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What Causes Income Inequality?
Progressive policies do.
Francis Menton
Photo by John Moore, Getty Images
New York City mayor Bill de Blasio has left no doubt that he wants to be the nation's preeminent spokesman for progressivism. Since unveiling his 13-point Progressive Agenda to Combat Income Inequality in May, de Blasio has campaigned on the message that income inequality is "the defining challenge of our time." His key policy prescriptions are well known: more "affordable" housing, generous welfare programs, higher minimum wages, and income-tax hikes on high earners. However, if such policies actually reduced the gap between rich and poor, jurisdictions that have already adopted them would experience less income inequality. They don't.
Last year, Bloomberg Rankings published a national study on income inequality, using U.S. Census Bureau income data to rank each of the 435 congressional districts by economists' standard measure of inequality, the Gini coefficient. The study found high levels of income inequality in areas of the country known for their political progressivism. Topping the inequality list was New York's tenth congressional district, which covers the West Side of Manhattan and Wall Street—including City Hall. Of the top 25 spots, 23 went to Democratic districts—and not just any Democratic districts. The five congressional districts covering some part of Manhattan earned the first, sixth, ninth, 13th, and 20th positions. Congressional districts in solidly liberal Chicago, Cambridge, Los Angeles, Santa Monica, and Berkeley placed in the Top 25. House minority leader Nancy Pelosi's San Francisco district ranked 14th on the list, while Democratic National Committee chairwoman Debbie Wasserman Schultz's Miami Beach district placed 23rd. All of those congressional districts have long been associated with progressive politics; most have long since adopted at least some of de Blasio's policy prescriptions, including extensive public and affordable-housing programs, generous welfare programs, relatively high and progressive state and local income taxes, and higher minimum wages.
Could it be that progressive policies intended to reduce income inequality actually cause it to increase? Quite likely, yes. The reason has to do with the effect of these policies on the low end of the income distribution. The government doesn't count the distribution of in-kind benefits as income—and means-tested handouts create incentives for recipients to keep their measured incomes low. Further, where higher minimum wages—another progressive agenda item—cause higher unemployment, we see even more zero-income earners. Having lots of low-earners or zero-earners doesn't help reduce inequality. Meanwhile, at the high end of the income distribution, taking money from high earners through higher marginal tax rates is not counted in official statistics as a cut in income—which means that income inequality again doesn't shrink.
New York City has far and away the most extensive public and "affordable" housing programs in the United States, and exceeds national averages in usage of welfare programs from Medicaid to welfare to food assistance. Gotham's highest-in-the-nation tax burden includes a top state income-tax rate of 8.82 percent on income exceeding $1.046 million, plus a top city income-tax rate of 3.876 percent, for a total of 12.696 percent on top earners. (California barely edges out New York City—which is something like a state within a state—for highest income-tax rate in the country, with a top rate of 13 percent.) New York City's minimum wage rose to $8 per hour in 2014 (thus exceeding the federal minimum of $7.25), and then to $8.75 this year, with a further scheduled increase to $9 for 2016; and de Blasio wants to raise the minimum to $15 (a state wage board voted last month to hike the minimum wage to $15 for workers in the fast food industry). Even with all this, the poverty rate in New York City is 20.3 percent, compared with 14.5 percent in the United States as a whole, according to the latest U.S. Census data.
Why does New York City have such a disproportionate poverty rate? A vast system of public and subsidized housing may be one reason. The Metropolitan Council on Housing lists some 178,000 units in New York City Housing Authority (NYCHA) low-income projects, approximately 100,000 "portable" Section 8 voucher recipients, another 90,000 "project-based" Section 8 voucher recipients, and yet another 100,000 or so units in so-called "Mitchell-Lama" buildings. That comes to almost 480,000 subsidized units—or more than 15 percent of the 3.1 million housing units in the city, far above national norms. And that's before the tens of thousands of new units now under construction as part of de Blasio's "inclusionary" affordable housing programs. The numbers also don't take into account the 1 million or so units covered by New York City's rent-stabilization program.
Housing subsidies don't count as income for government accounting purposes, nor are below-market rents taken into consideration. The exclusion of these subsidies and discounts makes a huge difference in the income-inequality calculation. The difference can be most dramatically illustrated by considering the effect on the measurement of income for residents of NYCHA projects in what are now gentrifying and expensive neighborhoods in Manhattan and Brooklyn. Many of the apartments in these projects would have market rental value in the range of $4,000 to $6,000 per month, but are rented to project residents for an average closer to $500 per month. Paying market-rate rent with after-tax dollars would cost tenants an extra $42,000 to $66,000 per year, but the gift of the same apartment to a subsidized tenant counts as zero income. So a family may be receiving a housing subsidy worth $42,000 or more, but it remains "poor" for want of spendable income.
The subsidized housing programs also entail powerful incentives for beneficiaries to keep their incomes low. NYCHA tenants must earn below certain income limits to qualify for housing and pay 30 percent of their household income toward rent, up to the maximum rent levels for the apartment size. In other words, if their income rises, so does their rent. In addition, most NYCHA and Section 8 tenants receive one or more government benefits, such as welfare or food stamps, also subject to income caps. To no one's surprise, these families structure their lives to keep their measured incomes low. NYCHA reports the poverty rate in its projects tops 51 percent.
Other means-tested, in-kind redistribution programs compound the disincentives to earn a living. Of course, progressive jurisdictions are strong supporters of these programs and have participation rates far above the national average. Medicaid is by far the largest such program—and another case where benefits don't show up in income-inequality measurements. After the recent expansions under Obamacare, Medicaid enrollment topped 71.1 million as of April 2015, or about 23 percent of the U.S. population. About 3.2 million New York City residents—or 38 percent of the population—were enrolled in Medicaid as of 2013, the latest data available. New York City goes to great lengths to maximize enrollment in all of these programs—including food stamps, WIC, energy-assistance programs, and free and reduced-priced school lunches—with little regard for the people trapped in poverty or near-poverty.
A much-higher minimum wage is a relatively recent addition to the progressive policy agenda, with Seattle, San Francisco, and Los Angeles County among the heavily Democratic areas of the country voting to phase in a $15-an-hour wage over the next few years. The progressive theory is that all or nearly all employees with wages below the mandatory level will see their pay increased, without adverse employment effects. But to the extent that much-higher minimum wages cause some people to become unemployable, zero-earners will be added to the income mix and measured income inequality will increase. This is likely to happen in New York if it embraces the $15 minimum wage. Today's $8.75 New York City minimum wage, if earned for full-time work for an entire year, yields an annual income around 35 percent of the city's median household income of slightly more than $50,000. A $15 minimum wage would push it to 60 percent of the median household income. What could possibly be wrong with that?
For the answer, look at Puerto Rico. The U.S. island territory is subject to the current federal minimum wage of $7.25 an hour, but its median household income is only about $20,000 per year. That means that the full-time minimum-wage work comes to over 70 percent of median household income. The negative effects have been massive. Puerto Rico's labor-force participation rate is only 42 percent—compared with 62 percent for the United States, which is the lowest it's been since the late 1970s. Even with that dramatically low labor-force participation, the unemployment rate on Puerto Rico exceeds 12 percent. And what effect do so many idle workers have on income inequality? Though Puerto Rico isn't exactly known for a large concentration of high-income earners, its Gini coefficient of 0.537 would put it in the top 10 congressional districts for income inequality.
The evidence indicates that de Blasio's policy prescriptions would actually make income inequality worse. The point of the mayor's progressivism may not be to improve the lot of the less fortunate or to ease poverty, but rather to broaden the base of voters permanently dependent on government programs.
Francis Menton is of counsel in the Litigation Department at a major New York-based international law firm. He blogs at www.manhattancontrarian.com.
The Food Stamp Pirouette
Francis Menton By Washington's own metrics, the program makes no dent whatsoever on poverty rates—except when it does.
Letter to a Manhattan Resident
Francis Menton A view from that big country outside New York
By Washington's own metrics, the program makes no dent whatsoever on poverty rates—except when it does. Francis Menton April 21, 2015


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Contest: Enter to Win an 11-Inch Apple MacBook Air [Deals Store]
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It uploads, and it distributes. Design thinking has allowed us to celebrate conventional solutions as breakthrough innovations and to continue with business as usual. Five links that caught my eye over the holiday. New Power: How it's Changing the 21st Century and Why you need to Know.


To train and test your agents, you will be given free access to Microsoft Azure cloud space, provided by our technological partner Microsoft Czech Republic and Slovakia.
b-toprok@microsoft.com and attach a copy of your registration confirmation e-mail.
You will need to register as a startup in our BizSpark program.
Please send me your startup name and I will generate an enrollment code for you.
BizSpark programに登録してからenrollment code を登録すりゃいいのかな?(ポチー -> REJECT!


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Global Protests & Candle Marches Call Out Pakistan-Based Terrorism On 26/11 Anniversary
Friday, November 29, 2019 by Indian Defence News
Growing protest against Pakistan-sponsored terrorism has taken center stage. In the last few days, there have been protests in London, Geneva and in the US
Growing protest against Pakistan-sponsored terrorism has taken centre stage. In the last few days, there have been protests in London, Geneva and in the US against Pakistan's visible non-action against terrorism emanating from their soil. On the anniversary of 2008 Mumbai terror attack, several Human rights activists gathered in front of the Pakistan embassy in Tokyo. Among the 166 innocent lives lost, a Japanese citizen, Hisashi Tsuda who was on a business trip during the fateful day, was also one of the victims of the dastardly terror attack.
According to the activists who gathered outside the embassy, Pakistan has deliberately not taken action against the likes of Hafiz Saeed despite India providing hard evidence against the most wanted terrorist. One of the Pakistani terrorists was captured alive by Indian security forces during this terror attack which turns out to be key evidence against the preparators. Investigations have pointed towards Hafiz Saeed - a co-founder of terror group Laskhar-e-Taiba and also a founder of Jamaat-ud-Dawah (JuD) operating mainly from Pakistan - as a mastermind behind this attack.
The US announced a bounty of $10 million on Saeed for his role in the 2008 Mumbai attacks. Saeed is listed on the NIA Most Wanted list and UN-designated terrorist. India has banned his organisations LeT and JuD as terrorist organisations. The United States, the United Kingdom, the European Union, Russia and Australia have also banned Lashkar-e-Taiba. However, Pakistan has been shy from taking any concrete action against Saeed, say the activists who gathered outside the Pakistan embassy.
Pakistan also faces strict scrutiny and serious warnings from FATF (Financial Action Task Force) for its lack of action to curb terror financing in Pakistan by pro-terror groups and individuals. As a tribute to the victims of 26/11 terror attack as well as a tribute to victims of terror attacks elsewhere, around 40-50 human rights activists demonstrated in front of the Pakistan embassy in Tokyo, demanding capital punishment for Saeed. Not just Tokyo, in London there have been continuous protests with around 100 demonstrators protesting against Pakistan. Muhajirs, Baloch, Gilgit/Baltistan, and Pashtuns gathered in London to protest against the Pakistani establishment. The protest lasted for over two hours.
Source>>
Labels: Army, Article, Border, Hizb, India, ISI, JeM, JuD, LeT, Pakistan, POK, Terrorism
Inside India's Large And Deadly Nuclear Weapons Program


