Be watchful- the grace of God appears suddenly. It comes without warning, to an open heart.
God's will can be a challenging concept; things like someone dying, someone enduring sickness, becoming handicapped, or losing a home or employment… Although these are negative things that happens, and can make God look like the bad guy- He only allows it for His own personal reasons. I can't help to think of Christ, who was despised, beaten and crucified for the redemption of many; but in his humanity wrestled with His own will even for just a second. He asked God the Father, Lord pass this bitter moment from me- and in the same prayer, then came to Himself and thought 'wait, never mind that God, please do your perfect will in my life." Someone who has not completely surrendered their will to God, will find themselves depressed, angry and faithless in the face of God's plans unfolding. We choose to die to ourselves, in order words we trade our plans and expectations for God's plans and expectations. But surrender is an ongoing daily process: many of us believe to be surrendered, until we face God's will with bitterness, restlessness, desperation or rebellion.
I remember Naomi, who lost her three sons and her husband- according to God's plan. He had orchestrated that Jesus Christ would come through the lineage of Boaz and Ruth. I remember Joseph, who was betrayed by his brothers and sold as a slave to Egypt. This was also according to God's plan for the people of Israel, otherwise they would starve to death and be lost as a people; God allowed his suffering for an appointed time. It's important to remember that God is only after our best interests, even if we have to wait for a future time. When we surrender all, we see God's mercy, favor and grace operating in the very areas which we surrender. The lord comforts us with His peace, His dreams and direction. We are no longer slaves, we are friends of God because He tells us what the Father says (revelation and strategy).
God had compassion on this widow woman, as well as many others in the bible. Scripture says that this boy was her only son, and the interesting thing about this story, is that the Lord told her 'don't cry'. if I'm going through a situation like this, and somebody tells me don't cry… I would be a little bothered. But I believe Jesus said this, because God cares about our emotional life and He is willing to have compassion on us. God's compassion changes our situations, which is why God asks us to live by faith and not by what we see. God has mercy on whomever He pleases, and ALL THINGS work for good for those who love God and who've been called according to His purpose. He is the God of suddenly, and can change our lives in an instant. The sweet thing about this story is that this woman did not ask God for a miracle! But Jesus helped her. Again, this reminds me about Joseph, Ruth and Naomi who did not ask anything from God, at least not according to scripture- but God treated them with favor. Have you been thinking about things, that may not be recorded anywhere for any to know- yet God does… do you have a situation where God can have compassion on you? God is the God of suddenly!
Jehovah, powerful and mighty God. I thank you for this word, and for the compassion you have on us such frail and weak vessels. May we be called to the resurrection of life, not to rebellion, bitterness, damnation, or shame. I pray for greater surrender in different areas of our lives, where we are able to give you what you require to excel in our purpose and calling. Just as the people witnessed do such a mighty work in the life of this family, may we be able to bring resurrection into people's lives. May those around us see your goodness, not your judgment. I pray for words in our mouths that may glorify you, and hearts that may yield to your love so that we may be able to love others. In Jesus name, Amen.


text:
Global Conversation
Covid Emergency
Fondazione Aurora
Our Historical Members
The Member Guide
City in metamorphosis
The Nagorno-Karabakh, 30 years lasting conflict, is still reality
30 September 2020 13 October 2020
chiaracriscenti
On Sunday the 27th of September, the heaviest clash since 2016 erupted, creating destruction and spreading fear.
The Nagorno-Karabakh conflict is an ethnic and territorial conflict that involves Armenia and Azerbaijan over the region of Nagorno-Karabakh and several other close districts, which are de facto controlled by the self-declared Republic of Artsakh.
Negotiations have so far failed to produce an enduring peace agreement and the fighting in the region remains one of Post-Soviet Europe's "frozen conflicts". While Nagorno-Karabakh is part of Azerbaijan according to international law, ethnic Armenians, who make up the large majority of the population at about 150,000 people, actually reject the Azerbaijani government. Since the collapse of the Soviet Union in the 1990s and the Azerbaijani advancement were pushed out during the Artsakh Liberation War, the region has been managing their own affairs, with support from Armenia.
At the end of this conflict, a ceasefire was signed in 1994, contributing to two decades of relative stability. However, after a four-day escalation in April 2016, another clash was initiated and is still ongoing. Both sides continue to accuse one another of starting the violence.
Where do other countries stand?
The international community is worried about the prospect of a greater war as Nagorno-Karabakh acts as a corridor for pipelines transporting oil and gas to world markets.
Turkey has already declared its support for Azerbaijan. In return, Armenia has accused Turkey of providing direct military support to help Azerbaijan gain control of the territory; a claim sternly rejected by Azerbaijan. Recently, Turkish President Recep Tayyip Erdogan, called on Armenia to immediately end its "occupation" of the region and withdraw, declaring that "this was the only way that would establish peace."
Many European countries such as; France, Germany, Italy, the United States, the European Union, and Russia urged for a ceasefire as they all have military bases situated in Armenia, nonetheless, maintaining friendly relations with Azerbaijan.
What is the situation like now?
The latest clash began on the morning of 27th September 2020, beside the Nagorno-Karabakh Line of Contact where both factions reported military and civilian casualties. In response to that, Armenia and the Republic of Artsakh introduced martial law and total mobilization, while Azerbaijan introduced martial law and a curfew, with Vilayet Eyvazov as the commander.
The press secretary of the President of the Republic of Artsakh, Vahram Poghosyan, stated that contrasts started when Azerbaijani armed forces launched artillery and aerial strikes against the civilian population including the capital city of Stepanakert. The authorities had to take action and strongly advised the population to seek cover in bomb shelters.
Following the Azerbaijani's use of force, the Armenian armed forces commenced shelling from mortars, weapons, and artillery on the positions of the Azerbaijan army along the frontline aiming at their settlements in the zone. By the afternoon Azerbaijani Ministry of Defence announced the retaking of seven of the villages in Nagorno-Karabakh. They also reported that the Azerbaijani Military Command proposed the Armenian one not to resist and to surrender so that the Armenian Army would not be completely destroyed and the death toll would not increase in the establishment of Aghdara, one of the main lines and not to multiply the number of deaths, also declaring that the treatment of prisoners of war and civilians would be realized in agreement with the conditions of the Geneva Convention and international humanitarian law. This has to be the worst Armenian-Azerbaijani crisis in recent years. It's a conflict that has left around 30,000 dead and with Armenian separatists taking control of the Azerbaijani territory, the number continues to grow.
© The Global Observer
GG Conversations
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TGO Guide

redpajama_set_name: RedPajamaCommonCrawl


text:
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Don't Hire a Veteran with Disabilities Because it Feels Good – Do it Because it's Smart Business
Nov 28, 2016 11:20 AM ET
Tweet This:
Hiring #veterans with disabilities is just good business. #MilFamStrong http://bit.ly/2gb2dZj
Hiring military veterans with disabilities isn't just feel-good altruism, it leads to strategic advantages. They're determined, self-aware, goal-oriented, determined, and inclusive. Today's veterans have led effective teams in ambiguous environments. They have managed large budgets, improvised around uncertainty and accomplished team objectives in lieu of personal ambitions.
Enter Eli Hinson, former Marine, and a wife and mother with two disabilities – dyslexia and deafness in her left ear.
Sit down and talk with Eli and you'll learn firsthand why hiring veterans makes good business sense. In her more than 16 years at global technology consulting firm, Booz Allen Hamilton, she has:
Managed backend cybersecurity for more than 40 internal websites.
Taken leadership roles in internal disability awareness programs.
Volunteered for local charity events.
Clearly, employing veterans has some obvious benefits. They're responsible, can execute commands in ambiguous professional environments, bring managerial skills to a team, and operate efficiently.
And yet veterans and the disabled are still stigmatized in the hiring process. Sometimes they are stereotyped for not being able to handle stress due to media portrayals of PTSD. Disabled veterans often get labeled charity cases for marketing purposes instead of being highlighted as valuable employees.
Recruiters and human resource professionals must realize employing veterans and people with disabilities is crucial to enterprise-wide growth. According to the I4CP study, high-performing companies are 37 percent more likely than low-performing companies to hire people with disabilities. That's a statistic that shows a clear correlation between success and hiring a diverse workforce.
Eli gained a host of technical skills during her time as a Marine field wireman in Okinawa, Japan.
She encoded, decoded, and set up complex cabling systems; identified sources of communications issues; and fixed multiple cables either hundreds of feet overhead on telephone poles or in muddy, bug-infested tunnels.
Eli utilized her technical experiences to land a job at Booz Allen. During her day-to-day, she is given the tools and resources to excel with her disabilities, like a read-back program on her laptop. Since joining the firm, she has earned three consecutive promotions.
In addition to going above and beyond in her day-to-day, she represented Booz Allen in the 2016 D.C. Dancing with the Stars charity event, a competition modeled off the eponymous reality TV show. She raised a remarkable $40k for the nonprofit Operation Renewed Hope (ORH), an organization that fights veteran homelessness. Her reasons for choosing ORH are intensely personal.
"I could have easily ended up homeless," Eli says. "When I returned from my service, my daughter was barely a month old and my husband had titanium rods in his back preventing him from working. I didn't have much in the way of savings and if I hadn't found employment with Booz Allen things could have ended up very differently for me."
Organizations should understand that diversity and inclusion (D&I) in recruiting practices helps acquire motivated employees, ultimately supporting the bottom line.
As Eli puts it, "These issues need to stop being so taboo. Once we bridge that cultural gap, and get people to actually start talking and familiarizing themselves with the disabled and veteran communities I think organizations will realize what a powerhouse talent pool they're missing out on."
Booz Allen Hamilton is a proud supporter of the more than 1.4 million Armed Forces service members, and their families. To read more about the firm's commitment to veterans, visit: http://www.boozallen.com/insights/better-our-world/supporting-veterans
Military Family Appreciation Month
More from Booz Allen Hamilton
How Booz Allen's Kathleen Featheringham Brings People and Machines Together
Jan 15, 2020 8:30 AM ET
Booz Allen CEO Wins Horatio Alger Award for Commitment to Education and Community Impact
Jan 9, 2020 7:00 AM ET
D.C. Tech Community Recognizes Booz Allen for Innovation Center, Entrepreneurial Support
Booz Allen Expands Its Colorado Footprint—and Space Capabilities
Year in Review: How Booz Allen Strengthened Its Commitment to STEM
Dec 31, 2019 7:00 AM ET
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Of all the ingenious qualities that have been nurtured within man, that of perception is immeasurably powerful and stimulating. Perception lures man to peer into the waters of inexhaustible possibilities.
Initially, there is perception, then comes a process of creative, exploratory, and conceptual thought, which leads to action. The original perception is neutral, then thought begins a search progression through old roots of embedded values, and supporting evaluations that lie deep within the mind and body, while at the same time striving to built new ones. In order for perception to be expressed, it must become charged either emotionally or intellectually. These two types of perception will have a dramatic impact on one's mental attitude. Through filtering your emotionally charged perception and balancing your intellectual perception, you will be able to grow toward self-mastery with an expanded, more harmonious, and infinite consciousness. Perception is an invisible field of overlapping realities that are engaged by one's own vibrational viewpoint. Most individuals are not aware of the overlapping of realities, until realities reveal themselves in relationship to changes in perception. This results in growth of mental attitudes and expanded awareness and possibilities. Conscious awareness expands and contracts. I expand and contract. The Earth expands and contracts reflecting a harmonious natural intelligence functioning in Oneness through divine expression. Perceptions are windows into realities, be they fixed or flexible. Thoughts are doorways to possible realities. Perception exists in perfect form and imperfect relationship with itself. The perceptions we have are the reality we create.
The obstacles to martial art's excellence lie within each of us. One of these obstacles is FEAR. Fear is real. Yet, fear is unique because in actuality, fear is an illusion and a misunderstood reflection of an individual's perception, experience, and lack of knowledge. For a martial artist, it is extremely detrimental. Fear has the capacity to choke out all creative expression, and can create doubt in the heart of every man. It can cause individuals to shake with the sheer thought of a thing. Fear expresses itself in many different locations throughout the body, and a martial artist must be aware of where fear resides. Fear expresses in the external body by freezing motion and instilling hesitation, locking down movement, creating rigidity in the muscles. Next, it can reside in the organs, creating extra stress, and it has the capacity to steal away the courage that resides in the heart, and causing anxiety and heart palpitations. Stress can also cause serious damage to organs and body cells. Very importantly, fear can attack one's mind, scattering thoughts, and leaving doubt and worry to enter. Thus, it gives the martial artist an altered reality or perception of what he is actually facing. This allows it to steal opportunity, and hinder growth and development.
So, a martial artist, should not give in to fear and indulge in fearful thoughts, but to master fear, and utilize it to his benefit. For a martial artist to dedicate his time trying to eliminate fear would be a life-long endeavor, for we are bombarded with illusions or perceptions that induce fear. Time is better spent trying to understand the mechanisms that stimulate fear and those connections that extend inwardly to himself from the physical world. We have ways to do this. Fear in the body can be released through relaxation techniques. Fear in the heart can be released through deep breathing exercises and Qi regenerating exercises. Fear in the mind can be relieved through meditation, visualization and self-analysis, and focused intentional thought. Remember, fear's effect is to keep you from accomplishing a thing, and your job is to forge your way through fear, coming out on the other side more determined and stronger than before.


