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Home > Blog > 2016 > October > An Interview with Professional Musher April Cox
Snow Driven: An Interview with Professional Musher April Cox
October 26, 2016 Timothy Dunn Lifestyle
All photos in this post courtesy of Facebook / Badas Musher - April Cox
If you're a fan of CKC mushing and follow any mushers on social media, there's a good chance that you've heard of BADAS Musher, a.k.a. April Cox.
Cox is a musher who's been enjoying the sport since the early 1990s. She entered her first official race in 2005 and never looked back. She's shown considerable skill, competing in a number of races and winning the top spot in the 2013 Chester Winterfest Sled Dog Races, while also earning second place in the 2013 Chemult, OR, Sled Dog Races and the 2013 Siskiyou Sled Dog Races.
Oh, and the name? According to the BADAS Musher herself, it actually goes back to a joke from her college days. "I was actually in college in my thirties taking business administration, and I found a sticker that said 'I have a BA in BS,' and I kind of played on that," she says. "So, it actually stands for Business Administration Degree Associate of Science."
Of course, mushing doesn't come easy for anyone, and it's taken a lot of patience, persistence, and hard work for Cox reach the level that she has. If a race is coming up, you better believe April Cox will be outside with her dogs, training for the next challenge.
"I go out, grab my harnesses, and throw them on the quad, [and then] pull out into the dog yard," Cox says, describing a typical training day. "I pull a quad into the dog yard, string out my gang line, then pull up at a drop chain stretched across the yard—like a staging area. I get each dog that I'm going to run, put them on the drop chain in order of how I'm going to hook them up. Then, go through down the line, harness each dog.
"[I] put any booties on if they need the booties, then hook them up to the gang line, open the gate, go get on the quad, and off I go! I make it sound so calm, but it's pretty nuts. Not that calm at all. Sometimes it can be a little chaotic."
Over the years, Cox learned how to handle the high-energy chaos of a team of dogs excited about doing what they do best, and she acquired that knowledge through first-hand experience. Cox has a better understanding of her dogs now, and they have come to know her better as well.
When she entered her first race at Shaver Lake, California, in 2005, both Cox and her dogs were excited, but also unsure of what to expect. She was new to training, so her dogs weren't as prepared as they could have been for the two-and-a-half-mile race. Nevertheless, Cox and her three-dog team showed up ready to go.
She remembers the dogs showing an odd calmness for their first time in the chute. They didn't stir much while waiting for the race to begin. In fact, Cox was worried the dogs looked so calm in the chute that they might not run when the time came.
"It was in Shaver Lake, California, and we're waiting in line, and I have helpers leading the dogs up to the chute," Cox recalls. "And, these helpers had been helping all these other teams who had raced before, so they're all banging their harnesses ready to go, and mine are standing all calm. And, someone made a comment about how calm my dogs were. And I said, 'Oh, yes. They're fine.' And he said, 'Well, wait until they get into the chute.' And I said, 'Well, they don't know what a chute is, they've never raced before either.'
"So, we get in the chute, and they're doing their countdown, because most races start in two-minute intervals. And I remember standing there, and people are lined on both sides of the chute, and I'm wondering, 'God, I hope they're gonna run,' you know, because they don't seem very excited, and you know it'd be kind of embarrassing if they say 'Go!' and we don't.
"Well, [I had] a family—a mom, her son, and her daughter that I had racing—and they were all mostly [piebald] white huskies. And I had Trixie in the lead, and I had Aladar and Suri in the back. And, I had just done weight pulling with Aladar at dog shows before that. And I said, 'Aladar, are you ready?' And he kind of wagged his tail. And I'm like, 'Trixie, are you ready?' And then I said, 'Well, I think they'll go . . .' because they watched the other dogs go.
"And they did. They took off pretty fast, so it was kind of exciting. But, needless to say, I came in dead last, because coming up the hill without the proper training and conditioning . . . it was just a lot. And it was pretty warm. It was like 45 [degrees], which is pretty warm for husky standards. And I was way over-dressed. I was dressed in my snowmobile suit that I had had from when I lived in Idaho, which is actually way too warm for California weather."
So, even though Cox didn't finish in one of the top spots for her first race, she learned a lot about what it would take to go further, faster. But, while that race was certainly difficult, one of the most challenging tests for Cox and her team would come from the 2016 Eagle Cap Extreme Dogsledding Race in Oregon, which runs through the Wallowa Mountains.
The weather proved to be one of the biggest obstacles in the Eagle Cap race. Cox and her team nearly lost their way when heavy snow drifts covered a section of the path. After it disappeared under the snow, Cox unintentionally veered off of the track for an unexpected three-foot drop. She later glimpsed a red sign through the heavy snowfall, which informed Cox that at some point the storm had turned her team around and led them off course.
Following a short rest for herself and her dogs, Cox miraculously managed to get her team back on track, despite the inclement weather. After she passed the affected area, officials caught up with Cox to tell her that the weather had become so harsh in that section of the race that it could be skipped altogether, but since she had already made her way through, she could continue on with the race. With the worst already behind them, the team pushed on and finished the race, showing off the skill and determination that led the Continental Kennel Club to become a sponsor of Cox and her team of hard-working dogs.
If you've read about April Cox's experiences as a musher and think you may have what it takes to get into the sport with your own team, the BADAS Musher offers some sage advice for newcomers:
"The best thing to do is to find a mentor, and there are mushers all over the country that do it recreationally or [for serious racing]. If you really want to do it, it's a lot of hard work.
"But, yeah. Find a mentor, hang out with them, and go on training runs. Everybody always appreciates help and an extra hand, and we love addicting people to the sport."
So, you've been warned. Once you get into mushing, there's a good chance you may be hooked for life. But, after hearing April Cox's exciting stories about the challenges she's faced and the bonding she's experienced with her team of dogs out on the trail, a mushing addiction doesn't sound like such a bad thing after all.
Tags: interviews exercise/activities
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Dogs instinctively curl up when they sleep because of their days living in the wild. Curling helps them keep their bodies warm while asleep, and it also protects their vital organs from would-be attackers.


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Q: Node js cors proxy results in forbidden I'm trying to communicate with an api with react. I was getting a cors error so i created a proxy api. I'm still getting a 403 error but this time it's different. I get a captcha error. I get data when i open the url in browser and also when i make a request in postman, but not through node js
const express = require('express');
const cors = require("cors");
const axios = require('axios');
const app = express();
const port = process.env.PORT || 8000;

app.use(cors({
  origin: '*'
}));

app.get("/", (req, res) => {
  const url = `URL_TO_EXTERNAL_API`;
  axios.get(url)
  .catch(error => console.log(error))
  .then(resp => {
    console.log(resp);
  });
});

app.listen(port, () => console.log(`http://localhost:${port}`));

here's a portion of the error i get
Here's the react request ->
fetch(local, {
  method: 'get',
  headers: new Headers({
    'Content-Type': 'application/json'
  })
}).then(response => {
  return response.json();;
}).then(data => {
  this.setState({ announcements: data })
});

where local is "http://localhost:8000?params..."
Do you know why the current solution is not working ?

A: The website or service behind URL_TO_EXTERNAL_API has some kind of captcha check to make sure that only humans can view the website. So basically you can't proxy that service, unless your proxy can bypass the captcha check.

redpajama_set_name: RedPajamaStackExchange


text:
Wellness Geeky
TOP Massage Chairs
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Home » Massage » Massage Chairs » OSIM uAstro2 Review TESTED [September 2018]
OSIM uAstro2 Review TESTED [September 2018]
December 26, 2019 By Jill Roberts
UPDATE: OSIM uAstro2 chair is no longer available. We recommend this chair instead
Massage has long been used to relieve stress, reduce muscle tension, and decrease pain.
Research also shows that it is effective in increasing circulation, improving mood, and promoting nervous system function.
If you want to reap these benefits on a regular basis, without the hassle and cost of booking sessions with a therapist, why not consider buying a massage chair?
OSIM – One of the leading producers of massage chairs
With inventive features such as voice controlled massage programs, music-massage synchronisation, and power-drum massager, OSIM have been on the cutting edge of technology since the company's inception in 1980. Unsurprisingly, it have won such accolades as Time Magazine's
"Invention of the Year," Forbes Asia's "Best 200 Under a Billion," and the Red Dot Design Award. In 2013, OSIM released their second generation of uAstro massage chairs. Its upgraded features and new aesthetic make it a popular product.
Let's take a closer look at some of the innovative features of the OSIM uAstro2
The Original uAstro Gets an Upgrade
OSIM have stepped it up a notch with the design of the uAstro2. Unlike its predecessor, the uAstro2 features footrest rollers, an upgrade to the neck and shoulder massage program, and (for the broader among us) a roomier seat-back. It also has the same great features of the original, including kneading massagers, preset programs, and built-in lumbar heat.
Total Body Relaxation
The OSIM uAstro2 aims to relax every muscle in your body and that's definitely what it does. Twelve preset programs with five levels of intensity ensure that your back, hips, legs, neck, and shoulders will get a thorough massage in just fifteen minutes.
It also includes built-in armrest and footrest massagers, so even your fingers and toes will get a time-out.
What's really special about the uAstro2 is the neck and shoulder massage which squeezes the muscles just like a masseuse. You can also choose from other targeted massages such as Back and Lumbar (Kneading) and Foot and Calf (Air), or automated programs like Stretch, Energize, and Relax.
State of the Art Technology: Full Body Scanning, Zero Gravity Recline, Color LCD Display
The uAstro2 utilizes the best in massage technology to melt away muscle tension, using optical pressure-point detection to customize each program to your body. The chair also features zero-gravity recline, which pivots the user to reduce pressure and evenly distribute weight throughout the body. Each session is completely customisable from the chair's remote, and you can monitor massage progress from the color LCD screen.
"Overall I like it very much and it feels like all my back pain has vanished and feel very flexible also. Highly recommended."
"The 2 has a MUCH improved neck massage and now has foot rollers. Also wider shoulder air bags for bigger guys. It's really a very good chair overall."
"I suffer from chronic pain and this chair is helping with pain immensely. I use it every day."
"My only complaint is I don't get to use it enough with everything (life) going on."
"Now onto the chair. WOW, describes it."
How much should you expect to pay for the OSIM uAstro2?
The uAstro2 retails at $3399 in stores and online. Brookstone also offer certified pre-owned massage chairs which have undergone repair and inspection. These are available at the lower price of $2899. The chair is currently only available in black.
With an average star rating of 4/5, reviews of the product itself are extremely positive. Some customers have found Brookstone's customer service less than satisfactory. However, there is always the option to buy the OSIM uAstro2 from other retailers.
Other users have commented that the chair takes up a lot of room – but they also thought the amazing results were worth the extra space!
Where can you read more"
A multitude of reviews are available online. Amazon.com provides detailed and unbiased reviews from product users around the world.
Where can you buy?
The OSIM uAstro2 is available online (with free delivery) at Amazon.com, and from OSIM distributers throughout the US, Europe, and Asia.
About Jill Roberts
Jill Roberts
Creator & Senior Editor at Wellness Geeky
Hi, my name is Jill Roberts, Owner Creator & Senior Editor at Wellness Geeky
My team and I are dedicated to provide you with reliable & uptodate Health & Wellness information.
My email is: [email protected]
I hold Master's Degree in Molecular Biology and PHD in Epidemiology and Clinical Research which makes me uniquely quality to provide you with well-researched information.
Find more About Us
Disclaimer: The information contained on this site is intended for educational purposes only and is not a substitute for advice, diagnosis, or treatment by a licensed physician. It is not meant to cover all possible precautions, drug interactions, circumstances, or adverse effects. You should seek prompt medical care for any health issues and consult your doctor before using alternative medicine or making a change to your regimen.
Jill Roberts Creator & Senior Editor at Wellness Geeky Hi, my name is Jill Roberts, Owner Creator & Senior Editor at Wellness Geeky My team and I are dedicated to provide you with reliable & uptodate Health & Wellness information. My...
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redpajama_set_name: RedPajamaCommonCrawl


