text:
package com.pso.bigquery.optimization.exceptions;

public class QueryPatternAnalyzerException extends RuntimeException {
  // TODO: Add more domain exceptions and handle errors
  //  throughout the project

  public QueryPatternAnalyzerException(String message) {
    super(message);
  }
}

redpajama_set_name: RedPajamaGithub


MKD Electric is pleased to announce that we have been awarded the Associated Builders and Contractors- Illinois Chapter 2017 Excellence in Construction: Award of Excellence for the Alt Fuels project. MKD Electric was selected as the electrical contractor to perform the electrical and instrumentation scope of work that would result in a clean technology-alternative fuel facility for heating limestone in kilns, in order to bake lime for steelmakers. The engineered fuel is recognized by the EPA as a non-hazardous fuel that will partially replace the use of coal as fuel for the plant and significantly reduce emissions.
It is significant that MKD Electric, an ABC-affiliated merit shop electrical and instrumentation contractor, was awarded and successfully completed this project in a predominately union controlled limestone plant. Due to the success of this project, MKD has been named the preferred electrical contractor for this company, serving multiple plants in the area.


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Rick Seibold
Band Pop Singer/Songwriter
Share Rick Seibold's EPK!
"Hard Work Pays Off For Rick Seibold And His New Album "New York""
New York, New York, May 4, 2009-
Though his years in the music industry add credit to his name, Rick Seibold's new album "New York" that debuted April 28th, is a product that appears to have spoken for itself. Within the first twenty-four hours of sale on iTunes, "New York" hit #16 on the iTunes Pop charts. Meanwhile, his previous EP album "That's The Day" also rose in the ranks to #41 on the iTunes Pop charts. This lands him amongst some of the most world-renown pop artists, and serves as further testament to the fact that Rick has reached great heights without any record label. The heights he has reached are nothing compared to what lies ahead. It is clear that Rick has already risen above the sea of independent musicians, and has even begun to crash the scene once only reserved for signed musicians.
His fan base continues to surpass such artists as well. From simple accolades to full reviews, it seems that the power of a good product is undeniable. One fan says via iTunes, "If you thought you were content with just listening to Rick Seibold via Facebook or Myspace players, then it is definitely time to upgrade to the new album…[He] has done it again with a stunning blend of chill-dance-pop with rhythmic instrumentals and smartly crafted lyrics." With such widespread acclaim this early on, it is undeniable that Rick will continue to rise in the ranks. - None
"Myxer Pick of the Day"
With a smooth swagger, and the natural ability to pen a perfect pop tune, today's featured artist, singer/songwriter Rick Seibold, is well on his way to the top! His upbeat, percussion-driven ditty, "Mary," is today's featured ringtone, and for good reason, it's a serious jam! Rick sounds as if he could pull off the perfect duet with Jack Johnson or Dave Matthews, and that my friends is a compliment of the highest order! Download today's ringtone and have a listen to this amazing talent on the rise! - Myxer
"Rick Seibold's New Album "New York" Slated To Hit iTunes With a SMASH!"
After months of toiling away in the studio, Rick Seibold's new album "New York" hits iTunes today. Sans record label, Rick creates a buzz unlike any other independent artist. His unique formula of online social networking, and a grueling schedule of shows have proven to pay off. His previous EP album "That's The Day" reached #3 on the iTunes pop charts, and garnered global attention. In the coming months, even greater success is expected of his new album, "New York."
This album has risen from the tenacity and passion Rick quickly found when he first took up residency in the city that never sleeps in 2005. In hopes of becoming a world-renowned musician, Rick gained knowledge of the industry while working for music giants MTV and BET. His nights were then spent performing at any venue possible, whether it was a bar, club, or street corner. In 2008, he returned to his roots in Nashville, TN to write and produce the album, and the release of "New York" marks his triumphant return to the Big Apple.
"I wanted to make a record that captured my experience of living in the bustling city of New York", Rick says. It seems evident that he has done that and more. From track to track it is as if Rick is not only offering an outlook on city-life, but also a fresh perspective on life as a whole. Perhaps the album's message will resonate with you, just as it has with his masses of loyal followers and fans. - Crowdsurf
"Rick Seibold Defies the Cliche"
Growing up in the small town of Wilmington, North Carolina, Rick had big dreams. But his story defies the cliche of such a story. With the help of social networking sites, viral videos, and no record label, Rick has generated an enormous fan-base. "It really comes down to people loving the music and then passing it on to their friends," Seibold says. "That's how this whole thing grew so fast." His debut album, "That's the Day," has spent twelve weeks in the iTunes Top 100 Pop Charts. This puts Seibold's first record among the top selling independent albums of 2008, reaching as high as 27 on Jan 26th and marking nearly 25,000 units since the beginning of the year. In addition, "That's the Day" reached as high as 3 in Canada and 18 in the United Kingdom. Rick's internet fan base flocked to him for his fun, easy going Jack Johnson or Colbie Caillat-like sound and lyrics inspired by his upbringing on the beaches of Wilmington, NC. More recently he has called Brooklyn, N.Y. and Nashville, Tenn. home as he hones his craft with mood-changing melodies and catchy choruses "Rick Seibold presents songs about love, learning and coming of age with a knack for smart and often catchy melodies... it's sure to make any young female listener swoon," says Alyssa Rashbaum (MTV, Spin, Vibe). "You know, I hope my music will be something comfortable for anyone listening to it," Seibold says. "Whether you're in the middle of a New York City blizzard or a Seattle rain, I hope you can put on my record and it'll make you feel like summer time." - Addicted to Noise
"Rick Seibold - That's the Day"
Rick Seibold hails from a small town in North Carolina, where he first got into music in middle school when he started playing the clarinet. By the age of 15 he was playing the guitar when a wrestling accident blew out his knee and he needed something to keep him occupied while he was in a cast. This led to him starting to write and record songs with his friends.
Taking his music from the things he has experienced and the places he has seen has only fuelled his creativity for creating infectious soulful songs. When a friend told him that the best songs came from just living life and that the more you live, the more you have to write about Rick decided that's what he would do.
The songs on his EP That's The Day are an example of this. Rick has put his life experiences and the songs of his heart out there for all of the world to hear only so that we may connect to our hearts in a much deeper way. Let's hope this MySpace budding artist does that for you. So far Canada seems to be letting him do that. His album peaked at number three on the iTunes Canada Pop Charts. - The 405
"Red Carpet Spectacle"
The red carpet. The flash of light bulbs. Hundreds of people straining to see the fashions.
The crush of people at Pennsbury High School's Saturday night prom was a lot like a prestigious awards show, with one significant difference.
Pennsbury's prom is so big that celebrity singers have entertained there in the past, including John Mayer. This year's big surprise entertainer was Rick Seibold, a singer from Nashville. - Courier Times Philadelphia
"Artists to look out for - Rick Seibold"
Thanks to modern technologies such as MySpace, musicians have been given a new output to show off their talents- most without even having to be signed!
My next planned series on My Two Pennies involves bringing to you highly recommended artists from these sites. These interviews are exclusive and are what the artist wants you to read about them, from where they came from to how they got into music.
First is Rick Seibold, with 7835 friends already on MySpace, he looks set to be riding 2009 to it's full extent.
"Rick Seibold presents songs about love, learning and coming of age with a knack for smart and often catchy melodies... it's sure to make any young female listener swoon." (Alyssa Rashbaum, MTV, SPIN, VIBE)
"Hey I'm Rick, I grew up in Wilmington, NC. Moved to New York City a few years ago and started playing around town in the local clubs. A little over a year ago I decided to make the move to Nashville to start recording a new record which I'm finishing up right now. It will be out this spring, probably in April."
"I started playing guitar when I was 15, and began writing songs as soon as I knew my first few chords. Where I grew up, there was a pretty vibrant film scene. When I was about 10 I played one of Anthony Hopkins's kids in a movie where I had to sing 'Silent Night', so I guess that's when I officially got my start!"
"Growing up at the beach in NC proved to be one of the early major influences on my music. I love the ocean, and have tried to write about my experiences growing up there as much as I can. Moving to New York City, and now to Nashville, has also had a major impact on my writing. Someone once told me if Jason Mraz and Chris Brown had a kid, it might look like Gary Coleman. But if they made an album, it would probably sound a lot like me."
"Look for the new record on iTunes in April."
Rick's first EP, 'That's The Day' is on sale on the iTunes Store and has achieved getting into the Top Three on the Pop Chart! - My Two Pennies
"you should know about… rick seibold"
rick seilbold is an indie artist originally from north carolina, but is now in nashville, by way of new york. follow that? :) rick writes great pop music that will make you smile and want to hit 'repeat' and if you're looking for cool new music, you should know about rick seibold.
his dope new ep "new york" releases april 28, but the first 100 people to pre-order can download the whole record now. check out the 'new york' ep sampler here, then befriend him on myspace, facebook and twitter. - The Gideon Chronicles
"That's the Day"
"On his debut four-song EP That's the Day, Rick Seibold presents a sort of grown-up college coffeehouse sound, with songs about love, learning and coming of age – no surprise since the singer has spent the last few years living from a suitcase, traveling, and, well, learning about life. "Mary," with its bongo-driven beat is about a girl who "claims shopaholic as a disease" and who will soon find "everything you need your cash can't buy." "Can't Go On" is an acoustic guitar-led track with observatory lyrics like "what good is life if it ain't with friends" and "summertime is a state of mind." Seibold also has a knack for smart and often catchy melodies as on "That's the Day," which is highlighted by lush instrumental and is by far the most textured song on the project with a smooth keys and swirling harmonies and the EP's closer, "Take it Slow." A fitting way to end the debut, "Take it Slow" is a sort of stripped-down radio-friendly pop song, a calling card to A&R men - with its raw strumming, guitar slaps and lyrics like "Baby would you tell me those secrets you've got locked inside / Cause baby I want to tell you some of mine," it's sure to make any young female listener swoon. "
- Alyssa Rashbaum (MTV, SPIN, VIBE)
Both Rick Seibold albums, 'New York' and 'That's the Day', available on iTunes and www.myspace.com/rickseibold
'Acoustic New York' Available July 28th, 2009!
Check out www.myspace.com/rickseibold for upcoming shows
For Rick Seibold, New York isn't just the name of his new EP, it's also the city that represents a pivotal time in the singer-songwriter's life. His sound has been shaped by struggle and his passion for his craft permeates through every note.
Raised on the East Coast in Wilmington, NC, Seibold's creativity began budding by the age of 10 when, at the suggestion of his parents, he began acting in films produced out of Wilmington's burgeoning film scene, including a part as one of Sir Anthony Hopkins' children in the 1994 film "The Road To Wellville". Spending many of his childhood summers on the North Carolina beaches, Seibold took up surfing and the guitar around age 15 where the sights and sounds of beach life quickly settled into his songs.
In winter of 2005, Seibold moved to New York City to take an internship at MTV. Over the course of following three years, he experienced more than his share of the requisite ebbs and flows that quite often characterize the life of a focused and passionate musician living out their dream on the mean streets of New York City. The bright lights and big city that first lured him soon gave way to a nomadic lifestyle that often found him without a place to live. When he was short on cash, you could find him performing in a nearby subway or park, playing and singing until he was hoarse. While his days would be spent working at MTV and BET, his nights were often filled with spot gigs at various bars and clubs, playing until the sun came up.
"Looking back, I'm not even sure how I made it," says Seibold, reflecting on this time in his life. "A friend of mine once said that the best songs are written from just living life, so that's what I was trying to do." And live he did.
Seibold channeled his struggles into a passionate pursuit of his music and, at the insistence of a friend, ultimately released his first collection of songs in EP form in 2007. His debut, That's The Day, would spend an impressive four months in the iTunes Top Pop Albums chart in 2008. Led by the first single, "Mary", his music found an audience with booth Canadian and UK listeners, garnering him an unexpected fan base north of the border and across the pond.
A 2008 move to Nashville has yielded a second collection of songs that marks not only the next chapter of Seibold's life, but also a stark stylistic departure from the acoustic simplicity of his previous release. The new project, New York, was written and largely produced by Seibold himself, and finds him settling firmly into a wildly hooky pop music sound. He has been compared sonically to that of One Republic, Rihanna, and even the Jonas Brothers.
Rick Seibold on iTunes
Rick Seibold's Facebook
Rick Seibold's Myspace

