Recently my lovely bloggy friend Rachel (who's no longer blogging at the moment, boo) asked me what products I love when my skin is going through a tough time. Now hey, I'm no Caroline Hirons, and I don't have perfect skin, but over the last few months I've really noticed a change in my skin, which in turn has made me feel more confident. So, I thought why not make a blog post and talk about the products I rated highly enough to recommend to one of my bestest friends, as I class you guys (if there's anyone reading this) as my friends too.
Origins Make A Differance Plus Rejuvenating Treatment - If you have oily, dehydrated skin and purchase one thing from this post then it has to be this. Make sure you pick up the treatment and not the moisturiser as they moisturiser isn't as great. I treat this as serum, putting it on after toner and before Hydraluron and moisturiser. It has a lovely cooling effect and really smooths out my skin. It works at hydrating whilst also fighting blemishes, how I don't know. But it's an amazing product!
REN Clear Calm 3 Clay Cleanser - This cleanser is sort of like a mud mask in consistency, due to the clay I guess. It works really well at exfoliating the skin (without any rough abrasive salts or sugars) whilst also deep cleansing. Whenever my skin is producing extra oil or producing extra blemishes I like to whip this guy out. It's also great if you want your makeup to last all day as it mattifies the skin (without drying it out), Super duper cleanser right here.
Origins High Potency Night A Mins - This moisturiser is great at rejuvenating the skin overnight. When I wake up in the morning my skin is left feeling hydrated, brightened and blemishes reduced. It has fruit acids in it which renew the skin whilst it's on, and the lovely thick consistency makes it feel super luxurious to apply.
Origins Drink Up Intensive Overnight Mask - So I actually tend to use this daily as a day moisturiser at work as it just feels so darn nice and makes my skin feel so soft. However, most normal people tend to use it a couple of nights a week overnight. When you wake up the next morning your skin is left feeling extremely hydrated, skin tone more even and just more plump. I would say no matter what your skin type this product is a fantastic.
An exfoliating toner - I've tried a couple of exfoliating toners and do find that they really do help clear up blemishes. I'm currently using the Boots Botanics Exfoliating Toner which I must say feels a bit too harsh to use everyday. However the previous exfoliating toners I've tried from Clarins and REN have both been amazing, which leads me to think I should stop being cheap and repurchase one of them. Anyway, Ms Hirrons is correct, don't miss this step.
A hydrating toner - There are two toners I love for their hydrating qualities; first up the Dr Weils for Origins Mega Mushroom Skin Relief. This is particularly good if you suffer from redness as it really does reduce it, especially blemishes. My current favourite is the Khiels Ultra Facial Toner. It has a really thick consistency that you can feel sink into your skin and boost your hydration levels.
So here are my top skincare products that I would recommend. Of course that's not my entire skincare routine, but it's the products I really love.
For the record my skin is blemish prone and dehydrated with oily patches in the centre of my face and dry skin on my cheeks. What might work for me may not work for you, and I'm not a skincare specialist.


Indian Bank has released a notification for the recruitment of 145 Specialist Officers for various posts of Assistant Manager, Manager, Senior Manager and More. Interested candidates may check the vacancy details and apply online from 10-04-2018 to 02-05-2018.
Educational Qualification: Candidate should have passed B.E/B.Tech. 4 year/MBA 2 year (Finance) or equivalent OR Post Graduation Degree/Diploma in relevant trade from a recognised university.
Application Fee: General/UR/OBC candidates have to pay Rs. 600 and SC/ST/PWD candidates have to pay Rs. 100 through online mode using Net-banking/Credit or Debit Card.
How to Apply: Interested and eligible candidates may apply online through Indian Bank website - http://www.indianbank.in/ - from 10-04-2018 to 02-05-2018.


When I was a kid, I wanted to be a nurse and marry a farmer. Politics never entered my mind except when it came to voting. I have never missed a vote since I reached the age of majority in this wonderful country of ours.
Today, I am putting myself forward for your consideration as a Councillor in Ward 4 of Wilmot Township.
Why? Because this town has given me and my family so much love and support that I want to offer to serve as your voice on the Council. It's the best idea I could come up with to let all of you know what a great place I think this is, and how grateful I am for the privilege of living here with all of you.
My day job is running Pfenning's Organic Vegetables Inc. with my husband and brother and sister-in-law. It is a job I love. In the past, I have also served on volunteer boards: seven years at Foodshare in Toronto, several years on the Organic Council of Ontario board, as well as the National Farmers Union Local and the Waterloo Region Food System Roundtable.
If you choose to elect me as councilor, I will work hard to consider everyone's voice, to listen, and to act in the best interest of the whole community.
In the coming weeks, I will be making myself available to hear what's on your mind by knocking on your doors and of course, attending many of the wonderful events we have.


ElderQuest for Education - ElderQuest: Adventures in Geriatrics!
ElderQuest harnesses the immersive aspects of 3D gameplay, in order to help players to better understand their patients' illnesses. At the "Bridge of Eyes," players must unlock a magical bridge-gate while experiencing sensory deficits that mimic real-world pathologies -- including diabetic retinopathy (right), and cataracts (below).
 Players must work through several challenges -- navigating the scene, identifying their own pathologies even as they experience them in the game, and finally clearing the bridge -- in order to escape a surge of deadly germ-clouds. Later on, players will experience hearing loss plus balance and gait problems as well.
Learning in Five Dimensions (or more) . . .
Within the world of immersive 3D games, reality itself is almost infinitely malleable. This freedom -- to shape the game world and the players' experience -- may be the most powerful capability of game-based education.
Unlike simulations or training videos, games grant enormous powers to creative instructors. By controlling gravity, we can help our students to fly; with a line of code, we can slow the passage of time, or summon a blizzard on a sunny summer day. Information can be presented and reinforced in numerous forms, facilitating multiple learning modalities and styles. And through judicious use of character and storyline, we can magnify our student-players' immersion and focus while heightening their motivation -- to continue to play the game, to overcome obstacles, and to learn what's necessary to "win."
 ElderQuest's drug-harvesting mission illustrates many benefits of immersive, story-based design: Drugs actually grow on trees in the Pharmlands of Jerissay. To save a "retirement village" from a devastating plague, players must harvest several types of medications from a magical orchard.
Harvesting pills from trees may seem simple, but it's actually quite dangerous. Throughout the Pharmlands, time-twisting magic accelerates the growth and ripening of pharmaceutical crops. Time speeds up for players as well: they'll age a full decade per minute spent in the orchard.
Those age-related changes quickly tax a player's game-world abilities -- sapping their strength, slowing their gait, and dulling their vision and hearing. Some meds are enormously heavy, slowing the player even further. Other drugs are more precious than gold, attracting the attention of thieving trolls who stalk unsuspecting players.
To save the plague-stricken retirement community, players must harvest their drugs wisely -- and quickly. Infirmity quickly takes its toll on players who age a century or more, and some unlucky players never leave the orchard. There's a bright side, though: time moves so quickly that players' bones turn to dust just a few seconds after death. The Pharmland's orchards are always very tidy!
The "Pharmlands Medications" chart (left) was created in order to guide the game-development team as they created the orchard. Players are sent into the orchard with a list of medication types which are required in order to treat patients in the retirement village. Given the village's elderly population, players must choose an appropriate drug from each category.
Within the orchard, each tree bears a specific category of medication (e.g., the "sleep-aid tree" has tryptophan, diphenhydramine, and zolpidem pills dangling from its limbs). Any one of those drugs will theoretically fulfill the sleep-aid requirement, but only one medication (tryptophan) is both cost-effective and safe for elderly patients.
Players who choose to harvest diphenhydramine will quickly discover that the pill is massively heavy: while it's possible to drag the pill to the gate, a lot of time will be wasted in the process. If a player plucks the zolpidem tablet, it's very likely to be stolen by a thieving drug-troll, especially if the player has aged a few decades and is unable hear the troll -- or outrun it!
The images/examples listed below will be explained in future updates to this page. Thanks!


Q: Have you found a way to route RS232 using GPIO nodes?
A: The nodes themselves do not use RS232 but the SA Port Router, which is an application that ships with PathfinederPRO software, can route RS232 data to and from RS232 and TCP. And the route changes can be dynamically changed by Pathfinder as well; for more information, see pages 189 to 200 in the Pathfinder User Manual.
If you do not have PathfinderPRO, there are many devices available that can convert RS232 into GPIO to interface a node (or vice-versa) and pull closures from a node to RS232. One such device is the Hall Research HR-4P, but a Google search will turn up many others.


The Way is fundamentally perfect. It penetrates everything. How could it depend upon practice and realization?
The vehicle of the Dharma is free and unobstructed. Why is concentrated effort necessary?
The truth is that the Great Body is well beyond the dust of this world. Who could believe there is a way to dust it?
It is never separate from anyone, always right there where you are. What good is it to go here or there to practice?
But if there is a gap, however narrow, the Way remains as distant as the sky from the earth. If you show the slightest preference or the slightest antipathy, the mind becomes lost in confusion.
Imagine someone who flatters himself on his own understanding and creates illusions about his own awakening, seeing the wisdom which penetrates all things, he joins the way, clarifies the soul, and awakens the desire to climb the sky.
This person has started on the limited outskirts: but he has hardly begun on the vital Way of absolute emancipation.
Do I need to speak of Buddha, who possessed innate knowledge? We still feel the influence of the six years he lived, seated in the lotus, in total immobility.
And the transmission of the seal has preserved the memory of Bodhidharma's nine years of zazen facing the wall.
Since it was thus ancient Buddhas and Patriarchs, how can anyone today excuse themselves from negotiating the Way?
So abandon all practice founded on intellectual comprehension, running after words and holding to the letter, and learn the about-face which directs your light to the inside, to illuminate your own true nature.
By themselves, body and mind drop off, and your original face appears.
If you want to attain awakening, you must practice awakening, without delay.
For zazen, find a quiet room. Eat and drink moderately. Throw off all obligations and give up all affairs.
Do not think: "This is good, that is bad." Do not take sides for or against. Stop all movement of the conscious mind. Do not judge thoughts and opinions. Have no desire to become a Buddha.
Zazen has absolutely nothing to do with the sitting posture or the lying posture.
In the place where you sit, spread out a thick mat and place a zafu on top of it. Sit down in the lotus or half-lotus posture.
In the lotus, place your right foot on your left thigh, then your left foot on your right thigh. In the half-lotus simply press your left foot against your right thigh.
Make sure that you loosen your clothes and your belt, and arrange them carefully.
Then place your right hand, palm up, on your left leg and your left hand on top of your right hand, palms up, the tips of the thumbs touching.
Sit straight in the correct posture, without leaning to the left or to the right, nor forwards or backwards.
Make sure your ears in line with your shoulders and your nose in line with your navel.
Place your tongue against the front of the palate: mouth closed, teeth just touch.
The eyes are always open, and always breathe gently through the nose.
Once you assume the correct posture, breathe once deeply in and out.
Sway your body from the right to the left, and sit, without moving, in this stable posture.
Think from the depths of non-thinking.
How do you think from the depths of non-thinking?
This in itself is the essential art of zazen.
This zazen is not step-by-step meditation, it is nothing other than the dharma of peace and happiness, the practice-realization of perfect awakening.
Zazen is the manifestation of ultimate reality. Snares and nets can never catch it.
Once you have grasped its heart, you are like the dragon when he enters the water and like the tiger when he enters the mountain.
For you must know that at this moment in zazen, true Dharma manifests itself. So from the start, drop off all physical and mental laziness and distraction.
When you get up again, move without hurrying, calmly and deliberately. Do not get up suddenly or clumsily.
Consider the past and see that transcending at once both illumination and non-illumination, that dying seated or standing, has always depended upon the vigor of zazen.
And further, the opening to illumination in the situations provided by a finger, a banner, a needle, a mallet; and the accomplishment of realization due to a fly-swatter, a fist, a stick, a cry-all this cannot be fully grasped by dualistic thinking. Nor any better understood through the practice of supernatural powers.
This is beyond that which man sees and hears, isn't it prior to knowledge and to perception?
So it doesn't matter if you are intelligent or not, and do not think whether you are superior or inferior. When you practice wholeheartedly, this in itself is following the Way.
Practice-realization cannot be defiled. Making the effort to follow the Way, is itself, its manifestation in daily life.
This world and the others (India and China) respect the seal of the Buddha. This school is unique in its devotion only to zazen: sitting immobile in total engagement.
Though it is said that there are as many souls as people, all negotiate the way in the same manner-through the practice of zazen.
Why give up your seat at home to wander in the dust of other lands? One false step and you stray from the Way traced out directly before you.
You have had the unique opportunity to take human shape. Do not waste it.
You can contribute to the essential work of the Way of the Buddha. Who could take vain pleasure from sparks of flint?
Form and substance are like the dew on the grass, the destined one similar to a flash of lightning. Your body will vanish soon, your life will be lost in an instant.
So I ask you please, honored disciples of Zen, for a long time so used to feeling for the elephant in the dark-do not fear the true dragon! Devote your energy to the way which points directly to the absolute. Respect those who realize their Self and no longer seek outside themselves. Put yourself in harmony with the illumination of the Buddhas, succeed to the legitimate dynasty of the satori of the Patriarchs. Practice like this always, and you can be as they. Your treasure house will open of itself, and you can use it as you please.


