text:
MassChallenge and The Joseph P. Kennedy, Jr. Foundation Partner to Accelerate Startups Serving the Intellectual and Developmental Disabilities Community
MassChallenge, the global network for innovators, and The Joseph P. Kennedy, Jr. Foundation, the leading organization for Intellectual Disabilities programs and public policy, are partnering to catalyze startup innovation for the Intellectual and Developmental Disabilities (IDD) community in collaboration with autism non-profit Blue Umbrella.
The partnership launched this summer with two startup showcases spotlighting the top startups impacting the IDD community. The latest showcase, on August 13 in Hyannis Port, MA, featured presentations from eight of the most promising IDD-focused startups in the world. The companies are tackling a diverse set of challenges including in-home support services, speech recognition, job training and employment, dating, and financial planning. Showcase attendees included leading early-stage investors, policymakers, and IDD advocates and experts.
"The showcase demonstrates that world-class founders are building venture-backable companies that will make a huge difference in the lives of people with intellectual disabilities and autism," said Christopher Kennedy McKelvy, Vice Chairman of the Foundation. "Our aim is to support those founders with the resources they need, and to draw more of them into this space, which is ripe for innovation and impact."
Startups featured at the showcase included:
Daivergent
Floreo
Hiki
Hope Trust
John's Crazy Socks
NeuroNav
Voiceitt
Beaming Health
"MassChallenge is honored to be a part of this meaningful initiative," said MassChallenge CEO Siobhan Dullea. "The excitement and energy around this showcase are palpable, and it is clear that true impact will be generated not only for the entrepreneurs in the room, but also for the individuals with intellectual and developmental disabilities whom they serve."
MassChallenge and the Foundation also hosted a virtual startup pitch competition on June 17. That event featured pre-seed companies, many of which are pre-revenue and still refining prototypes and customer discovery. Beaming Health, a service provider and resource directory for parents of children with autism, was selected as the first-prize winner at the virtual pitch event, receiving a $25K cash prize. Social Cipher, which builds social emotional learning games for neurodivergent youth, won a $10K cash prize. The June pitch competition also featured Advasys, Ability App, Making Authentic Friendships, Movia Robotics, and Valence Vibrations.
The showcases mark the first phase of the Foundation's plan to establish the premier IDD innovation community. As part of that vision, the Foundation will provide startups with the resources they need to scale, including capital, mentorship, and community.
"Founders focused on IDD need support from those who understand the community, the payor system, and the market." McKelvy said. "Given the Foundation's 75-year track record, we're uniquely positioned to fill that role."
"The Foundation has created an extraordinary opportunity to build an ecosystem that facilitates collaboration between founders, investors, and organizations dedicated to serving the special needs community," Hope Trust founder Josh Fishkind said. "Each of the Showcase presenters demonstrated cutting edge, innovative approaches to solving the largest challenges facing the ID and autism communities. Through the Foundation's efforts, the resources and funding needed to deliver these solutions to the mass market are within reach."
Approximately 6.5 million people in the United States, and as many as 200 million people globally, have an intellectual disability.
Aug.13 Startup Showcase:
Beaming Health (California): Service provider and resource directory for parents of children with autism
Daivergent (New York): A.I. Platform for maximizing IDD job readiness by building personalized, lifelong plans and providing a curated intervention library
Floreo (Washington D.C.): Telehealth-ready virtual reality therapy platform designed to teach important skills to neurodiverse individuals
Hiki (New York): Social eco-system for the neurodivergent community
Hope Trust (New Jersey): Revolutionary technology for financial and care planning for special needs loved ones
John's Crazy Socks (New York): E-commerce business spreading happiness through socks
NeuroNav (California): Technology-enabled concierge service that curates individualized support for adults with disabilities
Voiceitt (Florida): Automatic speech recognition for people with non-standard speech
June 17 Virtual Pitch Competition:
Ability App (Idaho): Helping people with disabilities find accessible spaces around the world
Advasys (Florida): Uses information technology to give a voice to those who can't communicate their health and safety needs
Making Authentic Friendships (New York): An app that enables individuals with special needs to make friends
Movia Robotics (Connecticut): Unlocking learning potential with robot-assisted instruction
Social Cipher (California): Game-based SEL learning for neurodiverse youth and their counselors
Valence Vibrations (California): Delivering real-time emotional feedback using haptic wristbands
About The Joseph P. Kennedy, Jr. Foundation:
Since its founding in 1946, The Joseph P. Kennedy, Jr. Foundation has been a leader in creating and supporting programs to benefit people with Intellectual Disabilities (ID), their families, and their communities. In its early years the Foundation was instrumental in founding Special Olympics. During President Kennedy's administration, the Foundation helped make ID research and programs a national priority, under the leadership of Eunice Kennedy Shriver. In 1980 the Foundation launched its Public Policy Fellowship Program, supporting rising leaders in key ID public policy roles. Senator Edward M. Kennedy was President of the Foundation until 2009. The Foundation is now led by Timothy Shriver, Chairman of Special Olympics.
About MassChallenge:
MassChallenge is the global network for innovators. Headquartered in the United States with seven locations worldwide, MassChallenge equips bold entrepreneurs to disrupt the status quo and to create meaningful change. Since launching in 2009, more than 2,900 MassChallenge alumni have raised $8.6B in funding, generated $3.6B in revenue, and created more than 186,000 total jobs. Learn more about MassChallenge at masschallenge.org.
About BlueUmbrella:
BlueUmbrella is a nonprofit innovation platform focused on autism and ID startups. Founded by successful entrepreneurs and investors with children on the spectrum, BlueUmbrella is connecting, convening, and catalyzing the emerging early-stage autism ecosystem.


text:
CAMBS: Nicky wows with jazz-inspired composition
Updated: Wednesday, March 2, 2011 (15:07)
Abbey College Young Musicians from left to right Ashley Jones (15), Marcus Bingham (14), Nicky Radford (14), Harry Branch (13), Mark Woodward (14).
A 14 YEAR OLD was given the top honour at Abbey College's first ever Young Musician of the Year competition.
Pianist Nicky was named overall winner of the competition after coming first in the intermediate solo, intermediate recital and composition categories.
She was also a member of the winning ensemble act along with Ashley Jones, 15, Marcus Bingham, 14, and Mark Woodward, 14.
Nicky wowed the audience with a performance of her own composition Atlantic Journey, inspired by the work of American jazz guitarist Pat Metheny.
After picking up her trophy Nicky said: "I am obviously very pleased but also slightly surprised. There were so many good musicians performing. This has certainly encouraged me to continue developing my musical ability. I am currently studying for GCSE music but would like to pursue this further through university."
Winner of the best junior musician was Ramsey Junior pupil Shona Drummond, best novice soloist was Harry Branch, 13, best senior soloist was Marcus Bingham, 14 and best senior recitalist was Natalie Davies, 17.
The judges were Sue Peacock, assistant headteacher, and Jenny Robertshaw, former head of music.
Kate Hamilton, subject leader for music said: "My congratulations go to all winners and competitors for such a superb evening and thanks to everyone who came to give their support."
All proceeds from the evening will go towards the Abbey College concert band's trip to EuroDisney in July, where they will perform to the crowds visiting the attraction.

redpajama_set_name: RedPajamaCommonCrawl


text:
Les collines d'Allan, plus rarement collines Allan, sont un groupe de collines à l'extrémité de la chaîne Transantarctique, situées dans les régions de la terre d'Oates et de la terre Victoria, en Antarctique. Elles sont généralement dépourvues de glace et mesurent  de long, juste au nord-ouest des collines de Coombs près des sommets du glacier Mawson et du glacier Mackay.

Elles ont été cartographiées en 1957-1958 par l'équipe néo-zélandaise de l'expédition Fuchs-Hillary et ont été baptisées en l'honneur du géologue néo-zélandais Robin Sutcliffe Allan.

En 2016, des carottes de glace bleue ont été extraites de ces collines, dans une zone où de la vieille glace gisait près de la surface. Ces carottes, chronologiquement discontinues, ont été datées par la méthode potassium-argon : la glace la plus ancienne date de 2,7 millions d'années, de loin la glace la plus ancienne jamais datée.

De nombreuses météorites ont été découvertes dans ces collines. Elles sont dénommées par le sigle ALH (pour Allan Hills), suivi d'un numéro. ALH 84001 notamment, une météorite martienne, a été médiatisée à la fin du  lorsqu'une micrographie électronique montrant des structures tubulaires évoquant des bactéries fossilisées a été largement diffusée dans la presse.

Notes et références

Allan
Terre de Oates
Allan
Éponyme d'une météorite

redpajama_set_name: RedPajamaWikipedia


text:
Judge declares mistrial in 1979 NY child murder case
09 May 2015 | 1:00 am
A New York judge declared a mistrial Friday for a man accused of murdering a "beautiful" six-year-old boy 36 years ago, in a case that has haunted millions of American parents.
A New York judge declared a mistrial Friday for a man accused of murdering a "beautiful" six-year-old boy 36 years ago, in a case that has haunted millions of American parents.
The jury informed Judge Maxwell Wiley that they could not agree to acquit or convict Pedro Hernandez, 54, after three weeks of deliberations.
Etan Patz vanished after leaving his Manhattan home to walk alone for the first time to the bus stop to go to school on May 25, 1979. His body has never been found.
Hernandez was ordered back into custody and the court will now reconvene on June 10, when, an official told AFP, a new trial date could be set.
Hernandez has been in custody since 2012, when he confessed to police that he killed Etan in the basement of a New York grocery store before dumping his body out with the trash.
He later retracted his confession and pleaded not guilty when the trial opened in January.
The jury reached deadlock when just one person on the 12-member panel refused to convict Hernandez, jurors later told reporters.
The dissenting juror said he was swayed by arguments over Hernandez's poor mental health and described the accused's confession as "very bizarre no matter how many times it happened."
Etan's disappearance awakened Americans to the dangers of child abduction and fueled a generation of hyper-vigilant child rearing by parents terrified of letting their offspring out of their sight.
Etan's father, Stan Patz, who attended each day of the trial, said he was convinced that Hernandez had kidnapped and murdered his son.
"We are frustrated and very disappointed the jury has been unable to come to a decision. Our long ordeal is not over," he told reporters, adding that he was confident there would be a retrial.
– 'He should pay' –
The defense also say they are preparing for Hernandez to be retried.
Defense lawyer Harvey Fishbein argued that Hernandez had an IQ of 70, which would put him in the bottom two percent of the population, and said that convicted sex offender Jose Ramos was the real culprit.
Patz said the family no longer believed Ramos was responsible.
"Maybe he's (Hernandez) a good man now but he did something impulsive and terrible and I think he should pay for it," Patz said.
He described his son as a "beautiful, outgoing, friendly, curious little kid" who "would have made a great adult."
Prosecutors had fought hard to build a case despite having no physical evidence to tie Hernandez to the crime.
New York district attorney Cyrus Vance did not immediately confirm he would seek a re-trial but said prosecutors believed there is "clear and corroborated evidence of the defendant's guilt beyond a reasonable doubt."
"The challenges in this case were exacerbated by the passage of time, but they should not, and did not, deter us."
Etan's parents were only alerted to his disappearance when he failed to come home from school at the end of the day.
His father was a photographer and his pictures of Etan were the first of a missing child featured on milk cartons as part of a national search.
In 1983, then US president Ronald Reagan declared the anniversary of his disappearance National Missing Children's Day.
Etan was declared legally dead in 2001.
A global mobility solution in COVID-19 pandemic
With a passport that provides visa-free travel to just 45 countries, Nigerian citizens are all too familiar with the downsides of owning a weak passport.
As tomorrow cometh
It is the time of the year we are immersed in the ritual of the momentous change in the calendar. One year becomes old and like a ripe fruit, falls off the bough and is replaced by a new year that begins its own count down from January 1.
APO Press Releases
Sudan: Food looted in Darfur, 730,000 people left without
Download logoThe UN chief condemned on Wednesday the looting and attacks against United Nations facilities, equipment and supplies gifted to the Sudanese authorities for civilian in El Fasher, Darfur. Last Tuesday evening, a World Food Programme (WFP) warehouse located in the Borsa area of El Fasher town, the capital of North Darfur State, came under attack from unknown armed groups. Over 1,900 metric tons of food…
Hundreds of homes feared lost in Colorado wildfires
Hundreds of homes are feared lost in fast-moving wildfires in the US state of Colorado, officials said Thursday, as flames tore through areas desiccated by a historic drought.
Arnold Schwarzenegger Donates 25 Tiny Houses To Homeless Veterans
Arnold Schwarzenegger has donated 25 tiny houses to veterans and their pets to celebrate the holiday season. Fox 11 reports that the former governor of California paid $250,000 to purchase the structures located in West Los Angeles. Taking to social media, on Dec. 23, Schwarzenegger took to social media to explain the project. "Today, I…


text:
Q: Not able to declare a function in the console when brackets on the same line In the Chrome console, I executed the following function:
function getData_working(){
  return [
    {Category: "Category 1", Key: "Key 10", Value: 1},
    {Category: "Category 2", Key: "Key 10", Value: 1}
  ];
}

It does work and executing getData_working() afterwards produces the expected array with two objects. However, when I execute the following function (and I've actually copied the former, only changing the placement of the brackets, to make sure it isn't a typo), I get undefined.
function getData2_failing(){
  return
    [{Category: "Category 1", Key: "Key 10", Value: 1},
    {Category: "Category 2", Key: "Key 10", Value: 1}];
}

What's happening here? How does someone explain this "bug" or "feature"?

A: You can't put return on its own line. Well, you can, but if you do, you will get a return value of undefined every time. This is because of how automatic semicolon insertion works.

*

*Forbidden LineTerminators: The following syntactic constructs forbid a newline ("LineTerminator") at a certain position. If there is a newline at that position, a semicolon is inserted. The ECMAScript standard calls the grammar rules below restricted productions.

  PostfixExpression
  LeftHandSideExpression [no LineTerminator here] ++
  LeftHandSideExpression [no LineTerminator here] --
ContinueStatement
continue [no LineTerminator here] Identifier? ;
BreakStatement
break [no LineTerminator here] Identifier? ;
ReturnStatement
return [no LineTerminator here] Expression? ;
ThrowStatement
throw [no LineTerminator here] Expression? ;
For PostfixExpression, the rationale is avoiding the modification of a value on the previous line. For continue, break, return and throw, the rationale is that if they are used without an argument, they should not refer to the next line if one forgets a semicolon.
source: http://www.2ality.com/2011/05/semicolon-insertion.html

If you want the brackets on the next line, you need
return (
  [ ... ]
);

Or do like you've done in your original code
return [
  ...
];

redpajama_set_name: RedPajamaStackExchange


text:
Police Board Clears Officer Brandon Ternand In 2012 Fatal Shooting Of Dakota Bright, An Unarmed Black Teen
Filed Under:Chicago, Chicago Police Board, Chicago Police Officer, Dakota Bright, Local TV, police shooting, Vi Nguyen
CHICAGO (CBS) — The Chicago Police Board has cleared a white police officer in the controversial shooting death of an unarmed black teenager six years ago.
The board's 5-3 vote on Thursday absolved Officer Brandon Ternand of any wrongdoing, despite a ruling by the Independent Police Review Authority calling the shooting of 15-year-old Dakota Bright unprovoked and unwarranted.
IPRA had recommended Ternand, who is white, be fired for using excessive force for shooting Bright, who is black, in the back of the head.
Bright was a freshman at Robeson High School when he was shot and killed while running away from police on Nov. 8, 2012. An autopsy revealed he was shot in the back of the head.
Dakota Bright (picture provided to CBS)
The shooting sparked controversy, and set off protests in the community. Bright's family said he had just left a friend's place, and was walking to his grandmother's house when he was shot and killed.
The night of the shooting, Ternand and his partner were responding to a call of a burglary that turned out to be a false alarm. Ternand told the Police Board they saw Bright walking in an alley with a gun in his hand.
IPRA had questioned Ternand's account of the shooting, in which he claimed he shot Bright after he reached for his waist and started turning toward him, after running away and scaling several fences.
Bright was unarmed when he was shot, although a .22-caliber revolver was found in a front yard near where Ternand and his partner began chasing the teen.
A majority of the Chicago Police Board found Ternand's testimony to be credible and persuasive. The board said he had been a tactical officer for five years, calling him a decorated officer, and noting his reputation for honesty was backed by character witnesses.
Bright's family sued the city over the shooting, and in 2016 the City Council approved a $925,000 settlement.
Ternand was placed on paid desk duty from 2012 to 2017 while the shooting was under investigation. He was suspended without pay last November. Having been cleared by the Police Board, Ternand will be reinstated to active duty, and given back pay for the time he was suspended.
The Chicago Police Department declined comment on the board's ruling.
"It would be improper for the Superintendent to comment on this specific case when there are other Police Board cases he has not commented on after the Police Board came to their decision," CPD News Affairs Officer Michael Carroll wrote in an email.