redpajama_set_name: RedPajamaCommonCrawl


A team of researchers has developed a new genetic barcoding technology which can be used to help identify and diagnose conditions and diseases within our genetic code.
Researchers have reportedly developed a new genetic barcoding technology that will enable the simultaneous analyzation of hundreds of genes.
For years, scientists have struggled with characterizing the individual functions of the 20,000 set of genes in the human body. Without this vital information, scientists remain clueless about how the human genome really works. To address this issue, a team of researchers from the Icahn School of Medicine at Mount Sinai developed a new genetic barcoding technology to determine the roles played by each gene in the body.
There is no doubt that the introduction of the gene-editing technology CRISPR has revolutionized the field of genetics. The CRISPR approach not only led to some groundbreaking development in curing genetic disorders but also helped researchers determine some gene functions. However, CRISPR has its limitations and is still not enough to track the tasks of the many thousands of genes in the body.
The new barcoding technology developed by the Mount Sinai researchers uses synthetic proteins known as epitopes to address the challenges of DNA-based barcoding like CRISPR. The researchers, led by postdoctoral fellows Aleksandra Wroblewska and Maxime Dhainaut, reported that they encoded triple combinations of linear epitopes into synthesized modules to produce over 100 unique protein barcodes which they called Pro-Codes.
"Using just 14 antibodies, we detected 364 Pro-Code populations; establishing the largest set of protein-based reporters. By pairing each Pro-Code with a different CRISPR, we simultaneously analyzed multiple phenotypic markers, including phospho-signaling, on dozens of knockouts," the researchers explained.
The team introduced the Pro-Code/CRISPR libraries to breast cancer cells, challenging the tumors with killer T cells specially engineered to recognize cancer cells. The experiment showed that most of the cancer cells were successfully eliminated by the T cells, but there were some that survived.
The Pro-Code barcoding technology helped the researchers identify the genes that were affected by the resistant cells, some of which were found to have unrecognizable roles in sensitizing cancer cells in the immune system.
At the moment, there are still fewer Pro-Codes produced as compared to DNA-based codes. However, the researchers claimed that more Pro-Codes could be created using their technique.
"Moreover, 1,000s of new Pro-Codes can be created by applying the principle we used here and introducing additional epitopes and epitope positions. Although generating genome-wide Pro-Code/CRISPR libraries cannot be done at the relative ease of DNA barcoded libraries, the Pro-Code will primarily be for more focused screens; concentrating on specific pathways or gene classes and targeting 100 to 500 genes," the team wrote.
The scientists also claimed that it is possible to produce Pro-Code/CRISPR libraries without overlapping epitopes that can be used in complex screenings for identifying redundant genes. The novel tool could also perform high-dimensional phenotyping of multiple proteins in pool screens, a feat that is reportedly not achievable with DNA barcoding.
Do you believe that developing genetic barcoding technologies would help us completely understand how our genome works?


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text:
Home Thoughts Opinions Obamacare and Keystone Pipeline At Top of Republican Agenda
Obamacare and Keystone Pipeline At Top of Republican Agenda
Dwight L. Schwab Jr.
Fri November 7, 2014
President Obama wasted no time showing his defiance following the election night slaughter of Democrats. He charged out Wednesday morning to declare he had no intention of signing a Republican bill to overturn his signature healthcare legislation (Obamacare) but would be willing to work with the GOP on other issues.
Although there has been no discussion on the subject following the election, the president wanted to take a stand after his party suffered one of the single biggest one-sided elections in history. Whether or not Obama plans to fight, or perhaps compromise parts of the Obamacare law, his ire was evident just hours after the humiliating defeat.
As the president said, "On healthcare, there are certainly some lines I'm going to draw. Repeal of the law I won't sign," he said at a hastily-called White House press conference. He went on to say he would resist efforts to weaken the law piecemeal, as he suspects the Republicans will do now emboldened with a stunning victory.
The press conference was almost surreal with the President of the United States providing a defiance that appeared shallow considering the Republicans had just taken the Senate majority and added to their majority in the House. However, a glimmer of compromise was provided when the president lamented he would not be adverse to "responsible changes" to "make it work better. There's no law that has ever been passed that is perfect."
Shortly before Obama's press conference, the new Senate Majority Leader, Mitch McConnell, made his feelings known on the subject saying, "If I had the ability, obviously I'd get rid of it. It's no secret that every one of my members thinks Obamacare was a huge legislative mistake."
The first few months after the Republicans take control of the Senate will be critical to establishing what they promised – a new order. McConnell is fully aware of the need for Republican unity and the ability to attract allies from across the aisle. The new majority leader said he would attempt to take out some of the more damaging parts of Obamacare "that are tremendously unpopular with the American people."
Obamacare aside, one of the first tests could be a bill to approve the Keystone XL crude oil pipeline from Canada, that has been stalled for years by the Obama administration. Republican Senator John Hoeven said in an interview on Wednesday that he has enough votes to pass a bill early in 2015 that would approve TransCanada's long-languishing $8 billion pipeline project.
It will be the first sign that gridlock is coming apart in Washington. Obama said at his news conference he would let the State Department-run process on Keystone play out. In actuality, there is nothing to "play out," but rather the votes that are now there to pass it.
It is no secret the Canadians supplying the oil are ecstatic about the election results. Energy markets look to the Republicans to lead the reform of crude and natural gas export laws and motivate the Obama administration to include those energy exports in new, or broader, trade agreements. It will be interesting to see how it all turns out now that the political environment has changed so dramatically.
Dwight L. Schwab Jr. is a moderate conservative who looks at all sides of a story, then speaks his mind. He has written more than 3500 national political and foreign affairs columns. His BS in journalism from the University of Oregon, with minors in political science and American history stands him in good stead for his writing.
Dwight has 30-years in the publishing industry, including ABC/Cap Cities and International Thomson. His first book, "Redistribution of Common Sense – Selective Commentaries on the Obama Administration 2009-2014," was published in July, 2014. "The Game Changer – America's Most Stunning Presidential Election in History," was published in April 2017.
Dwight is a native of Portland, Oregon, and now a resident of the San Francisco Bay Area.
Request Notification of Dwight's Stories


Official Drink up Grinches it's Christmas shirt, hoodie, sweater, longsleeve.
Because Grace wants to spend quality time with my sister Marlin, as feminist females, and as matriarchs, as they used their succubus's and Fallopian to attach to my kidneys and pancreas Marlin laughed at recalling a dream that she caused my diabetes by punching me in the pancreas, to turn me into a fag and die of AIDS. You keep voting Drink up Grinches it's Christmas shirt . And guess what. Your not getting any federal aide. You know why. Because reds don't believe in it. You get what you vote for. I live in Texas, this is the beginning of our rainy season. In 2014 it literally rained at least once a week from the beginning of November into mid March.
Drink up Grinches it's Christmas hoodie,ladies shirt, longsleeve and sweater.
Some weeks it would rain 5 of 7 days with a steady downpour for hours. This is why climate change and global warming does not make sense to a lot of people. Most people over the age of 60 have seen mass changes in the climate and weather. Most people born 1980 s and later to them this is normal. This is not going to matter. We are coming out in droves to vote for Stacey. You'll have to find Drink up Grinches it's Christmas shirt to win here in GA. We are all coming for these positions, to actually make America normal again. This guy is how the new Trump GOP operates. If you can't win on your own merits, cheat and lie. Vote everyone of these trump supporting crooks out of office.


Now that the weather is warming up, who has time to be slaving in the kitchen? Outdoors is officially the place to be because it is gardening season! While it is not quite *that* warm here is New England, I am eager to start my window boxes, and vegetable garden. I guess, I am a little ahead of things and will have to keep dreaming for a few more weeks.
In typical fashion, my cooking and baking (or lack there of) has been focused on lighter meals, and no-bake desserts. These Oreo Cookie Ball Carrot Truffles are a perfect example of a great no-bake dessert for Spring or Easter. By using a few common ingredients and some creativity, I was able to make the perfect treat for my kiddos.
I love no-bake desserts like these Oreo Cookie Ball Carrot Truffles because they are super easy and very quick to make. I literally just combined a package of Double Stuff Oreos and 1 package of cream cheese together and then formed the mixture into little balls. I then refrigerated them for one hour and then dipped them into orange melted chocolate.I had extra orange morsels from Halloween so that saved me the time of adding orange food gel to the chocolate. For the green, I used a little green sprinkles and that was it..easy peasy.
Heat Orange morsel, or white chocolate candies in he microwave for 2-3 2o second intervals; stir in-between each.
Remove cookie balls from the fridge and dip each one into the melted chocolate; place on baking sheet, and top each with a pinch of green sprinkles; repeat.
Even the official "Easter bunny" in our home approved!


Terraces at South Pasadena offer quaint boutique living in the heart of the city, with comfort and style at every turn. Terraces at South Pasadena's one- and two-bedroom floor plans were thoughtfully created with your convenience in mind...from oversized balconies, large windows, to let the light and nice breeze in and view of lush landscaping as well as the resort-style pool to give you a relaxing atmosphere, making you feel glad to be home.
Terraces at South Pasadena resides in a prime location in the very heart of South Pasadena and in close proximity to major freeways and some of the finest shops, restaurants, and entertainment of Old Towne Pasadena and Paseo Colorado Shopping Village...not to mention within reach of the well-renown Rose Bowl Parade. Living at Terraces at South Pasadena mean living with utmost convenience and fun at your very doorstep.
Whether they come with short hairs or long, from wagging tails and slobbering kisses to snuggly purrs, Terraces at South Pasadena understands that your pets are your family, so we do what we can to not only make Terraces at South Pasadena your home, but theirs as well.


text:
using namespace Syntax::Compiler;
using namespace Syntax::GraphicsCompiler;

EffectsCompiler::EffectsCompiler(void)
{
}

EffectsCompiler::~EffectsCompiler(void)
{
}


text:
Pattuglia acrobatica (Corn on the Cop) è un cortometraggio Merrie Melodies uscito nel 1965, diretto da Irving Spector e scritto da Friz Freleng. Questa è la missione poliziesca per Daffy Duck e Porky Pig.

Distribuzione

Edizione italiana

Esistono tre doppiaggi italiani per il corto. Il primo risale al 1966, distribuito in lingua italiana nelle sale cinematografiche. Il secondo realizzato negli anni ottanta dalla Effe Elle Due e nel 2003 il corto è stato ridoppiato per la TV. Da allora in poi viene usato sempre il ridoppiaggio.

Edizioni Home Video

VHS

Il corto è concluso in videocassetta, solo con il primo doppiaggio.

