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Tag: Loki series
"Loki" Is Lost In Space And Time In 1st Trailer!
I've had a great many ideas about how I would love for each of the upcoming Marvel Disney+ shows to look and feel since long before we saw anything official from any of them: and Loki has become my second most hotly-anticipated of the entire batch (just behind WandaVision, which seems like a technical masterpiece as well as a wildly entertaining story) thanks to the elaborate image I had concocted in my head of how it should look, ideally: like a mix of gritty science-fiction, Terry Pratchett absurdity, and fantasy horror. So you can imagine my shock and awe when the first full trailer for Loki revealed that this show is everything I was hoping it would be, and much, much more.
Loki | denofgeek.com
The last time Loki graced our screens was…well, that's a complex question. Technically that would be in 2018 when, both in our reality and in the main MCU timeline, he died pitifully attacking Thanos with a knife. But when the Avengers later staged their time heist and returned to the year 2012 to recover several Infinity Stones, they encountered the older, more feral and dangerous Loki that they had just battled and defeated in that year: and that Loki was able to escape with the Tesseract, opening a new rift in the Multiverse and ensuring that Captain America's best efforts to preserve the sanctity of each timeline wouldn't be entirely successful. The new series follows that Loki as he wanders throughout the Nine Realms, wreaking havoc and upsetting the natural order of things, while fighting characters and beings from the surreal side of Marvel comics.
That surrealism – already evident in the series – is precisely why I draw a comparison to Terry Pratchett: one of my favorite fantasy authors. By a complete coincidence, a trailer for the new BBC adaptation of Pratchett's The Watch dropped yesterday before the Loki trailer reveal, and in my opinion did a horrible job of conveying the author's characteristic brand of quirky, grammatical humor, or of capturing the colorful tone and style of his characters. This trailer (which has absolutely nothing to do with Pratchett) effortlessly achieves what any Pratchett adaptation should be aiming for – and if you don't believe me, just check out the trailer thumbnail above: Loki, dressed in a ratty old coat and splendid emerald green waistcoat, a self-congratulatory campaign button pinned to his breast, grinning from ear to ear, donning his horned helmet, and gleefully teasing us with the line "Come on? What did you expect?", all while standing in an abandoned arcade, surrounded by a group of absurdly-dressed misfits pointing spears and knives at him. It's not just a brilliant adaptation of the Loki comics and a striking visual that will entice audiences: it's a masterclass in absurd humor.
A large part of that has to do with Tom Hiddleston's deliciously entertaining performance, which is just the right amount of camp; just the right amount of Shakespearean villain; and more quintessentially British than ever before. Hiddleston is, in fact, channeling a number of iconic characters from around the globe, including James Bond and Good Omens' Crowley (a creation of Terry Pratchett and Neil Gaiman, by the way). He's also stepping into the shoes of real-life historical figures – most notably the legendary "D.B. Cooper".
Loki as "D.B. Cooper" | vulture.com
Those who love the thrill of trying to solve decades-old mysteries will enjoy this reference. D.B. Cooper, the unidentified man who somehow pulled off a mid-air robbery in 1971 before parachuting into a storm and disappearing from history makes an appearance in the Loki trailer: and the series finds a delightfully clever way of answering the questions surrounding Cooper's true identity, by suggesting that Cooper was none other than the God of Mischief. As for how he escaped an FBI manhunt, well, that's quite simple: he was snatched out of the sky by the Bifrost bridge of Asgard, of course! A few dollars drift out of Loki's attaché case as he disappears, and are scorched by the heat of the Bifrost – the same dollars, it would seem, that an eight-year old boy would find by a riverbank years after the event, mysteriously burned. The attention to detail here is simply staggering…and honestly, it's as good an explanation as any. Cooper's identity is still unknown, and the FBI officially gave up the search in 2016.
It's not the only unsolved mystery teased in the trailer. Near the end, a Polybius arcade game is also briefly visible in the background: a reference to the arcade game of the same name that sparked an urban legend in the early 2000's when it supposedly appeared out of the blue, hypnotizing or even brainwashing players, and attracting the attention of men in black. A crucial part of the legend was that the game would sometimes teleport players to other dimensions, and I expect that it will be used in the same way in Loki.
But why all this talk of historical mysteries? Well, the trailer confirms what we've long known: which is that the Time Variance Authority (or TVA) will employ Loki as one of their agents during the course of the show, and assign him various missions fixing the timeline and making sure human history proceeds as it's supposed to do. The TVA serves much the same function as the Commission in The Umbrella Academy, but with a much larger team of characters: including Mobius. M. Mobius, played by comedic actor Owen Wilson, and a severe-looking councilwoman with martial arts skills, played by Gugu Mbatha-Raw. In the case of Loki, it's unclear whether the TVA actually wanted him on their team, or were forced to recruit him because of the danger he poses as a rogue operative, wielding the Tesseract. The TVA may also need his help against a shadowy enemy we see in the trailer slaughtering their agents: a hooded figure who could be the time-traveler Kang, or an alternate, even more horrible, version of Loki.
Mobius M. Mobius and Loki | slashfilm.com
On the sidelines for now but sure to feature more prominently in time, there are hints of the Roxxon Energy Corporation – which, in the comics, tries to colonize space and harvest minerals on Mars. Interestingly, many theorists speculated that Roxxon would be the primary antagonist of Thor: Love And Thunder, with Christian Bale rumored to be playing the corporation's tyrannical minotaur leader, Dario Agger. Bale has now been confirmed to be playing a different villain, Gorr the God-Butcher, but it seems Roxxon will still play a part in the Loki series at least. We see both its inconspicuous façade on earth, as a grocery store, and its more secretive side in the form of a deep quarry on an alien planet. A red-haired woman is clearly visible in one shot, and Twitter immediately started asking if it might be Natasha Romanoff: but although I initially thought the same thing, the different hairstyle and sword at her hip makes me think this is someone else entirely.
If any major MCU characters are going to show up, I would bet on Chris Hemsworth's Thor and/or Idris Elba as Heimdall, since Loki's disguise as D.B. Cooper appears to be part of an ingenious plan to get in contact with both of his fellow Asgardians, and the Bifrost bridge that sucked him up has to have been summoned by somebody. Hopefully something happens by the end of the series that will allow this version of Loki to interact with the current version of Thor.
Trailer Rating: 10/10
"Loki" Could Expand Gender Diversity In The MCU!
While the Marvel Cinematic Universe has had an often uncomfortable relationship with the LGBTQ+ community (and recently, specifically the transgender community), things are looking up for the future, as new reports suggest that the upcoming Loki Disney+ series could feature not only a gender-fluid star, but a transgender supporting character. This would be a groundbreaking step forward if either rumor is accurate, and could pave the way for more diversity in future Marvel films and TV properties.
The Loki series recently began production, with star Tom Hiddleston (who, of course, plays the traditionally male trickster god and sorcerer, Loki) sharing a brief video of himself performing some high-flying stunts for an action sequence. In the MCU, Loki has only ever been a male, but while you may think that Loki's comics-canon gender-fluidity is something that would have come up in the movies already if it were also film-canon, there have been multiple rumors that suggest Loki could shift back and forth between genders throughout the duration of the show, with Hiddleston portraying their male side, and an actress (possibly Yesterday's Sophia Di Martino) taking on the role of Lady Loki. It's worth noting that the Loki we'll see in this show is the Loki who escaped into an alternate universe during the events of Avengers: Endgame, armed with the Space Stone and possibly a method of time-travel – this isn't the same Loki we've grown accustomed to in the years since 2012's The Avengers: this is an almost completely different iteration of the character, one who might be more comfortable with their gender-fluidity, and/or more likely to come out. The show, which is also rumored to include magical Marvel villains like Amora the Enchantress, could pose a problem that requires Loki to switch between genders in order to get out of harm's way. Who knows? There's any number of reasons why this crucial aspect of Loki's character is only just being seen onscreen now (and there's still no official confirmation that it is), and we'll just have to wait and see which one Marvel decides upon.
As for the other rumor, that a transgender actress will play an openly transgender character in the series, this is one that has been tossed around for a while: the character in question, a transwoman named Sera, was initially rumored to make her MCU debut in Thor: Love And Thunder, but new updates suggest that she will, instead, have a major supporting role in Loki – Sera is one of only a handful of transgender characters in Marvel comics, and her story is one that's always been linked to Marvel's Norse pantheon: born a male angel in the kingdom of Heven, Sera transitioned at a young age and was rescued from her dismal life by Thor's long-lost sister Angela, who soon became Sera's girlfriend. Considering that we've already had one of Thor's long-lost sisters show up in the MCU, it seems a bit of a stretch to assume that we're now going to introduce another one, which leads me to believe that either (a) Angela will not appear, and Sera's storyline will be very different from the comics, or (b) Marvel will replace Angela with Hela, and have the popular villain return, only to find true love. That might sound absurd, but Sera does have a connection to Hela in the comics that, with a lot of just a little tampering could be the base for a solid love-story, I guess? I mean, it's not like Hela tortured and imprisoned Sera or anything, right?
*she totally did, by the way*
Sera's inclusion in the series is said to be the stepping stone towards a role in future MCU movies, which probably includes Thor: Love And Thunder, where, completely coincidentally, she'd likely cross paths with the bisexual Queen of Asgard, Valkyrie (who, completely coincidentally, is said to be looking for a partner in the film). So maybe instead of falling for any of Thor's sisters, heroic, villainous or otherwise, Sera will end up with a very different but no less iconic member of the Thor mythos.
The other notable thing about this is that Sera is said to be played by a transgender actress – meaning that all that controversy earlier this year about Marvel president Kevin Feige tip-toeing around the subject of transgender representation, confirming it and immediately walking it back, could have a happy ending. All that being said, none of this is yet confirmed by Marvel or Disney. So keep your expectations measured, at least for right now.
What do you think? Will Loki and Sera be gender-fluid and transgender, respectively, in the Loki series, and how do you think the Marvel fandom will take this news, if it turns out to be true? Share your own thoughts, theories and opinions in the comments below!