After you have purchased the rolls of coins separate them and closely examine each coin in the roll.
Using this price guide will help you buy coins at a reasonable price.Visit us for to find about coin prices, coin value guide and coin values.Six decaying metal canisters filled with gold coins from the 1800s were unearthed in California by a couple out walking their dog.Whether you are buying or selling coins, knowing the value of your coins gives you the competitive edge.
Check with an professional organization like the Professional Numismatics Guild.
Coin Price Guide for Beginners (View the U.S. Currency Pricing Guide) How much is my coin worth.Coins also have value with collectors because of the rare and unique coins that are found throughout the world.
Coins, also known as Gold Pieces, are the most common form of currency in RuneScape.Our comprehensive values database is regularly updated by a team of expert analysts to guide the valuation of your favorite coins and collection.
If you do find a rare or valuable coin this way an added benefit is that you bought the coin for face value. 3.
There are many guides available that can help you determine the value of a coins. Determine Canadian Silver Dollar Values Here, discover how much you can expect to get for your silver dollar coins with a real transaction examples.
An article about the value of rare Australian 50 cent coins that you can find in your change.Search by name, location or numismatic specialty to find an ANA member dealer in your area.
I usually use a combination of coin books and online coin value guides to help me derive a price estimation for those who ask me what their coins are worth.
The last coin minted at Tower Hill, in 1975, was a sovereign.Look up your coin according to its date and origin, and you may be able to find its current value.


If you're looking to step into the wild, wild west, you're going to need the perfect costume and a horse. Lucky for you, this Carry Me Horse costume for kids includes both. This fun costume features 3D character pants that make it look like you're riding a horse. Just add your own cowboy shirt and hat and you're ready to ride off into the sunset.


After being licensed in 2001, Scott began his legal career with a single point of focus- helping his clients navigate the complex legal system. Scott has successfully handled hundreds of cases covering a wide variety legal issues ranging from child custody to multi-million dollar intellectual property disputes.
Scott prides himself on delivering the highest quality representation as well as an unsurpassed passion for his clients' ultimate success and satisfaction in every case.
Chris Oetjens is a Baton Rouge native who attended and graduated from Episcopal High School. Chris has merged his knowledge, experience and expertise of the emergency medical field with the skills and knowledge he received in law school to become one of the most aggressive and respected attorneys in the Baton Rouge area.
Chris not only believes, but actually provides, each client the respect, compassion and most importantly, the accessibility each client deserves.
Katherine Guilbeau Guillot was born and raised in Lafayette, Louisiana, and has been serving Lafayette and the surrounding parishes for nearly fifteen years.
Katherine has established a reputation for effectiveness and success as a seasoned and aggressive advocate for each of her clients. She spends individual time with every client to understand the details of their specific case. She then tirelessly works to develop a special plan of action to gain the most favorable possible results.


A set of 9 Easter flashcards for teaching young learners. This set includes 9 cards depicting 6 vocabulary words.
Use these bright pictures to explain a few traditional ways families celebrate the Easter holiday.


text:
Masonic lodges are located throughout the state. Visit their web sites for information about activities.
Rhode Island Lodges
Below is a list of our Lodges, locations, and the days they meet. You don't have to join a lodge in your town of residence. It may be more convenient to join closer to where you work, or one that meets on a day you consistently have free. You can discuss more by calling or visiting when our Lodges are open.
If you're interested in joining the Fraternity, submit an inquiry to our inquiry portal.
Daylight Lodge No.48 - Cranston (Lodge meets during the day) Submit Inquiry
Jenks Lodge No.24 - Pawtucket Submit Inquiry
Redwood Lodge No.35 - East Providence Submit Inquiry
Rising Sun Lodge No.30 - East Providence Submit Inquiry
St. John's Lodge No.1 Newport - Portsmouth Submit Inquiry
Washington Lodge No.3 - Warren Submit Inquiry
Adelphoi Lodge No.33 - Riverside Submit Inquiry
Franklin Lodge No.20 - Westerly Submit Inquiry
Harmony Lodge No.9 - Cranston Submit Inquiry
St. Paul's Lodge No.14 - Middletown Submit Inquiry
Overseas Lodge No.40 - Pawtucket (Military Service / Veterans Only) Submit Inquiry
Hope Lodge No.25 - Wakefield Submit Inquiry
Lafayette Lodge No.47 - Cumberland Submit Inquiry
St. Alban's Lodge No.6 - Bristol Submit Inquiry
St. Johns Lodge No.1 (Providence) - Cranston Submit Inquiry
Charity Lodge No.23 - Charlestown Submit Inquiry
Morning Star Lodge No.13 - Woonsocket Submit Inquiry
Mt. Vernon Lodge No.4 - North Kingstown Submit Inquiry
St. Andrews Lodge No.39 - Riverside Submit Inquiry
Manchester Lodge No.12 - Coventry Inquiry
Mount Moriah Lodge No.8 - Lincoln Submit Inquiry
Friendship Lodge No.7 - Chepachet Submit Inquiry
Ionic Lodge No.28 - Coventry Submit Inquiry
Roger Williams Lodge No.32 - East Providence Submit Inquiry
If you aren't sure what Lodge to consider, submit a general inquiry here.
Upcoming Lodge Meetings
Mount Vernon Lodge No. 4 Themed Table Lodge
Sat 2020 7:30 pm - 9:30 pm
http://rimasons.org/find-a-lodge Annual: Morning Star Lodge No. 13
Mon 2020 7:00 pm - 9:00 pm
http://rimasons.org/find-a-lodge Jenks Lodge No. 24
http://rimasons.org/find-a-lodge Annual: St. Paul's Lodge No. 14
http://rimasons.org/find-a-lodge St John's Lodge No. 1
Tue 2020 6:00 pm - 8:00 pm
http://rimasons.org/find-a-lodge Adelphoi Lodge No. 33
www.stpauls14.org St. Paul's Lodge No. 14
http://harmony9.org/ Harmony Lodge No. 9
Wed 2020 7:30 pm - 9:30 pm
http://rimasons.org/find-a-lodge Lafayette Lodge No. 47
Thu 2020 7:00 pm - 9:00 pm
http://ourgrouponline.org/MtVernonLodgeNo4/ Mount Vernon Lodge No. 4
Masonic Grand Lodge of Rhode Island
222 Taunton Ave., East Providence, RI 02914. P: (401) 435-4650 ~ E: This email address is being protected from spambots. You need JavaScript enabled to view it. Hours: 8:30am to 4:00pm
Copyright © 2019 ~ All Rights Reserved


text:
Allyene is a rural commune located in M'diq-Fnideq Prefecture, Morocco. The settlement has a population of 6126.

References

Rural communes of Tanger-Tetouan-Al Hoceima
Populated places in M'diq-Fnideq Prefecture


At Ross Milroy Realty, LLC ("RossMiami") your privacy is very important to us. This policy covers how RossMiami treats personal information that RossMiami collects and receives when you visit our web site (the "Web Site"). Personal information is information about you that is personally identifiable such as your name, email address and phone number, or that is otherwise not publicly available.
This policy only applies to information collected while you are on or using our Web Site.
We collect personally identifiable information including, but not limited to, name, email address and phone number, that you voluntarily provide to us. For example, when you send us an email or use certain features of the Web Site such as registering for or logging into the "Property Search" or the "Contact Agent" feature, you voluntarily provide certain personal information.
Additionally, during your visit we automatically collect certain information related to your Web Site visit. This information is non-personally identifiable or anonymous information about you, including the date and time of your visit, your IP address, your computer browser information, the Internet address that you visited prior to and after reaching the Web Site, the name of the domain and host you used to access the Internet, and the features or pages of our Web Site which you accessed.
If you request to be contacted by an agent of RossMiami or otherwise request information from us, we will disclose the personally identifiable information you provide to appropriate individual within the company to satisfy your request.
We do not rent or sell personally identifiable information to third parties who are not affiliated with RossMiami. We may disclose personally identifiable information to the appropriate affiliates of RossMiami, including any strategic affiliates and third party service providers, to respond to a request to provide information about available properties or services.
The Web Site may contain links to third party web sites and may redirect you to third party web sites. These include, but is not limited to, third party web sites of service providers to RossMiami.
Third party web sites are not under the control of RossMiami and we are not responsible for any third party web site, including without limitation the accuracy, reliability, completeness, correctness, sufficiency, or timeliness thereof, any link contained therein, or any changes or updates thereto. The inclusion of a link or access to a third party web site does not imply an endorsement thereof, or of the provider of any third party web site.


Extended enterprise training is any training an organization offers to people outside the "four walls" of the enterprise. Such as partners, customers and any other learning audience who is selling or using your product.
No organization is an island. Neither do they function in isolation. Adopting extended enterprise learning and development training ensures you're building, fostering and capitalizing on relationships that are vital to your company's success.
Offering extended enterprise training online pays off. Whether you want to train partners, customers or any other external learning audience, online training is the easiest, most cost-effective and practical way to do it.
Taking part in training will be voluntary for your extended enterprise. But using a powerful and easy-to-use enterprise learning management system like eFront, encourages learner engagement and increases completion rates.
Not only can you create, deploy and manage your extended enterprise training courses exactly the way you want, but eFront also allows you to design your environment to match your branding.
Keep track of your network's training and certifications with ease. And create effective reports with information that is important to you.
With eFront you can deploy separate but centrally managed training programs for different members of your extended enterprise i.e. partner, contractors or customers.
eFront is built to keep up with your hard work. A single eFront eLearning portal can accommodate tens of thousands of learners, regardless of their location. And can expand as your extended enterprise grows.
eFront is no ordinary extended enterprise LMS. Not only does it integrate seamlessly into other organizational infrastructure, but with access to the source code, you can truly make it yours.


The above dictum would hold true only for gravity-fed systems. With forced carbonation, the first draft is of course mostly foam.
simply distraction I've been craving?
a little choker with five decades of red.
lentils listen better, fall in line?
two, three, four at once! I point.


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text:
Battlefield Locations and Other Places »
Visiting The Somme »
Beaumont-Hamel Newfoundland Memorial »
Beaumont-Hamel Memorial Park was establish in memory of Newfoundlanders who fell in the First World War. The Parkcovers an area of 84 acres.
The memorial itself stands at the highest point of the park and consists of a great caribou cast in bronze, emblem of the Royal Newfoundland Regiment. At the base, three tablets of bronze carry the names of over 800 members of the Royal Newfoundland Regiment, the Newfoundland Royal Naval Reserve and the Newfoundland Mercantile Marine, who gave their lives in the First World War and who have no known grave.
Beaumont-Hamel was attacked by the 29th Division on 1 July 1916 and although some units reached it, the village was not taken. It was finally captured by the 51st (Highland) and 63rd (Royal Naval) Divisions on the following 13 November.
The attack on Beaumont-Hamel in July 1916 was the first severe engagement of the regiment, and the most costly. On the first day of the Battle of the Somme, no unit suffered heavier losses than the Newfoundland Regiment which had gone into action 801 strong. The roll call the next day revealed that the final figures were 233 killed or dead of wounds, 386 wounded, and 91 missing. Every officer who went forward in the Newfoundland attack was either killed or wounded. For this reason, the government of Newfoundland chose the hill south-west of the village, where the front-line trenches ran at the time of the battle, as the site of their memorial to the soldiers (and also to the sailors) of Newfoundland.
The memorial was designed by R.H.K. Cochius, with sculpture by Basil Gotto. It was unveiled by Earl Haig on 7 June 1925. The remains of the trenches of both sides remain in the Park.
The "Danger Tree" is the twisted remains of a tree which marked the spot where casualities were heaviest on the first day of battle - 1 July 1916.
GPS Location: Latitude: 50.072797 Longitude: 2,648048
Photographs from my visit to the Beaumont-Hamel Newfoundland Memorial - 29 October 2013
Photographs from my visit to the Beaumont-Hamel Newfoundland Memorial - 27 June 2016


text:
\section{Introduction}
\hspace{1.5em}The theory of free entropy and free entropy dimension
was developed by D. Voiculescu in the 1990's and it is one of the
most powerful and exciting new tools in the theory of von Neumann
algebras. D. Voiculescu \cite{Vo1} \cite{Vo} introduced the concept
of free entropy in relation to his free probability theory and the
concept of free entropy dimension, and he used them to prove that
the free group factors do not contain Cartan subalgebras, which
answered a long-standing open problem. Later this was generalized by
L. Ge \cite{Ge}, who showed that the free group factors do not
contain a simple masa. L. Ge \cite{Ge1} used free entropy to give
the first example of a separable prime II$_1$ factor. Later, L. Ge
and J. Shen \cite{Ge-Shen} computed the free entropy dimension of
some II$_1$ factors with property T, including ${\cal L}(SL({\Bbb
Z},2m+1))\ (m\leq 1)$. Recently, D. Hadwin and J. Shen
\cite{Don-Shen} introduced a new invariant, the upper free
orbit-dimension of a finite von Neumann algebra, which is closely
related to Voiculescu's free entropy dimension. Using their new
invariant, they generalized and simplified the proofs of most of the
applications of free entropy dimension to finite von Neumann
algebras.

Here we introduce a new invariant, $\mathfrak{K}_3$, which is a
modification of the upper free orbit-dimension, $\mathfrak{K}_2$;
when $\mathfrak{K}_2$ is defined, $\mathfrak{K}_3=\infty\cdot
\mathfrak{K}_2$. We then extend the domain of $\mathfrak{K}_3$ to
all finite von Neumann algebras that can be embedded into some
ultrapower of the hyperfinite II$_1$ factors. This includes algebras
acting on nonseparable Hilbert spaces.

 The organization of the paper is as follows. In section 2, we recall the definition of free orbit-dimension,
 and introduce a new invariant $\mathfrak{K}_3$
on von Neumann algebras. In section 3, we prove:

(1)$\mathfrak{K}_3({\mathcal S})=\mathfrak{K}_3({\mathcal G})$ when
$W^*({\cal S})=W^*({\cal G})$ (independence of the generators),

 (2) if ${\mathcal N}_1\cap{\mathcal N}_2$
is diffuse, then $\mathfrak{K}_3\left(({\mathcal N}_1\cup {\mathcal
N}_2)''\right)\leq \mathfrak{K}_3({\mathcal
N}_1)+\mathfrak{K}_3({\mathcal N}_2),$

(3) $\mathfrak{K}_3\left(W^*({\cal N}\cup
\{y\})\right)\leq\mathfrak{K}_3({\mathcal N})$ whenever there exist
normal operators $a$ and $b$ in $\mathcal N$ without common
eigenvalues such that $ay=yb\neq 0$.

