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WATCH LIVE: Guests arrive at Biden's inauguration
WATCH IT LIVE: President-elect Biden and VP-elect Harris sworn into office at 12 noon
Mother and son pummeled in random attack at NYC subway station
Baseball cannot be surprised by Jared Porter
President Biden's promise: America at another crossroads
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Suspect in Harlem shooting death nabbed in Virginia
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Mental health issues spike amid historic pandemic and social unrest
By Nancy Dillon
Several months into our coronavirus state of emergency, the state of our mental well-being keeps facing new challenges. (ShutterStock)
Feeling more anxious and worn out than usual? You're not alone.
Several months into our coronavirus state of emergency, the state of our mental well-being keeps facing new challenges.
Fear of an invisible danger, illness and a mounting death toll have been compounded by record job losses, a polarizing presidential election and the fight for racial justice in the wake of a series of black people dying at the hands of police.
Between April 23 and May 5, an estimated 23.5% of Americans reported feeling symptoms of depressive disorder, according to the new Household Pulse Survey conducted by the National Center for Health Statistics and U.S. Census Bureau.
That figure climbed to 25.1% between June 11 and June 16, the survey found.
A body is picked up from a temporary morgue outside of Brooklyn Hospital in April. (Theodore Parisienne/for New York Daily News)
The numbers are alarming, experts say. During January through June 2019, only 6.6% of Americans 18 and older had symptoms of depressive disorder, the Centers for Disease Control said.
"I think we can all agree we are probably in the biggest mental health crisis this country has seen in a very long time," Dr. Donna Demetri Friedman, executive director of Mosaic Mental Health in the Bronx, told the Daily News.
"There's enormous fear permeating our society right now, whether it's financial, physical or emotional," she said. "And black mental health is a big and important issue now, too. People are constantly seeing images without warning of black people being killed. It creates post-traumatic stress in people."
Indeed, 25.6% of black Americans reported feeling symptoms of depression between April 23 and May 5, with the figure jumping to 28.3% between June 11 and June 16, the Household Pulse Survey found.
Protestors march Wednesday, June 3, in Manhattan. (Barry Williams/for New York Daily News)
Dr. Friedman and other experts urged anyone in need of support to reach out for help. The NY COVID-19 Emotional Support Hotline at 844-863-9314 is one service connecting New York residents with free phone appointments with mental health professionals.
"It's very important to stay connected. Staying connected can help identify issues people can't identify by themselves," said Dr. Fernando Taveras, a psychiatrist with SOMOS Community Care and an assistant professor at Columbia University.
Pointing to data on medical workers in other countries, he said fear of the unknown during our historic pandemic can be very disorienting and "spark" symptoms of a mental illness in people who don't have the underlying diagnosis.
One study of 1,379 health care workers in Italy found that those who treated COVID-19 patients experienced high rates of psychological distress — with 49% reporting post-traumatic stress symptoms, 25% reporting symptoms of depression, 20% reporting symptoms of anxiety and 8% reporting insomnia.
Fear of an invisible danger, illness and a mounting death toll have been compounded by record job losses, a polarizing presidential election and the fight for racial justice in the wake of a series of black people dying at the hands of police. (ShutterStock)
Interestingly, a similar study of health care workers in Singapore found less than 15% reported symptoms of anxiety during the COVID-19 crisis, while 9% screened positive for depression and 8% screened positive for post-traumatic stress.
The study authors from Singapore's National University Health System said the scores were much better than expected, likely due to the respondents' previous experience with the 2003 SARS epidemic.
A previous study in Singapore found physicians and nurses reported an almost three times higher prevalence of PTSD during the 2003 SARS crisis.
"They had less mental health complaints because they experienced the crisis before. We don't have the experience in the U.S. We don't know what to do. People have not felt we're going in a good direction, and that increases the anxiety and depression here," Dr. Taveras said.
"For people who are particularly sensitive, we recommend they take a break from the news," he said. "And it's important to have a routine and exercise, even if it's just floor exercises or some yoga."
Interestingly, a similar study of health care workers in Singapore found less than 15% reported symptoms of anxiety during the COVID-19 crisis. (ShutterStock)
According to Dr. Eric Nestler, head of Hope for Depressions Research Foundation and a psychiatrist at Mount Sinai, demand for mental health services has gone up "significantly" in recent months.
And many people could be feeling the effects of this moment for years to come, he said.
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"We know based on our experience with traumatic events such as combat and 9/11, the mental health consequences don't typically occur during the time of peak trauma but when things start to ease," he said.
"We think we're at the very early stages of seeing a dramatic increase in the symptoms of PTSD, depression, anxiety and the associated conditions like substance abuse and suicide."
Like his colleagues, Dr. Nestler suggests "maintaining social ties as best you can."
He said it's important to practice "mindfulness" and recommended the meditation app Headspace.
"Believing in something also helps," he said, mentioning both organized religion and the burgeoning movement for social justice.
"Organizing around how to do better and how to make society better can be an extremely positive focus for people," he said.


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EXCLUSIVE: Royal Central travels to Thailand with Princess Tessy
Archie Harrison is christened at Windsor
The 93-year-old Queen conducted an impressive 40 engagements in June alone
The Queen and the Duchess of Gloucester to undertake a joint engagement in Cambridge
The Duke of Cambridge: 'absolutely fine' if children are LGBTQ
Grand Ducal Family of Luxembourg 23rd April 2019
Grand Duke Jean's funeral details are released
Laura Dekkers
Europe Correspondent
OBITUARY: Grand Duke Jean of Luxembourg
Grand Duke Jean's Knight of the Garter banner to be removed today
Copyright: Cour grand-ducale / Jochen Herling
This morning Grand Duke Henri of Luxembourg announced the passing of his father, Grand Duke Jean, aged 98. The Prime Minister of Luxembourg, Xavier Bettel has made a statement on the passing of "a hero" and has released the funeral details.
Prime Minister Xavier Bettel organised a press conference in which he made a live statement on the death of Grand Duke Jean. The Prime Minister called him "a hero, an example for us all, a symbol of our nation". He praised Grand Duke Jean for his loyal service and commitment to his country. Mr Bettel was clearly moved by the Grand Duke's death.
The Prime Minister also released more details on the passing and funeral arrangements of Grand Duke Jean. His Royal Highness passed away on Tuesday 23 April at 12.25 am at the age of 98. He was surrounded by his family.
Grand Duke Jean will, as expected, receive a State Funeral. The funeral will take place on Saturday, 4 May at 11 am at the Cathedral of Luxembourg. Further details are expected to be released in due course.
It is too early to tell who will be attending the funeral. However, it is expected that many members of the Belgian Royal Family will come and pay their respects to Grand Duke Jean as they are closely related. King Albert of Belgium was the brother-in-law of Grand Duke Jean. King Philippe of the Belgians and Grand Duke Henri are first cousins.
Grand Duke Jean reigned for nearly 36 years before he abdicated in favour of his son, Grand Duke Henri in 2000. In 1953, he married the Belgian Princess Josephine Charlotte, a sister of King Albert of Belgium. Grand Duchess Josephine Charlotte passed away aged 77 in 2005. The pair had five children together.
Last week, the Grand Duke was admitted to the hospital due to a pulmonary infection. On Sunday, the Grand Ducal Court announced that his condition had worsened significantly, and on Tuesday morning, Grand Duke Henri announced the passing of his "beloved father."
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redpajama_set_name: RedPajamaCommonCrawl


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Manchester United vs. Chelsea Pick – Betting Odds
Manchester United vs. Chelsea, England Premier League Soccer Match Day 5, Old Trafford Manchester, England 11 AM EST Sunday September 18, 2011 on FOX Soccer, Sky Sports HD1
by Jason Green, Soccer Handicapper, Predictem.com
Match Odds: Manchester United 1.80 / Chelsea 4.40 / Draw 3.40
Over/Under: 2.5 goals
Moneyline: Manchester United -125 / Chelsea +330 / Draw +250
In a titanic battle between the 2 teams that have the best odds to win the Premier League this season Manchester United plays host to Chelsea. Man U has shot out of the gates winning all 4 of their league matches and is tied atop the table with Manchester City while Chelsea sites tied for 2nd with 3 wins and 1 draw.
Manchester United's offense has been on fire in league play and they have a +15 goal differential while Chelsea is only at +4. However, the Man U offense did not look in sync in their 1-1 draw against Benfica in a Wednesday night Champions League match.
The last time these 2 HUGE rivals met was at Old Trafford and Manchester United beat Chelsea 2-1. The Red Devils have also won 3 of their last 4 matches when facing the Blues.
After a somewhat disappointing last season Man U striker Wayne Rooney has been superb so far this season with 8 goals in only 4 matches and he already has a couple of hat tricks. The Chelsea D really has to key on him and maybe even more so since the Blues main striker Didier Drogba is expected to miss the match due to a concussion.
Chelsea was impressive in winning their Champions League match on Tuesday night beating Bayer Leverkusen 2-0. They have star G Petr Cech back in goal and he will have to play well against Rooney and the solid Red Devils' offense.
Nani has been joined in the midfield by off-season signing Ashley Young and they are a formidable duo and if they can play well in this match helping Man U win the time of possession battle Chelsea will likely lose this match.
Manchester United Recent Form:
9/14/2011 Benfica 1 Manchester United 1 (Champions League)
9/10/2011 Bolton 0 Manchester United 5 (EPL)
8/28/2011 Manchester United 8 Arsenal 2 (EPL)
9/22/2011 Manchester United 3 Tottenham 0 (EPL)
9/14/2011 West Bromwich 1 Manchester United 2 (EPL)
Chelsea Recent Form:
9/13/2011 Chelsea 2 Bayer Leverkusen 0 (Champions League)
9/10/2011 Sunderland 1 Chelsea 2 (EPL)
8/27/2011 Chelsea 3 Norwich City 1 (EPL)
8/20/2011 Chelsea 2 West Bromwich 1 (EPL)
8/14/2011 Stoke City 0 Chelsea 0 (EPL)
Besides Didier Drogba MF Michael Essien will also miss this match due to injury. For Manchester United MF Tom Cleverley, F Danny Welbeck, and DF's Nemanja Vidic and Rafael will all miss this match.
Chelsea has started off well, but their once feared offense has not. There is not only player for the Blues that has more than 1 goal this season. Fernando Torres has not done much of anything since his big transfer from Liverpool last season and this is the perfect match for him to have his Blues coming out party.
Chelsea's MF duo of John Mikel Obi and Frank Lampard have to push the envelope on offense, but I think they really have to concentrate on thwarting the Man U offense. One thing the Blues have going for them is that Manchester United G David De Gea, as while he has only conceded 3 goals in 4 matches he has looked a little shaky.
Rooney has a history of playing well against Chelsea and even though he will be the main guy to stop the Blues will not be able to do it, as he will score at least one goal.
Man U is always stingy on defense at home and that trend will continue, as they will keep a clean sheet in this match.
Jason's Pick to Win the Match: I look for most of this match to be played in the midfield, as both teams will come out cautious. Chelsea will be happier with a draw and they will fall back on defense more, especially with their big gun of Didier Drogba not playing. However, I see Rooney scoring in the 2nd half for the only goal of the match, as the Red Devils will win this match 1-0 and keep their perfect EPL record in tact.