2Electives for Mathematics majors: Major electives are selected from Mathematics courses numbered above MATH 300; also PHYS 350 , PHYS 351 or CMPS 364 . Students must select as an elective at least one course from each of the following sets: (1) Applied Set: MATH 310 , MATH 320 , MATH 341 , MATH 360 , MATH 361 , MATH 368 , MATH 371 ; (2) Analysis Set: MATH 446 , MATH 460 , MATH 461 ; (3) Algebra Set: MATH 346 , MATH 448 . Students are encouraged to take their applied math elective in the second year.
3Students must complete either 30 credits in a separate discipline or 15 credits in two distinct majors from different departments. These courses must count towards a major or minor in the respective discipline.
4These EPW electives might also satisfy other curricular requirement(s). When an EPW elective satisfies other curricular requirements(s), an additional free elective will be required.
Mathematics majors are required to take the Math Major Field Test in order to graduate.


text:
Q: how to create a derived table from two other tables? I have 2 base tables TABLE_1 and TABLE_2. i want to derive a new table consisting ID of both table.
I NEED SQL QUERY TO CREATE THE THIRD TABLE CALLED NEW_TABLE in which ID_1 and ID_2 combines to form primary key. and if any of the ID in TABLE_1 or TABLE_2 is updated or deleted it should be reflected in NEW_TABLE also.
eg:
TABLE_1          TABLE_2
ID    XXX        ID     ZZZ
100   A          200     P
101   B          201     Q
102   N          202     R
103   F          203     S

NEW_TABLE
ID_1   ID_2
100    200
100    201
100    203
101    200
101    202
103    200
103    201
103    203

PLEASE HELP ME......thanks in advance..:-)

A: You can use SELECT INTO to select from some table(s) and insert the data into a new table.

A: you can use views for above case

A: Best way would be to use a view

A: Why dont you use just a query instead of table
SELECT
TABLE_1.ID,TABLE_2.ID
FROM TABLE_1,TABLE_2


Advancing awareness and understanding of the role of maritime affairs in the history of the United States. Through leadership, assistance, and expertise in maritime history, preservation, and archaeology, the Maritime Heritage Program helps to interpret and preserve our maritime heritage.
Vacationing, dining, shopping, transportation and entertainment in Massachusetts.
Links to sites that participate in the Massachusetts Visitor Information Network Service program.
Protecting, promoting and enhancing Massachusetts' wealth of natural, cultural and recreational resources for the well-being of all.
Statewide nonprofit organization dedicated to preserving the Commonwealth's historic and cultural heritage.
For more than two centuries the Massachusetts Historical Society has been collecting and preserving materials relating to the history of the commonwealth and Nation. The holdings of the Massachusetts Historical Society encompass millions of rare and unique documents and artifacts vital to the study of American history.
Encouraging the historic preservation and restoration of lighthouse, lightship and lifesaving station artifacts and documents across the Nation.
A non-profit historical and educational organization incorporated to educate, inform, and entertain those who are interested in America's lighthouses, past and present.
Founded at the Cape Cod National Seashore in 1995 to preserve the stations, history, boats and equipment of the US Life-Saving Service and US Coast Guard.
Links to maritime-related museums, historic sites and historic vessels in Massachusetts including, the Essex Shipbuilding Museum, the Gloucester Maritime Heritage Center and the Peabody Essex Museum.
Sits astride the historic shipping routes and fishing grounds for numerous ports around Massachusetts.
Boston Harbor Islands National Recreation Area, Cape Cod National Seashore, Salem Maritime National Historic Site, Boston National Historical Park, New Bedford Whaling National Historical Park, Essex National Heritage Area protect and preserve our maritime history.
The HABS/HAER program documents important architectural, engineering and industrial sites throughout the United States and its territories. Its collections, including a number of Massachusetts sites, are archived at the Library of Congress and available online. You can view these by clicking on the link above and entering the search term "Massachusetts."
This website, maintained by the U.S. Department of Transportation, Federal Highway Administration, includes information on state and nationally designated byway routes throughout America based on their archeological, cultural, historic, natural, recreational, and scenic qualities. Visit the America's Byways Connecticut River Byway website for more ideas.
Beattie, Mary Elizabeth. Obligation and Opportunity: Single Maritime Women in Boston, 1870-1930. Ithaca, New York: McGill-Queen's University Press, 2000.
Bither, Barbara A. and Boston National Historical Park. Charlestown Navy Yard. Images of America. Charleston, South Carolina: Arcadia Publishing, 1999.
Bourne, Russell. The View From Front Street: Travels Through New England's Historic Fishing Communities. New York: W.W. Norton & Company, 1989.
Cheney, Robert K. and Roland H. Woodwell. Maritime History on the Merrimac: Shipbuilding. Newburyport, Massachusetts: Newburyport Press, 1964.
Clafin, James. Lighthouses and Life Saving Along the Massachusetts Coast. Images of America. Charleston, South Carolina: Arcadia Publishing, 1998.
Clifford, J. Candace and Mary Louise Clifford. Nineteenth-Century Lights: Historic Images of American Lighthouses. Alexandria, Virginia: Cypress Communications, 2000.
Connolly, James B. The Port of Gloucester. New York: Doubleday, Doran & Company, Inc., 1940.
Fischer, Lewis R. The Market for Seamen in the Age of Sail. St. John's, Nfld.: International Maritime Economic History Association, 1994.
Garland, Joseph E. Down to the Sea: The Fishing Schooners of Gloucester. Boston: David R. Godine, Publisher, Inc., 1983.
German, Andrew W. Down on T Wharf: the Boston Fisheries as seen through the Photographs of Henry D. Fisher. Mystic, Connecticut: Mystic Seaport Museum, 1982.
Harrison, Tim and Ray Jones. The Golden Age of American Lighthouses: A Nostalgic Look at the U.S. Lights from 1850-1939. Guilford, Connecticut: The Globe Pequot Press, 2002.
Heyrman, Christine Leigh. Commerce and Culture: The Maritime Communities of Colonial Massachusetts, 1690-1750. New York: Norton, 1984.
Johnston, Paul Forsythe. The New England Fisheries: A Treasure Greater than Gold. Salem, Massachusetts: Peabody Museum of Salem, 1984.
Kimball, John R. H. Disasters &c: The Maritime World of Marblehead 1815-1865. Portsmouth, New Hampshire: Peter E. Randall, 2005.
Kraus, Theresa L. The Department of the Navy. Know Your Government. New York: Chelsea House Publishers, 1989.
Morison, Samuel Eliot. The Maritime History of Massachusetts, 1783-1860. Boston: Houghton Mifflin Company, 1941.
National Park Service. Salem: Maritime Salem in the Age of Sail. Washington, D.C.: U.S. Department of the Interior, 1987.
----------------------. Bowditch's Salem: A Walking Tour of the Great Age of Sail. Washington, D.C.: Department of the Interior, 2003.
Peckham, Courtney Ellis. Essex Shipbuilding. Charleston, South Carolina: Arcadia Publishing, 2002.
Shanks, Ralph and Wick York. The U.S. Life-Saving Service: Heroes, Rescues and Architecture of the Early Coast Guard. Petaluma: Costano Books, 1996; reprint, Petaluma: Costano Books, 2000.
Snow, Edward Rowe. The Lighthouses of New England, updated by Jeremy D'Entremont. Beverly, Massachusetts: Commonwealth Editions, 2002.
Story, Dana A. Growing up in a Shipyard: Reminiscences of Shipbuilding in Essex, Massachusetts. Mystic, Connecticut: Mystic Seaport Museum, 1991.
---------------. The Shipbuilders of Essex: A Chronicle of Yankee Endeavor. Gloucester, Massachusetts: Ten Pound Island Book Company, 1995.
Carse, Robert. The Young Mariners: A History of Maritime Salem. New York: W.W. Norton, 1966.
Knight, James E. and David Wenzel. Salem Days: Life in a Colonial Seaport. Mahwah, New Jersey: Troll Associates, 1982.


Computers can be upgraded in a variety of ways, making them fit for new tasks. One of the simplest variants is the expansion of the memory, which is also pleasant with the Micron MT8JTF25664AZ-1G4D1 2 GB DDR3 module.
Computers can be upgraded in a variety of ways, making them fit for new tasks. One of the simplest variants is the expansion of the memory, which is also pleasant with the Micron MT8JTF25664AZ-1G4D1 2 GB DDR3 module. Take advantage of this feature and increase the performance of your desktop PC.
The Micron MT8JTF25664AZ-1G4D1 module has the largest share of the 2 GB DDR3-SDRAM, but at the same time it is an important compatability feature. Your PC must support the storage technology in any case, which is why you should take a look at the manual or data sheet before purchase. There is often also the maximum possible equipment specified.
The Micron MT8JTF25664AZ-1G4D1 2 GB DDR3 RAM is the maximum memory clock frequency at 1333 MHz, which is often referred to as PC-10600. Furthermore, it is a memory without an ECC function, which is why it is intended in particular for conventional computers. RAM modules for servers and workstations are of course also available in the assortment.
To complete the upgrade of your PC, you must install the Micron MT8JTF25664AZ-1G4D1 2 GB DDR3 DIMM only in the corresponding slot on the main board. In most cases, it is sufficient to just open the housing for a short time to access the slots for the RAM memory. All other configurations will be carried out automatically at the next start.


text:
namespace Esthar.Data.Transform
{
    public sealed class AsmHardlinkBinding : AsmBinding
    {
        public readonly AsmSegment TargetSegment;

        public AsmHardlinkBinding(AsmSegment source, AsmSegment target)
            : base(AsmBindingType.Hardlink, source, source.Length - 1)
        {
            TargetSegment = target;
        }

        public override AsmSegment[] ResolveTargets()
        {
            return new[] { TargetSegment };
        }

        public override bool? ResolveCondition()
        {
            return true;
        }
    }
}

redpajama_set_name: RedPajamaGithub


As a Professional Organizer, we try to help you create new systems so your life moves along more smoothly and efficiently. We want you to have more time to do what you want to do and that means getting rid of your excess items that no longer serve your life (papers, clothes, makeup, toys, books, kitchen items, memorabilia, etc, etc.) And organizing what you do Need, Use, and Love. Time management can mean not being so involved in as many activities as you say 'yes' to. Time management can mean not feeling you/your home has to be 'perfect' all the time. Time management can mean having more time and energy to maintain your space so you can enjoy all the other things you want out of life.
Having clutter bogs you down and makes you have less time because you're always deliberating about the overwhelming projects you never have time to do. Please call for a de-clutter session. I can help.