Our monthly amazon haul coming in hot – like so hot that we are trying on swimwear when it is still chilly in the Midwest. Warning: these hauls are never super "organized" with our buying rational so we have a mix of everything from swimsuits to pillow cases!
I am not happy about it, but I am a leggings SNOB. I want them to stretch, but have compression. Be breathable, but not see-through. High-rise, but without the c-toe. These hit all the marks. We both are wearing size small in the leggings. This tank is everything we love- super soft, lightweight, and tieback. I would size down to a small. L is keeping the medium and it's only $15! DONE AND DONE – immediately ordered more colors in both!
I am sure that the majority of people think we have about 15 closets – trust me we don't. We are on a shopping budget just like everyone else so we LOVE when we find an item you can wear multiple ways. Today we are showing you how to style the Leith dress . If you have been hiding under a rock this is pretty much the most flattering and versatile dress. The dress comes in 8 colors too.
There are two categories your house can fall into…. with kids or without kids. That makes a huge difference on the best rug for your home. Trust me the pretty white rug from Pottery Barn that my sweet toddler decided to add purple paint too almost brought me to tears. Now I find rugs that work for our home and our price point in case we have another paint incident I tossed all the freaking non-washable paint too!
A key feature in every working gals arsenal is a finding the best work bag to fit her style. Over the years and as my job has evolved my needs have changed – so the bag changes too.
A few years ago I read that you should stay away from bags that have a well known label and are known to be expensive. It can send the wrong message walking into a meeting with a $1,500 bag. I am all for rewarding yourself with that bag and rocking it for hh and the weekends!
Now that Spring is here we've rounded up some of our favorite must-have accessories that should be in every gals closet. Lets be honest- who isn't excited to pack up their lived-in puffer coat and chunky heeled boots in exchange for cocktails on a patio (margs anyone?) in the perfect sundress and sandals.
It might seem like we are clothes fanatics and maybe we are, what we want pieces that you can transition from work, dates, soccer games etc. I hate buying a piece that I can only wear once.
Loft is killing it right now with all their versatile pieces…..
Our beloved Dr.Scholls shoes are on sale- all over 30% off! So comfy and cute.
Living Proof – This is my #1 pick from the sale!
Huge LOFT sale 40% + 10% extra for card holders… see everything we got from the sale here !
Christmas time is becoming more and more magical as my boys become a little older. I'd say this year was one of the best! Their ages are 3 and 4.5. Our top movies were Home Alone and the Grinch! Every car ride consisted of listening to the Home Alone soundtrack on repeat! Ernie the elf was a hit once again and no one cried to see Santa! It was all so perfect.


Motel 6 St. Robert MO is conveniently located off of Hwy Z. Visit Devil's Elbow River Lay Z Day Canoes & Camping and the St Robert Aqua Park all within 5 miles! Enjoy Free Wi-fi and Micro-Fridges in all guest rooms.
Motel 6 St. Robert, MO can be found at 14175 Hwy Z . The following is offered: Hotels & Motels . In St. Robert there are 24 other Hotels & Motels. An overview can be found here.


User Experience - Scalable Privacy - Internet2 Wiki for your Wiki!
1.	Pat has low vision, and needs text to be displayed in a large font, and prefers to have text read aloud as well being displayed visually. Pat is applying for a job, online, but does not want the potential employer to be aware of these accommodations. Pat is concerned that the employer might look for a reason to reject the application if they knew that it had been submitted by "a person with a disability".
2.	Pat is a child who wants to participate in an age-restricted chat room. Pat must establish that their age is between 6-12, but would like anonymity beyond that restriction, providing an avatar for that purpose. Pat's age must be established, and a consenting adult for Pat must attest willingness.
3.	Pat is a child who wants to subscribe to a newsletter from a toy company. Pat needs to supply an email address and a consenting adult for Pat must attest willingness.
4.	Newtown wants to host on-line "in my backyard" conversations on wikis, where on-line discussants are limited to those who are residents of a particular neighborhood, but are provided anonymity in those discussions.
5.	A scientific web site wants to ease the boarding process for potential users by using a standard "Research and Scholarship" attribute package. That package would include name, email address, user identifier, targeted ID, and user affiliation.
6.	A company wants to set up an internal "suggestions box", where authenticated employees of the company can then submit anonymous suggestions.


SolarGrid Tanzania is bringing off-grid solar appliances to the people of Tanzanian.
It is estimated that approximately 80% of Tanzania's population have no access to grid-based electricity at all. However, only 5% of this vast market are currently addressed by alternative solutions like off-grid solar power. The government and private companies have been neglecting these people for too long. In order to fill this gap, SolarGrid Tanzania was founded. We are able to successfully serve the energy needs of the poorest sector of society. Until now we have brought solar energy systems to more than 5000 families and the scaling-up process has just begun.
The high expectations of our customers with regards to price and quality as well as the rough weather conditions in rural Tanzania demand a thorough selection of products. It took us several months of intense testing till we were able to select suppliers that met our envisaged standards. Since then we continuously expanded our product range in order to fulfill the divers needs of our customers.
Currently, we are very happy to offer four different product configurations which stretch from high-quality lighting with four or six bulbs over a fully functional 24-inch flat TV to a business system that entails a 10 slot USB-hub and allows our customers to earn a living by charging mobile devices. With respect to the financial means of our customers we offer for all products a micro loan program. Payments are conducted through the state-of-the-art mobile payment mechanism "Pay-as-you-go".
A single kerosene lantern burning an average of four hours per day emits over 100kg CO2 per year. Thus, the usage of solar energy strongly contributes to climate protection.
The indoor use of kerosene imposes severe health risks on the poor such as lung problems, cancer and burn injuries. The World Health Organization estimates that a staggering 1.6 million people die every year from fatal fires caused by kerosene. Solar Energy completely prevents such incidents.
The dim kerosene light makes reading, studying and working at night extremely burdensome. Through the bright illumination solar products can therefore contribute to the educational and commercial success of the "off-grid families".
Daniel Uphaus studied in Tanzania for one year as an exchange student. After winning a business plan contest in 2010, he started implementing his vision to empower Africa with renewable energies. If he is not busy developing business strategies for SolarGrid he is a passionate magician, gardener and tea-drinker.
Edwin Kiemba conceptualized and supervised numerous programs for international non-profit organisation like the Dutch aid agency SNV before coming to SolarGrid. He is a priceless interpreter between the different cultures, but even he is sometimes puzzled by the German taste in fashion and music.


With Halloween coming up many people consider this to be a cursed time of year. Teenagers preparing to toilet paper and egg houses, witches holding seances, and curious people using ouija boards, it is no wonder that October is viewed as a cursed time. While it is not actually a cursed time of year it is indeed a very interesting one.
If we look beyond the commercialization of the holiday we can find a few people interested in celebrating Hallow's Eve. Some celebrate Halloween as a day of remembering the dead. Others use it for a chance to possibly contact the spirit world. What is Halloween, besides a day to give out candy?
Halloween originates as a derivative of the Celtic holiday called Samhain. Samhain, to the celts was known as Oíche Shamhna or night of samhain. The night was a celebration of the fall. People would gather around large bonfires to celebrate the end of spring and the coming of the new season. Houses were also adorned with carved pumpkins or turnips, this tradition references the legend of Jack of the Lantern.
Once in the land of Eire, there lived a blacksmith by the name of Jack. Quick to the drink but short on paying for it, Jack had few friends and was known to curse a lovely day as often as a miserable one.
One night late, when all civil folk were home abed, Jack was at a pub near the crossroads outside his village. A stranger in a long coat and hat sat near. Hoping for sympathy, Jack moaned aloud that he would give his soul to the Devil if he had the money for one last drink. Quick as can be, the stranger beside him disappeared and close to Jack's hand lay the coins for his drink.
Now although Jack was a drinking man, he was also a clever man and he knew now that stranger was no mere man. Being the thinking man, Jack snatched the coins from the counter and pocketed them in his purse where he kept a small silver cross. Trapped by the silver cross, the stranger, who was actually the Devil in disguise, had to agree not to claim Jack's soul for another ten years in order to escape Jack's pocket.
Now time went by and as time does and Jack's ways never changed. Without family or friend, Jack contented himself walking about late after the pub had closed and one night, ten years hence, he came across the Devil himself on a dark country road.
The Devil reminded Jack of their bargain and Jack, still not eager to leave life, agreed to go if the Devil would feed his terrible hunger with an apple from a nearby tree. Hopping upon Jack's shoulder's, the Devil hoisted himself into the tree to get the apple. Quickly Jack took his small knife and carved a cross in the tree. Now as we all know, the Devil has no power in the air and was trapped in the tree by the cross on its trunk. This time Jack made the Devil promise never to come for him again. Having no choice, the Devil agreed.
Some years later, Jack did meet his natural end. Because of his miserly and drinking ways, he was denied entrance to Heaven. The Devil himself turned him away at the entrance to Hell because he had promised years before never to take his soul. As the night was dark and the wind was howling, Jack begged the Devil for light to find his way. The Devil threw him an ember from the very fires of hell and Jack kept it lit inside a bit of rotten turnip that he found.
This tale was retold by C. Austin. The idea of the lantern on Halloween being that if a demon or evil spirit saw the visage of another Demon, it would be so afraid it would leave the home alone. It is this same reasoning that has children dressing up as ghouls, monsters, ghosts, and demons. Disguised as the evil spirits, the children go door to door, nuts and fruits for the festivities.
Halloween however has become an amalgam of many celebrated holidays. Samhain, being the primary but Halloween itself being short for Hallow's eve. The day before all saints day. Unlike other holidays such as Christmas which hold some foundation in Pagan tradition, Halloween has remained primarily pagan. It is a fantastic time of year where people can believe in the paranormal. Where things that go bump in the night have the possibility to be more than the cat knocking something over.
I myself have fond memories of sneaking into abandoned houses and forts at night in search of spirits or demons. With hopes that these tales of ghosts and goblins I heard were true. My heart racing, my mind over active as I walked down a dark damp tunnel. SNAP a branch breaks. What was that? I run outside before I can ever find out, the monsters have found me. I get home as quick as I can, thinking maybe just maybe the stories are true.
Where ever one finds them self on Halloween, it still seems to be the one time of year that continues to hold a mystique. While the imagery, and legend of other Holidays fades. Halloween, holds strong and true. Perhaps, it is because of the power of the idea of spirits, ghosts, and demons that helps us to hold on. Or maybe it is just myself that seems to recall with great fondness the excitement and mystery of the season.