Experience a day in the life of an AMDA student at an Open House. Events may include demonstration classes, a tour of classroom and housing facilities and a Q&A with alumni, faculty and current students. We look forward to meeting you!
Event begins at 9:00 AM. Please arrive 30 minutes early.
Prospective students experience what life is like at AMDA. Here, performing isn't just a hobby, but a profession that takes commitment and focus. At the AMDA Open House, you take part in two sample classes in your desired discipline with AMDA faculty, receiving feedback similar to enrolled students. You also hear from Admissions, faculty, current students, and featured alumni working in the industry—all in all, a day not to be missed.
If this open house date doesn't work for you, see our full list of AMDA open house dates.


Dorothy Braunwarth, 98, of Norwood Young America, died Monday, Jan. 15, 2018, at Ridgeview Medical Center in Waconia.
Mass of Christian Burial will be Friday, Jan. 19, at 11 a.m., at Church of the Ascension in Norwood Young America, with interment following in the church cemetery.
Visitation will be Friday, Jan. 19, an hour and a half prior to the service at the church.


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Prince Harry claims the palace approved Meghan Markle's ripped jeans
When Meghan Markle first appeared on the royal scene in ripped jeans and a white T-shirt, she made a lot of headlines — but not all for the right reasons.
In his new memoir, 'Spare,' Prince Harry claims his then-girlfriend's outfit at the 2017 Invictus Games in Toronto – consisted of the mother's 'Looker' frayed jeans ($174, originally $248), white "husband" shirt ($255) to her friend Misha Nonoo W Sarah Flint Flats "Natalie" ($ 450) – the palace "approved".
But after negative press and online trolls claimed Markle wasn't appropriately dressed for his royal girlfriend, the prince thought something more could be done.
"One ad defending Meg was enough to make a massive difference," he wrote, referring to the hate Markle was receiving online at the time.
Prince Harry said the 2017 Invictus Games were the "most controlled and anticipated venue" for their debut as a couple.
While Buckingham Palace is never in the habit of commenting on a family member's choice of dress, let alone defending it, Harry writes that "nobody" in the Palace defended the "Suits" actress.
Writing about the Invictus experience—which he described as the "perfect occasion" for their debut due to the games being held in Toronto, where Markle lived at the time—Harry said the pair focused their attention "on the game and the good cause" versus "clicking for the cameras."
Although both were "nervous" about being officially announced as a married couple, the duke said the couple had "hid from the world for far too long".
Markle wore controversial ripped denim during a wheelchair tennis match at the 2017 Invictus Games.
Markle carried a brown Everlane bag to the Games.
The couple discussed their experiences dealing with the press during their Netflix documentary, 'Harry & Meghan'.
In other Spear-related style finds, the Duke of Sussex details his wife's infamous Taj incident before the royal weddingHow Camilla Parker Bowles She turned his bedroom into her own dressing room After his departure and an "awkward moment" Markle shared She asks Kate Middleton to borrow her lip gloss (Which may or may not be the property of the royal family $28 from Clarins Lip Perfector).
Palace told Page Six that it would not comment on any allegations in the book.
Markle carried an Everlane gaming bag that became a quick sell-out.
Sameer Hussein
See also A UN body refuses to discuss China's treatment of Uyghur Muslims, a blow to the West
The NFL is set to make two major changes to the AFC playoffs after owners voted Friday in favor of the proposals
Pixel 7 is the latest smartphone with auto smash cameras

redpajama_set_name: RedPajamaCommonCrawl


Home - Cote's Auto Body Inc.
Welcome To Cote's Auto Body Inc.
Car owners know how important it is to find a reliable auto body shop. For over 50 years, Cote's Auto Body has been providing New Hampshire's seacoast with high quality repairs and friendly customer service to get our customers' cars up, running, and back on the road as quickly as possible. Whether it's a scratch in the paint or a ding in the door, with our expert color matching and attention to detail, it'll be like it never happened.
We're a family owned and operated auto body shop that thrives on value and precision. After an accident, few things are more important than making sure that when a car goes back on the road, it's safe and reliable. Our team of certified expert technicians aim for long-term solutions, not simple fixes. Our knowledgeable team of technicians take on a wide range of jobs, including auto body repair, dent repair, and painting.
At Cote's Auto Body, we know being without your car can be an inconvenience. Our staff and technicians are available to provide a quick response, prompt project completion, and exceptional service to solve your problem and get you on your way. With flexible appointment availability and financing options available, our goal is to eliminate the stress and make the repair process as easy on you as it could ever be.
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Kule is a small village in Raigad, Sangameshwar tehsil of district Ratnagiri.

Overview

The nearest bus stand is Sangameshwar S.T. Stand (M.S.R.T.C.) nearly 18 km.

References

Cities and towns in Ratnagiri district
Talukas in Maharashtra
Ratnagiri district

redpajama_set_name: RedPajamaWikipedia


I like to think that more people are bridging that international gap through G+ hangouts, Ravelry, Sykpe, etc.
Maybe it's just me and my online social circle, I don't know. There are a few folks that I absolutely love to "hangout" with that are not in the US, and this poses a problem when it comes to questions like, "Oh, that's a beautiful pattern, what size needle does it use?"
And then I say whatever US size it takes... and they don't know what the heck I'm talking about because they're on metric. I scoured the internet for a few minutes and found several needle conversion charts, but none of them were pretty, and they all had ads on the pages. I like my work/play space to be pretty, so I made up my own chart and thought that maybe someone else would like to have one handy for all their international knit together needs.


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Home > Loma Linda University Publications > Adventist Heritage > 20
Adventist Heritage - Vol. 11, No. 2
Adventist Heritage, Inc.
Arthur N. Patrick; Brian E. Strayer; Gary C. Jenlins; Madeline Steele Johnson; Patrick Allen; James R. Nix; Gary R. Land
Seventh-day Adventist History
2 | Editor's Stump
-- 3 | Founding Mothers: Women and the Adventist Work in the South Pacific Division
-- 16 | Sarah A. H. Lindsey: Advent Preacher on the Southern Tier
-- 26 | Mrs. Almira S. Steele and the Steele Home for Needy Children
-- 30 | Matilda Erickson Andross: The Original Missionary Volunteer
-- 48 | The Depression and the Role of Women In the Seventh-day Adventist Church
-- 55 | A Portrait Gallery: A Few Notable Women
-- 61 | The Millerites: An Advent Christian View
History | Religion
Department of Archives and Special Collections with the Department of History and the Division of Religion, Loma Linda University
Physical rights are retained by the institution. Copyright is maintained in accordance with U.S. Copyright laws.
Adventist Heritage, Inc., "Adventist Heritage - Vol. 11, No. 2" (1986). Adventist Heritage.
https://scholarsrepository.llu.edu/advent-heritage/20
History Commons, Religion Commons


Baptist Memorial Health Care believes strongly in corporate citizenship the importance of collaboration with local organizations to build stronger and healthier communities. To ensure our efforts have the greatest impact, we contribute strategic in-kind and financial contributions to causes and community health needs for which we have considerable expertise and where we can make a difference. We also conduct Community Health Needs Assessment (CHNA) to ensure our initiatives, activities and partnerships are aligned with the needs of our region.
Baptist makes contributions to IRS-designated 501(c)(3) organizations. We also support programs that add value to the scope of services offered by nonprofit organizations. We typically do not support operating costs or capital campaigns and give preference to organizations that have a broad base of funding sources.
Organizations that wish to submit a request for charitable contributions should first ensure that their requests align with our giving guidelines. They can then apply online between Jan. 1 and May 31. Applicants will be notified if their requests are approved after Sept. 1. Contributions will be awarded after Oct.1.
The organization and/or event's purpose fits within the Baptist defined focus areas of health care and/or education.
Proposals funded in any one year are not assured of future funding. If renewed funding is desired, the organization should submit documentation of outcomes and a new request the following year.
Baptist contributes in-kind gifts—goods or services instead of monetary contributions—to nonprofit organizations through programs, seminars, health fairs, board participation, events, employee engagement and other avenues. Requests for in-kind gifts must meet our giving guidelines and should be submitted online at least eight (8) weeks prior to the event/need by date.


Inflatable ball with d-handle Pony. Ideal for gymnastics, and fun !. It inflates easily with a normal bicycle pump thanks to the valve included in the equipment. It is machine, and in the event of a puncture it can be repaired with glue.


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Q: I'm finished == I'm ready? I've read about the usage of finished as an adjective and have a question about it. Does the sentence

I'm finished

express completely the same meaning as the sentence

I'm ready

A: You use "I'm finished" in situations when you know you are dead meat, to indicate you are in some serious trouble. A similar expression would be "I'm done for".
However, there are some situations in informal conversations, where I'm finished is used to indicate I'm ready. Obviously, this is incorrect usage.

A: It depends on the context. Largely, I'm finished' is not equal to 'I'm ready. But if the scene is something like what I will now mention, I'm finished' is understood as 'I'm ready.'

Come on buddy, let's go to movie ~ Sure, but let me complete my homework. Just 10 minutes.

And the friend waits in the room killing time by watching TV.
After 10 minutes...

Jack, I'm finished (with my homework). Let's go

Finished can be used to mean no longer doing something or having thing completed.
People do use this word that way as in here, here and here. See that here, in no example, it means someone is 'ready'.
True, it's used as an adjective and it's quite similar to -I'm done; I'm over with ... etc.

redpajama_set_name: RedPajamaStackExchange


The New York Times reports on the battle of Fort Davidson. Ewing's "defeat" is both a tactical victory in terms of the relative casualty numbers, and a strategic victory, in that Price's assault on St. Louis will never develop.
St. LOUIS, MO., THURSDAY, OCT. 6th, 1864.
The safe arrival at Rolla of Brig. Gen. EWING, with his command, is simply told, but when the circumstances of the campaign are detailed, and the mind dwells upon them for a space, a feeling of admiration for the skill and bravery of that Commander must find place in every breast, and cause a glow of patriotism to thrill the heart that truly feels for the brave men who are sacrificing so much to restore peace to the nation.
My communication of the 2nd inst., gave you a general idea of the battle of Arcadia Valley, which events have proven to be the struggle for the possession of the Iron Mountain Railroad, and hence the temporary possession of St. Louis by the rebels. A few more details of that affair are necessary to complete the account, and these I am enabled to give from the highest and most reliable sources.
came from PRICE, by the hands of a refugee, for EWING had persistently resisted every attempt of the rebels to gain on him by a flag of truce, a favorite dodge of theirs. The proposition was that if EWING would surrender his command, he would be permitted to march out of the fort, officers carrying their side arms, and private property of all held sacred from pillage. It is needless to say that this was disregarded, and the preparations for evacuation and retreat went on. At midnight these were completed and the garrison was withdrawn through a covered way, and the artillery removed by muffling the wheels and moving stealthily until the shadows of the mountain concealed all from view. The evacuation was a complete success, and the command moved rapidly out northward on the Caledonia road, as Gen. EWING hoped to reach Mineral Point, where he expected to find A.J. SMITH's brigade, but instead of this, he encountered in the streets of Caledonia the advance of SHELBY and drove it back. This circumstance was proof positive that the upper part of the road was in the hands of the rebels. He had little time to determine on his course, for SHELBY was now coming in force from Potosi, so EWING decided to strike for Rolla by way of Webster and Osage.
PRICE's proposition was accompanied by a threat that if it was not agreed to, he would take the fort by assault and kill every man found in it. SHELBY was ordered from Potosi for the purpose of making the assault, and it became his apparent duty now to follow EWING and either capture or demolish his command.
Wednesday night was very dark and stormy, succeeding a day of constant skirmishing with the enemy who seemed to spring from the ground on every side. The retreat was conducted under the most embarrassing circumstances, for a host of refugees, white and black, old men, women and children flocked to the Union banner and sought protection under its bright and friendly folds. At every attack and fire of the enemy they stampeded and imparted alarm to the raw soldiery, now no longer protected by earthworks, but exposed in the open field to an experienced and overwhelmingly superior foe. Nothing could have saved the entire force from utter demolition but the unflagging energy, coolness and courage of Gen. EWING and the battery of Rodman's guns.
This entry was posted in Missouri, Sterling Price, Thomas Ewing. Bookmark the permalink.