Collegamenti esterni

Cortometraggi di Daffy Duck
Cortometraggi di Porky Pig
Cortometraggi Merrie Melodies
Film su Halloween

redpajama_set_name: RedPajamaWikipedia


text:
Αlpha Ethniki – denominazione, fino al campionato 2005-2006, del primo livello del campionato greco di calcio, in seguito ridenominato Souper Ligka Ellada.
 Αlpha Ethniki – primo livello del campionato greco di calcio femminile


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Posted on 11th June 2018 by Duncan Pollard
Football, Identity and Suffering
Official YIFA merchandise!
They say it's the beautiful game. Certainly that's why I chose my football team. I wanted to watch attractive attacking football. It didn't matter that for more than 15 years of me first supporting them they did not win anything. Then it all changed and they took the league by storm and not only did the attractive football flourish, but so did success – all the way to the Champions League.
FROM ILLUSION TO IDENTITY
However, the "beautiful game" is just an illusion. Week in week out, the majority of football fans suffer. They stand on the terraces whilst at best their team grinds out a draw. For them the beauty of football is the expectation, the elation of the occasional, unexpected 3 nil win, or the avoidance of relegation on the last day of the season. Mostly it's the suffering of what might have been. The memory of past glories. At the other extreme too much success gets wearisome. Football is only authentic for the majority when the majority of it is infused with suffering. And the rest of it is improbable (like Leicester City winning the league).
So for the last few years I have been looking for a second team to support. One that once won the league or a cup back in the 1950's but now plays in a lower division. Week in week out. Where the supporters turn up in expectation, but go home with the realization that the whole football thing does not really matter. It's been 90 minutes of escape, and there is always next week, or next year.
My friends are helping me, with plenty of suggestions. One has even suggested that I stick with my current team as they don't seem to be winning things these days. Ouch! The thing is, that I am struggling to get excited with any of the suggestions. I want a team I can identify with.
CONIFA 3 – FIFA 0
With the 2018 FIFA World Cup about to start, the England national side could fit the bill. England had glory in 1966, and have done nothing since then (OK there was that save by Gordon Banks, Lineker in front of goal, and that 4-1 victory over the Netherlands in Euro96). England therefore fits the bill, but when international competitions come along my interest is also drawn to some of the other countries I have lived in. I especially identify with Chile, Uruguay and Switzerland. Especially Uruguay, who have won the World Cup twice, but a long time ago in an era when all you had to do was turn up and you'd get a game.
So I have been really intrigued by CONIFA. The third CONIFA World Football Cup has just finished in London. Kárpátalja beat Northern Cyprus in the final 3-2 on penalties. CONIFA comprises 47 Football associations representing states, minorities, stateless peoples and regions that are not affiliated with FIFA. Kárpátalja represents the Hungarian people living in western Ukraine. Identity is a big thing for these teams.
No one can force an identity upon you. It's your own. And that's the beauty of CONIFA. It uses football to bring peoples together. The football suffering pales into insignificance when the real world suffering of some of these nations is considered. Football is rightly put in its place, whilst at the same time it becomes a place for a better understanding of the very real issues of nationhood and identity.
The FIFA World Cup has cost US$10Bn to organize. I can think of better uses for this sum of money. Meanwhile the CONIFA World Football Cup games, featuring Ellen Vannin (the Isle of Man), Abkhazia, Tibet, Kiribati and others, were played at venues in London such as Gander Green Lane, Sutton, and Coles Park, Haringey.
CONIFA aims "to give football outsiders overseen by FIFA or left behind by their mothers country FA the chance to win their place on a global stage and advance football-wise and personally…. We give them the chance to play for the entity they feel part of in the bottom of their hearts". CONIFA wants "to bridge the world and bring people together from all over the world… [leaving] all politics behind". This feels more genuine than the highly commercialized, corrupt FIFA.
GOD'S OWN COUNTRY
Now it just so happens that Yorkshire is a member of CONIFA. You may remember that if Yorkshire was a country it would have come 12th at the London Olympics, and 14th at the Rio Olympics. Seven of the 23 players in the 2018 England World Cup squad were born in Yorkshire.
Yorkshire were not at the CONIFA World Football Cup this time around – they only joined the organisation this year. Nevertheless, I have my Yorkshire football shirt and hope to be able to support the team from God's Own Country get to the next competition, whilst providing some degree of solidarity with the other outsider nations present. With a bit of luck Raetia will organize it so I won't have to travel too far.
CategoriesOther Stuff TagsCONIFA, identity
2 Replies to "Football, Identity and Suffering"
KeighleyClaret says:
I agree with the sentiments and also have Yorkshire as my second team – however it is sad that CONIFA chose to expel Ellan Vannin over a dispute about an illegible player after they had already been eliminated. CONIFA needs teams like Ellan Vannin and Yorkshire should refuse to play until the ban is repealed.
Duncan Pollard says:
Yes it's a bit harsh. Hard to see FIFA doing that, for obvious reasons. Then again, hard to see Ellen Vannin getting into FIFA for the reasons that CONIFA exists…. hopefully CONIFA see sense
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Bits and pieces, opinions and things that don't fit elsewhere.
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The Way to Work
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2017 Booklist
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Tim Dalton Dunn; Kathleen M. Liuzzi; Dunn & Dunn; Attorneys for Appellees.
Edward P. Moriarity; Bradley L. Booke; Jacque M. Ramos; Terry Mackey; Moriarity, Badaruddin & Booke; Attorneys for Appellants.
Brief of Appellee, Donatelli v. Beaumont, No. 20080020 (Utah Court of Appeals, 2008).


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text:
[En images] Près de 300 personnes rassemblées contre les lois liberticides il y a 3 minutes par Dijoncter
Home Punk scene > Music, punk scene & subcultures >
'77 punk
Discussion in 'Music, punk scene & subcultures' started by punkdude, Sep 3, 2009.
punkdudeMember New Member
what do you guys think about '77 punk??
Sex pistols, ramones, the clash, sid vicious, ...
punkdude, Sep 3, 2009
AndersBActive Member Forum Member
I think it's great. It's always a good idea to check out what it all started with. They may not be to everyones liking, but they were pioneers in their own sense.
To survive in modern society is like jumping out of a skyscraper and be happy all the way down - Arne And
AndersB, Sep 3, 2009
ungovernableAutonome Staff Member Uploader Admin Team Experienced member
Sex pistols: Sellouts
Ramones: Sellouts
The Clash: Sellouts
Sid Vicious: A dumbass and a sellout. (never understood why so many co-called punks worship him like he was a hero). On top of all, he couldn't even play bass (he was playing like shit)
REAL punk started with bands like crass, conflict, flux of pink indians, ...
ungovernable, Sep 3, 2009
ghost in the voidExperienced Member Experienced member Forum Member
AndersB said:
what a curious thing to write. most historians generally agree that punk started with MC5 and THE STOOGES when refering to the modern punk "sound". a lot of people also believe that punk started in the 60s, not the late 70s. furthermore, the use of the word "punk" dates back to the inquisition, and has the same social implications as the word "faggot". faggot meaning fallen wood in forests, punk being the fungi that grew on such wood. this wood was used to burn heretics.
the word punk is generally used to describe outsiders to the mainstream, and as such, is very old. the culture of punk didn't start in the 1970s or the 1960s, but several centuries beforehand. i am really very suprised at how few people realise this.
ghost in the void, Sep 3, 2009
SpiderExperienced Member Experienced member
in terms of the punk of that era, i always liked the saints more than the pistols. I felt like the pistols were always putting forward what they percieved to be a punk persona, (ie "I am an Anarchist" hey so am i lets get behind this band sorta thing) i get a much stronger feeling of legitimacy and honesty from the saints. I suppose from a postmodern perspective where everything is derivative you could say that whilst these bands may not have been completely sincere with their percieved persona (see "Gangsta Rap") the bands that they helped to inspire took the ideologies and DIY ethics which they had glamourised, and put them into effect.
Vive le punx
The cracks are getting bigger
Spider, Sep 3, 2009
bulavkaActive Member Uploader Forum Member
ungovernable said:
earlier, a few years ago, I thought that the sex pistols - a real punk. in our country they are very popular among false punks. They are popular because the media made them so. and they are immoral. because of this group in our country punks - young people who are not interested in anything but a lot of drink, smoke, behave immorally, listen to bad commercial music with silly lyrics, and some of them are fascists. Sid - is a bad example for punk rock. and he did nothing to punk, to music, unlike other groups, such as Crass.
but, I like The clash. and like film about Joe Str.
I like punk 77 and respect the many groups, but I'm listening this music is not very often
"If you feel dirty, insignificant and unloved, then rats are a good role model. They exist without permission, they have no respect for the hierarchy of society, and they have sex fifty times a day." © (Banksy)
bulavka, Sep 3, 2009
The only song i like from The Clash is "Guns of brixtons", but Inner terresterials or Bolchoi cover is way better....
But oh well, The Clash is comparable to Sex Pistols... Anarcho bands hated them equally... Just read the lyrics of "punk is dead" by Crass
Worst thing is that sometimes i see punks who thinks the Sex Pistols are real anarchists just because they sing "anarchy in the UK"....
the funny thing is the 'PISTOLS actually inspired CRASS, because they wanted to take the play-acting and make it real. i actually don't mind SEX PISTOLS or THE CLASH at all, i just consider them major label rock bands rather than punk per se. THE EXPLOITED however are rubbish, because they actually believe they're a punk band and don't have the sense of irony of the other two. the SEX PISTOLS are actually more like an invented "boy band" like THE MONKEES before them, and NEW KIDS ON THE BLOCK and N-SYNC etc that followed. call him what you want, but their manager malcolm mclaren was very clever at marketing, and so was his partner vivian westwood, her brand now of course a major player in the fashion scene.
sociopop82Experienced Member Experienced member
i had my fun with it but 77 was a long time ago.
it's irrelevant to me.
the saints are great by the way
so are some older garage bands. namely
the sonics.
Escape from Hell.
sociopop82, Sep 4, 2009
professerslimeMember New Member
I also believe that "punk" predated Sex Pistols and the Ramones. Not just as a definition meaning outsider or a rebel. It also existed in late 60's bands that the corporations wouldn't touch because they were to radical or just unmarketable. Ever since they came up with the term latch key kid after WWII there has been a wild anti-authoritarian element to our social conscience. A lot of the mythology was inflated by all those exploitation films that came out in the 50's and the 60's. Fuking Hollywood. But in conclusion there always has been rebellion and dissent only now there is a music and a label attached to it.
professerslime, Sep 6, 2009
FredzillaNew Member New Member
I agree with the sex pistol things , but who are we to decide and believe what is real or what is fake ??? If someone is happy to be '' IS '' way , why not ???? I know it's bad for the scene and doesn't bring any good to the serious ''follower'' . But stil , beeing a '' PUNK ROCKER '' doesn't mean liberty of expression ???!!!??? The word '' POSEUR '' only apply to those who are deciding that you are a '' POSEUR '' , somewhere inside them , they are jealous of something ! About what ? I don't know , but they sure are jealous , they won't admit it but anyway it's how the human was made and it's how he keep growing the mentality ..... who's ever eard the sentence : live and let die .........So cheers and have a good life (don't worry i hate the human race more than i hate cockroachs or bed bugs...)
Fredzilla, Sep 7, 2009
mel_the_bellExperienced Member Experienced member
well the first punk record i ever got was a sex pistols single so i cant say they never influenced me or i never listened to em
also 70s punks great for say a jukebox in a pub............bound to find the adverts, or sex pistols or slf on it...but NOT crass, flux or rudimentary peni lol
Some say 'Get a Life' but I'd rather wait until I can download a pirated version of one with a working KeyGen
mel_the_bell, Sep 7, 2009
NGNM85Experienced Member Experienced member Forum Member
If the Pistols "sold out", which is debateable, it wasn't until the 1996 reunion, they didn't make any significant amount of money between the forming of the band, and the breakup in '79.
Umm, I might mention that the Clash forfeited potential earnings to keep the album prices low, they also wanted to be remembered for their political songs, not the poppier songs, many of which they wanted to be released as B-sides. I'd also question your definition of "sellout."
Since when does musicianship matter? Although, Glen was much better for the band,.
Negative. Thats' when ANARCHIST punk started. Punk is NOT Anarchist and Anarchism is not necessarily "punk." They have similar tendencies and commonalities but thats' it. Crass are no MORE punk than the Germs, or whoever, maybe less. I could care less, musical taste is pretty subjective, but Anarchists can't honestly claim any ownership of punk rock.
"Kick over the wall 'cause government's to fall
How can you refuse it?
Let fury have the hour, anger can be power
D'you know that you can use it?"-The Clash, "Clampdown"
NGNM85, Sep 9, 2009
I was going to let it go but this started bugging me, I was just going to amend my first post, but I can't find an "edit" feature. (If anybody knows how to do this I'd REALLY appreciate it.)
Again, not a personal attack, and musical taste is a persons' business, you could rock out to Britney Spears for all I care.
But of all the bands you could've gone after you choose THE CLASH?! Personally, I prefer the Pistols, even they produced very little material, and were much more limited musically, however...
The Clash forfeited so much money to keep their record prices low, they didn't see any real profits until they'd been together for SIX YEARS! I've never heard of a band doing that. They also supported the Anti-Nazi league and played a ton of Rock Against Racism gigs. The Clash dedicated a triple album to the socialist rebels (And democratically elected leaders until Reagan overthrew them.) the Sandinistas showing solidarity with them. Thats' not even mentioning all the anti-authoritarian, anti-corporate songs they wrote, (Even "Rock the Casbah' was a political song.) "Straight to Hell", "Clampdown", "White Riot", "Hate and War", "Police and Theives", "English Civil War", "Career Opportunities", "Police on My Back", "Washington Bullets"....etc., etc. If anything they established a formula for Crass to follow. The Pistols or the Ramones you can argue about, but the Clash,..it's absurd.
ExtinctionExperienced Member Experienced member Forum Member
That is a good point, but everybody is entitled to their opinion. Personally my favorite 77 band was The Damned. People never talk about them enough. They also got me into goth rock. Very underrated.
Everybody knows the bird is the word!
Extinction, Sep 9, 2009
Extinction said:
i think my fave 70s "punk" band would have to be the ruts...think i enjoy listening to them the most tbh
@NGNM85: Well i don't know much about The Clash, you seem to know it a lot better than me... I didn't know they did all this stuff, the band is better than i thought At least if we compare them to the Sex Pistols and Ramones, they are less stupid and more involved in activism...
i agree that THE CLASH are quite important. i don't think "gateway" major label bands should be completely dismissed by the underground. people can dis bands like them, BAD RELIGION, ANTI-FLAG, RANCID etc etc, but these bands offer a way for people to realize punk still exists, and on doing their research they can discover DIY and become involved.
"rock the casbah" is one of my favourite songs. i found it on a jukebox in a tiny opal mining town when i lived in the desert, and i danced the night away drunk as a satyr. if punk had to be strictly underground all the time, i would indeed have to have been rocking out to ms spears instead.
ghost in the void said:
This is a really good point.
I like to take it even further, being in a Pop-Punk band, I would say I'm probably producing the most entry-level punk music going around. Your average teenager who listens to Fall out boy can hear us and understand where we are going. By giving them this opening, we can hopefully start getting them to shows, and down the punk path. We've played shows with bands like Face-Eater, Collapsed, The Seduction... Heavy bands. People who came to see us were exposed to something completely foreign to them, I hope some of them decided to look further into it.
A big moment for me in my discovery of Heavy punk/Hardcore was when I was 16 I went to a For Amusement Only Show (A Melbourne Pop-Punk Band at the time) and Mindsnare( Melbournes Gods of Hardcore) Played right after. They said there was another hardcore show at a house after the gig and i was so blown away by the intensity of it that i ended up going and that started me on the path that eventually led me to anarchism.
Sorry to hijack thread I'm just really passionate about Punk at all levels being important for the survival of the scene.
Spider, Sep 10, 2009
Actually, Anti-Flag have yet to put out a major label record. Their previous efforts (I HIGHLY recommend "Die For the Government" it's AMAZING.) were on indie labels like their own AF Records, or Side One Dummy. Most of Bad Religion's catalog is also distributed by indie labels, like Epitaph. The Offspring didn't join the majors until "Americana",..which was also a great album, after releasing the self titled debut, "Ignition", their breakout album "Smash", and "Ixnay on the Hombre" all on indie labels. Rancid didn't go major until the mediocre "Indestructible." Rancid's "Out Come the Wolves", Offspring's "Smash', and Green Day's "Dookie" (Their major label debut.) came out around the same time, (All fantastic albums, btw.) with "Smash" setting the record for biggest indie record of all time, nearly killing Epitaph with sales, but also propelling punk rock onto the mainstream radar. "The year punk broke." What bothers me is that people can't seem to distinguish between selling records, and selling out. This thing with the Clash was a perfect example. Just because a band is well-liked does not automatically make them industry prostitutes. Theres' just this bogus reactive tendency in punk rock that if something outsells the Casualties latest it must be manufactured corporate crap.
Back to 70's punk…..
NGNM85, Sep 11, 2009
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text:
\section{Introduction}\label{intro}