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HomeNewsThey're Still "Looking" For MH370
They're Still "Looking" For MH370
August 5, 2014 Royce Christyn News 60
MH370 Missing Plane - yournewswire.com
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According to multiple news sources, officials are next week going to announce the next phase of the hunt for missing plane MH370. The plane, which went missing over five months ago, leaving hundreds of family members in grieving limbo, has yet to be found… so they say. First we will give you some information on this latest distraction update:
According to CNN:
"The Australian Transport Safety Bureau (ATSB), which is leading the search for the missing plane at the Malaysian government's request, is expected to engage a single, private contractor to search for the Boeing 777-200ER, and if successfully located, to positively identify and map the wreckage.
The operation, which the ATSB says is likely to begin in early September and will last up to a year, will include a number of underwater vehicles that will slowly scan some 60,000 square kilometers (23,000 square miles) of the southern Indian Ocean, looking for any sign of aircraft debris.
It's not yet clear exactly how the deep-sea search, which Australia has estimated will cost $56 million, will play out. While the ATSB has developed the overall search strategy, the independent contractor will be responsible for day-to-day operations in the search zone."
The Christian Post:
"Despite some reports online claiming that the missing plane may have been found, it has been confirmed that the mystery surrounding Malaysia Airlines Flight MH370 remains, and that the search for the missing plane and the passengers and crew is ongoing.
Deputy Prime Minister Warren Truss told the Australian Associated Press this week that the search for the missing Malaysia Airlines Flight MH370 has not been altered or hindered in any way by the recent crash incident of Malaysia Airlines Flight MH17 in Ukraine."
And finally, Upstream Online:
"Malaysia's Petronas is chipping in more than $20 million and deploying assets in the hunt for Malaysian Airlines flight MH370 that disappeared along with 239 people in March.
The state-owned oil giant is to deploy underwater search equipment and an offshore support vessel in what is says is a "new search area" for the plane that vanished while en route from Kuala Lumpur to Beijing.
Petronas said on Monday it is funding the 67 million Malaysian ringgits ($20.92 million) deployment of a Prosas towed side scan sonar unit and the anchor-handling tug supply vessel Go Phoenix. The Marshall Islands-flagged vessel is managed by Go Offshore Asia of Singapore.
The Fugro-owned survey vessel Fugro Equator and a Chinese survey vessel are said to be currently carrying out mapping of search areas, with a Malaysian unit to join the search this month."
BUT, HERE'S THE THING –
What if it's not in the water? There is a great read over here at this LINK, which we think our readers should take a look at. Here is a clip about an alleged passenger, Phillip Wood, who sent a picture from Diego Garcia Military Base – THE WHOLE ARTICLE, though long, is worth the read:
"I have been held hostage by unknown military personal after my flight was hijacked (blindfolded). I work for IBM and I have managed to hide my cellphone in my ass during the hijack. I have been separated from the rest of the passengers and I am in a cell. My name is Philip Wood. I think I have been drugged as well and cannot think clearly."
Let's dissect that. "I have been held hostage by unknown military personal after my flight was hijacked (blindfolded)." If this was a hoax given what was known by all the zit picking kids out there, the message would have instead said "I have been held hostage by American forces after my plane was hijacked. I am in Diego Garcia. But he does not know who his captors are, and that is a total green light for the story being true, because they sure as heck would never tell him. And he obviously did not know where he was.
He was blindfolded. The Army will blindfold people with a locked bag over the head that the people they put it on can't get off, to leave their hands free when they have been judged to not be a threat. Once in the cell, they took the handcuffs off because he was not a threat, he was the victim and they left the bag to avoid being identified. Philip, who may have been beta testing the latest IBM voice recognition software or using the voice recognition abilities of the Iphone, swiped the screen correctly to turn it on and then talked to it to get it logged in somewhere to get his message out because he could not see the screen. THE KICKER IS THAT IT IS OBVIOUSLY A VOICE RECOGNITION MESSAGE. IT'S CHOPPY AND THE WORD PERSONNEL WAS INTERPRETED AS THE WORD PERSONAL. THE TWO ARE TOO DIFFERENT FOR THAT TO BE A TYPO, VOICE ASSISTED SOFTWARE OBVIOUSLY DID THAT ERROR.
So, the first line of what he said FITS PERFECT.
"I work for IBM and managed to hide my cellphone in my ass during the hijack". Shills ripped this, saying he would not have bothered with where he worked. That is a load of BUNK. If he was flying to Bejing, he was most likely on a business flight for work, and wanted to make clear who he was so there would be no mistake. And where else would he put the cell phone other than his ass? in his ear? his pee hole? Come on now, that shillage was thin.
So the second line could not have been any different in a real scenario
"I have been separated from the rest of the passengers and am in a cell" The fact that he waited to be separated from the rest of the passengers to get the message out is absolutely the way it would have been, before separation they would have been closely watching all the passengers and he did not risk losing his phone while under close scrutiny. Once in the cell, the cell prevents escape and would not have immediate supervision. THAT would be the time to dial out."
READ THE WHOLE THING HERE
Royce Christyn
Journalist at News Punch
Documentarian, Writer, Producer, Director, Author.
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text:
Wake Up Charlotte
Queen's Feast wrapping up January run soon
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The Beagle Brigade: Formerly abused dog now shines with Homeland Security
There's no question that Murray is a good boy. He's working the international baggage claim area at Atlanta's airport, one of the busiest airports in the world.
Author: Michelle Boudin
Published: 9:09 PM EST November 6, 2019
Updated: 10:50 PM EST November 6, 2019
ATLANTA — Meet the Beagle Brigade. These dogs are on the frontlines of Homeland Security across the country and even at Charlotte's airport.
So how are they keeping you safe?
The dogs may be adorable, but they play an important role in keeping you and your family safe.
"We're what we call America's frontline, the dogs actually spearhead that," K9 Supervisor Ken Hodgkins with Customs and Border Protection said. "We try to keep any fruits or vegetables, meats out that have diseases."
Murray is a member of the Beagle Brigade, an elite team of detector dogs who sniff out plants, meats, and animal products. There's a whole lot of Murrays across the United States and in Charlotte that make up the Beagle Brigade.
The dogs work to keep us all safe from anything getting in the country that could destroy the food supply.
"It's very effective, these dogs can find up to 25 seizures a day from different passengers," Hodgkins said.
"You can tell they're very, very excited to do what they're doing," a traveler named Hope said.
Hope didn't even mind when Murray sold her out – he sniffed out the apple she'd packed as a snack, and of course, he got a treat for doing it.
It's a sweet life for a dog who almost didn't make it.
"He was very playful when I met him but I noticed that half of his left ear is missing and seems like tail a little bit shorter," his handler Amabele Gella said.
Gella is an Agriculture Specialist and K-9 Enforcement Officer with Homeland Security. She said Murray was in bad shape.
Someone had cut part of Murray's ear and tail and left him on the street. He was rescued from a kill shelter in Georgia and sent for special training to work with customs and border patrol.
"When he first came to us we had some reservations – is he gonna be able to make it back from that traumatic prior life that he had? He's one of our best dogs now," Hodgkins said.
In fact, almost all the dogs on the Beagle Brigade are rescues.
"He is living his best life," Gella said.
The dogs are retired when they turn 9, and their handlers have the option of taking them home as pets.
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January birthstones are made to display your qualities based on the month you were born, they also are the same color as your birth gem. Stones measure approximately 1 1/2" - some are slightly larger and some may be slightly smaller.
QUANTITY DISCOUNTS: Buy 100+ of any of the imprinted colored glass stones (found in Shop > Imprinted Glass Stone > Imprinted - Birthstones) and the price drops to $1.90/each (You may mix and match words).
Please note there is a handling time of up to 8 business days before shipment for these items. For large quantity orders please call for specific ship times.


In response to Trump's hostility toward immigrants, political leaders in New York and California vow to protect their most vulnerable.
The wake of the Presidential election has stirred an array of reactions—grief, denial, outrage, bewilderment—among the liberals and the progressives who supported Hillary Clinton. Her loss, scarcely predicted in the polls, and therefore all the more shocking to the people who voted for her (and who outnumber her opponent's supporters by more than a million and counting), has become a case study in political trauma. Donald Trump's victory inspired another, less noted reaction that may prove more politically significant than the current wave of demoralization: defiance.
Whether you are gay or straight, Muslim or Christian, rich or poor, black or white or brown, we respect all people in the state of New York.
It's the very core of what we believe and who we are. But it's not just what we say, we passed laws that reflect it, and we will continue to do so, no matter what happens nationally. We won't allow a federal government that attacks immigrants to do so in our state.
These announcements cannot be dismissed as simply the whiny objections of the coastal élites. Thirty-nine million people live in California—twelve per cent of the population of the United States. The state is home to the economic and cultural axes of Silicon Valley and Hollywood. Last year, its economy became the sixth largest in the world, a spot formerly held by France. Clinton beat Trump by twenty-eight points in California, and by twenty-one points in New York. Now the two states have triggered an uncommon development in a year that has offered us a great number of them: liberals invoking states' rights.
We have become accustomed to governors denouncing overreach by the federal government. Last year, more than thirty governors released statements opposing President Barack Obama's plan to resettle Syrian refugees in their states, even though federal authority clearly allowed him to do so. In 2012, Republican opponents of the Affordable Care Act based their challenge, at the Supreme Court, on the ground that it violated states' rights. Later that year, shortly after Obama was reëlected, more than a hundred thousand Texans signed a petition requesting that the White House respond to their demand for secession. (This presented a paradox in that the petitioners were recognizing federal authority in an attempt to claim that the federal government did not have authority over them.) It would seem ironic that two of the largest and most consistently Democratic states have now presented states'-rights arguments in their cause, were it not for history.
States' rights have a complicated genealogy, which ties the birth of the Confederacy to Southern resistance to civil rights. In 1980, when Ronald Reagan invoked the phrase in Philadelphia, Mississippi—where three civil-rights workers had been murdered, in 1964—he was offering a kind of absolution, if not for racial violence then for the principle that had been used to justify it. Today, the phrase is often shorthand for opposition to "big government," a term that carries racial implications of its own. Yet the argument for states' rights began in the early years of the republic, in an effort to confront exactly the type of threat to non-citizens that the incoming Administration poses.
In 1798, the passage of the Alien and Sedition Acts increased the residency requirement from five years to fourteen before immigrants could vote, and authorized the executive branch to summarily deport immigrants who were deemed dangerous or who had come from hostile nations. In response, James Madison and Thomas Jefferson, whose Democratic-Republican Party was favored by immigrants, wrote the Virginia and Kentucky resolutions, which held that individual states had the right to nullify unconstitutional laws within their borders. They further stipulated that states had the right to "interpose" themselves against the authority of the federal government. George Washington was among those who saw the resolutions as a cure that was worse than the illness of federal overreach. But the notion of states' rights survived.
Protesters plan to descend on Washington, D.C., for Donald Trump's Inauguration, and may number in the hundreds of thousands.


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A szlovén férfi kézilabda-válogatott Szlovénia nemzeti csapata, melyet a Szlovén Kézilabda-szövetség (Slovénül:Rokometna zveza Slovenije) irányít.

Eredmények nemzetközi tornákon

A szlovén-válogatott legnagyobb sikere egy ezüstérem, amit a hazai környezetben megrendezett 2004-es Európa-bajnokságon szerzett.