The Hannah History of Medicine and Medical Humanities Speaker Series is a yearly colloquium series open to the public featuring interdisciplinary speakers in the history of medicine and the medical humanities.
Assistant Professor, Department of Anthropology, Université de Montréal.
Location: T13 building, Room 107.
March 22, 2017: The Drowned, the Saved, and the Forgotten: Genocide Survivors and the Foundations of Modern Humanitarianism.
Click Here to Download Dr. Watenpaugh's Poster.
Title (Afternoon): The Drowned, the Saved, and the Forgotten: Genocide Survivors and the Foundations of Modern Humanitarianism.


Cycloserine belongs to the family of medicines called antibiotics. It is used to treat tuberculosis (TB). When cycloserine is used for TB, it is given with other medicines for TB. Cycloserine may also be used for other conditions as determined by your doctor.
Cycloserine is available only with your doctor's prescription.


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Mousa Madkhali  (; born 19 December 1987) is a Saudi Arabian professional footballer who currently plays for Hetten as a striker.

Honours
Al-Wehda
Prince Mohammad bin Salman League: 2017–18

References

External links

1987 births
Living people
Saudi Arabian footballers
Hetten FC players
Al-Wehda Club (Mecca) players
Al-Ain FC (Saudi Arabia) players
Al-Adalah FC players
Al-Entesar Club players
Saudi First Division League players
Saudi Professional League players
Saudi Second Division players
Association football forwards


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Editor's note: The crime news reported by the Southwest Chicago Post---taken directly from Chicago Police Department incident reports---is not by any means an exhaustive catalogue of all crime reported in the Chicago Lawn (8th) District. For example, it typically does not include news of crimes committed in the eastern and southern sectors of the district---because the Southwest Chicago Post's coverage area is primarily the neighborhoods that border Midway Airport and secondarily because including the relatively large volume of crime news from elsewhere in the district would be a logistical challenge. We make this note to offer a little helpful perspective and remind everyone that while crime is definitely a concern in all parts of the district (as it always has been), crime remains relatively low overall in Sector 1. May all of us work together diligently to keep it that way. May all of us also remember that a person charged with a crime is presumed innocent until proven guilty in a court of law.
* * *
A 32-year-old Westmont man told police that he was robbed of $10,000 cash as he sat in his vehicle (a gold-colored 2006 Chrysler Town & Country minivan)---parked near 52nd and Latrobe---minutes after obtaining the cash (in an unmarked white envelope) from his mother-in-law, who lives near 51st and Laramie and was loaning him the funds. The man told police he pulled over to talk on his cell phone at about 1:25 p.m. Wednesday, June 6, when he was approached by two men, one of whom allegedly put his hand in his waistband---implying he had a gun.
The man told police he handed over the envelope (which he said had 200 fifty-dollar bills in it), and the offenders fled on foot---south on Latrobe and then east on Archer.
The man told police the robbers were black men---the first being about 40-45 years old, 5'8 to 6'2 and 170 pounds, with brown eyes and black hair, wearing a white shirt, brown pants and a black baseball cap---and the second being about 45-50 years old, 5'8 to 6'2 and 200-250 pounds, wearing a blue shirt and brown pants.
Police reported that they reviewed footage from a security camera mounted on the Chicago City Clerk's satellite office at 5674 South Archer, but did not see the offenders described by the Westmont man.
A 20-year-old Back of the Yards man was shot in the stomach in an incident that occurred near Archer and Kilpatrick at 2:24 a.m. Friday, June 8.
Police responding to a call of a person shot on the street at Archer and Kostner arrived to find a Chicago Fire Department ambulance crew administering care to the victim.
Police on the scene received conflicting accounts of what occurred and where. The victims' friends, a 21-year-old man from the Belmont Cragin neighborhood and a 26-year-old man from suburban Schiller Park, told police the victim was injured in a drive-by shooting more than a block away. The victim told police that the 26-year-old shot him when a handgun he was showing him discharged. Police report that the victim gave his account with great difficulty because he was frightened.
The victim was taken by ambulance to Christ Hospital and Medical Center in Oak Lawn.
A 12-year-old Clearing boy using the bathroom in the fieldhouse at Hale Park said that a man looked at him in an inappropriate manner and threatened him with a raised fist.
Police said that the boy walked into the bathroom at about 7:25 p.m. Wednesday, June 6. He said he went to use a urinal and noticed a man looking at him. The boy reportedly asked, "Why are you looking at me?" The man reportedly answered, "No, I wasn't." The boy told police he then said, "Yes, you did look at me." The man then reportedly responded, "No, I didn't."
The boy said that the man then raised a fist in a threatening manner; and then boy then ran out to the baseball diamonds to alert his father---who immediately called police.
When police arrived, the man was not around. According to police, the boy described him as about 6'3 and 350 pounds, 40-50 years old, brown eyes, brown hair in a comb-over style, light complexion, beige polo shirt and blue jeans.
Burglars forced open the back door of a house near 66th and Central Park and stole two Sony Playstation game systems, one PlayStation video game, two Apple iPads, one MacBook and an external hard drive.
The crime was discovered by a next-door neighbor, a 30-year-old woman, who saw the door open at 11:00 a.m. Wednesday, June 6 and called police.
The victim, a 19-year-old man, told police that earlier in the morning, he saw a suspicious man going door to door in the neighborhood, claiming to be lost when people answered their doors. He described the man as a black man, 30-35 years old, wearing a baggy white t-shirt and blue jeans.
Burglars forced open a side window of a home near 59th and Komensky and stole a laptop computer, two stereo speakers, an amplifier, two pairs of Nike Jordan shoes, assorted jewelry and a tool box with tools. The burglars tried but failed to steal a TV that was secured in the home.
The crime was discovered by the victim, a 44-year-old man, at 2:00 p.m. Tuesday, June 5 when he came home from work.
Burglars opened an unlocked window in the rear of a home near 69th and Komensky and stole seven TVs, a laptop computer, two Sony PlayStation 3 video game systems, various PlayStation 3 games, a Nintendo Wii, two Amazon Kindles, five watches and two wedding rings.
The crime was discovered by the victim, a 34-year-old woman, at 8:10 a.m. Monday, June 4 when she came home from work. A neighbor told police that she had earlier seen a white van in the alley, which may have been connected to the crime.
Burglars kicked open the back door of a home near 55th and Melvina and stole an Xbox video game console.
The crime was discovered by the victim, a 46-year-old woman, when she returned from work at 6:50 p.m.Monday, June 4. The victim also noted that a rock had been thrown through a back window of the home.
A person who spoke with police at the scene said she had seen several men standing in front of the home at about 5:30 p.m., but apparently gave no other information.
Want to work directly with Chicago Police to prevent crime in your neighborhood? If you live in Beat 812 (see map above), come to the Clearing Branch Library at 7:00 p.m. Wednesday, June 13 and attend your monthly CAPS meeting. If you live in Beat 814, come to Vittum Park at 7:00 p.m. Thursday, June 14. Hear updates on crime in your neighborhood and learn how you can work with neighbors and police to make the community safer and better for all.
Posted by Southwest Chicago Post at 10:14 PM
Labels: Crime, Southwest Chicago Post
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I don't have to rush to finish up my previous backlog anymore! Yay!
Basically, my job now is answering calls, report to the engineers, and follow up every single case and return call to customers every one hour to keep them updated. At 1st I tot one hour cant do much thing leh. But now I know, in that 1 hour, many things can happen, but yet, unable to solve it in within a day.
As usual we will gather up at my late great-grandma's place. Although my my great-grandma already passed away. But we still gather up as usual to gamble and what-not… Previous years, my great-grandma is really old and she can't recognize most of us anymore.
from left:- 4th uncle, granduncle, eldest uncle.
aunt amy, cousin tracy, cousin wilson.
aunt amy, cousin swee jin, cousin tracy.


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Community Foundation grant sets financial clarity for Clark students
In News, News Releases
Community Foundation grant sets financial clarity for Clark students2015-07-092015-07-09https://www.clarkcollegefoundation.org/wp-content/uploads/2018/09/clark-college-logo-2.pngClark College Foundationhttps://www.clarkcollegefoundation.org/wp-content/uploads/2015/09/Community-Foundation_logo_horizontal_color500W.jpg200px200px
$30,000 grant to break generational cycles of poverty
Low-income, first-generation students, are the most likely to drop out of college, nearly four times more likely than those who have neither of these risk factors.
Clark College Foundation received a grant from the Community Foundation for Southwest Washington to assist in breaking the cycle of intergenerational poverty in the region. The $30,000 grant is designated for a full-time financial literacy coach within Clark College's Career Services department.
"This financial literacy coach will provide much-needed, one-on-one assistance to many of Clark's most vulnerable students to help them set practical and achievable financial goals so they can succeed in college," said Lisa Gibert, president/CEO, Clark College Foundation. "We are grateful to the Community Foundation for Southwest Washington's leadership in helping us serve the highest need families in our region."
At Clark, among pre-college and college-level students, 72 percent are first-generation, 47 percent come from low-income backgrounds and 39 percent have low incomes and are first-generation.
"Our grants committee saw this proposal as an opportunity to support students who are striving for a better life through higher education," said Jennifer Rhoads, president of the Community Foundation. "On behalf of our many supporters, we are honored to strengthen the efforts of our local community college as they work to ensure that every student is able to achieve their dream of a college degree."
The grant provides the funding to upgrade an existing Career Services position from part-time to full-time. The individual will conduct workshops and classroom presentations focusing on topics such as budgets, balancing a checkbook, understanding credit cards, credit scores and loan repayment, according to Edie Blakley, director of Career and Veteran Services.
"Working with students to teach them financial wellness strategies will result in self-sufficiency and increase the likelihood that they will attain their economic, education and career goals," said Blakley.
The coach will positively impact significant numbers of vulnerable Clark students, 54 percent of whom are from families below the poverty level and 60 percent are unemployed and/or on public assistance. Moreover, nearly three-quarters of Clark students are the first generation in their families to attend college.
career services, Clark College, clark college foundation, college, financial literacy
Ed Boston is new alumni director
Savoring the excellenceDonor Stories, Events, News
Bank of America grant funds new pre-apprenticeship edge programNews, News Releases


If Gestational Sac Is Seen On Your Ultrasound, Is It a Guarantee of a Normal Pregnancy?
What Does Empty Gestational Sac Indicate?
What Will Your Doctor Do, If He Sees an Empty Gestational Sac?
Pregnancy can progress differently for every woman. The conception of a baby can be a complex process. The initial phase of foetal development normally forms the basis for a safe and healthy pregnancy.
So the confirmation of pregnancy is usually followed up by conducting an ultrasound and blood tests. Your doctor may check the levels of HCG (Human Chorionic Gonadotropin) through a blood test. During an ultrasound, the doctor tries to find the gestational sac which helps him to ascertain the progression of the pregnancy.
The presence of a gestational sac by 3 to 5 weeks of gestation is normally considered a positive mark. It could also happen that a gestational sac is detected but without the presence of an embryo.
In some cases, during the ultrasound, the doctor is unable to see a gestational sac. Such a scenario can arise due to miscalculation of gestational dates which means it may be too soon to see a gestational sac. But if no gestational sac is spotted even during the follow-ups, it may indicate a possible ectopic pregnancy or a miscarriage.
The detection of gestational sac through an early ultrasound is usually the easiest way to gauge the advance of pregnancy. Gestational sac is the cavity containing the amniotic fluid which surrounds the developing embryo, is usually present in the uterus. It looks like a dark space with a white rim around when seen on an obstetric ultrasound.
A gestational sac is the only structural evidence available which may indicate the existence of an intrauterine pregnancy until an embryo can be detected. It usually develops about 5 to 7 weeks after the end of last menstrual cycle. It is normally noticeable by 3 to 5 weeks of gestation when the MSD (Mean Sac Diameter) measures about 2 to 3 mm in diameter by obstetric ultrasonography or when the HCG levels are around 1500 to 2000.
The shape of the gestational sac during initial days of pregnancy (before 8 to 10 weeks) is normally important. Although a round shaped gestational sac is usually considered ideal, there is no definite way of defining what other shapes mean. But in case the doctor finds the gestational sac to be unusually shaped, he may like to monitor it regularly to rule out any probable complications. Preferably, after 1 or 2 weeks of the last examination of the gestational sac, the doctor should be able to detect the fetal heartbeat which is a sure shot sign that pregnancy is advancing well.
Discovery of a gestational sac during an ultrasound is surely a positive pregnancy sign. However, the presence of a gestational sac alone may not guarantee a normal pregnancy. Apart from the gestational sac, it is important that a yolk sac, which forms within it, also gets developed. The yolk sac is vital as the developing embryo draws its nutrition from it till the placenta doesn't take over. The yolk sac normally becomes apparent by 5 to 6 weeks of gestation on a transvaginal ultrasound. Sometimes, a yolk sac is not found even though a gestational sac may be visible on ultrasound.
The most common reason can be a miscalculation of gestation. You may have to go for another follow-up ultrasound test. Comparing your HCG levels with the results of your ultrasound test can prove helpful here. If your HCG levels are less than 1500, then it may be too early to look for a gestational sac.
A miscarriage has taken place or is likely to happen. Dropping HCG levels may be an indication of a miscarriage.
If a gestational sac is not visible and the HCG levels are around 1500 to 2000, it may be a sign of an ectopic pregnancy. Ectopic pregnancy or extrauterine pregnancy means the embryo gets implanted outside the uterus. Such an occurrence is a medical emergency and may require further testing for proper diagnosis and treatment.
It is possible to detect an embryo in the gestational sac in 6 weeks of gestation. But sometimes, an anembryonic pregnancy or gestational sac miscarriage may occur. In such an instance the early gestational sac is empty and does not hold an embryo. It simply means that an embryo was unable to develop. The reasons can range from abnormal cell division to poor quality egg or sperm. Such a pregnancy loss normally happens during the initial days of pregnancy and often without a woman even realizing that she was pregnant.
In many cases, a woman undergoes natural miscarriage without intervention due to chromosomal irregularities, while in some cases a woman may have to opt for a D&C (Dilation and Curettage) procedure.
An empty gestational sac can be because of reasons like early intrauterine pregnancy, anembryonic pregnancy or blighted ovum. In case your doctor notices an empty gestational sac he may test your HCG levels and suggest another ultrasound. Because sometimes, it can be difficult to calculate the gestational age that early in pregnancy and the follow-up ultrasound tests may reveal an embryo later on.
But if the results are unfavourable, he may prepare you for a miscarriage as the pregnancy is not advancing normally and is not likely to result in the birth of a child. The doctor may also declare a pregnancy non-viable (failed pregnancy) if he observes an irregular gestational sac which simply means that the gestational sac does not possess the well-defined, consistent structure of a normal gestational sac.
Different women can have varied pregnancy experiences. It is important to remain stress-free and enjoy your pregnancy as much as possible. You can always seek emotional and physical support from family and friends. It is sensible to discuss with your doctor in detail for necessary guidance and treatment in case of any complications.
Next articleBaby Dropping – When Does It Happen and What Should You Do?
Is Eating Capsicum during Pregnancy Safe?