The Comptroller- General of the Nigerian Prisons Service (NPS), Mr Ja'afaru Ahmed, has ordered immediate redeployment of 32 senior officers who were recently elevated to the position of Deputy Comptrollers General of the Prisons Service.
A statement by the Public Relations Officer of the NPS, Mr Francis Enobore on Wendesday in Abuja said this was contained in a posting instruction issued on the 1st of August, 2017 and signed by the CGP.
He said this followed the need to have the recently promoted Deputy Controllers General take charge as substantive Directorate heads and also address some administrative gaps pursuant to the present administration's drive for excellent service delivery.
In the posting, DCG Shehu Kangiwa heads the Finance & Account Directorate while DCG Dorathy Atagiri takes charge of the Directorate of Health & Social Welfare Services.
DCG MBE Shemfe heads the Directorate of Operations and DCG Charles Ahaotu heads Works & Logistics Directorate. DCG Ogundana Jerome who was the Zonal Coordinator in charge of Zone A Lagos now takes charge of the Directorate of Inmates Training & Productivity.
Others include, ACG Emmanuel Ogundele mni who takes over Prisons Academy, Ijebu-Igbo as Commandant, ACG KA Babalola moves to Zone A, Lagos as Zonal Coordinator. ACG AI Omale, CP Bukar Shettima Hauwa and CP KN Okuori proceed on retirement.
The redeployment also affected 21 other senior officers across board.
Recall that on 14th July, 2017, 5 Assistant Controllers General promoted to the rank of Deputy Controllers General of Prisons were formally decorated by the Honourable Minister of Interior, Lt. Gen. Abdulraman Bello Dambazau (retd), CFR, PhD.
"Ja'afaru while wishing the retiring officers well in their future endeavours charged the Directorate heads and other command officers to deploy all available resources at their disposal in order to add value to their output," the statement said.


text:
Accountability flaws could hinder IMO's climate goals
Several transparency and accountability
flaws could hinder the International Maritime Organization's (IMO) ability to
deliver on its own climate goals in reducing carbon emissions, a new report
by Transparency International showed.
The report entitled, "Governance at the
International Maritime Organisation: The Case for Reform," outlines several key
policy issues and recommendations that the IMO, the United Nation's leading
shipping agency, must address in order to meet international standards for
transparency, accountability, and integrity.
The Berlin-based non-governmental
organization said that these changes are essential if the IMO is going to
honour its environmental and climate mandates and reach a reduction of
greenhouse gas emissions of at least 50 per cent by 2050.
"Our biggest recommendation is to
transform the IMO's accountability policies, which are currently hindering
policymaking and leaving the agency susceptible to private influence. While the
IMO's initial strategy adopted in April is a big step forward for the
international shipping sector, more must be done to ensure the agency meets its
targets," said Rueben Lifuka, vice
chair of Transparency International.
Currently, if left unchecked, shipping
emissions could grow from 2.5 percent to 17 percent by 2050. However, to limit
the rise in global temperatures by one-and-a-half-degree, as outlined in the
Paris Agreement, some research suggests this number must actually reach zero by
As the principal international body
tackling shipping issues, the IMO has a significant role to play in curbing
emissions. However, in its current set-up, the IMO is at risk of severely
under-delivering on its targets.
The report finds several specific flaws
in the IMO governance structure, including a disproportionate influence of
private industry over the IMO and an unequal influence of certain Member States
in the policymaking process. In addition, the agency suffers from an increase
in privately-operated registries in states that serve as tax havens and a lack
of delegate accountability.
As such, Transparency International has
urged the IMO and its Member States to take the following immediate actions:
– Engage in open dialogue with civil
society, industry and other key stakeholders to improve transparency and remove
current restrictions on journalists covering IMO meetings and speakers.
– Ensure decision-making processes are transparent and reflect the public
interest by developing rules for the appointment of Member State delegations
and third-party representatives.
– Ensure decision-makers are subject to robust integrity rules by expanding the
mandate of the Internal Oversight and Ethics Office to detect and investigate
potential breaches in the code of conduct and provide subsequent sanctions, if
necessary, or refer to national authorities, as appropriate. In addition,
extend the IMO's whistleblowing and complaint policies to include Member State
representatives.
World Maritime News.
AccountabilityInternational Maritime OrganisationmaritimeNewsspotlightTransparencyWhistleblowing
SIFAX Group sponsors The Next Titan
Tanker Hazards: NRC calls for patronage of its Tank Wagons for movement of petroleum products


text:
Official Guidebook of the 1939 New York World's Fair
World's Fairs Collection, 1-25, Hofstra University Library Special Collections
The Official Guidebook of the World's Fair was the go-to informational guide for visitors. It included maps, advertisements, ticket information, show times, and virtually everything else a patron may need to navigate the fairgrounds. The descriptions of the attractions and amusements provide a first-hand look at what the fair must have been like to experience. One interesting entry in the book describes the "Town of Tomorrow" provided by Crane Plumbing and Heating in the Community Interests Zone. The book lists the average home price of around $20,000. The guidebooks were updated each season.
← The 1939 World's Fair Comes to Hofstra College
Official Guidebook of the 1940 New York World's Fair →
The New York World's Fairs: The "World of Tomorrow" Celebrated Today.
"The Flushing Meadow Improvement," November, 1936
1939 World's Fair Certification of the Budget
New York World's Fair Inc. Press Release, 1939
Long Island Tourism and the 1939 World's Fair
Forest Hills-Kew Gardens Post, "Michael Radoslovich A Designer of the Fair"
The 1939 World's Fair Comes to Hofstra College
1939/40 New York World's Fair Official Guides
The Trylon and Perisphere
"Your World of Tomorrow" inside the Perisphere
The George Washington Statue at the 1939 World's Fair
The Government Zone at the 1939 World's Fair
Salvador Dali's "Dream of Venus" at the 1939 World's Fair
"Bring 'Em Back Alive" Frank Buck at the 1939 World's Fair
"108 World's Fair Recipes from Borden's!"
The Ford Exposition at the 1939 World's Fair
The Story of Lucky Strike
The Pennsylvania Railroad at the Fair
Postcards from the 1939 New York World's Fair
Tickets for the 1939 and 1940 World's Fair
Souvenir Spoons, 1939 New York World's Fair
Mr. Peanut Bowls, 1939 World's Fair
Publications from the 1939 and 1940 New York World's Fair
The 1939 and 1940 New York World's Fair in Pictures
Maps of the 1939 and 1940 New York World's Fair
"The Fair: Four Months to Go"
The Unisphere: Symbol of the 1964-1965 New York World's Fair
Parker Pen Pavilion Construction Photographs
1965 New York World's Fair Official Guidebook
"Our Fair City: a map guide to New York City prepared for visitors to the 1964-65 World's Fair as a service of the Chase Manhattan Bank"
The Jack Pickens Coble Home and the "House of Good Taste" Pavilion
Walt Disney and Technological Innovation at the 1964 New York World's Fair
The AMF Monorail at the 1964 New York World's Fair
Tickets and Stamp, 1964 New York World's Fair
"A Visit to the New York World's Fair," Official Storybook
Barbara Kingsley at the 1964 New York World's Fair
Commemorative Coin, Coaster, and Ashtray from the 1964 New York World's Fair
Wooden Shoes from the Sweden Pavilion, 1964 World's Fair
1964 New York World's Fair Wooden Bowl
1964 World's Fair License Plates
Hofstra University at the 1964 New York World's Fair