With the season half over and the Tigers now getting back to playing AL rivals, it's time for a mid-season statistical summary. One thing to keep in mind as you go through the tables below is that offense is down in the American League so far this year. The average runs scored per game is down from 4.9 in 2007 versus to 4.6 so far this year. It was even lower before the AL beat up the NL in interleague play. Anyway, where a team ranks among other teams in the league is more relevant than comparing raw numbers between 2007 and 2008.
The Tigers are third in the league in runs scored per game (4.9) after being second in 2007. Much has been made of those two 19 run games throwing off the offensive numbers and they do skew things a little but they have played enough games now where it's not a big factor. Their inconsistent offense (9 shutouts) was a problem earlier in the year but they have become more steady over the past month. run prevention has been their weakness overall this year. They have allowed 4.8 runs per game which is 12th in the league. The positive thing is that they did improve to 4.2 runs allowed per game in June.
Table 2 shows that the Tigers are above average in most offensive categories including 3rd in batting average, 3rd in isolated power, 4th in OBP and 4th in slugging. The one area where they are only average is walks. However, they have improved from 2.9 per game in 2007 (12th in the league) to 3.4 this year (7th).
Table 3 shows that their pitching is bad again this year. Looking at the starter and reliever splits, we can see that their starters ERA is 9th and their relievers ERA ranks 12th. Their starting pitching has improved remarkably since the beginning of the season though. After compiling a 5.44 ERA in their first 38 games, they have had a 3.67 ERA in their last 43.
Fielding is harder to measure but Table 4 illustrates that the Tigers are not doing nearly as well defensively this year. Using The Hardball Times plus/minus stat, they ranked 3rd in fielding last year and have dropped to 22nd this year. In this system, they break down fielding by looking at types of balls hit (ground ball, fly ball, pop up, line drive) . These stats are explained further in an article by Dave Studeman. This system is denoted +/- in the table.
Looking at Revised Zone Rating (RZR), they were 4th in 2007 and 18th so far this year. Each position on the field has a zone around it and a player's RZR is the proportion of balls hit into his zone which he converts into outs. Team RZR is the proportion of balls hit into all fielding zones which are converted into outs. Balls not hit into any zone are not considered in the calculation. For a further discussion of RZR and other fielding statistics, check out Fielding Stats at The Hardball Times by Dave Studeman.
The chart below also breaks RZR into infield and outfield. The table shows that both their infield and outfield defenses have slipped substantially this year. The infield ranked 6th in 2007 and is now 24th. Similarly, the outfield has gone from 1st to 17th.
In summary, both pitching and defense have slowed the Tigers this year. The offense was inconsistent early but they have become more balanced as the season has gone along. Scoring has not been a big problem overall. I think the offense will actually do a little better in the second half. The defense will probably remain average to below average. The pitching will likely be the key to whether they stay in the race. Their starting pitching has been solid over the last seven weeks. If they can keep that up (and that's a big if with two rookies in the rotation), they should be in good shape.
Wow. I had no clue the defense was down that much this season. I am really surprised. Nice work Lee.


Valentines Day for us doesn't mean flowers or chocolates or gifts- it means we have an excuse to buy shrimp. Woo! I always feel like it's better quality around the holidays. It's so hard for us to buy shrimp out here since we are spoiled New Englanders; it's just not as good as fresh off the Maine Coast shrimp.
This Valentines Day, we found a Spicy Sriracha Shrimp and Zucchini Lo Mein Recipe (see full recipe here). A few years ago there was no way I would have been able to eat this, but lately I've been really into spicy foods because it helps me sloooow down and not eat so much. And it helps me get in my daily water intake!! Win all around!
We did alter the recipe just a tiny tiny bit, and I thought I would give a shot at recipe writing.
In a small bowl, combine ingredients to make spicy sauce: brown sugar, soy sauce,srircha, and sesame oil.
In a wok or pan, heat up the butter and red pepper. Then add the eggs and scramble. When cooked thoroughly, remove and put on small plate.
Heat up olive oil, and throw in asparagus and zucchini until cooked through. Then add in eggs, shrimp, sauce, and noodles.
It was very good and very spicy! I drank about 3 glasses of water and probably only consumed (not even) a cup of food. Also, whenever I eat spicy foods my eyelids sweat, which everyone thinks is weird. Does anyone else's body react like that?
I'm not lyin'! That's my mascara melted off after dinner.


WASHINGTON — Gary D. Cohn is a Democrat, but his resignation as director of the White House National Economic Council on Tuesday actually underscored the fundamental divide between President Trump and his fellow Republicans as the president seeks to raise barriers to foreign trade.
From his perch in the West Wing, Mr. Cohn in effect served as a proxy for the business wing of the Republican Party as it fought what may be a losing battle against new tariffs on steel and aluminum imports. His departure deprives free trade proponents of perhaps their strongest voice inside Mr. Trump's inner circle.
Trade has always been the iron curtain that divided Mr. Trump and establishment Republicans. The president argues that other countries have run roughshod over the United States when it comes to the exchange of goods and services while most Republican leaders in Washington maintain that lower barriers have helped keep the United States as the world's leading economic powerhouse.
Until Mr. Trump's signature tax cut plan was passed in December, that schism was largely papered over in the name of unity as Republicans focused on lowering rates for corporations, wealthy Americans and middle-class families. But the rift over trade has taken center stage in the days since Mr. Trump roiled markets and international relations by declaring that he would impose tariffs of 25 percent on foreign steel and 10 percent on foreign aluminum.
Mr. Trump finds himself at odds with his party's leaders in Congress, like Speaker Paul D. Ryan of Wisconsin and Senator Orrin G. Hatch of Utah, the chairman of the powerful Senate Finance Committee, both of whom have long records supporting free trade. Mr. Hatch, who once delighted Mr. Trump by saying that he was on the way to making his "the greatest presidency" in history, has broken sharply with the White House over tariffs.
In a sharply worded letter on Tuesday to Mr. Trump, Mr. Hatch said that the proposed tariffs would be paid by American manufacturers and consumers, not foreign companies, and would undercut "the overwhelming and immediate success" of the tax overhaul. "Put simply, the proposed tariffs on steel and aluminum imports threaten to undermine that success," he wrote.
Mr. Trump even faced criticism of his proposed tariffs in person on Tuesday as he hosted Prime Minister Stefan Lofven of Sweden. "Swedish prosperity is built on cooperation, competitiveness and free trade, and I am convinced that increased tariffs will hurt us all in the long run," the prime minister told reporters as the president stood next to him at the White House.
Mr. Cohn, a former Goldman Sachs banker, has been the central protagonist in the trade war inside Mr. Trump's administration. As he sought to keep the president from blowing up the international trade consensus precipitously, Mr. Cohn was targeted by trade skeptics on the right, like Stephen K. Bannon, the president's former chief strategist. Breitbart News, once overseen by Mr. Bannon, regularly included an illustration of a globe next to Mr. Cohn's name to mock him as a globalist.
Skeptics of free trade did not mourn the news of Mr. Cohn's departure and suggested that Mr. Trump's Republican critics like Mr. Ryan and Mr. Hatch were on the wrong side of history.
While Mr. Trump's views on many issues have been flexible or evolved over the years, his belief that the United States was being shafted has been a core conviction going back decades, long before he first ran for president. It is an article of faith that has informed his opinion of trade pacts, tariffs and manufacturing and that helped him tap into populist discontent during the presidential campaign in 2016, particularly in traditionally Democratic states in the Midwest like Wisconsin, Michigan and Pennsylvania.
The disagreement on trade was obscured through much of Mr. Trump's tenure because he talked more about the issue than took action on it, meaning Republican free-traders could turn aside for a time.
He pulled out of the Trans-Pacific Partnership, a 12-nation trade deal negotiated by President Barack Obama, but the agreement had never been ratified by Congress and seemed unlikely to be. Mr. Trump threatened to scrap the North American Free Trade Agreement with Mexico and Canada, but backed off in favor of negotiations that so far have not yielded consensus. He has likewise suggested he would withdraw from a trade deal with South Korea but has not followed through.
In one of the most tangible actions he has taken, Mr. Trump slapped steep tariffs on foreign washing machines and solar energy panels in January, generating cheers from domestic manufacturers and protests from other business leaders. But he followed up days later with a reassuring speech to investors and corporate chieftains gathered in Davos, Switzerland, the ground zero for globalization and free trade, a speech written largely by Mr. Cohn.
In the end, though, Mr. Cohn could not head off the steel and aluminum tariffs, suggesting that the "America First" side of the White House had finally prevailed over his resistance.
Lori Wallach, the director of Public Citizen's Global Trade Watch and a longtime critic of trade deals, said the divide over trade crossed all sorts of traditional lines.
But Jason Furman, who was the chairman of the President's Council of Economic Advisers under Mr. Obama, said people on both sides of the issue were investing too much importance in it.
Even so, as Mr. Cohn takes his leave, it has now become the defining fault line between a president and his party.


Since campaigning began in March for Pakistan's upcoming elections, 28 people have been reportedly killed in a total of 14 separate attacks.
The run up to Pakistan's elections have taken a bloody turn, with a spate of attacks targeting candidates and campaign rallies.
The violence threatens to tilt the political landscape in favor of more conservative parties ahead of historic elections in May. The upcoming election will mark the first democratic transition of power from one civilian government to another.
Since campaigning began in March, 28 people have been reportedly killed in a total of 14 separate attacks. The attacks have mostly targeted the secular Awami National Party in Khyber Pakhtunkhwa province.
The coalition government that ruled in Khyber Pakhtunkhwa for the last five years was dominated by the ANP and the PPP. In recent years, they battled the Pakistani Taliban for control of parts of the province.
The Taliban has directly claimed responsibility for five of the attacks, and it is likely that affiliated groups were behind many of the other attacks. The attacks come as the Taliban have been operating with increasing impunity across the country.
Militants have carried out four bombings and one grenade attack against the Awami National Party in the past 10 days.
"These are planned and premeditated attacks. The idea is to scupper the political process and rig it in favor of a narrative and a political system that does not reject extremism," said Raza Rumi, director of policy and programs at the Jinnah Institute, an Islamabad-based think tank. "For sure this is the larger [Tehrik-i-Taliban Pakistan] network at play," he added.
The other major parties have not escaped unscathed. The ruling Pakistan Peoples Party, the main opposition Pakistan Muslim League-Nawaz and the Muttahida Qaumi Movement, have also been targeted. The PML-N is widely expected to form the next government.
Afrasiab Khattak, the president of the ANP in Khyber Pakhtunkhwa, believes the wave of attacks on his party – the largest in the province – is a way to preventing it from coming to power. He called on the Election Commission of Pakistan and the caretaker government to do more to prevent the attacks.
Mr. Khattak's calls have been echoed across the political spectrum, with the attacks being widely described as pre-poll rigging.
"There needs to be a unanimous position from across political lines otherwise the elections will be a fake," said retired General Talat Masood. "We need preventive measures to deal with this threat," he added.