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\section*{Introduction}

Estimating the restriction of a Laplace eigenfunction $f$ on a Riemannian manifold $X$ to a compact submanifold $Y$ is a classical problem in partial differential equations and global analysis. There are various types of restriction problems, for example, one may estimate $L^p$-norms ($0<p\leq\infty$) of the restriction $f|_Y$ in terms of the eigenvalue of $f$. A more refined quantity in the case $p=2$ are the Fourier coefficients of $f|_Y$ with respect to an orthonormal basis of Laplace eigenfunctions $g$ in $L^2(Y)$. Each Fourier coefficient is given by an integral $P_Y(f,g)$ of $f|_Y$ against $g$, and one can ask for estimates in terms of the eigenvalues of $f$ and $g$.

In the context of locally symmetric spaces, the integrals $P_Y(f,g)$ are also called \textit{periods} and they carry important arithmetic information of $f$ and $g$. One instance of this is when $X$ is an arithmetic hyperbolic surface and $Y$ a closed geodesic, in which case the period integrals $P_Y(f,g)$ are proportional to special $L$-values via Waldspurger's formula. In a series of papers, J.~Bernstein and A.~Reznikov~\cite{BR04,BR05,BR10,Rez08,Rez15} studied various types of period integrals for automorphic forms on hyperbolic surfaces in connection with the representation theory of the corresponding isometry groups of the universal coverings of $X$ and $Y$. More recently, their techniques were generalized to the case of higher-dimensional real hyperbolic manifolds by the authors~\cite{MO13,Su16}, where estimates for $P_Y(f,g)$ either in terms of the eigenvalue of $f$ or in terms of the eigenvalue of $g$ were obtained.

In this work we try to extend the techniques by Bernstein--Reznikov even further and study period integrals for arbitrary locally symmetric spaces $X$ of rank one. In this context, we obtain estimates for $P_Y(f,g)$ in terms of the eigenvalue of $f$ and $g$, for compact totally geodesic cycles $Y$ of a particular form. This includes real, complex and quaternionic hyperbolic manifolds $X$ as well as quotients of the $16$-dimensional octonionic plane in which case $Y$ is an $8$-dimensional real hyperbolic submanifold.

\subsection*{Statement of the results}

Let $X$ be a connected locally symmetric space of rank one. Then the universal cover $\widetilde{X}$ of $X$ is isomorphic to the hyperbolic space $H_\F^n$ over $\F$, where $\F=\R,\mathbb{C}}\def\D{\mathbb{D},\H$ ($n\geq2$) or $\F=\mathbb{O}}\def\bS{\mathbb{S}$ ($n=2$). The real dimension of $X$ is equal to $N=nd$, where $d=\dim_\R\F$. We can write $X=\Gamma\backslash H^n_\F$ for some discrete torsion-free subgroup $\Gamma\subseteq\Isom(H^n_\F)$ of the isometry group of $H^n_\F$. The space $X$ is a complete Riemannian manifold and we denote by $dx$ the Riemannian volume element of $X$ and by $\Delta_X$ the Laplacian on $X$. Then $\Delta_X$ is a self-adjoint operator on $L^2(X)=L^2(X,dx)$.

Consider a compact totally geodesic cycle $Y\subseteq X$ of the form $Y=(\Gamma\cap\Isom(H^m_\F))\backslash H^m_\F$ ($1\leq m<n$), where $H^m_\F$ is naturally embedded into $H^n_\F$. Write $M=md$ for the real dimension of $Y$, $dy$ for its Riemannian volume element and $\Delta_Y$ for the Laplacian on $Y$.

Let $f$ be a normalized square integrable Laplace eigenfunction on $X$ with eigenvalue $a\geq0$, i.e. $\Delta_Xf=af$. Then $f$ is smooth by elliptic regularity. Integrating the restriction $f|_Y$ of $f$ to $Y$ against a normalized eigenfunction $g\in L^2(Y)$ of $\Delta_Y$ with eigenvalue $b\geq0$, i.e. $\Delta_Yg=bg$, defines a period integral
$$ P_Y(f,g) = \int_Y f(y)g(y)\,dy,$$
called \emph{geodesic period}. Since $Y$ is compact, the integral always converges absolutely. We are interested in the asymptotic behavior of $P_Y(f,g)$ as the eigenvalues of $f$ and $g$ grow. Let $\{g_j\}_j$ be an orthonormal basis of $L^2(Y)$ consisting of Laplace eigenfunctions with eigenvalues $0=b_0<b_1\leq b_2\leq\ldots$. Then the period integrals $P_Y(f,g_j)$ are the Fourier coefficients of the expansion of $f|_Y$ into the basis $\{\bar{g}_j\}$ of $L^2(Y)$.

To simplify the statement of our main result, define the modified period
$$ \P_Y(f,g) = |a-b|^{-\frac{N-M-2}{4}}e^{\frac{\pi}{2}\sqrt{|a-b|}}P_Y(f,g). $$

We shall prove the following:

\begin{thmalph}\label{thm:Main}
There exists a constant $C=C_\Gamma>0$ such that
$$ \sum_{b_j\leq T} |\P_Y(f,g_j)|^2 \leq C\cdot a^{-\frac{N-2}{2}}e^{\pi\sqrt{a}}\cdot T^{\frac{N-1}{2}} \qquad \forall\,f\mbox{ with }a\leq T. $$
\end{thmalph}

For the case where one of the two eigenfunctions $f$ and $g$ is fixed, Theorem~\ref{thm:Main} implies the following bounds for the original geodesic periods $P_Y(f,g)$:

\begin{coralph}\label{cor:Main}
\begin{enumerate}[(1)]
\item\label{cor:Main1} For a fixed Laplace eigenfunction $f\in L^2(X)$ there exists a constant $C=C_{\Gamma,f}>0$ such that
$$ |P_Y(f,g_j)| \leq C\cdot b_j^{\frac{2N-M-3}{4}}e^{-\frac{\pi}{2}\sqrt{b_j}} \qquad \mbox{as $j\to\infty$.} $$
\item\label{cor:Main2} For a fixed Laplace eigenfunction $g\in L^2(Y)$ there exists a constant $C=C_{\Gamma,g}>0$ such that
$$ |P_Y(f,g)| \leq C\cdot a^{\frac{N-M-1}{4}} \qquad \forall\,f. $$
\end{enumerate}
\end{coralph}

We remark that our results hold for all locally symmetric spaces of rank one (not necessarily compact), in particular also for quotients $X$ of the octonionic plane $H^2_\mathbb{O}}\def\bS{\mathbb{S}$ in which case $Y$ is a compact $8$-dimensional hyperbolic manifold.

\subsection*{Relation to previous work}

Let us compare our results to the existing literature:
\begin{itemize}
\item
Burq--G\'{e}rard--Tzvetkov~\cite{BGT07}
estimated the $L^p$-norm ($2\leq p\leq\infty$) of $f|_Y$ for general compact Riemannian manifolds $X$ and compact submanifolds $Y$. The resulting bound for a single period is weaker than Corollary~\ref{cor:Main}~\eqref{cor:Main2}, which is to be expeceted as our results are for a very special family of Riemannian manifolds.
\item
Zelditch \cite{Ze92} estimated asymptotics of the period $P_Y(f,1)$ (i.e. $g$ is a constant function) where $X$ is a compact Riemannian manifold and $Y$ is any compact submanifold. The resulting bound for $P_Y(f,1)$ is the same as in Corollary~\ref{cor:Main}~\eqref{cor:Main2}.
\item
M\"{o}llers--{\O}rsted~\cite{MO13} estimated the period $P_Y(f,g)$ where $X$ is a real hyperbolic manifold (not necessarily compact) and $Y$ a special totally geodesic cycle of any dimension. In their work the form $f$ is fixed and $g$ varies, and the obtained bound is a special case of Corollary~\ref{cor:Main}~\eqref{cor:Main1}.
\item
Su~\cite{Su16} estimated the period $P_Y(f,g)$ where $X$ is a hyperbolic manifold (not necessarily compact) and $Y$ a special totally geodesic cycle of codimension 1. Here the form $f$ varies and $g$ is fixed, and the obtained bound is a special case of Corollary~\ref{cor:Main}~\eqref{cor:Main2}.
\end{itemize}

\subsection*{Method of proof}

Each automorphic form $f\in L^2(\Gamma\backslash H^n_\F)$ corresponds to an irreducible unitarizable representation $\pi_f$ of the isometry group $G=\textup{U}}\def\SU{\textup{SU}(1,n;\F)$ of the hyperbolic space $H^n_\F$. Then $H=\textup{U}}\def\SU{\textup{SU}(1,m;\F)\times\textup{U}}\def\SU{\textup{SU}(n-m;\F)\subseteq G$ is the subgroup of isometries leaving the hyperbolic subspace $H^m_F\subseteq H^n_\F$ invariant, and automorphic forms $g\in L^2((\Gamma\cap H)\backslash H^m_\F)$ correspond to certain irreducible unitarizable representations $\tau_g$ of $H$. The period integral $P_Y(f,g)$ can be interpreted as the special value of an $H$-invariant bilinear form on the product of $\pi_f$ and $\tau_g$. In this context, invariant bilinear forms are unique up to scalars and can therefore be related to invariant bilinear forms on explicit models of the representations $\pi_f$ and $\tau_g$. Such model forms are constructed in \cite{MOO16} for principal series representations in terms of explicit integral kernels, so that the period integrals $P_Y(f,g)$ can be written as the product of a certain explicit model integral and a proportionality scalar relating the two invariant bilinear forms (an idea due to Bernstein--Reznikov). Whereas the proportionality scalar can be estimated by standard techniques (see Section~\ref{estimate}), the evaluation of the model integral is the key computation in this paper (see Section~\ref{special}).