Ranil Wickremesinghe, Sri Lanka's four-times Prime Minister and the futurist largely credited with forming a national government with arch rival, SLFP - a first of its kind - has been in politics since the late '60s.
He got into active politics as a member of the Youth League of the UNP while in the university. Wickremesinghe, as a young university student, like many others, was not too happy with the educational reforms proposed by the Education Minister of the Dudley Senanayake Government. During that time, Wickremesinghe liaised between Prime Minister Senanayake and university students, by working closely with the Prime Minister, an experience he fondly remembers to date.
Although entering politics was not his childhood dream, the major factor that drew him deep into politics was the deleterious policies of the then Sirima Bandaranaike government.
The young lawyer in the making, his compatriots and other young people of the UNP wanted to rejuvenate the party to face future political challenges.
The sentiment expressed by many that the policies of Prime Minister Bandaranaike were intruding into the democratic rights of the people more or less rekindled their inspiration to explore the right path through democratic means.
"It was disastrous. Gradually, the restrictions on democratic rights became worse - we had to fight back," Wickremesinghe recalls.
People expected President J.R. Jayewardene's brother, H.W. Jayewardene QC and erudite civil lawyer to come forward. When he refused the mantle automatically fell on Wickremesinghe. When Wickremesinghe was first offered Kelaniya, his reply was in the negative.
In a recent interview with the Sunday Observer at the Temple Trees, a confident and futurist Prime Minister answered our questions on a wide range of topics from politics to good governance and literature.
Q: Do you see yourself as a veteran in statecraft?
A: A veteran in politics; 40 years in Parliament and 45 years in active politics. I suppose that is the spirit of a politician.
Q: What were the most crucial political decisions you had to take in the past four decades?
A: First, I was involved in some of the crucial decisions that were made by other leaders such as President J. R. Jayewardene and President R. Premadasa. On my own - yes, I had to take some crucial decisions as well.
One of the earliest decisions in my political career was to press for the free trade zones (FTZ), and the decision to build an FTZ in Biyagama.
As the Minister of Education I took decisions on improving the quality of education, not allowing politics to enter into the Ministry of Education and emphasizing on teacher development.
I am quite happy about my decision to set up the National Youth Services Council and the National Apprenticeship Board - the two institutions that continue to support thousands of youth to launch their careers here and abroad.
Another crucial decision was to stand for national unity when the riots took place in 1983 and to look at the political solution that President Jayewardene proposed. I was involved in most of the major decisions taken by President Jayewardene.
During the time of President Premadasa, the foremost step that impelled me to take was to stand by him during efforts by some to impeach him through Parliamentary procedures. My efforts were successful which prevented a breakup of the party – that itself was a very important achievement in my political life.
Elaborating on his political achievements Wickremesinghe said that as the Minister of Industries, he actively supported President Premadasa's economic agenda by getting the IMF and other donors to stand by Sri Lanka, thus ensuring economic growth.
Treading on contemporary politics and its fallout the Prime Minister recalled one of the most poignant and critical moments in his political life. It was when President Premadasa was assassinated in 1993. "We were heading for another crisis," he said.
The Prime Minister vividly explained how he had to take another major step to stabilize the country - that was to appoint then Prime Minister D.B. Wijetunga as the President after consulting then Leader of the Opposition Sirimavo Bandaranaike. And I was chosen to be the Prime Minister during the Presidency of D. B. Wijetunga.
Our next challenge at the time was to turn around the economy which had negative growth. We brought the economy forward, while giving my best to run the cohabitation government.
Amongs the more recent successes, the Prime Minister said that one of the most crucial decisions in his political life was to support a common candidate on a common platform for the Presidency in 2015 where I was made prime minister of the national government. After the general elections in 2015 I became the Prime Minister again under President Maithripala Sirisena.
Q: In 1987 when the Indo-Lanka agreement was signed and after the 13th Amendment, what were the problems the government faced? The LTTE refused to surrender in terms of the Indo-Lanka accord.
A: President Jayewardene expected normalcy somewhere in 1988. Then the LTTE u-turned. But Indian defense minister K C Pant was here when it happened. So we spoke to him and he committed the IPKF to take over the role of ensuring that the LTTE was brought under control.
Q: Looking back, how do you analyze the 2002-ceasefire Agreement? And what support did you get from the government?
A: I think it was a good decision still. It was a necessary step, given the time and circumstances. Our economy had virtually broken down; there were military setbacks. We needed that ceasefire. It was an opportunity for Tamil people to realize how a peace process could work - for both sides there were reasons for the ceasefire. The LTTE couldn't win the war anyway in those circumstances. Before that, even former president Chandrika Kumaratunga had tried it in 2001 after the LTTE attack on the airport, but it didn't work.
Party support - We had a mandate to go for the peace accord - we had a majority in Parliament and a lot of people in the country wanted it.
Q: How did former President J.R. Jayewardene influence you to become a political leader?
A: I worked very closely with President Jayewardene, in the opposition and the government. We talked a lot about the country's issues. I learned a lot from him, on how to handle state affairs. I was not a contender for a post or a successor. He faced some of the most difficult issues during his tenure. He had a very successful program under the government, to take the economy forward. But when the 1983 riots broke out, everything got disrupted and we picked up the pieces and started again. I remember him at that stage giving advice to us, repeating a verse from Rudyard Kipling's famous poem - 'IF'.
Rebuilding - And from '83 onwards, we learned the art of rebuilding. Then under President Premadasa we learned how to revive the economy at times of war. Through the Ministry of Industries throughout the war, we launched a huge economic development program.
The focus on industrial promotion, infrastructure development and industrial estates in rural areas, 200 garment factories and the promotion of tourism were key areas.
I also must mention that I had the privilege of working with Prime Minister Dudley Senanayake during the last part of his life, Presidents J. R. Jayewardene, R. Premadasa and D.B. Wijetunga.
I must say that President Premadasa's is yet another experience - his approach is different. I am quite happy that I backed him as the presidential candidate and supported him throughout. He had a very grassroots approach. His policies were people-oriented. I think he, together with President JR Jayewardene, were proponents of the social market economy, but he was more into explaining that to the people than President Jayewardene.
I also worked with President Chandrika Kumaratunga in the cohabitation government. I certainly had no issue from my part although there were ups and downs until November 2003. Apart from that, I also had the opportunity to know former political leaders such as S.W.R.D. Bandaranaike, Sirimavo Bandaranaike and Dr. N.M. Perera, Colvin R. de Silva and Bernard de Soysa. Some of them were my father's friends as well. I also knew Mahinda Rajapaksa for a long time.
Q: As a political veteran, what lessons would you like to share with young people and would-be politicians?
A: I am still learning in politics, and keep learning. Whether you are in politics or in any other profession you are in the University of Life - you never can graduate from it.
You must learn to face adversity in life - not to go under. There has to be a certain amount of detachment in yourself. This applies to everyone. The main thing is to 'hang on' - what I have learned.
Q: People describe you as a resilient leader. Do you agree?
A: I think in the past 30 - 35 years, the whole country has become resilient - now we have to build up- grab opportunities. The national government is a new experience for all Lankans.
Q: Is it difficult to run a coalition government?
A: Running a government is difficult even with your own party; so it is more difficult to run a coalition government. It is also about how you work with Parliament. If you are willing to consult Parliament and work alongside with them, it is fair. Running a government is not only about forming a government. It is time to ensure that Parliament gets involved, participate in the process of government.
Q: Prime Minister, you were proposing oversight committees - what are the latest developments?
A: That also is a recommendation we have been making for a long time. I am happy we brought it in, because it is working well now after initial setbacks. Then we have also held that the Public Accounts Committee and the Committee on Public Enterprises should not be headed by members of the government and the UNP has adhered to it. We also created a Committee on Public Finance. Now we want to bring in legislation to establish the parliamentary budget office. Also, we have agreed in principle that the J. R. Jayewardene Centre will have the added task for parliamentary research, and running legislative information service for MPs; the centre will do research for them and we are amending the Act and the centre will be under the supervision of Parliament. It also gives us the opportunity to have the centre separate from Parliament which means a lot of things can be done without having the security checks in Parliament which makes many people reluctant to come into the building now. The Speaker and I are taking some of these additional responsibilities and we need more space outside Parliament. We decided to undertake the present Agriculture Ministry for this work and the Ministry has put in a Cabinet paper to move elsewhere which decision has been much criticized. The reason is, we needed buildings near Parliament.
Besides, we got independent commissions working; the Constitutional Assembly (CA) is working well and the basic framework for democracy is established now.
Q: We are a country burdened with debts. What is your next step to increase revenue?
A: When we took over we could not even pay the interest on public debt; we had to borrow. Now with macroeconomic stabilization, the revenue has increased. At least we are able to pay the interest on our debts.
We have to increase revenue - again we are criticized for that. Our revenue has dropped to 10 or 11 percent of the GDP; taxes are not collected; people with political influence are getting away. Our plan is to have the revenue to become 15% of GDP by 2020. I remember earlier we had revenue of 20% and then came down to 18 %. With this you can see the loss.
We will have to reschedule our debts or at least go in for new borrowings to repay old loans. That's why we are going in for an investment strategy. Once the investments are coming in, we will be earning more foreign exchange. So our strategy has to be an investment-driven, outward-looking, export-oriented one. After the next five years we will be in a position even to repay our debts. When GDP expands, and the economy stabilizes hopefully by 2020, our burden will be less. But the fact is we will have to pay our debts. We do not want to keep this burden for the next generation. Now we are starting on our investments - then from 2020, to 2025 the economy should be picking up so we can repay our debts.
Q: Prime Minister, you've been positive about the Indian port development sector.
A: We are the transshipment point of India; if India develops its ports, it is beneficial to us. Besides we are working with India to develop the Colombo and Trincomalee ports. Japan is also supporting us to develop these two ports. We want to make it the hub of the Indian Ocean with three harbours and two airports - maybe three airports later with another in the Eastern or the North Central province - but, that's a long way off. I think we should be able to get a Private Public Partnership (PPP) agreement on Mattala soon.
Q: How are the other key development projects progressing?
A: There are plans for every district. Through the Hambantota port agreement the government expects USD. 1.1 billion - that will reduce our debts from USD 26 b to 25 b.
We have two corridors - South West Corridor from Kandy to Kurunegala, Kuliyapitiya, Chilaw, Galle, Matara, Hambantota, up to Moneragala, connecting two harbours and two airports.
Surbana Jurong of Singapore is working on the Eastern Development plan, including the Trincomalee port activities and the tourism industry, down to the Southern part of Arugam Bay, to Kuchchaveli, then under the President's supervision to Polonnaruwa , Moragahakanda, Malwathu Oya and the cultural triangle.
Among the other major plans are Colombo metropolis which is underway; Kandy city development project which is being done with Japanese assistance; Kurunegala will have a major development program with Central Expressway, Chilaw tourism development project and the Matara/Dedduwa tourism zone. We want to develop Hingurakgoda as the main regional airport and the Aviation Minister Nimal Siripala de Silva is currently working on this project. The development of heritage properties in Matara and Galle are being looked at. Kalutara will be developed as a tourism and industrial district. And economic development projects for North and East are another priority. Colombo will be the main financial services centre.
The development process needs two major ingredients; one is reconciliation and peace, which we are moving towards, putting the past behind us. Secondly, the democratic structure for a stable government.
Q: What are your views on the media's responsibility?
A: The problem is most people in the media have got the background experience of a different era (Rajapaksa era) and government intervention in economics. It got inculcated with the Rajapaksa theory of development. None of them today talk about democracy and doing away with corruption - I wonder what they were doing during that era!
I hope now that new people are coming in, it is going to be different. There are several new publications that have come up after 2015; but you have to give them time and space to change and we are not asking them to give uncritical support to the government.
But there are key areas and certainly you should not go in for extremism in race and religion. The country voted against it, we cannot go back.
Q: Your decision to implement the Right to Information Act (RTI) is a great leap towards democratizing society. What compelled you to do so?
A: The RTI was the draft we prepared in 2003; cabinet approved it; went to Parliament. Then the new government which came in after 2004 promised to go ahead with it, but they didn't.
As far as the UNP is concerned we are committed to the RTI; so was the civil society. When we won this election, we went for it. Many were interested in it, but hesitant to implement it. So when we came to power we wanted to do it; there are shortcomings - my approach was 'let us implement it - and let us amend it' after looking at it for two years. Had we started looking at every possible scenario we wouldn't have got the legislation. We have now identified the shortcomings of the RTI - the fact is that we got it - it is an achievement.
Q: You are also the longest serving party leader of the UNP. What is the secret?
A: There's no secret. Remember, in 1994 all the party leaders were pruned; I had to hang on and wait, and to build up. Some of them left politics. We went through a lot of party issues - our people got killed. But I took the challenge of keeping the party together. I think no party leader had to go through the experience I have had.
Q: For Sri Lankan women, irrespective of their achievements in many other fields, entering politics is still a distant dream.
A: Young people and professional women did not enter into politics due to the preferential voting system which costs money. From 1977, the focus was lost, especially in the past 10 years. The new system will ensure that more women are entering the political system, beginning from Pradeshiya Saba level.
Q: You are an avid reader - who are your favourite authors?
A: The best book I have read is the Dhammapada. I read on politics, history and world affairs. I used to read Sinhala books as well. But today I don't find many good books of Sinhala literature.
But I think this is due to the suppression of democracy over the past decade. Unless you have a really free society, peoples' creativity skills would not come out freely.
Art suffered under this political patronage of the previous regime.
We have to catch up and I think in the next few years we would see a revival of art, theatre and music. Once you've killed, suppressed democracy - the right to expression - it takes time to revive.
Q: How do you take criticism?
A: You reply when you have to. You are not what you were five minutes ago.
Q: Prime Minister, in five years from now, where would you see Sri Lanka?
A: My belief is a better nation and a strong economy; gradually build up our saying in international affairs. More than that, I want to see a Sri Lanka where younger people and the newer generation will be confident of their future. The world is transforming. And to know how to capitalize on it and how to play our cards well is important.
But, Shakesphere is the best!


text:
\section{introduction}

Let $ (\Sigma , g ) $ be a compact oriented Riemannian surface of negative curvature
and denote by $ \mathcal G $ the set of primitive closed geodesics on
$ \Sigma $ (counted with multiplicity). For $ \gamma \in \mathcal G $ denote by $ \ell_\gamma $ its length. The Ruelle zeta
function \cite{Ru1} is defined
by the analogy with the Riemann zeta function,
$ \zeta ( s ) = \prod_p ( 1 - p^{-s} )^{-1} $, replacing primes
$ p $ by primitive closed geodesics:
\begin{equation}
  \label{eq:defRuz}
\zeta_R ( s ) := \prod_{ \gamma \in \mathcal G } ( 1 - e^{ -s  \ell_\gamma } ) .
\end{equation}
The infinite product converges for $ \Re s \gg 1 $ and the meromorphic
continuation of~$ \zeta_R $ to $\mathbb C$ has been a subject of extensive study.

Thanks to the Selberg trace formula the order of vanishing of $ \zeta_R ( s ) $
at $ 0$ has been known for a long time
in the case of {\em constant curvature} and it is given by $ - \chi ( \Sigma ) $ where $ \chi ( \Sigma ) $ is the Euler characteristic. We show that the
same result remains true for {\em any} negatively curved oriented surface:

\vspace{0.04in}

\noindent
{\bf Theorem.} \emph{ Let $ \zeta_R ( s ) $ be the Ruelle zeta
function for an oriented negatively curved $C^\infty$ Riemannian surface
$ ( \Sigma, g ) $ and
let $ \chi ( \Sigma ) $ be its
Euler characteristic.
Then $s^{\chi(\Sigma)}\zeta_R(s)$ is holomorphic at $s=0$ and
\begin{equation}
\label{eq:t1}
s^{\chi(\Sigma)}\zeta_R(s)|_{s=0}\neq 0.
\end{equation}
}

\vspace{-0.05in}

\Remarks
1. The condition that the surface is $ C^\infty $ can be replaced by $ C^k $ for a sufficiently
large $ k $~-- that is an automatic consequence of our microlocal methods.

\noindent
2. As was pointed out to us by Yuya Takeuchi,
our proof gives a stronger result in which the cosphere
bundle $ S^* \Sigma = \{ ( x , \xi ) \in T^* \Sigma :
| \xi|_g = 1 \} $ is replaced by a connected contact 3-manifold~$ M$ whose
contact flow has the Anosov property
with orientable stable and unstable bundles (see \S\S \ref{s:resonances},\ref{s:gf}).
If $ \mathbf b_1 ( M ) $ denotes the first Betti number of $ M $
(see \eqref{e:Betti}) then $ s^{ 2 - \mathbf b_1 ( M ) } \zeta_R ( s ) $
is holomorphic at $ 0 $ and
\begin{equation}
\label{eq:b1M}
s^{ 2 - \mathbf b_1 ( M ) } \zeta_R ( s ) |_{s=0} \neq 0 .
\end{equation}
Theorem above follows from the fact that for
negatively curved surfaces $ 2 - \mathbf b_1 ( S^* \Sigma ) = \chi( \Sigma ) $ (see
Lemma \ref{l:topo} for the review of this standard fact).  For the
existence of contact Anosov flows on 3-manifolds which do not arise from
geodesic flows see \cite{HF}.

\noindent
3. Our result implies that for a negatively curved connected oriented Riemannian surface,
its length spectrum (that is, lengths of closed geodesics counted with multiplicity)
determines its genus. This appears to be a previously unknown inverse result~-- we refer
the reader to reviews~\cite{RBM,Wilkinson,Zelditch} for more information.