Homelessness among women and children is growing. With your support, our Center for Women & Children Expansion will help meet the growing demand for shelter and services — that's hundreds more women & children helped each day!
See how many more women & children our center expansion will help!


text:
JOHN CAGE WITH DAVID TUDOR – VARIATIONS IV
Sello: Modern Harmonic
It could be argued that there is no more controversial figure in music history as avant-garde electronic composer John Cage. Perhaps best known for his composition "4'33,"" which consisted of Cage sitting at a piano for four-plus minutes of total silence, Cage was both loved and loathed in the 60s and 70s as a leading light in avant-garde music and as an entertainingly weird guy who used radios, televisions, live dancers and his own Adam's apple as instruments in his live performances. Cage's music blurred the line between music, performance art and visual art in a way that no other composer has before or since.
Excerpts—7PM To 8PM
Excerpts—9PM To 10PM
Excerpts—10PM To 11PM
Disponible: Agotado
Categories: ESENCIALES, VINILOS
ALGODÓN NATURAL
Ramo incluye 10 cabezas de algodón.
PRINT CURVAS
Dimensiones: 42 cm Ancho / 60 cm Alto
Print: Papel bond 120 mg, impresión offset
Impreso en Perú
JUANECO Y SU COMBO – EL GRAN CACIQUE
OFFICIAL REISSUE OF FLEETWOOD MAC'S RUMOURS pressed up on virgin audiophile vinyl. Not the rarest record by any means (it is after all the second biggest-selling record of all time), but still heavily sweated amongst djs, collectors, and music lovers alike for good reason. This is the one with all hits – "Dreams," "Don't Stop," "Go Your Own Way," "You Make Loving Fun," "The Chain," etc. Stevie Nicks and company were at the top of their game with this one making it the top album of 1977 (catching a Grammy for best album and Diamond award from R.I.A.A.). Classic and essential.
Second Hand News
Never Going Back Again
Don't Stop
You Make Loving Fun
I Don't Want To Know
Oh Daddy
Gold Dust Woman
SUN RA – THUNDER OF THE GODS SMOKY RED
A new Sun Ra LP touches ground in the earthly bins, 50 years after its creation! Sun Ra is still trying to get our attention 50 years after dispatching this transmission. Humanitys path since then makes his message even more urgent today. Years after Herman Poole LeSonyr Ra Blount left the planet hes still trying to reach us, to wake us up and to change our destiny. Sun Ra and the Arkestra werent a traditional studio band, and every star in the vast galaxy of their discography reflects this. The origins of these records can be hard to pinpoint at times, but when it comes to Thunder Of The Gods, its a bit easier. Calling Planet Earth We'll Wait For You was discovered on tapes from Ras, Universe In Blue, believed to be recorded in 71. The raucous title track and Moonshots Across the Sky are unearthed from the 66 Strange Strings sessions. Modern Harmonic has once again paired Ras sonic art with the visual art of The Father of Modern Space Art, Chesley Bonestell, whose 1952 work Formation of the Earths Continents spralls across the front and back covers.
Calling Planet Earth- We'll Wait For You
Moonshots Across The Sky (Mono)
Thunder Of The Gods
DAVID BOWIE – AKA SPACE ODDITY
Unwashed And Somewhat Slightly Dazed
Letter To Hermione
Cygnet Committee
An Occasional Dream
The Wild Eyed Boy From Freecloud
God Knows I'm Good
Memory Of A Free Festival


Level 1 – Classes are suitable for students with no little or no previous experience. Classes include slow paced instruction with a focus on proper alignment. Designed to develop a basic foundation and understanding of yoga postures, and to build baseline strength while gently increasing flexibility and improving circulation.
Level 2 – Classes are suitable for students with previous yoga experience who feel comfortable moving through basic yoga postures and sun salutations.
Level 3 – Classes are suitable for those with an moderate to strong personal yoga practice and fitness level.
In hot vinyasa style yoga, students fluidly move from one pose to the next while connecting their breathing to their movements. This type of yoga is taught in a heated studio (85-90 degrees) which makes it a vigorous and revitalizing form of physical fitness. It incorporates strength, flexibility, balance, and physical and mental stamina in one session for the perfect blend of sweat and serenity!
Without the heat, this vinyasa style class provides students the space to deepen their yoga practice as they are guided through a carefully planned set of asanas designed to promote strength, focus, endurance, and tuning in to one's mind, body, breath connection. This is a vigorous class featuring advanced postures and deep stretching. Suitable for students with previous yoga experience.
Restorative Yoga is a slower-paced, relaxing version of the popular physical and spiritual discipline of yoga. This class serves to relax the body and mind. We begin with a gentle warm-up and move into seated and reclined poses, using props for optimal support. The intention of this class is to improve circulation in the joints, revitalize tissues, improve range of motion, and relax!
In gentle yoga flow we move slowly through asanas (poses) with the intention of evenly building strength and flexibility in the body. This class involves comprehensive queuing and one on one alignment coaching, making this a great class for beginners! Regular practice develops core strength, balance, flexibility, and increased mobility. Set to soothing traditional yoga music and practiced in a dimly lit room to promote relaxation.
Buti is an Indian Marathi word that means "a cure that has been hidden or kept secret." We believe EVERY woman has the power to transform and heal her body from within. Buti Yoga focuses on activating the Spiral Structure of the core to engage the deep abdominal muscles, stimulate the endocrine system, and activate kundalini energy. Classes utilize ancient pranayama and core engagement techniques seamlessly woven into a power yoga flow with bursts of cardio-intensive tribal movement. Experience Twisted Sister Yoga Studio's unique set up and awesome sound system designed to "Rock the Mat" in these high-energy, transformative workouts set to rap, hip-hop, and tribal dance music! For more about Buti Yoga check out this video: What is Buti Yoga?
Based on the elements of ballet warm-up exercises, yoga, and HIIT. The movements in Barre FusionFlow consist of high energy, low impact exercises, combined with traditional vinyasa. Designed to develop muscle tone and strength, while providing maximum calorie burning benefits. Burn up to 300 calories each 45 minute class. Bring water and a towel. Efficient, effective, and fun!
Based on the Bikram Chouldry hot yoga style synthesized from traditional hatha yoga techniques and popularized beginning in the early 1970s. This 60 minute class consists of the same series of 26 postures, including two breathing exercises with the studio heated to 95 degrees.
This class integrates kundalini yoga kriyas and breathing techniques into tradtional vinyasa flow to stimulate movement in the subtle energy body or chakra system while improving strength, flexibility, balance, mental focus, increasing oxygen in the body and brain, and calming the central nervous system. Following the movement portion of the class, students are guided into meditation with the option of working with mudras and mantras, and encouraged to sustain the meditation for twenty minutes. This is a healing practice that meets students where they are physically, mentally, and emotionally, and is suitable for all levels.
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text:
\section{Introduction}
Let $V$ be an $n$-dimensional vector space $V$ (over a division ring).
Denote by ${\mathcal G}_{k}(V)$ the Grassmannian formed by $k$-dimensional subspaces of $V$.
Suppose that $1<k\le n-k$.
For every integer $i$ satisfying $1\le i\le k$ we define the following relation
$${\mathfrak R}_{i}:=\{(X,Y)\in{\mathcal G}_{k}(V)\times{\mathcal G}_{k}(V): \dim_{l}(X\cap Y)=k-i\}.$$
We write $\dim_{l}$ for the linear dimension (the dimension of vector spaces and their subspaces),
since we want to distinguish it from the projective dimension
(the dimension of projective spaces and their subspaces) which will be denoted by $\dim_{p}$.
Note that $(X,Y)\in {\mathfrak R}_{i}$ if and only if
the distance between $X$ and $Y$ in the {\it Grassmann graph}
$$\Gamma_{k}(V)=({\mathcal G}_{k}(V), {\mathfrak R}_{1})$$
is equal to $i$.

The same relations can be defined on dual polar spaces.
Let $\Pi$ be a polar space of rank $n$.
Denote by ${\mathcal G}_{k}(\Pi)$ the polar Grassmannian formed by
singular subspaces of $\Pi$ whose projective dimension is equal to $k$.
The associated dual polar space is formed by maximal singular subspaces,
i.e. singular subspaces of dimension $n-1$.
For any integer $i$ satisfying $1\le i\le n$ we define
$${\mathfrak R}_{i}:=\{(X,Y)\in{\mathcal G}_{n-1}(\Pi)\times{\mathcal G}_{n-1}(\Pi): \dim_{p}(X\cap Y)=n-1-i\}.$$
As above, we have $(X,Y)\in {\mathfrak R}_{i}$ if and only if
the distance between $X$ and $Y$ in the {\it dual polar graph}
$$\Gamma_{n-1}(\Pi)=({\mathcal G}_{n-1}(\Pi), {\mathfrak R}_{1})$$
is equal to $i$.

By \cite{Chow}, every automorphism of the Grassmann graph $\Gamma_{k}(V)$
is induced by a semilinear automorphism of $V$ or a semilinear isomorphism of $V$ to
the dual vector space $V^{*}$ (the second possibility can be realized only in the case when $n=2k$).
Similarly, every automorphism of the dual polar graph $\Gamma_{n-1}(\Pi)$
is induced by an automorphism of the polar space $\Pi$.
The latter was proved by Chow \cite{Chow} for classical polar spaces only,
but Chow's method works in the general case \cite[Section 4.6]{Pankov-book}.
Some results closely related to these statements were obtained
\cite{Havlicek1,Havlicek2,HavlicekPankov,Huang1,Huang2,Huang3,Kreuzer,Pankov1,Pankov2}
and we refer \cite{Pankov-book} for a survey.

Every bijective transformation of ${\mathcal G}_{k}(V)$ preserving ${\mathfrak R}_{k}$
is an automorphism of $\Gamma_{k}(V)$ \cite{BH,HavlicekPankov}.
For the relation ${\mathfrak R}_{i}$ with $1<i<k$ the same is not proved.
However, all bijective transformations of ${\mathcal G}_{k}(V)$ preserving
${\mathfrak R}_{1}\cup\dots\cup {\mathfrak R}_{m}$
are automorphisms of $\Gamma_{k}(V)$ for every integer $m<k$ \cite{Lim}.
This is a generalization of the previous result; indeed, if $m=k-1$ then
the considered above transformations preserve ${\mathfrak R}_{k}$.
The same statement is proved for some dual polar spaces \cite{Huang5}.
Results of similar nature were established for other objects \cite{AbHVM, HVM1, HVM2, HH, Huang4}.

Now suppose that $k$ is an integer not greater than $n-2$
and consider the relation ${\mathfrak R}_{i,j}$, $0\le i\le j\le k+1$
formed by all pairs
$$(X,Y)\in {\mathcal G}_{k}(\Pi)\times {\mathcal G}_{k}(\Pi)$$
satisfying the following conditions
$$\dim_{p}(X^{\perp}\cap Y)=k-i\;\mbox{ and }\;\dim_{p} (X\cap Y)=k-j$$
($X^{\perp}$ consists of all points of $\Pi$ collinear to every point of $X$).
All automorphisms of the {\it polar Grassmann graph}
$$\Gamma_{k}(\Pi)=({\mathcal G}_{k}(\Pi), {\mathfrak R}_{0,1})$$
are described in \cite[Section 4.6]{Pankov-book}.
They are induced by automorphisms of $\Pi$,  except the case when $n=4$, $k=1$
and our polar space is of type $\textsf{D}$
(every $(n-2)$-dimensional singular subspace is contained in precisely two maximal singular subspaces).
Also, every automorphism of so-called {\it weak Grassmann graph}
$$\Gamma^{w}_{k}(\Pi)=({\mathcal G}_{k}(\Pi),{\mathfrak R}_{0,1}\cup {\mathfrak R}_{1,1})$$
is induced by an automorphism of $\Pi$ \cite{PPZ}.

We show  that every bijective transformation of ${\mathcal
G}_{k}(\Pi)$ preserving ${\mathfrak R}_{1,1}$ is induced by an
automorphism of $\Pi$ (Theorem \ref{theorem1}). Our second result
(Theorem \ref{theorem2}) states that the same holds for the relation
${\mathfrak R}_{0,t}$ if $n\ge 2t\ge 4$ and $k=n-t-1$. The latter
relation is equivalent to the fact that two elements with the
relation ${\mathfrak R}_{0,t}$ span a maximal singular subspace.