\subsection*{Structure of the paper}

In Section \ref{sec:GeodesicPeriods} we give a group-theoretic description of locally symmetric spaces of rank one and their special totally geodesic cycles. Also we introduce the automorphic invariant bilinear forms arising from geodesic periods. In Section \ref{representation} we recall the structure and representation theory of rank one groups. The non-compact models of the representations and invariant bilinear forms on these models are discussed. The geodesic periods are expressed as a product of a special value of the model invariant bilinear form and a proportionality scalar. The latter two objects are dealt with in Section \ref{special} and \ref{estimate}, respectively. The necessary integral formulas used in Section \ref{special} are summarized in the Appendix.\newpage

\section{Geodesic periods on rank one locally symmetric spaces}\label{sec:GeodesicPeriods}

\subsection{Locally symmetric spaces of rank one}\label{geometry}

Let $X$ be a connected locally symmetric space of rank one. Then its universal covering $\widetilde{X}$ is a globally symmetric space of rank one and hence of the form $\widetilde{X}=G/K$, where $G=\textup{U}}\def\SU{\textup{SU}(1,n;\F)$ with $\F=\R,\mathbb{C}}\def\D{\mathbb{D},\H$ ($n\geq2$) or $\F=\mathbb{O}}\def\bS{\mathbb{S}$ ($n=2$) and $K=\textup{U}}\def\SU{\textup{SU}(1;\F)\times\textup{U}}\def\SU{\textup{SU}(n;\F)$ the standard maximal compact subgroup of $G$. The group $G$ acts on $\widetilde{X}$ by isometries, and conversely every isometry of $\widetilde{X}$ is given by an element of $G$. We can therefore identify the fundamental group $\Gamma=\pi_1(X)$ of $X$ with a torsion-free discrete subgroup of $G$ so that $X=\Gamma\backslash G/K$.

The symmetric space $\widetilde{X}$ is a Riemannian manifold with the metric induced by the Killing form $\kappa$ on the Lie algebra $\mathfrak{g}}\def\frn{\mathfrak{n}$ of $G$ which we normalize to be
$$ \kappa(X,Y) = \frac{1}{2}\Tr(XY^*), \qquad X,Y\in\mathfrak{g}}\def\frn{\mathfrak{n}.$$
This metric descends to $X$ and gives rise to a Riemannian measure $dx$ of $X$. Denote by $\Delta_X$ the Laplacian on $X$ determined by the Riemannian metric. Then $\Delta_X$ extends to a self-adjoint operator on $L^2(X)$.

\subsection{Automorphic representations}

Every eigenfunction $f\in L^2(X)$ of $\Delta_X$ is automatically smooth since $\Delta_X$ is an elliptic operator. In view of the isomorphism $L^2(X)\simeq L^2(\Gamma\backslash G)^K$ we can regard $f$ as a right-$K$-invariant $L^2$-function on $\Gamma\backslash G$. The space $\Gamma\backslash G$ is equipped with a $G$-invariant Radon measure that descends to $dx$ (so that the quotient integral formula holds). Thus the action of $G$ given by right translation induces a unitary representation of $G$ on $L^2(\Gamma\backslash G)$. The closed subspace generated by all translates of $f$ is an irreducible subrepresentation of $L^2(\Gamma\backslash G)$, in which $f$ is the (up to scalar multiples) unique $K$-invariant vector.

Write $V_f\subseteq L^2(\Gamma\backslash G)$ for the space of smooth vectors of the subrepresentation generated by $f$, and $\pi_f$ for the action of $G$ on $V_f$. Since the Lie algebra $\mathfrak{g}$ of $G$ acts on $\Gamma\backslash G$ by vector fields, we have $V_f\subseteq C^\infty(\Gamma\backslash G)$ by the Sobolev Embedding Theorem.

\subsection{Geodesic cycles}

For $G=\textup{U}}\def\SU{\textup{SU}(1,n;\F)$ we consider subgroups of the form $H=\textup{U}}\def\SU{\textup{SU}(1,m;\F)\times\textup{U}}\def\SU{\textup{SU}(n-m;\F)$ ($1\leq m\leq n-1$). Note that for $G=\textup{U}}\def\SU{\textup{SU}(1,2;\mathbb{O}}\def\bS{\mathbb{S})=F_{4(-20)}$ we have $H=\textup{Spin}(8,1)$. Then $(G,H)$ forms a symmetric pair. More precisely, $H=G^\sigma$ is the fixed point subgroup of the involution $\sigma$ of $G$ given by conjugation with the element $\diag(\bm{1}_{m+1},-\bm{1}_{n-m})$. The intersection $K_H=K\cap H=\textup{U}}\def\SU{\textup{SU}(1;\F)\times\textup{U}}\def\SU{\textup{SU}(m;\F)\times\textup{U}}\def\SU{\textup{SU}(n-m;\F)$ is a maximal compact subgroup of $H$ and $H/K_H$ a rank one Riemannian symmetric space. The natural embedding $H/K_H\hookrightarrow G/K$ identifies $H/K_H$ with a totally geodesic submanifold of $G/K$.

Assume that $\sigma\Gamma=\Gamma$, then the intersection $\Gamma_H=\Gamma\cap H=\Gamma^\sigma\subseteq\Gamma$ is a torsion-free discrete subgroup of $H$ and the rank one locally symmetric space $Y=\Gamma_H\backslash H/K_H$ can be viewed as a totally geodesic submanifold of $X$. We denote by $dy$ the corresponding Riemannian measure on $Y$ and by $\Delta_Y$ the Laplacian.

Like in the case of $X$, every eigenfunction $g$ of $\Delta_Y$ generates an irreducible subrepresentation of $L^2(\Gamma_H\backslash H)$ whose subspace of smooth vectors is denoted by $W_g\subseteq C^\infty(\Gamma_H\backslash H)$. We write $\tau_g$ for the action of $H$ on $W_g$.

\subsection{Geodesic periods}

Let $f\in L^2(X)$ be a Maass form on $X$, i.e. $f$ is a square integrable eigenfunction of $\Delta_X$. Write $\Delta_Xf=af$ where $a\geq0$. Let $g\in L^2(Y)$ be a Maass form on $Y$ such that $\Delta_Yg=bg$ where $b\geq0$. Assuming that $Y$ is compact, the period integral
$$ P_Y(f,g) = \int_Yf(y)g(y)\,dy$$
converges absolutely. The purpose of this paper is to find estimates for $P_Y(f,g)$ in terms of $a$ and $b$. Throughout the paper we normalize $f$ and $g$ such that $\|f\|_{L^2(X)}=\|g\|_{L^2(Y)}=1$.

\subsection{Automorphic invariant bilinear forms}

We can view $P_Y(f,g)$ as the special value of an invariant bilinear form on $V_f\times W_g$. More precisely, define
$$ \ell^\aut_{f,g}:V_f\times W_g\to\mathcal{C}}\def\DD{\mathcal{D}, \quad (\phi,\psi)\mapsto\int_{\Gamma_H\backslash H}\phi(h)\psi(h)\,dh, $$
where $dh$ denotes the $H$-invariant measure on $\Gamma_H\backslash H$ that descends to $dy$, then
$$ P_Y(f,g) = \ell^\aut_{f,g}(f,g). $$
The integral above always converges since $\phi$, $\psi$ are smooth and $\Gamma_H\backslash H$ is compact. As a consequence of the $H$-invariance of $dh$, the bilinear map $\ell^\aut_{f,g}$ is invariant under the action $\pi_f|_H\otimes\tau_g$ of $H$, i.e.
$$ \ell^\aut_{f,g} \in \Hom_H(\pi_f|_H\otimes\tau_g,\mathbb{C}}\def\D{\mathbb{D}). $$

\section{Representation theory of rank one reductive groups}\label{representation}

We recall the classification of spherical unitary representations of the rank one groups $G=\textup{U}}\def\SU{\textup{SU}(1,n;\F)$, $\F=\R,\mathbb{C}}\def\D{\mathbb{D},\H,\mathbb{O}}\def\bS{\mathbb{S}$. For the subgroups $H=\textup{U}}\def\SU{\textup{SU}(1,m;\F)\times\textup{U}}\def\SU{\textup{SU}(n-m;\F)$ we describe the $H$-invariant bilinear forms on products of such representations of $G$ and $H$ as obtained in \cite{MOO16}.

\subsection{Group decompositions}

Let $\mathfrak{a}}\def\frb{\mathfrak{b}=\R H_0\subseteq\mathfrak{g}}\def\frn{\mathfrak{n}$ with
$$ H_0 = \left(\begin{array}{ccc}0&1&\\1&0&\\&&\bm{0}_{n-1}\end{array}\right)\in\mathfrak{g}}\def\frn{\mathfrak{n}, $$
then $\ad(H_0)$ acts on $\mathfrak{g}}\def\frn{\mathfrak{n}$ with eigenvalues $\{0,\pm1\}$ for $\F=\R$, and eigenvalues $\{0,\pm1,\pm2\}$ for $\F=\mathbb{C}}\def\D{\mathbb{D},\H,\mathbb{O}}\def\bS{\mathbb{S}$. Let $\alpha\in\mathfrak{a}}\def\frb{\mathfrak{b}_\mathbb{C}}\def\D{\mathbb{D}^*$ be given by $\alpha(H_0)=1$ and put
$$ \frn = \mathfrak{g}}\def\frn{\mathfrak{n}_\alpha+\mathfrak{g}}\def\frn{\mathfrak{n}_{2\alpha}, \qquad \overline{\frn} = \mathfrak{g}}\def\frn{\mathfrak{n}_{-\alpha}+\mathfrak{g}}\def\frn{\mathfrak{n}_{-2\alpha}. $$
Note that $\mathfrak{g}}\def\frn{\mathfrak{n}_{\pm2\alpha}=\{0\}$ for $\F=\R$. Denote by $A=\exp(\mathfrak{a}}\def\frb{\mathfrak{b})$, $N=\exp(\frn)$ and $\overline{N}=\exp(\overline{\frn})$ the corresponding analytic subgroups of $G$. We identify $\F^{n-1}\oplus\Im\F\simeq\overline{N}$ by
$$ (x,z)\mapsto\overline{n}_{(x,z)} = \exp\left(\begin{array}{ccc}z&z&x^*\\-z&-z&-x^*\\x&x&\bm{0}_{n-1}\end{array}\right)=\left(\begin{array}{ccc}1+z+\frac{1}{2}|x|^2&z+\frac{1}{2}|x|^2&x^*\\-z-\frac{1}{2}|x|^2&1-z-\frac{1}{2}|x|^2&-x^*\\x&x&\bm{1}_{n-1}\end{array}\right). $$
Then
$$ \overline{n}_{(x,z)}\cdot\overline{n}_{(x',z')}=\overline{n}_{(x,z)\cdot(x',z')}, $$
where
$$ (x,z)\cdot(x',z')= (x+x',z+z'+\tfrac{1}{2}(x^*x'-x^{\prime\,*}x)). $$
In particular, $\overline{n}_{(x,z)}^{-1}=\overline{n}_{(-x,-z)}$. Further, let
$$ M=Z_K(A)=\{\diag(z,z,k):z\in\textup{U}}\def\SU{\textup{SU}(1;\F),k\in\textup{U}}\def\SU{\textup{SU}(n-1;\F)\}\simeq\textup{U}}\def\SU{\textup{SU}(1;\F)\times\textup{U}}\def\SU{\textup{SU}(n-1;\F), $$
then $P=MAN$ is a (minimal) parabolic subgroup of $G$. Let $\rho=\frac{1}{2}\Tr\ad|_{\frn}$ be the half sum of positive roots.

We have the Iwasawa decomposition $G=KAN\simeq K\times A\times N$ and therefore, for every $g\in G$ there exists a unique $H(g)\in\mathfrak{a}}\def\frb{\mathfrak{b}$ such that $g\in Ke^{H(g)}N$. An easy computation shows that
$$ H(\overline{n}_{(x,z)}) = ((1+|x|^2)^2+4|z|^2)^{\frac{1}{2}}H_0, \qquad (x,z)\in\F^{n-1}\oplus\Im\F. $$

The subset $\overline{N}MAN\simeq\overline{N}\times M\times A\times N$ is open and dense in $G$, according to the Bruhat decomposition.

\subsection{The symmetric pair}

The parabolic subgroup $P$ is compatible with the subgroup $H$ in the sense that $P_H=P\cap H$ is a (minimal) parabolic subgroup of $H$. In fact, $P_H=M_HAN_H$ with $M_H=M\cap H$ and $N_H=N\cap H$. Let $\rho_H=\frac12\Tr\ad|_{\frn_H}$. The opposite nilradical $\overline{N}_H$ of $\overline{N}$ identifies with $\F^{m-1}\oplus\Im\F$ under the isomorphism $\F^{n-1}\oplus\Im\F\simeq\overline{N}$, where $\F^{m-1}\subseteq\F^{n-1}$ as the first $m-1$ coordinates.