A planar curve is an immersion of $S^1$ in ${\mathbb R}^2$.
The study of invariants of planar curves has been initiated by V. I. Arnold in \cite{a1},\cite{a2},
where he presented the three basic order 1 invariants $J^+,J^-,St$.
Various explicit formulas for these and other invariants appear in \cite{cd},\cite{p1},\cite{p3},\cite{s}.
The study of order 1 and higher order invariants has in general split into the study of $J$-invariants,
as in \cite{g1},\cite{g2},\cite{l},\cite{p2},
and $S$-invariants as in \cite{k},\cite{t},\cite{va},
where $J$-invariants are invariants which are unchanged when passing a triple
point, and $S$-invariants are invariants which are unchanged when passing a tangency.
But as we shall see, the space of invariants spanned by the order 1
$J$- and $S$-invariants is much smaller than the full space of order 1 invariants.

In this work we give a complete presentation of all order 1 invariants of planar curves.
Since the vector space in which the invariants take their values is arbitrary, the presentation
is in terms of a universal order 1 invariant.
The corresponding description for spherical curves, i.e. immersions of $S^1$ in $S^2$, appears in \cite{n6}.
Analogous study for immersions of surfaces in 3-space has been carried out in \cite{n1}--\cite{n5}.

The structure of the paper is as follows. In Section \protect\ref{state} we
present the basic definitions, and construct our invariant $F$ of planar curves.
In Section \protect\ref{abs} we construct an abstract invariant, as opposed to the explicit invariant $F$,
and show that the abstract invariant is universal (Theorem \protect\ref{absf}).
In Section \protect\ref{prf} we prove our main result, namely, that a slight modification $\widehat{F}$ of
the explicit invariant $F$ is universal (Theorem \protect\ref{main}).
In Section \protect\ref{ex} we prove the main algebraic fact used in the proof of Theorem \protect\ref{main}.

\section{Definitions and statement of results}\label{state}
By a \emph{curve} we will always mean an immersion $c:S^1 \to {\mathbb R}^2$.
Let ${\mathcal{C}}$ denote the space of all curves.
A curve will be called \emph{stable} if its only self intersections are transverse double points.
The generic singularities a curve may have are either a tangency of first order between two strands,
which will be called a $J$-type singularity, or three strands meeting at a point,
each two of which are transverse, which will be called an $S$-type singularity.
Singularities of type $J$ and $S$ appear in Figures \protect\ref{p2} and \protect\ref{p3}.
A generic singularity can be resolved in two ways,
and there is a standard way for considering one resolution positive, and the other negative,
as defined in \cite{a1}.
We denote by ${\mathcal{C}}_n \subseteq {\mathcal{C}}$ ($n\geq 0$) the space of all curves which have precisely
$n$ generic singularity points (the self intersection being elsewhere stable).
In particular, ${\mathcal{C}}_0$ is the space of all stable curves.
An invariant of curves is a function $f:{\mathcal{C}}_0 \to W$,
which is constant on the connected components of ${\mathcal{C}}_0$, and where $W$ in this work
will always be a vector space over ${\mathbb Q}$.

Given a curve $c\in {\mathcal{C}}_n$, with singularities located at $p_1,\dots,p_n \in {\mathbb R}^2$,
and given a subset $A\subseteq \{p_1,\dots,p_n\}$,
we define $c_A \in C_0$ to be the stable curve obtained from $c$ by resolving all singularities
of $c$ at points of $A$ into the
negative side, and all singularities not in $A$ into the positive side.
Given an invariant $f:{\mathcal{C}}_0 \to W$ we define the ``$n$th derivative'' of $f$ to be the function
$f^{(n)}:{\mathcal{C}}_n \to W$ defined by
$$f^{(n)}(c)=\sum_{ A \subseteq \{p_1,\dots,p_n\} } (-1)^{|A|} f(c_A)$$
where $|A|$ is the number of elements in $A$.
An invariant $f:{\mathcal{C}}_0\to W$ is called \emph{of order $n$} if
$f^{(n+1)}(c)=0$ for all $c\in {\mathcal{C}}_{n+1}$.
The space of all $W$ valued invariants on ${\mathcal{C}}_0$ of order $n$ is denoted $V_n=V_n(W)$.
Clearly $V_n \subseteq V_m$ for $n \leq m$.
In this work we give a full description of $V_1$. We will construct a ``universal''
order 1 invariant, by which we mean the following:

\begin{dfn}\label{uni}
An order 1 invariant $\widehat{f} :{\mathcal{C}}_0 \to \widehat{W}$ will be called \emph{universal}, if
for any $W$ and any order 1 invariant $f:{\mathcal{C}}_0 \to W$, there exists a unique linear map
$\phi: \widehat{W} \to W$ such that $f = \phi \circ \widehat{f}$. In other words, for any $W$, the natural map
$Hom_{\mathbb Q}(\widehat{W},W) \to V_1(W)$ given by $\phi \mapsto \phi \circ \widehat{f}$, is an isomorphism.
\end{dfn}

\begin{dfn}\label{gen2}
A \emph{standard} annulus will be an annulus $A$
of the form $A = {\mathbb R}^2 - U$, where $U \subseteq {\mathbb R}^2$ is a circular open disc
(i.e. $U$ is a disc geometrically, not just topologically.)
An immersion $e:[0,1] \to A$ will be called a \emph{simple arc} if
\begin{enumerate}
\item $e(0),e(1) \in {\partial} A$.
\item The ends of $e$ are perpendicular to ${\partial} A$.
\item The self intersections of $e$ are transverse double points.
\end{enumerate}
\end{dfn}

Let ${\Z^o} \subseteq {\mathbb Z}$ denote the set of odd integers. Let $D{\mathbb Z}$ denote the set of all columns ${a_1 \choose a_2}$
with $a_1 \in {\mathbb Z}$ and $a_2 \in {\mathbb Z}^o$.

\begin{dfn}\label{di}
Let $A$ be a standard annulus, and let $e$ be a simple arc in $A$.
\begin{enumerate}
\item For a double point $v$ of $e$
we define $i(v) \in \{1,-1\}$, where $i(v)=1$ if the orientation at $v$ given by the two tangents to $e$ at $v$,
in the order they are visited, coincides with the orientation of $A$ (restricted from ${\mathbb R}^2$).
Otherwise $i(v)=-1$.
\item We define the \emph{top} index of $e$, $I_1(e)\in{\mathbb Z}$ by $I_1(e) = \sum_v i(v)$
where the sum is over all double points $v$ of $e$.
\item We define the \emph{bottom} index of $e$, $I_2(e)$, to be the index defined in \cite{k}, that is,
$I_2(e) = (\phi - \omega)/\pi$ where $\phi$ is the total angle of rotation  of the radius vector
(from the center of $U$ to $e$)
along $e$, and $\omega$ is the total angle of rotation of the tangent vector along $e$.
Since by definition of simple arc the ends of $e$ are perpendicular to ${\partial} A$,
we have that $I_2(e)$ is an odd integer, that is, $I_2(e) \in {\Z^o}$.
\item We define the \emph{double} index of $e$ to be $I(e) = {I_1(e) \choose I_2(e)} \in D{\mathbb Z}$.
\end{enumerate}
\end{dfn}

Define ${\mathbb X}$ to be the vector space over ${\mathbb Q}$ with basis all symbols $X^{a_1,b_1}_{a_2,b_2}$ where
$a_1, b_1 \in {\mathbb Z} , a_2, b_2 \in {\Z^o}$.
For $c \in {\mathcal{C}}_0$, let $v$ be a double point of $c$, and let $u_1, u_2$ be the two tangents at
$v$ ordered by the orientation of ${\mathbb R}^2$.
Let $U$ be a small circular disc neighborhood centered at $v$,
and let $A={\mathbb R}^2 - U$. Assume the strands of $c$ cross ${\partial} U$ perpendicularly, so
$c|_{c^{-1}(D)}$ defines two simple arcs $c_1,c_2$ in $A$. The ordering $c_1,c_2$ is chosen so that
the tangent $u_i$ leads to $c_i$, $i=1,2$. They will be called the \emph{exterior arcs} of $c$.
We denote $I(c_1)$ by $a_1(v) \choose a_2(v)$ and $I(c_2)$ by
$b_1(v) \choose b_2(v)$.
We define $F : {\mathcal{C}}_0 \to {\mathbb X}$ as follows:
$$F(c) = \sum_v X^{a_1(v),b_1(v)}_{a_2(v),b_2(v)}$$
where the sum is over all double points $v$ of $c$.

For $c \in {\mathcal{C}}_0$ let $\omega(c)$ denote the Whitney winding number of $c$,
and let $G : {\mathcal{C}}_0 \to {\mathbb X}$ be the order 0 invariant defined by $G(c)=X^{\omega(c),0}_{1,-1}$.
Let $\widehat{F} : {\mathcal{C}}_0 \to {\mathbb X}$ be defined by $\widehat{F}(c) = F(c) + G(c)$.
In this work we will prove that $\widehat{F}$ is a universal order one invariant.