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Q: Contingency table and confidence intervals Let me describe my problem by a simplified example
I have a contingency table
datatable<-array(c(1,2,3,4,5,6,dim=c(3,2))
datatable<-cbind(datatable,rowSums(datatable))
datatable<-rbind(datatable,colSums(datatable))

here, m=3,n=2 This is a (m+1)x(n+1)=4x3 table. Now, i want a new mxn array, where the ij'th entrance is a confidence interval (a list of size 2 if you will) for the estimate datatable[i,j]. Below is a function that computes an interval for the estimate datatable[i,j] from the arguments a<-datatable[i,j] and b<-datatable[i,n+1]
CIfunction<-function(a,b) c(-1,1)+a*b

I hope it is not too messy, i couldn't think of an easier example.
How do i create such a table in a elegant way? In the real example, this is a function that should take arbitrary contingency tables and return the corresponding CI table.
I already found a way, using double loop over i resp. j, but it really smells like something that could be done in an elegant way using mapply/outer or thelike.

A: The code below involves some trickery because it relies on how R recycles vectors (under what circumstances does R recycle?). array and matrix objects are just folded vectors (stored column-wise). It's not hard to come up with tricks like this once you understand the behavior
datatable <- array (1:6, dim=c(3,2))
datatable <-cbind(datatable,rowSums(datatable))
datatable <-rbind(datatable,colSums(datatable))

# last column recycled as necessary
lower <- -datatable[1:3, 1:2] * datatable[1:3, 3]
upper <- -lower
CIlist <- list(lower, upper)

I do suggest, however, that you store datatable without the row and column totals and compute them only on print.

A: Thank for your comments and answers. I had abit of trouble generalizing the (elegant) method provided by ilir to arbitrary CI functions. I ended up just doing the double loop  mentioned above. My solution to the above problem would be something like
CItable<-function(datatable,CIfunction)
{
 m<-dim(datatable)[1]-1
 n<-dim(datatable)[2]-1

 CItable<-array(NA,dim=c(m,2*n))
 for(i in 1:m)
   {
    for(j in 1:n)
     {
      tempint<-CIfunction(datatable[i,j],datatable[i,n+1])
      CItable[i,2*j]<-tempint[2]
      CItable[i,2*j-1]<-tempint[1]
     }

   }
return(CItable)
}

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This standard is managed by the Ministry of Housing and Urban-Rural Development (MOHURD) and edited by the China Academy of Building Research and reviewed by the Expert Committee of the Organization. The "Design Standard for Energy Efficiency of Residential Buildings in Severe Cold and Cold Zones" is now approved as the industry standard, numbered JGJ26-2018, which will be implemented from August 1, 2019, which means the energy conservation standard has been further improved in these areas.
This standard is managed by the Department of Housing and Urban-Rural Construction and the Quality Supervision Bureau of Qinghai Province and compiled by the Qinghai Provincial Survey and Design Institute and reviewed by the Expert Committee of the organization. The trial version of the "Design Standard for Energy Efficiency of Residential Buildings in Qinghai Province" is now approved as the local standard of Qinghai Province, and the record number DB63T 1626-2018. This standard has been implemented since March 1st, 2018. In order to achieve the strategies on energy conservation and environmental protection, this energy-saving standard has further improved the target to 75%. According to the distribution of the province's climatic zones in the severe cold regions (A), (B) and (C), the heat conductivity coefficient of the external window is limited from 1.4 to 1.6 K [W / (m2 • K)], the level of air tightness requirement for outsides window (door) and the open balcony door is class 7.
This guide is managed by the Sichuan Provincial Department of Housing and Urban-Rural Development and completed by the China Southwest Architectural Design and Research Institute with the relevant departments. The Practical Guideline for 65% of Energy Efficiency in residential buildings in Sichuan Province covers four climate zones including of severe cold, cold, hot summer and cold winter, mild areas in Sichuan Province. Strive to improve the building energy efficiency standards in these regions and lay a solid foundation for the full implementation of the 65% energy conservation standards in the future. According to the different window-wall ratios, the thermal heat transfer coefficient of the external window is between 1.8-2.2 K [W/(m2·K)] in severe cold and hot summer and cold winter regions.