 In section 4 we prove:

(1) if $\{{\mathcal{M}}_{\lambda }\}_{\lambda \in
\Lambda }$ is a family of von Neumann algebras such that each $\mathcal{M}%
_{\lambda }$ has a central net of Haar untiaries, and $\alpha $ is a
nontrivial ultrafilter on $\Lambda $, then $
\mathfrak{K}_{3}(\prod^{\alpha }\mathcal{M}_{\lambda })=0, $

(2) if $\mathcal{M}$ is a von Neumann algebra with a central net of
Haar unitaries, then $\mathfrak{K}_{3}({\mathcal{M}})=0$,

(3) if $\mathbb{F}$ is a free group with the standard generating set
$G$ satisfying $|G|\geq 2$, then $\mathfrak{K}_{3}({\mathcal{L}}_{\mathbb{F}%
})=\infty$,

(4) if ${\mathcal N}_1$ and ${\mathcal N}_2$ are mutually commuting
diffuse subalgebras of $\mathcal M$, then $\mathfrak{K}_3(W^*({\cal
N}_1\cup {\cal N}_2))=0$,

(5) if $\cal M$ is a II$_1$ factor and $\mathfrak{K}_3({\cal
M})=\infty$, then $\cal M$ is prime (i.e., cannot be written as a
tensor product of two II$_1$ factors).

In section 5, we show how our invariant leads naturally to a
canonical decomposition of torsion-free groups into a union of
certain self-normalizing subgroups so that the intersection of any
two of them is $\{e\}$. We completely describe this decomposition
for free groups, and we present a related question for the free
group factor ${\cal L}_{{\Bbb F}_2}$.

All of the free entropy concepts require the von Neumann algebra
$\cal M$ under consideration can be tracially embedded into an
ultrapower of the hyperfinite II$_1$ factor. Throughout this paper,
we assume that all the von Neumann algebras we consider can be
embedded.
\section{Preliminaries}

\hspace{1.5em} First we recall the definition of $\mathfrak{K}_2$
introduced by D. Hadwin and J. Shen \cite{Don-Shen}, then we
introduce our new invariant $\mathfrak{K}_3$.

Let ${\cal M}_k({\Bbb C})$ be the $k\times k$ full matrix algebra
with entries in $\Bbb
 C$, and $\tau_k$ be the normalized
 trace on ${\cal M}_k({\Bbb C})$, i.e., $\tau_k=\frac{1}{k}Tr$,
 where $Tr$ is the usual trace on ${\cal M}_k({\Bbb C})$. Let
 ${\mathcal U}_k$ be the group of all unitary matrices in ${\cal M}_k(\Bbb
 C)$ and ${\cal M}_k({\Bbb C})^n$ denote the direct sum of $n$
 copies of ${\cal M}_k({\Bbb C})$. Define $\|\cdot\|_2$ on ${\cal M}_k(\Bbb
 C)^n$ by
 $$\|(A_1,\ldots,A_n)\|^2_2=\tau_k(A_1^*A_1)+\ldots+\tau_k(A_n^*A_n)$$for
 all $(A_1,\ldots,A_n)$ in ${\cal M}_k({\Bbb C})^n$.

 For every $\omega>0$, define the {\it $\omega$-ball}
 $Ball(B_1,\ldots,B_n;\omega)$ in ${\cal M}_k({\Bbb C})^n$ to be the subset of ${\cal M}_k(\Bbb
 C)^n$ consisting of all $(A_1,\ldots,A_n)$ in ${\cal M}_k(\Bbb
 C)^n$ such that $$\|(A_1,\ldots,A_n)-(B_1,\ldots,B_n)\|_2<\omega.$$
 Define the {\it $\omega$-orbit-ball} ${\mathcal
 U}(B_1,\ldots,B_n;\omega)$ in ${\cal M}_k({\Bbb C})^n$ to be the subset of ${\cal M}_k(\Bbb
 C)^n$ consisting of all $(A_1,\ldots,A_n)$ in ${\cal M}_k(\Bbb
 C)^n$ such that there exists some unitary matrix $W$ in ${\mathcal U}_k$ satisfying
  $$\|(A_1,\ldots,A_n)-(WB_1W^*,\ldots,WB_nW^*)\|_2<\omega.$$

Suppose $E\subseteq {\cal M}_k({\Bbb C})^n$, $\omega>0$. Define the
{\it covering number} $\nu_2(E,\omega)$ to be the minimal number of
$\omega$-balls that cover $E$ with the centers of these
$\omega$-balls in $E$; define the {\it $\omega$-orbit covering
number} $\nu(E,\omega)$ to be the minimal number of
$\omega$-orbit-balls that cover $E$ with the centers of these
$\omega$-orbit-balls in $E$.

Let $\mathcal M$ be a von Neumann algebra with a tracial state
$\tau$ and $x_1, x_2, \ldots, x_n$ be elements in ${\mathcal M}$.
For any $R,\varepsilon>0$, and positive integers $m$ and $k$, define
$\Gamma_R(x_1,\ldots,x_n;m,k,\varepsilon)$ to be the subset of
${\cal M}_k({\Bbb C})^n$ consisting of all $(A_1,\ldots,A_n)$ in
${\cal M}_k({\Bbb C})^n$ such that $\|A_j\|\leq R$ for $1\leq j\leq
n$, and
$$|\tau_k(A_{i_1}^{\eta_1}\cdots
A_{i_q}^{\eta_q})-\tau(x_{i_1}^{\eta_1}\cdots
x_{i_q}^{\eta_q})|<\varepsilon,$$ for all $1\leq i_1,\ldots,i_q\leq
n$, all $\eta_1,\ldots,\eta_q\in\{1,*\}$ and all $q$ with $1\leq
q\leq m$.

Define
$$\mathfrak{K}(x_1,\ldots,x_n;m,\varepsilon, \omega,R)=
\limsup_{k\rightarrow\infty}
\frac{\log(\nu(\Gamma_R(x_1,\ldots,x_n;m,k,\varepsilon),\omega))}{-k^2\log\omega}$$
$$\mathfrak{K}(x_1,\ldots,x_n;\omega,R)=
\inf_{m\in\Bbb N,\varepsilon>0}
\mathfrak{K}(x_1,\ldots,x_n;m,\varepsilon, \omega,R)$$
$$\mathfrak{K}(x_1,\ldots,x_n;\omega)=\sup_{R>0}\mathfrak{K}(x_1,\ldots,x_n;\omega,R)$$
$$\mathfrak{K}_2(x_1,\ldots,x_n)=\sup_{0< \omega< 1}\mathfrak{K}(x_1,\ldots,x_n;\omega)$$

D. Hadwin and J. Shen \cite{Don-Shen} also defined
$\mathfrak{K}_2(x_1,\ldots,x_n:y_1,\ldots,y_p)$ for all $x_1,
\ldots, x_n$, $y_1, \ldots, y_p$ in the von Neumann algebra $\cal M$
as follows. Let
\begin{eqnarray*}&&\Gamma_R(x_1,\ldots,x_n:y_1,\ldots,y_p;m,k,\varepsilon)\\
&=&\left\{(A_1,\ldots,A_n)\in {\cal M}_k({\Bbb C})^n:\ \mbox{there
exist}\ B_1,\ldots,B_p\ \mbox{in}\ {\cal M}_k({\Bbb C})\right.\\
&&\left.\mbox{such that}\
 (A_1,\ldots,A_n,B_1,\ldots,B_p)\in \Gamma_R(x_1,\ldots,x_n,y_1,\ldots,y_p;m,k,\varepsilon)
\right\},\end{eqnarray*}
$$\mathfrak{K}(x_1,\ldots,x_n:y_1,\ldots,y_p;m,\varepsilon,\omega,R)=
\limsup_{k\rightarrow\infty}
\frac{\log(\nu(\Gamma_R(x_1,\ldots,x_n;m,k,\varepsilon),\omega))}{-k^2\log\omega}$$
$$\mathfrak{K}(x_1,\ldots,x_n:y_1,\ldots,y_p;\omega,R)=
\inf_{m\in\Bbb N,\varepsilon>0}
\mathfrak{K}(x_1,\ldots,x_n:y_1,\ldots,y_p;m,\varepsilon,\omega,R)$$
$$\mathfrak{K}(x_1,\ldots,x_n:y_1,\ldots,y_p;\omega)=\sup_{R>0}\mathfrak{K}(x_1,\ldots,x_n;\omega,R)$$
$$\mathfrak{K}_2(x_1,\ldots,x_n:y_1,\ldots,y_p)=\sup_{0< \omega< 1}\mathfrak{K}(x_1,\ldots,x_n;\omega)$$

\begin{remark}\label{remark,x:y>x:y,z}From the definition, it is clear that

(1) $\mathfrak{K}_2(x_1,\ldots,x_n:y_1,\ldots,y_p)\geq
\mathfrak{K}_2(x_1,\ldots,x_n:y_1,\ldots,y_p,y_{p+1}),$

(2) if $\mathfrak{K}_2(x_1,\ldots,x_n:x_1,\ldots,x_{n+j})=0$ ($j\geq
0 $), then
$$\mathfrak{K}_2(x_1,\ldots,x_{n-1}:x_1,\ldots,x_{n+j})=0.$$
\end{remark}

Let $\infty\cdot 0=0$. For any subset $\mathcal G$ of $\mathcal M$,
define
$$\mathfrak{K}_3(x_1,\ldots,x_n:{\mathcal
G})=\inf\left\{\infty\cdot\mathfrak{K}_2(x_1,\ldots,x_n:A):\ A\
\mbox{is a finite subset of\ }{\mathcal G}\right\},$$
$$\mathfrak{K}_3({\mathcal
G})=\sup_{\begin{array}{l}E\subseteq {\mathcal G}\\
E\ \mbox{is finte}\end{array}}\inf_{\begin{array}{l}F\subseteq {\mathcal G}\\
F\ \mbox{is finte}\end{array}}\infty\cdot\mathfrak{K}_2(E:F).$$

 When
$\mathcal G$ is finite, it is not difficult to see that
$$\mathfrak{K}_3(x_1,\ldots,x_n:{\mathcal
G})=\infty\cdot\mathfrak{K}_2(x_1,\ldots,x_n:{\mathcal G})$$ and
$$\mathfrak{K}_3({\mathcal G})=\infty\cdot\mathfrak{K}_2({\mathcal
G}).$$

Note that the value of
$\mathfrak{K}_3(x_1,\ldots,x_n:y_1,\ldots,y_p)$ or
$\mathfrak{K}_3(x_1,\ldots,x_n)$ is always 0 or $\infty.$

\section{Key properties of $\mathfrak{K}_3$}
\begin{theorem}\label{equivalence}
If $\mathcal M$ is a von Neumann algebra with a tracial state
$\tau$, then the following are equivalent:

(1) $\mathfrak{K}_3({\mathcal M})=0$;

(2) if $x_1,\ldots,x_n\in\mathcal M$, then there exist
$y_1,\ldots,y_t\in\mathcal M$ such that
$\mathfrak{K}_2(x_1,\ldots,x_n:y_1,\ldots,y_t)=0$;

(3) for any generating set $\mathcal G$ of $\mathcal M$,
$\mathfrak{K}_3({\mathcal G})=0$;

(4) there exists a generating set $\mathcal G$ of $\mathcal M$ such
that $\mathfrak{K}_3({\mathcal G})=0$;

(5) if $\mathcal G$ is a generating set of $\mathcal M$, and $A_0$
is a finite subset of $\mathcal G$, then, for any finite subset $A$
with $A_0\subseteq A\subseteq \mathcal G$, there exists a finite
subset $B$ of $\mathcal G$ so that $\mathfrak{K}_2(A:B)=0$;

(6) there is an increasing directed family $\{{\mathcal
M}_\iota:\iota\in \Lambda\}$ of von Neumann subalgebras of $\mathcal
M$ such that

\ \ (a) each ${\mathcal M}_\iota$ is countably generated,

\ \ (b) $\mathfrak{K}_3({\mathcal M}_\iota)=0$,

\ \ (c) ${\mathcal M}=\cup_{\iota\in \Lambda}{\mathcal M}_{\iota}$.
\end{theorem}

Proof. It is clear that (1)$\Leftrightarrow$(2), (3)$\Rightarrow$(4)
and (3)$\Rightarrow$(5).

(4)$\Rightarrow$(2) Suppose $\cal G$ is a generating set of $\cal M$
and $\mathfrak{K}_3({\cal G})=0$. Let $\omega>0$ and $x_1, \ldots,
x_n$ be any elements in ${\cal M}$. Then there exist polynomials
$p_1, \ldots, p_n$ and elements $y_1,\ldots, y_s$ in $\cal G$ such
that
$$\|(x_1, \ldots, x_n)-(p_1(y_1,\ldots, y_s),\ldots, p_n(y_1,\ldots, y_s))\|\leq
\frac{\omega}{4}.$$

For any given $R>0$, if $(A_1, \ldots, A_s), (B_1, \ldots, B_s)$ in
${\cal M}_k({\Bbb C})^s$ and $\|A_j\|,\|B_j\|\leq R$ for all $1\leq
j\leq s$, then there exists a positive integer $N$ such that
\begin{eqnarray*}&&\|(p_1(A_1, \ldots, A_s), \ldots, p_n(A_1, \ldots,
A_s))-(p_1(B_1, \ldots, B_s),\ldots, p_n(B_1, \ldots, B_s))\|_2\\
& \leq & N\|(A_1, \ldots, A_s)-(B_1, \ldots,
B_s)\|_2.\end{eqnarray*}

Since $\mathfrak{K}_3({\cal G})=0$ and $y_1, \ldots, y_m$ are in
$\cal G$, there exist $y_1, \ldots, y_t$ ($t\geq s$) in $\cal G$
such that
$$\mathfrak{K}_2(y_1,\ldots, y_s:y_1, \ldots, y_t)=0.$$

Let $R> \max\{\|y_i\|, \|x_j\|: 1\leq i\leq n, 1\leq j\leq t\}$.
Note that if $(A_1,\ldots, A_n,B_1, \ldots, B_t)\in \Gamma_R(x_1,
\ldots, x_n,y_1, \ldots, y_t;m,k,\varepsilon)$, then, for
sufficiently small $\varepsilon$ and sufficiently large $m$, we have
$$\|(A_1, \ldots, A_n)-(p_1(B_1,\ldots, B_s),\ldots, p_n(B_1,\ldots, B_s))\|_2<\frac{\omega}{4}$$
and $$(B_1, \ldots, B_s)\in\Gamma(y_1, \ldots, y_s:y_1, \ldots,
y_t;m,k,\varepsilon).$$ It follows that there is a set $\Lambda$ and
a subset $\{(B_1^{\lambda}, \ldots,
B_s^{\lambda}):\lambda\in\Lambda\}$ of $\Gamma_R(y_1, \ldots,
y_s:y_1, \ldots, y_t;m,k,\varepsilon)$ with
$$\mbox{card}(\Lambda)\leq \nu(\Gamma(y_1, \ldots, y_s:y_1, \ldots,
y_t;m,k,\varepsilon),\frac{\omega}{4N}).$$ That means, for every
$(A_1, \ldots, A_n,B_1, \ldots, B_t)\in \Gamma_R (x_1, \ldots,
x_n,y_1, \ldots, y_t;m,k,\varepsilon)$, there is a
$\lambda\in\Lambda$ and a unitary $k\times k$ matrix $U$ such that
$$\|(B_1, \ldots, B_s)-U^*(B_1^{\lambda}, \ldots,
B_s^{\lambda})U\|_2\leq \frac{\omega}{4N}.$$ That gives
$$\|(A_1, \ldots A_n)-U^*(p_1(B_1^{\lambda}, \ldots, B_s^{\lambda}),\ldots, p_n(B_1^{\lambda}, \ldots, B_s^{\lambda}))U\|_2<\frac{\omega}{2}.$$

It follows that, if $\varepsilon$ is sufficient small and $m$ is
sufficient large, then for any $k\in\Bbb N$,
\begin{eqnarray*}&&\frac{\nu(\Gamma_R(x_1, \ldots, x_n:y_1, \ldots, y_t;m,k,\varepsilon),\omega)}{-k^2\log \omega}\\
&\leq&
\left(\frac{\log\omega}{\log(\omega/4N)}\right)\frac{\nu(\Gamma(y_1,
\ldots, y_s:y_1, \ldots,
y_t;m,k,\varepsilon),\omega/4N)}{-k^2\log(\omega/4N)}.\end{eqnarray*}
If we take $k\rightarrow \infty$ and take the infimum over $m$ and
$\varepsilon$, we get
\begin{eqnarray*}&&\mathfrak{K}_2(x_1, \ldots, x_n:y_1, \ldots, y_t;\omega)\\
&\leq&
\left(\frac{\log\omega}{\log(\omega/4N)}\right)\mathfrak{K}_2(y_1,
\ldots, y_s:y_1, \ldots, y_t)\\
&=&0.\end{eqnarray*}

Hence, we have $\mathfrak{K}_2(x_1, \ldots, x_n: y_1, \ldots,
y_t)=0$.