Often one is interested in order 1 invariants only up to order 0 invariants, and in that case the addition of
$G$ is not needed. Our definition of universal order 1 invariant however requires that \emph{all}
order 1 invariants be obtained by composition with linear maps (in a unique way).
For this a correction such as $G$ is needed, as will be explained after the proof of Theorem \protect\ref{main}.
We will see (Lemma \protect\ref{xw}) that if $F(c) \neq 0$ then
$\omega(c)$ can be computed from $F(c)$, so in terms of separating curves, the only difference between $F$
and $\widehat{F}$ is that $\widehat{F}$ separates the two different embedded curves, whereas $F$ vanishes on both.

\section{An abstract universal order one invariant}\label{abs}

In this section we will construct an ``abstract'' universal order 1 invariant, as opposed to the ``concrete''
invariant described in the previous section.
This will be an intermediate step in proving that the concrete
invariant is universal.

\begin{figure}
\scalebox{0.8}{\includegraphics{p1.eps}}
\caption{Representatives for the regular homotopy classes of simple arcs in $A$.}\label{p1}
\end{figure}

\begin{lemma}\label{omega}
Let $e$ be a simple arc in $A$ with initial point $i$ and final point $f$,
with $i \neq f$.
Let $\theta$ be the angle from $i$ to $f$,
in the positive direction, see Figure \protect\ref{p1} . Let $\phi$ be the total angle of rotation
of the radius vector (from the center of $U$ to $e$)
along $e$, and $\omega$ the total angle of rotation of the tangent vector along $e$.
Then $I_1(e) = (2\phi - \omega - \theta + \pi)/2\pi$.
\end{lemma}

\begin{pf} Both sides of the equality are invariant under regular homotopy in $A$,
keeping the initial and final points and tangents fixed (for the invariance of $I_1$
note that $i \neq f$). So,
it is enough to check the equality on representatives of the various regular homotopy classes
of simple arcs, such as those of the form appearing in Figure \protect\ref{p1}.
They are arcs that begin by spiraling around $U$ a number of times
and may then include a number of ``curls''. To verify the equality on such arcs, one can first
check it for embedded arcs, and then check that it is preserved
when adding a round tour around $U$ and when adding a curl.
\end{pf}

\begin{lemma}\label{li}
If $e,e'$ are two simple arcs in $A$ with the same initial point $i$ and final point $f$, with $i \neq f$,
then $I(e)=I(e')$ iff $e$ and $e'$ are regularly homotopic in $A$, keeping the initial and final
points and tangents fixed.
\end{lemma}

\begin{pf}
Let $I_1=I_1(e),I_2=I_2(e)$. Let $\theta,\phi,\omega$ be as in Lemma \protect\ref{omega}.
Then by Lemma \protect\ref{omega}, $I_1 = (2\phi - \omega - \theta + \pi)/2\pi$,
and by definition, $I_2 = (\phi - \omega)/\pi$.
These formulas are invertible, giving
$\phi = 2\pi I_1 - \pi I_2 + \theta - \pi$
and $\omega= 2\pi I_1 - 2\pi I_2 + \theta - \pi$.
Since the pair $\phi,\omega$ characterizes the regular homotopy class of $e$, so does the pair
$I_1,I_2$.
\end{pf}

Two curves $c,c' \in {\mathcal{C}}_1$ will be called \emph{equivalent}
if there is an ambient isotopy of ${\mathbb R}^2$ bringing a neighborhood $U$ of
the singular point of $c$ onto a neighborhood $U'$ of the singular point of $c'$, such that the configuration
near the singular point precisely matches and such that the exterior arcs in $A = {\mathbb R}^2-U$ are
regularly homotopic in $A$.
We always assume, as before,
that the strands of $c$ cross ${\partial} U$ perpendicularly, and so the exterior arcs are simple arcs.
Note that the crossing points with ${\partial} U$ are all distinct (even for $J$ type singularities).
By Lemma \protect\ref{li} a pair of exterior arcs with same initial and final points, are regularly homotopic
in $A$ iff their double indices are the same. So we have:

\begin{prop}\label{eq1}
Two curves $c,c' \in {\mathcal{C}}_1$ are equivalent
iff  they have the same singularity configuration and the same corresponding double indices of exterior arcs.
\end{prop}

The following is clear from the definition of order 1 invariant:

\begin{lemma}\label{eqv}
Let $f : {\mathcal{C}}_0 \to W$ be an invariant, then $f$ is of order 1 iff for any two equivalent $c,c' \in {\mathcal{C}}_1$,
$f^{(1)}(c) = f^{(1)}(c')$.
\end{lemma}

We will attach a symbol to each equivalence class of curves in ${\mathcal{C}}_1$ as follows. For $J$ type singularities there are
three distinct configurations which we name $J^+,J^A,J^B$, see Figure \protect\ref{p2}.
The $J^A,J^B$ singularities are symmetric with respect to a $\pi$ rotation, which interchanges the two exterior arcs,
and so, by Proposition \protect\ref{eq1},
the equivalence class of a $J^A$ or $J^B$ singularity is characterized by a symbol $J^A_{{a_1 \choose a_2},{b_1 \choose b_2}}$ and $J^B_{{a_1 \choose a_2},{b_1 \choose b_2}}$
(${a_1 \choose a_2},{b_1 \choose b_2} \in D{\mathbb Z}$), where ${a_1 \choose a_2},{b_1 \choose b_2}$ is an \emph{unordered} pair, registering the double indices of the two exterior arcs.
The $J^+$ configuration, on the other hand, is not symmetric, and so characterized by a symbol $J^+_{{a_1 \choose a_2},{b_1 \choose b_2}}$ with
${a_1 \choose a_2},{b_1 \choose b_2}$ an \emph{ordered} pair, with say, ${a_1 \choose a_2}$ corresponding to the lower strand in Figure \protect\ref{p2}.
(We will shortly see that this ordering may however be disregarded.)

As to $S$ type singularities, there are four types, as seen in Figure \protect\ref{p3},
depending on the relative orientations of the three strands participating in
the triple point, and the way they are connected
to each other. The distinction between them will be incorporated into the
way we register the double indices of the exterior arcs.
We will have a cyclicly ordered triple
${a_1 \choose a_2},{b_1 \choose b_2},{c_1 \choose c_2} \in D{\mathbb Z}$ registering the double indices of the three exterior arcs, in the cyclic
order they appear along $S^1$, and each may appear with or without a hat, according to the following
rule. For given exterior arc $e$ with double index ${a_1 \choose a_2}$,
let $u_1$ be the initial tangent of $e$ and $u_2$ the initial tangent
of the following segment. Then if $u_2$ is pointing to the right of $u_1$ then ${a_1 \choose a_2}$ will appear with a hat,
and if $u_2$ is pointing to the left of $u_1$ then ${a_1 \choose a_2}$ will appear unhatted. Since the ordering is cyclic, this gives
four types of $S$ symbols,
$S_{{a_1 \choose a_2},{b_1 \choose b_2},{c_1 \choose c_2}} , S_{\widehat{{a_1 \choose a_2}},{b_1 \choose b_2},{c_1 \choose c_2}} , S_{\widehat{{a_1 \choose a_2}},\widehat{{b_1 \choose b_2}},{c_1 \choose c_2}}, S_{\widehat{{a_1 \choose a_2}},\widehat{{b_1 \choose b_2}},\widehat{{c_1 \choose c_2}}}$, which
correspond to the four types of $S$ singularities.

\begin{figure}
\scalebox{0.8}{\includegraphics{p2.eps}}
\caption{Singularities of type $J^+$, $J^A$, $J^B$.}\label{p2}
\end{figure}

\begin{figure}
\scalebox{0.8}{\includegraphics{p3.eps}}
\caption{Singularities of type $S_{{a_1 \choose a_2},{b_1 \choose b_2},{c_1 \choose c_2}}$, $S_{\widehat{{a_1 \choose a_2}},{b_1 \choose b_2},{c_1 \choose c_2}}$, $S_{\widehat{{a_1 \choose a_2}},\widehat{{b_1 \choose b_2}},{c_1 \choose c_2}}$,
$S_{\widehat{{a_1 \choose a_2}},\widehat{{b_1 \choose b_2}},\widehat{{c_1 \choose c_2}}}$.}\label{p3}
\end{figure}

Let ${\mathbb F}$ denote the vector space over ${\mathbb Q}$ with basis all the above symbols, $J^+_{{a_1 \choose a_2},{b_1 \choose b_2}}$, $J^A_{{a_1 \choose a_2},{b_1 \choose b_2}}$,
$J^B_{{a_1 \choose a_2},{b_1 \choose b_2}}$,
$S_{{a_1 \choose a_2},{b_1 \choose b_2},{c_1 \choose c_2}}$, $S_{\widehat{{a_1 \choose a_2}},{b_1 \choose b_2},{c_1 \choose c_2}}$, $S_{\widehat{{a_1 \choose a_2}},\widehat{{b_1 \choose b_2}},{c_1 \choose c_2}}$, $S_{\widehat{{a_1 \choose a_2}},\widehat{{b_1 \choose b_2}},\widehat{{c_1 \choose c_2}}}$.
Let $\gamma$ be a generic path in ${\mathcal{C}}$, that is,
a path (regular homotopy) $[0,1] \to {\mathcal{C}}$, whose image lies in ${\mathcal{C}}_0 \cup {\mathcal{C}}_1$ and
which is transverse with respect to ${\mathcal{C}}_1$.
We denote by $v(\gamma) \in {\mathbb F}$ the sum of symbols of the singularities $\gamma$ passes,
each added with $+$ or $-$
sign according to whether we pass it from its negative side to its positive side, or from its positive side to its
negative side, respectively. Let $N \subseteq {\mathbb F}$ be the subspace generated by all elements $v(\gamma)$ obtained from
all possible generic \emph{loops} $\gamma$ in ${\mathcal{C}}$ (i.e. \emph{closed} paths), and let ${\mathbb G} = {\mathbb F} / N$.

For an order 1 invariant $f:{\mathcal{C}}_0 \to W$, since by Lemma \protect\ref{eqv} $f^{(1)}$ coincides on equivalent curves in ${\mathcal{C}}_1$,
it induces a well defined linear map $f^{(1)}:{\mathbb F} \to W$.

The following is clear:
\begin{lemma}\label{vgm}
If $\gamma$ is a generic path in ${\mathcal{C}}$, from $c_1$ to $c_2$, then $f^{(1)} ( v(\gamma) ) = f(c_2) - f (c_1)$.
\end{lemma}

From Lemma \protect\ref{vgm} it follows that $f^{(1)}$ vanishes on the generators of $N$,
so it also induces a well defined linear map $f^{(1)} : {\mathbb G} \to W$.

Let $\Gamma_m \in {\mathcal{C}}_0$ be a curve with Whitney number $m$, chosen once and for all as
base curve for its regular homotopy class.
Let $\widehat{{\mathbb G}} = {\mathbb G} \oplus \bigoplus_{m \in {\mathbb Z}} {\mathbb Q} u_m$,
where $\{ u_m \}_{m \in {\mathbb Z}}$ is a set of new independent vectors.
We define an order 1 invariant $\widehat{f} : {\mathcal{C}}_0 \to \widehat{{\mathbb G}}$ as follows:
For any $c \in {\mathcal{C}}_0$ there is a generic path $\gamma$
from $\Gamma_m$ to $c$, where $m$ is the Whitney number of $c$.
Let $\widehat{f}(c)=u_m + v(\gamma) \in \widehat{{\mathbb G}}$.
By definition of $N$,
$v(\gamma) \in {\mathbb G}$ is indeed independent of the choice of path $\gamma$.
From Lemma \protect\ref{eqv} it is clear that $\widehat{f}$
is an order 1 invariant, and we will now see that it is \emph{universal}:

\begin{thm}\label{absf}
$\widehat{f} : {\mathcal{C}}_0 \to \widehat{{\mathbb G}}$ is a universal order 1 invariant.
\end{thm}

\begin{pf}
For an order 1 invariant $f: {\mathcal{C}}_0 \to W$, define the linear map $\Phi_f:\widehat{{\mathbb G}}\to W$
by $\Phi_f|_{{\mathbb G}} = f^{(1)}$ and $\Phi_f(u_m)=f(\Gamma_m)$, $m \in {\mathbb Z}$.
We claim $\Phi_f \circ \widehat{f} = f$ and that $\Phi_f$ is the unique linear map satisfying this property.
Indeed, let $c \in {\mathcal{C}}_0$ and let $\gamma$ be a generic path from $\Gamma_m$ to $c$ where $m$ is the
Whitney number of $c$.
Then by Lemma \protect\ref{vgm}
$f(c) = f(\Gamma_m) + f^{(1)}( v(\gamma) ) = \Phi_f(u_m) + \Phi_f ( v(\gamma)) = \Phi_f ( \widehat{f} (c))$.