Érmek

Európa-bajnokság
  : 2004
Világbajnokság
  : 2017

Szövetségi kapitányok
Ljubomir Vranjes (2019–2022)
Uroš Zorman (2022–)

Jegyzetek

Külső hivatkozások
 A szlovén Kézilabda-szövetség honlapja

Szlovénia sportélete
Férfi kézilabda-válogatottak

redpajama_set_name: RedPajamaWikipedia


text:
Home»Gossip»PRODUCTION FRAME Did Big Ed Brown say goodbye to everything in the drama?
PRODUCTION FRAME Did Big Ed Brown say goodbye to everything in the drama?
About Part 2 of 90 Day Fiancé: Happily Ever After? On the Season 7 Tell All Special, Big Ed Brown demanded Liz Woods return the ring.
reason? Well, he felt humiliated because everyone caught him in a silly lie.
That might sound like typical Big Ed toxicity, but some fans aren't sure.
On Tell All, some people focused on the all-important details. Did the production set Ed to fail?
As we recall, early in part two of the four-part Tell All, Rosemary Vega was debunking Big Ed's lies.
Sean asked the 90 Day Fiancé villain if he was in touch with his ex Rose. It was a simple question and should have been easy to answer.
Instead, Ed fumbled and lied. He started with a silly "I don't know" and settled for a disingenuous "no."
Rose pointed out that this is not true. Not only did Ed keep in touch with her, he asked her to video chat on multiple occasions.
In fact, he even went so far as to ask her if she would fly to the Philippines to visit her.
The production had already taken transcripts of those conversations and put Ed's request on screen for Liz to read.
The main problem with the message was that Big Ed apparently lied to Liz.
But this is also why some 90 Day Fiancé fans think the drama is fake, or that the production set up Ed.
why? Late August 2021 due to the date of the message. we actually covered this Unfortunately some people don't follow me Hollywood gossipAnalysis of 90 day fianceeHard to believe, but it's true.
So what are fans worried about? timing. Between social media cues and Elizabeth's pregnancy timeline for her potstone, Tell All was taken in September 2022 of her.
Ed's text with Rose dates back to late August 2021. That was over a year before her on Tell All.
So are these messages just juicy empty burgers set up by the production to provoke fights?
Obviously, stirring the pot is the production's literal job (sometimes).
Wouldn't it be absolutely wild if a producer's idea of showing inflammatory text out of context led to on-screen divisiveness?
Some fans wonder, is that really what happened to Big Ed and Liz?
But it's not.
First, Big Ed is a grown man in his fifties. He's not in fifth grade. Exactly one year ago isn't a "long time" when it comes to getting in touch with his ex.
(It is also possible that some of his contacts with Rose were more recent than late August 2021.)
But second… this is not what it was before I met Liz. In fact, Ed's request to visit Rose in the Philippines week After he proposes to Liz.
And for those of us who paid attention in Part 2 (to be fair, there was a lot of drama in that two-hour episode), Liz also explained the timeline.
She and Ed basically have to think hard about whether or not they were actually together when he sent that message to Rose. Because the Part 2 split was their 11th breakup.
Getting in touch with your ex-boyfriend may or may not be normal. It depends on many factors.
Even if you're "on a break," it's a big deal to ask a new person to pop out and visit your ex about a month after proposing.
Ed didn't have to use his well-deserved humiliation as an excuse to attack Liz.
Also, if your ex asks if you're still in touch with him, you should honestly say yes, even if it's been a year.
Especially if the question was raised on Tell All.
Don't let Sean ask if the production knows the answer. It's like a lawyer asking a question in court, Ed.
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Welcome! Since 2004, Dr. John Sigle, DPM, FACFAS has been treating patients to provide the highest level of podiatric care in Springfield and central Illinois. Dr. Sigle's advanced training and experience in foot and ankle reconstructive surgery is coupled with his passion and dedication to his patients. The Foot & Ankle Center Of Illinois staff is also committed to provide excellent service and to provide you with prompt attention and comfort.
Dr. Sigle specializes in all areas of foot and ankle care including trauma, ankle fractures, Achilles ruptures, ankle and foot pain, heel pain, arthritis, ingrown toenails, hammer toes, toenail fungus, diabetic ulcers, bunions, fusions, revisions, and total ankle replacement. Dr. Sigle holds the most advanced certifications from the American Board of Podiatric Surgery and he has trained under leading experts all over the world.
We strive to develop a collaborative relationship with every patient. A better informed patient is able to understand the problem, diagnosis, treatment options, and work with me us to formulate a medical plan to heal, restore, and rejuvenate their life. Our website has an extensive patient library on foot care, including self-help tips, instructional videos, television and radio interviews, and articles.
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Nominations Sought for the Lindberg Award for Outstanding Scholar-Teacher in the College of Liberal Arts
The Lindberg Award goes to "a truly outstanding scholar and teacher in the Liberal Arts." Nominees should be tenured faculty members of the COLA faculty who have substantial experience at UNH and a distinguished record of teaching and scholarship. This award carries a $5,000 stipend, and the winner is expected to present the Lindberg Lecture on a topic of their choice in the subsequent academic year.
Nominations may be submitted by any faculty member and you may nominate more than one faculty member. Nominations from the prior two years (2020-21 and 2021-22) can be updated and carried over to this year's pool.
One letter of nomination (must address scholarship and teaching)
Nominee's current/up-to-date CV
Please send your nomination letter with current CV as one PDF to Kayla Laurie by March 24, 2023 (kayla.laurie@unh.edu).


As the AAFP celebrates the 10th anniversary of its Vaccine Science Fellowship program, the Academy is seeking applicants to be 2019-2020 fellows.
Past and current AAFP Vaccine Science Fellows gather during the 2018 Family Medicine Experience meeting in New Orleans: Front row (from left): Margot Savoy, M.D., M.P.H.; Amra Resic, M.D.; and Mina Khan, M.D. Back row (from left): David Cope, M.D.; Laura Morris, M.D., M.S.P.H.; John Epling, M.D., M.S.Ed.; Pamela Rockwell, D.O.; and Paul Hunter, M.D.
Two fellowship positions are available for family physicians interested in working with mentors to become more familiar with the complexities of vaccine science and policy. The deadline to apply is March 22.
The selected applicants will have the opportunity to gain education and practical hands-on experience through participating in meetings with leading experts in public health and immunization, federal and state vaccine policy groups, and vaccine manufacturers.
As the deadline to apply for the 2019-2020 fellowship program approaches, 2018 Vaccine Science Fellows Mina Khan, M.D., of Trussville, Ala., and Laura Morris, M.D., M.S.P.H., of Columbia, Mo., described their experience with the program to AAFP News.
For the past few years, Morris has served on the Missouri Advisory Council on Childhood Immunizations, a group of clinicians and public health experts who review and advise the governor and state legislature on immunization-related policy.
Family physicians interested in learning more about the science behind vaccine development and production and vaccine policy should consider applying for the 2019-2020 AAFP Vaccine Science Fellowship.
The deadline to apply is March 22.
Last year's fellows recently sat down with AAFP News to speak about their experience in the program.
"I also have an academic interest in evidence-based medicine, and the Vaccine Science Fellowship offered an opportunity to further develop that interest in a specific clinical application," she said. "Past fellows have used their expertise to advise state and national agencies that develop vaccine policy, and I hope to do the same."
Khan said one of her best memories as a fellow was getting to attend the National Foundation for Infectious Diseases' 2018 Clinical Vaccinology course in Bethesda, Md., where she was able to learn firsthand from some of the biggest names in the field of vaccine science.
Being mid-career and well-established in her clinical practice, Khan said she was interested in adding some skills to her résumé that could help her in creating health policy, a special interest for her.
"I felt that the Vaccine Science Fellowship would help me develop more knowledge in this field," she said. "In addition, with the recent resurgence in antivaccine sentiment among some fringe groups, which has now become more widespread, I wanted to be armed with the skills to address this issue and maintain a good level of immunization coverage in the community that I serve."
2018-2019 Vaccine Science Fellow Laura Morris, M.D., M.S.P.H.
In fact, said Morris, she learned (and remembered) more vaccine science in that two-day course than she has since medical school.
For her part, Khan said attending the National Vaccine Advisory Committee meeting in Washington, D.C., helped her "learn how some of the policy decisions are made in the field of vaccinology, and how they are then disseminated through different agencies at the national level, and eventually influence policy at state and local levels."
Additionally, Morris said that attending the CDC's Advisory Committee on Immunization Practices policy meetings at CDC headquarters in Atlanta "inspired great confidence to see the thoughtful and thorough work that goes into creating the national vaccine recommendations and vaccine schedule."
Along with basic sanitation practices such as handwashing and the advent of clean drinking water, Morris said vaccines represent one of the most effective ways to protect public health; they effectively save lives in a way that the most expensive, most exotic new medications on the market will never be able to touch.
"It seems like the recent U.S. cases of measles are far away from my practice in Missouri, but in reality, all communities are vulnerable -- some much more so when vaccination rates fall below the threshold needed for herd immunity," she said. "I have a responsibility to keep all families in my community as healthy as possible, and that means investing the time to counsel patients and parents about vaccinations."
2018-2019 Vaccine Science Fellow Mina Khan, M.D.
Khan said that with the current environment of vaccine hesitancy and resurgence of communicable diseases that once were only read about in medical textbooks, it's important to be well equipped to address these issues.
"I believe that the Vaccine Science Fellowship provides these skills and can be particularly useful for physicians who are involved in teaching medical students, residents and even support staff, such as registered nurses, licensed practical nurses and medical assistants," she said.
Another group of physicians that can specifically benefit from the fellowship, Khan said, are those directly involved in influencing policy, such as those who may serve on a state or local advisory board.
"Since joining the fellowship, I have personally taught lectures about vaccine confidence to medical assistant students who are often the first point of contact in a medical practice for patients and caregivers who may have questions or reservations about vaccines," Khan said.
Morris said family physicians who are interested in learning more about the science behind vaccine development and production should apply for the fellowship because it offers an educational opportunity unlike anything she's encountered since residency.
"I couldn't have arranged these experiences on my own even if I would have tried!" she said. "I've seen the insider's view of how vaccines are created, tested and deployed, as well as the policy development process."
Khan credited the fellowship with giving her a great opportunity to network and make contacts that she said will be highly valuable for the rest of her medical career.
"The mentors, many who are graduates of the Vaccine Science Fellowship themselves, have been extremely helpful in shaping the fellowship for me and providing valuable personal insight and knowledge when needed."
Fellowship applications should be emailed to Pamela Carter-Smith, M.P.A., along with a curriculum vitae, letter of interest, completed AAFP conflict of interest form, a photo and a letter from the physician's institution or department agreeing that 10 percent of the applicant's time will be devoted to the fellowship.


Welcome To TOYO Asian Cuisine!
Sticed chicken with mushroom, snow peas, waterchestnuts, carrot in white sauce.
Stir fried chicken with peanut, pepper, baby corn, waterchestnuts, snow pea, scallion in spicy kung pao sauce.
Stir fried chicken with bell pepper, bamboo shoot, celery, carrot in spicy chili sauce.
Crispy chicken with orange peel in sweet & spicy sacue.
Crispy chicken in sweet and spicy sauce on the bed of steam broccoli.
My order was ready precisely when promised! The front hostess/cashier was very kind, soft spoken and hospitable! I notice a peaceful environment whenever I visit this restaurant. The gyoza that we ordered was spot on perfect. In fact, the gyoza are my 13 year old daughter's favorite! They are becoming a favorite of mine, too!
TOYO Asian Cuisine (the "Company") provides this website https://ez2eat.com/toyoasiancuisine(the "website" or "site") for its customers and other permitted users. We've worked hard to give you the essential information here and elsewhere on our site to get you comfortable with our site and to make it useful to you. These Terms and Conditions of Use (the "Terms and Conditions") help enable us to provide you with the products and services, information, other content and features available through our site.
9. Terms and Conditions: Please also visit our Terms and Conditions section, which sets forth permitted application and website use, disclaimers, and limitations of liability governing the use of this website.


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text:
\section{Introduction}

A remarkable property of von Neumann entropy is the strong subadditivity (SSA): For a state $\rho$
on the 3-fold tensor product $B(\mathcal H_A\otimes\mathcal H_B\otimes \mathcal H_C)$, we have
\[
S(\rho)+S(\rho_B)\le S(\rho_{AB})+S(\rho_{BC})
\]
Here $\mathcal H_A$, $\mathcal H_B$ and $\mathcal H_C$ are finite dimensional Hilbert spaces and
$\rho_B$, $\rho_{AB}$, $\rho_{BC}$ are the restrictions of $\rho$ to the respective subsystems.
This was first proved by Lieb and Ruskai in \cite{liebruskai}.