(2)$\Rightarrow$(3)\ \ Suppose (2) is true, and suppose
$x_1,\ldots,x_n$ are elements of some generating set $\mathcal G$ of
$\mathcal M$. Then there exist $y_1,\ldots,y_t$ in $\mathcal M$ such
that $\mathfrak{K}_2(x_1,\ldots,x_n:y_1,\ldots,y_t)=0$.

Suppose $\varepsilon_0>0$, $m_0\in \Bbb N$, and $0< \omega< 1$. We
can choose $w_1,\ldots,w_s$ in $\mathcal G$ and polynomials
$p_1,\ldots,p_t$ so that each $\|y_j-p_j(w_1,\ldots, w_s)\|_2$
($1\leq j\leq t$) is small enough to make
$$\left|\tau(q(x_1,\ldots,x_n,y_1,\ldots,y_t))-
\tau(q(x_1,\ldots,x_n,p_1(w_1,\ldots,w_s),\ldots,p_t(w_1,\ldots,w_s)))\right|<\frac{\varepsilon_0}{4},$$
for every monomial $q$ with length at most $m_0$.

When $m$ is sufficient large, $\varepsilon$ is sufficient small,
if
$$(A_1,\ldots,A_n,B_1,\ldots,B_s)\in\Gamma_R(x_1,\ldots,x_n,w_1,\ldots,w_s;m,k,\varepsilon),$$
 then $$(A_1,\ldots,A_n,p_1(B_1,\ldots,B_s),\ldots,p_t(B_1,\ldots,B_s))\in
 \Gamma_R(x_1,\ldots,x_n,y_1,\ldots,y_t;m_0,k,\varepsilon_0).$$
Hence
$$\Gamma_R(x_1,\ldots,x_n:w_1,\ldots,w_s;m,k,\varepsilon)\subseteq
\Gamma_R(x_1,\ldots,x_n:y_1,\ldots,y_t;m_0,k,\varepsilon_0).$$ Since
$$\nu(\Gamma_R(x_1,\ldots,x_n:w_1,\ldots,w_s;m,k,\varepsilon),\omega)\leq
2\nu(\Gamma_R(x_1,\ldots,x_n:y_1,\ldots,y_t;m_0,k,\varepsilon_0),\omega),$$
we have,
$$\mathfrak{K}(x_1,\ldots,x_n:w_1,\ldots,w_s;\omega)\leq
2\mathfrak{K}(x_1,\ldots,x_n:y_1,\ldots,y_t;m_0,\varepsilon_0,\omega).$$
Then we get
$$\mathfrak{K}(x_1,\ldots,x_n:w_1,\ldots,w_s;\omega)\leq
2\mathfrak{K}(x_1,\ldots,x_n:y_1,\ldots,y_t;\omega).$$ Therefore
$$\mathfrak{K}_2(x_1,\ldots,x_n:w_1,\ldots,w_s)\leq 2\mathfrak{K}_2(x_1,\ldots,x_n:y_1,\ldots,y_t)=0.
$$Thus
$\mathfrak{K}_2(x_1,\ldots,x_n:w_1,\ldots,w_s)=0$. From the
definition, $\mathfrak{K}_3({\mathcal G})=0$.

(5)$\Rightarrow$(6)\ \ Suppose $A_0\subseteq
\{x_1,\ldots,x_n\}\subseteq \mathcal G$. From (4), there exists a
family $\{B_0, B_1, \ldots\}$ of finite subsets of $\mathcal G$ such
that $$\mathfrak{K}_2(x_1, \ldots, x_n: B_0)=0,$$ and for any
positive integer $n$, $$\mathfrak{K}_2(x_1, \ldots, x_n, B_0,
\ldots, B_{n-1}: B_n)=0.$$ Let $\cal G$ be the set $\{x_1,\ldots,
x_n\}\cup\cup_{n=0}^{\infty}B_n$ and $\cal N$ be the von Neumann
subalgebra generated by $\cal G$. Then $\cal N$ is countably
generated.

Let $A\subseteq \cal G$ be a finite subset. Then there exists a
positive integer $m$, so that $A\subseteq \{x_1, \ldots, x_n\}\cup
B_0\cup\cdots \cup B_m$. Since
$$\mathfrak{K}_2(x_1, \ldots, x_n, B_0, \ldots, B_{m}: B_{m+1})=0,$$
by Remark \ref{remark,x:y>x:y,z}, we have
$$\mathfrak{K}_2(A: B_n)=0.$$It follows that
$\mathfrak{K}_3\left({\cal G}\right)=0.$ Therefore
$\mathfrak{K}_3({\cal N})=0$ by the equivalence of (1) and (4).

It is not difficult to see that the union of all such $\mathcal N$'s
is $\mathcal M$.

(6)$\Rightarrow$(2)\ \ Suppose $x_1,\ldots,x_n$ are elements of
$\mathcal M$. From (5), there exists $\{\iota_1,\ldots,
\iota_n\}\subseteq \Lambda$ such that $x_1\in {\mathcal
M}_{\iota_1},\ldots, x_n\in{\mathcal M}_{\iota_n}$. Since
$\{{\mathcal M}_\iota:\iota\in \Lambda\}$ is an increasing directed
family, there exists $\iota\in\Lambda$, such that ${\mathcal
M}_{\iota_1},\ldots,{\mathcal M}_{\iota_n}\subseteq {\cal
M}_{\iota}$ and $\mathfrak{K}_3({\cal M}_{\iota})=0$. Therefore
there exist $y_1,\ldots,y_m\in {\mathcal M}_{\iota}\subseteq
\mathcal M$ such that $\mathfrak{K}_2(x_1,\ldots, x_n:y_1,\ldots,
y_m)=0$.\hfill $\Box$

\begin{remark}If $\cal M$ is finite generated, then $\mathfrak{K}_3({\cal M})=0$ is equivalent to $\mathfrak{K}_2({\cal M})=0$.
\end{remark}

\begin{corollary}\label{corollary, k3 is independent on generatoer}Suppose $\mathcal M$
 is a von Neumann algebra with a tracial state $\tau$, $\mathcal G$ is
a generating set of $\mathcal M$. Then $\mathfrak{K}_3({\mathcal M
})=\mathfrak{K}_3({\mathcal G})$.
\end{corollary}

\begin{corollary}\label{corollary, increasingly directed}Suppose $\{{\mathcal M}_{\iota}\}_{\iota\in\Lambda}$
is an increasingly directed family of von Neumann algebras. Then
$\mathfrak{K}_3(\cup_{\iota}{\mathcal M}_{\iota})\leq
\liminf_{\iota}\mathfrak{K}_3({\mathcal M}_{\iota}).$
\end{corollary}

\begin{remark}To see that Corollary \ref{corollary, increasingly directed} gives the best estimate, note that ${\mathcal L}_{{\Bbb F}_2}\otimes
{\mathcal R}=\cup_n\left( {\mathcal L}_{{\Bbb F}_2}\otimes{\mathcal
M}_{2^n}({\Bbb C})\right)$ and $\mathfrak{K}_3({\mathcal L}_{{\Bbb
F}_2}\otimes {\mathcal R})=0$, but $\mathfrak{K}_3({\mathcal
L}_{{\Bbb F}_2}\otimes {\mathcal M}_{2^n}(\Bbb C))=\infty$ for every
$n$.
\end{remark}

\vspace{0.2cm}

To prove Theorem \ref{theorem, intersection contains Haar unitary},
we need the following lemmas.
\begin{lemma}\label{add generator is the same} Let $\mathcal M$ be a von Neumann algebra with a tracial state $\tau$. Suppose
$x_1$, $\ldots$, $x_n$, $y_1$, $\ldots$, $y_p$, $w_1,\ldots,w_t$ are
elements of $\mathcal M$ and $x_1,\ldots,x_n\in
W^*(y_1,\ldots,y_p)$. Then, for $\omega>0$,
$$\mathfrak{K}(y_1,\ldots,y_p:w_1,\ldots,w_t;\omega)=\mathfrak{K}(y_1,\ldots,y_p:x_1,\ldots,x_n,w_1,\ldots,w_t;\omega).$$
\end{lemma}

Proof. It is not hard to get the ``$\geq$" part.

Assume $\varepsilon_0>0$, $m_0\in\Bbb N$, $R>1$. Since
$x_1,\ldots,x_n\in W^*(y_1,\ldots,y_p)$, there exist $m_1\in \Bbb N$
and $\varepsilon_1>0$ and a family of noncommutative polynomials
$q_1,\ldots, q_n$ such that
\newline $\|\left(q_1(y_1,\ldots,y_p),\ldots,
q_n(y_1,\ldots,y_p)\right)-(x_1,\ldots,x_n)\|_2$ is so small that
for any $m\geq m_1$ and $0<\varepsilon\leq \varepsilon_1$, we have,
for any $k\in\Bbb N$,
\begin{eqnarray*}&&\{(A_1,\ldots,A_p,q_1(A_1,\ldots,A_p),\ldots, q_n(A_1,\ldots,A_p),C_1,\ldots,C_t):\\
&&(A_1,\ldots,A_p,C_1,\ldots,C_t)\in\Gamma_R(y_1,\ldots,y_p,w_1,\ldots,w_t;m,k,\varepsilon)\}\\
&\subseteq&
\Gamma_R(y_1,\ldots,y_p,x_1,\ldots,x_n,w_1,\ldots,w_t;m_0,k,\varepsilon_0),
\end{eqnarray*}
which implies
$$\Gamma_R(y_1,\ldots,y_p:w_1,\ldots,w_t;m,k,\varepsilon)\subseteq
\Gamma_R(y_1,\ldots,y_p:x_1,\ldots,x_n,w_1,\ldots,w_t;m_0,k,\varepsilon_0)
.$$ Therefore
$\mathfrak{K}(y_1,\ldots,y_p:w_1,\ldots,w_t;\omega)\leq\mathfrak{K}(y_1,\ldots,y_p:x_1,\ldots,x_n,w_1,\ldots,w_t;\omega)$.\hfill
$\Box$

\vspace{0.2cm}

The following lemma is a slight extension of Theorem 1 in
\cite{Don-Shen}; the proofs are similar.
\begin{lemma}\label{change generator is the same} Let
$x_1,\ldots,x_n,y_1,\ldots,y_p,w_1,\ldots,w_t$ be elements in a von
Neumann algebra $\mathcal M$ with a tracial state $\tau$, and
$W^*(x_1,\ldots,x_n)=W^*(y_1,\ldots,y_p)$. Then
$$\mathfrak{K}_3(x_1,\ldots,x_n:w_1,\ldots,w_t)=\mathfrak{K}_3(y_1,\ldots,y_p:w_1,\ldots,w_t).$$
\end{lemma}

Proof. If $\mathfrak{K}_3(x_1,\ldots,x_n:w_1,\ldots,w_t)$ and
$\mathfrak{K}_3(y_1,\ldots,y_p:w_1,\ldots,w_t)$ are both infinity,
then they are equal. If one of them is zero, say it
$\mathfrak{K}_3(x_1,\ldots,x_n:w_1,\ldots,w_t)$, then we need to
prove that $\mathfrak{K}_3(y_1,\ldots,y_p:w_1,\ldots,w_t)=0$.

For every $0<\omega<1$, there exists a family of noncommutative
polynomials $q_1,\ldots, q_p$, such that
$$\|(y_1,\ldots,y_p)-\left(q_1(x_1,\ldots,x_n),\ldots,q_p(x_1,\ldots,x_n\right))\|_2\leq\frac{\omega}{4}.$$
For such a family of polynomials $q_1,\ldots,q_p$ and every $R>0$,
there always exists a constant $D\geq 1$, depending only on
$q_1,\ldots,q_p$ and $R$, such that
\begin{eqnarray*}&&\|\left(q_1(A_1,\ldots,A_n),\ldots,q_p(A_1,\ldots,A_n)\right)-\left(q_1(B_1,\ldots,B_n),\ldots,q_p(B_1,\ldots,B_n)\right)\|_2\\
&\leq& D\|(A_1,\ldots,A_n)-(B_1,\ldots,B_n)\|_2,\end{eqnarray*} for
all $A_1,\ldots,A_n,B_1,\ldots,B_n$ in ${\mathcal M}_k({\Bbb C})$
satisfying $\|A_j\|, \|B_j\|\leq R$, for $1\leq j\leq n$.

For $R>1$, $m$ and $k$ sufficiently large, $\varepsilon$
sufficiently small, if
$$(B_1,\ldots,B_p,A_1,\ldots,A_n)\in
\Gamma_R(y_1,\ldots,y_p,
x_1,\ldots,x_n:w_1,\ldots,w_t;m,k,\varepsilon),$$ then
$$\|(B_1,\ldots,B_p)-\left(q_1(A_1,\ldots,A_n),\ldots,q_p(A_1,\ldots,A_n)\right)\|_2\leq \frac{\omega}{4}.$$
It is clear that $(A_1,\ldots,A_n)\in \Gamma_R(
x_1,\ldots,x_n:w_1,\ldots,w_t;m,k,\varepsilon)$.