For uniqueness, it is enough to show that $\widehat{f} ( {\mathcal{C}}_0 )$ spans $\widehat{{\mathbb G}}$.
We have $u_m = \widehat{f}(\Gamma_m)$ for  $m \in {\mathbb Z}$ so it remains to show that
${\mathbb G} \subseteq span \widehat{f} ( {\mathcal{C}}_0 )$.
Indeed for any generating symbol $T$ of ${\mathbb G}$, $T$ is the
the difference $\widehat{f}(c)-\widehat{f}(c')$ for two
curves $c,c' \in {\mathcal{C}}_0$, namely, the two resolutions of a curve in ${\mathcal{C}}_1$
whose symbol is $T$.
\end{pf}

The following is also clear:

\begin{lemma}\label{iso}
Let $f: {\mathcal{C}}_0 \to W$ be an order 1 invariant.
Then $f$ is a universal order 1 invariant iff
$\Phi_f: \widehat{{\mathbb G}} \to W$ (appearing in the proof of Theorem \protect\ref{absf}) is an isomorphism.
\end{lemma}

We present two specific subfamilies of the set of generators of $N$.
They appear in Figures \protect\ref{p4},\protect\ref{p5}.
The figures are schematic, only indicating the way the strands
involved in the local configuration are
connected to each other. The two exterior arcs for the intersection point appearing in the left most curves,
labeled $a$ and $b$, are assumed to have double indices ${a_1 \choose a_2}$ and ${b_1 \choose b_2}$ respectively.
To obtain the double indices for the tangency points, note that exterior arcs that may be obtained from each
other by sliding their ends along ${\partial} A$ have the same bottom index.
The element obtained from Figure \protect\ref{p4} is $J^+_{{a_1 \choose a_2},{b_1 \choose b_2}} - J^+_{{b_1 \choose b_2},{a_1 \choose a_2}}$,
that is, in ${\mathbb G}$ we have the relation $J^+_{{a_1 \choose a_2},{b_1 \choose b_2}} - J^+_{{b_1 \choose b_2},{a_1 \choose a_2}} = 0$,
and so from now on we will simply regard the double indices of $J^+_{{a_1 \choose a_2},{b_1 \choose b_2}}$ as being \emph{un-ordered}
(as is true by definition for the $J^A$ and $J^B$ symbols).
The relation obtained from Figure \protect\ref{p5} is $J^B_{{a_1+1 \choose a_2},{b_1 \choose b_2}} - J^A_{{a_1 \choose a_2},{b_1-1 \choose b_2}} = 0$,
or after index shift:
$J^B_{{a_1 \choose a_2},{b_1 \choose b_2}} = J^A_{{a_1-1 \choose a_2},{b_1-1 \choose b_2}}$.

\begin{figure}
\scalebox{0.8}{\includegraphics{p4.eps}}
\caption{$J^+_{{a_1 \choose a_2},{b_1 \choose b_2}} - J^+_{{b_1 \choose b_2},{a_1 \choose a_2}} = 0$}\label{p4}
\end{figure}

\begin{figure}
\scalebox{0.8}{\includegraphics{p5.eps}}
\caption{$J^B_{{a_1+1 \choose a_2},{b_1 \choose b_2}} - J^A_{{a_1 \choose a_2},{b_1-1 \choose b_2}} = 0$}\label{p5}
\end{figure}

\section{Proof of main result}\label{prf}

Returning to our invariant $F:{\mathcal{C}}_0 \to {\mathbb X}$, we look at $F^{(1)}$.
It is easy to see that indeed the value of $F^{(1)}$ depends only on the symbol of a
given curve $c \in {\mathcal{C}}_1$, which by Lemma \protect\ref{eqv} proves
that indeed $F$ is an order 1 invariant.
We will need the value of $F^{(1)}$ only on the symbols of type $J^+$ and $J^A$, which are demonstrated in
Figures \protect\ref{p6},\protect\ref{p7}. Again, the figures are schematic, only indicating the way the strands
involved in the local configuration
are connected to each other. The two exterior arcs for the tangency point appearing in the middle curves,
labeled $a$ and $b$, are assumed to have double indices ${a_1 \choose a_2}$ and ${b_1 \choose b_2}$ respectively.
In Figure \protect\ref{p6} the two new intersection points $u$ and $v$ contribute additional terms
$X^{a_1,b_1}_{a_2,b_2}$ and $X^{b_1,a_1}_{b_2,a_2} $ respectively, and in Figure \protect\ref{p7} $u$ and $v$ contribute
$X^{a_1,b_1+1}_{a_2,b_2}$ and $X^{b_1,a_1+1}_{b_2,a_2}$ respectively.
So we have:
\begin{itemize}
\item $F^{(1)}(J^+_{{a_1 \choose a_2},{b_1 \choose b_2}}) =
X^{a_1,b_1}_{a_2,b_2} + X^{b_1,a_1}_{b_2,a_2} $
\item $F^{(1)}(J^A_{{a_1 \choose a_2},{b_1 \choose b_2}}) =
X^{a_1,b_1+1}_{a_2,b_2} + X^{b_1,a_1+1}_{b_2,a_2} $
\end{itemize}

\begin{figure}
\scalebox{0.8}{\includegraphics{p6.eps}}
\caption{$F^{(1)}(J^+_{{a_1 \choose a_2},{b_1 \choose b_2}}) =
X^{a_1,b_1}_{a_2,b_2} + X^{b_1,a_1}_{b_2,a_2} $}\label{p6}
\end{figure}

\begin{figure}
\scalebox{0.8}{\includegraphics{p7.eps}}
\caption{$F^{(1)}(J^A_{{a_1 \choose a_2},{b_1 \choose b_2}}) =
X^{a_1,b_1+1}_{a_2,b_2} + X^{b_1,a_1+1}_{b_2,a_2} $}\label{p7}
\end{figure}

We define ${\mathbb Y}$ to be the vector space over ${\mathbb Q}$ generated by the symbols $Y^n_{k,l}$, $n \in {\mathbb Z}$,
$k,l \in {\Z^o}$ and with the relation $$Y^n_{k,l} = -Y^n_{l,k}.$$
We define the map $\Psi: {\mathbb X} \to {\mathbb Y}$ by $$\Psi(X^{a,b}_{k,l}) = Y^{a+b}_{k,l}.$$

For each $m \in {\mathbb Z}$ define the linear functional $g_m :{\mathbb Y} \to {\mathbb Q}$ as follows:
$$g_m(Y^n_{k,l}) =
\begin{cases}
k-l         & \ \ \ \  \text{if} \ \ \ n-k-l=m \ \ \ \text{and} \ \ \ k+l =0       \\
(l-k)/2     & \ \ \ \  \text{if} \ \ \ n-k-l=m \ \ \ \text{and}  \ \ \ k+l = \pm 2       \\
0           & \ \ \ \  \text{otherwise}
\end{cases}$$
Let $\overline{{\mathbb Y}} \subseteq {\mathbb Y}$ denote the subspace defined by the family of all equation $g_m = 0$, $m \in {\mathbb Z}$,
that is, the intersection of kernels of all $g_m$.
Let $\overline{{\mathbb X}} = \Psi^{-1}(\overline{{\mathbb Y}})$.

We now look at the order 1 invariant $K = \Psi \circ F : {\mathcal{C}}_0 \to {\mathbb Y}$.
The invariant $K$ is closely related to the invariant defined by M. \`{E}. Kazarian in \cite{k}.
Let ${\mathbb J}$ denote the subspace of ${\mathbb G}$ spanned by all symbols of type $J$, then by the relation
$J^B_{{a_1 \choose a_2},{b_1 \choose b_2}} = J^A_{{a_1-1 \choose a_2},{b_1-1 \choose b_2}}$
it is also spanned by all symbols of type $J^+,J^A$.
It is clear from our explicit formulas for $F^{(1)}$ on the symbols of type $J^+,J^A$, that
$K^{(1)} = (\Psi \circ F)^{(1)} = \Psi \circ F^{(1)}$ vanishes on ${\mathbb J}$, and so induces a map
$K^{(1)} : {\mathbb G} / {\mathbb J} \to {\mathbb Y}$. As we shall now explain, the result
in \cite{k} may be stated, in the language of the present work, as follows:

\begin{thm}[M. \`{E}. Kazarian]\label{kaz}
$K^{(1)} : {\mathbb G} / {\mathbb J} \to {\mathbb Y}$ is injective, with image $\overline{{\mathbb Y}}$.
\end{thm}

To make precise the relation between our $K$ and the invariant of \cite{k} we first prove:

\begin{lemma}\label{xw}
If $c \in {\mathcal{C}}_0$ is a curve with Whitney winding number $m$, then for each intersection point $v$ of $c$,
the two double indices ${a_1 \choose a_2},{b_1 \choose b_2}$ attached to $v$ satisfy $a_1+b_1-a_2-b_2=m$.
\end{lemma}

\begin{pf}
For the two exterior arcs $c_1,c_2$ of $v$, let $\theta_1,\phi_1,\omega_1$ and $\theta_2,\phi_2,\omega_2$
be the corresponding angles as in Lemma \protect\ref{omega},
then $\theta_1+\theta_2=2\pi$ and $(\omega_1+\omega_2)/2\pi=m$.
So $I_1(c_1) + I_1(c_2) - I_2(c_1) - I_2(c_2) =
(2\phi_1 - \omega_1 - \theta_1 + \pi)/2\pi + (2\phi_2 - \omega_2 - \theta_2 + \pi)/2\pi
-(\phi_1 - \omega_1)/\pi -(\phi_2 - \omega_2)/\pi = m$.
\end{pf}

Accordingly, we define ${\mathbb X}^m \subseteq {\mathbb X}$ to be the subspace spanned by all $X^{a,b}_{k,l}$ with $a+b-k-l=m$ and
${\mathbb Y}^m \subseteq {\mathbb Y}$ to be the subspace spanned by all $Y^n_{k,l}$ with $n-k-l=m$.
We have ${\mathbb X}=\oplus_m {\mathbb X}^m$, ${\mathbb Y} = \oplus_m {\mathbb Y}^m$, $F$ maps all curves with
Whitney number $m$ into ${\mathbb X}^m$ and $\Psi$ maps ${\mathbb X}^m$ into ${\mathbb Y}^m$.
In a similar way each generating symbol
of ${\mathbb G}$ is related to a specific regular homotopy class and accordingly
${\mathbb G}$ and ${\mathbb J}$ split into $\oplus_m {\mathbb G}^m$ and $\oplus_m {\mathbb J}^m$.
Finally, define $\overline{{\mathbb Y}}^m \subseteq {\mathbb Y}^m$ to be the subspace defined by the equation $g_m=0$,
then $\overline{{\mathbb Y}} = \oplus_m \overline{{\mathbb Y}}^m$.
Each ${\mathbb Y}^m$ is naturally isomorphic to  $\Lambda^2 A$ of \cite{k} Remark 1,
where $Y^{m+k+l}_{k,l} \in {\mathbb Y}^m$ is identified with $e_k \wedge e_l \in \Lambda^2 A$.
This induces a projection $p: {\mathbb Y} \to \Lambda^2 A$, mapping any $Y^n_{k,l}$ to $e_k \wedge e_l$.
The invariant $S$ of \cite{k} Remark 1 is precisely $p \circ K$.
Now \cite{k} Theorem 2 relates to each regular homotopy class separately,
the space $I_1/I_0$ appearing there is the space of $S$-invariant on a given regular homotopy class, up to constants.
An $S$-invariant is by definition an invariant $f$ such that $f^{(1)}$ vanishes on ${\mathbb J}$, and so $I_1/I_0$ is
precisely the dual to our ${\mathbb G}^m / {\mathbb J}^m$. So, the statement of \cite{k} Theorem 2 is that
the dual to our $K^{(1)} : {\mathbb G}^m / {\mathbb J}^m \to \overline{{\mathbb Y}}^m$ is an isomorphism. It follows that
this map itself is an isomorphism, and so $K^{(1)} : {\mathbb G} / {\mathbb J} \to \overline{{\mathbb Y}}$ is an isomorphism.