For $ ( \Sigma , g) $ of constant curvature the meromorphy of $ \zeta_R $ follows
from its relation to the Selberg zeta function:
$$
\zeta_S ( s ) := \prod_{ \gamma \in \mathcal G } \prod_{ m=0}^\infty
( 1 - e^{- ( m + s ) \ell_\gamma } ) , \quad
\zeta_R ( s ) = \frac{ \zeta_S ( s  ) }{ \zeta_S ( s + 1  ) } ,
$$
see for instance \cite[Theorem~5]{Mark} for a self-contained presentation.
In this case the behaviour at $ s = 0 $ was analysed by Fried \cite[Corollary 2]{Fr1}
who showed that
\begin{equation}
\label{eq:fried}    \zeta_R ( s ) = \pm ( 2 \pi s )^{ |\chi ( \Sigma )| } ( 1 + \mathcal O ( s ) ), \end{equation}
where $ \chi ( \Sigma ) $ is the Euler characteristic of $ \Sigma $. A far reaching
generalization of this result to locally symmetric manifolds
has recently been provided by Shen
\cite[Theorem 4.1]{Sh} following earlier contributions
by Bismut~\cite{Bismut}, Fried~\cite{Fr-Anator}, and Moskovici--Stanton~\cite{Moskovici}.

For real analytic metrics the meromorphic continuation of $ \zeta_R ( s ) $
is more recent and
follows from results of Rugh \cite{Ru} and Fried \cite{Fr2}
proved twenty years ago.
In the $ C^\infty $ case (or $ C^k $ for sufficiently large $ k $)
that meromorphic continuation is very recent.
For Anosov flows on compact manifolds it was first established
by Giulietti--Liverani--Pollicott~\cite{glp} and then by Dyatlov--Zworski~\cite{DZ}.
See these papers and
\cite[Chapter 4]{ZS} for more references and for background information.
 { Here we only mention two particularly
relevant contributions:  \cite{DG} where the more complicated non-compact
case is considered
and \cite{River} where microlocal methods are used to
describe the correlation function of a Morse--Smale gradient flow.}

The value at zero of the dynamical zeta function for
certain two-dimensional hyperbolic open billiards was computed by Morita~\cite{Morita}
using Markov partitions. It is possible that similar methods could work in our
setting  because of the better regularity of stable/unstable foliations
in dimensions 2. However, our spectral approach is more direct and,
as it does not rely on regularity of the stable/unstable foliations,
can be applied in higher dimensions.

The first step of our proof is the standard factorization of $ \zeta_R $
which shows that the multiplicity of the zero (or pole) of $ \zeta_R $
can be computed from the multiplicities of Pollicott--Ruelle resonances
of the generator of the flow, $ X$,  acting on differential forms~-- see~\S\S\ref{s:resonances},\ref{s:zeta}. The resonances are defined as eigenvalues
of $ X $ acting on microlocally weighted spaces
-- see~\eqref{e:mercon} which we recall from the
work of Faure--Sj\"ostrand \cite{fa-sj} and \cite{DZ}.
The key fact, essentially from \cite{fa-sj} -- see \cite[Lemma 5.1]{DGF}
and Lemma \ref{l:res-states} below~-- is that
the generalized eigenvalue problem is equivalent to solving the equation
$ ( X + s )^k u = 0 $ under a {\em wavefront set condition}.
We should stress that the origins of this method lie
in the works on anisotropic Banach spaces by
Baladi~\cite{bal},
Baladi--Tsujii~\cite{bats},
Blank--Keller--Liverani~\cite{b-k-l},
Butterley--Liverani~\cite{but-liv},
Gou\"ezel--Liverani~\cite{go-liv}, and
Liverani~\cite{liverani,liverani2}.

Hence we need to show that the multiplicities of generalized eigenvalues at $s=0$
are the same as in the
case of constant curvature surfaces (for detailed analysis of
Pollicott--Ruelle resonances in that case we refer to \cite{DGF} and \cite{g-w}).
For functions and 2-forms that is straightforward. For 1-forms
the dimension of the eigenspace turns out to be easily computable using
the behaviour of $ ( X + s)^{-1} $ near $ 0 $ acting on functions
and is given by the first Betti number. That
is done in \S \ref{s:1fo} and it works for any
contact Anosov flow on a 3-manifold.
In the case of orientable stable and
unstable manifolds that gives holomorphy of
$ s^{ 2 - \mathbf b_1 ( M ) } \zeta ( s ) $ at $ s = 0$.

To show \eqref{eq:b1M}, that is to see that the order of vanishing is exactly $ 2 - \mathbf b_1 ( M )  $, we need
to show that zero is a semisimple eigenvalue, that is its algebraic and geometric
multiplicities are equal. The key ingredient
 is a regularity result given in Lemma~\ref{l:mystery}. It holds
for any Anosov flow preserving a smooth density and could be of independent
interest.

\medskip\noindent\textbf{Acknowledgements.}
We gratefully acknowledge partial support by a Clay Research Fellowship
(SD) and by the National Science Foundation grant DMS-1500852 (MZ).
We would also like to thank Richard Melrose for suggesting the proof of
Lemma \ref{l:de-rham}, Fr\'ederic Naud for informing us of
reference \cite{Morita} and the anonymous referee for helpful comments. We are particularly grateful to Yuya Takeuchi
for pointing out that a topological assumption made in an earlier version was
unnecessary -- that lead to the stronger result described in Remark 2 above.

\section{Ingredients}
\label{gepr}

\subsection{Microlocal analysis}

Our proofs rely on microlocal analysis, and we briefly describe microlocal tools
used in this paper providing detailed references to~\cite{ho1,ho3,ev-zw,DZ} and
\cite[Appendix E]{res}.

Let $M$ be a compact smooth manifold and $\mathcal E,\mathcal F$ smooth vector bundles over $M$.
For $k\in\mathbb R$, denote by $\Psi^k(M;\Hom(\mathcal E,\mathcal F))$ the class of pseudodifferential
operators of order~$k$ on~$M$ with values in homomorphisms $\mathcal E\to\mathcal F$
and symbols in the class $S^k$; see for instance~\cite[\S 18.1]{ho3}
and~\cite[\S C.1]{DZ}.
These operators act
\begin{equation}
  \label{e:acting}
C^\infty(M;\mathcal E)\to C^\infty(M;\mathcal F),\quad
\mathcal D'(M;\mathcal E)\to \mathcal D'(M;\mathcal F)
\end{equation}
where $C^\infty(M;\mathcal E)$ denotes the space of smooth sections
and $\mathcal D'(M;\mathcal E)$ denotes the space of distributional sections
\cite[\S 6.3]{ho1}.
For $k\in\mathbb N_0$, the class $\Psi^k$ includes all smooth differential operators of order $k$.
To each $\mathbf A\in \Psi^k(M;\Hom(\mathcal E;\mathcal F))$ we associate its principal symbol
$$
\sigma(\mathbf A)\in S^k(M;\Hom(\mathcal E;\mathcal F))/S^{k-1}(M;\Hom(\mathcal E;\mathcal F))
$$
and its wavefront set
$
\WF(\mathbf A)\ \subset\ T^*M\setminus 0,
$
which is a closed conic set. Here $T^*M\setminus 0$ denotes the cotangent
bundle of $M$ without the zero section. In the case of $\mathcal E=\mathcal F$ we use the
notation
$
\End(\mathcal E)=\Hom(\mathcal E;\mathcal E).
$
For a distribution $\mathbf u\in \mathcal D'(M;\mathcal E)$, its wavefront
set
$$
\WF(\mathbf u)\subset T^*M\setminus 0
$$
is a closed conic set defined as follows: a point $(x,\xi)\in T^*M\setminus 0$
does \emph{not} lie in~$\WF(\mathbf u)$ if and only if there exists an open conic
neighborhood $U$ of $(x,\xi)$ such that
$\mathbf A \mathbf u\in C^\infty(M;\mathcal E)$ for each
$\mathbf A\in \Psi^k(M;\End(\mathcal E))$ satisfying $\WF(\mathbf A)\subset U$.
See \cite[Theorem 18.1.27]{ho3} for more details.

{ The above abstract definition is
useful in this paper but for the reader's convenience we recall the more intuitive local definition in the case of distributions on $ {\mathbb R}^n $ (see \cite[Definition 8.1.2]{ho1}): if $ u \in \mathcal D' ( {\mathbb R}^n ) $ and $ ( x, \xi )
\in T^* {\mathbb R}^n \setminus 0={\mathbb R}^n \times ( {\mathbb R}^n \setminus \{0\} ) $ then
\[  ( x, \xi ) \notin \WF ( u ) \ \Longleftrightarrow \
\left\{ \begin{array}{c}
\exists \, \varphi \in C^\infty_{\rm{c}} ( {\mathbb R}^n ), \ \varphi ( x  ) \neq 0, \
 \varepsilon > 0 \ \text{ such that } \\
 | \widehat {\varphi u } (\eta ) | = \mathcal O ( \langle \eta \rangle^{-\infty} )
 \ \text{ for } \ \big| \eta/|\eta | - \xi/|\xi | \big| < \varepsilon .
\end{array} \right.
\]
Here $ \langle \eta \rangle := ( 1 + |\eta|^2 )^{\frac12} $ and
$ \mathcal O ( \langle \eta \rangle^{-\infty} ) $ means that the left hand side is
bounded by $ C_N \langle \eta \rangle^{-N } $ for any $ N $. Since the decay of
the Fourier transform, $ \hat v $, corresponds to regularity of a distribution $ v $,
this provides ``localized" information both in the position variable $ x $ (thanks to the cutoff $ \varphi $) and in the frequency variable $ \eta $ (thanks to the localization to the cone $ \big| \eta/|\eta| - \xi/|\xi| \big| < \epsilon $).
}

The wavefront set is preserved by
pseudodifferential operators: that is,
\begin{equation}
  \label{e:module-1}
\mathbf A\in \Psi^k(M;\Hom(\mathcal E,\mathcal F)),\
\mathbf u\in \mathcal D'(M;\mathcal E)\ \Longrightarrow\
\WF(\mathbf A\mathbf u)\subset \WF(\mathbf A)\cap\WF(\mathbf u).
\end{equation}
Following~\cite[\S 8.2]{ho1}, for a closed conic set $\Gamma\subset T^*M\setminus 0$ we consider
the space
\begin{equation}
  \label{e:hormander-space}
\mathcal D'_\Gamma(M;\mathcal E)=\{\mathbf u\in \mathcal D'(M;\mathcal E)\colon \WF(\mathbf u)\subset\Gamma\}
\end{equation}
and note that by~\eqref{e:module-1} this space is preserved by pseudodifferential operators.

We also consider the class $\Psi^k_h(M;\Hom(\mathcal E;\mathcal F))$ of semiclassical pseudodifferential
operators with symbols in class $S^k_h$.
The elements of this class are families of operators on~\eqref{e:acting} depending
on a small parameter $h>0$. To each $\mathbf A\in\Psi^k_h(M;\Hom(\mathcal E;\mathcal F))$
correspond its semiclassical principal symbol and wavefront set
$$
\sigma_h(\mathbf A)\in S^k_h(M;\Hom(\mathcal E;\mathcal F))/
hS^{k-1}_h(M;\Hom(\mathcal E;\mathcal F)),\quad
\WFh(\mathbf A)\subset \overline T^*M
$$
where $\overline T^*M$ is the fiber-radially compactified cotangent bundle,
see for instance \cite[\S E.1]{res}. For a tempered $h$-dependent family of distributions
$\mathbf u(h)\in \mathcal D'(M;\mathcal E)$, we can define its wavefront set
$
\WFh(\mathbf u)\subset \overline T^*M.
$

We denote by $\Psi^{\comp}_h(M)\subset \bigcap_k \Psi^k_h(M)$
the class of compactly microlocalized semiclassical pseudodifferential operators,
see \cite[Definition E.29]{res}.

\subsection{Differential forms}

Let $M$ be a compact oriented manifold.
Denote by $\Omega^k$ the complexified vector bundle of differential $k$-forms
on $M$.
The de Rham cohomology spaces are defined as the quotients
of the spaces of closed forms by the spaces of exact forms, that is
$$
\mathbf H^k(M;\mathbb C)={\displaystyle\{\mathbf u\in C^\infty(M;\Omega^k)\colon d\mathbf u=0\}
\over \displaystyle\{d\mathbf v\colon \mathbf v\in C^\infty(M;\Omega^{k-1})\}}.
$$
These are finite dimensional vector spaces over $\mathbb C$, with the dimensions
\begin{equation}
  \label{e:Betti}
\mathbf b_k(M): =\dim\mathbf H^k(M;\mathbb C)
\end{equation}
called $k$-th \emph{Betti numbers}. (It is convenient for us to study cohomology
over $\mathbb C$, which is of course just the complexification of the cohomology over $\mathbb R$.)