Note that for   finite symplectic and hermitian polar spaces,
the first result under some conditions was proved in
\cite{liu,zeng}.

For a finite classical polar space, the valency of every
${\mathfrak R}_{i,j}$ was given by Stanton \cite{Stanton}.  We
establish that the valencies of ${\mathfrak R}_{i,j}$ and
${\mathfrak R}_{i',j'}$ are distinct if $(i,j)\ne (i',j')$
(Proposition \ref{prop-val}).

\section{Polar spaces}
We recall  some basic properties of polar spaces
and refer \cite{BC, Pankov-book,Ueberberg} for their proofs.

Let ${\mathcal P}$ be a non-empty set whose elements are called {\it points}
and ${\mathcal L}$ be a family formed by proper subsets of ${\mathcal P}$ called {\it lines}.
Two distinct points joined by a line are said to be {\it collinear}.
Let $\Pi=(P,{\mathcal L})$ be a {\it partial linear space}, i.e.
each line contains at least two points and
for any distinct collinear points $p,q\in {\mathcal P}$ there is precisely one line containing them,
this line is denoted by $pq$.

We say that $S\subset {\mathcal P}$ is a {\it subspace} of $\Pi$
if for any distinct collinear points $p,q\in S$
the line $pq$ is contained in $S$.
A {\it singular} subspace is a subspace where any two distinct points are collinear.
Note that the empty set and a single point are singular subspaces.
Using Zorn lemma, we show that every singular subspace is contained
in a certain maximal singular subspace.

From this moment we suppose that $\Pi$ is a {\it polar space}.
This means that the following axioms hold:
\begin{enumerate}
\item[(P1)] each line contains at least three points,
\item[(P2)] there is no point collinear to all points,
\item[(P3)] if $p\in {\mathcal P}$ and $L\in {\mathcal L}$
then $p$ is collinear to precisely one point or all points of the line $L$,
\item[(P4)] every flag formed by singular subspaces is finite.
\end{enumerate}
If there is a maximal singular subspace of $\Pi$ containing more than one line
then all maximal singular subspaces of $\Pi$ are projective spaces
of the same finite dimension $\ge 2$.
We say that the {\it rank} of $\Pi$ is $n$ if this dimension is equal to $n-1$.

In the case when the rank of $\Pi$ is not less than $4$,
every maximal singular subspace $M$ can be identified with
the projective space associated to a certain $n$-dimensional vector space $V$ (over a division ring).
Then every non-empty singular subspace $S\subset M$ will be identified with
the corresponding subspace of the vector space $V$.

The collinearity relation on $\Pi$ is denoted by $\perp$.
We write $p\perp q$ if $p,q\in {\mathcal P}$ are collinear points and $p\not\perp q$ otherwise.
If $X,Y\subset {\mathcal P}$ then $X\perp Y$ means that every point of $X$ is collinear to all points of $Y$.
For every subset $X\subset {\mathcal P}$ satisfying $X\perp X$
the minimal singular subspace containing $X$ is called {\it spanned} by $X$
and denoted by $\langle X \rangle$.
For every subset $X\subset {\mathcal P}$ we denote by $X^{\perp}$
the subspace of $\Pi$ formed by all points  collinear to all points of $X$.

\begin{fact}\label{fact1}
Let $X$ be a subset of ${\mathcal P}$ satisfying $X\perp X$ and spanning a maximal singular subspace $M$.
Then $p\perp X$ implies that $p\in M$.
\end{fact}

\begin{fact}\label{fact2}
If $M$ is a maximal singular subspace of $\Pi$
then for every point $p\in {\mathcal P}$ such that $p\not\in M$ we have
$$\dim_{p} (p^{\perp}\cap M)=n-2.$$
\end{fact}

\begin{fact}\label{fact3}
For every singular subspace $S$ there are maximal singular subspaces $M$ and $N$
such that  $S=M\cap N$.
\end{fact}

\section{Results}
Let $\Pi=({\mathcal P},{\mathcal L})$ be a polar space of   rank
$n\ge 3$. Recall that the polar Grassmannian formed by all
$k$-dimensional singular subspaces of $\Pi$ is denoted by ${\mathcal
G}_{k}(\Pi)$. Suppose that $k\le n-2$. We consider the relation
${\mathfrak R}_{i,j}$, $0\le i\le j\le k+1$ is formed by all pairs
$$(X,Y)\in {\mathcal G}_{k}(\Pi)\times{\mathcal G}_{k}(\Pi)$$
satisfying
$$\dim_{p} (X^{\perp}\cap Y)=k-i\;\mbox{ and }\; \dim_{p} (X\cap Y)=k-j.$$
Since the equality
$$\dim_{p} (X^{\perp}\cap Y)=\dim_{p} (Y^{\perp}\cap X)$$
holds for any pair $X,Y\in {\mathcal G}_{k}(\Pi)$,
this relation is symmetric.
Every automorphism of $\Pi$ (a bijective transformation of ${\mathcal P}$ preserving ${\mathcal L}$)
induces a transformation of ${\mathcal G}_{k}(\Pi)$ which preserves all ${\mathfrak R}_{i,j}$.

First, we determine all automorphisms of the graph
$$\Gamma'_{k}(\Pi)=({\mathcal G}_{k}(\Pi),{\mathfrak R}_{1,1}).$$

\begin{theorem}\label{theorem1}
Every automorphism of $\Gamma'_{k}(\Pi)$ is induced by an automorphism of $\Pi$.
\end{theorem}

\begin{rem}{\rm
For $k=0$ this statement is trivial.
The edges of $\Gamma'_{0}(\Pi)$ are pairs of non-collinear points of $\Pi$
and every automorphism of this graph is an automorphism of
the collinearity graph $\Gamma_{0}(\Pi)$.
It is well-known that the class of automorphisms of $\Gamma_{0}(\Pi)$ coincides with
the class of automorphisms of $\Pi$.
}\end{rem}

In the case when $k\in \{1,\dots, n-3\}$,
the distance between $S,U\in {\mathcal G}_{k}(\Pi)$ in the Grassmann graph $\Gamma_{k}(\Pi)$
is equal to $2$ if and only if $(S,U)$ belongs to ${\mathfrak R}_{1,1}\cup {\mathfrak R}_{0,2}$.
The distance between $S,U\in {\mathcal G}_{n-2}(\Pi)$ in $\Gamma_{n-2}(\Pi)$
is equal to $2$ if and only if $(S,U)\in{\mathfrak R}_{1,1}$
(if $k=n-2$ then ${\mathfrak R}_{0,2}$ is empty).
Theorem \ref{theorem1} gives the following.

\begin{cor}
Let $f$ be a bijective transformation of ${\mathcal G}_{n-2}(\Pi)$
satisfying the following condition:
the distance between $S,U\in {\mathcal G}_{n-2}(\Pi)$ in $\Gamma_{n-2}(\Pi)$ is equal to $2$
if and only if the distance between $f(S)$ and $f(U)$ in $\Gamma_{n-2}(\Pi)$ is equal to $2$.
Then $f$ is an automorphism of $\Gamma_{n-2}(\Pi)$.
\end{cor}

Suppose that $k=n-t-1$, where $t$ is an integer satisfying $n\ge
2t\ge 4$. Then $(S,U)\in {\mathfrak R}_{0,t}$ is equivalent to the
fact that $S$ and $U$ span a maximal singular subspace of $\Pi$. Our
second result describes all automorphisms of the graph
$$\Gamma''_{k}(\Pi)=({\mathcal G}_{k}(\Pi),{\mathfrak R}_{0,t}).$$

\begin{theorem}\label{theorem2}
Every   automorphism of $\Gamma''_{n-t-1}(\Pi)$ is induced by an
automorphism of $\Pi$.
\end{theorem}

\section{Cliques}
From this moment we suppose that $k\in\{1,\dots,n-2\}$.
For every singular subspace $N$ such that $\dim_{p} N<k$
we denote by $[N\rangle_{k}$ the set of all elements of ${\mathcal G}_{k}(\Pi)$ containing $N$.
This subset is said to be a {\it big star} if $N\in {\mathcal G}_{k-1}(\Pi)$.

If $N$ and $M$ are singular subspaces satisfying
$$N\subset M\;\mbox{ and }\;\dim_{p} N<k<\dim_{p} M$$
then we denote by $[N,M]_{k}$ the set of all $S\in {\mathcal G}_{k}(\Pi)$
such that $N\subset S\subset M$.
This subset is called a {\it star} if
$$N\in {\mathcal G}_{k-1}(\Pi)\;\mbox{ and }\;M\in {\mathcal G}_{n-1}(\Pi).$$
In the case when $N=\emptyset$, we write $\langle M]_{k}$ instead of $[N,M]_{k}$.
We say that $\langle M]_{k}$ is a {\it top} if $M\in {\mathcal G}_{k+1}(\Pi)$.

All maximal cliques of the Grassmann graph
$$\Gamma_{k}(\Pi)=({\mathcal G}_{k}(\Pi),{\mathfrak R}_{0,1})$$
and the weak Grassmann graph
$$\Gamma^{w}_{k}(\Pi)=({\mathcal G}_{k}(\Pi),{\mathfrak R}_{0,1}\cup {\mathfrak R}_{1,1})$$
are known \cite[Subsection 4.5.1 and Subsection 4.6.2]{Pankov-book}.
Every maximal clique of $\Gamma^{w}_{k}(\Pi)$ is a big star or a top.
Every maximal clique of $\Gamma_{k}(\Pi)$ is a star or a top.
In the case when $k=n-2$, every star is contained in a certain top
and all maximal cliques of $\Gamma_{k}(\Pi)$ are tops.

\begin{prop}\label{prop-cliq1}
Every clique of  $\Gamma'_{k}(\Pi)$ is contained in a big star.
\end{prop}

{\noindent\it Proof.\quad} If ${\mathcal C}$ is a clique of $\Gamma'_{k}(\Pi)$ then it
is a clique of $\Gamma^{w}_{k}(\Pi)$. Hence ${\mathcal C}$ is
contained in a big star or a top. Since any two distinct elements of
a top are non-adjacent vertices of $\Gamma'_{k}(\Pi)$, ${\mathcal
C}$ is a subset in a big star. $\hfill\Box\medskip$

Let $N\in {\mathcal G}_{k-1}(\Pi)$.
For every $M\in {\mathcal G}_{k+1}(\Pi)$ containing $N$
the subset $[N,M]_{k}$ is said to be a {\it line}.
The big star $[N\rangle_{k}$ together with all lines defined above is a polar space of rank $n-k$
\cite[Lemma 4.4]{Pankov-book}.
This polar space will be denoted by $\Pi_{N}$.

\begin{lemma}\label{lemma-cliq1}
Let $N\in {\mathcal G}_{k-1}(\Pi)$.
Two distinct elements of the big star $[N\rangle_{k}$
are adjacent vertices of the graph $\Gamma'_{k}(\Pi)$ if and only if
they are non-collinear points of the polar space $\Pi_{N}$.
\end{lemma}

{\noindent\it Proof.\quad} Easy verification. $\hfill\Box\medskip$

\iffalse
By Proposition \ref{prop-cliq2}, every clique of  $\Gamma'_{k}(\Pi)$ is contained in a big star.
Lemma \ref{lemma-cliq1} implies that
a subset of the big star $[N\rangle_{k}$, $N\in {\mathcal G}_{k-1}(\Pi)$
is a clique of $\Gamma'_{k}(\Pi)$ if and only if it is formed by mutually non-collinear points of $\Pi_{N}$,
in other words, it is a clique of $\Gamma'_{0}(\Pi_{N})$.
\fi

\begin{prop}\label{prop-cliq2}
Suppose that $k=n-t-1$, where $t$ is an integer satisfying $n\ge
2t\ge 4$. If $S$ and $U$ are adjacent vertices of
$\Gamma''_{k}(\Pi)$, then $M:=\langle S\cup U\rangle$ is a maximal
singular subspace of $\Pi$ and every clique of $\Gamma''_{k}(\Pi)$
containing $S$ and $U$ is a subset in $\langle M]_{k}$.
\end{prop}

{\noindent\it Proof.\quad} It is clear that $S\perp U$ and $M$  is a maximal singular
subspace. Let ${\mathcal C}$ be a clique of $\Gamma''_{k}(\Pi)$
containing $S$ and $U$. Then for every $A\in {\mathcal C}$ we have
$A\perp S$ and $A\perp U$. By Fact \ref{fact1}, this implies that
$A\subset M$. Hence ${\mathcal C}$ is contained in $\langle M]_{k}$.
$\hfill\Box\medskip$

\section{Proof of Theorem \ref{theorem1}}

\begin{lemma}\label{lemma1-1}
Let $p$ and $q$ be non-collinear points of $\Pi$
and let $t$ be a point of $\Pi$ collinear to at least one of the points  $p,q$.
Then there exists a point of $\Pi$ non-collinear to $p,q,t$.
\end{lemma}

{\noindent\it Proof.\quad} Suppose that  the statement fails and every point of $\Pi$ is
collinear to at least one of the points $p,q,t$, in other words,
\begin{equation}\label{eq1}
{\mathcal P}=p^{\perp}\cup q^{\perp}\cup t^{\perp}.
\end{equation}
First we show that every maximal singular subspace of $\Pi$
contains at least one of the points $p,q,t$.