We now look at the following commutative diagram:

$$\begin{CD}
0 @>>> {\mathbb J} @>inc>> {\mathbb G} @>quot>> {\mathbb G} / {\mathbb J} @>>> 0 \\
& &      @V Id VV     @VV F^{(1)} V       @VV K^{(1)} V \\
0 @>>> {\mathbb J} @>>F^{(1)}|_{{\mathbb J}}> {\mathbb X} @>>\Psi> {\mathbb Y} @>>> 0
\end{CD}$$

The algebraic portion of our work is the following theorem, whose proof we defer to Section \protect\ref{ex}:

\begin{thm}\label{exact}
The sequence \ \ $0 \to {\mathbb J} @>F^{(1)}|_{{\mathbb J}}>> {\mathbb X} @>\Psi>> {\mathbb Y} \to 0$ \ \ is exact.
\end{thm}

It follows from Theorem \protect\ref{kaz}, Theorem \protect\ref{exact}, and the commutative diagram, that:

\begin{thm}\label{f1}
$F^{(1)} : {\mathbb G} \to {\mathbb X}$ is injective, with image $\overline{{\mathbb X}}$.
\end{thm}

We may now prove our main result:

\begin{thm}\label{main}
For $c \in {\mathcal{C}}_0$ let $\omega(c)$ denote the Whitney winding number of $c$,
and let $G : {\mathcal{C}}_0 \to {\mathbb X}$ be the order 0 invariant defined by $G(c)=X^{\omega(c),0}_{1,-1}$.
Let $\widehat{F} : {\mathcal{C}}_0 \to {\mathbb X}$ be defined by $\widehat{F}(c) = F(c) + G(c)$, then $\widehat{F}$
is a universal order one invariant.
\end{thm}

\begin{pf}
By Lemma \protect\ref{iso} it is enough to show that $\Phi_{\widehat{F}} : \widehat{{\mathbb G}} \to {\mathbb X}$ is an isomorphism.
Indeed, by Theorem \protect\ref{f1},
$\Phi_{\widehat{F}}|_{{\mathbb G}} = \widehat{F}^{(1)} = F^{(1)}$ is injective, and onto $\overline{{\mathbb X}}$.
Now, the values $X^{\omega(c),0}_{1,-1}$ for $G$ were precisely chosen so that
$\{ \widehat{F}(\Gamma_m) \}_{m \in {\mathbb Z}} = \{ \Phi_{\widehat{F}}(u_m) \}_{m \in {\mathbb Z}}$  will be a basis  for ${\mathbb X} / \overline{{\mathbb X}}$.
This is verified by evaluating the functionals $g_m \circ \Psi$
(which are the defining equations of $\overline{{\mathbb X}}$ in ${\mathbb X}$)
on $\widehat{F}(\Gamma_{m'})$. Indeed, from \cite{k}
and our definition of $g_m$ we know
$g_m(\Psi( F(\Gamma_{m'}))) = g_m(K(\Gamma_{m'} ))= 2\delta_{m,m'}\delta_{m',0}$
and so $g_m \circ \Psi( \widehat{F}(\Gamma_{m'}) ) =  2\delta_{m,m'}\delta_{m',0} + 2\delta_{m,m'}
= 2(1+\delta_{m',0})\delta_{m,m'}$.
It follows that $\Phi_{\widehat{F}} : \widehat{{\mathbb G}} \to {\mathbb X}$ is an isomorphism.
\end{pf}

We now see why we needed to add to $F$ a correcting order 0 invariant such as $G$.
From the proof of Theorem \protect\ref{main} we see that $\Phi_F : \widehat{{\mathbb G}} \to {\mathbb X}$
is non-injective and non-surjective.
$\Phi_F$ being non-injective means that not all order 1 invariants may be realized as $\phi \circ F$
($\phi : {\mathbb X} \to W$ a linear map).
Indeed $g_m \circ \Psi( F(\Gamma_{m'})) = 2\delta_{m,m'}\delta_{m',0}$ implies that any order 0 invariant of
the form $\phi \circ F$ vanishes on all curves with Whitney number $m \neq 0$.
$\Phi_F$ being non-surjective means that the choice of $\phi$ is non-unique, that is, there are $\phi \neq \psi$ with
$\phi \circ F = \psi \circ F$. Adding $G$ to $F$ solved both injectivity and surjectivity problems.
(If non-surjectivity were the only problem, then it would be fixed simply by diminishing the target space ${\mathbb X}$.)

We conclude this section by explaining the remark made in the introduction, that the space of
invariants spanned by all order 1 $J$- and $S$-invariants is much smaller than the full space of order 1 invariants.
Indeed, let ${\mathbb S} \subseteq {\mathbb G}$ be the subspace spanned  by all symbols of type $S$. By definition, an order 1 invariant $f$
is a $J$- or $S$-invariant, iff $f^{(1)}$ vanishes on ${\mathbb S}$ or ${\mathbb J}$ respectively.
Therefore an invariant $f$ is the sum of $J$- and $S$-invariants iff $f^{(1)}$ vanishes on ${\mathbb S} \cap {\mathbb J}$.
By observing relations in $N$ coming from loops as in \cite{n6} Figures 6,7,
one can show that ${\mathbb S} \cap {\mathbb J}$ is the subspace of ${\mathbb G}$
spanned by all symbols of the form
$J^+_{{a_1 \choose a_2},{b_1 \choose b_2}} - J^+_{{a_1+2 \choose a_2+2},{b_1 \choose b_2}}$ and $J^A_{{a_1 \choose a_2},{b_1 \choose b_2}} -J^A_{{a_1+2 \choose a_2+2},{b_1 \choose b_2}}$,
$a_1, b_1 \in {\mathbb Z} , a_2, b_2 \in {\Z^o}$.
Given that the symbols of type $J^+,J^A$ are independent in ${\mathbb G}$ (a fact that will be proved in Section \protect\ref{ex}),
we see ${\mathbb S} \cap {\mathbb J}$ is a fairly large subspace of ${\mathbb G}$.

\section{The exact sequence}\label{ex}

In this section we prove Theorem \protect\ref{exact}, stating that the sequence
$0 \to {\mathbb J} @>F^{(1)}|_{{\mathbb J}}>> {\mathbb X} @>\Psi>> {\mathbb Y} \to 0$ is exact.
For $n \in {\mathbb Z}$ and $k \leq l \in {\Z^o}$
let $A^n_{k,l}$ be the following set of symbols, viewed as elements in ${\mathbb G}$.
$$A^n_{k,l} = \{ J^+_{ {a \choose k} , {b \choose l} } \ : \  a+b = n \} \cup
\{ J^A_{{a \choose k},{b \choose l}} \ : \ a+b=n-1 \}.$$
Let ${\mathbb J}^n_{k,l} = span A^n_{k,l} \subseteq {\mathbb J}$. We point out that since the pair ${a \choose k} , {b \choose l}$ is unordered,
the meaning of the pair $a,b$ in the conditions $a+b=n$ and $a+b=n-1$ depends on whether $k=l$ or $k<l$.
If $k=l$ then $a,b$ runs over all unordered pairs $\{a,b\}$, whereas if $k<l$ then the pairing of $a$ with $k$
and $b$ with $l$ serves to distinguish them and so in this case $a,b$ runs over all \emph{ordered} pairs $( a,b )$.

For $n \in {\mathbb Z}$ and $k \leq l \in {\Z^o}$ define ${\mathbb X}^n_{k,l}$
to be the span of all symbols of the form  $X^{a,b}_{k,l}$ and $X^{a,b}_{l,k}$ with $a+b =n$.
Then ${\mathbb X}$ is the direct sum of all ${\mathbb X}^n_{k,l}$.
Here there is also a difference between the cases $k=l$ and $k<l$. When $k=l$ a basis for ${\mathbb X}^n_{k,k}$ is
$\{ X^{i,n-i}_{k,k} \}_{i \in {\mathbb Z}}$, whereas when $k < l$ a basis for ${\mathbb X}^n_{k,l}$
is $\{ X^{i,n-i}_{k,l} \}_{i \in {\mathbb Z}} \cup \{ X^{i,n-i}_{l,k} \}_{i \in {\mathbb Z}}$.
Let $B^n_{k,l} = F^{(1)}(A^n_{k,l}) \subseteq {\mathbb X}$. That is
$$B^n_{k,l} = \{ X^{a,b}_{k,l} + X^{b,a}_{l,k} \ : \ a+b=n \}
\cup \{ X^{a,b+1}_{k,l} + X^{b,a+1}_{l,k} \ : \ a+b=n-1 \}.$$
We see  $B^n_{k,l} \subseteq {\mathbb X}^n_{k,l}$, and so $F^{(1)}({\mathbb J}^n_{k,l}) \subseteq {\mathbb X}^n_{k,l}$.
$B^n_{k,l}$ being simply the image of $A^n_{k,l}$, the meaning of the pair $a,b$ here is
the same as noted above for $A^n_{k,l}$.

The main step in proving Theorem \protect\ref{exact} is the following:

\begin{prop}\label{ankl}
\begin{enumerate}
\item For any $n\in {\mathbb Z}, k \in {\Z^o}$, $B^n_{k,k}$ is a basis for ${\mathbb X}^n_{k,k}$.
\item For any $n\in {\mathbb Z}, k < l \in {\Z^o}$, $B^n_{k,l}$ is an independent set spanning a subspace
of codimension 1 in ${\mathbb X}^n_{k,l}$.
\end{enumerate}
\end{prop}

\begin{pf}
We first look at the following model space.
Let ${\mathbb E}$ be the vector space with basis all symbols $E_i$, $i \in {\mathbb Z}$.
Let $D$ be the following set of element in ${\mathbb E}$:
$$D = \{ E_{-i} +E_i \}_{i \geq 0} \cup \{ E_{-i} +E_{1+i} \}_{i \geq 0}$$
We claim that $D$ is a basis for ${\mathbb E}$. Indeed,
order the symbols $E_i$ as follows:
$$E_0, \ \ E_1, \ \ E_{-1}, \ \ E_2, \ \ E_{-2}, \ \ E_3, \ \ E_{-3}, \ \ \dots.$$
Order the elements of
$D$ by alternatingly taking an element from the first and second set, that is
$$2E_0, \ \ E_0 + E_1, \ \ E_{-1} + E_1, \ \ E_{-1} + E_2, \ \ E_{-2}+E_2 , \ \ E_{-2}+E_3, \ \ E_{-3}+E_3, \ \ \dots.$$
By induction, the span of the first $m$ elements in the first list coincides with that of the second list.
This establishes our claim, and we now prove (1) and (2) of the proposition.

\emph{Proof of (1):} Assume first that $n$ is even, and let $n=2m$.
Define an isomorphism $\varphi : {\mathbb E} \to {\mathbb X}^n_{k,k}$ by $\varphi(E_i) = X^{m-i,m+i}_{k,k}$ for all $i \in {\mathbb Z}$.
Then for all $i \geq 0$
\begin{itemize}
\item $\varphi(E_{-i} + E_i) = X^{m+i,m-i}_{k,k} + X^{m-i, m+i}_{k,k} = F^{(1)}(J^+_{{ m+i \choose k }{ m-i \choose k }})$
\item $\varphi(E_{-i} + E_{1+i}) = X^{m+i,m-i}_{k,k} + X^{m-1-i, m+1+i}_{k,k}
= X^{m+i,(m-1-i)+1}_{k,k} + X^{m-1-i, (m+i)+1}_{k,k} = F^{(1)}( J^A_{{m+i \choose k  }{ m-1-i \choose k }} )$
\end{itemize}
That is, $\varphi$ maps $D$ bijectively onto
$B^n_{k,k}$, and so $B^n_{k,k}$ is a basis for ${\mathbb X}^n_{k,k}$.