De Rham cohomology is typically formulated in terms of smooth differential forms.
However, the next lemma shows that one can use instead the classes $\mathcal D'_\Gamma$:
\begin{lemm}
  \label{l:de-rham}
Let $\Gamma\subset T^*M\setminus 0$ be a closed conic set.
Using the notation \eqref{e:hormander-space}, assume that
$
\mathbf u\in \mathcal D'_\Gamma(M;\Omega^k)$,
$d\mathbf u\in C^\infty(M;\Omega^{k+1}).
$

Then there exist
$
\mathbf v\in C^\infty(M;\Omega^k)$ and
$\mathbf w\in \mathcal D'_\Gamma(M;\Omega^{k-1})$ such that
$\mathbf u=\mathbf v+d\mathbf w.
$
\end{lemm}
\begin{proof}
Fix a smooth Riemannian metric on $M$ { and recall that
it defines an inner product on $ C^\infty ( M ; \Omega^k ) $. Since
$ d : C^\infty ( M ; \Omega^k ) \to C^\infty ( M ; \Omega^{k+1} ) $, we obtain the adjoint operator $ \delta : \mathcal D' ( M ;
\Omega^{k+1} )  \to \mathcal D' ( M ; \Omega^k ) $.}
 We use Hodge theory,
in particular the fact that the Hodge Laplacian
$
\mathbf\Delta_k:=d\delta+\delta d:\mathcal D'(M;\Omega^k)\to \mathcal D'(M;\Omega^k)
$
is a second order differential operator with scalar
principal symbol $\sigma(\mathbf\Delta_k)(x,\xi)=|\xi|^2_g$.
By the elliptic parametrix construction (see \cite[Theorem 18.1.24]{ho3}) there exists
a pseudodifferential operator
$
\mathbf Q_k\in \Psi^{-2}(M;\End(\Omega^k))
$
such that
\begin{equation}
  \label{e:paramet}
\mathbf Q_k\mathbf \Delta_k-I, \ \mathbf \Delta_k\mathbf Q_k-I:\mathcal D'(M;\Omega^k)\to C^\infty(M;\Omega^k).
\end{equation}
Using~\eqref{e:module-1} we now take
$
\mathbf w:=\delta \mathbf Q_k\mathbf u\in\mathcal D'_\Gamma(M;\Omega^{k-1}).
$

Then by~\eqref{e:paramet}
$$
\mathbf u-\delta d \mathbf Q_k\mathbf u-d\mathbf w=\mathbf u-\mathbf \Delta_k\mathbf Q_k \mathbf u\in C^\infty(M;\Omega^k).
$$
Since $d\mathbf u\in C^\infty(M;\Omega^{k+1})$, we have
$$
\mathbf \Delta_{k+1}(d\mathbf Q_k\mathbf u)=d(\mathbf\Delta_k \mathbf Q_k\mathbf u)\in C^\infty(M;\Omega^{k+1}).
$$
By~\eqref{e:paramet} this implies that $d\mathbf Q_k\mathbf u\in C^\infty(M;\Omega^{k+1})$
and thus $\delta d\mathbf Q_k\mathbf u\in C^\infty(M;\Omega^k)$,
giving $\mathbf v:=\mathbf u-d\mathbf w\in C^\infty(M;\Omega^k)$.
\end{proof}

\subsection{Pollicott--Ruelle resonances}
  \label{s:resonances}

We now follow \cite{fa-sj,DZ} and recall  a microlocal approach to Pollicott--Ruelle resonances.
Let $M$ be a compact manifold and $X$ be a smooth vector field
on $M$. We assume that $e^{tX}$ is an Anosov flow, that is each tangent
space $T_xM$ admits a stable/unstable decomposition
$$
T_xM=\mathbb RX(x)\oplus E_u(x)\oplus E_s(x),\quad
x\in M,
$$
where $E_u(x),E_s(x)$ are subspaces of $T_xM$ depending continuously on $x$
and invariant under the flow and for some constants $C,\nu>0$ and
a fixed smooth metric on $M$,
\begin{equation}
  \label{e:Anosov}
|de^{tX}(x)\cdot v|\leq Ce^{-\nu|t|}\cdot |v|,\quad
\begin{cases}t\geq 0,& v\in E_s(x),\\
t\leq 0,& v\in E_u(x).\end{cases}
\end{equation}
We consider the dual decomposition
$$
T_x^*M=E_0^*(x)\oplus E_u^*(x)\oplus E_s^*(x),
$$
where $E_0^*(x),E_u^*(x),E_s^*(x) $ are dual to $\mathbb RX( x ) ,E_s( x ) ,E_u( x ) $. In particular,
$E_u^*(x)$ is the annihilator of $\mathbb RX(x)\oplus E_u(x)$
and $E_u^* := \bigcup_{ x \in M } E_u^* ( x ) \subset T^*M$ is a closed conic set.

Assume next that $\mathcal E$ is a smooth complex vector bundle over $M$ and
$$
\mathbf P:C^\infty(M;\mathcal E)\to C^\infty(M;\mathcal E)
$$
is a first order differential operator whose principal part is given by $-iX$,
that is
\begin{equation}
  \label{e:principal-part}
\mathbf P(\varphi \mathbf u)=-(iX\varphi)\mathbf u+\varphi(\mathbf P\mathbf u),\quad
\varphi\in C^\infty(M),\quad
\mathbf u\in C^\infty(M;\mathcal E).
\end{equation}
For $\lambda\in\mathbb C$ with sufficiently large $\Im\lambda$, the integral
\begin{equation}
  \label{e:uhp}
\mathbf R(\lambda):=i\int_0^\infty e^{i\lambda t}e^{-it\mathbf P}\,dt:L^2(M;\mathcal E)\to L^2(M;\mathcal E)
\end{equation}
converges and defines a bounded operator on $L^2$, holomorphic in $\lambda$; in fact,
$\mathbf R(\lambda)=(\mathbf P-\lambda)^{-1}$ on $L^2$.

\renewcommand\thefootnote{\dag

The operator $\mathbf R(\lambda)$ admits a meromorphic continuation to the entire complex plane,
\begin{equation}
  \label{e:mercon}
\mathbf R(\lambda):C^\infty(M;\mathcal E)\to \mathcal D'(M;\mathcal E),\quad
\lambda\in\mathbb C,
\end{equation}
and the poles of this meromorphic continuation are the \emph{Pollicott--Ruelle resonances}%
\footnote{To be consistent with \cite{DZ} we use the spectral parameter $\lambda=is$ where
$s$ is the parameter used in~\S1. Note that $\Re s \gg 1 $ corresponds to
$ \Im \lambda \gg 1 $.}
of the operator $\mathbf P$. See for instance~\cite[\S3.2]{DZ}
and~\cite[Theorems~1.4,1.5]{fa-sj}.

To define the multiplicity of a Pollicott--Ruelle resonance $\lambda_0$,
we use the Laurent expansion of $\mathbf R$ at $\lambda_0$ given by~\cite[Proposition~3.3]{DZ}:
\begin{equation}
\label{eq:proj}
\mathbf R(\lambda)=\mathbf R_H(\lambda)-\sum_{j=1}^{J(\lambda_0)} {(\mathbf P-\lambda_0)^{j-1}\Pi\over (\lambda-\lambda_0)^j},\quad
\mathbf R_H(\lambda),\Pi:\mathcal D'_{E_u^*}(M;\mathcal E)\to\mathcal D'_{E_u^*}(M;\mathcal E),
\end{equation}
where $\mathbf R_H(\lambda)$ is holomorphic at $\lambda_0$,
$\Pi$ is a finite rank operator,
and $\mathcal D'_{E_u^*}(M;\mathcal E)$ is defined using~\eqref{e:hormander-space}.
The fact that $\mathbf R_H(\lambda),\Pi$ can be extended to continuous operators
on $\mathcal D'_{E_u^*}$ follows from
the restrictions on their wavefront sets given in~\cite[(3.7)]{DZ} together with~\cite[Theorem~8.2.13]{ho1}.
The multiplicity of $\lambda_0$, denoted $m_{\mathbf P}(\lambda_0)$, is defined as the dimension of the range of $\Pi$.

The multiplicity of a resonance can be
computed using generalized resonant states.
Here we only need the following special case:
\begin{lemm}
  \label{l:res-states}
Define the space of resonant states at $\lambda_0\in\mathbb C$,
$$
\Res_{\mathbf P}(\lambda_0)=\{\mathbf u\in \mathcal D'_{E_u^*}(M;\mathcal E)\colon
(\mathbf P-\lambda_0)\mathbf u=0\}.
$$
Then $m_{\mathbf P}(\lambda_0)\geq\dim\Res_{\mathbf P}(\lambda_0)$. Moreover
we have $m_{\mathbf P}(\lambda_0)=\dim\Res_{\mathbf P}(\lambda_0)$ under the following
semisimplicity condition:
\begin{equation}
  \label{e:rs-condition}
\mathbf u\in \mathcal D'_{E_u^*}(M;\mathcal E),\quad
(\mathbf P-\lambda_0)^2\mathbf u=0\quad\Longrightarrow\quad
(\mathbf P-\lambda_0)\mathbf u=0.
\end{equation}
\end{lemm}
\begin{proof}
We first assume that~\eqref{e:rs-condition} holds and prove that $m_{\mathbf P}(\lambda_0)\leq\dim\Res_{\mathbf P}(\lambda_0)$.
We have $(\mathbf P-\lambda)\mathbf R(\lambda)=I$ and thus
$(\mathbf P-\lambda_0)^{J(\lambda_0)}\Pi=0$. Take $\mathbf u$ in the range
of $\Pi$, then $\mathbf u\in\mathcal D'_{E_u^*}(M;\mathcal E)$ by
the mapping property in~\eqref{eq:proj} and $(\mathbf P-\lambda_0)^{J(\lambda_0)}\mathbf u=0$.
Arguing by induction using~\eqref{e:rs-condition} we obtain $\mathbf u\in\Res_{\mathbf P}(\lambda_0)$,
finishing the proof.

It remains to show that $\dim\Res_{\mathbf P}(\lambda_0)\leq m_{\mathbf P}(\lambda_0)$.
For that it suffices to prove that
\begin{equation}
  \label{e:reproduce}
\mathbf u\in\Res_{\mathbf P}(\lambda_0)\quad\Longrightarrow\quad \mathbf u=\Pi\mathbf u.
\end{equation}
We recall from~\cite[\S\S3.1,3.2]{DZ} that $\mathbf R(\lambda)$
is the restriction to $C^\infty$ of the inverse of the operator
\begin{equation}
  \label{e:direct}
\mathbf P-\lambda:\{\mathbf v\in H_{sG}(M;\mathcal E)\colon \mathbf P\mathbf v\in H_{sG}(M;\mathcal E)\}
\to H_{sG}(M;\mathcal E),
\end{equation}
where $H_{sG}(M;\mathcal E)\subset\mathcal D'(M;\mathcal E)$ is a specially constructed anisotropic Sobolev space
and we may take any $s>s_0$ where $s_0$ depends on $\lambda$.
Take $s>s_0$ large enough so that $\mathbf u$ lies in the usual Sobolev space $H^{-s}(M;\mathcal E)$.
Since $H_{sG}$ is equivalent to $H^{-s}$
microlocally near $E_u^*$ (see~\cite[(3.3),(3.4)]{DZ}),
we have $\mathbf u\in H_{sG}$.
We compute $(\mathbf P-\lambda)^{-1}\mathbf u=(\lambda_0-\lambda)^{-1}\mathbf u$ for
$\mathbf u\in\Res_{\mathbf P}(\lambda_0)$
and the space $C^\infty$ is dense in $H_{sG}\cap\mathcal D'_{E_u^*}$, thus~\eqref{e:reproduce} follows
from the Laurent expansion~\eqref{eq:proj} applied to $\mathbf u$.
\end{proof}
We finish this section with the following analogue of Rellich's uniqueness theorem in scattering
theory: vanishing of radiation patterns implies rapid decay. { To see
the connection we refer to the discussion around \cite[(3.6.15)]{res}:
an outgoing solution $ u = R_0 ( \lambda ) f $, $ R_0 ( \lambda )
= ( - \Delta - \lambda^2-i0)^{-1} $, $ f \in C^\infty_{\rm{c}} ( {\mathbb R}^n ) $, $\lambda>0$, has to have a {\em nonnegative} quantum
flux $ - \Im \langle ( - \Delta - \lambda^2 ) u , u \rangle = \Im \langle
R_0 ( \lambda ) f , f \rangle $. If that flux is nonpositive (and thus equal to zero), it follows that
$ u$ is rapidly decaying. In Lemma~\ref{l:mystery} below, the analogue of $ ( - \Delta - \lambda^2 ) u
$ is $ \mathbf P \mathbf u $ and rapid decay is replaced by smoothness.
Technically the proof is also different but the commutator argument is
related to the commutator appearing on the left hand side of \cite[(3.6.15)]{res}.}

\begin{lemm}
\label{l:mystery}
Suppose that there exist
a smooth volume form on $M$ and a smooth inner product on the fibers of $\mathcal E$,
for which $\mathbf P^*=\mathbf P$ on $L^2(M;\mathcal E)$.
Suppose that $ \mathbf u \in \mathcal D'_{E_u^*} ( M;\mathcal E ) $ satisfies
\[
\mathbf P \mathbf u \in C^\infty ( M;\mathcal E ) ,\quad \Im\, \langle \mathbf P \mathbf u , \mathbf u \rangle_{L^2} \geq 0 .
\]
Then $ \mathbf u \in C^\infty ( M ;\mathcal E) $.
\end{lemm}
\Remark Lemma~\ref{l:mystery} applies in particular when $\mathbf u$
is a resonant state at some $\lambda\in\mathbb R$ (replacing $\mathbf P$ by $\mathbf P-\lambda$),
showing that all such resonant states are smooth. This represents a borderline case
since for $\Im\lambda>0$ the integral~\eqref{e:uhp} converges and
thus there are no resonances.
\begin{proof}
We introduce the semiclassical parameter $h>0$ and use the following statement
relating semiclassical and nonsemiclassical wavefront sets of an $h$-independent
distribution $\mathbf v$, see~\cite[(2.6)]{DZ}:
\begin{equation}
  \label{e:wfrel}
\WF(\mathbf v)=\WFh(\mathbf v)\cap (T^*M\setminus 0).
\end{equation}
Since $\mathbf u\in \mathcal D'_{E_u^*}$ and $\mathbf P\mathbf u\in C^\infty$ we have
\begin{equation}
\label{e:wf-1}
  \WFh(\mathbf u)\cap (T^*M\setminus 0)\ \subset\ E_u^*,\quad
  \WFh(\mathbf P\mathbf u)\cap (\overline T^*M\setminus 0)\ =\ \emptyset.
\end{equation}
(The last statement uses the fiber-radially compactified cotangent bundle
and it follows immediately from the proof of~\cite[(2.6)]{DZ} in~\cite[\S C.2]{DZ}.)