The structure of states that saturate the strong subadditivity of entropy, called strongly
additive states,  was studied in \cite{hjpw}. In was shown that a state $\rho$ is strongly additive
if and only if it has the form
\begin{equation}\label{eq:ssaeq_hrpw}
\rho=\bigoplus_n A_n\otimes B_n,
\end{equation}
where $A_n\in B(\mathcal H_A\otimes \mathcal H_n)$ and $B_n\in B(\mathcal K_n\otimes \mathcal H_C)$ are positive operators
and $\mathcal H_B$ has a decomposition $\mathcal H_B=\bigoplus_n \mathcal H_n\otimes\mathcal K_n$ (see also \cite{japetz}, where
this was proved also for the infinite  dimensional case). Equivalently,
\begin{equation} \label{eq:ssaeq_ja}
\rho= (D_{AB}\otimes I_C)(I_A\otimes D_{BC})
\end{equation}
where $D_{AB}\in B(\mathcal H_A\otimes \mathcal H_B)$ and $D_{BC}\in B(\mathcal H_B\otimes \mathcal H_C)$ are positive
matrices.

The Markov property for states in the quantum (non-commutative) probability was introduced by Accardi \cite{accardi} and Accardi and Frigerio \cite{acfrig},
in terms of completely positive unital maps, so-called quasiconditional expectations. For tensor products, it was shown
that the Markov property is equivalent to strong additivity of the states \cite{ohyapetz}.

The definition of the Markov property does not require the tensor product structure and can be applied in much more general
situations. We are interested in the case of CAR algebras. The Markov states for
CAR algebras were studied in \cite{afimu}. The strong subadditivity of entropy on
CAR systems was recently shown and it was proved that strong additivity is
equivalent to Markov property in the case of even states, see \cite{moriya}.
For noneven states, a necessary and sufficient condition for equality in (SSA) was given in \cite{belpit}.

The aim of the present paper is to find the structure of strongly additive states  and Markov triplets on the CAR algebra. We find an analogue of
(\ref{eq:ssaeq_ja}) for any states and of (\ref{eq:ssaeq_hrpw}) for even states. This is done by a similar method as in \cite{japetz}, using the results of the theory of
sufficient subalgebras.

The paper is organized as follows. The preliminary section summarizes the most important results on the CAR algebra and on sufficient subalgebras. The main tool used in the sequel is the factorization Theorem \ref{thm:factorization} in Section 2.1. Section 3 shows the relation between strong additivity and Markov property for any states on the CAR algebra. Section 4 contains the main results.

\section{Preliminaries}

\subsection{Sufficient subalgebras}

We first recall the definition and some characterizations of a sufficient
subalgebra, which is a generalization of the classical notion of a sufficient statistic, see \cite{petz,ohyapetz} for details.

Let $\mathcal A$ be a finite dimensional algebra and let $\varphi,\psi$ be states on $\mathcal A$. Let $\mathcal B\subset \mathcal A$ be a subalgebra and let $\varphi_0$, $\psi_0$ be the
restrictions of the states to $\mathcal B$. Then $\mathcal B$ is sufficient for $\{\varphi,\psi\}$ is there is a completely positive, identity preserving map $E:\mathcal A\to \mathcal B$, such that $\varphi_0\circ E=\varphi$, $\psi_0\circ E=\psi$.

For simplicity, let us further assume that the states are faithful. Let $\rho_\varphi$, $\rho_\psi$ be the densities of $\varphi$, $\psi$ with respect to a trace ${\rm Tr}$:
\[
\varphi(a)={\rm Tr} \rho_\varphi a, \quad \psi(a)={\rm Tr} \rho_\psi a,\qquad a\in \mathcal A
\]
The relative entropy $S(\varphi,\psi)$ is defined as
\[
S(\varphi,\psi)=S(\rho_\varphi,\rho_\psi)={\rm Tr} \rho_\varphi(\log\rho_\varphi-\log \rho_\psi)
\]
It is monotone, in the sense that we have $S(\varphi,\psi)\ge S(\varphi_0,\psi_0)$ for any subalgebra $\mathcal B\subseteq \mathcal A$. We will also need the definition of the
generalized conditional expectation $E_\psi: \mathcal A\to \mathcal B$ with respect to the state $\psi$ \cite{acccec}
\[
E_\psi(a)=E_{\rho_\psi}(a)=\rho_{\psi_0}^{-1/2}E_\mathcal B(\rho_\psi^{1/2}a\rho_\psi^{1/2})\rho_{\psi_0}^{-1/2}
\]
where $E_\mathcal B:\mathcal A \to \mathcal B$ is the trace preserving conditional expectation. Then $E_\psi$ is a completely positive identity preserving map, such that
$\psi_0\circ E_\psi=\psi$ and it is a conditional expectation if and only if
$\rho^{it}_\psi\mathcal B\rho^{-it}_\psi\subseteq \mathcal B$ for all $t\in \mathbb R$.

The following theorem gives several equivalent characterizations of sufficiency.

\begin{thm}\label{thm:sufficiency} \cite{ohyapetz} The following conditions are
equivalent.
\begin{enumerate}
\item[(i)] The subalgebra $\mathcal B$ is sufficient for $\{\varphi,\psi\}$.
\item [(ii)] $S(\varphi,\psi)=S(\varphi_0,\psi_0)$.
\item[(iii)] $\rho_\varphi^{it}\rho_\psi^{-it}\in \mathcal B$, for all $t\in \mathbb R$.
\item[(iv)] $E_\varphi=E_\psi$.
\end{enumerate}

\end{thm}

Our results below are based on the following generalization of the classical
factorization criterion for sufficient statistics.

\begin{thm}\label{thm:factorization} \cite{japetz} Let $\varphi$, $\psi$ be
faithful states on $\mathcal A$ and let $\mathcal B\subseteq \mathcal A$ be a subalgebra, such that
$\rho_\psi^{it}\mathcal B\rho_\psi^{-it}\subseteq \mathcal B$ for all $t\in \mathbb R$. Then
$\mathcal B$ is sufficient for $\{\varphi,\psi\}$ if and only if
\[
\rho_\varphi=\rho_{\varphi_0}D,\qquad \rho_\psi=\rho_{\psi_0}D
\]
where $\varphi_0=\varphi|_\mathcal B$, $\psi_0=\psi|_\mathcal B$ and $D$ is a positive element
in the relative commutant $\mathcal B'\cap \mathcal A$.
\end{thm}

\subsection{The CAR    algebra}

We recall some basic facts about the CAR algebra, for details see \cite{armo, bratrob}.

The CAR algebra $\mathcal A$ is
the $C^*$- algebra generated by elements $\{a_i, i\in \mathbb Z\}$,
satisfying the  anticommutation relations
\begin{equation}\label{eq:car}
a_ia_j+a_ja_i =0,\quad   a_ia_j^*+a_j^*a_i=\delta_{ij},\qquad i,j\in \mathbb Z
\end{equation}
For a subset  $I\subset \mathbb Z$, the $C^*$-subalgebra generated by
$\{a_i, i\in I\}$  is denoted by $\mathcal A(I)$. If $I$ is finite,
$\mathcal A(I)$ is isomorphic  to the full
matrix algebra $M_{2^{|I|}}(\mathbb C)$ by the so-called Jordan-Wigner
isomorphism. Since
\[
\mathcal A=\overline{\bigcup_{|I|<\infty}\mathcal A(I)}^{\, C^*},
\]
there is a unique tracial state $\tau$ on $\mathcal A$, obtained as an
extension of the unique tracial  states on $\mathcal A(I)$, $|I|<\infty$. It has
the following product property:
\begin{equation}\label{eq:product}
\tau(ab)=\tau(a)\tau(b),\qquad a\in \mathcal A(I),\ b\in \mathcal A(J),\quad I\cap J=\emptyset
\end{equation}

\subsubsection{Graded commutation relations}

For $I\subseteq \mathbb Z$, we denote by $\Theta^I$ the (unique)
automorphism of $\mathcal A$, such that
\begin{equation}\label{eq:thetaI}
\Theta^I(a_i)=-a_i, \ i\in I,\qquad \Theta^I(a_i)=a_i,\ i\notin I
\end{equation}
in particular, we denote $\Theta^{\mathbb Z}$ by $\Theta$. The even  and odd
parts  of $\mathcal A$ are defined as
\[
\mathcal A_+:=\{ a\in \mathcal A,\ \Theta(a)=a\},\
\mathcal A_-:=\{ a\in \mathcal A,\ \Theta(a)=-a\}
\]
and  $\mathcal A(I)_+:=\mathcal A(I)\cap \mathcal A_+$, $\mathcal A(I)_-:=
\mathcal A(I)\cap \mathcal A_-$. Let  $I\cap J=\emptyset$
 and $a\in \mathcal A(I)_\sigma$, $b\in \mathcal A(J)_{\sigma'}$,
 $\sigma,\sigma'\in \{+,-\}$. Then we have the graded commutation
relations
\begin{equation}\label{eq:graded}
ab=\epsilon(\sigma,\sigma')ba
\end{equation}
where
\begin{eqnarray*}
\epsilon(\sigma,\sigma')&=& -1\quad  \mbox{if } \sigma=\sigma'=-\\
			&=& +1\quad  \mbox{otherwise}
\end{eqnarray*}

If $I$ is finite,
 then there is a self-adjoint unitary $v_I\in \mathcal A(I)$, such that
 $\Theta^I(a)=v_Iav_I$ for $a\in \mathcal A$ and
 \begin{equation}\label{eq:vI}
v_I=\Pi_{i\in I}v_i,\qquad v_i=a_i^*a_i-a_ia_i^*
 \end{equation}
Note that $v_iv_j=v_jv_i$ if $i\neq j$ and $\tau(v_i)=0$. Moreover,
$v_I\in \mathcal A(I)_+$ and $\mathcal A(I)_+=\mathcal A_I\cap \{v_I\}'$.