There exists a set $\{\mathcal
U(A_1^{\lambda},\ldots,A_n^{\lambda};\frac{\omega}{4D})\}_{\lambda\in\Lambda_k}$
of $\frac{\omega}{4D}$-orbit-balls that cover $\Gamma_R(
x_1,\ldots,x_n:w_1,\ldots,w_t;\\ m,k,\varepsilon)$ with the
cardinality of $\Lambda_k$ satisfying $|\Lambda_k|=\nu(\Gamma_R(
x_1,\ldots,x_n:w_1,\ldots,w_t;m,k,\varepsilon),\frac{\omega}{4D})$.
Thus there exists some $\lambda\in\Lambda_k$ and $U\in {\mathcal
U}_k$ such that
$$\|(A_1,\ldots,A_n)-U(A_1^{\lambda},\ldots,A_n^{\lambda})U^*\|_2\leq
\frac{\omega}{4D}.$$ It follows that
\begin{eqnarray*}&&\|(B_1,\ldots,B_p)-U\left(q_1(A_1^{\lambda},\ldots,A_n^{\lambda}),\ldots,q_p(A_1^{\lambda},\ldots,A_n^{\lambda})\right)U^*\|_2\\
&=&\|(B_1,\ldots,B_p)-\left(q_1(U(A_1^{\lambda},\ldots,A_n^{\lambda})U^*),\ldots,q_p(U(A_1^{\lambda},\ldots,A_n^{\lambda})U^*)\right)\|_2\\
&\leq & \frac{\omega}{2}.\end{eqnarray*} That is,
$$(B_1,\ldots,B_p)\in{\mathcal
U}\left(q_1(A_1^{\lambda},\ldots,A_n^{\lambda}),\ldots,q_p(A_1^{\lambda},\ldots,A_n^{\lambda});\omega\right).$$
Hence, we get \begin{eqnarray*} 0&\leq &\mathfrak{K}(y_1,\ldots,y_p:x_1,\ldots,x_n,w_1,\ldots,w_t;\omega, R)\\
&\leq & \inf_{m\in\Bbb N,\varepsilon>0}\limsup_{k\rightarrow \infty}\frac{\log(|\Lambda_k|)}{-k^2\log\omega}\\
&=&\inf_{m\in\Bbb N,\varepsilon>0}\limsup_{k\rightarrow
\infty}\frac{\log(\nu(\Gamma_R(
x_1,\ldots,x_n:w_1,\ldots,w_t;m,k,\varepsilon),\frac{\omega}{4D}))}{-k^2\log\omega}\\
&=&0.
\end{eqnarray*}
Therefore
$\mathfrak{K}_2(y_1,\ldots,y_p:x_1,\ldots,x_n,w_1,\ldots,w_t)=0$.
From Lemma \ref{add generator is the same}, we get
$\mathfrak{K}_2(y_1,\ldots,y_p:w_1,\ldots,w_t)=0$. So
$\mathfrak{K}_3(y_1,\ldots,y_p:w_1,\ldots,w_t)=0$.\hfill $\Box$

\begin{definition}A unitary matrix $A$ in ${\mathcal M}_k({\Bbb C})$
is called a Haar unitary matrix if the eigenvalues of $A$ are the
$k$-th roots of unity; equivalently, if $\tau_k(A^i)=0$ for $1\leq
i<k$ and $\tau_k(A^k)=1$.
\end{definition}

\begin{lemma}[\cite{Vo}]\label{Vo} Let $V_1,V_2$ be two Haar unitary matrices in ${\mathcal M}_k(\Bbb
C)$. For every $\delta>0$, let $$\Omega(V_1,V_2;\delta)=\{U\in
{\mathcal U}_k: \|UV_1-V_2U\|_2\leq \delta\}.$$ Then, for every
$0<\delta<r$, $\nu_2(\Omega(V_1,V_2;\delta),\frac{4\delta}{r})\leq
(\frac{3r}{2\delta})^{4rk^2}$.
\end{lemma}

\begin{definition}Suppose $\mathcal M$ is a von Neumann algebra
 with a tracial state $\tau$. Then a unitary $u$ in $\mathcal M$
 is called a Haar unitary if $\tau(u^m)=0$ when $m\neq 0$. In
 addition, $\cal M$ is called diffuse if $\cal M$ contains a Haar
 unitary.
\end{definition}

The following lemma is a slight extension of Theorem 6 in
\cite{Don-Shen}, and the proofs are similar.
\begin{lemma}\label{haar unitary} Let $x_1,\ldots,x_n, y_1,\ldots,y_p, v_1,\ldots,v_s, w_1,\ldots,w_t$
 be elements in a von Neumann algebra $\mathcal M$ with a
tracial state $\tau$. If $W^*(x_1,\ldots,x_n)\cap
W^*(y_1,\ldots,y_p)$ is diffuse, then
\begin{eqnarray*}&&\mathfrak{K}_3(x_1,\ldots,x_n,y_1,\ldots,y_p:v_1,\ldots,v_s,w_1,\ldots,w_t)\\
&\leq & \mathfrak{K}_3(x_1,\ldots,x_n:v_1,\ldots,v_s)+
\mathfrak{K}_3(y_1,\ldots,y_p:w_1,\ldots,w_t). \end{eqnarray*}
\end{lemma}

Proof. If one of $\mathfrak{K}_3(x_1,\ldots,x_n:v_1,\ldots,v_s)$ and
$\mathfrak{K}_3(y_1,\ldots,y_p:w_1,\ldots,w_t)$ is infinity, then we
are done.

Now suppose
$\mathfrak{K}_3(x_1,\ldots,x_n:v_1,\ldots,v_s)=\mathfrak{K}_3(y_1,\ldots,y_p:w_1,\ldots,w_t)=0$.
Since $W^*(x_1,\ldots,x_n)\cap W^*(y_1,\ldots,y_p)$ is diffuse, we
can find a Haar unitary $u$ in $W^*(x_1,\ldots,x_n)\cap
W^*(y_1,\ldots,y_p)$.

For $R>1+\max_{1\leq i\leq n, 1\leq j\leq p}\{\|x_i\|,\|y_j\|\}$,
$0<\omega<\frac{1}{2n}$, $0<r<1$ and $\varepsilon>0$, $m,k\in\Bbb
N$. Suppose
$$(A_1,\ldots,A_n, B_1,\ldots,B_p,U)\in\Gamma_R(x_1,\ldots,x_n,
y_1,\ldots,y_p,u: v_1,\ldots,v_s, w_1,\ldots,w_t;m,k,\varepsilon).$$
Then
$$(A_1,\ldots,A_n,U)\in\Gamma_R(x_1,\ldots,x_n,u:v_1,\ldots,v_s,w_1,\ldots,w_t;m,k,\varepsilon)$$
and
$$(B_1,\ldots,B_p,U)\in\Gamma_R(y_1,\ldots,y_p,u:v_1,\ldots,v_s,w_1,\ldots,w_t;m,k,\varepsilon).$$

Let $\{{\mathcal
U(A_1^{\lambda},\ldots,A_n^{\lambda},U^{\lambda})};\frac{r\omega}{24R}\}_{\lambda\in\Lambda_k}$
be a set of $\frac{r\omega}{24R}$-orbit-balls that cover
$\Gamma_R(x_1,\ldots,x_n,u: v_1,\ldots,v_s,\\
w_1,\ldots,w_t;m,k,\varepsilon)$ with the cardinality of $\Lambda_k$
satisfying
$$|\Lambda_k|=\nu(\Gamma_R(x_1,\ldots,x_n,u:
v_1,\ldots,v_s,w_1,\ldots,w_t;m,k,\varepsilon);\frac{r\omega}{24R}).$$
Also let $\{{\mathcal
U(B_1^{\sigma},\ldots,B_p^{\sigma},U^{\sigma})};\frac{r\omega}{24R}\}_{\lambda\in\Sigma_k}$
be a set of $\frac{r\omega}{24R}$-orbit-balls that cover
$\Gamma_R(y_1,\ldots,y_p,u:v_1,\ldots,v_s,\\
w_1,\ldots,w_t;m,k,\varepsilon)$ with the cardinality of $\Sigma_k$
satisfying
$$|\Sigma_k|=\nu(\Gamma_R(y_1,\ldots,y_p,u:v_1,\ldots,v_s,w_1,\ldots,w_t;m,k,\varepsilon);\frac{r\omega}{24R}).$$
When $m$ is sufficiently large and $\varepsilon$ is sufficiently
small, by Theorem 2.1 in \cite{Don1}, we can assume that all
$U^{\lambda}, U^{\sigma}$ to be Haar unitary matrices in ${\cal
M}_k(\Bbb C)$.

For any $$(A_1,\ldots,A_n,
B_1,\ldots,B_p,U)\in\Gamma_R(x_1,\ldots,x_n, y_1,\ldots,y_p,u:
v_1,\ldots,v_s, w_1,\ldots,w_t;m,k,\varepsilon),$$ there exist some
$\lambda\in \Lambda_k$, $\sigma\in \Sigma_k$ and
$W_1,W_2\in{\mathcal U }_k$ such that
$$\|(A_1,\ldots,A_n,U)-W_1(A_1^{\lambda},\ldots,A_n^{\lambda},U^{\lambda})W_1^*\|_2\leq \frac{r\omega}{24R},$$
$$\|(B_1,\ldots,B_p,U)-W_2(B_1^{\sigma},\ldots,B_p^{\sigma},U^{\sigma})W_2^*\|_2\leq \frac{r\omega}{24R}.$$
Therefore
$$\|W_1U^{\lambda}W_1^*-W_2U^{\sigma}W_2^*\|_2=\|W_2^*W_1U^{\lambda}-U^{\sigma}W_2^*W_1\|_2<\frac{r\omega}{12R}.$$

From Lemma \ref{Vo}, there exists a set
$\{Ball(U_{\lambda,\sigma,\gamma},\frac{\omega}{3R})\}_{\gamma\in
\Delta_k}$ in ${\mathcal U}_k$ which cover
$\Omega(U^{\lambda},U^{\sigma};\frac{r\omega}{12R})$ with
cardinality $|\Delta_k|\leq (\frac{18R}{\omega})^{4rk^2}$. This
implies
\begin{eqnarray*}&&\|(A_1,\ldots,A_n,B_1,\ldots,B_p,U)\\
&-&
(W_2U_{\lambda,\sigma,\gamma}A_1^{\lambda}U_{\lambda,\sigma,\gamma}^*W_2^*,
\ldots,W_2U_{\lambda,\sigma,\gamma}A_n^{\lambda}U_{\lambda,\sigma,\gamma}^*W_2^*,
W_2B_1^{\sigma}W_2^*,\ldots,W_2B_p^{\sigma}W_2^*,W_2U^{\sigma}W_2^*)\|_2\\
&\leq& n\omega.\end{eqnarray*}
 Then we get
\begin{eqnarray*}&&\mathfrak{K}(x_1,\ldots,x_n,
y_1,\ldots,y_p,u: v_1,\ldots,v_s, w_1,\ldots,w_t;2n\omega,R)\\
&\leq&\inf_{m\in\Bbb N,\varepsilon>0}
\limsup_{k\rightarrow \infty}\frac{\log(|\Lambda_k||\Sigma_k||\Delta_k|)}{-k^2\log(2n\omega)}\\
&\leq &4r\frac{\log (18R)-\log \omega}{-\log(2n\omega)}.
\end{eqnarray*}

Because $r$ is an arbitrarily small positive number, we have
$$\mathfrak{K}_3(x_1,\ldots,x_n,y_1,\ldots,y_p,u:v_1,\ldots,v_s,w_1,\ldots,w_t)=0.$$
Note that
$W^*(x_1,\ldots,x_n,y_1,\ldots,y_p,u)=W^*(x_1,\ldots,x_n,y_1,\ldots,y_p)$,
by Lemma \ref{change generator is the same}, we have
$$\mathfrak{K}_3(x_1,\ldots,x_n,y_1,\ldots,y_p:v_1,\ldots,v_s,w_1,\ldots,w_t)=0.$$\hfill $\Box$

\vspace{0.2cm}

Now we are ready to prove the following theorem.

\begin{theorem}\label{theorem, intersection contains Haar unitary}Suppose $\mathcal M$ is a von Neumann algebra with a
tracial state $\tau$, ${\mathcal N}_1$ and ${\mathcal N}_2$ are von
Neumann subalgebras of $\mathcal M$. If ${\mathcal N}_1\cap{\mathcal
N}_2$ is diffuse, then
$$\mathfrak{K}_3(({\cal N}_1\cup {\cal N}_2)'')\leq
\mathfrak{K}_3({\mathcal N}_1)+\mathfrak{K}_3({\mathcal N}_2).$$
\end{theorem}

Proof. If one of $\mathfrak{K}_3({\mathcal N}_1)$ and
$\mathfrak{K}_3({\mathcal N}_2)$ is infinity, then we are done.

Now suppose $\mathfrak{K}_3({\mathcal N}_1)=\mathfrak{K}_3({\mathcal
N}_2)=0$ and $u$ is a Haar unitary in ${\mathcal N}_1\cap{\mathcal
N}_2$.  Let ${\cal G}={\mathcal N}_1\cup{\mathcal N}_2$ and
$A_0=\{u\}$. Then $\cal G$ is a generating set of $\mathcal M$.
Suppose $A_0\subseteq A\subseteq {\cal G}$ and $A$ is finite, write
 $A=\{x_1,\ldots,x_n,u,y_1,\ldots,y_p\}$ with $x_1,\ldots,x_n\in{\mathcal
N}_1$ and $y_1,\ldots,y_p\in{\mathcal N}_2$. Since
$\mathfrak{K}_3({\mathcal N}_1)=\mathfrak{K}_3({\mathcal N}_2)=0$,
there exist $v_1,\ldots,v_s\in {\mathcal N}_1$,
$w_1,\ldots,w_t\in{\mathcal N}_2$ such that
$\mathfrak{K}_2(x_1,\ldots,x_n,u:v_1,\ldots,v_s)=0$ and
$\mathfrak{K}_2(y_1,\ldots,y_p,u:w_1,\ldots,w_t)=0.$

 Because $u\in
W^*(x_1,\ldots,x_n,u)\cap W^*(y_1,\ldots,y_p,u)$, then from Lemma
\ref{haar unitary}, we know that
\begin{eqnarray*}&&\mathfrak{K}_2(A:v_1,\ldots,v_s,w_1,\ldots,w_t)\\
&=&\mathfrak{K}_2(x_1,\ldots,x_n,u,y_1,\ldots,y_p:v_1,\ldots,v_s,w_1,\ldots,w_t)\\
&=&0.\end{eqnarray*}

Therefore, by Theorem \ref{equivalence}, $\mathfrak{K}_3(({\cal
N}_1\cup {\cal N}_2)'')=0$.\hfill $\Box$

\begin{lemma}\label{don} (\cite{Don}, Lemma 17) Suppose $\mathcal M$ is a von Neumann algebra with a
tracial state $\tau$, $x$ is a normal element in $\mathcal M$ such
that $x$ has no eigenvalues. Then there is a selfadjoint element $y$
with the uniform distribution on $[0,1]$ such that $W^*(x)=W^*(y)$.
\end{lemma}

\begin{lemma}\label{don, lemma 18} (\cite{Don}, Lemma 18) Suppose $n,n_1,p\in\Bbb
N$, $1\leq n_1\leq n$ and $p\geq 2$. Suppose $A$ is a diagonal
matrix whose diagonal entries are $\frac{1}{n_1}, \frac{2}{n_1},
\cdots, \frac{n_1}{n_1}, -1, \cdots, -1$, and $B$ is any selfadjoint
$n\times n$ matrix with $0\leq B\leq 1$. Suppose $0\leq
\varepsilon\leq \frac{1}{4}$, and
$\frac{1}{n_1}<4\varepsilon^{p-1}$. Let
$$\Sigma(A,B,\varepsilon^p)=\{W\in {\cal M}_n: \|W\|\leq 1, \|AW-WB_2\|_2<\varepsilon^p\}.$$
Then $\nu_2(\Sigma(A,B,\varepsilon^p),\varepsilon)\leq
(\frac{6}{\varepsilon})^{16n^2\varepsilon^{p-1}}$.
\end{lemma}

\begin{theorem}\label{theorem, without comon eigenvalues} Let $\mathcal M$ be a von Neumann algebra with a
tracial state $\tau$. Suppose $\mathcal N$ is a von Neumann
subalgebra of $\mathcal M$ and $y$ is an element in $\mathcal M$. If
$a$, $b$ are normal operators in $\mathcal N$ such that $a$ and $b$
have no common eigenvalues and $ay=yb\neq 0$, then
$\mathfrak{K}_3\left(W^*({\cal N}\cup
\{y\})\right)\leq\mathfrak{K}_3({\mathcal N})$.
\end{theorem}

Proof. There is no loss in assuming that $y, a, b$ have norm at most
1.