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Understanding, teaching, and living the faith
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Jonah, Patron of Catechists
For too long the prophet Jonah has been overshadowed (or swallowed up?) by the great fish that swam into his life. Granted, this fish with the cast-iron stomach is an important supporting character in the tale. But when measured by instruments other than yardsticks, a man is bigger than a fish.
Anyone not distracted by the whale will realize that Jonah is a prime candidate for the role of patron of catechists. He is no superhero like Abraham, Moses, and David. He does not stride across the pages of the Bible in sandals too large for the average Christian to wear. He does not possess a secret weapon like Samson's hair. Jonah does not have to raise twelve sons as Jacob did. Nor does he have to withstand a hundred and one forms of persecution as Jeremiah did. All in all, Jonah is an ordinary man living an ordinary prophet's life.
Jonah, utterly and pitifully human in his faults and foibles, goes about his noble task of prophesying, showing fear, anger, obstinacy, and unpredictability. And yet, God uses him to accomplish the impossible. Because of Jonah's flawed but successful service, he offers hope to every aspiring catechist who is less than perfect.
The story of Jonah opens with God's commanding him to march into Nineveh, the capital of Israel's dreaded enemy, Assyria. His mission: to preach against Nineveh and to predict that in forty days it would be wiped off the face of the earth—a job comparable to condemning communism is the streets of China. Any catechist assigned to spread the Good News to a class that has a reputation of being bad news should have no trouble identifying with Jonah.
Jonah does not go through the process of discernment or consult his spiritual director. He does the natural, cowardly thing. He runs away. Jonah is so terrified that simply refusing to cooperate is not enough. He boards a ship bound for the faraway city of Tarshish.
Jonah's goodness, however, redeems him. First of all, as the Bible points out, he pays the faire on the escape ship. He is not a stowaway. Further, he honestly and humbly, if not prudently, admits to the crew that he is fleeing from his God.
All goes well for the runaway prophet until a raging storm attacks the ship. So violet is this specially brewed storm that the sailors toss the cargo over board to lighten the ship. Somehow, in the midst of keeping the ship afloat, the crew finds time to cast lots to see who is to blame for the weather. Of course, the lots accuse J=nah. Unaware that the Lord is closing in on him, our hero lies fast asleep in the hold. quiet a feat considering the thunder, the waves, and his conscience, but characteristically human. Once prodded awake and apprised of the situation, Jonah proves himself a really good Joe. He suggests that the sailors dump him overboard to save themselves and the ship. They take his advice and what follows is a conversion story of the Guinness Book of Records. When the storm clears, all the sailor worship Jonah's God, the God who made the sea and the land.
Leaving John in the sea for awhile, let us consider the manner of his shipboard witness. He does not preach a work to the sailors, but surrenders his life for their sakes. Through his self-sacrifice the tough pagan sailors come to a knowledge of God. Ironically, when Jonah is a coward, a jinx, and an outcast—most looks and feels like a total failure, he succeeds.
Now for the only "fish" part of the Jonah tale. While Jonah flounders in the swirling waters, seaweed clinging about his head, the Lord sends a large fish to catch him. Notice that the whale's obedience to the Lord in the face of an unpleasant task exceeds that of Jonah. But who can surpass Jonah's trust in the Lord when he prays a prayer of thanksgiving—nota n act of contrition—from the belly of this fish. Again, Jonah shows himself the perfect model for catechists. Although in dire straits, half-drowned, and eaten by a sea monster (and what catechist has never been in a similar position?), Jonah is still full of living faith and praise for God. Living up to Jonah's confidence in him, the Lord delivered Jonah, via whale express, from ship to shore. There Jonah is spewed out safe and sound, through probably a bit shaken and seasick.
After his little game of hide-and-seek, Jonah is ready for Nineveh. His drenching has called him to his senses, and he realizes that the Lord's will for him might well be the least of many possible evils.
Once in Nineveh, after only a single day of prophesying doom, Jonah surely merits the prophet-of-the-year award. For everyone in the city, including the king, fasts and sits in sackcloth and ashes. Even the sheep and cattle fast and wear sackcloth in sympathy with their masters, making the fields of Nineveh a sight to behold
What was Jonah's secret? He was not a profuse prophet or the book named for him would be longer than four chapters. Under the circumstances, he was not likely to be a dynamic speaker. According to the Bible , his words were not accompanied by spine-tingling miracles. And he certainly did not bombard the Ninevites through the media. Except in a passive way, the mass conversion of Nineveh was not Jonah's doing at all. It was God's When Jonah finally let God use him great things happened. He literally was an overnight success. God was able to work through Jonah to save an entire city. Jonah learns that being a witness if being a tool in God's hands. God speaks and acts through people who turn over their lives to him.
The proper place for the story to end would be right here with God so moved by the people's repentance that he c=decides to cancel their punishment and with Jonah covered in glory. But the foolish unpredictable side of Jonah's human nature crops up and prolongs the tale.
Instead of being thrilled at Nineveh's overwhelmingly positive response to his message, Jonah is angry that God changed his mind about destroying the city. He had been rather look forward to seeing a spectacular fire and brimstone demolition of his enemy. Therefore, Jonah breaks into a complaining speech in which he pretends that he fled before only because he knew that God was going to forgive the Ninevites. Strangely, in the course of his tantrum Jonah pays God a memorable complement. He says, "I knew that you are a gracious and merciful God, slow to anger, rich in clemency loathe to punish." Jonah concludes by declaring that he is so angry he wants to die. After his outburst, the disgruntled man parks himself on the outskirts of the town to see if God will comply with his wishes and consume Nineveh.
What happens next is a lesson for catechists as well as for Jonah The divine Teacher prepares a concrete learning experience for his thickheaded and hardhearted prophet. In one day a shady gourd plan springs up to protect Jonah from the hot sun. The next day a hungry worm chops on the plant and kills it. For the second time Jonah insists he is so angry he could die. With that, God draws the moral from the withered plant. Jonah is concerned about one gourd plant. Softening his lesson with gentle humor, God asks, "And should I not be concerned over Nineveh, the great city, in which there are more than a hundred and twenty thousand persons who cannot distinguish their right hand form their left, not to mention the many cattle?"
Jonah apparently is speechless for the story abruptly ends with that question. No doubt, knowing the value of personal life stories in his work, Jonah repeated the story of Nineveh over and over during the remaining years of his ministry. Now almost three thousand years later, Jonah's tale, as a book of the Bible, is still influencing people.
Although he is called a minor prophet and sometimes even the comic relief of the Bible, Jonah is undeniably a man of no small stature. Later another prophet (who could also sleep through storms at sea) compared Jonah's three-day imprisonment in the whale to his own confinement in a tomb. Ever since than Jonah has been a sign and witness of the Resurrection. What more could you ask for in a patron of catechists?
(Originally published in The Catechist)
About Sister Kathleen Glavich, SND
Jesus ordered us to make disciples of all nations and teach them. Mary Kathleen, a Sister of Notre Dame from Chardon, Ohio, responds to this call through writing, speaking, giving retreats, and teaching. Her motto, adopted from Eddie Doherty's gravesite, is "All my words for the Word."
About Catholic Faith Corner
A warm welcome to Catholic Faith Corner! May its reflections help you know and live the Catholic faith, inspire you, and give you hope.
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Newest Book
In "Heart to Heart with Mary: A Yearly Devotional" Our Blessed Mother speaks to us every day of the year and a suggestion is provided for a responsse. It is similar to the popular book "Jesus Calling" and can be found on Amazon or purchased from me.
Totally Catholic! A Catechism for Kids and Their Parents and Teachers
Do you wish you had a simple, child-friendly summary of what Catholics believe? This book took first place in the Association of Catholic Publishers awards 2014 in the category of Children's Books. It can be purchased from Pauline Books and Media or from me. ($14.95)
A Heavenly Book
Do you believe in angels? Most people do. The Catholic Companion to Angels offers heavenly facts about these majestic creatures, confirming that you can turn to them for help. Uniquely, The Angels comes with three covers (statue, stained-glass, or painting). Choose the one you like best. Available from ACTA Publications and me. ($10.00)
My First Novel!
The Fisherman's Wife is the story of St. Peter's spouse, the long-suffering but delightful woman who puts up with Peter and his obsession with the itinerant preacher named Jesus. Through her we meet Peter, Jesus, and other biblical characters. In Capernaum she witnesses Jesus' healings and hears his words.The book is based on the Gospels, legends, and what is known about first-century Jewish women. The rest is sheer imagination. Order directly from me FOR AN AUTOGRAPHED COPY at kglavich@ndec.org.
A Bit of Humor
A Slow Student During indoor recess, the third and fourth graders were playing school. Taking part in the game, the teacher, Sister Janet, sat at a child's desk. Adam, who was playing the teacher, came up to her and asked, "And just how many years have you been kept back?" (from "Why Is Jesus in the Microwave?")
The Fisherman's Wife
100 Catholic Blogs by Catholic Sisters and Nuns
Ideas for Church fundraising
Some Uplifting Thoughts for Trying Times
Joseph, the Silent Saint Gets His Year
A Mentally Healthy New Year Review
Season of Love: Divine Intimacy
God's Amazing Love at Christmas: Divine Intimacy
The Fisherman's Wife
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When you need to work, Dickies will do the job every time. We love stocking this brand at West Brothers because they are so practical and so useful in everyday life. From cargo shorts with belts to loose fit work pants, and to the original work pants, there's everything you need here to get you through the working week. Whether your job involves sitting in an office, walking around, or even driving a vehicle, Dickies Pants have you covered – literally. Shop the Dickies Pants range in the West Brothers online store and make your work wardrobe that much more practical, comfortable, and reliable. You won't doubt the quality.