\subsubsection{Matrix units}

Let $A\subset \mathbb Z$ be a finite set, $A=\{i_!,\dots,i_n\}$. The relations
\begin{eqnarray*}
e_{11}^{(i_j)}:=a_{i_j}a_{i_j}^*,\qquad e_{12}^{(i_j)}:=V_{i_{j-1}}a_{i_j}\\
e^{(i_j)}_{21}:=V_{i_{j-1}}a_{i_j}^*,\qquad e_{22}^{(i_j)}:=a_{i_j}^*a_{i_j}
\end{eqnarray*}
with $V_{i_j}=\Pi_{k=1}^j(I-2a_{i_k}^*a_{i_k})$ define a family of mutually
commuting $2\times 2$ matrix units. The Jordan-Wigner isomorphism is then given by\[
e^{(A)}_{k_1l_1\dots k_nl_n}:=e^{(i_1)}_{k_1l_1}\dots e^{(i_n)}_{k_nl_n}\mapsto
e_{k_1l_1}\otimes\dots\otimes e_{k_nl_n}
\]
where $e_{kl}$ are standard matrix units in $M_2(\mathbb C)$. The elements
$\{e^{(a)}_\alpha,\alpha\in \mathcal J(A):=(\{1,2\}\times \{1,2\})^n\}$  span $\mathcal A(A)$. Note that $e^{(A)}_\alpha$ are either even or odd, we denote the set of
indices of the even resp. odd elements by $\mathcal J(A)_+$, resp.
$\mathcal J(A)_-$. Moreover, the elements $p_\alpha^{(A)}:=e_\alpha^{(A)}(e_\alpha^{(A)})^*$ and $q_\alpha^{(A)}:=(e^{(A)}_\alpha)^*e^{(A)}_\alpha$ are even
projections in $\mathcal A(A)$ and
\begin{equation}\label{eq:munits}
p_\alpha^{(A)}e^{(A)}_\beta q_\alpha^{(A)}=\delta_{\alpha,\beta}e^{(A)}_\alpha,
\qquad \alpha,\beta \in \mathcal J(A)
\end{equation}

\subsubsection{Conditional expectations}

Let $I\subseteq \mathbb Z$ be any subset. Then there is a unique conditional
expectation $E_I: \mathcal A\to \mathcal A (I)$, satisfying
\begin{equation}\label{eq:condexp}
\tau(ab)=\tau(E_I(a)b),\qquad a\in \mathcal A,\ b\in \mathcal A(I)
\end{equation}
This implies that $\Theta E_I=E_I\Theta$. If $J\subseteq \mathbb Z$, then
$E_I(a)\in \mathcal A(I\cap J)$ for $a\in \mathcal A(J)$ and $E_IE_J=E_JE_I=E_{I\cap J}$. Note
also  that the product property (\ref{eq:product}) implies that for $a\in \mathcal A(J)$ with $I\cap J=\emptyset$, $E_I(a)=\tau(a)$.

\section{Strong additivity and Markov property}

Let $A$, $B$, $C$ be disjoint finite subsets in $\mathbb Z$. Let us denote $\mathcal A=\mathcal A_{ABC}=\mathcal A(A\cup B\cup C)$, $\mathcal A_{AB}=\mathcal A(A\cup B)$ etc.
Let $\varphi$ be a faithful state on $\mathcal A$ and let $\rho$ be its density, that is,
$\varphi(x)={\rm Tr} \rho x$ for $x\in \mathcal A$.

Let $\varphi_{AB}$ denote  the  restriction of $\varphi$
to $\mathcal A_{AB}$, similarly $\varphi_{BC}$ and $\varphi_B$. Then the density of $\varphi_{AB}$ in $\mathcal A_{AB}$ is
\[
\rho_{AB}=E_{AB}(\rho),
\]
where $E_{AB}=E_{A\cup B}$. As an element in $\mathcal A$, $\rho_{AB}$ is the density of the state $\varphi\circ E_{AB}$.

\subsection{Strong subadditivity of entropy}

Let $\rho$ be the density of the state $\varphi$. Let
\[
S(\varphi)= -{\rm Tr} \rho(\log (\rho))
\]
be the von Neumann entropy of $\varphi$. The strong subadditivity for CAR
algebras\begin{equation*} \tag{SSA}
S(\varphi)-S(\varphi_{AB})-S(\varphi_{BC})+S(\varphi_B)\le 0
\end{equation*}
 was proved in \cite{moriya}.
This inequality is equivalent with
\[
S(\rho,\rho_{BC})-S(\rho_{AB},\rho_B)\ge 0.
\]
Since $\rho_{AB}=E_{AB}(\rho)$, $\rho_B=E_{AB}(\rho_{BC})$ are restrictions of $\rho$ and $\rho_{BC}$ to $\mathcal A_{AB}$, this  holds by monotonicity of the relative
entropy. Theorem \ref{thm:sufficiency} (ii)  then implies the  following.

\begin{thm}\label{thm:SSAsuff} The equality in (SSA) is attained if and only if the subalgebra
$\mathcal A_{AB}$ is sufficient for $\{\varphi,\varphi\circ E_{BC}\}$.

\end{thm}

\subsection{Markov triplets and strong additivity}

The state $\varphi$ is a Markov triplet if there exists a completely positive, identity preserving map $E: \mathcal A \to \mathcal A_{AB}$, such that
\begin{enumerate}
\item[(i)] $E(xy)=xE(y)$, for all $x\in \mathcal A_A$ and $y\in \mathcal A$.
\item[(ii)] $\varphi\circ E=\varphi$
\item[(iii)] $E(\mathcal A_{BC})\subseteq \mathcal A_B$
\end{enumerate}
The map $E$ is called a quasi-conditional expectation with respect to the triplet $\mathcal A_A \subset \mathcal A_{AB}\subset \mathcal A$.
Let us now define the subalgebras $\mathcal B\subset \mathcal C$ in $\mathcal A_{AB}$ by
\[
\mathcal C=\{ x\in \mathcal A_{AB}, \rho_{BC}^{it}x\rho_{BC}^{-it}\in \mathcal A_{AB}\}, \qquad
\mathcal B=\{ y\in \mathcal A_B, \rho_{BC}^{it}y\rho_{BC}^{-it}\in \mathcal A_B\}
\]
Note that $\mathcal C$ is the fixed point subalgebra of the generalized conditional
expectation $E_{\rho_{BC}}:\mathcal A \to \mathcal A_{AB}$ with respect to $\rho_{BC}$
\cite{acccec}.
We also have $E_{BC}(\mathcal C)= \mathcal B$. Indeed, if $x=E_{BC}(y)$ for
some $y\in \mathcal C$, then
\[
\rho_{BC}^{it}x\rho_{BC}^{-it}=E_{BC}(\rho_{BC}^{it}y\rho_{BC}^{-it})\in \mathcal A_B,
\]
so that $E_{BC}(\mathcal C)\subseteq \mathcal B$, the converse inclusion is clear.

\begin{thm}\label{thm:SSAmarkov} The state $\varphi$ is a Markov triplet if and only if $\varphi$
satisfies  equality in (SSA) and $\mathcal A_A\subseteq \mathcal C$.
\end{thm}

{\it Proof.} Let $\varphi$ be a Markov triplet and let $E$ be the
quasi-conditional expectation.
Then $E$ is a completely positive identity preserving map
$\mathcal A\to\mathcal A_{AB}$  and $\varphi \circ E=\varphi$.
Moreover, let $x\in \mathcal A_A$, $y\in \mathcal A_{BC}$, then
\[
\varphi\circ E_{BC}\circ E(xy)=\varphi\circ E_{BC}(xE(y))=\tau(x)\varphi(E(y))=
\tau(x)\varphi(y)=\varphi\circ E_{BC}(xy)
\]
Since by the commutation relations (\ref{eq:car}) $\mathcal A$ is spanned by elements of the form $xy$, the above equality implies that $E$ preserves
$\varphi\circ E_{BC}$ as well, so that $\mathcal A_{AB}$ is sufficient
 for $\{\varphi,\varphi\circ E_{BC}\}$ and equality in (SSA) holds by Theorem \ref{thm:SSAsuff}. Let
 \[
F=\lim_n \frac 1n \sum_{k=0}^{n-1}E^k
 \]
By the ergodic theorem, $F$ is a conditional expectation with range $\mathcal R(F)$ the fixed point subalgebra of $E$. By the property (i) of Markov triplets,
$\mathcal A_A\subseteq \mathcal R(F)$. Since $F$ also preserves $\varphi\circ E_{BC}$, we have by Takesaki theorem that $\rho_{BC}^{it}\mathcal R(F) \rho_{BC}^{-it}\subseteq \mathcal R(F)$, hence also $\rho^{it}_{BC}\mathcal A_A\rho_{BC}^{-it}\subseteq \mathcal R(F)\subseteq \mathcal A_{AB}$. It follows that $\mathcal A_A\subseteq \mathcal C$.

Conversely, suppose equality in (SSA) and $\mathcal A_A\subseteq \mathcal C$. Let
$E=E_{\rho_{BC}}:\mathcal A\to\mathcal A_{AB}$ be the generalized conditional expectation.
By Theorem \ref{thm:SSAsuff}, $\mathcal A_{AB}$ is sufficient for for $\{\varphi,\varphi\circ E_{BC}\}$ and by Theorem \ref{thm:sufficiency} (iv), $E_{\rho_{BC}}=E_\rho$,
 hence $\varphi\circ E=\varphi$. By the assumptions, $\mathcal A_A\subseteq \mathcal C$ the
 fixed point subalgebra of $E$. The property (iii) of Markov triplets is clear
 from the definition of $E_{\rho_{BC}}$.

{\hfill $\square$}

The following Corollary was already proved in \cite{moriya}.

\begin{coro}\label{coro:even} Let $\varphi$ be an even state. Then $\varphi$ is a Markov triplet if and only if it satisfies equality in (SSA).
\end{coro}

{\it Proof.} Since $\rho$ is even, $\rho_{BC}$ is even as well and we always have
$\mathcal A_A\subseteq \mathcal C$, by the graded  commutation relations. The proof now
follows  from Theorem \ref{thm:SSAmarkov}.

{\hfill $\square$}

\section{Characterization of strongly additive states and Markov triplets}

\begin{thm}\label{thm:SSAfactor} The state $\varphi$ satisfies equality in (SSA) if and only if there
are positive elements $x\in \mathcal A_{AB}$, $y\in \mathcal A_{BC}$, such that
\begin{equation}\label{eq:factor}
\rho=xy
\end{equation}

\end{thm}

{\it Proof.} Suppose that $\varphi$ satisfies equality in (SSA). Then $\mathcal A_{AB}$
is a sufficient subalgebra for $\{\varphi,\varphi\circ E_{BC}\}$. By Theorem \ref{thm:sufficiency}, this implies that $u_t:=\rho^{it}\rho_{BC}^{-it}\in \mathcal A_{AB}$ for all $t$.  Since
$\rho_{BC}^{it}u_s\rho_{BC}^{-it}=u_t^*u_{s+t}$ for $s,t\in \mathbb R$, this
implies that $u_t\in \mathcal C$ for all $t$. Hence, $\mathcal C$ is a sufficient subalgebra as well,  such that $\rho_{BC}^{it}\mathcal C\rho_{BC}^{-it}\subseteq
\mathcal C$. By Theorem \ref{thm:factorization},
\begin{eqnarray*}
\rho&=&xy\\
\rho_{BC} &=& x_0y
\end{eqnarray*}
where $x,x_0\in \mathcal C\subseteq \mathcal A_{AB}$ are  the densities of
the restrictions $\varphi|_\mathcal C$ and $\varphi\circ E_{BC}|_\mathcal C$ and
$y$ is a positive element in $\mathcal C'$.  Note also that $\varphi\circ E_{BC}|_\mathcal C$
is
the restriction of $\varphi$ to $E_{BC}(\mathcal C)=\mathcal B\subseteq \mathcal A_B$, so that
$x_0\in \mathcal A_B$.

By the graded commutation relations, we  have $(\mathcal A_A)_+ \subseteq \mathcal C$, so that
$\mathcal C'\subseteq ((\mathcal A_A)_+)'= \mathcal A_{BC}+v_A\mathcal A_{BC}$, \cite{armo} (all commutants are taken in the algebra $\mathcal A$).
Let $y\in \mathcal C'$, then  $y=d_1+v_Ad_2$, where $d_1,d_2\in \mathcal A_{BC}$. We have
\[
x_0y=E_{BC}(x_0y)=E_{BC}(x_0(d_1+v_Ad_2))=x_0d_1.
\]
Since $\varphi$, and therefore also its restriction to $\mathcal B$ is faithful, $x_0$ is invertible, so that $y=d_1\in \mathcal A_{BC}$.