For $n$ odd, let $n=2m+1$. This time we define the isomorphism
$\varphi : {\mathbb E} \to {\mathbb X}^n_{k,k}$ by $\varphi(E_i) = X^{m+i,m+1-i}_{k,k}$ for all $i \in {\mathbb Z}$.
We have for all $i \geq 0$,
\begin{itemize}
\item $\varphi(E_{-i} + E_i) = X^{m-i,(m+i)+1}_{k,k} + X^{m+i, (m-i)+1}_{k,k} = F^{(1)}(J^A_{ { m-i \choose k }{ m+i \choose k } })$
\item $\varphi(E_{-i} + E_{1+i}) = X^{m-i,m+1+i}_{k,k} + X^{m+1+i, m-i}_{k,k} = F^{(1)}(J^+_{{ m-i \choose k }{ m+1+i \choose k }})$.
\end{itemize}
That is, again, $\varphi$ maps $D$ onto $B^n_{k,k}$, and so $B^n_{k,k}$ is a basis for ${\mathbb X}^n_{k,k}$.

\emph{Proof of (2):} Let ${\mathbb E}'$ be another copy of ${\mathbb E}$ defined with symbols $\{E'_i\}$ and construct the corresponding
basis $D'$, then $D \cup D'$ is a basis for ${\mathbb E} \oplus {\mathbb E}'$.
Assume first that $n$ is even and $n=2m$.
Let $\varphi:{\mathbb E} \oplus {\mathbb E}' \to {\mathbb X}^n_{k,l}$ be the isomorphism defined as follows:
\begin{itemize}
\item For $i \leq 0$: $\varphi(E_i) = X^{m-i,m+i}_{k,l}$ and for $i \geq 1$: $\varphi(E_i) = X^{m-i,m+i}_{l,k}$
\item For $i \leq 0$: $\varphi(E'_i) = X^{m-i,m+i}_{l,k}$ and for $i \geq 1$: $\varphi(E'_i) = X^{m-i,m+i}_{k,l}$
\end{itemize}
Then the value of $\varphi$ on $D \cup D'$ is as follows:
\begin{itemize}
\item $\varphi(2E_0) = 2X^{m,m}_{k,l} \not\in B^n_{k,l}$
\item $\varphi(2E'_0) = 2X^{m,m}_{l,k} \not\in B^n_{k,l}$
\end{itemize}

For $i \geq 1$:
\begin{itemize}
\item $\varphi(E_{-i} + E_i) = X^{m+i,m-i}_{k,l} + X^{m-i, m+i}_{l,k} = F^{(1)}(J^+_{{ m+i \choose k },{ m-i \choose l }})$
\item $\varphi(E'_{-i} + E'_i) = X^{m+i,m-i}_{l,k} + X^{m-i, m+i}_{k,l} = F^{(1)}(J^+_{{ m-i \choose k },{ m+i \choose l }})$
\end{itemize}
(Note that $F^{(1)}( J^+_{ { m \choose k },{ m \choose l  } }  )$ is skipped here.)

For all $i \geq 0$:
\begin{itemize}
\item $\varphi(E_{-i} + E_{1+i}) = X^{m+i,(m-1-i)+1}_{k,l} + X^{m-1-i, (m+i)+1}_{l,k}
         = F^{(1)}(J^A_{{ m+i \choose k },{ m-1-i \choose l }})$
\item $\varphi(E'_{-i} + E'_{1+i}) = X^{m+i,(m-1-i)+1}_{l,k} + X^{m-1-i, (m+i)+1}_{k,l}
         = F^{(1)}(J^A_{{ m-1-i \choose k },{ m+i \choose l }})$
\end{itemize}

That is, the set $\varphi(D \cup D')$, which is a basis for ${\mathbb X}^n_{k,l}$,
does not precisely coincide with $B^n_{k,l}$. The only difference is that the two elements
$2X^{m,m}_{k,l}, 2X^{m,m}_{l,k}$ that appear in $\varphi(D \cup D')$, are replaced in
$B^n_{k,l}$ by the one element $X^{m,m}_{k,l} + X^{m,m}_{l,k} = F^{(1)}( J^+_{ { m \choose k },{ m \choose l  } }  )$.
It follows that $B^n_{k,l}$ is independent and $spanB^n_{k,l}$ is
of codimension 1 in ${\mathbb X}^n_{k,l}$.

For $n$ odd let $n=2m+1$.
Let $\varphi:{\mathbb E} \oplus {\mathbb E}' \to {\mathbb X}^n_{k,l}$ be the isomorphism defined as follows:
\begin{itemize}
\item For $i \leq 0$, $\varphi(E_i) = X^{m+i,m+1-i}_{k,l}$ and for $i \geq 1$: $\varphi(E_i) = X^{m+i,m+1-i}_{l,k}$
\item For $i \leq 0$, $\varphi(E'_i) = X^{m+i,m+1-i}_{l,k}$ and for $i \geq 1$: $\varphi(E'_i) = X^{m+i,m+1-i}_{k,l}$
\end{itemize}
Then the value of $\varphi$ on $D \cup D'$ is as follows:
\begin{itemize}
\item $\varphi(2E_0) = 2X^{m,m+1}_{k,l} \not\in B^n_{k,l}$
\item $\varphi(2E'_0) = 2X^{m,m+1}_{l,k} \not\in B^n_{k,l}$
\end{itemize}

For $i \geq 1$:
\begin{itemize}
\item $\varphi(E_{-i} + E_i) = X^{m-i,(m+i)+1}_{k,l} + X^{m+i, (m-i)+1}_{l,k} = F^{(1)}(J^A_{{ m-i \choose k },{ m+i \choose l }})$
\item $\varphi(E'_{-i} + E'_i) = X^{m-i,(m+i)+1}_{l,k} + X^{m+i, (m-i)+1}_{k,l} = F^{(1)}(J^A_{{ m+i \choose k },{ m-i \choose l }})$
\end{itemize}
(Note that $F^{(1)}( J^A_{ { m \choose k },{ m \choose l  } }  )$ is skipped here.)

For all $i \geq 0$:
\begin{itemize}
\item $\varphi(E_{-i} + E_{1+i}) = X^{m-i,m+1+i}_{k,l} + X^{m+1+i, m-i}_{l,k}
= F^{(1)}(J^+_{{ m-i \choose k },{ m+1+i \choose l }})$
\item $\varphi(E'_{-i} + E'_{1+i}) = X^{m-i,m+1+i}_{l,k} + X^{m+1+i, m-i}_{k,l}
= F^{(1)}(J^+_{{ m+1+i \choose k },{ m-i \choose l }})$
\end{itemize}

Again, the set $\varphi(D \cup D')$, which is a basis for ${\mathbb X}^n_{k,l}$,
does not coincide with $B^n_{k,l}$. The two elements
$2X^{m,m+1}_{k,l}, 2X^{m,m+1}_{l,k}$ that appear in $\varphi(D \cup D')$, are replaced in $B^n_{k,l}$
by the one element $X^{m,m+1}_{k,l} + X^{m,m+1}_{l,k} = F^{(1)}( J^A_{{m \choose k },{m \choose l}})$. Again
it follows that $B^n_{k,l}$ is independent and $spanB^n_{k,l}$ is
of codimension 1 in ${\mathbb X}^n_{k,l}$.
\end{pf}

Proposition \protect\ref{ankl} states in particular that for any $n \in {\mathbb Z}, k \leq l \in {\Z^o}$,
$F^{(1)}(A^n_{k,l})$ is an independent set.
This implies that $A^n_{k,l}$ itself is independent, and so a basis for ${\mathbb J}^n_{k,l}$,
and that $F^{(1)}|_{{\mathbb J}^n_{k,l}} : {\mathbb J}^n_{k,l} \to {\mathbb X}^n_{k,l}$ is injective.
We have already noted that the symbols of type $J^+,J^A$ span ${\mathbb J}$
and so ${\mathbb J}$ is the sum of all ${\mathbb J}^n_{k,l}$. Since ${\mathbb X}$ is the direct sum of all ${\mathbb X}^n_{k,l}$,
and each $F^{(1)}|_{{\mathbb J}^n_{k,l}}$ is injective, we obtain that ${\mathbb J}$ is the \emph{direct} sum of all ${\mathbb J}^n_{k,l}$.
(And since we have seen $A^n_{k,l}$ is a basis for ${\mathbb J}^n_{k,l}$, we also obtain
that the set of all symbols of type $J^+,J^A$ is a basis for ${\mathbb J}$.)

For $n \in {\mathbb Z}$ and $k \leq l \in {\Z^o}$ define ${\mathbb Y}^n_{k,l} = span\{ Y^n_{k,l} \}\subseteq {\mathbb Y}$.
Then for $k=l$ we have $dim {\mathbb Y}^n_{k,l}=0$ and for $k < l$ we have $dim {\mathbb Y}^n_{k,l} =1$.
${\mathbb Y}$ is the direct sum of all ${\mathbb Y}^n_{k,l}$, and $\Psi({\mathbb X}^n_{k,l}) \subseteq {\mathbb Y}^n_{k,l}$.

Now, to prove Theorem \protect\ref{exact} it is enough to show that for each $n \in {\mathbb Z}$ and $k \leq l \in {\Z^o}$,
the restricted sequence $0 \to {\mathbb J}^n_{k,l} @>F^{(1)}>> {\mathbb X}^n_{k,l} @>\Psi>> {\mathbb Y}^n_{k,l} \to 0$ is exact.
Clearly $\Psi \circ F^{(1)} |_{\mathbb J} = 0$, and  $\Psi|_{{\mathbb X}^n_{k,l}} : {\mathbb X}^n_{k,l} \to {\mathbb Y}^n_{k,l}$ is surjective,
and we have also established that $F^{(1)}|_{{\mathbb J}^n_{k,l}} : {\mathbb J}^n_{k,l} \to {\mathbb X}^n_{k,l}$ is injective.
Since $F^{(1)}({\mathbb J}^n_{k,l}) = span B^n_{k,l}$, and since by Proposition \protect\ref{ankl}
$codim ( span B^n_{k,l} ) = dim {\mathbb Y}^n_{k,l}$, indeed
$0 \to {\mathbb J}^n_{k,l} @>F^{(1)}>> {\mathbb X}^n_{k,l} @>\Psi>> {\mathbb Y}^n_{k,l} \to 0$ is exact, and so
$0 \to {\mathbb J} @>F^{(1)}|_{{\mathbb J}}>> {\mathbb X} @>\Psi>> {\mathbb Y} \to 0$ is exact.

redpajama_set_name: RedPajamaArXiv


text:
David Sparks indicted on murder charge in Spurlock case
Sparks Indicted On Murder Charge
(LEX 18) — According to court documents, David Sparks has been indicted on a charge of murder in connection to the death of Savannah Spurlock.
Kentucky State Police say that the indictment warrant was served on Sparks Wednesday at the Lincoln County Detention Center. Court documents say that Sparks committed "the offense of murder when he intentionally caused the death of Savannah Spurlock, or, while manifesting extreme indifference to the value of human life, he wantonly engaged in conduct which created a grave risk of death to her and thereby caused her death."
Sparks was initially charged with abuse of a corpse and tampering with evidence. He was arrested in July after the remains of the missing mother of four were found buried in a shallow grave on property belonging to Sparks' family.
Spurlock disappeared from the Other Bar in Lexington in January of this year. She was seen on surveillance video leaving with two men, one of which was later identified as Sparks.
Officers say her nude body was found in a 19-inch deep grave and her feet were bound with shipping tape. She was placed in multiple black bags. Police say that she was found buried with a rug from Sparks' bedroom, which he replaced with a similar looking rug.
Initially, Sparks had told police officers that Spurlock had gone back with him to his Lancaster home with two other men the night she disappeared, but that when he woke up the next morning she was gone.
Officers also testified that blood belonging to Spurlock was found inside a closet door at Sparks' Price Court rental.
A cause of death for Spurlock has not been released.
Sparks remains in the Lincoln County Detention Center. He will be arraigned Friday, September 6.