Gus Malzahn decided to take over play-calling duties for Auburn moving forward following the departure of former offensive coordinator Chip Lindsey.
Nearly two years ago, Gus Malzahn sat at a table inside Auburn Arena as he introduced Chip Lindsey as the Tigers' new offensive coordinator and announced that he would be stepping away from play-calling for good.
It will be the first time since early in the 2016 season that Malzahn will have full control of his offense. That season, after the Tigers got off to a terrible start offensively, he handed over the keys to then-offensive coordinator Rhett Lashlee.
That hiatus is now over, beginning with Auburn's upcoming appearance in the Music City Bowl on Dec. 28 against Purdue in Nashville, Tenn.
Malzahn admitted Thursday that when he introduced Lindsey as offensive coordinator two years ago, he believed at the time that it was "very accurate" that he would be retiring his playsheet and taking a step back as a head coach. Circumstances change, however, and after a disappointing 2018 season offensively, Malzahn felt in his heart that he needed to get back to what made him so successful as a coach and what facilitated his rise through the coaching ranks.
Malzahn first made a name for himself as a playcaller in Arkansas as a high school coach at Hughes, Shiloh Christian and Springdale, where his offensive scheme helped him reach the pinnacle of football in the state. He parlayed that into a college gig, serving as offensive coordinator at Arkansas in 2006 before leaving for the same position at Tulsa in 2007 and 2008.
While at Tulsa, Malzahn helped produce the nation's most prolific offense for two seasons. He then left for Auburn to become offensive coordinator under Gene Chizik, helping the Tigers win a national championship in 2010 with Heisman winner Cam Newton. That year, Auburn had the nation's No. 7 scoring offense and No. 5 rushing attack.
Malzahn's decision to assume playcalling played into his decision to hire Dillingham, who did not call plays at Memphis under Mike Norvell, leading Malzahn to say it was "the perfect fit" from that standpoint. Now, following a season in which Auburn's offense hit several new statistical lows, Malzahn expects to see a bounceback performance in 2019 while he calls the plays.
"I expect us to be quite a bit better," Malzahn said. "We'll operate like we have in the past when I called the plays, everything as far as that goes with it."


The Canadian National Vimy Memorial is located on route D55, approximately 10 km north of Arras, 15 km south of Lens, 135 km southeast of Calais and 175 km north of Paris. Visitors are advised that there is currently no access to the site by public transportation. Parking is available free of charge.
If you are travelling by car, please follow the directions below or view our interactive map.
When it comes to reaching the Canadian National Vimy Memorial, all roads lead to the N17 motorway. Select your starting point below and follow the instructions to reach the N17. Then, skip to step two for instructions on reaching the memorial.
Please note that tolls may apply on parts of these routes.
Take the A1 motorway toward Lille for approximately 160 km. Follow signs for Bethune/Calais.
Watch for signs to join the A26 towards Bethune/Calais/Cambrai/Reims. Continue for approximately 10 km.
Take exit 7 for Arras-Centre/Vimy and follow signs to join the N17 towards Lens/Thélus/Vimy/Liéven.
Take the N17 towards Lens/Thélus/Vimy/Liéven and drive for approximately 10 km.
Take the A26 motorway and continue on this road for approximately 130 km.
Take exit 7 for Arras Center/Vimy and follow signs to join the N17 towards Lens/Thélus/Vimy/Liéven.
Take the A1 motorway toward Paris for approximately 45 km. (Follow signs for E17 to Paris/Reims/Aéroport Lille Lesquin).
Watch for signs to join the A26/E15 toward Calais. Continue for approximately 15 km.
Once you have joined the N17, drive for approximately 2 km (10 km for those travelling from Arras) until you reach a roundabout.
At the roundabout, take the exit toward Thélus/Vimy onto Rue des Artilleurs Canadiens. Drive for 1.5 km.
As you leave Thélus, watch for a large black and white sign on your right. It will indicate an upcoming left turn toward the site.
For additional information on travel in the region, visit Explore Arras.
Note: Veterans Affairs Canada does not control the relevance, timeliness, or accuracy of materials produced by other organizations, nor does Veterans Affairs Canada endorse their views, products or services.


