поставила под сомнение заключение эксперта Шварзиммера о вменяемости сестёр
и ходатайствовала (требовала) назначить новую судебно-психиатрическую экспертизу.
Но суд (в том числе по требованию прокурора) отклонил это ходатайство защиты, удовлетворившись заключением Шварзиммера.

Приговор
Присяжные вынесли вердикт о виновности сестёр Кристины и Леи Папен.

29 сентября 1933 года сёстрам был вынесен приговор:
Кристина была приговорена к смертной казни,
Лея — к 10 (к десяти) годам каторги и 20 (двадцати) годам ссылки.

22 января 1934 года президент Франции Альбер Лебрен помиловал Кристину,
и приговор заменили каторгой.
И Кристина была переведена в  (в городе Ренн, что в Бретани).

Дальнейшая жизнь
Через несколько недель после приговора и перевода Кристины в реннскую тюрьму — в 1934 году — у неё обнаружилась (проявилась) психическая болезнь — депрессия;
она была помещена в психиатрическую больницу Ренна,
где умерла три года спустя — в 1937 году, в возрасте 32 года, от кахексии.

Лее за хорошее поведение снизили срок на 2 (два) года,
в 1941 году она освободилась,
разыскала мать и поселилась вместе с ней.
Много десятилетий после освобождения она работала горничной в различных гостиницах запада Франции.
Она умерла 24 июля 2001 года, в возрасте 89 лет,
замужем ни разу не была, детей не имела.
Похоронена в Нанте на .

Через год после суда над сёстрами Папен состоялся суд над Виолеттой Нозьер.
Судебно-психиатрическую экспертизу Виолетты Нозьер проводил тот же психиатр-эксперт, который освидетельствовал сестёр Папен,
и так же дал заключение, что Виолетта Нозьер психически вменяема и ответственна за своё преступление.
Заключение его экспертизы по Виолетте Нозьер было оспорено её (Виолетты) адвокатом,
в том числе на том основании, что год назад он (психиатр-эксперт) освидетельствовал сестёр Папен,
дал заключение, что они вменяемы,
а вскоре в тюрьме у одной из сестёр — Кристины — обнаружилась (проявилась) психическая болезнь.

В психоанализе
Молодой французский психиатр, будущий знаменитый психоаналитик Жак Лакан посвятил разбору преступления сестёр Папен статью «Мотивы параноидального преступления — преступление сестёр Папен» (), опубликованную в журнале сюрреалистов «» в 1933—1934 годах.

В искусстве
Дело сестёр Папен вдохновило многих писателей и деятелей кино и театра,
которые создали о них немало литературных произведений, пьес и фильмов.
В частности:
 В 1947 году Жан Жене написал пьесу «Служанки» (). Несмотря на бросающееся в глаза сходство сюжета его пьесы с историей сестёр Папен, Жан Жене всегда яростно отрицал, что на создание пьесы его вдохновило дело сестёр.
 В 1963 году французский режиссёр  снял фильм .
 В 1994 году американский (США) режиссёр  сняла фильм .
 1995: Клод Шаброль снял фильм «Церемония», экранизировав роман Рут Ренделл «Каменный приговор».
 2000:  снял фильм , экранизировав книгу Paulette Houdyer о деле сестёр Папен.
 2000: режиссёр документального кино  снял фильм .
 2005: французская певица Жюльет Нуреддин записала песню «Проклятый колокол» (альбом «Mutatis mutandis»).
 В 2019 году был снят фильм «Паразиты». В интервью режиссёр Пон Чжун Хо сообщил, что основными источниками вдохновения для создания фильма послужили картина «Служанка» (1960) Ким Ги Ёна и дело сестёр Папен.

Не путать с
 Сёстры Паппини

Примечания

Источники

Литература
 Фредерик Поттешер, книга «Знаменитые судебные процессы» (фр. «Les Grands Proces de L'Histoire»), в СССР издана издательством «Прогресс» в 1985 году, 302 стр.