It suffices to prove that for each
$A\in \Psi^{\comp}_h(M)$ with $\WFh(A)\subset T^*M\setminus 0$,
there exists $B\in\Psi^{\comp}_h(M)$ with $\WFh(B)\subset T^*M\setminus 0$
such that
\begin{equation}
  \label{e:mystery-step}
\|A\mathbf u\|_{L^2}\leq Ch^{1/2}\|B\mathbf u\|_{L^2}+\mathcal O(h^\infty).
\end{equation}
Indeed, fix $N>0$ such that $\mathbf u\in H^{-N}$, then $\|A\mathbf u\|_{L^2}\leq Ch^{-N}$
for all $A\in\Psi^{\comp}_h(M)$. By induction~\eqref{e:mystery-step} implies that
$\|A\mathbf u\|_{L^2}=\mathcal O(h^\infty)$. This gives $\WFh(\mathbf u)\cap (T^*M\setminus 0)=\emptyset$
and thus by~\eqref{e:wfrel} $\WF(\mathbf u)=\emptyset$, that is  $\mathbf u\in C^\infty$.

To show~\eqref{e:mystery-step}, note that $h\mathbf P\in\Psi^1_h(M;\End(\mathcal E))$ and
its principal symbol is scalar and given by
$$
\sigma_h(h\mathbf P)=p,\quad
p(x,\xi)=\langle \xi,X(x)\rangle.
$$
We now claim that there exists $ \chi\in C_c^\infty(T^*M;[0,1]) $ such that
$$
\begin{gathered}
\supp(1-\chi)\subset T^*M\setminus 0,\quad
H_p\chi \leq  0 \text{ near }E_u^*,\quad
H_p\chi <0  \text{ on }E_u^*\cap \WFh(A).
\end{gathered}
$$
To construct $\chi$, we first use
part~2 of~\cite[Lemma~C.1]{DZ} (applied to $L:=E_u^*$ which is
a radial source for $-p$) to construct
$f_1\in C^\infty(T^*M\setminus 0;[0,\infty))$ homogeneous of degree~1,
satisfying $f_1(x,\xi)\geq c|\xi|$ and $H_pf_1\geq cf_1$ in a conic neighborhood
of $E_u^*$, for some $c>0$. Next we put $\chi:=\chi_1\circ f_1$ where
$ \chi_1\in C_c^\infty(\mathbb R;[0,1]) $ satisfies
$$
\begin{aligned}
\chi_1=1\text{ near }0,\quad
\chi'_1\leq 0 \text{ on }[0,\infty), \quad
\chi'_1<0 \text{ on }f_1(\WFh(A)).
\end{aligned}
$$
It is then straightforward to see that $\chi$ has the required properties.

We now quantize $\chi$ to obtain an operator
$$
F\in \Psi^{\comp}_h(M),\quad
\sigma_h(F)=\chi,\quad
\WFh(I-F)\subset \overline T^*M\setminus 0,\quad
F^*=F.
$$
Since $H_p\chi\leq 0$ near $E_u^*$ and $H_p\chi<0$ on $E_u^*\cap \WFh(A)$ there exists
\begin{gather*}
A_1\in\Psi^{\comp}_h(M), \ \ \WFh(A_1)\subset T^*M\setminus 0, \ \
\WFh(A_1)\cap E_u^*=\emptyset,
\end{gather*}
such that
\begin{equation}
\label{eq:ineq}
-{\textstyle{ 1\over 2}}H_p\chi +|\sigma_h(A_1)|^2\geq C^{-1}|\sigma_h(A)|^2
\end{equation}
where $C>0$ is some constant.

Fix $B\in\Psi^{\comp}_h(M)$ with
$ \WFh ( B ) \subset T^* M \setminus 0 $ so  that
\begin{equation}
\label{eq:defB}
\left( \WFh\big([\mathbf P,F]\big)\cup\WFh(A_1)\cup\WFh(A) \right) \cap
\WFh ( I - B ) = \emptyset .
\end{equation}
By the second part of \eqref{e:wf-1} we have
$(I-F)\mathbf P\mathbf u=\mathcal O(h^\infty)_{C^\infty}$.
Since $\Im\langle \mathbf P\mathbf u,\mathbf u\rangle_{L^2}\geq 0$ this gives
\begin{equation}
  \label{e:varnish}
-\Im\langle F\mathbf P\mathbf u,\mathbf u\rangle_{L^2}\leq \mathcal O(h^\infty).
\end{equation}
On the other hand, since $\mathbf P$ and $F$ are both symmetric, we get
\begin{equation}
  \label{e:varnish2}
-\Im\langle F\mathbf P\mathbf u,\mathbf u\rangle_{L^2}={\textstyle{1\over 2i}}\langle [\mathbf P,F]\mathbf u,\mathbf u\rangle_{L^2}.
\end{equation}
We now observe that
$$
{\textstyle{1\over 2i}}[\mathbf P,F]\in \Psi^{\comp}_h(M;\mathcal E),\quad
\sigma_h\left({\textstyle{1\over 2i}}[\mathbf P,F]\right)=-{\textstyle{1\over 2}}H_p\chi.
$$
Using \eqref{eq:ineq} we can apply the sharp G\r arding inequality
(see for instance \cite[Theorem 9.11]{ev-zw})
to the operator ${1\over 2i}[\mathbf P,F]+A_1^*A_1-C^{-1}A^*A$
and the section $B\mathbf u$
to obtain
$$
\|A B \mathbf u\|_{L^2}^2\leq C\|A_1 B \mathbf u\|_{L^2}^2+{\textstyle{C\over 2i}}\langle B^*[\mathbf P,F]B\mathbf u,\mathbf u\rangle_{L^2}
+Ch\|B\mathbf u\|_{L^2}^2.
$$
From \eqref{eq:defB} we see that
$ A B \mathbf u \equiv A \mathbf u $,
$ A_1 B\mathbf u \equiv A_1 \mathbf  u $ and
$  B^* [\mathbf P,F] B\mathbf u \equiv [\mathbf P,F] \mathbf u $, modulo $ \mathcal O ( h^\infty )_{C^\infty}$.
Also, the first part of \eqref{e:wf-1} shows that $A_1\mathbf u=\mathcal O(h^\infty)_{L^2}$. Using~\eqref{e:varnish}
and~\eqref{e:varnish2}
we obtain~\eqref{e:mystery-step}, finishing the proof.
\end{proof}

\subsection{Contact flows and geodesic flows}
\label{s:gf}

Assume that $M$ is a compact three dimensional manifold and
$\alpha\in C^\infty(M;\Omega^1)$ is a contact form, that is
\[
d\vol_M:=
\alpha\wedge d\alpha\neq 0\quad\text{everywhere.}
\]
Then $d\vol_M$ fixes a volume form and an orientation on $M$.
The form $\alpha$ determines uniquely the \emph{Reeb vector field} $X$
on $M$ satisfying the conditions (with $\iota$ denoting the interior product)
\begin{equation}
\label{eq:dReeb}
\iota_X\alpha=1 \,,\quad
\iota_X (d\alpha)=0.
\end{equation}
We record for future use the following identity
which can be checked by applying both sides to a frame containing $X$:
\begin{equation}
  \label{e:identity}
\mathbf u\wedge d\alpha=(\iota_X \mathbf u)\,d\vol_M\quad\text{for all }
\mathbf u\in\mathcal D'(M;\Omega^1).
\end{equation}
We now consider the special case when $M$ is the unit cotangent bundle
of a compact Riemannian surface $(\Sigma,g)$:
\begin{equation}
  \label{e:circle-bundle}
M=S^*\Sigma=\{(x,\xi)\in T^*\Sigma\colon |\xi|_g=1\}.
\end{equation}
Let $ j: S^* \Sigma \hookrightarrow T^* \Sigma $ and put $ \alpha := j^* ( \xi dx ) $.
Then $\alpha$ is a contact form and the corresponding vector field $X$ is the generator
of the geodesic flow.

We recall a standard topological fact which
will be used in passing from the Betti number of $ M = S^* \Sigma $ to
the Euler characteristic of $ \Sigma $. It is an immediate consequence
of the Gysin long exact sequence; we provide a direct proof for the reader's convenience:
\begin{lemm}
  \label{l:topo}
Assume that $(\Sigma,g)$ is a compact connected oriented Riemannian surface of nonzero Euler characteristic,
$M$ is given by~\eqref{e:circle-bundle}, and $\pi:M\to\Sigma$ is the projection map.
Then for any
$
\mathbf u\in C^\infty(M;\Omega^1)$ with
$ d\mathbf u=0$
there exist $\mathbf v,\varphi$ such that
\begin{equation}
  \label{e:topo}
\mathbf u=\pi^* \mathbf v+d\varphi,\quad
\mathbf v\in C^\infty(\Sigma;\Omega^1),\quad
d\mathbf v=0,\quad
\varphi\in C^\infty(M).
\end{equation}
In particular, $\pi^*:\mathbf H^1(\Sigma;\mathbb C)\to \mathbf H^1(M;\mathbb C)$ is an isomorphism.
\end{lemm}
\begin{proof}
For computations below, we will use positively oriented local coordinates $(x_1,x_2)$ on $\Sigma$
in which the metric has the form
$g=e^{2\psi}(dx_1^2+dx_2^2)$,
for some smooth real-valued function $\psi$. The corresponding coordinates
on $M$ are $(x_1,x_2,\theta)$ with the covector
given by $\xi=e^{\psi}(\cos\theta,\sin\theta)$.
Let $V$ be the vector field on $M$ which generates rotations in the fibers of $\pi$.
In local coordinates, we have $V=\partial_\theta$.
To show~\eqref{e:topo} it suffices to find $\varphi\in C^\infty(M)$ such that
\begin{equation}
  \label{e:topo-p1}
V\varphi=\iota_V\mathbf u.
\end{equation}
Indeed, put $\mathbf w:=\mathbf u-d\varphi$. Then $d\mathbf w=0$
and $\iota_V\mathbf w=0$. A calculation in local coordinates
shows that $\mathbf w=\pi^*\mathbf v$
for some $\mathbf v\in C^\infty(\Sigma;\Omega^1)$
such that $d\mathbf v=0$.

A smooth solution to~\eqref{e:topo-p1} exists if
$\mathbf u$ integrates to 0 on each fiber of $\pi$.
Since~$\mathbf u$ is closed
and all fibers are homotopic to each other, the integral
of $\mathbf u$ along each fiber is equal to some constant $c\in\mathbb C$,
thus it remains to show that $c=0$.