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An replace to Google Assistant on Android gadgets will permit it to present higher visible responses and extra full info.
When answering queries, Assistant returns a display screen that extra resembles what you'd see in Google search outcomes.
Another new factor Google Assistant will do is present accompanying visuals when potential.
There's now lots much less empty display screen house when returning search outcomes.
In the instance beneath, you'll see that Assistant used to show outcomes horizontally in a carousel, leaving practically half the display screen empty.
Now, Assistant fills the display screen vertically which seems higher and is extra practical.
This replace is rolling out now on Android telephones solely. Android usually receives updates to Google software program earlier than iOS.
It wasn't explicitly said, however an iOS model is probably going not far behind.


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LifestyleHealth
In First Executive Orders, Biden Makes Clear Break With Trump Era
President Joseph R. Biden Jr. signed 17 executive orders, proclamations, and reviews just hours after his inauguration, signaling an abrupt change of course for the new administration, including rejoining the World Health Organization and Paris Climate Accord.
Despite his calls for unity and bipartisanship, Biden immediately reversed some of Trump's most controversial executive orders — in particular, those aimed at immigration, including building the southern border wall and bans on entry for Muslims from several countries. He also swiftly addressed pandemic response, infrastructure, and climate issues.
Among the most notable of Biden's orders was the move to halt building of the border wall between Mexico and the U.S., and to immediately halt the national emergency declaration Trump used to redirect billions of dollars to the wall. The legality of Trump's actions will now also be reviewed.
Biden also sought to bolster the Deferred Action for Childhood Arrivals program, which is most commonly referred to as "DACA" and limits the deportation of immigrants brought to the U.S. as children, often called "Dreamers." Biden encouraged legislation allowing for permanent status and providing a path to citizenship. The Trump administration tried to end it in 2017 and it was suspended in July 2020. (The program was reinstituted on December 4, 2020, under a federal judge's order, and the Department of Homeland Security was ordered to quickly begin accepting applications.)
Biden also ended what was termed the "Muslim-country travel ban", the previous administration's orders preventing certain individuals from entering the United States — first from primarily Muslim countries, and later, from more African countries. Biden called the ban a "stain on our national conscience … inconsistent with our long history of welcoming people of all faiths and no faith at all."
The Biden administration also reinstituted a plan to halt deportations of those who sought asylum from war-torn Liberia since 1991. The order ensures visa screening and vetting is not prejudicial, and strengthens interagency communication.
In a cautionary move, Biden enacted a regulatory freeze on any last-minute legislation from the Trump administration. The freeze, which makes such legislation subject to a 60-day review period, gives the new administration opportunity to review actions and analyze the facts and laws involved.
Another order makes the U.S. once again party to the Paris Climate Accord of 2015, which the U.S. exited in 2019. The agreement's goals aim to reduce emissions and use of fossil fuels, which contribute to climate change.
You will find more infographics at Statista
Dr. Anthony Fauci, Director of the National Institute of Allergy and Infectious Diseases, listens during an event at the State Dining Room of the White House. President Biden delivered remarks on his administration's COVID-19 response, and signed executive orders and other presidential actions. (Alex Wong/Getty Images)
Biden also reentered the U.S. into partnership with the World Health Organization and appointed Dr. Anthony S. Fauci of the National Institutes of Health to the organization's executive board as head of the U.S. delegation. This overturns a Trump administration decision to withdraw, which breached a decades-long partnership supporting vaccinations for common diseases globally.
The U.S. was the top donor, with voluntary contributions comprising 76% of the funds.
The pandemic response team disbanded by the Trump administration prior to the COVID-19 outbreak was reinstituted by a separate order. The National Security Counsel Global Health Security and Biodefense directorate was originally a monitoring and response task force for outbreaks and biological threats. Biden created several positions to coordinate with National Security Council.
He appointed Jeffrey D. Zients as official COVID response coordinator of the federal government response, and a second position reporting directly to the president for the national biodefense response. Biden bolstered data collection, information sharing, and transparency in another executive order Thursday.
An executive order now requires mask wearing and social distancing on all federal property. Biden stopped short of a national mandate, but launched a 100-day masking challenge. He urged state and local officials to enact measures to stop the spread of the coronavirus.
Additional orders Thursday require masks in airports and public transportation systems. International travelers must now provide proof of a negative Covid test for admittance to the U.S.
Among the actions was also an "Executive Order on Preventing and Combating Discrimination on the Basis of Gender Identity or Sexual Orientation," citing Title VII in the original Civil Rights act of 1964.
The order "will fully enforce Title VII and other laws that prohibit discrimination on the basis of gender identity or sexual orientation," according to the White House. The order applies to a broad spectrum of activities including sports, education, housing, and immigration.
President Donald Trump and First Lady Melania Trump acknowledge waiting supporters at Joint Base Andrews before boarding Air Force One for his last time as President. (Pete Marovich – Pool/Getty Images)
Biden ended Trump's 1776 commission, which released a controversial report on slavery in the U.S. and other history on Jan. 18, the day of the federal holiday celebrating Martin Luther King Jr.
Biden also revoked Trump's limits on diversity and inclusion training within the federal government and contractor ranks. He appointed Susan E. Rice as director of the Domestic Policy Council to organize an interagency effort to root out racism in governance, and tasked all federal agencies to provide report on equity within the ranks within the first 200 days.
In continued response to the coronavirus pandemic, he acted to extend moratoriums on evictions and asked agencies to do the same for federally guaranteed mortgages — all lasting until the end of March.
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text:
Papantoniou second 'pothen esches' trial postponed
The trial of former minister Yannos Papantoniou and his wife, Stavroula Kourakou, for submitting inaccurate "pothen esches" (declaration of source of income) forms for 2008 was postponed to June 8, 2015, an Athens court ruled on Wednesday. The decision was reached following a strike by lawyers.
Last week the couple was sentenced to four years in prison for submitting inaccurate source of wealth forms for 2009 but were allowed to pay off their terms for 14,560 euros each, pending appeal. Discrepancies in their stated and actual assets were detected when Kourakou's name appeared on the Lagarde list of Greeks with bank accounts in a Geneva HSBC branch.
(ekathimerini.com)


text:
Q: Change background-image after hover on submenu item Hello I got question regarding setting a background-image whenever I hover on an element of my submmenu. I have the following HTML:

            <button type="button" class="navbar-toggle collapsed">
            </button>

                        My data

                    Menu 1

                        Submenu1
                        Submenu2

                    Menu 2

                        submenu1
                        submenu2

                FAQ
                Contact
                Logout

I know this is not the proper way to generate a menu with submenu items but despite this, how could I get my grey home button (which becomes visible after hovering on the home button) to stay visible whenever I hover on the "my data" button? Is this possible with pure CSS?
I have a demo here: JSFIDDLE

A: Change this:
.not_active:hover {
    background-image: url(https://jira.transvar.org/secure/attachmentzip/unzip/14790/12882%5B1%5D/Home_Icon_for_IGB_pack/home_icon_grey_128x128_more_space_16x16.pn);
}

to this:
    .not_active:hover, .menuitem_wrapper:hover .not_active {
        background-image: url(https://jira.transvar.org/secure/attachmentzip/unzip/14790/12882%5B1%5D/Home_Icon_for_IGB_pack/home_icon_grey_128x128_more_space_16x16.pn);
}

I just added .menuitem_wrapper:hover .not_active to the same rule.

A: Use jQuery mouseover() and mouseout()
$(document).ready(function(){
     $(".dropdown-menu a").mouseover(function(){
         $("a.home_button.not_active").css("background-image", "url('https://jira.transvar.org/secure/attachmentzip/unzip/14790/12882%5B1%5D/Home_Icon_for_IGB_pack/home_icon_grey_128x128_more_space_16x16.png')");
     });

$( ".dropdown-menu a" ).mouseout(function() {
$("a.home_button.not_active").css("background-image", "url('https://upload.wikimedia.org/wikipedia/commons/3/34/Home-icon.svg')");
     });
});

Also make changes to CSS
.not_active:hover {
    background-image: url(https://jira.transvar.org/secure/attachmentzip/unzip/14790/12882%5B1%5D/Home_Icon_for_IGB_pack/home_icon_grey_128x128_more_space_16x16.png)! important;
}

Working Example JSFIDDLE

redpajama_set_name: RedPajamaStackExchange


text:
\section{Introduction}
In 1985, Arjeh Cohen and Jacques Tits proved the existence and uniqueness of $\widetilde{Q}_n$, a girth six, 2-fold cover of the hypercube $Q_n$. A 2-fold cover of a graph can be specified by a $\pm1$ signing of its adjacency matrix, and the eigenvalues of this cover are  the union of the eigenvalues of the original graph and of the signed matrix. Last year, Hao Huang resolved the Sensitivity Conjecture by proving that every $(2^{n-1}+1)$ vertex induced subgraph of $Q_n$ has maximum degree at least $\sqrt{n}$. His proof relies on constructing a $\pm1$ signing of the adjacency matrix of $Q_n$ which specifies exactly the cover $\widetilde{Q}_n$ described above (see Section \ref{cttohh}). Motivated by this connection, we develop a natural framework for both results. This framework extends to covers of larger index, or equivalently, adjacency matrices whose ``signings'' take values in larger cyclic groups represented by complex roots of unity.  As in the $\pm1$ signing case, the spectra of these complex adjacency matrices encode the new eigenvalues of our covering graphs. Our key definition is that of a \textbf{two-eigenvalue cover}, (or 2ev-cover) which is a covering graph whose adjacency spectra differs (as a multiset) from that of the graph it covers by exactly two eigenvalues. It is worth noting that the only symmetric or hermitian matrices with exactly one eigenvalue are scalar multiples of the identity matrix. Consequently, there is no analogous notion of a ``1ev-cover,'' so we are studying the extremal case.

2ev-covering graphs occur frequently in the study of distance-regular graphs, where this two eigenvalue condition is necessary for a lifted graph to retain the distance-regularity of its quotient. However, this condition is not sufficient: $Q_n$ is distance-regular, while $\widetilde{Q}_n$ is not. This begs the question of which combinatorial properties are necessarily preserved by 2ev-covers.  By employing ideas of Godsil and Hensel \cite{GH} we answer this question when the structure of the cover in question is governed by an abelian group. In particular we show walk regularity, a weaker version of distance-regularity, is preserved in abelian two eigenvalue covers. We then show that any cyclic 2ev-cover of the complete graph is a distance-regular and antipodal. Any cyclic 2ev-cover of a complete bipartite graph is distance-regular with diameter four. More generally, we give a necessary and sufficient condition for a cyclic 2ev-cover of a strongly regular graph to be distance-regular. We conclude with some examples and directions for future work.

The theory of covering graphs was first formally developed in 1974 by Biggs \cite{NB} and independently by Gross \cite{JG}. Biggs devoted a chapter of his book to setting down a categorical definition and explicating several interesting examples. Meanwhile, Gross developed an analogous theory as a tool for studying graph embeddings. This topological approach was echoed by Cohen and Tits \cite{CT}, who described $\widetilde{Q}_n$ as the cover corresponding to a particular index 2 subgroup of the fundamental group of the (1-skeleton of) the hypercube.