If $\mathfrak{K}_3({\mathcal N})$ is infinity, then we are done. So
we can assume $\mathfrak{K}_3({\mathcal N})=0$.

Since $ay=yb$, by the Putnam-Fuglede theorem, we have $a^*y=yb^*$.
Then for any polynomial $p$, $p(a,a^*)a=ap(b,b^*)$.
 Therefore for every bounded Borel
function $\varphi:{\Bbb C}\rightarrow{\Bbb C}$,
$\varphi(a)y=y\varphi(b)$. If $\lambda\in\sigma_p(a)$, then the
spectral projection $\chi_{\{\lambda\}}(b)$ is 0. If $a$ is
diagonal, then
$y=Iy=\sum_{\lambda\in\sigma_p(a)}\chi_{\{\lambda\}}(a)y=y\sum_{\lambda\in\sigma_p(a)}\chi_{\{\lambda\}}(b)=0.$
So $a$ can  not be diagonal, therefore we can write $a=d\oplus c$,
where $d$ is diagonal and $c$ has no eigenvalues. From Lemma
\ref{don}, there is a self-adjoint element $c_0$ with the uniform
distribution such that $W^*(c)=W^*(c_0)$, thus there is some Borel
function $\psi$ such that $c_0=\psi(c)$ and $\psi(d)=-1$, $0\leq
\psi(b)\leq 1$. Since $\psi(a)y=(-1\oplus c_0)y=y\psi(b)$, we can
replace $-1\oplus c_0$ with $a$, $\psi(b)$ with $b$. Hence we can
assume $b$ is selfadjoint and $a=-1\oplus c_0$ with
$\tau(a^n)=(-1)^n(1-\alpha)+\alpha\int_0^1t^ndt$, where
$\alpha=1-\tau(1\oplus 0).$

For each $n\in\Bbb N$, define the diagonal matrix $A_n$ with
eigenvalues
$$\frac{1}{[n\alpha]},\frac{2}{[n\alpha]},\ldots,\frac{[n\alpha]}{[n\alpha]},-1,-1,\ldots,-1,$$
where $[n\alpha]$ denotes the greatest integer function of
$n\alpha$. It is not hard to show that
$\tau_n(A_n^m)=(-1)^m\frac{n-[n\alpha]}{n}+
\frac{[n\alpha]}{m}\left(\frac{1}{[n\alpha]}\sum_{s=1}^{[n\alpha]}(\frac{s}{[n\alpha]})^m\right)$.
When $n\rightarrow\infty$,
$\frac{1}{[n\alpha]}\sum_{s=1}^{[n\alpha]}(\frac{s}{[n\alpha]})^m$
is a Riemann sum converging to $\int_0^1t^mdt.$ Hence
$\tau_n(f(A_n))\rightarrow \tau(f(a))$ as $n\rightarrow \infty$.
Choose matrix $B_n\in{\mathcal M}_n({\Bbb C})$ such that $0\leq
B_n\leq 1$ and $B_n$ converges in distribution to $b$ as
$n\rightarrow \infty$.

Let $x_1,\ldots,x_n$ be elements in $\mathcal N$. Then there exist
$y_1,\ldots,y_p$ in $\mathcal N$ such that
$$\mathfrak{K}_2(x_1,\ldots,x_n:y_1,\ldots,y_p)=0.$$ For any
$0<\omega<1$, $0<r<1$, $R>1$, $m\in\Bbb N$, $\varepsilon>0$ and
$k\in\Bbb N$, there exists a set $\{{\mathcal
U}(T_1^{\lambda},\ldots,T_n^{\lambda},A^{\lambda},B^{\lambda};\\\frac{r\omega}{64})\}_{\lambda\in\Lambda_k}$
 of $\frac{r\omega}{64}$-orbit-balls in ${\mathcal M}_k(\Bbb
 C)^{k+2}$ that cover
 $\Gamma_R(x_1,\ldots,x_n,a,b:y_1,\ldots,y_p;m,k,\varepsilon)$ with the
 cardinality of $\Lambda_k$ satisfying $|\Lambda_k|=\nu(\Gamma_R(x_1,\ldots,x_n,a,b:y_1, \ldots, y_p;m,k,\varepsilon),\frac{r\omega}{64})$.
When $m$ is sufficiently large and $\varepsilon$ is sufficiently
small, we can assume that $A^{\lambda}$ to be $A_k$ and
$B^{\lambda}$ to be $(U^{\lambda})^*B_kU^{\lambda}$ for some unitary
matrix $U$.

For $m$ is sufficiently large and $\varepsilon$ ($\leq
\frac{r\omega}{64}$) is sufficiently small, when
$(T_1,\ldots,T_n,A,B,C)\in \Gamma_R(x_1,\ldots,x_n,\\
a,b,y:y_1,\ldots,y_p;m,k,\varepsilon)$, it follows from Lemma 4 in
\cite{Don} that we can assume that $\|C\|\leq 1$. In addition, it is
clear that $\|AC-CB\|_2\leq \varepsilon$ and
$(T_1,\ldots,T_n,A,B)\in
\Gamma_R(x_1,\ldots,x_n,a,b:y_1,\ldots,y_p;m,k,\varepsilon)$. So
there exist some $\lambda\in\Lambda_k$ and $V\in{\mathcal U}_k$ such
that
$$\|(T_1,\ldots,T_n,A,B)-(VT_1^{\lambda}V^*,\ldots,
VT_n^{\lambda}V^*,VA_kV^*,V(U^{\lambda})^*B_kU^{\lambda}V^*)\|_2\leq
\frac{r\omega}{64}.$$ Hence
$$\|A_kV^*CV-V^*CVU^*B_kU\|_2=\|VA_kV^*C-CV(U^{\lambda})^*B_kU^{\lambda}V^*\|_2\leq
\frac{r\omega}{16}.$$  Then, by Lemma \ref{don, lemma 18}, there
exists a set
$\{Ball(C_{\sigma};\frac{\omega}{4})\}_{\sigma\in\Sigma_k}$ of
$\frac{\omega}{4}$-balls that cover $\{W\in {\mathcal M}_k:
\|W\|\leq 1,\
\|WA_k-(U^{\lambda})^*B_kU^{\lambda}W\|_2<\frac{r\omega}{16}\}$ with
$|\Sigma_k|\leq (\frac{24}{\omega})^{32rk^2}$, i.e., there exists
some $C_{\sigma}$ such that
$$\|V^*CV-C_{\sigma}\|_2=\|C-VC_{\sigma}V^*\|_2\leq \frac{\omega}{4}.$$
Thus
$$\|(T_1,\ldots,T_n,C)-(VT_1^{\lambda}V^*,\ldots,VT_n^{\lambda}V^*,VC_{\sigma}V^*)\|_2\leq
\frac{\omega}{2}.$$ Therefore
$$\nu(\Gamma_R(x_1,\ldots,x_n,y:a,b,y_1,\ldots,y_p);\omega)\leq
|\Lambda_k|\cdot |\Sigma_k|.$$

Hence, we get
\begin{eqnarray*} 0&\leq &
\mathfrak{K}(x_1,\ldots,x_n,y:a,b,y_1,\ldots,y_p;\omega,R)\\
&\leq & \inf_{m\in \Bbb
N,\varepsilon>0}\limsup_{k\rightarrow\infty}\frac{\log(|\Lambda_k||\Sigma_k|)}{-k^2\log\omega}\\
&\leq& \inf_{m\in \Bbb
N,\varepsilon>0}\limsup_{k\rightarrow\infty}\left(\frac{\log(|\Lambda_k|)}{-k^2\log\omega}+\frac{32rk^2(\log
24-\log\omega)}{-k^2\log\omega}\right)\\
&=& \inf_{m\in \Bbb
N,\varepsilon>0}\limsup_{k\rightarrow\infty}\frac{\log(|\Lambda_k|)}{-k^2\log\omega}+32r\frac{\log 24-\log\omega}{-\log\omega}\\
&=&32r\frac{\log 24-\log\omega}{-\log\omega}.
\end{eqnarray*} Since $r$ is an arbitrarily small positive number,
we have
$\mathfrak{K}(x_1,\ldots,x_n,y:a,b,y_1,\ldots,y_p;\omega,R)=0$,
whence, $\mathfrak{K}_2(x_1,\ldots,x_n,y:a,b, y_1,\ldots,y_p)=0$.
Therefore, by theorem \ref{equivalence}, $\mathfrak{K}_3({\mathcal
M})=0$.\hfill $\Box$

\begin{corollary}\label{corollary, x:u=0}Let $\cal M$ be a von Neumann algebra with a faithful trace $\tau$. Suppose $a, x_1, \ldots, x_n$
are elements in $\cal M$ such that $a$ is a normal element without
eigenvalues, and $ax_i=x_ia$ for all $1\leq i\leq n$. Then
$$\mathfrak{K}_3(x_1, \ldots, x_n:u)=0$$
\end{corollary}

Using the similar idea in the proof of Theorem \ref{theorem, without
comon eigenvalues}, we can prove the following theorem.

\begin{theorem}\label{add one unitary}Suppose $\mathcal M$ is a von Neumann algebra with
a faithful trace $\tau$ and ${\mathcal M}=\{{\mathcal N},u\}''$,
where $\mathcal N$ is a von Neumann subalgebra of $\mathcal M$, $u$
is a unitary element of $\mathcal M$. Let $\{v_1,v_2,\ldots\}$ be a
family of Haar unitary elements and $\{w_1,w_2,\ldots\}$ be a family
of unitary elements in $\mathcal N$ such that
$\|v_nu-uw_n\|_2\rightarrow 0$. Then $\mathfrak{K}_3({\mathcal
M})\leq \mathfrak{K}_3({\mathcal N})$.

 In
particular, if $v$ and $w$ are Haar unitary elements in $\mathcal N$
such that $vu=uw$, then $\mathfrak{K}_3({\mathcal M})\leq
\mathfrak{K}_3({\mathcal N})$.
\end{theorem}

\section{Applications}

\hspace{1.5em}Suppose $\Lambda$ is an infinite set. An {\it
ultrafilter} $\alpha$ on $\Lambda$ is a collection of subsets of
$\Bbb N$ such that the empty set $\emptyset\not\in\alpha$, $\alpha$
is closed under finite intersections, and, for each subset $A$ of
$\Lambda$, either $A\in\alpha$ or ${\Bbb N}\setminus A\in \alpha$.
One example of an ultrafilter is obtained by choosing an $\iota$ in
$\Lambda$ and letting $\alpha$ be the collection of all subsets of
$\Lambda$ that contain $\iota$. Such an ultrafilter is called {\it
principal}; ultrafilters not of this form are called {\it free}. We
will call an ultrafilter $\alpha$ {\it nontrivial} if it is free and
there exists a decreasing sequence in $\alpha$ whose intersection is
empty. Free ultrafilters on an countable set are always nontrivial.

Suppose $\mathfrak{X}$ is another set, $f:\Lambda\rightarrow
\mathfrak{X}$ is a mapping and $E\subseteq \mathfrak{X}$. we say
that $f(\iota)$ is {\it eventually in $E$} along $\alpha$ if
$f^{-1}(E)=\{\iota\in\Lambda: f(\iota)\in E\}\in \alpha$. If
$\mathfrak{X}$ is a topological space, we say that $f(\iota)$ {\it
converges to $x$} (in $\mathfrak{X}$) along $\alpha$, denoted by
$\lim_{\iota\rightarrow \alpha}f(\iota)=x$, if $f(\iota)$ is
eventually in each neighborhood of $x$. It is well known that if
$\mathfrak{X}$ is a compact Hausdorff space, the
$\lim_{\iota\rightarrow \alpha}f(\iota)$ always exists in
$\mathfrak{X}$ for every $f:\Lambda\rightarrow \mathfrak{X}$ and
every ultrafilter $\alpha$ on $\Lambda$.

Let $\alpha$ be a nontrivial ultrafilter on $\Lambda$. Suppose
${\mathcal M}_\iota$ is a finite von Neumann algebra with a faithful
trace $\tau_\iota$. Let $\prod_\iota{\mathcal M}_\iota$ be the
$l^\infty$-product of the ${\mathcal M}_\iota$'s, ${\mathcal
J}=\{\{x_{\iota}\}: \lim_{\iota\to\alpha}
\tau_{\iota}(x_{\iota}^*x_{\iota})=0\}$. Then define the {\it
ultraproduct} $\prod^\alpha{\mathcal M}_\iota$ of ${\mathcal
M}_\iota$ to be $\prod_{i\in\Bbb I}{\mathcal M}_i/\mathcal J$.

When ${\mathcal M}_\iota=\mathcal M$ for all $\iota$, then
$\prod^\alpha{\mathcal M}_\iota$ is called the {\it ultrapower} of
$\mathcal M$, denoted by ${\mathcal M}^\alpha$.

 Let $\mathcal M$ be a II$_1$ factor with the faithful trace
$\tau$. For every $\varepsilon>0$, and any elements
$x_1,x_2,\ldots,x_n$ in $\cal M$, if there exists a unitary
$u\in\cal M$ with $\tau(u)=0$ such that $\|ux_i-x_iu\|_2\leq
\varepsilon$ for every $i$, then we say that $\mathcal M$ has {\it
property $\Gamma.$}

It is well-known that if $\cal M$ is a II$_1$ factor with property
$\Gamma$, then there exists a central sequence $\{v_n\}$ of Haar
unitary elements in $\mathcal M$, i.e., $\|v_nx-xv_n\|_2\rightarrow
0$ for every $x\in{\mathcal M}$.