Let $M$ be a maximal singular subspace of $\Pi$.
If each of the points $p,q,t$ does not belong to $M$ then
$$p^{\perp}\cap M,\;q^{\perp}\cap M,\; t^{\perp}\cap M$$
are $(n-2)$-dimensional subspaces of $M$ (Fact \ref{fact2})
and \eqref{eq1} implies that $M$ is the union of these subspaces.
The latter is impossible,
since a projective space cannot be presented as the union of three hyperplanes.

By our assumption, $t$ is collinear to at least one of the points  $p$ and $q$.
Suppose that $q\perp t$.
Since $p\not\perp q$,
the line $qt$ contains the unique point $s$ collinear to $p$.
One of the following possibilities is realized:
\begin{enumerate}
\item[(1)] $s=t$,
\item[(2)] $s\ne t$.
\end{enumerate}
In the case (1), we take any point $v$ on the line $qt$ different from $q$ and $t$.
It is clear that $p\not\perp v$.
By Fact \ref{fact3}, there exist maximal singular subspaces $M$ and $N$ such that
$M\cap N=\{v\}$.
Since $p\not\perp v$, they do not contain $p$.
Then one of these subspaces contains $q$ and the other contains $t$.
So, each of these subspaces contains two distinct points of the line $qt$.
This means that this line is contained in $M\cap N$ which is impossible.

In the case (2), we take any point $w$ on the line $ps$ different
from $p$ and $s$. As in the previous case, we consider maximal
singular subspaces $M$ and $N$ such that $M\cap N=\{w\}$. One of
these subspaces contains $q$ or $t$. Then at least one of the points
$q,t$ is collinear to $w$. Since $q$ and $t$ both are collinear to
$s$ and $w$ is on the line $ps$, one of the points $q,t$ is
collinear to all points of the line $ps$. Thus $p$ is collinear to
$q$ or $t$  which is impossible. $\hfill\Box\medskip$

\begin{lemma}\label{lemma1-2}
Suppose that $N\in {\mathcal G}_{k-1}(\Pi)$.
Let $P,Q\in [N\rangle_{k}$ be adjacent vertices of $\Gamma'_{k}(\Pi)$
and let $T\in [N\rangle_{k}$ be a vertex of $\Gamma'_{k}(\Pi)$
non-adjacent to at least one of the vertices $P,Q$.
Then there exists $S\in [N\rangle_{k}$ adjacent to $P,Q,T$ in $\Gamma'_{k}(\Pi)$.
\end{lemma}

{\noindent\it Proof.\quad} By Lemma \ref{lemma-cliq1}, $P$ and $Q$ are non-collinear
points of $\Pi_{N}$ and $T$ is a point of $\Pi_{N}$ collinear to at
least one of the points $P,Q$. We apply Lemma \ref{lemma1-1} to the
polar space $\Pi_{N}$ and get the claim. $\hfill\Box\medskip$

Let $f$ be an automorphism of $\Gamma'_{k}(\Pi)$.

Show that $f$ transfers big stars to subsets of big stars.
We take any $N\in {\mathcal G}_{k-1}(\Pi)$.
Let $P$ and $Q$ be adjacent vertices of $\Gamma'_{k}(\Pi)$
contained in the big star $[N\rangle_{k}$.
Then $f(P)$ and $f(Q)$ are adjacent vertices of $\Gamma'_{k}(\Pi)$
contained in the big star $[N'\rangle_{k}$, where
$$N'=f(P)\cap f(Q).$$
We assert that $f(T)\in [N'\rangle_{k}$ for every $T\in [N\rangle_{k}$
and prove this statement in several steps.

(i). First, we consider the case when
$T$ is a vertex of  $\Gamma'_{k}(\Pi)$ adjacent to both $P$ and $Q$.
Then $f(P),f(Q),f(T)$
form a clique in $\Gamma'_{k}(\Pi)$ which,
by Proposition \ref{prop-cliq1}, is contained in a certain big star $[N''\rangle_{k}$.
We have
$$N'=f(P)\cap f(Q)=N''$$
which gives the claim.

(ii). Consider the case when $T$ is a vertex of  $\Gamma'_{k}(\Pi)$
adjacent to precisely one of the vertices $P,Q$.
Suppose that $T$ is adjacent to $P$.
Lemma \ref{lemma1-2} implies the existence of
a vertex $S\in [N\rangle_{k}$ in the graph $\Gamma'_{k}(\Pi)$ adjacent to $P,Q,T$.
By (i), $f(S)$ belongs to $[N'\rangle_{k}$.
Then
$$f(P)\cap f(S)=N'$$
and $f(P),f(S),f(T)$ form a clique of $\Gamma'_{k}(\Pi)$.
As in (i), we show that $f(T)\in [N'\rangle_{k}$.

(iii). Suppose that $T$ is a vertex of $\Gamma'_{k}(\Pi)$ non-adjacent to both $P$ and $Q$.
As above, we consider $S\in [N\rangle_{k}$ which is a vertex of $\Gamma'_{k}(\Pi)$ adjacent to $P,Q,T$
and obtain that $f(S)\in [N'\rangle_{k}$.
We apply the arguments from (ii) to $P,S,T$ and establish that $f(T)\in [N'\rangle_{k}$.

So, $f$ transfers big stars to subsets of big stars.
The same arguments show that $f^{-1}$ sends big stars to subsets of big stars.
This means that $f$ and $f^{-1}$ both map big stars to big stars, i.e.
there exists a bijective transformation $g$ of ${\mathcal G}_{k-1}(\Pi)$
such that
$$f([N\rangle_{k})=[g(N)\rangle_{k}$$
for every $N\in {\mathcal G}_{k-1}(\Pi)$.

Let $U\in {\mathcal G}_{k}(\Pi)$.
Then $g$ transfers the top $\langle U]_{k-1}$ to the top $\langle f(U)]_{k-1}$.
Indeed, we have
$$N\in \langle U]_{k-1}\Leftrightarrow U\in [N\rangle_{k}\Leftrightarrow f(U)\in [g(N)\rangle_{k}\Leftrightarrow
g(N)\in \langle f(U)]_{k-1}.$$
Similarly, $g^{-1}$ sends $\langle U]_{k-1}$
to the top $\langle f^{-1}(U)]_{k-1}$.

Therefore, $g$ and $g^{-1}$ both transfer tops to tops.
Since for any two adjacent vertices of the Grassmann graph $\Gamma_{k-1}(\Pi)$ there is a top containing them,
$g$ is an automorphism of $\Gamma_{k-1}(\Pi)$.
In some special cases,
there are automorphisms of $\Gamma_{k-1}(\Pi)$ which are not induced by automorphisms of $\Pi$.
However, every automorphism of $\Gamma_{k-1}(\Pi)$ transferring tops to tops is
induced by an automorphism of $\Pi$ \cite[Section 4.6.1]{Pankov-book}.
Thus $g$ is induced by an automorphism of $\Pi$.
An easy verification shows that this automorphism also induces $f$.

\section{Proof of Theorem \ref{theorem2}}

In this section we suppose that $k=n-t-1$, where $t$ is an integer satisfying $n\ge 2t\ge 4$.

\begin{lemma}\label{lemma2-1}
Let $S,U$ be adjacent vertices of $\Gamma''_{k}(\Pi)$. Let also $M$
be the maximal singular subspace spanned by $S$ and $U$. If $T\in
\langle M]_{k}$ is a vertex of  $\Gamma''_{k}(\Pi)$ non-adjacent to
at least one of the vertices $S,U$, then there exists a vertex  $Q$
of $\Gamma''_{k}(\Pi)$ adjacent to $S,U,T$.
\end{lemma}

{\noindent\it Proof.\quad}
We suppose that $T$ is non-adjacent to $S$ (the case when $T$ is non-adjacent to $U$ is similar).
First of all, we show that the general case can be reduced to the case when
$N:=S\cap U$ and $T$ are disjoint.
Indeed, if the projective dimension of $W:=N\cap T$ is equal to $w\ge 0$
then $S,U,T$ can be naturally identified with vertices of
the graph $\Gamma''_{k-w-1}(\Pi_{W})$ which is isomorphic to
the subgraph of $\Gamma''_{k}(\Pi)$ induced on $[W\rangle_{k}$.

So, we suppose that $N$ and $T$ are disjoint.
Since $n\ge 4$, we identify $M$ with
the projective space associated to an $n$-dimensional vector space $V$.
Every non-empty singular subspace of $M$ will be identified with the corresponding subspace of
the vector space $V$.
We set
$$m:=\dim_{l} N=n-2t$$
and suppose that
$$\dim_{l}(S\cap T)=i,\;\;\;\dim_{l}(U\cap T)=j.$$
Note that $i>m$, since $S$ and $T$ are non-adjacent vertices of $\Gamma''_{k}(\Pi)$.

First we consider the case when $i+j=n-t$, i.e. $T$ is spanned by $S\cap T$ and $U\cap T$.
Since $\dim_{l}(S/N)=\dim_{l}(U/N)=t$, we can choose vectors
$$
x_{1},\dots,x_{t}\in S\setminus N\;\mbox{ and }\; y_{1},\dots,y_{t}\in U\setminus N$$
such that
$$S=N+\langle x_{1},\dots,x_{t}\rangle,\;\;\;
U=N+\langle y_{1},\dots,y_{t}\rangle,\;\;\; T=\langle x_{1},\dots,
x_{i}\rangle+\langle y_{t-j+1},\dots,y_{t}\rangle.
$$ The vectors
$x_{1}+y_{1},\dots,x_{t}+y_{t}$ are linearly independent. We define
$$Q:=N+\langle x_{1}+y_{1},\dots,x_{t}+y_{t}\rangle.$$
An easy verification shows that
$$S+Q=U+Q=T+Q=M$$
which implies that $Q$ is a vertex of $\Gamma''_{k}(\Pi)$ adjacent to $S,U,T$.