Conversely, suppose $\rho=xy$ as above.
Then  $\rho_{AB}=xy_0$, $\rho_{BC}= x_0y$ and $\rho_B=x_0y_0$,
where $y_0=E_{AB}(y)\in \mathcal A_B$, $x_0=E_{BC}(x)\in \mathcal A_B$. Clearly, both $x$ and $x_0$ must commute with both $y$ and $y_0$. Then
$\rho^{it}\rho_{BC}^{-it}= x^{it}x_0^{-it}\in \mathcal A_{AB}$. By Theorem \ref{thm:sufficiency} (iii), $\mathcal A_{AB}$ is sufficient for $\{\varphi,\varphi\circ E_{BC}\}$, so that $\varphi$ satisfies equality in (SSA).

{\hfill $\square$}

\begin{thm}\label{thm:Markovfactor} The state $\varphi$ is a Markov triplet if and only if there are
positive elements $x\in \mathcal A_{AB}$ and $y\in (\mathcal A_{BC})_+$, such that
\[
\rho=xy
\]

\end{thm}

{\it Proof.} Let $\varphi$ be a Markov triplet. By Theorem \ref{thm:SSAmarkov},
 $\varphi$ satisfies equality in (SSA) and  by Theorem
 \ref{thm:SSAfactor} and its proof, there are positive elements $x\in \mathcal C$, $y\in \mathcal C'$, such
 that  $\rho=xy$. Since  $\mathcal A_A\subseteq \mathcal C$, $\mathcal C'\subseteq \mathcal A_A'=(\mathcal A_{BC})_++v_A(\mathcal A_{BC})_-$, \cite{armo}. This implies that $y=d_++v_Ad_-$, where $d_+\in (\mathcal A_{BC})_+$ and
 $d_-\in (\mathcal A_{BC})_-$. By the same reasoning as in the proof of Theorem \ref{thm:SSAfactor}, we get that $y=d_+\in (\mathcal A_{BC})_+$.

Conversely, let $\rho=xy$ as above, then $\varphi$ satisfies equality in (SSA) by
Theorem \ref{thm:SSAfactor}, and $\rho_{BC}=x_0y$, $x_0=E_{BC}(x)$. For $a\in \mathcal A_A$,
\[
\rho_{BC}^{it}a\rho_{BC}^{-it} =x_0^{it}ax_0^{-it}\in \mathcal A_{AB}
\]
by the graded commutation relations, so that $\mathcal A_A\subseteq \mathcal C$. By Theorem
\ref{thm:SSAmarkov}, $\varphi$ is a Markov triplet.

{\hfill $\square$}

\subsection{Even Markov triplets}

\begin{thm}\label{thm:evenmarkovfactor}
Let $\varphi$ be an even state. Then $\varphi$ is a Markov triplet if and only if there are positive elements $x\in \mathcal A_{AB}$ and $y\in \mathcal A_{BC}$, such that
\[
\rho=xy.
\]
Moreover, $x$ and $y$ can be chosen even.

\end{thm}

{\it Proof.} Follows easily from Corollary \ref{coro:even},
Theorems \ref{thm:SSAfactor} and \ref{thm:Markovfactor} and the fact that $\rho$ is even.

{\hfill $\square$}

We will now describe the subalgebras $\mathcal C$ and $\mathcal C'$ for even states.
Since $\rho_{BC}$ is even, both $\mathcal C$ and $\mathcal B$  and  their  commutants
$\mathcal C'$ and  $\mathcal B'$ are invariant under $\Theta$.

\begin{lemma}\label{lemma:c} If $\varphi$ is even, then
\[
\mathcal C= \mathcal A_A\bigvee \mathcal B
\]
\end{lemma}

{\it Proof.} Since  $\mathcal A_A\subseteq \mathcal C$ and clearly
also $\mathcal B\subseteq \mathcal C$, we have  $\mathcal A_A\bigvee \mathcal B\subseteq \mathcal C$.

Conversely, any element $x\in \mathcal C\subseteq \mathcal A_{AB}$ has the form
$x=\sum_\alpha e^{(A)}_\alpha
b_\alpha$  for some $b_\alpha \in \mathcal A_B$, where $e^{(A)}_\alpha$ are the matrix
units in  $\mathcal A_A$. By (\ref{eq:munits}), we have for any $\alpha$,
\[
p^{(A)}_\alpha x q^{(A)}_\alpha=
e^{(A)}_\alpha b_\alpha,
\]
since $q^{(A)}_\alpha$ is always even. As $\mathcal A_A\subseteq \mathcal C$,
this implies that
$e^{(A)}_\alpha b_\alpha\in \mathcal C$ for all $\alpha$. It follows
 that
 \[
 \rho^{it}_{BC}e^{(A)}_\alpha b_\alpha\rho^{-it}_{BC}=e^{(A)}_\alpha
 \rho_{BC}^{it}b_\alpha \rho_{BC}^{-it}\in \mathcal A_{AB},
 \]
 hence $b_\alpha\in \mathcal B$, so that $\mathcal C \subseteq  \mathcal A_A\bigvee \mathcal B$.

{\hfill $\square$}

\begin{lemma}\label{lemma:ccom} If $\varphi$ is even, then
\[
\mathcal C'=(\mathcal B'\cap \mathcal A_{BC})_++(\mathcal B'\cap \mathcal A_{BC})_- v_A
\]
\end{lemma}

{\it Proof.}  Since $\mathcal A_A\subseteq \mathcal C$, we have $\mathcal C'\subseteq \mathcal A_A'=
(\mathcal A_{BC})_++ (\mathcal A_{BC})_-v_A$, by \cite{armo}. Let $d_++v_Ad_-\in \mathcal C'$ and
let $x\in \mathcal B\subset \mathcal C$. Then we must have
$xd_+-d_+x=v_A(d_-x-xd_-)$. Applying $E_{BC}$ on both sides, we get
$xd_+-d_+x=d_-x-xd_-=0$, hence $d_+\in \mathcal B'\cap (\mathcal A_{BC})_+=(\mathcal B'\cap \mathcal A_{BC})_+$, $d_-\in \mathcal B'\cap (\mathcal A_{BC})_-=(\mathcal B'\cap \mathcal A_{BC})_-$.

Conversely, let $d_+ \in (\mathcal B'\cap \mathcal A_{BC})_+$, $d_-\in  (\mathcal B'\cap \mathcal A_{BC})_-$
and let $a\in \mathcal A_A$, $b\in \mathcal B$. Then by the graded commutation relations,
\[
ab(d_++v_Ad_-)=d_+ab+ v_Ad_-a_+b+v_Ad_-a_-b=(d_++v_Ad_-)ab
\]
so that $d_++v_Ad_-\in \mathcal C'$.
{\hfill $\square$}

\begin{lemma}\label{lemma:b}
Denote
$\tilde {\mathcal B} = \mathcal B'\cap \mathcal A_B$. Then
\[
\mathcal B'\cap \mathcal A_{BC}=\tilde {\mathcal B}\bigvee ((\mathcal A_C)_++v_B(\mathcal A_C)_-)
\]

\end{lemma}

{\it Proof.} It is easy to see that both
$\tilde {\mathcal B}$ and  $(\mathcal A_C)_++v_B(\mathcal A_C)_-$ are subsets in $\mathcal B'\cap \mathcal A_{BC}$.
Conversely,
any $y\in \mathcal A_{BC}$ has the form
$y=\sum_\beta b_\beta e^{(C)}_\beta$, for some $b_\beta\in \mathcal A_B$. Let $x\in \mathcal B$,
then
\begin{eqnarray*}
yx&=&\sum_\beta b_\beta e^{(C)}_\beta(x_++x_-)=\sum_\beta b_\beta x_+e^{C}_\beta+
\sum_{\beta\in \mathcal J(C)_+}b_\beta x_-e^{(C)}_\beta -
\sum_{\beta\in \mathcal J(C)_-}b_\beta x_-e^{(C)}_\beta\\
&=& \sum_{\beta\in \mathcal J(C)_+} b_\beta xe^{(C)}_\beta +
\sum_{\beta\in \mathcal J(C)_-}b_\beta\Theta(x)
e^{(C)}_\beta
\end{eqnarray*}
It follows that $yx=xy$  only if
$xb_\beta=b_\beta x$ for  $\beta\in \mathcal J(C)_+$ and $xb_\beta=b_\beta\Theta(x)$ for $\beta\in \mathcal J(C)_-$. This is true for all $x\in \mathcal B$ if and only if $b_\beta\in \mathcal B'$ for
$\beta\in \mathcal J(C)_+$ and $b_\beta v_B\in \mathcal B'$ for $\beta\in \mathcal J(C)_-$, this implies the statement of the lemma.

{\hfill $\square$}

Let us now look at the algebra $\mathcal B$. Let $P_1,\dots, P_m$ be the minimal central projections in $\mathcal B$. Since  $\mathcal B$ is  invariant under $\Theta$, we must have for each $i$, $\Theta(P_i)=P_j$ for some $j$. Suppose that
$\Theta(P_i)=P_i$, $i=1,\dots,k$ and $\Theta(P_i)=P_{i+1}$ for $i=k+2l+1$, $l=0,\dots, \frac {m-k}2-1$.

\begin{lemma} \label{lemma:central} Let us denote $P_A=\frac12 (1+v_A)$.
The minimal central projections in $\mathcal C$ are
\begin{eqnarray*}
Q_i&:=&P_i, \qquad i=1,\dots,k\\
Q_{k+1}&:=& P_AP_{k+1}+(1-P_A)P_{k+2}, \quad Q_{k+2}:=(1-P_A)P_{k+1}+P_AP_{k+2}\\
\dots\\
Q_{m-1}&:=& P_AP_{m-1}+(1-P_A)P_{m},\quad Q_m:=(1-P_A)P_{m-1}+P_AP_{m}
\end{eqnarray*}

\end{lemma}

{\it Proof.} Clearly,
$\mathcal Z(\mathcal C)\subset \mathcal A_A'\cap \mathcal A_{AB}=(\mathcal A_B)_++v_A(\mathcal A_B)_-$ and it is easy to see that if $x_++v_Ax_-\in \mathcal Z(\mathcal C)$, then $x_+,x_-$ must be in $\mathcal Z(\mathcal B)$. Therefore, $x_+=\sum c_j P_j$ and $x_-=\sum_j d_jP_j$, for some
$c_j,d_j\in \mathbb C$. Since $x_+$ is even, we must have
$c_j=c_{j+1}$ for $j=k+2l+1$, $l=0,\dots,\frac{m-k}2-1$. Similarly, we get
$d_j=0$ for $j=1,\dots,k$ and $d_j=-d_{j+1}$ for $j=k+2l+1$, $l=0,\dots,\frac{m-k}2-1$.

Suppose now that $P=x_++v_Ax_-$ is a projection, then
we must have $x_+^*x_++x_-^*x_-=x_+$ and $x^*_+x_-+x_-^*x_+=x_-$. This implies
that $c_j=|c_j|^2$ for $j=1,\dots,k$, $c_j=|c_j|^2+|d_j|^2\ge0$ and $c_j(d_j+\bar d_j)=d_j\in \mathbb R$, for $j>k$.
Hence $2c_jd_j=d_j$, so that either $d_j=0$ and then $c_j=c_j^2$, or $c_j=\frac12$ and then $d_j=\pm
\frac12$.