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An Ode to Nana
By Girlwiththebook Press
An Ode to an Inspirational Grandmother
In "Because You're the First" my character Kassandra has an amazing relationship with her grandmother.
I readily admit the inspiration for the character was my maternal grandmother, Tess Edgerton.
Nana was a first generation American. Her mother and father came to the U.S. from Bohemia (now the Czech Republic) in 1906; my great-grandmother newly married and pregnant with my grandmother. The young family settled in Mansfield, Connecticut.
Nana's childhood was atypical of those living in rural New England at the turn of the century. She turned out to be the oldest of the 13 children my great-grandparents had. They were poor immigrants; but they all sang and played an instrument, and from what I heard from my mother, knew how to give great parties. Lots of food and liquor flowed, and everyone had wonderful times.
At 13, Nana left formal schooling, and went to work in a button factory to help support the family. Long before there were adequate child labor laws in America, it was not one her fondest experiences. But that ended when she married my grandfather, Frederick T. Edgerton. Ten years her senior, his descendants are traceable back Plymouth Plantation. The Edgerton family came from Massachusetts to Coventry, Connecticut in the 1720s. My grandfather was a World War I veteran, and quite the looker and the catch. They first settled in a small house for a few years before moving into the family farmhouse, which still stands in Coventry today. My aunt was born there, and my mother grew up there.
My grandfather and me, when I was two years old
In the book, I create sort of what I call a "wishful thinking" scenario. By that I mean Kassandra, the main character, goes and visits her grandmother at the family farm as a teenager; years after her own family moved to Los Angeles. She gets to experience farm life, spending an entire summer there, and getting through one of the toughest experiences of her life with her grandmother's support.
I was personally not as lucky. My memories of my grandparent's farm are sketchy. This picture above with my grandfather is one of the few I have of myself on the farm. That is because he passed away in 1965. I was four years old. Since he died in the house, my grandmother, who was rather superstitious, would not return. My father and uncle took Nana's personal effects out of the house, and the rest of the contents of the farm; furniture, animals, you name it, was auctioned off. The farm was sold, and Nana moved in with my aunt and uncle who lived down the street from the farm.
Two years later, my parents, my brother and I left Connecticut for Southern California. I was seven years old.
I did visit Nana at my aunt's Connecticut homes during my childhood, first in Coventry, and later in Vernon. In 1990, my aunt and uncle retired, moving to northwestern Florida, where they still reside. She also came and visited us in California when I was growing up, which is where I truly got to know how wonderful she really was.
She passed away in 2002 just one month shy of her 96th birthday. To say I miss her would be an understatement. But she is always with me. Our relationship has evolved from physical to spiritual, and I feel her presence with me always. There is a funny (and rather crude) story my grandmother used to tell (she loved bathroom humor, as I admit, does pretty much everyone in our family) that Kassandra's grandmother tells her and her brother Matthew during their farm visit. That story was my grandmother's story. She had a hearty laugh I can still hear; loved a good joke, a "highball" as they used to call it, and was filled with compassion and love for other people.
Needless to say, I would have loved to have been that teenage Kassandra from "Because You're the First," getting to visit her grandmother on the family farm. But luckily, through our trips back and forth over the years, my grandmother and I did develop a wonderful connection that I have found comforting and inspirational throughout my life. One of the greatest things about writing is you can use the things that are wonderful about your life in your fiction. My grandmother was only one of many people that inspired me to write. But I admit, she is probably my favorite!!


text:
Q: Why does pressing Alt + Tab cause a freeze? Lately my computer is freezing a lot when I do the Alt + Tab combination in order to change window focus.
I don't have Compiz or anything like that.  What could be causing the system to lock up like this?

A: It is more likely to be a bug with Xorg/Xorg driver, if you can reproduce the freeze please open a bug report and if you have the skills try to debug it as described at https://wiki.ubuntu.com/DebuggingXorg .


text:
\section{Introduction}

The intermediate long wave (ILW) equation was introduced in \cite{Jos77} to describe the propagation of waves in a two-layer fluid of finite depth. The model represents the natural interpolation between the Benjamin--Ono deep water and Kortweg--de Vries shallow water theories. The ILW equation for the time $s$ evolution of a function $w=w(x)$ defined on the real line has the following form:
\begin{gather*}
w_s +2 w w_x + T(w_{xx})= 0, \qquad T(f) := \mathrm{p.v.}\!\int_{-\infty}^{+\infty}\! \frac{1}{2\delta}\left( \operatorname{sgn}(x-\xi)-\coth\frac{\pi (x-\xi)}{2\delta} \right) f(\xi) {\rm d}\xi,
\end{gather*}
where $\delta$ is a parameter and $\mathrm{p.v.}\int\cdot \, {\rm d}\xi$ represents the principal value integral. This equation can be rewritten in the formal loop space formalism (see for instance \cite{BDGR16b}, from which we borrow notations for the rest of the paper) as
\begin{gather}\label{eq:ILW equation}
u_t = u u_x + \sum_{g\geq 1} \mu^{g-1} \varepsilon^{2g} \frac{|B_{2g}|}{(2g)!}u_{2g+1},
\end{gather}
where $B_{2g}$ are the Bernoulli numbers. Indeed the action of the operator $T$ on a function $f$ can be written in terms of the derivatives of $f$ as $T(f) = \sum\limits_{n \geq 1} \delta^{2n-1} 2^{2n} \frac{|B_{2n}|}{(2n)!} {\partial}_x^{2n-1}f$, and, by setting $w = \frac{\sqrt{\mu}}{\varepsilon} u$, $s = - \frac{\varepsilon}{2 \sqrt{\mu}} t$ and $\delta = \frac{\varepsilon\sqrt{\mu} }{2}$, we see how the above equations are equivalent.

In \cite{SAK79} the integrability of the ILW equation was established, finding an infinite number of conserved densities and giving a corresponding inverse scattering problem. From this inverse scattering problem a Lax representation can of course be deduced. Also, through a simple Hamiltonian structure, the conserved densities generate an infinite number of commuting flows. However the explicit relation between the Lax representation of the ILW equation, its higher flows and the Hamiltonian structure was never, to the best of our knowledge, clarified in the literature.

In this paper we give an explicit and remarkably simple Lax description of the full hierarchy of commuting flows of the ILW equation (the ILW hierarchy) and their Hamiltonian structure in terms of a single Lax mixed difference-differential operator.

As it turns out, this Lax description corresponds to the equivariant bi-graded Toda hierarchy of \cite[Section~B.2]{MT11} in the somewhat degenerate case of bi-degree $(1,0)$, hence a reduction of the 2D-Toda hierarchy. In fact, in \cite{MT11}, because of the geometric origin of their problem, the authors always work with strictly positive bi-degree, but this is an unnecessary restriction. What we do here is prove that the bi-degree $(1,0)$ case corresponds to the ILW hierarchy in its Hamiltonian formulation. A similar question is raised in \cite[Remark~31]{Car03}, where the author notices that the definition of all but the extended flows of the extended bi-graded Toda hierarchy survives when one of the bi-degrees vanishes. Since in the equivariant case there is no need of extended flows, this problem does not arise in the Lax representation of the equivariant bi-graded Toda hierarchy. Of course this means that equivariant bi-graded Toda hierarchies of any bidegree~$(N,0)$, with $N\geq 1$, are well defined (see also \cite{BCRR14} for their relation with the rational reductions of the 2D-Toda hierarchy). We will study the relation of such hierarchies with the geometry of equivariant orbifold Gromov--Witten theory in an upcoming publication.