Even before these formal treatments, many interesting covering graphs were known to mathematicians studying distance-regular graphs. This connection can be described concretely by a result of Smith which states that every distance-regular graph is either primitive, bipartite, or antipodal (and hence a cover). See \cite{BCN}, Chapter 4. In 1990, Godsil and Hensel developed the theory of the distance-regular antipodal covers of the complete graph (drackn's). They described the relation between such covers and regular two-graphs, as well as a connection to sets of equiangular lines in euclidean space. \cite{GH}, \cite{LS}. This work was extended by Coutinho et. al. \cite{CGSZ}  to relate drackns to sets of complex lines, and by Iverson and Mixon \cite{IM}, to a broader class of lines called roux.

Backtracking to 1987, Gross and Tucker \cite{GT} expanded Gross's work and defined the permutation voltage graphs, from which all covering graphs can be constructed.  Zaslavsky  \cite{TZ} studied these graphs from a matroidal perspective under the name gain graphs. This study has blossomed in many different directions whose references are recorded by Zaslavsky in the dynamic bibliography \cite{TZ2}. One recent example is a paper of Cavaleri, D'Angeli, and Donno \cite{CDD}, wherein the authors answer a question of Zaslavsky on the algebraic characterization of balanced gain graphs.  In the final section of this paper, the authors call upon the connection between gain graphs and covering graphs to translate their results to the latter setting. In particular, they show that a certain condition on the balancedness of cycles is equivalent to the associated cover being isomorphic to a disjoint union of copies of the underlying graph. This is closely related (but of an opposite flavor) to our Theorem \ref{wreg1} on the walk regularity of 2ev-covers.  Cavaleri, D'Angeli, and Donno have given a spectral characterization of the covering graphs that are in some sense trivial while we are interested in those that have interesting combinatorial structure.

Running parallel to the inquiries above, the spectral theory of $\pm1$ signings of adjacency matrices has been well developed. The recent advances in this area have, at least partially, been inspired by Bilu and Linial's \cite{BL} proposed construction of expander graphs via an iterative sequence of 2-fold covers, and by Marcus, Spielman, and Srivastava's \cite{MSS} great success in implementing this proposal. Interest has again been renewed by Hao Huang's \cite{HH} resolution of the sensitivity conjecture, which is predicated on signed adjacency matrices with exactly two distinct eigenvalues. Even prior to Huang's proof, the taxonomy of two-eigenvalue signed graphs had begun to emerge, see \cite{ZS}, \cite{GF},\cite{MS}. Moreover, see \cite{BCKW} for a survey of results on two-eigenvalue signed graphs and some interesting open problems.

\section{Structure of covering graphs}
We begin with a connected undirected graph $Y=(V,E)$ and a permutation group $G$. Define a \textbf{symmetric arc function} to be a function $f: V\times V \rightarrow G\cup\{0\}$ for which $f(u,v)=f(v,u)^{-1}$ whenever $f(u,v)\in G$ and $f(u,v)=0$ if and only if $\{u,v\}\notin E(Y)$. The pair $(Y,f)$ is called a \textbf{gain graph}, and can be used to define a cover of $Y$ as follows. Let $R=\{1,2,\dots, r\}$ be a set on which $G$ acts. The \textbf{cover} $Y^f$ of $Y$ is the graph with vertex set $V\times R$ and $(v,j)\sim_{Y^f}(u,k)$ exactly when $v\sim_Yu$ and $f(u,v)j=k$.

Many structural characteristics of $Y^f$ are immediate. If $Y$ is a regular graph, then $Y^f$ is regular with the same valency. This can be considered as a prototypical example of the ``combinatorial regularity'' of covering graphs that we are interested in studying.

There exists a graph homomorphism $\gamma:Y^f\rightarrow Y$ defined by $\gamma(v,j)\mapsto v$; it is called the \textbf{covering map}. The preimages $\gamma^{-1}(v)$ are called the \textbf{fibers} of $Y^f$, and they induce cocliques (edgeless subgraphs). The subgraph induced by any two fibers $\gamma^{-1}(u),\gamma^{-1}(v)$ is a perfect matching if $\{u,v\}\in E$ and is a coclique otherwise.

Two distinct symmetric arc functions can yield isomorphic covering graphs. In particular, permuting the vertices of each fiber $\gamma^{-1}(u)$ of $Y^f$ by some $\sigma_u$ in the symmetric group $\sym(r)$ gives rise to an isomorphic cover $Y^g$ with symmetric arc function $g(i,j)=\sigma^{-1}_i f(i,j)\sigma_j$. Whenever we choose these permutations so that $g(i,j)=id$ for all edges $\{i,j\}$ of some spanning tree of $Y$ we call $g$ a  \textbf{normalized} symmetric arc function.

\section{Representations of finite groups}
Let $Y^f$ as above with $f$ a normalized arc function. Denote by $\langle f \rangle$ the subgroup of $\sym(r)$ generated by the image of $f$. The linear representations of this group are a valuable tool in studying the cover $Y^f$. To keep this paper self contained we will briefly mention the basic representation theory that is relevant to our work. The following discussion is essentially identical to (a subset of) that in Sections 8 and 9 of \cite{GH}. A more detailed account can be found in any text on the representation theory of finite groups, e.g, \cite{JL}.

  A \textbf{representation} $\phi$ of a group $G$ is a homomorphism into the general linear group $GL(r,\bb{C})$. We say that $r$ is the \textbf{degree} of the representation. For each $A\in GL(r,\bb{C})$ the map $\phi^A(g)$ defined by \[\phi^A(g)=A\phi(g)A^{-1}\] is also a representation of $G$, and is said to be \textbf{equivalent} to $\phi$. The \textbf{trivial} representation is the map from $G$ onto the identity matrix $I_r$. If $\phi$ and $\psi$ are representations of $G$ with degrees $r$ and $s$ respectively then the map $\phi+ \psi$ defined by \[g\mapsto \pmat{\phi(g)&0\\0&\psi(g)}\] is a representation of degree $r+s$. We say that $\phi+ \psi$ is the sum of $\phi$ and $\psi$. The set of representations of $G$ is closed under non-negative integral linear combinations. A representation $\phi$ is \textbf{irreducible} if there is no nontrivial subspace of $\bb{C}^r$ which is invariant under $\phi(g)$ for all $g\in G$. It can be shown that a representation of a finite group is reducible if and only if it is equivalent to a nontrivial positive integral linear combination of representations of $G$.

Define a vector space $\mc{V}$ to be the $\bb{C}$-span of the elements of $G$ and define a $G$-action on this vector space where each $g\in G$, acts on the basis vectors by left multiplication (and the action is extended linearly). In the basis of group elements of $G$, each $g$ acts as a permutation matrix $\rho(g)$. The map $\rho: G\rightarrow GL(|G|,\bb{C})$ is called the (left) \textbf{regular representation} of $G$.

  \begin{theorem}
  Let $G$ be a finite group. Then
 \begin{enumerate}
\item[(a)] $G$ has finitely many inequivalent irreducible representations $\phi_i$ for $(i=0,1,\dots)$.
\item[(b)] If $r_i$ denotes the degree of $\phi_i$ and $\rho$ is the regular representation of $G$ then $\rho$ is equivalent to $\sum_{i}r_i\phi_i$.
\item[(c)] If $G=H_1\times H_2$, then the regular representation of $G$ is the tensor product of the regular representations of $H_1$ and $H_2$.
\item[(d)]  If $G$ is abelian then each $r_i=1$ and $\phi_i(g)$ is a (not necessarily primitive) $m$th root of unity where $m$ is the order of $G$.\qed

 \end{enumerate}
  \end{theorem}

We have described above how to construct $Y^f$ from $(Y,f)$. When $\langle f\rangle$ acts regularly on the vertices of a fiber, the regular representation of $\langle f \rangle$ encodes this process as follows.  Let $A=A(Y)$ denote the adjacency matrix of $Y$, and suppose  $\phi$ is a degree $r$ representation of $\langle f \rangle$, define $A^{\phi(f)}$ to be the matrix obtained by replacing each non-zero entry $A_{u,v}$ with the matrix $\phi(f(u,v))$ and each zero entry with the $r\times r$ matrix of zeros.

 \begin{theorem}\label{rep1} (\cite{GH} Sections 7, 8)
 If $f$ is a normalized arc function then $\langle f \rangle$ acts regularly on the vertices of each fiber of $Y^f$ if and only if $A(Y^f)=A(y)^{\rho(f)}$.\qed
 \end{theorem}

If $Y^f$ is connected then $\langle f \rangle$ acts transitively on $\{1,\dots, r\}$ and $|\langle f \rangle|=r\cdot |stab(x)|$ for any $x\in \{1,\dots r\}$. If $\langle f \rangle$ is abelian we may quotient out this stabilizer, considering the resulting abelian subgroup acting regularly on $\{1,\dots r\}$. In this paper we will be concerned with abelian covers, and we will thus assume that our groups $\langle f \rangle$ act regularly on $\{1,\dots, r\}$. Hence we will assume that $A(Y^f)=A(Y)^{\rho(f)}$.

This issue is more subtle if $\langle f \rangle$  is not abelian: We certainly may always work with the regular representation of $\langle f \rangle$, but we will often miss out on certain covers of small index in the process. This is essentially the distinction between ordinary voltage graphs and permutation voltage graphs as explicated by Gross and Tucker \cite{GT}. See Section \ref{nonab} for an example of a cover that is not regular.

 \section{Two eigenvalue covers}
Since $\rho$ is equivalent to $\sum_{i=0}^{|\langle f \rangle |}\phi_i$, it follows that $A(Y)^\rho$ is equivalent to a block diagonal matrix whose blocks are the $A(Y)^{\phi_i(f)}$. The first of these blocks is $A(Y)^{\phi_0(f)}=A(Y)$, from which we deduce that the spectrum of $A(Y)$ is a subset of the spectrum of $A(Y^f)$. Let $\mathrm{Spec}(A)$ denote the spectrum of $A$ as a multi-set.

We say $Y^f$ is a \textbf{two-eigenvalue cover} (2ev-cover) if $\mathrm{Spec}((A(Y^f))\setminus \mathrm{Spec}((A(Y))$ has exactly two distinct elements.  Suppose $Y^f$ is a 2ev-cover, then each $A(Y)^{\phi_i(f)}$ for $i>1$ must be roots of the same minimal polynomial (of degree 2). Our primary means of studying 2ev-covers is via these $A(Y)^{\phi_i(f)}$.

\begin{prop}
Let $\phi$ be some non-trivial irreducible representation of $\langle f \rangle$ and let $Y^f$ be a cover for which $S=A(Y)^{\phi(f)}$ has exactly two distinct eigenvalues $\theta$ and $\tau$. Then there exist real numbers $\lambda, \mu$ so that  \[S^2=\lambda S+\mu I,\] Moreover, $Y$ is regular with valency $\mu=-\theta\tau$.
\end{prop}

\begin{proof}
The minimal polynomial of $S$ has degree 2 and may be written as $x^2-\lambda x-\mu$. Since $f$ is a symmetric arc function, $S$ is hermitian, and has real eigenvalues $\theta, \tau$. We have $\mu=-\theta\tau$ and $\lambda=-\theta-\tau$. Moreover $S$ satisfies the equation $S^2=\lambda S+\mu I$. Since $Y$ is loopless, $S$ has zero diagonal. Finally, the diagonal entries of $S^2$ are equal to the diagonal entries of $A(Y)^2$ which are the valency of the vertices of $Y$.\qed
\end{proof}

 \textbf{Remark:}  In the event that $\langle f \rangle$ is abelian, its irreducible representations have degree 1, and for any non-trivial irreducible representation $\phi$, the matrix $S=A(Y)^{\phi(f)}$ is particularly familiar. Here $S$ is the group valued adjacency matrix of a complex unit-gain graph. This is a direct generalization of the signed adjacency matrix, a setting where the 2ev condition has already been extensively investigated.