Ссылки

Персоналии по алфавиту
Убийцы Франции
Сёстры-преступницы
Женщины-убийцы
Приговорённые к смертной казни во Франции
Преступники, признанные невменяемыми


text:
E tū celebrates largest ever minimum wage increase
Wednesday, 19 December 2018, 12:13 pm
Press Release: E tu
The minimum wage is set to increase by $1.20 to $17.70 in April 2019 – the largest increase in the adult minimum wage in New Zealand history in dollar terms.
This is the biggest leap yet towards the Coalition Government's promise to increase the minimum wage to $20 by 2021.
E tū National Secretary Bill Newson says the increase is another clear demonstration of the Coalition Government's commitment to working people.
"This Government continues to prove that they really care about workers and their families," Bill says.
"Lifting the minimum wage is relatively straight forward, and the evidence shows that bringing wages up is the clear path out of poverty in New Zealand.
"Together with the recent Employment Relations Act changes and the ongoing work on Fair Pay Agreements, the Government is taking us in the right direction. This is another good step forward."
Mareta Sinoti, a cleaner at the National Library in Wellington, says that while the increase is welcome, it's not going to solve all the problems.
"The thing is, we need a Living Wage," Mareta says.
"Everything is just too expensive. Rent, food, and transport costs are increasing all the time. When you add up the 10-trip for the train, the costs of clothes for our families, and everything else, it's just too much.
"It's great that the minimum wage is going up to $17.70, but how long have we waited for it to get there?"
Find more from E tu on InfoPages.


text:
Poloroid is a former stage name of Essex-born singer-songwriter Danielle "Dan" Rowe, best known for her 2003 single "So Damn Beautiful". Originally referring to a trio, the name "Poloroid" was adopted by Rowe for her solo material in 2001, following the death of her bandmate John Horrocks. Her song "So Damn Beautiful" was featured on compilation albums, such as Global Underground 012: Bueno Aires, and was also used in the soundtrack to an episode of American drama Nip/Tuck. After her record label, Telstar Records, was dissolved in 2004, Rowe retired the "Poloroid" moniker and formed a new band named No Silence.

History
Poloroid were originally a trio consisting of Danielle Rowe, Lee Milleare and John Horrocks. After Horrocks committed suicide in May 2001, Rowe continued to use the Poloroid moniker as a solo artist. Her song "So Damn Beautiful", which she had co-written with both Milleare and Horrocks, was first brought to the attention of nightclubs after British disc jockey Dave Seaman included it on the mix album Global Underground 012: Bueno Aires in 1999. Because of this attention for the song, Rowe was invited to perform a live set for BBC Radio 1 in Ibiza. As well as interest from the clubs, the track was also used in the UK as part of soundtracks to various victories of the England national football team, such as their 5–1 defeat of Germany in 2001 and their 1–0 victory over Argentina in 2002.

"So Damn Beautiful" was officially released by Telstar Records in the UK on 29 September 2003, in a radically remixed form by trance DJs Dogzilla. The song was playlisted on BBC Radio 1's C list and received airplay, where Australian singer Holly Valance expressed a like for the song on The Chris Moyles Show. "So Damn Beautiful" peaked at number 28 on the UK Singles Chart and was Poloroid's biggest chart hit. It was also used in the episode "Nanette Babcock" of the American drama Nip/Tuck and was included on the soundtrack album Nip/Tuck: Original TV Soundtrack. On reviewing the album, UK-based website musicOMH described the track as "lovely" and called it "a truly haunting song".

Telstar Records was declared bankrupt in 2004, three days before the release of Poloroid's debut album. As a result of this, Rowe decided to retire the name Poloroid and chose instead to form a new band named No Silence, with whom she currently performs.

Discography

Singles

Other appearances
The following songs have been officially released, but do not feature on an album by Poloroid.

References

External links

Poloroid on Myspace

English women singers
English songwriters
Living people
Musical groups established in 1998
Musical groups disestablished in 2001
English dance music groups
British musical trios
Trip hop groups
Telstar Records artists
Year of birth missing (living people)


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