If a von Neumann algebra acts on a very large Hilbert space, it may
not contain any nontrivial central sequences, but it may contain a
{\it central net}, i.e., a net $\{x_{\lambda}\}$ in $\cal M$ such
that $\|x_{\lambda}a-ax_{\lambda}\|_2\rightarrow 0$ for every
$a\in\cal M$. Equivalently, $\cal M$ has a central net if and only
if, for every $\varepsilon>0$ and for every finite subset
$F\subseteq {\cal M}$, there is a Haar unitary $u$ such that
$\|ua-au\|_2<\varepsilon$ for every $a\in F$.

\begin{theorem}
\label{ultraproduct}Suppose $\{{\mathcal{M}}_{\lambda }\}_{\lambda
\in
\Lambda }$ is a family of von Neumann algebras such that each $\mathcal{M}%
_{\lambda }$ has a central net of Haar untiaries. Let $\alpha $ be a
nontrivial ultrafilter on $\Lambda $. Then
$$
\mathfrak{K}_{3}(\prod^{\alpha }\mathcal{M}_{\lambda })=0.
$$
\end{theorem}

Proof. Suppose $x_{1},\ldots ,x_{n}$ are any elements in $\prod^{\alpha }%
\mathcal{M}_{\lambda }$, and $x_{i}=\{x_{\lambda }^{i}\}_{\alpha }$. Since $%
\alpha $ is nontrivial, there exists a decreasing sequence $%
\{A_{k}\}_{k=1}^{\infty }$ in $\alpha $ whose intersection is empty, and $%
A_{1}=\Lambda $. If $\lambda \in A_{k}/A_{k-1}$, since ${\mathcal{M}}%
_{\lambda }$ has a central net of Haar unitaries, then there exists
a Haar unitary $u_{\lambda }\in \mathcal{M}_{\lambda }$ such that
$\Vert x_{\lambda
}^{i}u_{\lambda }-u_{\lambda }x_{\lambda }^{i}\Vert _{2}<\frac{1}{k}$ for $%
1\leq i\leq n$. Then $u=\{u_{\lambda }\}_{\alpha }$ defines a Haar
unitary
that commutes with $x_{1},\ldots ,x_{n}$. By Corollary \ref{corollary, x:u=0}%
, $\mathfrak{K}_{2}(x_{1},\ldots ,x_{n}:u)=0.$ Therefore, by Theorem \ref%
{equivalence}, $\mathfrak{K}_{3}\left( \prod^{\alpha
}\mathcal{M}_{\lambda }\right) =0$.\hfill $\Box $

\begin{remark}
Suppose $\cal M$ is a diffuse finite von Neumann algebra with a
faithful trace $\tau$. We can define a numerical invariant $\gamma
\left( \mathcal{M} \right) $ by
$$\gamma \left( \mathcal{M}\right) =\sup_{\begin{array}{c}F\subset
\mbox{ball}\left( \mathcal{M}\right)\\
F\mbox{ finite}\end{array}}\inf_{\begin{array}{c}u\ \mbox{is a}\\
\mbox{Haar unitary}\end{array}}\max_{a\in F}\| au-ua\|.$$
It is clear that $\gamma ( \mathcal{M}) =0$ if and only if $%
\mathcal{M}$ has a central net of Haar unitaries and that $\gamma (
\mathcal{M}) \leq 2$. F. Murray and J. von Neumann \cite{Mv} found a lower bound for $%
\gamma \left( \mathcal{L}_{\mathbb{F}_{n}}\right) $ for $n\geq 2$.
(Of course, they did not use our terminology.) It is not difficult
to modify the proof of Theorem \ref{ultraproduct} to prove
that if $\gamma ({\cal M}_{\lambda})\rightarrow 0$ along the ultrafilter $\alpha $, then $%
\mathfrak{K}_{3}(\prod^{\alpha }\mathcal{M}_{\lambda })=0.$
\end{remark}

\begin{lemma}\cite{Don-Li}
\label{goober}Suppose $\mathcal{M}$ is diffuse, countably generated
and has
a central sequence of Haar unitaries. Then there is a central sequence $%
\{u_{n}\}$ of mutually commuting Haar unitaries in $\mathcal{M}$.
\end{lemma}

\begin{theorem}
If $\mathcal{M}$ is a von Neumann algebra with a central net of Haar
unitaries, then $\mathfrak{K}_{3}({\mathcal{M}})=0$.
\end{theorem}

Proof. Suppose $x_{1},\ldots ,x_{n}\in \mathcal{M}$. Then there is a
sequence $\left\{ u_{n}\right\} $ of Haar unitaries so that if $a\in
\{ x_{1},\ldots ,x_{n},u_{1},u_{2},\ldots\} ,$ then $$\|
au_{n}-u_{n}a\| _{2}\rightarrow 0, $$ i.e., inductively choose
$u_{n}$ so that $\| au_{n}-u_{n}a\|_{2}<1/n$ for $a\in \left\{
x_{1},\ldots ,x_{n},u_{1},u_{2},\ldots u_{n-1}\right\} $. Hence
$\left\{ u_{n}\right\} $
is a central sequence in the von Neumann algebra $\mathcal{N}$ generated by $%
\{ x_{1},\ldots ,x_{n}$, $u_{1},u_{2},\ldots \}$. It follows from
Lemma \ref{goober} that there is a central sequence $\{
v_{n}\}$ of commuting Haar unitaries in $\mathcal{N}$. Then we get $\mathfrak{K}%
_{3}\left( \left\{ v_{1},v_{2},\ldots \right\} ^{\prime \prime
}\right) =0.$ We can choose unitaries $\left\{ w_{1},\ldots
,w_{m}\right\}$ that generate $W^{\ast }\left( x_{1},\ldots
,x_{n}\right),$ and, using Theorem \ref{add one unitary}, we
inductively get  $\mathfrak{K}_{3}\left( \left\{ w_{1},\ldots
,w_{j},v_{1},v_{2},\ldots \right\} ^{\prime \prime }\right) =0$ for
$1\leq j\leq m$. Hence, there exist $y_{1},\ldots ,y_{p}\in
\mathcal{N}$ such that $$ \mathfrak{K}_{2}\left( x_{1},\ldots
,x_{n}:y_{1},\ldots ,y_{p}\right) =0.
$$
Therefore $\mathfrak{K}_{3}({\mathcal{M}})=0.$\hfill $\Box $

\begin{theorem}
Suppose $\mathbb{F}$ is a free group with the standard generating
set $G$ satisfying $|G|\geq 2$ and let $\mathcal{L}_{\mathbb{F}}$ be
the group von Neumann algebra
generated by $\mathbb{F}$. Then $\mathfrak{K}_{3}({\mathcal{L}}_{\mathbb{F}%
})=\infty.$
\end{theorem}

Proof. For any $g\in\Bbb F$, we can view $g$ as a unitary in $\mathcal{L}_{\mathbb{F}}$. Note that $G$ generates ${\mathcal{L}}_{\mathbb{F}}$. Let $%
g_{1},\ldots ,g_{n}\in G$ ($n\geq 2$), and $y_{1},y_{2},\ldots
,y_{N}\in G.$ We will prove that $n\leq \delta \left( g_{1},\ldots
,g_{n}:y_{1},y_{2},\ldots ,y_{N}\right) $. Since $$\delta \left(
g_{1},\ldots ,g_{n}:y_{1},y_{2},\ldots ,y_{N}\right) \leq
1+\mathfrak{K}_{2}\left( g_{1},\ldots ,g_{n}:y_{1},y_{2},\ldots
,y_{N}\right) ,$$ we will conclude that $n-1\leq
\mathfrak{K}_{2}\left( g_{1},\ldots
,g_{n}:y_{1},y_{2},\ldots ,y_{N}\right) .$ From this it follows that  $%
\mathfrak{K}_{3}\left( {\mathcal{L}}_{\mathbb{F}}\right) =\mathfrak{K}%
_{3}\left( G\right) =\infty $.

It follows from Theorem 13 in \cite{Don} that when we compute
$\delta $ we
can replace the $\Gamma _{R}$-sets with the set of unitary elements in the $%
\Gamma _{R}$-sets. Let $$\Omega _{m,k,\varepsilon
}=\mathcal{U}_{k}^{n+N}\cap \Gamma_R (g_{1},\ldots
,g_{n},y_{1},y_{2},\ldots ,y_{N};m,k,\varepsilon ),$$  $$\Delta
_{m,k,\varepsilon }=\mathcal{U}_{k}^{n}\cap \Gamma _{R}(g_{1},\ldots
,g_{n}:y_{1},y_{2},\ldots ,y_{N};m,k,\varepsilon ).$$ If
$(U_{1},\ldots
,U_{n},W_{1},\ldots ,W_{N})\in \Omega _{m,k,\varepsilon }$, then  $%
(U_{1},\ldots ,U_{n})\in \Delta _{m,k,\varepsilon }$.  So $$\Omega _{m,k,\varepsilon }\subseteq \Delta _{m,k,\varepsilon }\times {\mathcal{%
U}}_{k}^{N}.
$$
Let $\mu _{k}$ denote Haar measure on ${\mathcal{U}}_{k}$, $\mu
_{k}^{n}$ denote the corresponding product measure on
${\mathcal{U}}_{k}^{n}$. Then $$ \mu _{k}^{n+N}\left( \Omega
_{m,k,\varepsilon }\right) \leq \mu
_{k}^{n}(\Delta _{m,k,\varepsilon })\cdot \mu _{k}^{N}\left( {\mathcal{U}}%
_{k}^{N}\right) \leq \mu _{k}^{n}(\Delta _{m,k,\varepsilon })\leq 1.
$$
We know from Theorem 3.9 in \cite{V02} that $\mu _{k}^{n+N}\left(
\Omega _{m,k,\varepsilon }\right) \rightarrow 1$ and thus $\mu
_{k}^{n}(\Delta _{m,k,\varepsilon })\rightarrow 1 $ as $k\rightarrow
\infty $. It follows that $$ \mu _{k}^{n}(\Delta _{m,k,\varepsilon
})\leq \nu _{2}(\Delta _{m,k,\varepsilon },\omega )\mu
_{k}^{n}(ball(1,\omega ))\leq \nu _{2}(\Delta _{m,k,\varepsilon
},\omega )(\omega )^{nk^{2}},
$$
so $\nu _{2}(\Delta _{m,k,\varepsilon },\omega )\geq \mu
_{k}^{n}(\Delta _{m,k,\varepsilon })(\frac{1}{\omega })^{nk^{2}}$.

By Lemma 1 in \cite{Don-Shen}, $$ \mathfrak{K}_{2}(g_1, \ldots,
g_n:y_1, \ldots, y_N)\geq  \delta (g_1, \ldots, g_n:y_1, \ldots,
y_N) -1.$$ Note that \begin{eqnarray*} \delta (g_1, \ldots, g_n:y_1,
\ldots, y_N)&=&-1+\limsup_{\omega \rightarrow
0^{+}}\inf_{m,\varepsilon }\limsup_{k\rightarrow \infty }\frac{\log
\left( \nu _{2}(\Delta
_{m,k,\varepsilon },\omega )\right) }{-k^{2}\log \omega }\\
& \geq &-1+\limsup_{\omega \rightarrow 0^{+}}\inf_{m,\varepsilon
}\limsup_{k\rightarrow \infty }\frac{\log \left( \mu _{k}^{n}(\Delta
_{m,k,\varepsilon })\right) -nk^{2}\log \omega }{-k^{2}\log \omega
}.\end{eqnarray*} Since $\mu _{k}^{n}(\Delta _{m,k,\varepsilon
})\rightarrow 1$, we have $$\mathfrak{K}_{2}(g_1, \ldots, g_n:y_1,
\ldots, y_N)\geq n-1.$$ Thus
$\mathfrak{K}_{3}({\mathcal{L}}_{G})=\infty .$\hfill $\Box $

\begin{theorem}Suppose $\mathcal M$ is a von Neumann algebra with
a faithful trace $\tau$, ${\mathcal N}_1$ and ${\mathcal N}_2$ are
mutually commuting diffuse subalgebras of $\mathcal M$. Then
$\mathfrak{K}_3(W^*({\cal N}_1\cup{\cal N}_2))=0$.
\end{theorem}

Proof. Since ${\mathcal N}_1$ and ${\mathcal N}_2$ are diffuse, we
can assume that ${\mathcal N}_1=\{u_{\lambda}:\lambda\in\Lambda\}''$
and ${\mathcal N}_2=\{v_{\sigma}:\sigma\in\Sigma\}''$ where
$u_{\lambda}, v_{\sigma}$ are all Haar unitaries. For any finite
subset $E$ of $\Lambda$ and finite subset $F$ of $\Sigma$, let
${\mathcal M_{E,F}}=\{u_{\lambda}, v_{\sigma}: \lambda\in E,
\sigma\in F\}''$. Then $\mathfrak{K}_3\left({\mathcal
M_{E,F}}\right)=0$ by Theorem \ref{add one unitary}. Let
$${\mathcal S}=\bigcup\left\{{\mathcal M_{E,F}}: E\ \mbox{is a
finite subset of}\ \Lambda, F\ \mbox{is a finite subset of}\
\Sigma\right\}.$$ It is clear that ${\mathcal S}$ generates
$W^*({\cal N}_1\cup{\cal N}_2)$. By Corollary \ref{corollary,
increasingly directed}, $\mathfrak{K}_3({\mathcal S})=0$. Thus
$\mathfrak{K}_3(W^*({\cal N}_1\cup{\cal N}_2))=0$ by Corollary
\ref{corollary, k3 is independent on generatoer}.\hfill $\Box$

\begin{corollary}If ${\mathcal N}_1$ and
${\mathcal N}_2$ are diffuse von Neumann algebras, then
$\mathfrak{K}_3({\mathcal N}_1\otimes {\mathcal N}_2)=0$.
\end{corollary}

The following corollary was proved by S. Popa \cite{popa2} and L. Ge
\cite{Ge2}.
\begin{corollary}\label{corollary, free group factor k_3=infty}If ${\mathcal M}$ is a II$_1$ factor and $\mathfrak{K}_3({\cal
M})=\infty$, then $\cal M$ is prime. In particular,
${\mathcal{L}}_{\mathbb{F}}$ is prime for every free group $\Bbb F$
with the standard generating set $G$ satisfying $|G|\geq 2$..
\end{corollary}

Let $\mathcal M$ be a von Neumann algebra with a faithful trace,
${\cal U(M)}$ be the set of all unitary elements in $\cal M$. For
any subset ${\cal S}\subseteq {\cal M}$, define $${\cal N}({\cal
S})=\{u\in {\cal U(M)}: u{\cal S}u^*\subseteq {\cal S}\}'',$$ and
\begin{eqnarray*}{\cal I}({\cal S})&=&W^*(\{y\in{\cal M}: \exists\  \mbox{two normal elements}\ a,b \\
&&\mbox{without common eigenvalues such that}\ ay=yb\neq
0\})\end{eqnarray*} Suppose $\mathcal A$ is a diffuse von Neumann
subalgebra of $\mathcal M$, and $\alpha$ is an ordinal. Then define
$${\mathcal N}_{\alpha}({\mathcal
A})=\left\{\begin{array}{lll}\mathcal A& \alpha=0\\
\left(\bigcup_{\beta<\alpha}{\mathcal N}_{\beta}(\mathcal
A)\right)''& \alpha\ \mbox{is a limit ordinal}\\
{\cal N}\left({\cal N}_{\beta}({\cal A})\right) & \mbox{if}\
\alpha=\beta+1,\end{array}\right.$$ and
$${\mathcal I}_{\alpha}({\mathcal
A})=\left\{\begin{array}{lll}\mathcal A& \alpha=0\\
\left(\bigcup_{\beta<\alpha}{\mathcal I}_{\beta}(\mathcal
A)\right)''& \alpha\ \mbox{is a limit ordinal}\\
{\cal I}\left({\cal I}_{\beta}({\cal A})\right) & \mbox{if}\
\alpha=\beta+1.\end{array}\right.$$

The following theorem is a easy consequence of Theorem \ref{add one
unitary}, Theorem \ref{theorem, without comon eigenvalues} and
Corollary \ref{corollary, increasingly directed}.

\begin{theorem}\label{thereom, normalizer}Let $\mathcal M$ be a von Neumann algebra with
faithful trace $\tau$ and $\mathcal A$ be a diffuse subalgebra of
$\mathcal M$ with $\mathfrak{K}_3({\mathcal A})=0.$ Then for any
ordinal $\alpha$, $$\mathfrak{K}_3\left({\mathcal
N}_{\alpha}({\mathcal A})\right)=\mathfrak{K}_3\left({\mathcal
I}_{\alpha}({\mathcal A})\right)=0.$$
\end{theorem}

\section{Applications to group theory}

\hspace{1.5em}Suppose $\cal M$ is a von Neumann algebra with a
faithful trace $\tau$ and $u$ is a Haar unitary in $\cal M$. Define
$${\cal N}_u=\left(\cup\{ {\cal N}: {\cal N}\subseteq {\cal M}, u\in
{\cal N}, \mathfrak{K}_3({\cal N})=0\}\right)''$$ to be the von
Neumann subalgebra generated by the union of those subalgebras $\cal
N$ containing $u$ such that $\mathfrak{K}_3({\cal N})=0$. It follows
from Theorem \ref{theorem, intersection contains Haar unitary} that
$\mathfrak{K}_3\left({\cal N}_u\right)=0.$ Therefore ${\cal N}_u$ is
the unique largest subalgebra containing $u$ with
$\mathfrak{K}_3({\cal N}_u)=0$.