\section{Covers of walk regular graphs}

A graph is \textbf{walk regular} if the number of closed walks of length $k$ starting from (and ending at) vertex $v$ depends only on $k$. Equivalently, a graph with adjacency matrix $A$ is walk regular if $A^k$ has constant diagonal for all positive integers $k$. Walk regular graphs have been studied due to their ``spectral regularity:'' The subgraphs $W/v$ have identical characteristic polynomials for all $v\in V(W)$, (see \cite{GM}). Moreover, it can be shown that all distance-regular graphs are walk regular. In Section 6 we will be able to say a fair amount about cyclic two-eigenvalue covers of distance-regular graphs. Working instead with walk regularity we can say more.

\begin{theorem}\label{wreg1} If $Y$ is a walk regular graph and $X=Y^f$ is a cyclic 2ev-cover of $Y$ then $X$ is walk regular.\end{theorem}

\begin{proof}
Let $\phi_0,\dots \phi_{r-1}$ be the irreducible representations of $\langle f \rangle$, with $\phi_0$ the trivial representation, and let $S_1=A(Y)^{\phi_1(f)}$.
Let $A_g$ denote the 01 matrix \[(A_g)_{u,v}=\begin{cases} 1& \text{if } f(u,v)=g \\ 0& \text{otherwise}\end{cases}.\]

\noindent Identifying our index set with the (additive) cyclic group on $\{0,\dots,r-1\}$ we may write \[S_1=\sum_{i=0}^{r-1} \omega^iA_i\] where $r$ is the index of the cover, and $\omega$ is a primitive $r$th root of unity. By replacing $\phi_1$ with $\phi_{i}$ for $i\in \{2,\dots r-1\}$ we exchange $\omega$ for the other $r$th roots of unity and obtain matrices $S_j=A(Y)^{\phi_j(f)}$ defined by \[S_j=\sum_{i=0}^{r-1} \omega^{j\cdot i}A_i.\]

\noindent Let $\circ$ denote the entry-wise matrix product, and let $\ell$ be a non-negative integer.

\textbf{Claim.} \[A(X)^\ell\circ I_{nr}=(\frac{1}{r}(\sum_{j=0}^{r-1}S_j^\ell)\otimes I_r)\circ I_{nr}\]

This will complete our proof since each of the individual $S_j^\ell$ terms have constant diagonal:  $S_1$ has the same support as $A$, so $S_1$ has zero diagonal.  By Proposition 4.1, any power of $S_1$  is polynomial in $\{I,S_1\}$, hence $S_1^\ell$ has constant diagonal. As we have noted in Section 4, the matrices $S_j$ for $j\geq 1$ must be roots of the same minimal polynomial. So each $S_j$ is a root of the minimal polynomial of $S_1$, and $S_j^\ell$ has constant diagonal as well. By assumption $Y$ is walk regular, so $S_0^\ell=A(Y)^\ell$ has constant diagonal. Hence the diagonal of $A(X)^\ell$ is the diagonal of a sum of matrices with constant diagonal and we are done.

\textbf{Proof of Claim.} From Theorem \ref{rep1} we have \[A(X)=\sum_{i=0}^{r-1}A_i\otimes \rho(\omega^i).\] So $A(X)^\ell$ is a sum of products of the form \[(A_{m_1}A_{m_2}\dots A_{m_\ell})\otimes \rho( \omega^{\sum_k m_k})\] where $m_i\in \{0,\dots r-1\}.$  Since $\langle f \rangle$ is regular, the only summands that have non-zero diagonal entries are those for which the righthand tensor factor is the identity i.e., those terms where $\sum m_k=0 \mod r$. Let $\mc{M}$ be the set of $\ell$-tuples $(m_1,\dots m_\ell)$ of elements of $\langle f \rangle$, and let $\mc{M}_0$ the subset of $\mc{M}$ for which  $\sum m_k=0\mod r$. The previous remark shows that $A(X)^\ell$ has the same diagonal as

\begin{equation} \sum_{M\in\mc{M}_0}\big(\prod_{m_i\in M}A_{m_i}\big)\otimes I_r=\Big(\sum_{M\in\mc{M}_0}\prod_{m_i\in M}A_{m_i}\Big)\otimes I_r\end{equation}

\noindent On the other hand, notice that \[S_j^\ell=(\sum_{i=0}^{r-1} \omega^{j\cdot i}A_i)^\ell=\sum_{M\in\mc{M}}\prod_{m_i\in M}\omega^{j\cdot m_i}A_{m_i}\]

\noindent Now, summing over all such $S_j$ we have \[\sum_{j=0}^{r-1} S_j^\ell=\sum_{j=0}^{r-1} \sum_{M\in\mc{M}}\prod_{m_i\in M}\omega^{j\cdot \sum m_i}A_{m_i} =\sum_{M\in\mc{M}}\sum_{j=0}^{r-1}\prod_{m_i\in M}\omega^{j\cdot \sum m_i}A_{m_i} \]

\noindent For any $M\in \mc{M}\backslash \mc{M}_0$ the $\omega^j$ term is independent of the product, hence \[\sum_{j=0}^{r-1}\prod_{m_i\in M}\omega^{j\cdot \sum m_i}A_{m_i}=\sum_{j=0}^{r-1}\omega^j\prod_{m_i\in M}\omega^{\sum m_i}A_{m_i}=0.\] And so \begin{equation}\sum_{j=0}^{r-1} S_j^\ell=\sum_{M\in\mc{M}_0}\sum_{j=0}^{r-1}\prod_{m_i\in M}\omega^{j\cdot \sum m_i}A_{m_i} =\sum_{M\in\mc{M}_0}\sum_{j=0}^{r-1}\prod_{m_i\in M}A_{m_i}=r\sum_{M\in\mc{M}_0}\prod_{m_i\in M}A_{m_i}.\end{equation}
Putting together equations $(1)$ and $(2)$ proves the claim.\qed
\end{proof}

\begin{corollary} If $Y$ is a walk regular graph and $X$ is an abelian 2ev-cover of $Y$ then $X$ is walk regular.\end{corollary}

\begin{proof}
The regular representation of the abelian group $G=H_1\times H_2$ is the tensor product of the regular representations of $H_1$ and $H_2$. Hence for each $g\in G$ we have $\rho(g)=\rho_{H_1}(h_1)\otimes \rho_{H_2}(h_2)$, and $\rho(g)$ still has zero diagonal unless $g=id$. So the only terms in the expansion of $(\sum_{g\in G} A_g\otimes \rho(g))^\ell$ that contribute to the diagonal are those whose first tensor factor has subscripts whose sum is zero mod $|G|$. The irreducible representations of $\rho$ are still of degree one and the argument  proceeds exactly as above. \qed
\end{proof}

\section{Covers of strongly regular graphs}
Let $X$ be a connected regular graph and $v\in V(X)$. The \textbf{distance partition} of $X$ with respect to  $v$ is a partition of $V(X)$ into cells $\{\Gamma_0(v),\Gamma_1(v),\dots,\Gamma_d(v)\}$ where $\Gamma_i(v)$ consists of the vertices at distance $i$ from $v$.  A partition $\pi=(\pi_1,\dots,\pi_k)$ of $V(X)$ is \textbf{equitable} if for each pair $(i,j)\in \{0,\dots d\}\times \{0,\dots d\}$ the number of edges of $X$ with one end in $\pi_i$ and the other end in $\pi_j$ depends only on $i$ and $j$.

A graph whose distance partition is equitable is \textbf{distance-regular}. Each distance-regular graph of diameter $d$ has an \textbf{intersection array} $\{b_0,b_1,\dots b_{d-1};c_1,c_2\dots c_d\}$ where $b_i$ is the number of neighbors $u\in \Gamma_{i+1}$ of any $v \in \Gamma_i$ and $c_i$ is the number of neighbors $u\in \Gamma_{i}$ of any $v \in \Gamma_{i+1}$.

A distance-regular graph with diameter 2 is \textbf{strongly regular}. The \textbf{parameters} of a strongly regular graph are the 4-tuple $(n,k,a,c)$ where $n$ is the number of vertices, $k$ is the valency, $a:=k-b_1-c_1$ is the number of common neighbors of a pair of adjacent vertices and $c:=c_2$ is the number of common neighbors of a pair of non-adjacent vertices. See Chapter 10 of \cite{CG2} for a thorough introduction.

A graph $Y$ of diameter $d$ is \textbf{antipodal} if ``$v$ is at distance $d$ from $u$'' is an equivalence relation on $V(Y)$. If $X$ is antipodal and distance-regular, we consider the graph $Y$ whose vertices are the antipodal classes with two classes adjacent if there is an edge of $X$ with one end in each class. $Y$ is called the \textbf{antipodal quotient} of $X$, and $X$ is a cover of its antipodal quotient. See Theorem 2.1 of \cite{GH}, \cite{BCN}, \cite{AG}. In the case that this antipodal quotient $Y$ is the complete graph, $X$ is a distance-regular antipodal cover of $K_n$, or \textbf{drackn} for short.\label{drackn}

Distance-regular graphs are a well studied topic in algebraic graph theory. They are related to combinatorial designs, linear codes, association schemes, and orthogonal polynomials. The literature is vast, but a good starting place would be \cite{BCN},\cite{NB},\cite{CG2}.

We take a quick detour to offer a few interesting examples.
\begin{enumerate}

\item The hypercube is distance-regular and antipodal with antipodal classes of size 2. Its antipodal quotient is also distance-regular. In the case of $Q_3$ this antipodal quotient is the complete graph $K_4$, and $Q_3$ is a drackn.
\item The Hoffman-Singleton graph $H$ is strongly regular with parameters (50,7,0,1), For any fixed vertex $v\in V(H)$ the subgraph induced by vertices at distance 2 from $v$ is a 6-fold distance-regular antipodal cover of $K_7$.
\item The Johnson graph $J(n,k)$ has as its vertices the $k$ element subsets of $\{1,\dots, n\}$. Two vertices are adjacent if  their intersection has size $k-1$. Any Johnson graph is distance-regular.
\end{enumerate}

Now we will give a characterization of when a cyclic 2ev-cover of a strongly regular graph is distance-regular.