Let $K\in C^\infty(\Sigma)$ be the Gaussian curvature of $\Sigma$ and
$d\vol_\Sigma\in C^\infty(\Sigma;\Omega^2)$ the volume form
of $(\Sigma,g)$, written in local coordinates as
$d\vol_\Sigma=e^{2\psi}dx_1\wedge dx_2$.
With $\chi(\Sigma)\neq 0$ denoting the Euler characteristic of $\Sigma$, we have
by Gauss--Bonnet theorem
$$
\int_M\mathbf u\wedge \pi^*(K\,d\vol_\Sigma)=2\pi \chi(\Sigma)\cdot c.
$$
It then remains to prove that $
\int_M \mathbf u\wedge \pi^* (K\, d\vol_\Sigma)=0$.
This follows via integration by parts from the identity
$\pi^*(K\,d\vol_\Sigma)=-dV^*$,
where $V^*\in C^\infty(M;\Omega^1)$ is the connection form, namely the unique
1-form satisfying the relations
$$
\iota_V V^*=1,\quad
d\alpha=V^*\wedge \beta,\quad
d\beta=-V^*\wedge \alpha,
$$
where $\alpha$ is the contact form
and $\beta$ is the pullback of $\alpha$
by the $\pi/2$ rotation in the fibers of $\pi$. This can be checked
in local coordinates using the formulas $\alpha=e^\psi(\cos\theta\,dx_1+\sin\theta\,dx_2)$,
$\beta=e^{\psi}(-\sin\theta\,dx_1+\cos\theta\,dx_2)$,
$V^*=\partial_{x_1}\psi\,dx_2-\partial_{x_2}\psi\,dx_1+d\theta$,
$K=-e^{-2\psi}\Delta\psi$; see also~\cite[\S3]{g-k}.

Having established~\eqref{e:topo}, we see immediately
that $\pi^*:\mathbf H^1(\Sigma;\mathbb C)\to\mathbf H^1(M;\mathbb C)$
is onto. To show that $\pi^*$ is one-to-one, assume
that $\mathbf v\in C^\infty(\Sigma;\Omega^1)$ satisfies
$\pi^*\mathbf v=d\varphi$ for some $\varphi\in C^\infty(M)$.
Then $V\varphi=\iota_Vd\varphi=0$, therefore $\varphi=\pi^* \chi$
for some $\chi\in C^\infty(\Sigma)$ and $\mathbf v=d\chi$ is exact.
\end{proof}

\section{Proof}
\label{proof}

In this section we prove the main theorem in a slightly more general
setting~-- see Proposition \ref{l:main}. We assume throughout that
$M$ is a three-dimensional connected compact manifold, $\alpha$ is a contact form on $M$, and
$X$ is the Reeb vector field of $\alpha$ generating an Anosov flow (see~\S\S\ref{s:resonances},\ref{s:gf}).
For the application to zeta functions we also assume that the corresponding
stable/unstable bundles $E_u,E_s$ are orientable.

\subsection{Zeta function and Pollicott--Ruelle resonances}
\label{s:zeta}

For $k=0,1,2$, let $\Omega^k_0\subset\Omega^k$ be the bundle of exterior $k$-forms $\mathbf u$
on $M$ such that $\iota_X \mathbf u=0$.
Consider the following operator satisfying~\eqref{e:principal-part}:
$$
\mathbf P_k:=-i\mathcal L_X:\mathcal D'(M;\Omega^k_0)\to\mathcal D'(M;\Omega^k_0).
$$
Note that by Cartan's formula
$$
\mathbf P_k\mathbf u=-i\,\iota_X(d\mathbf u),\quad
\mathbf u\in\mathcal D'(M;\Omega^k_0).
$$
As discussed in~\S\ref{s:resonances} we may consider Pollicott--Ruelle resonances
associated to the operators $\mathbf P_k$, denoting their multiplicities
as follows:
$$
m_k(\lambda):=m_{\mathbf P_k}(\lambda)\in\mathbb N_0,\quad
\lambda\in\mathbb C.
$$
The connection with the Ruelle zeta function comes from the following standard
formula (see \cite[(2.5) and~\S 4]{DZ})
for the meromorphic continuation of $\zeta_R$:
$$
\zeta_R(s)={\zeta_1(s )\over \zeta_0( s )\zeta_2(s  )},\quad
s\in\mathbb C.
$$
(It is here that we the assumption that the stable and unstable bundle are orientable.)
Here each $\zeta_k(s )$ is an entire function having a zero of multiplicity
$m_k( i s )$ at each $ s \in\mathbb C$. Therefore, $\zeta_R(s )$ has a zero at
$s =0$ of multiplicity
\begin{equation}
  \label{e:mul-ultimate}
m_R(0):=m_1(0)-m_0(0)-m_2(0).
\end{equation}
By Lemma~\ref{l:res-states} the multiplicity $m_k(0)$ can be calculated as
\begin{equation}
  \label{e:dim-mult}
m_k(0)=\dim\Res_k(0),
\end{equation}
where $\Res_k(0)$ is the space of resonant states at zero,
\begin{equation}
\label{eq:Resk}
\Res_k(0)=\{\mathbf u\in \mathcal D'_{E_u^*}(M;\Omega^k)\colon \iota_X\mathbf u=0,\
\iota_X(d\mathbf u)=0\}
\end{equation}
provided that the semisimplicity condition~\eqref{e:rs-condition} is satisfied:
\begin{equation}
  \label{e:algsim}
\mathbf u\in \mathcal D'_{E_u^*}(M;\Omega^k),\quad
\iota_X\mathbf u=0,\quad
\iota_X(d\mathbf u)\in \Res_k(0)\quad\Longrightarrow\quad
\iota_X(d\mathbf u)=0.
\end{equation}
The main result of this section is
\begin{prop}
\label{l:main}
In the notation of \eqref{eq:Resk} we have
\begin{enumerate}
\item $\dim\Res_0(0)=\dim\Res_2(0)=1$;
\item $\dim\Res_1(0)$ is equal to the Betti number $\mathbf b_1(M)$ defined in~\eqref{e:Betti};
\item the condition~\eqref{e:algsim} holds for $k=0,1,2$.
\end{enumerate}
\end{prop}
It is direct to see that Proposition~\ref{l:main} implies the main theorem when
applied to the case $M=S^*\Sigma$ discussed in~\S\ref{s:gf} (strictly speaking,
to each connected component of $\Sigma$). Indeed, $X$ generates an Anosov flow
since $\Sigma$ is negatively curved (see for example~\cite[Theorem~3.9.1]{k}),
the stable/unstable bundles are orientable since $\Sigma$ is orientable and $m_R(0)=\mathbf b_1(M)-2$ equals to $-\chi(\Sigma)$ by Lemma~\ref{l:topo}.

\subsection{Scalars and 2-forms}

We start the proof of Proposition~\ref{l:main} by considering the cases $k=0$
and $k=2$:
\begin{lemm}
  \label{l:scalars}
We have
\begin{equation}
\Res_0(0)=\{c\colon c\in\mathbb C\},\quad
\Res_2(0)=\{c\,d\alpha\colon c\in\mathbb C\},
\end{equation}
and~\eqref{e:algsim} holds for $k=0,2$, that is the resonance at 0 for
$ k=0,2$ is simple.
\end{lemm}
\Remark The argument for $\Res_0(0)$ applies to any contact Anosov flow on
a compact connected manifold. It can be generalized
to show that $\Res_0(0)$ consists of constant functions and $\Res_0(\lambda)$ is trivial
for all $\lambda\in\mathbb R\setminus 0$. This in particular
implies that the flow is mixing.
\begin{proof}
We first handle the case of $\Res_0(0)$. Clearly this space contains constant functions,
therefore we need to show that
\begin{equation}
  \label{e:const}
u\in\mathcal D'_{E_u^*}(M),\quad
Xu=0\quad\Longrightarrow\quad u=c\quad\text{for some }c\in\mathbb C.
\end{equation}
By Lemma~\ref{l:mystery} we have $u\in C^\infty(M)$.
Since $Xu=0$ we have $u=u\circ e^{tX}$ and thus
$$
\langle du(x),v\rangle=\langle du(e^{tX}(x)),de^{tX}(x)\cdot v\rangle\quad\text{for all }
(x,v)\in TM,\ t\in\mathbb R.
$$
Now if $v\in E_s(x)$ then taking the limit as $t\to \infty$ and using~\eqref{e:Anosov}
we obtain
$\langle du(x),v\rangle=0$. Similarly if $v\in E_u(x)$ then
the limit $t\to -\infty$ gives
$\langle du(x),v\rangle=0$. Therefore $du|_{E_u\oplus E_s}=0$. However
$E_u\oplus E_s=\ker \alpha$, thus we have for some $\varphi\in C^\infty(M)$,
$$
du=\varphi\alpha.
$$
We have $0=\alpha\wedge d(\varphi\alpha)=\varphi\alpha\wedge d\alpha$,
thus $du=0$, implying~\eqref{e:const} since $M$ is connected.

Next, \eqref{e:algsim} holds for $k=0$. Indeed, if $u\in\mathcal D'_{E_u^*}(M)$
then
$$
\int_M (Xu)\,d\vol_M=0,
$$
implying that $Xu$ cannot be a nonzero element of $\Res_0(0)$.

Now, assume that $\mathbf u\in \mathcal D'_{E_u^*}(M;\Omega^2)$
satisfies $\iota_X\mathbf u=0$. Then $\mathbf u$ can be written as
$$
\mathbf u=u\,d\alpha, \quad
u\in \mathcal D'_{E_u^*}(M);\quad
\iota_X(d\mathbf u)=(Xu)d\alpha.
$$
Therefore the case of $\Res_2(0)$ follows immediately from that of $\Res_0(0)$.
\end{proof}
Lemma~\ref{l:scalars} implies solvability
of the equation $Xu=f$ in the class $\mathcal D'_{E_u^*}$:
\begin{prop}
  \label{l:scalar-invertibility}
Assume that $f\in C^\infty(M)$ and $\int_M f\,d\vol_M=0$. Then
there exists $u\in \mathcal D'_{E_u^*}(M)$ such that
$Xu=f$.
\end{prop}
\begin{proof}
It follows from Lemma~\ref{l:scalars} and the proof of Lemma~\ref{l:res-states} that the
resolvent $\mathbf R_0(\lambda)$ of the operator $\mathbf P_0=-iX$ defined in~\eqref{e:mercon} has the expansion
$$
\mathbf R_0(\lambda)=\mathbf R_H(\lambda)-{\Pi\over\lambda}
$$
where $\mathbf R_H(\lambda)$ is holomorphic at $\lambda=0$ and the range of $ \Pi $ consists of constant functions.
By analytic continuation from~\eqref{e:uhp}, we see that $\mathbf R_0(\lambda)^*=-\mathbf R_{-\mathbf P_0}(-\bar\lambda)$
where $\mathbf R_{-\mathbf P_0}(\lambda)$ is the resolvent of $-\mathbf P_0$. Applying
Lemma~\ref{l:scalars} to the field $-X$, we see that the range of $\Pi^*$ also consists
of constant functions. By~\eqref{e:reproduce} we have $\Pi(1)=1$, therefore
$$
\Pi u={1\over\vol(M)}\int_M u\,d\vol_M.
$$
Now, put
$u:=-i\mathbf R_H(0)f$, then $u\in\mathcal D'_{E_u^*}(M)$ by~\eqref{eq:proj}.
Since $\Pi f=0$ and $(\mathbf P_0-\lambda)\mathbf R_0(\lambda)=I$, we have
$Xu=f$.
\end{proof}

\subsection{1-forms} \label{s:1fo}

It remains to show Proposition~\ref{l:main} for the case $k=1$,
that is to analyse the space
$$
\Res_1(0)=\{\mathbf u\in \mathcal D'_{E_u^*}(M,\Omega^1)\colon
\iota_X\mathbf u=0,\
\iota_X(d\mathbf u)=0\}.
$$
The next lemma shows that the $\dim\Res_1(0)=\mathbf b_1(M)$:
\begin{lemm}
  \label{l:1-forms}
Assume that $\mathbf u\in \Res_1(0)$. Then there exists $\varphi\in \mathcal D'_{E_u^*}(M)$
such that
\begin{equation}
  \label{e:de-rham-iso-0}
\mathbf u-d\varphi\in C^\infty(M;\Omega^1),\quad
d(\mathbf u-d\varphi)=0.
\end{equation}
The cohomology class $[\mathbf u-d\varphi]\in \mathbf H^1(M;\mathbb C)$
is independent of the choice of $\varphi$. The map
\begin{equation}
  \label{e:de-rham-iso-1}
\Res_1(0)\ni\mathbf u\ \mapsto\ [\mathbf u-d\varphi]\in \mathbf H^1(M;\mathbb C)
\end{equation}
is a linear isomorphism.
\end{lemm}
\begin{proof}
We first show that
\begin{equation}
  \label{e:it-is-closed}
\mathbf u\in\Res_1(0)\ \Longrightarrow\ d\mathbf u=0.
\end{equation}
Definition~\eqref{eq:Resk} shows that $ d\mathbf u\in \Res_2(0)$ and
therefore by Lemma~\ref{l:scalars} we have
$d\mathbf u=c\,d\alpha$ for some $c\in\mathbb C$.
From~\eqref{e:identity} and $\iota_X\mathbf u=0$ we also have $\mathbf u\wedge d\alpha=0$,
thus Stokes's theorem gives~\eqref{e:it-is-closed}:
$$
c\vol(M)=\int_M \alpha\wedge d\mathbf u =
 \int_M \mathbf u\wedge d\alpha
=0.
$$
Lemma~\ref{l:de-rham} and \eqref{e:it-is-closed} imply the existence
of $\varphi\in\mathcal D'_{E_u^*}(M)$ such that~\eqref{e:de-rham-iso-0} holds.
Moreover, if $\tilde \varphi\in \mathcal D'_{E_u^*}(M)$ is another function satisfying~\eqref{e:de-rham-iso-0}
then $d(\varphi-\tilde \varphi)\in C^\infty(M;\Omega^1)$, implying by Lemma~\ref{l:de-rham} that
$\varphi-\tilde \varphi\in C^\infty(M)$. Therefore $\mathbf u-d\varphi$ and $\mathbf u-d\tilde \varphi$
belong to the same de Rham cohomology class, thus the map~\eqref{e:de-rham-iso-1} is well-defined.