\subsection{Principal series representations}

Identify $\mathfrak{a}}\def\frb{\mathfrak{b}_\mathbb{C}}\def\D{\mathbb{D}^*\simeq\mathbb{C}}\def\D{\mathbb{D}$ by $\lambda\mapsto\lambda(H)$ so that $\rho=\frac{1}{2}((n-1)d+2(d-1))=\frac{1}{2}(nd+d-2)$, where $d=\dim_\R\F$. For $\lambda\in\mathfrak{a}}\def\frb{\mathfrak{b}_\mathbb{C}}\def\D{\mathbb{D}^*$ we consider the principal series representations (smooth normalized parabolic induction)
$$ \pi_\lambda = \textup{Ind}}\def\Hom{\textup{Hom}_P^G(\bm{1}\otimes e^\lambda\otimes\bm{1}). $$
Since $\overline{N}MAN$ is open dense in $G$, any smooth right-$P$-equivariant function on $G$ is uniquely determined by its values on $\overline{N}$. We can therefore realize $\pi_\lambda$ on a space $I(\lambda)$ of smooth functions on $\overline{N}\simeq\F^{n-1}\oplus\Im\F$. More precisely, we have $\cS(\overline{N})\subseteq I(\lambda)\subseteq C^\infty(\overline{N})$ where $\cS(\overline{N})$ denotes the space of Schwarz functions on $\overline{N}$. The $K$-spherical vector in $I(\lambda)$ is given by the function
$$ \phi_\lambda(x,z) = ((1+|x|^2)^2+4|z|^2)^{-\frac{\lambda+\rho}{2}}, \qquad (x,z)\in\F^{n-1}\oplus\Im\F. $$

The representation $\pi_\lambda$ is irreducible and unitarizable if and only if $\lambda\in i\R\cup(-R,R)$, where the constant $R$ is given by
$$ R = \begin{cases}\rho&\mbox{for $\FF=\R,\mathbb{C}}\def\D{\mathbb{D}$,}\\\frac{1}{2}((n-1)d+2)&\mbox{for $\FF=\H,\mathbb{O}$.}\end{cases} $$
For $\lambda\in i\R$ the invariant inner product on $\pi_\lambda$ is the $L^2$-inner product of $L^2(\F^{n-1}\oplus\Im\F)$ with respect to the Lebesgue measure on $\F^{n-1}\oplus\Im\F\simeq\R^{(n-1)d+(d-1)}=\R^{nd-1}$. The irreducible unitary representations of $G$ with a non-zero $K$-spherical vector are precisely the unitary completions of $\pi_\lambda$ (where $\lambda\in i\R\cup(-R,R)$) together with the trivial representation. Note that $\pi_\lambda\simeq\pi_{-\lambda}$ for those parameters.

If now $f\in L^2(X)$ is an eigenfunction of the Laplacian $\Delta_X$, then $\pi_f$ is an irreducible unitarizable representation with $K$-spherical vector $f$, hence $\pi_f\simeq\pi_\lambda$ for some $\lambda\in i\R\cup(-R,R)$ or $\pi_f$ is the trivial representation. The parameter $\lambda$ is related to the eigenvalue $a$ of $f$ by
$$ a = \rho^2-\lambda^2. $$

For the subgroup $H$ we denote by $\tau_\nu$ ($\nu\in\mathfrak{a}}\def\frb{\mathfrak{b}_\mathbb{C}}\def\D{\mathbb{D}^*\simeq\mathbb{C}}\def\D{\mathbb{D}$) the corresponding principal series representation, realized on a space $J(\nu)$ of smooth functions on $\overline{N}_H\simeq\F^{m-1}\oplus\Im\F$ containing the spherical vector
$$ \psi_\nu(y,w) = ((1+|y|^2)^2+4|w|^2)^{-\frac{\nu+\rho_H}{2}}, \qquad (y,w)\in\F^{m-1}\oplus\Im\F, $$
where $\rho_H=\frac{1}{2}(md+d-2)$

\subsection{Model invariant bilinear forms}\label{sec:InvBilinearForms}

In \cite{MOO16} intertwining operators in $\Hom_H\big(\pi_\lambda|_H,\tau_\nu\big)$ were constructed and shown to be generically unique (see also \cite{KS15} for a much more thorough treatment of the special case where $\F=\R$ and $m=n-1$). We briefly review the results and show how they give rise to invariant bilinear forms on $I(\lambda)\times J(\nu)$. Let $K_{\lambda,\nu}$ be the function on $\overline{N}\times\overline{N}_H$ given by
$$ K_{\lambda,\nu}((x,z),(y,w)) = \NN((-x,-z)\cdot(y,w))^{-2(\nu+\rho_H)}|x_2|^{(\lambda-\rho)+(\nu+\rho_H)}, $$
where
$$ \NN(x,z)=(|x|^4+4|z|^2)^{\frac{1}{4}} $$
and $x=(x_1,x_2)\in\F^{m-1}\oplus\F^{n-m}=\F^{n-1}$. Then the operator $A_{\lambda,\nu}:I(\lambda)\to J(\nu)$ given by
$$ A_{\lambda,\nu}u(y,w) = \int_{\overline{N}} K_{\lambda,\nu}((x,z),(y,w))u(x,z)\,d(x,z) $$
is intertwining for the action of $H$ and depends meromorphically on $(\lambda,\nu)\in\mathbb{C}}\def\D{\mathbb{D}^2$. For $(\lambda+\rho)+(\pm\nu-\rho_H)\notin(-2\mathbb{N}}\def\P{\mathbb{P})$ it was shown in \cite[Theorem 4.1]{MOO16} that $\dim\Hom_H(\pi_\lambda|_H,\tau_\nu)=1$ and the space of intertwining operators is spanned by $A_{\lambda,\nu}$ or its regularization.

In view of the isomorphism $\Hom_H\big(\pi_\lambda|_H,\tau_\nu\big)\cong \Hom_H\big(\pi_\lambda|_H\otimes\tau_{-\nu},\mathbb{C}}\def\D{\mathbb{D}\big)$, the bilinear form $\ell^\mod_{\lambda,\nu}:I(\lambda)\times J(\nu)\to\mathbb{C}}\def\D{\mathbb{D}$ given by
\begin{align*}
 \ell^\mod_{\lambda,\nu}(u,v) &= \int_{\overline{N}_H} A_{\lambda,-\nu}u(y,w)\cdot v(y,w)\,d(y,w)\\
 &= \int_{\overline{N}}\int_{\overline{N}_H}K_{\lambda,-\nu}((x,z),(y,w))u(x,z)v(y,w)\,d(y,w)\,d(x,z),
\end{align*}
is invariant for the diagonal action of $H$ by $\pi_\lambda|_H\otimes\tau_\nu$. Further, the uniqueness statement for intertwining operators implies the following uniqueness statement for invariant bilinear forms:

\begin{thm}\label{thm:UniquenessInvBilinearForms}
Let $(G,H)=(\textup{U}}\def\SU{\textup{SU}(1,n;\F),\textup{U}}\def\SU{\textup{SU}(1,m;\F)\times\textup{U}}\def\SU{\textup{SU}(n-m;\F))$, then the space $\Hom_H(\pi_\lambda|_H\otimes\tau_\nu,\mathbb{C}}\def\D{\mathbb{D})$ of invariant bilinear forms is one-dimensional if $(\lambda+\rho)+(\pm\nu-\rho_H)\notin(-2\mathbb{N}}\def\P{\mathbb{P})$ and spanned by $\ell^\mod_{\lambda,\nu}$ or its regularization.
\end{thm}

In particular, $H$-invariant bilinear forms are unique if $\lambda,\nu\in i\R$. In this case, the integral converges absolutely and no regularization is necessary.

\subsection{Proportionality}

Let $f\in L^2(X)$ and $g\in L^2(Y)$ be Laplace eigenfunctions and let $\pi_f$ and $\tau_g$ be the corresponding irreducible representations of $G$ and $H$. We fix equivariant isometric isomorphisms
$$ \theta:I(\lambda)\to V_f,\,u\mapsto\theta_u \qquad \mbox{and} \qquad \eta:J(\nu)\to W_g,\,v\mapsto\eta_v $$
which map the spherical vectors $\phi_\lambda$ and $\psi_\nu$ to $f$ and $g$. Assuming that $\lambda,\nu\in i\R$, both $\ell^\aut_{f,g}$ and $\ell^\mod_{\lambda,\nu}$ are $H$-invariant bilinear forms on $\pi_f\otimes \tau_g\simeq\pi_\lambda\otimes\tau_\nu$, so that Theorem~\ref{thm:UniquenessInvBilinearForms} implies the existence of a proportionality constant $b_{f,g}\in\mathbb{C}}\def\D{\mathbb{D}$ such that
$$ \ell^\aut_{f,g}(\theta_u,\eta_v) = b_{f,g}\cdot\ell^\mod_{\lambda,\nu}(u,v) \qquad \forall\,u\in I(\lambda),~v\in J(\nu). $$
In particular,
\begin{equation}
 P_Y(f,g) = \ell^\aut_{f,g}(f,g) = b_{f,g}\cdot\ell^\mod_{\lambda,\nu}(\phi_\lambda,\psi_\nu).\label{eq:Proportionality}
\end{equation}
In the Section~\ref{special} we compute explicitly the special value $\ell^\mod_{\lambda,\nu}(\phi_\lambda,\psi_\nu)$. In Section~\ref{estimate} we derive estimates for the proportionality constants $b_{f,g}$. Together these work implies Theorem~\ref{thm:Main}.

\section{Special values of model invariant bilinear forms}\label{special}

In this section we compute the special value $\ell^\mod_{\lambda,\nu}(\phi_\lambda,\psi_\nu)$ of the model invariant bilinear form $\ell^\mod_{\lambda,\nu}$ at the spherical vectors $\phi_\lambda\in I(\lambda)$, $\psi_\nu\in J(\nu)$.

\begin{thm}\label{thm:SpecialValue}
There exists a constant $c>0$ depending only on $m$, $n$ and $d$ such that
$$ \ell^\mod_{\lambda,\nu}(\phi_\lambda,\psi_\nu) = c\cdot\frac{\Ga(\frac{\lambda+\rho+1}{2})\Ga(\frac{\lambda+\rho-\nu-\rho_H}{2})\Ga(\frac{\lambda+\rho+\nu-\rho_H}{2})}{\Ga(\frac{\lambda+\rho-d+2}{2})\Ga(\lambda+\rho)}. $$
\end{thm}

\begin{rmk}
For $\F=\R$ the above integral was first computed in \cite[Proposition 3.1]{MO13} (see also \cite{KS15} for the special case where $m=n-1$).
\end{rmk}

From the well-known Stirling formula
$$ \Ga(a+ib) = \sqrt{2\pi}|b|^{a-1/2}e^{-\frac{\pi}{2}|b|}(1+\OO(|b|^{-1})), \qquad \text{as}~|b|\to\infty, $$
it follows that for fixed $a>0$ there exist $c_1,c_2>0$ such that
$$ c_1(1+|b|)^{a-1/2}e^{-\frac{\pi}{2}|b|} \leq |\Ga(a+ib)| \leq c_2(1+|b|)^{a-1/2}e^{-\frac{\pi}{2}|b|} \qquad \forall\,b\in\R. $$
Applied to Theorem~\ref{thm:SpecialValue} this yields the following asymptotics for $\ell^\mod_{\lambda,\nu}(\phi_\lambda,\psi_\nu)$:

\begin{cor}\label{cor:AsymptoticsModelPeriod}
For $\lambda,\nu\in i\R$ we have
$$ |\ell^\mod_{\lambda,\nu}(\phi_\lambda,\psi_\nu)| \sim (1+|\lambda|)^{1-n\frac{d}{2}}(1+|\lambda^2-\nu^2|)^{(n-m)\frac{d}{4}-\frac{1}{2}}e^{\frac{\pi}{2}|\lambda|}e^{-\frac{\pi}{4}(|\lambda+\nu|+|\lambda-\nu|)}. $$
\end{cor}

\subsection{Reformulation using intertwining operators}

As noted in Section~\ref{sec:InvBilinearForms} we have
$$ \ell^\mod_{\lambda,\nu}(\phi_\lambda,\psi_\nu) = \int_{\overline{N}_H} A_{\lambda,-\nu}\phi_\lambda(y,w)\cdot\psi_\nu(y,w)\,d(y,w). $$
The intertwining operator $A_{\lambda,-\nu}:I(\lambda)\to J(-\nu)$ maps the $K$-invariant vector $\phi_\lambda\in I(\lambda)$ to a scalar multiple of the $K_H$-invariant vector $\psi_{-\nu}\in J(-\nu)$, and since $\psi_{-\nu}(0,0)=1$ we have
$$ A_{\lambda,-\nu}\phi_\lambda = A_{\lambda,-\nu}\phi_\lambda(0,0)\cdot\psi_{-\nu}. $$
Hence,
\begin{align*}
 \ell^\mod_{\lambda,\nu}(\phi_\lambda,\psi_\nu) &= A_{\lambda,-\nu}\phi_\lambda(0,0)\cdot\int_{\overline{N}_H} \psi_\nu(y,w)\cdot\psi_{-\nu}(y,w)\,d(y,w)\\
 &= A_{\lambda,-\nu}\phi_\lambda(0,0)\cdot\int_{\overline{N}_H} ((1+|y|^2)^2+4|w|^2)^{-\rho_H}\,d(y,w),
\end{align*}
the latter integral being a positive constant depending only on $m$, $n$ and $d$.