Our key lemma shows that the 2ev condition forces the arc function to take a very particular form.

\begin{lemma}\label{ML}
Let $Y$ be a connected distance-regular graph with parameters $a:=k-b_1-1$ and $c:=c_2$. Let $\phi$ a non-trivial irreducible representation of $\langle f \rangle$, $S=A(Y)^{\phi(f)}$, and $x^2-\lambda x-\mu$ the minimal polynomial of $S$. Assume that $f$ is normalized so that $S$ is of the form \[S=\pmat{0&\bb{1}&\bb{0}&\dots\\\bb{1}^T&N_1 &B\\ \bb{0}^T&B^T & N_2&\\\vdots& &&\ddots}.\] There exist constants \[t=\frac{a-\lambda}{r},~~s=\frac{c}{r}\] so that \begin{itemize}\item For each $i\in \{1,\dots, r-1\}$, each column of the submatrix $N_1$ contains exactly $t$ entries equal to $\omega^i$. \item For each $j\in \{0,\dots, r-1\}$ each column of the submatrix $B$ contains exactly $s$ entries equal to $\omega^j$.\end{itemize}
\end{lemma}

\begin{proof}
For a submatrix $M$ of $S$, let $C_m(M)$ denote the sum of the entires of column $m$ of $M$. Matrix multiplication shows that the first row of $S^2$ is \[\pmat{k&C_1(N_1)&\dots& C_k(N_1)&C_1(B)&\dots&C_{\ell}(B)&0&\dots&0}.\]
On the other hand, $S^2 = \lambda S + \mu I$ so the first row of $S^2$ is \[\pmat{\mu& \lambda &\dots& \lambda& 0 &\dots& 0}\]

We have three immediate consequences. \begin{itemize}
\item $\mu=k$.
\item For each $1\leq m\leq k, C_m(N_1)=\lambda$.
\item For each  $1\leq n \leq \ell$, $C_n(B)=0.$
\end{itemize}

For some fixed column $m'$ of $N_1$, let $t_i$ denote the number of occurrences of $\omega^i$ in that column. We may write $C_m'(N_1)=\sum_{i=0}^{r-1}t_i\omega^i$. Note that $S$ is hermitian and that $\lambda$ is (minus) the sum of the two distinct eigenvalues of $S$, hence $\lambda$ is real. So the above expression for $\lambda$ must be symmetric under any permutation of the $rth$ roots of unity which fixes the rationals. It follows that $t_1=t_2=\dots =t_{r-1}$. Denote this common value by $t$. We have \[\lambda=C_{m'}(N_1)=t_0+t\Big(\sum_{i=1}^{r-1}\omega^i\Big)=t_0-t.\] Since $Y$ is distance-regular, we have $\sum_{i=0}^{r-1}t_i=a$ which we may now write as \[t_0+t(r-1)=a.\] Combining our equations we obtain \[t=\frac{a-\lambda}{r}.\] Note that $t$ is independent of the chosen column $m'$, this proves the first claim.

For some fixed column $n'$ of $B$, let $s_j$ denote the number of occurrences of $\omega^j$ in that column. We play the same game with the equations \[C_{n'}(B)=\sum_{j=0}^{r-1}s_j\omega^j=0,~~~~~\sum_{j=0}^{r-1}s_j=c\] and find that \[s:=s_0=s_1=\dots=s_{r-1}=\frac{c}{r}.\] is independent of $n'$. This proves the second claim. \qed
\end{proof}

Now we will employ this lemma to characterize cyclic 2ev-covers of strongly regular graphs. We begin with the degenerate case.

\begin{theorem}\label{drg1}
If $Y$ is the complete graph on $n$ vertices and $Y^f$ is a connected cyclic 2ev-cover of $Y$, then $Y^f$ is a drackn.
\end{theorem}

\begin{proof}
Note that $t>0$, since $t=0$ implies $Y^f$ is the disjoint union of $r$ copies of $K_n$.

Label the vertices of $Y$ as $v_0,\dots, v_{n-1}$, and let $\gamma$ be the covering map from $Y^f$ to $Y$. For each fiber $\gamma^{-1}(v_i)$  let $(v_i,0),\dots, (v_i,r-1)$ denote the vertices of that fiber. Let $v_0$ be the vertex of $Y$ indexing the first row and column of $S$, and choose (arbitrarily) some vertex of $\gamma^{-1}(v_0)$ to be $(v_0,0)$. We consider the distance partition of $Y^f$ with respect to $(v_0,0)$.

$Y^f$ has at most 4 distinct eigenvalues ($n-1$, $-1$, and the eigenvalues of $S$), hence it has diameter at most $3$. Any two vertices of $Y^f$ from the same fiber are not adjacent and do not have any common neighbors, hence the diameter of $Y^f$ is exactly 3. The assumption that $S$ is normalized along the first row and column is equivalent to the claim that \[\Gamma_{1}(v_0,0)=\{(v_i,0):i\in \{1,\dots n-1\}\}.\]

For any $j\neq 0$ and  $(v_j,q)$, the result of Lemma \ref{ML} and the fact that $t>0$ imply that, in the  $j$th column of $S$ there is some entry $S_{i,j}=\omega^q, i\neq 0$. Hence $(v_j,q)\sim(v_i,0)$ and the distance between $(v_0,0)$ and $(v_j,q)$ is at most (in fact, exactly) 2.  It follows that each vertex that is neither a neighbor of $(v_0,0)$ nor in $\gamma^{-1}(v_0)$ is at distance two from $(v_0,0)$. Moreover for any such vertex  $(v_j,q)$ there are exactly $t$ entries of the $j$th column equal to $\omega^q$. So $(v_0,0)$ and $(v_j,q)$ have exactly $t$ common neighbors. It now follows from Lemma 3.1 in \cite{GH} that $X$ is an $(n,r,t)$-drackn.  \qed
\end{proof}

Now we consider the case where $Y$ is strongly regular. As illustrated by Example 7.1, it is not true that all cyclic 2ev-covers of strongly regular graphs are distance-regular, but we can give an exact spectral characterization of those which are.

\begin{theorem}\label{drg2} Let $Y$ be a connected strongly regular graph with parameters $(n,k,a,c)$, $\phi$ a non-trivial irreducible representation of $\langle f \rangle$, $S=A(Y)^{\phi(f)}$, and $x^2-\lambda x-\mu$ the minimal polynomial of $S$. Then $Y^f$ is distance-regular if and only if $a=\lambda$.
\end{theorem}

\begin{proof}
Suppose $a=\lambda$. Then $t=0$ and  $N_1$ is a 0-1 matrix. We construct the distance partition of $Y^f$ with respect to $(v_0,0)$ and find that the neighborhood of $(v_0,0)$ induces a subgraph isomorphic to $Y[N_1]$, call this subgraph $H_0$, and note that, since the arc function is trivial on $N_1$, there are $r-1$ other subgraphs $H_1,\dots H_{r-1}$ contained in $y^f$, each of which is the neighborhood of some $(v_0,j)$ for $j\in \{1,\dots r-1\}$. Moreover, there are no edges between any of the distinct $H_i$ (again, since the arc function is trivial on $N_1$).

Since $Y$ is connected, we have $c>0$. By Lemma \ref{ML}, each column of $B$ contains $\frac{c}{r}$ entries equal to each $\omega^i$. Hence each vertex $(v_j,i)$ for $v_j\in N_2$ is adjacent to $\frac{c}{r}$ vertices of each $H_i$. This shows that  $\Gamma_2(v_0,0)$ is the union of fibers of vertices in $N_2$, and $\Gamma_3(v_0,0)$ is the union of $H_i$ for $i>0$. Finally,  $\Gamma_4(v_0,0)$ is the remaining $r-1$ vertices of the $v_0$ fiber. It follows that the distance partition is equitable, with intersection array \[\{k,k-a-1,\frac{c(r-1)}{r},1;1,\frac{c}{r},k-a-1,k\}.\]

Now suppose $a\neq\lambda$. So $t>0$ and each column of $N_1$ other than the first contains each $\omega^i$ for $i\in \{0,\dots, r-1\}$. Since $Y$ is connected, $c>0$, and each column of $B$ also contains at least one entry equal to each $\omega^i$.

As in the proof of Theorem \ref{drg1}, this means that when we construct the distance partition of $Y^f$ with respect to $(v_0,0)$, we have \[\Gamma_1(v_0,0)=\{(v_j,0) : j\in \{1,\dots, n-1\}\} \]  and \[\Gamma_3(v_0,0) \supseteq \{(v_0,j): j \in \{1,\dots r-1\}\}.\]
There are two types of vertices remaining. Those of the form $(v_\alpha,i)$ for $v_\alpha$ a neighbor of $v_0$ and those of the form $(v_\beta,i)$ for $v_\beta$ at distance 2 from $v_0$.  By Lemma \ref{ML}, each $v_\alpha$ is adjacent to $t>0$ neighbors of $v_0$ for which the edge joining them is assigned the value $\omega^j$. Since this holds for all $j\in\{1,\dots, r-1\}$ each $(v_\alpha,i)$ for $i>0$ is at distance 2 from $(v_0,0)$. Similarly, by Lemma \ref{ML} all vertices of the fiber $\gamma^{-1}(v_\beta)$ are at distance two from $(v_0,0)$, and none of these vertices are incident with any vertex of $\gamma^{-1}(v_0)$, so some vertices in $\Gamma_2$ have neighbors in $\Gamma_3$ while others do not, and $X$ is not distance-regular.\qed\end{proof}

As a special case of the previous theorem, we show that every cyclic 2-ev cover of a complete bipartite graph is distance-regular.

\begin{corollary}
Suppose $Y$ is complete bipartite and $Y^f$ is a cyclic $r$-fold 2-ev cover of $Y$. Then $Y=K_{n,n}$ with $r|n$, and $X$ is bipartite distance-regular with diameter 4.
\end{corollary}

\textbf{Proof}. It follows from Proposition 4.1  that $Y$ is regular.
 Hence $Y$ is strongly regular with parameters $(2n,n,0,n)$ for some $n$. Normalizing as in Theorem \ref{drg2} and noting that $a=0$ we have
\[S=\pmat{0&\bb{1}&\bb{0}\\\bb{1}^T&\bb{0}_{n\times n} &B\\ \bb{0}^T&B^T &N_2}\]

Hence $\lambda=0$ as well, and $Y^f$ is distance-regular. Moreover, each column of $B$ contains $\frac{c}{r}=\frac{n}{r}$ entries equal to $\omega^i$, hence $r|n$. It is well known that a graph is bipartite if and only if its adjacency spectrum is symmetric about 0. $Y$ is bipartite, and $S$ has spectrum symmetric around zero. Hence $Y^f$ has spectrum symmetric around zero and is bipartite.\qed

\section{Examples}
In this section we provide a few examples of 2ev-covers.

\subsection{The Cohen-Tits cover of $Q_n$}\label{cttohh} We wish to construct a 2-fold cover of $Q_n$ which contains no 4-cycles. We employ the inductive construction of $Q_n$ as two copies of $Q_{n-1}$ joined by a perfect matching.
Consider a $\pm1$ signing of the edges of $Q_{n-1}$, or equivalently, a symmetric arc function \[f:V\times V \rightarrow \bb{Z_2}\cup\{0\}.\]  The condition that the resulting 2-fold cover $Q_{n-1}^f$ has no 4-cycles  is equivalent to the condition that each 4-cycle of $Q_n$ has an odd number of edges $e$ with $f(e)=-1$. It follows that the opposite signing, $f'=-f$ will also give rise to a 4-cycle free 2-fold cover. Joining these oppositely signed copies of $Q_{n-1}$ by a perfect matching whose edges are all given the same sign (say, $+1$) yields a signing of $Q_n$ in which each 4-cycle has an odd number of negative edges, and thus gives rise to the desired cover.