\section{ILW hierarchy}

Here and in what follows we will use the formal loop space formalism in the notations of \cite{BDGR16b}. The Hamiltonian structure of the ILW equation~\eqref{eq:ILW equation} is given by the Hamiltonian
\begin{gather*}
\overline h_1^\mathrm{ILW}=\int\left(\frac{u^3}{6}+\sum_{g\ge 1}\mu^{g-1}\varepsilon^{2g}\frac{|B_{2g}|}{2(2g)!}u u_{2g}\right){\rm d}x
\end{gather*}
and the Poisson bracket $\{\cdot,\cdot\}_{{\partial}_x}$, associated to the operator ${\partial}_x$. The Hamiltonians $\overline h_d^{\mathrm{ILW}}\in\widehat\Lambda_u^{[0]}$, $d\ge 2$, of the higher flows of the ILW hierarchy are uniquely determined by the properties
\begin{gather*}
\left.\overline h_d^\mathrm{ILW}\right|_{\varepsilon=0}=\int\frac{u^{d+2}}{(d+2)!}{\rm d}x,\qquad\left\{\overline h_d^\mathrm{ILW},\overline h_1^\mathrm{ILW}\right\}_{{\partial}_x}=0.
\end{gather*}
For example,
\begin{gather*}
\overline h^{\mathrm{ILW}}_2=\int\left(\frac{u^4}{4!}+\frac{\varepsilon^2}{48}u^2 u_{xx}+\sum_{g\ge 2}\frac{|B_{2g}|}{(2g)!}\varepsilon^{2g}\left(\mu^{g-2}\frac{g+1}{2}u u_{2g}+\mu^{g-1}\frac{1}{4}u^2u_{2g}\right)\right){\rm d}x.
\end{gather*}
We refer the reader to the paper~\cite[Section 8]{Bur15}, which explains a relation between the Hamilto\-nians~$\overline h^\mathrm{ILW}_d$ and the conserved quantities of the ILW equation, constructed in~\cite{SAK79}. It is convenient to introduce an additional Hamiltonian $\overline h_0^\mathrm{ILW}:=\int\frac{u^2}{2}{\rm d}x$, which generates spatial translations. So the flows of the ILW hierarchy are given by
\begin{gather*}
\frac{{\partial} u}{{\partial} t_d}={\partial}_x\frac{\delta\overline h_d^{\mathrm{ILW}}}{\delta u},\qquad d\ge 0,
\end{gather*}
where we identify the times $t_1$ and $t$.

\section{Lax description of the ILW hierarchy}

Our Lax description of the ILW hierarchy is presented in Section~\ref{subsection:Lax description}, see Theorem~\ref{main theorem}. Before that, in Section~\ref{subsection:shift operators}, we recall necessary definitions from the theory of shift operators.

\subsection{Shift operators}\label{subsection:shift operators}

Let $\Lambda:={\rm e}^{{\rm i}\varepsilon{\partial}_x}$. We will consider formal series of the form
\begin{gather*}
A=\sum_{n\le m}a_n\Lambda^n,\qquad a_n\in\widehat{\mathcal A}_u,\qquad m\in\mathbb Z.
\end{gather*}
Via the operation of composition $\circ$, the vector space of such formal operators is endowed with the structure of a non-commutative associative algebra. The positive part~$A_+$, the negative part~$A_-$ and the residue~$\res A$ of the operator $A$ are defined by
\begin{gather*}
A_+:=\sum_{n=0}^m a_n\Lambda^n,\qquad A_-:=A-A_+,\qquad \res A:=a_0.
\end{gather*}
Let $z$ be a formal variable. The symbol $\widehat{A}$ of the operator $A$ is defined by
\begin{gather*}
\widehat{A}:=\sum_{n\le m}a_n {\rm e}^{nz}.
\end{gather*}

For an operator $L$ of the form
\begin{gather*}
L=\Lambda+\sum_{n\ge 0}a_n\Lambda^{-n},\qquad a_n\in\widehat{\mathcal A}_u,
\end{gather*}
one can define the dressing operator $P$,
\begin{gather*}
P=1+\sum_{n\ge 1}p_n\Lambda^{-n},
\end{gather*}
by the identity
\begin{gather*}
L=P\circ\Lambda\circ P^{-1}.
\end{gather*}
Note that the coefficients $p_n$ of the dressing operator do not belong to the ring $\widehat{\mathcal A}_u$, but to a~certain extension of it~(see, e.g.,~\cite[Section~2]{CDZ04}). The dressing operator $P$ is defined up to the multiplication from the right by an operator of the form $1+\sum\limits_{n\ge 1}\widehat{p}_n\Lambda^{-n}$, where $\widehat{p}_n$ are some constants.

The logarithm $\log L$ is defined by
\begin{gather*}
\log L:=P\circ {\rm i}\varepsilon{\partial}_x\circ P^{-1}={\rm i}\varepsilon{\partial}_x-{\rm i}\varepsilon P_x\circ P^{-1},
\end{gather*}
where $P_x=\sum\limits_{n\ge 1}(p_n)_x\Lambda^{-n}$. The ambiguity in the choice of dressing operator is cancelled in the definition of $\log L$ and, moreover, the coefficients of $\log L$ do belong to $\widehat{\mathcal A}_u$ (see the proof of Theorem~2.1 in~\cite{CDZ04}). To be more precise, one has the commutation relations
\begin{gather}\label{eq:computation of log L}
\big[\log L,L^m\big]=0,\qquad m\ge 1,
\end{gather}
which imply that
\begin{gather*}
\res \big[{\rm i}\varepsilon P_x\circ P^{-1},L^m\big]={\rm i}\varepsilon{\partial}_x\res L^m,\qquad m\ge 1.
\end{gather*}
These relations allow to compute recursively all the coefficients of the operator ${\rm i}\varepsilon P_x\circ P^{-1}$. As a result, if we write
\begin{gather*}
\log L={\rm i}\varepsilon{\partial}_x+\sum_{n\ge 1}f_n\Lambda^{-n},
\end{gather*}
then the coefficient $f_n$ can be expressed as a differential polynomial in the coefficients $a_0,a_1,\ldots$, $a_{n-1}$ of the operator~$L$, $f_n=f_n(a_0,\ldots,a_{n-1})\in\widehat{\mathcal A}^{[0]}_{a_0,\ldots,a_{n-1}}$. For example,
\begin{gather}\label{eq:formula for f_1}
f_1=\frac{{\rm i}\varepsilon{\partial}_x}{\Lambda-1}a_0=\sum_{n\ge 0}\frac{B_n}{n!}({\rm i}\varepsilon{\partial}_x)^n a_0.
\end{gather}