Suppose $\cal B$ is a diffuse von Neumann subalgebra of $\cal M$
with $\mathfrak{K}_3({\cal B})=0$. Since $\cal B$ is diffuse, there
exists a Haar unitary $u\in \cal B$. It is clear that ${\cal N}_u$
is the largest subalgebra of $\cal M$ that contains $\cal B$ whose
$\mathfrak{K}_3$ is 0.

If $u, v$ are two Haar unitaries in $\cal M$ such that ${\cal
N}_u\cap {\cal N}_v$ is diffuse, then, by Theorem \ref{theorem,
intersection contains Haar unitary},
$\mathfrak{K}_3\left(\left({\cal N}_u\cup {\cal
N}_v\right)''\right)=0.$ Therefore ${\cal N}_u={\cal N}_v$ by the
maximality of ${\cal N}_u$ and ${\cal N}_v$. Therefore, if ${\cal
N}_u\neq {\cal N}_v$, then ${\cal N}_u\cap {\cal N}_v$ is not
diffuse.

By Theorem \ref{thereom, normalizer}, it is clear that, for any Haar
unitary $u\in \cal M$, the normalizer
 of ${\cal N}_u$ is ${\cal N}_u$.

The above ideas have an interesting interpretation in discrete
groups. Let $G$ be a group and ${\cal L}_G$ be the corresponding
group von Neumann algebra.  For any $g\in G$, we can view $g$ as a
unitary in ${\cal L}_G$. If $g$ is an element in $G$ with infinite
order, then $g$ is a Haar unitary in ${\cal L}_G$. If $H$ is a
subgroup of $G\subseteq {\cal L}_G$, then $H''\cong {\cal L}_H.$

Suppose $G$ is a discrete torsion-free group (i.e., the only element
of finite order is identity) and ${\mathcal L}_{G}$ can be embedded
into an ultrapower of the hyperfinite II$_1$ factor. F. Radulescu
(see Proposition 2.5 in \cite{R}) proved that this is equivalent to
$G$ being algebraically embeddable in the unitary group of such an
ultrapower. For any $g\in G\setminus \{e\}$, let $H_g$ be the
subgroup generated by the set $\{H\leq G: g\in H,
\mathfrak{K}_3({\cal L}_{H})=0\}.$

By Theorem \ref{theorem, intersection contains Haar unitary} and
Theorem \ref{add one unitary}, we get the following theorem.
\begin{theorem}Let $G$ be a torsion-free group with the unit $e$.
The following statements hold:

(1) $\{H_g\setminus\{e\}: g\in G\setminus\{e\}\}$ is a partition of
$G\setminus\{e\}$,

(2) for any $g\in G\setminus \{e\}$, if $hH_gh^{-1}\subseteq H_g$,
then $h\in H_g$,

(3) for any $g\in G\setminus\{e\}$, if $h\in G\setminus H_g$, then
$hH_g\cap H_gh=\{h\}$.

\end{theorem}

We call $\{H_g: g\in G/\{e\}\}$ the {\it
$\mathfrak{K}_3$-decomposition} of the torsion-free group $G$.
Determining the $\mathfrak{K}_3$-decomposition of a particular group
involves both algebra and the theory of von Neumann algebras. In
\cite {Don-Shen}, there are many examples of groups $G$ whose
$\mathfrak{K}_3$-decomposition is $\{G\}$. We now provide a
particular example.

\begin{proposition}If $G$ is a free group, then, for any $g\in G\setminus \{e\}$,
$H_g$ is a maximal cyclic subgroup of $G$ containing $g$. In
particular, if $g$ is one of the free generators, then $H_g$ is the
subgroup generated by $g$.
\end{proposition}

Proof. Since every subgroup of free group is free, $H_g$ is free. By
definition of $H_g$ we see that $\mathfrak{K}_3({\cal L}_{H_g})=0$.
Therefore, by
 Corollary \ref{corollary, free group factor k_3=infty}, $H_g$ cannot have more than one generator. That implies $H_g$ is cyclic. By Theorem
 \ref{add one unitary}, $H_g$ must be a maximal abelian subgroup in
 $G$.\hfill $\Box$
\vspace{0.2cm}

Let ${\Bbb F}_2$ be a free group generated by two standard
generators $u,v$. We know that, in ${\Bbb F}_2$, $H_u$ is the
subgroup generated by $u$. This naturally raises the question.

\vspace{0.3cm}

\noindent \textbf{Question 1}.  In ${\cal L}_{{\Bbb F}_2}$, is
${\cal N}_u=W^*(u)$? In other words, is $W^*(u)$ a maximal
subalgebra of ${\mathcal L}_{{\Bbb F}_2}$ whose $\mathfrak{K}_3$ is
$0$?

 S. Popa \cite{popa3} proved that $W^*(u)$ is maximal injective. An
 affirmative answer to the question above would imply Popa's result,
 since $\mathfrak{K}_3({\cal M})=0$ whenever ${\cal M}$ is
 injective. This means that answering the question above is likely
 to be difficult. However, there are natural subquestions based on
 Theorem \ref{theorem, intersection contains Haar unitary} and Theorem
 \ref{theorem, without comon eigenvalues}, respectively.

\vspace{0.3cm}

\noindent \textbf{Question 1a} If $\cal M$ is a subalgebra of
${\mathcal L}_{{\Bbb F}_2}$ with $\mathfrak{K}_3({\cal M})=0$, and
${\cal M}\cap W^*(u)$ is diffuse, then must we have ${\cal
M}\subseteq W^*(u)$?

\vspace{0.3cm}

\noindent \textbf{Question 1b } If $y\in {\mathcal L}_{{\Bbb F}_2}$,
and $a, b\in W^*(u)$ without common eigenvalues, such that,
$ya=by\neq 0$, then must y be in $W^*(u)$?

\vspace{0.2cm}

We can give a partial solution to Question 1b by showing that if $w$
is a unitary in ${\mathcal L}_{{\Bbb F}_2}$ that conjugates a Haar
unitary in $W^*(u)$ into $W^*(u)$, then $w\in W^*(u)$.

 Suppose that $\cal M$ and $\cal N$ are von Neumann algebras and
${\cal N}\subseteq {\cal M}$. By a {\it conditional expectation}
from $\cal M$ onto $\cal N$, we mean a positive linear mapping $E:
{\cal M}\rightarrow {\cal N}$ such that

(1) $E(I)=I$,

(2) $E(x_1yx_2)=x_1E(y)x_2$, for any $x_1, x_2\in\cal N$ and
$y\in\cal M$.

 Define ${\mathcal
A}\bot{\mathcal B}$ to be $\tau(ab)=\tau(a)\tau(b)$ for any
$a\in\cal A$ and $b\in\cal B$.

\begin{theorem} Let $u,v$ be standard generators of ${\cal L}_{{\Bbb F}_2}$. If $\mathcal B$ is a diffuse von Neumann subalgebra of
$W^*(u)$, then $$\{w:\ w\ \mbox{is a unitary in}\ {\cal L}_{{\Bbb
F}_2}, w^*{\cal B}w\subseteq W^*(u)\}\subseteq W^*(u).$$
\end{theorem}

Proof. Suppose $\mathcal B$ is a diffuse von Neumann subalgebra of
$W^*(u)$. Define $${\cal N}=\{w:\ w\ \mbox{is a unitary in}\ {\cal
L}_{{\Bbb F}_2}, w^*{\cal B}w\subseteq {W^*(u)}\}.$$ It is
sufficient to prove $W^*(u)^{\perp}\subseteq {\cal N}^{\perp}$. The
proof is a modification of Lemma 2.5 in \cite{popa2}.

Let $g$ be en element in $W^*(u)^{\perp}$. It follows that
$gW^*(u)g^*\bot W^*(u)$.

Since $\cal B$ has no atoms, then, for any given $\varepsilon>0$,
there exists an orthogonal family of projections $e_1,\ldots,e_n$ in
$\cal B$ such that $\tau(e_i)<\varepsilon$ for $1\leq i\leq n$. Let
${\mathcal A}_{\varepsilon}$ be the von Neumann subalgebra generated
by $e_1,\ldots,e_n$, $\tau$ be the unique trace on ${\mathcal
L}_{{\Bbb F}_2}$, and $E_{{\cal A}_{\varepsilon}'\cap {\cal
L}_{{\Bbb F}_2}}$ be the unique $\tau$-preserving conditional
expectation from ${\mathcal L}_{{\Bbb F}_2}$ onto ${\cal
A}_{\varepsilon}'\cap {\cal L}_{{\Bbb F}_2}$ (i.e., $\tau\circ
E_{{\cal A}_{\varepsilon}'\cap {\cal L}_{{\Bbb F}_2}}=\tau$).

For any $w\in {\cal N}$, $g{\cal A}_{\varepsilon}g^*\bot w{\cal
A}_{\varepsilon}w^*$ since $g{\cal A}_{\varepsilon}g^*\bot W^*(u)$
and $w{\cal A}_{\varepsilon}w^*\subseteq w{\cal B}w^*\subseteq
W^*(u)$. Therefore, for $1\leq i\leq n$,
$$\tau(w^*ge_ig^*we_i)=\tau(ge_ig^*we_iw^*)=\tau(ge_ig^*)\tau(we_iw^*)=\tau(e_i)^2.$$
Summing up over $i$, we get
\begin{eqnarray*}|\tau(wg)|^2&\leq & \|E_{{\cal A}_{\varepsilon}'\cap {\cal
L}_{{\Bbb F}_2}}(wg)\|_2^2\\
&=&\|\sum_{i}e_iwge_i\|_2^2=\sum_{i}\|e_iwge_i\|_2^2\\
&=&\sum_{i}\tau(wge_ig^*w^*e_i)=\sum_i\tau(e_i)^2\\
&\leq& (\max_j\tau(e_j))\sum_i\tau(e_i)<\varepsilon.
\end{eqnarray*}
Since $\varepsilon>0$ is arbitrarily small, $\tau(wg)=0$. Therefore
for any $x\in {\cal N}''$, $\tau(x)=0$. Thus $g\bot {\cal N}''$.
\hfill$\Box$

redpajama_set_name: RedPajamaArXiv


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Home > india > My grandfather, Field Marshal Manekshaw
My grandfather, Field Marshal Manekshaw
'After he died, the phrase that struck me most about him was that 'Sam restored the izaat of the Army'.'
Field Marshal Sam Manekshaw's grandson Raoul-Sam Daruwala remembers the man behind the uniform.
Field Marshal Sam Hormusji Framji Jamshedji Manekshaw led India to its biggest military victory in the 1971 War.
A brilliant leader of men, he was India's first field marshal and a legendary soldier.
Two people who knew him closely were his grandsons Raoul-Sam Daruwala and Jehan Manekshaw.
"He really loved the Army. I don't think anyone who hasn't been in the army can understand what that feels like," says grandson Raoul-Sam Daruwala, a PhD in computer science who joined a start-up 16 years ago and lives in the United States.
"Sam was a Field Marshal and that meant a lot to him. The rank was a tremendous honour, the product of a career that he was so proud of."
"He knew that it was a form of recognition for all who served under him in 1971."
"I'm sure he would have been humbled to have received the Bharat Ratna, because so many great Indians had received one before him. But, to those who feel he's been slighted, I don't think he took it personally at all," he tells Rediff.com's Archana Masih in an interview over WhatsApp where he shares memories of his grandfather.
Did Field Marshal Manekshaw's home in Coonoor have a lot of memorabilia from his days as a soldier?
Not really. My grandmother was the decorator. Most of the memorabilia went to people who were interested in keeping it.
He also didn't really want to put a lot of weaponry in front of his grand-children. We never once saw a gun in the house.
I once found an official album of pictures from the 1971 War, some of the images weren't for younger eyes. It disappeared the next day.
Did men who had served with him or men who joined service after him come to see him? How often did you hear him speak in Gorkhali, Hindi, Punjabi — languages other than English?
I was three years old when he was promoted to Field Marshal so I never really knew him as 'the Chief'.
Over the years I'd spent time with him in Coonoor, New Delhi, Mumbai.
We'd bump into servicemen all the time and during those conversations he spoke Hindi, Punjabi and Gorkhali fluently.
He was always joking and laughing in languages I didn't know. I suspect I was the punchline in some of the jokes.
If someone had served under him directly, he'd remember just about everything about their careers in the army as well as the little details about their families.
I think he also spoke Pashtu (to some degree of fluency) from his time with the Frontier Force.
The evenings with the family were full of Gujarati stories.
What was his routine like? What were his interests?
He'd wake up at 5 am, sit down at his typewriter and then do all of his correspondence.
He had a few portable typewriters that he loved to work on. He'd open and reply to every letter sent his way. He'd get his work done early and then spend the rest of his time reading, gardening, telling stories and being a family man
He is known as the soldier's soldier — what was about him that made him connect so beautifully with that soldier on the ground or that soldier in the trench?
He really loved the army. I don't think anyone who hasn't been in the army can understand what that feels like.
I would get glimpses of it from the usual stories that people told me about him, or him about that life. But most of all you'd just see a difference in how totally engaged he was with people who had served with him.
After he died the phrase that struck me most about him was that 'Sam restored the izaat of the Army.'
I had to look that up, when I did it made perfect sense to me.
He was honourable, fair and respectful of his troops.
What are the lessons you have learnt from him?
Lessons taught by example:
Speak truth to power. He wasn't afraid of telling people that they were wrong. It isn't always easy but that doesn't mean you shouldn't do it. In my life, it's nearly always paid dividends.
Be respectful of those who are less privileged than you are. He always had time for the enlisted men.
Endeavour to get things 100% right, 100% of the time.
He was a perfectionist.
Do everything with a smile. It's so much easier than picking a fight. He also had a very disarming smile.
Enjoy your family. Enjoy yourself. Enjoy your friends.
Did he come to visit you in America? What did he say about your career?
No, he never came to the US.
He didn't really care one way or another about what I did for a living as long as it would make me happy.
What is it that you miss most about him?
I miss his company all the time. I also really miss going to every dhaba in sight in search of the best puris or rogan josh.
Do you think he got his due in later years? Do you think he should be awarded the Bharat Ratna?
I'm sure he would have been humbled to have received the Bharat Ratna, because so many great Indians had received one before him. But, to those who feel he's been slighted, I don't think he took it personally at all.
I took an awfully long time to complete my doctorate. He'd tease me all the time about how long it was taking me.
He'd been offered an honorary doctorate a number of times and turned them down saying, 'Son, how can I get my PhD before you do? Hurry up and finish it, won't you?'
I still took the slow route, but I did get there.
After I completed my PhD, he wrote to tell me that he accepted his.
I have both of our certificates. I plan to frame them together.
Feature Presentation: Aslam Hunani/Rediff.com
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