Now suppose that $$l:=n-t-(i+j)>0.$$
Then $t=i+j-m+l$. Since $i>m$, we have $i+j-m>0$.
We choose linearly independent vectors
$$x_{1},\dots,x_{i+j-m},x'_{1},\dots,x'_{l}\in S\setminus N$$
and linearly independent vectors
$$y_{1},\dots,y_{i+j-m},y'_{1},\dots,y'_{l}\in U\setminus N$$
such that
$$T=\langle x_{1},\dots,x_{i}\rangle+\langle y_{i-m+1},\dots, y_{i+j-m} \rangle+
\langle x'_{1}+y'_{1},\dots,x'_{l}+y'_{l}\rangle.$$
 We denote by $Q$ the subspace spanned by $N$ and the  vectors
$$x_{1}+y'_{1},x'_{1}+y'_{2},\dots,x'_{l-1}+y'_{l}, x'_{l}+y_{1}, x_{2}+y_{2},\dots, x_{i+j-m}+y_{i+j-m}.$$
Using the equalities
$$S=N+\langle x_{1},\dots,x_{i+j-m},x'_{1},\dots,x'_{l}\rangle,$$
$$U=N+\langle y_{1},\dots,y_{i+j-m},y'_{1},\dots,y'_{l}\rangle,$$
we establish that $S+Q=U+Q=M$.
To complete the proof we need to check that $T+Q=M$.

The conditions
$$x_{1},\dots,x_{i},y_{i-m+1},\dots, y_{i+j-m}\in T\;
\mbox{ and }\;x_{2}+y_{2},\dots, x_{i+j-m}+y_{i+j-m}\in Q$$ imply
that $x_{p}\in T+Q$ for every $p$ and $y_{p}\in T+Q$ if $p\ge 2$.
Since $x_{1}\in T$ and $x_{1}+y'_{1}\in Q$, we have $y'_{1}\in T+Q$.
Then $x'_{1}+y'_{1}\in T$ implies that $x'_{1}\in T+Q$. Step by
step, we  establish that all $x'_{q}$ and $y'_{q}$ belong to $T+Q$.
The conditions $x'_{l}\in T+Q$ and $x'_{l}+y_{1}\in Q$ guarantee
that $y_{1}\in T+Q$. Therefore, $Q+T$ coincides with $S+U=M$. $\hfill\Box\medskip$

Let $f$ be an automorphism of $\Gamma''_{k}(\Pi)$.
Show that for every $M\in {\mathcal G}_{n-1}(\Pi)$ there exists $M'\in {\mathcal G}_{n-1}(\Pi)$
such that
$$f(\langle M]_{k})\subset \langle M']_{k}.$$
We choose $S,U\in \langle M]_{k}$ such that $M$ is spanned by $S\cup
U$. Then $S$ and $U$ are adjacent vertices of $\Gamma''_{k}(\Pi)$
and the same holds for $f(S)$ and $f(U)$. Hence
$$M':=\langle f(S)\cup f(U)\rangle$$
is a maximal singular subspace.
We assert that $f(T)\in \langle M']_{k}$ for every $T\in \langle M]_{k}$
and prove this statement in several steps.

(i). If $T\in \langle M]_{k}$ is a vertex of $\Gamma''_{k}(\Pi)$
adjacent to $S$ and $U$ then $f(S)$, $f(U)$, $f(T)$ form a clique in
$\Gamma''_{k}(\Pi)$ and, by Proposition \ref{prop-cliq2}, we have
$f(T)\in \langle M']_{k}$.

(ii). Now, let $T\in \langle M]_{k}$ be a vertex of  $\Gamma''_{k}(\Pi)$
adjacent to precisely one of the vertices  $U$ and $S$.
Suppose that $T$ is adjacent to $U$ and non-adjacent to $S$.
Let $Q\in \langle M]_{k}$ be a vertex of $\Gamma''_{k}(\Pi)$ adjacent to $S,U,T$
(Lemma \ref{lemma2-1}).
By (i), $f(Q)$ belongs to $\langle M']_{k}$.
Then
\begin{equation}\label{eq2}
\langle f(U)\cup f(Q)\rangle=M'.
\end{equation}
We apply (i) to $U,Q,T$ and establish that $f(T)$ belongs to $\langle M']_{k}$.

(iii). Consider the case when $T$ is a vertex of $\Gamma''_{k}(\Pi)$
non-adjacent to both $S$ and $U$.
As above, we consider $Q\in \langle M]_{k}$ which is a vertex of $\Gamma''_{k}(\Pi)$
adjacent to $S,U,T$.
Then $f(Q)\in \langle M']_{k}$ and \eqref{eq2} holds.
We apply (ii) to $U,Q,T$ and obtain that $f(T)\in \langle M']_{k}$.

We apply the above arguments to $f^{-1}$ and
establish the existence of a bijective transformation $g$ of ${\mathcal G}_{n-1}(\Pi)$
such that
$$f(\langle M]_{k})=\langle g(M)]_{k}$$
for every $M\in {\mathcal G}_{n-1}(\Pi)$.

Let $U$ be a singular subspace of $\Pi$ such that $k<\dim_{p} U<n-1$.
Then there exist
$M_{1},M_{2}\in {\mathcal G}_{n-1}(\Pi)$ satisfying $U=M_{1}\cap M_{2}$.
We have
$$\langle U]_{k}=\langle M_{1}]_{k}\cap \langle M_{2}]_{k},$$
$$f(\langle U]_{k})=\langle g(M_{1})]_{k}\cap \langle g(M_{2})]_{k}=
\langle g(M_{1})\cap g(M_{2})]_{k}.
$$ We set
$$g(U):=g(M_{1})\cap g(M_{2})$$
and get an extension of $g$ to a transformation of
\begin{equation}\label{eq3}
{\mathcal G}_{k+1}(\Pi)\cup\dots\cup {\mathcal G}_{n-1}(\Pi)
\end{equation}
such that
$$f(\langle U]_{k})=\langle g(U)]_{k}$$
for every $U$ belonging to \eqref{eq3}.
We apply the same arguments to the pair $g^{-1},f^{-1}$
and show that $g$ is bijective.
It is clear that $f$ is inclusion preserving, i.e.
$$S\subset U\;\Longleftrightarrow\;f(S)\subset f(U)$$
for any $S,U$ belonging to \eqref{eq3}.
The latter guarantees that $f$ sends ${\mathcal G}_{i}(\Pi)$ to itself
for every integer $i$ satisfying $k<i\le n-1$.
Therefore, $f$ and $f^{-1}$ both send tops to tops
which implies that $f$ is induced by an automorphism of $\Pi$.

  \section{Valency of the relation for finite classical polar spaces}
Let $V$ be an $n$-dimensional vector space over the finite field
${\mathbb F}_{q}$ together with a non-degenerate symplectic,
hermitian or symmetric form of Witt index $d\ge 2$. For every $m\in
\{1,\dots,d\}$ we denote by ${\mathcal N}_{m}$  the set of all
$m$-dimensional totally isotropic subspaces. The associated polar
space $\Pi$ is defined as follows: the point set is ${\mathcal
N}_{1}$ and the lines are defined by elements of ${\mathcal N}_{2}$
(the line corresponding to $S\in {\mathcal N}_{2}$ consists of all
elements of ${\mathcal N}_{1}$ contained in $S$). This is a polar
space of rank $d$ and every ${\mathcal G}_{i}(\Pi)$ can be naturally
identified with ${\mathcal N}_{i+1}$.  For more information, see
\cite{taylor, wan1}.

 For $0\leq i\leq j\leq m,$ denote by $n_{i,j}$ the valency of
the relation
\begin{equation}\label{eq4}
\{(S,U)\in{\mathcal N}_{m}\times{\mathcal N}_{m}: \dim_l
(S^{\perp}\cap U)=m-i,\; \dim_l (S\cap U)=m-j\},
\end{equation}
where $S^{\perp}$ is the orthogonal complement to $S$.
If $m<d$ then \eqref{eq4} coincides with the relation ${\mathfrak R}_{i,j}$.
In the case when $m=d$,
we have $S^{\perp}\cap U=S\cap U$ for any $S,U\in {\mathcal N}_{m}$
and \eqref{eq4} is the relation ${\mathfrak R}_{i}$ defined on the dual polar space.

The valency $n_{i, j}$ is computed in \cite{Stanton} as follows:
$$
n_{i, j}=
q^{j^2+i(n-2m-2j+\frac{3}{2}i+\frac{1}{2}-\mu-\nu)}{m\brack j}{j\brack i}\\
\prod\limits_{s=0}^{j-i-1}(q^{\frac{n}{2}-m-\mu-s}-1)
(q^{\frac{n}{2}-m-\nu-s}+1)(q^{s+1}-1)^{-1},
$$
where
$\mu=\frac{1}{2}n-d$ and $\nu$ is a number such that $\mu+\nu$
equals $0,\frac{1}{2},1$ in the symplectic, hermitian and symmetric
cases, respectively.

\begin{prop}\label{prop-val}
All valencies $n_{i, j}$  are pairwise distinct.
\end{prop}

\begin{lemma}\label{5le7}
Let $f(x)$ be a polynomial of  degree at least $1$ over the rational
number field $\mathbb Q$.

{\rm (i)} If $f(x)$ has the following two factorizations:
\begin{eqnarray*}\label{4a0}
f(x)=(x^{r_1}-1)^{i_1}(x^{r_2}-1)^{i_2}\cdots
(x^{r_l}-1)^{i_l}=(x^{t_1}-1)^{j_1} (x^{t_2}-1)^{j_2}\cdots
(x^{t_h}-1)^{j_h},
\end{eqnarray*}
where $r_1>\cdots >r_l$ and $t_1>\cdots >t_h$, then $h=l$, $r_s=t_s$
and $i_s=j_s$, $s=1,\ldots,l$.

{\rm (ii)} If $f(x)$ has the following two factorizations:
\begin{eqnarray*}\label{4a0'}
f(x)=(x^r+1)(x^{r_1}-1)^{i_1}\cdots
(x^{r_l}-1)^{i_l}=(x^{t_1}-1)^{j_1}\cdots (x^{t_h}-1)^{j_h},
\end{eqnarray*}
then there   exists an $s\in \{1,\ldots,h\}$ such that $t_s=2r$.

{\rm (iii)} If $f(x)$ has the following two factorizations:
\begin{eqnarray*}\label{4a0''}
f(x)=(x^{k_1}+1)\cdots(x^{k_f}+1)(x^{r_1}-1)^{i_1}\cdots
(x^{r_l}-1)^{i_l}=(x^{t_1}-1)^{j_1}\cdots (x^{t_h}-1)^{j_h},
\end{eqnarray*}
where $k_1,\ldots,k_f$ are distinct odd  integers, then there  exist
 $s_1,\ldots s_f\in \{1,\ldots,h\}$ such that $t_{s_1}=2k_1, \ldots,
t_{s_f}=2k_f$.
\end{lemma}

{\noindent\it Proof.\quad} (i)  Consider the term  of second minimum degree of $f(x)$,
we have
$$
(-1)^{i_1+\cdots+i_l-1}i_lx^{r_l}=(-1)^{j_1+\cdots+j_h-1}j_hx^{t_h},
$$
which implies that  $j_h=i_l$ and $t_h=r_l$.  So
$$
(x^{r_1}-1)^{i_1}\cdots
(x^{r_{l-1}}-1)^{i_{l-1}}=(x^{t_1}-1)^{j_1}\cdots
(x^{t_{h-1}}-1)^{j_{h-1}},
$$
 By induction, (i) holds.