\subsection{Computation of $A_{\lambda,-\nu}\phi_\lambda(0,0)$}

We now compute the integral
$$ A_{\lambda,-\nu}\phi_\lambda(0,0) = \int_{\overline{N}} (|x|^4+4|z|^2)^{\frac{\nu-\rho_H}{2}}|x_2|^{(\lambda-\rho)-(\nu-\rho_H)}((1+|x|^2)^2+4|z|^2)^{-\frac{\lambda+\rho}{2}}\,d(x,z). $$
To simplify notation we abbreviate
$$ p=(n-1)d, \qquad p'=(m-1)d, \qquad p''=(n-m)d, \qquad q=d-1, $$
then the integral becomes
\begin{multline*}
 = \int_{\R^{p'}}\int_{\R^{p''}}\int_{\R^q} ((|x_1|^2+|x_2|^2)^2+4|z|^2)^{\frac{2\nu-p'-2q}{4}} |x_2|^{\lambda-\nu-\frac{p''}{2}}\\
 ((1+|x_1|^2+|x_2|^2)^2+4|z|^2)^{-\frac{2\lambda+p+2q}{4}}\,dz\,dx_2\,dx_1.
\end{multline*}
Using polar coordinates on $\R^{p'}$, $\R^{p''}$ and $\R^q$ as well as the volume formula $\textup{vol}}\def\len{\textup{len}(S^{k-1})=\frac{2\pi^{\frac{k}{2}}}{\Gamma(\frac{k}{2})}$ this simplifies to
\begin{multline*}
 = \frac{\pi^{\frac{p+q}{2}}}{2^{q-3}\Gamma(\frac{p'}{2})\Gamma(\frac{p''}{2})\Gamma(\frac{q}{2})}\int_0^\infty\int_0^\infty\int_0^\infty((r^2+s^2)^2+t^2)^{\frac{2\nu-p'-2q}{4}}\\
 ((1+r^2+s^2)^2+t^2)^{-\frac{2\lambda+p+2q}{4}} r^{p'-1}s^{\lambda-\nu+\frac{p''}{2}-1}t^{q-1}\,dr\,ds\,dt.
\end{multline*}
With \eqref{eq:IntFormulaMellinTrafoOfSphericalProduct} we first calculate the integral over $t$:
\begin{multline*}
 = \frac{\pi^{\frac{p+q}{2}}\Gamma(\frac{2\lambda-2\nu+2p'+p''+2q}{4})}{2^{q-2}\Gamma(\frac{p'}{2})\Gamma(\frac{p''}{2})\Gamma(\frac{2\lambda-2\nu+2p'+p''+4q}{4})} \int_0^\infty\int_0^\infty r^{p'-1}s^{\lambda-\nu+\frac{p''}{2}-1}(r^2+s^2)^{\nu-\frac{p'}{2}-q}(1+r^2+s^2)^{-\lambda-\frac{p}{2}}\\
 \times {_2F_1}\left(-\tfrac{2\nu-p'-2q}{4},\tfrac{q}{2};\tfrac{2\lambda-2\nu+2p'+p''+4q}{4};1-\tfrac{(1+r^2+s^2)^2}{(r^2+s^2)^2}\right)\,dr\,ds.
\end{multline*}
Substituting $(r,s)=(x\cos\theta,x\sin\theta)$, $x>0$, $0<\theta<\frac{\pi}{2}$ gives
\begin{multline*}
 = \frac{\pi^{\frac{p+q}{2}}\Gamma(\frac{2\lambda-2\nu+2p'+p''+2q}{4})}{2^{q-2}\Gamma(\frac{p'}{2})\Gamma(\frac{p''}{2})\Gamma(\frac{2\lambda-2\nu+2p'+p''+4q}{4})} \int_0^\frac{\pi}{2} \cos^{p'-1}\theta \sin^{\lambda-\nu+\frac{p''}{2}-1}\theta\,d\theta\\
 \times \int_0^\infty x^{\lambda+\nu+\frac{p''}{2}-2q-1} (1+x^2)^{-\lambda-\frac{p}{2}} {_2F_1}\left(-\tfrac{2\nu-p'-2q}{4},\tfrac{q}{2};\tfrac{2\lambda-2\nu+2p'+p''+4q}{4};1-\tfrac{(1+x^2)^2}{x^4}\right)\,dx.
\end{multline*}
Evaluating the first integral using \eqref{eq:SineCosineIntegral} and using the Euler transformation formula \eqref{eq:EulerTransformationFormula} on the hypergeometric function we obtain
\begin{multline*}
 = \frac{\pi^{\frac{p+q}{2}}\Gamma(\frac{2\lambda-2\nu+2p'+p''+2q}{4})\Gamma(\frac{2\lambda-2\nu+p''}{4})}{2^{q-1}\Gamma(\frac{p''}{2})\Gamma(\frac{2\lambda-2\nu+2p'+p''+4q}{4})\Gamma(\frac{2\lambda-2\nu+2p'+p''}{4})} \int_0^\infty x^{-\lambda+\nu-p'-\frac{p''}{2}-2q-1}\\
 \times {_2F_1}\left(\tfrac{2\lambda+p+2q}{4},\tfrac{2\lambda-2\nu+2p'+p''+2q}{4};\tfrac{2\lambda-2\nu+2p'+p''+4q}{4};1-\tfrac{(1+x^2)^2}{x^4}\right)\,dx.
\end{multline*}
Substituting $y=\frac{1+x^2}{x^2}$ we obtain
\begin{multline*}
 = \frac{\pi^{\frac{p+q}{2}}\Gamma(\frac{2\lambda-2\nu+2p'+p''+2q}{4})\Gamma(\frac{2\lambda-2\nu+p''}{4})}{2^q\Gamma(\frac{p''}{2})\Gamma(\frac{2\lambda-2\nu+2p'+p''+4q}{4})\Gamma(\frac{2\lambda-2\nu+2p'+p''}{4})} \int_1^\infty (y-1)^{\frac{2\lambda-2\nu+2p'+p''+4q-4}{4}}\\
 \times {_2F_1}\left(\tfrac{2\lambda+p+2q}{4},\tfrac{2\lambda-2\nu+2p'+p''+2q}{4};\tfrac{2\lambda-2\nu+2p'+p''+4q}{4};1-y^2\right)\,dy.
\end{multline*}
Using the Euler integral representation \eqref{eq:EulerIntegralRepresentation} we further find
\begin{multline*}
 = \frac{\pi^{\frac{p+q}{2}}\Gamma(\frac{2\lambda-2\nu+p''}{4})}{2^q\Gamma(\frac{p''}{2})\Gamma(\frac{q}{2})\Gamma(\frac{2\lambda-2\nu+2p'+p''}{4})} \int_0^\infty t^{\frac{2\lambda-2\nu+2p'+p''+2q-4}{4}}(1+t)^{\frac{2\nu-p'-2q}{4}}\\
 \times\int_1^\infty (y-1)^{\frac{2\lambda-2\nu+2p'+p''+4q-4}{4}} (1+ty^2)^{-\frac{2\lambda+p+2q}{4}}\,dy\,dt
\end{multline*}
The inner integral can be calculated using \eqref{eq:IntRepresentation3F2}:
\begin{multline*}
 = \frac{\pi^{\frac{p+q}{2}}\Gamma(\frac{2\lambda-2\nu+p''}{4})\Gamma(\frac{2\lambda-2\nu+2p'+p''+4q}{4})\Gamma(\frac{2\lambda+2\nu+p''}{4})}{2^q\Gamma(\frac{p''}{2})\Gamma(\frac{q}{2})\Gamma(\frac{2\lambda-2\nu+2p'+p''}{4})\Gamma(\frac{2\lambda+p+2q}{2})} \int_0^\infty t^{-\frac{2\nu-p'+4}{4}}(1+t)^{\frac{2\nu-p'-2q}{4}}\\
 \times{_2F_1}\left(\tfrac{2\lambda+2\nu+p''}{8},\tfrac{2\lambda+2\nu+p''+4}{8};\tfrac{2\lambda+p+2q+2}{4};-t^{-1}\right)\,dt.
\end{multline*}
Substituting $x=t^{-1}$ we find
\begin{multline*}
 = \frac{\pi^{\frac{p+q}{2}}\Gamma(\frac{2\lambda-2\nu+p''}{4})\Gamma(\frac{2\lambda-2\nu+2p'+p''+4q}{4})\Gamma(\frac{2\lambda+2\nu+p''}{4})}{2^q\Gamma(\frac{p''}{2})\Gamma(\frac{q}{2})\Gamma(\frac{2\lambda-2\nu+2p'+p''}{4})\Gamma(\frac{2\lambda+p+2q}{2})} \int_0^\infty x^{\frac{q-2}{2}}(1+x)^{\frac{2\nu-p'-2q}{4}}\\
 \times{_2F_1}\left(\tfrac{2\lambda+2\nu+p''}{8},\tfrac{2\lambda+2\nu+p''+4}{8};\tfrac{2\lambda+p+2q+2}{4};-x\right)\,dx.
\end{multline*}
Finally the last integral can be calculated using \eqref{eq:IntegralHypergeometric}:
\begin{multline*}
 = \frac{\pi^{\frac{p+q}{2}}\Gamma(\frac{2\lambda-2\nu+p''}{4})\Gamma(\frac{2\lambda-2\nu+2p'+p''+4q}{4})\Gamma(\frac{2\lambda+2\nu+p''}{4})\Gamma(\frac{2\lambda-2\nu+2p'+p''}{8})}{2^q\Gamma(\frac{p''}{2})\Gamma(\frac{2\lambda-2\nu+2p'+p''}{4})\Gamma(\frac{2\lambda+p+2q}{2})\Gamma(\frac{2\lambda-2\nu+2p'+p''+4q}{8})}\\
 \times{_3F_2}\left(\tfrac{2\lambda+2\nu+p''}{8},\tfrac{2\lambda-2\nu+2p'+p''+4q}{8},\tfrac{q}{2};\tfrac{2\lambda+p+2q+2}{4},\tfrac{2\lambda-2\nu+2p'+p''+4q}{8};1\right).
\end{multline*}
Simplifying the hypergeometric function ${_3F_2}$ to ${_2F_1}$, evaluating it with \eqref{eq:SpecialValueHypergeometric} and simplifying the whole expression using the duplication formula $\Gamma(z)\Gamma(z+\frac{1}{2})=2^{1-2z}\sqrt{\pi}\Gamma(2z)$ we get
\begin{equation*}
 = \frac{\pi^{\frac{p+q}{2}}\Gamma(\frac{2\lambda-2\nu+p''}{4})\Gamma(\frac{2\lambda+2\nu+p''}{4})\Gamma(\frac{2\lambda+p+2q+2}{4})}{\Gamma(\frac{p''}{2})\Gamma(\frac{2\lambda+p+2q}{2})\Gamma(\frac{2\lambda+p+2}{4})}
\end{equation*}
which is the desired identity.

\section{Estimating the proportionality constants}\label{estimate}

In this section we estimate the size of the scalar $b_{f,g}$ relating the automorphic and the model invariant bilinear form. The technique we use is due to Bernstein--Reznikov~\cite{BR04} and was applied in two other situations in \cite{MO13,MS17}. We therefore omit the details and only point out, at which steps one has to modify the arguments slightly.