Performing this process at the level of the adjacency matrix is exactly the construction given by Huang in \cite{HH}. Define \[A_1=\pmat{0&1\\1&0},~~~A_{n}=\pmat{A_{n-1}&I_{n-1}\\I_{n-1}&-A_{n-1}}.\] Then taking $f(u,v)=(A_n)_{u,v}$ defines a symmetric arc function on the hypercube $Q_n$. The resulting cover of index 2 is the unique girth six 2-fold cover of $Q_n$ determined by Cohen and Tits in \cite{CT}. Note that this cover is not distance-regular, however its distance distribution diagram is a tree which is not too far away from being a path. (See \cite{BCN} section 9.2E).

\subsection{Folded cubes}
The folded $n$-cube, denoted $\square_n$ is the antipodal quotient of $Q_{n}$. Equivalently, $\square_n$ is the graph obtained from $Q_{n-1}$ by adding a perfect matching between antipodal vertices. In \cite{BCN} the authors remark, without proof, that a 4-cycle free 2-fold cover of $\square_n$ exists if and only if $n$ is congruent to $0, 1$ mod 4 and $n>4$. (See \cite{BCN} section 9.2E). These covers are necessarily 2ev. A construction of these covers, as well as a proof of the non-existence of such covers for the other congruence classes can be found in the recent arxiv preprint of Alon and Zheng \cite{AZ}.

Alon and Zheng have constructed what they call unitary signings of Cayley graphs for elementary abeliean 2-groups. The graphs in question are the folded $d$-cubes and the cartesian products of folded $(k+1)$-cubes with $(d-k)$-cubes. And unitary signings can be viewed as symmetric arc functions on the underling graphs which give rise to cyclic covers of index 4 and girth $>4$.

\subsection{The 3-fold cover of $Kn(7,2)$}
The Kneser graph $Y=Kn(7,2)$ has as vertices the 21 two element subsets of $\{0,1,\dots 6\}$. Two such subsets are adjacent if they are disjoint. Its spectrum is \[\{10^{(1)},1^{(14)},-4^{6}\}.\] There is a unique antipodal distance-regular graph $X$ of diameter 4 on 63 vertices. It's spectrum is \[\{10^{(1)},5^{(12)},1^{(14)},-2^{(30)},-4^{6}\}.\] As suggested by the spectrum, $X$ is a 3-fold 2ev-cover of $Y$, (See \cite{BCN} 13.2B).

\subsection{Drackn's and dracknn's}

As seen in Section 6, the drackn's are precisely the cyclic 2-ev covers of the complete graph, and the distance-regular antipodal covers of the complete bipartite graph (dracknn's) are precisely the cyclic 2-ev covers of $K_{n,n}$. In particular, the distance-regular 2-fold covers of $K_{n,n}$ are the Hadamard graphs, (See \cite{BCN} Section 1.8). More generally, if $H$ is a complex Hadamard matrix of Butson type (meaning the entries are $q$th roots of unity), with \[B=\pmat{0& H\\H^T&0}\] then $f(u,v)=B_{u,v}$ defines a symmetric arc function which gives rise to a distance-regular 2ev-cover of $K_{q,q}$.

\subsection{A non-abelian 2-ev cover}\label{nonab}
Many of our proof techniques are predicated on the cyclic or abelian nature of the group $\langle f \rangle$. In Theorem \ref{wreg1}, It was important that $\langle f \rangle$ acted regularly $\{1,\dots, r\}$. In Theorems \ref{drg1} and \ref{drg2}, we made (implicit) use of the fact that the irreducible representations of $\langle f \rangle$ were one-dimensional.

However, we do not lose all of our tools when considering non-abelian covers. If we work with any permutation representation $\tau$ for $\langle f \rangle$, the the spectrum of the resulting cover will decompose as the union of the spectra of the covers defined by the irreducible representations of $\langle f \rangle$. The regular representation is no longer the only interesting choice. As a first example of a non-regular 2ev-cover, consider the line graph of the Petersen graph.  This graph can be obtained as a 2ev-cover of $K_5$ by taking the 3-dimensional permutation representation of $S_3$, and the following assignment with $\tau_1=(1,2),\tau_2=(2,3),\tau_3=(1,3)$. \[\pmat{0&1&1&1&1\\1&0&\tau_1&\tau_2&\tau_3\\1&\tau_1&0&\tau_3&\tau_2\\1&\tau_2&\tau_3&0&\tau_1\\1&\tau_3&\tau_2&\tau_1&0}\]

\subsection{An informative non-example}
There are several ways to interpret the statement ``a cover with two new eigenvalues.'' A priori, it is not clear that we have chosen the correct interpretation. One alternative is to consider covers where the number of distinct eigenvalues has increased by 2. The following example shows that our definition is better suited to certain combinatorial investigations.

We consider a $\pm 1$ signing of $A(K_{3n})$ with three distinct eigenvalues, one of which is $-1$ (and so coincides with the spectrum of $K_{3n}$).

Construct an arc function with $f(u,v)=-1$ for edges $\{u,v\}$ of a complete bipartite subgraph $K_{n,n}$, and $f(u,v)=1$ on all other edges. This signed matrix has spectrum \[\{(2n-1)^{(2)},-1^{(3n-3)},(-n-1)^{(1)}\},\] compared to the spectrum \[\{(3n-1)^{(1)},(-1)^{(3n-2)}\}\] of $K_{3n}$. We see that the cover is possessed of two new and distinct eigenvalues and one new eigenvalue coinciding with an eigenvalue of the base graph. However it can be shown that this cover is not distance-regular.

\section{Further questions}
We have explicated the relationship between 2ev-covering graphs and the representations of their group valued adjacency matrices. Moreover we have shown that this is a natural setting in which to study combinatorial regularity of covering graphs. This study sits at the intersection of several research areas, and there are many directions in which it could progress. We name just a few that are most closely related to our present work.
\begin{itemize}
\item It may be possible to extend Theorem \ref{drg2} to start with graphs of larger diameter. However, we expect that there are no connected 2-ev covers in this case, since a result of Gardiner implies that any such cover could not be antipodal. See \cite{AG}.

\item If $Y^f$ is a 2ev-cover of $Y$ we have very strong conditions on the balancedness of  cycles in the gain graph $(Y,f)$. In Theorem \ref{drg2} the condition that $\lambda=a$ implies that for any triangle in $Y$, with vertices $a,b,c$, we have $f(a,b)f(b,c)f(c,a)=id.$ Perhaps there are cases where these conditions imply the existence of some interesting biased graphs.

\item It is possible that Theorem 5.1 can be extended to non-abelian 2ev-covers. However our proof does not seem to extend in any straightforward way. Alternatively, there may be non-abelian 2ev-covers of walk regular graphs which fail to be walk regular. We would be curious to hear of any examples.

\item There are already a number of constructions of $\pm 1$ signings of adjacency matrices which have exactly two eigenvalues. It could be interesting to consider the associated covering graphs for some of these constructions. Moreover, it could be very interesting to extend existing methods for constructing 2ev-signings of adjacency matrices to construct cyclic 2ev-covers of larger index.

\end{itemize}

redpajama_set_name: RedPajamaArXiv


text:
Melodifestivalen 1999 var den 39:e upplagan av musiktävlingen Melodifestivalen och samtidigt Sveriges uttagning till Eurovision Song Contest 1999.

Finalen hölls på Victoriahallen/Stockholmsmässan i Stockholm den 27 februari 1999, där melodin "Tusen och en natt", framförd av Charlotte Nilsson, vann, genom att ha fått högst totalpoäng av jurygrupperna och tv-tittarna. För andra gången i festivalens historia fick tv-tittarna vara med och påverka resultatet genom telefonröstning, något som varje år sedan dess använts i festivalen. Finalkvällen blev återigen bara körd i en omgång istället för i två och systemet med att bjuda in kompositörer på förhand ströks. Dessutom återupptogs samarbetet med Svenska musikförläggareföreningen igen. För första gången någonsin spelades de tio finalbidragen upp i radio samma eftermiddag som finalen skulle äga rum.

Tusen och en natt fick sedan representera Sverige i ESC 1999 som arrangerades i Jerusalem i Israel den 29 maj 1999.

Tävlingsupplägg

Efter många år med att ha haft finalkvällen uppdelad i två omgångar och endast haft jurygrupper som enda resultat valde Sveriges Television att förändra och förnya konceptet. Finalkvällen behölls men likt många tidigare år (ej åren 1958-65, 1971, 1982-88 och 1991-96) tävlades bara finalen i en enda omgång. Skillnaden blev nu att makten delades mellan elva jurydistrikt runt om i Sverige samt telefonröster. Telefonröster hade förvisso använts redan år 1993 (och 1962 avgjordes hela finalen med hjälp av poströster) men det var först det här året som tittarna kunde rösta på alla tio melodierna och fick avlägga lika många poäng som alla de elva jurygrupperna tillsammans.

Systemet med att ha förinbjudna kompositörer togs bort och därmed gick alla tio finalplatser till bidrag som skickats in genom en allmän inskickning av bidrag. Totalt skickades det in 1 315 bidrag, vilket blev en ökning med 194 bidrag jämfört med året innan. Men Sveriges Television valde delvis att ta tillbaka det system som använts flitigt under 1980-talet, med att låta bidragen granskas av Svenska musikförläggareföreningen (SMFF) innan urvalsjuryn fick börja lyssna och bedöma. Till skillnad från 80-talets system behövde inte de inskickade bidragen gå via ett musikförlag innan de nådde Sveriges Television, utan bidragen skickades in till televisionen och granskades därefter av SMFF och Sveriges Television. Efter granskningen tog urvalsjuryn vid som tog ut tio tävlande bidrag.

Det här året blev det första året som festivalen direktsändes via internet (genom Sveriges Televisions hemsida), och dessutom var det första året som de tio finalbidragen spelades upp i Sveriges Radio P4 samma eftermiddag som finalen skulle hållas.

Återkommande artister

1 1994 tävlade Roger Pontare tillsammans med Marie Bergman.

Övrigt
 Martin Svensson var innan tävlingen favorittippad att vinna, medan den riktiga vinnaren (Charlotte Nilsson) enbart fick en tvåa (av fem) i betyg av Aftonbladets kritiker Annika Sundbaum-Melin.
 I tjejgruppen Ai ingick bland annat Josefine Sundström som senare samma år blev anställd på Sveriges Television som programledare för musikprogrammet Voxpop. Sundström skulle också komma att vara programledare för Melodifestivalen 2001 och senare även för Melodifestivalen 2013, då som webbprogramledare.
 På grund av tekniska problem fick Arvingarna ta om sitt nummer. Felet bestod i att bandets mikrofoner inte var påslagna. Trots sin "andra" chans lyckades inte bandet vinna.
 Thomas G:son fick med sin första låt i Melodifestivalen. I senare festivaler har han fått med över 25 låtar, där tre vunnit. Han har även tävlat med andra bidrag i andra europeiska länder (och lyckats vinna där ett antal gånger).
 Cleo Nilsson fick starta sist det här året och blev därmed den sista artisten som framförde ett bidrag i Melodifestivalen på 1900-talet (ej inräknat att vinnarlåten spelades igen).
 "Tusen och en natt" fick knappt 125000 röster.
 "Det gäller dej och mej" fick över 50000 färre röster än "Tusen och en natt":s 125000 röster.