\subsection{Lax description}\label{subsection:Lax description}

Let $\tau$ be a formal variable and
\begin{gather*}
\mathcal{L}:=\Lambda+u-\tau {\rm i}\varepsilon{\partial}_x.
\end{gather*}
From the discussion of the construction of the logarithm $\log L$ in the previous section it is easy to see that there exists a unique operator $L$ of the form
\begin{gather*}
L=\Lambda+\sum_{n\ge 0}a_n\Lambda^{-n},\qquad a_n\in\widehat{\mathcal A}^{[0]}_u[\tau],
\end{gather*}
satisfying
\begin{gather}\label{eq:definition of L}
L-\tau\log L=\mathcal{L}.
\end{gather}
Since $\big[L^{d+1},\mathcal{L}\big]=0$, $d\ge 0$, the commutator $\big[\big(L^{d+1}\big)_+,\mathcal{L}\big]$ doesn't contain terms with non-zero powers of $\Lambda$. Consider the following system of PDEs:
\begin{gather}\label{eq:Lax equations}
\frac{{\partial} u}{{\partial} T_d}=\frac{{\partial}\mathcal{L}}{{\partial} T_d}=\frac{1}{\tau {\rm i}\varepsilon(d+1)!}\big[\big(L^{d+1}\big)_+,\mathcal{L}\big],\qquad d\ge 0.
\end{gather}
The following theorem is the main result of our paper.
\begin{theorem}\label{main theorem}\quad
\begin{enumerate}\itemsep=0pt
\item[$1.$] The flows $\frac{{\partial}}{{\partial} T_d}$, given by~\eqref{eq:Lax equations}, pairwise commute.
\item[$2.$] The system of Lax equations~\eqref{eq:Lax equations} possesses a Hamiltonian structure given by the Hamiltonians
\begin{gather*}
\overline h^\mathrm{Lax}_d=\int\left(\frac{\res L^{d+2}}{(d+2)!}-\frac{\tau}{d+1}\frac{\res L^{d+1}}{(d+1)!}\right){\rm d}x,\qquad d\ge 0,
\end{gather*}
and the Poisson bracket associated to the operator ${\partial}_x$.
\item[$3.$] Let $y$ be a formal variable and define polynomials $P_d(y)\in{\mathbb Q}[y,\tau]$, $d\ge 1$, by
\begin{gather*}
P_d(y):=y\prod_{i=1}^{d-1}\left(y+\frac{\tau}{{\rm i}}\right)=\sum_{j=1}^d P_{d,j}y^j\tau^{d-j},\qquad P_{d,j}\in{\mathbb Q}.
\end{gather*}
The ILW hierarchy is related to the hierarchy~\eqref{eq:Lax equations} by the following triangular transformation:
\begin{gather}\label{eq:relation between Hamiltonians}
\overline h_d^\mathrm{Lax}=\sum_{j=0}^d P_{d+1,j+1}\tau^{d-j}\left.\overline h_j^\mathrm{ILW}\right|_{\substack{\mu=-\tau^{-1}\\\varepsilon\mapsto\varepsilon\sqrt{-\tau}}},\qquad d\ge 0.
\end{gather}
\end{enumerate}
\end{theorem}
\begin{proof}1. Hereafter, for simplicity, we will use $L^m_+$ to denote $(L^m)_+$. Let
\begin{gather*}
H_d:=\frac{1}{\tau {\rm i}\varepsilon(d+1)!}L^{d+1}_+,\qquad d\ge 0.
\end{gather*}
Let us first check that
\begin{gather}\label{eq:formulas for derivatives of L and log L}
\frac{{\partial} L}{{\partial} T_d}=[H_d,L],\qquad\frac{{\partial}\log L}{{\partial} T_d}=[H_d,\log L].
\end{gather}
Equations~\eqref{eq:computation of log L} and~\eqref{eq:definition of L} imply that
\begin{gather}
\frac{{\partial} L}{{\partial} T_d}-\tau\frac{{\partial}\log L}{{\partial} T_d}=[H_d,L]-\tau[H_d,\log L],\label{eq:equation1 for derivatives}\\
\res\left[\frac{{\partial}\log L}{{\partial} T_d},L^m\right]+\sum_{a+b=m-1}\res\left[\log L,L^a\circ\frac{{\partial} L}{{\partial} T_d}\circ L^b\right]=0,\qquad m\ge 1.\label{eq:equation2 for derivatives}
\end{gather}
We consider these equations as a system of equations for the pair of operators $\frac{{\partial} L}{{\partial} T_d},\frac{{\partial}\log L}{{\partial} T_d}$. Similarly to the discussion of the computation of the logarithm $\log L$ in the previous section, equations~\eqref{eq:equation1 for derivatives} and~\eqref{eq:equation2 for derivatives} allow to compute recursively all the coefficients of the operators $\frac{{\partial} L}{{\partial} T_d}$ and~$\frac{{\partial}\log L}{{\partial} T_d}$. Then it remains to note that the operators $\frac{{\partial} L}{{\partial} T_d}=[H_d,L]$ and $\frac{{\partial}\log L}{{\partial} T_d}=[H_d,\log L]$ satisfy system~\eqref{eq:equation1 for derivatives}--\eqref{eq:equation2 for derivatives}. This completes the proof of equations~\eqref{eq:formulas for derivatives of L and log L}.

When we know formulas~\eqref{eq:formulas for derivatives of L and log L}, the commutativity of the flows $\frac{{\partial}}{{\partial} T_d}$ is proved by a standard computation:
\begin{gather*}
 \tau {\rm i}\varepsilon(d_1+1)!\tau {\rm i}\varepsilon(d_2+1)!\left(\frac{{\partial}}{{\partial} T_1}\frac{{\partial} u}{{\partial} T_2}-\frac{{\partial}}{{\partial} T_2}\frac{{\partial} u}{{\partial} T_1}\right) \\
\qquad{} = \big[\big[L^{d_1+1}_+,L^{d_2+1}\big]_+,\mathcal{L}\big]+\big[L^{d_2+1}_+,\big[L^{d_1+1}_+,\mathcal{L}\big]\big]\\
\qquad\quad{} -\big[\big[L^{d_2+1}_+,L^{d_1+1}\big]_+,\mathcal{L}\big]-\big[L^{d_1+1}_+,\big[L^{d_2+1}_+,\mathcal{L}\big]\big]\\
\qquad {} \stackrel{\substack{\text{Jacobi}\\\text{identity}}}{=} \big[\big[L^{d_1+1}_+,L^{d_2+1}\big]_+,\mathcal{L}\big]-\big[\big[L^{d_2+1}_+,L^{d_1+1}\big]_+,\mathcal{L}\big]+\big[\big[L^{d_2+1}_+,L^{d_1+1}_+\big],\mathcal{L}\big]\\
\qquad{} = \underbrace{-\big[\big[L^{d_1+1}_-,L^{d_2+1}_+\big]_+,\mathcal{L}\big]-\big[\big[L^{d_2+1}_+,L^{d_1+1}\big]_+, \mathcal{L}\big]}_{=-\big[\big[L^{d_2+1}_+,L^{d_1+1}_+\big]_+,\mathcal{L}\big]}+\big[\big[L^{d_2+1}_+,L^{d_1+1}_+\big],\mathcal{L}\big]=0.
\end{gather*}
2. Note that the flows $\frac{{\partial}}{{\partial} T_d}$ can be written as
\begin{gather*}
\frac{{\partial} u}{{\partial} T_d}=\frac{1}{(d+1)!}{\partial}_x\res L^{d+1}.
\end{gather*}
Let us compute the flow $\frac{{\partial}}{{\partial} T_1}$. For the coefficients of the operator $L$, one can immediately see that $a_0=u$ and then, using formula~\eqref{eq:formula for f_1}, we get
\begin{gather*}
a_1=\tau\frac{{\rm i}\varepsilon{\partial}_x}{\Lambda-1}u.
\end{gather*}
This allows to compute
\begin{gather*}
\frac{{\partial} u}{{\partial} T_1}= \frac{1}{2}{\partial}_x\res L^2={\partial}_x\left(\frac{u^2}{2}+\frac{\tau {\rm i}\varepsilon{\partial}_x}{2}\frac{\Lambda+1}{\Lambda-1}u\right)=uu_x+\tau u_x-\tau\sum_{g\ge 1}\frac{|B_{2g}|}{(2g)!}\varepsilon^{2g}u_{2g+1}\\
\hphantom{\frac{{\partial} u}{{\partial} T_1}}{} = {\partial}_x\frac{\delta}{\delta u}\left(\frac{u^3}{6}+\tau\frac{u^2}{2}-\tau\sum_{g\ge 1}\frac{|B_{2g}|}{2(2g)!}\varepsilon^{2g}u u_{2g}\right).
\end{gather*}