It follows that any projection in $\mathcal Z(\mathcal C)$ is a sum of some of
the following  projections: $P_i$, $i=1,\dots,k$, $P_j+P_{j+1}$, $j=k+2l+1$, and
$\frac 12 (P_j+P_{j+1}\pm v_A(P_j-P_{j-1}))$, $j=k+2l+1$. Since the last
projection is equal to $Q_j$ or $Q_{j+1}$ and $Q_j+Q_{j+1}=P_j+P_{j+1}$, the
Lemma follows.

{\hfill $\square$}

\begin{thm} Let  $\varphi$ be an even faithful state on $\mathcal A$. Then $\varphi$
is a Markov triplet if and only if there is an orthogonal family of
projections
$P_1,\dots,P_m\in \mathcal A_B$ and  decompositions
$P_j\mathcal A_BP_j=\mathcal B_j\otimes \tilde \mathcal B_j$, where $\mathcal B_j$ and $\tilde \mathcal B_j$ are
full matrix algebras,  such that
\begin{enumerate}
\item $\Theta(P_j)=P_j$ and $\mathcal B_j$ and $\tilde \mathcal B_j$ are invariant under $\Theta$ for
$j=1,\dots,k$
\item $\Theta(P_j)=P_{j+1}$ and $\Theta(\mathcal B_j)=\mathcal B_{j+1}$, $\Theta(\tilde \mathcal B_j)=
\tilde \mathcal B_{j+1}$ for $j=k+2l+1$, $l=0,\dots,\frac{m-k}2-1$
\item Let us denote
\begin{eqnarray*}
 V_j&=&P_jv_B\\
 \mathcal C_j &=& \mathcal A_A\bigvee \mathcal B_j\\
 \tilde \mathcal C_j &=& \tilde\mathcal B_j\bigvee ((\mathcal A_C)_++V_j(\mathcal A_C)_-)
\end{eqnarray*}
  for $j=0,\dots,k$ and
\begin{eqnarray*}
  U_l&=&(P_{k+2l+1}+P_{k+2l+2})v_B\\
  \mathcal D_l &=& \mathcal A_A\bigvee (P_A\mathcal B_{k+2l+1}+(1-P_A) \mathcal B_{k+2l+2})\\
  \tilde \mathcal D_l &=&  (P_A\tilde \mathcal B_{k+2l+1}+(1-P_A)\tilde \mathcal B_{k+2l+2})\bigvee ((\mathcal A_C)_+ +U_l(\mathcal A_C)_-)
\end{eqnarray*}

  for $l=0,\dots,\frac{m-k}2-1$, then there is a decomposition
\begin{equation} \label{eq:evendecomp}
\rho=\bigoplus_{j=1}^k x_j\otimes y_j\oplus \bigoplus_{l=0}^{\frac{m-k}2-1}
(z_l\otimes w_l\oplus \Theta(z_l\otimes w_l)),
\end{equation}
where $x_j\in \mathcal C_j$ and
$y_j\in \tilde\mathcal C_j$  are positive and even
for $j=1,\dots,k$, and $z_l\in \mathcal D_l$,  $w_l\in \tilde \mathcal D_l$ are positive for
$l=0,\dots,\frac{m-k}2-1$.

\end{enumerate}

\end{thm}

{\it Proof.} Suppose that  $\rho$ has the form (\ref{eq:evendecomp}). Let us
define  $Q_1,\dots, Q_m$ from $P_1,\dots,P_m$
as in Lemma \ref{lemma:central}. Then $Q_j$ are mutually orthogonal projections
and it is easy to see that $Q_jx_j=x_j$, $Q_jy_j=y_j$ and
$Q_{k+2l+1}z_l=z_l$, $Q_{k+2l+1}w_l=w_l$, $Q_{k+2l+2}\Theta(z_l)=\Theta(z_l)$,
$Q_{k+2l+2}\Theta(w_l)=\Theta(w_l)$. Put
\[
x=\bigoplus_j x_j\oplus \bigoplus_l (z_l\oplus \Theta(z_l)), \qquad
y=\bigoplus_j y_j\oplus \bigoplus_l (w_l\oplus \Theta(w_l))
\]
then $x\in \mathcal A_{AB}$ and $y\in \mathcal A_{BC}$ are positive even elements and
$\rho=xy$. By Theorem \ref{thm:evenmarkovfactor}, this implies that $\varphi$ is a Markov triplet.

Conversely, suppose that $\varphi$ is an even Markov triplet. Then we have seen
that $\rho=xy$, where $x\in \mathcal C$ and $y\in \mathcal C'$ are positive and even. By Lemmas
 \ref{lemma:c}, \ref{lemma:ccom} and \ref{lemma:b}, $x\in \mathcal A_A\bigvee \mathcal B$ and
 $y\in \tilde \mathcal C:= \tilde \mathcal B\bigvee ((\mathcal A_C)_++v_B (\mathcal A_C)_-)$.

Let $P_1,\dots,P_m$ be the minimal central projections in $\mathcal B$ and let
$\mathcal B_j:= P_j\mathcal B$, $\tilde \mathcal B_j:=P_j\tilde \mathcal B$.  Then $\mathcal B_j$ and $\tilde \mathcal B_j$ are full matrix algebras and $P_j \mathcal A_BP_j=\mathcal B_j\otimes \tilde \mathcal B_j$. Moreover,
 we may suppose that there is some $k\le m$ such that 1. and 2. are fulfilled.

The minimal central projections $Q_1,\dots,Q_m$  in $\mathcal C$
are given by Lemma \ref{lemma:central}.
Let us denote $\mathcal C_j=Q_j\mathcal C_j$, $\mathcal C'_j =Q_j\mathcal C'$.
Then each $\mathcal C_j$, $\mathcal C'_j$ is isomorphic to a full matrix
algebra and we have a decomposition
\[
\mathcal C=\bigoplus_j \mathcal C_j\otimes \tilde I_j,\qquad \mathcal C'=\bigoplus_j I_j\otimes
\mathcal C'_j
\]
Since we are interested only in even elements in $\mathcal C'$, we take the algebra
$\tilde \mathcal C_j:=Q_j\tilde \mathcal C\subset \mathcal C'_j$.
For $j=1,\dots,k$, $\mathcal C_j$ are invariant under $\Theta$. For $l=0,\dots, \tfrac{m-k}2-1$, let us denote
$\mathcal D_l:=\mathcal C_{k+2l+1}$, $E_l:= Q_{k+2l+1}+Q_{k+2l+2}=P_{k+2l+1}+P_{k+2l+2}$. Then
$E_l$ is an even projection, the algebra $E_l\mathcal C= \mathcal D_l\oplus \Theta(\mathcal D_l)$ is
invariant under $\Theta$ and even elements in $E_l\mathcal C$ are of the form
$x\oplus \Theta(x)$, for some $x\in \mathcal D_l$. Similar relation hold for $\Tilde \mathcal C_j$ and $\tilde \mathcal D_l:=\tilde \mathcal C_{k+2l+1}$.

Let us denote $x_j:=Q_jx$, $y_j:=Q_jy$ for $j=1,\dots,k$ and
$x_j:=E_lx$, $y_j:=E_ly$ for $j=k+2l+1$, $l=0,\dots, \frac{m-k}2-1$.
Then all $x_j$, $y_j$ are positive and even and
\[
\rho=\bigoplus_{j=1}^k xj\otimes y_j \oplus \bigoplus_{l=0}^{\frac{m-k}2-1}
x_{k+2l+1}y_{k+2l+1}
\]
Moreover,  for $j=k+2l+1$ we must have
$x_j=z_l\oplus \Theta(z_l)$ for some positive $z_l\in \mathcal D_l$ and similarly
$y_j=w_l\oplus \Theta(w_l)$ for positive $w_l\in \tilde \mathcal D_l$, $l=0,\dots,\frac{m-k}2-1$. The rest of the proof now follows from Lemmas \ref{lemma:c} and \ref{lemma:b}.

{\hfill $\square$}

redpajama_set_name: RedPajamaArXiv


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text:
Sparkle shine and fade
Posted on December 1, 2012 December 17, 2012 by coolpapae
Sparkle – 2012
Directed by Salim Akil
Starring Jordin Sparks, CeeLo Green, Whitney Houston, Derek Luke, Mike Epps, Tamela Mann, Carmen Ejogo, Tika Sumpter, Omari Hardwick
Screenplay Mara Brock Akil, Howard Rosenman based on the film screenplay by Joel Schumacher and Rosenman
The first vision of Whitney Houston in the remake of the 1976 film, Sparkle is breathtaking. Not so much for anything she does, but for how much she has aged. It's not that it was a bad look. Some of it was makeup. She was now at the point where she was beyond the precious look that she was naturally graced with in her youth. This look was more earned than that of her more successful years. The ensuing scenes with her are a welcome sight. She is a protective mother, for sure. It's a role that feels like she lived it.
The tale is a loose approximation of a Supremes-type group. Instead of taking place in Harlem in the '50's, this one is in Detroit during the Motown heyday of the 60's. This film exists along side of those groups, and indeed inspired by them. It gives the story more of a logical basis and really feels right.
Whitney plays the mother of three girls, Sister (Ejogo), Dee (Sumpter) and Sparkle (Sparks) who form a girl band based around the singing talent of one and the writing talent of another. They are managed by a man named Stix (Luke) who loves the writer (Sparks) and they find success. This is countered by Sister's bad decisions to the point where tragedy strikes all three. In a striking missed opportunity, this tragedy is quickly, conveniently disposed of.
The soundtrack for the film, a mixture of songs from the original film as well as those from the new. For the most part, the work is fantastic, even incorporating Houston's fading skills to the maximum extent. Her best days were behind her, but she still could inspire, especially with the production of the movie behind her.
There is a real force behind the combination of Ejogo, Sumpter and Sparks. They really do come off as sisters with distinct personalities and an unbreakable bond. This helps the film to overcome the familiar plot points. Each is given a distinct relationship with their strict, loving and flawed mother. The different paths give the film a power that emanates from the mostly silent authority from Houston. It cannot be underestimated how important this work was to her. Just like it can't be overstated how important her presence is to this film.
Given that the film is named after Sparks' character, her performance is key to the film. She is uneven as an actor, but she her singing comes from a real sense of real joy. Luke is a good companion for her. His performances are never showy and this is no exception. He is the rare actor that can sink into every role and be just enough to act as a glue, but not necessarily standing out. He could benefit every movie.
The film's last 10 minutes are its best. The song "One Wing" is as good as anything R. Kelly ever wrote, but Sparks performs the hell out of it. Add to this moments of wounded joy shared between mother and daughter and you have the makings of a solid, emotionally honest film.
As the duet, "Celebrate" between Sparks and Houston rolls over the credits, thoughts of how little I had even pondered Whitney from the first time I had seen her on film, since her début in the atrocious The Bodyguard. As the world celebrated what would become her biggest song, I thought there had been unearthed an acting skill to match…let's say, Diana Ross in The Wiz. I stopped looking forward to her, like, apparently, most everyone else. Now, after this triumph, we can only look back and wonder what we missed.
(**** out of *****)
Tagged Derek Luke, Jordin Sparks, Sparkle, Whitney Houston, Whitney's final film
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2 thoughts on "Sparkle shine and fade"
Phil Grant says:
Great review CPE
coolpapae says:
Thanks Buddy.