(ii) The polynomial $(x^r-1) f(x)$ has the following two
factorizations:
$$
(x^r-1) f(x)=(x^{2r}-1)(x^{r_1}-1)^{i_1}\cdots
(x^{r_l}-1)^{i_l}=(x^r-1)(x^{t_1}-1)^{j_1}\cdots (x^{t_h}-1)^{j_h}.
$$
 So (ii) holds by (i).

 The proof of (iii) is similar to that of (ii), and will be
 omitted.$\hfill\Box\medskip$

\medskip{\em Proof of Proposition \ref{prop-val}:}
Suppose $n_{a, b}= n_{a', b'}$. Since $q$ is a prime power,
we obtain
\begin{eqnarray}\label{4a1}
b^2+a(n-2m-2b+\frac{3}{2}a+\frac{1}{2}-\mu-\nu)=b'^2+a'(n-2m-2b'+\frac{3}{2}a'+\frac{1}{2}-\mu-\nu),
\end{eqnarray}
and
\begin{eqnarray}\label{4a2}
&{m\brack b}{b\brack
a}\prod\limits_{s=0}^{b-a-1}(q^{\frac{n}{2}-m-\mu-s}-1)(q^{\frac{n}{2}-m-\nu-s}+1)
(q^{s+1}-1)^{-1}\nonumber \\
=&{m\brack b'}{b'\brack
a'}\prod\limits_{s=0}^{b'-a'-1}(q^{\frac{n}{2}-m-\mu-s}-1)(q^{\frac{n}{2}-m-\nu-s}+1)
(q^{s+1}-1)^{-1}.
\end{eqnarray}

Simplifying (\ref{4a1}), we have
\begin{eqnarray}\label{4a3}
(b-b')(2b+2b'-4a')=(a'-a)(2l+3a'+3a-4b),
\end{eqnarray}
where $l=n-2m+\frac{1}{2}-\mu-\nu$. Write
$$
\begin{array} {rcl}
f_{i,j}(x)&=&\prod\limits_{s=0}^{j-i-1}(x^{n-2m-2\mu-2s}-1)(x^{n-2m-2\nu-2s}+1),\\
g_{i,j}(x)&=&\prod\limits_{s=1}^{m-j}(x^{2s}-1)\prod\limits_{s=1}^{i}(x^{2s}-1)\prod\limits_{s=1}^{j-i}(x^{2s}-1)^2.
\end{array}
$$
The equality (\ref{4a2}) implies that
$f_{a,b}(q)g_{a',b'}(q)=f_{a',b'}(q)g_{a,b}(q)$  for all prime
powers $q$, so
\begin{eqnarray}\label{4a4}
f_{a,b}(x)g_{a',b'}(x)=f_{a',b'}(x)g_{a,b}(x).
\end{eqnarray}
Computing the degree of this polynomial, we have
\begin{eqnarray}\label{4a5}
(b-b')(2l-5b-5b'+m+8a')=(a'-a)(-2l-5a-5a'+8b).
\end{eqnarray}
Equalities  (\ref{4a3}) and (\ref{4a5}) imply that
\begin{eqnarray}\label{4a6}
(b-b')(2l+2m-b-b')=(a'-a)(2l+a+a').
\end{eqnarray}

Suppose $b\neq b'.$ Without loss of generality, we assume that
$b>b'$. Since $2l+2m-b-b'>0$ and $2l+a+a'>0$, by  (\ref{4a6}) one
gets $a'>a$ and
\begin{equation}\label{ineq}
b-a-b'+a'\geq 2.
\end{equation}
Write
$$\begin{array}{rcl}
k(x)&=&\prod\limits_{s=b'-a'+1}^{b-a}(x^{2s}-1)^2,\\
h(x)&=&\prod\limits_{s=b'-a'}^{b-a-1}(x^{n-2m-2\mu-2s}-1)(x^{n-2m-2\nu-2s}+1)\prod\limits_{s=m-b+1}^{m-b'}(x^{2s}-1)
\prod\limits_{s=a+1}^{a'}(x^{2s}-1).
\end{array}
$$
The equation (\ref{4a4}) implies that $k(x)=h(x)$.

Set
$$
h_1(x)=x^{2n-4m-4\nu-4b'+4a'}-1,\quad h_2(x)=x^{2n-4m-4\nu-4b+4a+4}-1.
$$
By Lemma~\ref{5le7}  there exist
$s_1,s_2\in\{b'-a'+1,\ldots, b-a\}$ such that $h_1(x)=x^{2s_1}-1$
and $h_2(x)=x^{2s_2}-1$. It follows that
$$
s_1=n-2m-2\nu-2b'+2a',\quad s_2= n-2m-2\nu-2b+2a+2.
$$
By (\ref{ineq}) we have $s_1-s_2=2b-2a-2b'+2a'-2>b-a-(b'-a'+1),$ a
contradiction. So $b=b'$ and $a=a'$, as desired. $\hfill\Box\medskip$

\begin{rem}{\rm  It is well known that $\mathcal N_m$
has a structure of a symmetric association scheme, each relation of
which has form (\ref{eq4}). By Theorem~\ref{theorem1} every
automorphism of this scheme is induced by an automorphism of the
associated polar space $\Pi$. }\end{rem}

\section*{Acknowledgement}
W. Liu was supported by NSFC(11271004). K. Wang  was supported by
NSFC(11271047) and the Fundamental Research Funds for the Central
University of China.

redpajama_set_name: RedPajamaArXiv


Are you feeling to have low luck? Here you can change or improve your luck by simple Astrology remedy suggested by Astro Uncle.
Some children have good luck and some have problems with in their luck. One can know about their luck by seeing their hand.
Luck line starts from the life line and goes towards the mountain of Saturn. If it is dark then it gives you success easily but if it is light then you have to work hard to get success.
Mostly male's have dark line comparatively female. If the luck line is straight and it has no cuts then child have good luck. However if it has cuts and break then you have to do such remedy. Luck line has part such as 18, 36 and so on.
If such luck line goes toward mountain of Jupiter then you get good result even after doing less work. If one small line from the luck line goes toward mountain of Jupiter then also you get good success and popularity.
If such luck line goes toward mountain of Sun then you will definitely get success in any profession. It gives you not only money but also prestige, respect, and fame.
You should do remedies for Sun, Moon and Jupiter for life time. You have to see sunrise. You can chant Adityahriday Strot. You can also do Tratak along with Sun.
You can also chant Gayatri mantra for 21 times and 108 times on Sunday. You have to donate gram. You can eat Kheer which is made from sugar cane juice. You should avoid salt on Sunday. You should serve parents, elders and leper.
You should eat less in many times. You can eat jaggery. Your luck will increase by drinking lukewarm water in morning.
Extra Remedy: Many people have problems after taking lunch/dinner. They have to drink water 2-3 times in a day in copper glass. They can wear copper coin in green thread on throat.


Michael Frankel is the chair of the Compilation Engagements Task Force. He has been a member of the Auditing and Assurance Standards Board since 2013. In addition, Michael serves on numerous local, national, and international not-for-profit boards.
Michael is a retired partner of Richter LLP, where he provided business advisory services to a wide range of companies, both private and public, in multiple industry sectors.


text:
TITLE INVESTIGATION OF HIGH-RESOLUTION CLIMATE PROJECTIONS OVER TURKEY AND ITS SURROUNDING REGIONS USING REGCM4.4
AUTHORS Mustafa Tufan Turp (1,2), Tugba Ozturk (2,3), Murat Türkeş (2), Mehmet Levent Kurnaz (2,4)
Center For Climate Change And Policy Studies, Bogazici University Istanbul (Turkey)
ABSTRACT In this study, projected future changes for the period of 2021–2050 in mean air temperature and precipitation climatology and year-to-year variability with respect to the control period of 1971–2000 were investigated for the domain of Turkey and its surrounding regions via regional climate model simulations. In order to investigate the projected changes in near future climate conditions, Regional Climate Model (RegCM4.4) of ICTP (the Abdus Salam International Centre for Theoretical Physics) was driven by two different global climate models. HadGEM2-ES global climate model of the Met Office Hadley Centre and MPI-ESM-MR global climate model of the Max Planck Institute for Meteorology were dynamically downscaled to 10 km for Turkey and its surrounding regions. The projections were performed based on the RCP4.5 and the RCP8.5 emission scenarios of the Intergovernmental Panel on Climate Change (IPCC). This research has been supported by Bogazii University Research Fund Grant Number 10421.


Mentorship is one of the best ways to promote diversity in the c-suite and at the board level, but some leaders need an extra push.
Despite the fact that countless reports have shown correlations between financial performance and mentoring new diverse and inclusive teams, some senior-level executives still need convincing. For leaders who are resistant, there may only be one solution: tie sponsorship and mentorship of staff to their own money and advancement.
Like IBM, to promote sponsorship and mentoring The Boston Consulting Group (BCG) ties sponsorship of a diverse workforce not just to compensation of its leaders, but to their own advancement. "Sometimes it's money, but more to succeed and advance and get to be a senior partner, explicitly, we ask 'who are you sponsoring and promoting and putting into business?' Without that you don't advance," said Matt Krentz, Global People Team Chair at BCG.
Nationwide strives for diversity and inclusion by requiring qualified diverse candidate pools for all director-and-above positions, before interviewing can begin.
"It helps discourage the process of saying 'I've got my two or three go-to candidates, that's who I want to talk to and I know I'm going to pick one of them' because you don't get a green light to pass go and start the hiring process until we're sure that we have looked broader and deeper than we otherwise might out of habit," said Shore.
In the midst of global transformation, IBM has strategically focused on technical women's career development and advancement. "If culture is what culture does, we have to start putting in place actions that reflect the culture we want, that, in turn, generates the culture down the road," said Gilbert.


Massive Family home conveniently located in a great pocket of Southern River. This beautiful family home is close to schools, shops and transport.
Very rarely does a home with such presentation become available for rent.
- Theatre room / Formal Living.
- Outdoor Entertaining Alfresco area.
- 2 x Split System Air Conditioning.


text:
Biografia
Cresciuto in una zona di campagna distante dai cinema, fin da giovane si appassiona al cinema. Ai tempi della scuola superiore frequenta il club scolastico di teatro in cui svolge il ruolo di sceneggiatore e direttore, durante un festival della cultura proietta un suo cortometraggio.

Poco dopo aver compiuto 20 anni si trasferisce a Tokyo in autostop, dopo varie difficoltà economiche riesce nel 2009 a fondare la compagnia Panpokopina.

Filmografia parziale

Regista e sceneggiatore
 - anche montatore (2011)
 (2017)
 - Co-diretto e scritto con Yuya Nakaizumi e Naoya Asanuma (2019)
 (2019)

Note

Altri progetti

Collegamenti esterni

redpajama_set_name: RedPajamaWikipedia


Blue Jasmine, Woody Allen's 2013 film, is just over three months away from it's US premiere, and we are slowly getting more images and information. We have a new image of lead Cate Blanchett, the ad that ran in the LA Times that could give us a clue to the poster and new release dates for other markets.
First is a new image of Cate Blanchett. she's the lead for the new drama, and it's certainly the darkest and broodiest image we have so far. It is the seventh photo we've seen of from the new film.
Another new image from the film was released this week, but only in print. Sony Pictures Classics took out full page ads in the LA Times this week to promote two of their summer films – Pedro Almodovar's 'I'm so Excited' and 'Blue Jasmine'. The image used to promote 'Blue Jasmine' is our first hint at what a poster will look like.
Or is it? It's the standard Woody Allen film font, and the centred look and cast name in alphabetical order are all design tricks we've seen before. They've used a blue colour to tie in with the film's name (we presume), and it seems like otherwise all is normal in Allen's world. This style is so embedded that it would have been pretty easy for an artwork person to mock this up. We are expecting to see a full poster very soon.
Bobby Cannavale was recently interviewed by Newsday, and in the interview he discussed Woody Allen.
And how was Woody Allen? Some find him intimidating.
About three weeks later, he sent over my scenes in a manila envelope. I had an hour to read them, then give them back. He sent a note saying, "Take a look — if you don't like it, you can pass. But I'd love for you to do it." It was amazing watching him work. I just wanted to remember it.
We also have some new release dates. The US limited release is 26th July, in New York and Los Angeles only at this point. Now the Seattle Times has announced it will open in that part of the Us on 16 August. It makes sense – a limited release that rolls out across the US over the coming weeks.
On the international side, IMDB has added Portugal to it's release dates, as September 5, earlier than the UK or France.
Finally, the MPAA has released ratings for 'Blue Jasmine' – PG13. Which means there's a finished cut doing the round! And a serious drama it may be, but it's not the R rating that was on films such as Husbands And Wives.
We are still waiting for that proper poster and a trailer. They usually do not come any later than three months out, so we are expecting it to hit any day now. We will post it up as soon as we get it.
Check out all our Blue Jasmine coverage.
What do you think of the new pic?