The local functionals $\overline h_d^\mathrm{Lax}$ are conserved quantities for the flow $\frac{{\partial}}{{\partial} T_1}$. Indeed,
\begin{gather*}
\frac{{\partial}}{{\partial} T_1}\int\res L^d {\rm d}x=\frac{1}{2\tau {\rm i}\varepsilon}\int\res\big[L^2_+,L^d\big]{\rm d}x,
\end{gather*}
which is zero because
\begin{gather*}
\int\res\big[f\Lambda^m,g\Lambda^n\big]{\rm d}x=\delta_{m+n,0}\int\big(f\cdot\Lambda^m g-g \cdot\Lambda^n f \big){\rm d}x=0,\qquad f,g\in\widehat{\mathcal A}_u,\qquad m,n\in\mathbb Z.
\end{gather*}
Therefore, the local functionals $\overline h_d^\mathrm{Lax}$ together with the Poisson bracket $\{\cdot,\cdot\}_{{\partial}_x}$ generate the flows which commute with the flow $\frac{{\partial}}{{\partial} T_1}$. Then these flows are uniquely determined by their dispersionless parts (see~\cite[Lemma 3.3]{LZ06} or~\cite[Lemma~4.14]{BR18}). Hence, it is sufficient to check the equation
\begin{gather*}
{\partial}_x\frac{\delta\overline h_d^\mathrm{Lax}}{\delta u}={\partial}_x\frac{\res L^{d+1}}{(d+1)!}
\end{gather*}
at the dispersionless level.

Denote $\widehat{L}_0:= \widehat{L}\big|_{\varepsilon=0}$. We see that it is sufficient to check that
\begin{gather}\label{eq:Hamiltonians at the dispersionless level}
{\partial}_x\frac{{\partial}}{{\partial} u}\left(\frac{\res\widehat{L}_0^{d+2}}{(d+2)!}-\frac{\tau}{d+1}\frac{\res\widehat{L}_0^{d+1}}{(d+1)!}\right)={\partial}_x\res\left(\frac{\widehat{L}_0^{d+1}}{(d+1)!}\right),\qquad d\ge 0.
\end{gather}
For this we compute
\begin{gather*}
\widehat{L}_0-\tau\log\widehat{L}_0={\rm e}^z+u-\tau z \ \Rightarrow \ \frac{{\partial}\widehat{L}_0}{{\partial} u}-\tau\frac{\frac{{\partial}\widehat{L}_0}{{\partial} u}}{\widehat{L}_0}=1 \ \Rightarrow \ \frac{{\partial}\widehat{L}_0}{{\partial} u}=\frac{1}{1-\tau\widehat{L}_0^{-1}}.
\end{gather*}
Therefore,
\begin{gather*}
\frac{{\partial}}{{\partial} u}\left(\frac{\res\widehat{L}_0^{d+1}}{(d+1)!}\right)=\frac{1}{d!}\res\left(\widehat{L}_0^d\frac{{\partial}\widehat{L}_0}{{\partial} u}\right)=\frac{1}{d!}\sum_{j=0}^d\tau^j\res\big(\widehat{L}_0^{d-j}\big),\qquad d\ge 0,
\end{gather*}
which gives
\begin{gather}\label{eq:recursion}
\frac{{\partial}}{{\partial} u}\left(\frac{\res\widehat{L}_0^{d+1}}{(d+1)!}\right)=\frac{\res\widehat{L}_0^d}{d!}+\frac{\tau}{d}\frac{{\partial}}{{\partial} u}\left(\frac{\res\widehat{L}_0^d}{d!}\right),\qquad d\ge 1.
\end{gather}
This implies equation~\eqref{eq:Hamiltonians at the dispersionless level}.

3. We see that
\begin{gather*}
\frac{{\partial} u}{{\partial} T_1}=\left.\frac{{\partial} u}{{\partial} t_1}\right|_{\substack{\mu=-\tau^{-1}\\\varepsilon\mapsto\varepsilon\sqrt{-\tau}}}+\tau u_x.
\end{gather*}
Using again the result of~\cite[Lemma~3.3]{LZ06} (see also~\cite[Lemma~4.14]{BR18}), we conclude that it is sufficient to prove equation~\eqref{eq:relation between Hamiltonians} at the dispersionless level, namely,
\begin{gather*}
\res\left(\frac{\widehat{L}_0^{d+2}}{(d+2)!}-\frac{\tau}{d+1}\frac{\widehat{L}_0^{d+1}}{(d+1)!}\right)=\sum_{j=0}^d P_{d+1,j+1}\tau^{d-j}\frac{u^{j+2}}{(j+2)!},\qquad d\ge 0.
\end{gather*}
Using formula~\eqref{eq:recursion} and the property $\widehat{L}_0\big|_{u=0}={\rm e}^z$, the last equation can be equivalently written as
\begin{gather}\label{eq:main elementary identity}
\frac{\res\widehat{L}_0^{d+1}}{(d+1)!}=\sum_{j=0}^d P_{d+1,j+1}\tau^{d-j}\frac{u^{j+1}}{(j+1)!}.
\end{gather}
Recursion~\eqref{eq:recursion} implies that
\begin{gather*}
\frac{\res\widehat{L}_0^{d+1}}{(d+1)!}=\left(\prod_{j=1}^d\left({\partial}_u^{-1}+\frac{\tau}{j}\right)\right)u,
\end{gather*}
where we define the action of the operator ${\partial}_u^{-1}$ in the polynomial ring ${\mathbb Q}[u,\tau]$ by ${\partial}_u^{-1}u^j:=\frac{u^{j+1}}{j+1}$, $j\ge 0$. Since we obviously have
\begin{gather*}
\sum_{j=0}^d P_{d+1,j+1}\tau^{d-j}\frac{u^{j+1}}{(j+1)!}=\left(\prod_{j=1}^d\left({\partial}_u^{-1}+\frac{\tau}{j}\right)\right)u,
\end{gather*}
identity~\eqref{eq:main elementary identity} becomes clear. This completes the proof of the theorem.
\end{proof}

\subsection*{Acknowledgements}

We would like to thank Andrea Brini, Guido Carlet, Oleg Chalykh, Allan Fordy, Alexander Mikhailov and Vladimir Novikov for useful discussions.
The work of the first author (Theorem~\ref{main theorem}, parts~1 and~3) was supported by the grant no.~16-11-10260 of the Russian Science Foundation.

\pdfbookmark[1]{References}{ref}

redpajama_set_name: RedPajamaArXiv


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