\subsection{Hermitian forms and construction of test functions}

For a Maass form $f\in L^2(X)$ with Langlands parameter $\lambda\in i\R$ let
$$ H^\aut_f(\phi) = \int_{\Gamma_H\backslash H} |\phi(h)|^2\,dh, \qquad \phi\in V_f, $$
then $H^\aut_f$ is an $H$-invariant Hermitian form on $V_f$. As explained in\cite{BR04,MO13,MS17}, for an orthonormal sequence $\{g_j\}_j\subseteq L^2(Y)$ of Maass forms on $Y$ with Langlands parameters $\nu_j\in i\R$ the following inequality holds:
\begin{equation}
 \sum_j |b_{f,g_j}|^2 H^\mod_{\lambda,\nu_j}(u) \leq H^\aut(\theta_u) \qquad \forall\,u\in I(\lambda),\label{eq:InequalityHermitianForms}
\end{equation}
where the $H$-invariant Hermitian form $H^\mod_{\lambda,\nu}$ on $I(\lambda)$ is given by
$$ H^\mod_{\lambda,\nu}(u) = \|A^\mod_{\lambda,-\nu}u\|_{L^2(\F^{m-1}\oplus\Im\F)}^2, \qquad u\in I(\lambda). $$

\begin{lem}\label{lem}
There exist $C_1,C_2>0$ and $\xi\in C_c^\infty(G)$, $\xi\geq0$, such that for $T\gg0$ and all Maass forms $f\in L^2(X)$ and $g\in L^2(Y)$ with Langlands parameters $\lambda,\nu\in i\R$, $|\lambda|,|\nu|\leq T$, there exists a test function $u_T\in C_c^\infty(\F^{n-1}\oplus\Im\F)\subseteq I(\lambda)$ of $L^2$-norm one such that
\begin{enumerate}[(1)]
\item $\int_G H^\mod_{\lambda}\def\La{\Lambda,\nu}(\pi_\lambda(k)u_T)\xi(k)\,dk \geq C_1\cdot T^{-2\rho}$,
\item $\int_G H^\aut_f(\theta_{\pi_\lambda(k)u_T})\xi(k)\,dk \leq C_2$.
\end{enumerate}
\end{lem}

\begin{proof}
The same statement in the context of $(G,H)=(\textup{GL}}\def\PGL{\textup{PGL}(3,\R),\textup{GL}}\def\PGL{\textup{PGL}(2,\R))$ is proven in \cite[Lemma 4.1]{MS17}, and we only provide the necessary details to translate the proof to our setting. First, we note that $K_{\lambda,\nu}((x,z),(0,0))=1$ for $(x,z)=(e_{n-1},0)\in\F^{n-1}\oplus\Im\F$. Let $u\in C_c^\infty(\F^{n-1}\oplus\Im\F)$, $u\geq0$, with support in a sufficiently small neighborhood around $(0,0)$, normalized to have $L^2$-norm one. Then
$$ u_T(x,z) = T^{\frac{p+2q}{2}}u(T(x-e_{n-1}),T^2z), \qquad (x,z)\in\F^{n-1}\oplus\Im\F, $$
defines a function $u_T\in C_c^\infty(\F^{n-1}\oplus\Im\F)\subseteq I(\lambda)$ of $L^2$-norm one with support around $(e_{n-1},0)$. As in \cite[Lemma 4.1]{MS17} one shows that $(\pi_\lambda(k)K_{\lambda,\nu})((x,z),(0,0))\sim1$ whenever $(x,z)\in\supp u_T$, $|\lambda|,|\nu|\leq T$ and $k$ is contained in some identity neighborhood $U\subseteq G$. This implies
$$ A_{\lambda,-\nu}\pi(k)u_T(0,0) \sim \int_{\F^{n-1}\oplus\Im\F} u_T(x,z)\,d(x,z) = \const\cdot T^{-\rho}  $$
for $k\in U$, $|\lambda|,|\nu|\leq T$ and $T\gg0$. Then for any left-$K_H$-invariant function $\xi\in C_c^\infty$, $\xi\geq0$, with $\xi|_U\equiv1$ we obtain
$$ \int_G H^\mod_{\lambda,\nu}(\pi_\lambda(k)u_T)\xi(k)\,dk = \int_G |A_{\lambda,-\nu}^\mod\pi_\lambda(k)u_T(0,0)|^2\xi(k)\,dk \geq C_1\cdot T^{-2\rho}, $$
which shows (1). For (2) the argument is the same as in \cite[Lemma 4.1]{MS17}.
\end{proof}

\begin{prop}\label{thm-1}
There exists a constant $C_3>0$ such that
$$ \sum_{|\nu_j|\leq T}|b_{f,g_j}|^2 \leq C_3\cdot T^{2\rho} \qquad \forall\,f\text{ with }|\lambda|\leq T. $$
\end{prop}

\begin{proof}
Apply Lemma~\ref{lem} to the inequality \eqref{eq:InequalityHermitianForms}, then the claim follows (see \cite[Proposition 4.2]{MS17} for details).
\end{proof}

\subsection{Proof of the main results}

Theorem~\ref{thm:Main} now follows from \eqref{eq:Proportionality} together with Corollary~\ref{cor:AsymptoticsModelPeriod} and Proposition~\ref{thm-1} after replacing the Langlands parameters $\lambda$ and $\nu$ by the eigenvalues $a=\rho^2-\lambda^2$ and $b=\rho_H^2-\nu^2$. Corollary~\ref{cor:Main} easily follows from Theorem~\ref{thm:Main} by fixing one of the Maass forms.

redpajama_set_name: RedPajamaArXiv


Challenge # 186 continues on the House Mouse and Friends Monday Challenge blog. Pauline, our talented teammate has chosen the theme of "MAKE IT MASCULINE".
For more inspiration and challenges, check out their official Blog HERE.
your blog and see what you come up with.
Fun Dad card, Sue. Love that shade of blue. Your scallops add so much to the look of your card. A winner in a Dad's eyes for sure. We are in the path of the big Blizzard today. I didn't notice if you were or not, but please be careful if you are. Keeya will have a ball for herself. Mine two won't even look out the door. LOL Enjoy your week.
Super Dad's card Sue !!
A wonderful card Sue, such a beautifully coloured image and pretty blues - a fabulous male card.
Wonderful Sue (can't say beautiful for a male card...can I??)!!! Your clouds are so perfectly done...I so struggle with them...you've created a beautiful day for these sweet little bunnies (also perfectly coloured by the way)to be taking a stroll!! Hope that you don't get hit with the storm...we are having some of the effects but more so along the lake...it's actually sunny where I am!!
Another fab card. I love the bright blue.
This is gorgeous Sue, I love the cute image, beautiful colours and wonderful detailing too.
Absolutely gorgeous and I love, love, love that brilliant blue color!
A gorgeous card Sue, I love your stunning colour combo and great design......brilliant colouring of cute little Hoppers!!
Love all of the blue on your manly mousey card, Sue. Hope if you are expecting snow, that it fizzled much like ours.
Hi Sue awwwwww love this sweet image gorgeous blues very striking design lovely work.
BSS MONDAY IDEA - IT DOESN'T HAVE TO BE RED!


text:
4 – Alan Bibey & Grasstowne
Posted on November 3, 2015 December 13, 2016 by John Lawless
What's up with Grasstowne? That's a question I've heard more than a few times this past few years.
Founder and mandolinist Alan Bibey has seen a lot of changes with this band in the nine years since they formed as a partnership between he, Steve Gulley, and Phil Leadbetter. Bibey's partners peeled off over the years to follow their own solo efforts, but Al has consistently found new members to keep his vision of a contemporary/traditional bluegrass outfit in clear view.
In fact, about three years ago he had such a strong group together that all the guys made a pact to stick together for at least five years to see how far they could go as a team. But you know what they say about best laid plans. Soon fiddler Adam Haynes was gone to The Grascals, singer Blake Johnson moved to IIIrd Tyme Out, and bass man Kameron Keller was with Junior Sisk. Suddenly it was just Alan and ace banjo picker Justin Jenkins.
But through all this time, Grasstowne had been recording, and the results can be seen in Grasstowne 4, undoubtedly the finest album they've ever made under this name. The tracks on this new Mountain Fever release cover several years of the band's history, with contributions from former members Haynes, Keller, and Johnson, plus now-departed vocalists Dustin Pyrtle and Shannon Slaughter, and guests Ronnie Bowman, Dwight McCall, and Ron Stewart.
Alan Bibey has now put his name on the band, in both the literal and figurative senses of the word. Rebranded now as Alan Bibey & Grasstowne, Al tackles the bulk of the lead singing on this project, and his voice comes across as strong and mature, something he says he has worked on consistently this past few years. He's also a primary songwriter here, with four of the twelve songs being his co-writes. I guess it really is his band.
Bibey's voice was always far better than serviceable, but the bump in his confidence is on display throughout Grasstowne 4. Right from the opening track, an Elmer Burchette number with a nasty groove called Cry Baby Cry, Alan delivers the mean right on cue. He's just as strong on more traditional fare, taking the lead on You Can Feel It Your Soul, which Flatt & Scruggs had cut in the early '60s. Justin Jenkins lays down a solid Scruggs-style guitar on that one as well.
I'm Country reunites Alan with a song he had recorded in his first high-profile gig with New Quicksilver in 1985, one written by Wes Golding who had recorded it previously with the Shenandoah Cut Ups.
The first of Bibey's originals to appear is Look What I've Done To Me, a co-write with Ronnie Bowman. It's an upbeat song that takes a hard look at the mistakes the singer has made. Dwight McCall adds a fine tenor vocal. Cold Dark Ground, from Bibey, Bowman and Mark Collie, was released two years ago as a single and was prominent on our Bluegrass Today Weekly Airplay chart, ending up as the #20 track for all of 2014. Blake Johnson sings it here, a beautiful rendition in every way.
Alan adds swingy singing to his repertoire on You Let Me Down, yet another written with Bowman. This is a fun song where Bibey's mandolin also shines, as does Jenkins' Crowe-inflected banjo. Former guitarist Shannon Slaughter sings high lead on Can't Stop The Rain, a fast moving trio arrangement with another wicked mandolin break.
The album's other prior release is This Old Guitar and Me, sung by the inimitable Ronnie Bowman, as smooth a male voice as bluegrass has ever seen. Grasstowne cut this one back when Haynes and Pyrtle were still in the band. An easy-going, mid-tempo ballad, it also did quite well with bluegrass radio in 2013 when it hit.
Heaven comes next, a lovely Gospel waltz from north Alabama songwriter Stacey Peek. It's performed by Alan and Shannon and Heather Slaughter, with just mandolin and guitar accompaniment. The simplicity of the arrangement serves this heartfelt testimony just right.
The last two tracks may be my favorites on the record, both well familiar to those who have followed bluegrass for years. Worried Man Blues gets a super driving delivery from Jenks' opening break to the last note of the ending. This is straight bluegrass played by people who have it deep in their bones. Bibey sings lead with McCall again on tenor. If anyone can find a fault with this one, your ears are better than mine. It's textbook.
Drawing this to a close is a grassy version of Kris Kristofferson's Darby's Castle. Shannon takes the lead and entirely sells this distinctly tragic story of a man so obsessed by his vision to create something for the woman he loved, that he forgot to be present in her life. The band gives the song a reading reminiscent of the '70s New South, and again, is note perfect.
We can't finish without some mention of the picking on this album. Though most comment here has been about Bibey's strong vocal work, one could never forget that it was his stellar mandolin playing that first brought him to wide attention. Each track shows again why he is held in such high regard, with carefully constructed solos appropriate to every song. Banjo man Jenkins is also a standout, on the CD's lone instrumental, the old timer Old Hen, often known as Cluck Old Hen, and on the various vocal numbers in turn.
Special kudos to Ron Stewart, whose fiddle is as spot on as ever, but even more so for his rhythm and lead guitar which drives the band on nearly every track. We all know what a superb fiddler and banjo player Ron is from his work with Lynn Morris, J.D. Crowe, and The Boxcars, though many fewer recognize his expertise on every other bluegrass instrument as well. Flatpickers will find a reason to resent him on this album.
Grasstowne 4 is available now from digital resellers, from the band web site, and at any of their live shows. Radio programmers can download tracks now from Airplay Direct.
They will be celebrating the release this Friday night (11/6) at Nashville's Station Inn with a performance by the current edition of the band. Bibey will be on mandolin, with Gena Britt on banjo, Zak McLamb on bass, Courtney Rorer on fiddle, and Greg Luck on guitar (Justin Jenkins is on vacation). Look for them to pick and sing songs from the new CD, with some special guests expected.
Anyone who appreciates contemporary bluegrass played with precision, soul, and authority will want this one in their library.
Duets & Ballads – The Cadleys
Landmark – Jake Workman
Seven True Stories – The High & Wides
Home To Stay – South Carolina Broadcasters


Air India's turnaround plan may be heading into fresh turbulence with a section of its pilots not cooperating with the management. Tushar Srivastava reports.
Air India's turnaround plan may be heading into fresh turbulence with a section of its pilots not cooperating with the management.
While the government has in-principle agreed to an equity infusion of R30,000 crore into the ailing national carrier by 2020, it comes with several riders.
While the government will play its part by releasing the funds in stages, the airline also has to prove that it means business.
Improving on-time performance and cutting costs are some of the benchmarks that would have to be implemented.
What has irked aviation ministry officials is the latest directive issued by Indian Commercial Pilots Association (ICPA), the union of former Indian Airlines pilots, directing its members not to follow a particular flight procedure introduced to reduce fuel consumption.
"The airline had recently modified the acceleration altitude in the take-off segment for the Airbus fleet," an AI official said. This, the official said, is in line with international practice and is followed by many reputed carriers.
But ICPA refused to accept the procedure citing safety concerns.
"If other airlines worldwide find it safe to follow this procedure, we wonder why only some pilots of AI are finding it unsafe?" said an aviation ministry official.
The airline's management pilots who fly Airbus planes have already started following the new procedure that saves fuel and improves engine life. "This procedure has been adopted by many airlines worldwide including Lufthansa. ICPA is resisting it for no valid reason. If they fly a new-generation aircraft the old fashioned way, what is the point?" asked a senior management pilot.
"Our objections are safety related," said an ICPA pilot.