I personally find it ironic that the word chosen by IESO is Danish. Denmark is where electricity prices for residential homes is the most expensive in Europe* at EURO per kWh of 0.3126 or Canadian 0.48 cents per kWh. Doesn't that make all Ontario residents feel cosy!
Denmark is home to VESTAS and their product line is exclusively wind turbines. Vestas employs over 24,000 people which makes them one of the 10 largest employers in the country. Vestas's website claim they have installed 97 GW (97,000 MW) of industrial wind turbines (IWT) globally. All those noise-emitting, bird- and bat-killing, intermittent and unreliable wind turbines might make the Danes "cosy" but somehow I doubt it, with the price they are paying for electricity.
The IESO post suggests we: turn off the phone, unplug appliances and devices, eat comfort food and use energy-efficient cooking methods like a pressure cooker! ** The message to the reader goes on to suggest pulling on wool socks and using our favourite blanket to get cosy and then to "get lost in the moment" by reading our favourite book!
IESO should stop the wasted spending on conservation efforts of this ilk. Does IESO not understand we are all billed monthly for our cost of electricity usage and have been doing our best to "stay cosy"? For many it has been an effort to simply avoid energy poverty.
Stop lecturing us, stop wasting our money and focus your efforts on managing the grid in a manner that will reduce the costs of electricity.
*Demark has the highest prices for residential electricity out of 41 European countries listed by Eurostat.
**Full disclosure—my wife's pressure cooker recently blew up and created a mess in our kitchen which we now must repair.
Using a "private prosecution," Wind Concerns Ontario has served the Honourable Chris Ballard, Minister of MOECC with a summons for violating the EPA (Environment Protection Act).
The issue is related to the lack of proper follow-up on the thousands of noise complaints filed by individuals throughout the province in locations where industrial wind turbines have been erected.
The complaints date back to 2006 and have continued without any significant attention from the Ministry and have even caused families to abandon their homes in order to seek relief from the noise effects.
Mr Ballard has been Environment Minister since August 2017, and complaints have continued, without resolution.
How much does wind power cost us?
Being asked to do a presentation at Wind Concerns Ontario's annual conference this past Saturday to describe the costs associated with industrial wind turbines was something I relished!
The presentation I developed used IESO information for 2017.
Discovered in the preparation of my presentation was the fact that nuclear and hydro power alone could have supplied over 100% of all grid-connected consumption for 2017, at a average cost of about 5.9 cents per kilowatt hour.
The cost for Class B ratepayers in 2017 however, was almost double, coming in at 11.55 cents per kwh.
So why the big jump? Have a look at the presentation to see why and look at Slide 6 in particular where you get an inkling of how IESO views the reliability of industrial wind generation in their forward planning process!
Author parkergallantenergyperspectivesblogPosted on March 5, 2018 March 13, 2018 Categories electricity bills, energy policy, hydro bills, wind energyTags IESO, Ontario electricity costs, Ontario hydro Bills, Wind Concerns Ontario, wind energy, wind farms, wind power, wind turbines4 Comments on How much does wind power cost us?
A simple Google search "wind power is cheapest energy" will generate 1.2 million hits.
If you search "wind power is most expensive energy" you get 2.1 million hits.
Two days last week in Ontario are real-world proof of the cost of wind power, no matter what the government or wind power industry spin tells you. Tuesday, December 5th and Wednesday December 6th were two very windy days, an excellent opportunity to examine both the power generation from industrial wind turbines in Ontario and their delivered cost of power to the grid.
$$$ IESO forecasts indicated that wind could have delivered 23.8% (177,100 MWh) of total Ontario demand (755,200 MWh) via the 4,200 MW of grid-connected wind capacity.
But wind turbines have a bad habit of generating power when it's not needed (middle of the night, spring and fall) so the intermittent power must often be curtailed (constrained/wasted but paid for). It was!
$$$ The IESO curtailed 41.8% of their forecast generation meaning 74,000 MWh were not used!
Via the contracts in place with wind power companies, IESO is obliged to pay for both delivered and curtailed power at prices for grid-accepted power at $135/MWh and $120/MWh for curtailed power.
$$$ Quick math: the cost for grid-accepted wind on those two days meant Ontario ratepayers got charged approximately $22.8 million or $221.14/MWh for grid-accepted wind. That means it cost ratepayers 22.11cents/kWh (kilowatt hour), well above what the average time-of-use rates would be for the average Ontario ratepayer! The cost of the delivered wind power for those two days was almost three times the current levied* "average" cost of 8.22 cents/kWh, and 3.7 times the off-peak cost of 5.9 cents/kWh.
There's more (sorry): be assured IESO instructed OPG to spill water over the hydro dams and Bruce Nuclear to steam off nuclear power — so power from our two reliable, emissions-free sources of power generation was also wasted. OPG and Bruce will be paid for that waste and the cost will be added to our bills. At the same time gas plants (backing up wind and solar) were being paid for idling.
Those two December days also saw sales of surplus power of 93,700 MWh to our neighbours in New York, Michigan, and others for pennies of the actual cost. In all probability, we recovered around 15% of their generation costs meaning, we bit the bullet for another $10/11 million.
Just the cost of the curtailed and grid-accepted wind and the losses on our surplus exports for those two days was $32/33 million for absolutely no benefit to any of us ratepayers. If every day of the year was like those two days last week, Ontario's ratepayers would be shelling out over $6 billion annually, due to the abysmal planning and management of the electricity sector by the current Ontario government.
Imagine how far $6 billion would go to improve our health care system.
* This price reflects the 17% deferral under the Fair Hydro Act.
With only one month left in the current year, the bad news on the electricity sector keeps getting worse.
Well before the official sources such as IESO report on how much power industrial wind turbines generated and how much was curtailed (constrained, or paid for but not added to the power grid), my friend Scott Luft has published his estimates for both the former and the latter for the month of November.
As he reports (conservatively), curtailed wind in November was over 422,000 megawatt hours (MWh) — that could have supplied 562,000 average Ontario households with free power for the month.
Instead, no one got free power; the cost of the 422,000 MWh of undelivered wind power to Ontario ratepayers was $120/MWh. That $50.7-million cost for the month was simply added to the costs of the electricity bills ratepayers will be obliged to pay, while some of it will deferred to the future as part of the Fair Hydro Plan.
No doubt the wasted wind power presented itself when it wasn't needed; if it had been accepted into the grid, that extra power could have caused blackouts or brownouts, so it was curtailed. At the same time, much of the grid-accepted wind was exported to our neighbours in New York, Michigan and elsewhere, at discount prices! Curtailed wind for November 2017 compared to 2016 was almost 55% higher.
How bad is it? Let's review the first 11 months of the current year, compared to 2016.
So far in 2017, curtailed wind is about 786,000 MWh higher (+33.8%) at just over 3.1million MWh. The cost of all the curtailed wind so far in 2017 is approximately $373.6 million, or $94.3 million more than 2016 costs.
Translation: ratepayers will pick up the approximately $165 million cost of that waste via their electricity bills.
Not only are we curtailing wind and spilling hydro, but we also steamed off nuclear power generated by Bruce Nuclear at the same time we pay for idling gas plants to back up intermittent wind and solar power generation.
The cost of "greening" Ontario with unreliable and intermittent wind and solar keeps climbing, no matter what their proponents or politicians say. As ratepayers and taxpayers we should reflect on why 25% of the waste of the noted 7.6 TWh of undelivered power and its cost of $539 million (so far this year) is being deferred via the Fair Hydro Plan. And at the same time, we should recognize that we have experienced the worst possible planning in the Energy Ministry in the history of the province.
The energy sector in Ontario needs real planning by experts that will provide real value for money and save ratepayers from paying more than $500 million a year … for nothing!
That statement is included in CanWEA's recently released brochure "The Secret is Out, Wind is in".
Had the unknown author(s) at CanWEA simply looked at the Ontario Energy Board's (OEB) semi-annual Regulated Price Plan they would have noted Table 2 on page 21 of the April 20, 2017 report that the cost of a wind-generated kilowatt hour (kWh) in Ontario is shown as 17.3 cents ($178/MWh), as the cost of "curtailed" (not added to the grid) wind is also included as a cost input.
Had the author(s) also simply looked at IESO data they might also have noticed that maybe wind energy costs are not continuing to fall! Saturday, November 25th was an example: it was a very windy day in Ontario with an especially windy night. Unfortunately for the wind power cheerleaders, our demand for power from 12 AM until 7 or 8 AM was relatively low, but the wind was really blowing. That meant the 4,200+ MW of wind capacity were running at 90% (approximately) of their capacity, at the same time as Ontario's demand for power was hovering mid-way between 11,000 and 12,000 MW. That's very close to what our nuclear plants can provide on their own without help from other generation sources.
As a result, IESO ordered wind's curtailment, hydro's spilling and nuclear steam-off. At the same time, they were exporting whatever the market would take.
So, all together on November 25, the IESO curtailed 35,600 MWh of grid-connected wind and accepted 30,600 MWh into the grid, while scrambling to prevent brownouts or blackouts by exporting about 50,000 MWh over the day.
Industrial-scale wind power developers get paid $120/MWh for curtailed wind and $135 MWh for grid-accepted wind.
Ontario's ratepayers picked up the costs of almost $8.6 million for curtailed and grid-accepted wind power produced when it wasn't needed.
The cost of the grid-accepted wind (30,600 MWh) was therefore just over $280/MWh or 28 cents per kWh or, 10.7 cents more than the OEB reported back in April. On top of that, we ratepayers also ate the costs of spilled hydro, steamed off nuclear and the losses on the 50,000 MWh exported at a price close to zero.


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At only one year old, North West is officially the youngest person ever to be enlisted in the army—the Balmain army, that is.
Her mom, Kim Kardashian, and her aunt/model-of-the-moment, Kendall—hold the Jenner—were recruited by Oliver Rousteing long ago, and now the "future of fashion" is joining the ranks, because of course she is.
Yesterday before being spotted at LAX, Kim shared a snap of North's new uniform: three jackets made especially for her little mini-me from the creative director himself.
Nori wasn't wearing one of her new jackets yesterday, though. Instead, she put on for Papa West in pint-sized clothing from his Yeezus tour. A black bomber jacket covered her concert tee with a Grim Reaper graphic. With it, she also wore a tiny pair of Dr. Martens.
Mrs. West kept up with an all-black ensemble, wearing thigh-high boots beneath a skirt, turtleneck top and coat that stopped right at her ankles. See more shots of their stroll through the airport below.


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