Next, assume that~\eqref{e:de-rham-iso-0} holds and
$\mathbf u-d\varphi$ is exact. By changing $\varphi$ we may assume that
$\mathbf u=d\varphi$. Since $\iota_X\mathbf u=0$ we have $X\varphi=0$,
which by Lemma~\ref{l:scalars} implies that $\varphi$ is constant
and thus $\mathbf u=0$. This shows that~\eqref{e:de-rham-iso-1}
is injective.

It remains to show that~\eqref{e:de-rham-iso-1} is surjective.
For that, take a closed $\mathbf v\in C^\infty(M;\Omega^1)$. We need to find $\varphi\in\mathcal D'_{E_u^*}(M)$
such that $\mathbf v+d\varphi\in \Res_1(0)$. This
is equivalent to $\iota_X(\mathbf v+d\varphi)=0$, that is
$
X\varphi=-\iota_X\mathbf v.
$
A solution $\varphi$ to the above equation exists by Proposition~\ref{l:scalar-invertibility}
since \eqref{e:identity} implies
$$
\int_M \iota_X\mathbf v\,d\vol_M= \int_M \mathbf v\wedge d\alpha=\int_M \alpha\wedge d\mathbf v=0.
$$
This finishes the proof.
\end{proof}
To prove Proposition~\ref{l:main} it remains to show the
semisimplicity condition:
\begin{lemm}
  \label{l:a-s}
Suppose that
$$ \mathbf u\in\mathcal D'_{E_u^*}(M;\Omega^1),\quad
\iota_X\mathbf u=0,\quad
\iota_X(d\mathbf u)=\mathbf v\in \Res_1(0).
$$
Then $\mathbf v=0$, that is, condition \eqref{e:algsim} holds for $k=1$.
\end{lemm}
\begin{proof}
We have $\alpha\wedge d\mathbf u=a\,d\vol_M$
for some $a\in \mathcal D'_{E_u^*}(M)$. By~\eqref{e:identity},
$$
\int_M a\,d\vol_M=\int_M \mathbf u\wedge d\alpha =\int_M \iota_X\mathbf u\,d\vol_M=0.
$$
Moreover since $\mathcal L_X(\alpha)=0$, $\mathcal L_X(d\alpha)=0$, and $d\mathbf v=0$ by~\eqref{e:it-is-closed},
we have
$$
(Xa)\,d\vol_M=\mathcal L_X(\alpha\wedge d\mathbf u)=\alpha\wedge d\mathbf v=0.
$$
Thus $Xa=0$ and Lemma~\ref{l:scalars} gives that $a=0$ and thus
$\alpha\wedge d\mathbf u=0$. This implies
$d\mathbf u= \alpha \wedge \iota_X d \mathbf u = \alpha\wedge \mathbf v$.
Now by Lemma~\ref{l:1-forms} there exist
$$
\mathbf w\in C^\infty(M;\Omega^1),\quad
\varphi\in \mathcal D'_{E_u^*}(M),\quad
\mathbf v=\mathbf w+d\varphi,\quad
d\mathbf w=0.
$$
Since $\iota_X\mathbf v=0$ we have $X\varphi=-\iota_X\mathbf w$.
Integration by parts together with~\eqref{e:identity} gives
\begin{equation}
\label{eq:zeroprod}
\begin{aligned}
0&=\Re \int_M d\mathbf u\wedge\overline{\mathbf w}
=\Re \int_M\alpha\wedge d\varphi\wedge\overline{\mathbf w}\\
&=\Re \int_M\varphi \,\overline{\mathbf w}\wedge d\alpha
=-\Re \langle X\varphi,\varphi\rangle_{L^2}.
\end{aligned}
\end{equation}
By Lemma~\ref{l:mystery} with $\mathbf P=-iX$ this implies $\varphi\in C^\infty(M)$
and thus $\mathbf v\in C^\infty(M;\Omega^1)$.

We can now use the same argument as in the proof of Lemma \ref{l:scalars}:
$ ( e^{tX } )^* \mathbf v = \mathbf v $ and hence
\[  \langle \mathbf v (x ) , z \rangle = \langle
\mathbf v ( e^{tX } x ) , d e^{tX } ( x ) \cdot z \rangle , \ \ ( x, z ) \in T M, \ \ t \in {\mathbb R} . \]
The right hand side tends to zero as $ t \to \infty $ for $ z \in E_s ( x ) $,
and as $ t \to - \infty $ for $ z \in E_u ( x ) $. Since $ \iota_X \mathbf v = 0 $
it follows that $ \mathbf v = 0 $.
\end{proof}

redpajama_set_name: RedPajamaArXiv


text:
'America's Got Talent': 13-Year-Old Courtney Hadwin Slays Some James Brown (Video)
Do you see why she earned Howie Mandel's golden buzzer?
Tony Maglio | August 15, 2018 @ 7:39 AM
Courtney Hadwin's got a brand new song in her bag of tricks. The 13-year-old "America's Got Talent" standout slayed James Brown's "Papa's Got a Brand New Bag" Tuesday night on NBC.
While her talents are no longer a surprise to viewers, they're still more than enough to bring the "AGT" audience and judges to their feet.
Watch Hadwin's first go-around here, when she earned Howie Mandel's golden buzzer.
Also Read: 'AGT' Trapeze Act Goes Horribly Wrong When Husband Can't Hold on to Wife (Video)
Hit the horns, and watch her new performance via the video above — it ain't no drag.
Cool area rugs.
10 Stars Who Just Need an Emmy to EGOT, From Elton John to Stephen Sondheim (Photos)
The EGOT -- an acronym for Emmy, Grammy, Oscar, Tony -- is among the greatest and most elusive honors in entertainment. These stars are (or were) close to achieving it.
Elton John, composer and musician (1947-)
Grammy: Best Pop Performance by a Duo or Group, "That's What Friends Are For" (1986); Best Instrumental Composition, "Basque" (1991); Best Male Pop Vocal Performance, "Can You Feel the Love Tonight" (1994); Best Male Pop Vocal Performance, "Candle in the Wind" (1997); Best Show Album, "Aida" (2000)
Oscar: Best Original Son, "Can You Feel the Love Tonight" from "The Lion King" (1994)
Tony: Best Score, "Aida" (2000)
Stephen Sondheim, composer and lyricist (1930-)
Grammy: Best Show Album, "Company" (1970); Best Show Album, "A Little Night Music" (1973); Song of the Year, "Send in the Clowns" (1975); Best Show Album, "Sweeney Todd" (1979); Best Show Album, "Sunday in the Park With George" (1984); Best Cast Show Album, "Into the Woods" (1988); Best Show Album, "Passion" (1994);
Oscar: Best Original Song, "Sooner Or Later (I Always Get My Man)" from "Dick Tracy" (1990)
Tony: Best Musical, "A Funny Thing Happened on the Way to the Forum," (1963); Best Score and Best Lyrics, "Company" (1971); Best Score, "Follies" (1972); Best Score, "A Little Night Music" (1973); Best Score, "Sweeney Todd" (1979); Best Score, "Into the Woods" (1988); Best Score, "Passion" (1994)
Joel Grey, actor (1932 - )
Grammy (2): Best Musical Theater Album, "Cabaret" (1968) and "Chicago" (1998)
Oscar: Supporting Actor, "Cabaret" (1972)
Tony: Featured Actor in a Musical, "Cabaret" (1967)
Benj Pasek and Justin Paul, composers (both 1985 - )
Grammy: Best Musical Theater Album, "Dear Evan Hansen" (2018)
Oscar: Best Original Song, "La La Land" (2017)
Tony: Best Original Score, "Dear Evan Hansen" (2017)
Henry Fonda, actor (1905-82)
Grammy: Best Spoken Word Album, "Great Documents" (1977)
Oscar: Best Actor, "On Golden Pond" (1981)
Tony: Best Actor, "Mister Roberts" (1948); Best Actor, "Clarence Darrow" (1975)
Oscar Hammerstein II, lyricist and producer (1895-1960)
Grammy: Best Original Cast Album, "The Sound of Music" (1960)
Oscar: Best Original Song, "The Last Time I Saw Paris" from "Lady Be Good" (1941); "It Might As Well Be Spring" from "State Fair" (1945)
Tony: Three awards for "South Pacific" (1950); Best Musical, "The King and I" (1952); Best Musical, "The Sound of Music" (1960)
Alan Jay Lerner, lyricist and writer (1918-86)
Grammy: Best Original Cast Album, "On a Clear Day" (1965)
Oscar: Best Original Screenplay, "An American in Paris" (1951); Best Adapted Screenplay and Best Original Song, "Gigi" (1958)
Tony: Best Musical, "My Fair Lady" (1957); Best Original Score, "Gigi" (1974)
Frank Loesser, composer (1910-69)
Grammy: Best Cast Album, "How to Succeed..." (1961)
Oscar: Best Song, "Baby, It's Cold Outside" from "Neptune's Daughter" (1949)
Tony: Best Musical, "Guys and Dolls" (1951); Best Musical, "How to Succeed..." (1962)
Jule Styne, composer and songwriter (1905-94)
Grammy: Best Cast Album, "Funny Girl" (1964)
Oscar: Best Song, "Three Coins in the Fountain" (1954)
Tony: Best Musical and Best Score, "Hallelujah Baby" (1968)
A select group of entertainers can round out their trophy cases with a competitive win from the Television Academy
'America's Got Talent' Judges Get Totally Freaked by Magic Show: 'I Want This to End' (Video)
By Beatrice Verhoeven | July 27, 2018 @ 8:25 AM
'America's Got Talent': Olivia Munn Learns That Magic Is Real (Video)
Neal E Boyd, 'America's Got Talent' Season 3 Winner, Dies at 42
By Ashley Boucher | June 11, 2018 @ 10:09 AM


text:
By Bloomberg February 9, 2015 22:32
Permira Cuts Hugo Boss Investment Again With $1.5 Billion Stake
Permira Holdings Ltd. is cutting its stake in Hugo Boss AG below 14 percent, boosting the clothier's free float in a move that may make it more attractive to institutional investors.
Abbey Lee Kershaw for BOSS Womenswear spring/summer 2015 | Source: Hugo Boss
METZINGEN, Germany — Permira Holdings Ltd. is cutting its stake in Hugo Boss AG below 14 percent, boosting the clothier's free float in a move that may make it more attractive to institutional investors.
Permira is selling a 10.4 percent stake to the market, and a further 500 million euros to Italy's Marzotto family, Hugo Boss said in a statement. The combined value of those divestments is about 1.3 billion euros ($1.5 billion), based on today's closing price of 109.60 euros per share.
Permira has over the last two years been cutting its holding in the fashion house, in which it acquired a majority stake in 2007. When the free float rose to 66 percent after a stake sale Dec. 12, Metzingen, Germany-based Hugo Boss said it would make the stock more attractive to institutional investors.
Gaetano Marzotto already sits on Hugo Boss's supervisory board. Permira acquired control of Valentino Fashion Group SpA from the Marzottos in 2007.
Hugo Boss shares have gained 7.8 percent this year, valuing it at 7.7 billion euros. That's even after reporting fourth- quarter sales and profit that missed estimates, as it contended with sluggish demand for high-priced fashion in economically stagnant Europe.
By Alex Webb. Editors: Simon Thiel, James Kraus, Jim Silver.
Li Edelkoort Reads Her Anti-Fashion Manifesto


text:
Q: Enable conversion between 64 bit integer types Trying to port some Windows C++ code to Linux + clang.
Here's a function:
inline __m256i XM_CALLCONV maskload_epi64( const int64_t *mem_addr, __m256i mask )
{
    return _mm256_maskload_epi64( mem_addr, mask );
}

Here's a compiler output:
no matching function for call to '_mm256_maskload_epi64'
    return _mm256_maskload_epi64( mem_addr, mask );
           ^~~~~~~~~~~~~~~~~~~~~
/usr/include/clang/4.0.0/include/avx2intrin.h:989:1: note: candidate function
      not viable: no known conversion from 'const int64_t *'
      (aka 'const long *') to 'const long long *' for 1st argument
_mm256_maskload_epi64(long long const *__X, __m256i __M)

Is there a compiler switch to make it cast between different 64 bit built-in types without errors?

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