Finalkvällen
Finalen av festivalen 1999 direktsändes i SVT2 (och SR P4) den 27 februari 1999 kl. 21.00-23.00 från Victoriahallen (Stockholmsmässan) i Stockholm. Programledare var Anders Lundin och Vendela Kirsebom Thomessen medan Anders Berglund var kapellmästare. Det här är den enda gången som en Melodifestival organiserades i Stockholm sändes i tvåan istället för ettan.

Det här året blev det första året som de elva jurydistrikten fick dela avgörandet med tv-tittarna. De bägge grupperna (jury respektive tittarna) gavs 50 procents makt vardera att avgöra finalen. Förvisso hade telefonröstning använts även år 1993, men då fick tittarna agera jury i den andra röstningsomgången. Jury respektive tittarna hade 50 totalt 473 poäng vardera att dela ut. Varje jury delade ut poängen 1, 2, 4, 6, 8, 10 och 12 poäng. Tittarnas andel delades ut i juryns poäng multiplar av elva: 1 poäng blev 11 poäng, 2 poäng blev 22 poäng, 4 poäng blev 44 poäng, 6 poäng blev 66 poäng, 8 poäng blev 88 poäng, 10 poäng blev 110 poäng och 12 poäng blev 132 poäng. Sju av tio bidrag gavs därför poäng per jurygrupp och telefonröster. Därför fick de tre bidragen med minst antal röster per jurygrupp (och även för telefonrösterna) noll poäng. Jurygrupperna representerade städer och bestod enbart av musiker, musikproffs etc.

Klockan 14.00 samma eftermiddag som finalen skulle hållas spelades bidragen upp i Sveriges Radio P4, samtidigt som man påbörjade telefonröstningen. Varje samtal kostade 60 öre och det fanns ingen begränsning på hur många samtal man fick ringa. Vart och ett av de tio bidragen hade dessutom två telefonnummer: ett där man avlade sin röst och ett där man kunde lyssna på låten i telefonen. Valde man dock att ringlyssna på låten kostade samtalet 4,90 kr per minut, pengar som gick oavkortat till Radiohjälpen. När sändningen sedan startade spelades de tio bidragen upp, med fortsatt röstning. Efter att alla bidragen hade framförts och spelats upp i en snabbrepris stängdes slussarna och rösterna räknades. Samtidigt som det skedde avlade jurygrupperna sina poäng. När alla jurygrupper hade sagt sitt adderades deras poäng med tittarnas poäng och man fick ut ett totalresultat. Vinnaren blev den som efter jurygruppernas och tittarnas poängutdelning hade fått högst totalpoäng. Samtliga Melodifestivalsfinaler åren efter det här året har avgjorts på samma sätt, dvs. med lika delar jurypoäng som telefonpoäng. Det kan noteras att Sverige använde telefonröstning som resultat i Eurovision Song Contest redan år 1997 och 1998, men i Melodifestivalen 1997 och 1998 avgjorde en jury.

I fem låtar användes bara orkester. De andra låtarna hade också en förinspelad band, låtar 2 och 8 utförde mestadels med förinspelad musik, och i låt nummer 4, "Stjärna på himmelen" av Drömhus, användes inte orkestern alls.

Startlista
Nedan listas bidragen i startordningen:

Poäng och placeringar

 Totalröster: 630 339 röster.
 Tittarsiffror: 3 212 000 tittare.

Juryuppläsare
Liksom år 1997 läste juryuppläsarna i första omgången upp de lägre poängen (1, 2, 4, 6, 8, 10) och tolvorna i andra omgången.

 Luleå: Christer Holmqvist
 Umeå: Anita Färingö
 Sundsvall: Peter Nässén
 Falun: Tinna Edlund
 Karlstad: Jenny Eklund
 Örebro: Päivi Kotka
 Norrköping: Per Dahlberg
 Göteborg: Henrik Johnsson
 Växjö: Henrik Olsson
 Malmö: Pernilla Månsson
 Stockholm: Sten Söderberg

Eurovision Song Contest

Efter Dana Internationals seger året innan fick Israel stå som värdland. Samma arena som stod värd 1979 användes (Binyanei Ha'ouma ICC i Jerusalem) även det här året och för första gången i ESC-historien var det tre programledare som ledde finalen, som hölls den 29 maj 1999.

EBU gjorde ett antal förändringar inför det här årets festival. För det första slopades språkregeln helt och hållet, vilket gjorde att alla tävlande länder själva fick bestämma vilket språk de ville sjunga på. Dessutom togs orkestern bort helt och hållet, vilket gjorde att man bara använde förinspelad musik till alla länders bidrag. Dock behölls regeln om att all huvud- och körsång skulle göras live och spärren för antalet länder behölls. Därför fick de sämst placerade året innan avstå tävlan det här året, till förmån för de länder som fick avstå 1998 års ESC.

Tjugotre länder kom till starten i och med att Bosnien och Hercegovina, Danmark, Island, Litauen och Österrike återkom igen. På grund av dålig placering året innan fick dock Finland, Grekland, Makedonien, Rumänien, Schweiz och Slovakien avstå tävlan. Egentligen skulle Lettland ha gjort debut, men drog sig ur och tillät då Ungern att tävla istället. Dock drog sig även Ungern ut vilket gjorde att Portugal fick deltaga istället.

Sverige tävlade som nummer femton (av tjugotre länder) och hade efter varje lands omröstning slutat på totalt 163 poäng och därmed också vunnit hela tävlingen. Det blev således Sveriges fjärde vinst i tävlingen. Det var också det sista seger under 13 år, fram till 2012. Island var länge nära att slå Sverige, men slutade på andra plats med 146 poäng (dock Islands dittills bästa resultat) och Tyskland slutade trea med 140 poäng. Allra sist blev Spanien som bara fick en poäng. Dock kan det tilläggas att man från nästföljande års Eurovision Song Contest införde den så kallade "Big Four"-gruppen som helt enkelt innebar att Frankrike, Spanien, Storbritannien och Tyskland inte längre behövde bekymra sig över sina nästkommande placeringar eftersom de från då började betala mest till tävlingen och därmed fick en automatisk finalplats oavsett hur bra eller dåligt de placerar sig.

Detta var den allra sista Eurovision Song Contest som sändes på 1900-talet. Estlands Diamond of Night, framförd av Evelin Samuel & Camille, fick därmed avsluta 1900-talets Eurovision Song Contest.

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TV-sändningarna i SVT:s öppna arkiv

1999 i Sverige
1999
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Musikevenemang i Älvsjö

redpajama_set_name: RedPajamaWikipedia


This collection includes memorabilia collected by the two brothers during their World War I service.
One photograph album containing photographs taken by a Canadian soldier between 1917 and 1919.
This fonds contains materials relating to Beach's academic career and interests in psychology, human behaviour, and counseling; as well as material relating to World War II, conflict in general, and behaviour under various forms of stress.


Cristiano Ronaldo scored his 20th Juventus goal as the unbeaten champions beat Sassuolo to go 11 points clear at the top of Serie A.
Sami Khedira scored from a rebound after a Ronaldo effort was saved.
The Portugal star then headed home Miralem Pjanic's corner in the second half – the ninth consecutive away league game in which he has scored.
Former Liverpool midfielder Emre Can scored a late third with his first touch after coming off the bench.
Ronaldo – who also had a goal correctly disallowed for offside – is Serie A's top scorer with 18 goals, and has scored 20 in 31 appearances for Juve in all competitions.
Juventus, who have won 20 and drawn three of their 23 league games, are chasing an eighth consecutive Serie A title.
Earlier, Atalanta briefly moved into the Champions League places with a 2-1 win over SPAL. Duvan Zapata scored their winner with his 18th goal in 12 games.
However, AC Milan retook fourth spot with a 3-0 win over Cagliari. Krzysztof Piatek became the first player to score four goals in his first four games for Milan since Mario Balotelli in 2013.


text:
Alan Hansen: Liverpool are facing an uncertain future and need an overhaul
A title race may never be over until it is mathematically decided, but Chelsea can certainly head into the final game of the season knowing they have done all of the hard work.
Troubled times: Steven Gerrard looks on in frustration during Liverpool's 2-0 home defeat to Chelsea on Sunday which confirmed they could not finish fourth Photo: GETTY IMAGES
By Alan Hansen
8:00AM BST 03 May 2010
Carlo Ancelotti's side have faced the toughest test of their run-in and emerged as convincing winners.
The question, in reality, was never likely to be whether Liverpool had the right attitude to beat Chelsea, but whether they had the right attributes.
In the end, they offered as bad a performance as they have produced this season.
That should not detract from the fact that Chelsea played extremely well at Anfield, inspired by the two players who have added to their squad most this season, Florent Malouda and Salomon Kalou. Both excelled on Sunday. Malouda has had an excellent season and Kalou has had a great four or five weeks as the campaign reaches its climax.
There was a stage when neither looked like they would be a Premier League player, when you assumed both would be shipped out, judged to be of insufficient quality. Malouda, though, has blossomed this year, and Kalou's recent contribution towards Chelsea's title push has been hugely important.
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Manchester United had a similar situation at the start of their run of three league championships in as many years with Patrice Evra and Nemanja Vidic.
They struggled to impose themselves on the side when they arrived, but with time they have developed into great players. Both were awesome at Sunderland, as Sir Alex Ferguson's team secured the win which takes the title race to the final weekend.
Whatever happens next week, the differences between Chelsea and Manchester United are not vast by any means. While it has been a terrific campaign in terms of the interest and intrigue across the league, it has not been a great one for quality, as has been proved in the performance of the English teams in the Champions League.
United, Chelsea and Arsenal have all been a lot easier to beat this season than in previous years, as their defeat tallies show. It is among Europe's elite that such a weakness is truly exposed, though. If you are easy to beat in the Champions League, then you will get beaten.
That is why, regardless of whether Chelsea maintain their lead on Sunday to win the title or United manage to overhaul them, both need a considerable revamp in the summer.
Chelsea will no doubt look to strengthen their midfield, but it is their squad that has been weakened most noticeably over the years.
Under Jose Mourinho, it was always said that Chelsea's second string would finish in the top six in the Premier League, but that is simply not the case any more. They require four or five players to return them to that sort of level. Winning titles and, particularly, winning the Champions League is not about 11 players, it is about 20, and there is no question both sides need new personnel if they are to continue to be successful.
At least Ancelotti and Ferguson will only have to look to buy a handful of players this summer. Liverpool, on the other hand, may need to go into double figures before they have a squad worthy of returning them to where everyone at the club feels they deserve to be.
The worry now that their exclusion from next season's Champions League has been confirmed is whether they will be able to attract the right players to help rebuild the club.
Liverpool's supporters are the best in the world, but players do not join clubs because of the fans. They come first for money, and then for what is on offer. Without the lure of the world's biggest club competition, Liverpool simply cannot compete.
The only solace is that Liverpool are still Liverpool; they still have the history and tradition to attract players. That will only last so long, though. Should they go three or four seasons without Champions League football, even that lustre will fade.
That is why this is the most crucial period in the club's history. At the end of last season, it would have seemed impossible that they would be in the position they are now. That is the nature of football, though. It is easy to fall from grace, but extremely difficult to rise to prominence again.
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